Physics For Jee

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Gujarat Secondary and Higher Secondary Education Board, Gandhinagar

QUESTION BANK PHYSICS

Price

: ` 85.00

Published by : Secretary Gujarat Secondary and Higher Secondary Education Board, Gandhinagar

I

Contribution 1

Dr. Hasmukh Adhiya (IAS)

Principal Secretary , Education Department Gandhinagar

2

Shri R. R. Varsani (IAS)

Chairman , G.S&H.S.E. Bord, Gandhinagar

3

Shri H. K. Patel (G.A.S)

Dy. Chairman, G.S&H.S.E. Bord, Gandhinagar

4

Shri M. I. Joshi (G.E.S)

Secretary , G.S&H.S.E. Bord, Gandhinagar

Coordination 1

Shri B. K. Patel

O.S.D., G.S&H.S.E. Bord, Gandhinagar

2

Shri D. A.Vankar

Assistant Secretary (Retd.), G.S&H.S.E. Bord, Gandhinagar

5

Shri G. M. Rupareliya

Assistant Secretary, G.S&H.S.E. Bord, Gandhinagar

Expert Teachers 1.

Shri J. M. Patel

Shree J. M. Chaudhary Sarvajanik Vidhyalaya, Mehsana

2.

Shri K. D. Patel

J. N. Balika Vidhyalaya, Saraspur

3.

Shri Mayur M. Raval

P. J. Vakharia High School, Kalol

4.

Shri S. G. Patel

Sarkari Schook, Sector-12, Gandhinagar

5.

Shri J. P. Joshi

Diwan Ballubhai High School, Ahmedabad

6.

Shri Vasudev B. Raval

Vidhya Mandir High School, Palanpur

7.

Shri Surendrabhai M. Rajkutir

Convent of Jesus And Merry

8.

Shri Sureshchandra H. Patel

Alambic Vidhyalaya, Vadodara

9.

Shri C. D. Patel

Lalbahadur Shastri Vidhyalaya, Vadodara

10. Shri Mukesh N. Gandhi

New English School, Nadiad

11. Shri Dineshbhai V. Suthar

Retired Teacher

12. Shri S. S. Patel

J. M. Chaudhary Sarvajanik Vidhyalaya, Mehsana

13. Shri Jayesh M. Purohit

Ankur Vidhyalaya, Ahmedabad

14. Smt. Asha M. Patel

Shree M.B. Vamdot Sarvajanik High School, Bardoli

15. Shri Maheshbhai Dhandhla

Bhavnagar

16. Shri Mukesh M. Bhatt

Bhavnagar

17. Shri Anilkumar Trivedi

Anand

18. Shri Anand Thakkar

Navchetan High School, Ahmedbad

19. Shri Sudhirkumar G. Patel

Nutan High School, Visnagar

20. Smt. Anita Pillai

Surat

II

P R E FA C E Uptil now , the Students had to appear in various entrance examinations for engineering and medical courses after std-12. The burden of examinations on the side of the students was increasing day-by-day. For alleviating this difficulty faced by the students, from the current year, the Ministry of Human Resource Development , Government of India, has Introduced a system of examination covering whole country. For entrance to engineering colleges, JEE(Main) and JEE(Advanced) examinations will be held by the CBSE. The Government of Gujarat has except the new system and has decided to follow the examinations to be held by the CBSE. Necessary information pertaining to the proposed JEE (Main) and JEE(Advanced) examination is available on CBSE website www.cbse.nic.in and it is requested that the parents and students may visit this website and obtain latest information – guidance and prepare for the proposed examination accordingly. The detailed information about the syllabus of the proposed examination, method of entrances in the examination /centers/ places/cities of the examinations etc. is available on the said website. You are requested to go through the same carefully. The information booklet in Gujarati for JEE( Main) examination booklet has been brought out by the Board for Students and the beneficieries and a copy of this has been already sent to all the schools of the state. You are requested to take full advantage of the same also However, it is very essential to visit the above CBSE website from time to time for the latest information – guidance . An humble effort has been made by the Gujarat secondary and Higher Secondary Education Boards, Gandhinagar for JEE and NEET examinations considering the demands of the students and parents , a question bank has been prepared by the expert teachers of the science stream in the state. The MCQ type Objective questions in this Question Bank will provide best guidance to the students and we hope that it will be helpful for the JEE and NEET examinations. It may please be noted that this “Question Bank” is only for the guidance of the Students and it is not a necessary to believe that questions given in it will be asked in the examinations. This Question Bank is only for the guidance and practice of the Students. We hope that this Question Bank will be useful and guiding for the Students appearing in JEE and NEET entrance examinations. We have taken all the care to make this Question Bank error free, however, if any error or omission is found, you are requested to refer to the text – books.

Date: 02/ 01/ 2013

M.I. Joshi Secretary

III

R.R. Varsani (IAS) Chairman

INDEX Unit Unit Name No.

GSEB

NEET

JEE

Page No.

PART - I 1

Phisycs & Measurement

Y

Y

Y

1

2

Kinematics

Y

Y

Y

21

3

Laws of Motion

Y

Y

Y

51

4

Work, Energy and Power

Y

Y

Y

78

5

Rotationl Motion

Y

Y

Y

107

6

Gravitation

Y

Y

Y

147

7

Properties of Solids and Liquids

Y

Y

Y

195

8

Thermodynamics

Y

Y

Y

258

9

Kinetic Theory of Gases

Y

Y

Y

288

10

Oscilations and Waves

Y

Y

Y

321

IV

Unit-1 Important Formula

1

SUMMARY •

Measurement of large distance (Parallax Method) equation D =

b 

where

D = distance of the planet from the earth.

where

•

 = parallax angle. b = distance between two place of observation. Measurement of the size of a planet or a star. equation  

d D

where

D = distance of planet from the earth, d = diameter of planet.   angular diameter of planet.

•

•

Measurement of mass The gravitational force on an object, of mass m, is called the weight of the object. 1 amu = 1.66  1027 kg = 1u Estimation of Error Absolute Error - Suppose the values obtained in several measurement of physical quantity a are a1, a2, ............... an If their arithmetic mean is a then a 

n

a1  a2 + ....  an 1  n n

a

i

i 1

 a1 = a - a1,  a2 = a - a2 ,----------  an = a - an  a2,  a2 -------  an are called absolute error

•

Average absolute error

a 

a1  a2  ...  an n

•

Fractional Error  a =

•

Percentage Error



1 n  ai n i 1

a a

Percentage error = a  100 % = •

a  100 % a

Combination of errors Addition Z = A + B  Z  A  B Substraction Z = A – B  Z  A  B Division Z =

A Z A B    B Z A B

Multiplication Z = A • B  Power Z = An 

Z A B   Z A B

Z A n Z A

2

•

•

•

Rule for determining number of significant figures •

All the non - zero digits are significant

•

All the zeros between two non zero digits are significant no matter where the decimal point is it at all.

•

If the number is less then 1 then zeros on the right of decimal point but to the left of the first non - zero digit are not significant.

•

In a number without decimal point the zeros on the right side of the last non zero digit are not significant.

Dimensions and Dimensional formulas. •

The expression of a physical quantity with appropriate powers of M, L, T, K, A etc is called the dimensional formula of that physical quantity.

•

The power of exponents of M, L, T, K, A are called dimensions of that quantity.

•

Some important units of distance 1 fermi (fm)  10 –15 m o

1 A  10 –10 m

1 AU  1.496  1011 m 1 light year  9.46  1015 m 1 par sec  3.08  1016 m

3

MCQ Questions For the answer of the following questions choose the correct alternative from among the given ones. Physics - scope and Excitement - Physics, Technology and society. - Fundamental sources of nature. - Nature of Physical laws 1. Physics is one of the basic disciplines in the category of ............... sciences. (A) Astro (B) Natural (C) Space (D) Genetic 2. ‘Physics’ comes from a ............... word meaning nature (A) Hindi (B) German (C) Greek (D) Sanskrit 3. Mechanics and newton’s motion laws as ............... laws dependad. (a) liner momentum (b) Energy conservation (c) Gravitational (d) Charge conservation 4. What is the approximate value of the Radious of a nucleus ? (d) 10 –15 m

5.

(a) 10 –14 m (b) 10 –31 m (c) 10 –19 m The scope for ratio of length is in order to ...............

(d) 1030

6.

(a) 10–40 (b) 1040 The range of time scale is about ...............

(c) 1020

(a) 10–10 sec to 1026 sec (c) 10 –15 sec to 1015 sec (b) 10 –22 sec to 1018 sec (d) 1020 sec to 1025 sec 7.

8. 9.

10.

Birth, evolution and death of stars etc. are studid in branch of physics known as ............... (a) Thermodynamics (c) Astro physics (b) Quantam physics (d) Electronics ............... is a branch of physics in wich heat engine and refrigeratior efficiency is studied. (a) optics (b) Thermodynamics (c) Mechanics (d) Quantom physics What is full name of LHC (a) Large hadron collider (c) Large heavy cullent (b) Large hadron cullent (d) Light heavy cullent The range of mass varies from ............... (a) 10–15 kg to 1026 kg

11.

(d) 10–20 kg to 1020 kg

(b) 1036 m

(c) 1028 m

(d) 10 –14 m

The approximate value of charge of an electron is ............... (a) 10 –18 c

13.

(c) 10–30 kg to 1055 kg

Length of Galaxies is in order of ............... (a) 1026 m

12.

(b) 10–20 kg to 1028 kg

(b) 1015 c

(c) 10 –38 c

The universe is made up of ............... (a) matter only (b) radiation only

(c) vaccum 4

(d) 10–19 c (d) matter and radiation

14.

15.

Nucleus of molecule is made up of wich fundamental constituents ? (a) only Electron

(c) Electron and Proton

(b) Proton and neutron

(d) Electron and neutron

In the development of nenotechnology and biotechnology ............... have played a vital role. (a) ECG

16.

17.

18.

19.

20.

(c) Atomatic fire microscope

(b) Atomic force mirror

(d) Atomic force microscope

What is full name of ECG ? (a) Electron cardiograph

(c) Electron colour gram

(b) Electro cardiograph

(d) Electric colour graph

What is full name of ESR ? (a) Electric space Radar

(c) Electron spin Resonance

(b) Electron space Range

(d) Electric spin Resonance

What is full name of NMR ? (a) Nuclear magnetic Resonance

(c) Nuclear mega Radar

(b) Neutron mega Resonance

(d) Nuclear micro Radar

............... deals with electric charge and magnatic phenomenna

26.

(b) four

(d) five

(c) Ampere’s

(d) Faraday’s

(b) Electromagnetic

(c) Nuclear

(d) Gravitational

The ............... force is the strongest of all fundamental forces. (b) Electromagnetic

(c) Gravitational

(d) Weak nuclear

Electromagnetic force is ............... (a) attractive force only

(c) repulsive force only

(b) attractive and repulsive force

(d) a short range force

Which of the following force binds The particle in the nucleons ? (b) Strong force (c) Gravitational force

(d) Weak force

Electromagnetic force is ............... range force (a) Short

28.

(c) two

(b) Newton’s

(a) Electromagnetic force 27.

(d) Mechanis

The ................ force is the force of mutual attraction between any two objects by virtue of their masses.

(a) nuclear 25.

(c) Themodynamic

When charges are at rest the force is given by ............... law.

(a) Weak 24.

(b) Electro dynamic

At present state, there are .............. fundamental forces in nature.

(a) coulomb’s 23.

(d) AFM

(a) Atomatic force mioroscope

(a) six 22.

(c) NMR

What is full form of AFM ?

(a) Dynamics 21.

(b) ESR

(b) long

(c) medium

(d) very short

Quarks - Quarks force is produced between (a) Proton - neutron (b) proton - proton

(c) neutron - neutron 5

(d) (a),(b), (c) are true

29.

30.

31.

Which partical are emitted during the  decay from the nucleus ? (a) neutron and proton

(c) electron and neutrino

(b) electron and neutron

(d) electron and proton

............... and ............... law’s are called inverse square law (a) Gravitation and weak

(c) Coulomb’s and strong

(b) Gravitation and coulomb’s

(d) Electromagnetic and coulomb’s

Which property of object is responsible for the electric force ? (a) electric charge

32.

(c) 10–13

(d) 10 –2

(b) 10–13

(c) 1036

(d) 10–36

(b) Maxwell

(c) Coulomb

(d) Faraday

(b) Maxwell

(c) Coulomb

(d) Farady

(b) C > A < B

(c) B > A > C

(d) C < A < B

(b) 10–14 km

(c) 10–18 km

(d) 10–20 km

(b) nuclear

(c) neutron

(d) electron

The force acting between two point charges kept at a certain distance is F1 Now magnitude of charge are double and distance between them is double. The force acting between them is F2 find out the ratio of F2/F1 = ............... (a) 16 : 1

42.

(b) 102

Strong nuclear force close not exist on ............... (a) Proton

41.

(d) 10 –2

Range of weak nuclear force is ............... (a) 10–15 km

40.

(c) 10–13

The weak nuclear force, Gravitational force and electromagnatic force are A, B and C Respectively then ............... (a) C > A > B

39.

(b) 102

Who has unified terrestrial and celestial domains under a common law of Gravitational (a) Newton

38.

(d) volume

Who has unified electromagnetism and optics ? (a) Newton

37.

(c) pressure

How much times is the electromagnatic force stronger then Gravitational force (a) 1013

36.

(b) mass

How much times is the strong nuclear force stronger then electro magnatic force ? (a) 1013

35.

(d) mass

How much times is the strong nuclear force stronger then weak nuclear force ? (a) 1013

34.

(c) volume

Which property of object is responsible for the Gravitational force. (a) electric charge

33.

(b) pressure

(b) 1: 16

(c) 1: 1

(d) 1: 8

If the resulting external force acting on system is zero then ............... of the system is constant and if the resultant external torque acting on a system is zero then ............... of the system is constart. (a) total energy, angularmomentum

(c) linermomentam, energy

(b) liner momentam, angularmomentum

(d) angular and linear momentam 6

43.

44.

Space is homogeneous and isotropic so ............... law of servation is the result of this (a) linear and angular momentum

(c) energy and charge

(b) angular and linear momentum

(d) charge and energy

Time is homogeneous so ............... law of conserbation is the result of this (a) angular momentum (b) linear momentum

45.

(c) energy

(d) charge

The basic reason behind existance of which conseration of law is still not known ? (a) angular momentum (c) energy (b) linear momentum (d) charge

46.

The Gravitational force between any two body charges with distance as F  r n where n = .......... (a) –1

47.

(b) 2

(c) –3

(d) –2

Match the column Column - I

Column - II

(1) space is isotropic

(P)

conservation of linear momentum

(2) space is homogeneous

(Q)

conservation of energy

(3) Time is homogeneous

(R)

conservation of charge still not known

(4) Time is isotropic

(S)

conservation of angular momentam

(a) 1- (S), 2-(P), 3-(R), 4-(Q)

(c) 1-(P), 2-(S), 3-(R), 4-(Q)

(b) 1-(S), 2-(P), 3-(Q), 4-(R)

(d) 1-(R), 2-(Q), 3-(P), 4-(S)

Measurement and system of units •

Units of physical quantities, system of units, SI system of units, fundamental or Base units. precision in measurement. Error in measurement and significant figures.

•

Dimensions and Dimensional formula, Dimensional analysis and its uses.

48.

Which of the following unit is not of length ? o

(a) light year 49.

(b) fermi

(c) A

becquerel is a ............... unit and its symbol is ............... (a) supplementary, Bq (b) fundamental, Bq

50.

(b) 7

(c) 6

(d) 4

(b) velocity

(c) force

(d) time

(b) velocity

(c) force

(d) time

Poise is the unit of (a) viscosity

53.

(d) derived, Bv

Which of the following physical quantity is fundamental ? (a) viscosity

52.

(c) derived, Bq

How many fundamental units are there in SI system ? (a) 5

51.

(d) becquerel

Which unit of physical quantity remains same for all unit system ? (a) meter

(b) second

(c) ampere 7

(d) kilogram

54.

Which of the following system of unit is not based on only units of mass length and time. (a) SI

55.

(b) MKS (b) kg. (b) Cd

(c) 6.67  10 –8

then value of G in

(d) 6.67  10 –5

(b) 30 m / s 2

(c) 18 m / s 2

(d) 20 m / s 2

(b) 95  10 –5

(c) 950  10 –7

(d) 9.5  10 –4

(b) 10 –7 m

(c) 10  10–6 m

(d) 10  10 –8 m

(b) 3.6  106 J

(c) 36  105 J

(d) 36  106 J

(c) x–2

(d) x–1

1 Mev = ............... ev (b) 104

(c) 105

(d) 106

Wave length of light radiation 0.000015 m = ............... (a) 15 micron

(b) 1.5 micron

(c) 150 micron

(d) 0.15 micron

(b) 3600 ''

(c) 180 ''

(d) 3600 '

(b) 3.14

 180  (c)     

10 = ...............

(a) 600 '' 68.

(b) 6.67  10 –7

(b) x2

(a) 107

67.

dyn cm gm 2

2

Nm 2 kg 2

If x meter is a unit of length then area of 1m2 = ............... (a) x

66.

(d) 10–24

100 walt hour = ............... joule. (a) 3.6  105 J

65.

(c) 1024

100 picometer = ............... (a) 10 –8 cm

64.

(b) 10 –8

950 dyne = ............... newton (a) 9.5  10–3

63.

(d) 1015

A partical has an acceleration of 72 km / min2 find acceleration in SI system. (a) 0.5 m / s 2

62.

(c) 1016

–11 If value of gravitational constant in MKS is 6.67  10

(a) 6.67  10 –9

61.

(d) CD

1 fem to m eter 1 0 0 n en o m eter = ...............

CGS = ...............

60.

(c) cd.

(b) 103

(a) 10 –6 59.

(d) A

What is the ratio of 10 micron to 1 nenometer ? (a) 104

58.

(c) k

Why derive luminous intensity simbol form of SI system ? (a) cd

57.

(d) FPS

Which of the following symbol of unit does not follow practical norms for the use of SI system ? (a) Kg

56.

(c) CGS

1 rad = ............... (a) 180

0

0

8

0

   (d)    180 

0

69.

1 g = ............... amu (a) 6.02  1023

70.

72.

(c) (ampere) 2  ohm

(c) joule/second

(d) ampere  volt

(b) rad / (sec)2

(c) m/sec

(b) N m –1 sec

(c) N m2 kg –2

(b) w m2 k –3

(c) density

(d) Distance

(b) angular momentum (c) Pressure

(d) Energy

(b) kg m –2s 2

(c) kg m –1

(d) kg ms –1

(b) viscosity

(c) Electric fild intensity (d) velosity

The force F is represented by equation F = P 1 + Q , where  is the length. The unit of P is same as that of ............... (b) velocity

(c) force

(d) momentum

Write the unit of surface tension in SI system. N m2

(b)

N m

(c)

dyne cm2

(d)

dyne cm

Which physical quantity has unit of pascal - secod ? (a) Velocity

83.

(b) volume

Volt/meter is the unit of ...............

(a) 82.

(d) Jule/second

The SI unit of momentum is ...............

(a) Surface tension 81.

(c) Jule

Joule/seed is the unit of ...............

(a) Work 80.

(d) w m –2 k –4

Light year is a unit of ...............

(a) kg  newton 79.

(c) w m –2 k 4

Unit of momentum physical quantity ?

(a) Work 78.

(d) N m kg –1

The unit of stefen Boltzman constant (  ) is ...............

(a) Mass 77.

(d) N/kg

unit of universal gravitational constant is ...............

(a) newton - second (b) newton/second 76.

(d) 3.08  1016 m

Write the unit of angular acceleration in the SI system.

(a) w 2 m –2 k –1 75.

(c) 1.496  1015 m

(a) ampere/volt

(a) kg m sec –1 74.

(b) 1.496  1011 m

Which of the following unit does not represent the unit of power ?

(a) N.Kg 73.

(c) 1.66  10 –27 (d) 1.66  1027

1 parsec = ............... (a) 10 –15 m

71.

(b) 6.02  10 –23

(b) viscocity

(c) energy

(d) coefficient of viscocity

Which physical quantity has unit of joule - second ? (a) velocity

(b) plank’s constant

(c) energy 9

(d) vescocity

84.

What is the least count of vernier callipers ?

85.

(a) 10 –4 m (b) 10–5 m What is the least count of screw gauge ?

86. 87.

(c) 10 –2 m

(d) 10–3 m

(a) 10 –4 m (b) 10–5 m (c) 10 –2 m (d) 10 –6 m For measurement of astronomical distance ............... is used. (a) vernier callipers (b) spherometer (c) screwgauge (d) indirect method Which mictoscope is used to measure the dimension of particle having dimension less than 4000 A0 ?

88. 89. 90.

(a) electron microscope (b) simple microscope (c) optical microscope (d) none of above In electron microscope electron behave like ............... (a) charge (b) mass (c) particles (d) wave Which wave length of light is used in an optical microscope ? (a) radiowave (b) X - ray (c) infrared (d) visible 2 The intercepted area of the spherical surface about the center is 0.25m having diameter 50 cm what will be solid angle ? (a) 4  10 –1 sr

91.

(b) 1  103 sr

(c) 10–1 sr

(d) 5  10–1 sr

One planet is observed from two diametrically opposite point A and B on the earth the angle subtended at the planet by the two directions of observations is 1.8o. Given the diameter of the earth to be about 1.276  107 m . What will be distance of the planet from the earth ? (a) 40.06  108 m

92.

(b) 4.06  108 m

(c) 400.6  1013 m

(d) 11  108 m

Find the distance at which 4 AU would subtend an angle of exactly 1" of arc. [1AU  1.496  1011 m,1"  4.85  1016 rad]

(a) 1.123  105 m 93. 94.

95.

(c) 1.123  1017 m

(d) 11.23  1017 m

The percentage error in the distance 100  5 cm is ... (a) 5 % (b) 6% (c) 8 % (d) 20 % In an experiment to determine the density of a cube the percentage error in the measurement of mass is 0.25 % and the percentage error in the measurement of length is 0.50 % what will be the percentage error in the determination of its density ? (a) 2.75 % (b) 1.75 % (c) 0.75 % (d) 1.25 % If A  b4 the fractional error in A is ............... (a)

96.

(b) 11.23  105 m

 b 

If P 

b

4

(b)

b    b 

 (c) 4 

b b

(d)  b 

4

A2 B where percentage error in A , B and C are respectively 2 %  3% and 5 % then C3

total percentage error in measurement of p (a)18 %

(b) 14 %

(c) 21 % 10

(d) 12 %

97.

In the experiment of simple pendulum error in length of pendulum () is 5 % and that of g is 3 % then find percentage error in measurement of periodic time for pendulum (a) 4.2 %

98.

(b) 1.2 %

Acceleration due to gravity is given by g 

(c) 2 %

(d) 4 %

GM what is the equation of the fractional error g / g R2

in measurement of gravity g ? [G & M constant] (a) – 99.

R R

(b) 2

R R

(c) –2

R R

(d) 

1 R 2 R

The period of oscillation of a simple pendulum is given by T  2 g what is the equation of

T in measurement of period T ? T 1   1   (a) (b) 2 (c) (d) 4 2   4   100. The length of a rod is (10.15  0.06) cm what is the length of two such rods ?

the relative error

(a) (20.30  0.06) cm

(b) (20.30  1.6) cm

(c) (10.30  0.12) cm

(d) (20.30  0.12) cm

4 3

3 101. For a sphere having volume is given by V r What is the equation of the relative error

V in measurement of the volume V ? V

(a) 3

r r

(b) 4

r r

(c)

4 r 3 r

102. Kinetic energy K and linear momentum P are related as K  relative error p

(a) p

(d)

1 r 3 r

p2 . What is the equation of the 2m

k in measurement of the K ? (mass in constant) k p

p

(b) 2 p

p

(d) 4 p

(c) 2p 2

103. Heat produced in a current carrying conducting wire is H = I Rt it percentage error in I, R and t is 2 % , 4 % and 2 % respectively then total percentage error in measurement of heat energy ............... (a) 8 % (b) 15 % (c) 5 % (d) 10 % 104. The resistance of two resistance wires are R 1 (100  5) and R 2  (200  7) are connected in series. find the maximum absolute error in the equivalent resistance of the combination. (a) 35  (b) 12  (c) 4  (d) 9  105. The periodic time of simple pendulum is T  2

 relative error in the measurement of T and g

 are a and  b respectively find relative error in the measurement of g

(a) a + b

(b) 2b + a

(c) 2a + b 11

(d) a - b

1 4 4 106. A physical quantity x is given by x = A3 B4 due to which physical quantgity produced the CD3

maximum percentage error in x (a) B

(b) C

(c) A

(d) D

V

107. The resistance R  where V 100  5 volts and I  10  0.3 anperes calculate the percentage I error in R. (a) 8 %

(b) 10 %

(c) 12 %

(d) 14 %

108. The number of significant figures in 0.000150 is ............... (a) 3

(b) 5

(c) 2

(d) 4

109. Which of the following numerical value have significant figure 4 ? (a) 1.011

(b) 0.010

(c) 0.001

(d) 0.100

110. What is the number of significant figures in 5.50  103 ? (a) 2

(b) 7

(c) 3

(d) 4

111. The mass of substance is 75.5 gm and its volume is 25 cm2. It’s density up to the correct significant figure is ............... (a) 3.02 gm / cm3

(b) 3.200 gm / cm3

(c) 3.02 gm / cm3

(d) 3.1 gm / cm3

112. The area of a rectangle of size 1.25  2.245 cm in significant figure is ............... (a) 2.80625 cm 2

(b) 2.81 cm 2

(c) 2.806 cm2

(d) 2.8062 cm2

113. The significant figures in 500.5000 are ............... (a) 5

(b) 3

(c) 7

(d) 6

114. Addition of measurement 15.225 cm, 7.21 cm and 3.0 cm in significant figure is ............... (a) 25.43 cm

(b) 25.4 cm

(c) 25.435 cm

(d) 25.4350 cm

115. Substract 0.2 J from 7.36 J and express the result with correct number of significant figures. (a) 7.160 J

(b) 7.016 J

(c) 7.16 J

(d) 7.2 J

116. After rounding of the number 9595 to 3 significant digits the value becomes ............... (a) 9600

(b) 9000

(c) 9590

(d) 9500

117. How many significant numbers are there in (2.30  4.70) 105 ? (a) 3

(b) 4

(c) 2

(d) 5

118. The radius of circle is 1.26 cm. According to the concept of significant figures area of it can be represented as (a) 4.9850 cm2

(b) 4.985 cm 2

(c) 4.98 cm 2

(d) 9.98 cm2

119. If A = 3.331 cm B = 3.3 cm then with regard to significant figure A + B = ............... (a) 6.6 cm

(b) 6.31 cm

(c) 6.631 cm 12

(d) 6 cm

120. If the length of rod A is (2.35  0.01) cm and that of B is (5.68  0.01) cm then the rod B is longer than rod A by ............... (a) (2.43  0.00) cm

(b) (3.33  0.02) cm

(c) (2.43  0.01) cm

(d) (2.43  0.001) cm

121. In acceleration, The dimensions for mass ............... for length .. and for time (a) 0,1,–2

(b) 1,0,–2

(c) –2,0,1

(d) –2,1,0

122. Dimensional formula for power is ............... (a) M2 L–2T –3 (b) M1L2T –2 (c) M1L3T –1 123. Dimensional formula for calories is ............... (a) M1L1T –2

1 –2 (b) M2LT

(c) M1L2T –2

(d) M0L2T –2 (d) M2 L2T –2

124. Dimensional formula for thermal conductivity (k) is .. 1 –2 –1 (a) M2LT K

1 –2 1 (b) M1LT K

(c) M1L0T –3K –1

(d) M1L1T –3K –1

125. Dimensional formula for Resistance (R) is ............... (a) M1L1T –3A –1

1 0 –1 (b) M1LT A

(c) M1L2T –3A –2

(d) M1L0T –3A –1

126. Dimensional formula for conductance is ............... (a) M –1L2T –3A 2

(b) M1L2T –2A1

(c) M1L–2T3A2

(d) M –1L–2T3A 2

127. Which physical quantity is represented by M1L3T –3 A 2 ? (a) Resistivily

(b) Resistance

(c) conductance

(d) conductivity

128. Which physical quantity is represented by M –1L–3T 3 A 2 ? (a) Resistivity

(b) Resistance

(c) conductance

129. Which physical quantity is represented by M1L1T –3A –1 ? (a) Stress (b) Resistance (c) Electricfield

(d) conductivity

(d) potential Difference

130. The dimensional formula of plank’s constant is ............... 1 –1 (a) M3L2T –1 (b) M1L2T –1 (c) M 2 LT 131. Dimensional formula of latent heat is ...............

(a) M0L2T –2

(b) M2 L0T –2

(c) M1L2T –1

(d) M1L2T –3 (d) M2 L2T –1

132. Dimensions of impulse are. 1 –1 1 1 (a) M –1L–1T1 (b) M1LT (c) M1LT 133. Write dimensional formula of coefficient of viscosity

(a) M1L2T –1

1 1 (b) M –1LT

(d) M1L2T –2

(c) M1L–1T –1

1 –1 (d) M1LT

(c) M1L1T –2

(d) M1L2T –2

(c) M –1L–2T3A1

(d) M3L1T –1A –2

134. Dimensional formula for torque is 1 –2 (a) M 2 L2T –3 (b) M 2 LT 135. Dimensional formula for capisitance (C)

(a) M –1L–2T 4A 2

(b) M1L–2T4A2 13

136. Dimensional formula for Boltzmann’s constant is ............... 1 –2 –1 (a) M1L1T –2 K –1 (b) M2LT (c) M1L2T –2K –1 K 137. Dimensional formula for electromotive force (emf)

(d) M2L2T1K –2

1 –1 –3 (a) M2LT (b) M1L2T –3K –1 (c) M1L1T –3K –1 (d) M1L2T3K –1 K 138. Which physical quantity has dimensional formula as CR where C - capisitance and R - Resistance ?

(a) Frequency

(b) current

(c) Time period

(d) acceleration

139. Write the dimensional formula of the ratio of linear momentum to angular momentum. 1 0 (a) M0 L–1T0 (b) M1L1T0 (c) M0LT (d) M0L1T1 140. If L and R are respesented as the inductance and resistance respectively then the dimensional

R will be ............... L

formula of

(a) M –2L1T –2 A1 (b) M0L0T –1A0 (c) M1L–1T0 A1 141. Write the dimensional formula of r.m.s (root mean square) speed. (a) M1L2T –2

(b) M0L2T –2

1 –1 (c) M0LT

142. One physical quantity represented by an equation as

(d) M1L3T1A0 (d) M1L0T –1

 (p – q)c where p, q and c are length 2

then quantity is .. (a) length

(b) velocity

(c) Area

(d) volume

143. The dimensional formula of magnetic flux is ............... (a) M1L2T –2A –1 (b) M1L2T1A2 (c) M1L2T –2A2 144. Which physical quantity has unit of pascal - second ? (a) Force

(b) Energy

(d) M –1L–2T1A2

(c) Coefficient of viscocity (d) velocity

145. Dimensional formula of CV ? where C - capacitance and V - potential different (a) M1L–2 T 4 A 2

(b) M1L2 T –3A1

(c) M 0 L0 T1A –1

(d) M 0 L0 T1A1

x 146. The equation of a wave is given by Y  A sin   – k  where  is the angular velocity and v  v is the linear velocity. Write the dimensional formula of K

(a) M0L0T1 (b) M1L0T –1 (c) M0L1T1 (d) M1L–1T1 147. If P and q are diffrent physical quantities then which one of following is only possible dimensionally ? p

(a) p + q 

(b) q

(c) p – q

(d) p = q

a 

148. From  p  2   v – b   constant equation is dimensionally correct find the dimensional formula v   for b ? where P = preasure V = volume (a) M0 L3T 0

1 3 (c) M 0 LT

(b) M1L3T 0 14

1 –1 (d) M1LT

149. Pressure P = A cosBx + c sinDt where xin meter and t in time then find dimensional formula of

D B

1 –1 1 –1 (a) M1LT (b) M0LT (c) M1L1T0 (d) M –1L0T1 150. Find the dimensional formula for energy per unit surface area per unit time

(a) M1L0T –2

1 –1 (b) M0LT

(c) M1L0T –3

(d) M1L–1T1

151. Equation of force F  at  bt 2 where F is force in Newton t is time in second, then write unit of b. (a) Nm –1 152. Pressure P 

(b) Nm 2

(d) Nm –2

a at 2 where x = distance, t= time find the dimensional formula for b bx

(a) M1L0T –4 153.

(c) Nm

1 –1 (b) M1LT

(c) M1L0T –2

(d) M1L0T –2

2

F  A 0 (1 – e – Bxt ) where F is force and x is desplacement. write the dimension formula of B 1 –1 (a) M2LT

(b) M0L–1T –2

(c) M1L0T –2

(d) M1L2T –1

154. Equation of physical quantity v  at  bt 2 where v = velocity t = time so write the dimensional formula of a in this equation 1 –1 (a) M0LT

1 –1 (b) M1LT

1 –2 (c) M0 LT

(d) M1L2T0

155. Density of substance in CGS system is 3.125 gm / cm3 what is its magnitude is SI system ? (a) 0.3125

(b) 3.125

(c) 31.25

(d) 3125

AR where L = length of wire A = Area of wire and L R is resistance of wire find dimension formula of 

156. The resistivity of resistive wire is  

(a) M1L3T –3A –2 (b) M1L2T –3A –2 (c) M2L3T1A2 (d) M2L3T –3A –2 157. A cube has numerically equal volume and surface area calculate the volume of such a cube. (a) 2000 Unit

(b) 216 Unit

(c) 2160 Unit

(d) 1000 Unit

158. Which out of the following is dimensionally correct. (a) p2 = hg

(b) p = h2g

(c) p = hg

(d) p = h2g

159. If energy E  G p h q cr where G is the universal gravitational constant. h is the plank’s constant and c is the velocity of light, then the values of p, q and r are respectively 1 1 5 2 2 2

(a) – , ,

(b)

1 1 5 , , 2 2 2

(c)

5 1 1 , ,– 2 2 2

(d)

1 1 5 ,– , 2 2 2

160. If the centripetel force is of the form mavbrc find the values of a, b and c (a) 1,2,1

(b) 1,2,–1

(c) 1,3,–2 15

(d) –1,3,–1

161. equation of  t   0 [1   (T2 – T1 )] find out the dimensions of the coefficient of linear expansion  suffix. (a) M0L0T1K1

1 1 1 (b) M0LT K

1 0 1 (c) M1LT K

(d) M0L0T0K –1

162. Test if the following equation are dimensionally correct (S = surface tension  = density P = pressure v = volume n = coefficient of viscocity r = redious) 2Scos 

(a) h  rg

p

(b) v  

(c) v 

pr4 t 8n

(d) all correct

163. Match list - I with list - II List - I

List - II

(1) Joule

(a) henry  ampere/sec (b) coulomb  volt

(2) Walt (3) volt

(c) metre  ohm (d) (ampere)2  ohm

(4) Resistivity (a) b,d,c,a

(b) c,a,b,d

(c) b,d,a,c

(d) b,c,a,d

164. Match column - I with column - II Column -I

Column - II

(1) capacitance

(a) M1L1T –3A –1

(2) Electricfield

(b) M1L2T –1

(3) planck’s constant

(c) M –1L–2T 4A 2

(4) Angular momentum

(d) M1L2T –1

(a) a,c,b,d

(b) c,a,d,b

(c) c,a,b,d

(d) a,b,d,c

– z

 kB  165. In the relation P  e , P is pressure, z is distance, k is boltz mann constant and  is 

the temperature. The dimensional formula of B will be (a) M 0 L2 T 0

(b) M1L0T1

1 –1 (c) M1LT

16

(d) M1L1T0

KEY NOTE 1(B) 2(C) 3(C) 4(A) 5(B) 6(B) 7(C)

26(B) 27(B) 28(A) 29(C) 30(B) 31(A) 32(B)

51(D) 52(A) 53(B) 54(A) 55(B) 56(A) 57(A)

76(D) 77(B) 78(D) 79(C) 80(A) 81(B) 82(D)

101(A) 102(B) 103(D) 104(B) 105(C) 106(C) 107(A)

126(D) 127(A) 128(D) 129(C) 130(B) 131(A) 132(B)

151(D) 152(A) 153(B) 154(C) 155(D) 156(A) 157(B)

8(B) 9(A) 10(C) 11(A) 12(D) 13(D)

33(A) 34(B) 35(C) 36(B) 37(A) 38(A)

58(B) 59(C) 60(D) 61(A) 62(C) 63(A)

83(B) 84(A) 85(B) 86(D) 87(A) 88(D)

108(A) 109(A) 110(C) 111(D) 112(B) 113(C)

133(C) 134(C) 135(D) 136(A) 137(B) 138(C)

158(C) 159(A) 160(B) 161(D) 162(D) 163(C)

14(B) 15(D) 16(D) 17(B) 18(C) 19(A) 20(B)

39(C) 40(D) 41(C) 42(B) 43(A) 44(C) 45(D)

64(C) 65(D) 66(A) 67(B) 68(C) 69(A) 70(D)

89(D) 90(A) 91(B) 92(C) 93(A) 94(B) 95(C)

114(B) 115(D) 116(C) 117(A) 118(C) 119(A) 120(B)

139(A) 140(B) 141(C) 142(C) 143(A) 144(C) 145(D)

164(B) 165(B)

21(B) 22(C) 23(D) 24(A) 25(B)

46(D) 47(B) 48(D) 49(A) 50(B)

71(A) 72(B) 73(C) 74(D) 75(A)

96(C) 97(D) 98(B) 99(A) 100(D)

121(A) 122(B) 123(C) 124(D) 125(C)

146(A) 147(B) 148(A) 149(B) 150(C)

17

HINT 91

  1.80  0.01  rad

33

Strong nuclear force 1 = = 102 Electronmagnaticforce 102

34

Strong nuclear force 1 = = 1013 Weak nuclear force 10-13

D

Electronmagnetic force 10-2 = = 1036 94 Gravational force 10-38

density ( ) =

35

41

F1 

kq1q 2 r12

F2 

10  10 –6 104 –9 10

58

10 –15  10–8 –9 100  10

60

km 72  1000 m 72   20 2 (min ) 3600 (sec) 2

64

Area = 

= [

96

A2 B C3

= 21 % 97

A 1  2  x –2 2 x x

–24

l g

gm

= 4 % 100

= (20.30  0.12) cm 103

p  F |1

length of two rods  2l

= 2(10.15  0.06)

F  p 1  q –1

heat energy H = I2 RT H I R T  100  [2   ]  100 H I R T

F N(Neuton) P   surface tension  (meter)

A  0.025 m

T  2

T 1  l 1 g  100  [    ]  100 T 2 l 2 g

1 amu = 1.66  10 –27 kg

= 10 %

2

105

2r  0.5m

Solid angle =

M  l  + 3  ]  100 M  l 

P A B C %  [2  3 ] P A B C

1gm  6.023  1023 amu

90

P

2

= 1.66  10

80

mass( m) volume( l 3 )

= 1.75 %

A = x 2m2

69

b  4.06  108 m 

percentage error in density

kq1' q '2 r22

49

1 m2 

b  1.27  107 m

g  4 2

l T2

g l T   2 g l T

A  0.4 Sr r2

= b + 2q

= 4  10 –1 Sr

18

107

Resis tan ce R

V I

148

R V I   R V I

PV – Pb 

R 5 0.3   R 100 10

 V  b  M 0 L3T 0

149

Bx  M 0 L0 T 0

mass 75.5 density =  volume 25

B

Re sistivity 

force electric ch arg e

Electricfield 

130

mass  (dis tan ce) 2 plank 's cons tan t  time

latent heat(Q) =

heat energy mass

coefficientof vis cos ity 

141

Urms 

142

If p = q = c = L then (p - q)c = L2 = Area

144

If F = nA n =

146

D  M 0 L0 T –1 

151

F  at  bt 2

b

153

F N  t 2 m2 2

F  A (1 – e – Bxt )

Bxt 2  dimensional less

u 2  root mean square speed

B

155

dv dx

F = pascal second dv A dx

D  M 0 L1T –1 B

F  bt 2  at

Force  time (length) 2

133

M 0 L0T 0  M 0 L–1T 0 X

Same as

Resis tan ce  Area length

129

131

cos Bx  dim ensionl less

R %  8% R

= 3.02 = 3.1 g / cm3 128

a ab –  constant v v2

 PV – Pb

R 8  R 100

111

a    P  2  (v – b)  cons tan t v  

M 0 L0T 0  M 0 L–1T –2 xt 2

Density = 3.125 

=

gm cm3

3.125  10 –3 kg 10 –6 m 3

= 3125 kg / m3 157

x y  A sin ( – k) v

volume of cube V  a 3 total surface area of cube A = 6a 2 V  A

x  k v

a 3  6a 2

x k   M 0 L0 T1 v

 V  (6)3  216 unit

a 6

19

159

E  G P hqcr E  M1L2 T –2

z  M 0 L0 T 0 kB

G  M –1L3T –2

 

kB uand P =   z

c  M 0 L1T –1 take it

 

 kB  p pz

(M1L2 T –2 )  (M –1L3T –2 ) p (M1L2T –1 )q

 M0L2T0

165

h  M1L2T –1

(M 0 L1T –1 ) c

= M –pq L3p2qr T –2p–q–r 1 1 5 P  , q  , r  2 2 2

160

F  ma v b r c 1 –2 F  M1LT

v M 0 L1T –1 1 0 r  M 0 LT

m  M1L0 T 0 take it (M1L1T –2 )  (M1 )a (L1T –1 ) b (L1 )c

= Ma LbcT – b  a  1, b  2, c  –1

20

Unit - 2 Kinematics

21

SUMMARY •

speed =

distance x time t

Average speed =

• •

Total distance Total time

Instantaneous speed = lim

t 0

x t

 displacement r Velocity v =  time t    r r Ins tan eous velocity υ  lim  t  0  t dt  t

•

Average acceleration Gave 

•

    υ dυ Ins tan tan eous acceleration a  lim  t   t dt Equation for Uniformally accelerated motion

•

• •

1 2 at 2

(1) υ = υ 0 + at

(3) d = o t +

 Vo  V  (2) s   t  2 

(4) V2 = Vo2 + 2ad

a Distan ce covered in n th Second Sn  Vo  (2n –1) 2 About Vectors

  A . B = AB cosq

 A

 ´ B = AB sinq nˆ

  2 A . A =|A|



iˆ . iˆ = ˆj . ˆj = kˆ . iˆ = 1

iˆ ´ iˆ = ˆj ´ ˆj = kˆ ´ kˆ = 0

iˆ . ˆj = ˆj . kˆ = kˆ . kˆ = 0

iˆ ´ ˆj = K ˆj ´ kˆ = iˆ kˆ ´ iˆ = ˆj

 

cosq

. =A B AB



A ´ A =0

ˆj iˆ kˆ   Ax Ay Az A ´ B = Bx By Bz

22

    A ^ B then A . B = 0

    A ^ B then | A ´ B | = AB



    A || B then A ´ B = 0

A

   || B then A . B = AB

    | A | = | B | and A and B is Q the angle between then

  | A + B | = 2A

(1)

θ =0

(2)

  θ =180 then | A + B | = 0

(3)

  θ =90 then | A + B | = 2 A

(4)

θ =60 then

(5)

  θ =120 then | A + B | = A

  | A + B | = 3A

For projectile - Time to reach the highest point tm  - Maximum height H = - Range R =

υ o sin  g

υ 20 sin 2 θ 2g

υ 20 sin2 θ g

- Maximum Range R 

- Flight time T 

υ 20 g

2υ o sin  g

- Equation of trajectory y  x tan  –

gx 2 2 2o cos 2 

- R  4H cot 

23

MCQ For the answer of the following questions choose the correct alternative from among the given ones. (1) A branch of physics dealing with motion without considering its causes is known as .... (A) Kinematicas (B) dynamics (C) Hydrodynemics (D) mechanics (2) Mechanics is a branch of physics. This branch is ... (A) Kinematics without dynamics (B) dynamics without Kinematics (C) Kinematics and dynamics (D) Kinematics or dynamics (3) To locate the position of the particle we need ... (A) a frame of referance (B) direction of the particle (C) size of the particle (D) mass of the particle (4) Frame of reference is a ... and a ... from where an obeserver takes his observation, (A) place, size (B) size, situation (C) situation, size (D) place, situation (5)

0

B -2 -1

A 1

2

3

(m)

As shown in the figure a particle moves from 0 to A, and then A to B. Find pathlength and displacement. (A) 2m, –2m (B) 8m, –2m (C) 2m, 2m (D) 8m, –8m (6)

A particle moves from A to B and then it moves from B to C as shown in figure. Calculate the ratio between path lenghth and displacement.

