Rrb Je Cbt1_2 Mechanical.pdf

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Preface Railway Recr uit ment Boar d (RRB) Junior Engineer Examinat ion 2019 is a combined t wo-st age examinat ion followed by Document Ver ificat ion conduct ed by t he r espect ive RRBs for r ecr uit ment of Junior Engineer in I ndian Railways (I R). I n ever y t wo year s, a lar ge number of candidat es appear for t his exam, compet i ng for a limit ed number of post s. Thus RRB(JE) is consider ed one of t he most sought exams in I ndia due t o it s low select ion r at io and t echnical nat ur e. Unlike befor e, t he RRB(JE) 2019 exam pat t er n and syllabus has been complet ely changed. The old pat t er n consist ed of single st age examinat ion wher ein t he candidat es wer e allot t ed depar t ment s in I ndian Rai lways aft er clear ing t he exam. But t he r evised pat t er n includes t wo st ages – CBT-I and CBT-I I followed by document ver ificat ion, t he candidat e is r equir ed t o qualify each st age in or der t o move on t o t he next st age. The pr elims stage includes Gener al I nt elligence and Reasoning, Quantit at ive Apt it ude, Gener al Science and Gener al Awar eness. H er e t he CBT-I is common for all t he br anches. The second st age, CBT-I I is of t he object ive t ype t o t est t he t echnical abilit y of t he r espect ive engineer ing discipline. GK Publicat ions has been t he ‘‘publisher of choice’’ t o student s pr epar ing for GATE, ESE and ot her technical t est pr ep examinat ions in t he count r y. GK P's RRB(JE) 2019 ser ies pr ovides a wide r ange of st udy mat er ial which is classified into guides and object ive solved paper books t o simplify t he ent ir e pr epar at ion. These books have been t hor oughly updat ed as per t he lat est pat t er n and syllabus t o pr ovide ever yt hing you need t o per fect your scor e. GK P has also launched an andr oid app t o pr ovide you wit h an updat e on all upcoming vacancies in t he technical segment and it also has a lot of added cont ent t o aid your pr epar at ion. We hope t his lit t le effor t of our s will be helpful in achieving your dr eams. I f you have any suggest ions for impr ovement of t his book, you can wr it e t o us at gkp@gkpublicat ions.com.

All the Best! Team GKP

Contents 

SYL L ABU S

CBT – I MATHEMATICS 1. N umber Syst em

1.1 - 1.9

2. Per cent age and Rat i o & Pr opor t i on

2.1 - 2.16

3. Pr obl ems On Age

3.1 - 3.4

4. Al l i gat i ons & M i xt ur es

4.1 - 4.5

5. Ti me and Wor k

5.1 - 5.11

6. Ti me and Di st ance

6.1 - 6.10

7. Boat s and St r eams

7.1 - 7.3

8. Si mpl e I nt er est and Compound I nt er est 9. Pr ofi t , L oss and Di scount

8.1 - 8.10 9.1 - 9.9

10. Aver age

10.1 - 10.12

11. Algebr a

11.1 - 11.10

12. Tr i gonomet r i c Rat i os and H ei ght & Di st ance

12.1 - 12.13

13. Cl ock and Cal endar s

13.1 - 13.10

14. Geomet r y

14.1 - 14.18

15. M ensur at i on

15.1 - 15.12

GENERAL INTELLIGENCE & REASONING 1. Al phabet i cal and N umber Ser i es 2. Anal ogy 3. Odd One Out

1.1 - 1.7 2.1 - 2.10 3.1 - 3.4

4. Codi ng Decodi ng

4.1 - 4.10

5. Bl ood Rel at i ons

5.1 - 5.7

6. Di r ect i on Sense Test

6.1 - 6.4

7. Venn Di agr ams

7.1 - 7.7

8. Syl l ogi sm 9. St at ement Concl usi on

8.1 - 8.10 9.1 - 9.7

10. St at ement Assumpt i on

10.1 - 10.6

11. M at hemat i cal Puzzl es

11.1 - 11.6

12. Cubes and Di ce

12.1 - 12.4

13. Anal yt i cal Reasoni ng

13.1 - 13.6

14. Dat a Suffi ci ency

14.1 - 14.9

GENERAL SCIENCE* 1. Physi cs

1.1 - 1.32

2. Chemi st r y

2.1 - 2.22

3. Bi ol ogy

3.1 - 3.43

GENERAL AWARENESS* 1. Science and Technology

1.1 - 1.7

2. Spor ts

2.1 - 2.7

3. I ndian Hist or y

3.1 - 3.8

4. Geogr aphy

4.1 - 4.14

5. I ndian Polit y

5.1 - 5.6

6. Cur r ent Affair s

6.1 - 6.21

CBT – II BASICS OF ENVIRONMENT AND POLLUTION CONTROL • Basics of Envir onment and Pollut ion Cont r ol

1 - 16

BASICS OF COMPUTERS AND APPLICATIONS • Basics of Computer s and Applicat ions

1 - 15

MECHANICAL 1. Engi neer i ng M echani cs

1.1 - 1.17

2. Engi neer i ng M at er i al s

2.1 - 2.36

3. St r engt h of M at er i al s

3.1 - 3.30

4. M anufact ur i ng Engi neer i ng (M achi ni ng, Wel di ng and Fi ni shi ng Pr ocess)

4.1 - 4.51

5. M et r ol ogy and I nspect i on

5.1 - 5.41

6. Fl ui d M echani cs and H ydr aul i c M achi ner y

6.1 - 6.50

7. I ndust r i al Engi neer i ng

7.1 - 7.34

8. Ther mal Engi neer i ng

8.1 - 8.64

PRACT I CE PAPE R • CBT-I

1 - 12

• CBT-II

1 - 10

*

Common for CBT-I and CBT-I I

SYL L ABU S RECRUITMENT PROCESS Only single online applicat ion {common to all the notified posts in opted RRB - Junior Engineer (JE), Junior Engineer (I nformat ion Technology) [JE(I T )], Depot M aterial Superintendant (DM S) and Chemical & M etallurgical Assistant (CM A)} has t o be submit t ed by t he candidat e t hr ough t he link pr ovided on t he official websit e of RRBs.

The ent ir e r ecr uit ment pr ocess shall involve, 1st st age Comput er Based Test (CBT), 2nd st age CBT, and Document Ver ificat ion/M edical Examinat ion as applicable. Select i on is made st r i ct ly as per mer it , on t he basis of CBTs. The dat e, t ime and venue for al l t he act ivi t i es vi z CBTs and DV or any ot her addit ional act ivi t y as applicable shall be fixed by the RRB and shall be intimated to the eligible candidates in due cour se. Request for postponement of any of t he above act ivity or for change of venue, date and shift will not be enter tained under any cir cumstances.

1ST STAGE CBT (COMMON FOR ALL NOTIFIED POSTS OF THIS CEN) Duration

:

90 minutes (120 Minut es for eligible PwBD candidat es accompanied wit h Scr ibe)

No. of Quest ions

:

100

The 1st st age CBT is of scr eening nat ur e and t he st andar d of quest ions for t he CBT wi ll be gener all y in confor mit y wit h t he educat ional st andar ds and/or minimum t echnical quali fi cat ions pr escr ibed for t he post s. The nor mali zed scor e of 1st st age exam shall be used only for shor t l ist ing of candi dat es for 2nd st age exam as per t heir mer i t . Candidat es who ar e shor t li st ed for 2nd st age CBT avail ing t he r eser vat i on benefi t s of a communi t y, PwBD and ExSM shall cont i nue t o be consider ed onl y against t hat communi t y for all subsequent st ages of r ecr ui t ment pr ocess. The Quest ions wi ll be of object ive t ype wit h mul t i pl e choices and ar e li kely t o i ncl ude quest ions per t aining t o:

a. Mathematics : N umber syst ems, BODM AS, Deci mal s, Fr act i ons, L CM and H CF, Rat i o and Pr opor t i on, Per cent ages, M ensur at i on, Time and Wor k , Time and Dist ance, Simpl e and Compound I nt er est , Pr ofit and L oss, Algebr a, Geomet r y, Tr igonomet r y, El ement ar y St at ist i cs, Squar e Root , Age Cal cul at ions, Calendar & Clock , Pipes & Cister n.

b. General Intelligence and Reasoning : Anal ogi es, Alphabet ical and Number Ser ies, Coding and Decodi ng, M at hemat ical oper at i ons, Rel at ionshi ps, Syll ogism, Jumbl ing, Venn Di agr am, Dat a I nt er pr et at i on and Suffi ci ency, Conclusions and Decision M ak ing, Si mi lar it ies and Di ffer ences, Analyt ical r easoni ng, Classifi cat ion, Dir ect ions, St at ement – Ar gument s and Assumpt ions et c.

c. General Awareness : K nowledge of Cur r ent affair s, I ndian geogr aphy, cult ur e and histor y of I ndia including fr eedom st r uggle, I ndian Polit y and const itution, I ndian Economy, Envir onmental issues concer ning I ndia and t he Wor ld, Spor ts, Gener al scient ifi c and t echnologi cal development s et c.

d. General Science : Physics, Chemi st r y and L ife Sci ences (up t o 10t h St andar d CBSE syll abus).

The sect i on wi se Number of quest i ons and mar k s ar e as bel ow : Subjects

N o. of

M arks for each

Questions

Section

Stage-I

Stage-I

M athematics

30

30

Gener al I nt ell igence & Reasoni ng

25

25

Gener al Awar eness

15

15

Gener al Sci ence

30

30

Tot al

100

100

Ti me i n M inut es

90

The sect ion wise dist r ibut ion given in t he above t able is only indicat ive and t her e may be some var i at ions in t he act ual quest ion paper s. M inimum per cent age of mar ks for eligibilit y in var ious cat egor ies: U R -40%, OBC-30%, SC-30%, ST -25%. These per cent age of mar ks for eli gibil it y may be r elaxed by 2% for PwBD candi dat es i n case of shor t age of PwBD candidat es against vacanci es r eser ved for t hem.

2nd Stage CBT Shor t l ist i ng of Candidat es for t he 2nd St age CBT exam shall be based on t he nor mali zed mar ks obt ained by t hem in t he 1st St age CBT Exam. Tot al number of candi dat es t o be shor t li st ed for 2nd St age shall be 15 t imes t he communi t y wi se t ot al vacancy of Post s not ifi ed agai nst t he RRB as per t heir mer i t i n 1st St age CBT. H owever, Railways r eser ve t he r ight t o i ncr ease/decr ease t his li mi t i n t ot al or for any speci fi c cat egor y(s) as r equir ed t o ensur e avai labil it y of adequat e candi dat es for al l t he not i fi ed post s. Dur ation

:

120 minut es (160 M inut es for eligible PwBD candidat es accompanied wit h Scr ibe)

No of Quest ions

:

150

Syllabus The Quest ions wi ll be of object ive t ype wi t h mult i pl e choices and ar e l ik ely t o i nclude quest ions per t aining t o Gener al Awar eness, Physics and Chemist r y, Basics of Comput er s and Applicat ions, Basics of Envir onment and Pollution Contr ol and Technical abilities for the post. The syllabus for Gener al Awar eness, Physics and Chemistr y, Basi cs of Comput er s and Appl icat i ons, Basics of Envir onment and Pol lut ion Cont r ol is common for all not i fi ed post s under t his CEN as det ai led bel ow:-

a) General Awareness K nowledge of Cur r ent affair s, I ndian geogr aphy, cult ur e and histor y of I ndia including fr eedom st r uggle, I ndian Polit y and const itution, I ndian Economy, Envir onmental issues concer ning I ndia and t he Wor ld, Spor ts, Gener al scient ifi c and t echnologi cal development s et c.

b) Physics and Chemistry Up t o 10t h st andar d CBSE syllabus.

c) Basics of Computers and Applications Ar chit ect ur e of Comput er s; i nput and Out put devi ces; St or age devi ces, Net wor k ing, Oper at i ng Syst em l i ke Windows, Unix, L inux; M S Office; Var ious dat a r epr esent at ion; I nt er net and Email; Websit es & Web Br owser s; Comput er Vir us.

d) Basics of Environment and Pollution Control: Basi cs of Envi r onment ; Adver se effect of envi r onment al poll ut i on and cont r ol st r at egies; Ai r, wat er and N oi se poll ut i on, t heir effect and cont r ol; Wast e M anagement , Gl obal war mi ng; Aci d r ai n; Ozone deplet ion.

e) Technical Abilities: The educat i onal quali fi cat ions ment ioned agai nst each post shown in Annexur e-A, have been gr ouped int o di ffer ent exam gr oups as bel ow. Quest i ons on t he Technical abili t i es wi ll be fr amed i n t he syll abus defined for var ious Exam Gr oups gi ven at Annexur e-VI I -A, B, C, D, E, F & G. The sect i on wi se Number of quest i ons and mar k s ar e as bel ow : Subjects

N o. of

M arks

Questions

for each Section

Stage-I I

Stage-I I

Gener al Awar eness

15

15

Physics & Chemi st r y

15

15

Basi cs of Comput er s and Applicat ions

10

10

Cont r ol

10

10

Technical Abilit ies

100

100

Tot al

150

150

Ti me i n M inut es

120

Basi cs of Envi r onment and Pollut i on

The sect ion wise dist r ibut ion given in t he above t able is only indicat ive and t her e may be some var i at ions in t he act ual quest ion paper s. M i nimum per cent age of mar k s for eligibi lit y in var ious cat egor i es: UR -40%, OBC-30%, SC-30%, ST -25%. This per cent age of mar ks for el igibi li t y may be r elaxed by 2% for PwBD candidat es, i n case of shor t age of PwBD candidat es against vacanci es r eser ved for t hem. Vi r t ual cal cul at or wi ll be made availabl e on t he Comput er M onit or dur ing 2nd St age CBT. D iscipline M apping Tables:(I ) Sl. N o.

T hree years Diploma in Engineering

E xam Gr oup

or Bachelor ’s D egr ee in Engineering/Technology 1.

M echanical Engineer ing Pr oduct ion Engi neer ing Aut omobi le Engineer ing M anufact ur ing Engineer ing

M echanical and Allied

M echat r onics Engineer ing

Engineer ing

I ndust r ial Engineer ing M achining Engineer ing Tool s and M achini ng Engi neer i ng Tool s and Die M aki ng Engineer i ng Combinat i on of any sub st r eam of basic st r eams of above di scipl ines 2. 3.

Elect r ical Engineer ing

Electr ical and Allied

Combinat i on of any sub st r eam of basic st r eams of Elect r ical Engi neer i ng

Engineer ing

Elect r oni cs Engineer ing I nst r ument at i on and Cont r ol Engineer i ng

Communicat i on Engineer i ng Comput er Science and Engineer i ng

Elect r onics and Allied

Comput er Engineer ing

Engineer ing

Comput er Sci ence I nfor mat i on Technology Combinat i on of sub st r eams of basic st r eams of above discipli nes. 4.

Civil Engineer ing Combinat i on of any sub st r eam of basic st r eams of Civil Engineer ing

Civi l and Allied Engineer ing

B.Sc., in Civi l Engineer i ng of 3year s dur at i on 5.

Pr int ing Technology/Engi neer i ng

Pr int ing Technology

(I I ) Sl. N o.

E ducat ional Qual ificat ions

E xam Gr oup

1.

B.Sc., Chemi st r y and Physi cs

CMA

Sl. N o.

E ducat ional Qual ificat ions

E xam Gr oup

1.

BE/B.Tech., (Comput er Science)

(I I I )

BE/B.Tech., (I nfor mat ion Technol ogy) PGDCA

Comput er Science and

B.Sc. Comput er Sci ence

I nfor mat ion Technology

BCA DOEACC “ B” L evel Cour se of 3 year s dur at i on or equivalent Al l t he candidat es wi t h t he above qual ificat i on shall be t est ed i n t he Exam Gr oup mapped as per t he above char t . H owever, candi dat es wit h educat ional qualificat i on of BE/B.Tech (Comput er Science) or BE/B.Tech (I nfor mat ion Technology), appl ying for bot h t he post s of JE(S& T Depar t ment ) and JE (I T), have t o opt for ei t her El ect r oni cs and All ied Engineer ing Exam Gr oup or Comput er Science and I nfor mat ion Technology Exam Gr oup. The educat i onal quali fi cat ion for t he post of DM S (Depot M at er ial Super int endent ) i s Thr ee Year s Diploma in Engi neer i ng i.e a candidat e wit h Thr ee Year s Diploma in any of Engi neer i ng di sci pl ines, can appl y for t hese post s as appli cable. Candi dat es wi t h educat ional qual ificat i ons not figur i ng in t he above char t and el igi bl e for DM S post s have t o choose any one of t he above l ist ed Exam Gr oups ot her t han CM A Exam Gr oup, dur ing t he r egist r at ion for onl ine appl icat i ons of t his CEN. A candidat e possessing mor e t han one minimum educat i onal quali fi cat ion, mapped t o differ ent Exam Gr oups, can choose any one Exam Gr oup. These candi dat es woul d be el igibl e for all t he post s for which t hey possess minimum educat ional qualificat ions. SH ORTLI ST I N G OF CAN DI DAT ES FOR 2 nd STAGE CBT, DV AN D EM PAN ELM EN T: Shor t l ist i ng of Candidat es for t he DV shal l be based on t he nor mal ized mar ks obt ained by t hem i n 2nd St age CBT. The nor mali zat ion scheme t o be adopt ed for shor t l ist ing t he candidat es fr om 1st St age CBT t o 2nd St age CBT and for DV on t he basis of per for mance i n 2nd St age CBT is det ail ed below: N ORM ALI SAT I ON OF TH E M ARKS: Whenever CBT is conducted in multiple sessions for t he same syllabus, the r aw mar ks obtained by the candidat es in differ ent sessions wi ll be conver t ed t o nor malized mar k s. The r aw mar ks for single session paper and nor malized mar ks for mult i session paper will be used for comput ing M er i t I ndex, whi ch is a common benchmar k for gener at ing mer it for candi dat es fr om differ ent Exam Gr oups.

CALCU LATI ON OF N ORM ALI ZED M ARKS FOR M U LT I -SESSI ON PAPERS: I n 1st St age CBT and for some Exam Gr oups i n 2nd St age CBT, t he exami nat ion may be conduct ed in mult isessions. H ence, for t hese mult isession paper s, a sui t able nor mal izat i on is appl ied t o t ake int o account any var i at i on i n t he di fficult y l evel s of t he quest i on paper s acr oss differ ent sessi ons. The for mula for cal culat ing t he nor mal ized mar ks for t he mul t i-session paper s is det ail ed below:

ˆ is given by : Nor mal izat i on mar k of j t h candidat e i n i t h sessi on M ij ˆ  M ij

M tg  M qg M ti  M i q

 M i j  M i q   M qg

M ij : is t he act ual mar ks obt ained by t he j t h candidat e i n i t h session. M tg : is t he aver age mar k s of t he t op 0.1% of t he candidat es consider i ng al l sessi ons.

M qg : is t he sum of mean and st andar d devi at i on mar k s of t he candidat es in t he paper consider i ng al l sessi ons. M ti : is t he aver age mar k s of t he t op 0.1% of t he candidat es in t he i t h session or mar ks of t opper i f sessi on st r engt h is less t han 1000.

M iq : is t he sum of t he mean mar ks and st andar d deviat ion of t he i t h session. CALCU LATI ON OF M ERI T I N DEX FOR ALL PAPERS : I n or der t o gener at e a common mer it l ist compr i si ng of candidat es who gave exami nat ion fr om differ ent Exam Gr oups, but el igi bl e for a common post , mer it i ndex wil l be comput ed. For all paper s for which t her e is only one session, act ual mar ks obt ained by t he candi dat es will be used for calcul at ing M er i t I ndex, whi le for paper s in mult i-sessi ons, nor mali zed mar ks wi ll be calcul at ed cor r esponding t o t he r aw mar ks obt ai ned by a candi dat e and t he M er it I ndex will be cal culat ed based on t he nor mal ized mar k s. The M er it I ndex wil l be comput ed usi ng t he for mula gi ven bel ow :

M er i t I ndex = Sq + (St – Sq)

M  Mq Mt  Mq

(M er it I ndex i s t he r el at i ve scor e of a candi dat e wi t hi n t he discipli ne.) M : M ar ks obt ained by t he candi dat e (act ual/r aw mar ks for single session exam and nor malized mar ks for mult i session exam. M q: The qualifyi ng mar k s for gener al cat egor y candidat e in t he paper (40).

M t : The mean mar ks of t op 0.1% or t op 10 whi chever is lar ger of t he candi dat es who appear ed in t he paper (in case of mult i sessi on exam incl udi ng all sessi ons) Sq: 350 i s t he scor e assigned t o M q. St : 900 i s t he scor e assi gned t o M t . The quali fying mar ks (M q) for gener al cat egor y candi dat e is 40. The M er it I ndex wi ll be calculat ed for UR, OBC, SC, ST candi dat es whose act ual mar k s for singl e session exam and nor mali zed mar ks for mul t i session exam ar e equal or above t he communi t y qual ifying mar ks pr escr ibed in Par a 13.2. Based on t he M er it I ndex gener at ed, a combined mer it list of t he candidat es of differ ent disciplines/ Exam Gr oup wil l be pr epar ed in t he descendi ng or der of mer i t and t he al lot ment of t he pr efer ence wil l be done on t he basi s of t his mer it l ist .

CBT – I



1

Number System

CHAPTER N U M BER SYSTEM I t divided int o r eal number and imaginar y number. The number which we can put on a number line ar e 1 1 1 r eal number , e.g. , 2, , , 2 et c. 2 3 4 And ot her number s ar e i magi nar y number s, e.g. 4 , 6 et c.

H er e 1 is denot ed by i . REAL N U M BERS Real number s ar e divided int o r at ional number s and ir r at ional number s. 1. Rat ional N umbers. p Rat ional number s can be put in t he for m of q wher e q is not equal t o 0. 2 3 e.g. , et c. 3 5 Again r ational number s ar e divided into terminating decimals and non-t er minat ing decimals. e.g. 1/2 = 0.5 i s a t er mi nat i ng deci mal whi l e 1/3, 1/6 ar e non-t er minat ing decimal. Ter minat ing decimals ar e being identified by t he fact t hat t her e i s n o pr i m e f act or ot h er t h an 2 or 5 i n t h e denominat or of t he lowest fr act ion while if t her e is any ot her pr ime fact or ot her t han 2 or 5, t han non-t er minat ing but r epeat ing. We can put a bar on it for defi ni ng t he same. For exampl e, 2. 3 —

means 2.333........, 4. 67 means 4.676767........ 2. I rr at ional N umbers. The number s which ar e non-t er minat ing and non r epeat ing ar e ir r at ional number s. e.g. 2,3 et c. Classificat ion of Rat ional numbers. Rat ional number s ar e fur t her divided int o decimals and i nt eger s. Ther e ar e posi t i ve i nt eger s, 0 and negat ive integer s. Posit ive int eger s ar e called natur al number s. whi le posi t i ve i nt eger s wit h 0 ar e cal led whole numbers. Nat ur al number s fur t her classified int o t wo differ ent categor ies : ( i ) Even and Odd numbers : This concept of even and odd looks t o be ver y simple but certain t hings need t o be under st ood i.e. E + E = E, E + O = O, O + O = E, E × E = E, O × E = E, O × O = O 0 (zer o) is neit her posit ive nor negat ive.

( ii ) P r i m e an d Com posi t e n u m ber s : Pr i m e number s ar e t hose, which does not have any fact or except 1 and it self like 2,3,5,7,11,13 et c. Composit e number s ar e number s ot her t han pr ime. Note : (1) 1 is neit her pr ime nor composit e. Note :(2) 2 is t he only even pr ime. Note : (3) Ther e ar e 25 pr ime number s fr om 1 t o 100 and 15 pr ime number s fr om 1 t o 50. U N I T’S DI GI T I N TH E PRODU CT E xampl e. F i n d t h e u n i t ’s di gi t i n t h e pr odu ct (256 × 27 × 159 × 182). Solution : Pr oduct of unit ’s digit s in given number s = (6 × 7 × 9 × 2) = 756 H ence, unit digit in t he given pr oduct is 6. E xampl e. F i n d t h e u n i t ’s di gi t i n t h e pr odu ct (367 × 639 × 753). Solution : Cl ear ly, unit di git i n 34 i s 1  Unit di gi t i n 364 i s 1.  U nit digit in 367 is 7. (U ni t digit in 1 × 3 × 3 × 3 i s 7) Cl ear ly, uni t di git i n ever y power of 6 i s 6.  U nit digit in 639 is 6. Cl ear ly, unit di git i n 74 i s 1.  Unit di gi t i n 752 i s 1.  U nit digit in 753 is 7. (U ni t digit in 1 × 7 i s 7) Unit di git i n given pr oduct = Unit digit in (7 × 6 × 7) = 4 FACTORI AL N U M BERS The hi ghest power of a pr i me number ‘a’ which i s cont ained in n ! is

n  n   n   a    2    3   ............   a  a  wher e [ x ] r epr esent s t he gr eat est int eger less t han or equal t o x. Example. What is t he highest power of 2 cont ained in 70! ? Sol ut i on.

 70   70   70   70   70   70   70   2    2    3    4    5    6    7  ...........   2  2  2  2  2  2  = 35 + 17 + 8 + 4 + 2 + 1 + 0... = 67

1.2

Number System

I DEN TI T I ES (General Formulae)  (a + b)2 = a2 + 2ab + b2  (a – b)2 = a2 – 2ab + b2  (a + b) (a – b) = a2 – b2  (a + b)2 + (a – b)2 = 2(a2 + b2)  (a + b)2 – (a – b)2 = 4ab  (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca  (a + b)3 = a3 + b3 + 3ab (a + b) = a3 + b3 + 3a2b + 3ab2  (a – b)3 = a3 – b3 – 3ab (a – b) = a3 – b3 – 3a2b + 3ab2  a3 + b3 = (a + b) (a2 – ab + b2) = (a + b)3 – 3ab(a + b)  a3 – b3 = (a – b) (a2 + ab + b2) = (a – b)3 + 3ab(a – b)  a2 + b2 = (a + b)2 – 2ab or (a – b)2 + 2ab  a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)  I f a + b + c = 0, t hen a3 + b3 + c3 = 3abc  (x + a) (x + b) = x 2 + x (a + b) + ab  (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4  a4 + a2b2 + b4 = (a2 + ab + b2) (a2 – ab + b2)  a4 – b4 = (a + b) (a – b) (a2 + b2)  (a + b)2 = (a – b)2 + 4ab  (a – b)2 = (a + b)2 – 4ab  a3 + b3 + c3 = (a + b + c)3 – 3(a + b) (b + c) (c + a)  (a + b) (b + c) (c + a) = (a + b + c) (ab + bc + ca) – abc  a2 + b2+ c2 – ab – bc – ca = 





1 [( a – b) 2  (b – c) 2  ( c – a) 2 ] 2

I f a2 + b2 + c2 – ab – bc – ca = 0, t hen

FG a  1 IJ H bK FG a – 1 IJ H bK

2

= a2 + 2

= a2 +

1

b

2

+

a=b=c

2a b

1 2a – b b2

BOD M AS To do t he simplificat ion, we should always car r y out t he oper at ions in t he or der of each let t er of t he wor d ‘BODM AS’, wher e B br acket s [{( )}] O of or D division  M multiplication  A addition + S subtr action – e.g. 12 – 4  5 + 36  9 + 8 = 12 – 4  5 + 4 + 8 = 12 – 20 + 4 + 8 = 24 – 20 = 4.

e.g. 1  [1 + 1  {1 + 1  (1 + 1  2) } ] = 1  [1 + 1  {1 + 1  (1 + = 1  [1 + 1  {1 + 1  = 1  [1 + 1  {1 + = 1  [1 + 1  = 1  [1 + =1 =

1 )}] 2

3 }] 2

2 }] 3

5 ] 3

3 ] 5

8 5

5 8

RATI ON AL I SAT I ON I f pr odcut of t wo sur ds is r at ional, t hen one of t hem is called r at ionalising fact or (R.F) of t he ot her. I RRATI ON AL N U M BERS TI PS : When doing t hese sort of problems, remember : (1) When you conver t a fr act ion int o a decimal, you end up wi t h eit her a t er mi nat i ng decimal or a r ecu r r i n g deci m al . N u m ber s w h i ch can be expr essed as a fr action ar e called rational numbers. 1/4 = 0.25 (Ter minat ing decimal) 1/3 = 0.33333..... (Recur r ing decimal) (2) Number s which cannot be expr essed as a fr act ion ar e called ir r at ional number s. These number s go on for ever wit hout any appar ent pat t er n e.g :

, 2 , 7 et c.   has been calculat ed by comput er s t o millions of decimal places and st ill no pat t er n has been found!  Squar e r oot s of all non-squar es ar e ir r at ional.  Cube r oot s of all non-cubes ar e ir r at ional. (3) The pr oduct of 2 ir r ational number s can sometimes be a r at i on al num ber. Remem ber on e of t he examples below: e.g., 3  12  36  6

5  5  25  5 (4) Expr essi ons which i nclude at l east 1 ir r at ional number ar e called sur ds.

e.g., 2, 5  7 et c. I N DI CES an is t he pr oduct of n fact or s each of which is ‘a’ called base of t he power and ‘n’ is any nat ur al number called index or exponent of t he power . e.g. I f a  a  a  a is wr it t en as a4, t hen a4 is called indices of base.

Number System

LAW OF I N DI CES  ao = 1  a1 = a  am  an = am+n  am  an  ap  ... = am + n + p + ...  (am )n = amn  (ab)m = am bm 

m n

 a  

p

= am  n  p  ...



am  an = am – n I f am = an, t hen m = n I f am = bm , t hen a = b I f am = 1, t hen m = 0



a– m =

  



FG a IJ H bK am



1 am

m

=

am bm

= am– n if m > n

bn am

=

an

1 a



n

a .n b 

n



m

a .n a 

mn



mn n



ab am  n

a  mn a

a

m

if n > m

n– m

a



mn

am  n



p n a . q n b  pq n ab



I f x n = y, t hen x = y 1/n  x = n y



e aj







n

=a

a p = ap/n

n m

n

2

p m

a 

=

mn

a pm

a n b  an b

H I GH EST COM M ON FACTOR (H CF) I t is t he gr eat est fact or common t o t wo or mor e given number s. I t is also called GCM (Gr eat est Common Measur e). e.g. 4 is t he GCM of 12 and 16.

1.3

M ethods to find H CF. 1. By met hod of fact orizat ion : 36 = 2 × 2 × 3 × 3 64 = 2 × 2 × 2 × 2 × 2 × 2  HCF = 2 × 2 = 4 (Pr oduct of common fact or s) 2. By division method : Suppose t wo number s ar e given. Divide t he gr eat er number by t he lesser ; t he l esser by t he r emai nder ; di vi de t he fi r st r emainder by t he new r emainder and so on t ill t her e is no r emainder. The last divisor is t he H CF r equir ed. Example. Find t he H CF of 64 and 36. Solution : 36)64(1 36 32)36(1 32 4)32(8 32 0 The H CF of 64 and 36 is 4. Example. Find t he H .C.F. of 126, 396 and 1080. Solution : Expr essing the number s in pr ime fact or s. 126 = 2 × 32 × 7 396 = 22 × 32 × 11 1080 = 23 × 33 × 5 The highest power of 2, which will divide 2, 22, and 23, is 2. The highest power of 3, which will divide 32 and 33, is 32, and t her e ar e no ot her common fact or s. Thus H .C.F. is 2 × 32, or 18. Example. Find t he H .C.F. of 440, 1800, 2800. Solution : 440 = 10 × 11 × 4, and of t hese t hr ee fact or s 10 and 4 divide all t hr ee number s 11 does not .  H .C.F. = 10 × 4 = 40 Some I mport ant Result s  The pr oduct of t wo number is equal t o pr oduct of t heir H CM and L CM .  I f t wo number s divided by a t hir d number gives t he same r emainder, t heir differ ence is exact ly divisible by t hat number.  I f ther e ar e mor e than 2 number s, say 4 number s. Find t he H CF of any 2 number s and t he H CF of the other 2 number s. The HCF of their HCFs gives t he H CF of all t he 4 number s. LEAST COM M ON M U LTI PLE (LCM ) L east common mult iple (L CM ) of t wo or mor e given number is the least number which is exactly divisible by each of t hem e.g. 30 is t he L CM of 2, 3, 5, 6.

1.4

Number System

M ethod to find LCM . 1. By met hod of fact orizations: Resol ve each one of t he given number s int o pr ime fact or s, t hen t heir L CM is t he pr oduct of highest power s of all fact or s, t hat occur in t hese number s. Example. Find t he L CM of 36 and 64. Solut ion : 36 = 2 × 2 × 3 × 3 64 = 2 × 2 × 2 × 2 × 2 × 2  L CM = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 = 576

2. U si n g t h e f or m u l a : Pr odu ct of n u m ber s = H CF × L CM I f t wo number s ar e given, t heir L CM is given by LCM =

Pr oduct of t wo number H CF

For t he above example LCM =

36  64 = 576 4

DI VI SI BI LI TY RU LES. N umber divisible by 2 3 4 5 6

Conditions Last digi t is 0 or an even number . Sum of all t he digit s of t he number is divisible by 3. Last t wo digit s of t he number is divisible by 4 or 00. Last digi t of t he number is 0 or 5. Last digit i s 0 or an even number , and sum of all t he digit s of t he number is divisible by 3. Differ ence bet ween digit /digit s in fr ont and doubled value of t he last digit is 0 (or ) is divisi ble by 7. Last t hr ee digit s of t he number is divi sible by 8. Sum of all t he digit s of t he number is divisible by 9. Last digi t is 0. I f ever y second digit is added and t hen subt r act ed sum of all ot her digit s, t he answer is 0, or divisible by 11. Number is divisi ble by bot h 3 and 4.

7 8 9 10 11 12

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. I f t he following gr oups of fr act ions is ar r anged in ascending or der ?

5 7 6 , , 16 18 17 5 6 7 , , (c) 16 17 18

7 6 5 , , 18 17 16 6 7 5 , , (d) 17 18 16 9 2 8 5 , , , 2. If f r act i on s ar e ar r an ged i n 13 3 11 7 ascendi ng or der, t hen t he cor r ect sequence is (a)

(b)

9 2 8 5 2 9 5 8 , , , , , , (b) 13 3 11 7 3 13 7 11 2 8 5 9 5 8 2 9 , , , , , , (c) (d) 3 11 7 13 7 11 3 13 3. Which one of t he following is t he lar gest ? (a)

2 5 , 6 3 , 3 7 and 8 2 (a) 8 2

(b) 2 5

(c) 6 3

(d) 3 7

4. I f n umer at or an d den omi nat or of a pr oper fractions ar e incr eased by the same quantity, then t he r esult ing fr act ion is (a) always gr eat er t han t he or iginal fr act ion (b) always less t han t he or iginal fr act ion (c) always equal t o t he or iginal fr act ion (d) none of t hese 5. I f x + y > 5 and x – y > 3, t hen which of t he following gives all possible values of x ? (a) x > 3

(b) x > 4

(c) x > 5

(d) x < 5

6. I f x an d y ar e n egat i ve, t hen whi ch of t he following st at ement s is/ar e always t r ue ? I . x + y is posit ive I I . xy is posit ive I I I . x – y is posit ive (a) I only

(b) I I only

(c) I I I only

(d) I and I I I only

Number System

7. The value of

3.

0.000064 is

3

(a) 0.02 (b) 0.2 (c) 2.0 (d) N one 8. I f 11, 109, 999 is divided by 1111, then what is t he r emainder ? (a) 1098 (b) 11888 (c) 1010 (d) 1110

1 1  of 2 2 9. The value of 1 1  of 2 2 2 (a) 2 3 10. Taking

(b) 1

10 = 3.1623(appr ox.). What is the appr ox, value 1 of ? 10 (a) 0.333 (b) 0.3162

(c) 0.3221

4. Find the value of (2744)1/3 : (a) 24

(b) 14

(c) 34

(d) 16 [RRB JE 2014 RED SH I FT ]

(d) 3

2 = 1.414, 3 = 1.732, 5 = 2.236 and

5. Find the L.C.M. of 148 and 185. (a) 680

(b) 740

(c) 2960

(d) 3700 [RRB JE 2014 RED SH I FT ]

6 = 2.449, t hen t he value of

9 2 5 3 11. 12.

13.

14.

15.

+

9– 5–

3

to three places of decimals is

(a) 9.2321 (b) 13.716 (c) 10.723 (d) 15.892 The cube r oot t o 1.061208 (a) 1.022 (b) 10.22 (c) 0.102 (d) 1.02 The least number having four digit s which is a per fect squar e is (a) 1004 (b) 1016 (c) 1036 (d) None of t hese The missing number in t he ser ies 8, 24, 12, 36, 18, 54,  is (a) 27 (b) 108 (c) 68 (d) 72 What is t he eight h t er m of t he sequence 1, 4, 9, 16, 25,........ ? (a) 8 (b) 64 (c) 128 (d) 200 Which of t he following is t he best appr oximat ion for t he following expr ession, (7.9986 / 0.115)  19.97 ?

(a) 15

(b) 10

(a) 3

1 , then the value of 'n' is : 8n  3 (b) 2

(c) 0

(d) –2

6. If 22n 1 

2

(c) 1.0

(d) 1.3

LEVEL-1 1. The val ue of (1 +0.1 +0.11 +0.111) is (a) 1.321

(b) 1.211

(c) 1.111

(d) 1.331 [RRB JE 2014 GREEN SH I FT ]

2. When a number is divided by 5, it gives r emainder 3. What i s t he r emainder when squar e of t hat number is di vi ded by 5? (a) 9

(b) 3

(c) 4

(d) 1

(d) 0.3437 [RRB JE 2014 GREEN SH I FT ]

1 2 is 1 2 1 (c) 1 3

1.5

[RRB JE 2014 GREEN SH I FT ]

[RRB JE 2014 RED SH I FT ]

7. What is the largest possible length of a scale that can be used to measure exactly the lengths 3 m, 5 m 10 cm and 12 m 90 cm ? (a) 10 cm

(b) 20 cm

(c) 25 cm

(d) 30 cm [RRB JE 2014 YEL L OW SH I FT ]

8. After measuring 120 metres of a rope, it was discovered that the metre rod was 3 cm longer. The true length of the rope measured is : (a) 116 m 40 cm

(b) 121m 20 cm

(c) 123 m

(d) 123 m 60 cm [RRB JE 2014 YEL L OW SH I FT ]

9. Solve

3

0.000064 = ?

(a) 0.4

(b) 0.04

(c) 0.004

(d) 0.0004 [RRB JE 2014 YEL L OW SH I FT ]

10. The HCF of two numbers is 6 and their LCM is 72. If one number is 24, the other number is (a) 12

(b) 18

(c) 36

(d) 72 [RRB JE 2015 26 th AU G 1 st SH I FT ]

11. The largest number which divides by 72 and 125, leaving remainders 7 and 8 respectively is (a) 13 (c) 65

(b) 56 (d) 900 [RRB JE 2015 26 th AU G 1 st SH I FT ]

1.6

Number System

12. The HCF of two numbers is 12 and their LCM is 72. If one number is 36, the other number is

5. Value of  (appr ox. value 3.14) is : (a) Ter minat ing decimal

(a) 12

(b) 24

(b) Recur r i ng decimal

(c) 36

(d) 48

(c) Non-t er minat ing non-r epeat ing decimal

[RRB JE 2015 26 th AU G 2 nd SH I FT ]

13. The largest number which divides 81 and 108, leaving remainders 6 and 3 respectively is

(d) I ndet er minat e [RRB SSE 2014 GREEN SH I FT ]

6. Which one of the following is not a pr ime number ?

(a) 9

(b) 15

(a) 71

(b) 91

(c) 18

(d) 515

(c) 61

(d) 31

[RRB JE 2015 26 th AU G 2 nd SH I FT ]

14. The HCF of two numbers is 15 and their LCM is 270. If one number is 45, the other number is (a) 18

(b) 90

(c) 81

(d) 675 [RRB JE 2015 26 th AU G 3 rd SH I FT ]

15. The largest number which divides 247 and 319, leaving remainders 7 and 4 respectively is (a) 15

(b) 30

(c) 45

(d) 56 [RRB JE 2015 26 th AU G 3 rd SH I FT ]

LEVEL-2 3

 

1. 22  22

3

is equal t o:

(a) 22

(b) 21

(c) 2– 2

(d) 2– 1 [RRB SSE 2014 GREEN SH I FT ]

2. Ar r ange t he fol l owi ng fr act i ons i n ascendi ng or der.

7 3 4 , , 10 8 5

[RRB SSE 2014 RED SH I FT ]

7. Fi nd t he val ue of :

 489  3752   489  375 2  489  375 (a) 144

(b) 864

(c) 2

(d) 4 [RRB SSE 2014 RED SH I FT ]

8. Fi nd t he L .C.M . of

1 5 2 4 , , , : 3 6 9 27

(a)

1 54

(b)

(c)

20 3

(d) None of t hese [RRB SSE 2014 RED SH I FT ]

9. Ar r ange t he fr act ions

(a)

16 7 5 3 13     29 12 8 4 16

3 7 4 , , 8 10 5

(b)

3 4 7 , , 8 5 10

(b)

16 5 7 13 3     29 8 12 16 4

(c)

4 3 7 , , 5 8 10

(d)

7 3 4 , , 10 8 5

(c)

3 13 7 5 16     4 16 12 8 29

(d)

3 5 7 13 16     4 8 12 16 29

3. By what l east number shoul d 192,000 be divided so as t o become a per fect cube? (a) 2

(b) 5

(c) 3

(d) 7 [RRB SSE 2014 GREEN SH I FT ]

[RRB SSE 2014 RED SH I FT ]

 a 10. I f    b

4. Fi nd t he value of: 3 + 0.03 + 0.003 + 0.0003 (a) 12

(b) 3.0333

(c) 3.3333

(d) 6.0333 [RRB SSE 2014 GREEN SH I FT ]

5 7 13 16 3 , , , and in 8 12 16 29 4

ascendi ng or der of magnit ude :

(a)

[RRB SSE 2014 GREEN SH I FT ]

10 27

(a)

1 2

(c) 2

x 1

 b    a

x 3

, t hen t he val ue of 'x ' i s :

(b) 1 (d) – 1 [RRB SSE 2014 RED SH I FT ]

Number System

11. Fi nd t he val ue of (0.000216)3 : (a) 0.6

14. H CF of t wo number s, each consi st i ng of four di gi t s, i s 103 and t h ei r L CM i s 19261. The di ffer ence of t he number s is

(b) 0.06

(c) 0.006

(d) 0.0006

(a) 6/8

[RRB SSE 2014 RED SH I FT ]

c = 2126 × 382 × 5128 and d 2125 × 382 × 5129 t hen H CF of a, b ,c and d is (a) 2125 × 381 × 5129

(b) 2125 × 381 × 5128

(c) 2

(d) 2129 × 382 × 5129

×3 ×5 82

128

15. L et a  3129  5128  7 22 ,b  3128  5129  722

c  3128  5128  722 andd  3129  5128  723 H CF of a, b, c and d is

13. L et x be a l east number which when divided by 21,33,35 and 55 leaves i n each case a r emai nder 3, but is exact ly divisible by 67. The sum of digit s of x is (b) 10

(c) 12

(d) 15

(d) 4/8 [RRB SSE 2015 1 st SEP 2 nd SH I FT ]

[RRB SSE 2015 1 st SEP 1 st SH I FT ]

(a) 8

(b) 5/9

(c) 5/8

12. I f a = 2129 × 381 × 5128 , b = 2127 × 381 × 5128

125

1.7

128

 7 22

129

 723

(a) 3129  5128  722

(b) 15 

(c) 3128  5128  723

(d) 15

[RRB SSE 2015 1 st SEP 2 nd SH I FT ]

[RRB SSE 2015 1 st SEP 1 st SH I FT ]

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (c) 11. (d)

2. (b) 12. (d)

3. (a) 13. (a)

4. (a) 14. (b)

5. (b) 15. (b)

6. (b)

7. (b)

8. (d)

9. (a)

10. (c)

LEVEL-1 1. (a)

2. (c)

3. (a)

4. (b)

5. (b)

11. (a)

12. (b)

13. (b)

14. (b)

15. (a)

6. (b)

7. (a)

8. (d)

9. (b)

10. (b)

7. (d)

8. (c)

9. (a)

10. (c)

LEVEL-2 1. (a)

2. (a)

3. (c)

4. (b)

5. (b)

11. (b)

12. (b)

13. (d)

14. (a)

15. (d)

6. (b)

EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S 1.

2 5 = 2  2.236 = 4.472

5 = 0.312 16

6 3 = 6  1.732 = 10.392

6 = 0.352 17

8 2 = 8  1.414 = 11.312

3 7 = 3  2.646 = 7.938

7 = 0.388 18 2.

3.

9 = 0.692, 13

2 = 0.666 3

8 = 0.727, 11

5 = 0.714 7

H ence ascending or der is

2 9 5 8 , , , 3 13 7 11

Clear ly 8 2 is lar gest . 2 4. L et pr oper fr act ion = 3

 Result ing fr act ion = H ence



21 3 = 31 4

2 3 < 3 4 1 2 < 2 3 3 31 < 5 51

1.8

Number System

3 4 < et c. 5 6 5. Solving x + y > 5 and x – y > 3 we get , x > 4. 6. Pr oduct of – ve number s is also +ve. 7. Given expr ession =

3

9 2 5 3 =

+

6–

10 2 5 Requi r ed scal e has t o be of l engt h 10 cm because 10 cm is t he shor t est lengt h in i n gi ven quest ion. Act ual lengt h has t o be 120m + (120 × 3) cm = 120 m + 360 cm  123 m 60 cm.

0.008 = 0.2

n

7.

8.

4 8 2 = =2 3 3 3

= 2

2

4 4 4   = .04 100 100 100 We know that, x × y = LCM × HCF (x, y are the two distinct numbers)  x × 24 = 72 × 6  x = 18. In such a questions it is better to check the options. In this case 13 satisfies the given condition. Rest of the numbers are large enough to be eliminated easily. set the number be x  x × 36 = 12 × 72  x = 24 Checking the options we get the correct answer as 15. HCF × LCM = Product of numbers 3

9.

5 3

1 9 5  9 3  10 – 53

6

10.

6 5  6 3  10  3 1 15 5 – 3 5  2 6 2 1 = [15  2.236 – 3  1.732 – 2  2.449] 2 1 = [33.540 – 5.196 – 7.898] 2 = 10.732 =

11. H er e

1.061208 = (1.02)3

Requir ed cube r oot = 1.02 12. Requir ed number =1024 = (32)2 13. Second t er m is 3 t imes of t he fir st t er m and t hir d t er m i s hal f of t he second t er m, r epeat t hi s pr ocess t he missing t er m is half of 54, i.e. 27. 14. Given sequence can be wr it t en as 12, 22, 32, 42, 52, ... H ence it ’s eight h t er m = 82 = 64 15. Rounding off, we get

1 3n  9

 22n 1  23n  9  20  5n – 10 = 0

1 1 1 1 4    2 2 2 9. Given expr ession = 1 1 1 = 2 1 3 +  2 2 2 4

10.

22n 1 

6.



11.

12.

13. 14.

0.000064  3

15  270 = 90 45 Numbers = 247, 319  remainders = 7 & 4,  other number =

15.

247  7  240  319  4  315 are divisible 

8  20  100 0.1

HCF (240, 315) = 15

LEVEL-2

LEVEL-1 1. 2.

3. 4.

5.

3

1 + 0.1000 + 0.110 + 0.111 = 1.321 Let the number be 8. Thus, when 82 = 64 will be divided by 5, then remainder will be 4.

1. 22  22

1 will be less than 0.3333. 3.1623 2744 i s a mul t i ple of 7. H ence, t he answer has t o be 14. 148 = 37 × 4 and 185 = 37 × 5 LCM = 37 × 4 × 5 = 740

2.

 

3

= 28  26 = 22.

7  0.7 10 3  0.375 8 

3 7 4   8 10 5

Number System

4  0.8 5 3.

192000  96,000  Not a perfect cube. 2

 a 10.    b

x 1

 b    a

x 3

 a   b

x 1

 a    b

3 x

192000  64,000  A perfect cube 3

  x  1  3  x

 Ans = 3

 2x = 4

4. 3 + 0.03 + 0.003 + 0.0003

 x=2

= 3.0333

1

6. 91 is not a prime number

1.9

1



11.  0.000326  3   216  3  216 6



1 3

91 = 7 × 13 = 6 × 10–2 = 0.06 7.

 489  3752   489  3752  489  375 i.e



 a  b 2   a  b 2 ab

4ab 4 ab

8. LCM 

LCM 1,5,2,4  HCF  3,6,9,27 

12. Highest power of 2 common in a1 b1 c1 d = 125 Highest power of 3 common in a1 b1 c1 d = 81 Highest power of 5 common in a1 b1 c1 d = 128  HCF  2125  381  5128

x x 13. A/Q x x

 21k1  3   33k 2  3   35k 3  3   x  55k 4  3

= LCM (211 331 351 55)k + 3 = 1155k + 3



9.

20 3

A/Q x = 67 k5

 67k 5  1155k  3

5  0.625 8

at k = 4

7  0.583 12

satisfies all conditions.

Number = 4623 which  sum of digits = 4 + 6 + 2 + 3 = 15 14. Product of two Numbers = LCM × HCF

13  0.8125 16 16  0.551 29 3  0.75 4  option (a) is correct

 x × y = 103 × 19261 = 103 × 103 × 11 × 17 = 1133 × 1751  Difference =

6 (Ans) 8

15. Highest power of 3 common answer a1 b1 c1 d = 129 Highest power of 5 common answer a1 b1 c1 d = 129 Highest power of 7 common answer a1 b1 c1 d = 23 

2

Percentage and Ratio & Proportion

CHAPTER

Percentage PERCEN T Per cent means out of hundr ed. For example if we say 50% 50 25 , 25% m ean s . Or w e can say 100 100 per cent age is a fr act ion in which denominat or is 100.

t h at m ean s

50% 1 33 % 3

1/10

1/4

25%

1/12

1/5

20%

1/13

divide t he number by 100.

1/6

16

I n conver sion of fr act ion int o per cent age mult iply by 100.

1/7

Fir st lear n how to change per centage into fr action or fr action int o per cent age. 



Con ver si on t abl e of per cen t age i n t o fract ion and vice-versa. (To learn)

25% means

25 or i n conver sion of % int o fr act i on 100

¼ × 100 = 25% (fr act ion) (per cent age)

1/2 1/3

2 % 3 2 14 % 7

1/8

12 ½%

1/9

11

1/11

1/14 1/15 1/16

1 % 9

10% 1 9 % 11 1 8 % 3 9 7 % 13 1 7 % 7 2 6 % 3 1 6 % 4

Nor mally a st udent get s quest ions based on per cent ages given in per cent age t able. N ote : I f a fr act ion is given mor e and you have t o find for less per cent age, t hen decr ease t he fr act ion and a lower fr act ion is t he answer or vice-ver sa. U SE OF PERCEN TAGE TABLE TYPE I Example 1. I f t he salar y of A is 20% mor e t han B, t hen by what %, B's salar y is less t han A? 1 Solution : I f we see t he per cent age t able, 20% = 5 1 2 We have t o find a fr act ion less t han t his, i.e. , t he answer is 16 %. 6 3 Example 2. I f t he r at e has r educed by 10%, by what % t he consumpt ion has t o be incr eased, so t hat expendit ur e r emains t he same ?

1 1 1 . We have to find increased fraction, i.e. or 11 % is the answer.. 10 9 9 Example 3. I f lengt h of a r ect angle is r educed by 40%, t hen by what % t he br eadt h has t o be incr eased so t hat ar ea r emains t he same ? Solution : I n percentage table, 10% =

Solution :

FG x  100%IJ H 100  x K

As 40% is not given in per cent age t able, we ar e asking for mor e, so denominat er must be less.

FG 40  100IJ = 66 2 %. H (100 – 40) K 3

2.2

Percentage and Ratio & Proportion

TYPE I I Ther e ar e t hr ee differ ent cases. We have t o solve in t he example :

A + x % of A = B

Case 1. Value of A and x are given and we have to find B. Example. Add 20% of a number t o a given number 600. So, 600 + 20% of 600 = 600 +

20 × 600 = 720. 100

Case 2. Value of A and B are given, we have to find the value of x. Example. What % has t o be added t o 200 t o get 250 ? H er e, A = 200, B = 250. Fir st of all , see t he change, i.e. 50 ( 250 – 200 ) I mportant  On which value it is changing on 200, or on 250. I n t his case it is 200.

50 1 = or 25% 200 4



Case 3. Value of x and B are given and we have to solve for A. Example. Adding 20% t o a number gives 480, what is t he number ? H er e, x = 20%, B = 480, 20  A = 480 A+ 100 This is fr equent ly asked quest ion. Adding 20% or 1/5 t o a number gives B, means subt r act a lower fr act ion

FG H

IJ K

i.e. 1/6 (lower t han 1/5) fr om t he number 480 (B) or 480 –

1 × 480 = 400. 6

I f %age asked is not given in per cent age t able t hen. Example. A's salar y 30% mor e t han B, by what % B salar y less t han A ? Formulae : ( x /100 ± x ) × 100 % As we ar e asking for a lower per cent age, so denominat or must be mor e or (+) has t o be done.

FG 30 IJ × 100 H 130 K

=

300 1 = 23 %. 13 13

TYPE I I I A×B=D I n t his case A and B ar e decr easing or incr easing hence changing t he value of C or find % change in C. I n quest ion of per cent age we can assume any t hing or ever yt hing t o be 100 inst ead of assuming x and y . Example. The lengt h and br eadt h of a r ect angle ar e incr eased by 30% and 20% r espect ively. What is %age change in Ar ea? Solution. I nst ead of assuming x and y as lengt h and br eadt h, assume 100 for lengt h and 100 for br eadt h. Ar ea = l × b L engt h incr eased by 30% means fr om 100 it became 130, Br eadt h incr eased by 20% means fr om 100 it became 120, so t he ar ea became 15600, which is 5600 mor e t han 10000, 56 5600 or = = 56%. 100 10000 TYPE I V We gener alize t he concept ,

A% of B = C

Percentage and Ratio & Proportion

I n t his concept : (i ) Eit her A and B ar e given and we have t o find t he value of C. (ii )

A and C ar e given and we have t o find t he value of B.

(iii ) B and C ar e given and we have t o solve for A

SOLVED EXAM PLES 1. (0.756 × ¾ ) is equivalent t o. Solution.

FG0.756  3IJ × 100 = 56.7%. H 4K

2. Find (12 ½) % of 88. Solution.

FG12 1 IJ % means 1 and 1 of 88 is 11. H ence t he answer is 11.1. H 2K 8 8

FG 2 IJ % of 75. H 3K FG 6 2 IJ % means 1 and 1 of 75 = 5. Solution. 15 15 H 3K F 1I Find GH 37 JK % of 48 2 FG12 1 IJ % is 1 . Solution. H 2K 8 F 1I 3 3 I f we mult iply by 3, is GH 37 JK % ,so × 48 = 18. 2 8 8

3. Find 6

4.

5. Find 43.5% of 20 Solution. 43.5% doesn't have any easy fr act ional r epr esent at ion x % of y = y % of x leads t o conclude 43.5% of 20= 20% of 43.5, i.e. 6.

1 × 43.5 = 8.7 5

FG 11 1 IJ % of which number is 12 ? H 9K F 1 I 1 and C is 12 Solution. A = GH 11 JK % = 9 9 1 × B = 12 9 B = 108.

or 7. 16% of ------- is 40. Solution. A = 16%,



C = 40

16 × B= 40, 100

or B = 250.

8. 24 is ------- % of 36. Solution. C = 24, B = 36. A t o find out .

2.3

2.4

Percentage and Ratio & Proportion

A × 36 = 24, 100 24 is hence 66

or A = 66

2 %. 3

24 2 of 36, i.e. of 36, 36 3

2 1 1 as 33 % = mult iply by 2. 3 3 3

9. What is 3% of 5%? Solution. A = 3%, B = 5%, C t o find.

or

3 5  =C 100 100 C = .0015

10. What % is 3% of 5% ? Solution. A t o be find out , B = 5%, C = 3%.

or

A 5 3  = 100 100 100 A = 60%.

11. 3% of what number is 5% ? Solution. A = 3%, B = t o be find out , C = 5%.

3 5 ×B = 100 100 5 = 1.6 3 At t imes we have t o find t he fr act ion like 15%, 18% and so on. or

B =

12. Find 15% of 480 (concept for ment al calculat ion) Solut ion. Fi r st fi nd 10% of 480 = 48 and 5% sh oul d be hal f of t he val u e (48) whi ch i s 10% . So 5% = 24 add t he t wo, i.e. 72. 13. Find 18% of 450. Solution. 10% of 450, i.e.

1 of 450 = 45 mult iply by 2 it becomes 20% = 90 10 of 20% is 90 t han 2% is 9 as ( 2 is

1 of 20) 10

I f one subt r act ion 2% fr om 20%, one get s 18%, so 90 – 9 = 81.

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. What per cent of 72 is 6 ? (a) 10% (b) 20%

1 (c) 8 % 3

(d) N one

2. I f 30% of a num ber is 12.6, t hen t he num ber is (a) 41 (b) 51 (c) 52 (d) 42

3. What will be 160% of a num ber whose 200% is 140 ? (a) 200 (c) 140

(b) 160 (d) 112

1 % mor e t han t hat of B , t hen 3 how mu ch per cent is B 's in come less t han t hat of A?

4. I f A's in come is 33

1 % 3

(a) 25%

(b) 33

(c) 40%

(d) N one

Percentage and Ratio & Proportion

5. 63% of 3

4 is 7

(a) 2.25 (c) 2.50

(b) 2.40 (d) 2.75

6. What per cent age is 3% of 5% ? (a) 60% (b) 50% (c) 15%

(d) 30%

7. A number i n cr eased by 37 number is (a) 27 (c) 24

1 % gi ves 33. The 2

(b) 25 (d) 22

8. 40 quint als is what per cent of 2 m t ones ? (a) 20% (b) 2% (c) 200% (d) 150% 9. I f 15% of 40 is gr eat er t han 25% of a number by 2, t hen number is (a) 16 (c) 24

(b) 20 (d) 32

10. 12.5 % of 192 = 50% of ? (a) 48 (b) 96 (c) 24 (d) N one 11. x% of y is y% of ? (a) x (c) 100x

x 100 y (d) 100 (b)

12. A st udent has t o secur e 40% mar ks t o pass. H e get s 178 m ar k s and f ai l s by 22 m ar k s. Th e maximum mar ks ar e (a) 200

(b) 500

(c) 800

(d) 1000

13. I n an exam i n at i on 1100 boys an d 900 gi r l s appear ed. 50% of t he boys and 40% of t he gir ls passed t he exam. The per cent age of candidat es failed is (a) 45

(b) 45.5

(c) 54.5

(d) 59.2

14. Fr om a container having pur e milk, 20% is replaced by water and the process is r epeated thrice. At the end of the third oper ation, the milk is (a) 40% pur e (b) 50% pur e (c) 51.2% pur e (d) None of t hese

2.5

15. 300 gms of sugar solut ion has 40% sugar in it . H ow much sugar should be added t o make it 50% in t he solut ion? (a) 10 gms.

(b) 40 gms.

(c) 60 gms.

(d) 30 gms.

LEVEL-1 1. The salar ies of A and B t ogether amount to ` 2000. A spends 95% of his salar y and B 85% of his. I f now, t heir savings ar e t he same, t hen what is A's salar y ? (a) ` 1500 (b) ` 1250 (c) ` 750 (d) ` 2600 2. I f t he r adius of a cir cle is decr eased by 50%, it s ar ea is r educed by (a) 25% (b) 50% (c) 75% (d) N one 3. The populat ion of a t own is 18000. I t incr eases by 10% during first year and by 20% during the second year. The populat ion aft er 2 year s will be (a) 19800 (b) 21600 (c) 23760 (d) N one 1 4. By mixing wat er a man gains t he pr ofit of 11 %. 9 Find t he quant it y of wat er mixed by him. (a)

1 lit er s 9

(b)

1 lit er s 8

(c)

1 lit er s 10

(d)

1 lit er s 11

5. M ohan’s expendit ur e and saving ar e in t he r at io of 4:1. H is income incr eases by 20% if his saving i ncr eases by 12%. By how much % shoul d hi s expendit ur e incr eases ? (a) 21% (b) 20% (c) 42% (d) none 6. A candidat e has t o secur e 40% of t he t ot al mar ks to pass. He gets 190 mar ks and fails by 190 mar ks. Find t he t ot al mar ks. (a) 900 (b) 800 (c) 700 (d) 950 7. M ukesh ear ned ` 4000 per mont h. Fr om t he last mont h his income incr eased by 8%. Due t o r ise in pr ice, his expendit ur e incr eased by 12%and his sav i n g decr eased by 4%. F i n d h i s i n i t i al expendit ur e and init ial saving. (a) 3000, 1000 (b) 1000, 3000 (c) 12000, 4000 (d) 6000, 2000

2.6

Percentage and Ratio & Proportion

8. A vessel of 80 lit r es is filled wit h milk and wat er. 70% of milk and 30% of wat er is t aken out of t he vessel. I t is found t hat t he vessel is vacat ed by 55%, find t he init ial quant it y of wat er. (a) 50 L (b) 60L (c) 40 L (d) 30 L 9. I mixed some wat er in pur e milk and t he mixt ur e is sold at t he cost pr ice of t he milk. I f I gained

2 16 %, in what r atio did I mix water in the milk ? 3 (a) 1:5

(b) 5 :1

(c) 7:1

(d) none

10. I n a cer t ai n camp, 20% of t he boys ar e fr om M ahar asht r a st at e and 30% of t hose ar e fr om Bombay cit y. I f t her e ar e 49 boys in t he camp who ar e fr om M ahar asht r a stat e but not fr om Bombay city, what is the t ot al number of boys in the camp ? (a) 70 (b) 245 (c) 163 (d) 350 11. A r ise of 600m is r equir ed t o get a r ail r oad over a m ou n t ai n t h e gr ade can be k ept dow n by lengt hening t he t r ack and cur ving it ar ound t he mount ain peak. The addit ional lengt h of the tr ack r equir ed t o r educe t he gr ade fr om 2% t o 3% is appr oximately (a) 10000 m (b) 20000 m (c) 120000 m (d) 30000 m 12. An insur ance collect or r eceives 15% commissions on t he pr emi um coll ect ed. H ow much must he collect per week in or der t hat his annual income may be ` 6500/- ?

1 3 2 (c) ` 720 (d) ` 650 3 13. A man t axed at a r at e of 5% as income t ax and 10% special sur char ge on t he amount of income t ax pays alt oget her ` 1110. Find his gr oss income if he is allowed a deduct ion of 20%. (a) ` 1250

(a) 20000 (c) 30000

(b) ` 833

(b) 25000 (d) none

14. A gar den has only r ed, gr een and whit e flower s. 60% of t he flower s have r ed colour s, 30% have gr een colour and 50% have whit e colour. I f no flower has all t he t hr ee colour s, what per cent age of t he flower s have only one colour ? (a) 40 (b) 50 (c) 60 (d) None of t hese

15. I n school X, t her e ar e 600 pupils of whom one quar t er ar e gir ls. I n school Y, t her e ar e 500 pupils of whom t hr ee-fift h ar e gir ls, and in school Z ther e ar e 700 pupi l s of whom hal f ar e gi r l s. What per cent age of t he t ot al number of pupils in t he t hr ee schools is compr ised of gir ls ? (a) 44.44% (b) 45% (c) 50% (d) 55.55%

LEVEL-2 1. What st r engt h of 20 ml alcohol should be added t o 10 ml of 50% alcohol t o get an aver age st r engt h of 20% alcohol ? (a) 0.5% (c) 50%

(b) 5% (d) 5.5%

2. A dishonest dealer sells as a weight of 800 gm in place of 1 kg and adds 20% impur it ies in sugar. What would be his profit % if he claims to be selling at cost pr ice? (a) 50% (b) 40% (c) 45.5% (d) None of t hese 3. The populat ion of a t own gr ows at t he r at e of 20% ever y 5 year s. I n how many year s will it double it self? (appr oximately) (a) 12 (b) 15 (c) 16 (d) 20 4. 30% of t he st udent s in a class wear glasses, 10% of t he st udent s who wear gl asses, wor e gol d r immed glasses. I f t he class has 200 students, how many wear gold r immed glasses ? (a) 6 (b) 7 (c) 8 (d) None of t hese 5. I n a village, 18% of t he populat ion ar e childr en and 10% of childr en ar e female. I f t he number of female childr en is 90, what is t he populat ion ? (a) 500 (b) 5000 (c) 600 (d) 6000 6. For annual income in t he slab of ` 1200000 t o 2000000, a per son pays tax at 20% over the sur plus on ` 1200000. Wini's annual salar y is ` 1500000. H ow much t ax does she pays per annum? (a) ` 300000 (b) ` 240000 (c) ` 120000 (d) ` 60000 7. The pr ice of a book includes 10% pr int ing cost , 20% paper cost , and anot her 15% labour cost . I f t he r emainder is ` 110, what is t he t ot al pr ice of t he book ? (a) ` 200 (b) ` 300 (c) ` 400 (d) ` 500

Percentage and Ratio & Proportion

8. The populat ion of a t own has incr eased t o 110% of it s value last year. I f t he populat ion t his year is 2.2. lakhs, what is last year 's populat ion? (a) ` 2.32 L akhs (b) ` 2 L akhs (c) ` 2.5 L akhs (d) None of t hese 9. The t ot al sales of a company amount s t o ` 24 cr or es and t he volume sold is 20000 unit s. What is t he aver age pr ice/ unit ? (a) ` 1000 (b) ` 1200 (c) ` 120 (d) ` 12000 10. Six fr iends have an aver age height of 167 cms. Penshu wit h height 162 cms leaves. What is t he new aver age height ? (a) 166 cm (b) 167 cm (c) 168 cm (d) 169 cm 11. M adhu got mar r ied 6 year s ago. Today her age is

2.7

13. A sum of money doubl es i t sel f at compound inter est in 15 yr s. I n how many yr s it will become eight t imes ? (a) 30

(b) 45

(c) 50

(d) 60

14. A man buys 6 dozen eggs for ` 10.80. 12 eggs ar e found t o be r ot t en and t he r est ar e sold at 5 for ` 1. Find his % gain or % loss. 1 (a) gain of 11 % yr s. 9 1 (b) loss of 11 % yr s. 9 1 (c) gain of 9 % yr s 11 1 (d) loss of 9 % 11 15. A bor r owed ` 800@ 6% and B bor r owed ` 600 @10%. Aft er how much t ime will t hey bot h have equal debt s ?

1 t imes her age at t he t ime of mar r iage. H er 4 son's age is 1/10 her age. H er son's age is (a) 3 yr s (b) 2.5 yr s (c) 2 yr s (d) 4 yr s 1

(a) 15

1 yr s. 3

(b) 16

2 yr s. 3

(c) 18

1 yr s 3

(d) none of t hese

12. I n a village t he cur r ent bir th r at e per t housand is 55 wher eas cor r esponding deat h r at e is 34 per t h ou san d. T h e n et gr ow t h r at e i n t er m of populat ion incr ease will be (a) 0.021% (b) 0.0021% (c) 21% (d) 2.1%

AN SWE RS OBJECTI VE TYPE QU ESTI ON S 1. (c) 11. (a)

2. (d) 12. (b)

3. (d) 13. (c)

4. (a) 14. (c)

5. (a) 15. (c)

6. (a)

7. (c)

8. (c)

9. (c)

10. (a)

7. (a)

8. (d)

9. (d)

10. (d)

7. (a)

8. (b)

9. (b)

10. (c)

LEVEL-1 1. (a) 11. (a)

2. (c) 12. (b)

3. (c) 13. (d)

4. (a) 14. (d)

5. (d) 15. (a)

6. (d)

LEVEL-2 1. (b)

2. (a)

3. (d)

4. (a)

5. (b)

11. (a)

12. (d)

13. (b)

14. (a)

15. (b)

6. (d)



2.8

Percentage and Ratio & Proportion

Ratio & Proportion RAT I O Rat io is t he r elat ion which one quant it y bear s t o anot her of t he same kind. The r at io of A t o B is usually wr it t en: A : B. The fir st t er m is called t he ant ecedent and t he second t er m t he consequent .

a ma = b mb i.e., t he value of a r at io r emains unalt er ed if t he ant ecedent and t he consequent ar e mult iplied or divided by t he same quant it y.

e.g., A and B have income in t he r at io of 5 : 4 does not mean t hat t he incomes ar e ` 5 and ` 4, but it means any income which is a mult iple of 5 and 4, i.e. 5x and 4x , wher e x is a var iable. Example 1. Divide ` 80 in t he r at io of 3 : 5 ? Solution. Divide 80 by 8 (sum of t he r at ios) we get 10 and have 10 is t he k fact or. H ence mult iply t he r at io 3 : 5 by 10 or 30 : 50. Example 2. Divide ` 98 among a, b, c, d in t he r at io in t he r at io 2 : 3 : 4 : 5 ? Solution. Divide ` 98 by 14 (sum of t he r at ios), we get 7 t hat is k fact or. The amount in ` ar e 7  2 ,7  3,7  4, 7  5 or 14, 21, 28, 35.

 Rat ios ar e compounded by mult iplying t oget her t he fr act ions which denot e t hem. Example 3. Find t he r at io compounded of t he t hr ee r at ios ? Solution.

Requir ed r at io =

 

2 a : 3b, 6 ab : 5 c2, c : a

2a 6 ab c 4 a    3b 5c2 a 5c

When t he r at io a: b is compounded wit h it self t he r esult ing r at io is a2:b2 and is called t he duplicat e r at io of a:b. Also a½ : b½ is called t he sub duplicat e r at io of a:b. (i )

The duplicat e r at io of 2a: 3b is 4a2 : 9b2.

(ii )

The sub duplicat e r at io of 49 : 25 is 7 : 5

(iii ) The t r iplicat e r at io of 2x : 1 is 8x 3 : 1 Rat io of Equalit y, Great er inequat lit y and L ess inequalit y. A r at io is said t o be a r atio of gr eat er inequalit y, of less inequality, or of equality, accor ding as the ant ecedent is gr eat er t han less t han or equal t o t he consequent .

 

A r at i o of gr eat er i n equ al i t y i s di m i n i sh ed an d a r at i o of l ess i n equ al i t y i s i n cr eased, by adding t he same quant it y t o bot h it s t er ms. When a ser ies of fr act ions ar e equal, each of t hem is equal t o t he sum of all t he numer at or s divided by t he sum of all t he denominat or s.

An import ant t heorem.

a I f a1 , a2 , a3 ,..............., n be unequal fr act ions, of which t he denominat or s ar e all of t he same sign, bn b1 b2 b3 t hen t he fr act ion

a1  a2  a3  ...  an lies in magnit ude bet ween t he gr eat est and least of t hem. b1  b2  b3  ...  bn

PROPORT I ON When t wo r at ios ar e equal, t he four quant it ies composing t hem ar e said t o be pr opor t ionals. Thus if

a c = , t hen a, b, c, d ar e pr opor t ionals. This is expr essed by saying t hat a is t o b as c is t o d, and t he b d pr opor t ion is wr it t en a:b::c:d

Percentage and Ratio & Proportion

2.9

The t er ms a and d ar e called t he extremes, b and c t he means.

 L et a, b, c, d be t he pr opor t ionals. a c Then by definit ion =  ad = bc b d H ence if any t hr ee t er ms of a pr opor t ion ar e given, t he four t h may be find out . ad Thus if a, c, d ar e given, t hen b = . c  Quant it ies ar e said t o be in cont inued pr opor t ion when t he fir st is t o t he second, as t he second is t o t he thir d. I f t hr ee quant it ies a, b, c, ar e in cont inued pr opor t ion, t hen

a : b= b : c

ac = b2.

and

 I f thr ee quant it ies ar e pr opor t ional, t he fir st is t o t he t hir d is t he duplicat e r at io of t he fir st t o t he second. L et t hr ee quant it ies be a, b, c, t hen

a b = b c

a2 a a b a a =  =  = 2 b c b c b b

Now

i.e.

a : c = a2 : b2

I f a : b = c : d,

and

e : f = g : h,

t hen

ae : bf = cg : dh .

 I f four quant it ies a, b, c, d, for m a pr opor t ion, many ot her pr opor t ions may be deduced by t he pr oper t ies of fr act ions. (1) I f a : b = c : d, t hen

b : a = d : c.

[I nver tendo]

(2) I f a : b = c : d, t hen

a:c= b:d

[Alternando]

(3) I f a : b = c : d, t hen

a + b : b = c + d : d.

[Componendo]

(4) I f a : b = c : d, t hen

a– b : b = c – d : d

[Dividendo]

(5) I f a : b = c : d, t hen

a+ b:a– b= c+ d :c– d

Example. Find second pr opor t ion of 2, 4 and 8. Sol ut i on.

a:b =c:d

or

2:b =4:8

or

4 b = 16

or

b =4

Example. Find t he t hir d pr opor t ion of .5, 12 and 8. Solution. I n t his quest ion, we have t o find t he value of c

a:b =c:d or or

.5 : 12 = c : 8 12c = 4

1 3 COM PARI SON OF TWO RATI OS Example. I f a : b = 2 : 3 and b : c = 4 : 5 , t hen what is t he r at io of a : b : c ?

or

c=

Sol ut i on.

a = (2  4) = 8

a:

b:

2

3

So t he r at io

b = (3  4) = 12 c = (3  5) = 15. a : b : c = 8 : 12 : 15

4

c

5

2.10

Percentage and Ratio & Proportion

 I f t he r at ios among 4 differ ent quant it ies has t o be found out . Example. What is t he r at io of a, b, c, d. I f r atio of a : b = 2 : 3, b : c = 4 : 5 and c : d = 6 : 7? Solution. a : b : c : d

2 : 3 : 3 : 3 4 : 4 : 5 : 5 6 : 6 : 6 : 7 48 : 72 : 90 : 105 a : b : c : d = 16 : 24 : 30 : 35 Example. Divide ` 1300 among a, b, c, d such t hat Solution.

a

b

2

3 2

c

c 2 a b = = = . b c d 3

d

3 2

3

Then, r at io of a, b, c, d = 8 : 12 : 18 : 27 so, sum of r at ios of a, b, c, d = 65. Divide 1300 by 65 t o get t he key fact or as 20. So t he division of money among a, b, c, d ar e 160, 240, 360, 540, i.e 8  20, 12  20, 18  20, 27  20. SACRI FI CI N G RATI O I f t wo or mor e par t ner s give some par t of t heir shar es t o incoming par t ner, t hen t he loss in t he shar es of exist ing par t ner s is called sacr ificing r at io. Sacr ificing r at io = Old r at io – New r at io. Example. What is t he sacr ificing r at io of t wo par t ner s who wer e dividing t heir shar e in t he r at io of 2 : 3 and aft er joining by incoming par t ner t heir r at io changed t o 1 : 1 : 1 ? Sol ut i on.

Sacr ificing r at io = Old r at io – New r at io



1 4 2 1 3 1  5  3  :  5  3  = 15 : 15  1 : 4    

E xample. Thr ee men wer e t r avell ing in a r ail way compar t ment . One of t hem was a ver y weal t hy businessman. The ot her t wo had food wit h t hem and invit ed t he businessman t o join t hem in t heir meal. One man had br ought 5 chapat his and t he ot her had 3. The t hr ee of t hem at e t oget her and t he food was shar ed equally. On par ting, the businessman gave 8 gold coins to the t wo men, as a token of his appreciation of t heir hospit alit y, and t old t hem t o shar e t he coins in pr opor t ion of t heir cont r ibut ion of food. What number of coins will each of t hem get ?

1 7 3 1 5 1 Solut ion. Sacr i fici ng r at io    :    = : 1:7  8 3   8 3  24 24 ` 8 has t o be divided in t he r at io of 1 : 7, i.e. ` 1 t o A and ` 7 t o B. FRACTI ON AL RATI O Cer t ain quest ion have t o be solved accor ding t o t heir fr act ional r at io t hat means t he r at io should be divided aft er solving fr act ion. Example 1. A and B can complet e a wor k in 2 & 3 days r espect ively. They complet e a wor k t oget her and get ` 50 as t he wages. H ow should t hey divide it ? Solution. ` 50 is not being divided in t he r at io of 2 : 3 as A who can complet e t he wor k in 2 days is mor e

Percentage and Ratio & Proportion

2.11

efficient t hen B who complet ed t he wor k in 3 days. Divide ` 50 in t he r at io of 3:2 1 1 =3:2 : = 6 2 3 So A should get ` 30 and B should get ` 20

Efficiency of A : Efficiency of B =

Example 2. A, B and C can complet e a wor k in 2, 3 and 4 days r espect ively. I f t hey complet e t he wor k t oget her, in what r at io t hey should divide t he money ? 1 1 1 : : = 6:4:3 = 6 : 4 : 3 2 3 4 12 N ote : Ever yt hing which is inver sely pr opor t ional should be divided in fr act ional r at io.

Solution. Efficiency of A : Efficiency of B : Efficiency of C =

Example 3. The M ahar aja of an I ndian St at e invit ed t hr ee ar t ist s t o per for m at his cour t . H e was much impr essed, and honor ed t hem wit h t he gold and gift s. On t he day of t heir depar t ur e he pr esent ed t he t hr ee of t hem wit h sevent een elephant s, expr essing a wish that t he oldest of t he ar t ists should get one-half of t he number of elephant s, t he middle one one-t hir d of t he number, and t he last one one-nint h. H ow t he elephant s will be divided among t hem? Solutilon. Rat io in which t hr ee ar t ist s should divide 17 elephant s

1 1 1 9: 6: 2 : : = 2 3 9 18 So oldest ar t ist s should get 9 elephant s middle one 6 and t he last 2 elephant s r espect ively. =

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. I f (4x 2 – 3y 2) : (2x 2 + 5y 2) = 12 : 19, t hen x : y is (a) 2 : 3

(b) 1 : 2

(c) 3 : 2 (d) 2 : 1 x x 2. I f = , t hen (x+ 5) : (y + 8) is equal t o 5 8 (a) 3 : 5 (b) 13 : 8 (c) 8 : 5

(d) 5 : 8

3. I f x : y = 6 : 5, then ( 5x + 3y) : ( 5x – 3y) is equal to (a) 2 : 1

(b) 3 : 1

(c) 5 : 3

(d) 5 : 2

4. What same number must be added t o each t er m of t he r at io 7 : 3 so t hat t he r at io becomes 2 : 3? (a) 1

(b) 2

(c) 5

(d) Can’t be det er mined

5. The r at io of t he t wo number s is 3 : 4 & t heir sum is 420. The gr eat er of t he t wo number s is (a) 175

(b) 200

(c) 240

(d) 315

6. Five bananas and four apples cost as much as three bananas and seven apples. The r at io of t he cost of one banana t o t hat of one apple is (a) 3 : 2

(b) 4 : 3

(c) 3 : 4

(d) 1 : 3

7. T h e speeds of t h r ee car s ar e i n t h e r at i o 5 : 4 : 6. The r at io bet ween t he t imes t aken by t hem t o t r avel t he same dist ance is (a) 5 : 4 : 6

(b) 6 : 4 : 5

(c) 10 : 12 : 15

(d) 12 : 15 : 10

8. A dog t akes 3 leaps for ever y 5 leaps of a har e. I f one leap of t he dog is equal t o 3 leaps of t he har e, t he r at io of t he speed of t he dog t o t hat of t he har e is (a) 8 : 5

(b) 9 : 5

(c) 8 : 7

(d) 9 : 7

9. ` 180 cont ained in a box consist s of one r upee, 50 paise and 25 paise coins in t he pr opor t ion of 2 : 3 : 4. What is t he number of 50 paise coins? (a) 120

(b) 150

(c) 180

(d) 240

10. I n a school, 10% of t he boys ar e same in number as

1 of t he gir ls and 10% of t he gir ls ar e same in 4

number as 1/25 of t he boys. What is t he r at io of boys t o gir ls in t hat school? (a) 3 : 2

(b) 5 : 2

(c) 2 : 1

(d) 4 : 3

2.12

Percentage and Ratio & Proportion

11. Two number s ar e in t he r at io 3 : 4 and t he pr oduct of t heir L .C.M . & H .C.F is 10800. The sum of t he number s is (a) 180

(b) 210

(c) 225

(d) 240

12. The ages of x and y ar e in t he r at io of 3 : 1. Aft er 15 year s t he r at io will be 2 : 1. Their pr esent ages ( in year s) ar e (a) 30, 10

(b) 45, 15

(c) 21, 7

(d) 60, 20

6. In two vessels A and B, milk and water are in the ratio of 4 : 3 and 3 : 5 respectively. The ratio in which these are to be mixed to obtain new mixture which contains half milk and half water is (a) 7 : 4

(b) 7 : 8

(c) 1 : 2

(d) 4 : 5 [RRB JE 2015 26 th AU G 1 st SH I FT ]

7. If a : b = 4: 5 and b : c = 2 : 3, then c : a is

13. Gold is 19 t imes as heavy as wat er and copper is 9 times as heavy as water. I n what r atio should ther e be mixed to get an alloy 15 times as heavy as water

(a) 15 : 8

(b) 3 : 4

(c) 8 : 15

(d) 3 : 5 [RRB JE 2015 26 th AU G 2 nd SH I FT ]

8. If a : b = 5 : 6 and b : c = 3 : 4, then c : a is

(a) 1 : 1

(b) 2 : 3

(a) 4 : 5

(b) 5 : 4

(c) 1 : 2

(d) 3 : 2

(c) 8 : 5

(d) 5 : 8

14. 85 lit er s of a mixt ur e cont ains milk and wat er in t he r at io 27 : 7. H ow much mor e wat er is t o be added t o get a new mixt ur e cont aining milk and wat er in t he r at io 3 : 1? (a) 5 lt .

(b) 6.5 lt .

(c) 7.25 lt .

(d) 8 lt .

[RRB JE 2015 26 th AU G 3 rd SH I FT ]

9. If a : b = 2 : 3 and a + b = 45. then a is equal to (a) 27

(b) 25

(c) 18

(d) 9 [RRB JE 2015 27 th AU G 3 rd SH I FT ] 2

10. If 2, x, x , 4 are in proportion, then x is equal to

LEVEL-1 1. Sachin is younger than Rahul by 4 years. If their ages are in the ratio of 7 ; 9, then how old is Sachin ? (a) 14 years

(b) 21 years

(c) 18 years

(d) 25 years [RRB JE 2014 RED SH I FT ]

2. The sum of two numbers is 40 and the difference of these two numbers is 4. Find the ratio of these two numbers.

(a) 2

(b) 4

(c) 8

(d) 16 [RRB JE 2015 27 th AU G 3 rd SH I FT ]

11. In two vessels A and B, spirit and water are in the ratio 2 : 1 and 2 : 3 respectively. The ratio in which these ire mixed which contains half water is (a) 5 : 3

(b) 3 : 5

(c) 2 : 3

(d) 3 : 2

(a) 11 : 9

(b) 11 : 18

[RRB JE 2015 27 th AU G 3 rd SH I FT ]

(c) 22 : 9

(d) 17 : 13

12. The average of 20 observations is 18 and average of 30 observations is 25. The average of all 50 observations is

[RRB JE 2014 RED SH I FT ]

3. IfA exceeds B by 40% and B is less than C by 20%. then A : C = ? (a) 3 : 1

(b) 3 : 2

(c) 26 : 25

(d) 28 : 25

4. If a : b = 3 : 5 and b : c = 2 : 3, then a : c is (b) 2 : 5

(c) 5 : 2

(d) 5 : 8

(b) 21.8

(c) 22.2

(d) 22.5 [RRB JE 2015 27 th AU G 3 rd SH I FT ]

[RRB JE 2014 YEL L OW SH I FT ]

(a) 3 : 2

(a) 21.5

13. If 4, x, 2x, 32 are in proportion, then x is equal to (a) 8 2

(b) 8

(c) 16

(d) 16 2 [RRB JE 2015 28 th AU G 2 nd SH I FT ]

[RRB JE 2015 26 th AU G 1 st SH I FT ]

5. Which of the following numbers are in proportion (a) 12, 27, 54, 24

(b) 27, 12, 24, 54

(c) 12, 27, 24, 54

(d) 54, 24, 12, 27

[RRB JE 2015 26

th

14. If 8, 3x, 6, 27 are in proportion, then x is equal to (a) 6 (c) 12

AU G 1 SH I FT ] st

(b) 9 (d) 15 [RRB JE 2015 28 th AU G 3 rd SH I FT ]

Percentage and Ratio & Proportion

15. In two vessels A and B, the milk and water are in the ratio 5 : 4 and 3 : 5 respectively. The ratio in which these are mixed to obtain new mixture which Contains half milk and half water is

2.13

7. X is 40 year s ol d and Y i s 60 year s old H ow many year s ago was t he r at io of t hei r ages 3:5? (a) 5 year s

(b) 10 year s

(c) 20 year s

(d) 37 year s

(a) 9 : 4

(b) 4 : 9

[RRB SSE 2014 YEL L OW SH I FT ]

(c) 4 : 3

(d) 3 : 4

8. Rs. 680 i s divided among A, B, C such t hat A get s

[RRB JE 2015 28 th AU G 3 rd SH I FT ]

LEVEL-2 1. I f a : b = 4 : 3 and b : c = 7 : 9, t hen a : b : c : ? (a) 24 : 21 : 30 (b) 12 : 15 : 21 (c) 8 : 6 : 12 (d) 28 : 2l : 27 [RRB SSE 2014 GREEN SH I FT ]

2 3 2. Ravi spends of his salar y on H ouse Rent ; 5 10 1 of hi s sal ar y on Food and of hi s Sal ar y on 8 Con v ey an ce. A f t er t h i s, h e i s l ef t w i t h Rs. 1400. Find his expendit ur e on Food. (a) Rs. 8000 (b) Rs. 3200 (c) Rs. 2400 (d) Rs. 1000

2 1 of what B get s and B get s of what C get s. 3 4 Then t hei r shar es ar e r espect i vely (a) Rs. 75, Rs. 325, Rs. 280 (b) Rs. 80, Rs. 120, Rs. 480 (c) Rs. 90, Rs. 210, Rs. 380 (d) Rs. 100, Rs. 200, Rs. 380 [RRB SSE 2014 YEL L OW SH I FT ]

9. X. Y and Z st ar t a busi ness X invest s 3 t i mes as much as Y invest s and Y invest s

2 r d of what Z 3

invest s. Then t he r at io of capit als of X. Y, Z is 3

[RRB SSE 2014 RED SH I FT ]

(a) 3 : 9 : 2

(b) 6 : 10 : 15

3. A sum of Rs. 312 was di vi ded among 60 boys and some gi r l s in such a way t hat each boy get s Rs. 3.60 and each gir l Rs. 2.40. The number of gi r ls is (a) 35 (b) 40 (c) 60 (d) 65

(c) 5 : 3 : 2

(d) 6 : 2 : 3

[RRB SSE 2014 YEL L OW SH I FT ]

4. Some st udent s planned a t r ip. The budget for food was Rs. 500 But , 5 of t hem fai led t o go and t hus t he cost of food for each member incr eased by Rs. 5. H ow many st udent s at t ended t he t r ip ? (a) 15 (b) 20 (c) 25 (d) 30 [RRB SSE 2014 YEL L OW SH I FT ]

5. I n a cl ass, t her e ar e t wo sect i ons A and B. I f 10 st udent s of sect ion B shift over t o sect i on A, t he st r engt h of A becomes t hr ee t i mes t he st r engt h of B. But , if 10 st udent s shift over fr om A t o B, bot h A and B ar e equal in st r engt h. H ow many st udent s ar e t her e in sect ions A and B ? (a) 50 and 30 (b) 45 and 15 (c) 90 and 40 (d) 80 and 40 [RRB SSE 2014 YEL L OW SH I FT ]

6. Si x per sons went t o a hot el for meal s. Fi ve of t hem spent Rs. 32 each on t heir meal s while t he 6t h per son spent Rs. 80 mor e t han t he aver age expendit ur e of all t he si x. Tot al money spent by al l t he per sons is : (a) Rs. 192 (b) Rs. 240 (c) Rs. 288 (d) Rs. 336 [RRB SSE 2014 YEL L OW SH I FT ]

[RRB SSE 2014 YEL L OW SH I FT ]

10. Ent r y fee t o an exhibit ion was Rs.80. L at er, t his was r educed by 25% which incr eased t he sal e by 20%. The per cent age i ncr eased in t he number of vi si t or s is (a) 30

(b) 40

(c) 60

(d) 80 [RRB SSE 2015 1 st SEP 1 st SH I FT ]

11. I f a : b = 8 : 15 , b : c = 5 : 8 and c : d = 4 : 5, t hen b : d is (a) 1 : 2

(b) 1 : 3

(c) 4 : 15

(d) 5 : 8 [RRB SSE 2015 1 st SEP 1 st SH I FT ]

12. A per son gave

1 par t his income t o hi s son and 5

40 % par t of his i ncome t o his daught er. H e lent out t he r emai ni ng money i n t hr ee t r ust s A, B and C i n t he r at io of 5 : 3 : 2. I f t he differ ence bet ween t he amount got by son and daught er is Rs. 50,000, how much amount di d he invest in t r ust B? (a) Rs. 20000

(b) Rs 30000

(c) Rs 40000

(d) Rs 50000 [RRB SSE 2015 1 st SEP 1 st SH I FT ]

2.14

Percentage and Ratio & Proportion

13. Two all oys A and B cont ain zinc and copper in t he r at io 5 : 6 and 7 : 8 r espect ively. I f equal quant it ies of all oys ar e melt ed t o for m a t hir d all oys C, t hen t he r at io of copper and zinc i n C wil l be (a) 76 : 89

(b) 89 : 76

(c) 48 : 35

(d) 35 : 48

15. A sum of Rs.16500 is t o be divided among A, B, C and D i n such a way t hat t he r at i o of shar es of A and B i s 3:4, t hat of B and C i s 1:3 and t hat of C and D is 6:7. Sum of shar es of A and D is (a) Rs.8000

(b) Rs.7500

(c) Rs.8500

(d) Rs.9000 [RRB SSE 2015 2 nd SEP 1 st SH I FT ]

[RRB SSE 2015 1 SEP 1 SH I FT ] st

st

14. I f A : B = 2 : 3, B : C = 5 : 6 and C : D 8 : 9, t hen A: D is (a) 2 : 9

(b) 20 : 81

(c) 20 : 27

(d) 40 : 81 [RRB SSE 2015 1 st SEP 3 rd SH I FT ]

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (c) 11. (b)

2. (d) 12. (b)

3. (b) 13. (d)

4. (d) 14. (a)

5. (c)

6. (a)

7. (d)

8. (b)

9. (a)

10. (b)

7. (a)

8. (c)

9. (c)

10. (a)

7. (b)

8. (b)

9. (d)

10. (c)

LEVEL-1 1. (a) 11. (b)

3. (d) 13. (b)

2. (a) 12. (c)

4. (b) 14. (c)

5. (c) 15. (a)

6. (a)

LEVEL-2 1. (d) 11. (a)

2. (c) 12. (b)

3. (b) 13. (b)

4. (b) 14. (d)

5. (a) 15. (c)

6. (c)

EXPLAN ATI ON S LEVEL-1 1.

6.

Applying allegation for milk we get:

S : R = 7X : 9x

A

 9x – 7x = 4

4 7

or x = 2

B

3 8

Hence, S or age of Sachin

1 2

= 7 × 2 = 14 years. 2.

3.

 a = 22 and b = 18

1 8

Hence, a : b = 11 : 9

Hence, required ratio =

a +b = 40 and a – b = 4

L et C = 100, such t hat B = 80 and A = 80  1.4 = 112

7.

Given: a : b = 3 : 5 and b : c = 2 : 3  a : b = 6 : 10 and b : c = 10 : 15

5.

1 1 : = 7 : 4. 8 14 a : b = 4 : 5 and b : c = 2 : 3 a : b = 8 : 10 and b : c = 10 : 15

H ence, A : C = 112 : 100 = 28 : 25

4.

1 14

 8.

c : a = 15 : 8

a : b = 5 : 6; b : c = 3 : 4 = 15 : 18

= 15 : 20

(making the value of ‘b’ equal in both the ratios)

 c : a = 24 : 15

 a : c = 6 : 15 = 2 : 5.

=8:5

Among the given options only 12, 27, 24 and 54 are in proportion i.e. 12 : 27 = 24 : 54 = 4 : 9.

= 18 : 24

Percentage and Ratio & Proportion

9.

a 2  a + b = 45 b 3 a : b = 2x : 3x  5x = 45

 Money left = 1 

33x 7x   1400 40 40

 x = 8000

 a = 2x = 18 10.

2, x, x2, 4 are in proportion  2x2 = x × 4 x=2

11.

spirit 

A

B

2 3

2 5

1 10

 ratio =

13.

1 6

1 1  10 6

500 500  =5 x 5 x

 No. of students who attended the trip = x – 5 = 20 Let the no. of students in class B = y According to question

20  18  30  25 Average = = 22.2 50 4, x, 2x, 32 are in propertion  4 × 32 = 2x × x x=8

15.

3  60  3.6 3. No. of girls  2 = 40 2.4 4. Let the no. of students initially = x

5. Let the no. of students in class A = x

 Product of extremes = product pf weavs

14.

3  8000  2400 10

 x = 25

= 6 : 10 = 3 : 5 12.

 Money spent on food 

According to question

1 2

Product of means = product of extremes

(x + 10) = 3(y – 10)

… (i)

also (x – 10) = y + 10

… (ii)

Solving eq. (i) and (ii), we get x = 50 and y = 30 6. Let avg expenditure be x.  x

32  5   80  2 6

 3x × 6 = 8 × 27

 x = 48

 x = 12

 Total money spent = (32 × 5) + (80 + 48)

Let quantity of liquid of A m mixture = x

= 288.

Let ________________ B __________ = y  Milk from A + Milk from B = total milk

5 3 1  x  y  x  y 9 8 2

x 9   9:4 y 4

7. According to question,

40  k 3  60  k 5  k = 10 years 8. According to question,

LEVEL-2 43  47 : 37 1. a : b   4 : (3)  7    b : c   7 : (9)  3   and 7 : 9  7  3 : 9  3

A

2 B 3

B

3 A 2

 a : b = 28 : 21 b : c = 21 : 27

and B 

 a : b : c = 28 : 21 : 27 2. Let the salary of Ravi be Rs. x Total expenditure 

2.15

2 3 1 33 x x x  x 5 10 8 40

1 C 4

C  4B  4 

3 A 2

 C  6A

Now A + B + C = 680

2.16

Percentage and Ratio & Proportion

A 

13. Let quantity of alloy A = x

3 A  6A  680 2

Let quantity of alloy B = y

 A = 80

Zinc in alloy (A + B) =

 B = 120

5x 7y  11 15

 C = 480 copper in alloy (A + B) 

9. According to question, x = 3y  x: y=

y

3 2 6   1 2 2

6x 8x  copper in C 11 15 89  5x 7y  Zinc in C 76  11 15

2 z 3

 x = y (quantities are same)

 y:x=2:3  x:y=6:2 14.

 x:y:z=6:2:3 10. Entry fee × No. of visitors = sales. multiplying factor 

x

6x 8y  11 15

3 6 x  4 5

3 8  3 in no. of   1    increase of 5   5 5

persons  60 % increase

A 2 B 5 C 8  ,  ;  B 3 C 6 D 9

A B C 2 5 8      B C D 3 6 9 

A 40  D 81

15. A : B  3 : 4  B : C  1 : 3   A : B : C  3 : 4 : 12 C : D  6 : 7  A : B : C = 3 : 4 : 12

a 8 b 5 c 4  ;  ;  11. b 15 c 8 d 5 

b c 5 4 1     c d 8 5 2

Amount received by son = 20%  12. Amount received by daughter = 40%  

Amount given to trust = 40 % Difference = 20% is Rs. 50,000 40% = Rs. 1,00,000. A:B:C=5:3:2  B gets

3  1,00,000  Rs. 30,000 10

C : D = 12 : 14  A : B : C : D = 3 : 4 : 12 : 14 sum of shares of A and D is  3  14     16500  8500  33 



Problems On Age

3

Problems On Age

CHAPTER Quicker M at hemat ical Appr oach FORM U LA 1. t 1 year s ear lier, t he fat her 's age was  t imes t hat of his son. At pr esent t he fat her 's age is  t imes t hat of his son. What ar e t he pr esent ages of t he son and t he fat her ? Son's age =

t1 (  1) (  )

Example. At pr esent t he age of fat her is five t imes t hat of t he age of his son. Thr ee year s hence, t he father 's age would be four times that of his son. Find t he pr esent ages of fat her and t he son. Sol ut i on.

3.1

Son's age =

3  (4  1) = 9 year s 54

and Fat her 's age = 5  9 = 45 year s. FORM U LA 2. The pr esent age of t he fat her is  t imes t he age of his son aft er t 2 year s, t he fat her 's age becomes  t imes t he age of his son. What ar e t he pr esent ages of t he fat her and his son ? Son's age =

(  1) t 2 

Example. Thr ee year s ear lier fat her was 7 t imes as ol d as hi s son. Aft er t hr ee year s t he fat her 's age woul d be four t i mes t hat of hi s son. What ar e t he pr esent ages of the father and son ? Solution. Son's age = =

3 (4  1)  3 (7  1) 74 9  18 = 9 yr s. 3

FORM U LA 3. t 1 year s, ago t he age of t he fat her was  t i mes t he age of hi s son. Aft er t 2 year s, t he age of t he fat her becomes  t i mes t he age of hi s son. What ar e t he pr esent ages of t he son and t he fat her ? t (  1)  t 1 (  1) Son's age = 2  Fat her ’s age =

 ( y  1) (t1  t 2 ) 

Example. Thr ee year s ago t he fat her was 7 t imes as old as his son. Aft er t hr ee year s t he fat her 's age would be four t imes t hat of his son. What ar e t he pr esent ages of t he fat her and t he son ? 3 (4  1)  3 (7  1) Solution. Son's age = 74 9  18 = = 9 years. 3 7(4  1)(3  3) Fat her ’s age = 74 736 = 3 = 42 years. Example. The age of man is 4 t imes t hat of his son. 5 year s ago, t he man was nine t imes as old as his son was at t hat t i me. What i s t he pr esent age of t he man ? Solution. Son's age =

b

g b9  4 g

5 91

= 8 year s

 Fat her 's age = 4  8 = 32 years Example. 10 year s ago, Asha's mot her was 4 t imes older t han her daught er. Aft er 10 year s, t he mot her wi ll be t wice older t han t he daught er. What i s t he pr esent age of Asha ? Solution. Asha's age =

10  2  1  10 (4  1) 42

= 20 years FORM U LA 4 : (DI RECT FORM U LA). Daught er 's/ son's age

Total age  Number of years ago (Times  1) Times  1 Example. The sum of t he age of a mot her and her daught er is 60 year s. Also 6 year s ago, t he mot her 's age was 8 t imes t he age of t he daught er. What ar e t he pr esent ages of t he mot her and t he daught er ? =

Solut ion. Daught er 's age =

60  6 (8  1) 81

= 11.33 years and M ot her 's age = 48.67 years.

3.2

Problems On Age

Example. The sum of t he age of son and fat her is 56 year s. Aft er 4 year s t he age of t he fat her will be t hr ee t imes t hat of t he son. What is t he age of t he son ? Solution. Son's age =

Total age  Number of years after (Times  1) Times  1

=

56  4 (3  1) 31

48 4 = 12 years. N ote : When the question deals wit h ‘ago’ a ‘+’ ve sign exists and when it deals with ‘after’ a ‘– ’ ve sign exists in t he numer at or s in for mula 4. FORM U LA 5 : Pr esent age = Fat her : son = x : y Aft er t year s =  :  =

t hen, Son's age = y  Fat her 's age = x 

t (  ) difference of cross product

t (  ) difference of cr oss product

E xample. The r at i o of t he age of t he fat her and t he son at pr esent is 6 : 1, aft er 5 year s t he r at io will become 7 : 2. What is t he pr esent age of t he son ? Sol ut i on Pr esent age = Fat her : 6, Son : 1 Aft er 5 year s = 7 : 2 Son's age = 1 

5 (7  2) = 5 year s. 6  271

Fat her 's age = 6 

5 (7  2) = 30 years. 6  271

N ote : While calculat ing t he differ ence t he cr oss pr oduct always t akes t he ‘+ ’ ve sign.

PRACTI CE EXERCI SE 1. The r at io of t he ages of t he fat her and t he son at pr esent is 7:1. After 4 year s, the r atio will become 4 : 1. What is t he sum of t he pr esent ages of t he fat her and t he son ? (a) 29 year s (b) 35 year s (c) 32 year s (d) None of t hese 2. I f 10 year s ar e subt r act ed fr om t he pr esent age of Ram and t he r emainder divided by 14, t hen you would get t he pr esent age of his gr andson Shyam. I f Shyam is 9 year s younger t o Sunder whose age is 14, t hen what is t he pr esent age of Ram? (a) 80 year s (b) 70 year s (c) 60 year s (d) None of t hese

1 of B’s age. B’s age will be t wice of 6 C’s age aft er 10 year s. I f C’s eight h bir t hday was celebr at ed t wo year s ago, t hen t he pr esent age of A must be (a) 5 year s (b) 10 year s (c) 15 year s (d) 20 year s 4. Sur esh is half his fat her ’s age. Aft er 20 year s, hi s fat her ’s age wi l l be one and a hal f t i mes Sur esh’s age. What is his fat her ’s age now? (a) 40 year s (b) 20 year s (c) 26 year s (d) 30 year s 3. A’s age is

1 t he age of his fat her K eit h now,, 3r d th and was 1/4 t he age of his fat her 5 year s ago, then how old will his fat her Keith be 5 year s fr om now ? (a) 20 year s (b) 45 year s (c) 40 year s (d) 50 year s 6. The ages of t he t wo per sons differ by 20 year s. I f 5 year s ago, t he older one be 5 t imes as old as t he younger one, t hen t hei r pr esent ages, i n year s, ar e (a) 25, 5 (b) 30, 10 (c) 35, 15 (d) 50, 30 7. Rajan got mar r ied 8 year s ago. His pr esent age is 6/5 times his age at t he t ime of his mar r iage. Rajan’s sister was 10 year s younger to him at the time of his mar r iage. The age of Rajan’s sister is (a) 32 year s (b) 36 year s (c) 38 year s (d) 40 year s 8. A fat her ’s age is t hr ee t imes t he sum of t he ages of his t wo childr en, but 20 year s hence his age will be equal t o t he sum of t heir ages. Then t he fat her ’s age is (a) 30 year s (b) 40 year s (c) 35 year s (d) 45 year s

5. I f Dennis is

Problems On Age

9. Rat io of Ashok’s age t o Pr adeep’s age is 4 : 3. Ashok will be 26 year s old aft er 6 year s. H ow old is Pr adeep now? (a) 18 year s (b) 21 year s (c) 15 year s (d) 24 year s

3.3

10. Sonu is 4 year s younger t han M anu while Dolly 1 is four year s younger t han Sumit but t imes 5 as old as Sonu. I f Sumit is eight year s old, how many t imes as old is M anu as Dolly? 1 (a) 6 (b) 2 (c) 3 (d) None of t hese

AN SWERS 1. (c)

2. (a)

3. (a)

4. (a)

5. (d)

6. (b)

7. (c)

8. (a)

9. (c)

10. (a)

EXPLAN ATI ON S 1.

Fat her : Son = x : y = 7 : 1 After t = 4 year = :  = 4 : 1 Son’s age =

y  t ( – ) differ ence of cross pr oduct

Now C’s eight bir t h day is celebr at ed t wo year s ago is pr esent age of C = 10 year s = 2x  x=5  Pr esent age of A = 1  x = 1 5 = 5 year s 4. L et fat her ’s pr esent age be x and t hat of Sur esh be y.

=

1  4(4 – 1) 7 1– 4 1

=

143 = 4 year s 3



x  t (  ) differ ence of cr oss product

and

and Fat her ’s age = =

x  10 =y ...(1) 14 Now, Shyam is 9 years younger to sunder whose age  Pr esent age of Shyam (y ) = 14 – 9 = 5 year s. Put t ing t he value of y in equat ions (1) will get x – 10 =5 14  x = 80 year s. H ence pr esent age of Ram is 80 year s. 3. A B C 10 : 60 : 20  5x : 30x : 10x  1x : 6x : 2x

1 x 2

...(i )

3 (y + 20) 2 Fr om (i ) and (ii ), we have

7  4(4  1) 7  1 4  1

743 = = 28 year s 3 H ence sum of t he ages of fat her and son = (28 + 4) = 32 year s 2. L et t he Ram’s pr esent age be x year s. and Shyam’s pr esent age be y year s. Now accor ding t o quest ion

y= x + 20 =

5.

IJ K

3 x  20 2 2 4x + 80 = 3x + 120 x = 40.

x + 20 =

 

FG H

...(ii )

Dennis = Dennis – 5 =

1 (Keith) 3 1 (K eit h – 5) 4

5 1 1 K eit h – 5 = K eit h – 4 3 4  Keit h = 45 H en ce K ei t h i s 50 year s ol d, 5 year s f r om now. 6. L et ages of t wo per sons be x and y .  x – y = 20 ...(i ) [ x > y ] x – 5 = 5(y – 5)  x – 5y = – 20 ...(ii ) Solving equat ions (i ) and (ii )  y = 10, x = 30 

3.4

7.

Problems On Age

L et t he age at t he t ime of mar r iage be x . Then

6 x+8= x 5 Solving, we get x = 40. Age of his sist er = 40 – 10 + 8 = 38. 8. L et t he sum of ages of t wo childr en = x year s, L et t he fat her ’s age = y year s  y = 3x ...(i ) 20 year s hence, y + 20 = x + 20 + 20 ...(ii ) Solving (i ) and (ii ), we get y = 30 year s.

9. Ashok’s pr esent age = 26 – 6 = 20 year s

Ashok' s age 4  Pr adeep' s age 3 H ence Pr adeep’s age = 15 year s 10. Age of Sonu + 4 year s = Age of M onu Age of Dolly+ 4 year s = Age of Sumit I f Sumit is 8 year old, t han Dolly is 4 year old and Sonu is 20 year old. M onu = 24 year 

t hen

M onu 24 = =6 Dolly 4 

4

Alligations & Mixtures

CHAPTER

ALLI GATI ON Wor d Alligation lit er ally means ‘linking', This r ule enables us t o find t he mean or aver age value of mixtur es when t he pr ices of t wo or mor e ingr edient s which may be mixed t oget her and t he pr opor t ion in which t hey ar e mixed ar e given. Rule of Alligat ion. I f t wo ingr edient s ar e mixed, t hen C.P. of dear er – Mean pr ice Quant it y of cheaper = M ean pr ice – C.P. of heaper Quantity of dearer We represent it as under

C.P. unit quant it y of cheaper (c)

C.P. unit quant it y of dear er (d) mean pr ice (m )

(d – m )

(m – c)

Cheaper quant it y : dear er quant it y = (d – m ) : (m– c) Solut ion of all t he quest ions given above can be done by using fundament als of t he same chapt er s but Alligat ion r ule incr eases t he speed and accur acy. Simult aneously enhances your I .Q. Alligat ion is not a chapt er but it 's a met hod or r ule t o solve quest ions based on weight ed aver age. Some examples ar e given below. So all t he above quest ions ar e fr om differ ent chapt er s. The fir st quest ion is fr om Simple I nt er est , second quest ion fr om t he Per cent ages and so on but t he met hod we use is t he same ‘Alligat ion'.

PRACTI CE EXERCI SE 1. Two vessels A and B ar e filled wit h dilut e sulfur ic aci d (i .e. m i xt u r e of su l f u r i c aci d an d wat er ). Vessel A h as t h e r at i o of aci d an d w at er i n 1 : 2, whi l e t he vessel B has t he r at io of aci d and wat er i n 3 : 1. To pr epar e 5 l i t r e mi xt ur e, di l ut e sulfur ic acid containing equal amount of acid and water (a) 2 and 3 lit r e (b) 3 and 2 lit r e (c) 1 and 4 lit r e (d) 4 and 1 lit r e 2. The wei ght of 1 l i t r e wat er i s 1 k g and 1 l i t r e of anot her liquid in 1.340 kg. The mixtur e weight of t he t wo l i qui ds i s 1.270 k g/l i t r e. The r at i o of t he two in 1 litr e mixture is (a) 27/34 : 7/34 (b) 7 : 27 (c) 27 : 7 (d) none of t hese

3. The cost of pur e milk is Rs. 12 per lit r e. Wat er is added t o it and t hen mixt ur e is sold at Rs. 13.75 per lit r e and t hus 25% of pr ofit is r ealized. The r at io of milk and wat er in t he mixt ur e is (a) 12 : 1 (b) 10 : 1 (c) 11 : 1 (d) none of t hese 4. I n br and A t he r at io of t wo t ypes of oils ar e 4 : 3 and in br and B t he r at io of t hese t wo oils is 2 : 3 H ow much kg of br and A should be mixt ur e wit h 5 kg of br and B t o gain t he r at io of t wo oils in 5:4 (a) 80 (b) 90 (c) 49 (d) None of t hese

4.2

Alligations & Mixtures

5. A milkman sells milk on CP aft er adding wat er. I f he had bought it at 85 paise per lit er and pr ofit 1 r ealized is 11 %, t hen quant ity of wat er added is 9 (a) 1 lit r e (b) 2 lit r e (c) 2.5 lit r e (d) none of t hese 6. A mer chant buys tea A @Rs.8 per kg and t ea B @ Rs. 14 per kg. The two br ands of tea, he mixed and sells the mixtur e @Rs. 12 per kg. I f pr ofit r ealized is 20%, then r atio of two br ands of tea in mixture is (a) 15 : 2 (b) 1 : 16 (c) 2 : 1 (d) none of t hese 7. A, B , C and D br ands of t ea ar e pur ch ased @Rs. 12, 15, 18 and 21 per kg r espectively mixt ur e i n t he r at i o 4 : 3 : 2 : 1. What i s t he SP per k g of t he mi xed, so t hat 10% of pr ofi t can be obt ai ned ? (a) 6 (b) 7

11. An alloy contains 90% copper and 10% tin. Another alloy has 93% copper and 4% t in. I n what r at io t hese t o be mixed so t hat t he mixt ur e must have 9% t in ? (a) 5 : 1 (b) 1 : 5 (c) 2 : 7 (d) None of t hese 12. Gold in 19 t imes as heavy as wat er. Copper is 9 t imes as heavy as wat er. I n what r at io t hese t wo be mi xed, so t hat wei ght of t he mi xt ur e be 15 t imes as heavy as wat er ? (a) 2 : 3 (b) 1 : 5 (c) 3 : 2 (d) None of t hese 13. I n a m i xt u r e, t h e r at i o of wi n e t o wat er i n 3 : 2. I n anot her mi xt ur e, t he r at i o of wi ne t o wat er is 4 : 5. What quantit y (in gallons) of second mixt ur e shoul d be mi xed wit h t he gal l ons of fir st mi xt ur e, so t hat t he r at i o of wi ne and wat er be 1 : 1 ?

(c) 16.50 (d) None of t hese 8. Two alloys of silver and copper cont ain silver and copper in the r atio 5 : 1 and 7 : 2. What quantities of each should mixed to pr epar e silver being 80% in 5 lbs ? (a) 1 : 4 (b) 2 : 3 (c) 3 : 2 (d) None of t hese 9. To pr epar e 21 lit r e of a 95% pur e solut ion 90% and 97% of pur e acidic solut ion ar e mixed.What quant it y of each solut ion is r equir ed ? (a) 6 and 15 lit r e (b) 10 and 11 lit r e (c) 5 and 16 lit r e (d) none of t hese 10. Two vessels cont ain milk and wat er in t he r at io 3 : 1 and 5 : 3 r espect ively. I n what r at io t hese t o be mixed t o pr epar e a t hir d mixt ur e cont aining mild and wat er in t he r at io 2 : 1 ? (a) 1 : 2 (b) 2 : 1 (c) 1 : 1 (d) None of t hese

(a)

22 5

(b) 5

2 5

(c) 5 (d) None of t hese 14. How much kg of tea (cost Rs. 10.40 per kg) should be mixed wit h tea (cost Rs. 8.8.per kg.) t o pr epar e 15 kg to sell at Rs. 146.40 ? (a) 7 kg (b) 8 kg (c) 9 kg (d) None of t hese 15. 4 li t r es wi ne is wi t hdr awn fr om a vessel ful l of wi ne and agai n fi l l ed wi t h wat er. 4 l i t r es i s w i t h dr awn an d r epl aced w i t h w at er t o f i l l t he vessel . I f r at i o of wi ne and wat er i n t he vessel n ow i s 36 : 13, t h en capaci t y of t h e vessel s i s (a) 23 gallons (b) 49 gallons (c) 28 gallons (d) none of t hese

Alligations & Mixtures

16. 10 lit r e milk is wit hdr awn fr om a bucket full wit h milk and r efilled wit h wat er. The same pr ocess is r epeat ed for t he 4 t imes. I f t he r at io of milk t o wat er is found t o be 16 : 65m t hen capacit y of t he bucket is (a) 81 lit r es 17.

18.

19.

20.

(b) 49 lit r es

(c) 30 lit r es (d) none of t hese I n what r at io, a gr ocer can mix t ea leaves of Rs. 10 per kg and Rs. 15 per kg t o make a mixt ur e of Rs. 12 per kg ? (a) 3 : 2 (b) 2 : 3 (c) 3 : 4 (d) None of t hese A bottle contains mixtur e of spir it and water. Spir it is 18%, 8 lit r es of mixt ur e is t aken out fr om t he bot t le which is again filled wit h wat er. I f pr esent per cent age of spi r i t i s 15%, t hen quant i t y of mixt ur e in t he bot t le is (a) 43 lit r e (b) 47 lit r e (c) 48 lit r e (d) none of t hese Fr om a vessel cont aining 72 lit r e of milk, 9 lit r e of milk is t aken out and wat er is added in place of it . The pr ocess was r epeat ed t wice again. What is t he quant it y of milk r emained in t he vessel ? (a) 3087/64 lit r es (b) 2087/64 lit r es (c) 4087/64 lit r es (d) None of t hese Thr ee ident ical vessels cont ain mixt ur e of milk and wat er. I n t he fir st vessel, r at io of milk and wat er is 5 : 4, in t he second vessel it is 4 : 3 while in t he t hir d vessel it is 3 : 2, The t hr ee mixt ur e wer e mi xed i n a bi gger vessel t o mak e a new mixt ur e. What is t he r at io of milk and wat er in t he new mixt ur e ? (a) 544 : 401 (b) 401 : 544 (c) 501 : 544 (d) 544 : 501

4.3

21. H ow many kg t ea wor t h Rs. 3.60 per kg should be mixed wit h 8 kg of t ea wor t h Rs 4.20 per kg t o get a pr ofit of 10% by selling t he mixt ur e @Rs. 4.40 per kg ? (a) 2 kg (b) 3 kg (c) 4 kg (d) 5 kg 22. I n what r atio must a gr ocer mix t wo var ieties of pulses costing Rs. 15 and Rs. 20 per kg r espectively so as to get a mixtur e wor th Rs. 16.50 per kg ? (a) 3 : 7 (b) 5 : 7 (c) 7 : 3 (d) 7 : 5 23. I n what r at io must a gr ocer mix t wo var iet ies of t ea wor t h Rs. 60 a kg and Rs. 65 a kg so t hat by selling t he mixt ur e at Rs. 68.20 a kg he may gain 10% ? (a) 3 : 2 (b) 3 : 4 (c) 3 : 5 (d) 4 : 5 24. I n what r at io must wat er be mixed wit h milk t o 2 gain 16 % on selling t he mixt ur e at cost pr ice ? 3 (a) 1 : 6 (b) 6 : 1 (c) 2 : 3 (d) 4 : 3 25. Tea wor t h Rs. 126 per kg and Rs. 135 per kg ar e mixed wih a t hir d var iet y in t he r at io 1 : 1 : 2. I f mixt ur e is wor t h Rs. 153 per kg, t hen pr ice of t he t hir d var iet y per kg will be (a) Rs. 169.50 (b) Rs. 170 (c) Rs. 175.50 (d) Rs. 180

AN SWERS 1. (b)

2. (a)

3. (c)

4. (c)

5. (d)

6. (c)

7. (c)

8. (b)

9. (a)

10. (a)

11. (a)

12. (c)

13. (b)

14. (c)

15. (c)

16. (c)

17. (a)

18. (c)

19. (a)

20. (a)

21. (c)

22. (c)

23. (a)

24. (a)

25. (c)

4.4

Alligations & Mixtures

EXPLAN ATI ON S 1 3

1.

H ence 2 pounds and 3 lbs.

3 4

9. 90

1.27 1 4 2.

97

i .e. 3, 2 1 6

95 2

1000

1340

I =

1 2

5

2 5  21  6, I I =  21  15 7 7

H ence (6, 15)

.7

.27

3 4

10. 3.

12

0

11 4 7

4.

7 45

5 9

10 100

11.

2 5

15

10 2

7.

13.

12  4 = 48 15  3 = 45

6

4

3 5

4 9

1 18

18  2 = 36 21  1 = 21

Again,

Cost pr ice = 150 Pr ofit = 15

SP =

1 45

80 4 = 100 5

7 9 1 30

165  Rs. 16.50 10

14.

1 10 3 gallons 5 = ? 9

10.40

8.80

.96

.64

Again



1 2

2 ? = 5 gallons 5



165



1 100

9

19

14

4

4 100

9 100

5 100

1 63

8

8. 5 6

1 12

1

12. 6.

2 3

1 24

11

5 8

3 ? = 32 15 ? = 9 kg

Alligations & Mixtures

15. Let total volume of wine be ‘a’ gallon and ‘b’ gallon is drawn out and refilled with water. Then aft er n t imes. Volume of l eft wine  a  b =   Tot al volume of wine  a 

By t he r ule of alligat ion :

Cost of 1 k g t ea Cost of 1 k g t ea of 2nd t ype of 1st t ype Rs.65 Rs. 60 M ean pr i ce

n

Volume left wine 36 = Volume of wat er in vessel 13 Volume of left wine  Volume of left wine  Volume of wat er

Rs.62

Given:

= 2



 a  b  a 



a4 6 = a 7 a = 28 gallons



36 49

16 10   =  1   65 V V = 30 lit r e

16.

 17.

=

10

36 36 = 36  13 49

 Requir ed r at io = 3 : 2 24. L et C.P of 1 lit r e milk be Re. 1. S.P. of 1 lit r e of mixt ur e = Re. 1

50 %. 3

 C.P. of 1 lit r e of mixt ur e = 100 

2

2

21.

3.6

3 6  1  Rs. . 350 7

By t he r ule of alligat ion : C.P.of 1 l it r e of wat er 0

C.P. of 1 li tr e of mil k Re. 1

M ean pr i ce 6 Re. 7

12 3

2

3

Gain =

15



a = 30 lit r e

1 7

4.2

6 7

4 2 Again 

4.5

 Rat io of wat er and milk =

4 4.40 = 110% = 4 kg

25. Since fir st and second var iet is ar e mixed in equal pr opor t ions, hence t heir aver age pr ice

22. By t he r ule of alligat ion :

=

Cost of 1 k g pulses Cost of 1 k g pulses of 2nd t ype of 1st t ype Rs.15 Rs. 20 M ean pr i ce Rs.16.50

3.50

1.50

 Requir ed r ate = 3.50 : 1.50

126  135 = Rs. 130.50 2

So, mixt ur e i s for med by mixing t wo var iet i es, one at Rs. 130.50 per kg and t he ot her at say, Rs. x per kg in t he r at io 2 : 2 i.e., 1 : 1

By t he r ule of alligat ion : Cost of 1 k g t ea of 1st t ype Rs. 130.50

Rs.153

=7:3

 100   68.20   110 

Cost of 1 k g t ea of 2nd t ype Rs. x M ean pr i ce

= 35 : 15 23. S.P. of 1 kg of mixtur e = Rs. 68.20 Gain = 10%  C.P. of 1 kg of t he mixt ur e

1 6 :  1 : 6. 7 7

(x – 153)



= Rs. 



= Rs. 62.



22.50

x  153 =1 22.50 x – 153 = 22.50 x = 175.50.



5

Time and Work

CHAPTER

TI M E AN D WORK Wor k is defined as t he amount of job assigned or t he amount of job act ually done. Wor k is always consider ed as a whole or 1.



I f A can do a piece of wor k in ‘a' number of days, t hen in one day Con v er sel y, i f a m an does in 1 /

 

 



th

of a w or k i n 1 day, t h en h e can com pl et e t h e w or k

I f A is ‘x ' t imes as good a wor kman as B, t hen he will t ake

1 th of t he t ime t aken by B t o do t he same x

wor k . I f A and B can do a piece of wor k in ‘x ' and ‘y ' days r espect ively, t hen wor king t oget her, t hey will t ake xy x  y th days t o finish t he wor k and in one day, t hey will finish par t of t he wor k. x y xy To compar e t he wor k done by differ ent people, fir st find t he amount of wor k each can do in t he same t ime. I f t he number of men t o do a job is changed in t he r at io a : b, t hen t he t ime r equir ed t o do t he wor k will be in t he r at io b : a, assuming t he amount of wor k done by each of t hem in t he given t ime is t he same, or t hey ar e ident ical. I f t wo men A and B t oget her can finish a job in ‘x ' days and if A wor king alone t akes ‘a' days mor e t han A and B wor king t oget her and B wor king alone t akes ‘b' days mor e t han A and B wor king t oget her, t hen



1 = a days. a

1 a

1 th of t he wor k is done. a

x=

ab .

To do a piece of wor k, t he number of men employed and t he number of days r equir ed t o do t he wor k ar e in inver se pr opor t ion. Also, t he number of men employed and t he hour s wor ked per day ar e in inver se pr opor t ion.

EFFI CI EN CY When we say t hat A can complet e a wor k in 10 days we put 1 day wor k i.e., 1/10 ,but we have t o wor k on it as, an efficiency which is inver sely pr opor t ional t o number of days in which a per son can complet e a piece of wor k. Example 1 A can complet e a piece of wor k in 12 days and B is 60% mor e efficient t han A. I n how many days B wil l complet e t he same wor k ? Sol ut i on. A's efficiency or A's one day wor k = B's efficiency is 60% mor e t han A or

1 12

1 160 2  = 12 100 15

or

15 days = 7.5 days 2

5.2

Time and Work

Example 2 A is t hr ee t imes mor e efficient t han B and t her efor e can complet e one wor k in 60 days less t han B t akes. I n how many days t hey will complet e t he wor k wor king t oget her ? number of days Solution. As A is t hr ee t imes mor e efficient t han B and efficiency = 1 So, if number of days in which A will complet e t he wor k = x Then, B will complet e t he wor k in 3x days  3x – x = 60 or x = 30 days 1 1 1 31  Toget her A's and B's efficiency =    22 days 2 30 90 90 M AN DAYS I f we say a par t icular building of ar ea 1000 sq m has t o be const r uct ed ,what could be appr ox cost defined by t he engineer, he calculat es t he mat er ial cost and labour cost . Now in what for m he calculat es t he labour cost . (a) I n how many days will t he building be complet ed depends on t he number of labour s. (b) H ow many labour s ar e to be used depends on the number of days in which the building is to be completed. So, the engineer calculates mandays i.e. multiplication of labour needed and days in which it will be completed and mandays always r emains const ant for a specific wor k. e.g. A building can be complet ed in 2000 mendays means if t her e ar e 20 men, it t akes 100 days, if t her e ar e 50 men it t akes 40 days, if t her e ar e 100 men it t akes 20 days. Or t he mult iplicat ion of man and days r emain const ant . Example 3 40 men can complet e one wor k in 20 days. I n how many days 50 men can complet e it ? Solution. H er e, number of mandays t o complet e t he wor k = 40  20, i.e. 800 mandays. 800 So if t her e ar e 50 men, t hey can complet e t he wor k in , i.e. 16 days 50 Example 4 50 men can complet e t he wor k in 100 days. I n how many days t he wor k will be complet ed if 10 men st ar t t he wor k and 10 mor e men joined aft er ever y 10 days? Sol ut i on. Number of mandays r equir ed t o complet e t he wor k = 50  100 = 5000 I n fir st 10 days, t he number of mandays = 10  10 = 100 For next 10 days, t he number of mandays = 20  10 = 200 For next 10 days, t he number of mandays = 30  10 = 300 For next 10 days, t he number of mandays = 40  10 = 400 For next 10 days, t he number of mandays = 50  10 = 500 For next 10 days, t he number of mandays = 60  10 = 600 For next 10 days, t he number of mandays = 70  10 = 700 For next 10 days, t he number of mandays = 80  10 = 800 For next 10 days, t he number of mandays = 90  10 = 900 Tot al wor k which has been complet ed = 4500 mandays. For t he r emaining 500 mandays wor k, t her e ar e 100 men so t he number of days in which r emaining wor k will be complet ed =

500 = 5 days 100

So t ot al number of days = 95 days

Time and Work

5.3

FRACTI ON OF WORK DON E BY A + FRACTI ON OF WORK DON E BY B =1 I f A can complet e one wor k in 20 days, t hen in 10 days he can complet e in 5 days he can complet e

10 of t he wor k, 20

5 of t he wor k 20

and in 7 days he can complet e

7 of t he wor k . 20

 Fr act ion of wor k done by A =

Number of days for which A did t he wor k Number of days in which A can complet e t he wor k

Gener ally speaking if A can complet e a wor k in 50 days and did t he wor k for x number of days, x might be anyt hing bet ween 1 t o 50, t hen t he fr act ion of wor k complet ed by A =

x . 50

I f it is known t hat 3 per sons ar e complet ing one wor k, t hen Fr act ion of t he wor k complet ed by 1st + Fr act ion of t he wor k complet ed by 2nd + Fr act ion of t he wor k complet ed by 3r d = 1 Example 5 A can complete the wor k in 12 days while B can complete it in 15 days. They star t the wor k together but A left aft er 4 days. I n how many days t he wor k will be complet ed? Solution. A did t he wor k for 4 days while B did t he wor k for (4 + x ) days, or Fr act ion of t he wor k complet ed by A + Fr act ion of t he wor k complet ed by B = 1

4  x  4  1 12 15



 x  4

or

15

1 

4 12

or

=6

or x=6 so, t ot al number of days t hey t ake t o complet e t he wor k = 4 + 6 = 10 days

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. Rambabu can do a piece of wor k in 5 days and Sant osh can do t he same wor k in 7 days. I f bot h wor k t oget her, t hey will finish t he wor k in how many days? (a) 12 days

(b) 2 days

11 12 days (d) day 12 35 2. Rajan can do a piece of wor k in 10 days, Rakesh can do in 12 days and M ukesh do t he same wor k in 15 days. I n how many days t hey can finish t he wor k, wor king t oget her ? (c) 2

3. Sohan and Rohan can do a piece of wor k in 10 days, Rohan and Mohan can do in 12 days, Sohan and M ohan in 15 days. The t ime t aken by t hem t o do t his wor k t oget her is (a) 16 days

(b) 15 days

(c) 12 days

(d) 8 days

4. Sunil and Sant osh t oget her can finish a wor k in 30 days. They wor ked for it for 20 days and t hen Sant osh left . The r emaining wor k was done by Sunil alone in 20 mor e days, Sant osh alone can finish t he wor k in (a) 52 days

(a) 3 days

(b) 5 days

(b) 30 days

(c) 4 days

(d) 2 days

(c) 48 days (d) 60 days

5.4

Time and Work

5. Rambabu, Santosh and Sunil can finish a piece of work in 10, 12 and 15 days respectively. I f Santosh st ops doing t he wor k aft er 2 days, Rambabu and Sunil will finish t he wor k in (a) 2 days (b) 3 days 1 1 (c) 2 days (d) 3 days 2 2 6. A man under t akes t o do a cer t ain wor k in 150 days. He employs 200 wor ker s. H e discover s that only a quar t er of t he wor k is done is 50 days. I n or der t o complet e t he wor k on shcedule, he must addit ionally employ (a) 50 wor ker s (b) 100 wor ker s (c) 150 wor ker s (d) 200 wor ker s 7. Rajan can do a pi ece of wor k i n 30 days and M ukesh in 20 days. M ukesh alone at it for 10 days and t hen leaves. Rajan alone can finish t he r emaining wor k in (a) 15 days (b) 12 days (c) 14 days (d) 18 days 8. A gr oup of 10 st udent s wor king one hour per day complet e a piece of wor k in 12 days. I f t her e ar e 12 st udent s in t he gr oup and t hey wor k one hour per day, t hen will be able t o complet e t he wor k in (a) 8 days (b) 9 days (c) 10 days (d) 11 days 9. Ram and Sunil can finish a work in 8 and 16 hours repectively. I f they work at it alternatively for an hour Ram beginning first, the work will be finish in 2 3 (a) 8 hour s (b) 10 hour s 3 4 2 1 (c) 10 hour s (d) 11 hour s 3 2 10. 12 men complet e a wor k in 18 days. 6 days aft er t hey had st ar t ed wor king, 4 men joined t hem. H ow many days will all of t hem t ake t o complet e t he r emaining wor k ? (a) 8 days (b) 12 days (c) 13 days (d) 9 days 11. 72 men can build a wall 280 m long in 42 days. The number of per sons who would t ake 36 days t o build a similar will 100 m in lengt h will be (a) 15 (b) 30 (c) 25 (d) 20 12. Sunil can do a wor k in 36 days, Ram in 54 days and Bal u i n 72 days. They st ar t ed wor k i ng together but befor e the wor k was to be over, Sunil left 8 days befor e and 12 days befor e Ram. Balu will complet e t he wor k alone in (a) 20 days (b) 24 days (c) 30 days (d) 34 days

13. Rajan can do a piece of job in 6 days, Sunil in 8 days and M ukesh in 12 days. Sunil and M ukesh wor k ed t oget her for 2 days and t hen Raj an r eplaces M ukesh. Now how long t he new par t ner will have t o wor k t o complet e t he job ? (a) 10 days (b) 7 days (c) 4 days (d) 2 days 14. 25 men wer e employed to do a piece of wor k which they could finish in 20 days. But t he men dr opped off by. 5 at t he end of ever y 10 days. I n what t ime will t he wor k be complet ed ? (a) 17 days

(b) 23

1 days 3

(c) 8 days (d) 26 days 15. I f 10 men can do a wor k in 6 days and 15 women can do t he same i n 5 days, t hen 8 men and 5 women can t oget her do t he wor k in (a) 7 days (b) 6 days (c) 5 days (d) 4 days

LEVEL-1 1. 4 men can complet e a piece of wor k i n 5 days. H ow many men ar e r equir ed t o complet e 3 t imes t he wor k i n 4 days? (a) 5 (b) 15 (c) 80 (d) 20 [RRB JE 2014 GREEN SH I FT ]

2. When Ram and Mohan work together, they complete a work in 4 days. If Ram alone can complete this work in 12 days then in how many days Mohan alone can complete this work ? (a) 10 days

(b) 8 days

(c) 6 days

(d) 16 days [RRB JE 2014 RED SH I FT ]

3. Pipe 'F can fill a tank in 36 hours and pipe 'Q' can fill this tank in 45 hours. If both the pipes are opened simultaneously, then how much time will be taken to fill this tank ?

1 hours 2

(a) 20 hours

(b) 40

(c) 9 hours

(d) 42 hours [RRB JE 2014 RED SH I FT ]

4. To complete a work P takes 50% more time than Q. If together they take 18 days to complete the work, how much time shall Q take to do it? (a) 30 days

(b) 35 days

(c) 40 days

(d) 45 days? [RRB JE 2014 YEL L OW SH I FT ]

Time and Work

5. Two pipes can fill a tank in 20 minutes and 30 minutes respectively. If both the pipes are opened simultaneously, then the tank will be filled in (a) 10 minutes (c) 15 minutes

5.5

12. A and B can do a piece of work in 24 days. If efficiency of A is double than B, then in how many days B alone can do the same work?

(b) 12 minutes

(a) 72

(d) 25 minutes

(b) 60

[RRB JE 2014 YEL L OW SH I FT ]

(c) 36

6. A and B can do a piece of work in 8 days. A alone can do the same work in 12 days. The number of days in which B alone can do the same work is

(d) 30 [RRB JE 2015 27 th AU G 1 st SH I FT ]

(c) 40

13. P and Q can do a piece of work in 12 days, Q and in R in 15 lays and R and P in 20 days. In how many days R alone can do the same work? (a) 70 (b) 60 (c) 45 (d) 30

(d) 48

[RRB JE 2015 27 th AU G 1 st SH I FT ]

(a) 20 (b) 24

[RRB JE 2015 26 th AU G 1 st SH I FT ]

7. P and Q can do a piece of work in 12 days, Q and R in 15 days and R and P in 20 days. In how many days P alone can do the same work? (a) 15 (c) 23.5

(b) 30 (d) 35 [RRB JE 2015 26 th AU G 1 st SH I FT ]

8. A and B can do a piece of work in 24 days. If efficiency of A is double than B, then in how many days, A alone can do the same work? (a) 30

(b) 36

(c) 60

(d) 72 [RRB JE 2015 26

(b) 30

(c) 47

(d) 60

[RRB JE 2015 27 th AU G 2 nd SH I FT ]

15. P and Q can do a piece of work in 10 days. Q and R in 12 days and R and P in 15 days. In how many days R alone can do the same work?, (a) 70 (b) 60 (c) 40 (d) 30 [RRB JE 2015 27 th AU G 2 nd SH I FT ]

th

AU G 2

nd

SH I FT ]

9. P and Q can do a piece of work in 12 days, Q and R in 15 days and R and P in 20 days. In how many days Q alone can do the same work? (a) 20

14. A can do a piece of work in 20 day and B can do the same work in 30 days. If they work together th number of days required to do the same work is (a) 18 (b) 16 (c) 14 (d) 12

LEVEL-2 1. A, B and C can do a piece of wor k in 12, 15 and 20 days r espect i vely. H ow long will t hey t ak e t o fi ni sh t he wor k t oget her ? (a) 10 days (b) 5 days (c) 8 days (d) 12 days

[RRB JE 2015 26 th AU G 2 nd SH I FT ]

[RRB SSE 2014 GREEN SH I FT ]

10. A can do a piece of work in 15 days and B ran do the same work in 10 days. If they work together, number of days required to complete the same work is

2. Pi pe 'P' can fill a t ank i n 10 hour s and Pi pe 'Q' can fil l t hi s t ank in 12 hour s. Pi pe 'R' can empt y t he ful l t ank in 20 hour s. I f all t he t hr ee pipes ar e oper at ed simult aneoulsy, t hen in how much t i me t his t ank wi ll be fill ed ?

(a) 5

(b) 6

(c) 7

(d) 8 [RRB JE 2015 26 th AU G 3 rd SH I FT ]

11. P and O can do a piece of work in 10 days, Q and R in 12 days and R and P in 15 days. In how many days P alone can do the same work? (a) 24

(b) 40

(c) 6

(d)

40 3

[RRB JE 2015 26 th AU G 3 rd SH I FT ]

(a) 7 hour s

(b) 7

1 hour s 2

(c) 8 hour s

(d) 8

1 hour s 2

[RRB SSE 2014 RED SH I FT ]

3. 36 men can compl et e a wor k in 18 days. I n how many days will 27 men complet e t he same wor k? (a) 24 days (b) 12 days (c) 30 days (d) 42 days [RRB SSE 2014 RED SH I FT ]

5.6

Time and Work

4. A man, a woman and a boy can t ogether complet e a pi ece of wor k in 3 days. I f a man alone can do it in 6 days and a boy alone in 18 days, how long wi ll a woman t ak e t o complet e t he wor k? (a) 9 days (b) 21 days (c) 24 days (d) 27 days [RRB SSE 2014 YEL L OW SH I FT ]

5. A t ap can fill a cist er n in 8 hour s and anot her t ap can empt y i t i n 16 hour s. I f bot h t he t aps ar e open, t he t i me t aken t o fi ll t he t ank wil l be (a) 8 hr s. (b) 10 hr s. (c) 16 hr s. (d) 24 hr s. [RRB SSE 2014 YEL L OW SH I FT ]

6. A t eam of 100 wor k er s i s supposed t o do a wor k in 40 days. After 35 days, 100 mor e wor ker s wer e empl oyed and t he wor k was fi ni shed on. Time. H ow many days would have i t been del ayed if addi t ional wor ker s wer e not employed? (a) 1 (b) 2 (c) 3 (d) 5 [RRB SSE 2015 1 st SEP 1 st SH I FT ]

7. A, B and C can do a wor k 6,12 and 24 days r espect ively. They all begin t oget her. A cont inues t o wor k t il l i t is fi nished, C leaves off 2 days and B one day befor e complet ion. I n what t ime (in days) is t he wor k finished? (a) 2 (b) 4 (c) 6 (d) 8 [RRB SSE 2015 1 st SEP 1 st SH I FT ]

8. Two per sons wor ki ng 2 hour s a day assembl e 3 machines in 3 days. The number of machines assembled by 5 per sons wor king 4 hour s a day in 4 days is (a) 10 (b) 12 (c) 15 (d) 20 [RRB SSE 2015 1 st SEP 2 nd SH I FT ]

9. Two pipes A and B can fi ll a t ank wi t h wat er in 2 hour s and 32 hour s r espect ively. Bot h t he pipes ar e open t oget her. I f t he t ank i s fi l l ed up i n 14 hour s, t hen t he fir st pipe must be t ur ned off after (a) 10 hour s

1 (b) 12 hour s 2

1 (c) 13 hour s 2

(d) 12 hour s [RRB SSE 2015 1 st SEP 2 nd SH I FT ]

10. Two pipes A and B r unning t oget her can fi ll a

2 mi nut es. Pipe B t ak es 3 minut es 3 mor e t han A t o fill it . The t ime, in minut es, t aken by B al one t o fil l t he cist er n is (a) 15 (b) 16 (c) 18 (d) 12 ci st er n in 6

[RRB SSE 2015 1 st SEP 3 rd SH I FT ]

11. 2 men and 7 women complet e a wor k in 14 days, whil e 3 men and 8 women complet e t he same wor k i n 11 days. I n how many days will 8 men and 6 women compl et e 3 t imes of t his wor k ? (a) 14 (b) 21 (c) 24 (d) 28 [RRB SSE 2015 1 st SEP 3 rd SH I FT ]

12. To do a cer t ain pi ece of wor k B t ak es four t imes as l ong as A and C t oget her and C t akes t hr ee t i mes as long as A and B t oget her. I f all t he t hr ee wor k ing t oget her complet e t he wor k in 12 days, how long (in days) would A alone t ake to complete t he wor k? (a) 21 (c) 48

9 11

(b) 26

2 3

(d) 60 [RRB SSE 2015 2 nd SEP 1 st SH I FT ]

13. I f 6 men and 8 women can do a pi ece of wor k in 10 days whil e 5 men and 15 women can do t he same wor k in 8 days? I n how many days wi ll 15 men and 20 women do t he same wor k? (a) 2 (b) 3 (c) 4 (d) 5 [RRB SSE 2015 2 nd SEP 1 st SH I FT ]

14. 8 men and 12 women can do a pi ece of wor k in 5 days, whi le 6 men and 8 women can do i t in 7 days. The t i me (i n days) t aken by t wo women and one man t o do t he same wor k is (a) 21 (b) 28 (c) 35 (d) 40 [RRB SSE 2015 2 nd SEP 2 nd SH I FT ]

15. Two pi pes X and Y can fi l l a t ank i n 15 and 20 minut es r espect i vel y. I f bot h t he pi pes ar e opened t oget her aft er how much t i me pi pe y should be t ur ned off so t hat t he t ank i s ful l in 12 minut es? (a) 3 minut es (b) 4 minut es (c) 5 minut es (d) 6 minut es [RRB SSE 2015 2 nd SEP 2 nd SH I FT ]

Time and Work

5.7

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (c)

2. (c)

3. (d)

4. (d)

5. (b)

11. (b)

12. (b)

13. (d)

14. (b)

15. (c)

6. (b)

7. (a)

8. (c)

9. (c)

10. (d)

LEVEL-1 1. (b)

2. (b)

3. (a)

4. (a)

5. (b)

11. (a)

12. (a)

13. (b)

14. (d)

15. (c)

6. (b)

7. (b)

8. (b)

9. (a)

10. (b)

7. (b)

8. (d)

9. (c)

10. (c)

LEVEL-2 1. (b)

2. (b)

3. (d)

4. (a)

5. (c)

11. (b)

12. (a)

13. (c)

14. (c)

15. (b)

6. (d)

EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S 1.

5.

57 35 11 2 = days 57 12 12

Balance wor k t o be done by Rb and SL = 1

1 1 1 654 1    2. = 10 12 15 60 4 H ence t hey can finish t he wor k in 4 days 3.

(S + R)’s 1 days wor k = (R + M )’s 1 days wor k = (M + S)’s 1 days wor k = 2(S + R + M )’s 1 day wor k =

 (S + R +M )’s 1 days wor k = H ence t i me t ak en by t hem t oget her = 8 days

1 10 1 12 1 15 1 4 1 4 t o do t hi s wor k

(Rb + SL )’s 1 day wor k

1 1 32 1    10 15 30 6 1  (Rb + SL ) will t ake 3 days to do of t he wor k 2 1 of t he wor k in 50 days 4 1  1 person does of work in 50200 days 4 3 1 per son does of wor k in 350200 days 4 3 3  50  200 n per sons do of the wor k in days 4 n

6. 200 per sons do

3  50  200 = 100 n  n = 300 Alr eady 200 wor ker s ar e t her e, hence 100 mor e per sons ar e t o be employed. 

Balance wor k =

1 3

20 x =

1 3

 Again,



1 x = 60

x+y =

7.

1 30

1 1 y = 30 60

1 60 i.e. Sant osh can finish in 60 days. 

1 1  2 2

=

4. Aft er 20 days,



FG 1  1  1 IJ  1 H 10 12 15 K 2

2(R b  S A  SL ) = 2

y =

8.

10 1  20 2 1 1  Balance wor k = 1   2 2 1 30 H ence wor k Rajan can do in 2 2 = 15 days M ukesh’s 10 day’s wor k =

Students Days 10

12

12

x

5.8

Time and Work

 12 : x :: 12 : 10,

 5 women = 4 men  8 men + 5 women = 8 men + 4 men = 12 men N ow, 10 men can finish t he wor k in 6 days  12 m en can f i n i sh t h e w or k i n 6  10 = 5 days 12

12 12 = x 10  x = 10 days 

9.

1 1 1 1 2  1  2  1...    ... = 1  =1 8 16 8 16 16 i .e.

10

1 days 2

LEVEL-1

6 1 10. 12 men’s 6 days wor k =  18 3 1 12  18  = 2 16  ? 3 ? = 9 days

 11.

1.

 2.



3.

FG 1  1 IJ  2FG 3  2 IJ  5 H 8 12 K H 24 K 12

5 7 = 12 12 1 1 (R + S)’s 1 day wor k =  6 8 43 7   24 24 H ence balance wor k will be done by R and S in 7 12 = 2 days 7 24 25  20 1 14. = 25  10  20  10  15  D 1  Balance wor k = 1–

4.

15. 10 men can do t he wor k in 6 days. I f t he wor k is t o be finished in 5 days, t hen number of women r equir ed = 15 I f t he wor k is t o be finished in 6 days, t hen number of women r equir ed 25 15  5 = = 2 6 25  10 men = women 2

Let the capacity of the tank be 180 units i.e LCM of 36 and 45 such that efficiencies of the two pipes = 5 units and 4 units.

180  20hrs. 54

L et t he t ime t ak en by P and Q t o complet e t he wor k alone be 3x and 2x such t hat t ot al w or k be 6x u n i t s an d t h ei r r espect i v e effi ci encies be 2 unit s and 3 unit s. 

6x  18 32

 x  15

5.

1 days 3 H ence t ime t o complet e t he wor k 1 1 = 23 days. 3 3

12  1 = 2 unit s. 4

 required time =

 D = 3

= 10  10  3

L et t he t ot al wor k be 12 unit s such t hat wor k done by Ram in one day be 1 uni t .

=

x  8 x  12 x   =1 36 54 72 x = 24 days

13. 2(S  M ) = 2

453  15 4

 Wor k done by M ohan i n one day

72  42 280 = ? 36 100  ? = 30 men.

12.

Required number of men will be

H ence, Q wi ll compl et e t he wor k i n 2x = 2 × 15 = 30 days L et t he capaci t y of t he t ank be 60 uni t s i.e. L CM of 20 and 30 such t hat t hei r r espect i ve effi ci encies be 3 unit s and 2 unit s. H ence, r equir ed t i me =

6.

60 = 12 minut es. 32

Let the total work be 24 units i.e. LCM of 8 and 12.

24 8 = 3 units

 work done by A and B in one day =

24 = 2 units 12 Hence, work done by B in one day = 1 unit and that by A alone in one day =

and required time =

24 = 24 days. 1

Time and Work

7.

Let the total work be 60 units i.e. LCM of 12, 15 and 20.  work done by P and Q in one day =

60 12

1  1   x 1  15 10  x=6 11.

Let P do work in P days

= 5 units A/Q

60 and that by Q and R in one day = 15 = 4 units

(2)

P = 24

(5  4  3) = =6 2

12.

Efficiency ratio A : B = 2 : 1 Time ratio A : B = 1 : 2

and that by P in one day = 6 4(work done by Q and R in 1 day) = 2 units

i.e x & 2x

1 1 1   x 2x 24  x = 36

60 Hence, required time = = 30 days. 2

A/Q

Set the work done by A and B value in one day be 2 units and 1 units respectively

 B can do work in 2x days = 72 13.

Let work done in 1 day by P, Q, R

 Total work = 24 × (2 + 1) = 72 units and

1 1 1 , , p q r

72  36 units required time = 2



Set the total work be 60 unit i.e. LCM of 12, 15 and 20



 Work done in one day

1 1 1   q r 15

....(2)

1 1 1   r p 20

...(3)

by P and Q = 5 units ....(1) by Q and R = 4 units ....(2) and by R and P = 3 units ....(3)  Adding (1), (2) and (3) we get



12  6 units  P+Q+R= 2

= 6 – 3 = 3 units

Work done by A in 1day 

1 15

1 10 Let number of days required = x Work done by in B in 1day 

1 1 1 1    r p 15 12

....(4)

2 1 1 1     r  60 r 15 12 20  Number of days in which r can do work = 60

= (P + Q + R) – (P + R)

60  20units. 3

....(1)

solving (3) & (4) :

Thus, work done y Q in one day

Hence, required time =

1 1 1   p q 12

By solving (4) & (2),

2(P + Q + R) = 5 + 4 + 3

10.

(1)

1 1 1   (3) P R 15 Solving (1), (2), (3)

Thus, work done by P, Q and R in 1 day

9.

1 1 1   P Q 10

1 1 1   Q R 12

60 and that by P and R in one day = 20 = 3 units

8.

5.9

14.

Work done by A in 1 day =

___________ B _______ =

1 20

1 30

Let no. of days required = D

1  1   D1  20 30 

5.10

Time and Work

 D = 12 15.

emptying speed =

Let R do the work in R days  A/Q,

1 1 1   P Q 10

1 1 1   Q R 12

....(1)

 Time =

16  1 l / hr 16

16  16 hours. 2 1

6. 35 × 100 + 5 × (100 + 100) = 100 × N ....(2)

 N = 45 days  No. of days object =(45 – 40) days = 5 days

1 1 1   P R 15

....(3) 7. one days work of A 

solving (1), (2), (3):R = 40

LEVEL-2

one days work of B 

1 12

one days work of C 

1 24

1. Let time taken by A, B, C combined is D days.

1 1  1     D1  12 15 20  

1 1 1 1 1 1      x      1   1 6 12 24 6 12 6

1 D1 5

x = 2 days Ans = 2 + 1 + 1 = 4 days

 D = 5 days 2. Let capacity of tank be 120 l  filling speed of

P

8.

120  12 l / hr 10

 filling speed of

120 Q  10 l / hr 12

9. Portion of tank filled by (A +B) + position filled by B =1

1 1  1  T    14  T  1  24 32  32

120  6 l / hr 20

 Time 

120 1  7 hours. 12  10  6 2

3. M1 D1 = M2 D2 36 × 18 = 27 × D2  D2 = 24 days 4. Let the work be completed by a woman in x days. According to the question

1 1 1       3  1 6 2 18  x = 9 days 5. Let capacity of tank be 16 litres

16  2 l / hr  filling speed = 8

M1D1H1 M2 D2 H2  w1 w2 232 544   x = 20 machines 3 x

emptying speed of

R

1 6

T 10.

27 1  13 Ans 2 2

1 1 3 + = x x + 3 20 x+x+3 3 = x(x + 3) 20

2x + 3 3  2 x  3 x 20 40x + 60 = 3x2 + 9x 3x2 – 31x – 60 = 0 x = 12 minutes Pipe A takes 12 minutes Pipe B takes 15 minutes 11. Let work done by a man in a day = x units Let work done by a woman in a day = y units  (2x + 7y) × 14 = (3x + 8y) 11

Time and Work

 28x + 98 y = 33x + 88y  5x = 10 y

 Time 

5.11

 8x  12y   5 x  2y

 x = 2y.  work (2x + 7y) × 14 = (2 × 2y + 7y) × 14 = 154 y. 3 times work will be done in



15. Let Tank capacity be 300 litres.  filling speed of x

154y   3 8x  6y

 =

28  5  y  35 days 4y

154y  3  21 days 22y

12. Let A alone takes x days to do the work. Let B alone takes y days to do the work.

300  20 l / min 15

 filling speed of y



300  15 l / min 20

Let C alone takes 2 days to do the work.

Let pipe of be turned off after + minutes

According to the question

According to the question 20 × 12 + 15 × t = 300

1 1 4   x 2 y

 t = 4 mins 

1 1 3   x y 7 Also

1 1 1 1    x y 2 12

Solving for x,

x

4  60 9  21 11 11

13. Let work done by a man = x units in one day Let work done by a woman = y units in one day According to the question (6x + 8y) 10 = (5x + 15y)8  60x + 80y = 40x + 120y  20x = 40y  x = 2y

 Time 



 6x  8y   10 15x  20y

200y  4days 50y

14. Let x units of work be done by a man in one day. Let y units of work be done by a woman in one day  (8x + 12y)5 = (6x + 8y)7  40x + 60y = 42x + 56y  2x = 4y  x = 2y

6

Time and Distance

CHAPTER RELATI VE M OTI ON

I n t his chapt er t he r elat ive mot ion is most impor t ant concept . Ther e is only one for mula applicable D = ST Some I mport ant Conversions :

 1 hour = 60 minut es = 60  60 seconds.  1 kilomet er = 1000 met r es  1 kilomet r e = 0.6214 mile; 1 mile = 1.609 kilomet r e, i.e.

8 kilomet r es  5 miles

 1 yar d = 3 feet



5m km = 18 s hr km 5 miles  hr 8 hr



m 18 km = s 5 hr





miles 22 ft = hr 15 sec

 Aver age Speed =

Tot al dist ance t r avelled Tot al t ime t aken

Relative M ot ion can be defined in t he following ways : (i ) Dependent as in t he case of boat s and st r eams. (ii ) I ndependent as in t he case of t r ains. When one body which is moving inside anot her moving body, t he mot ion is known as dependent , e.g. When a boat is moving inside a st r eam, t he speed of boat depends on t he speed of st r eam. I f boat is going in t he same dir ect ion as t he st r eam, t hen t he boat will move fast er t han t he speed at which boat is moving in st ill wat er, but when t he boat is moving in opposit e dir ect ion, t he speed of boat is slower t han speed of boat in st ill wat er. SPE E D

Dist ance Time d v= t d t= v d=vt

Speed =

 or or

H er e, v is speed, d is dist ance and t is t ime AVERAGE SPEED Tot al dist ance cover ed Tot al t ime t ak en I f a cer t ain dist ance is cover ed at t wo differ ent speeds i.e., if a dist ance x is cover ed at u km/hr and t he same dist ance is cover ed at v km/hr, t hen 2uv Aver age speed of t he jour ney = km/hr.. u  v 

Aver age speed =

6.2

Time and Distance

Example. For a man cover ing a cer t ain dist ance at 6 km/hr and r et ur ning at 4 km/hr, calculat e aver age speed of t he man. Given : u = 6 km/hr. v = 4 km/hr. 2uv Aver age speed = u  v 

2  6  4 48 = = 4.8 km/hr.. 6  4  10 I f r at io of t he speeds of A and B is a : b, t hen r at io of t he t ime t aken by t hem t o cover t he same 1 1 dist ance is : or b : a. a b Somet imes, in or der t o solve cer t ain pr oblems, we need t o conver t t he unit s of speed fr om km/hr. t o m/sec and vice ver sa. 5   1000   k m/hr = x  x  m/sec. =  x   m/sec.  18   3600  (since 1 km = 1000 m and 1 hour = 3600 sec.) 3600  18    m/sec =  y   km/hr. =  y   km/hr.. 1000  5  =



 (1) x

(2) y

STRATEGI ES AN D SOLVED PROBLEM S (1) I f t wo bodies ar e moving in t he same dir ect ion at u km/hr and ‘v ’ km/hr, such t hat u > v t hen t heir r elat ive speed = (u – v ) km/hr. E xample. I f t wo car s ar e moving t owar ds Nor t h wit h t he speeds 40 km/hr and 25 km/hr r espect ively, find t heir r elat ive speed. Ans. H er e, t wo speeds ar e 40 km/hr and 25 km/hr.

u = 40 km/hr and v = 25 km/hr. Relative speed = (u – v ) km/hr. = (45 – 25) km/hr. = 15 km/hr. (2) I f t wo bodies ar e moving in t he opposit e dir ect ions at u km/hr and v km/hr, t hen t heir r elat ive speed = (u + v ) km/hr. E xample. I f t w o bu ses ar e m ov i n g i n t h e opposi t e di r ect i on s, t h ei r speed h av i n g 35 km/hr and 30 km/hr r espect ively, find t heir r elat ive speed ? Ans. H er e, t wo buses ar e moving in t he opposit e dir ect ions. Their speeds ar e u = 35 km/hr and v = 30 km/hr.

 Relative speed = (u + v ) km/hr. = (35 + 30) km/hr. = 65 km/hr. (3) I f two bodies are moving in the same direction with different velocities, initiating earlier, then the distance cover ed by bot h t he bodies at t he point of over t ake is same for bot h t he bodies. Example. A man leaves a t own at 8 a.m. on his bicycle moving at 10 km/hr. Another man leaves the same town at 9 a.m. on his scooter moving at 30 km/hr. At what t ime does he over take the man on the bicycle? Ans. L et ‘t ’ be t he t ime t aken for t he scoot er t o t r avel t he dist ance, so as t o over t ake t he bicycle. The time taken by the bicycle to cover the same distance will be(t + 1). Speed of the bicycle is 10 km/hr. and t hat of t he scoot er is 30 km/hr. Now, using t he st r at egy, dist ance t r avelled by t he scoot er as well as t he bicycle t ill t he point of over t ake will be t he same. i .e.

30  t = 10  (t + 1)

  

30t = 10t + 10 30t – 10t = 10 20t = 10

Time and Distance



t =

6.3

1 hour = 30 mins 2

H ence, t he scoot er will over t ake t he bicycle at 9:30 a.m. (4) I f t wo bodi es init i at e at t wo opposi t e point s and st ar t movi ng t owar ds each ot her, t hen sum of t he di st ances cover ed by t hem t i ll t hei r cr ossi ng point is t he t ot al di st ance bet ween t he t wo ext r eme point s. E xample. I f t wo car s A and B move t owar ds each ot her, wher e car A st ar t s at 9 a.m. and car B at 10 a.m.. The speeds of t he car A and B ar e 40 k m/hr and 50 km/hr r espect i vel y. What dist ance does car t r avel when t he t wo car s meet , if t he init ial dist ance bet ween A and B is 400 km ? Ans. L et t hr s be t he t ime t aken by t he car B t o r each t he cr ossing point . H ence, A will t ake (t + 1) hr s. t o r each t he same point . Speed of car B = 50 km/hr. Dist ance cover d befor e t he cr ossing = 50  t Now, using t he st r at egy, we know t hat t ot al dist ance cover ed by t he t wo car s is t he t ot al dist ance bet ween t heir st ar t ing point s. i .e. 40 (t + 1) + 50t = 400

   

40 t + 50 t + 40 = 400 90 t = 360

t = 4 hr s. Dist ance cover ed by car B = 50  4 = 200 km.

(5) I f an object changes its speed in the ratio m : n , then the ratio of times taken becomes n : m E xample. The r at i o bet ween r at es of wal ki ng of A and B is 2 : 3. I f t he t ime t aken by B t o cover a di st ance is 24 minut es, fi nd t he t ime t aken by A t o cover t he same di st ance. Ans. Rat io of t he speeds of A and B = 2 : 3. Using st r at egy, we get t he t ime t aken by t hem in t he r at io 3 : 2 i .e. A : B = 3 : 2. I t is given t hat B t akes 24 minut es t o cover a dist ance, we need t o find t he t ime t aken by A. L et x be t his t ime t aken by A

 

3 x = 2 24 24 3 x= 2 = 36 minut es

GEN ERAL QU ESTI ON S  I f t he r at io of t he speeds of A and B is a : b, t hen t he r at io of t he t imes t aken by t hem t o cover t he same 1 1 : or b : a. a b Suppose a man cover s a cer t ai n di st ance at x k mph and an equal dist ance at y k mph. Then,

dist ance is



aver age speed dur ing t he whole jour ney =



FG 2xy IJ kmph H x  yK

Suppose a man cover s a cer t ain dist ance at x kmph, equal dist ance at y kmph and same dist ance at z kmph, t hen aver age speed dur ing t he whole jour ney =

FG 3xyz IJ H xy  yz  zx K

6.4

Time and Distance

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. Two mot or car s st ar t fr om one point and move along t wo r oads at r ight angles t o each ot her. I f t h ei r speeds be r espect i v el y 36 k m /h r an d 48 km/hr, t hen aft er half and hour t hey will be x km apar t . What is x ? (a) 30 km (b) 40 km (c) 12 km (d) 84 km 2. A man goes fr om Delhi t o Dehr adun. H alf t he dist ance he cover s by bus at an aver age speed of 50 km/hr and the other half by t r ain at an aver age speed of 70 k m/hr. The aver age speed for t he whole jour ney was (a) 56 km/hr (b) 64 km/hr (c) 60 km/hr (d) none of t hese 3. A man cover s a dist ance of 100 km par t ly by bus at 40 km/hr and par t ly by scoot er at 30 km/hr. H i s aver age speed for t he whol e jour ney was 32 km/hr. H ow far did he go by bus? (a) 20 km (b) 25 km (c) 30 km (d) 40 km 4. A per son shoot ing at a t r aget fr om a dist ance of 450 metres finds that the sound of the bullet hitting t he t ar get comes 1/2 seconds aft er he fir ed. A per son equidist ant fr om t he t ar get and shoot ing point hear s t he bullet hit 3 seconds after he hear d t he gun. The speed of sound is (a) 300 m/s (b) 350 m/s (c) 250 m/s (d) 400 m/s 5. A monkey climbs a slipper y pole 12m high. I t r ises 1met er in ever y one minut e and slips 1/2 met er in ever y next minut e. Find how soon it will r each t he t op? (a) 45 min (b) 40 min (c) 35 min (d) 48 min 6. A boy walks at 4 km/hr. and r eaches his school 5 minut es lat e. Next day he walks at 5 kms/hr and st ar t at t he same t ime, r eaches his school 10 minut es. ear ly. H ow far is t he school fr om his house? (a) 4 km (b) 5 km (c) 6 km (d) 3 km 7. A t r ain star ts fr om Delhi with a cer tain number of passenger s. At Ghaziabad 110 get down and 100 get in. At Aligr ah 50% get down and 25 get in. At K anpur 5 half of t hem get down and 50 get in. Wh en t h e t r ai n l eaves K an pu r t h er e ar e 200 passenger s. How many did boar d the tr ain at Delhi? (a) 300 (b) 400 (c) 500 (d) 600

8. The radius of a cir cular wheel is 35 cm and it moves at t he r at e of 500 r evolut ions per minut e. The speed of wheel is (a) 50 km/hr (b) 60 km/hr (c) 63 km/hr (d) 66 km/hr 9. Diamet er of t he fr ont wheel and hind wheel of a van ar e 70 cm and 98 cm r espect ively. How many mor e r evolutions the fr ont wheel will make when the hind wheel in going over a distance of 154 kms? (a) 10,000 (b) 15,000 (c) 20,000 (d) 25,000 10. Two men X and Y st ar t fr om a place P, walking at 3 km/hr, and 4 km/hr. By how much distance apar t t hey will be aft er 4 hour s if t hey ar e walking in t he same dir ect ion? (a) 4 km (b) 3 km (c) 2 km (d) 1 km 11. I n t he above quest ion, what will be t he dist ance bet ween t hem aft er 5 hour s i f t hey wal k i n opposit e dir ect ions? (a) 35 km (b) 14 km (c) 21 km (d) 28 km 12. A per son has t o make a jour ney of 72 km. H e r ides a cycle at 12 kms/hr. Aft er going cer t ain dist ance, t he cycle got punct ur ed and he walks 1 t he r emaining dist ance at 4 km/hr. Wher e did 2 t he cycle got punct ur ed if t ot al t ime t aken for t he 1 jour ney was 8 hour s? 2 (a) 18 km (b) 54 km (c) 36 km (d) 48 km 13. A car is stolen at 4.00 a.m the thief drives it towar ds n or t h at a speed of 50 k m /h r . T h e t h ef t i s discover ed at 4.30 a.m. and a police jeep is set t owar ds nor t h at 60 kms/hr. At what t ime did t he jeep will over t ake t he car ? (a) 6.00 a.m.

(b) 7.00 a.m.

(c) 7.30 a.m.

(d) 6.30 a.m.

14. A student goes to his school fr om his house walking at 4 k m /h r . an d r each es h i s sch ool 10 min. lat e Next day st ar t ing at same t he t ime he wal k s at 6 k m/ hr and r eaches hi s school 15 minut es ear ly. H ow far is t he school fr om t he house? (a) 3 km

(b) 5 km

(c) 9 km

(d) 24 km

Time and Distance

15. A t r ain tr avelled fr om A t o B and back in a cer t ain t ime at t he r at e of 60 km/hr. But if t he t r ain had t r avelled fr om A t o B at t he r at e of 80 km/hr. and back fr om B t o A at t he r at e of 40 km/hr it would t ak e t wo hour l onger. The di st ance bet ween A and B is (a) 480 km (b) 320 km (c) 540 km (d) 180 km

LEVEL-1 1. The length of two trains are 140 m and 160 m respectively. If they run at the speed of 60 km/h and 40 km/h respectively in opposite directions on parallel tracks, then find the time in which they will cross each other. (a) 10 sec

(b) 10.8 sec

(c) 9 sec

(d) 9.6 sec [RRB JE 2014 RED SH I FT ]

2. Ravi runs 200 metres in 24 seconds. Find his average speed : (a) 20 km/h

(b) 24 km/h

(c) 28.5 km/h

(d) 30 km/h [RRB JE 2014 RED SH I FT ]

3. A man completes 30 km of a journey at 6 km/hr and the remaining 40 km of the journey in 5 hours His average speed for the whole journey is : (a) 6

4 km/hr 11

1 (c) 7 km/hr 2

(b) 7 km/hr

7. Speed of a boat in still water is 3 km/hr. the speed of the stream is 1 km/hr. The time taken to go 4 km upstream in minutes is (a) 48 (c) 96

4. A train with a speed of 60 kmph crosses a pole in 30 seconds. The length of the train is (a) 500 m

(b) 750 m

(c) 900 m

(d) 1000 m [RRB JE 2014 YEL L OW SH I FT ]

5. Speed of a boat in still water is 3 km/hr. the speed of the stream is 2 km/hr. The time taken to go 4 km downstream in minutes is

(b) 60 (d) 120 [RRB JE 2015 26 th AU G 2 nd SH I FT ]

8. A train 100 m long is moving at 40 km/h. The time in seconds, it will take to pass another train 150 m long which is moving at 50 km/h in the opposite direction from the moment they meet is (a) 6

(b) 8

(c) 10

(d) 15 [RRB JE 2015 26 th AU G 2 nd SH I FT ]

9. The speed of a boat m still water is 15 km/hr. The speed of the stream is 3 km/hr. The time taken to go 12 km downstream in minutes is (a) 40

(b) 45

(c) 50

(d) 55 [RRB JE 2015 26 th AU G 3 rd SH I FT ]

10. A train 200 m long is moving at 40 km/hr . The time in seconds, it will take to pass another train 150 long which is moving at 50 km/h in the opposite direction from the moment they meet is (a) 8

(b) 12

(c) 14

(d) 117 [RRB JE 2015 26 th AU G 3 rd SH I FT ]

(d) 8 km/hr [RRB JE 2014 YEL L OW SH I FT ]

11. A person cycles from one place to another in 100 minutes. If his speed is 18 km/h, the distance between two places is (a) 20 km

(b) 30 km

(c) 15 km

(d) 25 km [RRB JE 2015 27 th AU G 1 st SH I FT ]

12. A train 200 m long is moving at 68 km/h. The time in seconds, it will take to pass another train 150 m long which is moving at 50 km/h in the same direction from the moment they meet is

(a) 120

(b) 96

(a) 40

(c) 80

(d) 48

(b) 50

[RRB JE 2015 26 th AU G 1 st SH I FT ]

(c) 60

6. A train 100 m long is moving at 58 km/h. The time in seconds, it will take to pass another train 150 m long which is moving at 50 km/h in the same direction from the moment they meet is

(d) 70 [RRB JE 2015 27 th AU G 1 st SH I FT ]

13. Speed of a boat in still water is 15 km/h. the speed of the stream is 3 km/h. The time taken to go 6 km upstream in minutes is

(a) 20

(b) 30

(a) 20

(b) 30

(c) 50

(d) 70

(c) 24

(d) 25

[RRB JE 2015 26 th AU G 1 st SH I FT ]

6.5

[RRB JE 2015 27 th AU G 2 nd SH I FT ]

6.6

Time and Distance

14. A train 120 in long is moving at 40 km/h. The time in seconds, it will take to pass another train 100 m long which is moving at 32 km/h in the opposite direction horn the moment they meet is (a) 11

(b) 22

(c) 44

(d) 88 [RRB JE 2015 27 th AU G 2 nd SH I FT ]

15. The speed of a car is 72 km/h. The time taken by it to cover a distance of 500 m in seconds is (a) 20 (b) 25 (c) 30 (d) 40 [RRB JE 2015 27 th AU G 3 rd SH I FT ]

LEVEL-2 1. What is t he r at io of angular speed of second's needle and hour 's needl e of a cl ock? (a) 1 : 60 (b) 60 : 1 (c) 3600 : 1 (d) 720 : 1 [RRB SSE 2014 GREEN SH I FT ]

2. A man dr ives a car and r eaches hi s dest inat ion in 4 h ou r s. H ad h e i n cr eased h i s speed by 10 k m/hr, he woul d have r eached i n 3 hour s, 12 minut es. What dist ance di d t he man cover ? (a) 80 km (b) 120 km (c) 160 km (d) 210 km [RRB SSE 2014 GREEN SH I FT ]

3. The speed of a boat in downst r eam dir ect ion is 14 km/hour and in upst r eam di r ect i on is 8 k m/hour. Fi nd t he speed of t hi s boat in st il l wat er : (a) 22 k m/hour (b) 6 km/hour (c) 3 km/hour (d) 11 k m/hour [RRB SSE 2014 RED SH I FT ]

4. M anoj can complet e a jour ney i n 10 hour s. H e t r avels fir st half of t he jour ney at t he speed of 21 k mph and second half of t he jour ney at t he speed of 24 kmph. Find t he t ot al jour ney : (a) 230 km (b) 234 km (c) 220 km (d) 224 km [RRB SSE 2014 RED SH I FT ]

5. A t r ain, 270 met r es l ong, is r unning at t he speed of 120 kmph. I t cr osses anot her t r ai n, which is r unni ng at t he speed of 80 k mph i n opposi t e di r ect i on on par all el t r ack, in 9 seconds. What is t he lengt h of anot her t r ain ? (a) 230 met r es (b) 100 met r es (c) 250 met r es (d) 330 met r es [RRB SSE 2014 RED SH I FT ]

6. Find t he angle bet ween the Hour hand the Minute hand of a clock when t he t ime is 03 : 40 t hat i s 40 mi nut es past 3 ? (a) 120° (b) 125° (c) 130° (d) 135° [RRB SSE 2014 RED SH I FT ]

7. Two t r ains appr oach each ot her at 30 km/hr and 27 k m/hr fr om t wo places 342 k m apar t . Aft er how many hour s wi ll t hey meet ? (a) 5 hr s. (b) 6 hr s. (c) 7 hr s. (d) 12 hr s. [RRB SSE 2014 YEL L OW SH I FT ]

8. The speed of a 150 m long t r ain i s 50 kmph. H ow much t i me wi l l i t t ak e t o pass a 600 m l ong pl at for m ? (a) 50 sec (b) 54 sec (c) 60 sec (d) 64 sec [RRB SSE 2014 YEL L OW SH I FT ]

9. A per son t r avels a cer t ai n dist ance on a bicycle wi t h a cer t ain speed. H ad he moved 3 k m/hour fast er, he would have t ak en 40 minut es less. H ad he moved 2 km/hour slower, he would have t ak en 40 minut es mor e. Sl ower speed of t he per son, in km/hour, is (a) 8 (b) 10 (c) 12 (d) 15 [RRB SSE 2015 1 st SEP 1 st SH I FT ]

10. A and B can wal k ar ound a ci r cul ar pat h i n 4 minut es and 9 minut es r espectively. I f they star t fr om t he same point in t he same dir ect i on, aft er how much time will they meet again for fir st time? (a) 4 mi nut es 12 seconds (b) 4 mi nut e 30 seconds (c) 7 mi nut e 12 seconds (d) 36 minut es [RRB SSE 2015 1 st SEP 1 st SH I FT ]

11. A t r ain l eaves a st at ion A at 5 am. And r eaches B at 9 a.m. Anot her t r ain l eave B at 6 : 30 am and r eaches A at 10 am on t he same day. On par all el t r acks. They wil l meet each ot her at (a) 7 : 40 a.m. (b) 7 : 50 a.m. (c) 8 : 40 a.m. (d) 8 : 50 a.m. [RRB SSE 2015 1 st SEP 2 nd SH I FT ]

12. A per son can r ow 9 k m/hour i n st ill wat er and he fi nds t hat it t akes him t wice as much t i me t o r ow upstr eam as to row downstream the same distance. The speed of t he cur r ent , in km/hour, is (a) 2 (b) 3 (c) 2.5 (d) 3.5 [RRB SSE 2015 1 st SEP 2 nd SH I FT ]

Time and Distance

1 t h of her jour ney 4 wit h a speed of 60 km/hour, 40% her jour ney wit h a speed of 50 km/hour, and t he r est wit h a speed of 42 k m/hour s. The aver age speed (k m/hour ) for t he whole jour ney i s about (a) 46.8 (b) 48.8 (c) 49.7 (d) 50.1

13. Anu t r avel by car and cover s

[RRB SSE 2015 1 SEP 2 st

nd

SH I FT ]

14. Two places A and B ar e 220 km apar t on a highway H ar i st ar t s fr om A and Juhi fr om B at t he same t i me on t he same day usi ng car s. I f t hey t r avel in t h e sam e di r ect i on (A t o B ), t h ey m eet i n

5

6.7

1 hour s, bu t i n 2 hour s 45 mi nut es when 2

t r avell ing t owar ds each ot her. Speed of t he car of Juhi , in km/hour, is (a) 20

(b) 30

(c) 50

(d) 60 [RRB SSE 2015 1 st SEP 3 rd SH I FT ]

15. A cover s a cer t ain di st ance in a cer t ain t ime, if B cover s half of t hi s dist ance i n t he double t ime, t he r at io of speeds of A and B is (a) 2 : 1

(b) 4 : 1

(c) 1 : 2

(d) 1 : 4 [RRB SSE 2015 1 st SEP 3 rd SH I FT ]

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (a) 11. (a)

2. (d) 12. (b)

3. (b) 13. (b)

4. (a) 14. (b)

5. (a) 15. (a)

6. (b)

7. (c)

8. (d)

9. (c)

10. (a)

LEVEL-1 1. (b)

2. (d)

3. (b)

4. (a)

5. (d)

11. (b)

12. (d)

13. (b)

14. (a)

15. (b)

6. (c)

7. (d)

8. (c)

9. (a)

10. (c)

7. (b)

8. (b)

9. (b)

10. (c)

LEVEL-2 1. (d)

2. (c)

3. (d)

4. (d)

5. (a)

11. (a)

12. (b)

13. (b)

14. (a)

15. (b)

6. (c)

EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S x2 = 242 + 182 = 576 + 324 = 900

1.

 2. v =

3.

x = 900 = 30 km 2uv 2  50  70 1 = = 58 kph u +v 50 + 70 3

 x km (100 – x) km  @40 kph @30 kph Aver age speed =





32 =

Tot al dist ance Tot al t ime 100 x 100  x + 40 30

x = 25 km

4. v =

450 m = 300 m/s 1 1 s 2

5. Aver age speed = 1 met er / 4minut es For 11 met r es, t ime t aken = 44 minut es For last 1 met r e jump add 1 minut es  Time t aken = 45 minut es 6. Shor t Tr ick :

d d 15 – = 4 5 60 

 

5d – 4 d 15  20 60

d 1 = 20 4 d = 5 km

6.8

Time and Distance

7. L et t her e wer e n passanger s at Delhi. n +10 10 

n  Delhi

n 75 n  50 + 25 40 + 2 + 50 20

Ghaziabad

3d  6 d  4 d  4 d =2 240



d = 480 km

LEVEL-1



 Aligar h K anpur (200)

1.

Relat i ve speed of t he t wo t r ai ns

n 75 + + 50 = 200 40 2



 (60  40) 

 n = 500 8. Speed = 500 r pm = 500  2 

22 35 1   60 =66 km/hr 7 100 100

22  70 cm = 15400000 cm 7



n 1 

 and

n 1 = 70000 n 1 c2 = 154 km



n2 

22  98 cm = 15400000 cm 7

 n 2 = 50000  n 1 – n 2 = 20000 cm 10. d = v  t = (4 – 3)  4 = 4 km 11. d = (4 + 3)  5 = 35 km

8

12.

72  x 1 x = + 1 2 12 4 2

200 18  = 30 km/hr. 24 5

2.

Average speed =

3.

Aver age speed =

4.

L engt h of t he t r ain = 60 

5.

Speed of the boat downstream = (3 + 2) km/hr

30  40 70  = 7 km/hr.. 30 10 5 6 5  30 18 = 500 met er.

Distance to be travelled = 4 km Hence, required time in minutes = 6.

4 × 60 =48 minutes. 5

Relative speed of the two trains with respect to each other = 68 – 50 = 18 km/hr or 18 ×

d d 1 – = 50 60 2



d 1 = 300 2 d = 150



t=



150 = 3 hr s. 50

d d 25 – = 4 6 60

14.



= 250 meters Hence, required time = 7.

Required time 

4

 3  1

250 = 50 seconds. 5

 60

= 120 minutes. 8.

3d – 2d 5 = 12 12

Relative speed = (40 + 50) 

5  25m / s 18

Hence, required time

d = 5 km

d  d d  d  80  40  –  60  60  = 2

5 = 5 m/s. 18

Distance to be covered = 150 + 100

H ence jeep will over t ake t he car at 7.00 a.m.

15.

140  160 = 10.8 sec. 250 9

x = 54 km





5 250  m/s 18 9

Thus, r equir ed t ime =

n 1 c1 = 154 km

9.

13.



 9.

100  150  10 seconds. 25

Relative Speed of boat (down stream) = 15 + 3 = 18 km/pr Distance = 12 km

Time and Distance

 Time (in minutes) =

12  60 18



= 40 min 10.



Total distance covered by train of length 200 m to cross another train of length 150 m = 350 m Relative speed = (40 + 50) km/pr = 90 km/pr  Time (in seconds) =

11.

350  18  14 seconds 90  5

100 hr. Time given = 100 min = 60 speed = 18 km/hr

100  30km 60 Total distance = length of 2 trains  Dis tan ce  18 

12.

= 200 + 150 = 350 m Relative speed = 68 – 50 = 18 km/hr = 5 m/s  Time taken to meet 

Dis tan ce speed

350  70sec onds 5 Relative speed of boat (upstream) = 15 – 3 

13.

speed of sec onds needle speed of hours needle 6  720 : 1  1    120 

2. Let distance be D km Let initial speed be 5 km/hr According to the question

D 4 S

...(1)

D 1 3 S  10 5

...(2)

Divide (1) by (2)

S  10 5  S 4  S = 40 km/hr  D = 40 × 4 = 160 km 3. Let the speed of boat in still water be x km/hr Let the speed of stream be y km/hr According to the question, x + y = 14 x – y = 18

= 12 km/hr

2x = 22

Distance = 6 km

6  60 = 30 12 Distance covered by train to pass = 120 + 100

 x = 11 km/hr

 Time (minutes) = 14.

= 220 m

4. Average speed =

21  24  2  22.4 21  24

 Total Distance = 22.4  10 = 224 km

Relative speed = 40 + 32 = 72 km/hr.

220  18  11  Time (seconds)  72  5 15.

5  speed =  72   m / s = 20 m/s  18   Time tabeu =

500 = 25 seconds 20

LEVEL-2 1. speed of hour needle = 0.5°/min 

0.5  1     / sec 60  120  speed of seconds needle = 360°/min 

 360   / sec  60 

6.9

5. Relative speed = 120  80 

 200 

5 m /s 18

5 18

Let the length of other train be x m. 270  x 9  According to question, 200 5 18  x = 230 m 6. Q  30h 

11 11  40  130. m  30  3  2 2

7. Time after the trains meet



342km  6hours 30  27 km / hr

6.10

Time and Distance

8. 50km / hr   50  5  m / s  18  



25  5 m /s 9

12. speed in still water = 9 km/ pr Time up : Time Down = 2 : 1  speed up : speed Down = 1 : 2 speed in still water 

 125   m/s  9 

 x = 6 km/ pr

150  600  m Time taken 

speed of current 

 125    m/s 9 

13. Distance 

= 54 seconds

2 9. 3 km / hr faster ... hr less 3 2 2 km / hr slower ... hr more 3 Let initial speed = V v



v1 v2 (t1  t2 ) 2v1 v2   t  t2  40 min  v1 t 2  v2 t1 v1  v2  1

232 2 32

speed of B 

36  9 m/min 4

36  4 m/min 9

36m  Time  9  4 m / min = 7 min 12 sec   11. T1 : T2 = 4 : 3.5  V1 : V2 = 3.5 : 4 Let total distance = 4 × 3.5x Distance covered by train 1 till 6 : 30 am = 1.5 × 3.5x  Meeting time = 1.16 hours i.e 7 : 40 am

1  2x  x   3 (Ans) 2

1 2 35 x x x 4 5 100

x 2x 35x   4  60 5  50 100  42

Avg speed 

Total Dis tan ce  48.8km / pr Total Time

14. Let speed of A be x km/hr let speed of B be y km/hr



220 1 11 5  xy 2 2

220 3 11 2  xy 4 4

 slower speed =(12 – 2) km/hr = 10 km/hr 10. Let length of track = LCM (4 and 9) = 36 m  speed of A =

Time 

1  x  2x   9 2

Distance left 14x  5.25x  Relative speed 75x

 x = 60 km/hr y = 20 km/hr  20 km/hr 15. According to question,

D T SA

 D   2  2T SB  SA : SB = 4 : 1 

7

Boats and Streams

CHAPTER

L et v 1 be speed of boat and v 2 be speed of t he st r eam.

 and

speed in up st r eam = v 1 – v 2 speed in down st r eam = v 1 + v 2

I f boat goes a cer t ain dist ance ‘d’ down st r eam and r et ur ns upst r eam and t ot al t ime t aken be t , t hen

t=

d d  v1  v2 v1  v2

Speed of boat in st ill wat er = Speed of st r eam =

1 speed in up st r eam + speed in down st r eam 2

1 speed in down st r eam – speed i n up st r eam 2

PRACTI CE EXERCI SE 1. A boat goes 20 k m upst r eam i n 2 hour s and downst r eam in 1 hour. H ow much t ime t his boat will t ake t o t r avel 30 km in all st ill wat er ? (a) 1 hr (b) 2 hr s (c) 1.5 hr s (d) 2.5 hr s 2. I n t he above quest ion, t he speed at which t he st r eam is flowing is (a) 10 km/hr (b) 20 km/hr (c) 15 km/hr (d) 5 km/hr 3. A boat t r avels 10 km in 1 hr downst r eam and 14 kms in 2 hr s upst r eam. H ow much t ime t his boat will t ake t o t r avel 17 kms in st ill wat er ? (a) 1 hr

(b) 2

1 hr s 2

(c) 2 hr s

(d) 2

1 hr s 2

4. A man goes by mot or boat a cer t ai n di st ance upst r eam at 15 k m/hr and r et ur n t he same downstr eam at 20 km/hr. The total time taken for the jour ney was 7 hr s. Find how far did he go. (a) 60 km (b) 50 km (c) 40 km (d) 120 km

2 km in 3 10 m i nu t es an d r et u r n s t h e same di st ance downst r eam in 5 minut es. Rat io of man’s speed in st ill wat er and t hat of t he st r eam will be (a) 3 : 1 (b) 1 : 3 (c) 2 :3 (d) 3 : 2

5. A man can r ow upst r eam a dist ance of

6. A man can r ow a cer tain dist ance down st r eam in 6 hour s and r et ur n t he same dist ance in 9 hour s. I f st r eam flows at t he r at e of 2 km/hr, t hen what will be man’s speed if he r ows in st ill wat er ? (a) 10 km/hr (b) 12 km/hr (c) 14 km/hr (d) 15 km/hr 7. A boat against t he cur r ent of wat er goes 9km/hr and in t he dir ect ion of t he cur r ent 12km/hr. The boat t ak es 4 hour s and 12 mi nut es t o move upwar ed and downwar d dir ect ion fr om A t o B. What is t he dist ance bet ween A and B? (a) 21.6 km (b) 21.0 km (c) 22 km (d) 30 km 8. A man t ak es 3 hour s and 45 minut es t o boat 15 km wit h t he cur r ent in a r iver and 2 hour s 30 minut es t o cover a dist ance of 5 km against t he cur r ent . Speed of t he boat in st ill wat er and speed of t he cur r ent r espect ively will be (a) 3 km/hr, 1 km/hr (b) 1 km/hr, 3 km/hr (c) 2 km/hr, 5 km/hr (d) none of t hese 9. A boat can be r owed 6km/hr along t he cur r ent and 4 km/hr against t he cur r ent . Speed of t he cur r ent and speed of t he boat i n st i l l wat er, r espect ively will be (a) 1 km/hr, 5 km/hr (b) 5 km/hr, 1 km/hr (c) 2 km/hr, 4 km/hr (d) none of t hese

7.2

Boats and Streams

10. A boat moves down t he st r eam at t he r at e of 1 km in 6 minut es and up t he st r eam at t he r at e of 1 km in 10 minut es. The speed of t he cur r ent is (a) 2 km/hr (b) 1 km/hr (c) 1.5 km/hr (d) 2.5 km/hr 11. A man can r ow 5 km per hour in st ill wat er. I f t he r iver is flowing at 1km per hour, it t akes him 75 minut es t o r ow t o a place and back. H ow far is t he place? (a) 3 km (b) 2.5 km (c) 4 km (d) none of t hese

12. Speed of a boat i n st i l l wat er i s 7 k m/hr and speed of t he st r eam is 1.5 km/hr . H ow much time will it take to move up is st r eam of a distance 7.7 km? (a) 75 minut es (b) 84 minut es (c) 72 minut es (d) none of t hese 13. A motor boat takes 2 hour s to tr avel a distance of 9 km down the cur r ent and it takes 6 hour s to travel the same distance against the current. What is the speed of the boat in still water in kmph? (a) 3 (b) 2 (c) 1.5 (d) 1

AN SWERS 1. (b) 11. (a)

2. (d) 12. (b)

3. (c) 13. (a)

4. (a)

5. (a)

6. (a)

7. (a)

8. (a)

9. (b)

10. (a)

EXPLAN ATI ON S 1. L et v 1 be t he speed of boat in st ill wat er and v 2 be t he speed of cur r ent



20 v1 + v2 =  20 1

20 v1 – v2 =  10 2 Fr om equat ions (i ) and (ii ) we get v 1 = 15 km/hr 

t

5.

...(i )

 ...(ii ) and

v1 + v2 =

10  10 1

14 7 2 Adding equat ions (i ) and (ii ), we get

v1 – v2 =

v1 = 

4.



t

17 km/hr 2

d 17 =  2hr s. 17 v1 2

d d  =7 20 15 d = 60 km

2 km min 30 2 km V1 + V2 = 3 5 min V1 – V2 =

2 km 15 min Solving equat ion (i ) and (ii ), we get

30 d =  2hrs. 15 v1

=

2. Fr om t he above t wo equat ion, we get v 2 = 5km/hr 3.

2 km V1 – V2 = 3 10 min

...(i ) and

...(ii )



 



d d 12 = 4  9 12 60 d = 21.6 km

7.

...(ii )

2 2  1 km 24 6  V1 = 30 15 = = 10 min 60 60 2 2 2  42 2 1 km  V2 = 15 30 = = 60 60 30 mi n 2 1 30 V1  = 10 1 V2 =3:1 (v 1 + v 2) t 1 = (v 1 – v 2) t 2 (v 1 + 2)  6 = (v 1 – 2)  9 v 1 = 10 km/hr

6.

...(i )

Boats and Streams

v1 + v2 =

8.

v1 – v2 =

15  4 k m/hr 3 3 4 5  2 k m/hr 1 2 2

...(i )

 ...(ii )

Solving equat ions (i ) and (ii ), we get v 1 = 3 km/hr and v 2 = 1 km/hr 9.

v1 + v2 = 6 ...(i ) and v1 – v2 = 4 ...(ii ) Fr om equat ions (i ) and (ii ), we get v 1 = 5 km/hr and v 2 = 1 km/hr

10. and

v1 + v2 =

1km  10 k m/hr 6 min

...(i )

v1 – v2 =

1km  6 k m/hr 10 min

...(ii )

Subt r act ing equat ion (ii ) fr om (i ), we get v 2 = 2 km/hr

11.

 12. t =

7.3

d d 75 =  5 1 5 1 60 2d  3d 5 = 12 4 d = 3 km 7.7 7 d 7.7  hrs. =  5.5 5 v1  v2 7  1.5 =

7  60 min = 84 minut es 5

13. L et t he speed of boat in st ill wat er and speed of cur r ent ar e x and y km/h r espect ively.  Downwar d speed of boat = (x + y ) km/h. Accor ding t o quest ion, 9 x+y =  2x + 2y = 9 ...(i ) 2 9 and x– y =  2x – 2y = 3 ...(ii ) 6 On solving equat ions (i ) and (ii ), we get 3 x = 3, y = 2 

8

Simple Interest and Compound Interest

CHAPTER

I N TEREST (I ) I t is t he money paid for t he use of money bor r owed. I t is gener ally a per cent age of t he sum bor r owed. I t is paid quar t er ly, half year ly or annually as agr eed upon. Pr inciple (P). The act ual money bor r owed is called pr inciple. Amount (A). The pr inciple t oget her wit h it s int er est is called amount . Rate percent per annum (r). I t is t he sum paid on ` 100 of t he loan for a year Time (n). Time for which money is bor r owed n is expr essed in number of per iods, which is nor mally one year. SI M PLE I N TEREST (S.I .) I f t hr oughout t he loan per iod, int er est is char ged on t he or iginal sums bor r owed (i.e. pr incipal), it is called simple inter est. F or mulae.



Simple int er est , S.I . =



Pr inciple, P =



Time, n =



Rat e, r =

Pnr 100

S.I .  100 r n

S.I .  100 Pr



S.I .  100 Pn Amount , A = Pr inciple + Simple int er est



When amount is given :

Pr incipal, P =

When pr incipal is given :

 Amount A = P  I +  100 

A  100 100 + nr

nr

COM POU N D I N TEREST (C.I .) M oney is said t o be lent at compound int er est when t he int er est due aft er a given t ime is added t o t he pr incipal goes on incr easing at t he end of ever y year by an amount t o t he int er est for t hat year. Differ ence bet ween final amount and t he or iginal pr inciple is called compound int er est .  When int er est is compounded annually 

r

 

r 2  100 

 

r 4  100 



r



r





r



Amount , A = P 1  1  1  2  1  3  ... 1  n  100   100   100   100   wher e, n = t ime in year s.  When int er est is compounded half-year ly Amount , A = P  1 +

 When inter est is compound quar t er ly

Amount , A = P  1 +

2n

4n

8.2

Simple Interest and Compound Interest

 When t ime is in fr act ion of a year, say 2

1 5

1   2 r r    5  Amount , A = P  1 +  1     100  100      Compound inter est = Amount – Pr inciple  Pr inciple (P) =

A n

n   = P  1  r   1 100   

r    1   100   When rates are different for different years, say r 1%, r 2%, r 3%, ..... r n% for 1st , 2nd, 3rd, ... nth year respectively, then r  r  r   r   Amount , A = P 1  1  1  2  1  3  ... 1  n  100 100 100 100       Remember. (1) Compound int er est for one year is equal t o t he simple int er est for one year. (2) The difference between C.I. and S.I. on the same sum for 2 years is one year interest on the S.I. for 1 year. COM PARI SON OF SI M PLE I N TEREST AN D COM POU N D I N TEREST I n case of simple int er est t he pr inciple r emains t he same for any fixed t ime per iod whet her for 1st year, 2nd year or 3r d year. While in case of compound int er est , t he pr inciple keeps on incr esing as t he amount aft er 1 year becomes t he pr inciple for t he second year, amount aft er t wo year s becomes t he pr inciple for the t hir d year and so on. The differ ence bet ween 2nd year and 2 year s is t hat 2nd year is one year i.e. t he second one and t wo year s means fir st and second year combined.

e.g. For t he t ime per iod 1st Jan 1991 - 1st Jan 1992 - 1st Jan 1993, fr om 1st Jan 1992 t o 1st Jan 1993 is 2nd year while 1st Jan 1991 - 1st Jan 1993 is 2 year. I t is always ver y easy t o calculat e t he simple int er est as compar ison t o Compound int er est . So we come acr oss a t able S.I . C.I .

1st year

2nd year

3rd year

2 years

3 years

I1 I1

I1 I 1+I 2

I1 I 1+2I 2+I 3

2I 1 2I 1+I 2

3I 1 3I 1+3I 2+I 3

The differ ence bet ween SI and CI for 2 year s is I 2 and for 3 year s is 3I 2+I 3. Now we have t o see what is I 1, I 2 and I 3. I f P is t he Pr inciple and r is t he r at e of int er est , t hen for one year Pr t Pr SI = or as t = 1 year 100 100 Pr H ence for fir st year t he SI is I 1 = . 100 As t he pr inciple r emains same in case of SI for ever y year weat her 2nd or 3r d, t he SI r emains I 1. I n case of CI , as the pr inciple keeps on changing t he amount (= pr inciple + I nter est for 1 year ) aft er 1year will become the principle for 2nd year. So P2 = P + I 1, for which we have to calculate interest for 2nd year. SI =

P2 r t (P+I 1 ) r  t = 100 100

wher e, P2 = Amount aft er 1 year or pr inciple for 2nd year



S.I . =

As t = 1 year, i.e. 2nd year.

I r Pr  1 = I 1 + I 2 (supposed t o be I 2) 100 100

Simple Interest and Compound Interest

8.3

As in 2nd year t he int er est is I 1 + I 2, so for t he 3r d year t he pr inciple will be equal t o amount aft er 2 year s, i.e. P + 2I 1 + I 2. So t he Compound I nt er est for t he 3r d year is P+ Pr inciple

I1 I nt er est for 1st year SI =

+

I1 + I2 I nt er est for 2nd year

 P+2I 1 +I 2  rt 100

As for 3r d year, t = 1 year SI =



Pr = I1 100 I 1r = I 2. 100 I 2r = I3 100

As alr eady define

So or we can say, and

2I r I r Pr  1  2 100 100 100

int er est for 3r d year = I 1 + 2I 2 + I 3 t ot al int er est for 3 year s = 3I 1 + 3I 2 + I 3.

Gener ally you would be asked t he quest ion on compar ison of SI and CI . To find t he CI we go for Pascal t r iangle. PASCAL TRI AN GLE I st Year

1 I1

2nd Year

1 I1

3rd Year

1 I1

4th Year

1 I1

So compound int er est

1 I2

+

1 I3

2 I2

3 I2

3 I3

1 I4

for 2 year s = 2I 1 + 1I 2, for 3 year s = 3I 1 + 3I 2 + I 3, for 4 year s = 4I 1 + 6I 2 + 4I 3 + I 4 ... and so on.

TI PS OF COM POU N D I N TEREST When doing t hese sor t of pr oblems, r emember : 1. Compound int er est will be of “ int er est ” t o you if you have a building societ y account , it wi ll t ell you how much you savings wil l be wor t h next year or in 5 year ’s t i me et c. The t able below shows how an init ial investment of £1,000 can gr ow in 5 year s, if t he int er est r ate is 6%: Year Amount at star t of year (`)

I nter est

Amount at end of year

T otal amount at end of year

1

1,000

1,000  0.06

1,000 + 60

1,060

2

1,060

1,060  0.06

1,060 + 63.60

1,123.60

3

1,123.60

1,123.60  0.06 1,123.60 + 67.42 1,191.02

4

1,191.02

1,191.02  0.06 1,191.02 + 71.46 1,262.48

5

1,262.48

1,262.48  0.06 1,262.48 + 75.75 1,338.23

Compound int er est gi ves you ` 338.23 for doi ng not hing! 2. Si mple int er est gi ves you a sl ight l y wor se deal: ` 1,000 invest ed for 5 year s wit h 6% si mpl e int er est wil l just gi ve you 6% of ` 1,000 ever y year (i .e. ` 60) mul t i pl y t hi s by 5 and we get ` 300.

8.4

Simple Interest and Compound Interest

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. Ram u w an t ed t o bor r ow Rs. 1000 f r om a moneylender for a per iod of one year. H owever, t he moneylender deduct ed Rs. 200 as int er est ch ar ges an d gave Rs. 800 t o Ram u . Ram u r et ur ned Rs. 1000 at t he end of t he year. The r at e of int er est char ged by t he moneylender is (a) 20

(b) 22.5

(c) 25

(d) 27.5

2. Rajan bor r owed Rs. 50000 fr om Rakesh at simple int er est . Aft er 3 year s, Rakesh got Rs. 3000 mor e t han what he had given t o Rajan. What was t he r at e of int er est per annum ? (a) 2 %

(b) 5 %

(c) 8 %

(d) 10 %

3. Rakesh t ook a loan for 7 year s at t he r at e of 6 % p.a. S.I . I f t ot al int er est paid was Rs. 2100, t hen pr incipal was (a) Rs. 4400 (b) Rs. 4800 (c) Rs. 5000 (d) Rs. 5200 4. H ow much should money lender lend at simple r at e of int er est of 15% in or der t o have Rs. 3234 1 at t he end of 1 year s ? 2 (a) Rs. 1640 (b) Rs. 2620

(a) 9

(b) 10

(c) 12

(d) 11

8. A sum of money was lent at simple int er est at 1 1 year s and 4 year s r espectively.. 2 2 I f differ ence in int er est s for two per iods was Rs. 5500, then the sum is

11% p.a. for 3

(a) Rs. 50050

(b) Rs. 55000

(c) Rs. 50000

(d) Rs. 50500

9. Rajan lent Rs. 1200 t o Rakesh for 3 year s at a cer t ain r at e of simple int er est and Rs. 1000 t o M ukesh for t he same t ime at t he same r at e. I f he gets Rs. 50 mor e fr om Rakesh t han fr om M ukesh, t hen t he r at e per cent is

1 (a) 8 % 3

2 (b) 6 % 3

1 (c) 10 % 3

2 (d) 9 % 3

10. The differ ence bet ween int er est s r eceived fr om Canar a Bank and Punjab & Sind Bank on Rs. 500 for 2 year s is Rs. 2.50. The differ ence bet ween t heir r at es is (a) 1 % (b) 0.5 % (c) 0.25 % (d) 2.5 % 11. I f compound int er est for t wo successive year s is Rs. 110 and Rs. 121 r espect ively, t hen t he r at e of int er est is

(c) Rs. 2610 (d) Rs. 2640 5. A n am ou n t Rs. 8000 becom es Rs. 9200 in 3 year s at simple int er est . I f r at e of int er est is incr eased by 3%, it would amount t o (a) Rs. 9920

(a) 10 %

(b) 8 %

(c) 6 %

(d) 4 %

12. A money lender finds t hat due t o a fall in t he r at e 3 %, his year ly income 4 diminishes by Rs. 615. H is capit al is (a) Rs. 260000 (b) Rs. 246000 (c) Rs. 238000 (d) Rs. 224000

of int er est fr om 8% t o 7

(b) Rs. 10560 (c) Rs. 11120 (d) Rs. 11820 6. A sum of money deposited at compound interest doubles itself in 4 years. I t will amount to sixten times at the same rate in (a) 12 year s

(b) 16 year s

(c) 24 year s

(d) 30 year s

7. For how many year s should Rs. 1200 be invest ed at 10% p.a. i n or der t o ear n t he same si mpl e int er est as is ear ned by investing Rs. 1800 at 12% p.a. for 5 year s ?

13. M r. M it t al finds t hat an incr ease in t he r at e of

7 1 inter est fr om 4 % t o 5 % per annum incr eases 8 8 his year ly income by Rs. 250. H is invest ment is (a) Rs. 1,00,000 (b) Rs. 1,20,000 (c) Rs. 1,50,000 (d) Rs. 2,00,000

Simple Interest and Compound Interest

14. I n how many year s will a sum of money double it self at 4% per annum ?

6. The simple interest on rupees 200 for 3 years at 6% per annum in rupees is

(a) 8 year s

(b) 16 year s

(a) 36

(b) 18

(c) 12 year s

(d) 25 year s

(c) 24

(d) 48

15. At a cer t ain r at e of simple int er est , a cer t ain sum doubles it self in 10 year s. I t will t r iple it self in (a) 12 year s (b) 15 year s (c) 20 year s (d) 30 year s

LEVEL-1 1. Find the simple interest on Rs. 4800 at the rate

1 of 8 % per annum for a period of 2 years 3 2 months. (a) Rs. 796

(b) Rs. 816

(c) Rs. 918

(d) Rs. 990 [RRB JE 2014 RED SH I FT ]

2. Simple Interest on Rs. 500 for 4 years at 6.25% per annum is equal to the Simple Interest on Rs.400 at 5% per annum for a certain period of time. The period of time is (a) 4 years (c) 6

1 years 4

(b) 5 years (d) 8

2 years 3

[RRB JE 2014 YEL L OW SH I FT ]

3. A sum becomes Rs. 2916 in 2 years at 8% per annum compound interest. The sum is (a) Rs. 2750

(b) Rs. 2500

(c) Rs. 2625

(d) Rs. 2560 [RRB JE 2014 YEL L OW SH I FT ]

4. If `200 becomes `240 in 4 years, then the rate of simple interest per annum is (a)

25 % 6

(b)

25 % 3

(c)

25 % 2

(d) 5% [RRB JE 2015 26 th AU G 1 st SH I FT ]

5. A sum of money doubles itself in 5 years when the interest is compounded annually. The number of years when it will become eight times is

[RRB JE 2015 26 th AU G 2 nd SH I FT ]

7. A sum of money doubles itself in 4 years when the interests is compounded annually. The number of years when it will become eight times is (a) 32

(b) 16

(c) 12

(d) 8 [RRB JE 2015 26 th AU G 2 nd SH I FT ]

8. The simple interest on rupees 800 for 7 years at 5% per annum is (a) `100

(b) `125

(c) `150

(d) `200 [RRB JE 2015 26 th AU G 3 rd SH I FT ]

9. The compound interest on rupees 12000 for 1 year at 10% per annum compounded half yearly is (a) `1200

(b) `1230

(c) `2520

(d) `2680 [RRB JE 2015 26 th AU G 3 rd SH I FT ]

10. The simple interest on rupees 800 for 3 years at 5% per annum in rupees is (a) 24

(b) 40

(c) 120

(d) 140 [RRB JE 2015 27 th AU G 1 st SH I FT ]

11. Compound interest on rupees 8000 for 1 year at 10% per annum compounded half yearly is (a) 800

(b) 1680

(c) 840

(d) 820 [RRB JE 2015 27 th AU G 1 st SH I FT ]

12. In how many years rupees 500 will amount to rupees 800 at simple interest of 10% per year? (a) 6

(b) 8

(c) 10

(d) 16 [RRB JE 2015 27 th AU G 2 nd SH I FT ]

13. Compound interest `16000 for 1 year at 10% per annum compounded half yearly is

(a) 10

(b) 12

(a) 1600

(c) 15

(d) 20

(b) 1640

[RRB JE 2015 26 th AU G 1 st SH I FT ]

8.5

(c) 1680 (d) 3360

[RRB JE 2015 27 th AU G 2 nd SH I FT ]

8.6

Simple Interest and Compound Interest

14. In how many years `500 will amount to `700 at simple interest of 5% per annum? (a) 4

(b) 5

(c) 6

(d) 8 [RRB JE 2015 27 th AU G 3 rd SH I FT ]

15. In how many years `2000 will amount to `2100 at 10% per annum compounded half yearly? (a) 2 (c) 1

(b) 1.5 (d) 0.5 [RRB JE 2015 27 th AU G 3 rd SH I FT ]

LEVEL-2 1. Fi nd compound int er est on Rs. 7,300 at t he r at e of 4% per an n u m f or 2 year s, compou nded annually?

com poun ded an nu al l y at 12

1 %, t hen each 2

inst al lment s wil l be of (i n Rs.) (a) 26736

(b) 26244

(c) 25736

(d) 24244 [RRB SSE 2015 1 st SEP 2 nd SH I FT ]

7. Two equal sum ar e l ent out at 6 % and 5% simple inter est per annum r espectively at the same time. The fi r st i s r ecover ed 24 year s ear li er t han t he second one and t he amount r eceived in each case was Rs. 28800. Each sum (in Rs.) was (a) 18000

(b) 20000

(c) 20500

(d) 2200

(a) Rs. 612

(b) Rs. 300

[RRB SSE 2015 1 st SEP 2 nd SH I FT ]

(c) Rs. 600

(d) Rs. 630

8. A comput er i s avai l abl e for Rs. 22750 cash payment or for Rs. 6200 cash down payment and t hr ee equal annual i nst al lment s of Rs. x. I f t he inter est char ged is 10% per annum. Compounded annual ly, t he value of x is

[RRB SSE 2014 RED SH I FT ]

2. I n how many year s, a sum wil l be t hr i ce of i t at si mple int er est @10% per annum ? (a) 15 year s

(b) 20 year s

(c) 30 year s

(d) 40 year s [RRB SSE 2014 YEL L OW SH I FT ]

3. A sum of money amount s t o Rs. 9680 i n 2 year s and Rs. 10648 i n 3 year s. The r at e of i nt er est per annum on compounded basi s is (a) 5% (c) 15%

(b) 10% (d) 20% [RRB SSE 2014 YEL L OW SH I FT ]

4. A man buys a TV by maki ng cash down payment of Rs. 4945 and agr ees t o pay t wo mor e year ly inst allment s of equivalent amount s at t he end of fi r st year and second year. I f t he r at e of int er est

1 is 7 % per annum, compounded annual ly, t he 2 cash value of t he TV (in Rs.) i s near est t o (a) 12840

(b) 13804

(c) 13824

(d) 14835 [RRB SSE 2015 1 st SEP 1 st SH I FT ]

5. A sum of Rs.5000 amounts to Rs 8640 at compound i nt er est i n a ai n t i mes, t hen t he same sum amount s t o what in one-t hir d of t he t ime? (a) Rs 5886

(b) Rs 6000

(c) Rs 6214

(d) Rs 7000 [RRB SSE 2015 1 st SEP 1 st SH I FT ]

6. A loan of Rs.62496 is t o be paid back in t hr ee equal annual i nst al l ment s. I f t he i nt er est i s

(a) 5517

(b) 5578

(c) 6565

(d) 6655 [RRB SSE 2015 1 st SEP 3 rd SH I FT ]

9. A sum of money at simpl e i nt er est amount s t o Rs.13800 in 3 year s . I f r ate of inter est is incr eased by 30%,t he same sum amount s t o Rs.14340 in t he same t ime. The r at e of int er est per annum is (a) 3%

(b) 4%

(c) 5%

(d) 8% [RRB SSE 2015 1 st SEP 3 rd SH I FT ]

10. A per son bor r owed some money on compound int er est and r et ur ned i t in t hr ee year s in equal annual inst allment s. I f the r ate of inter est in 15% per an n u m an d t h e an n u al i n st al l m en t i s Rs.48668, t hen t he sum bor r owed was (i n Rs ) (a) 101020

(b) 111050

(c) 111120

(d) 146004 [RRB SSE 2015 2 nd SEP 1 st SH I FT ]

11. A sum of Rs. x at si mpl e int er est amount s t o Rs. 14160 i n 3 year s. I f t h e r at e of i nt er est i s i ncr eased by 25 % t he same sum amount s t o Rs.14700 in t he same t i me. The value of x is (a) 12000

(b) 12400

(c) 13000

(d) 13400 [RRB SSE 2015 2 nd SEP 1 st SH I FT ]

Simple Interest and Compound Interest

12. A cer t ain sum of many is bor r owed at compound i n t er est for 3 year s at 5% per ann um. T he int er est for t he t hir d year is gr eat er t han t hat of secon d year by Rs.642.60. t h e sum (i n Rs.) bor r owed is (a) 24480

(b) 185400

(c) 244800

(d) 368400 [RRB SSE 2015 2

nd

14. A loan of Rs. 26480 i s t o be paid back i n t hr ee equal year l y i nst al l ment s. I f t he i nt er est i s compounded year ly at 10% per annum, then each inst all ment is of Rs.

SEP 2

compounded half year l y at t he r at e of 13

SH I FT ]

(b) 10864

(c) 10648

(d) 8827

15. A sum of Rs.78060 is divided bet ween A and B, so t hat t he amount of A aft er 3

1 % per 3

1 year s is equal t o 2

1 year, t he int er est is 2 compounded hal f year ly at 8% per annum. The shar e of B in t he given sum is (i n Rs.) t he amount of B aft er 4

annum, t hen each inst all ment is of Rs. (b) 45960

(c) 46080

(a) 11548

[RRB SSE 2015 2 nd SEP 3 rd SH I FT ] nd

13. A sum of Rs. 129780 is be paid back in thr ee equal h al f year l y i n st al l m en t s. I f t h e i n t er est i s

(a) 44690

8.7

(d) 49152 [RRB SSE 2015 2 nd SEP 2 nd SH I FT ]

(a) 40560

(b) 38560

(c) 37800

(d) 37500 [RRB SSE 2015 2 nd SEP 3 rd SH I FT ]

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (c) 11. (a)

2. (a) 12. (b)

3. (c) 13. (a)

4. (d) 14. (d)

5. (a) 15. (c)

6. (b)

7. (a)

8. (c)

9. (a)

10. (c)

7. (c)

8. (a)

9. (b)

10. (c)

7. (a)

8. (d)

9. (c)

10. (c)

LEVEL-1 1. (c)

2. (c)

3. (b)

4. (d)

5. (c)

11. (*)

12. (a)

13. (b)

14. (d)

15. (d)

6. (a)

LEVEL-2 1. (a)

2. (b)

3. (b)

4. (c)

5. (b)

11. (a)

12. (c)

13. (d)

14. (c)

15. (d)

6. (b)

EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S 1. Effect ively he bor r owed Rs. 800 and r et ur ned Rs. 1000 aft er one year. So he paid Rs. 200 as int er est on Rs. 800

 Rat e of int er est =

FG H

200  100 = 25% 800

IJ K

2. Rat e = 100  300 % = 2% 5000  3 3. Pr incipal = Rs.

FG 2100  100 IJ = Rs. 5000 H 76 K

4. L et r equir ed money be x .

Then



FG x  x  15  3 IJ H 100 2 K

= 3234

49 x = 3234 40

x=



FG 3234  40 IJ = 2640 H 49 K

5. Pr incipal = Rs. 8000, S.I . = Rs. 1200, Time = 3 year s.



Rate =

FG 100  1200 IJ % = 5%. H 8000  3 K

New r at e = 8%, Pr incipal = Rs. 8000,

8.8

Simple Interest and Compound Interest

Time = 3 year s.

FG H

8000  8  3 S.I . = Rs. 100

IJ K

= Rs. 1920.

6. L et sum = Rs. 100, Time = 4 year s, Amount due in 4 year s = Rs. 200

FG H

100 1 

10 100

IJ K

4

= 200

FG1  10 IJ = 2 H 100 K FG1  10 IJ = 2 H 100 K 4





14

...(i )

L et t he amount becomes 16 t imes in n year s

FG H

10 100



100 1 



FG1  10 IJ H 100K

IJ K

n

= 1600

...(ii )

n

= 16

14

n

= 16 = 24

n =4 4



500  r1  2 500  r2  2  = 2.50 100 100

10.



1000 (r 1 – r 2) = 250



r 1 – r 2=

7. S.I . r equir ed = Rs.

1800  12  5 100

= Rs. 1080. Ti me =

100  1080 = 9 year s. 120  10

8. L et t he sum Rs. x . Then

9 1 7 1    x  11    x 11    = 5500 2 100 2 100  

22 x = 5500 200



11x = 550000



x = 50000

250    0.25% 1000 4

11. I n t he fir st year, int er est = Rs. 110 I n t he second year, int er est = Rs. 121 Thus an additional inter eset of Rs. 11 is ear ned in the second year. This additional inter est is ear ned on the inter est ear ned in fir st year i.e., on Rs. 110.

 Rat e of int er est =

11  100 = 10% 110  1

12. L et t he capit al be Rs. x . Then

x  8 1 31 1 x  = 615 100 4 100 32x – 31x = 61500  4 x = 246000

 

13. L et t he invest ment be Rs. x . Then

x

n = 16



1 r = 8 % 3



Fr om equat ion (i ) and (ii )

e2 j

6r = 50



 New amount = Rs. (8000 + 1920) = Rs. 9920.



1200  r  3 1000  r  3  = 50 100 100

9.

41 1 39 1  x  = 250 8 100 8 100 2x = 20000 x = 100000

 

14. L et t he sum be x . Then S.I . = x

 Time =

100  S.I .  100  x  =   years Sum  Rat e  x  4 

= 25 year s 15. L et t he sum be x . Then S.I . = x Time = 10 year s.

 Rat e =

FG 100  x IJ % = 10% H x  10 K

Now, sum = x , S.I . = 2x , Rat e = 10%

 Time =

FG 100  2 x IJ years = 20 years. H x  10 K

Simple Interest and Compound Interest

LEVEL-1

10      CI  8000  1  2  100  

4800  8.S  2.2S = Rs. 918 100

1.

SI 

2.

500  4  6.25%  400  5  t

This is given in option 4.

L et t he r equir ed sum be Rs. x.

12.

108 108  x   2916 100 100  x = Rs. 2500

4.

PRT 10  300  500  T 100 100  T = 6 years SI 

Given that Rs. 200 becomes Rs. 240 in 4 years, thus it would have become Rs. 210 at the end of first year.

13.

10 × 100 = 200

7.

8.

The money gets doubled in 5 years which means it becomes twice of itself after every 5 years. Hence, it will be increased to 4 times in 10 years and 8 times in 15 years.

200  30  6  Rs.36 100 If the money gets doubled in 4 years then it will become 5 times in 8 years and 8 times in 12 years.

 800  9.

1 1   840 10 2  Total I = 800 + 840 = 1640 P = Rs. 500 A = Rs. 700  Interest = 200 

15.

1 PRT year = 2 100

10 1  100 2

= 600 CI for next

14.

1 10 1  year  600  600  2 100 2

r   A = P 1    100 

5    1.05   1    100 

11.

T

LEVEL-2 2

4   1. A  7500  1    812  100 

= Rs. 1230

PRT 100

 SI  800 

T

 No. of years = 0.5

 Total CL = 600 + 630

SI 

T

10      2100  2000  1  2  100  

= 630

10.

500  5  T 100

 T = 8 years

5 5   100 100 2

 12000 

1 year 2

 800  800 

PRT 100

Compound interest for

PRT 1 1  16000   = 800 100 10 2

CI for 2nd

S.I 

Simple interest =

1 year 2

CI for first



5%.

6.

Simple interest = Amount – Principle = 800 – 500 = 300

Hence, rate of simple interest = 5.

 8000

= 820

 t  6.25 years 3.

1 2

5  3  120 100

r     2 CI  P  1  100   

T 2

 CI = 8112 – 7500 = 612 2. According to the question sum becomes thrice.  If Rs. P is invested, it becomes 3P  Interest earned = 2P

P

 2P 

P  10  T  T  20years 100

8.9

8.10

Simple Interest and Compound Interest

3. According to the question

r   9680  P  1    100 

10. Let money borrowed be Rs. x According to the question

2

3 10648  r  r     1  10648  P  1       9680 100 100

r 100  r = 10% P.a 4. Let c be the cost of T.V.  1.1  1 

 x = 111120. 11. According to the question

x

xr3  14160 100

....(1)

r  x   r   x3  4 x  14700. ....(2) 100

  1 1 c  4945 1    1.075 (1.075)2  

solving (1) & (2)  x =12000.

= 4945 + 4600 + 4279 = 13824

r   5. 8640  5000  1    100 

2 3 48668 1.5  1.15  1  x 1.15  

T

12. I = Interest ...(1)

I1st year  T

r 3  k  5000  1    100 

P 20

I2nd year 

P P 21P   20 400 400

I3rd year 

P P 21P   20 400 20  400

cubing both sides

r  3 k 3   5000  1    100 

T

...(2)



Divide (1) and (2)  k3 = (5000)2 × 8640  k = 6000 6. Let equal instruments be of Ps. x. 2

 P = 244800 13. Let each installment be Rs. x 3

x 1.125  1.125  1  62496 1.125    x = 26244 (Ans) 7. Let each sum be Rs. x According to the question x  6   t  24  2  5  t  100 100

x5f  28800 Also, x  100 x = 18,000 (Ans) 8. (22750 – 6200) (1.2)3 = x[(1.1)2 + (1.1) + 1]  x = 6655

P  1.3r   3 P  r  3   540 9. 100 100  Pr = 60000 Also, P 

Pr3  13,800 100

 r = 5%

4P  642.6 8000

 129780 1.0665 

3

2 x 1.0665  1.0665  1  

 x = 49152 14. Let early equal installments be x. (26480) (1.1)3 2  1.1  1.1  1 x  

 x = 10648 15. Let sum with B = x  sum with A = (78060 – x) According to the question 9

4  4    x 1    78060  x   1    100 100  2

4    x 1    78060  x  100   x = 37500.

7

Profit, Loss and Discount

9

9.1

Profit, Loss and Discount

CHAPTER

PRI CE When one per son ent er s int o a t r ansact ion, he buys t he ar t icle or manufact ur es t he ar t icle at a cer t ain pr ice known as Cost Price. Then he t r ies t o sell at pr ice mor e t han t he cost pr ice, t he pr ice at which he sells t he ar t icle is known as Selling Price. I f he sells t he ar t icle at mor e t han t he cost pr ice, he get s t he pr ofit . Pr ofit = Selling pr ice – Cost pr ice But if he sells t he ar t icle at lower pr ice t han t he cost pr ice t han he get s t he loss. L oss = Cost pr ice – Selling pr ice Same way when a per son decides t o sell t he ar t icle at a cer t ain pr ice he wr it es a pr ice on t he ar t icle above t he selling pr ice or t he pr ice at which he want s t o sell it . The pr ice t hat he mar ks is known as M arked Pr ice. H e wr it es mor e pr ice on t he ar t icle becouse ever ybody asks for decr easing t he pr ice, which is wr it t en or mar ked. So decr ease in pr ice is known as discount . Discount = M ar ked pr ice – Selling pr ice At t imes 2 cust omer s come for t he same ar t icle; t he shopkeeper sells t he ar t icle above t he mar ked pr ice, t he pr ices above t he mar ked pr ices in known as Premium. Premium = Selling price – M arked price +Pr ofit CP

+Pr emium SP

– L oss

MP – Discount

PROFI T% AN D LOSS% Pr ofit is t he differ ence bet ween selling pr ice and cost pr ice and is t aken on t he cost pr ice. E xample. A man buys an ar t icle for ` 2000 and sells it for ` 2500. H e get s a pr ofit of 500 ` (2500 – 2000) on invest ing ` 2000. Per cent means upon hundr ed as t old in per cent age chapt er. So by unit ar y met hod On ` 2000

  or

pr ofit made is ` 500

500 2000 500 On ` 100 pr ofit made is `  100 2000 On ` 1 pr ofit made is `

Pr ofit % =

Pr ofit , (i.e. 500)  100 Cost price, (i.e. 2000)

Similar ly, if a per son get s some loss, t hen L oss% =

L oss  100 Cost pr ice

9.2

Profit, Loss and Discount

DI SCOU N T% AN D PREM I U M % Pr ofit on t he cost pr ice is t he same as Pr emium on t he M ar ked pr ice and loss on t he cost pr ice is same as Discount on t he M ar ked pr ice, i.e. (i) A man pur chases an ar t icle for ` 2000 and sells for ` 2500 get s a pr ofit of ` 500. Same way a man mar ks t he ar t icle for ` 2000 and sells for ` 2500 get s a pr emium of ` 500. (ii) A man pur chases an ar t icle for ` 2000 and sells for ` 1600 get s a loss of ` 400. Same way a man mar ks t he ar t icle for ` 2000 and sells for ` 1600 gives a discount of ` 400. So Pr emium% = Pr emium/M ar ked pr ice  100 and,

Discount % = Discount /M ar ked pr ice  100

Ther e is no dir ect r elat ionship bet ween cost pr ice and M ar ked pr ice. But bot h ar e r elat ed t o selling pr ice in some way. CP + Pr ofit = SP

FG P%  CPIJ H 100 K FG100  P% IJ  CP H 100 K

CP +

Same way if it is loss t han

FG100 – L % IJ CP H 100 K

= SP = SP

= SP

Selling pr ice and mar ked pr ice ar e r elat ed as M P + Pr emium = SP or or

FG100  Pr emium%IJ M P H K 100 FG100 – Discount %IJ M P H K 100

= SP = SP

Not e : I n quest ion of pr ofit and loss, a gener al mist ake made by st udent s is -

Finding t he pr ofit %, loss% or Discount % on t he selling pr ice. So adher e t o t he fact t hat P%, L % has t o be calculat ed on t he cost pr ice and D% has t o be calculat ed on t he mar ked pr ice.

PRACT I CE E XE RCI SE OBJECTI VE TYPE QU ESTI ON S 1. I f C.P. of 21 or anges i s equal t o t he S.P. of 18 or anges, t hen pr ofit per cent is

1 ( a) 7 % 7

5 (b) 5 % 9

2 1 (c) 16 % (d ) 7 % 3 3 2. A man sold t wo hor ses for Rs. 3000 each gaining 25% on t he one and losing 25% on t he ot her. H is loss per cent is 1 ( a) 7 % 3

(b) 6

1 % 4

1 (c) 7 % 3

5 (d ) 5 % 9

3. A man pur chases some or anges @Re. 1 for 6 and an equal number @ Re. 1 for 4. H e mixed t hem and sol d @ 20 pai se each. H i s gai n or l oss i n per cent is (a) loss 5%

(b) loss 4%

(c) pr ofit 5%

(d ) pr ofit 4%

4. Rajan buys lemons at t he r at e of 9 for 80p and sells t hem at 11 for 120p. H is gain per lemon is ( a)

200 99

(b)

11 99

(c)

120 99

(d )

11 9

Profit, Loss and Discount

5. By selling an ar t icle at Rs. 1250, a gain of 25% is made on t he CP. At what pr ice should t he ar t icle be sold in or der t hat a loss of 20% is made on t he selling pr ice ? (a) Rs. 800

(b) Rs. 850

1 2 (c) Rs. 833 (d ) Rs. 833 3 3 6. Raj an sol d h i s w at ch f or Rs. 75 an d got a per cent age of pr ofit equal t o t he cost pr ice. The cost pr ice of t he wat ch is (a) Rs. 40 (b) Rs. 60

13. Pr ofit ear ned by selling an ar t icle for Rs. 1,060 is 20% mor e t han t he loss incur r ed by selling t he ar ticle for Rs. 950. At what pr ice should the ar t icle be sold t o ear n 20% pr ofit ? (a) Rs. 1,800

(b) Rs. 1,080

(c) Rs. 1000

(d ) Rs. 1200

14. Even aft er r educing t he mar ked pr ice of a T.V. by Rs. 320, a shopkeeper makes a pr ofit of 15%. I f t he cost pr i ce be Rs. 3200, what per cent age of pr ofit would he have made if he had sold t he T.V. at t he mar ked pr ice ? (a) 10%

(c) Rs. 50 7. By sel ling 15 mangoes, a fr uit sel ler gai ns t he selling pr ice of 3 mangoes. H is gain is (a) 25%

(b) 16%

(c) 24 %

(d ) 27 %

8. Ram sol d a cow t o Rah i m at 20% pr of i t . Rahim sold it t o Rober t at 25% pr ofit . I f Rober t paid Rs. 900, t hen Ram had pur chased t he cow (in r upees) for (a) 600

(b) 700

(c) 750

(d ) 800

9. By selling a t owel for Rs. 126.90, a dr aper loses 6% For low much should he sell t he t owel t o gain 4% ? (a) Rs. 130

(b) Rs. 140.40

(c) Rs. 145.05

(d ) Rs. 160

10. By selling an ar t icle for Rs. 3640, a man loses 9%. H is gain or loss per cent if he sells it for Rs. 4200, is (a) no loss no again

(b) loss 5%

(c) gain 5%

(d ) gain 7.5%

11. Rajan buys mangoes at t he r at e of 3 kg for Rs. 21 and sells t hem at 5 kg. for Rs. 50. To ear n Rs. 102 as pr ofit , he must sell (a) 34 kg

(b) 52 kg

(c) 26 kg

(d ) 32 kg

12. Rajan pur chased a r efr i ger at or wi t h a mar k ed pr ice of Rs. 6000 in a sale wher e 25% discount was being offer ed on t he mar ked pr ice. H e was given a fur t her discount of 10% on t he amount ar r ived at aft er giving 25% discount . What was t he final amount paid by t he Rajan ? (a) Rs. 3900

(b) Rs. 4050

(c) Rs. 5400

(d ) Rs. 5650

(b) 20%

2 % 3 15. A mer chant blends t wo var iet ies of t ea fr om t wo differ ent t ea gar dens, one cost ing Rs. 45 per kg and ot her Rs. 60 per k g i n t he r at i o of 7 : 3 r espect i vel y. H e sel l s t he bl ended var i et y at Rs. 54.45 per kg. H is pr ofit per cent is (a) 5% (b) 10% (c) 25%

(d ) Rs. 52.50

9.3

(c) 9

1 % 11

(d ) 16

1 (d ) 11 % 9

LEVEL-1 1. A man sel ls hi s t wo car s at t he same pr ice. I n one car he makes a pr ofi t of 10%. I n ot her car he loses 10% over t he cost pr i ce. H i s t ot al gai n or loss per cent is (a) 1% loss (c) 2% loss

(b) 1% gain (d) No l oss no gain [RRB JE 2014 GREEN SH I FT ]

2. A man buys an article for Rs. 490 and sells it for Rs. 465.50. Find his loss percentage. (a) 4% (b) 4.5% (c) 5% (d) 5.5% [RRB JE 2014 RED SH I FT ]

3. A dealer professing to sell his goods at cost price, uses 900 gm weight for 1 Kg. His gain percent is (a) 9% (b) 10% (c) 11%

1 (d) 11 % 9 [RRB JE 2014 YEL L OW SH I FT ]

4. The setting price of a table is price. The gam percent is

1 (a) 20 % 3

1 (b) 20 % 2

1 (c) 25 % 4

1 (d) 33 % 3

4 times its cost 3

[RRB JE 2014 YEL L OW SH I FT ]

9.4

Profit, Loss and Discount

5. X, Y, Z started a business by investing Rs. 27000, Rs. 81000 and Rs. 72000 respectively. At the end on one year. Y's share of total profit was Rs. 36000. What was the total profit ? (a) Rs.108000 (b) Rs. 116000 (c) Rs. 80000 (d) Rs. 92000 [RRB JE 2014 YEL L OW SH I FT ]

6. Successive discounts of 20% and 10% are equivalent to a single discount of (a) 25% (b) 26% (c) 28% (d) 30% [RRB JE 2015 26 th AU G 1 st SH I FT ]

7. If CP of 10 articles is equal to SP of 8 articles, then in the whole transaction there is a (a) Profit of 20%

(b) Loss of 20%

(c) Loss of 25%

(d) Profit of 25% [RRB JE 2015 26 th AU G 1 st SH I FT ]

8. To gain 25% after allowing a discount of 10%, the shopkeeper should mark the price of the article which cost him `360 as (a) `450

(b) `460

(c) `486

(d) `500 [RRB JE 2015 26 th AU G 1 st SH I FT ]

9. A shopkeeper mixes 26 kg of tea which costs him `800 per kg with 30 kg of tea which costs him `1440 per kg. He sells the mixed tea at `1200 per kg. His gain is (a) 5%

(b) 8%

(c) 9%

(d) 10% [RRB JE 2015 26 th AU G 1 st SH I FT ]

10. Successive discounts of 30% and 10% are equivalent to a single discount of (a) 20%

(b) 35%

(c) 36%

(d) 37% [RRB JE 2015 26 th AU G 2 nd SH I FT ]

11. If C.P. of 12 articles is equal to S.P. of 10 articles, then in the whole transaction there is a profit of 50 % 3 (c) 25%

(a)

25 % 3 (d) 20%

(b)

[RRB JE 2015 26 th AU G 2 nd SH I FT ]

12. To gain 25% after allowing a discount of 20%, the shopkeeper should mark the price of the article which cost him `400 as (a) `525

(b) `580

(c) `625

(d) `650 [RRB JE 2015 26 th AU G 2 nd SH I FT ]

13. A shopkeeper mixes 40 kg of sugar which costs him `36 per kg with 27 kg of sugar which costs him `30 per kg. He sells the mixture at `35 per kg. His gain percent is (a)

38 % 9

(b)

400 % 49

(c)

100 % 9

(d) 10% [RRB JE 2015 26 th AU G 2 nd SH I FT ]

14. Successive discounts of 40% and 20% are equivalent to a single discount of (a) 60%

(b) 55%

(c) 54%

(d) 52% [RRB JE 2015 26 th AU G 3 rd SH I FT ]

15. If CP of 25 articles is equal to SP of 20 articles, then in the whole transaction there is a profit of (a) 10%

(b) 20%

(c) 25%

(d) 30% [RRB JE 2015 26 th AU G 3 rd SH I FT ]

LEVEL-2 1. A , B & C i n vest Rs. 26000, Rs. 34000 an d Rs. 10000 r espect ivel y in a business. They ear n a pr ofit of 3500. B 's shar e i n t he pr ofit i s: (a) Rs. 1200

(b) Rs. 1500

(c) Rs. 1700

(d) Rs. 1900 [RRB SSE 2014 GREEN SH I FT ]

2. A shop r educed t he pr ice of an ar t icle by 25%. I t s sale for t hat ar t icle incr eased by 25%. What is t he net effect on sal es i n r upees? (a) No change (b) I ncr ease by 5.75% (c) Decr ease by 5.75% (d) Decr ease by 6.25% [RRB SSE 2014 GREEN SH I FT ]

3. A mer chant is mi xing t wo qualit ies of r i ce, one which fir st pr ocur es at Rs. 70/kg and second at Rs. 40/kg i n t he r at io of 7 : 3 r espect ivel y. At what pr ice shoul d he sel l t he mixt ur e t o ear n a pr ofit of 20%? (a) Rs. 73.20/kg (b) Rs. 74/kg (c) Rs. 74.6/kg (d) Rs. 75.4/kg [RRB SSE 2014 GREEN SH I FT ]

Profit, Loss and Discount

4. A shopkeeper pur chased 100 or anges for Rs. 330 and t hen sold t hese or anges at t he r at e of Rs. 48 per dozen. What is hi s per cent age pr ofit ?

1 (a) 12 % 2 1 (c) 10 % 2

2 (b) 14 % 7 (d) 15%

5. I f a fr ame is sold at Rs. 60. t her e i s a l oss of 15%. For a pr ofi t of 2%, t he fr ame is t o be sol d at (a) Rs. 70

(b) Rs. 72

(c) Rs. 75

(d) Rs. 85 [RRB SSE 2014 YEL L OW SH I FT ]

6. On sel ling 100 pens, a shopk eeper gai ns pr ice of 20 pens. H is gai n per cent is (c) 15%

2 % 3

(b) gain of 14

2 % 7

(c) loss of 14

2 % 7

(d) gain of 16

2 % 3

[RRB SSE 2015 1 st SEP 2 nd SH I FT ]

[RRB SSE 2014 RED SH I FT ]

(a) 25%

(a) loss of 16

9.5

(b) 20% (d) 12% [RRB SSE 2014 YEL L OW SH I FT ]

7. A man bought an ar t icle for Rs. 240 and sol d i t at a loss of x% .H ad he pur chased it at 10% lesser cost pr ice and sold it or Rs.42 mor e, t hen he would

1 have had a gai n of of t he new cost pr ice. 4 The value of x is (a) 4

(b) 5

(c) 6

(d) 8 [RRB SSE 2015 1 st SEP 1 st SH I FT ]

8. Sur bhi makes a pr ofi t of 25% by sel li ng a pen at a cer t ain pr ice. I f she char ges Rs 1 mor e on each pen, she would gain 40 % .The or i gi nal cost pr i ce of one dozen pen i s(i n Rs.) (a) 60

(b) 72

(c) 80

(d) 84 [RRB SSE 2015 1 st SEP 1 st SH I FT ]

9. The sel l i ng pr i ce of a cer t ai n commodi t y was r educed by 20%. As a r esul t of it , t he sale was incr eased by 30%. What was t he t ot al effect of it on cash col lect ed by daily sal e? (a) 4% i ncr ease

(b) 4% decr ease

(c) 2.5% decr ease

(d) 2% i ncr ease [RRB SSE 2015 1 st SEP 2 nd SH I FT ]

10. M ar k ed pr ice of a washing machi ne is Rs.7200.

2 % of t he mar k ed 3 pr ice, t he gai n is 25%. I f i t i s sold for Rs.1600 below mar ked pr i ce, t hen t her e is a I f i t is sold at a di scount of 16

11. H ami d sol d a chair at a pr ofit of 6.5%. I f he had sold it for Rs.687.5 mor e, he would have gained x%. I f t he cost pr ice of t he chai r is Rs.12500 t hen t he value of x is (a) 10

(b) 12

(c) 14

(d) 16 [RRB SSE 2015 1 st SEP 2 nd SH I FT ]

12. A t r ader bought 864 ar t icles and sold 800 of t hem for t he pr i ce he paid for 864 ar t icle . H e sold t he r emaining ar t icl es at t he same pr ice per ar t icle as t he ot her 800. The per cent age gai n on ent i r e t r ansact ion is

1 (a) 7 % 2

(b) 8%

1 (c) 8 % 2

(d) 9% [RRB SSE 2015 1 st SEP 3 rd SH I FT ]

13. Anki t pur chased an ar t i cle for Rs.600 and sol d it at t h e gai n of 30% . F r om t hat am ou nt , h e pur chased anot her ar t icl e and sold i t at a loss of 30% . I n t he ent ir e t r ansact i on he has a (a) L oss of 9%

(b) Gain of 9%

(c) L oss of 6.9 %

(d) Gain of 6.9% [RRB SSE 2015 1 st SEP 3 rd SH I FT ]

14. A shopkeeper mar ks his goods at such a pr ice t hat aft er all owing a discount of 15% on t he mar k ed pr ice, he st il l ear ns a pr ofi t of 15 %. The mar k ed pr i ce of an ar t i cl e w h i ch cost s h i m Rs. 8500 is (a) Rs. 11000

(b) Rs. 11500

(c) Rs. 12000

(d) Rs. 12500 [RRB SSE 2015 2 nd SEP 1 st SH I FT ]

15. A per son bought an ar t icle at

4 of it s selling pr ice 5

and sol d at 10% mor e t han it s or i gi nal sell ing pr ice. H i s gai n per cent is (a) 20

(b) 10

(c) 18.75

(d) 37.5 [RRB SSE 2015 2 nd SEP 1 st SH I FT ]

9.6

Profit, Loss and Discount

AN SWE RS OBJECTI VE TYPE QU ESTI ON S 1. (c) 11. (a)

2. (b) 12. (b)

3. (b) 13. (c)

4. (a) 14. (b)

5. (c) 15. (b)

6. (c)

7. (a)

8. (a)

9. (b)

10. (c)

7. (d)

8. (d)

9. (a)

10. (d)

7. (b)

8. (c)

9. (a)

10. (d)

LEVEL-1 1. (a)

2. (c)

3. (d)

4. (d)

5. (c)

11. (d)

12. (c)

13. (a)

14. (d)

15. (c)

6. (c)

LEVEL-2 1. (c)

2. (d)

3. (a)

4. (b)

5. (b)

11. (b)

12. (b)

13. (a)

14. (b)

15. (d)

6. (b)

E XPL AN AT I ON S OBJECTI VE TYPE QU ESTI ON S 1.

3 2 SP – CP  100  16 %  100% = 18 3 SP

2.

1 2.5  2.5 = 6 % 4

 Rober t ’s cost pr ice = Rs. 150 I f Rober t ’s cost pr ice is Rs. 150, t hen Ram’s cost pr ice = Rs. 100 H ence if Rober t ’s cost pr ice is Rs. 900, t hen Ram’s cost pr ice = Rs.

3. CP for 12 eggs = 2 ; SP (24 eggs) = 4.80 N ow,

3 CP (for 12 eggs) = 5 CP for 24 eggs

 % L oss = 4. CP =

9. CP = Rs.

new SP =

80 120 ; SP = 9 11

10.

7.

 i.e.

125 1250 = 100 CP  CP = 1000 N ow, 120% of SP = 1000, 

SP = 833

FG 100  126.90IJ = Rs. 135 H 94 K

H ence t o gain 4%,

20  100%  4% 5

120 80 200    Gain per lemon = 11 9 99 5.

100  900  Rs. 600 . 150

91 = ? ?= Gain =

135  104 = Rs. 140.40 100

3640 4200 105% 5%

11. P = 10 – 7 = 3

 SP =

1 3

102 = Rs. 34 per kg 3

()25%  Rs. 4500 12. Rs.6000  ()1500

3  100% = 25% 12

()10%   Rs. 4050 ()450

8. L et Ram’s cost pr ice = Rs. 100  Ram’s selling pr ice = Rs. 100 

13. L et CP = K

100  20 20

= Rs. 120  Rahim’s cost pr ice = Rs. 120  Rahim’s selling pr ice = Rs. 120  = Rs. 150

100  25 100

 1960 – K =

120 (K  950)  K = Rs 1000 100

14. 15% of 3200 = 480 M .P. = 3200 + 480 + 320 = 4000 P = 4000 – 3200 = 800

 P% =

800  100%  20% 4000

Profit, Loss and Discount

15.

7  45 = 315 3  60 = 180 CP of 10 kg = 495 CP of 10 kg = 49.5 Pr ofit = 54.54 – 49.5 = Rs. 4.95 4.95 % pr ofit =  100% = 10% 49.5





If the selling price and profit +/ loss percent is same then there is always a loss of  10  10    % i.e. 1%. 100 

x

11.

Hence, there would be a net discount of 37% given; CP × 12 × = SP × 10



L oss % =

3.

Requir ed per cent =

4.

SP =

100 1  100  11 % 900 9

12.

13.

4  CP 3

43 1  100  33 % 3 3 4 Y's shar e = Rs. 36000 whi ch is =  81000 9 H ence, t ot al pr ofit

8.

 MP = Rs. 625. CP of (per kg) sugar after mixing

Hence, gain percent

2250 67  100 2250 67

35  



67  35  2250 95  100   10 2250 225

Successive discounts of 20% and 10% will be equal to a single discount of:



38 % 9

14.

 SP : CP = 5 : 4 = 1.25

Let initial value = 100 After 1st discount of 40%, value = 60 Discount1 = 40

Hence, there would be a profit of 25% in the entire transaction.

After 2nd discount of 20%, value 60 

Given that: 10 × CP = 8 × SP

Given that: MP × 0.9 = 360 × 1.25

Discount2 = 12 Single discount = (40 + 12) = 52%

Hence, MP or marked price will be Rs. 500. 9.

25  400 16

4  27000  81000  72000  9 = Rs. 80,000

– 20 – 10 + (– 20 × – 10)/100 = – 28 or 28%. 7.

4 5   400 5 4

40  36  27  30 2250  Rs. 40  27 67

=

6.

MP 

 MP 

 Requir ed pr ofit =

5.

SP 6  CP 5

 SP = 1.2 CP Hence, there will be a profit of 20%

490  465.5  100 = 5% 490

2.

7 9 63   x 10 10 100

10.

LEVEL-1 1.

Per kg cost price of the tea after mixing the

(26  800  30  1440) two varieties = (26  30) 8000 = Rs. 7 and selling price is given as Rs. 1200 Hence, required profit percent 8000    1200   7    100  5% . 8000 7

9.7

15.

Let CP of 1 article = x  CP of 25 articles = 25 x 3P of 25 articles 

 Profit =

25x  25 20

SP  CP  100  25% CP

625x  25x 20  100  25% 25x

4  48 5

9.8

Profit, Loss and Discount

LEVEL-2 1. The ratio is which profit distributed will be ratio of their investments.  B’ share

34000    3500    26000  34000  10000  = 1700 2. Let initial price be Rs. 100  New price 

3  100  Rs.75 4

Let initial sales = 100 units  New sales 

3. x =

5  100  125 4

 1  x 7. S.P initial  240   100  New C.P = 0.9 (240) = 216

 1  x New S.P  240    42 100 

x    240  1    42  1.25  216   100  x=5 8. old C.P = x  old S.P = 1.25 x New S.P = (1.25 x + 1) According to the question 1.25 x + 1 = 1.4 (x)

70 × 7 + 40 × 3 10

x

x = Rs.610

1 0.15

with a profit of 20% New x = 610 

20  610 100

Rs.73.20/kg

 lost of dozen 

1  12 = Rs. 80 0.15

9. Initial cash collected = SP × Quality Sold =x×y

4. C.P of one orange =

50  Rs.3.5 100

Final cash collected = 0.8x × 1.3y = 1.04 xy  4% Increase (Ans) 10. According to the question

S.P one orange =

 Pr ofit % 

48 4 12

Pr ofit  100 CP

0.5 2   100  14 % 3.5 7

SP  7200

50  7200  6000 3  100

 gain = 25 % CP=

4  6000  4800 5

if SP = 7200 – 1600 i.e 5600

5. According to the question

60 CP  0.85

 gain % 

5600  4800  100 4800

 for a profit of 2%

 gain % 

SP  CP 2  100%  16 % CP 3

 60  = 72 S.P  1.02   0.85  6. According to the question Profit = 20 C.P;

C.P  cost price

11. SP = 6.5 % of 12500

 6.5  x% =   12500  687.5  100 

100 SP – 100 P = 20 CP  100 SP = 120 CP  SP = 1.2 CP

Pr ofit% 

1.2CP  CP  100  20% CP

x = 6.5 +

12500

6875  12 (Ans) 125

Profit, Loss and Discount

12. Let price of 1 article = Rs. x  C.P = 864, Total SP 

864  864x 800

 864  864  x  864x   800    100 = 8%  % gain  864x 13. At the end of transaction, he lead

1.3  600   0.7  546  loss = 600 – 546 = 54  loss % 

54  100  9% 600

9.9

14. M. P = x

 S.P  0.85x 

0.85x  8500  100  15 8500

 x = 11500 15. Let original SP = x  CP = 0.8x  SPnew = 1.1x

gain  

.1.1x  0.8x  100% = 37.5% 0.8x 

10

Average

C H A P TE R

AV ERAGE The Aver age (ar i t hmet ic mean) of a gr oup or set of N number s is defined as t he sum of t hose number s di vi ded by N. H er e N is t he number of values or obser vat ions in a set . Aver age =

Sum of N number s/values Numbers of values/observat ions

Sum or Sum = N  Aver age N e.g. I f wei ght s of t hr ee chi ldr en ar e 80, 90 and 76 pounds, t hen aver age of t he wei ght s of chil dr en can be calcul at ed as : A=

Aver age =

80  90  76 Sum of weight s of childr en = = 82 Pounds. 3 Number of childr en

U sing Average to find a N umber Somet imes aver age wil l be given and we have t o fi nd t he mi ssi ng number / obser vat ion. I n t his t ype of case, t hey give us. (1) Aver age value and t he number of obser vat ions, aver age of which is given. They also give t he sum of obser vat ions. (2) Except t he missing obser vat ions, (I f mor e t han one obser vat ions ar e missing) e.g. I f aver age of 3, 4, 5 and x i s 5, what i s t he value of x ? We have given, N = 4 obser vat ions, and aver age of t hese obser vat i ons i s 5.



Aver age =

Sum N

345x  20 = 12 + x  x = 8. 4 H ence, t he mi ssi ng obser vat ion i s 8. e.g. I f aver age of fi ve number s i s 10 and sum of t hr ee number s is 16, t hen what is t he aver age of ot her t wo number s. L et t he five number s a, b, c, d and e. 

5=

a  b c d  e = 10 5 Al so, gi ven, a + b + c = 16. d  e 16  = 10 5 de  8 = 14   2 So, t he aver age of ot her t wo number s is 17. So as gi ven,

dc 2

= 17

Average in Different Cases 1. I f all t he number s in a set ar e t he same, t hen t hat number is t he aver age of t hat set . e.g. Aver age of set of values 5, 5 and 5 is 5. Also by t heor y,

Aver age =

15 Sum 555 = = =5 3 N 3

10.2

Average

2. I f t he number s in a set ar e not all t he same, t hen t he aver age must be gr eat er t han t he smallest number and less t han t he gr eat est number in t hat set . e.g.

Aver age of set of values 83, 84, 87, 97, 99

Sum 83  84  87  97  99 450 = 90 = = N 5 5 90 > 83 and 90 < 99. Aver age =

H er e

Weight ed Average To calcul at e t he weight ed aver age of a set number s, mul t i ply each number i n t he set by t he number of t i mes i t appear s, add all pr oduct s and divi de by t he t ot al number of number s in t he set . e.g. On one day, 20 out of 25 st udent s t ak e t est and t heir aver age was 80. On anot her day, t he r est 5 st udent s t ake t est and t heir aver age i s 90. What was t he aver age for t he ent ir e class ?

20(80)  5 90

1600  450 = = 82. 25 25 Fact : The aver age bet ween t wo set s of number s is closer t o t he set wi t h mor e obser vat i ons. Aver age of ent ir e class =

Some import ant F act s (1) I f each one of t he gi ven number s is incr eased by const ant K , t hen t hei r aver age is i ncr eased by K . (2) Aver age of n consecut ive nat ur al number =

n 1 2

(3) Aver age of squar es of n consecut ive nat ur al number = (4) Aver age of cubes of n consecut ive nat ur al number =

(n  1) (2n  1) 6

n (n  1) 2 4

(5) Aver age of n consecut ive even number s = n + 1 (6) Aver age of squar es of n consecut ive even number s = 2(n  1) (2n  1) 3 (7) Aver age of n consecut ive odd number s = n (8) I f each one of t he gi ven number s is mult ipl ied by const ant K , t hen t heir aver age is mult i pl ied by K .

t ot al dist ance dist ance (9) I f a car cover s some jour ney fr om A t o B at u k m/hr, and t he r et ur n t r i p at v km/hr. t hen t he aver age speed dur ing t he whol e jour ney i s, Aver age speed =

 2uv    k m/hr . u v (10) I f aver age of a set of number s is A, and i f anot her number is added t o t he set and a new aver age is calcul at ed, t hen new aver age will be l ess t han, equal t o, or gr eat er t han A, depending on whet her x is less t han, equal t o, or gr eat er t han A, r espect ivel y. (11) Aver age of t hr ee differ ent speeds u , v , w t o t r avel equal dist ance =

3uvw uv  vw  uw

(12) I f a per son is r eplaced by anot her per son by which (i ) Aver age is incr eased, t hen Age of t he new comer = Age of per son left = Number of per sons  I ncr ease in aver age age (ii ) Aver age is decr eased, t hen Age of new comer = Age of per son left – Number of per sons  Decr ease in aver age age. (13) I f a per son joins a gr oup wit hout r eplacing any per son by which (i ) Aver age incr eased, t hen age of t he new comer =Pr evious aver age age + Number of per sons including new comer  I ncr ease in aver age age

Average

10.3

or , age of t he new comer = I ncr eased aver age age + Number of per sons or iginally in t he gr oup  I ncr ease in aver age age (ii ) Aver age decr eased, t hen Age of t he new comer = Pr evious aver age age – Number of per sons (including new comer )  Decr ease in aver age age. (14) I f a per son leaves t he gr oup but no body joins t he gr oup by which (i ) Aver age incr eased, t hen Age of man left = Pr evious aver age age + Number of pr esent per sons  incr ease in aver age age (ii ) Aver age decr eased, t hen Age of man left = Pr evious aver age age + Number of pr esent per sons  Decr ease in aver age age (15) Let aver age marks obtained by x candidates in an examination is n . I f average mar ks of passed candidates is p and t hat of t he failed candiat es is q, t hen (i ) Number of passed candidat es = (ii ) Number of failed candidat es =

Total candidates (Passed average  Failed average) Passed average  Failed average

Total candidates (Passed average  Total average) Passed average  Failed average

(16) Geomet r ic mean of number s x 1, x 2, ............. x n = n x1  x2 ... xn

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. A m an cov er s a cer t ai n di st an ce at 90 km/hr and r et ur ns back t o t he st ar t ing point at 60 k m/hr. H is aver age speed dur ing t he whol e jour ney is (a) 60km/hr (b) 65km/hr (c) 72km/hr (d) 75km/hr 2. A man goes up hi l l wi t h an aver age speed at 20 kmph and comes down wit h an aver age speed of 30 kmph. The distance tr avelled in both the cases being the same, the aver age speed for t he ent ir e jour ney is (a) 18 kmph (b) 22 kmph (c) 24 kmph (d) 26 kmph 3. A man t r avels fir st 80 km at 20 kmph, next 30km at 15 kmph and then 80km at 20 kmph. His average speed for t he whole jour ney (in kmph) is (a) 20 (b) 22 (c) 17 (d) 19 4. A car cover s four successi ve 4 km di st ances at speeds of 10 k m ph , 20 k m ph an d 60 k m ph r espect ively. I t s aver age speed over t ot al dist ance is (a) 10 kmph (b) 20 kmph (c) 25 kmph (d) 30 kmph 5. Fir st 150 miles of his t r ip, John dr ove at 50 miles per hour, t hen due t o t r affic, he dr ove at only 20 miles per hour for t he next 120 miles. H is aver age speed, in miles per hour, for t he ent ir e t r ip is (a) 20 (b) 28 (c) 30 (d) 32

6. A car t r avels 70 kilometer s in one hour befor e some fault happens, t hen it t r avels for 120 km at 30 kmph. For t he ent ir e t r ip, aver age speed is (a) 33 (b) 36 (c) 38 (d) 40 7. A mot or t r avels 100 miles at t he r at e of 40 miles per hour. I f it r et ur ns t he same dist ance at a r at e of 50 miles per hour, t he aver age speed for t he ent ir e t r ip, in miles per hour is

200 (b) 40 3 50 400 (c) (d) 9 9 8. At a const ant velocict y of 30 miles per hour, car A l eaves point X  at 1 p.m.. Car B l eaves ‘X’ at 3 p.m. at const ant vel ocit y and over t ak es A at 5 p.m. The aver age speed of B will be (a) 20 (b) 25 (c) 40 (d) 60 9. A car cover s a distance in 36 minut es. I t r uns at 50 kmph on an aver age. The speed at which t he car must r un t o r educe t he t i me of jour ney by 6 minut es will be (a) 30 kmph (b) 42.86 kmph (c) 55 kmph (d) 60 kmph (a)

10.4

Average

10. A car t r avelled fr om city P t o cit y R in 30 minut es. The fir st half of t he dist ance was cover ed at 50 miles per hour, and t he second half was cover ed at 60 miles per hour. What was t he over age speed of t he car ?

200 400 (b) 11 11 500 600 (c) (d) 11 11 A man dr ives 150 km fr om A t o B in 3 hour s 20 minut es. H e comes back fr om B t o A in 4 hour s 10 minut es. Then, aver age speed fr om A t o B exceeds t he aver age speed for t he ent ir e t r ip by (a) 4 km/hr (b) 4.5 km/hr (c) 5 km/hr (d) 3 km/hr The aver age of fir st 50 nat ur al number s is (a) 22.25 (b) 24.25 (c) 25 (d) 25.5 M ean of 12, 22, 32, 42, 52, 62, 72 ............ 502 is (a) 850.5 (b) 858.5 (c) 854.5 (d) 852.5 Aver age of all odd number s upt o 100 is (a) 51 (b) 50 (c) 49.5 (d) 49 Aver age of 7 consecut i ve number s i s 33, t hen lar gest of t hese number s is (a) 106 (b) 105 (c) 104 (d) 107 (a)

11.

12.

13.

14.

15.

LEVEL-1 1. Find the average of all prime numbers between 30 and 50 : (a) 48

(b) 39

(c) 39.8

(d) 38 [RRB JE 2014 RED SH I FT ]

2. The average age of 5 members of a committee is the same as it was 3 years ago, because an old member has been replaced by a new member. The difference between the ages of old and new member is (a) 12 years

(b) 4 years

(c) 8 years

(d) 15 years

4. The average of 35 observations is 30. Out of these observations the average of first 18 observations is 30 and the last 18 observations is 40. The 18th observation is (a) 180 (b) 190 (c( 200 (d) 210 [RRB JE 2015 26 th AU G 1 st SH I FT ]

5. The average of first five prime numbers is (a) 3.6 (b) 4.5 (c) 5.5 (d) 5.6 [RRB JE 2015 26 th AU G 2 nd SH I FT ]

6. The average weight of 100 students is 46 kg. The average weight of boys is 50 kg. If the number of boys is 60, the average weight of girls in kilograms is (a) 35 (b) 40 (c) 45 (d) 50 [RRB JE 2015 26 th AU G 2 nd SH I FT ]

7. The average of first six even whole numbers is (a) 5 (b) 6 (c) 7 (d) 8 [RRB JE 2015 26 th AU G 3 rd SH I FT ]

8. The average marks of students of section A an B are respectively 60 and 70. The number of students in section A is 47 and in section B is 53. The average marks of both sections taken together are (a) 65.5 (b) 65.3 (c) 65.6 (d) 65.8 [RRB JE 2015 26 th AU G 3 rd SH I FT ]

9. The average weight of 100 students is 46 kg. The average weight of girls is 40 kg. If the number of girls is 40, the average weight of boys in kilograms is (a) 40 (c) 50

(b) 45 (d) 60 [RRB JE 2015 27 th AU G 1 st SH I FT ]

10. If 27, 12. 24, x are in proportion, then x is equal to (a) 9

(b) 30

(c) 54

(d) 60

For the above question, User had specified ‘ignore’ during keys upload.

[RRB JE 2014 YEL L OW SH I FT ]

[RRB JE 2015 27 th AU G 2 nd SH I FT ]

3. The average of 5 numbers is 40. If one number is excluded then the average becomes 38. The excluded number is

11. The average of squares of first five whole numbers is

(a) 48

(b) 58

(c) 60

(d) 78 [RRB JE 2015 26 th AU G 1 st SH I FT ]

(a) 6

(b) 7.5

(c) 9

(d) 11 [RRB JE 2015 27 th AU G 2 nd SH I FT ]

Average

12. The average of first 5 odd prime numbers is (a) 5.4

(b) 5.8

(c) 6.8

(d) 7.8 [RRB JE 2015 28 th AU G 2 nd SH I FT ]

13. The average of 18 observations is 30 and average of 22 observations is 40. The average of all combined observation is (a) 35.5

(b) 35

(c) 36

(d) 36.5 [RRB JE 2015 28

(b) 7

(c) 6

(d) 3

th

AU G 2

nd

SH I FT ]

[RRB JE 2015 28 th AU G 3 rd SH I FT ]

15. The average 16 observations is 25 and average of 24 observations is 30. The average of all 40 observation is (a) 26

(b) 27

(c) 27.5

(d) 28

5. I f aver age of 7 consecut ive number i s 203, t hen t he aver age of t he small est and second number is (a) 203 (b) 201 (c) 202.5 (d) 203.5 [RRB SSE 2015 1 st SEP 1 st SH I FT ]

14. The average of first seven whole numbers is (a) 8

[RRB JE 2015 28 th AU G 3 rd SH I FT ]

LEVEL-2 1. I f 12a + 6b = 54, what i s t he aver age of a & b? (a) 2.25

(b) 4.5

(c) 6

(d) Data insufficient

6. The aver age of n number s is x . I f the fir st number is i ncr eased by 1, second by 2, t hi r d by 3 and so on, t hen t heir aver age is y. The value of

(c) 3 : 5

(b) 5 : 2 (d) 5 : 3

(a) X

(c)

n  n  1 2

(b)

n 1 2

(d)

n 1 2

[RRB SSE 2015 1 st SEP 2 nd SH I FT ]

7. I n a mat hemat ics t est in a class t he aver age scor e of boys i s 33 and t hat of gi r l s is 42. I f t he aver age scor e of all t he st udent s in t he class is 38, t hen what is t he per cent age of gir ls in t he cl ass? (a) 44

4 9

(b) 46

5 9

(c) 55

5 9

(d) 55

4 9

[RRB SSE 2015 1 st SEP 3 rd SH I FT ]

8. Aver age scor e of gi r ls of class x in an examinat ion is 73 and t hat of boys is 71. I f t he aver age scor e of bot h boys and gi r ls i s 71.8, t hus per cent age of boys i n t he cl ass is

[RRB SSE 2014 GREEN SH I FT ]

(a) 35

(b) 40

3. The aver age of six number s is 3.95. I f t he aver age of fir st t wo number s is 3.40 and t he aver age of next t wo number s is 3.85, t hen fi nd t he aver age of t he r emaining t wo number s.

(c) 60

(d) 80

(a) 4.6

(b) 4.7

(c) 4.8

(d) 4.5

[RRB SSE 2015 2 nd SEP 1 st SH I FT ]

9. The aver age of t he t est scor es of a cl ass of x st udent s is 70 and t hat of y st udent s is 91. I f t he scor es of bot h t he cl asses ar e combi ned, t he aver age is 80.

[RRB SSE 2014 RED SH I FT ]

4. The aver age of mar ks of 28 st udent s in M at hs was 50. 8 st udent s left t he school and t hen t he aver age i ncr eased by 5. What i s t he aver age of mar k s obt ai ned by t he st udent s who l eft t he school? (a) 37.5 (b) 42.5 (c) 45 (d) 50.5 [RRB SSE 2014 YEL L OW SH I FT ]

yx

is.

[RRB SSE 2014 GREEN SH I FT ]

2. The aver age scor e of gir ls in a cl ass is 75 mar k s. T h e av er age scor es of boy s i n t h e cl ass i s 65 mar k s. I f t he aver age of t he cl ass i s 68.75 mar k s, what i s t he r at i o of boys t o gir ls in t he class? (a) 2:5

10.5

The val ue of

2x  5y is 5x  4y

(a)

11 10

(b)

33 10

(c)

24 5

(d)

26 5

[RRB SSE 2015 2 nd SEP 2 nd SH I FT ]

10.6

Average

10. The aver age of n number s is x .I f t he fir st number is decr eased by 1, second is decr eased by 2,t hi r d is decr eased by 3 and so on, t hen t hei r aver age is y. Then find t he value of x – y. (a)

(c)

n  n  1

(b)

2 n  n  1

(d)

2

13. The aver age scor e of t he st udent s i n a class is 65. The aver age scor e of boys is 60 and t hat of gir ls is 68. The per cent age of gir l s in t he class i s:

n 1 2 n 1 2

(b) 72 (d) 89 [RRB SSE 2015 2 nd SEP 3 rd SH I FT ]

12. The aver age of 100 number s i s 100. I f t he fir st number is i ncr ease by 1, second by 2, t hir d by 3 and so on , t hen t he aver age of t he number s so obt ained exceeds t he or i gi nal aver age by (a) 25.5

(b) 50

(c) 50.5

(d) 60

(c) 60

(d) 65

14. The aver age of 30 posit ive i nt eger s i s x. I f t he fir st jmber i s incr eased by 2. second by 4. t hir d by 6. four t h by 8. and so on, t hen t he aver age of t he r esult i ng number s i s y. The val ue of y-x is

11. Four positive integer s, when added thr ee at a t ime gi ve t he sums 180,197,208 and 219. Aver age of t hese four int eger s is (c) 73

(b) 62.5 [RRB SSE 2015 3 rd SEP 2 nd SH I FT ]

[RRB SSE 2015 2 nd SEP 2 nd SH I FT ]

(a) 67

(a) 37.5

(a) 27

(b) 29

(c) 31

(d) 33 [RRB SSE 2015 3 rd SEP 3 rd SH I FT ]

15. The aver age of mar k s obt ai ned by 150 st udent s was 40. I f t he aver age mar ks of passed st udent s wer e 50 and t hat of fai led st udent s was 20. Then t h e n u m ber of st u den t s w h o p assed t h e examinat i on was (a) 80

(b) 90

(c) 100

(d) 120 [RRB SSE 2015 3 rd SEP 3 rd SH I FT ]

[RRB SSE 2015 3 rd SEP 1 st SH I FT ]

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (c)

2. (c)

3. (d)

4. (b)

5. (c)

11. (c)

12. (d)

13. (b)

14. (b)

15. (a)

6. (c)

7. (d)

8. (d)

9. (d)

10. (d)

7. (a)

8. (b)

9. (c)

10. (*)

7. (c)

8. (c)

9. (c)

10. (b)

LEVEL-1 1. (c) 11. (a)

2. (d) 12. (d)

3. (a) 13. (a)

4. (d) 14. (d)

5. (d) 15. (d)

6. (b)

LEVEL-2 1. (d)

2. (d)

3. (a)

4. (a)

5. (a)

11. (a)

12. (c)

13. (b)

14. (c)

15. (c)

6. (d)

Average

10.7

EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S 1. H er e man cover s t he same dist ance, but wit h t wo differ ent speeds. When t he dist ance is same and t wo differ ent speeds ar e given for cover ing t hat distance, t hen aver age speed dur ing t he whole jour ney is given by, Aver age speed =

2uv uv

H er e, u = 90 km/hr and v = 60 km/hr.

 Aver age speed =

2  90  60 10800  90  60 150

= 72 km/hr. 2. H er e as given, man goes uphill wit h an aver age speed of 20 k mph and comes down wi t h an aver age speed of 30 kmph. Thus, u = 20kmph and v = 30 kmph I t is given that dist ance in both the jour ney is same. So, aver age speed for t he ent ir e jour ney

 2uv  km/hr =   u  v  2  20  30 = kmph 20  30  1200  =   k m/hr = 24 kmph. 50  3. M an t r avels t ot al dist ance of 80 + 30 + 80 = 190 kilomet er s. Times t aken for 1st phase =

80 k m = 4h r 20 k m /hr

for 2nd phase =

30km  2hr 15k m/hr

for 3r d phase =

80k m  4hr 20k m/hr

4. H er e a car cover s four successive 4 km dist ances at 4 di ffer ent speeds of 10 k mph, 20 k mph, 30 kmph and 60 kmph So, t ime for 1st 4 km =

4 km 24 hr s. = 10 K mph 60

2nd 4 km =

4km 12  hrs 20kmph 60

3r d 4 km =

4 km 8  hrs 30 K mph 60

4t h 4 km =

4 km 4  hrs 60 K mph 60

 Tot al di st ance = 4+4+4+4 = 16 km. H ence, aver age speed of car over 16 km =

=

16 k m 4   24 12 8      hr 60 60 60 60 

16  60 = 20 km./hr. 48

5. As given in dat a, John dr ove fir st 150 miles at speed of 50 mile per hour. So, for fir st 150 miles he t ook

150 miles = hour s 50 miles/hr Next 120 miles, he dr ove at speed of 20 miles per hour. So for t hat 120 miles, he t ook,

120 miles = 6 hour s 20 miles Thus, in t ot al he dr ove 150 + 120 = 270 miles in 6 + 3 = 9 hr s. So, his aver age speed for t he ent ir e t r ip =

Tot al dist ance John dr ove Tot al t ime he t ook

So, aver age speed for t he whole jour ney =

Tot al dist ance Tot al t ime t aken

=

150+120 miles 3+6 hrs

=

(80  30  80) 190 k m = = 19 km/hr.. (4  2  4)hr 10 hr .

=

270 miles / hr = 30 miles /hr.. 9

10.8

Average

6. Here given that a car travels 70 km in one hour befor e fault happens means it has cover ed 70 km distnace for first phase. Aft er t he faul t happened, i t has cover d t he r emaining 120 km at speed of 30 km/hr H ence, t ot al t ime for t r ip becomes = (t ime for 1st phase) + (t ime for 2nd phase) = 1 hr + 4 hr s = 5 hr s. And, t ot al dist ance becomes = (Dist ance in 1st phase) + (Dist ance in 2nd phase) = 70 km + 120 km = 190 kilomet er s.  Aver age speed =

Tot al dist ance Tot al t ime t aken

=

(120  70) km (1  4) hr

= 38 km/hr. 7. I n fir st phase, a mot or t r avels 100 miles at r at e of 40 miles per hour. So, t ime t aken dur ing 1st phase =

Dist ance Speed

=

100 miles 1 = 2 hr s. 40 miles/hr 2

Same as t hat of 1st phase, but speed t he mot or is differ ent . So t he t ime for 2nd phase will be differ ent. So t he dist ance dur ing 2nd phase is 100 miles and speed of mot or given is 50 miles/hr.

 Time (2nd phase) = =

(100  100) miles 9 hr s. 2

400 mi les/hr 9 8. As given, car A leaves point X at 1 p.m. and t r avels wit h const ant velocit y of 30 miles per hour. Car B L eaves point X at 3 p.m. and t r avels wit h const ant velocit y and over t akes car A at 5 p.m. So, at 5 p.m. t he dist ance cover ed by bot h car A and car B will be equal. L et D 1 and D 2 would be t he dist ances cover ed by car A and car B r espect ively, H ence 5 p.m. D1 = D2  (Time taken by car A  Speed of by A = Time t aken by car B  Speed of car B  30 km/hr  4 hr s = S2 (Speed of car B)  2 hr s  120 km = S2  2 hr s. =

 S2 = 60 km/hr 9. The dist ance cover ed by a car at 50 kmph is in 36 minut es. To pr oceed fur ther, we fir st find the total distance cover ed by it . Now, Dist ance = Speed  Time H er e, Speed = 50 kmph = 50 

 Dist ance =

1000 500 = 3600 36

500  36 = 500 km 36

Fur t her, we ar e gi ven t he t i me r educed by 6 minut es, i.e. 30 minut es t o complet e t he same dist ance.

Dist ance Speed

 Speed =

100 miles 50 miles/hr

Dist an ce Time



=

50 500 = 3 30

=

50 3600  km/hr 3 1000

= 2 hr s. H ence t ot al dist ance for t he ent ir e t r ip = (100 +100) = 200 miles and t ime t aken for t he t r ip

 Aver age speed (ent ir e t r ip) =

=

9 1 = 2  2 = hr s. 2 2

Tot al dist ance of t r ip Tot al t ime t aken for t r ip

= 60 km/hr 10. H er e, we apply t he for mula Dist ance = speed  t ime. The t ot al dist ance cover ed in 30 mins. The fir st half of t he dist ance was cover ed at 50 miles per hour.

Average

 Time Taken for fir st half distance =

x 2  50

wher e, x = t ot al dist ance cover ed. The second half of t he dist ance was cover ed at 60 miles per hour.  Time Taken for second half dist ance x = 2  60





x x   30 2  50 2  60 x 2

  x

= 40 km/hr. ...(3) Fr om r esult s (i ) and (ii ), we get t he differ ence of aver age speeds of (jour ney fr om A t o B) and t he whole jour ney.  Differ ence = (Average speed for A to B) – (Aver age speed for whole Jour ney) = (45 – 40)km/hr = 5 km/hr H ence, aver age speed fr om A t o B exceeds t he aver age speed for whole t r ip by 5 km/hr.

n (n  1) 2 So, aver age of fir st n nat ur al number s

12. Sum of fir st n nat ur al number s =

1   1     30 50 60

x 6  5    30 2  300 

=

Tot al Dist ance Tot al Time

=

1800 30

13. 12 + 22 + 32 + ..... +n 2 =

11

Dist ance 150 km = Time   20   3+  60   hr  

Dist ance 150 km = Time   10   4   60   hr   150 km = 36 km/hr 1 4 hr 6

...(ii )

2(Avg.Speed for A t oB) (Avg.Speed for B t oA) (Avg.Speed for A toB)  (Avg. Speed for B t oA)

=

2(45)(36) (45  36)

FG 2500 IJ = 50 H 50 K

15. L et t he number s be x, x + 1, x + 2, x + 3, x + 4, x + 5 and x + 6.



x  x 1  x  2  x  3  x  4  x  5  x  6 7

= 103

Aver age speed for t he whole jour ney =

14. Sum of odd number s upt o 100 = 1 + 3 + 5 + 7 +... + 95 + 97 + 99 = (1+99)+(3+97)+(5+95) +... upt o 25 pair s = 100 + 100 + 100 + ... (25 t imes) = 2500

 Aver age=

= 45 km/hr ...(i ) Now, we get aver age speeds for t he t r ips fr om A t o B and B t o A. Aver age speed fr om B t o A

=

FG 50  51  101 IJ = 42925. H 6 K F 42925 IJ  858.5  Requir ed aver age = G H 50 K =

600 1800 = 11 11  30 11. H er e dist ance bet ween A and B is 150 km and a man dr ives t his dist ance in 3 hour s 20 min. So his speed for t he t r ip fr om A t o B. Aver age speed fr om A t o B

=

n (n  1) (2n  1) 6

 12 + 22 + 32 + ... + 502

=

=

n (n  1) n  1  2n 2

 50  1  51  Aver age =  = = 25.5  2  2

1800 30  300 miles. 2= 11 11

Now, aver age speed =

10.9



7 x  21 = 103 7



7( x  3) = 103 7

 x + 3 = 103  x = 100  L ar gest number = x + 6 = (100+6) = 106

10.10

Average

LEVEL-1 1.

3000 Average weight of boys  number of boys

Prime numbers between 30 and 50 are: 31, 37, 41, 43, 47 

3000  50 100  40 27 : 12 : : 24 : x 

required average 10.

31  37  41  43  47  5  2.



288  10.67 27 correct ans is not available in the options.

199  39 5

x

Aver age age of t he five member s should have been incr eased by 3 year s.

11.

Squares of first 5 whole numbers = 0, 1, 4, 9, 16

12.

0  1  4  9  16 =6 5 Average of first 5 add prime Numbers

H ence, r equir ed di ffer ence = 3 x 5 = 15 year s.

3.

Let the number be x.  5 × 40 = x + 4 × 38

 Average 

 x = 48. 4.

=

Let the 18th observation be x.  18 × 30 + 18 × 40 = 35 × 30 + x

13.

 x = 210. 5. 6.

14.

First six whole numbers form on AP  0, 2, 4, ...

n (2a + (n – 1)d) = 30 2

6 = (2 × 0 +(6 – 1)2) = 30 2 30 5  Average = 6 Total marks in Section A = average marks × numbers of students = 60 × 47 = 2820 Similarly, total marks in Section B = 70 × 53 = 3710

2820  3710  65.3 47  53 Total weight of class = average weight × number of students

Average =

15.

Average =

= 46 × 100 = 4600 similarly, total weight of girls = 40 × 40 = 1600  Total weight of boys = 4600 – 1600 = 3000

sum of first 7 whole numbers 7

16  25  24  30 40

= 28

LEVEL-2 1. 12a + 6b = 54 As different values of a and b satisfies the given equation.  Data is insufficient to find average. Boys

2. Gir ls 75

 Average = 9.

18  30  22  40 18  22

0 1 2 3 4  5 6 7 =3

x = 40 kg.

Sum of Ap =

8.

Combined Average 



 50  60  x  40  100  46

7.

3  5  7  11  13 = 7.8 5

= 35.5

3  5  7  11  5.6 Required average  5 Set the average weight of girls be x kg.



7 24  12 x

65

68.75

3.75

6.25

 Boys : Girls = 6.25 : 3.75 = 5 : 3 3. Let avg of last two number be x  3.95 × 6 Let the average of last 2 number be x.

Average 10.11

According to question, 3.95  6  2  3.40  2  3.85  2x

9.

X

Y

70

91

 2x  23.7  14.5

x = 4.6

80

4. Let avg of students who left = x

55 

28  50  8x 20

11

 x : y = 11 : 10

 x = 37.5 5. Assume smallest No. = a



a  1  2  3  4  5  6  7

x 2 5 2x  5y y   5x  4y 5 x  4 y

 203

 a = 200  Avg of 200 and 206 = 203 Ans 6. Increase in sum of numbers 

n  n  1 2

Distributing the sum to n numbers 

n  n  1  n  1  2n 2

7. Boys

Girls 42

33

10

11 5 10  11 5 4 10 2

11 5  5 11 4 2 

38

36 2 24   5 3 5

10. According to question, 4

5

4:5

5  5   100  %  55 %  percentage of girls   5  4  9 8. Age of Boys 71

nx  (1  2    n) y n

nx  

n(n  1) 2 y n

Age of Girls 73

x  y 

n 1 2

11. Let the integers be a, b, c, d a + b + c = 180

71.8

b + c + d = 197 c + d + a = 208 1.2

0.8

d + a + b = 219

B:G

__________________

 3:2

3(a + b + c + d) = 804  a + b + c + d = 268

 3   100  % Percentage of boys in class   3  2 

= 60%

 Avg 

abcd  67 4

10.12

Average

 100 100  1    2 12. The increase in average = 100

(till 30 numbers) =

= 50.5 13.

14. Increase in the numbers = 2 + 4 + 6 + 8 + …

Boys 60

Girls 68

30 [2 × 2 + (30 – 1)2]= 930 2

Since total increase is 930, the increase in average of 30 numbers =

930 = 31 30

15. Let number of passed students = P

65

Let number of failed students = 150 – P

3

5

 150 × 40 = 50P + 20(150 – P)  5P + 2(150 – P) = 600

 5   100  %  62.5%  Percentage of girls   35 

 P = 100

11

Algebra

CHAPTER

QU ADRATI C EQU ATI ON S I f atleast one of the var iables in an equation is r aised to the power of two, then it is called a Quadratic equation . No var iable i n t he equat ion is r aised t o t he power higher t han 2. e.g ax 2 + bx + c H er e, x is var iable and a, b, c ar e r eal number s. As x is r aised t o power 2, so it is a quadr at ic equat ion. SOLU TI ON OF QU ADRATI C EQU ATI ON To sol ve t he quadr at i c equ at i ons t her e ar e t wo met hods: 1st M et hod (1) For solving a Quadr at ic Equat ion, set one side of t he equat ion equal t o zer o i .e. ax 2 + bx + c = 0 (2) Then split t he bx t er m in t wo such for ms t hat addit ion / subt r act ion of t hese t wo t er ms would be bx and t heir pr oduct is equal t o t he pr oduct of a and c. (3) Take common fact or s fr om t he fir st t wo t er ms of t he split ed equat ion and t hen again t ake common fact or s fr om 3 r d and 4 t h t er ms of t he spl i t ed equat ion. (4) Following t his pr ocedur e, we get t wo fact or s as a r esult . (5) Equat e t hese t wo fact or s t o zer os. H ence, we get t wo values of x . (6) These t wo values of x will be t he r oot s of the given quadr atic equation. Consider an example : (1) x 2 – 5x + 6 = 0 [by equat ing t o zer o] (2) Split 5x in t wo t er ms 3x and 2x . Addit ion of 3x and 2x = 3x + 2x = 5x and pr oduct of 3x and 2x = 3x  2x = 6x 2 which is equal t o (6 x 2) so, t he equat ion becomes : x 2 – 3x – 2x + 6 = 0 (3) Take common fact or s fr om 1st and 2nd and 3r d and 4t h t er ms.  x (x – 3) – 2(x – 3) = 0  (x – 3) (x – 2) = 0  x – 3 = 0 or x – 2 = 0 So, t wo r oot s of t he given equat ions ar e 3 and 2. H er e, t he fir st met hod is complet ed.

2nd M et hod H er e values of x can be det er mined by t he for mulae of x. For equat ion ax 2 + bx + c = 0,   b  b2  4 ac   x =  2a     By compar i ng gi ven quadr at i c equat i on wi t h t he st andar d quadr at ic equat ion ax 2 + bx + c = 0, we get values of a, b and c; then put t ing values of a, b and c in t he equat i on of x , we get r oot s of gi ven quadr at i c equat ion. Consider an example: x 2 – 5x + 6 = 0 Compar e given equat ion wit h ax 2 + bx + c = 0, we get a = 1, b = – 5, and c = 6. Put t hese values of a, b and c in t he equat ion, we get   b  b2  4 ac   x = 2a       (5)  (5) 2  4(1) (6)   =  2(1)    

5 1 =    2  or x=2  x=3 H ence second met hod is complet ed. I mport ant F act s About Quadr at ic Equat ion 1. Gen er al f or m of a Qu adr at i c E qu at i on i s ax 2 + bx + c = 0, wher e a, b, c ar e r eal number s.  I f  ,  ar e r oot s of ax 2 + bx + c = 0, t hen  =

 b  b2  4 ac 2a

 b  b2  4 ac 2a  A quadr atic equation has exactly two r oots, may be r eal or imaginar y. and

 =

11.2

Algebra

 x 1= T 2 = a + d ; x 2 = T 3 = a + 2d; x n = T n+1 = a + nd

q is a r oot of a quadr at ic equat ion, t hen it s ot her r oot is p – q .  ax 2 + bx + c = 0 is fact or izable int o t wo linear fact or s only when b2 – 4ac  0.  If p +

Put t ing for d, we get

x1 = a +

2. Relat ion bet ween Root s and Coefficient s L et ‘a’ and ‘b’ be t he r oot s of t he equat ion ax 2 + bx + c = 0

Sum of n ar it hmet ic means = n (single AM .) i.e. x 1 + x 2+ x 3 +............ + x n =

b Then, sum of r oot s =  a c and pr oduct s of r oot s = a

=

3. To form a Quadrat ic Equation Quadr at ic equat ion whose r oot s ar e  and  , is given by





x– a =b– x 2x =a+b

a+b 2 n – arithmetic means. I f bet ween t wo given quant it ies, a and b we have t o inser t ‘n ’ ar it hmet ic means x 1, x 2,...... x n, t hen a, x 1, x 2, ....x n, b will be in A.P. I n or der t o find t he values of t hese means we r equir e t he common differ ence. The above ser ies consist of (n + 2) t er ms and t he last t er m is b and fir st t er m is a.  b = T n+2 = a + (n + 2 – 1) d 



x =

d =

b a n +1

Any t hr ee number i n AP shoul d be t ak en as a – d, a, a + d



I n an A.P., t he sum of t er ms equidist ant fr om t he begining and end is const ant and equal t o sum of fir st and last t er m.



Any t er m of A.P. (except t he fir st ) is equal t o half t he sum of t er ms which ar e equidist ant fr om it

an =

Sn =



a+b =nx 2

Any four number, i n A.P. shoul d be t ak en as a – 3 d , a – d , a + d , a + 3d Similar ly five number in A.P. should be t aken as a – 2d, a – d, a, a + d, a + 2d An y si x n um ber i n A.P. sh ou l d be t ak en as a – 5 d , a – 3 d , a – d , a + d , a + 3d , a + 5d

ARI T H M ET I CAL PROGRESSI ON

n n (a + l ) = [2a + (n – 1) d] 2 2 Arithmetic mean (A. M .). The ar it hmet ic mean bet ween t wo given quant ies a and b is x, so t hat a, x, b ar e in A. P.

[  b – x n = d]

wher e, x is t he AM of a and b

A.P., G.P. AN D H .P. I f cer t ain quant it ies incr eases or decr eases by t he same const ant , t hen such quant it ies for m a ser i es whi ch i s cal l ed an ar it hmet ical pr ogr ession . Thi s const ant called common differ ence. e.g. 1, 4, 7, 10, 13 ,............; 9, 6, 3, 0, – 3 ,...........; a, a+d, a+2d, a+3d, ............ N ot at ion. The fir st t er m of t he ser ies is denot ed by ‘a’ common differ ence by ‘d’, t he last t er m by l , t he number of t er ms by n , sum of it s n t er ms by Sn, and t he n t h t er m be T n. T n = a + (n – 1) d

n (x + x n ) 2 1

n (a + d + b – d ) 2

=n

x2 – (  +  ) x +   = 0 i .e.x 2 – (sum of r oot s) x + (pr oduct of r oot s) = 0

b  a na + b nb + a = ; xn = n 1 n +1 n +1

and

for k = 1, an =

1 (a + an+k ), k n 2 n– k 1 (a + an+1) 2 n– 1



T n = Sn – Sn– 1 (n > 2)



Sum and differ ence of cor r esponding t er ms of t wo AP’s will for m a ser ies in A.P.

GEOM ET RI CAL PROGRESSI ON A ser ies in which each t er m is same mult iple of t he pr eceding t er m is called a G.P.. I n ot her wor ds a ser ies in which the r atio of successive t er ms is const ant is called a G.P.. Thi s const ant r at i o i s cal l ed common r at i o and i s denot ed by ‘r ’. e.g. 1, 4, 16, 64 ....; 9, 6, 4, ....; a, ar , ar 2.... nth term of a G.P. L et t he ser ies be, a, ar , ar 2, ar 3, ...  T 1 = a = ar 1– 1 ; T 2 = ar 2– 1 = ar ; T 3 = ar 2 = ar 3– 1; ......T n = ar n– 1

Algebra

Sum of n terms of a G.P.



L et S = a + ar + ar 2 + ar 3 + ...............+ ar n– 1 ........(i ) r  S = [ ar + ar 2 + ar 3 + ...............+ ar n– 1] + ar n ........(ii ) Subt r act ing (ii ) fr om (i ), S (1 – r ) = a – ar n



 

a (1  r n ) S= 1 r Sum of an infinit e number of t er ms of a G.P., when r < 1

n 

Common r at io, (r ) =

a and 1r

I f a1, a2, a3................and b1, b2, b3 ................be t wo G P’s of common r at io r 1, and r 2 r espect ively t hen

a 1 b 1 , a 2 b 2, a 3 b 3. ................an d

r at io will be r 1 r 2 and



I f a1, a2, a3 ..... be a G.P of +ve t er ms, t hen l og a 1 , l og a 2 , l og a 3 w i l l be an A .P. an d conver sely.

A =

ab , 2

G  ab,

Single geometric mean between a and b.





a a a , , r r3 r5

a a four t er ms as, ar 3, ar , , 3 . r r

A G  G H

ab 1  ab  ( a  b)2 2 2

A G  1   1 or G  H G H

and

I n par t icular, t hr ee t er ms ar e as, ar , a,

2ab ab

As A – G is always posit ive hence, A – G > 0,  A>G

ab E v en n u m ber of t er m s i n a G.P. m u st be

ar 5, ar 3, ar ,

H

A, G, H are in G.P. A– G=

x =

 

A H = a b = G2 and

x b  = common r at io of t he G.P. a x x2 = ab



r1 r espect ively.. r2

RELATI ON S BETWEEN A AN D G

Tn Tn 1

L et x be t he singl e geomet r ic mean bet ween t wo given quant it ies a and b. Then a, x , b ar e in G.P.

a1 a2 a3 , , b1 b2 b3

................ wi ll also for m G.P., wher e common

Since r  < 1, t he number of t er ms ar e infinit e

lim r n = 0 and hence in t his case S =

11.3

a ; and r 

 G>H  A>G>H A, G, H ar e in descending or der of magnit ude.

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. Which of t he following equat ions has r eal r oot s? (a) 3x 2 + 4x + 5 = 0

(b) x 2 + x + 4 = 0

(c) (x – 1) (2x – 5) = 0 (d) 2x 2 – 3x + 4 = 0 2. 3 chair s and 2 t ables cost ` 700, while 5 chair s and 3 t abl es cost ` 1100. What is t he cost of 2 chair s and 2 t ables ? (a) ` 300 (b) ` 350 (c) ` 450 (d) ` 600 3. I f x =

x  2a x  2b 4 ab , t hen value of + is x – 2a x – 2b ab

equal t o (a) 0 (c) 2

(b) 1 (d) None of t hese

x y ab is equal and b = , t hen x y x– y a– b

4. I f a = to (a)

xy x  y2

(b)

(c)

x x y

(d)

2

x2  y2 xy

FG y IJ H x – yK

y (y + x ) x   , det er mine t he r at io of (x – z) z y x : y : z. (a) 1 : 2 : 3 (b) 3 : 2 : 1 (c) 4 : 2 : 3 (d) 2 : 4 : 7

5. if

11.4

6.

Algebra

x y z =a   (2x + y + z) (x + 2y + z) (x + y + 2z)

14. The t er m independent of x in t he expansion of

FG 2x  1 IJ H 3x K

find a if x + y + z = 0. (a)

1 3

(b)

1 4

(c)

1 8

(d)

1 2

(a 2  bc)



b2 (b 2 – ca)



160 9 160 (c) 27 15.

c2 (c2  ab)

(a) 4 (b) 3 (c) 1 (d) 0 8. I f p, q, r , s ar e in har monic pr ogr ession and p > s, t hen (b) q + r = p + s

1 1 1 1    q p r s

(d) None of t hese

9. I f 2x 2 – 7xy + 3y 2 = 0, t hen t he value of x : y is (a) 3 : 2 (b) 2 : 3 (c) 3 : 1 and 1 : 2 (d) 5 : 6 10. The differ ence bet ween t he logar it hms of sum of t he squar es of t wo posit ive number s A and B and t he sum of logar ithms of t he individual number s is a const ant C. I f A = B, t hen C is (a) 2 (b) 1.3031 (c) log 2 (d) exp (2) 11. The ar it hmet ic mean of t he ser ies 1, 2, 4, 8, 16, ... 2n is (a)

(c)

n

2 –1 n

(b)

2n  1 n 1

(d)

2

n 1

–1 n 1

F Find t he t er m independent of x in G 3 x H (a) (b) (c) (d)

2

–

2

x2

IJ K

20

126070  318  29 184756  310  210 120000  320  219 320  221  (34)20

LEVEL-1 (a) 3200

(b) 3400

(c) 3146

(d) 3143 [RRB JE 2014 GREEN SH I FT ]

2. Gi ven t hat log 2 = 0.3 appr ox., one bi ll ion would be appr ox. (a) 29

(b) 210

(c) 2

(d) 230

20

[RRB JE 2014 GREEN SH I FT ]

3. I n how many differ ent ways can 3 ident ical whit e balls and 2 identical r ed balls be ar r anged besides each ot her, in a st r ai ght l ine? (a) 6 (c) 12

2n – 1 n 1

A.P. S2n = 3Sn. Then t he r at io (b) 6

(c) 8

S3 n is equal t o Sn (d) 10

(b) 10 (d) 120 [RRB JE 2014 GREEN SH I FT ]

4. What is t he pr obabi li t y of get t ing 3 aces i f t hr ee car ds ar e dr awn fr om a set of 52 playi ng car ds? (a) 523

12. I f t he sum of t he 6t h and t he 15t h elements of an ar it hmet ic pr ogr ession is equal t o t he sum of t he 7t h , 10t h an d 12t h el em en t s of t h e sam e pr ogr ession, t hen which element s of t he ser ies should necessar ily be equal t o zer o ? (a) 10th (b) 8th (c) 1st (d) None of t hese 13. L et Sn denot e t he sum of t he fir st ‘n’ t er ms of an

(a) 4

80 9 80 (d) 3 (b)

1. Fi nd t he val ue of 672 – 332.

1 1  (a) ps qr (c)

is

(a)

7. I f a + b + c = 0, find t he value of

a2

6

(c)

1 52!

(b)

1 523

(d)

432 52  51  50

[RRB JE 2014 GREEN SH I FT ]

5. I n a class of 40 st udent s, 25 ar e spor t s per sons an d 25 ar e m at h em at i ci an s. W h at i s t h e pr obabi lit y t hat t he monit or of t he class is bot h a spor t s per son and a mat hemat ician? (a)

1 40

(b)

1 25

(c)

1 4

(d)

1 50

[RRB JE 2014 GREEN SH I FT ]

Algebra

6. Sum of t wo number s i s 15 and sum of t hei r r eci pr ocals is

15 . The t wo number s ar e 56

(a) 4, 11

(b) 5, 10

(c) 6, 9

(d) 7, 8

14. In an examination, a student gets 4 marks for every correct answer and loses 1 mark for even' wrong answer. If he attempts in all 60 questions and secures 130 marks, then find the number of questions he attempted correctly. (a) 42

[RRB JE 2014 GREEN SH I FT ]

(b) 48

7. I f ,  ar e t he r oot s of quadr at i c equat ion x 2 + x +

(c) 36

1 = 0 , t hen

1 1  is  

(a) – 1

(d) 38 [RRB JE 2014 RED SH I FT ]

(b) 1

(c) 0

(d) None of t hese

15. If

x 6 x2  y2  , then find the value of 2 : y 5 x  y2

[RRB JE 2014 GREEN SH I FT ]

8. Value of (a)

6  6  6  ...... i s

5 2

(c) 3

(b) – 2

(a) 11

(c)

11 5

9. I f a, b, c, d, e and f ar e i n ar it hmet ic pr ogr ession, t hen e – c is equal t o (a) 2(b – a)

(b) c – b

(c) 2 (f – d)

(d) 2(d – b) [RRB JE 2014 GREEN SH I FT ]

10. Fi nd t he median of t he fol lowing number s : 14, 23, 20, 12, 11, 15, 24, 17, 9, 21, 25 (c) 17

(b) 20 (d) 14 [RRB JE 2014 GREEN SH I FT ]

11. A st udent was asked t o mult iply a number by 12. By mist ak e he mult i pl ied t he number by 21 and got t he answer 63 mor e t han t he cor r ect answer. What i s t he cor r ect answer ? (a) 9 (c) 7

(b) 8 (d) 84 [RRB JE 2014 GREEN SH I FT ]

12. Find the value of

7683   2323 : 768 2  768  232   232 2

(a) 1000

(b) 536

(c) 500

(d) 268 [RRB JE 2014 RED SH I FT ]

13. Find the value of (1+ 2 + 3 + 4+..........+45) : (a) 2140

(b) 2070

(c) 1035

(d) 1280 [RRB JE 2014 RED SH I FT ]

(b)

61 11

(d) 6 [RRB JE 2014 RED SH I FT ]

(d) 4

LEVEL-2

[RRB JE 2014 GREEN SH I FT ]

(a) 15

11.5

1. log4 5 × l og56 × l og67 is equal t o:

 7 (a) log   4

(b) log4 7

 4 (c) log   7

(d) log7 4 [RRB SSE 2014 GREEN SH I FT ]

2. The sum of fir st n odd nat ur al number s i s: (a) n 2 – 1 (b) n2 (c) (n + 1)2 (d) (n – l )2 [RRB SSE 2014 GREEN SH I FT ]

3. A per son put s 1 gr ain of r ice in t he fir st squar e of a chess boar d. I n the subsequent squar es, he puts t wice t hat of t he pr evi ous squar e. H ow many gr ai ns woul d he need t o put on al l t he squar es of t he chess boar d? (a) 64! (b) 264 – 1 63 (c) 2 – 1 (d) p(64, 2) [RRB SSE 2014 GREEN SH I FT ]

4. Which of t he fol lowing st at ement is cor r ect ? (a) n li near equat ions wit h n var iables may have a unique solut i on (b) n li near equat ions wit h n var iables may have no solut i on (c) Bot h (a) & (b) ar e cor r ect (d) Bot h (a) & (b) ar e wr ong [RRB SSE 2014 GREEN SH I FT ]

11.6

Algebra

5. Fi nd t he next number in t he ser ies. 1,2,6,24, 120,___________. (a) 240 (b) 480 (c) 560 (d) 720

10. The sum of t he squar es of t hr ee consecut ive odd number s is 2531. Find t hese t hr ee odd number s: (a) 21, 23, 25 (b) 25, 27, 29 (c) 27, 29, 31 (d) 23, 25, 27

[RRB SSE 2014 GREEN SH I FT ]

[RRB SSE 2014 RED SH I FT ]

6. A coin is t ossed t wo t i mes. On bot h occasi ons, t he r esul t is heads. When t he coin is t ossed a t hir d t ime, what i s t he pr obabil it y of get t i ng a head?

11. The pr esent age of a fat her is 3 year s mor e t han t hr ee t imes t he age of his son. Aft er t hr ee year s, fat her 's age wil l be 10 year s mor e t han t wice t he age of t he son. Find t he pr esent age of t he Father. (a) 30 year s (b) 33 year s (c) 36 year s (d) 39 year s

(a) 1 (c)

(b)

1 1  4 2

(d)

1 2

[RRB SSE 2014 RED SH I FT ]

3 1  4 2

[RRB SSE 2014 GREEN SH I FT ]

7. I f 3x + 7 = x 2 + p = 7x + 5, t hen t he value of 'p' wi ll be : (a)

1 2

(c) 2 2

(b) 8

1 2

(d) 8

1 4

12. Fi nd t he val ue of 10C3: (a) 720 (b) 240 (c) 120 (d) 1000 [RRB SSE 2014 RED SH I FT ]

13. 1250 ar t icles wer e dist r ibut ed among st udent s of a cl ass. Each st udent got t wice as many ar t i cl es as t he number of st udent s in t hat gr oup. The number of st udent s in t he gr oup was : (a) 25 (b) 45 (c) 50 (d) 100

[RRB SSE 2014 RED SH I FT ]

[RRB SSE 2014 YEL L OW SH I FT ]

 1 8. I f log10 7 = x, t hen t he value of logl0   i s equal 70

14. The sum of t wo number s is 2490. I f 6.5% of one number is equal t o 8.5% of t he ot her, t he gr eat er number is : (a) 1079 (b) 1380 (c) 1411 (d) 1250

t o: (a) – (1 + x) (c)

x 10

(b) (1 + x)– 1 (d)

1 10x

[RRB SSE 2014 YEL L OW SH I FT ]

[RRB SSE 2014 RED SH I FT ]

9. Fi nd t he value of: (51 + 52 + 53+ 54 +.... +100) (a) 3775 (b) 5050 (c) 1275 (d) 2525

15. L et  and  be t he r oot s of x 2 + k x + 8 = 0, such t hat  –  = 2, t hen t he val ue of k ar e (a) ±3 (b) ±6 (c) ±4 (d) ±8 [RRB SSE 2015 1 st SEP 1 st SH I FT ]

[RRB SSE 2014 RED SH I FT ]

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (c)

2. (d)

3. (a)

4. (a)

5. (c)

6. (b)

7. (b)

8. (d)

9. (c)

10. (b)

11. (b)

12. (b)

13. (b)

14. (c)

15. (b)

16. (c)

17. (b)

18. (b)

7. (a)

8. (c)

9. (a)

10. (c)

7. (d)

8. (a)

9. (a)

10. (c)

LEVEL-1 1. (b)

2. (d)

3. (b)

4. (d)

5. (c)

11. (d)

12. (a)

13. (c)

14. (d)

15. (b)

6. (d)

LEVEL-2 1. (b)

2. (b)

3. (b)

4. (c)

5. (d)

11. (b)

12. (c)

13. (a)

14. (c)

15. (b)

6. (b)

Algebra

11.7

EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S (x – 1) (2x – 5) = 0

1.



x = 1, x =

5 2

Bot h r oot s ar e r eal. For r eal r oot s of quadr at ic equat ion ax 2 + bx + c = 0 we have b2 – 4ac  0 For 3x 2 + 4x + 5 = 0 b2 – 4ac = 16 – 4  3  5 = – 44 2 For x + x + 4 = 0 b2 – 4ac = 1 – 4  1  4 = – 15 2 For 2x – 3x + 4 = 0 b2 – 4ac = 9 – 4  2  4 = – 23 2. L et cost of chair = ` x and cost of t able = ` y  3x + 2y = 700 (i ) and 5x + 3y = 1100 (ii ) Solving (i ) and (ii ) we get x = 100 and y = 200 Now 2x + 2y = 200 + 400 = ` 600.

3.

6 ab  2a 2 6b 2  2ab x  2b x  2a ab ab + = + 2 x  2b x  2a 2ab  2a 2 ab  2b 2 (a  b) ab =

2b(3b  a) 2a(3b  a) + 2b(a  b) 2a(b  a)

3b  a 3b  a = – =0 ba ba x y  xy ab x  y x  y = 4. = 2 x  y2 ab x y  x y x y 5. Subst it ut e in t he choices. Using x = 4, y = 2, z = 3, we find t he equat ion is satisfied. 6. Adding all t hr ee, we get 1 x yz =a= . 4 4 x yz

b

7.

=

g

a2 b2 c2   a2 – bc b2 – ca c2 – ab

a  2a2 – bc   2b2 – ca  b  2c2 – ab  c      a  bc   ca  b  ab  c [Applying componendo and dividendo on each t er ms]

=

2a3 – abc 2b2 – abc 2c3 – abc   abc abc abc 2 a3 + b3 + c3 – 3  abc





abc

6abc – 3abc =3 abc  (a  b  c)3  a3  b3  c3 – 3abc    if a  b  c  0   p, q, r , s ar e in H . P.

=

8.



1 1 1 1 , , , ar e in A.P.. p q r s



1 1 1 1 – = – r s p q



1 1 1 1  =  q r p s

9.

2



x2 x – 7 +3= 0 y y2

x 75 1 7  49  24 = = = 3, y 4 2 22

Thus x : y = 3 : 1 and 1 : 2 11. A.M . of t he ser ies 1, 2, 4, 8, 16, .... 2n =

Sum of (n + 1) t er ms of t he given ser ies n 1

1.(2 n 1 – 1) 2 1 = n 1

2n 1  1 n 1 Since ser ies is in G.P. wit h fir st t er m a = 1 and common r at io r = 2 Number of t er ms = n + 1 =

Sum of t his ser ies, Sn + 1 =

a.(r n  1) r 1

r 21 12. L et a be t he fir st t er m and d be common r at io of an A.P.  (a + 5d) + (a + 14d) = (a + 6d) + (a + 9d) + (a + 11d)  a + 7d = 0 H ence 8t h t er m = 0 13. L et a be t he fir st t er m and d be t he common differ ence of an A. P.

11.8

Algebra

H en ce t h e t er m

n {2a + (n – 1) d} 2

Sn =



FG 3 x 2 – 2 IJ H x2 K

2n S2n = {2a + (2n – 1) d} 2

=

3n {2a + (3n – 1) d} 2 Given : S2n = 3Sn

RS T

14.

1.

FG 2x  1 IJ H 3x K

2.

2

UV = W

67 – 33 = (67 + 33) (67 – 33) 1 billion = 109  Taking log both sides we get,

2an 2 n 1

9 lot 10 = n log 2  9 = n × 0.3  n = 30 3.

Required number of ways

5!  10. 3!  2!

6

4.

FG 1 IJ + C (2x) FG 1 IJ H 3x K H 3x K F1I F1I + C (2x ) G J + C (2x ) G J H 3x K H 3x K F1I F1I + C (2x ) G J + C G J H 3x K H 3x K

Required probability

2

= (2x )6 + 6C1 (2x )5

6

3

6

4

6

5

4

2

5

6

4



4

2

3

3

= 6C3  23 

FG H 20

IJ K

6.

33

Number of boys who are both sports person and a mathematician = 254 + 25 – 10 = 40

a + b = 15 and



654 1 160 8  3 2 1 27 27

2

1 1 15   a b 56

a  b 15  ab 56

 (a, b) = (7, 8) 7.

1 1       given, + = –1 &  = 1 Hence,

FG – 2 IJ Hx K

4  3a2 52  51  50

1

is

r

C3



Hence, required probability 

20

Cr (3x 2)20 –

C3

52

6

15. (r +1)t h t er m i n t he expansi on of ( x + a)n i s n Cr x n– r ar H ence ( r + 1)t h t er m i n t h e ex pan si on of 2 2 3x – 2 x

5.

6

H ence t er m independent of x

=

2

Let 109 = 2n

2

6

20  19 ...  11  310  210 10  9  ...  1

= 100 × 34 = 3400.

Sn = 6. S3n



C10  310 210

LEVEL-1

RS2a  (3n  1) 2a UV = 12an n  1W T n 1

3n and S3n = 2

x in

= 184756  310  210

2a d = n 1

n 2 a  (n  1) 2a n 1 2

20

=

n  n {2a + (2n – 1) d} = 3  {2a + (2n – 1) d} 2  4a + (4n – 2) d = 6a + (3n – 3) d  (n + 1) d = 2a

 Sn =

of

is (r + 1)t h i.e. 11 1t h t er m

S3n =



i n depen den t

20

   1   1.  1

r

= 20Cr 320 – r  2r  x 40 – 4r ,

which is independent of x if 40 – 4r = 0 i.e. r = 10

8.

6  6  6  ......  x

squaring both sides we get, 6 + x = x2 x=3

10 1  . 40 4

Algebra

9.

e – c = (e – d) + (d – c)

10.



1  264  1

= 2(b – a)



 2  1

14, 23, 20, 12, 11, 15, 24, 17, 9, 21, 25 When written in ascending order becomes:  sn

9, 11, 12, 14, 15, 17, 20, 21, 23, 24, 25 set the number be x.

 G.P 



1,

(768)  (232)

12.

3

2

(768)  (768  232)  (232)2 

(768  232)  (768)2  (768  232)  (232)2  (768)2  (768  232)  (232)2

= 768 + 232 = 1000

13.

= 23 × 4S - 1035

2,

 x × 4 – (60 – x) × 1 = 130

x2  y 2

2

2

6 5 61  2  2  2 11 x  y 6 5

LEVEL-2 1. log 54  log 65  log 76



4 log10

×5

×6

= 720 6. Probability of getting head in third occasion will

1 , since it is independent of the first two 2 tosses. be

3x + 7 = 7x + 5  4x = 2

x

log 65 5 log10

1 2

 1  1  3    7     p.  2  2

x 6  y 5



12,

2

 x = 38

log 54

24,

×2 ×3 ×4  Ans = 120 × 6

 Sx – 60 = 130



6,

5.

set the no. of correct questions be x

15.

 r  1

7. 3x + 7 = x2 + p = 7x + 5

Required sum = (1 + 4S) ×45

14.



 Ans = option (c)

x=7 3



1

4. Both statement A and statement B are correct.

 x × 21 = x + 12 + 63 Hence, correct answer = 12 × 7 = 84

64

a rn  1

Hence, median or mid-term = 17 11.

2



3 1  7   p. 2 4 p8

1 4

8. log107 = x log 76 6 log10

7 log10

 log 74 4 log10

2. the sum of first n odd natural numbers. = n2. 1 = 12

 1  1  log10    log10   70   7  10 

= –log107 – log1010 = –x – 1 9. S = (51 + 52 + … 100)



50 51  100 2

1 + 3 = 22

 sum A.P 

1 + 3 + 5 = 32.

= 3775

3. Number of grains = 1+2+4+8+… 20 + 21 + 22 + … 263.

11.9

n a  l 2

10. Let consecutive odd numbers be x, (x + 2), (x + 4)

11.10

Algebra

According the question: x2 + (x + 2)2 + (x + 4) = 2531 Buy hit and trial, 27, 29, 31 satisfies. 11. Let present age of father be x Let age of son be y According the question: x = 3 + 3y Also, (x + 3) = 10 + 2 (y + 3)

14. x + y = 2490

6.5 8.5 x y 100 100  13x = 17y  x = 1411 15. x2 + kx + 8 = 0   =

8 =8 1

 y = 10  Product of root 

 x = 33 12. n Cr 





10

n! r!  n  r  ! C3 

10! 3!  7!

10  9  8  7!  120 3  2  1  7!

13. Let no. of students be x.  According the question: x × (2x) = 1250  x = 25

c for (ax2 + bx + c) a

   2

4   2 or 2   4 k 

b = sum of roots = ± 6 2 

12

CHAPTER

Trigonometric Ratios & Height and Distance

Syst em of M easur ement of Angles Ther e ar e t hr ee syst em of measur ement 1. Sexagesimal syst em 2. Cent esimal syst em 3. Cir cular syst em

N ote: I f measur e of an angle is given in degr ee t hen to conver t it into radians, multiply the measure of angle  by and if t he measur e of an angle is given in 180 r adian, t hen t o conver t it int o degr ee, wr it e 180º at t he place of .

Sexagesimal Syst em

Tr i gonomet r i c Rat i os

I n t his system each angle is divided int o 90 equal par ts and each par t is know as a degr ee. Thus a r ight angle is equal t o 90 degr ees. One degr ee is denot ed as 1º. Each degr ee is divided int o 60 equal par t s each of which is known as one minut e one minut e is denot ed as 1’ each minut e is consist of 60 par t s, each par t is known as a second. One second is denot ed by 1” . H ence, 1 r ight angle = 90º (90 degr ee) 1 = 60º (60 min) 1 = 60 (60 seconds)

I n r ight angled t r iangle ABC, C e us ten o p Hy

 A

I n this system each angle is divided into 100 equal par ts and one par ts is known as a gr ader. Thus one r ight angle is equal t o 100 gr ade. One gr ade is denoted as 1g. One gr ade is divided int o 100 equal par t s one par t is known as a minut e and is denot ed as one minut e is also divided into 100 equal par t s, one par t is known as a second which is denot ed by 1’. H ence 1 r ight angle = 100g (100 gr ade) 1g = 100(100 minut es) 1 = 100(100 seconds)

BC Perpendicular = AC Hypot enuse

cos  =

AB Base = AC Hypot enuse

BC Perpendicular = AB Base 1 AB Base  cot  = = t an  BC Perpendicular t an  =

1 AC H ypot enuse =  cos AB Base 1 AC Hypot enuse  cosec  = = sin  BC per pendicular sec  =

Cir cular Syst em

T he rat io of t he circumference of t he circle t o the diameter of the circle is denoted by a Greek letter  and it is a constant quantity

Cir cumfer ence of ci r cle  D iameter of cir cle Relat ion among degree, radian and gr ade 180º = c = 100g and 1 r adian = 57º 17’ 44.8’.

B

sin  =

Cent esimal Syst em

l r I f t he angle subt ended by an ar c of l engt h (t o t he cent r e of ci r cl e of  r r adius r , is  t hen l = r I f t he lengt h of ar c is equal t o t he r adius of t he cir cle, t hen t he angle subt ended at t he cent r e of t he cir cle will be one r adian one r adian is denot ed by 1c.

Base

if < CAB = q, t hen

Fundamental Relation Among Trigonometric Rat i os I t s is clear fr om t he definit ion of t r igonomet r ic r at ios that 1 sin  1 sec  = cos 1 cot  = t an  sin  cos  and cot   . t an  = cos  sin  cos2  + sin 2  = 1 1 + t an 2  = sec2  1 + cot 2  = cosec2 

1. cosec  = 2. 3. 4. 5. 6. 7.

12.2

Trigonometric Ratios & Height and Distance

Tr igonomet r ic Rat ios of D iffer ent Angles  – 0 90º –  90º +  180º –  180º +  360º + 

sin  cos  tan  – sin  cos  – tan  cos  sin  cot  cos  – sin  – cot  sin  – cos  – tan  – sin  – cos  – tan  – sin  cos  – tan 

360º +  sin 

cos 

tan 

cot  sec  – cot  sec  tan  cosec  – tan  – cosec  – cot  – sec  – cot  – sec  – cot  sec  cot 

sec 

cosec  – cosec  sec  sec  cosec  – cosec  – cosec  cosec 

Val ues of Tr i gonomet r i c Rat i os of Some I mpor tant Angles Between 0  and 180 

CD  CD  cos  3. cosC  cosD  2 cos     2   2  CD  D C sin  4. cosC  cosD  2sin     2   2 

Trigonometric Ratios of M ultiples of an Angles 2 t an  1  t an 2  2. cos2  cos2   sin 2   1  2sin 2  1. sin 2  2sin  cos  

 2 cos2   1 

1 1 (1  cos2 ), sin 2   (1  cos2) 2 2 2 t an  4. t an 2  1  t an 2  cot 2   1 5. cot 2  2cot  6. sin 3  3sin  4 sin 3  2 3. cos  

Angle

0º

30º

45º

60º

90º

sin

0

1 2

1

3 2

1

cos

1

2

1 2

0

tan

0

1

3



7. cos3  4 cos3   3 cos 

cot



0

8. t an 3 

sec

1

cosec



 3 2 1 3 3

2 3 2

2 1

1 2 2

1 3 2

2 3

I ts Trigonomet ric Rat ios of Combined Angles 1. sin (A + B) = sin A cos B + cos A sin B. 2. sin (A– B) = sin A cos B – cos A sin B. 3. cos (A + B) = cos A cos B – sin A sin B. 4. cos (A – B) = cos A cos B + sin A sin B.

t an A  t an B 1  t an A t an B t an A  t an B t an (A – B) = 1  t an A t an B cot A cot B  1 cot (A + B) = cot A  cot B cot A cot B  1 cot (A – B) = . cot B  cot A 2sin A cosB  sin(A  B)  sin(A  B) A  B

5. t an (A + B) = 6. 7. 8. 1.

2. 2cos A sin B  sin(A  B)  sin(A  B) 3. 2cos A cosB  cos(A  B)  cos(A  B) 4. 2sin A sin B  cos(A  B)  cos(A  B)

C, D Formula

1  t an 2  1  t an 2 

3 t an   t an 3  1  3 t an 2 

Some I mpor t ant Result s  1. sin 15º  1 2. cos15º 

3 1 2 2 3 1 2 2

3. t an 15º  cot 75 º  2 3 4. cot 15 º  t an 75 º  2  3 5. sin 22

1º  2

2 2 2

1º 2 2  2 2 1º 7. t an 22  2  1 2 1º 8. cot 22  2  1 2

6. cos22

9. sin 18º  cos72º 

5 1 4

10. cos18º  sin 72º 

10  2 5 4

10  2 5 4 5 1 12. cos36º  sin 54 º  4

11. sin 36 º  cos54 º 

1. sin C  si n D  2sin  C  D  cos  C  D   2   2 

13. sin 9 º  cos81º 

3 5  5 5 4

CD  CD  2. sin C  sin D  2cos   sin  2  2    

14. cos9 º  sin 81º 

3 5  5 5 4

Trigonometric Ratios & Height and Distance

Angle of E levat ion t

I mpor t ant F act s

P

h I f ‘O’ be t he obser ver ’s eye and Sig of e OX be t h e h or i zon t al l i n e Lin  t hr ough O. I f P is at a higher O X Horizontal line level t han eyes. Then POX is called the angle of elevation. I n figur e, OX is a hor izont al line and OP is line sight and POX =  is angle of elevat ion.

Angl e of D epr essi on

Horizontal line

O

12.3

X

I f ‘O’ be t he obser ver ’s eye and  Lin OX is a hor izont al line. I f object eo fS igh P i s at a l ower l evel t han O. t Then POX is called t he angle P of depr ession. I n fig., OX is a hor izont al line and POX =  is angle of depr ession. Note: Angle of elevation of an object from an observer is same as angle of depression of observer from the object.

1. Angle of elevation and angle of depression are always acut e angle. 2. Any per pendicular line t o a plane is per pendicular t o all lines lying in t he plane. 3. I n si mi l ar t r i angl e, t he cor r espondi ng si des ar e pr opor tional 4. The ext er ior angle of a t r iangle is equal t o t he sum of int er ior opposit e angles A 2

1

 X

B

C

5. I n isosceles t r iangles, t he median is per pendicular t o t he base. I n Fig. ABC is isosceles and AD  BC. A

Appol oni us T heor em If in ABC, AD is median, then AB2 + AC2 = 2 (AD2 + BD2) A B

C

D

6. I n a t r i angl e, t he i nt er nal bi sect or of an angl e divides t he opposit e side in t he r at io of t he ar ms of t he angles. A

D

C

B

M -N Theorem (Trigonometric T heorem) I n a t r iangle ABC, D divides BC in t he r at io m : n

A 2 c

A 2

b

A  

B

D

C

I n fig. AD is bisect or of  A

BD c = DC b 7. I n a r ight angled t r iangle ABC t he mid point D of hypot enuse. AC is equidist ant fr om it s ver t ices A, B and C. i.e., AD = BD = CD 

B

1. 2.



B

D m:n

C

C

 m  n  cot   n cot B  m cot C  m  n  cot   m cot   n cot .

Pt ol emy’s T heor em I n cyclic quadr ilat er al ABCD AB . CD + AD . BC = AC . BD.