1 (D)  2 A particle moves from A to P and then it moves from P to B as shown in the figure. Find path length and dispalcement.

(A) 2 (7)

(B) 1

P

(A)

2l

3

, l

(C)

600

(B)

l 3

, l

(C) 2l, l

24

(D) l,

2l 3

(8)

A car goes from one end to the other end of a semicircular path of diameter ‘d’. Find the ratio between path legth and displacement. (A)

(9)

3 2

(B) 

(C) 2

(D)

π 2

A particle goes from point A to B. Its displacement is X and pathlength is y. So

x ..... y

(A) > 1 (B) < 1 (C)  1 (D)  1 (10) As shown in the figure a partricle statrs its motion from 0 to A. And then it moves from A to B. AB is an arc find the Path length



O

3

r

  π  (C) r  1 +  (D)  r  1 3 3 3  (11) Here is a cube made from twelve wire each of length l. An ant goes from A to G through path A-B-C-G. Calculate the displacement.

(B) r 

(A) 2r

E

H

F

G 

l 3 (12) As shown in the figure particle P moves from A to B and particle Q moves from C to D. Desplacements for P and Q are x and y respectivey then (A) 3l

(B) 2l

(C)

(B) x < y

(C) x = y

3l

(D)

5 4 3 2 1 O 1

2

(A) x > y

3

4

5

25

(D) x  y

(13) Shape of the graph of position  time given in the figure for a body shows that x

o t

(A) The body moves with constant acceleration (B) The body moves with zero velocity (C) The body returns back towards the origin (D) nothing can be said (14) The graph of position  time shown in the figure for a particle is not possible because ... x

o

t

(A) velocity can not have two values on one time (B) Displacement can not have two values at one time (C) Acceleration can not have two values at one time (D) A, B and c are true (15) An ant goes from P to Q on a circular path in 20 second Raidus OP = 10m. What is the average speed and average velocity of it ? R

0

0 12

O

p

(A)

 ms –1 , 3 ms –1 6

(B)

 3 ms –1 , ms –1 3 2

 ms –1 , 3 ms –1 (D)  ms –1 , 6 ms –1 3 (16) A particle is thrown in upward direction with initial velocity of 60 m/s. Find average speed and average velocity after 10 seconds. [g = 10 ms–2] (A) 26ms–1, 16ms–1 (B) 26ms–1, 10ms–1 –1 –1 (C) 20ms , 10ms (D) 15ms–1, 25ms–1

(C)

26

(17) The ratio of pathlength and the resepective time interval is (A) Mean Velocity (B) M ean speed (C) intantaneous velocity (D) intantaneous speed (18) A car moving over a straight path covers a distance x with constant speed 10 ms–1 and then

the same distance with constant speed of V2. If average speed of the car is 16ms–1, then V2 = .... (A) 30 ms–1 (B) 20 ms–1 (C) 40 ms–1 (D) 25 ms–1 (19) A bus travells between two points A ans B. V1 and V2 are it average speed and average velocity then (A) v1 > v2 (B) v1 < v2 (C) v1 = v2 (D) depends on situation (20) A car covers one third part of its straight path with speed V1 and the rest with speed V2. What is its average speed ?

3 v1v 2 (A) 2 v  v 1 2

2 v1 v 2 (B) 3v  v 1 2

3 v1 v 2 3v1v 2 (D) v1  2 v 2 2 v1  2 v 2 (21) Rohit completes a semicirular path of radius R in 10 seconds. Calculate average speed and average velocity in ms–1.

(C)

R R πR 2R 2R R 2R 2R (B) (C) (D) , , , , 10 10 10 10 10 10 10 10 (22) A particle moves 4m in the south direction. Then it moves 3m in the west direction. The time taken by the particle is 2 second. What is the ratio between average speed and average velocity ?

(A)

5 7 14 5 (B) (C) (D) 7 5 5 14 –1 (23) A particle is projected vertically upwards with velocity 30ms . Find the ratio of average speed and instantaneous velocity after 6s. [g = 10ms–1]

(A)

1 (B) 2 (C) 3 (D) 4 2 (24) The motion of a particle along a straight line is described by the function x = (3t – 2)2. Calculate the acceleration after 10s. (A) 9ms–2 (B) 18mls (C) 36ms–  (D) 6ms–  (25) Given figure shows a graph at acceleration  time for a rectilinear motion. Find average acceleration in first 10 seconds.

(A)

m a 2  S 

10

o 5

10

15

t

(A) 10ms–2 (B) 15ms–2 (C) 7.5ms–2 (D) 30ms–2 (26) A body starts its motion with zero velocity and its acceleration is 3m/s2. Find the distance travelled by it in fifth second. (A) 15.5m (B) 17.5m (C) 13.5m (D) 14.5m 27

(27) A body is moving in x direction with constant acceleration  . Find the difference of the displacement covered by it in nth second and (n–1)th second.  3 (C) 3 (D)  2 2 What does the speedometer measure kept in motorbike ? (A) Average Velocity (B) Average speed (C) intantaneous speed (D) intantaneous Velocity The displacement of a particle in x direction is given by x = 9 – 5t + 4t2. Find the Velocity at timt t = 0 (A) –8 ms–1 (B) –5 ms–1 (C) 3 ms–1 (D) 10 ms–1 A freely falling particle covers a building of 45m height in one second. Find the height of the point from where the particle was released. [g = 10ms–2] (A) 120m (B) 125m (C) 25m (D) 80m 2 The distance travelled by a particle is given by s = 3 + 2t + 5t The initial velocity of the particle is ... (A) 2 unit (B) 3 unit (C) 10 unit (D) 5 unit A particle is thrown in upward direction with Velocity V0. It passes through a point p of height h at time t1 and t2 so t1 + t2 = ....

(B)

(A) α (28)

(29)

(30)

(31)

(32)

2 v0 2h h v0 (B) (C) (D) g g 2g g (33) A particle is thrown in upward direction with initial velocity V0. It crosses point P at height h at time t1 and t2 so t1t2 = _______ (A)

V0 2 (B) 2g

2h (A) g

2V0 2 (C) g

(D)

h 2g

(D)

h 2g (34) Ball A is thrown in upward from the top of a tower of height h. At the same time ball B starts to fall from that point. When A comes to the top of the tower, B reaches the ground. Find the the time to reach maximum height for A. (A)

h g

(B)

2h g

(C)

4h g

(35) In the figure Velocity (V)  position graph is given. Find the true equation. V Vo X

xo

(A) v 

v0 x – v0 x0

(B) v  –

v0 x  v0 x0

28

(C) v 

v – v0 x – v 0 (D) v = 0 x + v 0 x0 x0

(36) In the figure there is a graph of a  x for a moving particle. Hence da = .... V dt a

X

o ao

xo

–x 0 –a 0 x0 a (B) (C) (D) 0 a0 x0 a0 x0 –1 (37) A particle is moving in a straight line with intial velocity of 10 ms . A graph of acceleration  time of the particle is given in the figure. Find velocity at t = 10 s.

(A)

a 5

o

10 –1

(A) 25 ms (B) 35 ms–1 (C) 45 ms–1 (D) 15 ms–1 (38) A graph of moving body with constant acceleration is given in the figure. What is the velocity after time t ? D V B

A o

(A) 0A +

C E t1

t

DC  0E BC

(B) 0A +

DC BC  DE (C) AB +  0E BC DC

(D) 0A + DC  AD BC

(39) V

o

t

The graph given in the figure shows that the body is moving with ..... (A) increasing acceleration (B) decreasing acceleration (C) constant velocity (D) increasing velocity 29

(40) Slope of the velocity-time graph gives of a moving body.. (A) displacement (B) acceleration (C) initial velocity (D) final velocity (41) The intercept of the velocity-time graph on the velocity axis gives. (A) initial velocity (B) final velocity (C) average velocity(D) instanteneous velocity (42) Here are the graphs of velocity  time of two cars A and B, Find the ratio of the acceleration after time t. B

V

A

60 130 t

1 1 (B) (C) 3 (D) 3 3 3 (43) Here is a velocity - time graph of a motorbike moving in one direction. Calculate the distance covered by it in last two seconds. (A)

V C

10m 5 o

t 2 3 4 5

1

(A) 5 m

(B) 20 m

(C) 50 m

(D) 25 m

a

(44)

t

In the above figure acceleration (a)  time (t) graph is given. Hence V  ..... (A) a (45)

(B)

(C) a2

a

(D) a3

X A o

B

t

The graph of displacent (x)  time (t) for an object is given in the figure. In which part of the graph the acceleration of the particle is positive ? (A) OA (B) AB (C) 0 - A - B (D) acceleration is not positive at any part. 30

(46) In a uniformly accelerated motion the slope of velocity - time graph gives .... (A) The instantaneous velocity (B) The acceleration (C) The initial velocity (D) The final velocity (47) The area covered by the curve of V – t graph and time axis is equal to magnitude of .... (A) change in velocity (B) change in acceleration (C) displacement (D) final velocity (48) An object moves in a straight line. It starts from the rest and its acceleration is 2ms–2. After reaching a certain point it comes back to the original point. In this movement its acceleration is -3ms-2. till it comes to rest. The total time taken for the movement is 5 second. Calculate the maximum velocity. (A) 6 ms–1 (B) 5 ms–1 (C) 10 ms–1 (D) 4 ms–1 (49) The relation between time and displacement of a moving particle is given by t  2x 2 where  is a constant. The shape of the graph x  y is ... (A) parabola (B) hyperbola (C) ellips (D) circle (50) Here are the graphs of x  t of a moving body. Which of them is not suitable ? x

x

(A)

(B) o

o

t

t

(D) x

(C) x

t

t

(51) Here are the graphs of v  t of a moving body. Which of them is not suitable ? (A)

V

V

(B)

t

t

(C)

V

V

(D)

t

t

V A o

Comprehension type questions

B C D

t

In the figure there is a graph of velocity  time for a particle. 31

(52) Which area shows the displacement covered by the particle after time t (A) closed fig AODCA (B) closed fig. ABCA (C) closed fig. AODCBA (D) none of above (53) Which part shows initial velocity of the particle ? (A) OA (B) AB (C) AC (D) AOA (54) How will you calculate the acceleration of the particle ? (A) taking length of AB (B) taking magnitude of BC (C) taking slope of AC (D) taking slope of AB X

(55) o

t

2

4

6

8

Given graph shows relation between position and time. Find correct graph of acceleration  time (A)

(B) a

a t 2

4

6

8

X

(C)

2

6

4

2

4

6

8

o

t 4

2

X

(D)

o

t

o

8

t 8

B

(56) X

A

60 130 t

Here are displacement  time graphs of particle A and B. If VA and VB are velocities of the particles respectively, then

(A)

1 3

VA = ..... VB

(B) 3

(C) 32

1 3

(D) 3

(57) X t

2

4

6

8

Given graph shows relation between position (x)  time (t) Find the correct graph of velocity  time. V

(A)

(B)

o

2

t 4

6

V

2

8

(C) V

t 4

6

8

(D) V t 2

4

6

2

8

4

6

8

t

(58) Particles A and B are released from the same height at an interval of 2 s. After some time t the distance between A and B is 100m. Calculate time t. (A) 8 s (B) 6 s (C) 3 s (D) 12 s (59) As shown in the figure a particle is released from P. It reachet at point Q at time t1 and reaches at point R at time t2 so

t1 = .... t2

2 4 1 1 (B) (C) (D) 1 1 3 2 (60) A particle moves in stright line. Its position is given by x = 2 + 5t – 3t2. Find the ratio of intial velocity and initial acceleration.

(A)

6 5 5 6 (B) – (C) (D) – 5 6 6 5 (61) A particle is moving in a circle of radius R with constant speed. It coveres an angle  in some time interval. Find displacement in this interval of time.

(A) 

 (B) 2Rsin θ (C) 2Rcos (D) 2Rsin 2 2 (62) A particle is moving in a straight line with initial velocity of 200 ms–1 acceleration of the particle is given by a = 3t2 – 2t. Find velocity of the particle at 10 second. (A) 1100ms–1 (B) 300ms–1 (C) 900ms–1 (D) 100ms–1

(A) 2R cos

33

(63) Angle of projection, maximum height and time to reach the maximum height of a particle are  , H and tm respectivley. Find the true relation. (A) (B) (C) (D) H 4H

tm 

tm =

2g

2H g

tm =

tm =

g

H 4g

(64) Particle A is projected vertically upward from a top of a tower. At the same time particle B is dropped from the same point. The graph of distance (s) between the two particle varies with time is. (B) S

(A) S

P t

t

(C)

S

h Q

(D) S 3h

R

t

t

(65) A car is moving with speed 30m. Due to application of brakes it travells 30m before stopping. Find its acceleration. m m m m (B) –15 2 (C) 30 2 (D) 10 2 2 s s s s 2 A particle moves with a constant acceleration 2m/s . Its intial velocity is 10m/s. Find velocity after t second. (A) (10 + t) ms–1 (B) 5(2 + t)ms–1 (C) 2 (5 + t)ms–1 (D) (10 + t2) ms–1 A particle moves in a straight lime with constant acceleration. At t = 10s velocity and displacement of the particle are 16ms–1 and 39m respectively. What will be the velocity after 10 s ... (A) 22 ms–1 (B) 18 ms–1 (C) 20 ms–1 (D) 28 ms–1 A particle moves with constant acceleration 2m/s2 in x direction. The distance travelled in fifth second is 19 m. Calculate the distance travelled after 5 second. (A) 50 m (B) 75 m (C) 80 m (D) 70 m Two bodies of masses m1 and m2 are dropped from heights H and 2H respectively. The ratio of time taken by the bodies to touch the ground is ...

(A) 15 (66) (67)

(68)

(69)

1 1 (B) 2 (C) (D) 2 2 2 1 (70) A freely falling stone crashes through a horizontal glass plate at time t and losses half of its velocity. Af(A)

t it falls on the ground. The glass plate is 60 m high from the ground. Find the total distance travelled 2 by the stone. [g = 10ms–2] (A) 120 m (B) 80 m (C) 100 m (D) 140 m (71) A freely falling object travells distance H. Its velocity is V. Hence, in travelling further distance of 4H its velocity will become ....

ter time

(A)

3V

(B)

(C) 2V

5V 34

(D) 3V

(72) A ball is thrown vertically upward direction. Neglacting the air resistance velocity of the ball in air will (A) zero (B) decrease when it is going up (C) decrease when it is coming down (D) remain constant (73) Two particles P and Q get 5 m closer each second while travelling in opposite direction. They get 1 m closer each second while travelling in same direction. The speeds of P and Q are respectively ... (A) 5 ms–1, 1 ms–1 (B) 3 ms–1, 4 ms–1 (C) 3 ms–1, 2 ms–1 (D) 10 ms–1, 5 ms–1 1

(74) Motion of a porticle is described by an equation υ =  A + y  2 where v, y and A are velocity distance and a constant respectively. Find the acceleratrion of the particle. 1 (D) 3 unit unit 2 (75) The minumum distance in which a car can be stopped is x. The velocity of the car is V. If the velocity is 2V then find the stopping distance.

(A) 1 unit

(B) 2 unit

(C)

1 x 2 (76) A particle moves in one direction with acceleration 2 ms–2 and initial velocity 3 ms–1.After what time its displacement will be 10 m ? (A) 1 s (B) 2 s (C) 3 s (D) 4 s (77) A goods train is moving with constant acceleration. when engine passes through a signal its speed is U. Midpoint of the train passes the signal with speed V. What will be the speed of the last wagon ?

(A) 2x

(A)

(78)

(79)

(80)

(81)

(B) 4x

(C) 3x

V2 – U2 2

(B)

(D)

V2  U2 2

2V 2  U 2 (C) (D) 2V 2 – U 2 2 Displacement of a particle in y direction is given by y = t2 – 5t + 5 where t is in second. Calculate the time when its velocity is zero. (A) 5 s (B) 2.5 s (C) 10 s (D) 3 s The area under acceleration versus time graph for any time interval represents... (A) Intial velocity (B) final velocity (C) change invelocity in the time interval (D) Distance covered by the particle A ball is thrown vertically upward. What is the velocity and acceleration of the ball at the maximum height ? (A) –gt ms–1, 0 (B) 0, –9 ms–2 (C) g ms–1, 0 (D) 0, –gt ms–2 The relation between velocity and position of a particle is V = Ax + B where A and B are constants. Acceleration of the particle is 10 ms–2 when its velocity is V, How much is the acceleration when its velocity is 2V. (A) 20 ms–2 (B) 10 ms–1 (C) 5 ms–2 (D) 0

35

(82) A particle moves on a plane along the path y = Ax3 + B in such a way that

dx  c . c, A, dt

B are constans. Calculate the acceleration of the particle. ^

^

(A) 3Axc jms –2

(B) 5Axc 2 jms –2 ^ (D)  c i  3Axc2 

^

(C) 3Axc 2 jms –2

^

 j  ms –2 

(83) The relation between velocity and position of a particle is given by V   –  x . Its initial velocity is zero. Find its velocity at time t 

(84)

(85) (86)

(87)

1 B

1 –1 (D) e2 ms–1 ms e An object moves in x - y plane. Equations for displacement in x and y direction are x = 3sin2t and y = 3cos2t Speed of the particle is (A) zero (B) constant and nonzero (C) increasing with time t (D) decreasing with time t Motion of a particle is decribed by x = (t – 2)2 Find its velocity when it passes through origin. (A) 0 (B) 2 ms–1 (C) 4 ms–1 (D) 8 ms–1 To introduce a vector quantity .... (A) it needs magnitude not direction (B) it needs direction not magnitude (C) it need both magnitude and direction (D) nothing is needed Which pair of two vectors is antiparallel.

(A) e ms–1

(B) 0 ms–1

(C)

(A) A

(B)

(C)

B

A B

(D)

A B

A B

  (88) In the above figure P and Q are two vectors. What from followings is true P Q

    (A) P and Q are equal (B) P and Q are perpendicular     (C) P and Q are antiparallel (D) P and Q are in same direction (89) Which from the following is a scalar ? (A) Electric current (B) Velocity (C) acceleration (D) Electric field   (90) P and Q are equal vectors what from the followings is true.     (A) P and Q are antiparallel (B) P and Q are parallel     (C) P and Q may be perpendicular (D) P and Q may be free vectors

36

  (91) P = Q is true, if ... (A) their magnitudes are equal (B) they are in same direction (C) their magnitudes are equal and they are in same direction (D) their magnitudes are not equal and they are not in same direction   (92) A and B are in opposite direction so they are (A) parallel vectors (B) anti parallel vector (C) equal vector (D) perpendicular vector ^ ^  ^  (93) A  1  2 j  2 k . Calculate the angle between A and Y axis.

(A) sin –1

5 3

1 3

–1 (B) sin

(C) sin –1

10 3

(D) cos –1

5 3

  (94) A and B are nonzero vectors. Which from the followings is true ?   2   2   2   2 (A) A + B – A – B = 2 A 2 + B2 (B) A  B – A – B  2 A 2 – B2









  2   2   2   2 (C) A  B – A – B  A 2  B2 (D) A  B – A – B  A 2 – B2      (95) C  A  B and A = B = C Find the angle between A and B   2 (B) (C) (D) 0 3 6 3 (96) The resultant of two vectors is maximum when they. (A) are at right angles to each other (B) act in oppsite direction (C) act in same direction (D) are act 1200 to each other   (97) The resultant of two veetors A and B (A) can be smaller than A – B in magnitude (B) can be greater than A + B in magnitude (C) can’t be greater than A + B or smaller than A – B in magnitude (D) none of above is true (98) The resultant of two forces of magnitude 2N and 3N can never be.

(A)

(A) 4N

(B) 1N

(C) 2.5N

(D)

1 N 2

  (99) The sum of P and Q is at right agnles to their difference then (A) A = B (B) A = 2B (C) B = 2A (D) A = B = A – B         (100) Magnitudes of A, B and C are 41, 40 and 9 respectively, A  B  C Find the angle between A and B

(A) sin –1

9 40

(B) sin –1

9 41

(C) tan –1

9 41

(D) tan –1

41 40

 (101) If A  3i  4j  9k is multiplied by 3, then the component of the new vector along z direction is ... (A) –3 (B) + 3 (C) –27 (D) + 27 37

        (102) A  B is perpendicular to A and B  2 A  B What is the angle between A and B  5π 2π π (B) (C) (D) 6 6 3 3 (103) Out of the following pairs of forces, the resultant of which can not be 18N (A) 11 N, 7 N (B) 11 N, 8 N (C) 11 N, 29 N (D) 11 N, 5 N     (104) A  3i  2j – 5k and B  i  j  2k Find A  2B

(A)

(A)

(B) 42 (C) 39 (D) 2      (105) What is the angle between Q and the resultant of P  Q and Q – P (A) 900 (B) 600 (C) 0 (D) 450     (106) A  2i  2j – k and B  2i – j – 2k Find 3A – 2B 40

(A) 2i  7j  k

(B) 2i  8j – k

(C) 2i + 8j + k

(D) i + 7j + k

(107) Linear momentajm of a particle is 3i + 2j – k kgms–1. Find its magnitude.





(A) 14 (B) 12 (C) 15 (D)     (108) A × B = C, Then C is perpendicular to   (A) A only (B) B only   (C) A and B both when the angle between them is ...   (D) A and B both whatever to be the angle between them   (109) If A  B  0 then   (A) A must be zero (B) B must be zero   (C) either A  0, B  0 or   0

11

  (D) either A = 0, B = 0 or θ = π 2

  (110) y component of A  B is ....  A  Axi + Ayj  Azk  B  Bxi + Byj  Bzk (A) AxBy – AyBx (B) AzBx – AxBz (C) AxBz – AzBx (D) AzBy – AyBz     (111) If A  B  0 then A  B = ...

(A) AB

(B)

A B

(C)

1 AB 2

(D) 0

    (112) If A  4i + 3j – 2k and B  8i + 6j – 4k the angle between A and B is (A) 450 (B) 0 (C) 60 (D) 90 38

    (113) A  2i – 3j  k and B  8i + 6j – 4k then A  B = .... (A) 28 (B) 14 (C) 0 (D) 7      (114) If A  2i + 5j – k and B  3i – 2j – 4k the angle between A and B is ...

(A) 0

(B)

π 2

(C)

(115) Which statement is true ?     (A) A  B  B  A     (C) A  B  – B  A (116) Which from the following is true ?   AB (A) cos   AB   A×B (C) tanθ =   A-B     (117) A  B  A  B  ....





π 4

(D)

π 6

    (B) A × B = – B × A   (D) A  B  AB

  AB (B) sin   AB

AB (D) cot     AB



(B) A2 + B2

(A) 0

(C)

A 2  B2

(D) A2B2

(118) The angle between i + j and z axis is .... (A) 0 (B) 45 (C) 90 (D) 180     (119) A  2i – j  2k and B  – i – 2j  2k what is the angle between A and B (A) cos–1 0.8888 (B) cos–1 0.4444 (C) sin–1 0.4444 (D) sin–1 0.8888   (120) A  Pi – 2Pj – k and B  – 3i + 2j   14k are perependcular to each other. Then p = ... (A) 3 (121)

(B) 4

(C) 2

In the above triangle AC = 5, BC = 8 and B =

(D) 1

 Find the value of angle A. 6

5  6

8

(A) sin–1 0.6 (B) sin–1 0.8 (C) sin–1 0.12 (D) sin–1 0.4    (122) A  – l + j – 2 k and B  2i – j  k Find the unit vectior in direction of A  B (A)

1 –i – 5j – 2k 23



(B)

1 –i + 5j – 3k 35

(C)

1 –i – 5j – 3k 29



(D)

1 –i – 5j – 3k 35





39





 

(123) What is unit vector along i + j ?   (A) i + j 2

(B)

i + j 2

(C)

i + j 3

(D)

i – j 2

    (124) Unit vector of A  B is k . Unit vector of A is i Then what is the unit vector of B (A) j (B) – j (C) any unit vector in xy plane (D) any unit vector in xz plane (125) Find a unit vector in direction of i + 2j – 3k

(A)

1   i + 2j – 3k 7

(C)

1  i + 2j – 3k 14





(126) Find a unit   A B (A) AB (127) If resultant



(B) –

1  i + 2j – 3k 2





1   i + 2 j – 3k 5   vector perpenduicular to both A and B       A×B A × B (B) (C) (D) A  B ABcosθ ABsinθ ABsinθ      of A  2i + j – k , B  i – 2j + 3k and C is unit vector in y direction, then C is (B) 3i – 2j + 2k (C) j (D) 2i + 3k



(D)





(A) – j   A = U  B Now find the true option. (128) A and B are two vectors U     (A) A and B equal vectors (B) A and B are in opposite direction     (C) A and B are in same direction (D) A  B  (129) Unit vector in direction of A is      A A (A) A (B)  (C) A A (D)  A A (130)

2 2 i + j + pk is a unit vector so p = ..... 3 3

2 1 (B) – 3 3 (131) Find a unit vector from the followings.

(A)

(C) 1

(D)

1 9

1  1 1 1  i– j (D) i+ j 2 2 2 2 (132) Train A is 56 m long and train B 54 m long. They are travelling in opposite direction with velocity (A) i + j

(B) i – j

(C)

m m and 5 respectively. The time of crossing is. s s (A) 12 s (B) 6 s (C) 3 s

15

40

(D) 18 s

(133) Graphs of velocity  time for two cars A and B moving in a straight line are given in the fig. The area covered by PQRS gives. S

V

B

P

A R

Q

t1

(A) (B) (C) (D) (134) V

t2

t

distance from A to B at time t2 distance from A to B at time t1 distance from A to B in time interval t2 – t1 change in distance from A to B in time interval t2 – t1 B A

o

Graphs velocity  At time t = 0 they (A) They will meet (C) They will meet

t

time is given for cars A and B moving in a straight line in same direction. are moving in the direction from A to B, then. once (B) They will never meet twice (D) none of above is true

(135) Velocity of particle A with respect to particle B is 4 m while they are moving in same direction. s m while they are in opposite direction. What are the velocities of the particles with s respect to the stationary frame of reference. (A) 7 ms–1, 3 ms–1 (B) 4 ms–1, 5 ms–1 (C) 7 ms–1, 4 ms–1 (D) 10 ms–1, 4 ms–1 (136) Stone A is thrown in horizontal direction with velocity of 10 ms–1 at the same time stone B freely falls vertically in downword direction. Calculate the velocity of B with respect to A after 10 second.

And it is 10

(A) 10 ms–1

(B)

(C) 10 101ms–1

101 ms –1

(D) 0

(137) A car moves horizontally with a speed of 3 ms–1. A glass wind screen is kept on the front side of the car. Rain drops strike the screen vertically. With the Velocity of 5 ms-1 Calculate the velocity of rain drops with respect to a ground. (A) 6 ms–1 (B) 4 ms–1 (C) 3 ms–1 (D) 1 ms–1

41

 (138) A man crosses a river through shortest distance D as given in the figure. VR is velocity of   water and V m is velocity of man in still river water. If VmR is relative velocity of man w.r.t. river, then find the angle made by swimming man with the shortest distance AB B

D

A

(A) tan –1

(C) tan –1

VR Vm – VR VR Vm2 – VR2

(B) tan –1

Vm VR2 – Vm2

(D) tan –1

VR VR2 – Vm2

(139) A particle has initial velocity 2i + 3j ms –1 and has acceleration i + j ms –2 . Find the velocity





 

of the particle after 2 second. (A) 3i + 5j ms –1

(B) 4i + 5j ms –1

–1

–1

(C)

  3i + 2j ms

  (D)  5i + 4j ms

 (140) Motion of a body in x - y plane is described by r   2ti   5t 2  6t  j m Then find velocity   of the body at t = 1 second. (A)

144ms –1

(B)

(C)

148ms –1

150ms –1

(D)

260ms –1

 (141) A particle moves in x - y plane. The position vector of the particle is given by r  3ti – 2t 2 j m Find





the rate of change of  at t = 1 second. Where  is the anghe betheen direction of motion and x (A)

16 25

(B)

12 25

(C) –

12 25

(D)

16 9

(142) x and y co-ordinates of a particle moving in x-y plane at some instant are x  2t 2 and y  Calculate y co-ordinate when its x coordinate is 8m. (A) 3 m (B) 6 m (C) 8 m (D) 9 m (143) A particle in xy plane is governed by x = A cos t, y = A (1 – sin  t). A and What is the speed of the particle. (A) A ω t

(C) A  cos  t

(B) A ω 2t 42



2 (D) A  sin

3 2 t 2

are constants.

t 2

(144) A particle is moving in a xy plane with y = 2x and Vx = 2 – t. Find Vy at time t = 3 second. (A) 2 ms–1 (B) –3 ms–1 (C) +3 ms–1 (D) –2 ms–1 (145) t1 and t2 are two values of time of a projectile at the same height t1 + t2 = ... (A) Time to reach maximum height (B) fight time for the projectile 3 3 time of the fight time. (D) time of the fight time. 4 2 (146) Eequetion of a projectile is given by y = Ax – Bx2. Find the range for the particle.

(C)

A A A 2A (B) (C) (D) B 4B 2B B (147) Angle of projection of a projectile with horizonal line is  at time t = 0, After what time the angle will be again  ?

(A)

(A)

V cos  g

(B)

Vsin  g

(C)

V0 sin  2g

(D)

2V0sinθ g

(148) A particle is projected with initial speed of V0 and angle of  . Find the horizontal displacement when its velocity is perpendicular to initial velocity.

V02 (A) gtanθ

(B)

V02 gsinθ

(C)

V0 sin  g

(D)

V02 tan

(149) Intial anlge of a projectile is  and its initial velocity is V0. Find the angle of velocity with horizontal line at time t.

  g (A) sin –1 1 – t  V0 cos  

  g (B) tan –1 1 – t  V0 cos  

  g (C) tan –1  tanθ – t V0cosθ  

  g (D) sin –1  tanθ – t V0cosθ  

(150) A stone is projected with an angle θ and velocity V0 from point P. It strikes the ground at point Q. If the both P and Q are on same horizontal line, then find average velocity. (C) V0 cos θ (D) V0sin θ 2 2 –1 (151) An object is projected with initial velocity of 100 ms and angle of 60. Find the verticle velocity when its horizontal displacement is 500 m. (g = 10 ms–1) (A) 93.35 ms–1 (B) –93.35 ms–1 (C) –8.65 ms–1 (D) 98 ms–1 (152) Angle of projection of a projectile is changed, keeping initial velocity constant. Find the rate of change of maximum height. Range of the projectile is R. (A) V0cosθ

(B) V0sinθ

R R R (B) (C) (D) R 4 3 2 (153) A cartasian equation of a projectile is given by y = 2x – 5x2. Calculate its initial velocity.

(A)

(A)

10ms –1

(B)

(C)

5ms –1

43

2ms –1

(D) 4 ms–1

(154) A body travelling in a circle at constant speed. (A) has a constant velocity (B) is not accelerated (C) has an outward radial acceleration (D) has an inward radial acceleration (155) The speed of a particle moving in a circle of radius r = 4 meter is 10 ms–1. What is its radial acceleration ? (A) 25 ms–2 (B) 20 ms–2 (C) 10 ms–2 (D) 15 ms–2 (156) If f is the frequency of a body moving in a circular path with constant speed. a is its centrifugal acceleration, so. (A) a  f

•

(C) a α f 3

(B) a α f 2

(D) a α

1 f

Following question is Assertion - Reason type question. choose

(A) If both Assertion - Reason are true, reason is correct explanation of Asserton. (B) If both Assertion - Reason are true but reason is not correct explanation of Asserton. (C) If Asserton is true but Reason is false. (D) If Reason is true but Asserton is false. (157) Asserton : At the highest point of projectile motion the velocity is not zero. Reason : Only the verticle component of velocity is zero. Where as horizontal component still exists. (A) a (B) b (C) c (D) d Match coloum type question. (158) A balloon rise up with constant net acceleration of 10 ms–2. After 2 second a particle drops from the balloon. After further 2 s match the following (take g = 10 ms–2) Table - 1 Table 2 (A) Hight of the particle from ground (P) zero (B) Speed of particle (Q) 10 SI unit (C) Displacement of particle (R) 40 SI unit (D) Accelereation of the particle (S) 20 SI unit (A) A - P, B - Q, C - R, D - S (B) A - Q, B - R, C - S, D - P (C) A - R, B - P, C - S, D - Q (D) A - R, B - S, C - P, D - Q Comprehensions type questions. A particle is moving in a circle of radius R with constant speed. The time period of the particle is T Now T . 6 (159) Average speed of the particle is

after time t =

R 2R (B) 6T 3T (160) Average velocity of the particle is

(A)

(C)

2R T

(D)

R T

3R 6R 2R 4R (B) (C) (D) T T T T (161) Range of a projectile is R and maximum height is H. Find the area covered by the path of the projectile and horizontal line.

(A)

(A)

2 RH 3

(B)

5 RH 3

(C) 44

3 RH 5

(D)

6 RH 5

KEY NOTE 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 30 31

A C A D B A A D C C C A C D B B B C D A C B A B C C A C B B A

32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62

B A C D C C A B B A B C C D B C A B C D C A D D A A B B B B A

63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93

B A B B C B C D B B C C B B D B C B A B C B A C D B A B C B A

45

94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124

A C C C D A B C B D B C C A D D B A B C B B C A C B C B D B C

125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161

C B B C B B C B C A A C B C B D C B C D B A D A C A B C B D A B A C C B A

• • •

HINT (9)

Pathlenth is always greater or equal to displacement

(18) Average speed =

2V1V2 V1 + V2

(23) h =

V0 2 2V0 2 speed = 2g 2g × 6

(24) V =

dx dv and a = dt dt

change in velocity time change in velocity is the area covered by the graph.