A

A

D D

B

C

B

C

8. Angles in t he same segment of a cir cle ar e equal.

12.4

Trigonometric Ratios & Height and Distance

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. I f sin  + cosec  = 2, t hen sin  + cosec  is equal t o: (a) 1 (b) 4 (c) 2 (d) none of t hese 2

2

cot 2   1 is equal t o: cot 2   1 (a) sin 2 (b) cos 2 (c) cosec 2 (d) sec 2 3. I f t an A = 2 t an B + cot B, t hen 2 t an (A – B) is equal t o: (a) t an B (b) 2 t an B (c) cot B (d) 2 cot B 4. I f t an A – t an B = x and cot B – cot A = y , t hen cot (A – B) is equal t o:

2. The value of

(a)

1 y x

(b)

1 xy

(c)

1 1  x y

(d)

1 1  x y

4 5. I f sin  =  and  lies in t he t hir d quadr ant , 5  t hen cos is equal t o: 2

1 2 2 (b)  (c) (d)  5 5 5 5 6. The value of cos 1º cos 2º cos 3º... cos 100º is equal t o: (a) 1 (b) – 1 (c) 0 (d) none of t hese 7. The value of sin 12º sin 48º sin 54º is equal t o: (a)

13. The cir cular wir e of diamet er 10 cm is cut and pl aced al ong t he ci r cumfer ence of a ci r cl e of diamet er 1m. The angle subt ended by t he wr it e at t he cent r e of t he cir cle is equal t o:

    rad (b) rad r ad (c) rad (d) 4 3 5 10 14. The gr eat est and least value of sin x cos x ar e: (a)

(a) 1, – 1

1 1 , (d) 2, – 2 4 4 15. I f A = sin 2  + cos4 , t hen for all r eal values of  (c)

(a) 1  A  2 (c)

13  A 1 16

3  A 1 4 3 13 (d)  A  4 16 (b)

LEVEL-1 1. The value of sin 2 30° + sin 2 60° is (a) 1

(b)

3 2

(c) 2

(d)

3 4

1

1 1 1 1 (a) (b) (c) (d) 16 32 8 4 8. The value of 2 cos x – cos 3x – cos 5x is equal t o: (a) 16 cos3 x sin 2 x (b) 16 sin 3 x cos2 x (c) 4 cos3 x sin 2 x (d) 4 sin 3 x cos2 x

1 1 1 2 1   x   , t hen  x  2  is equal t o: 2 x 2 x  (a) sin 2 (b) cos 2 (c) t an 2 (d) sec 2 10. T h e v al u e of x f or t h e m ax i m u m v al u e of 3 cos x  sin x is: (a) 30º (b) 45º (c) 60º (d) 90º 11. The equat ion (a + b)2 = 4ab sin 2 is possible only when (a) 2a = b (b) a = b (c) a = 2b (d) none of t hese 12. The value of t he expr ession 9. I f cos  =

2

sin y 1  cos y sin is equal t o: 1   1  cos y sin y 1  cos y (a) 0 (b) 1 (c) sin y (d) cos y

1 1 , 2 2

(b)

[RRB JE 2014 GREEN SH I FT]

2. t an 90° is undefined. As  is incr eased fr om 89° t owar ds 90°, value of t an  t ends t o (a) 0

(b) +oo

(c) 1

(d) undefined [RRB JE 2014 GREEN SH I FT]

3. The angle of elevation of a ladder leaning against a wall is 60° i.e. ladder makes an angle of 60° with the ground. The foot of the ladder is 4.6 metres away from the wall. What is the length of this ladder ? (a) 9.2 m

(b) 2.3 m

(c) 6.9 m

(d) 7.8 m [RRB JE 2014 RED SH I FT]

4. A kite is flying at a height of 75 m from the level ground, attached to a string inclined at 60° to the horizontal. The length of the string is (a) 50 2m (c)

50 2

m

(b) 50 3m (d)

50 3

m

[RRB JE 2014 YELLOW SH I FT]

Trigonometric Ratios & Height and Distance

5. A tower stands vertically on the ground. From a point on the ground which is 30 m away from the foot of the lower, the angle of elevation of the top of the tower is 60°. The height of the tower is:

10. From a point on the bridge across a river the angle of depressions of the bank on the opposite side of the river are 60° and 45° respectively. If the bridge is at a height of 3 m from the bank, the width of the river is

(a) 30 3 m

(b) 10 3 m

(a) 2  3  1 m

(b) 3  3  1 m

(c) 15 m

(d) 6 m

(c) 3  3 m

(d) 3  3 m

For the above question, User had specified ‘ignore’ during keys upload. [RRB JE 2015 26 th AU G 1 st SH I FT]

6. From a point on the bridge across a river the angles of depression of the banks on the opposite side of the river are 30° and 45° respectively. If the bridge is at a height of 4 m from the bank, the width of the river is

[RRB JE 2015 26 th AU G 3 rd SH I FT]

11. A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, the distance of the foot of the ladder from the wall is (a) 7.5 m (b) 5 3 m (c) 10 3 m

(d)

(a) 4  3  1 m

15 3 m 2

[RRB JE 2015 27 th AU G 1 st SH I FT]

(b) 2  2 3  1 m (c) 4  3  1 m (d) 2  2 3  1 m [RRB JE 2015 26 th AU G 1 st SH I FT]

7. A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, the height of the wall is (a) 7.5 m (c)

12.5

15 3 m 2

12. From a point on the bridge across a river the angles of depression of the bank on the opposite side of the river are 60° and 45° respectively. If the bridge is at a height of 4 m from the bank, the width of the river is (a)

4 3  3  m 3

(b) 5 3 m

(b) 4  3  1 m

(d) 10 3 m

(c)

2 1  3  m 3

[RRB JE 2015 26 th AU G 2 nd SH I FT]

8. From a point on the bridge across a river the angles of depression of the banks on the opposite side of the river are 30° and 45° respectively. If the bridge is at a height of 3m from the bank, the width of the river is (a) 2  3  1 m

(d) 2  2 3  1 m [RRB JE 2015 27 th AU G 1 st SH I FT]

13. The angle of elevation of the top of a tower from a point on the ground Inch is 30 m away from the foot of the tower is 45°. The height of the tow t is

(b) 3  3  1 m

(a) 15 m

(b) 10 3 m

(c) 4  3  1 m

(c) 30 m

(d) 30 3 m [RRB JE 2015 27 th AU G 2 nd SH I FT]

(d) 2  2 3  1 m [RRB JE 2015 26 th AU G 2 nd SH I FT]

9. A ladder just reaches the top of a wall. The foot of the ladder is 8 m away from the foot of the wall. The ladder makes an angle of 60° with the ground. The length of the ladder is (a) 4 m (c)

16 3 m 3

14. From a point on the bridge across a river the angle of depressions of the bank on the opposite side of the river are 30° and 60° respectively. If the bridge is at a height of 3 m from the bank, the width of the river is

(b) 16 m

(a) 4 3 m

(b) 2  3  1 m

(d) 16 3 m

(c) 2  3  3  m

(d) 2 3 m

[RRB JE 2015 26 th AU G 3 rd SH I FT]

[RRB JE 2015 27 th AU G 2 nd SH I FT]

12.6

Trigonometric Ratios & Height and Distance

15. The angle of elevation of top of a tower from a point on the groin 20 m away from the foot of the tower is 60°. The height of tower is

20 3 m 3

(a)

(c) 20 m

(b)

40 3 m 3

(d) 20 3 m

5. The lengt h of t he shadow of a pol e i s 90m, when t he sun's el evat i on i s 30 °.The l engt h of t he shadow of the pole is x meter when Sun's elevation is 60°. The value of x is (a) 15 3

(b) 30

(c) 45

(d) 20 3 [RRB SSE 2015 1 st SEP 2 nd SH I FT ]

[RRB JE 2015 27 th AU G 3 rd SH I FT]

6. I f 3 si n  + 4 cos = 5, t hen t he value of 4si n  – 3cos i s

LEVEL-2 1. The angle of elevat ion of a ladder leaning against a wall is 60° and t he foot of t he ladder is 7.5 m away fr om t he wall. The lengt h of t he ladder is (a) 15 m

(b) 14.86 m

(c) 15 64 m

(d) 15.8 m [RRB SSE 2014 YEL L OW SH I FT ]

2. I f si n  + cos =

3 ,t hen t he value of

3 (t an  + 4

cot ) is (a) 1

(c)

(b)

(c) 1

(d) 0 [RRB SSE 2015 1 st SEP 3 rd SH I FT ]

7. Fr om t he t op of a t ower 50 m hi gh t he angles of depr ession of t he t op and bot t om of a pole ar e found t o be 45° and 60° r espect i vely. The height of t he pole, i n met r e, is

3 1



(b) 50



3 1



(c) 50 3  3



(d) 50



3 1



(a) 50



[RRB SSE 2015 1 st SEP 3 rd SH I FT ]

8. I f x = (1 + cot  – cosec)(1 + t an  + sec),

3 2

(d) 3

0°<< 90°, t hen t he value of x is

3. A ship is appr oachi ng a li ght home, 100m hi gh above t he sea-level . The angle of depr ession of t he ship as obser ved fr om t he t op of t he l ight home, changes fr om 30° t o 45°. The dist ance, in m, t r avel l ed by t he shi p dur i ng t he per i od of obser vat i on , in m , is (a) 100( 3 + l)

(b) 100( 3 – l)

(c) 100 3 + l

(d) 100 3 + l [RRB SSE 2015 1 st SEP 1 st SH I FT ]

sec   t an  5  0   90 t hen t he value of sec   t an  3

cosec i s (a) 2

(c)

(b) – 2



3 4

[RRB SSE 2015 1 st SEP 1 st SH I FT ]

4. I f

(a) – 5

15

(b) – 1

(c) 1

(d) 2 [RRB SSE 2015 2 nd SEP 1 st SH I FT ]

9. The angle of elevat ion of t he t op of an unfini shed t ower at a poi nt 20m away fr om i t s base is 45°. H ow much higher must t he t ower be r aised so t hat it s angle of elevat ion of t he t op at t he same point be 60° (a) 20 3m (b) 20 3  1m (c) 20  3  1 m (d)

(b)

4

(a) – 2

15 4

(d) 4 [RRB SSE 2015 1 st SEP 2 nd SH I FT ]

20 3  1m 3 [RRB SSE 2015 2 nd SEP 1 st SH I FT ]

10. The expr ession tan 2+ cot 2– sec2cosec2 is equal to (a) 0

(b) 1

(c) – 1

(d) – 2 [RRB SSE 2015 2 nd SEP 2 nd SH I FT ]

Trigonometric Ratios & Height and Distance

11. Two pol es of equal height st and on eit her side of a r oadway which is 100m wi de. At a poi nt in t he r oadway bet ween t he poles, t he el evat i on of t he t ops of t he poles ar e 60°and 30°. t he hei ght of each pole, in met er is (a) 25 3 (c) 25





(d) 25



3 1

[RRB SSE 2015 2

nd

SEP 2

nd

(a) 2, 3

5 2

(d) 3, 4 [RRB SSE 2015 2 nd SEP 3 rd SH I FT ]

13. Fr om t he t op of a t ower 90 m high, t he angles of depr ession of t he t op and bot t om of a buil di ng ar e obser ved t o be 30° and 60° r espect i vely. The height of t he buil ding, in m, is (a) 40

(b) 45 3

(c) 60

(d) 40 3

(a)

1 2

(b)

(c)

2 4

(d) 1

SH I FT ]

(b) 1, 3

l 2  2m is 2

value of



12. I f 2si n 2 – 5si n cos +7cos2 =1, t hen possible values of t an  ar e

(c) 2,

14. I f cos – si n  = l and t an  = m sec2, t hen t he

(b) 50 3

3 1

12.7

1 4

[RRB SSE 2015 3 rd SEP 1 st SH I FT ]

15. A per son is st anding on t he gr ound and flyi ng a ki t e wi t h a st r i ng of l engt h 140 m at an angl e of 30°. Anot her per son is st andi ng on t he r oof of a building 20 m high and is flying a kit e at an angle of 45°. I f bot h per sons ar e on opposi t e sides of bot h t he ki t es, t he lengt h (in m ) of t he st r i ng t hat t he second per son must have so t hat t he t wo kit es meet s, (a) 70

(b) 60 2

(c) 50 2

(d) 50 [RRB SSE 2015 3 rd SEP 1 st SH I FT ]

[RRB SSE 2015 2 nd SEP 3 rd SH I FT ]

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (c)

2. (d)

3. (c)

4. (d)

5. (b)

11. (a)

12. (d)

13. (c)

14. (b)

15. (b)

6. (c)

7. (c)

8. (a)

9. (d)

10. (a)

7. (a)

8. (b)

9. (b)

10. (d)

LEVEL-1 1. (a) 11. (d)

2. (b) 12. (a)

3. (a) 13. (c)

4. (b) 14. (a)

5. (a) 15. (d)

6. (c)

LEVEL-2 1. (a) 11. (a)

2. (b) 12. (a)

3. (b) 13. (c)

4. (d) 14. (a)

5. (b) 15. (c)

6. (d)

7. (a)

8. (d)

9. (c)

10. (d)

12.8

Trigonometric Ratios & Height and Distance

EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S 1. Given t hat , sin + cosec = 2 On squar ing bot h sides, we get sin 2  + cosec2  + 2 = 4  sin 2  + cosec2  = 2 1  t an   cot 2   1 2. = 1  t an 2  cot 2   1 1 1  sec2 = = cos2   sin 2  cos2 3. Given t hat t an A = 2 t an B + cot B ...(i )

 t an A  t an B  N ow, 2 t an (A – B) = 2    1  t an A t an B   2 t an B  cot B  t an B  = 2 [Fr om (i )] 1   2 t an B  cot B  t an B cot B  t an 2 B  1  t an B  cot B  cot B = 1  t an 2 B  2 1  t an 2 B  4. Given t hat t an A – t an B = x ...(i ) and cot B – cot A = y ...(ii ) 1 N ow, cot (A – B) = t an  A  B  1  t an A t an B = t an A  t an B 1 t an A tan B  = t an A  tan B t an A  tan B 1 1 =  [fr om (i ) and (ii )] x y 5. Given t hat , 4 sin  =  and  lies in the IIIrd quadrent 5 16 3   cos  =  1  25 5 3 1  1  cos  5  1 N ow, cos =   5 2 2 2 1  But we take cos =  . Since, if  lies in I I I r d 2 5  quadr ant , t hen will be in I I nd quadr ant . 2 1  H ence, cos =  2 5 6. cos 1º cos 2º cos 3º ... cos 90º... cos 100º = cos 1º cos 2º cos 3º ... 0 ... cos 100º = 0 [  cos 90º = 0] 7. Now, sin 12º sin 48º sin 54º 1 =  cos36 º  cos60º  cos36 º 2

= 2

1  5  1 1   5  1 5  1 4 1      = 32 32 8 2 4 2  4  8. 2 cos x – cos 3x – cos 5x = 2 cos x – 2 cos x cos 4x = 2 cos x (1 – cos 4x ) = 2 cos x 2 sin 2 2x = 4 cos x (2 sin x cos x )2 = 16 sin 2 x cos3 x 1 1 1 9. Given t hat cos =  x    x  = 2 cos  x 2 x

=

2

1  1 We know t hat x  2   x    2 x x  = (2 cos )2 – 2 = 4 cos2  – 2 = 2 cos 2 [fr om (i )] 2

1 2 1  1   x  2    2cos2  cos2 2 x  2 10. L et f(x) = 3 cos x + sin x  3  1   cos x  sin x  = 2sin  x   f(x) = 2   2 2  3    Since, 1  sin  x    1 3    H ence, f (x ) is maximum if x   3 2   x =  30º 6 11. We have (a + b)2 = 4ab sin 2



 a  b 2 4 ab



sin 2  =

Since,

sin 2   1



 a  b 2 1 4 ab (a + b)2 – 4ab  1 (a – b)2  0 a = b

  12. The given expr ession can be wr it t en as

1  cos y  sin 2 y 1  cos2 y   sin  y  1  cos y sin y 1  cos y  cos y 1  cos y   0  cos y 1  cos y 13. Given t hat , diamet er of cir cular wir e = 10 cm, L engt h of wir e = 10 

H ence, r equir ed angle =

10   rad 50 5 1 f(x) = sin x cos x = si n 2 x 2 – 1  sin 2x  1 1 1 1   sin 2 x  2 2 2 =

14. L et We know



lengt h of arc radius of big cir cle

Trigonometric Ratios & Height and Distance

A

8.

3m

Thus, t he gr eat est and least value of f (x ) ar e 1 1 and  r espect ively.. 2 2 15. We have, A = sin 2  + cos4  = sin 2  + cos2 cos2   sin 2  + cos2  (Since, cos2   1)  sin 2  + cos4  1  A  1 Again, sin 2  + cos4  = 1 – cos2  + cos4  = cos4  – cos2  + 1

30°

45°

According to the figure.

AC  tan 45 CD AC = CD

In ACD,

2

1 3 3  2 =  cos       2 4 4



or CD = 3 m

3 A1 4

H ence

D

C

B

Again in ACB,

LEVEL-1

AC  tan 30 BC

1.

sin2 30° + sin2 60° = sin2 30° + cos2 30° = 1

 BC  3 3

2.

Value of tan q will ten A to +  as tan 90° is + .

Hence, BD  3 3  3 3  3  1 m.

3.

9.

4.2  cos60 L

l

 L = 4.6 × 2 = 9.2 m 4.

75  sin 60 L

L  5.

75  2 3

Let length of Ladder = l

 50 3 m

cos 60° =

Figure



tan 60° = Height / 30  Height of the tower = 30 3 meter. 6.

60°

8m

Figure In triangle ABD, tan 30° =

8 l

1 8   l  16m 2 2

10.

3m

DB AB

y

45°

From the figure, tan 60° =

3  3 x

60°  AB = 4 3 meter Similarly, in triangle CBD, tan 45° =

x river

DB BC

 BC = DB = 4 meter

7.

Hence, required distance AC = 4 3 + 4

x 3

= 4( 3 + 1) meter.

tan 45° =

cos 60 

h 15

 h  15 

1  7.5m 2

3 1 y  3 y

 Width of river = n + y = 3  3 m

12.9

12.10

Trigonometric Ratios & Height and Distance

tan 45° =

11.

60°

15m

h 1 30 h = 30 m



bridge

14.

x

30° 45° 4m

Let distance of foot of ladder from wall = x

sin 60 

60°

x 3  15 2

 x  15

tan 60 

tan 45  4m

river

3 1 y y=3



y

x

3  3 x

x 3

bridge

60°

 width of river = x + y  4 3

45° 15.

From the figure,

tan 60 

h

4  3 x

60° 20

4

x

3

tan 45 

tan 60° =

4 1 y

3

n  n  20 3 20

LEVEL-2 1. According the question: y=4



A

width of river = x + y



4 3

4



4 3 3 

45°

river

3 2

12.

y

x

3

m

60°

B

C

13.

7.5m cos 60 

h 30 m

45°

BC AC

1 7.5  2 AC  A = 15 m

Trigonometric Ratios & Height and Distance 12.11

2. (sin + cos)2 = sin 2   cos2   2cos  sin   3 2

 sin. cos = 1, since sin  + cos  = 1 

h x x

tan 60° 

2

3 3  sin2   cos2   (tan + cot) =   4 4  sin  cos  

and tan 60° =

3 3

 30 m (Ans)

1

Since tan 30° 

3 (Ans) 4

90

3 3

6. 3 sin  + 4 cos = 5

3.

3 4 sin   cos   1 5 5

30° 45°

 comparing with sin . sin + cos  – cos  = 1

100 we get sin  

45°

60°

cos  

x

100

3 5

4 5

 4 sin  – 3 cos 

100 3

 4

3 4 3  0 5 5

 x  100 3  100 4.

7.

sec   tan  5  sec   tan  3

45° h

 8 tan  = 2sec  

4sin  1 1   sin   cos  cos  4

50

60°

 cosec = 4 (Ans)

x

According the question:

5.

tan 45  1 

h

tan 60 

x

60°

x  x  50  h 50  h

x 50

 3  50  50  h

30°  h  50

90







3 1

 tan 60  3



8. (1 + cot  – cosec ) (1 tan  + sec )

h 90 h tan 30°  90 3

cos  1  sin  1    1      1    sin  sin  cos  cos  

12.12

Trigonometric Ratios & Height and Distance

 2

sin 2   cos2  1 sin  cos 

tan 30 

1 3

ED CD



 CD  x 3

=2+0=2

A/Q

9. h

x 3

 x 3  100

 x  25 3 12. 2sin 2   5 cos .sin   7 cos 2   1

x

60°

Divide by cos2.

45°

 2 tan 2   5 tan   4

20

sin2   cos2 

 tan 2   1

x tan 45  1  20



 x = 20

 tan 2   tan   6  0

tan 60  3 

cos2 

  tan   3 tan   2  0

hx 20

 tan   2 or 3

 h  20 3  20  20



13.



A

3 1

60°

30° 30°

10. tan2 + cot2 – sec2 . cosec2 put  = 45°, since all '' values should give same value of expression. =1+1 

2

 2  2

D

90 3

60° C

A

90 3

E x

x

tan 60  3  60°

D

 BC 

90

100 m

AB tan 60  3  BC

 BC 

x 3

90 BC

30° C

B

E

2

= 2 – 4 = –2 11.

90

30°

tan 30 

3 1 3



 AE = 30  BE = 60 m

AE 90 3

B

Trigonometric Ratios & Height and Distance 12.13

14. cos– sin = l ; tan = m sec2 sin 30° =

put  = 45° l=0;m=

l  2m   2

1 2

 AB = 70m  AF = (70 – 20) m = 50m

 AE  50 2

1 2 1 2 2

02

15. According to the question

A

140

50 2

50

45

E

F

20 C

30

1 AB  2 AC

B

D

13

Clock and Calendars

CHAPTER

Cl ock s The dial of a clock is a cir cle whose cir cumfer ence is divided int o 12 par t s, called hour spaces. Each hour space is fur t her divided int o 5 par t s, called minute spaces. This way, t he whole cir cumfer ence is divided int o 12 × 5 = 60 minut e spaces. The t ime t aken by t he hour hand (smaller hand) t o cover a dist ance of an hour space is equal t o t he t ime t aken by t he minut e hand (longer hand) t o cover a dist ance of t he whole cir cumfer ence. Thus, we may conclude t hat in 60 minutes, the minute-hand gains 55 minutes over the hour-hand. Note: The above statement (given in bold) is very useful in solving pr oblems in t his chapt er.I t can be r est at ed as : “I n an hour, the hour-hand moves a distance of 5 minute spaces wher eas t he minut e-hand moves a dist ance of 60 minut e spaces. Thus, t he minut e-hand r emains 60 – 5 = 55 minut e spaces ahead of t he hour -hand.” Some other facts : 1. I n ever y hour, bot h t he hands coincide once. 2. When t he t wo hands ar e at r ight angle, t hey ar e 15 minut e spaces apar t . This happens t wice in ever y hour. 3. When t he hands ar e in opposit e dir ect ions, t hey ar e 30 minut e spaces apar t . This happens once in ever y hour. 4. The hands ar e in t he same st r aight line when t hey ar e Coincide or opposit e t o each ot her. 5. T h e h ou r h an d m ov es ar ou n d t h e w h ol e cir cumfer ence of t he clock once in 12 hour s. So, t he minut e-hand is t welve t imes fast er t han t he hour -hand. 6. T h e cl ock i s di vi ded i n t o 60 equ al m i n u t e divisions. 7. 1 minut e division =

360  6 apar t . 60

8. T h e cl ock h as 12 h ou r s n u m ber ed f r om 1 t o 12 ar r anged ser ially. 9. E ach h ou r n u m ber i s ev en l y an d equ al l y separ at ed by fi ve mi nut e di vi si ons (= 5 × 6°) = 30° apar t .

10. I n one mi nut e, t he mi nut e-hand moves one minut e division or 6°. 

 1 11. I n one minut e, t he hour hand moves   .  2 

 1 12. I n one minut e, t he minut e-hand gains  5  .  2 mor e t han t he hour -hand. 13. When t he hands ar e t oget her, t hey ar e 0° apar t . H ence,

Formed in Formed in 12 hours 24 hour s

 0° or 180° 90° or any ot her angle

11

22

22

44

As per t he r equi r ed angl e, di ffer ence bet ween t he mi n ut e-hand an d t he hour -hand and t h e i ni t i al (or st ar t i ng) posi t i on of t he hour -hand, di ffer ent for mulae ar e used t o find out t he r equir ed t ime. Now consider t he Rules (Quicker M ethods) given in t he following pages. Variants in a Clock: I t is evident t hat t he t wo hands of a clock subt end an angle ‘’ bet ween t hem. At any t ime, t he same can be find out using t he foll owing for mula :



11 11   m  30 h  when m  30 h    2 2

or   30 h 

11 m 2

11   m  when 30 h  2 

(H er e, m = minut es and h = hour s) Gain or Loss : I n a cor r ect clock, hands of a clock coincide in ever y 65

5 min . 11

5 min . , 11 t hen clock gains t ime and if hands of a clock coincide I f hands of a clock coincide in less t han 65

in mor e t han 65

5 min . , t hen clock loses t ime. 11

13.2

Clock and Calendars

Too Fast And Too Slow: I f a watch indicat es 9.20, when the cor r ect t ime is 9.10, it is said t o be 10 minut es t oo fast . And if it indicat es 9.00, when t he cor r ect t ime is 9.10, it is said t o be 10 minut es t oo slow.

Cal endar s Based on t he given infor mat ion, if you ar e asked t o f i n d ou t , wh at day of t h e week was Sept em ber 15, 2008, then you can easily calculat e it on your finger t ips. But , if week-day on Januar y 28, 2008 is t o be find out , t hen it may be a t ough job. The questions on this topic ar e ver y common in var ious compet i t i ve exams. T he met hod of sol vi ng su ch quest ions lies in t he concept of obt aining t he number of odd days. Befor e jumpi ng ont o t he t opi c, l et us r eview some of t he basic concept s.

What is a Leap year and an Ordinary year? Consider t he year 1997. I s t he number 1997 exact ly divisible by 4? The answer is: No, it is not . L et us t ake t he year 1996. I s t he number 1996 exact ly divisible by 4? The answer is: Yes, it is. Now, extending t he given concept onto the following year s such as 1980, 1900, 1987, 1964, 1600, 1951, 1300, 1988, 2000, 1990, 1992, et c., we can conclude : (1) Wh enever t he num ber of year i s exact l y divisible by 4 (except t he cent ur y year s), t hen it is a L eap year. (2) Whenever t he number of year is not divisible by 4, t hen it is an Or dinar y year. (3) I n case of t he cent ur y year s, if t he number of t he year is exact ly divisible by 400, t hen it is a L eap year. (4) Whenever t he number of year is not divisible by 400, t hen it is an Or dinar y year. Ordinary year: An or dinar y year can be defined as t he year having 365 days which is equal t o 52 weeks and an ext r a day. Century year: A year is a cent ur y year if it is divisible by 100. N on-Century year: A year is a non-cent ur y year if it is not a cent ur y year, or not divisible by 100. Leap year: A year is a leap year if it is a non-cent ur y

year t hat is divisible by 4, or a cent ur y year t hat is divisible by 400. H ow to find out number of odd days? The t ot al number of days for a specific per iod of t ime when divided by 7 gives a r emainder. That r emainder is t er med as t he odd day(s). Count ing of Odd days: i . 1 or di nar y year = 365 days = 52 week s + 1 odd day An or dinar y year has 1 odd day. ii. 1 l eap y ear = 366 day s = 52 w eek s + 2 odd days A leap year has 2 odd days. iii. 1 centur y year = 100 year s = 76 or dinar y year s + 24 l eap y ear s = 76 + 2 × 24 = 124 odd days = 5 odd days Now, based on t he above fact , we can conclude t hat : i . Number of odd days in 100 year s = 5 ii. Number of odd days in 200 year s = 10 days = 1 week + 3 odd days = 3 iii. Number of odd days in 300 year s = 15 days = 2 weeks + 1 odd day = 1 i v. Number of odd days in 400 year s = (20 + 1) = 3 weeks = 0 T he following point s have t o be observed: (1) The fol l owi ng t abl e i s based on t he fact t hat 1st Januar y, 1 A.D. was a M onday. This t able is helpful in solving t he quest ions which assume t he given infor mat ion.

No. of odd days Days

1

2

4

5

6

7 or 0

Mon. Tue. Wed. Thu. Fri. Sat. Sun.

(2) I n an Or dinar y year, t he calendar for t he mont h of Januar y is t he same as t he calendar for t he mont h of Oct ober. I n shor t , in an Or dinar y year, January = Oct ober. (3) I n a L eap year, t he calendar for t he mont h of Januar y is t he same for t he mont h of July. I n shor t , in a Leap year, January = July.

Sol ved E xamples Example 1 At what t ime bet ween 3 O’clock and 4 O’clock, will the minute-hand and the hour-hand of a clock coincide wit h each ot her ? Sol ut i on When t he t wo hands of a clock coincide wit h each ot her, t he angle bet ween t hem is 0°.

3

11 m  30h 2 H er e, 

  0 and h = 3

Clock and Calendars





11 m  30h 2

90  30  5 

11 m  30  3 2

11 m 2

120 10 min .  10 min . 11 11 Ther efor e, t he angle bet ween t he t wo hands of t he m 

90  2 4  16 min 11 11 Ther efor e, t he t wo hands of t he clock will coincide m 

cl ock i s 90°, when t he t i me i s 5 hr s. 10 and 5 hr s. 43

11 m  30h 2 H er e,

m = minut es h = hour s H er e, m = 20 and h = 3



h = 4 and   180

11 180  m  30  4 2

st r aight line at 4 hour s 54

6 min . 11

Example 3 At what t ime bet ween 5 O’clock and 6 O’clock will the hands of a clock ar e per pendicular t o each other ? Sol ut i on

11 m  30h 2

H er e,   90 and h = 5

90 



11 m  30  5 2

11 m  240 2

m 

480 7 min .  43 mi n . 11 11

Also,   30 h 

11 m 2

Example 4 What is t he angle bet ween t he minut e-hand and t he hour -hand of a clock at 3 hr s. 20 min.? Sol ut i on



11 m  30 h 2

  angl e



(180  30  4)  2 6 m   54 mi n . 11 11 Ther efor e, t he hands of t he clock will be on t he same

10 min . 11

7 min . 11

4 min . past 3. 11 Example 2 At what t ime bet ween 4 O’clock and 5 O’clock will t he hands of a clock be in t he same st r aight line but not t oget her ? Sol ut i on When t he t wo hands of t he clock ar e in t he same straight line but not together, then the angle between t hem is 180°. at 16



13.3

11  20  30  3 2

 110  90  20    20 Example 5 The minut e-hand of an in-accur at e clock over t akes t he hour -hand in ever y 65 minut es int er val. H ow much in a day does t he clock gain or lose? Sol ut i on I n an accur at e clock, the hands of a clock coincide

5 minutes. But in this case, they are together 11 again after 65 minutes, hence t he clock gains time. every 65

5 5    65  Gain in 65 minut es =  65 minut es.  11  11 

Gain in one day (24 × 60 min.)



5 60 5  288 10   24  min . min.  10 11 65 143 143

Example 6 A wat ch, which gains unifor mally, was obser ved t o be 5 minut es slow at 10 a.m. on a Tuesday. On t he next day at 11 a.m., it was not iced t hat t he wat ch was 5 minut es fast . When did t he wat ch show t he cor r ect t ime?

13.4

Clock and Calendars

Sol ut i on Tot al hour s fr om 10 a.m. Tuesday t o 11 a.m. on t he next day = 25 hour s The wat ch gains (5 + 5) = 10 minut es in 25 hour s.

Example 10

The wat ch gains 5 min. in

1992 being a leap year, it had 2 odd days. So, t he fi r st day of t he year 1993 was t wo days beyond Wednesday. i.e., it was Fr iday.

1  25  125  10  5  10 hr s.  12 2 hrs.

Januar y 1, 1992 was a Wednesday. What day of t he week was Januar y 1, 1993? Sol ut i on

Example 11 Januar y 12, 1980 was a Sat ur day. The day of t he week on Januar y 12, 1979 was:

1  12 hours fr om 10 a.m. Tuesday 2

Sol ut i on

= 10 : 30 p.m. Tuesday. Example 7 Ther e ar e t wo clocks, bot h set t o show t he cor r ect t ime at 10 p.m. One clock gains one minut e in an hour while t he ot her gains 2 minut es in an hour, t hen by how many minut es will t he t wo clocks differ at 10 a.m. on t he next day? Sol ut i on Differ ence in minut es bet ween t he t wo clocks in one hour = 1 mi nut e. Tot al number of hour s (10 p.m t o 10 a.m. on next day) = 12 hour s.

The year 1979 being an or dinar y year, had 1 odd day. 12t h Januar y 1980 was a Sat ur day. But it was one day beyond 12th Januar y 1979. Ther efor e, 12t h Januar y 1979 was a Fr iday. Example 12 Febr uar y 20, 1999 was a Sat ur day. What day of t he week was December 30, 1997? Sol ut i on

Example 8

The year dur ing t his int er val was 1998 and it was not a leap year. Now, we calculat e t he number of odd days in 1999 upt o Febr uar y 19:

I f t he t ime in a clock is 8 hour s 20 minut es, t hen what t ime does it show on t he mir r or ?

19 Febr uar y 1999 gives 5 odd days

The t wo clocks differ by = 1 × 12 = 12 minut es.

Januar y 1999 gives 3 odd days

Sol ut i on

1998, being an or dinar y year gives 1 odd day

The t i me shown by t he cl ock , when seen i n t he mir r or = 12 hour s – 8 hour s 20 minut es

I n 1997, December 30 and 31 give n 2 odd days

= 3 hour s 40 minut es. Example 9 What day of t he week was 16t h July, 1776? Sol ut i on (1775 year s + 6 mont hs + 16 days) 100 year s have 5 odd days. 75 year s contain 18 L eap and 57 Or dinar y year s and t her efor e, (36 + 57) or 93 or 2 odd days. 1775 year s give 0 + 5 + 2 = 7 and so, 0 odd days.

Also, number of days fr om 1st Jan, 1776 to 16th July, 1776 M ar.

Apr.

M ay

Jun. Jul.

31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days = 28 weeks + 2 days = 2 odd days. 

Example 13

Sol ut i on

Now, 1600 year s have 0 odd days.

Jan. Feb.

Ther efor e, December 30, 1997 was 4 days befor e Sat ur day i.e., a Tuesday. The year next t o 1987 having t he same calendar as t hat of 1987 is:

16t h July, 1776 means



 Tot al n u m ber of odd day s = 3 + 5 + 1 + 2 = 11 days = 4 odd days.

Tot al number of odd days = 0 + 2 = 2.

H ence, 16t h July, 1776 was a Tuesday.

St ar t ing wit h 1987, we go on count ing t he number of odd days t ill t he sum is divisible by 7. Number of odd days = 1(1987) + 2(1988) + 1(1989) + 1(1990) + 1(1991) + 2(1992) + 1(1993) + 1(1994) + 1(1995) + 2(1996) + 1(1997) = 14/7 = 0 odd days. So, t he year next t o 1987 having t he same calendar as t hat of 1987 is 1998.

Clock and Calendars

13.5

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. An accur at e clock shows 8 O’clock in t he mor ning. Thr ough how many degr ees will t he hour -hand r ot at e when t he cl ock shows 2 O’cl ock i n t he after noon? (a) 30° (b) 180° (c) 90° (d) 150° 2. At 3 : 40, t he hour -hand and the minut e-hand of a clock for m an angle of: (a) 120° (b) 125° (c) 130° (d) 140° 3. The angle between the minute-hand and the hour hand of a clock when t he t ime is 4 : 20, is: (a) 10° (b) 15° (c) 5° (d) 20° 4. H ow many t imes do t he hands of a clock coincide in a day? (a) 24 t imes (b) 22 t imes (c) 20 t imes (d) 23 t imes 5. H ow many t imes in a day, do t he hands of a clock for m a r ight angle? (a) 24 t imes (b) 48 t imes (c) 22 t imes (d) 44 t imes 6. H ow many t imes in a day, bot h t he hands of a clock for m a st r aight line? (a) 22 t imes (b) 24 t imes (c) 44 t imes (d) 48 t imes 7. H ow many degr ees does an hour -hand move in 15 minut es? (a) 1.5° (b) 7.5° (c) 18° (d) 12° 8. H ow many degr ees will t he minut e-hand move in 20 minut es? (a) 90° (b) 150° (c) 120° (d) 180° 9. I f t he t ime in a clock is 6 hour s 45 minut es, t hen what t ime does it show on t he mir r or ? (a) 6 hr s. 45 min (b) 4 hr s. 15 min (c) 7 hr s. 45 min (d) 5 hr s. 15min 10. H ow many odd days ar e t her e in 249 days? (a) 5 days (b) 2 days (c) 3 days (d) 4 days 11. Which will be t he next leap year aft er 1896? (a) 1900 (b) 1904 (c) 1908 (d) 1898 12. Today is Fr iday. After 62 days, what day will it be? (a) Thursday (b) Saturday (c) Friday (d) Sunday

13. I f a year st ar t s and ends on M onday, t hen how many M ondays ar e t her e in t hat year ? (a) 52 (b) 53 (c) 51 (d) 50 14. A clock is st ar t ed at noon. By 10 minut es past 5, t he angle t hat t he hour -hand has t ur ned t hr ough is: (a) 145° (b) 150° (c) 155° (d) 160° 15. A t w h at t i m e bet w een 9 O’cl ock an d 10 O’clock will t he hands of a wat ch coincide? (a) 10 hrs. 49

1 min . 11

(b) 9 hr s. 49

1 min . 11

(c) 11 hr s. 49

1 min . 11

(d) 9 hr s. 59

1 min . 11

LEVEL-1 1. At what t ime bet ween 4O’clock and 5 O’clock will t he hands of a wat ch point in opposit e dir ect ions? (a) 4 hr s. 52

6 min . 11

(b) 4 hr s. 32

6 min . 11

(c) 4 hr s. 54

6 min . 11

(d) 4 hr s. 34

6 min . 11

2. The angle between the minute-hand and the hour hand of a clock when t he t ime is 8 : 30, is: (a) 80° (b) 75° (c) 60° (d) 105° 3. At what angle ar e t he hands of a clock inclined at 15 minut es past 5? (a) 58

1 2



1 2

(b) 64° 

1 2 4. H ow many degr ees will t he minut e-hand move, in t he same t ime in which t he second-hand moves 240°? (a) 4° (b) 5° (c) 6° (d) 8° (c) 67

(d) 72

13.6

Clock and Calendars

5. Ther e ar e t wo clocks, bot h set t o show t he cor r ect t ime at 10 a.m. One clock gains t wo minut es in one hour while t he ot her gain one minut e in one hour. I f t he clock which gains 2 minut e shows t he t ime as 22 minut es past 9 p. m., t hen what t ime does t he ot her wat ch show? (a) 9 hr s. 33 min. (b) 9 hr s. 12 min. (c) 9 hr s. 11 min. (d) 9 hr s. 23 min. 6. Which among t he following year s is a leap year ? (a) 1900 (b) 1800 (c) 1700 (d) 2800 (e) 2600 7. The fir st Republic Day of I ndia was celebr at ed on 26th Januar y, 1950. What was the day of the week on t hat dat e? (a) Wednesday (b) Thursday (c) Friday (d) Saturday (e) Tuesday 8. M ahat ma Gandhi was bor n on 2nd Oct ober, 1869. The day of t he week was (a) Wednesday (b) Thursday (c) Friday (d) Saturday (e) Sunday 9. I ndia got I ndependence on 15th August 1947. What was t he day of t he week on t hat day? (a) Wednesday (b) Thursday (c) Friday (d) Saturday (e) Sunday 10. I f t oday is Sat ur day, t hen what day of t he week will be on t he 338t h day fr om t oday? (a) Monday (b) Friday (c) Sunday (d) Saturday (e) Tuesday 11. I f t oday is M onday, which day of t he week will it be aft er one year ? (a) Tuesday (b) Wednesday (c) Monday (d) Eit her ‘(a)’ or ‘(b)’ (e) Sunday 12. I f 23r d Apr il, 1984 was a Monday, which day of the week was 15t h August in t hat year ? (a) Monday (b) Wednesday (c) Tuesday (d) Thursday (e) Friday 13. I f 3r d M ar ch, 1984 was a Sunday, t hen which day of t he week was 13t h July, 1987? (a) Monday (b) Sunday (c) Saturday (d) Tuesday (e) Friday

14. I f 10t h Apr il, 1883 was a Wednesday, t hen which day of t he week was 23r d August , 1879? (a) Sunday (b) Tuesday (c) Monday (d) Friday (e) Saturday 15. On Januar y 2, 1985, it was Wednesday. The day of t he week on Januar y 2, 1984 was: (a) Monday (b) Sunday (c) Tuesday (d) Wednesday (e) Saturday

LEVEL-2 1. Today is 1st August and t he day of t he week is M onday. This is a leap year. The day of t he week on t his day aft er 3 year s will be (a) Wednesday (b) Friday (c) Thursday (d) Sunday (e) Tuesday 2. M onday falls on 4t h Apr il, 1988. What was t he day on 3r d November, 1987? (a) Tuesday (b) Sunday (c) Monday (d) Wednesday (e) Saturday 3. The year next t o 1990 having t he same calendar as t hat of 1990 is ____. (a) 1998 (b) 2001 (c) 2002 (d) 2004 (e) 2000 4. Which dat es of Apr il, 2012 will be Sunday? (a) 1, 8, 15, 22, 29 (b) 3, 10, 17, 24, 31 (c) 2, 9, 16, 23, 30 (d) 5, 12, 19, 26 (e) 4, 11, 18, 25 5. I f 30 Januar y 1989 is M onday t hen what was t he day of week on 26 Oct 2003. (a) Monday (b) Sunday (c) Tuesday (d) Saturday (e) Friday 6. I f 9t h of t he mont h fal ls on t he day pr ecedi ng Sunday, on what day will 1st of t he mont h fal l? (a) Friday (b) Saturday (c) Sunday (d) Monday 7. Anil r eached a place on Fr iday. H e came t o know t hat he was t hr ee days ear lier t han t he scheduled day. I f he had r eached t her e on t he fol l owi ng Sunday, how many days l at e/ear ly he would have been? (a) One day ear li er (b) One day lat e (c) Two days lat e (d) Two days ear li er

Clock and Calendars

8. I f t he day befor e yest er day was ' Sunday, what day wi l l i t be t hr ee days aft er t he day aft er t omor r ow? (a) Sunday (b) Monday (c) Wednesday (d) Saturday 9. I f t he day aft er t omor r ow is Sunday, what day was t omor r ow's day befor e yest er day ? (a) Friday (b) Thursday (c) Monday (d) Tuesday 10. Su r esh w as bom on 4t h Oct ober 1999. Shashikant h was bor n 6 days befor e Sur esh. The I ndependence Day of t hat year fell on Sunday. Which day was Shashik ant h bom? (a) Tuesday (b) Wednesday (c) Monday (d) Sunday 11. Reaching a place of appointment on Fr iday, I found t hat I was t wo days ear li er t han t he schedul ed day. I f I had r eached on t he following Wednesday, how many days lat e woul d I have been? (a) One day (b) Two days (c) Thr ee days (d) Four days

13.7

12. lf t he 23r d of a mont h i s a Sunday, what day it woul d have been t wo weeks and four mor e days ear l ier ? (a) Monday (b) Tuesday (c) Wednesday (d) Thursday 13. I f t wo days aft er day aft er t omor r ow i s Sunday, what day was t he I day befor e yest er day? (a) Wednesday (b) Tuesday (c) Monday (d) Sunday 14. lf t hr ee days aft er t oday wi ll be Tuesday, what day was four days befor e yest er day? (a) Tuesday (b) Sunday (c) Monday (d) Wednesday 15. I f t he day befor e yest er day was Thur sday, when wi ll Sunday be? (a) Day aft er t omor r ow (b) Today (c) Tomor r ow (d) Two days aft er t oday

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (b) 11. (b)

2. (c) 12. (a)

3. (a) 13. (b)

4. (b) 14. (c)

5. (d) 15. (b)

6. (a)

7. (b)

8. (c)

9. (d)

10. (d)

7. (b)

8. (d)

9. (c)

10. (a)

7. (a)

8. (a)

9. (b)

10. (b)

LEVEL-1 1. (c) 11. (d)

2. (b) 12. (b)

3. (c) 13. (c)

4. (a) 14. (a)

5. (c) 15. (a)

6. (d)

LEVEL-2 1. (c) 11. (d)

2. (a) 12. (c)

3. (b) 13. (c)

4. (a) 14. (c)

5. (b) 15. (c)

6. (a)

EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S 1. Angle t r aced by hour -hand in 6 hour s

  30h 

11 m 2

H er e, m = 20 and h = 4

0

 360   6 = 1800. =   12 

   30  4 

2. The angle bet ween t he t wo hands of a clock at 3:40 is  

   10.

11 m  30h 2

H er e, h = 3 and m = 40   

11  20    120  110 2

11  40  30  3 2

   130.

3. The angle bet ween t he t wo hands of a clock at 4 : 20 is

4. The hands of a clock coincide 11 t imes in ever y 12 hour s (Since bet ween 11 and 1, t hey coincide only once, i.e. at 12 O’clock).  The hands of a clock coincide 22 t imes in a day.. 5. I n 12 hour s, t hey ar e at r ight angles, 22 t imes.  I n 24 hour s, t hey ar e at r ight angles, 44 t imes.

13.8

Clock and Calendars

6. The hands of a clock point in opposit e dir ect ions (in t he st r aight line) 11 t imes in ever y 12 hour s (because bet ween 5 and 7 t hey point in opposit e dir ect ions at 6 O’ clock onl y). So, in a day, t he hands of a clock point in t he opposit e dir ect ions 22 t imes. 7. I n 12 hour s t he hour -hand moves 360°. H ence, in 15 minut es it moves 

 360 15   12  60   7.5. 8. The minut e-hand moves 360° in 1 hour. H ence, i n 20 mi nut es t he mi nut e-hand moves 360  20  120. 60

9. The t ime shown by t he clock when seen in t he mir r or = 12 hr s. – 6 hr s. 45 min. = 5 hr s. 15 min. 249 10.  35 weeks + 4 odd days. 7

11. A leap year comes aft er ever y 4 year s. 1896 + 4 = 1900 But 1900 i s not a l eap year, because i t i s not divisible by 400. (for a cent ur y t o be leap year, it should be exact ly divisible by 400.) H ence, 1900 + 4 = 1904 will be t he next leap year aft er 1896. 12. Each day of t he week is r epeat ed aft er 7 days.  Aft er 63 days, it would be Fr iday.. So, aft er 62 days, it would be Thur sday. 13. As t he given year st ar t s and ends wit h M onday means t he next year wi l l st ar t wi t h Tuesday. H ence, t he given year is a non-leap year. Ther e will be 53 M ondays in t he year. 14. Angle t r aced by hour -hand in 12 hr s. = 360 Angle t r aced by hour -hand in 5 hr s 10 min. i.e.

m 

270  2 1 min .  49 min . 11 11

Ther efor e, t he hands of t he clock ar e t oget her at 9 hr s. 49

1 min . 11

LEVEL-1 1. When t he t wo hands of t he cl ock ar e faci ng opposit e dir ect ions, t hen t he angle bet ween t hem is 180°. 11  m  30h 2 H er e, h = 4 and   180. 11 180  m  30  4 2

m 

600 6 min .  54 min . 11 11

Ther efor e, t he hands of t he cl ock ar e faci ng 6 opposit e dir ect ions at 4 hr s. 54 min . 11 2. The angle bet ween t he t wo hands of a clock at 8:30 is H er e,

11 m  30h 2

   30h 

11 m 2

   30  8     75

3. The angl e bet ween t he hands of a cl ock at 15 minut es past 5 is 11   30h  m 2 H er e, h = 5 and m = 15

11 1  15    67 . 2 2  240  4. Second-hand moves 240° in  40 sec. 6  360 I n 60 sec t he minut e-hand cover s  6 . 60 H ence, in 40 sec. t he minut e-hand cover s 4°. 5. Differ ence in minut e bet ween t he t wo clocks in one hour = 1 minut e. Number of hour s = 11 hour s. I n 11 hour s, one of t he clock gains 22 minut es and shows t he time as 9:22 p.m. The ot her clock which gains 1 minut e per hour shows t he t ime as 9:11 p.m.    30  5 

31 hr s. 6 

 360 31      155.  12 6 15. When t he t wo hands of t he clock coincide, t hen t he angle bet ween t hem is 0°. 11  m  30h 2 H er e,   0 and h = 9 11  m  30  9 2

11  30 2

Clock and Calendars

6. A cent ur y year is a leap year only if it is exact ly divisible by 400. Only 2800 is exact ly divisible by 400. H ence, 2800 is a leap year. 7. Tot al number of odd days = 1600 year s have 0 odd day + 300 year s have 1 odd day + 49 year s (12 leap + 37 or dinar y) have 5 odd days + 26 days of Jan have 5 odd days = 0 + 1 + 5 + 5 = 4 odd days So, t he day was Thur sday. 8. 1600 year s have 0 odd day 200 year s have 2 × 5 = 10, i.e., 3 odd days. 68 year s cont ain 17 leap year s and 51 or dinar y year s. That is, 17 × 2 + 51 = 85 days, i.e., 1 odd day. I n 1869, upt o 2nd Oct ., t ot al number of odd days = 31(Jan.) + 28(Feb.) + 31(M ar.) + 30(Apr.) + 31(May) + 30(Jun.) + 31(Jul.) + 31(Aug.) + 30(Sep.) + 2(Oct .) = 275 days = 2 odd days.  Tot al odd days = 0 + 3 + 1 + 2 = 6 odd days.  The day was Sat ur day.. 9. 15 Aug., 1947 = (1600 + 300 + 46) year s + 1 Jan. t o 15 Aug. of 1947 = (1600 + 300 + 46) year s + 365 - 16 Aug. t o 31 Dec 1947 = (1600 + 300 + 46) year s + (365 - 138) days Number of odd days = 0 + 1 + 1 (fr om 11 leap year s and 35 or dinar y year s) + 3 = 5 odd days.  The day was Fr iday.. 338  48 10. N umber of odd days i n 338 days  7 complet e weeks +2 odd days. 2nd day aft er Sat ur day is a M onday. 11. The given year is a leap year or a non-leap year is not given. So, t he answer is eit her (a) or (b). 12. Count ing the number of days after 23rd Apr il, 1984 we have: Apr il may June July August days: 7 + 31 + 30 + 31 + 15 = 114 days Number of odd days in 114 days =

114  16 weeks +2 odd days 7

2nd day aft er M onday is Wednesday. 13. I t is given t hat 3r d M ar ch, 1984 was a Sunday So, 3r d Mar ch, 1987, was thr ee days after Sunday, i.e., on Wednesday. Number of days fr om 3r d M ar ch, 1987 t o 13t h July, 1987: M ar ch Apr il M ay June July Days: 28 + 30 + 31 + 30 + 31 = 150 =

150  3 odd days 7

3r d day aft er Wednesday is Sat ur day.

13.9

14. I t is given t hat 10t h Apr il, 1883 was a Wednesday. N umber of days fr om 10t h Apr i l , 1883 t o 23r d August 1883. Apr il may June July August Days: 20 + 31 + 30 + 31 + 23= 135 Number of odd days in 135 days =

135 7

= 2 odd days. 2 days aft er Wednesday is Fr iday. Number of odd days fr om 23r d August 1879 t o 23r d August 1883 ar e five. So, 23r d August , 1879 is five days back t o Fr iday is Sunday. 15. The year 1984 being a leap year, it has 2 odd days. So, t he day on 2nd Jan., 1984. But , 2nd Jan., 1985 was Wednesday.  2nd Jan., 1984 was M onday..

LEVEL-2 1. This being a leap year none of t he next 3 year s is a leap year. So, t he day of t he week will be 3 days beyond M onday i.e., it will be Thur sday. 2. Counting t he number of days after 3r d November, 1987 we have : Nov. Dec. Jan. Feb. M ar. Apr. Days: 27 + 31 + 31 + 29 + 31 + 4 = 153 days, cont aining 6 odd days. i.e., (7 – 6) = 1 day beyond t he day on 4t h Apr il, 1988. So, t he day was Tuesday. 3. We go on count ing t he no. of odd days fr om 1990 onwar d t ill t he sum is exact ly divisible by 7. The number of such days ar e 14 upt o t he year 2000. So, t he calendar for 1990, was r epeat ed in t he year 2001. N ote: Number of odd days = 1(1990) + 1(1991) + 2(1992) + 1(1993) + 1(1994) + 1(1995) + 2(1996) + 1(1997) + 1(1998) + 1(1999) + 2(2000) = 14 odd days. 4. 1st Apr il, 2012: 2000 + 11 + Number of days fr om 1st Januar y 2012 t o 1st Apr il, 2012. Number of odd days in 2000 year s = 0 Number of odd days in 11 year s = 13 Januar y Febr uar y M ar ch Apr il Odd days: 3 + 1 + 3 + 1 =8 Tot al number of odd days = 8 + 13 + 0 = 21 = 0 odd days. H ence, 1st Apr il, 2012 is a Sunday. 1st , 8 t h , 15 t h , 22 nd and 29 t h of Apr i l , 2012 ar e Sunday’s.

13.10

Clock and Calendars

5. Tot al number of odd days fr om 30 Januar y 1989 t o 30 Januar y 2003 = odd days in 3 leap year + 11 or dinar y year s = 3 odd days Total odd days fr om 30 Januar y 2003 to 26 October 2003 = Jan 1

F eb 28

M ar ch 31

A pr i l 30

Ju n e 30

Ju l y 31

A u gu st 31

Sept 30

= 3 odd days So, t ot al odd days = 6 odd days So, day on 26 Oct ober 2003 will be Sunday. 6. Accor ding t o quest ion, 9t h  Sat ur day Ther efor e, 9-7 = 2nd  Sat ur day  1st  Fr iday 7. Anil r eached t he place on Fr iday and he was t hr ee days ear l ier t han t he scheduled day. Ther efor e, t he scheduled day = Fr iday + 3 days = M onday I f he had r eached on Sunday t hen he woul d have ear l ier t han one day. 8. Day befor e yest er day was Sunday. Ther efor e, t oday is Tuesday. Day aft er t omor r ow will be Thur sday. Thur sday + 3 = Sunday 9. The day aft er t omor r ow i s Sunday. Ther efor e, t oday i s Fr i day. The day on t omor r ow's day befor e yest er day = Fr iday – 1 = Thur sday

10. Shashik ant was bor n on 29t h Sept ember 1999. 15t h August , 1999 was Sunday. Days upt o 29t h Sept ember fr om 15 August 16 + 29 = 45 days = 6 week s 3 old days Sunday + 3 = Wednesday. 11. Fr iday  2 days ear li er Ther efor e, scheduled day = Fr iday + 2 = Sunday, Sunday + 3 = Wednesday Ther efor e, I woul d have been l at e by 4 days including Sunday. 12. Two weeks ear l ier t han 23r d was also Sunday. 23 – 7 = 16 16 – 7= 9 4 days ear l ier t han 9 means 5t h, 9t h  Sunday 8t h  Sat ur day 7t h  Fr iday 6t h  Thur sday 5t h  Wednesday 13. The day aft er t omor r ow would be Fr i day. Today i s Wednesday. The day befor e yest er day was M onday. 14. Today + 3 = Tuesday  Today = Tuesday – 3 = Sat ur day Yest er day = Sat ur day – 1 = Fr iday Fr iday – 4 = M onday 15. Today is Thur sday + 2 = Sat ur day Ther efor e, t omor r ow will be Sunday. 

14

Geometry

CHAPTER

LI N E A line segement AB when extened indefinitely in both  t he dir ect ions is called line AB . A B Collinear point s Thr ee or mor e than thr ee points ar e said to be collinear , if t her e is a line which cont ains all of t hem. H er e, A, B, C ar e collinear point s.

A B C I nt er sect ing lines Two having a common point ar e called int er sect ing lines. H er e, line AB and CD int er sect at a point O. The point common t o t wo given lines is called t heir point of int er sect ion.

Classificat ion of angle : 1. Acut e angle : I t is an angle whose measur e is mor e t han 0 but less t han 90 2. Right angle : I t is an angle whose measur e is 90 3. Obt use angle : I t is an angle measur e mor e t han 90 but less t han 180. 4. St r aight angle : I t i s an angle whose measur e is 180 5. Reflex angle : I t is an angle whose measur e is mor e t he 180 but less t han 360. 6. Complet e angle : I t is an angle whose measur e is 360. Bisector of an angle A r ay OP is said t o be t he bisect or of AOB if AOB = BOP B

D

A

P O B

C

Concur r ent lines Thr ee or mor e lines ar e said t o be concur r ent if t hey meet at t he same point . l m

o n

Par allel lines Two lines l and m in a plane ar e said t o be par allel, if t hey have no common poi nt and we wr i t e l  m . The distance between two par allel ines aways r emains same. PLAN E A plane is a sur face such t hat ever y point of t he lines joining any t wo point s on it , lies on it . AN GLE Two r ays having a common init ial point is called an a angle.

  AOP =  BOP =

A

1  AOB 2

Complement ar y angles Two angles ar e said t o be complement ar y, if sum of t h ei r m easu r e i s 90  an d each an gl e i s cal l ed complement of t he ot her s. Supplement ar y angle Two angles ar e said t o be supplement ar y, if sum of t hei r measur es i s 180  and each angl e i s cal l ed supplement of t he ot her s. Adjacent angles Two angles having t he same ver t ex and a common ar m ar e said t o be adjancent , if t her e non-common ar ms ar e opposit e side of t he common ar m. H er e,  AOC and  COB ar e adjacent angles.

A C

B

O

A

O

O

B

14.2

Geometry

Angle of linear pair Two adjacent angle ar e said t o for m a linear pair of angles, if t her e non-common ar ms ar e t wo opposit e r ays.

C

T RAN SV ERSAL A st r aight line t hat cut s t wo or mor e st r aight lines at differ ent point s is called a t r ansver sal . Angles formed when a transversal cuts two lines L et AB and CD be t wo lines, cut by a t r ansver sal l, t hen following angles ar e for med :

l A

O

A

1

B

D

B O

C

A

3

4

H er e,  AOC and  BOC for m a linear pair of angles. V ert ical opposit e angle   I f t wo lines AB and CD int er sect at a point ‘O’, t hen pai r s of angl e for med by t hem i s cal l ed ver t i cally opposite angles.

H er e,  AOD and  BOC ;  DOB and  AOC ar e ver t ically opposit e angle. (i ) I f a r ay st ands on a line, t hen sum of t he adjacent angle, so for m is equal t o t he r ight angles. (ii ) I f two lines inter sct each-other , then t he ver tically opposit e angles so for med ar e equal.

8

C

5

7

B

2

6

D

(i ) Pair s of cor r esponding angles : ( 1,  5) and ( 4,  8) ; ( 2,  6); ( 3, 7) (ii ) Pair s of alt er nat e int er ior angles : ( 3, 6) and ( 3, 5) (iii ) Pair s of consecut ive int er ior angles : ( 3, 6) and ( 4, 5) (iv) The angles labelled 1, 2, 7, 8 ar e called ext er ior angles. (v) The angles labelled 3, 4, 5, 6 ar e called int er ior angles. Cor r esponding Angles Axiom (i ) I f a t r ansver sal cut s t wo par allel lines t hen each pair of cor r esponding angles ar e equal. (ii ) I f a t r ansver sal cut s t wo lines, making a pair of cor r esponding angles equal , t hen t he l i nes ar e par allel.

SOLVED EXAM PLES 1. In the given figure, three lines AB, CD and EF intersect in a common point, for ming angles as shown. I f  – 50  – 90, and  – , find the value of 

F

A

C

2. I f AB  CD EF as shown in t he given figur e, find  CEF. A

D

 

L

E

F

36

C

 

B

66

B

  E

Solution : Given  =  – 50,  = – 90,  =  Since  +  +  +  +  = 360  50 + 90 +  + 50 + 90 +  = 360  280 + 2 = 360  2 = 80   = 40

D

Solution : Refer r ing figur e, BLE =  ABL = 66  CL E = 180 – 66 = 144 x =  L CE + CL E = 36 = 144 = 150 3. I f AB and CD ar e par allel st r aight lines in t he given figur e, t hen find  RSD. P R

A C

3 x S Q

2 x

B

D

Geometry

Solution :  CSR +  SRA = 180   CSR +  BRP = 180  3x  + 2x  = 180  5x = 180  x = 36   RSD = 2x  = 72

x  x = 90   

x

B

C

x ar e

 

O

....(i )

x = 90 – x x = (90 – x )2 = (90)2 + x 2 – 180x 2 x – 181 x + 8100 = 0

181  (181) 2  4  8100 2 181  32761  32400 = 2 181  19  100,81 = 2 Since, x = 100 does not sat isfy (i ), Ther efor e x = 81 5. I f differ ence of t wo supplement ar y angles is 50, find smaller angle. Solution : I f x and y supplement ar y angles, t hen x + y = 180 and x – y = 50 ...(i ) Solving equat ions (i ) and (ii ), we get x = 115, y = 65 ...(ii ) 6. I n t he given figur e, AB and CD ar e par allel and PQ is per pendicular t o L M . I f  BNM = 50, find  PQD.



2z  3z  4 z = 180 12 9z  = 180 12 180  12  z= = 240 9 8. I f t he gi ven fi gur e, st r ai ght l i nes AB and CD int er sect at O. I f magnit ude of  is for t imes t hat of , find  

4. I f an angle and it s complement ar e x  and r espect ively, t hen find angle x . Sol ut i on : Si n ce t h e an gl es x an d complement ar y, t her efor e

14.3

x =

A

D

Solution : Given :  = 4 Fr om t he figur e, it is clear t hat  +  = 180     = 180 4 5  = 180  4   = 180 4   = 180   144 5 T RI AN GL ES A plane figur e bounded by thr ee line segments is called a triangle. A tr iangle has basically six par ts, three sides and t hr ee angles. The sum of t he t hr ee angles of a t r iangle is 180 Perimeter of a Triangle. Sum of t he lengt hs of t hr ee sides of a t r iangle is called per imet er of a t r iangle. E xt er i or Angl es of a T r i angl e. L et A B C be a t r iangle, whose ar e sides BC is pr oduced t o D. Then,  BCD is called an ext er ior angle of a t r iangle. A

M N

A

50

Q

C L

B D

P

Solut ion : 

 DQN =  BNM = 50  PQD =  PQN – DQN = 90 – 50 = 40 7. Find value of z in (degr ees), in t he given figur e

B

C

D

 ABC and  BAC are called its nterior opposite angles. M edians. The line segment which joins midpoint of t he side wit h t he opposit e ver t ex is called median of the triangle. A F

E G

B

Solution : Fr om t he figur e, z z z   = 180 6 4 3

D

C

Her e, AD, BE and CF ar e medians of the tr iangle ABC. Cent r oid. The poi nt of i nt er sect i on of al l t he t hr ee medians of a t r iangle is called cent r oid of t he t r iangle. I n the above figur e, G is the centr oid of the tr iangle ABC.

14.4

Geometry

Altitudes. The lengt h of per pendicular dr awn fr om t he opposit e ver t ex t o t he cor r esponding side is called alt itudes of t he t r iangle. A R

Q O

B

P

C

H er e, AP, BQ and CR ar e alt i t udes of t he t r iangle cor r sponding t o base BC, CA and AB. Or t hocent r e. The point of int er sect ion of all t he t hr ee alt it udes of a t r iangle is called or t hocent r e of t he t r iangle. I ncent r e. The point of int er sect ion of t he int er nal bisect or s of the angles of a tr iangle is called incentre of the triangle H er e I is t he incent r e of  ABC. A

I B

D

C

L et I D  BC. Then, a cir cle wit h cent r e I and r adius I O is calledf b incir cle of t he ABC. Circumcent re of a t riangle. The point of int er sect ion of t he Per pendicular bisect or s of t he sides of a t r iangle is called cir cumcent r e of t he t r iangles. A

O B

C

H er e ‘O’ is t he cir cumcent r e  ABC. Then, a cir cle wit h cent r e ‘O’ and r adius equal t o OA = OB = OC is called cir cumcir cle of  ABC.s N ot e: (i ) I f a side of a t r iangle is pr oduced, t hen ext er ior angle so for med is equal t o sum of t he t wo int er ior opposit e angle. (ii ) Any ext er ior angle of a tr iangle is gr eater t han eit her of the int er ior opposite angle. Congruence of Triangle. Two t r iangles ar e said t o be congr uent if and only if one of t hem can be made t o super pose on t he ot her , so as t o cover it exact ly. Thus, congr uent t r iangles ar e exact ly ident ical.

Ther e ar e Var ious cr it er ia for t he t wo t r iangles t o be congr uent : (i ) S– A– S : Two t r iangles ar e congr uent if t wo sides and included angle ar e cor r espondingly equal. (ii ) A– S– A : Two t r iangles ar e congr uent if t wo angles and one side ar e cor r esponding equal. (iii ) S– S– S : Two t r iangles ar e congr uent if all the t hr ee sides ar e cor r espondingly equal. (iv) R– H – S : Two r ight angled tr iangled t r iangles ar e congr uent if hypotenuse and one side ar e cor r espondingly equal. SI M I L AR T RI AN GL ES Two t r iangles ar e siad t o be similar t o each-ot her , if (i ) t heir cor r esponding angles ar e equal ; and (ii ) t heir cor r esponding sides ar e pr opor t ional. Charact er ist ic Pr oper t y of Similar it y (i ) A– A– A– Si mi l ar i t y : I f t wo t r i angl es ar e equiangular , t hen t he t r iangles ar e similar . (ii ) S– S– S– Similar it y : if cor r esponding sides of t wo t r iangles ar e pr opor t ional, t hen t hey ar e similar . (iii ) S– A– S – Similar it y : I f in t wo t r iangles, one pair of cor r esponding sides ar e pr opor t ional and t he included angles ar e equa, t hen t he t wo t r iangles ar e similar . Result s on Similar T riangle (i ) I f a l ine is dr awn par allel t o one side of a tr iangle inter secting the other two sides, t hen it divides t hese sides in t he same r at io. I t s conver se is also t r ue. (ii ) I nt er nal bi sect or of an angle of a t r iangl e divides the opposite side inter nally in the ratio of t he sides cont aining t he angle. (iii ) Rat io of t he ar eas of t wo similar t r iangles is equal t o t he r at io of t he squar es of any t wo cor r esponding sides. (iv) I f ar eas of t wo similar t r iangles ar e equal, t hen t r iangles ar e congr uent , i.e. equal and similar t r iangles ar e congr uent . Pyt hagor as’ T heor em. I n a r ight -angled tr iangle, squar e of t he hypot enuse is equal t o sum of t he squar es of t he ot her t wo sides. B

A

C

For a given r ight angled t r iangle ABC, wher e  B = 90 2 AB + BC2 = AC2

Geometry

14.5

SOLVED EXAM PLES 1. I n two similar tr iangles  ABC and DEF, DE = 3cm, EF = 5 cm, DF = 4 cm and BC = 20 cm, then find length of AB. A

 

D 3

4

E

I f s be t he side of the maximum squar e, t hen 1 1 s2  ( x  s) s  ( x  s) s = 800 2 2 s2 = x s – s2 = 800 x s = 800

F B

5

C

20

H ence, lengt h of t he diagonal of t he squar e

Solution : Fr om similar t r iangles ABC and DEF AB DE = BC EF AB 20



=

AB =



3 5 3  20  12 cm. 5

2. I f a t r aingle and a r ect angle have equal ar eas and equal al t i t u de, fi nd r el at i on bet ween base of r ect angle and base of t r iangle. Solution : L et h be t e common height of t r iangle and r ect angle. I f a and b be r espect ively t he bases of t r iangle and r ect angle, t hen 1 ah = b  h By hypot hesis, 2 1 a =b  2  a = 2b 3. Area of an isosceles right-angled tr iangle is 800 sq. metr es. The greatest possible squar e has been cut out from it. Find length of the diagonal of this square. Solution : L et x be t he base and side of isosceles ABC. By hypot hesis, 1 . x x = 800 2  x2 = 1600  x = 40 m

= 20 2 m 4. Per imeter of a t r iangle is 100 m and its sides ar e in t he r at io 1 : 2 : 2. Find ar ea of t he tr iangle (in m 2). Solution : L et t he sides be x , 2x and 2x . Then x + 2x + 2x = 100  x = 20 H ence sides ar e 2, 40 and 40 and s = 50 ( 2s = a + b + c) Ar ea of t he t r iangle

=

50(50  20) (50  40) (50  40) 2 50  30  10  10 = 100 15 cm

AB 2  BD 2

Solution : AD =

= x

H ypot enuse,

y

BC =

x

3 2

x h

x 2

D x

Ar ea of ABC =

B

2 x  40 2 m

x2 4

A

= s

x2 

=

B

90



s(s  a) (s  b) (s  c)

5. Find ar ea of an equilat er al t r iangle of side x.



A

=

=

C

x

800  20 m 40

s =



=

C

1 . BC. AD 2

1 x 3 . x. 2 2 3 2 x 4

14.6

Geometry

6. I n t he gi ven fi gur e,  ABC i s an equi l at er al t r i angl e. O i s t he poi nt of i nt er sect i on of t he medians. I f AB = 6 cm. Find OB.

8.  PQR and L M N ar e similar. I f 3 PQ = L M and M N = 9 cm, find QR. L

P

A

M

Q

R M

N

Solution : Fr om similar Ds PQR and L M N O B

C

Solution : Gi ven

AB = 6 cm, AM = 3 cm BM =



62  32

= 3 3 cm and OB =

2  BM  2 3 cm 3

7. An isosceles right triangle has an area of 200 sq. cm. Find area of a square dr awn on its hypotenuse. Solut ion : L et AB = BC = a  Ar ea of  ABC A

a a

2

90 B

=



C

a

1 2 a  200 2

a2 = 400 Ar ea of squar e on AC 2 a 2 a

=

= 2. a2 = 800 sq.cm.

LM MN = PQ QR 9 3PQ  = QR PQ 9 3  = QR 1 9  QR =  3 cm 3 QU AD RI L AT E RAL A plane figur e enclosed by four line segment is called a quadrilateral. PARAL L E L OGRAM A quadr ilat er al having opposite sides par allel is called a parallelogram. I n a par allelogram (i ) opposit e sides ar e equal. (ii ) opposit e angles ar e equal. (iii ) t he t wo diagonals of a par allelogr am bisect each-ot her . RECT AN GL E A par allelogr am each of whose angles is 90, is called a r eact angle. Diagonals of a r ect angle ar e equal, and it s conver se is also t r ue. SQU ARE A r ect angle having all sides equal is called a squar e. Diagonals of a squar e and bisect at r ight angle. RH OM BU S a par allelogr am having all sides equal but diagonals ar e not equal is called rhombus. Diagonals of a rhombus bisect at r ight angle

SOLVED EXAM PLES 1. Find side of a r hombus whose diagnals ar e 16 cm and 12 cm r espect ively. Solution : Given : d1 = 16 cm, 6 cm d2 = 12 cm



1 d12  d22 2 1 = 16 2  122 2 1 1 = 400 =  20 2 2

Side of r hombus =

 10 cm

2. Find number of sides of a r egular polygon each of whose angles measures 156 Solut i on : Sum of t he i nt er i or angl es of an n -sided r egular polygon = (2n – 4)  90



(2n – 4)  90 = 156  n



180n – 360 = 156 n

 

180n – 156n = 360

n =

360  15 24

Geometry

3. O is t he point of int er section of the diagonals AC and BD of a r hombus ABCD, P, Q, R ar e point s on OC, OB and OA r espectively such t hat OP = 1 unit , OQ = 2 unit s and OR = 4 unit s. Find angle PQR . D

D

C M 0 27

C

P

A

B

1

=

0 1

R

Q

A

B

Solution : Fr om  OPQ, PQ2 = 12 + 22 =5 Fr om  OQR, OR2 = 22 + 42 = 20  PQ2 + QR2 = 5 + 20 = 25 = PR2  PQR = 90 4. The ar ea of a r hombus is 120 cm 2. I f one of it s diagonals is of lengt h 10 cm, find lengt h of one of it s sides. D

C

1 1  70  27   70  27 2 2

= 1890 cm 2 6. The ar ea of a t r apezium is 275 cm 2. I f it s par allel isdes ar e in t he r at io 2 : 3 and t he per pendicular dist ance bet ween t hem is 5 cm, find smaller of t he par allel sides. Solution : L et t he par allel sides be 2x and 3x , t hen ar ea of t r apezium = 275 cm 2



1 (2 x  3 x)  5 = 275 2



x=

2  275  22 cm 25

Smaller of t he par allel sides = 2x = 44 cm 7. I f each angle of a polygon is 165, find number of sides of t he polygon. Solution : Ext er ior angle = 180 – 165 = 15

0 5

Number of sides =



12 A

B

Solution : Ar ea of a r hombus = diagonals.

1 pr oduct of it s 2

1  10  2nd diagonal 2  2nd diagonal = 24 cm Fr om t he figur e, 

=

122  52

= 169  13 5. One diagonal of a par al lel ogr am is 70 cm and per pendicular diatance of this diagonal fr om either of t he out l yi ng ver t i ces i s 27cm. Fi nd ar ea of par allelogr am (in cm 2). Solution : Given : AC = 70 cm and BM = 27 cm  Ar ea of par allelogr am ABCD = Ar ea opf ABC + Ar ea of  ACD

360 Ext erior angle 360  24 15

8. ABCD i s t r apezi um wi t h AB par al l el t o DC. I f AB = 10 cm, AD = BC = 4 cm and DAB = CBA = 60, find lengt h of CD.

120 =

AB =

14.7

C

D 4

4

60 A

60 P

Q

Solution: Fr om D ADP,

AP = cos 60 AD

  

x

x 1  4 2 x =2 CD = AB – 2x = 10 – 4 = 6 x =

B

14.8

Geometry

LOCU S The pat h t r aced out by a moving point under some geometrical conditions is called its locus. The plur al fr om of locus is called loci . e.g. A Acir cle wit h r adius or is t he locus of a point which moves is such a way t hat it s dist ance fr om t he fixed point O is always equal t o r .

A

r

Properties : 1. The locus of a point , equaidist ant fr om t he given fixed point s, is t he per pendicular bisect or of t he line segment joining t he given point s. 2. T h e l ocu s of a poi n t equ i di st an t f r om t w o int er sect ing lines is t he pair of lines beisect ing t he angles for med by t he given lines. CON CU RREN T LI N ES I f t hr ee or mor e lines pass t hr ough t he same point t hen t hey ar e said t o be concur r ent and t he common point is called point of concur r ency of t he given lines. T heor ems. 1. T h e an gl e bi sect or s of a gi ven t r i an gl e ar e concur r ent . 2. The per pendi cul ar bi sect or s of t he si des of a t r iangle ar e concur r ent . 3. The alt it udes of a t r iangle ar e concur r ent . CI RCL E S A cir cle is a set of all t hose point s in plane t hat ar e at a given const ant dist ance fr om a given fixed point in t he plane.

r

B P

chord

B

When a chor d passes t hr ough t he cent r e of t he cir cle, it is called diamet er of t he cir cle. Segment of circle : The chor d AB divide t he cir cle in t wo par t s. Each par t of t he cir cle is called segment of t he cir cle. The segment cont aining t he minor ar c is cal l ed m on or segm en t w h er eas t h e segm en t cont aining t he major ar c is called major segment of t he cir cle. Congruence of circle : Two ci r cl es ar e sai d t o be congr uent if eit her of t hem can be supper posed on t he ot her so as t o cover it exact ly. This is possible only when t he r adii of t he t wo cir cles ar e equal. M ajor segment

M inor segment

Sector of a circle : A sect or is t hat r egion of a cir cular disc C (o, r ) which lies bet ween as ar c and t he t wo r adii joining t he ext r emit ies of t he ar c and t he cent r e. Central angle : L et (O, r ) be any cir cle. Then, any angle wit h vr t ex ‘O’ is called cent r al angle of a cir cle. Tangent to a circle : A line t hat int er sect s t he cir cle in exact ly one point is called a t angent t o cir cle.

A

The fixed point is called centr e and t he given constant dist ance is called r adius of t he cir cle. H er e OA = r Concentric circle : Cir cles having t he same cent r e but wit h differ ent r adii ar e called concent r ic cir cle.

A

Arc of a circle : A cont inuous piece of a cir cle is called an ar c of a cir cle. H er e, ar c AB = l = ar c Chord of a circle : A line segment joining t wo point s on a cir cle is called chor d of t he cir cle.

Sect or

A

B

The point wher e t he t angent int er sect s t he cir cle is called point of cont act . Secant : A line which intersects circle in two distinct points is called secant of the circles. N ormal : A line per pendi cular t o a t angent at t he point of cont act is called nor mal t o a cir cle. Cyclic quadr ilat er al. I f all t he four ver t ices of a quadr ilat er al lie on a cir cle, such a quadr ilat r al is called a cyclic quadr ilat er al. I f four points lie on a cir cle, t hey ar e said to be concylic.

Geometry

Alternate segment : Segment opposit e t o t he angle for med by t he chor d of a cir cle wit h t he t angent at a point is called alt er nat e segment for t hat angle.

B

A

C P

A

T

I n t he given figur e, PAT is a t angent t o t he cir cle and AB i s a chor d, whi ch di vi des t he ci r cl e i nt o t wo segment s namely ACB and BDA, called t he alt er nat e segments. Fr o  BAT, t he alt er nat e segment is BPA. Fr o  BAP, t he alt er nat e segment is ACB. Theorem : 1. I f t wo ar cs of a cir cle ar e congr uent , t hen t heir cor r esponding chor ds ar e equal. 2. I f t w o ch or ds of a ci r cl e ar e equ al , t h en cor r esponding ar cs ar e congr uent . 3. The per pendicular fr om t he cent r e of a cir cle on a chor d bisect t he chor d. I t s conver se is also t r ue. 4. Ther e is one and only one cir cle passng t hr ough t hr ee non-collinear point s.

14.9

5. Equal chor ds of congr uent cir cles ar e equidist ant fr om t he cor r esponding cent r es. I t s conver se is also t r ue. 6. The angle subt ended by an ar c of a cir cle is t wice the angle subtended by it at any point of the alternate segment of t he cir cle wit h r espect t o t he ar c. 7. Any t wo angles in t he same segment of a cir cle ar e equal. 8. An angl e i n a semi ci r cl e i s a r i ght angl e. I t s conver se is also t r ue. 9. The sum of either pair of opposite angles of a cyclic quadr ilat er al is 180. TAN GEN TS TO A CI RCLE T heor em. 1. A t angent at any point of a cir cle is per pendicular t o t he r adius t he point of cont act . I t s conver se is also t r ue. 2. The length of two tangents dr awn fr om an exter nal point t o a cir cle ar e equal. 3. I t t wo chor ds of a cir cle int er sect inside or out side t he cir cle when pr oduced, t he r ect angle for med by t he t wo segment s of one chor d is equal in ar ea t o t he r ect angle for med by t he t wo segment s of anot her chor d. 4. I t t wo ci r cl es t ouch each ot her i nt er nal l y or ext er nally, t he point of cont act lies on t he l ine t hr ough t he cent r e.

SOLVED EXAM PLES 1. I f O is t he cent r e of a cir cle of r adius R and OA and OB ar e 2 r adii of t he cir cle, such t hat t he angle AOB is  is in r adians, find ar ea of the bigger segment of t he cir cle cm off by t he chor d AB. Solution : Ar ea of smaller segment OACB

FG H

IJ K

FG H

=

R 2      R sin  R cos   2 2 2 2 2

=

R 2    R 2 cos sin 2 2 2 2

R  R  sin  2 2 Ar ea of bigger segment =

ADB = R

=

= 1  108   7 2 2 180

1 108 22   77 2 180 7 = 46.2 cm 2 =

Solution : Requir ed ar ea

= Area of square – 4  ar ea of one sect or

LM R   R sin OP MN 2 2 PQ 2

Solution : Requir ed ar ea of t he sect or

3. Find ar ea of the shaded por tion in the given figur e, wher e t he ar cs ar e quadr ant s of a cir cle.

2

2

IJ K

2. I n a cir cle of r adius 7 cm, ar e subt ends an angle of 180° at t he cent r e° Find ar ea of t he sect or.

2

R2 2    sin  2

1   72 4 1 22 = 196  4   77 4 7 = 196 – 154 = 42 m 2 = 14  14  4 

14.10

Geometry

4. I n t he given figur e, PQ is t angent A. BC is t he diamet er. I f ABC = 42, find  PAB.

A

P

Q

Fr om t he given figur e, x2 + x2 = r 2  2x2 = r 2  side of t he squar e = 2x

= 2

42°

B

O

...(ii )

r2  2 r  10 2 cm. 2

7. Find ar ea of the shaded por tion in the given figur e

22   x  7  .   Sol ut ion : AP i s t angent t o t he ci r cl e at A. Ther efor e OA and AP ar e pr ependicular t o each ot her.  OAP = 90 Again OA = OB   OAB =  OBA = 42  PAB = PAO – BAO = 90– 42 = 48 5. I f lengt h of t he t angent fr om t he or igin t o t he cir cle x 2 + y 2 – 26x + K = 0 is 5, find K . Solution : Given cir cle is, (x – 13)2 + y 2 = 169 – K  Cent r e C is (13, 0) and

A

B m 7c

7c m

C

D

Solution : Fr om t he figur e, it is clear t hat AB = 14 cm and ABCD is a squar e. Also 2 semi-cir cles make a cir cle of r adius 7 cm.  Ar ea of shaded por t i on = Ar ea of squar e ABCD – Ar ea of cir cle of r adius 7 cm. = 14  14    7 2

T = 14  14 

C (13,0)

O

radius = Given :

= 196  154  43 cm 2 8. A cir cle A has a r adius of 3 cm, two cir cles B and C have a r adius each equal t o t he diamet er of cir cle A. Find r adius of a cir cle D which has an ar ea equal t o t he t ot al ar ea of A, B and C.

169  K

OT = 5, CT =

169  K

Fr om OCT, OC = OT + CT  132 = 52 + 169 – K 6. I f ar ea of t he given cir cle is 100  squar e cm, find side of t he squar e inscr ibed in t he cir cle. Solution : I f r is t he r adius of t he ci r cle, t hen 2

2

22  72 7

2

ar ea of t he cir cle =  r 2  100

Solution : Given : r adius of cir cle A = 3 cm, t hus diamet er of A = 6 cm Also, r adius of cir cle B = r adius of cir cle C = 6 cm. I f r is t he r adius of cir cle D, t hen

 32   . 6 2   . 6 2

= r2

r 2 = 81 r = 9 cm.

 

9. I f st r aight line y = x + C is a t angent t o t he cir cle x 2 + y 2 = 1, find C. Solut ion : L engt h of per pendi cul ar fr om t he cent r e (0, 0) on x – y + c = 0 must be equal t o t he r adius of t he cir cle.

r x

 

r 2 = 100 r = 10

 ...(i )



00C 1 1

=±1

C =  2

Geometry 14.11

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. The cir cumcentr e of a tr iangle is always the point of int er sect ion of t he (a) medians (b) per pendicular s bisect or s (c) alt it ude bisect or (d) per pendicular s dr opped fr om t he ver t ices on t he opposit e sides of t he t r iangle 2. The Qut ab M inar cast s a shadow 150 m long at t he same t i me when t he Vi kas M i nar cast s a shadow of 120 m long on the gr ound. I f the height of t he Vikas M inar is 80 m, t hen t he height of t he Qut ab M inar is (a) 180 m (b) 100 m (c) 150 m (d) 120 m 3. The number of t angent s t hat can be dr awn t o t wo non-int er sect ing cir cles is (a) 4 (b) 3 (c) 2 (d) 1 4. An angle is equal t o 1/3 r d of it s supplement . Find it s measur e (a) 60 (b) 80 (c) 90 (d) 45 5. A man goes 10 m due east and t hen 24 m due nor t h. Find t he dist ance fr om t he st ar t ing point . (a) 26 m (b) 24 m (c) 28 m (d) 30 m 6. The sum of t he int er ior angles of a polygon is 1620. The number of sides of t he polygon ar e (a) 9 (b) 11 (c) 15 (d) 12 7. H ow many sides a r egular polygon has wit h it s int er ior angle eight t imes it s ext er ior angle ? (a) 16 (b) 24 (c) 18 (d) 20

5 is t he measur e of each int er ior angle of a 6 r egular convex polygon, t hen it must be a (a) octagon (b) hexagon (c) dodecagon (d) pentagon 9. I n a qu adr i l at er al A B CD ,  B = 90  an d AD 2 = AB 2 + BC2 + CD 2, t hen ACD is equal t o (a) 90 (b) 60 (c) 30 (d) none of t hese 10. I n a t r i an gl e A B C,  A = x  ,  B = y  an d C = (y + 20). I f 4x – y = 10, t hen t he t r iangle is (a) Right-angled (b) Obtuse-angled (c) Equilateral (d) None of t hese 11. The per imet er s of t wo similar t r iangle ABC and PQR ar e 36 cm an d 24 cm r espect i vel y. I f PQ = 10 cm, t hen t he lengt h of AB is (a) 16 m (b) 12 m (c) 14 m (d) 15 m 8. I f

12. I n t he following figur e, find ADC.

(a) 55 (b) 27.5 (c) 60 (d) 30 13. Two isosceles tr iangles have equal ver tical angles and t heir ar eas ar e in t he r at io 9 : 16. The r at io of t heir cor r esponding height s is (a) 3 : 4 (b) 4 : 3 (c) 2 : 1 (d) 1 : 2 14. A 25 m long ladder is placed against a ver t ical wall inside a r oom such that the foot of t heladder is 7 m fr om t he foot of t he wall. I f t he t op of t he ladder slides 4 m downwor ds, t hen t he foot of t he ladder will slide by (a) 2 m (b) 4 m (c) 8 m (d) 16 m 15. The ar ea of a fi eld i n t he shape of a t r apezi um measur es 1440 m. The per pendi cular dist ance bet ween i t s par all el si des i s 24 m. I f t he r at i o of t he par al lel sides is 5 : 3, t he lengt h of t he longer par allel side is (a) 45 m (b) 60 m (c) 75 m (d) 120 m

LEVEL-1 1. Suppose it is 3 o’clock. After 20 minutes the angle between the smaller and bigger hands will be (a) 20 (b) 30 (c) 110 (d) 120 2. The r adius of t he cir cumcir cle of an equilat er al t r iangle of side 12 cm is (a)

4 3 cm 3

(b) 4 3 cm

4 2 cm 3 3. I f t he per imet er of an isosceles r ight t r iangle is (c) 4 2 cm

(d)

e6  3 2 j m, t hen t he ar ea of t he t r iangle is (a) 4.5 m 2 (b) 5.4 m 2 (c) 9 m 2 (d) 81 m 2 4. I f t wo diamet er s of a cir cle int er sect each ot her at r ight angles, t hen the quadr ilat er al for med by joining t heir end point s is a (a) Rhombus (b) Rectangle (c) Squar e (d) Par allelogr am 5. Of all t he chor ds of a cir cle passing t hr ough a given point in it , t he smallest is t hat which (a) is t r isect ed at t he point (b) is bisect ed at t he point (c) passes t hr ough t he cent r e (d) none of t hese

14.12

Geometry

6. I n a t r i angl e ABC, t he l engt hs of t he si des AB, AC and BC ar e 3, 5 and 6 cm r espect ively. I f a point D on BC is dr awn such t hat t he line AD bisect s t he angle A int er nally, t hen what is t he lengt h of BD ? (a) 2 cm (b) 2.25 cm (c) 2.5 cm (d) 3 cm 7. Wit h t he ver t ices of a ABC as cent r es, t hr ee cir cles ar e descr ibed each t ouching t he other t wo ext er nally. I f t he sides of t he t r iangle ar e 4, 6 and 8 cm r espect ively, t hen t he sum of t he r adii of t he t hr ee cir cles equals (a) 10 (b) 14 (c) 12 (d) 9 8. I n t he figur e given below, O is t he cent r e of t he cir cle. I f OBC = 37, t he BAC is equal t o A

12. I f P and Q ar e t he mid point s of t he sides CA and CB r espect ively of a t r iangle ABC, r ight -angled at C. Then t he value of 4 (AQ2 + BP2) is equal t o (a) 4 BC2 (b) 5 AB 2 (c) 2 AC2 (d) 2 BC2 13. I f one of t he diagonals of a r hombus is equal t o it s side, t hen diagonals of t he r hombus ar e in t he r at io (a) 3 : 1 (b) 2 : 1 (c) 3 : 1 (d) 2 : 1 14. I f t he sides of a r ight t r ingle ar e x , x + 1 and x – 1, t hen t he hypot enuse is (a) 5 (b) 4 (c) 1 (d) 0 15. ABCD is a squar e, F is mid point of AB and E is a point on BC such t hat BE is one-t hir d of BC. I f ar ea of FBE = 108 m 2, then the length of AC is (a) 63 m

O B

C

(a) 74 (b) 106 (c) 53 (d) 37 9. I f, in t he following figur e, PA = 8 cm, PD = 4 cm, CD = 3 cm, t hen AB is equal t o

(c) 63

2 m

(b)

3 2

2 (d) none of t hese 3 11. The t wo sides of a r ight t r iangle cont aining t he r ight angle measur e 3 cm and 4 cm. The r adius of t he incir cle of t he t r iangle is (a) 3.5 cm (b) 1.75 cm (c) 1 cm (d) 0.875 cm (c)

(b) 3 2 m

(c) 2 3 cm (d) 4 2 m 2. The pr oduct of t he lengt hs of t hr ee si des of a triangle is 196 cm and the radius of its circumcircle is 2.5 cm. The ar ea of the tr iangle is (a) 39.2 cm 2 (b) 19.6 cm 2 2 (c) 122.5 cm (d) 61.25 cm 2 3. I n t he figur e below, if t he per imet er of ABC is p, t hen t he per imet er of t he r egular hexagon is 3p B (a) 2

2p 3 C A 3p (c) 2 2p (d) 3 4. The ar ea of a r hombus is 2016 sq cm and it s side is 65 cm. The lengt hs of t he diagonals (in cm) r espect ively ar e (a) 125, 35 (b) 126, 32 (c) 132, 26 (d) 135, 25 (b)

(a) 5

2 m (d) 72 2 m LEVEL-2

1. Two cir cles touch each other inter nally. Their radii are 2 cm and 3 cm. The biggest chor d of the outer cir cle which is outside the inner cir cle is of length (a) 2 2 cm

(a) 3.0 cm (b) 3.5 cm (c) 4.0 cm (d) 4.5 cm 10. Two cir cles of unit r adius t ouch each ot her and each of t hem t ouches int er nally a cir cle of r adius t wo, as shown in t he following figur e. The r adius of t he cir cle which t ouches all t he t hr ee cir cles is

(b) 36

Geometry 14.13

5. Two cir cles with r adii ‘a’ and ‘b’ r espectively touch each ot her ext er nally. L et ‘c’ be t he r adius of a cir cle t hat t ouches t hese t wo cir cles as well as a common t angent t o t he t wo cir cles. Then 1 1 1 1 1 1 –  –  (a) (b) a b c b a c 1 1 1   (c) (d) None of t hese a b c 6. A r hombus OABC is dr awn inside a cir cle whose cent r e is at O in such a way t hat t he ver t ices A, B and C of t he r hombus ar e on t he cir cle. I f t he ar ea of t he r hombus is 32 3 m 2, t hen r adius of t he cir cle is (a) 64 m (b) 8 m (c) 32 m (d) 46 m 7. The sides of a t r iangle ar e 6 cm, 11 cm and 15 cm. The r adius of it s incir cle is (a)

5 2 cm 4

8. I n t h e gi v en d i agr a m , t w o ci r cl es p ass t h r ou gh each ot h er ’s cen t r e. I f t h e r adi u s of each ci r cl e i s 2, t h en wh at i s t h e per i m et er of t h e r egi on m ar k ed B ?

(a) (b)

FG 8 IJ  H 3K FG 4 IJ  H 5K

(c) 4 (d)

FG 5 IJ  H 3K

(b) 3 2 cm

(c) 6 2 cm

(d)

4 2 cm 5

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (b)

2. (b)

3. (a)

4. (d)

5. (a)

11. (d)

12. (b)

13. (a)

14. (c)

15. (c)

6. (b)

7. (c)

8. (c)

9. (a)

10. (a)

7. (d)

8. (c)

9. (d)

10. (c)

7. (a)

8. (a)

LEVEL-1 1. (a)

2. (b)

3. (a)

4. (c)

5. (d)

11. (c)

12. (b)

13. (a)

14. (a)

15. (b)

6. (b)

LEVEL-2 1. (d)

2. (b)

3. (d)

4. (b)

5. (c)

6. (b)

EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S 1. Ci r cu m cen t r e of a t r i an gl e i s t h e poi n t of intersection of per pendicular bisector s of its sides. 2. L et height of Qut ab M inar be x met r es 150 x  = 120 80 150  80  x= = 100 120

4. L et t he angle be x .

 H ence

1 × (180 – x ) ; 3 x = 45

x=

5.

26

24

T1

3.

10

T2 T3 T4

H ence 4 t angent s can be dr awn.

6. Sum of t he int er ior angles of a polygon of n sides  = (2n – 4)  2

14.14

Geometry

  = 1620  2 180  n– 2=9  n = 11 7. L et n be t he number of sides of t he polygon.  I nt er ior angle = 8  Ext er ior angle 

(2n – 4) 

 2 = 8  2 n n n – 2 = 16 n = 18

Also,

55 = 27.5 2 13. Given t r iangles ar e congr uent . 

(2n – 4)  

  5 8. = 150. 6 Ext er ior angle = 30 

Number of sides =

9.

360 = 12. 30

A

CAD = ADC =



h12 Area of I st t riangle = 2 , h2 Area of I I nd t riangle



h12 9 = 2 16 h2

h1 3 = h2 4 14. AB = 25  BC = 24  B C = 20 A B  = 25  A C = 15  A A = 8 m 15. Ar ea of t r apezium 

360 Ext er ior angle

=

AC = CD

D

B B

A

A

7

C

1 (sum of par allel sides) × height 2 1 = (5x + 3x ) × 24 = 1440. 2 Solving, we get x = 15 and lengt h of longer side = 5 × 15 = 75. =

90 C

B

AD2 = AB 2 + BC2 + CD 2 = AC2 + CD 2  ACD = 90 10. x + y + (y + 20) = 180  x + 2y = 160 (i ) and 4x – y = 10 (ii ) Solving (i ) and (ii ) we get  y = 70, x = 20 Hence the angles of the triangle ar e 20, 70, 90 Thus, t he t r iangle is r ight angled. 11.

P

A 10

B

 

12.

 

C Q

R

AB 36 = PQ 24 3 AB = 2 3 PQ =  10 = 15 2 AB = BC 110 BAC = BCA = = 55 2 ACD = 125

LEVEL-2 1.

Degr ees at 3 : 20

1 (5 min) dist ance 3 = 30 – 10 = 20 degr ees. 2. Ci r cu m cen t r e of a t r i an gl e i s t h e poi n t of int er sect ion of t he r ight bisect or s of it s sides. P QM  sin 60 = 60° QC C 6 3 60°  = Q R QC 6 M 6 2 62  QC = = 4 3 cm 3 = 5 min dist ance –

3.

Per imet er = (6 + 3 2 ) m i .e.

 

x+x+

2x = 6+3 2

2x +

2x = 6+3 2

x=

63 2

2 2 =3

m 2x

xm



3 2+ 2 

xm

2+ 2



Geometry 14.15

CC1 = 1, OC1 = r + 1 OC = AC – AO = CD – AO, [AC and CD ar e r adius of t he bigger cir cle]  OC = 2 – r  OC12 = CC12 + OC2  (r + 1)2 = 12 + (2 – r )2  r 2 + 2r + 1 = 1 + 4 – 4r + r 2  6r = 4

 Requir ed ar ea of t r iangle 1 1 =  x  x = x2 2 2 1 2 =  (3) = 4.5 m 2 2 6.

A

3

5

r =

 B

C

6

D

Divide BC = 6 in t he r at io 3 : 5 3 9 Thus BD = 6 = = 2.25 cm 35 4



7.

r1

A

BAC =

5 r3 r3

r2

C

10.

A r r

D

1

 AC – AB  CD – BD I n our case, BC  BD + DC  CD  BD

= = = = = = =

C

4

2 2 4 4 and – BD + DC = 2 3 1 = OE Radius of t he incir cle

12.

A

r

1

1 c2 C

Q

C

AQ2 = AC2 + QC2 BP2 = BC2 + CP2 2 AQ + BP2 = (AC2 + BC2) + (QC2 + CP2) = AB 2 + PQ2 1 O F1 I L = AB G ABJ , M PQ  AB P 2 H2 K N Q 2

5 AB2 4  4(AQ2 + BP2) = 5AB 2 13. Ar ea of a r hombus 1 = (pr oduct of t heir diagnonals) 2 =

1 1

90

2

O r

c1

P

B

AB = AP – BP = 8 – 3.5 = 4.5 cm.

1

D

1 1 BOC = 106 = 53 2 2

28 PB = = 3.5 cm 8



O

B

9. Fr om t he given figur e, PC  PD = PA  PB  (4 + 3)  4 = 8  PB



A

3E

AB = r 1 + r 2 = 4 BC = r 2 + r 3 = 6 CA = r 3 + r 1 = 8  2(r 1 + r 2 + r 3) = 4 + 6 +8  r1 + r2 + r3 = 9 8. OBC = 37 OB = OC = r adius  OCB = OBC = 37  BOC = 180 – (37 + 37) = 106 N ow,

11. I f incir cle of a tr iangle ABC touches BC at D,then BD – CD  = AB – AC I n our case, AC = 5, AB = 3

r1

r2 B

4 2 = 6 3

B

14.16

Geometry

1 xy ...(i ) 2 Again, Ar ea of a r hombus = 2  Ar ea of equilat er al t r iangle ABD

= (36)2 + (36)2 = 2 (36)2

=



AC = 36 2 m

LEVEL-1

A x

1.

x

A 3

y B

2 2

D

x

C x

O O 2

1

x

2 2 C

= 2

B

3 2 x 4

3 2 x 2 Fr om (i ) and (ii ), we have

...(ii )

=

1 xy = 2

3 2 x 2

abc 196 = 4R 4  2.5 = 19.6 cm 2 3. L et a be t he side of t he r egular hexagon L et ED  AC =

y 3 = x 1



AB is t he chor d of t he out er cir cle t ouching t he inner cir cle. 2. I f a, b, c ar e t hr ee sides of a t r iangle and R is r adius of it s cir cumcir cle, t hen  = Ar ea of t he t r iangle

 y:x = 3 :1 14. L et (x + 1) be t he hypot enuse

C E

x+

30° D

1

x–1

x

 (x +1) = x 2 + (x – 1)2 2  x + 2x + 1 = x 2 + x 2 – 2x + 1  x 2 – 4x = 0  x (x – 4) = 0  x = 4, as x  0  H ypot enuse = 4 + 1 = 5 2

15.

D

6k

60°

A

AD =

 Fr om ADE,

B

p = CD 6

AD = cos 30 AE



p 6 = a



6a =

C

3 2 2p 3

 Per imet er of r egular hexagon = 6k

E 2k

A

F

3k

B

L et t he side of t he squar e be 6K

   

1  3K  2K = 108 2 K =6 Side of t he squar e = 36m AC2 = CD 2 + DC2

4. Ar ea =



2p 3

1  (Pr oduct of t he diagonals) = 2016 2

1 (2x  2y ) = 2016 2 65 x 65

65

y x y

65

Geometry 14.17

xy = 1008 x + y 2 = 652 = 4225 (x – y )2 = x 2 + y 2 – 2xy = 4225 – 2016 = 2209  x – y = 47 Now (x + y )2 = (x – y )2 + 2xy = 6241  x + y = 79 Solving (i ) and (ii ) we get  x = 63, y = 16 Thus, diagonals ar e 126 and 32.  Also 

5.

2

PR = M C =

AC  AM

90 P

a

A

b



...(ii )



1  Ar ea of t he r hombus OABC 4

1 1 xy =  32 3 , 2 4

B

QR = 2 bc

N ow,

PQ = PR + RQ = 2 ac + 2 bc

PQ = 2 ab

...(i )

...(ii )

15

=

2 ac + 2 bc = 2 ab

=

Dividing it by 2 abc , we get +

b 6.

1

a

A

=

1

c

B x

y

y D O

11

6

Fr om (i ) and (ii ), we have

1

 s

wher e  = ar ea s = semi-per imet er of t he t r iangle ABC. H er e, 2s = 6 + 11 + 15 = 32  s = 16

Dr aw PN par allel t o AB  PN = AB = a + b, QN = BQ – BN = b – a  PQ2 = PN 2 – QN 2 = (a + b)2 – (a – b)2 = 4ab



...(ii )

y = 4, x = 4 3  Radius of t he cir cle = 2y = 8 7. I ncent r e of a t r iangle is t he point of int er sect ion of t he bisect or s of it s angles. I f r is r adius of t he incir cle, t hen r =

Similar ly

r =

 8.

16(16 – 6)(16 – 11)(16 – 15) 16  10  5  1 = 20 2

20 2 5 2 = cm 16 4

PQ = 2 (Radius of t he cir cle) PX = PY = QX = QY = 2 (Being r adii of cir cles)

x

X C

P

L et AC = 2x , OB = 2y  Radius = OC = 2y = OB Fr om  ODC, OC2 = OD 2 + CD 2

[Given]

 xy = 16 3 Fr om (i ) and (ii ), we have

2

N b

C

aM

...(i )

3y

Also ar ea of t he ODC

...(i )

90 Q

R

x=



(a  c) 2  (a  c) 2 = 2 ac

=

4y2 = y 2 + x 2 x2 = 3y2

 

2

 O

Q

B A

C Y

14.18

Geometry

H ence PYQX is a r hombus.  Thus PO = 1 = OQ [  Diagonal of a r hombus bisect each ot her ] cos  =

 

OP 1 = PX 2

 = 60 XPY = 120 Ar c XQY = =

Cir cumfer ence  120 360 2r 4 2  2 = = 3 3 3

Similar ly, Ar c XPY =

4 3

 Requir ed per imet er =

8 3



15

Mensuration

CHAPTER

TWO DI M EN SI ON Rectangle and Squar es Per imet er of a r ect angle = 2 (lengt h + br eadt h) Per imet er of a squar e = 4  lengt h Ar ea of a r ect angle = (lengt h  br eadt h) 2 Ar ea of a squar e= l (As in square, length = breadth) N ote : Ar ea is expr essed in squar e unit s M easurements of area in the M etric System. 100 squar e millimet r es = 1 squar e cent imet r e 100 squar e cent imet r es = 1 squar e decimet r e 100 squar e decimet r es = 1 squar e met r e 100 squar e met r es = 1 squar e decamet r e 100 squar e decamet r es = 1 squar e hect amet r e 100 squar e hect amet r es = 1 squar e kilomet r e N ote : One squar e decamet er which is equal t o 100 squar e met er s is called an acre. One squar e hectameter which is equal to 10000 squar e met er s is called a hect ar e. ‘5 square meters’ is quit e a differ ent t hing fr om ‘5 met ers square’. ‘5 squar e met er ’ r epr esent a ar ea which cont ains one squar e met er 5 t i mes wher eas ‘5 met er s squar e’ denot es t he ar ea of a squar e whose side is 5 met er s long and cont ains 5  5 or 25 squar e met er s. Diagonal of a Rectangle. A D

AC =

AB 2  BC2 B

Ar ea of a t r iangle =

b

gb

gb g

s s– a s– b s– c

wher e a, b, c ar e t he sides of a t r iangle while

s=

1 (a + b + c) or semi per imet er.. 2 A

B

D

C

A t r iangle whose all t he t hr ee sides ar e equal is called an equilat eral t riangle. I n an equilat er al t r iangle ABC, a = b = c.



s =



ar ea =

3a 2

3a 1 1 1 3 2  a a a = a 2 2 2 2 4

3 2 2 a 3 2 Area 4 a H eight = = = and 2 a Base PARAL L E L OGRAM Ar ea of a par allelogr am = Base  H eight L et CDEF be a r eact angle on t he same base DC and of t he same height FC. Then since par allelogr ams on t he same base and of t he same height ar e equal in ar ea.

C

Diagonal of a Square.

AC = AB 2 A

D

B

C

T RI AN GL E

1 1 Base  H eight =  BC  AD 2 2 wher e AD must be a per pendicular t o BC. Ar ea of a t r iangle=

Par alellogr am ABCD = Rect angle CDEF = CD  FC = Base  H eight H ence t he ar ea of a par al l el ogr am i s equal t o t he pr oduct of it s base and height .

15.2

Mensuration

RH OM BU S A r hombus is a par allelogr am all of whose sides ar e equal. I n a r hombus t he diagonals bisect each ot her at r ight angles. A B

= ADE + Rect . ABFE + BFC =

1 1 DE . AE + AE . EF + BF . FC 2 2 (BF = AE)

1 AE [DE + 2EF + FC] (EF = A(B) 2 1 = AE [DE + EF + FC + AB] 2 1 = AE [DC + AB] 2 1 = H eight  sum of par allel sides 2 H ence ar ea of a t r apenzium is equal t o t he pr oduct of half t he sum of par allel sides and height . REGU LAR POLYGON Regular polygon is a n sided enclosed figur e wher e all t he sides ar e equal. =

O

C

D

AB = BD = CD = DA, and AO = OC, BO = OD BOA = 1 r t . angle Ar ea of r hombus ABCD = 2 ar ea of BCD = 4 ar ea of BOA =4×

1 BO × OA 2

Ar ea of a hexagon =

BC AD 1  = BC × AD 2 2 2 H ence t he ar ea of a r hombus i s equal t o hal f t he pr oduct of it s diagonals. T RAPE ZI U M A t r apezium is four -sided figur e having a pair of opposit e sides par allel. Thus ABCD is a t r apezium in which AB DC Dr aw AE and BF per pendicular s fr om A and B t o DC Area of a trapezium ABCD =2×

D

A

B

E

F

C

3 3a 2 ; 2

Ar ea of oct agon = 2a2

e

j

2 1

Ar ea of Regular polygon of n sides having given lengt h of one side and r adius of cir cumscir bed cir cle

n 2 = a R – 2

FG a IJ H 2K

2

wher e R is r adius of cir cumscr ibed cir cle, a lengt h of one side, and n number of sides CI RCL E 2 Ar ea of a cir cle= r Cir cumfer ence of cir cle = 2r Si nce ar c i s t he par t of a ci r cumfer ence, t her efor e   2r lengt h of an ar c = 360 Since, segment is t he par t of ar ea, t her efor e   r 2 ar ea of a segment = 360

SOLVED E XAM PLES 1. Find t he diagonal of a r ect angle whose sides ar e 12 met er s and 5 met er s. Solution : L engt h of t he diagonal = =

12 2  5 2

169 = 13 met er s. 2. H ow many paving st ones, each measur ing 2½ met er s by 2 met er s, ar e r equi r ed t o pave a r ect angular cour t yar d 30 met er s long and 16½ met er s br oad ?

Solution : Ar ea of cour t yar d= 30  16½ sq. m = 495 sq. met er s Ar ea of each paving st one= 2½  2 sq. m Ar ea of each paving st one= 2½  2 sq. m  Number of st ones r equir ed= 495 5 = 99 3. A hall r oom, 39 m 10 cm long and 35 m 70 cm br oad, is to be paved with equal squar e tiles. Find t he l ar gest t i le which wil l exact ly fit and t he number r equir ed. Solution : 39 m 10 cm = 3910 cm; 35 m 70 cm = 3570 G.C.M . of 3910 and 3570 = 170

Mensuration

H ence, si de of t he l ar gest squar e t i l e i s 1 m 70 cm

 Number of t iles =

3910  3570 = 483 cm 170  170

4. Find t he widt h of a r oller which t r aver ses 128 km while cut t ing 6.4 hect ar es of gr ass. Solut ion : 128 km = 128000 m. 6.4 hect ar es = 6.4  10000 sq. m. I magining t he gr ass ar ea t o be 128000 m long, and as wide as t he r oller, we have Widt h r equir ed=

6.4  10000 1 m = m = 50 cm 128000 2

5. An oblong piece of gr ound measur es 19 met er s 2.5 dm by 12 met er s 5 dm. Fr om t he cent r e of each side a pat h 2 met er s wide goes acr oss t o t he centr e of the opposite side. Find the cost of paving these paths at the r ate of Rs 12.32 P per sq. meter. Solution : Area of the path AB = 19¼  2 sq. meters Ar ea of t he pat h CD = 12½  2 sq. met er s Ar ea of t he common por t ion = 2  2 sq. met er s C

H ence we must find t wo number s whose pr oduct is 540 and differ ence (37 – 30) or 7. Now, if t wo number s be a and b, t hen 2 2 (a + b) = (a – b) + 4ab = 49 + 4  540 = 2209  a + b = 47. But a– b =7 Solving, we get a = 27, b = 20



Requir ed width

=

119 sq. met er s 2

119  12.32 2 = ` 733.04 = ` 733.4 P 6. A mar gi nal wal k al l - r ound t he i nsi de of a r ect angular space 37 m by 30 m ocupies 570 sq. m. Find t he widt h of t he walk. Solution : Area of inner rectangle = (37  30 – 570) = 540 sq. m. A lit t le r eflect ion will show t hat t he differ ence bet ween si des of t he i nner r ect angl e must be equal t o di ffer ence bet ween si des of t he out er r ect angle. 

Cost = `

37 – 27 m =5m 2

O

A

B

Then AB =

D Area of be paved = (19¼  2 + 12½ v 2 - 2  2) sq. meters = (19¼ + 12½ – 2)  2

=

7. The per imet er of a r hombus is 146 cm and one of it s diagonals is 55 cm. Find t he ot her diagonal and t he ar ea of t he r hombus. Solution : L et ABCD be t he r hombus in which AC = 55 cm. D C

B

A

15.3

 BO =

146 55 = 36.5 cm and AO = = 27.5 cm 4 2

b36.5g – b27.5g 2

2

cm = 24 cm

H ence,ot her diagonal BD = 48 cm Ar ea of t he r hombus =

1 . AC . BD 2

1  55  48 sq. cm 2 = 1320 sq. cm. 8. Find ar ea of t he t r apezium whose par allel sides ar e 6 cm and 10 cm long and the distance between t hem is 4 cm 1 Solut ion : Area = height  (sum of parallel sides) 2 1 =  4  [6 + 10] sq. cm 2 = 32 sq. cm. 9. Find ar ea of t he a r egular hexagon whose side measur es 9 cm. =

Solut ion : Ar ea of a r egular hexagon = her e,



a = 9 cm

Area=

3 3a 2 2

3 3a 2 3 3  81 =  210.4 sq. cm. 2 2

15.4

Mensuration

10. Find t o t he near est met er, t he side of a r egular oct agonal enclosur e whose ar ea is 1 hect ar e. Solution : Area of a regular octagon = 2(1+ 2 ) a2



2(1+ 2 ) a2 = 1 hect ar e



a2 =

10000

e

2 1 2

j

= 2071 sq. m

Right ci r cul ar cylinder i s a sol i d descr i bed by t he r evolut ion of a r ect angle about one of it s sides or we can say a pr ism wit h t he infinit e sides on t he base. Volume of a cylinder = area of the base  height = r 2 h Cur ved sur face= cir cumfer ence  height = 2r h Tot al sur face ar ea of a cylinder = 2r h + r 2 = 2 r (r + h) PYRAM I D AN D CON E Pyr amid is a solid whose sides ar e t r iangle, having a common ver t ex and whose base is plane r ect ilinear fi gur e. A pyr ami d i s cal l ed t et r ahedr on, squar e pyr amid, pentagonal pyr amid, hexagonal pyramid etc. accor di n g t o i t s base bei n g a t r i an gl e, squ ar e, pent agon, hexagon et c. r espect ively.

 a = 46 met er s (apr r ox) TH REE DI M EN SI ON S Anyt hing which occupies space is called a solid or a three dimensional figure e.g., cube, cuboid, cylinder, cone, pr ism, spher e et c. CU BE AN D CU BOI D A solid figur e wit h six faces all of which ar e r ect angle is known as cuboid while a figur e with all the six faces, squar e, is called cube.

1 Volume of a Pyr amid =  base  height (Base means 3 ar ea of t he base) Ri ght ci r cul ar cone i s a sol i d gener at ed by t he r evolut ion of a r ight angled t r iangle about one of t he sides cont aining t he r ight angle as axis.

S

--Ph -R-

Volume of a cuboid = L  B  H (Length  Breadth  Height) Volume of a cube= L 3 wher e all t he sides ar e equal. Sur face ar ea of a cuboid = 2 (L B + BH + H L ) Sur face ar ea of a cube = 6 L 2 Diagonal of a cuboid = L 2  B 2  H 2 Diagonal of a cube = 3 L PRI SM AN D CYLI N DER Right pr ism is a solid bounded by plane faces of which t wo ar e congr uent par allel figur es and t he ot her s ar e r ect angles.

Volume of a r ight pr ism = Ar ea of t he base  H eight L at er al sur face of a r ight pr ism = Per imet er of t he base  H eight

1  area of base  perpendicular height 3 1 = r 2 h 3 Cur ved sur face= r l

Volume of a cone =

H 2  R2 Tot al sur face ar ea= r (r + l ) SPH E RE A spher e is a solid bounded by one sur face and is such that all str aight lines dr awn fr om a cer tain point within t he solid t o t he bounding sur face ar e equal. Spher e can be hollow as well as solid. e.g. t ennis ball is a hollow spher e while Cr icket ball is a solid spher e. wher e, l = slant height =

4 r 3 3 Sur face ar ea of a spher e = 4r 2 Volume of a spher e =

Mensuration

15.5

SOLVED E XAM PLES 1 meters 3 in length, 12 meters in breadth and 8 meters in depth, is full of water. Find the weight of water in metr ic tons, given that one cubic meter of water weighs 1000 kilogr ams. Solution: Volume of water 1 = 37  12  8 cub. meters 3 112 Weight of wat er =  12  8  1000 3 = 3584000 kg = 3584 metric tons A br ick measur es 20 cm by 10 cm by 7½ cm.H ow many br icks will be r equir ed for a wall 25 m long, 2 m high and ¾ m t hick? 3 Solution: Volume of wall = 25  2  cub. meters 4 20 10 15   Volume of one br ick = cub. m 100 100 200 3 = cub. m 2000  Requir ed number of br icks 3 3 = 25  2  = 25000  4 2000 3 The sur face of a cube is 30 sq. met er s. Find 8 it s volume. 3 243 Solution : 6 (edge)2 = 30 = sq. met er s 8 8 243 1 81  (edge)2=  = sq. met er s 8 6 16 81 9  edge= met er s = met er s 16 4 9 9 9 25   Volume = = 11 cub. met er s 4 4 4 64 The annual r ainfall at a place is 43 cm. Find t he weight in metr ic t onnes of t he annual r ain falling t her e on a hect ar e of land, t aking t he weight of wat er t o be 1 met r ic t onne t o t he cubic met er. Solution: Ar ea of land = 10000 sq. m 10000  43 Volume of r ainfall = = 4300 cub. m 100 Weight of wat er = 4300  1 = 4300 m tonnes A cubic met er of gold is ext ended by hammer ing so as t o cover an ar ea of 6 hect aes. Find t he t hickness of t he gold in decimals of cm, cor r ect t o t he fir st t wo significant figur es. Solut ion :1 cub. m = 1000000 cub. cm 6 hect ar es= 6  10000  10000 sq. cm 1000000  Thickness of gold = 6  10000  10000 1 = = .0017 cm. 600

1. A rectangular tank measuring internally 37

2.

FG H

3.

4.

 5.

IJ K

6. A ci st er n con t ai n i n g 8000 l i t r es m easu r es exter nally 3 m 3 dm by 2 m 6 dm by 1m 1 dm, t he sides being 5 cm t hick. Find t he t hickness of t he bot t om, supposi ng one cubi c met er t o cont ain 1000 lit r es. Solution : Volume of t he cist er n = 8000  1000 = 8 cub. m. I nt er nal lengt h = 3 m dm – 2  5 cm = 3 m 2 dm I nter nal br eadth = 2 m 6 dm – 2  5cm = 2 m 5 dm 8  I nt er nal height = m=1m 3.2  2.5 But ext er nal height = 1m 1 dm  Thickness of bot t om = 1m 1dm – 1 m = 1dm 7. The base of a pr ism is a t r iangle of which t he sides ar e 17 cm, 25 cm and 28 cm r espect ively. The volume of t he pr ism is 4200 cubic cm. What is t he height ? Find it s lat er al ar ea also. Solution: L et t he sides be a = 17 cm, b =25 cm, c = 28 cm ab c 17  25  28 Then, s = = = 35 cm 2 2 s – a = (35– 17)cm = 18 cm. s – b = (35– 25) cm = 10 cm s – c = (35– 28)cm = 7 cm

b

gb

gb g

s s– a s– b s– c

Ar ea of a t r iangle =

H ence ar ea of t he base = 35  18  10  7 = 210 sq. cm. 4200 H eight of t he pr ism = cm = 20 cm. 210 L at er al ar ea = per imet er of t he base  height = (17 + 25+ 28)  20 sq. cm = 1400 sq.cm 8. A hollow cylindr ical t ube open at bot h ends is made of ir on 2cm t hick. I f t he ext er nal diamet er be 50 cm and t he lengt h of t he t ube be 140 cm., find t he number of cubic cm of ir on in it . Solut ion :H eight = 140 cm Ext er nal diamet er = 50 cm  Ext er nal r adius, OB = 25 cm

O

A

B

Also, int er nal r adius, OA = OB – AB = (25 – 2) cm = 23 cm I t is easy t o see t hat volume of ir on will be found by subt r act ing volume of t he cylinder of r adius OA fr om volume t he cylinder of r adius OB. 22 22 Volume of ir on =  25  25  140 –  23 7 7  23  140 = 42240 cub. cm.

15.6

Mensuration

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. The lengt h of t he longest r od t hat can be placed in a r oom which is 12 m long, 9 m br oad and 8 m high is (a) 27 m (b) 19 m (c) 17 m (d) 13 m 2. A cir cular well is dug t o a dept h of 14 met r es wit h a diamet er of 2 met r es. What is t he volume 22   of t he ear t h dug out ?  Use   7   (a) 32 cubic met r es (b) 36 cubic met r es (c) 40 cubic met r es (d) 44 cubic met r es 3. The per imet er of one face of a cube is 20 cm. I t s volume must be (a) 8000 cm 3 (b) 1000 cm 3 (c) 125 cm 3 (d) 400 cm 3 4. The cur ved sur face of a r ight cir cular cone of height 15 cm and base diamet er 16 cm is (a) 120 cm 2 (b) 60 cm 2 (c) 136 cm 2 (d) 68 cm 2 5. The volume of a cube is V. The t ot al lengt h of it s edges is 6.

7.

8.

9.

10.

(a) 6V1/3 (b) 8 V (c) 12 V 2/3 (d) 12 V 1/3 The cr oss-sect ion of a canal is in t he for m of a t r apezium. I f canal t op is 10 m wide, t he bot t om is 6 m wide and ar ea of t he cr oss sect ion is 72 m 2, t hen dept h of t he canal is (a) 10 m (b) 7 m (c) 6 m (d) 9 m A hor se is t et her ed t o one cor ner of a r ect angular gr assy field 40 m by 24 m wit h a r ope 14 m long. Over how much ar ea of t he field can it gr aze ? (a) 154 m 2 (b) 308 m 2 (c) 150 m 2 (d) None of t hese The sides of a r ect angular field ar e in t he r at io 3 : 4 wit h it s ar ea as 7500 sq. m. The cost of fencing t he field @25-paise per met er is (a) ` 87.50 (b) ` 86.50 (c) ` 67.50 (d) ` 55.50 A t r ack i s i n t he for m of a r i ng whose i nner ci r cu m f er en ce i s 352 m an d t h e ou t er cir cumfer ence is 396 m. The width of the tr ack is (a) 44 m (b) 14 m (c) 22 m (d) 7 m A st eel wir e bent in t he for m of a squar e of ar ea 121 cm 2. I f t he same wir e is bent in t he for m of a cir cle, t hen t he ar ea of t he cir cle is (a) 130 cm 2 (b) 136 cm 2 (c) 154 cm 2 (d) none of t hese

11. The diamet er of a gar den r oller is 1.4 m and it is 2 m l ong. H ow much ar ea wi l l i t cover i n 5 22 r evolut ions ? use   7

FG H

IJ K

(a) 40 m 2 (b) 44 m 2 (c) 48 m 2 (d) 36 m 2 12. The number of r evolut ions made by a wheel of diamet er 56 cm in cover ing a dist ance of 1.1 km

22   is  Use   7   (a) 31.25 (b) 56.25 (c) 625 (d) 62.5 13. A bicycle wheel makes 5000 r evolutions in moving 11 km. Find diamet er of t he wheel. (a) 55 cm (b) 60 cm (c) 65 cm (d) 70 cm 14. The perimeter of a r ectangular field is 52 m. I f the length of the field is 2m more than thrice the breadth, then what is the breadth of the field? (a) 6.5 m (b) 6.25 m (c) 13 m (d) 6 m

1 1 1 : : . 2 3 4 I f t he per imet er is 52 cm, t hen t he lengt h of t he smallest side is (a) 9 cm (b) 10 cm (c) 11 cm (d) 12 cm

15. The sides of a tr iangle ar e in the r atio of

LEVEL-1 1. The ar ea of four walls of a r oom is 120 m 2. The lengt h is t wice t he br eadt h. I f t he height of t he r oom is 4 m, find ar ea of t he floor. (a) 48 m 2 (b) 49 m 2 (c) 50 m 2 (d) 52 m 2 2. A wir e is looped in t he for m of a cir cle of r adius 28cm. I t is r e-bent into a squar e for m. Deter mine lengt h of t he side of t he squar e. (a) 44 cm (b) 45 cm (c) 46 cm (d) 48 cm 3. A cir cular gr assy plot of land, 42 m in diamet er, has a pat h 3.5 m wide r unning r ound it on t he out side. Find t he cost gr avelling t he pat h at ` 4 per squar e met r e. (a) ` 2002 (b) ` 2003 (c) ` 2004 (d) ` 2000 4. A r ect angul ar t ank measur i ng 5 m × 4.5 m × 2.1 m is dug at the centr e of the field measur ing 13.5 m × 2.5 m. The ear th dig out is spr ead evenly over t he r emai ni ng por t i on of t he fi el d. H ow much is t he level of t he field r aised? (a) 4.0 m (b) 4.1 m (c) 4.2 m (d) 4.3 m

Mensuration

5. H ow many met r es of cl ot h 5 m wi de wi l l be r equir ed t o make a conical t ent , t he r adius of whose base is 7 m and whose height is 25 m? (Take  = 22/7) (a) 108 m (b) 110 m (c) 112 m (d) 115 m 6. A met all ic sheet i s of r ect angular shape wi t h di mensions 48 × 36 cm. Fr om each one of it s cor ner s, a squar e of 8 cm is cut off. An open box is made of t he r emaining sheet . Find volume of t he box. (a) 5110 cm 3 (b) 5130 cm 3 3 (c) 5120 cm (d) 5140 cm 3 7. A room 5 m 8 m is to be carpeted leaving a margin of 10 cm for m each wall. I f cost of t he car pet is ` 18 per m 2, t hen cost of car pet ing t he r oom will be (a) ` 702.60 (b) ` 691.80 (c) ` 682.46 (d) ` 673.92 8. A cir cle r oad r uns ar ound a cir cular gar den. I f differ ence bet ween cir cumfer ence of t he out er cir cle and t he inner cir cle is 44 m, t hen widt h of t he r oad is (a) 4 m (b) 7 m (c) 3,5 m (d) 7,5 m 9. A gar den is 24 m long and 14 m wide. Ther e is a pat h 1 m wide out side t he gar den along it s sides. I f t he pat h i s t o be const r uct ed wi t h squar e mar ble t iles 20 cm 20 cm, t he number of t iles r equir ed t o cover t he pat h is (a) 1800 (b) 200 (c) 2000 (d) 2150 10. A lawn is in t he for m of an isosceles t r iangle. The cost of t ur fing it came t o ` 1200 at `4 per m 2. I f t he base is 40 m long, t he lengt h of each side is (a) 120m (b) 25 m (c) 7.5 m (d) None of t hese 11. The r at io of t he r at e of flow of wat er in pipes var ies inver sely as t he squar e of t he r adius of t he pipes. What is t he r at io of t he r at es of flow in t wo pipes of diamet er s 2 cm and 4 cm ? (a) 1 : 2 (b) 2 : 1 (c) 1 : 8 (d) 4 : 1 12. A cir cle and a r ect angle have the same per imeter. The sides of t he r ect angle ar e 18 cm and 26 cm. What is t he ar ea of t he cir cle? (a) 88 cm 2 (b) 154 cm 2 2 (c) 1250 cm (d) 616 cm 2 13. A r ect angular lawn 80 met r es by 60 met r es has t wo r oads each 10 met r es wide r unning in t he middle of it , one par allel t o t he lengt h and t he ot her par allel t o t he br eadt h. Find t he cost of gr avelling t hem at ` 30 per squar e met r e. (a) ` 39,000 (b) ` 3, 900 (c) ` 3,600 (d) ` 36,000

15.7

14. The l engt h of a r oom i s doubl e t he br eat h. The cost of col our i ng t he cei l i ng at ` 25 per sq m i s ` 5,000 and t he cost of pai nt i ng t he four wal l s at ` 240 per sq m i s ` 64,800. Fi nd t he hei ght of t he r oom. (a) 4.5 m (b) 4 m (c) 3.5 m (d) 5 m 15. The t r unk of a t r ee is a r ight cylinder 1.5 m in r adius and 10 m high. The volume of t he t imber which r emains when t he t r unk is t r immed just en ou gh t o r edu ce i t t o a r ect an gu l ar par allelopiped on a squar e base is (a) 44 m 3 (b) 46 m 3 (c) 45 m 3 (d) 47 m 3

LEVEL-2 1. A hemispher ical bowl is filled t o t he br im wit h a bev er age. T h e con t en t s of t h e bow l ar e tr ansfer r ed into a cylindr ical vessel whose r adius is 50% mor e t han it s height . I f t he diamet er is same for bot h t he bowl and t he cyli nder , t he volume of t he bever age in t he cylindr ical vessel, as a per cen t age of t h e v ol u m e i n t h e hemispher ical bowl, is (a) 66

2 % 3

(b) 78

1 % 2

(c) 100% (d) mor e t han 100% 2. I n t he accompanyi ng fi gur e, AB is one of t he diamet er s of t he cir cle and OC is per pendicular t o it t hr ough t he cent r e O. I f AC is 7 2 cm, t hen what is t he ar ea of t he cir cle in sq. cm.? C

A

O

B

(a) 24.5 (b) 49 (c) 98 (d) 154 3. A wir e is in t he for m of a cir cle of r adius 35 cm. I f it is bent int o t he shape of a r hombus t hen what is t he side of t he r hombus? (a) 32 cm (b) 70 cm (c) 55 cm (d) 17 cm 4. A per son wishes t o make a 100 sqm r ect angular gar den. Since he has only 30 m bar bed wir e for fencing, he fences only t hr ee sides let t i ng t he house wall act as t he four t h side. The widt h of t he gar den is (a) 10 m (b) 5 m (c) 50 m (d) 100 m

15.8

Mensuration

5. The weight of a solid cone having diamet er 14cm and ver t ical height 51 cm is ......, if t he mat er ial of solid cone weighs 10 gr ams per cube cm. (a) 16.18 kg (b) 17.25 kg (c) 26.18 kg (d) 71.40 kg 6. I n a speci al r aci n g even t , t h e per son wh o en cl osed t h e m axi m u m ar ea wou l d be t h e wi nner and would get ` 100 for ever y squar e met r e of ar ea cover ed by him/her. Johnson, who su ccessf u l l y com pl et ed t h e r ace an d w as t h e ev en t u al w i n n er , en cl osed t h e ar ea shown in figur e below. What is t he pr ize money won ? (N ot e: ar c f r om C t o D m ak es a com pl et e semi-cir cle). AB = 3m, BC = 10m, CD = BE = 2m. C

B

D

E

A

(a) ` 2914 (b) ` 2457 (c) ` 2614 (d) ` 2500 7. ABCD is a four -sided figur e wit h AB par allel t o CD and AD par allel t o BC. ADE is r ight angle. I f t he per imet er of ABE is 6 cm, t hen t he ar ea of t he figur e ABCD is A

D

B

(a) 2 3 sq. cm. (b) 4 3 sq. cm. (c) 3 sq. cm. (d) none of t hese 8. I t is r equir ed to fix a pipe such t hat wat er flowing t hr ough it at a speed of 7 met r es per minut e fills a tank of capacity 440 cubic metr es in 10 minutes. The inner r adius of t he pipe should be (a)

2 m

(c) 1/2 m

12 2 cm, t hen t he ar ea of t he t r iangle is (a) 24 3 cm 2

(b) 24 2 cm 2

(c) 64 3 cm 2 (d) 32 3 cm 2 12. Semi -ci r cul ar l awns ar e at t ached t o bot h t he edges of a r ect an gu l ar f i el d m easu r i n g 42 m 35m. The ar ea of t he t ot al field is (a) 3818.5 m 2 (b) 8318 m 2 2 (c) 5813 m (d) 1358m 2 13. What is t he ar ea of t he inner equilat er al t r iangle if t he side of t he out er most squar e is ‘a’ ? (ABCD is a squar e) (a)

3 3a 2 32

(b)

3 3a 2 64

(c)

5 3a 2 32

A

B

C

2

(b) 2 m (d)

1 2

m

9. Wat er flows out t hr ough a cir cular pipe whose int er nal diamet er is 2 cm, at t he r at e of 6 met r es per second int o a cylindr ical t ank, t he r adius of whose base is 60 cm. By how much will be level of wat er r ise in 30 minut es? (a) 2 m (b) 4 m (c) 3 m (d) 5 m

D

3a 64 14. An edge of a cube measur es 10 cm. I f t he lar gest possible cone is cut out of t his cube, t hen volume of t he cone is (a) 260 cm 3 (b) 260.9 cm 3 3 (c) 261.9 cm (d) 262.7 cm 3 15. I n t he f i gur e, A BCD i s a squ ar e wi t h si de 10. BFD is an ar c of a cir cle wit h cent r e C. BGD is an ar c of a cir cle wit h cent r e A. What is t he ar ea of t he shaded r egion ?

(d)

C

E

10. A plot of land in t he for m of a r ect angle has a dimension 240 m × 180 m. A dr ainlet 10 m wide is dug all ar ound it (one the outside) and the earth dug out is evenly spr ead over the plot, incr easing i t s sur face l evel by 25 cm. The dept h of t he dr ainlet is (a) 1.225 m (b) 1.229 m (c) 1.227 m (d) 1.223 m 11. A squar e and an equilat er al t r iangle have t he same per imet er. I f t he diagonal of t he squar e is

(a) 100 – 50  (c) 50 – 100

(b) 100 – 25  (d) 25  – 100

Mensuration

15.9

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (c) 11. (b)

2. (d) 12. (c)

3. (c) 13. (d)

4. (c) 14. (d)

5. (d) 15. (d)

6. (d)

7. (a)

8. (a)

9. (d)

10. (c)

7. (d)

8. (b)

9. (c)

10. (b)

7. (c)

8. (b)

9. (c)

10. (c)

LEVEL-1 1. (c)

2. (a)

3. (a)

4. (c)

5. (b)

11. (d)

12. (d)

13. (a)

14. (a)

15. (c)

6. (c)

LEVEL-2 1. (c) 11. (c)

2. (d) 12. (a)

3. (c) 13. (a)

4. (b) 14. (c)

5. (c) 15. (c)

6. (b)

EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S

=

14 m

7.

1. L engt h of t he longest r od = Diagonal of t he r oom

14 m 24m

12 2  9 2  8 2

=

289 = 17 2. Requir ed volume = Volume of r ight cir cle cylinder = r 2h

=

40m

 Requir ed ar ea = shaded ar ea

22  1  14 = 44 cubic met r es 7

3. Volume of cube = (Side of t he cube)3

F 20I = G J H4K

= (5)3 = 125 cm 3

=

4. H er e, h = 15 cm, r = 8 cm 2

2

=

2

2

= 17

h r 15  8  Cur ved sur face of t he cone =  r l = 8  17 = 136 cm 2 5. Ther e ar e 12 edges in t he cube Given : Vol ume = V Each edge = V1/3 Tot al lengt h of t he edges = 12 V 1/3 6. L et h t he height of t he t r apezium H ence ar ea of cr oss-sect ion of t he channel in t he for m of t r apezium =

 

1 (10 + 6)  h = 72 2

8h = 72 h = 9m

1 (14)2 4



1 22   14  14 = 154 m 2 4 7

9. Widt h of cir cular t r ack = r 1 – r 2

3

 Slant height , l =

=

2r1  2r2 396  352 = = 7m 2 2

10. Side of a squar e = 121 = 11 cm  Per imet er of cir cle = 4  11 = 44 cm  2r = 44 wher e r is r adius of t he cir cle

22  r = 44 7  r = 7 cm  Ar ea of t he cir cle = r 2 

2

22  (7)2 = 154 cm 2 7 11. Gar den r oller is in t he for m of a cylinder whose r adius is 0.7 m and height is 2 m.  Ar ea cover ed in 5 r evolut ions = 5  2r h 22 = 5  2   0.7  2 7 2 = 44 m =

12. Number of r evolut ions =

1.1  1000  100 = 625 22 2  28 7

15.10

Mensuration

13. Cir cumfer ence of t he wheel 11000 met res = 2r = 5000 11  r = 10



r =

7. Ar ea of car pet = 7.8m × 4.8m = 37.44m 2 Cost of Car pet = ` 18 × 37. 44 = ` 673.92 H ence opt ion (d) is cor r ect .

11 7 7  = met r e 10 20 22

7  100 = 35 cm 20 Diamet er of t he wheel = 70 cm. 14. Given, 2x + 2y = 52 and x = 3y + 2 Solving (i ) and (ii ) we get x = 20, y = 6

0.1m

=

15.

0.1m

(i ) (ii )

8. or

2(R1 – R2) = 44 R1 – R2 = 7m

1 1 1 : : 6a : 4a : 3a 2 3 4  13a = 52 a = 4, smallest side = 3a = 12.

R1 R2

LEVEL-1 1. Using t he for mula, we get 2 × 4 (2b + b) = 120  b=5  l = 10 Ar ea = 50 2. Given : R = 28 Cir cumfer ence = 2(28) = 176 Side of t he squar e =

9. Ar ea of pat h = (26 × 16) – (24 × 14) = 80m 2 Ar ea of t ile = 0.2 × 0.2m 2



Number of t iles =

47.25 H eight r aised = = 4.2 11.25

5. Given : l = 25 cur ved sur face = r l = × 7 × 25 = 550 m 2.

 L engt h of t he clot h =

550 = 110. 5

6. Dimensions of t he box : lengt h = 48 – 16 = 32 widt h = 36 – 16 = 20 and height = 8  Volume = 32 × 20 × 8 = 5120

ar ea of pat h ar ea of one t ile

= 2000 t iles

1m

1m

176 = 44 4

3. Radius = 21 Requir ed ar ea = (24.52 – 212) = 500.5 Cost = 500.5 × 4 = 2002 4. Ar ea of field – Area of tank = 13.5 × 2.5 – 5 × 4.5 = 11.25 Volume of t ank = 41.25



0.1m

CARPET

1m

10. h 40 m

ar ea of t r iangle =



1 bh 2

1 120  40  h = h = 15 m 2 4 side =

202  152 = 625 = 25m 11. Rat io of t wo pipes ar e 1 cm and 2 cm squar es of t he Rat io of t he t wo pipes ar e 1 cm and 4 cm.  Rat es of flow of t he t wo pipes ar e in t he r at io 1 , i.e. 4 : 1. 4 12. Per imet er of t he r ect angle = 2(18 + 26) = 88 = 2r . 1:

Mensuration 15.11

H ence r = 14. Ar ea of t he cir cle = × 142 = 616. 13. Cost = Ar ea  r at e = [(80  10 + 60  10 ) – (10  10]  30 = 39000. 14. L et t he dimensions of t he r oom be 2x and x . Ar ea of ceiling = 2x  x =

5000 = 200, 25

5. Volume of t he solid cone =

1 22   49  51 = 2618 cm 3 3 7  Weight of t he solid cone = (2618  10) gm. =

=

 x = 10.  Ar ea of 4 walls = 2(l h + bh) = 2 (20h + 10h) =

64800 = 270. 240

Solving, we get 15. 2x2 = 32, 9  x2 = 2



h = 4.5 m

V = lbh =

9 × 10 = 45 2

2 × × r 3 3 2 Volume of cylinder = × r 2 × r 3 Since t he t wo ar e equal, t he per cent age is 100%. 2. L et OA = OC = r adius = r  OA 2 + OC2 = AC2

1.Volume of hemispher e =

 

e j

r2 + r2 = 7 2

2

2r 2 = 2  49 r =7

22  49 = 154 cm 2 7 3. Cir cumfer ence of t he cir cle = 2r  Ar ea of t he cir cle = r 2 =

22  35 = 200 cm 7 = Per imet er of t he r hombus

=2

220 = 55 cm 4 4. L et widt h of t he gar den be x met r es  Side of t he r hombus =

 L engt h of t he gar den =

100 x

100 + x + x = 30 x  x 2 – 15x + 50 = 0  x = 10, x = 5 x = 10 is omitted, because in that case, the gar den will become squar e which is cont r ar y t o what is given  Widt h of t he gar den = 5 met r es 

26180 kg = 26.18 kg 1000

6. Ar ea of t he semi-cir cle = Ar ea of t he t r iangle =

 = 1.571 sq. m. 2

1  AB  BE 2

1  3  2 = 3 sq. m. 2 Ar ea of t he r ect angle = 10  2 = 20 sq. m.  Tot al ar ea cover ed = 24.571 sq. m.  Pr ize money won by t he per son = 24.571  100 = ` 2457 8. Volume of t he pipe in one minut e =

LEVEL-2



1 (7)2 51 3

= r 2 (7) =

440 10

 r = 2 9. Volume of wat er in t he t ank = (60)2 × x wher e, x = level of t he wat er Wat er r eleased by pipe in 30 minut es = × (1)2 × 600 × 30 × 60 Equat ing t he t wo and solving, we get  (60)2  x =   600  30  60  x = 300 cm = 3 m 10. Volume of t he ear t h t aken out = Volume of t he ear t h on plot  [{260 × 200} – {240 × 180}] × x = 240 × 180 ×

1 4

Solving, we get x = 1.227. 11. L et x be t he side of t he squar e.



e

x 2 + x 2 = 12 2

j

2

 x = 12 Now, per imet er of t he equilat er al t r iangle = per imet er of t he squar e = 4x = 48 cm  Side of t he equilat er al t r iangle =

48 = 16 cm 3

15.12

Mensuration

 Ar ea of equilat er al t r iangle

3  (16)2 = 64 3 cm 2 4 Ar ea B = Ar ea D Ar ea C = Ar ea E

Dr aw per pendicular fr om cent r e t o one side of t r iangle.

=

12.

B

A

C

FG IJ H K

35 1 1 = 42  35 + 2   (21)2 + 2   2 2 2 = 1470 + 1386 + 962.5 = 3818.5 m 2

D

E

F C

L engt h of out er squar e = a

 L engt h of inner squar e =  Radius of cir cle =

l=

3a 2 2

10 cm = 5 cm 2 H ence volume of t he lar gest possible cone cut out of t his cube

a 2 2

2

1 22   (5)2  10 = 261.9 cm 3 3 7 15. Ar ea of t he por t ion DFBC =

1  (10)2 = 25 4 1  Ar ea of BCD =  10  10 = 50 2  Ar ea of t he por t ion DFBOD = Ar ea of t he por t ion DFBC – Ar ea of BCD = 25 – 50  Ar ea of t he por t ion DFBGD = 2 Ar ea of t he por t ion DFBOD = 2(25 – 50) = 50 – 100  =

A G

30°

l 3 2 2  = 2a 2

Radius of t he base =

 Tot al ar ea of t he field = Ar ea of A + 2  Ar ea of B + 2  Ar ea of C

B

l 2 si de 2 = a 2 2 Radius

3 3a2 3 2 l = 32 4 14. H eight of t he cone = 10 cm

D

13.

cos30=

 ar ea of t r iangle =

42 E 35

t hen

a 2

 

1

Alphabetical and Number Series

CHAPTER

I n ver bal ser i es, wor ds, l et t er s or di gi t s ar e given in a specific sequence/or der and we have fi nd out next wor d, let t er or di git t o compl et e t he gi ven ser ies. Number s or alphabet ical let t er s, ar e gener all y cal led t er ms of t he ser ies. These t er ms follow a cer t ain pat t er n t hr oughout . I n t he quest i ons we have t o ident ify last one or t wo t er ms t o cont inue t he ser ies or t o fi nd a mi ssing t er m i n bet ween gi ven t er ms t o cont inue t he sequence foll owed in t he quest ion. Ther e i s no set pat t er n and each quest ion may foll ow a di ffer ent pat t er n or sequent i al ar r angement of l et t er s or di gi t s, whi ch have t o det ect using common sense and r easoni ng abil it y. TYPES OF SERI ES COM PL ET I ON QU ESTI ON S. M ainly t her e ar e following four t ypes of ver bal ser ies complet ion quest ions : 1. AL PH ABET SERI ES. I n t hi s ser i es, given alphabet s foll ow a par t i cul ar sequence or or der. We have t o det ect t he pat t er n fr om t he given alphabet s and fi nd mi ssing alphabet or t he next al phabet t o cont i nue t he pat t er n.  Ther e ar e no set r ules.  Ther e can be omi ssi on of alphabet s i n an or der.  Al phabet s may al so be omit t ed i n an incr easing/decr easi ng or der, which may be di r ect incr ease or decr ease.  Ther e can also be alt er nat e or der.  Ther e may also be alt er nat e sequences  Ther e may be sever al ot her pat t er ns in t he let t er ser ies.  To t ack le l et t er ser i es quest ions, var y posit i on of t he al phabet and i t s posi t ion number in bot h for war d and backwar d sequences.  I n solving t hese quest ions pat t er n of t he alphabet ser ies should be not ed. Patt erns of L ett ers

I t shows var iat ion based on following : ( i ) Dist ances bet ween Alphabet s and Repet it ion of let t ers. To cont inue the series aft er Z, we again begin wit h A. I n ot her words, t he sequence is kept in a circular order. 

On r eaching Z, t he ser ies r est ar t s fr om A and on r eaching A, it r est ar t s fr om Z. Alphabet s in N atural series A B C D E F   1st 5th

G H

I

J K  10th

L

M N

O P  15th

Q

R S

T U  20th

O N M L K  15t h

J

I H G F  20th

V W

X Y Z  25th

Alphabet s in Reverse series Z Y  1st

X

W V U  5th

T

S

R Q P  10th

E

D

( ii ) Posit ion of Alphabet s. 1

A26

10

J17

19

S8

2

B25

11

3

C24

4

D 23

K 16

12

13

20

21

22

T7

L 15 U6

M 14

V5

5

6

E 22

7

F 21

14

W4

16

O12

24

X3

9

H 19

15

N 13

23

8

G20

P11

25

Y2

I 18

17

Q10

26

Z1 .

18

R9

C

B A  25th

1.2

Alphabetical and Number Series

N ote : I n m x n ; m = posit ion fr om left ,

n = posit ion fr om r ight .

m + n = 27

Some Skipping Pat t er ns (i ) Regular Order : N umber of alphabets skipped r emai ns t he same.

Example. A , D , G , J , ? Ans. M (ii ) I ncreasing Order : Each t ime t he number of alphabet s ski pped incr eases in a gi ven pat t er n.

Example. A C F J O ? Ans. U H er e, each t ime number of let t er s sk ipped i ncr eases by one. (iii ) Decreasing Order : Each t ime t he number of let t er s sk ipped decr eases i n a given pat t er n.

Example. A G L P S ? Ans. U H er e number of l et t er s ski pped decr eases by one each t ime, i.e., fi r st 5, t hen 4, t hen 3, and so on. (iv) I nterlinked Series : I n t his t wo or mor e differ ent ser i es ar e at t ached t oget her. These differ ent ser i es foll ow t heir own di ffer ent r ules.

Example. A D F J M R ? Ans. V H er e, t her e ar e t wo i nt er linked ser ies. E xamples 1. Which of t he given opt i ons will complet e t he given ser i es ? BDFH J? Solution : The ser ies follows t he pat t er n of moving t he let t er s t wo st eps for war d. B

D +2

F +2

J

H +2

L

+2

+2

2. A Z B Y C ? Solution : Ther e ar e t wo alt er nat e ser ies. +1

A

Z

+1

B

Y

C

X

–1

–1

Ser ies I : A B C (consecut ive let t er s in nat ur al ser ies) Ser ies I I : Z Y X (consecut ive iet t er s in r ever se ser ies)

3. I f each consonant in t he wor d ‘EXPL ORATI ON’ is r eplaced by t he pr evious alphabet and each vowel is r eplaced by t he alphabet following it of t he english alphabet ic ser ies and t hen t he or der of t he alphabet s t hus, for med is r ever sed, which of t he following will be t he sevent h fr om t he r ight ? Solut ion : E X P L O R A T I O N























F W O K P New for mat ion = F W O K P Q B S J P M

Q

B

S

J

P

M

On r ever sing = M P J S B Q P K O W F H ence, sevent h let t er for m t he r ight end = B

Alphabetical and Number Series

(v) addi t i on /su bt r act i on /m u l t i pl i cat i on / division by some number (vi ) m an y m or e com bi n at i on s of abov e r elationships

2. L ET TER SERI ES. This t ype of quest i ons usuall y consi st of a ser i es of small let t er s whi ch follow a cer t ain pat t er n. H owever, some l et t er s ar e mi ssi ng fr om t he ser i es. Then t hese mi ssing l et t er s ar e given in a pr oper sequence as one of t he alt er nat i ves.

Rules : 1. Difference between consecutive numbers

Example. aab... aaa.... boa.... (a) baa bab (e) bbb

(b) abb (d) aab

(i ) is same.

( c )

(ii ) ar e in ar it hmet ic pr ogr ession (A.P). (iii ) is a per fect squar e. (iv) ar e mult iples of a number.

Ans. (a)

(v) ar e pr ime number s.

M et hod : 



 



(vi ) is a per fect cube.

Fir st blank space should be filled in by ‘b', so t hat we have t wo a's followed by t wo b's.

(vii )ar e in geomet r ic pr ogr ession (G.P.) 2. Rat io between each consecutive number

Second blank space should be filled in eit her by ‘a', so t hat we have four a's followed by t wo b's, or by ‘b' so t hat we have t hr ee a's followed by t hr ee b's.

(i ) is same. (ii ) is in ar it hmet ic pr ogr ession (A.P) (iii ) is per fect squar e number.

L ast space must be filled in by ‘a'.

(iv) is t he mult iple of a number.

Thus, we have t wo possible answer s : ‘baa' and ‘bba'. But, only ‘baa' appear s in t he alt er natives. So, t he answer is (a). I n case, we had bot h t he possible answer s in t he alt er nat ives, we would have chosen t he one t hat for ms a mor e pr omi nent pat t er n, which is aabb/ aaabbb/aa. Thus, our answer would have been “ bba” .

(v) is a pr ime number. (vi ) is a per fect cube number. (vii )ar e in geomet r ic pr ogr ession (GP.) Types of N umber Series ( i ) Pure Series. I n this type of number ser ies, the number i t sel f obeys cer t ai n or der so t hat t he char act er of t he ser ies can be found out .

Number may be : Per fect squar e ; Per fect cube, Pr ime, and combinat ion :

3. N U M BER SERI ES. I n t he number ser i es, some number s ar e ar r anged i n a par t icul ar sequence. Al l t he number s for m a ser ies and change in a cer tain or der. Somet imes, one or mor e number s ar e wr ongl y put in t he number ser i es. We ar e r equ i r ed t o obser ve t h e t r en d i n w h i ch number s change in t he ser ies and find out which number /number s misfit int o t he ser ies t hat number /number s is odd number of t he ser i es.

( ii ) Difference series. I n t his t ype of number ser ies change in or der for t he di ffer ence bet ween each consecutive number of t he ser ies is found out . ( iii ) Rat io ser ies. I n t his t ype of number ser ies change in or der f or t h e r at i os bet w een each consecutive number of t he ser ies is found out .

I n number ser ies, set of given number s in a series are r elated to one another in a par ticular pat t er n or manner.

Relationship bet ween number s may be (i ) consecut ive odd/even number (ii ) consecut ive pr ime number s (iii ) squar es/cubes of some number s wi t h/ without var iat ion of addit ion or subtr action of some number s (iv) su m /pr odu ct /di f f er en ce of pr ecedi n g number (s)

1.3

( iv) M ixed series. I n t his t ype of number ser ies number s obeyi ng var ious or der s of t wo or mor e di ffer ent t ypes of ser i es ar e ar r anged alt er nat ely in a single number ser ies. Some I mport ant Pat t er ns



a, a ± d, a ± 2d, a ± 3d, ........



a, ak, ak 2, ak 3, ........ a a a, , a , 3 , ........ 2 k k k



1.4

Alphabetical and Number Series

   

(a)n , (a ± d)n , (a ± 2d)n , (a ± 3d)n ........ an + k, (a + 1)n + k, (a + 2)n + k , ........ an + k, (an + k )n + k , [(an + k )n + k ] n + k , ........ 1n, 1n + 2n, 1n + 2n + 3n, 1n + 2n + 3n + 4n, ........



Ser ies of pr ime number s i.e. 2, 3, 5, 7, 11, 13, 17, 19, 23 et c.

E xamples 1. Which number will complet e t he given ser ies ? 2, 4, 6, 8, ? Solution : Series is made of numbers which are multiples of 2. Difference between two consecutive numbers is 2. 2 4 6 8 10 2

3

+2

+2

8 +2

10 +2

2. Which number will complet e t he ser ies ? 12, 72, 432, ? Solution : Number s ar e mult iplied by 6 t o obt ain t he next number : 2

12

72

Solution : I n t his ser ies, fir st set compr ises of squar es of odd number s in descending or der and second set compr ises of squar es of even number s in ascending or der. 7

2

49

5 16

3

25

42

3 36

6

2

9

2

64

8

2

Ser ies I : 49, 25, 9 (squar es of 7, 5, 3) Ser ies I I : 16, 36, 64 (squar es of 4, 6, 8) This t ype of ser ies consist s of bot h number and let t er s.

5

OR 6

4

49, 16, 36, 9, ?

4. M I XED SERI ES.

4

2

3. Find missing number in t he ser ies given below.

432

2592

6 6 6 6 Given ser ies may also compr ise of t wo alt er nat e ser ies mer ged as one.

5. CORRESPON DEN CE SERI ES. This t ype of ser ies consist s of t hr ee sequences wi t h t hr ee di ffer ent element s (usual l y capit al let t er s, digit s and small let t er s). On t he basis of t he similar it y in positions in t he t hr ee sequences, a capi t al l et t er i s found t o cor r espond wi t h a unique digit and a unique small let t er, whenever i t occur s. We ar e r equi r ed t o t r ace out t hi s cor r esponden ce an d accor di n gl y ch oose t h e element s t o be filled in at t he desir ed places.

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S Direction (Q. 1 - 23). Find t he missing let t er s. 1. A, B, D, G, ..... (a) L (b) M (c) J (d) K

5. AZ, BY, CX, DW, ....... (a) EU (b) EV (c) EW (d) FW 6. DF, GJ, K M , NQ, RT, ..... (a) WV (b) UX (c) XZ (d) YZ

2. A, D, H , M , S ..... (a) U (c) Y

(b) V (d) Z

3. X, U, R, O, L , ..... (a) I (c) K

7. ABZ, BCY, CDX, DEW, ...... (a) EFV (b) FEV (c) DEF (d) DEV

(b) J (d) L

4. R, M , ....., F, D, C (a) F (c) H

8. NEQ, M GO, L I M , K K K , ..... (a) PRI (b) YAM (c) AAJ (d) JMI

(b) G (d) I

9. AYD, BVF, DRH , ....., K GL (a) FMI (b) GMJ (c) H L K (d) GLJ

Alphabetical and Number Series

10. A, CD, GH I , ......, UVWXY (a) LM NO (b) MNO (c) NOPQ (d) MNOP 11. DCBA, WXYZ, H GFE, STUV, L K JI , ..... (a) OPQR (b) MNOP (c) NOPQ (d) ONML 12. shg, r if, qje, pkd, ..... (a) ole (b) olc (c) nmc (d) nlb 13. ab __ ba __ b.a __ (a) bab

(b) aba

(c) bbb

(d) abb

14. abaa __ aab __ a __ a (a) abb

(b) aba

(c) bab

(d) aab

15. aa __ b __ abb ____ bb (a) abaa

(b) bbaa

(c) baaa

(d) baba

LEVEL-1 1. a __ ab __ baa __ bb __ (a) baba

(b) abab

(c) aabb

(d) bbaa

2. abc __ bc __ c __ (a) acbbab

(b) aabbca

(c) aabacb

(d) aababc

3. aabc ____ bca __ b __ a __ bc (a) aabca

(b) aacba

(c) aaaca

(d) abaac

4. a __ abbb __ ccccd __ ddccc __ bb __ ba (a) abcda

(b) abdbc

(c) abdcb

(d) abcad

5. adb __ ac __ da __ cddcb __ dbc __ cbda (a) bccba

(b) cbbaa

(c) ccbba

(d) bbcad

6. b __ abbc __ bbca __ bcabb __ ab (a) acaa

(b) acba

(c) cabc

(d) cacc

7. ac __ cab __ baca __ aba __ acac (a) aacb

(b) acbc

(c) babb

(d) bcbb

8. b __ b __ bb ____ bbb __ bb __ b (a) bbbbba (b) bbaaab (c) ababab (d) aabaab 9. a __ bc __ c __ abb __ bca __ (a) cccbc (b) cbbac (c) accba (d) abbba Directions (Q. 10- 15). Find t he missing number s. 10. 18, 21, 24, 27, ..... (a) 32 (c) 31

(b) 30 (d) 29

11. 885, 870, 855, 840, ...... (a) 835 (b) 855 (c) 825 (d) 815 12. 3, 9, 27, 81, ...... (a) 243 (c) 342

(b) 324 (d) 432

13. 1, 4, 9, 16, 25, ...... (a) 35 (b) 36 (c) 38 (d) 34 14. 216, 221, 231, 246, 266, ...... (a) 289 (b) 271 (c) 281 (d) 291 15. 11, 16, 23, 32, 43, ..... (a) 56 (b) 55 (c) 57 (d) 54

LEVEL-2 1. 1, 6, 15, 28, 45, ...... (a) 56 (b) 66 (c) 57 (d) 63 2. 64, 144, 256, 400, ..... (a) 529 (b) 484 (c) 676 (d) 576 3. 2, 5, 10, 17, ...... (a) 21 (c) 27

(b) 26 (d) 29

4. 2, 10, 30, 68, ..... (a) 130 (c) 142

(b) 140 (d) 138

1.5

1.6

Alphabetical and Number Series

5. 31, 35, 43, 59, 91, ...... (a) 135 (b) 145 (c) 155 (d) 165

11. Q – 49, P – 36, O – 25, N – 16, ..... (a) N – 9 (b) M – 9 (c) L – 9 (d) K – 9

6. F – 1, G – 3, J – 5, O – 8, ...... (a) W – 3 (b) L – 14 (c) U – 10 (d) Z – 26

12. B2D, E3H , I 4M , ...... (a) N5R (b) N5S (c) N5Q (d) N5T

7. D – 4, G – 7, I – 9, M – 13, ...... (a) P – 14 (b) O – 15 (c) Q – 16 (d) M – 17

13. A1E, F4J , K 9O, ....... (a) Q4U (b) W1X (c) P16T (d) T2V

8. B – 3, D – 6, F – 9, H – 12, ...... (a) L – 14 (b) M – 13 (c) J – 15 (d) I – 9

14. C4X, F9U, I 16R, ..... (a) K25P (b) L25P (c) L 25O (d) L27P

9. 3F, 6G, 111, 18L , ...... (a) 21O (b) 25N (c) 27P (d) 27Q

15. A3B, C7D, E11F, G15H , ..... (a) I20J (b) I19K (c) J19K (d) I19J

10. C3, H 8, O15, ...... (a) X24 (c) T 26

(b) I 35 (d) J 35

AN SWE RS OBJECTI VE TYPE QU ESTI ON S 1. (d) 11. (a)

2. (d) 12. (b)

3. (a) 13. (d)

4. (d) 14. (c)

5. (b) 15. (c)

6. (b)

7. (a)

8. (d)

9. (b)

10. (d)

7. (a)

8. (c)

9. (c)

10. (b)

7. (b)

8. (c)

9. (c)

10. (a)

LEVEL-1 1. (b) 11. (c)

2. (d) 12. (a)

3. (c) 13. (b)

4. (c) 14. (d)

5. (b) 15. (a)

6. (c)

LEVEL-2 1. (b)

2. (d)

3. (b)

4. (a)

5. (c)

11. (b)

12. (b)

13. (c)

14. (c)

15. (d)

6. (a)

E XPL AN AT I ON S OBJECTI VE TYPE QU ESTI ON S

5. Pat t er n is 1AZ1, 2BY 2, 3CX 3, 4DW 4, 5EV 5, .....

 M issing let t er = EV. 1

1

2

4

7

1. Pat t er n is A, A + 1, B + 2, D + 3, G +4,.....

 M issing let t er = K .

6. Pat t er n is 4

D 6F, 7G 10J, 11K

1

4

8

A, A + 3, D + 4, H + 5,

13

M + 6,

19

S + 7, ......

24

X,...

A 2B Z1, 2B 3C Y 2, 3C 4D X 3, 4D 5E W 4, 5E 6F V 5,....

 M issing let t er = EFV.

24

X, 24X – 3, 21U – 3, 18R – 3, 15O – 3, 12L – 3, ......

8. Pat t er n is 14N 5E

Q, 13M 7G 15O , 12L 9I 13M , 11 11 11 K K K , 10J 13M 9I , ....  M issing let t er = JM I .

 M issing let t er = I . 4. Pat t er n is R,

Q, 18R 20T, 21U

1

3. Pat t er n is

18

17

7. Pat t er n is

 M issing let t er = Z.

18

M , 14N

 M issing let t er = UX.

2. Pat t er n is 1

13

R – 5,

17

9. Pat t er n is 13

9

6

4

M – 4, I – 3, F – 2, D – 1, .....

 M issing let t er = I .

1

AY 2 4D, 2BV 5 6F, 4DR9 8H , 7GM 14 10J, 11K G20 12L ,...

 M issing let t er s = GM J.

Alphabetical and Number Series

10. Pat t er n is 1A,3C4D, 7G8H 9I , 13M 14N 15O16P, 21 22 23 24 25 U V W X Y, ......

 M issing let t er = M NOP.

13. Pat t er n is 12, 22, 32, 42, 52, .....

 M issing number = 62 = 36. 14. Pat t er n is + 5, + 10, + 15, + 20, ....  M issing number = 266 + 25 = 291.

11. Pat t er n is 4

D 3C2B 1A, W 4X 3Y 2Z1, 8H 7G6F 5E, S8T 7U 6V 5, 12 11 10 9 L K J I ,O12P11Q10R9, ......

15. Pat t er n is + 5, + 7, + 9, + 11, ....  M issing number = 43 + 13 = 56.

 M issing let t er = OPQR.

LEVEL-2

12. Pat t er n is

s88h 7g, r 99i 6f, q1010j 5e, p1111k 4d, o1212l 3c, .......  M issing let t er = olc.

1. Pat t er n is + 5, + 9, ......, + 21, + 25  M issing number = 45 + 21 = 66. 2. Pat t er n is 82, 122, 162, 202, .....

13. Pat t er n i s ab/ab/ab/ab/ab.

 M issing number = (20 + 4)2 = 576.

14. Pat t er n i s aba/aba/aba/aba.

3. Pat t er n is 12 + 1, 22 + 1, 32 + 1, 42 + 1, .....

15. Pat t er n i s aa/bb/aa/bb/aa/bb.

 M issing number = 52 + 1 = 26.

LEVEL-1

4. Pat t er n is 13 + 1, 23 + 2, 33 + 3, 43 + 4, .....

1. Pat t er n i s aaa/bbb/aaa/bbb.

 M issing number = 53 + 5 = 130.

2. Pat t er n i s abc/abc/abc/abc. 3. Pat t er n i s aa/bc/aa/bc/aa/bc/aa/bc. 4. pat t er n is aaa/bbbb/cccc/dddd/cccc/bbbb/a

5. Pat t er n is + 4, + 8, + 16, + 32, ....

i.e. + 22, + 23, + 24, + 25, .....  M issing number = 91 + 26 = 91 + 64 = 155. 6. Pat t er n var ies as 6

5. Pat t er n is

adbc/acbd/abcd/dcba/dbca/ebda H er e let t er s ar e equidist ant fr om t he begining and t he end of ser ies ar e t he same. 6. Pat t er n i s bcab/bcab/bcab/bacb/bcab. 7. Pat t er n i s com bi n at i on acac and abab

1.7

of

t w o ser i es

acac/abab/acac/abab/acac 8. Pat t er n i s babb/bbab/bbba/bbbb. Thus, in each sequence, a moves one step for war d and b t akes it s place and finally in t he four t h sequence, it is eliminat ed. 9. Pat t er n i s aa/b/cccc/a/bbbb/c/aa. 10. Pat t er n is a, a + d, a + 2d, ....... i.e. 18, 18 + 3, 21 + 3, 24 + 3, ......  M issing number = 27 + 3 = 30. 11. Pat t er n is a, a – d, a – 2d, .....  M issing number = 840 – 15 = 825. 12. Pat t er n is 3, 3  3, 9  3, 27  3, .......  M issing number = 81  3 = 243.

F – 1, 6 + 1G – 3, 7 + 3J – 5, 10 + 5O – 8, 15 + 8W i.e. numer als r epr esent t he places which var ies. 7. I n t he given ser ies, all t he alphabet s wit h t heir positions. 8. Pat t er n i s com bi n at i on of t w o ser i es 2 4 6 8 B, D, F, H ,10J, ...... and 3, 6, 9, 12, 15, ....... 9. Pat t er n is combinat ion of t wo ser ies 3, 6, 11, 18, 27, ..... and 6F,7G,9I ,12L ,16P, ......  M issing t er m = 27 P. 10. L et t er s wit h t heir posit ions. 11. Pat t er n var ies as Q, P, O, N, M , ........ and 49, 36, 25, 16, 9, 4, ...... 12. Numer al indicat es t he places bet ween let t er s. 13. Pat t er n is combinat ion of t wo ser ies 1 5 A E, 6F 10J, 11K 15O, 16P20T, ...... and 1, 4, 9,16, ...... 14. Pat t er n is combinat ion of t wo ser ies 3 CX 3, 6FU 6, 9I R9, 12L O12, ... and 4, 9, 16, 25, ... 15. Pat t er n is A3B, C7D, E11F, G15H , I 19J, K 23L , ....

2

Analogy

CHAPTER

‘Analogy ' means ‘cor r espondence'. I n quest ions based on analogy, a par t icular r elat ionship is given and anot her similar r elat ionship has t o be ident ified fr om t he given alt er nat ives. I n Analogy, r elat ionship bet ween t wo given wor ds is est ablished and t hen applied t o ot her wor ds. The t ype of r elationship may ver y, so while at tempt ing such questions fir st st ep is t o identify the type of r elationship. Analogy deals wit h t he cor r espondence of a par t icular r elat ion. I n t his chapt er we have given a r elat ion and have t o choose t he similar r elat ion wit h given choices. Such pr oblems can be expr essed eit her simply or in r at io-pr opor t ion for m. KI N DS OF RELATI ON SH I PS WI TH EXAM PLES 1. I nst rument and M easurement 

Ther momet er : Temper at ur e (Ther momet er is an inst r ument used t o measur e t emper at ur e).



Bar omet er : Pr essur e



Anemomet er : Wind vane



Odomet er : Speed



Scale : L engt h



Balance : M ass



Sphygmomanomet er : Blood Pr essur e



Rain Gauge : Rain



H ygr omet er : H umidit y



Ammet er : Cur r ent



Scr ew Gauge : Thickness



Seismogr aph : Ear t hquakes



Taseomet er : St r ains

2. Quant it y and U nit 

M ass : K ilogr am (K ilogr am is unit of mass).



L engt h : M et r e



For ce : Newt on



Ener gy : Joule



Resist ance : Ohm



Volume : L it r e



Angle : Radians



Time : Seconds



Pot ent ial : Volt



Wor k : Joule



Cur r ent : Amper e



L uminosit y : Candela



Pr essur e : Pascal



Ar ea : H ect ar e



Temper at ur e : Degr ees



Power : Wat t



Conduct ivit y : M ho



M agnet ic field : Oer st ed

3. I ndividual and Gr oup  Soldier s : Ar my (Gr oup of soldier s is called ar my).  Flower s : Bouquet  Singer : Chor us  Fish : Shoal  Rider s : Cavalcade  M an : Cr owd  Nomads : H or de

     

Cat t le : H er d Gr apes : Bunch Ar t ist : Tr oupe Sheep : Flock Bees : Swar m Sailor s : Cr ew

2.2

Analogy

4. Animal and Young one  Cow : Calf (Calf is young one of cow)  Cat : K it t en  But t er fly : Cat er pillar  Dog : Puppy  L ion : Cub  M an : Child 5. M ale and Female  H or se : M ar e (M ar e is female hor se).  St ag : Doe  L ion : L ioness  Dr one : Bee  Nephew : Niece 6. I ndividual and Class  M onkey : M ammal (M onkey belongs t o class of mammal).  Snake : Rept ile  Whale : M ammal  L izar d : Rept ile 7. I ndividual and D welling Place  Dog : K ennel (Dog lives in a kennel).  Cat t le : Shed  Poult r y : Far m  Fish : Aquar ium  H or se : St able 8. St udy and Topic  Or nit hology : Bir ds (Or nit hology is st udy of bir ds).  Onomat ology : Names  Et hnology : H uman Races  H er pet ology : Amphibians  Ast r ology : Fut ur e  Palaeogr aphy : Wr it ings  Semant ics : L anguage  Concology : Shells  Cr aniology : Skull  Ent omology : I nsect s 9. Tool and Act ion  Needle : Sew (Needle is used for sewing).  Swor d : Slaught er  Filt er : Pur ify  Pen : Wr it e  Spoon : Feed  Gun : Shoot  Chisel : Car ve  Axe : Gr ind  Spade : Dig

    

    

   

   

          

       

H or se : Pony Sheep : L amb I nsect : L ar va H en : Chicken Duck : Duckling

Dog : Bit ch Son : Daught er Sor cer er : Sor cer ess Gent leman : L ady Tiger : Tigr ess M an : M ammal Ost r ich : Bir d But t er fly : I nsect Rat : Rodent

Bee : Apiar y L ion : Den M onk : M onast er y Bir ds : Aviar y

Seismology : Ear t hquakes Bot any : Plant s Occult ism : Super nat ur al Ont ology : Realit y Pat hology : Diseases Ant hr opology : M an I cht hyology : Fishes Nephr ology : K idney H aemat ology : Blood M ycology : Fungi Zoology : Animals K nife : Cut M at t ock : Dig St eer ing : Dr ive Spanner : Gr ip M icr oscope : M agnify Shovel : Scoop Oar : Row Auger : Bor e

Analogy

10.            

Worker and Working Place Chef : K it chen (Chef wor ks in a kit chen). War r ior : Bat t lefield Sailor : Ship Beaut ician : Par lour Act or : St age L awyer : Cour t Teacher : School Cler k : Office Dr iver : Cabin Paint er : Galler y Wor ker : Fact or y Gambler : Casino

11.Wor ker and Tool  War r ior : Swor d (Swor d is t ool used by a war r ior ).  Tailor : Needle  Soldier : Gun  Chef : K nife  Far mer : Plough 12.         

13.          

14.    

Wor ker and Pr oduct M ason : Wall (M ason builds a wall). Teacher : Educat ion Chef : Food Judge : Just ice Chor eogr apher : Ballet Pr oducer : Film Ar chit ect : Design Tailor : Clot hes Dr amat ist : Play Product and Raw M at er ial Pr ism : Glass (Pr ism is made of glass). Clot h : Fibr e Road : Asphalt Book : Paper Sack : Jut e Omelet t e : Egg Jeweller y : Gold L inen : Flax Oil : Seed Paper : Pulp Part and Whole Relat ionship Pen : Nib (Nib is a par t of a pen). H ouse : K it chen Class : St udent Aer oplane : Cockpit

          

     

       

         

   

Far mer : Field Engineer : Sit e Pilot : Cockpit Ar t ist : Theat r e M echanic : Gar age Scient ist : L abor at or y Doct or : H ospit al Ser vant : H ouse Gr ocer : Shop Wait er : Rest aur ant Umpir e : Pit ch

Car pent er : Saw Woodcut t er : Axe L abour er : Spade Sculpt or : Chisel Doct or : St et hoscope Aut hor : Pen Far mer : Cr op H unt er : Pr ey Car pent er : Fur nit ur e Aut hor : Book Goldsmit h : Or nament s But cher : M eat Cobbler : Shoes Poet : Poem

But t er : M ilk Wall : Br ick Fur nit ur e : Wood Shoes : L eat her Pullover : Wool M et al : Or e Rubber : L at ex Jagger y : Sugar cane Wine : Gr apes Fabr ic : Yar n Pencil : L ead Fan : Blade Room : Window Book : Chapt er

2.3

2.4

Analogy

15. Wor d and I nt ensit y  Anger : Rage (Rage is of higher int ensit y t han Anger ).  K indle : Bur n  Quar r el : War  Famous : Renowned  Refuse : Deny 16. Wor d and Synonym  Abode : Dwelling (Abode means almost same as Dwelling).  Ban : Pr ohibit ion  Vacant : Empt y  Dear t h : Scar cit y  Sedat e : Calm  H ouse : H ome  Pr esage : Pr edict  Flaw : Defect  Fallacy : I llusion  M end : Repair  Pr esume : Assume 17. Word and Ant onym  At t ack : Defend (Defend means opposit e of At t ack).  Cr uel : K ind  Fr esh : St ale  I nit ial : Final  Chaos : Peace  Gr adual : Abr upt  Robust : Weak  Deep : Shallow  Affir m : Deny  L et har gy : Aler t ness

   

         

        

Wish : Desir e Sink : Dr own Er r or : Blunder Unhappy : Sad

Blend : M ix Assign : Allot Abduct : K idnap Dissipat e : Squander Br im : Edge Solicit : Request H aught y : Pr oud Fier ce : Violent Subst it ut e : Replace Alight : Descend

Advance : Ret r eat Best : Wor st I gnor e : Not ice Condense : Expand Cr eat e : Dest r oy Sink : Float Gent le : H ar sh Cor dial : H ost ile M our n : Rejoice

2. Choosing Analogous Pair I n t his t ype of quest ions, a pair of wor ds is given, followed by four pair s of wor ds as alt er nat ives and we ar e r equir ed t o choose t he pair in which wor ds bear t he same r elat ionship t o each ot her as t he wor ds of t he given pair bear. Example. Sink : Float : : ? (a) Br im : Edge (b) M end : Repair (c) Gent le : H ar sh (d) Flow : defect (e) Solicit : Request Ans. (b) Just as Sink is ant onym of float , so also gent le is opposit e of har sh. Example. Gossip : Exagger at ion (a) Gir l : Sist er

(b) Dish : Food

(c) Climb : Tr ek

(d) Smoke : Fir e

Ans. (d) Fir e expr esses a height ened magnit ude of smoke just as exagger at ion does of gossip. 3. Simple Analogy I t is the r elationship between words in these type of questions is easily per ceptible from general experiences

Analogy

2.5

of day-t o-day life and discer nible fr om t he given for m, pat t er n or combinat ions, yet aspir ing candidat es would appr eciat e t he fact t hat t he pr esent day life has become much mor e complex t han it was ever befor e. Wit h development of Science and Technology and incr ease in compet it ion, life is t hr owing bigger and higher chal lenges. One shoul d have adequat e k nowl edge and i nfor mat ion t o deal wi t h mor e complicat ed combinations. Example. M ouse is r elat ed t o M onit or as ...?... is r elat ed t o canvas. (a) Cat

(b) Br ush

(c) UPS

(d) Paint er

Ans. (b) M ouse put s t he cur sor wher e t he comput er oper at or want s t o t ype. The br ush is put wher e t he paint er want s t o wor k on a canvas. 4. Analogy of Words and Expression I n t his t ype of quest ions, a gr oup of t hr ee wor ds is given, followed by four ot her wor ds as alt er nat ives. The candidat e is r equir ed t o choose t he alt er nat ive, which is similar t o t he given t hr ee wor ds. Example. Sit ar : Guit ar : Tanpur a (a) Tr umpet (c) H ar monium (e) Flut e

(b) Violin (d) Mridanga

Ans. (b) Sit ar, Guit ar and Tanpur a ar e all st r ing inst r ument s. Violin is also a st r ing inst r ument . 5. D et ect ing Analogies I n t his t ype of quest ions, it is r equir ed t o t r ace out hidden analogy or common char act er ist ic among t he given wor ds or t o choose t he wor d which possesses same char act er ist ic as t he given wor d. Example. Judo : K ar at e : Taekwando (a) They ar e names of mar t ial ar t s. (b) They can be per for med by obese per sons. (c) They ar e per for med on st age. (d) They ar e impor t ant it ems of Asian Games. (e) They have or igin in K er ala Ans. (a) Judo, K ar at e and Taekwando ar e mar t ial ar t s and (a) is most suit able descr ipt ion for all t he t hr ee. Example : Java : Or acle : Cobol (a) They ar e t he names of Gr eek alphabet s. (b) They ar e Comput er languages. (c) They ar e t he names of islands. (d) They all ar e M ult i-Nat ional Companies (M NC’s) The r ight opt ion is (b) as it cor r ect ly expr esses t he analogy bet ween t he given wor ds. 6. T hr ee word Analogy I n t his t ype of quest ions, a gr oup of t hr ee int er -r elat ed wor ds is given and we ar e r equir ed t o t r ace out t he r elat ionship among t hese t hr ee wor ds and choose anot her gr oup wit h similar analogy, fr om among t he alt er nat ives pr ovided. Example. Pen : Pencil : I nk (a) Or ange : Banana : Juice (b) Table : Chair : Wood (c) Cow : M ilk : Cur d (d) Fish : Shar k : Wat er (e) Car : Engine : Cir cle Ans. (a) Pen cont ai ns ink and pencil belongs t o t he same cat egor y as pen i.e. st at ioner y. Similar ly, or ange cont ains juice and banana belongs t o t he same cat egor y as or ange, i.e. fr uit s.

2.6

Analogy

Example : Comput er : Floppy : I nfor mat ion (a) Camer a : Film : Pict ur e (b) DVD : CD : M usic (c) Book : Chapt er : K nowledge (d) Pen : Riffle : Essay. Ans. Floppy contains I nformation and is used in a Comput er, Similar ly, CD cont ains music and is used in t he DVD. H ence (b) bear s a simil ar anal ogy as t he wor ds gi ven i n t he question. Other opt ions ar e slight ly differ ent. I n (a), film cont ains t he Pict ur e but is not used again in t he Camer a. I t is used only t he fir st t ime when it is blank . I n (c) Chapt er cont ains knowledge but is not separ at e fr om t he book as Floppy is fr om a Comput er. I n (d),Riffle is used in a ball point Pen but does not cont ain Essay as such. 7. Double Analogy I n t wo wor ds indicat ed by I and I I ar e left out in each pair. Ther e ar e four alt er nat ives for eit her of t hem, mar ked (A), (B), (C), (D) for I and (P), (Q), (R), (S) for I I One of t hese opt ions in each set bear s a r elat ionship wit h t he given wor d in such a way t hat t her e is an analogy bet ween pai r s. The combi nat i on of r i ght r opt ions is indi cat ed by (a), (b), (c), (d ). The answer is to be indicated by ticking one of these f ou r com bi n at i on s. A cl ose st u dy of t h e fol l owing exer ci se woul d mak e t he concept mor e clear. 8. N umber analogy I n number analogy t he r elat ionship bet ween given number s is det ect ed and t hen applied t o t he second par t t o find missing number s.

This r elat ionship bet ween number s can be based on following pat t er ns: (1) Number can be odd/even/ pr ime number s (2) Number s can be mult iples of some number (3) Number s can be squar es/ cubes of differ ent number (4) Some number s can be added t o /subt r act ed fr om/mult iplied t o/ divided int o t he fir st number t o get t he second number (5) Second number can be t he sum/pr oduct / differ ence of t he dight is of fir st number

(6) Combinat ions of any mat hemat ical calculat ions given above can apply t o t he r elat ionship between two given number s. Types of Quest ions.

I t includes t wo t ypes of quest ions : 1. Choosing a similar ly r elat ed pair as t he given number pair on t he basis of r elat ion bet ween t he number s in each pair. Example. 582 : 194 :: 258 : ? (a) 82 (b) 86 (c) 92 (d) 58 Ans. (b) Just as 582 is r elat ed t o 194 as 582 is 3 t imes 194, in t he same way 258 is 3 t imes t he 86. Example. Find appr opr iate number for the four t h place 25 : 81 : : 36 : ? Solut ion.

25 :

81 :: 36







: 121 

52

92

62

112

H ence appr opr iat e number will be 121. 2.Choosing a number similar t o a gr oup of number s on t he basis of cer t ain common pr oper t ies t hat t hey possess. Example. Which number belongs t o given set of number s? 2, 3, 5, 7, 11, 13, 17, 19, 23...... (a) 15 (b) 9 (c) 21 (d) 29 (e) 27 Ans. (d) Given set of number s belong t o t he pr ime number s. I n t he given opt ion, only pr ime number is 29.

Analogy

2.7

PRACTI CE1EXERCI SE OBJECTI VE TYPE QU ESTI ON S I n each of t he following quest ions t he t wo wor ds given on one si de of doubl e col on i .e., :: bear a cer t ai n r elat ionship. The ot her side shows an incomplet e pair mar ked by a mi ssing wor d whi ch bear s a si mi lar r elat ionship wit h t he given wor d. Find t he missing wor d out of t he given opt ions (a) (b) (c) or (d). 1. H ill : M ount ain :: ? : Pain (a) Dist r ess (b) Discomfor t (c) Headache

(d) Fear

2. Raipur : Chhat isgar h :: ? : Ut t ar anchal (a) Nainital

(b) Ranchi

(c) Dehr adun

(d) Bhopal

3. Eye : Cat ar act :: ? : H epat it is (a) L iver

(b) Pancr eas

(c) St omach

(d) Lungs

4. Sculpt or : St at ue :: ? : Poem (a) Paint er

(b) Wr it er

(c) Singer

(d) Poet

5. Reading : K nowledge :: ? : Exper ience (a) Job (b) Wor king (c) Tr aveller (d) Tr aining I n each of the following questions, the two words given in t he salut ar y pair expr ess an explicit or implicit relationship with each other. This is followed by four pairs of words marked (a), (b), (c) and (d). The two words in one of these pair s bear s a r elationship similar to the words in t he main pair. Choose that analogous pair. 6. Boxing : Rink (a) Badmint on : H all (c) Swimming : Wat er

(b) Cr icket : Pit ch (d) Cr icket : Gr ound

7. H aemoglobin : Blood (a) Chlor ophyll : L eaf (c) M icr ot in : Lymph

(b) Chlor oplasm : Cell (d) Bile : H or mone

8. H ot : St eaming (a) Wat er : Flood (c) Clot h : Dr ess

(b) Cold : Chilly (d) Dr y : Famine

9. M onolit h : Rock (a) Cont inent : Ocean (c) Gr ain : Sand

(b) Tor : L ea (d) Cat ar act : Wat er fall

10. Theft : Confess (a) Fault : Admit (c) M ist ake : Agr ee

(b) Fight : Abet (d) H ar m : Do

Choose appr opr iat e wor ds fr om four choice given . 11. M ir r or is t o Reflect ion as Wat er is t o (a) Conduction (b) Disper sion (c) I mmer sion (d) Refr act ion 12. L ife is t o Deat h as H ope is t o (a) Sad (b) Despair (c) Pain (d) Cry 13. Black is t o Coal as H ar d is t o : (a) Rock (b) Br ick (c) Board (d) St one 14. Taj M ahal is t o L ove as Jallianwala Bagh is t o (a) Amr itsar (b) Mar tyr dom (c) War (d) Punjab 15. Eye is t o Tear as M out h is t o (a) Teet h (b) Tongue (c) Lips (d) Saliva

LEVEL-1 1. K ings is t o H ist or y as Rocks is t o (a) Geomet r y (b) Cutler y (c) Chemist r y (d) Geogr aphy 2. Gr apes is t o Wine as Wheat is t o (a) Flake (b) Barley (c) Bread (d) Cake 3. Cub is t o Tiger as Fawn is t o (a) Stag (b) M onkey (c) Ass (d) Sheep 4. Wat er is t o Ocean as Snow is t o (a) Peaks (b) Hail (c) Glacier (d) Mountain 5. Writer is to Reader as Producer is to (a) Seller (b) Consumer (c) Cr eator (d) Cont r actor

Fir st t wo wor ds ar e r elat ed in a par t icular manner. Select t he wor d fr om given choices which is r elat ed t o t hir d one in same manner. 6. Assumpt ion : L ogic :: Guess : ? (a) Estimation (b) Analysis (c) Basis (d) Study 7. Or chest r a : Conduct or :: Team : ? (a) Captain (b) Senior member (c) Manager (d) Coach 8. Judges : Bench :: Teacher s : ? (a) School (b) College (c) Team (d) Faculty

2.8

Analogy

9. Solidify : I ce :: L iquefy : ? (a) Vapour

(b) Wat er

(c) St eam

(d) M er cur y

10. St ar : St ellar :: L ight : ? (a) Lightening

(b) Lighting

(c) L ighten

(d) Lightning.

11. Speech : Sight : : Dumb : ? (a) Eyes

(b) Mouth

(c) Tongue

(d) Blind

containing three words. Only one of these options bears an analogy similar to the main words. Find t he cor r ect option. 4. St ar : Const ellat ion : Galaxy (a) Flower : Gar land : Bouquet (b) M oon : Sat ellit e : Eclipse (c) Sun : Ear t h : Venus (d) Clot h : Design : Dr ess . 5. M obile : Call : M essage

12. Tast e : Tongue : : L ight : ?

(a) FAX : Paper : Confir mat ion

(a) Film (b) Camer a (c) Eye (d) Brain 13. Red For t : Delhi : : Taj M ahal : ? (a) Agra (b) Kanpur (c) Haryana (d) Punjab 14.Pigeon : Peace : : Whit e Flag : ?

(b) Pencil : Wr it e : Dr awing

(a) Victor y (b) Sur r ender (c) Peace (d) L iber t y 15. Pear l : Necklace : : Flower : ? (a) Gar den (b) Bouquet (c) Pet al (d) Plant

(c) Calculat or : Abacus : Number s (d) Camer a : Pict ur e : L ense. 6. Count r y : St at e : Capit al (a) Shop : Cust omer : Seller (b) Cit y : Colony : Flat (c) H ospit al : Pat ient : M edicine (d) Office : Cabin : Boss. 7. Fact or y : M anager : Wor ker (a) School : Teacher : St udent (b) H ouse : Owner : Ser vant

LEVEL-2

(c) M ar ket : M all : Agent

Given below ar e t hr ee wor ds in each quest ion which ar e analogous t o one anot her in some way. Det ect t he analogy under lying t hem by choosing t he r ight opt ion out of (a), (b), (c) and (d) :

(d) Cour t : Case : L awyer.

1. K elvin : Celcius : Fahr enheit (a) They ar e t ypes of t her mo-met er. (b) They ar e SI unit s of t emper at ur e. (c) They ar e scales of t emper -at ur e. (d) They are used to measure body temperatur e. 2. Sahar a : M odiluft : I ndian Air lines

Directions : H er e t wo wor ds indicat ed by I & I I have been left out for which four alternat ive choices is given for each. Read wit h t he cor r ect wor ds t her e is some r elat ionship bet ween t he t wo wor ds t o t he left of t he sign:: and same r elat ionship exist s bet ween t he t wo wor ds t o t he r ight of ::. The cor r ect combinat ion is given as (a), (b), (c) & (d). Find t he cor rect combinat ion in each case. 8. I : Wheat :: Br ick : I I I . (A) Weed

(B) Field

(C) Bread

(D) Fanner

(b) They all r un on losses.

I I . (P) Building

(Q) Mason

(c) They ar e local Air lines.

(R) Clay

(a) They r epr esent joint vent ur es in air lines.

(d) They ar e air lines cont r olled by St at e Govt . 3. Palace : Thr one : Empir e (a) They ar e inher it ed by t he king. (b) They ar e a bone of cont ent ion. (c) They ar e places cont r olled by t he king. (d) People aim t o possess t hem.

I n each of the following questions three words are given which ar e int er-connect ed in some way. Below t hem, four opt ions mar ked (a), (b), (c) and (d) ar e given, each

(S) Kiln

(a) AS

(b) BR

(c) CQ

(d) DP

9. I : Night :: Sun : I I I . (A) Birds

(B) Fight

(C) Star s

(D) Dawn

I I . (P) Dusk

(Q) N oon

(R) Bright

(S) Day

(a) AP

(b) CD

(c) BQ

(d) CP

Analogy

Choose appr opr iat e number for four t h place. 12. 582 : 194 : : 2.58 : ?

10. I : Squar e :: Ar c : I I I . (A) L i ne

(B) Diagonal

(C) Rectangle I I . (P) Chord

(D) Per imet er

(a) 82

(b) 86

(Q) Cir cle

(c) 92

(d) 58

(R) Diamet er

2.9

(S) Cir cumfer ena

13. 108 : 27 : : 776 : ?

(a) BR

(b) CP

(a) 162

(b) 194

(c) DS

(d) AQ

(c) 137

(d) 147

11. I : Disease :: Psychiat r ist : I I

14. 36 : 18 : : 72 : ?

I . (A) Charlatan

(B) Sur geon

(a) 164

(b) 134

(C) Paediatrician I I . (P) Maladjustment (Q) Tr ut h (R) Medicine (S) Cur e (a) BD (c) AQ

(D) Doct or

(c) 94

(d) 14

15. 36 : 64 : : 81 : ? (a) 25

(b) 16

(c) 121

(d) 49

(b) CQ (d) DP

AN SWE RS OBJECTI VE TYPE QU ESTI ON S 1. (b)

2. (c)

3. (a)

4. (d)

5. (b)

11. (d)

12. (b)

13. (a)

14. (b)

15. (d)

6. (d)

7. (a)

8. (b)

9. (d)

10. (a)

7. (a)

8. (d)

9. (b)

10. (b)

7. (b)

8. (c)

9. (b)

10. (d)

LEVEL-1 1. (d)

2. (c)

3. (a)

4. (c)

5. (b)

11. (d)

12. (c)

13. (a)

14. (b)

15. (b)

6. (a)

LEVEL-2 1. (c)

2. (c)

3. (c)

4. (a)

5. (a)

11. (d)

12. (b)

13. (b)

14. (d)

15. (c)

6. (b)

E XPL AN AT I ON S OBJECTI VE TYPE QU ESTI ON S

10. Fault is admit t ed just as t heft is confessed.

1. Second is t he lar ger for m of t he fir st .

11. I n wat er light r ay r efr act s.

2. Fir st is t he capit al of t he second which is a st at e.

12. Ant onyms of hope is despair.

3. Second is a disease affect ing the fir st which is an or gan.

13. Rock is har d.

4. Second is a cr eat ion of t he fir st .

15. M out h pr oduced Saliva.

14. Mar tyr dom.

5. Fir st br ings t he second.

LEVEL-1

6. Boxing is played in a r ink while cr icket is played in a gr ound.

1. Geogr aphy.

7. H aem ogl obi n gi v es t h e bl ood r ed col ou r , chlor ophyll gives t he leaf gr een colour.

3. Fir st is young one of t he second.

8. St eaming is ext r emely hot as chilly is ext r emely cold. 9. A cat ar act is a lar ge wat er fall just as monolit h is a lar ge r ock.

2. Second is made of fr om fir st . 4. Glacier. 5. Pr oducer pr oduce for t he consumer. 6. Guess is made on Est imat ion as Assumpt ion is made on logic. 7. Second leads t he fir st .

2.10

Analogy

8. Second is a collect ion of t he fir st . 9. Second is pr oduced by the fir st which is a pr ocess. 10. Second is adject ive t o t he fir st . 11. A Dumb who cannot speak wher e blind cannot see i.e. sight pr oblem. 12. As t ast e is r elat ed t o t ongue, light is t o eye. 13. Taj M ahal is in Agr a. 14. Whit e Flag is sign of sur r ender. 15. Flower s put in bouguet .

LEVEL-2 4. A col l ect i on of t he fi r st mak es t he second. A collect ion of second makes t he t hir d. 5. Fir st can be used for bot h t he Second and t he Thir d.

6. Fir st cont ains t he Second and Second cont ains t he Thir d. 7. Owner cont r ols t he H ouse as well as t he Ser vant just as t he M anager does t he Fact or y and t he Wor k er. 8. Second is used t o make fir st . 9. Fir st is found is second. 11. Fir st deals wit h second. 12. 258  3 = 86. 13. 776  4 = 194. 14. Second number i s pr oduct of digit of t he fi r st number 15. (9 + 2)2 = 121.



3

Odd One Out

CHAPTER

‘Classificat ion' means ar r angement of given it ems on t he basis of some common char act er. I n t his t est , a gr oup of cer t ain it ems ar e given, out of which some ar e similar in some manner and one is differ ent fr om t he r est . We ar e r equir ed t o choose t his one it em which does not fit int o t he given gr oup. TYPES OF CLASSI FI CATI ON QU ESTI ON S. 1. ODD ON E OU T-WORDS. I n t his t ype of quest ions, four (or five) wor ds ar e given out of which one is dissimilar t o t he ot her s and st udent s have t o find t hat odd one fr om t he given set . Example. Choose t he wor d which is least like t he ot her wor ds in t he gr oup. I. (a) Chair (b) Cupboard (c) Table (d) Paper weight Ans. (b) H er e, all except paper weight ar e fur nit ur es. (a) H at (b) Bag (c) Pur se (d) Basket Ans (a) H at is used on head so odd one out is H at . I I I . (a) T (b) Z (c) Q (d ) H Ans. (c) Q is 17t h let t er of t he alphabet and all t hr ee opt ions ar e on even places wher e 17 is an odd number. II.

2. ODD ON E OU T-WORD GROU P OR PAI R OF WORDS Classificat ion of gr oup of wor ds, let t er s or number s is not much differ ent fr om single wor ds, let t er s or number s. H er e, we have t o obser ve t he t ype of r elat ion bet ween t wo r elat ed wor ds and find out t he pair which does not follow t he r elat ionship pat t er n of t he ot her gr oups. The r elat ed wor ds can be opposi t e or ident ical in meaning or nat ur e, or can var y in t heir degr ees of r epr esent at ion or can be r hyming wor ds. Example. Choose t he odd pair of wor ds. (a) L oom : Clot h (b) Table : Dr awer (c) Book : Page (d) Car : Wheel Ans. (a) I n all ot her pair s, second is par t of t he fir st . 3. CH OOSI N G ODD N U M ERAL I n t his t ype of quest ions, cer t ain number s ar e given, out of which all except one ar e alike in some manner while one is differ ent and t his number is t o be chosen as t he answer. Example. Choose t he number which is differ ent fr om ot her s in t he gr oup. (a) 49

(b) 64

(c) 121

(d) 156

Ans. (d) Each of t he number s except 156 is complet e squar e. 4. ODD ON E OU T-N U M ERAL PAI R/GROU P I n this type of classification, differ ent number s ar e given as options. These number s have some commonness; except one which is t he odd one. One has t o ident ify t he similar it y and t hen find t he odd one out as answer opt ion. The number can be odd/even/ consecut ive pr ime number s, mult iple of some number, single squar e or cubes of differ ent number s, plus/minus of some ot her number or combinat ions of any mat hemat ical calculation. Example. Choose t he numer al pair /gr oup which is differ ent fr om ot her s. (a) 83 – 75 (b) 58 – 50 (c) 49 – 42 (d) 25 – 17

3.2

Odd One Out

Ans. (c) I n each of t he pair s fir st number is eight mor e t han t he second. 5. ODD ON E OU T-LETTER GROU P I n t his t ype of quest ions, usually five gr oups of let t er s ar e given. Four of t hem ar e similar t o each ot her in some manner while one is differ ent and we have t o find it as t he answer. Example. Choose t he gr oup of let t er s which is differ ent fr om ot her s. (a) SU

(b) PN

(c) I K

(d) BD

Ans. (b) All ot her s ar e t wo alt er nat e let t er s.

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S Directions (Q. 1 – 10). Find out t he odd one fr om given four wor ds. 1. (a) Book (c) Pencil

(b) Paper (d) Pen

2. (a) Star (c) M oon

(b) Sun (d) Univer se

3. (a) Pineapple (c) Malta

(b) Or ange (d) Banana

4. (a) Kidney (c) Eye

(b) H ear t (d) Lung

5. (a) Engineer (c) Car penter

(b) Blacksmith (d) Bar ber

6. (a) Newt on (c) Mar coni

(b) Faraday (d) Beethovan

7. (a) I nch (c) Yard

(b) Foot (d) Quar t

8. (a) Comput er (c) Radio

(b) Televisi on (d) X-ray

9. (a) Dollar (c) Pound

(b) Fr anc (d) Ounce

10. (a) Panipat (c) Plassey

(b) Haldighati (d) Sarnath

Directions (Q. 11-15). Choose t he odd pair of wor ds.

15. (a) Car — Engine (c) I nk— Pen

(b) Ticket— Tr ain (d) St amp— Lett er

LEVEL-1 Directions (Q. 1-3). Choose t he odd pair of wor ds. 1. (a) MeetingChairman (b) Tomat oPot at o (c) Ar my Gener al (d) CracheI nfant 2. (a) MangoFr uit (c) Shir t Dr ess

(b) Boy Gir l (d) TableFur nit ur e

3. (a) GoodBett er (b) TepidHot (c) Gainr Pr ofit (d) Whisper Shout Directions. (Q. 4-13). I n each of the following questions, there are four options. The numbers, in these options, are alike in certain manner. Only one number does not fit in. Choose the one which is different from t he r est . 4. (a) 64 (c) 121

(b) 96 (d) 144

5. (a) 9 (c) 12

(b) 7 (d) 18

6. (a) 43 (c) 63

(b) 53 (d) 73

7. (a) 3730 (c) 5568

(b) 6820 (d) 4604

8. (a) 24 (c) 54

(b) 90 (d) 36

9. (a) 3215 (c) 4721

(b) 9309 (d) 2850

11. (a) Light— Heavy (c) Big— Lar ge

(b) Broad— Wide (d) Tiny— Small

12. (a) M ot her — Father (c) Mast er — Ser vant

(b) Sister — Br ot her (d) Uncle— Nephew

10. (a) 7654 (c) 9876

(b) 4567 (d) 4321

13. (a) Oil— Lamp (c) Power — M achine

(b) Water — Tap (d) Oxygen— Life

11. (a) 1472 (c) 2683

(b) 3848 (d) 4210

14. (a) Knife— Dagger (c) Car— Bus

(b) Pist ol— Gun (d) Engine— Tr ain

12. (a) 325 (c) 711

(b) 207 (d) 423

Odd One Out

13. (a) 1365 (c) 3175

(b) 5713 (d) 7531

Directions. (Q. 8-12). Choose t he gr oup of let t er s which is differ ent fr om ot her s.

LEVEL-2

8. (a) SU

Directions. (Q. 1-7). I n each of the following questions, there are four options. The numbers, in these options, ar e alike in cer t ain manner. Only one number does not fit in. Choose t he one which is differ ent fr om t he r est . 1. (a) 13 – 21

(b) 15 – 23

(c) 16 – 24

(d) 19 – 27

2. (a) 12 – 144 (c) 15 – 180

(b) 13 – 156 (d) 16 – 176

3. (a) 13 – 31 (c) 16 – 61

(b) 45 – 54 (d) 71 – 88

4. (a) 6 – 16 (c) 10 – 27

(b) 7 – 19 (d) 11 – 31

5. (a) 95 – 82 (c) 55 – 42

(b) 69 – 56 (d) 48 – 34

6. (a) 67 – 19 (c) 41 – 19

(b) 71 – 11 (d) 61 – 15

7. (a) 3 – 5 (c) 6 – 2

(b) 5 – 3 (d) 7 – 3

3.3

(b) PN

(c) IK

(d) BD

9. (a) H GF

(b) XWV

(c) NM L

(d) OPQ

10. (a) PSVX

(b) JMPS

(c) ORUX

(d) CFI L

11. (a) TRQPS

(b) KJHMF

(c) FCGDE

(d) KHGJI

12. (a) JMG

(b) PSM

(c) WZT (d) EIB Directions. (Q.13– 15). Ther e ar e four wor ds wit h t he let t er s jumbled up. Thr ee of t hem ar e alike. Find t he odd one out . 13. (a) NI TK (b) TI K (c) TI H (d) I TS 14. (a) LI TYAQU (c) TEAUBY

(b) TI TYUANQ (d) TEDUCAED

15. (a) FI WE (c) BUSHDNA

(b) FLAMEE (d) OMAWN

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (a)

2. (d)

3. (d)

4. (b)

5. (d)

11. (a)

12. (c)

13. (b)

14. (d)

15. (a)

6. (d)

7. (d)

8. (d)

9. (d)

10. (d)

7. (d)

8. (b)

9. (b)

10. (b)

7. (d)

8. (b)

9. (d)

10. (b)

LEVEL-1 1. (d)

2. (b)

3. (c)

11. (c)

12. (a)

13. (a)

4. (b)

5. (b)

6. (c)

LEVEL-2 1. (c)

2. (d)

3. (d)

4. (d)

5. (d)

11. (b)

12. (b)

13. (b)

14. (b)

15. (c)

6. (d)

EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S

7. Ot her s ar e t he unit of measur ing dist ances.

1. Ot her s ar e st at ionar y it ems.

8. Ot her s ar e elect r onic devices.

2. Ot her s for m a par t of t he univer se.

9. Ot her s ar e cur r encies.

3. Ot her s ar e juicy fr uit s.

10. Ot her s ar e famous bat t le field.

4. Ot her s pr esent in body in pair.

11. Ot her s ar e synonyms.

5. Ot her s want r aw mat er ials t o wor k.

12. Ot her s have familiar r elat ion.

6. Ot her s ar e sci ent i st wher e Beet hovan was a musician.

13. Fir st par t of pair ar e necessit ies of second.

3.4

Odd One Out

14. Others are of the same purpose but different in size.

6. Ot her pair s ar e of odd and pr ime number s.

15. I n ot her s second par t of t he pair holds fir st par t .

7. Sum of t wo number s is 8.

LEVEL-1 1. Ot her pair s second is t he head of fir st .

8. Ot her s ar e t wo alt er nat e let t er s. 9. Ot her s ar e in r ever se or der.

2. I n ot her pair s second is t he class t o which fir st belongs.

10. I n ot her gr oups, each l et t er m oves 3 st eps for war d.

3. Ot her s ar e differ ent in degr ees of compar ison t hough having same meaning.

11. Ot her gr oup consist s of five consecut ive let t er s but not in or der.

4. Ot her s ar e complet e squar es.

12. I n other s, places between the alphabets ar e same.

5. Ot her s ar e divisible by 3.

13. Wr it ing in t he meaningful way we get

6. Ot her s ar e pr ime number s. 7. Ot her s have t wo digit s same. 8. Ot her s have sum of bot h t he digit s . 9. I n ot her s no digit is r epeat ed. 10. I n ot her gr oups number is in descending or der. 11. I n all ot her t er ms digit 4 is used.

(a) KNI T (b) KI T (c) HI T (d) SI T So, odd one out is K I T. 14. Rear r anging let t er of t he given wor ds, we get (a) Quality

(b) Quantity

12. Rest all ar e divisible by 3.

(c) Beauty

(d) Educated

13. I n all the r est numbers, the sum of first and four th digits and sum of second, thir d digit is 8.

All thr ee options except (b) ar e used for explaining t he per son.

LEVEL-2

15. Rear r anging let t er s of t he given wor ds, we get

1. Ot her s pair s cont ain odd number s.

(a) WI FE

2. Ot her s pair s cont ain a number and it s pr oduct wit h 12.

(c) HUSBAND

3. Ot her s for med by t he int er change of number s. 4. Ot her pair s ar e copr imes. 5. Rest all pair s have differ ence of 13.

(b) FEMALE (d) WOMAN So, odd one out is H usband.



4

Coding Decoding

CHAPTER

A CODE is a ‘syst em of signals'. Ther efor e, coding is a met hod of t r ansmit t ing a message bet ween t he sender and t he r eceiver wit hout a t hir d per son knowing it . Coding and Decoding test is set up to judge t he candidat e's ability to decipher t he r ule that codes a par t icular wor d/ message and br eak t he code t o decipher t he message. TYPES OF CODI N G DECODI N G QU ESTI ON S

This t ype of quest ion involves following : 1. L et t er Coding 2. Number Coding 3. Substitution 4. M ixed lett er Coding 5. M ixed number Coding 6. Decipher ing individual let t er codes by Analysis 7. Gr oup Coding 8. Decoding 1. LETTER CODI N G. I n t hese quest ions, r eal alphabet s in a wor d ar e r eplaced by cer t ain ot her alphabet s accor ding t o a specific r ule t o for m it s code and we ar e r equir ed t o det ect t he common r ule and answer t he quest ions accor dingly.

L et t er coding is a secr et ive language used t o change r epr esent at ion of t he act ual t er m/wor d/value. Coded language can be fr amed by following : (i ) M oving let t er s one or mor e st eps for war d or backwar d (ii ) Subst it ut ing number s for let t er s and vice-ver sa (iii ) Wr it ing let t er s of t he given wor d in r ever se or der in par t of in whole (iv) Replacing let t er s in t heir nat ur al ser ies by t he same posit ioned let t er in t heir r ever se ser ies. Alphabet s in N atural series A B C D E F G H

 1st

I

J

K

L M N O P Q R S T U V W X Y











5th

10th

15th

20th

25th

Z

Alphabet s in Reverse series Z Y X W V U T

 1st

S R Q P O N M L

K

J

I

H G F

E D C B A











5th

10th

15th

20th

25th

4.2

Coding Decoding

E xamples 1. I f FACE is coded as GBDF; t hen how BADE will be coded? Solution : Wor d is coded by moving t he let t er s one st ep for war d. F A C E

G B D F

Similarly,

B A D E

C B E F

+1

+1

+1

+1

+1

+1

+1

+1

2. I f TAP is coded as SZO, t hen how is FREEZE coded? Solution : Each let t er in t he wor d TAP is moved one st ep backwar d t o obt ain t he cor r esponding let t er of t he code.

Thus, in FREEZE, F will be coded as E, R as Q, E as D and Z as Y. So, t he code becomes EQDDYD. E xamples. 1. I f DEL H I is coded as FGNJK , how is M ADRAS coded ? Sol ut i on:

4

5

12

8

9

D

E

L

H

I

+2

+2

M

A

D

R

A

S

|

|

|

|

|

|

6

7

14 10 11













F

G

N

O

C

F

T

C

U

J

K

2. I f BH OPAL is wr it t en as EERM DI , how is POWDER wr it t en ? Sol ut i on: 2

8

15

16

1

12

16

15

B +3 E 5

H – 3 E 5

O +3 R 18

P – 3 M 13

A +3 D 4

L – 3 I 9

P O +3 – 3

W D +3 – 3

E R +3 – 3

S

Z

H

L

23

4

5

A

18

O

3 I f AH M EDABAD is wr it t en as ASM VDZBZD, how is ASSASSI NATI ON coded ? Sol ut i on: 1

8

13

5

4

1

2

1

4

A

H

M

E

D

A

B

A

D



(+)



(+)



(+)



(+)



A

S

M

V

D

Z

B

Z

D

1

19

13

22

4

26

2

26

4









27

27

27

27

Coding Decoding

4.3

I t follows t hat , alt er nat e let t er ar e coded as follows : A B C D E F G H

I

J K L M N O P Q R S T U V W X Y Z

Z Y X W V U T S R Q P O N M L K J So ;

A S S A S S I N A T

I

I H G F E D C B A

O N

                          A H S Z S H

I M A G I

L N

 AH SZSH I M AGI L N

4. I f ZXV st ands for ACE; what will YZW st and for ? Solution : Letter s of the wor d ACE ar e decipher ed by decoding ZXV.The letter ar e decoded by substituting t heir r epr esent ed let t er s in t he nat ur al or der. (i.e., ‘Z’ is 1st in t he r ever se ser ies and ‘A’ is 1st in nat ur al ser ies) Z X V  let t er s in r ever se ser ies A C E  let t er s is nat ur al ser ies    1st 3rd 5th  posit ion of let t er s 2. N U M BER CODI N G. I n t he t ype of coding, digits and it s coded let t er s or vice ver sa ar e alr eady given. We have t o find out answer t o t he given quest ion t allying t he given code. Examples 1. I f CH I NA is wr it t en as 38126 and NEPAL is 25769, how is PL AI NE is coded ? Solution : Clear ly Alphabet s ar e coded as shown :

C H I N 3 8 1 2 H ence, PL AI NE is coded as 796125.

A 6

E 5

P 7

L 9

2. I f RESULT is coded as 798206, t hen how L ET will be coded ? Solution : L et t er s ar e coded by number s, and t o code t he given wor d, select r espect ive coded number s. R E S 7 9 8 So, code for L ET will be

U 2

L 0

T 6

 

let t er s code

L

E

T



let t er s

0

9

6



code

3. I f Y = 50 and BAT = 46, t hen GOA = ? Solution : Y = 50, and BAT = 46

A = 2, B = 4, C = 6 . . . . . T = 40, ...... Y = 50, Z = 52 GOA = 14 + 30 + 2 = 46

I n t he given code,



4. I n t he cor r ect ly wor ked out mult iplicat ion pr oblem at t he below, each let t er r epr esent a differ ent digit s. What is t he value of B ? A A A B — — — — B B A A C — — — — A 3 B

4.4

Coding Decoding

Solution :

B+C =B:C=0 AA ×B = BB



(10 A + A ) B = 10 B + B

AB = B , A = 1 B+A =3 B + 1 = 3,  B=2 11 12 — — 22 110 — — 132 5. I n a cer t ain code language 24685 is wr itten as 33776. How is 35791 wr itten in that code ? Solutions. L et t er s ar e coded as follows : 2 4 6 8 5 +1 – 1 +1 – 1 +1 3 3 7 7 6 So, 3 5 7 9 1 +1 – 1 +1 – 1 +1 4 4 8 8 2



6. I n a cer t ain code ‘256’ means ‘you ar e good’ ; ‘637’ means ‘we ar e bad’ and ‘358’ means ‘good and bad’. Which of t he fol l owi ng r epr esent s ‘and’ in t hat code ? Solutions : Given : ‘You ar e good’  ‘256’ ...( i ) ‘We ar e bad’  ‘637’ ...(ii ) and ‘Good and bad’  ‘358’ ...(iii ) Fr om equations (i ) and (iii ), ‘good’ is coded as ‘5’. Fr om equations (ii ) and (iii ), ‘bad’ is coded as ‘3’. Fr om equat ion (iii ), ‘and’ is coded as ‘8’. 3. SU BSTI TU TI ON . I n t his t ype of quest ions, some par t icular object s are assigned code names. Then a question is asked t hat is t o be answer ed in t he code language. Example. I f cook is called but ler , but ler is called manager, manager is called t eacher, t eacher is called cler k and cler k is called pr incipal, who will t each in a class ? Solutions : A ‘t eacher ' t eaches in a class and as given ‘t eacher ' is called ‘cler k'. So, a ‘cler k' will t each in t he class. 4. M I X E D L E T T E R CO D I N G (SE N T E N CE CODI N G). I n t his t ype of quest ions, t hr ee or four complet e messages ar e given in t he coded language and the code for a par t icular wor d is asked. To analyse such codes, any t wo messages bear ing a common

wor d ar e picked up. The common code wor d will mean t hat wor d. Pr oceeding similar ly by picking up all possible combinat ions of t wo, t he ent ir e message can be analysed. Example. I f ‘tee see pee' means ‘Dr ink fr uit juice', ‘see kee lee' means ‘Juice is sweet ' and ‘fee r ee mee' means ‘H e is int elligent ', which wor d in t hat language means ‘sweet ' ? Solut ions : I n fi r st and second st at ement s, common wor d is ‘Juice' and common code wor d is ‘see'. So, ‘see' means ‘Juice'. I n second and thir d statements, common wor d is ‘is' and common code i s ‘l ee'. So, ‘l ee' means ‘i s'. Thu s, i n second st at ement , r emaining wor d ‘sweet ' i s coded as ‘k ee'. Example. I n a cer t ain language "t a ki" means "come her e"; "po r ui mo" means "ver y far away"; and "r ui t a ju" means "far fr om her e". Which find t he wor ds st and for wor d "far "? Sol ut i on : Obser ve t h e f ol l owi n g coded sentences : ta ki  come her e po r ui mo  ver y far away r ui ta ju  far fr om her e H er e wor d ‘far ' i s r epeat ed i n 2nd and 3r d sent ences and so t he code is ‘r ui’. 5. M I XED N U M BER CODI N G. I n this type of questions, a few gr oups of number s each coding a cer t ain shor t message, ar e given. T h r ou gh a com par i son of t h e gi ven coded messages, taking two at a time, we ar e is r equir ed t o find t he number code for each wor d and t hen for mulat e t he code for t he message given. Example. I n a cer t ain code, ‘786' means ‘st udy very har d', ‘958' means ‘hard work pays' and ‘645' means ‘st udy and wor k '. Which of t he following is t he code for ‘ver y ' ? Solut ions : I n fi r st and second st at ement s, common wor d is ‘har d'and common code digit is ‘8’. So, ‘8' means ‘har d’. In the first and third statements, common word is ‘study' and the common code digit is ‘6'. So, ‘6' means ‘study'. Thus, in t he fir st st at ement , ‘7' means ‘ver y'. 6. D E CI P H E RI N G I N D I V I D U AL L E T T E R CODES BY AN ALYSI S. I n t his t ype of quest ions, cer t ain sample wor ds ar e gi ven al ong wi t h t hei r codes. We ar e i s r equir ed to decipher individual codes for different lett er s by compar ing, taking two wor ds at a t ime, and then answer the given questions accor dingly.

Coding Decoding

4.5

Example. I n a cer t ain coding syst em ETTPI st ands for APPL E. What is t he code for DEL H I ? Solutions : Each let t er of Apple moves + 4 st eps, so r equir ed code for DEL H I is H I PL M . 7. GROU P CODI N G. I n this type of coding, capital letters A to Z are coded by sing the small letters a to z. With the help of the coding pattern, we have to find the right code from the columns which is equivalent to the group of capital letters. 8. DECODI N G I t is t he r ever se of coding. H er e code needs t o be decipher ed wit h t he help of given wor ds. Each let t er in a wor d ar e r eplaced by cer t ain ot her alphabet s accor ding t o a specific r ule.

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. I f TAP is coded as SZO, t hen how is FREEZE coded? (a) EQDFYG

(b) ESDFYF

(c) GQFDYF

(d) EQDDYD

2. I n a cer t ain code, SI K KI M is wr it t en as TH L JJL . H ow is TRAI NI NG wr it t en in t hat code ? (a) SQBHOHOH

(b) UQBHOHOF

(c) UQBJOHHO

(d) UQBJOHOH

3. I n a cer t ai n code, M EN TI ON , i s wr i t t en as L N EI TN O. H ow i s PATTERN wr i t t en i n t hat code? (a) APTTREM

(b) PTAETNR

(c) OTAETNR

(d) OTAETRN

4. I n a cer t ain code, FORGE is wr it t en as FPTJI . H ow is CUL PRI T wr it t en in t hat code ? (a) CSJNPGR

(b) CVMQSTU

(c) CVNSVNZ

(d) CXOSULW

5. I n a cer t ai n code, T RI PPL E i s w r i t t en as SQH OOK D. H ow is DI SPOSE is wr it t en in t hat code ? (a) CHRONRD

(b) DSOESPI

(c) ESJTPTF (d) ESOPSI D Directions (Q. 6-10) : I n each of t hese quest ions a gr oup of let t er s is given fol l owed by four combi nat i ons of number /symbol number ed (a), (b), (c) and (d). L et t er s ar e t o be coded as per t he scheme and condit ions given below. You have t o find out the serial number of the combinat ion, which r epr esent s t he let t er gr oup. Ser ial number of t hat combinat ion is your answer. L et t er s : Q M S I N G D K A L P R B J E Number : 7 @ 4 # % $ 6 1 2

£

59 83

/Symbol Condit ions : (i ) I f the first letter is a consonant and the last a vowel, bot h ar e to be coded as the code of t he vowel.

(ii ) I f t h e f i r st l et t er i s a vowel an d t h e l ast a consonant , t he codes for t he fir st and t he last ar e t o be int er changed. (iii ) I f no vowel is pr esent in t he gr oup of let t er s, t he second and t he fift h let t er s ar e t o be coded as . 6. BARNIS (a) 9 2 # % 4 (b) 9 2 4 # % (c) 9 2 # % 9 7. DMBNIA (a) 6 @9 % # 2 (c) 2 @9 % # 2 8. IJBRLG

(d) None of t hese (b) 2 @9 % # 6 (d) 2 9 % # 2

(a) # 8 9 £ S

(b) # 8 9 £ #

(c) S 8 9 £ #

(d) S 8 9£ S

9. BKGQJN (a) 9 S 7 % (c) 9 1 S 7 8 % 10. EGAKRL

(b) 9 S 7 % (d) % 1 S 7 8 9

(a) # £ S 2 1 

(b)

(c)

(d)

£ S 2 1 #

£S213 #£S21#

11. I f PAI NT is coded as 74128 and EXCEL is coded as 93596, t hen how would you encode ACCEPT? (a) 455978 (b) 547978 (c) 554978 (d) 735961 12. I f D = 4 and COVER = 63 t hen BASI S = ? (a) 49 (b) 50 (c) 54 (d) 55 13. I f DEL H I is coded as 73541 and CAL CUTTA as 82589662, how can CAL I CUT be coded ? (a) 5279431 (b) 5978213 (c) 8251896 (d) 8543691 14. I f ROSE is coded as 6821, CHAI R is coded as 73456 and PREACH is coded as 961473, what will be t he code for SEARCH ? (a) 246173 (b) 214673 (c) 214763 (d) 216473

4.6

Coding Decoding

15. I f t he let t er s in PRABA ar e coded as 27596 and T H I L A K ar e coded as 368451, h ow can BH ARATH I be coded ? (a) 37536689 (b) 57686535 (c) 96575368 (d) 96855368

LEVEL-1 1. I f ENGLAND is written as 1234526 and FRANCE is wr it t en as 785291, how is GREECE coded ? (a) 381171 (b) 381191 (c) 832252 (d) 835545 2. I f SH ARP is coded as 58034 and PUSH as 4658 t han RUSH is coded as (a) 3568 (b) 3658 (c) 3685 (d) 3583 3. I n a cer t ain code GARI M A is wr it t en as 725432 and TI NA as 6482. H ow is M ARTI NA wr it t en in t hat code ? (a) 3256482 (b) 3265842 (c) 3645862 (d) 3658426 4. If PALAM could be given the code number 43, what code number can be given to SANTACRUZ ? (a) 75

(b) 85

(c) 120

(d) 123

5. I f Z = 52 and ACT = 48, t hen BAT is equal t o (a) 39

(b) 41

(c) 44

(d) 46

6. I f ‘light ’ is called ‘mor ning’, ‘mor ning’ is called ‘dar k ’, ‘dar k ’ i s cal l ed ‘ni ght ’, ‘ni ght ’ i s cal l ed ‘sunshine’ and ‘sunshine’ is called ‘dusk’, when do we sleep ? (a) Dusk (b) Dar k (c) Night (d) Sunshine 7. I n a cer t ain language, (i ) ‘sun shines br ight ly’ is wr it t en as ‘ba lo sul’ (ii ) ‘houses ar e br ightly lit’ as ‘kado ulo ar i ba’ and (iii ) ‘light comes fr om sun’ as ‘dopi kup lo nor ’. What wor ds ar e wr it t en for sun and ‘br ight ly’ ? (a) lo, ba (b) balo (c) sul, lo (d) ba, sul 8. I f whit e is called blue, blue is called r ed, r ed is call ed yel l ow, yell ow is cal led gr een, gr een i s called black, black is called violet and violet is called or ange, t hen what would be t he colour of human blood ? (a) Red (b) Gr een (c) Yel low (d) Violet

9. I f ‘TV is called Radio’, ‘Radio is called Aer oplane’, ‘Aer oplane i s call ed H el icopt er ’, ‘H el icopt er is called Bus’, ‘Bus is called Bike’, ‘Bike is called Wat er ’, ‘Wat er i s cal l ed Fr og’, ‘Fr og i s cal l ed Tom at o’, ‘Tom at o i s cal l ed F u r n i t u r e’ an d ‘Fur nitur e is called Cigar ett e’, t hen what Tomato sauce made up of ? (a) Wat er (b) Fur nit ur e (c) Fr og (d) None of t hese D ir ect ions : I n the following questions study the coded patterns and select the right option from given alternatives. 10. I n a cer t ain code ‘415’ means ‘milk is hot ’; ‘18’ means ‘hot soup’ and ‘895’ means ‘soup is tasty’. What number will indicat e t he wor d ‘t ast y’ ? (a) 3 (b) 6 (c) 7 (d) 9 11. I n a cer t ain code language : ‘dugo hui mul zo’ stands for ‘wor k is ver y har d’; ‘hui dugo ba ki’ stands for ‘Bingo is ver y smar t’; ‘nano mul dugo’ st ands for ‘Cake is har d’, ‘mul ki qu’ st ands for ‘Smar t and har d’. Which of the following wor ds st and for ‘Bingo’? (a) jalu (b) dugo (c) ki (d) ba 12. I f in a cer tain language ‘mu mit es’ means ‘who is she’ and ‘elb mu es’ means ‘wher e is she’, t hen what i s t he code for ‘wher e’ i n t hi s language ? (a) es (b) elb (c) mu (d) mit 13. I n a cer t ain code language ‘r oi ja kyo t wa’ means ‘M oody is wr iting letter s’ ‘pok ju ja t wa’ means ‘Woody is wr it ing car ds’, ‘t r n kyo pos un’ means ‘They ar e wr it ing essays’, and ‘r oi r us pok’ means ‘let t er s and car ds’. What is t he code wor d for ‘M oody’ ? (a) ja (b) twa (c) r oi (d) k yo 14. I n a cer t ain code ‘gr i chr i’ means ‘br and new’, ‘gyp t woh’ means ‘ver y old’, gr i bur t woh’ means ‘old and new’ and ‘chr i deh gyp’ means ‘old br and car ’. What is t he code wor d for ‘new car ’ ? (a) chr i gr i (b) gr i deh (c) deh gyp (d) t woh deh

Coding Decoding

LEVEL-2 1. I n a cer t ain code, ‘247’ means ‘spr ead r ed carpet ’; ‘236’ means ‘dust one car pet ’ and ‘234’ means ‘one red carpet ’. Which digit in that code means ‘dust ’? (a) 2 (b) 3 (c) 5 (d) 6 2. I n a cer t ai n code ‘975’ means ‘Thr ow away gar bage’; ‘528’ means ‘Give away smoking’ and ‘213’ means ‘Smoking is har mful ’. Which digit in t hat code means ‘Give’ ? (a) 5

(b) 8

(c) 2

(d) 3

3. I n a cer t ain code language, 134 means ‘Good and Tast y ’ , 478 means ‘see good pict ur es’ and 729 means ‘pictures are faint ’. Which of t he following numer ical symbols st ands for ‘see’ ? (a) 1

(b) 2

(c) 7

(d) 8

Directions (Q. 4 – 7) :

Read t he following infor mat ion car efully and answer these questions. I n a cer t ain coding syst em, 816321 means ‘The br own dog fr ight ened t he cat ’, 64851 means ‘The fr ight ened cat r an away’ 7621 means ‘The cat was br own’ 341 means ‘The dog r an’ 4. What is t he code for ‘t he dog was fr ight ened’ ? (a) 8263

(b) 8731

(c) 5438

(d) None of t hese

5. What is t he code for ‘fr ight ened’ ? (a) 2

(b) 6

(c) 3

(d) 8

6. What is t he code for ‘away’ ? (a) 5

(b) 7

(c) 6

(d) 1

7. What is t he code for ‘br own’ ? (a) 2 (b) 4 (c) 6 (d) 8 8. I n a code language, 256 means “ You ar e good” , 637 means “ We ar e bad” , 358 means “ Good and bad” . Then, what is t he code of ‘and’ ? (a) 8 (b) 2 (c) 3 (d) 5

4.7

9. In a certain code, ‘467’ means ‘Leaves are green’ ; ‘485’ means ‘green is good’ and ‘639’ means ‘they are playing’. Which digit stands for ‘leaves’ in that code ? (a) 4 (b) 6 (c) 7 (d) 3 10. I n a cer t ai n code l anguage ‘851’ means ‘good sweet fr uit ’ ; ‘783’ means ‘good r ed r ose’ and ‘341’ means ‘r ose and fr uit .’ Which of t he following digit s st ands for ‘sweet ’ in t hat language ? (a) 8 (b) 5 (c) 1 (d) 3 11. I n a cer t ain code language, ‘479’ means ‘fr uit is sweet ’ ; ‘248’ means ‘ver y sweet voice’ and ‘637’ means ‘eat fr uit daily ’. Which digit st ands for ‘is’ in t hat code ? (a) 7 (b) 9 (c) 4 (d) can’t det er mined. D ir ect ion : Below ar e given let t ers A t o Z. Under each capit al let t er a small let t er is wr it t en which is t o be used as a code for t he capit al let t er. A B C D E F G H I p t x q v y r z e J K L M N O P Q R n a s g c j l u m S T U V W X Y Z o w k d i b h f I n each of t he following quest ions, a gr oup of six capit al let t er s is given and it s code equivalent is given in one of t he columns (a), (b), (c) or (d). Your cor r ect answer is (a), (b), (c) or (d) accor ding t o t he code equivalent of the gr oup of let t er s is found in it . (a) (b) (c) (d) muvgt o zcuysm wlhxct ribgpq xfwlmd r sqwui zcyums mvquot gwdnje qvlmzi xfdwlm dgjtlq muqvot dgtjql rigbqp r qswui qwjvr x muvqt o r sqiwu wlhzbt whlcxt gwndej qwjr vx zcsmuy 12. CZVTPR 13. DEPRHW 14. GWXMAD 15. RQEMBS

4.8

Coding Decoding

AN SWE RS OBJECTI VE TYPE QU ESTI ON S 1. (d) 11. (a)

2. (b)

3. (c)

4. (c)

5. (a)

1 2. (b)

13. (c)

14. (b)

15. (c)

6. (d)

7. (c)

8. (c)

9. (a)

10. (b)

7. (a)

8. (c)

9. (b)

10. (d)

7. (a)

8. (a)

9. (c)

10. (b)

LEVEL-1 1. (b) 11. (d)

2. (b) 12. (b)

3. (a) 13. (c)

4. (d) 14. (b)

5. (d)

6. (d)

LEVEL-2 1. (d)

2. (b)

3. (d)

4. (b)

5. (d)

11. (b)

12. (c)

13. (b)

14. (d)

15. (a)

6. (a)

EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S 1.

2. S

T

A

P

F

R

E

E

Z

E

– 1

– 1

– 1

– 1

–1

–1

–1

–1

–1

S

Z

O

E

Q

D

D

Y

I

K

K

I

M

T

+1– 1 +1 – 1 +1 – 1 T

H

3.

4.

L

I

N

I

N

 +1  – 1  +1  – 1  +1  – 1

L

U

Q

B

H

O

G

 +1  – 1

H

O

F

EN

TI

ON

P

AT

TE

RN

– 1 L

N E

I T

NO

– 1 O

TA

ET

NR

O

R

G

E

C

+1 +2 +3 +4

F 5.

J

A

M

F



J

R

D  EQDDYD

P

T

J

I

T R – 1 – 1 S Q

I – 1 H

P – 1 O

P – 1 O



U

E – 1 D

6. 7. 8. 9. 10.

9 2 * %# 4 Condit ion (i ) is applicable Condit ion (ii ) is applicable Condition (iii ) is applicable Condit ion (ii ) is applicable

11.

P

A

I

N

T

E

X

C

E

L

7

4

1

2

8

9

3

5

9

6

P

R

I

T

+1 +2 +3 +4 +5 +6

C L – 1 K

L

V

N

S

D – 1 C

I – 1 H

S P – 1 – 1 R O

ACCEPT = 455978 12. D = 4 COVER = 63  3 + 15 + 22 + 5 + 18 = 63. C = 3, O = 15, V = 22, E = 5 and R = 18, BASI S = 2 + 1 + 19 + 9 + 19 = 50

V

N

Z

O – 1 N

S E – 1 – 1 R D

Coding Decoding

13. Alphabet s ar e coded as follows : D E L H I C A U 7 3 5 4 1 8 2 9 CAL I CUT  8251896. 14. Alphabet s ar e coded as follows : R O S E C H A I 6 8 2 1 7 3 4 5 SEARCH 214673. 15. Alphabet s ar e coded as follows : P R A B T H I L 2 7 5 9 3 6 8 4 BH ARATH I  96575368.

T 6

P 9

K 1

LEVEL-1 1. Alphabet s ar e coded as follows : E N G L A D F R C 1 2 3 4 5 6 7 8 9 GREECE  381191. 2. Alphabet s ar e coded as follows : S H A R P U 5 8 0 3 4 6 RUSH  3658. 3. Alphabet s ar e coded as follows : G A R I M T N 7 2 5 4 3 6 8 MARTI NA  3256482. 4. L et t er s ar e coded as follows : A =1, L = 12, M = 13, P = 16 H ence, A =1, B = 2, C = 3 ...... Z = 26 PAL AM =16 + 1 + 12 + 1 + 13 = 43 So, SANTACRUZ = 19 + 1 + 14 + 20 + 1 + 3 + 18 + 21 + 26 = 123 5. L et t er s ar e coded as follows : A = 2, C = 6, T = 40, Z = 52 So, A = 2, B = 4, C = 6 ...... Z = 52 ACT = 2 + 6 + 40 = 48 BAT = 4 + 2 + 40  46. 6. I n t he quest ion, night is called sunshine and we sl eep at n i gh t , t h er ef or e cor r ect an swer i s sunshine. 7. By compar ing all the given sentences, we get sunlo and br ightly-ba. 10. Code Sent ence 415 18 895

M ilk is hot H ot soup Soup is t ast y

4.9

From 3rd code and its sentence neither number ‘9’ is r epeat ed nor t he wor d ‘t ast y’. 11. Code

Sent ence

dugo hui mul zo Wor k is ver y har d hui dugo ba ki Bingo is very smart nano mul dugo Cake is har d mul ki qu Smar t and har d Fr om 2nd code and it s sent ence, neit her ‘ba’ nor ‘Bingo’ is r epeat ed. 12. Code Sent ence mu mit es who is she elb mu es wher e is she Code words ‘mu’ and ‘es’ ar e repeated in 2nd sentence. Only code left is ‘elb’which means ‘where’. 13. Code

Sent ence

r oi ja kyo t wa Moody is writing letters pok ju ja t wa Woody is wr iting car ds t r n kyo pos un They are writing essays r oi r us pok let t er s and car ds ‘M oody’ is in 1st sent ence only. Code wor ds ‘Ja’ and ‘twa’ ar e r epeat ed in 2nd sent ence and ‘k yo’ i n 3r d sent ence. Onl y code wor d ‘r oi ’ r emains which st ands for ‘M oody’. 14. Code

Sent ence

gr i chr i br and new gyp t woh ver y old gr i bur t woh old and new chr i deh gyp old br and car Wor d ‘new’ is pr esent in 1st and 3r d sent ence and so t he code ‘gr i’. Wor d ‘car ’ is only in t he 4t h sent ence and code ‘deh’ is not r epeat ed in any ot her sent ence.

LEVEL-2 1. H er e 2, 3 and 4 ar e r ecur r ing, so 6 must be dust . 2. Same as above 3. By compar ing, we get 4 = good, 7 = pict ur es, 8 = see So, 8 st ands for see. 4. I n t he pr oblem ‘t he’ is common; so ‘t he’  1. I n 1st and 4t h sent ence, dog is common, so dog  3 Similar ly r an  4, cat  6 was  7 br own  2 away  5 and fr ight ened  8 So, code for ‘t he dog was fr ight ened’ is 8731. 5. Code for fr ight ened is 8. 6. Code for away is 5. 7. Code for br own is 2.

4.10

Coding Decoding

8. ar e = 6,

5 = good, 2 = you,

bad = 3, 7 = we, 8 = and 9. Given : ‘L eaves ar e gr een’ ‘467’ ‘Gr een is good’  ‘485’ ‘They ar e playing’  ‘639’ Fr om equat ions (i ) and (iii ), ‘ar e’ is coded as ‘6’ Fr om equat ions (i ) and (ii ), ‘gr een’ is coded as‘4’ Fr om equat ion (i ), ‘L eaves’ is coded as ‘7’. 10. Given : ‘good sweet fr uit ’  ‘851’ ‘good r ed r ose’  ‘783’ ‘Rose and fr uit ’  ‘341’ Fr om equations (i ) and (ii ), ‘good’ is coded as ‘8’. Fr om equations (i ) and (iii ), ‘fruit’ is coded as ‘1’. Fr om equat ion (i ), ‘Sweet ’ is coded as ‘5’.

11. Given : ‘Fr uit is sweet ’  ‘479’ ...(i ) ...(ii ) ...(iii )

‘Ver y sweet voice’  ‘248’ ‘Eat fr uit daily’  ‘637’

...(i ) ...(ii ) ...(iii )

Fr om equations (i ) and (ii ) ‘sweet’ is coded as ‘4’. From equations (i ) and (iii ), ‘fruit’ is coded as ‘7’. Fr om equat ion (i ), ‘is’ is coded as ‘9’.

 ...(i ) ...(ii ) ...(iii )

5

Blood Relations

CHAPTER

Blood-r elat ion pr oblems ar ises t hr ough t he par t icular r elat ions among family member s. I n gener al, t he pr oblems cont ain t hr ee st ages of family. I t explains t hr ough a st at ement and we have t o analyse t he r elat ion bet ween t hem.

 To solve such pr oblems, we st ar t fr om gr and father or mother and classify into thr ee stages, we r epresent t he male member by

and female by

. I n case of siblings we use —

(dash) bet ween t hem.

Example. I n t he family of A, B, C, D, E and F . Ther e ar e t wo mar r ied couple. D is t he gr and mot her of A and mot her of B . C is t he wife of B and mot her of F . F is gr and daught er of E and A is br ot her of F . What is C t o A ? Solution : We st ar t fr om D as follows : — — — — — — — — — — — — — — — — — — — —

i.e. C is t he mot her of A .  Some pr oblems ar e also based on r elat ionships bet ween imaginar y people. Sever al st at ement s ar e given which may confuse a per son and at t imes may also appear cont r adict or y and false. The point is t o look at each par t of a given st at ement car efully, dr aw a diagr am or t able based on t he st at ement (s) and ar r ive at a conclusion. Example. ‘A’s fat her -in-law and B’s fat her is t he husband of B’s mot her. H ow is A r elat ed t o B?’ Solution : They ar e husband and wife. Example. ‘Point ing t o a lady in t he phot ogr aph Punkuj said, “ She is t he daught er of my gr andfat her ’s only son.” H ow is she r elat ed t o Punkuj ? Solution : She is his sist er. SOM E I M PORTAN T BLOOD RELATI ON S.



M ot her 's or fat her 's son

—

Br ot her



M ot her 's or fat her 's daught er

—

Sister



M ot her 's or fat her 's br ot her

—

Uncle



M ot her 's or fat her 's sist er

—

Aunt



M ot her 's or fat her 's fat her

—

Gr and fat her



M ot her 's or fat her 's mot her

—

Gr and mot her



Son's wife

—

Daughter -in-Law



Daughter 's husband

—

Son-in-Law



H usband's or wife's sist er

—

Sister -in-Law



H usband's or wife's br ot her

—

Br ot her -in-Law



Br ot her 's son

—

Nephew

 

Br ot her 's daught er Uncle or aunt 's son or daught er

— —

N i ece Cousin

5.2

Blood Relations

  

Sist er 's husband Br ot her 's wife Gr andson's or Gr and daught er 's daught er

— — —

Br ot her -in-Law Sister -in-Law Gr eat gr and daught er

Some I mpor t ant I nfor mat ion for Blood-r elat ion t est .



M ot her 's sist er - M at er nal aunt



Fat her 's sist er - Aunt



Son of mot her or fat her - Br ot her



Daught er of mot her or fat her - Sist er



Son of br ot her - Nephew



Daught er of br ot her - Niece



Wife of elder br ot her - Sist er -in-law



H usband of daught er - Son-in-law



Wife of son - Daught er -in-law



Sist er of wife - Sist er -in-law



Br ot her of wife - Br ot her -in-law



Fat her of fat her -Gr andfat her



Fat her of mot her - M at er nal gr andfat her



M ot her of fat her - Gr andmot her



M ot her of mot her - M at er nal gr andmot her



Br ot her of fat her -Uncle



Br ot her of mot her - M at er nal uncle



Son of fat her 's br ot her - Cousin



Son of fat her 's sist er - Cousin



Son of M ot her 's br ot her - Cousin



Son of M ot her 's sist er - Cousin



Son of M at er nal gr andfat her - M at er nal uncle



The only daught er of M at er nal gr andfat her - M ot her



Wife of mot her 's br ot her or wife or M at er nal gr and fat her 's son - M at er nal aunt



The only son of gr andfat her - Fat her



M ot her of wife - M ot her -in-law



Fat her of wife - Fat her -in-law



The only daught er of Fat her -in-law or M ot her -in-law - Wife



H usband of sist er - Br ot her -in-law



Son of sist er or br ot her -in-law - Nephew



Daught er of sist er or br ot her -in-law - Niece



Son of daught er or son-in-law - Gr and son



Daught er of daught er or son-in-law - Gr anddaught er



Son of Son or daught er -in-law - Gr andson



Daught er of son or daught er -in-law - Gr anddaught er



Only daught er -in-law of Gr andfat her - Mot her



Only daught er -in-law of fat her - Wife



H usband of daught er -in-law - Son

Blood Relations

5.3

TYPES OF BLOOD RELATI ON QU ESTI ON S. 1. D ecipher ing jumbled up D escr ipt ions I n t his t ype of quest ions, a r ound-about descr ipt ion is given in t he for m of cer t ain small r elat ionships and dir ect r elat ionship bet ween t he per sons concer ned is t o be decipher ed. Example. Point ing t o a phot ogr aph, a man said, “ I have no br ot her or sist er but t hat man's fat her is my fat her 's son.” Whose phot ogr aph was it ? Solution : Since t he man has no br ot her, his fat her 's son is he himself. So, t he man who is t alking is t he fat her of t he man in t he phot ogr aph or t he man in t he phot ogr aph is his son. 2. Relat ion Puzzle I n t his t ype, mut ual blood r elat ions or ot her infor mat ions of mor e t han t wo per sons ar e ment ioned and infor mat ion about any t wo is ment ioned. E xample. A and B ar e brother s. C and D are sister s. A’s son is D's br other. How is B related to C. Solution : B is t he br ot her of A; A's son is D's br ot her. This means D is t he daught er of A. Since C and D ar e sist er s, C is also t he daught er of A. So, B is t he uncle of C. 3. Coded Relat ions I n such quest ions, r elat ionships ar e r epr esent ed by cer t ain codes or symbols such as +, – , , ,* ,  et c. Then r elat ionships bet ween cer t ain per sons, given in t he for m of t hese codes, ar e t o be analysed. E xample. 1. I f A + B means A is t he sist er of B; A – B means A is t he br ot her of B; A  B means A is the daughter of B, which of t he following shows t he r elat ion t hat E is t he mat er nal uncle of D? Solution : E is the maternal uncle of D means D is the daughter of the sister (say F) of E i.e. D  F + E. 2. I f A + B means - ‘A is t he mot her of B'. A B mean - ‘A is t he br ot her of B', A  B means‘A is t he son of B' and A – B means - ‘A is t he sist er of B' t hen which of t he following means - ‘C is t he sist er of D' ? (a) D – C (b) D  P – C (c) C – P  D (d) P + D  C Ans. (c) (a) D – C means D is t he sist er of C Fr om figur e, D is t he sist er of C but it is not necessar y t hat C is t he sister of D. C may also be br other of D. H ence, opt ion (a) is nor cor r ect . (b) D  P – C (1) D  P means D is t he son of P (2) P – C means P is t he sist er of C I f C is supposed t o be a male, t hen he will be t he mat er nal uncle of D and if C is supposed t o a female t hen she will t he mat er nal aunt of D. H ence, opt ion (b) is not cor r ect . (c) C – P  D (1) C – P means C is t he sist er of P (2) P  D means P is t he br ot her of D Fr om Figur e, if P is br ot her of D, t hen eit her D will be a male or a female. I f D is a male, t hen C is t he sist er of D and if D is a femal e t hen also C is t he sist er of D. H ence opt ion (c) is cor r ect .

5.4

Blood Relations

(d) P + D  C (1) P  D means P is t he mot her of D. (2) D  C means D is t he br ot her of C. Fr om figur e, D is t he br ot her of C, so C may be a male or female. H ence, opt ion (d) is not cor r ect .

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. A’s mot her is the only daught er of B’s father. How is B’s husband r elat ed t o A ? (a) Uncle (b) Br ot her (c) Fat her (d) Grandfather 2. A is br ot her of B. B is t he son of C. D is C’s fat her t hen what is A of D ? (a) Br ot her (b) Son (c) Gr andson (d) Grandfather 3. A is t he fat her of C and D is son of B. E is br ot her of A. I f C is sist er of D, how is B r elat ed t o E ? (a) Br ot her (b) Sister (c) Br other-in-law (d) Sister-in-law 4. A and B ar e br ot her s. C and D ar e sist er s. A’s son is D’s br ot her. H ow is B r elat ed t o C ? (a) Fat her (b) Br ot her (c) Grandfather (d) Uncle 5. A woman int r oduces a man as t he son of t he br ot her of her mot her. H ow is t he man r elat ed t o t he woman ? (a) Nephew (b) Uncle (c) Son (d) Cousin 6. A man said t o a lady, “ Your mot her ’s husband’s sist er is my aunt ” . H ow is t hat lady r elat ed t o t hat man ? (a) Daught er (b) Gr and-daughter (c) M other (d) Sister 7. I f X is br ot her of t he son of Y’s son, how i s X r elat ed t o Y ? (a) Son (b) Br ot her (c) Cousin (d) Grand-son 8. I f B says t hat his mot her is t he only daught er of A’s mot her, how is A r elat ed t o B ? (a) Son (b) Grand-father (c) Uncle (d) Br ot her

LEVEL-1 1. A is B’s wife and C is A’s sister. D is the father of C, while E is D’s son. What is the r elat ion of E to B ? (a) Br ot her (b) Br other-in-law (c) Cousin (d) Father-in-law 2. Point ing t o a phot ogr aph, a man said, “ I have no br ot her or si st er but t hat man’s fat her i s my fat her ’s son” . Whose phot ogr aph was it ? (a) H is son’s (b) H is fat her ’s (c) H is nephew’s (d) H is own 3. K is t he br ot her of X, Z is t he son of X, P, t he daughter of K , is mar r ied to N, G and X ar e sister s to one another. Then state how is G r elated to Z ? (a) Sister (b) Aunt (c) M other (d) Mother-in-law 4. A is mot her of B and C. I f D is husband of C, t hen what is A for D ? (a) Aunty (b) M other (c) Sister (d) Mother-in-law 5. I f F is br ot her of A, C is daught er of A, K is sist er of F and J is br other of C, then who is uncle of J ? (a) F (b) C (c) K (d) A 6. K al yan i i s m ot h er -i n -l aw of Veen a, wh o i s si st er -i n-l aw of Ashok . Dheer aj i s f at her of Sundeep, t he onl y br ot her of Ashok . H ow i s K alyani r elat ed t o Ashok ? (a) Mother-in-law (b) Aunt (c) M other

(d) Wife

7. Pointing to M anju in the photogr aph, Rajesh said, “ She is t he daught er of my gr andfat her ’s only son” . H ow is M anju r elat ed t o Rajesh ? (a) Sister (b) Br other-in-law (c) Son (d) M other

Blood Relations

LEVEL-2 Direction (Q. 1-5) : Fr om t h e gi ven st at emen t s sol ve t he fol l owi ng quest ions : (1) I n a family, t her e is six member s A, B, C, D, E and F. (2) C is sist er of F. (3) B is t he br ot her of E’s husband. (4) D is t he fat her of A and gr andfat her of F. (5) Ther e ar e t wo fat her s, t hr ee br ot her s and a mot her in t he family. 1. H ow F is r elat ed t o E ? (a) Daught er (b) Husband (c) Son (d) Uncle

5. Which of t he following is gr oup of br ot her s ? (a) ABD (b) ABF (c) BDF (d) BFC 6. I ntr oducing a man, a woman said, “ His wife is t he only daught er of my mot her ” . H ow is t he woman r elat ed t o t he man ? (a) Sister-in-law (b) Wife (c) Mother-in-law (d) Aunt 7. I f A’s mot her is B’s mot her ’s daught er t hen how is B’s mot her r elat ed t o A. (a) M other

(b) Aunt

(c) N i ece

(d) Gr andmot her

8. Ther e ar e t hr ee women in t he family. I f t wo of t hem ar e mot her s and t wo ar e daught er s t hen how is t he youngest r elat ed t o t he eldest ?

2. Who is E’s husband ? (a) A (b) B (c) C (d) F 3. Who is t he mot her ? (a) A

(b) B

(c) D (d) E 4. H ow m any m al e m ember s ar e t h er e i n t h e gr oup ? (a) 1 (b) 2 (c) 3 (d) 4

(a) Daught er

(b) M other

(c) Gr and-daughter

(d) Gr and-mother

9. Point ing t o a boy a woman says, “ H is fat her is fat her -i n-l aw of t hat per son whose fat her i s father-in-law of mine” . Then how boy is r elat ed t o t he woman ? (a) Son (c) Son-in-law

(b) Br ot her (d) Br other-in-law

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (c)

2. (c)

3. (d)

4. (d)

5. (d)

6. (d)

7. (d)

8. (c)

LEVEL-1 1. (b)

2. (a)

3. (b)

4. (d)

5. (a)

6. (c)

7. (a)

LEVEL-2 1. (c)

2. (a)

3. (d)

4. (d)

5. (b)

6. (b)

7. (d)

8. (c)

E XPL AN AT I ON S OBJECTI VE TYPE QU ESTI ON S 1.

2. — — — — — — — — — — C — — — — — — — — — —

— — B’s fat her — — — — — — — — — —

–

— — B’s husband — — — — — — — — — — A

i.e. B’s husband is fat her of A.

5.5

i.e. A is gr andson of D. 3.

– — — — — — — — — — —

9. (b)

5.6

Blood Relations

3.

– — — — — — — — —

i.e. B is sist er -in-law of E. 4.

– — — — — — — — — — — –

–

i.e. G is aunt of Z. 4. — — — — — — — —

i.e. B is uncle of C. 5.

B–

– — — — — — — — — — —

i.e. A is mot her -in-law of D. 5.

–

— — — — — — — —

i.e. M an is cousin of woman. 6.

–

– — — — — — — — — — —

i.e. L ady is sist er of M an. 7.

– A

i.e. F is t he uncle of J. 6. — — — — — — — —

Y — — — — — — — —

–

Y’s son

i.e. K alyani is mot her of Ashok.

— — — — — — — — — — – son i.e. X is gr andson of Y. 8.

7.

Ragesh’s gr andfat her — — — — — —

A’s mot her — — — — — — — — — — — — –

— — — — — —

B’s mot her Rajesh i.e. M anju is sist er of Rajesh.

— — — — — — — — — — — — B i.e. A is uncle of B.

LEVEL-1

LEVEL-2 (1 – 5).

1. — — — — — — — — —

— — — — — — — — — — –

–

–

— — — — — — — — —

i.e. E is br ot her -in-law of B.

–

2. — — — — — —

6. — — — — — —

— — — — — —

i.e. Woman is wife of M an. i.e. F shows t he phot ogr aph of his son M .

Blood Relations

7.

B’s mot her — — — — — — — — —

5.7

9. — — — — — — — — — –

— — — — — — — — — A

i.e. B’s mot her is gr andmot her t o A. 8. — — — — — — — —

i.e. Youngest is gr anddaught er of t he eldest .

i.e. Boy is br ot her of woman. 

6

Direction Sense Test

CHAPTER

I n t his t ype of t est , t he quest ions consist of a sor t of dir ect ion puzzle. A successive follow-up of dir ect ions is for mulat ed and we ar e r equir ed t o ascer t ain final dir ect ion or t he dist ance bet ween t wo point s. The t est is meant t o judge abilit y t o t r ace and follow cor r ect ly and sense t he dir ect ion cor r ect ly. 0 30 45 60 90

N NW

NE 45° 45° 45° 45°

W

SW

E

SE

si n

0

cos

1

t an

0

1 2

1

3 2 1

1

S

3

2 2 1

3 2

1

1 2

0

3



Above figur e shows four main dir ect ions : Nor t h N, Sout h S, East E, West W Four car dinal dir ect ions : Nor t h East (NE), Nor t h West (NW), Sout h East (SE), Sout h West (SW) Pyt hagor as T heor em This t heor em is applicable only in r ight angle t r iangles, in which one angle is of 90. AC2 = AB 2 + BC2 wher e, AC is H ypot enuse

A

AB is Per pendicular (H eight ) BC is Base Shor t cut M et hod 1. I f 2 opposit e t ur ns ar e coming t oget her, t hey will cancel each ot her. B C e.g. left , r ight and r ight , left . 2. I f similar t ur ns ar e coming t oget her t hey will be opposit e t o t he ear lier dir ect ion. e.g. if a per son is going in Nor t h and t aking t wo left t ur ns it means now he is moving in opposit e of Nor t h i.e. Sout h. 3. St ar t fr om eit her left side or r ight side. 4. Two opposit e t ur ns will make no change.

PRACTI CE EXERCI SE 1. I go 10 m t o t he east , t hen I t ur n left and go 5 m, I t ur n left again and go 10 m and t hen again I t ur n left and go 10 m. I n which dir ect ion am I fr om t he st ar t ing point ?

2. A boy walks a while facing t owar ds t he sun he t ur ns t o his r ight and cont inues t o walk. L at er, he t ur ns left and finally, t ur ning t o his r ight , he st ops. Which dir ect ion is he facing now ?

(a) Nor t h

(a) Nor t h

(b) South

(b) South

(c) East

(c) East

(d) West

(d) West

6.2

Direction Sense Test

3. A man tr avels 7 kms towar ds East, then he tur ns left and travels 8 kms, again he turns left and travels 10 kms. Finally, he turns left and travels 2 kms. I n which direction is he from his star ting point?

7. A is 40 m Sout h-West of B, C is 40 m Sout h-East of B t hen, C is in which dir ect ion of A ? (a) South (b) West

(a) Nor t h-West

(c) East

(b) West

(d) Nor th-East

(c) East (d) Nor th-East 4. Ravi dr ove 6 km t owar ds East . H e t hen t ur ned r ight and dr ove 10 km. H e again t ur ned t o his r ight and dr ove 6 km. H e t hen t ur ned t o his left and dr ove 15 km. At what dist ance is he fr om t he st ar t ing point and in which dir ect ion ?

8. A man st ar t ed wal k i ng posi t i oni ng hi s back t owar ds t he sun. Aft er somet ime, he t ur ned left , then turned r ight and then towar ds the left again. I n which dir ect ion is he going now ? (a) Nor t h or Sout h (b) East or West (c) Nor t h or West

(a) 25 km, Sout h

(d) Sout h or West

(b) 21 km, Sout h

9. P, Q, R and S ar e playing a game of car r om. P, R and S, Q ar e par t ner s. S is t o t he r ight of R who is facing West . Then, Q is facing

(c) 15 km, Sout h (d) 21 km, Nor t h 5. I f Sou t h -E ast becom es N or t h , N or t h -E ast becomes West and so on, what will West ecome ?

(a) South (b) East

(a) Nor th-East

(c) West

(b) South-East

(d) Nor t h

(c) Nor t h-West (d) South-West 6. A man is facing Nor t h-west . H e t ur ns 90.in t he cl ock w i se di r ect i on an d t h en 135  i n t h e ant iclockwise dir ect i on. Whi ch di r ect i on i s he facing now ?

10. A man went 10 kms t owar ds Sout h. Then t ur ned East and cover ed 10 kms and tur ned to t he r ight . Again aft er 10 kms he t ur ned t o left and cover ed 10 kms t o r each t he dest inat ion. H ow far and in which dir ect ion is he t o his st ar t ing point ? (a) 20 2 km, Sout h-East

(a) West (b) Nor t h

(b) 20

(c) South

2 km, Nor t h-East

(c) 20 2 m, Sout h-East

(d) East

(d) 20 km, Sout h-East

AN SWE RS 1. (a)

2. (c)

3. (a)

4. (a)

5. (b)

6 (a)

7. (c)

8. (a)

9. (d)

10. (a)

Direction Sense Test

6.3

E XPL AN AT I ON S 5.

10

1.

5 Starting point

10 10

End point

2.

Finishing point

6.

Starting point

3.

8 kms

2 kms

10 kms

Finishing point

Starting point

4. Starting

7.

7 kms

6km

point

10km

6km

8.

I n M or ni ng

15km

Finishing point

t owar ds sout h

I n Eveni ng t owar ds nort h

6.4

Direction Sense Test S

9.

P

10.

R

Requir ed distance =

( 20)2  ( 20)2

= 20 2 km Q



7

CHAPTER

Venn Diagrams

This t ype of quest ions aim at analysing abilit y t o r elat e a cer t ain given gr oup of it ems and illust r at e it diagr amatically. I . LOGI CAL DI AGRAM S I n t his type a few t ypes of Venn diagr ams with t heir implicat ions made clear. L et a gr oup of t hr ee it ems ar e given. 1. I f items evidently belong t o t hr ee differ ent gr oups, then Venn diagr am r epr esent ing it would be as shown below. e.g. Doct or s, Engineer s, L awyer s These t hr ee it ems bear no r elat ionship t o each ot her. So, t hey ar e r epr esented by 3 disjoint figur es as shown. 2. I f one item belongs to t he class of the second and second belongs t o t he class of t hir d, t hen r epr esent at ion is in t he for m of t hr ee concent r ic cir cles, as shown below. e.g. Seconds, M inut es, H our s Seconds ar e a par t of minut es and minut es ar e a par t of hour s. So, Venn diagr am would be as shown in the figur e with cir cle A representing Seconds, circle B r epr esenting M inut es and cir cle C r epr esent ing H our s. 3. I f t wo separ at e it ems belong t o t he class of t he t hir d, t hey ar e r epr esent ed by t wo disjoint cir cles inside a bigger cir cle as shown below

e.g. Table, Chair, Fur nitur e Table and chair ar e separ at e it ems but bot h ar e it ems of fur nit ur e. So, t hey would be r epr esent ed as in t he f i gur e wi t h ci r cl e A r epr esent i n g Tabl e, ci r cl e B r epr esenting Chair and cir cle C r epr esenting Fur nitur e. 4. I f t wo separ at e t hings having somet hing common belongs t o t he class of t hir d, t hey ar e r epr esent ed by t wo joint figur es inside one bigger figur e is shown below.

5. I f t wo set s A and B ar e par t ly r elat ed and t hir d is differ ent .

7.2

Venn Diagrams

6. I f A is par t ly r elat ed t o B and B is par t ly r elat ed t o C.

7. I f A, B and C par t ly r elat ed t o each ot her.

I n t his t ype of quest ions, gener ally a Venn diagr am is given. 8. I f A belongs t o B and bot h of t hem is par t ly r elat ed t o C.

2.VEN N DI AGRAM S Each geomet r ical figur e in t he diagr am r epr esent s a cer t ain class. The candidat e is r equir ed t o st udy and analyse t he figur e car efully and t hen answer cer t ain quest ions r egar ding t he given dat a. Example. I n t he following diagr am, squar e r epr esent s gir ls, t he cir cle st ands for t all per sons, t he t r iangle is for t ennis player s and t he r ect angle st ands for t he swimmer s.

On t he basis of t he above diagr am, answer t he following quest ions : 1. Which let t er r epr esent s t all gir ls who ar e swimmer s but don't play t ennis ? Solutions : Tall gir ls, who ar e swimmer s ar e r epr esent ed by t he r egion common t o t he squar e, cir cle and t he r ect angle i.e., G and H . But , accor ding t o t he given condit ions, t he gir ls shouldn't be t ennis player s. So, r equir ed r egion should not be a par t of t he t r iangle i.e., H should be excluded. Thus, r egion r epr esent ing t he per sons sat isfying given condit ions is G. 2. Which let t er r epr esent s gir ls who ar e swimmer s, play t ennis but ar e not t all? Solutions : Gir ls who ar e swimmer s and play t ennis ar e r epr esent ed by t he r egion common t o t he squar e, t r iangle and r ect angle L e., H . But , it is given t hat t he gir ls shouldn't be t all. So, r equir ed r egion should not be a par t of t he cir cle. Since H is a par t of t he cir cle.

Venn Diagrams

7.3

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. Choose t he diagr ams which r epr esent boy, gir l and a dog.

(xv) Cancer, Aids, Disease. The diagr ams ar e (a) (b)

(a) (c) (b)

(c)

(d) (e)

(d)

2. The venn di agr am whi ch r epr esent s St at e, Count r y and Blanket is

Directions (Q. 4-9) : Each it em is r epr esent ed by a cir cle. M at ch t he r ight par t s on t he basis of t he r elat ionship among t he wor ds given in each of t hese quest ions.

(a)

(b)

(c) (d) 3. Choose the most suitable diagr am for given items. (i ) Dog, cat , animal. (ii ) Sist er, mot her, women. (iii ) Town, st at e, count r y. (iv) Rhombous, Quadr ilat er als, Polygons. (v) English, H indi, L anguage. (vi ) Women, mot her, widows. (vii ) Flower s, clot hes, whit e. (viii ) Kadamkuan, Pat na, Bihar. (ix) Diesel, pet r ol, fuel. (x) Dogs, cat s, pet s. (xi ) Table, st udy r oom, house. (xii ) H ockey, Foot ball, Games. (xiii ) Tennis fans, Cr icket player s, St udent s. (xiv) Boy, st udent , cr icket er.

4. Wat er : At mospher e : H ydr ogen (a) A (b) B (c) D (d) E 5. Shir t : Collar : Pocket (a) E (b) D (c) A (d) B 6. English : L at in : Gr eek (a) A (b) F (c) C (d) D 7. Week : Day : Year (a) F (c) C

(b) E (d) A

8. Fr ame : M odel : Paint ing (a) A (b) B (c) E (d) F 9. Cit y : Childr en : St at e (a) F (b) E (c) D (d) C

7.4

Venn Diagrams

LEVEL-1 Directions (Q. 1-5) : I n t he diagr am given below, cir cle r epr esent s pr ofessor s in a M edical college, t r i angl e st ands f or Su r gi cal Speci al i st s whi l e r ectangle represent s t he M edical Specialist s :

1. Pr ofessor s who ar e also sur gical specialist s ar e r epr esent ed by (a) Y (b) C (c) D (d) X 2. Su r gi cal speci al i st s w h o ar e al so m edi cal specialist s but no pr ofessor s ar e r epr esent ed by (a) X (b) Y (c) Z (d) B

6. I n t he fol lowi ng di agr am, t r i angle st ands for ‘Bihar i', squar e st ands for ‘English', and cir cle st ands for ‘Sanskr it '. Find how many ‘Bihar is' ar e such who r ead English as well as Sanskr it ?

(a) (b) (c) (d)

3 l,3,7 3,4,5 1,3,8

LEVEL-2 Directions (Q. 1-8) : St aff employed in a UNESCO office in Par is ar e r epr esent ed by four int er sect ing circles as in the given diagram. Each circle represents people who can r ead and wr it e english, Fr ench, Spanish and Russian. St r engt h of people in each cir cle is also shown alongside. St udy t he diagr am t o answer t hese quest ions.

3. Col l ege pr ocessor s w h o ar e al so m edi cal specialist s ar e r epr esent ed by (a) Y (b) X (c) Z (d) A 4. ‘B’ r epr esent s (a) Pr ofessor s who ar e not medical specialist s (b) Pr ofessor s who ar e not sur gical specialist s (c) Pr ofessor s who ar e neither medical specialists nor sur gical specialist s (d) Medical specialists who ar e neither pr ofessor s nor sur gical specialist s 5. ‘C’ r epr esent s (a) Pr ofessor s (b) Medical Specialists (c) Sur gical Specialists (d) M edical and Sur gical Specialist s

1. People who can r ead and wr it e English, Fr ench and Spanish ar e r epr esent ed by (a) A (b) D (c) K (d) F 2. Peopl e who can r ead and wr i t e al l t he four languages can be r epr esent ed by (a) J (b) M (c) L (d) K

Venn Diagrams

3. People who can r ead and wr it e all t he languages except Spanish ar e r epr esent ed by (a) K

(b) G

(c) B

(d) I

4. Peopl e who cannot r ead and wr i t e Russi an, English and Fr ench ar e r epr esent ed by (a) L (b) J (c) H (d) E 5. People who cannot r ead and wr it e Spanish and Fr ench but ar e conver sant wi t h Engl i sh and Russian ar e r epr esent ed by (a) B (b) J (c) M (d) K

7. H ow many people know only Spanish ? (a) 10 (b) 20 (c) 40 (d) 60 8. H ow many people can r ead and wr it e any one language except Fr ench ? (a) 100 (b) 160 (c) 140 (d) 120 9. I n t h e f ol l ow i n g di agr am , par al l el ogr am r epr esent s women, t r i angl e r epr esent s subi n spect or s of pol i ce an d ci r cl e r epr esen t s gr aduat es. Whi ch number ed ar ea r epr esent s women gr aduat e sub-inspect or s of police ?

6. Which of t he following languages is known by t he maxi mum number of peopl e as per t he diagr am ? (a) Spanish only (b) Fr ench only (c) English only (d) Russian only

(a) 5 (c) 8

(b) 3 (d) 13

AN SWE RS OBJECTI VE TYPE QU ESTI ON S 1. (a)

2. (c)

3. (i ) (e),(ii ) (e), (iii ) (c), (iv) (c), (v) (e), (vi ) (d), (vii ) (a), (viii ) (c), (ix) (e), (x) (e), (xi ) (c), (xii ) (e), (xiii ) (b), (xiv) (b), (xv) (e) 4. (d)

5. (a)

6. (d)

7. (a)

8. (a)

9. (a)

LEVEL-1 1. (c)

2. (c)

3. (b)

4. (d)

5. (c)

6. (a)

LEVEL-2 1. (c)

2. (b)

3. (d)

4. (d)

5. (a)

6. (c )

7. (b)

EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S

(iii)

1. All t he t hr ee ar e fr om differ ent gr oups. 2. St at e belongs t o t he count r y wher e blanket is fr om differ ent gr oup. 3. (i )

(ii)

7.5

(iv)

8. (c)

9. (b)

7.6

(v)

Venn Diagrams

(xiv)

(vi) (xv)

(vii)

(viii)

4. At mospher e cont ains bot h H ydr ogen and Oxygen. 5. Shir t has bot h Collar and Pocket . 6. English, Gr eek and Latin ar e differ ent languages. 7. Year cont ains week, Week cont ains day.

(ix)

8. Painting is fr amed in a fr ame while model is used for paint ing. 9. Childr en line in a cit y and cit y is a past of st at e.

LEVEL-1 (x)

(xi)

1. Rect an gl e r epr esen t s ar ea f or t h e medi cal specialist s wher eas t r iangle r epr esent s sur gical specialist s and cir cle r epr esent s pr ofessor s in a medical college. Thus ar ea wher e pr ofessor s who ar e al so sur gi cal speci al i st s but not medi cal specialist s is r epr esent ed by D. 2. Su r gi cal speci al i st s w h o ar e al so m edi cal specialist s but not pr ofessor s is common ar ea of t r iangle and r ect angle and not cir cle which is r epr esent ed by Z.

(xii)

3. Col l ege pr of essor s w h o ar e al so m edi cal speci al i st s i s t he common ar ea of ci r cl e and r ect angle r epr esent ed by X. 4. B r epr esent medical specialist s who ar e neit her pr ofessor s nor sur gical specialist s. 5. C r epr esent only sur gical specialist s.

(xiii)

6. Since t he number of Bihar is who r ead English and Sanskr it bot h, will be the number which will be enclosed by t r iangle, squar e and cir cle and such number is 3.

LEVEL-2 1. I n t his gr oup, we consider common ar ea of t he t r ee cir cles one r epr esent ing t hose people who

Venn Diagrams

can r ead and wr it e English, second Fr ench and t hir d Spanish. This is shown by t he let t er K in t he diagr am. I t r epr esent s t hose people who can r ead and wr it e Fr ench, English and Spanish. 2. Thi s ar ea i s common t o al l four ci r cl es and r epr esent ed by M . 3. I n t his gr oup we consider common ar ea of t he t hr ee cir cles except Spanish so subr act ing t he cir cle of Spanish fr om r est of t hr ee cir cles. We get t he ar ea of peopl e who speak s al l t hr ee languages except spanish is I .

6. Given :

7.7

A = 40 C = 2 A = 80 E=

1 A = 20 2

G = 2 E = 40 = A So, maximum number of people used English. 7. Fr om t he diagr am only spanish language is used by t he people who belongs t o E and E = 20 8. Number of people who can r ead and wr it e any one language except fr ench

4. Peopl e who cannot r ead and wr i t e Russi an. English and Fr ench ar e r epr esent ed by E.

= A+C+E

5. B is common ar ea of t hese people who can r ead and wr it e spanish and Fr ench.

= 140

= 40 + 80 + 20 9. The r equir ed r egion is t he one common t o t he par allelogr am, t r iangle and cir cle i.e. 3.



8

Syllogism

CHAPTER

I nt r oduct i on I t helps t o t hink of deduct i ve r easoni ng in t er ms of syll ogi sms. A syl l ogi sm i s a deduct i ve ar gument r el at i ng t wo pr emises and a conclusion, all of which ar e quant ified pr opositions, i.e. pr opositions joining concepts by using wor ds such as 'some' and 'all '. Deduct i ve l ogi c i s used t o der i ve concl usi ons fr om pr emi ses wher e t he t r ut h of t he concl usi on must al ways be cont ai ned i n t he t r ut h of t he pr emi ses.

L ear ning Object ives  To under st and t he st r uct ur e of an ar gument .  To i dent i fy ar gument s involvi ng deduct ive and induct ive r easoning. L et u s st ar t by f i r st u n der st an di n g t er m s l i k e ar gument , pr emise and conclusion. A r gu m en t — A n ‘ar gu m en t expr esses a si n gl e compr ehensive act of t hought which gives judgement and is suppor t ed by t wo st at ement s. So it consist s of t hr ee pr oposit ions. I t is divided int o t wo par t s: Pr emises and Conclusion Pr emises — Two pr oposit ion which suppor t t he t hir d pr oposit ion giving judgement ar e called ‘Pr emises’. Conclusion — The t hir d pr oposi t i on whi ch gives a judgement and is suppor t ed by t wo pr emises is called a ‘Conclusion’. L et ’s consider an ar gument . 1. All st udent s ar e wise. 2. Ravi is a st udent . 3. Ravi is wise. H er e obviously pr oposit ions 1 and 2 ar e t he pr emises and t he pr oposit ion 3 which follows fr om t he fir st t wo pr oposit ions, is called t he conclusion.

D educt ive and I nduct ive L ogic Consider t he following example. 1. All mammals have lungs. 2. Cow is a mammal. 3. Cow has lungs. I r r espect ive of any ot her char act er ist ics of a cow, i.e. it gives milk, it is whit e, et c., t he conclusion t hat a cow has lungs holds good, if (1) and (2) ar e t r ue.

T his is deduct ive logic. Some char act er ist ics of deduct ive logic: • I t i s a vali d ar gument . • Not hing can fur t her st r engt hen t hi s ar gument , i.e. no st at ement can mak e t he conclusion mor e valid. • Anot her way of l ook ing at deduct i ve logi c i s t hat when a ‘specific’ conclusi on is der i ved fr om a set of gener al st at ement s, t hen it is deduct ive logic. I n t he pr evious example, if we alt er t he sequence of sent ences, i.e. 1. Cow is a mammal. 2. Cow has lungs. 3. All mammals have lungs. The conclusion t hat all mammals have lungs is not necessar ily valid. However, it is not necessar ily invalid. Ther e is a pr obabilit y of t his conclusion being t r ue. T his is induct ive logic. Some char act er ist ics of induct ive logic. • The conclusion cannot be said t o be eit her t ot ally invalid or valid. • Addit ion of cer t ain ot her pr emises may make t he conclusion eit her mor e valid or invalid. • Anot her way of looking at induct ive logic is t hat when a ‘gener al’ conclusion is der ived fr om a set of specific st at ement s, t hen it is induct ive logic. D educt ive L ogic: T h e pr em i ses i n dedu ct i v e l ogi c can be ei t h er affir mat ive or negat ive and can also be univer sal or par t i cul ar. Thus we have basi cal l y four t ypes of pr emises as summar ized below Affirmative N egative U niver sal All P ar e Q No P is Q Par t icul ar Some P ar e Q Some P ar e not Q Let us under stand each of these four statement in their en t i r et y …

All P are Q This st at ement means t hat t her e is no P which is not a Q. For t hose who under st and t he language of set s, it means t hat P is a sub-set of Q. On e of t h e m ost com m on m i st ak es m ade i n

8.2

Syllogism

int er pr et ing t his stat ement is t hat t her e ar e cer t ain Q which ar e not P. This is false r easoning. I t is just a possibilit y t hat t her e ar e cer t ain Q t hat ar e not P but we cannot be sur e of it . I t is quit e possible t hat t he ent ir e set of P and Q over lap wit h each ot her.

I n fact , t her e ar e FOUR possible venn diagr ams for t his st at ement .

Q P

P Q

Similar ly we also cannot conclude t hat ‘All Q ar e P’ as t her e is a possibilit y of some Q not being P Thus t her e ar e t wo possible venn diagr ams for t his st at ement :

P,Q

Q P=Q

P

Q

P

Some P are not Q Again her e all t hat we know t hat t her e is at least one P which is not a Q.

N o P is Q Thi s st at ement i s t he si mpl est and has j ust one int er pr et at ion i.e. t he t wo set s of P and Q ar e disjoint or do not over lap.

Please not e t he ‘at least ’. I t is quit e possible t hat no P may be a Q. Thus t he conclusion t hat “ some P ar e Q’ cannot follow t his st at ement . To summar ize: Correct conclusion

P

Q

Thus we can conclude t hat no P is a Q and also t hat no Q is a P.

Some P are Q The meaning of t his st at ement is t hat t her e is at least one P which is a Q. Thus we can also conclude fr om t his st at ement t hat ‘some Q ar e P’ because ther e is at least one entity which is bot h a P and Q. This st at ement is most pr one t o be misunder st ood. One of t he most common misunder st andings of t his st at ement is t he conclusion ‘Some P ar e not Q’. Please not e ‘Some P ar e Q’ does not mean ‘Some P ar e not Q’. Under st and it this way : the facult y of a class knows t wo individuals of t he class per sonally and know t hat t hey ar e int elligent . Thus he is r ight in saying ‘some student s of t his class ar e int elligent ’. At t he same t ime all t he st udent may also be int elligent , it is just t hat he does not know t hem well. The facult y when he says t hat some st udent s of t his class ar e int elligent is just r efer r ing t o t hose t wo individuals. The facult y is not making a r emar k on t he ot her s. The ot her s may also be int elligent .

Commonly made wrong conclusion

All P ar e Q Some Q ar e P Q No Q is P — —

All Q ar e P

Some P ar e Q Some P ar e not Q

Some P ar e not Q Some P ar e Q Some Q ar e not P

Some Q ar e P — —

No P is

Now let s move on t o find t he infer ences dr awn fr om t wo such st at ement s. I deally we will t r y t o eliminat e t he use of venn di agr ams and t r y t o deci pher t he conclusions ver bally. St r ive har d t o visualize t he venn diagram if r equir ed and not to draw it as venn diagrams ar e not hing but cr ut ches and unnecessar ily t ake t ime.

St r at egy t o Solve Syllogisms 1. St ar t by wor king backwar ds, i.e. fr om t he given choices. 2. Always t ake t he given pr emises t o be t r ue. Do not use gener al knowledge her e. Tr ees may not be blues and r eds may not be blacks, but t he idea i s t o fi nd t he l ogi cal connect i on bet ween t he pr emises and select t he conclusion. Do not let t he pr emise dist r act you int o gener al knowledge and vice ver sa. 3. Eliminat e choices wher e a st at ement negat es a pr evious st at ement . 4. Pr act ice enough quest ions, so t hat you don’t need t o dr aw Venn di agr ams. Avoi d dr awi ng Venn di agr ams dur i ng t he exam, as t her e i s a t i me constr aint except in case you wish to validat e your answer and of cour se, if t ime allows.

Syllogism

8.3

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S

12.Statements: Some t ins ar e pens.

Directions for questions 1 to 10: You ar e r equir ed t o choose fr om among t he given four diagr ams — (a), (b), (c) and (d) — t he diagr am t hat best illust r at es a r elat ionship among t he t hr ee given classes i n t he questions.

Some pens ar e r ods. Conclusions: I . No r od is t in. I I . Some t ins ar e r ods. 13.Statements:

(a)

(b)

Some sabr es ar e bombs. K ey is a sabr e. Conclusions:

(c)

I . Some bombs ar e sabr es.

(d)

I I . Some keys ar e bombs. 1. Eat ables, M eat , Venison

14.Statements: All lenses ar e hor ns.

2. Cr iminals, Thieves, M ur der er s

No hor n is colour ful.

3. Bachelor s, Doct or s, H usbands

Conclusions:

4. Food, Spaghet t i, Oven

I . Some lenses ar e colour ful.

5. Food, Rice, Belt 6. Bever ages, L iquor, Juice 7. M at t er, Solid, Gas

I I . No lens is colour ful. 15.Statements: All flir t s ar e smar t .

8. Chair, Table, Fur nit ur e

All smar t ar e r ot t en.

9. Cut ler y, Spoon, Chair

Conclusions:

10. Cr oissant s, Baked eat ables, Br own br ead Directions for questions 11 to 15: I n each quest ion, t wo st at ement s ar e fol l owed by t wo concl usi ons, I and I I . You have t o t ake t he given t wo st at ement s t o be t r ue even if t hey seem t o be at var iance wit h commonly known facts. Read the conclusions and then decide which of the given conclusions logically follows f r om t h e t w o gi v en st at em en t s, di sr egar di n g commonly known fact s. M ark t he answer (a)

if only conclusion I follows

(b)

if only conclusion I I follows

(c)

if bot h I and I I follow

(d)

if neit her I nor I I follows

11. St at ement s: Some clouds ar e t hunder. All clouds ar e r ain. Concl usions:

I . All r ot t en ar e flir t s. I I . All smar t s ar e flir t s.

LEVEL-1 Directions for questions 1 to 15: I n each quest ion, t wo st at ement s ar e fol l owed by t wo concl usi ons, I and I I . You have t o t ake t he given t wo st at ement s t o be t r ue even if t hey seem t o be at var iance wit h commonly known facts. Read the conclusions and then decide which of the given conclusions logically follows f r om t h e t w o gi v en st at em en t s, di sr egar di n g commonly known fact s. M ark t he answer (a)

if only conclusion I follows

(b)

if only conclusion I I follows

(c)

if bot h I and I I follow

(d)

if neit her I nor I I follows

1. St at ement s:

I . Those clouds which ar e not t hunder, ar e also r ain.

All cot s ar e t ablet s.

I I . Those clouds which ar e not t hunder, ar e not necessar ily r ain.

Concl usions:

Some chimps ar e t ablet s. I . Some cot s ar e chimps. I I . Some chimps ar e cot s.

8.4

Syllogism

2. Statements:

9. Statements:

All willows ar e net t les.

All gr eat scient ist s ar e college dr opout s.

Some t r ees ar e willows.

Some cr iminals ar e college dr opout s.

Conclusions:

Conclusions:

I . Some t r ees ar e net t les.

I . Some cr iminals ar e gr eat scient ist s.

I I . Some t r ees ar e not net t les.

I I . All gr eat scient ist s ar e cr iminals.

3. Statements:

10.Statements:

Some dolls blink.

Sampr as is a good spor t sman.

All dolls bur ble.

Spor t smen ar e st r ong.

Conclusions:

Conclusions:

I . Those dolls which do not blink, also bur ble.

I . All st r ong per sons ar e spor t smen.

I I . T h ose dol l s w h i ch do n ot bl i n k , don ’t necessar ily bur ble. 4. St at ement s:

I I . Sampr as is st r ong. 11. Statements: All of my classmat es ar e int elligent .

Some fools ar e int elligent .

Ravinder is not int elligent .

Some fools ar e gr eat .

Conclusions:

Concl usions:

I . Ravinder is not my classmat e.

I . Some int elligent ar e gr eat .

I I . Ravinder needs t o shar pen up his br ains.

I I . All gr eat ar e int elligent . 5. St at ement s:

12. Statements: All hippopot ami have skin.

Some pumpkins ar e fat .

Rojo is a hippopot amus.

Some fat ar e pur ple.

Conclusions:

Concl usions:

I . Rojo has skin.

I . Some fat ar e pur ple and pumpkins.

I I . Animals other t han hippopotamus do not have ski n.

I I . Some pur ple ar e pumpkins. 6. St at ement s:

13. St at ement s:

Some bosses ar e punct ual.

Some sunglasses ar e blue.

Some punct ual ar e genial.

Ray-Ban is a sunglass.

Concl usions:

Concl usions:

I . No genial is a boss.

I . Some blue ar e sunglasses.

I I . Some bosses ar e genial.

I I . Some Ray-Bans ar e blue.

7. St at ement s:

14. St at ement s:

Some sea-men ar e foolish.

Some men ar e br oad-minded.

Some sea-men ar e illit er at e.

Br oad-minded people ar e educated.

Concl usions:

Concl usions:

I . All illit er at e ar e foolish.

I . All educat ed people ar e br oad-minded.

I I . No sea-men ar e foolish.

I I . Some men ar e educat ed.

8. St at ement :

15. St at ement s:

All boys ar e her oes.

Some pious people ar e pr iest s.

Sandy is a her o.

Joseph is a pr iest .

Concl usions:

Concl usions:

I . Sandy is a boy.

I . Some pr iest s ar e pious.

I I . All her oes ar e boys.

I I . Joseph is pious.

Syllogism

LEVEL-2 Direct ions for quest ions 1 t o 15: I n each of t he following quest ions, t hr ee st at ement s 1, 2 and 3 ar e followed by four conclusions I , I I , I I I , I V. You have t o t ake t he given st at ement s t o be t r ue even if t hey appear t o be at var iance wit h commonly known facts, and t hen decide which of t he conclusions logically f ol l ow(s) f r om t h e gi ven st at em en t s. F or each quest ion, mar k t he answer choice t hat you t hink is cor r ect . 1. St at ement s: 1. All bibs ar e r ed. 2. All r ed ar e flower s. 3. No flower s ar e t ablet s. Concl usions: I . Some flower s ar e bibs. I I . No t ablet s ar e bibs. I I I . Some r ed ar e bibs. IV. Some t ablet s ar e r ed. (a) I , I I and I I I follow (b) I I , I I I and I V follow (c) Only I and I I follow (d) Only I and I I I follow 2. St at ement s: 1. Some books ar e bibs. 2. All biscuit s ar e pencils. 3. No bibs ar e biscuit s. Concl usions: I . Some books ar e not biscuit s. I I . Some bibs ar e not pencils. I I I . Some books ar e not pencils. IV. Some pencils ar e not bibs. (a) I and I V follow (b) I I and I V follow (c) I and I I I follow (d) I I and I I I follow 3. St at ement s: 1. All big ar e novels. 2. Some novels ar e willows. 3. No willows is pencil. Concl usions: I . Some willow ar e big. I I . Some novels ar e pencils. I I I . No willow is big. IV. Some novels ar e not pencils. (a) I and eit her I I or I V follow (b) I , I I I and I V follow (c) I , I I and I I I follow (d) Only I V follows

4. Statements: 1. Some whit es ar e Canadians. 2. Some Amer icans ar e whit es. 3. No Amer ican is a black. Conclusions: I . Some Canadians ar e Amer icans. I I . Some whit es ar e blacks. I I I . Some blacks ar e not Canadians. IV. Some whit es ar e not blacks. (a) I and I V follow

(b) I I and I I I follow

(c) Only I I I follows

(d) Only I V follows

5. Statements: 1. All knot s ar e t ight . 2. No collar is t ight . 3. Some bush-shir t s ar e knot s. Conclusions: I . Some bush-shir t s ar e collar s. I I . Some bush-shir t s ar e t ight . I I I . Some collar s ar e not bush-shir t s. IV. Some bush-shir t s ar e not collar s. (a) Only I I and I I I follow (b) Only I I and I V follow (c) Bot h I & I V follows (d) Only I V follows 6. Statements: 1. All cr ooked ar e spades. 2. All big ar e jaded. 3. Some big ar e cr ooked. Conclusions: I . Some jaded ar e cr ooked. I I . Some spades ar e cr ooked. I I I . Some cr ooked ar e jaded. IV. Some spades ar e big. (a) I , I I and I I I follow (b) I I , I I I and I V follow (c) I , I I I and I V follow (d) All follow 7. Statements: 1. Some people ar e nice. 2. Some for eigner s ar e people. 3. No for eigner is a Canadian. Conclusions: I . Some for eigner s ar e nice. I I . Some people ar e Canadians. I I I . Some for eigner s ar e not nice.

8.5

8.6

Syllogism

IV. Some people ar e not Canadians.

11. Statements:

(a) Bot h I I and I I I follow

1. No foods ar e mangoes.

(b) Bot h I I I and I V follow

2. No or anges ar e pot at oes.

(c) Bot h I V and I follow

3. All mangoes ar e or anges.

(d) Only I V follows

Conclusions:

8. Statements:

I . Some foods ar e not or anges.

1. All goat s ar e boxes.

I I . Some or anges ar e not foods.

2. Some goat s ar e flower s.

I I I . No mangoes ar e pot at oes.

3. No chocolat e is box.

IV. Some or anges ar e foods.

Conclusions:

(a) Only I I I and I follow

I . Some flower s ar e chocolat es. I I . No goat s ar e chocolat es. I I I . Some flower s ar e boxes. IV. Some flower s ar e not chocolat es. (a) I I , I I I and I V follow (b) I , I I I and I V follow (c) Only I I and I I I follow (d) Only I I I follows 9. Statements: 1. Some I ndians ar e not Amer icans. 2. All Amer icans ar e Asians. 3. Some Asians ar e Amer icans. Conclusions: I . Some I ndians ar e not Asians. I I . All Amer icans ar e not I ndians. I I I . All Amer icans ar e I ndians. IV. Some Amer icans ar e I ndians. (a) Only I follows (b) Only I I follows (c) Only I I I follows (d) None follows 10. Statements: 1. Some candies ar e paper s. 2. All paper s ar e vanillas. 3. Some vanillas ar e t r ams. Conclusions: I . Some candies ar e t r ams. I I . Some paper s ar e t r ams. I I I . Some t r ams ar e paper s. IV. Some vanillas ar e candies. (a) Only I I and I I I follow

(b) Only I I and I V follow (c) Only I and I I follow (d) Only I I and I I I follow 12. Statements: 1. Some cot s ar e docks. 2. All cot s ar e clocks. 3. No fr ock is clock. Conclusions: I . No fr ock is a dock. I I . No fr ock is a cot . I I I . Some docks ar e not fr ocks. IV. Some docks ar e fr ocks. (a) Bot h I I I and I V follow (b) Bot h I and I I follow (c) Bot h I I and I I I follow (d) I , I I and I I I follow 13. Statements: 1. Some t umbler s ar e capsules. 2. No capsule is a medicine. 3. All medicines ar e syr ups. Conclusions: I . Some t umbler s ar e not medicines. I I . No medicine is a t umbler. I I I . Some syr ups ar e capsules. IV. Some syr ups ar e not capsules. (a) I , I I and I I I follow (b) I , I I I and I V follow (c) I I , I I I and I V follow (d) I and I V follow 14. Statements: 1. All coat s ar e paint s. 2. No paint s ar e shir t s.

(b) I and I V follow

3. Some shir t s ar e vest s.

(c) Only I V follows

Conclusions:

(d) I I , I I I and I V follow

I . Some vest s ar e shir t s. I I . Some coat s ar e shir t s.

Syllogism

I I I . No coat is a shir t .

8.7

15. Statements:

IV. Some vest s ar e not coat s.

1. All chimps ar e monkeys.

(a) I , I I and I I I follow

2. No monkeys ar e mammals.

(b) I and I V follow

3. Some hor ses ar e mammals.

(c) I and I I I follow

Conclusions:

(d) I , I I I and I V follow

I . Some hor ses ar e not chimps. I I . Some hor ses ar e not monkeys. I I I . Some monkeys ar e not chimps. IV. No chimp is a mammal. (a) I , I I I and I V follow (b) I I , I I I and I V follow (c) I I and I V follow (d) I , I I and I V follow

AN SWE RS OBJECTI VE TYPE QU ESTI ON S 1. (a)

2. (b)

3. (d)

4. (c)

5. (c)

11. (a)

12. (d)

13. (a)

14. (b)

15. (d)

6. (b)

7. (b)

8. (b)

9. (c)

10. (b)

7. (d)

8. (d)

9. (d)

10. (b)

7. (d)

8. (a)

9. (d)

10. (c)

LEVEL-1 1. (d) 11. (a)

2. (a) 12. (a)

3. (a) 13. (a)

4. (d) 14. (b)

5. (d) 15. (a)

6. (d)

LEVEL-2 1. (a)

2. (a)

3. (d)

4. (d)

5. (b)

11. (d)

12. (c)

13. (d)

14. (d)

15. (d)

6. (d)

EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S 11.

S

13. S

B

K1

R1 T

C

B

T

C

K2

R2

Fr om t he above venn diagr am we conclude only I follows.

Hence, clouds which ar e not thunder ar e also r ain. 12.

T

14.

P

H L

T

H L

P R2 R1

Some r ods may or may not be t i ns. Some t ins may or may not be r ods. H ence, none of t he conclusions definit ely follows.

K3

H ence, no lens is colour ful.

C

8.8

Syllogism

15.

6.

R

S

G2 B

F

P

B

S F

P

H ence, all flir t s ar e r ot t en.

G1

Some genial may or may not be bosses.

LEVEL-1 1.

Ch2

T

I2

7.

T

S

F

S F

Co

Co

Ch1 I1

H ence, some chimps may or may not be cot s. 2.

N

Some illit er at e may or may not be foolish.

N

8.

T2 W

W

H

H S1 B

B S2

T1

Fr om t he above venn diagr am we can conclude t hat some t r ees ar e definit ely net t les

None of t he conclusions follow.

. 3.

9.

Bu2

C2

D

D

Bl

Bl

D

Bu1

S

S

D

C1

None of t he conclusions follows.

All dolls, whet her t hey blink or not , bur ble. 10.

4. F

I

Sp Sp

I

F

Sa

St

Sa

G2

Thus, we can conclude t hat Sampr as is st r ong.

G1

Some int elligent may or may not be gr eat . H ence, neit her conclusion I nor I I follows. 5.

11.

I

I

R

Pur2 CM

Pum Fat

CM

Pum Fat

Pur1

Some pur ple may or may not be pumpkins.

H ence, Ravinder is definit ely not my classmat e.

Syllogism

12.

S

S

3.

H

I.

P2

H

N

W2

II.

B

Rojo

III. P1

SG

13.

RB1

H ence, opt ion (d) is t he cor r ect answer.

B

B SG

IV. W1

H ence, we can definit ely concludes t hat Raj has ski n. 4.

RB2

B1

I. II.

A1

H ence, some blue ar e sunglasses. M

C

III. IV.

B2

A2

E1 E2 B

M

B

W

H ence, opt ion (d) is t he cor r ect answer. 5.

14.

C3

I.

T C1

H ence, some men ar e educat ed. Pi

Pr

Pi

K

II. III.

Pr

C2

IV.

BS

J1

15. J2

Joseph may or may not be pious. H ence, onl y conclusion I follows.

H ence, opt ion (b) is t he cor r ect answer. 6.

J

II.

LEVEL-2 1.

I.

S

B

C

F

III. IV.

I.

R

II. T

B

III.

H ence, opt ion (d) is t he cor r ect answer. 7. C1

IV.

C2

H ence, opt ion (a) is t he cor r ect answer.

I. II.

N

P

III.

F2

2.

IV.

I. Bi

Bo

F1

II. III.

H ence, opt ion (d) is t he cor r ect answer.

P1 Bis2

P2

Bis1

IV.

8.

C2

I.

F

H ence, opt ion (a) is t he cor r ect answer.

B

II.

G C1

III. IV.

H ence, opt ion (a) is t he cor r ect answer.

8.9

8.10

Syllogism

9. I4

I1

As

13.

I.

S1 M1

II. Am

III.

I2

I.

T

IV.

C

II.

I3

III.

H ence, opt ion (d) is t he cor r ect answer.

M2

IV. S2

10. V T2

C

I.

P

Only conclusions I and I V follow. The set of ‘syr ups’ which ar e ‘medicines’ ar e definit ely not ‘capsules’. H ence, opt ion (d) is cor r ect .

II. III. IV.

T1

14. I.

V

S

II.

H ence, opt ion (c) is t he cor r ect answer. 11.

P III.

C

O

IV.

I.

F1

II.

M

III. IV.

P

F2

H ence, opt ion (d) is t he cor r ect answer. 15. H1

Ma

I. II.

H ence, opt ion (d) is t he cor r ect answer. 12.

I.

ClK

II. C

D

F1 F2

III. IV.

H ence, opt ion (c) is t he cor r ect answer.

III. Ch

H2

IV.

Mon

H ence, opt ion (d) is t he cor r ect answer.



9

Statement Conclusion

CHAPTER

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S D ir ect i ons for quest ions 1 t o 10: Each of t he following quest ions consist s of statements followed by two conclusions. Read the given statements car efully and ident ify which of the conclusions dir ectly follow. M ark your answer as: (a) if only conclusion I follows (b) if only conclusion I I follows (c) if bot h conclusions I and I I follow (d) if neit her conclusion I nor I I follows 1. St at ement s: Shei l a i s an M BA. Some M BAs speci al i ze i n I nfor mat ion Syst ems M anagement . Concl usi on: I . Shei l a speci al i zed i n I nfor mat i on Syst ems Management . I I . Not all MBAs specialize in I nfor mation System Management . 2. St at ement s: Asphalt is ver y har d. Oil is of t he same densit y as asphalt. Concl usi on: I . Oil is ver y har d. I I . The density of asphalt and oil can be measur ed. 3. St at ement s: Mur der is a mor al offence. You have committed a mor al offence. Concl usi on: I . You ar e a mur der er. I I . M ur der er s ar e immor al. 4. St at ement s: Ear t h i s a par t of our sol ar syst em. Our solar syst em is par t of t he milky way galaxy. Concl usi on: I . Ear t h is par t of t he milky way galaxy. I I . M ilky way galaxy is in t he solar syst em.

5. St at ement s: Wat ching cr icket is one of t he major r ecr eat ional act ivit ies in I ndia. H e has gone for r ecr eat ion. Concl usi on: I . H e has gone t o wat ch cr icket . I I . Cr icket is somet imes played for pleasur e in I ndia. 6. St at ement s: For t he past 10 year s people of K ashmir have lost t heir peace of mind because of fear of t er r or ist at t ack. M y uncle who is r esiding in K ashmir has also lost his peace of mind. Concl usi on: I . M y uncle may be fear ing a t er r or ist at t ack. I I . People staying in Kashmir can not lead a peaceful life. 7. St at ement s: Sachin owns a Fer r ar i. Some Fer r ar is have left hand dr ive. Concl usi on: I . Sachin had bought a Fer r ar i for himself. I I . Sachin’s car may have a left -hand dr ive. 8. St at ement s: The r eason for t he upsur ge in t he incidence of car di ac di sease among young pr ofessi onal s i s occu pat i on al st r ess. M r . Si n gh , a y ou n g pr ofessional r ecently succumbed t o car diac ar r est. Concl usi on: I . M r. Si ngh's deat h was due t o occupat i onal st r ess. I I . A l ot of y ou n g pr of essi on al s ar e f aci n g occupat ional st r ess now days. 9. St at ement s: CET for ms can be pr ocur ed eit her by paying cash or by submit t ing a DD. M any people pr efer cash. She has pr ocur ed a CET for m. Concl usi on: I . She submit t ed a DD for t he CET for m. I I . Pu r ch asi n g t h e f or m by m ak i n g cash payment i s mor e conveni ent t han payi ng t hr ough a DD.

9.2

Statement Conclusion

10. St at ement s: Polar bear s do not have fur on t hem. Animal X doesn’t have fur on it . Concl usi on: I . Animal X is fr om t he poles. I I . Polar bear ’s skin is t hick enough t o pr ot ect it fr om cold. Directions for questions 11 to 15: I n t he following quest ions, one/t wo st at ement s ar e followed by t wo concl u si on s 1 and 2. You h ave t o con si der t h e st at ement s t o be t r ue, even i f t hey seem t o be at var iance fr om commonly known fact s. You have t o decide which of t he given conclusions, if any, follow fr om t he given st at ement s. M ark your answer as: (a) if neit her conclusion 1 nor conclusion 2 follows (b) if bot h conclusions 1 and 2 follow (c) if only conclusion 1 follows (d) if only conclusion 2 follows 11. St at ement s: 1. Alexander visit ed Egypt . 2. Cleopat r a was t he queen of Egypt . Concl usions: 1. Alexander met Cleopat r a. 2. Egypt was a monar chy. 12. St at ement s: 1. I ndia is a democr at ic count r y. 2. Pakist an is a democr at ic count r y. Concl usions: 1. I ndia and Pakist an ar e bot h democr at ic. 2. I ndia and Pakist an ar e neighbor s. 13. St at ement s: 1. Cr icket is played wit h r ed balls. 2. Snooker uses many r ed balls. Concl usions: 1. Snooker can be played wit h cr icket balls. 2. Cr icket can be played wit h snooker balls. 14. St at ement s: 1. M any people ar e r ich and educat ed. 2. M any educat ed people ar e poor. Concl usions: 1. Rich people ar e educat ed. 2. Poor people ar e educat ed. 15. St at ement : The gover nment pr ovides subsidies t o help r ich people. Concl usions: 1. Rich people need help. 2. Rich people get subsidies.

LEVEL-1 Directions for questions 1 to 10: I n t he following quest ions, one/t wo st at ement s ar e followed by t wo concl u si on s 1 and 2. You h ave t o con si der t h e st at ement s t o be t r ue, even i f t hey seem t o be at var iance fr om commonly known fact s. You have t o decide which of t he given conclusions, if any, follow fr om t he given st at ement s. M ark your answer as: (a) if neit her conclusion 1 nor conclusion 2 follows (b) if bot h conclusions 1 and 2 follow (c) if only conclusion 1 follows (d) if only conclusion 2 follows 1. St at ement s: 1. The company r epor t ed l osses i n t he fi r st quar t er. 2. The company has been pr ofit able in t he past . Concl usions: 1. The company will make a pr ofit in t he second quar t er. 2. The company is going to be bought by a bigger company. 2. St at ement s: 1. A flag has many color s. 2. I ndia’s flag has 4 color s, including whit e. Concl usions: 1. I ndia may not be a nat ion. 2. I ndia has t o be a nat ion. 3. St at ement s: 1. Rich people have lot s of fr ee t ime. 2. Rich people buy many car s. Concl usions: 1. H aving many car s saves t ime. 2. M any r ich people pur chase car s. 4. St at ement s: 1. Air t r avel is mor e expensive t han t r ain t r avel. 2. Tr ain t r avel t akes less t ime t han air t r avel. Concl usions: 1. To save t ime, one should use t r ain t r avel. 2. To save money, one should use t r ain t r avel. 5. St at ement s: 1. At hlet es like David exer cise for mor e t han a quar t er of ever y day. 2. For anyone t o become an at hlet e, t hey must do mor e t han 6 hour s of exer cise ever yday. Concl usions: 1. D av i d ex er ci ses f or m or e t h an 6 h ou r s ever yday. 2. Anyone can be an at hlet e.

Statement Conclusion

6. St at ement s:

9.3

2. The sympt oms wer e of Typhoid.

1. Wat ches ar e used t o keep t r ack of t ime.

Concl usions:

2. Clocks ar e made all over t he wor ld t o keep t r ack of t ime.

I . Cont aminat ion of wat er may lead t o Typhoid.

Concl usions:

(a) Only conclusion I is t r ue

1. Cl ock s and wat ches ar e subst it ut es of each ot her.

(b) Only conclusion I I is t r ue

2. Wat ches ar e made all over t he wor ld. 7. St at ement s: 1. The gover nment holds exams t o select people for var ious post s lying vacant . 2. Many people give these exams t o get their fir st jobs. Concl usions: 1. Ther e ar e many post s t hat need t o be filled. 2. M any people ar e unemployed. 8. St at ement s: 1. Nor t h K or ea and Sout h K or ea ar e separ at ed by wat er. 2. They t hr eat en war on each ot her. Concl usions:

I I . Typhoid is a cont agious disease.

(c) Bot h conclusions I and I I ar e t r ue (d) Bot h conclusions I and I I ar e false 12. St at ement s: 1. 60% of t he gover nment empl oyees went on st r i ke. 2. M r. Gopal is a gover nment employee. Concl usions: I . M r. Gopal went on st r ike. I I . M r. Gopal did not par t icipat e in t he st r ike. (a) Only conclusion I follows (b) Only conclusions I I follows (c) Bot h conclusions I and I I follow (d) Eit her conclusion I or I I follows 13. St at ement s:

1. Nor t h K or ea and Sout h K or ea ar e islands.

1. Temple is a place of wor ship.

2. They ar e fight ing over who get s t o keep t he wat er bet ween t hem.

Concl usions:

9. St at ement s: 1. T h e I n t er n et i s a web of i nt er con n ect ed comput er s.

2. Chur ch is also a place of wor ship. I . H indus and Chr ist ians use t he same place for wor ship. I I . All chur ches ar e t emples.

2. All comput er s have input and out put devices.

(a) Neit her conclusion I nor I I follows

Concl usions:

(b) Bot h conclusions I and I I follows

1. The I nt er net is an out put device.

(c) Only conclusion I follows

2. A single comput er cannot make an I nt er net .

(d) Only conclusion I I follows

10. St at ement s:

14. St at ement s:

1. Wat er is necessar y for life t o sur vive.

1. I ndia is becoming indust r ialised.

2. Food is necessar y for life t o sur vive.

2. Pol l u t i on i s a pr obl em associ at ed w i t h industr alisat ion.

Concl usions: 1. Wat er and food ar e life. 2. L ife cannot sur vive wit hout wat er and food. Directions for questions 11 to 15: I n t he following quest ions, t wo st at ement s ar e given, followed by t wo concl u si on s I and I I . You h ave t o consi der t he st at ement s t o be t r ue even i f t hey seem t o be at var iance fr om commonly known fact s. You have t o decide which of t he given conclusions, if any, follow fr om t he given st at ement s. 11. St at ement s: 1. Due t o cont aminat ion of wat er, lar ge number of people wer e admit t ed t o hospit als.

Conclusi ons I . Pollut ed nat ions ar e indust r ialised. I I . I ndia may become pollut ed. (a) All ar e appr opr iat e (b) Only I is appr opr iate (c) Only I I is appr opriate (d) None is appr opr iate 15. St at ement s: 1. H appiness der ived fr om ext er nal mat er ials is moment ar y. 2. Ever lasting happiness has to come fr om within.

9.4

Statement Conclusion

Concl usions:

Concl usions:

I . Nobody can exper ience happiness fr om outside.

I . Cont r ast s hinder int egr at ion.

I I . H appi ness exper i enced fr om ci nema i s not last ing.

I I . Nat ional int egr at ion is essent ial for I ndia.

(a) Only I follows (b) Only I I follows (c) Neit her I nor I I follows (d) Bot h I and I I follow

LEVEL-2 D ir ect i ons for quest ions 1 t o 10: Each of t he following quest ions consist of st at ement s followed by t wo conclusions. Read t he given st at ements car efully and ident ify which of t he conclusions dir ect ly follow. M ark your answer as: (a) if only conclusion I follows (b) if only conclusion I I follows (c) if bot h conclusions I and I I follow (d) if neit her conclusion I nor I I follows 1. St at ement s: The Pl anni ng Commi ssi on i s opposed t o t he pr oposal . I t feel s t hat t he cost of subsi di zi ng helicopt er oper at ions will be exor bit ant . Concl usions: I . T h e Pl an n i n g Com m i ssi on w an t s t h e gover nment t o spend less on it self. I I . H elicopt er oper at ions must be subsidized. 2. St at ement s: The population below the pover t y line is computed on the basis of minimum daily calor ie r equir ement of food and act ual consumpt i on. For t y-ei ght per cent of I ndi a's popul at i on l i ves bel ow t he pover t y line. Concl usions: I . For t y-eight per cent of t he people in I ndia do not get t he r equir ed calor ies of food. I I . Pover t y is I ndia's biggest pr oblem. 3. St at ement s: The policy of liber alisation will make the rich richer and t he poor poor er. The dispar it y bet ween t he r ich and t he poor will widen. Concl usions: I . L iber alisat ion is not good for I ndia. I I . The r ich believe in liber alisat ion. 4. St at ement s: Nat ional int egr at ion cannot be achieved because t her e ar e cont r ast s bet ween t he r ich and t he poor and differ ent r eligious gr oups.

5. St at ement s: A simple DNA t est can pr edict whet her someone is mor e likely t o lose weight on a low fat or a low car bohydr ate diet, say US resear chers. Their study looked at how well people wit h differ ent genes far ed on differ ent weight loss diet s. Concl usions: I . Peopl e’s bodi es r eact t o cer t ai n nut r i ent s differ ent ly and t his is r elat ed t o t heir genet ic makeup. I I . A bet t er u n der st an di n g of on e’s gen et i c makeup helps in charting out one’s diet patter n t o lose weight . 6. St at ement s: Ther e ar e glar ing inequalit ies bet ween man and man. Laws can ensur e an or der ed society but only har d wor k wi l l hel p us t o achi eve t he soci al object ive of economic goals. Concl usions: I . On l y h ar d w or k can r em ov e econ om i c inequalities. I I . Economi c equal i t y i s mor e i mpor t ant t han discipline. 7. St at ement s: The deadlock per sist ed for t he four t h consecut ive day despite hectic efforts. Then, a meeting between t he Pr ime M inist er and t he far mer s' leader s was ar r anged. Concl usions: I . The M eet ing wit h t he Pr ime M inist er ended t he deadlock. I I . The meet ing bet ween t he Pr ime M inist er and t he far mer s' leader was t he r esult of t he hectic effor t . 8. St at ement s: Beaut y is God’s gift and ever yt hing and ever yone has beaut y, but not ever yone ident ifies it . Concl usions: I . Some people ident ify beaut y. I I . Beaut y is ubiquit ous. 9. St at ement s: The funct ion of science is t o supply r eliable and r elevant infor mation to the society. I ts infor mation i s r el i abl e due t o i t s el abor at e t ech ni que of ver ificat ion and also because such infor mat ion is capable of sur viving for cent ur ies.

Statement Conclusion

Concl usions:

9.5

13. St at ement :

I . Scient ific infor mat ion is beyond t he scope of r evision.

M inist er s ar r ived at t he public funct ion in t heir car s.

I I . Scient ific infor mat ion is ver ifiable.

Concl usions:

10. St at ement :

I . All minist er s ar e r ich

The human or ganism gr ows and develops thr ough st imulat ion and act ion.

I I . M inist er s have car s.

Concl usions:

(a) Only I I and I I I ar e implicit in t he st at ement .

I . I n er t hu m an or gan i sm can not gr ow an d develop.

(b) Only I is implicit in t he st at ement

I I . H uman or ganisms do not r eact t o st imulat ion and act ion.

(d) Only I I I and I ar e implicit in t he st at ement .

11. ‘All beggar s ar e poor.’

I I I . M inist er s came t o t he public funct ion.

(c) Only I and I I ar e implicit in t he st at ement 14. St at ement : Philant hr opes wit h t heir human compassion and zeal t o help t he needy have cont r ibut ed t o human welfar e in ever y societ y.

I f t he above st at ement i s t r ue, whi ch of t he following conclusions can be dr awn? (a) All t hose who ar e poor ar e beggar s.

Concl usions:

(b) I f A is r ich, t hen A is not a beggar

I . Rich per sons ar e philant hr opes

(c) I f A is not r ich, t hen A is not a beggar.

I I . Poor people cannot act as philant hr opes.

(d) I f A is a beggar, t hen A is not r ich.

(a) Only I is implicit in t he st at ement .

12. St at ement :

(b) Bot h I and I I ar e implicit

No childr en ar e vot er s.

(c) Neit her I nor I I is implicit .

Concl usions:

(d) Only I I is implicit .

I . All adult s ar e vot er s.

15. Fer t ilizer consumpt ion in I ndia in 1984-85 was 8.21 MT. By 1990 it was 13.75 M T and by 2000 it is expected to r each 16 MT. What is your conclusion?

I I . No vot er s ar e childr en. (a) Only conclusion I follows (b) Only conclusion I I follows

(a) Fer t ilizer consumpt ion is st eady.

(c) Bot h conclusions I and I I follow

(b) T h er e i s a st eady decr ease i n f er t i l i zer consumpt ion.

(d) Neit her conclusion I nor I I follows

(c) Fer t i l i zer consumpt i on does not show any t r end. (d) Ther e i s a st eady i ncr ease i n s fer t i l i zer consumpt ion.

AN SWE RS OBJECTI VE TYPE QU ESTI ON S 1. (d) 11. (d)

2. (b) 12. (c)

3. (d) 13. (b)

4. (a) 14. (a)

5. (d) 15. (a)

6. (a)

7. (b)

8. (d)

9. (d)

10. (d)

7. (a)

8. (a)

9. (d)

10. (d)

7. (d)

8. (b)

9. (b)

10. (d)

LEVEL-1 1. (a) 11. (a)

2. (c) 12. (d)

3. (a) 13. (a)

4. (b) 14. (c)

5. (c) 15. (b)

6. (a)

LEVEL-2 1. (d) 11. (d)

2. (a) 12. (b)

3. (d) 13. (a)

4. (a) 14. (c)

5. (c) 15. (d)

6. (d)

9.6

Statement Conclusion

EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S 1. None of t he conclusions follow as ‘some’ may mean al l and t he given st at ement does not gi ve any cl ar i t y on t he same. Shei l a may or may not specialize in I nfor mat ion Syst em M anagement . H ence, opt ion (d) is cor r ect . 2. Conclusion I is incor r ect because t he r eason for asphalt being har d is not known. I t could be t he pr esence of fact or s like asphalt ’s composit ion, it s t ext ur e et c. t hat make it har d. So, even if oil and asphalt ar e of t he same densit y, oil may not be as har d as asphalt. Only conclusion I I follows because t he st at ement clear ly indicat es t hat oil and asphalt ar e of t he same densit y and such a st at ement can only be made aft er measur ing t he densit y of t hese t wo subst ances. 3. None of t he concl usions is t r ue. Commit t i ng a single act of moral offence cannot account for being cal l ed i mmor al . So concl usi on I I i s i ncor r ect . Conclusion I does not follow because mur der is one of t he immor al act s but may not be t he only immor al act . 4. Only conclusion I follows as ‘ear t h’ is a subset of ‘sol ar syst em’ whi ch i s a subset of ‘mi l ky way galaxy’. Conclusion I I is incor r ect as ‘solar system’ is a subset of ‘milky way galaxy’ and not vice ver sa. 5. None of t he conclusion follows. Conclusion I is incor r ect because wat ching cr icket is one of t he r ecr eational activities but not the only recreational act ivit y. H e may or may not be wat ching cr icket . Conclusion I I is incor r ect because in I ndia, cr icket is watched for pleasu re and not played for pleasure. 6. Only conclusion I follows as it t alks of a possibility (may). Conclusion I I is incor r ect because the given statement s r efer to a specified t ime fr ame (over the past ten years) but conclusion II makes a general statement which cannot be definitely concluded. 7. Conclusion I is incor r ect because if Sachin owns a Fer r ar i, it does not necessar ily mean t hat he had bought one. H e could have got it as a pr esent fr om his fat her. Only conclusion I I follows as it t alks about a possibilit y. Since some Fer r ar is have left hand dr i ve, i t i s qui t e possi bl e t hat Sachi n’s Fer r ar i is also left -hand dr ive. 8. Conclusion I does not follow because occupational st r ess is one of t he r easons for car diac ar r est but not the only r eason for it. Conclusion I I is incor r ect because though ther e is an incr ease in the number of young pr ofessionals affected by car diac diseases, but it cannot be definit ely said t hat a lot of t hem have been pr ey t o occupat ional st r ess. The t er m

‘lot ’ cannot be defined as we do not have number s her e. 9. Conclusion I is incor r ect because t he st at ement s clear ly say t hat t her e ar e t wo ways of pr ocur ing t he for m, so she could have chosen eit her of t he two. Conclusion I I is incorr ect as people may prefer cash because it is safer or fast er or cheaper. 10. N one of t he concl usi ons fol l ow. I i s i ncor r ect because t her e could be anot her var iet y of animal which doesn’t have fur t oo. So, we cannot say that animal X is fr om t he poles. I I is out of scope as not hing has been said about polar bear ’s skin. 11. Only conclusion 2 follows, since a queen r eigns over a monar chy. 12. Onl y concl usi on 1 fol l ows, si nce geogr aphi cal pr oximit y is not ment ioned. 13. Bot h conclusions 1 and 2 follow, since t her e is no di ffer en ce bet ween r ed bal l s as used i n t he st at ement s. 14. Neit her conclusion follows, since not all r ich or poor people may be educat ed. 15. Neit her of t he conclusions follow. Conclusion 1 is incor r ect because if gover nment helps someone, i t does not i mply t hat t hat someone definit el y needs help. Conclusion 2 is incor r ect because we cannot say t hat all r ich people get subsidies.

LEVEL-1 1. Neit her conclusion follows, since we don’t know t he second quar t er r esult s or any plans of bigger companies. 2. Only conclusion 1 follows, since it is not mentioned t hat ever yt hing wit h a flag must be a nat ion. 3. Ther e is no r elat ion bet ween car s and t ime in t he st at em en t s. So con cl u si on 1 i s i n cor r ect . Con cl u si on 2 does n ot f ol l ow becau se t h e st at ement mer ely says t hat r ich people pur chase ‘m an y’ car s. Bu t wh et her a few r i ch peopl e pur chase many car s or many r ich people buy many car s is not hint ed at . 4. Bot h conclusi ons fol low, si nce accor di ng t o t he st at ement s, air t r avel is mor e expensive as well as slower. 5. Only conclusion 1 follows because quar ter of a day m ean s m or e t h an 6 h ou r s. Con cl u si on 2 i s incor r ect because not hi ng has been ment i oned about who all can become an at hlet e. 6. N one of t he concl usi ons fol l ow. 1 i s i ncor r ect because though a watch and a clock serve the same pr imar y pur pose but a wat ch can be used on t he move while a clock is st at ionar y. So, t hey cannot

Statement Conclusion

be used int er changeably. 2 is incor r ect because nothing has been said about pr oduction of watches. 7. None of the conclusions follow as nothing has been ment ioned about t he number of post s t hat ar e vacant and about t he number of people who ar e unemployed. So ‘many post s’ in conclusion 1 and ‘many people’ in conclusion 2 cannot be deduced fr om t he given st at ement s. 8. Neit her conclusion follows, since separ at ion fr om each ot her by wat er is not t ant amount t o being an island. Also, t he r eason for war t hr eat s is not given. 9. Only concl usion 2 follows, since I nt er net must have inter connected comput er s, or mor e t han one comput er at least . A comput er having input and output devices does not make the I nter net become an input or out put device. 10. Only conclusion 2 follows fr om t he st at ement s, since bot h food and wat er ar e given as necessar y for sur vival of life. 11. Conclusion I is cor rect as it talks about a possibility. Conclusion I I is out of scope. 12. M r. Gopal may or may not par t icipat e in st r ike. So, bot h conclusions ar e possible and one of t hem must be t r ue. 13. Templ e and Chur ch may or may not over l ap. Further, no infor mation is given about Hindus and/ or Chr ist ians. Thus, neit her conclusion follows. 14. Conclusion I I t alks about a ‘possibilit y’ of I ndian becoming pollut ed due t o indust r ializat ion and hence is cor r ect . Conclusion I cannot be dr awn as not hi ng has been ment i oned about pol l u t ed nat ions. 15. Onl y concl usi on I I fol l ows because happi ness exper ienced fr om cinema cannot be cat egor ized as happiness coming fr om wit hin one’s soul and st at em en t 1 cl ear l y st at es t h at ever l ast i n g happiness comes fr om wit hin.

LEVEL-2 1. I is neither stated nor implied. So, I does not follow. I I cont r adict s t he idea of planning commission. So, I I does not follow. 2. We cannot infer fr om the passage whether pover ty is I ndia’s biggest pr oblem or not . Only conclusion I follows. 3. The st at ement s do not say anyt hing about I ndia. H ence, I is negat ed. I I t alks about t he beliefs of t he r ich and is beyond t he scope of t he ar gument .

9.7

4. The st at ement implies t hat t he wide gap between the r ich and the poor hinder s national integr ation. So, I can be concluded. The st at ement does not make any ment ion of I ndia, so I I is beyond t he scope of t he ar gument . 5. I t ’s st at ed t hat a DNA t est can pr edict whet her someone is mor e likely t o lose weight on a low fat or l ow car bohydr at e di et . Thi s means t hat a per son’s genet ic makeup det er mines how his/her body r eact s to cer tain nutr ients and t his may play a r ole in char t ing out one’s diet pat t er n. H ence, conclusions I & I I follow. 6. When t he ar gument t alks about inequalit ies, it does not specify t hat it is mentioning the economic or t he social ones. We have no clar it y about t he social object ives of economic goals. Thus I is quit e f ar f et ch ed an d can n ot be con cl u ded. I I i s ir r elevant . 7. Neither of the conclusions follow because one does not know if t he meet ing ended t he deadlock and whet her it was t he r esult of t he hect ic effor t . 8. ‘Ubiquit ous’ means pr esent ever ywher e. So, I I definitely follows as t he main st at ement says t hat “Beauty is God’s gift and ever yt hing and ever yone has beaut y.” Not ever yone includes the possibilit y t hat no one ident ifies it . H ence, I does not follow. 9. Only I I follows. I is vague as t he ‘scope of r evision’ cannot be deduced. 10. Both conclusions are incor r ect as nothing has been said about iner t human or ganisms and about t he condit ions in which t hey r eact . 11. I f A is a beggar, t hen A is not r ich. 12. The gi ven st at ement i s U ni ver sal N egat i ve. Conclusion I I is Conver se of it . 13. The use of t er m ‘All’ in t he conclusion I makes it invalid. We know t hat , definit e conclusion cannot be dr awn fr om gener al st at ement . M i ni st er s ar r ived at t he public funct ion in t heir car s. Thus, mi ni st er s have car s and t hey at t ended publ i c funct ion. Ther efor e, only I I and I I I ar e implicit in t he st at ement . 14. Any per son can be philant hr ope whether he is r ich or poor. 15. The given dat a indicat es t hat t her e is a st eady incr ease of fer t ilizer consumpt ion.



10

Statement Assumption

CHAPTER

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S Direct ions for questions 1 t o 15: I n each quest i on bel ow i s gi v en a st at em en t f ol l ow ed by t w o assu m pt i on s/i n f er en ces n u m ber ed I an d I I . A n assumption is something supposed or taken for granted and an infer ence i s somet hing whi ch can be di r ect ly infer r ed fr om t he given fact s. You have t o consider t h e st at em en t an d t h e f ol l ow i n g assu m pt i on s/ infer ences and decide which of t hose is/ar e implicit in t he st at ement . Give answer : (a) if only I is i mpl icit (b) if onl y I I is impl icit (c) if eit her I or I I is impl icit (d) if neit her I nor I I is i mpl icit (e) if bot h I and I I ar e impl icit 1. St at ement : 'Do not ent er -avoi d t he r i sk of get t ing i nfect ed wi t h t he ABC disease'- wr i t t en out side t he quar ant ine war d no. 2 (meant only for ABC disease) of a hospi t al. Assumpt ions : I . Di sease ABC is cont agi ous. I I . Al l t he pat i ent s i n war d no. 2 suffer fr om disease ABC. 2. St at ement : I n cit y Z, peopl e pr efer t o buy Car X inst ead of Car Y as Car X has Ger man t echnology which i s ver y advanced. Assumpt ions : I . Car s wit h Ger man t echnology ar e per ceived t o be bet t er t han ot her car s i n cit y Z. I I . Had German technology been pr esent in Car Y also, its sales would have cr ossed car X's sales. 3. Statement : Railway does not pr ovide concession t o an y on e for t r avel l i ng t o cer t ai n hol i day dest inat ions. Assumpt ions : I . Rail way ser vices ar e avail abl e for t r avel li ng t o t hese holiday dest inat i ons.

I I . Rai l way s pr ovi des con cessi on t o cer t ai n per sons for t r avel l i ng t o pl aces ot her t han t hese hol iday dest inat i ons. 4. St at ement : "Tr avel l er s wi t h a t i ck et for t he second cl ass i f found t r avel li ng in t he fi r st class compar t ments would be penalized" - Not ice in the compar t ment s of a t r ai n. Assumpt ions : I . Tr avel ler s wi t h a t i cket for t he fi r st class ar e al so not al lowed t o t r avel i n t he second class compar t ment s. I I . I nspect i ons ar e car r i ed out i n t he t r ai n t o check t he t ick et s. 5. St at ement : The pr ices of pet r ol and di esel have r emained unchanged only in N iger i a since t he past t hr ee year s. Assumpt ions : I . Pet r ol an d -di esel pr i ces h av e ch an ged el sewher e i n t he wor l d dur i ng t hese t hr ee year s. I I . Befor e t hi s t hr ee year s per i od, pet r ol and di esel wer e avai lable at a pr ice differ ent fr om t he pr esent r at es. 6. St at ement : I f par ki ng space i s not availabl e in offi ce, par k your vehicl es in t he mall and wal k t o t he office. Assumpt ions : I . The mal l is at a walkabl e dist ance fr om t he office. I I . The office does not allow vi sit or s' vehi cles in it s pr emi ses. 7. St at ement : Far mer s must immediat ely swit ch ov er t o or gan i c f er t i l i zer s f r om ch em i cal fer t il izer s for bet t er yi el d. Assumpt ions : I . Al l t he far mer s use onl y chemi cal fer t il izer s. I I . Or ganic fer tilizer s ar e r eadily available t o t he far mer s.

10.2

Statement Assumption

8. St at ement : An adver t i sement by bank X-'Our int er est r at es for educat i on loans ar e l ower t han any ot her bank '. Assumpt ions : I . Some ot her bank s al so pr ovi de educat i on loans. I I . I nt er est r at es char ged on educat ion loans ar e di ffer ent for di ffer ent bank s. 9. St at ement : For any ki nd of pr oblem wit h your mobile phone, cont act our help desk immediately. Assumpt ions : I . H el p desk h as a sol u t i on t o al l k i nds of pr oblems r elated to mobile phones or will guide accor dingly. I I . Unless t he pr oblem is r epor t ed immediat el y, it cannot be sol ved. 10. St at ement : U se ou r m edi ci n e t o f i gh t t h e pr oblem of obesit y. Assumpt ions : I . Ot her sl i mming medi ci nes avail abl e i n t he mar k et do not r educe wei ght . I I . Obesi t y can n ot be con t r ol l ed w i t h ou t medi ci nes. 11. St at ement : The number of people l iving bel ow pover t y l ine i n ur ban ar eas has i ncr eased since last year. Assumpt ions : I . People li vi ng in r ur al ar eas ar e not bel ow t he pover t y l ine. I I . A si mi lar sur vey was conduct ed l ast year. 12. Statement : Tr avelling by t r ains r ather t han cabs is mor e convenient and economical in Par i s. Assumpt ions : I . Par i s is an expensive cit y. I I . Tr ai n ser vi ces ar e r easonabl y good in Par i s. 13. St at ement : A gover nment adver t i sement i n public int er est -For a child's bet t er ment al healt h, admi t him/ her t o a school onl y aft er fi ve year s of age. Assumpt ions : I . A chil d cannot l ear n befor e he/she t ur ns five. I I . Some school admi t chi l dr en who ar e below fi ve year s of age. 14. St atement : Alt hough he has done M BA t hr ough a cor r espondence cour se, he i s as smar t as a per son fr om coll ege X. Assumpt ions : I . St udent s fr om col lege X ar e k nown for t heir smar t ness.

I I . St u den t s sh ou l d do M B A t h r ou gh cor r espondence i n or der t o become smar t . 15. St at ement : Conveyance faci l i t y pr ovi ded by or gani sat ion hel ps empl oyees r epor t t o wor k on t ime. Assumpt ions : I . The conveyance faci lit y which i s pr ovide by t he or ganisat i on al ways r eaches on t ime. I I . I t i s not possible t o r epor t t o wor k on t i me unless conveyance facili t y is pr ovided by t he or ganisat ion.

LEVEL-1 Direct ions for questions 1 t o 15: I n each quest i on bel ow i s gi v en a st at em en t f ol l ow ed by t w o assu m pt i on s/i n f er en ces n u m ber ed I an d I I . A n assumption is something supposed or taken for granted and an infer ence i s somet hing whi ch can be di r ect ly infer r ed fr om t he given fact s. You have t o consider t h e st at em en t an d t h e f ol l ow i n g assu m pt i on s/ infer ences and decide which of t hose is/ar e implicit in t he st at ement . Give answer : (a) if only I is i mpl icit (b) if onl y I I is impl icit (c) if eit her I or I I is impl icit (d) if neit her I nor I I is i mpl icit (e) if bot h I and I I ar e impl icit 1. St at ement : A ver y lar ge number of people st ood in t he queue for buying t ickets for t he one-day inter national cr icket match scheduled to be played in the cit y on the next day. Assumpt ions : I . No ot her one-day int er nat ional cr ick et mat ch may be pl ayed i n t he ci t y for t he next si x mont hs. I I . M ajor i t y of t hose who st ood in t he queue may be abl e t o get t i ck et f or t h e on e-day int er nat i onal cr ick et mat ch. 2. St at ement : The hi ghway poli ce aut hor it y put up lar ge boar ds at r egular inter vals indicating t he speed l imit and danger s of over -speedi ng on t he highways. Assumpt ions : I . M ost of t h e m ot or i st s m ay dr i v e t h ei r v eh i cl es w i t h i n t h e speed l i m i t on t h e highways. I I . M ot or i st s gener all y i gnor e such caut ions and over -speed on t he highways.

Statement Assumption

3. St at ement : The empl oyees' associ at i on ur ged its member s to stay away fr om the annual function as many of t heir demands wer e not met by t he management .

10.3

8. Statement : The lar gest computer manufactur ing company slashed the pr ices of most of t he deskt op model s by about 15 per cent wi t h i mmedi at e effect .

Assumpt ions :

Assumpt ions :

I . M ajor i t y of t he member s of t he associat i on may not at t end t he funct ion.

I . The company may incur heavy losses due t o r educt i on i n pr i ces of t he desk t op.

I I . The management may cancel t he annual funct ion.

I I . The sal es of desk t op manufact ur ed by t he company i ncr ease subst ant i al ly in t he near futur e.