(25) Find

1 1 g (t + 1) 2 – gt 2 = 45 2 2

(30) h 2 – h1 = (31) V =

ds = 2 + 10t dt

(32,33) h = V0 t – 

1 2 gt 2

1 2 gt – V0 t + h = 0 2

 t2 +

2V0 2h + =0 g g

There are two real values of t. (35) slope is

V0 x0

(37) Aera covered by a  t graph gives change in velocity.. (42) acceleration = slope = tanθ (44)

da = constant dt 

da dv  =k dv dt



da k = dv a

  ada =  kdv



a2 = kv 2 46

d2x (45) a = 2 dt (48) If maximum velocity is V. V = a1t1 and V = a2t2 T = t1 + t 2 = V=

(59) h =

v v + a1 a 2

a1a 2 T a1 + a 2 1 2 1 gt1 and (h + 3h) = gt 2 2 2 2

(61)

R

S

Q R

S s =



R 2  R 2  2R 2 cos θ

2R 2 -2R 2 Cosθ 

2R 2 (1-Cosθ)

= 2Rsin

(65) stopping distance ds =

θ 2

V0 2 2g

(66) V = V0 + at d = V0 t +

1 2 at 2

1 2 gt H α t 2 2 (70) At time t velocity V = gt

(69) H =

(70) y = V0 t +

1 2 gt 2

t 1 t y = gt × – g   2 2 2

2

(73) V1 + V2 = 5 V1 – V2 = 1 So 2V1 = 6 V1 = 3m/s 1

(74) V = (A + y) 2 47

a =

dv dv dy   dt dy dt

v2 2a (81) V = Ax + B a = Av aαV (75) s =

dx dx 2 = c and 2 = 0 (82) dt dt 3 y = Ax + B 

dy dx = 3Ax 2 dt dt

and

d2y dx dx = 6Ax × 2 dt dt dt

d2 y  2 = 6Axc 2 dt  d2 x ^ d2 y ^ Now a = 2 i + 2 j at dt    (95) C  A  B C2 = A 2 + B2 + 2AB cosθ   (116) A – B = ABcosθ   A  B = ABsinθ    ^ (127) A + B + C = j

4 4 + + p2 = 1 9 9 (131) Velocity of A wrt ground ^  VAG = 10 l (130)

Velocity of B w.r.t. ground ^  VBG = –gt j    VBA = VBG  VGA

48

 VωB

(138)

V MB

V Mω

   V MW + V WB = V MB  V MW = Velocity of man w.r.t. water  V WB = Velocity of water w.r.t. Bank  V MB = Velocity of man w.r.t. Bank

tanθ =

VWB VMB

 (141) r = 3ti – 2t 2 j





  dr V = = 3i – 4t j dt –4 t 3 (142) x = 2t2 = 8  t = 2s tanθ =

3 2 3 2 t = (2) = 6m 2 2 (143) x = Acoswt y = A (i – sinwt) y=

Vα =

dx = –Aωsinwt Vy = –A ω coswt dv

V = Vx 2 + Vy 2

(145) A

C

o

t1 t2 t1 t1

B

= = + +

t (0A) = t (BC) t (0B) t2 = t (0A) + t (0B) t2 = t (BC) + t (0B) 49

(146) y = Ax – Bx2 dy = A – 2Bx dx At maximum height A – 2Bx = 0 

x =

A 2B

(147) Q

 (148) V0 = V0 cosθi + V0sinθj  V = V0 cosθi +  V0smθ – gt  j   V0  V = 0

V0 Now find x gsinθ

t =

(149) A t time t V x = V0 = V0 cosθ Vy = Visinθ – gt

tanα =

Vy V0 sinθ – gt = Vx V0 cosθ

2 (150) Range R = V0 sin2θ g

Flight time t f =

2V0sinθ g

Average velocity =

R tf

(156) Cenrifugal acceleration 2

ar =

v 2  2πr  1 2 2 =  = 4π rf r  T  r R

(161) A =  ydx 0

gx 2 take y = xtanθ – 2V0 xcosθ

50

Unit - 3 Laws of Motion

51

SUMMARY 

Important Points The law of inertia given by Galileo was represented by Newton as the first law of motion :" If no external force acts on a body, the body at rest remains at rest and a body in motion continues to move with the same velocity." This law gives us definition of force. 





The momentum of a body p  m v is a vector quantity.It gives more information than the velocity.It's



unit is kgms1 or Ns and dimensional formula M1L1T-1. Newton's second law of motion : The time-rate of change in momentum of a body is equal to the resultant external force applied on the body and is in the direction of the exterrnal force.   dP  ma is the vector relationship F= dt



The SI unit of force is newton (=N) . 1 N = 1 kg ms–2 . This law gives the value of force.It is consistent 





with the first law.( F =0 indicates that a =0). In this equation the acceleration of the body a is that 



which it has when the force is acting on it.(Not of the Past!) . F is only the resultant external force. The impulse of force is the product of force and the time for which it acts. when a large force acts for 

a very small time, it is difficult to measure F and t but the change in momentum can be measured, 



which is equal to the impulse of force ( F t ) Newton's third law of motion: " To every action there is always an equal and opposite reaction." 







Forces always act in pairs, and, FAB= – FBA. The action and the reaction act simultaneously.They act on different bodies,hence they cannot be cancelled by adding. But the resultant of the forces between different parts of the same body becomes zero. The law of conservation of momentum is obtained from Newton's second law and the third law.It is written as-"The total momentum of an isolated system remains constant." The concurrent forces are those forces of which the lines of action pass through the same point. For 

-

equilibrium of the body, under the effect of such forces,  F must be = 0 Moreover, the sum of the corresponding components also should be zero. (  Fx = 0,  Fy=o,  Fz = 0) Friction is produced due to the contact force between the surfaces in contact.It opposes the impending or the real relative motion. Static frictional force fs  fs (max)

=

s N and the kinetic friction is fk =  k N

52

µS  coefficient of static friction

µ k  coefficient of kinetic friction and  k   s

*

*

The reference frame , in which Newton's first law of motion is obeyed is called the inertial frame of reference and the one in which it is not obeyed is called non-inertial frame of reference. The frame of reference with constant velocity is an inertial frame of reference and one which has acceleration is noninertial frame of reference. On a body performing uniform circular motion a force equal to mv2 / r acts towards the centre of the circular path. This is called the centripetal force. The maximum safe speed on level curved road is vmax = µ s rg The maximum safe speed on a banked curved road is vmax =

*

 µ  tanθ   rg s 1  µ tanθ s  

Motion of a body on a friction less inclined plane: As shown in figure body of mass m is placed on a smooth inclined plane making an angle θ with the horizontal. Here, 1 mg cos   N †ÃÜå 2 mg sin   ma  acceleration of body θ = gsin θ .

*

Motion of a body on a Rough inclined plane: (1) Body moving down : If body moving down with acceleration a then, mg sin θ – µ s N  ma and N = mg cos θ  mg sinθ  µ s mg cosθ   ma

 a  g sinθ  µ s cosθ 

(2) 

 

Body moves up : If body is moving upwards with acceleration a then a  g  sin θ  μ s cos θ  . Pseudo Force : In non-inertial frame of reference due to acceleration one more additional force acting on a body in the opposite direction of acceleration of frame of reference is called pseudo force (FP) . when a man of weight m climbs on the rope with acceleration a then tension in the rope is T = m(g + a). When man sliding down with acceleration a then tension in the rope is T = m(g – a).

53

*

When masses in contact: In figure masses m1 ,m2 and m3 are placed in contact on a surface. F

acceleration a  m  m  m 1 2 3 Force on m1 : F1 = F Force on m2 : F2 = (m2 + m3) a 

Force on m3 : *

F3  m3a 

 m2  m3  F

m3

m2 F

m1  m2  m3

a

m1

m3 F  m1  m2  m3 

Masses connected by strings: Consider two masses m1 and m2 placed on a frictionless horizontal surface connected by a light inextensibule string. a If is pulled by a force F then a tension T is developed, m1

in the string.

m2

For m1 : F – T = m1a , For m2 : T = m2 a m2 F F  Tension in string T  m  m , acceleration a  m  m . 1 2 1 2

54

TT

m1

F

MCQ For the answer of the following questions choose the correct alternative from among the given ones. 1.

The velocity of a body of mass 20 kg decreases from 20 ms–1 to 5 ms–1 in a distance of 100 m. Force on the body is (A) -27.5 N

2.

(B) -47.5 N

(B) 20 N

(C) 22 N

(D) 4 N

Formula for true force is (A) F  ma

4.

(D) -67.5 N

A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes upto 2 m height further, find the magnitude of the force. (Consider g = 10 ms–2) (A) 16 N

3.

(C) -37.5 N

(C) F 

d  mv

(B) F  m

dt

dv dt

(D) F = m

d2x dt 2

A particle moves in the X–Y plane under the influence of a force such that its linear momentum is 

^

^

P (t)  A[ i cos (kt)  j sin(kt)] where A and k are constants. The angle between the force and

momentum is (A) 00 5.

`

(C) 450

(D) 900

Force of 5 N acts on a body of weight 9.8 N. what is the acceleration produced in ms-2 (A)49.00

6.

(B) 300 (B) 5.00

(C) 1.46

(D) 0.51

A lift is going up. The total mass of the lift and the passenger is 1000 kg The variation in the speed of the lift is as given in the graph. The tension in the rope pulling the lift at t= 10.5 sec will be (B) Zero

1

(A) 8000 N 3.6

(C) 12000 N (D) 17400 N 7.

10

12

Same force acts on two bodies of different masses 2 kg and 4 kg initially at rest. The ratio of times required to acquire same final velocity is (A) 2:1

8.

t (sec) 2

(B) 1:2

(C) 1:1

(D) 4:16

Which of the following quantities measured from different inertial reference frames are same (A) Force

(B) Velocity

(C) Displacement

55

(D) Kinetic Energy

9.

10,000 small balls, each weighing 1 g strike one square cm of area per second with a velocity 100 ms–1 in a normal direction and rebound with the same velocity. The value of pressure on the surface

will be (A) 2 103 Nm 2

(B) 2 105 Nm 2

(C) 107 Nm2

(D) 2 107 Nm 2

10. When the speed of a moving body is doubled

(A) Its acceleration is doubled (B) Its momentum is doubled (C) Its kinetic energy is doubled (D) Its potential energy is doubled 11. A particle moves in the XY Plane under the action of a force F such that the components of its linear momentum P at any time t are Px = 2 cost, Py = 2 sint. The angle between F and P at time t is (A) 900 (B) 00 (C) 1800 (D) 300 12. A player caught a cricket ball of mass 150 g moving at the rate of 20 ms-1 . If the catching process be completed in 0.1 s the force of the blow exerted by the ball on the hands of player is (A) 0.3 N (B) 30 N C) 300 N (D) 3000 N ^ ^  13. A body of mass 5 kg starts from the origin with an initial velocity u=30 i +40 j ms-1 . If a constant



^

^

  Force F= –  i +5 j  N acts on the body, the time in which the y-component of the velocity becomes 

zero is (A) 5 s



(B) 20 s

(C) 40 s

(D) 80 s

Third law of motion 14. Swimming is possible on account of

(A) First law of motion (B) second law of motion (C) Third law of motion (D) Newton's law of gravitation 15. A cold soft drink is kept on the balance.When the cap is open, then the weight (A) Increases (B) Decreases (C) First increase then decreases (D) Remains same

56

Conservation of linear momentum and impulse 16. A wagon weighing 1000 kg is moving with a velocity 50 km h-1 on smooth horizontal rails. A mass of 250 kg is dropped into it. The velocity with which it moves now is

(A) 2.5 km h 1

(B) 20 km h 1

(C) 40 km h 1

(D) 50 km h 1

17. The Figure shows the Position-time (x-t) graph of one dimensional motion of a body of mass 0.4 kg .

The magnitude of each impluse is x (m)

(A) 0.2 Ns (C) 0.8 Ns

(B) 0.4 Ns (D) 1.6 Ns 0

2

4

6

8

10

12

14 16 t (s)

Equllibrium of Forces 18. Three Forces F1, F2 and F3 together keep a body in equilibrium. If F1 = 3 N along the positive X- axis, F2 = 4N along the positive Y-axis ,then the third force F3 is

  4 3  with negative y-axis 3 4  with negative y-axis 4 3  with negative y-axis

(A) 5 N -making an angle θ = tan–1 3 4 with negative y-axis (B) 5 N - making an angle θ = tan–1 (C) 7 N - making an angle θ = tan–1 (D) 7 N - making an angle θ = tan–1

19. A solid sphere of mass 2 kg is resting inside a cube as shown in the figure. The cube is moving with ^   ^  –1 a velocity v =  5t i+ 2t j  ms Here t is the time in second. All surfaces are smooth. The sphere





is at rest with respect to the cube. what is the total force exerted by the sphere on the cube (g = 10 ms–2) y A B (A) 29 N

(B) 29 N

(C) 26 N

(D) 89 N

D

0

57

C

x

20. A particle of mass 2 kg is initially at rest. A force acts on it whose magnitude changes with time. The F (N)

force time graph is shown below. 20 The velocity of the particle after 10s is 10 (A) 10 ms-1 (B) 20 ms-1 0 (C) 75 ms-1 (D) 50 ms-1 t (s) 21. A block of mass 4 kg is placed on a rough horizontal plane. A time dependent force F = Kt 2 acts on a block , where k = 2 N s 2 , co-efficient of friction µ  0.8 .Force of friction between the block and the plane at t = 2 S is..... (A) 32 N (B) 4 N

(C) 2 N

(D) 8 N

^ ^  ^ ^  22. A 7 kg object is subjected to two forces (in newton) F1 = 20i+30 j and F2 = 8i-5 j The magnitude of

resulting acceleration in ms–2 will be (A) 5 (B) 4 (C) 3 (D) 2 23. A car travelling at a speed of 30 km/h is brought to a halt in 8 metres by applying brakes. If the same car is travelling at 60 km/h it can be brought to a halt with the same breaking power in (A) 8 m (B) 16 m (C) 24 m (D) 32 m 0 24. A given object takes n times more time to slide down 45 rough inclined plane as it takes to slide down a perfactly smooth 450 incline. The coefficient of kinetic friction between the object and the incline is (A)

1 2  n2

(B) 1  1 n 2

(C) 1  1 n 2

(D)

1 1 - n2

25. Two bodies of equal masses revolve in circular orbits of radii R1 and R2 with the same period Their

centripetal forces are in the ratio.  R2  (A)  R   1

2

26. Two masses M and

(B) M

R1

R (C)  1 R   2

R2

2

(D) R 1R2

2 are joined together by means of light inextensible string passed over a

frictionless pulley as shown in fig. When the bigger mass is released, the small one will ascend with an acceleration 3g 2

(A) g 3

(B)

(C) g

(D) g 2

M

2 M

58

27. A 0.5 kg ball moving with a speed of 12 ms-1 strikes a hard wall at an angle of 300 with the wall.

It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for 0.025 S the average force acting on the wall is (A) 96 N (B) 48 N (C) 24 N (D) 12 N 28 A shell of mass 200g is ejected from a gun of mass 4 kg by an explosion that generates 1.05 KJ of energy. The initial velocity of the shell is (A) 100 ms-1 (B) 80 ms-1 (C) 40 ms-1 (D) 120 ms-1 29. A gramophone record is revolving with an anguler velocity  A coin is placed at a distance r from the centre of the record. The coefficient of static friction is µ . The coin will revolve with the record if 2

(B) r   µg

(A) r = µ g2

(C) r 

µg 2

(D) r 

µg 2

30. A stone of mass 2 k g is tied to a string of length 0.5 m It the breaking tension of the string is 900N,

then the maximum angular velocity the stone can have in uniform circular motion is (A) 30 rad/s (B) 20 rad/s (C) 10 rad/s (D) 25 rad/s 31. A body of mass 6 kg is hanging from another body of mass 10 kg as shown in fig. This conbination is being pulled up by a string with an acceleration of 2 ms–2. the tension T1 is (g = 10 ms–2 ) (A) 240 N (B) 150 N (C) 220 N (D) 192 N

T1 10 kg

T2

a

6 kg

32. A sparrow flying in air sits on a stretched telegraph wire. If the weight of the sparrow is W ,which of the

following is true about the tension T produced in the wire? (A) T = W (B) T < W (C) T = 0 (D) T >W 33. Fig. shows the displacement of a particle going along X-axis as a function of time. The force acting on the particle is zero in the region (A) AB (B) BC (C) CD (D) None of these

Displacement

Y

B

C

D A

X Time

59

34. A Force (F) varies with time (t) as shown in fig. Average force over a complete cycle isF

F0

(A) Zero

(B)

(C) F0 

(D) 2F0

F0

2 0

t

35. A body of mass 0.05 kg is falling with acceleration 9.4 ms–2 . The force exerted by air opposite to motion is N (g=9.8 ms–2)

(A) 0.02 (B) 0.20 (C) 0.030 (D) Zero 36. The average force necessary to stop a hammer with 25 NS momentun in 0.04 sec is ___________N (A) 625 (B) 125 (C) 50 (D) 25 37. Newton's third law of motion leads to the law of consrevation of (A) Angular momentum (B) Energy (C) mass (D) momentum 38. A ball falls on surface from 10 m height and rebounds to 2.5 m. If duration of contact with floor is 0.01 sec. then average aceleration during contact is _______________ms -2 (A) 2100 (B) 1400 (C) 700 (D) 400 39. A vehicle of 100 kg is moving with a velocity of 5 m s . To stop it in 110 sec, the required force

in opposite direction is _______________N (A) 50 (B) 500 (C) 5000 (D) 1000 40. The linear momentum P of a particle varies with the time as follows. P  a  bt 2 Where a and b are constants. The net force acting on the particle is _____________ (A) Proportional to t (B) Proportional to t2 (C) Zero (D) constant 41. A vessel containing water is given a constant acceleration a towards the right, along a straight horizontal path. which of the following diagram represents the surface of the liquid ? (A) (B) (C) (D) a

a

a

a

42. A body of 2 kg has an initial speed 5 m/s. A force act on it for some time in the directine of motion.  The force ( F ) ---------time (t) graph is shown in figure. The final speed of the body is _________ F (N)

(A) 9.25 ms-1 (C) 14.25 ms-1

(B) 5 ms-1 (D) 4.25 ms-1

4 2.5

2

60

4

4.5

6.5

t (s)

43. which of the following statement is correct?

(A) A body has a constant velocily but a varying speed. (B) A body has a constant speed but a varying value of acceleration. (C) A body has a constant speed and zero accelaration. (D) A body has a constant speed but velocity is zero. 44. A force of 8 N acts on an object of mass 5kg in X- direction and another force of 6 N acts on it in Y direction. Hence, the magnitude of acceleration of object will be (A) 1.5 ms–2 (B) 2.0 ms–2 (C) 2.5 ms–2 (D) 3.5 ms–2 45. Three forces are acting simultaneously on a particle  moving with velocity V .These forces are represented in magnitude and direction by the three sides of a triangle ABC. The particle will now move with velocily __________



(A) Less than v

C

A

B



(B) greater than v



(C) v in the direction of the largest force BC



(D) v remaining unchanged.

46. A plate of mass M is placed on a horizontal frictionless surface and a body of mass m is placed on this plate, The coefficient of dynamic friction between this body and the plate is . If a force 2 mg. is

applied to the body of mass m along the horizonal direction the acceleration of the plate will be ________________ (A) (C)

µm

µm g M

(B) M  m  g

2m g M

(D) M  m  g

47. On the horizontal surface of a truck

m M

2mg 

2m

µ  0.6 , a block of mass

1 kg is placed. If the truck is

accelerating at the rale of 5 m / s 2 then frictional force on the block will be_____________N (A) 5

(B) 6 (C) 5.88 (D) 8 48. Two blocks of nass 8 kg and 4 kg are connected a heavy string Placed on rough horizontal Plane, The 4 kg block is Pulled with a constant force F.The co-efficient of friction between the blocks and the ground is 0.5, what is the value of F, So that the tension in the spring is constant throughout during the motion of the blocks ? (g=10 ms–2) (A) 40 N (B) 60 N 8 kg F 4 kg (C) 50 N (D) 30 N 61

49. Seven blocks, each of mass 1 kg are arranged one above the other as

7

shown in figure. what are the values of the contact forces exerted on the third block by the forth and the second block respectively ?

6

( g = 10 ms-2 )

4

(A) 40 N, 50 N

(B) 50 N, 40 N

(C) 40 N, 20 N

(D) 50 N, 30 N

2

50. A man is standing on a spring balance. Reading of spring balance is 60 kgf. If man jumps outside

balance, then reading of spring balance____________ (A) First increase than decreases to zero

(B) Decreses

(C) Increases

(D) Remains same

51. A car turns a corner on a slippery road at a constant speed of 10 m/s. If the coefficient of friction is 0.5, the minimum radius of the arc at which the car turns is_____________meter.

(A) 20

(B) 10

(C) 5

(D) 4

52. A person standing on the floor of a lift drops a coin. The coin reaches the floor of the lift in time to if

the lift is stationary and the time t2 if it is accelerated in upward direction. Than (A) t1  t 2

(C) t1  t 2

(B) t1 > t2

(D) Cannot say anything

53. A lift of mass 1000 kg is moving with an acceleration of 1 ms–2 in upward direction Tension developed

in the rope of lift is _________________N (g = 9.8 ms-2 ) (A) 9800

(B) 10,000

(C) 10,800

(D) 11,000

54. Three blocks of masses m 1,m2 and m3 are connected by massless strings as shown in figure, on a

frictionless table. They are Pulled with a force T3  40 N . If m1  10kg , m2  6 kg and m3  4 kg the tension T2 will be =________________N

(A) 20

(B) 40

(C) 10

(D) 32

T1

m1

m2

T2

m3

T3 = 40 N

55. If the surfaccs shown in figure are frictionless, the ratio of T1 and T2 is __________

(A) 3 : 2

(B) 1 : 3

(C) 1 : 5

(D) 5 : 1

F T2 3kg

62

T1 12kg

15kg

300

56. Three masses 1 kg, 6 kg and 3 kg are connected to each

other with threads and are placed on a table as shown in figure. The alcceleration with which the system is moving is ______ ms-2 (g=10ms-2)

T1

T1 1 kg

6 kg

T2

T2 3 kg

(A) zero (B) 1 (C) 2 (D) 3 57. A rope which can withstand a maximum tension of 400 N hangs from a tree. If a monkey of mass 30 kg climbs on the rope in which of the following cases-will the rope break? (take g =10 ms-2 and neglect the mass of rope) (A) When the monkey climbs with constant speed of 5 ms–1 (B) When the monkey climbs with constant acceleration of 2 ms–2 (C) When the monkey climbs with constant acceleration of 5 ms–2 (D) When the monkey climbs with the constant speed of 12 ms–1 58. An object of mass 3 kg is moving with a velocity of 5 m/s along a straight path. If a force of 12 N is applied for 3 sec on the object in a perpendicuiar to its direction of motion.The magnitude of velocity of the particle at the end of 3 sec is____________m/s. (A) 5 (B) 12 (C) 13 (D) 4 59. Same forces act on two bodies of different mass 2 kg and 5 kg initialy at rest. The ratio of times required to acquire same final velocity is ____________ (A) 5:3 (B) 25:4 (C) 4:25 (D) 2:5 ^ ^  60. A body of mass 5 kg starts motion form the origine with an initial veiocily  0 =30 i +40 j m/s If a



^

^

constant force F = – ( i +5 j)N acts on the body, than the time in which the Y-component of the velocity becomes zero is __________________ (A) 5 s (B) 20 s (C) 40 s (D) 80 s 61. A Block of mass 300 kg is set into motion on a frictionless horizontal surface with the help of frictionless pulley and a=1 ms-2 F a rope system as shown in figure. What horizontal force F should be applied to produce in the block an aeceleration of 1 ms-2 ? (A) 150 N (B) 100 N (C) 300 N (D) 50 N

63

62. A body of mass m rests on horizontal surface. The coefficient of friction between the body and the surface is µ . If the body is Pulled by a force P as shown in figure, the limiting friction between body

and surface will be__________ P

  (B) µ mg  2   

A) µmg



P

  (C) µ mg  2   

3P 

(D)  mg  2   

63. Three blocks A , B, and C of equal mass m are placed one over

A B C

the other, one on a smooth horizontal ground as shown in figure. Coefficient of friction between any two blocks of A, B and C is 0.5. What would be the maximum value of mass of block D so that the blocks A, B and C move without slipping over each other ? (A) 3 m (B) 5 m D (C) 6 m (D) 4 m 64. A train is moving along a horizontal track. A pendulum suspended from the roof makes an angle of 40 with the vertical, The acceleration of the train is ___________ms-2 (g = 10 ms-2 ) (A) 0.6 (B) 0.7 (C) 0.5 (D) 0.2 65. A bag of sand of mass m is suspended by rope. a bullet of mass

m is fired at it with a velocity υ 30

and gets emmbedded into it. The velocity of the bag finally is _________ (A)

31υ 30

(B)

30υ 31

(C)

υ 31

(D) υ 30

66. Three blocks having equal mass of 2 kg are hanging on a string passing over a

pulley as shown in figure. what will be the tension produced in a string connecting the blocks B and C

T2 A

B

(A) zero (B) 13.1 N T1 C (C) 3.3 N (D) 19.6 N 67. A partly hanging uniform chain of length L is resting on a rough horizontal table.  is the maximum possible length that can hang in equilibrium The coefficient of friction between the chain and table is ____________ (A)

lL L+l

(B)

L l

(C) 64

l L

(D)

l Ll

68. As shown in figure,the block of 2 kg at one end and the other of 3 kg at the other end of a light string

are connected. It the system remains stationary find the magnitude and direction of the frictional force (g = 10 ms-2 ) (A) 20 N, downward on slope (C) 10 N Downward on slope

(B) 20 N, upward on slope (D) 10 N upward on slope

2

kg

3 kg

30

0

69. A particle is resting over a smooth horizontal floor, At t = 0, a horizontal force starts acting on it. Magnitude of the force increses with time according to law F = t , where  = is constant Match the

column after seeing the figure. Column-1 (a) curve (i) shows. (b) curve (ii) shows

y

(ii) Column-2 (I) (p) velocity against time (q) velocity against acceleration (r) acceleration against time x (A) (i)-p (ii)-q (B) (i)-q, (ii)-r (C) (i)-r, (ii)-p (D) (i)-q, (ii)-p 70. A car of mass 1000 kg travelling at 32 m/s clashes into a rear of a truck of mass 8000 kg moving in the same direction with a velocity of 4 m/s. After the collision the car bounces with a velocity of 8 ms–1. The velocity of truck after the impact is ____________m/s (A) 8 (B) 4 (C) 6 (D) 9 71. A Block of mass m = 2 kg is resting on a rough inclined plane of inclination 300 as shown in fignre. The coefficient of friction between the block and the plane is µ =0.5 . What minimum force F shuld be applied perpendicular to the plane of block so that block does not slip on the plane ? (g=10ms–2) N = F + mg cos 30 (A) zero (B) 6.24N (C) 2.68 N (D) 4.3 N 0

30 0

72. The upper half of an inclined plane of inclination θ is perfectly smooth while the lower half is rough A

body starting from the rest at top come back to rest at the bottom, then the coefficient of friction for the lower half is given by____________ (A) μs  sinθ (B) μ s  cot θ (C) μ s  2cos θ (D) μ s  2 tan θ 73. A Block of mass m = 4 kg is placed over a rough inclined plane as shown in figure, The coefficient of friction between the block and plane is µs = 0.6. A force F = 10 N is applied on the block of an angle at 300. The contact force between the block and the plane is ___________ F

(A) 27.15 N (C) 10.65 N

(B) 16.32 N (D) 32.16 N

45 0

65

30 0

74. The motion of a particle of a mass m is describe by y  ut 

1 2 gt . Find the force acting on the 2

particle. (A) F = ma (B) F = mg (C) F = 0 (D) None of these 75. A balloon has a mass of 10 g in air, The air escapes from the balloon at a uniform rate with a velocity of 5 cm s and the balloon shrinks completely in 2.5 sec. calculate the average force acting on the balloon. (A) 20 dyne (B) 5 dyne (C) 0 dyne (D) 10 dyne 76. Two bodies A and B each of mass m are fixed together by a massless spring A force F acts on the mass B as shown in figure. At the instant shown, a body A has an acceleration a. what is the accelaration of B? F



(A)  m  a   

A m

(B) F–T

F

  (C)  a  m   

B m

F

(D) a

77. With what acceleration (a) should a box descend so that a block of mass M placed in it exerts a force Mg on the floor of the box? 4

(A)

4g 3

(B)

3g 4

(C) g 4

(D) 3g

78. A mass of 6 kg is suspended by a rope of length 2 m form the ceiling. A force of 50 N in the horizontal dircction is applied at the mid Point

P of the rope- as shown in figure. what is the angle the rope makes with the vertical in equilibrium ? (g = 10 ms-2) Neglect mass of the rope. (A) 400 (B) 300 (C) 350 (D) 450

T1 P

50 N

1m T2 w 60 N

79. The minimum force required to start pushing a body up a rough (cofficient of µ ) inclined plane is F1.

While the minimum force needed to prevent it from sliding down is F2. If the inclined plane makes an F1

angle  from the horizontal. such that tan   2 than the ratio F is 2 (A) 4

(B) 1

(C) 2 66

(D) 3

80. When forces F1, F2, F3 are acting on a particle of mass m such that F2 and F3 are mutually perpendicular,

then the particle remains stationary. If the force F1 is now removed than the acceleration of the particle is (A)

Fb 0 m

(B)

FF 1 2 m

(C)

F2 - F3 m

(D)

F2 m

Asseration and reason type question

81.

82.

83.

84.

Asseration and reason are given in following question. Each question have four options. One of them is correct select it. (a) Asseration is true. Reason is true and reason is correct explanatin for Assertion. (b) Asseration is true. Reason is ture but reason is not the correct explanatin of assertion. (c) Asseration is ture. Reason is false. (d) Asseration is false. Reason is true. Asseration : Frictional forces are conservative forces. Reason : Potential energy can be associated with frictional forces. (A) a (B) b (C) c (D) d Asseration : A body of mass 1 kg is moving with an accelaration of 1ms-1. The rate of change of its momentum is 1 N. Reason : The rate of change of momentum of body = force applied on the body. (A) a (B) b (C) c (D) d Asseration : A body falling freely under gravity becomes weightless. Reason : R = m(g – a) = m(g – g) = 0 (A) a (B) b (C) c (D) d Asseration : It is difficult to move bike with its breaks on. Reason : Rolling friction is converted into sliding friction, which is comparatively larger. (A) a (B) b (C) c (D) d

Comprehension :According to newton's second low of mation, F= ma, where F is the force required to produce an accelaration a in a body of mass m. If a = 0 than F = 0. If a force acts on a body for t seconds, the effect of the force is given by impulse = F  t = change in linear momentum of the body. 67

with the help of the passage given above, choose the most appropriate alternative for each of the following questions. 85 A cricket ball of mass 150 g. is moving with a velocity of 12 m/s and is hit by a bat so that the ball is turned back with a velocity of 20 m/s. If duration of contact between the ball and the bat is 0.01 sec. The impulse of the force is (A) 7.4 NS (B) 4.8 NS (C) 1.2 NS (D) 4.7 NS 86. Average force exerted by the bat is (A) 480 N (B) 120 N (C) 1200 N (D) 840 N 87. The force acting on a body whose linear momentum changes by 20 kgms-1 in 10 sec is (A) 2 N (B) 20 N (C) 200 N (D) 0.2 N 88. An impulsive force of 100 N acts on a body for 1 sec What is the change in its linear momentum ? (A) 10 N-S (B) 100 N-S (C) 1000 N-S (D) 1 N-S 89. Match the column Column - I Column - II (a) Body lying on a horizontal surface (p) is a self adjusting force (b) Static friction (q) is a maximum value of static friction (c) Limiting friction (r) is less than limiting friction (d) Dynamic friction (s) force of friction = 0 (A) a-s, b-p, c-q, d-r (B) a-p, b-q, c-r, d-s (C) a-s, b-r, c-q, d-p (D) a-r, b-q, c-p, d-s 90. A block B, placed on a horizontal surface is pulled with initial velocity V. If the coefficient of kinetic friction between surface and block is  , than after how much time, block will come to rest ? v

(A) gµ

(B)

gµ v

(C)

g v

v

(D) g

91. As shown in figure a block of mass m is attached with a cart. If the coefficient of static friction between the surface of cart and block is  than what would be acceleration  of cart to prevent the falling of

block ? mg

(A) α  µ g

(C) α  µ

g

(B) α  µm g

(D) α  µ

68

92. A particle of mass m is at rest at t = 0. The force exerting on it in x-direction is F(t) = Foe-bt . Which one

of the following graph is of speed V  t   t (A)

(B)

F0 b m

vt

vt

(C)

F0 b m

t

(D)

F0 b m vt

t

F0 b m vt

t

t

KEY NOTE 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 30

C C C D B A B A D B A B C C C C C C C D D A D B B A C A C A

31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

D D A A A A D A C A C C C B D A A B B A A C C D D C C C D C

61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90

69

A C A B C B D A C D C D A B A A B A D A D A A A B A A B A A

91 92

C B

HINT 1.

 v2  u2  F = ma = m  2S   

2.

H max 

u2 2g

 u  2g H max

This velocity is supplied to the ball by hand and initially the hand was at rest. It acquires this velocity in distance of 0.2 meter.  a 

3.

u2 2S

So upword force F = m (g+a) According to newton's second law force = rate of change of linear momentum. 

4.

^ ^  P  t   A  i cos kt  j sin kt   



F

d    P( t )  dt  

^  ^   i sin kt  j cos kt  = Ak     F.P  A 2 k   cos kt.sin kt  sin kt.cos kt 



F. P  0 5.

As weight W = mg = 9.8 N  m = 1 Kg a Fm

6.

At 10.5 sec the lift is moving upword with acceleration a

0  3.6  1.8 ms 2 2

Tension in rope T = m (g-a) = 1000 (9.8 - 1.8) = 8000 N

70

7.

t 

V 1 t  a a

(V is constant)

t1 a 2 m1 2 1     t 2 a1 m 2 4 2

9.

P= F

11.



A

=

n  mv -  -mv   A

^

[ a =

1 as F is constant] m

2nmv A



^

P  Px i  Py j , F  dp dt 

 

Now F . P  0

 θ  90o

 dv  12. Force exerted by ball, F  m    dt 

13. vy = 40 ms–1 Fy = –5 N

So ay =

Fy m

m = 5 kg

 1 ms 1

as v y  u y  at 0 = 40 –1t  t = 40s 14. Swimming is a result of pushing water in the opposite direction of the motion 15. Gas will cane out with sufficient speed in forward direction, So recetion of this forward force wil change

the reading of the spring balance. 16. According to the principle of conservation of linear momentum 1000× 50 = 1250× v \ v = 40 km h 17. impulse = P  m  v f  vi  

^

ay

^

19. As υ  5t i  2t j ^ ^  a  5i  2 j 

^

ax ^

F  ma x i  m  g  a y  i 

1F 1  m ax 2   g  a y 

max

2

m(g+ay)

= 26 N 71

20. Velocity after 10 sec. υ

1 (Area enclosed between F - t graph) = 50 ms-1 2

21. Fmax = µ mg = 0.8  4  10 = 32 N

applied force at t = 2S is F = kt2 = 2(2)2 = 8 N The body fails to move. Hence the force of friction at t = 2s = applied force = 8 N 22.







F  F1  F 2  a Fm

23. Here distance d2

as υ is doubled, d becomes 4 times. 24.

1 2S S  ut  at 2 as u = 0 t  2 a

for smooth plane a = gsin θ For rough plane a1  g (sinθ-μ cos θ) 

t1 

2S g  sin θ-μ cos θ 

 t1  nt  n

 n 2g  sin θ-μ cos θ   g sin θ

When θ = 45o sin θ = cos θ = 1 2 2

25.

 µ  1- 1

2

mv 2 m  R 42 mR  2  F    mR    R R T2 T 

As m and T are same for two bodies F R 26.

a

 m1  m 2  g m1  m 2

 MM  2 g g  MM   3  2

72

n2

2S g sinθ

27. As impulse = change in linear momentun

28.

Ft = 2mvsin300

 F

m1v1 +m2v2 = 0

v2 =

Now E =

2msin30 t

m1 v v1 = - 1 m2 20

1 1 2 2 m1v1 + m2 v2 2 2

By solving weight v1 = 100 ms-1 29. The coin will revolve with the record if centrifugal force mrw2 < f  mr2 <µR<µmg

r<

30.

µg 2

F  mr 2    F

mr

 30 rad

s

31. Total mass pulled up m = 10 + 6 = 6 kg

 T1  mg  a  33. d-t graph AB is linier  Velocity is uniforce a = 0  F = 0

35.

Fair  mg  a 

36.

F

38.

v0  2 gh1 and v  2 gh1

P 25   625 N t 0.04

madt  m  v2  v1  

a 

v2  v1 dt

dp  0  2bt  Ft dt

40.

F

42.

F .t  m  v2  v1  and F.t  Area of lower part of graph F  t

73

44.

2

F  F1  F2

and a  F M

2

46. Resultant force F = 2 µ mg- µ mg = µ mg

accleration in plate a  F m 48.

a 0

51.

v  µrg

52.

d

T  8µg

r v

 F  T  4µg 2

µg

1 2 1 gt1 and d   g  a  t 22 2 2

By comparing both eqn. 1 2 1 gt1   g  a  t 22 2 2

 t1  t 2 53. Tensile force T = m(g+a) T3 m1  m 2  54. Tensile force T2  m  m  m 1 2 3 F 55. acceleration a  m  m  m 1 2 3

T2  m1a T1  m1  m2  a  3 1   10 56. acceleration a    1 3  6 

57. Calculate F = m (g+a) 58.





P  F  t

m (vy–v0y ) = F  t

v y  12 j

and resultant velocity of t = 3 Sec v  vx 2  v y 2 59.

Ft1  m1  v-v0   m1υ1 Ft2  m2  v -v0   m2 v 

t1 m1  t2 m2

74

 Vy  V0   60. Fy = m  t   61. Tension T = 2 F

 ma

62. N + P sin θ = mg



=2F

F

N = mg - P sin θ

= ma 2

= mg - P 2

limiting friction Fs = µ s N 63.

a max 

Fs (max)  µ s M A g m g

D and a max  3m  m D

OR

Fmax g  2 mA

 M D  3m

T1  T2  T3  W 1  M D  6 µm  6  m  3 m 2

µmg  2 µmg  3 µmg  M D .g 64. ma = mg sin θ

 a = 10 sin 4o

 a = 0.7 ms-2

65. According to the low of conservation of momentum m m  v  m  v 30 30  

v 

v 31

66. for Block A

T2 -2g = 2a ....(1) for block B T1 +2g-T2= 2a ....(2) for block C 2g - T1= 2a ....(3) Adding (1) & (2) T1 = 4a ....(4) from (3) & (4)

m m  v  m  v  a  g 3 30 30  

2g-4a = 2a

g  T1  4a  4  3  13.1 N 67.