4. St at ement : The sar panch of t he vi llage call ed a meet ing of all t he heads of t he famil ies t o discuss t he pr obl em of acut e shor t age of dr i nk ing wat er in t he vi ll age. Assumpt ions : I . The sar panch had ear lier called such meetings t o discuss about var i ous pr obl ems.

9. St atement : The school aut hor i t y decided t o r ent out t he school pr emi ses dur i ng weekends and hol i days for or gani zi n g var i ous funct i ons t o augment i t s r esour ces t o meet t he gr owing needs of t he school. Assumpt ions :

I I . M ost of t he heads of famil ies may at t end t he meet ing call ed by t he sar panch.

I . The par ents of the school students may pr ot est agai nst t he deci sion of t he school aut hor it y.

5. St at ement : The muni ci pal cor por at ion advised all t he people l i vi ng in t he shant i es along t he beaches t o move t o higher places dur ing monsoon.

I I . Ther e may not be enough demand for hi r i ng t he school pr emi ses for or ganizing funct i ons.

Assumpt ions : I . M any people li vi ng in t he shant ies may l eave t he ci t y and r elocat e t hemsel ves elsewher e in t he st at e. I I . M ajor it y of t he peopl e l iving i n t he shant i es al ong t he beach may t r y t o r elocat e t o hi gher pl aces dur i ng monsoon. 6. St at ement : T h e l ar gest dom est i c ai r l i n es corpor ation has announced new summer schedules in which mor e number of flight s in tr unk r outes ar e int r oduced. Assumpt ions :

10. St at ement : The local civic body has ur ged all t he r esident s t o volunt ar i ly r educe consumpt i on of pot abl e wat er by about 30 per cent t o t ide over t he wat er cr i si s. Assumpt ions : I . M any r esi dent s may r educe consumpt ion of pot able wat er. I I . M any act i vist s may wel come t he civic body's move and spr ead awar eness among r esident s. 11. St at ement : The dr iver of t he huge t r uck pull ed t he emer gency br ak es t o avoi d hi t t i ng t he aut o r ick shaw whi ch suddenl y came i n fr ont of t he t r uck.

I . Mor e number of passenger s may tr avel by this ai r l ines cor por at i on dur i ng summer mont hs in t r unk r out es.

Assumpt ions :

I I . Ot her air li nes companies may al so i ncr ease t he number of fl ight s i n all t he sect or s.

I I . The t r uck dr iver may be able t o st op t he t r uck befor e it hi t s t he aut o r i ckshaw.

7. St at ement : The Chai r man of t he company decided to hold a gr and funct ion to celebr ate silver jubi lee dur i ng t he next weekend and invit ed a lar ge number of guest s.

12. St at ement : The doct or war ned t he pat i ent agai nst any fur t her consumpt ion of alcohol, i f he desi r ed t o get cur ed fr om t he ai lment and li ve a longer l ife.

I . The aut o r ickshaw dr iver may be abl e t o st eer hi s vehicle away fr om t he oncomi ng t r uck.

Assumpt ions :

Assumpt ions :

I . The company offi ci als may be abl e t o make al l t he necessar y pr epar at ions for t he si lver cel ebr at i on.

I . The pat ient may follow t he doct or 's advice and st op consumi ng alcohol.

I I . Major ity of the guests invited by the Chair man may at t end t he funct ion.

I I . The doct or may be abl e t o cur e t he pat ient f r om t h e ai l m en t , i f t h e pat i en t st ops consumi ng alcohol.

10.4

Statement Assumption

13. Statement : The Chair man of t he company ur ged al l t he empl oyees t o r efr ain fr om maki ng long per sonal calls dur i ng wor k ing hour s in or der t o boost pr oduct i vit y. Assumpt ions : I . M aj or i t y of t h e em pl oyees m ay r espon d posi t i vel y t o t he Chair man's appeal. I I . M ost of t he empl oyees may cont i nue t o make long per sonal call s dur ing wor ki ng hour s. 14. St at ement : The local cul t ur al club decided t o or gani se a musi cal event t o r ai se money t he const r uct ion of t he club bui lding. Assumpt ions : I . The local r esi dent s may not all ow t he club t o or ganise t he musi cal event in t he local it y. I I . The money collected by or ganizing the musical event may be subst ant ial enough for t he cl ub t o st ar t const r uct ion. 15. St at ement : The t r affi c pol ice depar t ment has put huge not ice boar ds at al l t he major junct ions of t he ci t y war ning dr i ver s t o r efr ain fr om usi ng cell phones whil e dr ivi ng or else t heir li cences wi ll be i mpounded. Assumpt ions : I . The dr i ver s of t he vehi cles may i gnor e t he war ning and cont inue using cell phones while dr iving. I I . The t r affic poli ce depar t ment may be abl e t o nab most of t he offender s and i mpound t heir l i cences.

LEVEL-2 D irect ions for quest ions 1 t o 10: Each quest i on below has a st at ement foll owed by t hr ee assumpt ions number ed I , I I and I I I . An assumpt ion is somet hi ng supposed or t aken for gr ant ed. You have t o consider t he st at ement and t he assumpt i ons and decide whi ch of t he assumpt ions is i mplicit i n t he st at ement . Then deci de which of t he answer s (a), (b), (c), (d) and (e) is t he cor r ect answer. 1. St at ement : "The r et ur n of count r y X as a ful lf l ed ged m em ber of t h e Com m on w eal t h i s dependent on t he 'cr edibility' of t he election which w i l l be i n pr ocess n ex t y ear ." - H ead of Commonwealt h Obser ver s' Gr oup (COG). Assumpt ions : I . The 'cr edibil it y' of t he elect ion pr ocess can be measur ed in t angible t er ms. I I . The el ect i on pr ocess i n count r y X al ways r emai ns a mat t er of debat e for t he wor l d communi t y.

I I I . The wor ld communit y has a common desi r e t o r est or e democr acy and i t s et hi cs al l over t he wor l d. (a) Al l I , I I and I I

(b) None

(c) Onl y I

(d) Only I I

(e) None of t hese 2. St a t em en t : " T h e I n di an M et eor ol ogi cal Depar t ment has pr oposed t o hold a br ainst or ming session of weat her exper t s next mont h t o t r y and st udy t he causes for t he fai l ur e of t he j ust concl uded summer monsoon i n t he count r y."Spok esper son of I n di an M et eor ol ogi cal Depar t ment . Assumpt ions : I . The behaviour of t he just -concluded monsoon was int r igui ng. I I . Thor ough scient i fi c i nvest igat ion i s lik el y t o r eveal t he causes of monsoon's fail ur e. I I I . Br ainst or mi ng sessions or gani sed in t he past didn't pr ove fr uit ful. (a) Al l I , I I and I I I (b) None (c) Only I I

(d) Only I and I I

(e) None of t hese 3. St at ement : "Why ar e you look ing sad? Did you not get bonus t his year too''? - M r. X said t o M r. Y. Assumpt ions : I . M r. X is t he wel l-wisher of M r. Y. I I . The expr ession of sadness on one's face is a si gn of one's financial cr unch. I I I . I f one get s bonus, one does not r emai n sad. (a) All

(b) Only I

(c) Only I and I I

(d) Only I I and I I I

(e) None of t hese 4. St at ement : "A non-Br ahmi n well-ver sed wi t h r i t uals could be appoint ed as a pujar i as wel l" r uli ng of t he Supr eme Cour t (SC). Assumpt ions : I . B r ah m i n s, don 't h av e a m on opol y ov er per for ming puja in a t empl e. I I . M er e el igi bil i t y for a post i s enough t o l ay cl ai m t o candidat ur e for t he post . I I I . People wi ll comply wi t h t he ver di ct gi ven by t he SC. (a) All I , I I and I I I (b) Only I and I I (c) Only I I and I I I (d) Onl y I and I I I (e) None of t hese

Statement Assumption

5. Statement : "Ever y successful per son who claims t o have come fr om a village has achieved success aft er he or she left t he village." - A leader of par t y X. Assumpt ions :

10.5

8. Statement : "A cour t can convict an accused solely on t he basi s of a dyi ng decl ar at ion but such a decl ar at i on should be fr ee fr om any doubt and t h e vi ct i m mak i ng t he st at ement sh oul d be ment all y fit ." - Supr eme Cour t .

I . The aver age I ndian village is a place with little capit al, l ow t echnol ogy and l i mi t ed mar ket access.

Assumpt ions :

I I . Oppor tunity for gr owth is mor e in metr o cities.

I I . Declar ation made by t he dying per son is likely t o be t r ue.

I I I . Oppor tunities for gr owth are scarce in villages. (a) All I , I I and I I I (b) Only I and I I I (c) Only I I and I I I (d) Only I and I I (e) None of t hese 6. Statements : "Today I am rejecting your proposal to play a cr icket match against your team because of the absence of Mr. Z but tomor r ow I am ready to play against your team at any cost." - Mr. X says to Mr. Y. Assumpt ions :

I . One who i s not ment al l y fi t al ways mak es st at ement s unt r ust wor t hy i n nat ur e.

I I I . I t i s possi bl e t o di st i n gu i sh w h et h er a decl ar at i on is doubt ful or not . (a) Only I

(b) Only I and I I

(c) Only I I and I I I (d) Only I I I (e) None of t hese 9. St at ement : "The 'X' st at e cabi net endor sed our pr oposal t o st agger shoppi ng t i mi ng by allowi ng mar k et s t o r emai n open t il l 10 PM t o mak e it mor e dynami c." - A leader of t r ader s.

I . M r. Z wil l be avail able t omor r ow.

Assumpt ions :

I I . M at ch wil l be pl ayed t omor r ow i r r espect i ve of avai labi lit y of M r. Z.

I . I f i t i s t ak en wel l by t r ader s, t h e st at e gover nment will per manently alter the closing t i me for all pr ominent mar ket s in t he cit y.

I I I . M r. Y wi ll be r eady t o pl ay a mat ch against M r. X t omor r ow. (a) All I , I I and I I I (b) Eit her I or I I and I I I (c) Only I I and I I I (d) Onl y I and I I I (e) None of t hese 7. Stat ement : "An ast er oi d whi ch bur ned up i n t h e ear t h 's at m osph er e i n J u n e cou l d h ave t r igger ed a mi st ak en nucl ear war bet ween I ndia and Pakist an had it det onat ed over Sout h Asia." St at ement of M r X i n a r epor t i n The N ew Yor k Ti mes. Assumpt ions : I . I ndia and Pakist an ar e equipped wit h nuclear weapons. I I . N ei t h er I n di a n or Pak i st an h ad t h e sophisticated sensor s that could deter mine the differ ence between a natur al NEO (near -ear th object ) i mpact and a nuclear det onat ion.

I I . Ther e is a need t o boost commer cial act i vi t i es in t he cit y. I I I . K eeping t he mar k et open t ill l at e i s lik el y t o enhance sales as well as mak e shopping mor e convenient for people with late wor king hour s. (a) Only I

(b) Only I and I I

(c) Only I I and I I I (d) Onl y I and I I I (e) None of t hese 10. St at ement : The hi gh cour t of st at e 'X' has di r ect ed t he st at e gover nment 'X' t o i ssue a not i fi cat ion mak ing speed gover nor s compul sor y for al l four -wheel er s i n t he cit y t o check highspeed dr iving. Assumpt ions : I . Speed gover nor s wil l put an end t o accident s. I I . Speed gover nor s wi l l hel p i n r educi ng t he possibili t i es of r oad acci dent s.

I I I . I ndia and Pakist an have no good r elat ionship wi t h each ot her.

I I I . H i gh speed of vehicles on t he r oad causes accident s.

(a) Only I

(a) Only I

(b) Only I and I I

(b) Only I and I I

(c) Only I I and I I I

(c) Only I I and I I I

(d) Onl y I and I I I

(d) Onl y I and I I I

(e) All I , I I and I I I

(e) None of t hese

10.6

Statement Assumption

Directions for questions 11 to 15: I n each quest i on below is given a statement followed by two assumptions number ed I and I I . An assumpt i on i s somet hi ng supposed or t aken for gr ant ed. You have t o consider t he st at ement and t he fol l owi ng assumpt i ons and deci de whi ch of t he assumpt i ons i s i mpli ci t i n t he st at ement . Gi ve answer :

I I . I f a differ ent cr op i s gr own i n t he successi ve season , n o addi t i on al n u t r i en t s su ch as fer t ili zer s ar e r equir ed t o be added t o t he soil. 13. St at ement : I f far mer s wai t t o i mpr ove t hei r yiel d, t hey must use or ganic fer t ili zer s i n place of chemical fer t i lizer s. Assumpt ions :

(a) if onl y Assumpt i on I is impl icit

I . Chemical fer t i lizer s have cer t ain ill effect s on health.

(b) if only Assumpt ion I I is i mpl icit (c) if eit her Assumption I or Assumpt ion I I is implicit (d) i f nei t her Assumpt i on I nor Assumpt i on I I i s implicit (e) if bot h Assumpt i ons I and I I ar e impl icit

I I . Chemical fer t i lizer s do not pr oduce as much yi el d as t he or gani c fer t i lizer s. 14. St at ement : St or e eat ables in t he deep fr eeze in or der t o pr eser ve t hese for a l ong t ime.

11. St at ement : A l eading uni ver si t y has begun a pr actice of displaying r esult s only on t he I nter net r at her t han on t he main not ice boar ds.

Assumpt ions : I . Food mater ial r emains eat able even aft er deep fr eezing for a l ong t ime.

Assumpt ions : I . Al l t he st udent s enr olled wit h t he uni ver si t y have access t o I nt er net at home. I I . M ost of t he st udent s r efer r ed t o t he r esul t s di splayed on bot h t he int er net as well as t he not i ce boar ds ear l ier.

I I . I t i s not possible t o st or e any eat abl e at r oom t emper at ur e even for a shor t er per iod of t ime. 15. St at ement : A l eading NGO decided t o open a li br ar y cont ai ni ng book s and newspaper s of all major publi sher s in a r emot e vi ll age. Assumpt ions :

12. St at ement : I n or der t o r epleni sh t he nut r i ent s in t he soil , i t is impor t ant t o gr ow differ ent t ypes of cr ops ever y alt er nat e season.

I . All ot her near by villages alr eady have similar li br ar ies. I I . Ther e i s adequat e number of l it er at e people in t he vi ll age.

Assumpt ions : I . A cr op can never be gr own for t he second t ime in t he same fi el d.

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (e) 11. (b)

2. (a) 12. (b)

3. (e) 13. (b)

4. (b) 14. (a)

5. (e) 15. (d)

6. (a)

7. (b)

8. (e)

9. (a)

10. (d)

7. (e)

8. (b)

9. (d)

10. (b)

7. (e)

8. (c)

9. (c)

10. (c)

LEVEL-1 1. (b) 11. (e)

2. (e) 12. (c)

3. (a) 13. (a)

4. (b) 14. (d)

5. (b) 15. (c)

6. (a)

LEVEL-2 1. (b) 11. (d)

2. (d) 12. (d)

3. (d) 13. (b)

4. (a) 14. (a)

5. (e) 15. (d)

6. (c)

11

Mathematical Puzzles

CHAPTER

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S

6. In the following question find the missing number.

1. Find the missing number:

4

3 11

144

9 15

4 5 3 17

6

?

9801

(a) 2250

(b) 8100

(c) 11036

(d) 1216

12 5

16 9

64 81

? 49

(a) 75

(b) 60

(c) 30

(d) 90

(b) 6

(c) 7

(d) 10

7. In the following question find the missing number: 2 3 5 25

15 7

(c) 8

(d) 13

(b) 15

(c) 18

(d) 20 17 6 3 34

9

35

7

36

3

4

1

2

? (b) 4

(c) 5

(d) 6

9.

4

2

(a) 3

10.

5. Select the missing number from the given responses.

9

6

4

15

8

5

12

16

8 ?

8

6

12

23 8 2 92

25 18 ? 150

(a) 1

(b) 3

(c) 5

(d) 6

Directions for questions 9 and 10: Some equations are solved on the basis of certain system. Find out the correct answer for the unsolved equation on that basis.

4. Find the missing number:

27

3 2 ? 100

8. In the following question find the missing number:

6 5 12 27 20 18 16 13 12 10 ? 11 13 9 4 (b) 12

1 4 10 50

(a) 12

3. Find the missing number:

(a) 9

8 9 ? 65

(a) 5

2. Find the missing number:

48 25

6 7 4 38

14

11

3 8 5 853

1 4 7 471

? 5 9 593

(a) 2

(b) 3

(c) 4

(d) 0

3 4 7 426

5 8 4 643

6 2 8 ?

(a) 554

(b) 693

(c) 184

(d) 717

11. If 92 + 53 = 72 and 31 + 33 = 20, then 52 + 91 = ?

(a) 12

(b) 10

(a) 38

(b) 143

(c) 8

(d) 6

(c) 39

(d) 75

11.2

Mathematical Puzzles

12. If 37 + 17 = 20 and 68 + 42 = 26, then 96 + 63 = ? (a) 74

(b) 159

(c) 33

(d) 62

6. In the following question, select the number which can be placed at the sign of question mark (?) from the given alternatives.

13. If 53 + 21 = 73 and 45 + 32 = 106, then 29 + 63 = ? (a) 150

(b) 49

(c) 202

(d) 178

3 10 6 186 9 5 3 138 5 3

14. 5 × 2 × 7 = 752, 4 × 7 × 3 = 347,1 × 4 × 9 = ? (a) 914

(b) 441

(c) 419

(d) 941

15. If 83 ÷ 42 = 5, 63 ÷ 21 = 6, then 56 ÷ 23 = ? (a) 11

(b) 6

(c) 9

(d) 17

(c) –1

(d) 2

2. In the following question, select the number which can be placed at the sign of question mark (?) from the given alternatives.

3 2

31 1 5

4

2

2 145 6 3

1

?

7

5

(a) 43

(b) 49

(c) 59

(d) 71

3. If 3 @ 3 * 3 = 3 and 48 @ 4 * 3 = 36, then 91 @ 13 *2=? (a) 4

(b) 8

(c) 10

(d) 14

4. In the following question, select the number which can be placed at the sign of question mark (?) from the given alternatives.

(c) 45

(d) 95

7. In the following question, select the number which can be placed at the sign of question mark (?) from the given alternatives.

2 5 14 9

(a) 11

(b) 16

(c) 21

(d) 31

5. If (3)2 @ 1 * 7 = 98 and (4)2 @ 2 * 16 = 178, then (5)2 @3*9=?

7 10 19 14

8 11 20 ?

(a) 14

(b) 15

(c) 17

(d) 19

8. If 6 @ 4 @ 7 = 101 and 2 @ 5 @ 11 = 150, then what is the value of A in A @ 8 @ 9 = 289? (a) 5

(b) 8

(c) 12

(d) 17

9. In the following question, select the number which can be placed at the sign of question mark (?) from the given alternatives.

20 72

90 110

2 3 6

3

56 ?

7 4

1 7 4

(a) 112

(b) 144

(c) 156

(d) 186

10. In the following question, select the number which can be placed at the sign of question mark (?) from the given alternatives.

3 2

12 13 156 14 ? 154 15 13 195

36 ? (b) 42

1. If 52 ÷ 24 = 1, and 83 ÷ 45 = 2, then 41 ÷ 33 = ? (b) 0

1 5

(a) 35

LEVEL-1 (a) 1

7 2

625

5 3 4096 4 2 ? (a) 216

(b) 1024

(c) 1296

(d) 2024

11. In the following question, select the number which can be placed at the sign of question mark (?) from the given alternatives.

(a) 218

3 2

1 2

2

(b) 262

5 4 1 7

7 6 4 2

3 11 1 ?

(c) 253 (d) 259

(a) 2

(b) 4

(c) 6

(d) 8

1

Mathematical Puzzles

12. If 9 * 2 * 5 = 23 and 1 * 4 * 8 = 29, then 1 * 6 * 3 = ? (a) 19

(b) 21

(c) 31

(d) 39

13. In the following question, select the number which can be placed at the sign of question mark (?) from the given alternatives.

9

6

8

117

2

7

68

(a) 48

3

14. In the following question, select the number which can be placed at the sign of question mark (?) from the given alternatives.

4

1

5 7

3 5

5 6

(c) 6

(d) 36

?

(d) 58

2

4. Find the missing number. 2 7 9 7 3 4 9 8 ? 126 168 216 (a) 8 (b) 3 5. Find the missing number.

(b) 52

(c) 55

39 17 ? (a) 11

(b) 31

(c) 32

(d) 37

18

27

75

41

60

19

45

45

?

(a) 20

(b) 30

(c) 15

(d) 35

6. Find the missing number.

52

55

63

69

49

79

57

?

(a) 117

(b) 120

(c) 107

(d) 110

7. Find the missing number shown by the question mark (?) in the center of the third figure.

22

11

19

11

LEVEL-2 13 (a) 8

1. Find the missing number.

7

12

22

9

45

5

4

5

93

99

4 11

7

?

3

(a) 3

(b) 33

(c) 22

(d) 9

20

7 13

4

(c) 12

6

5

3

30

1

4

46

3 1

6

2

4

? 8

(a) 111

(b) 113

(c) 36

(d) 288

2

7 12 (b) 10

1 2

 576

2 3 4 5

(d) 14

(a) 3

(b) 9

(c) 1

(d) 2

9. Find the missing number.

25 16 49 9 27 ? 64 125

?

(a) 14400

(b) 15600

(c) 23040

(d) 17400

14

8. Which of the following number can replace the question mark in the figure given below?

3. Find the missing number.

3 4

10 ?

2. Find the missing number.

2

11.3

(a) 7

(b) 343

(c) 216

(d) 6

11.4

Mathematical Puzzles

10. Find the missing letters. C D C E ? G

A

13. Find the missing number: 13 27

12

?

24 27

I 25

51

64

(a) F, G

(b) F, F

(a) 35

(b) 36

(c) G, H

(d) H, F

(c) 37

(d) 38

11. Find the missing number: 2

3

5

14. Find the missing number: 6

7

8

9

3

7

45

20

40

5

25 27 35

60 30 40

25 ? 35

30

30

30

?

41

63 6

8

(a) 82

(b) 83

(c) 86

(d) 26

12. Find the missing number: 7

19

4

9

9

?

?

3

6

(a) 36

(b) 33

(c) 45

(d) 60

15. Find the missing number: 8

28

22 69

46

5

38

42 79

12

(a) 27 (b) 21

24

(c) 28

(a) 40

27 (b) 41

(d) 17

(c) 31

(d) 51

?

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (b)

2. (b)

3. (b)

4. (c)

5. (d)

6. (c)

7. (d)

8. (b)

9. (b)

10. (d)

11. (a)

12. (c)

13. (c)

14. (a)

15. (b)

1. (c)

2. (d)

3. (d)

4. (a)

5. (b)

6. (a)

7. (b)

8. (c)

9. (c)

10. (c)

11. (b)

12. (a)

13. (d)

14. (b)

1. (c)

2. (b)

3. (a)

4. (c)

5. (b)

6. (a)

7. (a)

8. (d)

9. (b)

10. (a)

11. (a)

12. (b)

13. (c)

14. (b)

15. (b)

LEVEL-1

LEVEL-2

Mathematical Puzzles

11.5

EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S

9.

3

8

5

8

5

3

1

4

7

4

7

1

?

5

9

5

9

3

2

1. 4 × 3 = 12 and (12) = 144 11 × 9 = 99 and (99)2 = 9801 15 × 6 = 90 and (90)2 = 8100 2. 12 × 4 = 48 25  5

16 × 4 = 64 81  9

 ? = 3.

Similarly, 15 × 4 = 60

10. In column 1, (3 + 1) (4  2) (7 – 1) = 426.

49  7

In column 2,

3. (27 + 18) – (12 + 13) 

(5 + 1) (8  2) (4 – 1) = 643.

45 – 25 = 20

Similarly, in column 3,

(16 + 12) – (6 + 9) 

28 – 15 = 13

(10 + 11) – (5 + 4) 

(6 + 1) (2  2) (8 – 1) = 717. 11. Expression 92 + 53 = 72



9 – 2 = 7 and 5 – 3 = 2

Expression 31 + 33 = 20



3 – 1 = 2 and 3 – 3 = 0.

21 – 9 = 12

4. 27 = 9 × (2 + 1) 35 = 7 × (3 + 2)

 52 + 91 = (5 – 2) = 3 and (9 – 1) = 8 

3 – 1 = 2 and 7 – 7 = 0.

Expression 68 + 42 = 26



6 – 4 = 2 and 8 – 2 = 6.

5. First figure

 96 + 63 = (9 – 6) = 3 and (6 – 3) = 3

12 – 6 = 6; Second figure 12 – 4 = 8; 16 – 8 = 8

13.



33.

Expression 53 + 21 = 73  (5 + 3)2 + (2 + 1)2 = 82 + 32 = 73. Expression 45 + 32 = 106

Third figure

 (4 + 5)2 + (3 + 2)2 = 92 + 52 = 106.

11 – 5 = 6;

 29 + 63

14 – 8 = 6

 (2 + 9)2 + (6 + 3)2 = 112 + 92 = 202.

6. In column 1, 4 × 5 – 3 = 17

38.

12. Expression 37 + 17 = 20

36 = 4 × (4 + 5)

15 – 9 = 6



14. 5 × 2 × 7 = 7 5 2

In column 2, 6 × 7 – 4 = 38  In column 3, 8 × 9 – ? = 65  ? = 7.

4 × 7 × 3

=

3

4

7

1 × 4 × 9

=

7. In column 1, (2 + 3) × 5 = 25 In column 2, (1 + 4) × 10 = 50  In column 3, (3 + 2) × ? = 100  ? = 20.

9

1

4

Similarly,

8. In column 1, 17 × 6  3 = 34 In column 2, 23 × 8  2 = 92

15. 83 ÷ 42 = 5 (8 – 4) + (3 – 2) = 4 + 1 = 5.

 In column 3, 25 × 18  ? = 150

63 ÷ 21 = 6 (6 – 2) + (3 – 1) = 4 + 2 = 6.

 ? = 3.

Similarly, 56 ÷ 23 (5 – 2) + (6 – 3) = 3 + 3 = 6.

11.6

Mathematical Puzzles

LEVEL-1 1. 52 ÷ 24 = 1



(5 – 2) + (2 – 4) = 3 – 2 = 1

83 ÷ 45 = 2



(8 – 4) + (3 – 5) = 4 – 2 = 2

Similarly, 41 ÷ 33



8. {(5)2 + (4)2} – {(3)2 + (2)2} = (25 + 16) – (9 + 4) = 41 – 13 = 28 Similarly, {(8)2 + (4)2} – {(8)2 + (x)2} = 12 (Let x = ?)  (64 + 16) – (64 + x2) = 12  x2 = 4

(4 – 3) + (1 – 3) = 1 – 2 = – 1.

LEVEL-2

52 = 25, 53 = 125 and 32 = 9, 33 = 27

7 × 5 + (6 + 4) = 45

42 = 16, 43 = 64 and 72 = 49, 73 = 343.

12 × 7 + (5 + 4) = 93 10.

A

 x = 22. 2. We have

x = 2.

9. Corresponding sector have square and cubes of a number.

1. We have

Similarly, 22 × 3 + (11 + x) = 99



C +2

C

E +2

D

G +2

E

I +2

F

G

12 + 22 + 32 + 42 = 30 12 + 22 + 42 + 52 = 46 Similarly, 22 + 82 + 62 + 32 = 113. 3. 1 × 2 × 3 × 4 = 24

4.

+1

+1

+1

+1

11. Take the sum of the products of the upper and lower numbers respectively to get the central number.

24 × 24 = 576

5 × 3 = 15 and 6 × 8 = 48

2 × 3 × 5 × 4 = 120

15 + 48 = 63

120 × 120 = 14400.

2 × 7 = 14 and 3 × 9 = 27

Column  1 2 × 7 × 9 = 126 Column  2 7 × 3 × 8 = 168 Column  3  9 × 4 × 6 = 216

Hence, 6 should replace the ‘question mark’. 5. The sum of the numbers in a row is equal to 120. i.e. 18 + 27 + 75 = 45 + 75 = 120 41 + 60 + 19 = 79 + 41 = 120 Similarly, 45 + 45 + 30 = 120 Hence, 30 should replace the question mark. 6. Difference of the numbers in the bottom row is 10 times the difference of the number in the upper row.

14 + 27 = 41 Therefore, 6 × 7 = 42 and 8 × 5 = 40 42 + 40 = 82 12. 7 × 4 = 28;

19 + 9 = 28.

8 × 5 = 40; 28 + 12 = 14 9 × 3 = 27;

27 – 6 = 21

13. 12 + 13 = 25 27 + 24 = 51  ? = 64 – 27 = 37 14. 25 + 45 + 35 + 30 = 135 135  27 5

60 + 20 + 40 + 30 = 150

 ? – 57 = 10(69 – 63) = 10 × 6  ? = 57 + 60 = 117 Hence, 117 should replace the question mark. 7. In each figure, sum of the numbers at the right hand side vertices of the square is subtracted from the sum of the numbers at the left hand side vertices of the square to get the number in the center.

150  30 5

25 + 40 + 35 + 65 = 165 165  33 5

15. 24 + 22 = 46

Therefore, the required number is

27 + 42 = 69

(20 + 12) – (10 + 14) = 8.

 ? = 79 – 38 = 41 

12

Cubes and Dice

CHAPTER CU BES To t est t he abilit y of t he candidat es t o t hink about 3-dimensional figur es, quest ions ar e oft en asked on t he cube. Two t ypes of quest ions ar e gener ally asked on t he cube. 1. Var ious faces of the cube having differ ent number s (usually 1 t o 6) on each face ar e shown side by si de. The candi dat e i s t hen ask ed t o fi nd t he number s on some of t he hidden faces. Somet imes some objects or geomet r ical figur es may be dr awn on t he faces of t he cube. e.g. These ar e t he faces of a cube : 6 1

5 4

3

4 5

6

1 6

2

4

Q. Which number appear s on t he face opposit e t o t he face wit h number 6 in t he cube ? A. 2. As 1, 4, 3 & 5 ar e neighbour s of 6, only 2 is t he one opposit e t o it . 2. The faces of a given cube ar e paint ed, eit her in t he same colour or in differ ent colour s. The cube is t hen cut int o smaller cubes and one has t o find out the number of cubes with different specifications. H er e, one should r emember t hat t he number of cubes of side 1 unit t hat can be obt ained when a cube of side L unit s is cut , is L 3 = L L  L . One has t o visualise t he colour s on t he faces of t he smaller cubes as t he lar ger cube is cut . e.g. Given a cube of side 3cm. Opposit e faces of t he cube ar e paint ed r ed, gr een and blue, each a differ ent colour. The cube is cut into smaller cubes of side 1 cm each. Find t he number of cubes wit h (i ) 3 faces paint ed; (ii ) one face paint ed; and (iii ) no face paint ed.

Red

Green

Tot al number of smaller cubes = 3  3  3 = 27 (i ) So, as t he cube has eight cor ner s, only 8 small cubes will have 3 faces paint ed. (ii ) The cubes in t he middle of each side will have one face paint ed. As t her e ar e 6 sides, so t her e ar e 6 such cubes. (iii ) Only t he cube in t he exact middle shall have no side paint ed so, t he answer is 1.

I f a cube of dimension n  n  n is paint ed on all six suface, t hen smaller cubes for med will have  Tot al numebr of cubes = n 3  Tot al numebr of cubes paint ed on t hr ee sides = 8  Cubes paint ed on t wo sides = (n – 2)  4  3  Cubes paint ed on single side = (n – 2)2  6  Cubes paint ed on no sides = (n – 2)3. Case I . When n = 3

In   

t he above figur e Tot al number of cubes = n 3 = 33 = 27 Tot al number of cubes paint ed on t hr ee sides = 8 Tot al number of cubes paint ed t wo sides = (n – 2)  4  3 = (3 – 2)  4  3 = 12  Cubes paint ed on single side = (n – 2)2  6 = (3 – 2)2  6 = 6  Cubes paint ed on no sides = (n – 2)3  (3 – 2)3 = 1. Case I I . When n = 4

12.2

Cubes and Dice

In 1. 2. 3.

t he above figur e Tot al number of cubes = n 3 = 43 = 64 Tot al number of cubes paint ed on t hr ee sides = 8 Tot al number of cubes paint ed on t wo sides = (n – 2)  4  3 = (4 – 2)  4  3 = 24 4. Tot al numebr of cubes paint ed on single side = (n – 2)2 6 = (4 – 2)2 6 = 24 5. Cubes paint ed on no sides. = (n – 2)3 = (4 – 2)3 = 8. Case I I I . When n = 5

In 1. 2. 3.

t he above figur e Tot al number of cubes = n 3 = 53 = 125 Tot al number of cubes paint ed on t hr ee sides = 8 Cubes paint ed on t wo sides = ( n – 2)  4  3 = (5 – 2)  4  3 = 36 4. Cubes paint ed on single side = (n – 2)2 6 = (5 – 3)2 6 = 56 5. Cubes paint ed on no sides. = (n – 2)3 = (5 – 2)3 = 1.

2.

num ber of 3 cubes = n Thr ee si des pai nt ed cubes =8

n=3

n=4

n=5

n=6

27

64

125

216

8

8

8

8

24

36

48

3.

Two si des pai nt ed cubes = (n – 2)  4  3

12

4.

Si ngl e sides pai nt ed cubes 2 = (n – 2)  6

6

24

54

96

5.

No si de pai nt ed cubes = (n – 2)3

1

8

27

64

DI CE S Dices ar e cubi cal st r uct ur es in which number s or point s fr om 1 t o 6 ar e mar ked on sides. Pr oblems based on dices ar e ver y simple in nat ur e. Nor mally t her e ar e t wo cases: 1. Sum of number s on opposit e sides is seven. Nat ur ally t he faces opposit e t o each ot her will be 1– 6 2– 5 3– 4 2. When t hr ee differ ent posit ions of dice is given. I n such cases, fir st of all we will det er mine t he face which is not adjacent t o t he given number. That number will naturally be the number s marked on opposit e face. Example. 1 A dice wit h it s face number ed 1 t o 6, is shown in t hr ee differ ent posit ions x , y and z.

Find opposit e faces. Sol ut ions. Faces adjacent t o 6 ar e

Nat ur ally 1 will be mar ked opposit e t o 6. Faces adjacent t o 4, ar e

Nat ur ally 2 will be mar ked opposit e t o 4. Thus opposit e faces ar e 6– 1 4– 2 3– 5

Cubes and Dice 12.3

PRACTI CE EXERCI SE Directions (Q. 1 – 5): A cube is colour ed r ed on one face, green on the opposite face, yellow on another face and blue on a face adjacent t o t he yellow face. The ot her t wo faces ar e left uncolour ed. I t is t hen cut int o 125 smaller cubes of equal size. Answer t he following quest ions based on t he above statement. 1. H ow many cubes ar e uncolour ed on all t he faces? (a) 27 (b) 36 (c) 48 (d) 64 2. H ow many cubes ar e colour ed blue on one face, r ed or gr een on anot her face and have four uncolour ed faces ? (a) 8 (b) 12 (c) 16 (d) 23 3. H ow many cubes ar e colour ed r ed on one face, and have t he r emaining faces uncolour ed? (a) 8 (b) 10 (c) 12 (d) 16 4. H ow many cubes have at least one gr een face ? (a) 4 (b) 5 (c) 16 (d) 25 5. How many cubes have at least two colour ed faces ? (a) 23 (b) 21 (c) 20 (d) 19 6. Two posit ions of a dice ar e shown below. I f 1 is at t he bot t om which number will be on t he t op ? (a) 4

Directions (Q. 9 – 10) : Six sides of a cube are coloured

with different colours. Yellow is opposite white and blue is between red and green. One side is black. 9. Which colour is opposit e blue ? (a) r ed

(b) black

(c) gr een

(d) None of t hese

10. Which colour does not t ouch r ed side ? (a) gr een

(b) yellow

(c) whit e

(d) blue

Directions (Q. 11– 12) : A cube paint ed blue on all the faces is cut into 125 cubes of equal size. Now answer t he following quest ions : 11. H ow many cubes ar e not paint ed on any face ? (a) 8

(b) 16

(c) 27

(d) 54

12. H ow many cubes ar e paint ed on one face only ? (a) 8

(b) 16

(c) 36

(d) 54

D i r ect i on s ( Q . 13– 17) : St u dy t h e f ol l ow i n g infor mat ion and answer t he quest ions given below: (i ) A r ectangular wooden block is having lenght 6 cm, br eadt h 4 cm and height 1 cm. (ii ) Bot h sides having dimensi on 4 cm × 1 cm ar e paint ed wit h black colour. (iii )Both sides having dimension 6 cm × 1 cm ar e painted in r ed colour. (iv) Both sides with dimension 6 cm × 4 cm ar e painted in gr een colour.

(b) 3 (c) 2 (d) 5 7. I f the cube is tur ned twice to t he r ight , which will be t he hidden number s? (a) 1, 2, 5

5

(b) 3, 4, 6

2

(c) 1, 2, 6

3

1

(d) 2, 3, 5 8. Twent y-seven cubes ar e ar r anged in a block as sh ow n bel ow. H ow m an y cu bes w i l l be sur r ounded by ot her cubes on all sides ?

(v) The block is cut int o six equal pair s of 1 cm each (fr om 6 cm side) and int o 4 equal par t s of 1 cm each (fr om 4 cm side). 13. How many cubes will have all thr ee colour s black, gr een and r ed each at least on one side? (a) 6

(b) 12

(c) 10

(d) None of t hese

14. H ow many cubes will be for med ? (a) 6

(b) 12

(c) 16

(d) 24 of t hese

15. I f cubes having only “black as well as gr een” colour ar e r emoved t hen how many cubes will r emain? (a) 4

(b) 8

(c) 12

(d) None of t hese

16. H ow many cubes will have t wo sides wit h gr een colour and r emaining 4 sides wit hout any colour ? (a) 3

(b) 1

(a) 12

(b) 10

(c) 9

(d) 6

(c) 8

(d) 4

12.4

Cubes and Dice

17. H ow many cubes will have 4 colour ed sides and 2 sides wit hout colour ? (a) 8

(b) 4

(c) 16

(d) 10

Directions (Q. 18– 22) : Six dices wit h t heir t op faces erased have been given. The opposite faces of the dices have dot s which add up t o t hir t een. Wor k out t he number of dot s on t he t op faces, accor ding t o t he quest ion spot your answer fr om amongst t he given alternatives.

18. I f t he odd number ed dices have even number of dot s at t heir bottom faces, what would be the total number of dot s? (a) 20 (b) 22

19. I f dices I , I I I and I V have odd number of dot s at t heir t op faces, what would be t he t ot al number of dot s? (a) 15 (b) 18 (c) 21 (d) 17 20. I f even number ed dices have odd number of dot s at t heir t op faces, what would be t he t ot al number of dot s? (a) 19 (b) 18 (c) 17 (d) 16 21. I f dices I , I I and I I I have odd number of dot s on their upper faces and dices I V, V and VI have even number of dots on their bottom faces, then what would be the difference in the total numbers of t op face dot s bet ween t hese t wo set s? (a) 8 (b) 4 (c) 0 (d) 1 22. I f dices I I , V and VI have even number of dot s at t heir bot t om faces, what would be t he t ot al number of dot s? (a) 18 (c) 16

(c) 24 (d) 18

(b) 20 (d) 24

AN SWERS 1. (c)

2. (a)

3. (d)

4. (d)

5. (b)

6. (b)

7. (d)

8. (b)

9. (b)

10. (a)

11. (c)

12. (d)

13. (d)

14. (d)

15. (d)

16. (c)

17. (b)

18. (a)

19. (d)

20. (a)

21. (b)

22. (a)



13

Analytical Reasoning

CHAPTER

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. In a row of students, Ajay is 12th from the left and Vijay is 17th from the right. When Ajay and Vijay interchange their positions Vijay becomes 27th from the right. How many students are there between Vijay and Ajay? (a) 9

(b) 12

(c) 7

(d) 10

2. Modi is fourteenth from the right end in a row of 40 boys. What is his position from the left end? (a) 25th

(b) 27th

(c) 24th

(d) 26th

3. In a row of boys, Jayati is seventh from the start and a eleventh from the end. In another row of boys, Bharat is tenth from the start and twelfth from the end. How many boys are there in both the rows together?

7. In a row of girls facing North, Rajni is 10th to the left of Paro, who is 21st from the right end. If Meera, who is 17th from the left end, is fourth to the right of Rajni, how many girls are there in the row? (a) 37

(b) 43

(c) 44

(d) Data inadequate

8. Students line up in a queue in which Ashwani stands fifteenth from the left and Sandeep is seventh from the right. If they interchange their places, Sandeep would be fifteenth from the right. How many students are there in the queue? (a) 21

(b) 22

(c) 29

(d) None of these

9. In a row of children, Hardipak is eleventh from the left and Manoj is seventeenth from the right. When they interchange their places, Hardipak will be thirteenth from the left. Which of the following will be the new position of Manoj from the right?

(a) 36

(b) 37

(a) Eleventh

(b) Twenty-first

(c) 39

(d) None of these

(c) Nineteenth

(d) Twenty-ninth

4. Seema is 8 ranks ahead of Sajan, who ranks twentysixth in a class of 42. What is Seema’s rank from the last? (a) 9th

(b) 24th

(c) 25th

(d) 34th

5. In a class, among the passed students, Amita is twenty-second from the top and Ramnik, who is 5 ranks below Amita, is thirty-fourth from the bottom. All the students from the class have appeared for the exam. If the ratio of the students who passed in the exam to those who failed is 4 : 1 in that class, how many students are there in the class? (a) 60 (c) 90

(b) 75 (d) Data inadequate

6. In a queue, Sameer is ninth from the back. Wasim’s place is eighth from the front. Neetesh is standing between the two. What could be the minimum number of boys standing in the queue? (a) 8

(b) 10

(c) 12

(d) 14

10. Neeta is taller than Sania. Neelima is taller than Neeta. Nitika is taller than Neelima. Mehak is the tallest of all. If they stand according to their height, who will be in the middle? (a) Neeta

(b) Neelima

(c) Sania

(d) Nitika

11. In a team of five players, Vani is older than Rani. Sita is younger than Vani but older than Rani. Nita is younger than Mary and Rani. Rani is older than Mary. Whose age is middle of all the five in this team? (a) Vani

(b) Rani

(c) Sita

(d) Mary

12. Five persons – R, S, T, U and V – are in a queue facing a reservation counter. Immediately behind S is U. T is standing between R and V. In between R and U, no one is there. Then, where is S standing in the queue? (a) First

(b) Second

(c) Last

(d) Last but one

13.2

Analytical Reasoning

13. In a row of girls, Kamla is 9th from the left and Veena is 16th from the right. If they interchange their positions. Kamla becomes 25th from the left. How many girls are there in the row?

6. In a classroom, there are 5 rows, and 5 children – A, B, C, D and E – are seated one behind the other in 5 separate rows as follows: A is sitting behind C, but in front of B.

(a) 34

(b) 36

C is sitting behind E. D is sitting in front of E.

(c) 40

(d) 41

The order in which they are sitting from the first row to the last is

14. In a class of boys Rajan got the 11th rank and he was 31st from the bottom of the list of boys passed. Three boys did not take the examination and one failed. What is the total strength of the class? (a) 32

(b) 42

(c) 45

(d) 46

15. B is twice as old as A but twice younger than F. C is half the age of A but is twice older than D. Who is the second oldest? (a) B

(b) F

(c) D

(d) C

LEVEL-1 1. Ramesh ranks 13th in a class of 33 students. There are 5 students below Suresh rankwise. How many students are there between Ramesh and Suresh? (a) 12

(b) 14

(c) 15

(d) 16

2. In a row of trees, a tree is 7th from the left end and 14th from the right end. How many trees are there in the row?

(a) DECAB

(b) BACED

(c) ACBDE

(d) ABEDC

7. Suresh is 7 ranks ahead of Ashok in the class of 39 students. If Ashok’s rank is 17th from the last, what is Suresh’s rank from the start? (a) 16th

(b) 23rd

(c) 24th

(d) 15th

8. Of the six members of a panel sitting in a row, X is to the left of Q but on the right of P. Y is on the right of Q but is on the left of Z, Z is to the left of R. Find the members who are at the extreme ends? (a) QZ

(b) XZ

(c) PR

(d) QY

9. A goldsmith has five rings. Each having a different weight. Ring D weighing twice as much as Ring E. Ring E weighing four and one-half times as much as Ring F. Ring F weighing half as much as Ring G.

(a) 18

(b) 19

Ring G weighing half as much as Ring H.

(c) 20

(d) 21

Ring H weighing less than Ring D but more than Ring F.

3. Five girls – M, N, O, P and Q – are standing in a row. P is on the right of Q, N is on the left of Q, but N is on the right of M, P is on the left of O. Who is standing on the extreme right end? (a) Q

(b) N

(c) O

(d) P

4. Sita is elder than Swapna. Lavanya is elder than Swapna but younger than Sita. Suvarna is younger than both Hari and Swapna, Swapna is elder than Hari. Who is the youngest?

Which of the following represents the descending order of weights of the rings? (a) D, E, G, H, F

(b) D, E, F, H, G

(c) E, G, H, D, F

(d) D, E, H, G, F

10. Five athletes – A, B, C, D and E – participated in a race. 

B finished before 3 athletes exactly.



C and A finished neither 1st nor 5th.



D did not finish at last.

(a) Sita

(b) Lavanya



C finished before A.

(c) Suvarna

(d) Hari

Who came third?

5. In a row of boys, Srinath is 7th from the left and Venkat is 12th from the right. If they interchange their positions, Srinath becomes 22nd from the left. How many boys are there in the row? (a) 19

(a) A

(b) C

(c) D

(d) E

11. Rahul is taller than Saurav, Sachin is taller than Rahul and Dhoni is taller than Kaif but shorter than Saurav. Who is the tallest?

(b) 31

(a) Sachin

(b) Rahul

(c) 33

(c) Dhoni

(d) Saurav

(d) 34

(e) Kaif

Analytical Reasoning

12. Five friends – Ravi, Suhail, Mayur, Aakash and Krishna – participated in a race. Aakash finished the race before Krishna and after Suhail. Ravi finished the race before Suhail and Mayur. Who won the race?

3. Among five boys, Amit is shorter than Mandar. Prashant is taller than Vinod but shorter than Amit. Mandar is shorter than Pankaj. Who is the tallest in the group?

(a) Suhail

(b) Aakash

(a) Prashant

(b) Amit

(c) Ravi

(d) krishna

(c) Vinod

(d) Pankaj

(e) Mayur

(e) Mandar 4. In a class, Vasu is ranked 16th from the top and 15th from the bottom of a list. How many students are there in the class?

13. Five books are lying in a pile. E is lying on A, C is lying under B, A is lying above B, and D is lying under C. Which book is lying at the bottom? (a) A

(b) B

(a) 29

(b) 31

(c) C

(d) D

(c) 30

(d) 32

(e) None of these

(e) 28

14. There are seven flights – A, B, C, D, E, F and G. A lands after F. C lands before G and after B. D lands after E and before B. E lands after A. Which flight landed at last? (a) B

(b) C

(c) A

(d) G

5. Five persons - Keshav, Mahesh, Sameer, Ravi and Piyush– are sitting in a row facing North. Piyush is on right side of Keshav and Mahesh. Ravi is to the right side of Sameer. Only Ravi is sitting between Keshav and Sameer. Which of the following pairs could be sitting at the extreme ends? (a) Sameer and Piyush (b) Ravi and Mahesh

(e) None of these

(c) Sameer and Mahesh (d) Ravi and Keshav

15. There are five persons – P, Q, R, S and T – sitting in a row facing North. Q is sitting to the left of S and to the right of R. No two persons among R, Q and T are sitting adjacent to each other. Who is sitting at the extreme ends of the row?

(e) Ravi and Piyush 6. In a row of five, A is next to B. E is on right side of A, and D is to the immediately left of A. C and E do not sit together. Who is E’s neighbour?

(a) P and R

(b) P and Q

(a) A

(b) B

(c) Q and S

(d) R and T

(c) C

(d) D

(e) None of these

(e) Cannot be determined

LEVEL-2

7. B is heavier than D, who is heavier than A. E is heavier than C but lighter than D. Who is the lightest?

1. There were 6 friends – A, B, C, D, E and F. •

B is heavier than only E.

•

F is heavier than 4 friends exactly.

•

A or C is not heaviest.

•

A is heavier than C.

(a) A

(b) C

(c) E

(d) D

(e) Cannot be determined Directions for questions 8 and 9: Answer the questions on the basis of the information given below.

Who is 4th heaviest? (a) A

(b) C

(c) F

(d) D

(e) None of these 2. In the following colour sequence, R stands for Red, Y for Yellow, G for Green, B for Blue and W for White. If the sequence is continued, then which colour will come next?

I.

There are five friends – Alok, Jayesh, Bhagat, Subodh and Pramod.

II.

They are standing in a row facing South.

III.

Jayesh is to the immediate right of Alok.

IV.

Pramod is between Bhagat and Subodh.

V.

Subodh is between Jayesh and Pramod. 8. Who is at the extreme left end?

BBRBRWBRWGBRWGYBBRBRWB RW

(a) Alok

(a) Red

(b) Blue

(c) Subodh

(c) Green

(d) Yellow

(d) Jayesh

(e) White

13.3

(b) Bhagat

(e) Data insufficient

13.4

Analytical Reasoning

9. Who is in the middle? (a) Bhagat

(b) Jayesh

(c) Pramod

(d) Subodh

(e) Alok 10. Six friends A, B, C, D, E and F are sitting in a row facing East. Only ‘C’ is between ‘A’ and ‘E’, ‘B’ is just to the right of ‘E’ but left of ‘D’. ‘F’ is not at the right end. Who are to the right of ‘E’?

13. There are five persons – Aman, Baman, Chaman, Raman and Saman. Each of them is of a different weight and height. Aman is the tallest but not the heaviest. Saman, who is the lightest, is taller than exactly two persons. Chaman, who is shorter than at least two persons, is not the second lightest. Raman is the shortest as well as the heaviest. If Aman is the third heaviest, who is the second heaviest among them?

(a) A and B

(b) A and D

(a) Chaman

(b) Raman

(c) B and D

(d) D and E

(c) Baman

(d) Aman

11. Six persons – F, G, H, K, L and M – are of different heights. F, G and H are males, while K, L and M are females. G is taller than F, who is taller than exactly two persons. K is taller than H, who is taller than only one female. M is taller than L but smaller than G. The tallest person is not a female.

14. Five girls are sitting in a row. Sudha is sitting next to Padma. Krishna is sitting next to Rama who is sitting on the extreme left. Tapti is sitting on the extreme right. No body is sitting between Padma and Krishna. Who is sitting in the middle?

Who is the smallest female? (a) K

(b) M

(c) L

(d) Cannot be determined

12. Five books—A, B, C, D and E—are placed on a table one above another. B is placed below C, A is placed above E, D is placed below B and E is placed above C. Which of the following books touches the table? (a) B

(b) D

(c) A

(d) E

(a) Krishna

(b) Padma

(c) Sudha

(d) Tapti

15. Five policemen are standing in a row facing south. Shekhar is to the immediate right of Dhanush. Bala is between Basha and Dhanush. David is at the extreme right end of the row. Who is standing in the middle of the row? (a) Bala

(b) Basha

(c) Shekhar

(d) Dhanush

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (a) 2. (b) 3. (d) 4. (c) 11. (b) 12. (a) 13. (c) 14. (c) LEVEL-1 1. (b) 2. (c) 3. (c) 4. (c) 11. (a) 12. (c) 13. (d) 14. (d) LEVEL-2 1. (b) 2. (c) 3. (d) 4. (c) 11. (c) 12. (b) 13. (a) 14. (b)

5. (b) 15. (a)

6. (d)

7. (b)

8. (c)

9. (c)

10. (b)

5. (c) 15. (d)

6. (a)

7. (a)

8. (c)

9. (d)

10. (b)

5. (a) 15. (d)

6. (b)

7. (e)

8. (a)

9. (d)

10. (c)

EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S 1. From right end (i) Vijay’s present position: 27th (ii) Vijay’s former position: 17th Now, total number of students between Ajay and Vijay = 27 – 17 – 1 = 9. 2. Modi’s position from the left end = 40 + 1 – 14 = 27th.

3. Clearly, total number of boys in both the rows = (Number of boys in Jayati’s row) + (Number of boys in Bharat’s row) = (6 + 1 + 10) + (9 + 1 + 11) = 17 + 21 = 38. 4. Sajan rank 26th and Seema is 8 ranks ahead of Sajan. So, Seema ranks 18th. Number of students behind Seema in rank = (42 – 18) = 24. So, Seema ranks 25th from the last.

Analytical Reasoning

5. Amita is 22nd from the top and Ramnik is 5 ranks below Amita. So, Ramnik is 27th from the top. Also, Ramnik is 34th from the bottom.

14. Rajan’s rank in the class = 11th Also, Rajan’s rank in the class from bottom = 31st  Number of boys passed = 41

Number of students passed = (26 + 1 + 33) = 60 Total number of students in the class = 60 + 15 = 75.

Total number of boys = 41 + 31 + 1 = 45. 15. B = 2A

6. Minimum number of students will be possible in the following case. Sameer

Neetesh

F = 2B = 4A A = 2C  C 

Wasim





9

C  2D  D 

8



= 9 + 6 – 1 = 14.

LEVEL-1

1. Ramesh’s rank in the class = 13th Total number of student = 33

7. Here, Rajni’s position from the right end = 21 + 10

 Suresh’s rank in the class = 33 – 5 = 28th

= 31st

 Number of student between Ramesh and Suresh

Meera’s position from the right end = 31 – 4 = 27th Meera’s position from the left end = 17th (given) Thus, the total number of persons in the row = 27 + 17 – 1 = 43. 8. Sandeep’s new position is 15th from the right as well as from the left end of the row. Number of students in the queue = (14 + 1 + 14) = 29. 9. Manoj’s earlier position from the right end = 17th

= 28 – (13 + 1) = 14. 2. A tree’s rank from the left = 7th and that of from right= 14th.  Total number of trees = 14 + 7 – 1 = 20. 3. The given standing position is Left  M N Q P O  Right 4. The order of eldest to youngest is given below. Sita > Lavanya > Swapna > Hari > Suvarna.

Manoj’s earlier position from the left end = 13th Total number of persons in the row = 17 + 13 – 1 = 29

C A  2 4

Hence, the second oldest is B.

7

Thus, in the above case, total number of students

A 2

The correct order is F > B > A > C > D.



6

5.

7th

22nd

Srinath

Manoj’s new position from the right

Venkat 12th

= 29 + 1 – 11= 19th. 10. On the basis of the given information, we get

13.5

Total number of boys in the row = 22 + 12 – 1 = 33.

Mehak > Nitika > Neelima > Neeta > Sania

6. From the given information the order is DECAB.

Obviously, Neelima will occupy the middle position.

7. Ashok's rank from last = 17th

11. Vani > Rani Vani > Sita > Rani Mary, Rani > Nita. Rani > Mary  Vani > Sita > Rani > Mary > Nita.

12.  Reservation counter.. S U R T V 13. Kamal’s position from left = 9th Veena’s position from right = 16th After interchanging their positions,

 Suresh's rank from last = 17 + 7 = 24th  Suresh's rank from the start = 39 – 24 + 1 = 16th. 8. The given information can be shown as: P

X

Q

Y

Z

R

9. Let weight of the ring D be x gm. x

 Weight of E = 2 gm. x

2

x

Weight of F = 2  9  9 gm. 2x

Weight of G = 9 gm. 4x

So, that Kamal’s position from right = 16th

Weight of H = 9 gm. Weight of D > Weight of H > Weight of F.

 Total number of girls = (25 + 16) – 1 = 40.

Descending order of weight of the rings

Kamal’s position from left = 25th

D > E > H > G > F.

Analytical Reasoning

10. Order of athletes to finish race is D, B, C, A and E.

Hence, Sameer and Piyush could be sitting at the extreme ends.

Hence, C will come 3rd. 11. The order of the heights of the persons according to the given information will be

6. The arrangement is C D A B E

Sachin > Rahul > Saurav > Dhoni > Kaif

So, B is E’s neighbour.

Hence, Sachin is the tallest. 12. Aakash finished the race before Krishna and after Suhail. Ravi finished the race before Suhail and Mayur.

7. Arranging the given information in decreasing order we get B > D > A and D > E > C, but we don’t know whether A < C or C < A. Hence, we cannot decide.

Hence, Ravi won the race.

For questions 8 and 9:

13. The sequence of books according to the given information will be

Bhagat Pramod Subodh Jayesh Alok South

E A B C D so

10.

LEFT F A C

Hence, D is at the bottom of the pile.

E

14. From the given information, order of landing is:

B

FAE D BC G

D RIGHT

Hence, G landed last. 15. Since Q is sitting to the left of S and to the right of R, their places are R Q S Since R, Q and T are not sitting adjacent to each other, T must be to the right of S. P is sitting between R and Q. The final arrangement is R P Q S T

Hence, B and D are to the right of E. 11. The order of persons from the tallest to the smallest is as follows: G

K/M

M/K

F

H

L

12. The order of the books from bottom to top is shown below.

Hence, R and T are sitting at the extreme ends.

A

LEVEL-2

E C

1. The order of weight is D > F > A > C > B > E.

B

Hence, C is the 4th heaviest.

D

2. B | BR | BRW | BRWG | BRWGY | B | BR| BRW | BRWG Therefore, in the above sequence G will be the next letter. Hence, Green will come next in the series.

Hence, D touches the table. 13. The information for weight can be summarised as shown below. Raman > Chaman > Aman> Baman >Saman

3. Arranging the given information, we get

Hence, Chaman is the second heaviest.

Pankaj > Mandar > Amit > Prashant > Vinod

T  RT  RB  1

RIGHT

Tapti

Sudha

Padma

Krishna

LEFT

4. Using the formula,

Rama

14.

Hence, Pankaj is the tallest.

LEFT

Basha

Bala

Hence, there are 30 students in the class.

David

RIGHT

15.

T = 16 + 15 – 1 = 30.

Dhanush

Hence, Padma is in the middle.

We get,

Shekhar

13.6

5. Given information can be shown as: Sameer Ravi Keshav Mahesh Piyush

Hence, Dhanush is standing in the middle.  North

OR Mahesh

Sameer

Ravi

Keshav

Piyush



14

Data Sufficiency

CHAPTER

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S

6.

Directions for questions 1 to 15: Each quest ion is followed by t wo st at ement s, I and I I . Answer each quest ion using t he following inst r uct ions. M ar k: (a)

(b)

(c)

if the question can be answer ed using statement I alone, but cannot be answer ed using statement I I alone.

(d)

if t he quest ion can be answer ed using eit her st at ement I or I I individually.

(e)

if t he quest ion cannot be answer ed even using bot h t he st at ement s t oget her.

1.

I s M anish t aller t han Chandr a? I . Suhas is of t he same height as M anish and Chandr a. I I . Chandr a is not shor t er t han Suhas.

2.

3.

4.

Buses ar e always punct ual in Delhi cit y. H ow long, at t he most , will M r Dhir en have t o wait for t he bus?

I I . Sujit r ead t he last 40 pages of t he book in t he mor ning of M onday. 7.

I n a cer t ain language, ‘pit nac nit ’ means ‘r ed pant shi r t ’. Whi ch wor d means ‘pant ’ in t he language? I . ‘nit t im nac sir ’ means ‘he wor e r ed pant ’. I I . ‘nee jic pit ’ means ‘shir t is dir t y’.

8.

I n a code, ‘l ee pee t i n’ means ‘al ways k eep smiling’. What is t he code for ‘smiling’. I . ‘t in lut lee’ means ‘always keep left ’. I I . ‘dee pee’ means ‘r ose smiling’.

9.

Among t he five fr iends who is t he t allest ? I . Dinesh is t aller t han At ul and Char les. I I . Basu is shor t er t han Ena but t all er t han Dinesh.

10.

When is Tat a’s bir t hday? I . Tat a’s fat her was bor n on 27t h M ay, 1948. I I . Tat a is 25 year s younger t han his mot her.

11.

On what day in Apr il is H r it hik’s bir t hday?

I . M r Dhi r en has come t o t he bus st and at 9 a.m.

I . H r it hik was bor n exact ly 28 year s aft er his mot her was bor n.

I I . Ther e is a bus at 10 a.m. and possibly another bus even ear lier.

I I . H is mot her will be 55 year s 4 mont hs and 5 days on August 18 t his year.

H ow is Rajesh r elat ed t o Anish?

12.

At what t ime will t he plane leave t oday?

I . Bikr am is t he br ot her of Anish.

I . The plane nor mally leaves on t ime.

I I . Bikr am is Rajesh’s son.

I I . The scheduled depar t ur e is at 14.30.

H ow is Shahr ukh r elat ed t o Salman? I . Aami r ’s wi fe Reena i s pat er nal aunt of Salman. I I . Shahr ukh is the br ot her of a fr iend of Reena.

5.

I . The book has 300 pages out of which t wot hir ds wer e r ead by him befor e Sunday.

if the question can be answer ed using statement I I alone, but cannot be answer ed using statement I alone. if t he quest ion can be answer ed using bot h t he st at ement s t oget her, but cannot be answer ed using eit her st at ement alone.

H ow many pages of t he book – The M an Who Saw Tomor r ow did Sujit r ead on Sunday?

Tendulkar r anks t ent h in a class. H ow many st udent s ar e t her e in t he class? I . H is fr iend got 58t h r ank which is t he last . I I . Tendulkar ’s r ank fr om t he last is 49t h.

13.

On whi ch day i n Januar y, Ronal do l eft for Ger many? I . Ron al do h as so f ar spen t 10 y ear s i n Ger many. I I . Ronaldo’s fr iend Rivaldo left for Ger many on 15t h F ebr u ar y an d j oi n ed Ron al do 20 days aft er Ronaldo’s ar r ival.

14.2

14.

Data Sufficiency

I n which year was Sangit a bor n?

5.

I . Only the childr en of wor ker s wer e admitt ed.

I . Sangit a at pr esent is 25 year s younger t o her mot her. I I . Sangita’s br other, who was bor n in year 1964, is 35 year s younger t o his mot her. 15.

H ow many lect ur es wer e deliver ed in t he t wo days’ pr ogr amme? I . 18 speaker s wer e invit ed t o give at least one l ect u r e, ou t of w h i ch on e-si x t h of t h e speaker s could not come. I I . One-t hir d of t he speaker s gave t wo lect ur es each.

I I . Q was not a wor ker. 6.

(a)

if the question can be answer ed using statement I alone, but cannot be answer ed using statement I I alone.

(b)

if the question can be answer ed using statement I I alone, but cannot be answer ed using statement I alone.

(c)

if t he quest ion can be answer ed using bot h t he st at ement s t oget her, but cannot be answer ed using eit her st at ement alone.

(d)

if t he quest ion can be answer ed using eit her st at ement I or I I individually.

(e)

if t he quest ion cannot be answer ed even using bot h t he st at ement s t oget her.

1.

Who is t he shor t est st udent in t he class, a boy or a gir l?

I I . D is t he daught er -in-law of B. 7.

2.

I I . Ram was bor n in year 1988. 8.

I I . Year ‘k’ was a leap year. 9.

Do t he car s P and Q t r avel along t he same r out e for at least par t of t heir jour neys? I . P goes fr om A t o C via B. I I . Q goes fr om A t o D via B.

4.

H as any one of t hem seen t he pict ur e t wice? I . A few of t hem saw t he pict ur e t wice. I I . M ost of t hem saw t he pict ur e only once.

Does M r X belong t o Gujar at ? I . Gujar at is a st at e in I ndia. I I . M r X is not an I ndian.

10.

All who had got medals sat in t he fir st r ow for t he gr oup phot o. But none who was less t han five feet in height sat in t he 2nd r ow. (Ther e wer e only t wo r ows.) Did Amar sit in t he fir st r ow? I . H e had not got any medal. I I . H is height was 4 feet 10 inches.

11.

A, B, C and D st ay in a four -st or ey building. Who st ays on 3r d floor fr om t op? I . A and C st ay on odd-number ed floor. I I . D st ays above B.

12.

Find code for ‘Jamaica’. I . ‘Jamaica is wonder ful place’ st ands for ‘Raj is sheet al’s fr iend’. I I . ‘Tok y o pl ace w on der f u l i s’ st an ds f or ‘Ramesh’s fr iend is sheet al’s’.

13.

H ow many wives does Sheikh have? I . I n his fami l y t her e ar e five peopl e of hi s gener at ion. I I . Sheikh has t wo br ot her s.

14.

A was bor n in which year of 19t h cent ur y? I . H is br ot her was bor n in t he year 1972.

I I . I t was not A who lost his deposit . 3.

H ow many Sundays wer e t her e in t he mont h of M ar ch of a par t icular year ‘k’? I . 10th Febr uary of that year ‘k’ was a Saturday.

Four candidat es (A, B, C and D) cont est ed an elect ion in year 2000. One lost his deposit . Who won? I . C got mor e vot es t han A, but less vot es t han B.

What is Ram’s dat e of bir t h? I . Ram is exact ly 2 year s elder t o Gaur av who was bor n on 22-7-1982.

I . Eit her Ram or Shyam is t he shor t est boy in t he class. I I . Ram is shor t er t han Shyam, but t aller t han Sit a in t he class.

H ow is A r elat ed t o D? I . A is t he son of B.

LEVEL-1 Directions for questions 1 to 15: Each quest ion is followed by t wo st at ement s, I and I I . Answer each quest ion using t he following inst r uct ions. M ar k:

P is t he only son of Q. Was P admit t ed?

I I . Thr ee digit s of t he year ar e same. 15.

What is t he code for ‘economy’? I . ‘Opening of the economy’ stands for ‘economy is in doldr ums’. I I . I n t he above st at ement one wor d has not been coded.

Data Sufficiency

I I . Anu walks 2 km t owar ds Sout h fr om st ation A and t hen, befor e moving 8 k m t owar ds nor t h, walks 3 km t owar ds West .

LEVEL-2 D ir ect ions for quest ions 1 t o 5: I n each of t he f ol l ow i n g qu est i on s, a qu est i on i s f ol l ow ed by infor mat ion given in t hr ee st at ement s. You have t o study the question along with the statements and decide t h e i nf or m at i on gi ven i n wh i ch st at em en t (s) i s necessar y and sufficient t o answer t he quest ion. 1.

I I I . Befor e movi ng 3 k m t owar ds west , Anu walk s 4 k m t owar ds Nor t h fr om st at ion A. (a) Only I and I I (b) Only I I and I I I

H ow many br ot her s does Akash have? (Akash is a boy) I . N eel i m a i s A k ash 's m ot h er Su dh i r i s N eel i m a's h u sban d an d h as on l y on e daught er Riya.

(c) Eit her I and I I or I and I I I (d) All ar e necessar y (e) Any t wo of t he t hr ee 5.

I I . Riya has t wo younger and one elder br ot her. I I I . Ak ash is not t he youngest chil d. (b) Only I and I I

I . B and C ar e si t t i ng opposi t e t o each ot her.

(c) Only I and I I I

(d) All I , I I , and I I I

I I . E may sit ei t her on t he immediat e left of B or on t he immediat e r i ght of C. I I I . D cannot si t opposit e E, and A cannot sit opposit e E.

Who among P, Q, R, S and T was t he fir st t o r each t he st at ion?

(a) Only I and I I

I . Q r eached ear l ier t han T; P and R wer e not t he fir st t o r each.

(b) Only I I and I I I (c) Only I and I I or I I I

I I . P r eached ear l ier t han bot h R and T, but coul d not r each ear li er t han S, who was at t he st at i on befor e Q.

(d) All I , I I and I I I (e) Data inadequate

(c) Only I I and I I I

Direct ions for quest ions 6 t o 10: I n each of t he following questions, a question is followed by information gi ven i n t hr ee st at ement s. You have t o deci de t he i nfor mat i on gi ven i n whi ch of t he st at ement s i s necessar y and sufficient to answer the given question.

(d) All I , I I and I I I

6.

I I I . R di dn't r each just aft er P. (a) Only I and I I (b) Only I and I I or I I I

(e) None of t hese 3.

I s t he t ime 6 o'clock now? I . Aft er fift een mi nut es, t he minut e and t he hour hands of t he clock will mak e a r ight angle.

H ow is A r elat ed t o C? I . M has t wo daught er s. One of t hem is Z, who is mar r ied t o A.

I I . The t r ain which is r unning l at e by exact ly t wo hour s fr om it s scheduled t ime of ar r ival, ie 3 pm, has r eached now.

I I . C is t he mot her of V, t he younger si st er of Z.

I I I . 6 o'clock is t he t i me of depar t ur e of a t r ain and it is st il l on t he plat for m.

I I I . M is C's husband. (a) Only I and I I

(a) Only I

(b) Only I and I I I

4.

A, B, C, D, E and F ar e sit t ing ar ound a cir cular t able facing t he cent r e. Who is sit t ing opposit e t o A?

(a) Only I I and I I I

(e) Any t wo of t he t hr ee 2.

14.3

(b) Eit her I I or I

(c) Only I and eit her I I or I I I

(c) Eit her I I I or I I

(d) Any t wo of t he t hr ee

(d) Eit her I or I I or I I I

(e) All ar e necessar y

(e) All I , I I and I I I

Yogendr a is in which dir ect ion wit h r espect t o Anu? I . Yogendr a wal ks 1 km t owar ds N or t h-east fr om st at ion A and t hen, befor e wal ki ng 2 km towar ds south, walks 2 km towar ds East.

7.

H ow many sist er s does Neha have? (Neha is a gir l.) I . K r ishna, Neha's mot her, has t hr ee childr en. I I . Vinod is t he fat her-in-law of K r ishna and he has only one son and one gr andson.

14.4

Data Sufficiency

I I I . Pooja, Neha's si st er, has t wo sibli ngs.

(b)

if t he dat a in st at ement I I alone is sufficient t o answer t he question while the data in statement I alone is not sufficient t o answer t he quest ion.

(c)

i f t he dat a eit her i n st at ement I alone or i n st at ement I I alone is sufficient t o answer t he quest ion.

(d)

if t he dat a in bot h st at ement s I and I I t oget her is not sufficient t o answer t he quest ion.

(e)

if t he dat a in bot h st at ement s I and I I t oget her is necessar y t o answer t he quest ion.

11.

What colour will be on t he opposit e sur face of t he br own sur face of a cube? The sur faces of cube have differ ent colour s, namely r ed, black, gr een, br own, whit e and blue.

(a) Only I and I I (b) Only I I and I I I (c) Only I I and eit her I I I or I (d) All I , I I and I I I (e) Data inadequate 8.

Who is t he t allest among P. Q, R, S and T? I . R is t all er t han Q but not as t al l as T. I I . P is not t he shor t est . I I I . P is t all er t han onl y S. (a) Only I and I I (b) Only I and I I I (c) Only I and eit her I I or I I I

I . The gr een sur face i s bet ween r ed and black sur faces whi le t he blue sur face is adjacent t o t he whit e. I I . The br own sur face i s adjacent t o t he blue.

(d) All I , I I and I I I (e) Data inadequate 9.

On which day of t he week did Anu ar r ive? I . H er sist er, Tanu, cor r ect l y r emember s t hat she di d not ar r i ve on Wednesday. I I . H er fr iend, M anu, cor r ectly r emember s t hat she ar r ived befor e Fr i day.

12.

I I I . H er mot her cor r ect l y ment i ons t hat she ar r i ved befor e Fr iday but aft er Tuesday.

I . The tall per son is on the left of the fair person and t he weak per son i s sit t i ng bet ween t he int elli gent per son and t he fat per son, who is sit t ing on t he r ight of t he weak per son. I I . One of t he t wo per sons at t he ext r eme ends is i nt ell igent , and is second t o t he left of t he fat per son, who is on t he immediat e r ight of t he weak per son.

(a) Only I and I I (b) Only I I and I I I (c) Only I and I I I (d) All I , I I and I I I (e) Data inadequat e 10.

A, B, C, D and E ar e sitting in a row facing Nor th. Who among t hem is in t he middle? I . E is at t he r i ght end of t he r ow. I I . D si t s bet ween A and C. I I I . Neit her A nor C si t s at an ext r eme end.

13.

(b) Only I I and I I I (c) Any t wo of t he t hr ee (d) All I , I I , and I I I

I I . I n t he same code l anguage "what he want t o be" is wr it t en as jai kal i aaj gai koko; "I want sweet s what Renu needs" i s wr i t t en as baaj saaj pal i kal i aaj t ik a; and "PO ar e gent le" i s wr i t t en as bogo mai al i.

(e) Data inadequate

(a)

if t he dat a in st at ement I alone is sufficient t o answer t he question while the data in statement I I alone is not sufficient to answer t he question.

I f t ika pika mai is t he code for "Renu is PO" in a code language used in t er r it or y X? I . I n t he same code language "I want to be PO" i s wr it t en as jai k ali gai pal i mai , "Renu needs money" i s wr i t t en as saaj t ik a sik a; and "H e needs sweet s" is wr it t en as baaj koko saaj.

(a) Only I and I I

D ir ect ions for quest ions 11 t o 15: Each of t he quest ions below consist s of a quest ion fallowed by t he t wo st at ement s number ed I and I I given below it You have t o deci de whet her t he dat a pr ovi ded i n t he st at ement s ar e sufficient to answer the quest ion. Read bot h t he st at ement s and give answer

What char acter istic is possessed by t he per son who is on the left of the weak per son and who is sitt ing in a r ow of five per sons? I t is given that other s possess t he following qualities: fat, t all, fair and intelligent. And each individual possesses only one quality.

14.

H ow many car ds does B have? B is playing a game of car ds wit h A, C, D and E. I t is given t hat t he t ot al number of car ds is 158. I . A says t o B, "I f you give me t hr ee car ds, you wi ll have as many as E has and if I gi ve you t hr ee car ds, you wil l have as many as D has."

Data Sufficiency

I I . A and B t oget her have 10 car ds mor e t han what D and E t oget her have. And B has t wo car ds mor e t han what C has. 15.

14.5

I . Star ting fr om above ther e is an English book bet w een a H i st or y an d M at h s book , a H i st or y book bet ween a M at h s and an Engl i sh book , a H i ndi book bet ween an Engl ish and a M at hs book , a M at hs book bet ween t wo H i ndi book s and t wo H i ndi book s bet ween a M at hs and a H i st or y book.

The book of which subject is at the sixth position fr om t he t op in a pile of t en books, including 3 books of H ist or y, 3 of H indi, 2 of M at hs and 2 of English?

I I . I f we count fr om t he bot t om, t he book which is at t he fi ft h posit ion i s nei t her M at hs nor English.

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (a) 11. (c)

2. (c) 12. (e)

3. (c) 13. (e)

4. (e) 14. (c)

5. (d) 15. (e)

6. (c)

7. (e)

8. (d)

9. (c)

10. (e)

7. (a)

8. (c)

9. (c)

10. (b)

7. (c)

8. (b)

9. (c)

10. (b)

LEVEL-1 1. (e)

2. (c)

3. (e)

4. (a)

5. (e)

11. (c)

12. (c)

13. (e)

14. (e)

15. (c)

6. (e)

LEVEL-2 1. (b) 11. (e)

2. (e) 12. (c)

3. (c) 13. (d)

4. (c) 14. (d)

5. (d) 15. (a)

6. (d)

EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S 1. a

2. c

3. c

4. e 5. d

From statement I, we can conclude that Suhas, Manish and Chandra are of the same height. So, Manish is not taller that Chandra. Thus, only statement I is sufficient to answer the question. Statement II gives no relation between Manish and Chandra. From both the given statements, we find that Dhiren reached the bus stand at 9 a.m. and a bus is sure to arrive at 10 a.m. So, Dhiren has to wait for at the most one hour. From both the statements together, we find that Rajesh is the father of Bikram and Bikram is the brother of Anish. So, Rajesh is the father of Anish. Thus, both the given statements are needed. Clearly, both the statements together are not sufficient to answer the question. Statement I reveals that 58th rank is the last rank in the class. This means that there are 58 students in the class. So, statement I alone is

6. c

sufficient. Also, from statement II, we find that Tendulkar’s rank in the class is 10th and 49th from the last. So, there are (10 + 49 – 1) = 58 students in the class. Thus, statement II alone also is sufficient. From statements I and II, we find that Sujit read (300 x

7. e

8. d

), i.e. 200 pages before Sunday

and the last 40 pages on Monday. This means that he reads (300 – (200 + 40)), i.e. 60 pages on Sunday. Clearly, from each of the statements, we find that the code for ‘pant’ is either ‘nit’ or ‘nac’. So, none of them is sufficient to answer the question. Comparing the information in the question with statement I, we find that ‘tin’ and ‘lee’ are the codes for ‘always’ and ‘keep’. So, ‘pee’ represents ‘smiling’. Thus, statement I alone is sufficient. Again, comparing the information in the

14.6

9. c

Data Sufficiency

question with statement II, we find that the common code word ‘pee’ stands for the common word ‘smiling’. Thus, statement II alone is also sufficient. From statement I, we have: D > A, D > C. From statement II, we have: E > B > D. Combining the above two, we get: E > B > D > (A and C). So, E is the tallest.

10. e Clearly, even both the statements together do not reveal Tata’s birthday. 11. c Clearly, the birthday of Hrithik’s mother can be found out from statement II and then Hrithik's birthday can be determined using the fact given in statement I. Thus, both the statements are required. 12. e Clearly, even both the statements together do not reveal the exact time of departure of the plane today. 13. e Clearly, even from both the given statements, we cannot conclude the exact date of Ronaldo leaving for Germany. 14. c From both the given statements, we find that Sangita is (35 – 25) = 10 years older than her brother, who was born in year 1964. So, Sangita was born in 1954. Thus, both the given statements are needed to answer the question.

3. e

4. a

5. e

6. e

7. a

8. c

15. e From statement I, we find that the no. of speakers who attended the programme = 18 –

1 6

of 18 = 15.

From statement II, we find that (

1 3

x 15), i.e.

five speakers gave two lectures each, But we don’t know about number of lectures given by remaining 10 speakers. Because they can deliver one lecture each or more than two also.

LEVEL-1 1. e

2. c

Statement I does not mention the girls at all, so it is insufficient to answer the question. Statement II does not mention anything about the height of Sita. Hence, statement II is also insufficient to answer the question. Putting the two statements together, we are able to say that Ram is the shortest boy but it is not yet comparable to the height of the shortest girl. From statement I, we get the order of candidates getting votes as B > C > A. From statement II, we find that it was D who has lost his deposit (as A, B, C can’t have lost their deposit). Hence, the final order of candidates becomes B > C > A > D. Hence, B has got the maximum number of votes.

9. c

Combining both statements, both cars P and Q travel from A to B. But we don’t know the number of roads connecting A and B. Hence, the question cannot be answered. From statement I, we infer that a small number of people saw the picture twice. So, from statement I alone, we get the answer ‘yes’. Statement II alone is insufficient to answer the question. The father of P was not a worker. But the mother of P might be a worker. No information in this regard is available either from the given statements or from the introductory part. So, this question cannot be answered. Statements I and II taken alone are not sufficient to answer the question. Even when considered together, the two statements are not sufficient to answer the question, as D could be either the wife or sister-in-law of A. From statement I, it is clear that Ram’s date of birth is 22-7-1980. Hence, statement I is sufficient to answer the question. From statement II, though we get the year in which Ram was born, we cannot find out his date of birth. We get from statement I that 11th February was a Sunday. Hence, 18th and 25th February were also Sundays. The next Sunday can be looked upon as ‘32nd February’ [which is 3rd March for a leap year, as mentioned in statement II]. Since statement II mentions that year ‘k’ is a leap year, 3rd, 10th, 17th, 24th and 31st March are Sundays, giving us 5 Sundays in that March. In a normal year, 4th, 11th, 18th, 25th March only would have been Sundays. Hence, the answer for a normal year would be 4. Hence, we require both statements to answer the question. Since Mr X is not an Indian, it follows that Mr X does not belong to any state in India. As Gujarat is a state in India, we infer that Mr X does not belong to Gujarat.

10. b From statement II, we note that Amar’s height was less than five feet. Hence, we get from the introductory part that he did not sit in the 2nd row. As there were only two rows, this means that he sat in he first row. (Note: All those who had got medals sat in the first row. But this does not mean that only those who had got medals sat in the first row.) 11. c Combining both statements I and II, B and D stay on 2nd and 4th floors respectively. Since D is above B, B is on 2nd floor, i.e. 3rd floor from top.

Data Sufficiency

12. c Combining both statements I and II, when we compare between statements given, in first half other than Jamaica, remaining three words are repeated. Also in the other half of the statements, three words are repeated. Hence Jamaica stands for Raj. 13. e This is because we don’t have any information about his sisters. Hence, using both statements together we cannot answer the question. 14. e Statement I gives no clue. And statement II gives indication to 1811 and 1888. If brother was younger or elder, than it would have been (e) as the answer. 15. c Statement I in itself is not sufficient, but when we consider second statement, it is clear that economy stands for economy only.

From III: M is C’s husband . But III alone is not sufficient to answer the question. Combining I and III, we get A is C’s son in law.  The question can be answered using I and either II or III. 4. c

LEVEL-2 1. b

2. e

3. c

From I: Neelima and Sudhir have only one daughter Riya. Also, Akash is their son. No information regarding other children is given, so I is not sufficient to answer the question. From II: Riya has two younger brothers and one elder brother. No information regarding sisters is given, so II alone is not sufficient to answer the question. From III: no information on Akash’s brothers Combining I and II  Akash has two brothers  Only I and II are necessary to answer the question. From I: Q came before T and P and R did not reach first. No information regarding S is given, so I alone is not sufficient to answer the question. From II: P came before R, T but not before S (who reached before Q)  S reached first. So II alone is sufficient to answer the question. From III: no information regarding who reached first is given. Only statement II is required to answer the question.  None of these. From I: M(parent) has two daughters . one daughter is Z whose husband is A. We know that A is the son in law but no information on C is given, so I alone is not sufficient to answer the question. From II: C (mother) of V and Z. But no mention of A is made, so II alone is not sufficient to answer the question. Combining I and II, we get A is C’s son in law.

14.7

5. d

From I: Yogendra is to the South East of station A. But no information regarding Anu is given, so I alone is not sufficient to answer the question. From II: Anu is to the North West of station A. But no information regarding Yogendra is given, so II alone is not sufficient to answer the question. Using the position of Yogendra from I and the position of Anu from II the question can be answered. From III: Anu is to the North West of station A. But no information regarding Yogendra is given , so III alone is not sufficient to answer the question. Using the position of Yogendra from I and the position of Anu from III the question can be answered.  The question can be answered using I and either II or III From I: B and C are opposite to each other I is not sufficient to answer the question since no information regarding the others is given. From II: E is either to the left of B or to the right of C. II is not sufficient to answer the question since no information regarding the others is given. From III: Both D and A are not opposite E so F is opposite E. III is not sufficient to answer the question since no information regarding the others is given. Combining I , II and III  D is opposite A.  All I , II and III are necessary to answer the question.

6. d

From I: it is not 6 o'clock now because at 6:15 the hands are not at an exact right angle. I is sufficient to answer the question. From II: it is not 6 o'clock now because the 3 pm train which was 2 hours late has reached now. II is sufficient to answer the question. From III: it is not 6 o'clock now because the 6 o'clock train is still in the platform. III is sufficient to answer the question.

14.8

7. c

8. b

Data Sufficiency

The answer is: it is not 6 o'clock now.  Either I or II or III is sufficient to answer the question. From I and III it is clear that Neha has two other siblings. But I and III are not sufficient to answer the question since no information regarding brother or sister is given.

sitting at the extremes. No information regarding others is given, so III alone is not sufficient to answer the question. Combining II and III the answer is: among 5 seats the middle 3 are occupied by A, D and C and D is in the middle.  Only II and III are sufficient to answer the question.

From II it is clear that Neha has only one brother. II is not sufficient to answer the question since the other siblings are not mentioned. The answer is: the second sibling is a sister.  Only II and either III or I are sufficient to answer the question. From I it is clear that T is taller than R and Q. But I alone is not sufficient to answer the question.

11. e From I: More than one combinations of the given faces are possible, so I alone is not sufficient to answer the question. From II: brown is adjacent to blue But no information regarding other colours is given so II alone is not sufficient to answer the question. Combining statements I and II together we get

From II no information regarding anyone other than P is given So II is not sufficient to answer the question. From III it is clear that P is taller than S only  P is shorter than Q, R and T. But III alone is not sufficient to answer the question. Combining I and III , we get T is the tallest.

9. c

 Only I and III are sufficient to answer the question. From I it is clear that it is not Wednesday. But I alone is not sufficient to answer the question. From II it is clear that it is before Friday. But II alone is not sufficient to answer the question. From III it is clear that it is before Friday and after Tuesday.

But III alone is not sufficient to answer the question. Combining I and III : Anu arrived on Thursday (which is after Tuesday , it is not a Wednesday and it is before Friday.)  Only I and III are sufficient to answer the question. 10. b From I: E is at the right end. No information regarding others is given, so I alone is not sufficient to answer the question. From II it is clear that D sits between A and C. No information regarding others is given, so II alone is not sufficient to answer the question. From III it is clear that A and C, both are not

 white is opposite to brown.  Data in both statements I and II put together is necessary to answer the question 12. c From I: __, Intelligent, weak, fat,__ From II: Intelligent, weak, fat,__, __ From both the statements we can infer that the person sitting on the left of the weak person is intelligent.  Data in statement I alone or statement II alone is sufficient to answer the question 13. d From I: {I, want, to, be, PO} = { jai, kali, gai, pali, mai} {Renu ,needs, money} = {saaj, tika, sika} {He ,needs, sweets} = {baaj, koko, saaj} No code for 'is' given, so I is not sufficient to answer the question. From II: {what, he, want, to, be} = {jai, kali, aaj, gai, koko} {I, want, sweets, what, Renu, needs} = { baaj, saaj, pali, kali, aaj, tika} {PO, are, gentle} = {bogo, mai, ali} No code for 'is' is given, so II is not sufficient to answer the question. Both I and II do not have a code for "is". So data from both I and II together is not sufficient to answer the question.

Data Sufficiency

14. d Both statements I and II give the relation between the cards in each person's hand but do not give the number of cards in any of the five persons' hands. So data in both statements I and II put together is not sufficient to answer the question.

14.9

15. a From I: we can clearly see that the pile of books is as follows (from top to bottom) History English Mathematics History English Hindi Mathematics Hindi Hindi History So data in statement I alone is sufficient to answer the question. From II: the fifth book from bottom is neither mathematics nor English. II is not sufficient to answer the question.  Data in statement I alone is sufficient whereas data in statement II alone is not sufficient to answer the question.



1

Physics

CHAPTER PH YSI CAL QU AN TI TI ES U N I TS  L engt h C.G.S. : cent imet er M .K .S. : met er F.P.S. : foot  M ass C.G.S. : gr am M .K .S. : kilogr am F.P.S. : pound  T ime C.G.S. : second M .K .S. : second F.P.S. : second  Cur r ent S.I . : amper e (A)  Temper at ur e S.I . : kelvin (K )  Ar ea C.G.S. : cm 2 M .K .S. : m 2  Vol ume C.G.S. : cm 3 or cc. M .K .S. : met er 3  D ensit y C.G.S. : gm/cm 3 M .K .S. : kg/met er 3  Speed C.G.S. : cm/sec M .K .S. : met er /sec  Vel ocit y C.G.S. : cm/sec M .K .S. : met er /sec S.I . : km/sec  M oment um C.G.S. : gm cm/sec M .K .S. : kg.m/sec  F or ce C.G.S. : dyne M .K .S. or SI unit : newt on (N)  T hr ust C.G.S. : dyne or gm wt M .K .S. : newt on or kg.wt

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Pr essur e C.G.S. : dyne/cm 2 or g.wt /cm 2 M.K.S. : newton/m 2 or kg.wt/m 2 M .K .S. or SI Syst em: km/pascal Wor k C.G.S. : er g or dyne cm M .K .S. : joule or Nm P ow er C.G.S. : er g/sec M .K .S. : wat t E ner gy C.G.S. : er g M .K .S. : joule Acceleration due to gravity (g): M .K .S. : met er /sec2 M oment of F orce: C.G.S. unit : dyne cm S.I . unit : newt on met r e (Nm) D ensit y: C.G.S. : gr ams/cm 3 M .K .S. : kilogr am/m 3 H eat C.G.S. : calor ie S.I . : joule (J) Specific heat C.G.S. : cal/gm °C M .K .S. : J/kg°C Cal or i f i c Val u e of Speci f i c ener gy C.G.S. : cal/gm M .K .S. : J/kg F r equency her t z or Cycles/sec Char ge coulombs (C) Capaci t y far ads (F) Pot ent i al volts (V) Least Count of Screw Gauge mm or cm U n i ver sal G r a vi t a t i on al Const ant (G) Nm 2/kg2

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Angular D isplacement radians Angular Velocit y r adians/sec L uminous F lux lumen(L n) Solid angle steradians(Sr ) L uminous I nt ensit y Candela(cd) or L umens Candle Power Candela(cd) M agnet ic Pole St r engt h M .K .S. : weber S.I . : amper e-met r e (A-m) M agn et i c I n du ct i on or M agnet ic F lux Densit y newt on/amper e-met r e I nt ensit y of M agnet ic field C.G.S. : Gauss S.I . : Wb/m 2 or t esla E lect r ic Power joule/sec or wat t Resist ance Ohm Specific Resist ance Ohm-met r e Quant it y of heat calor ies Specific heat Cal/gm °C H ouse hol d consumpt i on of elect r ical ener gy Kilowat t-hour s Electro Chemical Equivalent gm/coulomb Self induct ance henr y M ut ual induct ance henr y Elect ro M ot ive F or ce volt s

1.2

Physics

VALU ES  Light year(ly) = 9.46 × 1012 km  Distance of moon from Earth = 3.85 × 105 km  Distance between Earth and the Sun = 1.5×1011 km  N umber of Satellites of Jupitor = 14  1 Astronomical U nit (1 AU ) = 1496 × 1011 met r es  Per sec(largest unit of distance) = 3.26 light year  Velocity of light = 3 × 108 met r e/sec 1  Time period of earth = 365 days 4  1 metre = 100 cm  1 kg = 1000 gr ams  Least count of Vernier callipers = 0.01 cm or 0.1 mm  1 cubic meter = 1000 lit r es  1 Litre = 1000 cubic cms  1 gm/cubic cm = 1000 kg/met r e  Relative density of mercury = 13.6  Density of water = 1 gm/cm 3  1 Square metre = 10,000 cm 2  N umerical value of G = 6.67 × 10– 11Nm 2/kg2 5  1 km/hour = met r es/sec 18  1 newton = 105 dynes  1 gm. weight = 980 dynes  1 kg. weight = 9.8 newt ons  N ormal temperature of H uman body = 36.9°C or 98.4°F  M elting point of water = 0°C  Boiling point of pure water = 100°C  M elting point of wax = 60°C  Freezing point of mercury = – 39°C  Velocity of sound in vaccum = 0  Time period for a seconds pendulum = 2 seconds  1 H orse Power = 746 wat t s  1 Kilo watt = 1.34 H or se Power  1 Watt = 107 er gs/second  Rat io bet ween coefficient of linear superficial cubical expansions  :  :  :: 1 : 2 : 3  Audible frequency range = 20 H z t o 20,000 H z  Value of g on M oon = 1.67 m/s2  Value of g on Sun = 27.4 m/s2  Value of g on Earth = 9.8m/s2  For a freely falling body initial velocity = Zer o  2  radians = 360°  1 radian = 57°18 or 57°29  Wave length of ruby laser = 6943Å  Wave length of H e-N e laser = 6328Å  Relat ive permeability of air vaccum = 1

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Permeability of free space (0) = 4× 10– 7 henry/metre Rel at i ve per m ea bi l i t y of a di am agn et i c substance r  1 Rel at i ve per m eabi l i t y of a pa r am a gn et i c substance r  1 Rel at i ve per m eabi l i t y of a f er r om a gn et i c substancer  1 1 kilo ohm = 103 ohm

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1 M ega ohm = 106 ohms

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Specific resistance of copper at 20°C = 1.7  10– 8

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Value of M echanical equivalent of heat

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= 4.18 joules/calor ies 

1 Watt H our = 3600 wat t -second

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1 Kilowatt Hour = 36  105 watt-second = 3.6 × 106 Joule

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1 M ega watt = 106 wat t

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Electro-chemical equivalent of gold = 0.0006812 gm/Coulomb

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1 M eV = 1.6  10– 12 joules

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1 a.m.u = 931.5 M eV

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1 Energy gap of a semi conductor = 1 eV

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Velocity of sound in air at 0°C = 330 m/sec

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Velocity of sound is air at 25°C = 351 m/sec

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1 farad = 106 M icr ofar ads

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Wave-lengt h ranges (i ) Visible spect r um: 0.4 m t o 0.7 m (ii ) I nfr ar ed spect r um: 0.7 m t o 100 m (iii ) M icr o waves: 10 m t o 10m (iv) Radio waves: 1 m t o 10 km (v) Ult r a violet spect r um: 0.4 m t o 1 nm (vi ) X r ays: 0.001 nm t o 10 nm (vii ) Gamma r ays: 0.0001 nm t o 0.1 nm Ratio of Specific heat of air Cp   1.4 Cv Dielect ric constant value of rubber = 3 H alf life period of radium = 1600 year s Dielect ric const ant value of paraffin = 2

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Time int er val   Displacement s Velocit y, V = Time t  For ce, F = M ass(m)  Acceler at ion (a)

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Weight = M ass(m)  Acceler at ion due t o gr avit y (g) Thr ust T  Pr essur e, P = Ar ea  A  Thr ust , T = Pr essur e (P)  Ar ea (A) Boyle's law: PV = Const ant Number of images for med by t wo plane mir r or s at an angle  360 n= 1  Size of t he image M agnificat ion = Size of t he object sin i Snell's law 12 = sin r Relat ion bet ween u, v and f 1 1 1   f u v Equat ions of mot ion 1 (i ) v = u + at (ii ) s = ut + at 2 (iii ) v 2 – u 2 = 2as 2 Wor k, W = For ce (F)  Displacement (S) Wor k, W = F  S cos  Work W  Power, P = Time  T  Pot ent ial Ener gy (P.E.) = mgh 1 2 K inet ic Ener gy (K .E.) = mv 2 Moment of force, M = Force(F)  Perpendicular distance M echanical advant age of wheel and axle Radius of t he wheel = Radius of t he axle M echanical advant age of scr ew jack Cir cumference of t he scr ew = Pit ch of t he scr ew M ass Densit y = Volume Densit y of t he subst ance Specific gr avit y = Densit y of Wat er Quant it y of heat , Q = mst Heat pr oduced Calor ific value = Unit mass H eat ut ilised Qu  Ther mal efficiency = Tot al heat pr oduced Q1  Fundamental pr inciple of calorimetr y H eat lost by hot body = H eat gained by cold body Velocit y of sound, V = m   Dist ance bet ween a cr est and at r ough = 2 Distance between two consecutive rests or tr ough =  Newt on's for mula : E Velocit y of sound in a gas, V = P

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 A  Dm  sin   2  Refr act ive index of t he pr ism,  = A sin 2 1 Power of a L ense, P = Focal lengt h in mt sf 

Relat ion bet ween I , V, H ; Char ge, Q = CV Q Cur r ent , i = t

I 2 = V2 + H 2

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r2 GM Equat ion of mot ion of a body under gr avit y 1 (i ) v = u ± gt (ii ) h = ut ± gt 2 (iii ) v 2 – u 2 = ±2gh 2 Equat ion of mot ion for a fr eely falling body

(i ) v = gt ...(u = 0) 

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1.3

(ii ) h =

1 2 gt 2

(iii ) v 2 = 2gh

M aximum height r eached by a body t hr own up, u2 h= 2g u Time of ascent = g u Time of descent = g 2u Time of flight = g Velocit y on r eaching t he gr ound (i ) v = 2gh (ii ) u = 2gh Angular displacement Angular velocit y,  = Time int erval Also, v = r 2 M agnit ude, L = mvr = m r v2 Cent r ipet al acceler at ion, a = r mv 2 2 Cent r al for ce, F =  m r r  v2  Banking angle,  = t an – 1    rg  Elect r ic pot ent ial, V =

Wor k done W  Char ge Q 

Potent ial differ ence of cells connect ed in ser ies E = E 1 + E 2 + E 3 + ... Ohm's law : V = iR

1.4 

Physics

Resist ance, R = 

l A

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A Specific r esist ance,  = R l Resist ances connet ed in ser ies, R = R1 + R2

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R1 R 2 R1  R 2

V2 Elect r ic power, P = V.i = i R = ...(V = iR) R Elect r ical wor k done, W = Vq Elect r ical wor k done, W = Vit Elect r ical wor k done, W = Vq = i 2Rt i 2 Rt H eat developed, Q = J Far aday's fir st law : M = Z.i.t Far aday's second law : M 1 : M 2 : M 3 = E 1 : E 2 : E 3 = Z 1 : Z2 : Z3  i M agnet ic induct ion, B = 0 2 r For ce act ing on a cur r ent car r ying conduct or placed in an ext er nal magnet ic field, F = ilB V1 n 1  V2 n 2 V1 i 2 n 1   V2 i 1 n 2 2

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Alber t Einstein's mass, ener gy equivalence, E = mc2 

DEVI CES AN D I TS U SES   

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Sonometer: To measur e fr equency of a t unning for k Vernier callipers: To measur e lengt hs accur at ely M easuring Jar: To measur e volume of liquids in millilit r es. M easur ing flask and pipet t e: To obt ai n fi xed amount of liquid Burette: To deliver any required volume of a liquid upto its maximum capacity Common balance: To measur e mass of a body Spring balance: To find weight of an object Compression spring balance: To find weight of a per son Postal balances: Used in Post offices Table or scale balances: Used in fancy shops and sweet shops Platform balances: Used in Railway st at ions, I r on and har dwar e shops and in par cel offices t o weight heavy weight s M icro balances: To measur e mass of a subst ance upt o one micr ogr am Electronic balances: Used in jewellar y shops Single pan analytical balance: Used in laboratories Clocks: To measur e t ime

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Beam Balance: To measur e mass of a subst ance Density bottle: To det er mine r elat ive densit y or specific gr avit y of liquids H ydraulic machine: To pr ess bales of cot t on and t o pr ess oil seeds for get t ing oil Barometer: To measur e at mospher ic pr essur e Altimeter: Used in air cr aft t o measur e alt it udes Clinical T hermometer: To measur e t emper at ur e of human body Six maximum and minimum T hermomet er: To measur e maximum and minimum t emper at ur e at a place Pressure cooker: House hold it em or appliance used for cooking Periscope: U sed in submar ines t o see object s on t he sur face of wat er Kaleidoscope: To obser ve a number of images wit h wonder ful designs and colour s Concave mirror: Used as a r eflect or in head light s of vechicles; Dent ist s and ENT specialist s used t o see smal l i nner par t s of t hr oat , nose. used as a shaving mir r or Convex mirror: U sed in t el escopes; Ar r anged i n fr ont of t he dr iver of a vechicle M agnetic compass: Used t o know posit ion of ship in mid sea and t o dr ive it in r equir ed dir ect ion for r eaching t heir dest inat ions Electro magnets: Used in tape r ecorders, speaker s, dynamos and motors Volt aic cell: Chemi cal ener gy i s conver t ed i nt o elect r ical ener gy Dry cell: Used in radios, torch lights, tape records etc. Telegraph: Used in sending messages t o a dist ance place in a ver y shor t t ime Electric iron: To pr ess t he clot hes Soldering gun: To connect var ious element s in a cir cuit Electric stove: For cooking or boiling of wat er, milk et c. H ydrometer: To deter mine specific gr avity of liquids Calorimeter: To measur e quant it ies of heat Bomb calorimeter: To det er mine calor ific value of fuels and foods Astronomical telescope: To view dist ant st ar s and planets Terrestrial telescope: To view dist ant object s on ear t h Dip circle: To det er mine dip of a place Gold leaf electroscope: To det ect pr esence of st at ic elect r icit y over a body Coolidge tube: I n pr oduct ion of X -r ays Screw Gauge: To measur e t hickness of a t hin glass plat e and diamet er of a t hinwir e or a small spher e

Physics               

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Gravity meters: To find value of ‘g’ at a given location Spring balance: To det er mine weight of a body Cent rifuge: To separ at e par t icles of higher mass fr om t hose of lower mass in a given mixt ur e Simple pendulum:: To det er mine ‘g’ value Ripple tank: Used for demonst r at ion of waves. Ammeter: To measur e cur r ent in amper es. Voltmeter: To measur e pot ent ial differ ence in volt s Rheostat: Used t o r egulat e t he value of a cur r ent in a cir cuit Resitance: Opposes t he flow of elect r ons t hr ough t he conduct or Tap-key: Used t o make or br eak an elect r ic cir cuit Cell, battery: Power sour ces Voltameter or Elect rolytic cell: Vessel i n which elect r olysis t ake place Elect ric mot or: Devi ce whi ch conver t s el ect r i cal ener gy int o mechanical ener gy AC D ynamo or AC Gener at or : Devi ce whi ch conver t s mechanical ener gy int o elect r ical ener gy Tr ansfor mer : Devi ce whi ch ei t her i ncr eases or decr eases magnit ude of an alt er nat ing volt age and ior n cor e t o minimise power losses N uclear reactor: Used in Nuclear Power St at ions t o dr ive t ur bines of t he elect r ic gener at or syst em Diodes: Used in r ect ifier cir cuit s P-n junction diode: Used as an elect r onic swit ch Light emitting diodes: Used in digit al clocks and digital calculat or s Junciton transistor: Used to stabilise power supplies Computer: Used in banki ng, i ndust r y, commer ce science, educat ion, weat her pr edict ion, war far e et c. St op wat ch: U sed in r unning r aces, l abor at or i es t ype inst itut es

TERM S AN D TH EI R EXAM PLES  Opaque bodies: st one, met als, wood, human body  Transparent bodies: glass, wat er  N atural fibres: cot t on, jut e, wool  Artificial fibres: nylon, dacr on, or lon  Self-luminous body: sun, st ar s, bur ning candle  Fundamental units: lengt h, mass and t ime  Derived units: ar ea, volume, densit y  Scal ar quant i t y: l engt h , mass, t i m e, vol um e, t emper atur e  Vector quantity: displacement , velocit y, for ce  Transverse waves: waves pr oduced on str ings, light waves  Longit udinal waves: sound waves i n a medium, vibr at ion in spr ing  M usical inst r ument s: (i ) St r inged inst r ument s: sit ar, violen, veena (ii ) Dr um inst r ument s: mr idangam, tabala ((iii ) Blow t ype inst r ument s: flut e, clar inet

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1.5

N on-luminous bodies: ear t h, moon, chair M edium: air, glass, wat er, vaccum Transparent subst ances: gr ound gl ass, l ayer of par affin wax, oiled paper N atural magnets: ear t h, bar magnet , H or se shoe magnet , r ing magnet M agnetic substances: ir on, st eel, nickel, cobalt N on-magnet ic subst ances: paper, wood, br ass, aluminium Physical effort: pushing, pulling, t ur ning, bending M ental effort: memor ising a poem, solving a pr oblem Renewable source of energy: sol ar ener gy, wind ener gy, wat er power, biomass ener gy N on-renewable source of energy: coal, oil, nat ur al gas Energy crisis: solar ener gy, t idal, nuclear ener gy Fuels: pet r ol, ker osene, diesel, coal Simple machines: wheel and axle, scr ew-jack Wheel and axle: winches, capstans, drills, tap handles Primary colours: r ed, gr een, blue Secondary colours: yell ow, magent a, cyan. Lenses: convex, concave Optical syst ems: t elescopes, micr oscopes Conductors of electricity: copper, aluminium, metals I nsulators: r ubber, glass, wood Capacit or s: L eydenj ar capaci t or, par al l el pl at e capacit or, hor izont al t ype capacit or Cat hode r ay-t ubes: t el evi si on pi ct u r e t u bes, comput er display t ubes, t ube light s in t he houses Flourescent substances: zinc sulphide Gravity meters: boliden gr avit y met er, gulf gr avit y met er L inear mot ion or Tr anslat or y mot ion: mot i on of t he bus Oscillatory or Vibratory motion: pendul um of a wall clock Gaseous laser: helium- neon laser Solid laser: r uby laser Diamagnetic substances: air, wat er, bismut h, gold Par amagnet ic subst ances: oxygen, mangenese, aluminium, plat inum, chr omium Ferromagnet ic substances: ir on, cobalt , ni ck el Power sources: cell, bat t er y Power consumer: bul b Connectors: conduct ing wir es H eating effects of electric current: elect r ic ir on, immer sion heat er, elect r ic st ove, elect r ic bulb N at ur al r adi o act i ve subst ances: u r ani u m , t hor ium, r adium, act inium I sotopes: 2010Ne, 2110Ne, 2210Ne; 1H 1 1H ,21H 3 I sobars: 4019K , 4020Ca, 136C, 137N

1.6 

  

Physics

Radio isotopes: r adio act ive sodium, r adio act ive cobalt , r adio act ive iodine Semi-conductors: gr aphite, pur e ger manium, silicon Trivalent atoms: gallium, indium, aluminium, boron Pentavalent atoms: ar senic, ant imony, phosphor us

M OT I ON A body if changes it s posit ion wit h r espect t o anot her, t hen it is called mot ion . Ever y one knows t hat even ear t h i s not st at i onar y. H owever, for al l par t i cal applicat ions, ear t h is asssumed as st at ionar y. Thus a body can be said in mot ion if it changes it s posit ion wit h r espect t o ear t h. Types of M otion 1. U nifor m mot ion. I f a body cover s equal dist ances in equal int er vals of t ime, it is said t o be moving wit h unifor m mot ion or mot ion of body is unifor m. 2. N on-uniform mot ion or Var iable mot ion I f a body cover s unequal distances in equal int er vals of time, then it is said to be moving with non-unifor m or var iable mot ion. ORI GI N Any ar bit r ar ily select ed fixed point wit h r espect t o which t he posit ion changes is known as or igin. DI STAN CE  I t is t he act ual lengt h cover ed by a body dur ing t he whole jour ney.  I t is a scalar quant it y.  I n SI unit s, it is measur ed in met r es(m). DI SPLACEM EN T  I t can be defined as t he dist ance t r avelled by a body in a par t icular dir ect ion.  I t is a vect or quant it y.  I t is t he shor t est dist ance bet ween any t wo point .  I n SI unit s, it is measur ed in met r es(m). SPE E D  The rate of change of movement (distance) is called speed.  I t is a scalar quant it y.  I n SI units it is measur ed in metr es per second(m/s). di st ance d   Mathematically, Speed = t ime t  Aver age speed I t is t he dist ance t r avelled by body per unit t ime. M at hemat ically, Aver age speed Tot al dist ance t ravelled by t he body = Time t aken by t he body t o cover t he t ot al dist ance U nifr om speed I f a body cover s equal dist ances in equal int er vals of t ime, t hen it is said t o be moving wit h unifor m speed.

SCALAR AN D VECTOR QU AN TI TI ES Scalar Quant it ies Physi cal quant i t i es possessing magni t ude onl y ar e called scalar quant it ies. e.g. ar ea, lengt h, volume, mass and speed ar e scalar quantities. Pr essur e, elect r ic cur r ent s ar e scaler quant it y. Vect or Quant it ies Physical quantities that can be r epr esented completely by magnitude and dir ection ar e called vector quantities. e.g. velocit y, acceler at ion, weight , momentum, cur r ent densit y and for ce ar e vect or qunt it ies. VELOCI TY The r at e of change of displacement i n a par t icul ar dir ect ion is called velocit y . I t is a vect or quant it y. I t s unit is m/s.   Displacement s m /s Mathematically, Velocity, V = Time t 

 

U nifor m velocit y I t a body moving in a par ticular dir ection, cover s equal dist ances in equal int er vals of t ime, it is said t o be moving wit h unifor m velocit y. Alt er nat ively a body is said t o be moving wit h unifor m velocit y if it s speed and dir ect ion does not change wit h t ime. Var iable velocit y A body is said t o be moving wit h var iable velocit y if it cover s unequal dist ance in equal int er vals of t ime. A body is said t o be moving wit h var iable velocit y if (1) dir ect ion of mot ion changes, or (2) speed changes or (3) bot h dir ect ion and speed changes. Physi cal Quant ity Dist ance Displacement M ass Time Speed Velocit y Acceleration Angular veloci ty

S.I . U nit s met r e met r e kg second met ers per second met ers per second met ers per second square radians per second

Symbol m m kg s ms– 1 ms– 1 ms– 2 rad s

–1

ACCELERAT I ON The r at e of change of velocit y is called acceler at ion .    Change in velocit y v  u  Acceler at ion, a = Time  t  Types of Accelerat ion 1. U nifor m acceler at ion. I f velocity of a moving body increase by equal amount in equal inter vals of time, then it is said to be moving wit h unifor m acceler ation .

Physics

2. Var iable acceler at ion. I f vel ocit y of a moving body changes by unequal amount s in equal inter vals of t ime, t hen it is said t o be moving wit h var iable acceler at ion . RE TARD AT I ON I f final velocity of a body is less than the initial speed, then it is called deceleration and the body is said t o be u n der goi n g r et ar dat i on . F or al l m at h em at i cal calculations, it is taken as acceleration with minus sign. GRAPH S Gr aphs ar e used for conveni ent ly r epr esent ing t he given dat a on t wo mut ually per pendicular axes. For t he pr oper ly plot t ed gr aphs, we can find out t hose values which ar e not given on t he dat a. Average velocit y ( vavg) It is defined as the velocity of a moving body per unit time. S Tot al dist ance t ravelled M at hemat ically, v avg = = t Tot al t ime t ak en I n t he case of a unifor mly acceler at ed mot ion, u+v v avg = 2 wher e, u = init ial velocit y, v = final velocit y Angular velocit y (  )  The r ate of change of angular displacement is called angular velocit y .  The velocity of a body moving in a cir cle is measur ed in r adians/ seconds (r ad s– 1). Angular displacement i n radians  = or  = Time t ak en t wher e,  = angular displacement E quat ions of mot ion Consi der mot i on of a body, movi ng wi t h uni for m acceler at ion. L et u is init ial velocit y and a is acceler at ion of t he body. Then Final velocit y of t he body v = u + at ...(i ) Since body is moving wit h unifor m acceler at ion, t hen uv Aver age velocit y of t he body, v ar g = 2 1 2 Dist ance cover ed by t he body, s = ut  at 2 1 Dist ance cover ed in n t h second s = u  a 2n  1 2 M ot ion under gr avit y When a body falls t owar ds t he ear t h, it s velocit y goes in incr easing continuously. The body is said to be falling under gr avit y. When a body falls under gr avit y, it cont inuously gains acceler at ion g. Thus equat i ons of mot i on of a body fal l i ng under gr avit y ar e wr it t en as 1 v = u + gt ; s = ut + gt 2 ; v 2 = u 2 + 2gs = u 2 + 2gh 2 I n t hese equat ions, ‘a’ is r eplaced by g.

1.7

When body goes away fr om t he ear t h, t hen its velocit y goes on decr easi ng cont i nuousl y. The val ue of g i s t ak en as negat i ve for such cases for sol vi ng t he pr oblems of physics 1 2 Thus v = u – gt , s = ut – gt , and v 2 – u 2 = – 2gs 2 or v 2 = u 2 – 2gh M OM EN T OF A FORCE The turning effect produced by a force is called its moment . F1

F2

F 1 l = F 2 l A

l

l

B

wher e, F 1 = F 2 = for ces act ing on t he ends of a bar l = ar m of moment Unit s : N-m in S.I syst em. Newt on is a unit of for ce. I t is equal t o t he for ce whi ch pr oduces accel er at i on of 1 m/s2 i n a mass of 1 k g  1 N = 1 kg  m 2; 1 N = 105 dyne KI N EM ATI CS I t is t he st udy of mot ion of object s independent of t he causes of mot ion. I t is also gener ally independent of t he object t hat is moving. M oment um I t is equal t o pr oduct of mass and velocit y of a body. I t is a vect or quant it y and is r epr esent ed by ‘P’. P=mv Unit s : kg-m/sec in SI syst em L aw of Conservat ion of M oment um M oment um can neit her be cr eat ed nor dest r oyed but can be changed fr om one for m t o anot her for m. When t wo or mor e bodies int er act mut ually, t hen sum t ot al of moment um r emains unchanged unless t hey ar e act ed upon by some unbalanced ext er nal for ce. F ORCE For ce is agency which pr oduces or t ends t o pr oduce, destr oys or tends to destr oy, the state of r est or unifor m mot ion of a body. I t also changes shape and dir ect ion of a body. I t SI syst em it is measur ed in newt on(N). For example a ball on a table moves when it is pushed. It shows that the ball needs a force to change its state of rest. Types of Force I n nat ur e we obser ve a number of for ces. 1. M uscular force : I t is exer ted in lifting heavy bodies. 2. Force of friction : It helps us to move on smooth roads. 3. M agnetic force of at t ract ion : I t i s r esponsi bl e for an ir on r od at t r act ed by a magnet . 4. Gravitational force : I t pulls all bodies towards the centr e of the earth. 5. Cohesion and Adhesion : Ther e ar e for ces of at t r act ion bet ween molecules. 6. N uclear force : Ther e ar e ver y shor t r ange for ces t hat keep t he nuclear par t icles t oget her.

1.8

Physics

I N ERT I A I ner t ia is a measur e of mass of a body. Gr eat er t he mass of a body gr eat er will be it s iner t ia or vice ver sa when any vechicle st ar t s suddenly, the passenger falls backwar d. I t is due t o iner t ial pr oper t y of t he body. I M PU L SE The pr oduct of for ce and time for which it acts is called impulse. I t is a vect or quant it y. I mpulse = For ce  Time The unit s of impulse is Ns. N EWTON ’S LAWS OF M OTI ON F irst law A body cont inues t o r emain in st at e of r est or unifor m mot ion in t he same dir ect ion in a st r aight line unless act ed upon by some ext er nal unbalanced for ces. Second law The r at e of change of moment um of a body is dir ect ly pr opor t ional t o t he implied for ce and t akes place in t he dir ect ion of t he for ce mv F i.e or F = ma t T hir d law To ever y act ion t hr er e is equal and opposit e r eact ion i .e. F1 = – F2 F or mulae of M ot ion  For ce = mass  acceler at ion  M oment um = mass  velocit y  Newt on is a unit of for ce. One newt on is t hat much f or ce w h i ch pr odu ces i n a m ass of 1 k g an acceler at ion of 1 m/s2  SI unit of moment um is kgm/s, velocit y is m/s  S.I unit of acceler at ion is m/sec2 RE ACT I ON When t wo for ces act ing on t wo bodies in cont act ar e equal and opposit e, t hen one of the ot her for ce is called react ion .

1 r adian =

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

A body is said t o possess mot ion of r ot at ion if (i ) all t he par t icles of t he body move in cir cles wit h t heir cent r es lying on t he axis of r ot at ion, and (ii ) all t he posit ion vect or s sweep out t he same angle in a given t ime int er val. Angular D isplacement I f a par t icle moving in a cir cle, t hen it may be defined as the angle swept out by t he posit ion vect or in a given t ime int er val. The S.I . unit of measur ement of angular displacement is r adians. Radian is t he angle subt ended at t he ent r e of a cir cle by an ar c of lengt h equal t o r adius of cir cle. S ; r

TI M E PERI OD The time per iod of a body r evolving about a fixed points is defined as t he t ime t aken by it t o complet e one full r evolut ion. I t is denot ed by T. FREQU EN CY ( n) Fr equency of a body r evolving about a fixed point (or a fi xed axi s) i s defi ned as t he number of compl et e r evolut ions made by it in a unit t ime. n = 1/T Relat ion bet ween Angular velocit y, Time per iod and F r equency 1  = 2  = 2 n T Relat ion bet ween L inear and Angular velocit y v=r M agnit ude of t he linear velocit y of a par t icle moving i n a ci r cl e i s pr oduct of t he angul ar vel oci t y and dist ance of t he par t icle fr om t he axis of r ot at ion. CEN TRI PETAL FORCE I t is t he for ce t hat comples a body t o keep moving in a cir cular path with a constant speed and is directed along r adius of t he cir cle t owar ds it s cent r e. mv 2 Cent r ipet al for ce F = = mr 2 ...(v = r ) r wher e m is mass of a par t icle moving along a cir cle of r adius r wit h a unifor m speed v. CEN TRI FU GAL FORCE I t is defined as t he for ce of r eact ion v exer t ed by a body moving unifor mly v F along a cir clular pat h on t he ext er nal F O agen t w h i ch i s pr ov i di n g t h e v cen t r i pet al f or ce, by vi r t u e of i t s const ant t endency t o t r avel along a st r aight line pat h. The cent r ifugal for ce act s along t he r adius but in t he out war d dir ect ion. M OM EN T OF I N ERTI A (M .I .) M .I . of a r igid body, about a given axis of r ot at ion i s su m of t h e pr odu ct s m r 2 t ak en f or al l t h e par t icles const it ut ing t he body, wher e m is mass of a par t icle and ‘r ’ is it s nor mal dist ance fr om t he axis of r ot at ion  I = mr 2 Moment of iner t ia is a scalar quantit y. I ts unit is kg m 2. 

ROTATORY M OTI ON

(in r adian) =

Angular velocit y (  ) I t i s def i n ed as t h e r at e of ch an ge of si n gu l ar displacement of a body. I f a body descr ibes a small angle  in small int er val of t ime t , t hen  = t Unit of angular velocit y is r adians per second(r ad s– 1).

360  57.3 2

Physics

WORK, POWER AN D EN ERGY W ORK When a body is displaced by t he applicat ion of a for ce, wor k is said t o be done by t he for ce. Wor k = For ce  Displacement , or W = F  S Unit s : I n S.I syst em – newt on-met r e or joule and I n C.G.S syst em – er gs; 1 joule = 107 er gs POW E R W The r at e of doing wor k is called power. P = t 1 joule Unit s : wat t and kilo wat t and 1 wat t = 1 sec H orse power is anot her unit of power, 1 H .P = 746 wat t EN ERGY The capacit y t o do wor k is called ener gy. Differ ent for m of ener gy ar e int er nal ener gy, kinet ic ener gy, pot ent ial ener gy, chemi cal, t her mal ener gy, elect r ical ener gy et c. Kinet ic ener gy It is the energy possessed by a body by virtue of its motion. 1 M at hemat ically, K .E = mv 2 2 where, m is mass of the body and v is velocity of the body Pot ent ial ener gy I t is t he ener gy possessed by t he body due t o it s some/ r est posit ion. P.E. = mgh wher e, g = acceler at ion due t o gr avit y h = height of the body above the r efer ence point L aw of Conservat ion of Energy The total energy of a closed system is constant. I t can neither be created nor destroyed, but can be converted into one form to other form. Relat ion bet ween M ass and Energy (E inst ein’s equat ion) Tot al ener gy of body, E = mc2 wher e, m = r elat ive mass of body

GRAV I TAT I ON

Al l obj ect s possessi ng mass have t he pr oper t y of gr avitation. Gr avit at ion is t he for ce of at t r act ion bet ween any t wo object s of t he univer se. A chair lying in a r oom att r act s all ot her object s including t he ear t h. GRAV I T Y I t is the force of attr action between object and the ear th. K epler’s laws 1. Each planet moves in an ellipt ical or bit wit h t he sun at one of it s foci. 2. The line joining t he planet t o t he sun sweeps equal ar eas in equal int er vals of t ime (or ) Velocit y of t he r adius vect or joining planet and t he Sun is const ant .

1.9

3. Square of the per iod of revolution of the planet r ound t he sun is dir ect ly pr opor t ional t o t he cube of t he aver age dist ance bet ween planet and t he sun, i.e. T2  r 3 N ewt on’s law of Gravit at ion Ever ybody in t his univer se at t r act s ever y ot her body wi t h a for ce, whi ch i s di r ect l y pr opor t i onal t o t he pr oduct of t heir masses and inver sely pr opor t ional t o t he squar e of t he dist ance bet ween t hem. m 1m 2 m 1m 2 M at hemat ically, F  ; or F = G d2 d2 wher e m 1 and m 2 = mass of two bodies and d is distance between them G = const ant called uni ver sal gr avi t at i onal const ant . Gravitational constant G between sun and earth N  mt 2 is 6.7  10– 11 kg 2 ACCELERATI ON DU E TO GRAVI TY (g) I t is t he acceler at ion with which a body falls fr eely it is i n depen den t of m ass of t h e body. T h e val u e of acceler at ion due t o gr avit y is not const ant at all par t s on t he ear t h’s sur face. GM The aver age value of g is 9.81 m/sec2. g  2 R Thus we find t hat ‘g’ is independent of mass of a body. I t depends on mass and r adius of t he ear t h. M ass of the Earth gR 2 M= G wher e, M = 5.91  1024 kg R = r adius of ear t h = 6.38 106 m G = univer al gr avit at ional const ant G = 6.67  10– 11 Nm 2/kg2 Variat ions in value of g 1. Var iat ions wit h alt it ude (height ). gh = g 1  2h   g  where, gh = acceleration due to gr avity at a height ‘h’. The value of ‘g’ decr eases wit h incr easing height (r adius). 2. Var iat ion wit h dept h. Acceler at ion due t o gr avit y at dept h ‘d’ g = g 1  d   R The value of ‘g’ decr eases wit h incr easing d. 3. Variation due t o shape of the eart h. Ear t h is not a per fect spher e. I t is slight ly bulging at t he equat or and flat t ened at t he poles, whi ch means t he pol ar r adi us of ear t h i s smal l er as compar ed t o it s equat or ial r adius. Thus value of ‘g’ at poles will be mor e t han it s value at equat or, i.e. gpolar > gequait or ial

1.10 Physics

Gr avit at ional pot ent ial ener gy

GM m P.E = – R Negative sign of gravitational potential at a point shows that for ce is applied opposite to the dir ection of mot ion as to move the body without acceler ation due to gr avity. SATELLI TE Sat ellit e is like a body t hat is r evolving in an or bit ar ound a compar at ively much lar ger body. e.g. moon is a sat ellit e of ear t h. Geost at ionar y sat ellit e. I t is t hat sat ellit e which r evolves ar ound t he ear t h wit h t he same angular speed in t he same dir ect ion as t hat of t he ear t h ar ound it s axis. Such a sat ellit e is also called geosynchronous sat ellit e. M ovement of Planet and Satellit es 4 r 2 G.M . = T2 wher e, r = r adius of t he or bit T = per iod of r evolut ion M = mass of t he sun (for planet ) or of ear t h (for an ear t h sat ellit e) H eight of Geost at ionary sat ellit e h = R0 – R wher e, R0 = r adi us of fi xed or bi t ar ound t he ear t h (42.250 km) wher e, R = r adius of ear t h(6380 km) h = 35,870 km Or bit al velocit y (v0) I t is t he speed of a sat ellit e in it s or bit v0 =

gR = 6.4  106  9.8 = 7.92 km/s Escape velocit y (v e) The minimum velocit y wit h which a body should be pr oject ed t o over come the ear t h’s gr avit ational field is called t he escape velocit y. v e = 11.2 km/s

SI M PLE H ARM ON I C M OTI ON PERI ODI C M OT I ON I t is a mot ion which r epeat it self in definit e int er val of t ime. e.g. mot ion of sun ar ound ear t h, mot ion of ar ms of a clock, mot ion of simple pendulum et c. Oscillat or y or Vibr at or y mot ion When a body moves t o and fr o on eit her side of a point in definit e t ime int er val, t hen t his mot ion is called oscillat or y or vibr at or y mot ion . T ime per iod I t is time taken by the body t o complete one oscillation. e.g. mot ion of a mass suspended fr om a spr ing mot ion of simple pendulum et c.

F r equency The time after which the body r etr aces its path is called t ime per iod and t he number of vibr at ions made in one second is called frequency . 1 1 or  Fr equency, n = Time per iod T D isplacement and Amplit ude The physcial quantity which varies uniformly with time in an oscillat or y mot ion is called displacement . The maximum value of displacement is called amplitude. SI M PLE H ARM ON I C M OTI ON A m ot i on i n w h i ch accel er at i on of t h e body i s pr opor t i on al t o i t s di spl acem ent fr om t he mean posi t i on and i s al ways di r ect ed t owar ds t he mean posit ion is called simple har monic mot ion . or Th e m ot i on of f oot of t he per pendicular dr opped fr om t h e par t i cl e m ov i n g i n a cir cle on t he hor izont al and ver t i cal di amet er s i s call ed simple har monic mot ion . Simple H ar monic mot ion is r epr esent ed by  2  y = A sin  t   ; y = A sin (t + ) T  wher e, A = amplitude  = angular fr equency  = phase differ ence PH ASE I t is t hat pr oper t y of wave mot ion which t ells posit ion of the par t icle at any instant. Phase is measur ed eit her by t he angle which t he par t icle makes wit h t he mean posit ion or by fr act ion of t ime per iod. Energy of a particle execut ing S.H .M . E=

1 m2 A 2 2

SI M PLE PEN DU LU M The metal sphere is called bob and the point fr om which t h e pen du l u m i s su spen ded i s cal l ed cen t r e of suspension . The dist ance between cent r e of suspension and cent r e of gr avi t y of t he bob i s cal l ed effect i ve l engt h of pendulum . Time per iod of t he simple pendulum, T = 2

l g

Seconds pendulum I t is the simple pendulum having a time per iod at 2 seconds. Its effective length is 99.992 cm or approximately one meter. On ear th, time period of the pendulum in mines or up the hills is mor e than that on the sea level because at these places ‘g’ is less than that on sea level.

Physics 1.11

GEN ERAL PROPERTI ES OF M ATTER At mospher ic pr essur e Air exerts pressure which is called atmospheric pressure. I f ‘h’ is height of t he at mospher ic air, ‘d’ is densit y of t he air and ‘g’ is acceler at ion due t o gr avit y, t hen Atmospheric pr essur e = hdg = 76  13.6  980 dyne/cm 2 D ensit y Densit y of a subst ance is defined as t he r at io of it s mass t o it s volume. m = v Unit s : kg/m 3 Relat ive densit y I t is defined as t he r at io of densit y of t he subst ance t o t he densit y of t he wat er at 4C. F l ui d A fluid may be defined as t hat st at e of mat t er which cannot indefinit ely or per manent ly oppose or r esist a shear ing st r ess. L iquid I t is a fluid which alt hough has no shape of it s own, occupies a definit e volume which cannot be alt er ed, however gr eat t he for ce applied t o it . Cent er of pressure The point of t he plane (immer sed in a liquid) at which the r esult ant pr essur e act s is called center of pressur e. F luid pr essur e A liquid cont ained in a vessel exer t s a for ce on t he bot t om of t he vessel and on t he sides of t he vessel. This for ce is nor mal t o t he sur face. The pr essur e is defined as for ce per unit ar ea. Unit s : newt on/m 2 Ar chimedes’s pr inciple When a body (t ot ally or par t ially) is immer sed in a fluid it appears to lose a par t of its weight and appar ent loss of weight is equal t o weight of fluid displaced. Appar ent weight of t he body = Act ual weight of t he body – Upt hr ust L aws of F loat at ion L et a body of weight W is immer sed in a fluid and W  is t he upt hrust . 1. I f W > W , t hen body will sink. 2. I f W = W , t hen body float s wit h whole of it s volume inside t he liquid. 3. I f W < W , t hen body will float wit h some of it s par t out side t he liquid Pascal’s law Pressure applied at any point in the fluid is tr ansmitted equally and undiminished t o all par t s of t he fluid. Boyle’s law Volume of a given mass of a gas var ies inver sely as pressure of gas provided temperatur e remains constant. 1 P or PV = const ant . V

BU OYAN CY The pr oper t y by vir t ue of which a body immer sed in a liquid exper ience an upwar d thr ust is called buoyancy . Buoyant for ce Whenever a body is immer sed in liquid an upwar d force act s on t he body by t he liquid. This upwar d for ce is called buoyant for ce. Buoyant for ce depends upon (i ) volume (V) of the solid body immersed in the liquid, (ii ) densi t y of t he l i qui d(  ) i n whi ch t he body i s immer sed, and (iii ) acceler at ion due t o gr avit y (g) M at hemat ically, Buoyant For ce = V..g Buoyant force = Weight of the liquid displaced by the body. Cent r e of Buoyancy The point at which t he buoyant for ce act s is called cent r e of buoyancy and is defined as cent r e of gr avit y of t he displaced liquid.

KI N ETI C TH EORY OF GASES M OLECU LE I t is t he smallest par t icle of t he subst ance which has al l t he pr oper t i es of t hat subst ance and whi ch can r emain in fr ee st at e. AT OM I t is the smallest unit of the substance which is invisible and which can not be dest r oyed.

 3A  Radius of at om, r at om =   4  N  

13

Avogadr o’s number I t r epr esent s t he number of molecules in one mole of any subst ance. I t s value is 6.02  1023 per gm mole.

m  6.02  1023 M wher e, m = mass of t he subst ance M = moleculer weight Number of molecules 

STATES OF M ATTER Ther e ar e mainly t hr ee st at es of mat t er viz., solid, liquid and gas 1. Solids : The molecules of solid ar e ver y close t o each ot her, t her efor e int er nal for ces of at t r act ion ar e ver y st r ong. M olecules of t he solid can not move fr om one place t o another place but they can vibr at e simple har monically about t heir mean posit ions. 2. Liquids : The molecules of liquids ar e much closer i n compar i son t o t he mol ecul es of a sol i d . The inter molecular for ces of at t r act ion ar e smaller than t hat of solid. The molecules of a liquid ar e fr ee to move with the volume of the liquid but velocit y of the molecules is much less t han that of the gases.

1.12 Physics

3. Gases : I n t his stat e of matt er, t he distance bet ween t he mol ecules i s ver y l ar ge. The int er molecular for ces of attr action acting between the molecules ar e ver y small, ther efor e gases do not have a definite shape (or ) definite volume. Kinet ic t heory and Gas pressure The pr essur e of a gas i s t he r esul t of cont i nuous bombar dment of t he gas molecules against t he walls of t he cont ai ner and i s equal t o t ot al moment um impar t ed per second per unit ar ea of t he walls of t he cont ainer by t he bombar ding molecules. TEM PERATU RE SCALE Kelvin scale T = (273 + t )  K I DEAL GAS I deal gases obey Boyl es and Char l es l aw for al l condit ions of pr essur e and t emper at ur e. Ther e is no int er molecular for ce of at t r act ion act ing bet ween t he molecules of gases. Gas equat ion : PV = nRT wher e, P = pr essur e ; T = t emper at ur e ; V = volume n = number of molecules R = universal gas constant = 8.31 joule/mole-kelvin Degree of F reedom The t ot al number of co-or di nat es or i ndependent quantities which must be known in order to describe completely the position of an object or the state of a system is called degrees of freedom of the object or system. 3 M onoatomic gas : U = RT 2 5 Diatomic gas : U = RT 2 7 6 Polyatomic gas : U = RT or RT  3RT 2 2 Vander Waal equation (Equation of state for real gases) a    P  2   V  b  = RT V

K i localor i e. I t i s t h e am ou n t of h eat r equ i r ed t o r ai se t h e t emper at ur e of 1 kg of wat er t hr ough 1C. 1kcal = 1000 calor ies T her mal E quilibr ium When t wo bodies come in cont act in such a way t hat no t r ansfer of heat t ak es pl ace fr om one body t o an ot h er , t h en t h e bodi es ar e sai d i n t h er m al equilibr ium. T her mal Ener gy The sum of i nt er nal k i net i c ener gy and i nt er nal pot ent i al of t he mol ecules of gas i s call ed t her mal ener gy (or ) int er nal ener gy of t he gas. T E M PERAT U RE Temper at ur e of a subst ance is t he degr ee of its hot ness or coldness. Following t hr ee t emper at ur e scales ar e commonly used for measur ement of t emper at ur e : 1. Celsius scale (  C) : Thi s scal e was gi ven by Cel si us. On cel si us scal e t he t emper at ur e of melt ing ice, i.e. melt ing point of ice is given t he value 0° and t emper at ur e of st eam i s given by 100°C. This scale has been divided int o 100 equal par t s of degr ees. Since t her e ar e 100 divisions or degr ees on t he cel si us scal e, i t i s al so cal l ed cent igrade scale (cent i = 100 and gr ade = division). Celsius scale is used par ticular ly in scientific wor k. 2. Fahrenheit scale ( F) : This scale was given by Fahr enheit. On Fahr enheit scale, ice point is given t he value of 32° and steam point is given a value of 212°, so t hat t her e ar e 212 – 32 = 180 degr ees between the t wo fixed points. The Fahr enheit scale is gener ally used for household t her momet er s. 3. Kelvin scale (K) : This scale was given by Kelvin. On t his scale of temper atur e, ice point has a value of 273 K and steam point has a value of 373 K and ther e ar e 372 – 273 = 100 divisions between two fixed points. This is also called absolute scale of temper at ur e. Conversion of Celsius to Fahrenheit to Kelvin scale

H E AT TH ERM OM ET RY

C F  32 K  273  = 100 180 100 Also, K = C + 273 Types of T hermomet ers 1. L iquid t her momet er s 2. Gas t her momet er 3. Plat inum t her momet er 4. Ther moel ect r i c t her momet er 5. M agnet ic t her momet er 6. Opt ical pyr omet er. Specific H eat I t is defined as t he amount of heat in calor ies r equir ed t o r aise t he t emper at ur e of a unit mass of a subst ance by 1°C(or 1°K).

H EAT I t is an agent which pr oduces t he sensation of war mt h. When an object i s heat ed, i t s mol ecul es begi n t o move fast er. H eat al ways fl ows fr om a hot t er body t o a colder body. H eat is measur ed in calor ie or K ilocalor ie. SI unit of heat is joule (J). H eat ing and Cooling of subst ances On cooling gases become liquid. When liquids are heated they change to gases and when solids ar e heated, t hey change to liquids. Liquids solidify on cooling. Calor ie. I t i s t h e am ou n t of h eat r equ i r ed t o r ai se t h e t emper at ur e of 1 gm of wat er t hr ough 1C.

Physics 1.13

SI unit : joules per kilograme per kelvin, i.e. J- kg– 1 K – 1.  Specific heat of wat er is 4200 Jkg– 1 K – 1.  Specific heat of wat er is maximum.  Specific heat of copper is 0.093 cal/gm °C, which means that 0.093 calor ie of heat is r equir ed to r aise the temperature of 1 gm of copper by 1 degree centigrade. T her mal capacit y I t i s t h e am ou n t of h eat r equ i r ed t o r ai se t h e t emper at ur e of whole body t hr ough 1°C. Ther mal capacit y = M ass of t he body  Specific heat S.I unit : jouls per kelvin(J/K ). The common unit of t her mal capacit y is calor ies per degr ee, which is wr it t en as cal/ °C or cal °C– 1. Principle of H eat M easurement When t wo bodies at differ ent t emer at ur es ar e placed in cont act wit h each ot her, t hen heat will pass fr om t he body at higher t emper at ur e t o t he body at lower t em per at u r e u n t i l bot h r each t o a com m on t emper at ur e in t his pr ocess. H eat lost by one body = H eat gained by t he ot her body. H eat lost or gained by a body is given by Q = m  s  (T 2 – T 1); Q = m.s.t Subl imat ion Dir ect conver sion of solid int o gases st at e i s cal led sublimat ion . H oar F rost Dir ect conver sion of vapour s int o solid st at e is called hoar fr ost . M elt ing Conver si on of sol i d i nt o l i qui d st at e at const ant t emper at ur e is called melt ing. Boil i ng Evapor at ion wit hin t he whole mass of t he liquid is cal l ed boi l i ng. Boi l i ng t ak es pl ace at a const ant t emper at ur e called boiling point . E vapor at i on Conver sion of liquid int o vapour s at all t emper at ur es is called evapor at ion . I t is a sur face phenomenon. Effect of pressure on melting point of a solid The var iat ion of melt ing point wit h pr essur e is given by t he for mula dL JL  dt T v 2  v 1  EXPAN SI ON OF SOLI DS Solids expand on heat ing. Dur ing t he expansion of solids, t he dist ance bet ween it s molecules incr eases. The expansion of solids does not depend on it s mass. Types of Expansion of solids I t is of t hr ee t ypes. 1. L inear expansion. Change in lengt h due t o change of t emper at ur e is called linear expansion .

Coefficient of linear expansion ( ) : I t i s t he incr ease in lengt h per unit lengt h of a solid when it s t emper at ur e is r aised by 1°C. 2. Super ficial expansion. The expansion in ar ea of an object due t o change in t emper at ur e is called super ficial expansion. Coefficient of superficial expansion(  ) : I t i s t he incr ease in ar ea per unit ar ea of solid when it s t emper at ur e is r aised by 1C. 3. Volume expansion or Cubical expansion(  ). T h e ex pan si on of v ol u m e du e t o ch an ge i n t emper at ur e is called cubical expansion. EXPAN SI ON OF LI QU I DS L iquids do not have definit e shape and size. Ther efor e t hey have cubical expansion alone. Types of Expansion of liquids I t is of t wo t ypes. 1. Appar ent expansion. I t is the expansion of a liquid in which the expansion of it s cont ainer has not been t aken int o account . Coefficient of apparent expansion of a liquid (a) : I t is incr ease in it s volume per unit volume which appear s to have taken place when it is heated t hr ough 1C in a expandable vessel.

Apparent expansion of t he liquid Or iginal volume  Temper at ure differ ence 2. Real expansion or Absolut e expansion. I t is the expansion of a liquid in which the expansion of it s cont ainer has also been t aken in account . Coefficient of real expansion of a liquid ( r ): I t is t he incr ease in it s volume per unit volume which act ually t akes place when it is heat ed t hr ough 1C a =

Real expansion of t he liquid Or iginal volume  Temper at ure differ ence Anomalous Expansion of Wat er Wat er ex pan ds an om al ou sl y ar ou n d 4  C. I f t emper at ur e of wat er at 0C i s i ncr eased, t hen it s volume decr ease upt o 4C, becomes minimum at 4C and t hen incr eases. The peculiar behaviour of wat er ar ound 4C is called anomalous expansion of wat er . Thus volume of wat er at 4C minimum while densit y at 4C is maximum. EXPAN SI ON OF GASES Ther e ar e t wo coefficient s of expansion in gases. 1. Volume coefficient (  v) At constant pr essur e, the change in volume per unit v ol u m e per degr ee cel si u s i s cal l ed vol u m e coefficient . 2. Pr essure Coefficient (  P) At const ant volume t he change in pr essur e per unit pr essur e per degr ee cel si us i s cal l ed pr essur e coefficient . r =

1.14 Physics

BOYLE’S LAW At const ant t emper at ur e, t he pr essur e of a definit e mass of gas is inver sely pr opor t ional t o it s volume. 1 At const ant t emper at ur e, P  or PV = const ant . V

2. Convect ion I n t his mode of t r ansfer ence, heat is t r ansmit t ed fr om one par t of body t o anot her par t by t he act ual movement of heat ed par t icles.

T H E RM OD YN AM I CS It is the branch of physics which deals with the conversion of heat int o mechanical ener gy and vice ver sa. TH ERM ODYN AM I CAL STATE I t r efer s t o a st at e of a body (or t he syst em) t hat is com pl et el y def i n ed by pr essu r e, v ol u m e an d t emper at ur e of t he body. LAWS OF TH ERM ODYN AM I CS Zer ot h law of T her modynamics Wh en ev er t wo bodi es A an d B ar e i n t h er m al equilibr ium wit h anot her body ‘C’, t hen bodies A and B will also be in t her mal equilibr ium wit h each ot her. F ir st law of T hermodynamics The heat ener gy given t o a syst em is equal t o incr ease in int er nal ener gy of t he syst em and wor k done. dQ = dU + dW or dU = dQ – dW wher e, dQ = heat ener gy dU = change in int er nal ener gy dW = wor k done I n case of a cyclic pr ocess, U = 0; t hus dQ = dW Second law of T her modynamics H eat can not flow fr om a colder body t o a hot t er body wit hout some wor k being done by an ext er nal ener gy. T hir d law of T her modynamics The absolute entr opy of a perfectly crystalline substance becomes zero at absolute zero temperature(0K). M ODES OF TRAN SFER OF H EAT Ther e ar e t hr ee modes of t r ansfer of heat . 1. Conduct i on I n t his mode of t r ansfer ence, heat is t r ansmit t ed fr om on e par t i cl e t o t h e ot her par t i cl e i n t he dir ect ion of fall of t emper at ur e. Ther e is no act ual movement of par t icles. U nder st abl e condi t ons, t he amount of heat ‘Q’ flowing in a time ‘t’ at r ight angles to the faces of a wall of which one face is at temper atur e T 1 while the other face is at temper atur e T 2 (T 2 > T 1) is given by T2  T1 Q = K .A.  t joules f wher e, K = t her mal conduct ivit y of t he mat er ial A = ar ea of faces of t he wall f = t hickness of t he wall. Coefficient of thermal conductivity(K). I t is t he amount of heat flowing i n one second acr oss t he opposi t e faces of a 1 cm cube, mai nt ai ned at a t emper at ur e differ ence of 1C.

Cold Convection cur rent Hot St and

Burner

3. Radi at i on I n t his mode of t r ansfer ence, heat is t r ansmit t ed fr om one place t o t he ot her dir ect ly wit hout heating t he int er vening medium. e.g. heat fr om Sun r eached ear t h. Power t r ansmit t ed, P =  A(T 24 – T 14) wat t wher e A = sur face ar ea  = emissivit y of t he sur face T 2 = t emper at ur e of t he body T 1 = t emper at ur e of t he sur r oundings  = St ef an s con st an t (5.67  10 8 wat t m et r e2  K – 4 ) Black body A per fect ly black body is one t hat absor bs complet ely all t he r adiat ions falling on it . K ir choff 's law At any t emper at ur e and for par t icular wavelengt h, r at io of t he emissive power t o t he absor pt ive power of all subst ances is same and is equal t o t he emissive power of a per fect ly black body. e i.e. = E a Now, E  = 1; t hus a = e wher e, E  = emissive power ;, a = absor pt ive power K ir choff 's law signifies t hat good absor ber s ar e good r adiator s. TH ERM ODYN AM I C PROCESS Reversible changes for perfect gas A reversible change always consists of a succession of states of equilibrium in the absence of any dispersive process. I sochor ic or I sovolumet r ic pr ocess I t r efer s t o a pr ocess, in which t her e is no change in t he volume of t he syst em pr essur e and t emper at ur e may change in such a pr ocess. Adiabat ic pr ocess I n t his pr ocess, t her e is change in t he heat cont ent (or ent halpy H ) of t he syst em, i.e. syst em neit her gains nor losses heat . Also for a per fect gas, equat ion for r ever sible adiabatic change is Tv – 1 = const ant (or ) T 1v 1 – 1 = T 2v 2 – 1 and Tp1– / = constant (or ) T 1p11– / = T 2p21– / wher e  = cp/cv

Physics 1.15

I sot her mal pr ocess I n t his pr ocess, t he t emper at ur e r emains const ant , i.e. dT = 0 I sobar ic pr ocess This pr ocess occur s at const ant pr essur e, i.e. dp = 0 Thus w = pdv = nRdT Q = n.cp dT Efficiency of H eat engine Q  = 1 2 Q1 Coefficient of per for mance in case of r efr iger at or Q2 Q2 C.O.P. =  W Q1  Q2 CARN OT CYCLE I t is a r ever sible cycle and consist s of t wo isot her mal (A  B and C  D) and t wo adiabat ic (B  C, D  A) changes.

A  B i s an i sot h er m al expr essi on at con st an t t emper at ur e T 2. V Q2 = w 2 = nRT 2 log B  0 VA B  C is an adiabat ic change in which t emper at ur e change for m T 2 and T 1 w = – n.c dT = – n.c (T 1 – T 2) > 0 C  D i s an i sot her mal compr essi on at const ant t emper at ur e T 1. V Q1 = w 1 = nRT 1 log d  0 Vc D  A is an adiabat ic compr ession. w = – n.c.dT = – n.c (T 1 – T 2) < 0 For t he complet e cycle, du = 0 w BC + w DA = 0 Qcycle = Q2 + Q1 = w 2 + w 1 = w cycle V Area of cycle, w cycle = nR(T 2 – T 1)log 2 V1 w Q 2  Q1 Q Efficiency,  =  = 1 1 Q Q2 Q2 T2  T1 T1  1 or = T2 T2

 is less t han one for a car not cycle. E nt r opy(s). dq r ev ds = T For a cyclic pr ocess, ds = 0 Unit s: J/K

For a r ever sible pr ocess, change in entr opy is given by B dq ds = SB – SA =  r ev T A For adiabat ic changes, ds > 0 This is valid for r ever sible changes only.

SOU N D WAVE The differ ent shaped vehicle which is r esponsible for t r ansmi ssi on of ener gy fr om one pl ace t o anot her t hr ough a medium wit hout any t r anslation of medium is called a wave. Wave mot ion Pr opagat ion of dist ur bance fr om one place t o anot her is called wave mot ion . M ECH AN I CAL WAVES The waves or iginat ed in an elast ic mat er ial (air, st eel and wat er ) ar e called mechanical waves. Types of M echanical waves These ar e of t wo t ypes. 1. Tr ansver se waves. T h e par t i cl es of m edi u m vibr at e at r ight angles in t r ansver se waves e.g. pr opagat ion of waves t hr ough a r ope. 2. L ongit udinal waves. The par t i cl es of medi um vibr at e t o and fr o in longit udinal waves e.g. sound waves. E lect r omagnet ic wave El ect r om agnet i c waves ar e f or med by n at ur al l y per pendicular vibr at ing elect r ic and magnt ic fields. Pulse I t is a wave of shor t dur at ion. I t can also called a wave of single dist ur bance. Ampl i t ude I t is t he maximum displacement of the par ticles of t he medium fr om t heir mean posit ion. Wavelengt h (  ) I t is t he dist ance bet ween near est par t icles which ar e in t he same st at e of vibr at ion. T ime period (T ) I t is defined as t he t ime t aken by any par t icle of t he medi um t o compl et e one osci l l at i on. I t i s al ways measur ed in seconds. F requency (f) Number of vibr at ions per second is called fr equency . 1 Mathematically, f= T wher e T is in seconds. Phase The phase of an oscillating particle at any instant denotes the position and dir ection of motion of the par ticle at that instant. I t is r epr esented either by the angle which the particle makes with the mean position or by fr action of wavelength or by fr action of time per iod.

1.16 Physics

E poch I f the particle does not start from mean position at t = 0, t hen it is said t o possess an init ial phase called epoch . I nt ensit y of wave The amount of ener gy flowing per unit t ime t hr ough unit ar ea per pendicular to t he dir ection of pr opagation of t he wave is called int ensit y of t he wave. Unit s : joule/m 2-sec or wat t /m 2 Pi t ch I t i s t he char act er i st i c of sound t hat depends on fr equency. I t det er mines t he shr illness or gr aveness of sound. A gr ave not e is called low pitched note while a shr ill note is cal led high pit ched not e. Smaller t he fr equency; smaller is t he pit ch. Gr eat er t he fr equency, gr eat er is t he pit ch. Wave velocity (c) I t is t he dist ance t r avelled by t he wave in one second. I t is also called velocit y of wave. M at hemat ically, velocit y of wave, V = n  wher e,  = wavelengt h of t he wave n = fr equency of t he wave I nfr asonic waves These ar e t he waves wit h fr equencies below audible r ange, i.e. less than 20 Hz. I nfrasonic waves ar e usually pr oduced by lar ge sour ces. e.g. waves or iginat ing fr om ear t hquakes Audible waves Sound waves in t he fr equency r ange 20 H z t o 20,000 H z which pr oduces t he sensat ion of hear ing is called audible waves. The audible frequency range of dogs is 15 Hz to 50,000 Hz. U lt r asonic waves Waves with fr equencies above audible r ange ar e called ult r asonic waves. Waves wit h fr equency above 20,000 H z is ult r asonic wave. Refr act ion of sound When sound waves travel fr om one medium to another, t hey ar e deviat ed fr om t heir pat h. Refr action of sound follows t he same laws as t hat of light v sin i i .e. = 1 v2 sin r wher e v 1 and v 2 ar e vel oci t i es of sound in t he t wo mediums. D iffr act ion of sound Sound waves easily bend r ound t he edges of t he holes and t he obst acles, t his is called diffr act ion of sound. F act ors affect ing Velocit y of sound in air 1. Effect of pressure. The change of pr essur e have no effect on t he velocit y of sound in air (or any ot her gas). 2. Effect of densit y. I f densit y of a gas incr eases, t hen velocit y of sound in it decr eases.

3. Effect of t emper at ure. The velocit y of sound in air incr eases on r aising t he t emper at ur e. 4. Effect of humidit y. Sound tr avels fast er in humid air and slower in dr y air. SH OCK WAVES I f speed of a body i n ai r is gr eat er t han t he speed of sound, t hen i t is cal led super sonic speed. Such a body leaves behind it a conical r egion of dist ur bance whi ch spr eads cont i nousl y. Such a di st ur bance i s cal l ed shock wave. RESON AN CE When a body i s set int o vi br at i ons by an ext er nal per iodic for ce whose fr equency is equal t o t he nat ur al fr equency of t he body, t hen ampl it ude of vibr at i on i ncr eases at each st ep and becomes l ar ge. Such vi br at i ons ar e cal l ed r esonant vibr at ions and t he phenomenon is called r esonance. Condit ion for Resonance Because the sound has to tr avel down the tube and back dur ing one half vibr at ion of the pr ong, length of the air column for r esonance must be one four th of the wavelength of the sound emitt ed by the tuning for k.  2 3 I f lengt h of t he air column is incr eased by , , 2 2 2 ... et c, wher e  is wavelengt h of sound in air, we will again obt ain r esonance in each case. SU PERPOSI TI ON OF WAVES Tw o or m or e pr ogr essi v e w av es can t r av el simult aneously in t he medium wit hout effect ing t he m ot i on of on e an ot h er . T h er ef or e r esu l t an t displacement of each par t icle of t he medium at any inst ant is equal t o vect or sum of t he displacement s pr oduced by t wo waves separ at ely. This pr inciple is called principle of superposition . Applicat ion of Pr inciple of Super posit ion 1. I nt er fer ence. When t wo waves of same fr equency t r avel in a medium simultaneously in the same dir ection, then due t o t heir super posit ion, t he r esult ant int ensit y at any point of t he medium is differ ent fr om t he sum int ensit ies of t he t wo waves. At some point , int ensit y of t he r esult ant wave is ver y lar ge while at some ot her point s it is ver y small or zer o. This phenomenon is called int er fer ence of waves. 2. Beat s. When two sound waves of near ly equal fr equencies ar e pr oduced simultaneously, t hen int ensit y of t he r esult ant sound pr oduced by t heir super posi t on incr eases and decr eases alter nately with t ime. This r ise and fall int ensity of sound is called Beat s. The number of maxima (or ) minima heard in one second is called beats fr equency.

Physics 1.17

Applications of Beats : (i ) Fr equency of unk nown t unni ng for k can be calculat ed. (ii ) M usical inst r ument s can be t uned by r educing beats. 3. St at ionar y waves. Stationar y wave is for med when two waves of same fr equency t r avel l i ng i n opposi t e di r ect i ons ar e super imposed on each ot her. This is a par t icular phenomenon of inter fer ence of waves. I n stationar y waves, t her e is no flow of ener gy in eit her dir ection. The medi um get s spl i t up i nt o segment s, each segement vibr at ing up and down a whole.Ther e ar e some par t icles which ar e per manent ly at r est , called nodes ‘N’ while others which suffer maximum displacement fr om t he mean posit ion ar e called antinodes. L ongit udinal st at ionar y waves These can be pr oduced in a long flexible spr ing or in air column inside a closed end or open end pipe. E lect r omagnet ic waves These can be pr oduced by r apid vibr at ion of cur r ent in a conduct or. I f r esist ance of conduct or is ver y small, t hen fr equency of t he oscillat ion is given by 1 f= ...  f  n  2 L C Spect r um of E lect r omagnet ic waves Fr equency (Hz)  104 106 108 1012 1014 1016 1018 1020 1022 Wavelengt h (m). 104 102 100 10 – 2 10 – 4 10 – 6 10 – 8 10 – 10 10 – 12 10 – 14 Radio micro U.V, V.L X-rays,  -r ays Visible range is comprised of radiations with frequency / wavelengt h r ange : Fr equency :

3.84  1014 H z < f < 7.69  1014 H z

Wavelengt h : 7.80  10 7 m > l > 3.90  10 7 r ed violet DOPPLER EFFECT Whenever t her e is a r elat ive mot ion bet ween sound sour ce and t he obser ver, t her e is an appar ent change in fr equency of t he sound sour ce. This effect is called Doppler effect .

L I GH T I t is an agent which pr oduces in us t he sensat ion of sight . I t it self is invisible but makes t he ot her object s visible. I t may be defined as t he r adiant ener gy which pr oduces t he sensat ion of light . M I RROR I t is a highly polished sur face fr om which most of light is r eflect ed. RAY OF LI GH T I t is a st r aight line pat h along which t he t r ansfer of light ener gy t akes place.

Pencil of light rays I t is a gr oup of inclined r ays of light diver ging fr om a point sour ce or conver ging t o anot her point .

(a)

(b)

Beam of light As shown in t he figur e t he gr oup of par al l el r ays i s called beam of light . D iver gent r ays I f r ays of l i ght ar e di ver gi ng fr om a poi nt sour ce so t hat di st ance bet ween r ays goes on i ncr easi ng as t hey m ove f or war d, t hen t he gr ou p i s cal l ed diver gent r ays as shown in fi gur e (a) above. Conver gent rays I f r ays of light ar e conver ging t o a point so t hat t he dist ance bet ween t he r ays goes on decr easing as t hey move for war d, then t he gr oup is called convergent rays as shown in figur e (b) above. Parallel r ays I f successive light r ays keep equal dist ance t hr ough, t hen t hey ar e par allel r ays. I M AGE I f a pencil of diver ging fr om a point ‘O’ is caused by r eflect ion (or r efr act ion) t o conver ge or t o appear t o diver ge fr om some ot her point I , t hen I is called image of t he object ‘O’. Real image I f r eflected (or r efr act ed) r ays fr om fir st point act ually meet at the second point, then the second point is called r eal image of fir st point . Real image can be t aken on t he scr een. Virt ual image I f r eflect ed (or r efr act ed) r ays fr om fir st point appear t o meet at t he second point , t hen it is called vir t ual image of fir st point . I t can not be t aken on t he scr een. REFLECTI ON AN D REFRACTI ON Refl ect i on I t is bending of l ight t o t he fir st medium fr om t he sur face of separ at ion of t he t wo media. The r ays ar e sen t back by t h i s pr ocess. T h e ph en om en on of r eflect ion of light is shown in t he figur e. I ncident ray

Reflected ray Plane mirr or

Refr act i on I t is bending of light fr om it s st r aight pat h as it ent er s fr om one medium t o anot her. II I l

1.18 Physics

M EDI U M Opt imal medium A subst ance or any por t ion of space t hr ough which light can pass is called opt ical medium . I t can be solid, liquid or gas. H omogeneous or I sot r opic medium M edium possessing same opt ical pr oper t ies in all t he dir ect ions is called homogeneous medium . Tr anspar ent body I t is a body t hr ough which light can pass easily, e.g. air, glass et c. Tr anslucent body I t is the body through which light can pass only partially, so t hat object can be seen only indist inct ly, e.g. oiled paper. Opaque body I t is a body which does not allow light t o pass t hr ough i t , e.g. br i ck s, wood et c. N o subst ance i s per fect l y t r anspar ent or per fect ly opaque. L uminous body A body that emits light itself is called as luminous body, e.g. st ar, sun, fir e et c. N on-luminous body A subst ance t hat does not emit light it self is called non-luminous body. REFLECTI ON OF LI GH T The pr ocess of sending back t he light r ays which fall on t he sur face of an object is called r eflect ion of light . N N or mal

A I nci dent r ay

i 

M

r

O N

B

Refl ect i ng r ay M ir r or  M

I ncident r ay The r ay of light or iginat ing fr om t he sour ce and falling on t he sur face of a mir r or OA is t he incident r ay. Point of incidence I t is the point at which the incident r ay comes in contact wit h t he mir r or. O is t he incident point . Reflect ed r ay The r ay of light which is sent back by t he mir r or is called r eflect ed r ay . OB is t he r eflect ed r ay. N or mal The nor mal is a line at r ight angles to the mirr or sur face at t he point of incidence. ON is t he nor mal. Angle of incidence (i) I t is the angle made by the incident r ay with the nor mal at t he point of incidence. Angle of reflect ion (r) I t is t he angle made by t he r efl ect ed r ay wi t h t he nor mal at t he point of incidence.

Regular r eflect ion When a beam of par allel light r ays falls on a shining but plane sur face, t he light r ays ar e r eflect ed back in t he same or der. I t is r egular r eflect ion. I nci dent r ay

Reflect i ng r ay

A

B

I r r egular r eflect ion When light beam falls on r ough but uneven sur face t he light r ays r eflect ed back in many dir ect ions. This is known as ir r egular r eflect ion. I t gives scat t er ed or diffused light. I nci dent r ay

Reflect i ng r ay

A

B

LAWS OF REFLECTI ON The r eflect ion of light fr om a plane sur face like t hat of a plane mir r or takes place accor ding to two laws which ar e called t he laws of r eflect ion . F irst law of Reflect ion The incident r ay t he r eflect ed r ay, and t he nor mal all lie in t he same plane. Second law of Reflect ion The angle of r eflect ion is always equal t o t he angle of incidence. i=r TYPES OF M I RRORS 1. Plane mir r or When a plane mir r or is r ot at ed t hr ough an angle , t hen t he image is r ot at ed t hr ough an angle 2. 2. Spher ical mir r or I t is t hat mir r or whose r eflect ing sur face is t he par t of a hollow spher e of glass. Ther e ar e t wo t ypes ( i ) Concave mir r or I t is that spher ical mir r or in which the r eflection of light t akes place at t he concave sur face (or bent in sur face). IR

M i r r or

R.R Concave Sur face

( ii ) Convex mir r or I t is that spherical surface at which the r eflectioin of l i ght t ak es pl ace i s convex sur face or t he sur face bulges out war ds.

Physics 1.19

FOCAL LEN GTH Focal lengt h of a concave or convex mir r or is equal t o half of t he r adius of cur vat ur e r f= 2 Relat ion bet ween Conjugat e dist ances (M ir r or for mul a) 1 1 1   v u f 1 1 1   I mage dist ance Object dist ance Focal lengt h L inear M agnificat ion I t is defined as t he r at io of image dist ance t o object ive distance.

v u (or ) I t is the r atio which the size of image bear s to the size of object. m=

m=

I O

Ar eal magnificat ion M agnificat ion in ar ea is called ar eal magnificat ion. I t is given by

Area of image v 2  Area of object u 2 Sign convent ion used in M irrors (1) All dist ances ar e measur ed fr om t he pole of t he mi r r or. (2) Dist ances opposit e t o t he dir ect ion of incident r ay ar e t aken as negat ive. (3) Dist ances in t he dir ect ion of incident r ay ar e t aken as posit ive. (4) When image is r eal, v is t aken as negat ive. When image is vir t ual, v is t aken as posit ive. (5) Focal lengt h of a concave mir r or is negat ive. Focal lengt h of a convex mir r or is posit ive. (6) Downwar d dist ances ar e t aken as negat ive (– ve). Upwar d dist ances ar e t aken as posit ive (+ve). REFRACTI ON OF LI GH T When a r ay of light passes fr om one medium t o ot her, it suffer s a change in dir ect ion at t he boundar y of separ at ion of t wo media. This is called r efr act ion . AON = i = angle of incidence m2 =

I ncident ray

A

N Nor mal r ay

Medium I O M

M Medium I I

r i N B

Plane mir ror

Refr act ion r ay

NN  = nor mal t o t he sur face at point A BON = r = angle of r efr act ion

L aws of Refr act ion 1. The incident ray, nor mal to the sur face of separation at t he point of incidence and r efr act ed r ay lie in t he same plane. 2. Sine of angle of incidence bear s a const ant r at io t o t he sine of angle of r efr act ion, i.e

sin i Snell's law  sin r wher e, I I I is r efr act ion index of I I medium wit h r espect t o medium I . TOTAL I N TERN AL REFLECTI ON I f angle of incidence is incr eased beyond cr it ical angle, t hen t he r ay is r eflect ed back int o t he fir st medium. This phenomenon is called t ot al int er nal r eflect ion . Cr it ical angle At a par t i cul ar angl e of i nci dence, t he angl e of r efr act ion (r ) becomes 90. The angle of incidence(c) for which t he angle of r efr action is 90, is called cr itical angle. Angle of Disper sion The differ ence in angles of deviat ion of any t wo r ays is called angle of disper sion for t hose r ays. D i sper si on The separation of white light into its constituent colours by r efr act ion(or ) ot her means is called disper sion of light . I

I I =

A IR Whit e l i ght

B

Red Bl ue Yel l ow C

D isper sive power I t is r at io of t he disper sion bet ween any t wo colour s t o t he deviat ion suffer ed in t he same pr ism by t he mean r ay. v  R W =  1 y

CU RVED SU RFACES I f cur ved sur faces ar e spher ical, t hen t he lenses ar e called spher ical lenses. Types of curved surfaces Cur ved sur faces ar e of t wo t ypes. 1. Convex lens (convergent ). The dist inguishing char acter ist ic of a convex lens is t hat it is t hicker at t he cent r e t han at t he edges. Focal lengt h of convex lens is posit ive. 2. Concave lens (diver gent ). The dist inguishing char act er ist ic of a concave lens is t hat , it is t hinner at t he cent r e t han at edges. Focal lengt h of concave lens is negat ive.

1.20 Physics

LEN S Power of lens I t is it s abilit y t o conver ge or diver ge t he r ays of light . I t is measur ed as r ecipr ocal of t he focal lengt h of a lens expr essed in met r es. Power of lens (P) =

1 f  mt s

L ens formula The gener al for mula for connect ing object and image dist ance, for bot h t he convex and concave lens is

1 1 1 =  v u f L ens-M aker 's formula For mula for t he r efr act ion t hr ough a lens is  1 1  1  =    1  f  R1 R 2  wher e,  = r efr act ive index of t he lens N ewt on's formula for lenses 2 x 1x 2 = f wher e, x 1 = dist ance of object fr om t he fir st focus x 2 = dist ance of image fr om t he second focus. Combinat ion of t wo lenses in cont act Focal lengt h f is given by,

1 1 1  = f1 f 2 f

Power, P = P1 + P2 H U M AN EYE Human eye for ms r eal image on the r etina. The eyeball is near ly spher ical chamber and can be r ot at ed in t he socket by means of it s six muscles. The out er coat ing called scler ot ic consist s of fibr ous whit e t issues. The fr ont par t of scler ot ic is t r anspar ent . A nor mal eye has power of accommodat i on whi ch enables object s as far as infinit y and as close as 25 cm t o be focussed on r et ina. N ear point The shor t est dist ance at which an eye can see clear ly is called near point . L east point I t is t he dist ance at which an eye can see clear ly. I t is t aken as 25 cm for a nor mal eye. Visual angle The angle which an object subtends at our eye is called visual angle. The appar ent size of an object as seen by our eye depends upon t he visual angle. M yopia or Shor t sight edness I t is t he defect of t he eye when it cannot see far -off object s clear ly. M yopia occur s due t o (i ) elongat ion of t he eye ball, or (ii ) decr ease in focal lengt h of t he lens. M yopia is cor r ect ed by a concave lens.

H ypermet r opia or L ong sight edness I t is t he defect of t he eye when i t cannot see near object clear ly. This defect can be cor r ect ed by using a conver ging lense of cor r ect focal lengt h. At igmat i sm I t is a defect of t he eye when a st r aight object looks as cur ved. I t occur s due t o asymmet r ic cur vat ur e of eye lens. This defect is cor r ect ed by using cylindr ical lens. T ELESCOPES They ar e used t o br ing t he dist ance object s closer and hence incr eases t he visual angle (i). I t is an opt ical inst r ument . Types of Telescopes 1. Ast r onomical t elescope 2. Ter r est r ial t elescope 3. Galellion t elescope 4. Reflect ing t elescope. M I CROSCOPE S M icr oscope is an opt ical instr ument which for ms lar ge i mage of a cl ose and mi nut e obj ect . Th i s i m age subt ends a lar ge visual angle at t he eye so t hat t he object looks lar ge. Types of M icroscopes 1. Compound micr oscope 2. Simple micr oscope 3. Elect r onic micr oscope ABERRATI ON OF LEN SES The image for med by t he lense suffer fr om following t wo main dr awbacks : 1. Spher ical aber r at ion Aber r at i on of t he lens due t o which all t he r ays passes thr ough t he lense ar e not focussed at a single point and t he image of a point object placed on t he axis is blur r ed, is called spher ical aber r at ion . M ar ginal rays

Fm

Par axi al r ays

Fp Fm

I t can be r educed by using (i ) stops (ii ) lens of lar ge focal lengt hs (iii ) plano-convex lenses (iv) cr ossed lenses (v) combining convex and concave lenses 2. Chr omat ic aber r at ion I mage of a whit e object for med by lens is usually col our ed and bl ur r ed. Thi s defect of t he i mage pr oduced by lens is called chr omat ic aber r at ion . WAVE N ATU RE OF LI GH T Wavefr ont The focus of all such par t icles of t he medium which ar e vibr ating in the same phase at any inst ant is called wavefront .

Physics 1.21

I nt erference of light When two light waves of exactly equal fr equency having a phase differ ence which is const ant wit h r espect t o t ime t r avel in t he same dir ect ion and over lap each ot her, t hen t he int ensit y is not unifor m shape. This phenomenon is called int er fer ence of light . Polar isat ion of light I t is t he only phenomenon in physics which pr oves t hat light is a t r ansver se wave. Plane of vibrat ion The plane cont aining dir ect ion of vibr at ion and t he di r ect i on of pr opagat i on of l i ght i s cal l ed plane of vibr at ion . Plane of polar isat ion The plane passing thr ough the dir ection of pr opagation an d con t ai n i n g n o vi br at i on s i s cal l ed pl ane of polarization . Br ewest er ’s law Br ewest er discover ed t hat t her e is a simple r elat ion bet ween polar ising angle ‘i p’ and r efr act ive index  of the mater ial r elat ive t o the sur r ounding medium. This is called Brewest er ’s law .

 = t an i p =

sin i p cos i p

The polar ising angle for air glass is 57. DOU BL E REFRACTI ON I t was discover ed by Er asmous Bar t hdinous. A r ay of unpol ar i sed l i ght i nci dent on a cal ci t e (or quar t z) cr y st al , spl i t s u p i n t o t w o r ef r act ed r ays. T h e phenomenon is called double r efr act ion . DI FFRACTI ON OF LI GH T Bending of light ar ound t he edges of an obst acle, or encr oachment of light wit hin t he geomer t ical shadow is called diffr act ion of light . At mospher e is t r anspar ent t o visible r adiat ion, but almost opaque t o infr ar ed r adiat ions. L ow lying clouds pr event infr ar ed r adiat ions t o pass t hr ough t hem and t hey keep t he ear th’s sur face war m at night . This effect is called gr een house effect . The differ ence bet ween int er fer ence and diffr act ion is t hat super -posit ion effect bet ween wavelet s st ar t ing fr om t wo coher ent sour ces whi l e di r ect i on i s t he diffr act ion in t he super posit ion.

M AGN E T I SM

M agnet ic Poles (m) When a magnet is br ought near a heap of ir on filling t he ends of t he magnet show t he gr eat est at t r act ion. These ends wher e the magnetic attr action is maximum ar e called poles of a magnet . I n ever y magnet ther e ar e t wo poles 1. Sout h pole ; 2. Nor t h pole. S.I . unit of st r engt h of a magnet ic pole: amper e met er

M agnet ic axis I t is t he line joining t he t wo poles of a magnet inside it s body. M agnet ic mer idian I t is t he ver t ical plane passing t hr ough axis of fr eely suspended magnet . Geogr aphic mer idian I t is t he ver t ical plane passing t hr ough t he axis of r ot at ion of t he ear t h. D eclinat i on I t is t he angle bet ween geogr aphic mer idian and t he magnet ic mer idian at t he place. M agnet ic lengt h The dist ance bet ween poles of t he magnet is called magnet ic lengt h . I t is denot ed by ‘2 l ’.

Unit s: met er in S.I . syst em. M agnet ic moment (M ) I t is pr oduct of t he pole st r engt h (m) and lengt h of magnet (2l ) is called magnet ic moment .



M = m  2l = 2lm

Unit s: Amper e-met er 2 or joule/t esla M agnet ic flux ( ) The t ot al number of magnet ic lines of for ce nor mal t o a sur face is called magnet ic flux .

Unit s: weber I ntensity of M agnet ic field (H ) When a magnetic mat er ial is place in a magnetic field, it becomes magnetised. The capability of magnetic field t o magnet ise a mat er ial is called magnet ic intensit y of t he field. B H = 0 Unit s : amp/met r e. M agnet ic pot ent ial M agnet ic pot ent ial at a point is defined as t he wor k done car r ying a unit nor t h pole fr om infinit y t o t hat point against t he field. or Magnetic potential is defined as a quantity whose space r at e of var iation in any dir ect ion gives int ensit y of t he magnet ic field. Unit s : joule/weber M AGN ETI C SU BSTAN CES 1. D iamagnet ic subst ances These ar e subst ances which on being placed in a magnet ic field get feebly magnet ised in dir ect ion opposi t e t o t hat of t he magnet i si ng fi el d. Such substances get r epelled when br ought near a st r ong magnet. This pr oper t y of diamagnet ic subst ances is called diamagnetism . e.g. Bi smut h (BI ), H ydr ogen (H 2), Ni t r ogen (N 2), Water (H 2O), Common salt (NaCl), Diamond (C), Gold (Au), Silver (Ag), Copper (Cu), Zinc (Zn) et c

1.22 Physics

2. Par amagnet ic subst ances These ar e subst ances which on being placed in a di amagnet i c fi el d get feebl y magnet i sed i n t he dir ect ion of t hese magnet ic field. Such subst ances, get at t r act ed t owar ds t he magnet , when br ought near a str ong magnet. This pr operty of paramagnetic subst ances is called paramagnet ism . e.g. Aluminum (Al), Sodium (Na), Plat inum (Pt ), M anganese (M n), Copper (I I ) chlor ide (CuCl 2) et c 3. F er r omagnet ic subst ances These ar e those which on being placed in a magnet ic field get st r ongly magnet ised in t he dir ect ion of t he magnet ic field. Such subst ances when br ought near t he magnet get ver y much at t r act ed t owar ds t he magnet . This pr oper t y of fer r omagnet ic subst ances is called ferr omagnet ism . e.g. Iron (Fe), Nickel (Ni), Cobalt (Co), Magnetite (Fe3O4)

connect i ng wi r e, as shown i n Fi g. (b), t he posit i ve char ge passes on t he ear t h as shown in Fi g. (c). N ow i f we fi r st r emove t he connect i ng wi r e and t hen t he char ged r od, t he negat i ve char ge spr eads on t he whol e conduct or as shown i n Fi g. (d).

E L E CT ROSTAT I CS

Mathematically, F = Eq Unit s : newt on Coulomb's I nverse square law I t st at es t hat t he for ce of at t r act i on (or ) r epul sion bet ween two char ges is (i ) dir ect ly pr opor tional to the pr oduct of two char ges, (ii ) inver sely pr opor tional to the squar e of the distance bet ween t hem.

ELECTRI CI TY Amber, glass, ebonit e, sulphur et c. on being r ubbed at t r act l i ght bodi es. Thi s pr oper l y i n mat er i al s i s developed due to electrification by friction. On acquiring t his pr oper ty, the mat er ial is called elect rified and t his pr oper t y is called electr icity . The elect r icit y developed on bodies, when t hey ar e r ubbed wit h each ot her is called fractional elect r icity (or ) stat ic elect r icity . Types of E lect ricit y 1. Posit ive char ges 2. Negat ive char ges Rules : L ike char ges r epel and unlike char ges at t r act each ot her. Test : Repulsion is a super t est of elect r ificat ion. E lect r oscope I t is an inst r ument used t o det ect and det er mine t he kind of elect r icit y pr esent on a char ged body. Electric field (E) The r egion in t he neighbour hood of an elect r ic char ge wher e its influence can be exper ienced is called electric field. Unit s : newt on/coulomb (or ) volt /met r e CH ARGE I N DU CTI ON L et a conduct or AB is mount ed on an insulating st and as shown bel ow. Br i ng a posi t i vel y char ge i n t he conduct or is at t r act ed t owar ds t he glass r od while t he posit ive char ge is r epelled. Thus near end A of t he conduct or acquir es negat ive char ge and it s far end B acquir es posit ive char ge. When posit ively char ged r od is r emoved, t he conductor against becomes elect r ically neut r al. T h i s sh ow s t h at du r i n g i n du ct i on , equ al an d opposi t e char ges ar e i nduced at t he t wo ends of t he conduct or. Holding the positively char ged r od near t he conduct or, i f far end is connect ed t o ear t h wi t h a

+ + +

I nsulat or B

Insulator

A Glass r od

(a)

Glass r od

Insulator

A

(c)

A

(b)

+ + +

I nsulat or B

B

+ + +

B

Glass r od

A

(d)

ELECTRI C F ORCE I t is t he for ce exper ienced by a char ge ‘q’ placed at a point in an elect r ic field of int ensit y ‘E’.

i.e. F 

q1 q 2 r2

wher e,

1 k 4 0

or

F=

qq 1 q1 q 2  k 12 2 4 0 d 2 r

k = constant called permitivity or dielectric constant = 9  109 Nm 2/c2 Value of k depends upon (i ) unit s in which for ce, char ge and dist ance (ii ) nat ur e of int er vening medium Unit s : Nm 2/c2 SEM I CON DU CT OR I t allows t he cur r ent par t ially when t he t emper at ur e is incr easing conduct ivit y also incr easing r esist ance wi l l be decr easi ng. Thi s i s t he speci al pr oper t y of semiconduct or. e.g. Ge, Si ELECTRI C FI ELD I N TEN SI TY (E) I nt ensit y of t he elect r ic field at a point is defined as t he for ce exper ienced by a unit posit ive char ge when placed at t hat point . I f a small t est char ge q0 exper iences a for ce F at a point in a elect r ic field, t hen int ensit y E of t he elect r ic field at t hat point is given by E = F/q0

Physics 1.23

I t is a vect or quant it y and has t he same dir ect ion as t hat of t he for ce on a unit posit ive char ge. Unit : S.I unit of electr ic int ensity is newton/Coulomb. ELECTRI C LI N ES OF FORCE The pat h t r aced by a t est char ge fr ee t o move under t he effect of an elect r ic field is called a line of for ce. The line of for ce is a cur ve so dr awn t hat a t angent t o it at any point gives dir ect ion of t he r esult ant elect r ic field at t hat point . Pr oper t ies. (1) A line of for ce st ar t s fr om a posit ive char ge and ends on a negat ive char ge. (2) No t wo lines of for ce cr oss each ot her. (3) I t is always nor mal t o t he sur face of t he conduct or. (4) These do not pass int o a closed conduct or. (5) These cont r act lengt hwise and expand sidewise. (6) The dir ect ion of for ce is given by t he dir ect ion in which a fr ee posit ive char ge t ends t o move.

+Q

–Q

Electric int ensity at a point in an Electric field I t is equal t o t he for ce in dynes exper ienced by a unit positive charge when placed at that point. The dir ection of elect r ic int ensit y is t he same as t hat for ce. Elect r ic int ensit y at a point , E =

1 q  2 4 E d

Unit s : dynes/e.s.u (or ) ELECTRI C DI POLE I t is a pair of equal and opposite char ge separ at ed by a fixed dist ance is called elect r ic dipole. ELECTRI C POTEN TI AL The electric potential at a point in an electric field is measured by the amount of work done in taking a unit +ve charge from infinity to that point against electric forces. +Q

+Q

Unit : volt 1 volt = 1 joule/1 coulomb = 1/300 e.s.u. of pot ent ial Pot ent ial difference (V) The potential difference between two points in an electric field is defined as the amount of wor k done in moving a unit posit ive char ge fr om one point to t he other. I f W amount of wor k is r equir ed t o move a char ge Q fr om one point t o anot her in t he elect r ic field, t hen pot ent ial differ ence bet ween t wo point s is given by W V= or W = QV Q S.I . unit Volt

Elect ric pot ent ial energy (U ) Electr ic potential ener gy of a system of char ges is the wor k done in br inging these char ges fr om infinity to near each other to for m the syst em. Pot ent ial ener gy U of a syst em of char ges q1 and q2 separ at ed by a dist ance (r ) a par t is given by U=

1 q1q 2 . 4 0 r

Capacit or (or ) condenser I t is a device, which is used for st or ing elect r ic char ge. Two met al plat es separ at ed by an insulat or const it ut e a capacit or or condenser. Types of Condensers 1. Fixed condenser

2. Var iable condenser

3. Elect r olyt ic condenser

4. L eyden jar

Capaci t ance I t is t he r at io of char ge of a conduct or t o it s pot ent ial Q C= far ad V CAPACI TY (C) The capacit y of a conductor may be defined as t he r at io bet ween t he char ge ‘Q’ given t o t he conduct or and t he pot ent ial V t o which it is r aised.

Q far ad V Capacity of conductor is 1 far ad if a char ge of 1 coulomb is r equir ed t o r aise it s pot ent ial 1 volt i .e.

C=

1 coulomb 1 volt Unit : Pr act ical unit of capacit y is far ad. 

1 far ad =

Far ad is t he capaci t y of conduct or whose pot ent i al r aises by 1 volt when char ge of 1 coulomb is given t o it 3  109 st at coulomb 1 coloumb = 1 / 300 st at volt 1 volt 1 far ad = 9  1011 st at far ad Capacit y of a parallel plat e condenser

1 far ad =

X

t

d

K 0 A Far ad d I f t her e i s vacuum (or ai r ) bet ween plat es, t hen k = 1 C=



C0 =

0 A d

Y

1.24 Physics

I f t her e is a dielect r ic medium (inst ead of vacuum) bet ween t he plat es, t hen capacit ance of t he capacit or incr eases k t imes C = kC wher e, k = dielect r ic const ant of medium d = dist ance bet ween t he plat es in cm A = ar ea of t he plat e in sq. cm Capacit or s in Ser ies combinat ion When a number of capacitor s having capacit ies C1, C2, C3 ... Cn ar e joi ned i n ser i es, t hen t hei r combi ned capacity is

1 1 1 1 1 =    ...  C C1 C2 C3 Cn +q

+

+q

v1

+q

v2

I t is t he pot ent ial differ ence at t he poles of t he cell when no cur r ent is flowing (open cir cuit ). E lect r ic cir cuit I t is t he closed pat h along which an elect r ic cur r ent flow. Cu r r en t f l owi ng t hr ough t he ci r cui t ( cl osed ci r cui t ) When t he cell is in closed cir cuit , a par t of e.m.f is used up t o over come int er nal r esist ance ‘r ’ of t he cell. I n such a case, cur r ent flowing t hr ough cir cuit R

+q

v3

i

vn

V

Capacit or s in Par allel combinat ion When ‘n' number of capacit or s having C1, C2, C3 ... Cn ar e joined in par allels t heir combined capacit y is C = C1 + C2 + C3 + ... + Cn Wh en ‘n ’ capaci t or s ar e con n ect ed i n par al l el combinat ion, t hen pot ent ial r emains const ant and t he r esult ant capacit y should be incr easing.

A

E lect r omot ive for ce

+q

C1

q1

C2

q2

C3

q3

Cn

E Rr wher e, E = e.m.f of t he cell i=

r = int er nal r esist ance of t he cell R = ext er nal r esist ance of t he cir cuit I f V is pot ent ial differ ence bet ween t he poles, t hen V = E – ir Ammet er I t is an inst r ument used t o measur e elect r ic cur r ent in amper es.

–q

B

qn

LI GH TN I N G CON DU CTOR When a char ged cloud passes by a t all building, t he char ge on t he cloud passes t o t he ear t h t hr ough t he building. This causes a big damage t o t he building. Thus t o pr ot ect t he t all building fr om light ing, t he light ing conduct or s, (which ar e point ed met al r oads) passes over t he char ge on t he clouds t o ear t h, t hus pr ot ect ing the buildings.

CU RRE N T E L E CT RI CI T Y EL ECTRI C CU RREN T I t is defined as t he r at e of flow of char ge t hr ough any sect ion of a conduct or. I f a char ge ‘q’ passes t hr ough any sect i on of a conduct or i n t ime t , t hen cur r ent flowing t hr ough it is given by i=

E r

q t

Unit : amper e When one coulomb of char ge flow t hr ough any sect ion of a conduct or in one second, t hen cur r ent fl owi ng t hr ough it is called one amper e.

Vol t met er I t i s an i n st r u m en t u sed t o m easu r e pot en t i al differ ence bet ween t wo point s in volt s. Ohm’s law I t st at es t hat cur r ent flowing t hr ough a conduct or is direct ly pr opor t ional t o the pot ential differ ence across it s ends, if t emper at ur e and ot her physical condit ions r emain unchanged. Vi

or

V = iR

wher e, V = pot ential differ ence i = cur r ent , R = r esist ance

S.I . unit of r esist ance is ohm . L imit at ion of Ohm’s law (1) Only small cur rent should be allowed to flow through t he ci r cui t so t hat t emper at ur e shoul d r emai n const ant . (2) The conduct or should not be subject ed t o any kind of st r ess or st r ain or t ension. ELECT RI C RESI STAN CE The elect r ic r esist ance of a conduct or is t he pr oper t y of t he conduct or by vir t ue of which it opposes t he flow of cur r ent t hr ough it .

Physics 1.25

L aws of Resist ance

ELECTRI C EN ERGY (JOU LE’S LAW)

Resist ance of t he conduct or,

Joule found t hat t he amount of heat (H ) pr oduced in a conduct or is dir ect ly pr opor t ional t o

R  L engt h of conduct or (l)

l Ar ea of cr oss-sect ion a  1 R =  a wher e  = specific r esist ance. Specific r esist ance () is also called r esist ance and it s 1 r eciprocal   is called conductivity (k) of the mater ial.  R

1 k= 

(i ) squar e of cur r ent (i) flowing thr ough the conductor, (ii ) r esist ance (R) of t he conduct or and (iii ) t ime (t ) for which t he cur r ent flows.

 H  i 2RT This r elat ion is called Joule's law or

H =

I 2 Rt J

(in calor ies)

wher e J = 4.18 EL ECTRI C POWER

Specific resist ance (or) Resist ivit y (  ) I t is defined as t he r esist ance offer ed by 1m lengt h of the conductor having an area of cross-section of 1 square met er. Ra = l Unit s of  : ohm-met er or ohm-cm

Elect r ic power of an appliance is defined as t he r at e of consumpt i on of el ect r i c ener gy or as i t s r at e of doing wor k.

W t wher e, W = elect r ic wor k done in t ime t Unit : S.I . unit of power is wat t . P=

1 j oul e second 1 joule = 1 wat t

CON DU CTAN CE (C)

1 wat t =

I t is t he r ecipr ocal of r esist ance, i.e. C = Unit s : mho or ohm – 1

1 R

or

Commer cial unit of elect r ic ener gy is kilowat t hour 1 kilowat t hour = 1000J/s  3600 s = 3.6  108 J

Conduct ivit y (K )

Colour code for Carbon Resist ances T h e val u e of r esi st ances used i n el ect r i cal an d elect r onic cir cuit s var y over a ver y wide r ange. R

4 t em p

I t is r ecipr ocal of t he specific r esist ance (or ) r esist ivit y (), i.e. k=

1 

Unit s : M ho (or ) ohm – 1 Super conduct ivit y I n case of most of t he met als, t her e occur s a decr ease in r esist ance wit h decr ease in t emper at ur e and t he r esi st an ce appr oach es zer o as absol u t e zer o of temper atur e is appr oached. This phenomenon is called super conductivity . e.g. r esist ance of mer cur y becomes zer o at 4K . T her mi st or s T h ese al l ow sem i con du ct or m at er i al s, w h ose r esistance var ies appr eciably with r ise in temper atur e. Ther mist or s ar e used in elect r onics indust r y, e.g. t o safeguar d t he heat er of a t elevision t ube against t he var iat ions in cur r ent .

These r esi st ances ar e usual l y car bon r esi st ances and a col our code i s used t o i ndi cat e val ue of t he r esist ance. N ot e : You can l ear n t he or der of col our by t he sent ence, “ B B R Y of Gr eat Br it ain has Ver y Good Wife” L etter s as an aid to memor y

Colour

F i gur e

M ulti plier

B

Black

0

10

0

B

Br own

1

10

1

R

Red

2

10

2

O

Or ange

3

103

Y

Yellow

4

10

4

G

Gr een

5

10

5

B

Blue

6

10

6

V

Violet

7

10

7

G

Gr ay

8

10

8

W

Whit e

9

10

9

1.26 Physics

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. A Jet engine wor ks on the principle of conser vation of : (a) linear moment um (b) angular moment um (c) ener gy

(d) mass

2. The sur face t emper at ur e of t he Sun is near ly : (a) 2000K

(b) 4000K

(c) 6000K

(d) 8000K

3. I f t he elect r ical r esist ance of a t ypical subst ance suddenly drops to zero, then the substance is called : (a) super conduct or

(b) semiconduct or

(c) conduct or

(d) insulat or

4. A spher ical air bubble is embedded in a piece of glass. For a r ay of light passing thr ough the bubble, it behaves like a : (a) conver ging lens. (b) diver ging lens. (c) piano-conver ging lens. (d) piano-diver ging lens. 5. ‘The st ar s seem t o be higher on t he sky t han t hey act ually ar e'. This can be explained by :

9. Magnets attract magnetic substances such as ir on, nickel, cobalt et c. They can also r epel : (a) par amagnet ic subst ances. (b) fer r omagnet ic subst ances. (c) diamagnet ic subst ances. (d) non-magnet ic subst ances. 10. When a r ay of light is going fr om one medium t o anot her, it s : (a) wavelengt h r emains same. (b) fr equency r emains same. (c) fr equency incr eases. (d) wavelengt h incr eases. 11. A body init ially at r est is acted upon by a const ant for ce. The r at e of change of i t s ki net i c ener gy var ies : (a) linear ly wit h squar e r oot of t ime. (b) linear ly wit h t ime. (c) linear ly wit h squar e of t ime. (d) inver sely wit h t ime. 12. Whi ch one among t he fol l owi ng st at ement s i s cor r ect ?

(a) at mospher ic r efr act ion.

(a) Convex mir r or s ar e used by doctor s to examine or al cavit y

(b) disper sion of light .

(b) Concave mir r or s ar e used as r eflect or s

(c) t ot al int er nal r eflect ion.

(c) Convex mir r or s ar e used as r eflect or s

(d) diffr act ion of light .

(d) Convex mir r or s should be used for shaving

6. Which one among t he following is not a sour ce of r enewable ener gy ? (a) H ydr oelectr icit y (b) Solar ener gy (c) Fuel cell (d) Wind ener gy 7. M ass of B is four t imes t hat of A, B moves wit h a velocit y half t hat of A. Then B has : (a) kinet ic ener gy equal t o t hat of A. (b) half t he kinet ic ener gy of A. (c) t wice t he kinet ic ener gy of A. (d) kinet ic ener gy one-four t h of A. 8. I n a pr essur e cooker cooking is "fast er because t he incr ease in vapour pr essur e : (a) incr eases t he specific heat . (b) decr eases t he specific heat . (c) decr eases t he boiling point . (d) incr eases t he boiling point .

13. Bat s can ascer t ain dist ances, dir ect ions, nat ur e and size of t he obst acles at night . This is possible by r eflect ion of t he emit t ed : (a) ult r asonic waves fr om t he bat . (b) ult r asonic waves fr om t he dist ant object s. (c) super sonic waves fr om t he bat . (d) super sonic waves fr om t he dist ant object s. 14. L ight t r avels slower in glass t han in air because: (a) r efr act ive index of air is less t han t hat of glass. (b) r efr act ive index of air is gr eat er t han t hat of glass. (c) densit y of glass is gr eat er t han t hat of air. (d) densit y of glass is less t han t hat of air. 15. The lines of for ce of a unifor m magnet ic field : (a) must be conver gent . (b) must be diver gent . (c) must be par allel t o each ot her. (d) int er sect .

Physics 1.27

LEVEL-1 1. As t he speed of char ged par t icl e incr eases in a cycl ot r on, (choose Tr ue (T) or False (F)) (a) t he par t i cl e moves t o a lar ger ci r cle

(a) g (c)

(b) t her e i s r elat ivist ic change in t he mass of t he par t icle (c) fr equency of t he cycl ot r on has t o be adjust ed

m1  m 2 g m2

(b)

m1  m 2 g m1

(d)

m2 .g m1  m 2

[RRB JE 2014 GREEN SH I FT ]

7. Which of t he fol lowing st at ement s is cor r ect ?

(a) F, F, F

(b) T, T, T

(a) Speed of light in vacuum is 3 × 108 m/s

(c) T, F, T

(d) T, T, F

(b) Speed of light is differ ent for differ ent colour s

[RRB JE 2014 GREEN SH I FT ]

R1 R 2 2. The for mula R = r epr esent s R1  R 2 (a) ser i es connect i on (b) par allel connect ion (c) br idge connect i on [RRB JE 2014 GREEN SH I FT ]

3. The ear t h conduct or pr ovides a pat h t o gr ound for (a) ci r cui t cur r ent

(b) leak age cur r ent

(c) over cur r ent

(d) hi gh vol t age [RRB JE 2014 GREEN SH I FT ]

4. I f t he mass of sun, ear t h and di st ance bet ween t hem is r espect i vel y M , m and r ; wor k done by t he sun's gr avi t y on ear t h for one r evol ut i on r ound t he sun is

(c)

GM m 2 r

(d) Al l of t he above [RRB JE 2014 GREEN SH I FT ]

8. I n H ei sen ber g's U ncer t ai ni t y pr i nci pl e, t he uncer t ai ni t y of moment um and posi t i on of a par t icle can be (a) r educed usi ng smal ler wavelengt h of pr obing light

(d) linear connect i on

(a) zer o

(c) Speed of li ght i s differ ent in differ ent media

(b)

GM m r2

(d)

GM m 2 r2

[RRB JE 2014 GREEN SH I FT ]

5. The choke of a t ube light wor k s on t he pr i nciple of

(b) r educed usi ng lar ger wavel engt h of pr obi ng light (c) r educed usi ng hi gh ener gy pr obe par t i cl es acceler at ed by cyclot r on (d) can 't be r edu ced as i t i s f u n dam en t al l y inher ent [RRB JE 2014 GREEN SH I FT ]

9. The speed of sound in air is approximately equal to : (a) 3 × l08 m/sec

(b) 330 m/sec

(c) 5000 m/sec

(d) 1500 m/sec [RRB JE 2014 RED SH I FT ]

10. 'When a body is wholly or partially, immersed in a fluid, it experiences an upthrust equal to the weight of the fluid displaced'. This is known as: (a) Pascal's principle (b) Archimedes principle

(a) bi-met allic

(b) capacitance

(c) Stoke's law

(c) induct ance

(d) ionizat ion

(d) Newton's Laws of Motion

[RRB JE 2014 GREEN SH I FT ]

[RRB JE 2014 RED SH I FT ]

6. I n t he fi gur e below, what is t he acceler at ion of body wit h mass m 2, gi ven g i s t he acceler at i on due t o gr avit y (assume pulley and sur faces ar e smoot h)

11. Which one of the following is not a scalar quantity? (a) Volume

(b) Mass

(c) Force

(d) Length [RRB JE 2014 RED SH I FT ]

m1

12. The resultant of two forces P and Q acting at an angle 0, is given by : m2

(a)

P 2  Q 2  2PQt an 

(b)

P 2  Q 2  2PQsin 

1.28 Physics

(c)

P 2  Q 2  2PQcos 

(d) P  Q  2PQtan 

4. What is the boiling point of water in Kelvin Seale? (a) 100 K

(b) 273 K

(c) 373 K

(d) 300 K [RRB SSE 2014 GREEN SH I FT ]

[RRB JE 2014 RED SH I FT ]

5. Acid r ain is caused by:

13. A cyclotron is a : (a) Bunch of Gamma Rays

(a) CO & CO2

(b) SO2 & O2

(b) High Frequency Oscillator

(c) SO2 & N O2

(d) NO2 & O2 [RRB SSE 2014 GREEN SH I FT ]

(c) Particle Accelerator (d) None of these [RRB JE 2014 RED SH I FT ]

14. The nucleus of an atom generally, contains : (a) Protons and Neutrons (b) Protons and Electrons (c) Electrons and Neutrons (d) Only Neutrons [RRB JE 2014 RED SH I FT ]

15. A bullet is fired vertically upwards with a velocity of 196 m/sec. What is the maximum height reached by the bullet ? (Assuming g = 9.8 m/sec2) (a) 1960 m

(b) 196 m

(c) 980 m

(d) 490 m [RRB JE 2014 RED SH I FT ]

LEVEL-2 1. I n a cl assi cal bl ood pr essu r e m easu r i n g inst r ument in which t he doct or obser ves t he r i se and fall of mer cur y, the hand air pump is attached t o a(a) I sobar

(b) Tr ansducer

(c) M anomet er

(d) M er cur y column [RRB SSE 2014 GREEN SH I FT ]

2. Conser vat ion of ener gy cor r esponds t o which law of t her modynamics? (a) Zer ot h l aw

(b) Fi r st l aw

(c) Second l aw

(d) Thir d l aw [RRB SSE 2014 GREEN SH I FT ]

3. I n our house when we swi t ch on heavy l oad appl iances, we not i ce t hat t her e is sl ight di p in t he gl ow of t he bulb t hat was al r eady swit ched on. Thi s i s due t o(a) H eavy cur r ent dr awn by heavy load (b) Addi t ional r esist ance added t o t he cir cuit (c) Resi st ance of elect r ical wi r i ng (d) Resi st ance of par t of t he ci r cuit decr easi ng fr om infi nit y t o a posit ive val ue [RRB SSE 2014 GREEN SH I FT ]

6. Whi ch pl anet has hot t ur bul ent at mospher e dominat ed by car bon-di-oxide? (a) Venus

(b) Mars

(c) Jupiter

(d) Nept une [RRB SSE 2014 GREEN SH I FT ]

7. A tunic fork when sounded together with another t uni ng for k of k nown fr equency of 240 H z, emi t s 2 beat s. On l oadi ng t he t uning for k of known fr equency t he number of heat s hear d ar e one per second. The fr equency of the t uning for k is: (a) 241Hz

(b) 242 H z

(c) 239 H z

(d) 238 H z [RRB SSE 2014 GREEN SH I FT ]

8. Tachymet er (or Tacheomet er i s an i nst r ument for measur ing(a) rpm (b) Tor que (c) Rot at ional kinet ic ener gy (d) Dist ances [RRB SSE 2014 GREEN SH I FT ]

9. W h i ch of t h e f ol l ow i n g i s N OT u sed f or measur ement of t emper at ur e? (a) Ther mocoupl es

(b) Ther most at s

(c) Pyr omet er s

(d) Al l ar e used [RRB SSE 2014 GREEN SH I FT ]

10. Al um i n i u m i s comm onl y used as con du ct or material in tr ansmission lines compar ed to copper because: (a) I t i s mor e conduct i ve (b) I t s t ensi le st r engt h is mor e (c) I t i s cost li er (d) I t i s cheaper and l ight er [RRB SSE 2014 RED SH I FT ]

11. Find t he dist ance of object fr om a concave mir r or of focal lengt h 10 cm so t hat t he si ze of i t s r eal image i s four t imes t he si ze of t he object . (a) 7.5 cm

(b) 5 cm

(c) 2.5 cm

(d) 12.5 cm [RRB SSE 2014 RED SH I FT ]

Physics 1.29

12. A bar omet er measur es : 14. The r efr act ive i ndex of wat er is

(a) Absolut e pr essur e

4 . What is t he 3

(b) At mospher i c pr essur e

speed of li ght i n wat er ?

(c) Gauge pr essur e

(a) 2.25 × 108 m/sec

(b) 4 × l08m/sec

(c) 1.5 × l08m/sec

(d) 2.67 × 108 m/sec

(d) Vacuum

[ RR B

SSE

2014

RE D

SH I F T ]

13. Which one of t he foll owi ng has t he dimensions of pr essur e ? (a) MLT – 2 (b) ML – 1T – 2 (c) ML – 2T – 2 (d) ML – 1T – 1 [RRB SSE 2014 RED SH I FT ]

[RRB SSE 2014 RED SH I FT ]

15. I f t he electr on in hydr ogen or bit jumps fr om t hir d or bi t t o second or bit t hen t he wavelengt h ( o f t he emit t ed r adi at ion i s given by : (wher e R = Rydber g const ant ) (a)  

R 6

(b)  

R 5

(c)  

36 5R

(d)  

5R 36

[RRB SSE 2014 RED SH I FT ]

1.30 Physics

AN SWERS OBJECTI VE TYPE QU ESTI ON S

1. (a)

2. (c)

3. (a)

4. (b)

5. (a)

11. (c)

12. (d)

13. (c)

14. (a)

15. (c)

6. (c)

7. (a)

8. (d)

9. (c)

10. (b)

7. (d)

8. (d)

9. (b)

10. (b)

7. (* )

8. (d)

9. (d)

10. (b)

LEVEL-1 1. (b)

2. (b)

3. (b)

4. (a)

5. (c)

11. (c)

12. (c)

13. (c)

14. (a)

15. (a)

6. (d)

LEVEL-2 1. (c)

2. (b)

3. (a)

4. (c)

5. (c)

11. (d)

12. (b)

13. (b)

14. (a)

15. (d)

6. (* )

EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S 1. A jet engine wor ks on the pr inciple of conservation of linear moment um. 2. Sur face temper atur e of t he sun in near ly 6000K. 3. Wh en t he el ect r i cal r esi st an ce of a t ypi cal su bst an ce su dden l y dr ops t o zer o t h i s phenomenon is called as super conduct or s. 4. A spherical air bubble behaves like a diverging lens. 6. (c) Fuel cell is not a sour ce of r enewable ener gy. 7. Suppose mass of A = m  mass of B = 4m Velocit y of A = v

Velocit y of B =

V 2

K inet ic ener gy of A =

=

1  v 4m    2 2

13. Emission of r eflect ed super sonic waves fr om t he bat can ascer t ain distance, dir ect ions, nat ur e and size of obst acle at night . 14. L ight t r avels sloveen in glass t han in air because r efr act ive index of air is less t han t hat of glass. 15. The lines of for ce of a unifor m magnetic field must be par allel t o each ot her.

LEVEL-1

1 mv2 2

K int enic ener gy of B =

10. When a r ay of light is going fr om one medium t o an ot h er i t s f r equ en cy r em ai n s sam e an d wavelengt h decr eases. 11. A body init ially at r est is acted upon by a constant for ce. The r at e of change of it s ki net ic ener gy var ies linear ly wit h squar e of t ime. 12. Conver e lense of lar ge focal lengt h is used for shar ing. I t is also used in solar cooker.

2

1 v2  4m  2 4

1 mv2 2 So kinet ic ener gy B = K inet ic ener gy A 8. We know t hat Pr essur e Temper at ur e =

as t he incr ease in vapour pr essur e, incr eases t he boiling point . 9. Magnet s at tr act magnetic substance such as ir on, ni ck l e, cobal t et c t hey al so r epel di amagnet i c subst ances.

1. All 3 points are true about charged particle in cyclotron whose speed increases. Particle moves to a larger circle, its relativistic mass changes because of speed change and frequency of cyclotron has to be adjusted. 2. When two resistances are in parallel, the formula gives the value of combined resistance of the circuit. 3. Earth conductor allows path for the leakage current in the body of the equipment or tool to ground. 4. Work done is zero because force is always perpendicular to the direction of movement of the earth. Also another way to look at this is earth returns to the same point after one revolution so work done must be zero as there are no latent energy forms involved.

Physics 1.31

5. Choke works on inductance principle. The function of choke is to provide high voltage enough for ionization to take place in a tube light and after establishment and sustenance of ionization, limit the voltage across the tube. That is the reason why a tube fuses when the choke is shorted.

Second law of thermodynamics states hot the entropy of an isolated system always increase.

6. Force = m.a Here m = m1+ m2 But force applied is m2g. Therefore acceleration = F/m.

 Option (b) is correct.

7. All the statements are correct. Speed of light changes in different media and it is different for different colours in media other than vacuum. 8. The uncertainty of position and momentum of particle in Heisenberg's Uncertainly principle cannot be reduced because it is inherent. 10. Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces and acts in the upward direction at the center of mass of the displaced fluid. 11. Force is a vector quantity. F = m.a and the direction of acceleration will determine the direction of force.

Third law of thermodynamics states that entropy of a system approaches a confound value as the temperature approaches absolute zero.

3. a

Heavy current drawn by heavy load because it has low resistance and takes more power.

4. c

Boiling point of water is 373K.

5. c

Acid rain is caused by chemical reaction when compounds like sulphur dioxide and nitrogen oxides are released into air, these substance rize very high into atmosphere, where they mix and react with water, oxygen and other chemicals to form acid rain.  Option (c) is correct.

6. *

Planet Venus has hot-turbulent atmosphere dominated by CO2.

7. *

The frequency of tuning fork is = 240 – 2 = 238 Hz

8. a

12. Whenever two vector quantities are added, their resultant vector is given by this formula. 2

Tachymeter can measure rpm.  Option (a) is correct.

9. d

All devices are used to measure temperature.

10. b

Copper has highest conductivity but low tensile strength compare to aluminium as copper is must costly so it is no used for transmission line.

2

P  Q  2PQ cos  13. A cyclotron accelerates charged particles outwards from the center along a spiral path. The particles are held to a spiral trajectory by a static magnetic field and accelerated by a rapidly varying (radio frequency) electric field.

11. d

14. Both protons and neutrons are found in the nucleus and are together called nucleons. However, electrons revolve in orbits outside of the nucleus. 15. v2 – u2 = 2as  0 – (196)2 = – 2 × 9.8 × S

1 1 1   u 4u f 5u 4u

2



1 10

 u = 12.5 cm. 12. b

S = 1960

A barometer is an instrument used to measure atmospheric pressure.  Option (b) is correct

LEVEL-2 1. c 2. b

Manometer is an instrument in which doctor observes the rise and fall of mercury. First law also known as law of conservation of energy, states that energy can neither the created not be destroyed in an Isolated system.

13. b

Pr essure 

Force mass  acceleration  Area Area

M1  L1 T 2 L2 M1 L1 T 2  Option (b) is correct

1.32 Physics

14. a

Refractive index 

C V

8



4 3  10  3 V

8

V 

9  10 = 2.25 × 108 m/s 4

 Option (a) is correct

15. d

Emitted radiation 1  1 1 1   4R  2  2   4R    16 36  6  4

 36  16   20   4R   4R   16  36    16  36 



5R 36

 Option (d) is correct

2

Chemistry

CHAPTER F or mul a

Composit ion of M at t er Classificat ion of mat t er M at t er Pur e subst ances El ement s

Compounds

M i xt ur es H omogenous

H et er ogenous

1. M atter. Anything that has mass and occupies space is matt er . 2. Pure Substance. A subst ance is t he for m of mat t er w h i ch h as def i n i t e com posi t i on el em en t an d compounds and pur e subst ance. 3. M ixture. A combination of two or more substances in which the substances r etain their identity is mixture. Types of M ixt ures. ( i ) H omogeneous mixt ur e : I t has a uni for m composit ion t hr oughout it s mass and no visible bou n dr i es of separ at i on bet w een v ar i ou s const it uent s. e.g. solut ion of sugar in wat er. ( ii ) H et erogeneous mixt ure : I t i s a subst ances having different composition throughout its mass. I t has visible boundar ies of separ ation between various constituents. e.g. mixture of sand and cement 4. Element. A subst ance that cannot be separ at ed into simpler subst ances by chemical means is element . Types of Element s. ( i ) M etal: It is a good conductor of heat and electricity. ( ii ) Nonmetal: It is poor conductor of heat and electricity. ( iii ) M et alloid: I t has int er mediat e pr oper t ies of metals and nonmetals. 5. Compound. A subst ance composed of at oms of t wo or m or e el em en t s ch em i cal l y u n i t ed i n f i xed proportions is compound. I n compound elements loss t heir ident it ies.

Symbol

I t is an abbr eviat ion or shor t ened for m for t he full name of an element . This pr esent syst em of symbols was int r oduced by Ber zelius. Symbols and t heir L at in names English name Symbol L atin name Ant imony Sb St i bium Copper Cu Cupr um Gold Au Aur um I r on Fe Fer r um L ead Pb Plumbum Sodium Na Nat r ium

I t is a gr oup of symbols of element s which r epr esent s one molecule of a subst ance e.g. H y dr ogen H 2 ; Ox y gen O 2 ; N i t r ogen N 2 ; Chlor ine Cl 2 ; Car bondioxide CO2 et c.

Chemical E quat i on I t is a way of wr iting a chemical r eaction inter ms of chemical symbols and for mulae. The equat ion should r epr esent a tr ue chemical r eactionwhich can be done in an labor ator y. Reactant s ar e wr it t en on left hand side and pr oducts on r ight hand side and these ar e separ ated by an ar r ow (). Each r eactant and pr oduct ar e seper at ed by(+) si gn.The physi cal st at es ar e r epr esented in br acket s.  r epr esents heat changes. I f H is negat ive, t he r action is exothermic and if its is posit ive, t he r eact ion is endot her mic. The equat ion should be balanced. e.g. 2H 2(g) + O2(g)  2H 2O(l ); H = – 136 K .cal N ote: Unbalanced r eaction viocates low of conser vation of mass.

I on s When electr ons ar e r emoved fr om or added to a neutr al atom or molecule, a charged particle called ion is formed or ions ar e for med by the heter olytic fission of a covalent bond. An ion that bear s a ‘+’ ve char ge is called cations while those which bears a ‘-’ ve charge ar e called anions. Types of I ons 1. M onoatomic ions. These cont ain only one at om. e.g. M g+2, Fe+3 2. Polyatomic ions. These contain more than one atom. e.g. OH – , SO4– 2

I oni c Compounds Neut r al compounds containing cations and anions ar e called ionic compounds. Cat ions ar e most ly der ived fr om met al at oms. Monoatomic anions ar e named by the addit ion of suffix -ide t o a st em der ived fr om t he name of element . e.g. Car bide C– 4, Oxide O– 2, Nit r ide N – 3, Fluor ide F – et c Polyatomic ions containing oxygen are called oxyanions. Types of I onic Compounds. These ar e of t wo t ypes 1. suffix-ate : I t is used for ion wit h lar gest number of oxygen at oms. e.g. NO3 – nit r at e 2. Suffix -ite : I t is used for ion with smallest number of oxygen at oms. e.g. N O 2 – nit r it e

2.2

Chemistry

When an element for ms mor e than two oxyanions, the prefix per meaning more and hypo meaning less ar e used in addition to suffixes-ite and - at e. e.g.

Cl O 4 – per chlor at e; Cl O 2 – chlor it e

Cl O 3 – chlor at e; ClO– – hypochlor it e

When t wo or mor e ani ons di ffer i n number of hydr ogen at oms, t hey ar e named as hydr ogen or dihydr ogen . e.g. PO 43 phosphat e H PO42 hydr ogen phosphat e H 2PO4 dihyr ogen phosphat e

The subscr ipt of t he cat ion is numer ically equal t o t he char ge on t he anion and subscr ipt of t he anion is numer cially equal t o t he char ge on t he cat ion. e.g. sodium car bonat e Na + 1 N a2

CO3– 2 2 CO3

Common name

F or mula

Systematic name

Dr yice Ammonia Table salt Caust ic soda Caust ic potash Soda ash Pear l ash Quick lime M ar ble Slaked lime M ilk of managesia Baking soda

CO2 NH 3 NaCl NaOH K OH Na2 CO3 K 2 CO3 CaO CaCO3 Ca(OH )2 M g(OH)2

Bleacking powder L aughing gas Cane sugar Epsom salt

CaOCl 2

Solid car bondioxide Tr ihydr ogen nitr ide Sodium chlor ide Sodium hydr oxide Pot assium hydr oxide Sodium car bonat e Pot assium car bonate Calcium oxide Calcium car bonat e Calcium hydr oxide M agnesium hydr oxide Sodium hydr ogen ar bonate Calcium oxy chlor ide Nit r ous oxide Sucr ose M agnesium sulphate heptahydr at e Calcium sulphat e dihydr at e

Gypsum

NaH CO3

N 2O Cl 2 H 22 O11 M gSO4. 7H 2O CaSO4. 2H 2O

Types of Chemical React ions

1. Combinat ion React ions React ions in which t wo or mor e subst ances combine t o f or m a com pou n d ar e cal l ed com bi n at i on reactions. e.g. 2Mg(s) + O2(g)  2M gO(s) Synt hesis is t he for mat ion of a compound fr om it s const it uent element s. e.g. N 2(g) + 3H 2(g)  2NH 3(g) 2. D ecombinat ion React ions React i ons i n whi ch a compound decomposes t o pr oduce t wo or mor e differ ent subst ances ar e called decomposit ion r eact ions.

e.g. 2KCI O3(s)  2K CI (s) + 3O2(g) D i ssociat i on. I t is a r ever sible decomposit ion r eact ion. e.g. PCl 5(s)PCl 3 + Cl 2 3. D isplacement React ions React ions in which one element r eplaced anot her element for m a compound ar e called displacement reactions. e.g. Zn(s) + H 2SO4(eq)  ZnSO4(aq) + H 2(g) These ar e r edox r eact ions; i.e. t hey involve t r ansfer of elect r ons fr om one subst ance t o anot her. 4. M etathesis Reactions or Double Decomposition React ions in which t wo compounds r eact t o for m t wo new compounds and no changes in oxidat ion number t ake place ar e called met at hesis r eact ions. e.g. pr ecipit at ion r eact ions, neut r alisat ion r eact ion AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) H Cl(aq) + NaOH(aq)  NaCl(aq) + H 2O(l) 5. I somer i sat ion The i nt er conver si on of one for m of i somer i nt o anot her is called isomer isat ion.  e.g. NH 4CNO   NH 2CONH 2

L aws of Chemical Combinat ion 1. L aw of con ser va t i on of m ass or L aw of indest r uct ibilit y of mat t er I n a chemical change, t ot al mass of t he r eact ant s is equal t o t he t ot al mass of t he pr oduct s; or M at t er can neit her be cr eat ed nor dest r oyed; or Dur ing a chemical r eact ion, t her e is no det ect able loss or gain in t he t ot al mass. 2. Law of definite (or) constant proportions (Proust) A chemical compound al ways cont ai ns t he same element s combined t ogether in t he same pr opor t ion by weight or A chemical compound has a fixed composition by weight. 3. L aw of mult iple propor t ions (D alt on) When t wo element s combine t o for m t wo or mor e compounds, t he di ffer ent wei ght s of one of t he element s combining with the const ant weight of the ot her bear a simple r at io t o one anot her. 4. L aw of combining volumes (Gay-L ussac) When gases combine, t hey do so in volume which bear a simple r atio t o one anot her and to t he volume of the product pr ovided all gases are measur ed under t he same condit ions of t emper at ur e and pr essur e. 5. L aw of r ecipr ocal pr opor t ions (Rit cher ) When two elements combine separately with a fixed weight of a thir d element, the ratio of their weights in which they do so is either the same or a whole number of multiple of the r atio in which they react together.

Chemistry

At oms At om I t is the smallest par ticle of an element that maint ains chemical ident it y t hr ough all t he chemical changes. That cannot have a st able independent exist ance.

conditions of temperature and pressure (STP) or normal condit ions of t emper at ur e and pr essur e (NTP). At STP, 22.4 lit r es of all gases have t he same number of molecul es and mass of each vol ume i n gr ams is numer ically equal t o it s molecular mass.

Avogadr o’s H ypot hesi s

M ol ecul e I t is t he smallest par t icle of an element or compound t hat can have a st able independent exist ence.

At omi ci t y N umber of at oms const it ut i ng a mol ecul e is cal led atomicity . e.g. Nit r ic acid - 5 H NO3 ; H ydr ochlor ic acid - 2H Cl Ozone - 3O3

At omic M ass I t is t he r at io of mass of one at om of an element t o 1 12 par t of mass of car bon -12 at om. 1 amu = 1.667  10– 24g or Avogr am = 1.667  10– 31 kg

Aver age At omic M ass I t is an aver age based on t he abundance of isot opes of t hat element in nat ur e. e.g. Car bon – 12.01 amu; Clor ine – 35.46 amu

Gram Atomic M ass or Gram Atom I t is the numer ical value of t he atomic mass expr essed in gr ams. Number of gram atoms =

2.3

M ass of el ement s in gr ams At omic mass of t he el ement

M ol ecul ar M ass

M olecular mass of a substance (element or compound) is defined as t he r at io of t he mass of one molecule of t he subst ance t o 1/12 par t of mass of car bon-12 at om. M olecular mass of a subst ance M ass of one molecule of t he subst ance = 1 par t of mass of car bon -12 at om 12

M ol e

The numer i cal val ue of t he mol ecul ar mass of a subst ance when expr essed in gr ams is cal led gr am molecular weight or gr am molecule or gr am mole or molar mass or mole. Mass of subst ance in grams Number of moles = Molecular mass No. of at om on M olecules Number of moles = 6.023  1023 Volume of gas at STP (in lit re) Number of moles = 22.4

ST P Condi t ions 0 C or 273K t emper at ur e, 1 at mospher e or 760 mm of K g or 76cm of H g pr essur e ar e cal l ed st an dar d

Equal volumes of all gases under similar condit ions of t emper at ur e and pr essur e cont ain equal number of mol ecul es. 2  volume densit y = M olecular weight

Relat ive D ensit y I t is t he r at io of mass a cer t ain volume of a gas t o t he mass of t he same volume of hydr ogen under similar condit ions of t emper at ur e and pr essur e. Relative density =

Mass if cer t ain volume of t he gas Mass of same volume of hydrogen

M ass of 22.4 lit r es of any gas at S.T.P is equal t o it s molecular mass in gr ams.

Avagadr o’s N umber I t is t he number of at oms pr esent in one gr am at om of an element or I t is t he number of molecules pr esent in one gr am molecular mass of a subst ance or I t is t he number of molecules pr esent in one gr am molar volume of a gaseous sust ance. I t is denot ed by N A , N 0 or N and has a value of 6.023  1023 and it s unit is mol – 1.

L oschmi dt N umber The number of molecules pr esent in 1 ml of a gas or vapour at STP is called L oschmidt number . I t s value is 2.689  1019 ml – 1

M ol e I t is t he amount of subst ance t hat cont ains as many element ar y ent it ies as t her e ar e at oms in 0.012 kg of car bon -12 or I t i s t he amount of subst ance whi ch cont ai ns one Avagadr o’s number 6.023  1023 of par t icles or 1 mole r epr esent s 22.4 lit r es of a gas at STP Gram atomic mass (or gr am at om). M ass of 6.023  1023 at oms of an element . Gram molecular mass M ass of 6.023  1023 molecules of any subst ance.

2.4

Chemistry

At omic St r uct ur e F undament al Par t icles The t hr ee basic subat omic par t icles; elect r on, pr ot on and neutr on which for m the building blocks of all atoms ar e called fundamental par ticles. E lect r on I t is a subat omic par ticle which car r ies a unit negat ive char ge. I t is discover ed by Sir J.J Thomson dur ing t he st udy of cat hode r ays in a dischar ge t ube. The name was int r oduced by St oney. e/m of elect r on = 1.76  108 coulomb/g M ass of elect r on = 9.11  10-28 g or 9.11  10 – 31 kg I t is

1 the mass of hydr ogen, i.e. 0.0005486 amu 1837

M ass of moving elect r on =

Rest mass of elect ron v  1   c

2

wher e, v = velocit y of elct r on c = velocit y of light Pr ot on I t is a subat omic par t ilce which car r ies a unit post ive char ge. The name pr ot on was suggest ed by Rut her for d. Char ge = + 1.602  10– 19 coulomb M ass = 1.007276 amu e/m = 9.58  104 c/g Goldst ein discover ed t he posit ive r ays or canal r ays. N eut r on I t is a subat omic par t icle which car r ies no char ge. I t is discover ed by Chadwick. M ass of neut r on = 1.675  10– 24 g or 1.675  10– 27 kg or 1.008665 amu Pr ot ons and neut r ons ar e found in t he nucleus at t he cent r e of t he at om. The elect r ons r evolve out side t he nucleus in shells or ener gy levels called unit . At omic models Thomson pr oposed wat er melon model of an at om. H e assumed it to be a spher e of positive electr icity in which t he elect r ons ar e embedded like seed.

Rut her for d’s  -r ay Scat t er ing E xper iment

H is model of at om is called nuclear model . (1)At om is spher ical and has a lot of empt y space in it . (2)The entir e mass of at om is concentr ated in its cent r e which is post ively char ged and lies at t he cent r e. (3)The nucleus is sur r ounded by elect r ons which move in cir cular pat h called or bit s. (4)Number of elect r ons is equal t o number of pr ot ons. H ence t he at om is neut r al. (5)Diamet er of nucleus is 10– 13 t o 10– 12 cm and t hat of at om is 10– 8 cm.

(6)The empt y space ar ound nucleus i s cal led ext r a nucluear part . The volume of atom is about 1015 times t he volume of nucleus. (7)This model is called planetary model as it r esembles sloar syst em. At omic number, Z The at omic number of an element is defined as t he number of unit post ive char ges or t he pr ot ons pr esent in t he nuclues of an at om. Z = number of unit posit ive char ges = number of pr ot ons = number of elect r ons (I n case of at om) = ser ial number of elements in periodic table M oseley’s r elat ionship is,  (Nu) = a(Z – b)

wher e,  = fr equency of X - r ay Z = at omic number a and b = const ant s. M ass number (A) I t is t he t ot al number of pr ot ons and neut r ons pr esent in t he nucleus of an at om of an element . A = Z + n (atomic number + number of neutrons) or n =A– Z A Z

wher e, A = mass number Z = at omic number ‘  ’ = symbol of element e.g. 11H , 73 L i, 2311Na, 42H e

I sot opes These ar e atoms of same element which have t he same at omic number but differ ent mass number s. They di ffer i n number of neut r ons pr esent i n t he nucleus. e.g. 3517Cl, 3717Cl I sobar s The at oms of di ffer ent el ement s whi ch have same mass number but differ ent at omic number s ar e called isobars. They di ffer i n number of el ect r ons, pr ot ons and neut r ons. e.g. 146C, 147N I sot ones These ar e t he at oms of differ ent element s which have t he same number of neut r ons. e.g. 146C and 168O

L i ght L ight is a for m of ener gy. I t i s an el ect r omagnet i c r adiation.

Chemistry

N ewt ons cor puscular t heory of light Accor di ng t o N ewt on, l i ght i s composed of mi nut e par t icles or cor puscules which t r avel in st r aight lines in all dir ect ions. Wave theory of light Accor ding t o H uygens, l ight t r avels i n t he for m of waves fr om a luminous object in all dir ect ions.

T ype of r adiation Gama r ays

Wavelength in Å 0.01 t o 0.1

X-r ays

0.1 t o 150

Ultra violet rays Visi ble light I nfr ar ed r ays M icr o waves Raido waves

150 t o 3800

Gener at ion sour ce Nuclei of r adioact ive element s By placing a met al obst acle in pat h of fast moving e – Sun r ays

3800 t o 7600

St ar s, ar c l amps

7600 to 6 106

I ncandescent object ive K lyst r on t ube

Char act er ist ics of Wave Wavelengt h The dir ect distance between any two adjacent ident ical poi nt s of t he wave, i .e. t he di st ance bet ween t wo adj acent cr est s or t wo adj acent t r oughs i s cal l ed wavelengt h . I t is denot ed by . U nit s : m, cm, A 0 or nm; 1 A 0 = 10– 8 cm= 10– 10 m; 1nm=10– 7cm = 10– 9m F r equency I t is t he number of wave cr est s or t r oughs passing or t r ough a given point in one second. I t is denot ed by (Nu). Unit s : cycle per second or H er t z(H z) h = c or



c 

or  =

6 106 t o 3 109 3 1014 t o 3 1014

c 

wher e, c = speed of pr opagat ion of wave Fr equency and wavelength ar e inver sely pr opor t ional t o each ot her. Vel ocit y I t is t he dist ance t r avelled by a wave in one second.I t is denot ed by ‘c’. Unit s : cms– 1 or ms– 1 Velocit y of light is 3  1010 cms– 1 or 3 108 ms– 1. Wave number I t is t he number of waves in one uni t lengt h. I t is denoted by  (Nu bar ). 1    = c  Unit s : cm – 1 or m – 1 Ampl i t ude H eight of t he cr est or dept h of t he t r ough of a wave is called it s amplit ude. I t is denot ed by A. I t is a measur e of int ensit y or br ighness of a beam of light .

Spect r um Wh i t e l i gh t i s com posed of sev en di f f er en t colour s(VI BGYOR). When a beam of whi t e l ight i s passed t hr ough a pr ism, it split s int o seven colour s; t his is called disper sion . This ar r ay of colour s similar t o a r ain bow is called spect r um . The r ange of visible r egion is 3800 t o 7600Å. The r adi at i ons wi t h fr equenci es l ower t han r ed l i ght ar e cal l ed infr a r ed and t hose whose fr equenci es ar e mor e t han vi ol et ar e cal l ed ult r a viol et r adi at i on . The ar r angement of var i ous t ypes of el ect omagnet i c r adi at i ons i n t he i ncr easi ng or der of wavel engt h or d ecr easi n g or d er of f r equ en ci es i s cal l ed elect r omagnet ic spect r um .

2.5

Fr om an alt er nat ing cur r ent wit h high fr equency

Quant um t heor y A black body is a per fect absor ber of ener gy, i.e. it absor bs complet ely all t he r adiat ions falling on it . Phot o elect ric effect When a beam of light of suit able fequency falls on t he sur face of a met al, elect r ons ar e eject ed fr om it . This is called phot o elect r ic effect . This was fir st obser ved by H er t z. The l i ber at ed el ect ons ar e cal l ed phot o electrons. E inst ein’s concept L ight i s pr opagat ed in space in bundles or packet s called phot on . Phot on has no mass, t hus one quant um of light is called phot on . Ener gy of phot on, E = h  wher e, n = element const ent

Bohr ’s T heor y of H yr ogen At om 1. Electr ons r evolve r ound the nucleus in cer tain fixed closed cir cular pat hs called or bit s. The angular momentum of or bit becomes quantised. 2.

mvr =

nh 2

wher e, m = mass of elect r on v = velocit y of elect r ons r = r adius of or bit n = number of or bit in which t he e– pr esent . Angular moment um is an int egr al mult iple of h/2 3. As l ong as an el ect r on r evol ves i n an or bi t , i t nei t her gai n nor l oses ener gy; such or bi t s ar e cal l ed st at i onar y st at es. Ener gy l evel s ar e t he st at ionar y st at es associat ed wit h a difinit e amount of ener gy.

2.6

Chemistry

4. The most st able st at e of an at om is it s gr ound or nor mal st at e. 5. When an elect r on jumps fr om one st at ionar y or bit t o anot her, t he emission or absr opt ion of ener gy takes place. Emission of ener gy takes place in ter ms of light . E = E2 – E1 = h When an electr on moves for m inner to outer or bit by absorbing a definite amount of energy, the electron is said to be in an excited st ate. Radius of or bit , r = 0.529  10– 8 n 2 cm wher e, n = number of t he or bit When n = 1, r = 0.529 Å or 5.29  109 cm. This is called Bohr ’s r adius and is denot ed by a0. Energy of an electron, 2.179  10 11 erg per atom n2 13.6 or En = eV per at om n2 313.6 or En = k.cal per mol n2 13.12 or En = kJ per mol n2 The negat i ve val ue of ener gy E n wou l d k eep on incr easing as t he elect r on moves t o t he ener gy level near er t o t he nucleus.

En =

2 e2 nh Split t ing of spect r al lines in an elect r ical field is called st ar k effect and in appli ed magnet i c field is call ed Zeeman effect. When an elect r on is excit ed int o n t h ener gy level, t he number of spect r al l i nes for med i n t he emi ssi on spect r um is given by

Velocit y of elect r on in nt h or bit , V n=

n(n  1) 2 wher e, n = number of shell in which t he elect r on is present in excited state.

Sommer feld ext ension t o Bohr ’s M odel

Electrons revolve round the nucleus in an elliptical orbits. They have a major axis and a minor axis with different wavelengths AB and CD respcetively. The nucleus of atom is pr esent at one of t he focii of ellipse.The angular momentum of revolving electr on in an elliptical orbit is an integral multiple of h/2 and is given by mvr = k

h 2

K shell; n = 1, k = 1 cir cular L shell; n = 2, k = 1,2 M shell; n = 3, k = 1,2,3 ellipt ical N shell; n = 4, k = 1,2,3,4 The t ot al ener gy of t he elect r on r emains same in an at om.

Wave nat ure of elect ron M at t er has dual char act er and it behaves like a wave and a par t icle. This was pr oposed by de Br oglie. DeBr oglie equat ion wavelengt h of r evolving elect r on

h mv wher e,  = wavelengt h m = mass of par t icle v = velocit y of par t icle h = Planck’s const ant M oment um, p = mv  =

h  wher e, = wave nat ur e p = par t icle nat ur e Or bi t I t is a well defined closed pat h ar ound t he nucleus in which t he elect r on r evolves. p=



Quant um N umber s Electrons of an atom are characterised by four quantum numbers. 1. Principle quant um number (N ei ls Bohr ) I t is denot ed by’n’. Values of n = 1,2,3,4 .......  I f n = 1, K shell I t denot es t he ener gy of r evolving elect r on. I t also gives t he r elat ive dist ance of el ect r on fr om t he nucleus. The maximum number of elect r ons t hat can be pr esent in an ener gy shell is given by 2n 2 wher e, n = number of shells. H ence K shell has 2 e– n=1 L shell has 8 e– n=2 – M shell has 18 e n=3 N shell has 32 e– n=4 2. Azimuthal quant um number (Sommer feld) I t is denot ed by ‘l ’. Values r anges fr om 0 t o (n– 1) l = 0,1,2,..........(n – 1) The main shells ar e made up of ener gy shells called sub ener gy st at es. I t descr ibes shape of t he or bit al and hence called or bit al quant um number. Shell

Value of n

Value of l

K L M N O

1 2 3 4 5

0 0, 1 0, 1, 2 0, 1, 2, 3 0, 1, 2, 3, 4

 l  0, it r epresent ed s  orbit al   l  1, it r epr esent ed p  or bit al     l  2, it r epr esent ed d  orbit al     l =3, it r epresent ed f  orbit al 

Chemistry

3. M agnet ic quantum number (L ande) I t is denot ed by ‘m’. Values of m ar e all whole number s r anging fr om – l t o + l including zer o. m = – l , – l +1 ...... – 1, 0, +1 ......+ l – 1, + l . Tot al number of values of m = (2l + 1). This denot es the spat ial or ient ation of t he or bit als. 4. Spi n qu an t u m n u m ber (U h l en beck an d Gouldsmit) I t is denot ed by m s or s. I t indicat es dir ect ion of spin of e– . 1 1 I t has only t wo values + and – or  2 2 Par allel spin   : Elect r ons spinning in t he same dir ect ion. Two elect r ons of par allel spin never be accomodat e in an or bit al. Opposite spi n : Electr ons spinning in the opposite dir ect ions. Per mit ted values

Quantum Symbol

Total number of permitted values

Pr inciple n Azimut hal l

1,2,3,4,...... 0,1,2, ..... (n– 1)

n n

M agent ic Spin

– l......0....+l +1/2 and – 1/2

2l+1 2

m m s or s

2.7

E l ect r oni c Confi gur at i on Ar r angement of el ect r ons i n t he space ar ound t he nucleus in an at om is called elect onic configuar at ion . Pauli’s exclusion pr inciple No t wo elect r ons in an at om can have t he same set of value for all t he four quant um number s. An or bit al can hold a maximum of 2 e- wit h opposit e spins. Aufbau pr inciple The newly ent er ing elect r on of an atom enter s int o that orbital with lower ener gy among the available ones.

F i g. M ot her ’s or bit al ener gy di agr am

The or der of incr easing ener gies of or bit als is found t o be 1s < 2s < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5d < 6s......... (n+l) r ule is used t o know t he sequence of ener gies of or bit als.I t st at es t hat , t he or bit al wit h t he lowest (n+l) value is filled fir st . I f t wo or mor e or bit als have same (n+l) values, t hen one wit h lower ‘n’ is filled fir st .

Per iodic Classificat ion of E lement s Repr esent at ive element s A 1 H 1.00797 I I A 3 4 Li Be 6.939 9.0122 11 12 Na Mg 22.9898 24.312 19 20 K Ca 39.098 40.08 37 38 Rb Sr 85.47 87.62 55 56  Ba Cs 137.33 132.905 87 88 Fr Ra (223) (226)

Repr esent at ive element s

I ner t gases 0 2 He III A IV A VA VI A VI I A 4.0026 5 6 7 8 9 10 B C N O F Ne Transit ion met als 10.811 12.01115 14.0067 15.994 18.9984 20.183 13 14 15 16 17 18 Al Si P S Cl Ar VI I B VI I I VI I I VI I I IB I I B 26.9815 28.086 30.9738 32.064 35.453 39.948 25 26 27 28 29 30 31 32 33 34 35 36 Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 54.9380 55.847 58.9332 58.71 63.54 65.37 69.72 72.59 74.9216 78.96 79.909 83.80 43 44 45 46 47 48 49 50 51 52 53 54 Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe (99) 107.07 100.905 106.4 107.870 112.41 114.82 118.69 121.75 127.60 126.9044 131.30 75 76 77 78 79 80 81 82 83 84 85 86 Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 186.2 190.2 192.2 195.09 196.967 200.59 204.37 207.19 208.980 (210) (210) (222) L ON G F ORM OF PERI OD I C T ABL E

III B IV B VB VI B 21 22 23 24 Sc Ti V Cr 44.956 47.90 50.942 51.996 39 40 41 42 Y Zr Nb Mo 88.905 91.22 92.906 95.94 57 72 73 74 *La Hf Ta W 138.91 178.49 180.948 183.85 89 † Ac (227)

104 Rf (257)

105 Ha (260)

106 Unh

107 Uns

108 Uno

109 Une

110 Uun

I nner t r ansit ion met als * L ant hani de ser ies † Act ini de ser ies

58 Ce 140.12 90 Th 232.038

59 Pr 140.907 91 Pa (231)

60 Nd 144.24 92 U 238.03

61 Pm (147) 93 Np (237)

62 Sm 150.35 94 Pu (242)

63 Eu 151.96 95 Am (243)

64 Gd 157.25 96 Cm (247)

65 Tb 158.924 97 Bk (247)

66 Dy 162.50 98 Cf (249)

67 Ho 164.930 99 Es (254)

68 Er 167.26 100 Fm (253)

69 Tm 168.934 101 Md (256)

70 Yb 173.04 102 No (253)

71 Lu 174.97 103 Lr (257)

2.8 







Chemistry

L avoisier classified element s simply into met als and non-met als. Dober einer ar r anged gr oup of t hr ee element s such that the atomic mass of centr al element is ar ithmatic mean of ot her t wo el ement s. These wer e cal l ed t r aids. e.g. L i, Na, K . Newland pr oposed law of oct aves which st at es t hat when element s ar e ar r anged in incr easing or der of t heir at omic masses, t he pr oper t ies of ever y eight h element ar e r epet it ion of fir st . L ot har M eyer plot t ed a gr aph of at omic volumes ver sus at omi c wei ght s of di ffer ent el ement s and showed t hat si mi l ar el ement s occupi ed si mi l ar post ions on cur ve. At omic volume =

Mass number Densit y

M andeleef’s per iodic law The physical and chemical pr oper t ies of element s ar e a per i odi c funct i on t o t hei r at omi c wei ght s. The element s pr edict ed by him ar e called eka element s. e.g. Scandium is called Eka Bor on .

Per i odi c L aw The physical and chemical pr opet ies of element s ar e per iodic funct ion of t heir at omic number s. Per i odi ci t y I t is t he occur ance of element s wit h similar pr oper t ies at r egular int er vals when ar r anged in incr easing or der of t heir at omic number s. M odern Periodic law (M osel ey) The physical and chemical pr oper t ies of element s ar e a per iodic funct ion of t heir elect r onic configur at ion. L ong form of Periodic t able The table is called extended form or long form of periodic table. I t h as 18 ver t i cal col um n s cal l ed gr ou ps and 7 hor izontal r ows called per iods. 18 ver tical columns ar e made int o 16 gr oups or families. Shortest period. Fir st per iod wit h only 2 element s. Short period. The second and t hir d per iods ar e called shor t per iods as t hey cont ain 8 element s.

L ong Per i od The four t h and fift h per iods cont aining 18 element s each ar e called long per iods. Each of t hese per iods cont ains t wo s-block element s, t en- d-bolck element s and six pblock elements. Longest period. Sixt h per iod I ncomplet e priod. Sevent h per i od.

Classificat ion of E lement s int o Blocks s-block element s The element s in which differ enciat ing elect r on ent er s t he s-or bi t al of out er most shel l ar e cal l ed s-bl ock elements. Gener al elect r onic configur at ion of s-block element s ar e ns1-2. p-block element s The element s in which differ enciat ing elect r on ent er s t he p-or bit al of t he out er most shell ar e called p-block elements. Gener al elect r onic configur at ion of p- block element s ar e ns2 np1-6. d- block element s The elements in which differ enciat ing elect r on enter s int o d-or bit al of penult imat e shell ar e called d-block elements. Gener al electr onic configur ation of d-block element ar e ns1-2 (n-1)d1-10. f-block element s. The element s in which differ enciat ing elect r on ent er s int o t he ant ipenult imat e shell (n-2) of f or bit als ar e called f-block element s. General electronic configur ation of f-block element ar e ns2 (n-1)d 0 t o 1 (n-2)f 1 t o 14.

Classificat ion of E lement s Based on el ect r oni c confi gu r at i on, el ement s ar e classified int o four t ypes. 1. I nert gas element s Element s in which the out er most s and p subshells ar e complet ely filled ar e called iner t gas element s. Gener al el ect r oni c configur at i on is ns2 np 6. T h ese el em en t s ar e cal l ed r ar e gases. T h ey const it ut e zer o gr oup of per iodic t able. 2. Repr esent at ive element s. E l em en t s i n w h i ch t h e ou t er m ost sh el l i s incompletely filled are called representative elements. Gener al elect r onic configur at ion is ns1, ns2, ns2,np1 t o ns2 np5. M etals, non-metals and metalloids come under this cat egor y. These ar e also called nor mal element s. 3. Transit ion element s Elements which have incompletely filled outer most and penultimate shells are called transition elements. Gener al elect r onic configur at ion is ns1-2 (n-1)d1-9. 4. I nner t ransit ion element s El ement s i n whi ch t hr ee out er most shel l s, i .e. valence penultimate and the antipenutlimate shells ar e incomplet ely filled ar e called inner t r ansit ion elements. Gener al el ect oni c con fi gur at i on i s n s2(n -1)d 0-1 (n-2)f 1-14.

Chemistry

Per i odi c Pr oper t i es The pr oper t i es whi ch ar e r epeat ed aft er cer t ai n i n t er val of at om i c n u m ber s ar e cal l ed per i odi c properties. At omic size The di st ance bet ween nucl eus and t he out er most elect r on of an at om is called at omic r adius. At omic r adium does not det er mined dir ect ly. Cr yst al radius or M et allic radius I t is one-half of t he dist ance bet ween nuclei of t wo adjacent met al at oms. Covalent r adius I t is one-half of t he dist ance bet ween cent r es of nuclei of t wo similar at oms bonded by a covalent bond. V ander Waal’s r adius I t is one half of t he int er nuclear dist ance bet ween t wo at oms faci ng each ot her bel ongi ng t o t wo near est molecules of t he element in solid st at e. r covalent < r met allic ; r covalent < r vander waals I onic r adius I t is t he effective dist ance fr om nucleus of t he ion upt o which it has an influence on it s elect r on cloud.

Val ency N umber of hydr ogen at oms or number of chlor i ne at oms or double t he number of oxygen at oms t hat combi ne wi t h an at om of t he el ement . El ect r ons pr esent in out er most shell ar e called valence electrons.

Acids and Bases All substances ar e basically classified int o acids, bases and neutr al substances. Robert Boyle defined acids and bases based on t heir pr oper t ies.

Aci ds These substances ar e sour in taste, tur ns blue lit mus red liberate hydrogen when reacts with metals, conducts electr icity in solut ion st ate and neutr alized bases.

B ases These subst ances ar e bit t er in t ast e, t ur ns r ed lit mus blue, soapy t o t ouch, conduct elect r icit y in aqueous solut ions and neut r alized acids.

M oder n D ef i n i t i on of A ci ds a n d B a ses (Based on T hr ee T heor ies) 1. Ar r heni us t h eor y (T h eor y of i on i zat i on or elect r olyt ic dissociat ion.) Acid: I t is a subst ance which cont ains hydr ogen and ionizes in aqueous solut ion t o give H + ions. Their gener al for mula is H X. e.g. H Cl, H 2SO4, H NO3, H SO4, CH 3COOH Base: I t i s a subst ance whi ch i oni ses i n wat er pr oducing hydr oxyl (OH – ) ions and ar e r epr esented as M OH . e.g. NaOH , Ca(OH )2, NH 4OH et c.

2.9

Salt: I t is a subst ances which can neit her give H + or OH – ions in aqueous solut ions. e.g. NaCl, K 2SO4 Strengt h of Acids and Bases I t depends on t he degr ee of dissociat ion(). Degr ee of dissociat ion, Number of moles dissociat ed = Tot al number of moles of subst ance 

St r ong acid : I t pr oduces lar ge number of H + ions in aqueous solut ion. e.g. H Cl, H 2SO4, H NO3 Weak acid : I t pr oduces less number of H + ions. e.g. CH 3COOH , H 3BO3 [H + ][X – ] [HX] Strong base dissociates to a larger extent in aqueous solution and produces lar ger number of OH – ions while weak bases dissociates to a lesser extent. e.g.NaOH , K OH , Ba(OH )2 ar e st r ong bases. NH 4OH , Ca(OH )2, Al(OH )3 ar e weak bases.

Dissociat ion const ant of acid, K a =

[M + ][OH – ] [M OH ] K a and K b ar e lar ge for st r ong acids and bases and small for weak acids and bases. The K a values of H Cl , H N O3, H 2SO4, K b val ues of N aOH , K OH cannot be det er mined as they dissociat e t o a lar ger ext ent and do not exist in equilibr ium. N eut r alizat ion. When acids r eact s with equal quantit y of base, both l oose t hei r char act er i st i c pr oper t i es and gi ve neutr al solution. H er e H + ions of acid combine wit h OH – ions of base t o give undissociat ed wat er.   N aCl  H 2O e.g. H Cl  N aOH  

Dissociat ion const ant of a base, K b =

Acid

base

salt

wat er

About 13.67 K.cals of heat is evolved which is called heat of neut r alization . The opposite reaction of neutralisation is Hydrolysis. 2. Bronst ed- Lowry theory (Pr ot on t heor y ) An acid is a subst ance t hat exhibit s a t endency t o loose one or mor e pr otons and a base is a substance t hat exhibit s a t endency t o gain pr ot ons. Acid : pr ot on donor Base : pr ot on accept or A pr ot on t r ansfer r eact ion is called pr ot olysis. When an acid loses a pr ot on, t he r emaining par t has a t endency t o gain it and it behaves as base.

  H*  A H A   pr oton base   H   Cl  H Cl     H O*  A  H A  H 2 O   3 acid – 1 base– 2

acid – 2

base– 1

2.10 Chemistry

A – is called conjugat e base of acid, H A and H 3O+ is conjugat e acid of base H 2O. Conjugate acid-base pair : I t is acid– base pair differ ing by a pr ot on.   H 3O*  Cl  e.g. H Cl + H 2 O     NH *4  OH  H 2O + N H 3  

These ar e r ever sible r eact ions. Conjugat e base of st r ong acid is always weak. N eutralization: Tr ansfer of a pr ot on fr om acid t o a base is called neut r alizat ion .  H 2O + H 2O e.g. H 3O+ OH +  Classification of Slovents: These are of four types. (i ) Pr ot ophilic solvent s : These have t endency t o accept pr ot ons e.g. wat er, alcohol, liquid NH 3 (ii ) Pr ot ogenic solvent s : These have t endency t o pr oduce pr ot ons e.g. wat er, H Cl (iii ) Amphipr ot ic solvent s : These act s as ei t her pr ot ophilic or pr ot ogenic e.g. wat er, NH 3 (iv) Aprotic solvents : These solvents which neither donat e nor accept pr ot ons e.g. benzene, CCl 4 St rengt h of Acids and Bases : St r ong aci d i s one whi ch has a gr eat er t endency t o donat e a pr ot on and i f t he t endency i s l ess, t he aci d i s weak .   H 3O+ + Cl HCl+ H 2O   H Cl is a st r ong acid t hen wat er Cl – is conjugat e base which is a weak base as it has less t endency t o be at t ached t o a pr ot on.   H 3 O+ +CH 3 COO  CH 3 COOH +H 2 O   CH 3COOH weak acid. CH 3COO– conjugat e base is a st r ong base as it has gr eat er t endency t o accept a pr ot on . The conjugate base of a strong acid and the conjugate acid of a st r ong base ar e weak. The conjugat e acid of a weak base and t he conjugat e base of a weak acid ar e st r ong. L eveling effect : St r ong aci ds such as H N O3, H 2SO4, H Cl when pr esent in aqueous solut ion all of t hem have same st r engt h, because al l t hese aci ds compl et ely di ssoci at e and donat e H + i ons easily t o water which exists as H 3O+ in wat er. Since t hese acids pr oduces H 3O+ ions in wat er and H 3O+ is the st r ongest acid in wat er, the str engt h of above acids come down t o t he level of H 3O+ st r engt h in water. Similar ly str ong bases such as NaOH, KOH, Ba(OH )2 come down t o it s st r engt h of OH – ions which ar e st r ong bases i n wat er. This i s call ed levelling effect . The strength of acid and base depends on the natur e of t he solvent used.

Amphoteric substances or Ampholyte : These ar e t he subst ances whi ch can act as an aci d as w el l as a base, i .e. capabl e of donat i n g and accept ing a pr ot on. e.g. Be(OH )2, Al 2O3, ZnO, Cr 2O3 et c. 3. L ewis t heory or Elect ronic t heory Acid : An acid is any molecule or ion that can accept a pair of electr ons and for ms a co-or dinat e covalent bond. e.g. H +, AlCl 3, BF 3 Elect r on-pair acceptor. Base : A base is any molecule or ion that can donate a pair of electr ons and for ms a co-or dinat e covalent bond. e.g. NH 3H 2O, OH – Elect r on pair donor is a base. N eut r ali zat ion : F or mat i on of a co-or di nat e covalent bond bet ween an acid and a base is called neutralization . Types of Lewis acids. (i ) M ol ecul es i n whi ch t he cent r al at om has incomplet e oct et of elect r ons in valence shell (i.e. having a vacant or bit al). e.g. BF 3, SO3, AlCl 3; t he cent r al at oms ar e B,S and Al. (ii ) Posit ive heavy met al wit h incomplet e st able or bitals e.g. Ag+, Cu +2, Fe+2, Fe+2 et c. I n cr ease of ch ar ge an d decr ease of si ze incr eases acidic st r engt h of t he ion. (iii ) M olecules cont aining double bonds bet ween differ ent at om e.g. CO2 All Bronst ed– L owry’s acids are L ewis acids. Br onsted– Lowry’s acid is a pr oton donor, Lewis acid is an elect r on pair accept or. e.g. H Cl is an acid. H Cl + NH 3  [NH 4] + + Cl – H Cl fur nishes a pr ot on – Br onst ed L owr y’s acid The pr ot on of H Cl accept s a lone pair of elect r ons from NH 3 to for m a co-or dinate covalent bond-Lewis acid. Al l L ew i s a ci ds n eed n ot be Br on st ed – L owr y’s acids. L ewi s aci d i s an el ect r ophil e. I t i s an el ect r on deficient molecule or ion. Br ownst ed – L owr y’s acid is a pr ot on donor and elect r on pair accept r or but all elect r on deficient molecules cannot be defined as pr ot on donor s. BF 3 is a L ewis acid, but not a Br onst ed acid.

Chemistry 2.11

Types of Lewis bases. These are of t hr ee t ypes. (i ) All at oms cont aining negat ive char ge; lar ger t he char ge densit y on t he ion, gr eat er is it s abilit y t o donat e an electr on pair and st r onger is basic char act er. e.g. Cl – , F – , O– 2, et c. (ii ) At oms and molecules cont aining one or mor e lone pair of elect r ons. (iii ) M olecules having C = C. The cloud of C = C wi l l for m coor di nat e bond t o gi ve compl ex compound. All Br onsted bases ar e Lewis bases.  Br onst ed-L owr y’s base is a pr ot on accept or. L ewis base is an elect r on pair donor . e.g. br omide is a base H + + Br –  H Br  Br omide is pr oton acceptor – Bronsted-Lowr y’s base, br omide i s el ect r on pai r donor -L ewi s base. As it has 4 lone pair of e– on it and donat es a pair t o H +. All Lewis bases need not be Bronsted– Lowry’s bases. L ewis base is a nucleophile. I t is an accept or of all posit ive ions and elect r on deficient molecules. Br onst ed-L owr y’s base is a pr ot ophile. I t accept s onl y pr ot ons. CaO i s a L ewi s base, but not a Br onst ed base. N eut r alizat ion. An acid and a base r eact t oget her t o for m salt and w at er w i t h t h e l i ber at i on of h eat i s cal l ed neutralization . T h e am ou n t of h eat ev ol v ed du r i n g t h e neut r alizat ion of one gr am equivalent of a base in di l u t e aqu eou s sol u t i on i s cal l ed h eat of neutralization . The heat of neut r alizat ion of a st r ong acid and a st r ong base is 13.7 K . cal/mole (at 25C).

I onic Pr oduct of Wat er

Th e pr oduct of h ydr ogen i on concen t r at i on an d hydr oxyl i on concent r at i on of pur e wat er or any aqueous solut ion is called ionic pr oduct of wat er .

H 2O H 2O H 2O base

K 

acid

 H  OH   H 2 O 

  H   OH      H 3O  OH    acid



K 

base

 H 3O  OH   H 2OH 2O 2

K H 2O  H   OH  or K  H 2O  H 3 O  OH   wher e, K = equilibr ium const ant Si n ce t he i oni zat i on of wat er i s ver y smal l , t he concent r at ion of wat er can be t aken as const ant . Then K[H 2O] = K w or K [H 2O] 2 = K w (another constant) K w = [H +][OH – ] or K w = [H 3O+] [OH – ]

wher e, K w is called ionic pr oduct of wat er which is influenced by t emper at ur e only. I t s val ue incr eases wit h t emper at ur e. At 25 t he value of K w is 1.0  10– 14. [H +][OH – ] = 1.0  10– 14 I n pu r e wat er and neut r al sol ut i ons, t he mol ar concent r at i ons of hydr ogen i on [H +] and hydr oxide ion[OH – ] ar e equal. [H +] = [OH – ] =

1.0  10 14 = 1.0  10– 7 mole/lit r e [H +] = 1.0  10– 7 mole/lit r e [OH – ] = 1.0  10– 7 mole/lit r e The r el at i onshi p K w = [H +][OH – ] hol ds good i n al l aqueous solut ion. [H +] = [OH – ] neut r al solut ion + – [H ] > [OH ] acidic solut ion [H +] < [OH – ] basic solut ion [H +] = 10– 1 M  10 7 M  10 14 M acidic

neut r al

basic

The degr ee of acidic or basic nat ur e of a solut ion can be expr essed in t er ms of hydr ogen ion concent r at ion. Kw 1.0  1014 = [OH  ] [OH  ] Kw 1.0  1014  Similar ly, [OH – ] = + [H ] [H + ] Pr ot on concent rat ion of Acids and Bases I n solut ion of st r ong acids [H +] = nor malit y of t he solut ion + or [H 3O ] or [H +] = Molar ity of the solution  Protocity Pr ot ocit y of an acid : I t is defined as t he number of pr otons pr ovided by each acid molecule upon ionization. I t is also called basicit y . I n solut ions of st r ong bases [OH – ] = nor malit y of t he solut ion or [OH – ] = molar it y of t he solut ion  acidit y Acidi t y of a base : I t i s defi ned as t he number of hydr oxyl ions pr ovided by each base molecule upon in ionization. pH and pOH The concept was int r oduced by Sor ensen. p H : I t i s n egat i v e l ogar i t h m of h y dr ogen i on concent r at ion. 1 pH = – log10 [H +] = log10 [H + ]

[H +] =

p OH : I t i s n egat i v e l ogar i t h m of h y dr ox y l i on concent r at ion 1 pOH = – log10 [OH – ] = log10 [H – ] For pur e wat er at 25C, pH + pOH = 7 + 7 = 14 For pur e wat er and neut r al solut ions pH = pOH = 7 pH = 0  7

Acidic neut r al

 14

basic

2.12 Chemistry

The lower the pH , the mor e acidic is t he solution and higher concentr ation of [H +] the higher the pH , the mor e basic is t he solution and higher concent r ation of [OH]. The pH value can be deter mined exper imentally using a pH mat er. On heat i ng, t he i oni zat i on of wat er incr eases t he concent r at ion of bot h H + and OH – ions. The value of bot h pH and pOH decr eases. The pH of an acidic solut ion incr eases upon dilut ion.

2. Quinonoid t heor y pH r ange of indicat or s. + – I f indicat or is weak acid, H ln + H 2O   H 3O + ln Dissociat ion const ant of indicat or

K ln =

Buffer Solut i on Sol ut i ons whi ch can r esi st t he change of p upon addit ion of small amount of a st r ong acid or a st r ong base is called Buffer solut ion .

pH = pK ln + log

H

Types Buffer s ar e of t wo t ypes. 1. Acid buffer : The solut ions wit h const ant pH r ange of 0 t o 7 ar e called acid buffer s. An acid buffer consists of a weak acid and its salt with a str ong base. e.g. CH 3 COOH + CH 3 COO Na; H 3PO4 + NaH 2PO4 2. Basic buffer : The solut ion wit h const ant pH r ange of 7 t o 14 ar e called basic buffer s. A weak base and it s salt wit h a st r ong acid for ms basic buffer. e.g. NH 4OH + NH 4Cl Fe(OH )3 + FeCl 3 Buffer capacit y The number of moles of an acid or a base added per lit r e of buffer t o change it s pH value by one unit is called buffer capacity .

Acid – Base I ndicator s I ndicat or s ar e used t o find equivalence point bet ween an acid and a base in volumet r ic analysis. Types Ther e ar e t wo t heor ies of indicat or s. 1. Ost wald’s t heor y. (i ) All indicat or s ar e eit her weak or ganic acids or bases. (ii ) I n solut ion t hey dissociat e t o give ions which exist in equilibr ium. acid indicator H ln + H 2O  H O + ln  3   undissociated indicat or ion indicat or molecule base indicat or ln + H 2O  H in + + OH 

indicat or molecule indicat or ion (iii ) The mol ecul ar and i oni c for ms of indicat or have differ ent colour s. (iv) Depending on [H +], t he ionization equilibr ium shift s eit her t o r ight or left . (v) The colour changes with change in pH of the solution.

[H  ][ln  ] [Hln] [ln  ] [H ln]

I f pH of solut ion is equal t o or gr eat er t han Pk ln+1 t hen solut ion exhibit t he colour of indicat or ions and if pH of solut ion is equal t o or less t han pK n– 1 , t hen solut ion exhibit s t he colour of indicat or molecule; t hus aci d-base indicat or s funct i on in p H r ange of (pK ln+1) t o (pK ln– 1) I ndicator

Colour Acid

Alkali

pH r ange

M ethyl or ange

r ed

yellow

3.1 – 4.4

M ethyl r ed

r ed

yellow

4.2 – 6.3

L it mus

r ed

blue

4.6 – 8.3

Bromothy molblue

yellow

blue

6.0 – 7.6

Thymol blue

yellow

pur ple

8.0 – 9.6

Phenolpht halein

colour less

r ed

8.3 - 10

Select ion of I ndicat or I t depends on t he change in pH value at the equivalence poi nt i n an aci d-base neut r al i zat i on r eact i on. The change of pH just at t he end-point is called pH r ange of t it r at ion . An indicat or will show a shar p change in it s colour if pH r ange of the indicator is within t he pH r ange of t he t it r at ion. T it r at ion. 1. Tit r at ion of st r ong acid wit h a st r ong base : pH r ange is about 3.3 t o 10.5. All t he indicat or s in t he above t able can be used. 2. Tit r at ion of weak acid wit h a st r ong base : pH r ange 8 t o 10 phenolpht halein and t hymol blue only fall in t his r ange. 3. Tit r at ion of weak base wit h a st r ong acid : pH r ange is 6 t o 3 met hyl or ange. 4. Tit r at ion of weak acid wit h a weak base: pH change near the end point is not shar p and hence it cannot be det ect ed wit h an indicat or.

Chemistry 2.13

Types of salts 1. N ormal salt s These ar e obtained by complete neutralization of an acid with a base. e.g. NaCl / CaCO3 2. Acidic salt s These ar e obt ained by par t ial neut r alizat ion of a pol ybasi c aci d wi t h a base. Gi ves H + i on s on secondar y ionizat ion. e.g. NaH SO4, Na2H PO4 3. Basic salt s These ar e obt ai ned by par t ial neut r al i zat ion of polyacidic base. These give OH – ions on secondar y ionization. e.g. Mg(OH )Br, Zn(OH)I 4. M ixed salt s. These ar e obt ained by neut r alizat ion of a mixt ur e of t wo acids wit h a base or a mixt ur e of t wo bases wit h an acid. e.g. CaOCl 2, K (NH 4) C2O4 et c.

Salt H ydr olysis

The pr ocess in which t he cat ion or anion of t he salt r eacts with water to pr oduce acidic or alkaline solution. The fr act i on of t he t ot al salt t hat is hydr olyzed at equilibr ium is called degr ee of hydr olysis. 1. Salt of a weak acid and a strong base Salt s of t his t ype under go anionic hydr olysis. salt

–

+

+

wat er

acid + base

CH 3COO + Na + H 2O H ence, CH 3COO– + H 2O

+

CH 3COOH + Na + OH – CH 3COOH + OH

–

The aqueous solut ion cont ains fr ee OH ions, hence t heir solut ion is basic hydr olysis const ant . Ot her examples ar e K CN, Na2CO3, H COOK et c. 2. Salt of st rong acid with, weak base Salt s of t his t ype under go cat ionic hydr olysis. As t he aqueous solut i on cont ains fr ee pr ot ons, t he solut ion is acidic hydr olysis const ant . e.g. N H 4  Cl –  H 2O

NH 4 OH + H + + Cl

H ence, NH 4  H 2 O

NH 4 OH + H +

3. Salt of weak acid and weak base Salt of this type undergo both anionic and cationic hydr olysis. The nature of their solution is almost neutral. e.g. N H 4 + CH 3COO– + H 2O

NH 4OH + CH 3COOH

As bot h H 3O and OH i ons ar e r el eased, t he solut ion is neut r al wit h PH 7. e.g. NH 4CN, CaCO3, AlPO4 4. Salt of st rong acid and strong base Salt of t his t ype do not under go hydr olysis, hence nat ur e of t heir solut ion is exact ly neut r al. The hydrolysis constant (K h) in aqueous solution of these salts is equal to ionic product of water (K w), i.e. Kh = Kw Na+ + Cl – + H 2O Na+ + OH – + H + + Cl – + H ence, H 2O H + OH – e.g. NaNO3, N 2SO4, NaClO4 +

–

Sol ut i ons A sol ut i on i s defi ned as homogenous mi xt ur e of t wo or mor e subst ances. Subst ances whi ch mak e up a sol u t i on ar e cal l ed com ponen t s. Th e component havi ng t he same physi cal st at e as t he sol ut i on and pr esent i n excess over t he ot her component i s cal l ed solvent . Ot her s whi ch ar e i n same or di ffer ent physical st at es and pr esent i n smal l er por t i ons ar e cal l ed sol ut es. For mat i on of a sol ut i on i s a physical pr ocess.

Types of Sol ut i ons

These ar e of t hr ee t ypes. 1. Gaseous solut ion. These ar e sol ut i ons i n whi ch gases and vapour s mi x i n al l pr opor t i ons and for m homogeneous mi xt ur es. Sol vent Sol ut e E xampl e gas gas at mospher i c ai r liquid per fumes sol id smok e 2. L i qui d sol ut i on. T h ese ar e t h e sol u t i on s i n whi ch sol vent i s l i qui d and sol ut e i s gas, l i qui d or sol id. Sol vent Sol ut e E xampl e liquid gas aer at ed dr i nk s liquid al cohol i n wat er sol id sugar sol ut i on Sol ubi l i t y of a gas i n a l i qui d i s gover n ed by H enr y’s l aw, whi ch st at es t hat “ at a con st ant t emper at ur e, sol ubi l i t y i s di r ect l y pr opor t i on al t o pr essur e of gas” . 3. Solid solut ion. At oms or molecul es of one soli d r epl ace t hose i n t he second sol i d t o for m sol i d solut ions. Sol vent sol id

Sol ut e gas liquid sol id

E xampl e sol i d i ce-cr eam t oot h past e all oy

Types of Sol ut i ons (Based on r elat ive amount of dissolved solut e pr esent in a solut ion at a given t emper at ur e)

These sol ut i ons ar e of t hr ee t ypes 1. Sat urat ed solut ion. I t i s t he sol ut i on cont ai ni ng maxi mum amount of sol ut e i n di ssol ved st at e at a gi ven t emper at ur e. I n t his solut ions t wo phases ar e in equilibr ium. 2. U n sa t u r a t ed sol u t i on . I t i s t h e sol u t i on cont ai ni ng l ess t han t he maxi mum quant i t y of a sol ut e i n di ssol ved st at e at a gi ven t emper at ue. 3. Super sat ur at ed sol ut i on. I t i s t he sol ut i on cont ai ni ng mor e amount of sol ut e di ssol ved t han i n a sat ur at ed sol ut i on. I t i s met a st abl e and becomes sat ur at ed by shak ing or st i r r ing.

2.14 Chemistry

Sol ubi l i t y

I t is t he amount of solut e r equir ed t o sat ur at e 100 gr ams of solvent at a given t emper at ur e. Weight of solut e Solubilit y =  100 Weight of solvent F act or s affect ing Solubilit y of subst ance 1. N ature of Solvent and Solute. I onic subst ances ar e sol u bl e i n pol ar sol v en t s an d n on -i on i c subst ances ar e soluble in non-polar solvent s. Rat e of dissolut ion of a solid in a liquid depends on ( i ) Size of solute particles : I ncrease in surface area increases the rate of dissolution as dissolution is a surface phenomenon. ( ii ) Agitat ion of t he mixture : St ir r i ng speeds dissolution. ( iii ) Temperature : Rise of t emper at ur e speeds up dissolut ion as t he t emper at ur e is incr eased, t he k i net i c ener gy and hence t he r at e of diffusion of bot h solut e and solvent par t icles incr eases. Some t ime solubilit y decr eases on I nver se of t emp. 2. Pressure. The effect of pr essur e on solubilit y of a gas in a liquid is gover ned by H enr y’s law. 3. Temperat ur e. The sol ubi l i t y of gases i n wat er generally decreases with increase of temperature.

H eat of Solution

I t is t he amount of ener gy evolved or absor bed when one mole of solute is dissolved in large excess of solvent. H eat of solut ion = lat t ice ener gy + hydr at ion ener gy

H ydr at i on E ner gy

I t i s t he amount of ener gy r eleased when ions ar e pr oduced fr om one mole of ionic subst ance in wat er.

L at t ice E ner gy

I t is t he amount of ener gy r equir ed t o separ at e one mole of ionic cr yst al int o it s const it uent ions. If lat t ice ener gy > hydr at ion ener gy, t he syst em cools down . If lat t ice ener gy < hydr ation ener gy, the system heats up. If lat t ice ener gy = hydr at ion ener gy, t he syst em has little effect of t emper ature

Solubi li t y Cur ves

Cur ves which show t he var iat ion of solubili t y wit h t emper at ur e ar e called solubilit y cur ves.

Concent r at i on

2. Percentage by weight. I t is t he number of gr ams of solut e pr esent in 100 gr ams of solut ion % by weight =

weight of solut e  100 weight of solut ion

3. Volume fraction. I t is t he volume(in ml) of t he solut e pr esent in 1.0 ml of solut ion. volume of solut e Volume fr act ion = volume of solut ion 4. Percentage of volume. It is the volume (in ml) of the solute present per 100 ml of the solution is volume

volume of solut e  100 volume of solut ion 5. M ole fraction. I t is t he r at io of number of moles of a component t o t he t ot al number of t he moles of all component s pr esent in solut ion. n M ole fr act ion of slout e, X solut e = nN n M ole fr act ion of solvent , X solvent = nN percent % by volume =



X

solut e

+X

solvent

=1

6. M ole per cent age. M ole per cent age = M ole fr act ion  100 7. M olality. I t i s t he number of gr am mol es of a solute pr esent in one kilogr am(1000g) of a solvent. m= m=

weight of solut e 1000  molecular weight weight of solvent

10  solub ilit y gram molecular weight of solut e

Unit : mol kg-1 8. M olarity. I t is t he number of gr am moles of solut e pr esent in one lit r e of t he solut ion. weight of solut e  1000 M= gr am molecular weight  solut ion in ml weight %  densit y  10 M= molecular weight of solut e

Unit : mol lit – 1 9. N ormality. I t is t he number of gr am equivalent s of solut e pr esent in one lit r e of solut ion. N= N=

weight of solut e  1000 volume of solut ion  eq. (m 1 ) weight of solut e

percent weight equivalent weight

molecular wei ght n

The amount of solut e pr esent in definit e quant it y of solut ion is called concent r at ion .

Equivalent weight =

M et hods of expr essing Concent r at ions 1. Weight fr act ion. I t i s t he number of gr ams of solut e pr esent in one gr am of t he solut ion. W weight of solut e  Weight fr action = w  W weight of solut ion

Nor malit y = n  M olar it y U nit : gm eq. lit – 1 10. Formality. I t is t he number of for mula weight of solut e pr esent in 1lit r e of solut ion.

Unit s. for mula weight lit r e-1

Chemistry 2.15

Char act er ist i cs of E qul i br i um st at e 1. The equi l i br i um st at e can be obt ai ned fr om bot h di r ect i ons 2. Chemi cal equi l i br i um i s dynami c i n nat ur e. 3. Cat al yst cannot shi ft t he posi t i on of equi l i bur i m but hel ps i n at t ai ni ng i t qui ck l y. 4. T h e equ i l i br i u m can be sh i f t ed by ch an gi n g condi t i ons l i k e t emper at ur e, pr essur e et c. 5. The syst em whi ch i s in equil i br i um st at e wi l l be i n equl i br i um as l ong as i t r emains undi st ur bed. 6. At equi l i br ui m st at e t he change i n fr ee ener gy (G = 0) i s zer o.

Sur face Chemi st r y A dsor pt i on T h e pr ocess of accu m u l at i on of a su bst an ce i n hi gher condi t i ons at t he sur face whi ch separ at es t wo phases. Adsor ben t I t i s t he subst ance whi ch t ak es up or adsor bs t he gas or l i qui d. Adsor at e I t i s t he subst ance whi ch i s adsor bed. Types of adsor pt i on 1. Physi cal or Vander waal ’s adsor pt i on. When a gas i s adsor bed on t he sur face of a sol i d by weak vander waal ’s for ce, t he phenomenon i s cal l ed ph ysi cal adsor pt i on . M u l t i l ay er ed pr ocess i nvol ves heat of adsor pt i on 1-10 K cal / m ol e. 2. Chemical or Act i vat ed adsor pt i on. When a gas i s adsor bed ont o t he sur face of a sol i d by for ces si mi l ar t o t hose of a chemi cal bond, i t i s cal l ed chemi sor pt i on or L angmui r adsor pt i on . I t i nvol ves heat of adsor pt i on 10100 K cal /mol e. A dsor pt i on I t i s t he phenomenon of i ncr ease i n concent r at i on t hr oughout t he body al ong wi t h t he sur face. Sor pt i on I t i s t h e pr ocess i n wh i ch bot h absor pt i on an d adsor pt i on t ak es pl ace. I t i s al so cal l ed r ever se pr ocess.

F act or s affect i ng adsor pt i on (1) N a t u r e of a d sor ben t a n d a d sor ba t e: Per manent gases ar e adsor bed l ess and easi l y l i qui fi abl e gases ar e adsor bed much. (2) Pr essu r e (3) Temper at ur e (4) Sur face ar ea of adsor bent .

Col l oi dal St at e Tr ue sol ut i on I t i s a homogenous mi xt ur e of sol ut e and sovl ent f or m i n g on e ph ase i n wh i ch t h e m ol ecu l es ar e mi xed r andoml y. The sol ut e par t i cl es never set t l e down size of par t i cl es < 10– 7cm. e.g. sal t sol ut i on Suspen si on . The sol ut i on i n whi ch t he par t i cl es can be seen wi t h nak ed eye and whi ch set t l e on st andi ng ar e cal l ed suspensions. Si ze of par t i cl es > 10 – 5cm.

C ol l oi ds A col l oi dal syst em i s het r ogeneous and consi st s of at l east t w o p h ases t h e d i sp er sed p h ase an d di sper sed medi um possessi ng cer t ai n char act er i st i c pr oper t i es. T h e si ze of par t i cl es r an ges f r om 10 – 5 -10– 7 cm. D i sper sed phase Subst ances whose par t i cl es ar e di st r i but ed i n a medi um. I t i s al so cal l ed di scont i nuous phase or inner phase. D i sper si on medi um The medi um i n whi ch t he col l oi dal par t i cl es ar e di sper sed. I t i s al so cal l ed cont i nuous phase or out er phase. D i sper sed phase

M edium N ame

E xample

1. gas

liquid

foam

soap lather

2. gas

solid

solid foam

cor k

3. liquid

gas

aerosol

fog, cloud, mist

4. liquid

liquid

emulsion

milk

5. liquid

solid

gels

cheese

6. solid

gas

smoke

dust

7. solid

liquid

sols

proteins

8. solid

solid

solid sol or gel minerals

Sol I f disper sion medium is a fluid, it is called sol . H ydr osol I f disper sion medium is wat er, it is called hydr osol . I f alcohol or benzene ar e used, t hey ar e called alcosol or benzosol. Types of colloids 1. Lyophilic colloids. These ar e susbst ances which passes int o colloidal st ate simply by br inging it in contact with a solvent. I t is also called r ever sible colloid. e.g. gum, gelat in.

2.16 Chemistry

2. Lyophobic colloids. These ar e insoluble substances which do not r eadily yield colloidal solut ions when br ought in cont act wit h solvent . I r r ever sible colloids ar e also called suspensoids. I f wat er is used, t hey ar e called hydr ophilic and hydr ohobic colloids.

D e- emulsificat ion T he pr ocess of con ver t i n g an em u l si on i nt o i t s com pon en t s, i .e. oi l an d w at er i s cal l e deemulsificat ion . Coagul at ion The phenomenon of change of col l oi dal st at e t o suspension st at e is called coagulat ion or flocculat ion .

M i scel l s

The amount of electr olyte r equir ed to coagulate a fixed amount of a sol depends on t he valency of flocculat ing i on. H ardy-Schulze rule Gr eat er t he valency of flocculat ingion, higher is it s capacit y t o cause pr ecipit at ion. Gold number Weight in milligr ams of a pr otective colloid to be added t o pr event t he coagulat ion of 10ml of a given gold solut ion on adding 1ml of 10% solut ion of NaCl. D onnan membr ane equlibr ium The pot ent ial differ ence devel oped due t o unequal concent r at ion of ions on t wo sides of t he membr ane. E lect r ophor esi s When an elect r ic cur r ent is passed t hr ough a colloidal solut ion cont aining opposit ely char ged solid par t icles and liquid medium and t he pr ocess in which only t he sol i d par t i cl es can move i s cal l ed cat aphor esi s or electophoresis. Zet a pot ent ial or Elect okinet ic pot ent ial I t is t he differ ence of pot ent ial bet ween fixed and diffused par t s of t he double layer. Gel Colloidal solut ions cont aining a liquid disper sed in solid. The pr ocess of for mat ion of gel is called gelat ion .

Subst ances behave as nor mal, st r ong elect r olyt es at low concentr ations but at higher concentrations exhibit colloidal pr oper ties due to aggregation of par ticles; such aggr egat ed par t i cl es ar e call ed micelles and t hese subst ances ar e called associat ed colloids. e.g. soap or det er gent s These ar e amphit het ic, i.e. t hey have bot h lyophilic and lyophobic gr oups. Br ownian movement Th e r an dom an d con t i n uou s m ot i on of col l oi dal par t icles in a disper sion medium is called Br ownian movement . This er at ic mot ion is a r esult of const ant bombar dment of colloidal par t icles by t he molecules of disper sion medium in all dir ect ions. These impar t moment um t o colloidal solut ion.

Tyndall effect When a beam of light is passed t hr ough a colloidal solut ion, it becomes visible as a br ight st r eak. This phenomenon is called tyndall effect and the illuminated pat h is called t yndall cone.

e.g. cur d, soap, boot polish.

EM U L SON S Emulsions ar e for med by disper sion of one liquid in another liquid. Disper sion of tiny par ticles of one liquid in another liquid is called an emulsion and the pr ocess is called emulsificat ion.

Cat al ysi s

Types of Emulsions

Char act erist ics of Cat alyst 1. Cat alyst r emains chemically unchanged dur ing a r eact ion.

1. Oil in-wat er. Oi l i s di sper sed phase and wat er i s di sper si on medium. e.g. M ilk 2. Wat er-in-oil. Wat er is disper esed phase and oil is disper sion medium. e.g. But t er Emulsifier s (or) E mulsifying agent s To get st abl e emul si ons, smal l amount of anot her subst ances ar e added; t hese ar e called emulsifer s or emulsifiying agents.

Substances which alter s the r ate of a chemical r eaction wi t hout under goi ng a chemi cal change i s cal led a cat alyst and t he phenomenon is called cat alysis.

2. Small quant it y is enough t o br ing about a r eact ion. 3. I t does not effect equlibr ium of r ever sible r eact ions. 4. I t h as m ax i m u m ef f i ci en cy at i t s opt i m u m t emper at ur e. 5. I t does not init iat e a r eact ion t hat does not occur. 6. Cat al yst i c pr om ot er s : Su bst an ces w h i ch incr ease act ivit y of cat alyst ar e called pr omot er s. 7. Specificity : Ever y cat alyst is specific in act ion and act s only on a par t icular subst r at e.

Chemistry 2.17

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. The mat er ial used for elect r ic fuse is an alloy of t in and lead. This alloy should have:

8. Whi ch one among t he fol l owi ng i s t he mai n ingr edient in cement ? (a) Gypsum

(a) high specific r esistance and low melting point.

(b) L ime st one

(b) low specific r esistance and high melting point.

(c) Clay

(c) low specific r esist ance and low melt ing point .

(d) Ash

(d) high specific resistance and high melting point. 2. Silver war e tur ns black aft er a per iod of t ime due t o for mat ion of :

9. Glass is act ually : (a) a cr yst alline solid. (b) an ionic solid.

(a) nit r at e coat ing on silver.

(c) an elast ic solid.

(b) sulphide coat ing on silver.

(d) a vitr ified liquid.

(c) chlor ide coat ing on silver. (d) oxide coat ing on silver. 3. When concent r at ed H 2SO4 spilt s on t he sur face, it should be immediat ely cleaned : (a) wit h a piece of clot h. (b) by adding cold wat er. (c) by adding solid Na2CO3. (d) by adding solid BaCl 2. 4. A bee-sitting leaves an acid which causes pain and ir r it at ion. The inject ed acid is :

10. Solutions in test tubes containing H 2O and aqueous NaOH can be differ ent iat ed wit h t he help of : (a) r ed lit mus.

(b) blue lit mus

(c) Na2CO3

(d) HCl (aqueous)

11. H uman st omach pr oduces acid ‘X' which helps in digest ion of food. Acid ‘X' is : (a) acet ic acid.

(b) met hanoic acid.

(c) hydr ochlor ic acid.

(d) cit r ic acid.

12. Whi ch one among t he fol l owi ng i s used as a moder at or in nuclear r eact or s?

(a) acet ic acid.

(a) Ozone

(b) H eavy hydr ogen

(b) sulphur ic acid.

(c) H eavy wat er

(d) H ydr ogen per oxide

(c) cit r ic acid. (d) met hanoic acid. 5. I r on nails ar e dipped int o blue copper sulphat e solut ion. Aft er some t ime ir on nails ar e :

13. Which one of t he following cont ains maximum per cent age of nit r ogen by mass? (a) Ur ea (b) Ammonium cyanide

(a) dissolved and blue colour is dischar ged.

(c) Ammonium car bonat e

(b) dissolved but blue colour is not dischar ged.

(d) Ammonium nit r at e

(c) not dissolved and blue colour is not discharged. (d) not dissolved but blue colour is dischar ged. 6. A st udent by chance mixed acet one wit h alcohol. Thi s mi xt ur e of acet one and al cohol can be separ at ed by : (a) filt r at ion.

14. Oxygen and ozone ar e (a) Allotr opes

(b) isomer s

(c) isot opes

(d) isobar s

15. When applied t o t he affect ed ar ea, which one of t he following will r elieve t he pain due t o ant -bit e or bee-st ing?

(b) separ at ing funnel.

(a) L emon juice

(b) Vinegar

(c) fr act ional cr ystallization.

(c) Baking soda

(d) Caustic soda

(d) fr act ional dist illation. 7. Which one among t he following met hods is not effect ive in r emoving ar senic fr om cont aminat ed gr ound wat er ? (a) Boiling (b) Rever se osmosis (c) I on exchange (d) Coagulation-adsorption

LEVEL-1 1. H adr ons and Bar yons ar e (a) I ndust r ial chemicals (b) Types of subat omi c par t i cl es (c) Alkalies (d) Cyclot r ons [RRB JE 2014 GREEN SH I FT ]

2.18 Chemistry

2. Wh i ch of t h e f ol l owi n g i s a h et er ogen eou s mixt ur e? (a) Br ass (b) Sugar sol ut ion in wat er (c) Air (d) Milk

11. Major contributing activity towards Global Warming by Greenhouse gases (a) Agriculture (b) Deforestation (c) Energy (d) lndustry

[RRB JE 2014 GREEN SH I FT ]

3. A class of compounds which ar e used as fragr ances when molecular wei ght i s low and ar e nat ur al ly occur r i ng fat s when molecular wei ght i s high in t he ser ies, is call ed (a) amino aci ds (b) ar omat ic compounds (c) est er s (d) or ganic aci ds

[RRB JE 2014 YEL L OW SH I FT ]

12. Electrostatic Precipitators are devices for (a) Particulate Emission Control (b) Water Pollution Control (c) Noise Pollution Control (d) Energy Pollution Control

[RRB JE 2014 GREEN SH I FT ]

[RRB JE 2014 YEL L OW SH I FT ]

4. Disinfection of drinking water is done to remove:

13. Biochemical Oxygen Demand (BOD) is a measure of

(a) Odour

(b) Bacterias

(c) Turbidity

(d) Colour [RRB JE 2014 RED SH I FT ]

5. Global warming is caused by : (a) N2

(b) CO2

(c) Ozone

(d) None of these

(a) Oxygen utilized during oxidation of organic matters (b) Suspended particles in water (c) Suspended particles in air (d) Noise level in air [RRB JE 2014 YEL L OW SH I FT ]

[RRB JE 2014 RED SH I FT ]

6. What is the General formula of Alkanes ? (a) CnH2n+2

(b) CnH2n

(c) CnH2n – 2

(d) CnH2n + 4 [RRB JE 2014 RED SH I FT ]

7. The pollutant responsible for ozone holes is : (a) CO2

(b) CO

(c) SO2

(d) CFC [RRB JE 2014 RED SH I FT ]

8. Ammonia is prepared commercially by the : (a) Oswald process

14. Biodegradable pollutants are (a) quickly degraded by natural means (b) can not be degraded (c) can be degraded by burning only (d) disposed in flowing water only [RRB JE 2014 YEL L OW SH I FT ]

15. The state in which molecular attractions are very strong is (a) Solid (c) Gas

(b) Liquid (d) Vapour [RRB JE 2014 YEL L OW SH I FT ]

(b) Hall process

LEVEL-2

(c) Contact process [RRB JE 2014 RED SH I FT ]

1. What is t he common pr oper t y bet ween L iAlH 4, Sodium amalgam and NaBH 4?

9. The elements which have same mass number but different atomic numbers are know as :

(a) They ar e used in r emoving slag fr om molt en met als

(d) Haber process

(a) Isotones

(b) Isobars

(b) They ar e used in manufact ur ing est er s

(c) Isotopes

(d) Halogens

(c) They ar e r educing agent s

[RRB JE 2014 RED SH I FT ]

10. Which one of the following is not a Noble Gas ? (a) Helium

(b) Bromine

(c) Argon

(d) Neon [RRB JE 2014 RED SH I FT ]

(d) They ar e coat ed on welding elect r odes [RRB SSE 2014 GREEN SH I FT]

2. Soaps ar e manufact ur ed by: (a) React ion of alkalies wit h glycer ol (b) React ion of fat s wit h soluble hydr oxides

Chemistry 2.19

(c) React ion of calcium and magnesium ions wit h dilute sulphur ic acid

(c) su m of t h e n u m ber of pr ot on s an d t h e neut r ons in t he nucleus of an at om

(d) React ion of dodecyl benzene wit h H 2SO4 and t hen NaOH

(d) number of pr ot ons or elect r ons in one gr am of Sodium

[RRB SSE 2014 GREEN SH I FT]

[RRB SSE 2014 YELLOW SH I FT]

3. Chemical bonding which r esult s in for mat ion of molecules fr om at oms is basically-

10. I sot opes of t he same element have (a) Same number of neut r ons

(a) Nuclear for ce

(b) Same at omic mass

(b) Shor t r ange for ces

(c) Same number of pr ot ons

(c) Elect r ost at ic for ce

(d) Differ ent at omic number

(d) Gr avitat ional for ce

[RRB SSE 2014 YELLOW SH I FT]

[RRB SSE 2014 GREEN SH I FT]

4. Glycer ol can be r epr esent ed by chemical for mula:

11. I n a r eaction between Zinc and I odine. Zinc I odide is for med. What is being oxidised ?

(a) C2HSO2

(b) C3H 7OH

(a) Zinc ions

(b) I odide ions

(c) C3H 5OH

(d) C3H 8O3

(c) Zinc At om

(d) I odine

[RRB SSE 2014 GREEN SH I FT]

[RRB SSE 2014 YELLOW SH I FT]

5. The most ideal di si nfect ant used for dr i nki ng wat er is :

12. Whi ch of t he fol l owi ng hal ogens i s t he best oxidising agent ?

(a) Alum

(b) Chlor ine

(a) F 2

(b) Cl 2

(c) L i me

(d) Nit r ogen

(c) Br 2

(d) l 2

[RRB SSE 2014 GREEN SH I FT]

6. Which one of t he following is gener ally added t o Table Salt t o make it flow fr eely in r ainy season ?

[RRB SSE 2014 YELLOW SH I FT]

13. Nit r ogen is used t o fill elect r ic bulbs because it (a) is light er t han air

(a) Ca3(PO4)2

(b) Na3PO4

(b) makes t he bulb t o give mor e light

(c) KCI

(d) KI

(c) does not suppor t combust ion

[RRB SSE 2014 RED SH I FT]

(d) is non-t oxic

7. Valence elect r ons in t he element A ar e 3 and t hat i n el ement B ar e 6. M ost pr obabl e compound for med fr om A and B is :

[RRB SSE 2014 YELLOW SH I FT]

14. Fr ot h float at ion pr ocess for t he concent r at ion of Or es is an illustr ation of t he pr act ical applicat ion of

(a) A2B

(b) AB2

(c) A2B3

(d) A3B2

(a) Adsor ption

[RRB SSE 2014 RED SH I FT]

(b) Absor ption

8. At oms of t he element s bel ongi ng t o t he same gr oup of per iodic t able will have : (a) Same number of pr ot ons (b) Same number of neut r ons (c) Same number of elect r ons (d) Same number of elect r ons in t he valence shell [RRB SSE 2014 RED SH I FT]

9. Avogadr o's number, N A means (a) number of pr ot ons in nucleus of an at om (b) number of at oms i n one gr am at om of an element

(c) Coagulation (d) Sediment ation [RRB SSE 2014 YELLOW SH I FT]

15. The pr esence of ni t r ogen i n t he pr oduct s of combust ion ensur es t hat (a) Complet e combust ion of fuel t akes place (b) I ncomplet e combust ion of fuel t akes place (c) dr y pr oduct s of combust ion ar e analysed (d) air is used for t he combust ion [RRB SSE 2014 YELLOW SH I FT]

2.20 Chemistry

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (a)

2. (b)

3. (d)

4. (d)

5. (d)

11. (c)

12. (c)

13. (b)

14. (a)

15. (c)

6. (d)

7. (a)

8. (b)

9. (c)

10. (a)

7. (d)

8. (d)

9. (b)

10. (b)

7. (b)

8. (d)

9. (b)

10. (c)

LEVEL-1 1. (b)

2. (d)

3. (c)

4. (b)

5. (b)

11. (c)

12. (a)

13. (a)

14. (a)

15. (a)

6. (a)

LEVEL-2 1. (a)

2. (b)

3. (c)

4. (d)

5. (b)

11. (a)

12. (a)

13. (c)

14. (a)

15. (d)

6. (a)

EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S 1. The mat er ial should have high specific value and low melt ing point which is used for elect r ic fuse is an alloy of t in and lead. 2. Silver conver t s int o silver sulphide in pr esence of air and H 2S, which is black in colour

2Ag  H2S   Ag 2S  2H+ Black

3. When concent r ated H 2SO4 split s on t he sur face it should be immediat ely cleaned by adding solid Bacl 2 (Bar ium Chlor ide). 4. A bee-sting leaves methanoic acid (HCOOH) which causes pain and ir r it at ion. 5. I r on i s mor e r eact i ve t han copper, so r epl ace copper fr om copper sulphat e solut ion.

Fe  CuSO4   FeSO4  blue

6. 9. 10.

11. 12.

green

Cu

reddish brown

Copper submitted on iron nails by which the colour of solut ion become light . Acet one and alcohol have differ ent boiling point s so it can be separ at ed by fr act ional dist illat ion. Glass is an elast ic solid. Red lit mus is uneffect ed by wat er (H 2O) because it is neut r al in nat ur e while NaOH is basic in nat ur e so r ed lit mus t ur ns int o blue colour. H uman st omach pr oduces H ydr ochl or i c (H cl ) which helps in digest ion of food. H eavy wat er i s used as moder at or i n nuclear r eact or s which r educes t he velocit y of neut r ons.

13. I n Ur ea (NH 2 CO NH 2)

28  100  46.6% 60 I n Ammonium Cyanide (NH 4CN) Per cent age of Nit r ogen =

28  100  63.6% 44 I n Ammonium Car bonat e (NH 4)2 CO3 Per cent age of Nit r ogen =

28  100  29.2% 96 I n Ammonium nit r at e (NH 4)2 NO3 Per cent age of Nit r ogen =

28  100  42.8% 98 So, in ammonium cyanide per centage of nit r ogen is maximum 14. Oxygen (O2) and ozone (O3) ar e allotr opic for ms of oxygen. 15. I n ant bit e and bee st ing for mic acid (H COOH ) is pr esen t , wh i ch i s r espon si bl e f or pai n an d ir r it at ion. For mic acid is a weak acid it can be neut r al so bak i ng soda (sodi um bi -car bonat eNa2CO3) is used in r elieve in pain due t o ant i-bit e or bee st ing. Per cent age of Nit r ogen =

LEVEL-1 1. Hadrons and Baryons are types of subatomic particles. Baryons are heavy subatomic particles that are made up of three quarks. 2. Milk is an example of a heterogeneous mixture. Mixtures can be separated into two (or more)

Chemistry 2.21

individual substances by physical means. Our glass of ice water is a mixture because we can easily separate the ice from the liquid water by filtration. 3. A class of compounds which are used as fragrances when molecular weight is low and are naturally occurring fats when molecular weight is high in the series, is called esters. 4. Disinfection of drinking water is done to remove Bacteria. Water disinfection means the removal, deactivation or killing of pathogenic microorganisms. Microorganisms are destroyed or deactivated, resulting in termination of growth and reproduction. When microorganisms are not removed from drinking water, drinking water usage will cause people to fall ill. 5. CO2 causes green house effect trading to global warming. 6. The alkanes comprise a series of compounds that are composed of carbon and hydrogen atoms with single covalent bonds. This group of compounds comprises a homologous series with a general molecular formula of C n H 2 n+2.

12. Electrostatic Precipitators are devices for particulate emission control. 13. Biochemical Oxygen Demand (BOD, also called Biological Oxygen Demand) is the amount of dissolved oxygen needed (i.e. demanded) by aerobic biological organisms to break down organic material present in a given water sample at certain temperature over a specific time period. The BOD value is most commonly expressed in milligrams of oxygen consumed per litre of sample during 5 days of incubation at 20 °C. Biochemical Oxygen Demand (BOD) is a measure of Oxygen utilized during oxidation of organic matters. 14. Biodegradable pollutants are quickly degraded by natural means. Biodegradable pollutants: Such pollutants are quickly degraded by microbes (bacteria and fungi) in nature e.g. sewage, ... Examples of such pollutants are: DDT, mercury, lead, arsenic, some pesticides, radioactive substances, glass, plastic, aluminium pieces, etc. 15. Solid has the highest molecular attractions hence is dense and compact.

LEVEL-2 1. a

In organic chemistry we normally learn about two important reducing reagents, sodium borohydride (NaBH 4) and lithium aluminum hydride (LiAlH4 or LAH). We learn that NaBH4 is a “weak reducing agent” and can only take aldehydes and ketones to alcohols easily. NaBH4 can handle esters, but it is very slow at converting them and thus not preferable.

2. b

Fats and oils are composed of triglycerides; three molecules of fatty acids attach to a single molecule of glycerol. The alkaline solution, which is often called lye (although the term “lye soap” refers almost exclusively to soaps made with sodium hydroxide), brings about a chemical reaction known as saponification.

3. c

A chemical bond is a lasting attraction between atoms, ions or molecules that enables the formation of chemical compounds. The bond may result from the electrostatic force of attraction between oppositely charged ions as in ionic bonds or through the sharing of electrons as in covalent bonds.

7. Chlorofluorocarbons (CFCs) and other halogenated ozone depleting substances (ODS) are mainly responsible for man-made chemical ozone depletion. The total amount of effective halogens (chlorine and bromine) in the stratosphere can be calculated and are known as the equivalent effective stratospheric chlorine (EESC). 8. The Haber Process combines nitrogen from the air with hydrogen derived mainly from natural gas (methane) into ammonia. The reaction is reversible and the production of ammonia is exothermic. The catalyst is actually slightly more complicated than pure iron. 9. Atoms of chemical elements having same atomic mass but a different atomic number are called Isobars. The sum of the number of protons and neutrons together form the atomic mass. 10. Among the given options, Bromine is not a Noble Gas.The six noble gases that occur naturally are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and the radioactive radon (Rn). 11. A greenhouse gas is a gas that absorbs and emits radiant energy within the thermal infrared range. Increasing greenhouse gas emissions cause the greenhouse effect.

2.22 Chemistry

4. d

The glycerol chemical formula is C3H8O3 and its extended formula is CH2OH-CHOH-CH2OH. The IUPAC name for glycerol is 1, 2, 3- Trihydroxypropane or 1, 2, 3- Propanetriol.

5. b

The most ideal disinfectant used for drinking water is Chlorine. Chlorine is one of the most commonly used disinfectants for water disinfection.

6. a

Ca3(PO4)2 is generally added to Table Salt to make it flow freely in rainy season.

7. b

Valence electrons in the element A are 3 and that in element B are 6. Most probable compound formed from A and B is A2B3.

8. d

Atoms of the elements belonging to the same group of periodic table will have Same number of electrons in the valence shell.

9. b

In chemistry and physics, the Avogadro constant, named after scientist Amedeo Avogadro, is the number of constituent particles, usually atoms or molecules, that are contained in the amount of substance given by one mole.

10. c

The atoms of a chemical element can exist in different types. These are called isotopes. They have the same number of protons (and electrons), but different numbers of neutrons. Different isotopes of the same element have different masses.

11. a

Zinc powder is added to a solution of iodine in ethanol. An exothermic redox reaction occurs, forming zinc iodide, which can be obtained by evaporating the solvent. In reaction between Zinc and Iodine. Zinc Iodide is formed. Zinc atom is being oxidised.

12. a

Fluorine is such a powerful oxidising agent that you can’t reasonably do solution reactions with it.

13. c

Filling a bulb with an inert gas such as argon or nitrogen slows down the evaporation of the tungsten filament compared to operating it in a vacuum. This allows for greater temperatures and therefore greater efficacy with less reduction in filament life.

14. a

Froth floatation process for the concentration of Ores is an illustration of the practical application of Adsorption.

15. d

The presence of nitrogen in the products of combustion ensures that air is used for the combustion.

3

Biology

CHAPTER Classificat ion of Plant s PLANT KINGDOM Cryptogamae (Plants without seeds)

Thallophyta

Algae

Bryophyta

 Phanerogamae (Seed bearing plants)

Pteridophyta

Gymnospermae

Angiospermae

These have pr imit ive vascular syst em.

2. Phaner ogamae I t includes all flower ing plant s which bear seeds. These ar e also called sper mat ophyt es. Classificat ion of Phaner ogames

Fungi

( i ) Gymnosper ms. Hepaticae

Psilopsida

Anthocerotae

Lycopsida

Musci

Sphenopsida

 Pteropsida

 Cycadopsida

Coniferopsida

Gnetopsida

Monocotyledons

Dicotyledons



Depending on t he pr esence or absence of flower s and seeds, ent ir e plant kingdom is divided int o following.

1. Cr ypt ogamae I t includes all non-flower ing plant s such as algae, fungi, lichens, mosses and fer n. Classificat ion of Cr ypt ogamae ( i ) T hallophyt a.  Nut r it i on i s aut ot r ophi c i n algae, wher eas fungi exhibit het er ot r ophic nut r it ion. Algae usually sust ain it self in a var iet y of habit at s such as wet land as well as on ot her plant s and even ani mal s. Some gr ow i n mar i ne wat er and ar e called seaweeds. ( ii ) Br yophyt a.  This gr oup der ives it s name fr om t he mosses whi ch gr ow on t he r ock s, wal l s and t r ee t r unks in moist and shady places.  These lack r oot s, flower s and seeds.  M osses also r et ain moist ur e like the sponges.  M em ber s of t h i s gr ou p r epr odu ce bot h v eget at i v el y t h r ou gh f r agm en t at i on , gemmae and t uber s, as wel l as by sexual met hods. ( iii ) Pt er idophyt a. 

These occur in humid and t r opical climat es and usually gr ow on soil, r ocks, in ponds and as epiphyt es on ot her plant s.

These ar e r epr esent ed by cycas, pines and cedar. These have t he naked (uncover ed) seeds, i.e. seeds ar e not enclosed in a fr uit . These ar e r epr esent ed by coni fer s, which gr ow in cool climat e of hills, sometimes using melt ing snow as a sour ce of wat er. H owever, some like cycads and member s of Gnet ales t hr ive in war m dr y climat e.

Some common gymnosper ms ar e : Abi es, Cedr u s, Pi nu s and ot her t i m ber yielding species ( ii ) Angi osper ms.  These have cover ed seeds, i.e. they ar e always enclosed in an ovar y or fr uit .  These r ange i n si ze fr om mi nut e fl oat i ng duckweeds to giant eucalyptus and silk cotton t r ees and include plant s of gr eat var iet y and for m-cact i, wat er lillies, sunflower s, or chids, pit cher plant s, I ndian pipe et c.  These ar e seed-bear ing plant s well adapt ed to t he ter r estr ial life and they occur in diver se habit at s like cold t undr a t o hot t r opical and even deser t ar eas. They also t hr ive well in aquat ic habit at s.  These plants repr esented by trees, shrubs and her bs ar e eit her monocot yledons or dicot yledons. They have a body well differ ent iat ed int o r oot , st em and leaves.  These ar e mainly classified int o t wo main classes : (a) Dicot yledons (b) M onocotyledons 

3.2

Biology

Classificat ion of Animals Animal Kingdom

Chrodata

Metazoa (multi cellular)

Protozoa (unicellular/acellular)

Eumetazoa

Parazoa / parifera Diploblastic (having 2 germinal layers) e.g. Porifera and coelenterata

Acoelomata

Vertibrata / Craninta

Protochordata / Acraniata

Hemichordata

urochordata cephalochordata

Agnatha (Jawless)

Super class

Triploblastic (having 3 germinal layers) Cyclostomata

Ostacodermi (Extinct)

Pseudoeoelomata





Tetrapoda Classes

Eucoelomata Osteichthyes (Bony fishes)

Desstrostomia or, Enterocoelom (e.g. Echinodermata and chordata) Amphibia



Pisces Classes

Placodermi Chondrichthyes (Extinct)(cartilaginous fishes)

Protostomia or, Schizocoelom (e.g. Annelida, Arthropoda and Mollusca)

Gnathostomata (with true jaw)

Based on t he pr esence or absence of not ochor d at some st age in t heir life, animals ar e categorised int o t wo major gr oups : (1) Non-chor dat es ; (2) Chor dat es Th e ani mal k i n gdom i s di vi ded i nt o 35 Phyl a (singular : Phylum) of which 11 ar e consider ed as major Phyla. Ani mal K i ngdom (Ani mal i a) i ncl udes about 1.2 million known species.

  

   

S.N o. Char acter



Chor data

F un damental Char act er s

Aves

Mammalia

Phylum – Pr ot ozoa (Pr ot ista)

D i f f er en t i at i on bet w een N on - ch or da t a a n d Chor dat a N on Chor data

Reptilia



Wor d meaning is fir st animals. These protozoans ar e micr oscopic called animalcules. Common feat ures are : Encystment and Pher omer an of r egener at ion H umidit y is needed in life span. Exo– skelet on is absent except Pohyst omela. Endo– skelet on is absent . L ocomot ar y or ganellae ar e pseudopodia / flagella / cillia / absent . Nut r it ion may be holozoic / holophit ic / par asit ic. Digest ion is int r acellular. Respir at ion occur s by plasmalemma. Excr et ion t akes place by plasmalemma. Cont r act ile vacuole is pr esent for osmor egulat ion (however it is absent in parasitic pr otozoa and mar ine pr ot ozoa). Cir culat ion by cyclosis. Asexual r epr oduct ion by binar y fission / mult iple fission / spor ulat ion . H ologamy t ype of sexual r epr oduct ion is pr esent .

1

Notochor d or (chor da dor sali s)

Absent in ever y st age of li fe cycle

Pr esent i n any st age of l ife cycl e

2

Gil l Slit s

Absent in ever y st at e of l ife cycle

Pr esent i n any st at e of l ife cycl e

3

N er vous syst em

Vent r al, Double and Solid

N er ve – cor d Dor sal, Singl e and H oll ow ner ve – cor d

4

Post and t ai l

Absent

Pr esent



5

M outh

Gener ally in fi r st segment

Aft er fir st segment in ant er ior r egi on



6

Anus

Gener ally in l ast segment

Befor e l ast segment poster ior r egion



7

L i ver

Absent

Pr esent

8

H ear t

Dor sal

Vent r al

Phylum – Porifer a

9

R.B.C.

Absent

Pr esent

10

Ci r culat or y system

Open / closed / absent

Closed



11

Posit i on of ali ment ar y canal

Dor sal to nerve cor d

Vent r al t o ner ve cor d



12

Symmetr y

All t ype

Bil at er al

13

Gr ade

All t ypes

Or gan syst em

14

Body wall

Di pl obl ast ic / Tr ipl oblast ic

Tr i pl obl ast i c

15

Coelom

All t ypes

Ent er ocoeli c

16

Repr oduct i on

M ainly asexual

Sexual

17

L ar vae

Gener ally pr esent Gener all y absent

18

Regener at i on

M or e

L ess

19

Temper at ur e

Cold bl ooded

Col d blooded / war m blooded

20

Tot al animals

95-97%

3-5%

  







Animals ar e pores bear ing. I ncur r ent por es ar e ost ia, mor e in number, small in size, cont r olled by por ocyt e cells and equal t o mout h. Excur r ent por es ar e osculum, one or few in number, large in size, controlled by myocytes and equal to anus. Cells ar e loosely ar r anged. Out er wall is pinacoder m made up of pinacocyt es (pinacoderm cells) I nner wall is choanoder m made up of choanocyt es or collar ed flagellat ed cells. Bet ween t hese t wo layer s, mesenchyma (mesohyl layer ) is pr esent in which amoebocyte cells ar e ther e. Choanocyt e cells ar e r esponsible t o t r ansfer sper ms t o ovum.

Biology                  

A m oebocy t es ar e com i n g f r om bot h l ay er s (pinacoder m and choamoder ms). Cent r al cavit y is spongocoel or par agast r ic cavit y. Endoskeleton is present (secr eted by scleroblast cells). Animals ar e sessile or sedent ar y or not locomot e. Digest ion is int r acellular (in choanocyt e cells) Respi r at i on and excr et i on t hr ough gener al body sur face by difussion. Animals ar e mainly bisexual. Fer t ilizat ion is int er nal. I n body spicules pr esent , may be mega scler es or micr o scler es. Spicules ar e for med by act inoblast cells. Spicules ar e of monoaxon t o polyaxon t ype (diaxon is absent ). Commonly called sponges. M ost ly mar ine. Animals of t hese phylum ar e maximum act ive in animal kingdom. Nut r it ion is holozoic. Chr omat ocyt es cells pr esent . Food is st or ed in t hesocyt e cells. Olynt hus is ancient sponge, pr esent in life cycle of most of t he sponges.

Phyl um– Coel ent er at a Centr al cavity is coelenteron or gastrovascular cavity, which is helpful in nut r it ion and cir culat ion.  H et scheck named it Cnidar ia.  St inging cells or Nematoblast or Cnidocyt es pr esent for offence and defence.  Out er layer is epider mis I nner layer is gast r oder mis Bet ween t hem mesoglea is pr esent .  M out h is meant for ingest ion and egest ion.  Digest ion is int er – cellular and int r a– cellular bot h.  Respir at ion t hr ough gener al body sur face.  Ner vous system is diffused type, ner ves brain absent.  Asexual r epr oduct ion by budding (gemmat ion)  M ostly bisexual.  Fer t ilizat ion is ext er nal.  Development is indir ect .  Planula lar va is common.  M et agen esi s i s pr esen ce of gen er at i on , i .e. polymor phism is pr esent . Classificat ion These ar e mainly of t wo for ms : 1. Pol yp. I t is long, cylindr ical body, sedent ar y and colonial, hollow, wit hout velum t hin. M esoglea wit hout st at ocyst and as asexual for m. 2. M edusa. I t is umbr ella shaped, fr ee swimming and solit ar y velum. 

3.3

Reef Cor al r eefs (cor al + algae) ar e of t hr ee t ypes : ( i ) Fringing reefs : Closest to sea shor e and wit hout cagor m. ( ii ) Barrier reefs : 1/2 t o 10 miles a way fr om sea shor e and having lagoon. ( iii ) Atoll : I t is cir cular cor al r eef having lagoon. I t is called cor al island. N ote : The lar gest cor al r eef is The Gr eat Bar rier Reef of Aust r alia (about 1920 kms.)

Phylum – Plat yhelmint hes         

            

Body is dor so vent r ally flat t ened. Commonly called flat wor ms. Epider mis is syncit ial. M uscles pr esent in t he for m of bundle. Digest ion is int er cellular. I ncomplet e aliment ar y canal is pr esent or absent . Gener ally par asit es. M ainly anaer obic r espir at ion occur. Excr et or y or gan ar e Flame cells or pr ot onephr idia or solenocyt es. Ner vous syst em is ladder shape. Ganglions ar e for med for t he fir st t ime. Bisexual. Sex duct s pr esent . Sucker s or hooks or bot h pr esent . Power of fer t ilit y is mor e. L ife cycle is complicat ed. Digenet ic. Vitelline glands give nour ishment to zygotes and also helpful t o give out zygot es out of ut er us. Fer t ilizat ion is int er nal. Development indir ect . M any lar vae ar e pr esent . Polyembr yony pr esent , i.e. fr om one zygot e many embr yos ar e for med.

Phylum – Aschelmint hes This phylum is similar toplatyhelminthesin charactersexcept  Commonly called r ound wor m / t r ue wor m  Pseudocoelome is pr esent .  Unisexual in nat ur e  M onogenet ic / Digenet ic  Excr et or y or gans ar e coelomoduct s.  I n ner vous syst em r ing is for med.  M aximum adapted for par asit ism.  Cleavage is spir al and det er minat e.  Eut yly is comont at ur e. i.e. number of cells ar e fixed and show incr ease in size only (auxet ic gr owt h).  Complet e and st r aight aliment ar y canal is pr esent .

3.4

Biology

Phylum – Annelida                

Body is met amer ically segment ed. Segment at ion is ext er nal by annuli. Segment at ion is int er nal by coelosept a. Tr ue coelom is pr esent Complet e aliment ar y canalis pr esent L ocomot ar y or gan ar e set ae Sacker s ar e pr esent in leech. Digest ion is int er cellular. Respir at ion by body wall / gills / par apodia. Excr et or y or gans ar e nephr edia. Cir culat or y syst em is closed t ype. Ner ve r ing pr esent . Animals ar e bisexual / unisexual. Fer t ilizat ion is ext er nal. Development is indir ect . Haemoglobin or erythroqruorine is desolved in plasma.

Phylum – Ar thr opoda               

Fir st lar gest phylum r egar ding populat ion (consist s of mor e t han 80% of animal kingdom) Joint ed legs pr esent . H aemocoel is pr esent . Cir culat or y syst em is open t ype. Exoskelet on is made up of chit in. Endoskeelt on is pr esent . Respir at ion by t r achea / gills / book lungs. Excr et ion by malpighian t ubules / gr een glands or ant ennar y glands / coxal gland. Br ain pr esent . Unisexual in nat ur e. Fer t ilizat ion is int er nal. I n most of ani mal s (except i nsect ) haemocyni ne disolved in plasma. Cillia t ot ally absent . Ecdysis / moult ing / exuviae is common. Compound eyes ar e pr esent her e.

Phylum – M ollusca            

Soft bodied animals. Body i s cover ed by mant l e, whi ch i s hel pful i n r espir at ion and secr et ion of shell. I n shell conchalin pr ot ein pr esent . Second lar gest gr oup. Torsion and detorsion may pr esent i.e. twisting of viscera ar ound axis, so animals become asymmetrical. Cir culat or y syst em open. Coelom is maximum r educed. Exo skelet on is made up of calcium car bonat e. M uscular foot pr esent . I nt er cellular digest ion. Respir at ion by ct enidia / pulmonar y sac. Excr et ion by coxal gland / renel gland / keber 's gland / met anephr idia / coelomoduct / or gan of bajonus.

            

Br ain pr esent . Unisexual. Fer t ilizat ion ext er nal. Development indir ect . Glochidium / velliger / t r ocophor e lar vae ar e pr esent. Hepat opancr eas pr esent. Body is differ ent iat ed int o head, viscer a and foot . Balancing or gans ar e st at ocyst . Osphar idium is chemor ecept or. K idney is sac like. I n hear t 1– 2 aur icles and 1 vent r icle pr esent . Cleavage is spir al and det er minat e. Redula ar e t eet h like st r uct ur e, may pr esent .

Phylum – E chinoder mat a        

      

Animals ar e exclusively mar ine. Skin is spiny. Commonly called swimming pebbles. Exo sk el et on and endo sk el et on ar e made up of calcar ius ossicles. L ocomot ion by t ube feet . Respir at ion by tube feet / der mal pappulae/bursae. Excr et ion by wat er vascular syst em. Wat er vascular syst em consist s of (i ) Teidmann's bodies (ii ) S– shaped stone canals. No br ain, Unisexual. Fer t ilizat ion is ext er nal. Development indir ect . Lar vae pr esent ar e Pinnaria / bipinaria / oricularia / echinopluteus M out h pr esent on or al sur face. Anus pr esent on abor al sur face. Cover ing of body is echinocar dium.

Algae They ar e simple, t hallose, aut ot r ophic non– vascular pl ant s havi ng uni cel led sex or gans and no embr yo for mat ion. Algae occur in following var iet ies of habit at s: 1. F r esh wat er for ms They occur in r iver s, ponds, pools, lakes and ditches. Some for ms occur as passively float ing and dr ift ing flor a (phytoplanktons), e.g. diat oms 2. M arine for ms M ost of t he member s of br own algae, r ed al gae, some gr een and blue– gr een algae occur in sea. 3. Ter r est r ial for ms Sever al member s of gr een and bl ue– gr een al gae and a few ot her s occur on damp soi l . Osci l lat or i a and N ost oc occur on al k al i ne and cal car eous soi l .

Biology

3.5

4. U nusual or Specialised habit at s Algae Repr oduct ion M et hods involved in r epr oduct ion ar e 1. Vegetative reproduction 2. Asexual reproduction 3. Sexual r epr oduct ion.

E conomic I mpor t ance  Fungi like Agar icus, Clavat ia, M or chella ar e used as human food.  M or chella is commonly called sponge mushr oom.  Tr ufflels is edible fungus and pr oduces it s fr uit ing body embeded in soil.

F un gi

L i ch en s

The br anch dealing wit h t he st udy of fungi is called mycology.  Fungi do not have chlor ophyll and t her efor e t hey do not pr epar e t heir own food mat er ial. They have t o obt ain it fr om ext er nal sour ces: eit her from dead or ganic matter (sapr ophytes) or fr om other living plant s and animals (par asit es). Some fungi have symbiot ic r elat ionship wit h ot her plants.  The t hal l us i s made up of t hi n, l ong, br anched filaments called hyphae. The mass of hyphae is called mycelium . I n Basidiomycet es, t he sept a ar e br oad in t he cent r e sur r ounded by a double membr ane (par ent hesome) on each side. Such a bar r el– shaped sept um is called dolipor e sept um .  Except slime molds all ot her fungi have cell walls. The cel l wal l i s made up of cyt i n. I n t he cl ass Oomycet es, t he cell wall is of cellulose.  The fungi lack chlor ophylls but have car otenoids.  The hyphae may be septate or unseptate. Unseptate hyphae have nucl ei scat t er ed in cyt opl asm. Thi s condi t i on i s cal l ed coenocyt ic. H owever, i n such hyphae, sept a ar e for med at the t ime of for mat ion of r epr oduct ive or gans or in older por t ions of hyphae. N ut r it ion On t he basis of nut r it ion, fungi ar e divided int o t wo categor ies : (i) Parasites (ii) Saprophytes Some fungi live in associat ion wit h ot her gr oups of pl an t s f or m u t u al ben ef i t . T h ese ar e cal l ed symbiont s. e.g. lichen, mycor r hiza A plant body which consist s of a single cell and is conver t ed complet ely int o a r epr oduct ive st r uct ur e is called holocar pic e.g. Synchyt r ium . But if only a par t of t he t hallus is used up in t he for mat ion of r epr oduct ive st r uct ur e, it is called as eucarpic. Repr oduct i on I n fungi r epr oduct ion is of t hr ee t ypes : (i ) Veget at ive (ii ) Asexual (iii ) Sexual The r eser ve food mater ial in fungi is glycogen which is soluble polysacchar ide. Glycogen is also found in animal cells and hence called animal st ar ch. 

  











I t is a composit e t halloid st r uct ur e, made up of an alga and a fungus. Algal component of a lichen is called phycobiont , and fungal component is called mycobiont . T h e al gal com pon en t i s u su al l y a m em ber of cyan oph y ceae or ch l or oph yceae. T h e f u n gal component i s usual l y a member of ascomycet es, r ar ely of basidiomycet es. The algal and fungal par t ner s ar e benefit ed by t his associat ion (symbiosis or mut ualism). The fungus der i ves n ut r i t i on fr om t h e al ga, wh i l e al ga i s pr ot ect ed fr om desiccat ion by t he fungus. L ichens can t oler at e ext r emes of climat e and ar e found ever ywher e r anging fr om hot deser t s t o cool mountains. A sex u al r epr odu ct i on (i n cl u di n g v eget at i v e r epr oduct ion) t akes place by fr agment ation, sor edia, isidia and conidia. Sexual reproduction occurs only in the fungal partner. The female r epr oductive or gan is a car pogonium and male r epr oduct ive or gan is a sper mogonium. The fr uiting bodies ar e eit her apothecia or per ithecia containing asci. Each ascus develops eight ascospor es.

Gymnasper m  

        



Fr uit -less seed plant s. Char act er ized by pr esence of naked seeds pr oduced in cones. L eaves ar e dimor phic (scaly leaves and foliage leaves) Pollinat ion is anemophilous (by wind) Embr yo development is mer oblast ic. Pol y em br y on y i s ch ar act er i st i c f eat u r e of gymnosper ms. Endosper m is haploid, gamet ophyt ic t issue and is for med befor e fer t ilizat ion fr om megaspor e. Gymnosper ms and angiosper ms r esemble in having siphonogamy. Wood of Cycas is manoxylic and polyxylic while t hat of Pinus is Pycnoxylic. Cycas is also called sagoplam. I n r achis of cycas, vascular bundles ar e ar r anged in t he for m of inver t ed omega. Tr ansfusion t issue meant for conduct ion of wat er is pr esent in leaflet s of cycas and leaves (needles) of Pinus. Gr owt h of male plant in Cycas is sympodial while of female plant is monopodial.

3.6

Biology

T he Cell All life begins ar e a single cell. A number of or ganism ar e made of single cell, t hey ar e called uni/acellular. e.g. Amoeba, Chlamydomonas, Acet abular ia, Yeast .  Cells ar e gr ouped into t issue, tissue into or gans, into or gan syst em, r esult in Division of L abour.  All t he cells int r act and co-oper at e wit h each ot her and r emain par t ially dependent on each ot her.  The cells having a common or igin and per for ming a similar but specific funct ion const it ut es a t issue e.g. muscle  Sever al t ypes of t i ssues j oi n t o for m an or gan e.g. liver, kidney, leaf, r oot s, et c.  Sever al or gans const it ut e an or gan syst em e.g. digest ive syst em, excr et or y syst em, et c.  Cells var y widely in size and shape. I f bact er ia ar e among the or ganisms wit h smallest cells, lar gest cell of all is t he yolk of an ost r ich egg. Pr ok ar yot ic cell  Pr okar yot ic cells nuclear mater ial i.e. DNA, RNA, pr ot ei ns, et c. i s not bound by a defi ni t e nucl ear membrane.  Cytoplasm lacks well defined cyt oplasmic or ganelles e.g. en dopl asm i c r et i cu l u m , gol gi com pl ex , mit ochondr ia, cent r ioles et c.  These cells ar e r epr esent ed by bact er ia, blue- gr een algae, mycoplasma or PPL O (pleur o pneumonia-like or ganism), spir ochet e and r icket t siae.  These cells ar e supposed t o be far smaller t han t he euk ar yot i c cel l s. H owever, t hey mul t i pl y mor e r apidly as compar ed t o most of t he eukar yot ic cells.  These cells have r emar kable amount of var iat ion in shape and size. E ukar yot ic cell  These cells have a nucleus which cont ains nuclear mat er ial enclosed by a double layer ed membr ane. All plant and animal cells fall under t his cat egor y.  Component s of Eukar yot ic cells : Cell wall (absent in animals cells and some pr otist s), plasma membr ane, cyt oplasm and or ganelles  I mpor t ant cyt oplasmic or ganelles: Mitochondria, plastids, ribosomes, endoplasmic reticulum, lysosomes, vacuoles, golgi complex, centrioles, etc. 

D iffer ences Bet ween Animal and Plant Cel l s S . C h a r a ct e r s N o.

E u k a r y ot i c C e l l C el l w al l p r esen t on l y i n p l an t cel l (B act er i a con si d er as p l an t s d u e t o p r esen ce of CW )

1

C ell W a ll

C el l w al l absen t i n an i m al cel l .

2

P l a st i d s

P l ast i d s or ch l oP l ast i d s p r esen t s r op l ast s absen t i n i n p l an t cel l s. an i m al cel l . (ex cep t F u n gi ) N o-ch l or op h y l l

3

C e n t r osom e C en t r osom e fou n d C en t r osom e abi n an i m al cel l s absen t i n pl an t (ex cep t L ow er cel l p l an t s)

4

V a cu ol e s

A bsen t bu t sm al l , Sap / T r u e v acu ot em p or ar y v acu o- l es ar e p r esen t i n l es m ay p r esen t s. p l an t cel l s.

5

C ell D i v i si on

Cyt ok i nesis by const r i ct ion for m at ion or fu r r ow i n g.

C el l di v i si on by cel l pl at e f or m or p h r agm op l ast

6

R e se r v e f ood ce l l s

R eser v e food i s gl y cogen i n an i m al pl an t cel l s.

M ai n l y st ar ch i s r eser v e food i n .

Cell Wall       

Pr ovides shape t o plant cell r igidit y t o cells. Funct ions as a bar r ier t o ent r y of pat hogens int o cell s. Pr ovides pr otection to protoplasm against mechanical injur y. Cont r ol ent r y of exit of mat er ials. Funct ions as apoplast which is per meable t o wat er and miner als dissolved in it . Plasmodesmat a pr oduce a pr ot oplasmic cont inuum bet ween adjacent cells. Bur st ing of cells on endomit osis.

Pr ot opl ast Plastid

Plasmalemma

Cell T heor y Five fundament al facts: 1. Body of living organisms consists of different types of cells and their products, i.e. cell is structural unit of life. 2. Single cell per for m all t he vit al act ivit ies, i.e. cells is funct ional unit of life. All act ivit ies of living or ganisms occur due t o t he met abolic act ivit ies of t he cell. 3. Cell is her edit ar y unit of life. 4. Cells ar e living or cell has life. (Each cell is made of a small mass of pr ot oplasm cont aini ng a nucl eus i n i t s i nsi de and a pl asma membr ane with or without a cell wall on its outside). 5. New cells ar ise fr om pr e-exist ing cell.

P r ok a r y ot i c C ell

   

Cytosol

Nucleus

Cell wit hout cell membr ane (H enst ein) I t cont ains 1 or mor e vacuoles. I n mat ur e plant cells, pr ot oplast for ms a per ipher al layer called pr imor dial ut r icle. I n animal cells pr ot oplast fills t he whole cell.

M i t ochondr i a   

L ife span of mit ochondr ia is 5-10 days. Aver age number of mit ochondr ia in each cell is 1000 – 1600/cell U sual l y pl ant cel l s have fewer mi t ochondr i a as compar ed t o animal cell.

Biology

Shape and Size  M it ochondr ia ar e spher ical or filamentous in shape.  Smallest mit ochondr ia ar e pr esent in yeast (< 1).  L ar gest mit ochondr ia ar e pr esent in Rana pipens (20-40 )  I n human body lar gest mit ochondr ia ar e pr esent in pancreatic cells (10).  Si ze depen ds on act i v i t y of cel l . A ct i ve cel l s (mer ist emat ic cells) have lar ge mit ochondr ia.  M it ochondr i al DNA can code t he synt hesi s of 10 differ ent types of pr oteins (Membrane pr oteins). Rest of t he pr ot ei ns and enzymes of mit ochondr ia ar e synt hesized under t he cont r ol of nuclear genes.  Enzymes for r eplicat ion and t r anscr ipt ion of DNA like DNA- polymer ase and RNA polymer ase.  I n mit ochondr ia, 70s t ypes r ibosomes ar e pr esent , whi ch ar e cal led mit or ibosomes (M it ochondr ia of mammals have 55s r ibosomes)  I n scur vy disease, sever al mit ochondr ia fuse to for m lar ge bodies called chondr iospher e. F unct ions of M it ochondr ia 1. Cell oxidation. M it ochondr ia plays an impor t ant r ole in ATP synt hesis dur ing aer obic r espir at ion. M it ochondr ia ar e called Power house of cell because t hey st or e ener gy in for m of ATP. ATP is ener gy cur r ency. I n ATP t wo high ener gy bonds ar e pr esent which ar e called pyr ophosphat ase bonds. 2. M i t ochondr i a hel p i n Vi t el l ogenesi s i n oocyt es, M it ochondr ial K inase makes t he yolk viscous and insoluble for a longer duration stor age. Mitochondria of oocyt es called Yolk nuclei.

Golgi Appar at us  



Golgi in plant s is called lipochondr ion. Golgi body pr esent in all eukar yot es but absent in mat ur e RBCs of mammals, sieve t ubes, br yophyt es et c. Number of Golgi body : One t o many in a cell locat ed near nucleus. M aximum number of golgi body pr esent in r hizodial cells of char a (gr een algae = 25,000)

E ndoplasmic Ret iculum (E R)

I t is a diffused cell or ganelle.  I t consist s of t hr ee st r uct ur e : 1. Cisternae : L ong flatt ened and unbr anched unit s ar r anged in st acks. 40– 50 nm t hick. 2. Vesicles : 25– 500 nm in diamet er. 3. Tubules : L ong, isolat ed, br anched unit s. 50– 190 nm t hick  St r uct ur e of E.R. is like t he Golgi body but in E.R. ci st er nae, vesi cl es and t ubul es ar e i sol at ed i n cyt oplasm and t hese do not for m complex.  Golgi body is localised cell or ganelle while E.R. is widespr ead in cyt oplasm. F unct ions of E.R. 1. Mechanical support : Par t icipates in t he for mation of cytoskelet on with micr o-filaments, micr otubules and int er mediat e filament s. 2. I nt r acel l ul ar exchange : For ms i nt r acel l ul ar conduct i ng syst em. Tr anspor t of mat er i al s i n cyt oplasm fr om one place t o anot her may occur s t hr ough t he duct s of E.R. 3. Rough E.R. : Par t icipat es in pr ot ein synt hesis. 4. Lipid secretion : Lipid synthesized by the agranular por t ion of E.R. (smoot h E.R.), st or ed int o t he Golgi befor e finally being ext r uded int o t he cyt oplasm as lipid dr oplet and t hence t o t he out side of cell. 5. Release of Glucose from Glycogen : Polymer isation of glucose to for m glycogen granules probably occur in t he hyaloplasm not in t he wall of E.R. but t he r et iculum seems t o play a r ole in t he br eakdown of glycogen (glycogenolysis). 6. Cellular metabolism : Membr anes of the r eticulum pr ovi des an i n cr eased su r f ace for m et abol i c act ivit ies wit hin t he cyt oplasm. 7. For mat ion of nuclear membr ane : Fr agment ed vesicles of disint egr at ed nuclear membr ane and ER elements ar r ange ar ound chr omosomes to for m a new nuclear membr ane. 8. For mat ion of cell plat e. 9. For mat ion of lysosomes and Golgi body. 10. Det oxi f i cat i on : Sm oot h E R con cer n ed w i t h det oxificat ion of dr ugs and st er oids. 

Pl ast i ds These ar e pr esent in plants and few pr otists (Euglena). Classificat ion On t he basis of funct ion plast ids ar e of t hr ee t ypes: 1. L eucoplast ; 2. Chlor oplast ; 3. Chr omoplast  Al l t ypes of pl ast i ds have common or i gi n fr om pr oplast ids, sac like non-lamellar st r uct ur es.  Differ ent t ypes of plast ids may t r ansfor m fr om one for m t o anot her. But chr omoplast s never t r ansfor m t o chlor oplast s. Plastid Leucoplast

Clternae

Veslcles

Tubules

3.7

Chloroplast

Chromoplast

3.8

Biology

R i bosom e    

    



Also called pr ot ein fact or y and engine of t he cell . Ribosomes ar e smallest cell or ganelles (150  250 Å) Unit membr aneless cell or ganelle. Ri bosom e can be seen on l y t h r ou gh el ect r on micr oscope, i.e. after electr on micr oscopes invention, r ibosomes wer e discover ed. Having negative char ge due to pr esence of phosphate gr oup (H 3PO43– ). Palade coined t he t er m Ribosome. Except mammalian RBC all living cells have ribosomes. Pr ot eins in r ibosomes ar e negat ively char ged. r -RNA is synt hesized in Nucleolus. Nucleoulus is called Ribosome fact or y. Ribosomes ar e univer sal cell or ganelle, pr esent in all pr okar yot es and eukar yot es.

Cell Cycle A growing cell under goes a cell cycle t hat is comprised of essent ially t wo per iods 1. I nter phase. 2. Per iod of division. (M it osis/M eiosis)

Str uctur e of Cell N ucl eus 



 



I n major it y of t he cells, nucleus lies in t he cent r e. H owever, it may also be in t he per ipher y of t he cell as in t he case of plant cells. The liquid pr ot oplasm in t he nucleus is called nucleoplasm , which is bound by a membr ane called nuclear membr ane. I t cont ai ns a net wor k of fi br ous mat er i al cal l ed chr omat in . The condensed chr omat in is in t he for m of fibr e like st r uct ur es called chr omosomes, which help in inher it ance or t r ansfer of char act er s fr om t he par ent s t o next gener at ion.

L ar gest component of t he cell. N u cl eu s i s dou bl e m em br an e bou n d den se pr otoplasmic body that contr ols cellular metabolism, enclose all t he genet ic infor mat ion, and is able t o t r ansmit t he same t o t he next gener at ion. Except mat ur e mammalian R.B.C. and Sieve cell of phloem, ever y living cell have a nucleus.

Pr ocar yotes have incipient nucleus devoid of nuclear membr ane, nucl eol us and hi st one. Such t ype of nucleus is called nucleoid.  Nucleus of undividing cell is called interphase nucleus.  The pr ot oplasm bet ween nucleus and t he plasma membr ane is called cyt oplasm .  L ar gest among t he or ganelles in t he cell ar e plast ids found in plant cells. These cont ain pigment s.  Ther e ar e r od shaped or spher ical or ganelles called mit ochondr ia.  I n t he cyt opl asm t her e ar e dr opl et s of subst ances di ssol ved i n wat er. U nder t he mi cr oscope, t hese dr opl et s appear as empt y spaces and ar e t er med as vacuoles.  I n amoeba, t he vacuoles cont ain food par t icles and ar e called food vacuoles.  M ost of t he pl ant cel l s have l ar ger vacuol es as compar ed t o animal cells. N umber, Size, Position and Shape  Nor mally a cell has single nucleus, this condition is called Uninucleate / Mononucleate / Monokaryotic.  Som et i m es a cel l m ay h ave t wo n u cl ei cal l ed Binucleat e / Dikar yot ic cell  Nucleus usually occupies 20% of cells mass. I t s size depend upon ploidy, cell mat ur it y and cell funct ion.  N ucl eus l i es near geomet r i c cent r e of t he cel l , per iphar al nucleus pr esent in adipocyt es and plant cell s.  Nucleus is rounded in young plant s and polygonal in animal cells, oval or ellipt ical in mat ur e plant cells. 

Cel l D i visi on I t is t he pr ocess by which a cell called par ent cell , divides int o t wo cells, called daught er cells.  Cell is usually a small segment of a lar ger cell cycle.  I n meiosis however, a cell is permanently transformed.  Cell division is t he biological basis of life. For simple unicellular or ganisms such as t he amoeba, one cell division r epr oduces an ent ir e or ganism. On a lar ger scal e, cel l di v i si on can cr eat e pr ogen y f r om mult icellular or ganisms, such as plant s t hat gr ow fr om cut t ings. But most impor t ant ly, cell division enables sexually r epr oducing or ganisms t o develop fr om the one-celled zygote, which itself was pr oduced by cell division fr om gamet es. And aft er gr owth, cell division allows for cont inual r enewal and r epair of t he or ganism.  T h e pr i m ar y con cer n of cel l di v i si on i s t h e maintenance of t he or iginal cell’s genome. Befor e division, the genomic infor mation which is stor ed in chr omosomes must be r eplicat ed, and duplicated genome separ ated cleanly between cells. A gr eat deal of cel l ul ar i nfr ast r uct ur e i s i nvol ved i n k eepi ng genomic infor mation consistent between gener ations. M i t osi s I t i s t h e pr ocess of cel l di v i si on w h er eby t h e chr omosomes ar e duplicat ed and dist r ibut ed equally t o t he daught er cells. 

Biology

I t is a cont inuous pr ocess which t akes place in five phases : (i ) Pr ophase; ( i i ) Pr o-met aphase; ( i i i M et aphase; (iv ) Anaphase; (v ) Telophase M eiosis I t occur s in t he ger m cells, which ar e dest ined t o for m gamet es in sexually r epr oducing or ganisms. Binar y F ission Prokaryotic cell division process is called Binary fission . Difference bet ween M it osis and M eiosis M i t osi s 1. The cel l di vi des onl y once aft er one r ound of D N A r epl i cat i on.

2. M i t osi s t ak es pl ace i n t h e som at i c cel l s.

3. I t occur s i n bot h sexu al l y as wel l as asexual l y r epr oduci ng or gani sm s. 4. DN A r epl i cat es once for one cel l di vi si on. 5. Cel l di vi des onl y once and t he chr om osom es al so di vi de onl y once. 6. Chr om osom e num ber r em ai ns const ant at t he end of m i t osi s. 7. Gen et i c const i t ut i on of daught er cel l s i s i den t i cal t o t h at of par ent cel l . The t wo dau ght er cel l s ar e for m ed.

8. N o cr ossi ng over of genes. 9. I t r equi r es l ess t i m e.

M ei osi s 1. There are two successive cell divisions: Fi r st and t he second m ei ot i c di vi si ons. 2. M ei osi s t ak es pl ace i n t he ger m (r epr oduct i ve cel l s) cel l s. 3. I t occur s onl y i n sexual l y r epr oduci ng or gani sm s.

3.9

N ut r i t i on Aut ot r oph 



T h ese ar e l i v i n g or gan i sm s t h at sy n t h esi s or gan i c subst ances fr om i n or gani c m ol ecu l es usi ng l i ght . These ar e pr imar y pr oducer s, i.e. all gr een plant s and many plankt ons ar e aut ot r ophs.

 Some bact er i a use chemi cal ener gy of sul phar compounds t o synt hesis or ganic subst ances.

H et er ot r oph These ar e living or ganisms t hat do not synt hesis food, animal, fungi et c.

Phot osysnt hesi s 



Por e openings of plant leaves ar e called st omat a. These open in day and close at night . I t occur s in visible light 4000 t o 7500 Å (does not occur in I nfr a-r ed and ult r a-violet light ).



Red, blue and violet wavelengt hs ar e absor bed by t he pigment s.

Temperatur e range : 35C (for conifer s) t o + 75 C for (Algae) I t t akes place at r apid r at e bet ween 10 C t o 35 C.  Cer t ain bact er ia and algae can also capt ur e light ener gy and use it t o make food e.g. phot osynt het i c bact er i a cont ai ns chl or ophyl l i n t i n y bod i es cal l ed ch r om at op h or es . I n chr omat o-phor es component s ot her t han wat er ar e combi ned wi t h CO2 t o for m sugar. N o oxygen i s r el eased. Essent ial Element s in Plant s Car bon – 45% 

4. DN A r epl i cat es once for t wo cel l di vi si ons. 5. Ther e ar e t w o cel l di vi si ons bu t chr om osom e di vi de onl y once. 6 Chr om osom al num ber i s r educed fr om di pl oi d t o t he hapl oi d. 7. Genet i c const i t ut i on of t he daugh t er cel l s usual l y di ffer s fr om t hat of t he par ent cel l due t o cr ossi ng over . Each chr om osom e of daught er cel l s usual l y cont ai ns a m i xt ur e of m at er nal and pat er nal genes. 8. Cr ossi ng over of genes t ak es pl ace. 9. I t r equi r es l onger t i m e.

Oxygen

–

3%

Hydr ogen

–

6%

Nit r ogen

–

1.3%

et c.

Gr owt h St imulat or s I ndole acet ic acid It incr eases t he t endency of cut t ing t o t hr ough out r oot s. E t hylene I t acceler at es r ipening of fr uit s

3.10 Biology

Composi t i on of F ood Food can be classified int o 1. Car bohydr at es and fat s : ener gy giving food 2. Pr ot eins : Body building food 3. M iner als and Vit amins : Pr ot ect ive foods Carbohydrates Protein M acr o nut r ient s Fat

U| V| W

UV M icr o nut r ient s MineralsW

Vit amins

M iner als and Vit amins 1. Car bohydr at es Gener al for mula : (C,H , O  Cx (H 2O)y  These pr ovide ener gy t o body  B asi c un i t i n si m pl e su gar m ol ecu l e (m on o sachar ide) is GlucoseC6H 12O6.  One gm of glucose yields 4.2 k cal ener gy  Car bohydr at es cont r ibut e about 45% of calor ies in most diet s.  L iver conver t s Glucose t o Glycogen. Gl ycogen is cal led animal st ar ch because it i s for med in animals not in plant s. I t is st or ed in liver and muscles. Types of Carbohydr at es. (i ) M ono Sacharides (C6H 12O 6) : H uman body can make use of car bohydr at es only in t he for m of monosachar ides. e.g. Glucose, Galact ose, Fr uct ose (Fr uit sugar ) (ii ) Disacharides : C12H 22O11 Common disachar ides ar e maltose, sucr ose and lact ose. (iii ) Poly Sacharides : (C6H 10O5)n St ar ch, Glucose and Glycogen. 2. F at s  These ar e mixt ur e of L ipids [Tr iglycer ides].  L ipids mixt ur e is solid at 20C.  Veget able fat s ar e highly unsat ur at ed.  Animal fats ar e satur ated and cause blood pr essure and hear t disease.  I t is made of C, H , and O alt hough, O is ver y less. Gener al for mula  C57 H 110O6. One gm fat = 9.0 k.cal.  Fat s ar e st or ed in Adipose t issues. 3. Pr ot ei ns  T h ey con t ai n N i t r ogen , Car bon , H ydr ogen , Oxygen  These ar e pr imar y foods.  I t is essent ial for gr owt h

   

   

I t is made up of aminoacids These r egulat e met abolism These make 12 t o 15% of diet calor ies. Out of 22 Amino acids : 9 for adult s and 10 for childr en ar e t o be supplied in the diet fr om outside and r est ar e manufactur ed by t he body. 1 gm pr ot eins = 4.2 k. cal. I t cannot be st or ed in human body. Pr ot eins for m lar gest molecule in nat ur e These pr ovide base for manufact ur ing enzymes, har mones and ant ibodies et c. It's deficiency causes Kwashiorker (Red haired body).

4. Wat er  Wat er usually for ms 75-90% of cyt oplasm.  70 t o 90 per cent of living cells is wat er.  I n humans about t wo-t hir d of t he body is for med of wat er and of t hese, about 55 per cent (20-22 lit r es) is confined t o cells as int r acellular wat er. The r emainder is found in ext r acellular fluids like blood, t issur e fluid and lymph.  I t is good solvent and thus it is needed in the body 5. M iner als  Minerals are requir ed (i ) t o car r y out met abolic pr ocess. (ii ) maint ain pr oper osmot ic pr essur e. (iii ) for pr eser vat ion of physical shape.  M iner als ar e inor ganic subst ances  At least 29 element s ar e found in our body.  M iner als have no ener gy value but imbalance of miner al level in t hebody may cause diseases. I mpor tant M i ner als Cal ci um

R equi r ement

Calcium (Ca) & Phosphorus (P) Potassi um (K) & Magnesium (Mg) Sodi um (N a)

Pr ot ect ion of t eet h

I odi ne (I )

Pr oper clot t ing of blood

pr oper muscular cont r act ion M aint aining wat er bal ance in body. Healthy functioning of thyroid gland

Cobal t (Co)

Utilization of vitamin (B12 esp.)

Zi nc (Zn)

H ealt hy funct ioning of t ongue

I r on  I t is essent ial for haemoglobin.  I t gives r ed colour t o RBC  I t t r anspor t s oxygen t o human body

Biology 3.11

6. Vit amins

 VI TAM I N S

Wat er soluble B, C & P I f t hey ar e t aken in excess t hey ar e washed away





Fat soluble A, D, E, & K (cannot be washed) I f t aken in excess it causes a condit ion called hypervitaminous   L oss of hair  Enlar gement of liver  Cr acked mout h Glycogen in the liver is a short ter m stor e. I f no other glucose supply is available, it will last the body for about six hours. Excess glucose not stor ed as glycogen is conver t ed int o fat and st or ed in t he fat cells.

I f concent r at ion of glucose r ises above 160 mg per 100 cm 3 of blood glucose, it is excer et ed by t he kidneys in ur ine. I f it falls below 40 mg/100 cm 3 of blood, t he br ain cells may be affect ed.

Element Found in N ormal H uman Body E l em en t

P er cen t

E l em en t

P er cen t

Ox y gen Car bon

65.0 18.0

Sodi u m Ch l or i n e

0.15 0.15

H y dr ogen

10.0

N i t r ogen Cal ci u m Ph osph or ou s

M agn esi u m

0.05

3.0 2.0

I r on I odi n e

0.004 0.00004

1.1

M an gan ese

0.00013

Pot asi u m

0.35

Copper

0.00014

Su l ph u r

0.25

Cobal t

0.00000016

Communi cabl e D i seases N ame AI DS (Acquir ed I mmune Deficiency Syndr ome) Br ucellosis

Cause T r ansmi ssi on H uman I mmunodeficiency Sexual r elat ions; shar ing of vi ew (H I V) syr inges; blood t r ansfusion

I ncubati on P er i od Sever al year s

Choler a

Br ucellus abor t us or B cat t le or goat s 3-6 year s melit eusis bact er ia Var icella zost er vir us (U S) I nfect ed per sons; ar t icles 10-21 days H er pes zost er vir us (U K ) Cont aminat ed by dischar ge fr om mucous membr ances Vi br io choler ae bact er ium Cont aminat ed wat er and seafood a few hour s-5 days

Common cold

N umer ous vir uses

Diphit her ia

Cor nybact er ium Di pht her ia bact er ium Vi r uses Cl ost r idium Welchii bact er ium N eisser ia gonnor r hoeoe bact er ia H epat it i s A vir us

Chick enpox Var icella

Encephalit is Gas gangr ene Gonor r hoea H epat it i s A (I nect ious) H epat it i s B (Ser umt ype B) I nfluenza L epr osy M al ar ia M easles (r ubeola) M eningit is

M umps

H epat it i s B vir us N umer ous vir uses (t ypes A,B,C) M ycobact er ium lepr ae bacillus Pl asmodium pr ot ozoa Rubeola Vir t ual Var ious bact er ia and vi r uses (Vir al meni ngit is) Vi r us

Respir at or y dr oplet s of infect ed per son Respit ar y secr et ions and saliva of infect ed per sons or car r ier s Bit e fr om infect ed mosquit o Soil or soil-cont aminat ed ar t icles

1-4 days 2-6 days 4-21 days 1-4 days

U r et hr al or vaginal secr et ions of 3-8 days infect ed per sons Cont aminat ed food and wat er 15-50 days I nfect ed blood; par ent er al I nject ion Dir ect cont act ; r espir at or y dr oplet s, possi bly air bor ne Dr oplet infect ion (minimally cont agious) Bit e fr om infect ed mosquit o Dr oplet infect ion Respir at or y dr oplet s

Dir ect cont act wit h infect ed per sons; Respir at or y dr oplet s and or al secr et ions

6 week s-6 mont hs 1-4 days Var iable 6-37 days 10-15 days var ies wit h causat ive (bact er ial meningit is) agent 14-21 days

3.12 Biology

N ame

Cause

T r ansmission

I ncubation P er iod Par at yphoid fever s Salmonella bact er ia I ngest ion of cont aminat ed food and wat er 1-14 days St r ept ococcus Pneumoniae Dr oplet infect ion Pneumonia 1-3 week s Bact er ium Dir ect cont act wit h nasophar yngeal Poliomyeli t s Polio vir uses 7-21 days secr et ions of infect ed per sons; vomit 10 days-6 Rabies Vir us Bit e fr om r abid animal mont hs Rubella (Ger man Dir ect cont act or dr oplet spr ead of Rubella vir us 14-21 days measles) nasophar yngeal secr et ion Gr oup A haeolyt ic Dir ect or indir ect cont act wit h infect ed Scar let fever 1-5 days St r ept ococcus bact er ia per sons, or dr oplet infect ion Smallpox (Var iola) Poxvir us var iola Dir ect cont act ; dr oplet 7-14 days Tr eponema pallidum Sexual r elat ions; cont act wit h open Syphilis 10-90 days bact er ia lesions; blood t r ansfusion Tet anus (lockjaw) Clost r idium t et ani bacil lus Animal faeces and soil 3-21 days M ycobact er ium Dr oplet spr ead; ingest ion fr om Tuber culosi s var iable t uber culosis baci llus cont aminat ed mil k Typhoid fever Salmonella t yphi baci llus Cont aminat ed food and wat er 7-24 days Whopping Cough Bor det ella por t ussi s Dr oplet spr ead 10-21 days (per t ussis) bact er ium Yellow fever Ar bovir us Bit e fr om infect ed mosquit o 3-6 days

D efi ci ency and D iseases Vitamin

Function

D eficiency

Sour ces

A

Ret inol

B

Complex

B1

Thiamine

 1 vit amin t o be discover ed by Ezkman’ in 1897  Essent ial for healt hy ner ves & mucous membrane  (Dest r oyed by high t emperat ur e & baking soda)

Ber i-Ber i Checks gr owt h in childr en

Yeast , bacon, veget ables, eggs and liver s.

B2

Riboflavin

To obt ain st eady and cont inuous r elease of ener gy

 Checks gr owth  Skin becomes r ough and r ed  Diar r hoea & digest ive upset s

M ilk , cer eals, veget ables, yeast , meat

B3

Niacin ‘or ’ Nicot inous Acid

 Essent ial for healt hy digest ive funct ion.  H elps to cont r ol cholest er ol level.

 Diar r haea.  A condit ion known as ‘Pel l agr a’

 For normal gr owt h in childr en  To maint ain conjuctiva aidsnight vi sion  To maint ain skin and mucous membr ane. st

Fish, liver , oil, but t er , mil k,  Roughness and dr y skin. egg, veget able yellow  I nabili t y t o see in dim light . veget able and  Xer opht hlmia leading to blindness (cor nea becomes r ough and dr y)  N ycta l opi a (Night bl indness)

Cholestr ol : LDL ® Low Density lipo protein ® > 160 mg per 100 CmB of blood H DL (H igh densit y l ipo prot ein), > 40 mg per , 100 CmB of blood

Biology 3.13

V i t am i n B6

Pyr idoxine

Folic Acid

F unct i on  Pr ot ein met abolism

D efi ci ency

Sour ces

U seful dur ing r adio t her apy

Pr epar ed fr om Anem i a Biot in, cholin, mosit ol pant ot henic acid & par amino benzoic cid ar e al so st ept omycine cult ur e. member s of vit amin B complex For RBC M eat , l iver , fish B 12 Cyt amen P r eni ci ous Anaem i a  Aids in for mat ion of RBC

(cynocobalamin) C Ascor bic  For pr oper for mat ion of collagen in connect ive Acid t issues  For mat ion of bones and t eet h D Calcifer ol N ecessar y for absor pt ion D D and met aboli sm of calcium and phosphor us U sed for r at 2

3

poi soni ng ul tr a violet li ght nat ur al found in animal food

E

Tocopher ol

K

Philloquine

P

H esper idin

 Scu r vy  Capillar y bleeding  Check s gr owt h in childr en

 Cit r us fr uit s, t omat oes, Gr een leaf, vegetable and potato (newone)  Decr eases wi t h age.

 R i ck et s  Oest eom alaci a : Defective deposition of enamel, leadi ng t o dent al car r ies.

 Can be for med by sun l ight  But t er , cheese et c.

 Essent ial for maint aining fir mness of sk in  Blood t hinner For pr oper clot t ing of blood  Jaundice  Can be made by body only in t he pr esence of bile.  H elps in maint aining nor mal capillar y r esist ance.  Deficiency may cause int er nal bleeding.  Sour ces – Gr eens veget able and cit r ous fr uit s.

Wheat , Animal, Food.

Gr een plant s Gr een peas, cabbage



N ut r i t i on

Small I ntenstine (D uodenum + I lleum)

D i gest ive Syst em



I t consist of  Mouth  Oesophagus  St omach  Small int enst ine Duodenum + I lleum  L ar ge I nt enst ine  (Colon)

 

I t is 20 feet long. Villi pr ovides mor e sur face ar ea (600 times) for better absor pt ion. At t he place wher e small and lar ge int enst ine ar e joined, t wo or gans ar e found.

UV W



1. Caecum L ost t heir ut ilit y 2. Appendix Food in t he small int est ine is called chyle.

L ar ge I nt enst i ne  

I t is 5 feet long and 2-5 inches in diamet er I t has t hr ee par t s : 1. Coecum 2. Colon : I t i s t hi ck er t han smal l i nt est ine and t hinner t han coecum. — I t does not secr et any enzymes — Digest ed r esidue is st or ed. — I t absor bs wat er fr om r esidue.

3.14 Biology

3. Rect um : — I t secr et s no enzymes. I t st or es di gest i ve r esidue and absor bs wat er fr om r esidue. — I t is last par t of t he aliment r y canal.

D igest ion I n Buccal Cavit y I n buccal cavit y, food get s mixed wit h saliva fr om salivar y gland dur ing chewing.  Composit ion of Saliva : 99.4% H 2O, 0.6% ot her compounds which includes inor ganic (NaCl, K Cl, Na2H PO4, CaCO3 ), mucous, salivar y amylase or Pt yline, lysozyme. pH : 6.8  Tot al amount :1– 1.5 lit .  Wit h saliva, Pb, H g and t heir iodides ar e excr et ed out .  Secr et i on of sal i va i s under cont r ol of M edul l a oblongat a.  Par a sympat het i c syst em i ncr eases secr et i on of saliva.  Sympat het ic syst em decr eases secr et ion of saliva.  F unct ions of Saliva (i ) To t ie food par t icles. (ii ) To make food moist and slimy. (iii ) Lysozyme and Thiocyanat e dest r oy bact er ia. (iv) Pt yline conver ts st ar ch into maltose, isomaltose and limit dext r ines. (v) H elpful in det ect ing t est e. (vi ) H elpful in wat er r egulat ion. N ote : On feeling t hir st , O.P. of body fluid is incr eased. N ow food is li k e a ball call ed bolus whi ch r eaches st omach by perist alsis.

(iv) To avoid r ot t ening of food. (v) To act ivat e enzymes. Pepsinogen    Pepsin Pr or ennin   Rennin F unct ions of Pepsin Pepsin

Pr ot ein   Pept one+Pr ot eoses



D igest ion in St omach  







Spal l anzani expl ai ned di gest i on i n gener al and Beaumont explained it in man. As food comes in st omach, gast r ic juice is secr et ed by 350 l ak h gast r i c gl ands. Gast r i c gl ands ar e st imulat ed by gast r in hor mone secr et ed by G– cells of pylor ic par t . Quant i t y of gast r i c j ui ce becomes maxi mum i n 1 1 t o 1 hour s. 2 Composit ion of Gast r ic juice : 99% H 2O, 1% ot her s which include inor ganic salt s, mucous, H Cl (0.4%), pepsinogen, pr or ennin, gast r ic lipase, int r insic fact or of De Cest ello. pH : 1.0– 1.5 Tot al amount : 1– 3 lit ./24 hr s.

Functions of H CL (i ) To pr ovide acidic medium. (ii ) To kill har mful bact er ia. (iii ) To dest r oy living par t of food.

F unct ions of Rennin or Rennet or Cymosin or Cur ding enzyme

Caseinogen Renni n  Casein — Ca —  Par acasseinat e 

F unct ions of Gast r ic lipase I t is negligibly act ive. L I PASE Fat   Fat t y acids + Glycer ol Now food is called chyme, it comes in Duodenum. Her e ent er ogast r one is secr et ed which st ops digest ion in st omach M easur es against Aut odigest ion of St omach  Ther e is a basic layer of mucous on mucosa.  Tight junct ion ar e pr esent bet ween cells.  Cells of mucosa ar e r eplaced at t he int er val of 2– 3 days. N ote : A bact er ium H elicobact or pylor i is r esponsible for pept ic ulcer.

D igest ion in D uodenum     

I t r eceives bile juice and pancr eat ic juice. H epat ocr ine st imulat es synt hesis of bile juice. CCK st imulat es cont r act ion of gall bladder. Secr et i n (di scover ed by Beyl i s & St ur l i ng) and pancreozymine stimulate secretion of pancreatic juice. Fir st bile juice r eact s on food by making the medium alkaline

Composit ion Bile juice 89% H 2O, 11% others which includes mucous, inorganic, bi l e jui ce (Sodi um or Pot assi um t aur ochol at e and glycholat e) , bile pigment , cholest er ol, lecit hin. pH : 7.7 – 8.6  Tot al amount : 500 – 1000 ml / 24 H r.  Bi l e sal t s ar e hel pful i n emul si fi cat i on of fat s. Emulsified fat molecules ar e called micelles.  The r eact ion is called saponificat ion .  Bile salt s ar e helpful in absor pt ion of Vit amin A, D, E and K .  Bile pigment , cholest er ol and lecit hin ar e excr et or y. Pancreat ic juice – Complet e juice Composit ion : 98.5% H 2O and 1.5% ot her s which includes inor ganic salt s, mucous, chemot r ypsinogen, car boxypept idase, ami n opept i dase, t r ypsi n ogen, amyl ase, mal t ase, i somal t ase, l i mi t Dext r i nase, l i pase or st eapsi ne, cholest er ol est er ase, polynucleot idase. pH : 8.0

Biology 3.15

Total amount : 500 – 800 ml / 24 hr s.

Ent er ok inase Tr ypsinogen  Tr ypsin

Trypsin Chemot rypsinogen  Chemot r ypsin car boxypept idase Poly/Tr i/Di pept ide  Amino acid Aminopept idase

I somalt ose M alt ose

I somalt ase

  Glucose

M alt ase

 Gl ucose

L imit dext rin

Dext rinase

 Glucose

Sucr ase Sucrose  Glucose  Fruct ose I nver t ase L act ose

L act ase

 Glucose  Galact ose

I mulsified fat Cholest erol Nucleic acid

L ipase

 Fat t y

acid  glycer ol

Ch. est erase

 Est ers

of cholest erol

Polynuclect idase

  Nucleot ides

D igestion in Small I ntest ine

H er e f ood get s su ccu s en t er i cu s st i m u l at ed by ent er okr inine har mone. Composit ion of succus ent ericus : 98.5% H 2O, 1.5% ot her s whi ch i ncl udes mucous, i n or gan i c sal t s, er apsi n , m al t ase, i som al t ase, l i m i t D ex t r i n a se, su cr a se, l ect a se, l i p a se, polynucleot idase, phasphot ase, nucleosidase

Erapsin Tr i/Di/Pept ide   A.A. Aminopept idase dipept idase Nucleot idase Nucleot ides  Pent ose  Nit r ogenous base

D igest ion of Cellulose   



I n m an di gest i on of cel l u l ose t ak es pl ace i n her bivor ous. Cy st ase an d cel l u l ose con v er t cel l u l ose i n t o cellulobiose. I n r abbit it is not complet ed in one inst ance. Rabbit feeds on i t s eveni ng faeces. I t i s copr ophagy or pseudor umination. Such animals ar e copr ophagous. Food passes fr om aliment ar y canal t wo t imes, t his phenomenon is cest r ophy.

Absor pt i on of F ood Absor pt ion fr om M out h H er e negligible t obacco, alkohal and some medicines as isopr enaline glycer ol t r i-nit r it e is absor bed.

Absor pt ion fr om St omach H er e gl ucose, H 2O, al cohol and vi t ami ne B 12 ar e absor bed. Absor pt ion fr om I nt est ine  Villi ar e helpful.  Vilicr inine st imulat es it .  I n villi b.v. and lect eals pr esent .  Car bohydr at e and pr ot ein ar e absor bed by b.v.  L i pi d i s absor bed by l ect eal s, col our l ess l ymph becomes milky due t o for mat ion of chilomicr ones. I t r eaches t o hear t fir st .  Car bohydr at e and pr ot ein r each t o liver.  Fr uct ose is absor bed by facilit at ed diffusion.  M annose and pent ose by passive t r anspor t .  Glucose and Galact ose by act ive t r anspor t H 2O by osmosis. D.A.A. by passive t r anspor t . L .A.A. by act ive t r anspor t . Vit amins A and D by simple diffusion. I n columnar cells, acids phasphot ase enzyme pr esent , shows act ive pr ocess for absor pt ion.

E gest i on       

To give out undigest ed food fr om aliment ar y canal. Faeces can r emain in colon for 36 hr s t hen moving int o r ect um by gast r o-colic r eflex. Faeces consist s of 3/4 H 2O and 1/4 solid. I n solid par t , 3% bact er ia, 30% r oughage, 20% fat , 15% inor ganic and 3% pr ot ein ar e pr esent . Dead mu cosal cel l s, mucou s, chol est er oal al so pr esent . NH 3, CH 4 ar e negligible. Br own colour is due to stercobilin and stercobilinogen . Foul smell is due t o sket ol , I ndol and t r ept ophan .

Teet h Teet h ar e har d, yellow subst ance sur r ounding pulp. Types of Teeth Each half of t he jaw hold four t ypes of t eet h 1. I ncissor s 2. Canines 3. Pr e molar s 4. M olar s M ilk Teet h (Decidious t eeth) 1. I ncissors

N umber 8

2. Canines

4

3. Pre molars 4. Molar s Total

8 12 32

Age 7– 9 mont hs 18 mont hs 24 mont hs 6-20 year s

U sed for Slice and cut pieces Tear food it ems Rip t ough food Gr iding

3.16 Biology

Permanent t eet h

   

L ower t eet h come ear lier t han upper. Dentine Har d yellow substance that sur r r ounds the pulp. Other parts : Enamel, cementum, periodontal ligament. Enamel : Pr otects teeth fr om sensitive heat and cold.

Teeth in Animals  

 



Cats, dogs and most other mammals have heterodont t eet h which have differ ent uses. U nl i k e ot her mammal s fi sh an d r ept i l es h ave homodont t eet h. All ar e of same size and shape and have only one use). Fi sh and r ept i l es l oose and r epl ace t hei r t eet h cont inuously. Snakes have teet h that cur ve back t owar d thethr oat. I n poisonous snakes cer tain teet h have a canal or gr oove, thr ough which poison can be ejected. Bir ds, t oads, t ur t les, and some t ypes of insect s and whales do not have t eet h.

Skelet al Syst em   

The human endoskelet on is made up of bones and car t ilage of var ious t ypes. Bone is a har d connect ive t issue in which gr ound subst ance is ver y har d and cont ains calcium salt s. M ar r ow of t h e l on g bon es i s t h e si t e f or t h e haemopoiesis, i.e. for mation of blood and blood cells.

Classificat ion of Skelet al syst em 1. Axial skeleton (80 bones)  I t consist s of skul l, ver t ebr al column, r i bs and st er num, i.e. skelet al element s which ar e pr esent along t he longit udinal axis of t he body.  Ther e ar e 80 bones in axial skelet on.  M andible is t he only movable bone in t he skull of man. Ver t ebr al column.  I n humans, 26 ver tebr ae ar e pr esent serially along the length of the trunk starting behind the occipital bone of t he skull.  Ver t ebr al column is main axis of t he body, which ar ticulates with skull, pector al gir dle, pelvic gir dle and t he r ibs.  Each ver t ebr a is cent r ally hollow.  Ver t ebr al column gives suppor t t o t he t r unk and pr ovides places for t he at t achment of the r ibs and bones of pelvis as well as it per mit s movement and pr oject s t he spinal cor d.

2. Appendicular skelet on (126 bones)  The appendicular skelet on is made up of bones of t he ar ms and legs and t heir suppor t s.  Shoulder gir dle : I t consist s of (i ) Scapula (shoulder blade) (ii ) Clavicle (collar bone).  Skelet on of t he ar m : I t is divided int o (i ) H umer us (upper ar m) (ii ) Radius and Ulna (for ear m) : Ulna is sit uat ed t owar ds t he lit t le finger side, wher eas r adius t owar ds t he t humb side. (iii ) Car pals (wr ist bones) (iv) Metacarpals (palm) (v) Phalanges (finger s)  Thumb has t wo bones.  Ot her finger s have t hr ee bones : (i ) Pr oximal (ii ) Middle (iii ) Distal  Bones of t he leg : I t consist of (i ) Femur (t high) (ii ) Tibia and fibula (leg) (iii ) Tar sals (back of t he foot ) (iv) M et at ar sals (for efoot ) (v) Phalanges (toes)  L eg is at t ached t o t he t r unk by a pelvic gir dle made up of t wo hip bones; each consist s of t hr ee bones : (i ) I llum (ii ) I schium (iii ) Pubis. These bones ar e fused in adult s. Classificat ion. Bones of appendicular skeleton may be classified as : ( i ) U pper ext remit y : Shoulder gir dle (clavicle or collar bones, scapula) Humerus; Ulna; Radius; 8 car pals; 5 metacarpals; 14 phalanges. ( ii ) L ower ext remit y : Pelvis gir dle (2 hip bones divisions of each – ilium, ischium, pubis; anter ior joint is symphysis pubis) Femur of t high bone (longest bone) Pat el l a (sesamoi d); Ti bi a; F i bul a; 7 t ar sal s (cal caneus i s heel bone) ; 5 met at ar sal s and 14 phalanges. N ote : (i ) Femur i s t he longest and heaviest bone of t he body. (ii ) Femur, tibia and fibula bones together suppor t shank of t he leg. (iii ) Tibia is lar ger t han fibula and bear s major body weight .

Biology 3.17

E xosk elet ons The har d mat er ial is for med mat er ial is for med mainly on t he outside of the body and is oft en called exoskeleton . I n sect s su ch as beet l es or dr agon f l i es an d cr ust aceans l i k e cr abs or l obst er s have a har d cover ing t o t heir bodies called cut icle.

H uman Skelet on Syst em

H uman skelet on consist s of 206 pieces of bones. Ther e ar e 22 bones in human skull  Cranial bones  F acial bones  E ar s  T hr oat Ther e is one bone  Shoulder gir dle Ther e ar e four bones  T hor ax Ther e ar e 25 bones  Ver t ebr al column Ther e ar e 26 bones  Ar ms Ther e ar e 6 bones I mpor t ant bones Radius, ulna, H umer us.  H ands Ther e ar e 54 bones : Palm or metacarpal bines : Finger bones or phalanges :  Pelvis Ther e ar e t wo bones  L egs. Ther e are 8 bones I mport and bones Femur, Tibia, Fibala, Pat ella  Feet. Ther e r ar e 52 bones : Ankle (t arsal) bones : I nst ep bones : M et at ar sal boen (5  2) Toe bones :

Gland Syst em Gland is an or gan in animal's body which synt hesises su bst an ces such as har mones, wh i ch ar e ei t h er dischar ged int o blood st r eam by endocr ine glands or car r ied out side t hr ough duct s by exocr ine gland Gland can be divided as follows :

1. E xocr ine Glands Syst em

Secr et ions ar e car r ied t hr ough duct s. Secr et ion is called enzyems. Secr et es lar ge quant it y I t consist s of following or gans : ( i ) L iver  L iver is lar gest gland in t he body.  Pr oduct ion of bi le, a gr een yellow alkal ine fluid t akes place in t he liver.

I t manufact ur es ur ea as wast e pr oduct of pr oteins.  I t st or es ir on, vit amin A and vit amin D.  M anufact ur er of Fibr inongen {essent ial for clot t ing of blood}.  I t r egulat es sugar level in t he blood in r ange of 80 – 150 mg/100 cm 3 of blood.  I t act s as det oxificat ing agent in body.  Conver sion of stor ed fat for use by the tissues. ( ii ) M ucous membr ane  I t sit uat ed in r espir at or y glands.  I t secr et e st icky subst ances.  M ucous t r aps dust par t icles fr om air.  I t also moist ens air which is br eat hed. ( iii ) K i dneys  These develop fr om mesoder m of the embryo.  These r emoves a major par t of waste pr oducts fr om t he body.  T h ese ar e t wo bu t j u st on e k i dn ey can per for m all funct ions.  These ar e made up of special cells nephr ons (each one is 5 cm long) and t ubular in shape.  These r emoves foll owi ng subst ances fr om blood uricacid: Ur ea, NH 3, NaCl, Phosphat es, H ar mones et c. 

Renal medulla Posterior vena cava Rental artery and vein Aorta Ureter Urinary bladder

Renal pelvis Kidney

Renal cortex

Ureter

Urethra

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These help in maint aining alkaline nat ur e of blood by r emoving acidic pr oducts fr om blood. The di sease t hat effect s k i dneys i s cal l ed Nephrit is or Brightsdisease. K idney and sweat glands co-or dinat e t heir activities. I n summer, sweat gland sweat mor e and r ole of kidney is r educed, but in wint er it is r ever sed.

2. E ndocr ine Gland Syst em Secr et ion is dir ect ly dischar ged t o blood. Secr et ion ar e called har mones. Secr et e in small qunat it y. Chemical N at ure of H ormones

Ami nes. H or mones of pineal gland (melat onin) and adr enal medulla. M odified Amino acids. H or mones of t hyr oid ar e iodinat ed t hyr onine. e.g. t hyr oxine. Pept i des. Hormones of hypothalamus (ARH, TRH, GRH, GI H), int er mediate (MSH ) and poster ior lobes of pituit ar y (ADH , oxyt ocin). ACTH of ant er ior pit uit ar y and calcit onin of t hyr oid belong t o t his cat egor y.

3.18 Biology

These ar e fur ther classified as (i ) Shor t pept ides e.g. oxyt ocin, ADH (ii ) L ong pept ides e.g. calcit onin, ACTH Pr ot ei ns. H or mones of following ar e pr ot einaceons : (i ) Pancr eas (e.g. insulin, glucagon) (ii ) Gast r oint est inal t r act (iii ) Some female hor mones (e.g. r elaxi n of ovar y and hCG of placent a) (iv) Par at hor mone (PTH ) (v) M ost hor mones of ant er i or pi t ui t ar y except ACTH (e.g. TSH , FSH , L A, LTH , GH ) St er oi ds. Hormones are derived from cholesterol and other steroids e.g. al dost er one, cor t isol , sex cor t i coi ds (adr enal cor tex), testosterone, estr adiol, progester one (gonads except r elaxin, placent a except hCG).

8. M elanocyt e Releasing hor mone (M RH ) : I t stimulates intermediate lobe of the pituitary gland to secrete its melanocyte stimulating hormone (MSH). 9. M elanocyt e I nhibiting hor mone (M I H) : It inhibits secretion of melanocyte stimulating hormone from the intermediate lobe of the pituitary gland. Tar get cells. Neur ohor mones act on t he cells of t he pituit ar y gland. S.N .

1

T h yr ot r opi n r el easi n g h or m on es (T RH )

2

Gr ow t h h or m on e r el easi n g h or m on e (GH RH ) Gr ow t h h or m on e i n h i bi t i n g h or m on e (GH I H ) Gon adot r opi n r el easi n g h or m on e (Gn RH )

H ypot hal amus  

I t develops fr om t he ect oder m of t he embr yo. I t l i es bel ow or i nfer i or t o t he t hal amus. I t i s connect ed t o t he ant er ior lobe of pit uit ar y gland by hypophysial por t al veins, however it is connect ed t o t he poster ior lobe of pit uit ar y gland mainly by axons of neur osecr et or y cells.

R elea si n g of I n h i bi t i n g H or m on es

3

4

N eur o H or mones

The neur osecr etor y cells (neur ons) of hypothalamus secr et e hor mones call ed neur ohor mones (r el easi ng factor s). 1. Adrenocorticotropic Releasing hormone (ARH ) : I t st imulat es ant er ior lobe of t he pit uit ar y gland t o secr et e it s adr eno-cor t icot r opic hor mone (ACTH ). 2. Thyr ot r opin Releasing hor mone (TRH ) : I t st imulat es ant er ior lobe of t he pit uit ar y gland t o secr et e it s t hyr oid st imulat ing hor mone (TSH ) or thyr o-tr ophin. 3. Somat ot r opin Releasing hor mone (SRH ) : I t st imulat es ant er ior lobe of t he pihlit ar y gland t o r elease it s gr owt h hor mone (GH ) or somat ot r opin. 4. Growth I nhibiting hormone (GI H) somatostatin (SS) : I t inhibit s secr et ion of gr owt h hor mone fr om t he ant er ior lobe of t he pit uit ar y gland. 5. Gonadot r opin Releasing hor mone (GnRH ) : I t st imulat es ant er ior lobe of t he pit uit ar y gland t o secr et e t wo gonadot r opic hor mones : (i ) Follicle st imulat ing hor mone (FSH ) (ii ) L ut einising hor mone (L H ) 6. Pr olact in Releasing hor mone (PRH ) : I t st imulat es ant er ior lobe of t he pit uit ar y gland t o secr et e it s pr olact in. 7. Pr olact in I nhibit ing hor mone (PI H ) : I t inbihit s secr et ion of pr olact in fr om t he ant er ior lobe of pit uit ar y gland.

5 6 7

8

9

Pr ol act i n r el easi n g h or m on e (PRH ) Pr ol act i n i n h i bi t i n g h or m on e (PI H ) A dr enocor t i cot r opi c h or m on e r el easi n g h or m on e (CRH ) M el anocyt e st i m ul at i n g h or m on e r el easi n g h or m on e (M RH ) M el anocyt e st i m ul at i n g h or m on e i n h i bi t i n g h or m on e (M I H )

C on t r ol a n d R egu la t i on of Sp eci fi c H or m on es Secr et i on St i m u l at es t h yr ot r opi n st i m u l at i n g h or m one r el ease St i m u l at es gr ow t h h or m one r el ease I n h i bi t s gr ow t h h or m one r el ease St i m u l at es r el ease of fol l i cl e st i m u l at i n g h or m one an d l u t ei n i si n g h or m on e St i m u l at es pr ol act i n r el ease I n bi t i s pr ol act i n r el ease St i m u l at es adr en ocor t i cot r opi c h or m one St i m u l at es m el an ocyt e st i m u l at i n g h or m one r el ease I n h i bi t s m el an ocyt e st i m u l at i n g h or m one r el eas

Pit uit ar y Gland (H ypophysis Cer ebr i)  

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I t develops fr om ect oder m of t he embr yo. I t i s l ocat ed j ust bel ow t he hypot hal amus. The pit uit ar y gland is sit uat ed in a depr ession t he sella t ur cica of sphenoid bone of t he skull. I t is t he smallest endocr ine gland. I t is about 1.3 cm in diamet er and weighs about half a gr am. I t is attached to the br ain by a stalk the infundibulum which is continuous with the hypothalamus above Adenohypophysis or par s dist alis and t he post er ior lobe or neur ohypophysis or par s ner vosa.

Biology 3.19

H armones of Pit uit ary gland 1. H or mones of Ant erior lobe. The anteriorlobe of the pituitary gland secretes following hormones, most of them are trophic hormones. (i ) Somat ot r opi c hor mone or Gr owt h hor mone (STH or GH ) or Somat ot r opi n (Soma- body, t r ophe nour i shment ) (ii ) Thyroid stimulating hormone (TSH) or Thyrotropin (iii ) Adr enocor t icot r opic hor mone (ACTH ) (iv) Pr ol act i n hor mone (PRL ) or M ammot r ophi n hormone (MTH) or Luteotr ophic hormone (LTH) (v) Gonadot r opic hor mones : (a) Follicle-st imulating hormone (FSH ) (b) L uteinizing hor mone (L H ) 2. H ormone of t he I nt ermediat e lobe. Melanocyte stimulating hormone (MSH) or intermedin causes dispersal of pigment granules in the pigment cells, ther eby darkening the colour in certain animals like fishes and amphibians. I t is believed that it is associ at ed wi t h t he gr owt h and devel opment of melanocytes in man which give colour to the skin. Tar get cells. M elanocyt es in skin. 3. H ormones of t he Post erior lobe. ( i ) Oxyt ocin (OT; Pitocin) : Oxyt ocin pr omotes cont r action of the uter ine muscle or labour pain and cont r act ion of the myoepi t hel i al cel l s of t he l act at i ng br east , squeezing milk into the large ducts behind the nipple. In late pregnancy the uterus becomes very sensitive to oxytocin. The amount secr et ed is increased just befor e and dur ing labour and by sucking of the baby. Because of its r ole, oxytocin is called birth hormone and milk ejecting hormone. Tar get cells. Cells of mammar y glands. ( ii ) An t i di u r et i c h or m on e (AD H ) or Vasopressin or Pit r essin : This hor mone has t wo main funct ions : (a) Antidiuretic effect : I t incr eases r eabsor ption of wat er i n t he di st al convol ut ed t ubul e, collect ing tubules and collect ing duct s of the n eph r on s of t h e k i dn ey s. A s a r esu l t , r eabsor pt ion of wat er fr om t he glomer ular filt r at e is incr eased. (b) Pressor effect : I nvoluntar y muscles in the walls of the intestine, gall bladder, ur inar y bladder and blood vessels are stimulated to contr act by ADH. Contraction of the walls of the blood vessels raises the blood pressure and this may be its most impor tant pressor effect. Tar get cells. Cells of kidneys.

T hyr oid Gland  

I t develops fr om endoder m of t he embr yo. I t is the lar gest endocr ine gland located anter ior to the thyr oid car tilage of the lar ynx in the neck.

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I t is bilobed or gan. The t wo lobes ar e connect ed by a nar r ow st r uct ur e called ist hmus. M i cr oscopi c st r uct ur e of t he t hyr oi d gland shows t hyr oid follicles composed of cubical epit helium and filled wit h a hor monous mat er ial called colloid. Smal l amount of l oose connect i ve t i ssue for ms st r oma of t he gl and. Besi des cont ai ni ng bl ood capi ll ar ies, t he st r oma cont ai ns small clust er s of specialized par afollicular cells or ‘c’ cells. I t can st or e enough hor mones in t he for m of colloid t o supply t he body (for about t wo mont hs).

H ar mone Secr at ion Thyr oid gland secr et es following hor mones. 1. T hyroxine (T 4) and Tri-iodot hyronine (T 3).  These ar e secr et ed by t he t hyr oid follicular cells.  T 4 and T 3 cont ain four and t hr ee at oms of iodine.  T 3 is secr eted in smaller amounts (10%) but it is more active and several times more potent than T4  T 4 is conver t ed int o T 3 by r emoval of one iodine in t he liver, kidneys and some ot her t issues.  T 4 and T 3 have similar effect s on t he t ar get cells, t hey ar e gener ally consider ed t oget her under t he name, t hyr oid hor mone (TH ).  Thyr oid gland is t he onl y gland t hat st or es it s hor mones in lar ge quant it y.  T 4 and T 3 ar e synt hesised by at t aching iodine t o t he aminoacid t yr osine. 2. Calcit onin (CT ).  I t is secr et ed by C-cells of t he t hyr oid gland.  I t is secr et ed when calcium level is high in t he bl ood. I t t h en l ow er s t h e cal ci u m l ev el by suppr essing release of calcium ions from the bones. Thus calcitonin has an action opposite t o that of the par at hyr oid hor mone on calcium metabolism. T hyr oid D isor der s 1. H y per t h y r oi di sm (H y per secr et i on of t h y r oi d hor mone) 2. Hypothyroidism (Hyposecr etion of thyr oid hor mone) (i ) Cr etinism (in childr en) (ii ) M yxoedema or Gull’s disease (in adult s) (iii ) Simple Goit r e (iv) H ashimot o’s disease

Par at hyr iod Glands  



They develop fr om t he endoder m of t he embr yo. The parathyroid glands consist of four separate glands locat ed on t he post er ior sur face of t he lobes of t he t hyr oid gland. Cells of parathyroid glands are arranged in a compact mass and ar e of t wo t ypes : (i ) Small chief cells (or pr incipal cells) (ii ) L ar ge oxyphil cells (or eosinophil cells).

3.20 Biology 

The cells ar e enclosed by a delicate connective t issue capsule. The chief cells ar e much mor e numer ous t han t he oxyphil cells. The lat er ar e absent in t he young and appear a lit t le befor e t he age of puber t y.

H or mone Chief cells of the par athyr oids secr ete a hor mone called par at hyr oi d hor mone (PTH ) or par at hor -mone or Collip’s hor mone.  PTH r egul at es cal ci um and phosphat e bal ance bet ween blood and ot her t issues.  PTH inhibit s collagen synt hesis by ost eoblast s and bone r esor pt i on by ost eocl ast s. I t mobi l i ses t he r elease of calcium int o t he blood fr om t he bones.  PT H i n cr eases cal ci u m absor pt i on f r om t h e int est ines. I t incr eases calcium r esor pt ion fr om t he nephr ons (and inhibit s phosphat e r esor pt ion) of t he kidneys. Par at hyr oid D isor der s 1. H ypopar at hyr oidism (deficiency of PTH ). 2. H yper par at hyr oidism (excess of PTH ).

Pancr eas  

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I t is der ived fr om t he endoder m of t he embr yo. The pancr eas lies infer ior t o t he st omach in a bend of t he duodenum. I t is bot h an exocr ine and an endocr ine gland i.e. het er ocr ine gland. A l ar ge pancr eat i c duct r uns t hr ough t he gl and, car r yi ng enzymes and ot her exocr i ne di gest i ve secr et i ons fr om t he pancr eat ic acinar cells t o t he small int est ine. The t issue of t he pancr eas has in addit ion t o t he acinar cells, gr oups of cells called islet s of L anger hans. Types of cells ident ified in the islets 1. Alpha cells (about 20%) 2. Bet a cells (about 70%) 3. Delt a cells or D-cells (about 5%) 4. PP cells or F-cells (5%)

H or mones 1. Glucagon (secr et ed by – cells).  I t st imulat es t he liver t o conver t st or ed glycogen int o glucose.  Glucagon is cont r olled by feedback in accor dance wit h t he level of glucose in t he blood. When blood sugar r ises, secretion of the glucagon is suppr essed and when it dr ops, secr et ion of t he glucagon is st imulated. Tar get cells : Glucagon act s on t he cells of t he liver and adipose t issue. 2. I nsulin (secr et ed by  – cells).  I t was fir st pr epar ed / found by Benting and Best .

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A. F. sanger coined t he t er m insulin and pr oposed t he molecular st r uct ur e of insulin (cow's insuline) H uman insulin was synt hesized by Tsan. I nsulin was t he fir st pr ot ein t hat was ar t ificially synt hesized in labor at or y and is cr yst allized. One molecule of insulin is made up of 51 amino acids t hat has t wo chains : (i ) – chain : I t is made up of 21 aminoacids. (ii ) – chain : I t is made up of 30 amino acids. Both the br anches or chains ar e bind together with cr oss bonds of disulphide bonds. F unct ions : (i ) I t i s an t agon i st i c t o gl u cagon . I n su l i n decr eases t he level of glucose in t he blood. I t act s by incr easing t he r at e at which glucose is t r anspor t ed out of t he blood and int o cells and by st imulat ing muscle cells t o t ake up su gar f r om t h e bl ood an d con ver t i t t o glycogen. L ike glucagon, insulin is pr imar ily r egulat ed by f eedback f r om t h e bl ood gl u cose concentr at ion. When blood sugar level dr ops, t he secr et ion of insulin is suppr essed. When blood sugar level incr eases, t he secr et ion of insulin is st imulat ed. (ii ) I t pr omot es pr ot ein synt hesis in t issue fr om amino acids. (iii ) I t r educes cat aboli sm of pr ot ei ns. I t is an anabolic hor mone. (iv) I t incr eases synt hesis of .fat in t he adipose t issue fr om fat t y acids. (v) I nsulin r educes br eakdown and oxidat ion of fat.

3. Somat ost at in (SS). I t r egulat es secr et ion of insulin and glucagon. Tar get cells : Bot h somat ost at i n and pancr eat i c polypept ide act on t he cells of t he pancr eas. 4. Pancr eat ic Polypept ide (PP).  I t appear s that pancr eatic polypeptide inhibits the r elease of digest ive’ secr et ion of t he pancr eas.  Both somatostatin and pancr eatic polypeptide ar e r el at i vel y newl y di scover ed hor mones of t he pancr eas, and bot h ar e st ill being st udied.

D isor der s of t he Pancr eas D iabet es mellit us (H ypoinsulinism) The insuli n-dependent diabet es mel lit us (I DDM ) is caused by a failure of the Beta-cells to pr oduce adequate amounts of insulin while non-insulin-dependent diabetes mellitus (NI DDM ) appear s to involve failur e of insulin to facilit ate the movement of glucose into cells. 1. Hypoinsulinism (H yper glycemia) 2. H yper insulinism – H ypoglycemia

Biology 3.21

Pineal Gland (E piphysis Cer ebr i)  

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I t develops fr om t he ect oder m of t he embr yo. T h e pi n eal gl an d i s l ocat ed bet ween cer ebr al hemispher es, wher e it pr ot r udes fr om t he r oof of t hir d vent r icle. Pineal gland is a small rounded body which consist s of (i ) Pineal cells (ii ) Suppor t ing glial cells.

H or mone : 1. Melatonin 2. Ser ot onin

T hymus Gland  

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I t is der ived fr om endoder m of t he embr yo. Thymus gland is located in the upper part of the thorax near the hear t. I t is a soft, pinkish, bilobed mass of lymphoid tissue. I t is a pr ominent gland at the t ime of bir th but it gr adually at r ophies I n t he adult. H assal l ’s cor puscl es ar e spher i cal or oval bodi es pr esent i n t he t hymus. They ar e phagocyt i c i n funct ion.

H or mone  Thymus secr et es a hor mone named t hymosin which st imulates t he development of cer tain kinds of white blood cells involved in pr oducing immunit y. I t also hast ens at t ainment of sexual mat ur it y.  Thymosin hor mone stimulates the lymphocytes to destroy the antigens produced by bacteria or pathogen. I t self may be destr oyed by these lymphocytes.

Gonads  

These develop fr om t he mesoder m of t he embr yo. These ar e sex glands (ovar ies and t he t est es). These pr oduce ova and sper ms r espect ively, i.e. cyt ogenic in nat ur e but also secr et e hor mones. 1. Ovar ies Following hormones are produced by the ovaries : (i ) Oest r ogens (ii ) Pr ogest er one (iii ) Relaxin (iv) I nhibinlactin 2. Test es A pair of test es is situat ed in the scr otum of male. T h e con n ect i v e t i ssu e pr esen t bet w een seminifer ous t ubules in a t est is cont ain small clust er s of endocr ine cells called inter st itial cells or L eydig’s cells. These cells secr et e var ious male sex - hor mones cal l ed andr ogens. The pr i nci pal andr ogen i s testosterone.

Funct ions of Test ost er one : I t st imulat es gr owt h and development of male secondar y sex or gans l i k e semi nal vesi cl es, pr ost at e and penis. I t also helps t o maint ain t heir nor mal funct ions. Because t hese or gans do not pr oduce gamet es (sper ms) and only help in r epr oduct ion ar e, t her efor e called secondary sex or gans.  These also stimulates t he development of male secon dar y sex u al ch ar act er s l i k e bear d, moust ache and low-pit ch male voice in man.  Test ost er one al so st i mul at es for mat i on of sper ns in the seminifer ous tubules of the testes.  This hor mone pr omot es gr owt h of many body t i ssues such as bones and muscl es. I t i s due t o t his fact t hat male has a higher st atur e t han t he femal e. 

Disor der s of t he Gonads 1. H ypogonadi sm  I n adequ at e gon adal f u n ct i on i s cal l ed hypogonadism .  I t is due t o defect s in, or injur y t o t he hypothalamus, pituitary gland of the testes or ovary. 2. Pr ecocious Puber t y  Ear ly mat ur at ion of ovar ies and t est es wit h pr oduct ion of ova, befor e t he age of 9 year s in gir ls or sper ms befor e 10 year s in boys is called sexual pr ecocit y. 3. E unuchoi di sm F ai l u r e of t est ost er on e secr et i on cau ses eunuchoidism. A eunuch has undeveloped and non-functional secondar y sex or gans like pr ost r at e, seminal v esi cl es an d pen i s, l ack s ex t er n al sex char act er s such as bear d, moust ache and low pit ch voice and does not pr oduce sper ms. 4. Gynaecomast ia (Gr. gyne = woman, mast os = br east )  Excessive development of male mammar y glands is called gynaecomast ia.  Somet i mes t hese secr et e mi l k . I t r esul t s when t he secr et ion of oest r ogens i s mor e t han andr ogens.

Respi r at ion Syst em Respi r at i on  When body is at r est : aer obic r espir at ion.  When involved in har d physical wor k : aer obi c + anaer obic r espir at ion. Glucose  CO2 + L act ic acid + Ener gy.  The human r espir at or y syst em consist s of exter nal nar es or nostr ils, nasal cavity, nasophar ynx, lar ynx, t r achea, br onchi, br onchiole and lungs.

3.22 Biology   





The air enter s into t he nasal cavit y t hr ough nost r ils. L ungs ar e chief or gan in t he r espir at ion. Each lung cont ains millions of small air chamber s called alveoli . Pharynx (back of the noset and mouth), Larynx (voice box) and t r achea (wind pipe) ar e air passages t hat connect nose mout h t o lungs. Pharynx is also connected t o ear by Euschatian tube. I nhaled air

  

Exhaled ai r

N itr ogen

–

79%

N itr ogen

–

79%

Oxygen

–

20.9%

Oxygen

–

16%

CO2

–

0.04%

CO2

–

4.04%

Nor mal r at e of r espir at ion in adult = 18/minut e. Pulse r at e = 72/minut e. Nor mal r at e of r espir at ion in a baby = 40/minut e

Sequence of Breat hing I nhalation  Exhalation  Pause (Nor mal br eathing) I nhalat ion  Pause  Exhalat ion (sickness, i.e. inver se br eat hing) Capacit y of L ungs Air capacity of lungs of male is about 10%higher than female. Male lung capacity  4500 – 5000 ml. Female lung capacity  4000 – 4500 ml.

Types of Circulat or y Syst ems Ther e ar e t hr ee t ypes of cir culat or y syt ems : 1. No cir culat or y syst em 2. Open cir culat or y syt em 3. Closed cir culat or y syt em The cir culat or y syst em of all ver t ebr at es, as well as of annelids and cephalopods ar e closed, i.e. blood never leaves t he system of blood vessels consisting of ar ter ies capillar ies and veins. Ar t er ies br ing oxygenat ed blood t o t he t issues (except pulmonar y ar t er ies), and veins br ing deoxygenat ed blood back to the hear t (except pulmonary veins). Blood passes fr om ar ter ies to veins thr ough capillaries,which ar e t hinnest and most numer ous of t he blood vessels and these capillar ies helps t o join tissue with ar ter ioles for t r anspor t at ion of nut r it ion t o t he cells.

Component s in

1. Bl ood. Pumped body fluid, liquid medium. 2. H ear t . Cen t r al pu m pi n g or gan ; m ai n com pon en t of cir culat or y syst em. 3. Blood vessels. Syst em of channels which pr ovides pat hways for cir culation of pumped body fluid.

Tot al air The amount of air that is exchanged in lungs in a quiet

B l ood

1 or nor mal br eat hi ng i s equal t o of capaci t y of 10



lungs. Vital capacity (VC) of Lungs I t is t he volume of air t hat can be made t o pass int o and out of l ungs by t he most for ceful r espi r at i on (inspir ation and expir at ion). I n adult male it can be 4 t o 4.5 lit r e I n female it can be 3.5 t o 4.0 lit r es. Vit al capacit y of lungs is measur ed by an inst r ument cal l ed Spiromet er. Cellular Respir at ion  I t involves ser ies of chemical r eact ions t hat occur in t he pr esence of oxygen.  Cells can obt ain some ener gy wit hout oxygen by a chemical pr ocess called glycolysis.  Glycolysis conver t s glucose molecules in t o smaller r molecules called pyr uvic acid. This act ion r eleases ener gy whi ch i s capt ur ed i n a compound cal l ed adenosine triphasphate (ATP).

Ci r culat or y Syst em I t is an or gan syst em t hat moves subst ances t o and fr om cells. I t can also help st abilize body t emper at ur e and pH (par t of homeost asis).

Cir culat or y Syst em



  

Cir culat es body heat . Car r ies nut r ient s and t r anspor t oxygen. Removes wast e pr oduct s fr om body cell. Car r ies ant ibodies t o fight against infect ions. Gen er al l y bl ood m ak es 5% of body wei gh t i n human bei ngs. An adul t of 80 k g wi l l have 5 k g bl ood.

Composit ion of Blood 1. Pl asma.  St r aw colour ed.  50 t o 60% of t ot al blood by volume.  Plasma consist of 90% wat er and r est hundr eds of substances (pr oteins, har mones, nut r ient s etc.)  Cont ains fibr onogen (helps in blood clot t ing).  H aemoglobin if high, immune syst em is high. 2. R.B.C. (Red Blood Cells).  I t also called er yt hr ocyt es  L ar gest number of cells in blood.  I n one dr op of blood, 5 t o 6 million of R.B.C.  Cont ains mainly haemoglobin (gives r ed colour ).  Affinit y t o car bon mono oxide (CO) is 200 t imes mor e t han t hat of O2.  R.B.C. count in male > female (10% less)  L ife of R.B.C. is 120 days

Biology 3.23

Diseases associat ed wit h R.B.C. ( i ) Polycyt haemia or E ryt haemia :  R.B.C. count goes abnor mally high.  Blood becomes t hicker and moves slowly.  Skin becomes bluish. ( ii ) T halassemia [her edit ary disease]  Body is not able to make enough haemoglobin.  Regular blood tr ansfusion [ever y 3 to 4 weeks] common in I ndia in Sindis and Punjabis  Came in I ndia fr om Alexander ’s Ar my. ( iii ) Anaemia :  Deficiency of ir on, folic acid or pr ot iens 3. Whit e Blood Corpuscles (W.B.C.).  I t is also called L eucocyt es.  Types of W.B.C. Ther e ar e 3 t ypes of W.B.C. (i ) Gr anulocytes – 70%  manufact ur ed in  (ii ) M onocyt es – 5%  r edbone-mar r ow. (iii ) Lymphocytes – 25% : m an u f act u r ed i n lymphgland (Thymous gland).  These ar e non pigment al (colour less).  L ar ger in size t han RBC but small in number.  I n one dr op of blood – 7500 W.B.C. R.B.C. Rat io of = 600 : 1 W.B.C.  These dest r oy t he pat hogens.  Dur ing infect ion W.B.C. count incr eases.  Leukemia : I t is mor e common in childr en . (i ) Caused because of cancer of Red bone mar r ow. (ii ) Abnor mal incr ease in W.B.C. and decr ease in R.B.C. Remedies : (i ) Bone t r ansplantation (ii ) Radiation t r eat ment . 4. Plat elet s or T hr ombocyt es.  Non nucleat ed like R.B.C.  Non pigment ed like W.B.C.  L ife – 10 days.  Smaller in count .  M anufact ur ed in Red bone mar r ow.   

W.B.C. =2:1 Thrombocytes Essent ial for blood clot t ing. Vi t am i n K i s essen t i al f or pr odu ct i on of Pr ot hr ombin. Rat io =

Blood Groups I t is based on ant igens. 1. ABO syst em. S. No.

Blood Group

Antigen

Antibody

To take

To give

Genotype

1

A

A

B

A, O

A, AB

IA IA / IA i

2

B

B

A

B, O

B, AB

IB IB / IB i

3

AB

AB

—

A, B, AB, O

AB

IA IB

4

O

—

AB

O

A, B, AB, O

ii

        

2. Rh syst em.  Discover ed by L eindst einer and Winner. (1900)  Race and Taylor wor ked in it .  Rh fact or comes fr om Rhehsus monkey.  I f Rh fact or pr esent , Rh posit ive 97% in I ndia and 80% in wor ld.  I f Rh positive blood is given to Rh negative per son, first time no agglutination second time death occur.  I f Rh negat i ve mot her i s havi ng Rh posi t i ve embr ouy, disease is er yt hr oblast ois foet alis.  People who have Rh ant igen on the RBC ar e Rh +.  People who lack Rh ant igen an t he RBC ar e Rh – .  90% people ar e Rh +, and Ot her s Rh – .  Plasma has no natur al antibody to the Rh ant igen , i.e. Rh + cannot pr oduce only ant ibody. But Rh – may build up ant ibodies called Ant i Rh if t hey r eceive Rh + blood.  I f an Rh – pat ient r eceives Rh + blood, it causes ant i Rh t o at t ack Rh + blood . 

Bl ood Gr oupi ng  

Developed by K ar l L andst einer in 1900. Membr ane of RBC contains pr oteins called Antigens (mor e t han 300 ant igens have been ident ified). M ainly t wo ant igens ar e impor t ant . (i ) A fact or (ii ) B fact or

Discover ed by L eindst einer. O gr oup was discover ed by De cest ello and st ur li. A t t h e poi n t of i n j u r y, ser ot on i n e an d t hr omboplast in (lipopr ot ein) ar e r eleased. Ser ot onine is helpful in vaso const r ict ion. T h r om bopl ast i n h el ps i n t h e f or m at i on of pr ot hr ombinase. I n pr esence of pr ot hr ombinase, pr ot hr ombin is conver t ed int o t hr ombin. Thr ombin r eact s wit h fibr inogen in pr esence of air and Ca++ t o for m fibr in. Net wor k of fibr in for ms clot . I n pr esen ce of XI I f act or (H agm en f act or / glycopr ot ein) danse net is for med.

 

If

|RS M other T| Rh –

Foetus Rh

+

|UV W|

 Pr oblem

I f i t happens, no chance of chi l d bi r t h, onl y abor t ion in second or subsequent t imes. Remedy : vaccine ant i– D, so t hat mot her does not develop ant i Rh. D est r u ct i on of RB Cs of f oet u s becau se of unmat ched Rh fact or is called Er yt hr oblast osis. I n ot her cases t her e is no pr oblem.

3.24 Biology

H uman H eart (M ammalian H ear t) 

       

Situat ed in thor acic cavit y in per icar dial cavity close t o it s fr ont wall. I t s br oad base faces upwar d and backwar d. I ts nar r ow apex faces downwar d, for war d and slight ly t o t he left side and r est s on diaphr agm, below car diac not ch of left lungs. Shape : Some conical. Colour : Dar k r ed. Chamber : 4 (2 aur icles and 2 vent r icles) Size : Roughy 12 x 9 cms. Weight : Aver age weight is about 300 gm in male, 250 gm in female. H ear t is a hollow, muscular,somewhat conical four chamber ed for ce pump, enclosed in a fibr ous bag. Wall of t he hear t is made up of car diac muscle fibr e, connect ive t issues and t iny body vessels. H ear t is sit uat ed in t he chest bet ween t he lungs.



 





 



I nt er nal St r uct ur e 



Wall of aur icles ar e t hi n t han wal l of vent r icl es, because t hey have t o push t he blood t o vent r icles only sit uat ed close t o t hem. Walls of vent r icles ar e t hick as t hey have t o pump t he blood quit e far away. Wall of left vent r icle is 3 times thicker as it has to pump t he blood t o complete body.





 

  

 

H ear t has t wo pumps : ( i ) L eft pump : I t l eads oxygenat ed bl ood whi ch i s r ed in col our. ( ii ) Right pump : I t l eads deoxygenat ed bl ood (pur pl e in col our ). Bl ood bet ween t wo pumps does not mi x dir ect ly. H owever i n foet us t her e i s an openi ng, so t wo pumps meet as l ungs ar e not funct i oni ng. So foet us get s oxygen fr om mot her but just befor e bi r t h openi ng is cl osed. I f t he openi ng i s not cl osed, t hen condi t i on of blue baby i s r eached, and immediat e sur ger y i s r equir ed. Auricle : Smal l er i n si ze and t hi nner muscul ar valves. Vent ricle : L ar ger i n si ze and t hi ck er muscul ar walls.







L eft ventricle : Pumps oxygenat ed bl ood t o r est of t he body. – Taller t he per son, lower the blood pressure in adult. – M ale have l ower bl ood pr essur e t han femal e. I n wal l of vent r i cl e, t endi nous cor ds pr esent . I nner wal l of vent r i cle i s r ai sed int o l ow muscul ar r idges as columnae car neae and few lar ge muscular el avat i on as papi l l ar y muscl es. B et w een v en t r i cl es i n t er v en t r i cu l ar sept u m pr esent , bent t o r ight side. L umen of left vent r i cle i s l ar ge and l umen of r i ght vent r i cl e i s smal l and semi l unar shape. Bet ween aur icl es and vent r i cles A.V. node pr esent . Bet ween 2 aur icles, int er aur icular sept um pr esent . I n emr byo i n i t for amen oval i s aper t ur e pr esent , t hat conver t s int o a fossae oval is on bi r t h t i me. Right auricle : I n t hi s 2 openi ng pr esent pr e caval and post caval (i n r abbi t 3 openi ngs). On openi ng of pr e cavals eust achi an valve pr esent . Behi nd i t cor onar y si nus opens on i t Thebasi us val ve pr esen t . Ri gh t au r i cl e open s i n t o r i gh t vent r i cl e. L eft aur icle : I n t hi s pul menar y vei ns open by com m on aper t u r e (w i t h ou t v al v e). I t br i n gs oxygenat ed bl ood fr om l ungs i nt o l eft aur i cl e. Ri ght aur i cl e S.A. node or pace mak er or hear t of hear t or node of K eit h and Flack pr esent . I t is made up of car di ac muscl es and st i mul at e hear t beat . I t i s connect ed t o vagus ner ve. Bot h aur i cl es open i nt o vent r i cl es by aur i cul o vent r icular valve. I n r ight side, it is made up of t hr ee flaps : (i ) tr icuspid valve (ii ) lower edges of valve are fixed to papillar y muscles of vent r icle wall by chor dae t endinae fibr es. I n left side it is bicuspid or M it r al valve. Bot h valves ar e one way, i.e. allow blood fr om A t o V not vice ver sa. Some F act s Taller per son has lar ger hear t . H uman hear t is fully developed about 8 week aft er conception (star ts beating after 4 week of conception). Fish have a t wo chamber ed hear t : (i ) Atr ium (ii ) Vent r icle. Bir ds and animals have 4 chamber ed hear t (H ighly developed).

H ear t Beat Th e spont an eou s an d r hyt hm i c cont r act i on and r elaxat ion of hear t t o pump out and r eceive blood is called heart beat . So systole and diastole are collectively called hear t beat .

Biology 3.25

Types H ear t beat is of t wo t ypes. 1. N eurogenic : Cont r oled by ner ve 2. e.g. lower animals – Ar t hr opoda, M ollusca. 2. M yogenic : By own muscles 2. e.g. var t ebr at es including man and r abbit . Some F act s  Neur ogenic par t is less  I t is under cont r ol of upper par t of medulla.  Sympat hat ic acceler at e hear t beat .  Par a sympat hat ic or vagal inhibit hear t beat .  Adr enal, t hyr oxine incr eases hear t beat .  H ear t beat is 210/ minut e in r abbit . 72 / minut e in man. 500 / minut e in mice. 800 / minut e in shr ew. 25 / minut e in elephant . 8 / minut e in belenopt er a.  I n left vent r icle pr essur e is 115– 125 mm H g.  I n r ight vent r icle pr essur e is 25– 30 mm H g.  I n man car diac cycle is of 0.8 sec.  H ear t beat below nor mal is br achae car dia.  H ear t beat above nor mal is t achae car dia. Origin of heart beat. Mammalian hear t is myogenic (myo = muscle, genic = or iginating fr om) i.e. hear t beat or iginat es fr om a muscle (however, it is r egulated by t he ner ves). The hear t beat or iginat es fr om t he sinoat r ial node (SA node)— pace maker, which lies in t he wall of the r ight atrium near opening of the superior vena cava. SA node is a mass of neur omuscular tissue. Sometimes SA node may become damaged or defect ive, so beat does not function properly. This can be remedied by the sur gical gr afting of an art ificial pace maker in the chest of the patient. The ar tificial pace maker stimulates the hear t at r egular inter vals to maintain its beat. Conduction of H eart beat . A n ot h er m ass of n eu r om u scu l ar t i ssu e, A t r i oVentricular node (AV node) is situated in the wall of the r ight atr ium. AV node picks up the wave of contr act ion pr opagated by SA node. A mass of specialized fibr es, bundle of His, or iginat es fr om the AV node. Bundle of H is divides int o t wo br anches ; One going t o each vent r icle. Wit hin t he mycocar dium of t he vent r icles, br anches of bundle of H is divide int o a net wor k of fine fibr es cal l ed Pur k inje fibr es. The bundl e of H i s and t he Pur kinje fibr es convey impulse of cont r act ion fr om AV node t o t he myocar dium of t he vent r icles. Regulat ion of H eart beat . The rat e of hear t beat is regulated by two mechanisms. 1. N ervous regulat ion : Car diac cent r e lies in t he medulla oblongat a of t he br ain.

Car diac cent r e is for med of t wo par t s : (i ) Car dio-inhibit or (ii ) Car dio-acceler ator Car dio-inhibitor is connected with the heart through vagus ner ve (i t car r i es— par asympat het i c ner ve fibres) and car dio acceler ator through sympathetic n er ve f i br es. Sen sor y f i br es ext en d f r om t h e receptors pr esent in the vena cava, aorta and car otid si nuses t o t he car di ovascul ar cent r e of i n t he medul l a obl ongat a. I mpul ses r ecei ved fr om t he aorta and carotid sinuses decrease the hear t rate, whereas impulses fr om the vena cava increase the heart rate. 2. H or monal r egulat ion. A dr en al i n (epi n eph r i n e) an d N or adr en al i n (n or epi n eph r i n e) h or m on es ar e secr eat ed by m edu l l a of t h e adr en al gl an ds. N or adr en al i n acceler at es t he hear t beat under nor mal condit ions while adr enalin does t his funct ion at t he t ime of emer gency. These hor mones dir ect ly influence t he SA node. Thyr oxi ne hor mone secr eat ed by t hyr oi d gl and i ncr eases oxi dat i ve met aboli sm of t he body cel l s. Thi s r equi r es mor e oxygen and t hus i ndi r ect l y i ncr eases hear t beat . H eart rat e Pulse per mi nut e i s cal l ed hear t r at e. H uman hear t beat s about 72 t i mes per mi nut e i n an adul t per son at r est , t hi s is hear t r at e of t hat per son. H ear t r at e i ncr eases dur i ng exer ci se, fever, fear and anger. Because smal l er ani mal s have t he hi gher met abol i c r at e, t hei r hear t r at e i s hi gher t han l ar ger ani mal s. An el ephant has nor mal hear t r at e of about 25 per mi nut e wher eas mouse has a nor mal hear t r at e of sever al hundr eds per mi nut e. Car diac (H ear t ) Out put The amount of bl ood pumped by hear t per mi nut e i s cal led car diac out put . H ear t of a nor mal per son beat s 72 t i m es per m i n u t e an d pu m ps ou t abou t 70 mi ll i l i t r es of bl ood per beat . Thus car di ac out put i s 72  70 or 5040 mi ll i l i t er s per mi nut e, i.e. about 5 l i t r es per mi nut e whi ch i s equi val ent t o t ot al body bl ood vol ume (about 5.5 l i t r es). I n t he day hear t pumps about 7,600 l i t r e of blood.

P u l se

Pul se i s r hyt hmic cont r act i on and r el axat i on i n t he aor t a and i t s mai n ar t er i es. Thus pulse i s a wave of i ncr ease whi ch passes t hr ough ar t er i es as t he l eft vent r i cl e pumps i t s bl ood i nt o t he aor t a. Pul se is a r egul ar jer k of an ar t er y. Ther efor e, it i s al so cal l ed ar t er ial pulse. The pulse r at e i s exact ly same as t he hear t r at e i.e. 72 because an ar ter y pulses ever y t ime t he hear t beat s. Pul se is usual ly t ak en on t he r adial and ul ner ar t er y i n t he wr i st but i t can be t ak en on any ar t er y t hat fl ows near enough t o t he sur face of t he body t o be fel t l ik e neck , t empl es and ankl es.

3.26 Biology

F act ors affect ing Pulse rat e 1. Pulse r at e in childr en is mor e r apid t han in adult s. 2. Pulse r at e is mor e r apid in t he female t han in t he male. 3. When t he per son is st anding up, pulse r at e is mor e r apid t han when he/she is lying down. 4. When any str ong emotion is exper ienced, then pulse r at e is incr eased e.g. anger, exict ement , fear et c. 5. Any excer cie incr eases r at e of t he pulse.

Car di ac Cycl e

The car diac cycle consist s of one hear t beat or one cycle of contraction and r elaxation of the car diacmuscle. Dur ing a hear t beat, ther e is contr action and r elaxation of at r ia and vent r icles. The cont r act ion phase is called systole while r elaxation phase is called diastole. When bot h at r ia and vent r icles ar e in diast olic or r elaxed phase, t his is joint diast ole. Successive st ages of Cardiac cycle 1. At r ial syst ole. The at r i a cont r act due t o a wave of cont r act ion, st imulat ed by t he SA node. The blood is for ced int o t he vent r icles as t he bicuspid and t r icuspid valves ar e open. 2. Beginning of vent r icular syst ole. The vent r icles begin t o cont r act due t o a wave of contr action, stimulated by the AV node. The bicuspid and t r icuspid valves close immediat ely pr oducing par t of t he fir st hear t sound. 3. Complet e vent ricular syst ole. When vent r i cl es compl et e t hei r cont r act i on, t he blood flows int o t he pulmonar y t r unk and aor t a as t he semilunar valves open. 4. Beginning of vent ricular diast ole. The vent r icles r elax and t he semilunar valves ar e closed. This causes second hear t sound. 5. Complet e vent r icular diast ole. The tricuspid and bicuspid valves open when pressure in t he ventr icles falls and blood flows fr om atr ia into t he vent r icles. Cont r act i on of t he hear t does not cause t hi s bl ood fl ow. I t i s due t o t he fact t hat pr essur e wit hin t he r elaxed vent r icles is less t han t hat in t he at r ia and veins. H ear t Sounds The beating hear t produces characteristic sounds which can be hear d by placing the ear against the chest or by usi ng st et hoscope (an i nst r ument which magni fies sounds and conduct s t hem t o ear ). I n a normal person, two sounds ar e pr oduced per heart beat. 1. F irst sound. This is caused par t ly by t he closur e of t he bicuspid and t r icuspid valves and par tly by t he contr action of t he muscles in t he vent r icles.

The fir st sound lubb is low pit ched not ver y loud and of long dur at ion. 2. Second sound. This is caused by the closur e of the semilunar valves and mar ks t he end of vent r icular syst ole. The second sound dup is highly pit ched, louder, shar per and shor t er in dur at ion. Deffects of H eart 1. Blue Baby syndrome (Cyanosis). Due t o per sist ing for amen ovalis in at r ial sept um even aft er bi r t h, t he i mpur e bl ood fr om r i ght aur icles comes t o left aur icle and t hen int o left vent r icle fr om wher e it is supplied t o t he body incr easing t he bluishness of t he body. 2. H ear t block (Cardiac arrest ). I f S-A node becomes defective it cannot gener ate i mpul se pr oper l y and t he hear t beat becoems irr egular or may stop. This is cor r ected by implanting ar tificial electr onic pacemaker in t he chest. 3. M ur mur ing hear t . Due t o defect s in valves inst ead of the nor mal lubdup sound, mur mur i ng sound appear s as t he sympt om. 4. M it r al st enosis. Due t o defect of mit r al valve, passage of blood t hr ough A-V aper t ur e, becomes impr oper. 5. Regur git at ion. Due to improper closure of A-V aperture, some blood dur ing ventr icular systole r eenter s the aur icle. Car diac valves may be defect ive by bir t h or may get damaged due t o st aphyl ococcal i nfect i on (rheumatic fever ). This can be r epair ed or r eplaced sur gically. 6. Angina pect or is.  Due t o bl ock age or t hi ck eni ng of cor onar y vessel’s wall, t her e is inadequat e blood flow t o hear t muscles associ at ed wi t h pai n i n hear t dur ing exer t ion or emot ional t ension.  The bl ock age may be du e t o deposi t i on of ch ol est er ol , sm ok i n g, di abet es or var i ou s ot her fact or s. Si mi l ar defect i s al so cal l ed cor onar y t hr ombosi s when cor onar y vessel s ar e blocked by blood clot s. 7. Brown’s heart disease (Br own’s at rophy). Excess lipid (or lipofuschin) level in blood causes it s deposit ion in cor onar y vessels which at r ophy and causes sympt oms as above. 8. H ooping hear t . Sometimes impr oper r hythms in hear t beat occur, t his is gener ally t he indicat ion of hear t at t ack. 9. M yocar dial infarct ion (H eart at t ack). I nsufficient blood and O2 supply t o par t of car diac muscl es may cause deat h of t hat par t . Fr om damaged hear t muscle cells, cer tain enzymes leak

Biology 3.27

hence t heir pr esence i n plasma is ver y simpl e diagnostic feat ur e for hear t att ack, t hese enzymes ar e lact at e dehydrogenase and t ransaminase. 10. D ext r ocar di a. I f hear t is placed towar ds r ight side in the thoracic cavit y as bor n defect . 11. E ct opia cor dis. Somet i mes hear t becomes l ocat ed out si de t he t hor acic cavit y as bor n defect . Blood Pr essur e  I t is t he r esult of t he sum of (i ) Osmot ic colloidal pr essur e of blood (ii ) Elast ic r ecoil of blood vessel’s wall. (iii ) Pumping for ce of hear t .  First measured by British physiologist Stephan Hales (1733) in car ot id ar t er y of mar e.  I n human it was first measured by Vaivre (1856) using L udwig mer cur y manomet er.  Riva Rocci (1896) discovered sphygmomanometer, the inst r ument t o measur e B.P.  Nor mal r ange of B.P. is 120 / 80 mm H g. (120 m m H g = sy st ol i c pr essu r e; 80 m m H g = diast olic pr essur e)  The di ffer en ces bet ween syst ol i c and di ast ol i c pr essur e is called pulse pr essur e (40 mmH g).  

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I t is is calculated as, B.P. =

80 mmHg  age in year s 50 mmHg  age in year s

I n any age it shouldn’t exceed 150/100 mm H g and shouldn’t be below 80/50 mmH g. The abnor mal r ise in B.P. (hyper t ension) occur s due t o defect of any of t he above t hr ee component s. Wit h incr easing age, r igidit y of t he ar t er ial wall causes high B.P. H ypot ension (low B.P.) may be due t o (i ) Chr onic vasodilat ion (ii ) Anemia (iii ) Blood loss and impr oper hear t act ion. B.P. is slight ly lower in female t han male unt il t he menopause. Dur ing sleep, t he pr essur e falls by 20-30 mmH g and may r each 180-200 dur ing exer cise.

Blood Coagulat ion The pr oper ty of blood to change fr om fluid to gel st ate within a few minutes of its coming in contact with air is called blood coagulation or blood clotting or haemostasis. Ai m To pr event excessive loss of blood fr om an injur y. Blood Clot t ing  The clot begins t o develops in 15 t o 20 seconds but is ful l y for med wi t hi n 3 t o 6 mi nut es i n a nor mal per son.

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I t is a nat ur e's device t o check t he excessive loss of blood fr om an injur y. Cl ot t ing is a bio– chemical r eact ion, explained by H obel. Clot t ing t heor y is given by cascode. Vit amin K is essent ial for clot t ing. Bleeding t ime is 1– 3 minut es. Clot t ing t ime is 2– 6 minut es. Pr ocess of cl ot t i n g i s i n i t i at ed by pl at el et s / Thr ombocyt es. Aft er clot t ing r emaining par t is ser um.

Ant icoagulant s 1. H eparin (hepar = liver ). I t is synt hesized by mast cells of connect ive t issue and l i ver cel l s. I t i s a het er opol ysacchar i de. I t increases effectivity of antithrombin I I I (a – globulin) which inactivat es t hr ombin, so pr event s conver sion of fibr inogen int o fibr in. 2. H ir udin. I t is an anticoagulant pr esent in the saliva of salivar y glands of leech and is mixed with blood of host dur ing it s st or age in it s cr op. 3. War far in. I t is an ant icoagulant of plant or igin, which when gi v en t o a pat i en t , l ow er s t h e f or m at i on of pr ot hr ombin and fact or s VI I , I X and X fr om liver cells by lower ing t he act ivit y of Vit amin K . 4. Sodium oxalat e, sodium cit rat e and EDTA. (Et hylene diamine t et r a acet ic acid) These ar e used as ant icoagulant s in blood banks as these bind Ca++, so t hese ar e called chelating agent s. 5. Chilling of blood. Also delays blood clot t ing as it lower s act ivit y of t he enzymes involved in blood clot t ing. Role of Vit amin– K in Blood Clot ting  Vit amin K is also called anst i-haemor r hagic fact or.  I t is a fat soluble vit amin and is essent ial for t he for mat ion of pr ot hr ombin fr om t he liver.  Deficiency of vitamin K causes hypopr othrombinemia which int er fer es wit h blood clot t ing.  Vitamin K is also synthesized by intest inal bact er ia.

Bl ood Vessel s

These are of t hr ee t ypes. 1. Ar t er ies  Car r y blood away fr om hear t (oxygenat ed blood)  A l l ar t er i es car r y oxygen at ed bl ood except pulmonar y art ery , which car r ies blood fr om Right vent r icle t o lungs. L ar gest ar t er y  Aor t a ,which car r ies blood fr om L eft vent r ical t o ot her par t s of t he body.  Ar t er ies have capacit y t o cont r act and expand on t heir own, t her efor e also for ce t he blood for war d.

3.28 Biology   

They have thicker muscular wall and smaller diameter. Flow of blood in ar t er ies is int er mit t ent flow. Pulmonary artery and Aorta are called great vessels.

2. Veins  All veins car r y deoxygenat ed blood towar ds hear t.  These cannot cont r act and expand on t heir own.  Flow of blood is cont inuous.  These have thinner muscular wall of larger diameter.  These ar e pr ovided with valve to pr otect back flow.  A l l v ei n s car r y deoxy gen at ed bl ood ex cept pulmonar y vein, which car r y blood fr om lungs t o left at r ium.  L ar get vein is (i ) Super ior vena covae, which br ings t he blood fr om head and ar ms. (ii ) I nfer ior vena cavae, which car r ies blood fr om t r unk and legs. 3. Capill ar ies  Th ese ar e hai r l i k e, t hi n bl ood vessel s t h at connect s ar t er ies and veins.  These have no muscular wall.

Lymphat ic Syst em This system retur ns fluid from body tissues to the blood st r eam. I t consist s of (i ) Lymph fluid (ii ) Lymph gland (iii ) Lymph vessels. Lymph fluid  Lymph nodes look like bump with diameter 1 to 25 mm.  Pale fluid r esembling blood plasma.  Cont ains WBC but r ar ely RBC.  Cir culates in lymph vessels and bat hes t he body cell.  Lymph fr om all but upper r ight quar t er of t he body r eaches t he t hor aci cduct , t he l ar gest l ymphat i c vessel which lies along t he fr ont of spine.

E l ect r ocar diogr am (E CG) 

I t is a gr aphic r ecor d of the electr ic cur r ent pr oduced by excitation of the car diac muscles. The instr ument used to r ecor d t he changes is electrocar diogr aph. Waller (1887) first recorded the electrocardiogr am but Einthoven (1903) st udied ECG in details, t her efor e, he got Nobel Pr ize in 1924 for discover y of ECG. He is also consider ed "fat her of elect rocardiography".

Pacem ak er SA node is called natural pacemaker . When SA node does not send impulses to the atria, ventricles fail to receive atrial impulses. The conducting system of the heart is disrupted. I n such patients normal heart beat can be restored and maintained with an artificial pacemaker .

Tempor ar y Pacemaker I t is used in emergency such as arr ythmia contr ol (e.g. br adycar dia – slow hear t beat). Electr odes of temporary pacemaker is introduced from jugular, sub clavian and femoral veins as well as from veins of the upper extremity. Per manent Pacemaker I t i s used i n at r i ovent r i cul ar (AV) bl ock , SA node dysfunct i on, et c. I n t he pat i ent s of St ok es adams syndr ome (vent r icular escape or vent r icular asyst ole) at r ial impulse suddenly fails t o be t r ansmit t ed t o t he vent r icles. I n such pat ient s, per manent pacemaker is implanted. Ar t ificial Pacemak er I t is an artificial electr onic device which r egular ly sends small amount of elect r ical char ges t hat st imulat e t he hear t . I t consist s of (i ) a pulse – gener at or cont ai ning cel l (sol i d st at e lit hium cell) t o pr oduce elect r ical impulse, (ii ) lead in t he for m of a wir e which t r ansmit s t he impulse and (iii ) elect r ode, which is connect ed t o t he por t ion of t he hear t wher e impulse is t o be t r ansmit t ed.

Repr oduct i ve Syst em 1. M ale Repr oduct ive Syst em H uman male r epr oduct ive syst em is composed of a pair of testes, genital duct s, sever al accessor y glands and penis.

Test es (Test icles).  Test is is pr imar y male r epr oduct ive or gan. I t lies in small sac-like muscular st r uct ur e out side t he abdominal cavit y called scr ot um.  I n adult males, each testis is a small, pinkish and oval.  2 in number (Diar chic)  4– 5 cms long, 2.5 cm wide and 3 cm t hick.  M esoder mal.  I n em br yon i c st age at t ach ed t o k i dn ey by mesor chium membr ane.  I n embr yo at 7t h mont h of division descend down in scr ot el sac. Condit ion is ext r a abdominal, due t o it 3C less t emper at ur e is available, helpful in sper mat ogenesis.

Biology 3.29     

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Bot h scr ot el sac int er nally separ at ed by sept um scr ot i. Ext er nal by a scar -like r aphe. I n scr ot el sac fat layer i s absent , i nvol unt ar y muscles ar e der t os t unic pr esent . Lumen of scr otel sac is vaginal coelome, in it testes pr esent . Vaginal coelome is connect ed t o main coelome by inguinal canal. I n some cases t est es do not descend down in sac, condition is cr yptor chidism such animals ar e high flanker s (sper m ar e not pr oduced) Test es is at t ached t o pr e posit ion by sper mat ic cor d (gonadial ar t er y, gonadial vein, ner ve and connect ive t issue pr esent ). Test is is at t ached t o wall of scr ot el sac by elast ic st r uct ur e i.e. guber naculum cor dis. U ni t of t est i s i s semi ni fer ous t ubul e wh er e sper ms are manufactur ed by sper matogenesis. All t ubul es for m a cor d of 200– 400 met er. These t ubules ar e spr eaded in connect ive t issue. I n t i ssue i nt er st i t i al cel l s or cel l s of L eydi ng pr esent . T hese cel l s secr et e t est ost er on e or andr ogen hor mones t o cont r ol secondar y sexual char act er s. Out side each seminifer ous t ubule, t unica pr opr ia pr esen t . I n si de i t ger m i n al epi t h el i u m or basement membrane present. Made up of cuboidal epit helium. Epit helium consist s of P.G.C. Pr imor dial ger m cells and few ser t oli cells. P.G.C. ar e endoder mal in nat ur e. P.G.C. ar e r esponsible for sper mat ogenesis. Different stages of formation of sperms can be seen as Sper matogonia (i ) Pr imar y sper mat ocyt e (ii ) Secondar y sper mat ocyt e ar e (iii ) Sper matid (iv) Sper ms.

Seminal vesicles.  I t is sit uat ed behind t he bladder.  About 70% of the semen in seminal fluid pr oduced by seminal vesicles. Penis.  I nt r omit t ent or gan.  Er ect ile, it s apex is penis glanse.  Cover ed by pr epuce (par t of skin).  Main function is insemination (to tr ansfer sper ms in vagina in deep).  I n T.S. 3 cor ds pr esent , 2 on dor sal side cor pus cover nosus, 1 on ventr al side cor pus spongiosum.  Bet ween cor pus caver nosus sept um pr esent .  Thr ough cor pus spongiosum ur et hr a passes.  Wal l of cor pus spongi osum ar e spongy due t o t r abeculae, having power of collect ion of blood at t he t ime of er ect ion.  Penial ar t er y supply blood. Semen.  Sper m s an d secr et i on of accesor y gl an ds collect ively called seminal fluid or semen .  I t is milky, semi-solid in natur e having par ticular smel l.  pH : 7.35 – 7.5  Specific gr avit y : 1.028 2. F emale Reproduct ive Syst em Female Reproductive Systems

Main organ Ovaries

Accessory organs Fallopian tube Uterus Vagina Ext. genitilia Bartholins gland Breasts

Pr ost at e Gland.  I t i s si t u at ed ar ou n d t h e f i r st par t of t h e ur et hr a.  Englar gement of pr ost at e r esult s in pr ost at it is i.e. ur inat ion is difficult or impossible. U r et hr a.  I t is a thick walled muscular duct and is a common passage for bot h ur ine and semen.  Ur et hr a t r aver se and open at t he t ip of penis.  F or pr odu ct i on of sper m s t est es r equ i r e – t emper atur e 1°C less t han the body temper at ur e. Types of sperms: Sper ms ar e of t wo t ypes (1) (2) L ar ge sper ms Small sper ms car r y X cr omosomes car r y Y cr omosomes rd 23 pair of chromosomes is called sex chromosome.

Ovar ies  2 in number (didelphic).  whit e / pinkish, almond like.  3 cm long, 2 cm wide, 1 cm t hick.  L i e i n t he l ower par t of abdomen at t ached t o dor sal wall by mesovar ium.

3.30 Biology  

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At t ached t o ut er us by ovar ian ligament . I n wall of ovar y viscer al per it onium, ger minal epit helium, and t unica albuginea pr esent . Ger minal epit helium is made up of P.G.C. Gr ound par t of ovar y is st r oma differ nt iat ed in t o out er cor t ex and inner medulla. M esoder mal in or igin. P.G.C. of ger minal epit helium is endoder mal in or igin. Fir st of all fr om P.G.C. follicle cells ar e for med. Fuse to form egg nest. One is developing ovum, r est cells ar e destr oyed t o give nour ishment t o developing ovum. Destr oyed cells are atratic cells. This phenomenon is at r asia. Fir st it is single layer ed called pr imar y follicle. Soon it becomes double layer ed, i.e. secondar y follicle. I t is soon conver t ed int o gr affian follicles or mat ur e follicle. I t 's for mat ion is folliculogenesis under cont r ol of FSH and L H .

Graffian F ollicle.  I t appear s as knob or st igma in medulla.  I t is r ound, discover ed by Gr aff .  I t is cover ed by t wo layer s : (i ) Thecae ext er na (ii ) Theca inter na : Fr om t heca inter na, est r ogen i s secr et ed t o con t r ol secon dar y sexu al char at er s i n femal e. I nsi de t heca i nt er na, membr ana gr anulosa is pr esent . 

Secondar y oocyt e is at t ached t o cover ing layer by gr oup of cells, i.e. discus pr oliger ous. The point of discus pr oliger ous close t o secondar y oocyt e is H ill of follicular cells or cumulus oophor us. Secondar y oocyt e i s cover ed by zona pell uci da (pr i mar y egg membr ane) and cor ona r adi at a (secondar y egg memebr ane).

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Cavit y of follicle is ant r um of H imor or follicular cavit y filled wit h follicular fluid. Gr affian's follicle is r ept ur ed, secondar y oocyt e come out of ovar y fr om st i gma poi nt fi r st i n coelome. I t is ovulation. This t ime L.H. is secr eted mor e. After r eptur ing rest part of gr affian's follicle which i s yel l ow col our ed i s cal l ed cor pus lut eum . I t consist s of clot t ing, fibr in, lut ein cells. L .H . i s helpful in for mat i on of cor pus l ut eum. Cor pus lut eum secr etes pr ogest eron and r elaxin . Pr ogest er on is helpful in implant at ion, t o st op ovulat ion dur ing gest at ion per iod. Relaxin is helpful t o r elax pelvic muscles dur ing parturition .

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I f fer t ilisation occur s, t hen cor pus luteum var ium is used. I f not t hen cor pus lut eum spor er um is used. Aft er for mat i on of pl acent a, cor pus l ut eum i s degener ated and conver ted into white scar corpus albicanse.

F allopian T ube.  Fer t ilisat ion occur in it . 10– 12 cm long muscular t ube.  I t is suppor t ed by double fold of per it onium.  I t shows four r egions : (i ) I nfundibulum : Br oad, funnel shaped pr oximal par t .  I t mar gin bear s fimbr iae. I n funnel ost ium aper t ur e pr esent .  Fr om it eggs ent er int o duct . (ii ) Ampulla : L ong, wide, t hin walled. (iii ) I st hmus : Ver y shor t , nar r ow t hick wall ed str aight par t . (iv) Ut er ine part : I t is nar r ow, communicate with ut er ine cavit y. U terus (Womb).  L ar ge, pyr ifor m, highly elast ic.  Development of embr yo t akes place in it .  I t i s l ocat ed above and behi nd t he ur i nnar y bladder.  Att ached t o body wall by mesometr ium ligament .  I n nullipar ous woman, it is about 8 cm long 5 cm wide 2 cm thick, weighting 10– 15 gms, some what lar ger in mult ipar ous woman.  I t shows four r egions : (i ) Upper wide dom shaped fundus t hat r eceives fallopian t ubes. (ii ) Cor nuae t he upper cor ner s wher e oviduct s ent er int o ut er us. (iii ) Middle large body or cor pus which is the main par t. (iv) Lower nar r ow cer vix that pr ojects into vagina.  Cer vix communicat es wit h ut er us by int er nal os. and wit h vagina by ext er nal os.  Cavit y of cer vix bet ween ext er nal and int er nal os. is cer vical canal. I t is 2.5 cm long.  Cer vi x i s composed of t he bi ggest and most power ful sphinct or muscles i n t he body. I t i s st r ong enough t o hold about 7 k g of foet us fl ui d.  I n t he wall of ut er us out er layer is per imet r ium, middle myomet r ium and inner endomet r ium.  M yomet r ium consist s of inner and out er of longe muscles and middl e cir cular muscl es. L ongest involunt ar y muscles pr esent her e.  The cavit y of ut er us can be expanded 500 t imes dur ing pr egnancy fr om 10 cm 3 t o 5000 cm 3.

Biology 3.31

Vagina.  M edian, elast ic, muscular t ube 7.5 cm long. Open int o vest ibule by vaginal or ffice.  Uter us opens into an elast ic muscular tube called vagina.  Vagina is lined by a stratified squamous epithelium wit hout any glands.  D u r i n g r epr odu ct i v e l i f e,v agi n a con t ai n s Lactobacillus acidophilus which keeps vaginal pH bet w een 4.9 an d 3.5 by pr odu ci n g l act i c acid fr om glycogen.  The lining for ms vaginal r ugae.  Space bet ween vaginal wall and cer vix is for nix . Ext ernal Genit alia.  Vest ibule is cover ed by t wo pair s of lips : (i ) L abia major is (ii ) L abia minor is  At t he ant er ior junct ion of labia minor is, a small er ect ile clit or is pr esent , homologous t o penis.  A m em br an ou s f ol d f ou r ch et t e con ect s t h e post er ior ends of t he labia minor is.  Ar ea bet ween four chet t e and anus is per inium .  Ur et hr a and vagina open by separ at e aper t ur e, t he ur et hr al and vaginal or ffice int o vest ibule.  Vagi nal or ffice i s cover ed by a membr ane, i.e. hymen.  A slit in hymen allows menst r ual flow to pass out.  A fl eshy el evat i on above t he l abi a maj or i s i s monsvenesis or mons pubis. Ber t holian Glands.  1 pair in number.  Analogous t o Cowper 's gland of male. Br east s.  1 pai r, t hei r devel opment i s under cont r ol of pit uitar y.  Nipples pr esent (absent in pr ot ot heir a).  Ar ound ni pple ar eola mammar y pr esent , i n it melanin is maximum.  M ilk is synthesized by lectogenesis under contr ol of pr olect in.  M ilk is secr et ed under cont r ol of oxyt ocin.  Ear ly milk is r ich in miner als, i.e. colust r um .  M odified sweat gland.  Glands open on t he nipples by lact ifer ous duct s.  Just under nipple, lact ifer ous sinuses pr esent t o st or e milk. Onset of Pubert y in Female. At t ains at t he age of 13 by est r ogen hor mone. I t includes :  Gr owt h of br east s  Gr owt h of ext er nal genit ilia

Br oading of pelvis  Gr owt h of pubic hair  I ncr ease in sub cut aneous fat  St ar t ing of M .C., i.e. menar ch Disorder s of female r eproduct ive syst em.  St erility  Menstr ual irr egularit y : This may be amenor r hea (absen ce of M .C.) or ex cessi v e bl eedi n g (dysmenor r hea). 

Phases of menst rual cycle : M enst r ual cycle consist s of t hr ee phases : ( i ) F ollicular (Pr olifer at ing) phase : I t last s for about 14 days. I t compr ises following event s :  L evel of FSH is incr eased in t he blood ear ly in t he cycle.  FSH stimulates gr owth of the selected pr imar y ovar ian follicle.  FSH st imulat es follicle t o secr et e est r adiol.  E st r adi ol i n h i bi t secr et i on of F SH an d st imulat e secr et ion of L H .  L H induces gr affian follicle t o bur st and eject its egg in to fallopian tube. I t occurs after about 14 days.  Est r adiol st imult es t he ut er us t o pr epar e for implant at ion and nour ishment of foet us.  Lining of fallopian tube becomes t hick t o move ova int o ut er us. ( ii ) Luteal or secretory phase : This phase last s for 10 days. I t involves following event s :  L H st imulat es gr affian follicle t o conver t int o cor pus lut eum.  Cor pus lut eum secr et es pr ogest er on and lit t le amount of est r adiol.  Pr ogester on inhibits the r elease of FSH so that it may not develop additional follicle and eggs. ( iii ) M enst rual or Bleeding cycle : I t last s about for 4 days. I t involves following event s :  I f fer t ilization does not occur, secondar y oocyte under goes aut olysis.  I n some wall of ut er us secr et es a lut eolysine. I t r eaches t o ovar y wher e it causes aut olysis of cor pus lut eum.  St r atum functionalis of endomatr ium is given out wit h bleeding.  M enst r ual flow cont inues for 3– 5 days.  Endomatr ium tissue, unfer tilised egg and 500– 600 ml of blood ar e lost dur ing t his per iod.  M enst ur al i s oft en descr i bed as funer al of unfer t ilized egg or weeping of ut er us for t he last ovum.

3.32 Biology

Artificial I nsemination.  I t is a t echnique t o make a female pr egnant by ar t ificially int r oducing semen int o t he vagina.  Advant ages : (i ) Semen of good qualit y male is used. (ii ) Pr eser ved semen can be t r asnpor t ed t o distant place.  Ar t ificial inseminat ion can be used in human.  I t may be AI H (Ar t ificial insesment husband).  I t may be AI D (Ar t ificial inseminat ion donor ).  Childr en pr oduced by ar t ificial inseminat ion ar e called tube babies. Progest erone – Sex hormone  I t st imulat es ut er us t o develop blood capillar ies t o r eceive and nour ish, t he fer t ilized ovum and also suppr esses, r elease of fur t her pr oduct ion of ova.  Pr ogest er on e con t r ol s l ast t w o w eek s of menst r uat ion cycle.  Secr et ion of pr ogest er ones is st opped if ovum is not fer tilized.  T he dest r oyed capi l l ar i es of ut er u s r esul t s i n menst r uat ion blood. E est r ogen  I t is cont inuously secr et ed.  I t cont r ols fir st t wo weeks of menst r ual cycle.

Sex D iffer ent iat ion M echani sm

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Pr imit ive gonads ar e ident ical in bot h sexes upt o 6 weeks of gest at ion pr essur e. This gonad is called bipot ent ial pr imor dial gonad. I t is differ ent iat ed int o cor t ex and medulla. Formation and devlopment of gonad is gonadogenesis. I n mal e, devl opement of t est es i s cont r ol l ed by Y-chromosome. Test icular differ ent iat ion begins in t he 7t h week of gest at ion. Differ ent iat ion of ovar ies occur lat er in female in compar ision t o male or t est es. I t begins in t he 11t h or 12t h w eek of gest at i on u n der con t r ol of XX chr omosomes. Test es descend down in scr ot um in 7t h – 8t h mont h. Ext er nal genit ilia appear in male by t he 5t h mont h. Sper m

Ovum (egg)

1. X and Y chr omosomes.

1.Only X chr omosomes.

2. Sur vive 2– 3 days.

2. Survive for only 24 hrs.

3. Mobile, i.e. capable of moving on its own.

3.I mmobile.

4. Pr oduced in lar ge

4.Produce very few in life time numbers equal to (500 to 600 in whole life).

X

+

Y

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XY

M ale

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XX

Femal e

Fir st 22 pair s of chr omosomes bet ween M and F ar e same but 23r d pair is differ ent . M

F

X

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Y

X

X

Fat her det er mines t he sex. When an ovum is fertilized by sperm +

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F er t i l i sat i on

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Sex of Child

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Zygot e

30-hr s aft er fer t i l i zat i on, zygot e under goes cel l division (2 divisions) Embr oy

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I t is called embr yo, unt il t he human char act er ist ics ar e shown. When human char act er ist ics ar e shown, it is called foetus (Fr om 2nd and 3rd month of pr egnancy till bir th it is called foet us). The cont r act i on of ut er us aft er deliver y is call ed involution.

Pl acent a

A common t issue of foet us and mot her (ut er us) which is physical, physiological and endocr inal connect ion is called placenta. F unct ions 1. To pr ovide nut r ient s t o foet us. 2. To r emove met abolic wast e. 3. H elpful in r espir at ion. 4. St or e food in t he for m of glycogen. 5. I t pr oduces est r ogen, pr gest er on and r elaxin. N ote : A placent a consist s of 6 t issues. Thr ee fr om t he foet al walls and 3 fr om mot her 's ut er us. F oet al t issues  Foet al endomet r ium, i.e. blood capillar ies.  Connect ive t issue.  Chor ionic epit helium M at er nal t issue  M ucous epit helium of ut er ine walls  Connect ive t issue  M at er nal endot helial

N er vous Syst em H uman ner vous syst em consist s of t hr ee cont r ast ing funct ional subsystems.

Biology 3.33

1. Cent r al nervous syst em  I t is is called main swit ch boar d .  All t he ner ves ar e for med befor e t he bir t h.  Ner ves gr ow up t o t he age of 5 year s and st op by t he age 19– 20 year s.  Consumes 20% of oxygen i nhal ed and 20% of glucose int ake. Parts of central nervous System Br ain

Spinal cord  Medul a  Cerebrum

2. Aut onomic ner vous syst em  Oldest par t of ner vous syst em.  Contr ols basic activities of life like hunger, thir st, hear t beat , br eat hing, feat , anger, hot , cold et c.  I t cont ains 3 par t s : (i ) Medulla oblongata : I t contr ols br eathing and hear t beat s. (ii ) H ypot halamus : I t cont r ols - t emper at ur e, sleep, hunger, t hir st et c. (iii ) Thalamus:I t cont r ols level of concent r at ion.

C.S.F. (Cer ebr ospinal fluid).  Chor oid plexus is r esponsible t o for m C.S.F.  I t is for med 20 ml/hr mainly by chor oid plexus and ependyma of vent r icles.  I t comes out of medulla oblongat a by for amen of magendie and for amina of lushka.  I t is slight ly alkaline and has specific gr avit y 1.005.  Dir ect i on of movement of C.S.F i n br ai n is fr om ant er ior t o post er ior. St r uct ur e BRAIN

Fore Brain or Prosencephalon

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Cell body

Axon

M at t er  The br ain is composed of outer gr ey matter and inner whit e mat t er in t he ar eas of cer ebr um, r oof of midbr ain and cer ebellum.  I n t he r est , gr ey mat t er is inner and t he olfact or y lobes ar e made up of gr ey mat t er only. M eninges (Singular – M eninx)  The cover ings of br ain ar e called meninges. These ar e t hr ee in number. These menings ar e int er nally lined by squamous epit helium

2. Medulla oblongata or Myelencephalon

F or e Br ai n 1. Olfact or y L obes 

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1 pair, ver y small, solid, club shaped, separ at ed fr om each ot her. Fully cover ed by cer ebr al hemispher es on dor sal side, clear ly visible only in vent r al view. Each olfact or y lobes consist s of (i ) Ant er ior olfact or y bulb (ii ) Post er ior olfact or y t r act . I n sh ar k an d dog, ol f act or y l obes ar e wel l developed. Piscian br ain wit h lar ge olfact or y lobes is called nose br ain. Their funct ion is t o r ecieve sense of smell.

M i d Br ain Optics lobes ar e found in it which ar e four in number, t her efor e t hey ar e called Cor por a quadr igemina. I n fr og t hey ar e t wo in number, hence called Cor por a bigemina.  They ar e solid and do not have opt ocoel.  2 super ior coculi and 2 infer ior coculi ar e pr esent .  Opt i c l obes ar e at t ached t o each ot her by cr ur a cerebri.  Funct ions : (i ) Super i or cocul l i ar e concer ned wi t h sense of sight . (ii ) I nfer ior coculi ar e concer ned wit h hear ing. N ote : Br ain wit h lar ge opt ic lobes is avian br ain. 

Dendrites

1.Cerebellum or metencephalon

3. Diencephalon or Thalamencephalon

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Part of Neuron

1.Optic Lobes

2. Cerebrum or Telencephalon

Br ai n (E ncephel on) The br ain is sit uat ed in t he cr anial cavit y of skull. I t s shape i s oval and col our i s l i ght yel l ow or cr eamish yellow. I t weighs 1200 - 1400 gms (98% of C.N.S) and has 100 billion neur ons.

Hind Brain or Rhombencephalon

1. Olfactory lobe

3. Per ipher al ner vous syst em.  Car r ies all t he messages sent bet ween cent r al ner vous syst em and r est of t he body.  I t consist s of 12 pair s of ner ves t hat or iginat e in the br ain plus 31 pair s of ner ves of the spinal cord. These ser ve as telephone wir e that car ry message to and from every receptor and effector in the body. 

Mid Brain or Mesen cephalon

3.34 Biology

H ind Br ain or Rhombencephal on 1. Cer ebellum or M et encephalon  I t is a small par t which lies below t he post er ior par t of cer ebr u m an d abov e t h e m edu l l a oblongat a.  I t has a median lobe called vermis and t wo later al lobes called cer ebellar hemispher es.  There are three Arbor vit ae or t ree of life (made of whit e mat t er ) in t he cer ebellum, of which t wo ar e pr esent in t he t wo lat er al lobes and one is pr esent in t he median lobe.  I n cer ebellum flask shaped cells ar e pur kenje's cell s.  I t is sur r ounded by cer ebr al cor t ex.  Funct ions : I t s funct ions ar e equilibr ium and coor dinat ion of movement of muscles in walking. N ote : Cer ebellum is mor e developed in aves. 2. M edulla oblongat a or M yelencephalon  I t is t he lower most par t of t he hind br ain and is conical in appear ance.  I t s lumen is called met acoel or 4t h vent r icle.  An oval mass pons var ol i l i es above medul a oblongat a. I t connect s medulla t o higher br ain cent r es. I t cont r ols r espir at ion.  F unct ions : (i ) Car diac cent r e (ii ) Respir at or y cent r e. (iii ) Regulator of vasodilation and vasoconstr iction. (iv) Regulator of per ist alsis and deglutinat ion. (v) Regulat or of glandular secr et ion. Some I mpor t ant F act s  I n br ain glut amic acid met abolism is common  I n latimar ia br ain's size is smallest in r atio of body size.  I n man 500 ml C.S.F. per day is absor bed.  I n Br ain 750 ml. blood/ minut e flows.  I n br ain's vent r icle 125 ml C.S.F. is pr esent out of t ot al 150 ml.  Vausois's sphinctor connects optic lobes to cerebellum.  Bar r body is pr esent in t he neur ons in females.  Aft er r emoving cer ebr al hemispher e deat h t akes place.  Amnesia is a condit ion in which memor y is lost par t ially or complet ely.  L oss of sensat ion is called anaest hesia.  Neur in pr ot ein is at t ached t o t he inner sur face of br ain.  Br ain st em – medulla, pons var oli & mid br ain collect ively for m br ain st em. A diffuse net wor k of ner ve cel l s ext end t hr ough i t . I t i s r et i cul ar activating syst em (RAS)

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Amygdal a i s found on t he vent r al sur face of par acoel. I t is r elat ed t o r ecent and new memor y. L imbic syst em : Limbic lobe (area of temporal lobes) + hippocampus + hypot halamus + par t of t halamus + amygdala for m it . I t i s concer ned wi t h behavi or emot i on, r ecent memor y, food habit and sexual behaviour. Medulla oblongat a comes out of foramen magnum in t he for m of spinal cor d.

Spinal cor d (M yelon)  

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I t is a par t of t he CNS which is 45 cms long ( in women 43 cms) and weighs 35 gms. I t or iginat es fr om lower par t of medulla oblongat a and is pr esent on dor sal side in t he neur al canal of ver t ebr al column up t o 1st lumbar ver t ebr a. I t is long, cylindrical, convex dorsally and flat ventrally. I t s last par t is conical, i.e. conus t er minalis which in t he end becomes t hr ead like, i.e. filum t er minalis. I t is having br achial and lumbo sacr al swelling at t he point of or igin of for e leg and hind leg.

M eninges  The same t hr ee meninges cover t he spinal cor d also.  Similar to br ain, epidur al space is also pr esent which is filled wit h fat and connect ive t issue. M at t er I n cont r ast t o br ain, gr ey mat t er is found on t he inner side and whit e mat t er is found on t he out er side. C.S.F. Dir ect ion of cir culat ion of C.S.F is fr om post er ior end t o ant er ior end. St r uct ur e  Spinal cor d is hollow, it s lumen is cent r al canal or neur ocoel filled wit h C.S.F.  Canal is lined by ependyma of ependymal cells.  Gr ey mat t er is but t er fly or H -shaped.  Ther e ar e t hr ee pair s of cor nuae or hor ns : (i ) dorsal (ii ) ventr al (iii ) lat er al hor ns.  Dor sal, vent r al and lat er al funiculli lie close t o t he whit e mat t er ar ea.  Dor sal hor n for ms dor sal ar ch while vent r al hor n for ms vent r al ar ch.  On dor sal side dor sal sulcus wit h sept um is pr esent and on vent r al side vent r al fissur e is pr esent .  Funct ions : I t conduct s i mpul ses t o and fr om t he br ai n and cont r ols r eflex act ion .

Cr anial N er ves 

Ther e ar e 12 pair s of cr anial ner ves (t ot al weight 12 gms) in Amniotes and only 10 pairs in Anamniotes.

Biology 3.35 

Nat ur e of t hese ner ves may be sensor y or mot or or mixed. They can t he ident ified as follows: 1, 2, 8, – sensor y 3, 4, 6, 11, 12 – mot or 5, 7, 9, 10 – mixed 3, 7, 9, 10t h at t ached t o ANS also.

Spinal N er ves 

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These ar e 31 pair s in number and all ar e of mixed t ype. T hey wei gh a t ot al of 150 gms. and ar e composed of medullat ed ner ve fibr es. At t he base of or i gi n of spi nal ner ves, gl and of swammar dam or calcar ious ganglion is pr esent . Classificat ion 1. Cer vical – 8 pair s 4. Sacr al – 5 pair s 2. Thor acic – 12 pair s 5. Coccygeal – 1 pair 3. L umber – 5 pair s Each spinal ner ve or iginat es fr om spinal cor d fr om dor sal r oot and vent r al r oot . I n dor sal r oot , r oot ganglion is pr esent . Bot h r oot s fuse in neur al canal t o for m spinal ner ve. I t comes out of ver tebr al column fr om inter ver tebr al for amen. Out side vertebral column, each spinal nerve is divided into t hree par t s 1. Dor sal br anch or Ramus dor salis 2. Vent r al br anch or Ramus vent r alis 3. Viscer al br anch or Ramus communicans I n r abbit 37 pair s spinal ner ves pr esent : 1. Cer vical – 8 pair s 4. Sacr al – 4 pair s 2. Thor acic – 12 pair s 5. Caudal – 6 pair s 3. L umber – 7 pair s I n rabbit caudal nerves combine with filum terminalis t o for m cauda equina.

A.N .S.  The t er m A.N.S was given by L angley.  I t i s par t i al l y i ndependent syst em cont r ol l ed by hypot halamus and it contr ols involuntar y actions as hear t beat , br eat hi ng, homeost asi s, per i st al si s, secr et i on of gl ands. So a gangl i oni c syst em t hat cont r ols involunt ar y act ion is A.N.S.  I nt er nal or gans r eceive fibr es fr om t wo sour ces : 1. Sensor y 2. M ot or  Nerve of sensory region originates from dorsal root and nerve of motor region is originated from ventral root.  Viscer al mot or fibr es ar e divided int o t wo par t s (i ) Pr eganglionic : M edullat ed (whit e mat t er ) (ii ) Post ganglionic : Non– M edullat ed (gr ey mat t er )

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I mmune Syst em

Syst em of body which pr event t he body fr om diseases and cancer is called immune syst em. (I mmune = Exempt or Fr eedom) Types of I mmunit y I mmunit y is of t wo t ypes 1. Congenital immunity or I nnat e immunit y or Nonspecific immunit y. 2. Acquired immunity or Adaptive or specific immunity. Accuir ed immunit y is of t wo t ypes ( i ) Passive acquir ed immunit y. When readymade antibodies are given in immuno supressive individual this is called passive immunity. I n passive immunit y t r ansfer of ant ibodies t o a r ecipient in a r eady– made for m. ( ii ) Act ive acquir ed immunit y.  Thi s i mmuni t y devel op aft er i nfect i on or vaccinat ion.  I t is gener at ed due t o pr evious cont r act ion of disease.  I n many cases it is life long (against measles).  I t is involved in act ive funct ioning of per sons own immune system-leading to the synt hesis of ant ibodies.

Ant igen or Agglut inogen 

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Reflex Act i on 

I t is funct ional unit of C.N.S. which was discover ed by M ar shal H all and defined by Best and Taylor.

I t is sudden, immediat e involunt ar y act ion against ext er nal st imuli. I n man it is poly– synapt ic. I f we pr ick needle on skin, somatic sensor y ner ves br i ng i mpul ses t o gr ey mat t er. H er e t hese ar e analysed and or der is given to muscle by motor ner ve. The path fr om which impulse is passed is reflex arch . Minimum t ime is consumed in it since it is not under cont r ol of br ain. Some common Reflexact ions : Coughi ng, yawni ng, sneezi ng, k nee-j er k r efl ex, blinking of eyes, scr at ch r eflex, flow of bile fr om gall bladder, per ist alsis, hear t beat . Significance : (i ) I t enables animal t o r espond immediately t o t he har mful st imuli so t hat no har m is caused it . (ii ) I t gives mor e t ime t o br ain t o wor k.

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For eign subst ances like pr ot ein or polysacchar ide which stimulat es pr oduction of ant ibodies ar e called antigen. Toxin of pat hogen, white of egg, feather s, fr uit , meat, dr u g i n du ces t h e i m m u n e sy st em t o pr odu ce antibodies. Si t e over t h e ant i gens t h at ar e r ecogn i sed by ant ibodies and r ecept or s found on B and T cells ar e called ant igen deter minants/epit opes. An ant igen may have one to sever al t ypes of ant igen det er minants.

3.36 Biology 

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Epitopes can bined specifically to antigen binding site (par at ope) of t he ant i body mol ecul e and T cel l r ecept or s. Epit opes ar e smallest unit of ant igenicit y. Each det er minant can st imulat e t he for mat ion of antibody or effector cell. Thus a pur e pr ot ein antigen may give r ise to many distinct antibodies and effector cell s.

Ant i bodi es 

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RBCs of all A, B, O blood gr oup possess a common H ant igen which is pr ecur sor for for mat ion of A and B antigens. I t i s a com pl ex gl y copr ot ei n secr eat ed by B I ymphocyt es in r esponse t o an ant igen. I t is also called agglut inin. These ar e complex glycopr ot ein molecule made up of 4 polypept ide chain : Two light and t wo heavy chain. These t wo chain held t oget her by disulphide bond in shape of Y– molecule. Two t op t ips of t his molecule bind wit h ant igen like lock and key fashion and make ant i gen-ant ibody complex. Ant ibodies for ms t hir d line of defence. Ant i bodies ar e pr ot einaceous subst ances occur in blood plasma and pr oduced by immune syst em t o over come t oxic effect of ant igens. Each ant i body has at l east t wo heavy and l i ght polypeptide chains. H eavy chain is also called H chain wher eas light chain is also called L chain which may be eit her L ambda or K appa t ype. H eavy chain consist s of 440 amino acids wher eas light chain consist s of 220 amino acids. Ther e is a var iable por t ion for binding t o ant igen t hr ough el ect r ost at i c i nt er act i on and a const ant portion that deter mines its adher ence and diffusivity. Var iable r egion also called V– region / antigen binding fr agment / Fab. Const ant r egi on al so cal l ed const ant fr agment / cr yst alline fr agment / Fc. Fc lacks t he abilit y t o bind t o ant igen and can be cr ystallized. A n t i bodi es ar e al so cal l ed i m m u n ogl obu l i n s / gammaglobulins. Ant ibodi es ar e pr oduced in r esponse t o ant igenic stimulat ion. Antibodies ar e pr oduced by B lymphocyte and plasma cell s.

Cells of I mmune Syst em 

They ar e lymphocyt es and ant igen pr esent ing cells like macr ophages.

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

A healt hy human has about a t r illion lymphocyt es. Lymphocyt es ar e of t wo t ypes : (i ) T– lymphocyt es (T– cells) (ii ) B– lymphocyt es (B– cells). Both of these develop in bone mar r ow (in foetus fr om yolk cells and t hen fr om liver spleen complex) fr om lymphat ic st em cells. For mat ion of lymphocyt es is called haemat opoiesis. Some of young lymphocytes migr ate into t hymus for processing. They are called T– lymphocytes afterwards they pass on to all t he lymphoid tissues of the body. The ot her t ypes of l ymphocyt es r emai n i n bone mar r ow and get pr ocessed t her e. They ar e called B– lymphocyt es because in bir ds t hey ar e pr ocessed in l ymphoi d di ver t i cul um of cl oaca cal l ed bur sa of Fabr icius. Aft er pr ocessing B– lymphocyt es migr at e t o all t he lymphoid t issues of body.

Ant i body F unct i ons 1. N eut r alisat ion Some antibodies function as antitoxins and neutralise the toxins pr oduced by pat hogens/for eign chemicals. 2. Agglut i nat i on Ant ibodies called agglut inins cause immobilisat ion and clumping of antigens (pr ecipitation) and antigen cont aining pat hogens. 3. Opsoni sat ion/Adher ence Ant ibodies called opsonins (I gG) at t ach t hemselves t o sur faces of ant i gen cont ai ni ng cel l s, so as t o r ecognised by phagocyt es. 4. Complement mediat ed cell lysis Ant igen cont aining cells ar e per for ated by enzymes pr oduced with the help of lysin antibodies (I gM– I gG) and cyt ot oxic T– cells. 5. Phagocyt osi s The l ysed i mmobi l i sed cl umped pat hogens ar e engulfed by phagocyt es.

Vaccinat ion and I mmunisat ion 





I t is t he phenomenon of incr easing specific ant ibody pr oduct ion and development of memor y B– and T– cells against t he pot ent ial at t ack of a pat hogen. I t is car r ied out t hr ough vaccinat ion and inject ion of ant iser um. When an i mmuni sed per son i s at t ack ed by t he pat hogen, t hen exi st i ng ant i bodi es i mmedi at el y at t ack t he ant igen while memor y T and B cells give r ise to a massive cr op of lymphocyt es and anitbodies.

Vacci nat i on Vacci ne  I t i s t h e su spen si on of i n act i vat ed pat h ogen s or ant i geni c pr ot ei n of pat hogen whi ch i s t ak en or al l y or i nj ect ed t o pr ovi de i mmuni t y for t hat pat hogen.

Biology 3.37 



Second gener at ion vaccines : I nst ead of at t enuat ed pathogen, their antigenic polypeptides were separated and used as vaccines. Wit h t he help of genet ic engineer ing or r ecombinant DN A t echnol ogy, ant i geni c pol ypept i des of t he pat h ogen s ar e got sy n t h esi sed i n t r an sgen i c or ganisms, e.g. hepatit is– B vaccine fr om t r ansgenic yeast . Thir d generation vaccines : These ar e pur e synt het ic ant igenic polypeptides or their genes extr act ed fr om t he pat hogens.

I mmune Syst em D isor der s I mpr oper funct ioning of immune syst em may cause di scomfor t (al ler gy), di sease (AI DS) or even deat h (anaphylactic shock).

These ar e divided int o t hr ee classes : 1. H yper sensit ive disor der or aller gy  2 in number (didelphic).  When a per son show hyper r esponse or hyper sensitiveness for a common antigen or agent then it is called aller gy.  The agents which cause allergy are called allergens. Common aller gens ar e : pol l en gr ai n s, f ood (egg. f i sh ), m edi ci n es (penicilline), cold, heat , sunlight , fibr es. M anifest at ions (effect ).

I mpor t ant allergic r eact ions : Sneezing ; Coughing ;Wat er ing of Eyes ; Oedema : Accumulat ion of t issue fluid below skin ; Br onchial Ast hma ; H ay fever ; Anaphylat ic shock ; Eczema 2. Aut o immune disor der When i mmune syst em does not di scr i mi nat es bet ween self and non-self ant igen, ant ibodies ar e for med against t he self ant igen. These ant ibodies destr oy self ant igen and also self tissue of t he body. So, antibody for mation against self antigen is called aut o immunit y. Some subst ances/t i ssues of t he pat i ent 's body develop antigenic act ivit y and hence ar e called self ant igens /aut oant igens. E xamples. M yast heni a gr avi s ; Per ni ci ous (Dest r uct i ve) an aem i a ; H ash i m ot o di sease ; Rh eu m at oi d ar t hr i t i s ; I .D.D.M (I nsul i n dependent di abet es mallit us) ; M ult iple scler osis. 3. I mmuno D eficiency D isor der I t may be due t o gene mut at ion, gene deficiency, infect ion, nut r it ional deficiency and accident s.

1. S. C.I . D . ( Sever e C om bi n ed I m m u n o D eficiency)  This disor der is due t o gene mut at ion or gene deficiency of enzyme adenosine deaminidase.  Tr eat ment : Gene t her apy 2. A.I .D .S. ( Acqu i r ed I m m u n o D ef i ci en cy Syndr ome)  I t is characterised by reduction in the number of CD4 or hel per T 4– l ymphocyt es because of infection of HIV (human immunodeficiency virus).  I t i s al so call ed slim disease. ARC i s AI DS r el at ed compl ex whi ch i s char act er i sed by swollen lymph nodes, fever, night sweat s and weight loss. 



I t is of t wo t ypes : H I V– I (mor e common i n I ndia, Eur ope and Amer ica) and H I V– I I (mor e common in Afr ica). I n India the first AI DS case was reported in 1986.

 Spread of disease : The disease is rapidly spr eading thr oughout the wor ld. H i gh r i sk gr oups i ncl ude pr ost i t ut es, dr ug addicts, homosexual males, per sons with extr amar it al r elat ions and r ecipient s of unscr eened blood t r ansfusions. AI DS does not spr ead t hr ough M osqui t o bi t es, H uggi ng, K i ssi ng, Shar i ng meal s, Toi l et s t owel s or ut ensi l s, Shak i ng hands, Coughing, Sneezing, Looking after AI DS patients.

 Tr ansmission : Ther e ar e t hr ee r out es of t r ansmission : ( i ) Par ent er al r out e : I t i s t hr ough bl ood contact involving unscr eened tr ansfusion of blood, t at toeing, infect ed r azor s of bar ber s, poor l y st er i l i sed den t al i n st r u m en t s, shar i ng of i nj ect i on needl es and or gan tr ansplant. ( ii ) Sexual route : I t account s for 85% of H I V i nfect i on due t o mul t i pl e sex par t ner s, pr ost i t ut es, homosexual i t y and ar t i fi ci al inseminat ion. Vir us is pr esent in sufficient st r engt h in semen and vaginal secr et ions of infect ed per sons. ( iii ) Transplacental route : I nfect ion can occur fr om i nfect ed mot her t o foet us (ver t i cal transmission) across placenta and to infants t hr ough milk (per inat al t r ansmission).

3.38 Biology

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. I n whi ch of t he fol l owi ng ar e M i t ochondr i a absent? (a) Fungi

(b) Angiosper ms

(c) Gr een algae

(d) Blue-gr een algae

2. Eucar yot ic or ganisms ar e t hose t hat cont ain (a) DNA thr eads

(b) Plastids

(c) Tr ue nucleus

(d) Vacuoles

3. Wi t h whi ch of t he fol l owi ng i s Azot obact er associat ed ? (a) Car bon fixat ion

(b) Nit r ogen fixat ion

(c) Fer ment at ion

(d) Root nodules

4. Whi ch of t he fol l owi ng mi cr onut r i ent s pl ays an impor t ant r ole in t he nit r ogen met abolism of plant s, especially in t he r educt ion of nit r at e?

12. Cor k cells ar e imper vious t o wat er because of t he pr esence of (a) Cellulose

(b) Cutin

(c) Lignin

(d) Suber in

13. The biotic relationship between insects and plants wit h r efer ence t o pollinat ion is called (a) Commonsalism

(b) Mutualism

(c) Par asitism

(d) Sapr ophytism

14. Phot osynt hesis gener ally t akes place in which por t ions of t he plant ? (a) L eaf and ot her chlor oplast bear ing par t s (b) St em and leaf (c) Root s and chlor oplast bear ing par t s (d) Bar k and leaf 15. Opium is obt ained fr om

(a) Bor on

(b) Copper

(a) Poppy leaves

(c) Molybdenum

(d) Zinc

(b) L at ex juice

5. Plant development is influenced by (a) Qualit y of light only

(c) Tablet t ype lat ex (d) Seed capsule of opium poppy

(b) Qualit y and quant it y of light (c) Qualit y and dur at ion of light (d) Qualit y, quant it y and dur at ion of light 6. Edible par t of t omat o is (a) Endocarp

(b) Fleshy t halamus

(c) Mesocar p

(d) Whole fr uit

7. Xylem is a complex t issue, consisting of differ ent t ypes of cells. Which of t he following per for m t he funct ion of conduct ion of wat er and miner als? (a) Fiber s

(b) Par enchyma cells

(c) Tracheary elements (d) None of t hese 8. The edible por t ion of mango is (a) Embr yo

(b) Endocarp

(c) Endosper m

(d) Mesocar p

9. The r ice gr ain is (a) A seed

(b) One seeded fr uit

(c) M any seeded fr uit (d) Mult iple seeded fr uit 10. I n which of the following ar e plastids not pr esent? (a) Aer nechyma

(b) Collenchyma

(c) Par enchyma

(d) Schler enchyma

11. Bamboo is a (a) H er b

(b) Gr ass

(c) Shrub

(d) Tr ee

LEVEL-1 1. I n t he cont ext of genet ics, DNA st ands for (a) Di -Neur o Acid (b) Dai ly N ews Analysis (c) Det oxic N eur o Acid (d) Deoxyr ibo N uclei c Acid [RRB JE 2014 GREEN SH I FT ]

2. M at ch t he foll owing: (1) Cell wall

(a) Animal cell (b) Pl ant cell

(2) ATP

(a) M it ochondr ia (b) Genes

(a) l – (a), 2 – (a) (c) 1 – (b), 2 – (a)

(b) l – (a), 2 – (b) (d) l – (b), 2 – (b) [RRB JE 2014 GREEN SH I FT ]

3. Synapses and Dendr i t es ar e associ at ed wi t h (a) cor t ex (c) r etina

(b) epit helium (d) ner ve-cell s [RRB JE 2014 GREEN SH I FT ]

4. A tissue that connects muscle to bones in humans is call ed (a) Tendon (c) Axon

(b) Fibr e (d) Femur [RRB JE 2014 GREEN SH I FT ]

Biology 3.39

5. H aemat ology is t he st udy r el at ed t o

14. Nephrons are connected with

(a) Pl ant r epr oduct i on syst em

(a) Respiratory System

(b) Blood

(b) Nervous System

(c) Food habi t s of ani mals

(c) Circulatory System

(d) Bones

(d) Excretory System [RRB JE 2014 GREEN SH I FT ]

[RRB JE 2014 YEL L OW SH I FT ]

6. Which of the following is not a food bor ne disease?

15. Sight of delicious food usually makes mouth watery, it is a

(a) Amoebiasis (b) Choler a

(a) Hormonal response

(c) I nfluenza

(b) Neural response

(d) H epat it i s A

(c) Optic response [RRB JE 2014 GREEN SH I FT ]

(d) Olfactory response

7. A pher omone secr et ed by an animal (a) infl uences t he behaviour of ani mal s of same species (b) pr ot ect s it fr om pr edat or s (c) at t r act s t he vict ims for i t s food (d) none of t he above [RRB JE 2014 GREEN SH I FT ]

8. Secretion of Insulin Hormone is by : (a) Thyroid

(b) Pituitary

(c) Adrenal

(d) Pancreas [RRB JE 2014 RED SH I FT ]

9. Earthworm belongs to which of the following Animal Phyla ? (a) Arthropoda

(b) Mollusca

(c) Annelida

(d) Protozoa [RRB JE 2014 RED SH I FT ]

10. The total number of bones in the average adult human skeleton is :

[RRB JE 2014 YEL L OW SH I FT ]

LEVEL-2 1. Accor ding t o I PCC, t hr ee fact or s cont r ibut ing t o Global war ming ar e : 1) CO2 emissions 2) Change of land use defor est at ion 3) Non-veg food Place t hem in t he or der of t heir cont r ibut ion t o global war ming. (a) 1, 2, 3

(b) 1, 3, 2

(c) 3, 1, 2

(d) 2, 1, 3 [RRB SSE 2014 GREEN SH I FT]

2. St omat a ar e locat ed in(a) Red blood cells (c) St omach

(b) Chlor ophyll (d) L eaves [RRB SSE 2014 GREEN SH I FT]

3. Bile is secr et ed by-

(a) 350

(b) 206

(a) St omach

(b) L iver

(c) 115

(d) 540

(c) L ar ge int est ine

(d) Gall bladder

[RRB JE 2014 RED SH I FT ]

11. Identify the disease which is caused due to deficiency of Protein ? (a) Scurvy

(b) Beri-Beri

(c) Night-Blindness

(d) Kwashiorkor [RRB JE 2014 RED SH I FT ]

12. In humans, bile juice is secreted by (a) pancreas

(b) small intestine

(c) esophagus

(d) liver

[RRB SSE 2014 GREEN SH I FT]

4. I n t he cont ext of act ion of medicines on human body, mat ch t he following: 1. Recept or s

a. Catalysts

2. Enzymes

b. Neur ologically act ive

3. Tr anquilizer s

c. Pr ot ei ns

(a) 1-c, 2-a, 3-b

(b) 1-a, 2-c, 3-b

(c) 1-b, 2-a, 3-c

(d) 1-a, 2-b, 3-c

[RRB JE 2014 YEL L OW SH I FT ]

[RRB SSE 2014 GREEN SH I FT]

13. Which of the following is connected with transport of water in plants ?

5. Which of t he following t issues t r anspor t s wat er and mi ner als fr om r oot s t o ot her par t s of t he plant?

(a) Phloem

(b) Xylem

(c) Epidermis

(d) Cambium [RRB JE 2014 YEL L OW SH I FT ]

(a) Phloem

(b) Vessel

(c) Sieve t ube

(d) Xylem [RRB SSE 2014 GREEN SH I FT]

3.40 Biology

6. An eating disorder of excessive weight loss usually due to undue concer n about body - shape is known as: (a) Anor exia ner vosa

(c) H er bivor es, Car nivor es, Pr oducer s (d) H er bivor es, Pr oducer s, Car nivor es [RRB SSE 2014 RED SH I FT]

11. Ear t hwor m belongs t o following animal phyla ?

(b) Appetitis (c) Aut ot r ophic disor der (d) Aut otr ophic syndr ome [RRB SSE 2014 GREEN SH I FT]

7. M at ch t he following -

(a) Por ifer a

(b) Annelida

(c) Mollusca

(d) Arthropoda [RRB SSE 2014 RED SH I FT]

12. Which one of the following contains Human body 's t her most at ?

1. Lysosomes

a. Power H ouse

(a) Pineal

(b) Pit uitar y

2. DNA

b. Chr omosomes

(c) Thyr oid

(d) Hypothalamus

3. Mitochondr ia

c. Suicide bags

(a) 1-a, 2-c, 3-b

(b) l-c, 2-b, 3-a

(c) 1-b, 2-c, 3 a

(d) 1-c, 2-a, 3-b [RRB SSE 2014 GREEN SH I FT]

8. Nat ur e's cleaner s ar e :

[RRB SSE 2014 RED SH I FT]

13. Wh i ch on e of t h e f ol l ow i n g bl ood gr ou p i s consider ed Univer sal Donor ? (a) AB

(b) O

(c) A

(d) B

(a) Pr oducer s

(b) Consumer s

[RRB SSE 2014 RED SH I FT]

(c) Decomposer s

(d) Car nivor es

14. The number of chr omosomes in a nor mal human body cell is :

[RRB SSE 2014 RED SH I FT]

9. Tr iple Vaccine is administer ed to a new bor n child t o immunize against :

(a) 43

(b) 44

(c) 45

(d) 46

(a) Whooping Cough, Tet anus and M easles (b) Whooping Cough, Tet anus and Dipht her ia (c) Tet anus, Dipht her ia and Small pox (d) Tet anus, Typhoid and H epat it is [RRB SSE 2014 RED SH I FT]

10. A cor r ect food chain is : (a) Pr oducer s, H er bivor es, Car nivor es (b) Pr oducer s, Car nivor es, H er bivor es

[RRB SSE 2014 RED SH I FT]

15. Which one of t he following diseases is caused due t o deficiency of pr ot ein ? (a) K washior kor

(b) Ricket s

(c) Ber i - Ber i

(d) Scur vy [RRB SSE 2014 RED SH I FT]

Biology 3.41

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (d) 11. (b)

2. (c) 12. (d)

3. (b) 13. (b)

4. (c) 14. (a)

5. (d) 15. (d)

6. (d)

7. (c)

8. (d)

9. (a)

10. (d)

7. (a)

8. (d)

9. (c)

10. (b)

7. (b)

8. (c)

9. (b)

10. (a)

LEVEL-1 1. (d)

2. (c)

3. (d)

4. (a)

5. (b)

11. (d)

12. (d)

13. (b)

14. (d)

15. (b)

6. (c)

LEVEL-2 1. (a)

2. (d)

3. (b)

4. (a)

5. (d)

11. (b)

12. (d)

13. (b)

14. (d)

15. (a)

6. (a)

EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S 1. Mitochondr ia is pr esent in fungi, angiosper m and gr een algae whit e it is not found in blue-gr een algae. 2. Eucar yot i c or gani sms ar e t hose cont ai n t r ue nucleus. 3. Azat abact er is associat ed wit h nit r ogen fixat ion, which is found in t he r oot nodules of leguminous plants. 4. M olybdenum is a micr onut r ient s which plays an i mpor t ant r ol e i n t he ni t r ogen met abol i sm of plant s, especially in t he r educt ion of nit r at e. 5. Pl ant devel opment i s i nfl uenced by qual i t y, quant it y and also by dur at ion of light . 6. Whole fr uit is an edible par t of t omat o. 7. Xylem is a complex t issue, consist ing of differ ent t ypes of cel l i n whi ch t r achear y el ement s ar e per for m t he funct ion of conduct ion of wat er and miner als. 8. The edible por t ion of mango is mesocar p. 9. The r ice gr ain is a seed of paddy cr op. 10. Plastide ar e pr esent in aer nechyma, collenchyma an d par en ch y m a w h i l e i t i s n ot f ou n d i n schler enchyma. 11. Bomboo is for m of gr ass. 12. Cor p cells ar e imper vious t o wat er because of t he pr esence of suber in.

13. The biotic r elationship between insects and plants wit h r efer ence t o pollination is called mutualism. 14. Phot osynt hesis t akes place in leaves and ot her chlor oplast bear ing plant s in which chlor ophyle is pr esent . 15. Opium is obt ained fr om seed capsule of opium poppy.

LEVEL-1 1. Deoxyribonucleic acid (DNA) is a molecule composed of two chains (made of nucleotides) that coil around each other to form a double helix carrying the genetic instructions used in the growth, development, functioning and reproduction of all known living organisms and many viruses. 2. Cell walls are present in most prokaryotes (except mycoplasma bacteria), in algae, plants and fungi. Mitochondria are the energy factories of the cells. The energy currency for the work that animals must do is the energy-rich molecule adenosine triphosphate (ATP). The ATP is produced in the mitochondria using energy stored in food. 3. Synapses and Dendrites are associated with nerve-cells.In the nervous system, a synapse is a structure that permits a neuron (or nerve cell) to pass an electrical or chemical signal to another neuron or to the target effector cell. 4. A tendon is a tough yet flexible band of fibrous tissue. The tendon is the structure in your body

3.42 Biology

that connects a muscle to a bone. The structure that transmits the force of the muscle contraction to the bone is called a tendon.

LEVEL-2 1. a

According to IPCC, three factors contributing to Global warming are CO2 emissions , Change of land use deforestation as well as non-veg food. Global carbon emissions from fossil fuels have significantly increased since 1900. Since 1970, CO 2 emissions have increased by about 90%, with emissions from fossil fuel combustion and industrial processes contributing about 78% of the total greenhouse gas emissions increase from 1970 to 2011. Agriculture, deforestation, and other land-use changes have been the second-largest contributors.

2. d

In botany, a stoma (also stomate; plural stomata) is a tiny opening or pore that is used for gas exchange. They are mostly found on the under-surface of plant leaves. In a stoma, there is the chloroplast, a cell wall, a vacuole and a cell nucleus.

3. b

Bile or gall is a dark green to yellowish brown fluid, produced by the liver of most vertebrates, that aids the digestion of lipids in the small intestine. In humans, bile is produced continuously by the liver (liver bile), and stored and concentrated in the gallbladder.

4. a

The principal minor tranquilizers are the benzodiazepines, among which are diazepam (Valium), chlordiazepoxide (Librium), and alprazolam (Xanax). These drugs have a calming effect and eliminate both the physical and psychological effects of anxiety or fear. Enzyme r is a catalyst. Enzymes are proteins functioning as catalysts that speed up reactions by lowering the activation energy.In biochemistry and pharmacology, a receptor is a protein molecule that receives chemical signals from outside a cell.

5. d

Xylem and Phloem tissues are present throughout the plant. They begin at the root and then move up to the stem, branches, and leaves. The xylem tissue transports water and minerals from the roots to the leaves whereas the phloem tissue transports food from the leaves to the other parts of the plant.

6. a

Anorexia nervosa is a potentially lifethreatening eating disorder characterized by self-starvation, excessive weight loss and

5. Hematology, also spelled haematology, is the branch of medicine concerned with the study of the cause, prognosis, treatment, and prevention of diseases related to blood. 6. Among the given options, Influenza is not a foodborne disease.Foodborne illness (also foodborne disease and colloquially referred to as food poisoning) is any illness resulting from the food spoilage of contaminated food, pathogenic bacteria, viruses, or parasites that contaminate food as well as toxins such as poisonous mushrooms. 7. A pheromone secreted by an animal influences the behaviour of animals of same species. It is generally used for mate selection. 8. Insulin and glucagon are hormones secreted by islet cells within the pancreas. They are both secreted in response to blood sugar levels, but in opposite fashion! Insulin is normally secreted by the beta cells (a type of islet cell) of the pancreas. 9. Earthworms are segmented worms of the phylum Annelida, which encompasses about 9,000 species and three classes. Class Oligochaeta are the freshwater worms (including earthworms); class Polychaeta are the marine worms; and class Hirudinea are the leeches. 10. The total number of bones in the average adult human skeleton is 206. 11. Kwashiorkor is a form of severe protein malnutrition characterized by edema, and an enlarged liver with fatty infiltrates. Sufficient calorie intake, but with insufficient protein consumption, distinguishes it from marasmus. 12. Bile or gall is a dark green to yellowish brown fluid, produced by the liver of most vertebrates, that aids the digestion of lipids in the small intestine. In humans, bile is produced continuously by the liver (liver bile), and stored and concentrated in the gallbladder. 13. Xylem is one of the two types of transport tissue in vascular plants, phloem being the other. The basic function of xylem is to transport water from roots to shoots and leaves, but it also transports nutrients. 14. Nephrons are connected with excretory System. It is the microscopic structural and functional unit of the kidney. It is composed of a renal corpuscle and a renal tubule. 15. Sight of delicious food usually makes mouth watery, it is a neural response.

Biology 3.43

negative body image. Anorexia can affect individuals of all genders, races and ethnicities. While most common among females, about 10-15% of all individuals with anorexia are males. People of all ages develop anorexia but it is most common for onset to occur during adolescence. 7. b

8. c

9. b

Mitochondria are tiny organelles inside cells that are involved in releasing energy from food. This process is known as cellular respiration. It is for this reason that mitochondria are often referred to as the powerhouses of the cell.Lysosomes are tiny sacs mostly present in animal cells. They are filled with digestive enzymes. The released enzymes then digest their own cell and ultimately the cell dies. Hence, lysosomes are called suicide bags of the cell. Nature’s cleaners are decomposers. Decomposers and scavengers break down dead plants and animals. They also break down the waste (poop) of other organisms. Decomposers are very important for any ecosystem. If they weren’t in the ecosystem, the plants would not get essential nutrients, and dead matter and waste would pile up. Triple Vaccine is administered to a new born child to immunize against whooping Cough, Tetanus and Diphtheria.

10. a

A correct food chain is Producers, Herbivores, Carnivores

11. b

Phylum: Annelida. Annelids are segmented worms. Earthworms belong to this phylum because their bodies are sectioned, creating the ridged or ringed appearance that gives the “ringed worms” of this phylum their name.

12. d

Hypothalamus contains Human body’s thermostat. The hypothalamus is a small region of the brain. It’s located at the base of the brain, near the pituitary gland. While it’s very small, the hypothalamus plays a crucial role in many important functions, including: releasing hormones. regulating body temperature.

13. b

In transfusions of packed red blood cells, individuals with type O Rh D negative blood are often called universal donors. Those with type AB Rh D positive blood are called universal recipients.

14. d

There are 46 total chromosomes in a normal human body cell. They pair up, creating 23 pairs of chromosomes.

15. a

Protein malnutrition, or kwashiorkor, is mostly found in people living in geographical areas that have limited food resources. It’s most commonly seen in children whose diets are low in protein and calories.

 

1

CHAPTER

SCIENCE AND TECHNOLOGY

I N VEN TI ON S AN D DI SCOVERI ES I nvention

Year

I nventor

Country

Aer oplane Ball-point pen Bar omet er Bar omet er, Aner oid Bicycle Bicycle t yr e (air ) Calculating M achine Cent igr ade Scale Cinema Cir culat ion of blood Clock (mechanical) Clock (pendulum) Diesel Engine Dynamite Elect r ic ir on Elect r ic lamp Elect r ic mot or Theor y of Evolut ion Fount ain pen Glider Gr amophone I nsulin Jet engine L ight ening conduct or L ocomot ive, st eam M achine gun M atch, safet y M icr ophone M icr oscope M ot or car, pet r ol M ot or cycle Penicillin Photogr aphy (film) Radar Radium Radio Refr iger ator Revolver Rubber (vulcanised) Safety lamp Safet y pin St eam engine Submar ine Telegr aph code Ther momet er Televisi on Tel escope Typewr it er Wat ch X-ray

1903 1888 1644 1799 1839 1888 1642 1742 1895 1628 1725 1657 1892 1867 1882 1879 1834 1858 1884 1853 1878 1923 1937 1752 1804 1861 1855 1878 1590

Or ville and Wilbur Wr ight John J. L oud E.Tor r icelli W.J.Cant e K . M acmillan J.B.Dunlop Blaise Pascal A. Celsius A.L . and J. L umier e William H ar vey H sing and L ing-Tsan Chr ist ian H uygens Rudolf Diesel Alfr ed Nobel H .W. Seeley Thomas Alva Edison M or it z Jacobi Char les Dar win L .E. Wat er man Sir Ger or ge Cayley Thomas Alva Edison Sir Gr eder ick Bant ing Sir Fr ank Whit t le Benjamin Fr anklin Richar d Tr evit hic Richar d Gat ling J.E.L undst r om David Hughes Z. Janssen K ar l Benz Edwar d Butler Sir Alexander Fleming H ohn Car but t Dr. A.H . Taylor and L .C.Young M ar ie and Pier r e Cur ie Guglielmo M ar coni James H ar r ison Samuel Colt Char les Goodyear Sir H umphr y Davy William H ur st Thomas Saver y David Bushnell Samuel F.B. M or se H ans L ipper shey John L ogie Bair d Galileo M it t er hofer A.L . Br eguet Wilhelm Roent gen

U.S.A U.S.A. I taly I taly Br itain Br itain Fr ance Fr ance Fr ance England China Holland Ger many Sweden U.S.A U.S.A Ger many England U.S.A England U.S.A Canada England U.S.A England U.S.A Sweden U.S.A Holland Ger many England England U.S.A U.S.A Fr ance I taly Scotland U.S.A U.S.A England U.S.A Br itain U.S.A U.S.A Nether lands Scotland I taly Austria Fr ance Ger many

1884 1928 1888 1922 1898 1901 1851 1835 1841 1816 1849 1639 1776 1837 1608 1926 1593 1864 1791 1895

1.2

SCIENCE AND TECHNOLOGY

ELEM EN TS FOU N D I N N ORM AL H U M AN BODY Element

Percent

Element

Percent

Oxygen

65.0

Sodium

0.15

Car bon

18.0

Chlor ine

0.15

Hydr ogen

10.0

Magnesium

0.05

Nit r ogen

3.0

I r on

0.004

Calcium

2.0

I odine

0.00004

Phosphor ous

1.1

Manganese

0.00013

Pot assium

0.35

Copper

0.00014

Sulphur

0.25

Cobalt

0.00000016

VARI OU S SCI EN CES Acoustics : St udy of sound (or science of sound) Aerodynamics : Br anch of mechanics t hat deals wit h t he mot ion of air and ot her gases.

Ceramics : Ar t and t echnology of making object s fr om clay, et c.

Aeronautics : Science or ar t of flight .

Chemist r y : St udy of el ement s and t hei r l aws of combinat ion and behaviour.

Agronomy : Sci ence of soi l management and t he pr oduct ion of field cr ops.

Chemot her apy : Tr eat ment of di seases by usi ng chemical subst ances.

Agrostolgy : St udy of gr asses.

Conchology : Br anch of zool ogy deal i ng wi t h t he shells of molluscs.

Anat omy : Sci ence deal i ng wi t h t he st r uct ur e of animals, plant s or human body. Ant hr opology : Sci ence deal i ng wi t h t he or i gi ns, physical and cult ur al development of mankind. Archaeology : St udy of ant iquit ies. Ast rology : Ancient ar t of pr edict ing t he cour se of human dest inies wit h t he help of indicat ions deduced fr om t he posi t i on and movement of t he heavenl y bodies. Astronautics : Science of Space t r avel. Astronomy : St udy of heavenly bodies. Astrophysics : Br anch of ast r onomy concer ned wit h t he physical nat ur e of heavenly bodies. Bacteriology : St udy of Bact er ia. Biochemistry : St udy of chemical pr ocesses of living things. Biology : St udy of living t hings. Biometry : Applicat ion of mat hemat ics t o t he st udy of living t hings. Bionics : St udy of funct i ons, char act er i st i cs and phenomena observed in the living world and application of t his knowledge t o t he wor ld of machines.

Cosmology : St udy of univer se as a whole and of it s for m, nat ur e et c. Cryptography : St udy of Cipher s (secr et wr it ings) Crystallography : St udy of t he st r uct ur e, for ms and pr oper t ies of cr yst als. Cryogenics : Science deal i ng wi t h t he pr oduct ion, cont r ol and applicat ion of ver y low t emper at ur es. Cytology : St udy of cells, especially t heir for mat ion, st r uct ur e and funct ions. Dactylography : St udy of finger pr int s for the pur pose of ident ificat ion. Ecology : St udy of t he r elat ion of animals and plant s t o t heir sur r oundings, animat e and inanimat e. Economics : Sci ence deal i ng wi t h t he pr oduct i on, dist r ibut ion and consumpt ion of goods and ser vices. Embryology : St udy of development of embr yos. Entomology : St udy of insect s. E pidemiology : Br anch of medi ci ne deal i ng wi t h epidemic diseases. Epigraphy : St udy of inscr ipt ions.

Bionomics : St udy of t he r elat ion of an or ganism t o it s envir onment .

Ethnology : Br anch of ant hr opology dealing wit h t he or igin, dist r ibut ion and dist inguishing char act er istics of t he r aces of mankind.

Botany : A st udy of plant s.

Ethology : St udy of animal behaviour.

SCIENCE AND TECHNOLOGY

1.3

Eugenics : St udy of t he pr oduct ion of bet t er offspr ing by t he car eful select ion of par ent s.

Phycology : St udy of Algae.

Genealogy : St udy of family or igins and hist or y. Genesiology : The science of gener at ion.

Physiology : St udy of t he funct ioning of t he var ious or gans of living beings.

Genet i cs : B r an ch of bi ol ogy deal i n g w i t h t h e phenomena of her edit y and t he laws gover ning it .

Pomology : Science t hat deals wit h fr uit s and fr uit gr owing.

Geography : Development of science of t he ear t h’s sur face, physical feat ur es, climat e, populat ion et c.

Psychology : St udy of human and animal behaviour.

Geology : Science t hat deals wit h t he physical hist or y of t he ear t h.

Rheology : St udy of t he defor mat i on and fl ow of mat t er.

Geomorphology : St udy of t he char act er ist ics, or igin and development of landfor ms.

Sci en t ol ogy : St u dy of ear t h qu ak es an d t h e phenomena associat ed wit h it .

Ger ont ology : St udy of ol d age, i t s phenomena, diseases, et c.

Sericulture : Raising of silkwor ms for t he pr oduct ion of r aw silk.

H istology : St udy of t issues.

Sociology : St udy of human societ y.

H or t i cu l t u r e : Cu l t i v at i on of f l ow er s, f r u i t s, veget ables and or nament al plant s.

Telepathy : Communicat ion bet ween minds by some means ot her t han sensor y per cept ion.

H ydr ology : St udy of wat er wi t h r efer ence t o i t s occur r ence, and pr oper t ies in t he hydr ospher e and at mospher e.

Therapeutics : Ar t and science of healing.

M etallurgy : Pr ocess of ext r act ing met als fr om t heir or es.

Physics : St udy of t he pr oper t ies of mat t er.

Radiology : St udy of X-r ays and r adioact ivit y.

Topography : A special descr ipt ion of a par t or r egion. Virology : St udy of vir uses. Zoology : St udy of animal life.

M et eor ology : Sci ence of t he at mospher e and i t s phenomena.

SCI EN CES

M etrology : Scient ific st udy of weights and measur es.

L owest melt ing point

M icrobiology : St udy of mi nut e l i vi ng or gani sms, including bact er ia, moulds, and pat hogenic pr ot ozoa.

Since M er cur y melt s at – 38.8C ( – 38 F), it is liquid at r oom t emper at ur e.

M ycology : St udy of fungi and fungus diseases. N eurology : St udy of the ner vous system, it s funct ions and it s disor der s. N umerology : St udy of number s. Odontology : Scient ific st udy of t he t eet h. Optics : St udy of nat ur e and pr oper t ies of light . Ornithology : St udy of bir ds. Osteology : St udy of bones. Paleobotany : St udy of fossil plant s. Paleontology : St udy of fossils. Pathology : St udy of diseases. P h on et i cs : St u dy of speech sou n ds an d t h e pr oduct ion, t r ansmission, r ecept ion, et c. Phrenology : St udy of t he facult ies and qualit ies of minds fr om t he shape of t he skull. Phthisiology : Scient ific st udy of t uber culosis.

M er cur y is used in t her momet er s because it expands as it is heat ed. H ighest melt ing point The t emper at ur e must be 3,652C (6,606 F) befor e car bon wi l l mel t . Two-t hi r d as hot as t he Sun ’s sur face. Rar est element s Ast at ine is t he r ar est element on Ear t h Rhodium is t he scar cest metal in t he wor ld just 3 t ones ar e mined ever y year. H eaviest met al A 33-cm 3 (13-in 3) cube of osmium weighs 640 kgs. Acid and Alkalis The acid or alkali (or base) cont ent of a subst ance is measur ed in pH (pot ent ial H ydr ogen) on a scale of 0 14. Aci ds di ssol ve i n wat er t o for m shar p-t ast i ng solut ions, like lemon juice. Alkalis dissolve in wat er t o for m soapy solut ions.

1.4

SCIENCE AND TECHNOLOGY

SCI EN TI FI C I N STRU M EN TS Altimeter : I t is a special t ype of aner oid bar omet er, used in measur ing alt it udes.

M agnetometer : used t o compar e magnet ic moment s and fields.

Ammeter : used t o measur e st r engt h of an elect r ic cur r ent .

M egaphone : u sed f or car r y i n g sou n d t o l on g distances.

Anemometer : used t o measur e vel oci t y and find dir ect ion of t he wind.

M icrophone : used for conver t ing sound waves int o elect r ical ener gy which is t r ansmit t ed t hr ough wir es and then r ecover ed into sound in a magnified intensity.

Audiometer : used t o measur e differ ence in hear ing. Barometer : used for measur ing atmospher ic pressur e. Calorimeter : used for measur ing quant it ies of heat . Chronometer : I t is a clock t o det er mine longit ude of a vessel at sea. Clinical T her momet er : I t i s a t her momet er for measur ing t emper at ur e of human body. Colorimeter : used for compar ing int ensities of colour. Commutator : used t o change or r ever se t he dir ect ion of an elect r ic cur r ent . I n dynamo it is used t o conver t t he alt er nating cur r ent int o dir ect cur r ent . Dynamo : used for conver t ing mechanical ener gy int o elect r ical ener gy. Dynamometer : used for measur ing elect r ical power. Elect rocardiograph (ECG) : used for det ect i on of elect r ic pulses of t he hear t . I t gives a gr aphic pict ur e of hear t beat s. Electroencephalograph (EEG) : used for r ecor ding of change in elect r ic pot ent ial in var ious ar ea of t he br ain by means of electr ode on the scalp or in the br ain itself. Electrometer : used for measur ing elect r icit y. Electroscope : used for det ect ing pr esence of elect r ic char ge. Galvanometer : used for measur ing elect r ic cur r ent . H ydrometer : used for measur ing t he r elative densit y of liquids. H ydr oscope : opt i cal i nst r ument used for seei ng object s below t he sur face of wat er. H ygr omet er : u sed f or m easu r i n g t h e r el at i ve humidit y of t he at mospher e. H ygr oscope : u sed t o sh ow t h e ch an ges i n at mospher ic humidit y. Lactometer : used for measur ing t he r elat ive densit y of milk. M anometer : used measur e t he pr essur e of gases.

M icroscope : used for magnified view of ver y small object s. Photometer : used for compar ing luminous int ensit y of t he sour ces of light . Pyk nomet er : used t o measur e t he densi t y and coefficient of expansion of liquid. Pyrheliometer : used for measur ing solar r adiat ions. Pyrometers : used t o measur e high t emper at ur e. Radar : used for det ecting and finding r ange of moving object s by t r ansmit t ing beams of r adio waves. Radio micrometer : used for measuring heat radiations. Rain gauge : used for measur ing r ainfall. Refractometer : used t o measur e r efr act ive index of a subst ance. Resist ance t her momet er : used for det er mi ni ng elect r ical r esist ance of conduct or. Seismograph : used for r ecor ding int ensit y and or igin of ear t hquake shocks. Sextant : used for measur ement of angular dist ances bet ween t wo object s. Sphygmomanomet er : used for measur i ng bl ood pr essur e. Stethoscope : medical inst r ument used for hear ing and analysing t he sound of hear t and lungs. Tel escope : used f or vi ewi ng di st ant obj ect s as magnified. Television : used for t r ansmi t t i ng vi si bl e movi ng images by means of wir eless makes. Thermometer : used t o measur e t he t emper at ur e. T hermost at : devi ce used for r egul at i ng const ant t emper at ur es. Viscomet er : used for measur i ng vi scosi t y i .e., pr oper t y of r esist ance of a fl ui d t o r elat i ve mot ion wit hin it self. Volt met er : used t o measur e pot ent i al di ffer ence bet ween t wo point s.

SCIENCE AND TECHNOLOGY

1.5

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. The genome sequencing of which vir us has been r ecent ly done by I ndian scient ist s ? (a) Hepatitis C (b) Hepatitis B (c) HIV (d) Flu Vir us 2. Net gr owt h r at e of populat ion is det er mined by (a) gr oss r epr oduct ion r at e (b) t he bir t h r at e and t he deat h r at e (c) t he pr essur e of populat ion (d) t he bir t h r at e in a count r y 3. APSARA is t he name of I ndia’s fir st (a) Nuclear r eact or (b) H elicopt er (c) Gr ound bat t le t ank (d) Railway locomot ive 4. Which of the following parts of the sunlight makes t he solar cooker hot ? (a) Ult r aviolet (b) Red light r ays (c) I nfrared (d) Cosmic r ays 5. H uman Genome Pr oject endeavour s t o (a) Decode DNA (b) I nvent AI DS t r eat ment (c) St udy evolut ion of human (d) Study finger pr int s 6. The Gr een r evolut ion in cr ops, Yellow r evolut ion in oil seeds and Gol den r evolut ion i n ...... has been an ample t est imony t o t he cont r ibut ions of agr icul t ur al r esear ch and devel opment effor t s under t aken in t he count r y. (a) H or t icult ur e (b) Wheat (c) Pet r ol (d) None of t hese 7. Which of the following does not come as a pr oduct of vehicular pollut ion ? (a) Sulphur dioxide (b) Nit r ogen oxide (c) Car bon monoxide (d) H ydr ogen per oxide 8. Which one of t he fol lowi ng does a TV r emot e cont r ol unit use t o oper at e a TV set ? (a) L ight waves (b) Sound waves (c) Micr owaves (d) Radio waves

9. Under I ndian space pr ogr amme, the abbr eviation GSLV st ands for (a) Geosynchr onous Sat ellit e L aunch Vehicle (b) Global Sat ellit e L ink Vehicle (c) Geost at ionar y Solar L ight Vehicle (d) None of t hese 10. Cent r e for DNA Finger pr int ing and Diagnost ics (CDFD) is locat ed at (a) Delhi (b) Kolkata (c) Hyderabad (d) Chennai 11. The Russi an nucl ear submar i ne, whi ch sunk killing all on boar d, was called (a) Slovik (b) L enin (c) Brodigaya (d) K ur sk 12. Richter scale is a/an ......... scale to measur e ear th t r emor s. (a) Exponential (b) L ogar it hmic (c) Geomet r i c (d) Physical 13. Which of t he following compounds is commonly used as an ant isept ic in mout h washes and toot h past es ? (a) Bor ax (b) Salt Pet r e (c) H ydr ogen per oxide (d) Sodium chlor ide 14. Ant hr ax i s one of t he most pot ent bi ol ogi cal weapons i n t he wor l d. I t i s spr ead by si ngl e bacter ium called (a) Bacillus thur icide (b) Bacillus ant hr at um (c) Thiobacillus (d) Bacillus anthr acis 15. ‘Thr eat of global war ming’ is incr easing due t o incr easing concent r at ion of (a) Nit r ous oxide (b) Ozone (c) Sulphur dioxide (d) Car bon dioxide

LEVEL-1 1. Shant i Swar up Bhat nagar Awar d is given in t he field of (a) liter atur e (b) science and t echnology (c) jour nalism (d) communit y leader ship

1.6

SCIENCE AND TECHNOLOGY

2. D epar t m en t of A t om i c E n er gy (D A E ) w as est ablished on (a) 3rd August 1950 (b) 3rd August 1952 (c) 3rd August 1954 (d) 3rd August 1956 3. Dhr uva, CI RUS and Apsar a ar e (a) pr essur ized heavy wat er r eact or s (b) r esear ch r eact or s (c) boiling wat er r eact or s (d) advanced heavy wat er r eact or s 4. Whi ch one of t he fol l owi ng i s pr esent i n t he ur anium or e and is r ecover ed as a by pr oduct in t he ur anium pr ocess plant ? (a) H aemat it e

(b) Dolomi t e

(c) Magnetite

(d) Ber yllite

5. I n di an Pr essu r i zed H eavy Wat er React or s (PH WR) use (a) ur anium based fuel (b) t hor ium based fuel (c) bot h ur anium and t hor ium based fuel (d) ur anium and yt r ium based fuel 6. Which one of t he foll owi ng is used mai nly for r adiogr aphy of var ious mat er ials? (a) Dhruva (c) Kamini

(b) FBTR (d) Apsara

7. I REMON is (a) Fast br eeder r eact or (b) Pr essur ized heavy wat er r eact or (c) A n at i on wi de n et wor k of en vi r on m en t al r adiat ion monit or ing st at ion (d) Weat her monit or ing syst em 8. Biot echnology par k is sit uat ed in which of t he following cit ies of I ndia? (a) Bhopal (b) L uck now (c) Bengalur u (d) Nashik 9. H ydr omet er is an inst r ument (a) for measur ing sound under wat er (b) t o det ect t he pr esence of hydr ogen i n t he at mospher e (c) for measur ing t he specific gr avit y of liquids (d) to detect the changes in atmospher ic humidity

10. ‘Bar ’ is t he unit of (a) heat (b) t emper atur e (c) cur r ent (d) at mospher ic pr essur e 11. Elect r ic cur r ent is measur ed by (a) anemomet er (b) volt met er (c) ammet er (d) commut at or 12. Fat hom is t he unit of (a) sound (c) distance

(b) depth (d) fr equency

13. K ilowat t is a unit t o measur e (a) wor k (b) electr icit y (c) power (d) ener gy 14. Ver y sm al l t i m e i n t er v al s ar e accu r at el y measur ed by t he (a) Pulsar s (b) Quar t z clocks (c) At omic clocks (d) Whit e dwar fs 15. Decibel is t he unit used for (a) speed of light (b) int ensit y of heat (c) int ensit y of sound (d) r adio wave fr equency

LEVEL-2 1. The t er m ‘Black Box’ is mor e commonly used in r elat ion t o which of t he following ? (a) I t is a box in which high gr ade ur anium is kept t o pr event r adiat ion (b) I t i s a t i me capsul e i n whi ch r ecor ds of impor t ant event s ar e kept t o be opened at a lat er dat e (c) I t is a flight r ecor der in an aer oplane (d) None of t hese 2. The lightning conductor used in building, pr otects t he building by (a) dissipating the elect r ic char ge away fr om t he building (b) conduct ing t he light ning safely t o t he gr ound (c) absor bing t he elect r ic char ge (d) none of t hese 3. The pr inciple of wor king of per iscope is based on (a) r eflect ion only (b) r efr act ion only (c) r eflect ion and r efr act ion (d) r eflect ion and int er fer ence

SCIENCE AND TECHNOLOGY

4. The wor king of t he quar t z cr yst al in t he wat ch is based on (a) Johnson effect (b) Phot oelect r ic effect (c) Edison effect (d) Piezo elect r ic effect 5. Jet engines ar e (a) r ot ar y engines (b) t ur bine engines (c) ext er nal combust ion engines (d) r eact ion engines 6. The conver sion of elect r ical ener gy into chemical ener gy is obser ved in (a) fan (b) st or age bat t er y (c) heat er (d) incandescent bulb 7. Gr eenhouse is (a) a bu i l di ng chi efl y of gl ass i n wh i ch t he t emper at ur e is ver y low (b) a building in which gr een plants are cultivated (c) a bu i l di ng chi efl y of gl ass i n wh i ch t he temper atur e is maintained within the desir ed r ange (d) none of t hese 8. The mixed oxide fuel is used for which of t he following ? (a) Nuclear React or s (b) Aer oplanes (c) Cr yogenic Engines (d) PSLV r ocket s 9. The anode in a dr y cell consist s of (a) gr aphite (b) zinc (c) copper (d) cadmium

1.7

10. A comput er can be fr eely pr ogr ammable (a) if it is of a digit al t ype (b) if it is cont r olled synchr onously (c) if it cont ains a r ead only memor y (ROM ) (d) if it cont ains a r andom access memor y (RAM ) 11. L ife appear ed about (a) 2.6 billion year s ago (b) 3.6 billion year s ago (c) 4.6 billion year s ago (d) 5.6 billion year s ago 12. Alber t Einst ein was a (a) physician (b) chemist (c) physicist (d) biologist 13. The mass-ener gy r elat ion is t he out come of (a) Quant um t heor y (b) Gener al t heor y of r elat ivit y (c) Field t heor y of ener gy (d) Special t heor y of r elat ivit y 14. Who invent ed Radar ? (a) Rober t Wat son wat t (b) M ax Planck (c) H umphr y Davy (d) H enr y Becquer el 15. The I ndian scient ist whose name is associat ed wi t h t h e measur ement of gr owt h i n pl ant s is (a) J.C. Bose (b) H .G. K hor ana (c) Meghnad Saha (d) C.V. Raman

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (a)

2. (b)

3. (a)

4. (c)

5. (a)

11. (d)

12. (b)

13. (c)

14. (d)

15. (d)

6. (a)

7. (d)

8. (d)

9. (a)

10. (d)

7. (c)

8. (b)

9. (c)

10. (d)

7. (c)

8. (c)

9. (a)

10. (c)

LEVEL-1 1. (b)

2. (c)

3. (b)

4. (c)

5. (a)

11. (c)

12. (b)

13. (c)

14. (c)

15. (c)

6. (c)

LEVEL-2 1. (c) 11. (b)

2. (b) 12. (c)

3. (c) 13. (d)

4. (d) 14. (a)

5. (d) 15. (c)

6. (b)



2

SPORTS

CHAPTER

SPORTS AN D RELATED TERM S  Badmi nt on Angled dr ive, ser ve, bir d, deuce, double dr op, fault , let , lob, love all, smash

 Bask et ball Ball, basket , blocking, dr ibbling, fr ee t hr ow, held ball, holding, jump ball, mult iple t hr ows, pivot .

 Baseball Base, bunt i ng, bat t er y, di amond, hi t t er, home, pit cher, pullout , st r ike.

 Billiar ds Baulk line, br eak, bolting, Cannon, cue, hazar d, inoff, jigger, l ong jenney, pot , scr at ch, scr ew back, shor t jenney, spot st r oke.

 Boxi ng Auxiliar y point syst em, babit punch, defence, hook, jab, knock out , slam.

 Br i dge Auct ion, Chi cane, declar er, dummy, gr and sl am, nor t r umps, r evoke, r uff, suit .

 Chess Bishop, capt ur e, cast ling, checkmat e, en passant , gambit , king, knight , pawn, queen, r ook, stalemat e.

 Cr ick et Ashes, boundar y, bowling, cat ch, chinaman, cr ease, duck, follow on, googly, gully, hat -t r ick, hit wicket , l.b.w., no ball, off br eak, on dr ive, pitch, r ubber, silly point , squar e leg, st one walling, yor ker, wicket .

 Cr oquet H oops, mallet , peg out .

 D r aught s Huff.

 F oot ball Advant age clause, blind side, cor ner kick, dr ibble, fr ee kick, mar king, off side, penalt y kick, t hr ow in t r ipping.

 Golf Bogey, bunker, caddie, dor my, fair way, four some, gr eed holes, links, par, put t , t ee, t hr eesome.  Gymnast ics A-bar s, ar iel, bl ock s cone of swi ng, dish, giant s, inlocat e, kip, planche, t ar iff, t umble, wr ap.  H ock ey Bul ly, cor ner, fl ick, fr ee-hit , r oll i n, scoop, shor t cor ner, st ick, st r iking cir cle, t ackle, t ie br eaker.  H or se-r acing Jockey, punt er, st eeplechase.  J u do Chui, dan, dojo, gyaku, ippon, r andor i, yoshi, yuko.  K ar at e Dachi, gedan, jion, koka, shir o, ude, zen-no.  P ol o Bunker, chukker, mallet  Row i n g Bow, bucket , cow, feat her, paddle, r egat t a.  Ru gby Tr ackle, lines, scr um, t ouch, t r y.  Shoot i ng Bag, bull’s eye, mar ks- manship, muzzle, plug.  Swi mmi ng Back-st r oke, br east -st r oke, but ter fly-st r oke, cr awl, fr ee st r oke.  Table Tennis Ant i l oop, back spi n, chop, l oop, penhol d, gr i p, t widdle.  Tenni s Ace, backhand, st r oke, deuce, fault , let, love, volley.  Volleyball Ace, bl ock i ng, doubl i ng, heave, hol di ng, spi k e, ser vice.  Wr est ling Half nelson, head lock, heave, hold, r ebouts, scissor.

2.2

SPORTS

CU PS AN D TROPH I ES ASSOCI ATED WI TH SPORTS AN D GAM ES  Air Racing

 Cricket

Jawahar lal Challenge Tr ophy

Ant hony Demellow Tr ophy

K ing’s Cup

Ashes

Schneider Cup (Sea plane r ace in UK )

 Archery Feder ation Cup

 Athletics Char minar Tr ophy Wor ld Cup

 Badminton Agar wal Cup Amr it Diwan Cup Austr alasia Cup Chadha Cup

Asia Cup Benson and H edges Cup Champions Tr ophy Char minar Challenge Cup C.K . Nayudu Tr ophy Deodhar Tr ophy Duleep Tr ophy G.D.Bir la Tr ophy Gillett e Cup Ghulam Ahmed Tr ophy

Eur opean Cup

I r ani Tr ophy

Nar ang Cup

Jawahar lal Nehr u Cup

K onica Cup

M er chant Tr ophy

Thomas Cup

Nat west Tr ophy

Uber Cup

Pr udential Cup (Wor ld Cup)

Wor ld Cup

Ranji Tr ophy

Yonex Cup

 Basketball Basalat Jha Tr ophy B.C.Gupta Trophy Todd Memor ial Tr ophy William Jones Cup Feder ation Cup

 Boat Rowing Amer ican Cup (H acht r acing) Wellington Tr ophy (I ndia)

 Boxing

Shar jah Cup Sheesh M ahal Tr ophy Vijay H azar e Tr ophy Vijay M er chant Tr ophy

 Football Bandodkar Tr ophy B.C.Roy Tr ophy DCM Cup Dur and Cup Eur opean Cup

Aspy Adjahia Tr ophy

Feder ation Cup

Feder ation Cup

Gover nor ’s Cup

 Bridge

Gr eat Wall Cup

H olkar Tr ophy

I FA Shield

Ruia Gold Cup

Jules Rimet Tr ophy (Wor ld Cup)

Singhania Tr ophy

Kings Cup

 Chess

L al Bahadur Shast r i Tr ophy

Naidu Tr ophy

Mer deka Cup

K hait an Tr ophy

Nat ions Cup

L imca Tr ophy

Nehr u Gold Cup

Wor ld Cup

SPORTS

Nizam Gold Cup

Rangaswami Cup

Rover s Cup

Rene Fr ank Tr ophy

Santosh Tr ophy

Scindia Gold Cup

Staffor d Cup

Shr ir am Tr ophy

Subr ot o Cup

Wor ld Cup

Todd M emor ial Tr ophy

Yadavindr a Cup

UEFA Cup Vittal Tr ophy Air lines Cup Asia Cup Amer ica Cup Winner ’s Cup

 Kabaddi Feder ation Cup

 Kho-Kho Feder ation Cup

 Shooting

I ndir a Gandhi Tr ophy

Nor t h Wales Cup

Rajiv Gandhi Tr ophy

Welsh Gr and Pr ix

 Golf

 Table Tennis

Canada Cup

Asian Cup

Eisenhower Tr ophy

Ber na Bellack Cup

Muthiah Gold Cup

Cor billion Cup (Women)

Nomur a Tr ophy

Elect r a Gold Cup

Ryder Cup

Gasper -Giest Pr ize

Walker Cup

Jayalaxmi Cup (Women)

U.S. Open

Pet hapur am Cup (M en)

Br it ish Open

Swaythling Cup (M en)

 H ockey Agha Khan Cup Allwyn Asia Cup Azian Shah Cup Beight on Cup Bombay Gold Cup Champions Tr ophy Clar ke Tr ophy Dhyan Chand Tr ophy Gur u Nanak Cup I ndir a Gandhi Gold Cup

Wor ld Cup

 Tennis Davis Cup Feder ation Cup Wimbledon Tr ophy U.S.Open Fr ench Open Austr alian Open Hopman Cup

 Volleyball

I nt er continental Cup

Feder ation Cup

Kuppuswamy Naidu Cup

Shivant hi Gold Cup

M ahar aja Ranjit Singh Gold Cup

Wor ld Cup

M odi Gold Cup

 Yacht Racing

Nehr u Tr ophy

Amer ica Cup

Obaidullah Gold Cup

2.3

2.4

SPORTS

OLYM PI C GAM ES  The fir st moder n Olympic Games t ook place in 1896, founded by t he Fr enchman Bar on de couber t in.

 They ar e held ever y four year s.  Women fir st compet ed in 1900.  The fir st separ at e wint er games celebr at ion was in 1924, beginning in 1994, t he wint er games will t ake place bet ween summer games celebr at ion.  The Olympic mot t o is Cit ius, Allius, For t ivs meaning Sur fer H igher and st r onger composed by Fat her Dixon in 1897, and int r oduced in 1920 for t he fir st t ime.

Venues Year

Summer Games

Winter Games

Year

Summer Games

Wint er Games

1896

At hens, Gr eece

-

1968

M exico Cit y, M exico

Gr enoble, Fr ance

1900

Par is, Fr ance

-

1972

M unich, Ger many

Sapporo Japan

1904

St . L ouis, USA

-

1976

Montr eal, Canada

I nnsbr uck, Aust r ia

1908

L ondon, UK

-

1980

M oscow, Russia

L ake Placid, USA

1912

St ockholm, Sweden

-

1984

L os Angeles, USA

Sar ajevo Yugoslavia

1920

Ant wer p, Belgium

-

1988

Seoul Sout h K or ea

Calgar y, Canada

1924

Par is, Fr ance

Chamonix Fr ance

1992

Bar celona, Spain

Alber t ville, Fr ance

1928

Amst er dam,

St . M or t iz

The Net her lands

Switzerland

1932

L os Angeles, USA

L ake Placid, USA

1936

Ber lin, Ger many

Gar misch Ger many

1994 1996 1998 2000 2002 2004 2006 2008

At lanta, USA Sydney, Aust r alia At hens, Gr eece Beijing, China

L illehammer Nor way Nagano, Japan Salt L ake Cit y, USA Tur in, I t aly -

2010

-

Vancouver, Canada

2012

L ondon, UK

-

2014

-

Sochi, Russia

2016

Rio de Janeir o

-

2018

-

Pyeonghchang,

Par t enkir chen 1948

L ondon, UK

St. Moritz, Switzerland

1952

H elsinki, Finland

Oslo, Nor way

1956

M ellbour ne, Aust r alia Cor t ina, I t aly

1960

Rome, I t aly

Valley, USA

1964

Tokyo, Japan

I nnsbr uck, Aust r ia

Sout h K or ea

COM M ON WEALTH GAM ES

Emblem

 Fir st held as t he Br it ish Empir e Games in 1930.

A br ight full r ising Sun wit h int er locking r ings.

 They t ak e pl ace ever y four year s and bet ween Olympic celebr at ions.

WORLD CU P (FOOTBALL)

 T h ey becam e t h e B r i t i sh commonwealt h Games in 1954

E m pi r e

The cur r ent t it le was adopt ed in 1970. ASI AN GAM ES M ot t o

Ever onwar d (coined by Pt . Jawahar lal Nehr u)

an d

 The lar gest single spor t ing event in t he wor ld is t he Wor ld Cup Foot ball t our nament .  The fir st edit ion was held in 1930.  The 2018 Wor ld Cup, t he lat est edit ion was held in Russi a ; Fr ance emer ged wi nner by defeat i ng Cr oatia.  T h e Wor l d Cu p i s n ow of f i ci al l y cal l ed t h e Jules-Rimet Cup.

SPORTS

2.5

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. Who was t he fi r st I ndi an t o wi n t he Wor l d Amat eur Billiar ds t itle? (a) Geet Set hi (b) Wilson Jones (c) M ichael Fer r eir a (d) M anoj K ot har i 2. Rangaswami Cup is associat ed wit h which of t he following ? (a) H ock ey (b) Pol o (c) Badminton (d) Basket ball 3. Gr and Slam in Tennis means a player has t o win (a) Australian open, French open and Wimbledon (b) Aust r alian open, U.S. open and wimbledon (c) Austr alian open, Fr ench open, U.S. open and Wimbledon (d) None of t hese 4. Which of t he following is a pair of names of t he same game ? (a) Golf-Polo (b) Billiar ds-Snooker (c) Volleyball-Squash (d) Soccer -Football 5. Whi ch of t he fol lowi ng i s a set of fi ve event s included in M oder n Pent at hlon ? (a) Cycl i ng, Sk at i ng, Shoot i ng, Gymnast i cs, Running (b) Judo, Shooting, Swimming, Cycling, Running (c) Hor se r iding, Fencing, Shooting, Gymnastics, Running (d) Hor se Riding, Fencing, Shoot ing, Swimming, Running 6. The famous player Pele is associdated with which of t he following games ? (a) Table Tennis (b) Foot ball (c) H ock ey (d) Volleyball 7. H ow many pl ayer s ar e t her e i n a wat er pol o t eam ? (a) 4 (b) 5 (c) 6 (d) 7 8. Who is t he fir st I ndian woman t o win an Asian Games gold in 400m r un? (a) M.L.Valsamma (b) P.T.U sha (c) Kamaljit Sandhu (d) K.Malleshwar i 9. Which of t he following place is called t he M ecca of I ndian Foot ball ? (a) Delhi (b) Mumbi (c) Kolkata (d) Ambala

10. Tr ipping is associated wit h (a) Snooker (b) Foot ball (c) Cr icket (d) L awn Tennis 11. The Olympic Symbol compr ises of five r ings or cir cles linked t oget her t o r epr esent (a) spor t ing fr iendship of all people (b) five cont inent s (c) both (a) and (b) (d) none of t hese 12. The fi ve i nt er t wi ned r i ngs or ci r cl e son t he Olympic Flag made of whit e (fr om left t o r ight ) ar e (a) blue, yellow, black, gr een, and r ed (b) yellow, r ed, gr een, black and blue (c) r ed, gr een, black, yellow, and blue (d) yellow, gr een, black, blue and r ed 13. Which of the following cups/tr ophies is associated wit h foot ball ? (a) Davis Cup (b) Deodhar Tr ophy (c) Champions Tr ophy (d) Santosh Tr ophy 14. I ndia won it s fir st Olympic hockey gold in...? (a) 1928 (b) 1932 (c) 1936 (d) 1948 15. I f you scor ed a cannon, which game would you be playing ? (a) Foot ball (b) Cr icket (c) Billiar ds (d) H ock ey

LEVEL-1 1. Who r ecei ved t he fir st Nobel pr i ze in Physics in I ndia? (a) Dr. C.V. Raman (b) Dr. H ar gobind K hur ana (c) Pr of. C.N .R. Rao (d) Pr of. N ar l ik ar

2. Which country won the FIFA world cup, 2014 in Football ? (a) Germany (b) Argentina (c) Brazil (d) France 3. Who is the winner of Mens Singles Title in Tennis in US open, 2014 ? (a) Roger Federer (b) Kei Nishikori (c) Marin Cilic (d) Rafael Nadal 4. Who is the winner of Nobel Prize, 2014 in the field of Economics ? (a) Patrick Modiano (b) Malala Yousafzai (c) Jean Tirole (d) Kailash Satyarthi

2.6

SPORTS

5. UBER Cup is related to (a) International Badminton (Men) (b) International Volleyball (Men) (c) International Volleyball (Women) (d) International Badminton (Women) 6. First Sportsperson to be conferred with Award "Bhart Ratna" (a) Sachin Tendulkar (b) Dhyan Chand (c) Balbir Singh (d) Vijay Amritraj 7. Next Asian Games in2018 shall be held in (a) Seoul (b) Bangkok (c) Kualalumpur (d) Jakarta 8. Who among the following did not win a medal in Asian Games 2014 ? (a) Yogeshwar Dutt (b) Sushil Kumar (c) Abhinav Bindra (d) Jitu Rai 9. What is common amongst Mahesh Bhupathi, Ivan Lendl, Roger Federer ? (a) They are all Arjun Award winners (b) They all International Tennis players (c) They are all Social Activists (d) They are all Asian Games medal winners 10. Which of the following pairs was announced recently as Joint Noble Peace Prize winner. (a) Kailash Satyarthi and Malala (b) Amartya Sen and Benazir (c) Morkel and Hosni Mubarak (d) Anwar Sadat and Begin 11. The UN Public service award 2015 for eliminating open defecation has been given to (a) Surat in Gujarat (b) Gorakhpur in Uttar Pradesh (c) Nadia in West Bengal (d) Mahabaleshwar in Maharashtra 12. The Champion’s League Trophy is given for (a) the best club in Soccer (b) best player of team in Basket ball (c) maximum goal maker in Hockey (d) man of the match in Cricket 13. Cricket batsman has to leave the field for “hit wicket” when (a) Ball hits leg before bat (b) Keeper removes the bail (c) Bat hits any fielder (d) Wicket is touched by bat or body of the batsman

14. The latest Nobel Peace Prize was awarded to two people, a man and a woman. To which countries do they belong? (a) China (b) India and Pakistan (c) US and Greece (d) India and Indonesia 15. The tournament that takes place in Roland Garros in Paris is associated with (a) Lawn tennis (b) Table tennis (c) Basketball (d) Bowling

LEVEL-2 1.

Who is the Winner of Pr o Kabaddi league in 2014? (a) U M umba (b) Jai pur Pink Pant her s (c) Pat na Pir at es (d) Bengalur u Bulls 2. Who is t he wi nner of Nobel Pr i ze, 2014 in t he field of L it er at ur e ? (a) Phil ip Rot h (b) Pat r ick M odiano (c) H ar uki M ur akami (d) Ngugi Wa Thiong'o 3. Ji t u Rai won Gol d M edal i n t he r ecent Asi an Games i n t he fol lowing field : (a) Ar cher y

(b) Wr est li ng

(c) Boxing

(d) Shoot ing

4. Shant i Swar up Bhat nagar Awar d i s given for out st andi ng cont r i but ion in t he fol lowi ng fi el d: (a) Science

(b) L it er at ur e

(c) Economy (d) Per for ming Ar t s 5. Who i s t he wi nner of M en 's Si ngl es Ti t l e i n Wi mbledon, 2014 in Tennis ? (a) Roger Feder er (b) Rafael Nadal (c) M ar in Cilic (d) Novak Djok ovic 6. M er deka Cup is associ at ed wi t h (a) I nt er nat i onal Table Tennis (b) Badmint on (c) H ock ey (d) I nt er nat ional Foot ball 7. M at ch Col . X (Spor t sper son) and Col. Y (Spor t s): Col. X Col. Y P. Jitu Rai 1. Badmint on Q. H eena Sidhu 2. Wr est li ng R. Jwala Gut t a 3. Shoot ing S. Yogeshwar Dut t

SPORTS

(a) P-3; Q-3, R-1, S-2 (b) P-2. Q-3, R-1, S-2 (c) P-2. Q-2. R-1. S-3 (d) P-3, Q-1. R-1. S-2 8. The slogan of Asian Games I ncheon 2014 was (a) Gr een, Cl ean and Fr iendship (b) We Cheer, We Shar e, We Win (c) Di ver si t y Shines her e (d) The Games of Your L i fe 9. What ar e t he t hr ee val ues cher i shed by t he Commonweal t h Games? (a) Get Set , Go, and Pl ay . (b) Fast er, H igher, St r onger. (c) Di ver sit y Shi nes her e. (d) H umanit y, Equali t y, Dest iny. 10. For par t i ci pat ion in which i nt er nat ional games event has the Men's I ndian Hockey team qualified aft er t he Wor l d H ockey L eague? (a) Common weal t h Games (b) Olympics (c) Asian Games (d) Wi nt er Ol ympi cs

2.7

11. Wher e will Commonwealt h Games 2018 be held? (a) Glasgow, Scot l and (b) I nchon, Sout h K or ea (c) Gol d Coast , Queensl and ,Aust r al ia (d) Abuja, Niger ia 12. To par t icipat e i n t he Olympi cs at Rio de Janei r o in 2016 whi ch game di d I ndian t eam quali fy i n? (a) M en's H ock ey (b) Swimming (c) Ar cher y (d) Boxing 13. What was t he mascot for t he Commonweal t h Games hel d i n I ndia i n 2010? (a) Bison (b) Peacock (c) Deer (d) Tiger 14. For t he 2016 Ol ympics, whi ch t eam defeat ed t he M en's I ndian H ockey t eam t o secur e t he pl ace? (a) Fr ance (b) Ger many (c) England (d) I taly 15. Whi ch i nt er nat i onal spor t s event wi l l be hel d i n Gol d Coast , Qu een sl an d , A u st r al i a i n 2018? (a) Asian Games (b) Common wealt h Games (c) Asia Paci fi c Games (d) Wi nt er Ol ympics.

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (b) 11. (c)

2. (a) 12. (a)

3. (c) 13. (d)

4. (b) 14. (a)

5. (d) 15. (c)

6. (b)

7. (d)

8. (c)

9. (c)

10. (b)

7. (d)

8. (b)

9. (b)

10. (a)

7. (c)

8. (c)

9. (d)

10. (b)

LEVEL-1 1. (a)

2. (a)

3. (c)

4. (c)

5. (d)

11. (c)

12. (a)

13. (d)

14. (b)

15. (a)

6. (a)

LEVEL-2 1. (b)

2. (b)

3. (d)

4. (a)

5. (d)

11. (c)

12. (a)

13. (d)

14. (c)

15. (b)

6. (d)



3

INDIAN HISTORY

CHAPTER

I N DI A’S FREEDOM STRU GGLE 1885

F or m at i on of I n di an N at i on al Congr ess. Fi r st sessi on hel d at B om bay on 28t h D ecem ber at t ended by 72 delegat es.

1916, Apr il 28

H ome Rul e l eague f ou nded by Ti l ak wi t h i t s headquar t er s at Poona (I ndian home r ule league of I ndia).

1905

Par t it ion of Bengal announced by Cur zon.

1916, Sept . 25

A n ot h er H om e Ru l e L eagu e st at r ed by Annie Besant .

1906

Muslim league founded at Dacca by Aga K han t he Nawab of Dacca and N aw ab M oh si n -u l -M u l k , I st communal Par t y.

1916

F ou n dat i on of B an ar as H i n du U n i v er si t y by M adan M oh an M alaviya, Lucknow pact .

1917, Apr il

M ahat ma Gandhi l aunches t he Champar an satyagr aha.

1917, Aug. 20

M ont ague, t he secr et ar y of St at e for I ndia, declar es t hat t he goal of t he Br it ish gover nment in I ndia is t h e i n t r odu ct i n of r espon si bl e gover nment .

1918, M ar ch 23

Fir st All-I ndia Depr essed Classes confer ence was held.

1918, Apr il

Row l at t (sedi t i on ) com m i t t ee submi t s i t s r epor t . Rowl at t Bi l l int r oduced on Febr uar y 16, 1919.

1919, Apr il 6

All I ndia har t al over Rowlat t Bill cal l ed by Gandhi ji , I st N at i onal agitation.

1919, Apr il 13

Jallianwala Bagh t r agedy and t he gr eat Amr it sar massacr e.

1919, Dec. 25

M ont ague Chelmsfor d Refor m or t he Gover nment of I ndia Act, 1919 announced.

1920

Fir st meet ing of All I ndia Tr ade Union Congr ess (AI TUC) held at Bombay pr esi ded over by L al a Lajpat Rai.

1920, Dec.

I ndi an N at i onal Congr ess (I NC) adopt s t h e N on -cooper at i on r esolut ion.

1907

Congr ess split at Sur at session.

1907

L al a L aj pat Rai and Aj i t Si ngh depor t ed t o M andal ay fol l owi ng r iot s in t he canal colony of Punjab.

1908

K hudir am Bose execut ed.

1908, July 22

T i l ak sen t en ced t o si x y ear s i m pr i son m en t on ch ar ges of sedit ion.

1909, M ay 21

M or l ey M i n t o Ref or m s or t h e I ndian Council Act 1909 announced.

1911

Capit al of I ndia was shift ed fr om Calcut t a t o Delhi.

1912, Dec. 23

Bomb t hr own on L or d H ar dinge in Ch an di n i Ch ow k , D el h i by Rashbehar i Bose and Sachi ndr a Sanyal.

1913, Nov. 1

Gh adar par t y f or m ed at San Fr ancisco t o or ganise a r ebellion in I ndia t o over t hr ow t he Br it ish r ule.

1914

Fir st Wor ld War st ar t ed.

1915, Jan.

Gandhiji ar r ives in I ndia.

1916

Gandhi ji founded t he Sabar mat i Ashr am at Ahmedabad.

3.2

INDIAN HISTORY

1922, Feb. 5

1922

Chaur i Chaur a incident which led

1923, Jan. 1

1925, June 16

t r icolour of I ndia on t he banks of

Cooper at ion movement .

Ravi in L ahor e.

Second M oplah upr ising, M alabar

1927, Nov. 8

1930, Feb. 14

Wor king commit t ee of I NC meet s

by Rabindr a Nat h Tagor e.

Disobedience r esolut ion.

Swar ajist par ty founded by Motilal

1930, M ar ch 12

Civil Disobedience Movement with

Deat h of Deshbandhu

his epic Dandi mar ch. 1930, Nov. 12

begins in L ondon t o consider t he

r evolut ionar ies & K akor i

r epor t of Si mon Commi ssi on for

conspir acy case.

t he fut ur e const it ut ional set -up in

A ppoi n t m en t

of

I ndia.

t h e Si m on 1931, M ar ch 5

1931, M ar ch 23

1929, Oct . 31

1929, Dec. 31

Bhagat Singh, Sukh Dev and Raj Gur u execut ed (in L ahor e Case).

Al l Par t i es M usl i m Confer ence 1931, Sept . 7

Second Round Tabl e confer ence begins Mahatma Gandhi ar r ives in L ondon t o at t end it .

Bhagat Si ngh and Bat uk eshwar Dut t dr op bomb i n t he cent r al

Gandhi I r wi n pact si gned. Ci vi l Disobedience movement suspended.

Nehr u Repor t for a new

under t he leader ship of Jinnah.

1929, Sept . 13

F i r st Rou n d t abl e con f er en ce

L oot i ng of t r ai n near K akor i by

for mulat es t he “ four t een poi nt s”

1929, Apr il 8

M ahat ma Gandhi l aunches t he

Nehr u and ot her s.

const it ut ion of I ndia. 1929, M ar ch 9

Fir st I ndependence Day obser ved.

at Sabar mat i and passed t he Civil

Commission. 1928

1930, Jan. 26

Vishwa Bhar ati Univer sity st ar t ed

Chit t ar anjan Das. 1925, August

J aw ah ar l al N eh r u h oi st s t h e

t o t h e su spen si on of t h e N on -

coast , K er ala. 1922, M ay

1930, Jan. 1

1932, Aug. 16

Br i t i sh Pr i me M i ni st er Ramsay

l egi sl at i ve Assembl y t o pr ot est

M ac D on al d an n ou n ces t h e

against t he Public Safet y Bill.

Communal Awar d giving separ at e elect or at e t o H ar ijans.

Jat in Das dies in jain after 64 days of fast .

1932, Sept . 20

Gandhi’s fast unt o deat h.

L or d I r win’s announcement t hat

1932, Sept . 26

Poona pact signed by which t he

t he goal of Br it ish policy in I ndia

H ar i j ans get r eser ved seat s i n

was t he gr ant of dominion st at us.

place of separ at e elect or at e.

L ahor e sessi on of t he Congr ess

1932, Nov. 17

begins at L ondon.

under Jawahar L al Nehr u adopt s t he goal of complet e independence (Poor na swar aj) for I ndia.

Thi r d Round Tabl e Confer ence

1935, Aug. 4

Gover nment of I ndia Act passed.

INDIAN HISTORY

1937

1938, Feb. 19-20

Elect ions held in I ndia under t he

1943, Oct .

t h e l eader sh i p of I N A an d

minist r ies in seven provinces.

pr ocl ai ms t he for mat i on of t he ‘Pr ovi si onal Gover nment of fr ee

H ar i pur an sessi on of t he I N C. t he congr ess pr esident .

1939, Mar. 10-12

Tr ipur i session of I NC.

1939, Apr il

Subhash Chandr a Bose r esigns the pr esident ship of t he I NC.

I ndia’ at Singapor e. 1943, Dec.

‘Divide and Quit ’. 1944, June 25

pr ovinces r esign against t he war

of I ndian polit ical leader s. 1946 Feb. 1-8

Rat ing M utiny.

1946, M ar ch 15

B r i t i sh Pr i m e M i n i st er A t t l ee

policy of t he Br it ish gover nment . 1939, Dec. 22

M u sl i m L eagu e obser v es t h e

announces t he Cabinet M ission. 1946, Aug. 6

r esi gn at i on of t h e con gr ess minist r ies as Deliver ance day. 1940, M ar ch

1946, Dec. 9

Vicer oy Linlithglow announces the

boycot t ed by t he M uslim L eague. 1947, Feb. 20

1941

Deat h of Rabindr anat h Tagor e.

1941, Jan. 17

Subhash Chandr a Bose escapes fr om I ndia t o Ger many.

1942, M ar ch 11

1942, Aug. 8

I NC meet s at Bombay and adopt s

1947, M ar ch 24

Quit I ndia M ovement begins.

1942, Sept . 1

I ndian National Ar my (Azad H ind Fauj) is for med at Singapor e.

B r i t i sh

L oar d M ount bat t en, l ast Br it i sh vi cer oy and gover nor gener al of I ndia is swor n in.

1947, June 3

Mountbatten Plan for the partition of I ndia is announced.

1947, July 4

I n di an

I n depen den ce

Bill

i n t r odu ced i n t h e H ou se of

t he Quit I ndia r esolut ion. 1942, Aug. 11

t he

June 1948.

Chur chi l l announces t he Cr i pps M ission.

t h at

gover nment would leave I ndia by

Con gr ess l au n ch es I n di vi du al Sat yagr aha movement .

B r i si t h Pr i m e M i n i st er A t t l ee decl ar es

August offer . 1940, Oct .17

Fi r st session of t he Const it uent

Assembly of I ndia st ar t s, but it is

l eagu e passes t h e Pak i st an

1940, Aug. 10

Wavell invit es Nehr u t o for m an int er im gover nment .

L ah or e sessi on of t h e M u sl i m Resolut ion.

Wavell calls Simla confer ences in a bid to for m the Executive Council

Second wor ld war begins. Vicer oy

Con gr ess m i n i st r i es i n t h e

K ar achi sessi on of t he M usl i m L eagu e adopt s t h e sl ogan of

declar es t hat I ndia t oo is at war. 1939, Nov. 5

Subhash Chandr a bose t akes over

A ct of 1935. T h e I N C f or m s

Subhash Chandr a Bose el ect ed

1939, Sept . 3

3.3

Com m on s an d passed by t h e Br it ish par liament (July 18). 1947, Aug. 15

I ndia won fr eedom.

3.4

INDIAN HISTORY

I M PORTAN T ORGAN I SATI ON S AN D PARTI ES Parties and Organisations

Founders, year and Place

M uslim L eague

Aga K ha, t he Nawab of Dacca M ohsin ul M ulk (1906, Dacca)

H ome Rule L eague

Bal Gangadhar Tilak (July 1916) , Annie Besant (Sept . 1916)

Ant i Non-Cooper at ion Associat ion

Pur ushott am Das Thakur das (1920-21)

Johr at Sar vajanik Sabha

Rash Behar i Ghose (1893, Assam)

Raja M undar i Social Refor m Associat ion

Vir sallngam (1878)

Ant i Cir cular Societ y

K . K . M it r a

L ok Seva M andal

L ala L ajpat Rai, Punjab

I ndependent Congr ess Par t y

Madan M ohan Malviya, (1926)

Unit ed I ndia Pat r iot ic Associat ion

Sayyid Ahmed Khan

Br it ish Associat ion of Avadh

Raja Shiv Pr asad Sahu

L iber al Associat ion

Sapr u, Jayakar and Chint amani

I ndian L iber al Feder at ion

Sur endr a Nat h Baner jee and ot her s (1919)

H indust an Seval Dal

N. G. H ar dikar

I ndependence of I ndia league

Jawahar L al Nehr u and Subhas Chandr a Bose, (1928)

Pr aja Par t y

Akr am Khan, Fazi-ul-H uq and Abdur Rahim

H indu Associat ion

Annie Besant

K r ishak Pr aja Par t y

Faziul Haq

Sout h I ndia Feder at ion of peasant s & Agr icult ur al labour

N.G. Ranga and Namboodir ipad (1935)

Unionist Par t y

Fazl Hussain

Rasht r iya Swayam Sevak Sangh

Hedgewar (1925)

All I ndia Unt ouchabilit y league or H ar ijan Sevak Sangh

Gandhiji (1932)

H indu M aha Sabha

Founded in 1917, Revived by M adan M ohan M alviya in 1925.

Jana Sangh

Shyam Pr asad M ukher jee

National M ohammedan Association

Amir Ali, 1878, Calcut t a

Communit st Par t y of I ndia (CPI )

M .N. Roy, 1920 at Tashkent

L abour Swar aj Par t y

M uzaffar Ahmmed & Qazi Nazr ul 1925 and Wor ker ’s Par t y

K ir t i K isan Par t y

Sohan Singh Josh (Punjab)

Wor ker ’s & Peasant s Par t y

S.S. M ir ajkar, K .N. Joglekar & S.V. Ghat e, 1927, Bombay

Bihar Socialist Par t y

Jai Pr akash Nar ayan, Phulan Pr asad Ver ma, 1931

Congr ess Socialist Par t y

Nar endr a Dev, Jai Pr akash Nar ayan and M inoo M asani 1934

For war d Block

Subhas Chandr a Bose, 1939

Congr ess L abour Par t y

1926, Bombay

Revolut ionar y Socialist Par t y

1940

Radical Democr at ic Par t y

M .N. Roy, 1940

Bolshevik L eninist par t y

I ndr a Sen & Ajut Roy, 1941

Revolut ionar y Communist Par t y

Saumendr anat h Tagor e, 1942

H indust an Socialist Republican Association

Chandr a Shekhar Azad and ot her s, 1928

Punjab Socialist Par t y

1932

I ndependent Labour Part y

B.R. Ambedkar

INDIAN HISTORY

3.5

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. Which one of t he following leader s was elect ed as Pr esident of t he All I ndia K hilafat Confer ence held in Delhi in 1919? (a) Mahatma Gandhi (b) Shaukat Ali (c) M .A. Jinnah (d) None of t hese 2. Which movement was launched along wit h t he K hilafat M ovement ? (a) Swadeshi M ovement (b) H ome Rule M ovement (c) Civil Disobedience M ovement (d) Non-cooper at ion M ovement 3. The Swar aj Par t y st ood for (a) Captur ing Legislatur e Councils by contest ing elect ions (b) Boycot t ing t he elect ions (c) Over t hr ow of Gover nment by or ganising a count r y wide st r ike (d) None of t hese 4. Of t he followings who was not a member of t he K hilafat Committ ee? (a) Maulana Shaukat Ali (b) Maulana Muhammad Ali (c) M .A. Jinnah (d) Maulana Abdul Kalam Azad 5. Who of t he fol lowi ng was not i nvol ved i n t he incident r elating to thr owing of a bomb in Centr al L egislat ive Assembly on Apr il 18, 1929? (a) Sar dar Bhagat Singh (b) Khudiram (c) B.K . Dut t (d) All t hese 6. The All I ndia Muslim League was founded under t he leader ship of (a) M aulana M ohmmed Ali (b) H asan I mam and M azhar -ul H aque (c) Agha Khan and Mohsinul-M ulk (d) H akim Ajmal K han and Zafar Ali K han

7. Who was t he fir st Englishman t o pr eside over t he Congr ess session at Allahabad in 1908? (a) W. Wedder bur n

(b) A.O. H ume

(c) Geor ge Yule

(d) M r s. Annie Besant

8. Who was t he fir st M uslim pr esident of t he I NC? (a) H akim Ajmal K han (b) Rafi Ahmad Kidwai (c) Abul Kalam Azad (d) Badr uddin Tayabjee 9. Which of the following did not have an impor tant r ole in t he Swadeshi movement ? (a) Women (b) Peasants (c) Students

(d) Muslims

10. M.K. Gandhi applied his ‘Satyagr aha’ first against t he (a) Eur opean indigo plant er s. (b) M ill owner s of Ahmedabad. (c) Br i t i sh gover nment t o pr ot est against t he Rowlatt Act s (d) Racist aut hor it ies of Sout h Afr ica. 11. The car di nal pr i nci pl e of M ahat ma Gandhi ’s Satyagr aha was (a) Fear lessness

(b) Tr uthfulness

(c) Non-violence

(d) All of t hese

12. Fir st gr eat exper iment in sat yagr aha came in 1917 in (a) Ahmedabad (c) Sabar mati

(b) Champaran (d) Bardoli

13. Gandhiji’s Champar an Movement was for (a) The secur it y of r ight s of H ar ijans (b) Civil disobedience movement (c) M aint aining t he unit y of H indu societ y (d) Solving t he pr oblem of t he indigo wor ker s 14. Who was t he polit ical gur u of Gandhiji ? (a) Naur oji (b) Gokhale (c) Tilak (d) L ala Lajpat Rai 15. Who gave t he call “ Do or Die” ? (a) Jawahar lal Nehr u (b) Mahatma Gandhi (c) L ala Lajpat Rai (d) Subhash Bose

3.6

INDIAN HISTORY

LEVEL-1 1. A sh ok a i n t h e 13t h year of h i s cor on at i on , appoint ed a special t ype of offi cer who sur veyed the land, kept land r ecor ds and car r ied out justice. These officer s wer e call ed (a) Amatyas

(b) Samahar t as

(c) Rajukas

(d) Chalukyas

2. Who bui lt t he Jagannat ha t emple of Pur i? (a) Anantavar mana Chodaganga (b) Nar simahavar mana (c) Aadiyavar mana (d) Par meshwar avar mana 3. Sect i on 66 A has been i n medi a cont r over sy r ecent l y. The sect i on per t ains t o (a) Communal H ar mony (b) Sexual Aggr ession

11. Goutam Buddha delivered his first sermons at : (a) Kusinagar

(b) Sarnath

(c) Pataliputra

(d) Vaishali

12. The Governor General of India at the time of foundation of Indian National Congress was ? (a) Lord Chelmsford (b) Lord Dalhousie (c) Lord Dufferin

(d) Lord Canning

13. Who was the advocate at the famous trials of three INA Soldiers ? (a) Bhulabhai Desai (b) Asaf Ali (c) Subhash Chandra Bose (d) C. Rajagopalachari 14. Match Column A (Dance type) and Column B (State). Column A

(c) Company's Act

Column B

P. Bihu

1. Gujarat

Q. Garba

2. UP

R. Tamasha

3. Assam

(a) Ahmedabad strike, 1918

S. Nautanki

4. Maharashtra

(b) Rowlatt Satyagraha, 1919

(a) P-4, Q-1, R-2, S-3 (b) P-3, Q-1, R-4, S-2

(c) Swadeshi Movement, 1905

(c) P-3, Q-1, R-2, S-4 (d) P-1, Q-4, R-2, S-3

(d) I nfor mat i on Technology

4. In which of the following movement did Gandhiji make the first use of Hunger Strike as a weapon?

(d) Champaran Satyagraha, 1917 5. The famous Chinese pilgrim 'Hieun Tsang' visited India during the reign of: (a) Harshavardhan

(b) Chandragupta II

(c) Ashoka

(d) Kanishka

6. Jama Masjid at Delhi was built by : (a) Akbar

(b) Jahangir

(c) Shah Jahan

(d) Aurangzeb

7. The 'Quit India Movement' was launched in the year : (a) 1920 A.D.

(b) 1930 A.D.

(c) 1942 A.D.

(d) 1946 A.D.

8. Who wrote 'Indica' ? (a) Kautilya

(b) Kalidasa

(c) Shudraka

(d) Megasthenes

9. 'Giddha' is a folk dance of : (a) Punjab

(b) Uttar Pradesh

(c) Assam

(d) Maharashtra

10. "The Servants of India Society' was founded by : (a) Jyotiba Phule

(b) G.K. Gokhale

(c) B.G. Tilak

(d) B.R. Ambedkar

15. The first capital of British Colonial India as (a) Delhi

(b) Mumbai

(c) Kolkata

(d) Madras

LEVEL-2 1. "Khalsa" was founded by(a) Gur u Gobind Singh (b) Gur u Ramdas (c) Gur u Nanak (d) Gur u Ar jun Dev 2. "M ahabhar at a" t he epic was wr it t en by(a) Vyasa

(b) Kalidasa

(c) Tulsidasa

(d) Valmiki

3. The famous queen Chand Bibi who fought against Akbar, defended t he cit y of (a) Ber ar

(b) Ahmad nagar

(c) Golconda

(d) M ysor e

4. Ar ya samaj was founded by(a) Raja Ram M ohan (b) Gopal K r ishna Gokhale (c) Swami Dayanand Sar aswati (d) Anne Besant

INDIAN HISTORY

5. I ndi a's fi r st war of I ndependence (r el at ed t o M eer ut mut iny) was in: (a) 1835

(b) 1857

(c) 1892

(d) 1905

6. Fr ench power declined in I ndia aft er t he bat t le of-

11. The r uler of which of t he following St at es was r emoved fr om power by t he Br it ish on the pr etext of M isgover nance ? (a) Awadh

(b) Jhansi

(c) Satara

(d) Nagpur

12. Palit ana Temples ar e locat ed near :

(a) Plassey

(b) Buxar

(a) Bhavnagar, Gujar at

(c) Talikota

(d) Wandiwash

(b) Ujjain, Madhva Pr adesh

7. The fir st I ndian r ailway t r ain jour ney bet ween Bombay and Thane was in t he year -

3.7

(c) Nasik, M ahar asht r a (d) Var anasi, Ut t ar Pr adesh 13. K unwar Singh, a pr ominent leader of Upr ising of 1857, belonged t o :

(a) 1857 (b) 1853

(a) Punjab

(c) 1818

(b) Rajasthan

(c) Madhya Pr adesh (d) Bihar

(d) 1854 8. Wit h which one of t he following movement s is t he slogan "Do or Die" associat ed ? (a) Swadeshi M ovement (b) Non-Cooper at ionM ovement (c) Civil Disobedience M ovement (d) Quit I ndia M ovement 9. Which one of t he following places was associat ed wit h t he begi nni ng of Vinoba Bhave Bhoodan M ovement ?

14. Wher e was t he Fir st Session of I ndian Nat ional Congr ess held in 1885 A.D. ? (a) Delhi

(b) Calcutta

(c) Bombay

(d) Sur at

15. The 1929 session of I ndian Nat ional Congr ess is of si gni fi cance i n t he hi st or y of t he Fr eedom M ovement because t he (a) at t ainment of Self-Gover nment was declar ed as t he object ive of t he Congr ess

(a) Dandi

(b) Kheda

(b) at t ainment of Poor na Swar aj was adopt ed as t he goal of t he Congr ess

(c) Pochampalli

(d) Champaran

(c) Non-Cooper at ion M ovement was launched

10. I den t i f y t h e M u gh al E m per or w h o gav e per mission t o East I ndia Company t o est ablish t heir fact or y at Sur at : (a) Akbar

(b) Jahangir

(c) Shahjahan

(d) Aurangzeb

(d) deci si on t o par t i cipat e i n t he Round Tabl e Confer ence in L ondon was t aken

3.8

INDIAN HISTORY

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (a)

2. (d)

3. (a)

4. (c)

5. (b)

11. (c)

12. (b)

13. (d)

14. (b)

15. (b)

6. (c)

7. (c)

8. (d)

9. (b)

10. (d)

7. (c)

8. (d)

9. (a)

10. (b)

7. (b)

8. (d)

9. (c)

10. (b)

LEVEL-1 1. (b)

2. (a)

3. (d)

4. (a)

5. (b)

11. (b)

12. (c)

13. (a)

14. (b)

15. (c)

6. (c)

LEVEL-2 1. (a)

2. (a)

3. (b)

4. (c)

5. (b)

11. (a)

12. (a)

13. (d)

14. (c)

15. (b)

6. (d)



4

GEOGRAPHY

CHAPTER EARTH

I t is believed t hat near ly 150 million year s ago, t her e was a si ngl e cont i nent on ear t h cal l ed Pangaea. This super -cont inent br oke int o sever al pieces, which began t o dr ift apar t .

ORI GI N One of the theories suggests that it is an outcome of a star formation . About 4,600 million year s ago, a giant cloud of gas and dust in space, having a swirling motion, gr adually contracted under its own gr avity. As the gas cloud or nebula shr ank in size, it spinned fast er t o conser ve its angular moment um. The r apid r otat ion pr evented it from total collapse into one object and a large number of smaller units were formed. The core of nebula r otating slowly became new star– the Sun.



SU RFACE FEATU RES OF TH E EARTH  M ajor sur face feat ur es of t he ear t h ar e cont inent s and ocean basins.

L I T H OSPH E RE

REALM S OF TH E EARTH 1. L it hospher e (land masses) 2. At mospher e (air envelope) 3. H ydr ospher e (wat er bodies) 4. Biospher e (life for ms)

Layers of the Earth (according to volume) :



L ess t han 1/3r d of t he whole sur face of t he ear t h is landmass, t he r est is cover ed by wat er.

1. Crust. I t for ms only 0.5 per cent of t he volume of t he ear t h



The cont inent s and ocean basins ar e ir r egular ly ar r anged on t he ear t h.

2. M antle. I t for ms 16 per cent of t he volume of t he ear t h



The nor t her n hemispher e has mor e land t han t he sout her n hemispher e.

3. Core : I t for ms 83 per cent of t he volume of t he ear t h



For sever al million year s, lar ge par ts of cont inents wer e cover ed wit h t hick masses of ice. This per iod on t he ear t h is called I ce age.

Ear t h being a spher ical body, it has it s cent r e at 6,400 km viz., mean r adius of t he ear t h.

SOM E N U M ERI CAL FACTS ABOU T TH E EARTH L and

Greatest known height (in meters)

M ount Ever est

8863

M ar iana Tr ench

11035

Aver age height

840

Aver age dept h

3808

Size and shape (in km)

Oceans and Seas

Greatest known depth (in meters)

Ar ea (in millions sq. km.)

Equat or ial semi-axis, a

6378.2

L and (29.22 per cent )

149

Polar semi-axis, b

6356.8

I ce sheet s and glacier s

15.6

Mean r adius

6371.0

Oceans and seas (70.78 per cent )

361

Equat or ial cir cumfer ence

40076

L and plus cont inent al shelf

177.4

Polar (mer idian) cir cumfer ence 40009

Oceans and seas minus continental shelf

332.6

Ellipticit y, (a-b)/a

Tot al ar ea of t he Ear t h

510.0

1/298

4.2

GEOGRAPHY

ROCKS Cl assi f i cat i on of r ock s. On the basi s of the mode of formati on, r ocks ar e usuall y classifi ed into three major types : 1. I gn eou s r ock s At one t ime, all par ent r ock mat er ial was liquid, hot and a st icky t hing called magma. I t har dens below or upon r eaching the ear t h’s sur face. When it har dens, it for ms i gneous r ock. I gneous r ocks ar e par ent s of all ot her r ocks and ar e also called pr i mar y r ocks. 2. Sedi m en t ar y r ock s They ar e for med by deposit ion and consolidat ion of miner al and or ganic mat er ial fr om pr ecipat ion of miner als fr om solut ion. 3. M et am or p h i c r ock s When or iginal char act er of t he r ocks– t heir colour, har dness, t ext ur e and mi ner al composit ion i s par t l y or w h ol l y ch an ged, i t gi v es r i se t o met amor phic r ocks, under favour able condit ions of heat and pr essur e. EART H M OVEM ENT S About 120 mi ll ion year s ago, t he l andmass call ed Gondwana land st ar ted br eaking because of the eart h movement s, vol canic er upt i ons and sea fl oods al l wor king t oget her. T ect on i c M ovem en t I t is t he ear t h movement s which br ing about t hese vast changes. The concent r at ion of gr eat int er nal for ces wit hin t he ear t h r aise local areas upwar ds or cause them sinking downwards. Types of Tect on i c movem en t . (i ) Su d d en m ov em en t s : These are commonly noticed during an earthquake. (i i ) Sl ow or Secu l ar m ov em en t s: T h ese movemen t s con t i nu e much l on ger as compar ed t o our life span. L AN DFORM S Ty p es of L an d f or m s Ther e ar e thr ee major l andfor ms : 1. M OUNTAI NS An uplift ed por t ion of t he ear t h’s sur face is called hi l l or mountai n. Cl assi f i cat i on of m ou n t ai n s M ountains are classified on the basis of their or igin or mode of for mati on as fol l ows : (i ) St r u ct u r al m ou n t ai n s (Tect on i c) All gr eat mount ain syst ems of t he ear t h ar e of t his t ype. Bot h t he fold and t he block mount ains ar e included in t his t ype.

(i i ) Fold mount ains Major mount ains of the present day including Alps in Eur ope, Rockies of Nor t h Amer ica, Andes of Sout h Amer ica and H imalayas of Asia ar e fold mount ains. Fold is r ecognised by t he bending of r ockst r at a, up and down or sideways. (i i i ) B l ock mou n t ai n s These mount ains ar e for med when gr eat blocks of t he ear t h’s cr ust may be r aised or lower ed dur ing t he lat e st ages of mount ainbuilding. e.g. Vosges in Fr ance (i v Vol can i c m ou n t ai n s As these are formed by t he accumulat ion of vol can i c mat er i al , t h ey ar e al so cal l ed mountai ns of accumulation. e.g. Mt . Fuji Yama in Japan (v) Resi d u al or D i ssect ed m ou n t ai n s Since t hey owe t heir pr esent for m due t o er osion by differ ent agencies, so t hey ar e also cal l ed r el i ct moun t ai n s or moun t ai n s of ci rcumdenudati on e.g. Nilgir i, Par asnat h, Gir nar 2. PL ATEAUS A plateau is an elevated area generally in contrast to the nearby areas. It has a large area on its top unlike a mountain and has an extensively even or undulating surface. The great Deccan Plateau with itsslope towards east is a tilted plateau in our country. 3. PL AI NS A relat ively flat and a low-lying land sur face wit h least differ ence bet ween it s highest and lowest point s is called plain. A plain may be as low as 30 met r es t o t he east of M ississippi river near the Appalachian r ange and as high as 1,500 met r es above sea level t o the west of t he r iver. VOL CANOES Th e vol cani c ph enomen on i s a majest i c nat ur al phenomenon. Cl assi f i cat i on of v ol can oes On the basis of fr equency of erupti on, vol canoes ar e classi fi ed as : 1. Act i v e v ol can o These er upt fair ly frequent ly as compared to others 2. D or m an t or sl eepi n g vol can o I n t hese er uption has not occurr ed r ecent ly. e.g. one in Bar ren I sland t o t he east of Andamans in India 3. Ex t i n ct or An ci en t v ol can o These have r ecorded no er uption in hist or ic times ar e called extinct.

GEOGRAPHY

EARTH QUAK ES



Any sudden disturbance below the eart h’s sur face, may pr oduce vibr at ions or shakings at t he cr ust . When r ocks br eak, t he par t icles next t o t he br eak ar e set in mot ion. I t is t he movement of one r ock mass against anot her t hat causes vibr ations. Some of t hese vibrations r each t he sur face and ar e called earthquakes.



The place of or igin of an ear t hquake inside t he ear t h is called it s focus.



The point on t he ear t h’s sur face ver t ically above t he focus is called epi centr e.

SOI L S  Soil is a mixt ur e of many solid, liquid and gaseous subst ances.  I t for ms t he t opmost layer of ear t h’s cr ust .  I t has bot h non-living and t he living mat t er like miner al par t i cl es, decayi ng plant r emai ns and insect s living t oget her wit h count less bact er ia on it s or ganic mat t er.  Soil holds wat er. This moist ur e is t aken in by t he r oot s of plant s.  Ther e is ai r in t he open spaces wi t hin t he soil cont aining mor e of car bon dioxide and also oxygen and nit r ogen.  Soil is final pr oduct of t he int er act ions bet ween weat her ing of under lying r ock, climat e, plant s and act ivit ies of millions of insect s ear t hwor ms. U ND ERGROU ND WATER  Rain-wat er or snow-melt that neither runs off along the sur face nor evaporat es but sinks into the ground is called undergr ound water.

 Part of the surface water moves downwards thr ough pores between mineral grains or cracks between rocks.  The wat er -hol di ng capaci t y of a r ock mat er i al depends on t he por e spaces called it s por osi ty.  Per meabi l i ty is t he pr oper t y of r ocks which allow wat er t o pass t hr ough t hem. Wat er Tabl e Th e l evel bel ow whi ch t he r ock s ar e compl et el y sat ur at ed wit h wat er is called water tabl e. Wel l s Wells ar e man-made holes dug into t he ear t h’s sur face in or der t o get under gr ound wat er for ir r igat ion or for human consumpt ion. A special type of well in which water rises automatically under t he pr essure of a column of wat er t o t he ground sur face, eit her t hr ough a nat ur al or man-made hole, is called ar tesi an wel l .

4.3

Sp r i n gs Spr ings ar e places wher e a flow of wat er r ises t o t he sur face t hr ough nat ur al r ock opening under hydr aulic pr essur e fr om t he dept h. A spring or a chain of them are common at the junctions of per meable and imper meable r ocks. H ot sp r i n gs an d Gey ser s The gr ound wat er comes in cont act with t he heat ed or super heat ed st eam inside t he ear t h and emer ges at t he sur face as a hot spr i ng. Geyser is a hot spr ing in which wat er is for ced up by st eam pr essur e at int er vals. ATM OSPH ERE The at mospher e ext ends t o t housands of kilomet er s, but it has no clear upper limit and it gr adually mer ges wit h t he out er space. Comp osi t i on The at mospher e is a mixt ur e of many discr et e gases, in which var ying quant it ies of t iny solid par t icles ar e suspended. Pu r e d r y ai r con st i t u t es mai n l y of (by vol ume) Nit r ogen – 78 % Oxygen – 21% Remaining one per cent is account ed for by gases like ar gon (0.93 per cent ) car bon dioxide (0.03 per cent ) hydr ogen, helium and ozone. Wat er v ap ou r  Wat er vapour absor bs par t s of t he insolat ion fr om t he sun and t hus r educes it s amount r eaching t he ear t h’s sur face.



I t also pr eserves t he ear th’s r adiat ed heat . I t t hus, act s like a blanket allowing t he ear t h neit her t o become t oo cold nor t oo hot .

 Wat er vapour absor bs heat dur ing t he pr ocess of evapor at ion.  Winds transport latent heat along with water vapour fr om one place t o anot her, wher e t his heat may be r eleased t hr ough condensat ion and pr ecipit at ion. D u st p ar t i cl es  The movement s of the atmosphere ar e sufficient t o keep a large quantity of tiny solid particles suspended wit h in it .  The amount of dust par ticles is more in subt ropical and t emper at e ar eas because of dr y and wi ndy conditions t han in t he equatorial and polar r egions.  These dust par t icles are significant from met eorological st andpoint.  Many of these act as hygroscopic nuclei around which wat er vapour condenses t o produce clouds.

4.4

GEOGRAPHY

Ot h er Gases  Carbon dioxide constit utes only 0.03 per cent of t he volume of t he air. Even so i t is ver y impor t ant met eor ologically because it is t r anspar ent t o t he incoming solar r adiat ion but opaque t o out going t err est rial r adiat ion.  Ozon e i s ot h er i m por t an t compon en t of t h e at mospher e. I t act s as a filt er and absor bs ult r aviolet r adiat ion fr om t he sun. I t is concent r at ed mai nl y bet ween 10 t o 50 k i lomet r es above t he ear t h’s sur face. Atmospher e can br oadl y be di vi ded i nto four l ayer s :



(i ) Tr oposph er e I t is t he lowest layer of t he at mospher e.



I t ext ends r oughly t o a height of kilomet r es near t he poles and about 18 kilomet r es at t he equat or. (i i ) St r at osp h er e I t ext ends upt o a height of 50 kilomet r es. I n t he lower par t of t his layer, i .e. upt o a height of 20 kilomet r es, t emper at ur e r emains const ant . Above 20 km it gr adually incr eases upt o a height of 50 kilomet r es because of t he pr esence of ozone layer which absor bs sun’s ult r a-violet r ays. (i i i )M esosph er e I t ext ends upt o a height of 80 kilomet r es. Temper at ur e decr eases wi t h hei ght agai n and r each es u pt o – 100 ° C at t h e h ei gh t of 80 kilomet r es. (i v) Th er mosph er e I n it s lower par t , t her e is an elect r ically char ged layer called i onospher e Radi o waves t r ansmi t t ed f r om t he ear t h ar e r eflect ed back t o t he ear t h by t his layer. Temper at ur e again st ar t s incr easing wit h height because of r adiat ion fr om t he sun. (v) Exosp h er e H er e at mospher ic gases ar e ver y t hin. Thi s par t is ext r emely r ar efied and gr aduall y mer ges wit h t he out er space. ATM OSPH ERI C PRESSURE



I t is weight of t he column of air at any given place and t ime.



I t is measur ed by means of an inst r ument called bar ometer . I t is measur ed as a for ce per unit ar ea.



I t is a ver y impor t ant fact or in pr oducing changes in our weat her



Cont r ast s in t emper at ur e cause changes in air densit y, whi ch ar e r esponsible for var iat ions in pr essu r e. Th ese var i at i on s cau se h or i zon t al movement s of air called wi nds.

I sobar I t is is an imaginar y line dr awn t hr ough places having equal at mospher ic pr essur e r educed t o sea level. The spacing of isobar s expresses r ate and direct ion of t he pr essur e changes and is called pr essur e gr adi ents. Pr essur e gr adi ent may be defined as t he decr ease in pr essur e per unit dist ance in t he dir ect ion in which pr essur e decr eases most r apidly. Ther e ar e two types of pr essur e systems (i) H igh pr essur e (ii) L ow pr essur e

WI ND S Ty p es of Wi n d s (1) Pl an et ar y w i n d s.

 Winds which blow t hr oughout t he year fr om one latitude to the other in r esponse to the latitudinal differ ences in air pr essur e, ar e called prevai l ing wi nds or planetary wi nds.  These winds blow over vast area of the continents and oceans.  Winds for climat e and human act ivit ies Ty p es of p l an et ar y w i n d s. (i ) Tr ad e w i n d s : These winds blow fr om subt r opi cal hi gh pr essur e ar ea (30° N and S) t owar ds t he equat or ial low pr essur e belt and ar e ext r emely st eady winds. To blow t r ade means t o blow st eadily in t he same dir ect ion and in a const ant cour se. (i i ) West er l i es : These wi nds bl ow fr om subt r opical high pr essur e belt s t owar ds sub-polar low pr essur e belt s These blow fr om sout h-west t o nor t h-east in t he Nor t her n H emispher e and nor t h-west t o sout h-east in t he Sout her n H emispher e. (2) Per i od i c w i n d s.

 The winds changing t heir dir ect ion per iodically w i t h ch ange i n season ar e cal l ed per i odi c winds.  M onsoons ar e t he best example of lar ge scale modificat ion of t he planet ar y wind syst em. Ty p es of Per i od i c w i n d s : (i ) M o n so o n w i n d s : T h ese ar e season al modifi cat ion of t he gener al pl anet ar y wi nd syst em.

GEOGRAPHY

(i i ) L an d an d Sea br eezes : These affect only a nar r ow str ip along the coast . During day t ime, t he land get s mor e heat t han t he adjacent sea and develops low air pr essur e. The sea being cool, develops a comparatively higher pressur e. The war m air of t he land being lighter ascends and it s place is t aken by t he cooler air coming fr om t he sea, which is called sea breeze. At t he higher el evat ion, war m air get s cooled and moves t owar ds t he sea. H ence, sea br eezes blow during day at the lower level and moderate the weather of the coastal fringe. At night, rapid r adi at i on mak es t he l and cool er t han t he adjoining sea. This r esult s in high pr essur e over t he land and low pr essur e over t he sea. Air star ts blowing from land to sea and is called l and breeze. (i i i ) M ou n t ai n an d Val l ey br eezes : Dur ing day t ime, slope of t he mount ain is heat ed mor e t han t he valley floor. The air fr om t he valley flows up t he slope, and is called val l ey br eeze. After sunset t he patt er n is r eversed. Rapid loss of heat thr ough t er r est r ial r adiat ion along t he mountain slopes result s in sliding of cold dense air fr om higher elevat ions t o valleys. This is called mountai n br eeze. (3) L ocal w i n d s L ocal winds develop as a result of local differ ences in t emper at ur e and pr essur e.

H YD ROSPH ERE H UM I DI TY Humidit y is the gener al t erm which describes invisible amount of wat er-vapour pr esent in t he air. I t may be expr essed quanti tati vel y as fol l ows : 1. Absol u t e h u mi di t y Weight of act ual amount of wat er-vapour pr esent in a unit volume of air is called absolute humidi ty. I t is usually expr essed as gr ams per cubic met r e of air. 2. Sp eci f i c h u m i d i t y Weight of wat er-vapour per unit weight of air, or proport ion of t he mass of water -vapour t o t he t otal mass of air is called speci fi c humi di ty. I t is measur ed in unit s of weight (usually gr ams per kilogr am), Rel at i v e h u m i d i t y I t is t he r at io of air ’s act ual wat er -vapour cont ent t o it s wat er -vapour capacity at a given t emper at ur e.

4.5

EVAPORATI ON I t is t he pr ocess by which wat er is t r ansfor med fr om liquid t o gaseous for m. Ra t e of evapor at i on I t depends on (i) t emper atur e (ii) moist ur e cont ent or degr ee of dr yness of air (i i i ) movement of air CONDEN SATI ON  I t is t he pr ocess of change of st at e fr om gaseous t o liquid or solid st at e.  When moist air is cooled, it may r each a level when its capacity t o hold wat er -vapour is exceeded by the act ual amount pr esent in it .  I n free air, condensation r esult s from cooling around ver y small par t icles called condensation nucl ei .  Condensati on depends on (i) amount of cooling (i i ) r elat ive humidit y of t he air Dew When t he moist ur e is deposit ed in t he for m of wat er dr oplets on cooler surface of solid objects such as st ones, gr ass blades and plant leaves, it is called dew. Wh i t e F r ost When condensation takes place at a dew point which is at or below fr eezing point (0C), excess moist ur e is deposit ed in t he for m of minut e ice cr yst als inst ead of dr oplet s of wat er. I t is called whi te fr ost. Fog I t is defined as a cloud wit h its base at or ver y near t he gr ound. M i st I n t his t ype of fog, visibilit y is mor e t han 1 kilomet r e but less t han 2 kilomet r es. CL OUDS I t is a mass of minut e dr oplet s of wat er or t iny cr ystals of ice for med by condensat ion of t he wat er -vapour in fr ee air at consider able elevat ions. Cl assi f i cat i on of Cl ou d s on t h e basi s of t h ei r appear an ce (i .e. gener al shape, str uctur e and verti cal extent) (i ) Ci r r u s cl ou ds. T h ese ar e h i gh , w h i t e an d t h i n . T h ese ar e composed of ice cr yst als. (i i ) Cu mu l u s cl ou d s. These exhibit a flat base and have t he appear ance of r ising domes. (i i i ) Str atus cl ouds. These are best described as sheets of layers that cover much or all of the sky.

4.6

GEOGRAPHY

PRECI PI TATI ON  Condensat ion of wat er -vapour in air in t he for m of wat er dr oplet s and ice and t heir falling on t he gr ound is called precipi tati on.  The pr ecipit at ion in t he for m of dr ops of wat er is called rai nfal l.  When t emper at ur e is less t han 0C, pr ecipit at ion t akes place in the for m of fine flakes of snow and is called snow-fal l.  Sl eet is fr ozen raindr ops and re-fr ozen melt ed snow wat er. I t may be a mixt ur e of snow and r ain or mer ely par t ially melt ed snow.  Precipitat ion in t he for m of hard rounded pellet s is called hai l. OCEANS Su bm ar i n e Tr en ch es or D eep s  These are deepest par ts of the oceans with their bottoms far below the average level of the ocean floors.  A long, nar r ow and st eep sided depr ession on t he ocean bot t om is called tr ench. Su bmar i n e Can yon s  These ar e deep gor ges on t he ocean floor.  These ar e st rikingly deep valleys wit h st eep slopes t hat for m long, concave pr ofiles. Paci f i c Ocean  I t is lar gest of all t he wat er bodies.  Toget her wit h it s associat ed seas, it cover s about one-t hir d of t he ear t h’s sur face and exceeds t ot al land ar ea of t he wor ld in size.  I t is t he deepest of all oceans.  The major por t ion of t he basin has an aver age dept h of about 7,300 met r es.  This vast ocean is dot t ed wit h mor e t han 20,000 islands. At l an t i c Ocean  I t is r oughly half t he size of t he Pacific Ocean and cover s about one-sixt h of t he ear t h’s t ot al ar ea.  I t r esembles t he let t er S in shape.  I t has numer ous mar gi nal seas on bot h sides, especially in it s nor t her n par t .  I t 's most st r i k i n g feat u r e i s t h e pr esence of mid-At lant ic Ridge. I t ext ends fr om t he nor t h t o the south paralleling the S shape of the ocean itself. I n d i an Ocean  I t is smaller t han t he At lant ic Ocean.  I t aver age dept h i s 4,000 met r es w h i ch i s compar at ively lesser t han t hat of ot her oceans.  The floor has fewer ir regularities in compar ison to the other two oceans. Linear deeps are almost absent. The only except ion is the Sunda Trench, which lies south of t he island of Java and runs parallel t o it.

OCEAN WATERS  Temper at ur e and sal ini t y of t he ocean wat er, det er mine movement s of lar ge masses of wat er.  Temperature decreases according to the increasing dept h of t he ocean.  Gener ally, t emper at ur e in ocean wat er s var ies fr om below – 5°C t o over 33°C.



H eat i n g of ocean w at er There are two main pr ocess of heati ng ocean water: (i) By absor pt ion of r adiat ion fr om t he sun (ii) By convect i on of heat t hr ou gh t he ocean bot t om fr om int er ior of t he ear t h.



Cool i n g of ocean w at er Ther e ar e t hr ee main pr ocesses of cooling ocean wat er. (i) Back r adiat ion of heat fr om t he sea sur face (ii) Convect ion (iii) Evaporation

M OVEM ENT OF WATER Facts r esponsi bl e for movement of water ar e fol l ows. 1. Sal i n i t y  Sea wat er cont ains a number of dissolved salt s which r esult in t he pr oper t y of salinit y.  The salinity is expressed as the number of gr ams of dissolved salt s in 1,000 gr ams of sea wat er.  Aver age salinit y of t he sea wat er is about 35 per t housand or 35%, which means t hat in one kilogr am of sea wat er t her e ar e 35 gr ams of dissolved salts. 2. Wav es  Waves ar e osci ll at or y movement s i n wat er, manifest ed by an alt er nat e r ise and fall of t he sea sur face.  Top par t of a wave is called it s cr est and lower par t bet ween t wo waves is called tr ough.  The time t aken by two consecutive cr ests to pass any fixed point is called wave per i od.  The ver t ical dist ance bet ween a t r ough and a cr est is called wave hei ght. Wave length (L) Period (T) Si ze and for ce of waves depend on three factor s:

Velocit y of wave, C =





(i) Velocit y of t he wind (ii) Dur at ion of t he wind (i i i ) Dist ance over which t he wind can blow unhindered. As t he waves move away fr om t he winds t hat dist ur b smoot hness of sea sur face, t hey begin t o move in an unifor m pat t er n of equivalent per iod and height . These t r ains of waves ar e called swel ls.

GEOGRAPHY

3. Ocean cu r r en t s I t is gener al movement of a mass of wat er in a fair ly defined dir ect ion over gr eat dist ances. I t can br oadl y di vi ded i nto two cl asses : (i ) War m cu r r en t s : These flow from low latit udes in Tr opical zones t owar ds high lat it udes in t he Temper at e and sub-polar zones. (i i ) Col d cu r r en t s : These flow fr om high lat it udes t owar ds low latitudes. TI D ES  Sea-water rises regularly twice a day at constant intervals. This is periodic phenomenon of alternate rise and fall in the level of the seas is called tides.  These are produced as a gravitational interaction of the earth, moon and the sun.  The sun by virt ue of its bigger size should at tr act more but owing to its greater distance from the earth it is unable t o exer t much influence. The moon t hough much smal l er i n size t han t he sun, i s relatively very close t o the ear th, and is thus able to att ract more t han t he sun.  On the full moon and the new moon, moon and the sun are almost in a line with the earth. Hence they exert their combined pull on the earth. Therefore, on these two days tides are highest and are called as spring tides. When t he moon is at fir st and last quar t er, t he sun and t he moon make a r ight angle at t he ear t h’s cent r e. The at t r act ion of t he sun and t he moon t ends t o balance each ot her. As a r esult , t ides wit h l owest ampli t ude occur. These t ides ar e cal led neap ti des.

 Tides also occur in t he ar ms of t he sea called gulfs. Gulfs wit h wide fr om and nar r ow r ears exper ience higher t ides. The height of t hese t ides may be t en met r es or mor e. When a gulf is connect ed wit h t he open sea by a nar r ow channel, wat er flows int o t he gulf at t he t ime of high t ide and comes out of t he gulf at low t ide. This movement of wat er, inwar d and out war d is called ti dal cur rent.

4.7

B I OSPH ERE   

  



I t r efer s t o t hat par t of t he ear t h in which all life for ms exist . Organisms or life forms in the biosphere vary in size from minut e bact eria t o large whales or huge tr ees. Al l or gani sms may br oadl y be gr ouped under two categor i es : (i) Plant kingdom (i i ) Animal kingdom H u man bei ngs ar e one of t he speci es of t he biospher e called H omo sapi ens. St udy of t he int er act ions bet ween or ganisms and t heir envir onment is called ecol ogy. All l i fe for ms ar e made of pr i mari l y three most abundant elements Carbon, H ydrogen and Oxygen Ot her element s such as ni t r ogen, i r on, sulfur, phosphorus and manganese are required only in small quantities. These elements are also called nutrients. Tr ansfer of ener gy in t he ecosyst em t akes place in a ser ies of st eps or levels, called food chai n.

Fi r st l evel i n t he food chain ar e plant s call ed pr oducers. They use light ener gy t o convert carbon dioxide and wat er t o pr oduce car bohydr at es and eventually to ot her biochemical molecules required t o suppor t life. This pr ocess of ener gy conver sion is called photosynthesi s. Secon d l evel of t he food chai n ar e pr i mar y con su mer. T h ese ar e pl an t eat i n g an i mal s (her bivor es) such as insect s, mice and goat . Third l evel are the secondar y consumer s, who feed on t he pr imar y consumer s (car nivor es) such as owl and lion. Some of t he species ar e called omni vor es because t hey ar e bot h her bivor es and car nivor es. H uman beings come under t his cat egor y. Decomposer s (microscopic organisms and bacteria) feed on t he det r it us or decaying or ganic mat t er der ived fr om all levels. They help in r ecycling miner al nut r ient s int o ecosyst em and t hus food chain is complet ed.

4.8

GEOGRAPHY

I N DI AN GEOGRAPH Y L OCATI ONAL SETTI NG

– Mt. Everest (8,848 m), called Sagarmatha in Nepal and Chomol angma in China



I ndia is often described as a sub-conti nent and is a par t of the Asian continent.



I t sprawls between snowy height s of t he H imalaya and shor es of the I ndian Ocean.

– Nanga Par bat (8, 126 m)

The countr y get s an abundance of sunshine from t he t ropical sun and moist ure from t he splashing monsoon r ains.

– K2 (8611 m)





I ndia is sit uat ed in t he Nor t her n H emispher e.



The Tropic of Cancer passes approximately through t he middle r egion of t he count r y.



The nor t her n-most fr inge of I ndia consist s of a mount ain syst em which r adiat es fr om t he Pamir s, r oof of t he wor ld, in t he hear t of Asia.



The hot and humid K anyakumar i const it ut es t he sout her n-most t ip wher e t he I ndian peninsula, gett ing narr ower and nar rower, loses itself into t he ocean.



Distance from nor thern-most to the southern-most point s in t he mainland of I ndia is about 3,200 kil omet r es.



East t o w est , I n di a i s appr oxi mat el y 3,000 kilomet r es long.



Acr oss t he east er n bor der s of I ndia and t he Bay of Bengal lie Myanmar (Burma). M alaysia, I ndonesia, Th ai l and, Cambodi a (K ampu chea), L aos and Viet nam.

– K anchenjunga (8, 598 m) – Nanda Devi (7,817 m) – Namcha Bar wa (7,756 m). 2. PL AI N OF I N D I A I t occupies t he space bet ween Plat eau and t he nor t her n mountains. Evolut ion of t he plain is at t r ibut ed t o a pr ocess of gr adual filling of an init ial r ift valley st r et ched in fr ont of t he nor t her n mount ains and for med as a consequence of a fr act ur e in t he cr ust . (a) Bhabar (b) Tar ai (c) Bhangar (d) Khadar (e) Deltaic plain 3. PENI N SUL AR PL ATEAU I t is a block of old cr yst al r ocks lift ed above t he level of sea in which t hese r ocks wer e deposit ed in t he Pr e-Cambr ian t imes and never submer ged again.

I SL A N D S

Acr oss t he west er n bor der s of I ndia lie Pakist an, I r an, I r aq and t he Ar ab count r ies.



I ndian islands in t he Bay of Bengal consist of t he Andaman and t he Nicobar gr oups.

PH YSI OGRAPH I C DI VI SI ON S At the macr o l evel I ndi a may be di vi ded i nto thr ee physiographi c uni ts :



Ther e ar e as many as 200 islands in t he Andaman gr oup alone, ext ending for 350 kilomet r es.



Ther e ar e 19 islands in t he Nicobar gr oup. Some of the islands have a length of 60 to 100 km, forming a clust er sout h of t he Andaman gr oup.



Arabian Sea I slands consist of Lakshadweep group. They ar e for med on a cor al deposit off t he K er ala coast . The sout her most of t he t hese islands lies just t o t he nor t h of t he M aldives.



1. H I M AL AYAN M OUN TAI N CH AI N I t consist s of a ser ies of par allel mount ain r anges wit h bold r elief and ar e char act er ised by highly r ugged topography. H i m al ay an M ou n t ai n H i m al ayan m ou n t ai n ch ai n , al l al on g i t s longi tudinal axis, is arranged into three main series of a par al lel ranges : (i) Great Himalayas (ii) L esser Himalayas (iii) Sub-Himalayas Ot her s as inner, middle and t he out er H imalayas. I nner H i mal ayas which have an aver age alt it ude of 6,000 met r es h ave w i t hi n t hem al most al l pr ominent H imalayan peaks such as

REL I EF AN D D RAI N AGE  The land of I ndia is char act er ised by great diversit y in it s r elief and dr ainage.  The H imalaya link t hese diver se cult ur e-gr oups in a thread which runs invisibly all along the mountain r ange.  Nort h I ndian Plain is an ar ea of level and low r elief an d i t s compar at i vel y u n i f or m su r f ace i s as impr essive as it s vast ext ent .

GEOGRAPHY

DRAI NAGE

GEN ERAL I N F ORM AT I ON

On the basi s of or i gi n, the r i ver systems of I ndi a can be cl assi fi ed i nto two categor i es:

WORL D’S B I GGEST COU NT RI ES

1. Himalayan r iver s

 These have lar ge basins, t heir cat chment ar eas extending over hundreds of t housands of squar e kil omet r es.  The I ndus dr ains over an ar ea of appr oximat ely 250,000 sqk m wi t hi n t he H i mal ayan r egi on alone.  H imalayan r ivers is per ennial.  H i mal ayan r i ver s whi ch have evol ved t hr ough a l ong and chequer ed hi st or y, con si st of t he t hr ee pr i nci pal syst ems : (i) Indus (ii) Ganga (iii) Brahmaputra B asi c ar ea of H i m al ay an r i v er s. Ri ver s

4.9

(i n or d er of p op u l at i on ) N ame

Popu l at i on (i n Mi l l i on s)

Con ti n en t

China

1, 264.5

Asia

India

1, 002.1

Asia

USA

275.6

Nor t h Amer ica

I ndonesia

212.2

Asia

Brazil

170.1

Sout h America

Pakist an

150.6

Asia

Russia

145.2

Europe-Asia

Bangladesh

28.1

Asia

Japan

126.9

Asia

Nigeria

123.3

Africa

WORL D ’S SM AL L EST COU N TRI ES

Tot al ba si n (km 2)

Ba si n a r ea w i t h a r ea i n I n d i a (k m 2)

Indus

1,165,000

Ganga

(i n or d er of p op u l at i on ) N ame

Popu l at i on

L ocat i on

321,290

Vatican City

900

Europe

1,060,000

861,404

Tuvalu

10, 588

Sout h Pacific

580,000

187,110

Nauru

10, 605

Sout h Pacific

Palau

18, 467

West Pacific

 These flow t hr ough shallow valleys which ar e mor e or less complet ely gr aded in most cases.

San Marino

25, 061

Europe

L iecht enst ein

32, 057

Europe

 A lar ge number of t hem ar e seasonal as t heir flow is mainly dependent on r ainfall.

Manaco

32, 149

Europe

St. K it t s-Nevis

45, 000

Eastern

Brahmaputra

2. Pen i n su l ar r i ver s

 Many of the Peninsular rivers have, st raight and gener ally linear cour ses. East f l ow i n g p en i n su l ar r i v er s 1. Kaveri 2. Krishna 3. Godavari 4. Mahanadi West f l ow i n g p en i n su l ar r i v er s 1. Narmada 2. Tapi 3. Sabar mati 4. Mahi 5. Saravati

Caribbean WORL D’S B I GGEST COU NT RI ES (i n or d er of ar ea) N ame Russia

Con ti n en t Europe-Asia

Canada China

Nor t h Amer ica Asia

USA Brazil

Nor t h Amer ica Sout h America

Austr alia India

Austr alia Asia

Argentina Kazakhstan

Sout h America Asia

Sudan

Africa

4.10 GEOGRAPHY

WORLD SM ALLEST COU N TRI ES (in order of area)

N ame

Si ze(i n sq. k m.)

L oca tion

Vat ican Ci ty

0.44

Eur ope

M onaco

1.95

Eur ope

N aur u

21.10

Sout h Paci fi c

T uval u

26.00

Sout h Paci fi c

San M ar i no

61.00

Eur ope

L iech t enst ei n

160.00

Eur ope

M ar shal l I sl an ds

181.00

Cen t ral Paci fi c

St . K i t ts-N evi s

269.00

East ern Car i bbean

M al dives M al t a

298.00 316.00

I ndi an Ocean M edi t err an ean

LON GEST RI VERS OF TH E WORLD River

Continent

Approximate Length (miles)

River

Continent

k m.

Approximate Length miles

k m.

Ni le

Afr ica

4,000

6,400

Ob

Asia

2,110

3,400

Amazon

South

4,000

6,400

I ndus

Asia

1,900

3,060

Amer ica Asia

3,430

5,488

San Fr ancisco Rio Gr ande

South Amer ica Nor t h Amer ica

1,800

2,900

1,800

2,900

Br ahmaputr a

Asia

1,800

2,900

Danube

Eur ope

1,750

2,820

Yangtze (ChangJiang) Yel low

Asia

2,900

4,700

(Hwang Ho) Cango

Afr ica

2,900

4,700

Euphr ates

Asia

1,700

2,740

Missour i

Nor t h

2,710

4,360

Or inoco

South

1,700

2,740

Amer ica

Amer ica

Amur

Asia

2,700

4,345

Zambezi

Afr ica

1,600

2,575

L ena

Asia

2,650

4,260

Murr ay

Australia

1,600

2,575

Niger

Afr ica

2,600

4,180

Ganges

Asia

1,560

2,510

M ekong

Asia

2,600

4,180

Arkanas

1,450

2,330

Mackenzie Peace Mississippi

2,514

4,046

Dnieper

1,420

2,290

2,350

3,780

Color ado

2,250

2,350

3,780

I rrawaddy

Nor t h Amer ica Asia

1,400

St . L awr ence

Nor t h Amer ica Nor t h Amer ica Nor t h Amer ica

Nor t h Amer ica Eur ope

1,300

2,090

Volga

Eur ope

2,290

3,685

Tigr is

Asia

1,150

1,850

GEOGRAPHY 4.11

LARGEST I SLAN DS OF TH E WORLD I sland

Location

Ar ea (sq.mi.)

(sq.km.)

Gr eenland (int egr al par t of Denmar k)

Atlant ic Ocean

840.000

2,166,086

New Guinea (west er n sect ion is called I r ian Jaya and is

Pacific Ocean

312.000*

808,000

Pacific Ocean

287,000*

743,300

Madagascar (independent countr y)

I ndian Ocean

226,657

587,041

Baffin (par t of Canada)

Ar ct ic Ocean

183,810

476,067

Sumat r a (par t of I ndonesia)

I ndian Ocean

182,860

473,607

Gr eat Br it ain (consist ing of England, Scot land, and Wales)

Atlant ic Ocean

88,760

229,885

H onshu (one of t he main islands of Japan)

Pacific Ocean

88,000*

228,000

Ellesmer e (par t of Canada)

Ar ct ic Ocean

82,119

212,688

Vict or ia (par t of Nor t hwest Ter r it or ies, Canada)

Ar ct ic Ocean

81,930

212,199

Sulawesi or Celebes (par t of I ndonesia)

Pacific Ocean

72,987

189,036

Sout h I sland (par t of New Zealand)

Pacific Ocean

58,093

150,461

Java (par t of I ndonesia)

Pacific Ocean

48,000*

126,392

Nor t h I sland (par t of New Zealand)

Pacific Ocean

44,281

114,688

Cuba (independent countr y)

Car ibbean Sea

44,218

114,525

Newfoundland (par t of Canada)

Atlant ic Ocean

43,359

112,300

L uzon (one of t he main island of Philippines)

Pacific Ocean

41,000*

106,200

I celand (independent count r y)

Atlant ic Ocean

39,768

103,000

M indanao (one of t he main islands of Philppines)

Pacific Ocean

37,000*

95,800

I r eland (sout her n par t is an independent count r y; nor t her n

Atlant ic Ocean

32,597

84,426

H okkaido (one of t he main islands of Japan)

Pacific Ocean

30,077

77,899

H ispaniola (east er n par t is t he Dominican Republic;

Car ibbean Sea

29,530

76,484

Sr i L anka (independent count r y)

I ndian Ocean

25,332

65,610

Tasmania (Aust r alian st at e)

Pacific Ocean

24,450

63,326

par t of I ndonesia; east er n sect ion for ms most of Papua New Guinea, which gained it s independence in 1975) Bor neo (sout her n par t is par t of I ndonesia; nor t her n par t is made up of Sult anat e of Br unei and t wo st at es of M alaysia)

par t is par t of t he Unit ed K ingdom)

west er n par t is H ait i)

* Appr oximate ar ea.

4.12 GEOGRAPHY

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. Which of t he following count r ies is not adjacent t o Afghanist an ? (a) Uzbekistan (b) Tur kmenist an (c) Tajikistan (d) Russia 2. Which is not a neighbour ing st ate of Jhar khand ? (a) Madhya Pr adesh (b) Bihar (c) West Bengal (d) Or issa 3. Tamil Nadu coast r eceives r ains bot h in summer and wint er due t o t he influence of (a) L and and sea br eezes (b) Fr equent cyclones fr om t he Bay of Bengal (c) S. W. monsoon in summer and N. E. t r ade winds in wint er (d) Near ness t o equat or 4. H BJ pipelines car r y nat ur al gas fr om (a) H at hr as t o Bat hinda and Jhansi (b) Hoshangabad t o Bilaspur and Jabalpur (c) H azir a t o Bijaipur and Jagdishpur (d) H issar t o Bar mer and Jaisalmer 5. Whi ch one of t he fol l owi ng shi pyar ds bui l ds war ships for I ndian Navy ? (a) M azgaon Docks, M umbai (b) Cochin Shipyar d, Kochi (c) Hindustan Shipyar d, Visakhapatnam (d) Gar den Reach Wor kshop, K olkat a 6. The r ank of I ndia in t he cat egor y of The Best Educat ion Syst em in Asia is (a) 4 (b) 3 (c) 5 (d) 2 7. The fir st air line to allow flyer s to sur f the net was (a) United Air lines (b) Singapor e Air lines (c) Emir at es Air lines (d) Air Canada 8. Which country is geographically in Nor th America but polit ically a par t of Eur ope ? (a) I celand (b) Gr eenland (c) Canar y I sland (d) Cuba 9. Wi t h whi ch count r y woul d you associ at e t he r eligion Shint oism ? (a) Japan (b) Tibet (c) Malaysia (d) Vietnam 10. Which cit y is known for it s chiken embr oider y ? (a) Ujjain (b) L uck now (c) Pune (d) Sur at

11. To which count r y does t he t r adi t ional mar t ial for m Taekwondo belong ? (a) K or ea (b) Japan (c) China (d) Cuba 12. I ndian por t t hat bagged t he I SO 9002 awar d is (a) Paradip (b) Visakhapatnam (c) Haldia (d) Tut icor in 13. Which US rice gr owing company has been granted t he pat ent of I ndian Basmat i ? (a) Kasmati (b) Tax Rice (c) Rice Tec (d) Texmat i 14. Which of t he following count r ies has t he second lar gest r ail net wor k in t he wor ld ? (a) India (b) U.S.A. (c) Russia (d) China 15. Hindi is wr itten in the Devanagr i Scr ipt. I n which of t he following scr ipt s is Punjabi wr it t en ? (a) Sanskr it (b) I ndo-I ranian (c) Gur mukhi (d) Devnagr i

LEVEL-1 1. H i r akud dam has been buil t on t he r iver (a) Cauver y

(b) Mahanadi

(c) Kr ishna

(d) Yamuna

2. Wi t h r efer ence t o wat er poll ut ion, BOD means (a) Bi ochemical Oxygen Di lut i on (b) Bi ochemical Oxygen Demand (c) Bi o Or ganic Dissol ut es (d) Basic Or gani c Dissolut es 3. A pp r ox , p er cen t age of ox y gen i n E ar t h 's at mospher e is (a) 17%

(b) 21%

(c) 25%

(d) 33%

4. L unar Ecl ipse occur s only on a (a) Fi r st quar t er day (b) New moon day (c) Full moon day

(d) L ast quar t er day

5. M i r ages gener ally occur in (a) mount ai ns

(b) for est s

(c) deser t s

(d) sea

6. I n Oct ober 2014 a cyclone hit Vishak hapat nam. The name of t he cyclone was (a) K at r ina (b) Hudhud (c) L ai la (d) H el en

GEOGRAPHY 4.13

7. Which National Park is known for the 'Asiatic Lions' ?

2. Aver age Albedo (over all) of t he Ear t h is: (a) 5 × 106 candela/day

(a) Corbett National Park

(b) 5 × 107 candela/day

(b) Kanha National Park

(c) 30 t o 35%

(c) Bandipur National Park

(d) 60 t o 65%

(d) Gir National Park 8. The Indian Standard Time (I.S.T.) is ahead of Greenwich Mean Time (G.M.T.) by :

3. T he i l l u mi n at i on of a beam of l i gh t due t o scat t er ing on collision wit h par t icles suspended in a fluid, is called:

(a) 6 hours

(a) Raman effect

(b) Tyndall effect

(b) 5 hours

(c) Snell's effect

(d) H uygens effect

(c) 6 hours 30 minutes

4. I nt ensit y of ear t hquake is measur ed in -

(d) 5 hours 30 minutes 9. Red rot is a plant disease which affects : (a) Wheat

(b) Rice

(c) Sugarcane

(d) Cotton

10. Which one of the following is also known as Red Planet ? (a) Mercury

(b) Venus

(c) Earth

(d) Mars

11. Galena is an ore of : (a) Lead

(b) Copper

(c) Aluminium

(d) Iron

12. Identify the city which faced large scale destructions due to 'Hudhud' cyclone recently ? (a) Chennai

(b) Vishakhapatnam

(c) Kolkata

(d) Hyderabad

13. The most effective farming method for returning minerals to the soil is (a) Contour ploughing

(a) Bar omet er scale (b) Pyr omet er scale (c) Tachomet er scale (d) Richt er scale 5. Sever al nat ions ar e following a pr ot ocol which bi nds t hem t o r educe emi ssi on t ar get s. Thi s pr ot ocol was adopt ed in: (a) K yoto, Japan

(b) Geneva, Swit zer land

(c) New Yor k, USA

(d) Par is, Fr ance

6. Which of these r ocks would have alumina as their main component ? (a) Siliceous

(b) Ar gillaceous

(c) Calcar eous

(d) I gneous

7. Which of t he following phenomenon is r elat ed t o t he for mat ion of clouds? (a) Condensation (b) Evapor ation (c) Sublimation (d) Vulcanization 8. El Nino effect is:

(b) Terracing

(a) Development of low pr essur e ar eas in sout h east Asian r egion

(c) Crop rotation (d) Furrowing 14. Winter rains in North-Western India are caused by (a) Western Disturbances (b) South West Monsoon (c) South Easterly Disturbances (d) Easterly Disturbances 15. Kaziranga National Park is in (a) Uttar Pradesh

(b) Tamil Nadu

(c) Assam

(d) Kerala

LEVEL-2 1. Appr oximat e quant it y of CO2 in t he at mospher e in PPM (par t s per million) is:

(b) Reduct ion in ice caps r esult ing in var iat ion in in solat ion absor pt ion (c) Prolonged warming in the Pacific Ocean surface ar ea (d) Sust ained t or nados in t he east er n coast of Nor t h Amer ica 9. River Damoder is called t he 'Sor r ow of ' . (a) Assam

(b) Bengal

(c) Or issa

(d) Ut t ar Pr adesh

10. Woollen clot hes keep t he body war m in wint er because(a) Wool is a bad conduct or of heat (b) Wool is a good conduct or of heat

(a) 2

(b) 20

(c) Wool incr eases body t emper at ur e

(c) 200

(d) 400

(d) Wool decr eases body t emper at ur e

4.14 GEOGRAPHY

11. Age of a Tr ee may be ascer t ained by :

13. I dent i fy t he cycl one whi ch caused l ar ge scal e dest r uct i ons i n Vi shak hapat nam t hi s year i n Oct ober ?

(a) Radius of it s St em (b) Number of Annual Rings (c) Number of Br anches (d) Cir cumfer ence of it s St em 12. The ozone layer is useful for living beings because:

(c) I t maint ains t he Nit r ogen cycle of t he ear t h

(b) Katr ina

(c) Hudhud

(d) Nilofar

14. Which one of t he following is r enewable r esour ce?

(a) I t ser ves as t he sour ce of oxygen (b) I t maint ains t he t emper at ur e of t he ear t h

(a) Phailin

(a) Coal

(b) Pet r ol eum

(c) Natur al Gas

(d) Wind

15. Which of t hese will not he oxidised by Ozone ?

(d) I t pr otects t hem fr om har mful ult r aviolet r ays of t he sun

(a) KI

(b) FeSO4

(c) KMnO4

(d) K 2MnO4

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (d)

2. (a)

3. (c)

4. (c)

5. (a)

11. (b)

12. (d)

13. (c)

14. (a)

15. (c)

6. (c)

7. (b)

8. (b)

9. (a)

10. (b)

7. (d)

8. (d)

9. (c)

10. (d)

7. (b)

8. (c)

9. (b)

10. (a)

LEVEL-1 1. (b)

2. (b)

3. (b)

4. (c)

5. (c)

11. (a)

12. (b)

13. (c)

14. (a)

15. (c)

6. (b)

LEVEL-2 1. (d) 11. (b)

2. (c) 12. (d)

3. (b) 13. (c)

4. (d) 14. (d)

5. (a) 15. (c)

6. (b)

5

INDIAN POLITY

CHAPTER

CON STI TU EN T ASSEM BLY 

Accor ding t o t he r ecommendat ions of t he Cabinet M ission in 1946, t he Const it uent Assembly was elect ed indir ect ly by t he Pr ovincial Assemblies in July 1946.



Constit uent Assembly had 389 member s including 93 r epr esent at ives of t he Pr incely st at es.



Congr ess had 211 members and the Muslim League had 73 member s.



On 11t h December 1946, t he Assembly elect ed Dr. Rajendr a Pr asad as it s Pr esident .



Aft er par tit ion, the Constit uent Assembly was left wit h only 299 member s.



On 29t h August , 1947 it appoint ed a Dr aft ing Commit t ee under t he Chair manship of Dr. B.R. Ambedkar.



Ot her member s of t he Dr aft ing Commit t ee wer e N. Gopalaswami Ayyangar, Alladikr ishan Swami Ayy ar , K .M . M u n sh i , Sayy i d M d. Saadu l l a, N. M adhava Rao and D.P. K hait an.



Const i t ut i on was publ i shed i n Feb. 1948 and received President’s signature on 26 Nov. 1949 and was declar ed as passed



M eaning and Words in the Preamble  We the People of I ndia I t si gn i f i es t h at t h e con st i t u t i on of I n di a i s or di nat ed by t he peopl e of I ndi a t hr ough t hei r r epr esen t at i v es assem bl ed i n a sov er ei gn Const it ut ion Assembly. Thus it declar es t hat t he ult imat e sover eignt y lies wit h t he people of I ndia.



Sover eign I t means t hat t he count r y is fr ee or independent in i t s ext er nal and i nt er nal mat t er s, t her e i s no ext er nal subor dination.



Social ist I nser t ed in t he Pr eamble by t he 42nd Amendment Act 1976 it int ended t o give a posit ive dir ect ion t o the gover nment in for mulating its policies. I t meant r educing in equalit y in societ y.



Secular I t was inser t ed by t he 42nd Amendment Act 1976. I t means (i ) t her e is no official r eligion of I ndia (ii ) t he St at e wi l l not favour or pr omot e any par t icular r egion.



D emocr at i c I t en v i sages n ot on l y a dem ocr at i c f or m of gover nment but also a democr atic society. As a for m of gover nment it envisages r epr esent at ive for m of government.



Republ i c I t implies that head of the I ndian state shall neither be her edit ar y nor dict at or. H e shall be elect ed by t he people dir ect ly or indir ect ly.



L iber t y I t means fr eedom of t hough, expr ession, beli ef, fait h and wor ship.



E qualit y I t means making all discr imination illegal by stat e.



F r at er nit y I t means secur ing t he dignity of t he individual and unit y and int egr it y of t he nat ion.

Const it ut ion was officially implement ed on 26t h Januar y 1950 (dat ed of commencement ).

TH E PREAM BLE The Preamble as amended in 1976 reads

We t he people of I ndia having solemnly r esolved t o const it ut e I ndia int o a sover eign Socialist Secular Democr at ic Republic, and t o secur e t o all it s cit izens. Just ice social , economic and poli t i cal . L i ber t y of thought expression, belief, faith and worship. Equality of st at us and of oppor t unit y; and t o pr omot e among t hem al l . Fr at er ni t y assur i ng t he di gni t y of t he individual and t he unit y & int egr it y of t he Nat ion. “ I n our Constit uent Assembly t his twent y sixt h day of November 1949 do her eby adopt , enact and give t o our selves t his Const it ut ion” .

5.2

INDIAN POLITY

PRESI DEN TS OF I N DI A N ame

Term of Office

PRI M E M I N I STERS OF I N DI A N ame

Term of Office

Jawahar lal Nehr u

Aug 15, 1947 - M ay 27, 1964

Guljar i L al Nanda

M ay 27, 1964 - June 9, 1964

May-July 1969 (Acting)

L al Bahadur Shastr i

June 9, 1964 - Jan 11, 1966

Gulijar i L al Nanda

Jan 11, 1966 - Jan 24, 1966

Justice Mohammed H idayat ullah

July - Aug 1969 (Acting)

I ndir a Gandhi

Jan 24, 1966 - M ar 24, 1977

V.V.Gir i

1969-1974

M or ar ji Desai

M ar 24, 1977 - July 28, 1977

Fakhr uddin Ali Ahmed

1974-1977

Char an Singh

July 28, 1979 - Jan 14, 1980

B.D.Jatti

Feb-July 1977 (Acting)

I ndir a Gandhi

Jan 14, 1980 - Oct 31,1984

Rajiv Gandhi

Oct 31, 1984 - Dec 2, 1989

Neelam Sanjeev Reddy

1977-1982

Giani Zail Sing

1982- 1987

V.P.Singh

Dec 2 , 1989 - Nov 9, 1990

R. Venkat ar aman

1987-1992

Chandr asekar

Nov 10, 1990 - June 21, 1991

Dr. Shankar Dayal Shar ma

1992-1997

P.V. Nar asimha Rao

June 21, 1991 - May 15, 1996

K .R. Nar ayanan

1997-2002

A.B. Vajpayee

May 16, 1996 - M ay 31, 1996

A.P.J. Abdul K alam

2002-2007

H .D. Deve Gowda

June 1, 1996 - Apr il 20, 1997

Pr at ibha Patil

2007-2012

I .K.Gujr al

April 21, 1997 - March 19,1998

Pr anab M ukher ji

2012-2017

At al Bihar i Vajpayee

Ram Nat h K ovind

Fr om 2017

M ar ch 19, 1998 -Oct . 13, 1999

At al Bihar i Vajpayee

Oct . 13, 1999 - 22 M ay 2004

Dr. Rajendr a Pr asad

1950 - 1962

Dr. Sar vapalli Radhakr ishnan

1962-1967

Dr. Zakir H usain

1967-1969

Var ahagir i Venkat a Gir i

VI CE PRESI DEN TS OF I N DI A N ame

Term of Office

Dr. S. Radha K r ishnan

1952-1962

Dr. Zakir H usain

1962-1967

V.V.Gir i

1967-1969

Gopal Swar up Pat hak

1969-1974

B.D.Jatti

1974-1979

Mohammed Hidayatullah

1979-1984

R. Venkat ar aman

1984-1987

Dr. Shankar Dayal Shar ma

1987-1992

K .R. Nar ayanan

1992-1997

K r ishan K ant

1997-2002

Bhair on Singh Shekhawat

2002-2007

Shr i M ohammad H amid Ansar i, also r e-elect ed in 2012

2007 -2017

Shr i Venkaiah Naidu

2017 t ill dat e

Dr. M anmohan Singh M ay 22, 2004 – 26 M ay 2014 Nar endr a M odi

Fr om M ay 26, 2014

INDIAN POLITY

5.3

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. N am e t h e com m i t t ee f or t he fr amewor k of r est r uct ur ed r ailways. (a) Jain Commit t ee (b) Venkat achelliah Commit t ee (c) Rakesh M ohan Commit t ee (d) Dines Goswami Commit t ee 2. Who i s l egal l y compet ent under t he I ndi an Const it ut ion t o declar e war or conclude peace ? (a) The pr esident (b) The Pr ime M inist er (c) The Council of M inist er s (d) The Par liament 3. What i s t he maxi mum number of mi ni st er s allowed in t he union Cabinet ? (a) 15 (b) 28 (c) 39 (d) Ther e is no such limit 4. What is t he minimum age laid down for a per son t o seek elect ion t o t he L ok Sabha ? (a) 18 Year s (b) 21 Year s (c) 25 Year s (d) 30 Year s 5. Which ar t icle of t he Const it ut ion of I ndia gives pr ecedence t o const it ut ional pr ovisions over t he l aws m ade by t h e U n i on Par l i am en t /St at e L egislat ur es? (a) 13 (b) 32 (c) 245 (d) 326 6. The meet ings of Rajya Sabha ar e pr esided over by t he (a) Pr esident (b) Vice Pr esident (c) Pr ime M inist er (d) Speaker of L ok Sabha 7. Who is t he Supr eme Commander of t he I ndian Ar med For ces ? (a) Pr ime M inist er (b) Defence M inist er (c) Pr esident (d) Chief of t he Ar my St aff 8. Accor ding to t he I ndian Constitution, what is the minimum educat ional qualificat ion r equir ed for cont est ing t he L ok Sabha elect ions ? (a) Post Gr aduat ion (b) Graduation (c) H igher Secondar y (d) No such qualificat ion is r equir ed

9. I n which one of the following ar eas does the St ate Gover nment not have cont r ol over i t s l ocal bodies ? (a) (b) (c) (d)

Citizen’s grievances Financial mat t er s L egislat ion Per sonnel mat t er s

10. Dur ing I ndian Decennial Census Oper at ions (a) Only I ndians on I ndian soil ar e count ed (b) Both I ndians and for eigner s on I ndian soil ar e counted (c) Bot h I ndians by bir t h and by domicile only ar e count ed (d) None of t hese 11. As per t he last I ndian Decennial Census, ...... is t he most lit er at e St at e in t he count r y. (a) Delhi

(b) Mahar ashtr a

(c) Ker ala

(d) Tamilnadu

12. As per the Constitut ion of I ndia, a citizen of I ndia should not be less t han ......... of age t o become t he Pr esident of I ndia. (a) 30 year s

(b) 35 year s

(c) 40 year s

(d) 50 year s

13. Elect r onic Vot ing M achines (EVM s) wer e fir st used in t he elect ions held in t he year (a) 1996

(b) 1997

(c) 1998

(d) 1999

14. The official language as per t he const it ut ion is (a) Sanskr it

(b) Hindi

(c) English

(d) None of t hese

15. Who elect s t he Vice-pr esident of I ndia? (a) M ember s of Rajya Sabha (b) M ember s of L ok Sabha (c) The union Cabinet (d) M ember s of bot h H ouses of Par liament

LEVEL-1 1. An individual who is not a member of either house of t he par l iament can be appoi nt ed as a member of t he Council of M ini st er s, but he has t o become t he member of t he eit her house in (a) 3 mont hs (b) 6 mont hs (c) one year (d) 2 year s

5.4

INDIAN POLITY

2. The t er m 'Republ ic' used i n t he pr eamble of t he Const it ut i on of I ndia impli es

9. How long can a Presidential Ordinance remain in force ?

(a) That t he head of t he st at e is her edi t ar y

(a) One year

(b) That t he head of t he st at e i s a const it ut i onal r ul er

(b) Two months

(c) That t he h ead of t he st at e i s an el ect ed r epr esent at ive (d) None of t he above 3. I n I ndi a, what i s t he mi ni mum per missi ble age for employment i n a fact or y? (a) 14 year s

(b) 16 year s

(c) 18 year s

(d) 21 year s

4. Who is the speaker of present Lok Sabha ? (As on 01.11.2014) (a) Smt. Sumitra Mahajan (b) Smt. Sushma Swaraj (c) Smt. Meira Kumar (d) None of these 5. The Fundamental Duties of the Indian citizens are incorporated in the following Article of our constitution ? (a) Article 21 A (b) Article 51 A (c) Article 370 A (d) Article 1. A 6. To be eligible for elected as President, a candidate must be : (a) Over 25 years of age (b) Over 30 years of age (c) Over 35 years of age (d) Over 60 years of age 7. The Consolidated Fund of India is a fund in which (a) All taxes except Income Tax collected by the Union as well as State Governments are deposited (b) All money received by or on behalf of the Government of India is deposited (c) The Union as wel1 as state Governments make equal contribution to this fund (d) Savings of Union and State Governments are deposited 8. Which part of the lndian Constitution reflects the mind and ideals of the farmers?

(c) Till the President revokes it (d) Six months 10. A freedom not granted to citizens by the Indian Constitution is (a) to reside and settle in any part of India (b) move freely throughout Indian territory (c) assemble peacefully even with arms (d) form associations and Unions 11. Which out of the following is incorrect regarding Lok Sabha? (a) 530 members are elected from states (b) 20 members are elected from Union Territories (c) 2 members from Anglo lndian community are elected by the community (d) 2 members of Anglo Indian community are nominated by the President if there is no member from the Anglo Indian community 12. Which democratic country has an unwritten constitution? (a) United States (b) England (c) India (d) Canada and America 13. Who presides over the present Lok Sabha when parliament session is on? (a) President, Pranab Mukherji (b) Vice President, Hamid Ansari (c) Prime Minister, Narendra Modi (d) Speaker, Sumitra Mahajan 14. The Indian constitution was written on 26/11/ 1949 and came into force on (a) Same day

(b) 26.01.1950

(c) 15.08.1950

(d) 26.01.1952

15. A few children between ages 8 and 14 were rescued from a factory where they worked under inhuman conditions. Which fundamental right of the constitution made this possible? (a) Right to Education

(a) Preamble

(b) Right to Freedom of Speech

(b) Fundamental Rights

(c) Right against exploitation

(c) Directive Principles

(d) Right to Freedom of Religion

(d) Emergency Provisions

INDIAN POLITY

LEVEL-2 1. By whi ch const i t ut i onal amendment di d t he Par l i am en t acqu i r e t h e r i gh t t o am en d Fundamental Rights? (a) 23r d

(b) 24th

(c) 25

(d) 26

th

th

2. H ow many Fundament al Right s ar e guar ant eed by t he Const it ut ion of I ndia? (a) 7 (c) 5

(b) 3 (d)6

3. An int er pr et at ion of t he I ndian Const it ut ion is based on t he spir it of t he(a) Fundament al r ight s (b) Fundament al dut ies (c) Pr eamble (d) Dir ective pr inciples 4. To be eligible for member ship of t he L ok Sabha, a per son should be at least : (a) 18 year s of age

(b) 30 year s of age

(c) 35 year s of age

(d) 25 year s of age

5. Who is t he Chair man of Rajya Sabha ? (As on 01.11.2014) (a) Sumit r a Mahajan (b) Hamid Ansar i (c) Ar un Jait ley

(d) Thambi Dur ai

6. Who of t he following is r egar ded as t he ar chit ect of t he I ndian Const it ut ion ? (a) Pandit Nehr u

(b) B.R. Ambedkar

(c) Mahatma Gandhi (P) Rajendr a Pr asad 7. What was t he over all vot ing per cent age in t he r ecent l y hel d Gener al El ect i ons for 16t h L ok Sabha ? (a) About 60% (b) About 55% (c) About 66% (d) About 78% 8. M oney can be spent out of t he Consolidated Fund of I ndia (a) wit h t he appr oval of t he Pr esident (b) wit h t he appr oval of t he Par liament (c) wit h t he appr oval of t he CAG (d) wit h t he appr oval of t he above aut hor it ies 9. Whi ch of t he fol l owi ng i s not a condi t i on for becoming a Cit izen of I ndia ? (a) Birth (b) Descent (c) Acquiring property (d) Natur alisation

5.5

10. The Oat h of Office is conduct ed t o t he Pr esident of I ndia by (a) The Speaker of L ok Sabha (b) The Chief Just ice ofIndia (c) The Vice-Pr esident of I ndia (d) The Pr ime-M inist er of I ndia 11. Which one is a par t of t he Dir ect ive Pr inciple of St at e Policy ? (a) Right t o equalit y befor e law (b) Right t o adult fr anchise (c) Or ganisat i on of Tr ade U nions and wor ker s rights (d) Or ganisat ion of Village Panchayat s 12. The implement of which one of t he following does not need any legislat ion? (a) Fundament Right s and Dut ies (b) Dir ect ive pr inciples of St at e Policy. (c) Pr omot ion of communit y welfar e (d) Suggest ion t o St at es for cit izen welfar e. 13. Which special ar ea r elat ed t o childr en is included in t he Dir ective Pr inciples of St ate Policy in I ndia (a) Ear ly child ood car e and educat ion (b) Compulsor y educat ion for all (c) Educat ion upt o age of 14 (d) Fr ee edu cat i on for Economi cal l y weak er sect ions 14. Which amendment added Fundamental Duties to t he I ndian Const it ut ion? (a) 37t h Amendment act in 1975 (b) 41 Amendment act in 1976 (c) 38t h Amendment act in 1975 (d) 42nd Amendment Act in 1976 15. Which aspect would need r egulat ion if t her e is mor e For eign Dir ect I nvest ment in hor t icult ur e? (a) Consumpt ion of wat er and r ight s of far mer s t o wat er (b) Soil conser vat ion , and r enewal using nat ur al pr ocesses (c) Subsidy t o far mer s, for spending t ime away fr om t heir far ms (d) Education of far mer s on advantages of growing ot her cr ops

5.6

INDIAN POLITY

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (c)

2. (a)

3. (d)

4. (c)

5. (a)

11. (c)

12. (b)

13. (a)

14. (b)

15. (d)

6. (b)

7. (c)

8. (d)

9. (d)

10. (a)

7. (b)

8. (a)

9. (d)

10. (c)

7. (c)

8. (b)

9. (c)

10. (b)

LEVEL-1 1. (b)

2. (c)

3. (a)

4. (a)

5. (b)

11. (c)

12. (b)

13. (d)

14. (b)

15. (c)

6. (c)

LEVEL-2 1. (b) 11. (d)

2. (d) 12. (a)

3. (c) 13. (a)

4. (d) 14. (d)

5. (b) 15. (a)

6. (b)



6

CHAPTER I M PORTAN T DATES Jan 1 Jan 8 Jan 10 Jan 12 Jan 15 Jan 23 Jan 25 Jan 26 Jan 28 Jan 30 Feb 2 Feb 5 Feb 13 Feb 14 Feb 24 Feb 28 M ar 3 M ar 4 M ar 8 M ar 12 M ar 15 M ar 16 M ar 19 M ar 21 M ar 22 M ar 23 M ar 24 M ar 26 Apr il 1 Apr il 5 Apr il 7 Apr il 14 Apr il 18 Apr il 22 Apr il 23 M ay 1 M ay 3 M ay 8 May 11 May 15 May 17 May 24

Ar my M edical Cor ps Est ablishment Day Afr ican National Congr ess Foundation Day Wor ld L aught er Day Nat ional Yout h Day (Bi r t hday of Swami Vivekanand) Ar my Day N et aj i Subh ash Ch andr a B ose's bi r t h anniver sar y I nt er nat i onal Cust oms Dut y Day, I ndi a Tour ism Day Republic Day Bir t h anniver sar y of L ala L ajpat Rai (M ar t y r 's day ) M ah at m a Gan dh i 's Martyr dom Day; World Leprosy Er adication Day Nat ional day of Sr ilanka K ashmir Day (Or ganised by Pakist an) Sar ojini Naidu's Bir th Anniver sar y St . Valent ine's Day Cent r al Excise Day Nat ional Science Day Nat ional Defence Day Nat ional Secur it y Day I nt er nat ional Women's Day Maur itius Day; Cent r al I ndustr ial Secur ity For ce Day Wor ld Consumer Day National Vaccinat ion Day Wor ld Disabled Day Wor ld For est r y Day Wor ld Day for Wat er Wor ld M et er ological Day Wor ld TB Day Bangladesh L iber at ion Day Or issa Day Nat ional M ar it ime Day Wor ld H ealt h Day B .R. A m bedk ar Rem em ber an ce D ay ; Fir e Ext inguishing Day Wor ld H er it age Day Wor ld Ear t h Day Wor ld Books Day I nt er nat ional L abour Day (M ay Day) I nter nat ional Ener gy Day I nt er nat ional Red Cr oss Day Nat ional Technology Day I nt er nat ional Family Day Wor ld Telecom Day Commonwealt h Day

CURRENT AFFAIRS May 31 June 5 June 21 July 1 July 4 July 11 July 26 Aug 9 Aug 12 Aug 14 Aug 15 Aug 19 Aug 20 Aug 29 Sept 5 Sept 7 Sept 8 Sept 14 Sept 16 Sept 25 Sept 27 Oct 2 Oct 3 Oct 4 Oct 5 Oct 6 Oct 8 Oct 9 Oct 10 Oct 17 Oct 20 Oct 24 Nov 7 Nov 9 Nov 10 Nov 14 Nov 17 Nov 26 Nov 30 Dec 1 Dec 4 Dec 7 Dec 10 Dec 11 Dec 14 Dec 19 Dec 23

Wor ld No Tobacco Day Wor ld Envir onment Day I nt er nat ional Yoga Day Doct or 's Day Amer ican I ndependence Day Wor ld Populat ion Day K ar gil Vict or y Day Quit I ndia M ovement Day I nt er nat ional Yout h Day Pakist an's I ndependence Day I ndia's I ndependence Day Wor ld Phot ogr aphy Day Sadbhavna Divas Spor ts Day (Dhyanchand's bir thday) Teacher 's Day For giveness Day I nt er nat ional L it er acy Day H indi Day, Wor ld Fir st Aid Day Wor ld Ozone Day Social Just ice Day Wor ld Tour ism Day Gandhi Jayanti Wor ld Nat ur e Day Wor ld Animal Day Wor ld H abit at Day; Wor ld Teacher 's Day Wor ld Wildlife Day I ndian Air for ce Day Wor ld Post al Day Wor ld M ent al H ealt h Day, Nat ional Post Day I nt er nat ional Pover t y Er adication Day Nat ional Solidar ity Day Unit ed Nat ions Day I nfant Pr ot ect ion Day, Wor ld Cancer Awar eness Day Pravasiya Bharatiya Divas , Legal Services Day Tr anspor t Day Childr en's Day, Wor ld Diabet ics day Gur u Nanak Dev's Bir t h Anniver sar y L aw Day Flag Day Wor ld AI DS Day Navy Day Ar med For ces Flag Day H uman Right s Day UNI CEF Day Nat ional Ener gy Conser vat ion Day Goa's L iber at ion Day K isan Divas (Far mer 's day)

6.2

CURRENT AFFAIRS

FI RST I N TH E WORLD

The fir st per son t o r each M ount Ever est

Sher pa Tenzing, Edmund H illar y

The fir st per son t o r each Nor t h Pole

Rober t Pear y

The fir st per son t o r each Sout h Pole

Amundsen

The fir st r eligion of t he wor ld

H induism

The fir st count r y t o pr int book

China

The fir st count r y t o i ssue paper cur r ency

China

The fir st count r y t o commence compet it ive examinat ion in civil ser vices

China

The fir st Pr esident of t he U.S.A.

Geor ge Washingt on

The fir st Pr ime M inist er of Br it ain

Rober t Walpole

The fir st Gover nor Gener al of t he Unit ed Nat ions

Tr igveli (Nor way)

The fir st count r y t o win foot ball Wor ld cup

Ur uguay

The fir st count r y t o pr epar e a const it ut ion

U.S.A.

The fir st Gover nor Gener al of Paki st an

M ohd. Ali Jinnah

The fir st count r y t o host NAM summit

Belgr ade (Yugoslavia)

The fir st Eur opean t o at t ack I ndia

Alexander , The Gr eat

The fir st Eur opean t o r each China

M ar co Polo

The fir st per son t o fly aer oplane

Wr ight Br ot her s

The fir st per son t o sail r ound t he wor ld

M agellan

The fir st count r y t o send man t o t he moon

U.S.A.

The fir st count r y t o l aunch Ar t ificial sat ellit e in t he space Russia The fir st count r y t o host t he moder n Olympics

Gr eece

The fir st ci t y on which t he atom bomb was dr opped

H ir oshima (Japan) (6 Aug. 1945)

The fir st per son t o l and on t he moon

Neil Ar mst r ong followed by Edwin E. Aldr in

The fir st shut t le t o go in space

Columbia

The fir st spacecr aft t o r each on M ar s

Viking-1

The fir st woman Pr ime M i nist er of England

M ar gar et That cher

The fir st muslim Pr ime M inist er of a count r y

Benazir Bhut t o (Pakist an)

The fir st woman Pr ime M i nist er of a count r y

M r s. S. Bhandar nai ke (Sr i L anka)

The fir st woman t o climb M ount Ever est

M r s. Junko Tabei (Japan)

The fir st woman cosmonaut of t he wor ld

Velent ina Ter eshkova (Russi a)

The fir st woman Pr esi dent of t he U.N. Gener al Assembly

Vijaya Lak shmi Pandi t

The fir st man t o fly I nt o space

Yur i Gagar in (Russia)

The fir st bat sman t o scor e t hr ee t est cent ur y in t hr ee successive t est s on debut

M ohd. Azhar uddin

The fir st man t o have climbed M ount Ever est t wice

Nawang Gombu

The fir st U.S. Pr esident t o r esign Pr esidency

Richar d Nixon

th

CURRENT AFFAIRS

FI RST I N I N DI A M ale The fir st Pr esident of I ndian Republic The fir st Pr ime M inist er of fr ee I ndia The fir st I ndian t o win Nobel Pr ize The fir st Pr esident of I ndian Nat ional Congr ess The fir st M uslim Pr esident of I ndian Nat ional Congr ess The fir st M uslim Pr esident of I ndia The fir st Br it ish Gover nor Gener al of I ndia The fir st Br it ish Vicer oy of I ndia The fir st Gover nor Gener al of fr ee I ndia The fir st and t he last I ndian t o be Gover nor Gener al of fr ee I ndia The fir st man who I nt r oduced pr int ing pr ess in I ndia The fir st I ndian t o join t he I .C.S. I ndia's fir st man in space The fir st Pr ime M inist er of I ndia who r esigned wit hout complet ing t he full t er m The fir st I ndian Commander -in-Chief of I ndia The fir st Chief of t he Ar my St aff The first I ndian member of the Viceroy's executive council The fir st Pr esident of I ndia who died while in office The fir st Pr ime M inist er of I ndia who did not face t he Par liament The fir st Field M ar shal of I ndia The fir st I ndian t o get Nobel pr ize in Physics The fir st I ndian t o r eceive Bhar at Rat na awar d The fir st I ndian t o cr oss English channel The fir st per son t o r eceive Gynanpit h awar d The fir st Speaker of t he L ok Sabha The fir st Vice-pr esident of I ndia The fir st Educat ion M inist er The fir st H ome M inist er of I ndia The fir st I ndian Air Chief M ar shal The fir st I ndian Naval Chief The fir st judge of int er nat ional Cour t of Just ice The fir st per son t o r eceive Par amveer Chakr a The fir st per son t o r each M ount Ever est wit hout oxygen The fir st Chief Elect ion Commissioner The fir st per son t o r eceive M agsaysay Awar d The fir st per son of I ndian or igin t o r eceive Nobel Pr ize in M edicine The fir st Chinese t r aveller t o visit I ndia The fir st per son t o r eceive St alin Pr ize The fir st per son t o r esign fr om t he cent r al cabinet The fir st for eigner t o r eceive Bhar at Rat na The fir st per son t o r eceive Nobel Pr ize in Economics The fir st Chief Just ice of Supr eme Cour t F emale The fir st lady t o become ‘‘M iss Wor ld’’ The fir st woman judge in Supr eme Cour t The fir st woman Ambassador The fir st woman Gover nor of a St at e in fr ee I ndia The fir st woman Pr ime M inist er The fir st woman t o climb M ount Ever est The fir st woman t o climb M ount Ever est t wice

Dr. Rajendr a Pr asad Pt . Jawahar L al Nehr u Rabindr anath Tagor e W. C. Baner jee Badr uddin Tayyabji Dr. Zakir H ussain L or d William Bent inck L or d Canning L or d M ount bat t en C. Rajgopalachar i James H icky Sat yendr a Nat h Tagor e Rakesh Shar ma M or ar ji Desai Gener al Car iappa Gen. M ahar aj Rajendr a Singhji S. P. Sinha Dr. Zakir H ussain Char an Singh S. H . F. M anekshaw C. V. Raman Dr. Radhakr ishnan M ihir Sen Sr i Shankar K ur up Ganesh Vasudeva M avalankar Dr. Radhakr ishnan Abul Kalam Azad Sar dar Vallabh Bhai Patel S. M ukher jee Vice Admir al R. D. K at ar i Dr. Nagendr a Singh M ajor Somnat h Shar ma Sher pa Anga Dor jee Sukumar Sen Achar ya Vinoba Bhave Har govind K hur ana Fahein Saifuddin Kitchlu Shyama Pr asad M ukher jee Khan Abdul Ghaffar Khan Amar t ya Sen Just ice H ir alal J. K ania Rit a Far ia M r s. M eer a Sahib Fat ima Bibi M iss C.B. M uthamma M r s. Sar ojini Naidu M r s. I ndir a Gandhi Bachhendr i Pal Sant osh Yadav

6.3

6.4

CURRENT AFFAIRS

The fir st woman Pr esident of t he I ndian National Congr ess The fir st woman chief just ice of a H igh Cour t The fir st woman pilot in I ndian Air For ce The fir st woman Pr esident of t he Unit ed Nat ions Gener al Assembly The fir st woman Chief M inist er of an I ndian St at e The fir st woman chair man of Union Public Ser vice Commission The fir st woman Dir ect or Gener al of Police (DGP) The fir st woman L ieut enant Gener al The fir st woman Air vice M ar shal The fir st woman chair per son of I ndian Air lines The fir st woman I .P.S. Officer The fir st and t he last M uslim woman r uler of Delhi The fir st woman t o r eceive Ashoka Chakr a The fir st woman t o cr oss English Channel The fir st woman t o r eceive Nobel Pr ize The fir st woman t o r eceive Bhar at Rat na The fir st woman t o r eceive Gyanpit h Awar d

M r s. Annie Besant M r s. L eela Set h H ar it a Kaur Dayal M r s. Vijaya L axmi Pandit M r s. Suchet a Kr ipalani Roze M illian Bet hew Kanchan Chaudhar y Bhattachar ya Puneet a Ar or a P Bandopadhyaya Sushma Chawla M r s. K ir an Bedi Razia Sultan Nir ja Bhanot Aar ti Saha M ot her Ter esa M r s. I ndir a Gandhi Ashapur na Devi

POPU LAR N AM ES OF PERSON ALI TI ES Popular N ame L ady wit h t he lamp Gr and Old man of I ndia I r on Duke Gur u Ji John Bull CR K ing M aker JP L it t le Cor por al Mahamana Gurudev M aid of Or leans Desh Bandhu Deen Bandhu Yank ee L ion of t he Punjab (Sher -e-Punjab) Bar d of Avon Panditji M an of Blood Andhr a K esr i

Personality Flor ence Night ingale Dadabhai Naor oji Duke of Wellingt on M .S. Golwalkar England and t he English people Chakravarti Rajagopalachari Ear l of War wick Jayapr akash Nar ayan Napoleon Pt. Madan Mohan Malaviya Rabindr anath Tagor e Joan of Ar c C.R. Das C.F. Andr ews I nhabit ants of U.S.A L ala Lajpat Rai Shakespear e Jawahar lal Nehr u Bismar k T. Pr akasam

GEOGRAPH I CAL DI SCOVERI ES Discovery Discoverer Amer ica Sea r out e t o I ndia via Cape of Good H ope Solar syst em Planet s Sout h Pole Nor t h Pole China

Chr ist opher Columbus Vasco-de-Gama Coper nicus Kepler Amundsen Rober t Pear y M ar co Polo

Popular N ame L okmanya Bapu Apost le of Fr ee Tr ade Netaji Deser t Fox Night ingle of I ndia L al, Bal, Pal

Personality Bal Gangadhar Tilak Mahatma Gandhi Richar d Cobden Subhash Chandr a Bose Gen. Rommel Sar ojini Naidu L ala L ajpat Rai, Bal Gangadhar Tilak, Bipin Chandr a Pal Fat her of English Poetr y Geoffer y Chaucer Feuhr er H itler I r on man Sar dar Vallabh Bhai Patel I ke Gen. Eisenhower Tomy At kins English soldier Jawan I ndian soldier Pool u Fr ench soldier Vizzy Mahar aja K umar of Vizianagar a G.I . Amer ican soldier Wizar d of t he Nor t h Sir Walt er Scot Samuel Clemens M ar k Twain Spar r ow Major Gen.Rajender Singh Shastr iji L al Bahadur Shastr i Babuji Jagjiwan Ram Discovery New Foundland H udson Bay Sailor of t he wor ld M ount Ever est Fir st per son t o set foot on t he moon Tasmania island Cape of t he Good H ope

Discoverer Gobot Sebastian H enr y H udson Magellan Edmund Hillar y Neil Ar mst r ong Tasman Baur t ho Romeiodeis

CURRENT AFFAIRS

6.5

CAPI TALS AN D CU RREN CI ES OF COU N TRI ES Count r y ASI A Afghanistan Bahr ain Bangladesh Bhutan China India I ndonesia I r an I raq I sr ael Japan Jor dan Taiwan Philippines Qatar Saudi Ar abia Singapor e Sr i L anka

Capit al

Kabul Manama Dhaka Thimpu Beijing New Delhi Djakar ta Tehr an Baghdad Tel Aviv Tok yo Amman Taipei Manila Doha Riyadh Singapor e Sr i Jaya wordenapur a K ot le Syria Damascus Cambodia Phnom Penh Kazakhstan Akmola K or ea (Nor t h) Pyongyang K or ea (Sout h) Seoul L ebanon Beir ut Malaysia K uala L umpur Maldives Male Mangolia Ulan-Bator Myanmar Yangoon Nepal Kathmandu Oman Muscat Pakist an I slamabad Tajikistan Dushambe Thailand Bangkok Tur k ey Ankara Turkemenistan Ashkabad (TMM) United Ar ab Abu Dhabi Emir at es Uzbekistan Tashkent Vietnam Hanoi Yemen Sana

Cur r ency Afghani Bahr aini Dinar Taka Ngult r um Yuan I ndian Rupee Rupiah Riyal I r aqi Dinar Shekel Yen Jor dan Dinar New Taiwan Piso Qatar i Riyal Riyal (SAR) Singapor e Dollar Sr i L ankan Rupee

Syr ian Pound Riel Tenge Won (K PW) Won (K RW) L ebnanese Pound Malaysian Ringgit Rufiyaa Tugr ik Kyat Nepalese Rupee Omani Rial Pakistani Rupee Tajik Rouble Baht Tur kish L ir a Manat Dir ham Som (UK S) Dong Riyal (YER)

Count r y E U ROPE Andorr a

Capital

Cur r ency

Andorr a la- vella Ar menia Yer evan Austria Vienna Azerbaijan Baku Belar us Minsk Belgium Br ussels Bosnia Her jegovina Sar ajevo Nor way Oslo Croatia Zagr eb Por tugal L isbon Cyprus Nicosia Czech Republic Pr ague Denmar k Copenhagen Estonia Tallinn Finland H elsinki Fr ance Par is Geor gia Tbilisi Ger many Ber lin Gr eece Athens Hungar y Budapest I celand Reykavik I r eland Dublin Lithuania Vilnius L uxembur g L uxembur g Macedonia Skopje Malta Valletta Moldova Chisinau M onaco M onaco M ont enegr o Podogor ica Nether lands Amster dam Bulgar ia Sofia Poland Warsaw Romania Buchar est Russia M oscow San M ar ino San M ar ino Slovakia Br atislava Slovenia Ljubljana Spain Madrid Sweden St ock holm Switzerland Ber ne Ukr aine Kiev United Kingdom L ondon Vatican City Vatican City I taly Rome

Eur o Dr am Eur o Manat Russian Rouble Eur o Dinar Norwegian Krone Kuna Eur o Eur o Kor una Danish K r one Eur o Eur o Eur o L ar i Eur o Eur o For int Kr ona Eur o Lit as Eur o Dinar Eur o L eu Eur o Eur o Eur o L ev Zlot y L ei Rouble I talian L ir a Eur o Eur o Eur o Krona(SEK) Swiss Fr anc Kar bovanets Pound St er ling I talian L ir a Eur o

6.6

CURRENT AFFAIRS

Count r y AF RI CA Botswana Burkina Faso Burundi Camer oon Cape Ver de Centr al African Republic The Gambia Ghana Guinea Guinea Bissau Kenya L esot ho Chad Comor os Madagascar Congo I vor y Coast Djibouti Egypt Maur itius M or occo Mozambique Namibia Niger Niger ia Rwanda Senegal Seychelles Sier r a L eone Liber ia Libya Malawi Mali Maur itania Somalia Sout h Afr ica Sudan Swaziland Tanzania Togo Tunisia Uganda Zambia Zimbabwe

Capit al

Cur r ency

Gabor one

Pula

Ouagadougou Bujumbur a Yaoundi Pr aia Bangui

Fr anc (CFA) Bur undi Fr anc Fr anc (CFA) Cape Verde Escudo Fr anc (CFA)

Banjul Accr a Conakr y Bissau Nair obi Maser u N' D' Jamena M or oni Antananrivo Br azzaville Yamoussoukro Djibouti Cair o Por t L ouis Rabat Maputo Windhock Niamey L agos Kigali Dakar Victor ia Fr eet own M onor ovia Tr ipoli L ilongwe Bamako Nouakchot t Mogadishu Cape Town K har t oum Mbabane Dar -es-Salaam L ome Tunis Kampala Lusaka H ar ar e

Dalasi Cedi Guinean Fr anc Peso K enya Shilling L ot i Fr anc (CFA) Comor ian Fr anc Malagasy Fr anc Fr anc (CFA) Fr anc (CFA) Djibout i Fr anc Egyptian Pound Maur it ius Rupee Dir ham Metical Rand Fr anc Nair a Fr anc Fr anc Rupee L eone L iber ian Dollar L ibyan Dinar Kwacha Fr anc (CFA) Ouguiya Shilling Rand Pound Lilangeni Shilling Fr anc Dinar Shilling Kwatcha Zimbabwe Dollar

Count r y

Capit al

Cur r ency

N ORTH AM ERI CA Antigua and St . Johns Barbuda Bahamas Nassau Barbados Br idgetown Belize Belmopan Canada Ottawa Panama Panama Cit y Costa Rica San Jose Cuba Havana Dominica Roseau Unit ed St at es Washington D.C. of Amer ica Guatemala Guat emala City Haiti Por t -au-Pr ince Hondur as Tegucigalpa Jamaica Kingst on M exico M exico Cit y Nicar agua Managua St . K it s and Basset er r e Nevis Tr inidad and Por t-of-Spain Tobago El Salvador San Salvador Gr enada St . Geor ge SOU TH AM ERI CA Ar gentina Buenos Air es Bolivia L a Paz Brazil Br asilia Chile Santiago Colombia Bogota Ecuador Quit o Guyana Geor get own Par aguay Asuncion Per u L ima Sur inam Par amar ibo Ur uguay M ontevideo Venezuela Caracas Fr ench Guyana K oenne OCEAN I A Australia Fiji Naur u New Zealand

Canberr a Suva Naur u Willingt on

Papua Por t M or esby New Guinea Solomon I sland H oniar a Tonga Nukualofa

Dollar Dollar Dollar Dollar Canadian Dollar Balboa Colon Peso St er ling Dollar Quetzal Gour de Lempir a Dollar Peso Cor doba Dollar Dollar Colon Dollar Austr al (Pesu) Boliviano Cr uzeir o Peso Peso S-or e Dollar Guar ani Nuevosol Guilder Peso Boliver Fr anc Australian Dollar Fiji Dollar Dollar New Zealand Dollar Kina Dollar Panga

CURRENT AFFAIRS

6.7

U N I TED N ATI ON ORGAN I SATI ON • Wor ld's lar gest int er nat ional or ganisat ion and a successor of L eague of Nat ions. (L eague of Nat ions was for med aft er t he I Wor ld War, but it failed). • The Char t er of t he UN was signed at San Fr ancisco on June 26, 1945, at a meet ing of t he r epr esent at ives of 50 st at es, r epr esent ing 2/3r d of t he wor ld populat ion. • The Char t er or Const it ut ion was for med at Dumbar t on Oaks (Washingt on DC) Confer ence by USSR, UK , US and China. • For mally came int o exist ence on Oct 24, 1945. • Fir st r egular session was held in L ondon in Jan,1946 and Tr ygve Le (Nor way) was elect ed the fir st Secr et ar y Gener al. • H eadquar t er s locat ed at Fir st Avenue, UN Plaza, New Yor k Cit y, New Yor k, US. • The pr esent member ship of UN is 192. Swit zer land was The 190t h (in 2002), East Timor was t he 191st (in 2002), while M ont enegr o became t he 192nd member in 2006. • US, UK , China, Fr ance and Russia ar e t he per manent member s of t he Secur it y Council. • The pr esent Secr et ar y Gener al of UN is Ant onio Gut er r es of Por t ugal. SOM E I M PORTAN T U .N . AGEN CI ES N ame of Agency

E st d.in

H eadquar t er s

Pur pose

I nter national L abour Or ganisation (I L O)

1919

Geneva

To impr ove living condit ions and st andar d of wor ker s.

Wor ld H ealt h Or ganisat ion (WHO)

1948

Geneva

At t ainment of highest possible level of healt h by all people.

Unit ed Nat ions Educat ional, Scientific and Cultur al Or ganisation (UNESCO)

1946

Par is

To pr omot e collabor at ion among nat ions t hr ough educat ion, science and cult ur e.

I nter national At omic Ener gy Agency (I AEA)

1957

Vienna

To pr omot e peaceful Ener gy uses of at omic ener gy.

Unit ed Nat ions I nt er nat ional Childr en's Emer gency Fund (UNI CEF)

1946

New Yor k

To pr omot e childr en's welfar e all over t he wor ld.

Unit ed Nat ions Confer ence on Tr ade and Development . (UNCTAD)

1964

Geneva

Pr omot es int er nat ional t r ade t o acceler at e economic gr owt h of developing count r ies.

I nt er national Civil Aviat ion Or ganisation (I CAO)

1947

Mont r eal

Pr omot es safet y of int er nat ional aviation.

I nt er nat ional M onet ar y Fund (I MF)

1945

Washingt on D.C.

Pr omot es int er nat ional monet ar y cooper at ion.

Unit ed Nat ions Envir onment al Pr ogr amme (UNEP)

1972

Nair obi

Pr omot es int er nat ional co-oper at ion in human envir onment .

Unit ed Nations I ndust r ial Development Or ganisation (UNI DO)

1967

Geneva

Set s int er nat ional r egulat ions for r adio, t elegr aph, t elephone and space r adio communicat ion.

I nt er nat ional Bank for Reconst r uct ion and Development (IBRD)

1945

Washingt on D.C.

Development of economies of member s by facilit at ing invest ment of capit als by pr oviding loans.

Wor ld M et eor ological Or ganisat ion (WM O)

1950

Geneva

Pr omot e int er nat ional exchange of weat her r epor t s.

Wor ld Tr ade Or ganisat ion (WTO)

1995

Geneva

Set t ing r ules for wor ld t r ade t o r educe t ar iffs.

Unit ed Nat ions Development Pr ogr amme (UNDP)

1965

New Yor k

H elp developing count r ies incr ease t he wealt h pr oducing capabilit ies of t heir nat ur al and human r esour ces.

6.8

CURRENT AFFAIRS

OT H E R I N T E RN AT I ON AL ORGAN I SAT I ON S AN D GROU PS T he Common Wealt h • I t w as or i gi n al l y k n ow n as T h e B r i t i sh Commonwealt h of Nat ions'. I t is an associat ion of sover eing and independent st at es which for mally made up t he Br it ish empir e. • H eadquar t er s : L ondon. • M ember s : 54 • The Br it ish M onar ch (Queen Elizabet h I I ) is t he symbolic head of t he commonwealt h. • Com m on w eal t h h eads of gov er n m en t m eet (CH OGM ) is held in ever y 2 year s. Asia Pacific E conomic Co-operat ion (APEC) • Est ablished : Nov, 1989 • Object ive : To pr omot e t r ade and invest ment in t he Pacific basin. • M ember s : 21 Asian Development Bank (ADB) • Est ablished : Dec 19, 1966 • Obj ect i v e : To pr om ot e r egi on al econ om i c co-oper ation. • M ember s : 67 • H eadquar t er s: M anila Associ at i on of Sou t h - E ast Asi an N at i on s (ASEAN ) • Est ablished : Aug. 8, 1967 • Object ive : Regional, economic, social and cult ur al cooper at ion among t he non-communist count r ies of Sout h-East Asia. • M em ber s : B r u n ei , I n don esi a, M al ay si a, Philippines, Singapor e, Thailand, Viet nam, L aos, Myanmar, Cambodia. • H eadquar t er s : Djakar t a Group of 7 or G-7 (Formerly, G8) • Est ablished : Sept . 22. 1985 • Object ive : To pr omot e co-oper at ion among major non-communist economic power s. • M ember s : Fr ance, Ger many, Japan, U K , U S, Canada, I taly. Group of 15 (G-15) • Est ablished : 1989 • Obj ect i ve : To pr omot e economi c co-oper at i on among developing nat ions. • M ember s :18 Group of 77 (G-77) • Est ablished : Oct ., 1967 • Obj ect i ve: To pr omot e economi c co-oper at i on among developing nat ions. • M ember s : 132

I nt er nat i on al Cr i m i n al P ol i ce O r gan i sat i on (I N T ERPOL ) • Est ablished : 1914 • Object ive : To pr omot e int er nat ional cooper at ion among cr iminal police aut hor it ies • M ember s : 190 • H eadquar t er s : Fr ance I nt er nat ional Olympic Commit t ee (I OC) • Est ablished : June 23, 1894 • Object i ve: To pr omot e t he Ol ympi c i deal s and administ er Olympic games. • M ember s : 204 • H eadquar t er s : Switzer land I nt ernat ional Organisat ion F or St andardisat ion (I SO) • Est ablished : Feb: 1947 • Obj ect i v e : To pr om ot e t h e dev el opm en t of inter nat ional standar ds • M ember s : 163 • H eadquar t er s : Switzer land I n t er n at i on al Red Cr oss an d Red Cr escen t M ovement • Est ablished : 1928 • Objective : To promote wor ldwide humanitarian aid • H eadquar t er s : Geneva E ur opean U nion • Established : Apr. 8, 1965. Effective on July 1. 1967. • Object ive : To cr eat e a unit ed Eur ope in which m em ber cou n t r i es w ou l d h av e su ch st r on g economic and political bonds that war would cease t o be a r ecur r ing fact . • M ember s : 27 (The t en new count r ies which joined i n 2004 ar e Cypr us, Czech Republ i c, Est oni a, H u ngar y, L at vi a, L i t hu an i a. M al t a, Pol an d, Slovakia and Slovania). • H eadquar t er s : Br ussels (Belgium). The common Eur opean, cur r ency. Eur o, was launched on Jan. 1, 1999. N orth At lantic Treaty Organisat ion (N AT O) • Est ablished : Apr il 4, 1949 • Object ive : M ut ual defence and co-oper at ion • M ember s : 28 • H eadquar t er s : Br ussels Organisat ion of Pet roleum Export ing Count ries (OPE C) • Est ablished : Sept ., 1959 • Obj ect i ve : A t t em pt s t o set wor l d pr i ces by con t r ol l i ng oi l pr oduct i on an d al so per sues member int er est in t r ade and development . • Members : 12 (Algeria, Iran, Iraq, Kuwait, Libya, UAE, Nigeria, Qatar, Saudi Arabia, Ecuador and Venezuela). • H eadquar t er s : Vienna (Aust r ia)

CURRENT AFFAIRS

Sou t h Asi a n Associ at i on Co-oper at ion (SAARC) • Est ablished : Dec. 8, 1985 •

f or

• M ember s : 188 • H eadquar t er s : Geneva Shanghai Cooperat ion Or ganisat ion (SCO) • Est ablished : June 7, 2002. • Object ive : To develop mut ual cooper at ion. • M em ber s : Ru ssi a, Ch i n a, K azak h i st an , Uzbekistan, K yr ghizt an and Tajikistan. BRI CS (Brazil, Russia, I ndia, China, South Africa) • Established : 2010 befor e inclusion of Sout h Afr ica i t i s k n ow n as B RI C, w h i ch i s est abl i sh ed in 2009. • Object ive : To encour agi ng commer cial polit ical and cult ur al cooper at ion bet ween BRI CS nat ions.

Regi on al

Obj ect i ve : To pr om ot e econ omi c, soci al and cult ur al cooper at ion

•

M ember s : Bangladesh, Bhut an, I ndia, M aldives, Nepal, Pakist an. Sr ilanka & Afghanist an. • H eadquar t er s : K at hmandu Wor ld M et eor ological Or ganisat ion (WM O) • Est ablished : Oct . 11, 1947, Effect ive fr om Apr il 4, 1951. • Objective : Specialised UN Agency concer ned wit h met eor ological cooper at ion. WORLD: M I SCELLAN EOU S N at ional Emblems of F amous Count r ies

Countr y

Emblem

Countr y

Emblem

Austr alia

Kangar oo

I taly

White Lily

Bangladesh

Wat er Lily

Japan

Chr ysant hemum

Canada

White Lily

Nor way

Lion

Denmark

Beach

Pakistan

Cr escent

Fr ance

Lily

Spain

Eagle

Ger many

Corn Flower

Sr i L anka

Sword & L ion

I ndia

Lion Capit al

Russia

Sickle & H ammer

I r an

Rose

United Kingdom

Rose

I r eland

Shamrock

U.S.A.

Golden Rod

Parliament ’s N ame of t he F amous Count ries

Coun tr y

6.9

P ar liam en t

Cou ntr y

P ar lia ment

Afghan ist an

Shor a

Mal aysi a

M aj l is

Aust rali a

Par l iament

Mal dive

M aj l is

Bangl adesh

Jati a Par li am ent

Mangol ia

K hur al

Bhut an

Tasongadu

Nepal

Rast ht r iya Panchayat

Canada

Par l iament

Neth er l ands

States Gener al

Chi na

Nati onal People Congress

Norway

Stort i ng

Denm ar k

Folk eti ng

Pak i st an

Nati onal Assem bly

Egypt

People's Assembly

Pol an d

Scym

Fr ance

Nati onal Assembly

Spai n

Cr ot es

Ger man y

Bundest ag

Sweden

Ri k sdag

Great Br it ai n

Par l iament

Sout h Afr ica

Par l i ament

I ndi a

Par l iament (San sad)

Swi tzer land

Federal Assembl y

I r an

Maj l is

Russia

Duma

I r el an d

Dai l Ei r eann

Tai wan

Yuan

I sr ael

K nesset

Tur key

Grand Nat i onal Assembl y

Japan

Diet

U.S.A.