µ

M L

Weight of chain of lenght  = µg Weight of a chain of lenght ( L   ) = µ L -   g f S  µ s R  µ s µ(L-) g and f S  µg

 µg  µ s .µ  L-  g

 µs 

 L

75

68. for 3 kg body m1a = m1g-T

T  m1g  m1a......(1)

for 2 kg body T - m2 gsin θ = m2a

 T  m2a 

m2 g ......(2) 2

from (1) and (2) a = g = 10 ms-2 fS = m 2a=20 N 69.

F  t 

dp  t  P    tdt dt

 vα t 2

now

t 2 2

 mv 

Which shows graph of v  t 2 is parabola so (ii) - (P)

dv 1  dt 2

m (2t ) 

 a α t Which shows graph of a - t is linier so (i)....R 



1

1

70. according to eqn m1 v1  m2 v2  m1 v1  m2 v2 71.

1  v2  9 m s

N  F  mg cos 30 and mg sin 30  µ s N

72. For upper half v 2 - v02 = 2 ad 2  0  2g sin θ  µ 2

v 2  gl sin 

For lower half v12  v02  2a 'd 0  gl sin   2g sin    s cas.

73.

 2

 µ 2  2 tan θ

N  F sin 30  mg cos 45

fs max   µ s N †ÃÜå fc  N2 +fs2  max  75.

F

dp d  dm    mv     .v dt dt  dt  dm 10 4gram cm   sec and v  5 s dt 2.5

76.

ma'  F  T

ma'  F  ma

 a'  F m  a

77. As the box is descending  R  mg  a 

mg  mg  ma 4

a 

3g 4

76

78.

T1 sin θ  50 N and T1 cos θ  60 N  tan θ  5 4

79.

 θ  400

F1  mgsin θ  µ cos θ  F2  mgsin θ  µ cos θ  F1 tan θ  µ 2µ  µ  F  tan θ - µ  2µ - µ  3 2

80. Here F1  F2  F3  a  F1 a 90.

v  v0  at as v  0 and v0  v a

V t

µg 

V t

µmg  ma

f  ma t 

V µg

91. pseudo force exerted on block F = m α

force of friction f = µN  μmα when f W, bolck will not tall  μmα  mg

α  g

m

92. F(t) = F0e–bt –bt  ma = F0e

a =

F0 bt e m



dv Fo bt e dt  dt m

dv 

Integrating t

Fo bt v e dt m 0

t

Fo  e  bt  v     m  b  0

 v

which is exponential function So graph (b) is true.

77

Fo 1  ebt  mb

Fo  bt e dt m

Unit - 4 Work Energy And Power

78

SUMMARY 



The product of the magnitude of the displacement during the action of a force and the magnitude of the component of the force in the direction of displacement is known as work. Its unit in joule and its dimensional formula is M 1L2T-2. If angle between force and displacement is (i) for  = 0  W = Fd

 

(ii) for  =

 

 w = 0 (iii) for  w = -Fd

If  is an acute, work is positive and work is done by the force on the body. If  is an obtuse angle, work is negative and work is done by the body aganist the force. Work on a body lying on rough horizontal surface acted by force. As there is no displacement along Y-axis N + Fsin  = Mg  N = Mg - F sin Responsible resultant force for displacement along X-direction = F cos - N = F cos - (Mg - Fsin) Work W = [F cos - (Mg - Fsin]d = [F(cos - sin ) -  Mg] d   F  . dl f



The work done by a variable force is given by W =

i

 

If a variable force and displacement due to it are in the same direction the area below the F - x graph gives value of work. The ability of a body to do work by virtue of its motion is known as its Kinetic energy. If the velocity of a body of mass `m' is `v' its kinetic energy is K=



1 mv 2  p 2 / 2m . It is scalar quantity.. 2

Work Energy Theorem : Work done by a resultant force on a body is equal to  

1 2

2 2 the change in its kinetic energy. W  mv  mv0    K 0  

 



When a body has the ability to do work due to its position in a force field or its configuration. It is known as potential energy. It is scalar quantity. If the gravitational potential energy, due to the gravitational field of Earth, is randomly taken to be zero on its surface the potential energy of a body of mass m, at height h is mgh, where g is the gravitational acceleration. The value of `h' is negligible compared to the radius of the earth. The sum of the potential energy (U) and the kinetic energy (K) of a substance is called the mechanical energy. E = K + U. 79



Considering potential energy of a spring as zero in its normal state, if its langth is changed by x, the potential energy of the spring is U =



1 2 kx . Here k is the spring constant. Unit 2

of k is N/m and dimensional formula is M1L0T-2 The forces for which work done is independent of the path of motion of the body but depends only on initial and final positions, are called conservative force. The force of gravitation or the restoring force developed in a spring due to its compression or extension are conservative forces.

dU . dx



The relation between the conservative force and the potential energy is F = –



The time-rate of doing work is called power. Its unit is watt (J/s). Its dimensional formula is M1L2T-3.



 



The power P = W / t or P = F . v 1 horse power  746 watt Unit of electric energy for domestic use is 1 unit = 1 kWh = 3.6  106 J If during collision of two bodies the kinetic energy is conserved the collision said to be elestic. A body of mass m1 moving with velocity v1 undergoes elastic collision with a body of mass m2 moving with velocity v2 in the direction of v1 If after collision their velocities are v1' and v '2 . m1  m2 2m2 2m1 m  m1 v1  2 v2 v1' = m  m v1  m  m v2 and v '2 = 1 2 1 2 m1  m 2 m1  m 2



In case of complete inelastic collision bodies colliding move together after collision with a common velocity v. In this case v



m1v1  m2v2 m1  m2

A body of mass m1 moving with velocity v1 collides with a stationary body of mass m2 elastically. They move with velocities v1' and v '2 making angles 1 and 2 with direction of v1 then m1v1  m1v1 ' cos1  m2v2 ' cos2

0  m1v1 ' sin 1  m2 v1 ' sin 2 and m1v12  m1v1 ' 2  m2 v2 '2 

As Newton's low of impact, the co-efficient of restitution is, e =

v2 ' v1 ' , ' e ' depends on the types of materials of bodies colliding. v1  v2

For perfectly elastic collision e = 1 and for perfectly inelastic collision e = 0. Equations of velocities after collision of two bodies can be written as : v1 ' 

(1  e)m1 (m e  m 2 ) (m1  m2e) (1 e)m2 v1  v2 and v'2  v1  1 v2 m1  m2 m1  m2 m1  m 2 m1 + m 2 80

MCQ For the answer of the following questions choose the correct alternative from among the given ones. 1.

How much is the work done in pulling up a block of wood weighing 2KN for a length of 10m on a smooth plane inclined at an angle of 30 o with the horizontal ? (A) 1.732 KJ

2.

(D) 100 KJ

(B) 4 N

(C) 1 N

(D) 14 N

A 60 kg JATAN with 10 kg load on his head climbs 25 steps of 0.20m height each. what is the work done in climbing ? (g = 10 m/s 2) (A) 5 J

4.

(C) 10 KJ

A force of 7N, making an angle  with the horizantal, acting on an object displaces it by 0.5m along the horizontal direction. If the object gains K.E. of 2J, what is the horizontal component of the force ? (A) 2N

3.

(B) 17.32 KJ

(B) 350 J

(C) 100J

(D) 3500J

A ball of mass 5 kg is stiding on a plane with intial velocity of 10 m/s. If coefficient of friction between surface and ball is 1 2 , then before stopping it will describe...... (g = 10 m/s 2)

5.

(B) 5 m

(C) 7.5 m

The relationship between force and position is shown in the figure given (in one dimensional case) calculate the work done by the force in displacing a body from x = 0 cm to x = 5 cm (A) 30 ergs

(B) 70 ergs

(C) 20 ergs

(D) 60 ergs

(D) 10 m

20

Force (Dyne)

(A) 12.5 m

10

0

• 1

• 2

10

• 3

•

• 4 5

• 6 x (cm) 

20

6.

The force constant of a wire is K and that of the another wire is 3k when both the wires are stretched through same distance, if work done are W 1 and W2, then... (A) w 2  3w12

7.

(B) W2 = 0.33W1

(C) W2 = W1

(D) W2 = 3W1

A ball is released from the top of a tower. what is the ratio of work done by force of gravity in first, second and third second of the motion of the ball ? [hn  (2n – 1)] (A) 1 : 2 : 3

(B) 1 : 4 : 9

(C) 1 : 3 : 5 81

(D) 1: 5 : 3

8.

A spring of spring constatnt 10 3 N/m is stretched initially 4cm from the unstretched position. How much the work required to stretched it further by another 5 cm ? (A) 6.5 NM

9.

(B) 2.5 NM

(C) 3.25 NM

(D) 6.75 NM

The mass of a car is 1000 kg. How much work is required to be done on it to make it move with a speed of 36 km/h ? (A) 2.5  104 J

(B) 5  103 J

(D) 5  104 J

(C) 500 J

10. A body of mass 6 kg is under a force, which causes a displacement in it given by 2t3

3

(in m). Find the work done by the force in first one seconds.

(A) 2 J

(B) 3.8 J

(C) 5.2 J

11. A 8 kg mass moves along x - axis. Its accelerations as a function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from x = 0 to x = 6cm ? (B) 98  10-3 J (D) 9.8  10-3 J

12. The work done by a force acting on a body is as shown in the graph. what is the total work done in covering an intial distance of 15m ? (A) 50 J (C) 100 J

20 • 15 • 10 • 05 • 00 •

(D) 24 J

• 2

• 4

• 6

x (m)

Force (N)

(A) 48  10-3 J (C) 4.8  10-3 J

a (cm / sec2) 

S=

(B) 75 J (D) 25 J

13. A spring gun of spring constant 90 10 2 N / M is compressed 4cm by a ball of mass 16g. If the trigger is pulled, calculate the velocity of the ball. (A) 60 m/s

(B) 3 m/s

(C) 90 m/s

(D) 30 m/s

14. A uniform chain of length 2m is kept on a table such that a length of 50cm hangs freely from the edge of the table. The total mass of the chain is 5kg. What is the work done in pulling the entire chain on the table. (g = 10 m\s 2) (A) 7.2 J

(B) 3 J

(C) 4.6 J 82

(D) 120 J

15. A uniform chain of length L ans mass M is lying on a smooth table and one third of its is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the hanging part on to the table is (A) MgL

(B) MgL / 3

(C) MgL/9

(D) MgL / 18

16. A cord is used to lower vertically a block of mass M by a distance d with constant down-word acceleration 9 2 .work done by the cord on the block is (A) - Mgd/2

(B) Mgd/4

(C) -3Mgd/4

(D) Mgd

17. A block of mass 5 kg is resting on a smooth surface. At what angle a force of 20N be acted on the body so that it will acquired a kinetic energy of 40J after moving 4m (A) 300

(B) 450

(C) 600

(D) 1200

18. Natural length of a spring is 60 cm, and its spring constant is 2000N/m. A mass of 20kg is hung from it. The extension produced in the spring is..... (g = 9.8 m/s 2) (A) 4.9 cm

(B) 0.49 cm

(C) 9.8 cm

(D) 0.98 cm

19. The potential energy of a body is given by U = A - Bx 2 (where x is displacement). The magnitude of force acting on the particle is (A) constant

(B) proportional to x

(C) proportional to x2

(D) Inversely proportional to x

20. A uniform chain of length L and mass M is lying on a smooth table and ( 1 4 )th of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the hanging part on to the table is (A) MgL

(B) MgL/9

(C) MgL/18

(D) MgL/ 32

21. If Wa, Wb, and Wc represent the work done in moving a particle from X to Y along three different path a, b, and c respectively (as shown) in the gravitational field of a point mass m, find the correct relation between Wa, Wb and Wc (A) Wb > Wa > Wc

(B) Wa< Wb < Wc

(C) Wa > Wb > Wc

(D) Wa = Wb = Wc

22. An open knife edge of mass m is dropped from a height h on a wooden floor. If the blade penetrates upto the depth d into the wood, the average resistance offered by the wood to the knife edge is, (A) mg

(B) mg (1+ h d ) 83

(C) mg (1  h d ) 2

(D) mg (1 -

h

d

)

23. A toy car of mass 4 kg moves up a ramp under the influence of force F plotted against displacement x. The maximum height attained is given by (Take g = 10 m/s2) (A) ymax = 5m (B) ymax = 10m (C) ymax = 15m (D) ymax = 20m 24. A particle of mass 0.5kg travels in a straight line with velocity v = ax , Where a = 5 m s 1 . The work done by the net force during its displacement from x = 0 to x = 2m is (A) 50 J (B) 45 J (C) 25 J (D) None of these 25. Velocity time graph of a particle of mass 2kg moving in a straight line is as shown in figure. Wrok done by all forces on the particle is (A) 400 J (B) -400 J (C) -200 J (D) 200 J 26. A mass of M kg is suspended by a weight-less string, the horizontal force that is required to displace it until the string makes an angle of 60o with the intial varticale direction is 3

1

2

(A) Mg / 3 (B) Mg . 2 27. Given below is a graph between a variable force (F) (along y-axis) and the displaement (X) (along xaxis) of a particle in one dimension. The work done by the force in the displacement interval between 0m and 30m is (A) 275 J (B) 325 J (C) 400 J (D) 300 J

(C) Mg / 2

84

(D) Mg . 3

2

28. Force F on a particle moving in a straight line varies with distance d as shown in the figure. The work done on the particle during its displacement of 12m. (A) 27 J (B) 24 J (C) 36 J (D)26 J 29. A force F = Ay2 + By + C acts on a body in the y-direction. The work done by this force during a displacement from y = -a to y = a is (A)

2Aa 3 3

(B)

2Aa 3 Ba 2   ca (C) 3 2

2Aa 3  2ca 3

(D) None of these.

(ENERGY AND CONSERVATVE, NON CONSERVATIVE FORCE) 30. A spring with spring constant K when streched through 2cm the potential energy is U. If it is streched by 6cm. The potential energy will be...... (A) 6U

(B) 3U

(C) 9U

(D) 18U

31. If linear momentum of body is increased by 1.5%, its kinetic energy increases by.......% (A) 0%

(B) 10%

(C) 2.25%

(D) 3%

32. With what velocity should a student of mass 40 kg run so that his kinetic energy becomes 160 J ? (A) 4 m/s

(B)

8 m/s

(C) 16 m/s

(D) 8 m/s

33. A body of mass 1 kg is thrown upwords with a velocity 20 m/s. It momentarily comes to rest after a height 18m. How much energy is lost due to air friction. (g = 10 m/s2) (A) 20 J

(B) 30 J

(C) 40 J

(D) 10 J

34. Two bodies of masses m 1 and m2 have equal kinetic energies. If P 1 and P2 are their respective momentum, what is ratio of P 2 : P1 ? (A) m1: m2

(B)

(C)

m2 : m1 85

m1 : m2

(D) m12 : m22

35. A body having a mass of 0.5 kg slips along the wall of a semispherical smooth surface of radius 20 cm shown in figure.What is the velocity of body at the bottom of the surface ? (g = 10 m/s 2) (A)

0.5 Kg

R = 20cm

(B) 2 m/s

2 m/s

(C) 2 2 m / s

(D) 4 m/s

36. Two bodies of masses m and 3m have same momentum. their respective kinetic energies E1 and E2 are in the ratio..... (A) 1 : 3

(B) 3 : 1

(C) 1 :

(D) 1 : 6

3

37. What is the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to vertical height of 18 cm (Take g = 10 m/s2) (A) 0.4 m/s

(B) 4 m/s

(C) 6 m/s

(D) 0.6 m/s

18 cm

38. A particle is placed at the origin and a force F = kx is acting on it (Where K is positive constant) If Ucos = 0, which one of the following graph of U(x) versus x. (where U is the potential energy function) Y

U(x)

Y

U(x)

(A)

(B)

X

X

U(x)

Y

Y U(x)

(C)

(D)

X

86

X

39. A force  time graph for a linear motion is shown in figure where the segments are circular. what is linear momentum gained between zero and 8 second ? (A) -2  N S (C) 4  NS

Force (N) 

+2

(B) 0 NS (D) -6  NS

• • • • • • • •

0

2

4

6

8

t(s)

–2

40. A particle is dropped from a height-h. A constant horizontal velocity is given to the particle. Taking g to be constant everywhere kinetic energy E of the particle with respect to time t is correctly shown in.... E

(A)

(B)

E

t

t

E E

(C)

(D)

t

t

41. Which of the following graph is correct between kinetic energy (E), potential energy (U) and height (h) from the ground of the particle.

(B)

Energy

Energy

(A)

E

U

E

U

Height

Height U

(D)

Energy

Energy

(C)

E

E

U

Height

Height 87

42. An engine pump is used to pump a liquid of density  continuosly through a pipe of cross-sectional area A. If the speed of flow of the liquied in the pipe is v, then the rate at which kinetic energy is being imparted to the liquid is (A)

1 A  V3 2

(B)

1 A  V2 2

(C)

1 AV 2



(D) A  V



43. The velocity of a body of mass 400 gm is 3iˆ  4ˆj m/s. So its kinetic energy is ...... (A) 5 J

(B) 10 J

(C) 8 J

(D) 16 J

44. A particle is moving under the influence of a force given by F = kx, where k is a constant and x is the distance moved. What energy (in joule) gained by the particle in moving from x = 1m to x = 3m ? (A) 2 k

(B) 3 k

(C) 4 k

(D) 9 k

45. A spring is compressed by 1 cm by a force of 4 N. Find the potential energy of the spring when it is compressed by 10 cm. (A) 2 J

(B) 0.2J

(C) 20 J

(D) 200 J

46. when 2kg mass hangs to a spring of length 50 cm, the spring stretches by 2 cm. The mass is pulled down until the length of the spring becomes 60 cm. What is the amount of elastic energy stored in the spring in this condition, if g = 10 m/s 2 (A) 10 J

(B) 2 J

(C) 2.5 J

(D) 5 J

 

47. The potential energy of a projectile at its highest point is 1 2 initial kinetic energy. Therefore its angle of projection is ...... (A) 300

(B) 450

(C) 600

th

the value of its

(D) 750

48. Two bodies P and Q have masses 5 kg and 20 kg respectively. Each one is acted upon by a force of 4 N. If they acquire the same kinetic energy in times t P and tQ then the ratio tp tq = .......... (A)

1

(B) 2

2

(C)

49. A particle of mass 0.1kg is subjected to a force which varies with distance as shown in figure. If it starts its jounery from rest at x = 0. What is the particle's velocity squre at x = 6 cm ? (A) 0 (m/s)2

2

5

(D)

5

6

 F(N)

x(m) 

(B) 240 2 (m/s)2

(C) 240 3 (m/s)2

(D) 480 (m/s)2 88







50. The potential energy of 2kg particle, free to move along x axis is given by  x4 x2  U ( X )     J . If its mechanical energy is 2 J, its maximum speed is.... m/s 2   4

(A)

3

2

(B)

(C)

2

1 2

(D) 2

51. If the K.E. of a body is increased by 44%, its momentum will increase by....... (A) 20 %

(B) 22 %

(C) 2 %

(D) 120 %

52. A bullet of mass 0.10 kg moving with a speed of 100 m/s enters a wooden block and is stopped after a distance of 0.20m. what is the average resistive force exerted by the block on the bullet ? (A) 2.5 x 102 N

(C) 25 x 102 N

(B) 25 N

53. A simple pendulum is released from A as shown in figure. If 10 g and 100 cm represent the mass of the bob and length of the pendulum. what is the gain in K.E. at B ? ( g = 10 m/s2 ) (A) 0.5 J (C) 5 J 54. A rifle bullet loes

A 60

0

(B) 5  10-2 J (D) 0.5  10-3 J

  1 10

(D) 2.5 x 104 N

B

th

of its velocity in passing through a plank. The least

number of such planks required just to stop the bullent is (A) 5

(B) 10

(C) 11

(D) 20

55. A sphere of mass m moving the velocity v enters a hanging bag of sand and stops. If the mass of the bag is M and it is raised by height h, then the velocity of the sphere was (A)

mm 29h m

(B)

m

M 29 h m

56. A particle is acted upon by a force F which varies with position x as shown in figure. If the particle at x = 0 has kinetic energy of 20 J. Then the calculate the kinetic energy of the particle at x = 16 cm. (A) 45 J (C) 70 J

m

(C) M  m 29h (D) M

29 h

10 • 5•  F(N)

0• -5 • -10 •

(B) 30 J (D) 135 J

89









 





x(m) 

57. A frictionless track 12345 ends in a circular loop of radius R. A body slids down the track from point 1 Which is 6 cm. Maximum value of R for the body to successfully complete the loop is (A) 6 cm

(B)

15

(C)

(D)

12

5

12

cm

4 5

cm cm

58. If the water falls from a dam into a turbine wheel 19.6m below, then the velocity of water at the turbine is ...... ( g = 9.8 m/s 2) (A) 9.8 m/s

(B) 19.6 m/s

(C) 39.2 m/s

(D) 98.0 m/s

59. A bomb of 12 kg divedes in two parts whose ratio of masses is 1:4. If kinetic energy of smaller part is 288 J, then momentum of bigger part in kgm/sec will be (A) 48

(B) 72

(C) 108

(D) Data is incomplete

60. An ice-cream has a marked value of 700kcal. How many kilo-watt-hour of energy will it deliver to the body as it is digested (J = 4.2 J/cal) (A) 0.81 kwh

(B) 0.90 kwh

(C) 1.11 kwh

(D) 0.71 kwh

61. A spherical ball of mass 15 kg stationary at the top of a hill of height 82m. It slides down a smooth surface to the ground, then climbs up another hill of height 32m and finally slides down to horizontal base at a height of 10 m above the ground. The velocity attained by the ball is (A) 30 10 m/s

(B) 10 30 m/s

(C) 12 10 m/s (D) 10 12 m/s

62. A bomb of mass 10 kg explodes into 2 pieces of mass 4 kg and 6 kg. The velocity of mass 4 kg is 1.5 m/s, the K.E. of mass 6 kg is ....... (A) 3.84 J

(B) 9.6 J

(C) 3.00 J

(D) 2.5 J

63. A bomb of mass 3.0 kg explodes in air into two pieces of masses 2.0 kg and 1.0 kg. The smaller mass goes at a speed of 80m/s. The total energy imparted to the two fragments is (A) 1.07 KJ

(B) 2.14 KJ

(C) 2.4 KJ

(D) 4.8 KJ

64. The bob of simple pendulum (mass m and length l) dropped from a horizontal position strike a block of the same mass elastically placed on a horizontal frictionless table. The K.E. of the block will be (A) 2 mgl

(B) mgl /2

(C) mgl 90

(D) zero

65. A gun fires a bullet of mass 40 g with a velocity of 50 m/s. Because of this the gun is pushed back with a velocity of 1 m/s. The mass of the gun is (A) 1.5 kg

(B) 3 kg

(C) 2 kg

(D) 2.5 kg

66. The decreases in the potential energy of a ball of mass 25 kg which falls from a height of 40 cm is (A) 968 J

(B) 100 J

(C) 1980 J

(D) 200 J

67. If a man increase his speed by 2 m/s, his K.E. is doubled, the original speed of the man is (A) (2+2 2 ) m/s

(B) (2 +

2 ) m/s

(C) 4 m/s

(D) (1+2 2 ) m/s

68. The potential energy of a conservative system is given by U(X) = (x 2-5x) J. Then the equilibrium position is at....... (where x in m) (A) x = 1.5 m

(B) x = 2m

(C) x = 2.5 m

69. The potential energy of a particle varies with distance x as shown in the graph. The force acting on the particle is zero at (A) C

(B) B

(C) B and C

(D) A and D

(D) x = 5 m



 W( x)

D

 C x 

70. A nucleus at rest splits into two nuclear parts having same density and radii in the ratio 1:2. Their velocites are in the ratio (B) 4:1

(C) 6:1

71. A body of mass 1.5 kg slide down a curved track which is quadrant of a circle of radias 0.75 meter. All the surfaces are frictionless. If the body starts from rest, its speed at the bottom of the track

(D) 8:1 0.75m 0.75m

(A) 2:1

is ..... ( g = 10 m/s 2) (A) 3. 87 m/s

(B) 2 m/s

(C) 1.5 m/s

(D) 0.387 m/s

72. A single conservative force F(x) acts on a 2.5 kg particle that moves along the xaxis. The potential energy U(x) is given by U(x) = (10 + (x - 4) 2) where x is in meter. At x = 6.0m the particle has kinetic energy of 20J. what is the mechanical energy of the system ? (A) 34 J

(B) 45 J

(C) 48 J 91

(D) 49 J

POWER 73. A body initially at rest undergoes one dimensonal motion with constant acceleration. The power delivered to it at time t is proportional to..... (A) t

1

(B) t

2

(C) t

3

(D) t2

2

74. An electric motor develops 5KW of power. How much time will it take to lift a water of mass 100 kg to a height of 20 m ? ( g = 10 m/s 2 ) (A) 4 sec (B) 5 sec (C) 8 sec (D) 10 sec 75. Bansi does a given amount of work in 30 sec. Jaimeen does the same amount of work... in 15 sec. The ratio of the output power of Bansi to the Jaimeen is.... (A) 1:1 (B) 1:2 (C) 2:1 (D) 5:3 76. A rope-way trolly of mass 1200kg uniformly from rest to a velocity of 72 km/ h in 6s. What is the average power of the engine during this period in watt ? (Neglect friction) (A) 400 W (B) 40,000 W (C) 24000 W (D) 4000 W 77. A body of mass m is accelarated uniformly from rest to a speed v in time T. The instantaneous power delivered to the body in terms of time is given by..... mv 2 (A) 2 . t T

mv 2 2 (B) 2 . t T

(C)

mv 2 2T

.t

(D)

mv 2 2T

2

.t2

78. 1 kg apple gives 25 KJ energy to a monkey. How much height he can climb by using this energy if his efficiency is 40%. (mass of monkey = 25 kg and g = 10 m/s2) (A) 20m (B) 4m (C) 30m (D) 40m 79. A force of



 2iˆ  3jˆ  kˆ  N acts on a body for 5 second, produces a displacement 

of 3iˆ  5jˆ  kˆ . What was the power used ? (A) 4W (B) 20 W (C) 21 W (D) 4.2 W 80. If the force F is applied on a body and it moves with a velocity v, the power will be (A) Fv

(B)

F

(C)

V

F

2 V

(D) Fv 2

81. From an automatic gun a man fires 240 bullet per minute with a speed of 360 km/ h. If each weighs 20 g, the power of the gun is (A) 400 W (B) 300 W (C) 150 W (D) 600 W 92

82. A body of mass M is moving with a uniform speed of 10 m/s on frictionless surface under the influence of two forces F 1 and F2. The net power of the system is F1  M  F2 (A) 10 F1F2 M

(B) 10 ( F1  F2 ) M

(C) ( F1  F2 ) M

(D) Zero

83. A coolie 2.0m tall raises a load of 75 kg in 25 from the ground to his head and then walks a distance to 40m in another 25. The power developled by the coolie is ( g = 10 m/s2) (A)0.25 kw (B) 0.50 kw (C) 0.75 kw (D) 1.00 kw 84. A body is moved along a straight line by a machine delivering a constant power. The velocity gained by the body in time t is proportional to..... (A) t

3

4

(B) t

3

(C) t

2

1

4

(D) t

1

2

ELASTIC - INELASTIC (COLLISION) 85. The coefficient of restitution e for a perfectly elastic collision is (A) 1 (B) 0 (C)  (D) -1 86. Two balls at same temperature collide. What is conserved (A) Temperature (B) velocity (C) kinetic energy (D) momentum 87. A particle of mass m moving with horizontal speed 6 m/s as shown in figure. If m<<M then for one dimensional elastic collision, the speed of lighter particle after collision will be

(A) 1 m/s in original direction. (B) 2 m/s opposite to the original direction. (C) 1 m/s opposite to the original direction. (D) 2 m/s in original direction. 88. A rubber ball is dropped from a height of 5 m on a planet where the acceleration due to gravity is not known. On bouncing, it rises to 1.8 m. The ball losses its velocity on bouncing by a factor of (A) 16 25

(B)

9

(C)

25

3

5

(D)

2

5

89. Three objects A, B and C are kept in a straight line on a frictionless horizontal surface. These have masses m, 2m and m respectivelly. The object A moves towards B with a speed 9 m/s and makes an elastic collision with it. Thereafter, B makes compelety inelastic collision with C. All motion occur on the same straight line. Find final speed of the object C 93

(A) 3 m/s (B) 4 m/s (C) 5 m/s (D) 1 m/s 90. Two solid rubber balls P and Q having masses 200 g and 400 g respectively are moving in opposite directions with velocity of P equal to 0.3 m/s. After collision the two balls come to rest, then the velocity of Q is (A) 0.15 m/s (B) 1.5 m/s (C) -0.15 m/s (D) Zero 91. A sphere collides with another sphere of identical mass. After collision, the two sphere move. The collision is inelastic. Then the angle between the directions of the two spheres is 92.

93.

94.

95.

96.

(A) Different from 90 (B) 90 (C) O (D) 45 A ball is allowed to fall from a height 20m . If there is 30% loss of energy due to impact, then after one impact ball will go up to (A) 18 m (B) 16 m (C) 12 m (D) 14 m If a skater of weight 4 kg has intial speed 40 m/s and 2 nd one of weight 6 kg has 6 m/s. After collision, they have speed (couple) 6 m/s. Then the loss in K.E. is..... (A) 48 J (B) zero (C) 96 J (D) None of these A metal ball of mass 2 kg moving with a velocity of 36 km/h has a head on collision with a stationary ball of mass 3 kg. If after the collision, the two balls move togather, the loss in kinetic energy due to collision is (A) 40 J (B) 60 J (C) 100 J (D) 140 J A neutron having mass of 1.67  10-27 kg and moving at 108 m/s collides with a deutron at rest and sticks to it. If the mass of the deutron is 3.34  10-27 kg then the speed of the combination is (A) 3.33  107 m/s (B) 3  105 m/s (C) 33.3  107 m/s (D) 2.98  105 m/s A mass of 100g strikes the wall speed 5 m/s at an angle as shown in figure and is rebounds with the same speed. If the contact time is 5 103 sec, what is the force applied on the mass by the wall (A) 100 3 N to right (B) 100 N to right (C) 100 3 N to left (D) 100 N to left 94

97. Three identical sperical balls A, B and C are placed on a table as shown in the figure along a straight line. B and C are at rest initially. The ball A hits B head on with a speed of 10 m/s. Then after all collision A and B are brought to rest and C takes off with velocity of...... (elastic collision)

(A) 20 m/s (B) 2.5 m/s (C) 10 m/s (D) 7.5 m/s 98. A ball dropped from a height of 4m rebounds to a height of 2.4m after hitting the ground. Then the percentage of energy lost is (A) 40 (B) 50 (C) 30 (D) 600 99. A billiard ball moving with a speed of 8 m/s collides with an identical ball originally at rest. If the first ball stops after collision, then the second ball will move forward with a speed of..... (elastic collision) (A) 8 m/s (B) 4 m/s (C) 16 m/s (D) 1.0 m/s 100. A bullet of mass m moving with velocity v strikes a block of mass M at rest and gets embedded into it. The kinetic energy of the composite block will be 1

M

2 (A) 2 mv  (m  m)

1

m

2 (C) 2 MV  (m  M)

(B)

1 2 (m  m) mv  2 M

1

m

2 (D) 2 mv  (m  M)

101. A body of mass m1 is moving with a velocity v. It collides with another stationary body of mass m 1. They get embeded. At the point of collision, the velocity of the system (A) Increases (B) Decreases but does not become zero (C) Remains same (D) Become zero 102. Two small particles of equal masses start moving in opposite directions from a point. A in a horizontal circular orbit. Their tangential velocities are v and 2v, repectively, as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A, these two particles will again reach the point A (A) 1 (B) 2 (C) 3 (D) 4 95

103. Four identical balls are lined in a straight grove made on a horizontal frictionless surface as shown. Two similar balls each moving with a velocity v collide elastically with the row of 4 balls from left. What will happen

(A) One ball from the right rolls out with a speed 2v and the remaining balls will remain at rest. (B) Two balls from the right roll out speed v each and the remaining balls will remain statinary. (C) All the four balls in the row will roll out with speed v 4 each and the two colliding balls will come to rest. (D) The colliding balls will come to rest and no ball rolls out from right.

(ASSERTION & REASON) *

Assertion and Reason are given in following questions. Each question have four option. One of them is correct it. (1) If both assertion and reason and the reason is the correct explanation of the Assertion. (2) If bothe assertion and reason are true but reason is not the correct explanation of the assertion. (3) If the assertion is true but reason is false. (4) If the assertion and reason both are fare. 104. Assertion : In the elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision. Reason : In the elastic collision the linear momentum of the system is conserved. (A)1 (B) 2 (C) 3 (D) 4 105. Assertion : When a gas is allowed to expand, work done by gas is positive. Reason : Force due to gaseous pressure and displacement (of position) on in the same direction. (A) 1 (B) 2 (C) 3 (D) 4 96

106. Assertion : A light body and heavy body have same momemtum. Then they also have same kinetic energy. Reason : Kinetic energy does not depend on mass of the body. (A) 1 (B) 2 (C) 3 (D) 4 107. Assertion : Mountain roads rarely go straight up the slope. Reason : slope of mountains are large, therefore, more chances of vehicle to slip from roads. (A) 1 (B) 2 (C) 3 (D) 4 108. Assertion : The change in kinetic energy of a particle is equal to the work done on it by the net force. Reason : Change in kinetic energy of particle is equal to the work done only in case of a system of one particle. (A) 1 (B) 2 (C) 3 (D) 4 109. Assertion : Work done in moving a body over a closed loop is zero for every force in nature. Reason : Work done does not depend on nature of force. (A) 1 (B) 2 (C) 3 (D) 4 110. Assertion : A weight lifter does no work in holding the weight up. Reason : Work done is zero because distance moved is zero. (A) 1 (B) 2 (C) 3 (D) 4 Kinetic energy

111. Assertion : stopping distance = Stopping force Reason : Work done in stopping a body is equal to K.E. of the body. (A) 1 (B) 2 (C) 3 (D) 4 112. Assertion : The mass equivalent of 1000 kwh energy is 40 microgram. Reason : This follows from E = mc 2 where C  3 108 m s (A) 1 (B) 2 (C) 3 (D) 4 113. Assertion : Work done by centripetal force is zero. Reason : This is because entripetal force is always along the tangent. (A)1 (B) 2 (C) 3 (D) 4 114. Assertion : Two bodies of different masses have same momentum. Their kinetic energy are in the inverse ratio of their masses. Reason : K.E. = (A) 1

1 2 mv 2

(B) 2

(C) 3 97

(D) 4

115. Assertion : Linear momentum is conserved in both, elastic and inelastic collisions. Reason : Total energy is conserved in all such collisins. (A) 1 (B) 2 (C) 3 (D) 4 116. Assertion : Both, a stretched spring and a compressed spring have potential energy. Reason : Work is done against the restoring force in each case. (A) 1 (B) 2 (C) 3 (D) 4

(COLUMN) 117. A force F = kx (where k is positive constant) is acting on a particle. Match column-I and column-II, regarding work done in displacing the particle. Column - I Column - ii (a) From x = -4 to x = -2 (P) Positive (b) From x = -2 to x = -4 (Q) zero (c) From x = -2 to x = +2 (R) negative (A) a - R, b - P, c - Q (B) a - P, b - Q, c - R (C) a - R, b - Q, c - P (D) a - Q, b - P, c - R 118. A body falls freely under the action of gravity from a height h above the ground. Column - i Column - ii (a) P.E. = 2(K.E.) (P) constant at every point (b) P.E. = K.E. (Q) at height h 3 (c) P.E. =

1 (K.E.) 2

(R) at height

2h

3

(d) P.E.+ K.E. (S) at height h 2 (A) a - P, b - Q, c - R, d - S (B) a - Q, b - P, c - S, d - R (C) a - S, b - R, c - Q, d - P (D) a - R, b - S, c - Q, d - P 119. Two vehicles moving on a horizontal road are stopped by same retarding force. Column - I (a) When they have same K.E. (b) When they have different masses but same velocity (c) When both have same momentum (d) When both have same mass but different velocities

Column - ii (P) faster body stop in larger distance (Q) larger body stops in larger distance. (R) heavier body stops in larger distance. (S) stopped in same distance.

(A) a - P, b - Q, c - R, d - S (C) a - S, b - R, c - Q, d - p

(B) a - Q, b - P, c - S, d - R (D) a - R, b - S, c - Q, d - P 98

KEY NOTE 1

B

26

A

51

A

76

B

101

B

2

B

27

B

52

C

77

A

102

B

3

D

28

A

53

B

78

D

103

B

4

D

29

B

54

A

79

A

104

D

5

A

30

C

55

A

80

A

105

A

6

D

31

D

56

B

81

A

106

D

7

C

32

B

57

D

82

D

107

A

8

C

33

A

58

B

83

C

108

B

9

D

34

B

59

A

84

D

109

D

10

D

35

B

60

A

85

A

110

A

11

A

36

B

61

C

86

D

111

A

12

B

37

C

62

C

87

A

112

A

13

D

38

C

63

D

88

D

113

C

14

B

39

B

64

C

89

B

114

B

15

D

40

A

65

C

90

C

115

B

16

A

41

A

66

B

91

A

116

A

17

C

42

A

67

A

92

D

117

A

18

C

43

A

68

C

93

D

118

D

19

B

44

C

69

C

94

B

119

C

20

D

45

A

70

D

95

A

21

D

46

D

71

A

96

C

22

B

47

A

72

A

97

C

23

B

48

A

73

B

98

A

24

A

49

D

74

A

99

A

25

B

50

A

75

B

100

D

99

HINT 1.

W  Fd cos 

2.

 K  Fd cos  = w

3.

W  F d  ( Mg ) ( d ) ; where M = m1 + m2

4.

a  g and v  v   2gd  d 

5.

W = Area under curve of F  x graph.

6.

2 According W  kx

7.

hn  (2n  1)

8.

W

9.

According to work energy theorm W  k

10.

s

2

2 0

vo 2 2g

v  0 

1 2

1 k x 22  x12 2





l l 2t 3 d 2s  acceleration a  2  4t , work W   F ds   mads 3 dt 0 0

11. W = F x = max = (m) (Area covered by curve of a  x graph) 12. work W = Area covered by curve of F  x graph 13. Loss in P.E. of spring = gain in K.E. of ball 14.

W=

mgl , where n is, fraction of length of the chain hanging from the table. 2n 2 mgl ( n = 3 Given) 2n 2

15. W =

16. Tension in the cord, T = M (g17. Fd cos  = 18.  x  19.

F

g mg mgd )= . Work done by cord = -Td = 2 2 2

1 1 mv2 - mu2 2 2

F mg  K k

du dx

100

20.

w

mgl ( n = 4 given) 2n 2

21. Gravitational force is a conservative force and work done against it is a point function i.e. does not depends on the path. 22.

v 2  v 20  2ad , (v  0) and v0  2gh

o 



2gh



2

 2(g  a)d a 

23. Work done = gain in potential energy Area under curve = mgh 24. Intial velocity at x = 0; v1 = 0 Final velocity at x = 2; v2= 5  2 3 2 Work done =

1 m ( v 22  v12 ) 2

25. Intial velocity of particle v1= 20 m/s Final velocity of particle v2= 0 1 W  k  k 2  k1  m (v 22  v12 ) 2

26.