Congress

6.10 CURRENT AFFAIRS

•

Wonders of the World Ancient

M edieval

H anging Gar dens of Babylon

Colosseum of Rome

Temple of Diana at Ephesus (Rome)

Gr eat Wall of China

St at ue of Jupit er at Olympia

Por celain Tower of Nanking

M ausoleum of M ausolus (Rular of H alicar nassus)

St onehange of England

Pyr amids of Egypt

M osque at St . Sophia (Const ant inople)

L ight H ouse at Alexandr ia

Cat acombs of Alexandr ia

Colossus of Rhodes

L eaning Tower of Pisa, Taj M ahal (I ndia)

I N DI AN DEFEN CE •

The Pr esident of I ndia is t he supr eme commander of t he I ndian Defence Syst em.

•

The whole administ r at ive cont r ol of t he Ar med for ces lies in t he M inist r y of Defence.

•

I ndian Defence System has been divided into thr ee ser vices-Ar my, Navy and Air For ce.

•

T h e I n di an Ar m y i s or gan i sed i n t o sev en commands :

Commands

•

H eadquar ter s

1. West er n Command

Chandi M andir

2. East er n Command

K olk at a

3. Nor t her n Command

Udhampur

4. Sout her n Command

Pune

5. Cent r al Command

L ucknow

6. Ar my Tr aining Command

Shimla

7. Sout h West er n Command

Jaipur

I n di an ai r f or ce i s or gani sed i n t o seven commands :

Com m an ds 1. West er n Com m and 2. Cent r al Com m and 3. E ast er n Com m and 4. Sout h West er n Com m and 5. Tr ai ni ng Com m and 6. M ai nt enance Com m and 7. Sout her n Com m and

H eadqu ar t er s N ew Del hi A l l ahabad Shi l l ong Jodhpur B angal ur u N agpur Thi r uvanant h apur am

I n di an n avy commands :

i s or gan i sed

i n t o t h r ee

Commands

H eadquar ter s

East er n Command

Vishakhapat nam

Sout her n Command

Kochi

West er n Command

Mumbai

Ranks of Commissioned Officer s

Ar my

Air F or ces

N avy

Gener al

Air Chief M ar shal

Admir al

L t . Gener al

Air M ar shal

Vice Admiral

M ajor Gener al

Air Vice M ar shal

Rear Admir al

Br igadier

Air Commodor e

Commodore

Colonel

Gr oup Capt ain

Capt ain

L t . Colonel

Wing Commander

Commander

M ajor

Squadr on L eader

L t . Commander

Capt ain

Flt . L ieut enant

L ieut enant

L ieut enant

Flying Officer

Sub-L ieut enant

PARAM I LI TARY AN D RESERVED F ORCES I ndo-T ibet an Border Police (I T BP) • I t was established in 1962, after the Chinese attack. • I t is basically employed in t he Nor t her n bor der s f or mon i t or i n g t he bor der s and al so t o st op smuggling and illegal immigr at ion. N at ional Securit y Guar ds (N SG) • I t was est ablished in 1984. • I t has been est ablished t o count er t he sur ge of milit ancy in t he count r y. • I t is a highly t r ained for ce which deals wit h t he milit ant s effect ively. Cent ral I ndust rial Securit y F orce (CI SF ) • I t was set -up in 1969 aft er t he r ecommendat ions of Just ice B. M ukher ji. • I t s objective is to monitor t he indust r ial complexes of Cent r al Gover nment . Assam Rifles • I t was est abl i shed i n 1835 and i s t he ol dest par amilit ar y for ce in t he count r y. • I t s m ai n obj ect i v e i s t o k eep v i gi l an ce of inter national bor der s in Nor th East and counter ing i nsur gency oper at i ons i n Ar unachal Pr adesh, M anipur, M izor am and Nagaland. Border Securit y F orce (BSF ) • I t was est ablished in 1965. • I t keeps a vi gi l over t he int er nat ional bor der s against t he int r usion in t he count r y Cent ral Reserve Police F orce (CRPF ) • I t was set -up in 1939.

CURRENT AFFAIRS 6.11

•

I t s main object ive is t o assist t he St at e / Union Ter r it or y Police in maint enance of law and or der. • The 88t h Bat t alion of CRPF, known as ‘M ahila Batt alion' (commissioned on Mar ch 30,1986) is the wor ld's fir st par amilitar y for ce compr ising entir ely of women. N at ional Cadet Corps (N CC) • I t was est ablished in 1948. (came int o exist ence on 16t h july 1948) • I ts main objective is to stimulate interest among the youth in the defence of the country in order to build up a reserve man power to expand armed forces. Terr it orial Army (TA) • I t was est ablished in 1948. • I t is a volunt ar y, par t -t ime for ce (bet ween 18 and 35 year s), not pr ofessional soldier s, but civilians, who wish t o assist in defence of t he count r y. H ome Guards • I t was est ablished in 1962, t o assist t he police in maint aining secur it y, t o help defence for ces and t o help local author it ies in case of any event ualit y. Coast Guar d • I t was set -up in 1978. (18t h Aug. 1978) • I ts object ive is to pr otect the mar itime and other national inter ests in the mar itime zones of I ndia. I t was set-up in 1920. I nt elligence Bureau (I B) • I t s obj ect i ve i s t he col l ect secr et i nfor mat i on r elat ing t o count r y's secur it y. • I t was or iginally set-up as Cent r al Special Br anch (CSB) in 1887. Cent ral Bureau of I nvest igat ion (CBI ) • The CBI was est ablished in 1941 as t he special pol i ce est abl i sh m en t , t ask ed wi t h dom est i c secur i t y. I t was r enamed t he cent r al Bur eau of invest igat ion on 1 Apr il 1963. • I t s object ive is t o invest igat e cases of misconduct by public servants, cases of cheating, embezzlement and fr aud. • CBI i s also ent r ust ed wi t h t he invest igat i on of i nt er nat ional cr ime cases in coll abor at ion wit h I NTERPOL. N at ional Crime Records Bureau (N CRB) • I t was est ablished in 1986. • I t s object ive is t o collect cr i me st at ist i cs at t he nat i onal l evel , i nfor mat i on of i nt er st at e and i nt er nat i onal cr i mi nal s t o hel p i nvest i gat i on agencies. Rapid Act ion F orce (RAF ) • I t was est ablished in 11 december 1991. • Under t he oper at ional command of CRPF. • 10 bat t alions of t he CRPF have been r eor ient ed for t ackling communal r iot s in t he count r y.

FOLK AN D TRI BAL DAN CES St at e D ance Assam Bihu, Khel Gopal, Rakhal Leela, Tabal Chongli, Canoe, Nongkr em. AP Ghanta Mar dala, Bur r akatha, Veedhi Natakam, K uchipudi Bihar Jata Jatin, Chhau, Bakho, Kathaputli, J h i j h i y a, Sam o Ch ak w a, J at r a, K ar ma. Gujar at D an di y a Raas, Gar ba, Gom ph , Tippani, Bhavia, Zer iyun. Haryana Ghumar, Phag Dance, Daph, Dhamal, K hor ia. HP Jhor a, Thali, M ahasu, Jadda J& K Rouf, H ikat , Damali. Kar nataka Yakshagana, H ut t ar i Kerala Kaikottikali, Kaliyattam, Tappatikkali. MP M aach, Gher o, Gafa, K at ha-K eer t an. Mahar ashtr a L ezim, Tamasha Punjab Gi ddha (women), Bhangr a (men), Naqual, Bhand. Rajasthan Sui si ni , Chak r i , Ter aht aal , Gi nad, K hyal, Gangor e. Tamil Nadu K olat t am, Pinnal, K ar agam UP J h or a, K aj r i , K ar an , N ou t an k i , Chappeli, Raasleela and Ramleela WB K ahi, Jat r a. I M PORTAN T I N STRU M EN T PLAYERS I nst r ument Pl ayer Sitar I nvent ed by Ami r K husr au; N i k hi l Banner ji, Pt . Ravi Shankar, Vilayat K han, H ar i Shank ar Bhat t achar ya, Devbr at Choudhar y Tabla Allah Rakha K han, K ishan M ahar aj, Nikhil Ghosh, Zakir H ussain, Shaafal Ahmed K han Sarod Al i Ak bar K han, Al l auddi n K han, A m j ad A l i K h an , B u ddh adev Dasgupt a, Bahadur K han, Shar an Rani, Zar in S. Shar ma Violin Baluswamy Dikshit ar, Gujanan Rao Joshi, L al gudi G. Jayar aman, M .S. Gopal a-K r i sh n an , M y sor e T. Ch ow di ah , T.N . K r i sh n an , L . Subr amaniam Sant oor Pt . Shiv K umar Shar ma Shehnai Ust ad Bismillah K han Flut e Pt .H ar i Pr asad Chaur asiya, Pannalal Ghose, T.R. M ahalingam Veena K.R. Kumar aswami lyer, Dor aiswamy lyengar Dhrupad Har idas Swami Mandolin V. Sr inivas

6.12 CURRENT AFFAIRS

CABI N ET M I N I STERS N AM E Shr i Raj Nath Singh Smt . Sushma Swar aj Shr i Ar un Jait leyPiySush Goel Smt . Nir mala Sithar aman Shr i Nit in Jair am Gadkar i Shr i Mukht ar Abbas Naqvi Shr i Piyush Goyal Shr i D.V. Sadananda Gowda Smt . Uma Bhar at i Shr i Ramvilas Paswan Shr i Gir ir aj Singh Smt. Maneka Sanjay Gandhi Shr i Ravi Shankar Pr asad Shr i Jagat Pr akash Nadda Shr i Sur esh Pr abhu Shr i Anant Geet e Smt. H ar simr at K aur Badal Shr i Nar endr a Singh Tomar Shr i Chaudhar y Bir ender Singh Shr i Jual Or am Shr i Radha Mohan Singh Shr i Thaawar Chand Gehlot Smt . Smr i t i I r ani Dr. H ar sh Var dhan Shr i Pr akash Javadekar

PORTFOLI O M inister of H ome Affair s. Minist er of Exter nal Affair s. M inist er of Finance, M inist er of Cor por ate Affair s M inist er of Defence M inist er of Road Tr anspor t and H ighways; and M inist er of Shipping. River Development and Ganga Rejuvenation. Minist er of Minor ity Affair s M inister of Railways and M inist er of Coal M inist er of Statistics and Pr ogr amme I mplement ation and M inister of Chemicals & Fer t ilizer s M inister of Dr inking & Sanitation. M inist er of Consumer Affair s, Food and Public Distr ibution. M inister of Micr o, Small and Medium Enter pr ises. M inist er of Women and Child Development . M i n i st er of L aw an d J u st i ce; an d M i n i st er of E l ect r on i cs and I nfor mat ion Technology. M inist er of H ealth and Family Welfar e. M inister of Civil Aviation & Mini ster of Commer ce & I ndustr y M inist er of H eavy I ndustr ies and Public Enter pr ises. M inister of Food Pr ocessing I ndust r ies. M i ni st er of Rur al Devel opment ; M i ni st er of Panchayat i Raj and M inister of Par liament ar y Affair s. M inist er of Steel. Minist er of Tr ibal Affair s. M inister of Agr icult ur e and Far mer s Welfar e. M inist er of Social Just ice and Empower ment . M inister of Textiles M inist er of Science and Technology; and Minister of Ear t h Sciences, M inist er of Envir onment , For est and Climat e Change M inist er of H uman Resour ce Development .

M I N I STERS OF STATE (I N DEPEN DEN T CH ARGES) Rao I nder jit Singh Shr i Sant osh K umar Gangwar Shr i Gir ir aj Singh

M inister of State, Chemical & Fer t ilizer Minister of State (Independent Charge) of the Ministry of Labour and Employment. M inist er of State (I ndependent Char ge) of the M inist r y of M icr o, Small and Medium Enter pr ises. Shr i Shr ipad Yesso Naik M inist er of State (I ndependent Char ge) of the M inist r y of Ayur veda, Yoga and Natur opathy, Unani, Siddha and Homoeopathy (AYUSH ). Col. Rajyavardhan Singh Rathore M inister of State (I ndependent Char ge) of t he Ministr y of Yout h Affair s and Spor t s; and M i ni st er of St at e (I ndependent Char ge) of t he M i ni st r y of I nfor mat ion and Br oadcasting.. Dr. Jitendr a Singh M inist er of State (I ndependent Char ge) of the M inist r y of Development of Nor t h East er n Regi on; M inist er of St at e in t he Pr ime M inist er ’s Office; Minister of State in the Ministry of Personnel, Public Gr ievances and Pensions; M inister of State in t he Depar t ment of At omic Ener gy; and Minister of St at e in the Depar tment of Space.

CURRENT AFFAIRS 6.13

Sur esh R. K. Singh Dr. M ahesh Shar ma Shr i M anoj Sinha Shr i H ar deep Singh Pur i Shr i Alphons Kannanthanam

Minister of State (I ndependent Charge) of the Ministr y of Power; and Minister of State (I ndependent Char ge) of the Ministr y of New and Renewable Ener gy. Minister of State (I ndependent Charge) of the Ministry of Culture; and Minister of St ate (I ndependent Char ge) of t he Ministr y of Tour ism & Civil Aviation. M inister of State (I ndependent Char ge) of t he Ministr y of Communicat ions; and M inist er of St at e in the M inist r y of Railways. Minister of State (I ndependent Char ge) of the Ministr y of Housing and Ur ban Affairs.. M inist er of State (I ndependent Char ge) of the Ministr y of Tour ism.

M I N I STERS OF STATE Gener al (Ret d.) V.K . Singh

M inist er of State in t he Ministr y of Exter nal Affair s.

Shr i Shiv Pr atap Shukla Shr i S.S. Ahluwalia

M inist er of State in t he Ministr y of Finance. Minister of State in the Ministry of Electronics and Information Technology

Shr i Ramdas Athawale Shr i Ram K r ipal Yadav

M inister of State in t he Minist r y of Social Just ice and Empower ment . M inist er of State in t he Ministr y of Rur al Development .

Shr i Har ibhai Par thibhai Chaudhar y Shr i H ansr aj Gangar am Ahir

M inister of St ate for Coal and Mines M inist er of State in t he Ministr y of Home Affair s.

& Shr i K ir en Rijiju Shri Babul Supriyo

M i ni st er of St at e i n t he M inist r y of H eavy I ndust r ies and Publ ic Ent er pr ises.

Shr i Ramesh Chandappa Jigajinagi Shr i Rajen Gohain

M inist er of State in t he Ministr y of Dr inking Wat er and Sanit ation. M inister of State in t he M inist r y of Railways.

Shr i Par shott am Rupala

M inist er of State in t he Ministr y of Agr icultur e and Far mer s Welfar e; and M inist er of Stat e in the Ministr y of Panchayati Raj.

Shr i Radhakr ishnan P. Shr i K r ishan Pal

M inister of State in t he M inist r y of Road Tr anspor t and H ighways; and M inist er of St at e in the M inist r y of Shipping. M inister of State in t he Minist r y of Social Just ice and Empower ment .

Jaswantsingh Sumanbhai Bhabhor Shr i Vishnu Deo Sai

M inist er of State in t he Ministr y of Tr ibal Affair s. M inist er of Stat e in t he Ministr y of Steel.

Shr i Sudar shan Bhagat Shr i Jayant Sinha

M inist er of State in t he Ministr y of Agr icultur e and Far mer s Welfar e. M inist er of State in t he Ministr y of Civil Aviation.

Y. S. Chowdar y Sadhvi Nir anjan Jyoti

Minister of State in the Ministry of Science and Technology; and Minister of St ate in t he Minist r y of Ear t h Sciences. M inist er of State in t he Ministr y of Food Pr ocessing I ndust r ies.

Shr i Vijay Sampla Shr i Ar jun Ram Meghwal

M inister of State in t he Minist r y of Social Just ice and Empower ment . M inist er of State in t he Ministr y of Wat er Resour ces

Shr i Ajay Tamt a Smt . K r ishna Raj

M inister of State in t he M inist r y of Text iles. M inist er of State in t he Ministr y of Agr icult ur e & Far mer s Welfar e.

Shr i Anant K umar Hegde

M i n i st er of St at e i n t h e M i n i st r y of Sk i l l Devel opm en t an d Entr epr eneur ship. M inister of State in t he Minist r y of H ealth and Family Welfar e.

Smt . Anupr iya Patel Shr i C. R. Chaudhar y

Minister of State in the Ministr y of Consumer Affair s, Food and Public Distr ibution.

Shr i P.P. Chaudhar y

M inist er of State in t he Ministr y of Law and Just ice; and M inist er of State in t he Ministr y of Cor por at e Affair s. M inist er of State in t he Ministr y of Defence.

Dr. Subhash Ramr ao Bhamr e

6.14 CURRENT AFFAIRS

CH I EF M I N I STERS & GOVERN ORS OF I N DI AN STATES S. N o. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

N ame of States Andhra Pradesh Arunachal Pr adesh Assam Bihar Chhattisgarh Goa Gujarat Haryana Himachal Pr adesh Jammu Kashmir Jharkhand Karnataka Kerala Madhya Pradesh Maharashtra Manipur Meghalaya Mizoram Nagaland Odisha Punjab Rajasthan Sikkim Tamil Nadu Telangana Tripura Uttar Pr adesh Uttarakhand West Bengal

Chief M inister Shr i N. Chandr ababu Naidu Shr i Pema Khandu Shr i Sar bananda Sonowal Shr i Nitish Kumar Shr i Bhupesh Baghel Shr i Manohar Par r ikar Shri Vijay Rupani Shr i Manohar Lal Khat tar Shr i Jayr am Thakur – Shr i Raghubar Das Shr i H. D. Kumar aswamy Shr i Pinar ayi Vijayan Shr i Kamalnath Shr i Devendr a Fadnavis Shr i N. Bir en Singh Shr i Conrad Sangma Shri Zor amt hanga Shr i Neiphiu Rio Shr i Naveen Patnaik Captain Amarinder Singh Shr i Ashok Gehlot Shr i Pawan Chamling Shri Edappadi K. Palaniswami Shr i K. Chandr asekhar Rao Shr i Biplab Kumar Deb Shr i Yogi Adityanath Shr i Tr ivendr a Singh Rawat Shr i Mamat a Baner jee

Governor Shr i. E. S. L. Nar asimhan Brigadier BD Mishr a (Retd) Jagdish Mukhi L alji Tandon. Ram Nat h Kovind Anandiben Pat el Mridula Sinha Shr i Om Pr akash K ohli Satyadev Nar ayan Ar ya Achar ya Dev Vr at Sat ya Pal Malik Draupadi Murmu Vajubhai Rudhbhaivala P. Sathasivam Anandiben Patel C. Vidhyasagar Rao Najma Heptulla Tathagat a Roy K . Rajasekhar an Shr i Acharya Padmanabha Shr i Ganeshi L al V. P. Singh Badnor e Kalyan Singh Ganga Pr asad Banwar ilal Pur ohit Shr i. E. S. L. Nar asimhan K aptan Singh Solanki Shr i Ram Naik Baby Rani Maur ya Shr i Keshr i Nath Tr ipathi

Chief M inister

Lieutenant Governors/Administrators Admir al (Retd) Devendr a Kumar Joshi V. P. Sing Badnor e Shr i Pr aful Patel Shr i Pr aful Patel Shr i Far ooq K han Kir en Bedi Anil Baijal

U N I ON TERRI TORI ES Union Territory 1 2 3 4 5 6 7

Andaman & Nicobar Chandigarh Dadr a and Nagar Haveli Daman and Diu Lakshadweep Puducher r y New Delhi

Shr i V. Nar ayanswami Shr i Ar vind Kejar iwal

N EW JU DI CI ARY Mr. Ranjan Gogoi Mr. K.K. Venugopal

: Chief Justice of I ndia : Attor ney Gener al of I ndia

Mr. Ranjit Kumar Mr. Nar endr a M odi Mr. Sunil Ar or a Mr. Rajiv Mehr ishi Mr. Just ice H.L. Dattu

: : : : :

Solicitor Gener al of I ndia Chair man, NI TI Aayog Chief Election Commissioner Compt r oller and Audit or -Gener al of I ndia. Chair per son, National Human Right Commission (NHRC)

CURRENT AFFAIRS 6.15

Mr. Pr adeep Kumar Sinha Mr. Nr ipendr a Mishr a Mr. Justice Vangala Eshwaraiah Ms. Stuti Nar ain Kacker

: : : :

Dr. P.L. Punia Mr. Nand Kumar Sai Mr. Vinay Mittal Dr. M. S. Swaminathan Mr. Ajit Doval

: : : : :

Cabinet Secr etar y. Pr incipal Secr et ar y to Pr ime Minist er . Chair man, National Commission for Backwar d Classes. Chair per son, National Commission for Pr otection of Child Rights

Mr. Shar ad K umar

Chair man, National Commission for Scheduled Castes Chair man, National Commission for Scheduled Tr ibes. Chair man, UPSC. Chair man, National Commission on Far mer s (NCF). Nat i onal Secur it y Advi ser and Special Adviser t o PM (I nt er nal Security). : Dir ector -Gener al, Nat ional I nvest igation Agency (NI A).

Ashwani Lohani Mr. Desh Dipak Ver ma Mr s. Snehlata Sr ivastav Mr. Sanjeev Tr ipathi Mr. RC Tayal Mr. Rajiv Rai Bhatnagar

: : : : : :

Chairman, Railway Board. Secretar y-Gener al, Rajya Sabha Secr etar y-Gener al, L ok Sabha. Secr etar y, Resear ch and Analysis Wing. Dir ector -Gener al, NSG. Dir ector-Gener al, CRPF.

Mr. K.K. Shar ma Mr. Rajesh Ranjan Mr. Dhar mendr a K umar Mr. R.K. Pachnanda Mr s. Rajni Kant Mr. Rajendr a Singh

: : : : : :

Dir ector -Gener al, Bor der Secur ity For ce (BSF). Dir ector -Gener al, Centr al I ndust r ial Secur it y For ce (CI SF). Dir ector -Gener al, Railway Pr otection For ce. Dir ector -Gener al, I ndo-Tibetan Bor der Police. Dir ector -Gener al, Sashastr a Seema Bal. Dir ector -Gener al, I ndian Coast Guar d.

Lt. Gen. Avtar Singh Mr. D.P. Singh Mr. K. Vijay Raghavan Mr. K. Sivan Mr. Sekhar Basu

: : : : :

Dir ector-Gener al, Defence I ntelligence Agency. Chair man, UGC. Pr inciple Scient ific Adviser to the Gover nment of I ndia. Chair man, Space Commission and I SRO. Chairman, Atomic Energy Commission and Secretary, Dept. of Atomic Ener gy.

Mr. Gair ul Hasan Rizvi Mr. Radha Krishna Mathur Mr. Soumya Swaminathan Mr. C. Chandr amouli Mr. Just ic Balbir Singh Chauhan Mr. Baldev Raj

: : : : : :

Chair per son, National Commission for Minor ities. Chief I nfor mation Commissioner. Dir ector -Gener al, I ndian Council of Medical Resear ch. Registr ar-Gener al of I ndia and Census Commissioner. Chair man, Law Commission. Pr esident , I ndian National Academy of Engineer ing (I NAE).

Mr. Justice (Ret d.) B. N. Kir pal Mr. Dilip Rath Lt. Gen. Sanjeev Kumar Shr ivastav Ur jit Patel Mr. Justice Chandr amauli Kumar Pr asad Mr. Shashidhar Sinha

: : : : : :

Chair man, National For est Commission. Chair per son, National Dair y Development Boar d (NDDB). Dir ector -Gener al, Bor der Roads Or ganisat ion. Gover nor, RBI Chair man, Pr ess Council of I ndia. Chairman, Audit Bur eau of Cir culations (ABC).

Mr. Pr akash Chandr a Vanaja N. Seena Justice Sudhansu Jyoti Mukhopadhaya Balr aj Joshi Shashi Shanker

: : : : :

Chair per son, Centr al Boar d of Dir ect Taxes (CBDT). Chair man, Centr al Boar d of Excise & Cust oms. Chair per son, Competition Appellate Tr ibunal CMD, NH PC. CMD, ONGC.

6.16 CURRENT AFFAIRS

Mr. B.C. Tr ipathi Mr. Sanjiv Singh Mr. Utpal Bor a Anita Kar wal

: : : :

CMD, GAI L. Chair man, I OCL. CMD, Oil I ndia L td. Chair man, CBSE.

Ajay Tyagi Mr. Har sh Kumar Bhanwala Mr. Rajnish Kumar Mr. M S Raghavan Mr. Mahesh Mit tal Kumar Rajiv Kumar Chander

: Chair man, Secur ities & Exchange Boar d of I ndia. : Chair man, National Bank for Agr icultur e and Rur al Development (NABARD). : Chairman, SBI : Chairman, I DBI . : Chair man, Company Law Boar d. : I ndia's Per manent Repr esentat ive to UN.

Mr. S.K. Roy Mr. Ashwin B. Pandya Ms. Lalitha K umar amanglam Dr. Y.V. Reddy Mr. Pr onab Sen Ms. Pahlav Nihalani

: : : : : :

Chair man, LI C Chair man, Centr al Water Commission. Chair per son, Nat ional Commission for Women. Chair man, 14th Finance Commission. Chair man, National St atistical Commission. Chair per son, Centr al Boar d of Film Cer tification.

Dr. Budha Rashmi Mani Mr. Rashesh Shah Mr. R.S. Shar ma Mr. Hemant Mr. K.V. Chowdar y Mr. K.S. Vyas

: : : : : :

Dir ector -Gener al, Ar chaeological Sur vey of I ndia. Pr esident, FI CCI . Chairman, TRAI Chair man, Pension Fund Regulator y & Development Author ity. Centr al Vigilance Commissioner (CVC). Dir ector, Bhabha Atomic Resear ch Cent r e.

Nar inder Dhr uv Batr a Hr ushikesh Senapaty Ajay S. Shr ir am Mr. Sam Pit r oda Mr. Jawhar Sir car

: : : : :

Pr esident, I ndian Olympic Association. Dir ector, NCERT. Pr esident, CI I . Chair man, National K nowledge Commission. CEO, Pr asar Bhar ti Boar d.

Mr. Rishad Pr emji Mr. Sandeep Jajodia Mr. Sur esh Gopi Viveck Goenka N.K . Singh

: : : : :

Chair man, NASSCOM. Pr esident, ASSOCHAM. Chair man, National Film Development Cor por ation (NFDC). Chair man, PTI . Chair man, 15th Finance Commission

CURRENT AFFAIRS 6.17

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. Whi ch of t he fol l owi ng count r i es has began labor ator y tests of A(H1N1) flu vaccine? (a) China

(b) Canada

(c) The US

(d) Mexico

2. T h e Sou t h A si an A ssoci at i on f or Regi on al Cooper at i on (SA A RC) is r egi on al int er gover nment al or ganization and geopolitical union in South Asia. SAARC was founded in ______ in 1985. (a) Dhaka, Bangladesh (b) New Delhi, I ndia (c) Kathmandu, Nepal (d) Male, Maldives 3. Wor ld Senior Cit izen’s Day is celebr ated on (a) August 6

(b) August 8

(c) July 7

(d) July 9

4. Which business house has pr omoted Essar gr oup of companies ? (a) Ruias

(b) Ambanis

(c) Goenkas

(d) Kanor ias

5. Nor th Kor ea, officially the Democr atic people’s Republic of Kor ea, is a countr y in East Asia, in the nor ther n par t of the K or ean Peninsula what is t he capital of Nor t h Kor ea ? (a) Pyongyang

(b) Seoul

(c) Tokyo

(d) Beijing

6. The New identit y of softwar e ser vices pr ovider Satyam Computer Ser vices Ltd. is (a) Satyam Mahindra

(b) Mahindra Satyam

(c) Tech Sat yam

(d) Sat yam Tech

7. Cor sica is an island in t he M editer r anean Sea belonging to Fr ance. What is capital of Cor sica?

9. Which gas is r eleased fr om paddy fields? (a) Car bon dioxide

(b) Methane

(c) Ammonia

(d) Hydr ogen Sulphide

10. Which of the following companies has r olled out the Jaguar and Land Rover (JLR) in I ndia? (a) Tat a Mot or s

(b) For d Motor Co.

(c) Por sche

(d) Mer cedes

11. Who has been appoint ed as the FI FA pr esident ? (a) Sepp Blatter

(b) I ssa H ayatou

(c) Gianni I nfantino

(d) Joao Havelange

12. What is the main Chemical component of Vinegar? (a) Acetic acid

(b) Citr ic acid

(c) Tar t ar ic acid

(d) Nitr ic acid

13. ______ basically r efer s to a section of the financial mar ket wher e financial instr ument s wit h high liquidity and shor t-t er m matur it ies ar e tr aded. (a) Tr ade Mar ket

(b) Capital Market

(c) Money Mar ket

(d) Expor t Market

14. The new centr al vigilence commissioner of I ndia (a) Pr adeep Kumar

(b) KV Chowdar y

(c) Shar ad Kumar

(d) Shanta Sinha

15. Mont er o 09, the newly launched pr emium spor t utility vehicle by H indust an Motor s and Japanbased Mitsubishi Motor s, is (a) impor ted fr om Japan (b) made in I ndia (c) assembled in I ndia (d) None of these 16. I t is pr oposed t o set up an I I T in Jaipur with the collabor at ion of which count r y? (a) Germany

(a) Paris

(b) Ajaccio

(b) Fr ance

(c) Zagreb

(d) Hawana

(c) Australia

8. At which of the following places will Coimbator ebased Elgi Equipments limited set up a subsidiary?

(d) Japan 17. Who is the Pr esident of Wor ld Bank?

(a) Sao Paolo in Br azil

(a) Jim Yong Kim

(b) Rio de Janeir o in Br azil

(b) Pascal L amy

(c) Santiago in Chile

(c) Raul Castr o

(d) Car acas in Venezuela

(d) None of these

6.18 CURRENT AFFAIRS

18. Gr een Banking means

26. Which par t y in Russia does Mr. Put in heads ?

(a) financing of ir r igat ion pr oject s by bank

(a) One Russia

(b) Victor ious Russia

(b) development of for estr y by banks

(c) Unified Russia

(d) United Russia

(c) financing of envir onment fr iendly pr oject by banks (d) None of these

27. Wher e is Titicaca , the highest lake in the wor ld, located? (a) Russia and Cent r al Asian countr ies

19. Vijay Kelkar Committ ee r epor t deals with

(b) US and Canada bor der

(a) Centr al-State financial r elations

(c) South Afr ica

(b) Tax r efor ms

(d) Per u and Bolivia bor der

(c) Tr ade r efor ms (d) Refor ms in public sector enter pr ises 20. Which spor t is Lin Dan associated wit h ? (a) Table Tennis

(b) Mahesh Bhupati

(c) M ar k Knowles

(d) Tr avis Par r ot t

21. Jnanpith Awar d was pr esented t o whom among t he following? (a) H indi poet Sat ya Vr at Shastr i

28. The I ndian Space Resear ch Or ganisat ion (I SRO) is the space agency of the I ndian gover nment . I SRO headquar ter is locat ed in? (a) New Delhi

(b) Chennai

(c) Mumbai

(d) Bengaluru

29. Whi ch i s t he onl y st at e t hat has appl i ed t he elect r onic payment facilit y for it s tr ader s so t hat t hey coul d mak e t heir VAT payment t hr ough e-payment ?

(b) Sanskr it poet Sat ya Vr at Shastr i

(a) Gujar at

(b) Maharashtra

(c) Ur du poet Ghulam Azad

(c) Tamil Nadu

(d) Ker ala

(d) Hindi wr iter Ramakant Tr ipat hi 22. At which point of Ear t h t her e is no gr avity? (a) At Nor th and Sout h Pole (b) At equat or

(d) At cent r e of t he Ear t h 23. I n t he Panchayat s, women r eser vat i on has incr eased fr om 33% t o (a) 40 %

(b) 50 %

(c) 51 %

(d) 49 %

24. Who star ted the ‘Yellow Revolut ion’ in 1980’s in I ndia ? (a) Sam Pit r oda

(b) Nar ayan Mur this

(c) Nandan Nilekani

(d) Swaminathan

25. I nt er nat ional L iter acy Day is celebr at ed on (a)

Sept ember

(c) 8th Sept ember

(b)

(a) Red Rose

(b) Black Rose

(c) White Rose

(d) Gr een Rose

31. Oceansat -2 has been inject ed int o or bit at an alt it ude of about

(c) On the ocean sur face

6th

30. What was t he name of t he ship t hat sank near t he Par adip por t causing oil spill ?

1st

August

(d) 1st October

(a) 238 km

(b) 526 km

(c) 728 km

(d) 936 km

32. Th e I n t er n at i onal Day of Ol der Per son s i s obser ved on? (a) 2nd Oct ober

(b) 24t h October

(c) 16t h October

(d) 1st October

33. The fir st non-stop air -conditioned ‘Dur ant o’ tr ain was flagged off between (a) Sealdah-New Delhi (b) Mumbai-Howr ah (c) Bangalor e-Howrah (d) Chennai-New Delhi

CURRENT AFFAIRS 6.19

34. Which of t he foll owing is t he lat est Advanced Landing Gr ound (ALG) oper ationalised by I AF in L adakh ?

42. The Feder al Reser ve Syst em also known as the Federal Reserve or simply as the Fed is the central banking syst em of?

(a) Daulat Beg

(b) Fuk Che

(a) China

(b) Fr ance

(c) Nyoma

(d) None of these

(c) UK

(d) USA

LEVEL-1 35. Wit h which of t he following countr ies has I ndia si gn ed i t s si x t h ci v i l n u cl ear cor por at i on agr eement ? (a) Kazakhstan

(b) Namibia

(c) Mongolia

(d) Nepal

36. I n which of the following islands, massive tsunami wave killed at least 113 people ? (a) Java

(b) Samoan I slands

(c) Fiji

(d) Philippines

37. For t he fir st t ime, t he tot al number of account s held by women in public sect or banks in I ndia has cr ossed (a) 50 lakh

(b) 80 lakh

(c) 1 cr or e

(d) 1.5 cr or e

38. Which of the following count r ies has dr opped the wor d ‘communism’ fr om its Constitut ion ? (a) Cuba

(b) Nor th Kor ea

(c) China

(d) none of these

39. The MahabodhiVihar, a UNESCO World Heritage Sit e, is a Buddhist temple locat ed at which of the following place? (a) Guwahati, Assam (b) Deoghar, Jhar khand (c) Pat na, Bihar

(d) Dar bhanga, Bihar

40. Wh at i s t h e si gn i f i can ce of Ozon e l ayer i n atmospher e? (a) I t maint ains t he cycle of seasons. (b) I t causes timely ar r ival of M onsoon (c) I t filter s t he ult r aviolet r ays of t he sun (d) I t pr event s t he global war ming 41. The National Rur al Employment Guar ant ee Act (NREGA) has been r enamed after (a) Lal Bahadur Shastr i

43. Guwahat i is sit uated on which of t he following bank of the r iver ? (a) Sone

(b) H ooghly

(c) Brahmaputra

(d) Teest a

44. The second war shi p l aunched under Pr oj ect 15-A is named (a) I NS Kolkata

(b) I NS K ochi

(c) I NS Chennai

(d) I NS Coimbat or e

45. On Apr il 6, 1896, the first modern Olympic Games wer e hel d i n _________ wi t h at hlet es fr om 14 count r ies par ticipating. (a) L ondon, UK

(b) Athens, Gr eece

(c) Tokyo, Japan

(d) New Yor k, USA

46. Which of the following countries has the lowest ratio of external debt to Gross National I ncome ? (a) China

(b) India

(c) I ndonesia

(d) Malaysia

47. I CI CI Bank is an I ndian mult inat ional banking and financial ser vices company headquar ter ed in Mumbai, M ahar asht r a, I ndia, with it s r egist er ed office in Vadodar a. I CI CI st ands for ? (a) I ndustr ial Cr edit and I nvestment Corpor ation of I ndia Bank (b) I n du st r i al Cu st om er an d Cor por ation of I ndia Bank

I n su r an ce

(c) I n t er n at i on al Cr edi t an d I n v est m en t Cor por ation of I ndust r ial Bank (d) I ndian Cr edit and I ndustr ial Cor por at ion of I ndia Bank 48. The "God of Small Things" is t he debut novel narr ating a story about the childhood exper iences of fr ater nal twins whose lives ar e dest r oyed by t he "Love Laws" t hat lay down "who should be loved, and how. And how much". I t is wr it ten by?

(b) Sar dar Vallabhbhai Patel

(a) Chetan Bhagat

(c) Mahatma Gandhi

(c) Pr anab M ukher jee (d) Arundhati Roy

(d) Rajiv Gandhi

(b) Amish Tr ipathi

6.20 CURRENT AFFAIRS

49. What does NSAB stand for in I ndian context ? (a) Nat ional Societ y for Animal Br eeder s

58. Which count r y’s l ower house of Par l iament is known as Wolesi Jir ga?

(b) Nat ional Suppor t Activity for Bhutan

(a) Afghanistan

(b) Bangladesh

(c) Nat ional Spir itual Assembly of Buddhists

(c) Pakist an

(d) I r an

(d) National Secur it y Advisor y Boar d

LEVEL-2 50. Which of t he following cit ies will host t he 2020 Olympics ? (a)Chicago (c) Madr id

(b) Rio de Janeir o (d) Tokyo

51. 'Follow-on' is used which of the following game? (a) Badminton

(b) Tennis

(c) Foot ball

(d) Cr icket

52. Which family owns the H er o gr oup of Companies in I ndia?

59. Which of t he following companies is posing a stiff competit ion t o Sweden’s Er icsson t o become the wor ld’s lar gest telecom equipment manufactur er ? (a) Samsung

(b) Huawei

(c) Nokia

(d) Panasonic

60. What is the pr esent per missible power densit y value for mobile tower s oper ating at 1800 M Hz? (a) 4.5 wat t /met r e2

(b) 9 wat t/met r e2

(c) 13.5 wat t/met r e2

(d) 18 wat t/met r e2

61. I n which countr y is the headquar t er s of Nestle company located ?

(a) Hindujas

(b) Fir odias

(a) Sweden

(b) Switzerland

(c) Munjals

(d) Mansingh

(c) Ger many

(d) Nether lands

53. What is t he name of I ndia’s fir st indigenously developed nuclear submar ine? (a) I NS Sindhughosh (b) K 15 (c) I NS Ar ihant

(d) I NS Vikr amditya

54. Per iyar Tiger Reser ve locat ed in Thekkady at the distr ict of I dukki is one of t he 27 t iger r eser ves in I ndia is located in which state?

62. Wher e ar e Cent r e for Cell ul ar and molecular Biology, Indira Gandhi Centre for Atomic Research and Vik r am Sar abhai Space Resear ch Cent r e r espectively locat ed? (a) Hyder abad, Tar apur and Shr ihar ikota (b) Chennai, Tar apur and Thir u-vananthapur am

(a) Ker ala

(b) New Delhi

(c) H y der abad, K al pak k am vananthapuram

(c) Odisha

(d) Karnataka

(d) Chennai, Tar apur and Sr ihar ikota

55. Which of t he following countr ies in the M iddle East is facing a civil war between the Gover nment for ces and r ebed insur gent s ? (a) Syria

(b) Egypt

(c) Tunisia

(d) Jor don

56. SARFAESI Act is an I ndian law. I t allows banks and ot h er f i nan ci al i n st i t ut i on s t o au ct i on r esident ial or commer cial pr oper ties to r ecover loans. I n SARFAESI , F stands for ? (a) Fol low

(b) Financial

(c) Fill-up

(d) For um

57. H ow do you obser ve ever y year t he fir st M onday of October ? (a) Envir onment Day (b) Wor ld H abitat Day (c) Wor ld Diabet es Day(d) UNO Day

an d

Thir u-

63. What is Wakhan Cor r idor ? (a) Nor th East er n extr emit y of Afghanistan (b) Par t of the bor der bet ween I ndia and China in Ladakh r egion (c) A passage bet ween I r an and Afghanist an (d) N ar r ow st r i p j oi n i n g B an gl adesh w i t h Myanmar 64. Who is t he pr esent Gover nor of Kar nataka? (a) Hans Raj Bhar dwaj (b) Mridula Sinha (c) Vajubhai Rudabhai Vala (d) Sathasivam

CURRENT AFFAIRS 6.21

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (a)

2. (a)

3. (b)

4. (a)

5. (a)

6. (b)

7. (b)

8. (a)

9. (b)

10. (a)

11. (c)

12 (a)

13. (c)

14. (b)

15. (a)

16. (b)

17. (a)

18. (c)

19. (b)

20. (b)

21. (b)

22. (d)

23. (b)

24. (a)

25. (c)

26. (d)

27. (d)

28. (d)

29. (d)

30. (b)

31. (c)

32. (d)

33. (a)

34. (c) 41. (c)

42. (d)

43. (c)

44. (b)

56. (b)

57. (b)

58. (a)

59. (b)

LEVEL-1 35. (c)

36. (b)

37. (c)

38. (b)

39. (a)

45. (b)

46. (a)

47. (a)

48. (d)

49. (d)

40. (c)

LEVEL-2 50. (d)

51. (d)

52. (c)

53. (c)

54. (a)

60. (a)

61. (b)

62. (a)

63. (a)

64. (c)

55. (a)

CBT – II

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

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BASICS OF ENVIRONMENT AND POLLUTION CONTROL DEFINITION Environmental pollution is “the contamination of the physical and biological components of the earth/atmosphere system to such an extent that normal environmental processes are adversely affected”. Pollution is the introduction of contaminants into the environment that cause harm or discomfort  to humans or other living organisms, or that damage the environment” which can come “in the form of chemical substances, or energy such as noise, heat or light”. “Pollutants can be naturally occurring substances or energies, but are considered contaminants when in excess of natural levels.

TYPES OF POLLUTION: There are several types of pollution, and while they may come from different sources and have different consequences, understanding the basics about pollution can help environmentally conscious individuals minimize their contribution to these dangers.

AIR POLLUTION Air pollution is defined as any contamination of the atmosphere that disturbs the natural composition and chemistry of the air. This can be in the form of particulate matter such as dust or excessive gases like carbon dioxide or other vapors that cannot be effectively removed through natural cycles, such as the carbon cycle or the nitrogen cycle. Air pollution comes from a wide variety of sources. Some of the most excessive sources include:  Vehicle or manufacturing exhaust  Forest fires, volcanic eruptions, dry soil erosion, and other natural sources  Building construction or demolition Depending on the concentration of air pollutants, several effects can be noticed. Smog increases, higher rain acidity, crop depletion from inadequate oxygen, and higher rates of asthma. Many scientists believe that global warming is also related to increased air pollution.

Pollutant

Sources

Effects

Ozone. A gas that can be found in Ozone is not created directly, but is formed Ozone near the ground can cause a when nitrogen oxides and vo latile organic number of health prob lems. Ozone can two places. Near the ground (the compounds mix in sunlight. That is why ozone lead to more frequent asthma attacks in tropo sphere), it is a major part of is mostly found in the summer. Nitrogen oxides people who have asthma and can cause smog. The harmful ozo ne in the come from burning gasoline, coal, or other sore throats, coughs, and breathing fossil fuels. There are many types of volatile difficulty. It may even lead to premature lower atmosphere should not be organic compounds, and they come from death. Ozone can also hurt plants and confused with the protective layer sources ranging from factories to trees. crops. of ozone in the upper atmosphere (stratosphere), which screens out harmful ultraviolet rays Ca rbo n monoxide. A gas that comes from the burning of fossil fuels, mostly in cars. It cannot be seen or smelled

Carbon mono xide is released when engines Carbo n monoxide makes it hard for body burn fossil fuels. Emissions are higher when parts to get the oxygen they need to run engines are not tuned properly, and when fuel co rrectly. Exposure to carbon monoxid e is not completely burned. Cars emit a lot of the makes people feel dizzy and tired and carbon monoxide found outdoors. Furnaces gives them headaches. In high and heaters in the home can emit high co ncentrations it is fatal. Elderly people concentrations of carbo n monoxide, too, if they with heart disease are hospitalized more are not properly maintained. often when they are exposed to higher amounts of carbon monoxide.

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BASICS OF ENVIRONMENT AND POLLUTION CONTROL

Nitrogen dio xide.  A reddish- Nitrogen dioxide mostly comes from po wer High levels of nitrogen dioxide exposure brown gas that comes from the plants and cars. Nitrogen dioxide is formed in can give people coughs and can make burning of fossil fuels. It has a two ways-when nitrogen in the fuel is burned, them feel short of breath. People who are strong smell at high levels. or when nitrogen in the air reacts with oxygen exposed to nitrogen dioxide for a long at very high temperatures. Nitrogen dioxide time have a higher chance of getting can also react in the atmosphere to form ozone, respiratory infections. Nitro gen dioxid e acid rain, and particles. reacts in the atmosphere to form acid rain, which can harm plants and animals. Particulate matter. Solid or Particulate matter can be divided into two Particulate matter that is small enough liquid matter that is suspended in types-coarse particles and fine particles. can enter the lungs and cause health the air. To remain in the air, Coarse particles are formed from sources like prob lems. Some of these problems particles usually must be less than road dust, sea spray, and construction. Fine include more frequent asthma attacks, 0.1-mm wide and can be as small particles are formed when fuel is burned in respiratory problems, and premature automobiles and power plants. death. as 0.00005 mm. Sulphur dio xide.  A corrosive gas Sulfur dioxide mostly comes from the burning Sulfur d ioxide exposure can affect that cannot be seen or smelled at of coal or oil in power plants. It also comes people who have asthma or emphysema low levels but can have a “rotten from factories that make chemicals, paper, or by making it more difficult for them to egg“ smell at high levels. fuel. Like nitrogen dio xide, sulfur dioxide breathe. It can also irritate people's eyes, reacts in the atmosp here to form acid rain and noses, and throats. Sulfur dioxide can particles. harm trees and crops, damage b uildings, and make it harder for people to see long distances. Lead. A blue-gray metal that is Outside, lead comes from cars in areas where High amounts of lead can be dangerous very toxic and is found in a unleaded gasoline is not used. Lead can also for small children and can lead to lower number of forms and locations. come from power plants and other industrial IQs and kidney prob lems. For adults, sources. Inside, lead paint is an important exposure to lead can increase the chance source of lead, especially in houses where paint of having heart attacks or strokes. is peeling. Lead in old pipes can also be a source of lead in drinking water. Toxic air pollutants. A large number of chemicals that are known or suspected to cause cancer. Some imp ortant pollutants in this category include arsenic, asbestos, benzene, and dioxin.

Each toxic air pollutant comes from a slightly Toxic air pollutants can cause cancer. different source, but many are created in Some toxic air pollutants can also cause chemical plants or are emitted when fossil fuels birth defects. Other effects depend on the are burned. Some toxic air pollutants, like pollutant, but can include skin and eye asbestos and formaldehyde, can be found in irritation and breathing problems. building materials and can lead to indoor air problems. Many toxic air pollutants can also enter the food and water supplies.

Stratospheric ozone CFCs are used in air conditioners and If the ozone in the stratosphere is depleters.Chemicals that can refrigerators, since they work well as coolants. destro yed, people are exposed to more destroy the ozone in the They can also be fo und in aerosol cans and fire radiation fro m the sun (ultraviolet stratosphere. These chemicals extinguishers. Other stratospheric ozone radiation). This can lead to skin cancer include chlorofluorocarbons depleters are used as so lvents in industry. and eye problems. Higher ultraviolet (CFCs), halons, and other radiation can also harm plants and compounds that include chlorine animals. or bromine

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

Greenhouse gases. Gases that stay in the air for a long time and warm up the planet by trapping sunlight. This is called the “greenhouse effect“ b ecause the gases act like the glass in a greenhouse. Some of the important greenhouse gases are carbon dioxide, methane, and nitrous oxide.

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Carbon dioxide is the most important The greenhouse effect can lead to greenhouse gas. It comes from the burning of changes in the climate of the planet. fossil fuels in cars, p ower plants, houses, and Some of these changes might includ e industry. Methane is released during the more temperature extremes, higher sea processing of fossil fuels, and also comes from levels, changes in forest compositio n, natural sources like cows and rice paddies. and damage to land near the coast. Nitrous oxide comes from ind ustrial sources Human health might be affected by and decaying plants. diseases that are related to temperature or by damage to land and water.

WATER POLLUTION Water pollution involves any contaminated water, whether from chemical, particulate, or bacterial matter that degrades the water’s quality and purity. Water pollution can occur in oceans, rivers, lakes, and underground reservoirs, and as different water sources flow together through the water cycle the pollution can spread. Causes of water pollution include:  Increased sediment from soil erosion  Improper waste disposal and littering  Leaching of soil pollution into water supplies  Organic material decay in water supplies The effects of water pollution include decreasing the quantity of drinkable water available, lowering water supplies for crop irrigation, and impacting fish and wildlife populations that require water of a certain purity for survival. Ground water is being polluted by percolation of contaminated surface water through the layers of the earth. Release of raw sewage in unlined soak-pits and release of toxic effluents by the industries into surface water bodies, are the main causes of ground water pollution.

Major water pollutants, examples and sources Category Sources Examples I. Affecting health Infectious agents Bacteria, viruses and parasites Sewage, human and animal excreta Organic chemicals Pesticides, plastics, detergents. oil Agricultural, industrial and domestic wastes Inorganic chemicals Acids, caustics, salts, metals Industrial and domestic effluents Radioactive materials Uranium, thorium, randon, etc Mining, power plants, natural sources 2. Affecting ecossslcm Plant nutrients Nitrates, phosphates, etc Chemical fertilisers, sewage, manure Sediments Silt, soil Soil erosion Thermal Heat Industries, power plants Oxygen demanding Agricultural waster, manure Sweage, agricultural runoff Indiscriminate and overuse of fertilizers, chemicals and pesticides have also caused ground water pollution through the seepage of irrigation water into ground water reserves. The hazards of ground water pollution depend on several factors such as:  Concentration or toxicity of the pollutant  The level of ground water if the level is higher chances of contamination are more  Conditions of ground water recharge

Marine Pollution: Marine pollution is the matter of International concern from the point of view of conservation of living resources. All coastal nations dispose of millions of gallons of untreated sewage, millions of tonnes of garbage, unlimited amount of low level radioactive wastes etc. into the seas. In addition to the marine environment, areas along the coasts, such as, estuaries, reefs, wetlands, mangroves, etc. are adversely affected due to enormous dumping of pollutants into the ocean. This problem is further aggravated due to the fact that about 40% of the world’s population lives near the sea.

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BASICS OF ENVIRONMENT AND POLLUTION CONTROL

The main sources of marine pollution are: 1. Municipal wastes and sewage 2. Industrial effluents 3. Runoff agricultural wastes 4. Oil spills from tankers 5. Offshore drilling and mining 6. Submarine nuclear testing 7. Dumping of radioactive wastes

The consequences of marine pollution are as follows: 

The pollutants adversely affect the productive ocean regions, thus causing huge losses of fish populations and coral reefs. This results in economic losses amounting to billions of dollars per year.



Eutrophication, due to the influx of organic pollutants, results in the formation of red tides. These are blooms (massive growth) of red algae, which inhibit the movement of ships and also kill marine fauna.



Dumping of huge amounts of toxic wastes in a short duration of time, creates areas of oxygen-depleted zones in the coastal waters. In these zones, most of the aquatic lives die or migrate elsewhere.



Discarded garbage, sewage, plastic refuse, etc. that are dumped in the oceans sometimes accumulate in the beaches. This spoils the aesthetic beauty of the region and results in loss of tourism.

Water Pollutants and their Effects: Most of the rivers and fresh water streams in India are badly polluted by industrial wastes or effluents. The major sources of pollution of some Indian rivers are listed in table below:

Name of river 1. Kali 2. Yamuna 3. Ganga 4. Gomti 5. Dajora 6. Damodar 7. Hoogly

8. Sone Bhadra 9. Cooum, Adyar and Buckinghum canal (Chennai) 10. Kaveri 11. Godavari 12. Siwan 13. Kulu 14. Suwao

Indian rivers and sources of their pollution Sources of pollution Sugar mills: distilleries: paint, soap, rayon, silk. Yarn, tin and glycerine D.D.T. factory, sewage, Indraprastha Power Station, Delhi. Jute, chemical, metal and surgical industries: tanneries, textile mills and great bulk of domestic sewage of highly organic nature. Paper and pulp mills sewage. Synthetic rubber factories. Fertilizers, fly ash from steel mills, suspended coal particles from washeries, and thermal power station. Power stations: paper pulp, jute textiles, chemical mills, paint, varnishes, metal, steel, hydrogenated vegetable oil, rayon, soap, match, shellac, and polyethene industries and sewage. Cement, pulp and paper mills. Domestic sewage, automobile workshops. Sewage, tanneries, distilleries, paper and rayon mills. Paper, mills. Paper, sulphur, cement, and sugar mills. Chemical factories, rayon mills and tanneries. Sugar industries.

Contamination of water with industrial wastes is most dangerous. The sewage of big cities is often drained into rivers. This sewage promotes the growth of phytoplankton’s. The excessive growth depletes the oxygen of water. This reduction of oxygen and the presence of poisonous wastes affect the fish population. Besides these, rivers, lakes and ponds are also used directly by people for bathing or washing. This contaminates the water with the germs of various diseases- like cholera, dysentery and hepatitis.

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

Type or Industry Inorganic pollutants Mining Mine wastes : Chlorides, various metals, ferrous sulphate, sulpluric acid hydrogen sulphide, ferric hydroxide, surface wash offs, suspended solids, chlorides and heavy metals. Iron and steel Suspended solids, iron cyanide, thiocy-anate, sulphides, oxides, of copper, chromium, cadmium, and mercury. Chemical plants Various acids and alkalies, sulphates, nitrates of metals, phosphorus, fluorine, silica, and suspended particles. Pharmaceuticals —

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Organic pollutants — Oil, phenol and naptha

Aromatic compound solvents, organic acids, nitro compound dyes, etc.

Proteins, carbohydrates, organic solvents, intermediate products, drugs and antibiotics. Soap and detergent Tertiary ammonia compounds, alkalies. Fats and fatty acids, glycerol, polyphosphates, sulphonated hydrocarbons. Food processing Highly putrescible organic matter and — pathogens. Paper and pulp Sulphides, bleaching liquors Cellulose fibres, bark, wood, sugars and organic acids. Some pollutants produce only temporary effects in water whereas others have long standing effects. There are several types of physical and chemical effects produced by pollutants. These are:  Addition of poisonous substances  Addition of suspended particles  Addition of non-toxic salts  Water de-oxygenation  Heating of water.

SOIL POLLUTION Soil, or land pollution, is contamination of the soil that prevents natural growth and balance in the land whether it is used for cultivation, habitation, or a wildlife preserve. Some soil pollution, such as the creation of landfills, is deliberate, while much more is accidental and can have widespread effects. Soil pollution sources include:  Hazardous waste and sewage spills  Non-sustainable farming practices, such as the heavy use of inorganic pesticides  Strip mining, deforestation, and other destructive practices  Household dumping and littering Soil contamination can lead to poor growth and reduced crop yields, loss of wildlife habitat, water and visual pollution, soil erosion, and desertification.

NOISE POLLUTION Noise pollution refers to undesirable levels of noises caused by human activity that disrupt the standard of living in the affected area. Noise pollution can come from:  Traffic  Airports  Railroads  Manufacturing plants  Construction or demolition  Concerts

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BASICS OF ENVIRONMENT AND POLLUTION CONTROL

Some noise pollution may be temporary while other sources are more permanent. Effects may include hearing loss, wildlife disturbances, and a general degradation of lifestyle.

RADIOACTIVE POLLUTION Radioactive pollution is rare but extremely detrimental, and even deadly, when it occurs. Because of its intensity and the difficulty of reversing damage, there are strict government regulations to control radioactive pollution. Sources of radioactive contamination include: 

Nuclear power plant accidents or leakage



Improper nuclear waste disposal



Uranium mining operations

Radiation pollution can cause birth defects, cancer, sterilization, and other health problems for human and wildlife populations. It can also sterilize the soil and contribute to water and air pollution.

THERMAL POLLUTION Thermal pollution is excess heat that creates undesirable effects over long periods of time. The earth has a natural thermal cycle, but excessive temperature increases can be considered a rare type of pollution with long term effects. Many types of thermal pollution are confined to areas near their source, but multiple sources can have wider impacts over a greater geographic area. Thermal pollution may be caused by: 

Power plants



Urban sprawl



Air pollution particulates that trap heat



Deforestation



Loss of temperature moderating water supplies

As temperatures increase, mild climatic changes may be observed, and wildlife populations may be unable to recover from swift changes.

LIGHT POLLUTION Light pollution is the over illumination of an area that is considered obtrusive. Sources include: 

Large cities



Billboards and advertising



Night time sporting events and other night time entertainment

Light pollution makes it impossible to see stars, therefore interfering with astronomical observation and personal enjoyment. If it is near residential areas, light pollution can also degrade the quality of life for residents.

ENVIRONMENTAL PERFORMANCE INDEX 2018 In news 2018: The 2018 Environmental Performance Index (EPI) finds that air quality is the leading environmental threat to public health. Now in its twentieth year, the biennial report is produced by researchers at Yale and Columbia Universities in collaboration with the World Economic Forum. The tenth EPI report ranks 180 countries on 24 performance indicators across 10 issue categories covering environmental health and ecosystem vitality. Switzerland leads the world in sustainability, followed by France, Denmark, Malta, and Sweden.

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

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Key findings: 

Switzerland’s top ranking reflects strong performance across most issues, especially air quality and climate protection.



In general, high scorers exhibit long-standing commitments to protecting public health, preserving natural resources, and decoupling greenhouse gas (GHG) emissions from economic activity.



India and Bangladesh come in near the bottom of the rankings, with Burundi, Democratic Republic of the Congo, and Nepal rounding out the bottom five.



Low scores on the EPI are indicative of the need for national sustainability efforts on a number of fronts, especially cleaning up air quality, protecting biodiversity, and reducing GHG emissions.



Some of the lowest-ranking nations face broader challenges, such as civil unrest, but the low scores for others can be attributed to weak governance, they note.

EPI and Global Sustainability Data The EPI builds on the best available global data from international research entities, such as the Institute for Health Metrics and Evaluation, the World Resources Institute, and the Sea Around Us Project at the University of British Columbia, as well as international organizations such as the World Bank and the UN Food and Agriculture Organization. Nevertheless, serious data gaps limit the ability to measure results – and particularly changes in performance – on a number of important issues. “As the EPI project has highlighted for two decades, better data collection, reporting, and verification across a range of environmental issues are urgently needed,”. The world needs better data on sustainable agriculture, water resources, waste management, and threats to biodiversity. Supporting global data systems is one of the most important steps the world community can take to achieving sustainable development goals.

DUST MITIGATION PLAN Centre had notified dust mitigation norms.

The norms mandate that: 

No building or infrastructure project requiring Environmental Clearance shall be implemented without approved Environmental Management Plan inclusive of dust mitigation measures.



Roads leading to or at construction sites must be paved and blacktopped (i.e. metallic roads).



No excavation of soil shall be carried out without adequate dust mitigation measures in place.

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No loose soil or sand or Construction & Demolition Waste or any other construction material that causes dust shall be left uncovered,

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Wind-breaker of appropriate height i.e. 1/3rd of the building height and maximum up to 10 meters shall be provided.



Water sprinkling system shall be put in place.



Dust mitigation measures shall be displayed prominently at the construction site for easy public viewing.

How it works? (Steps taken) The teams are empowered to take on-the-spot action against violators and if necessary, issue “stop-work” orders. The campaign will also include enforcement of pollution-control measures for vehicles, driving discipline, inspection of power plants in Delhi to ensure compliance with the norms on pollution. Besides field surveys by empowered teams of officials, a series of seminars on mitigation of pollution will also be organised during the period. These include - a workshop on Environmental and Health; Air Pollution Abatement Technologies; enlisting support from NGOs, Civil Society, citizens; Clean Air Day in Universities, Colleges and Schools; a Mini Marathon for Clean Air; enhancing the role of PSUs and industries, apex industrial bodies; launching a national digital forum for discussions on air pollution; Indoor Air Pollution Management and a conference of Environment Ministers of States and Union Territories.

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BASICS OF ENVIRONMENT AND POLLUTION CONTROL

MINAMATA CONVENTION In news Recently, the Union Cabinet approved the proposal for ratification of Minamata Convention on Mercury enabling India to become a Party of the Convention.

About 

The approval entails ratification of the Minamata Convention on Mercury along with flexibility for continued use of mercury based products and processes involving mercury compound up to 2025.



The first Conference of the Parties (CoP) under the Minamata Convention took place in Geneva, Switzerland in 2017 which India attended as observer.

It is financed through Global Environment Facility.

Details about the convention 

The Minamata Convention on Mercury is first global legally binding treaty to protect human health and the environment from the adverse effects of mercury.



It was agreed in Geneva, Switzerland in January 2013 and came into force in August, 2017.

The Minamata Convention has put party nations to: 

Reduce and eliminate the use and release of mercury from artisanal and small-scale gold mining (ASGM).



Control mercury air emissions from coalfired power plants, coal-fired industrial boilers, certain non-ferrous metals production operations, waste incineration and cement production.

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Phase-out or take measures to reduce mercury use in certain products such as batteries, switches, lights, cosmetics, pesticides and measuring devices, and create initiatives to reduce the use of mercury in dental amalgam.

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Phase out or reduce the use of mercury in manufacturing processes such as chloralkali production, vinyl chloride monomer production, and acetaldehyde production.



It also puts a ban on new mercury mines.



The Convention also addresses interim storage of mercury and its disposal once it becomes waste, sites contaminated by mercury as well as health issues.

Waste management Waste management involves collecting, transporting, disposing, recycling and monitoring waste generated through human activities. General waste management techniques are: Landfill: It involves having the waste buried off in empty, deserted locations outside the city. Dumped waste is made to undergo compression to enhance the density and make the fill stable. It is later covered to discourage vermin growth. A gas extraction system is customarily installed to exact the gas (arising out of decomposition) through a burrow pit. Incineration: Waste is exposed to high temperature to trigger combustion and ultimately reduce to ash, gas and heat energy. Toxic wastes from industry are thermally treated in furnace and boiler to extract energy. This method is useful where land is scarce. Gasification and Pyrolysis methods involve heating waste in short supply of oxygen at high temperature inside a pressurized and sealed vessel. The resultant residue is used for energy generation. Recycling: Paper, plastic, PVC and other homogenous products can be recycled to put them in use in a new garb. This also rids the environment of non-biodegradable, chemical wastes that significantly disturb the ecological balance. Biological reprocessing: Wastes of organic origin are made to undergo biological decomposition and re-used as compost or mulch for agriculture and landscaping. Gas collected is used for electricity generation. Waste Reduction and Avoidance: The stress is on increased use of second hand products, repaired products and reducing the use of complex disposable items to keep a tab on waste generation in abundance.

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

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Recycling Solutions: Recycling is a superlative way to capitalize on accumulated waste by chemically treating it to make it fit for re-use. Recycling equipment make the waste processing method streamlined and cost-effective.

Global warming Global warming which is also referred to as climate change, is the observed rise in the average temperature of the Earth's climate system the global surface temperature is likely to rise a further 0.3 to 1.7 °C in the lowest emissions scenario, and 2.6 to 4.8 °C in the highest emissions scenario .These readings have been recorded by the “national science academies of the major industrialized nations”. Future climate change and impacts will differ from region to region. Expected effects include increase in global temperatures, rising sea levels, changing precipitation, and expansion of deserts. Causes: Global warming is a serious environmental issues. The causes are divided into two categories include "natural" and "human influences" of global warming. Natural Causes of Global Warming:  rotation of the sun that changes the intensity of sunlight and moving closer to the earth  greenhouse gases  Volcanic eruption. Human Influences on Global Warming:  industrial revolution  Mining  Deforestation Effects:  heat waves,  droughts,  heavy rainfall with floods,  heavy snowfall ,  ocean acidification,  species extinctions due to shifting temperature regimes

Acid rain Acid rain, or acid deposition, is a broad term that includes any form of precipitation with acidic components, such as sulfuric or nitric acid that fall to the ground from the atmosphere in wet or dry forms. This can include rain, snow, fog, hail or even dust that is acidic. Causes of Acid Rain : This image illustrates the pathway for acid rain in our environment.Acid rain results when sulfur dioxide (SO2) and nitrogen oxides (NOX) are emitted into the atmosphere and transported by wind and air currents. The SO2 and NOX react with water, oxygen and other chemicals to form sulfuric and nitric acids. These then mix with water and other materials before falling to the ground. While a small portion of the SO2 and NOX that cause acid rain is from natural sources such as volcanoes, most of it comes from the burning of fossil fuels. The major sources of SO2 and NOX in the atmosphere are:  Burning of fossil fuels to generate electricity. Two thirds of SO2 and one fourth of NOX in the atmosphere come from electric power generators.  Vehicles and heavy equipment.  Manufacturing, oil refineries and other industries. Winds can blow SO2 and NOX over long distances and across borders making acid rain a problem for everyone and not just those who live close to these sources.

Ozone depletion Ozone depletion, gradual thinning of Earth’s ozone layer in the upper atmosphere caused by the release of chemical compounds containing gaseous chlorine or bromine from industry and other human activities. The thinning is most pronounced in the polar regions, especially over Antarctica. Ozone depletion is a major environmental problem because it increases the amount of ultraviolet (UV) radiation that reaches Earth’s surface, which increases the rate of

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BASICS OF ENVIRONMENT AND POLLUTION CONTROL

skin cancer, eye cataracts, and genetic and immune system damage. The Montreal Protocol, ratified in 1987, was the first of several comprehensive international agreements enacted to halt the production and use of ozone-depleting chemicals. As a result of continued international cooperation on this issue, the ozone layer is expected to recover over time.

Important terminologies: ·

Garbage Pollution: Mismanagement of solid waste by households, waste collectors and waste disposal contractors.

·

Plastic Pollution: Waste of all types of non-biodegradable plastic of both hard and soft material.

· Pollution by Hospitals: Mismanagement of all types of waste generated by the hospitals instead of its environment friendly disposal. ·

Indoor Pollution: Kitchen emissions, smoking in home, loud music, spillage of sewerage.

·

Industrial Pollution: Smoke from chimney, waste and effluent from manufacturing process in factories.

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

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PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. Consider the following:

2.

3.

4.

5.

1. Carbon dioxide 2. Oxides of Nitrogen 3. Oxides of Sulphur Which of the above is/are the emission/ emissions from coal combustion at thermal power plants? (a) 1 only (b) 2 and 3 only (c) 1 and 3 only (d) 1, 2 and 3 Human activities in the recent past have Caused the increased concentration of carbon dioxide in the atmosphere, but a lot of it does not remain in the lower atmosphere because of : 1. Its escape into the outer stratosphere. 2. The photosynthesis by phyto-plankton in the oceans. 3. The trapping of air in the polar ice caps. Which of the statements given above is/are correct? (a) 1 and 2 (b) 2 only (c) 2 and 3 (d) 3 only Consider the following statements: Chlorofluorocarbons, known as ozone-depleting substances, are used 1. In the production of plastic foams 2. In the production of tubeless tyres 3. In cleaning certain electronic components 4. As pressurizing agents in aerosol cans Which of the statements given above is/are correct? (a) 1, 2 and 3 only (b) 4 only (c) 1, 3 and 4 only (d) 1, 2, 3 and 4 Acid rain is caused by the pollution of environment by (a) carbon dioxide and nitrogen (b) carbon monoxide and carbon dioxide (c) ozone and carbon dioxide (d) nitrous oxide and sulphur dioxide Photochemical smog is a resultant of the reaction among: (a) NO2, 03 and peroxyacetyl nitrate in the presence of sunlight (b) CO, 02 and0 peroxyacetyl nitrate in the presence of sunlight (c) CO, CO2 and N02 at low temperature (d) High concentration of N02, O3 and CO in the evening

6. There is some concern regarding the nanoparticles of some chemical elements that are used by the industry in the manufacture of various products. Why?

1. They can accumulate in the environment, and contaminate water and soil. 2. They can enter the food chains. 3. They can trigger the production of free radicals. Select the correct answer using the code given below. (a) 1 and 2 only (b) 3 only (c) 1 and 3 only (d) 1, 2 and 3 7. Which of the following are some important pollutants released by steel industry in India? 1. Oxides of sulphur 2. Oxides of nitrogen 3. Carbon monoxide 4. Carbon dioxide Select the correct answer using the code given below. (a) 1, 3 and 4 only (b) 2 and 3 only (c) 1 and 4 only (d) 1,2, 3 and 4 8. Brominated flame retardants are used in many household products like mattresses and upholstery. Why is there some concern about their use? 1. They are highly resistant to degradation in the environment. 2. They are able to accumulate in humans and animals. Select the correct answer using the code given below. (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 9. The scientific view is that the increase in global temperature should not exceed 2 °C above preindustrial level. If the global temperature increases beyond 3°C above the pre-industrial level, what can be its possible impact/impacts on the world? 1. Terrestrial biosphere tends toward a net carbon source 2. Widespread coral mortality will occur. 3. All the global wetlands will permanently disappear. 4. Cultivation of cereals will not be possible anywhere in the world. Select the correct answer using the code given below. (a) 1 only (b) 1 and 2 only (c) 2, 3 and 4 only (d) 1, 2, 3 and 4,

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BASICS OF ENVIRONMENT AND POLLUTION CONTROL

10. In the context of mitigating the impending global warming due to anthropogenic emissions of carbon dioxide, which of the following can be the potential sites for carbon sequestration?

(c) M axi mum acid i s due t o st r ong Car bonic Acid (d) Acid r ain affect s ecosyst em [RRB SSE 2014 YEL L OW SH I FT ]

15. The wor king pr inciple of tur bidimeter is based on

1. Abandoned and uneconomic coal seams

(a) r efl ect ion of l ight

2. Depleted oil and gas reservoirs

(b) r efr act ion of l ight

3. Subterranean deep saline formations

(c) scat t er ing of l ight (d) adsor pt ion of light [RRB SSE 2015 1 st SEP 1 st SH I FT ]

Select the correct answer using the code given below. (a) 1 and 2 only

(b) 3 only

(c) 1 and 3 only

(d) 1, 2 and 3

11. Biological Oxygen Demand (BOD) is a standard criterion for

LEVEL-1 1. The major sour ce of car ci nogenic hydr ocar bon, benzo () pyr ene pr esent i n ur ban at mospher e is (a) const r uct ion act ivit i es (b) r oad t r affic

(a) Measuring oxygen levels in blood

(c) bur st i ng of cr acker s

(b) Computing oxygen levels in forest ecosystems

(d) domest i c bur ni ng

(c) Pollution assay in aquatic ecosystems (d) Assessing oxygen levels in high altitude regions 12. In the context of solving pollution problems, what is/ are the advantage/advantages of bioremediation technique?

1. It is a technique for cleaning up pollution by enhancing the same biodegradation process that occurs in nature. 2. Any contaminant with heavy metals such as cadmium and lead can be readily and completely treated by bioremediation using microorganisms. 3. Genetic engineering can be used to create microorganisms specifically designed for bioremediation. Select the correct answer using the code given below:

[RRB SSE 2015 1 st SEP 1 st SH I FT ]

2. The pr escr i bed per missible noise level , L eq for commer cial ar ea at day t ime is (a) 75 dBA

(b) 50 dBA

(c) 55 dBA

(d) 65 dBA [RRB SSE 2015 1 st SEP 1 st SH I FT ]

3. The gl obal war mi ng i s caused by gr een house gases, which ar e (a) CO, N 2O, CH 4 and CFC (b) CO2, NO2, CH 4 and H 2O (c) CO2, N 2O, CH 4 and H 2O (d) CO2, N O2, CH 4 and CFC [RRB SSE 2015 1 st SEP 1 st SH I FT ]

4. Which of t he fol lowing r ol es fl y ash does not pl ay in concr et e (a) I mpr ovi ng t he wor kabi li t y (b) Acceler at ing t he st r engt h gain

(a) 1 only

(b) 2 and 3 only

(c) Delaying t he set t ing t i me of concr et e

(c) 1 and 3 only

(d) 1, 2 and 3

(d) H elps in long-t er m st r engt h gain

13. I n potable water, the dissolved oxygen is stipulated as-

[RRB SSE 2015 1 st SEP 2 nd SH I FT ]

5. One t ur bi di t y unit NTU is equal t o

(a) <6g/l

(b) >6g/l

(a) 1.0 mg/l far mazin (b) 1.0 meq/l Si O2

(c) <6mg/l

(d) >6mg/l

(c) 1.0 mg/l Si O2

(d) 1.0 meq/l kaol in

[RRB SSE 2014 GREEN SH I FT ]

[RRB SSE 2015 1 st SEP 2 nd SH I FT ]

14. I n r ef er en ce t o A ci d r ai n , w h at i s cor r ect st at ement

6. The pr escr i bed per missible noise level , L eq for r esi dent i al ar ea at day t ime is

(a) The pH value i s bel ow 5.6 (b) I t occur s due t o pr esence of sul phur i c aci d or ni t r ic acid i n t he at mospher e

(a) 65 dBA

(b) 45 dBA

(c) 50 dBA

(d) 55 dBA [RRB SSE 2015 1 st SEP 2 nd SH I FT ]

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

7. W h i ch of t h e f ol l ow i n g i s n ot u sed as a suppl ement ar y cement at ions mat er ial? (a) Fl y ash

(b) Gypsum

(c) Ri ce husk ash

(d) Si li ca fume

8. Accor di ng t o I S 456, if t he maximum aggr egat e si ze i s i ncr eased fr om 20 mm t o 40 mm, t he mi ni mum cement cont ent r equi r ement changes (i n kg/cum) by (b) 20

(c) – 30

(d) 30

(c) gast r oint est inal pr oblem (d) t he ir r it at i on in alveol i of t he l ungs [RRB SSE 2015 2 nd SEP 2 nd SH I FT ]

[RRB SSE 2015 1 st SEP 3 rd SH I FT ]

(a) – 20

13

15. The machi ne ‘A' and machine ‘B' pr oduce equal noise l evel s, i.e., 60 dBA each. The summat ion of t hese t wo noise level s is (a) 100 dBA

(b) 66 dBA

(c) 63 dBA

(d) 55 dBA [RRB SSE 2015 2 nd SEP 2 nd SH I FT ]

LEVEL-2

[RRB SSE 2015 1 st SEP 3 rd SH I FT ]

9. The t ur bi di t y in sur face wat er is due t o pr esence of (a) dissolved or ganics

1. I n st r at ospher e, t he t emper at ur e incr eases wi t h al t i t ude due t o pr esence of (a) r adicals

(b) chlor ofluor ocar bons

(c) HCFCs

(d) Ozone

(b) col loidal mat er ial

[RRB SSE 2015 2 nd SEP 2 nd SH I FT ]

(c) di ssolved i n or gani cs

2. N uclear densi t y guage can be used for al l t he foll owi ng pur poses, except

(d) di ssolved color s [RRB SSE 2015 1 st SEP 3 rd SH I FT ]

10. T h e pr escr i bed per m i ssi bl e n oi se l evel , f or r esi dent i al ar ea at night t ime is (a) 45 dBA

(b) 50 dBA

(c) 40 dBA

(d) 55 dBA

(a) M oist ur e cont ent (b) Wet densi t y (c) Dr y densi t y (d) St andar d penetr at ion r eading

[RRB SSE 2015 1 st SEP 3 rd SH I FT ]

11. The cont i nuous exposur e of hi gh concent r at i on of r epar able suspended par t icul at e mat t er may cause

[RRB SSE 2015 2 nd SEP 3 rd SH I FT ]

3. A wat er bor ne di sease pol iomyeli t is i s caused by (a) vir uses

(b) pr ot ozoa

(c) bact er ia

(d) hel mi nt hes

(a) eye ir r it at i on

[RRB SSE 2015 2 nd SEP 3 rd SH I FT ]

(b) kidney damage

4. I n pot abl e wat er, t he per missible li mi t of nit r at e ni t r ogen is

(c) fail ur e of r espi r at or y syst em (d) car diac disease [RRB SSE 2015 2

nd

SEP 1 SH I FT ] st

12. The pr escr i bed per missible noise level , L eq for commer cial ar ea at ni ght t ime is (a) 45 dBA

(b) 65 dBA

(c) 50 dBA

(d) 55 dBA [RRB SSE 2015 2 nd SEP 1 st SH I FT ]

13. The pH of acid r ai n shoul d always be less t han (a) 5.6 even aft er pr ecipit at i on (b) 7.0 aft er pr ecipit at ion (c) 6.5 aft er pr ecipit at ion (d) 4.2 aft er pr ecipit at ion [RRB SSE 2015 2 nd SEP 1 st SH I FT ]

14. The exposur e of gaseous pollut ant sulphur dioxide may cause (a) br onchi t is and pul monar y emphysema (b) lungs fai lur e and k idney damage

(a) 10 mg/l

(b) 25 mg/l

(c) 40 mg/l

(d) 15 mg/l [RRB SSE 2015 2 nd SEP 3 rd SH I FT ]

5. Car bon monoxi de for ms car boxyhemogl obi n i n human blood t hat may cause (a) incr eased oxygen car r yi ng capaci t y (b) decr eased oxygen car r ying capaci t y (c) damage in cent r al ner vous syst em (d) damage in ci r cul at or y syst em [RRB SSE 2015 2 nd SEP 3 rd SH I FT ]

6. Tw o m ach i n es ar e w or k i n g i n a n oi sy envi r onment and joint ly pr oduct 55 dBA noi se level. I f t he envir onment al noi se level i s also 55 dBA, t he summat i on of noise level s is (a) 110 dBA

(b) 56 dBA

(c) 55 dBA

(d) 58 dBA [RRB SSE 2015 2 nd SEP 3 rd SH I FT ]

14

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

(b) car bon dioxide, sulphur dioxide, water vapour s and chl or ofl uor ocar bons

7. The aver age concent r at i on of ozone pr esent i n t he st r at ospher e is appr oximat ely (a) 5 ppm

(b) 0.05 ppm

(c) 10 ppm

(d) 15 ppm

(c) car bon monoxide, nit r ous oxide, met hane and, hydr o-chlor ofluor ocar bons (d) car bon di ox i de, n i t r ogen di ox i de, w at er vapour s , met hane and ozone

[RRB SSE 2015 2 nd SEP 3 rd SH I FT ]

8. The t ot al col ifor m bact er ia ar e r epor t ed as most pr obabl e number (M PN) per (a) 10 ml of wat er (b) 1000 ml of wat er

[RRB SSE 2015 3 rd SEP 1 st SH I FT ]

12. The desir able amount of fluor i de ions i n pot able wat er s for opt imal dent al heal t h i s: (a) 1.5 mg/l

(c) 100 ml of wat er

(b) 1.0 mg/l

(c) 0.5 mg/l

(d) 1ml of wat er

(d) 0.05 mg/l [RRB SSE 2015 3 rd SEP 2 nd SH I FT ]

[RRB SSE 2015 3 rd SEP 1 st SH I FT ]

9. The ant hr opogenic sour ces of air pollut ion in well pl anned ci t y is

13. Wh i ch of t he f ol l owi ng i s n ot con si der ed as secondar y pollut ant ? (a) Phot ochemical smog

(a) constr uction activities, r oad tr affics, r ail tr affic, fugi t ive emissions

(b) Per oxy acet yl nit r at e (c) Acid mi st

(b) const r uct ion act ivi t i es, r oad t r affi c, domest ic burning (c) const r uct ion act ivit ies, r oad t r affi cs, bur st ing of cr acker s, dust st or ms (d) const r uct ion act ivit ies, r oad t r affics, domest ic bur ni ng, i ndust r ial emissions

(d) Car bon monoxi de [RRB SSE 2015 3 rd SEP 2 nd SH I FT ]

14. I n t he st at ist ical di st r ibut ion of noi se levels, t he back gr ound noise l evel is r epr esent ed by:

[RRB SSE 2015 3 rd SEP 1 st SH I FT ]

10. When t he measur ed and st andar d r efer ence pr essur e level becomes equal, t he sound pr essur e level (SPL ) is equi valent t o (a) 1 dBA

(b) 10 dBA

(c) 0 dBA

(d) 1.012 dBA

(a) L 90

(b) L 50

(c) L 10

(d) L 1 [RRB SSE 2015 3 rd SEP 2 nd SH I FT ]

15. Acid r ain i s caused due t o for mat ion of: (a) su l ph u r i c aci d an d car bon i c aci d i n t h e at mospher e (b) sulphur ic acid and nitric acid in the atmosphere

SEP 1 SH I FT ]

(c) nitr ic acid and car bonic acid in the atmospher e

11. The major gr een house gases cont r i but i ng i n gl obal war ming ar e

(d) sulphur ic acid, nit r ic acid and car bonic acid in t he at mospher e

(a) car bon di oxide, nit r ous oxide, met hane and wat er vapour s

[RRB SSE 2015 3 rd SEP 2 nd SH I FT ]

[RRB SSE 2015 3

rd

st

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (d) 11. (c)

2. (c) 12. (d)

3. (c) 13. (a)

4. (d) 14. (b)

5. (a) 15. (c)

6. (d)

7. (d)

8. (c)

9. (b)

10. (d)

7. (b)

8. (c)

9. (b)

10. (a)

7. (c)

8. (c)

9. (b)

10. (c)

LEVEL-1 1. (b) 11. (c)

2. (d) 12. (d)

3. (c) 13. (a)

4. (b) 14. (a)

5. (c) 15. (c)

6. (d)

LEVEL-2 1. (d) 11. (a)

2. (d) 12. (b)

3. (a) 13. (d)

4. (a) 14. (a)

5. (b) 15. (b)

6. (d)

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

15

EXPLAN ATI ON S LEVEL-1 1. Road traffic is the major source of carcinogenic hydrocarbon present in urban atmosphere. 2. The prescribed permissible noise level, Leq for commercial area at day time is 65 dBA. According to Central Pollution Control Board,

7. Gypsum is not used as a supplementary cementations material. 9. Turbidity in surface water is due to colloidal material. 10. The prescribed permissible noise level, Leq is 45 dBA. According to Central Pollution Control Board,

Area Limits in dB(A), Leq Category of Area Code Day time Night time A Industrial area 75 70 B Commercial area 65 55 C Residential area 55 45 D Silence Zone 50 40

3. Green house gases are the gas mixed in the atmosphere that absorbs the infrared radiation emitted by the earth’s surface.

Area Limits in dB(A), Leq Category of Area Code Day time Night time A Industrial area 75 70 B Commercial area 65 55 C Residential area 55 45 D Silence Zone 50 40

11. Continuous exposure to suspended particles are harmful to the lungs and can cause failure of respiratory system. 12. The prescribed permissible noise level, Leq for commercial area at night time is 55 dBA. According to Central Pollution Control Board, Area Limits in dB(A), Leq Category of Area Code Day time Night time A Industrial area 75 70 B Commercial area 65 55 C Residential area 55 45 D Silence Zone 50 40

We are not accustomed to these gases because neither nitrogen nor oxygen, the two most abundant gases of the atmosphere (78% and 21%, respectively), that many of us have heard of, have this ability to intercept infrared radiation. 4. Fly ash never improves strength of concrete. The advantages of using fly ash far outweigh the disadvantages. The most important benefit is reduced permeability to water and aggressive chemicals. Properly cured concrete made with fly ash creates a denser product because the size of the pores are reduced. 5. One turbidity unit NTU is equal to 1.0 mg/l SiO2. 6. The prescribed permissible noise level, Leq is 55 dBA. According to Central Pollution Control Board, Area Limits in dB(A), Leq Category of Area Code Day time Night time A Industrial area 75 70 B Commercial area 65 55 C Residential area 55 45 D Silence Zone 50 40

13. pH of acid rain should always be less than 5.6 even after precipitation. 14. The exposure of gaseous pollutant sulphur dioxide may cause bronchitis and pulmonary emphysema.

LEVEL-2 1. In the stratosphere, temperature increases with altitude. The reason is that the direct heat source for the stratosphere is the Sun. A layer of ozone molecules absorbs solar radiation, which heats the stratosphere. The amount of ozone present in the ozone layer is tiny, only a few molecules per million air molecules. 2. Nuclear density gauge can be used for all the following purposes except standard penetration reading. 3. A virus that may cause paralysis and is easily preventable by the polio vaccine. 4. The permissible limit of nitrate nitrogen in potable water is 10 mg/l . 7. The average concentration of ozone present in the stratosphere is approximately 10 ppm.

16

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

Sulfure dioxide (SO 2 carbon monoxide (CO 2 ),nitrogen oxides (No x ), and particulate matter (PM).

Photochemical oxidants (ozone, nitrogen dioxide, sulfur trioxide) and secondary particular matter.

Chemical reactants characterized with a direct pollution effect on living beings and ecosystems, and with an indirect effect through the formation of secondary pollutants.

Chemical products, highly reactive when photoactivation is involved in the chemical process of their formation

Complicated control Direct control through the process: understanding reduction of anthropogenic and interrupting the emissions. chemical reactions leading to their generation.

8. Total or fecal coliform bacteria are reported as most probable number per 100 mL 9. Anthropogenic sources of air pollution in well planned city are construction activities, road traffics, and domestic burning 10. When the measured and standard reference pressure level becomes equal, the sound pressure level (SPL) is equivalent to 0 dBA 11. The major green house gases contributing in global warming are carbon dioxide , nitrous oxide,methane and water vapor. 12. The desirable amount of flue le ions in potable waters for optimal dental health is 1.0 mg/l 13. Primary Pollutants Versus Secondary Pollutants Air pollutant formed in the atmosphere as a result of the chemical or physical Air pollutant emitted interactions between the direclty from a source into primary pollutants the atmosphere. themselves or between the primary pollutants and other atmospheric components.

14. L90 is frequently taken as the Lp of the background level. L10-L90 is often used to give a quantitative measure as to the spread or “how choppy” the sound was. L10 is the noise level exceeded for 10% of the time of the measurement duration. 15. Scientists have discovered that air pollution from burning of fossil fuels is the major cause of acid rain. The main chemicals in air pollution that create acid rain are sulfur dioxide (SO 2) and nitrogen (NOx). Acid rain usually forms high in the clouds where sulfur dioxide and nitrogen oxides react with water, oxygen, and oxidants. this mixture forms a mild solution of sulfuric acid and nitric acid. Sunlight increases the rate of most of these reactions. Rainwater, snow, fog, and other forms of precipitation containing those mild solutions of sulfuric and nitric acids fall to earth as acid rain.

Basics of Computers and Applications PERSON AL COM PU T ERS Per sonal comput er s can be cat egor ized by size and por t abilit y as : 1. Deskt op comput er s 2. L apt op or not ebooks 3. Per sonal Digit al Assist ants (PDAs) 4. Por t able comput er s 5. Tablet comput er s 6. Wear able comput er s U ses. Per sonal computers ar e normally operated by one user at a time to perform such general purpose tasks as word pr ocessing, I nter net br owsing, I nternet faxing, e-mail and other digit al messaging , multimedia playback, computer game play, computer pr ogr amming, etc. The user of modern personal computer may have significant knowledge of the operating environment and application pr ogr am s, bu t i s n ot n ecessar i l y i n t er est ed i n pr ogr amming not even able to wr ite pr ogr ams for the computer. Ther efore, most software written primarily for per sonal comput er s t ends t o be desi gned wi t h simplicity of use, or “user-friendliness” in mind. However, the software industry continuously provide a wide range of new products for use in personal computer s, tar geted at both the exper t and the non-expert user. COM PU T ER COM PON EN TS 1. Comput er case wit h power supply (usually sold t oget her ) 2. Mother boar d 3. Pr ocessor wit h fan (usually sold t oget her ) 4. At least one memor y car d 5. M ass st or age 6. K eyboar d and mouse for input 7. M onit or for out put The mot her boar d connect s ever t hing t oget her. The memory card(s), graphics card and processor are mounted directly onto the motherboard (the processor in a socket an the memory and graphics cards in an expansion slot). The mass storage is connected to it with cables. Same for keyboard and mouse, except that they are external and connect to the back plate. The monitor is also connected to the back plate, except not (usually) dir ectly to the motherboard, but to a connector in the graphics card.

M ass st orage. I t can be (i ) H ar d disk (ii ) Floppy dr ive or zip dr ive (bot h wit h media) (iii )Opt ical dr ive (CD or DVD) The oper at ing syst em (e.g., M icr osoft Windows, Linux or many ot her s) can be locat ed on eit her of t hese, but t ypically it son one of t he har d disks. Alive CD is also possible, but ver y slow and used for eit her inst allation of t he OS or pr oblem solving. A typical computer also has (i ) Sound car d (ii ) Net wor k car d (iii )M oder n and possibly r out er Common addit i ons, connect ed on t he out si de (per ipher als). Pr i nt er ; Scanner ; Webcam; Speak er s; M i cr ophone; Headset; Car d reader ; Gaming devices such as a joystick Sever al funct ions (implement ed by chipset s) can be integr ated into the mother boar d, such as typically USB and net wor k, but also gr aphics and sound. But even if these ar e pr esent, a separ ate car d can be added if what is available isn’t sufficient . The gr aphics and sound car d can have a br eak out box to keep the analog par ts away fr om the electr omagnetic r adiat ion inside t he comput er case. For r eally lar ge amount s of dat a, a t ape dr ive can be used or (ext r a) har d disks can be put t oget her in an ext er nal case. These component s can usually be put t oget her wit h lit t le k nowledge, t o build a comput er. I f somet hing shouldn’t go somewher e, it usually doesn’t fit (this used t o not always be t he case in t he past ) and if it does fit it can usually do lit t le har m. Most per sonal computer s ar e standar dized to the point t hat pur chased soft war e is expect ed t o r un wit h lit t le or no customizat ion for the par t icular comput er. M any PCs ar e also user -upgr adeable, especially desktop and wor kst at ion class comput er s. Devices such as main memor y, mass st or age, even t he mot her boar d and cent r al pr ocessing unit may be easily r eplaced by an end user. This upgr adeabilit y is, however, not idefinit e due t o r api d ch anges i n t h e per sonal compu t er indust r y, A PC that was consider ed t op-of-the-line five or six year s pr ior may be impr act ical t o upgr ade due t o changes in indust r y st andar ds. Such a comput er

2

Basics of Computers and Applications

usually must be t ot ally r eplaced once it ’s no longer suitable for its pur pose. This upgr ade and replacement cycle is par tially r elated to new r eleases of the pr imar y mass-mar ket oper at ed syst em, which t ends t o dr ive t he acquisition of new har dwar e and t ends of obsolet e pr ev i ou sl y ser v i ceabl e h ar dw ar e (see pl an n ed obsol escence). The har dwar e capabilit ies of per sonal comput er s can somet imes be ext ended by t he addit ion of expansion car ds connect ed via an expansion bus. Some standar d per ipheral buses often used for adding expansion cards in personal computers as of 2005 ar e PCI , AGP (a highspeed PCI bus dedicat ed t o gr aphics adapt er s), and PCI Expr ess. Most per sonal computer s as of 2005 have mul t i pl e physi cal PCI expansi on sl ot s. M any al so i ncl ude an AGP bus and expansi on sl ot or a PCI Expr ess bus and one or mor e explansion slot s, but few PCs cont ain bot h buses. M ot her boar d. The mother boar d (or mainboar d) is the pr imar y cir cuit boar d wi t hi n a per sonal com put er. M an y ot h er component s connect di r ect l y or i ndi r ect l y t o t he mot her boar d. M ot her boar ds usually cont ain one or mor e CPUs, suppor t ing cir cuit r y - usually int egr at ed cir cuits (I Cs) pr oviding the inter face between the CPU memor y and input /out put per ipher al cir cuit s, main memor y, and facilities for init ial setup of the computer immediat ely aft er being power ed on (often called boot fi r mwar e or, i n I BM PC compat i bl e comput er s, a BI OS). I n many por t abl e and embedded per sonal comput er s, t he mot her boar d houses near ly all of t he PC’s cor e component s. Oft en a mot her boar d will also cont ain one or mor e per ipher al buses and physical connect or s for expansi on pur poses. Somet i mes a secondar y daught er boar d i s connect ed wi t h t he mot her boar d t o pr ovide fur t her expandibilit y or t o sat isfy space constr aint s. M ain M emory. A PC’s main memor y (i.e., it s pr imar y st or e) is fast st or age t hat is dir ect ly accessible by t he CPU, and is used t o st or e t he cur r ent ly execut ing pr ogr am and i mmi di at el y needed dat a. PCs use semi conduct or Random Access Memor y (RAM ) of var ious kinds such as DRAM or SRAM as t heir pr imar y st or age. Which exact kind depends on cost /per for mance issues at any par t icular t ime. M ain memor y is much fast er t han mass st or age devices like har d disks or opt ical discs, but is usually volat ile, meaning it does not r et ain it s contents (inst r uctions or data) in the absence of power, and is much mor e expensive for a given capacit y t han is most mass st or age. M ain memor y is gener ally not suit able for long-t er m or ar chival dat a st or age. M ass st or age devices st or e pr ogr ams and dat a even when t he power i s off; t hey do r equi r e power t o per for m r ead/wr it e funct ions dur ing usage. Although

semiconductor flash memor y has dr opped in cost, the pr evailing for m of mass stor age in personal computers is st ill t he elect r omechanical har d disk. The disk dr ives use a sealed H ead/Disk Assembly (H D A ) w h i ch w as f i r st i n t r odu ced by I B M ’s “ Wi nchest er ” di sk syst em. The use of a seal ed assembly allowed t he use of t he positive air pr essur e t o dr ive out par t icles fr om t he sur face of t he disk, which impr oves r eliabilit y. Video Car d. The video car d- ot her wise cal led a gr aphics car d, gr aphics adapt er or video adapt er - pr ocessor s and r ender s t he gr aphics out put fr om t he comput er t o t he comput er display, also called t he Visual Display Unit (VDU), and is an essent ial par t of t he moder n comput er. M I CROPROCE SSOR. A micr opr ocessor is a mult ipur pose pr ogr ammable logic device t hat r eads binar y inst r uct ions fr om a st or age device called memor y, accept s binar y dat a as i n pu t an d pr ocess dat a accor di n g t o t h ose instr uctions and pr ovides r esults as output. A typical pr ogr ammable machine can be r epr esent wit h t hr ee component s : micr opr ocessor, memor y, and I /O.

Memory Micro processor

I/O

These t hr ee component s wor k t oget her or int er act wit h each ot her t o per for m a given t ask, t hus t hey compr ise a syst em H ARD WARE . The physical component s of t his syst em ar e called har dwar e. SOF TWARE A set of inst r uct ions wr it t en for t he micr opr ocessor t o per for m a t ask is called a pr ogr am and a gr oup of pr ogr ams is called soft war e. APPL I CAT I ON S. Th e mi cr opr ocessor appl i cat i ons ar e cl assi fi ed pr imar ily in t wo cat egor ies ( i ) I n Re-programmable syst ems. Such as micr ocomput er s, t he micr opr ocessor is used for comput ing and dat a pr ocessing. These systems, include gener al pur pose micr opr ocessor capabl e of handl i ng l ar ge dat a. M ass st or age device (disks), and per ipher als such as I /O device (pr inter ).

Basics of Computers and Applications

( ii ) Embedded syst em. I n embedded syst ems, t he micr opr ocessor is par t of a f i n al pr odu ct an d i s n ot av ai l abl e f or r epr ogr ammabl e t o t he end user. A copyi ng machine is a t ypi cal example of an embedded syst em . Th e mi cr opr ocessor s u sed i n t hese syst ems ar e gener ally cat egor ised as ( a ) M i cr ocon t r ol l er s t h at i n cl u de al l t h e components like micr opr ocessor, memor y and I /O. ( b) I ntegrated microprocessor that include various devices such as timers and various types of I /O on a chip. ( c) Gener al pur pose micropr ocessor with discr ete component s such as micr opr ocessor, memor y and I /O. Embedded syst ems can also be incr eased as pr oducts t hat use micr opr ocessor t o per for m t heir oper at ions t hey ar e called as micr opr ocessor based pr oduct s. e.g. washing machines, dish washer s, aut omobi le dashboar d cont r ol s, t r affi c l i ght cont r ol l er s, and aut omat ic t est ing inst r ument s. BI N ARY DI GI T. The micr opr ocessor oper at es in binar y digit s 0 and 1, also known as bit s. Bit is an abbr eviat ion for t he t er m binar y digit . Those digit s ar e r epr esent ed in ter ms of electr ical voltages in the machine : gener ally 0 r epr esent s one vol t age l evel and 1 r epr esent s anot her. The digit s 0 and 1 ar e also synonyms wit h low and high r espect ively. Each mi cr opr ocessor r ecogni zes and pr ocesses a gr oup of bit s called t he wor d and micr opr ocessor ar e classified accor ding t o t heir wor d lengt h. M emor y. M emor y is like t he pages of a not ebook wit h space for a fixed number of binar y number s on each line. H owever t hese pages ar e gener ally made of semiconduct or mat er ial. Each line in 8 bit r egist er t hat can st or e 8 bi t bi nar y bi t s, and sever al of t hese r egist er s ar e ar r anged in a sequence called memor y. I nput /Out put . The user can enter instructions and data into memor y thr ough devices such as keyboar d or simple switches. These devices ar e called input devices. The mi cr opr ocessor r eads i nst r uct i ons fr om t he memor y and pr ocesses t he dat a accor ding t o t hose inst r uct ions. The r esult can be displayed by a device such as seven segment L ED (light emit t ing diodes) or pr i nt ed by a pr i nt er. These devi ces ar e call ed out put devices. M I CROPROCESSOR AS A CPU . The cent r al pr ocessing unit (CPU) consist s of t he Ar ithmet ic Logic Unit (ALU) and Cont r ol Unit (CU).

3

The CPU cont ains var ious r egist er s t o st or e dat a, AL U t o per for m Ar it hmet ic and logical, oper at ions, inst r uction decoder s, count er s and cont r ol lines. The CPU r eads i n st r uct i on s fr om t h e memor y and per for m t he t asks specified. I t communicat es wit h input /out put devices eit her t o accept or to send dat a. These devices ar e also known as per ipher als. The CPU i s t h e pr i m ar y an d cen t r al pl ay er i n communicat ing wit h devices such as memor y, input an d ou t pu t . H ow ev er , t h e t i m i n g of t h e communicat ion pr ocess is cont r olled by t he gr oup of cir cuit called cont r ol unit . CPU on single chip called mi cr opr ocessor. Arithmetic Logic Unit (ALU)

Input

Control Unit

Output

Memory

Fig. (a) Traditional block diagram of

Input

Micro processor as CPU

Output

Memory

F i g. (b) Bl ock di agr am of a compu t er w i t h t he a comput er micr oprocessor as CPU

1. bit micr opr ocessor. The I nt el 4004 was t he fir st 4 bit pr ogr ammable device t hat was pr imar ily used in calculat or s. 2. bit micr opr ocessor. The int el 8008 is 8 bit micr opr ocessor, which was in t ur n super seded by t he I nt el 8080. I nt el 8080 wi dely used i n cont r ol applicat ions, and small comput er s also wer e designed using t he 8080 as t he CPU. Wit hin a few year s aft er t he emer gence of t he 8080, t he M ot or ola 6800 and Zilog Z80 and I nt el 8085 mi cr opr ocessor wer e devel oped as impr ovement s 3. 16 bit microprocessor. 8086/88 4. 32 bit microprocessor. 80380/486 and Pent ium 5. 64 bit microprocessor. M ot or ola 68000 ser ies

4

Basics of Computers and Applications

M I CROCOM PU T E RS. M icr ocomput er is classified in four gr oups. 1. Per sonnel comput er s. These micr ocomputer s ar e single user systems and being used for var iety of pur poses, such as payr oll, busi ness account s, wor d pr ocessi ng, l egal and medi cal r ecor d k eepi ng, per sonnel fi nance and inst r uctions. A typical configur ation includes a 16 or 32 bit micr opr ocessor, 2 to 4 M B (megabyte) of system memory, a video screen, a dot matrix pr inter. 2. Work st at ions. These ar e high per for mance cousi ns of t he PC. T hey ar e used i n en gi n eer i n g an d sci en t i fi c applications such as computer -aided design (CAD), computer aided engineer ing (CAE), and computer aided manufacturing (CAM), they generally include syst em memor y l ar ger t han 200 M B, st or age memor y in giga byt es and high r esolut ion scr een. The wor k st at i ons ar e desi gned ar ound RI SC pr ocessor s. The RI SC pr ocessor s t end t o be fast er and mor e effi ci ent t han t he pr ocessor s used i n per sonnel comput er. 3. Single boar d micr ocomput er s. These mi cr ocomput er s ar e pr i mar i l y used i n college, labor atories and industr ies for instr uctional pur poses or t o evaluate the per for mance of a given mi cr opr ocessor. They can al so be par t of some lar ger syst ems. Typical ly t hese micr ocomput er s include 8 or 16 bit micr opr ocessor.

e.g. These computers include such systems as Intel SDK85, SDK86, Motorola Evaluation kits, these are generally used to write and execute assembly language programs and to perform interfacing experiments 4. Si n gl e ch i p m i cr ocom pu t er s ( m i cr ocont r ol l er s) These micr ocomput er s ar e designed on a single chip, which t ypically includes a chip of 64 byt es of R/W memor y fr om 4K t o 2K byt es of ROM and sever al si ngl e l i nes t o connect I /Os. These ar e complet e micr ocomput er s on a chip, t hey ar e also k n ow n as m i cr ocon t r ol l er s. T h ese ar e u sed pr i m ar i l y f or su ch f u n ct i on s as con t r ol l i n g appliances and t r affic light s. e.g. Zilog Z8, I nt el M CS51 and 96 ser ies, and M ot or ola 68H C11. COM PU TER LAN GU AGES. N ibble. I t is a gr oup of four bit s M nemonic : A combinat ion of let t er s t o suggest t he oper at ion of an inst r uct ion. Compiler : A pr ogr am t hat t r anslat es english like wor ds of high language in t he machine language of a comput er. A complier r eads a given pr ogr am, called a sour ce code and t hen t r anslat es t he pr ogr am int o t he machine language which is called an object code. Assembler : A comput er pr ogr am t hat t r anslat es an assembly language pr ogr am fr om mnemonics t o t he binar y machine code of a comput er. M onit or program : A pr ogr am t hat int er pr et s t he input fr om a keyboar d and conver t s t he input int o it s binar y equivalent .

COM PU TER SYSTEM . 8085 Programming M odel. The 8085 pr ogr amming model includes six r egist er s, one accumulat or and one flag r egist er. I t has t wo 16 bit r egist er t he st ack point er and t he pr ogr am count er. Registers : The 8085 has six gener al pur pose r egist er s t o st or e 8 bit dat a. These ar e ident ical as B, C, D, E, H and L . They can be combined as Regist er s pair s BC, DE, and H L t o per for m some 16 bit oper at ions. The pr ogr ammer can use t hese r egist er s t o st or e or copy dat a int o t he r egist er s by using dat a copy inst r uct ions. Accumulator A

(8)

B

(8)

C

(8)

D

(8)

E

(8)

H

(8)

Flag Register

L

(8)

Stack pointer

(SP)

(16)

Program counter

(PC)

(16) Address Bus

Data Bus 8

Lines

Lines

Unidirectional

Bidirectional

D7

D6

S

Z

D5

F ig. (a) Progr amming model D4 D3 D2 AC P F ig. (b) F lab Register

D1

D0 CY

Basics of Computers and Applications

Accumulat or. The Accumulat or is an 8 bit r egist er t hat is par t of t he ar it hmet ic/logic unit (AL U). This r egist er is used t o st or e 8 bit dat a and t o per for m ar it hmet ic/logic unit (AL U). This r egist er is used t o st or e 8-bit dat a and t o per for m ar it hmet ic and logical oper at ions. The r esult of an oper at i on i s st or ed i n t he accumul at or. The accumulat or is also ident ified as r egist er A. FLAGS. The ALU includes five flip-flops, which ar e set or r eset aft er an oper at ion accor ding t o dat a condit ions of t he r esult in t he accumulat or and ot her r egist er s. They ar e called zer o (Z), car r y (CY); sign (S), par it y (P) and Auxiliar y car r y (AC) flags. The most commonly used flags ar e zer o, car r y, and sign. The micr opr ocessor uses t hese flags t o t est dat a condit ions. Aft er an addit ion of t wo number s, if t he sum in t he accumulat or is lar ger t han eight bits, the flip-flop used t o indicat e a car r y called car r y flag (CY) is set t o one. When an ar ithmetic oper ation is zer o the flip flop called t he zer o (Z) flag is set t o one. The fl ags have cr i t i cal i mpor t ance i n t he deci si on making pr ocess of t he micr opr ocessor. The condit ions (set or r eset ) of t he flags ar e t est ed t hr ough soft war e inst r uct ions. PROGRAM COU N TER (PC). T h i s 16 bi t r egi st er deal s wi t h sequ en ci n g t h e execut ion of inst r uct i on. Thi s r egi st er i s a memor y point er. M emor y locat ions have 16 bit addr esses, and that is why t his is a 16-bit r egister. The micr opr ocessor uses t hi s r egi st er t o sequence t he execut i on of t he inst r uct ions. The funct ion of t he pr ogr am count er is t o point t o t he memor y addr esses fr om which t he next byt e is t o be fet ched. When a byt e (machine code) is being fet ched, t he pr ogr am count er is incr ement ed by one t o point t o t he next memor y locat ion. STACK POI N TER (SP). The stack pointer is also a 16 bit register used as memory poi n t er. I t poi n t s t o a m em or y l ocat i on i n R/W memor y, called t he stack. The beginning of st ack is defined by loading a 16-bit address in the stack pointer. OPCODE FORM AT. I n t he desi gn of t he 8085 mi cr opr ocessor chi p, al l oper ations, registers and status flags are identified with a specific code. All int er nal r egist er s ar e ident ified as follows : Code Regist er s Code Regist er pair s 000 B 00 BC 001 C 01 DE 010 D 10 HL 011 E 11 A F or S P 100 F 101 G 110 Reser ved for M emor y Relat ed Oper at ion 111 A

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I N STRU CTI ON CODES. An inst r uct ion code is a gr oup of bit s t hat inst r uct t he comput er t o per for m a specific oper at ion. I t is usually divided i nt o par t s, each havi ng it s own par t i cular int er pr et at ion. The most basic par t of an inst r uct ion codes is it s oper at ion par t . Oper at ion Code. The oper at ion code of an inst r uct ion is a gr oup of bit s that define such oper ations as add, subt r act , mult iply, shift and complement . The number of bit s r equir ed for t he oper at ion code of an inst r uct ion depends on t he t ot al number of oper at i ons avai l abl e i n t he comput er. The oper at ion code must consist of at least n bit s for a given 2n (or less) dist inct oper at ions. OP Code. What t ask t o be per for med, called t he oper at ion code (OPcode). Oper and. When dat a t o be oper at ed on called t he oper and. The oper and (or dat a) can be specified in var ious ways. I t may include 8-bit or (16 bit) data, an int er nal r egist er, a memor y location, or 8 bit (or 16 bit) addr ess. I n some inst r uct ions, t he oper and is implicit . ADDRESSI N G M ODES. The control unit of a computer is designed to go through an inst r uct ion cycle t hat is divided int o t hr ee major phases. (i) Fet ch t he inst r uct ion fr om memor y (ii ) Decode t he inst r uct ion (iii) Execut e t he inst r uct ion I mplied mode. A l l r egi st er r ef er en ce i n st r uct i on s t h at u se an accumul at or ar e i mpl i ed mode i nst r uct i ons. Zer o addr ess inst r uct ions in a st ack-or ganized comput er ar e implied mode inst r uct ions since t he oper ands ar e implied t o be on t op of t he st ack. I mmediat e mode. The operand is specified in the instruction itself. In other words, an immediate mode instruction has an operand field r ather than an addr ess field. The operand field contains the actual operand to be used in conjunction with the operation specified in the instr uctions. Regist er mode. I n t his mode t he oper ands ar e in r egist er s, that r eside wit hin t he CPU. The par t icular r egist er is select ed fr om a r egist er field in t he inst r uct ion. A K bit field can specify any one of 2k r egist er s. Regist er indir ect mode. I n this mode t he inst r uct ion specifies a r egist er in t he CPU, whose cont ent gives t he addr ess of t he oper and in memor y. The advant age of a r egist er indir ect mode inst r uct ion is t hat t he addr ess field of t he inst r uct ion uses fewer bit s t o select a r egist er t han would have been r equir ed t o specify a memor y addr ess dir ect ly,

6

Basics of Computers and Applications

Aut o I ncrement or Decrement mode. This is similar t o r egist er indir ect mode except t hat t he r egist er is incr ement ed or decr ement ed aft er or befor e it s value is used t o access memor y. When t he addr ess st or ed in t he r egist er r efer s t o a t able of dat a in memor y, it is necessar y t o incr ement and decr ement t he r egist er s aft er ever y access t o t he t able D ir ect addressing mode. I n t hi s mode t he effect i ve addr ess i s equal t o t he addr ess par t of t he inst r uct ion. The oper and r esides in memor y and it s addr ess is given dir ect ly by t he addr ess field of t he inst r uct ion. I ndir ect addr essing mode. I n t his mode t he addr ess field of t he inst r uct ion gives t he addr ess wher e t he effect ive addr ess is st or ed in memor y. Effect ive addr ess = addr ess par t of inst r uct ion + cont ent of CPU r egist er I N PU T AN D OU TPU T SYSTEM I nput /out put devices ar e t he means t hr ough which t he M PU communicat es wit h t he out side wor ld. The M PU accept s binar y dat a as input fr om devices such as keyboar ds and A/D conver t er s and send dat a t o out put devices such as L EDs or pr int er s. Ther e ar e t wo differ ent met hods by which I /O devices can be ident ified. One uses an 8 bit addr ess and t he ot her uses a 16 bit addr ess. PERI PH ERAL M APPED I /O. I n this type of I /O, the MPU uses eight addr ess lines to identify an input or an output device; t his is known as per ipher al mapped I /O. The eight addr ess lines can have 256 (28 combinat ions) addr esses; t hus the MPU can identify 256 input devices and 256 output devices with addresses ranging fr om 00H to FFH input and output devices are differentiated by the contr ol signals. The MPU uses the I /O Read Contr ol Signal for input devices and the I /O Wr ite Contr ol Signal for output devices. The entir e r ange of I /O addr esses fr om 00 to FF is known as an I /O map, and individual addr esses ar e r efer r ed t o as I /O devices addr esses 8 I /O por t number s. M E M ORY M APPE D I /O (I /O W I T H 16 BI T ADDRESSES). I /O is connect ed as if it is memor y r egist er. This is known as memor y mapped I /O. The M PU uses t he same cont r ol signal (M emor y Read or M emor y Wr it e) an d i n st r u ct i on s as t h ose of m em or y. I n som e micr opr ocessor such as mot or ola 6800, all I /O have 16 bit addr esses. I /Os and memor y shar e t he same memor y map (64K). I n memor y mapped I /O, the MPU follows the same steps as if it is accessing a memor y r egist er. I N PU T DEVI CES These include a mouse, tr ack ball, space ball, joyst ick, digit izer s, dials and but t on boxes. Some ot her input

devices used in par t icular applicat ions ar e dat a glove, t ouch panels, image scanner s and voice syst ems. 1. K eyboar ds. An alphanumer ic keyboar d on a gr aphics syst em i s used pr i mar i l y as a devi ce for ent er i ng t ext st r i ngs. The k eyboar d i s an effi ci ent devi ce for input t ing such non gr aphics dat a as pict ur e labels associated with graphics display. Keyboar ds can also be pr ovided with featur e to facilitate entr y of scr een coor dinat es, menu select or s or gr aphics funct ions. 2. M ouse. A mouse is small hand held box used t o posit ion t he scr een cur sor. Wheels or r oller s on t he but t on of t he mouse can be used t o r ecor d t he amount and dir ect ion of movement s. Anot her met hod for det ect ing mouse mot ion is wit h an opt ical sensor. For t hese syst ems, t he mouse i s moved over a special mouse pad t hat has gr id of hor izont al and v er t i cal l i n es. T h e opt i cal sen sor det ect s movement s acr oss t he l i nes i n t he gr i d. Si nce mouse can be picked up and put down at anot her posit ion wit hout change in cur sor movement . I t is used for making r elat ive changes in t he posit ion of t he scr een cur sor. One, t wo or t hr ee but t ons ar e usual l y i ncl uded on t he t op of t he mouse for signalling t he execut ion of some oper at ion, such as r ecor ding cur sor position or invoking a funct ion. Most gener al pur pose graphics system now included a mouse and keyboar d as major input devices. 3. Track Ball and Space Ball. A t r ack ball is a ball t hat can be r ot at ed wit h t he fi nger s or pal m of t he hand t o pr oduce scr eencur sor movement , pot ent iomet er s at t ached t o t he ball measur e t he amount and dir ect ion of r ot at ion. Tr ackballs are often mounted on keyboar ds or other devices such as t he mouse. While a track ball is two dimensional posting device, a space ball pr ovides six degr ee of fr eedom. Unlike t he t r ack ball, space ball does not act ually move. St r ain gauges measur e t he amount of pr essur e applied to t he spaceball to pr ovide input for spat ial posit ioning and or ient at ion as t he ball is pushed or pulled in var ious dir ect ions. Space balls ar e used for t hr ee-di mensi onal posi t i oni ng and sel ect i on oper at ions in vir t ual r ealit y syst ems, modelling, animat ion, CAD, and ot her applicat ions. 4. Joyst i ck s. A joyst ick consist s of a small, ver t ical lever called st ick mount ed on a base t hat is used t o st eer and scr een cur sor ar ound. M ost joyst icks select scr een posi t i on wi t h act ual st i ck movem ent . Ot her s r espond t o pr essur e on t he st ick. Some joyst icks ar e mount ed on keyboar d, ot her s funct ion as st and alone unit s. Pot ent iomet er mount ed at t he base of joyst ick measur es t he amount of movement , and spr ings r etur n the stick to the center position when i t i s r el eased on e or m or e bu t t on s can be

Basics of Computers and Applications

5.

6.

7.

8.

pr ogr ammed t o act as input swi t ches t o si gnal cer t ai n act ions once a scr een posi t i on has been select ed. Dat a Glove. Dat a glove t hat can be used t o gr asp a “ vir t ual” object . The glove is const r uct ed wit h a ser ies of sensor s t hat det ect hand and fi nger mot i ons. Elect r omagnet ic coupl ing bet ween t r ansmi t t ing antennas and r eceiving antennas is used to pr ovide infor mat ion about t he posit ion and or ient at ion of the hand. The tr ansmitting and r eceiving antennas can each be st r uct ur ed as a set of t hr ee mut ually per pendicular coils, for ming a t hr ee dimensional car t esian co-or dinat e syst em. D igit izer s. A com m on dev i ce f or dr aw i n g, pai n t i n g or int er act ively select ing co-or dinat e posit ions on an object is a digit izer. These devices can be used t o i n pu t co-or di n at e v al u es i n ei t h er a t w o dimensional or t hr ee dimensional space. Typically, a digit izer is used t o scan over a dr awing or object and t o input a set of discr et e co-or dinat e posit ions, which can be joined wit h st r aight -line segment s t o appr oximat e the cur ve or sur face shapes. One t ype of digit izer is t he gr aphics t ablet , also r efer r ed t o as a dat a t abl et wh i ch i s u sed t o i n pu t t wo dimensional coor dinates by activating a hand cursor or st yles at select ed posit ions on a flat sur face. I mage Scanners. Dr awing, gr aphs, color and black and whit e photos or t ext can be st or ed for comput er pr ocessing wit h an image scanner by passing an opt ical scanning mechanism over t he infor mat ion t o be st or ed. The gr adat ions of gr ay scale or color ar e t hen r ecor ded and st or ed in an ar r ay. Once we have t he int er nal r epr esen t at i on of a pi ct u r e, w e can appl y t r ansfor mat ions t o r ot at e, scale or cr op the pict ur e t o a par t i cul ar scr een ar ea. We can al so appl y var ious image pr ocessing met hods t o modify t he ar r ay r epr esent at ion of t he pict ur e. For scanned t ext i nput , var i ous edi t i ng oper at i ons can be per for med on st or ed document s. Some scanner s ar e able t o scan eit her gr aphical r epr esent at ion or t ext , and t hey come i n a var i et y of si zes and capabilities. Touch Panels. Touch panel s al l ow di spl ayed object s or scr een posit ions t o be select ed wit h t he t ouch of a finger. A t ypi cal appl i cat i on of t ouch panels i s for t he selection of pr ocessing options that ar e r epr esented wit h gr aphical icons. Some syst ems, such as t he plasma panels ar e designed wit h t ouch scr eens. Ot her syst em can be adapt ed for t ouch input by fit t ing a t r anspar ent device wit h a t ouch-sensing mechanism over t he video monit or scr een. Touch

7

input can be r ecor ded using opt ical, elect r ical or acoust ical met hods. 9. L ight Pens. Light pen is pencil-shaped devices ar e used to select scr een posit ions by detect ing the light coming fr om point s on t he CRT scr een. They ar e sensit ive t o t he shor t bur st of light emit t ed fr om t he phosphor coat ing at t he inst ant t he elect r on beam st r ikes a par t icular point . Ot her light sour ces, such as t he back gr ound l i ght i n t he r oom, ar e usual l y not det ect ed by a light pen. PRI N T E RS. Pr inter s pr oduce output by either impact or non-impact methods. I mpact pr inter s pr ess for med char acter faces against an inked r ibbon ont o t he paper. A line pr int er is an example of impact device wit h t he t ype faces mount ed on bands, chai ns, dr ums or wheels. Noni mpact pr int er s and pl ot t er s use l aser t echniques, i nk j et spr ays, xer ogr aphi c pr ocesses as used i n phot ocopyi ng machi ne, el ect r ost at i c met hods and elect r ot her mal met hods t o get images on t o paper. Char acter impact printers often have a dot matrix print head containing a r ectangular ar r ay of pr otr uding wir e pins, with the number of pens depending on the quality of t he pr int er. I n a l aser devi ce, l aser beam cr eat es a ch ar ge di st r i but i on on a r ot at i n g dr um coat i ng wi t h a phot oel ect r i c mat er i al , such as sel eni um. Toner i s applied t o t he dr um and t hen t r ansfer r ed t o paper. I nkjet met hods pr oduce out put by squir t ing i nk in hor izont al r ows acr oss a r oll of paper wr apped on a dr um. The electr ically char ged ink st r eam is deflected by an elect r ic field t o pr oduce dot mat r ix pat t er ns. A deskt op inkjet plot t er wit h r esolut ion of 360 dot s per inch. An elect r ost at ic device places a negat ive char ge on t he paper, one complet e r ow at a t ime along t he lengt h of t he paper. Then t he paper is exposed t o a t oner. The t oner is posit ively char ged and so it is at t r act ed t o t he negat ively char ged, ar eas, wher e it adher es t o pr oduce the specified output . Elect r other mal met hods use heat in a dot mat r ix pr int head t o out put pat t er ns on heat sensi t i ve paper. We can get l i mi t ed col or out put on an impact pr int er by using differ ent color ed r ibbons. Non impact devices use var ious techniques to combine thr ee color pigments (cyan, magneta and yellow) to pr oduce a r ange of color pat t er ns. Par allel pr int er s use : (a) RS-232C interface (b) Cent r onics int er face (c) H andshake mode

8

Basics of Computers and Applications

STORAGE U N I T I t consist s of main memor y and secondar y memor y. M ain M emory : 1. A fl i p-fl op made of el ect r oni c semi conduct or devices is used t o fabr icat e a memor y cell. These memor y cel l s or gani zed as a Random Access Memor y (RAM). Each cell has a capabilit y to stor e one bit of infor mat ion. A main memor y or st or e of a comput er is or ganized using a lar ge number of cells. Each cell st or es a binar y digit . 2. A memor y cell, which does not loose the bit st or ed i n i t when no power is suppl ied t o t he cell , i s known as a non-volat ile cell. 3. A wor d is a gr oup of bit s, which ar e st or ed and r etr ieved as a unit. A memor y system is or ganized t o st or e a number of wor ds. 4. A Byt e consist s of 8 bit s. A wor d may st or e one or mor e byt es. 5. The st or age capacit y of a memor y is t he number of byt es it can st or e. 6. The addr ess of t he locat ion fr om wher e a wor d is t o be r et r i eved or t o be st or ed i s ent er ed i n a M emor y Addr ess Register (M AR). 7. The dat a r et r ieved fr om memor y or t o be st or ed in memor y ar e placed in a M emor y Dat a Regist er (MDR). 8. The t ime t aken t o wr it e a wor d is known as t he Wr it e t ime. 9. The t ime t o r et r ieve i nfor mat i on i s cal led t he Access t ime of t he memor y. 10. The t ime t aken t o access a wor d in a memor y is independent of the addr ess of t he wor d and hence it is know as a Random Access M emor y (RAM ).

 The main memor y used t o st or e pr ogr ams and dat a in a comput er is a RAM . 11. A RAM may be fabricated with per manently stor ed i nfor mat i on, whi ch cannot be er ased. Such a memor y is called a Read Only M emor y (ROM ).

 For mor e specialized uses, a user can st or e his won special funct ions or pr ogr ams in a ROM . Such ROM 's ar e called Pr ogr ammable ROM (PROM).

1. F loppy Disk Drive (F DD) : I n t hi s devi ce, t he medium used t o r ecor d t he dat a is called as floppy disk. I t is a flexible cir cular disk of diamet er 3.5 i nches made of pl ast i c coat ed wi t h a magnet i c mat er ial. This is housed in a squar e plast ic jacket . Dat a r ecor ded on a floppy disk is r ead and st or ed in a comput er 's memor y by a device called a floppy disk is r ead and st or ed in a comput er 's memor y by a device called a floppy disk dr ive (FDD). A floppy disk is inser t ed in a slot of t he FDD. Floppy Disks wit h var ious capacit ies ar e as follow:

 51/4 dr ive- 360KB, 1.2MB (1 KB= 210 = 1024 bytes)  31/2 dr ive- 1.44 M b, 2.88 M B (1M B= 220 byt es) 2. Compact Disk Drive (CDD) : CD-ROM (Compact Disk Read Onl y M emor y) used a l aser beam t o r ecor d and r ead dat a along spir al t r acks on a 51/4 di sk . A di sk can st or e ar ou n d 650 M B of infor mat ion. CD-ROM s ar e nor mally used t o st or e massive t ext dat a. Recent ly CD wr it er s have come in the mar ket . Using a CD wr it er, lot of infor mation can be wr it t en on CD-ROM and st or ed for fut ur e r efer ence. 3. H ard Disk Drive (H DD) : Unlike a floppy disk t hat is flexible and r emovable, t he har d disk used in t he PC is per manent ly fixed. The dat a t r ansfer r ate between the CPU and har d disk is much higher as compar ed to the between the CPU and the floppy disk dr ive. The CPU can use t he har d disk t o load pr ogr ams and dat a as well as t o st or e dat a. CLASSI FI CATI ON OF COM PU TERS Comput er s come in sizes fr om t iny t o monst r ous, in bot h appear ance and power. The size of a comput er t hat a per son or an or ganizat ion needs depends on t he comput ing r equir ement s. Supercomputers : The might iest comput er s-and, of cou r se, t h e m ost ex pen si v e-ar e k n ow n as super comput er s. Super comput er s pr ocess billions of inst r uct ions per second. One uses super comput er s for t asks t hat r equir e mammoth dat a manipulation, such as wor l dwi de weat h er f or ecast i ng and weapons r esear ch.

 Dat a is pr esent ed ser ially for wr it ing and is r et r ieved ser ially dur ing r ead.

M ainframes : I n t he jar gon of t he comput er t r ade, lar ge comput er s ar e called mainfr ames. M ainfr ames ar e capable of pr ocessing dat a at ver y high speedsmillions of inst r uct ions per second-and have access t o billions of char act er s of dat a. Their pr incipal use of it is for pr ocessing vast amount s of dat a quickly, some of t h e obvi ou s cu st om er s ar e ban k s, i n su r an ce companies, and manufact ur er s.

Secondary / Auxiliary storage devices : M agnet ic sur face r ecor ding devices used in comput er s as H ar d disks, Floppy disks, CD-ROM s and M agnet ic t apes.

Personal Computers : Per sonal comput er s ar e oft en called PCs. A PC usually comes wit h a tower that holds the main cir cuit boar ds and disk dr ives of the computer,

12. A ser ial access memor y is or ganized by ar r anging memor y cells in a linear sequence.

 I nfor mat ion is r et r ieved or st or ed in such a memor y by using a r ead/wr it e head.

Basics of Computers and Applications

and a collect ion of per ipher als, such as a keyboar d, mouse, and moni t or. The t er m "PC" oft en means machines t hat ar e compat ible t o I BM ot her t han a Macint osh. Personal Computers (PC) and M AC : A PC is based on a mi cr opr ocessor or i gi nal l y made by t he I nt el Company (I ntel's Pentium) with other companies such as AM D. The comput er s made by M acint oshes which uses, Power PC pr ocessor, made by M ot or ol a ar e r efer r ed as M ac. Also, t he oper at ing syst em soft war e t hat r uns t hese t wo kinds of comput er s is differ ent . PCs usual l y use an Oper at i ng Syst em made by M icr osoft , i.e., Windows. M acint oshes use oper at ing syst em, called M ac OS, made by Apple. N ot ebook Comput ers : A comput er t hat fi t s i n a br i efcase?. N ot ebook compu t er s, al so k nown as L apt op comput er s, ar e por t abl e and popul ar wit h tr aveler s who need a comput er that can go with them. M ost n ot ebook s accept di sk et t es or n et w or k connect i ons, so i t i s easy t o move dat a fr om one comput er t o anot her. I N TERN ET The I nt ernet i s a global syst em of int er connect ed comput er net wor ks t hat use t he st andar d I nt er net Pr ot ocol Sui t e (TCP/I P) t o ser ve bi l l i ons of user s wor ldwide. I t is a net wor k of net wor ks t hat consist s of millions of pr ivat e, public, academic, business, and gover nment net wor ks, of local t o global scope, t hat ar e li nked by a br oad ar r ay of elect r onic, wir eless and opt ical net wor king t echnologies. The I nt er net car r ies a vast r ange of infor mat ion r esour ces and ser v i ces, su ch as t h e i n t er -l i n k ed h y per t ex t document s of t he Wor ld Wide Web (WWW) and t he infr ast r uct ur e t o suppor t elect r onic mail. N ET WORKS A computer network, oft en simply r efer r ed t o as a net wor k , i s a col l ect i on of comput er s and devi ces i nt er connect ed by communi cat i ons channel s t hat faci l i t at e communi cat i ons and al l ows shar i ng of r esour ces and i nfor mat i on among i nt er connect ed devices. Computer networking or Data communicat ions (D at acom) i s t he engi neer i ng di sci pl i ne concer ned wit h t he comput er net wor ks.

The t hr ee t ypes of net wor ks ar e: (i ) t he I nt er net (ii ) t he int r anet (iii ) t he ext r anet . Examples of differ ent net wor k met hods ar e: 1. Local ar ea networ k (LAN), which is usually a small net wor k const r ained t o a small geogr aphic ar ea.

9

An exampl e of a L A N woul d be a comput er net wor k wit hin a building. 2. M et r opolit an ar ea net wor k (M AN), which is used for medi um si ze ar ea. exampl es for a cit y or a st at e. 3. Wide ar ea net wor k (WAN) t hat is usually a lar ger net wor k t hat cover s a lar ge geogr aphic ar ea. 4. Wir eless L ANs and WANs (WL AN & WWAN) ar e t he wir eless equivalent of t he L AN and WAN. I P ADDRESS An I nt ernet Prot ocol addr ess (I P addr ess) i s a n u m er i cal l abel assi gn ed t o each dev i ce (e.g., comput er, pr i n t er ) par t i ci pat i ng i n a comput er n et w or k t h at u ses t h e I n t er n et Pr ot ocol f or communicat ion.[1] An I P addr ess ser ves t wo pr incipal funct ions: host or networ k inter face identification and locat ion addr essing. I t s r ole has been char act er ized as follows: “ A name indicates what we seek. An address indicates where it is. A route indicates how to get there. I mport ant Devices U sed in N et work 1. M odem : A modem (modulat or -demodulat or ) is a device t hat modulat es an analog car r ier signal t o encode digit al infor mat ion, and also demodulat es such a car r ier signal t o decode t he t r ansmit t ed infor mat ion. The goal is t o pr oduce a signal t hat can be tr ansmitted easily and decoded to repr oduce the or iginal digit al data. M odems can be used over any means of t r ansmit t ing analog signals, fr om light emit t ing diodes t o r adio. 2. Router : A router is a device t hat for war ds dat a pack et s acr oss compu t er n et wor k s. Rou t er s per for m t he dat a “ t r affic dir ect ing” funct ions on t he I nt er net . A r out er is connect ed t o t wo or mor e dat a lines fr om di ffer ent net wor ks. When dat a comes in on one of t he lines, t he r out er r eads t he addr ess infor mat ion in t he packet t o det er mine it s ult imat e dest inat ion. 3. Bridge : A net work bridge connect s mul t i pl e net wor k segment s. Br idging i s a for war di ng t echni que used i n pack et -swi t ched comput er net wor k s. U nl i k e r out i ng, br i dgi ng mak es no assumptions about where in a networ k a par ticular addr ess is locat ed. I nst ead, it depends on flooding and examinat ion of sour ce addr esses in r eceived packet header s t o locat e unknown devices. Once a device has been locat ed, it s locat ion is r ecor ded in a t able wher e t he M AC addr ess is st or ed so as t o pr eclude t he need for fur t her br oadcast ing. 4. H ub : hub is a device for connecting multiple twisted pair or fiber optic Ethernet devices together and making them act as a single segment. The device is a for m of multiport repeater.

10

Basics of Computers and Applications

5. Repeater : A repeater is an elect r onic device t hat r eceives a signal and r et r ansmit s it at a higher level and/or higher power, or ont o t he ot her side of an obst r uct i on, so t hat t he si gnal can cover longer dist ances. 6. Server : A ser ver comput er i s a comput er, or ser ies of comput er s, t hat link ot her comput er s or el ect r oni c devi ces t oget her. They oft en pr ovi de essent i al ser vi ces acr oss a net wor k , ei t her t o private user s inside a lar ge or ganization or to public users via the inter net. For example, when you enter a quer y in a sear ch engine, t he quer y is sent fr om your computer over t he int er net to t he ser ver s t hat st or e all t he r elevant web pages. The r esult s ar e sent back by t he ser ver t o your comput er. EM AI L E l ect r on i c m ai l , com m on l y cal l ed em a i l or e-mail, is a met hod of exchanging digit al messages fr om an aut hor t o one or mor e r ecipi ent s. M oder n email oper at es acr oss t he I nt er net or ot her comput er net wor ks. Some ear ly email syst ems r equir ed t hat t he aut hor and t he r ecipient bot h be online at t he same t ime, a la i nst ant messaging. Today’s email syst ems ar e based on a st or e-and-for war d model . Email ser ver s accept , for war d, del iver and st or e messages. Neit her t he user s nor t heir comput er s ar e r equi r ed t o be onl i ne si mul t aneousl y; t hey need connect only br iefly, t ypically t o an email ser ver, for as long as it t akes t o send or r eceive messages. EM AI L ADDREES An email address ident ifies an email box t o which email messages ar e deliver ed. An example for mat of an email addr ess is lewis @ example .com which is r ead as lewis at example dot net . I t has two par t s. The par t befor e t he @sign is t he local-par t of t he addr ess, oft en t he user name of t he r ecipient lewis and the par t aft er t he @ sign is a domain name i.e. example.com t o which t he email message will be sent . M S OFFI CE M icrosoft Office is a proprietary commercial office suite of inter-related desktop applications, servers and services for the Microsoft Windows and Mac OS X oper ating systems, intr oduced by Micr osoft in 1989. I nitially a marketing term for a bundled set of applications, the first version of Office contained Microsoft Word, Micr osoft Excel, and Microsoft PowerPoint. W ORD Micr osoft Wor d is a wor d pr ocessor and was pr eviously consi der ed t o be t he mai n pr ogr am i n Offi ce. I t s pr opr i et ar y DOC for mat i s consi der ed a de fact o standard, although Word 2007 can also use a new XMLbased, Micr osoft Office-optimized format called .DOCX which has been st andar dized by Ecma I nt er nat ional

as Office Open XM L and it s SP2 updat e will suppor t ODF and PDF. Wor d is also available in some edit ions of M icr osoft Wor ks. I t is available for t he Windows and M ac platfor ms. The fir st ver sion of Wor d, r eleased in t he aut umn of 1983, was for t he M S-DOS oper at ing system and had the distinction of introducing the mouse t o a br oad populat ion. Wor d 1.0 could be pur chased wit h a bundled mouse, t hough none was r equir ed. Following t he pr ecedent s of L isaWr it e and M acWr it e, Wor d for Macintosh attempted to add closer WYSI WYG feat ur es int o it s package. Wor d for M ac was r eleased in 1985. Wor d for M ac was t he fir st gr aphical ver sion of M icr osoft Wor d. Despit e it s bugginess, it became one of t he most popular M ac applicat ions. EXCEL M i cr osoft Excel i s a spr eadsheet pr ogr am whi ch or iginally competed with the dominant Lotus 1-2-3, but eventually outsold it . I t is available for the Windows and Mac platfor ms. Micr osoft r eleased the fir st ver sion of Excel for the Mac in 1985, and t he fir st Windows ver sion (number ed 2.05 to line up with t he Mac and bu n dl ed w i t h a st an dal on e Wi n dow s r u n -t i m e envir onment) in November 1987. OU TLOOK M icr osoft Outlook (not t o be confused with Out look Expr ess) is a per sonal infor mation manager and e-mail communication softwar e. The replacement for Windows Messaging, Micr osoft M ail and Schedule+ star ting in Office 97, it includes an e-mail client, calendar, t ask manager and addr ess book. On t he M ac, M i cr osoft offer ed sever al ver si ons of Outlook in the late 1990s, but only for use with Microsoft Exchange Ser ver. I n Offi ce 2001, i t i nt r oduced an alter nat ive application with a slightly differ ent featur e set called Micr osoft Entour age. I t reintr oduced Outlook in Office 2011, r eplacing Entour age. POWE RPOI N T M i cr osof t Power Poi n t i s a popu l ar pr esent at i on pr ogr am for Windows and M ac. I t is used t o cr eat e slideshows, composed of t ext , gr aphics, movies and ot her object s, which can be displayed on-scr een and navigat ed t hr ough by t he pr esent er or pr int ed out on t r anspar encies or slides.

Basics of Computers and Applications

11

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. Which of t he following is t he fast est ? (a) CPU (b) magnet ic t apes and disks (c) video t er minal (d) sensor s, mechanical cont r oller s 2. The input unit of a comput er (a) feeds dat a t o t he CPU or memor y (b) r et r ieves dat a fr om CPU (c) dir ect s all ot her unit s (d) all of t hese 3. Offline device is (a) a device which is not connect ed t o CPU (b) a device which is connect ed t o CPU (c) a dir ect access st or age device (d) an I /O device 4. Which of t he following is a set of gener al pur pose int er nal r egist er s ? (a) Stack (b) Scratchpad (c) Addr ess r egist er (d) St at us r egist er 5. A single bus st r uct ur e is pr imar ily found in (a) main fr ames (b) super comput er s (c) high per for mance machines (d) mini-and micr o-comput er s 6. Which of t he following r egist er s is used t o keep t r ack of addr ess of t he memor y locat ion wher e t he next inst r uct ion is locat ed ? (a) M emor y Addr ess Regist er (b) M emor y Dat a Regist er (c) I nst r uct ion Regist er (d) Pr ogr am Count er 7. Which of t he following r egist er s is loaded wit h t he cont ent s of t he memor y locat ion point ed by t he PC ? (a) M emor y Addr ess Regist er (b) M emor y Dat a Regist er (c) I nst r uct ion Regist er (c) Pr ogr am count er

8. I n a gener ic micr opr ocessor, inst r uct ion cycle time is (a) shor t er t han machine cycle t ime (b) lar ger t han machine cycle t ime (c) exact ly double t he machine cycle t ime (d) exact ly t he same as t he machine cycle t ime 9. Pr ogr am St at us Wor d (PSW) cont ai ns var i ous (differ ent) st atus of (a) CPU (b) ALU (c) pr ogr am (d) r egist er s 10. When an int er r upt occur s, CPU saves t he value of— — — in a st ack, (a) accumulator (b) pr ogr am st at us wor d (PSW) only (c) I nstr uct ion Addr ess Counter (I AC)only (d) bot h PWS and I AC 11. Bus Ar bit r at ion is (a) clear ing t he bus (b) lat ching infor mat ion on t he bus (c) deciding t he cont r oller of t he bus (d) cont r olling t he bus 12. Cont r ol M emor y Addr ess Regist er is pr esent in (a) ALU (b) I nst r uct ion Regist er Unit (c) Cont r ol Unit (d) Disk Cont r ol I nt er face Unit 13. Which of t he fol lowi ng is not one of t he t hr ee pr i mar y funct i ons t hat on-l i ne di r ect access syst em can ser ve? (a) inquir y (b) backup (c) update (d) pr ogr amming 14. Which of the following is not tr ue of punched cards as dat a ent r y media? (a) They can be used as t ur n ar ound document s (b) They ar e inexpensive (c) I nput is slow compar ed wit h ot her media (d) They ar e easily damaged

12

Basics of Computers and Applications

15. M agnet ic t ape can ser ve as

7. ASCI I coding all ocat ed bi nar y codes t o Engli sh al phabet s and symbols for comput er use. M or e r ecent ly a new st andar d has been adopt ed which al locat es code t o almost all t he languages of t he wor l d and also t o symbols cover i ng mor e t han a lakh char act er s. The new st andar d is call ed

(a) input media (b) out put media (c) secondar y st or age media (d) all of t hese

LEVEL-1

(a) CCS

1. I n t he cont ext of I nfor mat ion Technology, OCR means (a) Opt ical Char act er Recognit i on (b) Oct agonal Cycl ic Rechar ge (c) Oct adecimal Cycli c Regener at i on (d) Opt ical Char act er Regener at i on [RRB JE 2014 GREEN SH I FT ]





2. I n Boolean algebr a 1  1  0  0 = ? (a) 0 (c) 2

(b) 1 (d) – 1 [RRB JE 2014 GREEN SH I FT ]

3. Which of t he fol lowi ngis not an I /O devi ce of t he comput er ? (a) Keyboar d (b) Joy st i ck (c) ALU (d) Pr int er [RRB JE 2014 GREEN SH I FT ]

4. W h at i s f l oat i n g poi n t w i t h r ef er en ce t o comput er s? (a) I t i s a soft war e subr out i ne ar ound whi ch ot her subr out i nes ar e bui lt (b) I t i s a r epr esent at i on of r eal number s t o facilit at e comput ing (c) I t i s t h e m ai n al gebr ai c f or m u l a of t h e soft war e (d) I t i s t h e vol t age poi n t gi v en t o var i ou s oper at i ng unit s of t he comput er [RRB JE 2014 GREEN SH I FT ]

5. A syst em of di gi t al r u l es for exch an ge and pr ocessi ng of dat a bet ween var i ous devi ces i s called (a) soft war e pr ogr amme (b) algor it hm (c) pr ot ocol (d) infor mat ion pr ocessing [RRB JE 2014 GREEN SH I FT ]

6. A t heor et ical comput er wi t h infini t e t ype and m em or y, u sed i n an al y si s of pr obl em s of comput at i on, is call ed (a) Tape calculat or (b) Babbage machine (c) Tur i ng machi ne (d) Theor et ical machi ne [RRB JE 2014 GREEN SH I FT ]

(b) Unicode (c) Standar d CCS code (d) Univer sal CCS code [RRB JE 2014 GREEN SH I FT ]

8. For using passwor ds on t he I nt er net a soft war e is used so t hat t he passwor d is not int er cept ed easi ly. I t is call ed (a) Coding

(b) Malwar e

(c) Virus

(d) Encr ypt ion [RRB JE 2014 GREEN SH I FT ]

9. A soft war e, codi ng of which is available fr eely on I nt er net and i s open for user s for fur t her use an d i m pr ov em en t an d w h i ch i s gen er al l y developed in a coll abor at i ve manner is call ed (a) open sour ce soft war e (b) unlicensed soft war e (c) fr ee soft war e (d) communi t y soft war e [RRB JE 2014 GREEN SH I FT ]

10. Wh i ch of t h e f ol l ow i n g ar e m ach i n e l ev el languages? (a) C++

(b) Java

(c) Python

(d) None of t hese [RRB JE 2014 GREEN SH I FT ]

11. Which of t he fol lowing st at ement s is i ncor r ect ? (a) M i cr osoft windows is GUI (b) L i nux is GUI (c) M or e t han 5000 k B dat a can be st or ed in a DVD (d) A 1 TB fl ash dr i ve can st or e 2 mi ll ion fil es each of size 1 M B [RRB JE 2014 GREEN SH I FT ]

12. The t er ms AL U, CPU, I /O devi ces per t ai n t o (a) comput er s (b) envi r onment al engineer i ng (c) di esel engi ne (d) en gi n eer i n g dr aw i n g an d or t h ogon al pr oject i ons [RRB JE 2014 GREEN SH I FT ]

Basics of Computers and Applications

13. I n a comput ing devi ce 'M H z' is ment i oned in t he specifi cat ions. I t r efer s t o (a) si ze of memor y

5. Who wr ot e/invent ed t he L inux soft war e? (a) M i cr osoft (b) Apple I NC (c) IBM (d) None of t hese

(b) speed of comput at i on (c) cl ock speed (d) none of t he above [RRB JE 2014 GREEN SH I FT ]

14. The value of binary 1111 is : (a) 23

(b) 23 – l

(c) 24

(d) 24 – l

[RRB SSE 2014 YELLOW SH I FT]

6. A t echnique of anonymous communicat i on over a computer networ k using encr yption of messages and spl it t ing bet ween t he nodes, i s cal led(a) Spice r out i ng (b) Onion r out i ng (c) Cabbage r out ing (d) Flower r out i ng

[RRB JE 2014 RED SH I FT ]

15. The term 'Operating System' means : (a) A set of programmes which controls computer working (b) The way a computer operator works (c) Conversion of high level language into machine level language (d) None of these [RRB JE 2014 RED SH I FT ]

LEVEL-2 1. Which of t hese i s N OT an Oper at ing Syst em? (a) Android (b) iOS (c) Linux (d) Power poi nt [RRB SSE 2014 YELLOW SH I FT]

2. A soft war e user i nt er face feat ur e t hat allows t he user t o view somet hing ver y si mi lar t o t he end r esul t whi l e t he document i s bei ng cr eat ed i s called(a) For mat cr eat or (b) For mat fideli t y (c) WYSI WYG (d) WYGI WYS [RRB SSE 2014 YELLOW SH I FT]

3. I n a comput er syst em t her e ar e soft war es and l anguages at var i ous l evel s, l i k e H i gh l evel L an gu age (H L ), M ach i n e L an gu age (M L ), Compiler (C). Which of the following is the cor r ect indi cat ive r epr esent at i on fr om user (U) t o t he comput er (COM P)? (a) U  H L  C  M L  Comp (b) U  C  M L  H L  Comp (c) U  C  H L  M L  Comp (d) U  M L  H L  C  Comp [RRB SSE 2014 YELLOW SH I FT]

4. Which of t hese devi ces per for ms t he funct ion of bot h i n pu t dev i ce an d ou t pu t dev i ce f or a comput er ? (a) Joy St i ck (b) M ouse (c) Modem (d) Pr int er [RRB SSE 2014 YELLOW SH I FT]

13

[RRB SSE 2014 YELLOW SH I FT]

7. Pr ocessing speed of comput er i s measur ed i n(a) M I PS(M i lli on I nst r uct i on Per Second) (b) M H z of clock (c) Bot h (a) and (b) (d) None of t hese

[RRB SSE 2014 YELLOW SHIFT]

8. To close a pr esent at ion and quit Power Point , one must click t he close but t on on t he : (a) menu bar (b) t i t le bar (c) st andar d t ool bar (d) common t ask s t oolbar [RRB SSE 2014 RED SH I FT]

9. Expr ession + + i is equi valent in ‘C’ t o : (a) i = i + l (b) i = i + 2 (c) i = 2i (d) None of t hese [RRB SSE 2014 RED SH I FT]

10. W h i ch of t h e f ol l ow i n g r at i on al r el at i on oper at ions in 'C means "not equal t o" ? (a) # (b) == (c) ! =

(d) < = [RRB SSE 2014 RED SH I FT]

11. M i cr osoft Windows is a/an : (a) Wor d-pr ocessing pr ogr am (b) Dat abase pr ogr am (c) Oper at ing syst em (d) Gr aphics pr ogr am [RRB SSE 2014 RED SH I FT]

12. __________wi ll t r ansl at e t he compl et e pr ogr am at once fr om a H i gh L evel L anguage t o t he M achine L anguage. (a) Compiler (b) Joy st i ck (c) Por t s (d) L i ght pen [RRB SSE 2014 RED SH I FT]

13. The wor d funct ion t hat cor r ect s t ext as we t ype is r efer r ed t o as : (a) Aut o inser t (b) Aut o cor r ect (c) Aut o summar i ze (d) Tr ack changes [RRB SSE 2014 RED SH I FT]

14

Basics of Computers and Applications

14. Pr imar y Stor age, in computer t er minology, r efer s to : (a) H ar d Disc Dr i ve (b) Random Access M emor y (RAM ) (c) Read Only M emor y (ROM )

15. What does an elect r onic spr eadsheet consist of ? (a) Rows (b) Columns (c) Cells (d) Al l of t he above [RRB SSE 2014 RED SH I FT]

(d) T he st or age devi ce wher e t h e oper at i n g syst em is st or ed [RRB SSE 2014 RED SH I FT]

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (a)

2. (a)

3. (a)

4. (b)

5. (d)

11. (c)

12. (c)

13. (d)

14. (b)

15. (d)

1. (a)

2. (a)

3. (c)

4. (b)

5. (c)

11. (d)

12. (a)

13. (c)

14. (d)

15. (a)

1. (d)

2. (c)

3. (a)

4. (c)

5. (d)

11. (c)

12. (a)

13. (b)

14. (b)

15. (d)

6. (d)

7. (c)

8. (b)

9. (a)

10. (d)

6. (c)

7. (b)

8. (d)

9. (a)

10. (d)

6. (c)

7. (b)

8. (b)

9. (b)

10. (c)

LEVEL-1

LEVEL-2

EXPLAN ATI ON S LEVEL-1 1. Opt i cal Char act er Reader i s ful l for m of OCR, which can r ead a char acter and conver t its bitmap image t o equival ent ASCI I codes. 2. I t i s equivalent t o 0.1 = 0 3. Ar it hmet i c L ogical Unit i s not an I nput devi ce whil e all t he ot her t hr ee ar e. 4. Fl oat i ng point number s ar e used i n comput er s t o r epr esent r eal number s. Si nce r eal number s can not be accur at ely r epr esent ed i n comput er s t hr ough binar y number s. 5. Pr ot ocols ar e r ules developed for exchanging and pr ocessing of dat a bet ween var i ous devices. Exampl es incl ude H TTP, I P, FTTP et c. 6. Tur i ng machi ne i s a mat hemat i cal model of a hypot het i cal comput ing machi ne which can use a pr edefi ned set of r ules t o det er mi ne a r esult fr om a set of input var iables. 7. The new st andar d which all ocat es codes t o almost al l languages and symbol s, t ot ali ng mor e t han a lakh i s cal led U ni code. I t mak es t r ansfer

8.

9.

10. 11. 12.

13.

and r euse of t r ansl at ed dat a et c. ver y easy. Al so it r epr esent s each char act er wi t h 16 bit s. To pr ot ect passwor ds et c. Encr ypt i on i s used which i s coding each dat a point i n a par t icul ar pat t er n whi ch is not easy t o decode. Open sour ce soft war es l i k e U ni x have t hei r sour ce code fr eely avai lable and t hese ar e developed t hr ough collabor at ion of coder s fr om acr oss t he wor ld. M any devel oper s as a pr inci ple use only open sour ce soft war es. Al l t he l anguages ment ioned her e ar e high l evel languages, in which i t i s easier t o wr it e code. 1 TB fl ash dr i ve can st or e appr oximat ely 1 mi lli on file sizes each of 1 M B. Al l t hese devi ces ar e r elat ed t o comput er s. CPU is Cent r al Pr ocessi ng Unit , whi le AL U is Ar it hmet i c and L ogi cal uni t , whil e I /O is I nput Out put devi ces. M H z is used t o measur e t he number of oper at i ons t hat can be done by t he CPU i n 1 second. So i t r efer s t o clock speed.

Basics of Computers and Applications

14. 1111 of binar y i s 24 – 1. I t i s 23 + 22 + 21 + 20 15. An oper at ing system (OS) is syst em soft war e that manages comput er har dwar e and soft war e r esour ces and pr ovides common ser vices for comput er pr ogr ams.

LEVEL-2 1.Power point is not an oper at ing syst em but an applicat ion t o make pr esent at ion slides. 2. When user s can see somet hing ver y similar t o end r esult while document cr eat ion, it is called WYSI WYG. 3. I t is t he cor r ect r epr esent at ion wher e user gives input in high level language, it is t hen compiled an d becom es m ach i n e l an gu age w h i ch i s comput ed and t he out put is similar ly pr ovided t o t he user. 4. The modem is an input and an out put device. I t is used for sending and r eceiving infor mation and dat a over t elephone lines. 5. L I NUX soft war e was invent ed by L inus Tor valds while st udying comput er science at Univer sity of H elsinki in 1991. 7. Pr ocessing speed is measur ed in MH z. Nowadays i t i s even measur ed i n GH z i .e. Gi ga H er t z. Basically it r epr esent s how many oper at ions can be pr ocessed in 1 second.

15

8. To close applications such as Powerpoint, MS-word et c. one must click t he close but t on on t he t it le bar which is at t he t op r ight hand cor ner. 9. ++I in C means i = i + 2. This was a way developed t o wr it e smaller lines in code. 10. != means not equal t o in C. 11. M icr osoft Windows is t he wor ld's most popular commer cial oper at ing syst em for PCs. 12. Com pi l er s ar e u sed t o t r an sl at e en t i r e pr ogr am s f r om H i gh l ev el l an gu age t o m ach i n e l an gu age, so t h at com pu t er can under st and and execut e it . 13. When we t ype t ext , Aut o cor r ect feat ur e cor r ect s t he wor d's spelling or any ot her por t ion of t he t ext . 14. Pr imar y st or age, also known as main st or age or memor y, is t he ar ea in a comput er in which dat a i s st or ed for qui ck access by t he comput er 's pr ocessor. The t er ms r andom access memor y (RAM ) and memor y ar e oft en used as synonyms for pr imar y or main st or age. 15. An elect r onic spr eadsheet like M S-Excel consist s of Cells which are ar r anged in Rows and Columns. So all ar e pr esent .

CBT – II

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

1

BASICS OF ENVIRONMENT AND POLLUTION CONTROL DEFINITION Environmental pollution is “the contamination of the physical and biological components of the earth/atmosphere system to such an extent that normal environmental processes are adversely affected”. Pollution is the introduction of contaminants into the environment that cause harm or discomfort  to humans or other living organisms, or that damage the environment” which can come “in the form of chemical substances, or energy such as noise, heat or light”. “Pollutants can be naturally occurring substances or energies, but are considered contaminants when in excess of natural levels.

TYPES OF POLLUTION: There are several types of pollution, and while they may come from different sources and have different consequences, understanding the basics about pollution can help environmentally conscious individuals minimize their contribution to these dangers.

AIR POLLUTION Air pollution is defined as any contamination of the atmosphere that disturbs the natural composition and chemistry of the air. This can be in the form of particulate matter such as dust or excessive gases like carbon dioxide or other vapors that cannot be effectively removed through natural cycles, such as the carbon cycle or the nitrogen cycle. Air pollution comes from a wide variety of sources. Some of the most excessive sources include:  Vehicle or manufacturing exhaust  Forest fires, volcanic eruptions, dry soil erosion, and other natural sources  Building construction or demolition Depending on the concentration of air pollutants, several effects can be noticed. Smog increases, higher rain acidity, crop depletion from inadequate oxygen, and higher rates of asthma. Many scientists believe that global warming is also related to increased air pollution.

Pollutant

Sources

Effects

Ozone. A gas that can be found in Ozone is not created directly, but is formed Ozone near the ground can cause a when nitrogen oxides and vo latile organic number of health prob lems. Ozone can two places. Near the ground (the compounds mix in sunlight. That is why ozone lead to more frequent asthma attacks in tropo sphere), it is a major part of is mostly found in the summer. Nitrogen oxides people who have asthma and can cause smog. The harmful ozo ne in the come from burning gasoline, coal, or other sore throats, coughs, and breathing fossil fuels. There are many types of volatile difficulty. It may even lead to premature lower atmosphere should not be organic compounds, and they come from death. Ozone can also hurt plants and confused with the protective layer sources ranging from factories to trees. crops. of ozone in the upper atmosphere (stratosphere), which screens out harmful ultraviolet rays Ca rbo n monoxide. A gas that comes from the burning of fossil fuels, mostly in cars. It cannot be seen or smelled

Carbon mono xide is released when engines Carbo n monoxide makes it hard for body burn fossil fuels. Emissions are higher when parts to get the oxygen they need to run engines are not tuned properly, and when fuel co rrectly. Exposure to carbon monoxid e is not completely burned. Cars emit a lot of the makes people feel dizzy and tired and carbon monoxide found outdoors. Furnaces gives them headaches. In high and heaters in the home can emit high co ncentrations it is fatal. Elderly people concentrations of carbo n monoxide, too, if they with heart disease are hospitalized more are not properly maintained. often when they are exposed to higher amounts of carbon monoxide.

2

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

Nitrogen dio xide.  A reddish- Nitrogen dioxide mostly comes from po wer High levels of nitrogen dioxide exposure brown gas that comes from the plants and cars. Nitrogen dioxide is formed in can give people coughs and can make burning of fossil fuels. It has a two ways-when nitrogen in the fuel is burned, them feel short of breath. People who are strong smell at high levels. or when nitrogen in the air reacts with oxygen exposed to nitrogen dioxide for a long at very high temperatures. Nitrogen dioxide time have a higher chance of getting can also react in the atmosphere to form ozone, respiratory infections. Nitro gen dioxid e acid rain, and particles. reacts in the atmosphere to form acid rain, which can harm plants and animals. Particulate matter. Solid or Particulate matter can be divided into two Particulate matter that is small enough liquid matter that is suspended in types-coarse particles and fine particles. can enter the lungs and cause health the air. To remain in the air, Coarse particles are formed from sources like prob lems. Some of these problems particles usually must be less than road dust, sea spray, and construction. Fine include more frequent asthma attacks, 0.1-mm wide and can be as small particles are formed when fuel is burned in respiratory problems, and premature automobiles and power plants. death. as 0.00005 mm. Sulphur dio xide.  A corrosive gas Sulfur dioxide mostly comes from the burning Sulfur d ioxide exposure can affect that cannot be seen or smelled at of coal or oil in power plants. It also comes people who have asthma or emphysema low levels but can have a “rotten from factories that make chemicals, paper, or by making it more difficult for them to egg“ smell at high levels. fuel. Like nitrogen dio xide, sulfur dioxide breathe. It can also irritate people's eyes, reacts in the atmosp here to form acid rain and noses, and throats. Sulfur dioxide can particles. harm trees and crops, damage b uildings, and make it harder for people to see long distances. Lead. A blue-gray metal that is Outside, lead comes from cars in areas where High amounts of lead can be dangerous very toxic and is found in a unleaded gasoline is not used. Lead can also for small children and can lead to lower number of forms and locations. come from power plants and other industrial IQs and kidney prob lems. For adults, sources. Inside, lead paint is an important exposure to lead can increase the chance source of lead, especially in houses where paint of having heart attacks or strokes. is peeling. Lead in old pipes can also be a source of lead in drinking water. Toxic air pollutants. A large number of chemicals that are known or suspected to cause cancer. Some imp ortant pollutants in this category include arsenic, asbestos, benzene, and dioxin.

Each toxic air pollutant comes from a slightly Toxic air pollutants can cause cancer. different source, but many are created in Some toxic air pollutants can also cause chemical plants or are emitted when fossil fuels birth defects. Other effects depend on the are burned. Some toxic air pollutants, like pollutant, but can include skin and eye asbestos and formaldehyde, can be found in irritation and breathing problems. building materials and can lead to indoor air problems. Many toxic air pollutants can also enter the food and water supplies.

Stratospheric ozone CFCs are used in air conditioners and If the ozone in the stratosphere is depleters.Chemicals that can refrigerators, since they work well as coolants. destro yed, people are exposed to more destroy the ozone in the They can also be fo und in aerosol cans and fire radiation fro m the sun (ultraviolet stratosphere. These chemicals extinguishers. Other stratospheric ozone radiation). This can lead to skin cancer include chlorofluorocarbons depleters are used as so lvents in industry. and eye problems. Higher ultraviolet (CFCs), halons, and other radiation can also harm plants and compounds that include chlorine animals. or bromine

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

Greenhouse gases. Gases that stay in the air for a long time and warm up the planet by trapping sunlight. This is called the “greenhouse effect“ b ecause the gases act like the glass in a greenhouse. Some of the important greenhouse gases are carbon dioxide, methane, and nitrous oxide.

3

Carbon dioxide is the most important The greenhouse effect can lead to greenhouse gas. It comes from the burning of changes in the climate of the planet. fossil fuels in cars, p ower plants, houses, and Some of these changes might includ e industry. Methane is released during the more temperature extremes, higher sea processing of fossil fuels, and also comes from levels, changes in forest compositio n, natural sources like cows and rice paddies. and damage to land near the coast. Nitrous oxide comes from ind ustrial sources Human health might be affected by and decaying plants. diseases that are related to temperature or by damage to land and water.

WATER POLLUTION Water pollution involves any contaminated water, whether from chemical, particulate, or bacterial matter that degrades the water’s quality and purity. Water pollution can occur in oceans, rivers, lakes, and underground reservoirs, and as different water sources flow together through the water cycle the pollution can spread. Causes of water pollution include:  Increased sediment from soil erosion  Improper waste disposal and littering  Leaching of soil pollution into water supplies  Organic material decay in water supplies The effects of water pollution include decreasing the quantity of drinkable water available, lowering water supplies for crop irrigation, and impacting fish and wildlife populations that require water of a certain purity for survival. Ground water is being polluted by percolation of contaminated surface water through the layers of the earth. Release of raw sewage in unlined soak-pits and release of toxic effluents by the industries into surface water bodies, are the main causes of ground water pollution.

Major water pollutants, examples and sources Category Sources Examples I. Affecting health Infectious agents Bacteria, viruses and parasites Sewage, human and animal excreta Organic chemicals Pesticides, plastics, detergents. oil Agricultural, industrial and domestic wastes Inorganic chemicals Acids, caustics, salts, metals Industrial and domestic effluents Radioactive materials Uranium, thorium, randon, etc Mining, power plants, natural sources 2. Affecting ecossslcm Plant nutrients Nitrates, phosphates, etc Chemical fertilisers, sewage, manure Sediments Silt, soil Soil erosion Thermal Heat Industries, power plants Oxygen demanding Agricultural waster, manure Sweage, agricultural runoff Indiscriminate and overuse of fertilizers, chemicals and pesticides have also caused ground water pollution through the seepage of irrigation water into ground water reserves. The hazards of ground water pollution depend on several factors such as:  Concentration or toxicity of the pollutant  The level of ground water if the level is higher chances of contamination are more  Conditions of ground water recharge

Marine Pollution: Marine pollution is the matter of International concern from the point of view of conservation of living resources. All coastal nations dispose of millions of gallons of untreated sewage, millions of tonnes of garbage, unlimited amount of low level radioactive wastes etc. into the seas. In addition to the marine environment, areas along the coasts, such as, estuaries, reefs, wetlands, mangroves, etc. are adversely affected due to enormous dumping of pollutants into the ocean. This problem is further aggravated due to the fact that about 40% of the world’s population lives near the sea.

4

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

The main sources of marine pollution are: 1. Municipal wastes and sewage 2. Industrial effluents 3. Runoff agricultural wastes 4. Oil spills from tankers 5. Offshore drilling and mining 6. Submarine nuclear testing 7. Dumping of radioactive wastes

The consequences of marine pollution are as follows: 

The pollutants adversely affect the productive ocean regions, thus causing huge losses of fish populations and coral reefs. This results in economic losses amounting to billions of dollars per year.



Eutrophication, due to the influx of organic pollutants, results in the formation of red tides. These are blooms (massive growth) of red algae, which inhibit the movement of ships and also kill marine fauna.



Dumping of huge amounts of toxic wastes in a short duration of time, creates areas of oxygen-depleted zones in the coastal waters. In these zones, most of the aquatic lives die or migrate elsewhere.



Discarded garbage, sewage, plastic refuse, etc. that are dumped in the oceans sometimes accumulate in the beaches. This spoils the aesthetic beauty of the region and results in loss of tourism.

Water Pollutants and their Effects: Most of the rivers and fresh water streams in India are badly polluted by industrial wastes or effluents. The major sources of pollution of some Indian rivers are listed in table below:

Name of river 1. Kali 2. Yamuna 3. Ganga 4. Gomti 5. Dajora 6. Damodar 7. Hoogly

8. Sone Bhadra 9. Cooum, Adyar and Buckinghum canal (Chennai) 10. Kaveri 11. Godavari 12. Siwan 13. Kulu 14. Suwao

Indian rivers and sources of their pollution Sources of pollution Sugar mills: distilleries: paint, soap, rayon, silk. Yarn, tin and glycerine D.D.T. factory, sewage, Indraprastha Power Station, Delhi. Jute, chemical, metal and surgical industries: tanneries, textile mills and great bulk of domestic sewage of highly organic nature. Paper and pulp mills sewage. Synthetic rubber factories. Fertilizers, fly ash from steel mills, suspended coal particles from washeries, and thermal power station. Power stations: paper pulp, jute textiles, chemical mills, paint, varnishes, metal, steel, hydrogenated vegetable oil, rayon, soap, match, shellac, and polyethene industries and sewage. Cement, pulp and paper mills. Domestic sewage, automobile workshops. Sewage, tanneries, distilleries, paper and rayon mills. Paper, mills. Paper, sulphur, cement, and sugar mills. Chemical factories, rayon mills and tanneries. Sugar industries.

Contamination of water with industrial wastes is most dangerous. The sewage of big cities is often drained into rivers. This sewage promotes the growth of phytoplankton’s. The excessive growth depletes the oxygen of water. This reduction of oxygen and the presence of poisonous wastes affect the fish population. Besides these, rivers, lakes and ponds are also used directly by people for bathing or washing. This contaminates the water with the germs of various diseases- like cholera, dysentery and hepatitis.

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

Type or Industry Inorganic pollutants Mining Mine wastes : Chlorides, various metals, ferrous sulphate, sulpluric acid hydrogen sulphide, ferric hydroxide, surface wash offs, suspended solids, chlorides and heavy metals. Iron and steel Suspended solids, iron cyanide, thiocy-anate, sulphides, oxides, of copper, chromium, cadmium, and mercury. Chemical plants Various acids and alkalies, sulphates, nitrates of metals, phosphorus, fluorine, silica, and suspended particles. Pharmaceuticals —

5

Organic pollutants — Oil, phenol and naptha

Aromatic compound solvents, organic acids, nitro compound dyes, etc.

Proteins, carbohydrates, organic solvents, intermediate products, drugs and antibiotics. Soap and detergent Tertiary ammonia compounds, alkalies. Fats and fatty acids, glycerol, polyphosphates, sulphonated hydrocarbons. Food processing Highly putrescible organic matter and — pathogens. Paper and pulp Sulphides, bleaching liquors Cellulose fibres, bark, wood, sugars and organic acids. Some pollutants produce only temporary effects in water whereas others have long standing effects. There are several types of physical and chemical effects produced by pollutants. These are:  Addition of poisonous substances  Addition of suspended particles  Addition of non-toxic salts  Water de-oxygenation  Heating of water.

SOIL POLLUTION Soil, or land pollution, is contamination of the soil that prevents natural growth and balance in the land whether it is used for cultivation, habitation, or a wildlife preserve. Some soil pollution, such as the creation of landfills, is deliberate, while much more is accidental and can have widespread effects. Soil pollution sources include:  Hazardous waste and sewage spills  Non-sustainable farming practices, such as the heavy use of inorganic pesticides  Strip mining, deforestation, and other destructive practices  Household dumping and littering Soil contamination can lead to poor growth and reduced crop yields, loss of wildlife habitat, water and visual pollution, soil erosion, and desertification.

NOISE POLLUTION Noise pollution refers to undesirable levels of noises caused by human activity that disrupt the standard of living in the affected area. Noise pollution can come from:  Traffic  Airports  Railroads  Manufacturing plants  Construction or demolition  Concerts

6

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

Some noise pollution may be temporary while other sources are more permanent. Effects may include hearing loss, wildlife disturbances, and a general degradation of lifestyle.

RADIOACTIVE POLLUTION Radioactive pollution is rare but extremely detrimental, and even deadly, when it occurs. Because of its intensity and the difficulty of reversing damage, there are strict government regulations to control radioactive pollution. Sources of radioactive contamination include: 

Nuclear power plant accidents or leakage



Improper nuclear waste disposal



Uranium mining operations

Radiation pollution can cause birth defects, cancer, sterilization, and other health problems for human and wildlife populations. It can also sterilize the soil and contribute to water and air pollution.

THERMAL POLLUTION Thermal pollution is excess heat that creates undesirable effects over long periods of time. The earth has a natural thermal cycle, but excessive temperature increases can be considered a rare type of pollution with long term effects. Many types of thermal pollution are confined to areas near their source, but multiple sources can have wider impacts over a greater geographic area. Thermal pollution may be caused by: 

Power plants



Urban sprawl



Air pollution particulates that trap heat



Deforestation



Loss of temperature moderating water supplies

As temperatures increase, mild climatic changes may be observed, and wildlife populations may be unable to recover from swift changes.

LIGHT POLLUTION Light pollution is the over illumination of an area that is considered obtrusive. Sources include: 

Large cities



Billboards and advertising



Night time sporting events and other night time entertainment

Light pollution makes it impossible to see stars, therefore interfering with astronomical observation and personal enjoyment. If it is near residential areas, light pollution can also degrade the quality of life for residents.

ENVIRONMENTAL PERFORMANCE INDEX 2018 In news 2018: The 2018 Environmental Performance Index (EPI) finds that air quality is the leading environmental threat to public health. Now in its twentieth year, the biennial report is produced by researchers at Yale and Columbia Universities in collaboration with the World Economic Forum. The tenth EPI report ranks 180 countries on 24 performance indicators across 10 issue categories covering environmental health and ecosystem vitality. Switzerland leads the world in sustainability, followed by France, Denmark, Malta, and Sweden.

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

7

Key findings: 

Switzerland’s top ranking reflects strong performance across most issues, especially air quality and climate protection.



In general, high scorers exhibit long-standing commitments to protecting public health, preserving natural resources, and decoupling greenhouse gas (GHG) emissions from economic activity.



India and Bangladesh come in near the bottom of the rankings, with Burundi, Democratic Republic of the Congo, and Nepal rounding out the bottom five.



Low scores on the EPI are indicative of the need for national sustainability efforts on a number of fronts, especially cleaning up air quality, protecting biodiversity, and reducing GHG emissions.



Some of the lowest-ranking nations face broader challenges, such as civil unrest, but the low scores for others can be attributed to weak governance, they note.

EPI and Global Sustainability Data The EPI builds on the best available global data from international research entities, such as the Institute for Health Metrics and Evaluation, the World Resources Institute, and the Sea Around Us Project at the University of British Columbia, as well as international organizations such as the World Bank and the UN Food and Agriculture Organization. Nevertheless, serious data gaps limit the ability to measure results – and particularly changes in performance – on a number of important issues. “As the EPI project has highlighted for two decades, better data collection, reporting, and verification across a range of environmental issues are urgently needed,”. The world needs better data on sustainable agriculture, water resources, waste management, and threats to biodiversity. Supporting global data systems is one of the most important steps the world community can take to achieving sustainable development goals.

DUST MITIGATION PLAN Centre had notified dust mitigation norms.

The norms mandate that: 

No building or infrastructure project requiring Environmental Clearance shall be implemented without approved Environmental Management Plan inclusive of dust mitigation measures.



Roads leading to or at construction sites must be paved and blacktopped (i.e. metallic roads).



No excavation of soil shall be carried out without adequate dust mitigation measures in place.



No loose soil or sand or Construction & Demolition Waste or any other construction material that causes dust shall be left uncovered,



Wind-breaker of appropriate height i.e. 1/3rd of the building height and maximum up to 10 meters shall be provided.



Water sprinkling system shall be put in place.



Dust mitigation measures shall be displayed prominently at the construction site for easy public viewing.

How it works? (Steps taken) The teams are empowered to take on-the-spot action against violators and if necessary, issue “stop-work” orders. The campaign will also include enforcement of pollution-control measures for vehicles, driving discipline, inspection of power plants in Delhi to ensure compliance with the norms on pollution. Besides field surveys by empowered teams of officials, a series of seminars on mitigation of pollution will also be organised during the period. These include - a workshop on Environmental and Health; Air Pollution Abatement Technologies; enlisting support from NGOs, Civil Society, citizens; Clean Air Day in Universities, Colleges and Schools; a Mini Marathon for Clean Air; enhancing the role of PSUs and industries, apex industrial bodies; launching a national digital forum for discussions on air pollution; Indoor Air Pollution Management and a conference of Environment Ministers of States and Union Territories.

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BASICS OF ENVIRONMENT AND POLLUTION CONTROL

MINAMATA CONVENTION In news Recently, the Union Cabinet approved the proposal for ratification of Minamata Convention on Mercury enabling India to become a Party of the Convention.

About 

The approval entails ratification of the Minamata Convention on Mercury along with flexibility for continued use of mercury based products and processes involving mercury compound up to 2025.



The first Conference of the Parties (CoP) under the Minamata Convention took place in Geneva, Switzerland in 2017 which India attended as observer.

It is financed through Global Environment Facility.

Details about the convention 

The Minamata Convention on Mercury is first global legally binding treaty to protect human health and the environment from the adverse effects of mercury.



It was agreed in Geneva, Switzerland in January 2013 and came into force in August, 2017.

The Minamata Convention has put party nations to: 

Reduce and eliminate the use and release of mercury from artisanal and small-scale gold mining (ASGM).



Control mercury air emissions from coalfired power plants, coal-fired industrial boilers, certain non-ferrous metals production operations, waste incineration and cement production.



Phase-out or take measures to reduce mercury use in certain products such as batteries, switches, lights, cosmetics, pesticides and measuring devices, and create initiatives to reduce the use of mercury in dental amalgam.



Phase out or reduce the use of mercury in manufacturing processes such as chloralkali production, vinyl chloride monomer production, and acetaldehyde production.



It also puts a ban on new mercury mines.



The Convention also addresses interim storage of mercury and its disposal once it becomes waste, sites contaminated by mercury as well as health issues.

Waste management Waste management involves collecting, transporting, disposing, recycling and monitoring waste generated through human activities. General waste management techniques are: Landfill: It involves having the waste buried off in empty, deserted locations outside the city. Dumped waste is made to undergo compression to enhance the density and make the fill stable. It is later covered to discourage vermin growth. A gas extraction system is customarily installed to exact the gas (arising out of decomposition) through a burrow pit. Incineration: Waste is exposed to high temperature to trigger combustion and ultimately reduce to ash, gas and heat energy. Toxic wastes from industry are thermally treated in furnace and boiler to extract energy. This method is useful where land is scarce. Gasification and Pyrolysis methods involve heating waste in short supply of oxygen at high temperature inside a pressurized and sealed vessel. The resultant residue is used for energy generation. Recycling: Paper, plastic, PVC and other homogenous products can be recycled to put them in use in a new garb. This also rids the environment of non-biodegradable, chemical wastes that significantly disturb the ecological balance. Biological reprocessing: Wastes of organic origin are made to undergo biological decomposition and re-used as compost or mulch for agriculture and landscaping. Gas collected is used for electricity generation. Waste Reduction and Avoidance: The stress is on increased use of second hand products, repaired products and reducing the use of complex disposable items to keep a tab on waste generation in abundance.

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

9

Recycling Solutions: Recycling is a superlative way to capitalize on accumulated waste by chemically treating it to make it fit for re-use. Recycling equipment make the waste processing method streamlined and cost-effective.

Global warming Global warming which is also referred to as climate change, is the observed rise in the average temperature of the Earth's climate system the global surface temperature is likely to rise a further 0.3 to 1.7 °C in the lowest emissions scenario, and 2.6 to 4.8 °C in the highest emissions scenario .These readings have been recorded by the “national science academies of the major industrialized nations”. Future climate change and impacts will differ from region to region. Expected effects include increase in global temperatures, rising sea levels, changing precipitation, and expansion of deserts. Causes: Global warming is a serious environmental issues. The causes are divided into two categories include "natural" and "human influences" of global warming. Natural Causes of Global Warming:  rotation of the sun that changes the intensity of sunlight and moving closer to the earth  greenhouse gases  Volcanic eruption. Human Influences on Global Warming:  industrial revolution  Mining  Deforestation Effects:  heat waves,  droughts,  heavy rainfall with floods,  heavy snowfall ,  ocean acidification,  species extinctions due to shifting temperature regimes

Acid rain Acid rain, or acid deposition, is a broad term that includes any form of precipitation with acidic components, such as sulfuric or nitric acid that fall to the ground from the atmosphere in wet or dry forms. This can include rain, snow, fog, hail or even dust that is acidic. Causes of Acid Rain : This image illustrates the pathway for acid rain in our environment.Acid rain results when sulfur dioxide (SO2) and nitrogen oxides (NOX) are emitted into the atmosphere and transported by wind and air currents. The SO2 and NOX react with water, oxygen and other chemicals to form sulfuric and nitric acids. These then mix with water and other materials before falling to the ground. While a small portion of the SO2 and NOX that cause acid rain is from natural sources such as volcanoes, most of it comes from the burning of fossil fuels. The major sources of SO2 and NOX in the atmosphere are:  Burning of fossil fuels to generate electricity. Two thirds of SO2 and one fourth of NOX in the atmosphere come from electric power generators.  Vehicles and heavy equipment.  Manufacturing, oil refineries and other industries. Winds can blow SO2 and NOX over long distances and across borders making acid rain a problem for everyone and not just those who live close to these sources.

Ozone depletion Ozone depletion, gradual thinning of Earth’s ozone layer in the upper atmosphere caused by the release of chemical compounds containing gaseous chlorine or bromine from industry and other human activities. The thinning is most pronounced in the polar regions, especially over Antarctica. Ozone depletion is a major environmental problem because it increases the amount of ultraviolet (UV) radiation that reaches Earth’s surface, which increases the rate of

10

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

skin cancer, eye cataracts, and genetic and immune system damage. The Montreal Protocol, ratified in 1987, was the first of several comprehensive international agreements enacted to halt the production and use of ozone-depleting chemicals. As a result of continued international cooperation on this issue, the ozone layer is expected to recover over time.

Important terminologies: ·

Garbage Pollution: Mismanagement of solid waste by households, waste collectors and waste disposal contractors.

·

Plastic Pollution: Waste of all types of non-biodegradable plastic of both hard and soft material.

· Pollution by Hospitals: Mismanagement of all types of waste generated by the hospitals instead of its environment friendly disposal. ·

Indoor Pollution: Kitchen emissions, smoking in home, loud music, spillage of sewerage.

·

Industrial Pollution: Smoke from chimney, waste and effluent from manufacturing process in factories.

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

11

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. Consider the following:

2.

3.

4.

5.

1. Carbon dioxide 2. Oxides of Nitrogen 3. Oxides of Sulphur Which of the above is/are the emission/ emissions from coal combustion at thermal power plants? (a) 1 only (b) 2 and 3 only (c) 1 and 3 only (d) 1, 2 and 3 Human activities in the recent past have Caused the increased concentration of carbon dioxide in the atmosphere, but a lot of it does not remain in the lower atmosphere because of : 1. Its escape into the outer stratosphere. 2. The photosynthesis by phyto-plankton in the oceans. 3. The trapping of air in the polar ice caps. Which of the statements given above is/are correct? (a) 1 and 2 (b) 2 only (c) 2 and 3 (d) 3 only Consider the following statements: Chlorofluorocarbons, known as ozone-depleting substances, are used 1. In the production of plastic foams 2. In the production of tubeless tyres 3. In cleaning certain electronic components 4. As pressurizing agents in aerosol cans Which of the statements given above is/are correct? (a) 1, 2 and 3 only (b) 4 only (c) 1, 3 and 4 only (d) 1, 2, 3 and 4 Acid rain is caused by the pollution of environment by (a) carbon dioxide and nitrogen (b) carbon monoxide and carbon dioxide (c) ozone and carbon dioxide (d) nitrous oxide and sulphur dioxide Photochemical smog is a resultant of the reaction among: (a) NO2, 03 and peroxyacetyl nitrate in the presence of sunlight (b) CO, 02 and0 peroxyacetyl nitrate in the presence of sunlight (c) CO, CO2 and N02 at low temperature (d) High concentration of N02, O3 and CO in the evening

6. There is some concern regarding the nanoparticles of some chemical elements that are used by the industry in the manufacture of various products. Why?

1. They can accumulate in the environment, and contaminate water and soil. 2. They can enter the food chains. 3. They can trigger the production of free radicals. Select the correct answer using the code given below. (a) 1 and 2 only (b) 3 only (c) 1 and 3 only (d) 1, 2 and 3 7. Which of the following are some important pollutants released by steel industry in India? 1. Oxides of sulphur 2. Oxides of nitrogen 3. Carbon monoxide 4. Carbon dioxide Select the correct answer using the code given below. (a) 1, 3 and 4 only (b) 2 and 3 only (c) 1 and 4 only (d) 1,2, 3 and 4 8. Brominated flame retardants are used in many household products like mattresses and upholstery. Why is there some concern about their use? 1. They are highly resistant to degradation in the environment. 2. They are able to accumulate in humans and animals. Select the correct answer using the code given below. (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 9. The scientific view is that the increase in global temperature should not exceed 2 °C above preindustrial level. If the global temperature increases beyond 3°C above the pre-industrial level, what can be its possible impact/impacts on the world? 1. Terrestrial biosphere tends toward a net carbon source 2. Widespread coral mortality will occur. 3. All the global wetlands will permanently disappear. 4. Cultivation of cereals will not be possible anywhere in the world. Select the correct answer using the code given below. (a) 1 only (b) 1 and 2 only (c) 2, 3 and 4 only (d) 1, 2, 3 and 4,

12

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

10. In the context of mitigating the impending global warming due to anthropogenic emissions of carbon dioxide, which of the following can be the potential sites for carbon sequestration?

(c) M axi mum acid i s due t o st r ong Car bonic Acid (d) Acid r ain affect s ecosyst em [RRB SSE 2014 YEL L OW SH I FT ]

15. The wor king pr inciple of tur bidimeter is based on

1. Abandoned and uneconomic coal seams

(a) r efl ect ion of l ight

2. Depleted oil and gas reservoirs

(b) r efr act ion of l ight

3. Subterranean deep saline formations

(c) scat t er ing of l ight (d) adsor pt ion of light [RRB SSE 2015 1 st SEP 1 st SH I FT ]

Select the correct answer using the code given below. (a) 1 and 2 only

(b) 3 only

(c) 1 and 3 only

(d) 1, 2 and 3

11. Biological Oxygen Demand (BOD) is a standard criterion for

LEVEL-1 1. The major sour ce of car ci nogenic hydr ocar bon, benzo () pyr ene pr esent i n ur ban at mospher e is (a) const r uct ion act ivit i es (b) r oad t r affic

(a) Measuring oxygen levels in blood

(c) bur st i ng of cr acker s

(b) Computing oxygen levels in forest ecosystems

(d) domest i c bur ni ng

(c) Pollution assay in aquatic ecosystems (d) Assessing oxygen levels in high altitude regions 12. In the context of solving pollution problems, what is/ are the advantage/advantages of bioremediation technique?

1. It is a technique for cleaning up pollution by enhancing the same biodegradation process that occurs in nature. 2. Any contaminant with heavy metals such as cadmium and lead can be readily and completely treated by bioremediation using microorganisms. 3. Genetic engineering can be used to create microorganisms specifically designed for bioremediation. Select the correct answer using the code given below:

[RRB SSE 2015 1 st SEP 1 st SH I FT ]

2. The pr escr i bed per missible noise level , L eq for commer cial ar ea at day t ime is (a) 75 dBA

(b) 50 dBA

(c) 55 dBA

(d) 65 dBA [RRB SSE 2015 1 st SEP 1 st SH I FT ]

3. The gl obal war mi ng i s caused by gr een house gases, which ar e (a) CO, N 2O, CH 4 and CFC (b) CO2, NO2, CH 4 and H 2O (c) CO2, N 2O, CH 4 and H 2O (d) CO2, N O2, CH 4 and CFC [RRB SSE 2015 1 st SEP 1 st SH I FT ]

4. Which of t he fol lowing r ol es fl y ash does not pl ay in concr et e (a) I mpr ovi ng t he wor kabi li t y (b) Acceler at ing t he st r engt h gain

(a) 1 only

(b) 2 and 3 only

(c) Delaying t he set t ing t i me of concr et e

(c) 1 and 3 only

(d) 1, 2 and 3

(d) H elps in long-t er m st r engt h gain

13. I n potable water, the dissolved oxygen is stipulated as-

[RRB SSE 2015 1 st SEP 2 nd SH I FT ]

5. One t ur bi di t y unit NTU is equal t o

(a) <6g/l

(b) >6g/l

(a) 1.0 mg/l far mazin (b) 1.0 meq/l Si O2

(c) <6mg/l

(d) >6mg/l

(c) 1.0 mg/l Si O2

(d) 1.0 meq/l kaol in

[RRB SSE 2014 GREEN SH I FT ]

[RRB SSE 2015 1 st SEP 2 nd SH I FT ]

14. I n r ef er en ce t o A ci d r ai n , w h at i s cor r ect st at ement

6. The pr escr i bed per missible noise level , L eq for r esi dent i al ar ea at day t ime is

(a) The pH value i s bel ow 5.6 (b) I t occur s due t o pr esence of sul phur i c aci d or ni t r ic acid i n t he at mospher e

(a) 65 dBA

(b) 45 dBA

(c) 50 dBA

(d) 55 dBA [RRB SSE 2015 1 st SEP 2 nd SH I FT ]

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

7. W h i ch of t h e f ol l ow i n g i s n ot u sed as a suppl ement ar y cement at ions mat er ial? (a) Fl y ash

(b) Gypsum

(c) Ri ce husk ash

(d) Si li ca fume

8. Accor di ng t o I S 456, if t he maximum aggr egat e si ze i s i ncr eased fr om 20 mm t o 40 mm, t he mi ni mum cement cont ent r equi r ement changes (i n kg/cum) by (b) 20

(c) – 30

(d) 30

(c) gast r oint est inal pr oblem (d) t he ir r it at i on in alveol i of t he l ungs [RRB SSE 2015 2 nd SEP 2 nd SH I FT ]

[RRB SSE 2015 1 st SEP 3 rd SH I FT ]

(a) – 20

13

15. The machi ne ‘A' and machine ‘B' pr oduce equal noise l evel s, i.e., 60 dBA each. The summat ion of t hese t wo noise level s is (a) 100 dBA

(b) 66 dBA

(c) 63 dBA

(d) 55 dBA [RRB SSE 2015 2 nd SEP 2 nd SH I FT ]

LEVEL-2

[RRB SSE 2015 1 st SEP 3 rd SH I FT ]

9. The t ur bi di t y in sur face wat er is due t o pr esence of (a) dissolved or ganics

1. I n st r at ospher e, t he t emper at ur e incr eases wi t h al t i t ude due t o pr esence of (a) r adicals

(b) chlor ofluor ocar bons

(c) HCFCs

(d) Ozone

(b) col loidal mat er ial

[RRB SSE 2015 2 nd SEP 2 nd SH I FT ]

(c) di ssolved i n or gani cs

2. N uclear densi t y guage can be used for al l t he foll owi ng pur poses, except

(d) di ssolved color s [RRB SSE 2015 1 st SEP 3 rd SH I FT ]

10. T h e pr escr i bed per m i ssi bl e n oi se l evel , f or r esi dent i al ar ea at night t ime is (a) 45 dBA

(b) 50 dBA

(c) 40 dBA

(d) 55 dBA

(a) M oist ur e cont ent (b) Wet densi t y (c) Dr y densi t y (d) St andar d penetr at ion r eading

[RRB SSE 2015 1 st SEP 3 rd SH I FT ]

11. The cont i nuous exposur e of hi gh concent r at i on of r epar able suspended par t icul at e mat t er may cause

[RRB SSE 2015 2 nd SEP 3 rd SH I FT ]

3. A wat er bor ne di sease pol iomyeli t is i s caused by (a) vir uses

(b) pr ot ozoa

(c) bact er ia

(d) hel mi nt hes

(a) eye ir r it at i on

[RRB SSE 2015 2 nd SEP 3 rd SH I FT ]

(b) kidney damage

4. I n pot abl e wat er, t he per missible li mi t of nit r at e ni t r ogen is

(c) fail ur e of r espi r at or y syst em (d) car diac disease [RRB SSE 2015 2

nd

SEP 1 SH I FT ] st

12. The pr escr i bed per missible noise level , L eq for commer cial ar ea at ni ght t ime is (a) 45 dBA

(b) 65 dBA

(c) 50 dBA

(d) 55 dBA [RRB SSE 2015 2 nd SEP 1 st SH I FT ]

13. The pH of acid r ai n shoul d always be less t han (a) 5.6 even aft er pr ecipit at i on (b) 7.0 aft er pr ecipit at ion (c) 6.5 aft er pr ecipit at ion (d) 4.2 aft er pr ecipit at ion [RRB SSE 2015 2 nd SEP 1 st SH I FT ]

14. The exposur e of gaseous pollut ant sulphur dioxide may cause (a) br onchi t is and pul monar y emphysema (b) lungs fai lur e and k idney damage

(a) 10 mg/l

(b) 25 mg/l

(c) 40 mg/l

(d) 15 mg/l [RRB SSE 2015 2 nd SEP 3 rd SH I FT ]

5. Car bon monoxi de for ms car boxyhemogl obi n i n human blood t hat may cause (a) incr eased oxygen car r yi ng capaci t y (b) decr eased oxygen car r ying capaci t y (c) damage in cent r al ner vous syst em (d) damage in ci r cul at or y syst em [RRB SSE 2015 2 nd SEP 3 rd SH I FT ]

6. Tw o m ach i n es ar e w or k i n g i n a n oi sy envi r onment and joint ly pr oduct 55 dBA noi se level. I f t he envir onment al noi se level i s also 55 dBA, t he summat i on of noise level s is (a) 110 dBA

(b) 56 dBA

(c) 55 dBA

(d) 58 dBA [RRB SSE 2015 2 nd SEP 3 rd SH I FT ]

14

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

(b) car bon dioxide, sulphur dioxide, water vapour s and chl or ofl uor ocar bons

7. The aver age concent r at i on of ozone pr esent i n t he st r at ospher e is appr oximat ely (a) 5 ppm

(b) 0.05 ppm

(c) 10 ppm

(d) 15 ppm

(c) car bon monoxide, nit r ous oxide, met hane and, hydr o-chlor ofluor ocar bons (d) car bon di ox i de, n i t r ogen di ox i de, w at er vapour s , met hane and ozone

[RRB SSE 2015 2 nd SEP 3 rd SH I FT ]

8. The t ot al col ifor m bact er ia ar e r epor t ed as most pr obabl e number (M PN) per (a) 10 ml of wat er (b) 1000 ml of wat er

[RRB SSE 2015 3 rd SEP 1 st SH I FT ]

12. The desir able amount of fluor i de ions i n pot able wat er s for opt imal dent al heal t h i s: (a) 1.5 mg/l

(c) 100 ml of wat er

(b) 1.0 mg/l

(c) 0.5 mg/l

(d) 1ml of wat er

(d) 0.05 mg/l [RRB SSE 2015 3 rd SEP 2 nd SH I FT ]

[RRB SSE 2015 3 rd SEP 1 st SH I FT ]

9. The ant hr opogenic sour ces of air pollut ion in well pl anned ci t y is

13. Wh i ch of t he f ol l owi ng i s n ot con si der ed as secondar y pollut ant ? (a) Phot ochemical smog

(a) constr uction activities, r oad tr affics, r ail tr affic, fugi t ive emissions

(b) Per oxy acet yl nit r at e (c) Acid mi st

(b) const r uct ion act ivi t i es, r oad t r affi c, domest ic burning (c) const r uct ion act ivit ies, r oad t r affi cs, bur st ing of cr acker s, dust st or ms (d) const r uct ion act ivit ies, r oad t r affics, domest ic bur ni ng, i ndust r ial emissions

(d) Car bon monoxi de [RRB SSE 2015 3 rd SEP 2 nd SH I FT ]

14. I n t he st at ist ical di st r ibut ion of noi se levels, t he back gr ound noise l evel is r epr esent ed by:

[RRB SSE 2015 3 rd SEP 1 st SH I FT ]

10. When t he measur ed and st andar d r efer ence pr essur e level becomes equal, t he sound pr essur e level (SPL ) is equi valent t o (a) 1 dBA

(b) 10 dBA

(c) 0 dBA

(d) 1.012 dBA

(a) L 90

(b) L 50

(c) L 10

(d) L 1 [RRB SSE 2015 3 rd SEP 2 nd SH I FT ]

15. Acid r ain i s caused due t o for mat ion of: (a) su l ph u r i c aci d an d car bon i c aci d i n t h e at mospher e (b) sulphur ic acid and nitric acid in the atmosphere

SEP 1 SH I FT ]

(c) nitr ic acid and car bonic acid in the atmospher e

11. The major gr een house gases cont r i but i ng i n gl obal war ming ar e

(d) sulphur ic acid, nit r ic acid and car bonic acid in t he at mospher e

(a) car bon di oxide, nit r ous oxide, met hane and wat er vapour s

[RRB SSE 2015 3 rd SEP 2 nd SH I FT ]

[RRB SSE 2015 3

rd

st

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (d) 11. (c)

2. (c) 12. (d)

3. (c) 13. (a)

4. (d) 14. (b)

5. (a) 15. (c)

6. (d)

7. (d)

8. (c)

9. (b)

10. (d)

7. (b)

8. (c)

9. (b)

10. (a)

7. (c)

8. (c)

9. (b)

10. (c)

LEVEL-1 1. (b) 11. (c)

2. (d) 12. (d)

3. (c) 13. (a)

4. (b) 14. (a)

5. (c) 15. (c)

6. (d)

LEVEL-2 1. (d) 11. (a)

2. (d) 12. (b)

3. (a) 13. (d)

4. (a) 14. (a)

5. (b) 15. (b)

6. (d)

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

15

EXPLAN ATI ON S LEVEL-1 1. Road traffic is the major source of carcinogenic hydrocarbon present in urban atmosphere. 2. The prescribed permissible noise level, Leq for commercial area at day time is 65 dBA. According to Central Pollution Control Board,

7. Gypsum is not used as a supplementary cementations material. 9. Turbidity in surface water is due to colloidal material. 10. The prescribed permissible noise level, Leq is 45 dBA. According to Central Pollution Control Board,

Area Limits in dB(A), Leq Category of Area Code Day time Night time A Industrial area 75 70 B Commercial area 65 55 C Residential area 55 45 D Silence Zone 50 40

3. Green house gases are the gas mixed in the atmosphere that absorbs the infrared radiation emitted by the earth’s surface.

Area Limits in dB(A), Leq Category of Area Code Day time Night time A Industrial area 75 70 B Commercial area 65 55 C Residential area 55 45 D Silence Zone 50 40

11. Continuous exposure to suspended particles are harmful to the lungs and can cause failure of respiratory system. 12. The prescribed permissible noise level, Leq for commercial area at night time is 55 dBA. According to Central Pollution Control Board, Area Limits in dB(A), Leq Category of Area Code Day time Night time A Industrial area 75 70 B Commercial area 65 55 C Residential area 55 45 D Silence Zone 50 40

We are not accustomed to these gases because neither nitrogen nor oxygen, the two most abundant gases of the atmosphere (78% and 21%, respectively), that many of us have heard of, have this ability to intercept infrared radiation. 4. Fly ash never improves strength of concrete. The advantages of using fly ash far outweigh the disadvantages. The most important benefit is reduced permeability to water and aggressive chemicals. Properly cured concrete made with fly ash creates a denser product because the size of the pores are reduced. 5. One turbidity unit NTU is equal to 1.0 mg/l SiO2. 6. The prescribed permissible noise level, Leq is 55 dBA. According to Central Pollution Control Board, Area Limits in dB(A), Leq Category of Area Code Day time Night time A Industrial area 75 70 B Commercial area 65 55 C Residential area 55 45 D Silence Zone 50 40

13. pH of acid rain should always be less than 5.6 even after precipitation. 14. The exposure of gaseous pollutant sulphur dioxide may cause bronchitis and pulmonary emphysema.

LEVEL-2 1. In the stratosphere, temperature increases with altitude. The reason is that the direct heat source for the stratosphere is the Sun. A layer of ozone molecules absorbs solar radiation, which heats the stratosphere. The amount of ozone present in the ozone layer is tiny, only a few molecules per million air molecules. 2. Nuclear density gauge can be used for all the following purposes except standard penetration reading. 3. A virus that may cause paralysis and is easily preventable by the polio vaccine. 4. The permissible limit of nitrate nitrogen in potable water is 10 mg/l . 7. The average concentration of ozone present in the stratosphere is approximately 10 ppm.

16

BASICS OF ENVIRONMENT AND POLLUTION CONTROL

Sulfure dioxide (SO 2 carbon monoxide (CO 2 ),nitrogen oxides (No x ), and particulate matter (PM).

Photochemical oxidants (ozone, nitrogen dioxide, sulfur trioxide) and secondary particular matter.

Chemical reactants characterized with a direct pollution effect on living beings and ecosystems, and with an indirect effect through the formation of secondary pollutants.

Chemical products, highly reactive when photoactivation is involved in the chemical process of their formation

Complicated control Direct control through the process: understanding reduction of anthropogenic and interrupting the emissions. chemical reactions leading to their generation.

8. Total or fecal coliform bacteria are reported as most probable number per 100 mL 9. Anthropogenic sources of air pollution in well planned city are construction activities, road traffics, and domestic burning 10. When the measured and standard reference pressure level becomes equal, the sound pressure level (SPL) is equivalent to 0 dBA 11. The major green house gases contributing in global warming are carbon dioxide , nitrous oxide,methane and water vapor. 12. The desirable amount of flue le ions in potable waters for optimal dental health is 1.0 mg/l 13. Primary Pollutants Versus Secondary Pollutants Air pollutant formed in the atmosphere as a result of the chemical or physical Air pollutant emitted interactions between the direclty from a source into primary pollutants the atmosphere. themselves or between the primary pollutants and other atmospheric components.

14. L90 is frequently taken as the Lp of the background level. L10-L90 is often used to give a quantitative measure as to the spread or “how choppy” the sound was. L10 is the noise level exceeded for 10% of the time of the measurement duration. 15. Scientists have discovered that air pollution from burning of fossil fuels is the major cause of acid rain. The main chemicals in air pollution that create acid rain are sulfur dioxide (SO 2) and nitrogen (NOx). Acid rain usually forms high in the clouds where sulfur dioxide and nitrogen oxides react with water, oxygen, and oxidants. this mixture forms a mild solution of sulfuric acid and nitric acid. Sunlight increases the rate of most of these reactions. Rainwater, snow, fog, and other forms of precipitation containing those mild solutions of sulfuric and nitric acids fall to earth as acid rain.

Basics of Computers and Applications PERSON AL COM PU T ERS Per sonal comput er s can be cat egor ized by size and por t abilit y as : 1. Deskt op comput er s 2. L apt op or not ebooks 3. Per sonal Digit al Assist ants (PDAs) 4. Por t able comput er s 5. Tablet comput er s 6. Wear able comput er s U ses. Per sonal computers ar e normally operated by one user at a time to perform such general purpose tasks as word pr ocessing, I nter net br owsing, I nternet faxing, e-mail and other digit al messaging , multimedia playback, computer game play, computer pr ogr amming, etc. The user of modern personal computer may have significant knowledge of the operating environment and application pr ogr am s, bu t i s n ot n ecessar i l y i n t er est ed i n pr ogr amming not even able to wr ite pr ogr ams for the computer. Ther efore, most software written primarily for per sonal comput er s t ends t o be desi gned wi t h simplicity of use, or “user-friendliness” in mind. However, the software industry continuously provide a wide range of new products for use in personal computer s, tar geted at both the exper t and the non-expert user. COM PU T ER COM PON EN TS 1. Comput er case wit h power supply (usually sold t oget her ) 2. Mother boar d 3. Pr ocessor wit h fan (usually sold t oget her ) 4. At least one memor y car d 5. M ass st or age 6. K eyboar d and mouse for input 7. M onit or for out put The mot her boar d connect s ever t hing t oget her. The memory card(s), graphics card and processor are mounted directly onto the motherboard (the processor in a socket an the memory and graphics cards in an expansion slot). The mass storage is connected to it with cables. Same for keyboard and mouse, except that they are external and connect to the back plate. The monitor is also connected to the back plate, except not (usually) dir ectly to the motherboard, but to a connector in the graphics card.

M ass st orage. I t can be (i ) H ar d disk (ii ) Floppy dr ive or zip dr ive (bot h wit h media) (iii )Opt ical dr ive (CD or DVD) The oper at ing syst em (e.g., M icr osoft Windows, Linux or many ot her s) can be locat ed on eit her of t hese, but t ypically it son one of t he har d disks. Alive CD is also possible, but ver y slow and used for eit her inst allation of t he OS or pr oblem solving. A typical computer also has (i ) Sound car d (ii ) Net wor k car d (iii )M oder n and possibly r out er Common addit i ons, connect ed on t he out si de (per ipher als). Pr i nt er ; Scanner ; Webcam; Speak er s; M i cr ophone; Headset; Car d reader ; Gaming devices such as a joystick Sever al funct ions (implement ed by chipset s) can be integr ated into the mother boar d, such as typically USB and net wor k, but also gr aphics and sound. But even if these ar e pr esent, a separ ate car d can be added if what is available isn’t sufficient . The gr aphics and sound car d can have a br eak out box to keep the analog par ts away fr om the electr omagnetic r adiat ion inside t he comput er case. For r eally lar ge amount s of dat a, a t ape dr ive can be used or (ext r a) har d disks can be put t oget her in an ext er nal case. These component s can usually be put t oget her wit h lit t le k nowledge, t o build a comput er. I f somet hing shouldn’t go somewher e, it usually doesn’t fit (this used t o not always be t he case in t he past ) and if it does fit it can usually do lit t le har m. Most per sonal computer s ar e standar dized to the point t hat pur chased soft war e is expect ed t o r un wit h lit t le or no customizat ion for the par t icular comput er. M any PCs ar e also user -upgr adeable, especially desktop and wor kst at ion class comput er s. Devices such as main memor y, mass st or age, even t he mot her boar d and cent r al pr ocessing unit may be easily r eplaced by an end user. This upgr adeabilit y is, however, not idefinit e due t o r api d ch anges i n t h e per sonal compu t er indust r y, A PC that was consider ed t op-of-the-line five or six year s pr ior may be impr act ical t o upgr ade due t o changes in indust r y st andar ds. Such a comput er

2

Basics of Computers and Applications

usually must be t ot ally r eplaced once it ’s no longer suitable for its pur pose. This upgr ade and replacement cycle is par tially r elated to new r eleases of the pr imar y mass-mar ket oper at ed syst em, which t ends t o dr ive t he acquisition of new har dwar e and t ends of obsolet e pr ev i ou sl y ser v i ceabl e h ar dw ar e (see pl an n ed obsol escence). The har dwar e capabilit ies of per sonal comput er s can somet imes be ext ended by t he addit ion of expansion car ds connect ed via an expansion bus. Some standar d per ipheral buses often used for adding expansion cards in personal computers as of 2005 ar e PCI , AGP (a highspeed PCI bus dedicat ed t o gr aphics adapt er s), and PCI Expr ess. Most per sonal computer s as of 2005 have mul t i pl e physi cal PCI expansi on sl ot s. M any al so i ncl ude an AGP bus and expansi on sl ot or a PCI Expr ess bus and one or mor e explansion slot s, but few PCs cont ain bot h buses. M ot her boar d. The mother boar d (or mainboar d) is the pr imar y cir cuit boar d wi t hi n a per sonal com put er. M an y ot h er component s connect di r ect l y or i ndi r ect l y t o t he mot her boar d. M ot her boar ds usually cont ain one or mor e CPUs, suppor t ing cir cuit r y - usually int egr at ed cir cuits (I Cs) pr oviding the inter face between the CPU memor y and input /out put per ipher al cir cuit s, main memor y, and facilities for init ial setup of the computer immediat ely aft er being power ed on (often called boot fi r mwar e or, i n I BM PC compat i bl e comput er s, a BI OS). I n many por t abl e and embedded per sonal comput er s, t he mot her boar d houses near ly all of t he PC’s cor e component s. Oft en a mot her boar d will also cont ain one or mor e per ipher al buses and physical connect or s for expansi on pur poses. Somet i mes a secondar y daught er boar d i s connect ed wi t h t he mot her boar d t o pr ovide fur t her expandibilit y or t o sat isfy space constr aint s. M ain M emory. A PC’s main memor y (i.e., it s pr imar y st or e) is fast st or age t hat is dir ect ly accessible by t he CPU, and is used t o st or e t he cur r ent ly execut ing pr ogr am and i mmi di at el y needed dat a. PCs use semi conduct or Random Access Memor y (RAM ) of var ious kinds such as DRAM or SRAM as t heir pr imar y st or age. Which exact kind depends on cost /per for mance issues at any par t icular t ime. M ain memor y is much fast er t han mass st or age devices like har d disks or opt ical discs, but is usually volat ile, meaning it does not r et ain it s contents (inst r uctions or data) in the absence of power, and is much mor e expensive for a given capacit y t han is most mass st or age. M ain memor y is gener ally not suit able for long-t er m or ar chival dat a st or age. M ass st or age devices st or e pr ogr ams and dat a even when t he power i s off; t hey do r equi r e power t o per for m r ead/wr it e funct ions dur ing usage. Although

semiconductor flash memor y has dr opped in cost, the pr evailing for m of mass stor age in personal computers is st ill t he elect r omechanical har d disk. The disk dr ives use a sealed H ead/Disk Assembly (H D A ) w h i ch w as f i r st i n t r odu ced by I B M ’s “ Wi nchest er ” di sk syst em. The use of a seal ed assembly allowed t he use of t he positive air pr essur e t o dr ive out par t icles fr om t he sur face of t he disk, which impr oves r eliabilit y. Video Car d. The video car d- ot her wise cal led a gr aphics car d, gr aphics adapt er or video adapt er - pr ocessor s and r ender s t he gr aphics out put fr om t he comput er t o t he comput er display, also called t he Visual Display Unit (VDU), and is an essent ial par t of t he moder n comput er. M I CROPROCE SSOR. A micr opr ocessor is a mult ipur pose pr ogr ammable logic device t hat r eads binar y inst r uct ions fr om a st or age device called memor y, accept s binar y dat a as i n pu t an d pr ocess dat a accor di n g t o t h ose instr uctions and pr ovides r esults as output. A typical pr ogr ammable machine can be r epr esent wit h t hr ee component s : micr opr ocessor, memor y, and I /O.

Memory Micro processor

I/O

These t hr ee component s wor k t oget her or int er act wit h each ot her t o per for m a given t ask, t hus t hey compr ise a syst em H ARD WARE . The physical component s of t his syst em ar e called har dwar e. SOF TWARE A set of inst r uct ions wr it t en for t he micr opr ocessor t o per for m a t ask is called a pr ogr am and a gr oup of pr ogr ams is called soft war e. APPL I CAT I ON S. Th e mi cr opr ocessor appl i cat i ons ar e cl assi fi ed pr imar ily in t wo cat egor ies ( i ) I n Re-programmable syst ems. Such as micr ocomput er s, t he micr opr ocessor is used for comput ing and dat a pr ocessing. These systems, include gener al pur pose micr opr ocessor capabl e of handl i ng l ar ge dat a. M ass st or age device (disks), and per ipher als such as I /O device (pr inter ).

Basics of Computers and Applications

( ii ) Embedded syst em. I n embedded syst ems, t he micr opr ocessor is par t of a f i n al pr odu ct an d i s n ot av ai l abl e f or r epr ogr ammabl e t o t he end user. A copyi ng machine is a t ypi cal example of an embedded syst em . Th e mi cr opr ocessor s u sed i n t hese syst ems ar e gener ally cat egor ised as ( a ) M i cr ocon t r ol l er s t h at i n cl u de al l t h e components like micr opr ocessor, memor y and I /O. ( b) I ntegrated microprocessor that include various devices such as timers and various types of I /O on a chip. ( c) Gener al pur pose micropr ocessor with discr ete component s such as micr opr ocessor, memor y and I /O. Embedded syst ems can also be incr eased as pr oducts t hat use micr opr ocessor t o per for m t heir oper at ions t hey ar e called as micr opr ocessor based pr oduct s. e.g. washing machines, dish washer s, aut omobi le dashboar d cont r ol s, t r affi c l i ght cont r ol l er s, and aut omat ic t est ing inst r ument s. BI N ARY DI GI T. The micr opr ocessor oper at es in binar y digit s 0 and 1, also known as bit s. Bit is an abbr eviat ion for t he t er m binar y digit . Those digit s ar e r epr esent ed in ter ms of electr ical voltages in the machine : gener ally 0 r epr esent s one vol t age l evel and 1 r epr esent s anot her. The digit s 0 and 1 ar e also synonyms wit h low and high r espect ively. Each mi cr opr ocessor r ecogni zes and pr ocesses a gr oup of bit s called t he wor d and micr opr ocessor ar e classified accor ding t o t heir wor d lengt h. M emor y. M emor y is like t he pages of a not ebook wit h space for a fixed number of binar y number s on each line. H owever t hese pages ar e gener ally made of semiconduct or mat er ial. Each line in 8 bit r egist er t hat can st or e 8 bi t bi nar y bi t s, and sever al of t hese r egist er s ar e ar r anged in a sequence called memor y. I nput /Out put . The user can enter instructions and data into memor y thr ough devices such as keyboar d or simple switches. These devices ar e called input devices. The mi cr opr ocessor r eads i nst r uct i ons fr om t he memor y and pr ocesses t he dat a accor ding t o t hose inst r uct ions. The r esult can be displayed by a device such as seven segment L ED (light emit t ing diodes) or pr i nt ed by a pr i nt er. These devi ces ar e call ed out put devices. M I CROPROCESSOR AS A CPU . The cent r al pr ocessing unit (CPU) consist s of t he Ar ithmet ic Logic Unit (ALU) and Cont r ol Unit (CU).

3

The CPU cont ains var ious r egist er s t o st or e dat a, AL U t o per for m Ar it hmet ic and logical, oper at ions, inst r uction decoder s, count er s and cont r ol lines. The CPU r eads i n st r uct i on s fr om t h e memor y and per for m t he t asks specified. I t communicat es wit h input /out put devices eit her t o accept or to send dat a. These devices ar e also known as per ipher als. The CPU i s t h e pr i m ar y an d cen t r al pl ay er i n communicat ing wit h devices such as memor y, input an d ou t pu t . H ow ev er , t h e t i m i n g of t h e communicat ion pr ocess is cont r olled by t he gr oup of cir cuit called cont r ol unit . CPU on single chip called mi cr opr ocessor. Arithmetic Logic Unit (ALU)

Input

Control Unit

Output

Memory

Fig. (a) Traditional block diagram of

Input

Micro processor as CPU

Output

Memory

F i g. (b) Bl ock di agr am of a compu t er w i t h t he a comput er micr oprocessor as CPU

1. bit micr opr ocessor. The I nt el 4004 was t he fir st 4 bit pr ogr ammable device t hat was pr imar ily used in calculat or s. 2. bit micr opr ocessor. The int el 8008 is 8 bit micr opr ocessor, which was in t ur n super seded by t he I nt el 8080. I nt el 8080 wi dely used i n cont r ol applicat ions, and small comput er s also wer e designed using t he 8080 as t he CPU. Wit hin a few year s aft er t he emer gence of t he 8080, t he M ot or ola 6800 and Zilog Z80 and I nt el 8085 mi cr opr ocessor wer e devel oped as impr ovement s 3. 16 bit microprocessor. 8086/88 4. 32 bit microprocessor. 80380/486 and Pent ium 5. 64 bit microprocessor. M ot or ola 68000 ser ies

4

Basics of Computers and Applications

M I CROCOM PU T E RS. M icr ocomput er is classified in four gr oups. 1. Per sonnel comput er s. These micr ocomputer s ar e single user systems and being used for var iety of pur poses, such as payr oll, busi ness account s, wor d pr ocessi ng, l egal and medi cal r ecor d k eepi ng, per sonnel fi nance and inst r uctions. A typical configur ation includes a 16 or 32 bit micr opr ocessor, 2 to 4 M B (megabyte) of system memory, a video screen, a dot matrix pr inter. 2. Work st at ions. These ar e high per for mance cousi ns of t he PC. T hey ar e used i n en gi n eer i n g an d sci en t i fi c applications such as computer -aided design (CAD), computer aided engineer ing (CAE), and computer aided manufacturing (CAM), they generally include syst em memor y l ar ger t han 200 M B, st or age memor y in giga byt es and high r esolut ion scr een. The wor k st at i ons ar e desi gned ar ound RI SC pr ocessor s. The RI SC pr ocessor s t end t o be fast er and mor e effi ci ent t han t he pr ocessor s used i n per sonnel comput er. 3. Single boar d micr ocomput er s. These mi cr ocomput er s ar e pr i mar i l y used i n college, labor atories and industr ies for instr uctional pur poses or t o evaluate the per for mance of a given mi cr opr ocessor. They can al so be par t of some lar ger syst ems. Typical ly t hese micr ocomput er s include 8 or 16 bit micr opr ocessor.

e.g. These computers include such systems as Intel SDK85, SDK86, Motorola Evaluation kits, these are generally used to write and execute assembly language programs and to perform interfacing experiments 4. Si n gl e ch i p m i cr ocom pu t er s ( m i cr ocont r ol l er s) These micr ocomput er s ar e designed on a single chip, which t ypically includes a chip of 64 byt es of R/W memor y fr om 4K t o 2K byt es of ROM and sever al si ngl e l i nes t o connect I /Os. These ar e complet e micr ocomput er s on a chip, t hey ar e also k n ow n as m i cr ocon t r ol l er s. T h ese ar e u sed pr i m ar i l y f or su ch f u n ct i on s as con t r ol l i n g appliances and t r affic light s. e.g. Zilog Z8, I nt el M CS51 and 96 ser ies, and M ot or ola 68H C11. COM PU TER LAN GU AGES. N ibble. I t is a gr oup of four bit s M nemonic : A combinat ion of let t er s t o suggest t he oper at ion of an inst r uct ion. Compiler : A pr ogr am t hat t r anslat es english like wor ds of high language in t he machine language of a comput er. A complier r eads a given pr ogr am, called a sour ce code and t hen t r anslat es t he pr ogr am int o t he machine language which is called an object code. Assembler : A comput er pr ogr am t hat t r anslat es an assembly language pr ogr am fr om mnemonics t o t he binar y machine code of a comput er. M onit or program : A pr ogr am t hat int er pr et s t he input fr om a keyboar d and conver t s t he input int o it s binar y equivalent .

COM PU TER SYSTEM . 8085 Programming M odel. The 8085 pr ogr amming model includes six r egist er s, one accumulat or and one flag r egist er. I t has t wo 16 bit r egist er t he st ack point er and t he pr ogr am count er. Registers : The 8085 has six gener al pur pose r egist er s t o st or e 8 bit dat a. These ar e ident ical as B, C, D, E, H and L . They can be combined as Regist er s pair s BC, DE, and H L t o per for m some 16 bit oper at ions. The pr ogr ammer can use t hese r egist er s t o st or e or copy dat a int o t he r egist er s by using dat a copy inst r uct ions. Accumulator A

(8)

B

(8)

C

(8)

D

(8)

E

(8)

H

(8)

Flag Register

L

(8)

Stack pointer

(SP)

(16)

Program counter

(PC)

(16) Address Bus

Data Bus 8

Lines

Lines

Unidirectional

Bidirectional

D7

D6

S

Z

D5

F ig. (a) Progr amming model D4 D3 D2 AC P F ig. (b) F lab Register

D1

D0 CY

Basics of Computers and Applications

Accumulat or. The Accumulat or is an 8 bit r egist er t hat is par t of t he ar it hmet ic/logic unit (AL U). This r egist er is used t o st or e 8 bit dat a and t o per for m ar it hmet ic/logic unit (AL U). This r egist er is used t o st or e 8-bit dat a and t o per for m ar it hmet ic and logical oper at ions. The r esult of an oper at i on i s st or ed i n t he accumul at or. The accumulat or is also ident ified as r egist er A. FLAGS. The ALU includes five flip-flops, which ar e set or r eset aft er an oper at ion accor ding t o dat a condit ions of t he r esult in t he accumulat or and ot her r egist er s. They ar e called zer o (Z), car r y (CY); sign (S), par it y (P) and Auxiliar y car r y (AC) flags. The most commonly used flags ar e zer o, car r y, and sign. The micr opr ocessor uses t hese flags t o t est dat a condit ions. Aft er an addit ion of t wo number s, if t he sum in t he accumulat or is lar ger t han eight bits, the flip-flop used t o indicat e a car r y called car r y flag (CY) is set t o one. When an ar ithmetic oper ation is zer o the flip flop called t he zer o (Z) flag is set t o one. The fl ags have cr i t i cal i mpor t ance i n t he deci si on making pr ocess of t he micr opr ocessor. The condit ions (set or r eset ) of t he flags ar e t est ed t hr ough soft war e inst r uct ions. PROGRAM COU N TER (PC). T h i s 16 bi t r egi st er deal s wi t h sequ en ci n g t h e execut ion of inst r uct i on. Thi s r egi st er i s a memor y point er. M emor y locat ions have 16 bit addr esses, and that is why t his is a 16-bit r egister. The micr opr ocessor uses t hi s r egi st er t o sequence t he execut i on of t he inst r uct ions. The funct ion of t he pr ogr am count er is t o point t o t he memor y addr esses fr om which t he next byt e is t o be fet ched. When a byt e (machine code) is being fet ched, t he pr ogr am count er is incr ement ed by one t o point t o t he next memor y locat ion. STACK POI N TER (SP). The stack pointer is also a 16 bit register used as memory poi n t er. I t poi n t s t o a m em or y l ocat i on i n R/W memor y, called t he stack. The beginning of st ack is defined by loading a 16-bit address in the stack pointer. OPCODE FORM AT. I n t he desi gn of t he 8085 mi cr opr ocessor chi p, al l oper ations, registers and status flags are identified with a specific code. All int er nal r egist er s ar e ident ified as follows : Code Regist er s Code Regist er pair s 000 B 00 BC 001 C 01 DE 010 D 10 HL 011 E 11 A F or S P 100 F 101 G 110 Reser ved for M emor y Relat ed Oper at ion 111 A

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I N STRU CTI ON CODES. An inst r uct ion code is a gr oup of bit s t hat inst r uct t he comput er t o per for m a specific oper at ion. I t is usually divided i nt o par t s, each havi ng it s own par t i cular int er pr et at ion. The most basic par t of an inst r uct ion codes is it s oper at ion par t . Oper at ion Code. The oper at ion code of an inst r uct ion is a gr oup of bit s that define such oper ations as add, subt r act , mult iply, shift and complement . The number of bit s r equir ed for t he oper at ion code of an inst r uct ion depends on t he t ot al number of oper at i ons avai l abl e i n t he comput er. The oper at ion code must consist of at least n bit s for a given 2n (or less) dist inct oper at ions. OP Code. What t ask t o be per for med, called t he oper at ion code (OPcode). Oper and. When dat a t o be oper at ed on called t he oper and. The oper and (or dat a) can be specified in var ious ways. I t may include 8-bit or (16 bit) data, an int er nal r egist er, a memor y location, or 8 bit (or 16 bit) addr ess. I n some inst r uct ions, t he oper and is implicit . ADDRESSI N G M ODES. The control unit of a computer is designed to go through an inst r uct ion cycle t hat is divided int o t hr ee major phases. (i) Fet ch t he inst r uct ion fr om memor y (ii ) Decode t he inst r uct ion (iii) Execut e t he inst r uct ion I mplied mode. A l l r egi st er r ef er en ce i n st r uct i on s t h at u se an accumul at or ar e i mpl i ed mode i nst r uct i ons. Zer o addr ess inst r uct ions in a st ack-or ganized comput er ar e implied mode inst r uct ions since t he oper ands ar e implied t o be on t op of t he st ack. I mmediat e mode. The operand is specified in the instruction itself. In other words, an immediate mode instruction has an operand field r ather than an addr ess field. The operand field contains the actual operand to be used in conjunction with the operation specified in the instr uctions. Regist er mode. I n t his mode t he oper ands ar e in r egist er s, that r eside wit hin t he CPU. The par t icular r egist er is select ed fr om a r egist er field in t he inst r uct ion. A K bit field can specify any one of 2k r egist er s. Regist er indir ect mode. I n this mode t he inst r uct ion specifies a r egist er in t he CPU, whose cont ent gives t he addr ess of t he oper and in memor y. The advant age of a r egist er indir ect mode inst r uct ion is t hat t he addr ess field of t he inst r uct ion uses fewer bit s t o select a r egist er t han would have been r equir ed t o specify a memor y addr ess dir ect ly,

6

Basics of Computers and Applications

Aut o I ncrement or Decrement mode. This is similar t o r egist er indir ect mode except t hat t he r egist er is incr ement ed or decr ement ed aft er or befor e it s value is used t o access memor y. When t he addr ess st or ed in t he r egist er r efer s t o a t able of dat a in memor y, it is necessar y t o incr ement and decr ement t he r egist er s aft er ever y access t o t he t able D ir ect addressing mode. I n t hi s mode t he effect i ve addr ess i s equal t o t he addr ess par t of t he inst r uct ion. The oper and r esides in memor y and it s addr ess is given dir ect ly by t he addr ess field of t he inst r uct ion. I ndir ect addr essing mode. I n t his mode t he addr ess field of t he inst r uct ion gives t he addr ess wher e t he effect ive addr ess is st or ed in memor y. Effect ive addr ess = addr ess par t of inst r uct ion + cont ent of CPU r egist er I N PU T AN D OU TPU T SYSTEM I nput /out put devices ar e t he means t hr ough which t he M PU communicat es wit h t he out side wor ld. The M PU accept s binar y dat a as input fr om devices such as keyboar ds and A/D conver t er s and send dat a t o out put devices such as L EDs or pr int er s. Ther e ar e t wo differ ent met hods by which I /O devices can be ident ified. One uses an 8 bit addr ess and t he ot her uses a 16 bit addr ess. PERI PH ERAL M APPED I /O. I n this type of I /O, the MPU uses eight addr ess lines to identify an input or an output device; t his is known as per ipher al mapped I /O. The eight addr ess lines can have 256 (28 combinat ions) addr esses; t hus the MPU can identify 256 input devices and 256 output devices with addresses ranging fr om 00H to FFH input and output devices are differentiated by the contr ol signals. The MPU uses the I /O Read Contr ol Signal for input devices and the I /O Wr ite Contr ol Signal for output devices. The entir e r ange of I /O addr esses fr om 00 to FF is known as an I /O map, and individual addr esses ar e r efer r ed t o as I /O devices addr esses 8 I /O por t number s. M E M ORY M APPE D I /O (I /O W I T H 16 BI T ADDRESSES). I /O is connect ed as if it is memor y r egist er. This is known as memor y mapped I /O. The M PU uses t he same cont r ol signal (M emor y Read or M emor y Wr it e) an d i n st r u ct i on s as t h ose of m em or y. I n som e micr opr ocessor such as mot or ola 6800, all I /O have 16 bit addr esses. I /Os and memor y shar e t he same memor y map (64K). I n memor y mapped I /O, the MPU follows the same steps as if it is accessing a memor y r egist er. I N PU T DEVI CES These include a mouse, tr ack ball, space ball, joyst ick, digit izer s, dials and but t on boxes. Some ot her input

devices used in par t icular applicat ions ar e dat a glove, t ouch panels, image scanner s and voice syst ems. 1. K eyboar ds. An alphanumer ic keyboar d on a gr aphics syst em i s used pr i mar i l y as a devi ce for ent er i ng t ext st r i ngs. The k eyboar d i s an effi ci ent devi ce for input t ing such non gr aphics dat a as pict ur e labels associated with graphics display. Keyboar ds can also be pr ovided with featur e to facilitate entr y of scr een coor dinat es, menu select or s or gr aphics funct ions. 2. M ouse. A mouse is small hand held box used t o posit ion t he scr een cur sor. Wheels or r oller s on t he but t on of t he mouse can be used t o r ecor d t he amount and dir ect ion of movement s. Anot her met hod for det ect ing mouse mot ion is wit h an opt ical sensor. For t hese syst ems, t he mouse i s moved over a special mouse pad t hat has gr id of hor izont al and v er t i cal l i n es. T h e opt i cal sen sor det ect s movement s acr oss t he l i nes i n t he gr i d. Si nce mouse can be picked up and put down at anot her posit ion wit hout change in cur sor movement . I t is used for making r elat ive changes in t he posit ion of t he scr een cur sor. One, t wo or t hr ee but t ons ar e usual l y i ncl uded on t he t op of t he mouse for signalling t he execut ion of some oper at ion, such as r ecor ding cur sor position or invoking a funct ion. Most gener al pur pose graphics system now included a mouse and keyboar d as major input devices. 3. Track Ball and Space Ball. A t r ack ball is a ball t hat can be r ot at ed wit h t he fi nger s or pal m of t he hand t o pr oduce scr eencur sor movement , pot ent iomet er s at t ached t o t he ball measur e t he amount and dir ect ion of r ot at ion. Tr ackballs are often mounted on keyboar ds or other devices such as t he mouse. While a track ball is two dimensional posting device, a space ball pr ovides six degr ee of fr eedom. Unlike t he t r ack ball, space ball does not act ually move. St r ain gauges measur e t he amount of pr essur e applied to t he spaceball to pr ovide input for spat ial posit ioning and or ient at ion as t he ball is pushed or pulled in var ious dir ect ions. Space balls ar e used for t hr ee-di mensi onal posi t i oni ng and sel ect i on oper at ions in vir t ual r ealit y syst ems, modelling, animat ion, CAD, and ot her applicat ions. 4. Joyst i ck s. A joyst ick consist s of a small, ver t ical lever called st ick mount ed on a base t hat is used t o st eer and scr een cur sor ar ound. M ost joyst icks select scr een posi t i on wi t h act ual st i ck movem ent . Ot her s r espond t o pr essur e on t he st ick. Some joyst icks ar e mount ed on keyboar d, ot her s funct ion as st and alone unit s. Pot ent iomet er mount ed at t he base of joyst ick measur es t he amount of movement , and spr ings r etur n the stick to the center position when i t i s r el eased on e or m or e bu t t on s can be

Basics of Computers and Applications

5.

6.

7.

8.

pr ogr ammed t o act as input swi t ches t o si gnal cer t ai n act ions once a scr een posi t i on has been select ed. Dat a Glove. Dat a glove t hat can be used t o gr asp a “ vir t ual” object . The glove is const r uct ed wit h a ser ies of sensor s t hat det ect hand and fi nger mot i ons. Elect r omagnet ic coupl ing bet ween t r ansmi t t ing antennas and r eceiving antennas is used to pr ovide infor mat ion about t he posit ion and or ient at ion of the hand. The tr ansmitting and r eceiving antennas can each be st r uct ur ed as a set of t hr ee mut ually per pendicular coils, for ming a t hr ee dimensional car t esian co-or dinat e syst em. D igit izer s. A com m on dev i ce f or dr aw i n g, pai n t i n g or int er act ively select ing co-or dinat e posit ions on an object is a digit izer. These devices can be used t o i n pu t co-or di n at e v al u es i n ei t h er a t w o dimensional or t hr ee dimensional space. Typically, a digit izer is used t o scan over a dr awing or object and t o input a set of discr et e co-or dinat e posit ions, which can be joined wit h st r aight -line segment s t o appr oximat e the cur ve or sur face shapes. One t ype of digit izer is t he gr aphics t ablet , also r efer r ed t o as a dat a t abl et wh i ch i s u sed t o i n pu t t wo dimensional coor dinates by activating a hand cursor or st yles at select ed posit ions on a flat sur face. I mage Scanners. Dr awing, gr aphs, color and black and whit e photos or t ext can be st or ed for comput er pr ocessing wit h an image scanner by passing an opt ical scanning mechanism over t he infor mat ion t o be st or ed. The gr adat ions of gr ay scale or color ar e t hen r ecor ded and st or ed in an ar r ay. Once we have t he int er nal r epr esen t at i on of a pi ct u r e, w e can appl y t r ansfor mat ions t o r ot at e, scale or cr op the pict ur e t o a par t i cul ar scr een ar ea. We can al so appl y var ious image pr ocessing met hods t o modify t he ar r ay r epr esent at ion of t he pict ur e. For scanned t ext i nput , var i ous edi t i ng oper at i ons can be per for med on st or ed document s. Some scanner s ar e able t o scan eit her gr aphical r epr esent at ion or t ext , and t hey come i n a var i et y of si zes and capabilities. Touch Panels. Touch panel s al l ow di spl ayed object s or scr een posit ions t o be select ed wit h t he t ouch of a finger. A t ypi cal appl i cat i on of t ouch panels i s for t he selection of pr ocessing options that ar e r epr esented wit h gr aphical icons. Some syst ems, such as t he plasma panels ar e designed wit h t ouch scr eens. Ot her syst em can be adapt ed for t ouch input by fit t ing a t r anspar ent device wit h a t ouch-sensing mechanism over t he video monit or scr een. Touch

7

input can be r ecor ded using opt ical, elect r ical or acoust ical met hods. 9. L ight Pens. Light pen is pencil-shaped devices ar e used to select scr een posit ions by detect ing the light coming fr om point s on t he CRT scr een. They ar e sensit ive t o t he shor t bur st of light emit t ed fr om t he phosphor coat ing at t he inst ant t he elect r on beam st r ikes a par t icular point . Ot her light sour ces, such as t he back gr ound l i ght i n t he r oom, ar e usual l y not det ect ed by a light pen. PRI N T E RS. Pr inter s pr oduce output by either impact or non-impact methods. I mpact pr inter s pr ess for med char acter faces against an inked r ibbon ont o t he paper. A line pr int er is an example of impact device wit h t he t ype faces mount ed on bands, chai ns, dr ums or wheels. Noni mpact pr int er s and pl ot t er s use l aser t echniques, i nk j et spr ays, xer ogr aphi c pr ocesses as used i n phot ocopyi ng machi ne, el ect r ost at i c met hods and elect r ot her mal met hods t o get images on t o paper. Char acter impact printers often have a dot matrix print head containing a r ectangular ar r ay of pr otr uding wir e pins, with the number of pens depending on the quality of t he pr int er. I n a l aser devi ce, l aser beam cr eat es a ch ar ge di st r i but i on on a r ot at i n g dr um coat i ng wi t h a phot oel ect r i c mat er i al , such as sel eni um. Toner i s applied t o t he dr um and t hen t r ansfer r ed t o paper. I nkjet met hods pr oduce out put by squir t ing i nk in hor izont al r ows acr oss a r oll of paper wr apped on a dr um. The electr ically char ged ink st r eam is deflected by an elect r ic field t o pr oduce dot mat r ix pat t er ns. A deskt op inkjet plot t er wit h r esolut ion of 360 dot s per inch. An elect r ost at ic device places a negat ive char ge on t he paper, one complet e r ow at a t ime along t he lengt h of t he paper. Then t he paper is exposed t o a t oner. The t oner is posit ively char ged and so it is at t r act ed t o t he negat ively char ged, ar eas, wher e it adher es t o pr oduce the specified output . Elect r other mal met hods use heat in a dot mat r ix pr int head t o out put pat t er ns on heat sensi t i ve paper. We can get l i mi t ed col or out put on an impact pr int er by using differ ent color ed r ibbons. Non impact devices use var ious techniques to combine thr ee color pigments (cyan, magneta and yellow) to pr oduce a r ange of color pat t er ns. Par allel pr int er s use : (a) RS-232C interface (b) Cent r onics int er face (c) H andshake mode

8

Basics of Computers and Applications

STORAGE U N I T I t consist s of main memor y and secondar y memor y. M ain M emory : 1. A fl i p-fl op made of el ect r oni c semi conduct or devices is used t o fabr icat e a memor y cell. These memor y cel l s or gani zed as a Random Access Memor y (RAM). Each cell has a capabilit y to stor e one bit of infor mat ion. A main memor y or st or e of a comput er is or ganized using a lar ge number of cells. Each cell st or es a binar y digit . 2. A memor y cell, which does not loose the bit st or ed i n i t when no power is suppl ied t o t he cell , i s known as a non-volat ile cell. 3. A wor d is a gr oup of bit s, which ar e st or ed and r etr ieved as a unit. A memor y system is or ganized t o st or e a number of wor ds. 4. A Byt e consist s of 8 bit s. A wor d may st or e one or mor e byt es. 5. The st or age capacit y of a memor y is t he number of byt es it can st or e. 6. The addr ess of t he locat ion fr om wher e a wor d is t o be r et r i eved or t o be st or ed i s ent er ed i n a M emor y Addr ess Register (M AR). 7. The dat a r et r ieved fr om memor y or t o be st or ed in memor y ar e placed in a M emor y Dat a Regist er (MDR). 8. The t ime t aken t o wr it e a wor d is known as t he Wr it e t ime. 9. The t ime t o r et r ieve i nfor mat i on i s cal led t he Access t ime of t he memor y. 10. The t ime t aken t o access a wor d in a memor y is independent of the addr ess of t he wor d and hence it is know as a Random Access M emor y (RAM ).

 The main memor y used t o st or e pr ogr ams and dat a in a comput er is a RAM . 11. A RAM may be fabricated with per manently stor ed i nfor mat i on, whi ch cannot be er ased. Such a memor y is called a Read Only M emor y (ROM ).

 For mor e specialized uses, a user can st or e his won special funct ions or pr ogr ams in a ROM . Such ROM 's ar e called Pr ogr ammable ROM (PROM).

1. F loppy Disk Drive (F DD) : I n t hi s devi ce, t he medium used t o r ecor d t he dat a is called as floppy disk. I t is a flexible cir cular disk of diamet er 3.5 i nches made of pl ast i c coat ed wi t h a magnet i c mat er ial. This is housed in a squar e plast ic jacket . Dat a r ecor ded on a floppy disk is r ead and st or ed in a comput er 's memor y by a device called a floppy disk is r ead and st or ed in a comput er 's memor y by a device called a floppy disk dr ive (FDD). A floppy disk is inser t ed in a slot of t he FDD. Floppy Disks wit h var ious capacit ies ar e as follow:

 51/4 dr ive- 360KB, 1.2MB (1 KB= 210 = 1024 bytes)  31/2 dr ive- 1.44 M b, 2.88 M B (1M B= 220 byt es) 2. Compact Disk Drive (CDD) : CD-ROM (Compact Disk Read Onl y M emor y) used a l aser beam t o r ecor d and r ead dat a along spir al t r acks on a 51/4 di sk . A di sk can st or e ar ou n d 650 M B of infor mat ion. CD-ROM s ar e nor mally used t o st or e massive t ext dat a. Recent ly CD wr it er s have come in the mar ket . Using a CD wr it er, lot of infor mation can be wr it t en on CD-ROM and st or ed for fut ur e r efer ence. 3. H ard Disk Drive (H DD) : Unlike a floppy disk t hat is flexible and r emovable, t he har d disk used in t he PC is per manent ly fixed. The dat a t r ansfer r ate between the CPU and har d disk is much higher as compar ed to the between the CPU and the floppy disk dr ive. The CPU can use t he har d disk t o load pr ogr ams and dat a as well as t o st or e dat a. CLASSI FI CATI ON OF COM PU TERS Comput er s come in sizes fr om t iny t o monst r ous, in bot h appear ance and power. The size of a comput er t hat a per son or an or ganizat ion needs depends on t he comput ing r equir ement s. Supercomputers : The might iest comput er s-and, of cou r se, t h e m ost ex pen si v e-ar e k n ow n as super comput er s. Super comput er s pr ocess billions of inst r uct ions per second. One uses super comput er s for t asks t hat r equir e mammoth dat a manipulation, such as wor l dwi de weat h er f or ecast i ng and weapons r esear ch.

 Dat a is pr esent ed ser ially for wr it ing and is r et r ieved ser ially dur ing r ead.

M ainframes : I n t he jar gon of t he comput er t r ade, lar ge comput er s ar e called mainfr ames. M ainfr ames ar e capable of pr ocessing dat a at ver y high speedsmillions of inst r uct ions per second-and have access t o billions of char act er s of dat a. Their pr incipal use of it is for pr ocessing vast amount s of dat a quickly, some of t h e obvi ou s cu st om er s ar e ban k s, i n su r an ce companies, and manufact ur er s.

Secondary / Auxiliary storage devices : M agnet ic sur face r ecor ding devices used in comput er s as H ar d disks, Floppy disks, CD-ROM s and M agnet ic t apes.

Personal Computers : Per sonal comput er s ar e oft en called PCs. A PC usually comes wit h a tower that holds the main cir cuit boar ds and disk dr ives of the computer,

12. A ser ial access memor y is or ganized by ar r anging memor y cells in a linear sequence.

 I nfor mat ion is r et r ieved or st or ed in such a memor y by using a r ead/wr it e head.

Basics of Computers and Applications

and a collect ion of per ipher als, such as a keyboar d, mouse, and moni t or. The t er m "PC" oft en means machines t hat ar e compat ible t o I BM ot her t han a Macint osh. Personal Computers (PC) and M AC : A PC is based on a mi cr opr ocessor or i gi nal l y made by t he I nt el Company (I ntel's Pentium) with other companies such as AM D. The comput er s made by M acint oshes which uses, Power PC pr ocessor, made by M ot or ol a ar e r efer r ed as M ac. Also, t he oper at ing syst em soft war e t hat r uns t hese t wo kinds of comput er s is differ ent . PCs usual l y use an Oper at i ng Syst em made by M icr osoft , i.e., Windows. M acint oshes use oper at ing syst em, called M ac OS, made by Apple. N ot ebook Comput ers : A comput er t hat fi t s i n a br i efcase?. N ot ebook compu t er s, al so k nown as L apt op comput er s, ar e por t abl e and popul ar wit h tr aveler s who need a comput er that can go with them. M ost n ot ebook s accept di sk et t es or n et w or k connect i ons, so i t i s easy t o move dat a fr om one comput er t o anot her. I N TERN ET The I nt ernet i s a global syst em of int er connect ed comput er net wor ks t hat use t he st andar d I nt er net Pr ot ocol Sui t e (TCP/I P) t o ser ve bi l l i ons of user s wor ldwide. I t is a net wor k of net wor ks t hat consist s of millions of pr ivat e, public, academic, business, and gover nment net wor ks, of local t o global scope, t hat ar e li nked by a br oad ar r ay of elect r onic, wir eless and opt ical net wor king t echnologies. The I nt er net car r ies a vast r ange of infor mat ion r esour ces and ser v i ces, su ch as t h e i n t er -l i n k ed h y per t ex t document s of t he Wor ld Wide Web (WWW) and t he infr ast r uct ur e t o suppor t elect r onic mail. N ET WORKS A computer network, oft en simply r efer r ed t o as a net wor k , i s a col l ect i on of comput er s and devi ces i nt er connect ed by communi cat i ons channel s t hat faci l i t at e communi cat i ons and al l ows shar i ng of r esour ces and i nfor mat i on among i nt er connect ed devices. Computer networking or Data communicat ions (D at acom) i s t he engi neer i ng di sci pl i ne concer ned wit h t he comput er net wor ks.

The t hr ee t ypes of net wor ks ar e: (i ) t he I nt er net (ii ) t he int r anet (iii ) t he ext r anet . Examples of differ ent net wor k met hods ar e: 1. Local ar ea networ k (LAN), which is usually a small net wor k const r ained t o a small geogr aphic ar ea.

9

An exampl e of a L A N woul d be a comput er net wor k wit hin a building. 2. M et r opolit an ar ea net wor k (M AN), which is used for medi um si ze ar ea. exampl es for a cit y or a st at e. 3. Wide ar ea net wor k (WAN) t hat is usually a lar ger net wor k t hat cover s a lar ge geogr aphic ar ea. 4. Wir eless L ANs and WANs (WL AN & WWAN) ar e t he wir eless equivalent of t he L AN and WAN. I P ADDRESS An I nt ernet Prot ocol addr ess (I P addr ess) i s a n u m er i cal l abel assi gn ed t o each dev i ce (e.g., comput er, pr i n t er ) par t i ci pat i ng i n a comput er n et w or k t h at u ses t h e I n t er n et Pr ot ocol f or communicat ion.[1] An I P addr ess ser ves t wo pr incipal funct ions: host or networ k inter face identification and locat ion addr essing. I t s r ole has been char act er ized as follows: “ A name indicates what we seek. An address indicates where it is. A route indicates how to get there. I mport ant Devices U sed in N et work 1. M odem : A modem (modulat or -demodulat or ) is a device t hat modulat es an analog car r ier signal t o encode digit al infor mat ion, and also demodulat es such a car r ier signal t o decode t he t r ansmit t ed infor mat ion. The goal is t o pr oduce a signal t hat can be tr ansmitted easily and decoded to repr oduce the or iginal digit al data. M odems can be used over any means of t r ansmit t ing analog signals, fr om light emit t ing diodes t o r adio. 2. Router : A router is a device t hat for war ds dat a pack et s acr oss compu t er n et wor k s. Rou t er s per for m t he dat a “ t r affic dir ect ing” funct ions on t he I nt er net . A r out er is connect ed t o t wo or mor e dat a lines fr om di ffer ent net wor ks. When dat a comes in on one of t he lines, t he r out er r eads t he addr ess infor mat ion in t he packet t o det er mine it s ult imat e dest inat ion. 3. Bridge : A net work bridge connect s mul t i pl e net wor k segment s. Br idging i s a for war di ng t echni que used i n pack et -swi t ched comput er net wor k s. U nl i k e r out i ng, br i dgi ng mak es no assumptions about where in a networ k a par ticular addr ess is locat ed. I nst ead, it depends on flooding and examinat ion of sour ce addr esses in r eceived packet header s t o locat e unknown devices. Once a device has been locat ed, it s locat ion is r ecor ded in a t able wher e t he M AC addr ess is st or ed so as t o pr eclude t he need for fur t her br oadcast ing. 4. H ub : hub is a device for connecting multiple twisted pair or fiber optic Ethernet devices together and making them act as a single segment. The device is a for m of multiport repeater.

10

Basics of Computers and Applications

5. Repeater : A repeater is an elect r onic device t hat r eceives a signal and r et r ansmit s it at a higher level and/or higher power, or ont o t he ot her side of an obst r uct i on, so t hat t he si gnal can cover longer dist ances. 6. Server : A ser ver comput er i s a comput er, or ser ies of comput er s, t hat link ot her comput er s or el ect r oni c devi ces t oget her. They oft en pr ovi de essent i al ser vi ces acr oss a net wor k , ei t her t o private user s inside a lar ge or ganization or to public users via the inter net. For example, when you enter a quer y in a sear ch engine, t he quer y is sent fr om your computer over t he int er net to t he ser ver s t hat st or e all t he r elevant web pages. The r esult s ar e sent back by t he ser ver t o your comput er. EM AI L E l ect r on i c m ai l , com m on l y cal l ed em a i l or e-mail, is a met hod of exchanging digit al messages fr om an aut hor t o one or mor e r ecipi ent s. M oder n email oper at es acr oss t he I nt er net or ot her comput er net wor ks. Some ear ly email syst ems r equir ed t hat t he aut hor and t he r ecipient bot h be online at t he same t ime, a la i nst ant messaging. Today’s email syst ems ar e based on a st or e-and-for war d model . Email ser ver s accept , for war d, del iver and st or e messages. Neit her t he user s nor t heir comput er s ar e r equi r ed t o be onl i ne si mul t aneousl y; t hey need connect only br iefly, t ypically t o an email ser ver, for as long as it t akes t o send or r eceive messages. EM AI L ADDREES An email address ident ifies an email box t o which email messages ar e deliver ed. An example for mat of an email addr ess is lewis @ example .com which is r ead as lewis at example dot net . I t has two par t s. The par t befor e t he @sign is t he local-par t of t he addr ess, oft en t he user name of t he r ecipient lewis and the par t aft er t he @ sign is a domain name i.e. example.com t o which t he email message will be sent . M S OFFI CE M icrosoft Office is a proprietary commercial office suite of inter-related desktop applications, servers and services for the Microsoft Windows and Mac OS X oper ating systems, intr oduced by Micr osoft in 1989. I nitially a marketing term for a bundled set of applications, the first version of Office contained Microsoft Word, Micr osoft Excel, and Microsoft PowerPoint. W ORD Micr osoft Wor d is a wor d pr ocessor and was pr eviously consi der ed t o be t he mai n pr ogr am i n Offi ce. I t s pr opr i et ar y DOC for mat i s consi der ed a de fact o standard, although Word 2007 can also use a new XMLbased, Micr osoft Office-optimized format called .DOCX which has been st andar dized by Ecma I nt er nat ional

as Office Open XM L and it s SP2 updat e will suppor t ODF and PDF. Wor d is also available in some edit ions of M icr osoft Wor ks. I t is available for t he Windows and M ac platfor ms. The fir st ver sion of Wor d, r eleased in t he aut umn of 1983, was for t he M S-DOS oper at ing system and had the distinction of introducing the mouse t o a br oad populat ion. Wor d 1.0 could be pur chased wit h a bundled mouse, t hough none was r equir ed. Following t he pr ecedent s of L isaWr it e and M acWr it e, Wor d for Macintosh attempted to add closer WYSI WYG feat ur es int o it s package. Wor d for M ac was r eleased in 1985. Wor d for M ac was t he fir st gr aphical ver sion of M icr osoft Wor d. Despit e it s bugginess, it became one of t he most popular M ac applicat ions. EXCEL M i cr osoft Excel i s a spr eadsheet pr ogr am whi ch or iginally competed with the dominant Lotus 1-2-3, but eventually outsold it . I t is available for the Windows and Mac platfor ms. Micr osoft r eleased the fir st ver sion of Excel for the Mac in 1985, and t he fir st Windows ver sion (number ed 2.05 to line up with t he Mac and bu n dl ed w i t h a st an dal on e Wi n dow s r u n -t i m e envir onment) in November 1987. OU TLOOK M icr osoft Outlook (not t o be confused with Out look Expr ess) is a per sonal infor mation manager and e-mail communication softwar e. The replacement for Windows Messaging, Micr osoft M ail and Schedule+ star ting in Office 97, it includes an e-mail client, calendar, t ask manager and addr ess book. On t he M ac, M i cr osoft offer ed sever al ver si ons of Outlook in the late 1990s, but only for use with Microsoft Exchange Ser ver. I n Offi ce 2001, i t i nt r oduced an alter nat ive application with a slightly differ ent featur e set called Micr osoft Entour age. I t reintr oduced Outlook in Office 2011, r eplacing Entour age. POWE RPOI N T M i cr osof t Power Poi n t i s a popu l ar pr esent at i on pr ogr am for Windows and M ac. I t is used t o cr eat e slideshows, composed of t ext , gr aphics, movies and ot her object s, which can be displayed on-scr een and navigat ed t hr ough by t he pr esent er or pr int ed out on t r anspar encies or slides.

Basics of Computers and Applications

11

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. Which of t he following is t he fast est ? (a) CPU (b) magnet ic t apes and disks (c) video t er minal (d) sensor s, mechanical cont r oller s 2. The input unit of a comput er (a) feeds dat a t o t he CPU or memor y (b) r et r ieves dat a fr om CPU (c) dir ect s all ot her unit s (d) all of t hese 3. Offline device is (a) a device which is not connect ed t o CPU (b) a device which is connect ed t o CPU (c) a dir ect access st or age device (d) an I /O device 4. Which of t he following is a set of gener al pur pose int er nal r egist er s ? (a) Stack (b) Scratchpad (c) Addr ess r egist er (d) St at us r egist er 5. A single bus st r uct ur e is pr imar ily found in (a) main fr ames (b) super comput er s (c) high per for mance machines (d) mini-and micr o-comput er s 6. Which of t he following r egist er s is used t o keep t r ack of addr ess of t he memor y locat ion wher e t he next inst r uct ion is locat ed ? (a) M emor y Addr ess Regist er (b) M emor y Dat a Regist er (c) I nst r uct ion Regist er (d) Pr ogr am Count er 7. Which of t he following r egist er s is loaded wit h t he cont ent s of t he memor y locat ion point ed by t he PC ? (a) M emor y Addr ess Regist er (b) M emor y Dat a Regist er (c) I nst r uct ion Regist er (c) Pr ogr am count er

8. I n a gener ic micr opr ocessor, inst r uct ion cycle time is (a) shor t er t han machine cycle t ime (b) lar ger t han machine cycle t ime (c) exact ly double t he machine cycle t ime (d) exact ly t he same as t he machine cycle t ime 9. Pr ogr am St at us Wor d (PSW) cont ai ns var i ous (differ ent) st atus of (a) CPU (b) ALU (c) pr ogr am (d) r egist er s 10. When an int er r upt occur s, CPU saves t he value of— — — in a st ack, (a) accumulator (b) pr ogr am st at us wor d (PSW) only (c) I nstr uct ion Addr ess Counter (I AC)only (d) bot h PWS and I AC 11. Bus Ar bit r at ion is (a) clear ing t he bus (b) lat ching infor mat ion on t he bus (c) deciding t he cont r oller of t he bus (d) cont r olling t he bus 12. Cont r ol M emor y Addr ess Regist er is pr esent in (a) ALU (b) I nst r uct ion Regist er Unit (c) Cont r ol Unit (d) Disk Cont r ol I nt er face Unit 13. Which of t he fol lowi ng is not one of t he t hr ee pr i mar y funct i ons t hat on-l i ne di r ect access syst em can ser ve? (a) inquir y (b) backup (c) update (d) pr ogr amming 14. Which of the following is not tr ue of punched cards as dat a ent r y media? (a) They can be used as t ur n ar ound document s (b) They ar e inexpensive (c) I nput is slow compar ed wit h ot her media (d) They ar e easily damaged

12

Basics of Computers and Applications

15. M agnet ic t ape can ser ve as

7. ASCI I coding all ocat ed bi nar y codes t o Engli sh al phabet s and symbols for comput er use. M or e r ecent ly a new st andar d has been adopt ed which al locat es code t o almost all t he languages of t he wor l d and also t o symbols cover i ng mor e t han a lakh char act er s. The new st andar d is call ed

(a) input media (b) out put media (c) secondar y st or age media (d) all of t hese

LEVEL-1

(a) CCS

1. I n t he cont ext of I nfor mat ion Technology, OCR means (a) Opt ical Char act er Recognit i on (b) Oct agonal Cycl ic Rechar ge (c) Oct adecimal Cycli c Regener at i on (d) Opt ical Char act er Regener at i on [RRB JE 2014 GREEN SH I FT ]





2. I n Boolean algebr a 1  1  0  0 = ? (a) 0 (c) 2

(b) 1 (d) – 1 [RRB JE 2014 GREEN SH I FT ]

3. Which of t he fol lowi ngis not an I /O devi ce of t he comput er ? (a) Keyboar d (b) Joy st i ck (c) ALU (d) Pr int er [RRB JE 2014 GREEN SH I FT ]

4. W h at i s f l oat i n g poi n t w i t h r ef er en ce t o comput er s? (a) I t i s a soft war e subr out i ne ar ound whi ch ot her subr out i nes ar e bui lt (b) I t i s a r epr esent at i on of r eal number s t o facilit at e comput ing (c) I t i s t h e m ai n al gebr ai c f or m u l a of t h e soft war e (d) I t i s t h e vol t age poi n t gi v en t o var i ou s oper at i ng unit s of t he comput er [RRB JE 2014 GREEN SH I FT ]

5. A syst em of di gi t al r u l es for exch an ge and pr ocessi ng of dat a bet ween var i ous devi ces i s called (a) soft war e pr ogr amme (b) algor it hm (c) pr ot ocol (d) infor mat ion pr ocessing [RRB JE 2014 GREEN SH I FT ]

6. A t heor et ical comput er wi t h infini t e t ype and m em or y, u sed i n an al y si s of pr obl em s of comput at i on, is call ed (a) Tape calculat or (b) Babbage machine (c) Tur i ng machi ne (d) Theor et ical machi ne [RRB JE 2014 GREEN SH I FT ]

(b) Unicode (c) Standar d CCS code (d) Univer sal CCS code [RRB JE 2014 GREEN SH I FT ]

8. For using passwor ds on t he I nt er net a soft war e is used so t hat t he passwor d is not int er cept ed easi ly. I t is call ed (a) Coding

(b) Malwar e

(c) Virus

(d) Encr ypt ion [RRB JE 2014 GREEN SH I FT ]

9. A soft war e, codi ng of which is available fr eely on I nt er net and i s open for user s for fur t her use an d i m pr ov em en t an d w h i ch i s gen er al l y developed in a coll abor at i ve manner is call ed (a) open sour ce soft war e (b) unlicensed soft war e (c) fr ee soft war e (d) communi t y soft war e [RRB JE 2014 GREEN SH I FT ]

10. Wh i ch of t h e f ol l ow i n g ar e m ach i n e l ev el languages? (a) C++

(b) Java

(c) Python

(d) None of t hese [RRB JE 2014 GREEN SH I FT ]

11. Which of t he fol lowing st at ement s is i ncor r ect ? (a) M i cr osoft windows is GUI (b) L i nux is GUI (c) M or e t han 5000 k B dat a can be st or ed in a DVD (d) A 1 TB fl ash dr i ve can st or e 2 mi ll ion fil es each of size 1 M B [RRB JE 2014 GREEN SH I FT ]

12. The t er ms AL U, CPU, I /O devi ces per t ai n t o (a) comput er s (b) envi r onment al engineer i ng (c) di esel engi ne (d) en gi n eer i n g dr aw i n g an d or t h ogon al pr oject i ons [RRB JE 2014 GREEN SH I FT ]

Basics of Computers and Applications

13. I n a comput ing devi ce 'M H z' is ment i oned in t he specifi cat ions. I t r efer s t o (a) si ze of memor y

5. Who wr ot e/invent ed t he L inux soft war e? (a) M i cr osoft (b) Apple I NC (c) IBM (d) None of t hese

(b) speed of comput at i on (c) cl ock speed (d) none of t he above [RRB JE 2014 GREEN SH I FT ]

14. The value of binary 1111 is : (a) 23

(b) 23 – l

(c) 24

(d) 24 – l

[RRB SSE 2014 YELLOW SH I FT]

6. A t echnique of anonymous communicat i on over a computer networ k using encr yption of messages and spl it t ing bet ween t he nodes, i s cal led(a) Spice r out i ng (b) Onion r out i ng (c) Cabbage r out ing (d) Flower r out i ng

[RRB JE 2014 RED SH I FT ]

15. The term 'Operating System' means : (a) A set of programmes which controls computer working (b) The way a computer operator works (c) Conversion of high level language into machine level language (d) None of these [RRB JE 2014 RED SH I FT ]

LEVEL-2 1. Which of t hese i s N OT an Oper at ing Syst em? (a) Android (b) iOS (c) Linux (d) Power poi nt [RRB SSE 2014 YELLOW SH I FT]

2. A soft war e user i nt er face feat ur e t hat allows t he user t o view somet hing ver y si mi lar t o t he end r esul t whi l e t he document i s bei ng cr eat ed i s called(a) For mat cr eat or (b) For mat fideli t y (c) WYSI WYG (d) WYGI WYS [RRB SSE 2014 YELLOW SH I FT]

3. I n a comput er syst em t her e ar e soft war es and l anguages at var i ous l evel s, l i k e H i gh l evel L an gu age (H L ), M ach i n e L an gu age (M L ), Compiler (C). Which of the following is the cor r ect indi cat ive r epr esent at i on fr om user (U) t o t he comput er (COM P)? (a) U  H L  C  M L  Comp (b) U  C  M L  H L  Comp (c) U  C  H L  M L  Comp (d) U  M L  H L  C  Comp [RRB SSE 2014 YELLOW SH I FT]

4. Which of t hese devi ces per for ms t he funct ion of bot h i n pu t dev i ce an d ou t pu t dev i ce f or a comput er ? (a) Joy St i ck (b) M ouse (c) Modem (d) Pr int er [RRB SSE 2014 YELLOW SH I FT]

13

[RRB SSE 2014 YELLOW SH I FT]

7. Pr ocessing speed of comput er i s measur ed i n(a) M I PS(M i lli on I nst r uct i on Per Second) (b) M H z of clock (c) Bot h (a) and (b) (d) None of t hese

[RRB SSE 2014 YELLOW SHIFT]

8. To close a pr esent at ion and quit Power Point , one must click t he close but t on on t he : (a) menu bar (b) t i t le bar (c) st andar d t ool bar (d) common t ask s t oolbar [RRB SSE 2014 RED SH I FT]

9. Expr ession + + i is equi valent in ‘C’ t o : (a) i = i + l (b) i = i + 2 (c) i = 2i (d) None of t hese [RRB SSE 2014 RED SH I FT]

10. W h i ch of t h e f ol l ow i n g r at i on al r el at i on oper at ions in 'C means "not equal t o" ? (a) # (b) == (c) ! =

(d) < = [RRB SSE 2014 RED SH I FT]

11. M i cr osoft Windows is a/an : (a) Wor d-pr ocessing pr ogr am (b) Dat abase pr ogr am (c) Oper at ing syst em (d) Gr aphics pr ogr am [RRB SSE 2014 RED SH I FT]

12. __________wi ll t r ansl at e t he compl et e pr ogr am at once fr om a H i gh L evel L anguage t o t he M achine L anguage. (a) Compiler (b) Joy st i ck (c) Por t s (d) L i ght pen [RRB SSE 2014 RED SH I FT]

13. The wor d funct ion t hat cor r ect s t ext as we t ype is r efer r ed t o as : (a) Aut o inser t (b) Aut o cor r ect (c) Aut o summar i ze (d) Tr ack changes [RRB SSE 2014 RED SH I FT]

14

Basics of Computers and Applications

14. Pr imar y Stor age, in computer t er minology, r efer s to : (a) H ar d Disc Dr i ve (b) Random Access M emor y (RAM ) (c) Read Only M emor y (ROM )

15. What does an elect r onic spr eadsheet consist of ? (a) Rows (b) Columns (c) Cells (d) Al l of t he above [RRB SSE 2014 RED SH I FT]

(d) T he st or age devi ce wher e t h e oper at i n g syst em is st or ed [RRB SSE 2014 RED SH I FT]

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (a)

2. (a)

3. (a)

4. (b)

5. (d)

11. (c)

12. (c)

13. (d)

14. (b)

15. (d)

1. (a)

2. (a)

3. (c)

4. (b)

5. (c)

11. (d)

12. (a)

13. (c)

14. (d)

15. (a)

1. (d)

2. (c)

3. (a)

4. (c)

5. (d)

11. (c)

12. (a)

13. (b)

14. (b)

15. (d)

6. (d)

7. (c)

8. (b)

9. (a)

10. (d)

6. (c)

7. (b)

8. (d)

9. (a)

10. (d)

6. (c)

7. (b)

8. (b)

9. (b)

10. (c)

LEVEL-1

LEVEL-2

EXPLAN ATI ON S LEVEL-1 1. Opt i cal Char act er Reader i s ful l for m of OCR, which can r ead a char acter and conver t its bitmap image t o equival ent ASCI I codes. 2. I t i s equivalent t o 0.1 = 0 3. Ar it hmet i c L ogical Unit i s not an I nput devi ce whil e all t he ot her t hr ee ar e. 4. Fl oat i ng point number s ar e used i n comput er s t o r epr esent r eal number s. Si nce r eal number s can not be accur at ely r epr esent ed i n comput er s t hr ough binar y number s. 5. Pr ot ocols ar e r ules developed for exchanging and pr ocessing of dat a bet ween var i ous devices. Exampl es incl ude H TTP, I P, FTTP et c. 6. Tur i ng machi ne i s a mat hemat i cal model of a hypot het i cal comput ing machi ne which can use a pr edefi ned set of r ules t o det er mi ne a r esult fr om a set of input var iables. 7. The new st andar d which all ocat es codes t o almost al l languages and symbol s, t ot ali ng mor e t han a lakh i s cal led U ni code. I t mak es t r ansfer

8.

9.

10. 11. 12.

13.

and r euse of t r ansl at ed dat a et c. ver y easy. Al so it r epr esent s each char act er wi t h 16 bit s. To pr ot ect passwor ds et c. Encr ypt i on i s used which i s coding each dat a point i n a par t icul ar pat t er n whi ch is not easy t o decode. Open sour ce soft war es l i k e U ni x have t hei r sour ce code fr eely avai lable and t hese ar e developed t hr ough collabor at ion of coder s fr om acr oss t he wor ld. M any devel oper s as a pr inci ple use only open sour ce soft war es. Al l t he l anguages ment ioned her e ar e high l evel languages, in which i t i s easier t o wr it e code. 1 TB fl ash dr i ve can st or e appr oximat ely 1 mi lli on file sizes each of 1 M B. Al l t hese devi ces ar e r elat ed t o comput er s. CPU is Cent r al Pr ocessi ng Unit , whi le AL U is Ar it hmet i c and L ogi cal uni t , whil e I /O is I nput Out put devi ces. M H z is used t o measur e t he number of oper at i ons t hat can be done by t he CPU i n 1 second. So i t r efer s t o clock speed.

Basics of Computers and Applications

14. 1111 of binar y i s 24 – 1. I t i s 23 + 22 + 21 + 20 15. An oper at ing system (OS) is syst em soft war e that manages comput er har dwar e and soft war e r esour ces and pr ovides common ser vices for comput er pr ogr ams.

LEVEL-2 1.Power point is not an oper at ing syst em but an applicat ion t o make pr esent at ion slides. 2. When user s can see somet hing ver y similar t o end r esult while document cr eat ion, it is called WYSI WYG. 3. I t is t he cor r ect r epr esent at ion wher e user gives input in high level language, it is t hen compiled an d becom es m ach i n e l an gu age w h i ch i s comput ed and t he out put is similar ly pr ovided t o t he user. 4. The modem is an input and an out put device. I t is used for sending and r eceiving infor mation and dat a over t elephone lines. 5. L I NUX soft war e was invent ed by L inus Tor valds while st udying comput er science at Univer sity of H elsinki in 1991. 7. Pr ocessing speed is measur ed in MH z. Nowadays i t i s even measur ed i n GH z i .e. Gi ga H er t z. Basically it r epr esent s how many oper at ions can be pr ocessed in 1 second.

15

8. To close applications such as Powerpoint, MS-word et c. one must click t he close but t on on t he t it le bar which is at t he t op r ight hand cor ner. 9. ++I in C means i = i + 2. This was a way developed t o wr it e smaller lines in code. 10. != means not equal t o in C. 11. M icr osoft Windows is t he wor ld's most popular commer cial oper at ing syst em for PCs. 12. Com pi l er s ar e u sed t o t r an sl at e en t i r e pr ogr am s f r om H i gh l ev el l an gu age t o m ach i n e l an gu age, so t h at com pu t er can under st and and execut e it . 13. When we t ype t ext , Aut o cor r ect feat ur e cor r ect s t he wor d's spelling or any ot her por t ion of t he t ext . 14. Pr imar y st or age, also known as main st or age or memor y, is t he ar ea in a comput er in which dat a i s st or ed for qui ck access by t he comput er 's pr ocessor. The t er ms r andom access memor y (RAM ) and memor y ar e oft en used as synonyms for pr imar y or main st or age. 15. An elect r onic spr eadsheet like M S-Excel consist s of Cells which are ar r anged in Rows and Columns. So all ar e pr esent .



1

Engineering Mechanics

CHAPTER FRE E-BODY DI AGRAM

W1

A fr ee-body diagr am consist s of a diagr ammat ic r epr esent at ion of a single body or a W2 1 subst yst em of bodies isolat ed fr om it s sur r oundings but shown under t he act ion of for ces and moment s due t o ext er nal act ions. 2 Consider, for example, a book lying flat on a t able. The book exer t s it s weight on t he R 1 t able and t he t able exer t s it s own weight as well as t r ansmit s t he weight of t he book on t he gr ound. A fr ee-body diagr am for t he book alone would consist of it s weight W act ing R 2A R 2B t hr ough t he cent r e of gr avit y and t he r eact ion exer t ed on t he book by t he t able t op as shown in t he figur e. F.B.D. A fr ee-body diagr am may be dr awn for any single member of a syst em, any subsyst em of t he syst em or t he ent ir e syst em ir r espect ive of whet her t he syst em is in equilibr ium: at r est , in unifor m mot ion or in a dynamic st at e of mot ion.

U ses of Free Body Diagrams in Statics (i )

The pr oblems involving equilibr ium of bodies under any syst em of for ces can be simplified by dr awing fr ee body diagr am of each body separ at ely. (ii ) All equat ions of equilibr ium can be applied t o each fr ee body diagr am. (iii ) The unknown for ces for equilibr ium of each body can be obt ained ver y easily.

TYPES OF STRU CTU RE S A st r uct ur e may consist of a t r uss or a fr ame pin-connect ed or r igidly secur ed. A t r uss is an assemblage of slender bar s fast ened t oget her at t heir ends by smoot h bolt s or ball-and socket joint s act ing as hinges. A t r uss, by definit ion, is a connect ed st r uct ur e. The bar member s, t her efor e, act as t wo-for ce member s which can eit her be in t ension or in compr ession: t her e can be no t r ansver se for ce in a member of a t r uss. A fr ame str uct ur e, on t he ot her hand, consist s of member s which may be subject ed t o a t r ansver se load in addit ion t o t he axial load. A simple st r uct ur e is t hus a pin connect ed fr ame or t r uss. A t r uss consist s of slender -bar member s which can car r y no t r ansver se loads. I t follows t hat t he loading in a t r uss must be at t he joint s only. A t r uss consist ing of member s which lie in plane and ar e loaded in t he same plane is called plane t r uss. I f a t r uss is made of noncoplanar member s, it is r efer r ed t o as space t r uss. Similar ly, a fr ame may be a plane fr ame or a space fr ame depending upon it s st r uct ur e. Tr usses ar e classified as just -r igid, over -r igid and non-r igid mechanism. I f t he member s ar e allowed any r elat ive movement , t hen t he assemblage of member s is called a non-rigid t r uss or mechanism. I f t he member s ar e not allowed any r elat ive movement t hen it is called a rigid truss. A j u st -r i gi d t r u ss i s t h at w h i ch , on t h e r em ov al of an y si n gl e m em ber , becom es n on -r i gi d. A n over -r i gi d t r uss i s t he one t hat has r edundant member s whi ch may be r emoved t o r ender t he t r uss just-r igid. F

C D

C D

A B (a) Non-Rigid Truss — A Mechanism

1.2

Engineering Mechanics

AN ALYSI S OF SI M PLE PLAN E TRU SS Assumpt ions (i ) Each t r uss is composed of r igid member s lying in one plane. (ii ) Weight of t he member s ar e neglect ed because t hey ar e small in compar ison wit h t he loads. (iii ) For ces ar e t r ansmit t ed fr om one member t o anot her t hr ough smoot h pins fit t ed in t he member s.

M et hods of Analysis of a F ramed St ruct ur e (1) Analytical met hod

(2) Gr aphical met hod

(1) Analyt ical M et hod This met hod includes (i ) M ethod of Joints : I n t his met hod, t he fr ee body diagr am of each joint is separ at ely analysed t o obt ain the magnitude of str esses in the tr uss member s. The unknown forces ar e then deter mined by equilibr ium equat ions viz. V = 0 and H = 0, i.e. sum of all t he ver t ical for ces as well as t he hor izont al for ces is equat ed t o zer o. The analysis also yields nat ur e of st r esses (whet her compr ession or t ension). The pr inciple of t r iangle of for ces, or r esolut ion of for ces is applied for analysis. L ogic is applied so t hat isolat ed joint is in equilibr ium under t he act ion of ext er nal for ces and int er nal st r esses (called for ces in t he member ). Car e should be t aken t o select a joint which does not have mor e t han t wo unknown for ces.

(ii ) M ethod of Sections : This met hod is convenient , when t he for ces in t he few member s of a t r uss is r equir ed t o be found out . Take such a sect ion t hat t he ext er nal for ces on one side of t he sect ion yi eld t he for ces in t he member s cut by t he sect ion. Use of Bow’s not at ion has been found convenient in t he analysis of fr ames. I n t his met hod, eit her left or r ight side of t he sect ion is consider ed. While t aking moment s at a point , t he appr opr iat e sign of t he moment should be used. e.g. consider ing left hand side of t he sect ion X-X, t he for ces ae, ab and hi should be used for comput at ion and not gf , dc, and hd N ote : Only that joint can be solved eit her analytically or gr aphically which has not mor e t han two known for ces. Bow’s N otation. Bow’s not at ions ar e used t o name t he differ ent member s of a t r uss. Ever y member s of t r uss is denot ed by t wo capit al let t er s, placed in t he spaces on eit her side of t he member of t he space diagr am as shown in t he figur e. Ther efor e t he load W will be named as PQ, t he r eact ion R1 as RP and r eact ion R2 as QR. The member s of t he fr ame will be named as PS, QS and RS r espect ively.

(2) Gr aphi cal M et hod This met hod is based on (a) L aw of t r iangle of for ces (b) L aw of polygon of for ces (c) Bow’s not at ion (d) L ink or funicular polygon : The funicular or link polygon is used for t he gr aphical det er minat ion of suppor t r eact ions. Consider a hor izont al st r ing AB suppor t ed at A and B. I f a for ce PQ applied at point ‘C’ of t he st r ing, t he st r ing will dist or t in t he shape shown by dot t ed lines. This distor ted shape of the str ing is called ‘funicular st ring or link polygon.’

Engineering Mechanics

1.3

VI RTU AL DI SPLACEM EN T AN D VI RTU AL WORK For the solution of problems of mechanics, it is necessar y to conceive an infinitesimal (extr emely small) hypothetical or imaginar y displacement of t he par t icle. Such an imaginar y infinit esimal displacement is known as Vir t ual Displacement . I f such a displacement is r ect ilinear, it is called Vir tual r ect ilinear displacement . The pr oduct of a for ce and vir t ual displacement caused by it is known as Vir t ual Wor k . The pr inciple of vir t ual wor k st at es t hat if a par t icle or r igid body or mor e gener ally, a syst em of connect ed r igid bodies, which is in equilibr ium under var ious ext er nal for ces, is given an ar bit r ar y displacement fr om t hat posit ion of equilibr ium, t he t ot al wor k done by t he ext er nal for ces dur ing t he displacement is zer o. Conversely, if the algebraic sum of the virtual works for every such displacement is zero, the system is in equilibrium.

Forces Omitt ed in t he F or mation of E quat ions for Virt ual Work (i ) Tension in an inext ansible st r ing. (ii ) React ions fr om t he sur faces. (iii ) React ion bet ween a body and t he sur face on which it r olls wit hout sliding. (iv) M ut ual act ion and r eact ion bet ween bodies whose equilibr ium is t o be consider ed t oget her.

Special Cases of Work Done Initial tension  Final tension  Extension 2 = M ean t ension × Ext ension

(i ) Work done by an elastic string during stretching =

(ii ) Wor k done in r aising a body by a height h = mgh (iii ) Wor k done in r ot at ion = Tor que × Angle t ur ned in r adian (iv) Wor k done by a couple = M oment of t he couple × Angle t ur ned in r adians

LAWS OF M ECH AN I CS Entity M ass

Law L aw of conser vat ion of mass

L inear moment um

Newt on's law

Angular moment um

Euler 's law

Energy

L aw of conser vat ion of energy

Statement M ass can neither by created nor destroyed by any physical or chemical means. The rat e of change of momentum of a body equals t he for ce impr essed upon it . The r at e of change of anular moment um of a body about an or igin O equals t he moment i m pr essed upon i t about t he origin. The r at e of change of int er nal ener gy and k inet ic ener gy of any mechanical syst em equals t he sum of t he r at es of wor k done by t he ext er nal for ces and t he ener gy flux acr oss t he boundar y as well as the energy developed wit hin t he syst em.

M at hemat ical F or mulat ion

d m =0 dt

b g

F=

d d p = p mV = dt dt

d r  mV dt d H = H = dt

M=

bg

b g b

g

–

E N E RGY Ener gy is t he capacit y t o do wor k. I nfact ener gy is possessed by a body, wher eas wor k is done by a for ce. Consider a par t icle which moves fr om a point A t o a neighbour ing point A (Fig.1)  F  I f r denot es t he posit ion vect or cor r esponding t o point A t hen dr is t he d r displacement of par t icle fr om A t o A  A  The wor k of t he for ce F cor r esponding t o t he displacement dr is defined as   r W  d F.dr This wor k done on t he body is st or ed in it in t he for m of ener gy and t he body in t his r  dr posit ion is r eady t o do t he wor k W h , if it is allowed t o fall fr eely on t he gr ound. Unit s : Unit s of ener gy is N.m and is same as t hat of wor k done.

A

1.4

Engineering Mechanics

F or ms of E ner gy (1) M echanical Energy The ener gy possessed by a body due t o it s posit ion or mot ion is called mechanical ener gy. Types. M echanical ener gy is of t wo t ypes ( i ) Kinetic Energy (K.E.). Ener gy possessed by a body by vir t ue of it s mot ion is called K inet ic Ener gy . I t is measur ed by t he amount of wor k which t he body can do due t o it s mot ion. The t wo t ypes of K .E. ar e as follows: (a ) Kinetic Energy of Translation: K inet ic ener gy due t o r ect ilinear mot ion of a body is called kinet ic ener gy of t r anslat ion. e.g., a car moving on a r oad, a moving bullet , wat er moving in a st r eam et c. Consider a mass ‘m ’ which is init ially at r est (i.e. u = 0). I f a const ant for ce F is applied, it st ar t s moving and let t he mass at t ains t he velocit y v aft er t r avelling a dist ances, t hen

mv2 2 This wor k done is st or ed in t he body in t he for m of kinet ic ener gy of t r anslat ion. mv2  K inet ic ener gy of t r anslat ion = 2 (b) Kinetic Energy of Rotation. The kinet ic ener gy due t o r ot ar y mot ion of t he body is called kinet ic ener gy of r ot at ion. e.g. a fly wheel r ot at ing about it s axis. Consider a body of mass m , r ot at ing about an axis O. L et t he body is composed of element al masses m 1, m 2, m 3, ..... et c. sit uat ed at dist ances r 1, r 2, r 3 ..... et c. fr om t he cent er O as shown in t he figur e and let t he linear velocit ies of t hese masses be v 1, v 2, v 3, ..... r espect ively. L et  be t he angular velocit y of t he body. Then 2 mr 2 K .E. of mass, m = 2 I 2  K .E. of r ot at ion = 2 wher e, I = mass moment of iner tia of the body about an axis passing t hr ough O. 4  2 N 2I But = 2 N, t her efor e K .E. of r ot at ion = 2 (ii ) Potential Energy. The ener gy possessed by a body by vir t ue of t o it s posit ion is called pot ent ial ener gy. The pot ent ial ener gy of a body is measur ed by t he wor k done against conser vat ive for ces act ing on t he body in br inging t he body fr om some r efer ence or dat um posit ion t o t he posit ion in quest ion. The most common for m of pot ent ial ener gy is t he gr avit at ional pot ent ial ener gy which is t he ener gy possessed by a body due t o it s posit ion wit h r espect t o t he sur face of ear t h. All bodies on t he sur face of ear t h have been suppossed t o possess zer o pot ent ial ener gy. I f m is t he mass of a body held at a height of h fr om ear t h’s sur face, t hen t he pot ent ial ener gy possessed by t he body is mgh. Wor k done on t he mass =

LAW OF CON SERVATI ON OF EN ERGY I f a par t icle (or body) is act ed upon by a conser vat ive for ce syst em, t he sum of t he kinet ic ener gy and pot ent ial ener gy is const ant .

E ner gy St or ed in t he Spr ing When a spr i ng i s st r et ched a for ce act i ng on a body does not r emai n const ant , but var i es wi t h t he displacement under gone by t he spr ing. L et a body on a fr ict ional hor izont al plane is at t ached t o a spr ing wit h st iffness k N/m.

(a)

(b)

Engineering Mechanics

1.5

When t he body is at S, a dist ance l fr om t he wall, t her e is no for ce in t he spr ing (kl = 0). This will be consider ed t he st andar d position. When t he body is at P, a dist ance (l + s) fr om t he wall, t her e is no change in t he pot ent ial ener gy of t he body which is always on t he hor izont al plane. The for ce act ing on t he body in t his posit ion is pr opor t ional t o t he defor mat ion of t he spr ing fr om t he st andar d unst r essed lengt h. Ther efor e F = – k.s. Wor k done by t his for ce for a differ ent ial change, ds = – k.s. ds

z

k s2 o 2 wher e k = spr ing const ant and s is change in lengt h of spr ing. The body possesses t his pot ent ial ener gy because of t he pull of t he spr ing which var ies wit h t he defor mat ion. 

Pot ent ial ener gy = –

s

k.s. ds =

I M PU L SE I f a const ant for ce P act s for a t ime t on a body, t hen impulse of t he for ce = P × t I t t is var iable over t ime, for ce-t ime plot can be pr epar ed and t he ar ea under t he cur ve gives t he impulse of t he var iable for ce. Unit s : Unit of impulse is N. sec.

L inear M oment um Pr oduct of t he mass m and t he velocit y v is t he linear moment um Fr om Newt on’s second law of mot ion, t he t ime r at e of change of t he pr oduct of t he mass and velocit y of a par t icle is pr opor t ional t o t he impr essed for ce. Ther efor e L inear moment um, G = mv Linear M otion. L et a const ant for ce F act on a body of mass ‘m ’ for t ime ‘t ’ and changes it s velocit y fr om u t o v under an acceler at ion a, t hen F = ma Fr om v = u + at , at = v – u  I mpulse = F × t = mv – mu I mpulse = Change of moment um This is called t he equat ion of moment um Fr om F × t = mv – mu , mv  mu F = t This value of F is called t he t ime aver age of for ce. I f t he for ce is var iable, t hen t t t dv L inear I mpulse = F .dt = m. dt = m.dv = m (v – u ) o o o dt Thus t he linear impulse is equal t o t he change in t he linear moment um dur ing t he t ime int er val t .

z z

z

Angular M ot i on L et a const ant t or que T act on a body of moment of iner t ia I for t ime t and changes it s angular velocit y fr om 1t o 2 under an angular acceler at ion , t hen Angular I mpulse = Tor que × t ime wher e Tor que,T = I  and 2 = 1 + t or t = (2 – 1)  Angular I mpulse = I × t = I (t ) = I (2 – 1) = change in angular moment um I f t or que T is var iable, t hen Angular I mpulse =

z

2

1

I .d = I (2 – 1)

Thus, angular impulse is t he pr oduct of M .I . about axis of r ot at ion and t he change in angular velocit y.

1.6

Engineering Mechanics

L inear I mpulse-M oment um Relat ion for an Assemblage of Par t icles The sum of the exter nal forces acting on an assemblage of a n particles equals the time rate of change of the linear momentum of a mass m , which is equal to the sum of the masses of n particles and which possesses a velocity equal to that of the centre of masses of n particles. Thus, sum of the external forces acting on the gr oup of particles. dF d (mv) F = = dt dt wher e m = mass of n par t icles v = velocit y of t he cent r e of mass t he gr oup of n par t icles. L inear I mpulse =

z

t

o

 F . dt =

z

y

u

d (mu ) = m (v – u )

So, t he linear impulse I of all t he for ces act ing in t he st at ed t ime int er val is equal t o t he change in linear moment um of a mass m.

KI N EM ATI CS OF A PARTI CLE I t is t he st udy of mot ion of a body or of a par t icle wit hout any r efer ence t o t he for ce syst em act ing on it .

D i spl acement Displacement of a body is defined as change of it s posit ion wit h r espect t o a given object . Displacement is a vect or quant it y and t he r efer ence object as well as dir ect ion of displacement should essent ially be st at ed. When t he movement is st r aight , t he displacement is r ect linear t r anslat ion and when t he mot ion is along a cur ved pat h, t he displacement is cur vilinear t r anslat ion .

Speed I t is change of posit ion per unit t ime, i.e., dist ance per unit t ime is called speed.

Vel oci t y Velocit y is speed associat ed wit h dir ect ion. Velocit y is a vect or quant it y having bot h magnit ude and dir ect ion. I f s is t he displacement in t ime t , t hen s veloci t y, v = t ds I n differ ent ial fr om, v= dt Velocit y is differ ent fr om speed, which r efer s t o t he magnit ude of t he velocit y ir r espect ive of it s dir ect ion.

Acceler at ion ( a or f )

v t dv dv d2s v I n differ ent ial for m, a = or 2 ds dt dt The unit of acceler at ion is m/sec2. Acceler at ion is a vect or quant it y I f velocit y incr eases, it s r at e of change is called acceler at ion and it decr eases, t he r at e of change of velocit y is called r elar dat ion or deacceler ation . I t is t he r at e of change of velocit y, i.e a =

RE CTI LI N EAR M OTI ON I f t he mot ion of a body is such t hat t he dir ect ion of it s r esult ant R r emains const ant t hen t he mot ion of t he body is along a st r aight pat h and is called r ect ilinear mot ion.

M ot ion U nder Const ant Acceler at ion I f init ial velocit y of a body is u, final velocit y is v, t ime int er val is t , acceler at ion is a and displacement is s, t hen 1 v2 = u 2 + 2 as and s = ut + at 2 2

M ot ion U nder Gr avit y

1 gt 2 2 wher e h is ver t ical displacement and g is acceler at ion due t o gr avit y t o be t aken as 9.81 m/s2. '+' sign t o be used for mot ions along gr avit y, wher eas ‘– ’ sign for mot ion against gr avit y.

v2 = u 2 + 2gh and h = ut +

Engineering Mechanics

1.7

Dist ance Tr avelled by a Par ticle in n th Second Consider a r ectilinear motion of a par ticle starts fr om or igin O and moving along x -axis wit h an init ial velocit y u . L et t he dist ance t r avelled in n t h second is s, final velocit y is v and const ant acceleration is a. The distance tr avelled in the (n - 1) seconds is sn – 1.  Dist ance t r avelled by t he par t icle in n t h second, a s = sn – sn – 1 = u + (2n – 1) 2

M ot ion U nder Var iable Acceler at ion L et , s be t he dist ance t r avelled by t he par t icle in t ime t wit h a velocit y v and acceler at ion a. dv a= v ds

Rel at i ve Velocit y (i )

Two bodies moving in the same direction will have their relative velocity equal to the difference of their velocities. Let u and v be the velocities of particles A and B r espectively. I f u > v and u & v have same direction, then Relat ive velocit y of A wit h r espect t o B = u – v in t he dir ect ion of u . Relat ive velocit y of B wit h r espect t o A = v – u (which is negat ive, opposit e t o t he dir ect ion of u ). (ii ) Two bodies moving in opposit e dir ect ions. Relat ive velocit y of A wit h r espect B = u + v in t he dir ect ion of u . Relat ive velocit y of B wit h r espect t o A = u + v in t he dir ect ion of v.

LAW OF PARALLELOGRAM OF VELOCI TI ES Parallelogram law st at es t hat, "if t wo velocities are r epresent ed in magnit ude and dir ection by two adjacent sides of a par allelogr am, t heir r esult ant will be r epr esent ed in magnit ude and dir ect ion by t he diagonal of t he parallelogram. w = r esult ant of velocit ies u, and v =

u 2  v2  2 uv cos 

v sin  u  v cos  I f t wo par t icles A and B ar e moving wit h velocit ies u and v and t he angle bet ween t heir dir ect ion of mot ion is  , t hen r elat ive velocit y of A wit h r espect t o B is obt ained by combining opposit e velocit y of B wit h A. This is shown in Fig. (a). Fig. (b) shows r elat ive velocit y of B wit h r espect t o A. and

t an  =

(a )

(b)

(c)

Thus

R1 = Relat ive velocit y of A wit h r espect t o B. = u 2  v2  2 uv cos (180   )

and

R2 = Relat ive velocit y of B wit h r espect t o A = u 2  v2  2 uv cos (180   )

But in opposit e dir ect ion of R1.

CU RV I LI N EAR M OTI ON I f t he dir ect ion of r esult ant of body R var ies t o maint ain it s posit ion t hr ough cent er of gr avit y, t hen body is r esult ing in a cur ved pat h and is called cur vilinear mot ion. Cur vilinear mot ion in a plane is mot ion along a plane cur ve (pat h). The velocit y and acceler at ion of a point on such a cur ve will be expr essed in (a) r ect angular component s (b) t angent ial as nor mal component s (c) r adial as t r ansver se component s

1.8

Engineering Mechanics

Angular Velocit y () I t is defined as t he angle in r adians t ur ned by a par t icle in unit t ime.

2 N 60 Tangent ial or linear velocit y of a par t icle under going a cur vilinear mot ion and having angular velocity is given by v = r  and t =

d v2  dt r Angular velocit y and displacement given by, Angular acceler at ion,=

1 t 2 2 2n  1 ( 2   20 )  and  = Angle t ur ned t hr ough in n t h second =  0  2 2    + t ,

t =

M OTI ON OF PROJE CTI LE When a par t icle is pr oject ed upwar ds at a cer t ain angle (but not ver t ical), it t r aces some pat h in t he air and falls back on at some ot her point . The velocit y of pr oject ion of par t icle has t wo component s : Ver t ical component which pr oject s t he body ver t ically upwar ds, and H or izont al component which move t he body hor izont ally in it s dir ect ion. Due to combined effect of both the components, the particle move along a par abolic pat h, t hen t he par t icle is called a pr oject ile. The pat h t r aced by a pr oject ile is called it s Tr aject or y. The angle at which t he pr oject ile is t hr own int o space is called t he angle of pr oject ion and is denot ed by as shown in figur e Velocity of Projection. The velocit y wit h which a pr oject ile is pr oject ed is known as t he velocit y of pr oject ion (u ). I t always act s along t he t angent t o t he cur ve. Range. M inimum dist ance bet ween t he point of pr oject ion and t he point wher e t he pr oject ile st r ikes t he gr ound is called t he r ange (R). Time of Flight. Tot al t ime by a pr oject ile t o r each maximum height and t o r et ur n back t o t he gr ound is called t he t ime of flight (t ). L et t = t ime of flight u = velocit y of pr oject ion, and  = angle of pr oject ion Equation of Trajectory. (Ver t ical dist ance t r avelled in t seconds) 1 gx 2 y  x tan   2 u 2 cos2  wher e,

x = hor izont al dist ance t r avelled in t seconds u cos  = init ial hor izont al velocit y of pr oject ile 2u sin  Time of flight, t = g Range, R =

u 2 sin 2 g

u2 M aximum value of R = g , wher e 2 = 90°,   = 45°

I f t he values of R and u ar e given, t hen t her e will be t wo values of sat isfying t he equat ion for R, say and  t her efor e  = 900 –  i.e. for a given r ange and velocit y of pr oject ion, t her e ar e t wo angles of pr oject ion, whose sum is 900. u 2 sin 2  M aximum height of pr oject ion, H max  , 2g

Engineering Mechanics

1.9

Ver t ical component of velocit y of hit t ing t he gr ound, v = gt wher e t = t ime of flight Result ant velocit y at t he gr ound, w  u 2 cos2  

1 2 2 g t 4

(1 / 2) gt gt  tan 1 u cos  2u cos  Vel oci t y at t i m e t : T h e h or i zon t al com pon en t of v el oci t y v x i s con st an t an d equ al t o u cos  . Ver t ical component of velocit y at t ime t is, v = u sin  – gt Angle wit h hor izont al at which t he pr oject ile hit s t he gr ound = tan 1



Result ant velocit y at t ime, t = vx2 + v2 =

u 2 cos2   (u sin   gt ) 2 =

I nclinat ion of r esult ant velocit y t o t he hor izont al = t an 1

vy

 t an 1

u 2  2 gut sin   g 2 t 2

u sin   g t u cos 

vx Velocity of H eight h : H or izont al component of velocit y v x = u cos  Ver t ical component of velocit y at height h is given by v 2y = u 2 sin 2 – 2 gh .......... ( fr om v 2 = u 2 + 2 as) H ence r esult ant velocit y v, at height h is given by v2 = v x2 + v y2 = u 2 cos2 + u 2 sin 2 – 2gh or v2 = u 2 – 2 gh

M OTI ON OF PROJECTI LE ON AN I N CLI N ED PLAN E L et a pr oject ile be pr oject ed on an inclined plane. The inclinat ion of t he plane and t he angle of pr oject ion ar e  and  r espect ively t o t he hor izont al. I n going fr om O to A t he pr oject ile has t r avelled zer o dist ance perpendicular t o plane. 2u sin (  )  Time of flight fr om O t o A, t = g cos  Range of pr oject ile on t he inclined plane, 2 u 2 sin ( – ) cos  OM OA = = cos  g cos2   For maximum r ange, –  = –  2 u2 u 2 (1 – sin ) M aximum Range, Rmax = = g (1  sin  g cos2  For pr oject ing t he par t icle down t he plane, R =

2 u 2 sin (  ) cos 

g cos2  N ote : Two angl es of elevat i on pr oducing t he same r ange ar e compl ement ar y angl es (e.g. 20° and 70°, 35° and 55° et c.)

KI N EM ATI CS OF A RI GI D BODY I N PLAN E M OTI ON

Y

Plane M ot ion of a Rigid Body

B

Plane mot ion of a r igid body t akes place if ever y point in t he body r emains at a const ant dist ance fr om a fixed plane. I n t he figur e, t he xy plane is a fixed r efer ence plane and lamina shown is r epr esent at ive of all laminas which compose t he r igid body. The z dist ance t o any point in t he lamina r emains const ant as t he lamina moves.

Tr ansl at i onal M ot i on

A

X Z

I n t r anslat ion r ot at ion as t he lamina moves, ever y st r aight link in t he lamina is always par allel t o is or iginal dir ect ion. I n t he given figur e t he line AB does not r ot at e.

Rot at i onal M ot i on The r otational motion takes place in the bodies like shafts, pulleys, flywheels etc. their an axis, usually the geometr ic axis of the body. This motion is considered as angular motion. Sometimes linear motion of any particle on the rotating body can be expressed in ter ms of its radius and the angular velocity or acceleration of the rotating body.

1.10

Engineering Mechanics

I f at any inst ant , t he angular velocit y of r ot at ing body is N r evolut ions/minut e, t hen it s angular velocit y 2 N revolution / second.  60 Angular velocit y is defined as t he angle in r adians t ur ned t hr ough by a par t icle in unit t ime. Angular displacement (  ) is given by d  r adians per second. dt Tangent i al or l i near vel oci t y of a par t i cl e under goi ng a cur vi l i near mot i on and havi ng angul ar vel oci t y is given by v = r and  = t d v2 Angular acceler at ion,  = = r dt Angular velocit y,  = t 1 Angular displacement ,  = t + t 2 2 2n  1  2   20   and = Angle moved t hr ough in n t h second = + 2 2

I M PACT I n impact , t he t ime int er val dur ing which t he for ces act ar e quit e small and usually indet er minat e. The sur face of t wo colliding bodies ar e along t he line of impact . (i ) Dir ect impact occur s if init ial velocit ies of t he t wo colliding bodies ar e along t he line of impact . (ii ) Dir ect cent r al impact occur s if mass cent r es in par t (i ) ar e also along t he line of impact . (iii ) Dir ect eccent r ic impact occur s if init ial velocit ies ar e par allel t o t he nor mal t o t he st r iking sur faces but ar e not collinear. (iv) Oblique impact occur s if init ial velocit ies ar e not along t he line of impact .

DI RE CT CEN TRAL I M PACT Energy Lost Due to I mpact Suppose a body of mass m falls on anot her body of mass M fr om a height h and let t he body of mass M penet r at e int o gr ound t hr ough a dist ance s befor e it comes t o r est . L et v be t he velocit y of mass m befor e impact and V be t he final of t he composit e mass (M + m ) aft er impact . Then

v = 2 gh Fr om conser vat ion of moment um mv = (M + m ) V m  V= V Mm L et E L be t he loss of kinet ic ener gy in impact . Then 1 1 EL = mv 2 – (M + m )V 2 2 2 Subst it ut e t he value of V fr om equat ion (i ), we have EL =

.....(i )

LM N

m m2 1 1 v 2 = 1 mv 2 1 – mv 2 – (M + m ). 2 M m ( M  m) 2 2 2

OP Q

m is always posit ve, t her e is always a loss of ener gy due t o impact . The r esist ance R offer ed by t he Mm gr ound can be calculat ed by equat ing t he wor k done t o K inet ic and Pot ent ial ener gies. 1 K .E. of impact = mv 2 2 Pont ent ial ener gy dur ing penet r at ion int o t he gr ound = (M + m ) gs Wor k done against r esist ance = R  s 1  R  s = mv 2 + (M + m ) gs 2 As

Engineering Mechanics

1.11

There can be following cases (1) When aft er t he impact bot h bodies move wit h common velocit y V.

LM N

OP Q

LM N

OP Q

m 1 mv 2 1 – M  m 2 (2) When mass M is penet r at ed t o an ext ent s and aver age r esist ance offer ed is R. Ener gy lost , E L =

m 1 mv 2 1 – M  m 2 (3) When mass M does not move aft er t he impact , i.e. V = 0 Rs=

1 mv 2 2 (4) When mass m penet r at es int o M and comes out wit h a velocit y V 2 and mass M also moves wit h a velocit y V 1. L et t be t hickness of mass M which m penet r at es and comes out . R1  s1 =

1 EL = R  t = mv 2 – 2

LM mV MN

2 2

 M V1 2 2

OP PQ

...(ii )

Collision of E last ic Bodies L et masses m and m  wit h init ial velocit ies u 1 and u 2 impact or collide. L et t heir velocit ies aft er collision be v 1 and v 2. Then fr om law of conser vat ion of moment um m 1v 1 + m 2v 2 = m 1u 1 + m 2u 2 ...(i )

N ewt on's E xper iment al L aw of Colliding E last ic Bodies This law st at es t hat , “ when t wo bodies impinge dir ect ly, t heir r elat ive velocit y aft er impact is in ‘Constant Rat io’ t o t heir r elat ive velocit y befor e impact , and is in t he opposit e dir ect ion. I f t he impact is not nor mal or dir ect but is oblique, t hen same fact holds for t he component s of t heir velocit ies along t he common nor mal at the point of cont act .” v1 – v2 Fr om t his law, we get =– e u1 – u 2

v1 – v2 ...(ii ) u1 – u 2 The const ant r at io ‘e’ is called coefficient of r est it ut ion ” or coefficient of r esilience. The value of ‘e’ depends upon t he nat ur e of t he mat er ial of t he colliding bodies, but is independent of t heir masses and magnit ude of t heir vel ocit i es. When bodies ar e per fect ly elast ic, e= 1 Fr om equat ion (ii ), m 1(v 1 – v 2) = e m 1(u 2 – u 1) e=

or

Adding equat ions (i ) and (ii ), and Now,

v1 = v2 =

(m1  e m2 ) u1  m2 (1  e)u2 m1  m2

m1 (1  e) u1  (m2 – e m1 )u2 m1  m2

kinet ic ener gy lost , EL = I nit ial K .E. – Final K .E. =

FG 1 m u H2

1 1

2



IJ FG K H

...(iii )

IJ K

m1 m 2 1 1 1 (1 – e2) (u 1 – u 2)2 m2u22 – m1 v1 2  m 2 v2 2 = 2(m1  m 2 ) 2 2 2

I nelastic impact : I n t his case, one body absor bs t he ot her or cl ings t o it , i.e. t hey have a common final veloci t y. I f e = 0, t hen equat ion (iii ) becomes

v2 = 

m1u1  m2u2 , m1  m2

v1 =

m2 u2  m1 u1 m2  m1

v1 = v 2

1.12

Engineering Mechanics

Purely Elast ic I mpact : When a movi ng body st r i k es a st at i onar y body on a smoot h sur face, t he nfi nal speed of t he movi ng body wi l l be t he i ni t i al speed of t he ot her body, whi l e fi nal speed of t he for mer l y st at i onar y body wi l l be t he i ni t i al speed of t he movi ng body, i .e. movi ng body st ops and t he st at i onar y one assumes i t s speed. For elast ic impact , put e = 1 in equat ion (iii ), we get

v1 = 2m 2u 2  u1 (m1 – m 2 ) m 2  m1

v2 =

and

2m1 u1  u 2 (m 2 – m1 ) m 2  m1

PRACTI CE EXERCI SE LEVEL-1 1. Two blocks wit h masses M and m ar e in cont act wit h each ot her and ar e r est ing on a hor izont al fr ict ionless floor. When hor izontal for ce is applied t o t he heavier, t he blocks acceler at e t o t he r ight . The for ce bet ween t he t wo blocks is (a) (M + m) F/m (b) M F/m (c) mF/M (d) mF/(M + m) 2. The combined motion of r ot ation and tr anslat ion may be assumed to be a mot ion of pur e r otat ion about some centr e which goes on changing fr om time to t ime. The centr e in quest ion is known as (a) Shear cent r e (b) M et a cent r e

(a)

t 3

(b)

t 3

t t (d) 2 3 (Take inclinat ion of plane = 30) (c)

5. AB is t he ver tical diamet er of a cir cle in a ver t ical plane. Anot her diamet er CD makes an angle of 600 wit h AB. Then t he r at io of t ime t aken by a par t icle t o slide along AB t o t he t ime t aken by it t o slide along CD is (a) 1 :

3

(b)

2 :1

(c) 1 :

(d) 3 : 2 2 6. Whi ch of t he fol l owi ng gr aphs r epr esent s t he mot i on of an obj ect s movi ng wi t h a l i near l y incr easing acceler at ion against t ime ? (a) (b)

(c) I nst ant aneous cent r e (d) Gr avit at ional cent r e 3. Two bodies of mass m 1 and m 2 ar e dr opped fr om differ ent heights h 1 and h 2 r espectively. Neglecting t he effect of fr ict ion, t he r at io of t imes t aken t o dr op t hr ough t he given height s would be (a)

m1 m2

(b)

m2 h2 m1 h1

Fh I (c) G J Hh K Fh I (d) G J Hh K

½

1

2

2

1

2

4. A block slides down a smoot h inclined plane in t ime t aft er having been r eleased fr om it s t op. An i dent ical bl ock on being r eleased fr om t he same point and falling freely will r each the gr ound in t ime

(c)

(d)

7. Two par t icles wit h masses in t he r at io 1 : 4 ar e m ov i n g w i t h equ al k i n et i c en er gi es. T h e m agn i t u de of t h ei r l i n ear m omen t u m s wi l l confor m t o t he r at io (a) 1 : 8

(b) 1 : 2

(c)

(d)

2 :1

2

8. I f two bodies one light and other heavy have equal k i n et i c en er gi es, w h i ch on e h as a gr eat er moment um? (a) H eavy body (b) L ight body (c) Bot h have equal moment um (d) I t depends on t he act ual velocit ies.

Engineering Mechanics

9. The angular moment um of a syst em is conser ved if t her e (a) ar e no for ces pr esent (b) ar e no magnet ic for ces pr esent

15. For maxi mum hor i zont al r ange, t he angl e of pr oject ion of a pr oject ile should be

(a) 30 (c) 60

(b) 45 (d) 90

(c) is no net for ce on t he syst em (d) ar e no t or ques pr esent 10. A body moves al ong a st r ai ght l i ne and t he var iat ion of it s kinet ic ener gy wit h t ime is linear as depict ed in t he figur e below

The for ce act ing on t he body is (a) dir ect ly pr opor t ional t o velocit y (b) inver sely pr opor t ional t o velocit y (c) zer o (d) constant 11. For per fectly elastic bodies, the value of coefficient of r est it ut ion is (a) 1 (b) 0.5 t o 1 (c) 0 t o 0.5 (d) zer o 12. A st eel ball is dr opped fr om a height h 1 ont o a st eel pl at e and r ebounds t o a hei ght h 2. The coefficient of restitution between the ball and plate will be h (a) 1 h2 (b)

h2 2 h1

LEVEL-2 16. Two st ones ar e pr oject ed wit h t he same velocit y (in magnitudes) but making differ ent angles with t he hor izont al. I f t heir r anges ar e equal and t he angle of pr oject ion of one is 600, t hen t he r at io of t he maximum height s at t ained (y 2/y 1) will be (a) 3 : 1 (b) 2 : 1 (c) 1 : 2 (d) 1 : 3 17. A st one is pr oject ed hor izont ally fr om a cliff at 10 m/sec and l ands on t he gr ound below at 20 m fr om t he base of t he cliff. Find t he height h of t he cliff. Use g = 10 m/sec2 (a) 18 m (b) 20 m (c) 22 m (d) 24 m 18. Two bal l s ar e pr oj ect ed fr om t he same poi nt making angles of 600 and 300 with the ver tical axis. I f both t he balls ar e to attain the same height, the r at io of t he speeds of pr oject ion V 1 / V 2 shoule be (a) 1 : 2 (b) 1 : 1 (c) 1 : 3 (d) 2 : 1 19. For a pr oject ile of r ange R, t he kinet ic ener gy is minimum aft er t he pr oject ile cover s (fr om st ar t ) a dist ance equal t o (a) 0.25R (b) 0.5R (c) 0.75 R (d) R 20. I f t he t ime of flight of a bullet over a hor izont al r ange R i s T seconds, t he i ncl i nat i on of t he dir ect ion of pr oject ion wit h t he hor izont al is

(c)

h2 2h1

(a) tan – 1

(d )

h2 h1

(c) tan – 1

13. Consi der a one-di mensi onal el ast i c col l i si on bet ween an incoming body A of mass m 1 and a body B of mass m 2 init ially at r est . For body B t o move wit h gr eat est kinet ic ener gy aft er collision (a) m 1 > m 2 (b) m 1 < m 2 (c) m 1 = m 2 (d) none of t hese 14. Dist ance of t he cent r oid of a semicir cle of r adius r fr om it s base is

4r 3 4 (c) 3r (a)

(b)

3 4r

(d)

2 3r

1.13

F gT I GH 2 R JK F gT I GH 4 R JK 2

2

(b) t an – 1 (d) tan – 1

FG gT IJ H 2R K FG 2 gT IJ H RK

21. The shaft of a mot or st ar t s fr om r est and at t ains full speed of 1800 r pm in 10 seconds. The shaft has an angular acceler at ion of (a) 3  r ad/sec2 (b) 6 r ad/sec2 2 (c) 2 r ad/sec (d) 18 r ad/sec2 22. A flywheel 200 cm in diamet er is r ot at ing at 100 r evolut ions per minut e. A point on t he r im of t he flywheel will not have (a) angular velocit y (b) cent r ipet al acceler at ion (c) t angent ial acceler at ion (d) all of t hese

1.14

Engineering Mechanics

23. Two bodi es of mass M and m ar e movi ng i n concent r ic or bit s of r adii R and r such t hat t heir per i ods ar e same. Then r at i o bet ween t hei r angular velocit y is (a) R : r

(b) m R : M r

(c) 1 : 1

(d)

m R : M r

24. A cylinder of r adius 10 cm. r olls without slipping on a hor izontal plane. At the inst ant as shown the magnit ude of t he velocit y of A r elat ive t o B is 40 cm/sec. The velocit y of t he cent r e O at this instant is (a) 10 2 cm/sec (b) 15 2 cm/sec (c) 20

2 cm/sec

(d) 25

2 cm/sec

25. A per son st anding on a unifor mly r ot at ing t ur n t able has his ar ms held close t o his chest . I f he out st r et ches his ar ms (a) t he moment of iner t ia will decr ease (b) t he angular moment um will incr ease

26. A boy t wi r ls a 15 l b bucket of wat er i n a ver t ical ci r cle. I f t he r adi us of cur vat ur e of t he pat h is 4 ft , det er mine t he minimum speed t he buck et must have when i t is over head at A so no wat er spil ls out . (a) 11.35 ft /s (b) 0 (c) 6.26 ft /s (d) 2.83 ft /s 27. A 0.6 kg br ick is t hr own into a 25 kg wagon which is i nit ially at r est . I f, upon ent er ing, t he br i ck has a vel ocit y of 10 m/s as shown, det er mi ne t he fi nal vel oci t y of t he wagon. (a) v = 0.203 m/s (b) v = 0.208 m/s (c) v = 0.240 m/s (d) v = 0.234 m/s 28. The fr i ct i on exper i enced by a body, when i n mot i on, is k nown as (a) r oll ing fr ict i on (b) dynamic fr ict ion (c) limit ing fr ict i on (d) st at ic fr ict i on 29. The bodies which r ebound after impact ar e called (a) Per fect ly inel ast ic bodi es (b) Per fect ly elast ic bodi es (c) n ei t h er Per f ect l y el ast i c n or Per f ect l y inelast i c bodi es (d) none of t hese 30. The moment of iner t ia of a soli d cone of mass m and base r adius r about it s ver t i cal axi s is

(c) t he speed of r ot at ion will decr ease

(a) 3mr 2/5 (c) 2mr 2/5

(d) t he angular velocit y will r emain const ant

(b) 3mr 2/10 (d) 4mr 2/5

AN SWERS LEVEL-1 1. (d) 11. (a)

2. (c) 12. (d)

3. (c) 13. (a)

4. (c) 14. (a)

5. (c) 15. (b)

6. (b)

7. (b)

8. (b)

9. (d)

10. (b)

22. (c)

23. (c)

24. (c)

25. (c)

LEVEL-2 16. (d) 26. (a)

17. (b) 27. (a)

18. (c) 28. (b)

19. (b) 29. (b)

20. (a) 30. (b)

21. (b)

EXPLAN ATI ON S 2.

LEVEL-1 1.

Fr om fr ee body diagr am F – N = Ma or N = ma Fr om equat ions (i ) and (ii ), we get N =

mF (m  M )

...(i ) ...(ii )

The inst ant aneous cent r e of a link is t he point about which all t he point s on t he link appear t o r ot at e at t hat inst ant .

s = ut +

3. and



u=0 1 h 1 = gt 12 , 2

1 gt 2 2

and h 2 =

1 2 gt 2 2

Engineering Mechanics

t1 t2

H ence

4.

=

FG h IJ Hh K

½

7.

1

(KE)2 =

1 m 1v 22 2

Appar ent ly r at io of t imes t aken is indepedent of t he mass of bodies.

and

For mot ion along t he plane

But kinet ic ener gies ar e equal and



h sin 

or

t2

h sin 

1 = (g sin  ) t 2 2 2h 1 = g sin 2 

For ver t ical mot ion : 1 h = gT 2 2 2h or T2 = g

Requir ed r at io =

...(ii )

8.

s = d (diamet er of cir cle) and a = g 1 d = gt 2  2 2d 2 t = ... (i ) g For mot i on al ong t he inclined diamet er CD, s=d and a = g cos 600 = g/2 1 g d =   T2  2 2 4d or T2 = ... (ii ) g Fr om equat ions (i ) and (ii ) we get t 1 = T 2 The acceler ation which var ies linear ly wit h time can be expr essed as dv a = = a0 + kt dt Hence no curve shows with acceleration versus time, kt 2 2 Cur ves (a), (d) ar e dr awn wit h velocit y against t ime. None of t hem sat isfies t he above r elat ions.

v = v 0 + a0 t +

(b) sat isfies t his r elat ion.

=2

v1 m1 m1 v1 =  v m2 2 m 2 v2

1 2=1:2 4 L et light body be L and heavy body H



t 2 For mot ion along t he ver t ical diamet er AB

s = s0 + v 0t +

½

=

T = t sin  = t sin 30 =

H owever,

FG IJ H K

m2 v1 = m1 v2

or

... (i )

m1 1  m2 4

m 1v 21 = m 2v 22

N ow,

Fr om equat ions (i ) and (ii ), we get

6.

1 m v2 2 1 1

2

a = g sin  and l =

5.

(K E)1 =

1.15

a0 t 2 kt 3 + 2 6

1 1 m L vL 2 = m H vH 2 2 2

FG IJ H K

mL v = H mH vL

or

2

<1

m L vL Momentum of light body v = = L . m H vH Momentum of heavy body vH

FG v IJ Hv K

2

H L

=

vH <1 vL

H ence moment um of light body is less t han t he moment um of heavy body or heavier body has gr eat er momentum as compar ed to light er body. 9.



Net torque on a system ,  n = time rate of change of t he syst em’s angular moment um,  dL i.e. dt   At const ant L , t her efor e,  n = 0. Thi s condi t i on may be sat i sfi ed i f t her e ar e tor ques pr esent, as long as ther e is no net tor que.

10.

FG IJ H K

dx 1 m dt 2 wher e k is const ant Differ ent iat ing, we get K inet ic ener gy =

m

dx dt

F d xI = k GH dt JK 2

2

2

= kt

1.16

Engineering Mechanics

For ce = mass  acceler at ion

FG IJ H K

2

20 = 2 sec. 10 Since hor izont al component of velocit y r emains const ant , t her efor e

17. Time of flight =

k v 11. Value of coefficient of r est it ut ion is unit y, if t he collision is elast ic, i.e. if no ener gy is dissipat ed dur ing collision. = m

d x dx = k/ 2 dt dt

=

1 2 gt 2 1 =  10  22 2 = 20 m

h =

Minimum value of coefficient of restitution would be zer o for plast ic collision, i.e., if ener gy get s ent ir ely dissipat ed. 12. Velocit y of ball just befor e impact ,

v1 =

2 g h1

I t s velocit y as it r ebound, v2 =

=

2 g h2 .

v2 – 0 0 – v1

=

2 gh 2

=

– (– 2 gh1

h2 h1

13. Aft er elast ic collision, t he velocit y of st at ionar y body, 2m1 v v2' = m1  m2 1

RS T

2m1 1 v1 K inet ic ener gy = m2 m 2 1  m2 2

=

2m1 m2

bm  m g 1

2

UV W

Ar ea =



V1 = 1: V2

20.

V02 sin 2  2 2g  2 2g V0 sin 2  1

=

sin 2 30  sin 2  2 = si n 2 60 sin 2  1

1 3

3

V 02 sin 2 2 sin  cos  = V02  g g

T =

2V0 sin  g T 2 g2 4 sin 2 

g 2T 2 4 sin 2 



R =

or

= t an – 1

21.

=1:3



1

R =

V02 =

16. Si nce r anges and vel oci t i es of pr oject i on ar e same and angl e of pr oj ect i on of one i s 60 0, t her efor e angle of pr oject ion of t he ot her would be (90º – 60º) = 30º

2

2

2V0 sin  V 2 sin 2 = 0 2g 2g which is half of t he r ange.

4r 3

LEVEL-2

FG V IJ HV K

= V 0 cos  

2

=

si n 2 60

=

= (hor izont al component of velocit y)  (half t he t ime of flight )

v12

V 2 sin 2 15. H or izont al r ange, R = 0 g For maximum hor izont al r ange, sin 2 = 1 or 2 = 90 or  = 45

H2 H1

H 1 = H 2;

 yc =

sin 2 30 

2

2

19. K i net i c ener gy i s mi ni mum at t he poi nt of maximum height , and t hat height is at t ained dur ing half of t he t ot al t ime of flight Dist ance cover ed

1 2 r 2

x c = r,

But

2

H ence k i net i c ener gy wi l l have t he gr eat est possible value when m 1 > m 2. 14.

FG V IJ HV K 1

By definit ion,

e=

V 2 sin 2  2 2g H2 = 2  2 2g H1 V1 sin 2  1

18.



F gT I GH 2 R JK

2 sin  cos  gT 2 = g 4 tan 

2

= t

2   1800 = 0 +  t 60 or  = 6 r ad/sec2 22. Since ther e is no change in magnitude of angular speed, t her e will be no t angent ial acceler at ion. 2 2 23. Given, T1 = , T2 = 1 2 T1   = 2 T2 1 or

Engineering Mechanics

Since t ime per iods of t wo bodies ar e st at ed t o be same, t heir angular velocit ies will confor m t o t he r at io 1 : 1. 24. Velocity = r adius  angular velocity



V A/B = r A/B  = 10 2  = 40 cm/sec  = 2 2 r ad/sec

Thus C is t he inst ant cent r e of velocit y.



V0 = 10 = 20 2 cm/sec

25. I n the absence of an exter nal tor que, the angular moment um is conser ved,

i.e., I  = const ant An incr ease in t he value of moment of iner t ia (due t o st r et ching of ar ms) will br ing about a decr ease in t he value of speed of r ot at ion of t he t ur n t able. 26. Cent r ifugal acceler at ion has t o be equal t o t he do gr avit y accel er at i on (g)

1.17

v2 =g r

v=

32.17 * 4 = 11.35 ft /s

27. T he m oment um i n t h e hor i zon t al di r ect i on r emain same. So

m 1v 1cos () = (m 1 + m 2)v 2 0.6 * 10cos (30) = (0.6 +25)v 2

 v = 0.203 m/s 28. Roll ing fr i ct i on - when body r oll s. Dynamic fr i ct i on - when body moves. L i mi t ing fr ict ion - when body just t o st ar t move. St at ic fr ict i on - when body doesn't move. 29. I f body r ebounds wit h same vel ocit y t hen i t is per fect ly elast i c body. I f body st i ck s wit h same velocit y t hen it is per fect ly i nel ast i c body.

2

Engineering Materials

CHAPTER M ATE RI AL STRU CTU RE Types of M at er ial St r uct ur e

(1) M acro Structure : This is seen by low-power magnificat ion of upt o 10x or t he naked eye. These may be obser ved dir ect ly on a fr act ur e sur face or on a for ging specimen. I nt er nal flaws open up under appli ed st r ess. (2) M icro Structure : The st r uct ur e is obser ved under an opt ical micr oscope at magnificat ions r anging fr om 20x to 2000x. Full infor mat ion r egar ding str uct ur e can be obtained only by examinat ion of pr epar ed sect ions of t he specimen. (3) Crystal Structure : This st r uct ur e descr ibes t he r egular it y and t ype of at omic ar r angement wit hin a cr yst al. (4) Electronic Structure : This r efer s t o t he elect r ons in t he out er most shells of individual at oms t hat for m t he solid. Spect r oscopy is used t o st udy t his t ype of st r uct ur e. (5) N uclear Structure : This is det er mined by nuclear spect r oscopic met hod. M ossbauer St udies and magnet ic r esonance (NM R) ar e t he common t echniques used for st udying.

CRYSTAL SYSTE M Solids can eit her be cr yst alline or non-cr yst alline (amor phous). (1) Cr yst alline Solids A cr ystal is a solid whose constituent molecules or atoms ar e arr anged in a systematic patt er n. Cr ystalline solids ar e usually built up of a number of cr ystals which may be similar or of widely var ying sizes and metallic or non-met allic. When descr ibing cr ystalline str uctur es, atoms (or ions) ar e t hought of as being solid spher es having well-defined di amet er s. Thi s i s t er med as t he "at omi c har d spher e m odel " i n F i g . C r y s t a l l i n e s t r u c t u r e wh i ch sph er es r epr esent ing near est neighbour atoms touch one another. ( h i g h l y o r d er ed st a t e) (2) N on-Cr yst alline (Amor phous) I n mat er ials (such as glass) which ar e non-cr yst alline, called amor phous, t he int er nal st r uct ur e is not based on a r egular r epet it ion pat t er n.

Fig. Amorphous structure (disordered state)

Cr yst al l ogr aphy I n cr yst allogr aphy t he st r uct ur e implies t he ar r angement and disposit ion of at oms wit hin a cr yst al.

( a)

Fig. Space lattice

( b)

2.2

Engineering Materials

Br avious L at t ice Ever y mat er ial is a building mat er ial block of same unit called br avious lat t ice.

Types of Br avious lat t ice Ther e ar e 7 differ ent t ypes of br avious lat t ices. Table : Charact erist ics of Basic crystal syst ems

S. N o.

Crystal System

Relat ion between primitives

I nt er face angles

Examples

1. Cubic

a=b=c

 =  =  = 90º

CaF 2, NaCl, Au, Cu, Al and Fe

2. Tet r agonal

a = b c

 =  =  = 90º

SnO2, NiSO4, Sn, TiO3

3. Or thogonal

abc

 =  =  = 90º

BaSO4, MgSO4, KNO3

4. Hexagonal

a=bc

 =  = 90º;  = 120º

SiO2, AgCl, Zn, Gr aphit e

5. Rhombohedral

a=b=c

 =  =   90

CaSO4, SiO2, CaCO3

6. M onoclinic

abc

 =  = 90º 

FeSO4, CaSO4, 2H 2O, NaSO4

7. Tr iclinic

abc

      90º

CuSO4, K 2Cr 2O7

U nit Cell Geometry

Engineering Materials

2.3

At omic Packing F act or (APF ) I t is defined as fr act ion of volume applied by spher ical at oms as compar ed t o t he t ot al available volume of t he st r uct ur e. I t is also called fr act ion or r elat ive densit y of packing. Volume of atoms in a unit cell At omic packing fact or = Volume of the unit cell

Cubic St r uct ur es ( i ) Simple Cubic St r uct ur e There is one atom at each corner and this atom is shar ed by B unit cells. So number of atoms per unit cell 1 n = 8=1 8 Size of unit cell (lat t ice const ant ) = a 4 1  r 3 3 APF = a3 But a = 2r  APF = 0.5 H ence coor dinat ion number of simple cubic is 6. ( ii ) Body-Cent red Cubic St ruct ure (BCC) Ot her met als possessing BCC st r uct ur e ar e M O, V, M n, Ta, -Cr and Nb. I r on has cubic st r uct ur e. At r oom t emper at ur e t he unit cell of ir on has an at om. 1  Tot al number of at oms =  8 = 1 at om 8 BCC cr yst al has one at om at t he cent r e = 1 at om Tot al at oms in BCC = 2 at oms Along wit h one at om at each cor ner t her e is at om at t he cent r e of each unit cell which is in physical cont act wit h all t he cor ner at oms 1 n = 8+1=2 8 4 2   r 3 3 APF = a3 But 4r = a 3  APF = 0.68 H ence coor dinat ion number of bcc is 8. (iii ) Face Centred Cubic st ructure (FCC) Along wit h one at om at each cor ner, t her e is an at om on each face. 1 1 n =8 +6 =4 8 2 4 3 4  r 3 APF = a3 But 4r = a 2

 APF = 0.74 H ence coor dinat ion number of FCC is 12. I n this type of lattice structur e, atoms are located at the corners of the cube and at the centre of each face. This t ype is t ypical of t he met als Cu, Al, Pb, Ag, Au, -Fe, Ca,  -Co, et c. 1 1  Tot al at oms in FCC unit cell =  8 +  6 = 1 + 3 = 4 at oms 2 8 ( iv) H exagonal close-packed st ruct ur es A lat t ice st r uct ur e of t his t ype has an at om at each cor ner of t he hexagon, one at om each at t he cent r es of t he t wo hexagonal faces and one at om at t he cent r e of t he line connect ing t he per pendicular s in t he t hr ee r hombuses which combine and for m t he hexagonal close-packed st r uct ur e. The at omic packing fact or for an H CP met al is found t o equal 0.74. This is ident ical t o t he packing fact or of an FCC met al because each has a co-or dinat e number of 12. H CP is found in such met als as Be, M g, Ca, Zn, Cd, Ti and ot her s.