W  Fl sin  U  mgl(1  cos )  W  U; (  600 )

27. Work = Area under (F - d) graph 28. Work = Area under (F - d) graph 29.

W   Fdy 2

=

  Ay

2



 By  C dy

a

a

 Ay3 By 2     cy  2  3  a



2Aa 3  2Ca 3

30. Energy U  x2 ( k constant)

101

31.

K 

32.

P2 2m

K  P 2 ( m constant)

dk dp 2 k p

1 k  mv 2 2

v 

2k m

33. Energy lost due to air friction =

34.

P m

35. mgR =



1 mv2 - mgh 2

m1 p1  p2 m2

1 2 mv 2

p2 2m

36.

E

37.

V  2gh

38.

U

 E

1 m

kx 2 2

39. F. t = momentum, The resulatant fore is zero, therefore F. t = 0 40. Kinetic Energy (K.E)  v 2 and v 2  t 2 .... 41. P.E (U)  h and E = K + U = constant 42.

1 mv 2 1 Al  V 2 1   Av 2 2 t 2 t 2 

43. Find the value of

v

and than, K.E =

1 mv 2 2

3

44.

U   kx dx. 1

45. Spring constant K 

F 1 , U  kx 2 x 2

46. Spring constant K 

F 1 , U  kx 2 x 2

102

v2o Sin 20 mv02 sin 2 0 1 , U  mg H max   Ko 47. Hmax = 2g 2 2

48. According to equation

1 1 2 2 mp v p  mQ vQ  Vp VQ , 2 2 m v

FΔt

p p P Using impulse momentum, FΔT = m v Q Q Q

49.

1 2 mv  area covered by curve of F  x , graph 2

50. K.E. is maximum than P.E. minimum. so

du 0  0 dx

OR 1 4

For x   4 U ( x )    U min . 51.

E1  E and E 2 1.44E . Than P E

52. F d =

1 m  v 22  v12  . 2

53. Loss of potential energy = Kinetic energy gained. 54. Let the thickness of one planks be s and n planks are arranged just to stop bullet. v 2  v 0 2  2as , put v 

9 v0 10

 Now

v0 2 100 v2  n  0 2as 19 2as

55. By conservation of linear momentum  mv = (m+M) Vsys By convervation of low of energy.

1 (m+M) v2 sys = (m+M) gh 2

Find Vsys and put it sin egn.(1) 56. Work done = Area under F  X graph with proper sign. 5 2

57. Condition for up direction loop (Parabola) h = R 58. P.E. of water = K.E. at turbine. mgh =

1 2 mv 2

m

1

1 59. The bomb into two masses m1 + m 2 =12kg and m  4 , K.E. 2

energy of smaller part =

1 m1 v12 2

103

60. 1 k cal = 103 calorie = 4200J = 700 K cal 

4200 kwh 3.6106

700  4200 kwh 3.6  10 6

61. From the conservation of energy, 1 2

mg (h1  h 2 )  mv

2

62.

m 1 v 1 + m 2 v 2 = 0 and than K.E. =

63.

m1v2 , = m2 v2 ,

Total energy of system =

1 m 2 v 22 2

1 1 m1 v12  m 2 v 22 2 2

64. The collision between bob and block is elastic  P.E.  K.E. 65.

mG vG  mB vB

66.

U  mgh

67. Initial kinetic energy, E =

1 2 mv 2 1 2

2 Final kinetic energy, 2E = m( v  2)

68. For a conservation field. Force, F  

du , dx

At equilibrium position, F = 0 69.

F 

du it is clear that slope of U - x curve is zero at point B and C. dx 3

M1  R1  70. Given 1 = 2  M =   2  R2 

V1  2    According to low conservation of linear momentum, V2  1  1 2 71. Conservation of energy, mgh = mv 2

72. P.E. at X = 6m is U = 10 + (6 - 4)2 = 14 J Mechanical energy = K.E. + P.E. = 20 + 14 = 34 J

104

3

73. Power, P = Fv = mav 74. Power, P =

mgh t

75. Power, P 

1 t

2

76.

EP

77.

P  2mE ; P E

78.

P  2mE ; P E

2m

 

79.

P  Wt

P

F.d t

80. ............. 81. Power of gun =

2 Total K.E.of fired bullet 1  n  mv t time 2

82. Work done by forces = 0 ; W 

p t

mgh (75)(10)(2)   0.75 KW t 2

83.

P

84.

1 mv 2  kt (K, Cons tan t) , 2

v

2k 12 2k 12 t , dx  t dt. m m

Take integration.

85. ............ 86. ............ 87.

 m  m2   2m2 u 2   u1    v1   1  m1  m2   m1  m2 

where m1 = m and m2 = M; m << M. m1 = 0, v1 = -u1+ 2u2

88.

v v2 h  2 ; i.e. fractional loss in velocity  1 2 v1 v1 h1

89.

v2 

2m1u1  6m / s m1  m 2

i.e. After elastic collision B strike to C with velocity of 6 m/s, Now for B and C 2m –– 6m/s

m  3m  Vsys Rest  (2m) (6) + 0 = (3m) (V sys)

90.

m p v p  m Q vQ  0

105

91. Angle will be 90o if collision is perfectly elastic. 92.

mgh ' = 70% of mgh

1

1

1

93. Loss in K.E. = (initial K.E. - Final K.E.) of system  m1u12  m 2 u 22  (m1  m 2 )v 2 2 2 2 94. By low of conservation of momentum 2  10 = (2+3) v v  4 m s Loss in K.E. =

1 1 (2)(10)2 - (5)(4)2 2 2

95. Momentum of neutron = Momentum of combination 96. Force = Rate of change of momentum. 

Intial momentum p1  mvsinˆj  mv cos ˆj 

Final momentum p 2   mvsinˆj  mv cos ˆj 

P F t



97. In elastic head on collision velocity gets interchanged. 98. U1 = mgh1 and U2= mgh2 % energy lost =

U1  U 2  100 U1

99. In elastic head on collision velocites gets interchanged. 100. By conservation of momentum, mv + M(o) = (m+M) v  m  v mM

Velocity of composite block V = 

1 (M  m) v 2 2 101. By conservation of momentum, m1V  m2 (0)  (m1  m2 )v

K.E. of composite block =

102. Let initially particle x is moving in anticlock-wise direction and y in clockwise direction. As the ratio of velocities of x and y Vx

1

particles are Vy  2 , therefore ratio of their distance covered will be in the ratio of 2:1. It means they collide at B. so, after two collision these two particles will again reach the point A. 103. Momentum and kinetic energy is conserved only in this case. 106

Unit - 5 Rotational Motion

107

SUMMARY * 1.

Important Formula, Facts and Terms Centre of mass of system of particles

R cm 

n m1 r1  m1 r 2      m n r n m rn  n m1  m 2      m n n 1 m n

for rigid body M.R cm   m n r n

for Two body System

m1 r1  m 2 r 2

 r1 m2 +1 OR  +1= m1 r2  r m +m   = 1 2 m1 r2 2.

OR

  r2 =

 r1 +r2 m 2 +m1  = m1 r2

rm1 m1 +m 2

OR

 rm 2 r1 = m1 +m 2

    m1 v1 + m2 v 2 + .........+ mn v n V cm = M      P  M vcm  p1  p 2  .........  p n similarly

a cm 

m1 a1  m 2 a 2     m n a n M

F  M a cm  F1  F 2     Fn 3.

Torque = T = r  F = I      rF sin = product of force and perpendicular distance between point of rotation and line of action.



 



Angular momentum = L  r  P  I  | L | = rpsin  = product of linear momentum and perpendicular distance between point of rotation and line of action. 2

2

Moment of inertia = I = m1r1  m 2 r2      m n rn n

=

m r

2

n n

n 1

108

2

4.

Low of conservation of angular momentum

    dL d I  I   p dt dt

  As L  I 

  dL    when  = then L remains constant dt Its geometrical representation in planetary motion dA is an areal velocity dt

Let Then 5.

m

dA L = dt 2

OR

dA L = dt 2m

Radius of gyration {K} As I = m1r1 2  m 2 r2 2      m n rn 2 If all particles have same mass then



= m r12  r2 2     rn 2

r I  nm

2

1



2

 r2     rn n

2



Here (nm = M)

 mk 2 2

k

6.

2

2

(r1  r2     rn ) n

Some relations between linear and rotational motion. v = rw

   v  w r

Here

w =

  d w d 2   2 dt dt    where a  a r  aT

d dt

ar  wv aT  r

 a  ar 2  at 2  2v 2  2 r 2  2 r 2 2   2 r 2  r 4   2 7.

Equilibrium of a rigid body.

8.

When F  F1  F2     Fn  0 it is in linear equilibrium     When P  P1  P 2     P n  0 it is in rotational equilibrium Two theorm for moment of inertia

IZ  IX  IY I  I cm  Md 2

Theorm of perpendicular axis. Theorm of Parallel axis. 109

9.

Rolling down of body on an inclined plane.    2 gh  V  2 , K 1  R 2 

a

2 sin  1  K 2   R 2 

Condition for rolling without sliding

s 

tan  , 2 1 R 2 K





For ring  s 

disc

1 tan  2

(K = R)

1 R    s  tan   K   3 2 

2 solid sphere  s  7 tan 

10.

 2  K  R   5  

Rotational Kinetic Energy. 1 2 R.K.E = I  2

2 1 2V  MK 2 2 R

V  2  I  MK &    R 

L2  2I Total kinetic Energy

= Rotational K.E. + Linear K.E.



Now

(1)

K2  1 V2 1 1 MK 2 2  MV 2  MV 2  2  1 2 2 2 R R 

2 R.K.E K 2 R LinearK .E

K2

Rotational K .E K2 R2  2  (2) 2 2 Total K .E R K 1 K 2 R

 K2 R 2  Percentage rotational K.E. =  K 2   100%  1 R2 

110

(3)

Translatio nal K.E R2 1  2  2 2 Total K.E R K 1 K R2

Comparison between physical quantities of linear motion and rotational motion Translational motion Linear displacement,

Rotational motion Angular displacement, 

d

Linear velocity, V Linear acceleration, a 

Angular velocity, w

dv dt

Angular acceleration,  

dw dt

Mass, m

Moment of inertia, I

Linear Momentum, P  m v

Angular momentum, L  I w

Force,

Torque,  

F  ma

Newton's Second Law of Motion,

F

A result similar to newtown's Second Law,

dP dt



1 2 Translational kinetic energy K = mv 2



dL dt

dL dt

Rotational kinetic energy K =

p2 2m



1 2 I 2

L2 2I

Work, W = F d

Work, W =

Power, P = Fv Equations of linear motion taking place with constant linear acceleration

Power, P = w Equations of rotational motion taking place with constant angular acceleration :

v = VO + at

d = v O t+



w = WO  t

1 2 at 2

1   W0 t  at 2 2

2ad = v 2 -v 0 2

2a = W 2  W0 2

Law of conservation of linear momentum when F  0 then P is constant Impulse linear  F  t  P 2  P1 111

Law of conservation of angular momentum  when P  0 then L is constant Impulse    rotational   t  L 2  L1

Value of

V, a, t for some Rolling Bodies

Shape of Body

velocity

velocity

acceleration 1 2

Ring/Hollow cylinder

gh

gl sin 

Disc/Solid cylinder

4 gh 3

4   gl sin   2 3 

Solid Sphere

10 gh 7

 10   gl sin   2 7 

Shell/Hollow spher

6 gh 5

6   gl sin   2 5 

1

3

g sin 

3 g sin 

1

7

g sin 

14 g sin 

2

5

g sin 

10 g sin 

2

1 3

R2

1

g sin 

1

5

 g sin 

2

2

1 2

K2

Time

2

5

3

Moment of inertia an radius of gyration for some symmetric bodies Body

Axis

Thin rod of

Passing through its

Length L

centre and per pendicular to its length

Ring of radius R

Any diameter

Ring of radius R

Passing through its centre and perpendicular to its plane

Circular disc radius R

Passing through its centre and perpendicular to its plane of

Circular disc

Any diameter of radius R

Figure

I

K

For rolling body K 2 R 2

L

1 ML2 12

2 3

1 MR 2 2

R 2

MR 2

R

-

-

1

1 MR 2 2

R 2

1 2

1 MR 2 4

R 2

-

Body

Axis

Figure

Hollow cylinder of radius R

Geometrical axis of the cylinder

Any diameter of radius R

Hollow sphere of radius R

Any diameter

K

For rolling body K 2 R 2

MR 2

R

1

1 MR 2 2

Solid cylinder Geometrical axis of radius R of the cylinder Solid sphere

I

R 2

1 2

2 MR 2 5

2 R 5

2 5

2 MR 2 3

2 R 3

2 3

MCQ For the answer of the following questions choose the correct alternative from among the given ones. 1.

2.

The centre of mass of a systems of two particles is (A) on the line joining them and midway between them (B) on the line joining them at a point whose distance from each particle is proportional to the square of the mass of that particle. (C) on the line joining them at a point whose distance from each particle inversely propotional to the mass of that particle. (D) On the line joining them at a point whose distance from each particle is proportional to the mass of that particle. Particles of 1 gm, 1 gm, 2 gm, 2 gm are placed at the corners A, B, C, D, respectively of a square of side 6 cm as shown in figure. Find the distance of centre of mass of the system from geometrical centre of square. {A} 1 cm {B}

2 cm

{C} 3 cm

3.

{D} 4 cm Three particles of the same mass lie in the (X, Y) plane, The (X, Y) coordinates of their positions are (1, 1), (2, 2) and (3, 3) respectively. The (X,Y) coordinates of the centre of mass are {A} (1, 2)

{B} (2, 2)

{C} (1.5, 2) 113

{D} (2, 1.5)

4.

Consider a two-particle system with the particles having masses M1 , and M 2 . If the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle be moved so as to keep the centre of mass at the same position? {A}

5.

M 2d {B} M  M 1 2

{C}

M1d M2

M2d {D} M 1

Four particles A, B, C and D of masses m, 2m, 3m and 5m respectively are placed at corners of a square of side x as shown in figure find the coordinate of centre of mass take A at origine of x-y plane. {A}

{C} 6.

M1d M1  M 2

7x    2x ,  10    x 10x   ,  2 7 

{B}

10 x    2 x,  7  

{D}

 x 7x   ,   2 10 

From a uniform circular disc of radius R, a circular disc of radius R 6 and having centre at a distance + R 2 from the centre of the disc is removed. Determine the centre of mass of remaining portion of the disc. R R R R {B} {C} {D}  70 70 7 7 A circular plate of uniform thickness has a diameter of 56 cm. A circular portion of diameter 42 cm. is removed from +ve x edge of the plate. Find the position of centre of mass of the remaining portion with respect to centre of mass of whole plate. {A} - 7 cm {B} + 9 cm {C} - 9 cm {D} + 7 cm Two blocks of masses 10 kg an 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the centre of mass is : {A} 30 m/s {B} 20 m/s {C} 10 m./s {D} 5 m/s A particle performing uniform circular motion has angular momentum L., its angular frequency is doubled and its K.E. halved, then the new angular momentum is : {A} ½ {B} ¼ {C} 2L {D} 4L A circular disc of radius R is removed from a bigger disc of radius 2R. such that the circumferences of the disc coincide. The centre of mass of the remaining portion is R from the centre of mass of the bigger disc. The value of  is. {A} ½ {B} 1/6 {C} ¼ {D} 1/3 Three point masses M1, M2 and M3 are located at the vertices of an equilateral triangle of side 'a'. what is the moment of inertia of the system about an axis along the attitude of the triangle passing through M1, ?

{A} 7.

8.

9.

10.

11.

2

{A} M1  M 2  a 4

2

{B} M 2  M 3  a 4

114

2

2

{C} M1  M 3  a 4 {D} M1  M 2  M 3  a 4

12.

13.

A body of mass m is tied to one end of spring and whirled round in a horizontal plane with a anstant angular velocity. The elongation in the spring is one centimeter. If the angular velocity is doubted, the elongation in the spring is 5 cm. The original length of spring is… {A} 16 cm {B} 15 cm {C} 14 cm {D} 13 cm A cylinder of mass 5 kg and radius 30 cm, and free to rotate about its axis, receives an angular impulse of 3 kg M2S-1 initially followed by a similar impulse after every 4 sec. what is the angular speed of the cylinder 30 sec after initial imulse ? The cylinder is at rest initially. {A} 106.7 rad S1

14.

{B} 206.7 rad S1 {C} 107.6 rad S-1 {D} 207.6 rad S-1 Two circular loop A & B of radi r a and rb respectively are made from a uniform wire. The ratio of their moment of inertia about axes passing through their centres and perpendicular to their planes IB is I  8 A

15.

16.

then

rb Ra is equal to… ra

{A} 2 {B} 4 {C} 6 {D} 8 If the earth were to suddenly contract so that its radius become half of it present radius, without any change in its mass, the duration of the new day will be… {A} 6 hr {B} 12 hr {C} 18 hr {D} 30 hr





In HC1 molecule the separation between the nuclei of the two atoms is about 1.27 A 1A  10 10 m . The approximate location of the centre of mass of the molecule is   A ˆi with respect of Hydrogen atom ( mass of CL is 35.5 times of mass of Hydrogen) {A} 1 {B} 2.5 {C} 1.24 {D} 1.5

17.

18.

19.





Two bodies of mass 1kg and 3 kg have position vector ˆi  2ˆj  kˆ and (-3i-2j+k) respectively the center of mass of this system has a position vector…… {A}  2ˆi  2kˆ {B}  2ˆi  ˆj  kˆ {C} 2ˆi  ˆj  kˆ {D}  ˆi  ˆj  kˆ Identify the correct statement for the rotational motion of a rigid body {A} Individual particles of the body do not undergo accelerated motion {B} The center of mass of the body remains unchanged. {C} The center of mass of the body moves uniformly in a circular path {D} Individual particle and centre of mass of the body undergo an accelerated motion. A car is moving at a speed of 72 km/hr the radius of its wheel is 0.25m. If the wheels are stopped in 20 rotations after applying breaks then angular retardation produced by the breaks is …… {A} -25.5 rad s 2

{B} -29.5 rad s 2

{C} -33.5 rad s 2

{D} -45.5 rad s 2

20.

A wheel rotates with a constant acceleration of 2.0 rad sec 2 If the wheel start from rest. The

21.

number of revolution it makes in the first ten seconds will be approximately. {A} 8 {B} 16 {C} 24 {D} 32 Two discs of the same material and thickness have radii 0.2 m and 0.6 m their moment of inertia about their axes will be in the ratio {A} 1 : 81 {B} 1 : 27 {C} 1 : 9 {D} 1 : 3 115

22. A wheel of mass 10 kg has a moment of inertia of 160 kg m 2 about its own axis. The radius of gyration will be _______ m. {A} 10 {B} 8 {C} 6 {D} 4 23. One circular rig and one circular disc both are having the same mass and radius. The ratio of their moment of inertia about the axes passing through their centres and perpendicular to their planes, will be…… {A} 1 : 1 {B} 2 : 1 {C} 1 : 2 {D} 4 : 1 24. One solid sphere A and another hollow sphere B are of the same mass and same outer radii. The moment of inertia about their diameters are respectively I A and I B such that… {A} I A  I B 25.

26.

{B} I A  I B

A ring of mass M and radius r is melted and then molded in to a sphere then the moment of inertia of the sphere will be….. {A} more than that of the ring {B} Less than that of the ring {C} Equal to that of the ring {D} None of these A circular disc of radius R and thickness R/6 has moment of inertia I about an axis passing through its centre and perpendicular to its plane. It is melted and recasted in to a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation is … 2I I {C} 8 5 One quater sector is cut from a uniform circular disc of radius R. This sector has mass M. It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is…

{A} I 27.

{B}

{A} 1 2 MR

28.

I A dA {C} I A  I B {D} I  dB (radio of their densities) B

{D}

I 10

2 {B} 1 4 MR

2

2 {C} 18 MR {D} 2MR 2 A thin wire of length L and uniform linear mass density  is bent in to a circular loop with centre at O as shown in figure. The moment of inertia of the loop about the axis xx' is ….

2 {A}  L

8

3 {B}  L

2

3

{C} 5  L

2

16  2

3 {D} 3  L

16  8 2 29. Two disc of same thickness but of different radii are made of two different materials such that their masses are same. The densities of the materials are in the ratio 1:3. The moment of inertia of these disc about the respective axes passing through their centres and perpendicular to their planes will be in the ratio. {A} 1 : 3 {B} 3 : 1 {C} 1 : 9 {D} 9 : 1

116

30.

Let I be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle of Q with AB. The moment of inertia of the plate about the axis CD is then equal to….  2 A small disc of radius 2 cm is cut from a disc of radius 6 cm. If the distance between their centres is 3.2 cm, what is the shift in the centre of mass of the disc… {A} -0.4 cm {B} -2.4 cm {C} -1.8 cm {D} 1.2 cm A straight rod of length L has one of its ends at the origin and the other end at x=L If the mass per unit length of rod is given by Ax where A is constant where is its center of mass. {A} L/3 {B} L/2 {C} 2L / 3 {D} 3L / 4 A uniform rod of length 2L is placed with one end in contact with horizontal and is then inclined at an angle  to the horizontal and allowed to fall without slipping at contact point. When it becomes horizontal, its angular velocity will be…..

{A} I 31.

32.

33.

{A} w = 34.

3 g sin  2L

36.

38.

2L {B} w = 3 g sin 

{C} w =

6 g sin  L

{D}

I cos 2

{D} w =

L g sin 

{B} 3v 2a

{C}

3v

{D} zero 2a Consider a body as shown in figure, consisting of two identical bulls, each of mass M connected by a light rigid rod. If an impulse J = MV is imparted to the body at one of its ends, what would be its angular velocity. What is V ? {A} V / L {B} 2V / L {C} V / 3L {D} V / 4L A thin circular ring of mass M and radius r is rotating about its axis with a constant angular velocity w. Two objects each of mass m are attached gently to the opposite ends of a diameter of the ring. The ring will now rotate with an angular velocity….

M  M  2m 

  M  2m  M A smooth sphere A is moving on a frictionless horizontal plane with angular speed  and centre of mass velocity v. It collides elastically and head on with an identical sphere B at rest. Neglect friction everywhere. After the collision, their angular speeds are A and B respectively, Then {A} A < B {B} A = B {C} A =  {D}  = B Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of ….. {A} 0.4 m from mass of 0.3 kg {B} 0.98 m from mass of 0.3 kg {C} 0.7 m from mass of 0.7 kg {D} 0.98 m from mass of 0.7 kg {A}

37.

{C} I cos 2 

A cubical block of side a is moving with velocity V on a horizontal smooth plane as shown in figure. It hits a ridge at point O. The angular speed of the block after it hits O is …. {A} 3v 4a

35.

{B} I sin 2 

( M  2m) M  2m

{B}

{C}

117

M M m

{D}

39.

40.

In a bicycle the radius of rear wheel is twice the radius of front wheel. If rF and rr are the radius, vF and vr are speed of top most points of wheel respectively then... {A} vr = 2vF {B} vF = 2vr {C} vF = vr {D} vF > vr From a circular disc of radius R and mass 9M, a small disc of radius R/3 is removed from the disc. The moment of inertia of the remaining portion about an axis perpendicular to the plane of the disc and passing through O is…. 40 37 MR 2 {C} 10 MR 2 {D} MR 2 9 9 A child is standing with folded hands at the centre of a platform rotating about its central axis the kinetic energy of the system is K. The child now stretches his arms so that the moment of inertia of the system doubles. The kinetic energy of the system now is… {A} 2 K {B} K/2 {C} K/4 {D} 4K If the earth is treated as a sphere of radius R and mass M. Its angular momentum about the axis of rotation with period T is…..

{A} 4 MR 2 41.

42.

{B}

MR 3 MR 2  2MR 2 4MR 2 {B} {C} {D} T T 5T 5T If the angular momentum of any rotating body increases by 200%, then the increase in its kinetic energy will be….. {A} 400% {B} 800% {C} 200% {D} 100% The M.I. of a body about the given axis is 1.2 kgm2 initially the body is at rest. In order to {A}

43.

44.

45.

46.

produce a rotational kinetic energy of 1500 J. an angular acceleration of 25 rad sec 2 must be applied about that axis for duration of …. {A} 4 sec {B} 2 sec {C} 8 sec {D} 10 sec An automobile engine develops 100kw when rotating at a speed of 1800 r.p.m. what torque does it deliver ? {A} 350 Nm {B} 440 Nm {C} 531 Nm {D} 628 Nm The moment of inertia of two rotating bodies A and B are I A and I B . I A  I B  and their angular momentum are equal. If their K.E. be K A and K B respectively then…. {C} KA KB  1 {D} KB KA  1 2 The centre of mass of the disc undergoes S.H.M. with angular frequency  equal to.. {A} KA, KB

47.

{A} 48.

k m

{B} KB KA  1

{B}

2k m

{C}

2k 3m

Three rings, each of mass P and radius  are arranged as shown in the figure the moment of inertia of the arrangement about YY' axis will be. 2 2 2 2 {A} 7 2 P {B} 2 7 P {C} 2 5 P {D} 5 2 P

118

{D}

4k 3m

49.

50.

51.

52.

53.

If distance of the earth becomes three times that of the present distance from the sun then number of days in one year will be …. {A} 365  3

{B} 365  27

{A} 2

{B}

55.

57.

365

1 2

{C} ½

{D} 2

A player caught a cricket ball of mass 150 gm moving at a rate of 20 m/s If the catching process is Comlitad in 0.1 sec the force of the flow exerted by the ball on the hand of the player ….. N {A} 3 {B} 30 {C} 150 {D} 300 Two disc one of the density 7.2 gm/cc and other of density 8.9 gm/cc are of the same mass and thickness their moment of inertia are in the ratio of …… 1 {C} 8.9 7.2 8.9  7.2 Two identical hollow spheres of mass M and radius R are joined together and the combination is rotated about an axis tangential to one sphere and perpendicular to the line connecting their centers. The moment of inertia of the combination is ________.

{B}

{D} 8.9  7.2

2 2 2 {A} 10 MR 2 {B} 4 3 MR {C} 32 3 MR {D} 34 3 MR A rod of length L rotate about an axis passing through its centre and normal to its length with an angular velocity . If A is the cross-section and D is the density of material of rod. Find its rotational K.E.

{A} 56.

{D}

365 3 3

3 3 A solid sphere and a solid cylinder having same mass and radius roll down the same incline the ratio of their acceleration will be…. {A} 15 : 14 {B} 14 : 15 {C} 5 : 3 {D} 3 : 5 The ratio of angular momentum of the electron in the first allowed orbit to that in the second allowed orbit of hydrogen atom is ……

{A} 7.2 8.9 54.

{C}

1 2

AL3D

{B}

1 6

AL3 D

{C}

1 24

AL3 D

{D}

1 12

AL3 D

Initial angular velocity of a circular disc of mass M is w1 Then two spheres of mass m are attached gently two diametrically oppsite points on the edge of the disc what is the final angular velocity of the disc? Mm  M  4m   M   M  w1  w1  w1  w1 {A}  {B}  {C}  {D}   M   M   M  4m   M  2m  A circular disc x of radius R is made from an iron plate of thickness t. and another disc Y of radius 4R is made from an iron plate of thickness t/4 then the rotation between the moment of

inertia I X and I y is ___________ {A} I y  64 I x 58.

{B} I y  32 I x

{C} I y  16 I x

{D} I y  I x





A Pulley of radius 2 m is rotated about its axis by a force F  20 t  5t 2 N where t is in sec applied tangentially. If the moment of inertia of the Pulley about its axis of rotation is 10 KgM 2 , 119

59.

the number of rotations made by the pulley before its direction of motion is reversed is : {A} more than 3 but less then 6 {B}more than 6 but less then 9 {C} more than 9 {D} Less then 3 Two spheres each of mass M and radius R/2 are connected with a mass less rod of length 2R as shown in figure. What will be moment of inertia of the system about an axis passing through centre of one of the spheres and perpendicular to the rod? 21 2 5 5 2 2 2 MR 2 {B} MR {C} MR {D} MR 5 5 2 21 Four particles each of mass 'm' are lying symmetrically on the rim of the disc of mass M and radius R moment of inertia of this system about an axis passing through one of the particles and perpendicular to plane of disc is _____ R2 {A} 16 MR 2 {B} 3M  16M  2 R2 {C} 3M  12M  {D} Zero 2 A mass M is supported by a mass less string wound around a uniform cylinder of mass M and radius R as in figure. With what acceleration will the mass fall on release? {A} 2/3g {B} g/2 {C} g {D} 4g/3 Calculate rotational K.E. of earth due to its rotation about its own axis.

{A} 60.

61.

62.

M e =6×10 24 kg Re=6400 Km

63.

{A} 6.2  1029 Joule {B} 2.6 10 29 Joule {C} 62 1029 Joule {D} 26 1029 Joule A cord is wound round the circumference of wheel of radius r. the axis of the wheel is horizontal and moment of inertia about it is I A weight mg is attached to the end of the cord and falls from the rest. After falling through the distance h. the angular velocity of the wheel will be…. {A}

64. 65.

66.

2 gh I  mr

1  2mgh  2 {B}  2  I  mr 

1  2mgh  2 {C}  2  I  2mr 

{D}

2gh

If rotational K.E. is 50% of translational K.E. then the body is ….. {A} Ring {B} solid cylinder {C} Hollow sphere {D} Solid sphere A meter stick of mass 400 gm is pivoted at one end and displaced through an angle 600 the increase in its P.E. is ______ J. {A} 2 {B} 3 {C} Zero {D} 1 Tow uniform rod of equal length but different masses are rigidly joined to form L shaped body which is then pivoted as shown. If in equilibrium the body is in the shown configuration ratio M/m will be……. {A} 2 {B} 3 {C}

2

{D}

3 120

67.

A light rod carries three equal masses A, B and C as shown in the figure the velocity of B in vertical position of rod if it is released from horizontal position as shown in the figure is ……. {A}

68.

2g

{B}

18g {C} 7

4g {D} 3

8g 3

A gramophone record of mass M and radius R is rotating with angular speed W. If two pieces of wax each of mass M are kept on it at a distance of R/2 from the centre on opposite side then the new angular velocity will be…..

 m M M m {B} {C} {D} 2 M m M m M A solid cylinder rolls down a smooth inclined plane 4.8m high without slipping what is its linear speed at the bottom of the plane, if it starts rolling from the top of the plane? (take g = 10 m/S2) {A} 4 m/S {B} 2 m/S {C} 10 m/S {D} 8 m/S The M.I of a disc of mass M and radius R about an axis passing {A}

69.

70.

MR 2 through the centre O and perpendicular to the plane of disc is . 2 If one quarter of the disc is removed the new moment of inertia of disc will be….. {A} 71.

MR 2 3

MR 2 4

{C}

2

 {A} 3

I1

I 2 is.

4 2 {B} 3

82 {C} 3

74.

{D}

3 MR 2 2

162 {D} 3

A molecule consist of two atoms each of mass 'm' and separated by a distance of 'd' If 'K' is the average rotational K.E. of the molecule at particular temperature then its angular frequency is….

2 k d m {B} d 2 k m {C} 2d m k {D} m k d 4 A car is moving with a constant speed the wheels of the car make 120 rotations per minute the breaks are applied and the car comes to rest in 8 sec how many rotation are completed by the wheels before the car is brought to rest. {A} 4 {B} 6 {C} 8 {D} 10 The angular momentum of a wheel changes from 2L to 5L in 3 seconds what is the magnitudes of torque acting on it? {A} L {B} L/2 {C} L/3 {D} L/5

{A} 73.

3 MR 2 8

The moment of inertia of a uniform rod about a perpendicular axis passing through one of its ends is I1. The same rod is bent in to a ring and its moment of inertia about a diameter is I2, Then

72.

{B}

121

75.

76.

77.

A uniform disc of mass 500kg and radius 2 m is rotating at the rate of 600 r.p.m. what is the torque required to rotate the disc in the opposite direction with the same angular speed in a time of 100 sec ? {A} 600  Nm {B} 500  Nm {C} 400  Nm {D} 300  Nm The moment of inertia of a meter scale of mass 0.6kg about an axis perpendicular to the scale and passing through 30 cm position on the scale is given by (Breath of scale is negligible). ________ {A} 0.104 kg m 2 {B} 0.208 kg m 2 {C} 0.074 kg m 2 {D} 0.148 kg m 2 How much constant force should be applied tangential to equator of the earth to stop its rotation in one day ? {A} 1.3  10 22 N

78.

79.

80.

81.

82.

83.

84.

{C} 1.3  10 23 N

{B} 8.26  10 28 N

{D} None of these

A constant torque of 1500 Nm turns a wheel of moment of inertia 300 kg m 2 about an axis passing through its centre the angular velocity of the wheel after 3 sec will be…….... rad/sec {A} 5 {B} 10 {C} 15 {D} 20 A disc of mass M and radius R is rolling with angular speed w on a horizontal plane, as shown in figure. The magnitude of angular momentum of the disc about the origin O is ______ 2 2 {A} 1 2 MR  {B} MR 2  {C} 3 2 MR  {D} 2 MR2 A mass m is moving with a constant velocity along the line parallel to the x-axis, away from the origin. Its angular momentum with respect to the origin {A} Zero {B} remains constant {C} goes on increasing {D} goes on decreasing A body is rolling down an incline plane. If the rotational K.E. of the body is 40% of its translational K.E. then the body is …. {A} ring {B} Cylinder {C} solid sphere {D} hollow sphere A spherical ball rolls on a table without slipping, then the fraction of its total energy associated with rotation is . {A} 2/5 {B} 3/5 {C} 2/7 {D} 3/7 A binary star consist of two stars A (2.2 Ms) and B(mass 11Ms) where Ms is the mass of sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B. about the centre of mass is _____. {A} 6 {B} ¼ {C} 12 {D} ½ A small object of uniform density rolls up a curved surface with initial

2

velocity 'u'. It reaches up to maximum height of 3v 4 g with respect

85.

to initial position then the object is ____. {A} ring {B} solid sphere {C} disc {D} hollow sphere A particle of mass m slides down on inclined plane and reaches the bottom with linear velocity V. If the same mass is in the form of ring and rolls without slipping down the same inclined plane. Its velocity will be______. {A} V

{B}

{C}

2V 122

V 2

{D} 2V

GRAPHICAL QUESTIONS. 86.

Moment of inertia of a sphere of mass M and radius R is I. keeping mass constant if graph is plotted between I and R then its form would be.

{A}

87.

{B}

{C}

{D}

{B}

{C}

{D}

The graphs between loge L and loge P is ____ where L is angular momentum and P is linear momentum

{A}

90.

{D}

The graphs between angular momentum L and angular velocity w will be.

{A}

89.

{C}

According to the theorem of parallel axis I  I cm  md 2 the graph between I  d will be

{A}

88.

{B}

{B}

{C}

{D}

Let Er is the rotational kinetic energy and L is angular momentum then the graph between LogeEr and log e L can be

{A}

{B}

{C}

123

{D}

MACHING COLUMN TYPE 91.

Match list I with list II and select the correct answer. List - I

List - II Moment of inertia

System (x) A ring about it axis

(1)

MR 2 2

(y) A uniform circular disc about it axis

( 2)

2 MR 2 5

(z) A solid sphere about any diameter

(3)

7 MR 2 5

(w) A solid sphere about any tangent

(4) MR 2 s (5)

9 MR 2 5

Select correct option

92.

Option?

X

Y

Z

W

{A}

2

1

3

4

{B}

4

3

2

5

{C}

1

5

4

3

{D}

4

1

2

3

Match the shape of graph with given pair of physical quantities. Physical Quantities {X} Moment of inertia  distance {Z}Angular momentum(L)  letargen parallel axis  angular geloaty (w) {Y} log e E r  log e L Shape of graph

{P}

{W} log e L  log e P

{Q}

{R}

Select Correct Option Option?

X

Y

Z

W

{A}

R

S

P

Q

{B}

Q

P

S

R

{C}

S

Q

R

P

{D}

P

R

Q

S 124

{S}

ASSERTION - REASONING TYPE 92.

93.

94.

95. 96.

97.

98.

99.

In the following questions statement - 1 (Assertion) is followed by statement - 2 (Reason). Each question has the following four choices out of which only one choice is correct. {A} Statement - 1 is correct (true), Statement - 2 is true and Statement- 2 is correct explanation for Statement - 1 {B} Statement -1 is true, statement -2 is true but statement-2 is not the correct explanation four statement -1. {C} Statement - 1 is true, statement-2 is false {D} Statement-2 is false, statement -2 is true Statement -1 — The angular momentum of a particle moving in a circular orbit with a constant speed remains conserved about any point on the circumference of the circle. Statement -2— If no net torque outs, the angular momentum of a system is conserved. Statement -1— A sphere and a cylinder slide without rolling from rest from the top of an inclined plane. They will reach the bottom with the same speed. Statement -2 — Bodies of all shapes, masses and sides slide down a plane with the same acceleration. Statement -1— Friction is necessary for a body to roll on surface Statement -2— Friction provides the necessary tangential force and torque. Statement -1— A body tied to a string is moved in a circle with a uniform speed. If the string suddenly breaks the angular momentum of the body becomes zero. Statement -2 — The torque on the body equals to the rate of change of angular momentum. Statement -1 — If there is no external torque on a body about its centre of mass, then the velocity of the centre of mass remains constant. Statement -2 — The Linear momentum of an isolated system remains constant. Statement -1 —Two cylinder one hollow and other solid (wood) with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow will reach the bottom of inclined plane first. Statement -2 — By the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline. Statement -1 —A thin uniform rod AB of mass M and length L is hinged at one end A to the horizontal floor initially it stands vertically. It is allowed to fall freely on the floor in the vertical plane, The angular velocity of the rod when its ends B strikes the floor

3g L

Statement -2 — The angular momentum of the rod about the hinge remains constant through out its fall to the floor. 100 .Statement -1 —If the cylinder rolling with angular speed- w. suddenly breaks up in to two equal halves of the same radius. The angular speed of each piece becomes 2w. Statement -2—If no external torque outs, the angular momentum of the system is conserved.