2.4

Engineering Materials

Cr yst al st r uct ur es of some common met als bcc

fcc

I r on (except 900° t o 1400 °C) Chr omium Tungsten Vanadium Molybdenum Tantalum

I r on (except 900° t o 1400 °C) Aluminum Copper Lead Ni ck et Silver Platinum

hcp Titanium Zinc Zir conium Magnesium Cobalt

N ote : Pr oper t ies of solid cr yst alline depends on t he basic cr yst al st r uct ur e of t he solid e.g. fcc mor e duct ile while hcp is less duct ile. bcc is usually har der.

DEFECTS I N TH E M ATERI ALS 1. Point s D efect s ( i ) Vacancy defect. I t appear s due t o missing of at om fr om t he lat t ice.

( ii ) I nterstitial defect. When for eign at om occupies t he int er st it ial sit e, t he defect is called int er st it ial defect .

( iii )Subst it ut ional defect . I f r egul ar at om i s r epl aced by anot her for ei gn at om, t he defect i s cal l ed substitut ional defect .

( iv) Frenkel defect. When at om in t he lat t ice point goes and occupies int er st it ial void of ot her at om, t hen it is called Fr enkel defect .

(v) Schottky defect. I n t he combinat ion of cat ion and anion if t her e is a vacancy defect , it is called Schot t ky defect . 2. Sur face Defect s ( i ) Grain boundary defect. The bound lengt h is mor e at t he gr ain boundar y due t o or ient at ion mismat ch can easily be br oken. So, at mospher ic oxygen r eact s wit h t he at om at gr ain boundar y and cor r odes. Thus finer t he gr ain st r uct ur e, lower will be t he cor r osion r esist ance. Smal l-angl e gr ai n boundar y

Or ient at ion M ism at ch

1 ( ii ) Tilt boundary defect. When or ient at ion mismat ch at t he gr ain boundar ies is – 1, gr ain boundar ies 2 ar e called t ilt boundar ies.

Engineering Materials

2.5

(iii ) Twin boundary defect. When or ient at ions on one side ar e mir r or image of opposit e side, such gr ain boundar y defect s ar e called t win boundar y defect s. ( iv) St acking fault s. When gr ai n boundar i es on bot h t he sides ar e par al lel t o each ot her wi t h sl i ght dist ur bance of bonds in a ver y r egion, such gr ain boundar ies ar e called st acking fault s. 3. Line Defects ( i ) Edge dislocation. Alt hough millions of dislocat ions ar e alr eady t her e in t he mat er ial but assume t hat t her e is a per fect mat er ial and one face of t he mat er ial is fixed, i.e. r est r ict ed t o move. On t he opposit e face, a unifor m pr essur e is applied on 50% ar ea. Once t his pr essur e exceeds beyond a cer t ain value, slipping of at oms t akes place. This unit amount of plast ic defor mat ion which appear s in t he dir ect ion of applied load is called bur ger vect or. Ther e will be a boundar y ‘AB’ as shown in t he figur e below bet ween slipped and unslipped r egions called edge dislocat ion . I n t his case bur ger vect or is per pendicular t o t he dislocat ion line.

Ther e appear s an ext r a half plane at t he dislocat ion line. The int er st it ial void at t he dislocat ion line is lar ger in size. Upon applying load on any mat er ial, dislocat ions move in t he dir ect ion of applied load. I f dislocat i ons comes out of t he sur face, it is said t hat plast ic defor mat ion has begun in the mater ial. The r epr esentat ion of dislocat ion mot ion is phase in t he figur e below. I f ext r a half plane is above t he dislocat ion line it is called posit ive edge dislocat ions r epr esent ed by  and if half plane is below t he dislocat ion line it is called negat ive edge dislocat ion r epr esent ed by symbol T.

Fig. Dislocation M otion

( ii ) Screw dislocation. I t is for med by a shear st r ess t hat is applied t o pr oduce dist or t ion shown in figur e below. The scr ew dislocat ion is t r aced ar ound t he dislocat ion line by t he at omic planes of at oms. The bur ger vect or in t his case is par allel t o t he dislocat ion line.

PH ASE DI AGRAM S Phase diagr ams pr ovide a convenient way of r epr esent ing which st at e of aggr egat ion (phase or phases) is st able for a par t icular set of condit ions. I n addit ion, phase diagr ams pr ovide valuable infor mat ion about melt ing, cast ing, cr yst allizat ion, and ot her phenomena. I t helps us t o under st and and pr edict micr ost r uct ur e of alloy. Following t er ms ar e r equir ed t o under st and t he phase diagr ams :



Component : I t is a chemically r ecognizable species (Fe and C in car bon st eel, H 2O and sucr ose in sugar solut ion in wat er ). A binar y alloy cont ains t wo component s, a t er nar y alloy - t hr ee, et c.



Phase : I t is a por t ion of a syst em t hat has unifor m physical and chemical char act er ist ics. Two dist inct phases in a syst em have dist inct physical and/or chemical char act er ist ics (e.g. wat er and ice, wat er and oil) and ar e separated fr om each other by definite phase boundaries. A phase may contain one or more components. A phase is a st r uct ur ally homogeneous por t ion of mat t er.

2.6

Engineering Materials



A single-phase system is called homogeneous, systems with two or mor e phases ar e mixtur es or heter ogeneous syst ems. A solut ion (liquid or solid) is phase wit h mor e t han one component ; a mixt ur e is a mat er ial wit h mor e t han one phase.



Solvent : M ajor component in solut ion, while solute is minor component .



System : The ser ies of possibl e alloys consist ing of t he same component s but wit hout r egar d t o all oy composit ion.



Solid solution : I t consist s of at oms of at least t wo differ ent t ypes wher e sol ut e at oms occupy ei t her substitutional or interstitial positions in the solvent lattice and the cr ystal str uctur e of the solvent is maintained.



Solubility Limit of a component in a phase is t he maximum amount of t he component t hat can be dissolved in it (e.g. alcohol has unlimit ed solubilit y in wat er, sugar has a limit ed solubilit y, oil is insoluble). The same concept s apply t o solid phases: Cu and Ni ar e mut ually soluble in any amount (unlimit ed solid solubi lit y), while C has a limited solubilit y in Fe. I n gener al, solubility limit changes wit h t emper atur e. I f solute available is mor e t han t he solubilit y limit t hat may lead t o for mat ion of differ ent phase, eit her a solid solut ion or compound. The pr oper t ies of an alloy depend not only on pr opor t ions of t he phases but also on how t hey ar e ar r anged st r uct ur ally at t he micr oscopic level. Thus, t he micr ost r uct ur e is specified by t he number of phases, t heir pr opor t ions, and t heir ar r angement in space.



Phase equilibrium : I t is t he set of condit ions wher e mor e t han one phase may exist . I t can be r eflect ed by const ancy wi t h t i me i n t he phase char act er i st i cs of a syst em at const ant t emper at ur e, pr essur e and composit ion. The st at e of equilibr ium is never complet ely achieved because of ver y slow r at e of appr oach of equilibr ium in solid systems. This leads to non-equilibrium or meta-stable state, which may persist indefinitely and which has mor e pr act ical significance t han equilibr ium phases. I n t her modynamics t he equilibr ium is descr ibed as a st at e of a syst em t hat cor r esponds t o t he minimum of t her modynamic funct ion called t he free energy. A syst em at a met a-st able st at e is t r apped in a local minimum of fr ee ener gy t hat is not t he global one.



Gibbs phase rule : These include t wo ext er nal var iables namely t emper at ur e and pr essur e along wit h int er nal var iables composit ion (C) and number of phases (P) is r elat ed t o number of independent var iables among t hese gives t he degr ees of fr eedom (F) as follows :

F+P=C+2 The degr ees of fr eedom cannot be less t han zer o so t hat we have an upper limit t o t he number of phases t hat can exist in equilibr ium for a given syst em. For pr act ical pur pose, in met allur gical and mat er ials field, pr essur e can be consider ed as a const ant , and t hus t he condensed phase r ule is given as follows : F+P=C+1 Equilibrium Phase Diagrams : A diagr am t hat depi ct s exi st ence of di ffer ent phases of a syst em under equilibr ium is t er med as phase diagr am . I t is also known as equilibr ium or const it ut ional diagr am . Equilibr ium phase diagr ams r epr esent t he r elat ionships bet ween t emper at ur e and t he composit ions and t he quant it i es of phases at equilibr ium.The equilibr ium st at e of t he syst em is defined by t wo independent par amet er s (P and T), (T and V), or (P and V). These diagr ams do not indicat e t he dynamics when one phase t r ansfor ms int o anot her. H owever, it depict s infor mat ion r elat ed t o micr ost r uct ur e and phase st r uct ur e of a par t icular syst em in a convenient and concise manner. The obt ainable infor mat ion fr om a phase diagr am can be summar ized as follows : 

To show phases ar e pr esent at differ ent composit ions and t emper at ur es under slow cooling (equilibr ium) condit ions.



To indicat e equilibr ium solid solubilit y of one element /compound in anot her.



To indicat e t emper at ur e at which an alloy st ar t s t o solidify and t he r ange of solidificat ion.



To indicat e t he t emper at ur e at which differ ent phases st ar t t o melt .



Amount of each phase in a t wo-phase mixt ur e can be obt ained.

A phase diagr am is act ually a set of solubilit y cur ves t hat r epr esent s locus of t emper at ur es above which all composit ions ar e liquid ar e called liquidus, while solidus r epr esent s set of solubilit y cur ves t hat denot es t he locus of t emper at ur es below which all composit ions ar e solid. Ever y phase diagr am for t wo or mor e component s must show a liquidus and a solidus, and an int er vening fr eezing r ange, except for pur e syst em, as melt ing of a

Engineering Materials

2.7

phase occur s over a r ange of t emper at ur e. Whet her t he component s ar e met als or nonmet als, t her e ar e cer t ain locat ions on t he phase diagr am wher e t he liquidus and solidus meet . For a pur e component , a cont act point lies at t he edge of t he diagr am. The liquidus and solidus also meet at t he ot her invar iant posit ions on t he diagr am which r epr esents an invar iant r eaction t hat can occur only under a par ticular set of conditions between par ticular phases. Phase diagr ams ar e classified based on t he number of component s in t he syst em. Single component syst ems have unar y diagr ams, t wo-component systems have binar y diagr ams, t hr ee component systems ar e r epr esented by t er nar y diagr ams and so on. When mor e t han t wo components ar e pr esent , phase diagr ams become ext remely complicat ed and difficult t o r epr esent . U nary diagrams: I n t hese syst ems, t her e is no composit ion change (C=1), t hus only var iables ar e t emper at ur e and pressur e. Thus in r egion of single phase two var iables (temper atur e and pressur e) can be varied independently. I f t wo phases coexist t hen, accor ding t o Phase r ule, either t emper at ur e or pr essur e can be var ied independently, but not bot h. At t r iple point s, t hr ee phases can coexist at a par t icular set of t emper at ur e and pr essur e. At t hese point s, neit her t emper at ur e nor t he pr essur e can be changed wit hout disr upt ing t he equilibr ium i.e. one of t he phases may disappear. Figur e-1 depict s phase diagr am for wat er. The point A, B and C (in Figur e-1) shows t he single phase, t wo phases and t hr ee phases at a par t icular t ime r espect ively. Cr it i cal point

225 at m 1 at m

Pr essur e

I ce (sol id)

0.006 at m

Wat er (l iqui d) A B

C Water Vapor Tr iple (gas) point 0.01C 100C 374C Temper at ur e

Figure 1 : U nary phase diagram for water

Binary diagrams : These diagr ams const it ut es t wo component s, e.g.: t wo met als (Cu and Ni), or a met al and a compound (Fe and Fe3C), or t wo compounds (Al 2O3 and Si 2O3), et c. The condensed phase r ule is applicable in engineer ing aspect . The pr esent at ion of binar y diagr am becomes less complicat ed, by assuming t he pr essur e constant i.e. no var iation of pr essur e. Thus binar y diagr ams ar e usually dr awn showing var iations in temper atur e and composit ion only. I t is also t o be not ed t hat all binar y syst ems consist only one liquid phase i.e. a component is complet ely soluble in t he ot her component when bot h ar e in liquid st at e. H ence, binar y syst ems ar e classified accor ding t o t heir solid solubilit y. Ext ent solid solubilit y for a system of t wo met allic component s can be pr edict ed based on H ume-Rut her y condit ions, summar ized in t he following :



Cr yst al st r uct ur e of each element of solid solut ion must be t he same.



Size of at oms of each t wo element s must not differ by mor e t han 15%.



Element s should not for m compounds wit h each ot her i.e. t her e should be no appr eciable differ ence in t he elect r o-negat ivit ies of t he t wo element s.



Element s should have t he same valence.

All t he H ume-Rot her y r ules ar e not always applicable for all pair s of element s which show complet e solid solubilit y. I mpor t ant binar y syst ems ar e descr ibed below :

I som or ph ou s syst em : B ot h t h e com pon en t s ar e com pl et el y sol u bl e i n each ot h er , e.g.: Cu -N i , Ag-Au, Ge-Si, Al 2O3-Cr 2O3. Figur e-2 depict s a t ypical phase diagr am for an isomor phous syst em made of t wo met allic element s A and B. A set of coor dinat es – a t emper at ur e and a composit ion – is associat ed wi t h each point in t he diagr am. I f t he alloy composi t ion and t emper at ur e specified, t hen t he phase diagr am al l ows det er minat ion of t he phase or phases t hat will pr esent under equilibr ium condit ions. Ther e ar e only t wo phases in t he phase diagr am, t he liquid and t he solid phases. These single-phases r egions ar e separ ated by a two-phase r egion wher e bot h liquid and solid co-exist . The ar ea in t he Figur e-2 above t he line mar ked liquidus (A'bB') cor r esponds t o t he r egion of st abilit y of t he liquid phase, and t he ar ea below t he solidus line (A'dB') r epr esent s t he st able r egion for t he solid phase.

2.8

Engineering Materials

Tem p era tu re

L i q u i d u s -A b B  L iq u id

a

U

b c

Y L +S

B

V

d

A

S o l i d u s -A d B 

e S o lid

U 0 A

10

20

30

X 40

50 60 (% ) C o m p o sitio n

V 70

80

90 100 B

Figure 2 : Phase diagram for typical isomorphous binary system

For the inter pr etation of the phase diagr am, let's consider the ver t ical line ae dr awn cor r esponding to composit ion of 50%A + 50%B and assume t hat t he syst em is under going equilibr ium cooling. The point a on t he line ae signifies t hat for t hat par t icular t emper at ur e and composit ion, only liquid phase is st able. This is t r ue up t o t he point b which lies on t he liquidus line, r epr esent ing t he st ar t ing of solidificat ion. Complet ion of solidificat ion of the alloy is r epr esented by the point, d. Point e cor r esponds to single phase solid r egion up to the r oom temper atur e. Point c lies in t he t wo-phase r egion made of both liquid and solid phases. Cor r esponding micr o-str uctur al changes ar e also shown in figur e-2. As shown in figur e-2, above liquidus only a liquid phase exist s, and below t he solidus single solid phase exist s as complet ely solidified gr ains. Bet ween t hese t wo lines, syst em consist bot h solid cr yst als spr ead in liquid phase. I t is cust omar y t o use L t o r epr esent liquid phase(s) and Gr eek alphabet s (,  , ) for r epr esent ing solid phases. Bet ween t wo ext r emes of t he hor izont al axis of t he diagr am, cooling cur ves for differ ent alloys ar e shown in figur e-3 as a funct ion of t ime and t emper at ur e. Cooling cur ves shown in figur e-3 r epr esent A, U', X, V' and B cor r espondingly in figur e-2. Change in slope of t he cooling cur ve is caused by heat of fusion. 50%A+50%B

Pure B

Temper at ur e

Pure A

A+28%B

A+74%B

Ti me

Figure 3 : Cooling curves for isomorphous binary system

Anot her impor t ant aspect of int er pr et ing phase diagr ams along wit h phases pr esent is finding t he r el at ive amount of phases pr esent and t heir individual composit ion. Pr ocedur e t o find equilibrium concent rat ions of phases :  A t ie-line or isot her m (UV) is dr awn acr oss t wo-phase r egion t o int er sect t he boundar ies of t he r egion. Per pendicular s ar e dr opped fr om t hese int er sect ions t o t he composit ion axis, r epr esent ed by U' and V' in figur e-2, fr om which each of each phase is r ead. U' r epr esent s composit ion of liquid phase and V' r epr esent s composit ion of solid phase as int er sect ion U meet s liquidus line and V meet s solidus line. Procedure t o find equilibrium relat ive amount s of phases (lever rule) :  A t ie-line is const r uct ed acr oss t he t wo phase r egion at t he t emper at ur e of t he alloy t o int er sect t he r egion boundar ies.



Engineering Materials



2.9

The r elat ive amount of a phase is comput ed by t aking t he lengt h of t ie line fr om over all composit ion t o t he phase boundar y for t he ot her phase, and dividing by t he t ot al t ie-line lengt h. Fr om figur e-2, r elat ive amount s of liquid and solid phases is given r espect ively by cV Uc CL  , CS  , UV UV and it is not iced t hat CL  CS  1

Eutectic system : M any binar y syst ems have component s which have limit ed solid solubilit y, e.g.: Cu-Ag, Pb-Sn. The r egions of limited solid solubility at each end of a phase diagr am ar e called ter minal solid solut ions as t hey appear at ends of t he diagr am. M any of t he binar y systems wit h limited solubilit y ar e of eut ect ic t ype, which consists of specific alloy composition known as eutectic composition that solidifies at a lower temper atur e t han all ot her composit ions. This low t emper at ur e which cor r esponds t o t he lowest t emper at ur e at whi ch t he l i qui d can exi st when cooled under equil i br ium condit i ons i s known as eut ect i c t emper at ur e. The cor r esponding point on t he phase diagr am is called eut ect ic point . When t he liquid of eut ect ic composit ion is cooled, at or below eut ect ic t emper at ur e t his liquid t r ansfor ms simult aneously int o t wo solid phases (t wo t er minal solid solut ions, r epr esent ed by  and  ). This t r ansfor mat ion is known as eut ect ic r eact ion and is wr it t en symbolically as : L iquid (L )  solid solut ion-1 () + solid solut ion-2 ( ) This eut ect ic r eact ion is called invar iant r eact ion as it occur s under equilibr ium condit ions at a specific t emper at ur e and specific composit ion which cannot be var ied. Thus, t his r eact ion is r epr esent ed by a ther mal hor izontal ar r est in t he cooling cur ve of an alloy of eut ectic composit ion. A typical eut ectic type phase diagr am is shown in figur e-4 along wit h a cooling cur ve. As shown in figur e-4, t her e exist t hr ee single phase r egions, n am el y l i qu i d (L ),  an d  ph ases. T h er e al so ex i st t h r ee t w o ph ase r egi on s : L + , L +  and +  . These t hr ee t wo phase r egions ar e separ at ed by hor izont al line cor r esponding t o t he eut ect i c t emper at ur e. Bel ow t he eut ect i c t emper at ur e, t he mat er i al is ful ly sol id for al l composi t i ons. Compositions and r elative amount of the phases can be deter mined using tie-lines and lever r ule. Compositions t hat ar e on l eft -hand-si de of t he eut ect i c composit i on ar e k nown as hypo-eut ect ic composi t i ons whi l e compositions on r ight-hand-side of the eutectic composition ar e called hyper-eutectic compositions. Development of micr o-st r uct ur e and r espect ive cooling cur ves for eut ect ic alloys ar e shown in figur e-5, 6, 7 and 8 for differ ent composit ions. The phase t hat for ms dur ing cooling but befor e r eaching eut ect ic t emper at ur e is called pro-eut ect ic phase. c d g - L iqui dus c b d f g - Sol idus a b /e f - Sol vus

, -Terminal solid solutions

c

L Temperat ur e

L+

L+ f

b 

g



d Eut ect ic poi nt Eut ect ic i sot her m + H yper -eut ect i c composi t ions e Composit ion (%wt .)

H ypo-eutect ic composi t ions a A

Figure 4 : Typical phase diagram for a binary eutectic system L g 

b

L+ d



f

+ a

L  Composit ion (%wt .)

Temperat ur e

Temper at ur e

c

A

B

e B

Ti me

Figure 5 : Cooling curve and micro-structure development for eutectic alloy that passes mainly through terminal solid solution

Engineering Materials

Temperat ur e

c

L

g

L+

f +  L  

a

A

L+

b



Temperat ur e

2.10

Composit i on (%wt .)



e

Ti me

B

F igure 6 : Cooling curve and micro-structure development for eutectic alloy that passes through t erminal solid solution wit hout formation of eutectic solid

g

L

L+

L+ f 

b 

Pr o-eut ect ic phase Eut ect ic phase

L  

a A

Temper at ur e

Temper at ur e

c

+ e

Composit i on (%wt .)

B

Ti me

F igur e 7: Cooling curve and micr o-st r uctur e development for eutectic alloy that passes t hr ough hypo-eut ect i c r egion c Temperat ur e

Temperat ur e

g

L

L+

L+ f

b 

 + 

a A

L  

Eut ect ic phase

Composit i on (%wt .)

e B

Ti me

F igur e 8: Cooling curve and micr o-st r uctur e development for eutectic alloy that passes t hr ough eut ectic-point

I nvariant reactions : The eut ect ic r eact ion, in which a liquid t r ansfor ms int o t wo solid phases, is just one of t he possible t hr ee-phase invar iant r eact ions t hat can occur in binar y syst ems t hose ar e not isomor phous. I t r epr esent s t hat a liquid phase, L , t r ansfor ms int o t wo differ ent solids phases  and  upon cooling dur ing t he eut ect ic r eact ion. I n t he solid st at e analog of a eut ect ic r eact ion, called a eut ect oid r eact ion, one solid phase having eut ect oid composit ion t r ansfor ms int o t wo differ ent solid phases. Anot her set of invar iant r eact ions t hat occur oft en in binar y syst ems ar e per it ect ic r eact ion wher e a solid phase r eact s wit h a liquid phase t o pr oduce a new solid phase, and in per it ect oid r eaction , t wo solid phases r eact to for m a new solid phase. Perit ectic r eact ion is commonly pr esent as par t of mor e-complicat ed binar y diagr ams, par t icular ly if t he melt ing point s of t he t wo component s ar e quit e differ ent . Per it ect ic and per it ect oid r eact ions do not give r ise t o micr o-const it uent s as t he eut ect ic and eut ect oid r eact ions do. Anot her invar iant r eact ion t hat involves liquid phase is monot ectic r eact ion in which a liquid phase t r ansfor ms int o a solid phase and a liquid phase of differ ent composit ion. Over a cer t ain r ange of composit ions t he t wo liquids ar e immiscible like oil and wat er and so constit ut e individual phases, t hus monot ect ic r eact ion can said t o be associat ed wit h miscibilit y gaps in t he liquid st at e. Example syst em for monot ect ic r eact ion : Cu-Pb at 954C and 36%Pb. Analog t o monot ect ic r eact ion in solid st at e is monot ect oid r eact ion in which a solid phase t r ansfor ms t o pr oduce t wo solid phases of differ ent composit ions. Anot her not able invar iant r eact ion t hat is associat ed wit h liquid immiscibilit y is synt ect ic r eact ion in which t wo liquid phases r eact t o for m a solid phase. All t he invar iant r eact ions ar e summar ized in t he t able-1 showing bot h symbolic r eact ion and schemat ic par t of phase diagr am.

Engineering Materials

2.11

Table 1 : Summary of invariant react ions in binary systems R eaction

Symbolic equation

Eut ect ic

L  + 

Eut ect oid

 + 

Per it ect ic

L + 

Per it ect oid

 + 

M onot ect ic

L1  L2 + 

M onot ect oid

1 2 + 

Synt ect ic

L 1 + L 2 

I ntermediate phases : An int er mediat e phase may occur over a composit ion r ange (int er mediat e solid solution) or at a r elat ively fixed composit ion (compound) inside t he phase diagr am and ar e separ at ed fr om ot her t wo phases in a binar y diagr am by t wo phase r egions. M any phase diagr ams cont ain int er mediat e phases whose occur r ence cannot be r eadily pr edict ed fr om t he nat ur e of t he pur e component s. I nt er mediat e solid solut ions often have higher elect r ical r esistivit y and har dness than either of the t wo components. I nter mediate compounds for m r elat ively at a fixed composit ion when t her e exist s a st oichiomet r ic r elat ionship bet ween t he component s, for example: M g2Ni and M gNi 2 in M g-Ni syst em. These ar e called int er -met allic compounds, and differ fr om ot her chemical compounds in t hat t he bonding is pr imar ily met allic r at her t han ionic or covalent , as would be found wit h compounds in cer t ain met al-nonmet al or cer amic syst ems. Some met al-nonmet al compounds, Fe3C, ar e met allic in nat ur e, wher eas in ot her s, M gO and M g2Si, bonding is mainly covalent .

CON TROL OF M ATERI AL PROPERTI ES M etals and alloys may not possess all t he pr oper t ies r equir ed in a finished pr oduct . The mat er ial pr oper t ies can, however, be cont r olled by var ious met hods t o make t he mat er ial suit able for a given applicat ion. These met hods include 1. Alloying

2. H eat t r eat ment

3. M echanical wor king

4. Recr ystallization.

1. Al loyi ng I f for eign element occupies int er st it ial sit e it pr oduces compr essive st r ain field in t he host at oms. I t is because i nt er st i t ial voi d i s al ways smal l er t han t he si ze of i mpur i t y at om. I f i mpur i t y at om occupi es subst it ut ional sit e and if it s size is smaller, it pr oduces t ensile st r ain field. I f impur it y size i s bigger t han t he host at oms, it pr oduces compr essive st r ain field. These st r ain fields cr eat e an obst acle in t he movement of dislocat ion. This incr eases st r engt h of mat er ial. So, alloys ar e always st r onger t han t he pur e met al . Bot h car bon and nit r ogen occupies int er st it ial sit e in ir on but nit r ogen pr oduces har der and st r onger mat er ials because it pr oduces st r ain field of lar ger int ensit y.

Al l oys An alloy is a combinat ion of a met al wit h ot her mat er ials. The manufactur ing pr oper ties of an alloy depend on the pr oper ties, distr ibution, size, and shape of the var ious phases pr esent, and on the natur e of phase int er faces. The most commonly-used alloy in engineer ing is that of ir on and car bon, popular ly known as steel. The car bon pr esent in steel is in the for m of Fe3C called cementite containing 6.67% of C. Pur e ir on has t wo differ ent allot r opic for ms. Fig. (a) shows t he cooling cur ve of pur e ir on. Bet ween 1537 C and 1400C, t he solid ir on exist s in t he for m of bcc cr yst als and is commonly known as -ir on. Fr om 1400C t o 910C, t he cr yst al st r uct ur e is fcc, t he cor r esponding name being -ir on. Below 910C, t he st r uct ur e again changes back t o bcc, and t his phase is r efer r ed t o as -ir on. H owever, t her e is no basic st r uct ur al differ ence bet ween t he - and t he -phase. Fig. (b) shows t he ir on-car bon equilibr ium diagr am. I n t his figur e, t he por t ion involving t he -phase is not of much int er est so far as t he nor mal manufact ur ing pr ocesses ar e concer ned because the t emper atur e is ver y high. For casting pr ocesses, the

2.12

Engineering Materials

liquid-solid tr ansfor mation at 1125C is significant, wher eas for heat tr eatment of steels, the tr ansfor mations ar ound 723C play an impor t ant r ole. At 1125C, t he solubilit y of cement ite in -ir on is limited to 2% as indicat ed by t he point A in Fig. (b) This solid solut ion of -ir on and Fe3C is commonly t er med as aust enit e. I n t he bcc phases (i.e., - and -phase) of ir on, t he solubilit y of Fe3C is much smaller (ar ound 0.33% in t he -phase and 0.1% in t h -phase, as indicat ed by t he point s C and B in Fig. (b). The solid solut ion of Fe3C in -ir on is called fer r it e. The eut ect oid (E ') composit ion of fer r it e and cement it e is r efer r ed t o as pear lit e which consist s of alt er nat e t hin laminat es of cement it e and fer r it e.  B +L

Temperature (°C)

1537

Liquid

1537 1500

P

1400

+

-iron(bcc)

1400

L(liquid)

1492

Peritectic

1125

+L

L + Fe3C

E

A

Eutectic

(austenitc) 910 1200

-iron(fcc)

Pearlite (ferric) (eutectoid) Ferrite Cementite pearlite + pearlite

910 900

C%

-iron(bcc)

 + Fe3C

E

+ C

723

0 0.3

0.83

Cementite  + Fe3C

2

4.3

Steel

Time

Fig. (a ) Cooling curve of pure iron

Cast iron

6.67 (  100% Fe 3C)

Fig. (b) I ron-carbon equilibrium diagram

910

Austenite

Cementite °C

2

°C Temperature  (°C)

1

Ferrite Austenite

723

Ferrite

C

2

C

C%

0 0.03

Pearlite Cementite

1

Pearlite

Austenite

Pearlite C1

Hypoeutectoid steel

0.83

C2

2

Hypereutectoid steel

Fig. (c) M icrostructures of various phases of steel

The different str uctures for the various phases of steel are indicated in Fig. (c). As can be noticed, the structure of ferrite is thick and r ounded, wher eas that of cementite tends to be thin and needle-like. Ferrite is soft and cementite is very hard. The transformation of austenite into fer rite and cementite is achieved only when the cooling is slow. A rapid cooling rate transforms austenite into a metastable phase, known as martensite. Depending on the composition and temperature drop, there exists a minimum cooling rate for such a transformation. Martensite is br ittle and this proper ty limits its applicability. Theor et ical t emper at ur es acr oss which a change of phase occur s can be found out fr om Fig. (c). For a composit ion C1, t he lower and the upper cr itical temper atur es ar e l C and u C . Similar ly, for a composit ion 1 1 C2, t he cr it ical t emper at ur es ar e l C and u C . 2

2

Engineering Materials

2.13

Classificat ion of Alloy St eel Alloy st eels ar e br oadly classified int o t wo cat egor ies (a) L ow alloy st eels (b) H igh alloy st eels I n low alloy st eels, t ot al cont ent of t he alloying element s, such as Cr, Ni, M o, V, and M n, is kept wit hin 5%. Each alloying element impar t s a specific pr oper t y t o t he or iginal mat er ial. Table : Effects of alloying elements on steel Element Effect (s) Remark (s) Al

Pr omot es deoxidizat ion Pr omot es nit r iding Rest r ict s gr ain gr owt h

Typically low per cent age

B

I ncr eases har denabilit y I mpair s impact st r engt h slight ly

Typical per centage 0.001-0.003 Used in st eels wit h car bon cont ent less t han 0.6%

Co

Cont r ibut es t o r ed har dness Sust ains har dness dur ing t emper ing I ncr eases har denabilit y

Typical per cent age 0.5-2 t o incr ease har denabilit y

Cr

I ncr eases r esist ance t o cor r osion, abr asion and wear I ncr eases high t emper at ur e st r engt h

Typical per cent age 4-8 t o incr ease cor r osion and wear r esist ance

Cu

I ncr ease cor r osion r esist ance Count er act s br it t leness fr om S

Typical per centage 0.1-0.4 Typical percentage 0.25-0.40 to counteract brittleness from S

Mn

I ncr eases har denabilit y I ncr ease r esist ance Reduces ductility and weldabilit y I ncr eases har denabilit y significantly

Typical per cent age > 1 t o incr ease har denabilit y

Mo

I ncr eases st r engt h, t oughness, r ed har dnes, and hot st r engt h when used wit h Cr, M n, and V enhances cor r osion and abr asion r esist ance.

Used for H SS cut ting t ools, for ged cr ankshafts, tur bine r ot or s, high pr essur e cylinder s and boiler plat es, and gear s Typical percentage 2-5 to increase toughness and strength.

Ni

I ncrease toughness and impact strength Typical per centage 12-20 to increase cor rosion resistance I mpr oves cor r osion r esist ance Used in case har dened par t s such as high speed gear s and bear ings

P

I ncr ease har denabilit y Typical per cent age 0.2-0.9 t o incr ease st r engt h I mpr oves machinabilit y Typical per cent age 2 I ncr eases st r engt h in low car bon st eel Used in spr ing st eels I mpr oves cor r osion r esist ance St r engt hens low alloy st eels I ncr eases har denabilit y

Si

Act s as deoxidizer I mpr oves magnet ic pr oper t ies when pr esent in lar ge per cent age

Typically high percentage to improve magnetic properties

S

I mpr oves machinabilit y of ver y low car bon st eels

Typical per cent age 0.08-0.15 Nor mally consider ed an impur it y

Ti

I ncr eases aust enit ic har denabilit y Reduces martensitic hardness in Cr steels

Fixes car bon in iner t par t icles, r esult ing in r emar kable car bide for ming effect

V

I ncr eases st r engt h while r et aining duct ilit y Pr oduces fine gr ain size I ncr eases har denabilit y

Typical per cent age 0.15 For ms st able car bides t hat per sist at quit e high t emper atur e Nor mally used in combinat ion wit h chr omium

W

I mpar ts har dness and wear r esistance Typical per cent age 4 t o impar t wear r esist ance Significant ly impr oves r ed har dness Typical per cent age 18 t o impr ove r ed har dness I mpar t s st r engt h at high t emper at ur e Used in high speed t ool mat er ials.

Typical per cent age 0.2-5

2.14

Engineering Materials

2. H eat Treatment Cont r ol of mat er ial pr oper t ies can also be achieved wit hout t he addit ion of ot her element s. This is done by subject ing t he mat er ial t o a cont r olled cycle of heat ing and cooling. Take a simple example wher e aust enit e st eel (above 723C) is cooled at differ ent r at es. Fig. (a) shows t he var ious r esult ing st r uct ur es along wit h a few mechanical pr oper t ies. Transformation-begins curve 0

Lower critical temperature

1 700

Water quenched

Oil Air quenched cooled

Furnace cooled

600 1 Temperature  (°C)

Austenite Temperature > 723°C

Coarse pearlite A B Fine pearlite

Pearlite

500 Nose 400 2 300

Tr a

Austenite

ns

Bainite fo r

m

a ti

200 3

Martensite

Very tine pearlite

Fine pearlite

Coarse pearlite

on

-e

nd

sc

ur

In this region austenite-martensite transformation is instantaneous

ve

100 Transformation-completed curve Martensite 1

t1 10

t2102

103

104

105

Time t (sec)

Fig. (a) Effect of cooling rate on microstructure and properties

Fig. (b) TTT diagram for carbon steel

Obviously by changing only t he r at e of cooling, differ ent phases can be achieved. The infor mat ion on t he change of phase wit h t he cooling r at e can be convenient ly displayed wit h t he help of a t ime-t emper at ur et r ansfor mat ion diagr am (commonly known as t he TTT diagr am). I n such a diagr am, t he t emper at ur e is plot ted along the ver tical axis (using a linear scale), wher eas t he abscissa r epr esent s the t ime on a logar it hmic scale. The TTT diagr am for car bon st eel is given in Fig. (b). When aust enit e is br ought t o a t emper at ur e 1 fr om 0 (in essent ially zer o t ime) and t her eaft er held at 1, t he t r ansfor mat ion t o pear lit e begins aft er a lapse of t ime t 1, as shown by t he point A in t he figur e. Such a t r ansfor mat ion, t aking place at a const ant t emper at ur e, is known as an isot her mal t r ansfor mat ion . The point B indicat es a t ime t 2, aft er which t he t r ansfor mat ion is complet e. I n Fig. (b), the tr ansfor mations cor r esponding to other t emper atur es, viz., 2 and 3, ar e also shown. At about 600C, the tr ansfor mation star ts aft er a minimum lapse of t ime, and this par t of t he diagr am is called the nose. Below this temper atur e, austenite tr ansfor ms into bainite which is an intimate mixtur e of ferr ite and cementite (cementit e exists in the for m of tiny spher oids). Bainite cannot be pr oduced by cont inuous cooling. When the temper ature of isother mal tr ansfor mation is decr eased (above 600C), the time r equired for the transformation reduces. This results in a finer grain structure as less time is available for the growth of new nuclei. The cur ve, indicat ing t he beginning of t he t r ansfor mat ion, does not exist below about 220C. Below t his t emper at ur e, aust enit e inst ant aneously st ar t s t r ansfor ming int o mar t ensit e. I t impar t s high mechanical pr oper t ies t o st eels.

Time Temper at ur e Tr ansfor mat ion (T T T ) D iagr am Pr of. Bain heat ed t he st eel samples havi ng an eut ect oi d composi t i on at a t emper at ur e at which aust eni t e i s st abl e. These sampl es ar e subsequent l y quenched t o di ffer ent t emper at ur es bel ow 725C. The fir st sampl e was quenched t o 700C and i t was obser ved t hat for a subst ant i al per iod of t i me, t her e was no change in t he micr o st r uct ur e of ‘aust enit e’. Although aust enit e cannot be stable below 725C, t his per iod is called incubation period. Decr easing qenching temper atur e, incubation per iod decr eases and at 550C it is only ‘one sec’ decr easing quenching t emper at ur e fur t her, again t her e is incr ease in t he incubat ion per iod. On projecting % decomposition vs time graph to temperature vs time graphs two ‘C’ curves appears. This diagram is called Time-Temperature-Transformation curve (or) TTT Curve or C-Curve or S-curve or Bains Curve.

F act or s r esponsi bl e for cur ve ( i ) Driving force : G = Gfinal – Ginit ial I f G is +ve, it means final phase is st able. I f G is xer o, it means phase is neut r al. I f G is – ve, it means phase is unst able. As t her e is a decr ease in t emper at ur e, (G) dr iving for ce value incr eases and so incubat ion per iod decr ease fr om 725C t o 555C.

Engineering Materials

2.15

( ii ) At omic mobilit y (D iffusion) I t decr eases wit h decr ease in t emper at ur es slight ly below 725C, t he dr iving for ce is ver y low t hat is why incubat ion per iod is mor e. Below 550C alt hough value of dr iving for ce is ver y high but due t o lower t emper at ur e, diffusion is ver y low. This incr eases incubat ion per iod.

H eat Tr eat ment Pr ocesses used in engineer ing pr act ice 1. Aust emper i ng In the TTT-diagram, there appears two lines other than C-curves at 220C and 100C horizontal to time axis.

220C-M sStar t of mar t ensit e 100C-M f Finish of mar tensit e

Fig. Critical Cooling Rate

I f cooling r ate is such that, it just touches the nose of TTT diagr am, it is called critical cooling rat e (CCR). Any cooling r ate gr eater than (or ) equal t o CCR will not pr oduce pear lite and micr ostr uctur e car bon will fr eeze at its locat ion and it is like colloidal solution of car bon (or ) cementite int o fer r ite. This phase of ir on is called mar tensite, which is the har dest phase of ir on. All the lines on TTT diar am ar e that of decomposition of austenite into some other micr o str uctur e. Once the austenite conver t s into some other micr ostr uctur e, it never r econver t s again. The slow cooling pr ocess pr oduce coar se str uctur e and fast cooling pr ocess pr oduce fine str uctur e. Bainite cannot be pr oduced by continuous cooling. To pr oduce bainite, the sample has to be quenched below the nose of TTT diagr am but above mar tensite star t line (220C). Then sample is maintained at this t emper atur e for substant ial per iod of time till entir e austenite conver ts into bainite. The pr ocess is called Austempering. Advant ages of Aust emper ing (i ) Br it t le mar t ensit e does not for m. (ii ) Quenching cr acks bet ween cor e and sur face do not develop. (iii ) Duct ilit y is incr eased. (iv) I mpact st r engt h and t oughness ar e incr eased. Object s of heat t reatment . (i ) To incr ease t he har dness of met als. (ii ) To r elieve t he st r ess set up in t he mat er ial aft er hot or cold wor king. (iii ) To impr ove machinabilit y. (iv) To soft en t he met al. (v) To modify t he st r uct ur e of t he mat er ial t o impr ove it s elect r ical and magnet ic pr oper t ies. (vi ) To change t he gr ain size. (vii )To incr ease t he qualit ies of a met al t o pr ovide bet t er r esist ance t o heat , cor r osion and wear.

2. Anneal i ng I t is t he heat ing of st eel t o aust enit e t emper at ur e and t hen cooling slowly in t he fur nace. M ain objects of Annealing : (i ) To r educe har dness (ii ) To impr ove machinabilit y (iii ) To incr ease or t o r est or e duct ilit y (iv) To r elieve int er nal st r esses (v) To r educe or eliminat e st r uct ur al in homogeneit y (vi ) To r efine gr ain size (vii ) To pr epar e st eel for subsequent heat t r eat ment Slow cooling r esults in t he for mation of spher oidal car bide and coar se lamilar pear lite. These pr oducts ar e ver y soft. The cooling r at e dur ing annealing var ies fr om depending upon alloying element in the steel and lower r ate of cooling is used for alloy steels as compar ed to plain C-steels. Annealing r esult s in t he for mation of

2.16

Engineering Materials

fer r ite, spher oidal cementite and coar se pear lite. All t heses phases and micr o-const ituent s ar e r elatively soft and ther efor e called softening treatment and produces relatively lower hardness values while ductility increases.

Types of Annealing process Ther e ar e four t ypes of anneling pr ocess : ( i ) Full Annealing : Pr imar y object ive of t his pr ocess is t o r educe har dness and incr ease duct ilit y. The pr ocess involves (a) H eat ing t he st eel t o about 50 t o 75C above t he upper cr it ical t emper at ur e for hypoeut ect oid st eels and above t he lower cr it ical t emper at ur e for hyper eut ect oid st eel. (b) H olding it at t his t emper at ur e for a sufficient t ime depending upon t he t hickness of object t he holding t ime is 3-4 min/mm of t hickness of t he lar gest sect ions. (c) Slowly cooling in t he fur nace. The r at e of cooling var ies fr om 30C-200C per hour depending upon t he composit ion and st abilit y of aust enit e, when slow cooling in done aust enit e is decompose int o coar se pear lite and fer r it e st r uct ur es in hypoeut ect oid steels and in eutectoid st eels is conver ted int o pear lite. Hyper -eutectoid steels may under go full annealing after hot wor king like cooling. Austenite decomposes int o pear lit e and cement it e. ( ii ) Process Annealing : I t is usually car r ied out t o r emove t he effect s of cold wor king and t o soft er t o make i t sui t abl e for fur t her pl ast i c defor mat i on as i n t he case of sheet and wi r e i ndust r i es. I t i s t he r e-cr yst allizat ion of cold wor ked st eel by heat ing below cr it ical t emper at ur e. The exact t emper at ur e depends upon t he ext ent of cold wor king gr ain size composition and holding time Dur ing r e-cr ystallization t he defor med gr ains ar e r eor ient ed t o incr ease plast icit y and r emove int er nal st r esses. I t should be not ed t hat t his pr ocess pr oduces no change in micr ost r uct ur e and t he pr ocess is pr imar ily used for l ow car bon st eels. (iii ) Spheroidise Annealing : This process is applied to high carbon steels which are difficult to machine. I t causes formation of all carbides in the steel in the for m of ver y small globules or spheroids like spheres. The process consists of heating the steel near the lower critical temperature (730-770C), holding at this temperature and their cooling slowly to 600C. The rate of cooling in the furnace is 25-30C/hr. Varieties of heat treatment can be used to produce a spheroidized structure but all of them are relatively lengthy and costly. ( iv) Diffusion Annealing : Diffusion annealing or homogenizing is applied t o alloy st eel ingot s and heavy complex cast ings for eliminat ing t he chemical in homogeneit y wit hin t he separ at e cr yst als by diffusion. H omogenizing is car r ied out at t emper at ur e 1000-1200C. The opt imum t emper at ur e being 1150C at which diffusion pr oceeds quit e easily. The st eel is heat ed t o 1150C as t he highest r at e available for t he fur nace. H ol di ng t i me i s k ept at mi ni mum, fol l owed by cool i ng wi t h t he fur nace for 6-8 hour s t o 800-850C and t hen fur t her cooling in air. Aft er H omogenizing cast ings under go full annealing t o r efine t heir st r uct ur es. H igher t emper at ur es ar e select ed t o enable diffusion mor e and mor e.

3. N or mal i zi ng M ain object s of nor malising ( i ) To r efine the gr ain str uctur e of the st eel to impr ove machinability, tensile str ength and str uctur e of weld. ( ii ) To r emove st r ains caused by cold wor king pr ocesses like hammer ing, r olling, bending et c., which makes t he met al br it t le and unr eliable. ( iii ) To r emove dislocat ions caused in t he int er nal st r uct ur e of t he st eel due t o hot wor king. ( iv) To impr ove cer t ain mechanical and elect r ical pr oper t ies. Process : I t is a final heat t r eat ment pr ocess which is given t o a pr oduct , which ar e subject ed t o r elat ively high st r esses, t he pr ocess consist of heat ing st eel t o a t emper at ur e 40-50C above t he line which aust enit e is st able, holding at t hat t emper at ur e for a shor t per iod and subsequent ly cooling in air at r oom t emper at ur e. This is called air quenching, Nor malizing pr oduces micr o-st r uct ur es consist ing of fer r it e and pear lit e for hypo-eut ect ic st eel and pear lit e and cement it e for hyper et ect oid st eels. This alloy st eel st r uct ur e consist of fer r it e and sor bit e (medium pear lit e). Nor malizing r aises the yield point ultimate t ensile st r engt h and impact st r engt h in st eels. Nor malized st eels ar e har der and st r onger but less duct ile t hen annealed st eels wit h t he same composit ion.  The upper cr it ical t emper at ur e for a st eel depends upon it s car bon cont ent . I t is 900C for pur e ir on, 860C for steels with 2.2% carbon, 723C for steel with 0.8% car bon and 1130C for steel with 1.8% carbon.  St eel cont aining 0.8% car bon is known as eut ect oid st eel. St eel cont aining less t han 0.8% car bon is called hypoeut ect oid st eel and st eel cont aining above 0.8% car bon is called hyper eut ect oid st eel.

Engineering Materials

2.17

Compar ison bet ween N ormalizing and Annealing (i ) Nor malizing r equir es a heat ing r ange which is about 40C above t hat of annealing. (ii ) M echanical pr oper t ies of st eels ar e bet t er t han t hose pr oduced by annealing. (iii )H eat t r eat ment pr ocess is of shor t dur at ion due t o incr eased r at e of cooling of t he met al in air. (iv) I f mechanical pr oper t ies is not t he main aim of t he heat t r eat ment , bet t er machinabilit y and r emoval of int er nal st r esses is possible in annealing t hat obt ained by nor malizing.

4. H ar deni ng H ar dening is a pr ocess in which st eel is heat ed t o aust enit e t emper at ur e held at t his t emper at ur e and t hen quenched in wat er oil or molt en salt bat hs. H ypoeut ect oid st eels ar e heat ed fr om 30-50C above t he Upper Cr it ical Temper at ur e (UCT), while hyper eut ect oid st eels ar e heat ed above t he lower cr it ical t emper at ur e (LCT). Cool i ng at a r at e hi gher t hen t he cr i t i cal val ue enabl es t he aust enit e t o super cool ed t o mar t ensit e point . Due t o r api d cool i ng car bon fr eezes at i t s l ocat i on and t he mi cr ost r uct ur e appear s l i k e col l idal sol ut i on of cement i t e i s fer r it e. Thi s mi cr ost r uct ur e i s cal l ed mar t ensit e whi ch i s ver y har d and br it t l e. har dened st eel i s ver y br i t t l e and can not be used for pr act i cal pur poses. Aft er har deni ng st eel must be t emper ed t o r educe br it t leness r elieve t he inter nal st r esses caused by har dening and t o obt ain t he desir ed mechanical pr oper t ies of st eel s i n case of t ool st eel s i t i ncr eases t he har dness and wear r et ai ning t he t oughness at t he same t i me. I n case of st r uct ur al st eel , i t i mpr oves st r engt h, duct i l i t y and t oughness.

St ages of H ardening H ar dening is car r ied out in t hr ee st ages : (i ) H eat ing t he object t o a t emper at ur e above t he cr it ical point . (ii ) H olding t he object at t his t emper at ur e for a definit e per iod. (iii )Quenching in a suit able medium. Hardening process is based upon metallurgical reaction of eutectoid, which depend upon following factors : ( i ) Adequate C-content : I n or der t o pr oduce har d st r uct ur es like mar t ensit e, at least 0.5% C must be pr esent . Car bon incr eases har dness and upt o 1% wear r esist ance sur face is obt ained. These ar e some alloys which r emain dissolved in t he aust enit e and become ver y har d dur ing decomposit ion of aust enit e t o mar t ensit e. ( ii ) Austenite decomposition : TTT diagr am r elat es t he decomposit ion of aust enit e wit h r espect t o t ime and t emper at ur e condit ions. This diagr am explains t he decomposit ion of aust enit e t o var ious st r uct ur al component s like pear lit e, banit e and mar t ensit e. ( iii ) H eating rate and H eating time : These on composit ion of st eel, it s st r uct ur e, r esidual st r esses, for m and size of t he par t t o be har dened. I f heat ing r at e is t oo high t her e will be a t emper at ur e gr adient bet ween sur face and t he cor e, t he best way is t o fir st heat t o t he r equir ed t emper at ur e and hold it at t hat t emper at ur e t he heat ing t ime for car bon t ool st eels and medium alloy st r uct ur al st eel is 25-50% mor e t han for C-st r uct ur al st eels. For high alloy st r uct ur al st eels and t ool st eels it should be 50-100% higher. H eat ing in salt bat hs may be conduct ed mor e unifor mly and r apidly t han heat ing in box fur naces. ( iv) Quenching medium : Quenching must pr ovide for a cooling r ate above the cr itical value to prevent austenite decomposition into pear lite and intermediate regions. On the martensitic transfor mation temperature r ange, cooling should be slower to avoid high internal stresses, distortion and cr acking the most widely employed quenching media are water aqueous solution, oil, air and molt en salt . Wat er and aqueous solut ion, are most widely used as quenching media in hardening car bon and certain low alloy steels which have a high cooling rate (cr itical). (v) Quenching rate (200 C/min) : I t depends upon quenching medium. H igher t he quenching r at e, mor e is t he t emper at ur e gr adient bet ween cor e and t he sur face. ( vi ) Size and Rate : L ong ar t icles of cylindr ical and ot her cr oss-sect ion should be quenched wit h t heir main axis per pendicular t o t he bot h sur face t hin and flat ar t icle, should be immer sed an edge. The dir ect ion of movement dur ing cooling should coincide wit h t he dir ect ion of immer sion. H eavy maximum ar t icles should be held st at ionar y and t he liquid should be agit at ed. (vii ) Surface condition : When a par t of a surface is quenched in water only its surface is in contact and the heat is dissipated from the par t to the water thr ough surface only. Cooling rate depends upon the surface ar ea of the par t.

2.18

Engineering Materials

5. Tem per i n g Austenite

Temperature  (°C)

M ar t ensit e which is for med dur ing quenching is t oo br it t le and hence cannot be used in many cases, t he r esidual st r esses ar e also developed dur ing mar t ensit e for mat i on hence har deni ng should be fol lowed by t emper i ng. Temper i ng consi st i n heat i ng t he har dened st eel t o a temper atur e below the lower cr itical t emper at ur e holding it for sometime and t hen cooling slowly. I t is t he final oper at ion in heat t r eat ment . Temper ing r efer s t o secondar y heat ing of mar t ensit e obtained by a r apid cooling of aust enit e. Dur ing t his pr ocess, no change of phase t akes place because t he t emper at ur e i s never r ai sed beyond t he l ower cr i t i cal t emper at ur e (723C). This pr ocess har dens t he st eel wit h r educt ion in st r engt h; al so, i t adds t o t he t oughness and duct il i t y. The di ffer ent st r uct ur es, indicated in t he figur e r esult fr om t he disper sion of car bides.

723 700

Spheroidite

500

Sorbite

300 100

Troosite Martensite Original state

F i g . T e m p er i n g o f m a r t en si t e

Advantages of Tempering : (i ) Residual st r esses ar e r elieved (ii ) Duct ilit y is impr oved (iii )Toughness is incr eased Higher the temper ing temperature, more the r esidual str esses. Hardness is reduced and toughness is incr eased at high temper ing temperatur e. The work is cooled slowly after tempering. The cooling r ate consider ably affect the r esidual stresses. The slower the cooling, lesser will be the stresses. Rapid cooling in water develops new thermal stresses.

Types of Tempering Temper ing is classified accor ding t o t he t emper ing t emper at ur e, because it has much effect on pr oper t ies of st eel . ( i ) H igh temperature tempering (Sorbite) : I t is employed at 500-650C. The r esult ing st r uct ur e consist of sor bit e which gives good st r engt h and t oughness. Residual st r esses ar e complet ely r elieved if holding t emper at ur e is 100-120C t he holding t ime can be incr eased t o give desir ed pr oper t ies t o st eel. ( ii ) M edium temperature tempering (Troosite) : This t ype of t emper ing is employed at 350-500C t he r esult ing st eel st r uct ur e consist of t emper ed t r oost it e. This pr ocess incr eases endur ance limit and elast ic limit . Aft er t emper ing t he wor k is cooled in wat er t hen by incr easing t he endur ance limit in case of spr ings. This pr ocess is used for spr ing st eel and die st eels. (iii ) Low temperature tempering : The temper atur e employs at 250C and the holding time is 1-3 hr s. This pr ocess r educes int er nal st r esses incr eases st r engt h and t oughness and pr ovides high wear r esist ance. This met hod is employed in manufact ur e of measur ing t ools and cut t ing t ools. This is also employed t o t he component s which ar e sur face har dened by car bur izing nit r iding or car bo-nit r iding.

6. Case H ar deni ng H ar d sur faces ar e pr oduced on r elat ively soft cor es by case har dening. The par t s which ar e subject ed t o wear and impact ar e har dened by sur face t r eat ment , har d sur face has good wear r esist ance and t he soft cor e has good t oughness. Since mild st eels can not be har dened by quenching, so t her e st r engt h is incr eased by case har dening. A har d wear r esist ance sur face is called case and r elat ively soft and t ough inside is called cor e.

Case hardening Techniques I . F ow Low carbon st eel ( i ) Car bur izing I t i s appl i ed t o l ow Car bon st eel up t o 0.18% car bon. I n car bur i zi ng t he car bon cont ent of t he sur face l ayer i s i ncr eased t he pur pose i s t o obt ai n a har d layer on t he wor k pi ece suface aft er heat t r eat ment s. The sur face l ayer i s enr i ched wi t h car bon upt o 1%. Car bur i zi ng i s fol l owed by har deni ng and l ow t emper at ur e t emper i ng. L ow C-st eel i s heat ed upt o 870  C i n t he at mospher i c whi ch cont ai n car bon. Fe + 2CO  FeC + CO2 Fe3C i s aust eni t e wi t h di ssol ve car bon at 870C, maxi mum amount of car bon can be dissolved i n aust enit e.

Engineering Materials

2.19

Car bur izing is per for med in t he following ways : ( a ) Pack Carburizing : The par t s ar e sur r ounded by car bur izing mixt ur e and packed in a st eel box which is t ight ly cover ed and sealed wit h clay on all sides t o pr event ent r y or exit of gases. The cont ainer is heat ed t o pr oper t emper at ur e for t he r equir ed amount of t ime. The t emper at ur e is gener ally 9501000C and t ime is 6-8 hr s. higher t emper at ur es ar e select ed t o enable diffusion mor e and mor e. The car bon diffuses int o low C-st eel dur ing t his t ime t he car bur ising mixt ur e consist of 50% good Char coal 20% BaCO3, 5% CaCO3 and 5-12% Na2CO3. Bar ium car bonat e act s as ener gizer it incr eases t he act ion of car bon on low C-st eel. The car bur izing mixt ur e is in t he for m of coar se par t icles so t hat sufficient air is t r apped, visual t he cont ainer t o for m Co. I n 2 CO  CO2 + C (at omic st at e). This at omic st at e car bon diffuses int o aust enit e. The addit ion of car bonat es act ivat es t he car bur izer t o for m t he at mospher e of CO. When heat ing per iod is over t he boxes ar e cool down t o 450-500C and t hen opened. The car bur ized par t s ar e t aken out and quenched if r equir ed. This met hod is bat ch pr oduct ion met hod. Disadvant ages : (1) H igh labour cost for packing and unpacking. (2) I t is difficult t o quench dir ect ly fr om car bur ising t emper at ur e. (3) M uch t ime is needed in heat ing and cooling. ( b) Gas Carburising : The wor k is t r eat ed in a medium of gases cont aining CO and hydr ocar bon gases such as CH 4. Pr opane but ane et c. nat ur al gas is also used which gives ver y high qualit y of car bur ized case these gases ar e mixed with definite amount of air which pr ovides O2 and for ms CO fr om hydr ocar bon gases. Wor k pieces ar e heat ed in an at mospher e of r egulat ing car bur izing gases. The t emper at ur e, maint ained is gener ally 930-950C and t he holding t ime is 3-12 hr s depending upon t he t hickness of the case t he chief r eact ion of car bur ising is t he dissociat ion of met hane and CO. This at omic st age of C diffuses int o aust enit e t his pr ocess may be per for med in cont inuous fur naces. The wor k pieces ar e placed in t he fur nace such t hat t he gases will have easy access t o t he sur face to be car bur ized. The wor k pieces ar e t hen cool down t o r oom t emper at ur e and fur t her heat t r eat ment can be done aft er war ds. This pr cess has many advant ages over pack car bur izing and hence it is ext ensively used t he pr ocess can be made fully aut omat ic. Advant ages : (1) H igh qualit y car bur ized case can be pr oduced. (2) H eat consumed is less. (3) Time consume is also less. (4) Pr ocess is clean. (5) Pr oduct ion cost is low. (6) Closer qualit y limit s can be achieved. (7) Ther e is mor e flexibilit y of oper at ion. ( c) Liquid Carburising : I n this pr ocess, molt en stat e bath containing 20% NaCN is used. Cyanide pr ovides C and N. The t emper at ur e is maint ained at 950C. The pr ocess consist of heat ing t he wor k pieces in molt en bat h at 950C. At t his high t emper at ur e C and small amount of N ar e diffused in t he sur face of wor k pieces and t he case is har dened. The only disadvant ages t hat t he cost of car bur izing salt is ver y high. ( ii ) N i t r iding I n nit r iding nit r ogen cont ent of t he sur face is incr eased. This is done by heating t he st eel in t he at mospher e of NH 3 gas t he par t s t o be nit r ided ar e placed in an air t ight cont ainer. NH 3 is passed cont inuous t hr ough t he wor k pieces at a t emper at ur e of 500 t o 650C. This at omic st at e of nit r ogen diffuses int o st eel sur faces and r eact wit h alloys of st eel making a case cont aining t he alloy nit r ides. Thus pr esence of alloying element s such as Al, Cr, M o et c. is necessar y, in or der t o obt ain a case of maximum har dness medium. C-st eel cont aining 3% of alloying element s pr oduces har dest case on t he st eel sur face. M ost common size of nitr iding case is 0.3 to 5 mm thick, it requires 30-40 hrs at 500-520C. The time can be reduced by using a double state pr ocess in which the work is first heated at 500-520C and then at 550-600C for the nitr iding process. After nitriding the wor k is cooled in the furnace in the spring of ammonia. Advant ages : (a) I t incr eases har dness of sur face layer which is har der as compar ed t o car bur ising. (b) I t incr eases wear r esist ance, endur ance limit and r esist ance t o cor r osion. (c) No heat t r eat ment is r equir ed aft er nit r iding. U ses : I t is used dur ing manufact ur e of gear s component s of machine t ools cylinder s of power ful engines, cylinder liner s, gauges, cams, values et c.

2.20

Engineering Materials

Disadvant ages : (a) Case pr oduced is br it t le. (b) Pr ocess is cost ly compar ed t o car bur ising. ( iii ) Cyaniding I t is a case of high har dness and wear r esist ance is pr oduced on C-alloy st eels. Wor k is immer sed in molt en salt bat h cont aining NaCN, which is heat ed t o 820-860C. This is usually followed by wat er quenching. The cyanide bath consists of 20-30% NaCN, 25-50% NaCl and 25-50% Na2CO3. The time r equir ed is 30-90 minutes depending upon t he dept h of case which is 0.15-0.5 mm. At omic C diffused int o st eel. The wor k can be dir ect ly quenched as soon as it is t aken out of bat h, Then low t emper at ur e t emper ing is done at 200C. The case obtained by cyaniding is of high wear r esistants and endur ances limit compar ed to car bur ising. Case har dness var ies in t he following sequence N it riding > Cyaniding > Carburizing Cyaniding is used to produce light cases on small parts like small shafts, worms, gears, nuts, springs, pins etc. ( iv) H igh t emperat ure or Deep cyaniding A deeper or t hick case of 0.5-2 mm is obt ained by heat ing t he wor k piece in a salt bat h at 930-360C for 1.5-6 hrs depending upon the depth of case required. The bath consist of 82% BaCl 2, 8% NaCN and 10% NaCl. This atomic C and N diffuse into steel after deep cyaniding. The wor k is cooled in air to r efine the gr ains. The wor k is again har dened and quenched, this is followed by low temper at ur e temper ing at 200C. Advant ages : (a) I t r equir es less t ime t han car bur ising. (b) Ther e is less dist or t ion of t he wor k piece. (c) Resist ance t o cor r osion and wear is high. Disadvant ages : (a) H igh cost . (b) Cyaniding bat h is t oxic, hence wor ker needs pr ot ect ion. (v) Carbo-nit r iding (Gas Cyaniding) A mixt ur e of ammonia and hydr ocar bon gas is used. The wor k is heat ed t o 850C in t he mixt ur e for 2-10 hrs. This is followed by quencing and then tempering is employed at 180C. Martempering can also be employed instead of low temperature tempering. Both Carbon and Nitrogen diffuses simultaneously with Carbon diffusing at a higher rate than nitrogen. The alloying elements form car bides and nitrides with Carbon and Nitrogen. The car bides and nitr ides lower the stability of austenite and troostite is formed in the case. This reduces endurances limit, ductility and toughness of the steel. Advant ages : (a) I t is per for med at lower t emper at ur e (850-870C) as compar ed t o Car bur izing (930-950C). (b) H igher r esist ance t o cor r osion and wear. (c) Oper ation is clean and ther e is no soot on wor kpiece when compar ed with liquid cyaniding. Hence medium and lar ge size wor k can be pr ocessed. (d) Nit r ogen and Car bon cont ent s in t he case can be cont r olled. (e) Toxic salt s ar e not r equir ed. (f) The wor k can be machined. I I . For M edium carbon steel ( i ) I nduct ion H ar dening This pr ocess is employed t o incr ease har dness, wear r esist ance and endur ances limit of t he sur face of t he wor k piece. H eat t r eat ment is given t o t he sur face only by supplying excessive heat t o t he sur face followed by dr ast ic quenching. The sur face is heat ed t o t he aust enit e r ange and t hen quenched immediat ely t o for m mar t ensit e. The st r uct ur e of cor e r emains unchanged because it is not effect ed by heat . The component s should cont ain 0.4-0.5% car bon or sufficient alloying element s as chr omium Ni, or M o. The wor k is placed in H elical coils called induct or. The coil consist of sever al t ur ns of wat er cooled cut ubing. Alt er nat e cur r ent is passed t hr ough t he induct or and t hus alt er nat e magnet ic field sets up. The field induces eddy cur r ent s on t he sur face layer s and heat is gener at ed. The cur r ent densit y is not unifor m t hr ough out t he cr oss-sect ion of t he wor k piece. H eat ing r at e in t he r ange of t r ansfor mat ion t emper at ur es is upt o 300C/sec and if it is 250C/sec it is 900C. I nduct ion har dening can be per for med in t hr ee ways : ( a ) Whole wor k sur face is heat ed at one t ime and quenched. This pr ocedur e is used for har dening sur faces of small component s like shaft s and shar ps of t ools.

Engineering Materials

2.21

( b) Sect ions of wor k ar e heat ed and quenched consequent ly. This pr ocedur e is used for har dening gener als of Cank shaft t eet h of gear s and cams of cam shaft . ( c) Wor k in made t o t r avel wit h r espect t o st at ionar y induct or or vice ver sa and spr ay quenching is done. This pr ocedur e is used for har dening long shaft s and axles. Gener ally induct ion heat ing is followed by low t emper at ur e t emper ing at about 180C. Advant ages : (a) Time r equir ed is ver y small hence t he pr ocess is ver y quick and pr oduct ivit y is high. (b) Scale i s not for med hence machining t ime is saved mor eover mat er ial is not l ost due t o scale for mat ion. (c) Pr ocess can be aut omat ed. (d) Dept h of har dness can be cont r olled easily. (e) Dist or t ion is r educed. (f) Due t o high speed gr ain gr owt h decar bur ising do not occur. (g) Bot h ext er nal and int er nal sur faces can be har dened. (h) I t is an efficient mass pr oduct ion met hod. Almost all component s can be unifor mly har dened i.e. t he qualit y of all t he component s is unifor m. (i ) Where ever required the hardering can be localised and there is no need to protect the remaining surfaces. Disadvant ages : Each t ype of wor k piece r equir es differ ent fixt ur es for it s holding, wher eas differ ent wor k pieces can be t r eat ed at one t ime in car bur izing and nit r iding. U ses : I t i s most l y used i n i ndust r y for har deni ng sur faces of cam shaft , cr ank shaft gear s, aut omobil e component s, cold r olling mills, splined shaft , cr ane wheels, spindles, br ake dr ums et c. ( ii ) F lame H ardening The pr ocess consi st s of heat ing t he sur face of medium car bon st eel by t emper at ur e gas fl ame at 240C-3300C and immediat ely cooling in air or in wat er. H eat may be supplied by oxyacet ylene t or ch. The fuel used by t he flame may be oxyacet ylene, nat ur al gas or der osene t he flame r apidly impar t s lar ge amount of heat t o t he sur face. The heat is supplied so quickly t o t he sur face and for so shor t t ime t he cor e r emains unaffect ed. As soon as t he desir ed t emper at ur e is achieved, wat er is immediat ely spr ayed which cools t he sur face. By pr oper contr ol of heating and cooling oper ation the cor e is not affected by the tr eatment. The thickness of har dened l ayer is 2-4 mm and it s st r uct ur e i s mar t ensit e. The st r uct ur e of under l ine layer s is t r oost omar t ensit e. Advant ages : (a) Ther e is pr act ically no dist or t ion of t he wor k piece because only small sect ions of t he wor k piece ar e heat ed. (b) As heat ing r at e is high, t he wor k sur face r emains clean. (c) Pr ocess can easily be aut omat ed. (d) Pr ocess is mor e efficient for lar ge wor k as compar ed t o induct ion heat ing. I t s is ver y economical for lar ge wor k. Disadvant ages : (a) Ver y t hin sect ions may get dist or t ed ext ensively. (b) Over heat ing may cause cr ack. U ses : To incr ease wear r esist ance and sur face har dness of pist on pins, lar ge gear s, hand t ools shaft s, cams, mill r olls et c.

Age H ar dening and Pr ecipit at ion H ar dening The pr ocess of har dening is applicable only for t hose alloys t hat exist as a t wo-phase mat er ial at t he r oom t emper at ur e and can be heat ed up t o a single phase. The phase diagr am of one such alloy is shown in t he figur e below Assuming t hat t he composit ion is 3% Cu and 97% Al, t he alloy exist s as a t wo-phase mat er ial (+ ) below a t emper at ur e 1. H owever, bet ween 1and 2, it exist s as a single-phase solid solut ion (). A solut ion heat t r eat ment pr ocess consist s in heat ing t he alloy t o a t emper at ur e bet ween 1and 3. Also, a sufficient t ime is given at t his t emper at ur e for t he mat er ial t o homogenize. A subsequent r apid quenching does not allow all t he  -phase t o separ at e out . Thus, solut ion becomes super sat ur at ed. This super sat ur at ed  -phase pr ecipit at es slowly, t he t at e being dependent on t he final t emper at ur e aft er quenching. The pr ecipit at ion t akes place at t he gr ain boundar ies and cr yst allogr aphic planes, making t he slippage of at omic layer s mor e difficult . Thus, alloy becomes har der and st r onger.

2.22

Engineering Materials

I f pr ecipitation takes place at the r oom temper atur e, a longer time is necessar y for the completion of pr ecipitation, t his pr ocess is called age hardening. I f pr ecipitation r at e is incr eased by quenching the specimen to a temper atur e higher t han t he r oom temper at ur e, t hen pr ocess is cal led precipitation hardening.

Fig. Principle of precipitation hardening

Defects due to H eat Treatment of Steels (i ) Over heating (iii ) Decar bur ization (v) Black fr act ur e (vii ) Defor mation and volume change after hardening; (ix) I nsufficient har dness aft er quenching (xi ) Excessive har dness aft er t emper ing (xiii ) Er osion

(ii ) (iv) (vi ) (viii ) (x) (xii ) (xiv)

Bur ning Excessive har dness of hot wor ked annealed st eel Quenching cr acks Warping Soft spot s I nsufficient har dness aft er t emper ing Cor r osi on.

CLASSI FI CATI ON OF EN GI N EERI N G M ATERI ALS Engi neer i ng mat er i als may be classi fi ed int o t he fol lowi ng

1. M et als M et als ar e element subst ances t hat give up elect r ons t o for m met allic bonds and conduct elect r icit y. When t wo or mor e pur e met als ar e melt ed t oget her t o for m a new met al it s pr oper t ies ar e quit e differ ent fr om t hose of t he or iginal met als, it is called an alloy. M et als possess specific pr oper t ies like plast icit y, st r engt h, lust r e, har dness, r esist ance t o cor r osion, t her mal and elect r ical conduct ivit y, malleabilit y, st iffness, t he pr oper t y of magnet ism, et c.

M et al s Act ually pur e met als ar e r ar ely used but ar e allowed t o get desir able pr oper t ies for t he applicat ion. Proper t ies of met als ( i ) Tensile st rengt h I t is t he pr oper t y of a mat er ial t o wit hst and st r et ching for ce. I t can be found by st r et ching a specimen of a mat er ial on a t est ing machine by applying known values of t he tensile load and measur ing the r esulting ext ension of t he specimen unt il t he mat er ial fr act ur es. M at er ials may have differ ent values of st r engt h when t est ed in t ension, compr ession and shear. ( ii ) Toughness This is t he abilit y of a met al t o wit hst and shock loading. Toughness may be measur ed by subject ing a t est specimen t o sudden loads using a machine designed for t he pur pose. The machine is capable of measur ing t he amount of ener gy r equir ed t o br eak t he t est piece. ( iii ) D uct il it y This is t he abilit y possessed by a met al t o wit hst and dr awing (st r et ching) wit hout showing sings of fr act ur e. Ductility r efer s to t he amount a metal will extend befor e br eaking when under a tensile load. The manufactur e of wir e, wher e a metal has t o be dr awn down fr om one diameter to another depends on the ductility possessed by a met al. ( iv) M alleabilit y This is the abilit y possessed by a met al t o withst and compr essive for ces without showing signs of fractur e. M et als t hat have t o be pr essed or dr awn int o shapes must be duct ile. (v) H ar dness I t is t he abilit y of a mat er ial t o wit hst and sur face wear. H ar dness may be measur ed by subject ing t he sur face of t he met al t o pr essur e fr om a st eel ball or point er.

Engineering Materials

2.23

H ardness testing ( a ) Vickers diamond pyramid test : This method uses a squre based diamond pyramid to make an indentation in the surface of the metal. The square impression is measur ed using a micr oscope and a hardness number is calculated. applied load Area and impression Because of t he smallness of t he indent at ions component s wit h a polished sur face finish may be t est ed wit hout not iceable damage t o t he sur face. Above a har dness value of 500, t he Vicker s t est is mor e accur at e, because t her e is less chance of t he diamond defor ming under heavier loads as would be t he case when using a st eel ball. ( b) Rockwell test : I n the test either a steel ball or a diamond cone ar e used as the indentor. I f a steel ball is used, it is refer red to as scale B, if a diamond cone, scale C or A depending upon the load used. The har dness value is shown on a dial and as no measur ement of t he impr ession is necessar y, t his met hod is ver y useful for quick r out ine har dness checking of component s. ( c) Brinell test : I n t his t est , a har dness st ell ballis for ced int o t he specimen under t est by means of a suitable standar d load. The diamet er of t he impr ession made by t he ball is measur ed using a micr oscope and t he r esult ant har dness value is found by calculat ion. P Applied load kg mm Br iness har dness number, H = = 1 Area of impression D D – D 2 – d 2 2 wher e, P = load in kilogr ammes. D = diamet er of t he ball in maillimet er s d = diamet er of t he impr ession in millimet er s.

Vicker s pyr amid har dness number =

LM N

e

j OPQ

( d ) Shore scleroscope test : I n t his t est , an indent or weighing 25g is cont ained wit hin a glass t ube and allowed t o fall fr eely down t his t ube fr om a height of 250 mm. The r ebound of t he indent or up t he t ube is measur ed fr om gr aduat ions on t he t ube and t his is t aken as an index of har dness.

Classificat ion of M et als M et als may be fur t her subdivided as ( i ) Ferrous metals : e.g. cast ir on, wr ought ir on and st eel) and alloys (e.g. silicon st eel, high speed st eel, spr ing st eel et c. ( ii ) N on-ferrous metals : e.g. copper, aluminium, zinc, lead etc. alloys br ass, br onze, dur alumin etc. The ir on gr oup which includes all ir ons and st eels ar e called fer r ous met als (fer r ous ir on), whilst ot her s ar e specified as non-fer r ous.

N on M etals Commonly adopt ed non-met allic mat er ials ar e leat her, r ubber, asbest os and plast ic. Difference bet ween M et als and N on-met als

S. N .

Property

M etals

N on-M etals

1.

Str uct ur e

All solid metals have crystalline st ructure. Exist in amorphic or measomophic forms

2.

Excit at ion of valence Easy elect r on by E.M .F. (electronmotive force)

Difficult

3.

St at e

Gener ally solids at r oom t emper at ur e

Gases and solid at ordinary temperature

4.

L ust r e

Possess met allic lust r e

Do not posses met allic lust r e (except iodine and gr aphit e)

5.

Conduct ivit y

Good conduct or of heat and elect r icit y

Bad conduct or s of heat and elect r icit y

6.

M al l eabi l it y

M al leabl e

Not malleable

7.

Ductilit y

Duct ile

Not duct ile

8.

H ar dness

Generally har d

H ar dness var ies

9. 10.

Elect r olysi s Densit y

For m anion H igh densit y

For m anions L ow densit y

2.24

Engineering Materials

2. Cer amics Cer amics ar e compounds of met allic and non-met allic element s. M ost of t hem ar e oxides e.g., silica, Al 2O3, M gO, concr et e, silicon car bide, bor on nit r ide, fer r it es, gar net s, et c. Cer amic mat er ials ar e a diver se gr oup of non-met allic, inor ganic solids wit h a wide r ange of composi t ions and pr oper t ies. Their st r uct ur e may be eit her cr yst alline of glassy. Cer amic includes all pr oduct s made fr om fir ed clay, such a br icks, t iles, fir eclaly r efr act or ies and elect r ical por celain. Clay is r eally a t her moset t ing mat er ial. I n t he r aw st at e, it s molecules ar e ar r anged in layer s which ar e separ at ed by molecules of wat er. This allows t he layer s t o glide over each ot her, leading t o t he char act er ist ic plast icit y of unfir ed clay. Fir ing dr ives off t he wat er molecules and at t he same t ime, st r ong chemical links for m bet ween t he layer s, so t hat t he day becomes har d and r igid. Since t he change is not r ever sible, clay cannot be soft ened again once fir ed, and it is t hus a t her moset t ing mat er ial.

Advant ages of Cer amics ( i ) Refr act or iness, or abilit y t o wit hst and high t emper at ur es wit hout det er ior at ion ( ii ) St r engt h and r igidit y at high t emper at ur e ( iii ) Fr eedom fr om cr eep at high t emper at ur e ( iv) H ar dness r esist ance t o wear and br it t leness (v) Resist ance t o abr asion

Some high-temperature Ceramics Subst ance Chemical name

M elting point (C)

Characteristics and U ses

Alumina

Aluminium oxide

2050

Widely used in spar k-plugs, cutting-tools, cr ucibles, pyr ometer -sheaths, gauges (with a life twenty times t hat of st eel) sur face-plat es for pr ecision- checking equipment .

Ber yllia

Ber yllium oxide

2350

Cr ucible for special mat er ials, as a moder at or in high t emper at ur e nuclear r eact or s.

Magnesta

M agnesium oxide

2800

Refractory material, furnace-linings and crucibles.

Thor ia

Thor ium oxide

3050

Not used much, because it is ‘fissionable’

Zir coma

Xir conium oxide

2690

Used in t he ‘stablised’ for m, liner s for jet and r ocket motor tubes, facing high-temper atur e furnace walls.

Tool mat er i al s Some cut t ing t ools ar e der ived fr om a mor e or less pur e cer amic base. These subst ances possess gr eat har dness and good compr essive st r engt hs, even at high t emper at ur e, t hough, compar ed wit h met als, t heir t ensile st r engt h is low, and t ools made fr om t hem ar e r elat ively br it t le. Cer amic t ools can be used at higher cut t ing speeds t han can har d met als, but t hey ar e par t icular ly useful in cut t ing t ough mat er ials such as plast ics, r ubber and wood, as well as met als. Cer amic bodies used for t his pur pose include sint er ed highpur it y alumina (cor undum), bor on car bide, r ecr yst allised silicon car bide (car bor undum), alumina bonded wit h glass and ot her mat er ials. Gr inding wheels ar e usually composed of abr asive par t icles of alumina or silicon car bide, bonded, wit h such mat er ials as clay, quar t z, or felspar.

Cer met s To get mat er ials having high melt ing point , st r engt hened r igidit y at high t emper at ur e, shock r esist ance, a combinat ion of met als and cer amics in suit able combinat ion is obt ained by power -met allur y met hods. I n this powder s of suit able par t icle size ar e t hor oughly mixed in t he cor r ect pr opor t ions and ar e t hen compr essed in dir es, at high pr essur e. A degr ee of cold-welding occur s bet ween t he met al par t icles in t he mixt ur e. Gr aingr owt h occur s acr oss t he boundr ies of t he minut e welds and knit s t oget her t he met al par t icles, giving a t ough and cont inuous st r uct ur e. H ar d r igid cer amic par t icles ar e t hen bonded in a t ough met allic mat r ix. I f a mat er i al of consi der abl e r igi di t y at high t emper at ur es is r equir ed, t hen pr opor t ion of cer amic may need t o be high, in or der t hat t he r esult ant cer met will not defor m. ( i ) Low ceramic-content : Cer met defor ms in compr ession. ( ii ) H igh cer ami c-cont ent : Cer m et wi t hst an ds comp r essive loads and is also t ough ( iii )M et al fibr es impr ove t he t ensi l e st r engt h, but t he mat er ial is also r igid, due t o t he cer amic body. Fig. Composite materials

Engineering Materials

2.25

U ses : Cer met s gener ally ar e suit able for uses such as lamp filament s, air cr aft jet - engine par t s, gas-t ur bine par t s, r ocket -engine component s, cut t i ng, dr il li ng, and gr i nding t ools, fr ict i on par t s nclear -power applicat ions, heat ing- element s, bear ings and magnet ic-cor e mat er ials

Some Cer met s Par ticle mater ial Flake silver or copper Alumina (70%)

Bond Structural type Gaphit e L aminat ed Chromium (30%) Bonded par t icles

M agnesia

Nick el

Alumina (40 – 70%) M olybdenum bor ide

I r on (30 – 60%) Nick el

Tit anium car bide

Var ious alloys cont aining M o, AI and Cr Cobal t

Tungst en car bide or Tit anium car bide

Characteristics and uses Cur r ent br ushes-low fr ict ion Ver y suit able for high-t emper at ur e ser vice, Good r esist ance t o impact and t her mal shock . Flame sprayed heat- Applicable to stainless steels, alloy steels and incor esist ant coat ing nel-to raise the working temperature by about 80C. Bonded par t icles Tur bine blades Bonded pat icles Cut t ing-t ools for machining t it anium Bonded par t icles I n air craft engines where r efract oriness, t hermal shock -r esist ance and r esist ance t o oxidat ion ar e necessary. Bonded par t icles Cut t ing-t ools for most mat er ial, including masonr y and glass met al-for ming dies.

Cer m et s ar e t ool m at er i al s wh i ch con si st s of ext r em el y h ar d, abr asion-r esist ant par t icles held t oget her by a st r ong, shock-r esist ant bonding mat er ials. The most widely used and best known consist s of particles of tungsten car bide bonded with tough, strong cobalt, though car bides of t it anium or t ant alum ar e somet i mes used i nst ead of t ungst en car bide.

Fig. Type of structure in a tungsten carbide/cabalt cermet. Particles of tungsten carbide (white) in cobalt matrix (black ) (x 100)

F ilament -r einfor ced cer amics These ar e cer met s in which a har d, r igid cer amic body is r einfor ced wit h st r ands of t ough, shockr esist ant met al.

3. Or ganic M at er ials These mater ials are der ived directly from carbon. They usually consist of carbon chemically combined with hydrogen, oxygen or other non-metallic substances. I n many instances their structures ar e fair ly complex.

Common or gani c mat er ials Plast ics and Synt het ic r ubber s ar e called polymer s because t hey ar e ter med by polymer izat ion r eact ion in which r elat ively simple molecules ar e chemically combined int o massive long-chain molecules or t hr ee dimensional st r uct ur e.

Or ganic mat er ials : (i ) Plast ics : PVC, PTFE , polyt hene, (ii ) Fibr es : t er ylene, nylon, cot t on, nat ur al and synt het ic r ubber s, leat her, et c.

Composi t es (1) M et als and alloys and cer amics (i ) St eel r einfor ced concr et e (ii ) Disper sion har dened alloys (2) M et als and alloys and or ganic polymer s (i ) Vinyl- coat ed st eel (ii ) Whisker -r einfor ced plastics (3) Cer amics and or ganic polymer s (i ) Fiber -r einfor ced plastic (ii ) Car bon-r einfor ced r ubber.

2.26

Engineering Materials

N ON -M E TALLI C M ATERI ALS These ar e used in engineer ing pr actice due to their low density, low cost, flexibility, r esistant to heat and electr icity.

Commonly used N on-met allic mat er ials 1. Pl ast i cs These ar e synt het ic mat er ials which ar e moulded int o shape under pr essur e wit h or wit hout t he applicat ion of heat . These can also be cast , r olled, ext r uded, laminat ed and machined.

Types of Plastics (i ) Thermosetting plastics : The t her moset t ing plast ics ar e t hose which ar e for med int o shape under heat and pr essur e and r esult s in a per manent ly har d pr oduct . The heat fir st soft ens t he mat er ial, but as addit ional heat and pr essur e is applied, it becomes har d by a chemical change called phenol-for maldehyde (Bakelit e), phenol-fur fur al (Dur it e), ur ea-for maldehyde (Plaskon), et c. (ii ) Thermoplastic. The thermoplastic materials do not become hard with the application of heat and pressure and no chemical change occurs. They remain soft at elevated temperatures until they are hardened by cooling. These can be remelted repeatedly by successive application of heat. common thermoplastics are Cellulose nit r at e (Celluloid), polyt hene, polyvinyl acet at e, polyvinyl chlor ide (P.V.C.) et c. The plastics are extremely r esistant to cor rosion and have a high dimensional stability. U ses : (a) manufacture of aeroplane and automobile parts. (b) For making safety glasses, laminated gear s, pulleys, self-lubricating bearing etc., due to their resilience and strength.

Polymer s M ost solid subst ances-met als, salt s and miner als-exist in a pur ely cr yst alline for m, i.e. at oms or ions of which t hey ar e composed ar e ar r anged in some r egular geomet r ic pat t er n. This is gener ally possible because at oms (or ions) ar e small and ver y easily maneouor able. I n polymer s, lar ge molecules ent angle wit h each ot her and ar e consequent ly much less, man manoeuvr able. As a r esult only a limit ed degr ee of cr yst alline ar r angement is possible with most polymers as they cool, and only restricted r egions occur in which t he linear molecular chains ar r ange t hemselves in an or der ed pat t er n. These or der ed r egions F i g . ‘C r y s t a l l i t e s ’ i n a s o l i d p l a s t i c s m a t e r i a l ar e called cr yst allit es. Thus in a solid state, polymer s consist of both cr ystalline and amor phous r egions. Highly cr ystalline polymer s cont ain up t o 90 cr yst alline r egions whilst ot her s ar e almost complet ely amor phous.

M elting point of polymers Amor phous and cr yst alline plast ics mat er ials on heat ing become pr ogr essively less r igid and show no clear t r ansit ion fr om solid t o liquid, since amor phous become less viscous as t he t emper at ur e r ises. I n hi ghly cr yst alline polymer s, a sudden change in t he r at e of expansion occur s on heat ing t o a cer t ain t emper at ur e. At t hi s poi nt al l cr yst al li ne r egions have become amor phous when soli d cr yst al li ne met al s change t o amor phous liquids. I n plast ic mat er ials t his is designat ed as melt ing point (T m ). Gr eat er t he degr ee of cr yst allint y of polyt hene, t he higher it s melt ing point , Because Vander waal’s for ces ar e much gr eat er in t he cr yst alline r egions since t he molecules ar e closer t oget her t her e.

M elting points of some crystalline polymers Pol ymer

M elti ng poi nt (T m ) (C )

Polyt hene (50% cryst alline)

120

Polyt hene (80% cryst alline)

135

Polypr opylene

176

Polyvinylchior ide

212

Polyt etr afluoroethylene

327

Engineering Materials

2.27

Glass t ransition t emper at ur e At ambient temper atur e polymer s may be either soft and flexible or har d, br itt le and glassy. The temper atur e at which t his soft -t o-glassy change occur s is called glass-t r ansit ion t emper at ur es T g.

Vicat soft ening t emperature I t offer s a useful means of compar ison bet ween plast ic mat er ials as far as t heir r esponse t o t emper at ur e change is concer ned. I t measur es t emper at ur e at which a st andar d indent or penet r at es a specific dist ance int o t he sur face of t he plast ics mat er ial using a st andar d for ce. Dur ing t he t est , t emper at ur e is r aised at a specified unifor m r at e and penet r at ion (usually 1 mm) has been at t ained.

M echanical pr oper t ies of plast ic mat er ials The str ength of plast ics mater ials is gener ally much lower than that of most other constr uctional mater ials. Never theless plastics ar e light mater ials with a r elative densit y between 0.9 and 2.0 so that when consider ed in t er ms of str ength/weight r atio they compar e favour ably with some metals and alloys. The mechanical pr oper t ies of most engineer ing met als and alloys var y ver y litt le wit hin t he r ange of ambient t emper at ur es encount er ed i n ser vi ce. Thi s i s expect ed si nce no st r uct ur al changes occur unt i l t he r ecr ystallisation temper atur e is r eached and this is usually well above 100C. With many polymer -par ticular ly t her moplast ic-mat er ials, mechanical pr oper t ies var y consider ably wit h t emper at ur e in t he ambient r egion. Bot h T m and T g will affect mechanical pr oper t ies as will t he gr adual r educt ion in Van der Waals’ for ces wit h r ise in t emper at ur e. Thus a t her moplast ic polymer may have a t ensile st r engt h of say, 70 N/mm 2 at 0C, falling to 40 N/mm 2 at 25C, and to no mor e than 10 N/mm 2 at 80C. As tensile strength falls r ise in temper atur e t her e is a cor r esponding incr ease in % elongat ion.

2. R u bber I t is one of t he most impor t ant nat ur al plast ics. I t r esist s abr asion, heat , st r ong alkalis and fair ly st r ong acids. Soft r ubber is used for elect r ical insulat ions. I t is also used for power t r ansmission belt ing, being applied t o woven cot t on or cot t on cor ds as a base. U ses : H ar d r ubber is used for piping and as lining for pickling t anks.

3. L eat her I t is ver y flexible and can wit hst and consider able wear under suit able condit ions. U ses : For power t r ansmission belt ing and as a packing or as washer s.

4. F er r odo I t is a t r ade name given t o asbest os lined wit h lead oxide. U ses : As a fr ict ion lining for clut ches and br akes.

N ON -FE RROU S M E TALS Non-fer r ous met als ar e t hose which cont ain a met al ot her t han ir on as t heir chief const it uent . Non-fer r ous met als ar e usually employed in indust r y due t o (i ) ease of fabr icat ion (cast ing, r olling, for ging, welding and machining) (ii ) r esist ance t o cor r osion (iii ) elect r ical and t her mal conduct ivit y (iv) weight

N on-F er r ous M et als used in E ngineer ing Pr act ice 1. Al umi ni um I t is whit e met al pr oduced by elect r ical pr ocesses fr om oxide (alumina), which is pr epar ed fr om a cl ayey miner al called bauxit e. I t is a light met al having specific gr avit y 2.7 and melt ing point 658C. The t ensile st r engt h of t he met al var ies fr om 90 M Pa t o 150 M Pa. I n its pur e state, metal is weak and soft for most pur poses, but when mixed with small amounts of other alloys, it becomes har d and r igid. So, it is blanked, for med, dr awn, tur ned, cast, for ged and die cast . I ts good electr ical conductivity is an impor tant pr oper ty and is widely used for over head cables. The high r esistance to cor r osion and its non-toxicity makes it useful metal for cooking utensils under or dinar y condit ion and thin foils ar e used for wr apping food items. I t is extensively used in air cr aft and automobile components wher e saving of weight is an advantage.

2.28

Engineering Materials

Aluminium Alloys The al uminium may be alloyed wit h one or mor e ot her element s like copper, magnesium, manganese, silicon and nickel. The addit ion of small quant it ies of alloying element s conver t s t he soft and weak met al int o har d and st r ong met al, while st ill r et aining it s light weight .

M ain Aluminium Alloys (i ) Duralumin : I t is an impor t ant and int er est ing wr ought alloy. Composit ion : Copper = 3.5 - 4.5% Manganese = 0.4 - 0.7% Magnesium = 0.4 - 0.7% Remaining is Aluminium This alloy possesses maximum t ensile st r engt h (upt o 400 M Pa) aft er heat t r eat ment and age har dening. Aft er wor king, if t he met al is allowed t o age for 3 or 4 days, it will be har dened. This phenomenon is call ed age hardening. I t is widely used in wr ought condit ions for for ging, st amping, bar s, sheet s, t ubes and r ivet s. I t can be wor ked in hot condit ion at a t emper at ur e of 500C. H owever, aft er for ging and annealing, it can also be cold wor ked. Due t o it s high st r engt h and light weight , t his alloy may be used in aut omobile and air cr aft component s. I t is also used in manufact ur ing connect ing r ods, bar s, r ivet s, pulleys et c. (ii ) Y-alloy : I t is also called copper-aluminium alloy . The addit ion of copper t o pur e aluminium incr eases it s st r engt h and machinabilit y. Composit ion : Copper = 3.5 - 4.5% Manganese = 1.2 - 1.7% Ni ck el = 1.8 - 2.3% Silicon, M agnesium, I r on = 0.6% each Remaining is Aluminium. This alloy is heat t r eat ed and age har dened like dur alumin. The ageing pr ocess is car r ied out at r oom t emper at ur e for about five days. I t is mainly used for cast pur poses, but it can also be used for for ged component s like dur alumin. Since Y -alloy, has bet t er st r engt h (t han dur alumin) at high t emper at ur e, t her efor e, it is much used in air cr aft engines for cylinder heads and pist ons. (iii )M agnalium : I t is made by melt ing t he aluminium wit h 2 t o 10% magnesium in a vacuum and t hen cooling it in a vacuum or under a pr essur e of 100 t o 200 at mospher es. I t also cont ains about 1.75% copper. Due to its light weight and good mechanical pr oper ties, it is mainly used for air cr aft and automobile component s. (iv) H indalium : I t is an alloy of aluminium and magnesium wit h a small quant it y of chr omium. I t is t r ade name of aluminium alloy pr oduced by H indust an Aluminium Cor por at ion L t d., Renukoot . I t is pr oduced as a r olled pr oduct in 16 guage, mainly for anodized ut ensil manufact ur e.

2. C opper I t is most widely used non-fer r ous met al. I t is a soft , malleable and duct ile mat er ial wit h a r eddish-br own appear ance. I t s specific gr avit y is 8.9 and melt ing point is 1083C. The t ensile st r engt h var ies fr om 150 M Pa t o 400 M Pa under differ ent condit ions. I t is a good conduct or of elect r icit y. I t is lar gely used in making electr ic cables and wir es, for electr ic machiner y and appliances, in electr otyping and elect r oplating, in making coins and household ut ensils.

Alloys of Copper Alloy Br asses (i ) Cap copper (ii ) Gilding metal (iii ) (iv) (v) (vi )

Car t idge metal 63/37 br ass M unt z met al L eaded br ass

Composition 91% Cu, 2% Zn 85% Cu, 15% Zn dr ess jeweller y 70% Cu, 30% Zn 63% Cu, 37% Zn 60% Cu, 40% Zn 56-61%, copper

Uses Used as deoxidiser Bullet envelops, dr awn cont ainer s, Cartr idges Cold pr ess wor k Engineer ing applications H ot st ampings 1.5-3.5% lead, balance zincExt r uded r ods

Engineering Materials

Br onzes (i ) Coinage br onze (ii ) Gun met al (iii ) 86/7/5/2 br onze (iv) Phosphor br onze (v) Bear ing br onze N ick el-Copper -Alloys (i ) Alloys-1 (ii ) Alloys-2 (iii ) Alloys-3 (iv) Alloys-4 (v) M onel met al Aluminium Alloys Y-alloy

Al-Cu alloy L m-6 RR 77

95% Cu, 3.5% Zn, 1.5% Zn 88% Cu, 10% Sn, 2% Zn 86% Cu, 7% Sn, 5% Zn, 2% Pb 93.7% Cu, 6% Sn, 0.3% P 75% Cu, 5% Sn, 20% Pb

Coins for gings Casting Cast ing for medium pr essur es Scient ific wor ks Bear ings

Cu 98%, Ni 2% Cu 94%, Ni 5% Fe 1% Cu 80%, Ni 20% Cu 55.60%, Ni 40.45% Cu 33%, Ni 68%, Fe 2%

Fir e box st ays Ship’s copper smit h’s wor k Condenser t ubes Elect r ical r esist ance and t her mocuples Tur bine blades

Cu 3.5– 4.5%, Ni 1.8— 2.3% M ag 1.2— 1.7%, M g 1.2— 1.7% Balance aluminium 92% Al, 8% Cu Si 10— 13% Balance aluminium Zn 5.5%, M g 2.8%, Cu 0.4%

Castings and for gings

2.29

Gr avit y die cast ing I nst r ument cases Air cr aft

Classificat ion of copper alloys These ar e br oadly classified int o following t wo gr oups : (i ) Copper-zinc alloys (Brass) : The most widely used copper-zinc alloy is br ass. There ar e var ious types of br asses, depending upon propor tions of copper and zinc. This is fundament ally a binar y alloy of copper wit h zinc each 50%. By adding small quant it ies of ot her element s, pr oper t ies of br ass may be gr eat ly changed. e.g., addit ion of lead (1 t o 2%) impr oves t he machining qualit y of br ass. I t has a gr eat er st r engt h t han t hat of copper, but have a lower t her mal and elect r ical conduct ivit y. Br asses ar e ver y r esistant t o atmospher ic cor r osion and can be easily solder ed. They can be easily fabr icated by pr ocesses like spinning and can also be electr oplated with metals like nickel and chr omium. Table : Composit ion and U ses of Brasses according t o I ndian St andards

I ndian St andar d designat ion

Composition in per centages

Car t r idge br ass

Copper = 70 Zinc = 30

Yellow br ass (M unt z met al)

Copper Zinc Copper Zinc Lead Copper Zinc Tin Copper Zinc Tin Copper Zinc Ni ck el

L eaded br ass

Admir alr y br ass

Naval br ass

Nickel br ass (Ger man silver or Nickel silver )

= 60 = 40 = 62.5 = 36 = 1.5 = 70 = 29 =1 = 59 = 40 =1 = 60-45 = 35-20 = 5-35

Uses I t is a cold wor king br ass used for cold r olled sheets, wir e dr awing, deep dr awing, pr essing and t ube manufact ur e.I t is suit able for hot wor king by r olling, ext r usion and st amping.

These ar e used for plat es, t ubes et c.

2.30

Engineering Materials

(ii ) Copper-t in alloys (Bronze). The alloys of copper and tin are usually called bronzes. The useful range of composition is 75 to 95% copper and 5 to 25% tin. The metal is comparatively hard, resists surface wear and can be shaped or rolled into wires, rods and sheets very easily. I n corrosion resistant properties, bronzes are superior to brasses. Types of bronzes: ( a ) Phosphor bronze: When br onze cont ains phosphor us, it is called phosphor bronze. Phosphor us increases strength, ductility and soundness of castings. Tesile strength of this alloy when cast varies fr om 215 MPa to 280 MPa but increases upto 2300 MPa when rolled or drawn. This alloy possesses good wearing qualities and high elasticity. The metal is r esistant to salt water cor rosion. Composition of the metal varies according to whether it is to be for ged, wrought or made into castings. Composit ion (accor ding t o I ndian st andar ds) of common phosphor br onze: Copper = 87-90% Tin = 9-10% Phosphor us = 0.1-3%. U sed : For bear ings, wor m wheels, gear s, nut s for machine lead scr ews, pump par t s, linings and for many ot her pur poses. I t is also suit able for making spr ings. ( b) Silicon bronze : I t cont ains 96% copper, 3% silicon and 1% manganese or zinc. I t has good gener al cor r osion r esist ance of copper combined wit h higher st r engt h. I t can be cast , r olled, st amped, for ged and pr essed eit her hot or cold and it can be welded by all t he usual met hods. U sed : For boiler s, t anks, st oves or wher e high st r engt h and good cor r osion r esist ance is r equir ed. ( c) Beryllium bronze : I t is a copper base alloy cont aining about 97.75% copper and 2.25% ber yllium. I t has high yield point, high fatigue limit and excellent cold and hot cor r osion r esistance. I t is par ticular ly suit able mat er ial for spr ings, heavy dut y elect r ical swit ches, cams and bushings. U ses : Since wear r esist ance of ber yllium copper is five t imes t hat of phosphor br onze, t her efor e, it may be used as a bear ing met al in place of phosphor br onze. I t has a film forming and a soft lubricating property, which makes it more suitable as a bearing metal. ( d ) M anganese bronze : I t is an alloy of copper, zinc and lit t le per cent age of manganese. Composit ion : Copper – 60% Zinc – 35% Manganese – 5% This metal is highly resistant to corrosion. It is harder and stronger than phosphor bronze. U sed : For bushes, plungers, feed pumps, rods etc. Worm gears are frequently made from this bronze. ( e) Aluminium br onze : I t i s an al l oy of copper and al umi ni um. The al umi ni um br onze wi t h 6-8% aluminium has valuable cold wor king pr oper t ies. The maximum t ensile st r engt h of t his alloy is 450 M Pa wit h 11% of aluminium. They ar e most suit able for making component s exposed t o sever e cor r osion condit ions. When ir on is added t o t hese br onzes, t his mechanical pr oper t ies ar e impr oved by r efining t he gr ain size and impr oving t he duct ilit y. Aluminium br onzes ar e widely used for making gear s, pr opeller s, condenser bolt s, pump component s, t ubes, air pumps, slide valves and bushings et c. Cams and r oller s ar e also made fr om t his alloy. The 6% aluminium alloy has a fine gold colour which is used for imit at ion jeweller y and decor at ive pur poses.

3. Sol der Types of Engineering solders (i ) Soft solders. I t is based on lead and t in and melt bet ween 180° C and 250° C. The ideal t in man’s solder has 2 par t s t in t o 1 lead but shor t age and high cost of t in oft en causes 5% or even lower t in cont ent s t o be used. For some special elect r ical pur pose pur e t in must be used, and solder s cont aining lead should never be used on pr ecious met als, e.g. r esist ance wir es. Plumber s solder r equir es a long past y r ange and has 2 par t s lead and 1 par t t in.

Engineering Materials

2.31

(ii ) Brazing solders. These ar e high zinc br asses, and t hey melt at ar ound 850°C— 900°C. (iii ) Silver solders. These ar e alloys of silver wit h copper and zinc. They have 64 per cent copper and 43 per cent silver and t heir melt ing r angs ar e 720/725°C, 725/750°C. Silver solder s can give gr eat st r engt h t han br azing solder s and under some condit ions have bet t er cor r osion r esist ance in chemical plant s while t heir whit e colour may somet imes be advant ageous. (iv) White metals. Whit e met al bear ing alloys ar e usually eit her t in base or lead base. The for mer being t he babbit met als. Tin base alloys ar e in most ways super ior t o lead alloys but t hey ar e mor e cost ly, and ar e subject t o anist r opic expansion which may init iat e failur e. The har dening const ituents necessar y for t he wear -r esist ance ar e pr ovided by t he pr esence of antimony, upt o 15 per cent , and of copper, usually limit ed t o 4 per cent . I n t in base alloys it is desir ed t o pr oduce t wo well-defined compounds, which will be embeded in t he soft er t in-r ich mat r ix : (a ) A compound of t in and copper which cr yst allises fr om t he molt en alloy in t he for m of neeles and for ms a sor t of net wor k. (b) A har d compound of t in and ant imony which separ at es in t he for m of char act er ist ic cuboids. The melting pr ocedur e, casting temper ature etc. must be car efully contr olled so that these two compounds ar e cor r ect ly ar r anged or t he bear ing wi ll not wear uni for mly. For cheaper bear ings lead may be subst it ut ed for par t of t he t in. L ead base alloys cont ain anyt hing over 50% lead, and t hese ar e also har dened wit h ant imony and copper. Wher e loads ar e not excessive and speeds ar e slow t hey ar e except ionally good.

4. Gun M etal I t is an alloy of copper, t in and zinc, which usually cont ains 88% copper, 10% t in and 2% zinc. This met al is also called admir alt y gun met al . The zinc is added t o clean t he met al and t o incr ease it s fluidit y. I t is not suitable for being wor ked in the cold state but may be for ged when at about 600C. The met al is ver y str ong and r esistant to cor r osion by water and atmospher e. Or iginally, it was made for cast ing guns. U ses : For cast ing boiler fit t ings, bushes, bear ings, glands et c.

5. L ea d I t is a bluish gr ey met al having specific gr avit y 11.36 and melt ing point 326C. I t is so soft t hat it can be cut wit h a knife. I t has no t enacit y. I t is ext ensively used for making solder s, as a lining for acid t anks, cist er ns, wat er pipes, and as coat ing for elect r ical cables. Lead base alloys ar e employed wher e a cheap and cor r osion r esistant mater ial is r equir ed. An alloy containing 83% lead, 15% ant imony, 1.5% t in and 0.5% copper is used for lar ge bear ings subject ed t o light ser vice.

6. T i n I t is br ight ly shining whit e met al. I t is soft , malleable and duct ile. I t can be r olled int o ver y t hin sheet s. I t is used for making impor t ant alloys, fine solder, as a pr ot ect ive coat ing for ir on and st eel sheet s and for making t in foil used as moist ur e pr oof packing. A t in base alloy cont aining 88% t in, 8% ant imony and 4% copper is called babbit met al. I t is a soft mat er ial wit h a low coefficient of fr ict ion and has lit t le st r engt h. I t is t he most common bear ing met al used wit h cast ir on boxes wher e bear ings ar e subject ed t o high pr essur e and load. Those alloys in which lead and t in ar e pr edominat ing ar e designat ed as whit e met al bear ing alloys. U sed : For lining bear ings subject ed t o high speeds like t he bear ings of aer o-engines.

7. Bear i ng M et al s

Commonly used Bear ing met als (i )

Copper-base alloys. These ar e most impor t ant bear ing alloys, which ar e har der and st r onger t han the white metals (lead base and tin base alloys). These ar e used for bear ings subject ed to heavy pr essur es. (ii ) L ead-base alloys (iii ) Tin-base alloys (iv) Cadmium-base alloys : These cont ain 95% cadmium and 5% silver. These ar e used for medium loaded bear ings subject ed t o high t emper at ur e.

2.32

Engineering Materials

Select ion of Bearing met als I t depends upon t he condit ions under which it is t o be used. I t involves fact or s r elat ing t o bear ing pr essur es, r ubbing speeds, t emper at ur es, lubr icat ion et c. Required pr opert ies in bearing mat erial : (i ) I t should have low coefficient of fr ict ion. (ii ) I t should have good wear ing qualit ies. (iii ) I t should have abilit y t o wit hst and bear ing pr essur es. (iv) I t should have ability to oper ate satisfactor ily wit h suit able lubr ication means at the maximum r ubbing speeds. (v) I t should have a sufficient melt ing point . (vi ) I t should have high t her mal conduct ivit y. (vii ) I t should have good cast ing qualit ies. (viii )I t should have minimum shr inkage aft er cast ing. (xi ) I t should have non-cor r osive pr oper t ies. (x) I t should be economical in cost .

8. Zinc Base Alloys M ost of t he die cast ings ar e pr oduced fr om zinc base alloys. These alloys can be cast ed easily wit h a good finish at fair ly low t emper at ur es. These also have consider able st r engt h and low in cost . Usual alloying element s for zinc ar e aluminium, copper and magnesium and t hese ar e all held in close limit s. Composit ion of t wo st andar d die cast ing zinc alloys (i ) Aluminium 4.1%, Copper 0.1%, M agnesium 0.04% and r emainder is Zinc. (ii ) Aluminium 4.1%, Copper 1%, M agnesium 0.04% and r emainder is Zinc. Aluminium impr oves mechanical pr oper t ies and r educes t endency of zinc t o dissolve ir on. Copper incr eases t ensi l e st r engt h, har dness and duct i li t y. M agnesi um has t he benefi ci al effect of maki ng t he cast ings per manently st able. U ses : I n t he aut omot ive indust r y and for ot her high pr oduct ion mar ket s such as washing machines, oil bur ner s, r efr iger at or s, r adios, phot ogr aphs, t elevision, business machines et c.

9. N ickel Base Alloys These ar e widely used in engineer ing and indust r y on account of t heir high mechanical st r engt h pr oper t ies, cor r osion r esist ance et c.

Commonly used N ickel base Alloys (i )

M onel metal. I t is an impor t ant alloy of nickel and copper, which cont ains 68% nickel, 29% copper and 3% ot her const it uent s like ir on, manganese, silicon and car bon. I t s specific gr avit y is 8.87 and melt ing point 1360C. I t has a t ensile st r engt h fr om 390 MPa t o 460 M Pa. I t r esembles nickel in appear ance and is st r ong, duct ile and t ough. I t is super ior t o br ass and br onze in cor r osion r esist ing pr oper t ies. U ses : I t is used for making pr opeller s, pump fit tings, condenser t ubes, st eam tur bine blades, sea water exposed par t s, t anks and chemical and food handling plant s. (ii ) I conel. I t consist s of 80% nickel, 14% chr omium, and 6% ir on. I t s specific gr avit y is 8.55 and melt ing point 1395C. This alloy has excellent mechanical pr oper t ies at or dinar y and elevat ed t emper at ur es. I t can be cast , r olled and cold dr awn. U ses : (a) For making spr ings which have to withst and high temper at ur es and ar e exposed to cor r osive action. (b) For exhaust manifolds of air cr aft engines. (iii ) N ichrome. I t consist s of 65% nickel, 15% chr omium and 20% ir on. I t has high heat and oxidat ion r esist ance. U ses : I n making elect r ical r esist ance wir e for elect r ic fur naces and heat ing element s. (iv) N imonic. I t consist s of 80% nickel and 20% chr omium. I t has high st r engt h and abilit y t o oper at e under int er mit t ent heat ing and cooling condit ions. U ses : I n gas t ur bine engines.

Engineering Materials

2.33

STRESS-STRAI N DI AGRAM The st iffness of a mat er ial is of equal impor t ance in designing t he st r uct ur e along wit h t he st r engt h of a mat er ial. The ot her mechanical pr oper t ies such as har dness, t oughness and duct ilit y ar e lesser impor t ant in t he select ion of mat er ial. These pr oper t ies ar e det er mined by making t est s on mat er ials and compar ing t he r esult s wit h est ablished st andar ds. One of t he t est s (t ension t est of st eel) and it s r esult s will be consider ed, which helps t o develop sever al impor t ant basic concept s.

I n t ensile t est , a specimen in gr ipped bet ween jaws of a t est ing machine. The elongat ion in a specified lengt h called gauge lengt h is obser ved simult aneously. These dat a ar e t hen plot t ed on a gr aph wit h t he or dinat e r epr esent ing t he load and absscisa r epr esent ing t he elongat ion. Figur e r epr esent s such a gr aph for st r uct ur al st eel. I n t his st r ess is plot t ed against t he unit elongat ion (i.e. st r ain) only by r educing obser ved values t o a unit basis. The pr oper t ies of one specimen is compar ed wit h t hose of ot her specimens. The diagr am shown in t he above figur e is called st r ess st r ain diagr am.

H OOKE'S LAW Consider st r aight line por t ion of t he st r ess-st r ain diagr am. The slope of t hat line is t he r at io of st r ess t o st r ain. I t is called modulus of elast icit y is denot ed by E.  Slope of st r ess-st r ain cur ve = E = e or,  = Ee

PRACTI CE EXERCI SE LEVEL-1 1. Abil it y of a mat er i al t o r esi st defor mat i on due t o st r ess is call ed (a) t oughness (b) stiffness (c) plast icity (d) har dness 2. Abil it y of mat er ial t o r esist fr act ur e due t o hi gh impact load is call ed (a) t oughness (b) stiffness (c) plast icity (d) har dness 3. Etching solution used for medium and high car bon st eel, pear lit ic st eel and cast i r on, is (a) ni t al-2% H N O3 i n et hyl alcohol (b) pi cr al - 5% picr ic acid and et hyl alcohol (c) 1% hydr ofluor ic acid in wat er (d) 50% N H 2 OH and 50% wat er

4. Pr oper t y of mat er ial due t o whi ch t hey can be dr awn into wir es, is called (a) elast icit y (b) plast icity (c) stiffness (d) duct ilit y 5. Abil it y of a mat er ial t o under go lar ge per manent defor mat i ons in t ensi on, is call ed (a) t oughness (b) stiffness (c) plast icity (d) har dness 6. Abil it y of a mat er i al t o r et ain t he defor mat ion per manently is called (a) stiffness (b) duct ilit y (c) har dness (d) plast icity 7. Abi l i t y of a mat er i al t o r esi st penet r at i on by anot her mat er i al , is call ed (a) stiffness (b) duct ilit y (c) har dness (d) plast icity

2.34

Engineering Materials

8. Pr oper t y of mat er ial due t o which i t can be r oll ed or hammer ed int o t hin sheet s is call ed (a) br it t l eness (b) duct ilit y (c) malleability (d) fatigue 9. Pr oper t y of mat er i al due t o which i t br eak s wi t h li t t le per manent di st or t ion is call ed (a) br it t l eness (b) duct ilit y (c) malleability (d) fatigue 10. Pr oper t y of mat er ial due t o whi ch r ecovery after unloading is complete but instantaneous, is called (a) elast icit y (b) plast icity (c) anelasticity (d) enelast icit y 11. Wi t h incr ease in t he har dness of mat er ial elast ic r ecover y aft er defor mat i on (a) i ncr eases (b) decr eases (c) r emains same (d) none of t hese 12. Abil it y of a mat er ial t o under go lar ge per manent defor mat i on in compr ession, is known as (a) duct ilit y (b) malleabilit y (c) br it t l eness (d) har dness 13. Ability of a mater ial to exhibit consider able elastic r ecover y on r elease of l oad, is k nown as (a) t oughness (b) stiffness (c) r esi li ence (d) har dness 14. Si li con when added t o copper incr eases i t s (a) machinability (b) br it t l eness (c) elect r ical conduct ivit y (d) har dness and st r engt h 15. St r ess-concent r at i on occu r s wh en a body i s subject ed t o (a) extensive str ess (b) non-uni for m st r ess (c) r ever se st r ess (d) fluct uat ing st r ess 16. M et als which can be easi ly dr awn i nt o wir e is (a) tin (b) copper (c) lead (d) zinc 17. When a body r ecover s it s or i ginal di mensions on r emoving t he l oad, it is call ed (a) elastic (b) plastic (c) br it tle (d) none of t hese 18. Amount of external ener gy r equired t o defor m an el ast i c body is call ed (a) el ast ic ener gy (b) pl ast ic ener gy (c) st r ain ener gy (d) none of t hese 19. A met al which is br it t le in t ension can become ductile (a) in t he pr esence of not ches (b) under hydr ost at ic compr ession (c) in t he pr esence of embr it t le-ment agent s such as hydr ogen (d) all of t hese

20. A met al which is duct ile in t ension can become br it tle (a) in t he pr esence of not ches (b) under hydr ost at ic compr ession (c) in t he pr esence of embr it t le-ment agent s such as hydr ogen (d) all of t hese

LEVEL-2 21. Nit r iding is a pr ocess for (a) nor malising

(b) annealing

(c) case har dening

(d) t emper ing

22. Pr ocess of r eheat ing t he har dened st eel t o some t emper at ur e below t heor et ical r ange, followed by any r at e of cooling is called (a) normalising

(b) annealing

(c) t emper ing

(d) spher oidising

23. Pr ocess in which steel is heated in a molten salt bath having temper a-tur e 250C to 500C above t he cr it ical t emper at ur e, t hen quenched i nt o a molt en bat h at sufficient r ate bet ween 200C to 450C, held t her e for sufficient time and cooled to r oom temper atur e, is called (a) nor malising

(b) annealing

(c) t emper ing

(d) spher oidising

24. H eat t r eat ment pr ocess used t o soft en har dened st eel is (a) normalising

(b) annealing

(c) t emper ing

(d) spher oidising

25. The pr ocess in which t he st eel is heat ed slight ly above t he lower cr it ical t emper at ur e and t hen cooled slowly t o a t emper at ur e of 600C is called (a) nor malising

(b) annealing

(c) t emper ing

(d) spher oidising

26. H eat t r eat ment pr ocess usually applied t o high car bon t ool st eel which ar e difficult t o machine is (a) normalising

(b) annealing

(c) t emper ing

(d) spher oidising

27. Heat tr eat ment pr ocess which decr eases har dness and t ensile strength but increases machinability is (a) nor malising

(b) annealing

(c) t emper ing

(d) spher oidising

28. The process in which steel is heated at about C, where the structure consists of entirely austenite, then it is cooled suddenly at a temperature of about 250C to 525C is called (a) normalising

(b) annealing

(c) har dening

(d) austemper ing

Engineering Materials

29. Sor bit e is obt ained by (a) annealing of st eel (b) quenching st eel dur ing t r ansfor mat ion (c) bot h (a) and (b) (d) none of t hese 30. Age har dening gener ally applied t o (a) cast ir on (b) high alloy st eel (c) alloys of aluminium, magnesium, nickel et c. (d) alloys of chr omium, silicon et c. 31. Age har dening is r elat ed t o (a) cast ir on (b) dur alumin (c) st ainless st eel (d) brass 32. I nduct ion har dening have high (a) volt age (b) cur r ent (c) fr equency (d) power fact or 33. I nduct ion har dening is t he pr ocess of (a) unifor m har dening (b) har dening t he cor e (c) select ive har dening (d) har dening sur face for wear r esist ance 34. I n flame har dening, t he flame used is of (a) oil bur ner (b) gas bur ner (c) oxy-acetylene (d) none of t hese 35. Con st i t u en t s of H ay n ess st el l i t e, h av i n g super for mance t han H SS ar e (a) t ungst en, chr omium and vanadium (b) t ungst en, chr omium and cobalt (c) t ungst en, molybdneum and cobalt (d) cobalt , nickel and aluminium 36. A r ever sible changes in an at omic str uct ur e of the m et al wi t h a cor r espon di n g ch an ge i n t h e pr oper t ies of st eel, is called (a) allot r opic (b) polyt r opic (c) cr itical point (d) none of t hese 37. St eel can be har dened only if it is heat ed above (a) lowest cr it ical t emper at ur e (b) middle cr it ical t emper at ur e (c) highest cr it ical t emper at ur e (d) none of t hese 38. When a st eel cont aining less t han 0.8% car bon is cooled slowly fr om t emper at ur e above or wit hin t he cr it ical r ange, it cont ains (a) fer r it e mainly (b) pear lite mainly (c) fer r it e and pear lit e (d) pear lit e and cement it e

2.35

39. Temper atur e at which t he change ends on heating t he st eel, is called (a) lower cr it ical t emper at ur e (b) upper cr it ical t emper at ur e (c) point of r ecalescence (d) none of t hese 40. St eel cont aining 0.8% car bon has (a) no cr it ical point (b) one cr it ical point (c) t wo cr it ical point s (d) t hr ee cr it ical point s 41. L ower cr it ical point for all st eels is (a) 600C (b) 723C (c) 900C (d) 91400C 42. Gamma ir on exist s in t he t emper at ur e r ange of (a) 300C to 600C (b) 600C to 900C (c) 900C t o 1400C (d) 1400C to 1530C 43. Alpha ir on exist s at (a) below 768C (b) bet ween 768C t o 900C (c) bet ween 900C t o 1400 (d) between 1400C to 1530C 44. K noop har dness number (K H N) is equal t o (a)

P LC

(b)

(c)

2P LC

(d)

P L2 C 2P L2 C

where, L = length of long diagonal, mm C = constant related to length of pr oject ed ar ea 45. Delt a ir on exist s in t he t emper at ur e r ange of (a) 0C t o 768C (b) 768C to 900C (c) 900C t o 1400C (d) 1400C to 1530C 46. Alpha () ir on (a) has body cent r ed at omic ar r angement (b) is commonly called fer r it e (c) is magnet ic (d) all of t hese 47. Vicker 's Pyr amid Number (VPN) is equal t o (a)

2P sin  d2

P sin  2 (c) d2

(b)

P sin  d2

(d) None of t hese

wher e,P = load in kg  = angle between opposite faces of diamond pyramid

2.36

Engineering Materials

48. When the atoms ar e ar r anged r egular ly in some dir ection but not in other s, in a mater ial, it is called

50. Br inell har dess number is equal to (a)

(a) amor phous mat er ial (b) mesomor phous mat er ial (c) cr yst alline st r uct ur e

(b)

(d) none of t hese 49. When t he at oms ar e ar r anged i n defi nit e and or der ly manner in a mat er ial, it is called

(c)

(a) amor phous mat er ial (b) mesomor phous mat er ial (c) cr yst alline st r uct ur e

(d)

(d) none of t hese

P D  ( D2  d 2 ) D FD H

P (D 2  d 2 )

IK

2P D  D2  d 2

FH

2P

D D  (D 2  d 2 )

IK

wher e d = diamet er of the impression.

AN SWERS LEVEL-1 1. (b)

2. (a)

3. (b)

4. (d)

5. (c)

6. (d)

7. (c)

8. (c)

9. (a)

10. (c)

11. (a)

12. (b)

13. (c)

14. (d)

15. (b)

16. (a)

17. (a)

18. (c)

19. (b)

20. (a)

27. (d) 37. (a) 47. (c)

28. (d) 38. (a) 48. (b)

29. (a) 39. (b) 49. (c)

30. (c) 40. (b) 50. (d)

LEVEL-2 21. (c) 31. (b) 41. (b)

22. (c) 32. (c) 42. (c)

23. (b) 33. (d) 43. (a)

24. (c) 34. (c) 44. (b)

25. (d) 35. (c) 45. (d)

26. (d) 36. (a) 46. (d)

3

Strength of Materials

CHAPTER

STRESS, STRAI N AN D M OH R'S CI RCLE STRESS. I t is t he int ensit y of I nt er nal Resist ance offer ed by a body/ mat er ial against defor mat ion. I nt ensit y means per unit ar ea. P P St r ess = Resist ing for ce per unit ar ea.

N ormal Stress & Shear Stress Resist ing for ce nor mal t o t he plane or sect ion per unit ar ea is called nor mal st r ess. I nt ensit y of r esist ing for ce par allel t o t he sect ion/plane is called shear st r ess. (ii) Nor mal st r ess: F P Fr om figur e, aver age st r ess = = = avg A A This Aver age st r ess avg can be assumed t o be t he st r ess at any point on that plane if resisting force is uniform throughout the plane. I f Resist ance is not unifor m t hen st r ess at any point on t he sect ion can be defined as F = A dF F  = line = A  0 dA A (ii ) Shear St r ess : Resist ing for ce par allel t o plane is Q P Q H ence shear st r ess,  = A Fr om equilibr ium, F = 0  –P+Q=0  P=Q P

F Resi st ance

F.B.D P

Fig

f A

P

P

P

P So = A U nit of stress : N/m 2 (similar t o unit of pr essur e) 1M Pa = 1  106 N/m 2 = 1N/mm 2

P

Q

Poisson's Rat io

The r at io of lat er al st r ain t o t he longit udinal st r ain is a const ant quant it y and is called t he Poisson's r at io and 1 is denot ed by  or m L at eral st rain  = L ongit udinal st r ain For the most metals its value is between 0.25 to 0.33. For cement concr et e 0.15, for r ubber 0.5, for steal 0.3.

ST RAI N Similar t o st r ess it is a quant it y used t o measur e t he int ensit y of defor mat ion. I t is defined as change in dimension wit h r espect t o or iginal dimension.

N ormal Strain, 

l

M easur es t he change in size (Elongat ion or Cont r act ion) of an ar bit r ar y line segment on a body dur ing it s defor mat ion.

avg =

l l

P

P L

3.2

Strength of Materials

Shear ing St rain,  M easur es t he change in shape. (Change in angle bet ween t wo lines t hat ar e or t hogonal in t he undefor med Y shape of t he body under defor mat ion)

  area  Shear st r ain  =  =   L  radius  change in volume V Volumet r ic st r ain = = original volume V



 L 

X

H OOKE'S LAW I f st at es t hat t he st r ess is dir ect ly pr opor t ional t o st r ain upt o t he elast ic limit .    =E ( E is modulus of elast icit y)  Ratio of stress and strain is used to measure the stiffness of a material, this ratio is called Modulus of Elasticity.. 

TH E RM AL ST RE SSE S Ever y mat er ial expands when t emper at ur e r ises and cont r act s when t emper at ur e falls. The change in lengt h due t o change in t emper at ur e is found t o be dir ect ly pr opor t ional t o t he lengt h of member and also t o change in t emper at ur e. H ence if  is constant of pr opor t ionalit y, t is change in t emper at ur e and L is lengt h of t he member, t hen change in lengt h, = t L wher e,  = const ant of pr opor t ionalit y called coefficient of t her mal expansion. I t is defined as change in unit lengt h of t he mat er ial due t o unit change in t emper at ur e. This value is differ ent for differ ent mat er ials.

M ater ia l St eel Copper St ain less st eel Brass, Br on ze Alu mi ni um

Coeffi ci en t of th er m al ex pansion 12  10– 6/ °C 17.5  10– 6 /°C 18  10– 6/ °C 19  10– 6/ °C 23  10– 6/ °C

I f t he bar is fr ee t o ext end when t emper at ur e is incr eased by ‘t ’ degr ees, its ext ension (fr ee expansion) would have been  t L . But this extension is completely prevented in this case by for ces developed at suppor t s. This suppor t for ce P such t hat it causes shor t ening () of t he bar by  t L . H ence  =  tL PL  =  tL AE P or = E t A i.e.  = E t wher e, p = t emper at ur e st r ess, which is compr essive in t his case. I f fr ee expansion is prevented par tly as in case of member shown in the figur e, then shor tening caused by support reaction P is given by  = t L – 



PL = t L –  AE

Composit e Sect ion U nder T her mal St r ess (i) Consider a composit e bar r igidly fit t ed at suppor t s A and B. Then suppor t r eact ions can be found. Tot al expansion, l = a t l a + s t l s

L

 

tL P

Strength of Materials

3.3

or l = (a l a + s l s) t … (i ) L et p be t he suppor t r eact ion. Then Pl a Pl s l = + … (i i ) A aE a A sE s Fr om equat ions (i) and (ii), ls l (a l a + s l s) t = P ( a + ) A A aE a sE s (ii) Consider a steel bar of uniformly varying diameter is held between two unyielding support at room temperature. I f t he t emper at ur e r ises by t degr ees, t hen maximum st r ess induced in t he bar. L et P be t he for ce developed at t he suppor t . Then 4Pl l = t L = E d1 d2

P A Example. A Composite bar is r igidly fitted at the suppor ts A and B as shown in the figur e. Deter mine the reactions at t he suppor ts when temper at ur e r ises by 20 C. Take E a = 70GN/m 2, E s = 200 GN/m 2, a = 11  10– 6/C and s = 12  10– 6/C. Pmax =

Fr ee expansion,  = a t L a + s t L s = 11  10– 6  20  1000 + 12  10– 6  20  3000 = 0.94 PL s PL a I f P is suppor t r eact ion, t hen  = A E + A E a a s s H er e, E a = 70  1000 N/mm 2 and E s = 200  1000 N/mm 2 Sol ut i on.

 or

P

LM 1000  3000 OP = 0.94 N 600  70  1000 300  200  1000 Q P = 12735.48 N.

Temper at ur e St r ess in Compound Bar s When Temper at ur e r ises t he t wo mat er ials of a compound bar exper ience differ ent fr ee expansions. Since, t hey ar e pr evented fr om separ ating, the two materials will have common position which is possible only by development of t emper at ur e st r esses. Consider t he compound bar of lengt h L shown in t he figur e. L et t he r ise in t emper at ur e be t . L et 1, 2 be coefficient of t her mal expansion and E 1, E 2 be modulii of elast icit y of t he t wo mat er ials. L et 1 > 2. Fr ee expansion of bar (i) = 1 t L . Fr ee expansion of bar (ii) = 2 t L . Si nce 1 > 2, t he posi t i on of t he t wo bar s, i f fr ee expansi ons ar e per mit t ed t o expend at AA and BB as shown in t he figur e. But since two bar s are rigidly secured at the ends the position of the bars will be some wher e in between AA and BB say at CC (figure). To secure this position compr essive for ce P1 will act in bar (I ) so as to produce shortening by 1 and a tensile for ce P2 will act in bar (I I ) to pr oduce an extension 2 in bar (I I ). From equilibrium condition,

3.4

Strength of Materials

P1 = P2 = P. 2 t L + 2 = 1 t L – 1 1 + 2 = 1 t L – 2 t L

Fr om t he figur e, or

PL PL + = (1 – 2) t L A 1E1 A 2E 2

or or

PL

FG A E  A E IJ H AEAE K 2

2

1 1

1 1

or

2

2

= (1 – 2) t L

P=

H ence

st r ess in bar (I ) is, 1 =

and

st r ess in bar (I I ) is, 2 =

FG A A E E IJ H A E + A E K  ( –  ) t L 1

1 1

2

1

2

2

2

1

2

P (compr essive) A1 P (t ensile) A2

Example. A compound bar is made of a cent r al st eel plat e 60 mm wide and 10 mm t hick t o which copper plat es 40 mm wide by 5 mm t hick ar e connect ed r igidly on each side. The lengt h of t he bar at nor mal t emper at ur e is 1 met r e. I f t he t emper at ur e is r aised by 80C, det er mine t he st r esses in each met al and t he change in lengt h. Take : E s = 200 GN/m 2 ; E c = 100 GN/m 2 ; s = 12  10– 6/c ; c = 17  10– 6/c Solution.

t = 80 C, L = 1 m = 1000 mm . E s = 200 GN/m 2 = 200  109 / (103  103) = 2  105 N/mm 2 E c = 100 GN/m 2 = 100  109 / (103  103) = 1  105 N/mm 2 A s = 60  10 = 600 mm 2 A c = 40  5 = 200 mm 2 for each plat e. Fr om equilibr ium of for ces P s = 2 Pc I f s and c ar e t he changes in lengt h of st eel and copper plat es r espect ively, t hen fr om t he figur e s + c = c t L – s t L H er e,

PsL PL + c = (c – s) t L A sE s A cEc

or

F 2P GH A E

or

c

s s

or or



Pc



Pc A cE c

LM 2 N 600  2  10

5



I JK

= (c – s) t

1 200  1  105

OP Q

= (17 – 12)  10– 6  80

Pc = 6000 N and Ps = 2  6000 = 12000 N. St r ess in copper =

6000 = 30 N/mm 2. Ans. 200

Strength of Materials

and

St r ess in st eel =

3.5

12000 = 20 N/mm 2. Ans. 600

PsL s Change in lengt h of compound bar = s t L + A E s s = 12  10– 6  80  1000 +

1200  1000 = 1.06 mm 600  2  105

E LASTI C CON STAN TS M odulus of E last icit y I t is defined as t he r at io of linear st r ess t o linear st r ain wit hin elast ic limit . Fr om H ook’s law ,   e or  = eE  or E = e

M odulus of Ri gidi t y

I t is defined as t he shear ing st r ess t o t he shear ing st r ain wit hin elast ic limit .   G = 

Bulk M odul us

When a body is subject ed t o ident ical pr essur e p in t hr ee mut ually per pendicular dir ect ion, t hen t he body under goes unifor m changes in t hr ee dir ect ion wit hout under going dist or t ion of shape. The r at io of change in volume t o or iginal volume is called volumet r ic st r ain (ev).  Then Bulk modulus, K = e v V wher e, ev = volumet r ic st r ain = V

Relat ionship Bet ween E last ic Const ant s E = 2 G (1 + ) E = 3 K (1 – 2 ) and

E =

9K G 3K  G

V OLU M E TRI C STRAI N Consider a bar of r ectangular section subjected t o str es x,y and z in x , y and z dir ections r espectively. The str ess   pr oduces str ain of magnitude in its dir ection and a str ain  of opposite natur e at r ight angles to its dir ection. E E  y y  y   z  x  z  x z  ex= x – – ; ey =  + – ; ez= – – + E E E E E E E E E

STRESS-STRAI N DI AGRAM (M I LD STEEL)

×

( x   y   z ) dv = ( 1 – 2 v) v E

B × A F × C Stress

Volumet r ic st r ain, ev =

E

Figur e shows st r ess – st r ain diagr am for t he t ypical mild st eel specimen. The following salient point s ar e obser ved on st r ess st r ain cur ve. (i) Limit of proportionality (A) : I t is t he limit ing value of t he st r ess O F Strain upt o which st r ess is pr opor t ional t o st r ain. (ii) Elastic limit : I t is t he limit ing value of st r ess upt o which if t he mat er ial is st r essed and t hen r eleased (unloaded) st r ain disappear s complet ely and t he or iginal lengt h is r egained. This point is slight ly beyond t he limit of pr opor t ionalit y.

3.6

Strength of Materials

(iii) U pper yield point (B) : I t is the st r ess at which, the load star t s r educing and the extension incr eases. This phenomenon is called yielding of mater ial. At this stage str ain is about 0.125 and str ess is about 250 N/mm 2. (iv) Lower yield point (C) : At t his st age t he st r ess r emains same but st r ain incr eases for some t ime. (v) U ltimate stress (D) : I t is t he maximum st r ess t he mat er ial can r esist . This st r ess is about 370-400 N/ mm 2. At t his st age cr oss sect ional ar ea at a par t icular sect ion st ar t s r educing ver y fast . This is called neck formation. Aft er t his st age, load r esist ed and hence t he st r ess developed st ar t s r educing. (vi) Breaking point (E) : The st r ess at which finally t he specimen fails is called br eaking point . At t his st age st r ain is 20 t o 25 per cent . I f unloading is made wit hin elast ic limit , it follows a st r aight line par allel t o t he or iginal st r ai ght por t ion as shown by FF in t he figur e. Thus if it is loaded beyond elast ic limit and t hen unloaded a per manent st r ain (OF) is left in t he specimen. This is called per manent set .

St ress-St rain Relation in Aluminium and H igh Str ength Steel

y Stress

F

F O 0.2

 Strain

F ig. Stress-Strain relationship in Aluminium and high strength steel

Str ess st ain diagr am is shown in the figur e. This str ess  at which if unloading is made ther e will be 0.2 per cent per manent set is known as 0.2 percent pr oof str ess and this point is tr eated as yield point for all pr actical pur poses.

Str ess-St rain Diagr am (Br it tle M aterial)

Stress

x

Strain

F ig. Stress-Strain relations for brittle material.

I n br ittle mater ials t her e is no appr eciable change in r at e of str ain. Ther e is no yield point and no necking takes place. Ultimate point and br eaking point ar e one and the same. The str ain at failur e is ver y small, ver y less str ain is obser ved.

Behaviour of M at er ials under Compr ession As t her e is chance of buckling (lat er ally bending) of long specimen, for compr ession t est s shor t specimen ar e used. Hence, t his test involve measur ement of smaller changes in lengt h. I t r esult s into lesser accur acy. However pr ecise measur ement t s have shown t he followign r esult s. (i) I n case of duct ile mat er ials st r ess-st r ain cur ve follows exact ly same pat h as in t ensile t est upt o and even slight ly beyond yield point . For lar ger values t he cur ves diver ge. Ther e wil not be necking in case of compr ession t est s. (ii) For most br it t le mat er ials ult imat e compr essive st r ess in compr ession is much lar ger t han in t ension. I t is because of flows and cr acks pr esent in br it t le mat er ials which weaken t h emat er ial in t ension but will not affect t he st r engt h in compr ession.

SH EAR FORCE AN D BEN DI N G M OM EN T The shear r epr esent s t he t endency t o slide. The shear ing for ce at any point along a loaded beam is t he algebr aic sum of all t he ver t ical for ces act ing t o one side of t he point . Thus for t he beam shown in t he figur e, t he shear for ce at t he cr oss sect ion x  x as measur ed fr om t he left hand side is

Strength of Materials

3.7

F 1 = R1 – W 1 which is equal t o t he shear for ce, F 2 = W 2 – R2 as measur ed fr om r ight hand side.

Since F 1 = F 2, t her efor e beam is in equilibr ium. A t hin slice of t he beam at sect ion x – x subject ed t o t hese for ces as shown in fig (b). The shear for ce is assumed t o be posit ive when it pr oduces a clockwise moment and negat ive when it pr oduces an ant iclockwise moment or shear for ce will be consider ed posit ive when r esult ant of t he for ces t o t he left is upwar ds, or t o t he r ight is downwar ds.

Bending M oment Bending moment at any sect ion on t he loaded beam is algebr aic sum of moment s of all t he for ces act ing on eit her side of t his sect ion. Clockwise moment at sect ion x – x , M 1 = R1 x – W 1 (x – a) and ant iclockwise moment at sect ion x – x , M 2 = R2 ( l – x ) – W 2 (l – x – b) For t he equilibr ium of t he beam, M1 = M2 Clockwise moment s ar e assumed t o be posit ive and ant iclockwise negat ive due t o all t he loads act ing t o t he left of a sect ion. The shear for ce and bending moment s ar e vect or quant it ies and following sign convent ions ar e gener ally used. F

F

F

F

+ve shear force

–ve shear force

The bending moment is positive if it tries to sag the beam. I f the left portion of the beam is consider ed positive moment works out to be clockwise. Whereas the bending moment is negative if it tr ies to bag the beam.

+ve bending moment Sagging Bending M oment

– ve bending moment H ogging Bending M oment

Relat ionship Bet ween L oad I nt ensit y, Shear F or ce and Bending M oment Consider t he beam AB subject ed t o a gener al loading as shown in t he figur e. I n t his figur e, t he fr ee body diagr am of an element al lengt h x at dist ance x fr om t he left suppor t is dr awn wit h posit ive sense of all t he for ces and moment s.

(a) Position of element

(b) Positive sences of SF and BM

(c) Enlarged view of element

The int ensit y of loading on t he element al lengt h may be t aken as const ant .

3.8

Strength of Materials

Consider ing t he ver t ical equilibr ium of for ces, we get F + F – F – W x = 0 F or =W x dF I n t he limit ing case as x  0, = W.. dx Taking moment in equilibr ium of t he element about r ight face, we get x M + M – M + F x + W x =0 2 Neglect ing t he small quant it y of higher or der, we get M + F x = 0 M or = – F.. x M = – F. I n t he limit , as x  0, x

Shear For ce Diagrams (SFD) and Bending M oment Diagr ams (BM D). A diagr am in which or dinat e r epr esent s shear for ce and abscissa r epr esent s t he posit ion of t he sect ion is called shear force diagram. Similarly bending moment diagram may be defined as a diagram in which ordinate represents bending moment and abscissa r epr esent s t he posit ion of t he sect ion. I n t hese diagr ams, following infor mat ion is given: (i) Values at all salient point s, i.e. t he point s wher e values ar e maximum/minimum, and t he point s wher e nat ur e of var iat ion changes. (ii) Nat ur e of var iat ion bet ween t he salient point s. (iii) L ocat ion of t he point of cont r aflexur e, i.e. t he point wher e moment changes it s sign. Obviously at t his point bending moment is zer o.

Shear F or ce and Bending M oment D iagr ams ( i ) Cant ilever Subject ed to a Concentrated L oad at End Consider t he beam shown in fig. (a) At t he sect ion dist ance x fr om fr ee end, and consider ing t he for ces on left hand side por t ion. F =–W Thus shear for ce is constant at any section and its value is – W. H ence SFD is as shown in t he fig (b). At sect ion x – x , M = – W x , i.e. linear var iat ion At x = 0, M =0 At x = l , M = – Wl . H ence BM D is as shown in fig (c). ( ii ) Cant ilever Subject ed t o U nifor mly Dist r ibut ed L oad Consider the left hand por tion of the beam from the section x – x shown in t he fig. (a) F = – wx i.e., linear var iat ion At x = 0, F = 0. At x = l , F = – wl . H ence SFD is as shown in t he fig. (b) x wx 2 M = – wx =– , 2 2 i.e., par abolic var iat ion At x = 0, M =0 wl 2 At x = l , M =– . 2 H ence BM D is as shown in fig. (c).

(a) SPACE DI AGRAM

(b) S.F.D.

(c) B.M .D. w/unit lenght

X

X l

(a) SPACE DI AGRAM

(b) S F D

(c) B M D

Strength of Materials

( iii ) Cant ilever Subject ed t o U niformly Varying L oad

w / unit l engt h

L et t he load var y fr om zer o at suppor t A t o w/unit lengt h at suppor t B linear ly. At any dist ance x fr om left hand side, L oad,



x

x wx = w . l F =–

3.9

l

(a) SPACE DI AGRAM

x 1 wx2 w x=– l 2l 2

i.e., par abolic var iat ion At x = 0, F = 0.

(b) S.F.D.

w x F = 2l

2

w l . 2 H ence SFD is as shown in t he fig. (b) At x = l ,

M =–

=

x 1 x wx3 w x =– l 6l 2 3 (c) B.M .D.

i.e., cubic var iat ion At x = 0, M =0

wl 2 . 6 H ence BM D is as shown in t he fig. (c) W ( iv) Simply Support ed Beam Subject ed t o a Concent rat ed L oad A B C Consider t he beam AB of span l subject ed t o a concent r at ed a b load W act ing at a dist ance ‘a’ fr om left hand suppor t as Wb Wa l l l shown in t he figur e. Wb Wa Now, RA = , and RB = l l Consider por t ion AC at any sect ion dist ance x fr om A. Wb F = , i.e., const ant l (b) S.F.D. Wb M = x, l i.e., linear var iat ion At x = 0, M = 0. W ab At x = a, M = l For por t ion AC, SFD and BM D can be dr awn. (c) B.M .D. Consider the portion CB. At any section at a distance x from B and considering right hand side portion, we get Wa F =– , i.e. const ant l Wa M = x , i.e. linear var iat ion l At x = 0, M =0 At x = l ,

At x = b,

l If a = b = , t hen 2

M =–

M =

W ab l

l 2 = W, 2 l

W

F =

and

M=

W

l l 2 2 = Wl . l 4

3.10

Strength of Materials

(v) Simply Support ed beam subject ed t o U DL . Consider t he simply suppor t ed beam of span l subject ed t o unifor mly dist r ibut ed load w per unit lengt h as shown in t he fig. (a). The r eact ions developed ar e wl RA = RB = 2 At sect ion x – x dist ance x fr om A, F = RA – wx =

(a) SPACE DI AGRAM

wl – wx 2

i .e. linear var iat ion

wl 2 wl wl F = – wl = – 2 2

At x = 0,

F =

At x = L , i.e. linear var iat ion At sect ion x – x ,

M = RA x – wx

i.e. par abolic var iat ion At x = 0 and x = L , At x =

L , 2

M=

(b) S.F.D.

x wLx wx2 = – 2 2 2

M = 0.

(c) B.M .D.

wL L w L2 w L2  – = . 2 2 8 8

TYPI CAL STAN DARD CASES OF CAN TI LEVERS AN D BEAM S S. N o. D iagr am Shear force Bending moment Cantilevers. ( Shear for ce and bending moment ar e bot h maximum at fixed end of a cant ilever ). 1.

F max = – W

M max = – W l

2.

F max = – wl

M max = –

3.

F max = – wl 1

M max = – wl 1 l 

4.

F max = – wl 1

M max = –

wl 12 2

5.

F max = –

wl 2

M max = –

wl 2 6

6.

F max = –

wl 2

M max = –

wl 2 3

wl 2 2

FG H

l1 2

IJ K

Beams. (Shear for ce may be maximum at suppor t or under point loads. Bending moment may be maximum at t he point wher e shear for ce is zer o or changes signs.) 7.

F max =

W 2

M max = +

Wl at mid span. 4

Strength of Materials

8.

F max =

Wa fr om B t o C. l

M max = +

W ab at C. l

9.

F max =

wl at A and B. 2

M max = +

l wl 2 at 2 8

10.

F max =

wl at A and B. 4

M max = +

11.

F max = –

wl at A. 3

M max =

3.11

fr om A.

wl 2 at C. 12

wl 2 9 3

at

l 3

fr om B.

DEFLECTI ON , TORSI ON AN D COLU M N TH EORY OF SI M PLE BEN DI N G Following assumpt ions ar e made : (1) Befor e t he applicat ion of t r anser ve loads of t he beam it is init ially st r aight . (2) The mater ials of the beam is per fectly homogeneous and isotr opic (i.e. equal elastic pr oper ties in all dir ections). (3) The beam mat er ial is st r essed wit hin it s elast ic limit and t hus obeys H ook’s law. (4) The t r ansver se sect ion (i.e. AC or BD in Fig. 1) r emain plane befor e and aft er bending of t he beam. (5) The value of Young’s modulus of elast icit y E of t he mat er ials is t he same in t ension or compr ession. Consider a small element ABCD of beam subject ed t o posit ive bending moment M . L et us assume sect ion of beam r ect angular (it may have any shope). Due t o bending moment , t op fibr e AB cont r act s and t he bot tom most fiber CD ext ends. Aft er bending AB changes t o A B  and CD changes t o CD . Assume for t he small infint esimal lengt h , t he bent lengt h can be consider ed as a par t of a cir cle of a definit e r adius. L et t he cent r e of cir cle or cent r e of cur vat ur e be O, r adius of t he cir cle or r adius of cur vat ur e upt o t he net ur al axis is R. Now consider of layer EF at a dist ance y fr om t he net ur al axis which is r educed t o lengt h E F  aft er t he bending of t he beam. M

A E

M B

b

tC yC

F

N

L

C

D

y/N

A

d

N

A'

D (b) Cross Section

(c) Strain diagram

R

h

Ft

yt



FC

N

tt (a) Longitudinal Section

C

B

y

B' F'

N' E'

t (d) Stress diagram

C'

L' D'

(e)

Final lengt h – Original lengt h Or iginal lengt h EF = AB = NL = N L 

St r ain in layer EF = I nitially,

St r ain in layer EF , e = E F  – N L  N L  wher e, N L  = R and E F  = (R – y )  Distance between the layer s EF and NL i.e., y is changed t o y  but change is t his thickness is negligible and y = y  ( R  y) – R y  St r ain in layer EF , e = = R R  e y ...(i )

3.12

Strength of Materials

So, st r ain is dir ect ly pr opor t ioal t o dist ance of fibr e fr om net r ual axis. The st r ain is compr essive or t ensile depending upon posit ion of t he layer, i.e. whet her layer is above or below t he neut r al layer. L et , y c = dist ance of t he t op fibr e neut r al axis, and y t = dist ance of t he bot t om fibr e fr om neut r al axis. yt Then, maximum compr essive st r ain, et = R yt and maximum t ensile st r ain, et = R These var iat ion ar e shown in t he figur e. st ress  As we know elast icit y, E = = st rain e y E   =eE = R  E  = R y E   = y ...(ii ) R Since E and R ar e const ant s for par t icular sect ion, t her efor e st r ess at any point is dir ect ly pr opor t ional t o it s dist ance fr om t he neut r al axis.

Posit ion of N eut ral Axis I nt er sect ion of t he plane of neut r al layer wit h t he cr oss sect ion of t he beam is called neut r al axis. Now consider st r ess in layer, EF =  But  = eE wher e E = Young’s modulus of mat er ial For ce on element ar y ar ea at EF level, wher e a = ar ea of element ar y st r ip

E R F = b ×y = a =y

yE a = – E  ya R R y E c ya  Tot al for ce on t his sect ion, F =  F – R yt For equilibr ium, r esultant for ce on t he section should be zer o, i.e. total compr essive for ce F C acting on the sect ion above the neutr al axis is equal to total tensile for ce F t act ing on the section below the neutr al axis, i.e Fc – Ft = 0  F =0 y E c y  a = 0  R yt 1 y =  yd a A  Ay = 0 Ther efor e, fir st moment of ar ea about t he N.A. is zer o since A  0 y =0  This shows t hat , net ur al axis of t he beam passes t hr ough t he cent r iod of t he sect ion or neur t al axis passes t hr ough t he cent r oids of all t he sect ion along t he lengt h t he beam. 

F =

M oment of Resist ance of a Sect ion Refer ing t o Fig. (a), for ce on element ar y ar ea a, and,

F = –

E y  a R

moment of t his for ce about N .A , M = – F = – y



E c 2 M oment of for ce about N .A . = R  y a yt

E E 2 y  a  – y  = y a R R

Strength of Materials

3.13

M oment of r esist ance is due t o int er nal st r esses developed in sect ion, yc M r = E  y2a and t his must be equal t o applied moment M . R yt yc

wher e,

2

 y a

yt

= second moment of the ar ea about N.A. = M oment of int er t ia about N.A. = I NA

E I R NA Assume C.G. is cent r iod of t he sect ion and X -X and Y -Y ar e hor izont al and ver t ical axes passing t hr ough t he cent r iod.  I NA = I xx Y E  M = I X X R xx N A M E  = I xx R Y Using t his r eact ion and equat ion (ii ), we have  E M = = y R I xx This for mulat ion is called flexur e for mula. Since t ot al compr essive for ce on t he sect ion above t he NA is equal t o t ot al t ensile for ce below t he neut r al axis, hence t hey const it ut e a couple of ar m h.  Moment of resistance, M r = Fc  h = Ft  h = Applied moment, M I n t er ms of maximum compr essive st r esses or maximum t ensile st r esses M t = M yc =  Zt I M =



xx

M yc M c = = I xx ZC wher e Z stands for modulus of section and is equal to moment of iner tia of the section about NA divided by the extr eme value of y . d For symmet r ical cr oss sect ion, yc = y t = 2 wher e  = dept h of t he sect ion I  Zc = Zt = xx d/2 and, moment of r esist ance, M = Z

SLOPE AN D DEFLECTI ON OF BEAM Consider a beam AB which t ake cur ved shape as shown in t he figur e. Consider an element ar y lengt h CD equal t o ds of t he beam. L et t angent t o t he elast ic cur ve at C makes an angle  wit h x axis of t he beam. d2 y d 1 dx 2 Then = = 3 R ds 2 2 dy 1 dx

LM F I OP MN GH JK QP

d2 y dy 1 M is ver y small, t hen = = dx R EI dx2 Posit ive bending moment should cause posit ive cur vat ur e. When cur vat ur e is posit ive, angle  decr eases going on fr om C t o D. Ther efor e d2 y 1 d M = = = R ds EI dx2 d2 y  M = EI dx2 When

3.14

Strength of Materials

dy = dx

z zz

M dx + C1 EI M Deflect ion of beam, y = dx  dx + C1 x + C2 EI wher e C1 and C2 ar e const ant and t heir values can be obt ained fr om t he end condit ions of t he beam. Slope of t he beam,

d3 y dM = EI dx dx 3 d4 y Unifor m load, W = EI dx 4 Shear for ce, F =

M et hods for F inding Slope and Deflection at a Point of Beam 1. M acaulay's met hod 2. M oment ar ea met hod 3. Conjugat e beam met hod

W A

B

RA

a

b

RB

1. M acaulay's M et hod Point load on a simply suppor t ed beam : React ion, RA = Differ ent ial equat ion of bending becomes

l

Wb ab

and

Wa RB = ab

d2 y W bx = – M x = – RA  x + W (x – a) = – + W (x – a) 2 ab dx dy W( x  a)2 W bx 2 EI =– + + C1 dx 2 2(a  b)

EI I nt egr at ing, we get

EIy =–

I nt egr at ing again, we get

Applying end condit ions At x = 0, y = 0, which gives At

x = a + b,

y = 0, which gives

W bx 3 W( x  a)3 + + C1 x + C2 6 6(a  b)

C2 = 0 C1 =

W ab 6

W( x  a) 3 W ab Wbx 3 + + 6 6 6( a  b)

W( x  a)2 W ab W bx 2 dy =– + + 2 6 dx 2(a  b) Deflect ion under t he load is obt ained by put t ing x = a W a2 b2  yc = 3 EI l l I f concent r at ed load W is act ing at mid-span, t hen a = b = 2 Wl 3  (y )x = l/2 = 48 EI dy M aximum deflect ion is obt ained wher e =0 dx a(a  2b)  x = 3

and



EI

y max

W ab( a  2b) = 9( a  b) EI

...(ii ) ...(iii)

FG a  2bIJ H ab K

Subst it ut ing equat ions (ii) and (iii), we get EIy =–

...(i )

RS a (a  2bUV T3 W

12

FG a  2bIJ H ab K FG a  2bIJ H ab K

x

...(iv ) ...(v )

Strength of Materials

At

EI

x = 0,

FG dy IJ H dx K

a=b=

a=b=

FG a  2bIJ H ab K FG a  2bIJ H ab K

W ab 6EI Wl 2 B = – 16 EI

l , 2

W ab 6EI Wl 2 B = – 16 EI

At x = l, For

W ab 6

A =

 For

x 0

=

B = –

l , 2

FG 2a  bIJ H ab K

U nifor mly Dist r ibut ed L oad on Simply Suppor t ed Beam M oment at sect ion X – X

M x = RA  x – w

Now

RA = RB =

Differ ent ial equat ion of bending is

x2 2

wl 2

wlx wx 2 wx 2 d2 y = – M = – R  x + = – + x A 2 2 2 dx2 2 3 dy wl x wx EI =– + + C1 4 6 dx wl x 3 wx 4 EIy =– + + C1 x + C2

EI I nt egr at ing, we get and

Apply end condit ions : At x = 0, y = 0, which gives C2 = 0 wl 3 At x = l , y = 0, which gives C1 = 24 EIy =–

 At x =



l , t he deflect ion is maximum. 2 5wl 4 y max =

Sl ope At

x = 0,

 At

x = l,

12

24

wx 4 wl 3 x wl x 3 + + 12 24 24

384EI dy wx 3 wl 3 wl x 2 EI =– + + dx 6 24 4 wl 3 A = 24EI wl 3 E I A = 24 wl 3 B = – 24 EI

Cantilever Beam Car rying a Concentr at ed Load at F ree End Taking a sect ion at a dist ance x fr om t he fr ee end, we have M x = – Wx Thus differ ent ial equat ion of bending becomes,

d2 y = – M x = Wx dx2 dy W x2 EI = + C1 dx 2

EI I nt egr at ing, we get

3.15

3.16

Strength of Materials

EIy =

Again int egr ating

Apply end condit ions : At

dy Wl 2 Wl 3 = 0. Ther efor e C1 = – and C2 = dx 2 3 2 2 dy Wx Wl EI = – dx 2 2 3 Wx Wl 2 Wl 3 EIy = – x+ 6 2 3 2 Wl Wl 3 A = – and yA = 2 EI 3 EI

x = l , y = 0,

 and At

W x3 + C1 x + C2 6

x = 0,

Beams of Var ying Cr oss-sect ion For a beam of var ying cross-section and made of different materials, EI should be taken inside the integr ation sign.

d2 y = – Mx dx2 Mx d2 y  2 = – EI dx dy Mx  =– dx dx EI Consider a beam made of t wo mat er ials and differ ent cr oss-sect ion as shown in t he figur e. Then M x = RA  x – W 1 (x – a) – W 2 (x – b) W1 RA (l 1 + l 2) = W 1 (l 1 + l 2 – a) + W 2 (l 1 + l 2 – b) e.g.

EI

z

RA = W 1

FG1  H l

a  l2 1

IJ FG1  K +W H l 2

b  l2 1

IJ K

C

D A

RB = (W 1 + W 2) – RA

a

RA



l1

d y = – Mx dx2 d2 y Mx =– EI dx2 2 d y RA  x W ( x  a) W2 ( x  b) + 1 + 2 = – EI EI EI dx

I nt egr at ing, we get

dy = dx Then

y=

z

l1

0

0

b

l2

RB

x ×

RS R  x  W (x  a)  W (x  b) UV dx+ RS R  x  W (x  a)  W (x  b) UV dx + C E I E I E I E I T EI W T EI W RS R  x  W ( x  a)  W ( x  b) UV RS R  x  W ( x  a)  W ( x  b) UV dx . dx + E I E I E I EI EI T W T EI W dx. dx A

zz l1

z

B

E

2

EI

W2 ×

1

1 1

l1

0

A

1 1

2

1 1

1

1 1

2

1 1

1 1

l2

l1

A

1

2 2

zz l2

l2

l1

l1

2

2 2

A

2 2

1

2 1

1

2

2 2

2 2

+ C1 x + C2 Example. A cant ilever beam of lengt h L car r ies a concent r at ed load P at it s fr ee end. The beam for t he fir st half of it s lengt h (for fixed end t o cent r e) is made of diamet er D and for t he r emaining lengt h is D/2. Show t hat deflect ion at t he fr ee end  = Sol ut i on.

23 Pl 3 , wher e I 2 is moment of iner t ia of t he smaller sect ion. 384 EI 2

D 4 D 4 , and I 2 = 64 16  64  I 1 = 16 I 2 Consider a sect ion at a dist ance x fr om t he fr ee end as shown in t he figur e M x = – Px

I1 =

Strength of Materials

Now

EI

d2 y = – M x = Px dx2

I nt egr at ing t wice, we get

y =

zz z L 2

L 2

0

0

L 2

=

0

3.17

zz

Px EI 2 d x d x +

Px EI 2 . x dx + 3

z

L

L

3

L

L

L 2 L

Px dxdx 2 EI 1

Px 3 Px . x dx = 3EI 2 EI 1 2 3

3

L 2

0

Px 3 + 3EI 1

L

L 2

3

3 PL PL PL PL PL PL3 23 PL = 24EI + 3EI – 24EI = 24EI + 48EI – 384EI  y = 384 EI 2 2 1 1 2 2 2 Example. A cant ilever beam of lengt h l and car r ying a unifor mly dist r ibut ed load of int ensit y w per unit lengt h is made of st eel having modulus of elast icit y E s and moment of iner t ia I . I t is suppor t ed by an elast ic r od made of aluminium at t he fr ee end. The modulus of elast icit y of r od is E a, lengt h L and ar ea of cr oss-sect ion A. Calculat e t he for ce exer t ed by t he r od on t he cant ilever. Sol ut i on: L et F be t he for ce bet ween t he beam and r od at A.

Compr ession of r od, r = Deflect ion of beam due t o for ce F at A, 1 =

FL EaA Fl 3 (upwar d) 3E sI

Deflect ion of beam at A due t o unifor mly dist r ibut ed load w , 2 = Now

wl 4 (downwar d) 8E sI

2 – 1 = r Fl 3 FL wl – = 3E sI EaA 8E sI 4

F =



wl 4 8E sI

LM L  l OP N E A 3E I Q 3

a

2. M ohr 's M oment Area M ethod

s

Consider a par t AB of t he deflect ed beam as shown in t he figur e. A small element CD = ds. Dr aw t angent s at C and D t o t he deflect ed beam. Also dr aw nor mals t o t hese t angent s t o meet a point O, so t hat COD = d. I n limit ing case for small deflect ion,

1 d = R ds 1 M Also, = R EI M M dx d A  d = . ds  = EI EI EI wher e, dA = M dx is ar ea of t he bending moment diagr am bet ween point s C and D.

ds  dx ; and

Fr om t he figur e,

z z B

B

A

A

d =

M dx EI

3.18

Strength of Materials

z

M dx EI Now dist ance of point B above t he t angent AB 1 at A = BB 1. 

Change of slope fr om A t o B, A – B =

B

A

z

x  xB

Ver t ical dist ance bet ween t he point s C and D, BB 1 =

M (x – x C) dx EI B

x  xA

The expr ession within the integr al sign is moment of the M /EI diagr am between points A and B with r espect to point B and it is equal to the distance between the deflected posit ion of B above the tangent at A. Example. A cant ilever beam of lengt h l is loaded only one half of it s lengt h fr om t he fr ee end wit h a unifor mly dist r ibut ed load of w per unit r un. Der ive a for mula for t he deflect ion at t he fr ee end. w/unit length

B

A

C

A

w/unit length l 2

l 2

(a)

(b)

Sol ut i on: Unifor mly dist r ibut ed load is ext ended upt o t he fixed end B and an equal and opposit e load is applied bet ween BC as shown in Fig. (a). The bending moment diagr am is shown in Fig (b).

1 A x  A 2 x2 A = 1 1 = EI EI

LM 1  wl N3 2

2

l

FG H

3 1 wl 2 l l 3 l    l 4 3 8 2 2 8

IJ OP = 1 L wl K Q EI MN 8

4



7wl 4 384

OP = 41 Q 384

wl 4 EI

TORSI ON OF CI RCU LAR SH AFTS Assumptions (i ) The mat er ial of t he bar is homogeneous, per fect ly elast ic and obeys H ooke's law. (ii ) The st r ess does not exceed t he limit of pr opor t ionalit y. (iii ) Cr oss-sect ions r ot at e as if r igid, i.e. ever y diamet er r ot at es t hr ough t he same angle. T A 

B

B

C

C



x

dx

O d/2 T l

(a)

(b)

Consider a cir cular shaft t o lengt h l and diamet er d subject t o a couple T as shown in Fig. (a). A line AB on t he sur face of t he shaft , which is par allel t o t he axis befor e st r aining, t akes up t he for m of a helix AC aft er st r aining. L et  be t he angle of shear st r ain on t he sur face. Then BC = l 



=

BC l

Strength of Materials

But

3.19

 G  = shear st r ess in t he shaft G = modulus of r igidit y.  = G =

wher e,



 T G = = r J l

 wher e J =

d 4 is polar moment of iner t ia of t he shaft cr oss-sect ion 32

COM PARI SON OF H OLLOW AN D SOLI D SH AFTS I f shaft is hollow having inner r adius r and out er r adius R, t hen

eR

  . R 2   R3 = 2

T hol l ow = T sol i d

Thol l ow = Tsol i d



4

FR r I GH R JK 4

4

j

 r4 =

 J R hollow

1 R3

COM POSI TE CI RCU LAR SH AFTS (i) Shafts in Series Consider a shaft made of t wo mat er ials connect ed in ser ies and subject ed t o t he t or que T.

d2

2

T

1

d1

T l2

l1

G11 T2 1  G  = and = 2 = 2 2 d1 d2 l1 J2 l2 2 2 J G  J  J  JG T1 = 1 1 = 1 1 1 , or T2 = 2 2 = 2 2 2 d2 l2 d1 l1 2 2 Since bot h t he shaft s ar e subject ed t o t he same t or que, t her efor e T1 = T 2  3  d24  d1 J2 J G d2 1 d1 1 l1 32  =  = = and = 2 2  =  J1 l2 d1 2 d2 2 J 1G1  d14  d2 32 Then,

T1 J1

=

FG IJ H K

FG d IJ Hd K 2

1

4



l1 G  2 l2 G1

( ii ) Shafts in Parallel Consider a shaft made of t wo mat er ials connect ed in par allel and T be t he t or que applied t o t he composit e shaft . Then T = T1 + T2 T1 =



T =

G11 J 1 l1

and T 2 =

G11 J 1 G 2 2 J 2 + l1 l2

I f l 1 = l 2 = l and 1 = 2 = , t hen

G 2 2 J 2 l2

d1

d2

l2

l1

3.20

Strength of Materials

T = wher e J 1 =

d14 ; 32

 (G1 J 1 + G2 J 2); l

J2 =

 32

ed

4 2

 d14

and

=

Tl G1 J 1  G 2 J 2

j

TAPE RE D CI RCU LAR SH AFT Consider t aper ed cir cular shaft and T be t he t or que applied at t he ends. L et 1 and 2 be t he maximum shear st r esses at t he ends having diamet er s d1 and d2 r espect ively and  be t he shear st r ess at a dist ance x fr om t he end having diamet er d2. T

l  d

d2

d1 T dx

x

Since t or que is same t hr oughout t he shaft , t her efor e    3 d13 1 = d23 2 = d  16 16 16  d13 1 = d23 2 = d3  Consider a small lengt h dx of t he shaft at a dist ance x fr om t he lar ger end. L et d be diamet er of t he shaft at t his sect ion. I f d is angle of t wist of t he small lengt h dx , t hen 32T T d = dx = dx d 4G GJ d  d1 Now t an  = 2 2l d  d1 d2  d1  d = d2 – x = d2 – kx ......[ k = 2 ] l l 32 T dx  d = G(d2  kx)4

FG H



IJ K

32T l Tot al angle of t wist for t he lengt h l = 3G 16 T M aximum shear st r ess, max = d13

FG H

Fd GH

2 1

 d22  d1 d2 d13 d23

IJ K

I JK

STRAI N EN ERGY AN D RESI LI EN CE Ext er nal for ces when applied t o a member, causes defor mat ion of t he member. The amount of defor mat ion depends upon t he manner in which t he load has been applied. The member opposes t he defor mat ion and i n doing so it develops int er nal st r esses. These int er nal st r eses have t he capacit y t o do wor k and get s st or ed as st r ain ener gy in it . Only elast ic mat er ials get st r ained and develop st r esses and so st r ain ener gy can be st or ed only in elast ic mat er ials. The wor k done in pr oducing an elast ic st r ain in t he member is st or ed as st ain ener gy in t he mat er ial, which r eappear s on t he r emoval of load. We know, wit hin t he elast ic limit , st r ess is dir ect ly pr opor t ional t o st r ain

u  Normal strain

e

 (,) u  Shear strain

Volumetric strain

(,)

Shear strain

Normal strain

e

(,) u  Volumetric strain

Consider a bar car r ying a nor mal st r ess  (compr essive or t ensile) pr oducing a nor mal st r ain e. Then St r ain ener gy per unit volume, u =

1 .E (ar ea of t he shaded por t ion) 2

Strength of Materials

st r ain, e =

But

3.21

 E 2

 u = 2E



2

 Tot al st r ain ener gy, U =  volume of t he specimen 2E The st r ain ener gy absor bed by t he specimen is also called r esilience. U nits : St r ain ener gy has unit s of wor k done, i.e. N-mm or kg-cm et c.

Pr oof Resil i ence The maximum st r ain ener gy absor bed by t he body up t o it s elast ic limit is called pr oof r esilience. 2

2

e ( e ) (A.L )  Volume = = 2E 2E Pc  c . L 2 E

Pr oof r esilience = wher e, cA = load at elast ic limit = wher e

FG IJ FG H KH

IJ K

FG  A IJ FG L   IJ H 2 K H E K e

e

 e L change in lengt h up t o elast ic limit E Pr oof r esilience =

FG 1 IJ (P ) (L ) H 2K e

e

Pr oof r esilience per unit volume is called modulus of r esilience. 2

M odulus of r esilience =



e 2E

Similar ly, shear st r ain ener gy per unit volume, u s =

1 2 2  = 2 2G

and,

2  volume 2G

t ot al shear st r ain ener gy, U =

wher e, = shear st r ess G = modulus of r igidit y or shear modulus. Similar ly,

volumet r ic st r ain ener gy per unit volume, u v =

2 1  v . ev = v 2 2K 2

and,

t ot al volumet r ic st r ain ener gy, V v =

v  Volume 2K

wher e, v = volumet r ic st r ess K = bulk modulus. Example. A st eel specimen 1.5 cm 2 in cr oss-sect ion st r et ches by 0.005 cm over a 5 cm gauge lengt h under an axial load of 30 kN. Calculat e st r ain ener gy st or ed in t he specimen at t his st age. I f load at elast ic limit for t he specimen is 50 kN, calculat e elongat ion at elast ic limit and pr oof r esilience. Sol ut i on. Cr oss sect ional ar ea = 150 mm 2 ; Elongat ion, l = 0.005 mm ; L = 50 mm, and P = 30,000 N Tot al st r ain ener gy,U =

FG 1 IJ ( P) (l ) = FG 1 IJ (30,000) (0.05) = 750 N– m H 2K H 2K

As we know, elongat ion due t o 30,000 N load = 0.05 mm

0.05  50,000 = 0.0833 mm (it is in t he elast ic limit ) 30,000 Ther efor e, maximum st r ain ener gy at elast ic limit , i.e. Ther efor e, elongat ion due t o 50,000 N load =

Pr oof r esilience =

FG 1 IJ H 2K

(Pe) (l ) =

FG 1 IJ H 2K

(50,000) (0.08333) = 2083.33 mm

3.22

Strength of Materials

Example. A bar 100 cm in lengt h is subject ed t o an axial pull such t hat t he maximum st r ess is equal t o 15000 N/mm 2. I f ar ea of cr oss-sect ion is 2 cm 2 over a lengt h at 95 cm and for middle 5 cm lengt h, cr oss sect ional ar ea is 1 cm 2. I f E = 2 105 N/mm 2, calculat e st ain ener gy st or ed in t he bar. Sol ut i on. St r ess in par t I and I I I :

1 =

St r ess in par t I I : 2 = Tot al st r ain ener gy st or ed,

15,000 = 75 N/mm 2 200 15,000 = 150 N/mm 2 100

LM   Volume of par t I +   Volume of par t I I +   Volume of Part III 2E MN 2E 2E L (75)  (200) ( 475)OP + (150)  (100) (50) =2 M MN 2  2  10 PQ 2  2  10

U =

1

2

2

2

1

2

2

2

5

5

= 2671.825 + 281.25 = 2953.125 N– mm.

TH I N -WALLED PRESSU RE VESSELS Thin walled pr essur e vessels ar e used for t r anspor t at ion and st gor age of gases and fluids. I n t his t ype of vessels t hi ck ness of t he wal l , t i s ver y sm al l as compar ed t o di amet er of t he vessel s. Thi ck ness equal s t o 1/10 or less of inner r adius. A vessel car r ying a gas or liquid under a pr essur e p is subjected t o t ensile for ces, which r esist t he bur sting for ces developed acr oss longit udinal and t r ansver se sect ions. P t

P L

d

Z

N

 D

Now consider a cylinder ical vessel of diamet er D , t hickness t lengt h L subject ed t o int er nal pr essur e P. Consider equilibr ium of vessel along t he sect ion Z-Z. Element ar y for ce act ing nor mal t o an element of t he cylinder at an angle  fr om hor izont al diamet er,

pL D . d 2 Same for ce act s on t he symmet r ically placed element on t he ot her side of t he ver t ical cent er line. So hor izont al component of such for ces cancels out , and bur st ing for ce f is summat ion of t he ver t ical component of t hese element ar y for ces. dF = p.dA =



F

=

 0

pL D . d sin  (for ver t ical component only) 2

pL D  pL D   cos 0 sin d = 2 2 0  F = pDL Tot al bur st ing for ce F , act ing nor mal t o t he cut t ing plane z-z, is r esist ed by t he equal for ces P act ing on each cut sur face of t he vessel wall. Consider ing st at ic equilibr ium in ver t ical dir ect ion, i .e. V = 0  F = p DL = 2P 

F =





Strength of Materials

3.23

This equat ion indicat es t hat bur st ing for ce can be det er mined if int er nal gas or fluid pr essur e p is known, along wit h vessel diamet er and lengt h. St r ess in t he logit udinal sect ion t hat r esist s bur t ing for ce F is obt ained by dividing it by t he ar ea of t wo cut sur faces. This gives pDL F t = ... [for m  = ] 2t  L A

pD ...(i ) 2t St r ess obt ained by equat ion (i ) is called t angent ial st r ess because it act s t angent ially t o t he sur faces of t he cylinder. These st r esses ar e also called cir cumfer ent ial st r ess, hoop st r ess, and gir t h st r ess. N ote: St r ess comput ed by equat ion (i ) is aver age st r ess for t hin cylinder s only, which is pr act ically equal t o maximum st r ess at t he inside sur faces, for t he st r ess dist r ibut ion in t hick walled cylinder s. Now consider equilibr ium of t r ansver se sect ion. Bur sting for ce acting over t he end of t he cylinder is r esist ed by r esult ant P of t he t ear ing for ces act ing over t he t r ansver se sect ion. Since t is small as compar ed t o D, 2 Ar ea of t r ansver se sect ion =  (D + t ) t F=  Dt (closely appr oximat ed) D Now P =F D 2 P   Dt  l = 4 P=(Dt). L PD  l = ...(ii ) 4t wher e t is called longit udial st r ess, because it act s par allel t o t he longit udinal axis of t he cylinder. Fr om equat ions (i ) and (ii ), it is clear t hat logit udinal st r ess is one half t he value of t angent ial st r ess. Wit h t his, it is concluded t hat if pr essur e in a cylinder is r aised t o t he bur st ing point , failur e will occur along a logit udinal sect ion or longit udinal seam of t he cylinder . When a cylinder ical tank is composed of two sheets riveted together, then strength of t he logit udinal joint should be t wice t he st r engt h of gir t h joint . For cylinder, wit h r ounded or dished end, t he bur st ing for ce t r aver se sect ion may st ill be comput ed as t he pr oduct of t he int er nal pr essur e mult iplied by t he Girth Joint pr oject ed ar ea of t he t r ansver se sect ion. H er e it is assumed t hat t he volume bet ween t r ansver se sect ion A-A and t he r ounded end is filled wit h a fluid. The r esult ant longit udial for ce will equal t o pr oduct of t he pr essur e int ensit y mult iplied by t he shaded ar ea of t he t r aver se sect ion. Bur st ing for ce = Pr oduct of pr essur e and pr oject ed ar ea t =



D 2 4 Bur st ing for ce is r esist ed by t he for ce P dist r ibut ed acr oss t he t hickness of t he vessel. Again consider ing  (D + t ) t =  Dt Since, P =F

i .e.

 

F = p

D 2 4 pD = 4t

Rounded or Convexend

 Dt = p

which is same as equat ion (ii ) The t angent ial or cir cumt er ent ial st r ess t and longit udinal or axial st r esses l ar e quit e lar ge in compar ision t o t he r adial st r ess p, t her efor e, in st r ain det er minat ion p is not consider ed.  1 l Cir cumfer ential st r ain, ec = t – E m E

3.24

Strength of Materials

wher e, e = Young’s modulus of elast icit y of t he mat er ial and

1 = Poissons r at io m

pD 1 pD pD 1  = (2 – ) 2tE m 4 tE 4 tE m change in dimaet er, D = ec  D ec =

 Ther efor e,

...(iii )

pD 2  1 2   4 tE  m  1 t pD 1 pD Axial st r ain or longit udinal st r ain, ea = l – =  E m E 4 tE m 2tE pD  2 1–   ea = 4 tE  3 pDL  2 1– Change in lengt h, L = ea  L = 4 tE  m  D =



...(iv )

Final volume – I nit ial volume I nit ial volume   ( D  D ) 2  ( L  L ) – D 2 L 4 4 =  2 D L 4 2D L  = (neglect ing higher or der t er m of dD and dl ) D L = 2ec + ea

Volumet r ic st r ain =

=

2 pD 1 pD 2 pD 4 (2 – ) (1  ) = (5 – ) 4 tE m 4 tE m 4 tE m

...(v )

pDV 4 (5 – ) 4 tE m p Change in volume of t he liquid, V2 = V K wher e K is bulk modulus of t he liquid. N ote: Equat ions (iii ) t o (vii ) ar e applicable t o t he cylinder vessels wit h flat ends. Change in volume of t he cylinder, V1 =

...(vi ) ...(vii )

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. x, y and xy ar e r ect angular st r ess component s at a point . The r adius of M ohr ’s cir cle is

(c) span/lengt h of t he beam (d) r atio of over all depth to thickness of the flange 3. The value of t he maximum shear st r ess will be

2

(a)

 2x

(c)

2 y



 2y



(b)

 x   y  2     xy 2  

(d)

 x –  y  2     xy 2  

 2xy

2

2 x

  

2 xy

2. The maxi mum al l owabl e compr essi ve st r ess cor r esponding t o lat er al buckling in a discr et ely later ally suppor t ed symmet r ical I -beam, does not depend upon

(a) 25 5

(b) 50 5

(c) 100 5 (d) 200 5 4. The buckling load in a st eel column is

(a) modulus of elast icit y

(a) r elat ed t o t he lengt h

(b) r adius of gyr at ion about t he minor axis

(b) dir ect ly pr opor t ional t o t he slender ness r at io

Strength of Materials

(c) inver sely pr opor tional to the slender ness r atio (d) non-linear ly t o t he slender ness r at io 5. I f t he pr inci pal st r ess cor r esponding t o a t wodimensinal state of st r ess ar e 1 and 2 and 1 is gr eater than 2 and both ar e tensile, then which one of the following would be the cor r ect cr iter ion for failur e by yielding, accor ding to the maximum shear str ess cr iter ion ? (a)

 yp (1   2 )  2 2

(b)

 yp 1  2 2

(c)

 yp 2  2 2

(d) 1 = ± 2 yp

6. Consider t he following st at ement s : I n a uni-dimensional st r ess system, t he pr incipal plane is defined as one on which t he 1. Shear st r ess is zer o 2. Nor mal st r ess is zer o 3. Shear st r ess is maximum 4. Nor mal st r ess is maximum Of t hese st at ement s : (a) 1 and 2 ar e cor r ect (b) 2 and 3 ar e cor r ect (c) 1 and 4 ar e cor r ect (d) 3 and 4 ar e cor r ect 7. A M ohr ’s cir cle r educes t o a point when t he body is subject ed t o (a) pur e shear (b) uniaxial st r ess only (c) equal and opposi t e axi al st r esses on t wo mut ual l y per pendi cul ar planes, t he pl anes being fr ee of shear (d) equal axial st r esses on t wo mut ually per pendicular planes, t he planes being fr ee of shear. 8. The number of st r ai n r eadi ngs (usi ng st r ai n gauges) needed on a plane sur face t o det er mine t he pr incipal st r ains and t heir dir ect ions is (a) 1 (b) 2 (c) 3 (d) 4 9. When t wo mut ual l y per pendi cul ar pr i nci pal st r esses ar e unequal but al ik e, t he maximum shear st r ess is r epr esent ed by (a) t he diamet er of t he M ohr ’s cir cle (b) half t he diamet er of t he M ohr ’s cir cle (c) one-t hir d t he diamet er of t he M ohr ’s cir cle (d) one-four t h t he diamet er of t he M ohr ’s cir cle 10. I f t he val ue of Poi sson’s r at i o i s zer o, t hen i t means t hat (a) t he mat er ial is r igid (b) t he mat er ial is per fect ly plast ic (c) t her e is no longit udinal st r ain in t he mat er ial (d) none of t hese

3.25

LEVEL-1 11. The plane of maximum shear st r ess has nor mal st r ess t hat is (a) maximum

(b) minimum

(c) zer o

(d) none of t hese

12. When  and Young’s M odul us of El ast i ci t y E r emains constant, the ener gy-absor bing capacity of part subject to dynamic forces, is a function of its (a) length

(b) cr oss-sect ion

(c) vol ume

(d) none of t hese

13. The shear st r ess dist r ibut ion over a r ect angular cr oss-sect ion of a beam follows (a) a st r aight line pat h (b) a cir cular pat h (c) a par abolic pat h

(d) an ellipt ical pat h

14. When a col um n i s fi xed at bot h ends, cor r espondi ng Eul er ’s cr i t i cal l oad i s (a) (c)

 2 EI

(b)

L2

3 2 EI

2  2 EI L2 4  2 EI

(d) L2 L2 15. For the two shafts connected in par allel, find which st at ement is t r ue ? (a) Tor que in each shaft is t he same (b) Shear st r ess in each shaft is t he same (c) Angle of t wist of each shaft is t he same (d) Tor sional st iffness of each shft is t he same.

16. The slenderness ratio of a compression member is : (a)

Effect ivelengt h Least r adiusof gyrat ion

(b)

Act ual lengt h M oment of iner t ia

(c)

Moment of iner t ia Act ual lengt h

(d)

Act ual lengt h R adiusof gyrat ion

17. The length of a bar is L metres. It extends by 2 mm when a tensile force F is applied. Find the strain produced in the bar : (a)

0.002 L

(b)

2 L

(c)

0.2 L

(d)

L 0.002

3.26

Strength of Materials

18. For perfectly elastic bodies, the value of coefficient of restitution is : (a) zero

(b) 0.5

(c) 1.0

(d) 0.25

uniform load of W/L

(1)

19. Choose the option which correctly shows the relationship between Modulus of Elasticity (E); Modulus of Rigidity (c) and Bulk Modulus (K) : (a) E 

KC K C

(b) E 

2K C 2K  C

(c) E 

9K C 3K  C

(d) E 

3K C K  2C

20. The property of a material by which it can be rolled into sheets is called : (a) Elasticity

(b) Plasticity

(c) Ductility

(d) Malleability

(2) (3) (4) (a) 1 (b) 2 (c) 3 (d) 4 27. A si mply suppor t ed beam car r ies a var ying load fr om zer o at one end t o  N /m at t he ot her end (as under ).

N/m

21. A simply supported beam of length L is loaded with a uniformly distributed load of "  per unit length. The maximum bending moment will be : (a)

L 2 4

(b)

(c)

L 2 2

(d) L2

A

L The lengt h of t he beam is L . The shear for ce wi ll be zer o at a dist ance 'x ' fr om A. Fi nd 'x ' :

L 2 8

(a)

22. Which of the following property is generally NOT shown by metal ? (a) Electrical conduction

28.

(c) dullness (d) ductility 23. I n S.I syst em, unit of st r ess i s: (a) kg/cm 2 (b) N (c) N/m 2 (d) Watt 24. What i s t he funct i on of push r od i n a di esel engi ne? I t t r ansfer s for ce bet ween (a) Cam and r ocker ar m (b) Connect ing r od and pi st on (c) Cr ankshaft and pist on (d) None of t hese 25. I n C.G.S syst em, t he unit of st r ain i s: (a) cm/k g (b) m/k g (c) no unit (d) None of t hese 26. I n t he case of a uni for ml y dist r ibut ed load on a si mpl y suppor t ed beam, t he bendi ng moment diagr am would be-

29.

30.

31.

32.

L 2

L

(b)

L 4

L 3 3 Which one of t he following is t he most significant pr oper t y t o be consi der ed i n t he sel ect i on of mat er ial for the manufact ur e of locating pins and dr il l jig bushes used i n jigs and fixt ur es ? (a) Wear Resi st ance (b) Elasticity (c) Shear St r engt h (d) Tensile St r engt h What is t he main shaft of an engine t hat cont r ols t he movement of pi st on ? (a) axle (b) dr ive shaft (c) cr ank shaft (d) cam shaft The component of t he engine t hat connect s t he li nk bet ween t he small end of t he connect ing r od and t he pist on is k nown as (a) Cams (b) Fl y wheel (c) Gudgeon pin (d) Pi st on r i ng The negat ive r at io of t r ansver se t o axi al st r ain is cal led as (a) Young's modul us (b) Shear modul us (c) Poisson's r at io (d) Bulk modulus of elast i ci t y To con n ect pi st on t o t h e con n ect i n g r od t he_______ ar e used: (a) r od caps (b) cap bol t s (c) smal l end bear ings (d) gudgeon pi ns (c)

(b) Sonorous in nature

LEVEL-2

B

(d)

Strength of Materials

3.27

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (d)

2. (a)

3. (b)

4. (a)

5. (a)

6. (c)

7. (d)

8. (c)

9. (b)

10. (d)

17. (a)

18. (c)

19. (c)

20. (d)

29. (c)

30. (c)

31. (c)

32. (d)

LEVEL-1 11. (b)

12. (c)

21. (b)

22. (c)

13. (c)

14. (d)

15. (c)

16. (a)

LEVEL-2 23. (c)

24. (a)

25. (c)

26. (b)

27. (c)

28. (d)

EXPLAN ATI ON S 2.

3.

Si nce al l owabl e compr essi ve st r ess depends l D upon r at io and r r at io as per I .S. Code y T 800 : 1984. Ther efor e, it will not depend upon t he modulus of elast icit y. x = 200 , y = 100, xy = 100

p1 , p2 =

=

5.

y

±

2

 x –  y  2     xy 2   2

 200 – 100  2    100 2  

200  100 ± 2

50 2  100 2 = 150 ± 50 5 = 150  50 5 , 150 – 50 5

p2 = 150 – 50 5 A t hir d pr incipal st r ess will be t hese p = 0 3

150  50 5 – 0 = 130.9 2 2 This will be maximum shear str ess in any plane Also max =

=

 p1 –  p2

= 50 5 2 This is maximum in plane shear st r ess. 2 EI 2 E.Ar 2 4. Per = = 2 2 =

l 2 EA

lr

2

l

=

2 E.A

 slender ness r at io



=

1 – 0 1 = 2 2

Cr it er ion 

 yp 1 =± 2 2

6.

 p1 = 150 + 50 5

 pmax –  pmin

 max –  mi n 2

M aximum Shear St r ess =

= 150 ±

max =

1

1 > 2 M aximum Shear St r ess Theor y

2

x   y

2

2

Per could be non linear ly r elat ed t o slender ness r at io so bet t er t o avoid choice (d).

Only (1) can be cor r ect (4) also Sat isfies  cor r ect st at ement is (1) and (4) 7. A M ohr ’s cir cle r educes t o a point when t he body i s subj ect ed t o equal axi al st r esses on t wo mut ually per pendicular planes, the planes being fr ee of shear.

FG  IJ H 2 EK 2

12.

St r ain ener gy,

U =

 and E r emaining const ant , U is pr opor t ional t o (A.L .) which is volume. Also, since U is a funct ion of 2, t hat por t ion of the par t which is pr one to high localised str esses will absorb a high amount of energy, making it vulnerable to failure, such a part, therefor e, is designed to have such a contour that, when it is subjected to time-varying or impact loads or other types of dynamic for ces, the par t absorbs or less uniform stress distribution along the whole length of the part is whole length of the part is ensured.

3.28

13.

Strength of Materials

A par abolic path

21. SF and BM Formulas Simply supported with uniform distributed load Fx = Shear force at X Mx = Bending Moment at X

 2 EI

14. Euler ’s cr it ical load, P =

( L eff ) 2 wher e, L eff = effect ive lengt h of t he column. When bot h ends ar e fixed, L ef f = 0.5 L 2

2

P Cr =



 EI (0.5 L)

2

=

 EI 2

0.25 L

x B

A

2

or

4  EI L2

16. Slender ness r at io is t he r at io of t he lengt h of a column and the least r adius of gyr ation of its cr oss sect i on. Oft en denot ed by l ambda. I t i s used ext ensively for finding out t he design load as well as i n cl assi f yi n g var i ou s col u mn s i n sh or t / int er mediat e/long. 18. The coefficient of restitution (COR) is the ratio of the final to initial relative velocity between two objects after they collide. It normally ranges from 0 to 1 where 1 would be a perfectly elastic collision. 19. We know that, E = 2C(1 +  ) ...(i)

and E = 3K (1 – 2  )

w Per unit length

C L

BM wL/2



wL/2

...(ii)

wL2 8

E 2C

E 1 2C

Equating this value in (i) E   E    E  3K 1  2   1    3K 1   2 2C C     

E   3C  E  E  3K 3    3K  C    C  EC = 3K(3C – E) = 9KC – 3KE  EC + 3KE = 9KC  E(C + 3K) = 9KC

E 

B

C

A

From (i)

1 

RB

RA

9KC C  3K

20. Malleability is a substance's ability to deform under pressure (compressive stress). If malleable, a material may be flattened into thin sheets by hammering or rolling. Malleable materials can be flattened into metal leaf. Many metals with high malleability also have high ductility.

22. Dullness is the property which is not shown by the metal. 23. The S.I . uni t of st r ess is N/m 2. 24. A push r od is a component of t he val ve t r ai n of cer tain pist on engines. I t 's funct ion is essent ially t o push t he valve open. Rods used in an over head valve engine t o open and close t he valves. One end is pushed up by the cam and ot her end makes cont act wit h t he r ock er ar ms whi ch r ot at es and pushes t he valve open. 25. St r ain is unit less quant it y. I t i s dimensionl ess. 26. Si mply suppor t ed beam wi t h udl,

W N/m B

A l

Now FBD, t ake sagging (+ve) and hoggi ng (– ve),

Strength of Materials

x

3.29

By FBD,

l

WL/2 A

B

C

x

l 2

l 2

x

2L/3

RB

L oad at poi nt C =

l x 2 x 2 2

WL 2

Now,  M B = O

 M A = (A bx = 0) = 0 – 0 = 0

RA  L 

 M B (A bx = L )



L/3

RA

l x x – x. Now, M x  2 2 Mx =

C L

L2 L2  0 2 2

WL L  2 3

WL 6

 RA 

For maximum bending,

H er e, fr om simil ar AXX and AWB,

dM x l  – x  0 dx 2

WXX x Wx   WXX  W L L

l l , so bendi ng moment at x  t o be 2 2 maxi mum so,

Shear for ce at X– X cr oss sect ion,

Now, x 

2

l    2 l  l  Mc  .   2  2 2

Mc 

SF at



 l 2 l 2  l 2   4 8 8



WL 1   WXX  x 6 2

WL 1 Wx  x 6 2 L



l 8

H ence, BM D shoul d be

0 2

Parabolic curve

l 8

B

C

So, opt ion (b) is cor r ect .



WL Wx 2 – 6 2L

Wx 2 WL  2L 6

 x2 

‘W’ N/M Wxx

x

X

27.

A



WL 1 Wx 2  6 2 L SF at X-X sect ion = 0 [By quest ion]

2

A

XX

x

X L

B

2L2 L2  6 3

1 3

28. Tensile strength is the criteria for selection of material in manufacturing of locating pins and jug bushes. 29. Crankshaft is the main shaft that controls the movement of piston.

3.30

Strength of Materials

30. Gudgeon pin is defi ned as t he pin which hol ds pi st on r od and connect i ng r od t oget her. 31. Poi sson’s r at i o: Poi sson’s r at i o i s t he r at i o of t r ansver se cont r act i on st r ai n t o l ongi t udi nal ext ension st r ai n i n t he dir ect ion of st r et ching for ce. Tensile defor mat ion i s consi der ed posi t i ve and com pr essi ve defor mat i on i s con si der ed negat i ve.

32. Gudgeon pins are those pins which holds piston and connecting rod together.

4

CHAPTER

Manufacturing Engineering (Machining, Welding and Finishing Process)

M ETAL CU TTI N G The pr imar y objective in met al cutting is the pr oduction of chips; which ar e r emoved to obt ain a component and event ually t he chips ar e t hr own away. These chips may const it ut e mor e t han 50% of t he init ial wor kpiece mat er ial. M et al cut t ing pr ocesses in gener al should be car r ied out at high speeds and feeds wit h t he least cut t ing effor t at a minimum cost . F act or s affect ing met al cut t ing. (i )

Pr oper t ies of wor k mat er ial.

( ii ) Pr oper t ies and geomet r y of t he cut t ing t ool. ( iii ) I nt er act ion bet ween t ool and t he wor k dur ing met al cut t ing. M echanics of met al cut t ing. A cut t ing t ool exer t s a compr essive for ce on t he wor kpiece which st r esses t he wor k mat er ial beyond it s yield point and t her efor e t he mat er ial defor m plast ically and shear s off. Plast ic flow t akes place i n a localised r egion called shear plane. Chi p The shear ed mat er ial begins t o flow along t he cut t ing t ool face in Pl a n t he for m of small pieces called chips. eo Sh f s Tool ear h ea The applied compr essive for ce which leads t o t he for mat ion of r an gl e chips is called cut t ing for ce. The fl owing chips cause t ool wear. The heat pr oduced dur ing shear ing act ion r aises t emper at ur e of t he wor k mat er ial, cut t ing Wor k piece t ool and chips. Temper at ur e r ise in t he cut t ing t ool soft ens it and causes loss of keeness in t he cut t ing edge leading t o it s failur e. Cut t ing for ce, heat and abr asive wear ar e t he essent ial feat ur es of t he mat er ial cut t ing pr ocess. T ypes of met al cut t ing processes. ( i ) Or t hogonal cut t ing (t wo dimensional) : I n this t ype, major cutt ing edge of t he tool is pr esent ed t o the wor kpiece per pendicular to the dir ect ion of feed mot ion. I n or t hogonal cut t ing only t wo for ces ar e involved simplifying t he analysis of cut t ing mot i on. ( ii ) Oblique cut t ing : I n oblique cut t ing, major edge of t he cut t ing t ool is pr esent ed t o t he wor kpiece at an angle which is not nor mal (angle  90º) t o t he dir ect ion of feed mot ion. CH I PS FORM ATI ON Types of chipes. ( i ) Cont inuous chip. This t ype of chip is pr oduced when duct ile mat er ials such as aluminium, copper, wr ought ir on, and mild st eel ar e machined at nor mal cut t ing speeds. This t ype of chip is pr oduced when t her e is low fr ict ion bet ween chip and t he t ool face. This chip has t he shape of long st r ing or cur ls int o a t ight r oll. ( ii ) Cont inuous chip wit h built up edge. When high fr ict ion exist s bet ween chip and t he t ool, t he chip mat er ial welds it self t o t he t ool face. This welded mat er ial incr eases fr ict ion fur t her which in t ur n leads t o t he building up of layer upon layer of chip mat er ial. Thus t he built mat er ial is r efer r ed t o as a built up edge. The built up edge

4.2

Manufacturing Engineering

gr ows and br eaks down when it becomes unst able. Chips wit h built up edge ar c not desir ed as t hey r esult in higher power consumpt ion, poor sur face finish and lar ge t ool wear. ( iii )D iscont inuous chip (segment ed chip). Discont inuous or segment ed chips ar e plast ically defor med wor kpiece segment s which ar e eit her loosely connected to each other or completely unconnected. This type of chip is pr oduced while machining br it t le mat er ials such as cast ir on or cast br ass but may also be pr oduced when machining duct ile mat er ials at ver y low speeds and high feeds. For br it t le mat er ials discont inuous chip is associat ed wit h fair sur face finish, lower power consumpt ion and r easonable t ool life. H owever, t his t ype of chip is not desir able in t he case of duct ile mat er ials as t hey indicat e poor sur face finish and excessive t ool wear. ( iv) N on-homogeneous chips. These chips ar e char act er ized by not ches on t he fr ee side of t he chip and t hey ar e for med due t o nonunifor m st r ain in t he mat er ial dur ing chip for mat ion. This t ype of chip is for med of mat er ials whose yield st r engt h decr eases wit h t emper at ur e and which have poor t her mal conduct ivit y. CU TT I N G T OOLS Type of cut t ing t ools. (1) Single-point cut t ing t ool. This t ype of t ool has a effect i ve cut t ing edge and r emoves excess mat er ial fr om t he wor k mat er i al al ong t he cut t ing edge. Types of single-point cut t ing tool : ( i ) Gr ound type ( ii ) For ged t ype ( ii ) Tr ipped type ( iv) Bit t ype

Minor cutting edge

Major cutting edge Tool axis

Shank Minor flank Major flank

Base

Single-point cut ting t ools ar e commonly used in lathes, planer s, shaper s, bor ing machines and slott er s. These t ools may be left-handed or r ight-handed. A t ool is said t o be r ight /left-handed if t he cut ting edge is on t he r ight or left side when t he t ool is viewed fr om t he point end. (2) M ult i-point cut t ing t ools. These t ools have mor e t han one effect ive edge. e.g. milling cut t er s, dr ills, r eamer s, br oaches and gr inding wheels et c. TOOL M ATERI ALS. Properties of cutting tool materials. ( i ) Red hardness or H ot hardness: I t is t he abilit y of a mat er ial t o r et ain it s har dness at high t emper at ur es. ( ii ) Wear resist ance: I t enables t he cut t ing t ool t o r et ain it s shape and cut t ing efficiency. ( iii )Toughness: I t r elat es t o t he abilit y of a mat er ial t o r esist shock or impact loads associat ed wit h int er r upt ed cut s. ( iv) Ot her pr opert ies : Thermal conductivity, specific heat, coefficient of ther mal expansion, hardenability, dimensional stability, fr eedom fr om distortion after heat tr eatment, resistance to decar bur isation, grindability and weldability. Classificat ion of t ool mat erials. (1) Car bon-t ool st eels. These ar e plain car bon st eels wit h 0.6 t o 1.5% car bon. Ver y lit t le quant it ies of M n, Si, Cr or V ar e added t o incr ease har dness and r efine gr ain size. L ow car bon var iet ies possess good t oughness and shock r esist ance wher eas t he high car bon var iet ies ar e abr asion r esist ant wit h an abilit y t o maint ain a shar p cut t ing edge.

Manufacturing Engineering

4.3

Types of carbon tool steels. (i ) Wat er har dening t ype (ii ) Oil har dening t ypes Car bon t ool st eels possess good machinabilit y, high sur face har dness wit h a fair ly t ough cor e. They loose t heir har dness r apidly at t emper at ur es above 200°C. Their dimensional st abilit y dur ing heat t r eat ment is poor. They cr ack under dr ast ic dimensional changes. Car bon t ool st eels ar e used in t he manufact ur e of milling cut t er s, t wist dr ills, t ur ning and for m t ools for use on wood, br ass, aluminium and magnesium. I n t hese applicat ions, t emper at ur e at t he cut t ing edge is below 200°C, t her eby t he t ool r et ains it s shar p edge and har dness. (2) H igh speed steel. These ar e basically high car bon st eel with significant quantit ies of t ungst en, molybdenum, chr omium, vanadium and coabalt . These alloying element s impr ove har dness, har denabilit y, t oughness and wear r esist ance. They also impr ove high t emper at ur e pr oper t ies. These st eels r et ain t he keenness of t he cut t ing edge and har dness upt o 600°C, t her eby per mit t ing much higher cut t ing speeds. Types of high speed steel : (i ) T t ype for Tungst en pr edominant alloy. (ii ) M t ype for M olybdenum pr edominat e alloy. Accor ding t o AI SI designat ion (Amer ican I r on and St eel I nst it ut e), M olybdenum high speed st eels ar e slight ly t ougher at t he same level of har dness as compar ed t o t he t ungst en var iet y. Pr esence of chr omim impr oves har denabilit y, vanadium impr oves abr asion r esistance. Gr indability decr eases with higher per cent age of vanadium. Cobalt is added t o high speed st eel t o incr ease it s r ed har dness. H igh speed st eels ar e used for single point lat he t ools, milling cut t er s, dr ills, br oaches, hobs, shaver s, t aps and t ools bit s. (3) Cast cobalt base alloys. I t is combinat ion of t ungst en, chr omium and cobalt which for m an alloy wit h high r ed har dness, wear r esist ance and t oughness. This mat er ial is made by melt ing t he element s t oget her and t hen cast in moulds, hence t hey ar e called cast alloys. Pr imar y element s in cast alloys. (i ) Cobalt (ii ) Chr omium (iii ) Tungsten (iv ) Car bon. Ot her el ement s such as Vanadium, M ol ybdenum, Tant al um, et c, ar e added t o i mpr ove speci fi c pr oper t ies. A par t of chr omium and t ungst en can be subst i t ut ed by mol ybdenum, t ant alum and columbium t o for m car bides. I n cer t ain inst ances nickel and bor on ar e also added. These cast alloys have pr oper t ies in bet ween high speed st eels and cement ed car bides. These alloys ar e used for machining cast ir on, alloy st eels, non-fer r ous met als, cer t ain super alloys, gr aphite and plast ic. (4) Cement ed car bides. These ar e car bides of t ungst en, t it anium and t ant alum wit h cer t ain amount of cobalt . These car bides ar e pr oduced t hr ough t he powder met allur gy r out e. Types of cement ed carbides : ( i ) St r aight t ungst en car bide cobalt gr ade : They ar e char act er i sed by hi gh abr asi ve wear r esist ance and st r engt h. They ar e not as t ough as H SS but possess higher har dness, densit y and modulus of eleast icity. I n commer cial gr ades, the cobalt content var ies fr om 3 t o 12%. Wit h incr ease in cobalt cont ent har dness, elast ic modulus, compr essive st r engt h decr ease, however t r ansver se r upt ur e st r engt h incr ease. ( ii ) Alloyed tungsten carbide crade : The st r aight tungsten car bide tools fail r apidly due to for mat ion of cr at er while machining st eel at higher speeds. H owever, t his cr at er wear can significant ly r educed by t he addit ion of car bides of t it anium, t ant alum, neobium, et c.

4.4

Manufacturing Engineering

U se of carbide t ools. The st r aight car bides ar e used for machining no-duct ile mat er ials which ar e not pr one t o met allic adhesion bet ween chip and t he t ool face like C.I ., non fer r ous alloys, plast ics, wood, glass et c. Alloyed car bide ar e used for cut t ing all gr ades of st eel and cut t ing speeds 3 or 4 t imes t hose possible wit h H SS can be used. (5) Ceramic t ool mat er ials. (i ) Alumination oxide (ii ) Silicon car bide (iii ) Bor on car bide (iv) Tit anium car bide (v) Tit anium bor ide Best machining r esult s have been pr ovided by sint er ed alumina. A popular cer amic t ool cont ains 99% Al 2 O3 and t he r emaining 1% is shar ed by Cr 2 O3, M gO and NiO. Cer amic t ool cut t ing edge r et ains t he keenness even when t he chip is r ed hot . Cer amics ar e st able in air even at temper atur es ar ound 18800°C. They have good cor r osion r esist ance. They ar e non-magnet ic and do not conduct electr icit y. Har dness is high fr om ver y low t emper at ur es t o ver y high temper atur es. Wit h t hese t ools built up edge is never encount er ed. Cer amic t ools cut most met al and allow higher cut t ing speeds. Their t ool life is longer and t hey pr ovide super ior sur face finish. No coolant is r equir ed and coefficient of fr ict ion bet ween chips and t ool face is less. (6) D iamond t ools. I t is ext ensively used for t r uing t he gr inding wheels and t o a lesser ext ent for fine finishing of met als. Diamond is one of t he har dest mat er ials wit h excellent abr asion r esist ance. I t s t her mal conduct ivit y and melt ing point ar e high. I t pr ovides highest tool life and t ool can pr oduce mir r or like sur face finish, as defor mat ion dur ing cut t ing is minimum. Diamond t ools ar e used t o machine Al-Si alloys cont aining high silicon levels, abr asive non fer r ous alloys, silica, cement ed tungsten cabide, cer amics, etc. Diamond abr asvie belt s ar e used for machining t elevision scr eens. Poly cr yst alline diamond inser t s ar e br azed int o cut t ing edges of cir cular saws for cut t ing const r uct ion mat er ials like concr et e, r efr act or ies, st one et c. TOOL GEOM ETRY. Tool geomet r y r efer s t o t ool angles, shape of t he t ool face and for m of cut t ing edges. F act ors affect ing t ool geomet ry. (i ) Wor k mat er ial (ii ) M achining var iables (viz, cut t ing speed, feed and dept h of cut ) (iii ) Tool mat er ial (iv) Type of cut t ing E n d cu tting e dg e a n gle

S ide c le ara nc e

An gl es of tool H e el

S ide ra ke N o se ra dius

S ide c uttin g ed ge a ng le

B a ke ra k e

S ide re lie f End c le ara nce

Flan k

S h an k

End re lie f

Fig. Various angles of single point tool

Manufacturing Engineering

4.5

Back rake angle. I t measur es downwar d slope of t op sur face of t he cut t ing t ool fr om the nose t o the r ear along t he longitudinal axis. I t guides the dir ect ion of ch i p f l ow an d al so pr ot ect s t h e cu t t i n g t ool poi n t . Soft er t he mat er i al , gr eat er shoul d be t he posi t i ve r ak e angl e. Back r ake angle may be posit ive, neut r al or negat ive. F ig. Positive Back rake angle

Side rake angle. Side r ake angle measur es t he slope of t he t op sur face of t he t ool t o t he side in dir ect ion per pendicular t o t he longit udinal axis. This angle also guides t he dir ect ion of chip away fr om t he job. The amount by which a chip is bent depends upon t his angle. When the side r ake angle incr eases, magnitude of chip bending decr eases and t her efor e power needed t o par t and bend t he chip decr eases. Smoot her sur face fur nish is pr oduced by lar ger side r ake angle.

S ide ra ke

Vx

S ide re lief a ng le

F ig. Side rake and r elief angles

End cut ting edge angle. I t is t he angle bet ween minor flank and a plane per pendicular t o t he side of shank. This angle ser ves as a r elief angle allowing a small sect ion of t he end cut t ing edge t o cont r act t he machined sur face. This angle usually var ies fr om 5° t o 15°. Side cutt ing edge angle. The angle bet ween side cut t ing edge and longit udinal axis of t he t ool is called side cut t ing edge angle. I t contr ols t he dir ection of chip flow and avoids the for mat ion of built up edge. This angle also distr ibut es t he cut t ing for ce and heat pr oduced over lar ger cut t ing edge. Side relief angle. This angle is bet ween flank of t he t ool and a plane nor mal t o t he base just under t he side cut t ing edge. This angle allows t he t ool t o fed sideways int o t he job in or der t o cut t he wor k mat er ial wit hout r ubbing. L ar ger side r elief angle leads t o t he br eaking of t he cut t ing edge due t o insufficient suppor t , wher eas if t his angle is t oo small, t he t ool cannot be fed int o t he wor k mat er ial. When t he side r elief angle i s ver y small, t he t ool will r ub against t he job and t her efor e it will get over heat ed and become blunt and t he sur fae finish obt ained will be poor. End relief angle. The angle bet ween plane nor mal t o t he base and flank is called end r elief angle. This angle pr event s t he r ubbing bet ween wor k mat er ial and cut t ing t ool. Ver y lar ge end r elief angles lead t o t he br eaking of cut t ing edge, however, if t his angle is ver y small t he t ool will r ub on t he job leading t o poor sur face finish. This angle var ies fr om 6° t o 10°. N ose radius. Nose r adius incr eases finish and st r engt hen t he cut t ing t ip of t he t ool. Smaller r adii pr oduce smoot her sur face finish. Small r adii is mor e suit able for t hin cr oss-sect ion of wor k. Tool signat ur e. The t ool angles have been st andar dised by t he Amer ican St andar ds Associat ion. Cut t ing t ool signat ur e consist s of following seven impor t ant par amet er s. Back r ake, side r ake, end r elief, side r elief, end cut t ing edge, side cut t ing edge angles and nose r adius. A t ool of signat ur e 10, 10, 6, 6, 8, 8, 2 will have t he following angles. Back r ake angle = 10, Side r ake angle = 10, End r elief angle = 6, Side r elief angle = 6, End cut t ing edge angle = 8, Side cut t ing edge angle = 8, Nose r adius = 2 mm.

4.6

Manufacturing Engineering

TOOL LI FE. Excessive cut t ing speeds cause a r apid failur e of t he cut t ing edge of t he t ool; t hus, t he t ool can be declar ed t o have had a shor t life. F actors used t o evaluat e t ool life. ( i ) Change of t he qualit y of t he machined sur faces ( ii ) Change in the magnitude of the cutting force r esulting in changes in machine and workpiece dimensions t o change ( iii ) Change in t he cut t ing t emper at ur e. The select ion of t he cor r ect cut t ing speed has an impor t ant bear ing on t he economics of all met al-cut t ing oper at ions. The cor r ect cut t ing speed can be est imat ed wit h r easonable accur acy fr om t ool-life gr aphs or fr om t he Taylor Tool-life r elat ionships, pr ovided necessar y dat a ar e obt ainable. F act ors affect ing t ool life. ( i ) Tool geomet ry : A t ool wit h lar ge r ake angle becomes weak as a lar ge r ake r educes t he t ool cr osssect ion and amount of met al t o absor b t he heat . The effect of end cut t ing edge angle is t o impr ove sur face finish, r igidit y and equivalent cut t ing speed. The opt imum end cut t ing edge angle is 4° t o 10°. Clear ance is opt imum for 12° t o 15°. ( ii ) Tool mat erial : Effect of t ool mat er ial on t ool life is r emoving lar ge or maximum volume of met al by not only incr easing cut t ing speed but t he high r at e at which t he st ock will be r emoved per cut t ing edge or t ool life. An ideal t ool would r emove t he same amount of met al per cut t ing edge or any speed. Physical and chemical pr oper t ies of wor k mater ials influence-t ool life by affecting st abilit y and r ate of wear of t ool. ( iii ) Cutt ing fluid : I t affect t ool life t o a gr eat ext ent . Cut t ing fluids car r ies away t he heat gener at ed and keeps t he t ool, chip and wor k piece cool. Cut t ing fluids also r educes coefficient of fr ict ion at t he chip t ool int er face and incr eases t ool life. M ACH I N ABI LI TY I t is pr oper t y of t he mat er ial which gover ns t he ease or difficult y wit h which it can be machined under a given set of condit ions. Cr it er ia for machinabilit y. (1) Tool life. Tool life is a dir ect funct ion of cut ting speed. By incr easing the cutt ing speed, tool life may be decr eased and vice-ver sa. H ence speed is usually t aken as r efer ence t o expr ess machinabilit y r at ings. (2) M achinabilit y rat ings (based on cut t ing speed). (3) Cut t ing for ces. This is necessar y t o limit t he values of cut t ing for ce in keeping wit h t he r igidit y of t he machine and t o avoid vibr ations dur ing machining. I f cut t ing for ce is high and consequent ly t he power consumpt ion is also high, lar ger machine t ool may be r equir ed, t hus incr easing over head cost and unit cost of the par t pr oduced. H igher t he cut t ing for ces induced under a set of cut t ing condit ions dur ing t he machining of a mat er ial, lower is it s machinabilit y index. (4) Surface finish. I f given mat er ial allow higher cut t ing speed or induce lower cut t ing for ces, it may not pr oduce good sur face finish. Wher e a finish pr oduced on a par t is a cause for r eject ion, it has an impor t ant bear ing on t he cost . H igher t he sur face finish, bet t er is it s machinabilit y. (5) Tool r igidit y. Tool defect s due t o cut t ing for ces causing a det r iment al effect not t he t ool life, e.g. in end mill since it is clamped only at ends, consider able for ce may cause t he t ool t o deflect as it s slender ness r at io is lar ge. (6) Penet r at ion r at io (7) Case of chip disposal (8) Temper at ur e of cut t ing t ool (9) Wor k har dening.

Manufacturing Engineering

4.7

FACTORS AFFECTI N G M ACH I N ABI LI TY. (1) M achine var iables. Efficiency of machi ning oper at i ons depends on over all r igidi t y of machi ne t ool, cut t i ng t ool and wor kpiece. I f t he machine is not sufficient ly r igid and has less power, t he t ool life will be r educed in addit ion t o poor accur acy and sur face finish. L ower speed, dept h and t ool have t o be employed. Thus machinabilit y of a mat er ial is indir ect ly affect ed by t he machine var iables. (2) Tool var iables. ( i ) Tool materials : When t ools ar e made fr om t he pr oper t ool mat er ial and have pr oper geomet r y, a cut t ing oper at ion can be per for med efficient ly. I f cut t ing t ool is not opt imized, t he mat er ial r emoval r at es must be r educed t o obt ain a r easonable value of t ool life, ot her wise machining cost s will be incr easing. Cut t ing t ools ar e made of high car bon st eel, high-speed st eels, var ious t ype of sint er ed car bides, cast alloys, cer amics, diamonds. Specific cut t ing speeds, defined as t he speed pr oducing a pr edet er mined lengt h of t ool life, ar e lar gely influenced by t he t ype of t ool mat er ial. Select ion of a par t icular t ool mat er ial for a given job depends on t he wor k mat er ial, machining condit ions, wear r esist ance of t he t ool mat er ial and t he cost of t he t ool mat er ial. ( ii ) Tool geomet r y : Pr oper t ool geomet r y i s essent i al for effi ci ent cut t i ng. For ces ar e most l y independent of t ool angles except t he t r ue r ake. (For ce change for ot her var iables also but t he change is less significant e.g. nose r adius, side cut t ing edges angle).

Var iable incr eased Rake angle Relief angle Side cut t ing edge angle End cut t ing edge angle Nose r adius

Effect on surface finish L ar ge impr ovement Slight det er ior at ion Slight t o lar ge impr ovement L ar ge det er ior at ion L ar ge impr ovement

(iii )N ature of engagement of tool with work : Tool life is usually mor e wher e cont inuous cut t ing occur (t ur ning a full cylinder ) and is shor t for int er mit t ent and heavy impact loads (machining slot s or keyways) when t he speeds has t o be r educed. For milling, init ial cont act of t he cut t er wit h t he wor kpiece affect s t ool life consider ably in t he case of car bides and ot her br it t le cut t ing t ool mat er ials. Angle of engagement wit h t he wor k should be less t han about 20° and if it is incr eased above 35°, it may cause pr emat ur e failur e of t he cut t ing edge. (3)

Cut t ing condit ions. (i ) Cutt ing speed (ii ) Feed (iii ) Dept h of cut

(4)

Wor k mat erial variables. ( i ) H ardness : I t is t he r esult of chemical composit ion and heat t r eat ment of t he alloy. As har dness incr eases, machinabilit y decr eases. ( ii ) Tensile strength : Sur face impr oves wit h incr ease in t ensile st r engt h of mat er ial. ( iii )Chemical composit ion : Effect of var ious element s in t he mat er ial : ( a ) Carbon : L ow car bon st eels wit h 0.01 – 0.15% car bon have poor machinabilit y because of t heir pr edominant fer r i t e st r uct ur e, hi gh duct i lit y and t oughness. When car bon cont ent incr eases for 0.25– 0.3%, machinabilit y impr oves. Any fur t her incr ease in car bon det er ior at es t he machinabilit y. ( b) M anganese : M achinabilit y is impr oved t o a cer t ain ext ent by t he addit ion of manganese upt o 1%. Wit h fur t her incr ease, machinabilit y det er ior at es.

4.8

Manufacturing Engineering

( c) Phosphorus : Addit ion of phosphor us upt o 0.15% has a favor able effect . I t r eact s wit h ir on t o for m ir on phosphide which incr eases br it t leness of chips and impr oves sur face finish. ( d ) Sulphur : I t has similar effect as phosphor us. ( e) Lead : When lead is added upt o 0.2%, it has a favour able effect on chip br eaking and t her efor e fr ee-cut t ing st eels cont ain lead. ( f ) Si l i con ( g) N ickel : M achinabilit y det er ior at es wit h higher cont ent of nickel. ( iv) M icrostructure. H ar d const it uent s necessit at e low cut t ing speeds and vice ver sa. L ar ger gr ain size al lows higher cut t i ng speeds. St eel s having same chemi cal composit i on and har dness manufact ur ed in t he same plant but in differ ent shift s, can have mar ked differ ence in t heir machinabilit y r at ings. Car bide and oxide inclusions cause r apid wear on t he cut t ing t ool, hence cut t ing speed has t o be subst ant ially r educed. M icr ost r uct ur e can be var ied by heat t r eat ment . (v) Degree of cold work. Cold wor king incr eases t ool life and per mit s an incr ease in cut t ing speed. I n st eels cont aining 0.3 t o 0.4% car bon, t her e is pr act ically no differ ence in machinabilit y of hot r olled and cold-r olled bar s. ( vi ) Shape and dimensions wor k (vii ) Rigidit y of wor k piece (viii ) St r ain har denability DRI L L I N G. I n t his pr ocess, t he hole is pr oduced or enlar ged using a specific t ype of end cut t ing t ool. Dr illing involves t wo pr incipal cut t ing edges. SH API N G M ACH I N E. I t is a r ecipr ocat ing t ype int ended t o pr oduce flat sur faces. These sur faces may be ver t ical, hor izont al or inclined. Types of shaping machines. ( i ) H or izont al t ype ( ii ) Ver t ical t ype ( iii ) Cr ank t ype, Gear ed or H ydr aulic t ype ( iv) Univer sal t ype M echanisms involved. (i )

Cr ank and slot t ed lever mechanism

( ii ) Whit wor t h quick r et ur n mechanism ( iii ) H ydr aulic shaper mechanism PLAN I N G M ACH I N E. I t is int ended t o pr oduce plane and flat sur faces by a single point cut t ing t ool. I t is a lar ge machine. I n t his machine, t he job r ecipr ocat es, and aft er ever y double st r oke t he t ool is subject ed t o feed eit her in ver t ical or hor izont al dir ect ion. Types of planing machine. (i ) Open t ype planer (ii ) Pit planer (iii ) Double housing planer Types of mechanisms used for table drive. ( i ) Open and cr ossed belt dr ive ( ii ) Rever sible mot or dr ive ( iii ) H ydr aulic dr ive

Manufacturing Engineering

4.9

M I LLI N G M ACH I N E I n t his machine, t he wor k is fed against a r ot at ing mult ipoint cut t er. The cut t er r ot at es at a high speed since t he mult iple cut t ing edge r emoves met al at a fast er r at e. This machine finds applicat ions on a wider r ange, since it is possible t o hold many cut t er s on t he mandr el or ar bor. The accur acy and sur face finish ar e good in t his machining compar ed t o ot her pr ocesses. The milling pr ocess is dist inguished by a t ool wit h one or mor e t eet h which r ot at e about a fixed axis while the wor kpiece is fed into the tool. The chips pr oduced by milling ar e gener ally short, discontinuous segments. The milling pr ocess ut ilises small, light t ools which can be easily r ot at ed at high speeds t o pr oduce flat or cur ved sur faces on wor kpieces of a wide var iet y of size and shapes. T ypes of milling machines. ( i ) Plain milling : The plain milling cut t er which is used t o pr oduce flat sur faces, has cut t ing edges on t he per ipher y which ar e par allel wit h t he axis of r ot at ion. The per for mance of t his cut t er is impr oved if t he t ools ar e ar r anged in helical fashion r at her t han par allel t o t he axis of r ot at ion. The use of helical cut t er not only pr ovides smoot her cut t ing but chip for mat ion is also impr oved. Cut t er s wit h helix angles up 45º ar e nor mally called plain milling cut t er s, wher eas cut t er wit h helix angles gr eat er t han 45º ar e called helical mills. ( ii ) End or face milling : The end milling cut t er has t eet h on one end as well as on t he per ipher y. I n use it s axis of r ot at ion is per pendicular t o t he sur face pr oduced, while in plain million t he cut t er axi s is par allel t o the finishing sur face. The end milling cut t er is used for pr oducing flat sur faces and pr ofiles– ver y oft en simult aneously. M illing cut t er s used ar e, milling saws, angle mills, fly mills, T-slot mi lls, for med mills, et c. Compound indexing. The indexing met hod is called compound due t o t wo separ at e movement s of t he cr ank in t wo differ ent hole cir cles. GRI N D I N G I t is a met al cut t ing pr ocess in which t he met al is r emoved by a shear ing pr ocess just as in ot her cut t ing oper at ions. Gr inding pr ocesses employ an abr asive wheel containing many gr ains of har d mat er ial bonded in a mat r ix. The act ion of a gr inding wheel may be consider ed a mult iple-edge cut t ing t ool except t he cutting edges ar e ir r egular ly shaped and r andomly spaced ar ound the face of the wheel. Each gr ain r emoves a shor t chip of gr adually incr easing t hickness. Aft er ‘use’ t he wheel can become ‘loaded’ wit h dull gr ains and/or find adher ing chips. The wheel must be dr essed by passing a diamond t ip t ool acr oss t he wheel face t o gener at e new shar p cut t ing sur faces. Gr inding is gener ally car r ied out wit h wheel and wor k moving in opposit e dir ect ions. This is somet imes r efer r ed t o as up gr inding. I n down gr inding t he wheel and wor k move in t he same dir ect ion. Types of Grinding. ( i ) Sur face gr inding ( ii ) Ext er nal or cylindr ical gr inding ( iii ) I nter nal gr inding ( iv) Cent r eless gr inding (v) Off hand gr inding. F act ors affect ing performance of a grinding wheel. ( i ) Abrasive type : The abr asives gener ally used ar e aluminium oxide, silicon car bide, and diamond. Diamond is t he har dest subst ance known and is used for ver y har d wor k mat er ials such as glass, car bide, and cer amics. Aluminium oxide and silicon car bide ar e mor e commonly used for making t he gr inding wheels. Silicon car bide is har der t han aluminium oxide but dulls mor e r apidly. Gener ally, aluminium oxide abr asives ar e select ed for t he sur face gr inding of st eels and br onzes, wher eas silicon car bide is chosen for t he sur face gr inding of cast ir on, br ass, aluminium, har d alloys, and car bides. ( ii ) Grain size : The size of t he gr ains i s gener all y by t he gr it size. A 60 gr it si ze, for exampl e, i s appr oximat ely 1/60 inch squar e. The lar ger t he size of t he gr ains, t he mor e will be t he mat er ial r emoval capacit y, but t he qualit y of t he sur face finish det er ior at es. Thus, t he gr ain size is det er mined pr imar ily by t he sur face qualit y r equir ement s.

4.10

Manufacturing Engineering

( iii )Bonding material : The bond mat er ials commonly used ar e vit r ified clay, r esinoid mat er ials, silicat es, r ubber, shellac, and met als. The vit r ified bond is st r ong and r igid. I t is t he most common t ype of bond used. The r esin bonds ar e made fr om synt het ic or ganic mat er ials. Such bonds ar e st r ong and fair ly flexible. The silicat e bonds ar e essent ially t he silicat es of soda (wat er glass). These bonds ar e not as st r ong as t he vit r ified bonds. ( iv) Structure : Since t he gr inding wheel is similar t o a milling cut t er wit h a ver y lar ge number of t eet h r andomly or ient ed, it must have voids t o allow space for t he chips. I f t he voids ar e t oo small for t he chips, t he chips st ay in t he wheel, blocking t he voids. This is known as loading of t he wheel. L oading causes inefficient cut t ing. I f t he voids ar e t oo lar ge, again t he cut t ing act ion is inefficient since t her e will be t oo few cut t ing edges. I n an open st r uct ur e, t he gr ains ar e not t oo densely packed, and in a wheel wit h a closed st r uct ur e, t he gr ains ar e t ight ly packed. For gr inding duct ile wor k mat er ials, lar ger chips ar e pr oduced, and t o r educe t he t endency of loading, an open st r uct ur e is pr efer r ed. I n t he case of har d and br it t le wor k mat er ials, a closed st r uct ur e is pr efer r ed. I n t he case of har d and br it t le wor k mat er ials, a closed st r uct ur e is select ed. The st r uct ur e depends on t he r equir ed gr ade and also t he nat ur e of cut . For a r ough cut , an open st r uct ur e is mor e suit able. (v) Grade : The gr ade is det er mined by t he st r engt h of t he bonding mat er ial. So, a har d wheel means st r ong bonding and t he abr asive gr ains can wit hst and lar ge for ces wit hout get t ing dislodged fr om t he wheel. I n t he case of a soft wheel, t he sit uat ion is just t he opposit e. When t he wor k mat er ial is har d, t he gr ains wear out easily and t he shar pness of t he cut t ing edges is quickly lost . This is known as glazing of t he wheel. A glazed wheel cut s less and r ubs mor e, making t he pr ocess inefficient . To avoid t his pr oblem, a soft wheel should be used so t hat t he gr ains which lose t he shar pness get easily dislodged as t he machining for ce on t he individual gr ains incr eases. Thus, t he layer s of new gr ains ar e exposed, maint aining t he shar pness of t he wheel. When t he wor k mat er ial is soft, a har d wheel should be used since t he pr oblem of glazing will be absent and a longer wheel life will be achieved. So, for a wor k mat er ial, t her e exist s an opt imal gr ade– t oo har d a wheel causes glazing, wher eas t oo soft a wheel wear s out ver y fast . CEN TRELESS GRI N DI N G. Advant ages. (i )

L ess met al is r emoved.

( ii ) No t endency for chat t er and deflect ion of t he wor k. ( iii ) I t can be used for mass pr oduct ion. ( iv) No cent r e holes is necessar y. (v)

Size of wor k is cont r olled.

( vi ) Unskilled wor ker s can also be employed. D isadvant ages. ( i ) M ult iple diamet er s cannot be gr ound. ( ii ) H ollow wor kpieces cannot be cor r ect ly gr ound. ABRASI VE S. I t is a subst ance used for making gr inding wheel. Types of abrasives. ( i ) N atural abrasives : Solid quar t z, emer y, cor undum and diamond. ( ii ) Artificial abrasives : These ar e silicon car bide and aluminium oxide. Silicon car bide cont ains 56% silica sand, 34% powder ed coke, 2% salt and 12% saw dust . Sand pr oduces silica, coke gives car bon, salt helps t o fuse it . Aluminium oxide is pr oduced by heat ing bauxit e wit h coke and ir on bor ings in a fur nace.

Manufacturing Engineering

4.11

BON DS. A bond is employed t o hold abr asives t oget her. Types of bonds. (i ) Vitr ified bond ( ii ) Silicat e bond ( iii ) Shellac bond ( iv) Resinoid bond (v) Rubber bond ( vi ) Oxychlor ide bond N ON -TRADI TI ON AL M ACH I N I N G PROCESSES ABRASI VE JET M ACH I N I N G (AJM ). I t is t he r emoval of mat er ial fr om a wor kpiece by a high speed syst em of abr asive par t icles car r ied by gas fr om a nozzle. The pr ocess is used chiefly t o cut mat er ials t hat ar e sensit ive t o heat damage and t hin sect ions of har d mater ials that chip easily and t o cut int r icat e holes t hat would be mor e difficult t o pr oduce by ot her met hods. Pr ocess. Abr asive powder is fed fr om a mixing chamber t hat vibr at es at 50 H z int o an or ifice chamber wher e it is ent r ained in t he gas st r eam and t hen passed t hr ough a connect ing hose finally emer ging fr om a small nozzle at high velocit y. Power feed r at e is cont r olled by amplit ude of t he mixing vibr at ion and pr essur e r egulator contr ols the gas flow. Duplex units allow unint er r upted pr oduction while r eloading or can per for m independent oper at ions wit h differ ent abr asives. The pr essur e can be r eleased t hr ough blow off out let s in some unit s. Abr asive flow fr om t he nozzle can be st opped in 10 t o 15 milliseconds. The nozzle is mount ed in a fixt ur e for aut omat ic oper at ion eit her t he wor kpiece or t he nozzle being moved by cam dr ives, pant ogr aphs or ot her suit able mechanisms. Nozzles must be highly r esist ant t o abr asion and ar e made of t ungst en car bide or synt het ic sapphir e. For pr ecision cut t ing nozzles ar e made wit h an ext er nal t aper on t he t ip t o minimize secondar y r icochet of abr asive par t icles t hat r ebound fr om t he wor kpiece. Dist ance of nozzle fr om wor kpiece affect t he size of machined ar ea and t he r at e of mat er ial r emoval. The abr asive par t icles fr om any nozzle follow par allel pat hs for a shor t dist ance and t hen t he abr asive jet flar es out war d like a nar r ow cone. Abr asive mat er ials used in abr asive jet machining include aluminium oxide, silicon car bide, dolomit e (calcium magnesium car bonat e), sodium bicar bonat e and small glass beads. har dness, st r engt h, par t icle size and par t icle shape of t he abr asive affect it s cut t ing per for mance and life. Abr asives ar e not or dinar ily r e-used because of r educed cutt ing action they give and because contaminat ion wit h mat er ial abr aded fr om t he wor k can clog small or ifices in t he equipment and t he nozzle. Carrier gas. Or dinar ily air, nit r ogen or car bon dioxide is used as t he car r ier or pr opellant gas. Commet r ical cyl inder gases ar e usually of sat isfact or y pur it y. Nozzle pr essur e can r ange bet ween 2 and 8.5 bar but or dinar ily is about 5 bar. H igh pr essur e r esult s in r apid nozzle wear, low pr essur e gives slow r emoval r at es. Advant ages. (i ) Abilit y t o cut int r icat e hole in mat er ial of any har dness. (ii ) Abilit y t o cut fr agile, br it t le or heat sensit ive mat er ial wit hout damage. (iii ) Absence of mechanical cont act bet ween t ool and wor k. (iv) L ow capit al cost . D isadvant ages. (i ) L imit ed applicabilit y because only small amount s of met als can be r emoved and nozzle must be close t o t he wor kpiece.

(ii ) Slow r emoval r at e and inefficient use of power. (iii ) St r ay cut t ing. (iv) Embedding of abr asive in t he wor kpiece.

4.12

Manufacturing Engineering

Appl icat i ons. Abr ading and fr eest one glass, cleaning, cutt ing fine lines exposing an ar ea for elect r ical contact machining semiconductor s such as ger manium, silicon gallium, cutting and etching mater ials such as quar tz, sapphir e, mica and glass, debur r ing, mar king. WATER-JET M ACH I N I N G Pr ocess. When pur e wat er is abr asive-laden and shaped int o a small (0.08 t o 0.5 mm diamet er ) coher ent st r eam fir ed at velocit ies up t o 1000 m/s (fast er t han a 30.06) caliber r ifle bullet ) int o a mat er ial, t he r esult s ar e clean, smoot h cut s at sur pr isingly fast r at es. I n oper at ion, t he wat er is pr essur ized by a hydr aulic int ensifier, which is a lar ger pist on wor king on a smaller one. Diamet er r at ios var y fr om 10 : 1 t o 20 : 1 wit h a 20 : 1 r at io, a pr essur e of 200 bar int ensified becomes 4000 bar on t he wat er in t he smaller -diamet er pist on. Pr essur ized wat er is t hen r out ed t o an at t enuat or t hat act s as an accumulat or t o maint ain const ant flow and pr essur e t o t he cut t ing nozzle by modulating the hydr aulic system’s pulsating action into a smooth str eam of water under for ce. A maintained 4000 bar pr essur e behind a 0.5 mm office r epr esent s t he high end of wor king r ange. Abr asive can be anyt hing fr om sand t o salt depending on t he applicat ion. Abr asive is int r oduced int o t he st eam aft er it leaves t he or ifice. Appl icat i ons. ( i ) Wat er jet can be used t o cut fer r ous and nonfer r ous plat es, composit es of all t ypes, t her moplast ics, high-st r engt h alloys, elast omer s, gels, cer amics, pr int ed cir cuit boar ds, paper boar ds, and a wide r ange of food pr oduct s. ( ii ) I nt er nal cut s can be made. ( iii )No st ar t ing hole is necessar y. ( iv) I nside cor ner r adii as t ight as 1.5 t o 3 mm can be made. Advant ages.

(i ) Water jets ar e shar p in all dir ections. Robot ar ms can swing them quickly without r egar d to or ientation. (ii ) Cut s ar e made without heat , leaving a fine sur face finish. A wide var iet y of mat er ial including fer r ous and nonfer r ous plat e composit es of all t ypes, t her moplast ics, high-st r engt h alloys, elast omer s, glass, cer amics and pr int ed cir cuit boar ds can be cut . (iii ) A minimum of wast e and allowes line-t o-line cut t ing t o nest ed par t s. (iv) Air bor ne par t icles like, glass, fiber glass, and ot her t oxic mat er ials et c. associat ed wit h most cut t ing pr ocesses ar e eliminat ed. (v) Cut edge is ver y smoot h and r equir es no fur t her finishing. D isadvant ages.

(i )

A 0.08 mm st r eam pr oject ed int o t he at mospher e will complet ely vapor ize at about 600 mm fr om t he nozzle.

(ii ) Noise can easily r each a high-decibel l evel. Eit her t he wor king unit must be sound shi elded or wor ker s in t he vicinit y must wear ear pr ot ect ion. (iii ) Back pr essur es of r eact ive for ces ar e pr esent , however, t hey ar e not t r oublesome since t he wor king ener gy is concent r at ed int o a minut e cr oss-sect ional ar ea at any given moment . (iv) Spent wat er and abr asive must be cont ained. Typical consumpt ion of wat er is 5 t o 10 l/m, wit h 70% evapor at ion dur ing t he cut . Remaining wat er is dir ect ed int o a cont ainer under t he equipment .

Manufacturing Engineering

4.13

ELECTRO CH EM I CAL M ACH I N I N G (ECM ). This met hod r emoves met al wit hout t he use of mechanical or t her mal ener gy. Elect r ic ener gy is combined wit h a chemical t o for m a r eact ion of r ever se plat ing. Dir ect cur r ent at r elat ively high amper age and low voltage is continuously passed between anodic wor kpiece and cathodic tool (electr ode) thr ough a conduct ive elect r olyt e. At t he anode sur face, elect r ons ar e r emoved by t he cur r ent flow and t he met alic bonds of t he molecular st r uct ur e of t his sur face ar e br oken. These sur face at oms pr oceed t o go int o solut ion as met al ions. Simult aneously, posit ive hydr ogen ions ar e at t r act ed t o t he negat ively char ged sur face, and emit t ed at t he cat hode sur face t o for m hydr ogen molecules. Dissolved mat er ial is r emoved fr om t he gap bet ween wor k and t ool by t he flow of elect r olyt ic which also aids in car r ying away t he heat and hydr ogen for med. Exposur e of t he wor kpiece t o hydr ogen is t hus r educed. Advant ages.

(i )

H igh met al r emoval r at es for t ough or har d alloys.

(ii ) Rapid met al r emoval r at es when machining complex t hr ee dimensional sur faces. (iii ) Ability t o machine complex t hr ee dimensional sur faces without bur r s and without the str iation mar ks left by miling cut t er s. (iv) Fr eedom fr om met allur gical damage. (v) Accur acy and economy t hr ough t he inher ent char act er ist ics of t he pr ocess. Requirement s of ECM operat ions.

(i )

A cat hode t ool pr epar ed wit h an appr oximat e mir r or image of t he configur at ion t o be machined int o t he wor kpiece (wit h compensat ion for over cut ).

(ii ) A wor kpiece and means t o hold and locat e it in close pr oximit y t o t he t ool. (iii ) A means of supplying t he gap bet ween t ool and wor kpiece wit h pr essur ized flowing conduct ive liquid (el ect r ol yt e). (iv) A car efully cont r olled sour ce of d.c. elect r ical power of sufficient capacit y t o maint ain a cur r ent densit y bet ween t ool and t he wor kpiece. Appl icat i ons. (i ) This pr ocess is par t icular ly suit able for small deep holes. (ii ) Penet r at ion r at es fr om 0.75 t o 12.5 m per min or mor e ar e possible. (iii ) Air cr aft and aer ospace component s ar e fr equently pr oduced by t his method, because t he high st r engt h t emper at ur e r esist ant mat er ials used ar e difficult t o machine in ot her ways. Advant ages. (i )

Abilit y t o machine economically t he complex shaped wor kpiece.

( ii ) M achining abilit y is independent of t he mechanical pr oper t ies of t he wor k mat er ial. ( iii ) Can dispense wit h oper at ion like gr inding and hence fast er in some cases. ( iv) Pr oduct ion of st r ess fr ee sur face. D isadvant ages. (i )

Cor r osion and r ust for mat ion in t he machine t ool and wor k.

( ii ) M inimum cor ner r adii t hat can be machined is limit ed t o about 0.2 mm. ( iii ) Gr ain boundar y at t ack may t ake place under cer t ain condit ions. ( iv) L ar ge floor space r equir ement s and high capit al cost .

4.14

Manufacturing Engineering

U LTRASON I C M ACH I N I N G (U SM ). I t is t he r emoval of mat er ial by par t icles of abr asive t hat vibr at e in a wat er slur r y cir culat ing t hr ough a nar r ow gap bet ween wor kpiece and t he t ool t hat oscillat es at about 20,000 H z (cycles per second). The t ool r epr oduces it s shape in t he wor kpiece, gener ally t o an accur acy ± of 0.025 mm and somet imes t o t oler ance of 0.0125 mm or less wit hout bur r s. Accur acy depends on t he size of t he t ool, r igidit y of machine and t he t ool, t emper at ur e of t he slur r y, gr it size and pr ocedur e for r oughing and finishing. Ult r asonic machining is used chiefly on har d, br it t le mat er ials t hat do not conduct elect r icit y, however, it is used on bot h met als and nonmet als and on duct ile as well as br it t le mat er ials. I t par t icular ly well suit ed t o t he pr oduct ion of r elat ively shallow ir r egular cavit ies and is one of t he few pr ocesses suit able for machining ext r emely fr agile such as honeycomb. Appl icat i ons. (i ) Four nar r ow r ect angular holes wer e cut at t he same t ime in t he moving par t of hydr aulic ser vo valve. (ii ) A car bon flat 75 × 100 × 1 mm was machined by USM making 2176 holes t hr ough it , each 0.8 mm squar e, in less t han 10 min. (iii ) Dense cer amic 1.6 mm t hick was machined by USM making 27 t hr ough-holes in 3 r ows of 9, t he cent r es accur at e t o ± 0.025 mm. H oles about 4.75 mm in diamet er wer e made simult aneously by a t ool t hat r esembled 27 t ubes. The t ool made 150 par t s befor e it needed shar pening. Advant ages. (i ) H igh accur acy and good finish ar e easily at t ainable. (ii ) Equipment is safe t o oper at e. (iii ) L it t le or no heat gener at ion dur ing machining. (iv) L ow cost of met al r emoval. (v) H ar d br it t le mat er ials ar e easily machinable. D isadvant ages. (i ) L ow r at e of met al r emoval. (ii ) Soft er mat er ials ar e difficult t o machine. (iii ) H igh cost . ELECTRON BEAM M ACH I N I N G (EBM ). Elect ion beam machining (EBM) uses elect r ical ener gy t o gener at e ther mal ener gy for r emoving mat er ial. A pulsat i ng st r eam of high speed el ect r ons pr oduced by a gener at or i s focused by el ect r ost at ic and elect r omagnet ic feilds t o concent r at e ener gy on a ver y small ar ea of wor k. H igh power beams ar e used wit h elect r on velocit ies exceeding one half t he speed of light . As t he elect r ons impinge on t he wor k t heir kinet ic ener gy is t r ansfor med int o t her mal ener gy and vapor izes t he mat er ial locally. Beams can be concent r at ed on spot s as small as 0.01 mm in diamet er. The pr ocess is usually per for med in vaccum, t o pr event collisions of elect r ons wit h gas molecules which would scat t er or diffuse t he elect r on beam. L ead shielding is r equir ed t o pr ot ect t he oper at or fr om r adiat ion. EBM is gener ally limit ed t o dr illing ext r emely small holes and cut t ing nar r ow slot s or cont our s in t hin mat er ials t o close t oler ances. Ther e is no t ool wear or pr essur e on t he wor k. St ock r emoval r at e is gener ally about 1.6 mm 3/min wit h a penet r at ion r at e of about 0.25 mm/s or fast er. Appl icat i ons. ( i ) Dr illing gas or ifices for pr essur e differ ent ial devices, wher eby closely dimensioned holes ar e dr illed t hr ough t he par t . These holes r egulat e t he amount of gas t hat flows in a given amount of t ime. ( ii ) Pr oducing wir e dr awing dies, light r ay or ifices and spinner et s t o pr oduce synt het ic fiber s. ( iii ) Pr oducing met er ing holes eit her r ound or pr ofile shaped t o be used as flow holes on sleeve valve r ocket fuel inject or s or inject ion nozzles on diesel engines. Advant ages. ( i ) Good for micr o-machining. ( ii ) Can cut any mat er ial. ( iii ) No t ool wear.

Manufacturing Engineering

4.15

( iv) No t ool wor k cont act . (v)

L it t le physical and met allur gical damage t o t he wor kpiece.

( vi ) Concent r at ed sour ce of heat ing. D isadvant ages (i )

H igh cost .

( ii ) H igh level of skill needed. ( iii ) Can be applied for small cut s only. ( iv) Vacuum r equir ement s limit t he size of wor kpiece t hat can be handled. ELECTRI CAL DI SCH ARGE GRI N DI N G (EDG). I t is similar t o EDM except t hat t he elect r ode is a r ot at ing wheel (gener ally gr aphit e). Posit ively char ged wor kpiece ar e immer sed in or flooded by a dielect r ic fluid and fed past t he negat ively char ged wheel by ser vo cont r olled machine t able. M et al is r emoved by int er mit t ent high fr equency elect r ical dischar ges passing thor ough the gap bet ween wheel and wor k. The r ever se of the for m on t he wheel face is tr ansfer r ed t o t he wor kpiece. Chips ar e flushed away by fluid car r ied t hr ough t he cut t ing ar ea by wheel r ot at ion. Wheel never comes closer t o t he wor kpiece t han t he pr esent lengt h of spar k (usually 0.01 t o 0.075 mm) used for any par t icular gr inding oper at ion. Appl icat i ons. (i )

Gr inding st eel and car bide at t he same t ime wit hout wheel loading.

( ii ) Gr inding t hin sect ions wher e abr asive wheel pr essur es might cause dist or t ion. ( iii ) Gr inding br it t le mat er ials of fr agile par t s wher e abr sive mat er ials might cause far act ur ing. ( iv) Gr inding t hr ough for ms wher e diamond wheel cost s would be excessive. (v)

Gr inding cir cular for ms in dir ect compet it ion wit h abr asive wheel met hods.

Advant ages. (i )

Gr inding pr essur e is ver y low r esult ing int o st r ess and bur r -fr ee gr inding.

( ii ) H eat effect s such as gr inding cr acks, t emper ing of wor k and t r ansfor mat ion of sur face layer s ar e eliminat ed because elect r olyt ic gr inding is a cold gr inding pr ocess. ( iii ) I nsignificant wheel wear guar ant ee cut t er concent r icit y. D isadvant ages. (i )

Cannot be applied t o elect r ically non-conduct ing mat er ials.

( ii ) I n some cases t he sur face machined, has been found t o have micr o-cr acks. ( iii ) Tool wear can limit t he degr ee of accur acy at t ainable. ( iv) Nor mally applicable t o small sized jobs. CH EM I CAL M ACH I N I N G. I n t his pr ocess, t wo most impor t ant fact or s ar e et chant r esist ant mat er ial (called maskant or r esist ) and etchant used. Types of maskant s or resists. ( i ) Cut and peel maskants : These ar e used almost exclusively for chemical milling of air cr aft missile and st r uct ur al par t s ar e applied by flow dip or spr ay coat ing t o t hicknesses of 0.0125 mm t o 0.375 mm dr y film for m. The mat er ials ar e r emoved fr om ar eas t o be et ched by cut t ing and peeling away t he unwant ed ar eas gener ally wit h a t emplat e t o aid accur acy. Et ching dept hs of 12 mm or mor e can be at t ained and aft er a cer t ain ar ea has been et ched, addit ional maskant may be r emoved so t hat st ep et ching is possible. (ii ) Phot ogr aphic r esist s (iii ) Scr een r esist s.

4.16

Manufacturing Engineering

F act ors affect ing select ion of et chant . (i ) M at er ials t o et ched (ii ) Type of maskant or r esist used (iii ) Dept h of et ch (iv) Sur face finish r equir ed (v) Pot ent ial damage or alt er at ion of met allur gical pr oper t ies of t he mat er ial (vi ) Speed of mat er ial r emoval (vii ) Per missible oper at ing envir onment (viii ) Economics of mat er ial r emoval (ix) H eat t r eat condit ion of mat er ial. CH EM I CAL M I LLI N G. Chemical milling is t he pr ocess used t o shape met als t o an exact ing t oler ance by t he chemical r emoval of met al or deep et ching of par t s. Amount of met al r emoved or dept h of et ch is cont r olled by amount of immer sion t ime in t he et ching solut ion. L ocat ion of t he unet ched or unmilled ar eas of a par t is cont r olled by masking or pr ot ect ing t hese ar eas fr om t he act ion of et chant solut ion. Appl icat i ons. (i ) Remove met al fr om a por t ion of t he ent ir e sur face of for med or ir r egular ly shaped par t s such as for gings, cast ings, ext r usions, or for med wr ought st ock. (ii ) Reduce web t hicknesses below pr act ical machining, for ging, cast ing, or for ming limit s. (iii ) Taper sheet s and per for med shapes. (iv) Pr oduce st epped webs r esult ing in consolidat ion of sever al det ails int o one int egr al piece. Advant ages. (i ) Sur faces wit h complex pr ofiles of t hin sheet s can be pr epar ed. No bur r s ar e pr oduced. (ii ) Bot h sides of wor k sheet can be pr epar ed at a t ime. (iii ) No mechanical wor k is done on t he sur face which offer s higher fat igue. (iv) Pr ocess r educes t ooling t ime consider ably. (v) Tooling cost is low. (vi ) Pr ocess is flexible. D isadvant ages. (i ) L ar ger floor ar ea is r equir ed because of t he size and number of bat hs r equir ed in t he pr ocess. (ii ) Cost of manufact ur ing is high. (iii ) L ower met al r emoval r at e is available (0.2 mm per min). (v) Shar p cut s ar e not possible. PH OTOCH EM I CAL M ACH I N I N G Phot ochemical machining or chemical blanking is t he pr ocess of pr oducing met allic and non-met allic par t s by chemical act ion. The pr ocess consist s of placing a chemically r esist ant image of t he par t on a sheet of met al and exposing t he sheet t o chemical act ion which dissolves all t he met al except t he desir ed par t . M ost par t s pr oduced in t his way ar e similar t o t hin gauge st ampings and ar e gener ally flat and of complex design. I n phot ogr aphic r esist pr ocess of phot ochemical machining met al can be chemically cleaned in numer ous ways including decr easing, pumice scr ubbing, elect r o cleaning, or chemical cleaning. Cleaned met al is coat ed wit h phot ogr aphic mat er ial which exposed t o light of t he pr oper wavelengt h, will polymer ize and r emain on t he panel as it goes t hr ough developing st age. This polymer ized layer t hen act s as t he bar r ier t o t he et ching solut ion applied t o t he met al. M et hods of coat ing t he met al wit h t he phot o-r esist ar e dipping, spr aying, flow coat ing, dollar coating, or laminat ing. Most applicable method is det er mined by the t ype of r esist used and t he par t ’s physical for m.

Manufacturing Engineering

4.17

Appl icat i ons. The use of phot ochemical machining is gener ally limit ed t o r elat ively t hin mat er ials, fr om 0.002 mm t o 1 mm t hick. The limit of mat er ial t hickness is gener ally a funct ion of t he t oler ance desir ed on finished par ts. Advant ages.

(i )

Wor king on ext r emely t hin mat er ials wher e handling difficult ies and die accur acies pr eclude t he use of nor mal mechanical met hods.

(ii ) Wor king on har dened or br it t le mat er ials wher e mechanical act ion would cause br eakage or st r ess concent r at ion point s. Chemical blanking wor ks well on spr ing mat er ials and har dened mat er ials which ar e r elat ively difficult t o punch. (iii ) Pr oduct ion of par t s which must be absolut ely bur r -fr ee. (iv) Pr oduct ion of ext r emely complex par t s wher e die cost would be pr ohibit ive. (v) Pr oduct ing shor t r un par t s wher e r elat ively low set up cost s and shor t cut fr om pr int t o pr oduct ion offer advant ages. This is especially impor t ant in r esear ch and development pr oject s and in model shops. LASER BEAM M ACH I N I N G (LBM ). The wor d L ASER st and for “ L ight Amplicat ion using Simulat ed Emission of Radiat ion.” L aser pr ovides int ense and uni-dir ect ional beams of light ; t his light is coher ent in nat ur e. The mechanism by which a laser beam r emoves mat er ial fr om t he t he sur face being wor ked involves a combinat ion of t he melt ing and evapor at ion pr ocesses. H owever, wit h some mat er ials, t he mechanism is pur ely one of evapor at ion. Advant ages. (i ) No mechanical cont act bet ween t ool and t he wor k. (ii ) Beam can be pr oejct ed t hr ough a t r anspar ent window. (iii ) L aser can be used wit h mat er ials sensit ive t o heat shock such as cer amics. (iv) Wor kpiece is not subject ed t o lar ge mechanical for ces. (v) L aser oper at es in any t r anspar ent envir onment including air, iner t gas, vacuum and even cer t ain liquids. D isadvant ages. The laser syst em is quit e inefficient .



4.18

Manufacturing Engineering

F ORM I N G PROCE SS I t can be defined as a pr ocess in which desir ed size and shape ar e obt ained t hr ough t he plast ic defor mation of a mat er i al. The st r esses induced dur i ng t he pr ocess ar e gr eat er t han t he yield st r engt h, but less t han t he fr actur e str engt h of t he mat er ial. The type of loading may be tensile, compr essive, bending, or shear ing or a combinat i on of t hese. Types of forming processes. (i ) Col d for mi ng (ii ) H ot for mi ng I f wor king t emper at ur e is higher t han t he r ecr yst al lizat ion t emper at ur e of t he mat er i al, t hen t he pr ocess is call ed hot for ming. Ot her wi se t he pr ocess is call ed cold for ming. The flow stress behaviour of a mater ial is entirely different above and below its r ecr ystallization temper ature. Dur ing hot wor king, a lar ge amount of plastic defor mat ion can be impar t ed without significant str ain har dening. This is impor tant because a lar ge amount of st r ain har dening r ender s t he mater ial br itt le. The fr i ct ional char act er ist ics of t wo for mi ng pr ocesses ar e also ent ir ely differ ent . e.g. coeffi cient of fr ict i on in cold for ming is gener al ly of t he or der of 0.1, wher eas t hat in hot for ming can be as high as 0.6. Again hot for ming lower s down t he mat er ial st r engt h so t hat a machine wit h a r easonabl e capacit y can be used even for a pr oduct havi ng l ar ge di mensi ons. Typical for ming pr ocesses. ( i ) Rolling : The job is dr awn by means of fr i ct ion t hr ough a r egul at ed openi ng bet ween t wo power dr iven r olls.The shape and size of t he pr oduct ar e decided by t he gap bet ween r olls and t heir cont our s. This i s a ver y useful pr ocess for t he pr oduct ion of sheet met al and var ious common sect i ons.

e.g. r ai l, channel angle ( ii ) Forging : I n for gi ng, t he mat er ial is squeezed bet ween t wo t o mor e di es or alt er it s shape and size. Depending on t he sit uat ion, t he die may be open or closed. ( ii i ) Drawing : I n t his pr ocess, cr oss-sect ion of a wir e or t hat of a bar or t ube is r educed by pulli ng t he wor k piece t hr ough t he coni cal or i fice of a di e. When hi gh r educt ion i s r equir ed, it may be necessar y t o per for m t he oper at i on i n sever al passes. (iv ) Deep drawing : I n deep dr awi ng, a cup-shaped pr oduct i s obt ained fr om a fl at sheet met al wit h t he help of a punch and a die. The sheet met al is held over t he die by means of a blank holder t o avoid defect s i n t he pr oduct . (v) Bending : Thi s is a pr ocess of bending a met al sheet plast i call y t o obt ai n t he desi r ed shape. Thi s is achi eved by a set of suit abl y designed punch and di e. (vi ) Extrusion : Thi s i s a pr ocess basical ly si mi lar t o t he cl osed di e for gi ng. But i n t hi s oper at i on, t he wor kpiece is compr essed in a closed space, for cing t he mat er ial t o flow out t hr ough a suit able opening, call ed die. I n t his pr ocess, only t he shapes wi t h const ant cr oss-sect ions (die out l et cr oss-sect ion) can be pr oduced. WORKI N G PROCESSES

Temper at ur e of t he wor kpiece in met al wor king depends on following fact or s. (i ) I nit ial t emper at ur e of t he t ools and mat er ial (ii ) H eat gener at ion due t o plast ic defor mat ion (iii ) H eat gener at ed by fr ict ion at t he die/mat er ial int er face (iv) H eat t r ansfer bet ween defor ming mat er ial and dies and sur r ounding envir onment Usually t emper at ur e is highest at t he mat er ial/t ool iner face wher e fr ict ion gener at es t he heat and it falls off t owar d t he inside of t he wor kpiece and int o t he die.

Manufacturing Engineering

4.19

CLASSI FI CATI ON OF WORKI N G PROCESS. (I ) On t he basis of t eperat ure of workpiece. (1) H ot wor king. When pl ast ic defor mat i on of met al is car r ied out at t emper at ur e above t he r ecr yst al l isat ion t emper at ur e, t he pr ocesses per for med on met als ar e called hot wor king. H ot wor king pr ocess can be consider ed as simult aneous combinat ion of cold wor king and annealing. Any wor k har dening effect caused by plast ic defor mat ion is neut r alised immediat ely by t he effect of high t emper at ur e. H ot wor king pr ocess facilit at es met al shaping wit h low power r equir ement s though it is expensive t o handle hot mat er ials. Ther e is also loss of met al by scaling and fine dimensional t oler ance cannot be achieved in t his pr ocess. H ot wor king incr eases t he densit y since any por es or cavit y in t he cast met al disappear dur ing hot -wor king. Gr ain st r uct ur e become mor e r efined. H ot -wor king pr ocesses such as r olling, ext r ust ion, or for ging t ypically ar e used at fir st conver t a cast ingot i nt o a wr ought pr oduct . The st r ai n in hot -wor ki ng i s l ar ge ( e  2 t o 4) compar ed wit h t ension or cr eep t est s. Usually H ot -wor king is car r ied out at high st r ain r at es in t he r ange 0.5 t o 500 s– 1. H ot -wor king r esult in decr ease in t he ener gy r equir ed t o defor m t he met al and incr eased abilit y t o flow wit hout cr acking, but r apid diffusion at hot wor king t emper at ur e aids in decr easing t he chemical inhomogeneit ies of t he cast -ingot st r uct ur e. The st r uct ur e and pr oper t ies of hot -wor ked met als ar e gener ally not so unifor m over t he cr oss sect ion as in met als which have been cold-wor ked and annealed. Since defor mat ion is always gr eat er in t he sur face layer s, t he met al will have a finer r ecr yst allized gr ain size in t his r egion. Because int er ior will be at higher t emper at ur es for longer t imes dur ing cooling t han will be t he ext er nal sur faces, gr ain gr owt h can occur in t he int er ior of lar ge pieces, which cool slowly fr om t he wor king t emper at ur e. L ower t emper at ur e limit for t he hot -wor king of a met al is t he lowest t emper at ur e at which t he r at e of r ecr yst allizat ion is r apid enough t o eliminat e st r ain har dening in t he t ime when t he met al is at t hat t emper at ur e. For a given met al or alloy, t he lower hot -wor king t emper at ur e will depend on fact or s like amount of defor mat ion and t ime t hat t he met al is at t hat t emper at ur e. Since gr eat er i s t he amount of defor mat i on, lower is t he r ecr yst al l i zat ion t emper at ur e, t he lower t emper at ur e limit for hot -wor king is decr eased for lar ge defor mat ions. M et al which is r apidly defor med and cooled r apidly will r equir e a higher hot -wor king t emper at ur e for t he same degr ee of defor mat ion t han will a met al slowly defor med and slowly cooled. The upper limit for hot-wor king is det er mined by t emper atur e at which eit her melt ing or excessive oxidat ion occur s. Gener ally maximum wor king t emper at ur e is limit ed t o 50 t o 60C below t he melt ing point . This is t o allow for t he possibilit y of segr egat ed r egions of lower -melt ing-point mat er ial. Only a ver y small amount of a gr ain-boundar y film of a lower -melt ing const it uent is needed t o make a mat er ial cr umble int o pieces when it is defor med. Such condit ion is called hot shor t ness or bur ning. Most hot-wor king oper ations ar e car r ied out in a number of multiple passes, or steps. Gener ally wor k i ng t emper at ur e for t he i nt er medi at e passes i s k ept wel l above t he mi ni mum wor ki ng temper atur e in or der to take advantage of economics offer ed by lower flow str ess. Some gr ain gr owth will likely to occur subsequent to the r ecr ystallization at t hese t emper atur es. Since a fine gr ainsized pr oduct is usually desir ed, common pr act ice is to lower the wor king temper at ur e for the last pass to the point wher e gr ain gr owth dur ing cooling fr om the wor king temper atur e will be negligible. Finishing temper atur e is usually just above the minimum r ecer ystallization temper atur e. To ensur e a fine r ecr ystallized gr ain size, amount of defor mation in the last pass should be r elatively lar ge. Types of hot wor king processes. ( i ) Rolling ( ii ) For ging (iii ) Pipe welding ( iv) H ot pier cing (v) H ot dr awing ( vi ) H ot spinning (vii ) H ot ext r uding

4.20

Manufacturing Engineering

(2) Cold wor k ing. A metal is said to be cold worked, if it is mechanically processed below the cr ystallisation temper ature of the metal. Cold working produces an improved surface finish and closer dimensional tolerance and because of this char acter istic cold wor king pr ocesses ar e gener ally used in making end-use pr oducts. Since r ecr ystallisation does not take place in cold wor king, the gr ains ar e per manently distor ted. Residual stresses are set-up during the cold working. As their presence is undesir able, a suitable heat tr eatment is gener ally necessary to neutr alise these str esses and restor e the metal to its or iginal struct ur e. Cold-wor king of met al r esults in an incr ease in st r engt h or har dness and a decr ease in duct ilit y. When cold-wor king is excessive, met al will fr act ur e befor e r eaching t he desir ed size and shape. Ther efor e cold-wor king oper ations ar e usually car r ied out in sever al steps with inter mediate annealing oper ations int r oduced t o soft en t he cold-wor ked met al and r est or e t he duct ilit y. This sequence of r epeat ed coldwor king and annealing is fr equent ly called cold-wor k-anneal cycle. Alt hough need for annealing oper at ions incr eases t he cost of for ming by cold-wor king, it pr ovides a degr ee of ver sat ilit y which is not possible in hot -wor king oper at ions. By suit ably adjust ing cold-wor kanneal cycle, the par t can be pr oduced wit h any desir ed degr ee of str ain har dening. I f finished pr oduct is st r onger t han fully annealed mat er ial, t hen final oper at ion must be a cold-wor king st ep wit h t he pr oper degr ee of defor mat ion t o pr oduce t he desir ed st r engt h. This would pr obably be followed by a st r ess r elief t o r emove r esidual st r esses. Types of cold working processes : (i )

Drawing

( ii ) Squeezing ( iii ) Bending ( iv) Shear ing (v)

Hobbing

( vi ) Shot peening ( vii ) Cold ext r uding When t her e is excessive cold wor king, t he met al may fr act ur e befor e r eaching t he desir ed shapes and size, and in or der t o avoid it , cold wor king oper at ions ar e car r ied out in sever al st eps. Compar ison of Cold working and H ot wor king process. C old w or k in g pr ocess

H ot w or k in g pr ocess

1. D one at a t emper atur e below the val ue r equir ed for r ecr yst al l i sat i on, so no appl i cabl e r ecover y t ak es pl ace dur ing defor mat i on.

1. D one at a t emper at ur e above r ecr yst al l i sat i on t emper atur e, so i t can be r egar ded as a si mul t aneous occur ence of defor mat i on and r ecover y pr ocess.

2. H ar dening i s not el i mi nat ed as w or k ing i s done at a t emper at ur e bel ow r ecr yst al l i sat i on. So thi s is al w ays accompanied by st r ai n har dening.

2. H ar dening due t o pl ast i c defor mat i on i s compl et el y el i mi nat ed by r ecover y and r ecr yst al l i sat i on. Thi s i s t r ue, only i f r at e of cr yst al l i sat i on i s higher t han r at e of defor mat i on.

3. D ecr eases mechani cal pr oper t i es like el ongat i on, r educt i on of ar ea and i mpact val ues.

3. I ncr eases mechanical pr oper t i es like el ongat i on, r educt i on of ar ea and i mpact val ues.

4. Cr yst al l i sat i on does not t ak e pl ace. Gr ains ar e el ongat ed.

4. Refi nement of cr yst al s occur s. .

Manufacturing Engineering

4.21

5. U ni for mi ty of mat er i al i s l ost and pr oper t i es ar e gr eat l y effect ed.

5. Pr omot es uni for mi t y of mat er i al by faci l i t at i ng di ffusi on of al l oys, const i t uent s and br eak s br i t t l e fi l m of har d const i t uent s or i mpur i ty e.g. cement i t e in st eel .

6. Chances of cr ack pr opagat i on i s mor e.

6. Cr ack s and unoxi dised bl ow hol es ar e somet i mes w el ded up; al t er nat i vely, ser i ous cr ack s or faul t s ar e usual l y shown up at an ear l y st age.

7. I ncr eases ul t i mat e t ensi l e st r ength, yi el d poi nt har dness and fat igue st r engt h but decr eases r esi st ance t o cor r osi on, i f sever el y w or k ed yi el d point may coi nci de w i t h ul t i mat e st r engt h val ue.

7. U l t i mat el y t ensi l e st r engt h, yi el d poi nt , fat i gue st r ength, har dness and r esi st ance t o cor r osi on et c. ar e not effect ed i f hot w or k ing i s done pr oper ly.

8. I nt er nal and r esi dual st r esses ar e pr oduced.

8. I nt er nal and pr oduced.

r esi dual

st r esses

ar e

not

I I . On the basis of t ype of forces applied. (1) D ir ect compr ession t ype pr ocess. I n t hese pr ocess, for ce is applied t o t he sur face of wor kpiece and t he met al flows at r ight angles to t he dir ect ion of compr ession.

e.g. for ging and r olling (2) I ndir ect compr ession pr ocesses. These include wir e dr awing, t ube dr awing, ext r usion and deep dr awing of a cup. (3) Tension t ype processes. These pr ocesses ar e car r ied out under t he applicat ion of t ensile for ces

e.g. st r et ch for ming, wher e a met al sheet is wr apped t o t he cont our of a die. (4) Bending pr ocesses. These pr ocesses involves t he applicat ion of bending moment t o t he sheet . (5) Shear ing pr ocesses. These pr ocesses involves t he applicat ion of shear ing for ces of sufficient magnit ude t o r upt ur e t he met al in t he plane of shear. F ORGI N G I t is t he pr ocess of shaping met als by impact or ver y high pr essur e. When t he mt al is heat ed t o plast ic t emper at ur e and shaped, it is called hot for ging or simplify for ging. When t he met al is defor med and shaped at r oom temper at ur e r aising t he st r ess level beyond elast ic limit , t he pr ocess is called cold for ging. M ain for ging oper at ions. (i )

Upset t ing or jumping

( ii ) Dr awing or necking ( iii ) Sett ing down ( iv) Punching (v)

Bending

( vi ) Welding ( vii ) Cutting (viii )Swaging

4.22

Manufacturing Engineering

TYPES OF FORGI N G. (1) Smit h for ging. I n smit h for ging, t he for ging for ce is supplied by dr op hammer s, st eam hammer s or by hydr aulic pr esses, so t hat for gings var ying in size fr om less t han 0.5 kg up t o 200 t ons can be pr oduced. The shaping of component depends ent ir ely upon the manipulat ive skill of the smit h, and met hod used for small quant it ies of for gings or for pr eliminar y shaping pr ior t o dr op for ging. Tools and equipment s used in smit hy : H ear t h; Anvil; H ammer s; Tongs; Swages; Swage block; Fuller s; Flat t er (2) Closed die for ging. This pr ocesses include bot h dr op for ging wher ein some for m of hammer is used and pr ess for ging in which a mechanical or hydr aulic pr ess is used. I n each pr ocess, a closed impr ession die is used t o shape a piece of heat ed met al. I n such a die, met al is t ot ally enclosed in t he die cavit y, one half of t he die being at t ached t o t he r am and ot her half of t he anvil. Closed-die for ging uses car efully machined matching die blocks to pr oduce forgings to close dimensional t oler ances. The pr eshaped billet is placed in t he cavit y of the blocking die and r ough-for ged t o close t o t he final shape. The gr eat est change in t he shape of t he met al usually occur s in t his st ep. I t is t hen t r ansfer r ed t o t he finishing die, wher e it is for ged t o final shape and dimensions. Usually blocking cavit y and t he finishing cavit y ar e machined into the same die block. Fuller ing and edging impr essions ar e oft en placed on t he edges of t he die block. (3) U pset or machine forging. This pr ocess pr oduces a lar ge var it y of component s, t he mechanical pr oper t ies of whi ch benefit consider ably fr om t he for ged st r uct ur e obt ained. I n this type of for ging, machines used ar e essentially double acting and their dies move in the hor izontal plane. The fixed half of t he die cor r esponds to the anvil in a for ging hammer and has a pair of gr ipping jaws, t he wor k is held by t hem, when for ging t akes place. EXTRU SI ON By t his pr ocess, a block of met al is r educed in cr oss-sect ion by for cing it t o blow t hr ough a die or ifice under high pr essur e. Types of ext rusion pr ocess. (1)

I mpact ext r usion. I t is a cold ext r usion pr ocess used for t he manufact ur e of collapsible t ube and long shells.

(2)

H ydr ost at ic ext r usion. I n t his pr ocess, billet is sur r ounded by liquid on all sides except at it s fr ont end which is point ed and which bear s against a cone shaped die. The liquid t r ansmit s t he pr essur e applied t o t he r am pushing t he billet for war d, and compr esses t he billet cir cumfer ent ially. Since t her e is no fr ict ion bet ween billet and t he cont ainer, wor king loads ar e less and ext r usion is also defect fr ee.

(3) Direct ext rusion. I n dir ect ext r usion, a heat ed billet of met al is placed in t he cont ainer of pr ess and t hen for ced t hr ough t he die, t he ext r usion t akes place in t he dir ect ion of mot ion of t he r am. (4)

I ndir ect E xt r usion. I n indir ect ext r usion, ext r usion t akes place in t he dir ect ion opposit e t o t he movement of t he r am. Gener ally indir ect pr ocess is used t o employ a fixed r am and move cont ainer and t he billet . Since t her e is no r el at i ve movement bet ween bil let and cont ai ner, l ess mechanical ener gy i s used in over coming fr ict ional for ces bet ween t he t wo.

(5)

E xt r usion for ging. This is basically a for ging pr ocess which is allied t o ext r usion so far as par t of t he component is ext r uded whilst hot t hr ough a die.

D RAWI N G Cold dr awing is used ext ensively in t he pr oduct ion of r ods of var ious cr oss-sect ions, and wir e.

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4.23

Wir e dr awing. I t is a simple pr ocess of r educing diamet er of met al r ods by dr awing t hem t hr ough conical opening in die blocks. St eel, ir on or non-fer r ous r od is conver t ed int o wir e by dr awing it t hr ough a conical hole having an included angle of 8-24 degr ee. I n cont inuous wir e dr awing, t he wir e passes t hr ough a succession of holes of decr easing size in dies made of st eel, t ungst en car bide, r uby or diamond. Reduct ionin cr oss-sect i onal ar ea usually being about 30 per cent . The amount of r educt ion which can be achieved in a single pass t hr ough a die is limit ed by t he t ensile st r engt h of t he dr awn wir e. M aximum t ensile for ce = cr oss-sect ional ar ea of dr awn wir e  yield st r engt h aft er dr awing. D eep dr awing. I n t his pr ocess, annual por t ion of t he sheet met al wor kpiece bet ween blank holder and die is subject ed t o a pur e r adial dr awing, wher eas por t ions of t he wor kpiece ar ound t he cor ner s of t he punch and die ar e subject ed t o a bending oper at ion. Again, por t ion of t he job bet ween punch and t he die walls under goes a longit udinal dr awing. I n t his oper at ion var ying amount of t hickening and t hinning of t he wor kpiece is unavoidable. BL AN KI N G I t is a pr ocess of cut t ing or shear ing a blank fr om sheet or st r ip mat er ial. Refer r ing figur e, punch is of t he built -up t ype having a low-car bon-st eel shank, which is fit t ed in t he pr ess r am, and a har dened-st eel cut t ing edge held by scr ews on t o t he face. Aim of having a built -up punch of t his t ype is t wo fold. I t gives a st r ong edge t hat will wit hst and t he abr asive act ion for a consider able per iod wit hout losing it size, and it per mit s a gr eat economy in t he use of expensive st eel. I n t his par t icular design, cut t ing edge is held in posit ion by means of a spigot and hollow-head scr ews. To assist the clamping and make up shunt height of t he pr ess, t he bed is mount ed on a bolst er plat e. I t is locat ed by means of st r ong dowels and clamped in posit ion wit h hollow-head scr ews. I n or der t o guide t he mat er ial a guide plat e fit t ed wit h a st op is at t ached t o t he face of t he blank bed. The guide plat e also act s as a st r ipper and t akes t he scr ap webbing off t he punch. I t aids in get t ing a good blank, and a r easonable out put per hour, as t ime is not wast ed in having t o pull t he scr ap clear by hand.

Punch Hardened steel cutting face

Guide stripper plate

Blank bed

PI ERCI N G OR PU N CH I N G The main differ ence bet ween pier cing and blanking is that the metal por t ions t hat ar e cut out in t he for mer pr ocess ar e scar p. H oles of var ious shapes ar e pier ced. When t wo or mor e pier cing punches ar e employed t oget her in a die, t heir lengt hs should differ slight ly i n or der t o r educe t he for ce and i mpact r equi r ed at one t i me. Diamet er s of holes which ar e t o be pier ced should be at least t wice t he met al t hickness, in or der t o avoid excessive punch br eakage.

Bolster plate

Piercing rollers Mandrel Solid-billet Seamless

tubing Pier ced holes should not be locat ed t oo close t o adjacent holes. Dr illing should be used for smaller holes and for holes which must be quit e close t oget her.

This pr ocess is employed for t he pr oduct ion of seamless t ubes. I t offer s most economical mechanical wor king pr ocess for t he manufact ur e of seamless t ubes. I t consist s of passing t he hot r olled billet at 1100C bet ween t wo conical r oller s and over a mandr el which helps in pier cing and cont r olling t he size of bor e as t he billet is for ced over it . The solid billet is fir st centr e-punched and then heat ed to 1100 in a fur nace befor e being pier ced. I t is t hen pushed int o two-pier cing r oller s, which impar t both axial and r otat ional motion. Due t o alter nate squeezing and bulging oper at ions, a seam is for med in t he cent r e and pier cing mandr el cont r ol size and shape of t he hole. The fir st pass makes a r at her t hick-walled t ube which is again passed bet ween gr ooved r oller s over a plug held by a mandr el and is conver t ed int o a longer t ube wit h specified wall t hickness. The t ube is again passed t hr ough a r eeling machine which helps in st r aight ening and sizing t o t he final dimensions.

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Manufacturing Engineering

Tubes ar e t hen cooled and placed in t he pickling bat h of dilut e sulphur ic acid to r emove t he oxide and scale for mat ion. The one pass pr ocess is applicable t o scamless t ubes of 15 cent imet er s diamet er. L ar ger t ubes upt o 35 cent imet er s diamet er ar e t hen given a second pass or pier cing r oller s. SPI N N I N G I t means shaping a met al blank as it r evolved at a high speed in lat he. For small-bat ch pr oduct ion, is used because of t he (i ) simplicit y of equipment needed (ii ) ease wit h which a chuck can be pr epar ed (iii ) gives lower over all pr oduct ion cost s t han any alt er nat ive met hod. The pr ocess lends it self t o t he pr oduct ion of goods dir ect fr om a flat blank. I t is also a ver y useful adjunct t o t he pr ess and dr op stamp. Types of spinning. (1) H and spinning. Wit h hand spinning , t he blank as it r evolves at a high speed is subject ed, by a st eel t ool, t o pr essur e exer t ed by t he oper at or, who has t he t ool handle t ucked under his r ight ar m. The t ool is lever ed off a t ee t est and is pr event ed fr om slipping by means of a peg. The lever age obt ained is adjust ed t o suit t he class of wor k being handled. (2) M achine spinning. EXTRU SI ON OF TU BI N G Wit h moder n equipment , t ubing may be pr oduced by ext r usion t o t oler ances as close as t hose obt ained by cold-dr awing. To pr oduce t ubing by extr usion, a mandr el must be fast ened t o the end of t he ext r usion r am. The mandr el ext ends t o t he ent r ance of ext r usion die, and t he clear ance bet ween t he mandr el and t he die wall det er mines t he wall t hickness of t he ext r uded t ube.

(a) Pi erci ng

(b) Extr usi on Fi g. Tube extrusi on

Classificat ion of Pipe and Tubing depending on met hod of manufact ure. ( i ) Seamless pipes. Ext r usion is an excellent met hod of pr oducing seamless pipe and t ubing, especially for met als which ar e difficult t o wor k. ( ii ) Welded pipes.



Manufacturing Engineering

4.25

JOI N I N G PROCESS All joining pr ocesses involving at omic bonding ar e of per manent in natur e. I n mechanical bonding, str ength of t he joint i s less t han combined st r engt h of t he or i ginal member s. H owever i n at omi c bonding, t he si t uat ion i s not necessar i ly so. Classificat ion of joint ing pr ocess. Joining pr ocess ar e classified accor ding t o t he composit ion of t he joint . (i ) Autogeneous : I n aut ogeneous pr ocess, no fil ler mat er i al is added dur ing joi ni n. e.g. all t ypes of soli d pahse welding and r esi st ance wel di ng (ii ) H omogeneous : I n homogeneous pr ocess, fil ler mat er i al used t o pr ovide t he joint is same as t he par ent mat er ial. e.g. ar c, gas, and t her mi t wel di ng (iii ) H eterogeneous : I n heter ogeneous pr ocess, filler material differ ent from the par ent mater ial is used. e.g. solder ing and br azing . Two mat er i al s whi ch ar e i nsol ubl e i n each ot her, such as i r on and si l ver, can be j oi ned by a het er ogeneous pr ocess. Thi s may be achieved by usi ng a fi ll er mat er ial (e.g. copper and t i n) which is solubl e i n bot h t he par ent mat er i al s (i.e. i r on and silver ). WELDI N G I t is t he met hod of joining met als by applicat ion of heat , wit hout t he use of solder or any ot her met al or alloy having a lower melt ing point t han t he met als being joined. Types of welding. These may be divided int o t wo main gr oups. (1) Pr essur e welding. I n a pr essur e welding, t he met al joined is never br ought t o a molt en st age but it is heat ed t o a welding t emper at ur e and t he act ual union is br ought about by t he applicat ion of pr essur e. Types of pressure welding : ( i ) Forge welding ( ii ) Resist ance elect r ic welding : (a) Butt welding (b) Flash welding (c) Spot welding (d) Scam welding (e) Pr oject ion welding (2) F usion welding. I n fusion welding, t he met al being joined is act ually melt ed and union is pr oduced on subsequent soldification. Pr inciples of F usion (liquid st at e) welding : I n a fusi on pr ocess, t he mat er i al ar ound t he joi nt is melt ed in bot h t he par t s t o be joined. I f necessar y, a mol t en fi l l er mat er i al i s al so added. Thus, fusi on wel di ng pr ocess may by ei t her aut ogeneous or homogeneous. M et allur igically, t her e ar e t hr ee dist inct zones in a welded par t : (i ) Fusion zone (ii ) H eat affect ed unmelt ed zone ar ound t he fusion zone (iii ) Unaffect ed or iginal par t . Factors governing fusion welding process : (i ) Char act er i st ics of t he heat sour ce. (ii ) Nat ur e of deposit ion of t he fi ll er mat er ial in t he fusion zone cal led wel d pool. (iii ) H eat flow char act er i st ics i n t he joint . (iv) Gas met al or sl ag met al r eact ions in t he fusion zone. (v) Cooling of t he fusion zone with the associated cont r act ion, r esidual str esses, and metallur gical changes.

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Manufacturing Engineering

Types of fusion welding : (i ) Gas welding (ii ) Elect r ic ar e welding (iii ) Ther mit welding H eat sour ce. A heat sour ce, sui t able for weldi ng, shoul d r el ease t he heat in a isolat ed zone. M or eover, t he heat should be pr oduced at a high t emper at ur e and at a high r at e.

Sour ces of heat . (i ) Elect r ic ar c (as in var ious ar c weldi ngs) (ii ) Chemi cal flame (as i n gas welding) (iii ) Exot her mic chemical r eact ion (as i n t her mit welding) (iv) Elect r ic r esist ance heat i ng (as in el ect r osl ag and ot her r esi st ance wel di ng pr ocesses). GAS WELDI N G I n gas welding, t he heat r equir ed t o pr oduce fusion of t he met al being welded is obt ained by bur ning some gas issuing fr om t he nozzle of a blow pipe or t or ch. The gas fr om t he nozzle is alr eady mixed wit h t he r equir ed amount of oxygen t o pr oduce a flame. The flame may be oxidising, neut r al or r educing. The nat ur e of flame depends upon t he met al or alloy t o be welded.

Types of gas welding. ( i ) Oxy-acet ylene welding : Oxygen and acet ylene can be st or ed in cylinder s and t heir flow t o t he welding t or ch is cont r olled by valves. When acetylene and oxygen ar e mixed in pr oper pr opor tion, flame tempeatur es ar ound 3500C can be achieved. The oxygen to acet ylene pr opor tion can be var ied to pr oduce eit her oxidising, r educing or neut ur al flame. I t pr ovides differ ent t emper at ur es for differ ent met als. ( ii ) Air-acet ylene welding : I n t his welding, t emper at ur e available is pr obably t he lowest in compar ison t o all ot her gas welding pr ocesses. The acetylene gas is supplied to the tor ch fr om a cyliner and air is dr awn into the tor ch fr om atmospher e and it s quant it y can be adjust ed for pr oper combust ion by var ying t he opening of air inlet . This t ype of welding is used for lead welding, low t emper at ur e br azing and solder ing oper at ions. ( iii )Oxy-hydr ogen welding : The gas used in this pr ocess is hydr ogen instead of acetylene and temper atur e upto 2000C is obtained. I t is best suit ed for welding t hin sheet s, low temper at ur e melt ing alloys and for br azing. No oxides ar e for med in t he sur face if a r educing at mospher e is used. I n t his case flame adjust ment is difficult due t o absence of dist inguishing colour t o judge t he pr opor t ion of hydr ogen and oxygen. ( iv) Oxy-L PG process : This is similar t o t he above except t hat L iquefied Pet r oleum Gas is used wit h oxygen for pr oducing t he flame. The t or ches ar e designed for handling lar ger volumes of gas, because L PG r equir es lar ge quantity of oxygen (2 to 4 times) as compar ed to acetylene. The temper atur e obtained is about 25000C. ELECTRI C ARC WELDI N G Welding arc. An ar c is an electr ic dischar ge between two elect r odes which takes place thr ough an electr ically conducting hot ionised gas called plasma. An elect r ic ar c used for welding is called welding arc and is usually bet ween a t hin r od (or coir e) and a plat e, it is t her efor e bell shaped. A welding ar c is a high cur r ent low volt age elect r ic dischar ge oper at ing gener ally in t he r ange of 10 t o 2000 amper es and at 10 t o 50 volt s. I n a welding cir cuit , t he ar c act s as a load r esist or.

Welding ar c is usually divided int o five par t s

Manufacturing Engineering

4.27

(i ) Cat hode spot (ii ) Cat hode dr op zone (iii ) Ar c column zone (iv) Anode zone (v) Anode spot Ar c blow. Nor mally, welding ar c is or ient ed along t he axis of elect r ode r at her t han along t he shor t est dist ance bet ween elect r ode and t he wor kpiece. H owever, an elect r ic ar c may be deflect ed fr om it s int ended pat h by var iet y of fact or s, such as deflect ion of ar c dur ing welding is called an ar c blow.

Fact or s caussing ar c blow. (i ) I mpr oper posit ioning of gr ound (or ear t h) connect ion (ii ) L ar ge fer r omagnet ic mass close t o welding sit e (iii ) L ong ar c lengt h (iv) H igh welding cur r ent (v) Use of dc supply TYPES OF ARC WELDI N G (1) M et allic ar c welding. I n t his pr ocess, met al elect r ode ser ves t o car r y t he ar c t o act as t he for m filler r od which deposit s molt en int o t he joint . Coat ed elect r odes ar e used r esult ing in t he for mat ion of a layer of slag at t he sur face of t he weld. When an ar c is st r uck bet ween elect r ode and t he wor kpiece, bot h ends of elect r ode and t he wor kpiece dir ect ly under t he t ip of elect r ode melt and become mixed. Pr essur e pr oduced by t he st r eam of ions flowing fr om t he elect r ode causes a cr at er t o for m in t he molt en met al of t he wor kpiece, and molt en met al fr om t he elect r ode falls int o t his cr at er. As t he ar c moves along t he joint , met al flows back int o t he cr at er, t her efor e base met al and elect r ode met al get int imat ely mixed. M ost engineer ing met als and alloy may be welded by t his pr ocess. (2) Carbon ar c welding. Types of carbon arc welding : (i ) I ndir ect met hod : I t employs an ar c bet ween t wo car bon elect r odes. (ii ) Dir ect met hod : I n t his ar c is st r uck bet ween a single car bon elect r ode and t he wor k. (iii ) Car bon ar c pr ocess : I t is used for welding st eel sheet , copper and somet imes aluminium. I t is also used t o melt small ar eas of met al t o be gouged. (3) Submer ged ar c welding. This pr ocess is an aut omat ic for m of t he metallic ar c welding pr ocess which can be used in t he st r aight line joining of met als. A t ube which deliver s powder ed flux int o t he pr epar ed joint in advance of t he elect r ode is built int o t he elect r ode holder. This powder ed flux envelops melt ing end of t he elect r ode and complet ely cover s t he ar c. M uch of t he flux melt s and r ises t o t he t op of t he molt en weld met als, wher e it for ms a coat ing of pr ot ect ive slag. Any unmelt ed flux is r emoved by a suct ion syst em and can be r eused. The elect r ode is in t he for m of bar e coiled wir e and is gener ally copper plat ed t o ensur e low r esist ance elect r ical cont act bet ween wir e and t he cont act shoes. This pr ocess is used for welding low and medium car bon st eels, low alloy sheet s, copper, aluminium and t it anium. (4) F lux cored ar c welding. I n t his welding, t he flux is car r ied inside t he elect r ode which consist s eit her of a t ube or a st r ip of met al folded ar ound t he flux. Car bon dioxide is used as an auxiliar y gas. I f such a gas shield is not used, t hen flux should cont ain mat er ials which will decompose t o liber at e non-oxidising gases in t he r egion of weld. (5) E lect r o-slag welding. I n t his welding, heavy sect ions can be joined in a single r un by placing t he plat es t o be welded in a ver t ical posit ion so t hat molt en met al is deliver ed pr ogr essively t o t he ver t ical gap. I n t his pr ocess,

4.28

Manufacturing Engineering

t he ar c is used only t o init iat e melt ing and t her eaft er heat is gener at ed by elect r ical r esist ance offer ed by t he slag which is sufficient ly conduct ive t o allow t he cur r ent t o pass t hr ough it fr om t he elect r ode t o t he weld pool beneat h. As t he elect r ode wir e melt s, level of t he pool r ises and t he aut omat ic wat er cooled weld shoes ar e r aised at a suit able r at e t o keep pace wit h solidificat ion and deliver y of molt en met al. The molt en slag bat h above t he weld pool act s bot h as heat sour ce and shield t o pr ot ect t he weld fr om oxidat ion. Dur ing welding slag t emper at ur e upt o 2000C ar e pr oduced. This leads t o consider able melt ing at t he edges of t he wor kpieces so t hat int imat e mixt ur e of t he mat er ials of elect r ode and wor kpieces t akes place. This pr ocess is used for joining lar ge cast ing and for gings. M ild, low alloy and high alloy st eels and t it anium can be welded by t his pr ocess. SOLI D STAT E WEL DI N G This pr ocesses can be car r ied out bot h at t he r oom t emper at ur e and at an elevat ed t emper at ur e wit hout , melt ing any par t of t he joining sur faces. A solid phase weldi ng done at t he r oom t emper at ur e does not al low r ecr yst al li zat ion and gr ai n gr owt h at t he int er face. This r educes duct i lit y of t he joi nt t o some ext ent . An incr ease in wor k ing t emper at ur e not only incr eases t he duct i li t y but also elemi nat es some ot her defect s. The shape and size of voi ds at t he int er face ar e modified consi der ably dependi ng on t he amount of diffusion. TYPES OF SOLI D-STATE WELDI N G. 1. E lect r ical r esist ance welding. ( i ) Spot welding : I n t his welding pr ocess, t he par t s t o be joined ar e clamped fir mly bet ween a pair of heavy elect r odes which ar e connect ed t o t he secondar y cir cuit of a st ep down t r ansfor mer syst em. M aximum r esist ance in such cir cuit s exist s at t he sur face of t he t wo par t s t o be joined. (ii ) Seam welding : I n t his pr ocess, a cont inuous weld is pr oduced by passing t he wor k bet ween r otat ing wheel shaped electr odes which exer t t he welding pr essur e and conduct the welding cur r ent.

M et hods of seam welding. (a) Cont inuous mot ion welding : I n t his welding, elect r odes r ot at e at const ant speed and welding cur r ent eit her flows cont inuously or is int er r upt ed. (b) I nt er mi t t ent pr ocess : I n t hi s wel di ng, a ser i es of over l appi ng spot wel ds i s pr oduced. The elect r odes t r avel t he dist ance necessar y for each successive weld and t hen st op. Then cur r ent is swit ched ON and the weld is made; t he whole pr ocess being contr olled aut omat ically. ( iii )Flash welding : This pr ocess can be used for joining t oget her t he ends of sheet s, r ods, wir es or t ubes. The t wo wor k par t s ar e placed in t he clamping jaws of a machine, and as t he par t s ar e br ought t oget her int o light cont act , a volt age sufficient ly high t o cause ar cing bet ween t he ends is applied. Ar cing cont inues as t he t wo par t s advance, unt il t he wor k at t ains a welding t emper at ur e, suffi ci ent pr essur e bei ng appl i ed t o pr oduce a cont i nuous wel d. Flashing and upset t i ng ar e accompanied by t he expulsion of met al, slag and oxides fr om t he joint . I n t his pr ocess, t he moving plat es should advance at t he cor r ect speed in r elat ion t o t he cur r ent used. Too high speed st ops ar cing act ion bet ween wor k par t s r esult ing in inadequat e weld. I f speed is t oo low, t he ar cing act ion may be int er mit t ent , and again heat ing will be insufficient t o for m a good weld. ( iv)Butt welding : I n t his pr ocess, no ar cing t akes place bet ween sur face being joined; t he heat being pr oduced solely by t he elect r ical r esist ance at t he but t ing sur faces t o t he passage of a cur r ent . The pr essur e is applied befor e t he cur r ent begins t o flow and is maint ained t hr oughout t he cont act ar ea unt il a sufficient ly high t emper at ur e has been r eached t o per mit t he for ging of a weld. Best r esult s ar e gener ally obt ained if two pieces ar e equal in cr oss-sect ional ar ea and of equal specific r esist ance. The sur faces to be joined must be clean, par allel and r easonably smooth, other wise local over heating may occur in t he r egion of any pr oject ion.

Manufacturing Engineering

4.29

(v) Projection welding : This is modificat ion of spot welding pr ocess, in which cur r ent flow and t he r esult ant heat ing ar e localised t o a r est r ict ed ar ea by embossing one of t he par t s t o be joined. Wher e t hick sect ions have t o be joined, pr oject ion welding can be used since spot welding would not be possible due t o heavy cur r ent s and pr essur es r equir ed. I n pr oject ion welding, a bet t er heat balance can be obt ained in difficult t o weld combinat ion of composit ion and t hicknesses, and t he finished appear ance is oft en super ior, since elect r ode on t he finish side will have a layer cont act ing ar ea which consider ably r educes elect r ode indent at ion. Elect r ode life is longer because of t he lar ge cont act ar eas employed and use of har der high r esist ance alloys. Pr essur es and cur r ent densit it es ar e also lower than in spot welding and these factor s will r educe wear and distor tion of the electr odes. The desing of pr oject ion should for m easily wit hout causing any dist or t ion in t he par t dur ing for ming and it should be st r ong enough t o suppor t init ial elect r ode pr essur e befor e t he cur r ent begins t o flow. Dur ing welding, t he pr oject ion wit hout undue spr ead and should t hus leave t he par t s in int imat e cont act .

(vi)Percussion welding. I n t his pr ocess, elect r ic ar c is mainly used t o heat t he welding member s and only a pr essur e is ut ilized t o effect a weld. This pr ocess r elies on elect r ic ar c for heat ing r at her t han on t he elect r ic r esist ance in t he met al. The met al pieces t o be welded ar c held apar t ; one in a st at ionar y clamp and ot her in a sliding clamp backed up against a heavy spr ing pr essur e. Dur ing welding, t he movable clamp is set fr ee and it car r ies wit h it t he piece t o be welded. When t he pieces ar e about 1.5 mm apar t , t her e is a sudden dischar ge of elect r ic ener gy. This cr eates intense ar cing over t he sur faces t o be welded and r ises t hem t o a high t emper at ur e. The ar c get s ext inguished by t he per cussion blow of t he t wo par t s coming t oget her wit h sufficient for ce t o make t he weld. (2) Solder ing and Br azing. Solder i ng and br azi ng pr ocesses ar e car r i ed out by al lowing a mol t en fil ler mat er i al t o flow i n t he gap bet ween par ent bodi es. Obviousl y, t he fi ller mat er ial has t o have a mel t ing poi nt much l ower t han t h at of t h e par en t bodi es. Wh en f i l l er m at er i al i s a copper al l oy ( e.g. copper -zi n c an d copper -silver ), t he pr ocess is called brazing. A similar pr ocess with a lead-t in alloy as t he filler mat er ial is call ed solder ing. The most common heat sour ce for t hese pr ocesses is el ect r i cal r esist ance heat ing.

Advant ages of solder ing and br azing pr ocesses: (i ) H eat i ng of t he par ent mat er ials is negl igi bl e t o cause any change in t hei r st r uct ur e or pr oper t ies. (ii ) These can join mat er ials which ar e i nsolubl e in each ot her. To pr oduce a per fect joint , ent ir e gap between t he par ent bodies must be filled up by the filler mater ial. This i s achieved essent ial ly t hr ough a capill ar y act ion. (3) Adhesive Bonding. This pr ocess i s most commonl y used in t he air cr aft and aut omobil e indust r ies wher e sheet met als ar e joined in var i ous configur at i ons. Advant ages. I n t hi s pr ocess t her e i s no uncer t aint y about t he fl ow of adhesi ve in t he joint , because t he adhesive is fi r st appli ed on t he sur faces and t he joint i s made wi t h subsequent appli cat ion of heat and pr essur e. D isadvant age. I nher ent l ow st r engt h of t he r esul t i ng joint .

Types of bonding forces taking part in an adhesive joint : (i ) van der Waals for ce : I t is due t o const ant movement of t he posi t i ve and negat ive char ges of mol ecul es. (ii ) Polar force bet ween adhesive and r elatively br itt le oxide film : I t is due to dipole adhesive molecules. This for ce is nor mall y sever al t i mes i n magni t ude higher t han t he van der Waals for ce.

4.30

Manufacturing Engineering

DU TY CYCLE I t is t he per cent age in t ime a power sour ce supplies cur r ent in successive t en minut e int er vals wit hout exceeding a pr edet er mined r ise in t he t emper at ur e of it s vit al component s. Usually 60% dut y cycle is t he st andar d indust r ial r at ing. A power sour ce wit h such a r at ing can supply it s r at ed out put for 6 minut es in ever y t en minut e int er val of it s oper at ion. Dut y cycle is one of t he most impor t ant r at ing specificat ion of a welding power sour ce and it det er mines t he t ype of ser vice for which a power sour ce is designed. H eavy dut y indust r ial unit s for manual welding ar e nor mally r at ed at 60% dut y cycle while for aut omat ic and oft en for semi-aut omat ic welding pr ocesses, t he r at ing is 100%. L ight dut y power sour ces may be of as low as 20% dut y cycle. When a welding power sour ce is r equir ed t o supply mor e t han it s r at ed out put cur r ent , it is possible t o do so if it s dut y cycle is r educed. H EAT FL OW CH ARACTERI STI CS (1) Welding speed. I n t he fusi on weldi ng pr ocesses, t he heat sour ce i s moving, except in spot welding wher e t he sour ce is st at i onar y. Once t he st eady st at e i s r eached, even wi t h a movi ng heat sour ce, t he t emper at ur e di st r i but ion r el at ive t o t he sour ce becomes st at ionar y. The most convenient way of analyzi ng such a pr oblem wit h a movi ng sour ce is t o assume t he sour ce as st at ionar y and t he wor k pi ece t o move wi t h same veloci t y in t he opposit e dir ect i on. This speed is call ed welding speed. (2) Types of heat sources. ( i ) Plant heat source : The heat i s liber at ed in a small zone whi ch is i deali zed as a poi nt sour ce, and t he heat fl ow fr om t he sour ce is t hr ee-dimensional. ( ii ) Line heat source : The heat i s liber at ed al ong a li ne and t he heat sour ce i s ideal ized as a li ne sour ce in a few cases e.g. in but t wel ding of r elat ivel y t hi n plat es. 