125

PASSAGE BASED QUESTIONS Passage - I A Solid sphere of mass M and radius R is released from rest at the top of a frictionless inclined plane of length 'd' and inclination ?. In case (a) it rolls down the plane without slipping and in case (b) it slides down the plane 101. The ratio of the acceleration of the sphere in case (a) to that in case (b) is [A] 1[B] 2/3 [C] 5/7 [D] 7/9 102. The ratio of the velocity of the spheres when it reaches the bottom of the plane in case (a) to that in case (b) is

2 3

[C]

5 7

[D]

[C]

2

[D]

7 3 103. The ratio of time taken by the sphere to reach the bottom in case (a) to that in case (b) as

[A]

2

[A] 1

[B]

[B]

3 2

7

5

Passage - II A uniform disc of mass M and radius R rolls without slipping down a plane inclined at an angle  with the horizontal. 104. The acceleration of the centre of mass of the disc is 2g sin  3 105. The frictional force on the disc is

[A] g sin 

[B]

[C]

g sin  3

Mg sin  2 Mg sin  [B] [C] Mg sin  3 3 106. The magnitude of torque acting on the disc is

[A]

[D]

2g cos  3

[D] None

2MgR sin  MgR sin  [D] 3 3 107. If the disc is replaced by a ring of the same mass M and the same radius R, the ratio of the frictional force on the ring to that on the disc will be

[A] MgR

[B] MgR sin 

[A] 3/2

[C]

[B] 2

[C] 2 Passage - III A solid cylinder of mass M and R is mounted on a frictionless horizontal axle so that it can freely rotate about this axis. A string of negligible mass is wrapped round the cylinder and a body of mass m is hung from the string as shown in figure the mass is released from rest then___

[D] 1

108. The acceleration with which the mass falls is [A] g

[B]

 mg   [C]  Mm

mg M

126

[D]

2Mng M  2m 

109. The tension in string is [A] mg

[B]

Mmg M  m

[C]

2Mmg M  2m 

[D]

Mg  M  2m 

110. The angular speed of cylinder is proportional to hn, where h is the height through which mass falls, Then the value of n is___ [A] zero [B] 1[C] ½ [D] 2 111. The moment of inertia of a uniform circular disc of mass M and radius R about any of its diameter is ¼ MR 2 , what is the moment of inertia of the disc about an axis passing through its centre and normal to the disc? 1 3 2 MR 2 [C] MR [D] 2MR 2 2 2 112. A solid cylinder of mass M and radius R rolls down an inclined plane of height h. The angular velocity of the cylinder when it reaches the bottom of the plane will be.

[A] MR 2

[A]

2 gh R

[B]

[B]

2 gh R 2

[C]

2 gh R 3

[D]

1 gh 2R

113. A cylinder of mass m and radius r is rotating about its axis with constant speed v Its kinetic energy is _____ [A] 2mv2

[B] mv2

[C]

1 2

mv2

[D] mv2

114. A circular disc of mass m and radius r is rolling on a smooth horizontal surface with a constant speed v. Its kinetic energy is _____ [A]

1 4

mv2

[B]

1 2

mv2

[C]

3 4

mv2

[D] mv2

115. A solid sphere is rotating about a diameter at an angular velocity w. if it cools so that its radius reduces to 1/n of its original value. Its angular velocity becomes_____ [A]  n [B]  n 2 [C] n [D] n2 116. In above question (115) If the original rotational K.E. of the sphere is K, Its new value will be_____ [A] K n 2 [B] K n 4 [C] n 2 K [D] n 4 K 117. A solid sphere is rotating about a diameter due to increase in room temperature, its volume increases by 5%, If no external torque acts. The angular speed of the sphere will. [A] increase by nearly 1/3 % [B] decrease by nearly 1/3 % [C] increase by nearly ½ % [D] decrease nearly by ½ % 118. A cylinder of mass M has length L that is

3 times its radius what is the ratio of its moment of inertia about its own axis and that about an axis passing through its centre and perpendicular to its axis? [A] 1

[B]

1 3

[C]

127

3

[D]

3 2

119. A uniform rod of length L is suspended from one end such that it is free to rotate about an axis passing through that end and perpendicular to the length, what maximum speed must be imparted to the lower end so that the rod completes one full revolution? [A]

2gL

[B] 2 gL

[C]

6gL

[D] 2 2gL

120. The height of a solid cylinder is four times that of its radius. It is kept vertically at time t=o on a belt which is moving in the horizontal direction with a velocity v = 2.45t 2 where v in m/s and t is in second. If the cylinder does not slip, it will topple over a time t = ____ [A] 1 second [B] 2 sec. [C] 3 sec. [D] 4 sec. 121. The moment of inertia of a thin rod of mass M and length L about an axis passing through the point at a distance L/4 from one of its ends and perpendicular to the rod is _____

7 ML2 ML2 ML2 ML2 [B] [C] [D] 48 12 9 3 122. A thin uniform rod AB of mass M and length L is hinged at one end A to the horizontal floor. Initially it stands vertically. It is allowed to fall freely on the floor in the vertical plane. The angular velocity of the rod when its end B strikes the floor is ____ [A]

[A]

g L

[B]

2g L

[C]

3g L

[D] 2

g L

123. A circular disc of radius R is free to oscillate about an axis passing through a point on its rim and perpendicular to its plane. The disc is turned through an angle of 60? and released. Its angular velocity when it reaches the equilibrium position will be__ [A]

g 3R

[B]

2g 3R

[C]

2g R

[D] 2

2g R

124. The moment of inertia of a hollow sphere of mass M and inner and outer radii R and 2R about the axis passing through its centre and perpendicular to its plane is 3 13 31 62 MR 2 MR 2 MR 2 MR 2 [B] [C] [D] 2 32 35 35 125. If a is aerial velocity of a planet of mass M its angular momentum is [A] M [B] 2 MA [C] A 2 M [D] AM 2 126. A wheel having moment of inertia 2 kg M 2 about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheels rotation in one minute will be..

[A]

  2  Nm Nm Nm Nm [B] [C] [D] 15 18 15 12 127. A wheel is rotating at 900 rpm about its axis. When power is cut off it comes to rest in 1 minute, the angular retardation in rad / sec is ___

[A]

    [B] [C] [D] 2 4 6 8 128. What is the moment of inertia of a solid sphere of density and radius R about its diameter?

[A]

[A]

105 5 R 176

[B]

176 5 R 105

[C] 128

105 2 R 176

[D]

176 2 R 105

129. A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In the first two second it rotate through an angle 1, in the next 2 sec. it rotates through an angle 2, find the ratio

2 = ____ 1

[A] 1 [B] 2 [C] 3 [D] 4 130. A gramophone record of mass M and radius R is rotating at an angular velocity w. A win of mass M is gently placed on the record at a distance R/2. from its centre. The new angular velocity of the system is [A]

2 wM 2M  m 

[B]

2 wM M  2m 

[C] 

[D]

wm M

KEYNOTE Q.No. 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 30

Ans C A B C D A C C B D B B A A A C B B A B A D B C B C A D B A

Q.No. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Ans A C A A A B C B C A B D B B C B D A C A C B C D C D A A A B

Q.No. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90

Ans A B B B D D D C D C C A C A C C A C C B C C A C C D C A B B

129

Q.No. 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120

Ans D B D A A D D D C A C C D B A D A D D C B C D C D C B A C A

Q.No. 121 122 123 124 125 126 127 128 129 130

Ans A C B D B A A B C A

Hints 1.

Let Rcm is at origin

M R cm  m1 r1  m 2 r2   O = m1 r1 + m 2 r2    m1 r1  m 2 r2 r1 m = 2 r2 m1

2.

m1 r1

– ve sign ignore as distance

^ ^  rA = 0 i + 6 J

Here

^ ^  rB = 6 i + 6 J ^ ^  rC = 6 i + 0 J ^ ^  rD = 0 i + 0 J

M = 1 + 1 + 2 + 2 = 6 gm M R cm  m A rA  m B rB  m C rC  m D rD

3.

The x and y co.ordinates of centre of mass are x=

m1x1 + m 2 x 2 + m3 x 3 m1 + m 2 + m3

1  (x1 +x2 +x3) =2 3

as m1 = m2 = m3

similtaraly for y = 2

 (x, y) = (2, 2)

4.

m1x1 = m2x2 and m1(x1 – d) = m2 (x2 – d’)  m1d = m 2 d '

5.

 d' =

m2

m1d m2

Here at origin  A (0, 0), B (x, 0), C (x, x), D (0, x) xcm =

(m  0) + (2mx) + (3mx) + (4m  0) m + 2m + 3m + 4m

y cm =

(m × 0) + (2m × 0) + (3mx) + (4mx ) m + 2m + 3m + 4m

130

r2

6.

Let mass per unit area of disc = m Mass of disc = M = R 2  m 2

R 2 m R Mass of removed disc = M ' =     m = 36 6

from figrue 00' = M  0 = M'  M ' x = M'

R 2

R + (M – M')x 2

R + Mx 2

R  M' x=    2  M – M'

7.

Let mass per unit area of Plote = m 2

 56  Mass of whole Plote = M =    m  2  2

 42  Mass of removed part = M1 =    m  2 

Mass of remaining Portion M2 = M – M1 C.M of whole disc R = O at origin C.M of removed Plote = r1 = 28 – 21 = 7cm C.M of remaining Portion r2 = ?

M  O = Mi ri + M2r2 8.

The Velocity of C.M. is given by Vcm =

9.

10.

m1v1 + m 2 v2 m1 + m 2

1 2 1 1 I  = I    = L 2 2 2 2E 2E'  L=  L' =  ' E=

Let m is the mass of unit area then mass of big disc = (2R)2 m = M Let m is the mass of unit area then mass of small disc = R2m = M1 = Mass of remaining Portion = M2 = M – M1 M2 =

3M 4

131

M 4

Let G be the C.M of remaining Portion M2(OG) = M1(OO’) 3M M (R) = R 4 4 1 3 The moment of inertia about AD = ? I = m1 (Perpendicular distance of m1 from AD)2 + m2 (Perpendicular distance of m2 from AD)2 + m3 (Perpendicular distance of m3 from AD)2 =

11.

2

9 9  0 + m 2    + m3    2 2

12.

2

a2  (m 2 + m3 ) 4 Let L is original length & K spring anstant then m (L + x1) w12 = kx1 & M (L + x2) w22 = kx2 Taking ratio 2

 L + x1   1  x1      = 2 x  L + x2   2 

13.

given x1 = 1cm, x2 = 5cm and 2 = 1

Inital angular momentum = Li = Ii = I x 0 = 0 angular mumentum after initial inpulse = 3kgm2s–1 angular mumentum after initial 4 sec = 3 + 3 = 6kgm2s–1 angular mumentum after initial 8 sec = 6 + 3 = 9kgm2s–1 angular mumentum after initial 28 sec = 24 kgm2s–1 angular mumentum after initial 30 sec = 24 kgm2s–1

MR 2 1 I = 24 here I = =  5  (0.3)2 = 0.225kgm 2 2 2 24 24 = = 106.7 rad s–1 I 0.225 2 IA = ma ra , IB = mb rb2  =

14.

r  I m  B = b   b IA ma  ra 

2

Let K is the mass of unit length of the wire then ma = (2ra )k and mb = (2rb )k m r  b = b ma ra

2

m  r  r  I  B =8=  b  b =  b IA  m a   ra   ra 

132

3



rb =2 ra

15.

Let M be the mass and R1 the initial radius of the earth 1 is the angular veloalty of the rotation of the earth, the duration T1 of the day is T1 =

2 2 and T2 = 1 2

According to law of conservation of angular momentum I11 = I12 ^

16.

17.

m1 = 1

m2 = 35.5    m r + m r 1 1 2 2 rcm = m1 + m 2

r1 = 0 r2 = 1.27 i

   m r + m r 2 2 rcm = 1 1 m1 + m 2

18.

Theory [B] The centre of mass of lucky remains uncharged.

19.

o

Go 72  1000/3600 = = 80rad/sec r 0.25

As 2 = w 2 – w o 2

 = O,  = 2n = 2  20 = 40 rad 20.

 = wot +

21.

I=

1 1 MR 2 = ( R 2 t  )R 2 2 2

IR 4

22.

1 2 t   = 100 rad 2

R  I  1 =  1 I2  R2 

As t   are same

4

I = MK2 = 160  K2 =

160 160 = = 16 m 10

23.

I ring MR 2 2 = = 1 I Disc 1 MR 2 2

24.

I A = ISolid =

 K=4

 2:1

2 MR 2 = 0.4MR 2 5 2  IA < IB MR 2 = 0.66MR 2 3 As Volume and Mass remain same

and I B = I hollow = 25.

Iring = MR12 ISolid =

2 MR 22 5

R2 <<< R1

133

26.

2 Volume of disc = V1 = R1  t =

Volume of Sphere = V2 = I1 = M.I of disc = I =

27.

 R13 6



4  R 32 3

R 13 4 = R 32 6 3

R13 = 8R 32

 R 1 = 2R 2

1 MR 12 2

2 MR 12 I 2 MR 2 = = I2 = M.I of sphere = 5 5 2 5 Mass of the ontire disc would be AM and its moment of inertia about the given axis would be 12 (AM)R 2 . For the given section the moment of inertia about the since axis be one qvarter

28.

of this is 12 (AM)R 2 . Mass per unit length of the wire =  Mass of L length M = L When it is lent in form of circularring

2r = L

 R=

L 2

Moment of inertia of ring about given axis = 29.

3 MR 2 2

1 1  M  1 M2 2 MR = M M.I of disc = = 2 2  t  2 t As their mass & thickness are some

 M M  R2 =   =  2 R t t  

1  Let IZ be the M.I of square plote about the axis passing through the centre and perpendicular to the plane of square, hence according to Perfendicular axis theorm. IZ = IAB + IA'B' Also IZ = ICD + IC'D' I

30.

As axis are symmetric IAB = IA'B' =

IZ 2

IZ 2 So we can say that IAB = IA'B' = ICD = IC'D' = I

And ICD = IC'D' =

31.

Let the radius of complete disc is a & that of small disc is b After small disc is cut from complete disc let the C.M. shift to O2 at distance x2 flem original centre O. The Position of new C.M. is givenly let 6 is mass percunitarea.

X cm

–6b 2  1 = 6a 2 – 6b 2

134

32.

Let the mass of an element of length dx of the rod located at a distance X away from left M dx . the x cordinate of the C.M. is given by.. L

and is

1

Total mass of rod =  Ax  dx = 0

x cm

33.

1 1 = xdm =  M  AL2     2 

1

 x (Ax dx) 0

1 1 1 mL2 2 2  By Conservation of Energy P.E. of rod = Rotational K .E. M g sind = I = 2 2 2 3

 =

= 34.

AL2 2

3g sin L

As here l = 2L

3g sin 2L

Angular momentum of Block w.r.t. O before collision with a on collision the block will rotate about the side 2 passing through O. Now its angular momentum = Iw Acc. to law of conservation of angular momentum

O = MV

a  Ma 2 Ma 2    = 3 v MV   = I   +  4v 2 2   6

here I is the M.I of block about the axis perpendicular to the plane passing through O. 35.

36.

Given system of two purtides will rotate about its centre of mass. L initial angular momentum = Mv   2 2 L Final angular momentum  2I = 2M    2 By law of conservation of angular momentum 2 L V MV   = 2M  L     = 2 L   2 Initial angular momentum of ring L = I = MR2 Final angular momentum of ring and particles = (MR2 + 2mR2) ' As No external forque so According to Law of conservation of angular momentum. MR2 = (MR2 + 2mR2) '

 ' =

wM (M + 2m) 135

37.

As it is head-on elastic collision between two idential balls there fore they will exchange their linear vecocity is A comes to rest and B starts moving with linear velocity V. As there is no friction any where, forque on both the spheres about their centre of mass is zero and their angular velocities remains unchanged. Therefore A =  and B = 0

38.

I = 0.3x2 + 0.7 (1.4 – x)2 For minimum work moment of inertia of the system should be minimum is

39. 40.

dI = 0 = 03  2x – 0.7  2 (1.4 – x) = 0 dx  x = 0.98 m [B] 0.98 m from mass of 0.3 kg Angular speed for both wheels are different but lincar speed for both same so VF = Vr M.I of complete disc about O point

I Total = 12 (gM) R 2 Radius of removed disc =

R 3

i.e. M  R 2

As M = R 2

Mass of removed portion =

gM M g 2

1 R MR 2 M.I of removed disc about it own axis = M   = 2 3 18 2

M.I of removed disc about O. I removed = Icm

MR 2 MR 2  2R  + Mx = + M =  18 2  3  2

Itotal = I removed disc + I remaining part 41.

E=

L2 = K given K  1 If L is constant when child stretches his arms the moment of inertia 2I I

of system get doubled so kinetic energy will becomes half i.e. 42.

Moment of inertia of sphere I =

2 MR 2 about its axis 5

L = I

L2 2I

43.

E=

44.

Rotational K.E. = According to

E  L2

1 2 I = 1500 2 w = wo + t

136

K 2

45.

 = 1800 rpm 

2  1800 = 60 rad/sec 60

P = 

I2 2I

KA =

J2 2IA

As K =

47.

As M is the mass of disc, the force is producing angular accoration in the disc, therefore

– 48.

4kx = –m(2 x )   = 3

and

KB =

J2 2IB

46.

4k 3m

For Ring 1 and 2 I1 and I2 about YY ' =

3 1 MR 2 and for Ring 3. I3 about YY ' = MR 2 2 2

M.I. of system I = I1 + I 2 + I3 49.

T2 2 r23 = T12 r13

50.

a=

51. 52.

g sin (1 + k 2 / R 2 )

 a1 =

5g sin 7

and

a2 =

2g sin 3

nh 2 Impulse = Change in momentum = F  t L=

 F=

L t

53.

1 M1R 12 I1 R 12 2 = = 2 1 I2 R2 M 2R 2 2 2

54.

Apply theorm of parallel axis

55.

Rotational K.E. =

56.

If  = 0

57.

M ass = Volume   = M = R 2t

1 2 I 2 I1w1 = I2w2

1 m1R 12 2 Here direction of Motion will be reversed when force F = 0 or 20 t – 5r2 = 0 or t = 4sec.

M.I of x is I x = 58.

If  is angular accellaration then forque  = I = F.r OR 10   = (20t – 5t 2 )  2 OR d d  = 4t – t 2 and w = dt Also dt = t

137

 d = .t  dt = (4t – t 2 )  t dt = (4t 2  t 3 ) Taking intigration

=

4t 3 t4 – If n rotations are completed in As then Putting t = 4 3 4

4  64 64 – 64 = 3 3 which is 3 < 3.4 < 6  n = 3.4  = 2n =

2

59. 60.

2

2 2 21 R R I = M   + M(2R) 2 + M    MR 2 5 2 5 2 5     According to the Parallel axis theorm M.I of disc about an axis passing through particle (3) 1 3 MR 2 + MR 2 = MR 2 2 2

and perpendicular to plane of disc is  61.

In falling through a height h. P.E. = K.E mgh =

1 2 1 2 1 2 1 1 2 2 mv + I  = mv +  mr   2 2 2 2 2 

According to V2 – v2 = 2ad taking v = 0 & d = h, V2 = 2ah so mgh = 62.

I=

3 m  2ah 4

2 2 MR 2 =  6  1024  (6.4  106 ) 2 = 9.83  1037kgm 2 5 5

Angularvelocity of earth 

2 T

K.E. of rotation of the earth  2gh 2 1 + kr2

1 2 I 2

=

V 2gh = 2 2 r + k2

2mgh mr 2 + mk 2

63.

We know v =

64.

kr = 50% kT

65.

Centre of mass of stick is at midpoint when it is displaced by 600 Its c.m. rises up to height h from figure h =

l l l – cos  = (1 – cos) 2 2 2

so increase in P.E. = mgh = mg  0.4  10 

l (1 – cos) 2

1 (1 – cos60o ) = 1 Joule 2

138

66.

Net forque about O should be zero

67.

1  1  Hence Mg  sin 60o  = Mg  sin 30o  2  2  Loss in PE = gain in agular kinetic energy

Mg

l l  2l  + mg   + mgl = Iw2 3 2 3

Here I = Moment of inertia about fixed point 2

2

14 l  2l   M   + m    ml 2 = ml 2 9 3 3

From fig. 2mgl = 68.

1  14 2  2  ml     = 2 9 

36g 2 36 g As V  r  l = 14l 3 14l

I = I'' 

 MR 2  R 2  MR 2 =  + 2m    ' 2  4   2

 M = (M + m) '

2gh k2 1+ 2 r

69.

V =

70.

New Mass

3 M 4

M.I of disc =

MR 2 2

1 3  2 3 [C] M  R = MR 2  2 4  8 For a rod of mass M and length L., the MI about a perpendicular axis passing through one

New MI = I ' = 71.

and is I1 =

ML2 3  R=

when it is bent to form a ring, then L = 2R

L  2

MR 2 ML2 ML2 = = The M.I of the ring about its diameter is I2 = 2 4 2  2 82 2

72.

89l 7

2

d d The M.I of the molecule = M   + m   2 2 2 2  d  md I = 2m   = 2  4 1 2 The Rotational K.E. of the moldule ( K ) = I 2 2K = I

139

73.

2  120 = 4 rad/sec 60 Now  = 0 + t o =

Total angle descrited in 8 second is  = wot +

1 2 t 2

dL 5L – 2L 3L = = =L dt 3 3

74.

=

75.

For disc I =

MR 2 500  4 = 2 2 so angular momentum L = I

angular speed =  =

2  600 60

final angular momentum in opposite direction = –1000  20

kgm 2 sec

So change in angular momentum = L = 2  1000  20 kgm 2 / sec

76.

dL 2  1000  20 = = 400 N.m. dt 100 By Parallel axis therom solved problem.

77.

1 = 2 rad/day and 2 = 0, t = 1 day

=

w 2 – w1 t Forque required to stop the earth = T = I = F.R. =

F=

78. 79.

80.

81.

I. R

d dt The disc have two types of motion translational and rotational so there will be two types of angularmotion there fore total angular momentum should be total of both  = I = I

L = LT + LR L = Momentum x perpenlicular distance between point of rotation and line of action = m.V.y all remain constant L = remaing constant R.K.E. =

40 T.K.E 100

1 2 4 1 1 I =  mv 2 = mv 2 2 10 2 5



1 V2 2 mk 2  2 = mv 2 2 r 5

140

82. 83.

1 2 1 2 1 2 1 I  + mv =  mr 2 2 + mr 2 2 2 2 2 5 2 Here C.M. wrt A.

Total energy E =

  11M5 5d    d= 6  11M5 + 2.2M5  2

d2  5d  LA = 2.2 M 5 w   = 55M 5 w 36  6  2

d2 d LB = 11 M 5    = 11M 5  36 6

84.

At the highest point the whole energy is conserted to P.E. of the object. P.E. = mgh,

3v 2 h= 4g

Total K.E. of body =

 PE = mg

3v 2 3mv 2 = 4g 4

1 2 1 2 1 2 1 IV 2 mv + Iw = mv + 2 2 2 2 r2

Now, K.E. = P.E. 85.

Here in first case

1 mg 2 = mvh 2

 v = 2gh

1

 2gh  2  In second case v ' =  2 1 + K  2 R 

As for the ring K = R 86.

87. 88. 89.

90.

v ' = gh

2 MR 2  I  R2 5 This relation shows that the graph between I and R will be paracola symatric to I axis I=

Graph should be paracola symmetric to I - axis, but it should not pass from origin because there is a constant value Icm plesent for d = 0 L = Iw L  w If I constant so graph between L and w will be straight line with constant stop and passing from origin.     log e L = log e P + log e r L=r  P If graph is drawn between loge L and loge P then it will be straight line which will not pass through origin at loge P = 0 loge L = loge r

Er =

L2 so Log Er = 2 log L – log (2I) 2I

So the graph log E r  log L will be straightline with constant stop and when Log L = 0. Log Er = – log (2I) 141

93. 94.

The correct choice is [D] since the centrifetal force is radial. Forque is zero so L = constant. The correct choice is [A] If a body slides down an indined plane its acaleration is a = g (sin  –  cos ) which depends only on g,  and  .

98.

a=

99.

Icm of hollow cylinder is loss, so it will have more accderation and will take less time to reach bottom. Loss in P.E. is equal to gain in rotational K.E. As the centre of mass of the rod falls through

g sin I 1 + cm 2 mR

the distance

L 2

L  2   ML2  2 Mg  I      = 2 2  3  100.

3g L

If I is the M.I. of the complete cylinder, The M.I of each Riece becomes the angular speed of each Piece becomes 2.

101.

In case (a) accelaration down the plane is a1 =

g sin 5 = g sin I   7 1 + 2  MR  

a1 5 = a2 7

So

In ase (b) a 2 = g sin 102.

using V2 = 2ad we can find. 2

v1 a = 1 2 a2 v2



v1 = v2

5 7

d1 1 2 = at , we find that d2 2

103.

From d =

104.

Mg sin – f = Ma

7 5

a 1 and  = fR MR 2 ,  = R 2 1 a 1 Hence fR = MR 2  = MRa 2 R 2

 = I . 

1 Ma 2

105.

f =

106.

  fR =

As I =

 f=

Mg sin 3

MgR sin 3

142

I since L = I, 2

107.

For a ring I = M R2  Correct choice is

108.

Mg – T = Ma, T is Tension in string forque on cylinder  = TR = I I  1 a 1  =  MR 2   2 = Ma R R 2 2 

=

a=

109. 110.

2mg (M + 2m)

[D]

MMg M + 2M From conservation of energy we have T=

[D]

1 1 MV 2 + I2 2 2

Mgh =



1 11 1  MR 2 2 +  MR 2  2  (2m + M) 2 R 2 2 22 4  1

 2 4mgh  =  2   (M + 2m) R  1

As   h 2 111.

According to perpendicular axis theorm Ix = I y =

1 MR 2 4

(Considering two perpendicular diameter in x & y directions)

So Ix  I y = Ic = 112.

For solid cylinder I = As Mgh =

113.

1 1 1 MR 2 + MR 2 = MR 2 4 4 2 1 MR 2 2

1 1 MV 2 + I2 2 2

V 1 K.E. in Rotation  1 Iw2 Here I  mr 2 and  = r 2 2 2

therefore K.E =

114.

1 1 1 v  MR 2    = mr 2 2 2 4 r

Rotational K.E = K.E =

1 2 1 2 I & Translational K.E = mv 2 2

1 1 Mv 2 + I2 2 2

143

115.

2 2 mr 5

A solid sphere M.I =

2

2 2 r As Iw = I11 = mr 2 1 = m   1 5 5 n

116.

=

1 n2

K=

1 2 1 I and K1 = I112 2 2

 1 = n 2 

2

117.

K1 =

1 2 r 1 2 1 2  m    1 =  mr 2  2  (n 2) 2 2 5 n 2 5 n

K1 =

1 I  2  n 2  kn 2 2

V=



dv dr  4  =3 log V = log   + 3 log r differentiating we have v r  3 

4 3 r 3

dr 1 dv 1 1 = =  0.5% = % r 3 v 3 6

No external forque so I = constant   r 2  = constant

2 2 mr  = constant 5

 2 log r + log  = log c

dr d d dr 1 1 + =0  = –2 = –2  % = – % r   dt 6 3 -ve sign for decrensing  2

118.

 R2 1 L2  2 I1 = MR , I 2 = M  +  2 12   4

119.

In one full revolution the incrase in P.E = MgL, where M is the mass of rod.

 MgL =

120.

1 2 1  ML2  2 I =   2 2 3 

The cylinder will topple when the forque mgr equals the forque  ma 2gr g = h 2 dV Now V = 2.45t2 and a = dt d a = [2.45(t) 2 ] = 4.gt dt

so ma

h = mgr 2

 a=

144

h 2

121.

Using I = I cm + Md2 Parallel axis theorm

L L L ML2 – = and Icm = 2 4 4 12

d=

ML2 ML2 7ML2  I= + = 12 16 48 122.

Loss in P.E = gain in Rotational K.E. centre of Mass of rod is at

P.E =

 123.

MgL 2

1 2 1  1 ML2  2 I =   2 22 3 

Gain in R.E =

MgL ML22 = 2 6

L . 2

3g L

 =

P.E. at 600 = Mgh (1 – cos) 1  Eqvillibrium at K.E =  I 2  2  For I According to Parallel axis theorm (as d = R) I = I cm + MR 2 =

Here 124.

1 3 MR 2 + MR 2 = MR 2 2 2

1 2 I = Mgh (1 – cos 60o ) 2

We can obtain hollow sphere as it solid sphere of radius R is removed from a solid sphere of 2R mass of hollow sphere M = M1 – M2 It  is density M1 =

4 4 (2R )3  and M 2 = R 3 3 3

28 2 R  3 Moment of inertia of hollow sphere M =

I=

125.

2 2 M1 (2R) 2 – M 2R 2 By substituting the values of M1 and M2 5 5

Areal Velocity A =

A=

R 2 R 2 = 2 2

2 R 2 and T =  T

 =

2A R2

L = I

145

126.

Here I = 2kgm2 n1 = 60rpm = 1rps T = ? n2 = 0 t = 1mn = 60 sec T = I. =

I (2 – 1 ) t

2 – 1 t

127.

Angular retardation  =

128.

For sphere I =

2 2 4 3  2 MR 2 =  R   R 5 5 3 

129.

From  = wt +

1 2 t 2

1 = 0 +

1  (2) 2 = 2 2

1 + 2 = 0 +

130.

1  (4) 2 = 8 2

Angular momentum of Recoul L = I

Where I =

1 MR 2 2

Let w' be the angular velocity after putting coin of mass m at distance R2 from centre the angular 2 momentum of system L' ( I + mr2) ' since T = 0 so L' = L  (I + mr ) '  I 

 ' =

1 MR 2  R I  2 r= but = = 2 2 2 2 1 MR + mr 2 I + mr  2mr  2 1 + 2  MR  

 2 M   ' =    2M + m 

146

Unit - 6 Gravitational

147

SUMMARY 1.

Gravitational force between two point masses is

F

2.

Gm1 m2

r2 Acceleration due to Gravity (I) (II)

GM  9.81ms 2 R2 At a height h from surface of earth

on the surface of earth =

g

g1 

h  1  R    (III)

2

 g (1 

2h ) if h << R R

At a depth d form the surfce of earth

d ) R g1 = g if d = R i. e. on the surface of earth (IV) Effect of rotation of earth at latitude  g1 = g – R2 cos2  - at the equator  = 0 g1 = g – R2 = minimum value - At the pole  = 900 g1 = g – R2 = maximum value - At the equator effect of rotation of earth is maximum and value of g is minimum. - At the pole effect of rotation of earth is zero and value of g is maximum. Field Strength  Gravitational field strength at a point in gravirtational field is defined as, g 1  g /(1 

3.



F = gravitational force per unit mass m  Due to point mass 

E

GM 1 (towards the mass) E  2 2 r r  Due to solid sphere E

GM r R3 At r = 0, E = 0 at the center

inside points Ei 

At r = R, E 

GM i.e. on the surface R2

out side points Eo =

GM 1 or E o  2 2 r r

148

At r   , E  0  Due to a sphericell shell inside points E=0

GM r2

outside points E0 =

GM R2 on the surface E – r graph is discontinuous  on the axis of a ring

just outside surface E 

Er 

GMr (R2  r 2 )

3

2

At r = 0, E = 0 i.e. at the center If r >> R, E 

4.

GM , i.e. ring behaves as a points mass r2

As r   E  0 Gravitational potential :(i) Gravitational potential at a point in a gravitational field is defined as the negative of work done in moving a unit mass from infinity to that point per unit mass, thus

Vp  

w  wp

m (ii) Due to point mass V 

Gm rm

v   as r  0 and v  0 as r   (iii) Due to solid sphere

 inside points Vi  –

GM (1.5 R2  0.5r 2 ) R

GM i.e. on the surface R V - r graph is parabola for inside points

as r = R V  

GM r Due to sphercal shell

 out side points v   (iv)

inside points Vi  

GM R

outside points Vi  

GM R

149

(v)

on the axis of a ring Vr  

GM R2  r 2

GM i.e. at center R Gravitational potential Energy (i) This is the work done by gravitational forces in arranging the system from infinite sepration in the present position (ii) Gravitational potential energy of two point massess is

at r = 0, V  

5.

Gm1m2 r (iii) To find the gravitational points energy of more than two points masses we have to make pairs of masses. Neighter of the pair should be repeated. For example in case of four point masses. U 

 m4 m3 m4 m2 m4 m1 m3 m2 m3 m1 m2 m1  U G        r23 r41 r32 r31 r21  r43  for n point masses total number of pairs will be

(iv)

n( n  1) 2 If a point mass m is plaled on the surface of earth the potential energy here is Uo

Uo  

GMm R

and potential energy at height h is

Uh  

GMm (R  h)

the difference in potential energy would be mgh U = Uh – Uo = 1 + h/r If h << R, U = mgh 6.

 Relation between field strength E and potential V (i) if V is a function of only one variable (Say r) then dV E   slope of U  r graph dr (ii) If V is funtion at three coordinates variable; e x, y, and z then   v v v  E    iˆ  ˆj  kˆ   x y z 

150

7.

Escape velocity (i) From the surface of earth

Ve  2 gR 

8.

2GM GM as g  2 R R

=11.2 km / sec (ii) Escape velocity does not depend upon the angle which particle is projected form the surface and the mass of body Motion of satellites

GM r

(i) orbital speed Vo  (ii) time period T 

2 r 3/ 2 T 2  r 3 GM

(iii) Kinetic energy K 

GMm 2r

(iv) Potential energy U   

GMm ... r

(v) Total Mechanical energy. E   9.

GMm r

Kepler’s laws - First law : Each planet moves in an elliptical orbit with the sun at one focus of ellipse - Second law : The radius vectors drawn form the sun to a planet, sweeps out equal area in equal time interval i.e. areal Velocity is constant. this law is derived from the law of conservation of angular momentum dA L  dt 2m = constant here L is the angular momentum and m is mass of planet

- Third law

-

-

T 2  r3 where r is semi-major axis of elliptical path The gravitational force acting between two bodies is always attractive. It is independent of medium between bodies. It holds good over a wide range of distance. It is an action and reaction pair. It is conservative force. It is a central force and obey inverse square law as F  1/ r 2 The value of G is never zero any where but the value of g is zero at the center of earth. the acceleration due to gravity is independent of mass, shape, size etc of falling body. the rate of decrease of the acceteration due to gravity with height is twice as compared to that with depth. It the rate of rotation of earth increases the value of acceleration due to gravity decreases at all points on the surface of earth except at poles. 151

-

If the radius of planet decreasees by n% keeping its mas unchanged, the accelreotion due to gravity on its surface increases by 2n%. If the mass of a planet increases by n% keeping its radius unchanged the acceleration due to gravity on its surface increases by n%. The value of g at a location gives the value of intensity of gravitational field at the location. The orbital velocity of a satellite is independent of mass of the satellite but depends upon the mass and radius of planet around which the rotation is taking place The value of orbital velocity for a satellite near the surface of earth is 7.92 kms–1. The direction of orbital velocity of satellite at an instant is along the tangent to the orbital path at that instant. The work done by a satellite in a complete orbit is zero. For a satellite orbiting close to the surface of earth (h << R), the time period of revolution 2  g / R  0.001237rad / sec T A geostationary satellite revolves around the earth from west to east. Its period of revoluton is one day i.e. 24 hours. The orbital velocity of geostationary satellite is 3.08 kms–1. Its height above the surface of earth is about 36000 km. The relative angular velocity of geostationary satellite w.r.t earth is zero. When a satelilte is orbiting in its orbit, no energy is required to keep it in its orbit. When the total energy of a satellite is negative, it will be moving in either a circular or an elliptical orbit. When the total energy of a satellite is zero, it will escape away from its orbit and its path becomes parabolic. When the height of satellites is increased its potential energy will increase and K.E. will decrease. When the velocity of satellite is increased, its total energy will increase and it will start orbiting in a circular path of larger radius. For a satellite orbiting in a circular orbit, the value of potenial energy is always greater than its K.E. If the velocity of a satellite orbiting the earth is increased by 41.4% or its K.E. is doubled, then it will escape away from the gravitational field of earth

is 2 R / g  84.6 minutes and angular velocity.   -

-

Escape velocity =.

2  orbital velocity If the gravitational force is inversely proportional to the nth power of distance r, then the orbital velocity of a satellite V0  r

-

n/ 2

and time period is T  r

( n  2) 2

When a body is projected horizontally with velocity v, from any height from the surface of earth, then the following possibilities are there. (i) If v < v0, the body fails to revolve around the earth and finally falls to the surface of earth. (ii) If v = v0, the body will revolve around the earth in circular orbit. (iii) If v < ve the body will revolve around the earth in elliptical orbit. (iv) If v = ve, the body will escape from the gravitational field of earth. (v) If v > ve the body will escape, following a hyperbolic path.

152

-

1 th of its present size without any change in mass, the duration n of the day will be nearly 24/n2 hours. Force function F(r) is related with potential energy function U(r) by a relaion

-

dU dr A given planet will have atmosphere if the root mean square velocity of molecules in its atmosphere

-

If earth suddently contracts to

F 

(i.e. Vrms = -

In the weightlessness state, the bodies donot have weight but they do possess inertia on account of their mass. the bodies floating inside the space craft may collide with each other and crash. If a body is released from a height aqual to n times the radius of earth, then its striking velocity on the surface of earth is

 -

3RT ) is smaller than escape velocity for that planet. M

2ngR n 1

If polar ice caps melt then moment of inertia, Angular velocity will decrease and period of rotaiton of earth increase. The, line joining the places on earth having same value of g are called isogams. Gravity meter and Etvos gravity balance are used to measure changes in accelaration due to gravity.

153

Newton’s Law of Gravitation For the answer of the following questions choose the correct alternative from among the given ones. 1.

2.

3. 4.

5.

6.

7.

8.

9.

Two identical solid copper spheres of radius R are placed in contact with each other. The gravitational force between them is proportional to (A) R2 (B) R-2 (C) R-4 (D) R4 The gravitational force Fg between two objects does not depend on (A) sum of the masses (B) product of masses (C) Gravitational constant (D) Distance between the masses The atmosphere is held to the earth by (A) clouds (B) Gravity (C) Winds (D) None of the above Two sphere of mass m1 and m2 are situated in air and the gravitational force between them is F. The space around the masses is now filled with liquid of specific gravity 3. The gravitational force will now be (A) F (B) 3F (C) F/3 (D) F/9. A satellite of the earth is revolveing in a circular orbit with a uniform speed v. If the gravitational force suddennly disappears, the satellite will (A) Contineue to move with velocity v along the original orbit. (B) Move with a Velocity v, tangentially to the original orbit. (C) Fall down with increasing velocity. (D) Ultimately come to rest somewhere on the original orbit. Correct form of gravitational law is    Gm1m 2 Gm1m 2 Gm1m 2 Gm1m 2  ˆ F = – F = – r F =– r (A) F = – (B) (C) (D) 2 2 3 r r r r3 Mass M is divided into two parts xM and (1 - x) M. For a given separation, the value of x for which the gravitational force between the two pieces becomes maximum is (A) 1 (B) 2 (D) 1/2 (D) 4/5 24 The earth (mass = 6 x 10 kg) revolves around the sun with angular velocity 2  10–7 rad/ sec in a circular orbit of radius 1.5 x 108 km. The force exerted by the sun on the earth is = ................... N (A) 18  1025 (b) zero (C) 27  1039 (D) 36  1021 Two particle of equal mass go round a circle of radius r. Under the action of their mutual gravitational force. The speed of each particle is =.................. (A) 

10.

11.

1 1 2r Gm

(B)  

Gm 2r

(C)  

1 Gm 2 r

(D) 

4Gm r

The distance of the moon and earth is D the mass of earth is 81 times the mass of moon. At what distance from the center of the earth, the gravitational force will be zero (A) D/2 (B) 12D/3 (C) 4D/3 (D) 9D/10 One can easily “ Weight the earth ” by calculating the mass of earth using the formula (in usual notation) (A)

g Re G

(B) g/G Re2

(C) 154

G 2 Re g

(D)

G 3 Re g

12.

Three equal masses of m kg each are plced the vertices of an equilateral triangle PQR and a mass of 2m kg is placed at the centroid 0 of the triangle which is at a distance of 2 m from each of vertices of triangle. The force in newton. acting on the mass 2m is = ............

13.

14.

15.

(A) 2 (B) 1 (C) 2 (D) zero Which of the follwing statement about the gravitational constant is true (A) It is a force (B) It has no unit (C) It has same value in all system of unit (D) It depends on the value of the masses. Two point masses A and B having masses in the ratio 4 : 3 are seprated by a distance of lm. When another point mass c of mass M is placed in between A and B the forces A and C is 1/3rd of the force between band C, Then the distance C form A is = ............... m (A) 23 (B) 1/3 (C) 1/4 (D) 2/7 The gravitational force between two point masses m1 ans m2 at separation r is given by mm F  G 1 2 2 The constant k ............... r (A) Depends on system of units only. (B) Depends on medium between masses only. (C) Depends on both (a) and (b) (D) is independent of both (a) and (b)

Acceleration Due to Gravity 16.

As we go from the equator to the poles, the value of g ............... (A ) Remains constant (B) Decreases (C) I ncreases (D) Decreases upro latitude of 45

17.

If R is the radius of the earth and g the acceleration due to gravity on the earth’s surface, the mean density of the earth is = ............... (A) 4 G / 3gR

18.

19. 20.

22.

(B) 3 R / 4 gG

(C) 3g / 4 RG

(D)  RG / 12 g

g=10ms-2.

The radius of the earth is 6400 km and In order that a body of 5 kg weights zero at the equator, the angular speed of the earth is = ............... rad/sec (A) 1/80 (B) 1/400 (C) 1/800 (D) 1/600 The time period of a simple pendulum on a freely moving artificial satellite is ............... sec (A) 0 (B) 2 (C) 3 (D) Infinite A spherical planet far out in space has mas Mo and diameter Do. A particle of m falling near the surface of this planet will experience an acceleration due to gravity which is equal to (A) GM 0 / Do 2

21.

o

(B) 4mGMo / Do 2

(C) 4GMo / Do 2

(D) GmMo / Do 2

A body weights 700 g wt on the surface of earth How much it weight on the surface of planet whose mass is 1/7 and radius is half that of the earth (A) 200g wt (B) 1400g wt (C) 50 g wt (D) 300g wt. 2 The value of g on he earth surface is 980 cm/sec . Its value at a height of 64 km from the earth surface is ............... cm5–2 (a) 960.40 (B) 984.90 (C) 982.45 (D) 977.55 155

23.

24.

25.

If earth rotates faster than its present speed the weight of an object will. (A) increases at the equator but remain unchanged of the poles. (B) Decreases at the equator but remain unchanged at poles. (C) Remain unchanged at the equator but decreases at poles. (D) Remain unchanged at the equator but increases at the poles. The moon’s radius is 1/4 that of earth and its mass is 1/80 times that of the earth. If g represents the acceleration due to gravity on the surface of earth, that on the surface of the moon is .............. (A) g/4 (B) g/5 (c) g/6 (D) g/8 The depth of at which the value of acceleration due to gravity becomes 1/n the time the value of at the surface is (R = radius of earth)

26.

27.

28. 29. 30.

31.

32.

33.

34.

( n  1) n

 n  (D) R    n 1 If the density of smalll planet is that of the same as that of the earth while the radius of the planet is 0.2 times that of lthe earth, the gravitational acceleration on the surface of the planet is ............... (A) 0.2g (B) 0.4g (c) 2g (D) 4g If mass of a body is M on the earth surface, than the mass of the same body on the moon surfae is (A) M/6 (B) 56 (C) M (D) None of these An object weights 72 N on the earth. Its weight at a height R/2 from earth is = .............. N (A) 32 (B) 56 (C) 72 (D) zero If the radius of earth is R then height ‘h’ at which value of ‘g’ becomes one - fourth is (A) R/4 (B) 3R/4 (C) R (D) R/8 If the mass of earth is 80 times of that of a planet and diameter is double that of planet and ‘g’ on the earth is 9.8 ms-2 , then the value of ‘g’ on that planet is = ............... ms-2 (A) 4.9 (B) 0.98 (c) 0.49 (D) 49 Assuming earth to be a sphere of a uniform density, what is value of gravitational acceleration in mine 100 km below the earth surface = ............... ms-2 (A) 9.66 (B) 7.64 (C) 5.00 (D) 3.1 Let g be the acceleration due to gravity at earth’s surface and k be the rotational K.E. of earth suppose the earth’s radius decreases by 2% keeping alt other quantities same then (A) g decreases by 2% and K decreases by 4 % (B) g decreases by 4% and K increases by 2% (C) g increases by 4% and K increases by 4% (D) g decreases by 4% and K increases by 4% A body weight 500 N on the surface of the earth. How much would it weight half way below the surface of earth (A) 125N (B) 1250N (C) 500N (D) 1000N The radii of two planets are respectively R1 and R2 and their densities are respectively 1 and 2 the ratio of the accelerations due to gravity at their surface is ...............

(A) R/n

(A) g1 : g 2 

(B) R

1  2 . 2 2 R1 R2

(C) g1 : g2  R1 2 : R21

(C) R/n2

(B) g1 : g2  R1 R2 : 12 (D) g1 : g2  R1 1 : R22 156

35.

36. 37.

38.

39.

40.

41.

42.

43.

44.

45.

46.

47.

At what height over the earth’s pole, the free fall acceleration decreases by one percent = .................. km (Re = 6400 km). (A) 32 (B) 80 (C) 1.253 (D) 64 Weight of a body is maximum at (A) moon (B) poles of earth (C) Equator of earth (D) Center of earth At what distance from the center of earth, the value of aceeleration due to gravity g will be half that of the surfaces (R = Radius of earth) (A) 2R (B) R (C) 1.414 R (D) o.414 R The acceleration due to gravity near the surface of a planet of radius R and density d is proportional to (A) d/R2 (B) dR2 (C) dR (D) d/R The acceleration due to gravity is g at a point distance r from the center of earth R. if r < R then (A) g  r (B) g  r2 (c) g  r-2 (D) g  r-1 The density of a newly discoverd planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of earth. If the radius of the earth is R, the radius of planet would be ..................... (A) 2R (B) 4R (C) 1/4 R (D) ....... Density of the earth is doubled keeping its radius constant then acceleration, due to gravity will be .................. ms-2(g = 9.8 ms2) (A) 19.6 (B) 9.8 (C) 4.9 (D) 2.45 Weight of body of mass m decreases by 1% when it is raised to height h above the earth’s surface. If the body is taken to a depth h in a mine. change in its weight is (A) 2% decreases (B) 0.5% decreases (C) 1% increases (D) 0.5% increases If density of earth increased 4 times and its radius becomes half of then out weight will be... (A) Four times it present calue (B) doubled (C) Remain same (D) halved A man can jump to a height of 1.5 m on a planet A what is the height ne may be able to jump on another planet whose density and radius are respectively one- quater and one- third that of planet A (A) 1.5 m (B) 15 m (C) 18 m (D) 28 m If the value of ‘g’ acceleration due to gravity, at earth surface fis 10ms–2. its value in ms–2 at the center of earth, which is assumed to be a sphere of Radius ‘R’ meter and uniform density is (A) 5 (B) 10/R (C) 10/2R (D) zero A research satellite of mass 200 kg. circles the earth in an orbit of avrage radius 3R/2 where R is radius of earth. Assuming the gravitational pull 10 N, the pull on the satellite will be =..........N (A) 880 (B) 889 (C) 890 (D) 892 Acceleration due to gravity on moon is 1/6 of the acceleration due to gravity on earth. If the e ratio of densities of earth e and moon m is   5 3 then radius of moon Rm in terms m of Re will be ...............

(A)

5 Re 18

(B)

1 Re 6

(C)

157

3 Re 16

(D)

1 2 3

Re

48.

49.

50.

The acceleration of a body due to the attraction of the earth (radius R) at a distance 2R from the surface of the earth is = ............... (g = acceleration due to gravity at the surface of earth) (A) g/9 (B) g/3 (C) g/4 (D) 9 The height at which the weight of a body becomes 1/16 th its weight on the surface of (radius R) is (A) 3R (B) 4R (C) 5R (D) 15R A spherical planet has a mass Mp and diameter Dp A particle of mass m falling freely near the surface of this planet will experience an acceleration due to gravity, equal to (A)

51.

4 GMp Dp 2

(B)

GMpm Dp 2

(C)

GMp Dp 2

(D)

4GMpm Dp 2

Assuming the earth to have a constant density, point out which of following curves show the variation acceleration due to gravity from center of earth to points far away from the surface of earth ...............

(D) None of these

Gravitational potential, Energy and Escape Velocity 52.

53.

54.

In a gravitational field, at a point where the gravitational potential is zero (A) The gravitational field is necessarily zero (B) The gravitational field is not necessarily zero (C) Nothing can be said definetely, about the gravitational field (D) None of these The mass of the earth is 6.00  1024 kg and that of the moon is 7.40  1022 kg. The constant of gravitation G = 6.67  10-11 Nm2 kg–2. The potential energy of the system is -7.79  1028 Joules. the mean distance between the earth and moon is = ............. meter. (A) 3.80  108 (B) 3.37  108 (C) 7.60  108 (D) 1.90  102 The masses and radii of earth and moon are M1, R1 and M2, R2 respectively. Their centres are d distance of apart. The minimum velocity with which a particle of mass m should be projected from a point midway between their centres so that it esacapes to infinity is............... (A) 2

G (M1  M 2 ) d

(B) 2

(C) 2

Gm ( M1  M 2 ) d

(D) 2

2G (M1  M 2 ) d Gm (M 1  M 2 ) d ( R1  R2 ) 158

55.

A rocket is launched with velocity 10 kms-1. If radius of earth is R then maximum height attained

56.

by it will be = .............. (A) 2R (B) 3R (C) 4R (D) 5R What is the intensity of gravitational field at the center of spherical shell Gm (B) g (C) zero (D) None of these r2 Escape velocity of a body of 1 kg. on a planet is 100 ms-1. Gravitational potential energy of the body at the planet is =............... J (A) -5000 (B) -1000 (C) -2400 (D) 5000 A body of mass m kg starts falling from a point 2R above the earth’s surface. Its K.E. when it has fallen to a point ‘R’ above the Earth’s surface = ..................... [R - Radius of Earth, M-mass of Earth G-Gravitational constant]

(A) 57.

58.

1 GMm 1 GMm 2 GMm 1 GMm (B) (C) (D) 2 R 6 R 3 R 3 R The Gravitational P.E. of a body of mas m at the earth’s surface is -mgRe. Its gravitational potential energy at a height Re from the earth’s surface will be = ............ here (Re is the radius of the earth)

(A) 59.

1 1 mg Re (D)  mg Re 2 2 A body is projected vertically upwards from the surface of a planet of radius R with a velocity equal to half the escape velocity for that planet. The maximum height attained by the body is .......... (A) R/3 (B) R/2 (C) R/4 (D) R/5 Energy required to move a body of mass m from an orbit of radius 2R to 3R is ................

(A) 60.

61.

–2 mgRe

(B) 2 mgRe

(C)

GMm G m G m G m (B) (C) (D) 2 2 12R 3R 8R 6R Radius of orbit of satellite of earth is R. Its K.E. is proportional to

(A) 62.

1 1 1 (B) (C) R (D) R R R 3 /2 A particle falls towards earth from infinity. It’s velocity reaching the earth would be............. (A)

63.

(A) infinity 64.

66.

(C) 2 gR

2gR

(D) zero

The escape velocity of a sphere of mass m from earth having mass M and Radius R is given by (A)

65.

(B)

2GM R

(B) 2

GM R

(C)

2GMm R

(D)

GM R

The escape velocity for a rocket from earth is 11.2 kms–1 value on a planet where acceleration due to gravity is double that on earth and diameter of the planet is twice that of earth will be = ........... kms–1 (A) 11.2 (B) 22.4 (C) 5.6 (D) 53.6 –1 The escape velocity from the earth is about 11 kms . The escape velocity from a planet having twice the radius and the same mean density as the earth is =............ kms–1. (A) 22 (B) 11 (C) 5.5 (C)15.5 159

67.

If g is the acceleration due to gravity at the earth’s surface and r is the radius of the earth, the escape velocity for the body to escape out of earth’s gravitational field is................. (A) gr

68. 69.

(B)

(B) 

GM r

(C) 

Ve 2

(B)

Ve 2

(C)

(B) 12800

2GM r

Ve 2 2

(D) 

(D)

(C) 3200

2GM R2

4GM r

Ve 4

(D) 1600

The escape velocity of a planet having mass 6 times and radius 2 times as that of earth is.......... 3 Ve

(B) 3 Ve

(C)

2 Ve

(D) 2Ve

There are two planets, the ratio of radius of two planets is k but the acceleration due to gravity of both planets are g what will be the ratio of their escape velocity. 1/ 2

 kg 

(B)  kg 

1/ 2

(C)

 kg 

2

(D)

 kg 

2

The escape velocity of a body on the surface of the earth is 11.2 km/sec. If the mass of the earth is increases to twice its present value and the radius of the earth becomes half, the escape velocity becomes = ............... kms–1 (A) 5.6

77.

(D)  

An artificial satellite is revolving round the earth in a circular orbit. its velocity is half the escape velocity. Its height from the earth surface is = ............... km

(A) 76.

(C)   2GMR

The escape velocity of a body from earth’s surface is Ve. The escape velocity of the same body from a height equal to 7 R from earth’s surface will be

(A) 75.

8 8 GR G  (B)   M 3 3

Two small and heavy sphere, each of mass M, are placed distance r apart on a horizontal surface the gravitational potential at a mid point on the line joining the center of spheres is

(A) 6400 74.

(D) m-1

The escape velocity of an object from the earth depends upon the mass of earth (M), its mean density (), its radius (R) and gravitational constant (G), thus the formula for escape veloctiy is

(A) 73.

(C) m0

(B) m

(A) zero 72.

(D) 1120 kms–1

The escape velocity of a particle of mass m varies as...................

(A)   R 71.

(D) r/g

The escape velocity of a projectile from the earth is approximately (A) 11.2 kms–1 (B) 112 kms–1 (C) 11.2 ms–1 (A) m2

70.

(C) g/r

2gr

(B) 11.2 (C) 22.4 (D) 494.8 1 1 Given mass of the moon is of the mass of the earth and corresponding radius is of 81 4 –1 the earth, If escape velocity on the earth surface is 11.2 kms the value of same on the surface of moon is = ....... kms–1. (A) 0.14 (B) 0.5 (C) 12.5 (D) 5 160

78.

3 particle each of mass m are kept at vertices of an equilateral triangle of side L. The gravitational field at center due to these particies is ................ (A) zero

79.

3GM L2

(C)

9Gm L2

12 Gm (D) 3 L2

Escape velocity on the surface of earth is 11.2 kms–1 Escape velocity from a planet whose masses the same as that of earth and radius 1/4 that of earth is = ....... kms–1 (A) 2.8

80.

(B)

(B) 15.6

(C) 22.4

(D) 44.8

The velocity with which a projectile must be fired so that it escapes earth’s gravitational does not depend on .................. (A) mass of earth (B) Mass of the projectile (C) Radius of the projectiles’s orbit (D) Gravitational constant

81.

The escape velocity for a body projected vertically upwards from the surface of earth is 11 kms-1. If the body is projected at an angle of 450 with the vertical, the escape velocity will be ............kms-1. (A)

82.

11 2

84.

(B) 2 Ve

(C) 4 Ve

(D) Ve/2

(A) 6.67  10-9 J (B) 6.67  10-10 J (C) 13.34  10-10 J (D) 3.33  10-10 J A particle of mass M is situated at the center of a spherical shell of same mass and radius a the magnitude of gravitational potential at a point situated at a / 2 distance from the center will be 4GM a

(B)

GM a

(C)

2GM a

(D)

3GM a

The mass and radius of the sun are 1.99 x 1030 kg and R = 6.96 x 108 m. The escape velocity of rocket from the sun is =.......... km/sec (A) 11.2

86.

(D) 11

A particle of mass 10g is kept on the surface of a uniform sphere of mass 100 kg and radiius 10 cm. Find the work to be done aginst the gravitational force between them to take the particle is away from the sphere (G = 6.67  10-11 SI unit)

(A) 85.

(C) 22

The acceleration due to gravity on a planet is same as that on earth and its radius is four times that of earth. What will be the value of escape velocity on that planet if it is Ve on the earth (A) Ve

83.

(B) 11 2

(B) 12.38

(C) 59.5

(D) 618

The mass of a space ship is 1000 kg. It is to be lauched from earth’s surface out into free space the value of g and R (radius of earth) are 10ms-2 and 6400 km respectively the required energy for this work will be = ................. J (A) 6.4  1011

(B) 6.4  108

(C) 6.4  109 161

(D) 6.4  1010

87.

88.

The diagram showing th variation of gravitational potential of earth with distance from the center of earth is

(A)

(B)

(C)

(D)

A shpere of mass M and Radius R2 has a concentric cavity of Radius R1as shown in figure. The force F exerted by the shpere on a particle of mass m located at a distance r from the center of shhere varies as (O  r   )

(A)

(B)

(C)

(D)

162

89.

90.

91.

Which one of the following graphs represents correctly the variation of the gravitational field with the distance (r) from the center of spherical shell of mass M and radius a

(A)

(B)

(C)

(D)

Which of the following graphs represents the motion of a planet moving about the sun.

(A)

(B)

(C)

(D)

The curves for P.E. (U) K.E. (Ek) of two particle system ae shown in figure . At what points systemwill be bound. (A) Only at point D Ek (B) Only at point A (C) At point D and A (D) At points A, B and C

163

Energy

The correct graph representing the variation of total energy (E) kinetic energy (K) and potential energy (U) of a satellite with its distance from the centre of earth is...........

Energy

92.

Energy

(B)

Energy

(A)

(C)

93.

94.

(D)

A shell of mass M and radius R has a point mass m placed at a distance r from its center. The gravitational potential energy U(r) –v will be

(A)

(B)

(C)

(D)

Motion of satellite If Ve and Vo are represent the escape velocity and orbital velocity of satelllite correspoinding to a circular orbit of radius r, then (A) Ve = Vo

(B) 2 Vo  Ve

(C) Ve  Vo / 2 (D)ve and vo are not related 164

95.

If r represents the radius of the orbit of a saltellite of mass m moving around a planet of mas M, the velocity of the satellite is given by (A)  2 

96.

gM r

99.

100.

101

102

103.

(C)  

GM r

(D)  2 

GM r

(B) V1 < V2

V1 V2 (D) r  r 1 2

(C) V1 > V2

A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the centere of earth in new orbit is two times of the earlier orbit. The time period in second orbit is .........hours. (a) 4.8

98.

GMm r

Two satellites of mass m1 and m2 (m1 > m2 ) are revolving round the earth in cirular orbits of r1 and r2 ( r1 > r2) respectively. Which of the following statement is true regarding their speeds V1 and V2 (A) V1 = V2

97.

(B)  2 

(B) 48 2

(C) 24

(D) 24 2

As astronaut orbiting the earth in a circular orbit 120 km above the surface of earth,gently drops a spoon out of space-ship. The spoon will (A) Fall vertically down to the earth (B) move towards the moon (C) Will move along with sapace - ship (D) Will move in an irregualr way then fall down to earth The period of a satellite in cirular orbit around a planet is independent of (A) the mass of the planet (B) the radius of the planet (C) mass of the satellite (D) all the three parameters (A), (B) and (C) Two satellites A and B go round a planet p in circular orbits having radii 4 R and R respectively if the speed of the satellite A is 3V, the speed if satellite B will be (A) 12V (B) 6V (C) 4/3 V (D) 3/2 V A small satellite is revolveing near earth’s surface. Its orbital velocity will be nearly = .......... kms–1. (A) 8 (B) 4 (C) 6 (D) 11.2 A satellite revolves around the earth in an elliptical orbit. Its speed (A) is the same at all points in the orbit (B) is greatest when it is closest to the earth (C) is greatest when it is farthest to the earth (D) goes on increasing or decresing continuously depending upon the mass of the satellite If the height of a satellite from the earth is negligible in comparison of the radius of the earth R, the orbital velocity of the stellite is ............ (A) gR

(B) gR/2

(C)

165

g/R

(D)

gR

104

105.

106.

107.

108.

109.

110.

111.

A satellite is moving around the earth with speed v in a circular orbit of radius r. If the orbit radius is decreasd by 1% its speed will (A) increase by 1% (B) increase by 0.5% (C) decrrease by 1% (C) Decrrease by 0.5% orbital velocity of an artifiicial satellite does not depend upon (A) mass of earth (B) mass of satellite (C) radius of earth (D) acceleration due to gravity orbital velocity of eath’s satellite near the surface is 7 kms–1. when the radius of orbit is 4 times that of earth’s radius, then orbital velocity in that orbit is =........kms-1 (A) 3.5 (B) 17 (C) 14 (D) 35 Two identical satellites are at R and 7R away from each surface, the wrong statement is (R - Radius of earth) (A) ratio of total energy will be 4 (B) ratio of kinetic energes will be 4 (C) ratio of potential energies will be 4 (D) ratio of total energy will be 4 but ratio of potential and kinetic energies will be 2 Which one of follwoing sttements regarding artificial satellite of earth is incorrect (A) The orbital velocity depends on the mass of the satellite (B) A minimum velocity of 8kms-1 is required by a satellite to orbit quite close to the earth. (C) The period of revolution is large if the radius of its orbit is large (D) The height of geostationary satellite is about 36000 km from earth The weight of an astronuat, in an artificial satellite revolving around the earth is (A) zero (B) Equal to that on the earth (C) more than that on earth (D) less than that on the earth The distance of a geo-stationary satellite from the center of the earth (Radius R= 6400km) is nearest to (A) 5R (B) 7R (C) 10R (D) 18R A geo-stationary satellite is orbiting the earth of a height of 6R above the surface of earth, R being the radius of earth. The time period of another satellite at a height of 2.5 R from the surface fo earth is = ...............hr (A) 6

112.

(B) 6 2

(C) 10

(D) 6 / 2

1 1 (and not as 2 )where R R R is separation between them, then a particle in circular orbit under such a force would have its orbital speed v proportional to

If the gravitational force between two objects were proportional to

1 1 (B) R0 (C) R1 (D) 2 R R A satellite moves around the earth in a circular orbit of radius r with speed v, If mass of the satellite is M , its total energy is

(A) 113.

(A) 

1 MV 2 2

(B)

1 MV 2 2

(C) 166

3 MV 2 2

(D) MV2

114.

A satellite with K.E. Ek is revolving round the earth in a circular orbit. How much more K.E. should be given to it so that it may just escape into outerspace ? (A) Ek

115.

116.

117.

118.

119.

120.

121.

(B) 2 Ek

(D) 3 Ek

Potential energy of a satellite having mass m and rotating at a height of 6.4 x 106 m from the surface of earth (A) -0.5 mg Re (B) -mg Re (C) -2mgRe (D) 4 mgRe When a satellite going round the earth in a circular obrit of radius r and speed v loses some of its energy, then r and v changes as (A) r and v bothe will increase (B) r and v both will decease (C) r will decrease and v will increase (D) r will increase and v will decrease The time period of a satellite of earth is 5 hours If the sepration between the earth and the satellite is increased to four times the previous value, the new time period will become ...... hours (A) 10 (B) 120 (C) 40 (D) 80 A person sitting in a chair in a satellite feels weightless because (A) the earth does not attract the objects in a satellite (B) the normal force by the chair on the person balances the earth’s attraction (C) the normal force is zero (D) the person in satellite is not accelerated Two satellites A and B go round a planet in cirular orbits having radii 4R and R respectively If the speed of satellite A is 3v, then speed of satellite B is (A) 3v/2 (B) 4v/2 (C) 6v (D) 12v A satellite moves in a circle around the earth, the radius of this circlr is equal to one half of the radius of the moon’s orbit the satellite completes one revolution in .........lunar month (A) 1/2 (B) 2/3 (C) 2–3/2 (D) 123/2 The additional K.E. to be provided to a satellite of mass m revolving around a planet of mass M, to transfer it from a circular orbit of radius R1 to another radius R2 (R2 > R1) is  1 1   2 2  R2   R1

(A) G M m 

 1 1    R2   R1

(C) 2 G M m  122.

1 Ek 2

(C)

 1 1    R2   R1

(B) G M m 

(D)

 1 1 1  GMm    2 R2   R1

Rockets are launched in eastward direction to take advantage of (A) the clear sky on eastem side (B) the thiner atmosphere on this side (C) earth’s rotation (D) earth’s tilt 167

123.

124.

A satellite of mass m is orbiting close to the surface of the earth (Radius R = 6400 km) has a K.E. K. The corresponding K.E. of satellite to escape from the earth’s gravitational field is (A) k (B) 2 k (C) mg R (D) m k A planet moving along an elliptical orbit is closest to the sun at a distance r 1 and farthest away at a distance of r2. If v1 and v2 are the liner velocities at these points respectively, then the v1 ratio v is ............ 2 r1 (A) r 2

125.

 r1  (B)    r2 

2

r2 (C) r 1

 r1  (D)    r2 

2

The time period T of the moon of planet Mars(Mm) is related to its orbital radius R as (G = Gravitational constant)

4π 2GR 2 (B) T  Mm

4 R 2 (A) T  GMm

2

2

2 R 2G (d) T2 = 4 Mm GR2 Mm A geostationary satellite is orbiting the earth at a height of 5 R above that of surface of the earth. R being the radius of the earth. The time period of another satellite in hours at a height of 2R from the surface of earth is ............ hr

(C) T 2  126.

127.

128.

129.

130.

(A) 5 (B) 10 (C) 6 2 (D) 6 / 2 The figure shows ellipticall orbit of a planet m about the sun s. the shaded area SCD is twice the saded area SAB. If t1 is the time for the planet to move from C and D and t2 is the time to move from A to B then

(A) t1 > t2 (B) t1 > 4t2 (C) t1 = 2t2 (D) t1 = t2 The period of a satellite in a circular orbit of radius R is T. the period of another satellite in a circular orbit of radius 4R is (A) 4T (B) T/4 (C) 8T (D) T/8 If the earth is at one- fourth of its present distance from the sun the duration of year will be (A) half the pesent Year (B) one-eight the present year (C) one-fourth the present year (D) one-sixth the present year The orbital speed of jupiter is (A) greater than the orbital speed of earth (B) less than the orbital speed of earth (C) equal to the orbital speed of earth (D) zero 168

131.

132.

133.

134.

Kepler’s second law regarding constancy of aerial velocity of a palnet is consequence of the law of conservation of (A) energy (B) angular Momentum (C) linear momentum (D) None of these The largest and shortest distance of the earth from the sun are r1 and r2 its distance from the sun when it is at the perpendicular to the major axis of the orbit drawn from the sun r1 r2 2r1 r2 r1  r2 r1  r2 (A) (B) r  r (C) r  r (D) 4 3 1 2 1 2 According to keplar, the period of revolution of a planet (T) and its mean distance from the sun (r) are related by the equation (A) T 3 r 3  cons tan t

(B) T 2 r 3  cons tan t

(C) Tr 3  cons tan t

(D) T 2 r  cons tan t

A satellite of mass m is circulating around the earth with constant angular velocity. If radius of the orbit is Ro and mass of earth M , the angular momentum about the center of earth is (A) m GMRo (C) m

135.

136.

137.

138.

139

(B) M GMRo

GM Ro

(D) M

GM Ro

The earth E moves in an elliptical orbit with the sun s at one of the foci as shown in figure. Its speed of motion will be maximum at a point ..........

(A) C (B) A (C) B (D) D The period of revolution of planet A around the sun is 8 times that of B. The distance of A from the sun is how many times greater than that of B from the sun. (A) 2 (B) 3 (C) 4 (D) 5 The earth revolves round the sun in one year. If distace between then becomes double the new period will be .......... years. (A) 0.5 (B) 2 2 (C) 4 (D) 8 The maximum and minimum distance of a comet from the sun are 8 x 1012m and 1.6 x 1012 m. If its velocity when nearest to the sun is 60 ms-1, What will be its velocity in ms-1 when it is farthest ? (A) 6 (B) 12 (C) 60 (D) 112 The period of moon’s rotation around th earth is nearly 29 days. If moon’s mass were 2 fold its present value and all other things remained unchanged the period of moons’s rotation would be nearly .....days (A) 29 2

(B) 29 / 2

(C) 29  2 169

(D) 29

140.

141. 142.

143.

144.

145.

If the velocity of planet is given by   G a M b R c then (A) a =1/3 b = 1/3 c = -1/3 (B) a =1/2 b = 1/2 c = -1/2 (C) a =1/2 b = -1/2 c = 1/2 (D) a =1/2 b = -1/2 c = -1/2 The radius of orbit of a planet is two times that of earth. The time period of planet is.........years. (A) 4.2 (B) 2.8 (C) 5.6 (D) 8.4 If r denotes the distance between the sun and the earth, then the angular momentum of the earth around the sun is proportional to (A) r3/2 (B) r (C) r1/1 (D) r2 What does not change in the field of central force (A) potential energy (B) Kinetic energy (C) linear momentum (D) Angular momentum A thin uniform annular disc (See figure) of mass M has outer radius 4R and inner radius 3R. The work required to take a unit mass from point P on its axix to infinity is.............. (A)

2GM (4 2  5) 7R

(B) 

2GM (4 2  5) 7R

(C)

GM 4R

(D)

2GM ( 2  1) 5R

Suppose the gravitational force varies inversely as the nth power of distance the time period of planet in circular orbit of radius R around the sun will be proportional to  n  1    2 

146.

147.

Rn

n 2   2 

(D) R 

(A) R (B) R (C) If the radius of the earth were to shrink by 1% its mass remaing the same, the accelration due to gravity on the earth’s surface would (A) decrease by 2% (B) remain Unchanged (C) increase by 2% (D) increases by 1% A body of mass m is taken from earth surface to the height h equal to radius of earth, the increase in potential energy will be (A) mg R

148.

 n 1    2 

1 (B) 2 mgR

(C) 2 mgR

1 (D) 4 mgR

An artificial satellite moving in a circular orbit around earth has a total (kinetic + potential lenergy) Eo, its potential energy is (A) -Eo (B) 1.5 Eo (C) 2 Eo (D) Eo

170

149.

Two bodies of masses m1 and m2 are initially at rest at infinite distance apart. They are then

allowed to move towards each other under mutual agravitational attraction Their relative velocity of apporach at sepration distance r between them is  2 G ( m1  m 2 )  (A)   r  1



1 2

1

 2 r (C)    2 G m1 m 2 

150.

1

 2 G ( m1  m 2 )  2 (B)   r 

(D)

 2 G m1 m 2  2 r  

A geostationary satellite orbits around the earth in a circular orbit of radus 3600 km the time period of a satellite orbiting a few hundred kilometers above the earth’s surface (Rearth = 6400 km) will approximately be = ...... h (A) 1/2 (B) 1 (C) 2 (D) 4

KEY NOTE 1.C 2.A 3. B 4. A 5. B 6. C 7. A 8. D 9. C 10. D 11. B 12. D 13. A 14. A 15. A 16. C 17. C 18. C 19. D 20. C 21. B 22. A 23. B 24. B

1.C 25. B 2 . A 25. B 26. A 3. B 27. C 26. A 4. A 28. A 27. C 5. B 29. C 6. C 28. A 30. C 7. A 29. C 31. A 8. D 32. C 9. C 30. C 33. B 10. D 31. A 34. D 11. B 32. C 35. A 12. D 26. B 33. B 13. A 27. C 14. A 34. D 38. C 15. A 35. A 39. A 16. C 40. D 17. C 26. B 41. A 18. C 27. C 42. B 19. D 38. C 43. B 20. C 44. C 39. A 21. B 45. D 40. D 22. A 46. B 23. B 41. A 47. A 24. B 48. A

42. B 43. B 44. C 45. D 46. B 47. A 48. A

49. C 50.C A 49. 51. C 50. A 52. A 51. 53.C A 54.A A 52. 55. 53. AC 56. C 54. 57.A A 58.CB 55. 59.CD 56. 60. A 57. 61.A D 58. 62.B A 63. 59. DB 64. A 60. A 65.B 61. 66.D A 67.A B 62. 68. C 63. 69.BC 64. 70.A A 71. 65.BD 72. C

66. A 67. B 68. C 69. C 70. A 71. D 72. C

73. A 74.73. A A 75. A 74. A 76. C 77.75. C A 78.76. A C 79.77. C C 80. B 81.78. D A 82.79. B C 83.80. B B 84. D 85.81. D D 86.82. D B 87.83. C B 88. B 89.84. D D 90.85. C D 91.86. D D 92. C 93.87. C C 94.88. B B 95.89. D D 96. B

90. C 91. D 92. C 93. C 94. B 95. D 96. B 171

97. B 121 D 98. C97. B122. C 99. C 98. C123 B 100. B 124. C 99. C 101. A 125. A 102. B100.126. B C 103. D 127. 101. A C 104. B 128. C B B 105. B102.129. 106. A103.130. D B 107. D 104.131. B B 108. A 132. C B B 109. A105.133. 110. B106.134. A A 111. D107.135. D B 112. B 136. C A B 113. A108.137. A A 114. A109.138. 115. A110.139. B D 116. C 140. B D C 117. D111.141. B C 118. C112.142. 119. C113.143. A D 120. C 144.A

114. A 115. A 116. C 117. D 118. C 119. C 120. C

145. A 146.DC 121 147. B 122. C 148. C 123 149.BB 150.CC 124.

125. A 126. C 127. C 128. C 129. B 130. B 131. B 132. C 133. B 134. A 135. B 136. C 137. B 138. A 139. D 140. B 141. C 142. C 143. D 144.A

145. A 146. C 147. B 148. C 149. B 150. C

HINT 4 G (  R 3.  )2 4 3   2  2 R4 2 4R 3

(1)

) (c) F  G (m)(m  2 (2 R)

(4) (5) (6)

(A) Gravitational force does not depend upon the medium (B) Due to inertia to direction (D) the force exerted by sun on the earth F = m2R  (6  1024 ) (2  107 ) 2 (1.5  1011 )  36  1021 N

(7)

F  xm ( x  1) m  m 2 x ( x  1) for maximum force 

dF 0 dx

dF  m2  2 m2 x  0  x  1/ 2 dx (C) cenripetal force provided by the gravitational force of attraction between two particites 

(9)



m 2 G ( m )( m )  r (2 r ) 2

 

Gm r

1 2

(D - X)

(10)

(D) For will be zero at the point of zero intensity x

11.

12.

m1 m1 

m2



81m 9 D D 10 81m  m

B mg 

GM e m where Me and Re is the mass and radius of the earth respectivel Re 2

 Me 

g Re 2 G

(D) Here FOA = FOB = FOC =

    F  FOA  FOB  FOC

G (m) (2m) r2

172

=

14.

= 0 (A) let a point mass C is placed at a distance of x m from the point mass A as show in figure mA here

x

m C

(1–x)

mB

ma 4  3 mb

G (m)(mA) .........(i ) x2 G (m)(mB) FBC  .........(ii) (1  x )2 FAC 

According to given problem FAC 

1 FBC 4

with the help of equ (i) and (ii)



G (m)(mA) 1 G (m)(mB )  x2 3 (1  x) 2



mA x2 4 x2 x2     4  mB 3(1  x) 2 3 3(1  x) 2 (1  x) 2

2

x  2  2x  x 1 x

 3x  2 17.

(c) g 

 x  23 m

GM 4 and M   R 3  2 R 3

G 4 3g .  R3     2 R 3 4 RG (c) for condition of weight lessness at equator g 

18.

 g  19.

R

10 1  rad sec 3 6400  10 800

(D) Time peripd of simple pendylym T  2 In artificial satellite g’ = 0  T   173

l g'

20.

GM GMo 4GMo (C) g  R 2  ( Do 2) 2  Do 2

21.

(B) we know that g  on the planet gp 

GM R2

GM 7 4 7g R2 4

Hence the weight on the planet = 700 

4 = 400 gm t 7

2

2

22.

g '  R   6400  2    g '  960.400 ms   g  ( R  h)   6400  64 

23.

g 1  g   2 R Cos 2  Rotation of the earth results in the decreased weight apparently.this decrease in weight is not felt at the poles as the angle of latitude is 90’

24.

GM (B) using g  R 2

25.

1 (B) g  g (1 

26.

4 (A) g  πGR 3



we get gm = g/5

d (n 1) )d  R R n

and g'  4/3πGR'

g' R'   0.2  g'  0.2g g R

27.

(C) mass does not vary from place to place

28.

 R  4  R  (a) g '  g    g   g  Rh  R  R2  9

2

2

4 W'  4/9 w  (72)  32 N 9 2

29.

 R    9 / 4 by solving h  R (c) g '  g  Rh

30.

 Mp   Re   1  2 9.8  0.49ms 2 (c) g p  g e     9.8   (2)  Me Rp 80 20     

31.

d 100    2 (A) g'  g  1    9.8  1    9.66 ms  R  6400 

2

174

32.

(C) g 

GM 2 and k = L 2I 2 R

1 1 k 2 2 and R R i.e. if radius of earth decreases by 2% then g and k both increase by 4% (B) Weight on surface of earth, mg = 500N and weight below the surface of earth at

If mass of the earth and its angular momentum ramains constant then g 

33.

d =

R mg  d  250N , mg1 = mg 1    mg 1  1/2   2 2  R

34.

4 (D) g  π GR 3

35.

(A) g α

g1 g2



R1 1 R 2 2

GM g α 1 or r α 1 r2 g r2

If g decreases by one percent then r should be increase by 1/2 % i.e. R 

1  6400  32 km 2  100

2

37.

1 R  R    R  h  2 R  h    (C) g '  g  2 Rh Rh

Hence distance form center = R + 0.414R = 1.414R 4 π GR  g α d R (  d given in the problem) 3

38.

(C) g 

39.

(A) insided the earth g' 

40.

(D) g 

4 π Gr  g ' r 3

Rp  g p 4  π  GR  Re  g e 3

 Rp 

41.

(A) g  

42.

(B) For height

Re R  2 2

Δg 24 100%  100 % g 4

For depth 43.

  e  1   (1)     2   p 

Δg d h 100%    1/2  0.5 g R R

(B) g  R

175





2  1 R  0.414 R

44.

(C) H 

V2 1 H g  Hα  B  A 2g g HA gB

Now g B  

gA as g  R 12

HB g A   12  HB  12 HA  12 1.5  18 m HA gB

46.

2  R   R  2   (B) g'  g  R  h   g  3 R   4/9.g(g  10 ms )    2

47.

4 g e e Re (A) g  3 πGR  g α R  g   . Rm m m



6 5 Re  . 1 3 Rm

 Rm 

5 Re 18

48.

 R   R  (A) g '  g    g   g /9  R  2R   R  2R 

49.

(C) g ' 

g (1  h / R) 2

g g  16 (1  h / R ) 2 2

 h 1    16  R 1

h 4 R

h 3 R

h  3R 50.