Manufacturing Engineering

4.31

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. The cut t ing edge of t he t ool is per pendicular t o t he dir ect ion of t ool t r avel in (a) or t hogonal cut t ing of met als (b) oblique cut t ing of met al (c) bot h (a) and (b) (d) none of t hese 2. The cut ting edge of t he tool is inclined at an angle less t han 90° t o t he dir ect ion of t ool t r avel in (a) or t hogonal cut t ing of met als (b) oblique cut t ing of met als (c) bot h (a) and (b) (d) none of t hese 3. Two component s of t he cut t ing for ces which ar e per pendicular ar e act ing on t he cut t ing t ool in (a) or t hogonal cut t ing of met al (b) oblique cut t ing of met al (c) bot h (a) and (b) (d) none of t hese 4. Or t hogonal cut t ing syst em is also known as (a) one-dimensional cut t ing syst em (b) t wo-dimensional cut t ing syst em (c) t hr ee-dimensional cut t ing syst em (d) none of t hese 5. Thr ee component s of t he cut t ing for ces whi ch ar e mut ual l y per pendi cul ar ar e act i ng at t he cut t ing edge in (a) or t hogonal cut t ing of met als (b) oblique cut t ing of met als (c) bot h (a) and (b) (d) none of t hese 6. I n or t hogonal cut t ing syst em (a) cut ting t ool pr epar es a sur face par allel t o the wor k face (b) chip flows over t he t ool face and dir ect ion of t he chip flow velocit y is nor mal t o t he cutt ing edge (c) m axi m u m ch i p t h i ck n ess occu r s at t h e middle (d) all of t hese 7. I n oblique cut t ing syst em (a) cut t i ng edge of t he t ool may or may not pr epar e a sur face par allel t o t he wor k face (b) maximum chip t hickness may not occur at t he middle (c) mor e t han one cut t ing edges ar e in act ion (d) all of t hese

8. I n met al cut t ing oper at ions discont inuous chips ar e pr oduced while machining (a) br it t le mat er ial (b) ductile mat er ial (c) har d mat er ial (d) soft mat er ial 9. I n met al cut t ing oper at ions cont inuous chips ar e pr oduced while machining (a) br it t le mat er ial (b) ductile mat er ial (c) har d mat er ial (d) soft mat er ial 10. When material is ductile and cutting speed is high, t hen (a) cont inuous chips ar e for med (b) cont i nuous chi ps wi t h bui l t -up edges ar e for med (c) discont inuous chips ar e for med (d) none of t hese 11. When mater ial is ductile and cutting speed is low, t hen chips for med ar e (a) cont inuous (b) cont inuous chips wit h build-up edges (c) discontinuous (d) none of t hese 12. When mater ial is br it t le, non duct ile and cut t ing speed is low, t hen (a) cont inuous chips ar e for med (b) cont i nuous chi ps wi t h bui l t -up edges ar e for med (c) discont inuous chips ar e for med (d) none of t hese 13. I n metal cutting oper ations, chips ar e for med due to (a) linear defor mat ion (b) shear defor mat ion (c) linear t r anslation (d) none of t hese 14. Discontinuous chips are formed dur ing machining of (a) cast ir on (b) mild st eel (c) aluminium (d) none of t hese 15. Cont inuous chips ar e for med dur ing machining of (a) cast ir on (b) mild st eel (c) aluminium (d) none of t hese

4.32

Manufacturing Engineering

16. Wit h incr ease of cut t ing speed, t he built -up edge (a) becomes longer (b) smaller or finally does not for m (c) may or may not for m (d) none of t hese 17. M at er ial having highest cut t ing speed is (a) cast ir on (b) br onze (c) aluminium (d) high car bon st eel 18. M at er ial having lowest cut t ing speed is (a) cast ir on (b) br onze (c) aluminium (d) high car bon st eel 19. Chi ps wi t h bui l t -up edges ar e for med whi l e machining (a) br it t le mat er ial (b) ductile mat er ial (c) cast mat er ial (d) all of t hese 20. I n metal cutting operations, shear angle is defined as t he angle made by t he shear plane wit h t he (a) t ool axis (b) dir ect ion of t ool t r avel (c) cent r al plane of wor k piece (d) none of t hese 21. Angle made by t he face of t ool and t he plane par allel t o t he base of t he cut t ing t ool is called (a) lip angle (b) r ake angle (c) cut t ing angle (d) clear ance angle 22. Angle bet ween t he t ool face and t he gr ound end sur face of flank is called (a) lip angle (b) r ake angle (c) cut t ing angle (d) clear ance angle 23. Angle bet ween t he t ool face and a line t angent t o the machined sur face at the cutting point is called (a) lip angle (b) r ake angle (c) cut t ing angle (d) clear ance angle

24. Angle bet ween t he shear plane and wor k sur face is called (a) lip angle (b) r ake angle (c) cut t ing angle (d) shear angle 25. A n gl e on w h i ch t h e st r en gt h of t h e t ool depends is (a) lip angle (b) r ake angle (c) cut t ing angle (d) clear ance angle 26. For br it t le mat er ial like br ass, t he r ake angle pr ovided is (a) zer o (b) negative (c) positive (d) none of t hese 27. For copper, t he r ake angle pr ovided is (a) zer o (b) positive (c) negative (d) none of t hese 28. Car bide t ipped t ools gener ally have r ake angle (a) zer o (b) positive (c) negative (d) none of t hese 29. For machining br ass by high speed st eel t ool, t he r ake angle pr ovided is (a) zer o (b) positive (c) negative (d) none of t hese 30. Negat ive r ake angles ar e pr ovided t o (a) give bet t er finish (b) incr ease st r engt h of cut t ing t ool point (c) decr ease t emper at ur e r ise at t he t ool-t ip (d) all of t hese 31. Negat ive r ake angles ar e used for (a) heavy loads (b) car bide t ools (c) har d mat er ials (d) all of t hese 32. Cement ed car bide t ool ar e gener ally poor in (a) t or sion (b) t ension (c) compr ession (d) shear 33. Bi ndi ng mat er i al used i n cement ed car bi de t ools is (a) nickel (b) cobalt (c) chr omi um (d) silicon 34. Cut t ing t ool used on lat he, shaper and planer is (a) single point (b) t wo point (c) t hr ee point (d) mult i point 35. Cut t i ng t ool used on mi l l i ng and br eachi ng machine is (a) single point (b) t wo point (c) t hr ee point (d) mult i point

Manufacturing Engineering

36. A single point t ool has (a) lip angle (b) r ake angle (c) cut t ing angle (d) all of t hese 37. L ip angle of a single point t ool is gener ally in t he r ange (a) 20° t o 40° (b) 40° t o 50° (c) 50° t o 60° (d) 60° t o 80° 38. I n a single point t ool t he angle by which t he face of t he t ool is inclined t owar ds back is called (a) side r ake angle (b) back r ake angle (c) end r elief angle (d) side r elief angle 39. I n a single point tool, t he angle by which the face of t he t ool is inclined sideways, is called (a) side r ake angle (b) back r ake angle (c) end r elief angle (d) side r elief angle 40. Angle pr ovided in a single point t ool t o cont r ol chip flow is (a) side r ake angle (b) back r ake angle (c) end r elief angle (d) side r elief angle 41. To pr event t ool fr om r ubbing t he wor k , angl e pr ovided on t ools is (a) lip angle (b) r ake angle (c) clear ance angle (d) r elief angle 42. Relief angles on high speed t ools gener ally var y in t he r ange (a) 0° t o 5° (b) 5° t o 10° (c) 10° t o 20° (d) 20° t o 30° 43. I n a si ngl e poi nt t ool , t he angl e bet ween t he sur face of t he flank immediat ely below t he point and a plane at r ight angles t o t he cent r e line of t he point of t ool, is called (a) side r elief angle (b) end r elief angle (c) side r ake angle (d) back r ake angle

4.33

44. I n a si ngl e poi nt t ool , t he angl e bet ween t he sur face of t he flank immediat ely below t he point and a line dr awn fr om t he point per pendicular t o t he base, is called (a) side r elief angle (b) end r elief angle (c) side r ake angle (d) back r ake angle 45. Vel oci t y of t ool r el at i v e t o w or k pi ece i s called (a) cut t ing velocit y (b) chip velocit y (c) shear velocit y (d) aver age velocit y 46. Velocit y of t ool along t he t ool face is called (a) cut t ing velocit y (b) chip velocit y (c) shear velocit y (d) aver age velocit y 47. Vect or sum of cut t i ng and chi p vel oci t i es i s equal t o (a) shear velocit y (b) aver age velocit y (c) bot h (a) and (b) (d) none of t hese 48. Velocit y of chip r elat ive t o t he t ool is act ing (a) along t he t ool face (b) along t he shear plane (c) nor mal t o t he shear plane (d) nor mal t o t he t ool face 49. Velocity of chip r elative to the wor k-place is acting (a) along t he t ool face (b) along t he shear plane (c) nor mal t o t he shear plane (d) nor mal t o t he t ool face 50. Chip t hickness r at io is t he r at io of (a) cut t ing velocit y t o chip velocit y (b) dept h of cut t o chip t hickness (c) chip t hickness t o dept h of cut (d) none of t hese 51. Cut t ing r at io is t he r at io of (a) chip velocit y t o cut t ing velocit y (b) chip t hickness t o dept h of cut (c) cut t ing velocit y t o chip velocit y (d) none of t hese

4.34

Manufacturing Engineering

52. Chip compr ession fact or is equal t o (a) chip t hickness r at io

1 (b) chip thickness ratio (c) chip t hickness r at io × shear angle (d) none of t hese 53. I n met al cut t ing oper at i ons, chi ps ar e for med due t o (a) elast ic defor mat ion of met al (b) plast ic defor mat ion of met al (c) elast ic plast ic defor mat ion of met al (d) all of t hese 54. I n met al machining, due t o plast ic defor mat ion of met al maximum heat is gener at ed in t he (a) shear zone (b) fr ict ion zone (c) wor k-t ool cont act zone (d) none of t hese 55. I n met al machining, due t o fr ict ion bet ween t he moving chip and t he t ool face, heat is gener at ed in t he (a) shear zone (b) fr ict ion zone (c) wor k-t ool cont act zone (d) none of t hese 56. I n met al machining due t o bur nishing fr ict ion, heat is gener at ed in t he (a) shear zone (b) fr ict ion zone (c) wor k-t ool cont act zone (d) none of t hese 57. The coefficient of fr ict ion bet ween chip and t ool can be r educed by r educing t he (a) dept h of cut (b) widt h of t ool (c) effect ive r ank angle (d) any of t hese 58. The dept h of cut depends upon (a) cut t ing speed (b) t ool mat er ial (c) r igidit y of machine t ool (d) all of t hese 59. Dept h of cut for r oughing oper at ion as compar ed t o finishing oper at ion is (a) mor e (b) less (c) same (d) none of t hese

60. Feed depends upon (a) dept h of cut (b) r igidit y of machine (c) finish r equir ed (d) all of t hese 61. The aver age cut t ing speed for machining a cast ir on by a high speed st eel t ool is (a) 10 m / min (b) 22 m / min (c) 30 m / min (d) 300 m / min 62. The aver age cut t ing speed for machining a cast ir on by a high speed st eel t ool is (a) 10 m / min (b) 22 m / min (c) 30 m / min (d) 300 m / min 63. Cut t ing speed is gener ally low for (a) soft mat er ial (b) castings (c) r egular shaped mat er ials (d) none of t hese 64. The aver age cut t ing speed for machining copper alloys by a high speed st eel t ool is (a) 10 m / min (b) 22 m / min (c) 30 m / min (d) 200 t o 300 m/min 65. I n most high speed milling cutter s, positive r adial r ake angle is (a) 10 – 15° (b) 7 – 10° (c) 20 – 25° (d) 15 – 20° 66. I n machine t ools, chat t er is due t o (a) fr ee vibr at ions (b) r andom vibr at ions (c) for ced vibr at ions (d) self-excit ed vibr ations 67. The usual r at io of for war d and r et ur n st oke in shaper is (a) 2 : 1 (b) 1 : 2 (c) 2 : 3 (d) 3 : 2 68. I n pier cing and punching oper at ions, t he angle of shear is pr ovided on (a) die (b) punch (c) bot h on punch and die (d) not pr ovided t o all

Manufacturing Engineering

69. I n pi er ci n g oper at i on , t h e cl ear an ce i s pr ovided on (a) die (b) punch (c) half on die and half on punch (d) may be pr ovided on any member 70. I n dr awing oper at ion, incr ease of punch r adius (a) has much i nfl uence on punch l oad and i t decr eases (b) does not influence t he punch load much (c) punch load incr eases (d) punch load depends on ot her fact or s 71. Angul ar cl ear ance pr ovi ded on di es i s of t he or der of (a) 5 t o 10° (b) 3 t o 5° (c) 1/2 t o 1° (d) 0.1 t o 0.5° 72. Cent r e of pr essur e of a piece t o be blanked or pier ced in power pr ess lies at (a) c.g. of ar ea of piece (b) c.g. of per imet er of piece (c) cent r e of piece (d) cent r e of per cussion 73. I n dr awing oper at ion, incr ease of die r adius (a) has much i nfl uence of punch l oad and i t decr eases (b) does not influence t he punch load much (c) punch load incr eases (d) punch load depends on ot her consider at ions 74. Bending oper at ion should be per for med (a) par allel t o t he gr ain dir ect ion (b) at 30° t o t he gr ain dir ect ion (c) at r ight angle t o t he gr ain dir ect ion (d) t her e is no such cr it er ion 75. I n bending oper at ion, t he met al t akes shape of (a) die (b) punch (c) aver age of t wo (d) could t ake any shape 76. I n bl an k i n g oper at i on , t h e cl ear an ce i s pr ovided on (a) die (b) punch (c) half on die and half on punch (d) may be pr ovided on any member

4.35

77. Most practical method of taking care of springback dur ing bending is t o (a) tr y a sample, make the necessar y adjustment and t r y again (b) punching t he inside cor ner of bend (c) use hollow concave punches (d) under cut t ing t he punch so t hat t he mat er ial is fr ee t o over bend 78. Sher adising is (a) a zinc diffusion pr ocess (b) an oxidising pr ocess used for aluminium and magnesium ar t icles (c) a pr ocess used for maki ng t his phosphat e coat ings on st eel t o act as base (d) t he pr ocess of coat ing of zinc by hot dipping 79. Aver age cut t ing speed in machining mild-st eel by single point t ool of H .S.S. is (a) 10 m / mt (b) 20 m / mt (c) 30 m / mt (d) 40 m / mt 80. Tool life is said t o be over if (a) poor sur face finish is obt ained (b) sudden incr ease in power and cut t ing for ce wit h chat t er ing t ake place (c) over heat ing and fuming due t o fr ict ion st ar t (d) all of t hese 81. Tool life is most affect ed by (a) cut t ing speed (b) t ool geomet r y (c) feed and dept h (d) micr ost r uct ur e of mat er ial being cut 82. Flank wear occur s mainly on (a) nose par t, fr ont r elief face and side r elief face (b) nose par t and t op face (c) cut t ing edges (d) all of t hese 83. Pr oper t y essent ial for a t ool mat er ial used for high speed machining is (a) r ed har dness and impact r esist ance (b) r ed har dness and wear r esist ance (c) t oughness and impact r esist ance (d) r ed har dness, wear r esistance and toughness. 84. M ost machinable met al is one which (a) pr oduces discont inuous chips (b) per mit s maxi mum met al r emoval per t ool grind (c) r esult s in maximum lengt h of shear plane (d) r esult s in minimum value of shear angle

4.36

Manufacturing Engineering

85. I t is possible t o cor r elat e t ool life wit h which of t he following pr oper t y of t he met al ? (a) gr ain size (b) t oughness (c) har dness (d) micr o const it uent s

93. Cr at er wear occur s mainly due t o (a) abr asion (b) diffusion (c) oxidation (d) adhesion

86. The met al in machining oper at ion is r emoved by (a) t ear ing chips (b) dist or t ion of met al (c) shear ing t he met al acr oss a zone (d) cut t ing t he met al acr oss a zone

94. Chips wit h built up edge can be expect ed when machining (a) har d mat er ial (b) br it t le mat er ial (c) t ough mat er ial (d) ductile mat er ial

87. Aver age cut t ing speed in machining cast ir on by a single point cut t ing t ool of H .S.S. is (a) 6 m / mt (b) 11 m / mt (c) 22 / mt (d) 33 m / mt

95. Cr at er wear t akes place in a single point cut t ing t ool at (a) flank (b) side r ake (c) face (d) tip

88. Galvanising is (a) zinc diffusion pr ocess (b) an oxidising pr ocess used for aluminium and magnesium ar t icles (c) a pr ocess used for making t hin phosphat e coat ing on st eel t o act as a base or pr imer for enamels and paint s (d) t he pr ocess of coat ing of zinc by hot dipping 89. The fr ont r ak e r equi r ed t o machi ne br ass by H .S.S. t ool is (a) 15° (b) 10° (c) 5° (d) 0° 90. Best all-r ound coolant for car bide t ool is (a) soluble oil (b) k er osene (c) t ur pent ine oil (d) compr essed air 91. Anodising is (a) a zinc diffusion pr ocess (b) an oxidising pr ocess used for aluminium and magnesium ar t icles (c) a pr ocess used for making t hin phosphat e coat ing on st eel t o act as a base or pr imer for enamels and paint s (d) t he pr ocess of coat ing of zinc by hot dipping 92. Best coolant and lubr icant for st eel and wr ought ir on is (a) wat er sol ubl e oi l or su l ph u r -based an d miner al oils (b) miner al and fat t y oils (c) soluble oils (d) dry

96. Best coolant and lubr icant for cast ir on is (a) wat er sol ubl e oi l s or sul phur based and miner al oils (b) miner al and oils (c) soluble oils (d) dry 97. Which of t he following t ool mat er ials has highest cut t ing speed ? (a) car bon st eel (b) t ool st eel (c) HSS (d) carbide 98. M ain funct ion of t he cut t ing fluid is t o (a) pr ovide lubr icat ion (b) cool t he t ool and wor kpiece (c) wash away t he chips (d) impr ove sur face finish 99. Which of the following is used as cut t ing fluid for t h e t ur n i ng an d m i l l i n g oper at i on on al l oy st eels ? (a) CO2 (b) k er osene (c) soluble oil (d) sulphur ised miner al oil 100. Cont inuous chips will be for med when machining speed is (a) high (b) low (c) medium (d) ir r espect ive of cut t ing speed

Manufacturing Engineering

LEVEL-1 101. I n hot wor k ing pr ocess (a) gr ai n st r uct ur e of t he met al i s r efined (b) por osit y of met al i s lagel y eli minat ed (c) m ech an i cal pr oper t i es such as du ct i l i t y, t ougnes elongat ion and r educt i on in ar ea ar e impr oved (d) al l of t hese 102. Pr ocess of incr easing t he cr oss-sect i on of a bar and r educing i t s lengt h is call ed (a) spinning (b) upset ting (c) dr awing down (d) r eaming 103. I n hot wor k ing pr ocess (a) poor sur face fini sh is pr oduced (b) scal e is for med on met al sur face (c) cl ose t ol er ances can not be mai nt ained (d) al l of t hese 104. Cold wor ki ng (a) r equi r es much hi gher pr essur e t han hot wor king (b) incr eases har dness (c) di st or t gr ain st r uct ur e and does not pr ovi de an appr eciable r educt ion i n si ze (d) al l of t hese 105. Cold wor k ing of met al incr eases (a) t ensile st r engt h (b) har dness (c) yi el d st r engt h (d) al l of t hese 106. Col d wor k i ng pr ocess can be appl i ed on t he component s having diamet er up t o (a) 12 mm (b) 25mm (c) 49mm (d) 50mm 107. Parts of circular cross section which are symmetrical about the axis of rotation are made by (a) hot spi ni ng (b) hot for gi ng (c) hot ext r ust i on (d) hot h per ci ng 108. Plast ic defor mat ion of met al at high temper at ur e i n t o a pr edet er m i n ed si ze or sh ape u si n g compr essi ve for ces exer t ed t hr ough some t ype of di e by a hammer a pr ess or an upset t i ng machine, is call ed (a) ext r usion (b) pier cing (c) for ging (d) casting 109. For ging of st eel i s done at a t emper at ur e of (a) 400°C (b) 800°C (c) 1000°C (d) 1300°C

4.37

110. Pr ocess used for maki ng seaml ess t ube is (a) ext r usion (b) pier cing (c) for ging (d) casting 111. Pr ocess used for maki ng nut s and bolt s is (a) ext r usion (b) cold peeni ng (c) hot pi er ci ng (d) up set t i ng 112. Pr ocess consist s of pushi ng t he met al insi de a chamber t o for ce i t out by high pr essur e t hr ough an or ifice, which is shaped t o pr ovide t he desir ed for m of t he fini shed par t , is call ed (a) ext r usion (b) pier cing (c) for ging (d) swaging 113. Pr ocess used t o impr ove fat igue r esist ance of t he met al by set t i ng up compr essive st r esses in i t s sur face, is call ed (a) ext r usion (b) pircing (c) cold peeni ng (d) up set t i ng 114. Pr ocess, in which t he cr oss-sect ional ar ea of bar s, r ods or t ubes in t he desir ed ar ea is r educed by r epeat ed bl ows, is call ed (a) ext r usion (b) pier cing (c) r eaming (d) swaging 115. Pr ocess of sh api n g t h i n m et al sh eet s by pr ocessing t hem against a for m is call ed (a) spinning (b) upset ting (c) dr awing down (d) r eaming 116. I n r ol l i ng oper at i ons, t he r ol l r ot at es wi t h a sur face veloci t y (a) lower t han t he speed of incoming met al (b) exceedi ng t he speed of i ncomi ng met al. (c) equal t o speed of t he i ncoming met al (d) none of t hese

4.38

Manufacturing Engineering

117. H ot pr ess for gi ng (a) casues a st eadil y applied pr essur e i nst ead of impact for ce (b) is used t o for ce t he end of a heat ed bar int o a desi r ed shape (c) is a for ging oper at ion i n whi ch t wo halves of a r ot at ing die open and cl ose r api dl y whi le impact i ng t he end of t he heat ed t ube or shell (d) i s a f or gi n g m et h od f or r ed u ci n g t h e di met er of a bar and in t he pr ocess making it longer 118. I n hot wor ki ng (a) anneal ing oper at ion is not necessar y (b) power r equir ement s ar e l ow (c) sur face finish i s good (d) gr ai n r efinement is possible 119. For ging of plain cor bon st eel i s car r i ed out at (a) 750°C (b) 900°C (c) 1100°C (d) 1300°C 120. Swaging (a) causes a st eadil y applied pr essur e i nst ead of impact for ce (b) is used t o for ce t he end of a heat ed bar int o a desi r ed shape (c) is a for ging oper at ion i n whi ch t wo halves of a r ot at ing die open and cl ose r api dl y whi le impact i ng t he end of t he heat ed t ube or shell (d) i s a f or gi n g m et h od f or r ed u ci n g t h e di met er of a bar and in t he pr ocess making it longer 121. M echani cal pr oper t i es of t he met al i mpr ove in hot wor ki ng due t o (a) r ecover y of gr ai ns (b) r ecr yst allisat ion (c) gr ai n gr owt h (d) refinement of grain size 122. Roll for gi ng (a) causes a st eadil y applied pr essur e i nst ead of impact for ce (b) is used t o for ce t he end of a heat ed bar int o a desi r ed shape (c) is a for ging oper at ion i n whi ch t wo halves of a r ot at ing die open and cl ose r api dl y whi le impact i ng t he end of t he heat ed t ube or shell (d) i s a f or gi n g m et h od f or r ed u ci n g t h e maki ng it longer

123. Pr oduct ion of cont our s in fl at blank is call ed (a) blanking (b) pier cing (c) per for at ing (d) punching 124. Upset for gi ng (a) causes a st eadil y applied pr essur e i nst ead of impact for ce (b) is used t o for ce t he end of a heat ed bar int o a desi r ed shape (c) is a for ging oper at ion i n whi ch t wo halves of a r ot at ing die open and cl ose r api dl y whi le impact i ng t he end of t he heat ed t ube or shell (d) is a for ging method for r educing the dimeter of a bar and in the pr ocess making it longer 125. I n four high r olling mill the bigger rollers are called (a) guide r ol ls (b) back up r ol ls (c) main r ol ls (d) suppor t r ol ls 126. A polished and et ched sur face of the cr oss-sect ion of a hot wor ked pr oduct wi ll be havi ng (a) fi br e lik e st r uct ur e (b) gr ai n field li ke st r uct ur e (c) car bon pr ecipit at ed at boundar i es (d) car bon in t he for m of fl ak es 127. Not ching is t he oper at ion of (a) r emoval of excess met al fr om t he edge of st r ip t o m ak e i t su i t abl e f or dr aw i n g w i t h ou t wr inkling (b) cut t i ng of t he excess met al at edge which was r equir ed for gr i ppi ng pur pose dur ing pr ess wor k ing oper at i on (c) cut t ing i n a single l ine acr oss a par t of t he met al st r ip t o allow bendi ng or for ming in pr ogr essi ve di e oper at i on whi l e t he par t r emains at t ached t o t he st r ip (d) punching in which punch is st opped as soon as t he met al fr act ur e is complet e and met al is not r emoved but hel d in hole 128. L ancing i s t he oper at ion of (a) r emoval of excess metal fr om the edge of str ip t o mak e i t sui t abl e for dr awi ng wi t hout wr inkling (b) cut t i ng of t he excess met al at edge which was r equir ed for gr i ppi ng pur pose dur ing pr ess wor k ing oper at i on (c) cut t ing i n a single l ine acr oss a par t of t he met al st r ip t o allow bendi ng or for ming in pr ogr essi ve di e oper at i on whi l e t he par t r emains at t ached t o t he st r ip (d) punching in which punch is st opped as soon as t he met al fr act ur e is complet e and met al is not r emoved but hel d in hole

Manufacturing Engineering

129. L aser is pr oduced by (a) gr aphite (b) r uby (c) diamond (d) emer ald 130. I n j ect i on m ou l di n g i s t h e i deal m et h od of pr ocessing (a) plastics (b) t her mo-set t i ng plast ics (c) t her moplast ics (d) nonfer r ous mat er ials 131. Compr essi on moul di ng i s t he ideal met hod of pr ocessing (a) plastics (b) t her mo-set t i ng plast ics (c) t her moplast ics (d) nonfer r ous mat er ials 132. Sl ugging is t he oper at ion of (a) r emoval of excess metal fr om the edge of str ip t o mak e i t sui t abl e for dr awi ng wi t hout wr inkling (b) cut t i ng of t he excess met al at edge which was r equir ed for gr i ppi ng pur pose dur ing pr ess wor k ing oper at i on (c) cut t ing i n a single l ine acr oss a par t of t he met al st r ip t o allow bendi ng or for ming in pr ogr essi ve di e oper at i on whi l e t he par t r emains at t ached t o t he st r ip (d) punching in which punch is st opped as soon as t he met al fr act ur e is complet e and met al is not r emoved but hel d in hole 133. I n dr awing oper at ion t he met al flows due t o (a) duct ilit y (b) wor k har deni ng (c) plast icity (d) shear ing 134. Tr imming is t he oper at ion of (a) r emoval of excess metal fr om the edge of str ip t o mak e i t sui t abl e for dr awi ng wi t hout wr inkling (b) cut t i ng of t he excess met al at edge which was r equir ed for gr i ppi ng pur pose dur ing pr ess wor k ing oper at i on (c) cut t ing i n a single l ine acr oss a par t of t he met al st r ip t o allow bendi ng or for ming in pr ogr essi ve di e oper at i on whi l e t he par t r emains at t ached t o t he st r ip (d) punching in which punch is st opped as soon as t he met al fr act ur e is complet e and met al is not r emoved but hel d in hole

4.39

135. H emming i s t he oper at i on (a) in which t he edges of sheet ar e t ur ned over t o pr ovide st i ffness and a smoot h edge (b) of pr oducing cont r ous in sheet met al end of bending pr eviously r oll for med sect ions (c) i n w h i ch a ser i es of i m pact bl ow s ar e t r ansfer r ed on dies so t hat solid or t ubul ar wor k changes i n cr oss sect ion or geomet r ic shape (d) empl oyed t o expand a t ubul ar or cyli ndr ical part 136. Bulging i s t he oper at i on (a) in which t he edges of sheet ar e t ur ned over t o pr ovide st i ffness and a smoot h edge (b) of pr oducing cont r ous in sheet met al end of bending pr eviously r oll for med sect ions (c) i n w h i ch a ser i es of i m pact bl ow s ar e t r ansfer r ed on dies so t hat solid or t ubul ar wor k changes i n cr oss sect ion or geomet r ic shape (d) empl oyed t o expand a t ubul ar or cyli ndr ical part 137. Gear ar e shapi ng is r el at ed t o (a) template (b) for m t oot h pr ocess (c) hob (d) gener at ing 138. Gear s best mass pr oduced by (a) milling (b) hobbin (c) shapping (d) for ming 139. Wh i ch of t h e f ol l ow i n g i s a gear f i n i sh i n g oper at i on ? (a) Hobbing (b) Shaping (c) Milling (d) Saving or bur nishi ng 140. I n pr ess oper at oin, t he si ze of pi er ced hole i s independent of t he size of (a) punch (b) die (c) aver age of punch and die (d) punch and clear ance 141. For dr awi ng oper at i on, t he best suit ed pr ess is (a) knuckle joint pr ess (b) cr ank shaft and connect ing r od pr ess (c) t oggle pr ess (d) r ack and pinion pr ess

4.40

Manufacturing Engineering

142. Swaging i s t he oper at oin (a) in which t he edges of sheet ar e t ur ned over t o pr ovide st i ffness and a smoot h edge

148. Smal lest t hi ck ness whi ch can be measur ed by sl ip gauges is (a) 0.001 mm

(b) of pr oducing cont r ous in sheet met al end of bending pr eviously r oll for med sect ions

(b) 0.01 mm

(c) i n w h i ch a ser i es of i m pact bl ow s ar e t r ansfer r ed on dies so t hat solid or t ubul ar wor k changes i n cr oss sect ion or geomet r ic shape

(d) none of t hese

(d) empl oyed t o expand a t ubul ar or cyli ndr ical part 143. Blanking and piercing oper ation can be per for med simulat aneously in (a) si mple die (b) pr ogr essi ve die

(c) 1.001 mm 149. Which machi ne par t is not cold for med ? (a) Food cont ainer (b) St ai nl ess st eel vessel (c) Cr ank shaft (d) H eat i ng duct 150. Odd pair is (a) Dimpling and flar ing (b) Welding and solder ing

(c) compound die

(c) Thr eading and bor ing

(d) combinat i on die

(d) H obbing and swaging

144. Cut t ing and for ming oper at ions can be done in si ngle oper at ion on (a) si mple die (b) pr ogr essi ve die (c) compound die (d) combinat i on die 145. St r et ch for ming is t he oper at i on (a) in which t he edges of sheet ar e t ur ned over t o pr ovide st i ffness and a smoot h edge (b) of pr oducing cont r ous in sheet met al end of bending pr eviously r oll for med sect ions (c) i n w h i ch a ser i es of i m pact bl ow s ar e t r ansfer r ed on dies so t hat solid or t ubul ar wor k changes i n cr oss sect ion or geomet r ic shape (d) empl oyed t o expand a t ubul ar or cyli ndr ical part 146. I n pr ess oper at i on, t he si ze of t he blanked par t is dependent on t he size of (a) punch (b) die (c) aver age of punch and die (d) di e and clear ance 147. A mast er gauge i s used by (a) mast er s (b) exper ienced t echnician (c) al l machines (d) none of t hes

151. Effect t hat is associ at ed wi t h cold for ming is (a) st r ain har dening (b) shr inkage (c) sur face decolour i ng (d) sur face r oughness 152. Cl ose t ol er an ce of d i m en si on s i s easi l y maint ained in cold for med par t s because (a) di es ar e of exact diment i on (b) no shr i nkage occur s (c) st r engt h incr eases (d) hi gher defor mi ng for ce i s appli ed 153. I f t her e ar e bad effect s of st r ai n har dening on a cold for med par t , t he par t must be (a) anealed (b) t emper ed (c) har dened (d) nor malized 154. Advant age of col d for ming is (a) gr ai n r efinement t akes place (b) st r engt h and har dness i ncr ease (c) no consequent heat t r eat ment is needed (d) for ce r equi r ed i s r el at ively small 155. St r et ch for ming is a pr ocess in whi ch (a) al l defor mat i ons occur i n t he di r ect i on of st r et ch (b) all for ces ar e applied int he dir ect ion of str et ch (c) advant age is t aken of pl ast ic st at e induced due t o st r et ch (d) no dies ar e used

Manufacturing Engineering

156. Cor r ect combi nat ion in a col d bendi ng pr ocess is

163. Pr e-bendi ng is not possibl e in

(a) t hicker met al , small er bend angle, small er bend r adius

(a) t hr ee r ol l single pinch machi ne

(b) har der metal, smaller bend angle, lar ger bend radius

(c) four r oll double pi nch machi ne

(c) thinner metal, smaller bend angle lar ger bend radius (d) thicker metal, lar ger bend angle, smaller bend radius 157. Ver nier cal iper is used t o measur e (a) ext er nal and i nt er nal di amet er of shaft s (b) t hickness of par t s (c) dept h of hol es (d) al l of t hese 158. Bending pr ocess r equi r es highest for ce is (a) Bot t om bendi ng (b) Thr ee point bendi ng (c) Air bending (d) none of t hese 159. Bot t om of bend (t ensi on si de) does not mak e cont act wit h t he di e in (a) bot t om bendi ng

(b) t hr ee r ol l double pinch machi ne (d) pyr amid machine 164. Roll ing machine is amenabl e t o NC CN C is (a) pyr amid machine (b) t hr ee r ool single pinch machi ne (c) four r oll double pi nch machi ne (d) t hr ee r ool double pinch machi ne 165. Bar of sect ion can be col d bent by (a) r oll bendi ng (b) r ot at ing die and wi per block (c) ser i es of r ol ls (d) mat ching shape wheels 166. Dr awing pr ocess does not bel ong t o t he gr oup is (a) deep dr awing (b) stamping (c) pr essing (d) shallow dr awing 167. A cylindr ical vessel wi t h fl at bot t om can be deep dr awn by

(b) ai r bendi ng

(a) si ngle act ion deep dr awi ng

(c) t hr ee point bendi ng

(b) double act ion deep dr awi ng

(d) al l of t hese

(c) t r iple act ion deep dr awi ng

160. I n which cold bendi ng pr ocess, one set of punch and die can pr oduce only one angl e of bendi ng ? (a) Air bending (b) Thr ee point bendi ng (c) Bot t om bendi ng (d) None of t hese 161. Fl at t ening is a pr ocess in which a met al st r i p is (a) bent over 18° and t hen pr essed (b) pr essed t o r emove ki nks and wr inkl es (c) bent t o cr eat e a smal l ki nk (d) pr essed against r ubber pad 162. I n flexible pr ess br ake die (a) any ben d angl e i r r espect i ve of pun ch i s obtained (b) good sur face finish on t ensi on side of bend is ensur ed (c) smal ler bending for ce is r equir ed (d) none of t hese

4.41

(d) shallow dr awing 168. Ri ng r oll ing i s used (a) for pr oduci ng a seamless t ube (b) t o i ncr ease t he t hi ck ness of a r i ng (c) t o decr ease t he t hi ck ness of a r i ng (d) for pr oducing lar ge cyli nder 169. I n r ing r ol li ng (a) inner r ol ler i s r oll er is power (b) out er r ol ler i s r oll er is power

lar ger in di amet er but out er dr iven lar ger in di amet er but i nner dr iven

(c) bot h r oll er s ar e equal in di amet er but i nner r oll er is power dr iven (d) lar ger out er r ol ler i s power dr iven 170. Cold or hot r oll ing does not pr oduce (a) a hollow ci r cular sect i on (b) a T sect i on (c) an I sect i on (d) a channel sect i on

4.42

Manufacturing Engineering

171. Pr ocess of ext r usion is li ke (a) a vi scous l ubr i can t pour i ng t hr ou gh t he mout h of cont ainer (b) a t oot h past e comi ng out of it s t ube (c) har d par t icl es t hr own out of a nozzl e under ai r pr essur e (d) abr asive par t icle in t he for m of slur r y coming out of an openi ng 172. A t oot h past e t ube can be pr oduced by (a) solid for war d ext ur si on (b) sol id backwar d ext r usion (c) holl ow back war d ext r usi on (d) holl ow for war d ext r usi on 173. Seamless t ube can be pr oduced by (a) t wo high r ol li ng mi ll (b) r i ng r oll ing combined wit h st r et ch for mi ng (c) pier cing (d) st eam hammer for gi ng 174. Swaging i s opposit e of (a) (b) (c) (d)

for ging ext r usion pier cing none of t hese

175. Pr ocess pr oducing gr ain st r uct ur e wi t h gr ai ns aligned along geomet r ical shape of cr ank shaft is (a) casting (b) r olling (c) welding (d) none of t hese 176. Pr ocess cannot be used for pr oducing a cr ank shaft is (a) casting (b) r olling (c) welding (d) bending 177. A poppet val ue can be pr oduced by (a) r olling (b) swaging (c) combind for gi ng and ext r usi on (d) st r et chi ng 178. Needle is pr oduced by (a) swaging (b) ext r usion (c) machining (d) for ging

179. M et al not good for impact ext r usion is (a) al loys of Zn and Sn (b) st ai nl ess st eel (c) low car bon annealed st eel (d) al loys of al uminium and lead 180. M at er ial good for ext r usion is (a) st ai nl ess st eel (b) br ass cast i ng (c) low car bon annealed st eel (d) low car bon wor k har dened st eel 181. I n w h i ch f or gi n g m ach i n e an v i l on w h i ch wor k pi ece is pl aced moves t owar d descending punch ? (a) Boar d dr op hammer (b) Ai r li ft hammer (c) Tr ip hammer (d) H i gh ener gy r at e for ging machi ne 182. Upset t i ng or col d headi ng is a (a) r oll ing pr ocess (b) ext r uding pr ocess (c) bendi ng pr ocess (d) for ging pr ocess 183. For ext r usion pr ocess (a) complex sect oins ar e pr oduced fr om bar stocks (b) good sur face fi ni sh and cl ose t ol er ance i s gener at ed (c) t he st r engt h of fi nished pr oduct is impr oved due t o col d wor ki ng (d) al l of t hese 184. For ming pr ocess which does not involve r ot at oin of wor k piece is (a) spinning (b) t hr ead r ol li ng (c) upset ting (d) r i ng r ol li ng 185. Di ffer ent st eps in Bl ock ing, which i s fi nishi ng oper at i on i n f or gi n g ar e i n t h e f ol l ow i n g oper at i on i n for ging ar e i n t he foll owing or der (a) coining, t r imming, planishi ng (b) t r immi ng, planishi ng, coi ni ng (c) pl anishing, coining, t r immi ng (d) pl anishing, t r immi ng, coi ni ng 186. The oper at ion t hat r emoves fins and flashes fr om a for ged par t is (a) combi nat i on of t r i mmi ng, pl ani shi ng and coining (b) combinat ion of t r i mming and planishi ng (c) combi nat ion of planishing and coini ng (d) t r imming

Manufacturing Engineering

187. A hack saw blade cut on (a) for war d st r oke (b) r et ur n st r oke (c) bot h (a) and (b) (d) depends upon dir ect ion of for ces 188. The for ge hammer s used for pl ani shi ng and coining ar e (a) Ai r lift hammer s (b) H elve and t r i p hammer s (c) Boar d dr op hammer s (d) St eam hammer s 189. The major pr oblem i n hot ext r usion is (a) desi gn of punch (b) desi gn of die (c) wear and t ear of die (d) wear of punch 190. Ext ur sion pr ocess can effet i vely r educe t he cost of pr oduct t hr ough (a) mat er ial saving (b) pr ocess i n t ooli ng cost (c) savi ng in t ooling cost (d) savi ng in admini st r at ive cost

4.43

195. H ack saw blade pr efer r ed for cut t ing br ass have number of t eet h (a) 14 per inch (b) 25 per inch (c) 32 per inch (d) 40 per inch 196. Teet h of hacksaw blades ar e bend (a) al t er nat ely r i ght or left and ever y t hir d or four t h left st r aight (b) t owar ds r ight (c) t owar ds left (d) none of t hese 197. A fi le wi t h 20 t eet h per inch is known as (a) r ough fi le (b) smoot h fi le (c) bast ar d file (d) second cut fi le 198. A fi le r emoves met al dur ing i t s (a) for war d st r ock (b) r et ur n st r ok (c) bot h (a) and (b) (d) none of t hese

191. I n a soli d ext r usi on die, pur pose of k nock out pin is (a) shopping the par t to ext r ude t hr ough the hose (b) eject ing t he par t aft er ext r usi on (c) allowing t he job to have ber tt er sur face finish (d) r educing t he wast e of mat er i al

199. Fi le used for wood wor k is (a) si ngle cut fi le (b) double cut fi le (c) r asp cut fi le (d) none of t hese

192. Widt h of slot cut by a hacksaw blade as compar ed t o widt h of bl ade is (a) mor e (b) less (c) equal (d) none of t hese

201. When fi le i s moved t o and fr o over t he wor k i t is called (a) dr aw filing (b) cr oss fi li ng (c) pull nd push filing (d) none of t hese

193. L engt h of hack saw blade is measur ed fr om (a) cent er of hole at one end t o t he cent er of hole at t he ot her end (b) ext r eme end t o ext r eme end (c) 16 t imes t he widt h (d) none of t hese

202. Scr i bing bl ock i s used t o (a) locat e t he cent er of r ound bar s (b) check t he t r ueness of fl at sur faces (c) check t he accur ay of a li mit gauge (d) check t he diamet er of shaft s or st uds

194. H ack saw blade is specifi ed by i t s (a) width (b) length (c) mater ial (d) number of t eet h

200. Sl ip gauges ar e made of (a) cast ir on (b) aluminium (c) al loy st eel (d) copper

203. Sur face plat e i s used t o (a) locat e t he cent er of r oung bar s (b) check t he t r ueness of fl at sur faces (c) check t he accur acy of a li mi t gauge (d) check t he diamet er of shaft s or st uds

4.44

Manufacturing Engineering

204. Sl ip gauge i s used t o (a) locat e t he cent er of r ound bar s (b) check t he t r ueness of fl at sur faces (c) check t he accur acy of a li mi t gauge (d) check t he diamet er of shaft s or st uds 205. Chisel s used for met al cut t i ng ar e (a) har dened (b) annealed (c) t emper ed (d) al l of t hese

LEVEL-2 206. A single V and single U-but t welds ar e for sheet s of appr oximat e t hickness (a) 1 t o 5 mm (b) 5 t o 15 mm (c) 15 t o 25 mm (d) mor e t hat 25 mm 207. For welding plat es of t hickness less t han 5 mm, bevelling of it s edges is (a) done t o a singe V or U-gr oove (b) done to a double V or U-gr oove on one side (c) done to a double V or U-groove on both sides (d) not r equir ed

212. I n el ect r i cal r esi st ance wel di ng, cl eani ng of met als t o be welded is (a) immat er ial (b) impor tant (c) not effect ing t he welding (d) none of t hese 213. I n electr ical r esistance welding, when the cur r ent passes thr ough t he metal, the gr eatest r esistance is at t he (a) sur face (b) point of cont act of elect r ode and met al (c) point of cont act of met als t o be welded (d) none of t hese 214. Welding of chr omium molybdenum st eels cannot use (a) t her mit welding (b) elect r ical r esist ance welding (c) oxyacet ylene welding (d) all of t hese 215. I n elect r ical r esist ance welding (a) volt age is high and cur r ent is low (b) volt age is low and cur r ent is high (c) bot h volt age and cur r ent ar e high (d) bot h volt age and cur r ent ar e low

208. For welding plat es of t hickness mor e t han 12.5 mm, bevelling of it s edges is (a) done t o a singe V or U-gr oove (b) done t o a double V or U-gr oove on one side (c) done t o a double V or U-gr oove on bot h sides (d) not r equir ed

216. I n elect r ical r esist ance welding voltage, r equir ed for heat ing is in t he r ange (a) 1 t o 5 volt s (b) 6 t o 10 volt s (c) 11 t o 15 volt s (d) 16 t o 20 volt s

209. I n single V-but t welds, t he angle bet ween edges is kept about (a) 40 t o 50 (b) 50 t o 60 (c) 60 t o 70 (d) 70 t o 90

217. I n electr ical r esist ance welding, pr essur e applied var ies in t he r ange (a) 50 t o 100 kgf/cm 2 (b) 100 t o 200 kgf/cm 2 (c) 250 t o 550 kgf/cm 2 (d) 600 t o 900 kgf/cm 2

210. I n a welding a flux is used t o (a) per mit per fect cohesion of met als (b) r emove t he oxides of t he met als for med at high t emper at ur e (c) bot h (a) and (c) (d) none of t hese 211. Pl ai n and but t wel ds can be used for sheet shaving t hickness appr oximat ely (a) 25 mm (b) 40 mm (c) 50 mm (d) 100 mm

218. I n el ect r i c r esi st an ce w el di n g, t w o copper elect r odes used ar e cooled by (a) water (b) air (c) bot h (a) and (b) (d) none of t hese 219. Amount of cur r ent r equir ed in electr ic r esistance welding is r egulat ed by changing (a) input supply (b) pr imar y t ur ns of t he t r ansfor mer (c) secondar y t ur ns of t he t r ansfor mer (d) all of t hese

Manufacturing Engineering

220. Wel di ng in which t he met als t o be joi ned ar e heat ed t o a molt en st at e and allowed t o solidify in pr esence of a filler mat er ial, is called (a) plast ic welding (b) fusion welding (c) t her mit welding (d) none of t hese 221. An example of fusion welding is (a) ar c welding (b) gas welding (c) t her mit welding (d) for ge welding 222. Welding in which met als t o be joined ar e heat ed t o t he plast ic st at e and t hen for ged t oget her by ext er nal pr essur e wit hout t he addit ion t o filler mat er ial, is called (a) plast ic welding (b) fusion welding (c) t her mit welding (d) none of t hese 223. An example of plast ic welding is (a) ar c welding (b) gas welding (c) for ge welding (d) all of t hese 224. Spot -wel di ng, pr oj ect i on wel di n g an d seam welding ar e classificat ion of (a) elect r ic r esist ance welding (b) t her mit welding (c) ar c welding (d) for ge welding 225. Welding pr ocess in which t wo pieces t o be joined are over lapped and placed between two electrodes, is called (a) spot welding (b) pr oject ion welding (c) seam welding (d) but t welding 226. Spot welding is used for welding t op joint s in plat es having t hickness (a) 0.25 mm t o 1.25 mm (b) 1.25 mm t o 2.50 mm (c) 2.5 mm t o 3 mm (d) mor e t han 3 mm 227. M at er ial difficult t o be spot welded is (a) st ainless st eel (b) copper (c) mild st eel sheet (d) all of t hese

4.45

228. N umber of zones of heat gener at i on i n spot welding ar e (a) 2 (b) 3 (c) 5 (d) 8 229. I n spot welding, t he t ip of elect r odes is of (a) st ainless st eel (b) aluminium (c) copper (d) brass 230. I n spot welding, spacing bet ween t wo spot welds should not be less t han (a) d (b) 1.5 d (c) 3 d (d) 4.5 d wher e d is elect r ode t ip diamet er 231. I n spot welding, t he elect r ode t ip diamet er (d) should be equal t o (a)

(b) 1. 5 t t (c) 3 t (d) 4.5 t wher e t is t hickness of plat e t o be welded. 232. M ult ispot welding pr ocess is (a) seam welding (b) pr oject ion welding (c) t her mit welding (d) per cussion welding 233. A cont inuous spot welding pr ocess is (a) seam welding (b) pr oject ion welding (c) t her mit welding (d) per cussion welding 234. I f t wo pieces of differ ent met als ar e t o be welded by pr oject ion welding, t hen pr oject ion should be made on t he met al piece having (a) higher conduct ivit y (b) lower conduct ivit y (c) same conduct ivit y (d) none of t hese 235. Welding pr ocess used for joining r ound bar s is (a) spot welding (b) pr oject ion welding (c) seam welding (d) none of t hese 236. Seam wel di ng used for met al sheet s havi ng t hickness in t he r ange (a) 0.025 t o 3 mm (b) 3 t o 6 mm (c) 6 t o 10 mm (d) mor e t han 10 mm

4.46

Manufacturing Engineering

237. I n pr oject ion welding, diamet er of t he pr oject ion as com par ed t o t h i ck n ess of t h e sh eet i s appr oximately (a) half (b) equal (c) 1.5 t imes (d) double 238. Welding of st eel st r uct ur es on sit e of a building is done by (a) spot welding (b) pr oject ion welding (c) seam welding (d) ar c welding 239. I n ar c welding, elect r ic ar c is pr oduced bet ween t he wor k and t he elect r ode by (a) cur r ent flow (b) volt age differ ence (c) cont act r esist ance (d) all of t hese 240. I n ar c welding, t emper at ur e of heat of ar c is in t he r ange (a) 1000C to 2000C (b) 2000C to 4000C (c) 4000C to 6000C (d) 6000C to 7000C 241. I n ar c welding, cur r ent used is (a) A.C. cur r ent of high fr equency (b) A.C. cur r ent of low fr equency (c) D. C. cur r ent (d) all of t hese 242. I n ar c wel di ng, wi t h i ncr ease of t hi ck ness of mat er i al t o be wel ded, t he wel di ng cur r ent r equir ement (a) incr eases (b) decr eases (c) r emains same (d) none of t hese 243. I n ar c welding, t oo low welding speed r esult s in (a) wast age of elect r ode (b) excessive pilling up of weld met al (c) over hauling wit hout penet r at ion edges (d) all of t hese 244. Welding pr efer r ed for joining t hin foils is (a) gas welding (b) x-r ays and gamma r ays (c) micr o waves (d) all of t hese 245. Tr ansfer of met al fr om t he elect r ode occur s due t o molecular at t r act ion in (a) gas welding (b) D.C. ar c welding (c) metallic welding (d) t her mit welding

246. Welding pr ocess using a pool of molt en met al is (a) car bon ar c welding (b) submer ged ar c welding (c) TI C welding (d) MI G welding 247. Welding pr ocess in which flux is used in t he for m of gr anules is (a) gas welding (b) D.C. ar c welding (c) submer ged ar c welding (d) t her mit welding 248. An arc is pr oduced between a bar e metal electr ode and t he wor k in (a) gas welding (b) D.C. ar c welding (c) submer ged ar c welding (d) t her mit welding 249. Welded rod coated with fluxing mater ial is used in (a) gas welding (b) shielded ar c welding (c) D.C ar c welding (d) ar gon ar c welding 250. I n ar c welding, eyes need t o be pr ot ect ed against (a) int ense glar e (b) spar ks (c) infr a-r ed r ays (d) infr a-r ed and ult r a violet r ays. 251. I n which t ype of welding, a pool of molt en met al is used ? (a) Electr oslag (b) Submer ged ar c (c) MIG (d) TI G 252. Plain and but t welds may be used on mat er ials upto appr oximately (a) 25 mm t hick (b) 40 mm t hick (c) 50 mm t hick (d) 60 mm t hick 253. M ain cr it er ion for select ion of electr ode diameter in ar c welding is (a) mat er ials t o be welded (b) t ype of welding pr ocess (c) t hickness of mat er ial (d) volt age used

Manufacturing Engineering

254. Which of t he following cur r ent is pr efer r ed for welding of non-fer r ous met als by ar c welding ? (a) A.C. low fr equency (b) A.C. high fr equency (c) D.C. (d) All of t hese7 255. I n ar c welding, air is cr eat ed bet ween elect r ode and t he wor k by (a) flow of cur r ent (b) volt age (c) mat er ial char act er ist ics (d) cont act r esist ance 256. Open ci r cui t volt age for ar c wel di ng i s of t he or der of (a) 18 - 40 volt s (b) 40 - 95 volt s (c) 100 - 125 volt s (d) 130 - 170 volt s 257. M at er ial used for coat ing t he elect r ode is called (a) pr ot ect ive layer (b) binder (c) slag (d) flux 258. Plug weld joint is used (a) wher e longit udinal shear is pr esent (b) wher e sever e loading is encount er ed and t he upper sur faces of bot h pieces must be in t he same plane (c) t o j oi n t wo pi eces of met al s i n t he same manner as r ivet s join met als (d) t her e is not hing like plug weld joint 259. Welding pr ocess using non-consumable electr odes is (a) L aser welding (b) MI G welding (c) TI G welding (d) lon beam welding 260. When welding is going on, ar c volt age is of t he or der of (a) 18 - 40 volt s (b) 40 - 95 volt s (c) 100 - 125 volt s (d) 130 - 170 volt s 261. Gases used in t ungst en iner t gas welding ar e (a) hydr ogen and oxygen (b) CO2 and H 2 (c) ar gon and neon (d) ar gon and helium

4.47

262. T-joint weld is used (a) wher e longit udinal shear is pr esent (b) wher e sever al loading is encount er ed and the upper sur faces of bot h pieces must be in t he same plane (c) t o joi nt t wo pi eces of met al s i n t he same manner as r ivet s join met als (d) t o join t wo pieces per pendicular ly 263. I n M I G welding (a) no flux is r equir ed (b) welding speeds is high (c) incr eased cor r osion r esist ance (d) even unclean sur face can be welded t o obt ain sound welds 264. Copper is (a) easily spot welded (b) ver y difficult t o be spot welded (c) good for spot welded as any ot her mat er ial (d) pr efer r ed t o be welded by spot welding 265. I t is not possible t o ar c weld all t ypes of met als wit h t r ansfor mer set because it does not have pr ovision for (a) cont r ol of cur r ent (b) cont r ol of volt age (c) cont r ol of t ime dur at ion (d) change of polar it y 266. Two sh eet s of sam e m at er i al but di f f er en t t hickness can be but t welded by (a) adjust ing t he cur r ent (b) t ime dur at ion of cur r ent (c) pr essur e applied (d) changing t he size of one elect r ode 267. Pr oject ion welding is (a) mult i-spot welding pr ocess (b) cont inuous spot welding pr ocess (c) used t o for m mesh (d) used t o make cant ilever s 268. H alf cor ner weld is used (a) wher e longit udinal shear is pr esent (b) t o j oi n t wo pi eces of met al s i n t he same manner as r ivet s join met als (c) wher e efficiency of joint should be 50% (d) none of t hese 269. I n r esistance welding, volt age used for heat ing is (a) 1 V (b) 10 V (c) 100 V (d) 1000 V

4.48

Manufacturing Engineering

270. I n r esist ance welding, t he pr essur e is r eleased (a) just at t he t ime of passing t he cur r ent (b) just aft er complet ion of cur r ent (c) aft er t he weld cools (d) dur ing heat ing per iod 271. Gr ey cast ir on is best welded by (a) TI G (b) ar c (c) MIG (d) oxy-acetylene 272. Seam welding is (a) mult i-spot welding pr ocess (b) cont inuous spot welding pr ocess (c) used t o for m mesh (d) none of t hese 273. Upt o what t hickness of plat e, edge pr epar at ion for welding is not r equir ed ? (a) 4 mm (b) 5 mm (c) 8 mm (d) 10 mm 274. Pr eheat ing is essent ial in welding (a) high speed st eel (b) st ainless st eel (c) cast ir on (d) ger man silver 275. Gr ey cast ir on is usually welded by (a) gas welding (b) r esist ance welding (c) ar c welding (d) all of t hese 276. Br azing met als and alloy commonly used ar e (a) copper (b) copper alloy (c) silver alloys (d) all of t hese 277. For ge welding is best suit ed for (a) st ainless st eel (b) high car bon st eel (c) cast ir on (d) wr ought ir on 278. Two sheet s of di ffer ent mat er i al s but same t hickness can be spot welded by (a) adjust ing t he cur r ent (b) t ime dur at ion of cur r ent (c) adjust ing t he pr essur e applied (d) changing t he size of neelect r ode 279. Wel di n g pr ocess, i n wh i ch el ect r ode i s n ot consumed is (a) t ungst en-iner t gas (TI G) welding (b) MI G welding (c) D.C. ar c welding (d) ar gon ar c welding

280. A non-consumable elect r ode is used in (a) gas welding (b) D.C. welding (c) A.C. ar c welding (d) ar gon ar c welding 281. A consumable elect r ode is used in (a) TI G welding (b) MI G welding (c) submer ged ar c welding (d) car bon ar c welding 282. Vol t age dr op acr oss t he ar c bet ween t he t wo elect r odes known as ar c volt age, incr eases as t he ar c lengt h (a) incr eases (b) decr eases (c) r emains same (d) none of t hese 283. I n ar c welding using dir ect cur r ent , amount of u sef u l ar c h eat at t h e an ode an d cat h ode r espect ively ar e (a) one t hir d and t wo t hir d (b) t wo t hir d and one t hir d (c) equal (d) none of t hese 284. I n ar c welding using A.C. amount of useful heat at t he anode and cat hode r espect ively ar e (a) one t hir d and t wo t hir d (b) t wo t hir d and one t hir d (c) equal (d) none of t hese 285. I n TI G ar c welding, t he welding zone is shielded by an at mospher e of (a) helium gas (b) ar gon gas (c) eit her (a) or (b) (d) none of t hese 286. TI G welding is pr efer r ed for (a) mild st eel (b) aluminium (c) silver (d) all of t hese 287. Welding pr ocess pr efer red for cutting and welding for non-fer r ous met als is (a) iner t gas ar c welding (b) MI G welding (c) submer ged ar c welding (d) D.C. ar c welding

Manufacturing Engineering

288. When t he wor k i s connect ed t o t he negat i ve t er minal and t he elect r ode holder t o t he posit ive t er minal, t he welding set up is said t o have (a) str aight polar it y (b) r ever sed polar it y (c) both (a) and (b) (d) none of t hese 289. When t he wor k i s connect ed t o t he posi t i ve t er minal and t he elect r ode holder t o t he negat ive t er minal, t he welding set up is said t o have (a) str aight polar it y (b) r ever sed polar it y (c) bot h (a) or (b) (d) none of t hese 290. I n welding copper alloys wit h TI G ar c welding (a) dir ect cur r ent wit h r ever se polar it y (DCRP) is used (b) dir ect cur r ent wit h st r aight polar it y (DCSP) is used (c) A. C. is used (d) all of t hese 291. I n welding magnesium wit h TI G ar c welding (a) dir ect cur r ent wit h r ever se polar it y (DCRP) is used (b) dir ect cur r ent wit h st r aight polar it y (DCSP) is used (c) A.C. is used (d) all of t hese 292. I n welding aluminium wit h TI G ar e welding (a) dir ect cur r ent wit h r ever se polar it y (DCRP) is used (b) dir ect cur r ent wit h st r aight polar it y (DCSP) is used (c) A.C. is used (d) all of t hese 293. I n ar c welding, penet r at ion is deeper for (a) DCRP (b) DCSP (c) A.C. (d) none of t hese 294. I n ar c welding, penet r at ion is minimum for (a) DCRP (b) DCSP (c) A.C. (d) none of t hese 295. Welding pr ocess, in which heat is pr oduced for welding by chemical r eact ion, is called (a) gas welding (b) t her mit welding (c) for ge welding (d) r esist ance welding

4.49

296. M olt en met al is pour ed for joining t he met als in (a) ar c welding (b) t her mit welding (c) gas welding (d) TI G welding 297. I n t her mit welding, heat is gener at ed fr om (a) chemical r eact i on i nvol ving fi nel y divi ded aluminium and ir on oxide (b) combust ion of gases (c) elect r ic ar c (d) none of t hese 298. I n t her mit welding, t her mit used is mixt ur e of (a) aluminium and ir on oxide (b) aluminium and char coal (c) ir on oxide and char coal (d) none of t hese 299. I n t her mit welding, aluminium and ir on oxide ar e mixed in t he pr opor t ion (a) 1 : 1 (b) 1 : 3 (c) 3 : 1 (d) 2 : 1 300. Ther mit welding is used for (a) joining r ails, t r uck fr ames and br oken mot or castings (b) r epair ing br oken shear s (c) r eplacing br oken gear t eet h (d) all of t hese 301. I n t her mit welding (a) weld cools unifor mly (b) all par t of t he weld sect ion ar e molt en at t he same t ime (c) pr oblem wi t h i nt er nal r esi dual st r esses is minimum (d) all of t hese 302. Wel di ng pr ocess whi ch empl oys exot her mal ch em i cal r eact i on f or dev el opi n g h i gh t emper at ur e is (a) t her mit welding (b) gas welding (c) ar c welding (d) r esist ance welding 303. Gr ay ir on is gener ally welded by (a) gas welding (b) ar c welding (c) TI G welding (d) MI G welding 304. I n gas wel di ng, maximum fl ame t emper at ur e occur s at (a) inner cone (b) out er cone (c) next of inner cone (d) t ip of t he flame

4.50

Manufacturing Engineering

305. I n gas w el di n g, f l am e t em per at u r e of t h e oxyacet ylene gas used is (a) 1200C (b) 1800C (c) 2400C (d) 3200C

315. Oxidising flame as compar ed to neutral flame has inner cor e (a) less luminous (b) shor t er (c) mor e luminous (d) both (a) and (b)

306. Acet ylene gas is st or ed in cylinder s in (a) solid for m (b) liquid for m (c) gaseous for m (d) all of t hese

316. Neut r al flame is used t o weld (a) copper or br ass (b) cast ir on (c) st eel (d) all of t hese

307. Chemical used in pr oducing acet -ylene gas is (a) car bon (b) cr ushed bone (c) char coal (d) all of t hese 308. L ow pr essur e acet yl en e i s pr oduced at t h e welding sit e by chemical r eact ion bet ween wat er and (a) calcium chlor ide (b) calcium car bide (c) calcium car bonat e (d) car bon

317. Rate of welding steel by carbur ising flame as compar ed t o neut r al flame is (a) mor e (b) less (c) same (d) none of t hese

309. I n gas welding, maximum t hickness of mat er ial which can be welded wi t h a 30 mm diamet er welding r od is (a) 3 mm (b) 6 mm (c) 15 mm (d) 300 mm 310. Filler mat er ial is essent ially used in (a) gas welding (b) spot welding (c) seam welding (d) all of t hese 311. Gener ally the oxy-acet ylene welding is done with (a) neut r al flame (b) oxidising flame (c) car bur ising (d) all of t hese 312. Neut r al flame is obt ained by sup-plying (a) equal volumes of oxygen and acet ylene (b) mor e volume of acet ylene and less volume of oxygen (c) mor e volume of oxygen and less volume of acet ylene (d) none of t hese 313. Oxidising flame is obt ained by supplying (a) equal volumes of oxygen and acet ylene (b) mor e volume of acet ylene and less volume of oxygen (c) mor e volume of oxygen and less volume of acet ylene (d) none of t hese 314. Car bur ising flame is obtained by supplying (a) equal volumes of oxygen and acet ylene (b) mor e volume of acet ylene and less volume of oxygen (c) mor e volume of oxygen and less volume of acet ylene (d) none of t hese

318. Oxidising flame is used t o weld (a) brass (b) copper (c) br onze (d) all of t hese 319. Car bur ising flame is used t o weld (a) st eel, cast ir on, copper et c. (b) br ass and br onze/c (c) har d sufr acing mat er ials such as st ellit e (d) all of t hese 320. Rat i o of ox ygen t o acet y l en e f or com pl et e combust ion is (a) 1 : 1 (b) 1.5 : 1 (c) 2 : 1 (d) 2.5 : 1 321. I n gas wel di ng usi ng oxygen and acet yl ene cylinder s, pr essur e of gas is mor e in (a) acet ylene cylinder (b) oxygen cylinder (c) equal in bot h t he cylinder s (d) none of t hese 322. Temper at ur e of oxy-hydr ogen flame as compar ed t o oxy-acet ylene flame is (a) mor e (b) less (c) same (d) none of t hese 323. Ends of t wo pipes of unifor m cr oss-sect ion ar e welded by (a) spot welding (b) seam welding (c) pr oject ion welding (d) upset butt welding 324. The name key hole welding r efer s t o (a) Pr oject ion welding (b) Per cussion welding (c) Pulsed TI G welding (d) Plasma ar c welding 325. Black colour is gener ally paint ed on (a) oxygen cylinder (b) acet ylene cylinder (c) hydr ogen cylinder (d) none of t hese

Manufacturing Engineering

4.51

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (a)

2. (b)

3. (a)

4. (b)

5. (b)

6. (d)

7. (d)

8. (a)

9. (b)

10. (a)

11. (b)

12. (c)

13. (b)

14. (a)

15. (b)

16. (b)

17. (c)

18. (d)

19. (c)

20. (b)

21. (b)

22. (a)

23. (c)

24. (d)

25. (b)

26. (a)

27. (c)

28. (c)

29. (a)

30. (d)

31. (d)

32. (d)

33. (b)

34. (a)

35. (d)

36. (d)

37. (d)

38. (b)

39. (a)

40. (a)

41. (d)

42. (d)

43. (a)

44. (b)

45. (a)

46. (b)

47. (a)

48. (a)

49. (b)

50. (b)

51. (a)

52. (b)

53. (b)

54. (a)

55. (b)

56. (c)

57. (b)

58. (d)

59. (a)

60. (d)

61. (c)

62. (b)

63. (b)

64. (d)

65. (d)

66. (d)

67. (d)

68. (b)

69. (a)

70. (b)

71. (c)

72. (b)

73. (a)

74. (c)

75. (b)

76. (b)

77. (a)

78. (a)

79. (c)

80. (d)

81. (a)

82. (a)

83. (d)

84. (b)

85. (c)

86. (c)

87. (c)

88. (d)

89. (d)

90. (a)

91. (b)

92. (a)

93. (b)

94. (d)

95. (a)

96. (d)

97. (d)

98. (b)

99. (d)

100. (a)

LEVEL-1 101. (d)

102. (b)

103. (d)

104. (d)

105. (d)

106. (b)

107. (a)

108. (c)

109. (c)

110. (b)

111. (c)

112. (a)

113. (c)

114. (d)

115. (a)

116. (b)

117. (a)

118. (d)

119. (d)

120. (c,d)

121. (d)

122. (d)

123. (a)

124. (b)

125. (b)

126. (a)

127. (a)

128. (c)

129. (b)

130. (b)

131. (c)

132. (d)

133. (c)

134. (b)

135. (a)

136. (d)

137. (c)

138. (b)

139. (d)

140. (a)

141. (c)

142. (c)

143. (c)

144. (d)

145. (b)

146. (b)

147. (c)

148. (c)

149. (c)

150. (d)

151. (a)

152. (b)

153. (a)

154. (b)

155. (c)

156. (b)

157. (d)

158. (a)

159. (b)

160. (c)

161. (a)

162. (b)

163. (d)

164. (c)

165. (b)

166. (a)

167. (b)

168. (c)

169. (d)

170. (a)

171. (b)

172. (c)

173. (c)

174. (d)

175. (d)

176. (b)

177. (c)

178. (a)

179. (b)

180. (c)

181. (d)

182. (d)

183. (d)

184. (c)

185. (b)

186. (d)

187. (a)

188. (b)

189. (c)

190. (a)

191. (b)

192. (a)

193. (b)

194. (b)

195. (a)

196. (a)

197. (a)

198. (a)

199. (c)

200. (c)

201. (a)

202. (a)

203. (b)

204. (c)

205. (d)

LEVEL-2 206. (b)

207. (d)

208. (c)

209. (d)

210. (c)

211. (a)

212. (b)

213. (c)

214. (b)

215. (b)

216. (b)

217. (c)

218. (a)

219. (b)

220. (b)

221. (b)

222. (a)

223. (c)

224. (a)

225. (a)

226. (a)

227. (b)

228. (c)

229. (c)

230. (c)

231. (a)

232. (b)

233. (a)

234. (a)

235. (c)

236. (a)

237. (b)

238. (d)

239. (c)

240. (c)

241. (d)

242. (a)

243. (d)

244. (c)

245. (c)

246. (b)

247. (c)

248. (c)

249. (b)

250. (d)

251. (a)

252. (a)

253. (c)

254. (c)

255. (d)

256. (b)

257. (d)

258. (c)

259. (c)

260. (a)

261. (d)

262. (a)

263. (d)

264. (b)

265. (d)

266. (a)

267. (a)

268. (d)

269. (b)

270. (c)

271. (d)

272. (b)

273. (a)

274. (c)

275. (a)

276. (d)

277. (d)

278. (d)

279. (a)

280. (d)

281. (b)

282. (a)

283. (b)

284. (c)

285. (c)

286. (b)

287. (a)

288. (b)

289. (a)

290. (b)

291. (a)

292. (c)

293. (b)

294. (a)

295. (b)

296. (b)

297. (a)

298. (a)

299. (b)

300. (d)

301. (d)

302. (a)

303. (a)

304. (a)

305. (d)

306. (b)

307. (b)

308. (a)

309. (a)

310. (a)

311. (a)

312. (a)

313. (c)

314. (b)

315. (d)

316. (d)

317. (a)

318. (d)

319. (c)

320. (d)

321. (b)

322. (a)

323. (d)

324. (a)

325. (a)

5

CHAPTER

Metrology and Inspection

5.1

Metrology and Inspection

I N TERCH AN GEABI LI TY When a syst em has been wor ked out , so t hat any one component will assemble cor r ect ly wit h any mat ing component , bot h being chosen at r andom, t he met hod is called int er changeable syst em.

Concept of I nt er changeabi li t y I t is ver y difficult t o manufact ur e any component or par t wit h exact dimensions due t o human inaccur acy or machine or pr ocess incapabilit y. Thus for t he component s t o meet funct ional aspect as well as t o br ing down t he cost of pr oduct ion, pr inciple of int er changeabilit y concept is int r oduced. I nt er changeable pr oduct ion means pr oduct ion of par t s t o such a degr ee of accur acy t hat will ensur e an assembly which will meet t he funct ional r equir ement s.

LI M I TS AN D FI TS Advant ages (1) Cost of t he assembly decr eases because of holes and shaft s can be made at differ ent places wher e t he mat er ial and t he labour is cheap. (2) St andar dizat ion of hole and shaft is possible. (3) Qualit y of pr oduct incr eases. (4) maint enance of t he assemble becomes cheap and easy because when eit her hole or shaft fail s one does not have t o t hr ough away t he complet e assembly and t he failed par t s only needs t o be r eplaced.

Si ze

F ig. L imits and tolerances for hole and shaft

A number expr essing, in a par t icular unit , t he numer ical value of a lengt h. Act ual size I t is size of a par t as may be found by measur ement . Limits of size The t wo ext r eme per missible sizes bet ween which act ual size is cont ained. M aximum limit of size Gr eat er of t he t wo limit s of size is called maximum limit . M inimum limit of size Smaller of t he t wo limit s of size is called minimum limit . Basic size Basic size of component is fixed by t he designer fr om funct ional point of view. I t is ver y difficult t o achieve t his size. A lit t le var iat ion in t he basic size is r equir ed in pr oduct ion, which must be t oler able and is called t oler ance. L imit s L ar ger and smaller dimensions ar e called limit s, t her e being a high limit and a low limit . Differ ence between high and low limits, is mar gin allowed for var iation in wor kmanship, and is called tolerance. 0.02 I f all the tolerance is allowed on one side of the nominal diameter  e. g. 25.00 0.00  , the system is called unilateral (uni

e

j

0.01 = one), while if it is divide, some being allowed on either side of the nominal diamet er e. g. 25.00 –0. 01 , t he syst em is

called bilateral (bi = two). An int er changeable syst em is gener ablly called limit syst em or syst em of limit s and fit s. A syst em of limit s must pr ovide infor mat ion fr om which most usual t ypes of engagement bet ween t wo mat ing par t s may be obt ained.

5.2

Metrology and Inspection

Allowances

An int ent ional differ ence bet ween t he hole dimension and shaft dimension for any t ype of fit is called allowance. M aximum allowance is obt ained by subt r act ing minimum shaft size fr om t he lar gest hole size M inimum allowance is obt ained by t he differ ence bet ween lar gest shaft size and t he smallest hole size.

FI TS When t wo par t s ar e t o be assembled, t he r elat ion r esult ing fr om t he differ ence bet ween t heir sizes befor e assembly is called a fit . Basic size of a fit. I t is t hat basic size which is common t o t wo par t s of a fit . Var iat ion of fit . This is t he ar it hmet ical sum of t oler ances of t wo mat ing par t s of fit .

Classificat ion of F it s 1. Depending on Act ual limits of hole or shaft ( i ) Clearance fit : I f lower limit of hole is lar ger t han the upper limit of shaft it is consider ed as clear ance fit.

( ii ) Transition fit : This t ype of fit appear s when t her e is over lap in t he t oler ance zones. Physically it indicat es t hat when a par t is select ed r andomly fr om hole and a shaft lot , some of t he assemblies can be made wit hout appliat ion of for ce and for some of t he assemblies for ce will be r equir ed.

( iii ) I nterference fit : I f maximum size of hole is smaller t han t he minimum size of shaft , t hen for ce has t o be applied t o make t he assembly. Such fit s ar e called int er fer ence fit s.

2. Depending on value of clearance To pr ovide t he var ious t ypes of engagement bet ween shaft and t he hole, t her e must be var iat ions in t he differ ence bet ween t heir aver age sizes. ( i ) Running fit A smoot h easy (but not loose) fit for t he pur pose of a moving bear ing pair. Bet ween t he zones of int er fer ence and clear ance, t her e exist s a r ange of fit s which ar e neit her one nor t he ot her (e.g. a ‘push’ fit ). These ar e called t r ansit ion fit s. ( ii ) Push fit I t can be assembled wit h light hand-pr essur e (locat ing plugs, dowels, et c.). To obt ain a for ce fit , shaft must be slight ly lar ger t han t he hole (int er fer ence fit ). ( iii )Dr iving or Press fit I t can be assembled wit h hammer or by medium pr essur e. I t gives a semi-per manent fit such as is necessar y for a keyed pulley on a shaft . ( iv) F orce fit I t r equir es gr eat pr essur e t o assemble and gives a per manent fit . I t is used for wheels and hubs on shaft s fr om which t hey ar e never likely t o be r emoved. I n a r unning fit , t he shaft is slight ly less in diamet er t han t he hole (clear ance fit ). T hese classification may be furt her subdivided by making such fit s as (i ) Slack r un (ii ) Close r un (iii ) L ight dr ive (iv) H eavy dr ive et c.

Metrology and Inspection

5.3

CLEARAN CE I t is t he differ ence between t he sizes of the hole and shaft , befor e assembly, when this differ ence is posit ive. M inimum and M aximum Clearance I n a clear ance fit , minimum clear ance is t he differ ence bet ween minimum size of t he hole and maximum size of t he shaft . I n a t r ansit ion or clear ance fit , maximum clear ance is t he differ ence bet ween maximum size of t he hole and minimum size of t he shaft .

I N TE RF E RE N CE This is t he ar it hmet ical differ ence bet ween sizes of t he hole and shaft befor e assembly, when t his differ ence is negat ive.

M inimum and M aximum I nt er fer ence I n int er fer ence fit , t he minimum int er fer ence is ar it hmet ical differ ence bet ween maximum size of t he hole and minimum size of t he shaft befor e assembly.

Fig. Disposition of maximum interference and minimum interference

I n int er fer ence or t r ansit ion fit , maximum int er fer ence is ar it hmet ical differ ence bet ween maximum size of t he hole and maximum size of t he shaft befor e assembly.

DE V I ATI ON

I t is defined as algebr aic differ ence bet ween a size (act ual & maximum) and cor r esponding basic size.

U pper D eviat i on I t is t he algebr aic differ ence bet ween maximum limit of size and cor r esponding basic size.

L ower D evi at i on I t is t he algebr aic differ ence bet ween minimum limit of size and cor r esponding basic size.

F undament al D evi at ion I t is one of the two deviations which is conventionally chosen to define position of tolerance zone in r elation to the zero line.

5.4

Metrology and Inspection

TOLERAN CE Toler ance is equal t o algebr aic differ ence bet ween upper and lower deviat ions and has an absolut e value wit hout sign. I n t he cont ext of t his t er minology for limit s and fit s, t he differ ence bet ween maximum limit of size and minimum limit of size is called t oler ance. I n t he figur e given below, toler ance etc. ar e shown too much exagger ated. I t is assur ed that shaft touches at the bottom of hole and then ot her dimensions ar e consider ed. A line is assumed to be t her e in between upper toler ance on shaft and lower toler ance on hole. This line is called zer o line. This also r efer s to basic size fr om axis of the shaft or hole. All t oler ances and deviations ar e r efer r ed to fr om t his datum line. Set t ing of t oler ance value alone is not sufficient t o define par t icular limit s, but posit ion of t he t oler ance zone r elat ive t o t he basic size of t he feat ur e must also be specified.

Fig. Disposition of maximum clearance and minimum clearance

Tol er ance Zone I n a gr aphical r epr esentation of toler ance, the zone bounded by two limit s of size of the par t is called toler ance zone. I t is defined by its magnitude (i.e. toler ance) and by its posit ion in r elat ion to the zer o line.

Gr ades of Toler ance I n a st andar dised syst em of limit s and fit s, gr oup of t oler ances ar e consider ed cor r esponding t o t he same level of accur acy for all basic sizes.

St andar d Tol er ance I t is a t oler ance belonging t o any one of t he st andar d gr ades.

St andar d Tol er ance U nit A unit which is a funct ion of basic size and which is common to the for mula defining differ ent gr ades of toler ances.

U se of Limit s and Toler ances I t ensur es t hat component s made in one fact or y may be assembled wit h mat ing par t s made in anot her and, by eliminat ing necessit y for pr eliminar y t r ial of a fit . Par t s may be made and held in st ock unt il t hey ar e r equir ed. The applicat ion of a syst em of limit s and fit is not confined t o cir cular holes and shaft s but may be applied t o any set of condit ions wher e a par t icular t ype of fit is r equir ed. Ther efor e, it may be used for keys and keyways, squar e and flat fit t ing combinat ions and similar applicat ions.

SH AF T A term used by convention to designate all external features of a part, including those which are not cylindrical.

Basic Shaft A shaft whose upper deviat ion is zer o, e.g. shaft h

Basic H ole A hole whose lower deviat ion is zer o, e.g. hole H

Go Limit I t r efer s t o upper limit of shaft and lower limit of a hole. Thus it cor r esponds t o maximum mat er ial condit ion.

Metrology and Inspection

5.5

N o Go Limit I t refers to the lower limit of a shaft and upper limit of a hole. Thus it corresponds to minimum material condition. For shaft s a t o h , upper deviat ion is below t he zer o line and for shaft s j t o zc it is above t he zer o line. For holes A t o H , lower deviat ion is above t he zer o line and for J t o zc it is below t he zer o line. Upper deviat ion for shaft s is denot ed by es and lower deviat ion by ei . For holes t he cor r esponding deviat ions ar e denot ed by ES and EI r espect ively. I n t he specificat ions, for mulae ar e given t o det er mine fundament al deviat ion. e.g. for shaft s, fundament al devi at i on (upper devi at i on es t o l ower devi at i on ei ) i s det er mi ned by means of for mul ae gi ven i n t he Table below. Ot her deviat ions may be der ived dir ect ly using absolut e value of t he t oler ance I T by means of algebr aic r elat ionship. ei = es – I T es = ei + I T Formulae for fundamental deviations for shafts for sizes upto 500 mm : Upper Deviation (es) Lower Deviation (ei)

Shaft designation a

In microns (for D in mm) = – (265 + 1.3 D) for D  120 = – 3.5 D for D > 120

b

= – (140 + 0.85 D) for D  160

Shaft designation j 5 to 8 k 4 to k 8 k for grade  3 and  8

In microns (for D in mm) No formula = + 0.6

3

D

=0

= – 1.8D for D > 160 c

= – 52 D0.2 for D  40

m

= + (IT7 - IT6)

d

= – (95 + 0.8D) for D > 40 = – 16D0.44

n p r

e

= – 11D0.41

s

= + 5D0.34 = + IT7 + 0 to 5 = geometric mean of values ei for p and s = IT8 + 1 to 4 for D  50

f g

= – 5.5 D0.41 = – 2.5 D0.34

h

=0

t u v x y z za zb zc

= + IT7 to + 0.4D for D > 50 = IT7 + 0.63D = + IT7 + D = + IT7 + 1.25D = + IT7 + 1.6D = + IT7 + 2D = + IT7 + 2.5D = IT8 + 3 + 3.15D = + IT9 + 4D = + IT10 + 5D

The deviat ion given in t he t able is t hat cor r esponding in pr inciple t o t he limit close t o t he zer o line, i.e., upper deviat ion es for shaft s a t o h and lower deviat ion ei for shaft s j t o zc. For holes also, deviat ions ar e der ived fr om t hose of t he cor r esponding shaft s as follows : The gener al r ule is t hat , hole limit s ar e ident ical wit h t he shaft limit s of t he same symbol (let t er and gr ade) but disposed on t he ot her side of t he zer o line, i.e upper deviat ion of hole, EI = es of shaft of t he same let t er symbol but of opposit e sign. I n t he hole basis syst em, differ ent clear ances and int er fer ences ar e obt ained by associat ing var ious shaft s wit h a single hole whose lower deviat ion is zer o (H hole). This is st andar d pr act ice as it is ver y convenient t o make cor r ect holes. I n shaft basis syst em, upper deviat ion of shaft is zer o.

5.6

Metrology and Inspection

I n I .S. specificat ion, 18 gr ades of t oler ance ar e designat ed : I T01, I T0, I T1 t o I T16. These -ar e called standar d t olerances and t heir numer ical values have been det er mined in ter ms of t he st andar d t oler ance unit i , expr essed as i (micr ons) = 0.45 3 D + 0.001 D. D = geometr ic mean of lower and upper diameter s of a particular diameter step in which diameter lies. (in mm). This for mula has been empir ically calculat ed on t he basis of for mer nat ional st andar ds of ot her count r ies and consider s t he fact t hat in most usual cases, t he t oler ance var ies mor e or less par abolically in t er ms of diamet er for t he same manufact ur ing condit ions. Relative magnitude of each grade Gr ades

I T5

I T6

I T7

I T8

I T9

I T10

I T11

I T12

I T13

I T14

I T15

I T16

Values

7i

10i

16i

25i

40i

64i

100i

160i

250i

400i

640i

1000i

Factors affecting magnitude of tolerance (i ) Size : I t is accommodat ed by t he fundament al t oler ance limit (ii ) Qualit y.

i = 0.45

D + 0.001D.

LI M I T SYSTE M L imit s and allowance depend upon (i ) Nominal size; e.g. whet her 25 mm, 100 mm, et c. (ii ) Class of fit r equir ed (iii ) Qualit y of pr oduct .

Basic of Fit (or Limit) System A fit or limit syst em consist s of a ser ies of t oler ances ar r anged t o suit a specific r ange of sizes and funct ions, so t hat limit s of size may be select ed and given t o mat ing component s t o ensur e specific classes of fit . This syst em may be ar r anged on t he following basis : 1. H ole basis system I n t his syst em, limit s on t he hole ar e kept const ant and t he var iat ions necessar y t o obt ain t he classes of fit ar e ar r anged by var ying t hose on t he shaft .

‘s’ denotes shafts to give various fits with hole

2. Shaft basis syst em I n t his syst em, limit s on t he shaft ar e kept const ant and var iat ions necessar y t o obt ain t he classes of fit ar e ar r anged by var ying t he limit s on t he holes. I n indust r ial pr act ice, hole basis syst em is used because many holes ar e pr oduced by st andar d t ooling, e.g. r eamer s, dr ills et c., whose size is not adjust able. Subsequent ly t he shaft sizes ar e mor e r eadily var iable about t he basic size by means of t ur ning or gr inding oper ations. Thus hole basis syst em r esults in consider able r educt ion in r eamer s and ot her pr ecision t ools as compar ed t o a shaft basis syst em because in shaft basis syst em due t o non-adjust able nat ur e of r eamer s, dr ill et c. gr eat var iet y (of sizes) of t hese t ools ar e r equir ed for pr oducing differ ent classes of holes for one class of shaft for obt aining differ ent fit s.

‘H ’ denotes hole to give various fits with shafts

Metrology and Inspection

5.7

I SO System This syst em allows for 27 t ypes of fit and 18 gr ades of toler ance, cover ing a size r ange of 0 to 3150 mm. I n t his system , holes ar e designated by capital letter s ABCDE... etc., and shaft ar e designat ed by small letter s cover ing the same r ange. The 18 accur acy gr ades ar e cover ed by the numer als 01, 0, 1, 2, 3, .... 16. For specifying any par t icular hole or shaft , t he r ule is t o wr it e t he let t er followed by t he numer al denot ing t oler ance gr ade, e.g. H 8 for a hole and f6 for a shaft . A fit involving t hese t wo element s is wr it t en as H 8-f6 or H 8/f6.

LI N EAR AN D AN GU LAR M EASU REM EN T M easur ing inst r ument ar e designed for : (i ) L ine measur ement (for measur ing dist ance bet ween t wo lines) like st eel r ule et c. (ii ) End measur ement (for measur ing dist ance bet ween t wo sur faces) like micr omet er scr ew gauge et c. Classificat ion of M easur ing inst rument s Depending on t he accur acy which can be at t end measur ing inst r ument s ar e classified as follows : (i ) Non-pr ecision inst r ument s (ii ) Pr ecision inst r ument s. L ine gr aduat ed measur ing I nst rument s These incor por at e gr aduat ion spacing r epr esent ing known dist ance. These ar e used for dir ect measur ement of specific dist ance wit hin t heir capacit y r ange. These ar e used in a gr eat var iet y of measur ing inst r ument s. Types of L ine graduat ed mesuring inst rument s ( i ) L ine graduat ed rules and t apes These ar e used for dir ect lengt h compar ison and t hey have no auxilar y device. ( ii ) L ine gr aduat ed bar st andar ds

CAL L I PE RS For t he par t s which can’t be measur ed dir ect ly wit h the scale, the assist ance of calliper s is essent ial. Thus, it act as accessor ies t o scale. The calliper consist s of t wo legs hinged at t op, and ends of legs-span t he par t t oo be inspect ed. This span is maint ained and t r ansfer r ed t o t he scale. Classificat ion of Calliper s (i ) Fir m joint calliper (ii ) Fir m joint divider (iii )Spr ing calliper Ot her classificat ion of Calliper s (i ) Out side calliper (ii ) I nside calliper (iii )Tr ansfer calliper (iv) H ermo phrodite calliperrs : I t is also calle old leg calliper and is a main t ool consist ing of one divider and one calliper leg. Thus it is used in layout wor k. I t can have t wo t ypes of legs : (a) Not ched legs (b) Cur ved legs. D ial calliper. These pr ovide a t ypical dir ect - r eading capabilit y of 0.02 mm. These unit ar e quit e flexible offer ing a t ypical measur ement r ange of 150 mm wit h accessor ies available for r ange ext ension as well as specific measur ement accessor ies.

SU RFACE PLATE I t pr ovide dat um or r efer ence sur face for measur ement . Sur face plat e t er ms t he basis of measur ement and is a must for met r ology labor at or y wher e inspect ion wor k is car r ied out . I t is also used t o check flat ness of anot her sur face.

5.8

Metrology and Inspection

Angle Plate This is an accessor y needed wit h sur face plat e for measur ement pur poses. I t 's t wo sur faces ar e per pendicular t o each ot her. I t is available in var ious designs. Cast ir on angle plat es ar e widely used for wor kshop and inspect ion pur pose. Angle plat e ar e available in t wo gr ades depending upon t heir accur acy ( i ) Grade 1 angle plates These ar e of size 3 t o 6 when not finished on t heir int er ior faces ar e webbed. ( ii ) Grade 2 angle These plat es may or may not be webbed.

V-BL OCK These ar e widely used for wor kshop and inspect ion pur pose for checking out r oundness of t he cylinder ical wor kpieces and for mar king cent r es accur at ely et c. Gener ally t he angle of V is 90 and t hese ar e available in wide var iet y of shapes. I n case of V-Block, t he wor king sur faces ar e flanks of vees, base and face, t op and side faces. Depending upon accur acy, V-blocks specifies int o t wo gr ades : (i ) Gr ade A (ii ) Gr ade B. Two t ype of V-block commonly used one of t hem being deeper and wider t han t he ot her.

L ar ger vee block having one vee only for special pur pose such as checking t r iangle effect s or t aps and ot her t hr ee-flut ed t ools, 120 degr ee include angle vee -blocks ar e also available t o hold cylindr ical pieces.

STRAI GH T EDGE These ar e used for checking st r aight ness and flat ness of par t s in conjuct ion wit h t he sur face plat es and spir it level. These may be made of st eel or cast -ir on. I t is widely used for t est ing machine t ool slide ways. They ar e heavily gibbed and bow shaped t o pr event dist or t ion.

Types of Str aight Edges (i ) Tool-maker ’s st r aight edge (ii ) Wide-edge st r aight edge (iii ) Angle st r aight edge (iv) Box st r aight edge Tolerance in flatness : I t is defined as maximum per missible dist ance separ at ing t wo imaginar y par allel planes, wit hin which t he sur face under consider at ion can just be enclosed. Tolerance in squareness : I t is defined as maximum per missible dist ance separ at ing t wo imaginar y par allel planes, wit hin which t he sur face under consider at ion can just be enclosed.

SPI RI T LEV ELS These ar e used for measur ing small angle of inclination and also enable t he position of a sur face to be det er mined wit h r espect t o t he hor izont al. A spir it level consist s of a sealed glass t ube, gr ound on it s inside sur face t o a convex for m wit h a lar ge r adius of cur vat ur e R. A scale is engr aved on t he glass at t he t op of t he t ube. The t ube is near ly filled wit h et her such t hat only a small volume r emains at t he t op par t of t hese t ube, which cont ains et her vapour s in t he for m of a bubble.

FRAM E LEVEL I t is used for checking ver t ical sur faces. Side edges of t he fr ame level ar e made st r ict ly squar e wit h t he base. A glass t ube filled wit h et her is mount ed in t he base. For checking ver t ical sur faces, side edge of t he fr ame level is placed int o exact cont act wit h t he sur face and r eading of t he bubble not ed down. Posit ion of t he fr ame level in t r ansver se dir ect ion is checked by anot her less accur at e t ube.

Metrology and Inspection

Principle

B

l

5.9

B

Glass tube is set in t he base adjusted in such a way that when base is hor izont al, A bubble r ests at the centr e of the scale, which is engr aved on the glass. When base R of the level is moved out of the hor izont al, then bubble tr ies t o r emain at t he  h highest point of the tube and t hus moves along t he scale.  O Relat ions bet ween movement of bubble and ot her condit ions : L A L et B be t he t op of t he t ube r adius and posit ion of t he bubble when base is at OA (hor izont al). I f base is t ilt ed t hr ough an angle  and base occupies posit ion OA ', t hen bubble will move t hr ough dist ance L t o B ', wher e angle BOB ' =  ar c (l ) = .R  l = R h h = L .   = L R. h  l = L

R. (one r adian equals 206,365 seconds of an ar c) 206,265 Scale spacing or t he dist ance bet ween adjacent gr aduat ion is gener ally about 2mm. I f is t aken is seconds, t hen

l =

2206,265 = 2 sec 206,000 The inclinat ion of 2" causes bubble movement of 2 mm. This is sensit ive spir it level and is r ecommended for r esear ch labor at or y. Thus, for R = 206 m,  =

U N I VERSAL SU RFACE GAU GE This is most ver sat ile inst r ument used in non-pr ecision measur ement . I t is gener ally used wit h sur face plat e for layout wor k and inspect ion. M any t ime sur face gauge is ver y successfully used for cent r ing t he wor k at lat he. I t is capable of pr ecision measur ement also when used in conjuct ion wit h dial indicat or.

FEELER GAU GE These ar e used t o measur e widt h of t he gap bet ween t wo par allel flat faces e.g. in gauging of t he clear ance bet ween pist on and cylinder. I t consist of a nar r ow st r ip of sheet made t o a given t hickness. The complet e set consist s of a number of gauging blade of differ ent t hickness assembled t oget her. Their wor king depends ent ir ely on t he sense of feel. The feeler blade should neit her be for ced bet ween sur faces nor should it slide fr eely r at her t he cor r ect blade will give a char act er ist ic gauge bit t ype of feel. I t is necessar y t hat , t wo blades should be joined t oget her for not ing any dimension. Feeler gauge gener ally compr ise of a ser ies of gauging blades of differ ent gr ades and t hicknesses var ying fr om 0.03 t o 1 mm assembled in pr ot ect ive shealt h. M aximum var iat ion wit h t hickness of blade should not exceed 0.04 mm per blades upt o and including 0.3 mm t hick 0.006 mm for blades over 0.3 mm in t hickness.

AN GLE GAU GE These ar e similar t o feeler gauges. A set of angle gauge consist of 18 blades wit h t heir ends cut at var ious angle fr om 2 t o 45 degr ee. 43° 28 10°38  Angle gauge or angle t emplat e as shown in t he figur e is pieces of gauge st eel 33° which have been accur at ely for med t o t he angle engr aved on t hem. These ar e used by fit t ing t hem in t ur n on t he piece t o be gauged, unt il it is found t hat fit s per fect ly. The fit t ing is judged by holding wor k and gauge in fr ont of a light and obser ving if any light is visible bet ween gauge and wor k. Since a gap of 0.025 mm is easily seen, t his met hod is limit ed as r egar ds accur acy mainly by accur acy of t he gauges t hemselves and ext ent of t he r ange of angles r epr esent ed.

5.10

Metrology and Inspection

VERN I ER I N STRU M EN TS Principle of vernier : When t wo scales or division slight ly differ ent in size ar e used, t hen differ ence bet ween t hem can be ut ilised t o enhance t he accur acy of measur ement .

Ver nier Calliper I t essent ially consist of two st eel r ules and these can slide along each other. One of the scale, i.e. main scale is engr aved on a solid L-shaped fr ame. On this scale on gr aduation ar e divided into 20 par ts so that one small division equals 0.05 cm. One end of the fr ame contains a fixed jaw which is shaped into a contact tip at its extr emity.

Reading of Ver nier Scale To under st and t he wor king of ver nier scale, assume t hat each small division of t he main scale is 0.025 unit s. The ver nier scale cont ains 25 divisions and t hese coincide exact ly wit h 24 divisions of main scale. 1 1 So, one ver nier division = of 24 scale division, i.e. 24  0.025 = 0.024 unit s. 25 25 Therefor e, difference between one main scale small division and one ver nier division (least count of the instr ument) equals 0.025 – 0.024, i.e. 0.001 unit s, i.e., if zer o of main scale and zer o of ver nier coincide, t hen fir st ver nier division will r ead 0.001 unit less t han t he 1 small scale division. I f zer o ver nier scale lies in bet ween t wo small divisions on main scale, it s exact value can be judged by seeing as t o which ver nier division is coinciding wit h main scale division. Thus r ead a measur ement fr om a ver nier calliper, note the units, t en t h an d f or t een t h w h i ch t h e zer o on t h e v er n i er has moved for m t he zer o on t he main scale. Not e down t he ver nier division which coincides wit h a scale division and add t o pr evious r eading t he number of t housands of a unit indicat ed by t he ver nier divisions e.g. r eading in t he scale shown in t he figur e is : 3 unit s + 0.1 unit + 0.075 + 0.008 = 3.183 unit s. When using ver nier calliper for inter nal measur ements, width of the measur ing jaws must be taken into account.

Ver nier H eight Gauge This is also a sor t of ver nier calliper, equipped wit h a special base block and ot her at t achment s which make t he inst r ument suit able for height measur ement . The ver nier height gauge is mainly used in t he inspect ion of par t s and layout wor k.

Speci fi cat i on Ver nier height gauge is specified by specifying (i ) Range of measur ement (ii ) Type of scales desir ed (iii ) Any par t icular r equir ement in r egar d t o t he t ype of ver nier desir ed. Gener ally all par ts of the height gauges ar e made of good qualify steel, or stainless steel also in cer tain cases.

Test for Accur acy Er r or s in t he r eading at any por t ion which for measur ing r ange of height gauge should not exceed ± 0.02 mm for measur ing r ange of height gange of 250 mm ± 0.04 mm for above 250 mm and upt o 750 mm ± 0.05 mm for r ange of mor e t han 750 mm.

Ver nier Dept h Gauge I t is used for measur ing dept h of holes, r ecesses and dist ances fr om a plane sur face t o pr oject ion. I n t his gr aduat ed scale can be slide t hr ough t he base and ver nier scale r emains fixed.

Metrology and Inspection

5.11

M I CROM ETERS M icrometer Screw Gauge I t essent ially consist of an accur at e scr ew having about 10 t o 20 t hr ead per cm and r evolves in a fixed nut . End of t he scr ew for ms t one measur ing t ips and ot her measur ing t ip is const it ut ed by a st at ionar y anvil in t he base of t he fr ame. The scr ew is t hr eaded for cer t ain lengt h, t his por t ion is called sleeve and it s end is t he measur ing sur face. The spindle is advanced or r efr act ed by t ur ning a t himble connect ed t o spindle. The spindle is slide fit over the bar rel an d bar r el i s t he f i xed par t at t ached wi t h t h e fr ame. Th e bar r el i s gr aduat ed i n u ni t of 0.05 cm , i.e. 20 division per cm, which is lead of t he scr ew for one complet e r evolut ion. The t himble has got 25 devisions ar ound it s per ipher y on cir cular por t ion. Thus it sub-divides each r evolut ion of t he scr ew in 25 equal par t s i.e. each division cor r esponds t o 0.002 cm. A lock nut is pr ovided for locking a dimensions by pr event ing mot ion of t he spindle.

Rat chet st op is pr ovided at t he end of t himble cap maint ain sufficient and unifor m mesur ing pr essur e so t hat st andar d condit ion of measur ement ar e at t ained. Rat chet st op consist of an over iding clut ch held by a weak spr ing. When spindle is br ought in cont act wit h t he wor k at t he cor r ect pr essur e, t he clut ch st ar t slipping and no fur t her movement of t he spindle t akes palce by r ot at ion of r at chet . I n t he backwar d movement , it is posit ive due t o shape of r at chet . Backlash : I t is lack of mot ion of t he spindle when r ot at ion of t himble is changed in dir ect ion. M easuring range : I t is t ot al t r avel of t he measur ing spindle for a given micr omet er. Total error : I t cor r esponds t o t he maximum differ ence of or dinat es of t he cummulat ive er r or.

Cummulat ive E r r or I t is t he deviat ion of measur ement fr om t he nominal dimension det er mined at any opt imal point of measur ing r ange. I t includes effect of all possible individual er r or s such as er r or s of t hr ead, er r or of measur ing faces et c. I t can be det er mined by some t est wit h slip gauge. When t est ed of 20°C, t ot al er r or should not exceed following values : For gr ade 1,

t ot al er r or

=

For gr ade 2,

t ot al er r or

=

FG 4  L IJ m H 100 K FG10  L IJ m H 50 K

wher e L = upper limit of t he measur ing r ange in mm.

Reading of M icr omet er I n or der t o make possible t o r ead upt o 0.0001 inch in micr omet er scr ew gauge, a ver nier scale is gener ally made on t he bar r el. The ver nier scale has t wo st r aight lines on bar r el and t hese coincides wit h exact division on t he t himble. Thus one small division on t himble is fur t her subdivided int o 10 par t s and for t aking r eading one has t o see which of the ver nier scale division coincides wit h a division of t he t himble. Accor dingly r eading for a given ar r angement on main bar r el : 0.120" thumble : 0.014" ver nier scale : 0.0001"  Tot al r eading = 0.1342" Befor e t aking r eadings, anvil and spindle must be br ought toget her car efully and init ial r eading not ed down. I t s caliber at ion must be checked by using st andar d gauge block.

5.12

Metrology and Inspection

I n metric micr ometer, pitch of scr ew thr ead is 0.5 mm, so t hat one r evolution of the scr ew moves it axially by 0.5 mm. Main scale on bar r el has least divisions of 0.5 mm. The thimble has 50 divisions on its cir cumfer ence. 0.5 mm = 0.01 mm  One division on t he t himble = 50 I f vernier scale is also incorporated, then subdivisions of thimble can be estimated upto an accuracy of 0.001 mm. Reading of micr omet er in above figur e is 3.5 mm on bar r el and 7 divisions on t himble = 3.5 + 7 × 0.01 = 3.57

St ick M i cr omet er I t is designed for measur ement of longer int er nal lengt h. I t s accur acy is of t he or der of ± 0.005 mm t hr oughout t he r ange.

M icr omet er D ept h Gauge I t is used for measur ing dept h of holes slot s and r ecessed ar ea. The lengt h of micr omet er dept h gauge var ies fr om 0 t o 255 mm. The scale her e is caliber at ed in t he r ever se dir ect ion. The accur acy again depend upon sense of t ouch.

T hr ead M i cr omet er I t is used for t he measur ement of pit ch diamet er but accur acy is influenced by helix angle of t he t hr ead.

Scr ew T hr ead M icr omet er Calliper I t is used for accur at e measur ement of t he pit ch diamet er of scr ew t hr eads.

Out side M icr omet er Calliper I t is designed for fast , easy and pr ecise measur ment of lar ge wor k having dimension above 250 mm.

V-Anvil M icr omet er Calliper Any out of r oundness can be quickly checked in centr eless gr inding and machining oper ation. Dir ect r eading eliminates use of special featur es. I t can be used for measur ing odd-flut ed taps, milling cutter s, reamer s etc.

Blade Type M icr omet er I t is ideally suit ed for fast and accur at e measur ement of cir cular for med t ools, diamet er and dept h of all t ypes of nar r ow gr ooves, slot s, keyways, r ecesses et c. I t has non-r ot at ing spindle which advances t o cont act t he wor k wit hout r ot at ion.

M icr omet er for M easur ing T hickness of Cylinder Wall Or dinar y micr omet er can not be used for measur ing t hickness of wall of a t ube, sleeve or bush because of concavit y of the inter nal sur face. The micr ometer s meant for t his pur pose, fixed anvil is pr ovided wit h a spher ical measur ing sur face and fr ame is cut away on t he out side t o per mit t he anvil being int r oduced int o t ubes of diamet er as small as 7.5 mm. I n anot her design, anvil is made of cylindr ical for m, it s axis being per pendicular t o t he axis of t he spindle.

Gr oove M i cr omet er These micr omet er s ar e designed for measur ing gr ooves, r ecesses and shoulder s locat ed inside a bor e. These have st andar d (12.7 mm) diamet er disc 6.35 mm diamet er discs ar e used t o r each har d t o get at locat ion inside small bor e. All disc, having t hicknesses of 0.75 mm and ar e har dened and lapped t o minimise par allax and t o achieve a higher degr ee of accur acy. These gr oove micr omet er measur e not only t hickness and spacing of gr ooves' but also measur e fr om an edge t o a land or fr om shoulder t o gr oove. M icr omet er s ar e sat in-chr ome finished t hr oughout .

Appl i cat i ons

(a ) Outside measurement

(b) I nside measurement (add 1.5 mm to Reading)

(c) Edge to Edge measurement (add 0.75 mm to Reading)

Metrology and Inspection

(d) Edge to Edge measurement (add 0.75 mm to Reading)

(e) Edge to Edge measurement with locating disc. (add 1.5 mm to Reading)

5.13

(f) Edge to Edge measurement with locating disc. (Add 0.75 mm to Reading)

Digital M icrometer These offer direct reading to 0.0001 mm. and employ liquid display operating on a alkaline manganese batter y.

Differential Screw M icrometer A ver y high degr ee of accur acy can be obt ained in t he micr omet er scr ew gauges ut ilising t he pr incipl e of differ ent ial scr ew on t he oper at ing spindle. I n such micr omet er, t he scr ew has t wo t ypes of pit ch one smaller and one lar ger.

CYLI N DER GAU GE The cylinder gauge is ideal gauge for deter mining t aper ing, out of r ound or scor ed cylinder. Dial indicator shows inst antly condit ion of cylinder to two t housandth of a cm. Adjustable r ods ar e fur nished to make it mor e ver satile. The dial is mounted at r ight angles to the sled. Sled has two line cont act points which ar e at all times in alignment wit h the walls of the cylinder. Two har dened contacts points, which ar e independently cause the needle to tr avel over the dial featur e a unique double spr ing action making the gauge self centr ing and obsoletely non-collapsible. Handles of var ious for ms and locking mechanism ar e also pr ovided. I n case of deep bor es, i.e. cylinder of int er nal combust ion engines, st eam engine et c, it is fr equent ly necessar y not only t o measur e diamet er at a given dept h but also t o check t he hole (i ) cylindr ical shape (ii ) t aper effect .

F ollowing I nstr ument s are M ainly U sed for T his Pur pose K eilpar t Gauge This is a commer cial gauge and it s oper at ion is just similar t o t he gauge descr ibed aboved. I t consi st s of one measur ing head and one movable measur ing head. The movement of movable measur ing r od is t r ansmit t ed t o dial indicat or by push r od t hr ough a spr ing act uat ed hinged member. Thus hor izont al movement of t he r od is t r ansmit t ed int o ver t ical dir ect ion and diamet er indicat or gives indicat ion of var iat ion of size.

Aut omobile Cylinder Bor e Gauge This is another commer cially used gauge for estimating quickly the wear in automobile cylinder s. I t consists of ser ies of steel balls ar r anged in cir cular holder. These balls come in contact with t he sur face of the cylinder. This t ype of gauge will not measur e ovalit y or det ect sur face imper fect ions but it gives mean diamet er at t he plane of t he measur ement . I t can be r eadily used t o indicat e t aper ed cylinder bor e. Telescope int er nal gauge is used for measur ing int er nal diamet er dur ing machining oper at ion.

Ball Type Plug Gauge These ar e used for measur ement of diamet er of bor es. These also det ect s ovaling t aper effect and also indicat e sur face finish. This gauge is based on t he pr inciple of t hr ee equally spaced balls of same size which ar e moved out war d by spr ing loaded cone.

Taper Par allel Gauges These ar e used for measur ement of diamet er of ver y small hole which ar e t oo small t o per mit t he ent r y of int er nal micr omet er.

Pin Gauge The diamet er of lar ge bor es which cannot easily measur ed by micr omet er or ver nier et c. can be measur ed by means of a bar wit h spher ical ends, t he lengt h of which is slight ly less t han t he diamet er t o be measur ed. The bar is placed in t he bor e wit h it s ends in cont act wit h t he bor e sur face, it s axis t hus being a chor d of t he cir cle. K eeping one end in cont act wit h t he bor e sur face, ot her end is br ought t o make cont act at ot her side. Two posit ions ar e shown by full and dot t ed lines, t he dist ance bet ween t wo point s of cont act is measur ed and fr om t his dimension, and t he lengt h of t he pin gauge diamet er on t he bor e is det er mined.

5.14

Metrology and Inspection

B is cent r e of t he dot t ed cir cle along which fr ee end of t he gauge moves. AB is diamet er D. BP1 and BP2 r epr esent t wo posit ion of bar of lengt h L . By pr oper t y of int er sect ion, BP × PA = P1P × PP2 = l 2 Since BP2 = BP12 – P1P2 = L 2 – l 2



BP =



PA =

L2  l 2

l2 L2  l 2

D = BP + PA =

D  L+



Er r or in D, D =



L2  l 2 

l2 2L

l2 L2  l 2

F GH

F l I = L  G1  H L JK 2

2



1 2

F GH

 L 1

l2 2L2

I JK

I JK

2 D D  l   L = l  l  1 – l L l L L 2L

H ence, er r or in t he diamet er is only a small fr act ion

FG l IJ HLK

of t he er r or in t he measur ement of l.

SLI P GAU GE These ar e in the for m of r ectangular pr isms, ver y accur ately made in var ying lengths. They ar e made of har dened st eel having flat par allel sur face. These ar e called gauge block . Their specific use is in t est ing t he lengt h of a finished component .

Gr ade of Slipgauge Bur eau of I ndian St andar d for slip gauges specifies t hr ee gr ades of slip gauges : ( i ) Grade I : These ar e used for pr ecise wor k, such as t hat car r ied out in a good class r oom. Typical uses include set t ing up sine bar s and sine t ables, checking gap gauges and set t ing dial t est indicat or s t o zer o. ( ii ) Grade I I : This is t he wor kshop gr ade. Typical use include set t ing up machine t ools, posit ioning milling cut t er s and checking mechanical widt h. I t is int ended for use in sur face gauge. I t is also called height t r ansfer gauge. I t is used t o check accur acy or par al leli sm of sur face, and t o t r ansfer measur ement in l ayout wor k by scr ibing t hem on a ver t i cal sur face. ( iii )Grade 0 : These ar e commonly called inspect ion gr ade and it s use confined t o t ool r oom or machine shop inspect ion. Grade 00 : These ar e kept for wor k of highest pr ecision only. Caliberation grade. This is a special gr ade, wit h act ual size of t he slips st at ed or caliber at ed on a special char t supplied wit h t he set . Examples of maximum permissible error in mean length of gauge at 20 °C: N omi na l 10 20 30 100

Gr ade 0 mi cr on 0.02 0.04 0.10 0.20

Gra de I m icron 0.10 0.15 0.30 0.50

Gr a de I I mi cr on 0.35 0.45 0.75 1.20

Metrology and Inspection

5.15

OPTI CAL M ETH ODS Advantages. (i ) Simplicity (iii ) Ver satility (v) Feasible solut ions.

(ii ) Non-cont act measur ement s (iv) Cost effect iveness

Optical I nstruments for Angular M easurement 1. Aut o coll imat or : Thi s i s an opt i cal i nst r ument used for measur ement of smal l angul ar di ffer ence. For smal l angul ar measur ement , aut ocollimat or pr ovides a ver y sensit ive and accur at e appr oach. Aut ocollimat or is essent ially infinit y t elescope and collimat or combined int o one inst r ument . Pr inciple of wor king : O is a point sour ce of light placed at pr inciple focus of a collimat ing lens. The r ay of light fr om O incident on t he lens will now t r avel as a par allel beam of light . I f t his beam now st r ikes a plane r eflect or which is nor mal t o t he opt ical axis, it will be r eflect ed back along it s own pat h and focused at t he same point O. I f plane r eflect or be now t ilt ed t hr ough a small angle , t hen par allel beam will be deflect ed t hr ough t wice t his angle, and will be br ought t o focus at  in t he same plane at a dist ance x fr om O. Obviously OO' = x = 2 f wher e f = focal lengt h of t he lens.

For high sensitivity, i.e. for lar ge value of x for a small angular deviation , a long focal lengt h is r equir ed. F act or s gover ning specificat ion of Aut ocollimat or ( i ) F ocal lengt h. I t det er mines basic sensit ivity and angular r ange. Longer the focal length, lar ger is t he linear displacement for a given r eflect or t ilt but maximum r eflect t ilt which can be accommodat ed is consequent ly r educed. Sensit ivit y is t her efor e t r aded against measur ing r ange. ( ii ) Effect ive apert ur e. The maximum separ at ion bet ween r eflect or and aut ocollimat or or wor king dist ance, is gover ned by effect ive aper t ur e of t he object ive and angular measur ing r ange of t he inst r ument becomes r educed at long wor king dist ance. I ncr easing maximum wor king dist ance by t he effect ive aper t ur e t hen demands a lar ger r eflect or for sat isfact or y image cont r ast . Aut ocollimat or design t hus involves many conflict ing cr it er ia and for t his r eason a r ange of inst r ument is r equir ed t o opt ically cover ever y applicat ion. Air cur r ent in t he opt ical pat h bet ween aut ocollimat or and t he t ar get mir r or cause fluct uat ion in t he r eading obt ained. The effect is mor e pr onounced as dist ance fr om aut collimat or t o t ar get mir r or incr eases. Also er r or may occur due t o er r or s in flat ness and r eflect ivt iy of t he t ar get mir r or which should be of high qualit y. 2. Angle dekkor. I t is a t ype of collimat or which car r ies t wo scales : (i ) Fixed hor izont al scale : I t can be always seen t hr ough t he eyepiece (ii ) I lluminat ed ver t ical scale : I t is focussed on hor izont al scale The t wo scale ar e mut ually at 90° t o each ot her and t hese dir ect ly measur es angular deviat ions. I t is also possible t o view t he r eflect ion fr om t wo or mor e sur faces simult aneously, t hr ough t he eyepiece. I n or der t o compar e t hem, scale of t he inst r ument ar e so caliber at ed t hat each small division of bot h scale r epr esent s an angl e of one minut e. The t wo r eadi ng of t he scal es at t he point of t heir int er sect ion r epr esent r elat ive angular posit ion of t he sur face being t est ed.

5.16

Metrology and Inspection

FOCU SSI N G TELESCOPE I t is also called micro-alignment t elescope. I t is an impor t ant and power ful opt ical inst r ument used t o check and ensur e geomet r ical int egr it y of component and t heir assembly. I t is based on t he concept s of geomet r ic opt ics and is simple and st r aight for war d t o use. M icr o-alignment t elescope is used t o measur e deviat ion fr om st r aight line of sight t o set and check alignment , squar eness, st r aight ness, flat ness, par allelism, ver t icaling and level. M icrot elescope is used t o r ead dir ectly t o scale and is able t o focus down t o zer o dist ance fr om t he fr ont object ive. Pr imar y opt ical axis is concent r ic wit h par allel t o t he out side of t he t ube t o wit hin 6.4 µm and 3 seconds ar e r espect ivley. The t ube it self is cylindr ical t o wit hin 5µm. I n pr act ice set t ing accur acy achieved is 50 µm at a dist ance of 30 met r es ar e pr opor t ionally for longer and shor t er dist ances down t o 3 met r es.

I N T E RF E RROM E T RY I n engineer ing applicat ion, wavelengt hs of light s ar e used for ext r emely accur at e measur ement s of sur face flat ness, slip gauge and ot her end gauges. These light waves have wavelengt hs of t he or der of 0.000375 t o 0.000675 mm and by est imat ion of widt h of light s int er fer ence bands, measur ement t o wit hin 0.000025 or 0.00005 mm ar e possible. I nt er fer ence met hods ar e suit able for det er mining absolut e size of caliber at ion and r efer ance gr ade gauge I .

Gauge Requir ement s (i ) (ii ) (iii ) (iv)

Flat end faces must be flat and par allel t o each ot her. They must have high degr ee of sur face finish. Act ual size must agr ee wit h it s nominal size t o wit hin a ver y small t oler ance. Edges must be pr oper ly r ounded off and debur r ed.

I nt er fer r omet er s The int er fer r omet er is suit able for measur ing set s of slip gauge upt o 100 mm.

Type of I nt er fer r omet er s ( i ) N PL-H ilger gauge interferrometer : Among t hem most pr ominent is NPL -H I L GER gauge int er fer r ometer. I n the design of this inter fer r ometer, batches of upto 20 gauge of differ ent size setup ar e used for measur ement in oper at ion. (ii )Laser interferrometer : I t is used t o check and caliber at e machine t ool for var ious geomet r ic feat ur e dur ing assembly. Bet t er accur acy t han gages or indicat or s. I t is also used t o aut omat ically compensat e for posit ioning er r or s in coor dinat e measur ing machined and Comput er Numer ical Cont r ol machines.

ST RAI GH T N E SSS I t is defined as deviat ion of sur fae fr omidealst r aight line.

M easur ement of St r aight ness At may places it is r equir ed t hat t he sur face must be per fect ly st r aight , e.g. in a lat he it is desir ed t hat t ool must move in st r aight pat h t o gener at e per fect cylinder and it is possible only when t he cont r olling guideways ar e t hemselves st r aight . Also st r aight line or plane is t he basis of measur ement in most met hods. St r aight ness can be measur ed in following t hr ee ways : ( i ) Straight edge : I t is a piece of block of which one sur face is exact ly st r aight . By keeping t his sur face on t he machine par t under invest igat ion, t he amount of light coming fr om t he ot her side is being obser ved. I f ver y small amount of light is coming fr om t he int er face, it means measur ed sur face is st r aight . ( ii ) Spirit level : Sur face under examinat ion is divided int o number of segment s equal t o t he size of spir it level. Spir it level is t her e kept fr om one segment t o anot her and posit ion of bubble in it is not ed down. The deviat ions of bubble for m t he cent er posit ion ar e r ecor ded. Spir it is used in it due t o it s low viscosit y.

Metrology and Inspection

5.17

(iii ) Autocollimator : As shown in t he figur e below, r eflect or of aut ocollimat or is moved on t he sur face and deviat ions  ar e ar e r ecor ded in t er ms of S = 2f . H ence char act er ist ic can be plot t ed on t he paper. By joining fir st and last point , a st r aight line can be for med and deviat ions of t he sur face fr om t his ideal st r aight line can be not ed down.

Fig. Autocollimat or

FLATN ESS Flat ness is defined as t he depar t ur e of sur face fr om ideal flat sur face. I nit ially aut ocollimat or or spir it level is moved along t hr ee differ ent dir ect ions AB, BC and CA as shown in t he figur e. These movement s will have t hr ee r efer ence point s A, B and C for ident it y. Thr ough t hese point s a plane can be defined. Now aut ocollimat or move pr act ically in all t he possible dir ect ions and char act er ist ic of sur face is not ed down. The deviation of sur face fr om t he r efer ence plane is called flatness.

F ig. M ovem en t of a u toc o l l i m a t o r o v e r su r f a c e

F lat ness Test ing Simplest for m of flat ness t est ing is possible by compar ing t he sur face wit h an accur at e sur face. M at hemat ically, flat ness er r or of t he sur face st at es t hat depar t ur e fr om flat ness is minimum separ at ion of pair of par allel planes which will be in just cont act all point s on t he sur face. Flat ness deviat ion (er r or s of flat ness) ar e indicat ed as follows : (i ) µ or mm per met r e when convexit ies ar e allowed as well as concavit ies. (ii ) Concave t o .....µ or mm, when bet ween t he ends, only concave sur face ar e allowed. (iii ) Convex t o .... or mm, when bet ween t he ends, only convex sur faces ar e allowed. Flat ness of sur face can be t est ed by (i ) Sur face plat e (ii ) Optical flats (iii ) Pr ecision level (iv) aut ocol limet er (v) I nt er fer r omet er s (vi ) Beam compar at or. Deviat ion of lar ge sur face such as sur faces t able or machine t able fr om t he t r ue plane may be det er mined by t he use of eit her a spr it level or an aut ocollimat or.

SQU ARE N E SS Squar eness Test ing Squar ness of sur face can be t est ed by following means : ( i ) Square tester : Engineer s t r y squar e aut ocollimet er and some specially designed for t he pur pose, called square tester. ( ii ) I ndicator method : This met hod is suit able for checking t he squar eness of block whose opposit e faces ar e supposed to par allel. I t is assumed that squar eness of t he block has alr eady been assur ed to r easonable accur acy by t he use of squar e et c. as ot her wise full sensit ivit y of t he met hod can’t be obt ained. The inst r ument for t his pur pose is designed by N.P.L . and is ver y suit able for checking squar eness while manufact ur ing a squar eblock. The inst r ument consist of par allel st r ip (fr ame wor k) and a flat base a knife edge and some ot her for m of indicat or is mount ed on t he fr ame wor k.

Cor r ect ion of Squar eness E r r or The pur pose of det er mining er r or in squar eness of a wor kpiece is t hat it may be cor r ect ed, ot her wise knowing t he er r or will not be of much value. Assume t hat t wo r eading obt ained denot ed pr eviously ar e l 1 and l 2 and dist ance bet ween cont act point is l . L et lengt h of t he squar e block AD be L . Then

er r or in squar eness =

l1  l 2 L  2 l

5.18

Metrology and Inspection

ROU N DN ESS AN D CI RCU LARI TY Oft en t he t er ms r oundness and cir cular it y ar e used int er changeably. Roundness is defined as a condit ion of a sur face of r evolut ion (like cylinder, cone or spher e) wher e all point s of t he sur face int er sect ed by any plane per pendicular t o a common axis, in case of cylinder and cone (or passing t hr ough a common cent r e in case of spher e) ar e equidist ant fr om t he axis (or cent r e). Since axis and cent r e donot exist physically, measur ement s have t o be made wit h r efer ence t o sur faces of t he figur es of r evolut ion only, what ever is measur ed by r efer r ing t o t he sur face of r evolut ion is t he cir cular contour. I t may be under st ood t hat while r oundness expr esses a par t icular geomet r ic for m of a body of r evolut ion in all t he t hr ee dimensionns, cir cular cont our is t he char act er ist ic for m of t he ent ir e per ipher y of plane figur e. For measur ing r oundness, it is only t he cir cular it y of t he cont our which is det er mined.

M easur ement of Roundness 1. D iamet r al met hod I n t his met hod, measur ing plunger s ar e locat ed 180 apar t and diamet er is measur ed at sever al places. This met hod is suit able only when specimen is ellipit ical or has an even number of lobes. Diamet r al check does not necessar ily disclose effect ive size or r oundness. This met hod is unr eliable in det er mining r oundness. 2. Cir cumfer ent ial comfining gauge I t is a shaft confined in a r ing gauge and r ot at ed against a jet indicat or pr obe. I t is useful for inspect ion of r oundness in pr oduction. H owever t his met hod r equir e a separ at e highly accur at e mast er for each size par t t o be measur ed. The clear ance bet ween par t and gauge is cr it ical and r eliabily. This t echnique does not allow measur ement of ot her r elat ed geomet r ic char act er ist ic, such as concent r icit y, flat ness of shoulder s et c. This values obt ained ar e dependent on t he shape of t he specimen. SPECIMEN

CIRCUMFERENTIAL CONFINING GAUGE

CENTRE

CYLINDRICAL PART

CENTRE

INDICATOR

3. Rot at ing on cent r e Some par t s (such as shaft ) may be inspect ed for r oundness while mount ed on cent r es. I n t his case, r eliabilit y is dependent on many fact or s like angle of cent r es, alignment of cent r es, r oundness and sur face condit ions of t he cent r e and cent r e holes, and r un out of piece. Out of st r aight ness of t he par t will cause a doubling r unout effect and appear t o be r oundness er r or. 4. U sing a V-block Set up employed for assessing cir cular it y er r or (lobing) by using a V-block is shown in t he figur e. Vee-block is placed on a sur face plate and the wor k t o be checked is placed upon it . A sensit ive dial indicat or is fir mly fixed in a st and and it s feeler made t o r est against t he sur face of t he wor k. Wor k is r ot at ed t o measur e r ise and fall of wor kpiece. For det er ming number of lobes on t he wor kpiece, t he wor kpieces is fir st t est ed in a 60 V-block anot her in a 90 V-block. Then number of lobes is equal t o number of t imes t he indicat or point er deflect s dur ing r ot at ion of t he wor kpiece t hr ough 360. I dea of t est ing wor kpiece in t wo V-block is t hat when an ellipit ically shaped par t is r ot at ed on 4V-block is t hat when an ellipt ically shaped par t is r ot at ed on 9V-block of an angle 60, no change in r eading is indicated, wher eas if same par t is r ot at ed on a angle 90 angle an veeblock, t wo maximum and t wo minimum r eadings on t he indicat or.

Metrology and Inspection

5.19

Types of V-block (i ) F ixed angle V-block : Dependi ng on number of l obes on a par t , fol l owi ng angles of V-bl ock s ar e r ecommended for measur ement of cor r ect r oundness by V-block met hod :

L obes

Angl e of V-bl ock

Thr ee point out of roundness

60

Five lobed par t

108

Seven-lobed par t

12834

( ii ) Adjustable angle V-block : I t is usually difficult t o aser t ain number of lobes of a par t and have lar ge number of fixed angle V-blocks. V-Block which can be adjust ed t o cor r ect angle t o show out of r oundness is bet t er choice. V-block met hod is limit ed in t he det er minat ion of r oundness of par t s because it is suit able only when number of lobes is known and is unifor mly ar r anged, which is never t he case.

Adjustable V-Block

3-Jaw inside micrometer

For wor kshop pur pose, V-block method is quite accur at e as it is capable of indicat ing nor mal r equir ement of accur acy. H owever for ver y pr ecise job wher e mor e r eliable and mor e accur at e r esult s ar e desir ed, t he second met hod is r ecommended which is quicker and also eliminat es t he effect of angle of t he block and t he number of lobes on apar t , but of cour se, is ver y cost ly one. 5. T hree point probe Thr ee-pr obe wit h 120 spacing is ver y useful for det er mining effect ive size in cases of doubt ful geomet r y of par t . They per for m like a 60 V-block. I t will show no er r or for 5 and 7 lobes, magnify er r or for 3 lobes par t s. 6. Accur at e spindle (i ) Par t fixed, ext er ior spindle wit h pr obe r ot at es (ii ) Pr obe fixed, par t r ot at es wit h spindle.

M ETH OD OF M EASU RI N G SU RFACE FI N I SH Ther e ar e t wo met hods used for measur ing finish of machined par t : 1. Surface inspect ion by compar ison met hods I n compar ison method, sur face textur e is assesed by obser vation of the sur face. But these method ar e not r eliable as these can be misleading if compar ison is not made with sur face pr oduced by same techniques. ( i ) Touch inspect ion M ain limit at ion of t his met hod is t hat t he degr ee of sur face r oughness can't be assessed. This met hod can simply t ell which sur face is mor e t ough. I n t his met hod, finger is moved along t he sur face at a speed of about 25 mm per second and ir r egular it ies as small as 0.01 mm can be easily det ect ed. A modificat ion of it is possible by using a t able t ennis ball, which is r ubbed over t he sur face and vibr at ion fr om t he ball t r ansmit t ed t o hand and sur face r oughness judged t her e by. ( ii ) Visual inspect ion This method is limited to r ougher sur face and r esult var y fr om per son to per son. Mor e accur ate inspection can be done by using illuminat ed magnifier. ( iii ) Scr at ch inspect ion I n t his met hod, a soft er mat er ial like lead or plast ic is r ubbed over t he sur face t o be inspect ed.

5.20

Metrology and Inspection

(iv) M icroscopic inspection Thi s i s best met hod for exami ni ng t he sur face fi ni sh but i t suffer s due t o l i mi t at i on t hat onl y a smal l por t ion of t he sur face can be inspect ed at a t ime. Thus sever al r eadi ngs ar e r equi r ed t o get an aver age val ue. I n t hi s met hod, a mast er fi nished sur face is placed under t he mi cr oscope and compar ed wi t h t he sur face under i nspect i on. I n anot her met hod, a st r ai ght edge i s pl aced on t he sur face t o be inspect ed and a beam of light s pr oject ed at an angle of about 60° t o t he wor k piece. Thus shadows cast int o t he sur face scr at ches ar e magnified and sur face ir r egullar it ies can be studied. (v) Sur face phot ogr aphs I n t his met hod, magnified phot ogr aphs of t he sur face ar e t aken wit h differ ent t ypes of illuminat ion. I n case of ver tical illuminat ion, defect s like ir r egular it ies and sur face appear as dar k spots and flat por t ion of t he sur face appear as br ight ar ea. I n case of oblique illuminat ion, r ever se is t he case. Phot ogr aph wit h differ ent illuminat ion ar e compar ed and r esult s assessed. ( vi ) M icr o int er fer r omet er I n this method, an optical flat is placed on the sur face to be inspected and illuminated by a monochr omatic sour ce of light . I nt er fer ence bands ar e st udied t hr ough a micr oscope. Defect s, i.e. scr at ches in t he sur face appear as int er fer ence line ext ending fr om dar k bands int o t he br ight bands. The dept h of defect s is measur ed in t er ms of fr act ion of t he int er fer ence band. ( vii )Wallace surface dynamomet er This is a sor t of fr ict ion met er and consist s of a pendulum in which t est ing shoes ar e clamped applied. I n t his met hod, pendulum is lift ed t o init ial posit ion and allowed t o swing over t he sur face t o be t est ed. I f sur face is smoot h, t hen t her e will be loss sect ion and pendulum swings for a longer per iod. Thus t ime measur e is a dir ect measur e of sur face finish. ( viii )Reflect ed light int ensit y I n t hi s met hod, a beam of l i ght of k nown quant i t y i s pr oj ect ed upon t he sur face. Thi s l i ght i s r ef l ect ed i n sever al di r ect i on i n beam of l esser i nt en si t y an d ch an ge i n l i gh t i n t en si t y i n di ff er en t di r ect i on i s m easu r ed by ph ot ocel l . T he m easu r ed i nt en si t y ch anges ar e al r eady calli ber at ed by means of r eading t aken fr om sur face of k nown r oughness by some ot her sui t able met hod. 2. D ir ect inst r ument M easur ment s ( i ) St ylus pr ob inst r ument Skid

Pickup

Stylus

( a ) Pr ofi lomet er This is dynamic I nst r ument similar in pr inciple t o a gr amophone pick-up. A finely point ed st ylus mount ed in t he pick up unit is t r ansver sed acr oss t he sur face ( b) Tomlinson sur face met er This inst r ument uses mechanical-cum-opt ical mean magnificat ion. The di amond st yl us on t he sur face fi ni sh r ecor der i s hel d by spr i ng pr essur e agai nst t he sur face of a l apped st eel cyl i nder. The st yl us i s al so at t ached t o body i nst r ument by a l eaf spr i n g an d h ei gh t i s adj u st abl e t o en abl e t h e di am on d t o be posi t i on ed con v en i en t l y. The l apped cyl i nder i s suppor t ed on one si de by t he st yl us and on t he ot her si de by t wo fi xed r ol l er s. The st yl us i s r est r ai ned fr om al l mot i on except t he ver t i cal by t he t ensi on i n coil and leaf spr ing. The t ensile for ces in t hese t wo spr ings ar e also keep t he lapped st eel cylinder i n posi t i on bet ween st yl us and pai r of fi xed r oll er. A l ight spr i ng st eel ar m is at t ached t o t he hor izont al lapped st eel cylinder and it car r ies at it s t ip a diamond scr iber which wear s against a smoked glass.

Metrology and Inspection

5.21

When measur ing surface finish body is traver sed acr oss the surface by a screw r otated by a synchronous mot or. Any ver tical movement of t he st ylus caused by the sur face ir r egular ities, causes the hor izontal lapped st eel cylinder t o r oll. By it s r olling, t he light ar m at t ached t o it s end pr ovides a magnified movement on a smoked glass plat e. The ver t ical movement coupled wit h hor izont al movement pr oduces a t r ace on t he glass magnified in ver t ical dir ect ion and t heir being no magnificat ion in hor izont al dir ect ion. Then smoked glass tr ace is fur ther pr ojected at 50 or 100 magnification for examination. This instr ument is compar atively cheap one and gives r eliable r esult s. ( c) Taylor-H obson Taly surf Taly sur f is an elect r ic inst r ument wor king on car r ier modulat ing pr inciple. The inst r ument also gives same infor mat ion as t he pr evious inst r ument, but much mor e r apidly and accur at ely. I t r ecor ds st at ic displacement of t he st ylus and is dynamic inst r ument line pr ofilomet er.

(ii ) M – system and E– system ( a ) Centre line average method (CLA) or M ean line system (M – system) : I t is ver y-widely used and all the instr ument for measur ing sur face r oughness available in the count r y ar e designed t o measur e in t his syst em. I t is mor e useful and sat isfact or y means of cont r olling at t he point of pr oduct ion, t he consist ency of r esult s fr om a pr ocess when pr oduct ion par amet er have been est ablished. Sur face r oughness is expr essed by Ra. ( b) E (Envelope syst em) syst em I t has bet t er definit ion of peak measur e and it is mor e easily applied t o t he sur face finish inst r ument based on int er fer ence pr inciple. ( iii ) Ot her met hods of measuring sur face r oughness (a) Taper sect ioning (b) Gloss measur ement (c) Diffr act ion t echnique (d) Tact ile t est (e) Pneumat ic gauging (f) Ther mal compar at or.

5.22

Metrology and Inspection

Roughness Grade N umber Var ious r oughness gr ade Number s NI t o N12 in 5 gr oups ar e specified as under by I SP.

Roughness gr ade number

Roughness value, R a µ m

Roughness symbol

N12 N11 N10

50 25 12.5

–

N9 N8 N7 N6 N5 N4 N3 N2 N1

6.3 3.2 1.6 0.8 0.4 0.2 0.1 0.05 0.025

 





I N SPECTI ON OF SCREW TH READ 1. Effect ive D iamet er M easurement The effect ive diamet er or pit ch diamet er can be measur ed by following met hods : ( i ) M icr omet er met hod I f cor r ect ly adjusted, this micr ometer gives pit ch diameter. This value should match with given r elation. Pit ch dia = D – 0.6403 p (in case of whit wor t h t hr ead) wher e 0.6403 p = dept h of t hr ead D = out side diamet er p = pitch ( ii )Wir e met hods ( a ) One-wire method : I n t hi s met hod, one wi r e i s pl aced Micrometer bet ween t wo t hr eads at one side and on t he ot her side anvil Anvil of t he measur ing micr omet er cont act s wit h t he cr est . Act ual measur ement over wir e on one side and t hr eads on ot her side. = size of gauge ± differ ence in t wo micr omet er r eading. Thi s met hod i s used for measur i ng effect i ve di amet er of counter pit ch thr ead ‘p' and dur ing manufact ur ing of t hr eads. The difficulty of this method is that, micr ometer axis may not remain exactly at r ight angle to the thread axis. ( b) Two wire method : Effect ive diamet er of a scr ew t hr ead may be ascer t ained by placing t wo wir e or r ods of identical diameter between flankes of thr ead and measur ing distance over the outside. Effective diamet er is t hen calculat ed. E= T+p wher e, T = dimension under t he wir e = M – 2d M  dimension over t he wir e d = diamet er of each wir e. The value of p depends upon t he diamet er of wir e and pit ch of t he t hr ead. p = 0.9605 p – 1.1657d (for whit wor t h t hr ead) p = 0.866 p – d (for met er t hr ead)

Metrology and Inspection

5.23

(c) Three wire method : I t is most accur at e met hod for measur ing effect ive diamet er.

E

Dist ance over t he wir e, M = E + d

FG1  cosec x IJ H 2K

M

–

p x cot 2 2

wher e, E = effect ive diamet er r = r adius of t he wir e d = diamet er of wir es. 2. Checking “ Thread from and “ Angle” by optical projection of thread This met hod is applicable only t o ext er nal t hr eads because int er nal t hr eads cannot be pr oject ed. The st andar d t ype of pr oject or is used which consist ing of a pr oject or lamp, a condensor lens or collimat or, pr oject ion lens and t he scr een. The scr ew t hr ead t o be examined is placed in t he par allel beam of light bet ween condensor lens and t he pr oject or lens. The moder n pr oject or s ar e equipped wit h wor k holding fixtur e, pr oject ion lamp and t he lenses sit uat ed on t op of t he cabinet and scr een at t he fr ont . The light r ays fr om t he lens ar e dir ect ed downwar ds int o t he cabinet , and hence t o t he scr een by a syst em of pr isms and mir r or s, br inging ever y t hing wit hin t he r each of t he oper at or. Enlar ged image of t he t hr ead for m appear s on t he gr ound-glass scr een on which is mount ed t he t emplat e or dr awing of t he for m made t o scale equal t o magnificat ion of t he lens. I n t his way, t wo for ms (i.e. ideal and pr oject ed) ar e compar ed. One of the difficult ies in pr oject ing scr ew t hr ead is t he fact t hat for m is specified on an axial plane. So cor r ect ion for it must be consider ed. The nor mal pit ch is less t han axial pit ch p and is given by pN = p cos  wher e, = helix angle

M EASU REM EN T OF PI TCH 1. Ext ernal T hreads ( i ) Screw pitch or profile gauge : Such a gauge consist s of ser ies of t hr ead for ms wit h var ying pit ch. One which coincides per fect ly wit h t he t hr ead under t est given t he pit ch. The accur acy of measur ement depends upon t he met hod of slit t ing and is used t o judge t he per fect ness. ( ii ) M icr oscopic met hod : I t is a mor e accur at e met hod. Scr ew t hr eads can be inspect ed and t hen pr ofile angles and linear pit ches checked wit h t he aid of a go micr ometr ic micr oscope. Effect ive pit ch diamet er s can also be measur ed by t his met hod. ( iii ) For st ill mor e accur at e pur poses, it is necessar y t o employ a special scr ew pit ch measur ing machine by which act ual pit ch er r or of individual t hr eads can be measur ed. The pit t er and mat r ix ar e t ypical example of pit ch measur ing machine. 2. I nt er nal T hreads Pitch of internal thread can be measured on any standard pitch machine by using an adaptor. This adaptor carries a bar which can be inserted into r ing, the stylus being fitted to the bar end engaging with the thread in the usual manner. The ring gauge is mounted on a face plate or on the head stock of the machine, which will accomodate rings upto several indicators. For ver y large r ings, a special setup on a sur face plate is necessary, utilising an indicator and slip gauge. G — St ylus Q — Rod holding st ylus R L — L ine for adjust ing st ylus G. This is ar r anged opposit e t o point er K . M — L ocking scr ew for Q. T — I ndicat or point er which is ar r anged t o r ead zer o while t aking micr omet er r eading.

5.24

Metrology and Inspection

Types of T hr ead Gauges 1. Plug scr ew gauge For gauging nut or int er nal t hr eads it is obvious t hat full for m plug gauge made accur at ely t o minimum dimension of int er nal t hr ead which will ensur e t hat all dimensions of t he t hr ead ar e not less t hat minimum if it will assembled wit h t he t hr ead. M ajor, minor and effect ive diamet er will be checked and it will ensur e t hat size of pit ch, angle or for m of t he t hr ead is not r educing below minimum. I t is not possible t o judge individual er r or of a par t icular component . A plain NOT GO gauge is r equir ed for minor diamet er, effect ive diamet er and major diamet er. NOT GO major diamet er gauge has it s major diamet er on t he cor r esponding upper limit of t he wor k wit h it s flank and minor diamet er being well clear ed off t he wor k dimension. I n or dinar y pr act ice, NOT GO effect ive diamet er gauge is t he only one used apar t for m full for m GO gauge. This gauge is necessar y to ensur e that thr eads of t he wor k ar e not thin even though major and minor diameter ar e within the limits. Gauging parallel int ernal screw t hreads : Following t hr ee gauges ar e r ecommended : ( i ) GO and N OT GO plug gauge : To check t oler ance on t he minor diamet er. ( ii ) GO screw plug gauge : To check minimum effect ive diamet er. (iii ) N OT GO screw plug gauge : To check maximum effect ive diamet er. 2. Ring scr ew gauge For pr oduct ion gauging of bolt s, equivalent mat ing sur face of t he bolt t hr eads is called r ing guage. As in case of plug gauges of a syst em of limit , gauges can be pr ovided by full for m NOT GO and GO effect ive diamet er r ing gauge. The NOT GO r ing gauge is t r uncat ed in it s minor diamet er and clear ed on it s major diamet er. Fir st st ep in using a set of r ing t hr ead gauge is t o r ead legend st amped on t he r ings. The legend should give size and pit ch. e.g. 7/16-20 NF means t he wor kpiece should be of "7/16" out side or major diamet er having 20 t hr eads per inch (Nat ional fine t hr eads ser ies). The pit ch diamet er upper limit is 0.4050" which should also be st amped on t he gauge. NOT GO gauge member should also have same legend except diamet er would be 0.4024" (cor r esponding t o lower limit ). Somet ime class of fit also appear s on t he r ing. A t hr ead r ing cannot or dinar ily be used t o analyse the individual er r or pr esent in t he scr ew. I f pit ch diameter is over size, t he r ing will not engage of cour se neit her will it t ur n on an over size major diamet er, if minor diamet er is off (i.e. scr ew who be r oot s ar e filled wit h dir t or made fr om wor k die or lat he t ool) t he gauge will blind. Excessive scr ew t hr ead lead er r or s, like t aper will or dinar ily be det ect ed aft er a few t ur ns of r ing gauge, wher e fine t hr eads ar e being gauged. Car e must be exer cised not t o for ce t he t hr eads of t he r ing gauge for scr ew.

Taylor ’s Pr inciple Applied t o Scr ew T hr eads Gauge A GO gauge should check bot h geomet r ic feat ur e, size and t hus be of full for m, wher eas NOT GO gauge should check only one dimension. I n line wit h t his, GO gague is made of full for m t o t he full lengt h of t hr ead t o t he maximum diamet er of t hr ead. I f NOT GO t hr ead gauge is also made of full for m, t hen vir t ual r educt ion in effect ive diamet er due t o pit ch er r or s may give misleading r esult s. To over come t his difficult y and t o account for Taylor ’s pr inciple, NOT GO gauge for t he t hr eads ar e made t o check for major diamet er and t he effect ive diamet er which is not influenced by t he er r or s in pit ch or for m of t he thr ead.

Gauge Wear The wear t akes place mor e r apidly in scr ew t hr ead gauge t han in plain gauge. The allowance r ecommended by B.S.I for wear on gener ally used solid r ing and plug gauge is 0.05 mm. Ther efor e, fr equent checking of solid pat t er n gauge is necessar y.

F unct ions of Var ious Types of Gauges ( i ) Solid or adjust able GO screw ring gauge I t is employed for checking maximum mat er ial limit of t he vir t ual pit ch diamet er. The solid GO gauge has t o be inspect ed per iodically for wear wit h wear check plug. M aximum size of t he pit ch diamet er of t he Go scr ew r ing gauge is checked by NOT GO check plug gauge. Adjust able GO scr ew r ing gauge is t o set wit h t he set t ing plug and cont r olled by t he wear check plug.

Metrology and Inspection

5.25

( ii ) Wear check plug for solid and adjust able GO ring gauge I t checks t he maximum wear limit s of t he pit ch diamet er of t he GO scr ew r ing gauge and should not ent er by mor e t hen 1 t ur n when scr ewed fr om eit her side of t he GO scr ew r ing gauge. ( iii ) GO screw calliper gauge I t checks maximum limit of vir t ual pit ch diamet er in an axial plan t aking int o account er r or s of pit ch and flank angle. This gauge violat es Taylor ’s pr inciple in t he r espect t hat it does not det ect er r or s of for m, depar t ur e fr om soundness and per iodic er r or s of pit ch. ( iv) N OT GO screw calliper gauge I t checks minimum limit of pit ch diamet er checking of t he wor k by t his gauge, i.e. it is not passing by mor e t han t wo t hr eads over t he wor k should be car r ied in at least for t hr ee posit ions evenly dist r ibut ed over t he ci r cumfer ence.

M EASU REM EN TS OF GEAR I nspect ion of Gear I nspect ion of gear ar e mainly of t wo t ypes : 1. Analyt ical inspect ion Anal yt i cal i nspect i on of gear mean al l t he i ndi vidual el ement s of gear t eet h ar e check ed. Thi s met hod i s sl ow and t edi ous and not of much use for i ndust r y. The discr et e er r or of pi t ch , t oot h pr ofi l e et c can’t gi ve a t r ue over al l assessment of t he accur acy of a gear. I t i s not easy t o assess accur at ely how t hese el ement al values combi ne i n pr act i ce t o gi ve a pr escr i bed per for mance under oper at i onal condi t i on. Al l er r or s i n pi t ch pr ofi l e cause var i at i ons in t he uni for mi t y of r ot ar y mot i on and er r or s in t oot h al ignment or h el i x angl e r esul t i n t he concent r at i on at smal l ar eas inst ead of bei ng di st r i but ed uni for ml y. Analyt ical inspection of gear consist in det er mining following t eet h elements in which err or s ar e caused due t o manufact ur ing er r or s : (i ) Pr ofile (ii ) Spacing (iii ) Pit ch (iv) Runout eccent r icit y concent r icit y (v) Thickness of t oot h (vi ) Lead (vii )Backlash 2. F unct ional inspect ion I t consist of car r ying out r unning t est of gear wit h anot her gear which is moved accur at e and called cont r ol gear or mast er gear t o det er mine composit e vibr at ion, noise level or var iat ion in act ion. Rolling t est I t is most commonly used t est under pr oduct ion condit ion. I n r olling t est , gear t o be t est ed is act ually compar ed wit h a har dened and gr ound mast er gear. This t est r eveals t hat any er r or s in t oot h for m, pit ch and concent r icit y of pit ch line. This t est is gener ally per for med on a most commonly used machine Par kson gear t est er. M easurement of t oot h t hickness Toot h t hickness is gener ally measur ed at pit ch cir cle and is t her efor e pit ch line t hickness of t oot h. M ethods of measuring Gear tooth thickness (i ) By gear t oot h ver nier calliper (ii ) Const ant chor d met hod (iii ) Base t angent met hod (iv) M easur ement by dimension over pins. Toot h t hickness can be ver y convenient ly measur ed by a gear t oot h ver nier calliper. I t is gener ally measur ed C at pit ch cir cle and is t her efor e, r efer r ed as pit ch-line t hickness of t oot h. OA = R d E Chor dal thickness, w = AB = 2 AD A B AOD =  =

D

360 4N

wher e N = number of t eet hs.  w = 2AD = 2 OA sin  = 2R sin 360/4N.

P.C.D 2R Module = = Number of t eet h N N. m 2



R =



w = 2

O

(Wher e, P.C.D  Pit ch cir cle diamet er )

FG IJ = Nm sin  90  H K N 

360 Nm sin 4N 2



5.26

Also,

Metrology and Inspection

chor dal addendum, d = OC – OD

But

OC = OE + addendum = R + m =

and

OD = R cos =

Nm m 2

FG 90 IJ HNK Nm F 90 I N m  1  2 – cos  90   cosG J = H N K 2  N  N  2

Nm cos 2

Nm m– 2 I n case of helical gear s,above expr ession have t o be modified t o t ake int o account change in cur vat ur e along pit ch line.

d =



Ver t ical number of t eet h, for helical gear =

w =



Nm n 3

cos 

sin

FG 90 .cos IJ , HN K

N cos3

3

d=

These for mula is applied when back lash is ignor ed.

Nm n 3

cos 

LM1  2 cos  – cosFG 90 cos IJ OP HN K PQ MN N 3

3

Checking Pr ofile of I nvolut e Shape of Gear Pr ofile er r or is deviat ion of act ual t oot h fr om t he t heor et ical pr ofile in t he designed r efer ence plane of r ot at ion. For t est ing pr ofile, t ip r elief and any por t ion of t he t oot h sur face below t he act ive pr ofile is not consider ed.

M et hods of Checking 1. I nspect ion of profile using dividing head and height gauge This is a ver y t ime consuming met hod but best suit ed for caliber at ion of mast er involut e. I t is t her efor e useful only for ver y pr ecise component s and involut e mast er cams. 2. Gear involut e measur ing machine I t i s desi gned for check i ng i nvol ut e pr ofi l es of t he spur and ot her gear s. Thi s machi ne i s sui t abl e for i nspect i on of gear havi ng modul e fr om 1 t o 10 mm havi ng maxi mum out si de di amet er upt o 300 mm. Permitted tolerances on the profile error

Accuracy class or Grade of gear 1 2 3 4 5

Profile tolerance in microns 2.0 + 0.06 K 2.5 + 0.10 K 3.0 + 0.16 K 4.0 + 0.25 K 5.0 + 0.40 K

H er e K is t oler ance and given by wher e, m = module D = pit ch circle diamet er

K

= m + 0.1

D

M easur ement of Gear Pitch Gear pit ch can be measur ed in t he following ways. 1. Cumulat ive cir cular pit ch er r or over a span of t eet h 2. Adjacent pit ch er r or or pit ch var iat ion 3. Base pit ch var iat ion.

Composit e M et hod of Gear Checking Composit e t est ing of gear s consist in measur ing t he var iat ion in cent r e dist ance when a gear is r olled in t ight mesh (double flank cont act ) wit h specified or mast er gear.

Types of Checki ng 1. Tot al var iat ion Tot al composit e var iat ion is t he cent r e dist ance var iat ion in one complet e r evolut ion of t he gear being inspected;

Metrology and Inspection

5.27

2. Tooth to tooth composite variation Toot h t o t oot h composit e var iat ion in t he cent r e dist ance var iat ion as t he gear is r ot at ed t hr ough any 360 incr ement of . A unifor m t oot h t o t oot h var iat ion shows pr ofile var iat ion wher eas sudden jump N indicat es t he pit ch var iat ion. Composit e t ype a checking t akes car e of all t he er r or s in t he year.

Par kinson Gear Test er The pr inciple of t his cent r e is t o mount a st andar d gear on a fixed ver t ical spindle for t he gear t o be t est ed on anot her similar spindle mount ed on sliding car r iage, maint aining t he gear s in mesh by spr ing pr essur e. Test er ar e available for t est ing spur gear, Bevel gear, helical gear, and wor m gear.

Checking of Gears U sing Grat ings A gear could eit her be t est ed by checking involut e pr ofile, helix pr ofile and pit ch or wher e fast er speed of measur ement is desir ed by met hod using int er act ion of all of above par amet er s, i.e composit e t est (meshing gear under t est wit h a per fect ly mast er gear ). M echanical met hods of t est ing demands t he highest st andar ds of mechanical accur acy which some how is limit ed. H owever wit h gr at ing, it possible t o measur e lar ge absolut e angular and linear displacement t o great accuracy. A big advantage of the gr ating is its output in the for m of digital pulses which can be conveniently manipulat ed by digit al elect r onic t echniques. Gear t r ansmission er r or s ar e measur ed convenient ly using gr at ing.

Gear Toot h Calliper I t is used t o measur e t hickness of gear t eet h at t he pit ch line or chor dal t hickness of t eet h and dist ance fr om t op of a t oot h t o t he chor d. Thickness of a t oot h at pit ch line and addendum is measur ed by an adjust able t ongue, each of which is adjust ed in dependent ly by adjoining scr ew on gr aduat ed bar s. The effect of zer o er r or s showed so t aken int o consident ion.

Const ant Chor d M et hod Bot h chor dal t hickness and chor dal addendum ar e dependent upon t he number of teet h. H ence for measur ing a lar ge number a gear s for set , each having differ ent number of t eet h would involve separ at e calculat ion. Const ant chor d of a gear is measur ed wher e t oot h flanks t ouch t he flanks of t he basic r ack. The t eet h of t he r ack ar e st r aight and inclined t o t heir cent r e line at t he pr essur e angle.

ALI GN M EN T TEST 1. On L athe M achine Befor e var ious t est s on any machine t ool ar e car r ied out , it is essent ial t hat it should be inst alled in t r uly hor izont al and ver t ical planes. I n hor izont al plane, bot h longit udinal and t r ansver se dir ect ions ar e equally impor t ant . I f bed is not inst alled t r uely in hor izont al t r ansver se dir ect ion, t wist will be int r oduced. Thus movement of saddal can’t be in a st r aight line and t r ue geomet r ic cylinder can’t be gener at ed. The level of machine bed in longitudinal and tr ansver se dir ections is gener ally t ested by a sensitive spr it level. St r aight ness of bed in longit udinal dir ect ion for a long beds can also be det er mined by ot her met hod e.g. using st r aight edge, aut o collimat er s or by t wo wir e met hod. But t he t est in t r ansver se dir ect ion can be car r ied out only by spr it level.

I t is desir ed that por t guideway should be convex only as cutting forces and weight of carrige act downwar d on it. I f fr ont guideways are concave, then effect will be cumulative. The tendency of carriage, under cutting forces is to lift upwards fr om rears and this is traversed by a gib placed underneath the guideways with the result, an upward force acts on the rear guideway, which must therefore be made concave. Tr ansver se level may be in any dir ect ion, but no t wist can be t oler at ed.

5.28

Metrology and Inspection

( a ) Parallelism of Tail stock guideways with M ovement of carriage I f t ail st ock guideways ar e not par allel wit h t he car r iage movement , t her e will be some offset of t he t ail st ock cent r e and t his r esult in t aper t ur ning. To check par allelism of t he t ailst ock guideways in bot h t he planes i.e. hor izont al and ver t ical, a block is placed on t he guideways as shown in t he figur e and faler of t he indicat or is t ouched on t he hor izont al and ver t ical sur faces of t he block. The dial indicat or is held in t he car r iage and car r iage is moved.

Fig. Error indicated by the pointer of dial indicator.

( b) Pit ch accur acy of L ead screw The accur acy of t hr eads cut on any machine depends upon accur acy of it s lead scr ew. Test for t his is per for med by fixing a posit ive st op on the lat he bed. Against t he st op, lengt h bar s and slip gauges can be locat ed. A suit able met hod for r ecor ding pr ogr essive and per iodic er r or s is by using a suit ably divided scale, which is placed close to the line of centres. A microscope is rigidly mounted on the car riage in a conveniently posit ion t o not e r eading on t he scale. 2. Alignment Test on M illing M achine Machine is always carefully adjusted and aligned on the test stand or is assembly department of the manufacture. M achine should be car efully levelled up by means of spr it level befor e st ar t ing wit h act ual t r ial t est machine t ools for t he wor kshop must be able t o pr oduce wor k pieces of given accur acy wit h in pr escr i bed limit , consist ent ly and wit hout r equir ing ar t ist ic skill on t he par t of t he oper at or. Test to be applied 2 Cut t er spindle Axial Slip or Float

Eccent ricit y of Ext er nal Diamet er I nt er nal t aper for t r ue running

Wor k Table Table sur face par allel wit h ar bor r ising t owar ds over ar m

Test Diagram 3

Gauges and M ethod 4 (A) Clock I ndicat or (B) Test t wo places at 180°

Permissible Error 5 A. 0.01 mm B. 0.01 mm

Clock I ndicat or slowly move mandr el for maxi mum eccent ricity (A) Near est t o spindle nose (B) At a dist ance of 300 mm

A. 0.01 mm B. 0.025 mm

St at ionar y mandr el. Clock indicat or s Under mandr el

0.025 mm per 300 mm

Test t able sur face for max. t r avel

Metrology and Inspection

5.29

3. Alignment tests on pillar type drilling M achine (i ) Flatness of clamping surface of base This t est is per for med by placing a st r aight edge on t wo gauge block on t he base plat e in var ious posit ion and er r or not ed down by inser t ing t he feeler gauge. This er r or should not exceed

0.1 mm 1000

clamping sur face and sur face should be concave only. ( ii ) F lat ness of clamping surface of t able This t est is per for med in t he same manner on t est (i ) but on t he t able. ( iii ) Perpendicularity of drill head guide t o t ie base plat e Squar eness (per pendicular it y) of dr ill head guide t o t he base plat e is t est ed in (i ) ver t ical plane passing t hr ough axes of bot h spindle and column. (ii ) plane at 90 t o t he plane ( iv) Perpendicularit y of spindle sleeve wit h base plat e This t est is per for med in bot h t he planes specified in t est (iii ) and in t he similar manner wit h differ ence t hat fr ame levels ar e t o be placed on spindle sleeve and base plat e. (v) True running of spindle t aper For t his t est , t est mandr el is placed in t aper hole of spindle and dial indicat or is fixed on t he t able and t eet h made t o scan t he mandr el. Spindle is r ot at ed slowly and r eading of indicat or not ed down. Er r or should not exceed

0.03 mm for machine wit h t ape upt o mor se number 2 100

0.04 mm for machine wit h t aper lar ger t han mor se number 2. 300

( vi ) Parallelism of spindle axis wit h vert ical movement This t est is per for med in t wo plane (A) and (B) at r ight angle t o each ot her. This t est mandr el is fit t ed in t he t aper ed hole of t he spindle and dial indicat or is fixed on t he t able wit h it s feeler t ouching mandr el. (vii )Squareness of clamping surface of table to its axis For t his t est , dial indicat or is mount ed in t he t aper hole of t he spindle and it s feeler is made t o t ouch sur f ace of t he t abl e. Tabl e i s sl owl y r ot at ed an d r eadi ng of di al gau ge not ed down , wh i ch should not exceed

0.05 mm diamet er.. 300

5.30

Metrology and Inspection

TOLERAN CE AN ALYSI S I N M AN U FACTU RI N G AN D ASSEM BLY Toler ance can be defined as t he magnit ude of per missible var iat ion of dimension or ot her measur ed or cont r ol cr it er ia fr om t he specified value. Toler ances have t o be allowed because of t he inevit able human failing and machine limit at ions which pr event ideal achievement s dur ing fabr icat ion. I n or der t o maint ain economic pr oduct ion and facilit at e t he assembly of component s, it is necessar y t o allow a limit ed deviat ion fr om t he designed size. Toler ance const it ut es an engineer ing legalit y for deviat ion fr om t he ideal value and for mulat ion of t oler ance should be given due consider at ion.

F act or Affect ing Select ion of Toler ance ( i ) Funct ional r equir ement , i.e. per mit t ed deviat ion in size must per mit t he assembly t o funct ion cor r ect ly. ( ii ) Standar disation. ( iii ) M anufactur ing needs. Wher e t he funct i onal per for mance pr ovi des some l at i t ude, t hen t ol er ances choi ce may be infl uenced and det er mined by fact or s like st andar disat ion, met hods of t ooling and manufact ur ing equipment available. The numer ical values of t oler ances may r ange acr oss t he ent ir e spect r um of measur ement s and funct ional r equir ement is t aken as t he only cr it er ion t o decide t he value of t oler ance.

D i sadvant ages (i ) Too many differ ent tolerances which mean excessive amounts of calculations and studies to establish tolerance values. (ii ) Excessive amount of special t ooling. (iii ) Complicat ions in inspect ion et c. I f a limit ed number of st andar d t oler ances ar e est ablished and t oler ances ar e chosen fr om t hese, so t hat t hese ar e slight ly closer t han the funct ion dict ates, t hen advant ages of fewer var iat ions of t ooling, few calculations and incr eases unit quant it ies because of r epeat ed use of t he same designs ar e obt ained.

N eed for Toler ances ( i ) Var iat ions in t he pr oper t ies of mat er ial being machined which pr oduce er r or s. ( ii ) Pr oduct ion machines t hemselves have some inher ent inaccur acies built int o and have t he limit at ions t o pr oduce per fect par t s. ( iii ) I t is impossible for an oper at or t o make per fect set t ing. I n set t ing up t he machine, i.e. in adjust ing t he t ools and wor kpiece on t he machine, some er r or s ar e likely t o cr eep in. An at t empt t o ent ir ely over come t hese fact or s wit h a view t o obt ain ideal condit ions would r esult in exhor bit ant cost s. The par t s should, t her efor e, be made as inaccur at e as t oler able t o sat isfy t he funct ional r equir ement s. Thus t oler ances ar e specified t o t he dimensions of all manufact ur ed par t s and t hese should be just enough t o do t he int ended job and not bet t er. Thus toler ance is a compr omise between t he accur acy r equir ed for pr oper functioning and abilit y t o economically pr oduce t his accur acy.

Repr esent at ion of Toler ances Toler ances ar e basically specified in t wo for ms. ( i ) U nilat er al t oler ances I n t hese, t ot al t oler ance is r elat ed t o a basic dimension, and is in one dir ect ion only. This for m of t oler ance is usually indicat ed when t he machining of making par t s is called for, as t his gr eat ly assist s t he oper at or. The oper at or machines the upper limit of shaft (lower limit for a hole) knowing fully well t hat he st ill has t he whole t oler ance left for machining befor e t he par t s ar e r eject ed. M et hods of specifying unilat eral t oler ances ( a ) Specify as limit ing dimensions Diamet er of hole : 25.000 mm, 25.002 mm Diamet er of shaft : 24.999 mm, 24.997 mm

Metrology and Inspection

5.31

( b) One limit ing size may be specified wit h it s t oler ances Diamet er of hole : 25.000 + 0.002, – 0.000 mm Diamet er of shaft : 25.000 + 0.000, – 0.002 mm (c) Nominal size may be specified for bot h par t s, wit h a not at ion showing bot h allowance and t oler ance. Diamet er of hole : 25.000 + 0.002, – 0.000 mm Diamet er of shaft : 25.000 – 0.001, – 0.003 mm H er e dimension is allowed t o var y only in one dir ect ion. (ii ) Bilateral tolerances : I n bilat er al t oler ance, t ot al t oler ance is specified on bot h sides (plus and minus) of t he basic dimension. These t oler ances usually have plus and minus t oler ance of equal amount , but not necessar ily always. This syst em per mit s t he oper at or t o t ake full advant age of t he limit syst em especially in posit ioning a hole. Bilat eral t oler ances may be expr essed as : 25.000 + 0.001, – 0.002 mm This system is used in machining pr ocess like dr illing in which dimensions ar e most likely t o deviat e in one dir ect ion only (dr illing hole is always over size r at her t han under size). Bilat er al t oler ance syst em defines t he t heor et ically desir ed size of t he basic size. This syst em is used in mass pr oduct ion wher e machine set t ing is done for t he basic size. Under such condit ions, if t oler ances ar e specified and unilat er al, t hen t hese should be changed t o bilat er al t oler ances by changing t he basic size. The basic size should be midway bet ween upper and lower limit s.

ASSEM BLY LI N E BALAN CI N G Line balancing is the appor tionment of sequential wor k activities int o wor k stations in or der to gain a high utilisation of labour and equipment, and ther efor e, minimum idle t ime. Compatible wor k activities ar e combined int o appr oximat ely equal time gr oupings that do not violate the or der (or pr ecedence) in which they must be done. Once a line is oper at ing at full capacity, t he cycle t ime at which t he complet ed pr oduct s leave the pr oduct ion line will be given by AT available time / period Cycle Time (CT) = = output output units required / period I t t he t ime r equir ed at any st at ion exceeds t hat which is available t o one wor ker, addit ional wor ker may have t o be added t o t he st at ion. Theor et ical or ideal number of wor ker s needed on t he assembly line will be given by (worker time / unit)  (output units / period) Theor et ical minimum number of wor ker s = available time / period t output units / period = t = CT available time / period

COM PARAT ORS

FG H

IJ K

The compar at or is an inst r ument used for compar ing t he dimensions of a component wit h a st andar d of lengt h.

Essent ial Part s of a Comparator (1) A fixed sur face fr om which all measur ement s ar e t aken. (2) A ver y sensit ive indicat or which will show t he movement of a sliding piece usually t er minat ing in an anvil wit h a cur ved face. (3) Some means of set t ing t he cur ved face (r efer r ed t o in 2) at adjust able dist ance fr om sur face (1). This ar r angement is used t o measur e t he differ ence bet ween lengt h or diamet er of a component and a st andar d of lengt h, usually made up of slip gauges. Thus the pur pose of a compar ator is to detect and display the small differ ences between unknown linear dimension and lengt h of t he st andar d. The differ ence in lengt h is det ect ed as a displacement of a sensing pr obe. The impor tant and essent ial funct ion of the instr ument is t o M agnify or Amplify t he small input displacement so t hat it is displayed on an analog scale.

5.32

Metrology and Inspection

U ses of Comparators (i ) (ii ) (iii ) (iv) (v)

To inspect newly pur chased gauges. I n mass pr oduct ion, wher e component s ar e t o be checked at a ver y fast r at e. As labor at or y st andar ds fr om which wor king or inspect ion gauge ar e set and cor r elat ed. I n sel ect i ve assembl y of par t s, wher e par t s ar e gr aded i n t hr ee gr oups dependi ng upon t hei r t ol er ance. As wor king gauges, t o pr event wor k spoilage and t o maint ain r equir ed t oler ance at all impor t ant st ages of manufactur e.

Types of Compar at or s 1. M echanical Compar at or A mechanical compar at or employs mechanical means for magnifying t he small movement of t he measur ing st ylus br ought about due t o t he differ ence bet ween st andar d and t he act ual dimension being checked. The met hod of magnifying t he small st ylus movement in all t he mechanical compar at or s is by means of lever s, gear t r ains or by t heir combinat ions. 2. E lect r ical Compar at or These compar at or s have lit t le or no moving par t s and hence can r et ain t heir accur acy over per iods and also t he sensit ivit y can be adjust ed at will. A higher magnificat ion can be achieved as compar ed t o mechanical compar at or s. 3. Pneumat ic Compar at or Pneumat ic compar at or s wor k on t he pr inciple t hat of an air jet . Upon decr easing t he st and-off dist ance, t he pr essur e on t he back gr ound of jet will incr ease. This pr essur e is called back pr essur e and can be dir ect ly r elat ed t o t he measur ement .

F i g . P r in ci p l e of a n a ir jet

Types of Pneumat ic Comparat or s These compar at or s can be fur t her be divided int o t hr ee : ( i ) F low t ype

Fig. Rotameter

The design of head is decided by the geomet r ic feat ur e under measur ement . Depending upon t he dist ance t her e will be back pr essur e ‘P2’ and t he differ ence bet ween (P2 – P1) decides posit ion of t he float . This float is having aer ofoil shape blades so t hat due t o t he flow of air it r ot at es. I f it is st at ionar y, it will st ick to the wall of r otameter and due to static fr iction between float and tube, the system will become insensitive t o small change in back pr essur e ‘P2’. The t ube is slight ly t aper ed t o maint ain linear r elat ionship in t he measur ement because air is compr essible.

Metrology and Inspection

5.33

(ii ) Differential type This compar at or is used t o compar e t wo sur face i.e., which sur face is at a higher level and which is lower. Two pr essur e st r eams P1 and P2 does not mix wit h each ot her as shown in bellow. One st r eam goes inside t he bellow and ot her r emains on t he out er sur face. Depending upon posit ion of bellow fr om t he neut r al posit ion, it can be known t hat which sur face is higher and which is lower.

Bellow

( iii ) Back pressur e t ype Back pr essur e r eadings can be not ed down fr om t he manomet er.

4. M echanical Comparators (a) Usually cheaper t han t he ot her devices. (b) Does not r equir e any ext er nal power supply t o oper at e. (c) Accur acy in inspect ion is independent of accur acy in manufact ur ing differ ent linkages. (d) Usually it has linear scale. (e) M or e moving linkages which br ing down t he accur acy. (f) Range of t he inst r ument is limit ed. (g) Due t o t he iner t ia of moving linkages makes t he compar at or sensit ive t o machine vibr at ions. Air gauging has r ecent ly incr eased due t o it s ver y high amplicat ions as no physical cont act is made eit her wit h t he set t ing gauge or t he par t being measur ed.

Based on the physical phenomenon on which the operation of pneumatic gauges is based these may be classified as : (i ) Flow or velocit y t ype compar at or. I t oper at e by sensing t he manomet r y r at e of air flow. (ii ) Back pr essur e t ype compar at or. I t oper at e by sensing t he velocit y differ ent ial pr essur e acr oss a vent ur i chamber. 5. Opt ical Compar at or s I n optical compar at or s, magnificat ion is obt ained wit h t he help of light beams which have advant age of being st r aight and weight less. Opt ical compar at or s have t heir own built -in illuminating device which t ends to heat t he inst r ument and t hus accur acy is liable t o suffer. I n mechanical opt ical compar at or s a small displacement of t he measur ing plunger is amplified fir st by a mechanical syst em consist ing of pivot ed lever s and fur t her a simple opt ical syst em involving t he pr oject ion of an image is amplified. 6. E lect r onic Compar at or s I t is based on t he pr inciple of applicat ion of fr equency modulat ion or r adio oscillat ions. I t pr ovides a r eliable means of measur ing ext er nal and int er nal measur ement s wit h a r emar kable accur acy and ease.

5.34

Metrology and Inspection

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. The case wit h which obser vat ions can be made accur at ely is called (a) r eadability

(b) sensit ivity

(c) accur acy

(d) r eputability

2. Accur acy of measur ing equipment is (a) t he closeness wit h which a measur ement can be read directly from a measur ing instr ument. (b) a measur e of how close t he r eading is t o t he t r ue size. (c) the differ ence between measur ed value and act ual value (d) the smallest change in measur e that can be measured 3. Er r or s ar e gener ally di st r ibut ed in accor dance wit h t he Gaussian dist r ibut ion is (a) cont r ollable er r or s (b) calibr at ion er r or s (c) avoidable er r or s (d) r andom er r or s 4. Figur e given below shows the dimension obtained on a component by a certain instrument.

This inst r ument is (a) pr ecise but not accur at e (b) accur at e but not pr ecise (c) neit her pr ecise nor accur at e (d) sensit ive. 5. Toler ance ar e specified (a) t o obt ain desir ed fit s (b) because it is not possible t o manufact ur e in size exact ly (c) t o obt ain high accur acy (d) t o have pr oper allowance 6. The fact t hat how closely t he inst r ument r eading follows t he measur ed var iables is called (a) fidelity (b) accur acy (c) t hr eshold sensit ivit y (d) pr ecision

7. Pr ocess capabilit y of t he machine t ool is (a) t o pr oduce maximum number of component s in unit t ime. (b) equal t o t he mean value of component (c) equ al t o t h e t h r ee t i m es t h e st an dar d deviation (d) equal t o six t imes t he st andar d deviat ion. 8. A bor e 14.67 mm in a wor kpiece can be measur ed by (a) st eel r ule (b) pneumat ic gauges (c) mi cr omet er (d) plug gauge 9. Scale sensit ivit y is defined as t he r at io of (a) change i n scal e r eadi ng t o cor r espondi ng change in point er deflect ion (b) least r eading of scale t o r ange of scale (c) l east r eadi ng of scal e t o uni t measur abl e quantity (d) least count of scale t o t or ange of scale 10. A thr ee-lobed par t if checked on 60° V-block would pr ovide magnificat ion of t he r adial out -of-r ound char act er ist ics (a) 1 t ime (b) 2 t imes (c) 3 t imes (d) 4 t imes 11. A five-lobed par t , if gauged in t he V-block would pr oduce t he magnificat ion of t he r adial out -ofr ound char act er ist ics (a) 0 t ime (b) 1 t ime (c) 2 t imes (d) 3 t imes 12. A mast er gauge is a/an (a) new gauge (b) int er nat ional r efer ence standar d (c) st andar d gauge for check i ng accur acy of gauges used on shop floor s (d) gauge used by exper ienced t echnicians 13. Expr essing a dimension as 32.5/32.3 mm is t he case of (a) unilat er al t oler ance (b) bilat er al t oler ance (c) limit ing dimensions (d) none of t hese 14. M oir e fr inges ar e obser ved when (a) an opt ical flat is placed over smoot h sur face (b) a mi cr oscope i s used t o obser ve su r face t ext ur e (c) index gr at ing is moved over scale gr at ing (d) whit e light is diffused t hr ough a pr ism

Metrology and Inspection

15. T h e su r f ace r ou gh n ess on a dr aw i n g i s r epr esent ed by (a) cir cles (b) squar es (c) zig-zag lines (d) tr iangles 16. I n i nt er fer omet r i c met hods, pat h di ffer ence bet ween one br ight band and t he next is var ied by (a) half wave lengt h (b) t wo wave lengt h (c) one quar t er wavelengt h (d) t wo wave lengt hs 17. The ‘best si ze wi r e’ for measur i n g ef fect i ve diamet er of t hr eads is of diamet er p sec  p cos (a) (b) 2 2 p cos (c) (d) none of t hese 4 wher e, p = pit ch of t hr ead  = semi-angle of t hr ead 18. On a t r iple t hr ead scr ew (a) lead = pit ch (b) lead = 3 pit ch

1 pit ch (d) lead = 9 pit ch 2 T h e m et h od of f r act i on al coi n ci den ces i n int er fer omet r y t echniques is used for (a) measur ement of end gauges (b) flat ness of sur face (c) linear displacement measur ement s (d) sur face r oughness measur ement Aut ocollimat or is used for (a) par allelism measur ement (b) st r aight ness measur ement (c) flat ness measur ement (d) angular measur ement The pitch cir cle r adius r p and base cir cle r adius r b of a gear ar e r elat ed by t he r elat ionship ( = pr essur e angle) (a) r b = r p cos  (b) r b = r p sin  (c) r b = r p t an  (d) r b = r r (cos – ) Syst emat ic er r or s ar e (a) r andomly dist r ibut ed (b) r egular ly r epet it ive in nat ur e (c) distributed on both + ve and – ve sides of mean value (d) unpr edictable The maximum amount by which t he r esult differ s fr om t he t r ue value is called (a) cor r ect ion (b) discrepancy (c) er r or (d) accur acy (c) lead =

19.

20.

21.

22.

23.

5.35

24. To inscr ibe lines par allel t o t he edges of a par t , t he inst r ument used is (a) ver nier calliper s (b) scr ew gauge (c) her maphr odit e calliper s (d) combinat ion set . 25. Which of the following ar e not contr ollable er r or s? (a) Calibr at ion er r or s (b) Envir onment al er r or s (c) Avoidable er r or s (d) Random er r or s 26. A scal e w h ose gr adu at i on m ar k s ar e i n a di scont i nuou s manner and ar e composed of aligned number s indicating directly the numer ical value of t he quant it y measur ed is called (a) linear scale (b) equidist ant scale (c) r egular scale (d) digit al scale 27. The thr ead micr ometer measur es (a) major diamet er of t he t hr ead (b) minor diamet er of t he t hr ead (c) effect ive diamet er of t he t hr ead (d) r oot diamet er of t he t hr ead 28. V-block is used in t he wor kshop t o check (a) r oundness of a cylindr ical wor k (b) sur face r oughness (c) t aper on a job (d) none of t hese 29. Repeat abilit y of measur ing equipment is (a) t he closeness wit h which a measur ement can be r ead dir ectly fr om a measur ing inst r ument (b) a measur e of how close t he r eading is t o t he t r ue size (c) differ ence between measured value and actual value (d) t he capabilit y t o indicat e t he same r eading again and again for a given measur and 30. The pur pose of r atchet scr ew in micr omet er scr ew gauge is t o (a) lock a dimension (b) impar t blow mot ion (c) maint ain sufficient and unifor m measur ing pr essur e (d) t ake car e of wear of scr ew t hr eads.

LEVEL-1 31. Ext er nal t aper can be accur at ely measur ed wit h t he help of (a) sine bar and slip gauge (b) dividing head (c) combinat ion set (d) cl inomet er

5.36

Metrology and Inspection

32. St ick micr omet er s ar e designed for measur ing (a) bor e of cylinder s (b) longer ext er nal lengt hs (c) cylindr icity (d) longer int er nal lengt hs 33. A sine bar is specified by (a) it s t ot al lengt h (b) cent r e dist ance bet ween t he t wo r oller s (c) size of t he r oller s (d) weight of sine bar. 34. The number of slip gauges in a set ar e (a) 87

(b) 45

(c) 103

(d) 31

(e) all of t hese 35. T h e essen t i al r equ i r em en t f or accu r acy of measur ement wit h a sine bar is (a) flat ness of upper sur face (b) equalit y of size and r oundness of r oller s (c) exact distance between r oller axes and mutual par allelism. (d) par allelism of r ol ler s t o upper sur face and equalit y of axis dist ance as fr om sur face (e) all of t hese 36. I n a sine bar t he st andar d lengt h is measur ed (a) fr om edge t o edge (b) between inner cir cumfer ences of two r oller s (c) bet ween out er cir cumfer ence of t wo r oller s (d) bet ween t he cent r es of t wo r oller s 37. Accur acy of set t ing a sine bar (a) decr eases appr eciably wit h st eep angle (b) is poor for small angles (c) is maximum when angle of measur ement is 45C (d) none of t hese 38. Figur e below shows a case of er r or in r elat ive locat ion of sur faces. This case is for

(a) misalignment

(b) axial r unout

(c) r adial r unout

(d) squar eness er r or

39. The M -syst em and E-syst em i n met r ol ogy ar e r elat ed wit h measur e-ment of (a) gear s (b) scr ew t hr eads (c) flatness (d) sur face finish. 40. All t he t hr ead char act er ist ics can be measur ed pr ecisely wit h (a) scr ew pit ch gauge (b) micr omet er wit h V-anvil (c) t ool r oom micr oscope (d) t hr ead gauge 41. The advantage of ver nier caliper over micr o-meter is t hat it (a) is easier and quicker t o use (b) is mor e accur at e (c) can be used to make both inside and outside measurements over a range of sizes (d) all of t hese 42. I n layout wor k, a pencil should not be used t o dr aw lines on met al because (a) it will wipe off easily (b) t he line will be t oo wide for accur at e wor k (c) lines will smudge and be difficult t o see (d) all of t hese 43. The basic unit in angular measur ement is (a) degr ee (b) minut e (c) second (d) r ight angle 44. M illimet r e scale in a micr omet er is mar ked on (a) bar r el (b) thimble (c) spindle (d) anvil 45. Cir cular scale of t he micr omet er is mar ked on (a) anvil (b) bar r el (c) r at chet (d) thimble 46. The following t ype of gauges has gauging sections combined on one end (a) combinat ion gauge (b) limit gauge (c) Go and No Go gauge (d) pr ogr essive gauge 47. Accur acy is (a) r epeat abilit y of a measur ing pr ocess (b) er r or of j u dgem en t i n r ecor di n g an obser vat ion (c) abi l i t y of i nst r ument t o r epr oduce same r eading under ident ical sit uat ions (d) agr eement of t he r esult of a measur ement wit h t he t r ue value of t he measur e quant it y

Metrology and Inspection

48. The least accur at e measur ing device is (a) air gauge (b) micr omet er scr ew gauge (c) opt ical pr oject or

(a) L = H tan

 d  tan (45 –  ) 2 2

(b) L = H cot

 d   tan 45 – 2 2 2

(c) L = H tan

d  tan 45 – 2 2

FG H

(d) st eel scale 49. A compar at or for it s wor king depends on (a) accur at ely calibr at ed scale (b) compar ison with standar d such as slip gauges

FG H

5.37

IJ K

IJ K

(d) None of t hese 54. Following figur e shows t he measur ement of

(c) accur at e micr omet er gauge (d) opt ical devices 50. Pr ecision is (a) r epeat abilit y of a measur ing pr ocess (b) agr eement of t he r esult of a measur ement wit h t he t r ue value of t he measur ed quant it y (c) abilit y of an inst r ument t o r epr oduce same r eading under ident ical condit ions (d) all of t hese 51. Following figur e shows t he pr inciple of Hole tolerance

(a) r oundness (b) r adius of cur vat ur e (c) cylindr icity (d) flatness 55. The r adius of a pulley block is measur ed as shown in t he figur e is calculat ed by

Number of hole gradings Tolerance on shaft

(a) t r aceabilit y (b) int er changeability

(a) R =

l–d 8d

(b) R =

(l – d ) 2 4d

(c) R =

(l – d ) 2 8d

(d) R =

(l – d ) 2 2d

(c) matched fit s (d) select ive assembly 52. M ost accur at e inst r ument is (a) st eel scale (b) micr omet er scr ew gauge (c) ver nier caliper

56. The r adi us of concave sur face can be easi l y det er mined by a dept h micr omet er as shown in t he figur e. Radius of cur vat ur e R is equal t o

(d) opt ical pr oject or 53. A pl ug of di amet er d r est s i n an angl e as shown i n fi gur e. A n equat i on gi vi ng di st ance L i n t er ms of d, H and  woul d be

(a)

d2 h  8h 2

(b)

d2 h – 8h 2

(c)

d2 h 4h

(d)

d2 h 8h

5.38

Metrology and Inspection

57. The accessor y of slip gauges is

65. The cor r ect way of designat ing fit , is

(a) scr ibing and cent r e point s (a)

(b) measur ing jaws (c) holder (d) base

(c)

(e) all of t hese 58. The later al faces of slip gauges ar e at r ight angles cor r ect t o wit hin (a) ± 1 degr ee

(b) ± 30 minut es

(c) ± 10 minut es

(d) ± 1 minut e

59. According to accuracy, slip gauges are classified under which of the following number of accuracy classes (a) t wo

(b) t hr ee

(c) four

(d) five

60. A pr ot ect or in slip gauges is pr ovided t o (a) pr ot ect slip gauges when not is use (b) take up all the wear when in use (c) clean t he slip gauges (d) facilit at e wr inging of slip gauges

61. The r at io of t he sur face ar ea A and t he volume V of a cylinder of diamet er d and lengt h l is

(c)

A ld = V dl

A 4l (b) = V d (d)

g7 50 H g g7

(b)

g7 H8

(d)

H8 – 50 g7

66. Dr illed holes and honed holes, could be designated by following gr ades r espect ively (a) H 5, H 11

(b) H 6, H 10

(c) H 8, H 6

(d) H 10, H 5

67. Sensit ivit y of measur ing equipment is (a) closeness wit h which a measur ement can be r ead dir ect ly fr om a measur ing inst r ument (b) a measur e of how close t he r eading is t o t he t r ue size (c) differ ence between measured value and actual value (d) t he smallest change in measur and t hat can be measur ed 68. Expressing a dimension as 25.3+0.05 mm is the case of

LEVEL-2

A 6d (a) = V l

Hg

A 2d  4l = V dl

62. The t wo slip gauges in pr ecision measur ement ar e joint ed by (a) assembling

(b) sliding

(c) adhesion

(d) wr inging

63. Plug gauges ar e used t o (a) measur e t he diamet er of t he wor kpieces (b) measur e t he di amet er of t he hol es i n t he wor kpieces (c) check diamet er of t he holes in t he wor kpieces (d) check lengt h of holes in t he wor kpieces 64. Ter m “ Allowance” i n li mit s and fit s is usuall y r efer r ed t o (a) minimum clear ance bet ween shaft and hole (b) maximum clear ance bet ween shaft and hole

(a) unilat er al t oler ance (b) bilat er al t oler ance (c) limit ing dimensions (d) all of t hese 69. Sur face r oughness on a dr awing is r epr esent ed by (a) tr iangles (b) cir cles (c) squar es (d) r ect angles 70. The di amet er of fi ni sh t ur ned shaft can best be check ed wi t h a (a) combinat ion set (b) slip gauge (c) height gauge (d) micr omet er scr ew gauge 71. A ccu r at e cen t r i n g of w or k m ou n t ed i n an independent chunk can be deter mined by using a (a) cent r e gauge

(b) height gauge

(c) dial indicat or

(d) sur face gauge

72. I n limit s and fit s syst em, basic shaft syst em is one whose

(c) differ ence of t oler ance of hole and shaft

(a) lower deviat ion is zer o

(d) di f f er en ce bet w een m ax i m u m si ze an d minimum size of the hole

(b) upper deviat ion is zer o (c) minimum clear ance is zer o (d) maximum clear ance is zer o

Metrology and Inspection

73. To ch eck t h e di am et er a t wi st dr i l l wi t h a mi cr omet er, t he measur ement must be t ak en acr oss t he (a) mar gins of t he dr ill

80. For gr ade I T 7, value of t oler ance is equal t o (a) 8 i

(b) 10 i

(c) 16 i

(d) 24 i

81. Planer gauge is used for

(b) flut es of t he dr ill

(a) t est ing flat ness of sur face

(c) lips of t he dr ill

(b) adding t o ut ilit y of measur ement s on sur face plate

(d) web of t he dr ill  0.00

74. Expr essing a dimension as 18.3  0.02 mm is t he case of (a) unilat er al t oler ance (b) bilat er al t oler ance

(c) angular measur ement (d) all of t hese 82. I S specificat ions specify ver nier caliper s as t ype A, B and C. This classificat ion is based on

(c) limit ing dimensions

(a) accur acy

(d) none of t hese

(b) least count

75. I S : 919 on limit s and fit s specifies which of t he fol l owi ng number s of gr ades of fundament al t ol er an ces, an d f u n dam en t al dev i at i on s r espectively (a) 25, 18

(b) 25, 16

(c) 18, 22

(d) 18, 25

76. Basic shaft and basic hole ar e those whose upper deviations and lower deviations r espectively ar e (a) + ve, – ve

(b) – ve, + ve

(c) zer o, zer o

(d) none of t hese

77. The st andar d t oler ance unit is equal t o

e 0.45 e 0.45 e 0.45 e

(a) 0.45

3

(b)

4

(c) (d)

3

4

j D j + 0.001 D D j + + 0.01 D D j + 0.01 D D + 0.001 D

wher e D i s geomet r i c means of t he l ower and upper diameters of a particular diameter step. 78. Eden-Rolt compar at or is a popular instr ument for t he

(c) r ange (d) int er nal or ext er nal measu-r ement and for mar king pur pose 83. The cr oss-sect ion of st r aight edges upt o 180 mm lengt h is (a) r ect angular (b) cir cular (c) I -section (d) elliptical 84. Opt ical micr omet er is used t o (a) measur e small linear displacement s (b) measur e sur face pr ofiles (c) measur e sur face r oughness (d) set ver y small displacement by r ot at ing t he glass block t hr ough r elat ively lar ge angles 85. I n pr ecision polygon, a cent r al hol e and small holes ar e dr illed t hr ough t he t hickness (a) for mount ing pur poses (b) t o achieve high accur acy (c) for ease of manufact ur e (d) t o make t hem light 86. Pr eci si on pol ygons ar e cal i br at ed fr om fi r st pr inciples using

(a) calibr at ion of slip gauges

(a) one aut ocollimat or

(b) absolute measurement of length of slip gauges

(b) t wo aut ocollimat or s

(c) measur ement of flat ness

(c) t hr ee aut ocollimat or s

(d) measur ement of angles

(d) t wo pr ecision spir it levels

79. I t is desir able t o handle t he slip gauges wit h a clot h or chamois leat her in or der t o

87. Bevel pr ot r act or is used for (a) angular measur ement

(a) avoid injur y t o hands

(b) linear measur ement

(b) pr ot ect t he sur faces of slip gauges

(c) height measur ement

(c) insulat e t hem fr om t he heat of t he hand

(d) flat ness measur ement

(d) none of t hese

5.39

5.40

Metrology and Inspection

88. Clinomet er is r elat ed wit h (a) Engineer ’s par allels (b) angle gauges

(a) di mensi oi n i s al l owed t o var y onl y i n one dir ect ion (b) t ot al t oler ance is r elat ed t o a basic dimension

(c) spir it level (d) bevel pr ot r act or

(c) both (a) and (b)

89. A compar ator (a) needs t o be calibr at ed (b) need not be calibr at ed (c) cont ains a calibr at ed scale (d) i s h i gh l y accu r at e ov er i t s com pl et e measur ing r ange 90. T he essent i al con di t i on f or i nt er -f er om et r y measur ement is (a) an air gap (a wedge) of var ying thickness must exist bet ween t he t wo sur faces (b) an opt ical flat is r equir ed (c) wor k sur face must be r eflect ive (d) monochr omat ic sour ce of light is r equir ed (e) all of t hese 91. The minimum t oler ance t o which mechine can possi bl y be expect ed t o wor k adn pr oduce no defect i ve under t he speci fi ed condi t i ons, of a manufact ur ing pr ocess or a machine is called it s (a) pr ocess capability st udy (b) machine accur acy st udy (c) both (a) and (b)

(d) none of t hese 96. Accur acy of measur ing equipment is (a) t he closeness wit h which a measur ement can be r ead di r ect l y f r om a m easu r i n g inst r ument (b) a measur e of how close t he r eading is t o t he t r ue size (c) t he differ ence bet ween measur ed value and actual value (d) t he smallest change is measur ed t hat can be measur ed 97. Pr ecision of measur ing equipment is (a) t he closeness wit h which a measur ement can be r ead di r ect l y f r om , a m easu r i n g inst r ument (b) a measur e of how close t he r eading is t o t he t r ue size (c) the difference between measured value and actual value (d) t he smallest change in measur and t hat can be measur ed 98. Repeat abilit y of measur ing equipment is

(d) none of t hese 92. Toler ances ar e specified (a) t o obt ain desir ed fit s (b) because it is not possible t o manufact ur e a size exact ly (c) t o obt ain high accur acy (d) t o have pr oper allowance 93. Toler ances ar e basically specified in (a) unilat er al for m

(b) bilat er al for m

(c) both (a) and (b)

(d) none of t hese

94. Bilat er al t oler ance (a) usuall y have pl us and minus t oner ance of equal amount (b) specifies t he t ot al t oler ance on bot h sides of t he basic dimension (c) may be expr essed as 25.000 + 0.001 – 0.002 mm (d) syst em defines t he t heor et ically desir ed size of t he basic size (e) all of t hese

95. I n unilat er al t oler ance

(a) measur e of how close t he r eading is t o t he t r ue size (b) t he differ ence bet ween measur ed value and actual value (c) t he smallest change in measur and t hat can be measur ed (d) t he capabilit y t o indicat e t he same r eading again and again for a given measur and. 99. Advant age of ver nier caliper over micr omet er s is t hat it (a) is easier and quicker t o use (b) is mor e accur at e (c) can be used t o make bot h inside and out side measur ement s over a r ange of sizes (d) none of t hese 100. Combinat ion set can be used t o (a) check angular sur faces (b) dr aw cir cles and ar cs (c) scr ibe lines (d) all of t hese

Metrology and Inspection

5.41

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (a)

2. (b)

3. (d)

4. (a)

5. (b)

6. (a)

7. (d)

8. (c)

9. (a)

10. (c)

11. (a)

12. (c)

13. (c)

14. (c)

15. (d)

16. (b)

17. (a)

18. (b)

19. (a)

20. (d)

21. (a)

22. (b)

23. (d)

24. (c)

25. (d)

26. (d)

27. (c)

28. (a)

29. (d)

30. (c)

LEVEL-1 31. (a)

32. (d)

33. (b)

34. (e)

35. (e)

36. (d)

37. (a)

38. (c)

39. (d)

40. (c)

41. (c)

42. (a)

43. (d)

44. (a)

45. (d)

46. (d)

47. (d)

48. (d)

49. (b)

50. (a)

51. (d)

52. (d)

53. (c)

54. (b)

55. (c)

56. (a)

57. (e)

58. (c)

59. (d)

60. (b)

LEVEL-2 61. (d)

62. (d)

63. (c)

64. (a)

65. (c)

66. (d)

67. (d)

68. (b)

69 (a)

70. (d)

71. (c)

72. (b)

73. (a)

74. (a)

75. (d)

76. (c)

77. (a)

78. (a)

79. (c)

80. (c)

81. (b)

82. (d)

83. (a)

84. (d)

85. (a)

86. (b)

87. (d)

88. (c)

89. (a)

90. (e)

91. (c)

92. (b)

93. (c)

94. (d)

95. (c)

96. (b)

97. (a)

98. (d)

99. (d)

100. (a)

101. (d)

102. (d)

103. (c)

104. (d)

105. (b)

106. (a)

107. (c)

108. (d)

109. (a)

6

C H A P TE R

Fluid Mechanics and Hydraulic Machinery

F L U I D PROPERTI ES ( i ) M ass density () . I t is mass per unit volume.

 =

Mass of t he fluid Volume of t he fluid

Dimensionally,  = ML– 3 I n M K S unit , it is expr essed in met r ic slugs per met er cube (slug/m 3). I n SI unit s, it is expr essed in Newt ons per met er cube (N/m 3). Densit y of liquid may be consider ed as const ant but densit y of gases var ies wit h t he var iat ion of pr essur e and temperat ur e. ( i i ) Specific weight . I t is weight of fluid per unit volume. weight of fluid  = volume of t he fluid I n M K S syst em, it is measur ed in kg/m 3. I n SI syst em, it is measur ed in N/m 3. To conver t  in SI unit s mult iply  (M K S) by 9.81. The specific weight of wat er is 9810 N/m 3. Specific weight of a fluid var ies due t o

(a) change of gr avit y (b) effect of pr essur e and t emeper at ur e (i i i ) Specific volume ( vs). I t is defined as volume per unit mass. I n M K S unit it is measur ed in m 3/slug. I n SI syst em it is measur ed in m 3/kg. (iv) Specific gr avit y (S). I t is defined as t he r at io of weight densit y (or densit y) of a fluid t o t he weight densit y (or densit y) of a st andar d fluid. The st andar d liquid is wat er and t he st andar d gas is air. I t is a dimensionless quantit y. M at hemat ically, weight density (or density) of liquid For liquids, S = weight density (or density) of water For gases,

S =

weight density (or density) of gas weight density (or density) of air

(v) Viscosit y (D ynamic viscosit y). I t is defined as t he pr oper t y of fluid which offer s r esist ance t o t he movement of one layer of fluid over another adjacent layer of t he fluid. L et t wo fluid layer s at dist ances y and (y + dy ) fr om t he sur face have velocit y values u and (u + du ) as shown in t he figur e. The t op layer causes a shear st r ess on t he adjacent lower layer while t he lower layer causes a shear st r ess on t he adjacent t op layer. The shear st r ess () is pr opor t ional t o t he r at e of change of velocit y wit h r espect t o y .

F ig. Velocity variation near a solid boundary

6.2

Fluid Mechanics and Hydraulic Machinery

M at hemat ically,



du dy

du dy wher e,  is const ant of pr opor t ionalit y known as coefficient of dynamic viscosit y du is called velocit y gr adient . dy   = du dy Unit of viscosit y.

or

=·

FG IJ H K

I n M K S syst em : kgf – sec/m 2 I n C.G.S. syst em : dyne – sec / cm 2. 1 dyne-sec/cm 2 is called one poise. I n S.I . syst em : Newt on – sec / m 2 = Ns / m 2

(vi ) Kinemat ic viscosit y (  ). I t is t he r at io of t wo physical pr oper t ies of a fluid. I t is defined as t he r at io bet ween dynamic viscosit y and densit y of t he fluid  viscosity M at hemat ically, = = density  Unit of kinemat ic viscosit y . FT L T M 2  2 2 2 unit of  T L = L , wher e F = ML = = L = M M unit of  T2 T L3 L3 2 I n M K S and SI syst em : m /s. I n CGS syst em : cm 2/s. One cm 2/s is known as one St oke. One st oke = one cm 2/sec =

FG 1 IJ H 100 K

2

= 10–

4

m 2/sec

The pr act ical unit of kinemat ic viscosit y is cent i st oke. 1 1 cent ist oke = st oke. 100 Newt on’s law of viscosit y. I t st at es t hat , shear st r ess on a fluid element layer is dir ect ly pr opor t ional t o t he r at e of shear st r ain. M at hamat ically,

 = 

FG du IJ H dy K

CLASSI FI CI ATI ON OF FLU I DS

(i) I deal fluid: Thi s flui d is incompr essibl e and possesses no viscosit y. Such a fluid is only an imaginar y fluid. All exist ing fluids have some viscosit y. (i i ) Real fluid: A fluid t hat possesses viscosit y is called a r eal fluid. I n act ual pr act ice, all fluids ar e r eal fluids. (i i i ) N ewtonian fluids: A r eal fluid in which t he shear st r ess is dir ect ly pr opor t ional t o t he r at e of shear st r ain (or velocit y gr adient ) is called Newt onian fluid. (iv) N on-N ewtonian fluids: A r eal fluid in which t he shear st r ess i s n ot pr opor t i on al t o t h e r at e of sh ear st r ai n (or velocit y gr adient ) is called non-Newt onian fluid. (v) I deal plastic fluid: A fluid in which t he shear st r ess is mor e t han t he yield value and t he shear st r ess is pr opor t ional t o t he r at e of shear st r ain (or velocit y gr adient ) is called ideal plast ic fluid.

Fluid Mechanics and Hydraulic Machinery

6.3

RH EOLOGI CAL CLASSI FI CATI ON OF FLU I DS. Newt onian fluids

Non-Newt onian fluids :   

du dy

du dy e.g. air, wat er and many Pur ely viscous fluids ot her engineer ing fluids Time-I ndependent Time-Dependent behaveasNewtonianfluids under nor mal (i ) Pseudoplast ic fluids (i) Thixot r opic fluids cir cumst ances. n n du du     ,n<1 + f(t ) dy dy e.g. fine par t icle e.g. cr ude oils suspensions bent onit ic (ii ) Dilament fluids (ii) Rheopect ic fluids = 

FG IJ H K



FG IJ H K

FG du IJ H dy K

n

,n 1

e.g. ult r afine ir r egular (iii ) I deal plast ics or Bingham fluids   0  

FG du IJ H dy K



FG du IJ H dy K

Viscoelast ic fluids

Viscoelast ic fluids

du  E dy wher e E is modulus of elast icit y L iquid solid 

n

+ f(t ),

f(t ) incr eases par t icle suspensions

e.g. combinat ions in pipe flow and polymer ised fluids wit h dr ag r educt ion feat ur es.

(iii) Rar e liquid-solid suspensions.

n

e.g. wat er suspensions of clay and fly ash. VARI ATI ON OF VI SCOSI TY WI TH TEM PERATU RE. Viscosit y of liquids decr eases wit h t he incr ease of t emper at ur e while viscosit y of t he gases incr eases wit h t he incr ease of t emper at ur e.

LM N

For liquids,

1  = 0 1  t  t 2

For gases,

 = 0 + t –  t 2

OP Q

CLASSI FI CATI ON OF FLU I DS BASED ON DEN SI TY AN D VI SCOSI TY.

Type of fluid

Density

Viscosity

I deal fluid

const ant

zer o

I ncompr essible fluid

const ant

non-zer o

I nviscid fluid

const ant or var iable

zer o

Real fluid

var iable

non-zer o

Newt onian fluid

const ant or var iable

=

du dy

Non-Newt onian fluid

const ant or var iable

 

du dy

Per fect gas

p RT

zer o or non-zer o

6.4

Fluid Mechanics and Hydraulic Machinery

CLASSI FI CATI ON OF FLOWS BASED ON M ACH N U M BER.

Type of flow I ncompr essible flow Compr essible subsonic flow Tr ansonic flow Super sonic flow H yper sonic flow

M ach number, M M less t han 0.3 M bet ween 0.3 and 1 M r anging bet ween values less t han 1 and mor e t han 1 M gr at er t han 1 but less t han 5 M gr eat er t han 5

COM PRESSI BI L I T Y I t is t he r ecipr ocal of bulk modulus of elast icit y ‘k ’ and is defined as t he r at io of compr essive st r ess t o volumet r ic st r ain. V L et V = volume of gas enclosed in a cylinder dV

p = pr essur e of gas when t he volume is V

L et pr essur e be incr eased t o (p + dp), so t hat cor r esponding volume decr ease t o (V – dV)



FG dV IJ H VK

Volumet r ic st r ain = –

Cylinder

Piston

[negat ive sign is due t o decr ease of volume when pr essur e value incr eases] Bulk modulus, K =

Compr essibilit y =

Increase of pressure = Volumetric strain

FG H

dp dV  V

IJ K

=–

FG dp IJ V H dV K

1 K

SU RFACE TEN SI ON ( ) I t is defined as t he t ensile for ce act ing on t he sur face of a liquid in cont act wit h a gas or on t he sur face bet ween t wo immiscible liquids such t hat the contact sur face behaves as if it is a membr ane under t ension. The value of t his for ce per unit lengt h of t he fr ee sur face is same as t he sur face ener gy per unit ar ea. I t is expr essed in S. I . syst em as N/m. 

Sur face t ension on a liquid dr oplet :

p

d 2 =  d 4

p=



4 d

wher e, d = diamet er of dr oplet

p = pr essur e inside t he dr oplet . Wit h decr ease of diamet er of t he dr oplet , t he pr essur e int ensit y inside t he dr oplet incr eases. 

Sur face t ension on a hollow bubble : p =



Sur face t ension on a liquid jet :

p =

8 d 2 d

Fluid Mechanics and Hydraulic Machinery

6.5

CAPI L L ARI T Y. I t is defined as a phenomenon of r ise or fall of a liquid sur face in a small t ube r elat ive t o t he adjacent gener al level of liquid when t he t ube is held ver t ically in t he liquid. The ter m used is capillar y r ise or capillar y fall. I t s value depends upon specific weight of t he liquid, diamet er of t he t ube and sur face t ension of t he liquid. For equilibr ium,







Diameter of tube = 2r = d

h

d 2  h  w =   d cos  4

wher e  = angle made by sur face layer in t he glass t ube wit h glass t ube

w = specific weight of t he liquid 

h =

Capillary rise

4  cos  wd

For wat er, t he value of  is near ly zer o. I f a glass is dipped in mer cur y, t he level of mer cur y in t he t ube will be lower t han t he gener al level of t he out side liquid. h

For equilibr ium,

 2 d cos = wh d 4

h=

or

 Mercury

4  cos  wd

Value of  for mer cur y  FLU I D STATI CS I n fluid st at ics, fluids (liquids and gases) ar e at r est . Ther e is no r elat ive mot ion bet ween adjacent or neighbour ing fluid layer s. The velocit y gr adient , which is equal t o change of velocit y bet ween t wo adjacent fluid layer s divided by t he dist ance bet ween t he layer s i.e.

FG H

Shear st r ess which is equal t o 

u y

du will be equal t o zer o. dy

IJ will also be zer o. K

For ces act ing on t he fluid par t icles. (i ) Due t o pr essur e of fluid nor mal t o t he sur face (ii ) Due t o gr avit y (or self weight fo fluid par t icles) Vapour pr essur e. Pr essur e exer t ed by t he vapour for med at t he fr ee sur face of a liquid is called vapour pr essur e. This phenomenon of vapor isation is due to the sur face molecules escaping the liquid by over coming the molecular at t r act ive for ces by vir t ue of t heir t r anslat ional moment um. Difference bet ween vaporisat ion and boiling : Vapor isat ion occur s if vapour pr essur e < pr essur e above a liquid sur face at a t emper at ur e. Boiling occur s if vapour pr essur e = pr essur e above a liquid sur face at a t emper at ur e. Pascal’s law. This law st at es t hat int ensit y of pr essur e at a point in a st at ic fluid is same in all t he dir ect ions.

6.6

Fluid Mechanics and Hydraulic Machinery

Pressure variat ion in a fluid at rest.

p A –

FG p  p . zIJ A + w (A  z) = 0 H z K p =w z



p = wz

or

wher e, p = pr essur e above at mospher ic pr essur e

z = dept h of t he point fr om fr ee sur face z=

Fig. Forces on a fluid element

p and t his is known as pr essur e head. w

Absolut e pr essur e. I t is defined as t he pr essur e which is measur ed wit h r efer ence t o absolut e vacuum pr essur e. Gauge pr essure. I t is defined as t he pr essur e which is measur ed t aking at mospher ic pr essur e as t he dat um. Vacuum pr essur e. I t is defined as t he pr essur e below t he at mospher ic pr essur e. The dat um in t his case is t he at mospher ic pr essur e.

Mathematically, Absolut e pr essur e = At mospher ic pr essur e + Gauge pr essur e i.e.

pabs = pat m + pgauge Vaccum pr essur e = At mospher ic pr essur e – Absolut e pr essur e

F orce on plane and curved surfaces. H er e,

du = 0. dy

Thus for ces act ing on t he fluid par t icles will be (i) due t o pr essur e of fluid nor mal t o t he sur face (ii) due t o gr avit y (or self weight of fluid par t icles). Total pressure and Centre of pressure. Tot al pr essur e is defined as t he for ce exer t ed by a st at ic fluid on a sur face eit her plane or cur ved, when t he fluid comes int o cont act wit h t he sur face. This for ce act s always nor mal t o t he sur face. wher e, A = t ot al ar ea of t he sur face

x = distance of centr e of gravity of the area from fr ee sur face of fluid G = cent r e of gr avit y of t he plane sur face P = cent r e of pr essur e (a point ) h = dist ance of cent r e of pr essur e fr om fr ee sur face of fluid

w = specific weight of t he fluid

Fluid Mechanics and Hydraulic Machinery

6.7

Tot al pr essur e over t he ar ea, P = w A x Dist ance of cent r e of pr essur e fr om fluid sur face,

h = x +

k2 IG = x + G x Ax

wher e I G and K G ar e moment of iner t ia of t he sur face about an axis passing t hr ough t he cent r e of gr avit y of t he sur face lying on it s plane par allel t o t he fluid sur face and r adius of gyr at ion of t he body. Cent r e of pr essur e lies below t he cent r e of gr avit y of t he ver t ical sur face. I nclined plane surface submerged in a fluid.

F ig. (a)

F ig. (b)

Consider Fig.(a). Tot al pr essur e on one face of t he lamina, P = w A x Ver t ical dept h of t he cent r e of pr essur e fr om t he fluid sur face

h= x +

 When  = 0, t hen

I G sin2  k 2 sin 2  = x + G x Ax

kG2 sin 2  = 0, and cent r e of gr avit y (G) and cent r e of pr essur e (P) coincide. x

 When  = 90, t he value of h = x +

k G2 which is same as for ver t ical submer sion. x

Consider Fig. (b) t o find t ot al pr essur e and t he cent r e of pr essur e of a cur ved sur face submer ged in a fluid. The cur ved sur face AB lies submer ged inside t he fluid. Consider a small ar ea dA at a dept h of h fr om wat er sur face. Tot al pr essur e p over t his small ar ea dA act s nor mal t o t he sur face. Tot al pr essur e over t he whole cur ved ar ea, P =

z

wh  dA

area

Tot al for ce on t he cur ved sur face, p =

and

px2  p2y

inclinat ion of r esult ant wit h hor izont al is given by, t an  =

py px

6.8

Fluid Mechanics and Hydraulic Machinery

Geometric properties of plane figures. S. N o. 1.

2.

3.

F igur e

Figure with C.G. shown on it

Rectangle

Cir cle

Semicir cle

Area and M oment of I nertia A = bd IG =

bd 3 12

A=

d 2 4

IG =

d 4 64

A=

d 2 8

I G = 0.11r 4; I 0 = 4.

5.

6.

7.

Tr iangle

Tr apezium

Par abola

Ellipse

d 4 8

A=

1 bd 2

IG =

bd 3 bd 3 ; I0 = 36 12

A=

d (a + b) 2

IG =

a 2  4ab  b 2 a b

A=

2bd 8 ; IG = bd 3 3 175

I0 =

16 2 bd3; I 01 = bd3 105 7

A=

 bd  bd 3 ; IG = 4 64

Fd I GH 36 JK 3

Per imet er (appr oximat ely)

 (b  d ) 2

or

BU OYAN CY AN D FLOTATI ON . A body may float in liquid in fully immer sed st at e or in a par t ly immer sed st at e. Cent re of buoyancy. I t is the cent r oid of t he displaced fluid. I t is t he point wher e buoyant for ce is supposed t o act . (a)

 {1.5 (b  d ) – 2

bd }

Fluid Mechanics and Hydraulic Machinery

6.9

M et acent r e. When a float ing body is t ilt ed by a small angle, t he point of int er sect ion of line joining pr evious buoyant cent r e and cent r e of gr avit y G, and ver t ical t hr ough t he new buoyant cent r e is called M et acent r e. Thus met acent r e is t he int er sect ion of buoyant for ce and cent r eline of t he body. I t is t he point about which a float ing body oscillat es when it is t ipped or made unst able.

Not at ions used : F B = buoyant for ce V = volume of float ing body

x = x coor dinat e of cent r e of element ar y ar ea fr om r efer ence axis

x = dist ance fr om t he axis t o t he line of act ion of buoyant for ce Refer figur es (a) and (b)

(b)

BM = dist ance bet ween met acent r e and t he buoyant cent r e

GM = met acent r ic height , i.e. dist ance bet ween cent r e of gr avit y and met aceent r e. I I We have, BM = , and GM = – BG V V P0 x  GM = [for P0 and x r efer Fig. (b)] W tan  The body is in

(a) st able equilibr ium if M is above G or GM is posit ive. (b) unst able equilibr ium if M is below G or GM is negat ive. (c) neut r al equilibr ium if M coincides wit h G or GM = 0. L ar ger is t he met acent r ic height (GM ), mor e st able is t he body. M AN OM ET RY. M anomet er s ar e used for t he measur ement of fluid pr essur e by measur ing height of t he same or anot her const ant densit y fluid at r est . Refer r ing figur e below, a liquid under pr essur e is cont ained in a pipe of diamet er D. Anot her t ube is fit t ed on t o t his pipe wit h t he end of t ube opening t o at mpsher e. Patm

Patm 1 density

 density

 density h

D

h1

P

pressure 'P'

D

h

Fig. direct vertical tube manometer, i.e. piezometer P = gh + Patm

Fig. U tube manometer P + gh = P atm + 1gh 1  (P – P atm) = (1h 1 – h)g Patm

P

D h

h1 

F ig. I nclined tube manometer P + gh = P atm + 1gh 1 sin   (P – P atm) = (1h 1 sin  – h)g

6.10

Fluid Mechanics and Hydraulic Machinery

D ifferent ial manomet er s. These manomet er s ar e used fr equent ly in flow syst ems t o est imat e t he pr essur e at a point wit h r espect t o t hat at anot her by differ ence. Upr ight differ ent ial manomet er is convenient when 3 >1 and 2. This syst em will be adopt ed when t he fluids A and B ar e not allowed t o come in cont act wit h each ot her. M anomet er fluid wit h densit y 3 is per mit t ed t o have cont act wit h bot h t he fluids A and B. B1 P2 p3

h3

h1 P1 A p1

h2

p2 0

0

U pr ight differ ent ial manomet er Pressure balance across O – O P 1 + 1gh 1 + 2gh 2 = P 2 + 3gh 3  (P 1 – P 2) = (2h 2 – 1h 1 – 2h 3)g

I nver t ed different ial manometer Pressure balance across O – O P 1 – 1gh 1 =P 2 – 2gh 2 – 3gh 3  (P 1 – P 2) = (2h 2 – 1h 1 – 3h 3)g

FLU I D KI N EM ATI CS Types of Fluid Flow. (i )

L aminar and t urbulent flows: M ain differ ence in laminar and t ur bulent flows is t he r andomness and impr act icabilit y of t ur bulent flow as compar ed to t hat in laminar flow. A t ur bulent flow compr ises of mor e ir r ever sibilit y. Tur bulent losses ar e pr opor t ional t o V 2 wher eas laminar losses ar e pr opor t ional t o V.

( ii )

St eady and unst eady flows: I n st eady flow, t ime r at e of change of a par t icular pr oper t y (or flow par amet er ) at a given sect ion is equal t o zer o. Thus if u is point velocit y at a sect ion and for all inst ant s of t ime,

( iii )

if

du = 0, t hen flow is st eady; dt

if

du  0, t hen flow is unst eady.. dt

U nifor m and non-uniform flows: I n case of non-uni for m fl ow, t he change of say vel oci t y per uni t di st ance i.e.

du i s non zer o. dx

The val ues of vel oci t y at t wo di st ant sect i ons at t he same i nst ant of t i me ar e t o be seen. I f t hey ar e differ ent , it is non-unifor m ot her wise it is unifor m. ( iv)

Viscous and non-viscous flows: Viscous flow makes due consider at ion of t he viscosit y effect s (r eal fluids) and non viscous flows develop ideal fluid theor y. I deal fluid is fr ict ionless, incompr essible fluid, but a per fect gas should not be confused wit h an ideal fluid. I n ideal fluid, t he flow pr ocess is r ever sible. I n adiabat ic flow, no heat is t r ansfer r ed t o and fr om t he fluid. Rever sible adiabat ic flow is isent r opic.

Fluid Mechanics and Hydraulic Machinery

(v)

6.11

Compr essible and incompr essible flows: When var iat ion of mass densit y of a fluid is consider ed, t he t r eat ment is t hat of compr essible flows other wise it is incompr essible. Wat er at pr essur e mor e than 600 kPa is to be tr eat ed as compr essible.

( vi )

One t wo and t hree dimensional flows: I f changes of velocit y, pr essur e et c. ar e neglect ed in t r ansver se t o main flow dir ection, t he t r eatment is t hat of one-dimensional flow and if consider ed, it is t hat of t wo mensional flow, i.e. if

du = 0, it is one-dimensional dy

if

du  0 , it is t wo-dimensional. dy

A t hr ee dimensional flow is a flow in which t he velocit y component s u , v   in t hr ee mut ually per pendicular dir ect ions ar e funct ions of space coor dinat es x, y, z and t ime t . ( vii ) Rot at ional and ir r ot at ional flows: du dv dw dw dv du = , = , = ; t he flow is ir r ot at ional ot her wise it is r ot at ional. dy dx dy dx dz dz For r ot at ional flow, t he cr oss der ivat ives of velocit y component s in t wo mut ually per pendicular dir ect ions ar e unequal. Also symbolically, if V r efer s t o velocit y.

If

St eady

U nsteady

One dimensional flow

V = f (x )

V = f (x, t )

Two dimensional flow

V = f (x, y )

V = f (x, y, t )

Thr ee dimensional flow

V = f (x, y, z)

V = f (x, y, z, t )

( viii ) Sub-crit ical, cr it ical or super -crit ical (for open channel) flow ( ix)

Sub-sonic, sonic or super sonic (for compressible fluids) flow

N ot e: All t he above t ypes of flow can exist independent of each ot her, making it possible t o have var ious combinat ions. STREAM LI N ES. I t is an imaginar y line dr awn in a flow field such t hat a t angent dr awn at any point on t his line r epr esent s t he dir ect ion of velocit y vect or. Fr om t he definit ion, it follows t hat t her e can be no flow acr oss a st r eam line. Consider a par t icle moving along a st r eam line for a shor t dist ance ds, having it s component s dx, dy and dz along t he t hr ee mut ually per pendicular axes. L et velocit y component s of V s in x , y , and z dir ect ions be u, v and w r espect ively. Time t aken by a fluid par t icle t o move t hr ough a dist ance ds along a st r eam line wit h a velocit y V s is

which is t he same as

t =

ds Vs

t =

dz dy dx = = w v u

H ence differ ent ial equat ion for a st r eam line is,

dy dz dx = = v w u

6.12

Fluid Mechanics and Hydraulic Machinery

Forms of energy present in t he fluid. (i) Elevat ion (ii) K inetic (iii) Pr essur e head (iv) I nt er nal ener gy Gener al equat ion is,

P1 P V2 V2 + z1 + 1 + H m = 2 + z2 + 2 + J [I 2 – I 1 + q] 1 2 2g 2g

... (1)

wher e J = mechanical equivalent of heat (I 2 – I 1) = change of int er nal ener gy H m = ext er nal sour ce of ener gy supplied at a cer t ain r at e

q = heat unit s per unit weight of flowing fluid t r ansfer r ed out of t he fluid t o t he sur r oundings. I n equat ion (1) (i) I f r at e of loss of heat is such t hat t emper at ur e of t he syst em r emains const ant , t hen

I2 = I1

(ii) For well insulat ed syst ems, q = 0 (iii) I f t her e is no ext er nal supply of heat ener gy, H m = 0 (iv) For incompr essible flow, 1 = 2 =  (v) The quant it y [J (I 2 – I 1) + J q] r efer s, t o fr ict ional head loss = h L Equat ion (1) t hen r educes t o,

P1 P V2 V2 + z1 + 1 = 2 + z2 + 2 + h L   2g 2g

LM P  z  V OP N  2 g Q becomes a const ant value. 2

Somet imes, h L is neglect ed being small, t hen

The above gener al energy equat ion is applicable for (i) ideal and r eal fluids (ii) liquid and gases or vapour s and (iii) compr essible and incompr essible fluids

VELOCI TY AN D ACCELERATI ON S. L et V be t he r esult ant velocit y at any point in a fluid flow. L et u, v and w be it s component s in t he x, y and z dir ect ions. The velocit y component s ar e funct ions of space co-or dinat es and t ime. M at hemat ically, t he velocit y component s ar e given as : u = f 1 [ x, y, z, t ], v = f 2 [ x, y, z, t ], and w = f 3 [ x, y, z, t ] Result ant velocit y, V = | ui + vj+ wk| =

u 2  v 2  w2 du u dx u dy u . dz u ax = = . + . + + dt x dt y dt z dt t dx dy dz But = u, = v , and =w dt dt dt u du u u u  ax = =u + v +w +  y dt z t x dv u u u u Similar ly, ay = =u + v +w + dt y z t x dw w w w w v w  and az = = u dt x y z t V For st eady flow, =0 x wher e V is r esult ant velocit y u v w  = = =0 t t t

Fluid Mechanics and Hydraulic Machinery

6.13

L ocal acceler at ion. I t is defiend as t he r at e of change of velocit y wit h r espect t o t ime at a given point in a flow field. u v w , and ar e known as local acceler at ions. t t t Convect ive acceler at ion. I t is defined as t he r at e of change of velocit y due t o t he change of posit ion of fluid par t icles in a fluid flow. u v w I n t he above expr essions for ax, ay and az , it excludes t he fact or s , and . t t t

VELOCI TY POTEN TI AL. The velocit y pot ent ial is not a physical quant it y which could be dir ect ly measur ed and t her efor e, it s zer o posit ion can be ar ibt r ar ily chosen. I t is defined as a scalar funct ion of space and t ime such t hat it s negat ive der ivat ive wit h r espcet t o any dir ect ion gives t he fluid velocit y in t hat dir ect ion. I t is denot ed by t he symbol . M at hemat ically, velocit y pot ent ial is defined as  = f (x, y, z ) for st eady flow such t hat    ;v = – u = – and w = – x y z wher e u, v and w ar e component s of velocit y in x, y and z dir ect ions r espect ively. The cont iunit y equat ion for incompr essible st eady flow is       –  –  – =0 x x y y z z

FG H

or

2



FG H

IJ K

 2

x 2 y 2 V 2 = 0, which is L aplace equat ion for 



 2 z2

IJ K

FG H

IJ K

=0

Any value of  t hat sat isfies L aplace equat ion will cor r espond t o some case of fluid flow. St r eam F unct ion ( ). I t is defined as the scalar function of space and time, such that its partial derivative with r espect to any direction gives the velocity component at r ight angles to that dir ection. I t is defined for two dimensional flow. Proper t ies of st r eam funct ion. (i) I f st r eam funct ion  exist s, it is a possible case of fluid flow which may be r ot at ional or ir r ot at ional.

(ii) I f st r eam funct ion sat isfies t he L aplace equat ion, it is a possible case of an ir r ot at ional flow. The pr oduct of slope of equipot ent ial line and t he slope of st r eam line at t he point of int er sect ion is equal t o – 1. Thus equipot ent ial lines ar e or t hogonal t o t he st r eam lines at all point s of int er sect ion. Flow N et s E qu i pot en t i al l i n es an d st r eam l i n es i n t er sect each ot h er or t hogonally. The st r eamlines and equipot ent ial lines for m a net of mut ually per pendicular lines which is called flow net . The flow net i s a gr aphical r epr esent at ion of t wo dimensional, ir r otat ional flow and consists of a family of str eamlines inter sect ing or t hogonally a famil y of equi pot ent ial lines and in t he pr ocess for ming small cur vilinear squar es. Figure shows a flow net , in which  value decr eases in t he dir ect ion of flow.

– (

) 

(

)  –2 

 + 2 h

 + 

 90°

90° 

I f t he di st ance bet ween st r eaml i nes at some smal l r egi on of fl ow i s h, and t he di st ance bet ween equipot ent ial lines is s, t hen for a flow net , s  h. The smaller t hese dimensions, t he bet t er will be t he r epr esent at ion of flow. This concept may be explained fr om t he r elat ion   = s h I f  = , t hen  h = s

6.14

Fluid Mechanics and Hydraulic Machinery

M et hods of drawing F low net s. (i) By mat hemat ical analysis (ii) By hydr aulic models (iii) By gr aphical met hods (iv) By elect r ical analogy Equat ion of M ot ion. Accor ding t o Newt on’s second law of mot ion, t he net for ce F x act ing on a fluid element in t he x dir ect ion is equal t o mass m of t he fluid element mult iplied by t he acceler at ion ax in t he x -dir ect ion. F x = max

Thus mat hemat ically,

For ces pr esent in t he fluid flow. (i) F g, gr avit y for ce (ii) F p, t he pr essur e for ce (iii) F v, for ce due t o viscosit y (iv) F t , for ce due t o t ur bulence (v) F e for ce due t o compr essibilit y Thus in t he above equat ion, net for ce, F x = (F g)x + (F p)x + (F v)x + (F t )x + (F c)x

(i)

I f for ce due t o compr essibilit y F e, is negligible, t hen r esult ing net for ce F x = (F g)x + (F p)x + (F v)x + (F t )x and equat ions of mot ion ar e called Reynold equations of motion.

(ii ) For flow, wher e F t is negligible, t he r esult ing equat ions of mot ion ar e called Navier-St okes equat ions. (iii) I f t he fl ow i s assumed t o be i deal , vi scous for ce F v i s zer o and equat i ons or mot i on ar e cal l ed Euler ’s equat ions of mot ion . CON TROL VOLU M E AN ALYSI S FOR M ASS M OM EN TU M AN D EN ERGY CON SERVATI ON Cont r ol volume.

D

I t is an ident ified volume fixed in space. A cont r ol volume may eit her be infinit esimally small or as lar ge as desir ed. A cont r ol volume is bounded by a cont r ol sur face. An infinitesimal contr ol volume is called differential control volume. A differ ent ial cont r ol volume is assumed t o have ver y small but non-zer o dimension. I t s volume is denot ed by V and it s cont r ol sur face by s.

C A

Control volume ABCDA B Differential control

volume M ass equat ion. This is also called cont inuit y equat ion . Consider t he fluid between t wo sect ions A and B. Since t her e can be no flow acr oss t he walls of t he t ube, aft er a given t ime t , t he same fluid will be cont ained bet ween A ' and B '. Since under st eady condit ions of flow, mass of t he fluid bet ween A and B r emains const ant , t he mass ent er ing t he sect ion at A will be equal t o t he mass leaving at B.

Thus

1A 1s1 = 2A 2s2

Dividing by dt ,

1A 1

s s1 = 2 A 2 2 t t



s1 s  u1 , 2  u 2 t t 1A 1u 1 = 2A 2u 2 (cont inutiy equat ion)

and

A 1u 1 = A 2u 2 =

As t  0,

dV dt

A

A

B

B

A2 A1 2

1 u1

u2 s1

s 2



Fluid Mechanics and Hydraulic Machinery

6.15

M OM EN TU M EQU ATI ON . I t is based on t he second law of mot ion, i.e. net for ce act ing in any dir ect ion is equal t o r at e of change of moment um in t hat dir ect ion, i.e. linear impulse = change in linear moment um I n t he x dir ect ion, 

I nit ial linear impulse linear impulse = Final linear moment um Q. t . v x1 F x . t = Q. t . v x2 Q or F x = Q (v x2 – v x1) = vx2  vx1 g wher e v x2 and v x1 = component s of velocit y in x -dir ect ion of point s 1 and 2. F x = component for net for ce in x -dir ect ion. Q For pipe flow, p1 A 1 – p2 A 2 = v2  v1 g wher e suffix 1 indicat es upst r eam point and suffix 2 indicat es downst r eam point . V 1 and V 2 ar e aver age velocit ies at sect ions 2 and 1 of pipe.

b

b



g

g

M oment of moment um equat ion. I f a fluid par t icle of mass m moves along a cur ved pat h such t hat it s dist ance fr om t he axis of r ot at ion (i.e. fixed cent er ) changes wit h t ime, t hen r adial dist ance will be differ ent at differ ent posit ions of t he par t icle. I f t he flow ent er s t he cont r ol volume at a unifor m velocit y v 1 and at a st eady r at e, t hen moment um of flow ent er ing cont r ol volume = Q v 1 I f t his flows has a r adius r 1, t hen moment of moment um of flow ent er ing cont r ol volume = Q v 1 r 1 Similar ly, if v 2 and r 2 ar e velocit y and r adius at t he out let of cont r ol volume, t hen moment of moment um of flow leaving t he cont r ol volume = Q v 2 r 2 Accor ding t o moment of moment um pr incipal, t he r esult ing t or que is equal t o t he t ime r at e of change in moment of moment um. Ther efor e T = Q (v 2 r 2 – v 1 r 1) BERN OU LLI ’S EQU ATI ON . Assumpt ions.

(i) I deal fluid (ii) Cont inuit y of flow (iii) St eady and incompr essible flow. Euler ’s ener gy equat ion is,

p  gdz  v  v = 0 

I nt egr at ing, we get

z z z

dp  gdz  vdv = const ant  For an incompr essible flow, = const ant 

p v2  gz  = const ant (Ber noulli’s ener gy equat ion)  2

or

p v2 = const ant  z g 2g p p = = pr essur e ener gy per unit weight of fluid or pr essur e head g  v2 = kinet ic ener gy per unit weight or kinet ic head 2g

z = pot ent ial ener gy per unit weight or pot ent ial head

6.16

Fluid Mechanics and Hydraulic Machinery

APPLI CATI ON S OF BERN OU LLI ’S EQU ATI ON .

I t is applied t o t he pr oblems of (i) flow under a sluice gat e which is a const r uct ion in an open channel (ii) fr ee liquid jet (iii) r adial flow (iv) fr ee vor t ex mot ion. FLOW TH ROU GH AN ORI FI CES. F low t hr ough small orifices : Jet of wat er or fluid has minimum ar ea at vena cont r act a. Coefficient of cont r act at ion,CC =

Area of jet at vena cont r act a Ar ea of opening

= 0.61 for d << H 1 Ber noulli’s equat ion gives, U t =

2 gH 1 and

U a = Cv

2 gH 1

wher e U t and U a ar e t heor et ical and act ual velocit ies at vena cont r act a. I f t her e is no fr ict ion, Cv = 1. Other wise

Cv = 0.97 t o 0.99 Q = ar ea  velocit y = Cc a Cv

wher e,

2 gH 1 = Cd a 2 gH

Cd = dischar ge coefficient = Cc  Cv

Deter minat ion of coefficient of velocit y (Cv). By t he t r aject or y met hod Cv =

x 4 yH

F low t hrough large orifice : When d is not ver y small compar ed t o H 1 or H 2 , dQ = Cc B dH 2gH I nt egr at ing over H 1 and H 2 gives Q=

2 3/2 Cc 2g.B H 3/2 2  H1 3

Neglect ing velocit y of appr oach, we get Q=

2 Cc 2g.B 3

LMF MNGH H

2



U 20 2g

I JK

3/2

F GH

 H1 

U 20 2g

I JK

3/2

OP PQ

I f velocit y of appr oach is t aken int o account , value of Cc for lar ge or ifices used ar e obt ained fr om st andar d char t s.

Fluid Mechanics and Hydraulic Machinery

6.17

LAM I N AR FLOW I n laminar flow the fluid par ticles move along st r aight , par allel paths in layer s or laminae. The magnitudes of t he velocit ies of adjacent laminae ar e not same.

Characteristics of laminar flow. (i) The flow is gover ned by t he law r elat ing t o r at e of angular defor mat ion, i.e. = 

dv for flow in x dir ect ion. dy

(ii) Ther e is pr act ically no int er fer ence of fluid par t ilces of one layer over t hose of adjacent layer. (iii) Ther e is “ no slip” condit ion occur s at t he boundar y. (iv) The flow is r ot at ional. (v)

Ther e is continuous dissipation of ener gy due t o viscous, shear and ener gy must be supplied exter nally t o maint ain t he flow.

(vi) Ener gy loss is pr opor t ional t o fir st power of velocit y and viscosit y. (vii) Flow r emains laminar as long as

V. l  < Rcr it ical . 

Cr it ical velocit y. The velocit y at which t he flow changes fr om laminar t o t ur bulent is called cr it ical velocit y . I t is t he velocit y at lower cr it ical Reyolds number. I t is differ ent for differ ent fluids.

A flow is mor e likely t o be laminar if (i) velocit y v is low (ii) widt h or diamet er of passage is small (iii) densit y  is low (iv) viscosit y of t he flowing fluid is high. The above four var iables ar e gr ouped in t he for m of a non-dimensional par a called Reynolds number, Re. Reynolds number. Reynolds det er mined a non-dimensional quant it y called Reynolds number (R).  Vd Vd  For cir cular pipe flowing full, Re =   wher eV = mean velocit y

d = diamet er of pipe  = kinemat ic viscosit y of t he fluid  = mass densit y of fluid µ = absolut e viscosit y For non-cir cular cr oss-sect ion, r at io of t he cr oss-sect ional ar ea t o t he wet t ed per imet er is called h ydr aulic r adius (R). I t is used in t he Reynolds number, viz. Re = wher e,

b g

V 4R 

 =  = kinemat ic viscosit y 

Flow t hr ough a pipe (or closed conduit ) is L aminar at Re values less t han 2000 t ur bulent at Re values mor e t han 3000. I n t r ansit ion st age at Re values bet ween 2000 and 3000.  L aminar flow depends upon: vel oci t y of fl ow, ar ea of sur face i n cont act , t emper at ur e of fl ui d independent of nat ur e of sur face in cont act and pr essur e of flow.

 Tur bulent flow depends upon squar e of velocit y of flow, ar ea of sur face in cont act , t emper at ur e, densit y of fluid and nat ur e of cont act and is independent of pr essur e of flow.

6.18

Fluid Mechanics and Hydraulic Machinery

U pper and lower cr it ical Reynolds number : The Reynolds number above which t he flow changes fr om laminar t o t ur bulent is called lower cr it ical Reynolds number . Velocit y of t he fluid cor r esponding t o t he lower cr it ical Reynolds number called lower critical velocity . Similar ly, Reynolds number and t he velocit y which t ur bulent flow may change t o laminar flow ar e call ed upper cr it ical Reynolds number and upper cr it ical velocit y r espect ively . The cr it ical Reynolds number in differ ent sit uat ions may be differ ent depending upon choice of t he char act er lengt h l , in place of t he diamet er d chosen for pipe flow.

Case

Char acterist ic lengt h

U pper cr it ical Reyonlds N umber

1. Pipe flow 2. Open channel flow

(a) H ydr aulic mean dept h R (b) Flow dept h, y

4000 2000

3. Par allel plat es

Spaced ‘a’ apar t

300

4. M ovement of small spher e t hr ough st ationar y fluid

Spher e diamet er

3000

b

g

d p rz h = (– ) k dl l The law is valid for velocit ies upt o 3 t o 4.5 mm/sec. u = k.

Dischar ge velocit y is given by,

Relat ion bet ween Shear st r ess and Pressure gradient . For st eady t wo dimensional unifor m flows, But   



p 2u  2 =0 x y

u , t her efor e y

 p z   x y

i.e. pr essur e gr adient in t he dir ect ion of flow is equal t o t he shear gr adient nor mal t o t he dir ect i on of mot i on. St oke’s law. Resist ing for ce dr ag on a small spher e moving under laminar condition in a viscous fluid was fir st calculat ed by St okes and is known as St oke’s solut ion. Tot al dr ag, F D = 3  µ Vd Out of which  µ V d is cont r ibut ed by the pr essur e and r emaining 2 µVd is contr ibut ed by t he shear str ess on t he body. The spher e aft er a shor t dist ance of fall in t he viscous fluid at t ains a const ant velocit y called t er minal velocit y . To at t ain t his equillibr ium of t he spher e under t he st eady st at e condit ion, t he dr ag for ce and buoyant for ce must add up t o balance t he weight of t he falling spher e. At equilibr ium, and

F D = 3. . V. d . =

t er minal velocit y, V =

d2 s   18 

b

 d3 s  6

b

g

g

Above expr essions ar e valid for R < 1 and is known as Stokes’law. Dr ag for ce. The dr ag for ce exper ienced by t he spher e is gener ally expr essed as CD = FD wher e,

 V2 .A 2

A = ar ea of cr oss-sect ion of t he moving body nor mal t o t he dir ect ion of mot ion CD = known as t he coefficient of dr ag.

Fluid Mechanics and Hydraulic Machinery

6.19

St eady laminar flow in hor izont al circular pipe (H agen-Poisuile equat ion). For st eady laminar flow t o occur in t he pipe, a pr essur e gr adient is maint ained in t he dir ect ion of flow which over comes t he fr ict ional for ces (or shear for ces) opposing t he flow. I t is also assumed t hat t he pr essur e is dist r ibut ed unifor mly acr oss t he pipe cr oss-sect ion. Shear stress distribution : Figur e shows a hor izont al pipe of r adius conveying a viscous incompr essible fluid of viscosit y µ.  0

dr

ro

r dp = . 2 dl



U r

p r 2

Umax

(p +

p dr)  r2 r

This r elat ion holds goods for laminar as well as t ur bulent flow. The negat ive sign indicat es t hat t he pr essur e is decr easing in t he dir ect ion of flow.

r dp  0 = (– ) 0 . 2 dl and at r = 0, = 0 At

r = maximum,

Velocity distribution : Velocit y dist r ibut ion for laminar flow is given by

b g 41 FGH  pl IJK er

u = 

2 0

 r2

j

This is equat ion of a par abola, so t he maximum velocit y var ies par abolically along t he diamet er. M aximum velocity (U max) : M aximum velocit y will be obt ained, if r = 0, i.e. at t he cent er.

b g 41 FGH pl IJK . r 1 F p I Q = b g G J . r 8  H l K

U max =  Discharge :

2 0

2 0

Average or M ean velocity of flow through the pipe : I t is obt ained fr om cont inuit y equat ion, V 

Q 

r02

b g 81 FGH  pl IJK . r

2 0

 

L et aver age velocit y V occur s at r adius r 1, t her efor e equat ing u t o V, we get

bg 81 FGH  pl IJK .  r

2 0

b g 41  pl er

2 0

= 

 r2

j

H ead loss : Fr om H agen-Poiseulli’s equat ion for laminar flow, H ead loss, h l = Also,

hl =

p1  p2 8 V L 32  V L   .   r02  d2 128 VL   d4

Relat ion bet ween Darcy’s frict ion coefficient f and Re. For laminar flow t hr ough cir cular pipe (i.e. for Re < 2000), 16 f = Re Local velocity in t er ms of maximum velocit y,

F GH

u = U max 1 

r2 r02

I JK

6.20

Fluid Mechanics and Hydraulic Machinery

CORRECTI ON FACTORS. Kinet ic ener gy cor rect ion fact or ( ). Due t o viscosit y, t he velocit y dist r ibut ion acr oss any cr oss-sect ion in r eal fluid flow is non-unifor m and V2 does not r epr esent t he 2g t r ue kinet ic ener gy acr oss t he sect ion. To compensat e t he discr epancy, a coefficient known as kinet ic ener gy cor r ect ion fact or () is used. 1 v V dA  = A A V

boundar y r esist ance and t her efor e kinet ic ener gy per unit weight given by

z FGH IJK

wher e

v = velocit y at any point in t he cr oss-sect ion. V

 = 1.0 for unifor m velocit y dist r ibut ion

Vmax

 = 1.02 t o 1.05 for t ur bulent flows

Average velocity Velocity Distribution

 = 2.0 for laminar flow. M oment um cor rect ion fact or (  ). Due t o non-unifor m distr ibution of velocity, the actual momentum tr ansfer r ed acr oss the sect ion is differ ent t han t hat comput ed using t he aver age velocit y. I t is given by

z FGH IJK

1 v A V For laminar flow in pipe, = 1.33  =

2

dA

and for t ur bulent flow in pipe,  var ies fr om 1.01 t o 1.07. BOU N DARY LAYER The nar r ow r egion sit uat ed in t he vicinit y of a solid boundar y wher e t he effect s of viscosit y ar e confined is called boundar y layer . The dr ag for ce exper ienced by a solid body immer sed in a flowing fluid can be explained only wit h t he boundar y layer concept . The motion of fluids wit h very lit tle fr iction [ver y low viscosity] amounts t o fluid flow at ver y lar ge Reynolds number. For such flows, Pr andt l made t he following obser vat ions : (i) Viscous effect s ar e confined t o a ver y t hin layer, called boundar y layer , near t he solid sur face. (ii) The flow out side t he boundar y layer can be consider ed fr ict ionless or ideal. Boundary layer growt h over a flat plat e. U

U Y

U 

Parabolic distribution of velocity

X x Leading edge

F ig. Velocity distribution and growth of boundary layer over a flat plate at zero incidence placed i n parallel flow

U  denot es undist ur bed velocit y or fr ee st r eam velocit y. I n t he downst r eam dir ect ion fr om t he leading edge, t hickness of t he r et ar ded layer  incr eases cont inuously, i.e. t hickness of the boundar y layer incr eases along t he plat e in t he downst r eam dir ect ion. The viscous shear wit hin t he boundar y layer will be high due t o t he exist ence of high velocit y gr adient . Under cer t ain condit ions of flow and boundar y configur at ion, t hickness of t he boundr y layer incr eases consider able in t he downst r eam dir ect ion so much, so t hat flow in t he boundar y layer get s r ever sed. The decler at ed fluid par t icles ar e for ced away and boundar y layer is separ at ed fr om t he solid boundar y. This phenomenon is called boundar y layer separ at ion . I t occur s in case of flow past blunt bodies like cir cular cylinder s and spher es. A r egion of highly deceler at ed flow exist s downst r eam of such bodies. This r egion is called wake zone.

Fluid Mechanics and Hydraulic Machinery

6.21

BOU N DARY LAYER TH I CKN ESS. (1) N ormal t hickness ( ). The t r ansi t i on fr om zer o vel oci t y at t he sol i d st at i onar y boundar y t o t he fr ee st r eam vel oci t y U  out side t he boundar y layer t akes place asympt ot ically. Boundar y layer is usually defined as t hat dist ance fr om t he boundar y wher e velocit y differ s by 1% fr om t he fr ee st r eam velocit y,

i.e.

y = for u = 0.99 U 

(2) D isplacement t hickness (  ). The boundar y layer for mat ion and t he r esult ing velocit y dist r ibut ion has t he effect of displacing t he flow r at e slight ly out war d fr om t he solid boundar y as compar ed wit h t he pot ent ial flow (or ideal flow) t hat would exist in t he absence of t his boundar y layer.

Displacement t hickness may be defined in any one of t he following ways : (i) I t is the dist ance measur ed per pendicular t o t he boundar y by which t he fr ee st r eam is displaced on account of for mat ion of boundar y layer, or (ii) I t can be defined as an addit ional wall t hickness t hat would have t o be added t o compensat e for t he r educt ion in flow r at e on account of boundar y layer for mat ion. I t is obt ained fr om t he expr ession

 =

1 U

z (U



0



– u ) dy =

z FGH1  Uu IJK dy



0



The quant it y ( U  – u ) is called velocit y defect . (3) M oment um thickness ( ) . This is defined in a way similar t o t he flow r at e displacement t hickness. This is obt ained fr om t he expr ession

z

FG H

IJ K

u u 1 U  dy 0 U N ote: For par allel flow past a flat plat e held at zer o incidence, t he appoximat e r elat ionship among t he t hr ee t hickness of a boundar y layer descr ibed above is 

 =

= 3= 7.5  (4) Energy t hickness ( e). The r easoning in t he case is same as for t he case of momet um t hickness. This can be obt ained fr om t he expr ession

z

F GH

I JK

u u2 1 dy U U 2 0 With t he ener gy t hickness known, t he loss of ener gy per unit t ime can be det er mined fr om t he r elation 1 E L = bU 3 2 

e =

Charact er ist ies of boundary layer.

6.22

Fluid Mechanics and Hydraulic Machinery

Consider t he boundar y layer for med on a flat plat e kept par allel t o flow of fluid of velocit y U .  incr eases as dist ance fr om leading edge x incr eases, i.e. x  decr eases as U  incr eases, i.e. 

1 U

 incr eases as kinemat ic viscosit y incr eases, i.e.   0 

FG U IJ , t her efor e  HK 

0

decr eases as x incr eases.

 I f pr essur e incr eases in t he downst r em dir ect ion, i.e.

p is posit ive, flow near t he boundar y is fur t her x

r et ar ded, boundar y layer gr owt h is fast er and boundar y layer is suscept ible t o separ at ion.

 I f pr essur e decr eases in t he downst r em dir ect ion, i.e.

p  is negat ive, t he boundar y layer gr owt h is x

r educed.

 Var iat ion of , 0 and for ce F et c char act er ist ics ar e gover ned by iner t ial and viscous for ces, t her efor e t hey ar e funct ions of eit her

Ux UL or , i.e. Reynolds number..  

 When Rx < 2  102, boundar y layer is laminar velocit y dist r ibut ion is par abolic When Rx > 4  105, boundar y layer is t ur bulent When 4  105 > Rx > 2  105, boundar y layer is in t r ansit ion.

 Cr it ical value of Reynolds number at which boundar y layer changes fr om laminar t o t ur bulent depends upon tur bulence in fr ee steam flow, sur face r oughness, pr essure gr adient, plate cur vature and temperatur e differ ence bet ween fluid and boundar y.  Velocity dist r ibution in laminar boundar y layer follows par abolic law while that in tur bulent layer follows logar it hmic law or power law. L aminar boundary layers on a flat plate. Flow in t he init ial st ages of boundar y layer development shows char act er ist ics of laminar mot ion. Viscous for ces slow down t he par t icles near t he solid boundar y. The Reynolds number in t he boundar y layer is expr essed as Re = U 

FG x IJ H vK

wher e x = dist ance fr om leading edge. Dimensionless boundary layer thickness depends upon (i ) fr ee st r eam velocit y, U 

(ii ) dist ance x measur ed fr om t he leading edge (iii ) t he kinemat ic viscosit y v of t he fluid. For a t wo dimensional laminar boundar y layer, it is given by

 =5 x

v Ux

=

5 RE

Nominal t hickness of t he boundar y layer is given by

 =5

vx U

Fluid Mechanics and Hydraulic Machinery

6.23

LAWS OF FLU I D FRI CTI ON A fluid in mot ion is always subject ed t o a cer t ain r esist ance. Such a r esist ance is gener ally assumed t o be due t o fr ict ion. I n r ealit y, t his r esist ance is mainly due t o sliding of t he adjacent fluid layer s. A liquid will have a st eady st r eamline (viscous) flow at low velocit ies only. Aft er r eaching a cer t ain velocit y, t he flow changes int o a t ur bulent flow wit h t he appear ance of eddy cur r ent s. For fluids in mot ion, t he fr ict ional r esist ance obeys cer t ain laws.

Laws of fluid friction for steady streamline flow. (1) Fr ict ional r esist ance is pr opor t ional t o velocit y, sur face ar ea of cont act and t emper at ur e. (2) Fr ict ional r esist ance is independent of pr essur e and nat ur e of t he sur face of cont act . The flow est ablishment lengt h is given by L = 0.057 R d ... for laminar flow L = 0.7 R d ... for t ur bulent flow

and

wher e, d = diamet er of t he pipe R = Reynolds number of flow. L aws of frict ion for t urbulent flow. (1) (2) (3) (4)

Fr ict ional Fr ict ional Fr ict ional Fr ict ional

r esist ance is pr opor t ional t o squar e of t he velocit y. r esist ance is independent of t he pr essur e. r esist ance is pr opor t ional t o densit y of t he fluid. r esist ance slight ly var ies wit h t emper at ur e.

D ar cy’s fr ict ion fact or. I n smoot h pipes (due t o Blasius), and

64  for laminar flow R 0.3164 f = for t ur bulent flow R1 / 4

f =

e

j

H EAD LOSSES. (i ) M ajor head losses. H ead loss due t o fr ict ion falls in t his cat egor y and is given by Dar cy - Weisbach’s equat ion

hf = wher e,

Now,

l V  

32  V l f l V2 =  d2 2 gd

= lengt h of pipe = aver age velocit y of flow = specific waight of fluid = coefficient of dynamic viscosit y. 4 fl V 2 fl Q 2 hf = = d 2g 3.0258d 5

Chezy’s for mula : V = C mi hf wher e, i = l A d m= = for a cir cular sect ion P 4 Vw C = Chezy’s const ant = f'

F I GGsubst it ut ingV  Q JJ e / d / 4j K H 2

6.24

Fluid Mechanics and Hydraulic Machinery

Value of f in Dar cy Weisbach for mula.

d V 4 fl V 2 . is a funct ion of t he Reynolds number, Re =  2g d 16 (a) When Re < 2000, f= ... (H agen Poiseuille) Re 0.079 (b) When Re < 2000 and Re < 1,00,000, f = ... (Blasius) Re1/4 (c) When Re < 1,00,000 t hen value of f can be found fr om 1 = 2 log10 Re . 4f 0.8 ... (K ar man Pr andt l) 4f The coefficient f in t he for mula,

hf =

(i i ) M inor head losses. V2 2g (b) Due t o sudden cont r act ion : I f V 2 is velocit y in t he cont r act ed por t ion, t hen

(a) At the ent rance :

h l = 0.5

hl =

F 1  1I GH C JK

2

c

V 22 2g

wher e, CC = coefficient of cont r act ion is nor mally t aken as 0.62

Vs2 2g (c) Sudden expansion : I f V 1 and V 2 ar e velocit ies in t he pr e-enlar gement and enlar ged sect ion r espect ively, t hen

h l = 0.5

or

ht =

bV

1

 V1

g

2

2g (iii) Ot her head losses. Ot her head losses ar e t hose in bends, pipe fit t ing and joiner ies such as valves, r educer s, T-junct ions, et c.

I f V is velocit y of flow in t he pipe, t hen t hese losses ar e expr essed in t er ms of velocit y head t imes velocit y head

V2 as K 2g

KV 2 2g wher e, K is called loss coefficient .

i.e

hl =



Fluid Mechanics and Hydraulic Machinery

6.25

GAS TU RBI N ES The gas t ur bine obt ains it s power by ut ilizing t he ener gy of a jet or bur nt gases and air, velocit y of t he jet being absor bed as it flows over sever al r ings of moving blades which ar e fixed t o a common shaft . I t t hus r esembles a st eam t ur bine and it s blades ar e designed in t he same manner as shown in t he velocit y diagr ams of st eam t ur bines. The gas t ur bine r equir es an air compr essor which is usually dr iven off it s own shafting. This absor bs a consider able pr opor tion of the power pr oduced and thus lower s the over all efficiency. Gas t ur bines have been const r uct ed t o wor k on t he following fuels : coal gas, pr oducer gas, blast fur nace gas, oil and pulver ized coal TYPES OF GAS TU RBI N ES. (1) Const ant pressure gas t urbine. I n t his t ype, fuel is bur ned at const ant pr essur e and t he cycle used is same as t he Joule cycle. The t ur bine is of t he r eact ion t ype using oil fuel and is fit t ed wit h an axial flow air compr essor. The r ot or consist s of five r ings of moving blades. The mult i-st age r ot ar y air compr essor is coupled t o t he r ot or shaft . The t ot al air supply is dr awn fr om t he sur r ounding at mospher e by t he compr essor and is compr essed t o t he combust ion pr essur e of 1 t o 4 at omospher e, it is t hen for ced int o t he combusion chamber. Par t of t his air is used as combust ion air for t he oil which ent er s t he bur ner, t he r emainder is for ced t hr ough t he annular space bet ween wall of t he combust ion chamber and t he bur ner jacket . The air r eceives heat fr om t he bur ner jacket and also mixes wit h t he pr oduct s of combust ion chamber and t he bur ner jacket . This r aises t emper at ur e and volumeof t he air. The use of a ver y lar ge quant it y of air in excess of t he combust ion air pr event s t emper at ur e of t he mixt ur e fr om r eaching values which ar e t oo high for t he met al of t he r ot or blades. I t also pr event s t he bur ner fr om becoming t oo hot . (2) Simple open cycle gas t ur bine (const ant pr essure heat addit ion) or Air st andard Br ayt on (or joule) cycle.

F ig. Simple open cycle gas t ur bine

F ig. Representat ion of Joule cycle on h-s or (T-s) diagram

Refer r ing figur e, we can obt ain t her mal efficiency on t he basis of 1 kg of wor king fluid flow. Heat supplied = q = h 3 – h 2 = cp(T 3 – T 2) if cp is assumed const ant in pr ocess 2-3. H eat r eject ed = h 4 – h 1 = cp(T 4 – T 1) if cp is assumed const ant in pr ocess 4-1. Net wor k = H eat supplied – H eat r eject ed = cp{(T 3 – T 2) – (T 4 – T 1)}

6.26

Fluid Mechanics and Hydraulic Machinery

This net wor k may also be found fr om t he differ ence bet ween t ur bine and compr essor wor k.



Ther mal efficiency, t h = 1 

1

.... [since r p = pressure ratio =

(  1) rp 

p2 ] p1

Fr om t his equat ion, t her mal efficiency of t he Br ayt on cycle is same as t hat of t he Ot t o cycle. N ote. For t ur bine and compr essor wor k we have neglect ed t he change in kinet ic and pot ent ial ener gies. ACTU AL BRAYTON CYCLE. T – s diagr am for an act ual Br ayt on cycle is shown in figur e. Pr essu r e l oss i n t h e com bu st i on ch am ber i s r epr esent ed by p3 – p2'.

I n t his cycle 1– 2 is isent r opic compr ession. 1– 2' is act ual compr ession. 3– 4 is isent r opic expansion. 3– 4' is act ual expansion. Compr essor efficiency, c = Based on st at ic values, 



c =

w Isentropic compression work  c Actual compression work wca

h2  h1 c p (T2  T1 )  h2 '  h1 c p (T2 ' T1 )

H er e 1 kg of air passing t hr ough t he compr essor is assumed or

c =

T2  T1 T2 'T1

wta Actual turbine work Tur bine efficiency, t = Isentropic turbine work  w t or

t ha =

(w t )  t  c  w c qa. c

Fr om t his equat ion, t her mal efficiency is incr eased by impr oving eit her t or c or bot h. Air rate (AR). I t is defined as t he air flow r equir ed per kWh out put .

ma mass of air required  = k Wh out put k Wh out put

ma 3600  wnet wnet in kJ / kg ma 3600 The r ecipr ocal of air r at e is called specific power. Air r at e is act ually t he cr it er ion of size of t he plant , i.e. lower t he air r at e, t he smaller t he plant . Work rat io (WR). I t is defined as t he r at io of net wor k t o t he t ur bine wor k. 

AR =

WR =

wnet wt

Opt imum pr essur e r at io for maximum cycle t her mal efficiency.

p2 r p (opt ) = p  x

FG p IJ Hp K 1

2

FG H

IJ K

Fluid Mechanics and Hydraulic Machinery

6.27

M et hods t o improve efficiency and specific out put of simple cycle. (1) Regener at ion. This is done by pr eheat ing air wit h t he t ur bine exhaust , t hus saving t he fuel consumpt ion. (2) I mpr oving t ur bine out put .

This is done by following met hods. (i ) Reheating : The whole expansion in t he t ur bine is achieved in t wo or mor e st ages and r eheat ing is done aft er each st age. (ii ) I ncr easing t he value of maximum cycle t emper at ur e, i.e. t ur bine inlet t emper at ur e.

This r equires (a) bet t er qualit y of fuel (b) new mat er ials which can wit hst and high t emper at ur es (c) blade cooling met hods. (iii ) I mproved turbine efficiency. I t depends on design impr ovement s. (3) Reducing compr essor input .

This is done by following met hods. (i ) I ntercooling. Compr essor wor k is r educed by int er cooling t he air bet ween compr essor st ages. (ii ) By lowering inlet temperature to compressor. I t is not pr act icable because t his will incr ease t he pr essur e r at io. (iii ) By increasing compressor efficiency. This depends upon t he design impr ovement . RECI PROCAT I N G AI R COM PRESSORS Recipr ocat ing air compr essor s ar e used for pr oducing high pr essur e air. At omspher ic air is dr awn int o a cylinder dur ing suct ion st r oke of pist on and is compr essed by t he pist on dur ing t he r et ur n st r oke; t he pist on dr iven by power fr om a ext er nal sour ce. I f air is admit t ed t o one side of t he pist on being only, t he compr essor is called single act ing. A double-act ing compr essor admit s air t o each side of t he pist on al t er nat ely. Whi lst one face of t he pist on is per for ming t he suct i on st r oke, ai r on t he ot her face i s compr essed.

Use of compr essed air deliver ed by t he compr essor (i ) To dr iver a compr essed air engine (ii ) For pr oducing an air blast for wor k-shop (iii ) For spr aying t he fuel int o cylinder of t he diesel engine. Air compr essor s ar e also used for cooling lar ge buildings. Fr esh air is dr awn int o t he compr essor and compr essed adiabat ically, which causes a lar ge incr ease in t emper at ur e. H igh-pr essur e air leaving t he compr essor is cooled by means of cold wat er. This cool high pr essur e air is t hen par t ially expanded adi abat i cal l y i n anot her cyl i nder, whi ch r educes i t s t emper at ur e t o much bel ow t he sur r oundi ng at mospher e, t he cold air is t hen cir culat ed t hr ough t he duct s of building. P2 3

2 2 2

T

PV = c (isothermal) PV r = c (Adiabatic)

P2

PV n = c (Polytropic)

2

P1

2 2

P1 4

1 V2

V1

V

1 S

F ig. T heoretical p-V diagram for a single r eci pr ocat ing ai r compr essor.

6.28

Fluid Mechanics and Hydraulic Machinery

Sequence of operat ions represent ed on t he diagram (i ) Operation 4 – 1 : Volume of air V 1 aspir at ed int o t he compr essor at pr essur e p1 and t emper at ur e T 1. (ii ) Operation 1 – 2 : Air compr essed accor ding t o t he law pV n = C fr om pr essur e p1 t o p2. Volume decr eases fr om V 1 t o V 2. Temper at ur e incr eases fr om T 1 t o T 2. (iii ) Operation 2 – 3 : Compr essed air of volume V 2 and at pr essur e p2 wit h t emper at ur e T 2 deliver ed fr om t he compr essor. Work done (neglect ing clear ance volume). L et p1 and v 1 be t he init ial condit ion of air befor e adiabat ic compr ession, p2 and v 2 be t he final condit ion aft er compr ession. Then wor k done per cycle in dr iving t he compr essor, W = p1 V 1

R|F p I S|GH p JK T

FG  IJ H   1K

2

U| V| W

 1 

1

1

I f compr esson follow t he law, pV n = const ant , t hen wor k r equir ed t o dr ive t he compr essor will be a similar P expr ession except t hat n must be subst it ut ed for .



W = p1 V 1

FG n IJ R|SFG p IJ H n  1K |H p K T 2

1

n 1 n

U|  1V |W

P2 6 3

P1 5

I f p1 is in k Pa and V 1 is in m 3, t hen W will in k N-m.

F n IJ P (V Wor k done (wit h clear ance volume), W = GH n  1K 1

R|F p I – V ) SGH p JK |T 2

1

4

Clearance volume = v3 = v c

n 1 n

1

2

n 1 n

1

2

I f compr essions ar e adiabat ic,  should be used in place of n . I f int er cooling is per fect , point d will be on t he isot her mal line, t hen



W = p1 V1

FG n IJ LMM FG p IJ H n  1K M H p K N

WN 60 wher e N = number of r evolut ions per minut e. Power r equir ed =

2

1

n 1 n

Fp I  G J Hp K 3 2

n 1 n

d

pvn = c

Isothermal p1

OP LMF p I  1P  p V MG J PQ MNH p K 2

pvn = c c

p2

Tot al wor k r equir ed t o dr ive t he compr essor per cycle will be t he sum of wor k r equir ed in low pr essur e and high pr essur e cylinder s.

LMF p I MMGH p JK N

v

Swept volume = v1 – v 3 = v s Total volume = v1

e

p3

Let p1 and V 1 r epresents condition of air enter ing low pr essur e cylinder, p2 and V 2 r epr esent condit ion of air ent er ing high pr essur e cylinder and p3 be t he final pr essur e of t he air.

p1V 1 = p2V 2

v1

p

L et compr ession follow t he law, pV n = const ant , and int er cooling be incomplet e, so t hat point d has not r eached t he isot her mal

1

1

v4

U|  1V |W

TWO-STAGE COM PRESSOR.

1

4

volume = v1 – v4

WN kW 60 wher e, N = number of complet e cycles per minut e.

Tot al wor k r equir ed, W =

v3

pvn = c

Effective swept

Power r equir ed t o dr ive compr essor =

L FG n IJ MM p V H n  1K M N

2

OP 2P PQ

3

2

b

a

n 1 n

OPOP  1P P PQPQ

v2

v1

v

Fluid Mechanics and Hydraulic Machinery

M U LT I STAGE COM PRESSOR.

6.29

p

I n mul t i st age compr essor s, ai r i s compr essed i n sever al st ages inst ead of full compr ession being per for med in single cy l i n der. I n pr i n ci pl e, i t i s equ i val en t t o n u m ber of compr essor s in ser ies, t he air passing fr om one cylinder t o t he next , t he pr essur e incr easing in each cylinder.

6

9

3 5

Delivery Pressure (P3 or P d )

Without Inter cooling Perfect Inter cooling

H.P.

pvn = c

 Cycle 8156 is single st age compr essor.  Cycles 8147 and 7456 ar e t wo st age compr essor wit hout int er cooling bet ween cylinder s.

Intermediote Pressure (P 2)

 Cycles 8147 and 7236 ar e that a two-stage compr essor with per fect int er cooling bet ween cylinder s.

Intake Pressure (P 1 or P s)

7 2 LP 8

4 pv = c 1 v

Per fect int er cooling. I t means t hat aft er t he i nit i al compr essi on in t he L .P. cyl i nder, wi t h i t s consequent t emper at ur e r i se, ai r i s cool ed i n an i nt er cool er s back t o i t s or i gi nal t emper at ur e, i.e. T 2 = T 1, in whi ch case point 2 li es on i sot her mal t hr ough poi nt 1. Condit ion for maximum efficiency. I f inital pr essur e p1 and final pr essur e p3 ar e fixed, t hen best value of inter mediat e pr essur e p2 t hus found will denot e at what pr essur e t o exhaust t he fir st st age, so t hat wor k done on t he compr essor will be mini mum.

p1, V 1 and t er m

n 1 is a const ant . n

Since for maximum efficiency, pr essur e r at io in each cycle is same, t her efor e

p2 p = 3 p1 p2 For maximum efficiency, int er mediat e pr essur e is geomet r ic mean of t he init ial and final pr essur e, i.e.

p2 =

p1 p3

mn p1 V1 For m st age compr essor, wor k done per cycle, W = n 1

LMF p  1I MMGH p JK N m

1

n 1 mn

OP  1P PQ

This equat ion applies t o any t ype of compr essor or mot or and even t o vapour engines, pr ovided n  .

Condit ions for maximum efficiency. (i ) Air is cooled t o t he init ial t emper at ur e bet ween t he st ages. (ii ) Pr essur e r at io in each st age is same. (iii ) Wor k r equir ed for each st age is same. I sot her mal hor sepower compr essor. Theor et ical hor sepower of compr essor is calculat ed on t he assumpt ion t hat t he compr ession cur ve of t he p-v diagr am is an isot her mal. Then I sot her mal wor k done per cycle = ar ea of p - v diagr am = p1 V 1 loge r = p1V 1 log

p1 V1 log e r  N 60 wher e N = number of cycles per minut e I sot her mal hor sepower =

V2 V1

6.30

Fluid Mechanics and Hydraulic Machinery

I ndicat ed hor se power of a compr essor. I t is t he hor sepower obt ained fr om t he act ual indicat or car d t aken dur ing a t est on t he compr essor. Compr essor efficiency =

isot her mal horsepower indicat ed hor sepower

I sot her mal efficiency. I sot her mal efficiency =

isot her mal horsepower = 70% (gener ally) shaft hor sepower

wher e shaft hor sepower is t he br ake hor sepower r equir ed t o dr ive t he compr essor. Adiabat ic efficiency. I t is t he r at io of hor sepower r equir ed t o dr ive t he compr essor compar ed wit h t he ar ea of hypot het ical diagr am assuming adiabat ic compr ession.

FG  IJ p V H   1K

1 1

Adiabatic efficiency =

LMF p I MMGH p JK N 2

 1 

1

OP N  1P  PQ 60

power required t o dr ive t he compr essor

Volumet r ic efficiency. I t is t he r at io of volume of air inhled at STP t o t he swept volume of t he pist on. Volumet r ic efficiency =

volume of air inhaled at S.T.P. swept volume of pist on

Value of t he volumet r ic efficiency var ies bet ween 70 and 90 per cent accor ding t o t he t ype of compr essor. Compar ison bet ween Recipr ocat ing and Rot ar y air compr essor s. S.N .

Par t i cular s

Reci pr ocat i ng

Rot ar y

1.

Suitability

Suit able for low dischar ge

Suit able for handing lar ge

of air at high pr essur e

volumes of air at low pr essur es

2.

Oper ation speed

L ow

High

3.

Air supply

Pulsating

Continuous

4.

Balancing

Cyclic vibr at ions occur

L ess vibr at ions

5.

L ubr icat ing syst em

Gener ally complicat ed

Gener ally simple lubr icat ion syst ems ar e r equir ed

6.

Qualit y of air

Cont aminat ed wit h oil

Air deliver ed is r elat ively

deliver ed 7.

Air compr essor size

mor e clean L ar ge for given dischar ge m 3/min

Small of same dischar ge 2000-3000 m 3/min

8.

Fr ee air handled

250-300

9.

Deliver y pr essur e

800 t o 1000 bar

Nor mally below 10 bar

10.

Usual standar d of

I sot her mal compr ession

I sesnt r opic compr ession

compr ession

Fluid Mechanics and Hydraulic Machinery

6.31

Compar ison bet ween Axial flow and Cent rifugal compr essors. S.N .

Par t i cular s

Axial flow compr essor s

Cent r ifugal compr essor s

1.

Pr essur e r at io

1.2 : 1 (For high pr essur e) r at io mor e number of st ages ar e r equir ed)

4:1

2.

I sesnt r opic efficiency

85 t o 88% (wit h moder n) aer o-foil blades)

70%

3.

Flexibilit y of oper at ion

L ess

M or e (due t o adjust able per whir l and diffuser vanes)

4.

Fr ont al ar ea

L ess (main cause in adopt ing t he axial flow compr essor s for air cr aft )

M or e

5.

Effect of deposit for mat ion on t he sur face of impeller r ot or.

Per for mance affect ed

Per for mance not affect ed

6.

St ar t ing t or que

H igh

L ow compar at ively

7.

Suit abilit y

Used univer sally wit h lar ge gas t ur bines

Suit able for super -char ging I .C. engines and for comp r essor s for r efr iger ant s and indust r ial gases

8.

Efficiency vs. speed cur ve

L ess flat

M or e flat compar at ively

CEN T RI FU GAL COM PRESSORS These in it s simplest for m consist s of a number of cur ved vanes fit t ed symmet r ically. The r ot or r ot at es in an air t ight volut e casing wit h inlet and out let point s. The casing for t he compr essor is so designed t hat kinet ic ener gy of t he air is conver t ed int o pr essur e ener gy befor e it leaves t he casing as shown in t he figur e.

AXI AL FLOW COM PRESSOR I n axial flow compr essor s, t he gas essent ially flows par allel t o t he axis. I t consist s of a number of r ot at ing blade r ows fixed on a r ot at ing dr um and st at or blades ar e fixed on casing in alt er nat e r ows. E ach st age con si st s of on e m ovi n g r ow of bl ades an d on e r ow of f i xed bl ades. T h e en t h al py and pr essur e of gas r ises as it passes thr ough r otating blades. This happens at the expense of a r eduction in r elat ive velocit y, t he absolut e velocit y of t he gas incr easing along t he axis of r ot or due t o wor k input . This incr ease in kinet ic ener gy is par t ly conver t ed int o pr essur e ener gy as t he air passes t hr ough diver ging fixed blades. These fixed blades help t he gas t o r each next set of moving blades for fur t her compr ession. The blades are generally made of aerofoil section to reduce losses caused by turbulence and boundary separation.

6.32

Fluid Mechanics and Hydraulic Machinery

I n axial flow compr essor s, t he dr um wit h r ot or blades, r ot at es inside a casing wit h a fixed or st at or blades. H YDRAU LI C M ACH I N ES (TU RBI N ES) Efficiencies of Tur bines. (i )

H ydraulic efficiency ( h ). H ydr aulic efficiency of t ur bine is t he r at io of power developed by t he r unner, i.e. wat er hor sepower (W.H .P.) t o t he net power supplied by t he wat er at t he ent r ance t o t he t ur bine. These t wo power s differ by t he amount of hydr aulic losses. W. H.P. w Q  Q H 75

h =

b

g

wher e, Q = quant it y of wat er act ually st r iking t he r unner,

Q = quant it y of wat er dischar ged dir ect ly t o t he t ail r ace wit hout st r iking t he t ur bine r unner I f Q is negligibly small, t hen

h =

b

W.H.P. wQH 75

g

( ii ) M echanical efficiency ( m ). M ech ani cal effi ci en cy of t ur bi ne i s t he r at i o of power obt ai ned fr om sh aft of t he t ur bi ne, i.e. shaft or br ake hor sepower. (S.H .P. or B.H .P.) t o t he power developed by t he r unner (i.e. W.H .P.). These t wo power s differ by t he amount of mechanical losses viz. bear ing fr ict ion et c.

m =

b

g

B.H. P. or S.H. P. W.H. P

( iii ) Volumet ric efficiency ( v ). Volumet r ic efficiency is t he r at io of quant it y of wat er act ually st r iking t he r unner and t he quant it y of wat er supplied t o the t ur bine. These t wo quant ities differ by the amount of wat er t hat slips dir ect ly t o t he t ail r ace wit hout st r iking t he r unner.

v =

Q Q  Q

(iv)Overall efficiency o ). Over all efficiency of t he t ur bine is t he r at io of power available at t he t ur bine shaft t o t he power supplied by t he wat er at t he ent r ance t o t he t ur bine.

o =

b

g

B.H.P. or S.H.P.

bNet power available at the turbine entranceg

Over all efficiency of t he t ur bine, o = h × m  o = b × v × m

Fluid Mechanics and Hydraulic Machinery

6.33

CLASSI FI CATI ON OF H YDRAU LI C TU RBI N ES.

M oder n t ur bines classified on t he following basis. (1) Act ion of fluid on t he moving blades (2) Dir ect ion of flow of fluid in t he r unner (3) Specific speed. (4) Disposit ion of shaft (1) Classificat ion according t o act ion of fluid on moving blades. (i ) I mpulse turbine : These r equir es high head and small quant it y of flow. The fluid is br ought in ending in a nozzle. The whole pr essur e ener gy of wat er is t r ansfor med int o kinet ic ener gy. The fluid coming out of t he nozzle in t he for m of a fr ee jet is made t o st r ike on a ser ies of bucket s mount ed on t he per ipher y of a wheel. The fluid is deliver ed t o t he wheel on a par t of it s cir cumfer ence filling or st r iking only a few of t he bucket s at a t ime. M ost ly, in hydr aulic t ur bines, t he wheel r evolves in open air, i.e. t her e is no differ ence of pr essur e in t he wat er at t he inlet t o t he r unner and t he dischar ge. Ther efor e, casing of an impulse t ur bine has no hydr aulic funct ion t o per for m. I t is necessar y only t o pr event splashing and t o lead t he wat er t o t he exit and also act s as a safeguar d against accident s. This t ur bine is also called a fr ee jet t ur bine. ( ii ) Reaction turbine : These r equir es low head and high r at e of flow. The fluid befor e ent er ing t he t ur bine has pr essur e as well as kinet ic ener gy. All pr essur e ener gy is not t r ansfor med int o kinet ic ener gy as in case of impulse t ur bine. The moment on t he wheel is pr oduced by bot h kinet ic and pr essur e ener gies. Types of r eact ion tur bines: (a) For medium head and medium flow (b) For low head and lar ge flow. N ote : The fluid leaving t he t ur bine has st ill some of t he pr essur e as well as kinet ic ener gy. The pr essur e at t he inlet t o t he t ur bine is much higher t han t he pr essur e at t he out let . Thus t her e is a possibilit y of fluid flowing t hr ough some passage ot her t han t he r unner and escape wit hout doing any wor k. H ence a casing is absolut ely essent ial due t o t he differ ence of pr essur e in r eact ion t ur bine. (2) Classificat ion according t o direct ion of flow of fluid in t he runner. H ydr aulic Turbines Tangent ial flow t ur bine

Radial flow t ur bi ne

Axial flow tur bine

Out war d r adial flow t ur bine (cent r ifugal t ype)

Mixed (r adial and axial) flow t ur bine

I nwar d r adial flow t ur bine (centr ipet al t ype)

(3) Classificat ion accor ding t o specific speed.

Specific speed ( N s ) is defined as the speed of a gemet r ically similar hypothet ical fluid machine handling unit dischar ge against a unit head. Ns =

N Q

b gH g

3/ 4

wher e, Q = dischar ge m 3/s N

= speed in r pm

g = gr avit at ional const ant m/s2 H = head under which t he machine oper at es.

6.34

Fluid Mechanics and Hydraulic Machinery

For t ur bines, N s =

N P H

5/4

; and for pumps,

N Q

Ns =

H 3/4

QH  o k W ( = pq) 1000 The specific speed can be used t o compar e differ ent t ypes of pumps and t o compar e differ ent t ypes of t ur bines t o select pick t he t ype of pump or t ur bine most suit able for a given wor k. wher e power gener at ed by shaft ,

P=

For small specific speeds, r adial flow gives best efficiency and for high specific speeds axials flow machi nes ar e most effi ci ent . The mi xed fl ow machi nes ar e most effi ci ent at t he i nt er medi at e r ange of specific speeds. Specific speed of hydraulic t ur bines.

Type of turbine

Specific speed

Type of runner

Pelt on

Fr ancis

P in H P

P in kW

Slow

10 t o 20

8.5 t o 17

Nor mal

20 t o 28

17 t o 24

Fast

28 t o 35

24 t o 30

Slow

60 t o 120

50 t o 100

Nor mal

120 t o 180

100 t o 150

Fast

180 t o 300

150 t o 250

–

300 t o 1,000

250 t o 850

Kaplan

H eads of hydraulic t urbines.

The head act ing on a t ur bine may be defined in t wo ways as follows : ( i ) Gross head. I t is defined as t he differ ence bet ween head r ace level and t he t ail r ace level when no wat er is flowing. The gr oss head is oft en called st at ic head or t ot al head and it may be r epr esent ed by H 1. (ii )N et or effective head. I t is t he head available at t he ent r ance t o t he t ur bine. I t is obt ained by subt r act ing fr om t he gr oss head all t he losses of head t hat may occur as wat er flows fr om head r ace t o t he ent r ance of t he t ur bine. The losses of head ar e mainly due t o fr ict ion occur r ing in penst ocks, canal, et c. Thus, if H r epr esent s net head and h f is t ot al loss of head bet ween head r ace and t he ent r ance of t he t ur bine, t hen H = H g – hf For a r eact i on (or encased) t ur bine, net head is equal t o differ ence bet ween pr essur e head at t he ent r ance t o t he t ur bine plus velocit y head in t he penst ock at t his point plus elevat ion of t his point above t he assumed dat um and elevat ion of t he t ail wat er plus velocit y head in t he dr aft t ube at it s exi t . ...

FP V I F H = G w  2g  Z J – G Z H K H 1

2 1

1

2



V22 2g

I JK

For an impulse t ur bine, net head is equal t o differ ence bet ween pr essur e head at t he ent r ance t o t he nozzle plus velocit y head in t he penst ock at t his point plus elevat ion of t his point above t he assumed head dat um and elevat ion of t he t ail wat er. ...

H =

FP  V Z I GH w 2 g JK 1

2 1

1

– Z2

Fluid Mechanics and Hydraulic Machinery

6.35

PELTON WH EEL TU RBI N ES This t ur bine is of t he par allel flow impulse t ype. This is suit able for highe heads. The quant it y of wat er r equir ed t o wor k t he Pelt on wheel is less as t he ener gy head is lar ge. The wat er r eaching t he t ur bine comes t hr ough penst ocks fr om t he r eser voir. The nozzle issues a power ful jet whcih impinges on t he bucket s placed at t he out er per ipher y of t he wheel. These vanes ar e of t he for m of double hemispher i cal cups. The jet st r ikes t he cent r al diving edge of t he double cups. Thus, jet get s deflect s on each si de. The wat er aft er impar t ing it s ener gy t o t he t ur bine, is dischar ged int o t he t ail r ace. The needle valve pr ovides necessar y r egulat ion of t he dischar ge t o t he t ur bine.

F ig. Pelt on wheeel

REACT I ON TU RBI N ES The pr incipal dist inguishing feat ur es of a r eact ion t ur bine ar e t hat only a par t of t he t ot al head of wat er is conver t ed int o velocit y head befor e i t r eaches t he r unner, and t hat t he wat er complet ely fills all t he passages in t he r unner. Types. (1) Fr ancis t ur bine (2) K aplan tur bine. (1) F rancis t urbine. Wat er is led t o t he t ur bine t hr ough t he penst ock pipe whose end is connect ed t o t he spir al casing of t he t ur bine. This spir al casing dir ect s t he wat er evenly t o t he guide blades. The wat er t hen passes t hr ough t he r unner and finally goes t o t he t ail r ace t hr ough t he dr aft t ube. The closed type Francis turbine is shown in t he figur e. Types of closed type F rancis t urbines. ( i ) H orizontal type : I t is used for medium and high heads and ver t ical t ype for medium and low heads. ( ii ) Vertical types : I t is used for medium and low heads. (2) K aplan and Pr opeller t ur bines. I n t hese r unner i s essent i al l y a pr opel l er wor k i ng i n r ever se and blades ar e mount ed so t hat blade angles can be adjust ed t oget her by means of suit able gear ing while t he machine is in oper at ion. The guide vanes or wicket gates of entr y ar e also adjustable. This machine is much suit able for low heads, say upt o 36 m. An el ect r i c gener at or coupl ed t o a K apl an t ype r unner may be enclosed in a fair ing within a fair ed tunnel. H eads as low as 3.6m may be used. A number of vanes ar e fixed t o t he boss (or hub). When t he vanes ar e composit e wit h t he boss, t he t ur bine is called propeller t ur bine. When t he vanes ar e adjust able t he t ur bine is called K aplan t ur bine.

6.36

Fluid Mechanics and Hydraulic Machinery

DRAFT TU BE Dr aft t ube is a pipe or passage of gr adually incr easing cr oss-sect ional ar ea which connect s r unner exit t o t he t ail r ace. I t may be made of cast or plat e st eel or concr et e. I t must be air t ight and under all condit ions of oper at ion it s lower end must be submer ged below t he level of wat er in t he t ail r ace. U ses of draft tube. ( i ) I t per mit s negat ive or suct ion head t o be est ablished at t he r unner exit , t hus making it possible t o inst all t he t ur bine above t he t ail r ace level wit hout loss of head. ( ii ) I t conver t s a lar ge pr opor t ion of velocit y ener gy r eject ed fr om t he r unner int o useful pr essur e ener gy. PERFORM AN CE CH ARACTERI ST I C CU RVES FOR TU RBI N ES. The t ur bines ar e gener ally designed t o wor k at par t icular values of H , Q, P, N and 0 which ar e called designed condit ions. But oft en t ur bines ar e r equir ed t o wor k at condit ions differ ent fr om t hose for which t hey have been designed. Ther efor e, it is essent ial t o det er mine exact behaviour of t he t ur bines under var ying condit ions by car r ying out t est s eit her on t he act ual t ur bines or on t heir small scale model s. The r esul t s of t hese t est s ar e usual l y gr aphi cal l y r epr esent ed and t he r esul t i ng cur ves ar e cal l ed char act er ist ic cur ves. For t he sake of convenience, t he char act er ist ic cur ves ar e plot t ed in t er ms of unit quantities. U nit speed (N u). Speed of t he t ur bine under unit head is called unit speed. N Nu = H U nit discharge (Q u). The dischar ge flowing t hr ough t he t ur bine under a unit head is called unit dischar ge. Qu =

Q H

U nit power (P u ). Out put of t he t ur bine under a unit head is called unit power . Pu =

P H3 2

Specific speed (N s). The speed of specific r unner is called specific speed.

N P H5 4 For any ot her t ur bine also exact ly same r elat ionship for N s may be der ived. Ns =

CEN TRI FU GAL PU M PS Cent r ifugal pump is used t o r aise liquids fr om a lower t o a higher level by cr eat ing r equir ed pr essur e wit h t he help of cent r ifugal act ion. I n gener al it can be defined as a machine which incr eases t he pr essur e ener gy of a fluid, as a pump may not be used t o lift wat er at all, but just boost t he pr essur e in a pipeline. Whir ling mot ion is impar t ed t o t he liquid by means of backwar d cur ved blades mount ed on a wheel called impeller . Liquid enter s the impeller at the centr e (eye) of the pump and dischar ges into the case sur rounding t he impeller. The pr essur e head developed by cent r ifugal act ion is ent ir ely due t o velocit y impar t ed t o t he liquid by t he r ot at ing impeller and not due t o any displacement or impact . At t he ent r ance t o t he impeller, since t her e ar e no guide vanes (as in t he case of t ur bines), t he dir ect ion of absolut e velocit y of liquid at t his point of t he impeller is not dir ect ly known. H owever, for best efficiency of t he pump, it is commonly assumed t hat t he liquid ent er s t he impeller r adially, i.e. absolut e velocit y of t he liquid at t he ent r ance t o t he impeller (or at t he inlet t ip of t he impeller vane) is r adial in dir ect ion. Thus in t his case,  = 90 and velocit y of whir l V w at inlet is equal t o zer o. Again, it is desir ed t hat t he liquid ent er s and leaves t he vane wit hout shock. This can be ensur ed if inlet and outlet t ips of the vane ar e par allel t o t he dir ect ion of t he r elat ive velocit ies at t he t wo t ips.

Fluid Mechanics and Hydraulic Machinery

6.37

CAVI TAT I ON I t is defined as t he phenomenon of for mat ion of vapour bubbles of a flowing liquid in a r egion wher e pr essur e of t he liquid falls below it s vapour pr essur e and sudden collapsing of t hese vapour bubbles in a r egion of higher pr essur e. When vapour bubbles collapse, a ver y high pr essur e is cr eat ed. The met all ic sur faces, above which t he liquid is flowing, is subject ed t o t hese high pr essur es, which cause pit t ing act ion of t he sur face. Thus, cavities ar e for med on t he met allic sur face and also consider able noise and vibr at ions ar e pr oduced. Caviat ion includes for mat ion of vapour bubbles of t he flowing liquid and collapsing of t he vapour bubbles. For mat ion of vapour bubbles of t he flowing liquid t akes place only whenever pr essur e in any r egion falls below vapour pr essur e. When pr essur e of t he flowing liquid is less t han it s vapour pr essur e, t he liquid st ar t s boiling and vapour bubbles ar e for med. These vapour bubbles ar e car r ied along wit h t he following liquid to higher pr essur e zones wher e these vapour s condense and bubbles collapse. Due to sudden collapsing of t he bubbles on t he met allic sur face, high pr essur e is pr oduced and met allic sur faces ar e subject ed t o high local st r esses. Thus t he sur faces ar e damaged. Effect s of cavit at ion. (i )

The met allic sur faces ar e damaged and cavit ies ar e for med on t he sur faces.

( ii ) Due t o sudden collapse of vapour bubble, consider able noise and vibr at ions ar e pr oduced. ( iii ) Efficiency of a t ur bine decr eases due t o cavit at ion. Due t o pit t ing act ion, sur face of t he tur bine blades becomes r ough and t he for ce exer t ed by wat er on t he t ur bine blades decr eases. H ence wor k done by wat er or out put hor se power becomes less and t hus efficiency decr eases. The hydr aulic machines subject ed t o cavit at ion ar e cent r ifugal pumps and r eact ion t ur bines.



6.38

Fluid Mechanics and Hydraulic Machinery

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. I n Red wood viscomet er (a) absolut e value of viscosit y is det er mined (b) part of the head of fluid is utilised in overcoming fr iction (c) fluid dischar ges thr ough or ifice with negligible velocit y (d) compar ison of viscosit y is done. 2. Cent r e of buoyancy is (a) point of inter section of buoyant force and centre line of t he body (b) cent r e of gr avit y of t he body (c) cent r oid of displaced volume fluid (d) mid point bet ween C.G. and met acent r e. 3. Length of mer cur y column at a place at an altitude will var y wit h r espect t o t hat at gr ound in a (a) linear r elat ion (b) hyper bolic r elat ion (c) par abolic r elat ion (d) manner fir st slowly and t hen st eeply 4. When power is tr ansmitted through a considerable dist ance by means of wat er under pr essur e, t he maximum power is t r ansmit t ed when fr ict ional loss of head is (a) one t hir d of t he t ot al head supplied (b) half of t he t ot al head supplied (c) 10% of t he t ot al head (d) 17.7% of t he t ot al head. 5. A r ot amet er is a device used t o measur e (a) velocit y of fluid in pipes (b) velocit y of gauges (c) vot ex flow (d) flow of fluids 6. Reynolds number for pipe flow is given by

vD vD (b)   vD vD (c) (d)   7. Wit h r ise in gas t emper at ur e, dynamic viscosit y of most of t he gases (a) incr eases (b) decr eases (c) does not change significant ly. (d) none of t hese (a)

8. The flow of wat er in a pipe of diamet er 3000 mm can be measur ed by (a) Vent ur imet er

(b) Rot amet er

(c) Pilot t ube

(d) Or ifice plat e.

9. W h i ch of t h e f ol l ow i n g i s di m en -si on l ess par amet er ? (a) Pr essur e coefficient (b) Fr oude number (c) Dar cy Weisbach fr ict ion fact or (d) None of t hese 10. A small plast ic boat loaded wit h pieces of st eel r ods i s fl oat ing i n a bat h t ub. I f t he car go i s dumped int o t he wat er allowing t he boat t o float empt y, t he wat er level in t he t ub will (a) r i se

(b) fall

(c) r emains same

(d) r ise and t hen fall

11. St eady flow occur s when (a) pr essur e does not change along t he flow (b) velocit y does not change (c) condit ions change gr adually wit h t ime (d) condit ions do not change wit h t ime at any point 12. A flow in which each liquid par ticle has a definite pat h and t heir pat hs do not cr oss each ot her, is called (a) Steady flow

(b) Unifor m flow

(c) St r eamline flow

(d) Tur bulent flow

13. Buoyant for ce is (a) r esultant of upthrust and gr avity forces acting on t he body (b) r esult ant for ce on t he body due t o t he fluid sur r ounding it (c) r esultant of static weight of body and dynamic t hr ust of fluid (d) equal t o t he volume of liquid displaced by t he body 14. I n a r ect angular not ch, t he r at io of per cent age dischar ge er r or in is measur ement of head 1 (a) 1 (b) 2 3 3 (c) (d) 4 2

Fluid Mechanics and Hydraulic Machinery

15.

24.

Cavit at ion is caused by (a) high velocit y (b) low bar omet r ic pr essur e (d) low pr essur e

16.

17.

(b) Steady flow

(c) Tur bulent flow

(d) L aminar flow

A lar ge Reynold number is indicat ion of (a) smoot h and st r eamline flow (b) laminar flow (c) steady flow (d) highly t ur bulent flow

26.

Non unifor m flow occur s when (a) di r ect i on and magni t ude of vel oci t y at al l point s ar e ident ical (b) velocit y of successive fluid par t icles, at any point , is same at successive per iods of t ime

(c) head loss var ies linear ly wit h flow r at e (d) shear st r ess var ies linear ly wit h r adius

19.

20.

21.

The fr ict ion r esist ance in pipe is pr opor t ional t o V 2, accor ding t o (a) Fr oude-number

(b) Reynolds-Weber

(c) Dar cy-Reynolds

(d) Weber -Fr oude

I n laminar flow, maximum velocity at the centre of pipe is how many times to the aver age velocity ? (a) Two

(b) Thr ee

(c) Four

(d) None of t hese

Pitot tube is used to measur e the velocity head of (a) st ill fluid

(b) laminar flow

(c) t ur bulent flow

(d) flowing fluid

(c) magnit ude and dir ect ion of velocit y do not change fr om point t o point in t he fluid (d) velocit y, dept h, pr essur e, et c. changes point t o point in t he fluid flow 27.

I n st eady flow of a fluid, t he acceler at ion of any fluid par t icle is (a) constant (b) var iable (c) zer o (d) never zer o

28.

For measur ing flow by a vent ur imet er, it should be installed in (a) ver t ical line (b) hor izont al line (c) inclined line wit h upwar d flow (d) in any dir ect ion and in any locat ion

29.

Fr oude number is significant in (a) su per son i cs, as w i t h pr oj ect i l e an d j et pr opulsion (b) full immer sion or complet ely enclosed flow, as wit h pipes, air cr aft s wings, nozzles et c., (c) si mul t aneous mot i on t hr ough t wo fl ui ds wher e t her e i s a sur face of di scont i nui t y, gr avit y for ces, and wave maki ng effect , as wit h ship’s hulls (d) all of t hese

30.

The fluid for ces consider ed in t he Navier St okes equat ion ar e (a) gr avit y, pr essur e and viscous (b) gr avit y, pr essur e and t ur bulent (c) pr essur e, viscous and t ur bulent (d) gr avit y, viscous and t ur bulent

I n equilibr ium condit ion, fluids ar e not able t o sustain (a) shear for ce (b) r esist ance t o viscosit y (c) sur face t ension (d) geomet r ic similit ude

22.

23.

Flow occur r ing in a pipeline when a valve is being opened is (a) steady

(b) unsteady

(c) laminar

(d) vor t ex

Tot al pr essur e on 1 m  1 m gat e i mmer sed ver t ically at a dept h of 2 m below t he fr ee wat er sur face will be (a) 1000 kg (b) 2000 kg (c) 4000 kg

(d) 8000 kg

u v w   =0 x y z

25.

I n a t ur bulent flow in a pipe (b) fluid par t icles move in st r aight lines

(b)

u v w u v w = 1 (d ) = u.v.w..     x y z x y z wher e u, v and w ar e component s of velocit y in x, y and z dir ect ions r espect ively

I f t he par t icles of a fluid at t ain such velocit ies t hat var y fr om point t o point in magnit ude and dir ect ion as well as fr om inst ant , t he flow is (a) Unifor m flow

u v w =0   x y z

(c)

(a) Reynolds number is gr eat er t han 10000

18.

The gener al equat i on of cont inui t y for t hr eedimensional flow of a compressible fluid for steady flow is (a)

(c) high pr essur e

6.39

6.40

31.

32.

Fluid Mechanics and Hydraulic Machinery

The dept h of cent r e of pr essur e in r ect angular lamina of height h wit h one side in t he liquid sur face is at h (a) h (b) 3 2h h (c) (d) 3 2

37.

Reynolds number is significant in

38.

The magnit ude of wat er hammer depends on the (a) lengt h of pipeline (b) speed at which t he valve is closed (c) elastic proper ties of the liquid flowing thr ough t he pipe and pipe mat er ial (d) all of t hese

39.

A cylinder is kept on a hor izont al boundar y past which an ideal fluid flows per pendicular t o t he cylinder axis. I t will exper ience (a) no lift for ce

(d) none of t hese

(a) su per son i cs, as w i t h pr oj ect i l e an d j et pr opulsion (b) full immer sion or complet ely enclosed flow, as wit h pipes, air cr aft wings, nozzles et c. (c) si mul t aneous mot i on t hr ough t wo fl ui ds wher e t her e i s a sur face of di scont i nui t y, gr avit y for ces, and wave making effect , as wit h ship’s hulls (d) all of t hese 33.

(b) some lift for ce (c) lift for ce in ver t ically downwar d dir ect ion (d) lift for ce in ver t ically upwar d dir ect ion

Two dimensional flow occur s when t he (a) dir ect ional and magnit ude of t he velocit y at all point s ar e ident ical (b) velocit y of successive fluid par t icles, at any point , is same at successive per iods of t ime (c) magnit ude and dir ect ion of velocit y do not change fr om point t o point in t he fluid (d) fl ui d par t icles move in a plane or par all el pl an es and t he st r eaml i n e pat t er ns ar e ident ical in each plane

34.

35.

A st r eamline is defined as t he line (a) par allel t o cent r al axis flow (b) par allel t o out er sur face t o pipe (c) of equal velocit y in a flow (d) along which t he pr essur e dr op is unifor m M at ch number is significant in (a) su per soni cs, as wi t h pr oj ect i l es an d j et pr opulsion (b) full immer sion or complet ely enclosed flow, as wit h pipes, air cr aft s wings, nozzles et c. (c) si mul t aneous mot i on t hr ough t wo fl ui ds wher e t her e i s a sur face of di scont i nui t y, gr avit y for ce, and wave making effect s, as wit h ship’s hull (d) all of t hese 2

36.

2

     For an ir r otational flow t he equat ion  x 2  y2 is called (a) Ber noulli’s equat ion (b) Cauchy Riemann’s equat ion (c) Euler ’s equat ion (d) L aplace equat ion.

Separ at i on of fl ow occur s due t o r educt i on of pr essur e gr adient t o (a) zer o (b) negligibly low value (c) t he ext ent such t hat vapour for mat ion st ar t s

40.

Wh en t h e w at er f l ows ov er a r ect an gu l ar su ppr essed wei r, t he pr essur e beneat h t he nappe is (a) ver y high (b) slight ly above at mospher ic (c) at mospher ic (d) negat ive

41. The specific speed of a hydr aulic t ur bine is given by N P P N (a) (b) 5/4 H H 5/4 NP NP (c) (d) H 3/2 H 3/2 42. For supplying feed wat er t o a boiler which of t he following pump is not used ? (a) St eam inject or (b) Recipr ocat ing pump (c) M ultist age centr ifugal pump (d) Gear pump 43. I n hydr aulic coupling, as t he r at io of t he speed dr i ven sh aft t o dr i vi ng sh af t i n cr eases, t h e efficiency (a) decr eases (b) incr eases (c) r emains const ant (d) is independent of speed r at io

Fluid Mechanics and Hydraulic Machinery

6.41

44. Pump select ed for pumping sewage is (a) Recipr ocat ing pump (b) Open impeller cent r ifugal pump (c) M ultist age centr ifugal pump (d) Scr ew pump

51.The hydr aulic r adius is given by

45. I n a cent r ifugal pump when deliver y valve is fully closed, t he pr essur e of fluid inside t he pump will (a) become zer o (b) r educe (c) incr ease (d) r emain unalt er ed

52.Specific speed (met r ic) = k ×specific (FPS), wher e k has t he value of (a) 1 (b) 1.75

46. Under shot water wheels ar e those on which wat er acts (a) pur ely by impulse (b) par t ly by impulse and par t ly by r eact ion (c) pur ely by r eact ion (d) none of t hese 47. I n t h e case of pel t on t ur bi n e i n st al l ed i n a hydr aulic power plant , t he gr oss head available is t he ver t ical dist ance bet ween (a) for ebay and t ail r ace (b) r eser voir level and t ur bine inlet (c) for ebay and t ur bine inlet (d) r eser voir level and t ail r ace 48.A hydr aulic coupling belongs t o t he cat egor y of (a) power absor bing machines (b) power developing machines (c) ener gy gener at ing machines (d) ener gy t r ansfer machines 49.For pumping molasses, it is pr efer able t o employ (a) r ecipr ocating pump (b) cent r ifugal pump wit h double shr ouds (c) open impeller pump (d) mult ist age centr ifugal pump 50.I n t he case of a cent r ifugal pump, cavit at ion will occur if (a) it oper at es above t he minimum net posit ive suct ion head (b) it oper at es below t he minimum net posit ive suct ion head (c) t he pr essur e at t he inlet of t he pump is above t he at mospher ic pr essur e (d) t he pr essur e at t he inlet of t he pump is equal t o t he at mospher ic pr essur e

(a) wet t ed per imet er divided by ar ea (b) ar ea divided by squar e of wet t ed per imet er (c) squar e r oot of ar ea (d) ar ea divided by wet t ed per imet er

(c) 2.37 (d) 0.67 53.The cavit at ion and pit t ing can be pr event ed by cr eat ing which one of t he following condit ions? (a) Reducing t he pr essur e head (b) Reducing t he velocit y head (c) I ncr easing t he elevat ion head (d) Reducing t he piezomet r ic head 54.Pr essur e r ise due t o wat er hammer in a penst ock depends on (a) wat er l evel i n t he r eser vior, and el ast ici t y of wat er (b) densit y of wat er (c) r oughness of pipe (d) all of t hese 55.A pumped st or age plant is a (a) high head plant (b) r un-off r iver plant (c) peak load plant (d) base load plant 56.Whi ch of t he following st at ement s about gear pumps ar e t r ue? I . Gear pumps ar e best suit ed for high pr essur es and small dischar ges. I I . Gear pumps ar e self pr iming. I I I . Efficiency of a gear pump depends on t he slip. I V. The dischar ge is inver sely pr opor t ional t o t he axial lengt h of t he gear t oot h. (a) I and I V only (b) I I and I I I only (c) I I I and I only (d) I I and I V only 57.I n t he figur e below, which cur ve r epr esent s t he con di t i on f or back w ar d cu r v ed v an es i n a cent r ifugal pump? (a) Cur ve A (b) Cur ve B (c) Cur ve C (d) Cur ve D

6.42

Fluid Mechanics and Hydraulic Machinery

58.I n all r eact ion t ur bines, maximum efficiency is obtained, if (a) guide vane angle is 90° (b) blade angle of t he r unner s is 90° at t he inlet (c) blade angle of t he r unner s is 90° at t he out let (d) angle of t he absolut e velocit y vect or at t he out let is 90°

66.Jet pumps ar e oft en used in pr ocess indust r y for t heir (a) high efficiency (b) easy maint enance (c) lar ge capacit y (d) capaci t y t o t r an spor t gases, l i qui ds an d mixt ur es of bot h

59.The shut off head will be near ly (a) Zer o (b) 18.0 m (c) 25.5 m (d) 35 m

67.Spout ing velocit y is (a) ideal velocit y of jet (b) 50% of ideal velocit y of jet (c) act ual velocit y of jet (d) velocit y of jet under some specified condit ions

60.Double hemispher ical bucket s ar e used in (a) Kaplan tur bine (b) Fr ancis t ur bine (c) Pr opeller t ur bine (d) Pelt on wheel 61.A r ecipr ocating pump (a) r equir es pr iming (b) r equir es air vessel (c) r equir es high oper at ing speeds (d) is used for viscous fluids only 62.Air vessel in a r ecipr ocat ing pump is used t o (a) obt ain cont inuous supply of wat er at unifor m r at e (b) incr ease deliver y (c) r educe suct ion head (d) r emove any ent r apped air fr om wat er 63.A t aper ed dr aft t ube as compar ed t o a cylindr ical dr aft t ube (a) pr event s hammer blow and sur ges (b) r esponds bet t er t o load funct ions (c) conver t s mor e of kinet ic head int o pr essur e head (d) pr even t s cavi t at i on even u nder r edu ced dischar ges 64.The specific speeds of r adial vane pump, mixed flow pump and axial flow pump ar e NS1, NS2 and NS3 r espect ively. Then

68.The comparison between pumps operating in ser ies and in par allel is (a) Pumps oper ating in ser ies boost the dischar ge, wher eas pumps oper ating in par allel boost the head. (b) Pu m ps oper at i n g i n par al l el boost t h e dischar ge, wher eas pumps oper at ing in ser ies boost t he head. (c) I n bot h cases t her e woul d be a boost i n dischar ge only. (d) I n bot h cases t her e would be a boost in head onl y. 69.A pelt on t ur bine is consider ed suit able for which of t he following head ? (a) 10 t o 12 met r es (b) 20 t o 30 met r es (c) 35 t o 50 met r es (d) 100 t o 250 met r es 70.M at er ial for wat er t ur bine should have (a) high cr eep r esist ance (b) high t emper at ur e r esist ance (c) high cor r osion r esist ance (d) low duct ilit y

LEVEL-1 71.

The coefficient of dischar ge (Cd ) of an or ifice var ies wit h (a) Reynold number (b) Weber number (c) Fr oude number (d) M ach number

72.

The shear st r ess dist r ibut ion for a fluid flowing in bet ween t he par allel plat es, bot h at r est , is (a) const ant over t he cr oss-sect ion (b) maximum at the mid-plane and varies linearly wit h dist ance fr om mid-plane

(a) NS1 > NS2 > NS3 (b) NS3 > NS2 > NS1 (c) NS2 > NS3 > NS1 (d) NS3 > NS1 > NS2 65.Cen t r i f u gal pu m ps oper at i n g i n ser i es wi l l r esult in (a) higher dischar ge (b) r educed power consumpt ion (c) higher head (d) low speed oper at ion

(c) zer o at the mid-point and var ies linear ly with dist ance fr om mid-plane (d) zer o at t he plat es and var ies exponent ially t o mid-point

Fluid Mechanics and Hydraulic Machinery

73.

74.

75.

Fr iction dr ag is generally lar ger than the pr essure dr ag in (a) flow past a spher e (b) flow past a cylinder (c) flow past an air foil (d) flow past a t hin sheet .

80.

81.

Capillar it y is due t o (a) Cohension (b) Adhesion (c) Adhesion and cohesion (d) Gr avity

(c) Re1/2

(d) Re– 1/2

I f one of t he walls moves in t he dir ect ion of flow wi t h uni for m vel oci t y whil e t he ot her wal l i s st at i onar y, t hen t he r esul t i ng fl ow bet ween par allel walls is called

(d) Euler ’s flow

Pr inciple of similit ude for ms t he basis of 82.

I n ser ies-pipe applications

(b) designing models so t hat t he r esult can be conver t ed t o pr ot ot ypes

(a) head loss t hr ough each pipe added t o obt ain t he t ot al head loss

(c) compar i ng si mi l ar l y bet ween desi gn and act ual equipment

(b) head loss is same t hr ough each pipe (c) fr ict ion fact or is assumed for each pipe (d) flow incr eases

I f V is t he mean velocit y of flow, t hen accor ding t o Dar cy-Weisbach equation for pipe flow, ener gy loss over a lengt h of pipe is pr opor t ional

83.

For pipe flow, at const ant diamet er, capacit y is pr opor t ional t o

(a) V

(b) 1/V

(a)

(c) V 2

(d)

(c) head3/2

V

H ead loss in t ur bulent flow in a pipe

84.

head

(b) head (d) head2

The velocit y dist r ibut ion for flow bet ween t wo fixed par allel plates (a) is const ant over t he cr oss-sect ion

(b) var ies inver sely as squar e of velocit y (c) var ies appr oximat ely as squar e of velocit y

(b) is zer o at t he plat es and incr eases linear ly t o t he midplane

(d) var ies inver sely as velocit y.

(c) var ies par abolically acr oss t he sect ion

The hor izont al component of for ce on a cur ved sur face is equal t o t he

(d) i s zer o i n m i ddl e an d i n cr ease l i n ear l y t owar ds t he plat es

(a) pr oduct of pr essur e at it s cent r oid and ar ea

85.

(b) weight of liquid r et ained by t he cur ved ar ea (c) for ce on a ver t ical pr oject ion of t he cur ved sur face

79.

(b) Re– 1

(c) Couet t e flow

(a) var ies dir ect ly as velocit y

78.

(a) Re

(b) St oke’s flow

(d) hydr aulic design

77.

T he dr ag coefficient for laminar flow var ies as (wher e Re = Reyonlds number )

(a) Plug flow

(a) compar ing t wo ident ical equipment s

76.

6.43

The most economi cal sect ion of a r ect angul ar channel for maximum discharge is obtained when it s dept h is equal t o (a) half t he br eadt h

(d) weight of liquid ver t ically above t he cur ved sur face

(b) t wice t he br eadt h

Tot al pr essu r e on t h e t op of a cl osed cylindr ical vessel of r adius r filled wit h liquid is pr opor t ional t o 1 (a) r (b) r 1 (c) (d) r 2 r2

(d) one t hir d t he br eadt h.

(c) same as t he br eadt h 86.

Region downst r eam fr om t he st r eamline wher e separ at i on t ak es pl ace fr om t he boundar y i s known as (a) wake

(b) lift

(c) drag

(d) cavit ation

6.44

87.

Fluid Mechanics and Hydraulic Machinery

A low pr essur e of t he or der of 10– 10 t or r can be measur ed in a chamber wit h (a) M anomet er (b) Bour don vacuum gauge (c) Pir ani gauge (d) I onisat ion chamber

88.

To avoid a cor r ect ion for t he effect of capillar it y in manomet er s, diamet er of t ube should be (a) less t han 1 mm (b) less t han 3 mm (c) less t han 4.5 mm (d) all of t hese

89.

The r iver flow dur ing floods can be classified as (a) st eady unifor m flow (b) unst eady unifor m flow (c) st eady non-unifor m flow (d) unst eady non-unifor m flow

90.

Wake always occur s (a) befor e a separ at ion point (b) aft er a separ at ion point (c) befor e and aft er a separ at ion point (d) none of t hese

91.

I n a flow field, at t he st agnat ion point (a) pr essur e is zer o (b) t ot al ener gy is zer o (c) pr essur e head is equal t o velocit y (d) all the velocity head is conver ted into pr essur e head.

92.

Tot al dr ag on a body is t he sum of (a) pr essur e dr ag and velocit y dr ag (b) fr ict ion dr ag and velocit y dr ag (c) fr ict ion dr ag and pr essur e dr ag (d) pr essur e dr ag, velocity dr ag and fr iction dr ag.

93. Specific speed of a pump and specific speed of a t ur bine ar e (symbols have t he usual meaning) (a)

N Q

(b)

N Q

(c)

N Q

(d)

N Q

H 3/4 H H H

3/4

5/4

5/4

and

N

and

N

and

N

and

N

P

H 5/4 H H H

P 3/4

P 5/4

P 5/4

r espect ively r espect ively r espect ively r espect ively

94. One hor se power is equal t o (a) 102 watt s (b) 75 watt s (c) 550 watt s (d) 735 watt s 95. Recipr ocat ing pumps ar e no mor e t o be seen in i n du st r i al appl i cat i on s (i n com par i son t o cent r ifugal pumps) because of (a) high init ial and maint enance cost (b) lower dischar ge (c) lower speed of oper at ion (d) necessit y of air vessel 96. The maximum cont inuous power available fr om a h y dr oel ect r i c pl an t u n der m ost adv er se hydr aulic condit ions, is called (a) Base power (b) Fir m power (c) Pr imar y power (d) Unpr edictable 97. The movable wicket gat es of a r eact ion t ur bine ar e used t o (a) cont r ol t he flow of wat er passing t hr ough t he tur bine (b) cont r ol t he pr essur e under which t he t ur bine is wor king (c) st r engt hen t he casing of t he t ur bine (d) r educe t he size of t he t ur bine 98. A Four neyr on t ur bine is (a) out war d flow r eact ion t ur bine (b) inwar d flow r eact ion t ur bine (c) out war d flow impulse t ur bine (d) inwar d flow impulse t ur bine 99. The degr ee of r eact ion of a t ur bine is defined as t he r at io of (a) st at ic pr essur e dr op t o t ot al ener gy t r ansfer (b) t ot al ener gy t r ansfer t o st at ic pr essur e dr op (c) change of velocit y ener gy acr oss t he t ur bine t o t he t ot al ener gy t r ansfer (d) velocit y ener gy t o pr essur e ener gy. 100. I n a cent r ifugal pump inst allat ion while st ar t ing, t he posit ion of deliver y valve is (a) fully open (b) fully closed (c) half open (d) mor e t han half open

Fluid Mechanics and Hydraulic Machinery

101. For at t aining a non-over loading char act er ist ic in cent r ifugal pumps (a) back war d bent vanes ar e pr efer r ed over for war d bent vanes (b) f or w ar d ben t v an es ar e pr ef er r ed ov er backwar d bent vanes (c) for war d bent vanes ar e pr efer r ed over vanes r adial at out let (d) vanes r adi al at out l et ar e pr efer r ed over backwar d vanes 102. Which one of t he following st at ement s r egar ding r ecipr ocat ing pumps is cor r ect ? (a) Friction head is mainly responsible for causing cavi t at i on i n a r eci pr ocat i ng pump at t he beginning of t he suct ion st r oke. (b) E f f ect of accel er at i on pr essu r e on recipr ocating pumps appears parabolic and has t he maxi mum effect at t he mi ddl e of t he deliver y st r oke. (c) Air vessel r educes t he acceler at ion head and consequently reduces the effect of fr iction head also. (d) The maximum per mi ssible suct ion lift in a dou bl e act i n g r eci pr ocat i n g pu m p i s independent of vapour pr essur e. 103. On an immer sed body in a flowing fluid t he lift for ce is (a) due t o buoyant for ce. (b) always in t he opposit e dir ect ion t o gr avit y (c) due t o wake phenomenon (d) t he dynamic fluid-for ce component nor mal t o appr oach velocit y 104. A plot bet ween power gener at ed in M W and t ime is known as (a) L oad cur ve (b) L oad dur at ion cur ve (c) L oad fact or (d) Demand cur ve 105. A cent r ifugal pump t akes t oo much power, due t o (a) low speed (b) air in wat er (c) air leakage (d) heavy liquid 106. I n r eact ion t ur bine t he dr aft t ube is used t o (a) transpor t water to downstream without eddies (b) r econver t kinet ic ener gy t o flow ener gy by a gr adual expansion of t he flow cr oss-sect ion (c) incr ease t he effect ive head (d) pr event air fr om ent er ing 107. A dr um of r adius R full of a fluid of densit y d is r ot at ed at  r ad/sec. The incr ease in pr essur e at t he out er edge of t he dr um will be

2 R2 d 2 Rd (c) 2 (a)

(b)

2 R d 2

(d)

 R d2 2

6.45

108. I n axial flow t ur bines, wat er ent er s (a) r adially but leaves axially (b) axially but leaves r adially (c) at an angle but leaves axially (d) axially and leaves axially

LEVEL-2 109. When pr essur e p, flow r at e Q, diamet er D, and density d, a dimensionless gr oup is r epr esented by p pQ 2 (a) (b) 2 4 4 dQ D dD (c)

pD4d Q2

(d)

pD 4

dD 2 110. Vi scosi t y i s t he most i mpor t ant pr oper t y i n t he (a) t r avel of a bullet t hr ough air (b) wat er jet issuing fr om a fir e air (c) for mat ion of soap bubbles (d) flow of cast or oil t hr ough a t ube. 111. A t ype of flow in which t he fluid par t icles while moving in t he dir ect ion of flow r ot at e about t heir mass cent r e, is known as (a) steady flow (b) unifor m flow (c) laminar flow (d) r ot at ional flow 112. I f pr essur e at any point in t he liquid appr oaches the vapour pr essur e, liquid star ts vapor ising and cr eates pocket s or bubbles of dissolved gases and vapour s. This phenomenon is (a) sur face t ension (b) adhesion (c) vapor isation (d) cavit ation 113. Fir e hose nozzle is gener ally made of (a) diver gent shape (b) conver gent shape (c) cylindr ical shape (d) par abolic shape. 114. Dischar ge of br oad cr ested weir is maximum in the head of wat er on downst r eam si de of wei r as compar ed t o head on t he upst r eam si de of t he weir is (a) one-half (b) one-t hir d (c) two-thir d (d) t hr ee-four t h 115. F or si m i l ar l y, i n addi t i on t o m odel s bei n g geomet r ically similar t o pr ot ot ype, t he following in bot h cases should also be equal (a) r at io of iner t ial for ce t o for ce due t o viscosit y (b) r at i o of i n er t i al f or ce t o f or ce du e t o gr avit ation (c) r at io of iner t ial for ce t o for ce due t o sur face t ension (d) all t he four r at ios of iner t ial for ce t o for ce due t o viscosit y, gr avit at ion, sur face t ension and elast icit y

6.46

Fluid Mechanics and Hydraulic Machinery

116. Dischar ge over a shar p-edged r ect angular not ch of widt h w and dept h h is equal t o (a) (c)

2 C d w 2 gh 5/ 2 3

(b)

2 C d w 2 gh 3/ 2 3

(d)

2 C d w 2 gh 3 8 C d w 2 gh 3/ 2 15

117. I n a fr ee vor t ex mot ion, t angent ial velocit y of t he wat er par t icles is pr opor t ional t o

r2

(a) r

(b)

(c) 1/r

(d) 1/r 2

wher e, r = dist ance fr om t he cent r e. 118. To r eplace a pipe of diameter D by n par allel pipes of diamet er d, t he for mula used is D D (a) d = (b) d = 1/ 2 n n D D (c) 3/ 2 (d) 2 / 5 n n 119. Ver t i cal com pon en t of pr essu r e f or ce on a submer ged cur ve sur face is equal t o (a) weight of liquid ver t ically above t he cur ved sur face and ext ending upon t he fr ee sur face (b) for ce on a ver t ical pr oject ion of t he cur ved sur face (c) pr oduct of pr essur e at cent r oid and sur face ar ea (d) hor izont al component

(a) gr avit y velocit y and viscous (b) gr avit y, pr essur e and t ur bulent (c) pr essur e, viscous and t ur bulent (d) gr avit y, viscous and t ur bulent 121. For a flow t o be r ot at ional, velocit y nor mal t o t he plane of ar ea should be equal t o t he (a) angular velocit y vect or (b) half t he angular velocit y vect or (c) t wice t he angular velocit y vect or (d) zer o 122. Cont i n u i t y equ at i on f or an i n com pr essi bl e fluid is

(c)

A 1V1 A V = 2 2 1 2

(b) 1 A 1 V 1 = 2 A 2V 2 (d)

(a) Reynolds number (Re) (b)

1 Re

(c)

4 Re

(d)

16 Re

124. H ydr aul i c gr ade l i ne for any fl ow syst em as compar ed t o ener gy line is (a) above

(b) below

(c) at same level

(d) uncer t ain

125. The effect of negative pr essur e beneath the nappe i n case of fl ow of wat er over a r ect angul ar suppr essed weir is t o (a) decr ease t he dischar ge (b) incr ease t he dischar ge (c) incr ease fr ict ional r esist ance (d) r educe fr ict ional r esist ance 126. Gr adually var ied flow is (a) st eady unifor m flow (b) st eady non-unifor m flow (c) unst eady unifor m flow (d) unst eady non-unifor m flow

120. The fluid for ces consider ed in t he Navier -St okes equat ion ar e

(a) A 1 V 1 = A 2V 2

123. L oss of head due t o fr i ct i on i n a pi pe of uni for m di amet er wi t h vi scous fl ow i s equal t o

 A 1 A 1 = 2 2 2 V1

wher e V = velocit y,  = densit y and A = ar ea

127. A flui d i n whi ch r esi st ance t o defor mat i on i s independent of t he shear str ess, is known as (a) Bingham plastic fluid (b) Pseudo plast ic fluid (c) Dilatant fluid (d) Newt onian fluid 128. Pr essur e dr ag r esult s fr om (a) skin-fr iction (b) defor mation dr ag (c) development of a st agnat ion point (d) occur r ence of a wake 129. Tur bulent flow is hydr aulically smoot h if r at io

height of r oughness pr oject ion is less t han t hickness of laminar sub - layer (a) 1.00

(b) 0.75

(c) 0.50

(d) 0.25

Fluid Mechanics and Hydraulic Machinery

130. Bluff body sur face

(a) (b) (c) (d)

is smoot h so t hat fr ict ion can be neglect ed coincides wit h st r eamlines does not coincide wit h st r eamlines per pendicular t o st r eamlines

131. Cent r e of pr essur e on an inclined plane is (a) at t he cent r oid

(b) above t he cent r oid

(c) below t he cent r oid

(d) at met acent r e

132. Separ ation of flow occur s when pr essur e gr adient (a) t ends t o appr oach zer o (b) becomes negat ive (c) changes abr upt ly (d) r educes t o a value when vapour for mat ion star ts 133. The fr ict ion head lost due t o flow of a viscous fluids t hr ough a cir cular pipe of lengt h L and diamet er d wit h a velocit y v, and pipe fr ict ion fact or ‘f’ is (a)

4 f L v2  d 2g

v2 (c) 2g

(b)

4 f L v2  d 2 2 g

4 f L v2  (d) d 2g

134. The r ate of change of linear moment um is equals to (a) act ive for ce

(b) r eact ive for ce

(c) t or que

(d) wor k done

135. Component of the for ce of fluid on the body (which is gener ally inclined t o t he dir ect ion of mot ion of t he body) par allel t o t he dir ect ion of mot ion is called (a) drag

(b) lift

(c) wake

(d) thr ust

136. Ener gy loss in flow t hr ough nozzle as compar ed t o vent ur imet er is (a) same

(b) mor e

(c) less

(d) unpr edictable

137. The r at e of change of moment of moment um r epr esent s t he (a) for ce ext er t ed by fluid (b) t or que applied by t he fluid (c) wor k done by t he fluid (d) power developed by t he fluid

6.47

138. I f t he char act er ist ics of a pump ar e as shown in t he figur e, t hen abscissa r epr esent s (a) head (b) RPM (c) dischar ge (d) power 139. To obt ai n r el at i on for t he speci fi c speed, t he assumpt ion made is t hat all pumps (a) ar e similar (b) of a given t ype ar e similar (c) of a given t ype ar e geomet r ically similar (d) ar e hydr aulically similar 140. Water tur bines may be put in the decr easing or der of specific speed as (a) Pr opeller t ur bine, React ion t ur bine, I mpulse tur bine (b) Pelton wheel, Fr ancis tur bine, Kaplan t ur bine (c) React ion t ur bine, I mpulse t ur bine, Pr opeller tur bine (d) Pr opeller t ur bine, I mpulse t ur bine, React ion tur bine 141. M ixed flow t ur bines ar e (a) r adial inwar d flow type (b) r adial out war d flow t ype (c) par t ly r adial par t ly axial (d) par allel flow type 142. A foot valve is pr ovided on (a) Centr ifugal pumps (b) Kaplan tur bines (c) Pelt on wheels (d) All of t hese 143. Which one is differ ent fr om t he ot her s ? (a) Axial flow impeller (b) M ixed flow impeller (c) War ped vane impeller (d) Shr ouded impeller 144. I n a r eact ion t ur bine (a) it is possible t o r egulat e t he flow wit hout loss (b) it must be placed at t he foot of t he fall and above t he t ail r ace (c) wor k done i s pur el y by t he change i n t he kinet ic ener gy of jet (d) only par t of the head is conver t ed into velocity befor e wat er ent er s t he wheel

6.48

Fluid Mechanics and Hydraulic Machinery

145. The discharge thr ough a rectangular weir varies as (a) H (b) H 1/2 (c) H 2 (d) H 5/2 146. The hydr aulic gr adient line is always (a) below t he t ot al ener gy line (b) par allel t o t he bot t om (c) above t he t ot al ener gy line (d) none of t hese 147. I n involut e casing t he velocit y of wat er (a) decr eases but pr essur e incr eases (b) and pr essur e bot h incr ease (c) and pr essur e bot h decr ease (d) incr eases but pr essur e decr eases 148. The pump which is differ ent fr om ot her s is (a) Simplex pump (b) Plunger pump (c) Pist on pump (d) Cent r ifugal pump 149. I f a cent r ifugal pump is noisy in oper at ion, it may be due t o (a) faulty pr iming (b) suct ion head t oo high (c) air in wat er (d) mechanical defect 150. Fr ancis t ur bine is best suit ed for (a) medium head applicat ion fr om 24 t o 180 m (b) high head inst allat ion above 180 m (c) low head inst allat ion upt o 30 m (d) all t ypes of heads 151. H ydr aulic accumulat or is used for (a) accumulat ing oil (b) supplying lar ge quant it ies of oil for ver y shor t dur ation (c) supplying ener gy when main supply fails (d) accumulating hydr aulic ener gy 152. Run away speed of a hydr aulic tur bine is the speed (a) at full load (b) at which t ur bine r unner will be damaged (c) at whi ch t he t ur bi ne r unner i s al l owed t o r evol ve fr eel y wi t hout l oad and wi t h t he wicket gat es wide open (d) cor r espon di n g t o m ax i m u m ov er l oad per missible 153. Accor di ng t o t he l aws of pr opor t i on al i t y f or homologous t ur bines, speed is pr opor t ional t o (a)

H /D

(b)

H/D

(c)

H / D2

(d)

HD

154. A hydr aulic int ensifier nor mally consist s of (a) t wo cylinder s, t wo r ams and a st or age device (b) a cylinder and a r am (c) t wo co-axial r ams and t wo cylinder s (d) a cylinder, a pist on, st or age t ank and cont r ol valve 155. Accor di n g t o t he l aws of pr opor t i onal i t y for homologous turbines, dischar ge is pr opor tional to (a) D H

(b) D 2

H

(c) D 2 H 3/2 (d) D 2 H 156. Any change in load is adjust ed by adjust ing on tur bine (a) net head (b) absolut e velocit y (c) blade velocit y (d) flow 157. For pumping viscous oil, pump used will be (a) cent r ifugal pump (b) r ecipr ocating pump (c) t ur bine pump (d) scr ew pump 158. The cavitation in r eaction type hydr aulic tur bines is avoided by (a) using highly polished blade (b) using st ainless st eel r unner (c) installing t he t ur bine below the tail r ace level (d) all of t he above 159. Pr opeller t ur bine is best suit ed for (a) medium head applicat ion fr om 20 t o 180 m (b) low head inst allat ion up t o 30 m (c) high head inst allat ion above 180 m (d) all t ypes of heads 160. A hydr aul i c jump can occur under al l of t he following condit ions except : (a) on t he upst r eam side of t he sluices (b) at t he foot of t he spillways (c) wher e t he gr adient suddenly changes fr om a st eep slope t o a flat slope (d) when wat er moving in shoot ing flow impact s wi t h wat er h avi n g a l ar ger dept h wi t h st r eaming flow 161. As per the aer ofoil theor y of K aplan tur bine blade design, guide angle is as t he angle bet ween (a) lift and r esult ant for ce (b) dr ag and r esult ant for ce (c) lift and t angent ial for ce (d) lift and dr ag 162. Power r equir ed t o dr ive a cent r ifugal pump is pr opor t ional t o (D = impeller diamet er ) (a) D (b) D 2 3 (c) D (d) D 4

Fluid Mechanics and Hydraulic Machinery

163. Rat io of maximum load t o r at ed plant capacit y is known as (a) L oad fact or (b) Ut ilizat ion fact or (c) M aximum load fact or (d) Capacity fact or

6.49

165. For cent r ifugal pump (a) head  speed2 (c) power  speed

(b) head  diamet er 2 3

(d) none of t hese

166. For smal l di schar ge at hi gh pr essur e, pump pr efer r ed is

164. A double act ing r ecipr ocat ing pump compar ed t o single act ing pump will have near ly (a) double efficiency (b) double head (c) double flow (d) double weight

(a) centr ifugal

(b) axial flow

(c) pr opeller

(d) r ecipr ocat ing

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (d)

2. (c)

3. (d)

4. (a)

5. (d)

6. (c)

7. (a)

8. (c)

9. (d)

10. (b)

11. (d)

12. (c)

13. (d)

14. (d)

15. (d)

16. (c)

17. (d)

18. (a)

19. (a)

20. (d)

21. (c)

22. (b)

23. (a)

24. (a)

25. (d)

26. (d)

27. (c)

28. (d)

29. (c)

30. (a)

31. (c)

32. (b)

33. (d)

34. (c)

35. (a)

36. (d)

37. (c)

38. (d)

39. (a)

40. (d)

41. (a)

42. (d)

43. (b)

44. (b)

45. (d)

46. (a)

47. (b)

48. (d)

49. (c)

50. (d)

51. (d)

52. (d)

53. (d)

54. (d)

55. (a)

56. (b)

57. (c)

58. (d)

59. (d)

60. (d)

61. (b)

62. (a)

63. (c)

64. (b)

65. (c)

66. (b,d)

67. (a)

68. (b)

69. (d)

70. (c)

LEVEL-1 71. (a)

72. (c)

73. (b)

74. (c)

75. (b)

76. (b)

77. (d)

78. (c)

79. (d)

80. (d)

81. (a)

82. (d)

83. (c)

84. (a)

85. (c)

86. (d)

87. (a)

88. (d)

89. (d)

90. (b)

91. (b)

92. (c)

93. (a)

94. (d)

95. (a)

96. (b)

97. (b)

98. (a)

99. (a)

100. (b)

101. (a)

102. (c)

103. (b)

104. (a)

105. (b)

106. (b)

107. (a)

108. (a)

LEVEL-2 109. (a)

110. (c)

111. (c)

112. (a)

113. (c)

114. (d)

115. (* )

116. (c)

117. (c)

118. (d)

119. (a)

120. (a)

121. (c)

122. (a)

123. (d)

124. (b)

125. (b)

126. (b)

127. (d)

128. (d)

129. (d)

130. (c)

131. (c)

132. (c)

133. (a)

134. (c)

135. (a)

136. (a)

137. (b)

138. (c)

139. (c)

140. (a)

141. (c)

142. (a)

143. (d)

144. (d)

145. (d)

146. (a)

147. (a)

148. (d)

149. (d)

150. (a)

151. (c)

152. (c)

153. (a)

154. (c)

155. (b)

156. (d)

157. (d)

158. (d)

159. (b)

160. (a)

161. (a)

162. (d)

163. (b)

164. (c)

165. (a)

166. (d)

6.50

Fluid Mechanics and Hydraulic Machinery

EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S 42.

Boi l er feed pump has t o suppl y wat er at hi gher pr essur es– mor e t han t he pr essur e of st eam. Gear pumps ar e never used for such ser vi ces. N2 43. I n fluid coupling, efficiency = N1 Thus var iat ion of efficiency wit h speed r at io is linear. 48.A hydr aulic coupling consists of a pump unit fitted on t he dr iving shaft and a t ur bine unit on t o t he dr i ven shaft . These t wo uni t s ar e housed i n a common casing in such a manner t hat out put of t he pump is fed t o t he t ur bine unit which r ot at es and drives the driven shaft. The medium fluid used is gener ally oil. 49. Molasses is highly viscous ther efor e open impeller should be used. Nor mally a scr ew pump is used for pumping highly viscous fluids. 53. When absolut e local pr essur e at any point in a conduit car r ying a liquid appr oaches t he vapour pr essur e pv of the liquid, dissolved gases and liquid vapour emer ge out of t he liquid as bubbles. These bubbles may t r avel t o r egions of higher pr essur e and collapse. At t he point of bubble collapse, t he boundar y get s damaged. The phenomenon of for mat ion, t r avel and collapse of vapour bubbles is cal led cavitation. 56. Gear pumps ar e best sui t ed for l ow pr essur es (1000 k N /m 2) and smal l di schar ges (l ess t han 30 lps). These ar e self pr iming and ar e oft en used t o pr ime lar ge cent r ifugal pumps. Dischar ge,

Q = K .C (D– C). L

FG N IJ H 60K

wher e L = axial lengt h of t eet h, C = cent r e t o cent r e distance bet ween axes of gear s. D = out side diamet er of gear. 58. I n case of r eact ion t ur bines, maximum efficiency is obt ained when absolut e velocit y vect or at t he out let is 90°. 59. By ext r apolat ing Q vs H cur ve, t he shut off head which is cor r esponding t o Q = 0 is 35.5 m. 60. Bucket s ar e double hemispher ical in shape. These bucket s ar e made of cast ir on, br onze or st ainless st eel . 61. A r ecipr ocat ing pump is self pr iming. I t oper at es at low speeds. I t is not suit able for viscous liquids, paper pulp, molasses, et c. 62. Air vessel smoot hens wat er dischar ge which is ot her wise pulsat ing. Air vessel may be fit t ed on suct ion side and/or deliver y side. 63. A t aper ed dr aft t ube is used for low specific speed ver t ical shaft t ur bines. I t s maximum cone angle is 8°. I f angle is mor e t han t his, wat er does not t ouch t he inner walls, causing vor t ices and loss of head. M aximum efficiency obt ained in t his tube is 9%. The t ube must dischar ge sufficient ly below t he t ail wat er level. 65. Ser ies oper at ion is achieved by having one pump di schar ge i nt o t h e suct i on of t he next . Thi s ar r angement i s used pr imar il y t o incr ease t he dischar ge head.

LEVEL-1 97. The pur pose of guide vanes or wicket gat es is t o dir ect wat er t o ent er int o t he r unner vanes at a suit able angle t o avoid any wast age of ener gy due t o shock and t o conver t par t ly pr essur e ener gy of t he ent er ing wat er i nt o ki net i c ener gy. I t also r egulates supply of wat er accor ding t o t he load on t he t ur bine.

7

Industrial Engineering

7.1

Industrial Engineering

C H A P TE R PRODU CT I ON PLAN N I N G I n pr oduct ion, pr oduct s ar e manufact ur ed by t he t r ansfor mat ion of r aw mat er ial (int o finished goods). Planning looks ahead, ant icipat es possible difficult ies and decides in advance as t o how t he pr oduct ion, best , be car r ied out . Cont r ol phase makes sur e t hat t he pr ogr ammed pr oduct ion is const ant ly maint ained.

FU N CTI ON S OF PRODU CTI ON PLAN N I N G A pr oduct ion planning and cont r ol syst em has following funct ions t o per for m, some befor e t he ar r ival of r aw mat er ials and t ools, and ot her s while t he r aw mat er ial under goes pr ocessing. For ecast ing Pr ior planning (1) Planning phase Pr oduct design Active planning Pr ocess planning and r out ing M at er ial cont r ol Tool cont r ol Loading Scheduling (2) Action phase Dispatching Pr ogr ess r epor t ing Data pr ocessing (3) Cont r ol phase Expediting Cor r ect ive act ion Replanning (1) Planning phase ( i ) Pr ior planning : (a) For ecast ing : Est imat ion of t ype, quant it y and qualit y of fut ur e wor k. (b) Pr oduct design : Collect ion of infor mat ion r egar ding specificat ions, bill of mat er ials, dr awings, et c. ( ii ) Act ive planning : (a) Pr ocess planning and r out ing : Finding t he most economical pr ocess for doing a wor k and (t hen) deciding how and wher e t he wor k will be done. (b) M at er ial cont r ol : Det er minat ion of t he r equir ement s and cont r ol of mat er ials. (c) Tool cont r ol : Det er minat ion of t he r equir ement s and cont r ol of t ools used. (d) L oading : Assignment of wor k t o manpower, machiner y, et c. (e) Scheduling : I t is t he t ime phase of loading and deter mines when and in what sequence t he wor k will be car r ied out . I t fixes t he st ar t ing as well as t he finishing t ime for t he job. (2) Act ion phase Dispat ching : I t is t he t r ansit ion fr om planning t o act ion phase. I n t his phase t he wor ker is or der ed t o st ar t t he act ual wor k. (3) Cont r ol phase ( i ) Pr ogr ess r epor ting. ( a ) Dat a r egar ding t he job pr ogr ess is collect ed. (b) I t is int er pr et ed by compar ison wit h t he pr eset level of per for mance. ( ii ) Cor r ect ive act ion. ( a ) Expedit ing : I t means t aking act ion if t he pr ogr ess r epor t ing (j) indicat es a deviat ion of t he plan fr om t he or iginally set t ar get s. ( b) Replanning : Replanning of t he whole affair becomes essent ial, in case expedit ing fails t o br ing t he deviat ed plan t o it s act ual (r ight ) pat h.

7.2

Industrial Engineering

FU N CTI ON AL ORGAN I SATI ON OF PRODU CTI ON PLAN N I N G AN D CON T ROL . PRODUCTI ON PLANNI NG AND CONTROL

PRODUCTI ON PLANNI NG

I NVENTORY CONTROL

PRODUCTI ON CONTROL

TRAFFI C

Production Budget

Material Budget

Manufact uring Met hods and Processes

Stores Record keeping

Manufact uring and Subsidiary Orders

Receiving

Machinery and Equipment

Material Requirements

Routing

Shipping

Plant Layout

Stores-Keeping

Scheduling – Machine Loading

I nt ernal Transportat ion

Tool Design

Solveging

Dispatching

Operat ion Sheets and instraction Cords

Expediting Follow-up

Mot ion and Time Study

Tool-Keeping

Machines and Manpower Requirements Master Schedule

Types of Pr oduct ion Syst em 1. Cont inuous 2. I nt er mit t ent

CON T I N U OU S PRODU CTI ON SYSTE M I t involves a continuous or almost continuous physical flow of mat er ial. I t makes use of special purpose machines, and pr oduces st andar dized it ems in lar ge quant i t i es e.g. chemical pr ocessing, ci gar et t e manufact ur i ng, and cement manufact ur i ng

Classificat ion of Cont inuous Pr oduct ion Syst em ( i ) M ass and fl ow li ne pr oduct i on ( ii ) Cont inuous or pr ocess pr oduct ion.

I N TE RM I TTE N T PRODU CTI ON SYSTE M I t is specified by t he int er mit t ent or int er r upt ed flow of mat er ial t hr ough t he plant . I t makes use of gener al pur pose machines and pr oduces component s differ ent in nat ur e and in small quant it ies.

e.g. machine shops, r epair and maint enance shops, welding shops, et c.

Classificat ion of I nt er mit t ent Pr oduct ion Syst em ( i ) Bat ch pr oduct ion (ii ) Job pr oduct ion.

Industrial Engineering

7.3

ROU T I N G Rout ing pr escr ibes t he sequence of oper at ions wher e and by whom t he wor k will be done. I t for ms t he basis for loading and scheduled funct ion of planning. The r outing plan is usually made in advance of pr oduction in continuous manufact ur ing system of pr oduction, however, r outing decisions ar e often made in the shop floor which includes last minute changes. The design of for mat used for t he r out ing sheet depends upon t he t ype of manufact ur ing syst em. Usually it is made simpler by t he usage of differ ent symbols and not at ions for quick and easy ident ificat ion. Row pr ocedur es ar e evolved, based on t he nat ur e of t he pr oduct ion syst em and t he complexit ies of or ganisat ion. I n int er mit t ent t ype of pr oduct ion syst ems, r out ing has t o det er mine for ever y bat ch wher e t he oper at ion is t o be done and t hr ough which r out e. I n case of cont i nuous manufact ur i ng syst ems, t he manufact ur i ng l i ne l ayout i s cl ear and fi xed and t h e r out i ng becomes mechani cal . I t does not pose much of a pr obl em because al l fol l ows t he same r out e.

L OAD I N G I n planning pr oduct ion, shop or der s have t o be conver t ed int o wor k load on individual wor k machines or gr oups of machines. The pr oduct ion manager must know the loading on each equipment /machine in or der t o deter mine whet her capacit y ut ilisat ion is full or ot her wise. H e should also indent ify over loads and under loads on each machine t o assist pr oduct ion planning t o balance t he load. M anual syst ems gener ally used t o load oper at ions against act ual capacit y by cr eat ing a bar char t on a load boar d. The design of t he loading boar d is made t o suit t he specific needs of t he syst em. The boar d compr ises of a list of r esour ces (machines) on t he left , a t ime scale acr oss t he t op and bar s of differ ent lengt hs against each r esour ce t o indicat e t he ext ent of loading.

SCH E D U L I N G Under pr oduct ion cont r ol syst em usually t hr ee t ypes of schedules ar e pr epar ed. ( i ) M ast er schedules I t indicat es t he dat a on which var ious jobs in hand ar e r equir ed t o be complet ed. M ast er Schedule is like a cont r ol boar d and a key schedule about all pr oduct ion act ivit ies. ( ii ) Oper at ion schedule I n t his, t ime is fixed for doing a par t icular piece of wor k wit h a given machine. ( iii ) D et ailed oper at ion schedule I n t his, t ime r equir ed for doing each det ailed oper at ion of given job is shown. While doing so, machine t o be used or t he pr ocess t o be followed is t aken int o account .

I N VEN TORY (OR STOCK OF GOODS)

I nventory is a physical stock items that a production enterprise keeps in hand for efficient running of its production. I nvent or y consi st s of r aw mat er i al s, component par t s, suppl i es or fi ni shed assembl i es et c. whi ch ar e pur chased fr om an out side sour ce and t he goods manufact ur ed in t he int er pr ise it self. I nvent or y cont r ol may be defined as, t he syst emat ic locat ion, st or age and r ecor ding of goods in such a way t hat desir ed degr ee of ser vice can be made t o t he oper at ing shops at minimum ult imat e cost . I nvent or y ser ves as t he buffer or safet y against it s planning, sudden demand cont inuous pr oduct ion et c. Any for m of invent or y is a sign of inefficiency. L evel of invent or y may be r educed by : (i ) bet t er planning (ii ) cont inuous monit or ing of st ock (on-line) (iii ) r eliable vendor s (iv) use of Just in-Time (JI T) concept s.

PU RPOSE OF I N V EN TORY CON TROL

I nvent or y cont r ols ar e needed t o have pr oper t r ack of t he invent or y, i.e. t o find t hat t he available movable mat er ial and t o find out t hat mat er ial of r equir ed size and qualit y in r equir ed quant it y is availabl e so t hat manufact ur ing wor k does not suffer at all. Each depar tment has and is likely t o get t he desir ed mat er ial wit hout loss of t ime and wit h least labour. The cont r ols ar e obser ved and exer cised so t hat obsolet e it ems ar e separ at ed fr om used it ems and also space shor t age pr oblem is solved. I nvent or y has been divided in differ ent cat egor ies so t hat t her e is no pr oblem about laying hand on t he r equir ed mat er ial.

7.4

Industrial Engineering

Such a cont r ol aims at having per iodic invent or y check up and t hose who ar e supposed t o exer cise cont r ol ar e t o ensur e t hat t he needed mat er ial is pur chased and st ocked when t hat is available at economical pr i ces, so that manufactur e is not r equir ed to unnecessar ily wast e the money. I nventor y means keeping unto and accur ate r ecor d of t he available mat er ial and t he object ive of t his cont r ol is t o ensur e t hat t his r ecor d is pr oper ly and car efully kept .

TYPES OF I N VEN TORY 1. Raw I nvent or y I t includes all such it ems which ar e supplied by some ot her fir m in t he for m of r aw mat er ial, for t he concer n t o which t hese ar e supplied. These r aw mat er ials ar e used in t he finished pr oduct in some for m or t he ot her.

2. Pr ocess I nvent or y These ar e used in pr ocess of manufact ur e and as such t hese ar e neit her r aw mat er ial nor finished goods.

3. F i ni shed I nvent or y These ar e finished goods and r eady t o be t aken in st or e and st ock for sale pur pose.

QU ALI TY STAN DARDS Ther e ar e six impor t ant st andar ds used as a t ool t o cont r ol inver t or y :

1. St andar d or der I t means quantity to be purchased at any time. I t is the difference between maximum and minimum quantity.

2. L ead or pr ocur ement t ime I t is a t ime which is calculat ed by a fir m on t he basis of past exper ience. I t is t he t ime t hat lapses bet ween pr epar ing of invoices for t he placement of or der s and t he t ime t aken for t he supply of mat er ials.

3. M aximum st or e I t is upper limit of t he invent or y and is t he lar gest quant it y which should be kept in st or es t aking int er est s of t he company int o consider at ion.

4. M ini mum st or es I t indicat es lower limit of t he invent or y, t he safet y mar gin can be used in case of emer gency. This quant it y also includes abnor mal r equir ement s of t he fir m.

5. St ock holdi ng I t means holding of stock so that inventor y does not go out of stock. I t is also means a buffer stock which can be utilised when mater ial falls even below the minimum level. I t takes car e of shor tage of inventor y in the mar ket and helps in r educing the var iety of items to be handled. I t also helps in deciding quick mater ials at the time of need and necessity.

6. Or der i ng poi nt I t means a point which indicates time to initiate a purchase order. It is the quantity of material required between the exhaustion of available stocks and the supply during the interval between placement of an order and delivery of materials.

VARI ABLES I N I N VEN TORY CON TROL 1. Cont r ol l abl e (i ) H ow much quant it y acquir ed or or der ed (ii ) Fr equency and t iming of acquisit ion or or der ing

2. U ncont r ol l abl e (i ) H olding cost (C1) in Rs/unit pr oduct /unit t ime or I nvent or y car r ying cost This cost consist s of expendit ur e made for (a) I nsur ance (b) Stor age and handling (c) Obsolescence and Depr eciat ion (d) Det er i or at on (e) Taxes (f) I nt r est et c (ii ) Shortage cost or stock out cost (C2) in Rs/unit pr oduct /unit t ime or back or der cost This cost in associat ed wit h t he shor t age of goods.

Industrial Engineering

7.5

( iii ) Set up cost (M tg cost) or Order cost (Procurement) or Replenishment cost in Rs/order or setup Set up cost includes expendit ur es made on (a) Calling quotation (b) Pr ocessing quot at ions (c) Placing pur chase or der ed (d) Receiving and I nspection (e) Ver ifying and payment of bills (f) Ot her incident char ge. ( iv) Demand : Number of unit s r equir ed per or der I t is of t wo t ypes : (a) Det er minist ic — known demand (b) I ndet er minist ic — not fixed or not cer t ain (v) Amount delivered : H ow much quant it y of t he pr oduct /goods, supplied/or der ed at a t ime ( vi ) Lead time : Time elapsed (pass by) bet ween placing of or der and r eceiving t he goods (finished)

I nvent or y M odel Ext r eme M aximum A

K

N or mal M axi mum Aver age U sage

I nvent ory N umber of i t ems q B E O 0

t

10

5 t

N or mal M inimum

C 20

15

25 Ext r eme M i ni mum

D Ti me dur at i on (days)

R =

q t

q = Rt



wher e, R = r at e of demand q = quant it y t = cycle t ime Point B, C and E— Re-or der t ime Reor der quant ity — BK and CK

CODI FI CATI ON OF M ATE RI ALS I t helps in object ive descr ipt ion of var ious ar t icles and descr ibes an ar t icle fr om gener al t o par t icular. A good codificat ion has enough flexibilit y and can cover all t he mat er ials r eceived in t he concer n. Not only t his but it has enough pr ovision for unfor eseen cont ingencies. I t becomes easy t o specify mat er ial at ever y st age. These days codification is applied t o all component s and pr oduct s as well as r aw mat er ials, semi-finished and finished goods.

M N E M ON I CS or Alpha N umeric Syst em SS St ainless St eel N umer ic 0 2   Raw mat er ial Metal I nvent ory Turnover Rat io (I TR). ITR =

02

208

Diamet er of r od

L engt h of r od

6  M.S

5  Bar

3  Round

025  Dia in mm

Annual consumpt ion Aver age invent or y

PU RPOSE OF H OLDI N G I N VEN TORY 1. D i scont i nuit y fact or I nvent or y is int ended t o make pr ovision for discont inuit y of funct ion, such as pur chasing, pr oduct ion, dist r ibut ion and sales wit h r espect t o t ime. By having invent or y bet ween t he above st ages each funct ion could be decoupled and able t o per for m at opt imum levels wit hout a br eak.

7.6

Industrial Engineering

2. U ncer t aint y fact or I t pr ovides pr ot ection against unexpect ed var iation in t he funct ion with a view to have some safety chushion.

3. E conomic fact or I t pr ovides a means for obt aining economic lot sizes and gaining quant it y discount s.

DI SADVAN TAGE OF H OLDI N G I N VEN TORY 1. L ocked up wor king capit al The invest ment in invent or ies r epr esent a sizable sum common t o all indust r ies. As invent or ies r epr esent usuable but idle r esour ces, holding t he invent or ies involves locked up wor king capit al, which will not br ing any r et ur ns.

2. St or age The invent er ies ar e t o be st or ed pr oper ly wit hout any det er ior at ion and unaut hor ised r emoval.

3. Absol escen ce Wit h t he emer ging t echnology, many of t he it ems pur chased ear lier and st or ed may not be of any use. These it ems would be lost in value and had t o be disposed off.

I N VE N TORY FU N CTI ON COSTS I t r epr esent t he cost s associat ed wit h t he funct ions of t he invent or y syst em. While developing an opt imal invent or y policy, t he cr it er ion is t he minimisat ion of t ot al cost .

M ajor Component s of I nvent or y Cost s ( i ) Ordering cost (C o) This r epr esent s expenses involved in placing an or der wit h t he out side supplier. This includes t he cost s involved in pr ocessing and or der ing for pur chase, expedit ing over t he or der s, r ecei ving t he consignment and inspect ion. I t is expr essed as cost in r upees per or der .

l Cos Tota

t

ing rr y Ca

Cost

s Co

t

Orde ring cos

t

econ o cost mic

( ii ) Set up cost (CSC) Order of quantity (Q) —  The par allel of t he or der ing cost when t he it ems ar e pr oduced wit hin t he or ganisat ion, is t he set -up cost . This r epr esent s init ial pr oduct cost in changing over t he exist ing pr oduct ion r un t o pr oduce t he or der ed it ems. I t is expr essed as cost in r upees per or der . Bot h t he or der ing cost and set up cost ar e independent of t he quant it y or der ed. ( iii ) Carrying cost (C c) This r epr esent s cost of holding and st or age of invent or y. I t is pr opor t ional t o t he amount of invent or y and t he t ime over which it is held. ( iv) Shor t age cost This r epr esent s loss t o t he fir m due t o non-availabilit y of t he it em, when it is r equir ed. (v) U nit cost I t r efer s t o t he nominal cost of t he invent or y it em per unit . I t is pur chase pr ice of t he it em if it is bought fr om out side. I t is t he pr oduct ion cost of t he it em pr oduced wit hin t he or ganisat ion. I f it is assumed as const ant and independent of t he quant it y or der ed, it does not affect invent or y cont r ol decisions. Consider quantity discounts, wher e unit cost var ies for differ ent or der quantities, then the effect of quantity discount s separ at ely have t o be consider ed separ at ely. The unit cost is expr essed as r upees per unit . The inventor y decisions ar e based on t he t otal invent or y cost s as an out come of t he decisions. The car r ying cost is dir ect ly pr opor t ional t o t he or der quant it y (Q), i.e. when t he or der quant it y per or der is incr eased, t he car r ying cost is incr eased. When t he or der quant it y is incr eased, t he number of or der s per unit t ime will be less and hence t he or der ing cost would decr eases (i.e. inver sely pr opor t ional t o or der quant it y (Qs). Tot al invent or y cost = car r ying cost + or der ing cost .

E conomic or der quant it y (E OQ) The or der quant it y (Q) for which t he t ot al cost is minimum is called Economic Or der Quant it y (EOQ). This may be expr essed in t er ms of unit s or in t er ms of r upee value of t he unit s or der ed. I n t he case of pur chasing, it is called Economic or der Quant it y . I n t he case of pr oduct ion of it em wit hin t he or ganisat ion is called ‘Economic Bat ch Size'.

Industrial Engineering

7.7

DE TE RM I N I STI C M ODE LS While making invent or y decisions it is r equir ed t hat some for m of mat hemat ical analyt ical model is applied and used. I n t he det er minist ic models, t he dat a values ar e known and r emains same dur ing t he per iod under st udy.

Det er minist ic M odel I (A) Economic lot size for mulae for opt imum pr oduct ion quant it y ‘q’ per cycle for a single pr oduct , so as t o minimise t ot al aver age cost /unit t ime. t

0 qdt

=

t

0 Rt .dt

I nvent or y level

B

q = Rt St orage St ock O dt

t ime

A t ime (days)

Assumpt i ons 1. The demand is unifor m (consumpt ions) at a r at e (number of it em unit s t ime), i.e. demand R is fixed 2. Pr oduct ion/Supply r at e is unifor m Demand q = = 0 0 or Pr oduct ion is inst ant aneous. 3. L ead t ime is zer o 4. H olding cost is C1 Rs/unit pr oduct /unit t ime 5. Set -up cost is C3 Rs/unit pr oduct /unit t ime 6. St age cost ar e not per mit ive (C2 = 100) 7. These is no buffer st ock. Tot al cost of invent or y = H olding cost + Cost of C i .e. C = C1  (Ar ea of OAB) + C3 1  C = C1  qt + C3 2 C 1 Aver age cost /unit t ime, C(t ) = Rt .C1 + 3 ...(i ) 2 t For cost t o be minimum,

dC  t  d 2C  t  = 0 and = +ve dt dt 2



dC t  C 1 = RC1 – 23 = 0 dt 2 t



C 1 RC1 = 23 2 t



t =

d 2C  t 

2C3 RC1

2C3 = +ve t3 dt H ence equat ion (ii ) gives opt imum value of t he cycle and shielding t ime as

For minimisat ion,

2

=0+

t* =

2C3 RC1 = opt imal t ime int er val

...(ii )

7.8

Industrial Engineering

Also, Opt imum Or der Quant it y or Economic or der Quant it y (EOQ) EOQ = q* = Rt * = R

2C3 RC1

2c3 R C1 Subst it ut ing value of t * in equat ion (i ), we have

q* =



Cmi ni mum =

2C3 .C + RC1 1

1 R 2

= Cmi nmum =

RC1C3 + 2

C3 2C3 RC1

=

R 2  2  C3 .C12 + 4 R C1

RC1C32 2C3

RC1C3 2

2RC1C3

wher e, C1 = I  P = H olding cost P = cost of unit pr oduct . I = a fr act ion denot ing per cent age of pr oduct cost as car r ying cost . E xampl es

Quant it y D iscount Concept 1. A machine manaufact ur er r equir es 2500 met er /year st or age cost is 0.20 Rs/unit /year. I nvent or y cost char ge 20% and pur chase pr ices ar e as follows : (a) Rs 6.75 for an or der of 200 or less (b) Rs 6.50 for an or der of 201 t o 500 (c) Rs 6.25 for an or der of 501 t o 800 (d) Rs 6.00 for an or der of 801 and above Det er mine Economic Bat ch Quant it y (EBQ) for pur chase if cost of each pur chase or der is ` 25 Sol ut i on. Given : C3 = 425/St or age cost , s = 0.20/unit /year



Aver age cost per unit t ime, C(t ) =

For C(t ) t o be minimum,

C 1 Rt . C1 + 3 + s.R.t 2 t

dC  t  C 1 = RC1 – 23 + s.R = 0 2 t dt



C3 1  =  C1  s  R  t2 2 

t* =



q* = Rt * =

2C3R C1  2s

L et q* lie bet ween 0 t o 200. 

P = 6.75. 

2C3 R  C1  2s

C1 = I  P = 0.2  6.75 = 1.35

2  2.5  2500 = 267.26  267 1.35  2  0.2 which does not lie bet ween assumed quant it y r ange. Second t r ial Assume quant it y q  lie bet ween 201 t o 500  P = 6.5  C1 = 0.2 6.5 = 1.3

q* =

2  25  2500 = 271 1.30  2  0.2 which does lie in bet ween assumed r ange; hence EBQ is 271 

q* =

Industrial Engineering

7.9

E xt r a Tr anspor t at ion Cost 2. A factor y engaged in Casting of car bur ettor as an annual demand of 50,000 carburettor s from an automobile engine, the set up cost is Rs 1800/year and over head cost is 500/ day. Daily production capacity is 250, material cost for each carburettor is 20. There are 15 workers engagged on a wage rate of Rs 50/ day. Assume instantaneous supply, no storage, the cost of insur ance, taxes depriciation etc are 20% of the unit cost. Determine (i ) Economic lot size, Number of r uns of sum (pr oduct ion size) and Dur at ion of each r un (ii ) EBQ if supplier imposes t he condit ion of Rs 5/- as t r anspor t at ion char ge per 100 unit s over and above 5000 units. Sol ut i on. Unit cost , P = M at er ial cost + L abour cost + Over head

15  50 500 + = 20 + 3 + 2 = Rs 25/unit 250 250 H olding cost , C1 = I  P = 0.2  25 = Rs. 5/unit Demand r at e, R = 50,000/yr. Set up cost , C3 = 1800/yr. = 20 +

Economic lot size, q* =

2C3R C1 =

Number of pr oduct ion r un =

Dur at ion of each r un = Also,



2  1800  50000 = 6000 unit s 5

P 50,000 = = 8.33 q* 6000  9(8 or der s of 6000 each and 9t h or der of 2000) 1year 12 = = 1.33 Number of run 9

600 = 24 day 250 24 days = 16 day, set up t ime

q* = f = q* = f = q* =

2C3 R  2Rqt e C1 1  2 f 

0 B Buffer st ock = = =0 q q q 2C3 R  2Rqt e C1 1  2 f 

B Buffer st ock 0 = = =0 q q q 2C3 R  2Rqt e C1

wher e, t e = lot size wit hout ext r a t r anspor t cost q= L ot size wit hout ext r a t r nspor t cost



te = q* = n =

5 = 0.05 100 2  1800  50000  2  50,000  50,000  0.05 = 5568 unit s 5 50000  9 (8 t ur ns of 5568 and 9t h of 5456) 5500

7.10

Industrial Engineering

2. D et er mini st i c M odel I (B) Economic lot size wit h differ ent r at es of demand

R3

R2

R1

R4 Rn

q D

t1

t3

t2

tn

t4 T

Tot al demand = D Quant it y per cycle = q Number of cycles =

D = Number of pr oduct ion cycles q

D 1 qT. C1 +  q  C3 2   2 d C  q dC  q For cost t o be minimum, = 0, and = +ve dq 2 dq

Tot al cost (aver age) of I nvent or y, C(q) =

dC  q  D  1 = T.C1 +  2  .C3 = 0 dq 2  q 



D 1 TC1 = 2 .C3 q 2

 Cmi ni mum =



=

1 C .T 2 1

 C  D  2 3   +  C1   T 

2 2 2 C3 D .C1 T . . 4 C1 T +



q* =

2C3 D . C1 T

DC3 2

C3 .D / T  C1

D 2C3 2 .C1 T . = 2C3 D

C1C3 T.D + 2

3. D et er mini st i c M odel I (C) Economic lot size wit h finit e r at e of pr oduct ion or /and Replenishment .

K

R n io

Pr

pt

od u

um ns Co

Q

– R ct io n

q

t2

t1 t

H er e,

K = pr oduct ion r at e (finit e) R = consumpt ion r at e (no supply or pr oduct ion) K – R = r eplenishment r at e (boulding up of st ock) and alway K > R

1 Qt  C1 + C3 2 Q = (K – R) t 1 Q = Rt 2 t = t1 + t2

Tot al aver age cost , C(t ) = H er e, and

C1C3 T.D = 2

2C1C3  D.T 

Industrial Engineering

t1 =

But

Q Q and t 2 = K R R QR + QK – QR Q Q + = K R R K – R  R



t =

 Also

Q  K    R K – R q = Rt



t =

t =

q R

q Q  K  =   R R  K– R 

 

C(q, t ) =

1 K – R q   . t  C1 + C3 2  K 

Per unit t ime,

C(q, t ) =

C3 1 K – R q    C1 + 2  K  t

C(q) =

R 1 K – R q   C1 + C3 q 2  K 

q  ...  t   R 

For cost t o be minimum,

dC  q =0 dq



dC  q  R  1 K – R =  K  C1 +  q 2  C3 = 0 dq 2     R 1 K – R .C = C q2 3 2  K  1



R KR = R K R 1 K

N ow,



2C3  K .R  . C1  K  R 

q* =



I f K =  t hen

1 = 0 M odel I (a) K Cmi ni mum =

1 2

2

2C3  K .R  K  R   C1 + C1  K  R   K  2

=

2 C3C12  K   K  R  . . .R + 4 C1  K  R   K 

=

R  2.C1C3R  1   K 

For inst ant enasus pr oduct ion, K = 



7.11

q* = Cmi ni mum =

2C3R C1 2C1C3 R M odel I (a)

C3 R 2C3  K .R  C1  K  R  C3 2R 2  C1 . (K  R) 2C3 K .R

7.12

Industrial Engineering

4. D et er minist ic M odel I I (A) When shor t age ar e per mit t ed, pr oduct ion r at e is infinit e, lead t ime is zer o, sheduling per iod, t is var iable

R

z q

St ock t2 t1

q -z

C2 =  [I (a )]

Shor t age

t1 =

H er e,

z q z and t 2 = R R

Tot al cost , C(q) =

1 1 z.t C + (q – z)t 2 . C2 + C3 2 1 1 2

=

1 z2 1 (q – z) .C1 + .C2 + C3 2 R 2 R

2

Aver age t ot al cost , C(q, t ) =

2  1  q  z 1  1 z2  .C1  .C2  C3  2 R t 2 R  

2 (q  z)   0  2 R .C2  0   

1 dC = t dq t  cos t

I f C2 = t hen

q* =

2C3 R  C1  C3    C1  C2 

q* =

2C3 R M odel I (a) C1

Cmi ni mum =

 C2  2C1C3 R    C1  C2 

wher e, C2 = wor k or der or shor t age



5. D et er minist ic M odel I I (B) Single or der level syst em wit h const ant sheduling t ime per iod 2 2 1 z 1  q  z C C(z) = C + + 3 2 qp 1 2 qp tp



tp =

qp R

wher e, t p = sheduling per iod kept on const ant . For cost t o be minimum,



d C( z) =0 dz

z* =

Cmi ni mum =

C2 qp C1  C2 RC3 1 C1C2 . Rt p + qp 2 C2  C2

Industrial Engineering

7.13

M AKE AN D BY DECI SI ON (Pr i ce  Quali t y) FC + VC Var i able Cost (VC)

Cost M AKE

Fi xed Cost (FC)

8U Y QB

Buy t he pr oduct ion befor e QB and M ake t he pr oduct ion aft er QB in gr aph Example. The pr oduct ion, demand and cost dat a for a cer t ain pr oduct manufact ur ed on a machine is given below : Fixed cost /lot , C3 = Rs. 36 Var iable cost /unit , P = Rs. 10 Per cent age char ge for t axes and insur ance, I = 50% Pr oduct ion r at e, K = 100,000 per year Demand r at e, R = 10,000 per year Det er mine economic M anufact ur ing Quant it y (EM Q) Sol ut i on. C1 = I  P = 0.5  0.1 = 0.05

q* =

2C3  K R  = C1  K  R 

2  36  105  104    = 0.05  105  104 

 109  1440  = 4000 4   9  10 

PROBABI LI STI C M ODEL The ideal inventor y models such as Wilson and pr oduct ion models, do not account for the r ise and uncer t aint y in t heir for mulat ion. I n r ealit y t hese sit uat ions r ar ely occur. The major var iat ions ar e in t he demand r at e and lead t ime. Bot h t he demand r at e and lead t ime may var y fr om cycle t o cycle and t heir var iat ions ar e pr obabilistic in natur e. I n the pr obabilistic models, inventor y decisions ar e based on the pr obability distr ibution of demand and/or t hat of t he lead t ime.

STATI C I N VEN TORY M ODELS I n t his case, t he decision-making is based on a single cycle only. Nor mally, in t his model, sit uat ion of t he uncer t aint y of demand exist s and t he or der ing quant it y is t o be decided upon for a single per iod. This is also called one per iod model . This t ype of model is useful for det er mining quant it y for r et ail r at es of per ishable goods, seasonal r at es et c. St at ic invent or y models ar e based on t he following idealisat ions : (i ) Only a single pur chase or der is possible dur ing t he per iod under consider at ion. (ii ) Reor der ing dur ing t he per iod is not per mit t ed. (iii )I f it ems ar e not used dur ing t he specific per iod it is consider ed as wast e or loss. (iv) Whenever goods ar e used dur ing t he per iod it will have a gain of effect iveness (example a pr ofit ). (v) The pr oblem would be in t he for m of ‘Given t he pr obabilit y dist r ibut ion of t he demand dur ing t he per iod under consider at ion, det er mine t he or der quant it y (i.e. invent or y level of t he beginning of t he per iod). (vi ) Optimum value of the order quantity or initial level of inventory is based on the marginal analysis (otherwise called incr ement al analysis), i.e. or der quant it y is per mit t ed t o go on incr easing upt o a level when a fur t her incr ease would r esult in decr ease of effect iveness.

Per ishable Goods Pr oblem This is called classical Newspaper boy pr oblem. L et G be t he gr oss gain or pr ofit for each unit used (or sold) and L be t he loss for each unit not used (and hence wast ed). L et P(S) be t he pr obabilit y densit y of t he demand of S unit s. S

L et P(S) = cummulat ive pr obabilit y =

 P(S) .

S

Then pr obabilit y of demand for S or mor e = 1 – P(S).

7.14

Industrial Engineering

This is t he pr obabilit y of selling S unit s or mor e. I f S is t he amount or der ed dur ing t he per iod, t hen Pr obabilit y of r at e of all t he it ems = 1– P(S) Expect ed gr oss gain (or pr ofit ) = G [1 – P(S)] Pr obabilit y of 5t h unit not sold = 1 – [1 – P(S)] = P(S) Expect ed loss = L P(S) Expect ed net gain = G [1 – P(S)] – L P(S) = G – (G + L ) P(S) If G – (G + L ) P (S) = 0 t hen

P (S) =

G GL

Consider ing mar ginal analysis, t he sit uat ion has t o occur when P (S – 1) <

G  P (S). GL

H ence choose t he or der quant it y S, so t hat above condit ion is sat isfied. Example. A newspaper boy buys paper s for 5 pai se each and sells t hem for 6 paise each. H e cannot r et ur n unsold newspaper s. Daily demand R for newpaper s follows t he dist r ibut i on : R : 10 11 12 13 14 15 16 pR : 0.05 0.15 0.40 0.20 0.10 0.05 0.05 I f each day's demand is independent of the pr evious day's, how many paper s should be or der ed each day? Sol ut i on. L et I m be t he number of newspaper s or der ed per day and R be t he demand for i t , i.e. t he number t hat ar e act ual ly sold per day. Now C1 = ` 0.05 C2 = ` (0.06 – 0.05) = ` 0.01 Pr obabilit ies for demand ar e R : 10 11 12 13 14 15 16 PR : 0.05 0.15 0.40 0.20 0.10 0.05 0.05 Im

P

R

R0

:

0.05

0.20

0.60

0.80

0.90

0.95

1.00

The desir ed opt imum value of I m i s det er mi ned by double i nequali t y PR  I m 1 < C2 < PR  I m C1  C2 N ow,

C2 0.01 1 C1  C2 = 0.01  0.05 = 6 = 0.167

This suggest s t hat I m0 must l ie bet ween 10 and 11 because 0.05 < 0.167 < 0.20.



I m = 11. 0

SAFETY STOCK DETERM I N ATI ON Safet y st ock r efer s t o a lower level of st ock int ended t o be maint ained all t he t ime, as a safet y against st ock out . I n t he idealised Wilson and pr oduct ion models, it is assumed t hat t he r at e of deplet ion (or demand) is const ant . I n r eal life sit uat ions t her e will be var iat ion in t he lead t ime and pr oduct ion schedule. Safet y st ock act s as cushion against uncer t aint ies and is able t o pr ovide t he desir ed ser vice levels t o t he cust omer. Safet y st ock is also called Buffer st ock, Safet y cushion . I t is also pr act ice t o call ‘Reser ved st ock' which would t ake car e of demand fluct uat ion and t he st ock, which is pr ovided t o t ake car e of t he var iat ion in lead t i me, as safety stock .

Industrial Engineering

7.15

I mplicat ions of Safet y St ock (i ) I t act s as cushion against st ockout . (ii ) I t decr eases t he penalt y (or st ockout ) cost s; t he fir m may have t o face due t o uncer t aint y of st ock when needed. (iii )I t incr eases t he car r ying cost . H ence, an economic balance has t o be made bet ween st ockout cost s and t he incr ease in invent or y car r ying cost . (iv)The amount of safet y st ock would nor mally be higher for (a) higher st ockout cost s (b) bet t er ser vice levels (c) lower car r ying cost s (d) lar ger var iat ion in consumpt ion (e) lar ger var iat ion in lead t ime.

SERVI CE LEVEL Nor mally, Ser vice level is expr essed in following t wo ways : (i )

Ser vice level =

Number of unit sdemanded – Number of unit sshor t age Number of unit sdemanded

This is used in det er minist ic models. I f Q is number of unit s or der ed per cycle, it is assumed t hat t his is t he t ot al amount demanded dur ing t he cycle. QB Ser vice level = Q wher e, B = number of unit s shor t age per cycle. (ii ) Ser vice level =

Number of per iodswit hout st ockout = Fr act ion of or der per iods wit hout st ockout . t ot al number of per iods

This is expr essed in t he for m as per cent age of ser vice t ime = per cent age of t ime, t her e will be no st ockout .

E ST I M AT I ON OF SAF E T Y ST OCK RE QU I RE M E N T S (PROBABI L I ST I C M ODEL) Safet y st ock is det er mined by assuming same pr obabilist ic dist r ibut ion of demand (or consumpt ion) dur ing t he lead t ime. [The dist r ibut ion may be assumed as Nor mal dist r ibut ion or Poission dist r ibut ion or any ot her pr obabilist ic dist r ibut ion based on pr evious r ecor ds].

Case 1. N ormal Distribution • • • • If

The ser vice level is specified by t he management . The pr obabilit y dist r ibut ion of demand dur ing lead t ime is consider ed as Nor mal dist r ibut ion . The mean () and standard deviation () of the demand during the lead time is known from the previous data. The pr oblem is t o find Re-or der point (ROP) and safet y st ock level (S). R is r at e of consumpt ion (demand), L is lead t ime, t hen Consumpt ion dur ing t he lead t ime = RL I f s is safet y st ock, and Q is r e-or der quant it y, t hen M aximum st ock level, Q + S M inimum st ock level, S = s Re-or der point (ROP) = S + RL i.e. Reor der point (ROP) = Safet y st ock + consumpt ion dur ing lead t ime M ean lead t ime consumpt ion =  Reor der point (ROP) = s +  Re-or der point (ROP) = Safet y st ock + M ean lead t ime consumpt ion. Now consider nor mal dist r ibut ion of lead t ime consumpt ion . The ar ea under t he nor mal cur ve gives pr obabilit y of demand of Re-or der point or less dur ing t he lead t ime. I f Re-or der point is t he st ock level available at t he point of or der ing, t hen t his ar ea will indicat e fr act ion of t he or der per iods wit hout st ockout . Thus, ar ea under t he nor mal cur ve for x = ROP gives t he ser vice level . x ROP   Nor mal var iat e, z = =  

7.16

Industrial Engineering

Fr om t he t able, for ar ea under nor mal cur ve we can find value of ‘z'-cor r esponding t o t he ar ea equal t o t he ser vice level expr essed in decimal fr act ion. Then safet y st ockout = z and ROP =  + z. Example. The mean and st andar d deviat ion of t he demand dur ing t he lead t ime for an it em ar e 100 and 5 r espect ively. Find t he r ecor der point and safet y st ock, if policy of t he management is t o have 95% ser vice level. The values of ar eas for z = 1.64 is 0.9465 and z = 1.65 is 0.9505. Sol ut i on. Given : ser vice level = 95% Given values given : A = 0.9465, z = 1.64 A = 0.9505, z = 1.65 By int er polation,

z = 1.64 +

LM 0.9500  0.9465 (1.65  1.64)OP = 1.64875 N 0.9505  0.9465 Q

 = 100,  = 5 Safet y st ock = z  = 1.64875  5 = 8.2435 = 8.24 unit s Reor der point =  + z  = 100 + 8.24 = 108.24 unit s.

Case 2. Poission distr ibut ion • • •

I t is similar t o case 1, but poission dist r ibut ion is assumed for lead t ime consumpt ion. I t is good fit for small, infr equent demand, wher e demand r at e is fair ly const ant . Poission dist r ibut ion can be r epr esent ed by a single par amet er, namely mean (). St andar d deviat ion,  =  • Fr om t he t able of summat ion of poission dist r ibut ion, Reor der point (ROP) is dir ect ly obt ained. Example. The aver age daily demand of an it em is 2 unit s. The lead t ime demand is Poission dist r ibut ion. What should be t he r eor der point and safet y st ock, if ser vice level of 95% is desir ed and t he lead time in 4 days ? Sol ut i on. Aver age demand per day = 2 unit s L ead t ime = 4 days M ean (aver age) demand dur ing t he lead t ime = 2  4 = 8 unit s.  =8

x 

10

11

12

13

14

0.921 0.935 0.970 0.983 0.816 8 As t he value of x indicat es t he occur r ence or less x = Reor der point (ROP) Fr om above t able, value of x = 13 is fair ly sat isfying for 95% ser vice level. H ence ROP = 13 unit s Safet y st ock = Reor der Point – Aver age demand = 13 – 8 = 5 units

LEAD TI M E (L) I t is t he t ime per iod bet ween placement of an or der and ar r ival of mat er ial.

RE-ORDER LEVEL (ROL) I t is t hat stock level at which fr esh or der should be placed wit h t he supplier for pr ocur ing addit ional inventor y equal t o t he economic or der quant it y. R.O.L . is so fixed t hat t he cust omer s can be ‘r easonabl y’ ser ved fr om t his st ock unt il t he r epleni shment of si ze q ar r ives against t he or der placed. So t he pr obl em of when t o or der r educes t o fixing R.O.L . wit h t he oper at i ng poli cy t hat as t he st ocks cr oss t his l evel an or der i s placed (on t he vendor or wit hin t he wor ks, of which t he si ze is pr edet er mined and fixed). R.O.L . = L ead t i me demand + Safet y st ock. When demand pat t er n i s almost st at i onar y and depict s no t r end or seasonal var iat i ons, L ead t i me demand = L ead t i me  aver age demand. If  = st andar d deviat ion or demand r at e, and d = demand r at e per unit t ime Then

r e-or der level = M = dL + K  L

Industrial Engineering

7.17

L ead t ime demand is t he st ock level, which on t he aver age is sufficient t o fill cust omer 's or der as t he st ocks ar e being r eplenshied. ‘On t he aver age’ implies that dur ing this per iod of r eplenshment, 50% of t he cust omer s' or der s can be filled while 50% may be eit her r efused or back or der ed t o be filled lat er. I t is because act ual demand may be differ ent fr om t he pr edict ed value. When R.O.L . is equal t o lead t ime demand, only 50% of t he or der s ar e expect ed t o be filled dur ing t he r eplenishment per iod. This may not be accept able t o t he management who may like t o limit t he ‘disser ve’ t o t he cust omer s down t o 5% or 10% at t he cost of ext r a st ocking. This ext r a st ock in excess of t he lead t ime demand is called safet y st ock (or buffer st ock or cushion st ock ).

SAFETY STOCK Safet y st ock = (maxi mum lead t ime – nor mal l ead t ime) × aver age demand I f demand exceeds t he for ecast , it woul d r esul t in bad ser vi ce t o t he cust omer. I f demand fall s shor t of t he for ecast over st ock ing will r esult depending upon t he for ecast er r or. I f st andar d devi at i on of t he for ecast er r or i s comput ed, t he safet y st ock may be det er mine t o keep t he ser vice down t o 5% or 10%. safet y st ock = Z × st andar d devi at ion.

M ean Absolut e D eviat ion (M AD). I t may be calcul at e (far mor e easil y) and used t o det er mine t he safet y st ock. Safet y st ock = k . MAD wher e, k = ser vice fact or =

 Z and M AD = 2



Safet y st ock =

2 S. D . 

 Z . M AD 2

CLASSI FI CATI ON OF I N VEN TORY I nvent or y classificat ion r efer s t o t he classificat ion of it ems int o cat egor ies, based on t he nat ur e of impor t ance, wit h a view for select ive cont r ol.

1. ABC Analysis I nvent or y it ems ar e classified on t he basis of monet ar y (i.e. r upee) value of t he annual consumpt ion of differ ent it ems in t he st or es. This t echnique is pr opular ly called Always Bet t er Cont r ol . As a manufactur ing concer n incr eases its act ivities, it is r equir ed to pur chase mor e and mor e items and the r esponsibility of taking car e 100 of each it ems also incr eases. I t also becomes difficult to pay equal 90 attention to all the items t hus pur chased. I n case all the items ar e 70 pur chased at a t ime and that too in bulk quant ities which cannot be % of put to immediat e use, their cont r olling will involve lot of money and inventory cost manual oper ation. I n any industr ial sit uation it may be noted that a handful of top high A B C value items (say 10 t o 20 per cent ) would account for a substant ial por tion of the cost of tot al consumption value (say 75 to 80 per cent). 0 10 30 100 These items ar e classified as A it ems. Similar ly, a lar ge number of it ems, called as t r ivial many (about 70%) may account for small per centage (about 10 to 15%) of t he tot al annual consumpt ion value and these ar e called C items. The inter mediar y items ar e called B items.

S.N o. 1. 2. 3. 4. 5. 6. 7.

A items St r ict cont r ol No safet y st ock Fr equent or der ing Cent r al pur chasing M aximum follow up and expediting M aximum effor t t o r educe lead t ime Accur at e for ecast s in mat er ial planning

B items M oder at e cont r ol L ow safet y st ock L ess number of or der s per year Combination pur chasing Per iodic Follow up M oder at e effor t

C items L ess cont r ol H igh safet y st ock Bulk or der ing Decent r alised pur chasing Follow up only in except ional cases M inimum effor t

Gener al est imat e on post data

Rough est imat es based on usual pr act ice

7.18

Industrial Engineering

L imit at ions of ABC-analysis (i ) The analysis depends only on t he annual consumpt ion value and does not depend upon t he unit cost . Hence the item with high unit cost may be classified under C-items, when the annual consumption is low. (ii ) I t does not depend on ot her impor t ance of t he it em, such as whet her it is vit al t o t he r unning of act ivit ies in t he or ganisat ion or it s availabilit y is scar ce et c. (iii )The r esult of ABC analysis should be r eviewed per iodically and updat ed.

2. H M L Analysis (H M L Classificat ion) I n t his met hod of classificat ion, t he unit cost of it em ar e given impor t ance. The t ot al it ems ar e divided int o high, medium and low cost it ems. This t ype of analysis would be useful in det er mining t he safet y st ock levels. This analysis will also be helpful in cases of it ems wher e t he unit cost of it em is high, but t he annual consumpt ion is low, ot her wise ABC analysis is used, such it ems may fall under ‘C' cat egor y and one may not exer cise t he due cont r ol. I t is mainly used t o cont r ol t he invent or ies of pur chased mat er ial. Basis is unit pr ice of mat er ial. H  H igh cost it ems M  M edium cost it ems L  L ow cost it ems

3. FSN Analysis (FSN Classification) This r efer s t o t he cat egor isat ion of invent or y it em in t he for m as Fast , Slow and Non-moving it ems. This is based on t he consumpt ion r at e of t he invent or y. This analysis will be useful for ar r angement of it ems of st ocks in t he st or es and t heir dist r ibut ion and handling met hods. Basis is issues fr om t he st or es F  Fast moving S  Slow moving N  Not moving

4. XYZ Analysis (XYZ Classificat ion) This is similar t o ABC analysis, except t hat in t he case, t he closing invent or y values of differ ent it ems ar e t aken int o consider at ion inst ead of t he value of t he annual wage of it ems used in ABC analysis. x cat egor y r efer s t o t he it ems whose invent or y values ar e high. The z cat egor y r efer s t o t hose wit h low invest ment in them. The y categor y it ems ar e t he inter mediar y bet ween the t wo. I t depends upon t he value of the inventor y possessed by t he fir m, r at her t han what it has consumed or used. I t is used for classifying t he mat er ial in st or age. Analysis based on value of t he st ock X  I t ems wit h Right invent or y value Y  I t ems wit h M oder at e invent or y value Z  I t ems wit h L ow invent or y value

5. SDE Analysis This r efer s t o t he classificat ion of invent or y it ems as scar ce, difficult and easy t o obt ain it ems. I t is based on t he level of difficult y in t he pr ocur ement of invent or y. I t is used in lead t ime analysis and decisions r elat ed t o pur chasing st r at egies. This analysis is based on availabilit y, posit ion of each it ems. I n this analysis S SCARCE I TEM S, which ar e in shor t supply and t heir availabilit y is scar ce. This includes impor t ant it ems D DI FFI CULT I TEM S, which ar e available in mar ket but not easily pr ocur ed (available) E EASI LY available it em most ly local it ems.

6. VED Analysis This r efer s t o t he classificat ion of invent or y it ems as vit al. Essent ial and desir able it ems. This is wed in t he maint enance st or es for cont r ol of invent or y of spares. Per t ains t o classificat ion, maint enance spar es, denot ing essent ialit y of st ocked spar es as follows : V  VI TAL it ems, wit hout which pr oduct ion would come t o halt . E  ESSENTI AL it ems, wit hout which dislocat ion of pr oduct ion wor k occur s. D  DESI RABL E it ems. Remaining it ems which do not cause any immediat effect /loss in pr oduct ion fall under t his cat egor y. Note : Basis of this (VED) analysis is criticality of items whereas basis of ABC analysis, is consumption of values.

7. Break Even Analysis The effect of quant it y on t he pr ofit cont r ibut ion of t he pr oduct is shown in t he figur e.

Industrial Engineering

7.19

Profit

Cost and income

Break even point l Sa

i es

Loss

ta To

sts l co

F+

om nc

e

bl ria Va

ts os ec

Q ,A

AQ

Fixed cost Q1

Q2

Plant Activity (Quantity)

Fig. Break even chart

Sales income is r epr esent ed by t he st r aight line BQ, in which Q is t he quant it y sold and B is t he income per unit . Cost s t o t he fir m consist of fixed cost s F, which ar e independent of t he quant it y pr oduced and includes execut ive salar ies, depr eciat ion of plant and equipment et c. Var iable cost s AQ, wher e A r epr esent s const ant t ot al cost per unit , including mat er ials, labour and ot her dir ect cost s t hat var y wit h t he plant act ivit y. Var iable cost s ar e shown in t he given figur e by t he st r aight line AQ. The der ivat ion int o fixed and var iable cost s r epr esent s only an appr oximat e int er pr et at ion of t he cost s funct ion and may not be valid for a ver y wide r ange of Q. Tot al cost s ar e given by t he sum of fixed and var iable cost s (F + AQ) and t he point of int er sect ion of t his line wit h t hat of sales income is br eak-even point (BEP) cor r esponding t o a sales volume Q1. Act ivit y below Q1 r esult s in a loss act ivit y and Q1 gives pr ofit . At t he point of int er sect ion, F + AQ1 = BQ1 F  Q1 = B–A I f a plant is oper ating at a point Q2, it is wor king with a mar gin of safet y (denoted by ), defined as Q  Q1 Q  = 2 = 2 – 1 Q1 Q1 Z and it can be shown t hat  = F wher e, Z = pr ofit of t he plant The desir able level of plant act ivit y can be expr essed in t er ms of t he safet y mar gin or t he pr ofit as Z Q2 = Q1 [ 1 + ] = Q1 1  F

I N VE N TORY CON TROL SYSTE M S Any invent or y syst em consist s of t hr ee element s : (i ) Pr ocur ement (ii ) St or age

LM N

OP Q

(iii ) I ssue

The it ems held in st or age at any event of t ime is t he invent or y. All t he t hr ee element s ment ioned above for m an ever going pr ocess. The pur pose of invent or y (st or age element ) is t o delink t he pr ocur ement and t he use. The pr ocur ement causes the input and need for use (i.e. the demand) causes the outflow. When t he pr ocur ement, dur ing a per iod of time exceeds t he issue, then an invent or y is for med. When t he issue exceeds the pr ocur ement, shor t age would r esult . A good invent or y syst em is one which could pr ovide t he r ight mat er ial, t he r i ght quant it y at t he r ight t ime when it needed at a t ot al minimum cost . The invent or y contr ol system is intended to maintain the system based on the policies developed. The feedback indicator for the contr ol syst em is the invent or y level. The r eview of t he st ock level may be continuous or per iodic. The decision on ‘When t o or der ’ and ‘H ow much t o or der ' ar e based on t his r eview. The syst em of or der ing may be t hr ee fold : (i ) Per pet ual, i.e. place or der when invent or y r eaches t he r eor der point specified. (ii ) Per iodic, i.e. place or der on a t ime scale, wit h equally spaced t ime int er val. (iiii ) Preplanned, i.e. planned ordering of the inventory, to fit in to meet the preplanned production requirements.

7.20

Industrial Engineering

TYPE OF I N VEN TORY CON TROL SYSTEM S 1. Two Bin Syst em 

    

‘Bin' r efer s t o t he cont ainer (or t he place) wher e a par t icular it em is st or ed. As per t his met hod, ever y it em is kept in t wo bins. A pr edet er mined amount of st ock (equal t o Reor der Point ) of t he it em is set aside and kept in a second bin. The nor mal issues ar e made fr om t he fir st bin only when t he fir st bi n becomes empt y a r eplenishment or der (i.e. r eor der s) would be placed immediat ely. Aft er war ds t he it em would be issued fr om t he second bin unt il t he ar r ival of st ock, for which t he or der has been placed. Then, second bin is filled up t o t he pr edet er mined quant it y, and t he r est is put in t o t he fir st bin and fur t her issues would st ar t fr om t he fir st bin. Two bin met hod follows closely t he ideal Wilson M odel wit h safet y st ock. Amount of st ock t o be kept in t he second bin is t he level of Reor der point (ROP). The or der quant it y would be t he EOQ. The amount filled int o t he fir st bin is Q + S – ROP. Two bin met hod can be used for low value it ems.

2. F ixed Or der Quantit y Syst em (Q-System) • • • •

The or der quant it y for ever y or der is fixed. Or der s ar e t r igger ed whenever st ock level r eaches t he Reor der point (ROP). The number of or der s placed depends upon t he var iat ion in demand and lead t ime. Safet y st ock depends upon accur acy, r eliabilit y and hor izon of for war d pr ogr amme as well as lead t ime var iable. • The level of safet y st ock depends upon t he var iat ions dur ing t he lead t ime and t he ser vice levels fixed by t he policy. Q syst em is suit able for A-class it ems.

Buffer stock : This is int ended for aver age consumpt ion dur ing t he lead t ime. Buffer st ock = M ean r at e of consumpt ion dur ing lead t ime (R)  M ean r at e t ime (L ) = Reor der L avel Reserve stock : This is t he st ock r equir ed t o t ake car e of var iat ion in demand dur ing t he lead t ime. This depends upon t he ser vice level. Reser ve st ock  z = S Safety Stock : This is int ended t o make pr ovision, t o t ake car e of maximum delay consider at ion. Safet y st ock = Pr obabilit y of occur r ence of maximum delay in lead t ime  M ean r at e of consumpt ion (R) dur ing t he lead t ime.

Reor der point (ROP) = Buffer st ock + Reser ve st ock + Safet y st ock. Note : I n the absence of maximum delay consideration, the r eserve stock and safety stock are taken as one and the same.

Industrial Engineering

7.21

3. Per i odic Review Syst em (P-Syst em) This is also called per iodic r eplenishment syst em The r eview per iod is const ant . The or der quant it y would var y dur ing each or der. M aximum invent or y level (S) = Review t ime (T) +M ean lead t ime (b)  aver age demand r at e (R) + Safet y st ock. S = R (T + L ) + Safet y st ock • This syst em is used in t he cases wher e (i ) usage of t he it em is fair ly r egular (ii ) lead t imes ar e nor mally shor t (iii ) many it ems fr om t he same supplier on r egular basis • I nvent or y cost s ar e higher t han t he Q-syst em. • P-syst em is used for B and C it ems • • • •

R1

R2

R3

R4

4. S-S Syst em (Opt ional Replenishment Syst em) • Review per iod is fixed. • The maximum level ‘S' is fixed, similar t o P-syst em, as S = R (T + L ) + Safet y st ock • The or der quant it y or der ed Q dur ing t he t ime of r eview maximum invent or y (S) [pr esent st ock level (X) + pending or der quant it y not yet r eceived (Q)] Q = S – (X + Q), X  S = 0, X < S • This syst em has been found advant ageous in t he cost of bulk chemicals, pig ir on et c wher e physical assessment of st ock is cost ly and could be inaccur at e. 5 EOQ s

Safet y st ock R1

R2

R3

R4

Ti me 

At At At At

R1 – R2 – R3 – R4 –

st ock st ock st ock st ock

level level level level

is less t han S, hence an or der is placed. is equal t o S, t he or der quant it y placed is equal t o EOQ. is gr eat er t han S hence no or der is placed. is less t han S and an or der is placed.

7.22

Industrial Engineering

N E TWORK FLOW M ODE LS A net wor k consist s of a set of nodes, being connect ed by a set of ar cs. I f each ar c has a specified dir ect ion only, t hen t he net wor k is called dir ect ed net wor k . I n many sit uat ions net wor k act s as an int er mediar y for mat hs modeling. I t helps t o visualise t he pr obl em mor e clear ly t hr ough var ious st ages or si t uat ions of event s. Net wor k flow models consider t he net wor k such t hat , a flow can t ake place in t he ar cs (br anches) of t he net wor k .

N odes

Nodes indicat e a set of t wo or mor e ver t ices which belong t o t he net wor k. A node may r epr esent event, place, st age or a st or age point . I n a gener alised way, it may be consider ed as a st at e wit h values and pr oper t ies, and can t ake numer ical values based on t he at t r ibut es specified. A node may be ident ified by it s associat ed node number. I t act s as end point s of ar c (or br anch). Node ar e also called as ver t ices or point s.

Sour ce N ode

I t is one at which the fir st tr ansaction would take place aft er this node. I t indicates the completion of flow (or a single r un or path). The value of attr ibutes of this node may indicat e final effect of the actions being per for med on t he net wor k. I t is mostly a mer ge node. (The dir ection of ar cs/br anches connecting this node is towar ds t his node.)

Br anch (Ar c)

A line joining t wo differ ent nodes is called a br anch. A br anch is char act er ised by a pair of nodes (i, j ) which it joins. M or e t han one br anch can connect ed any t wo nodes. 2

Node

a Br Br

Node

Node

1

Or ient ed Branch

2

an

L oop

h

2

ch

1

1

nc

h

a Br

h nc

nc

ch

a Br

B

n ra

Node

Branch

3

(1, 2) Non-orient ed Br anch

Or ient ed Br anch An or ient ed br anch is one, in which t he nodes ar e an or der ed pair, i.e. sense of dir ect ion is specified. I t is defined as i , j or as ‘i– j ’. The node i is t he point of beginning and node j is t he point of t er minat ion of t he or ient ed br anch. A non-or ient ed br anch is usually defined as (i, j ).

e j

Or i ent ed N et wor k I f or ient at ions ar e assigned t o all t he br anches in a net wor k, it is called as or ient ed net wor k .

Pat h A pat h connect s t he sour ce node and t he sink node in a sequence of Br anches st ar t ing fr om t he sour ce t o sink. Ther e may be mor e t han one at t he same, t hen it is called loop. Ther e may be mor e t han t wo nodes and br anches. I t is called cycles.

Tr ee A t r ee i s a con n ect ed n et w or k w h i ch h as n o loops (cycles). The path joining any two nodes is unique. I f a t r ee has N nodes, t hen it has (N– 1) br anches. H ence, it must be not ed t hat t he spanning t r ee must have all t he pr oper t ies of a ‘t r ee’.

SPAN N I N G TREE I t is one which connects all the nodes in a network. The proper ty of the spanning tr ee is that, the addition of one more branch (arc) to the tree would r esult in a cycle. There ar e more than one spanning tree in a networ k.

Spanning Tree

Industrial Engineering

7.23

L engt h of t he Spanning Tr ees L engt h of t he spanning t r ee is defined as t he sum of t he dist ances of all br anches (ar cs) in t r ee.

M inimal Spanning Tr ees

The spanning tree, for a given networ k with the least (total) distance is called minimal spanning tree.

Spanning Tr ees Pr oblem

This is in t he for m as; given a Net wor k, det er mine t he M inimal Spanning Tr ee, i.e. minimise t he sum of dist ances of t he br anches connect ing all t he nodes. N ote : All the br anches in t he Spanning Tr ee ar e non-dir ectional; and each having an unique path connect ing all the nodes. Ther e is no such t hing as sour ce or sink nodes. One can st ar t the path fr om any node. Ther e ar e no loops.

Pr ocedur e t o D et er mine M inimal Spanning Tr ee

Step 1 : Select an ar bit r ar y node init ially. Step 2 : Identify a node that is closest to the selected node. The closest node must have following characteristics: (i) I t is dir ect ly connect ed t o t he selet ed node. (ii) The dist ance of t he br anch connect ing t hese t wo nodes is minimum. I nclude t his br anch and t he nodes in t he set of t he M inimal Spanning Tr ee. Step 3 : Out of r emaining unconnected nodes (call it as set G) deter mine the one that is closest to a node, in set G, alr eady selected in t he spanning tr ee. (whenever ther e is a tie, it is br oken ar bitr ar ily). Step 4 : Go t o set p 3 and r epeat unt il all t he nodes ar e select ed, in t he set G, of t he spanning t r ee. Now apply t he above pr ocedur e t o det er mine t he minimal spanning t r ee. Let G is the set of connected nodes (with branches) in the Minimal Spaning tree. Select node 1 as initial node. The branches directly connecting the node 1 are (1, 2) = 2 and (1, 3) = 3. The branch with minimum distance is (1, 2). Hence (1, 2) is taken into the set G. The remaining unselected nodes 3 and 4 are considered to be in the set G. I n t he next st ep, consider all t he br anches t hat ar e dir ect ly connect ed wit h any one node in set G t o any one node in set G, (ot her t han t he br anches alr eady select ed) Set G : (1, 2) and Set G : (3, 4) Dist ance for (1, 3) = 3 Dist ance for (2, 3) = 4 Dist ance for (2, 4) = 5 The minimum dist ance br anch is (1, 3) and node 3 is t he closest . H ence node 3 is added. To set G. Set G : (1, 2, 3) and Set G : (4). Dist ance for (2, 4) = 5 2 Dist ance for (3, 4) = 6 2 5 The minimum dist ance br anch is (2, 4), hence added t o set G. Now all t he nodes have been added t o set G. 4 Set G consist s of br anches : (1, 2), (1, 3) and (2, 4) which const it ut e t he 1 M inimal spanning t r ee. 3 M inimum lengt h = 2 + 3 + 5 = 10. 3 Cr it ical pat h met hod should be cont inue fr om her e and queing t heor y aft er t hat because Net wor k model, CPM , PERT ar e same M inimal spanning tree

SI M PLE QU EU I N G M ODELS A queuing model is a suitable model t o r epr esent a ser vice or ient ed pr oblem wher e cust omer s ar r ive r andomly t o r eceive some ser vice, t he ser vice t ime also being a r andom var iable.

Basic Queuing Pr ocess 1. I nput (a arrival pat t ern) Since t he cust omer s ar r ive in a r andom fashion, t her efor e t heir ar r ival pat t er n can be descr ibed in t er ms of pr obabilit ies. We assume t hat t hey ar r ive accor ding t o a poisson pr ocess, i.e. t he number of unit s ar r iving unt il any specific t ime has a poisson dist r ibut ion. 2. Queue (a wait ing line) 3. Service mechanism (a service pat t ern) The pat t er n accor ding t o which t he cust omer s ar e ser ved, we shall deal wit h t he queuing models in which t he ser vice t ime follows t he exponent ial dist r ibut ion.

7.24

Industrial Engineering

M odel I : N ot at ions L et n = number of cust omer s in t he syst em bot h wait ing and in ser vice  = aver age number of cust omer s ar r iving per unit of t ime  = aver age number of cust omer s being ser ved per unit t ime   = = t r affic int ensit y (also called ut ilizat ion fact or )  Expect ed number of unit s in t he syst em (wait ing + being ser ved), L s =

 – 

 2 Expect ed number of unit s in t he queue, L q = L s – =   ( –  ) Expect ed t ime per unit in t he syst em (expect ed t ime a unit spends in syst em) 1 Expect ed number of unit s in t he syst em Ws = = Ar r ival r at e –   1 Expect ed wait ing t ime per unit in t he queue, W q = W s – =   –  

b

g

 –  1 Aver age wait ing t ime in non-empt y queue (aver age wait ing t ime of an ar r ival who wait s) W n = –   Var iance of t he queue lengt h = 2 –   – –  t Pr obabilit y densit y funct ion of wait ing t ime (excluding ser vice) dist r ibut ion =  1 – e 

Aver age lengt h of non-empt y queue (lengt h of queue t hat is for med fr om t ime t o t ime)

b

g

b

FG H

g – b – gt

Pr obabilit y densit y funct ion of wait ing + ser vice t ime dist r ibut ion =  –  e

Ln =

IJ b g K

 

Pr obabilit y of queue lengt h being gr eat er t han or equal t o n , t he number of cust omer s (p  n ) =

FG  IJ n H K

Example. A self– ser vice st or e employs one cashier at it s count er. 8 Cust omer s ar r ive on an aver age ever y 5 minut es while t he cashier can ser ve 10 cust omer s in t he same t ime. Assuming poisson dist r ibut ion for ser vice r at e, det er mine ( i ) Aver age number of cust omer s in t he syst ems. ( ii ) Aver age number of cust omer s in queue a aver age queue lengt h. ( iii ) Aver age t ime a cust omer spends in t he syst em. ( iv) Aver age t ime a cust omer wait s befor e being ser ved. Sol ut i on. 8 Ar r ival r at e,  = = 1.6 cust omer /minut e 5 10  = = 2 cust omer s/minut e 5 ( i ) Aver age number of cust omer s in t he syst em  1.6 Ls = = =4 2 – 1.6 –  ( ii ) Aver age number of cust omer in t he queue

2 (1.6 ) 2 = = 3.2  ( –  ) 2 (2 – 1.6) ( iii ) Aver age t ime a cust omer spends in t he syst em 1 1 Ws = = = 2.5 minut es 2 – 1.6 –  ( iv) Aver age t ime a cust omer spends in t he queue  1.6  1  Wq = =   = 2 minut es  ( –  ) 2  2  1.6  Lq =

Industrial Engineering

7.25

Example. Cust omer s ar r ive at a sales count er manned by a single per son accor ding t o a Poisson pr ocess wit h a mean r at e of 20 per hour. The t ime r equir ed t o ser ve a cust omer has an exponent ial dist r ibut ion wit h a mean of 100 seconds. Find t he aver age wait ing t ime of a cust omer. Sol ut i on. Given,  = 20 per hour and  =

3600 = 36 per hour 100

 Aver age wait ing t ime of a cust omer in t he queue = E (W q) =

 20 = hour s = 125 seconds  ( –  ) 36 36 – 20

b

g

1 1 Aver age wait ing t ime of a cust omer in t he syst em = E (W s) = = hour s = 225 seconds 36 – 20 –  Example. A bank has only one t ypist . Since t he t yping wor k var ies in lengt h (number of pages t o be t yped), t he t yping r at e is r andomly dist r ibut ed appr oximat ing a Poisson dist r ibut ion wit h mean ser vice r at e of 8 let t er s per hour. The let t er s ar r ive at a r at e of 5 per hour dur ing t he ent ir e 8 hour wor k dut y. I f t he t ypewr it er is valued at Rs. 1.50 per hour, det er mine ( i ) Equipment utilizat ion, ( ii ) The per cent age t ime t hat an ar r iving let t er has t o wait . ( iii ) Aver age syst em t ime. ( iv) Aver age cost due t o wait ing on t he par t of t he t ypewr it er. Sol ut i on. H er e,  = 5 per hour and  = 8 per hour  5 ( i ) Equipment utilizat ion,  = = = 0.625 8  ( ii ) Per cent t ime an ar r iving let t er has t o wait = per cent t ime t he t ypewr it er r emains busy = 62.5% 1 1 1 ( iii ) Aver age syst em t ime, W s = = = hr = 20 min 8– 5 3 –  (iv) Average cost due to waiting on the part of the typewriter /day = [8 hours  (1 – 0.625)]  ` 1.50 = ` 4.50.

CRI TI CAL PATH M E TH OD CPM (Cr itical Path M et hod) : Pr oject management t echnique t hat is used when act ivit y times ar e det er minist ic Activity : Dist inct par t of a pr oject , involving some wor k, whose complet ion r equir es some amount of t ime. e.g. dr illing a hole, st ar t ing a bus, issuing t he wor k or der, float ing a t ender, et c. Activity duration : I t is t he physical t ime r equir ed t o complet e an act ivit y. I n CPM , it is t he best est imat e of t he t ime t o complet e an act ivit y. I n PERT, it is t he expect ed t ime or aver age t ime t o complet e an act ivit y. Critical activity : This act ivit y has no r oom for schedule deviat ion. I n case of deviat ion or slips, t he ent ir e pr oject complet ion will slip. An act ivit y wit h zer o slack is also same. Critical path : The sequence or chain of cr it ical act ivit ies for t he pr oject const it ut es cr it ical pat h. I t is t he longest dur at ion pat h t hr ough t he net wor k. Crashing : The pr ocess of r educing an act ivit y t ime by adding fr esh r esour ces and hence usually incr easing cost . Cr ashing is needed for finishing t he t ask befor e est imat ed t ime. Crash cost : Cost associat ed wit h an act ivit y when it is complet ed in t he possible t ime (cr ash t ime), which is lesser t han t he expect ed or nor mal t ime. Dummy activity : An act ivit y t hat consumes no t ime but and zer o r esour ce only funct ion of t hese act ivit y is t o designat e a pr ecedence r elationship shows pr ecedence among act ivities. I t is useful for pr oper r epr esent at ion in t he net wor k. Earliest Finish (EF) time : The ear liest t ime t hat an act ivit y can finish, fr om t he beginning of t he pr oject . Earliest Start (ES) time : The ear liest t ime t hat an act ivit y can st ar t , fr om t he beginning of t he pr oject . Event : I t is t he beginning, complet ion point ,or milest one accomplishment wit hin t he pr oject . An act ivit y begins and ends wit h event s. An event t r igger s an act ivit y of t he pr oject . Expected activity time : The aver age act ivit y t ime t hat is used in t he pr oject sheduling. Free slack (float) : The lengt h of t ime up t o which an act ivit y can be delayed for channelling r esour ces or r eadjust ment s, wit hout affect ing t he st ar t s of t he succeeding act ivit ies. I mmediate predecessor : An act ivit y, which should immediat ely pr ecede t he act ivit y under consider at ion.

7.26

Industrial Engineering

Latest Finish (LF) time : I t is t he lat est t ime t hat an act ivit y can finish, fr om t he beginning of t he pr oject , wit hout causing a delay in t he complet ion of t he pr oject . Latest Start (LS) time : I t is t he lat est t ime t hat an act ivit y can st ar t , fr om t he beginning of t he pr oject , wit hout causing a delay in t he complet ion of t he pr oject . M ost Likely time ( t m) : I t is t he t ime for complet ing an act ivit y t hat is t he best est imat e; under t he given condit ions (used in PERT). N ormal cost : Cost associat ed wit h an act ivit y when it is complet ed in nor mal t ime. Optimistic time ( t o) : I t is t he t ime for complet ing an act ivit y if ever yt hing in t he pr oject goes well (used in PERT). Pessimistic time (t p) : I t is t he t ime for complet ing an act ivit y if ever yt hing in t he pr oject goes wr ong (used in PERT). Predecessor activity : An act ivit y t hat must occur befor e anot her act ivit y in t he pr oject which is decided on pr ecedence r elat ionship. Project : Set of act ivit ies which ar e int er r elat ed wit h each ot her and ar e t o be or ganised for a common goal or object ive dur ing a specified t ime-fr ame. Project network : A visual r epr esent at ion of t he int er dependence bet ween differ ent act ivit ies of a pr oject which ar e nor mally associat ed wit h a t ime-wise sequencing. Resource allocation methods : Allocat ion of r esour ces t o t he act ivit ies so t hat pr oject complet ion t ime is as small as possible and r esour ces ar e well ut ilized. Slack : I t is t he amount of t ime t hat an act ivit y or a gr oup of act ivit ies can delay in get t ing complet ed wit hout causing a delay in t he complet ion of t he pr oject . An act ivit y having slack cannot be cr it ical act ivit y. Successor activity : I t is t he act ivit y t hat must occur aft er anot her act ivit y (which is pr edecessor ). Total slack (Float) : The time up to which an act ivity can be delayed wit hout affecting star t of the succceeding activities. U pdating : I t involves some r evision of t he pr oject schedule aft er par t ial complet ion wit h r evised infor mat ion. Variance : I t is t he measur e of deviat ion of t he t ime dist r ibut ion for an act ivit y.

FLOAT OR SLACK Earliest Start Time (ES) i : This is t he ear liest occur r eance t ime for t he event fr om which t he act ivit y ar r ow or iginat es. We call it (ES)i for node i of act ivit y i – j . Earliest F inish Time (EF) j : This is t he ear liest st ar t t ime for t he event fr om whi ch t he act ivit y ar r ow or iginates. (EF)j = (ES)i + t ij Latest Finishing Time (LF) j : This is t he lat est occur r eance t ime for t he node at which t he act ivit y ar r ow t er minat es. Latest Start Time (LS) i : This is t he lat est finish t ime for t he node at which t he act ivit y ar r ow t er minat es. (LS)i = (L F)j – t ij I t is defined as t he amount of t ime an act ivit y can be delayed wit hout affect ing t he dur at ion of t he pr oject . On a cr it ical pat h, t he float is zer o.

M easur es of F loat There ar e t hr ee measur es of float ( i ) Total float : I t is t he maximum t ime, which is available t o complet e an act ivit y minus t he act ual t ime which t he act ivit y t akes. Tot al float = [(L F)j – (ES)i ] – t ij = [(L F)j – t ij ] – (ES)i = (L S)i – (ES)i ( ii ) Free float : This is based on t he possibilit y t hat all event s occur at t heir ear liest t ime. I t is a sit uat ion when t he pr oject is or ganised on ear liest t ime t o give t he best possible chances of complet ion on t i me. Fr ee float = (Ear liest Finish t ime – Ear liest st ar t ing t ime – act ivit y dur at ion) = [(EF)j – (ES)i ] – t ij = (EF)j – [(ES)i + t ij ] = (EF)j – Earliest finish time for i – j ( iii ) I ndependent float : I t is impor t ant when t he net wor k of t he pr oject r uns on ear liest t ime. I f an act ivit y r eaches t he next st age at t he lat est t ime, independent float will indicat e if t he consider ed act ivit y (which is just next ) will r each at t he next st age so as t o allow t he following act ivit y t o begin at t he ear liest t ime I ndependent float = (EF)j – (L S)i – t ij

Industrial Engineering

7.27

PERT (PROGRAM EVALU ATI ON AN D REVI EW TECH N I QU E) PERT i ncor por at es pr obabi l i st i c t i me-est i mat es for each act i vi t y. I t empl oys Bet a-di st r i but i on for t he t ime-est imat es. The pr ocedur e for making t he net wor k and det er mining t he cr it ical pat h is same as CPM . I t is t he pr oject management t echnique used when act ivit y t imes ar e pr obabilist ic.

Time Estimate in PERT PERT allows uncer t aint y in t he est imat es for t ime of each act ivit y. Ther e ar e t hr ee t ime est imat es in PERT . (1) Opt imist ic t ime. ( t o). I t for an act ivit y is t hat est imat e for t he complet ion of t he act ivit y which happens when ever y best t hing happens t o facilit at e t he execut ion. Thus, when ever yt hing goes well, t he est imat e is opt imist ic t ime. (2) Pessimist ic t ime ( t p). When ever y t hing goes wor st , t he dur at ion of t ime-est imat e is t he pessismistic time. (3) M ost likely time. ( t m). I t is in bet ween t he opt imist ic and pessimist ic t imes. Under nor mal cir cumst ances, t his is t he pr obable t ime in which an act ivit y is complet ed. I n PERT, it is assumed t hat t he t hr ee t ime est imat es ar e r andom var iables, dist r ibut ed as Bet a dist r ibut ion. The pr obabilit y of most likely t ime is four t imes t hat of eit her of t he r emaining t wo. M at hemat ically,

te =

t o  4tm  t p 6

F t p – to I Var iance for t he act ivit y, V t e = G H 6 JK St andar d deviat ion for t he Net wor k, () = St andar d deviat ion for t he act ivit y, Nor mal deviat ion,

te =

2

sum of var iance along cr itical pat h =

 2

t p – to

 Ts – TB z = 

DI FFEREN CE BETWEEN CPM AN D PERT PERT and CPM or iginated near ly dur ing same per iod but in differ ent context. CPM or iginated fr om constr uction pr oject while PERT evolved t hr ough R and D pr oject s. Bot h CPM and PERT shar e same basic appr oach for const r uct ing t he pr oject net wor k and for det er mining t he cr it ical pat h of t he net wor k. Ther e ar e some differ ences in PERT and CPM .PERT is associat ed wit h uncer t aint y in t he t ime est imat es for act i vi t y, wh i l e i n CPM , t h ese est i m at es ar e t r eat ed as fai r l y det er mi ni st i c. PERT i s consi der ed a event -or ient ed while CPM is mainly act ivit y-or ient ed. 1. 2. 3. 4.

PE RT Time est imat es ar e pr obabilist ic Event or ient ed Focussed on t ime M or e suit able for new pr oject s

1. 2. 3. 4.

CPM Time est imat e det er minist ic Act ivit y or ient ed Focussed on t ime-cost t r ade-off M or e suit ed for r epet it ive pr oject s.

7.28

Industrial Engineering

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. Pr oduct ion planni ng consist s of (a) pr eplanning and r outing (b) scheduling and dispat ching (c) expediting

(a) sequencing

(b) scheduling

(c) r out ing

(d) dispatching

9. The pr ime i nput for aggr egat e pl anning is (a) act ive for ecast i ng (b) passive for ecast ing (c) demand for ecast ing

(d) al l of t hese 2. For ecast ing pr ovides guide lines on (a) how many of t he pr oduct s pr oduced ar e li kely t o be demanded by t he cust omer s (b) amount of business, t he fir m can expect dur ing t he planni ng per iod (c) mat er i als r equir ement s (d) al l of t hese 3. For ecast ing which assumes a st at ic envor onment in t he fut ur e is (a) passive for ecast ing

(d) none of t hese 10. Which of t he fol lowing st at ement i s cor r ect ? (a) Qualit at ive models for for ecast ing ar e based on subject ive opini ons and judgement s (b) Gant t char t s ar e r efi ned ver si on of t r avel char t s. (c) For ecast i ng is t he pr ime i nput for aggr egat e planning. (d) Al l of t hese 11. Sequencing is a subset of

(b) act ive for ecast i ng

(a) r out ing

(b) scheduling

(c) long t er m for ecast i ng

(c) expediting

(d) none of t hese

(d) shor t t er m for ecast i ng 4. The funct ion which aut hor ises pr oduct ion as well as cont r ol is (a) r out ing

(b) scheduling

(c) dispatching

(d) expediting

5. T h e qu al i t y m odel of f or ecast i n g based on t he consensus opi ni on of a panel of exper t s i s cal l ed (a) Composit e Sales Team M et hod (b) Exponent ial Smoot hening M et hod (c) Regr ession M et hod (d) Delphi M et hod 6. Char t which is useful for scheduling and contr ol is (a) Kanban (b) Gant t Char t (c) Fl ow Pr ocess char t (d) X and R Char t 7. The monit or ing and foll ow up funct i on int ended for compl et ion of job wit hin t he due dat e is call ed (a) scheduling

(b) expediting

(c) r out ing

(d) none of t hese

8. The pr oduct i on pl anni ng funct i on r efer s t o t he det er mi nat i on of t he sequence of oper at i ons t o be per for med for t he job and all ocat i on of facili t i es wher e t hese oper at i ons ar e t o be per for med i s called

12. L ong r ange of for ecast ing is (a) one year or less (b) mor e t han 1 year s (c) one t o t hr ee year s (d) none of t hese 13. Advant age of Del phi met hod i s t hat it (a) pr ovides int ui t i ve base for for ecast i ng (b) is useful for long r ange planning (c) is helpful for development of new pr oduct s (d) al l of t hese 14. Degr ee of r el at i on sh i p bet ween t wo or m or e var i abl es i s expr essed by (a) r egr essi on

(b) cor r el at ion

(c) seasonal i ndex

(d) none of t hese

15. Kanban (a) i s an infor mat ion car r yi ng t he expr ession of needs (b) is a simple way of cont r olli ng pr oduct i on (c) is a r equest (car d) for pr oduct ion or wit hdr awal (movement ) of par t s (d) al l of t hese 16. The pr epl anning of pr oduct ions, wor k for ce and invent or y at t he br oadest level is (a) r egr essi on

(b) aggr egat e planning

(c) for ecast ing

(d) none of t hese

Industrial Engineering

17. Cost comput at i ons for t he st r at egi es depends upon (a) str ategy applied (b) data available (c) common logi c used in comput at i on (d) al l of t hese 18. I n l inear pr ogr amming appr oach, t he demand for a par t i cular per iod can be full fi lled by (a) r egular t ime pr oduct i on (b) over t ime pr oduct i on (c) sub cont r act (d) al l of t hese 19. Regr essi on met hod of for ecast i ng i s appl i cabl e mainly for (a) Casual models (b) Qualitat ive for ecast ing (c) Ti me-Ser i es models (d) Delphi met hod 20. A ggr egat e Pl an n i n g r ef er s t o p l an n i n g of pr oduct ion (a) at t he br oadest long t er m per i od levels (b) at t he shor t t er m per i od levels (c) wor k for ce and invent or y at t he br oadest levels (d) wor k for ce and i nvent or y at t he shor t t er m per i od levels 21. Inventary can be in the form of

22.

23.

24.

(a) r aw mat er ials (b) supplies (c) br ought out par t , semi finished goods and sub-assemblies (d) all of t hese Annual car r ying cost, for a given annual demand (a) will incr ease with the incr ease in the number of or der s placed per annum (b) will decr ease with the incr ease in the number of or der s placed per annum (c) is independent of number of orders placed per annum (d) will increase with the decrease in lead time Annual or der ing cost , for a given annual demand (a) wi l l i ncr ease wi t h t he decr ease i n or der quantity (b) wi l l decr ease wi t h t he decr ease i n or der quantity (c) is independent of or der quant it y (d) will decrease with the decrease in lead time For a gi ven l evel of safet y st ock and E OQ or der ing, Reor der point (a) depends only on t he r at e of consumpt ion

25.

26.

27.

28.

29.

30.

31.

7.29

(b) is independent of t he r at e of consumpt ion (c) depends only on the lead t ime (d) depends upon the rate of consumption and lead time The monthly demand is Rs. 2000 of sales. Annual car r ying cost is Rs. 2400. The or der ing cost per or der is Rs. 600. The EOQ is (a) One mont h of sales (b) Two mont hs of sales (c) Thr ee mont hs of sales (d) Four mont hs of sales The mat hemat ical t echnique for finding t he best use of limit ed r esour ces in an opt imum manner is called (a) oper at ion r esear ch (b) linear pr ogr amming (c) net wor k analysis (d) queuing t heor y I n or der t hat l i near pr ogr ammi ng t echni ques pr ovide valid r esult s, t he (a) relation between factors must be linear (positive) (b) r el at i on bet ween fact or s m ust be l i near (negative) (c) bot h (a) or (b) (d) onl y one fact or shoul d change at a t i me, ot her s r emaining const ant Th e l i near pr ogr ammi ng t ech ni qu es can be applied successfully t o indust r ies like (a) ir on and st eel (b) food pr ocessing (c) oil and chemical (d) banking (e) all of t hese The simplex met hod is t he basic met hod for (a) value analysis (b) oper at ion r esear ch (c) linear pr ogr amming (d) model analysis Gr aphical method, simplex method ar e concer ned with (a) value analysis (b) linear pr ogr amming (c) br eak-even analysis (d) queuing t heor y I n si m pl ex m et h od of l i n ear pr ogr am m i n g, t he object ive r ow of t he mat r ix consist s of (a) names of t he var iables of t he pr oblems (b) coefficient of t he object ive funct ion, which is t he pr ofit cont r ibut ion per unit of each of t he pr oducts (c) slack var iables (d) none of t hese

7.30

32.

33.

34.

35.

Industrial Engineering

I n linear pr ogr amming, shadow pr ices ar e (a) cost of br ought out it ems (b) maximum cost per it em (c) value assigned t o one unit of capacit y (d) lowest sale pr ices I n simplex met hod, if all t he basic var iable ar e gr eat er t han zer o (>0), t he solut ion is called (a) non-degener at e (b) degener at e (c) basic solut ion (d) none of t hese I n linear pr ogr amming (a) it is applicable t o linear models only (b) uncer t aint ies in t he fut ur e cannot convenient ly be incor por at ed in model (c) no solut ion is available t o t ime spans shor t er t han per iods in model (d) all of t hese A l i near pr ogr ammi ng pr obl em i s cal l ed so, because in t hat pr oblem (a) all t he funct ions expr essing t he const r aint s ar e linear (b) object ive funct ion also should be linear (c) both (a) and (b) (d) none of t hese

LEVEL-1 1. The mast er schedule (a) is t he out come of t he aggr egat e planni ng (b) dr ives t he ent ir e pr oduct i on syst em (c) is a high l evel schedul e t hat speci fi es what end pr oducts or pr oduct modules ar e t o be pr oduced and t he t ime per iod dur ing whi ch t hey ar e t o be made (d) al l of t hese 2. M ast er schedule (a) det er mines t he over al l pr oduct ion plan for t he near fut ur e (b) pr ovides r ough t ime schedule for pr oduct ion of final pr oduct s (c) r efer t o t he 'end it ems' or final pr oduct s (d) al l of t hese 3. M ast er schedul ing consi der s (a) demand for ecast s and pending or der s (b) r esour ce availabi lit i es and capacit ies (c) st ock st at us and i nvent or y i nfor mat ions (d) feed back i nfor mat i on on act ual shop l oads, backlog and leadt imes (e) al l of t hese 4. Det er mi nat i on of t he sequence of oper at oins t o be per for med and t he al locat ion of facil it ies wher e t hese oper at oi ns ar e t o be per for med is

(a) mast er scheduli ng (b) aggr egat e planning (c) r out ing (d) for ecast ing 5. Rout ing pr ocedur e depends upon (a) t ype of manufact ur i ng-Job. Bat ch, M ass/Fl ow pr oduct ion (b) nat ur e of pr ocesses i nvolved (c) avail abili t y of pl ant , machiner ies and facili t i es (d) char act er ist ics of t he plant , machiner ies and facilities (e) al l of t hese 6. A r ou t e sh eet m ay be desi gn ed t o com bi n e act i vit ies such as aut hor it y t o (a) pr oduct ion (b) mat er ial r equi r ement s (c) t ool or der s and move or der s (d) al l of t hese 7. A decision mak ing pr ocess t o det er mi ne when a job is t o be st ar t ed in a machine and when it i s t o be complet ed, is (a) scheduling (b) r out i ng (c) mast er scheduli ng (d) aggr egat e planning 8. The or der i n whi ch di ffer ent jobs ar e being t ak en up i n a machi ne or pr ocess is call ed (a) scheduling (b) sequencing (c) r out ing (d) aggr egat e planning 9. I n sequenci ng (a) t he t asks bel ong t o t he var ious jobs (b) t he t asks ar e per for med on t he same faci li t y (c) act s as an aid t o det er mine t he scheduli ng (d) emphasi s is on t he economi c or der (e) al l of t hese 10. I n r out i ng (a) t he var ious t ask s bel ong t o t he same job (b) t ask s ar e per for med i n var ious faci li t i es (c) acts as a constraint for scheduling (d) emphasis is on t he t echnological or der (e) al l of t hese 11. The mai n differ ence bet ween flow shop and job shop models in scheduling is t hat (a) in flow shop t her e is a cont inuous flow wit hout idle t ime for any machi ne wher eas i n t he job shop t he fl ow i s i nt er mi t t ent wi t h i dl eness enfor ced on t he machine. (b) in flow shop t he r out ing for al l jobs t hr ough t he machi nes i s uni di r ect i onal wher eas in t he Job shop t he r out i ng may fol low any pat h.

Industrial Engineering

(c) i n fl ow shop al l t he jobs ar e pr ocessed i n a si ngle machi ne, wher eas in job shop, jobs ar e t o be pr ocessed in mor e t hen one machine. (d) in flow shop t he r out ing for al l jobs shoul d be such t hat t o minimise t he make span, wher eas in t he job shop t he r out ing for all jobs should be such t hat t o mi nimise t he mean flow t ime. 12. Objective of scheduling pr oblem is (a) efficient utilisation of the r esour ces (or facilities) (b) r apid r espond t o demands (c) confor mance t o t he pr escr i bed dead-l ines (d) completion of job at the minimum protracted (total) time (e) al l of t hese 13. Fl ow of wor k t hat is unidir ect ional i n t he sense t hat al l jobs r equi r e oper at i ons and pr ocessing in t he same or der, is call ed (a) sequencing (b) r out ing (c) scheduling (d) fl ow shop 14. The amount of t i me measur ed fr om t he zer ot h t i me, at which t he job is due t o be complet ed is (a) compl et ion (b) fl ow t i me (c) due dat e (d) pr ocessing t i me 15. The t ime at which t he pr ocessing of t he job in t he machine i s complet ed, is (a) complet ion t i me (b) fl ow t i me (c) due t at e (d) pr ocessing t i me 16. Ti me t ak en for t he j ob fr om i t s ar r i val t o t he syst em unt i l i t s depar t ur e, is (a) complet ion t i me (b) fl ow t i me (c) due dat e (d) pr ocessing t i me 17. The algebr aic differ ence bet ween the flow time and due dat e is (a) mean lat eness (b) job (c) r ange of lat eness (d) job t r adi ness 18. The lat eness of t he job, fai li ng t o meet t he due dat e is call ed (a) job lat eness (b) mean lat eness (c) job t ar di ness (d) r ange of lat eness 19. The mean number of jobs in the processing system at a time, is called (a) inpr ocess invent or y (b) mean t ar di ness (c) mean lat eness (d) weight ed mean fl ow t i me 20. M axi mum flow t i me (a) is t he maxi mum t ime at whi ch any job st ays int he syst em (b) is t he t ot al pr ot ect ed t i me

7.31

(c) is mak e-span (d) al l of t hese 21. A pur ch asi n g assi st an t has cal cu l at ed t h e car r yi ng cost Rs. per un i t annu m, and t he EOQ = 500 units for an item. He must have taken t hat t he annual or der ing cost for t his it em (a) Rs. 500 (b) Rs. 100 (c) Rs. 31.62 (d) Rs. 22.36 22. I n t he Pr oduct ion M odel for det er mi ni ng t he Economi c Bat ch Si ze, t he pr oduct i on r at e i s consider ed as (a) equal t o demand r at e (b) less t han demand r at e (c) gr eater t han demand r ate (d) independent of demand r ate 23. If R = demand r at e, K = pr oduct ion r at e, Cc = car r ying cost per unit t ime, Co = or der ing cost , and Cc = cost of penalt y per unit t ime, t hen economi c or der quant i t y (EOQ) can be expr essed as (a) (b) (c)

FG IJ FG K IJ H K H K  RK 2RC F C  C I F K I G JK GH K  R JK C H C 2RC F C  C I F K I G JK GH K  R JK C H C IJ FG K  R IJ 2RC F C G C HC C K H K K 2RC o Cs Cc Cc  C s o

s

s

c

c

o

s

c

(d)

o

c

24.

25.

26.

s

s

s

s

c

I f EOQ is within the range of the lowest discounted rate offer ed, then (a) accept t he discount offer and or der for t he minimum in t he r ange (b) r eject t he discount offer (c) consi der t he t ot al cost s of t he r anges of discount befor e t aking t he decision (d) accept t he discount offer and or der at EOQ level I n ABC analysis, t he C it ems ar e t hose which r epr esent s (a) sm al l per cent age of t h e t ot al annu al consumption value (b) h i gh per cen t age of t h e t ot al an n u al consumption value (c) small per cent age of closing invent or y value (d) high per cent age of closing invent or y value The method of classification of items to be adopted for spar e par t s invent or y is (a) ABC analysis (b) XYZ analysis (c) VED analysis (d) SDE analysis

7.32

27.

28.

29.

30.

Industrial Engineering

I n P-syst em of invent or y cont r ol (a) order quantity remains constant (b) t ime bet ween or der ing r emains const ant (c) Recor der point r emains const ant (d) pr oduct ion r at e r emains const ant MRP indicates (a) M at er ials Reor der ing Point (b) M ater ials Reor der ing Planing (c) M at er ials Requir ement s Planing (d) M at er ials Requir ement s Point I t em B r equir es four number s of it em C. Pr oduct P r equi r es t wo number s of i t ems B and fi ve number of it em C. I f five number of pr oduct P is t o be manufact ur ed, t hen number of i t em C r equir ed will be (a) 65 (b) 55 (c) 45 (d) 35 For a given annual consumption, the minimum total inventory cost is proportional to square root of the product of (a) or der ing cost per or der (b) car r ying cost per unit per year (c) bot h (a) and (b) (d) none of t hese

(c) mi ni mise t he maximum t r adi ness (d) mi ni mi se t he mean t r adi ness 5. EDD (Earliest Due Date) sequencing of jobs in a single facility, would (a) mi ni mi se t he maximum lat eness (b) be i ndependent of pr ocessi ng t i mes of t he job (c) mi ni mise t he maximum t r adi ness (d) al l of t hese 6. MWKR (Most wor k Remaining) pr ior ity r ule in t he job shop dynamic syst em would most oft en gi ve sat i sfact or y per for mance for mini mi si ng

(c) mean lat eness (d) mean wait ing t i me (e) al l of t hese 2. Shor t Pr ocessi ng Time (SPT) sequencing of jobs in a single faci li t y woul d minimi se t he (a) mean lat eness (b) maximum t r adiness (c) maximum t r adiness (d) mean t r adi ness 3. The sequencing of jobs in t he ascendi ng or der of t hei r pr ocessi ng t ime, is call ed (a) shor t est pr ocessing t i me (b) ear li et due dat e (EDD) (c) dispatching (d) none of t hese 4. EDD (Ear l iest Due Dat e) sequencing of jobs in a si ngl e faci lit y, would (a) mi ni mise t he mean l at eness (b) mi ni mise t he maximum t r adi ness

(b) mean lat eness

(c) make span

(d) fl ow t i me

7. Whenever some jobs ar e found wai t i ng for t he oper at i on equal t o or gr eat er t han maxi mum allowable waiting time, over -r iding pr ior ity is given and ar e dispatched as per (a) FCFS (Fir st Come Fir st Ser ved) (b) SPT (c) LWK R (L east Wor k Remaining) (d) None of t hese 8. S/OPN (Sl ack Per Oper at i ons) r ul e for pr i or i t y di spat chi ng i s useful for r educi ng (a) number of remaining oper ations

LEVEL-2 1.Sh or t est Pr ocessi n g T i m e (SPT ) sequ en ci n g mi ni mises t he (a) mean fl ow t i me (b) inpr ocess invent or y

(a) mean t r adi ness

(b) number of t ar dy jobs (c) make span (d) mean lat eness 9. Flow control is adopted for control of (a) pr oduct ion of lar ge vol umes of single or a few t ypes of pr oduct s (b) int er mit t ent pr oduct i on of small quant it ies of many i t ems (c) or der i ng r aw mat er ials (d) consumpt i on of new mat er ials 10. Or der cont r ol r efer s t o t he i ndi vi dual cont r ol exer cised over the (a) mat er ials (b) quantit y and qualit y (c) pr ocessing det ails (d) pr ogr ess of wor k on each job or l ot (e) al l of t hese 11.

T h e sy st em advan t ageou s i n case of bu l k chemicals, pig ir on etc, wher e physical asesment of st ock is cost ly is (a) P syst em

(b) Q syst em

(c) S – S syst em

(d) S – A syst em

Industrial Engineering

12.

I n t he S– S opt i onal r epl acement syst em, t he decision not t o place t he or der is t aken when t he st ock level dur ing t he r eview t ime is

19.

(d) none of t he above 13.

The coor dinat ing of det ailed pr oduct ion plans in mult i st age pr oduct i on syst em, wit h i nvent or y cont r ol of pl anned or der r el ease, so t hat t he dependent -demand it ems ar e made available in t he appr opr iat e t ime schedule is called (a) M RP (M at er ials Requir ement s Planning)

(d) maximum inventory level 20.

(c) Q = S + (X + Q) (d) Q = S – (X + Q) 21.

(c) Time phasing (d) None of t hese

15.

A fir m pr oduces and used 2400 it ems annually. The cost of set t ing up for pr oduct ion is Rs. 850 and t he weekly pr oduct ion r at e is 100 unit s. The pr oduct ion cost is Rs. 5 per it em. The annual stor age and car r ying is 10% of aver age inventor y. The t ime, each opt imum pr oduct ion r un would t ake, will be

22.

FG d IJ H vK F vI N G 1 J H dK

(b) N 1

(c)

(d)

I nvent or y management consist s of (a) effect ive r unning of st or es

(c) 6 mont hs

(d) 1 mont h

(b) st ock cont r ol syst em

The key feat ur es of M RP syst em ar e (a) Planned or der r eleases

(c) st at e of mer chandise met hod of st or ing and maint enance et c

(b) Time-phasing of r equir ement s

(d) all of t hese 23.

(e) All of t hese

I n t he Si mplex met hod, t he exi st ence of mor e t han one opt imum solut ion is indicat ed, when (a) values of t he index r ow, cj – zj under one or mor e of t he non - base decision var iables is/ ar e zer o

I t em s h avi ng l i mi t ed n um ber of suppl i er s, difficult t o get qualit y supplier s and one has t o go in for -off places for supply ar e called

(b) some of t he values in t he const ant column (bi ) ar e zer o b (c) all t he r eplacement r at ios, i (s is t he key a is column) ar e negat ive

(a) difficult it ems (b) scar ce it ems (c) vit al it ems (d) dead it ems

(d) the values of index row, cj – zj, indicate opmality, with artificial var iable in the base

Scar ce it ems ar e (b) cannot be pr ocur ed easily

A basic feasible solution in simplex method is one, when all t he

(c) of shor t supply or impor t ed it ems

(a) decision var iables ar e in t he base

(d) all of t hese

(b) decision variables and sur plus var iables ar e assigned zer o values

(a) most ly available in indigenous mar ket

18.

FG d IJ H vK F vI N G 1 J H dK

(a) N 1

(b) 9 mont hs

(d) Gener at ion of lower level r equir ement s

17.

I n a pr oduct ion model, N is the optimum number of unit pr oduced per or der, v is pr oduct ion r at e in unit s pr oduced per day and is demand r at e u n i t s per day. T h e l ev el of t h e m ax i m u m invent or y, will be

(a) 12 months

(c) Pr ovisions for r esheduling

16.

I f Q is or der quant it y, S is maximum invent or y, X is pr esent st ock level and Q is pending or der quant it y, t hen (a) Q = S – (X + Q) (b) Q = S + (X + Q)

(b) BOM (Bill Of M at er ials)

14.

(b) or der quant it y

(c) EOQ

(b) below t he level of S (c) at t he level of S

Sum of buffer stock, r eserve stock and safety stock is equal t o (a) r ecor der point

(a) above t he level of S

7.33

M ean r at e of consumpt ion dur ing lead t ime (R) mult iplied by mean lead t ime (L ) is equal t o (a) buffer st ock

(b) r eser ve st ock

(c) safet y st ock

(d) none of t hese

24.

(c) base var iables ar e non - negat ive (d) base var iables satisfy the constr aint equations

7.34

25.

Industrial Engineering

Ar t ificial var iable is int r oduced, in t he simplex met hod t o

26.

(a) det er mine t he init ial basic feasible solut ion, when sur plus var iable is pr esent (b) conver t t he inequat ion wit h t he sign gr eat er t han or equal t o, in t he for m of an equat ion (c) apply Big - M met hod for solut ion t o linear pr ogr amming pr oblems (d) indicate the sensitivity of the sur plus var iable

L imit at ions of linear pr ogr amming models ar e based on cr it er ia of (a) pr opor tionalit y (linear it y) (b) additivity (c) divisibility (d) det er ministic (e) all of t hese

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (d)

2. (d)

3. (a)

4. (b)

5. (d)

6. (b)

7. (b)

8. (c)

9. (c)

10. (a)

11. (b)

12. (b)

13. (d)

14. (b)

15. (d)

16. (b)

17. (d)

18. (d)

19. (a)

20. (c)

21. (d)

22. (b)

23. (a)

24. (d)

25. (c)

26. (b)

27. (c)

28. (e)

29. (c)

30. (b)

31. (b)

32. (c)

33. (a)

34. (d)

35. (c)

LEVEL-1 1. (d)

2. (d)

3. (e)

4. (c)

5. (e)

6. (d)

7. (a)

8. (b)

9. (e)

10. (e)

11. (b)

12. (e)

13. (d)

14. (c)

15. (a)

16. (b)

17. (b)

18. (c)

19. (a)

20. (d)

21. (a)

22. (c)

23. (c)

24. (d)

25. (a)

26. (c)

27. (b)

28. (c)

29. (b)

30. (c)

LEVEL-2 1. (e)

2. (a)

3. (a)

4. (c)

5. (d)

6. (c)

7. (a)

8. (b)

9. (a)

10. (e)

11. (c)

12. (a)

13. (a)

14. (b)

15. (e)

16. (a)

17. (c)

18. (a)

19. (a)

20. (d)

21. (b)

22. (d)

23. (a)

24. (c)

25. (a)

26. (e)

8

Thermal Engineering

C H A P TE R

BASI C CON CEPTS AN D LAWS OF TH ERM ODYN AM I CS System, Surrounding and Boundary A syst em is defined as qnant it y of mat t er or a r egion in space chosen for st udy. Ever yt hing ext er nal t o t he syst em is sur r ounding. Ther modynamic syst em and sur r ounding is always separ at ed by t he boundar y. Surrounding System Boundary

Univer se = Syst em + Sur r ounding Closed System. The syst em in which only ener gy t r ansfer can t ake place and no mass can cr oss t he boundar y.

Mass No Closed system Energy Yes Open System. The syst em in which bot h ener gy and mass can cr oss t he boundar y. Energy out

Mass in

Mass out

Control Volume

Energy in

N ote : M ost of t he engineer ing devices ar e gener ally open syst em. I solated System. The syst em in which neit her ener gy nor mass cr oss t he boundar y.

T her modynami c Pr oper t i es I nt ensive pr oper t ies ar e t hose pr oper t ies t hat ar e independent of mass of a syst em. Ext ensive pr oper t ies ar e t hose whose value depends upon t he mass of a syst em. m v T

Divide into Two equal parts

p E

m 2 V 2

m 2 V 2

Extensive property

E — 2 T p

E — 2 T p

Intensive property

N ote : Ext ensive Pr oper t y per unit mass is always I nt ensive pr oper t y

8.2

Thermal Engineering

State Process, and Cycle Consider a syst em not under going any change and at t his point all t he pr oper t ies can be measur ed and gives us a set of pr oper t ies, descr ibes state.

m = 2 kg

m = 2 kg

T2 = 300K

T1 = 300K

V2 = 3m3

V1 = 2m3 State 1

State 2

Any change t hat a syst em under goes fr om one equilibr ium st at e t o t he anot her equilibr ium st at e is called process. And ser ies of st at e t hr ough which a syst em passes is known as path of a system . 2

2 P

T

1

1

V

S

I f t he init ial and end st at e ar e ident ical in a ser ies of pr ocesses, it becomes t her modynamic cycle. 1 Q1

2

p 4 Q2 (b)

3

v

Zer ot h L aw of T her modynamics I f t wo bodies ar e in t her mal equilibr ium (TE) wit h t hir d body separ at ely t hen t hey will also be in t her mal equilibr ium. This law of t her modynamics defines t emper at ur e. A

TE

TE

C TE

B

H eat H eat is defined as t he t er m of ener gy t hat is t r ansfer r ed by vir t ue of a t emper at ur e differ enace bet ween t wo syst em or syst ems & sur r ounding. Unit of heat is J. I t is a pat h funct ion and a t r ansient phenomenon. We r epr esent heat by Q (I nexact differ ent ial) or Q. Q = m sT or Q



2

1

Q = 1Q2 or Q1 –

Qin(+)

Sign convent ion for H eat t r ansfer

2

System

Qout(–)

Work. Wor k is t he ener gy t r ansfer associat ed wit h a for ce and displacement . A r ising & lower ing of pist on, a r ot at ing shaft ar e associat ed wit h wor k t r ansfer s.

Thermal Engineering

8.3

Unit of wor k is J. This is also a pat h funct ion. We r epr esent wor k by W or W 2



1

W = 1W 2 or W 1 –

Win(–)

2

=



2

1

pdv

System

Wout(+)

Wor k done by t he syst em is posit ive and wor k is done on t he syst em is negat ive First law of thermodynamics St at e t hat ener gy can neit her be cr eat ed nor be dest r oyed dur ing a pr ocess, it can change t he for ms M at hemat ical expr ession : Q – W = dU for a pr ocess

 Q =  W

for a cycle

Wher e dU = change in int er nal ener gy, unit is J For an ideal gas t he int er nal ener gy is t he funct ion of t emper at ur e only, U = f (T) Enthalpy. Ent halpy is t he t ot al ener gy of a syst em, t his pr oper t y is int r oduced t o simplify t he pr oblem solving. Enthalpy H = U + PV kJ Specific enthalpy, h = u + pv kJ/kg

Thermodynamic Processes 1. Constant volume process (or isochoric process)

P = const ant . T

(a) PVT Relat ion V = const ant ,

2

1 – 2 Heating

P

(b) PV diagr am

2 – 1 Cooling

1 V

(c) Wor k done

(d) H eat t r ansfer



2

1

W =

1W 2



2

1

pdv = 0

=0 T2

Q = m s  d T T1

1Q2 = m cv (T 2 – T 1) wher e cv = Specific heat at const ant volume. (e) I nt er nal ener gy dU = Q – W U 2 – U 1 = mcv (T 2 – T 1) (f) Enthalpy H = U + PV O



dH = dU + pdv + vdp = mcvdT + m RdT PV = m RT Pdv + Vdl = m RdT H 2 – H 1 = mcp(T 2 – T 1) cp = cv + R

 W = 0

8.4

Thermal Engineering

,

, dh

cv

cv

cp

cp

cv

cp

cp

Processes

cv

cp

cv

cp

Adibatic

Adibat ic cp cv

Thermal Engineering

8.5

First law of thermodynamics applied to open system Steady flow energy system St eady flow pr ocess means at any point in t he cont r ol volume, pr oper t y of flow does not change wit h t ime.

Q

2

Mass — m1 Mass flow rate – m1

P2

m2

v2

m2

u2

2 1

Pressure P1

z2

W 2

Sp. Vol. v1

Q = Heat transfer rate W = Power developed.

Internal Energy u1 Potential head z1 Velocity, C1

1

Fr om I st law of t her modynamics. Ener gy in = Ener gy out Fr om mass balance equat ion

m1 = m2 = m

m 1(p1v 1 + u 1 + gz1 + We know I n kW units

 C12 C2   ) + Q = m 2 (p2v 2 + u 2 + gz2 + 2 ) + W 2 2

2 = m  1 = m p 1v 1 + u 1 = h 1 & m p2v 2 + u 2 = h 2, specific ent halpy

   c22  c12       g( z2 – z1 )  Q – W = m  h2  h1       2 

SFEE 



Wher e Q = H eat t r ansfer r at e and W = Wor k t r ansfer r at e or Power developed I n kJ units SFEE

  c22  c12 Q – W = m  h2  h1     2 

    g( z2 – z1 )   

Wher e Q = Wor k done and W = H eat t r ansfer

c22  c12 + g(z2 – z1) 2 Wher e w = Wor k done per unit mass and q = H eat t r ansfer per unit mass. We nor mally apply SFEE t o t he following st eady flow devices. 1. Nozzles and diffuser s 2. Tur bines (gas or st eam) 3. Compr essor s 4. Thr ot t ling valves 5. M ixing chamber s 6. H eat exchanger s 7. Pipe or duct flow I n kJ/kg unit, SFEE

c2

q – w = (h 2 – h 1 ) +

8.6

Thermal Engineering

Application of Steady Flow Processes (i ) N ozzle and Diffusor A nozzle is a device which incr eases t he velocit y or kinet ic ener gy of a fluid at t he expense of it s pr essur e dr op, wher eas a diffusor incr eases t he pr essur e of a fluid at t he expense of it s kinet ic ener gy 1

2

C.S.

m

Insulation

1

m

2

Fig. shows a nozzle which is insulat ed. The st eady flow ener gy equat ion of t he cont r ol sur face gives

V12 V22 + Z1g + q = h 2 + + Z2g + w 2 2 H er e q = 0, w = 0, and t he change in pot ent ial ener gy is zer o.

h1 +

h1 +

V12 V2 = h2 + 2 2 2

... (i )

The cont inuit y equat ion gives

 = m

A 1 V1 A V = 2 2 v1 v2

... (ii )

When t he inlet velocit y or t he ‘velocit y of appr oach’ V 1 is ver y small compar ed t o t he exit velocit y V 2, t hen V 1 is neglect ed and equat ion (i) becomes

h1 = h2 +

V22 2

b

or

g

V 2 = 2  1000 h1  h2 m / s

wher e (h 1 – h 2) is in kJ/kg. ( ii ) T hrot t ling D evice When a fluid flows t hr ough a const r ict ed passage, like a par t ially opened valve, an or ifice, or a por ous plug, t her e is an appr eciable dr op in pr essur e, and t he flow is said t o be t hr ot t led. Figur e shows t he pr ocess of t hr ot t ling by a par t ially opened valve on a fluid flowing in an insulat ed pipe.

F ig. F low through a valve

I n t he st eady-flow ener gy equat ion,

q = 0, w=0 and t he changes in P.E. ar e ver y small and ignor ed. Thus t he st eady flow ener gy equat ion r educes t o V22 V22 = h2 + 2 2 Oft en t he pipe velocit ies in t hr ot t ling ar e so low t hat t he kinet ic ener gy t er ms ar e also negligible. ... h 1 = h2 or ent halpy of t he fluid befor e t hr ot t ling is equal t o t he ent halpy of t he fluid aft er t hr ot t ling.

h1 +

Thermal Engineering

8.7

(iii ) Turbine and Compressor Tur bines and engines give positive power out put, wher eas compr essor s and pumps r equir e power input. m

C.S.

1

WT Tur bine

I nsulat ion 2

F ig . F low th r ou g h a tu r bin e

For a t ur bine (Fig.) which is well insulat ed, t he flow velocit ies ar e oft en small, and t he kinet ic ener gy t er ms can be neglect ed. The st eady flow ener gy equat ion t hen becomes h1 = h2 + w or w = (h 1 – h 2) I t is seen t hat in t ur bines wor k is done by t he fluid at t he expense of it s ent halpy. Similar ly, for an adiabat ic pump or compr essor, wor k is done upon t he fluid and W is negat ive. Thus t he st eady flow ener gy equat ion becomes W h1 = h2 – m W or = h2 – h1 m I n compr essor t he ent haply of t he fluid incr eases by t he amount of wor k input . ( iv) H eat exchanger A heat exchanger is a device in which heat is t r ansfer r ed fr om one fluid t o anot her. The st eady flow ener gy equat ion for t he cr oss-sect ion gives ms 2

Cont r ol Surface

2 Exhaust st eam

Cool ing wat er i n mc

1

3

1

3 4

mc

Cool ing wat er out

4

Condensat e F i g . S te a m c o n d en se r

m ch 1 + m sh 2 = m ch 3 + m sh 4 or m s (h 2 – h 4) = m c (h 3 – h 1) H er e t he kinet ic ener gy and pot ent ial ener gy t er ms ar e consider ed small, t her e is no ext er nal wor k done, and ener gy exchange in t he for m of heat is confined only bet ween t he t wo fluids, i.e., t her e is no ext er nal heat int er act ion or heat loss. ( v ) St eam desuper heat er w1 Wat er

1

2

H igh t emper at ur e st eam w2 2 C.S.

1

3 w3 3

L ow t emper at ur e st eam

8.8

Thermal Engineering

Figur e shows a steam desuperheater wher e the temper atur e of the super heated steam is r educed by spr aying wat er. I f w 1, w 2 and w 3 ar e t he mass flow of t he inject ed wat er, of t he st eam ent er ing, and of t he st eam leaving, r espect ively, and h 1, h 2 and h 3 ar e t he cor r esponding ent halpies, and if K .E. and P.E. t er ms ar e neglect ed as befor e, t he st eady flow ener gy equat ion becomes w 1h 1 + w 2h 2 = w 3h 3 and t he mass balance gives w 1 + w 2 = w3

SECON D LAW OF TH ERM ODYN AM I CS The fir st law of t her modynamic st at es t hat a cer t ain ener gy balance will hold when a syst em under goes a change of st at e or a t her modynamic pr ocess. But it does not pr ovide any infor mat ion on whet her t hat change of st at e or t he pr ocess is at all feasible or not . The second law of t her modynamic pr ovides t he cr it er ion t o t he pr obabilit y of var ious pr ocesses. Joul e’s exper i ment s shows t hat when ener gy suppl i ed t o a syst em i n t he for m of wor k , t hen i t can be conver t ed in heat as follows : Wor k t r ansfer  I nt er nal ener gy incr ease  H eat t r ansfer The complet e conver sion of heat t aken fr om a single t her mal r eser voir, int o wor k in a cycle is not possible. A syst em is t aken fr om st at e-1 t o st at e-2 by wor k t r ansfer W 1-2 and t hen t aken fr om st at e-2 t o st at e-1 by heat t r ansfer Q2-1 t o complet e t he cycle. t hen W 1-2 = Q2-1

But when t he syst em is t aken fr om st at e-1 t o st at e-2 by heat t r ansfer Q1-2 and t hen st at e-2 t o st at e-1 by wor k t r ansfer W 2-1 t o complet e t he cycle t hen it is always found t hat Q1-2 > W2-1 H ence heat can not be conver t ed complet ely and cont inuously int o wor k. Wor k is said t o be high gr ade ener gy and heat a low gr ade ener gy. The complet e conver sion of low gr ade ener gy t o high gr ade ener gy in a cycle is not possible. H eat Engine Cycle Cyclic heat engine wor ks under a t her modynamic cycle in which t her e is a net heat t r ansfer t o t he syst em and net wor k t r ansfer fr om t he syst em. The syst em which execut es under t his t her modynamic cycle is called a heat engi ne. L et Q is heat t r ansfer r ed t o t he syst em, W E is wor k done by t he syst em, t he wor k W C is done upon t he syst em t hen heat Q2 is r eject ed fr om t he syst em as shown in Fig. (a). The net heat t r ansfer in a cycle t o eit her of t he heat engines Qnet = Q1 – Q2 ...(i) Q1 WE System WC Q2

(a) H eat engine cycle per formed by a closed system undergoing four successive energy int er act ions wit h t he sur r oundings H2O Vapour Furnace

Q1

Turbiner

Boiler

WT Q2

Condenser

Sea or river Pump

H2O Wp

(b) H eat engine cycle performed by a steady flow system interacting with the surroundings as shown

Thermal Engineering

and t he net wor k t r ansfer in a cycle Fr om Fig (a) W net = W T – W P Fr om Fig (b) W net = W E – W C By t he fir st law of t her modynamics, we have

... (ii)

Q = W cycle

H2O(l)

cycle

Q1

8.9

H2O(g)

B WP

T

P C H2O

Q2

WT

H2O

F ig . C yclic h ea t en g in e w ith

 Qnet = W net en e r g y i n te r a c t i o n s r ep r ese n t ed in a block d ia gra m or Q1 – Q2 = W T – W P ... (iii) Figur e r epr esent s a cyclic heat engine in t he for m of a block diagr am indicat ing t he var ious ener gy int er act ions dur ing a cycle. Boiler (B), t ur bine (T), condenser (C), and pump (P), all four t oget her const it ut e a heat engine. The funct ion of heat engine cycle is t o pr oduce wor k cont inuously at t he expense of heat input t o t he syst em, t he efficiency of heat engine is

W Net work out put of t he cycle = net Q1 Tot al heat out put t o t he cycle Fr om equat ions (i), (ii), (iii), and (iv), we have  =

... (iv)

Wnet W  WP Q  Q2 Q2 = T = 1 =1– Q1 Q1 Q1 Q1 This is also known as t he t her mal efficiency of a heat engine cycle.  =

H eat r eser voir s A heat r eser voir is defined as a body of infinit e heat capacit y, which is capable of absor bing or r eject ing an unlimit ed quant it y of heat wit hout suffer ing any change in it s t emper at ur e. L et Q1 is t he heat t r ansfer r ed t o t he syst em fr om sour ce Q2 is heat r eject ed fr om syst em

Figur e show cyclic heat engine exchanging heat wit h sour ce and a sink and pr oducing W net = W T – W P

Kelvin-planck St at ement of Second L aw

The K elvin-Planck st at ement of t he second law st at es : I t is impossible for a heat engine t o pr oduce net wor k in a complet e cycle if it exchanges heat only wit h bodies at a single fixed t emper at ur e.

Wnet Q2 =1– Q1 Q1 I f Q2 = 0 (i.e., W net = Q1, or  = 1.00), t he heat engine will pr oduce net wor k in a complet e cycle by exchanging heat with only one r eser voir, t hus violat ing t he Kelvin-Planck st at ement. Such a heat engine is called a per petual mot ion machine of t he second kind, abbr eviat ed t o PM M 2. A PM M 2 is impossible. Efficiency of a heat engine =

Clausius’ Stat ement of t he Second L aw H eat always flows fr om a body at a higher t emper at ur e t o a body at a lower t emper at ur e. The r ever se pr ocess never occur s spont aneously. Clausius’ st at ement of t he second law st at es : I t is impossible t o const r uct a device which, oper at ing in a cycle, will pr oduce no effect ot her t han t he t r ansfer of heat fr om a cooler t o a hot t er body.

8.10

Thermal Engineering

Refrigerator

A refrigerator is a device which, operating in a cycle, maintains a body at a temper atur e lower than the temper atur e of t he sur r oundings. L et t he body A (Fig.) be maint ained at t 2, which is lower t han t he ambient t emper at ur e t 1. Even t hough A is insulat ed, t her e will always be heat leakage Q2 int o t he body fr om t he sur r oundings by vir t ue of t he t emper at ur e differ ence. I n body A, Q2 and W ar e of pr imar y int er est . t her e is a per for mance par amet er in a r efr iger at or cycle, called t he coefficient of per for mance abbr eviat ed t o COP, is defined as COP =

Desir ed effect Wor k input

Q2 W Q1 = Q2 + W =

since



[COP] ref =

Q2 Q1  Q 2

... (i )

H eat Pump

A heat pump is a device which oper at ing in a cycle, maint ains a body say B at a t emper at ur e higher t han t he t emper at ur e of t he sur r oundings. Due t o t emper at ur e differ ence, t her e will be heat leakage Q1 fr om t he body t o t he sur r oundings. COP =



[COP]H.P =

Q1 W Q1 Q1  Q 2

... (ii)

Fr om equat ions (i) and (ii), we get [COP]H.P. = [COP] ref. + 1 The COP of a heat pump is gr eat er t han COP of a r efr iger at or by unit y. At st eady st at e, t he elect r ical ener gy W supplied t o an elect r ic heat er is dissipat ed as heat t o t he space, but when supplied t o a heat pump dissipat es Q1 (>W) giving a t her mal advant age. For heat t o flow fr om a cooler t o a hot t er body, W cannot be zer o, and hence, t he COP (bot h for r efriger at or and heat pump) cannot be infinit y. Ther efor e W > 0, and COP < .

Thermal Engineering

8.11

Carnot Cycle Car not cycle is an ideal cycle in which all t he pr ocesses const it ut ing a r ever sible cycle. 1

T1

2

T T2

4

3 (a)

s

(i ) Reversible isothermal process (1-2) in which heat Q1 ent er s t he syst em at t emper at ur e T 1 r ever sibly fr om a const ant t emper at ur e at T 1. The int er nal ener gy of t he syst em incr eases. (ii ) Reversible adiabatic process (2-3) in which wor k W E is done by t he syst em adiabat ically and r ever sibly at t h e ex pen se of i t s i n t er n al en er gy an d t h e t em per at u r e of t h e sy st em decr eases f r om T 1 t o T 2, t he int er nal ener gy decr eases dur ing t his pr ocess. ( iii ) Rever si bl e i sot her mal pr ocess (3-4)-i n wh i ch h eat Q 2 i s r ej ect ed at con st an t t em per at u r e. I n t his pr ocess t he int er nal ener gy of t he syst em fur t her decr eases. ( iv) Rever sible adiabat ics pr ocess (4-1) i n whi ch wor k W P i s done upon t he syst em r ever si bl y and adiabat ically and t he int er nal ener gy of t he syst em incr eases, and t he t emper at ur e r ises fr om T 2 t o T 1.

Carnot H eat E ngine

Source at Temp T1 Q1 2

Heat Exchanger T1 1

Turbine

Wp

Heat Exchanger T2

Pump

4

WT

3

Q2 Sink at Temp T2

A cyclic heat engine oper at ing on t he car not cycle is called a carnot heat engine. The car not cycle is shown in t he figur e. The heat Q1 is t r ansfer r ed t o t he syst em r ever sibly and isot her mally at T 1 in t he heat exchanger. The wor k is done by t he syst em r ever sibly and adiabat ically in t he t ur bine, t hen heat Q2 is t r ansfer r ed fr om t he syst em r ever sibly and isot her mally at t emper at ur e T 2 in t he heat exchanger, and W P is done upon t he syst em r ever sibly and adiabat ically by t he pump.

Reser ved H eat E ngine Since all t he pr ocesses ar e r ever sible in t he car not heat engine, it is possible t o imagine t hat t he pr ocesses ar e individually r ever sed and car r ied out in t he r ever sed or der Source at t1

Source at t1

W

W

Sink at t2 (a) Carnot heat Engine

Sink at t2 (b) Reversed heat engine

The heat pump and r efr iger at or ar e applied t o t he r ever sed heat engine.

8.12

Thermal Engineering

Actual Brayton Cycle with I ntercooling Reheat and I ntercooling 1-2'-3'-4'-6-7'-8-9'10' shows t he such a act ual cycle as shown in t he figur e. Act ual t her mal efficiency, t ha =

=

(h6  h7 ')  (h8  h9 ' )  (h2 'h1 )  (h4 ' h3 ) (h6  h5 ' )  (h8  h7 ') (T6  T7 ' )  (T8  T2 ')  (T9 'T1 )  (T4 'T3 ) (T6  T5 ')  (T8  T7 ' )

I mprovent in efficiency and output of simple cycle Addition to simple cycle

E ffect on efficiency

Effect on output

1. Regener at ion

+ 50%

N il

2. I nt er cooling

– 6.5%

+ 10.2%

3. Reheat

– 10.4%

+ 24.5%

66.7%

+ 24.0%

68%

+ 10.2%

– 18.2%

+ 34.7%

+ 80%

34.7%

4. Reheat + Regener at ion 5. I nt er cooling + Regener at ion 6. Reheat + I nt er cooling 7. Reheat + I nt er cooling Regener at ion

I C EN GI N E 1. Air Standard Cycle 1. Wor king subst ance is air 2. Cycles ar e mat hemat ical models 3. Used in I .C. Engine and Gast ur bine Assumption 1. I t follows t he law pV = m RT 2. The physical const ant s of t he wor king medium ar e t he same as of air at st andar d ambient i.e., molecular weight = 29, Cp = 1.005, Cv = 0.718 and r = 1.4 3. Const ant specific heat s Cp and Cv does not t hr ough out t he cycle. 4. The wor king medium does not under go any chemical change t hr ough out t he cycle. 5. Compr esion and expansion pr ocesses ar e r ever sible adiabat ic i.e., t her e is no gain and loss of ent r opy. 6. The oper at ion of engine is fr ict ionless. 7. The K .E and P.E of t he wor king fluid air negligible.

Air Standard Cycle Parameters 1. Air st andar d Efficiency (or Ther mal Efficiency) W Net W.D t h = = H .S H eat supplied t o t he syst em 2. Specific Wor k done (or wor k t r ansfer ) I t is t he W.D per unit mass of wor king subst ance i .e., size of plant needed t o per for m t he t ask (or job)

w =

W W.D = wher e W.D is wor k done mass m

3. Specific Air consumpt ion I t is t he r ever se of specific W.D = 4. Wor k r at io (r w )

m mass = W W.D

Net W.D in t he cycle  veW.D in a cycle N ote: I deal cycle have high t h , wigh w and high r w .

rw =

Thermal Engineering

8.13

Cycles 1. Ot t o cycle (const ant volume cycle) (used for Pet r ol engine or Gas Engine) 2. Diesel cycle (const ant pr essur e cycle) (used for slow speed Diesel Engine) 3. Dual cycle (mixed or limit ed pr essur e cycles) (used for high speed disel engine) Dual combust ion cycle 4. Br ayt on cycle (Jouble cycle) (used for gas t asbine plant s)

2. Otto Cycle A ver y common t ype of int er nal combust ion engine is t he spar k ignit ion used in aut omobile. The ot t o cycle is t he air st andar d cycle of such an engine. 3 WE

Pv r =C

Q

Q 1

T

T

3

WE

C 1 V=

2 4

2

Wc Q

1

Wc

2

1

Q

v L et m be t he fixed mass of air under going t he cycle of oper at ions H eat supplied, Q1 = Q2 – 3 = m cv (T 3 – T 2) H eat r eject ed, Q2 = Q4 – 1 = m cv (T 4 – T 1)

Efficiency, = 1 –

FG V IJ HV K

or

2 S

 1

2

4

V =C

1

ot t o = 1 –

rk   1

1

wher e r k = compr ession r at io and given by

Volume at t he beginning of compression V = 1 Volume at t he end of compression V2 Efficiency of the air standard Otto cycle is thus a function of compr ession r atio only. The higher the compr ession r at io, t he higher t he efficiency.

rk =

3. Diesel Cycle I t consist of t wo r ever sible adibat ics one r ever sible isobar i.e. const ant pr essur e and one r ever sible isochor e, i.e. const ant -volume pr ocess. Air is compr essed in pr ocess 1-2. H eat is t hen added t o it fr om an ext er nal sour ce at const ant pr essur e pr ocess (2-3). Air expends in pr ocess (3-4). Then heat is r eject ed at const ant volume in pr ocess (4-1) and cycle r epeat s it self. 2

3

3 WE

Q r PV =C

P

Wc

1

r PV =C

T

Q

2 Wc

4 Q

1

Q

2

v

s

(a)

Efficiency,  = 1 –

4

V =C

2

1

Consider m kg of air in t he cylinder. H eat supplied, Q1 = Q2 – H eat r eject ed, Q2 = Q4 –

WE

1 p=C

(b) 3 1

= m cp (T 3 – T 2) = m cv (T 4 – T 1)

b b

g g

m cv T4  T1 Q2 T4  T1 =1– =1– m c p T3  T2 Q1  T3  T2

b

g

8.14

Thermal Engineering

Efficiency may be expr essed in t er ms of any t wo of t he following t hr ee r at ios. rk =

V1 V2

Expansion r at io, r e =

V4 V3

Compr ession r at io,

V3 V2 r k = re · rc

Cut -off r at io, r c = We know,

Diesel = 1 –

1 

Since r c > 1, t her efor e

1 1 rc   1 · ·  rc  1 rk   1

F r  1I GH r  1 JK  c

>1

c

Ther efor e, efficiency of t he Diesel cycle is less t han t hat of t he Ot t o cycle for t he same compr essi on r at io.

4. Limited Pr essur e Cycle, M ixed Cycle or Dual Cycle H eat is added r ever sibly par tly at const ant volume and par t ly at constant pr essur e. Remaining cycle is same as diesel cycle. H eat supplied, Q1 = m cp (T 3 – T 2) + m cp (T 4 – T 3) H eat r eject ed, Q2 = m cv (T 5 – T 1) Efficiency,

 =1– =1–

b g

g b

m cv T5  T1 Q2 =1– m cv T3  T2  m c p T4  T3 Q1

b

b

T5  T1 T3  T2   T4  T3

g b

g

g

Efficiency of t he cycle can be expr essed in t er ms of t he following r at ios Compr ession r at io, r k =

V1 V2

Expansion r at io, r e =

V5 V4

rc =

V4 V3

Cut -off r at io,

p3 p2 rk = rc · re

Const ant volume pr essur e r at io, r p =

 

Dual = 1 –

1

rp  re  1

b

g

rk   1 rp  1  rp rc  1

Thermal Engineering

8.15

Q For t he same compr ession r at io and for t he same amount of heat supplied, discuss t he  of Ot t o cycle, diesel cycle and Dual cycle.

W.D H .S – H .R H .R = = 1– H .S H .S H .S = T.ds = Ar ea under 2 – 3 = Ar ea under 2 – 3 = Ar ea under 2 – 3  Ot t o cycle  Diesel cycle  Dual cycle

3 3

 = 1 H .S

3

3 T

2

4

4

4

1– 2– 3– 4– 1 v=c 1 1 – 2 – 3 – 4 – 1 1 – 2 – 3 – 3 – 4 – 1 S H .S is same in all 3 cycles And below I f H .R is less, t hen  will be mor e fr om t he cur ve F rom t he curve Ot t o cycle – Ar ea under 1 – 4 is heat r eject ion which is less t han ot her cycles. So  of Ot t o cycle will be mor e. I I nd dual cycle  H .R less t han Diesel cycle but mor e t hen Ot t o cycle ot t o > Dual > Diesel

I .C. Engine

Engine Compoent s and Basic Engine Nomenclat ur es (Single Cylinder )

Ver t ical E ngine CYL I NDER H EAD SU CTI ON EXH AU ST or I N L ET VALVE or OU TL ET VALVE

CL EARAN CE VOL U ME

PI STON VC

GUDGEON PI N A WRI ST PI N

OP DEAD CEN TER (TDC) CYL I N DER WAL L

STROVE V B WATER A JACK ETS

CONNECTI ON ROD BDC (Bot tom Dead Cent r e) Cr ank pine Cr ank web M ain bear ings

BAL AN CE w.f. or count er wt .

Smal l end   Pi st on  Gudgeon pi n or Pi st on CONN ECTI NG ROD Bi g End  Cr ank pi n

H or izont al E ngine I DC

ODC (Out er Dead Cent re)

IV Cyli nder head EOV I nner Dead Cent r e (I DC)

8.16

Thermal Engineering

I C Engine Classification The I C engines can be classified on t he basis of cycle oper at ion in cylinder, t ype of fluid, t ype of ignit ion et c. 1. Basic engine design : Recipr ocat ing engines, Rot ar y engines 2. Wor king cycle : Engine wor king on Ot t o cycle (SI engines), wor king Diesel cycle (C I Engine) 3. No. of st r oke : Two st r oke and four st r oke Engine 4. Fuel : Pet r ol engine; CNG Engine, L PG (L iquefed Pet r oleum Gas) Engine, L DO (light Diesel oil), H SD (H igh speed Diesel) Engine 5. M et hod of I gnit ion : SI Engine (bat t er y ignit ion), CI Engine (Compr ession I gnit ion) 6. M et hod of Cooling : Wat er cooled and Air cooled Engine 7. Cylinder ar r angement : Ver t ical Engine, H or izont al Engine

FOU R STROKE PETROL EN GI N E SUCT I ON Spar k Plug or Out l et val ue I NL ET VAL U E or Exhaust val ve VC TDC VS

BDC Before st ar t ing cam Shaft posi t ion and TOP DEAD CE N TER (TDC)

1. SUCTI ON STROKE I NL ET valve open (or suct ion value pens) TDC  BDC, Char ge (Pet r ol and air ) is admit ed and t he car ge is supplied by Car bur et t or. BDC, Suct ion valve get s closed.

2. COMPRESSI ON STROKE

1 REV or 180 r ot at ion of shaft 2 BDC

Bot h t he valve r eamin closed BDC  TDC. Char ge is being compr esed (compr ession t akes place). Pr essur e and Temp. incr eased char ge par t ionspr epar s t hemselves for ignit ion.

1 Rev. or 180 r ot at ion. 2 At t he end of t he compr ession st r oke, spar k is at t ained at t he spar k plug and igniat ion occuss. Ther e is sor t of explosion in t he cylinder. So it is t he spak ignit ion engine (SI Engine) TDC

3. EXPANSI ON STROK E

TDC

Bot h t he valve r emain closed TDC  BDC, Gas expands. Wor k is obt ained wor king st r oke or power st r oke BDC

1 Rev. or 180 2

4. EXH AUST STROK E

BDC

Exhaust valve opens BDC  TDC Gases ar e exhaust ed out .

1 Rev. or 180 2 Exhaust value closed. TDC

Thermal Engineering

I t number of r evolut ions ar e N t hen t he number of cycles will be Range of compr ession r at io r = 6 – 9 r =9 r =

8.17

N 2

(is pr efer able)

V1 V  Vs V  1 s = c V2 Vc Vc

F our St r oke Diesel E ngine FU EL I N JECTOR I nlet valve

Exhaust valve

TDC

Befor e st ar t ing

BDC TDC

1. SUCTI ON STROK E TDC Suct ion value opens TDC  BDC, Char ge (pur e air ) sucked in

1 Rev. or 180 2

BDC Suct ion valve closes 2. Compr ession st r oke

BDC

Bot h t he value r emain closed BDC  TDC Pr ocess of campr ession t akes place, pr essur e and t emp. incr eases

1 Rev. or 180 r ot at ion of shaft 2 Range of compr ession r at io, r = 14 t o 22 (M used in ver yt st r ong for ext r a ondir ar y and engine mfg) I n buses and t r ucks r = 18 At t he end of t he compr ession, Diesel is inject ed int o t he cylinder at a pr essur e higher t han t he pr essur e insdie t he cylinder. As soon as diesel get s inject ed it ignit es. I t is t he compr ession ignit ion engine (CI Engine) fuel inject or and fuel inject ion pump is used. TDC

3. Expansion st r oke TDC Bot h t he valves r emains closed TDC  BDC Gas expands Wor k is obt ained wor king st r oke or power st r oke

1 Rev. or 180 2

BDC 4. EXH AUST STROK E

BDC

Exhaust valve opens BDC  TDC Gases ar e exhaust ed out TCD

1 Rev. or 180 2

8.18

Thermal Engineering

Exhaust valve closes.

Fuel Consumption Glass t ube

FU EL T EN K

100CC

Amount of fued compr esed in kg/sec

100C.C   gm/c.c kg/sec 1000  t wher e  for Pet r ol = 0.72 t o 0.78  0.75  for Diesel = 0.82 t o 087  0.85 t  t ime in seconds t aken for combiust ion of 100 c.c. fuel

mf =

Br ak e D ynamomet er Fl y wheel Br ak e dr um Engi ne

Net balancing for e Spr ing balance

Br ake dr um side view

weight (tends t o apply br ake)

W = (w – s) Newt ons effect ive r edius of br eak dr um = r met er s Br aking Tor que = W  r B.P = T  w = T  2N Br ake Power, B.P =

1. F or Solving t he Pr oblems

Nm s

2 N T kW 1000

2 N T kW 1000 wher e N  r .p.s T  Tor que in kN-m Br ake Power, B.P =

2. I ndi cat ed Power Fr om t he I ndicat ed diagr am I .P = Gr oss Power – Pumping power L et ar ea of t he diagr am = a cm 2 L engt h of t he diagr am = l cm Spr ing No. or Pr essur e scale or spr ing st iffinessd in bar /cm = s

Thermal Engineering



I ndicat ed mean effect ive pr essur e (i.m.e.p)

or wher e

Pmi =

a s l

a = Aver age ht . of t he diagr am l Wor k done/sec = for ce  Displacement = (Pmi  A)  L Wor k done/sec = Pmi  A  L 

N 2

(Four s st r oke)

Pmi L AN n 2 Aver age velocit y = L N n  No. of cylinder s I p = Pmi L A N n Two st r oke Pmi  i.m.e.p in k Pa L  st r oke or lengt h of t he st r oke in m

I ndicat ed power I p =

wher e

A  Cr oss-sect ional ar ea of t he cylinder =

d  bor e or dia of t he cylinder n  No. of cylinder s

 2 2 d m 4

3. Brake M ean Effective Pressure (b.m.e.p) = bbi Pbi  L A N 2 2 BP B.m.e.p = L AN B.P =

(four st r oke)

4. Friction Power (F.P)

F.P = I .P – B. P

5. M echanical Efficiency mach mech =

B.P I .P

6. I ndicated Thermal Efficiency ( th) Ener gy equivalent t o B.P Energy supplied by fuel Ener gy supplied by fuel = m f  C.V m f  M ass of fuel per second C.V = Calor ipic value or heat ing value in kJ/kg mech =

i t h =

Ip

m f  C.V

7. Brake Thermal Efficiency bth b t h =

B.P m f  C.V

8. M echanical Efficiency ( mech) mech =

bt h i t h

9. Volumetric Efficiency ( v) v =

Volumeof air at ambient condit ion Swept Volume

8.19

8.20

Thermal Engineering

=

v =

Mass of char ge sucked in Mass of charge represent ed by cylinder volume Va Vs

10. b.s.f.c (Brake Specific Fuel Consumption) m f kg fuel used in kg/hr · = b. p kw b. p kwn isfc mech = bsfc m i .s.f .c = f kg/kwh Ip

b.s.f .c =

11.

12.Fuel – Air (F/A) or Air – Fuel (A/F)Ratio mf F = , ma A

A ma  F mf

13. Relative Fuel – Air Ratio ( fr ) fr =

Acut al fuel– air r at io St oichiomet r ic fuel – air rat ion

14. Relative Efficiency or Efficiency Ratio ( rel ) rel = Br ake r elat ive  =

i th air starded bt h air star ded

15. Air Supply to Engine by Orifice is Given by Q = A V m 3/sec wher e

V = Cd 2 gH A  ar ea of or ifice Cd  Coeff. of dischar ge of or ifice H  H ead

Q = A.Cd 2 gH Weight of t he air supplied = Q.W a Wa  Weight of one m 3 of air in kg

Two Stroke Petrol Engine

Crown of the piston

spark plug

EXHAUST PORT

Transfer Port

crank case

INLET PORT

Balance wt.

TDC Suct ion on t he lower side of t he pist on spar k occur s, ignit ion t akes place on t he upper side of t he pist on

Thermal Engineering

8.21

Expansion I nlet par t closed pr elimaner y compr ession of t he char ge in t he cr ank case (cr ank case campr ession) Exhaust por t st ar t s get t ing opended. Tr ansfer par t also st ar t s get ting opened and t he char ge fr om t he cr ank case is being t r ansfer r ed in t he cylinder on t he upper side of t he pist on. BDC : Pist on posit ion Tr ansfer por t – get s closed Scavanging – wast age of char ge Exhaust por t get s closed Act ual compr ession in t he cylinder on t he upper side of t he pist on I nlet por t st ar t s get t ing opened fr om t he lower side of t he pist on Pist on r eaches TDC Q. A gas engine is wor king on cor slt Ot t o cycle volume cycle gave t he following r esult s dur ing 1 H our t est r un. Cylinder diameter = 24 cm St r oke = 48 cm Effcet ive dia of t he br ake wheel = 1.25 m Net load on t he br eak = 1236 N Aver age speed = 226 r pm Aver age explosions/ min = 77 m.e.p = 73.57 N/cm 2 Super Gas used = 13 m 3 at 15C and 771 min of H g C.V of t he gas = 16727 kj/m3 Cooling wat er based, mw = 625 kg I nlet t emp of t he wat er Twi = 25C Out let = Two D et er mine mech ,i t h = ? and H eat balance 26 1.25 2   1236  60 2 kw = 18.25W B.P = 2N T = 1000 236 Explosion = 77 but t heor t ically = = 113 2  113 – 77 = 36 explsions ar e missed 36  2 = 72 cycles ar e missed

N 2 N = Number of cycles ar e excut ed/second I p = Pm L A

REFRI GERATI ON AN D PSYCH ROM ETRY Refr i ger at ion Cycl es Refr iger at ion by N on-cycl ic Pr ocesses Refr iger at ion is t he cooling of a syst em below t he t emper at ur e of it s sur r oundings.

Anot her medium of r efr iger at ion is solid car bon dioxide or dr y ice. At at mospher ic pr essur e CO2 cannot exist in a liquid st at e, and consequent ly, when solid CO2 is exposed t o at mospher e, it sublimat es, i.e. it goes dir ect ly fr om solid t o vapour, by absor bing t he lat ent heat of sublimat ion (620 kJ/kg at 1 at m., – 78.5C) fr om t he sur r oundings. Thus dr y ice is suit able for low t emper at ur e r efr iger at ion.

8.22

Thermal Engineering

REVERSED H EAT EN GI N E CYCLE A r ever sed heat engine cycle, is visualised as an engine oper at ing in t he r ever se way, i.e. r eceiving heat fr om a low t emper at ur e r egion, dischar ging heat t o a high t emper at ur e r egion, and r eceiving a net inflow of wor k. Under such condit ions t he cycle is called heat pump cycle or r efr iger at ion cycle. For a heat pump,

(COP)H.P. =

Q1 Q1 =– W Q1  Q 2

For a r efr iger at or,

(COP)ref. =

Q2 Q2 = W Q1  Q 2

Q1 = T 1 (s2 – s3) Q2 = T 2 (s1 – s4)  W net = W C – W E = Q1 – Q2 = (T 1 – T 2) (s1 – s4) wher e T 1 = t emper at ur e of heat r eject ion, T 2 = t emper at ur e of heat absor pt ion. H er e,

and

(COPref.)rev. =

T2 Q2 = T1  T2 Wnet

(COPH.P.)rev. =

Q1 T1 = Wnet T1  T2

VAPOU R COM PRE SSI ON REFRI GERATI ON CYCLE I n an act ual vapour r efr iger at ion cycle, an expansion engine, as shown in t he figur e is not used, si nce power r ecover y is small and does not just ify t he cost of engine. A t hr ot t ling valve or a capillar y t ube is used for expansion in r educing t he pr essur e fr om p1 t o p2. Q1 2 Condenser

3

p1 p2

Expansion valve

WC 1

Compressor

4 Evaporater

Q2

Oper at ions Repr esent ed for an I dealized Plant The basic oper at ions involved in a vapour compr ession r efr iger at ion plant ar e shown in t he flow diagr am, and t he pr oper t y diagr ams.

(a)

(b)

(c)

(d)

F ig. Vapour compression refr iger at ion cycle proper ty diagr ams

(i )

Compression : A r ever sible adiabat ic pr ocess 1-2 or 1 - 2 eit her st ar t ing wit h sat ur at ed vapour (st at e 1), called dr y compr ession , or st ar t ing wit h wet vapour (st at e 1), called wet compr ession. Dr y compr ession (1-2) is always pr efer r ed t o wet compr ession (1- 2), because wit h wet compr ession t her e is a danger of t he liquid r efr iger ant being t r apped in t he head of t he cylinder by t he r ising pist on which may damage t he valves or t he cylinder head, and t he dr oplet s of liquid r efr iger ant may wash away t he lubr icat ing oil fr om t he walls of t he cylinder, t hus acceler at ing wear.

Thermal Engineering

8.23

(ii ) Cooling and Condensing : A r ever sible const ant pr essur e pr ocess, 2-3, fir st desuper heat ed and t hen condensed, ending wit h sat ur at ed liquid. H eat Q1 is t r ansfer r ed out . ( iii ) Expansion : An adiabat ic t hr ot t ling pr ocess 3-4, for which ent halpy r emains unchanged. St at es 3 and 4 ar e equilibr ium point s. Pr ocess 3-4 is adiabat ic (t hen only h 3 = h 4), but not isent r opic. T ds = dh – vdp

or

s4 – s3 = –

z vdp T

p2

p1

H ence it is ir r ever sible and cannot be shown in pr oper t y diagr ams. St at es 3 and 4 have simply been joined by a dot t ed line. ( iv) Evaporation : A const ant pr essur e r ever sible pr ocess, 4-1, which complet es t he cycle. The r efr iger ant is t hr ot t led by t he expansion valve t o a pr essur e, t he sat ur at ion t emper at ur e at t his pr essur e being below t he t emper at ur e of t he sur r oundings. H eat t hen flows, by vir t ue of t emper at ur e differ ence, fr om t he sur r oundings, which get s cooled or r efr iger at ed, t o t he r efr iger ant , which t hen evapor at es, absor bing heat of evapor at ion. The evapor at or t hus pr oduces cooling or r efr iger at ing effect , absor bing heat Q2 fr om t he sur r oundings by evapor at ion. I n r efr iger at ion pr act ice, ent halpy is t he most sought -aft er pr oper t y. The diagr am in p-h coor dinat es is found t o be t he most convenient . When st eady st at e has been r eached, for 1 kg flow of r efr iger ant t hr ough t he cycle, t he st eady flow ener gy equat ions (neglect ing kinet ic and pot ent ial ener gy and P.E. changes) may be wr it t en for each of t he component s in t he cycle as given below. For compr essor, h 1 + Wc = h2  Wc = (h 2 – h 1) kJ/kg For condenser, h 2 = Q1 + h 3  Q1 = (h 2 – h 3) kJ/kg For expansion valve, h3 = h4 or (h f)p1 = (h f )p2 + x 4 (h fg)p2



x4 =

d h i  dh i dh i f

f

p1

p2

fg p 2

This is quality of the r efr iger ant at t he inlet to t he evapor at or (mass fr action of vapour in liquid-vapour mixtur e). For evapor at or, h 4 + Q2 = h 1  W 2 = (h 1 – h 4) kJ/kg This is called refrigerating effect. I t is t he amount of heat r emoved fr om t he sur r oundings per unit mass flow of r efr iger ant . I f p-h char t for a par t icular r efr iger ant is available wit h t he given par amet er s, it is possible t o obt ain fr om t he char t t he values of ent halpy at all t he car dinal point s of t he cycle. Then for t he cycle COP = =

Q2 Wc h1  h4 h2  h1

If m  is mass flow of r efr iger ant in kg/sec, t hen r at e of heat r emoval fr om t he sur r oundings = m  (h 1 – h 4) kJ/sec = m (h 1 – h 4)  3600 kJ/hr

One Tonne of Refrigeration I t is defined as t he r at e of heat r emoval fr om t he sur r oundings equivalent t o t he heat r equir ed for melt ing 1 t onne of ice in one day. I f lat ent heat of fusion of ice is t aken as 336 kJ/kg, t hen 1 t onne is equivalent t o heat r emoval at t he r at e of 1000  336 or 14,000 kJ/hr.. 24

8.24

Thermal Engineering

b

g

 h1  h4  3600 m t onnes 14,000 Rat e of heat r emoval in t he condenser, Q1 = m (h 2 – h 3) kJ/s I f condenser is wat er -cooled, m c t he flow-r at e of cooling wat er in kg/s, and (t c2 – t c1) t he r ise in t emper at ur e of wat er, t hen Q1 = m (h 2 – h 3) = m c (t c2 – t c1) kJ/s pr ovided heat t r ansfer is confined only bet ween r efr iger ant and wat er, and t her e is no heat int er act ion wit h t he sur r oundings. Rat e of wor k input t o t he compr essor, Wc = m (h 2 – h 1) kJ/s = m (h 2 – h 1) kW wher e h 1 and h 2 ar e in kJ/kg.



Capacit y of t he r efr iger at ing plant =

Actual Vapour Compression Cycle I n or der to ascer tain that ther e is no droplet of liquid r efr iger ant being car r ied over int o t he compr essor, some super heat ing of vapour is r ecommended aft er t he evapor at or. A small degr ee of subcooling of t he liquid r efr iger ant aft er t he condenser is also used t o r educe t he mass of vapour for med dur ing expansion, so t hat t oo many vapour bubbles do not impede t he flow of liquid r efr iger ant t hr ough t he expansion valve. B ot h su per h eat i n g of vapou r at t h e evapor at or ou t l et an d su bcool i n g of l i qu i d at t h e con den ser out l et cont r i but e t o an i ncr ease i n t he r efr i ger at i ng effect , as shown i n t he fi gur e. The compr essor di schar ge t emper at ur e, however, i ncr eases due t o super heat , fr om t ' 2 t o t 2, and l oad on t he condenser also incr eases. Somet imes, a liquid-line heat exchanger is used in t he plant , as shown in t he figur e. The liquid is subcooled in t he heat exchanger, r educing t he load on t he condenser and impr oving t he COP. For 1 kg flow, Q2 = h 6 – h 5, Q1 = h 2 – h 3 Wc = h 2 – h 1 and h 1 – h 6 = h 3 – h 4 O 4

3

2

P

5

G1 h

(a) (b) F ig. Vapour compression cycle wit h a suction-line heat exchanger.

Volumet ric E fficiency of Compressor I n r efr iger ant compr essor s which may be single cylinder or mult i cylinder, due t o clear ance, leakage past t he pist on and valves t he t hr ot t ling effect s at t he suct ion and dischar ge valve, t he act ual volume of gas dr awn int o t he cylinder is less t han t he volume displaced by t he pist on, t he r at io of t hese t wo is called volumet ir c efficiency .

vol = =

Act ual volume of gas drawn at evaporat or pressur e and t emper at ur e Pist on displacement volume

mv1  2 N D L n 4 60

Thermal Engineering

wher e

8.25

m = r efr iger ant mass flow r at e D and L = diamet er and st r oke of compr essor n = number of cylinder s N = r.p.m.

AI R CON DI TI ON I N G Air condit ioning means t he aut omat ic cont r ol of an at omospher ic envir onment eit her for comfor t of human beings or animals or for t he pr oper per for mance of some indust r ial or scient ific pr ocess. The pur pose of air conditioning is to supply sufficient volume of clean air cont aining a specific amount amount of water vapour and at a t emper at ur e capable of maint aini ng pr edet er mi ned at mospher ic condit ions. The space may be small compar t ment such as r esear ch t est cabinet or a cinema hall.

Classificat ion of Air Condit ioning The air condit ioning syst ems ar e br oadly classified int o t wo gr oup : (1) Comfor t air condit ioning The comfor t air condit ioning syst ems ar e fur t her subdivided int o t hr ee gr oups. ( i ) Summer air conditioning : I t is used t o r educe t he sensible heat and wat er vapour cont ent of t he air by cooling and dehumidifying. (ii) Winter air conditioning : I t is used t o incr ease t he sensible heat and wat er vapour cont ent of t he air by heat ing and humdification. ( iii )Year-round air conditioning : This syst em assur es t he cont r ol of t emper at ur e and humidit y of air in an enclosed space t hr ought t he year when t he at omospher ic condit ions ar e changing as per t he season. (2) I ndust r ial air condit ioning

Pr oper t iesof M oist Air Dr y Air Dr y air is the mechanical mixtur e of gases like, oxygen, nitr ogen, car bondioxide, hydr ogen, ar gon, neon, kr ypton, helium, ozone and xenon. H owever O2 and H 2 make up t he major par t of t he combinat ion. Dr y air consist of 21% O2 and 79% N 2 by volume and 23% O2 and 77% N 2 by mass.

M oist Air

I t is a mixt ur e of dr y air and wat er vapour. The quant it y of wat er vapour pr esent i n t he ai r depends upon t emper at ur e of t he air and it s quant it y may change fr om zer o t o maximum.

Wat er Vapour The wat er vapour pr esent in air is called moist ur e and it s quant it y in air is an impor t ant fact or in all air condit ioning syst ems. The mixt ur e of air and wat er vapour at a given t emper at ur e is said t o be sat ur at ed when it cont ains maximum amount of wat er vapour t hat it can hold. I f t emper at ur e of mixt ur e of air and wat er vapour is above t he sat ur at ion t emper at ur e of t he wat er vapour, t he vapour is called super heat ed vapour .

D r y-bul b Temper at ur e The t emper at ur e of air measur ed by or dinar y t her momet er called dr y-bulb-t emper at ur e.

Wet -bul b Temper at ur e The t emper at ur e measur ed by t he t her momet er when it s bulb is cover ed wit h wet clot h and is exposed t o a cur r ent of moving air is called wet bulb temperat ur e. The differ ence between dr y-bulb and wet-bulb temper atur e is called wet -bulb-depr ession . Wet bulb depr ession becomes zer o when t he air is fully sat ur at ed.

D ew-bul b Temper at ur e The temper atur e of the air is r educed by continuous cooling than the water vapour in the air will star t condensing at a par t icular t emper at ur e. The t emper at ur e at which t he condensing st ar t s is called dew-point t emper at ur e. Dew point t emper at ur e is equal t o t he st eam t able sat ur at ion t emper at ur e cor r esponding t o t he act ual par t ial pr essur e of the water vapour in the air. The differ ence between dr y bulb-t emper atur e and dew-point temper at ur e is called dew-point depr ession .

Specific H umidit y (H umidit y Rat io) I t is t he mass of wat er vapour pr esent per kg of dr y air. I t is given in gr ams per kg of dr y air.

Absol ut e H umi dit y The weight of wat er vapour pr esent in unit volume of air is called absolut e humidit y .

8.26

Thermal Engineering

Degree of Saturation I t is defined as t he r at io of mass of wat er vapour associat ed which unit mass of dr y air t o mass of wat er vapour associat ed wit h unit mass of dr y air sat ur at ed at t he same t er mper at ur e.

Relative H umidity I t is defined as t he r at io of act ual mass of wat er vapour in a given volume t o t he mass of wat er vapour if t he air is sat ur at ed at t he same t emper at ur e.

Sensible H eat of Air The quant it y of heat which can be measur ed by measur ing t he dr y bulb temper at ur e of t he air is called sensible heat .

Total H eat The t ot al heat of t he himid air is t he sum of t he sensible heat of the dr y air and sensible and lat ent heat of wat er vapour associat ed wit h dr y air.

H umid Specific Volume The volume of the mixture per kg of dry air in the mixture, expressed in cubic metres, is called humid specific volume.

D alton’s Law of Partial Pressure I t st at es t hat , t ot al pr essur e of a mixt ur e of gases is equal t o sum of t he par t ial pr essur es exer t ed by each gas when it occupies t he mixt ur e volume at t he mixt ur e t emper at ur e. As per t he Dalt on’s law of par t ial pr essur e pt = pa + pb + pc I f t his is applied t o t he moist air which cont ains dr y air and wat er vapour, t hen pt = pa + pv wher e, pt = t ot al pr essur e of moist air pa = par t ial pr essur e of dr y air pv = par t ial pr essur e of wat er vapour

Specific H umidit y (w)

w =

mass of wat er vapour in mixt ure mv = massof dr y air in mixt ur e ma

M ass of t he mixt ur e is (m a + m v).

ma =

H er e,

pa V R aT

wher e, pa = par t ial pr essur e of dr y air V = volume of mixt ur e Ra = gas const ant for dr y air

pv V RvT wher e, pv = par t ial pr essur e of wat er vapour Rv = gas const ant for wat er vapour. and

 But

mv =

w = Ra =

pv V R a T R a pv    R v T pa V R v pa R Ma

and

Rv =

R Mv

R = univer sal gas const ant wher e M a and M v = molecular weight s of air and wat er vapour r espect ively 

w =

M v pv 18 pv p pv     0.622 v  0.622 M a pa 29 pa pa pt  pv

M asses of air and wat er vapour in t er ms of specific volumes ar e given by

ma =

V va

and

mv 

V vv

....(i)

Thermal Engineering

8.27

wher e v a and v v ar e specific volumes of dr y air and wat er vapour at t he given mixt ur e t emper at ur e and at r espect ive par t ial pr essur e.

w =



D egr ee of Sat ur at ion ( ) = wher e,



va vv

mass of wat er vapour associat ed wit h unit massof dr y air w  massof wat er vapour associat ed wit h sat urat ed unit massof dr y air ws

w s = specific humidit y of air when air is fully sat ur at ed. pv p 0.622 1  vs pt  pv p pt  pvs p pt  v  v  = p p p  p p pvs vs t v vs 1  v 0.622 pt pt  pvs

FG H FG H

IJ K IJ K

FG H

IJ K

LM MM MN

OP PP PQ

wher e, pvs = par t ial pr essur e of wat er vapour when air is fully sat ur at ed at t he same t emper at ur e of air.

Relat ive H umidit y (  )

Relative humidity can be defined as the r atio of partial pr essur e of water vapour in a given volume of mixtur e to the partial pressure of water vapour when same volume of mixture is saturated at the same temperatur e. pv V massof wat er vapour in a given volume R T p m = =  v  v massof wat er vapour in samevolumeif sat ur at ed at t hesamet emperat ure mvs pvsV pvs RvT   = p 1  (1   ) vs pt When pvs << pt  Relat ive humidit y plays mor e impor t ant par t in comfor t air condit ioning and indust r ial air -condit ioning t han specific humidit y. Relat ive humidit y signifies t he absor pt ion capacit y of t he air. M or e moist ur e will be absor bed by t he air if t he init ial r elat ive humidit y of t he air is less. pa w 1 p =   1.6 w  a 0.622 pvs pvs

Enthalpy of M oist Air

I t is sum of t he ent halpy of dr y air and ent halpy of t he wat er vapour assciat ed wit h dr y air.  Ent halpy of moist air = ent halpy of one kg of dr y air + ent halpy of wat er vapour associat ed wit h one kg of dr y air or h = 1.005 T db + w [2500 + 1.88 T db] kJ/kg of dr y air or h = (1 + 1.88 w ) T db + w [(h fg)dp + 2.3 T dp] = (1 + 1.88 w ) T db + w [(h fg)dp+ 2.3 T dp] =Cpma × T db + w [(h fg)dp + 2.3 T dp] wher e Cpma is called mean humid specific heat of air . As 1.88 w << 1, t her efor e value of Cpma is always t aken as 1.0216 kJ/kg of dr y air -K for all pr act ical pur poses in air -condit ioning calculations  h = 2.0216 t db + w [(h fg)dp] + 2.3 T dp]

AI R COM PRE SSORS Air compr essor s ar e used for supplying high-pr essur e air. Ther e ar e many uses of high-pr essur e air i n t he indust r y. The main uses of high-pr essur e (compr essed) air ar e :  t o dr ive compr essed air engines (air mot or s) used in coal mines,  t o inject or spr ay fuel int o t he cylinder of a Diesel engine (air inject ion Diesel engine),  t o oper at e dr ills, hammer s, air br akes for locomot ives and r ailway car nages, wat er pumps and paint spr ays,  t o st ar t lar ge (heavy) Diesel engines,  t o clean wor kshop machines, gener at or s, aut omobile vehicles, et c.,  t o oper at e blast fur naces, gas t ur bine plant s, Bessemer conver t or s used in st eel plant s, et c.,  t o cool lar ge buildings and air cr aft s, and  t o super char ge I .C. engines.

8.28

Thermal Engineering

Ther e ar e mainly t wo t ypes of air compr essor s viz. r ecipr ocat ing air compr essor s and r ot ar y air compr essor s. Recipr ocat ing air compr essor s ar e similar t o r ecipr ocat ing engines wher e a piston r ecipr ocat es inside a cylinder. I n r ot ar y air compr essor s, air is compr essed due t o r ot ation of impeller or blades inside a casing. Air compr essor s ar e dr iven by engines or elect r ic mot or s.

Reci pr ocat ing Ai r Compr essor The r ecipr ocat ing air compr essor may be single-act ing (air is admit t ed t o one side of t he pist on only) or doubleact ing (air is admit t ed t o each side of t he pist on alt er nat ively), and may be single-st age or mult i-st age. Singlest age compr essor s ar e used for deliver y pr essur es upt o 10 bar, t hr ee-st age compr essor s for pr essur e upt o 200 bar and t wo-stage compr essor s for pr essur es in bet ween 10 to 200 bar. The aver age piston speed of a r ecipr ocating air compr essor is limit ed t o about 300 t o 400 met r es per minut e t o r educe fr ict ion wear. Compressed air di schar ge t o r ecei ver Pi st on dri ven fr om ext ernal source

3

2

Pr essur e

Pi st on

P2

P v n = const ant 1

P1 V2

0

a

Ai r inl et

b

V1 Volume

Single-stage air compressor without clearance.

Single-St age Air Compr essor I ndicator Diagram: The event s descr i bed above can be convenient l y r epr esent ed by P-V di agr am shown above figur e. The diagr am is dr awn for a compr essor wit hout clear ance. line 4-1- suct ion st r oke line 1-2- compr ession st r oke line 2-3- deliver y of t he compr essed air The net wor k r equir ed for compr ession and deliver y of t he air per cycle is r epr esent ed by t he ar ea 1-2 -3 -4. Deliver y P2

3

2 P2

3

2 2 2 n

Act ual cur ve PV . C I sot her mal curve

P1

I sot r opi c cur ve

Pr essur e

Pressur e

Compr ession cur ve

Suct i on 4

0

P1 0

Volume

1

4 volume

The amount of wor k done on t he air will depend upon t he nat ur e of t he compr ession cur ve. I f t he compr ession occur s ver y r apidly in a non-conduct ing cylinder so t hat t her e is no heat t r ansfer, t he compr ession will be pr act ically isent r opic. I f it is car r ied out slowly so t hat t he heat of t he compr ession is ext r act ed fr om t he air by t he jacket cooling wat er, t he compr ession will appr oach isot her mal. H owever, in act ual pr act ice neit her of t hese condit ions can be fulfilled and t he act ual compr ession will be bet ween isent r opic and isot her mal as shown in figur e.

Let pressure P1 is in N/m 2 and volume v 1 is in m 3 r epr esent init ial condit ion of t he air befor e compr ession. I f t he compr ession is polyt r opic

I ndicat ed Wor k r equi r ed t o be done on t he ai r W, per cycle assuming compr ession cur ve t o be polyt r opic PV n = C, is given by ar ea 1-2-3-4 of indicat or diagr am. P v  P1v1  P1v1 W = p2 V2  2 2 n 1

Thermal Engineering

8.29

n  P v  P1v1  by ideal gas r elat ion we get n 1 2 2 n m R  T2  T1  W= n 1  P2v2  n W = n  1 P1v1  P v  1   1 1  W=



1

v2  P2  n   But for polyt r opic compr ession P1v 1n = P2v 2n . H ence v1  P1  Wor k r equir ed per cycle (or per r evolut ion, if compr essor is single-act ing), 1    n     P P n  2 2 W= P v       1  n  1 1 1   P1   P1    

n 1    P2  n n  W P1v1     1  Joule per cycle n 1   P1    

Th i s equ at i on gi ves t he wor k r equi r ed i n Joul es per cycl e (or per r evol u t i on, i f t h e com pr essor i s single-act ing) in compr essing and deliver ing t he air. I ndicat ed power of t he compr essor =

W N J /s or Wat t 60

wher e W = wor k r equir ed in Joules per cycle, and N – No. of cycles per for med per minut e ( r.p.m.) for single-act ing compr essor, if p1v 1 in above equat ion is subst it ut ed by mRT 1, t hen wor k r equir ed per cycle,

Wor k r equir ed per kg of air,

n 1    P2  n n  m RT1    1  W = n 1  P  Joule per cycle  1    

n 1   n   P n  2 RT    1  Joule per cycle W=  n  1 1   P1     I ndicat ed power of t he compr essor = W × mass of air deliver ed per second J/sec. or Wat t B. I f t he compr ession is isent r opic. 1    P2     P1v1    1  Joule per cycle W=  P   1  1    

Wor k r equir ed per kg of air, 1    P2     W= RT1    1  Joule per cycle  P   1  1     I f t he compr ession is isot her mal (pv = C), t hen wor k r equir ed

W = ar ea under 1-2-3-4

v  W = P1v1 log e  1   P2v2  P1v1  v2  v  W = P1v1 log e  1   v2 

8.30

Thermal Engineering

P  W = P1v1 log e  2  Joule per cycle  P1  Wor k r equir ed per kg of air

P  W = RT1 l og e  2  Joule per cycle  P1 

Work I nput to the Compressor with Clearance Volume 3

P2

2 Compression

Expansion

n

pV = const ant

n

pV = constant

P1

4

(V 1 – V 4) Fr ee ai r del iver y

1 I nt ake pr essur e

Swept volume Vr = V1 – V3

Vc

Fr om indicat or diagr am shown in figur e. W I nput = W compr ession (W c) – W expansion (W e) n 1   n   P n  2 P v    1 Wc =  n  1 1 1   P1     n  P v  P41v4  We = n 1 3 3

As

 P3v3  n W e = n  1 P4 v4  P v  1   4 4 

P3v 3n = P4v 4n

1/ n

p  v3 =  4 v4  p3 

1/ n

p   1  p2 

p   2  p1 

1 / n

n 1    P2  n n  P1v4    1  We = n  1  P   1     W I nput = W c – W e

W I nput

n 1   n   P n  2 = P v  v     1  n  1 1 1 4   P1    

n 1    P2  n n  W I nput = P1vactual    1   P  n 1  1     N ote : Clear ance volume does not affect t he wor k input .

M ulti-stage Air Compressor H igh Pr essur e r equir ed by Single – St age: 1. Requir es heavy wor king par t s. 2. H as t o accommodat e high pr essur e r at ios. 3. I ncr eased balancing pr oblems. 4. H igh Tor que fluct uat ions. 5. Requir es heavy Flywheel inst allat ions.

Thermal Engineering

8.31

This demands for M ult i-st aging. L.P. = L ow Pr essur e I .P. = I nt er mediate Air Del iver y Pr essure H .P. = H igh Pr essur e

I nter cooler

L.P. Cylinder

I .P. Cylinder

H .P. Cylinder

I n t er c o o l er : Compr essed ai r is c o o l ed bet ween cyl inder s.

I nt er cooler

Air I nt ak e

A mult i-st age air compr essor wit h int er cooler is shown in fig.. Ser ies ar r angement of cylinder s, in which t he compressed air from earlier cylinder (i.e. discharge) becomes the intake air for the next cylinder (i.e. inlet ). When cooling is per fect , i.e., when air is cooled t o int ake t emper at ur e in t he int er cooler (T1 = T2), t he point 2 will lie on t he isot her mal line 1 - 3" as shown in fig.. I t may be not ed t hat each st age will incr ease t he pr essur e of air while t he int ake t emper at ur e T1 (cor r esponding t o point 1) is maint ained same at t he end. The isot her mal line dur ing t he pr ocess has been appr oximat ed as shown by t he diagr am, and t he shaded ar ea 2-3-3’-2’ shows t he saving of wor k as a r esult of t his appr oximat ed isot her mal. P P3

33 3

Pr essur e

Saving i n wor k due t o i nt er cooli ng n

PV = C

P2

2

I sot hermal 1

P1 0

2

Volume

V

Combined ideal indicator diagram of two-stage compression with perfect-intercooling.

Per fect -i nt er cool i ng I f int er cooling is per fect or complet e, t he point 2 will lie on t he isot her mal line, i.e., point 2 will coincide wit h point 2", t hen P1V 1 = P2V 2. Tot al wor k r equir ed per cycle,

n 1 n 1    P2  n  P3  n n   P v   2 W=      Joule per cycle n  1 1 1  P1  P  2  

I ndicat ed power of t he compr essor =

W N J/s or Wat t 60

I f P1v 1 in above eqn. is subst it ut ed by mRT 1, t hen wor k r equir ed per kg of air may be wr it t en as n 1 n 1    P2  n  P3  n n  W = RT1       2 Joule per cycle  P  n 1  P2   1   I ndicat ed power of t he compr essor = W  mass of air deliver ed per second J/sec. or Wat t H eat r eject ed t o int er cooler per min. Q = mk p(T 2 – T 2) kJ wher e, m = mass of air compr essed per minut e, k p = specific heat of air at const ant pr essur e, T 2 = t emper at ur e of air befor e ent er ing t he int er cooler, and T 2 = t emper at ur e of air aft er leaving t he int er cooler

8.32

Thermal Engineering

CON DU CTI ON M ODES OF H EAT TRAN SFER H eat t r ansfer may be defined as t he t r ansmission of ener gy fr om one r egion t o anot her as a r esult of t emper at ur e gr adient and it t akes place by t hr ee modes : ( i ) Conduct ion : Conduct ion is the t r ansfer of heat fr om one par t of a subst ance t o anot her par t of t he same subst ance, or fr om one subst ance to anot her in physical cont act wit h it , wit hout appr eciable displacement of molecules for ming t he subst ance. ( ii ) Convect ion : Convect ion is t he t r ansfer of heat wit hin a fluid by mixing of one por t ion of t he fluid wit h anot her. ( a ) Free or natural convection : I t occur s when t he fluid cir culat ed by vir t ue of t he nat ur al differ ence in densit ies of hot and cold fluids, t he denser por t ions of t he fluid move downwar d because of t he gr eat er for ce or gr avit y, as compar ed wit h t he for ce on t he less dense. ( b) Forced convection : When work is done to blow of pump the fluid, it is said to be forced convection. ( iii ) Radiat ion : Radi at i on i s t h e t r ansfer of h eat t hr ou gh space or m at t er by m eans ot her t han conduct ion or convect ion. Radiat ion of heat is t hought of as elect r omagnet ic waves or quant a (as convenient ) an emanat ion of t he same nat ur e as light and r adio waves. All bodies r adiat e heat , so a t r ansfer of heat by r adiat ion occur s because hot body emit s mor e heat t han it r eceives and a cold body r ecives mor e heat t han it emit s. Radiant ener gy (being elect r omagnet ic r adiat ion) r equir es no medium for pr opagat ion, and will pass t hr ough a vacuum.

H EAT TRAN SM I SSI ON BY CON DU CTI ON F our ier ’s L aw of Conduct ion This is an empir ical law based on obser vat ion and st at es. That t he r at e of flow of heat t hr ough a single homogeneous solid is dir ect ly pr opor t ional t o t he ar ea of t he sect ion at r ight angles t o t he dir ect ion of heat flow, and t he change of t emper at ur e wit h r espect t o t he lengt h of t he pat h of t he heat flow. I t is be r epr esent ed by t he equat ion, Q  A.

dt dx wher e k = const ant of pr opor t ionlit y and is known as t her mal conduct ivit y of t he body The – ve sign of k is t o t ake car e of t he decr easing t emper at ur e along wit h t he dir ect ion of incr easing t hickness dt of t he dir ect ion of heat flow. The t emper at ur e gr adient is always negat ive along posit ive x dir ect ion and dx t her efor e t he value of Q becomes + ve. Q = – k.A

Thus,

Q dx . A dt W 1 m W 2  Unit of k : = mK K m Fr om t he above equat ion, met er ials wit h high t her mal conduct ivit ies ar e good conduct or s of heat , wher eas mat er ials wit h low t her mal conduct ivit ies ar e good t hemal insulat or s. Conduct ion of heat occur s most r eadily in pur e met als, less so in alloys, and mush less r eadily in non-met als. The ver y low t her mal conduct ivit ies of cer t ain t her mal insulat or s, e.g. cor k is due t o t heir por osit y, t he air t r apped wit hin t he mat er ial act ing as an insulat or. N ow,

k=

Sequence of M at er ials of Decr easing T her mal Conduct ivit y (i ) (ii ) (iii ) (iv)

Pur e met al M et al alloys Non-met allic cr yst alline and amor phous subst ance Liquid

Thermal Engineering

(v)

8.33

Gases

Thermal Conduction in Liquids H eat conduct ion in liquids and gases ar e based on t he movemoment of at oms in a molecule. I n case of liquids, the r andom tr anslator y motion is small and it appear s that tr anslation of energy occur s by longitudinal vibration, similar t o t he pr opagat ion of sound. Br idgman assumpt ions for t he conduct ivit y of liquids. ( i ) Adjacent molecules ar e ar r anged in a simple cubical fashion. ( ii ) When t he ener gy fr om one molecule t o anot her is t r ansfer r ed at t he velocit y of sound, t hen t her mal conduct ivit y in liquid is given by t he expr ession,

k=3 wher e

=

v R vs . 2 = 3.  2s A0  

R is known as Bolt zmann const ant per molecule A0

R = univer sal gas const ant A 0 = avogadr o’s number v s = sonic velocit y

T her mal Conduct ion in Gases H eat conduct ion in a gas is explained in t er m of t he net of t r anspor t of ener gy acr oss imaginar y sur face by molecule psssing in bot h dir ect ion t hr ough t he sur face. When t emper at ur e gr adient exist s in a gas as shown in t he figur e bet ween t wo par allel sur face, molecules in t he r egions of higher t emper at ur es must have higher r andom velocit y t han in t he lower t emper at ur e r egions.

k=

Ther mal conduct ivit y, wher e

1 . n. vs . f . .  6

n = number of molecules per unit volumes vs = ar it hmat ic mean molecular velocit y

f = number of degr ee of fr eedom  = bolt zmann’s const ant  = molecular mean dept h pat h Gases wi t h t he hi gher mol ecul ar wei ght s have smal l er t her mal conduct i vi t y t han t hose wi t h l ower molecular weight s. This is due t o fact t hat molecular pat h of t he gas molecules decr ease wit h incr ease in densit y and k is dir ect ly pr opor t ional t o mean fr ee pat h of t he molecule. This can be explained by t he compar ison of t wo gases as follows : At STP,

k for N 2  0.021 W / mk k for H 2  0.15 W / mk

Conduct ivit y of t he liquid can also be explained in t er ms of dynamic viscosit y  as

LM N

k = 1 wher e,

OP Q

4.5  cv 2n

n = number of at oms in a molecule

ON E DI M EN SI ON AL H EAT CON DU CTI ON Conduct ion of H eat T hr ough Slabs Consider t he t r ansmission of heat t hr ough ar ea A of a wall (t hickness x ) consist ing of a number of slabs. L et conductivities of differ ent slabs (wall mater ials) be k 1, k 2 and k 3, t hickness of slabs (also called as ‘path’ length)be x 1, x 2 and x 3 and t emper at ur es at t he wall sur faces be T 1, T 2, T 3 and T 4 (T 1 > T 4) r espect ively. Since quant it y of heat t r ansmit t ed per unit t ime t hr ough each slab is t he same, t her efor e

8.34

Thermal Engineering

b

g

b

g

b

k2 T2 – T3 k3 T3 – T4 Q k1 T1 – T2 = = = A x1 x2 x3 

T1 – T4 =

or

Q=

LM N

Q x1 x2 x3   A k1 k2 k3

A(T1 – T4 )

LM x Nk

1

1



x2 x3  k2 k3

g

OP Q

OP Q

Conduction of H eat Through Pipe Walls Ar ea t hr ough which heat is t r ansmit t ed, A = 2r l Pat h lengt h = dr Q = – kA

 or I ntegr ating

FG d T IJ H dr K

dr r r2 dr Q r1 r

dT per unit t ime dr

= – k. 2r l

Q

= – k. 2l .dT

z

= – k. 2l =



z

T2 T1

dT

T1 – T2 r2  1 1  log e  2l  k r1 

Condition of H eat Through a H ollow Sphere Refer r ing figur e, consider a hollow spher e of int er nal r adius r 1 and ext er nal r adius r 2. L et t he inside and out side t emper at ur es be T 1 and T 2 and let t he t her mal conduct ivit y be k. Consider a small element of t hickness of dr at any r adius r . Ar ea t hr ough which heat is t r ansmit t ed, A = 4r 2 Pat h lengt h = dr (over which t he t emper at ur e fall in t emper at ur e is dT) Q = – k. 4r 2

 I ntegr ating,

Q

or

z

r2

dr

r1

2

r

= – 4k

Q =

z

T2 T1

dT dr dT

b

4 kr1 r2 T1 – T2

b

r2 – r1

g

g

T1 – T2

FG r – r IJ H 4kr r K 2

1

1 2

The t er m

r2 – r1 is known as t her mal r esist ance of t he spher e. 4 kr1 r2

H eat Transfer at The I nt erface of Two Solids Fig. (a) shows t he t emper at ur e dist r ibut ion t hr ough a composit e wall which is linear t hr ough each of it s component . This means t hat at t he int er face of t wo solids, t her e is a discont inuit y in t he t emper at ur e gr adient alt hough t he t emper at ur e it self is cont inuous.

At t he int er face,

k1

FG d T IJ H dx K

1

FG IJ H K

dT = k 2 dx

2

Thermal Engineering

8.35

I N SU L AT I ON The insulation is a mater ial which r etar ds the heat flow with r easonable effectiveness. I t is not necessar y to have low ther mal conductivity for insulating mater ial. The pur pose of insulation is to pr event the flow of heat fr om the system to the sur r ounding as in case of steam and hot water pipes which ar e used for air conditioning in winter and to pr event t he flow of heat fr om the sur r ounding to the syst em as in the case of br ine pipes which ar e used for air conditioning in summer and domestic r efr iger ator s and water cooler s. U ses of insulations for industrial purposes. ( i ) Air condit ioning syst em ( ii ) Refr iger at or s and food pr eser ving st or es ( iii ) Pr eser vat ion of liquid gases ( iv) Boiler and st eam pipes

Cr it ical T hickness of I nsulat ion When a solid cylinder r adius R1 is insulat ed wit h an insulat ion t hickness R2 – R1 as shown in t he figur e, t hen heat flow fr om t he sur face of t he solid cylinder t o t he sur r ounding is given by Q =

b g FR I 1 1 log eG 2 J  k2 H R1 K R2 h0 2L T1 – Ta

wher e L = lengt h of t he cylinder k 2 = conduct ivit y of insulat ion h 0 = combined heat tr ansfer coefficient (convection and r adiation) on t he out er sur face of t he insulat ion. R 1 I n t he above expr ession as R2 incr eases, t he fact or log e 2 k2 R1

FG IJ H K

1 incr eases but t he fact or R h decr eases. 2 0

FG IJ + 1 H K Rh LM 1 log F R I  1 OP = 0 MN k GH R JK R h PQ 1 1 1 F 1 I  – = 0 or R k R h GH R JK

Ther for e Q becomes maximum when The r equir ed condit ion is

d dR2 

2

becomes minimum.

2 0

2

e

2

R2 1 log k2 R1

1

2

2 0

0

2 2

2

=

k2 h0

wher e R2 is known as critical radius of insulation as Q becomes maximum at t he value of R2c. The cr it ical t hickness of insulat ion = R2 – R1 The var iat ion of Q wit h r espect t o R2 is shown in t he figur e below When R2 = R1 it indicat es bar e cylinder wit hout insulat ion. The cr it ical value of R2 is r epr esent ed by point 2 and t he r adius at t his point is r epr esent ed by R2c, Whenever t he value of R2 > R2c t he value of Q decr eases.

Cr it ical T hickness of I nsulat ion for Spher e H eat flow t hr ough spher e wit h insulat ion, T1  T2 Q= = R 2 – R1 1  4 R 2 R1 k 4 R 22 h0



d dR2

LM 1 – MN R k 1

1 1  2 R 2 k R 2 h0

LM 1 – MN R k

OP = 0 PQ

b

4  T1  T2

1

g

1 1  R 2 k R 22 h0 or

R2 =

OP PQ

2k h0

8.36

Thermal Engineering

FI NS Fins ar e commonly used for incr easing t he heat t r ansfer r at es fr om sur face whenever it is not possible t o incr ease t he r at e of heat t r ansfer eit her by incr easing t he heat t r ansfer co-efficient on t he sur face or by incr easing t he t emper at ur e differ ence bet ween t he sur face and sur r ounding fluids. The fins ar e commonly used in engines used for scoot er s and mot or cycles as well as small capacit y compr essor s. The cir cumfer ent ial fins of r ect angular or t r iangular pr ofile ar e commonly used on t he engine cylinder of scoot er s and mot or cycles. The pin t ype fins ar e used on t he condenser of a domest ic r efr iger at or.

H eat Tr ansfer thr ough F in Figur e shows a r ect angular fin of lengt h L , which is per pendicular t o sur face fr om which heat is t o be r emoved, 2  is t he t hickness of t he fin, W is widt h of fin which is par allel t o t he sur face fr om which heat is t o be r emoved. cosh m (L – x)  ( h / mk) sin h m (L – x)   cos h m L  ( h / mk) sin h m L  = T – Ta = T0 – Ta = fin sur face temper atur e at a distance x measur ed fr om base of the fin. Ta = at mospher e t emper at ur e T 0 = base t emper at ur e

   0  wher e  0 T

Then 

T

Ph kA c = per imet er of t he fin = 2W + 4 = cr oss sect ional ar ea of t he fin = conduct ivit y of fin mat er ial = convect ion heat t r ansfer coefficient

H eat lost by t he fin, Q = k A c m 0

For r ect angular fin, m = For cir cular fin, m =

h. P  kA c

hP  kA c

h d =  2 k. d 4

Af

c

m 0

LM1  mk t an h mL OP MM mk h P  t an h m L P MN h PQ

4h kd

 = 0

LM cos h mbL – xg OP MN cos h m L PQ

and Q = k A c m 0 I f t he fin is sufficient ly long, t hen  – mx = cos h mx – sin h mx 0 = e Q = k A c m 0 H eat lost wit h fin Effect iveness of t he Fin (). = H eat lost wit hout fin

Effect iveness  should be mor e t han 1

Ac

k

wher e, d = diamet er of t he fin I f end of t he fin is insulat ed, t hen

W L

LM h cosh mL  km sinh mL OP = k A N km cos h mL  h sinh mL Q hb2 W  4 g h = k W.2

x

T0

m = P Ac k h

2

x

k A c m.  0 h cos h m L  mk sin h m L mk cos h m L  h sin h m L = hA c.0

km = h

LM1  mk t an h mL OP MM mk h P  t an h m L P MN h PQ

Thermal Engineering

8.37

h Ph 2W. h = = k  kA c k. W.2 When 2 << W, a non-dimensional number (Biot number ) is  I nt ernal r esist ance of t he fin mat er ial h k Bi = = 1 Ext er nal r esist ance of t he fluid on fin mat er ial k h wher e,

m=

LM Bi  t an he Bi . L j OP B i M 1 + B i  t an he B i . L j P MN PQ

1

 =

L = non-dimensional lengt h  Now t her e ar e t hr ee impor t ant cases : ( i ) I f B i = 1, t hen  = 1. This is possible when poor conduct ing mat er ial is used for fin. ( ii ) I f B i > 1, t hen  < 1. Ther e is an adver se effect on heat flow and fin wor ks as insulat or. This is possible if t he value of h is quit e high t his is t he main cause for not using t he fins on t he st eam condenser t ubes. ( iii ) I f B i < 1, t hen  > 1 which is t he most desir able case. This is possible in t he case of high conduct ivit y mat er ials like Cu, Al. Ther efor e fins ar e desir able in all t ypes of heat exchangess. wher e, L =

Variation of H eat L oss From Fins with Length of Fin I t can be seen fr om t he gr aph t hat t he r at e of incr ease of Q wit h incr ease of L is quit e less as compar ed wit h t he r at e of incr ease of Q wit h incr ease of

k . h

E fficiency of F in f =

Heat lost fr om t he fin Heat lost fr om t he fin, if t he whole sur face of t he fin is maint ained at base t emperat ure

1 k .....  m  . t an h m L L h Temper at ure er r or of t her momet er Temper at ur e dist r ibut ion along t he fin is given by =

h K

cosh m(L  x) T  Ta  = x = cosh m L 0 T0  Ta I n t his case, Ta = T g, T o = T w , T x = T L , At x = L

TL  T g  1 T  Ta = x = = Tw  T g o cos h m L To  Ta

CON V E CT I ON H EAT TRAN SFER BY CON VECTI ON A pr ocess of heat t r ansfer by the combined act ion of heat conduct ion and mixing mot ion is known as convect ion . H eat i s t r ansfer r ed by convect i on due t o mass movement and mi xi ng of macr oscopi c el ement or gas. The act ual pr ocess of ener gy t r ansfer fr om one fluid par t icle t o anot her is by conduct ion, but t he ener gy may be tr anspor t ed fr om one point t o anot her by t he displacement of fluid. As the mot ion of the fluid is involved, t he heat t r ansfer by convect ion is par t ly gover ned by t he law of fluid mechanics.

8.38

Thermal Engineering

Fr om t he Newt on's law of cooling, q = h A T = Q = h A(T w – T f ) wher e T w = t emper at ur e of wall T f = t emper at ur e of fluid h = heat t r ansfer co-efficient Unit of heat t r ansfer co-efficient : W/m 2K

h A(T w – T  )

Types of Convection H eat Transfer (i ) N atural convection :The flow of fluid t ake place when t he hot or cold body comes in cont act wit h t he fluid. The heat flow fr om t he hot body t o fluid due t o t emper at ur e differ ence bet ween t he body and t he fluid and fluid densit y in t he vicinit y of t he hot body decr eases. The differ ence in densit y of t he hot and cold fluid cause t he fluid t o flow in t he upwar d dir ect ion. The for ce causing t his flow is known as buoyancy for ce. Whenever the mot ion of t he fluid is caused only due t o t he differ ence in densit y r esult ing fr om temper at ur e gr adient wit hout t he use of pump or fan, t hen t he mechanism of heat t r ansfer is known as nat ur al or fr ee convect ion. ( ii ) F orce convect ion :I t may be caused by some ext er nal agency such as pump or a blower. The mot ion of flui d in for ce convect i on wil l be fast er t han in t he fr ee or nat ur al convect ion, t her efor e, r at e of heat t r ansfer incr eases. As t he convect ion pr ocess depends upon t he mot ion of t he fluid par t icles also, so t he heat t r ansfer by convect ion depends upon t he law of heat conduct ion, geomet r y of t he syst em and law of fluid dynamics.

SOM E U SEFU L RATI OS (i )

( ii )

Reynold number, (Re). I t is t he r at io of iner t ia for ce t o t he viscous for ce. vL Re =  – fluid densit y, v – mean velocit y  L – char act er ist ic lengt h,  – dynamic viscosit y Prandtl number, (P r ). I t is t he r at io of kinet ic viscosit y t o t her mal diffusit y.  cp

cp – sp heat , k – t her mal conduct ivit y k ( iii ) N usselt number, (N u). I t is a dimensionless heat transfer coefficient which gives a measure of the ratio of the heat tr ansfer r ate q to the rate at which heat would be conducted within the fluid under a temper ature  gr adient . L hL q L q Nu = = = . k   k .k L ( iv) Stant on number,(St ) . I t is t he r at io of wall heat t r ansfer r at e (h ) t o t he mass heat flow r at e (vcp). I t Pr =

gives a measur e of t he r at io of t he heat t r ansfer coefficient h or

FG q IJ H  K

t o t he flow of heat per unit

t emper at ur e r ise due t o t he velocit y and heat capacit y of t he fluid.

(v)

hL h Nu k St = = = vc p RePr vL  c p   k Peclet number (pe). I t is t he r at io of mass heat flow r at e t o t he heat flow r at e by t he conduct ion under k a unit t emper at ur e gr adient and t hr ough a t hickness L q  . L M ass flow r at e = Vcp

FG H

IJ K

Thermal Engineering

8.39

LV VC p LV  Pe = = = k = q k  c p L V c pL VL c p . Again, Pe = =  K K  P e = Re . Pr (vi ) Graetz number (Gr). I t is t he r at io of heat capacit y of t he fluid flowing t hr ough t he pipe per unit lengt h of t he pipe t o t he conduct ivit y of t he pipe. So t his number is r elat ed only for t he heat flow t o t he fluid passing t hr ough cir cular pipes. cp  2 2 D  V . mc p L L =  D V c p =  . VD . c p . D =  Re . Pr . D  Gr = = 4 4 kL k 4 L 4  k L k VC p

D 4L ( vii ) Grashaff number (G r ). I t is only used in t he nat ur al convect ion heat t r ansfer. I t is defined as t he r at io of pr oduct of I ner t ia for ce and buoyancy for ce t o t he squar e of viscous for ce. = Pe



Gr =

I ner t ia for ce  Buoyance for ce (Viscous for ce)

2

2

=

 g.L 

2

3

=

gL3 

2

FG H

...   

F or mulae for Solving t he Pr oblems of N at ur al and F or ced Convect ions (1)

N at ur al Convect ion ( i ) Nu = = ( ii ) Nu = = ( iii ) Nu = =

(2)

= 0.13 (Gr

1 Pr ) 3

for ver t ical sur faces when 108 < Gr Pr < 1012

1

= 0.56 (Gr Pr ) 4 for ver t ical sur faces when 105 < Gr Pr < 108 1

= 0.13 (Gr Pr ) 3 for hor izont al cylinder s when 109 < Gr Pr <1012 1

= 0.53 (Gr Pr ) 4 for hor izont al cylinder when 104 < Gr Pr < 109 1

for hor izont al hot cir cular plat e facing upwar d when Gr Pr > 109

= 0.17 (Gr Pr ) 3 1

= 0.71 (Gr Pr ) 4 for 103 < Gr Pr < 109 1

( iv) Nu = 0.667 Pr 2 (Pr  0.952) F or ced Convect ion



1 1 h aL = 0.664 Re2 Pr 3 k ha d ( ii ) Nu = = 0.023 Re0.8Pr 0.4 k M odel Test ing  p m L p v (i ) m = . . vp m  p L m L p m vm = . vp Lm p

( i ) Nu =

(3)

h aL k h aL k ha d k ha d k ha d k ha d k

1 4

1

(Gr ) 4

for ver t ical plat e

for t he flow over flat plat e for t he flow t hr ough t ube

 

IJ K

8.40

Thermal Engineering

vm vp

(ii )

F (iii ) m Fp

Lm Lp

=

(t est ing for r esist ance in t he same fluid)

F I FL I GH JK GH L JK

 v = m . m p vp

(iv) H .P. =

2

m

2

(t est ing for t hr ust for pr opeller

p

F pv p 75

Q = h A(T w – T f ) T w = t emper at ur e of wall T f = t emper at ur e of fluid h = heat t r ansfer co-efficient Unit of heat t r ansfer co-efficient : W/m 2K Under st eady heat flow conduct ion. H eat conduct ed = H eat convect ed fr om solid boundar y t o adjacent fluid wher e

Q = – kA



FG dT IJ H dx K

– k = h A(T w – T a) or h =

FG dT IJ H dx K

xL

Tw – Ta x L Fr om t he above equat ion, for a syst em in which t he wall t emper at ur e T w , t he ambient t emper at ur e T a and t her mal conduct ivit y of solids, k ar e const ant , t hen any change in h mean change in

FG dT IJ H dx K

x L

i.e. heat t r ansfer co-efficient can be consider ed fundament ally as a slope of t he t emper at ur e dest r ibut ion in t he fluid.

RE SI STAN CE CON CE PT Over all H eat Tr ansfer Co-Efficient I f U is over all co-efficient of t he heat t r ansfer, t hen Q = UA(T a – T b) =

U =

 H er e,

b

g

A Ta – Tb 1 x 1   ha k hb

1 1 x 1   ha k hb

h a = heat t r ansfer co-efficient fr om hot fluid t o met al sur face h b = heat t r ansfer co-efficient fr om met al sur face t o cold fluid k = t her mal conduct ivit y of met al wall.

H eat Tr ansfer T hr ough Composit e Cylinder When a hot fluid at a t emper at ur e T a flowing t hr ough a pipe is separ at ed by t wo layer s fr om cold fluid at t emper at ur e T b(say at mospher e). (not at ions car r y t he usual meaning), t hen



Q=

Ta – Tb

r r 1 L 1 1 1 1 O  log e 2  log e 3  M P 2l N r1 ha k1 r1 k2 r2 r4 hb Q

her e, r1 < r2 < r3 I f U a and U b ar e the over all co-efficients of heat tr ansfer r efer red to inside and the outside ar ea of tube r espectively, t hen Q = U aA a(T a – T b) = U bA b (T a – T b)

Thermal Engineering

8.41

H eat Transfer Through Composite Spherical Shell Pr oceeding in t he same way as in t he pr evious case, t he heat t r ansfer in t his case will be given by t he expr ession Q =

Ta – Tb r2 – r1 r – r 1 1   3 2  ha A a 4 k1 k2 r1 r2 4 k2 r2 r3 hb Ab

Prandtl N umber I f t he fr ee st r eam velocit y is low and t emper at ur e differ ence bet ween fr ee st r eam and t he plat e is small, t hen t he fr ict ional heat gener at ion can be neglect ed and ener gy equat ion becomes

u.

where,

=

T T 2 T  v. = . 2 x y y

... (i )

Thermal conduct ivit y k = c p Ther mal capacit y

and t his r at io is known as t her mal diffusivit y H ydr odynamic equat ion or moment um equat ion for boundar y layer is

u u 2 u  v. = . x y y 2 I n t his case t he t emper at ur e and velocit y dist r ibut ion would be ident ical

u.

T  Tw v u df  = = = wher e  = y = st r et ching fact or  x U m d T  Tw The r esult is of consider able pr act ical impor t ance because all t he gases have  .   1 , gr eat er will be t he differ ence bet ween t emper at ur e and velocit y pr ofile. Also if  C p   Viscous nat ur e of fluid = = = k H eat conduct ing capacit y k  cp H eat st orage capacit y



For t hr ee differ ent condit ions of Pr, t he t her mal and hydr aulic boundar y layer s ar e shown in t he figur e.

F or mulae U sed F or Solving Pr oblems For laminar flow (i )

 4.64 = (Von-K ar man equat ion) = x Rex

(ii )

 = (Pr ) 3 t

5 Rex

(Blasius equat ion)

1

(iii ) (M ass flow t hr ough boundar y layer at a sect ion x = m x), m x =

5 U  8

... (ii )

8.42

(iv)

Thermal Engineering

M ass flow bet ween t wo sect ions), m = 

1 2



1 2

(v)

Cf x = 0.664 (Rex )

(vi )

Cf a = 1.328 (Rel )

(vii )  a = Cf a

5 U( 2  1 ) wher e 2 > 1 8

U 2 2

(viii )  x = 0.332

U L

Rex = 0.332 U 2 (Rex )

(ix)

 x = 0.664

U L

Rel = 0.664 U 2 (Rel )

(x)

F =  a.A = 0.664 UW (Rel ) = 0.664

(xi )

F x = x.A 1





1 2

1 2

A U 2

(Rel )



1 2

1

(xii ) Nu x = 0.332 (Rex ) 2 (Pr ) 3 1

1

(xiii ) Nu a = 0.664 (Rel ) 2 (Pr ) 3 (xiv)

2 St x (Pr ) 3

2

(xv) St a (Pr ) 3

F h I (Pr ) = G H U c JK F h I (Pr ) = G H U c JK x

2 3

=

C fx 2

p

a

2 3

=

C fa

p

2

; wher e, Rex =

hx . x ha . L and Nu a = k k For turbulent flow:  0.39 (i ) = x (Rex )1 / 5

x.U LU and Rel =  

Nu x =

(iii ) Cf a = 0.072 (Rel )

(ii ) Cf x = 0.0576 (Rex )

1 5





1 5

2 (iv)  a = Cf a . U 2

1

(v)

(vi ) Nu x = 0.0288 (Rex)0.8 (Pr ) 3

F = a . A 1

2

(vii ) Nu a = 0.036 (Rel )0.8 (Pr ) 3 2 (Pr ) 3



(viii ) St x . (Pr ) 3 = 0.0288 (Rex )



1 5

1 5

(ix) St a . = 0.036 (Rel ) For laminar and turbulent flows: 1

(i )

1

Nu a = 0.664 (Rec ) 2 (Pr ) 3 wher e, Rec = 5 × 105 and L c =

5  105 .  U

1 ha (L  L c ) = 0.036 [(Re ) 0.8  (Re ) 0.8 ] (Pr ) 3 l c k 1 k (iii ) Over all heat t r ansfer co-efficient for bot h r egions h a = [0.036 (Rel )0.8 – 836] . (Pr ) 3 L

(ii )

Tur bulent r egion :

Thermal Engineering

8.43

RAD I AT I ON H EAT TRAN SFER BY RADI ATI ON The heat t r ansfer by r adiat ion does not r equir e any medium and it is mor e effect ive in vaccum r at her t han in a medium. Ther e ar e t wo t heor ies which explain t he t r ansfer of heat by r adiat ion. ( i ) Wave Theory Accor ding t o t his t heor y, it is assumed t hat all t he space in a univer se is filled up by a hypot het ical medium called ether. I t was stated that the r adiation heat transfer take place because of pr opagation of electr omagnetic wave t hr ough t he et her. A hot body emit s elect r omagnet ic wave due t o t he vibr at ion of molecules and gives out it s st or ed ener gy. These waves t r avel t hr ough space unt il t hey st r ike anot her body wher e par t of t heir ener gy is absor bed and r econver t ed back int o ext er nal ener gy. The r adiat on in t he for m of elect r omagnet ic wave cause a decr ease in int er nal ener gy of t he emit t ing body unless t he heat is gener at ed wit hin t hat body equivalent t o t he decr ease in int er nal ener gy. ( ii ) Quant um T heor y Some of t he atoms and molecules of a solid body ar e r aised t o excited st at es when t hey ar e heat ed and t her e is t endency for t hese at om and molecules t o r et ur n t o lower ener gy levels spont aneously. Dur ing t his pr ocess, ener gy in t he for m of elect r omagnet ic r adiat ion is emit t ed. This emit t ed ener gy is not cont inuous but in t he for m of successive and separ at e quant it ies called "quant a". M ax planck has given t he equat ion E = hf wher e h = Planck's const ant f = fr equency of vibr at ion

Reflect ion, Absor pt ion and Tr ansmission of Radiat ion Radiat ion is t he pr oper t y of all subst ances and each emit s ener gy and amount of ener gy emit t ed depends upon t he t emper at ur e level. I n gener al, r adiat ion ener gy falling on a body par t ly r eflect ed, par t ly t r ansmit t ed and par t ly absor bed. For a par t icular wave lengt h, t he amount of r eflect ed ener gy depends upon (i ) t ype of t he mat er ial (ii ) sur face finish (iii ) angle of incidence For a given finish, t he por t ion r eflect ed depends on wave lengt h. I f t he t ot al incident ener gy Qi is falling on an object as shown in t he figur e and Qa is absor bed, Qr is r eflect ed and Qt is t r ansmit t ed, t hen Qa + Qr + Qt = Qi

Qa Qr Qt   =1 Qi Qi Qi  + r + = 1 wher e  is known as absor bivit y r is known as r eflect ivit y  is known as t r ansmissivit y The values of , r ,  in above equat ion ar e dimensionless and var y fr om 0 t o 1. For most of t he solids and liquids,  = 0 except few as glass et c.  +r = 1 M ost solid bodiesas well as liquids ar e st r ong absor ber s of the ther mal r adiation. The absor pt ion occur s in a ver y t hin r egion near t he sur face of solids. For good conduct or s of elect r icit y t his is of t he or der of one micr on and in elect r ical insulat or s t his t hickness may be as much as t housands of micr ons. 

Opaque Body The bodies which do not t r ansmit any r adiat ion ener gy ar e known as opaque bodies. All engineer ing mat er ials ar e consider ed opaque.

Bl ack Body If r = 0 and  = 0, t hen =1 This means t hat incident ener gy is ent ir ely absor bed by t he body. Such body is defined as a "Black body".

8.44

Thermal Engineering

W hit e Body If  = 0 and  = 0, t hen r =1 This means t hat ent ir e incident r adiant ener gy is r eflect ed by t he body and such body is called "Whit e body".

Tr anspar ent Body If  = 0 and r = 0, t hen =1 This means t hat ent ir e incident r adiant ener gy passes t hr ough t he body and such bodies ar e called t r anspar ent bodies. N ote : Since for most of t he gases r = 0, t her efor e  +  = 1 Ther e ar e no absolut ely black, whit e and t r anspar ent bodies in nat ur e.

Refl ect i on of Radi at ion Specul ar Refl ect i on The r eflect ion of t he r adiat ion ener gy is known as “ specular reflect ion” when t he angle of r eflect ion ( 2) is equal t o t he angle of incidence  1 as shown in t he figur e.

D i ffused Refl ect i on The r eflect ion of t he r adiat ion ener gy is known as “ diffused r eflect ion” when it is r eflect ed in all dir ect ion as shown in t he figur e. SOURCE

SOURCE

REFLECTED RAYS

2

1

(a)

(b)

Secular Reflection

Diffused Reflection

Specular r eflect ion occur s on a highly polished sur face while t he diffused r eflect ion occur s on a r ough sur face.

Absor pt ion and Tr ansmission The non-r eflect ed par t of incident r adiat ion is par t ly absor bed and par t ly t r ansmit t ed t hr ough t he body. The absor pt ion and t r ansmission depend on t he mat er ial and geomet r y of t he body r eceiving t he ener gy as well as on t he wavelengt h of t he incident r adiat ion. Nor mal glass is t r anspar ent t o light r ays but non-t r anspar ent t o ult r aviolet r ays and it only t r ansmit s a small par t of t her mal r adiat ion. The wor d t r anspar ent is used wit hin some limit s because a t hin layer of wat er is t r anspar ent t o light r adiat ion but it does not r each t he bot t om of t he sea.

BLACK AN D GREY SU RFACE A black body which absor bs all t he incident r adiat ion and none is r eflect ed or t r ansmit t ed. I n ot her wor ds, a black body is one whose absor pt ivit y is 100%.

Plank's L aw of Radiat ion The t ot al emissive power of a gr ey body var ies wit h t he wavelengt h at a given t emper at ur e and also var ies wit h a t emper at ur e. Source  E b = f (, T) For t he monochr omat ic emissive power of black body, E b =

C1 5 C2

e T – 1 Circular caruity

Thermal Engineering

8.45

3.21  10 8

= 37.45 × 10– 17 J-m 2/sec 10 16 and C2 = 1.438 cm-K = 14380 – K. Tot al emissive power of t he black body is given by Bolt zmann as wher e

C1 =

z

z





E b(0 – ) =

E b d  =

0

0

C1 5

eC2 / T – 1

d

Solving t he above equat ion and subst it ut ing t he values of C1 and C2, we have E b =  T 4 (St efan Bolt zmann's law) – 8 wher e,  = 5.67 × 10 J/m 2-sec-K 4

Stefan– Boltzmann Law The maximum r at e of r adiat ion t hat can be emit t ed fr om a sur face at an absolut e t emper at ur e T s (in K ) is given by t he Stefan– Boltzmann law as

 Q = A sTs4 (W) emit , max wher e

 = 5.67  10-8 W/m 2 · K – 4 wher e ‘K ’ is t he St efan– Bolt zmann const ant . A s is t he sur face ar ea t hr ough which r adiat ion t akes place

The idealized sur face t hat emit s r adiat ion at t his maximum r at e is called a blackbody, and t he r adiat ion emit t ed by a blackbody is called blackbody radiation. The r adiat ion emit t ed by all r eal sur faces is less t han t he r adiat ion emit t ed by a blackbody at t he same t emper at ur e, and is expr essed as

 = A sTs4 (W) Q emit wher e  is t he emissivity of t he sur face. The pr oper t y emissivit y, whose value is in t he r ange 0  1 is a measur e of how closely a sur face appr oximat es a blackbody for which  = 1. When a sur face of emissivit y and sur face ar ea As at an absolut e t emper at ur e Ts is complet ely enclosed by a much lar ger (or black) sur face at absolut e t emper at ur e Tsur r separ at ed by a gas (such as air ) t hat does not int er vene wit h r adiat ion, t he net r at e of r adiat ion heat t r ansfer bet ween t hese t wo sur faces is given by shown in figur e. 4 Qr ad = A s (Ts4  Tsurr )

Sur r oundi ng sur faces at T sur r

Ai r . Qem i t t ed

. Qi nci dent

, A s , T s . 4 4 Qr ad = A s (T s – T sur r )

Gr ey Body

F ig : Radiat ion heat transfer between a sur face and the sur faces surrounding it

The body which absor bs a definit e per cent age of incident r adiat ion waves ir r espect ive of t heir wavelengt hs, t he body is known as grey body. When t he absor pt ivit y of a body var ies wit h wavelengt h of r adiat ion waves, t he body is known as coloured body.

E missive Power of Gr ey Body The emissive power of t he body is always less t han t he black body. The r at io of emissive power of gr ey body t o t hat of black body at given t emper at ur e is const ant for all wave lengt hs.

8.46

Thermal Engineering

Emissivit y of a gr ey body,  = For monochr omat ic r adiat ion,

Eg Emissive power of grey body at a t emperat ure = Emissive power of black body at same t emperat ure Eb

Emissivity,  =

z z z

E g E b



Eg =

0 



 =

z



E g . d  and E b =

z

E b . d 

0



E g . d 

0 

=

0

E b . d 

0

E b .  . d 

z



E b . d 

0

H ence emissivit y of sur face is pr oper t y of t he sur face and only depends on t he char act er ist ic of t he sur face and independent on t he wavelengt h of the incident r adiat ion waves, while absor pt ivit y of the sur face is not pr oper t y of t he sur face because it depends on t he wavelengt h of t he incident r adiat ion waves.

K ir choff 's L aw K ir choff's law st at es t hat t he r at io of t ot al emissive power t o absor pt ivit y is const ant for all t he subst ances which ar e in t he t her mal equilibr ium wit h t he sur r oundings. A ppl yi n g t h i s l aw on t h r ee bodi es wh ose em i ssi ve power s ar e E 1 , E 2, an d E 3 an d absor pt i vi t i es ar e 1, 2, and 3 r espect ively. E E1 E E = 2 = 3 = const ant = b 1 2 3 b Since absor pt ivit y for black body, b = 1 E1 E2 E3  = 1, = 2 and = 3 Eb Eb Eb But accor ding t o t he definit ion of emissivit y of a body E1 E2 E3 = 1, = 2 and = 3 Eb Eb Eb Compar ing bot h above equat ions, we get  1 = 1, 2 =  and 3 = 3 I n gener al for m we can wr it e,  =  H ence t his law st at e t hat t he emissivit y of a body is equal t o it s absor pt ivit y when t he body r emains in t her mal equilibr ium wit h it s sur r oundings.

Wien's Displacement L aw mT = 2.9 mm-K wher e m is t hat wavelengt h at which monochr omat ic emissive power of t he black body becomes maximum at a t emper at ur e T. As t he body is heat ed, t he maximum int ensit y is shift ed t o t he shor t er wave lengt hs and fir st visibl e sign of incr ease in t emper at ur e of t he body is a dar k r ed colour. Wit h fur t her incr ease in t emper at ur e, t he colour appear s as br ight r ed, t hen br ight yellow and finally whit e.

Sol id Angle I t is defined as t he angle cover ed by unit ar ea on a sur face of a spher e of unit r adius when joined wit h t he cent r e of spher e and it is measur ed in t he st er adians. A Solid angle,  = 2 R wher e A is ar ea on a sur face of a spher e of r adius R which is cut out fr om t he issuing cent r e of a spher e.

I nt ensit y of Radiat ion I t is defined as t he r at e of emission of r adiat ion in a given dir ect ion fr om a sur face per unit solid angle and per unit pr oject ed ar ea of a r adiat ing sur face on a plane per pendicular t o t he dir ect ion of r adiat ion. E b = I b

Thermal Engineering

8.47

Radi osi t y I t refers to all the radiant ener gy leaving surface. The radiation leaving the surface includes the r eflected portion as well as emission as shown in the figur e. Therefore, r adiosity is differ ent fr om the emissive power.

Types of Radi osit y (i )

Spectral radiosity. I t is t he r at e at which r adiat ion leaves per unit ar ea of the sur face at a wavelengt h  per unit wavelength. I t is r elated to the int ensity of r adiation associated wit h emission also shown in figur e. ( ii ) Total radiosity. I t is associat ed wit h t he ent ir e spect r um of wave lengt h. The int ensit y of r adiat ion is based on pr oject ed ar ea while t he r adiosit y is based on t he act ual sur face ar ea.

RADI ATI ON BETWEEN TWO I N FI N I TELY LON G CYLI N DERS Two con cen t r i c cyl i n der of r adi u s R 1 an d R 2 ar e sh own i n f i gu r e. T 1 and T 2 ar e t emper at ur er s of inner and out er sur face of inner and out er cylinder r espect ively and t heir r elat ive emissivit ies ar e 1 and 2. T h en n et h eat t r an sf er bet w een t h e i n n er an d ou t er con cen t r i c cylinder s is given by Qnet =

For concent r ic spher e,

A 1 (T14  T24 ) A 1  1 1 A 2

FG 1  1IJ H K

=

2

net heat flow, Qnet =

 A 1 (T14  T24 ) R 1  1 1 R 2

FG 1  1IJ H K 2

 A 1 (T14  T24 )

FG IJ FG 1  1IJ H K H K

R1 1  1 R2

2

2

N ETWORK AN ALYSI S

rG

Con si der t w o bodi es A an d B r adi at i n g h eat as sh ow n i n f i gu r e. I t emissivit y of body A is a, t hen amount of ener gy r adiated by A is a . ‘E ba and if G is t he amount of ener gy falling on A fr om B, t hen out of G, t he amount of ener gy r eflect ed by A is r aG, wher e r a is t he r eflect ivit y of t he sur face A. Tot al ener gy going away fr om t he body A is,  = aE ba + r aG .... (i ) Since t r ansmissivit y is assumed t o be zer o for solid bodies, t her efor e a + r a = 1  r a = 1 – a Fr om K ir chhoff 's law, a = a  J = a E ba + (1 – a)G Net ener gy lost by t he body A, q = J – G q = aE ba – aG ... (ii ) Fr om equat ion (i ) and (ii ), we get

G

A

q = E ba  J Joule per unit ar ea 1 a a Tot al heat lost fr om A is given by, Q =

wher e

E ba  J  1  a  1    a  A a

1  a is t he sur face r esist ance of t he body as it is r elat ed t o t he sur face pr oper t ies of t he body.. a A a

a Eba

8.48

Thermal Engineering

When t wo sur face r esist ance of t he t wo bodies and space r esist ance bet ween t hem is consider ed, t hen net heat flow can be given by an elect r ic cir cuit as shown in figur e. Eb1



(i )

E b1  E b2 1  1 1 2 1   A 1 1 A 1 F12 A 2 2

=

Q =

Eb2

1 – 2 A2 2

1 A1F1–2

A 1 (T14  T24 )

FG H

1  1 1  2 1   1 F1 2 2

I f t wo bodies ar e infinit e par allel plat e, t hen A 1 = A 2 and



( ii )

Q =

1 – 1 A11

IJ K

A1 A2

F1 – 2 = 1

A 1 (T14  T24 ) 1 1  1 1 2

I f t wo bodies ar e concent r ic cylinder or spher e, t hen F 1– 2 = 1



Q =

A 1 (T14  T24 )

FG H

1  2 1  1 2

( iii ) For concent r ic cylinder s,



1 2

D 1 L D R A1 = = 1 = 1 A2 D2L D2 R2 Q=

A1 (iv) For concent r ic spher e, = A2 

IJ A KA

Q=

A 1 (T14  T24 )

FG H

1  2 1  1 2

FG R IJ HR K

IJ R KR

1

2

2

1

2

A 1 (T14  T24 )

FG H

1  2 1  1 2

IJ FG R IJ K HR K

2

1

2

I n all the above case when the grey surface ar e r eplaced by the black body, the surface resistance becomes zer o.

Thermal Engineering

8.49

PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 1. A closed syst em is one, which (a) per mi t s t he passage of ener gy and mat t er acr oss t he boundar ies (b) does not per mit t he passage of ener gy and mat t er acr oss t he bandar ies (c) permits the passage of energy across the boundary but does not permit the passage of matter (d) per mi t s t he passage of mat t er acr oss t he boundary but does not per mit the passage of ener gy 2. An isolat ed syst em is one, which (a) per mi t s t he passage of ener gy and mat t er acr oss t he boundar ies (b) per mit s t he passage of ener gy only (c) does not per mit t he passage of ener gy and mat t er acr oss it (d) per mit s t he passage of mat t er only 3. A system comprising of a single phase, is known as (a) open syst em (b) closed syst em (c) homogeneous system (d) heter ogeneous system 4. The char act er ist ic of a cont r ol volume is/ar e (a) t he volume, shape and posit ion wit h r espect t o an obser ver ar e fixed (b) mat er ial flow acr oss t he boundar y (c) bot h (a) and (b) (d) none of t heses 5. Cont r ol volume r efer s t o a (a) specified mass (b) fixed r egion in t he space (c) closed syst em (d) none of t he above 6. Specific heat is t he amount of heat r equir ed t o r aise t he t emper at ur e (a) by unit degr ee of a subst ance (b) by unit degr ee of a unit mass (c) of a unit mass by 10° (d) none of t he above 7. The r at io of specific heat s of a gas at const ant pr essur e and at const ant volume (a) var ies wit h t emper at ur e (b) var ies wit h pr essur e (c) is always const ant (d) none of t he above

8. I nt er nal ener gy of a per fect gas depends upon (a) t emper at ur e only (b) t emper at ur e and pr essur e (c) t emper at ur e, pr essur e and specific heat s (d) none of t he above 9. With r ise of temper atur e, the specific heat of water (a) incr eases (b) decr eases (c) fir st decr eases t o minimum t hen incr eases (d) r emains const ant 10. For a closed syst em, differ ence betweeen the heat added t o t he syst em and wor k done by t he gas, is equal t o t he change in (a) enthalpy (b) entr opy (c) int er nal ener gy (d) t emper at ur e 11. Wh i ch of t h e f ol l owi n g i s n ot an ext en si ve pr oper t y of a t her mo-dynamic syst em ? (a) Tot al mass (b) Tot al int er nal ener gy (c) Tot al volume (d) Temper at ur e 12. Ther mal equilibr ium between two or mor e bodies exist s, when t hey ar e br ought t oget her, t her e is no change in (a) density (b) pr essur e (c) t emper at ur e (d) all of t hese 13. The sequence of process that eventually returns the working substance to its original state, is known as (a) event (b) t her modynamic cycle (c) t her modynamic pr oper t y (d) none of t hese 14. A system consisting of more than one phase is called (a) isolat ed syst em (b) open syst em (c) non-unifor m syst em (d) het er ogeneous syst em 15. K elvin-Plank’s law deals wit h (a) conver sion of wor k int o heat (b) conver sion of heat int o wor k (c) conser vat ion of wor k (d) conser vat ion of heat 16. A ccor di n g t o K el v i n -Pl an k ’s st at em en t , a per pet ual mot ion machine of (a) fir st kind is possible (b) fir st kind is impossible (c) second kind is impossible (d) second kind is possible

8.50

Thermal Engineering

17. A per pet ual mot ion machine of t he fir st kind i.e. a m ach i n e w h i ch pr odu ces pow er w i t h ou t consuming any ener gy is (a) possi bl e accor di ng t o fi r st l aw of t her modynamics (b) impossible accor ding t o fir st law of ther modynamics (c) impossible according to second law of thermodynamics (d) possible accor ding t o second law of t her modynamics. 18. According to kinetic theory of gases, at absolute zero (a) specific heat of molecules r educes t o zer o (b) kinet ic ener gy of molecules r educes t o zer o (c) volume of gas r educe t o zer o (d) pr essur e of gas r educe t o zer o. 19. I n an isot her mal pr ocess, int er nal ener gy (a) incr eases (b) r emains const ant (c) decr eases (d) none of t he above 20. For t he same expansion r at io, wor k done by t he gas in case of adiabat ic pr ocess as compar ed t o wor k done in case of isot her mal pr ocess is (a) same (b) mor e (c) less (d) none of t he above 21. I n i sot her mal expansi on, wor k done by gas depends upon (a) at omicit y of gas only (b) expansion r at io only (c) adiabatic index (d) bot h (a) and (b) 22. A pr ocess, in which no heat is supplied or r eject ed fr om t he syst em and ent r opy is not const ant , is called (a) isot her mal (b) isent r opic (c) polyt r opic (d) hyper bolic 23. A process, in which the wor king substance neither r eceives nor gives out heat t o it s sur r oundings dur ing it s expansion or cont r act ion, is called (a) isot her mal pr ocess (b) isent r opic pr ocess (c) polyt r opic pr ocess (d) adiabatic pr ocess 24. I f v 1 is volume at t he begining and v 2 is volume at t he end of expansion, t hen expansion r at io (r ) is given by

v1 (a) v 2 (c)

v1  v2 v1

(b)

v2 v1

(d)

v1  v2 v2 – v1

25. The net work done in a polytropic process is given by (a)

(c)

p1 V1  p2 V2

bn  1g

p1 V1  p2 V2 n

(b)

(d)

p2 V2  p1 V1

bn  1g

p2 V2  p1 V1 n

26. The polyt r opic index (n) is given by

FG p IJ Hp K (a) FV I log G J HV K

(b)

FG p IJ Hp K (c) FV I log G J HV K

(d) None of t he above

log

2

1

1

2

log

FG V IJ HV K Fp I log G J Hp K

log

1

2

1

2

1

2

1

2

27. I n a t hr ot t ling pr ocess (a) W = 0 (b) E = 0 (c) H = 0 (d) all of t he above 28. I n a r ever sible polyt r opic pr ocess (a) ent halpy r emains const ant (b) ent r opy r emains const ant (c) some heat t r ansfer occur s (d) int er nal ener gy r emains const ant 29. I n t he polyt r opic equat ion, pV n = const ant , if value of n is infinit e, t hen pr ocess is called (a) const ant pr essur e pr ocess (b) const ant volume pr ocess (c) const ant t emper at ur e pr ocess (d) none of 30. I n t h e pol y t r opi c pr ocess equ at i on , pV n = const ant , if n = 0, t he pr ocess is called (a) const ant pr essur e pr ocess (b) const ant volume pr ocess (c) const ant t emper at ur e pr ocess (d) none of t hese 31. I n t h e pol y t r opi c pr ocess equ at i on , pV n = const ant , if n = 1, t he pr ocess is called (a) const ant pr essur e pr ocess (b) const ant volume pr ocess (c) const ant t emper at ur e pr ocess (d) none of t hese

Thermal Engineering

32. I n t h e pol y t r opi c pr ocess equ at i on , pV n = const ant , t he pr ocess will be adiabat ic, if n in t he euqt ion is (a) zer o (b) one (c)  (d) infinite 33. I n a r ever si bl e adi ablat i c pr ocess, t he r at i o of T 1/T 2 is equal t o

Fp I (a) G J Hp K

 1 

FV I (c) G J HV K

 1 

1

(b)

2

1

FG p IJ Hp K

 1 

FG V IJ HV K

 1 

2

1

2

(d)

2

1

34. A r ever sible pr ocess (a) must pass t hr ough a cont i nuous ser i es of equlibr ioum st at es. (b) leaves no histor y of the events in surr oundings. (c) must pass t hr ough t he same st at es on t he r ever sed pat h as on t he for war d pat h. (d) all of t hese. 35. I sent r opic flow is (a) r ever sible adiabat ic flow (b) ir r ever sible adiabat ic flow (c) fr ict ionless fluid flow (d) none of t he above. 36. Non quasi-st at ic pr ocess is (a) fr ee expansion of gas (b) expansion of a gas in a cylinder under constant pr essur e (c) r apid compr ession of a gas in a cylindder. (d) gr adual compr ession of a gas in a cylinder. 37. For an ideal gas, the change in enthalpy (H) for an elemental change in temperature T is given by (a) Cp . T (b) Cv . T (c)

Cp Cv

T

Cv (d) C T p

38. One hundr edth of a kilogr am of air is compr essed in a pist on-cylinder device. At an inst ant of t ime when T = 400 K , t he r at e at which wor k is being done on t he air is 8.165 kW, and heat is being r em ov ed at a r at e of 1.0 k W, t h e r at e of t emper at ur e r ise will be (a) 10 K /s (b) 100 K /s (c) 1000 K /s (d) 10000 K /s. 39. For which of t he following sit uat ions, zer ot h law of t her modynamics will not be valid ?

8.51

(a) 50 cc of water at 25C ar e mixed with 150 cc of wat er at 25C (b) 500 cc of milk at 15C ar e mixed with 100 cc of wat er at 15C (c) 5 kg of wet st eam at 100C i s mixed wi t h 50 kg of dr y and sat ur at ed st eam at 100C (d) 10 cc of wat er at 20C ar e mixed wit h 10 cc of sulphur ic acid at 20C. 40. Fi gur e shows a vessel divided int o t wo par t s by a diaphr agm. The space A cont ains a gas and space B is evacuat ed. These t wo spaces ar e separ at ed by a d i aph r agm . N ow i f t h e d i ap h r agm i s punctur ed, the gas will expand into the evacuated space B . Whi ch of t h e fol l owi ng st at ement i s invalid ?

Gas Under Pressure

Diaphragm

A

B

Vacuum (a) A fr ee expansion is a non-quasist at ic pr ocess (b) The pr essure and volume of gas ar e not r elated t hr ough t he equat ion of st at es (c) The pr ocess is ir r ever sible (d) The ener gy is t r ansfer r ed t o t he syst em.

LEVEL-1 1. I n a t her modynamic syst em, a pr ocess i n whi ch volume r emains const ant i s cal led_____pr ocess. (a) isobar ic (b) i somet r i c (c) adiabatic (d) isent r opic [RRB JE 2014 GREEN SH I FT ]

2. Coefficient of per for mance of a commer cially used r efr iger at or would be close t o (a) 40% (b) 85% (c) 1.5 (d) 3.5 [RRB JE 2014 GREEN SH I FT ]

3. I n a ther modynamic system, ther mal equilibr ium is achi eved when t wo bodies r each (a) same t her mal ener gy (b) same ent r opy (c) same t emper at ur e (d) same molecul ar ener gy [RRB JE 2014 GREEN SH I FT ]

8.52

Thermal Engineering

(b)

Temp

(a)

Temp

4. A hot body follows Newton's law of cooling. Typical t emper at ur e-t i me gr aph of t he cool i ng body would be

Time

Time

9. Boyle's law states that: (a) The pressure of a gas varies directly with temperature at constant volume i.e. PT. (b) The product of pressure and volume of a given mass of a gas is constant at constant temperature i.e. PV = constant. (c) The volume of a gas varies directly with temperature at constant pressure i.e. VT. (d) The pressure of a gas varies directly with volume at constant temperature i.e. PV.

(d)

10. The entropy of universe tends to be : (a) Minimum (b) Zero (c) Average (d) Maximum

Temp

(c)

Temp

[RRB JE 2014 RED SH I FT ]

Time

Time [RRB JE 2014 GREEN SH I FT ]

5. Consider t he fol lowing gr aph :

Temp °C

C

D

A B 0

(Calorie) Heat

Whi ch por t i on r epr esent s t he 'L at ent heat of fusion'? (a) OA (b) AB (c) BC (d) CD [RRB JE 2014 GREEN SH I FT ]

6. Which of t he fol lowing does not sublimat e? (a) I ce (b) Ammoni um chlor i de (c) Naphthalene (d) Camphor [RRB JE 2014 GREEN SH I FT ]

7. The pressure exerted on the walls of a container by a gas is due to the fact that Gas molecules : (a) Stick to the walls of the container (b) Lose their kinetic energy (c) Get accelerated towards the wall (d) Change their momentum due to collision with the wall. [RRB JE 2014 RED SH I FT ]

8. Zeroth Law of thermodynamics forms the basis of measurement. (a) Pressure (b) Temperature (c) Work (d) Momentum [RRB JE 2014 RED SH I FT ]

[RRB JE 2014 RED SH I FT ]

11. At what temperature, both Celsius and Fahrenheit scales will show the identical readings ? (a) 100° (b) 0° (c) –40° (d) 40° [RRB JE 2014 RED SH I FT ]

12. A closed thermodynamic system is one in which (a) There is no energy or mass transfer across the boundary (b) There is no mass transfer, but energy transfer exists (c) There is no energy transfer, but mass transfer exists (d) Both energy and mass transfer take place across the boundary, but mass transfer is controlled by valves [RRB JE 2014 YEL L OW SH I FT ]

13. Pressure reaches a value of absolute zero (a) at a temperature of –273 K (b) under vacuum condition (c) at the earth's centre (d) when molecular momentum of system becomes zero [RRB JE 2014 YEL L OW SH I FT ]

14. Zeroth Law of Thermodynamics states that (a) Two thermodynamic systems are always in thermal equilibrium with each other (b) If two systems are in thermal equilibrium, then the third system will also be in thermal equilibrium (c) Two systems not in thermal equilibrium with a third system will also not be in thermal equilibrium with each other (d) When two systems are in thermal equilibrium with a third system they are in thermal equilibrium with each other [RRB JE 2014 YEL L OW SH I FT ]

Thermal Engineering

8.53

15. In fluid flow, the line of constant piezometric head passes through two points which have the same (a) Elevation (b) Pressure (c) Velocity (d) Velocity potential

23. Zeroth law of thermodynamics forms the basis of measurement of (a) Pressure (b) Temperature (c) Heat exchange (d) Work

[RRB JE 2014 YEL L OW SH I FT ]

[RRB JE 2015 26 th AU G 3 rd SH I FT ]

[RRB JE 2014 YEL L OW SH I FT ]

17. Relative Humidity is the percentage of the (a) absolute humidity value to the amount of humidity actually present (b) increase of humidity/absolute humidity (c) amount of humidity actually present to the absolute humidity (d) None of these [RRB JE 2014 YEL L OW SH I FT ]

18. Which of the following is an extensive property? (a) Pressure (b) Heat capacity (c) Temperature (d) Specific volume [RRB JE 2015 26 th AU G 1 st SH I FT ]

19. Kelvin Planck’s law deals with (a) Conservation of heat (b) Conservation of heat (c) Conservation of heat into work (d) Conservation of work into heat [RRB JE 2015 26 th AU G 1 st SH I FT ]

20. Turbine used for low head is (a) Kaplan turbine (b) Francis turbine (c) Pelton Wheel turbine (d) Propeller turbine [RRB JE 2015 26 th AU G 1 st SH I FT ]

21. A system and its environment put together constitute (a) An adiabatic system (b) An isolated system (c) A segregated system (d) A homogeneous system [RRB JE 2015 26 th AU G 2 nd SH I FT ]

22. Slope of constant volume line on temperature entropy diagram is given by (a) Cp/T (b) T/Cp (c) Cv/T (d) T/Cv [RRB JE 2015 26 th AU G 2 nd SH I FT ]

24. Internal energy is defined by (a) Zeroth law of thermodynamics (b) First law of thermodynamics (c) Second law of thermodynamics (d) Law of entropy [RRB JE 2015 26 th AU G 3 rd SH I FT ]

25. The given P-V diagram indicates following cycle 1

P1

I se

Pressure

16. A man is standing on a boat in still water. If he walks in the boat towards the shore, the boat will (a) move away from the shore (b) remain stationary (c) move towards the shore (d) Sink

P4 P2

4 I se n

P3

n. e

xp .

2 . co

mp

V1 = V4

.

3 V2 = V3

Volume

(a) Carnot cycle (c) Diesel cycle

(b) Otto cycle (d) Dual cycle

[RRB JE 2015 26 th AU G 3 rd SH I FT ]

26. Control volume refers to (a) A fixed region in space (b) A specified mass (c) An isolated system (d) A reversible process only [RRB JE 2015 27 th AU G 1 st SH I FT ]

27. All of the following are intensive properties EXCEPT (a) Mass (b) Density (c) Pressure (d) Temperature [RRB JE 2015 27 th AU G 1 st SH I FT ]

28. Kinematic viscosity of gases on increase of temperature (a) Decreases (b) Increases (c) Remains the same (d) First decreases then increases [RRB JE 2015 27 th AU G 1 st SH I FT ]

29. First law of thermodynamics deals with (a) Conservation of heat (b) Conservation of momentum (c) Conservation of mass (d) Conservation of energy [RRB JE 2015 27 th AU G 2 nd SH I FT ]

8.54

Thermal Engineering

30. An isothermal process is governed by (a) Boyle’s law (b) Charle’s law (c) Gay-lussac’s law (d) Avogadro’s law [RRB JE 2015 27

th

AU G 2

nd

SH I FT ]

31. Heat is closely related with (a) Energy (b) Temperature (c) Entropy (d) Enthalpy [RRB JE 2015 27 th AU G 2 nd SH I FT ]

32. An engine operates between temperature limits of 900 K and T2: and another between T2 and 400 K. For both the engines to be equally efficient. T2 should be equal to (a) 600 K (b) 650 K (c) 625 K (d) 700 K [RRB JE 2015 27 th AU G 2 nd SH I FT ]

33. Work done in a free expansion process is (a) Zero (b) Positive (c) Negative (d) Maximum [RRB JE 2015 27 th AU G 3 rd SH I FT ]

34. The internal energy of an ideal gas is (a) A function of temperature only (b) A function of pressure (c) A function of volume (d) Both pressure and volume [RRB JE 2015 27 th AU G 3 rd SH I FT ]

35. The process in which no heat enters or leaves the system is called (a) Isobaric (b) Isothermal (c) Isentropic (d) Isochoric [RRB JE 2015 27 th AU G 3 rd SH I FT ]

36. Economiser used in power plants is used to heat (a) flue gases (b) intake air (c) steam (d) feed water [RRB JE 2015 27 th AU G 3 rd SH I FT ]

37. Which of the following is a high head turbine (a) Propeller turbine (b) Francis turbine (c) Kaplan turbine (d) Pelton wheel turbine [RRB JE 2015 27 th AU G 3 rd SH I FT ]

38. Which one of following expression is true for Tds equation (a) du – PdV (b) du + PdV (c) du – VdP (d) du + VdP [RRB JE 2015 28 th AU G 1 st SH I FT ]

39. PMM1 is closely related with (a) First law of thermodynamics (b) Second law of thermodynamics

(c) Third law of thermodynamics (d) Zeroth law of thermodynamics [RRB JE 2015 28 th AU G 1 st SH I FT ]

40. Isolated system indicates (a) Mass of substance cross the boundary (b) Energy of substance cross the boundary (c) Both mass and energy of substance cross the boundary (d) Both mass and energy substance does not cross the boundary [RRB JE 2015 28 th AU G 1 st SH I FT ]

41. The thermal diffusivity of a substance is given by : (a)

K C

(b)

K C

(c)

KC 

(d)

C K

[Where K = Thermal conductivity; p = Mass density; C = Specific heat] [RRB JE 2014 RED SH I FT ]

42. Fins are provided on heat transferring surface in order to increase : (a) Heat transfer area (b) Heat transfer coefficient (c) Temperature gradient (d) Mechanical strength of the equipment [RRB JE 2014 RED SH I FT ]

43. Heat is mainly transferred by conduction, convection and radiation in (a) Insulated pipes carrying hot water (b) Refrigerator freezer coil (c) Boiler furnaces (d) Condensation of steam in a condenser [RRB JE 2014 YEL L OW SH I FT ]

44. Heat transfer in liquids and gases is essentially due to (a) Conduction (b) Convection (c) Radiation (d) Conduction and Radiation put together [RRB JE 2014 YEL L OW SH I FT ]

45. In which of the following mechanisms, is heat conducted in liquid and gas (a) Lattice vibration (b) Transportation of free electrons (c) Collisions and diffusion (d) No heat conducted [RRB JE 2015 26 th AU G 1 st SH I FT ]

Thermal Engineering

46. An essential requirement for the transfer of heat from one body to another body is that (a) Both bodies must be solids (b) Both bodies must be in contact (c) Temperatures of the two bodies must be different (d) Temperatures of the two bodies must be same [RRB JE 2015 26 th AU G 1 st SH I FT ]

47. Stefen Boltzman law is applicable to (a) Gray body (b) White body (c) Black body (d) Blue body [RRB JE 2015 26 th AU G 2 nd SH I FT ]

48. Heat is transferred by conduction, convection and radiation in (a) Insulated pipe carrying hot water (b) Refrigerator freezer coils (c) Melting of ice (d) Boiler furnaces [RRB JE 2015 26 th AU G 2 nd SH I FT ]

49. The concept of overall heat transfer is used in the heat transfer in the case of (a) Conduction (b) Convection (c) Radiation (d) Combined mode of heat transfer of conduction and convection [RRB JE 2015 26 th AU G 3 rd SH I FT ]

50. The unit of thermal conductivity in S.I. unit is (a) W/mK (b) W2/mK (c) W/m2K (d) W/m [RRB JE 2015 26 th AU G 3 rd SH I FT ]

51. All radiations in a black body are (a) Reflected (b) Refracted (c) Transmitted (d) Absorbed [RRB JE 2015 26 th AU G 3 rd SH I FT ]

52. Fourier’s law of heat conduction gives the heat flow for (a) Irregular surfaces (b) Non-uniform temperature surfaces (c) One dimensional cases only (d) Two dimensional cases only [RRB JE 2015 27 th AU G 1 st SH I FT ]

53. Absorptivity of a body is equal to its emissivity (a) For a polished body (b) Under thermal equilibrium (c) At one particular temperature (d) At shorter wavelengths [RRB JE 2015 27 th AU G 1 st SH I FT ]

8.55

54. Metals are good conductor of heat because (a) Their atoms collide frequently (b) Their atoms are relatively far apart (c) They contain free electron (d) They have high density [RRB JE 2015 27 th AU G 2 nd SH I FT ]

55. Heat conduction in gases is due to (a) Electromagnetic waves (b) Motion of electrons (c) Mixing motion of the different layers of the gas (d) Elastic impact of molecules [RRB JE 2015 27 th AU G 2 nd SH I FT ]

56. The heat flow through solids only by (a) Conduction (b) Convection (c) Radiation (d) Does not flow [RRB JE 2015 27 th AU G 3 rd SH I FT ]

57. A perfect black body is (a) One which absorbs total radiant energy (b) Black in colour (c) One which does not reflect the radiant energy (d) One which absorbs all radiant energy at all wavelengths [RRB JE 2015 27 th AU G 3 rd SH I FT ]

58. Heat transfer by molecular collision in (a) Conduction (b) Convection (c) Radiation (d) Scattering [RRB JE 2015 28 th AU G 1 st SH I FT ]

59. Which one of the following have a highest thermal conductivity (a) Boiling water (b) Steam (c) Solid ice (d) Rain water [RRB JE 2015 28 th AU G 1 st SH I FT ]

60. The ratio of thermal conductivity to electrical conductivity is equal to (a) Prandtl number (b) Schmidt number (c) Lorentz number (d) Lewis number [RRB JE 2015 28 th AU G 2 nd SH I FT ]

61. The effectiveness of a fin will be maximum in an environment with (a) Free convection (b) Forced convection (c) Radiation (d) Convection and radiation [RRB JE 2015 28 th AU G 2 nd SH I FT ]

8.56

Thermal Engineering

62. Which one of the following modes of heat transfer would takes place predominantly from boiler furnace to water wall (a) Convection (b) Conduction (c) Radiation (d) Conduction and Convection [RRB JE 2015 28 th AU G 3 rd SH I FT ]

63. Heat pipe is widely used now a days, use (a) It acts as an insulator (b) It acts as conductor and insulator (c) It acts as a superconductor (d) It acts as afin [RRB JE 2015 29 th AU G 1 st SH I FT ]

64. Ice (a) (b) (c) (d)

is very close to a Gray body Black body White body Specular body

65. In which one of the following materials is the heat propagation minimum due to conduction heat transfer? (a) Lead (b) Copper (c) Water (d) Air [RRB JE 2015 29 th AU G 2 nd SH I FT ]

LEVEL-2 1. Assume t hat a 1 t on air condit i oner is r equir ed t o cool a r oom of size 14' × 14' × 14'. H ow many 1 t on ACs woul d be r equir ed for a hall of size of 24' × 24' of t he same r oof hei ght as t hat of t he pr evious r oom? (a) 2 (b) 3 (c) 4 (d) 5 [RRB SSE 2014 GREEN SH I FT ]

2. Effi ciency of Car not cycle i s:

(c) 1 

T1 T1  T2

[RRB SSE 2014 RED SH I FT ]

5. I f a system in equilibr ium is subjected to a change of concent r at i on; t emper at ur e or pr essur e, t he equilibrium shifts in a dir ection that tends to undo t he effect of t he change i mposed. This i s k nown as: (a) L e Chat el ier 's Pr inciple (b) L aw of M ass Act i on (c) Van der Waal s Pr inciple (d) None of t hese [RRB SSE 2014 RED SH I FT ]

dQ for an i r r ever sible pr ocess i s: T (a) L ess t han zer o (b) Gr eat er t han zer o (c) Equal t o zer o (d) Any one of t hese

6. The val ue of 

[RRB SSE 2014 RED SH I FT ]

[RRB JE 2015 29 th AU G 1 st SH I FT ]

Q1 (a) 1  Q1  Q 2

4. Which law of t her modynamics defines Ent r opy? (a) Zer oth (b) Fir st (c) Second (d) Thir d

T (b) 1  1 T2 (d) 1 

Q1 Q2

[RRB SSE 2014 GREEN SH I FT ]

3. I n a buil di ng, wat er is t o be pumped t o a height of 10m at t h e r at e of 1 l i t r e/second. Power r equir ement woul d be appr ox: (Tak e ‘g’ 10 m/scc2) (a) 10 Wat t s (b) 100 Wat t s (c) 500 Wat t s (d) 1 kW [RRB SSE 2014 GREEN SH I FT ]

7. Which of t he fol lowi ng cycles is used i n t her mal power plant s ? (a) Rankine (b) Car not (c) Ott o (d) Joule [RRB SSE 2014 RED SH I FT ]

8. The char act er i st ic equat ion of Gases PV = nRT holds good for : (a) M onoat omi c Gases (b) Di at omi c Gases (c) I deal Gases (d) Real Gases [RRB SSE 2014 RED SH I FT ]

9. For an I deal Gas, t he change in Ent halpy (H ) for an el ement al change i n t emper at ur e (T) i s gi ven by: (wher e Cp = H eat capacit y at Const ant Pr essur e; CV = H eat capacit y at Const ant Vol ume) (a) CV.T (c)

CV .T CP

(b)

CP .T CV

(d) CP .T [RRB SSE 2014 RED SH I FT ]

10. A per fect gas at 27 °C i s heat ed at const ant pr essur e t i l l i t s vol ume i s doubl ed. The fi nal t emper at ur e is : (a) 54°C (b) 108°C (c) 327°C (d) 600°C [RRB SSE 2014 RED SH I FT ]

11. Zer ot h law of t her modynami cs i s t he basi s of (a) Pr essur e measur ement (b) Temper at ur e measur ement (c) Densit y measur ement (d) Vi scosi t y measur ement [RRB SSE 2015 1 st SEP 1 st SH I FT ]

Thermal Engineering

12. A pr ocess i n whi ch no heat cr osses t he boundar y of t he syst em is call ed (a) I deal pr ocess (b) Adiabat ic pr ocess (c) I sot her mal pr ocess (d) I sobar i c pr ocess [RRB SSE 2015 1 st SEP 1 st SH I FT ]

13. The ent halpy of a subst ance is defi ned as (a) h = pv+RT (b) h = u+pT (c) h = u+pv (d) h = u-pv [RRB SSE 2015 1 st SEP 2 nd SH I FT ]

14. Cycl ic int egr al of any pr oper t y is (a) Zer o (b) One (c) I nfinit e (d) Two [RRB SSE 2015 1 st SEP 2 nd SH I FT ]

15. The machi ne whi ch vi ol at es t he fi r st l aw of t her modynami cs is k nown as (a) PMM-I (b) PMM-II (c) PMM-I II (d) H eat engi ne [RRB SSE 2015 1 st SEP 2 nd SH I FT ]

16. The device in which t he wor k is done by t he fluid at t he expense of i t s ent hal py is k nown as (a) Compr essor (b) Thr ot t ling devi ce (c) Tur bi ne (d) H eat exchanger [RRB SSE 2015 1 st SEP 2 nd SH I FT ]

17. A device which incr eases t he vel ocit y of a fluid at t he expense of i t s pr essur e dr op is known as (a) Diffuser (b) Nozzle (c) Thr ot t li ng devi ce (d) H eat t r ansfer [RRB SSE 2015 1 st SEP 2 nd SH I FT ]

18. The Pr andtl number for liquid, which has velocity boundar y l ayer much t h i ck er t h an t her m al boundar y layer, is (a) Pr >> l (b) Pr <<1 (c) Pr  1 (d) Pr = 0 [RRB SSE 2015 1 st SEP 2 nd SH I FT ]

19. A fluid flow i n which t he densi t y of t he fluid does not change dur ing flowing is called as (a) I ncompr essible (b) Unifor m (c) Compr essible (d) Non-linear [RRB SSE 2015 1 st SEP 2 nd SH I FT ]

20. The for mat i on of bubbles of vapour i n a flowi ng fl ui d i s called as (a) Cavit ation (b) Cor r osi on (c) Gasification (d) Boiling [RRB SSE 2015 1 st SEP 2 nd SH I FT ]

21. A lar ge body of infinit e heat capacit y is known as (a) Ther mal Ener gy Reser voir (TER) (b) M echani cal ener gy r eser voi r (M ER)

8.57

(c) Ther momet er (d) H eat pump [RRB SSE 2015 1 st SEP 3 rd SH I FT ]

22. "H eat al ways fl ows fr om a body at a hi gher t emper at ur e t o a body at a lower t emper at ur e" is t he st at ement of (a) Fi r st l aw of t her modynami cs (b) Second law of t her modynami cs (c) Thir d l aw of t her modynami cs (d) Zer ot h law of t her modynami cs [RRB SSE 2015 1 st SEP 3 rd SH I FT ]

23. The net wor k out put for any heat engi ne i s given by (a) QL – QH (b) QL + QH (c) QH – QL (d) 2QL Wher e, QH = magnit ude of heat t r ansfer bet ween heat engi ne and sour ce QL = magnit ude of heat t r ansfer bet ween heat engi ne and si nk [RRB SSE 2015 1 st SEP 3 rd SH I FT ]

24. A cyclic device which pr oduces wor k continuously at t he expense of heat i nput is k nown as (a) Refr i ger at or (b) H eat pump (c) H eat engi ne (d) Thr ot t li ng [RRB SSE 2015 1 st SEP 3 rd SH I FT ]

25. A li qui d for ms an i nt er face wit h anot her li quid or gas; t he sur face ener gy per unit ar ea of t he int er face is k nown as (a) Sur face t ensi on (b) Specific ener gy (c) Specific heat (d) Suct ion ener gy [RRB SSE 2015 1 st SEP 3 rd SH I FT ]

26. Accor di ng t o Dar cy's law, for l aminar fl ow in a sat ur at ed soil , t he r at e of fl ow is pr opor t ional t o t he (a) Cr oss sect i onal ar ea (b) H ydr aulic gr adient (c) Coefficient of per meabi li t y (d) Por osi t y of t he soil [RRB SSE 2015 1 st SEP 3 rd SH I FT ]

27. The effi ci ency of per pet ual mot i on machine-l l (PMM -I I ) is (a) 0% (b) 50% (c) 75% (d) 100% [RRB SSE 2015 2 nd SEP 1 st SH I FT ]

28. A ther mal ener gy r eser voir t hat supplies the heat is k nown as (a) Refr iger at or (b) Sour ce (c) Sink (d) H eat engi ne [RRB SSE 2015 2 nd SEP 1 st SH I FT ]

8.58

Thermal Engineering

29. Ther mal efficiency of heat engine may be defined as (a) l – (Q2/Q1) (b) 1 – Q1/Q2) (c) (Q1/Q2) 1 (d) (Q2/Q1) – 1 Wher e, O1 = magnit ude of heat t r ansfer bet ween heat engi ne and sour ce Q2 = magnit ude of heat t r ansfer bet ween heat engi ne and sink. [RRB SSE 2015 2 nd SEP 1 st SH I FT ]

30. “ Th eor et i cal l i m i t s” f or t h e per for m an ce of com m on l y u sed en gi n eer i n g sy st em s i s det er mi ned by (a) Zer ot h law of t her modynami cs (b) Fi r st law of t her modynami cs (c) Second law of t her modynami cs (d) Thir d law of t her modynami cs [RRB SSE 2015 2 nd SEP 1 st SH I FT ]

31. Radi at i on of a bl ack body, i n t er m s of i t s t emper at ur e foll ows: (a) Newt on's law of cooli ng (b) Pl ank's l aw (c) St efan's l aw (d) Ei nst ei n Bose equat i on [RRB SSE 2014 GREEN SH I FT ]

32. Consi der t hat t wo sol i d bodi es A and B ar e t ou ch i n g each ot h er an d t r ansmi t t i ng h eat t hr ough conduct i on. I n t he gr aph bel ow, OX r epr esent s t he fir st body and XY r epr esent s t he second body.

T4

34. The bendi ng of bimet alli c st r ips dur ing r ise in t emper at ur e is due t o di ffer ence i n t heir : (a) Coefficient of li near expansi on (b) Thick ness (c) Ther mal conduct ivit i es (d) Elast i c pr oper t ies [RRB SSE 2014 RED SH I FT ]

35. St eady St at e H eat flow impli es (a) negl igi bl e flow of heat (b) no di ffer ence of t emper at ur e bet ween t he bodies (c) const ant heat fl ow r at e i .e. heat fl ow r at e independent of t i me (d) unifor m r at e i n t emper at ur e r ise of a body [RRB SSE 2014 YEL L OW SH I FT ]

36. Which of t he fol lowi ng fl uid flow condi t ions has hi gh heat t r ansfer coeffici ent ? (a) Fr ee convect i on in air (b) For ced convect i on in air (c) Fr ee convect i on in wat er (d) Condensat ion of st eam [RRB SSE 2015 1 st SEP 1 st SH I FT ]

37. The value solar const ant is (a) 1353 k W/m 2 (c) 1353 J/m 2

(b) 1353 W/m 2 (d) 135 k J/m 2 [RRB SSE 2015 1 st SEP 1 st SH I FT ]

38. "At t her mal equi libr i um, t he r at i o of t he t ot al em i ssi ve power t o t h e t ot al absor pt i vi t y i s const ant for all bodies" is known as (a) K ir chhoff's law

T3

(b) Wi en's displacement l aw

T2

(c) M axwell 's t heor y (d) St efan Bolt zmann law [RRB SSE 2015 1 st SEP 2 nd SH I FT ]

T1 X Y O St at e Tr ue (T) or Fal se(F). 1) Temper at ur e gr adient is mor e in A t han in B 2) The heat fl ow is det er mined by Four ier 's l aw 3) A r ea u n der t h e cu r v e r epr esen t s h eat dissipat i on r at e. (a) T, T, T (b) T, T, F (c) T, F, T (d) F, F, T [RRB SSE 2014 GREEN SH I FT ]

33. St efan Bol t zmann L aw i s appl i cabl e for heat Tr ansfer by : (a) Conduction (b) Convect i on (c) Radiation

(d) Al l of t hese [RRB SSE 2014 RED SH I FT ]

39. T h e su bst an ce f or w h i ch Pr an dt l n u m ber (Pr ) < < 1 is char act er ized by (a) H i gh r at e of heat diffusi on (b) L ow r at e of heat di ffusi on (c) H i gh r at e of mass diffusi on (d) L ow r at e of mass di ffusi on [RRB SSE 2015 1 st SEP 3 rd SH I FT ]

40. The cr itical radius of insulation of cylinder is given by (a) k /h (b) 2k/h (c) k/2h (d) k/4h [RRB SSE 2015 1 st SEP 3 rd SH I FT ]

Thermal Engineering

41. M ax i m u m spect r al em i ssi v e pow er t emper at ur e (T) is pr opor t ional t o (a) T 4 (b) T 5 5/2 (c) T (d) T 3

8.59

at

44. The peak fr equency at whi ch ener gy is r adi at ed fr om a bl ackbody r adi at or is dependent on (a) col or (b) distance (c) int ensit y (d) t emper at ur e

[RRB SSE 2015 1 st SEP 3 rd SH I FT ]

[RRB SSE 2015 2 nd SEP 2 nd SH I FT ]

42. The r adiation ener gy emit ted by the Sun between t he wavel engt h of 0.4 n t o 0.76 m is known as (a) ult r aviolet r adiat ion (b) infr ar ed r adiat ion (c) t her mal r adiat ion (d) visible r adiat ion [RRB SSE 2015 2

nd

45. H eat t r ansfer t ak es place accor ding t o (a) Zer ot h L aw of Ther modynami cs (b) Fi r st L aw of Ther modynami cs (c) Second L aw of Ther modynami cs (d) Thir d L aw of Ther modynamics. [RRB SSE 2015 2 nd SEP 3 rd SH I FT ]

SEP 1 SH I FT ] st

43. To r educe t he heat loss wit h i nsulat ion i n spher e t he fol lowing condi t ion shoul d be sat isfi ed (a) r i nsul at i on t hi ck ne >> r cr it ical (b) r i nsul at i on t hick ness << r cr i t i cal (c) r i nsulat ion t hi ck ness = r cr i t i cal (d) r insulat ion t hick ness = 0 [RRB SSE 2015 2 nd SEP 1 st SH I FT ]

AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (c)

2. (c)

3. (c)

4. (c)

5. (b)

6. (b)

7. (c)

8. (a)

9. (c)

10. (c)

11. (d)

12. (c)

13. (b)

14. (d)

15. (b)

16. (c)

17. (b)

18. (b)

19. (b)

20. (c)

21. (d)

22. (c)

23. (d)

24. (b)

25. (a)

26. (a)

27. (d)

28. (c)

29. (b)

30. (a)

31. (c)

32. (c)

33. (a)

34. (d)

35. (a)

36. (a)

37. (a)

38. (c)

39. (d)

40. (d)

LEVEL-1 1. (b)

2. (c)

3. (c)

4. (d)

5. (b)

6. (a)

7. (d)

8. (b)

9. (b)

10. (d)

11. (c)

12. (b)

13. (d)

14. (b)

15. (c)

16. (a)

17. (d)

18. (b)

19. (c)

20. (a)

21. (b)

22. (d)

23. (b)

24. (a)

25. (b)

26. (a)

27. (a)

28. (b)

29. (d)

30. (a)

31. (b)

32. (a)

33. (a)

34. (a)

35. (c)

36. (d)

37. (d)

38. (b)

39. (a)

40. (d)

41. (b)

42. (a)

43. (c)

44. (b)

45. (c)

46. (c)

47. (c)

48. (d)

49. (d)

50. (a)

51. (d)

52. (c)

53. (b)

54. (a)

55. (d)

56. (a)

57. (d)

58. (b)

59. (c)

60. (c)

61. (a)

62. (c)

63. (c)

64. (b)

65. (d)

LEVEL-2 1. (b)

2. (b)

3. (b)

4. (c)

5. (a)

6. (a)

7. (a)

8. (c)

9. (d)

10. (c)

11. (b)

12. (b)

13. (c)

14. (a)

15. (a)

16. (c)

17. (b)

18. (a)

19. (a)

20. (a)

21. (a)

22. (b)

23. (c)

24. (c)

25. (a)

26. (b)

27. (d)

28. (b)

29. (a)

30. (c)

31. (c)

32. (c)

33. (c)

34. (a)

35. (c)

36. (d)

37. (b)

38. (a)

39. (a)

40. (a)

41. (b)

42. (d)

43. (a)

44. (d)

45. (c)

8.60

Thermal Engineering

EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S 1. An isochoric process, also called a constantvolume process, an isovolumetric process, or an isometric process, is a thermodynamic process during which the volume of the closed system undergoing such a process remains constant. 2. Coefficient of performance of a commercially used refrigerator would be close to 1.5 3. Thermal equilibrium means that there is no heat transfer going on between the bodies, which simply means that the bodies are at the same temperature. Heat content is a whole different concept - it is the amount of heat contained in a body, which is crucially dependent on mass. 4. Newton's law of cooling applies to convective heat transfer; it does not apply to thermal radiation.Newton's law of cooling states that the rate of heat exchange between an object and its surroundings is proportional to the difference in temperature between the object and the surroundings. Temperature 100 80 60 40 20 0

10

20

30

40

50

60

Time

5. Latent heat process is a process in which temperature remains constant and phase change takes place. 6. Ice will not sublimate at atmospheric conditions. 7. Due to change in momentum. 8. The Zeroth law of thermodynamics is the basis for measurement of temperature and setting its scale. In simple word, Zeroth law of thermodynamics says that "When two bodies are separately in thermal equilibrium with the third body, then the two are also in thermal equilibrium with each other." 9. Boyle's law (sometimes referred to as the BoyleMariottelaw, or Mariotte's law) is an experimental gas law that describes how the pressure of a gas tends to increase as the volume of the container decreases.

10. Entropy seems to tend to some maximum value. 11. The formulas for converting between degree Celsius and degree Fahrenheit are: °F = (°C – 9/5) + 32 °C = (°F – 32) – 5/9 To find the temperature when both are equal, we use an old algebra trick and just set ºF = ºC and solve one of the equations. °C = (°C × 9/5) + 32 °C – (°C × 9/5) = 32 –4/5 × °C = 32 °C = –32 × 5/4 °C = –40 °F = (°F × 9/5) + 32 °F - (°F × 9/5) = 32 -4/5 – °F = 32 °F = – 32 – 5/4 °F = –40 So the temperature when both the Celsius and Fahrenheit scales are the same is -40 degrees. 12. There are mainly three types of system: Open System: Both mass transfer and energy transfer can take place. Closed System: Only energy transfer can take place. Isolated System: Both mass transfer and energy transfer can not take place. 13. Absolute pressure becomes zero when molecular momentum of the system becomes zero. 14. The zeroth law of thermodynamics states that if two thermodynamic systems each are in thermal equilibrium with a third, then they are in thermal equilibrium with each other. Accordingly, thermal equilibrium between systems is a transitive relation. 15. In fluid flow, the line of constant piezometric head passes through two points which have the same Velocity. 16. The boat will start moving away from the shore. 17. Relative humidity is the ratio of the partial pressure of water vapor to the equilibrium vapor pressure of water at a given temperature. Relative humidity depends on temperature and the pressure of the system of interest. The same amount of water vapor results in higher relative humidity in cool air than warm air.

Thermal Engineering

18. Pressure, temperature and specific volume are intensive property. Heat capacity is an extensive property. Extensive property is a property of matter that changes as the amount of matter changes. 19. Kelvin Planck's law deals with conservation of heat into work. 20. Kaplan turbine is used for low head. Low head turbines are those turbines with a head of 20 metres (66 ft) or less to produce energy. 21. An isolated system does not exchange energy or matter with its surroundings. A system and its environment put together constitute an isolated system.

to its volume (V), provided that the temperature of the gas remains constant. 31. Heat is closely related with temperature. T1  T2 T1  T2  32. T1 T1 [Both engines equally efficient]



33. 34. 35.

dT T 22. ds  C v 23. Zeroth law of thermodynamics forms the basis of measurement of temperature. It states that when two bodies are in thermal equilibrium with a third body, then they are also in thermal equilibrium with each other. 24. The first law of thermodynamics makes the use of the key concepts of internal energy, heat and system work.

U  Q  W U  Change in internal energy

25.

26.

27.

28. 29.

30.

Q  Heat added to the system W  Work done by system Otto cycle is an idealized thermodynamic cycle which consists of two isentropic (reversible adiabatic) processes and two isochoric (constant volume) processes. Control volume: Control volume is a volume fixed in space or moving with constant velocity through which the fluid (gas or liquid) flows. Mass and volume are extensive properties. Intensive properties are those properties that does not depend on the system size or the amount of material in the system, such as density, pressure etc. If temperature increases, the kinematic viscosity of gases increases. First law of thermodynamics deals with conservation of energy. The law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but can be neither created nor destroyed. Boyle's law: According to Boyle's law, the pressure (P) of a given mass of gas is inversely proportional

8.61

36.

37.

900  T2 T2  400  900 T2

 T2 = 600 Kelvin Work done in free expansion process is always zero. Internal energy of an ideal gas depends only on temperature, not on pressure and volume. An isentropic process is an idealized thermodynamic process that is both adiabatic and reversible. The work transfers of the system are frictionless and there is no transfer of heat or matter. Economiser: Economizers are generally heat exchangers which are designed to exchange heat with the fluid, generally water. Pelton wheel turbine, turgo turbines are high head turbine.

38. du =  Q –  W

 du  Tds – pdv  Tds  du  pdv q1 39. PMM1 is closely related with first law of thermodynamics. It violates the first law of thermodynamics. Heat

Heat

Work

Work (a) PMM-1

(b) reversed PMM-1

40. Isolated system is a system that can not exchange either energy or matter outside the boundaries of the system.

LEVEL-1 1. In heat transfer analysis, thermal diffusivity is the thermal conductivity divided by density and specific heat capacity at constant pressure. It measures the rate of transfer of heat of a material from the hot side to the cold side. It has the SI derived unit of m²/s.



heat conducted k  heat stored  cp

2. Fins are used to increase heat transfer rate.

8.62

Thermal Engineering

3. Heat is mainly transferred by conduction, convection and radiation in Boiler furnaces. 4. Heat can be transferred from one place to another by three methods: conduction in solids, convection of fluids (liquids or gases), and radiation through anything that will allow radiation to pass. The method used to transfer heat is usually the one that is the most efficient. 5. Collisions and diffusion. 6. Heat transfer between two bodies only takes place when temperature of one body is different from temperature of other body. 7. Stephan Boltzman law is applicable to black body, theoretical surfaces that absorb all incident heat radiation. 8. Heat transfer takes place in boiler furnaces through three modes, namely conduction, convection and radiation. 9. Overall heat transfer concept is used in the cobined mode of heat transfer of conduction and convection. 10. Unit of thermal conductivity in S.I. unit i W/mk. 11. An object that absorbs all radiation falling on it, at all wavelength is called a black body. 12. Fourier's law of heat conduction is applicable only for one dimensional cases. 13. When the body is under thermal equilibrium, then its absorptivity is equal to its emissivity. 14. Metallic bonds are made from a lattice of irons in a 'cloud' of free electrons. These free electrons are responsible for the ability of metals to conduct heat. 15. Heat conduction in gases is due to elastic impact of molecules. 16. In solids, flow of heat is through conduction heat transfer method. 17. Black body is that type of body which absorbs all radiant energy at all wavelengths. 18. Convection is the heat transfer due to bulk movement of molecules within fluids such as gases and liquids. 19. Solid ice has highest thermal conductivity than boiling water, steam and rain water. Thermal conductivity is evaluated primarily in terms of the Fourier's law for heat conduction. 20. Lorentz number is defined as the ratio of thermal conductivity to electrical conductivity at a given temperature.

K L T Where,

21.

22.

23.

24.

25.

L = Lorentz number  = electrical conductivity K = Thermal conductivity Effectiveness of fin: Effectiveness of fin is defined as the ratio of actual heat transfer that takes place from the fin to the heat that would be dissipated from the same surfaces area without fin. It will be maximum with free convection. Radiation is the heat transfer due to emission of electromagnetic waves. From boiler furnace to water wall, radiation modes of heat transfer takes place. Heat pipe acts as a superconductor. Heat pipes are transport mechanisms that can carry heat fluxes ranging from 10 w/cm2 to 20 kw/cm2 at nearly the speed of sound (340 m/sec.) Ice is very close to black body because ice is a very good absorber of radiation outside the visible region. Because air has lowest value of thermal conductivity amongst given options.

LEVEL-2 1. 1 t onne AC's r equir ed for size of r oom = 14 × 14 × 14 = 2744 unit Now, for size of r oom = 24 × 24 × 14 = 8064 unit So, number of t onne of AC's r equir ed

8064  2.938 = 3 uni t (Appr ox) 2744 2. T 2 i s sour ce t emper at ur e T 1 is sink t emper at ur e Car n ot cy cl e ef f i ci en cy i s a f u n ct i on of t emper at ur e l imit s onl y. for t his 

T car not  1  1 T2 3. Given: H eight = 10 m Di schar ge(Q) Rat e = 1 l it r e/second =

1 m 3/sec 1 1000

H er e, Power 



Wor k mgh   Qgh Time t

PVgh  PQgh t

 10000  = 100 wat t

1  10  10 1000

Thermal Engineering

4. The second law of t her modynami cs st at es t hat t he st at e of ent r opy of t he ent ir e uni ver se, as an isol at ed syst em, wi ll al ways incr ease over t ime. The second law also st at es t hat t he changes in t h e en t r opy i n t h e u n i v er se can n ev er be negat i ve. 5. L e-chat el ier 's pr inci ple: A pr i nci pl e st at i ng t hat i f a const r aint (such as a change i n pr essur e, t emper at ur e or concent r at ion of a r eact ant ) is appl i ed t o a sy st em i n equ i l i br i u m t h e equi libr i um wi ll shift so as t o t end t o count er act t he effect of t he const r aint . 6. Value of 

dQ for an i r r ever i sble pr ocess i s less T

t han zer o. 7. Rank ine cycle is used i n t her mal power plant s. 8. I deal Gas equat i on is given by PV  nRT 9. The change in ent halpy(  H ) is given by, Cp.  T.. 10. T 1 = 27 + 273 = 300K V1 = V V 2 = 2V T2 = ? By using,

P1 V1 P2 V2  T1 T2 Gi ven, P1 = P2 So,

V1 V1  T1 T2



V 2V  300 T2

 T 2 = 300 × 2 = 600K = (600 – 273) °C = 327°C

11. The zer oth law is incr edibly impor tant as it allows us t o define t he concept of t emper at ur e scale. I f two systems ar e each in ther mal equilibr ium with t hir d, t hey ar e al so in t her mal equil ibr ium wi t h each ot her. 12. I n t her modynamics, an adi abat i c pr ocess is one t hat occur s wi t hout t r ansfer of heat or mass of substances between a thermodynamic system and it s sur r oundings. I n an adiabat ic pr ocess, ener gy is t r ansfer r ed t o t he sur r oundi ng only as wor k. 13. Ent hal py of a syst em is equal t o t he syst em’s int er nal ener gy pl us t he pr oduct of i t s pr essur e and vol ume. h = u + pv

8.63

14. Cycl ic int egr al of any pr oper t y i s zer o. 15. PM M -I  PM M -I st ands for per pet ual mot i on machine-I . I t is a hypot het ical machine whi ch can pr oduce useful ener gy (wor k) wi t hout any sour ce or which can pr oduce mor e ener gy t han con su m ed. I t v i ol at es t h e f i r st l aw of t her modynami cs. 16. Tur bine i s a device designed t o do wor k at t he expense of a decr ease i n eit her t he ent halpy of a fl ui d or t he k inet i c ener gy of t he fluid or bot h. 17. Nozzle incr eases t he velocit y of t he fluid at t he expense of it s pr essur e dr op. 18. Pr andt l number for li qui d, whi ch has vel ocit y boundar y l ayer much t h i ck er t h an t her m al boundar y layer, i s gi ven by “ Pr >> 1” . 19. I ncompr essible fluid is a fluid whose density does n ot ch an ge wh en t h e pr essur e chan ges. I n i ncompr essi bl e fl ow, t he mat er i al densi t y i s const ant wit hi n a flui d par cel. 20. For mat ion of bubbl es of vapour in a fl owing fluid is cal led as cavit at ion. 21. Ther mal Ener gy Reser voi r : A t her mal ener gy r eser voi r (TER) i s defi ned as a l ar ge body of i n f i n i t e h eat capaci t y, wh i ch i s capabl e of absor bi ng or r eject ing an unlimit ed quant it y of heat without suffer ing appr eciable changes in it s t her modynamic co-or di nat es. 22. Second law of t her modynami cs st at es t hat heat f l ows n at u r al l y f r om an obj ect at a hi gh er t emper atur e t o an object at a lower t emper at ur e, and heat does not fl ow in t he opposit e dir ect i on of i t s own. 23. QH – QL wher e, QH = M agnit ude of heat t r ansfer bet ween heat engi ne and sour ce. QL = magnit ude of heat t r ansfer bet ween heat engi ne and sink. 24. H eat Engi ne: A heat engi ne i s a devi ce t hat conver t s chemi cal ener gy t o heat or t her mal ener gy and t hen t o mechani cal ener gy or t o el ect r i cal ener gy. I t i s a cycl i c devi ce whi ch pr oduces wor k cont i nuousl y at t he expense of heat input . 25. Sur face Tension: Sur face t ension is measur ed as t he ener gy r equi r ed t o incr ease t he sur face ar ea of a li quid by a unit of ar ea. The sur face t ensi on of a l i qu i d r esu l t s f r om an i m bal an ce of i nt er mol ecul ar at t r act i ve for ces, t he cohesi ve for ces bet ween t he molecules.

8.64

Thermal Engineering

26. H ydr aul ic Gr adient : The hydr aul ic gr adient is a vect or gr adi ent bet ween t wo or mor e hydr aul ic head measur ement s over t he lengt h of t he fl ow pat h. Accor di ng t o dar cy’s l aw, for l aminar fl ow in a sat ur ated soil, the r at e of flow is pr opor t ional t o hydr auli c gr adient . 27. P M M – II Machines are those machines which violate the second law of thermodynamics because such machines will absorb continuously heat energy from a single thermal reservoir and will convert the aborbed heat energy completely into work energy. P M M II efficiency is approximately 100%

Heat Source

th 

Q Wnet Qnet  1 2 Q1 Q1 Q1

where, Q2  Heat rejected Q1  Heat supplied Source

T1

Q Heat engine

W

Q2

Q

PMM-2

29. Thermal efficiency of Heat engine,

Sink

W=Q

28. Source is a thermal energy reservoir that supplies the heat and Sink is a thermal energy reservoir that receives the heat.

Also,  th (rev.cycle) = 1

T2

 T2 T1

30. Second law of thermodynamics is used in determining the thererical limits for the the performance of commonly used engineering systems, such as heat engines and refrigerators etc.

PRACTI CE PAPER CBT-I M ATH EM ATI CS 1. Ratio of the time taken by A alone to complete a certain piece of work while that by B and C to complete the same work together is 4 : 3. If the time taken by C to complete the work alone is 8 days and the ratio of the efficiencies of B and C is 1 : 3, then find the time in which A and B can complete the work while working together. (a) 4.5 days

(b) 6 days

(c) 7.5 days

(d) 8 days

2. The length of each side of a rhombus is 29 cm while that of one of the diagonals is 40 cm. What will be the area of the rhombus? (a) 840 cm2 (c) 820 cm2

(b) 880 cm2 (d) 800 cm2

9. If (cos2A – sinA)(sinA + cos2A) = –1, then find the value of 2 + cos2A. (a) 0 (b) sec2A (c) 1 (d) 2 10. If four interior angles of a pentagon are 140°, 90°, 70° and 80°, then find the value of the fifth interior angle. (a) 20° (b) 160° (c) 140° (d) 40° 11. Find the value of secA(secA + tanA + 2tan2A. (a) 0 (c) 2 12. Find the value of

3. Three successive discounts of 15%, 12.5%, and 20% are equivalent to a single discount of (a) 38.5% (c) 39.8%

(b) 41.5% (d) 40.5%

4. A and B together have Rs.2,448. If 25% of A's amount is equal to 35% of B's amount, then find the amount with A. (a) Rs. 1,020

(b) Rs. 1,428

(c) Rs. 1,462

(d) Rs. 1,468

5. If the cost price of 50 articles is equal to the selling price of 40 articles, then find the profit or loss percent. (a) 20% (c) 25%

(b) 16.67% (d) 33.33%

6. A’s income is 25% more than B’s income while C’s income is 20% more than B's income. By what percent is A’s income more or less than C's income? (a) 5.26% (c) 4.28%

(b) 4.16% (d) 4.46%

7. In what time will a 143 m long train running at the rate of 45 km/hr cross a man coming towards it at the rate of 4.5 km/hr? (a) 11.4 sec

(b) 10.4 sec

(c) 11.64 sec

(d) 10.64 sec

8. If B : A + (A 2 +B 2 ) = 2 : 3 then find the value of B : A. (a) 8 : 15 (c) 12 : 5

(b) 24 : 7 (d) 6 : 8

(a)

(c)

3 4 3 5

cosA )– 1+sinA

(b) 1 (d) 3 (2 × 8n+1 + 4 × 23n – 1 ) for n = 2. (4 × 22n +1 – 2 × 42n +1 )

(b) –

3 4

(d) 

3 5

13. If the average age of sixteen students is 18.5 years and the average age of these students along with their teacher is 22 years, then find the age of the teacher? (a) 58 years (b) 78 years (c) 68 years (d) 76 years 14. If b +

1 1 = –1 , then find the value of b9  15 . b b

(a) 2 (c) 1

(b) –1 (d) 0

15. Medians BE and CF of ABC intersects at point O. P and Q are the midpoints of BO and CO respectively.If PQ = 3.5 cm, then find the length of BC. (a) 3.5 cm (b) 7 cm (c) 10.5 cm (d) 4.5 cm 16. In triangle PQR, S and T are the points on PQ and PR respectively, such that ST is parallel to QR and PS : SQ = 3 : 2. If RT = 5 cm, then find the ratio of the areas of triangle PST and quadrilateral STRQ. (a) 9 : 4 (b) 9 : 25 (c) 16 : 25

(d) 9 : 16

2

PRACTICE CBT-I

17. If Rs. 14,600 amounts to Rs. 16,404.56 invested in compound interest (compounded annually) for two years, then the rate of interest is (a) 5% (b) 6% (c) 7% (d) 8%

25. The difference of a number consisting of three different digits from the number formed by reversing the digits is always divisible by

18. From the top of a tower 40 3 meters high the angle of depression of the top and bottom of a pole are observed to be 45° and 60° respectively. If the pole and tower stand on the same plane, then find the distance between the top of the tower and that of the pole.

26. Pipe A can empty a tank in 12 minutes, pipe B can empty the tank in 18 minutes while the pipe C can empty the tank in 36 minutes. In how much time the tank will be empty if all three pipes are opened together?

(a) 20 3 m

(b) 20 2 m

(c) 40 2 m

(d) 8.6 m

19. A person borrows `33,000 at 20% compound interest. How much he has to pay equally at the end of each year, to settle his loan in two years? (a) `22,450 (c) `17,000

(b) `21,600 (d) `19,600

20. A faulty clock shows correct time at 6 a.m. It lags 1sec at the end of every 10 sec. What will be the time shown by the clock at 9 p.m.? (a) 7:00 p.m.

(b) 7:30 p.m.

(c) 8:00 p.m.

(d) 8:30 p.m.

21. Given that x2 – y2 = 60. If x = 6 + y, then the average of x and y is (a) 10

(b) 6

(c) 5

(d) 3

1

22. What is the value of x, if x 

?

3

3 5

(a)

31 110

(b)

31 101

(c)

41 101

(d)

3 11

7 2

a 1 b 2 12a2  8c 2  and  , then is equal to b 4 c 3 33a2  c 2 (a) –4 (b) –1

23. If

2 17 24. O is the incentre of the ΔABC. From point O a perpendicular is drop on side BC such that (c) 0

(d)

perpendicular meets BC at point P. If BOP  39, then ABC is (a) 39°

(b) 78°

(c) 51°

(d) 102°

(b) 10

(c) 11

(d) Both (a) and (c)

(a) 22 minutes

(b) 10 minutes

(c) 6 minutes (d) 5 minutes 27. A circle is inscribed in an equilateral triangle of side 6 cm. A square is inscribed in this circle, then the area of the square (in cm2) is (a) 12 (b) 6 (c) 18 (d) 24 28. A conical vessel of base radius 9 cm and height 8 cm is filled with milk. If the milk leaks through a hole at the bottom of the conical vessel into a cylindrical jar of radius 3 cm, then find the level of the milk in the jar after the milk has leaked completly. (a) 12 cm (b) 15 cm (c) 24 cm (d) 27 cm

1 29. A shopkeeper sells an article at 12 % loss. If he 2 sells it for `51.80 more, then he gains 6%. What is the cost price of the article? (a) `210 (b) `240 (c) `280 (d) `300 30. Ramesh travelled from city A to city B, in which he

1 th of the total 4 distance by car. If the rest 40% of the distance is travelled in a bus, then what is the total distance? (a) 100 km (b) 120 km (c) 140 km (d) 160 km travelled 42 km by motorcycle and

4 5

(a) 9

GEN ERAL I N TELLI GEN CE & REASON I N G 31. A series is given, with one term missing. Choose the correct alternative from the given ones that will complete the series. DNP, ELS, FJV, ___?___ (a) GHZ (b) HHY (c) GIW (d) GHY 32. A series is given, with one term missing. Choose the correct alternative from the given ones that will complete the series. 8, 15, 29, 50, 78, ___?___ (a) 110 (b) 113 (c) 118

(d) 115

PRACTICE CBT-I

33. Pointing to a picture, Seetha said, “ The only son of this lady’s mother is my father”. How is the lady related to Seetha’s mother? (a) Daughter-in-law

(b) Sister-in-law

(c) Mother

(d) Sister

34. A, B, C, D and E are standing at distinct positions in a straight line with three of them facing North and the rest facing South. B is equidistant from C and E, who are at a distance of 60 m from each other facing South. E is equidistant from B and D. A is at a distance of 10 m from C’s right. A is not at any extreme end. What is the distance between A and E? (a) 40 m

(b) 30 m

(c) 60 m

(d) 50 m

(a) B

(b) Y

(c) O

(d) W

3

41. Which of the following figures correctly represents the relation between: e-payment modes, paytm and cash.

(a)

(b)

(c)

(d)

35. If UP = 37 and BUS = 42, then CUT = ? (a) 44

(b) 24

(c) 43

(d) 34

36. If + means ÷ , ÷ means –, – means × and × means + , what will be the value of the following expression: 5 × 3 – 12 + 4 ÷ 2 = ? (a) 5

(b) 9

(c) 12

(d) 20

Directions: In question nos. 42 and 43, select the related letter/word from the given alternatives.

37. In this question, some equations are solved on the basis of a certain system. On the same basis find out amongst the four alternatives for the unsolved equation.

42. F-L : I-O :: P-V : ? (a) R-Y

(b) S-X

(c) S-Y

(d) R-W

43. Baboon : Infant :: Beaver : ?

5 × 8 × 3 = 358, 9 × 4 × 0 = 094, 2 × 3 × 4 = ?

(a) Kitten

(b) Kid

(a) 243

(b) 423

(c) Joey

(d) Fawn

(c) 324

(d) 432

38. Select the missing number from the given responses. 23 54 70 1

27 55 85 15

a_bab_aab_bcaa_abc

31 56 100 ?

(b) 30

(c) 29

(d) 130

(a) 400 m

(b) 500 m

(c) 300 m

(d) 600 m

B

Y

B B

Y

(c) abac

(d) acab

Y O

(c) Pumpkin

G

W

(b) Gourd (d) Jackfruit

46. (a) 325

(b) 118

(c) 272

(d) 253

47. (a) Solar energy (c) Fossil fuel 48. (a) Man

40. Four positions of a cube are shown in the diagram. Which letter is opposite to ‘G’ in the given cube?

G

(b) bcab

45. (a) French Beans

39. Sundar started from his house at sunrise and cycled 100 m in the direction of the rising sun, then he turned left and rode for 400 m. He then turned westward and cycled for another 500 m and finally turning left rode 100 m. How far is Sundar from his house?

G

(a) bbab

Directions: In question nos. 45 to 48, find the odd word/ number from the given alternatives.

(a) 37

R

44. Which one set of letters when sequentially placed at the gaps in the given letter series will complete it?

(c) Hen

(b) Biomass (d) Radiant energy (b) Whale (d) Seal

49. Karan is taller than Mukesh and is shorter than Anil. Rakesh is shorter than Sunil, who in turn is shorter than Anil. Who is the tallest? (a) Anil

(b) Sunil

(c) Mukesh

(d) Karan

4

PRACTICE CBT-I

50. In the following question, the number of letters skipped in between adjacent letters in the series is consecutive odd numbers. Which of the following series observes this rule? (a) B D G J P

(b) C E I O V

(c) D F J P X

(d) E G K P X

51. A man and his wife have four sons and three daughters. All four sons are married and have five children each. Find the total number of members in the family. (a) 33

(b) 25

(c) 28

(d) 29

Directions: In question nos. 52 and 53, one/two statements are given followed by two/three conclusions I, II and III. You have to consider the statements to be true even if they seem to be at variance with commonly known facts. You are to decide which of the given conclusions can definitely be drawn from the given statements. 52. Statements: I.

All gloves are things.

Conclusions: All gloves are warm.

II. All warm are things. III. Some warm are things. (a) Only conclusion III follows (b) Only conclusion I follows (c) Only conclusion II follows

Answer figures: (a)

(b)

(c)

(d)

GEN ERAL AWAREN ESS 56. The first Indian foreign policy was formulated by: (a) Dr Bhimrao Ambedkar (b) Pt. Jawaharlal Nehru (d) Babu Jagjivan Ram 57. How many women members were there in the Indian Constituent Assembly? (a) 6

(b) 9

(c) 11

(d) 15

58. Which among the following was the port city during Indus Valley Civilization? (a) Kalibanga

(b) Lothal

(c) Banawali

(d) Ropar

59. Gautama Buddha preached his last sermon at:

(d) None follows 53. Statement: People succeed when they work hard.

(a) Rajgriha

(b) Pataliputra

(c) Kundalgrama

(d) Vaishali

60. Cairo, Dublin and Lima are examples of:

Conclusions: I.

Europe, London, United Kingdom

(c) Indira Gandhi

II. Some gloves are warm. I.

55. Choose from the following diagrams (A), (B), (C) and (D) the one that illustrates the relationships among three given classes:

Only hard work leads to success.

II. Honesty does not lead to success. (a) Only conclusion I follows (b) Only conclusion II follows (c) Both conclusions I and II follow (d) Neither conclusion I nor II follows 54. Which figure represents the relationship among Universe, Planets, Galaxies?

(a) Enclaves

(b) Edge Cities

(c) Primate Cities

(d) World Class Cities

61. Mughal Emperor Jahangir tomb is located at which of the following place? (a) Agra

(b) Karachi

(c) Kabul

(d) Lahore

62. Who among the following plays Mohan Veena (slide guitar)? (a) A.R.Rehman (b) Sushmit Sen

(a)

(b)

(c) Vishwa Mohan Bhatt (d) Ehsaan Noorani 63. ECOMARC is related with: (a) Goods Exported

(c)

(d)

(b) Goods Imported (c) Goods Safe for Environment (d) Only processed food

PRACTICE CBT-I

64. Maanch is a lyrical folk drama that is very popular in: (a) Odisha (b) Madhya Pradesh (c) Chhattisgarh (d) Jharkhand 65. Which soil types of Indialack fertility due to intensive leaching? (a) Red Soil (b) Laterite Soil (c) Black Soil (d) AlluvialSoil 66. The world's first Hindi speaking realistic humanoid robot has been developed by: (a) Ajay Srivastava (b) Pradeep Kumar (c) Ranjit Srivastava (d) Pankuj Sharma 67. Name the women cricket player, who is the world's highest wicket-taker in ODIs format. (a) Cathryn Lorraine Fitzpatrick (b) Jhulan Goswami (c) Betty Wilson (d) Diana Edulji 68. International Day for the Preservation of the Ozone Layer is observed every year on 16th September. The theme for International Day for the Preservation of the Ozone Layer 2018 was: (a) Keep Cool and Carry On: The Montreal Protocol (b) Protect Ozone layer for future generations (c) Unite for the protection of Ozone layer (d) No to pollutants and save to Ozone layer 69. What is the India rank in the Press Freedom Index 2018 released by Reporters Without Borders (RSF)? (a) 140 (b) 138 (c) 150 (d) 128 70. Who has become the first non-classical or jazz artist to win Pulitzer Prize for music recently? (a) Kendrick Lamar (b) Caroline Shaw (c) Kevin Puts (d) Henry Threadgill

GEN ERAL SCI EN CE 71. The disadvantages of liquid propellants are (a) Difficult to store and handle

73. Cracking in brass is caused by (a) Ammonia (b) Nitric acid (c) Hydrogen peroxide (d) Sulphuric acid 74. Which logic gate is a ‘universal gate’? (a) NAND gate

(b) AND gate

(c) XOR gate

(d) OR gate

75. Strip cropping helps to stop (a) Air pollution

(b) Soil erosion

(c) Water pollution

(d) Marine pollution

76. Which part of the human eye acts as an aperture? (a) Retina

(b) Pupil

(c) Sclera

(d) Cornea

77. The manufacture of chlorofluorocarbon compounds has been phased out under the (a) Kyoto protocol (b) Montreal protocol (c) Stockholm Convention on Persistent Organic Pollutants (d) Protocol to the Convention on Long-range Transboundary Air Pollution on the Control of Emissions of Volatile Organic Compounds 78. Which coating of mucins gives them considerable water-holding capacity and makes them resistant to proteolysis? (a) Fat coating

(b) Sugar coating

(c) Protein coating

(d) Glycoprotein coating

79. The scientific method of dating based on the analysis of patterns of tree rings is known as (a) Dendrology

(b) Dendrochronology

(c) Chronology

(d) Acanthochronology

80. Which fossil species/genus is transitional between feathered dinosaurs and modern birds?

(b) Extreme toxicity

(a) Vampyronassa

(b) Cladoselache

(c) Moderately cryogenic

(c) Archaeopteryx

(d) Rhyniognatha

(d) All of these 72. ‘ABO’ blood group is known to be a ‘Universal Recipient’ because (a) antibody A and B both are present in AB blood group

81. Which of the following is useful in ice-cream making? (a) Agar-Agar

(b) Algal extract

(c) Algal bloom

(d) Algal excreta

82. Which of the following ecological pyramid is not inverted?

(b) no antigen is present in AB blood group

(a) Energy pyramid

(c) antigen A and B both are present in AB blood group

(b) Pyramid of numbers

(d) no antibody is present in AB blood group

5

(c) Ecological pyramid of biomass (d) All of these

6

PRACTICE CBT-I

83. Which hormone regulates blood pressure and water (fluid) balance in human beings? (a) Rennin (c) Thyroid

92. Which among the following converts mechanical energy into electrical energy?

(b) Renin-angiotensin

(a) Tube Light

(b) Solar Cell

(d) Parathyroid

(c) Candle

(d) Dynamo

84. Dengue is caused by the (a) Virus

(b) Bacteria

(c) Nematode

(d) Protozoan

85. Parasitic plants have a modified root known as (a) Haustorium

(b) Vellum

(c) Prop roots

(d) Pneumatophore

86. The study of teeth is referred to as (a) Osteology

(b) Archaeology

(c) Opthalmology

(d) Otolaryngology

87. Photography Exposure is measured in (a) Lux seconds

(b) Candela

(c) Roentgen

(d) Decibels

88. The rise in temperature in the stratosphere is caused by the absorption of (a) Ultra violet radiations (b) Infra red radiations

93. Which of the following are also called as the amphibians of the plant kingdom? (a) Lichens

(b) Algae

(c) Mosses

(d) Ferns

94. Which of the following is not an example of Igneous rocks? (a) Granite

(b) Basalt

(c) Obsidian

(d) Limestone

95. Which of the following chemicals is used in food preservation? (a) Sodium benzoate

(b) Diclofenac

(c) Azadirachtin

(d) Sodium thiopental

96. Which of the following are examples of total internal reflection? (a) extreme shining in diamond (b) Looming (c) Mirage (d) All of the above

(c) Visible spectrum (d) Ions present in the stratosphere 89. Nasonov pheromone is emitted by the (a) Ants

(b) Queen bees

(c) Worker bees

(d) Drones

90. Which of the following gas is used as a popular recreational drug? (a) Neon

(b) Helium

(c) Nitrous oxide

(d) MIC

91. Which of the following gas is used for artificial ripening of fruits? (a) Ethylene

(b) Methane

(c) Propane

(d) Butane

97. When a comet travels close to the sun, the ice which melts in the direction of its propagation is called: (a) Coma

(b) Croma

(c) Nebula

(d) Parabola

98. Date of manufacture of food items fried in oil should checked before buying because oil becomes rancid due to: (a) Oxidation

(b) Reduction

(c) Hydrogenation

(d) Decrease in viscosity

99. Which of the following is not caused by any virus? (a) Measles

(b) Mumps

(c) Botulism

(d) Poliomyelitis

100. Which is the oldest technique of food preservation? (a) Refrigeration

(b) Sugaring

(c) Pickling

(d) Drying

PRACTICE CBT-I

7

AN SWERS 1. (b)

2. (a)

3. (d)

4. (b)

5. (c)

6. (b)

7. (b)

8. (c)

9. (d)

10. (b)

11. (c)

12. (d)

13. (b)

14. (a)

15. (b)

16. (d)

17. (b)

18. (c)

19. (b)

20. (b)

21. (c)

22. (a)

23. (a)

24. (d)

25. (d)

26. (c)

27. (b)

28. (c)

29. (c)

30. (b)

31. (d)

32. (b)

33. (b)

34. (d)

35. (a)

36. (c)

37. (b)

38. (c)

39. (b)

40. (c)

41. (d)

42. (c)

43. (a)

44. (d)

45. (d)

46. (c)

47. (c)

48. (c)

49. (a)

50. (c)

51. (a)

52. (a)

53. (d)

54. (d)

55. (c)

56. (b)

57. (d)

58. (b)

59. (d)

60. (c)

61. (d)

62. (c)

63. (c)

64. (b)

65. (b)

66. (c)

67. (b)

68. (a)

69. (b)

70. (a)

71. (d)

72. (d)

73. (a)

74. (a)

75. (b)

76. (b)

77. (b)

78. (b)

79. (b)

80. (c)

81. (a)

82. (a)

83. (b)

84. (a)

85. (a)

86. (a)

87. (a)

88. (a)

89. (c)

90. (c)

91. (a)

92. (d)

93. (c)

94. (d)

95. (a)

96. (d)

97. (a)

98. (a)

99. (c)

100. (d)

EXPLAN ATI ON S 1. (b) Let the total work be 8 × 3 = 24 units.

4. (b) 25% of A = 35% of B

 Work done by C in 1 day = 3 units and that by B = 1 unit.

A:B=7:5

Thus, time taken by B and C together = 6 days while that by A alone = 8 days.

Hence, required amount =

 Work done by A in 1 day = 3 units s

Hence, required time = D

2. (a)

C 20

20

A

24 = 6 days. 4

cm

= Rs. 1,428. 5. (c) Let the selling price of 1 article be Re.1  Cost price of 50 articles = Rs. 40 and their selling price = Rs. 50

Hence, there would be profit equivalent to

50 – 40 × 100 = 25%. 40 6. (b) Let the income of B = Rs. 100

cm O

29 cm

 A's income = Rs. 125 and C's income = Rs. 120

B

Hence, required percentage

AB2 = AO2 + OB2  292 = 202 + OB2  OB = 21 cm

Hence, required area 1

= 4 × 2 × 21 × 20 = 840 cm2. 3. (d) x ×

7 × 2448 12

17 7 4 119 × × = = 59.5% of x 20 8 5 200

Hence, required discount = 40.5%.

=

125 – 120 × 100 = 4.16%. 120

7. (b) Relative speed = 45 + 4.5 = 49.5 km/hr = 49.5 ×

5 = 13.75 m/s 18

Hence, required time =

143 = 10.4 seconds. 13.75

8. (c) Going through the options we get the required answer as 12 : 5.

8

PRACTICE CBT-I

9. (d) (cos2A - sinA)(sinA + cos2A) = –1 cos4A



sin2A

-

Using mid-point theorem: BC = 7 cm.

= –1

 cos4A + cos2A = 0

P

16. (d)

 cos2A (cos2A + 1) = 0

3h 5

Since cos2A can never be negative in the given range thus A = 90° and (cos2A + 2) = 2.

2h 5

Sum of all the exterior angles of any polygon is always 360°.

Q

 (180 – 140)° + (180 – 90)° + (180 – 70)° + (180 – 80)° + (180 – x)° = 360°

Required ratio

Hence, x = 160°. 11. (c) Putting A = 45° in secA(secA + tanA +

cosA ) 1 + sinA

– 2tan2A, we get the required value as 2. 3n–1

(2 × 8 12. (d)

(4 × 2

+4×2

2n +1

–2×4

)

2n+1

 3x  Area of  PST 9   . Area of  STRQ  5x  2   3x 2 16

r   17. (b) 14600 ×  1+  100 

(2

–2

4n + 3

= 16,404.56

E 45°

(23n + 4 + 23n + 1 ) 2n + 3

2

 r = 6%.

) 18. (c)

=

R

5x

2



n+1

T

3x

S

10. (b) Let the fifth angle be x°.

60°

) D

C

(210 + 27 )

3 = 7 =– . 11 5 (2 – 2 )

40 3 m

13. (b) Age of the teacher = 17 × 22 – 16 × 18.5 = 78 years. 14. (a) b +

1 = –1 b

In ABE ,

 b2 + b + 1 = 0

 (b – 1) (b2 + b + 1) = 0

tan30 

 b3 = 1

 b9 

1 15

b

AB 40 3

 AB  40 m

In EDC ,

 2.

sin 45 

A

15. (b)

B

A

40  EC  40 2 m EC

19. (b) Let the amount be `x. Then,   20  20    33000  1  100   x  1  100   x

E

F O

Q

P



3.5 cm B

C

x5 6  x  33000  6 5

PRACTICE CBT-I



x

11  39600 6

and

 x = `21,600.



= 15 hrs. = 15 × 60 × 60 = 54000 sec  Lag at the end of 54000 sec = 5400 sec = 1 hr 30 min

=

Hence, the time shown = 7:30 p.m. 2

=6

2

and x – y

b 2 = c 3

 a : b : c= 1 : 4 : 6

20. (b) Difference between the two time limits

21. (c) Given x – y

9

12a2  8c 2

12  12  8  62 = 33a2  c 2 33  12  62

12 1  24 3 11  12

24. (d)

 4.

A

= 60,

therefore, x + y= 10 O

x  y 10  Hence, the average of x and y = = 5. 2 2

C

P

ABO = OBP

1

22. (a) x =

B

3

3

4

5

=

7 2

5

ABC 2

and OBP = 90  BOP = 90  39  51

1

 x=

3

3

5

3

 x=

(100z + 10y + x) – (100x + 10y + z) = 99(z – x) = 9 × 11 × a number.

3 8 5 17

1 3

25. (d) Let the 3 digit number be 100x + 10y + z. Then,

26. (c) Let the capacity of the tank be (LCM of 12, 18 and 36) 36 units. Then,

1

 x=

 x=

4 17 2

Hence, ABC = 2  51  102.

3 93 17

1 51 3 93

93 31  .  x= 330 110

Pipe A empties

36 = 3 units, pipe B empties 12

36 36 = 2 units while pipe C empties = 1 unit 18 36 from the tank in 1 minute. Hence, required time =

36 = 6 minutes. 3  21

27. (b) Let ‘r’ be the radius of the circle and ‘a’ be the length of the side of the square. Then, ‘r’ is the inradius of the equilateral triangle.  r=

6 2 3

 3 cm

 Diameter= 2r = 2 3 cm

a 23. (a) b

1 = 4

 Diameter of the circle = Diagonal of the square  Diagonal of the square = 2 3 cm

10 PRACTICE CBT-I

2 3

 Side of the square =

2

Hence, area of the square 

36. (c) 12 

 6

6 cm 2

37. (b) 423 2

 6 cm .

28. (c) Let the level of the milk in the jar be ‘h’. Then,

38. (c) The middle column is the average of the other two columns in each row. 39. (b)

500 m

volume of the conical vessel

100 m

= volume of the cylinder jar 

50 0m

2 1 2   9   8 =   3  h 3

300 m 100 m 400 m house

 h = 24 cm. 29. (c) Let the cost price be `x. Then,

12.5  6 x 100

Using Pythogoras theorem, distance from house = 51.80 =

 x = 280.

3002  4002 = 500 m. +3

42. (c) 30. (b) Let the distance between city A and city B be x km. Then,

+3 L

F

42 



x 2x  4 5

42 

=x

 5  8 x 20

+3

=x

+3 P

V +6

2

2

3

3

2

N   L   J  H 3

P   S   V  Y 32. (b) 8 + 7 = 15 15 + 14 = 29 29 + 21 = 50 78 + 35 = 113 33. (b) The lady in the picture in Seetha’s father’s sister. So she is Seetha’s mother’s sister-in-law. E

B 30m

Y

S +6

43. (a) The young one of a baboon is called an infant while the young one of a beaver is called a kitten. Hence, option (A) is the correct answer. 44. (d) The series is a,ab,abc,a,ab,abc,… 45. (d) Gourd, Pumpkin and French Beans grow on vines while Jackfruit grows on trees. Hence, option (d) is the answer. 46. (c) Sum of the digits of 272 is 11. While in rest of the three options, sum of the digits of the numbers is 10.

50 + 28 = 78

30m

+6

Similarly,

1 1 1 31. (d) D  E  F  G

D

O

I

+6

 x = 120.

34. (d)

400 m

30 m A C 10m

(A who is not at the extreme end of the row) Distance between A and E = 50 m. 35. (a) UP = 37, B U S = 2 + 21 + 19 = 42  C U T = 3 + 21 + 20 = 44

47. (c) Fossil fuel is the answer since it is a non renewable source of energy, while energy resources in options (a), (b) and (d) are renewable energy sources. 48. (c) Animals in options (a), (b) and (d) are mammals while hen is a bird. Hence, option (c) is the correct answer. 49. (a) The order of the heights of five given persons will be: Anil > Karan >Mukesh and Anil > Sunil > Rakesh Hence, Anil is the tallest among the five.

PRACTICE CBT-I 11

50. (c) In the series D F J P X the number of letters skipped between adjacent letters is consecutive odd numbers starting from 1. 2

4

6

8

D   F   J   P  X 51. (a) Total members = 2 + 3 + 4 × 2 + 4 × 5 = 33. 52. (a)

Things Gloves

Warm

As is clear from the Venn diagram, only conclusion (III) follows. Hence, option (a) is the correct answer. 53. (d) The given statement shows a cause and effect relationship. It says that when people work hard, they succeed. Neither does the statement say anything about hard work being the only way to succeed nor does it say anything about any other way that will or will not lead to success. Therefore, we can say that none of the conclusions follow. Hence, option (d) is the correct answer. 54. (d) Universe contains millions of galaxies and each galaxy contains millions of stars and planets. 55. (c) London is a part of United Kingdom and United Kingdom is a part of Europe. 56. (b) India's first foreign policy was conducted under the guidance of Pt. Jawaharlal Nehru, the first Prime Minister of independent India. 57. (d) Among the 299 members of the assembly, 15 were women who had either been voted or chosen to represent their provinces, who left their mark on the making of the republic. 59. (d) Lord Buddha delivered his last sermon in Vaishali, informing his disciples about his impending Mahaparinirvana and his desire to go to Kushinagar. 60. (c) A primate city is a major city that works as the financial, political, and population center of a country and is not rivaled in any of these aspects by any other city in that country. Normally, a primate city must be at least twice as populous as the second largest city in the country. 61. (d) The tombof Jahangir is located in Shahdara, a suburb of Lahore to the northwest of the city. 62. (c) Vishwa Mohan Bhatt plays Mohan Veena (slide guitar). 64. (b) Maanch is a lyrical folk drama and a form of operatic ballet that is very popular in Malwa region of Madhya Pradesh.

66. (c) The world's first Hindi speaking realistic humanoid robot has been developed by Ranjit Srivastava. The robot has been named as Rashmi. The developer hailing from Ranchi claimed that Rashmi can speak Hindi, Bhojpuri, and Marathi along with English. The Humanoid Robot uses linguistic interpretation (LI), artificial intelligence (AI), visual data and facial recognition systems, the developer said. 67. (b) Veteran Indian women's team pacer Jhulan Goswami announced her retirement from T20 Internationals. The 35-year-old will only play ODIs (as India don't play Test cricket), in which format she is the world's highest wicket-taker with 200 scalps from 169 games. 68. (a) International Day for the Preservation of the Ozone Layer is observed every year on 16th September throughout the world. The theme for International Day for the Preservation of the Ozone Layer 2018 was 'Keep Cool and Carry On: The Montreal Protocol'. 69. (b) India's ranking in the Press Freedom Index has fallen two places to 138, in an annual report of Reporters Without Borders (RSF), blaming "physical violence" against journalists like Gauri Lankesh as the key reason behind the country's low ranking. Norway topped the list of having the world's freest press for the second year in a row, the Reporters Without Borders (RSF) said while North Korea remained the most repressive country followed by Eritrea, Turkmenistan, Syria and then China. 70. (a) Kendrick Lamar becomes the first rapper to win Pulitzer Prize for music. 71. (d) The main difficulties with liquid propellants are with the oxidizers. These are generally at least moderately difficult to store and handle due to their low reactivity with common materials, may have extreme toxicity (nitric acids), moderately cryogenic (liquid oxygen), or both. 73. (a) Brass is an alloy made of copper and zinc; the proportions of zinc and copper can be varied to create a range of brasses with varying properties. Cracking in brass caused by ammonia attack. 74. (a) The NAND gate is a logic gate used to build digital logic circuits. Given two inputs, A and B, A NAND B will be true if at most one of A and B is true. In other words, A NAND B is false if both A and B are true, and true otherwise. The NAND gate is a “universal gate.” 75. (b) Strip cropping is a method of farming used when a slope is too steep or too long, or otherwise,

12 PRACTICE CBT-I

when one does not have an alternative method of preventing soil erosion. 76. (b) The eye includes a lens not dissimilar to lenses found in optical instruments such as cameras and the same principles can be applied. The pupil of the human eye is its aperture; the iris is the diaphragm that serves as the aperture stop. 77. (b) A chlorofluorocarbon (CFC) is an organic compound that contains only carbon, chlorine, and fluorine, produced as a volatile derivative of methane, ethane, and propane. They are also commonly known by the DuPont brand name Freon. The manufacture of such compounds has been phased out under the Montreal Protocol. 78. (b) The dense “sugar coating” of mucins gives them considerable water-holding capacity and makes them resistant to proteolysis. 79. (b) Dendrochronology or tree-ring dating, is the scientific method of dating based on the analysis of patterns of tree rings, also known as growth rings. Dendrochronology can date the time at which tree rings were formed, in many types of wood, to the exact calendar year. 80. (c) Archaeopteryx was an early bird that is transitional between feathered dinosaurs and modern birds. 81. (a) Agar is derived from the polysaccharide agarose, which forms the supporting structure in the cell walls of certain species of algae, and which is released on boiling. 82. (a) The pyramid of energy is drawn after taking into consideration the total quantity of energy utilized by the trophic levels in an ecosystem over a period. As the quantity of energy available for utilization in successive trophic levels is always less because there is loss of energy in each transfer, the energy pyramid will always be upright.

83. (b) The renin-angiotensin system (RAS) or the reninangiotensin-aldosterone system (RAAS) is a hormone system that regulates blood pressure and water (fluid) balance. If the renin-angiotensinaldosterone system is abnormally active, blood pressure will be too high. 84. (a) Dengue fever also known as break bone fever, is an infectious tropical disease caused by the dengue virus. Dengue is transmitted by several species of mosquito within the genus Aedes, principally A. aegypti. 85. (a) Parasitic plants have a modified root, the haustorium, that penetrates the host plant and connects to the xylem, phloem, or both. Haustoria do not penetrate the host’s cell membranes. 86. (a) The study of bones and teeth is referred to as osteology. 87. (a) In photography, exposure is the quantity of light reaching a photographic film, as determined by shutter speed and lens aperture. Exposure is measured in Lux seconds, and can be computed from exposure value (EV) and scene luminance in a specified region. 88. (a) The stratosphere is the second-lowest layer of Earth’s atmosphere. The stratosphere contains the ozone layer, which is the part of Earth’s atmosphere that contains relatively high concentrations of ozone. Abrupt rise in temperature is caused by the absorption of ultraviolet radiation (UV) radiation from the Sun by the ozone layer, which restricts turbulence and mixing. 89. (c) Nasonov pheromone is emitted by the worker bees and used for orientation.

PRACTI CE PAPER CBT-I I GEN ERAL AWAREN ESS 1. "Khalsa" was founded by(a) Gur u Gobind Singh (b) Gur u Ramdas (c) Gur u Nanak (d) Gur u Ar jun Dev 2. "M ahabhar at a" t he epic was wr it t en by(a) Vyasa

(b) Kalidasa

(c) Tulsidasa (d) Valmiki 3. An int er pr et at ion of t he I ndian Const it ut ion is based on t he spir it of t he(a) Fundament al r ight s (b) Fundament al dut ies (c) Pr eamble (d) Dir ective pr inciples 4. To be eligible for member ship of t he L ok Sabha, a per son should be at least : (a) 18 year s of age (b) 30 year s of age (c) 35 year s of age (d) 25 year s of age 5. Sever al nat ions ar e following a pr ot ocol which bi nds t hem t o r educe emi ssi on t ar get s. Thi s pr ot ocol was adopt ed in: (a) K yoto, Japan

(b) Geneva, Swit zer land

(c) New Yor k, USA

(d) Par is, Fr ance

6. Which of these r ocks would have alumina as their main component ? (a) Siliceous

(b) Ar gillaceous

(c) Calcar eous (d) I gneous 7. M at ch Col . X (Spor t sper son) and Col. Y (Spor t s): Col. X Col. Y P. Jitu Rai 1. Badmint on Q. H eena Sidhu 2. Wr est li ng R. Jwala Gut t a 3. Shoot ing S. Yogeshwar Dut t (a) P-3; Q-3, R-1, S-2 (b) P-2. Q-3, R-1, S-2 (c) P-2. Q-2. R-1. S-3 (d) P-3, Q-1. R-1. S-2

8. The slogan of Asian Games I ncheon 2014 was (a) Gr een, Cl ean and Fr iendship (b) We Cheer, We Shar e, We Win (c) Di ver si t y Shines her e (d) The Games of Your L i fe 9. What ar e t he t hr ee val ues cher i shed by t he Commonweal t h Games? (a) Get Set , Go, and Pl ay . (b) Fast er, H igher, St r onger. (c) Di ver sit y Shi nes her e. (d) H umanit y, Equali t y, Dest iny. 10. What is t he li kely change t hat For ei gn Di r ect I n v est m en t w i l l br i n g abou t i n I n d i an H or t icul t ur e? (a) I ncr eased use of fer t ili zer and hybr id seeds. (b) Sl owing down of or ganic far mi ng . (c) M or e invest ment in fl or i cult ur e in hil l r eas. (d) Expansion of ar ea under plant at i on 11. Wh at i n for mat i on does t h e Gr oss Dom est i c Pr oduct not show? (a) Cont r ibut ion of industr y t owar ds t he national I ncome (b) Expendi t ur e on goods and ser vi ces by t he gover nment (c) I ncome di st r ibut ion acr oss di ffer ent sect ions of t he populat ion. (d) Pur chasing power of t he people in a count r y. 12. H ow do e- commer ce vent ur es bui ld up t r ust of t he buyer s i n t heir goods? (a) Cash on deli ver yfacility (b) Ensur e a flexi ble r et ur n poli cy (c) Bet t er adver t ising (d) Pr ompt deli ver y 13. Which of t he following celebr it ies was r ecent ly appoint ed as "Br and Ambassador " of Telengana? (a) Deepika Pallikal (b) WS L axman (c) Saina Nehwal (d) Sania Mir za 14. BK S I yengar, who di ed r ecent l y, was a wor l d r enowned (a) Yoga Gur u (b) Ar tist (c) Folk Singer (d) Film Dir ect or

2

PRACTICE PAPER CBT-II

15. Which Count r y has r ecent ly launched "Gandhi I nspir ed Tour ist At t r act ion Pr oject " ?

23. Which one of t he foll owing i s cor r ect ?

(a) England

(b) Sout h Afr ica

The speci fic vol ume of wat er when heat ed fr om 0°C

(c) USA

(d) Japan

(a) fi r st incr eases and t hen decr eases (b) fi r st decr eases and t hen incr eases

PH YSI CS & CH EM I STRY 16. Which of t he fol lowi ng is used as a moder at or in nucl ear r eact or s ? (a) H ar d wat er (b) M i ner al wat er (c) Deioni zed wat er (d) H eavy wat er 17. A particle moves along a circular path with constant speed. What is t he nat ur e of it s accel er at i on ? (a) I t i s zer o (b) I t i s U ni for m (c) I t s dir ect ion changes (d) I t s magnit ude changes 18. A body i s at r est on t he sur face of t he ear t h. Which of t he fol lowing St at ement s i s cor r ect ? (a) No for ce is act i ng on t he body (b) Only weight of t he body act s on it (c) Net downward force is equal to net upward force (d) None of t hese i s cor r ect 19. The Speci fic H eat of t he gas in an i sot her mal pr ocess is (a) Zer o (b) I nfinite (c) Negat ive (d) Remai ns const ant 20. I n a Simple H ar monic Oscil lat or, at t he mean position (a) Kinetic Ener gy is minimum, Potential Ener gy is maximum (b) Bot h K i net i c and Pot ent i al Ener gi es ar e maximum (c) K i n et i c E n er gy i s m ax i m u m . Pot en t i al Ener gy is minimum (d) Bot h K i net i c and Pot ent i al Ener gi es ar e minimum 21. M i r age is a phenomenon due t o (a) Refl ect ion of l ight (b) Refr act ion of l ight (c) Tot al I nt er nal r eflect i on of l ight (d Diffr act ion of li ght 22. Which of t he fol lowi ng cannot be speed-t ime (v-t gr aph of a body in mot i on ? v v (a)

o

t

t

v (c)

(d)

o

t

o

(d) decr eases st eadily 24. Whi ch of t he fol l owi ng i s t he most aci di c i n nat ur e? (a) Phenol (b) 2-Nit r ophenol (c) 3- Nit r ophenol (d) 2,4- Dinit r ophenol 25. What is t he coor dinat ion number of zinc in t he complex ion [ Zn (OH )4(H 2O)2] 2– ? (a) 6

(b) 4

(c) 2

(d) 0

26. Addi t i on of 5 gr ams of sol ut e A (mol ecul ar mass=100) t o 100 gr ams of solvent B causes t he same elevat ion of boiling point as t he addit ion of 10 gr ams of solut e C t o 200 gr ams of t he same sol vent B. What i s t he mol ecul ar mass of t he solut e C? (a) 200

(b) 50

(c) 150

(d) 100

27. Which of t he following pair of molecules of t he compounds wi l l r eact t oget her i n pr esence of sodi u m h y dr ox i de sol u t i on u n der su i t abl e con di t i on s t o pr odu ce a m i x t u r e of f ou r compounds? (a) CH 3CH 2CHO + CH 3CH 2CHO (b) CH 3CH O + CH 3CHO (c) CH 3CHO + CH 3CH 2CHO (d) CH 3CH O + C6H 5CHO 28. The cor r ect st at ement concer ning t he molecule of ammonia is (a) I t is a V- shaped molecule. (b) The H -N-H bond angle ar e less t han 109°-28´. (c) The r epulsion bet ween pair and bond pair s of electrons is less than that bond pair- bond pair. (d) I t s shape is t et r ahedr al wit h lone pair s in t wo positions.

(b

o

(c) incr eases st eadily

t

29. I n an indust r ial ar ea, a sample of 10 lit r es of air was found t o cont ain ml of sulphur dioxide gas. The concent r at ion of sulphur dioxide gas in air in ppm is (a) 300

(b) 400

(c) 500

(d) 600

PRACTICE PAPER CBT-II

30. For t he r eact ion 2NO, (g)  N 2O4(g) + 60kJ, an incr ease in t emper at ur e will (a) favour t he decomposit ion of N 2O4 (b) favour t he for mat ion of N 2O4 (c) r esult in t he for mat ion of a differ ent r eact ion pr oduct (d) st op t he r eact ion.

BASI C OF COM PU TERS & APPLI CATI ON S 31. Which of the following st atement s about Machine language is cor r ect ? (a) M achine l anguage is machine dependent (b) M achi ne language i s machine i ndependent (c) M achine l anguage is easier t han high-l evel language t o wr it e pr ogr ams (d) M ach i n e l an gu age pr ogr am s r equ i r e assembler 32. H TTP st ands for (a) H yper Text Tr ansmi ssi on Pr ot ocol (b) H yper Text Tr ansfer Pr ogr am (c) H yper Text Tr ansfer Pr ot ocol (d) H yper Text Tr ansmission Pr ogr am 33. A Vi r us can not ____________________ (a) St eal har d disk space (b) St eal CPU t i me (c) L og keyst r ok es (d) I ncr ease/decr ease t he wor d lengt h of CPU 34. Which of t he fol lowing is not a val id cat egor y of Read Only M emor y (ROM )? (a) PROM (b) EPROM (c) EE PROM (d) EEEPROM 35. The 16's compl iment of t he hexadecimal number (A10)16 is (a) (5FO)16 (b) (5EO)16 (c) (5EF)16 (d) (6FO)16 36. Which of t he fol lowi ng wor ks on t he pr i nci pl e of ‘l ocal it y of r efer ence'? (a) RAM (b) ROM (c) Cache memor y (d) Associat i ve memor y 37. Which of t he fol lowing i s a r eal t ime oper at i ng syst em? (a) MS-Windows (b) Linux (c) Unix (d) QNX 38. Consider i ng l's compl ement r epr esent at ion for negat ive number s, -126 wil l be st or ed int o an 8 bi t memor y space as (a) 10000001 (b) 11111111 (c) 10111110 (d) 11100001

3

39. Consi der i ng X and Y as bi nar y var i abl es, t he equi val ent Boolean expr ession for (X.Y)' is (a) X '+Y (b) X + Y ' (c) X ' + Y ' (d) X '.Y ' 40. ‘C’ i s a_________________ (a) L ow-level pr ogr ammi ng l anguage (b) H igh-level pr ogr ammi ng l anguage (c) Assembly language (d) M achine language

BASI CS OF EN VI RON M EN T & POL L U T I ON S 41. The green house gases as per their decreasing order of effectiveness are (a) CFC, N2O, CO2, CH4 (b) CO2, CH4, N2O and CFC (c) CH4, CO2, N2O and CFC 42.

43.

44.

45.

46.

(d) CFC, CH4, CO2, and N2O In villages, to disinfect the well water, the most common disinfectant used is (a) silver and bromine (b) potassium permanganate (c) iodine solution (d) chlorine and bromine The carbon monoxide causes (a) coughing and choking problem in respiratory system (b) broncho-constriction (c) headache, vomiting, slurring of speech convulsions, coma and death (d) chronic bronchitis followed by asthma An octave band is a frequency band with upper and lower cut-off frequencies having a ratio of (a) 3 (b) 4 (c) 2 (d) 5 The CO2 concentration in the atmosphere was 355 ppm in 1990 that is increasing at a rate of (a) 1.00 ppm (b) 1.5 ppm (c) 0.50 ppm (d) 0.20 ppm The amount of oxygen required to decompose the organics under strong acidic conditions is called (a) chemical oxygen demand (b) biochemical oxygen demand (c) biological oxygen demand (d) theoretical oxygen demand

4

PRACTICE PAPER CBT-II

47. The photochemical smog is (a) criteria pollutant (b) primary pollutant (c) secondary pollutant (d) carcinogenic pollutant 48. The A - weight scale covers sounds of frequencies from (a) 800 to 3000 HZ (b) 500 to 2000 HZ (c) 1000 to 3000 HZ (d) 1200 to 4000 HZ 49. The rain water turned acidic when its pH falls below (a) 7.0

(b) 6.5

(c) 5.6

(d) 5.2

50. In case of air pollution, the most affected part of vegetation is (a) stems of vegetation (b) roots of vegetation (c) leaves of vegetation (d) fruits of vegetation

4. T h e h or se pow er t r an sm i t t ed by a bel t i s dependent upon (a) t ensi on on t i ght and sl ack si de of t he bel t (T 1, T 2) (b) r adius of t he pulley (R) (c) speed of t he pulley (N) (d) all of t hese 5. Which of t he following has no unit (a) kinemat ic viscosit y (b) sur face t ension (c) bulk modulus (d) str ain 6. M I S st ands for (a) milit ar y inspect ion scheme (b) management infor mat ion syst em (c) management int elligence syst em (d) management infor mat ion syst em 7. Two spr i n gs h avi n g spr i n g con st an t k an d ar r angement as shown i n fi gur e. Equi val ent spr ing const ant keq is

M ECH AN I CAL

k

1. I n a t ensi l e t est on mi l d st eel speci men, t he br eaking st r ess as compar ed t o ult imat e t ensile st r ess is (a) mor e (b) less (c) same (d) may have any value 2. Cavit at ion occur s in cent r ifugal pumps when t he suct ion pr essur e is (a) equal t o t he vapour pr essur e of t he liquid at t he t emper at ur e (b) less t han t he vopour pr essur e of t he liquid at t he t emper at ur e (c) mor e t han t he vapour pr essur e of t he liquid at t he t emper at ur e (d) none of t he above 3. For applicat ion involving high dischar ge and low pr essur e (a) a cent r i fugal pump wit h bl ades of for war d cur vat ur e (b) a cent r ifugal pump wit h blades of backwar d cur vat ur e

k

m

(a) k/2

(b) k

(c) 2k

(d) 4k

8. ABC analysis deals wit h (a) analysis of pr ocess char t (b) flow of mat er ial (c) schedule of job (d) cont r olling invent or y cost s money 9. The r elat ion bet ween moduleus of Elast icit y (E), modulus of r igidit y (G) and Bulk modulus (K ) is given by (a) E =

6K G G  9K

(b) E =

9K G G  3K

(c) E =

G  9K 3K G

(d) E =

G  3K 9K G

10. The r at io of elongat ion in a pr ismat ic bar due t o its own weight (W) as compar ed to another similar bar car r ying an addit ional weight (W) will be

(c) an axial flow pump

(a) 1 : 2

(b) 1 : 3

(d) a r ecipr ocat ing pump

(c) 1 : 4

(d) 1 : 2.5

PRACTICE PAPER CBT-II

11. Pr oduction schedule does not pr ovide infor mation about (a) pr oduct ion schedule

(c) a fixed beam car rying a unifor mly var ying load (d) none of t hese 19. A st andar d ice point t emper at ur e cor r esponds t o t he t emper at ur e of

(b) mater ial handling (c) both (a) and (b)

(a) wat er at 0°C

(d) none of t hese

(b) ice at – 4°C

12. The buckling load for a given mat er ial depends on (a) salender ness r at io and ar ea of cr oss-sect ion (b) poisson’s r at io and modulus of elast icit y (c) slender ness r at io and modulus of elast icit y (d) slender ness r at io, ar ea of cr oss-secit on and modulus of elast icit y 13. Whi ch t ype of t hr ead can t r anhsmit power i n eit her dir ect ion (a) acme

(b) squar e

(c) but tr ess

(d) BSW

(c) solid and dr y ice (d) mixt ur e of ice and wat er under equilibr ium condit ions 20. Reheat fact or is always (a) gr eat er t han 1

(b) less t han 1

(c) equal t o 1

(d) none of t hese

21. The cr ippling load accor ding t o Euler 's t heor y of long column, when bot h ends of t he column ar e hinged is equal t o (a)

14. St r ess r elaxat ion is t he phenomenon (a) st r ess r educes on incr easing load

(c)

(b) in which par t s ar e not loaded (c) in which str ess r emains constant on increasing load (d) in which defor mat ion tends t o loosen the joint and pr educes a st r ess r educt ion 15. The cor iolis component of acceler ation exist s only whenever a point (a) moves in a st r aight line (b) m ov es al on g a st r ai gh t l i n e w h i ch h as r ot at ional mot ion (c) moves along a cir cular pat h (d) none of t hese 16. Tot al ener gy line is .......... t he hydr aulic gr adient line (a) above

(b) below

(c) same

(d) none of t he above

17. I f a shaft subject ed to bot h t ension T and bending moment M , t hen maximum shear st r ess is 16 (a) d 3

M T

32 (c) d 3

M 2  T2

2

2

5

16 (b) d 3

M T

32 (d) d 3

M 2  T2

2

2

18. The point of cont r aflexur e does not occur s in

 2 EI 4l 2

 2 EI

l2

(b)

(d)

4  2 EI

l2 2 2 EI

l2

22. Cur t is Tur bine is a (a) pr essur e compounded impulse t ur bine (b) velocit y compounded t ur bine (c) pr essur e-velocit y compounded tur bine (d) velocit y compounded impulse t ur bine 23. The Gr ubl er 's cr i t er i on for det er mi ni ng t he degr ee of fr eedom (n) of a mechanism having plane mot ion is (a) n = (l – 1) – j

(b) n = 2(l – 1) – 2j

(c) n = 3(l – 1) – 2j

(d) n = 4(l – 1) – 2j

24. Vapour compr ession r efr iger at ion is some what li ke (a) Car not cycle (b) Rankine cycle (c) r ever sed car not cycle (d) none of t he above 25. For mult iple clut ch, t he no of discs ont he dr iving shaft is n1 and no. of discs on t he dr iven shaft is n2 t hen no. of pair s of cont act sur faces (n) is (a) n = n 1 + n 2

(b) n = n 1 – n 2

(c) n = n 1 + n 2 – 1

(d) n = n 1 + n 2 + 1

26. St r ain is defined as t he r at io of

(a) a fixed beam car r ying a unifor mly distr ibuted load

(a) change in volume t o or iginal volume

(b) a simply suppor ted beam car r ying a unifor mly distr ibuted load

(c) change in cross-sectional ar ea to or iginal crosssect ional ar ea

(b) change in lengt h t o or iginal lengt h

(d) any one of t he above

6

PRACTICE PAPER CBT-II

27. I n r egen er at i v e ai r pr eh eat er , t h e h eat i s tr ansfer r ed

34. I n Tayl or 's Rel at i on for t ool l i fe i s gi ven by VT n = C, wher e n has t he highest value for

(a) fr om a met al wall t hr ough one medi um t o anot her

(a) high speed st eel t ools

(b) fr om heat ing on int er mediat e mat er ial and t hen heat ing t he air fr om t his mat er ial

(c) cer amic t ools

(c) by dir ect mixing (d) heat is t r ansfer r ed by bleeding some gas fr om fur nace 28. Ther mal r adiat ion ext ends over t he r ange of (a) 0.01 t o 0.1 

(b) 100 t o 250 

(c) 0.1 t o 100 

(d) 250 t o 1000 

29. Deflect ion on a simply suppor t ed beam having point load at t he cent r e is (a)

Wl 3 8EI

Wl 3 (c) 3EI

(b)

Wl 3 48EI

5Wl 3 (d) 384

30. H ooke’s law holds good upt o (a) yield point

(b) car bide t ools (d) none of t hese 35. Unifor m sand har dness is obt ained t hr oughout t he mould by following moulding pr ocess (a) jolt (b) sand slinger (c) diaphr agm moulding (d) squeezing 36. The vapour compr ession r efr iger at or employs t he following cycle (a) r ankine

(b) car not

(c) r ever sed r ankine (d) r ever sed car not 37. I n M ohr 's cir cle, point on t he cir le gives (a) hydr ost at ic compr ession (b) hydr ost at ic t ension (c) shear st r ess (d) all of t hese

(b) limit of pr opor t ionalit y (c) br eaking point (d) elast ic limit 31. For low clear ance volume in st eam engine, t he pr ocess most suit able is

38. M inimum number of t eet h t o avoid int er fer ence is .......... for a pr essur e angle of 20 (a) 16

(b) 17

(c) 18

(d) 32

39. M aximum fluct uat ion of speed is given by

(a) pV 1.2 = const ant

(b) pV = const ant

(a) r at io of maximum and minimum speed

(c) pV r = const ant

(d) pV 1.5 = const ant

(b) di f f er en ce bet w een t h e m ax i m u m an d minimum speed

32. For a r efr iger at ion syst em if T 1 and T 2 ar e t he l i mi t i ng t emper at ur e (T 1 > T 2 ) and W i s t he wor kdone on t he syst em, t hen heat input t o t he syst em is

T1 W (a) T  T 1 2 (c)

T1  T2 W T2

(b)

T1  T2 W T1

T2 W (d) T  T 1 2

33. Unifor m flow is given by t he r elat ion (a)

v 0 t

v  const ant (c) s

(b)

v 0 s

v  const ant (d) t

(c) r at io of maximum fluct uat ion of speed t o t he mean speed (d) di f f er en ce bet w een t h e m ax i m u m an d minimum ener gy 40. The condensing pr essur e due t o t he pr esence of non condemnabl e gases, as compar ed t o t hat act ually r equir ed for condensing t emper at ur es wit hout non condemnable gases. (a) will be higher (b) will be lower (c) will r emain unaffect ed (d) unpr edictable 41. The lowest speed is used in (a) knur ling

(b) t aper t ur ning

(c) bor ing

(d) thr eading

PRACTICE PAPER CBT-II

42. I n wir e dr awing t he pr ess used is (a) knukle pr ess (b) t oggle pr ess (c) double act ion pr ess (d) none of t hese 43. When t he vent ur imet er is i ncl ined t hen what happens t o t he r eading (a) incr eases (b) decr eases

51. A st eam pipe is t o be insulat ed by t wo mat er ials having ther mal conductivities K 1 and K 2 such that (K 1 < K 2 ) so, for bet t er insulat ion (a) K 1 should be put over pipe and K 2 over it (b) K 2 should be put over pipe and K 1 over it (c) K 1 and K 2 may be put in any or der (d) none of t hese 52. Viscous for ce is pr opor t ional t o (a) shear st r ess due t o viscosit y (b) coefficient of viscosit y

(c) r emains t he same

(c) ar ea of t he pipe

(d) it is not applicable t o vent ur imet er

(d) none of t hese

44. Dr aft t ube is used (a) to tr anspor t water downstream without eddies (b) t o conver t t he kinet ic ener gy t o flow ener gy by a gr adual expansi on of t he fl ow cr oss sect ion (c) for safet y of t he t ur bine (d) t o incr ease flow r at e 45. Ammonia-absor ption r efr iger ation cycle r equir es (a) ver y lit t le wor k input (b) maximum wor k input (c) zer o wor k input (d) none of t he above 46. A t her mal power plant designed to oper ate in cold climat e is oper at ed is hot climat e it will develop (a) less power (b) mor e power (c) same power (d) mor e or less depending on t he size 47. A r el at i on bet ween t her modynami c boundar y layer and hydr odynamic boundar y layer exists in (a) Reynold's number (b) Pr andt e’s number (c) Nusselt ’s number (d) Euler ’s number 48. The r elat ive coefficient of per for mance is (a) act ual COP/t heor et ical COP (b) t heor ect ical COP/act ual COP (c) act ual COP x t heor ect ical COP (d) 1-act ual COP x t heor et ical COP 49. M et hod of gover ning used in diesel engine is (a) quant it y gover ning (b) quality gover ning (c) combined gover ning (d) hit and miss gover ning 50. I n S.J. unit , one t on of r efr iger at ion is equal t o (a) 210 kJ/min (b) 21 kJ/min (c) 420 kJ/min (d) 840 kJ/min

7

53. A r efr iger at or cycle oper at es bet ween condenser t emper at ur e of + 27 C and evapor at or t emper at ur e of – 23C. The C.O.P. of t he cycle is (a) 0.2

(b) 1.2

(c) 5

(d) 6

54. I n cet ane number t he new r efer ence t aken is (a) hept amethylnonane (b) -methyl napt halene (c) n - hept ane (d) iso-oct ane 55. D om est i c r ef r i ger at or w or k i n g on v apou r compr essi on cycl e uses t he fol l owi ng t ype of expansion device (a) elect r ically oper at ed t hr ot t ling valve (b) manually oper at ed valve (c) t her most at ic valve (d) capillar y t ube 56. Fins ar e used t o (a) incr ease t he ar ea (b) heat accumulat ion (c) t o incr ease t he sur face ar ea of heat t r ansfer (d) none of t hese 57. Tds = dH - Vdp is applicat ion t o (a) closed syst em (b) open syst em (c) r ever sible and ir r ever sible syst ems (d) all of t hese 58. Ber noulli's equat ion is applicable for (a) the flow is unifor m, steady and incompr essible (b) t he flow is non-viscous, unifor m and st eady (c) the flow is steady, non-viscous, incompr essible and ir r ot ational (d) none of t hese

8

PRACTICE PAPER CBT-II

59. L ift in a body in fluid medium is pr opor t ional t o (a) pr essur e for ce (b) r esist ance for ce (c) pr essur e for ce – r esist ance for ce (d) pr essur e for ce + r esist ance for ce 60. I n N ewt on's l aw of vi scosi t y, shear st r ess i s pr opor t ional t o (a) velocit y

(b) velocit y gr adient

(c) shear st r ain

(d) viscosit y

61. For a liquid if t he velocit y is incr eased t wice and diamet r e of t he pipe is decr eased one four t h t hen r at io of Reynold's number will be (a) 2

(b) 1

(c) 1/2

(d) 1/4

62. Which st at ement is wr ong r elat ed t o cor e (a) cor e is dr awn aft er pour ing met al (b) cor e is dr awn befor e pour ing met al (c) cor e has no r elat ion wit h solidficat ion (d) bot h (a) and (c) 63. I n a centrifugal pump dischar ge Q is proportional to (a) N 2

(b) N

(c) N 3

(d) N 1/2

64. Toot h past e t ube is made by following pr ocess (a) for ging

(b) pier cing

(c) impact ext r usion (d) dr awomg 65. Pick up t he cor r ect st at ement about giving up of heat fr om one medi um t o ot her i n ammoni a absor pt ion syst em (a) st r ong solut ion t o weak solut ion (b) weak solut ion t o st r ong solut ion (c) st r ong solut ion t o ammonia vapour (d) ammonia vapour t o weak solut ion 66. Fr eon gr oup of r efr iger ant s ar e (a) inflammable

68. I n t her mit welding, ir on oxide and aluminium ar e mixed in following pr opor t ions (a) 1 : 1

(b) 3 : 1

(c) 1 : 3

(d) 1 : 2

69. I n vapour compr essi on r efr i ger at i on syst em, r efr iger ant occur s as liquid bet ween (a) condenser and expansion valve (b) compr essor and evapor at or (c) expansion valve and evapor at or (d) compr essor and condenser 70. Which of t he following or ganisation is best suited for milit ar y (a) functional or ganisat ion (b) line or ganisat ion (c) staff or ganisat ion (d) line and st aff or ganisat ion 71. The t er m ‘value’ in value engineer ing r efer s t o (a) t ot al cost of t he pr oduct (b) ut ilit y of t he pr oduct (c) selling pr ice of t he pr oduct (d) manufact ur ed cost of t he pr oduct 72. M icr omot ion st udy is (a) subdivision of an oper at ion int o t her bligs and t heir analysis (b) enlar ged view of mot ion st udy (c) analysis of one st age of mot ion st udy (d) minut e and det ailed mot ion st udy 73. CPM and PERT t echnique is used for (a) shor t t er m

(b) long t er m

(c) moder at e t er m

(d) none of t hese

74. A body having velocit y of 199 m/s. Find it s M ach number (a) 1.2

(b) 0.8

(c) 0.6

(d) 1.4

75. Tools fails suddenly due t o

(b) t oxic

(a) flank wear

(c) non-inflammable and t oxic

(b) cr at er wear

(d) non-toxic and non-inflammable

(c) both (a) and (b)

67. I n gr ey cast ir on, car bon is pr esent in t he for m of (a) cement i t e

(d) eit her (a) or (b) 76. Oper at ion r esear ch is used in

(b) fr ee car bon

(a) linear pr ogr amming

(c) spher oids

(b) queuing t heor y

(d) flakes

(c) t r anspor t ation pr oblems (d) all of t hese

PRACTICE PAPER CBT-II

77. L aser applicat ion has made applicat ion in (a) elect r onic indust r y

9

(c) 710 C and 0.69% C (d) 910° C and 4.30% C 85. Which gas is not har mful used in r efr iger at ion

(b) casting (c) machining

(a) CO2

(b) NH 3

(d) none of t hese

(c) Fr eon

(d) none of t hese

78. The advant age of dr y compr ession is t hat (a) it per mit s higher speeds t o be used (b) i t per mi t s compl et e evapor at i on i n t he

evaporator (c) it results in high volumetric and mechanical efficiency (d) All of the above 79. Selender ness Rat io of a column is t he r at io of (a) it s lengt h t o least r adius of gyr at ion (b) it s lengt h t o least lat er al dir ect ion (c) it s lat er al dimension t o r adius of gyr at ion (d) none of t he above 80. 3 - 2 - 1 pi n met hod check s .......... degr ees of fr eedom

86. Th e COP of a vapou r compr essi on pl ant i n compar ison t o vapour absor pt ion plant is (a) mor e

(b) less

(c) same

(d) unpr edictable

87. C language is a (a) high level

(b) low level

(c) middle level

(d) none of t hese

88. At omic packing fact or for body cent er ed cubic is (a) 0.69

(b) 0.63

(c) 0.58

(d) 0.71

89. H ar dness of wat er is due t o (a) impur ities

(b) Ca and M g

(c) Ca and Na

(d) none of t hese

90. Semless pipes ar e made by

(a) 9

(b) 6

(a) ext r usion

(b) drawing

(c) 3

(d) 12

(c) pier cing

(d) for ging

81. Wh en cool i n g of A u st en i t e t ak es pl ace i n t emper at ur e r ange of 205  C – 315  C t hen for mat ion of (a) spher odite

(b) t r ossit e

(c) mar t ensit e

(d) sor bit e

82. The cr it ical pat h of a net wor k r epr esent s (a) t he minimum t ime r equir ed for complet ion of pr oject (b) t he maximum t ime r equir ed for complet ion of pr oject

91. Power gener at ed by which sour ce is now most effect ive (a) wind

(b) t idal wave

(c) solar

(d) none of t hese

92. Conver t 1026 decimal t o oct al (a) 536

(b) 534

(c) 524

(d) 520

93. Guided Rocket missile is a (a) mass var iant

(c) maxi mum cost r equi r ed for compl et i on of pr oject

(b) t ime var iant

(d) mi ni mum cost r equi r ed for compl et i on of pr oject

(d) none of t hese

83. The evolut ion of heat of solut ion t akes place in ammonia absor pt ion plant when (a) ammonia vapour goes int o solut ion

(c) mass and t ime var iant 94. For a shaft having nor mal st r ess and bendi ng st r ess  t hen equi val ent shear st r ess  i s (a)

1 2

2  2

(b)

1 2

 2  4 2

(c)

1 2

 2  4 2

(d)

1 2

2  2

(b) ammonia vapour is dr iven out of solut ion (c) lit hium br omide mixes wit h ammonia (d) weak solut ion mixes wit h st r ong solut ion 84. The t emper at ur e and car bon cont ent at which eut ect oi d r eact i on occur i n Fe - C equat i on diagr am ar e (a) 723 C and 0.02% C (b) 723 C and 0.80% C

95. Octane number of petrol found is I ndian r efiner ies is (a) 40 - 45

(b) 50 - 55

(c) 80 - 85

(d) 20 - 30

10 PRACTICE PAPER CBT-II

96. I n a r ever sible heat engine the temper atur e limits ar e 100 k and 400 K . I f heat out put is 200 kJ and heat input is

99. For minimum axial t hr ust , t he t ype of gear used should be (a) wor m hear

(a) 400 kJ

(b) 200 kJ

(b) helical gear

(c) 100 kJ

(d) 800 kJ

(c) her r ingbone hear

97. Velocit y pot ent ial funct ion is applicable for (a) r ot at ional flow

(b) ir r ot at ional flow

(c) both (a) and (b)

(d) none of t hese

(d) hypr id gear 100.Shor t hor izont al lines on pr essur e-ent halpy char t show

98. Two plat es ar e joined by r ivet s. The pull r equir ed t o t ear off t he plat e acr oss a r ow of r ivet s per pit ch lengt h is equal t o (a) ( p – d)t × f t

(b) p × t × f t

(c) d × t × f t

(d) ( p + d) × t × f t

(a) const ant pr essur e lines (b) const ant t emper at ur e lines (c) const ant t ot al heat lines (d) const ant ent r opy lines

wher e d = diamet er of r ivet p = pit ch of r ivet t = t hickness of r ivet f t = per missible t ensile st r ess

AN SWERS 1. (a)

2. (a)

3. (c)

4. (d)

5. (a)

6. (b)

7. (c)

8. (c)

9. (d)

10. (a)

11. (c)

12. (b)

13. (d)

14. (a)

15. (b)

16. (d)

17. (c)

18. (c)

19. (b)

20. (c)

21. (c)

22. (b)

23. (b)

24. (d)

25. (a)

26. (d)

27. (c)

28. (b)

29. (c)

30. (a)

31. (a)

32. (c)

33. (d)

34. (d)

35. (a)

36. (c)

37. (d)

38. (a)

39. (c)

40. (b)

41. (a)

42. (b)

43. (c)

44. (c)

45. (b)

46. (a)

47. (c)

48. (a)

49. (c)

50. (c)

M ECH AN I CAL 1. (b) 11. (b) 21. (c) 31. (b) 41. (a) 51. (a) 61. (a)

2. (b) 12. (d) 22. (d) 32. (d) 42. (b) 52. (a) 62. (a)

3. (c) 13. (b) 23. (c) 33. (b) 43. (c) 53. (c) 63. (b)

4. (d) 14. (d) 24. (d) 34. (b) 44. (b) 54. (a) 64. (c)

5. (d) 15. (b) 25. (c) 35. (c) 45. (a) 55. (d) 65. (b)

6. (b) 16. (a) 26. (d) 36. (d) 46. (a) 56. (c) 66. (d)

7. (c) 17. (b) 27. (b) 37. (d) 47. (b) 57. (d) 67. (d)

8. (d) 18. (b) 28. (c) 38. (c) 48. (a) 58. (c) 68. (b)

9. (b) 19. (d) 29. (b) 39. (b) 49. (b) 59. (d) 69. (c)

10. (b) 20. (a) 30. (b) 40. (a) 50. (a) 60. (b) 70. (b)

71. (b) 81. (c) 91. (b)

72. (a) 82. (b) 92. (b)

73. (a) 83. (a) 93. (d)

74. (c) 84. (b) 94. ( d)

75. (b) 85. (a) 95. (c)

76. (d) 86. (a) 96. (d)

77. (a) 87. (c) 97. (b)

78. (d) 88. (c) 98. (a)

79. (a) 89. (d) 99. (c)

80. (a) 90. (c) 100. (a)