(A) Gravitational attraction fore on particle B

Fg 

GMpm (Dp / 2) 2

Acceleration of paritcle due to particle due to gravity a 

176

Fg 4GMp  m Dp 2

51.

(C) g α r it(r  r) and g α

52.

(A) I  

53.

(A) U  

1 (it r  R) r2

dv dx

GMm r

6.67 10 11  7 10 22  6 10 24 r

7.79 10 28 

r  3.8 108 m 54.

(A) V  

GM 1 GM 2  d /2 d /2

Now P.E  mv  

2GM (m1  m2 ) (m  mass of particie) d

K.E = P.E 

1 2 2GM m  (m1  m2 ) 2 d

  2 55.

(C) If the body is projected with velocity υ (υ  Ve) then height up to where it rises

h

57.

G(M1  M 2 ) d

R 2

Ve 1 V2



 11.2     10 

2

(A) Ve  2GM R

100  

58.

R

 4 R(approx)

P.E. U  

GMm  5000 J R

GM R

GM  5000 R

(B) P.E. U   Uinitial = –

GMm GMm  r Rh

GMm GMm and Ufinal = – 3R 2R

loss of P.E. = gain in K.E =

GMm GMm GMm – = 2R 3R 6R

177

59.

(D) ΔU  U 2  U1 

mgh mgRe mgRe   1  h/Re 1  Re 2 Re

 U 2  (  mg Re) 

60.

mg Re 2

(A) If body is projected with velocity υ (υ  Ve) then height up to which it will rise R

but u  ve/2  h 

2



 Ve    1  Ve/2 

61.

1 U 2   mgRe 2

R  R/3 4 1

(D) change in potential energy in displacing a body from r 1and r2 is given by 1 1  1  GMm  1  U  GMm     GMm    6R  2 R 3 R   r1 r2 

GMm 1  K.E. α 2R R

62.

(A) K.E =

63.

(B) this should be equal to escape velocity i.e. = 2gR

64.

(A) A Escape velocity does not depend on the mass of the projectile

65

(B) Ve 

Vp

g p Rp .  22  2 g e Re

 Ve  2Ve  2 11.2  22.4Kms 1

2GM R R

 G

66.

(A) Ve 

69.

since the planet is having double radius incomparision to earth therefore the escape velocity becomes twice i.e. 22 kms-1 (C) Because it does not depend on the mass of projetile

71.

(D)

8

3

 Ve α R if  constant

Gravitational potential of A at 0  

GM 2GM  r /2 r

of B at 0  

GM 2GM  r /2 r

Total potential at 0  

4GM r

178

72.

V1  V2

73.

1 where r is the position of body from the surface r

(C) Ve α

r2  r1

(A) V 

R  7R V  2 2  V2  1 R 2 2

GM 1 2GM  R h 2 R

4R  2(R  h)  h  R  6400 km

Vp  Ve

Mp Re 1   6   3 Ve  3 Ve Me Rp 2

74.

(A)

75.

(A)

76.

(C) Ve 

77.

(C)

78.

2GM 4 2 2GM 2 11.2 .    2.5 kms1 R R 9 R 9 (A) Due to three particles net intensity at the center

υ  2gR 

V1  V2

g1 R1 .  gk  (kg)1/2 g2 R2

2GM M  Ve α R R If M becomes double and R becomes half then escape velocity; becomes two times on earth Ve =

2GM  11.2 kms 1 R

on moon Vm 

    I  I A  IB  I C  0

because out of these three intensities ARE equal in magnitude and between each other is 120’

80.

1 1 if R becomes then Ve will be 2 times 4 R (D) Escape velocity does not depends upon the angle of projection

82.

(B)  

83.

Ve = 2Ve (B) potential energy of system of two mass

79.

(C)

Ve α

2gR 

g Rp Vp  p.  1 4  2 Ve g e Re

GMm  6.67 10 11 100 10 10 3 U   6.67 10 10 J 2 R 10 10 so, the amount of work done to take the particle up to infinte will be 6.67  10-10 J

179

84.

Vp = Vsphere + Vpartical 

(D)

85.

(D) Ve =

86.

(D)

2GM = R

GM GM 3GM   a a/2 a

2  6.67 1011  1.99  1030  618 kms1 6.98  108

m  GMm  GMm W  0-   gR 2  mgR  R R  R   1000  10  6400  103  64 109

 6.4  1010 J GM R

87.

(C)

Vin  

88.

(B)

F  0 When 0  r  R1

Vsurface 

GM R

Vout  

GM r

beccause intensity is zero inside Caviy Fincrease When R1  r  R 2 Fα

89.

(D)

1 when r  R 2 R

Intensity will zero inside the spherical shell I = upto r = a and I

1 when r > a r2

90. 91.

(C) kepler’s law T2 α r3 (D) System will be bound at points where total energy is negative. In the given curve at point A, B, and C the P.E. is more than K.E.

92.

(C)

93.

U-

GMm GMm GMm K and E   r 2r 2r

For satellite U, K, and E varies with r and also U and E remains negativw where K reamin always positive (C) Gravitational P.E. = m x gravitational potential U = mv so the graph of U will be same as that of V for a spherical shell

94.

(B) Ve  2gR

96.

(B) υ 

and V0  gR

 Ve 

GM it r1  r2 then V1  V2 r 180

2 V0

97.

(B) T α r

3

2

if r becomes double then time period will become (2)3/2 times so new time period

will be 24 x 2 2 hr i.e. = 48

2

98.

(C) The velocity of the spoon will be equal to the orbital velocity when dropped out of the space ship

100.

(B)

VA 3V  VB VA



104.

(B)

GM V  A  R VB

υ

υα

RB  RA

R 1  4R 2

 VB  6V

1 r

% increase in speed = =

1 % decrease in radius 2

1 (1%) = 0.5 % 2

105.

(B)

υ

GM r

106.

(A)

υα

1 it orbital radius becomes 4 times then orbital velocity will becomes halt r

107.

(D)

orbital rudius of satellites

U1   K1   

110. 111.

(B) (D)

r1 = R+R = 2R r2 = R+7R = 8R

GMm GMm and U 2   r1 r2

GMm 2r1

and K2  

GMm 2r2

U1 K 1 E 1   4 U2 K 2 E2

6R from the surface of earth and 7R from the center Distance of satellite from the center are 7R and 3.5R respectively 3/2

3/2

T2  r2   3.5R      T2  24   T1  r1   7R 

112.

(B

 6 24 hr

Gravitational force provides the required centripetal force for orbiting the satellite

mυ 2 K  R R

becuse F α

1 R

 υ α Ro 181

Total energy = -K.E. = 

1 2 2 m

113.

(A)

114.

(A) B.E. = -K.E. And it this amount of energy(Ek) given to satellite it will escape into outer space 1 GMm g Re 2 m GMm GMm P.E. = = – = – = – = – mg Re = 0.5 mg Re 2 Re  h 2 Re 2R e r

115.

(A)

116.

(C) B. E. =

117.

R  (D) T2  T1  2   R1 

118.

B (C) V  A

120.

time period of revolution of moon around the earth = 1 lunar month

GMm 1 if B.E. decreases the r also decreases and V increases as   r r 3/2

V

 r  Te  e  Tm  rm 

121

rA  rB

3 /2

 T1 (4)3/2  8T1  40 hr 4R 2 R

1   2

3 /2

 Te  2  2 /3 lunar month

GMm GMm (D)  2R  K.E.   2R 1 2

K .E 

GMm  1 1     2  R1 R2 

122.

Because Earth rotation from west to east direction

123.

(B) k 

124. 125.

GMm 2R (C) v1 r1 = v2r2 (angular momentum is constant)

(A) Time period T 

T 2 

3/ 2

4 2 R 3 GMm 2

126.

2R 2R  GMm GM R

3

T1 R1 (6R ) 3   8 (C) 2 3 (3R ) 3 T2 R2 2

 T2 

24  24  72 8

 T2  6 2

182

127.

(C)

ΔA L  ΔAαΔt Δt 2m

ΔASCD t1 2A    t1  2t 2 ΔASAS t 2 A



3

 R  T1   1  T2  R2 

2

 R      4R 

3

2

128.

(C)

129.

 T1   1  1 1 2 3  (B) since T α r  T    4   T  8 T    

130.

VJ re (B) orbital radius of Jupiter > orbital radius of Earth V  r As rj < re there fore Vj < Vee e j

131.

dA L (B) dt  2m = constant

132.

(C) The earth moves around the sun in elliptical path, so by using the properties of ellipse

2

 T2  8T1

3

r1  (1  e) a and r2 = (1-e) r2, a =

r1  r2 2

 r1r2  (1  e 2 ) a 2 where a= semi major axis b= semi minor axis e= eccentircity Now required distance = sem latysrectum = b2/9  a2 133. 134.

135.

(1  e 2 ) r1r2 2r r   12 a (r1  r2 )/2 r1  r2

(B) T2/ r3 = constant T2 r-3 = constant (A) Angular momentum = Mass  orbital velocity x Radius

 m

GM  R0 R0

 m

GMR 0

(B) speed at the earth will be maximum when its distance from the sun is minimum because m r = constant 3

136

(C)

TA  rA    TB  rB 

(B)

T2  r2    T1  r1 

2

3

137.

3

4   8   A  4B 

2

 (2)

3

2

2

 rA  4B  8

3

2

 4rB

 2 2  T2  2 2 years 183

138.

(A) By conservation of angular momentum (mr) = constant Vmin X rmax = Vmax X rmin

60 1.6 1012 60   12 ms1 12 8 10 5 (D) Time period does not depends upon the mass of satellite

Vmin  139.

1 1 1 GM  G 2 M 2R 2 R

140.

(B) υ 

141.

 R2   (B) T2  T1   R1 

3

142.

2

3

 1 2  2  2.8 yar

(C) Angular momentum of earth around the sun L = MEVo r

GMS r

 ME

2

.r  M E .GMS .r

ME = mass of earth MS = mass of Sun r = Distance between sun and the earth 143.

L α r (D) For central force toraqe is zero

  

144.

dL  0  L  Constant dt

(A) Wext U  U F  Gdm   0  (1)  x   G

M π  7R 2

2π r dr 16R 2  r 2



2 GM 7R2



GM 7R2



2 GM 7R2

 r2





2 GM 4 2R  5R 7R





2 GM 4 2 5 7R



π

rdr 16 R2  r2 d2 GM  2 a 7R

 16R

2

 

(2) 4R 3R

 184

145.

(A)

m 2 Rα

T α R (C) g 

147.

(B) ΔU 

α

1  T 2 α R n 1 Rn

n+1 2

GM R2

146.

148.

 4π 2  1  m  2  R Rn  T 

is mass ramains constant then  

mgh 1  mgR 2 1 h R (C) P.E. = 2. total energy = 2E0 GMM GMM and Eo = – r 2r (B) let velocities of these masses at r distance from each other be v1 and v2 respectively By conservation of momentum

Because we know U = – 149.

 m 1 V1  m 2 V2 ____________ 1 

m 1 V1  m 2 V2  0

By conservation of energy change in P.E. = change in K.E. Gm1m 2 1 1 2 2  m1V1  m 2 V2 r 2 2 2



2

m1V1 m 2 V2 2 Gm1m 2   _______ ii  m1 m2 r

on solving (i) and (ii)

1 

2Gm 2

2

r m1  m2 

and

V ap p  V1  V 2  3

150.

1 R2

T2  r2    (C) T1  r1 

2

υ2 

2Gm 1

2

r m1  m 2 

2G m 1  m 2  r 3

 6400  2  T2  24    2 hours  3600 

185

Assertion - Reason Type Questions

1.

2.

Direction (Read the following uestions and choose) (A) If both Assertion and Reason are true and the Reason is correct explanation of assertion (B) If both Assertion and Reason are true, but reason is not correcte explanation of the Assertion (C) If Assertion is true, but the Reason is false (D) If Assertion is faluse, but the Reason is true Assertion : The value of acc. due to gravity (g) does not depend upon mass of the body GM Reason : This follows from g  2 , where M is mass of planet (earth) and R is radius of R planet (earth) (a) A (b) B (c) C (d) D -2 Assertion : Unit of gravitational field intensity is N/kg or ms Reason : Gravitational field intensity

Force N kg.m/sec 2    ms  2 mass kg kg 3.

4.

5.

(a) A (b) B (c) C (d) D Assertion : The time period of a geostationary satellite is 24 hours Reason : Such a satellite must have the same time period as the time taken by the earth to complete one revolution about its axis (a) A (b) B (c) C (d) D Assertion : Even when orbit of a satellite is elliptical, its plane of rotaiion passes through the center of earth Reason : This is in accordance with the principle of conservation of angular momentum (a) A (b) B (c) C (d) D Assertion : The time Period of pendulum, on a satlellite orbiting the earth is infinity Reason : (a) A

6.

Time period of a pendulum is inversely proportional to (b) B

(c) C

g (d) D

Assertion : The escape velocity on the surface of a planet of the same mass but

1 times the 4

radius of earth is 5.6 kms-1

7.

8.

9.

Reason : (a) A Assertion : Reason : (a) A Assertion :

The escape velocity Ve = 2gR (b) B (c) C (d) D The comet does not obey kepler’s law of planetaty motion The comet does not have ellitical orbit (b) B (c) C (d) D The square of the period of revolution of a planet is proportional to the cube of its distance from the sun. Reason : Sun’s gravitational field is inversely proportional to the square of its distance from the planet (a) A (b) B (c) C (d) D Assertion : Space ship while entering the earth’s atmoshere is likely to catch fire 186

10

Reason : Temperature of upper atomosphere is very high (a) A (b) B (c) C (d) D Assertion : The earth is slowing down and as a result the moon is coming nearer to it Reason : The angular momentum of earth - moon system is not conserved (a) A (b) B (c) C (d) D

187

Assertion - Reason Type question : 1. A 2. A 3. A

Both the asseration and reason are true and the latter is correct expalnation of the former. Both the asseration and reason are true and the latter is correct expalnation of the former. As the satellite is to be stationary over a particular place, its time period of revolution = 24 hours= time peroid of revolution of earth about its axis.

4. A

T  2π  on a satellite, there is a weightless -ness, so g= 0 hence T   g

5. A

Thus both the asseration and reason are correct As no torque is acting on the planet, its angular momentum must stay constant in a magnitude as well as direction there for planet of rotation must pass throught the center of earth

6. D

Ve 

2 GM R

Ve' = Ve

R

i.e.

R

Ve α

1 R

 2 Ve  2  11.2  22.4 kms1 4

thus assertion is wrong but Reason is correct 7. A Both the asseration and reason are correct and the Reason is correct explanation of Asseration 8. A A Both the asseration and reason are correct and the Reason is correct explanation of Asseration 9. C Here Asseration is correct but Reason is wrong because the space ship while entering the earth’s atmosphere may catch fire due to atmoshperic air friction 10. D Here Both the asseration and reason are wrong because the angular momentum of earth moon system is conserved in the absence of extermal touque.

188

Comperehensions Type Questions (1)

If a smooth tunnel is dug across a diameter of earth and a particle is related from the surface of earth, the particle oscillates simple harmonically along it. (1) Time period of the particle is not equal to 3 2π R 2 GM (C) 84.0 min (D) Nome of these (2) Maximum speed of the these

(A) 2π R g

(B)

2 GM R

(B)

(A) 2.

3.

GM R

(C)

3 GM 2R

(D)

GM 2R

When a paricle is projected from the surface of earth, it mechanincal energy and angular momentum about center of earth at all time is constant (i) A particle of mass m is projected from the surface of earth with velocity vo at angle  with horizontal suppose h be the maximum height of particle from surface of earth and v its speed at that point them v is (A) v0coso (B) >v0coso (C)
Vo 2 sin 2θ 2g

(C) 

Vo 2 sin 2θ 2g

(B) 

Vo 2 sin 2θ 2g

Vo 2 sin 2θ (D) can be greater than or less than 2g

A solid sphere of mass M and radius R is surrounding by a spherical shell of same mass M and raius 2R as shown. A small particle of mass m is relased from rest from a height (h <
(A)

2 hR 2 GM

(C)

(B)

2hR 2 GM

(D) None of these

(ii) what time will it take to move from A to B ? (A) 

(B) 

R2

(C) 

GMh R2

R2 GMh

(D) None of these

GMh

189

(iii) with what apporximate speed will it colide at B ?

4.

5.

(A)

2 GM R

(C)

GM 2R

(B)

3 GM 2R

(D)

GM R

A planet is revolving round the sun in elliptical orbit. Velocity at perigee position (nearest) is v1| and at apogee position (farthest) is v2 Both these velocities are perpendicular to the joinging center of sun and planet r is the minimum distance and r2 the maximum distance. (i) when the planet is at perigee position, it wants to revolve in a circular orbit by itself. For this value of G (A) Should increase (B) Should decrease (C) data is in sufficeint (D) will not depend on the value of G (ii) At apogee position suppose speed of planer is slightly decreased from v2, then what will happen to minimum distance r1 in the subsequent motion (A) r1 and r2 bothe will dicreases (B) r1 and r2 bothe will increases (C) r2 will remain as it is while r1 will increase (D) r2 will remain as it is while r1 will decrease Garvitational potential at any point inside a spherical shall is uniform GM and is given by  where M is the mass of shell and R its R 3GM radius. At the center solid sphere, potential is  2R (i)

There is a concentric hole of radius R in a solid sphere of radius 2R Mass of the remaining portion is M What is the gravitation potential at center? (A)

 3GM 7R

(B)

 5GM 7R

(C)

 7GM 14

(D)

 9GM 14 R

190

Solution (1)

(i) T  2π

R g

putting g 

GM we have R2

T

3 2π R2 Gm

(ii)maximum speed is at centre from conservation mechanicall energy(from cerface to center) increase in K.E. = decrease in P.E. 1 mn 2  U i  U f  m(Vi - Vf ) 2

υ  2(vi  vf )  GM 3 GM   2    12   R (2)

(i)

GM R

From conservation of angular momentum at A and B

mVocosθ  m  R  h   R      V0 cosθ  R h 

   Vo cosθ

(ii)

From consercation of mechanical energy, Decrease in K.E. = increase in P.E.



1 2 m V0  V 2 2



2

 V0  V 2 





2 gh 1 h R

2

mgh 1 h R but here V  V0cos θ

2

2

 V0  V 2  V0 sin 2θ so let V0  V 2  x here x  V0 x

h 

2

sin 2θ

2gh 1 h R x 2g  x

R

2 x  O sin 2  i.e. h > i.e. h > 2g 2g

191

(3)

(i)

Acceleration due to gravity near the surface of shell can be assumed to be uiform

g

G(2M) GM  (2R) 2 2R 2

From h 

t =

(ii)

2h hR 2 2 g GM

GMh  GM  VA  2 gh  2  2  h  R2  2R  From A to B field due to shell is zero, but field due to sphere is not-zero hence tAB

(iii)

1 2 gt 2

R R2   VA GMh

KA=0 polential between A and B due to shell is From energy conservation KA + UA = KB + UB KB = (UA = UB) = m(VA – VB) 1 mVB2 = m(VA – VB) 2

 VB  (4)

2(VA  VB ) =

GM R

At perigee position v1 > v0 where v0 is the orbital velocity for circular motion

V0 

GM  G r

so value of G should incease , so that v0 will increase for this position and which will become equal to v1 (5)

(ii) path will becomemore elliptical , keeping r2 constant and r1 to decrease Density of given material 

M 3M  3 3 4 28πR 3 3 π[(2R)  R ]

Vwhole = Vhole + Vremaing Vremaing = Vwhole – Vhole

3  GM1 GM2     2  2R R  192

4 8 3 here M1  π (2R)  M 3 7 4 M M 2  ( ) R 3  3 7

Vremaining  

9GM 14R

Match the Column (1)

(2)

on the surface of earth acceleration due to gravity is g and gravitational potential is V match the following Table - 1 Table -2 (A)

At height h = R value of g

(P)

decreases by a factor

1 4

(B)

At height h = R/2 value of g

(Q)

decreases by a factor

1 2

(C) (D)

At height h = R value of v At depth h = R/2 value of V

(R) increases by a factor 11/8 (S) increases by a factor 2 (T) None Density of planet is two times the density of earth Radius of this planet is half (As compared to earth) Match the following Table-1 Table-2 (A) Acceleration due to gravity on (P) Half this planet’s surface (B) Gravitational potential (Q) same

on the surface (C)

(3)

Gravitational potential (R) Two times at centre (D) Gravitational field strength (S) four times at centre let V and E denote the gravitational potential and gravitational field at a point. Then the match the follwing Table-1 Table-2 (A) E = 0, V = 0 (P) At center of Spherical shell (B) E  0, V= o (Q) At centre of solid sphere (C) V  0, E -0 (R) At centre of circular ring (D) V  0, E  0 (S) At centre of two point masses of equal magnitude (T) None 193

(4)

(5)

Match the following Table-1 (A) time period of an earth satellite in circular orbit (B) Orbital velocity of satellite (C) Mechnical energy of stellite Match the following (for a satellite in circular orbit) Table-1 Table-2 (A)

kinetic energy

(P)

GMm 2r

(B)

potential energy

(Q)

GM r

(C)

Total energy

(R)



(D)

orbital velocity

(S)

GMm 2r

GMm r

solution : (1) A-p, B-Q, C-s D-t (2) A-Q, B-p, C-p D-o (3) A-t, B-t, C-p,q,r,s, D-t (4) A-q, B-t, C-r, D-s (5) A-s, s-B, c-p, D-q

• • •

194

Table -2 (P) Independent of mas of satellite (Q) independent of radius of orbit (R) independent of mass of earth (S) none

Unit - 7 Proporties of Liquid a Solid

195

SUMMARY Surface tension of a liquid is measured by the force acting per unit length on either side of an imaginary line drawn on the free surface of liquid, the direction of this force being perpendicular to the line and tangential to the free surface of liquid. So if F is the force acting on one side of imaginary line of length l. then T = Fl. 1. 2.

It depends only on the nature of liquid and is independent of the area of surface or length of line considered. It is a scalar as it has a unique direction which is not to be specified.

3.

Dimension M1 L0 T 2

4.

Unit :

5.

It is a molecular phenomenon and its root cause is the electromagnetic forces.

N SI  m

Force Due to Surface Tension If a body of weight W is placed on the liquid surface, whose surface tension is T, and if F is the minimum force required to pull it away from the water then value of F for different bodies can be calculated by the following table. Body Force 1.

Needle (length l)

F  2 T  w

2.

Hollow disc (inner radius r1 Outer radius r2)

F  2 r1  r2  T  w

3.

Thin ring (radius r)

F  4 rT  w

4.

Circular plate (or disc) (radius r)

F  2 rT  w

5.

Square frame (side l)

F  8T  w

6.

Square plate

F  4T  w

1.

Examples of Surface Tension : When mercury is split on a clean glass plate, it forms globules. Tiny globules are spherical on the account of surface tension because force of gravity is negligible. The bigger globules because force of gravity is negiligible. The bigger globules get plattened from the middle but have round shape near the edges.

196

2.

When a molten metal is poured into water from a suitable height, the falling stream of metal breaks up and the detached portion of the liquid in small quantity acquire the spherical shape.

3.

Rain drops are spherical in shape because each drop tends to acquire minimum surface area due to surface tension, and for a given volume, the surface area of sphere is minimum. Oil drop spreads on cold water. Whereas it may remain as a drop on hot water. This is due to the fact that the surface tension of oil is less than that of cold water and is more than that of hot water. If a small irregular piece of camphor is floated on the surface of pure water, it does not remain steady but dances about on the surface. This is because, irregular shaped camphor dissolves unequally and decreases the surface tension of the water locally. The unbalanced forces make it to move haphazardly in different directions. Take a frame of wire and dip it in soap solution and take it out, a soap film will be formed in the frame. Place a loop of wet thread gently on the film. It will remain in the form, we place it on the film according to figure. Now, piercing the film with a pin at any point inside the loop, It immediately takes the circular from as shown in figure. When a greased iron needle is placed genetly on the surface of water at rest, so that it does not prick the water surface, the needle floats on the surface of water despite it being heavier because the weight of needle is balanced by the vertical components of the forces of surface tension. If the water surface is pricked by one end of the needle, the needle sinks down. Hair of shaving brush / painting brush when dipped in water spread out, but as soon as it is taken out, its hair stick together.

4. 5.

6.

7.

8.

1.

Factors affecting surface tension Temperature : The surface tension of liquid decreases with rise of temperature. The surface tension of liquid is zero at its boiling point and it vanishes at critical temperature. At critical temperature, intermolecular forces for liquid and gases becomes equal and liquid can expand without any restriction. For small temperature differences, the variation in surface tenstion with temperature is linear and is given by the relation. 197

Tt  T0  1  t  Where Tt and TO are the surface tensions at tOc and 0OC respectively and  is the temperature coefficient of surface tension. Examples : (i) Hot soup tastes better than the cold soup. (ii)

Machinery parts get jammed in winter.

2.

Impurities : The presence of impurities either on the liquid surface or dissolved in it, considerably affect the surface tension, depending upon the degree of contamination. A highly soluble substance like sodium chloride when dissolved in water, increases the surface tension of water. But the sparingly soluble substances like phenol when dissolved in water, decreases the surface tension of water. Applications of Surface Tension

1.

2.

The oil and grease spots on cloths cannot be removed be pure water. On the other hand, when detergents (like soap) are added in water, the surface tension of water decreases. As a result of this, wetting power of soap solution increases. Also the force of adhesion between soap solution and oil or grease on the clothes increases. Thus, Oil, grease and dirt particles get mixed with soap solution easily. Hence, clothes are washed easily. Surface tension of all lubricating oils and paints is kept low so that they spread over a large area.

3.

A rough sea can be calmed by pouring oil on its surface.

4.

In soldering, addition of 'flux' reduces the surface tension of molten tin, hence, it spreads. Molecular Theory of Surface Tension The maximum distance upto which the force of attraction between two molecules is appriciable is called molecular range  109 m . A sphere with a molecule as centre and radius equal to molecular range is called the sphere of influence. The liquid enclosed between free surface (PQ) of the liquid and an imaginary plane (RS) at a distance r (equal to molecular range) from the free surface of the liquid form a liquid film. To understand the concept of tension acting on the free surface of a liquid, let us consider four liquid molecules like A, B, C and D. Their sphere of influence are shown in the figure. Molecule A is well within the liquid, so it is attracted equally in all directions. Hence the net force on this molecule is zero and it moves freely inside the liquid. Molecule B is little below the free surface of the liquid and it is also attracted equally in all directions. Hence the resultant force acting on it is also zero. Molecule C is just below the upper surface of the liquid film and the part of its sphere of influence is outside the free liquid surface. So the number of molecules in the upper half (attracting the molecules upward) is less than the number of molecule in the lower half (attracting the molecule downward). Thus the molecule C experiences a net downward force. Molecule D is just on the free surface of the liquid. The upper half of the sphere of influence has no liquid molecule. Hence the molecule D experiences a maximum downward force.



1. 2 3.

4.

198



Thus all molecules lying on surface film experiences a net downward force. Therefore, free surface of the liquid behaves like a stretched membrane. Surface Energy : The potential energy of surface molecules per unit area of the surface is called surface energy. Unit :

J Dimension : ML2 m2

Suppose that the sliding wire LM is moved through a small distance x, so as to take me position . L M  . In this process, area of the film increases by 2  x (on the two sides) and to do so, the work done is given by,

W  fx  T  2  x  T  2x   T  A T 

1.

W A

i.e. Surface tension may be defined as the amount of work done in increasing the area of the liquid surface by unity against the force of surface tension at constant temperature. Work done in Blowing a liquid drop or soap-bubble If the intial radius of liquid drop is r1 and the final radius of liquid drop is r2 then, W  T  increment in Area



 T  4 r22  r12 2.



In case of soap-bubble,





W  T  8  r22  r12 (Bubble has two free surfaces) Splitting of Bigger Drop : When a drop of radius R splits into n smaller drops, (each of radius r) then surface area of liquid increases. Hence the work is to be done against surface tension. Since the volume of liquid remains constant therefore 4 4 R 3  n  r 3 3 3

 R 3  nr 3



Work done  T  A  T  4nr 2  4R 2



Formation of Bigger drop : Amount of surface energy released = Initial surface energy - final surface energy





E  4 r 2 T n  4R 2T Excess Pressure : Due to the property of surface tension a drop or bubble tends to contract and so compresses the matter enclosed. This in turn increases the internal pressure which prevents further contraction and 199

equilibrium is achieved. So in equilibrium the pressure inside a bubble or drop is greater than outside and the difference of pressure between two sides of the liquid surface is called excess pressure. Excess pressure in different cases is given in the following table : Plane surface Concave surface

P =

2T R

P =

2T R

4T R

P =

2T R

Cylindrical liquid surface

Liquid film of unequal radii

P = 0 Convex surface

Drop

P =

2T R

Bubble in air

Bubble in liquid

P =

P 

 1 1  P  2T     R1 R 2 

I R

Shape of liquid meniscus : When a capillary tube is dipped in a liquid, the liquid surface becomes curved near the point of contact. This curved surface is due to the resultant of two force i.e. the force of cohesion and the force of adhesion. The curved surface of the liquid is called meniscus of the liquid. If liquid molecule A is in the contact with solid. (i.e. wall of capillary tube) then forces acting on molecule A are (i) Force of adhesion Fa (ii) Force of cohesion FC (iii) Resultant force FN makes an angle  with Fa. FC sin 135o tan    Fa  FC cos 135o

FC 2 Fa  FC

Example : Pure water in silver coated capillary tube.

200

Angle of contact : Angle of contact between a liquid and a solid is defined as the angle enclosed between the tangents to the liquid surface and the solid surface inside the liquid, both the tangents being drawn at the point of contact of the liquid with the solid. Capillarity : If a tube of very narrow bore is dipped in a liquid, it is found that the liquid in the capillary either ascends or descends relative to the surrounding liquid. This phenomenon is called capillarity. The root cause of capillarity is the difference in pressure on two sides of (concave and convex) curved surface of liquid. Examples : (i) Ink rises in the fine pores of bloting paper leaving the paper dry. (ii) A towel soaks water. (iii) Oil rises in the long narrow spaces between the threads of wick. Ascent Formula : hρg 

(1)

2T R

rhρg Rhρg  2cosθ 2

 T 

Useful Facts and Formulae : Formation of double bubble : r1 r2 Radius of the interface r  r  r 2 1

1 1  and P  4T     r1 r2  (2)

...1

...2

Formation of a single bubble Under isothermal condition two soap bubble of radii "a" and "b" coalesce to from a single bubble of radius C. Now Pa  P0  Va 

4T 4T 4T , Pb  P0  , PC  P0  a b C

4 3 a 3

Now as mass is conserved, Vc 

4 3 c 3

a   b   c

Vc 

4 3 c 3

Pa Va P V PV  b b  c c RTa RTb RTc

201

Temp is constant,  Ta  Tb  Tc 4T   4 3   4T   4T   4 3     P0    a    P0     P0    c  a 3 b   C  3    

 

P0 c 3  a 3  b 3 T 4 a 2  b2  c 2

 

Hooke's Law and Modulus of Elasticity : According to this Law, within the elastic limit, stress is proportional to the strain. i.e. stress  strain 

Stress = E = constant Strain

The constant E is called the modulus of elasticity. There are three modulii of elasticity namely (i) Young's modulus (y), (ii) Bulk modulus (B), and modulus of rigidity (), responding to three types of the strain. (1)

Young's modulus (Y) It is defined as the ratio of normal stress to longitudinal strain within limit of proportionality F Normal stress FL Y  A   longitudin al strain A L Force constant of wire, K

Y 

mgL  r 2

F YA   L

Bulk Modulus : When a solid or fluid (liquid or gas) is subjected to a uniform pressure all over the surface, such that the shape remains the same, then there is a change in volume. Then the ratio of normal stress to the volumetric strain within the elastic limits is called as Bulk modulus. This is denoted by K. K

Normal stress Volumetric strain

F  pV  A   v v V 202

Modulus of Rigidity : Within limits of proportionality, the ratio of tangential stress to the shearing strain is called modulus of rigidity of the material of the body and is denoted by  ,  

Tangential stress Shearing strain

In this case the shape of a body changes but its volume remains unchanged.

F F   A   A Pascal's Law : It states that if gravity effect is neglected, the pressure at every point of liquid in equilibrium of rest is same. Working of hydraulic lift, hydraulic press and hydroulic breaks : It is used to lift the heavy loads. If a small force f is applied on pistion of C then the pressure exerted on the liquid f a [a = Area of cross section of the pistion in C] This pressure is transmitted equally to piston of cylinder D. P

F  PA 

(1) (2)

(1) (2) (3) (4)

f A Af   a 

As A >> a, F << f

Thermal Expansion : When matter is heated without any change in its state it usually expands. According to atomic theory of matter, a symmetry in potential energy curve is responsible for thermal expansion. As energy of atoms increases, hence the average distance between the atoms increases. So the matter as a whole expands. Thermal expansion is minimum in case of solids but maximum in case of gases because intermolecular force is maximum in solids but minimum in gases. Solids can expand in one dimension (linear expansion) two dimensions (superficial expansion) and three dimensions (volume expansion) While liquids and gases usually suffer change in volume only. Heat : The form of energy which is exchanged among various bodies or system on account of temperature difference is defined as heat. We can change the temperature of a body by giving heat (temperature rises) or by removing heat (temperature falls) from body. Heat is a scalar quantity. It's units are joule, erg, cal, kcal etc. 1 kcal = 1000 cal = 4186 J and 1 cal = 4.18 J 203

(1)

Specific Heat : The amount of heat energy required to raise the temperature of unit mass of a body through (or K) is called specific heat of the material of the body.  If Q heat changes the temperature of mass m by   then specific heat C  m 2 2 1 Unit L T  Molar specific Heat : Molar specific heat of a substance is defined as the amount of heat required to raise the temperature of one gram mole of the substance through a unit degree, it is represented by C



(2)

C 



MQ M Q 1 Q   M m   

Unit : M1 L2 T 2  1 (3)

Latent Heat : The amount of heat required to change the state of the mass m of the substance while its temperature remaining constant is written as : Q = mL, where L is the latent heat Latent heat is also called as cal J Heat of Transformation, It's unit is or and dimension : L2 T 2 gm kg Elastic behaviour : The propertory of matter by virture of which a body tends to regain its original shape and size after the removal of deforming force is called elasticity. In solids, atoms and molecules are arranged in such a way that each molecule is acted upon by the forces due to heighbouring molecules. These forces are known as intermolecular forces. For simplicity, the two molecules in their equilibrium positions (at inter-molecular distance r = r0) are shown by connecting them with a spring. In fact, the spring connecting the two molecules represents the inter-molecular force between them on applying the deforming forces, the molecules either come closer or go far apart from each other and restoring forces are developed. When the deforming force is removed, these restoring forces bring the molecules of the solid to their respective equilibrium position (r = r0) and hence the body regains its original form. Thermal Capacity : It is defined as the amount of heat required to raise the remperature of the whole body (mass m) through 10C or 1K. Q  The value of thermal capacity of a body depends upon the nature of the body and its mass.

Thermal capacity  mc  c 

Dimensions M1 L2 T 2  1 Units :

Cal J , o C k

204

Principle of Calorimetry : Calorimetry means 'measuring heat'. When two bodies (one being solid and other liquid or both being liquid) at different temperatures are mixed, heat will be transfered from body at higher temperature to a body at lowe temperature till both acquire same temeprature. The body at higher temperature releases heat while body at lower temeprature absorbs it, so that Heat lost = Heat gained i.e., Principle of calorimetry represents the law of conservation of heat energy. Thermometry : A branch of science which deals with the measurement of temperature of a substance is known as thermometry. Phase change : We use the term phase to describe a specific state of matter, such as solid, liquid or gas. A transtion from one phase to another is called a phase change. For any given pressure a phase change takes place at a definite temperature, usually accompanied by absorption or emission of heat and a change of volume and density. In phase change 10ce at 00C melts into water at 00C. Water at 1000C boils to form steam at 1000C. Streamline flow : Stream line flow of a liquid is that flow in which each element of the liquid passing through a point travels along the same path and with the same velocity as the preceding element passes through that point. A streamline may be defined as the path, straight or curved, the tangent to which at any point gives the direction of the flow of liquid at that point. The two streamlines cannot cross each other and the greater is the crowding of streamlines at a place, the greater is the velocity of liquid particles at that place. Laminar Flow : If a liquid is flowing over a horizontal surface with a steady flow and moves in the form of layers of different velocities which do not mix with each other, then the flow of liquid is called laminar flow. In this flow, the velocity of liquid flow is always less than the critical velocity of the liquid. The laminar flow is generally used synonymously with streamlined flow. Turbulent flow : When a liquid, moves with a velocity greater than its critical velocity, the motion of the particles of liquid become disordered or irregular. Such a flow is called a turbulent flow. In a turbulent flow, the path and the velocity of the particles of the liquid change continuously and haphazardly with time from point to point. In a turbulent flow, most of the external energy 205

maintaining the flow is spent in producint eddies in the liquid and only a small fraction of energy is available for forward flow. Critical velocity : The critical velocity is that velocity of liquid flow upto which its flow is streamlined and above which its flow becomes turbulent. Reynold's number : Reynold's number is a pure number whichdetermines the nature of flow of liquid through a pipe. NR 

Inprtial force per unit area Viscous force per unit area

Now

dm  AV  3 dt

d V  dm    dt   V AV  V 2  dt Inertial force per unit area   A A A Viscous force per unit area 

F nv  A r

If 0  N R  2000,

If 2000  N R  3000

 NR 

Vr 

laminar or stream line Unstable flow Equation of continuity : The equation of continuity is derived from the principle of conservation of mass. A non-viscous liquid in streamline flow passes through a tube AB of varying cross section. Let the cross sectional area of the pipe at points A and B be a1 and a2 respectively. Mass of the liquid entering per second at A = Mass of the liquid leaving per second at B.

a1 v1 1  a 2 v 2 2

 a1 v1  a 2 v 2

If the liquid is incompressible  av = constant which is the equation of continuity.

206

If N R  3000 definitely turbulent flow

MCQ For the answer of the following questions choose the correct alternative from among the given ones. 1.

2.

Two wires are made of the same material and have the same volume. However, wire 1 has crosssectional area A and wire 2 has cross-sectional Area 3A. If the length of wire 1 increases by on appling force F. How much force is needed to stretch wire 2 by same amount. (A) F (B) 4F (C) 6F (D) 9F The increases in length in l of a wire of length L by longitudinal stress. Then the stress is propotional to.............. (A)

L 

(B)

L

(C)



(D)

2  L

3.

The dimensions of four wires of the same material are given below, in which wire the increase in length will be maximum when the same strain is applied. (A) Length 100 cm, Diameter 1 mm (B) Length 200 cm, Diameter 2 mm (C) Length 300 cm, Diameter 3 mm (D) Length 50 cm, Diameter 0.5 mm

4.

The young's modulus of a wire of length L and radius r is Y reduced to

N . If the length and radius are m2

L r and . Then what will be its young's modulus ? 2 2

Y (B) Y (C) 2Y (D) 4Y 2 A beam of metal supported at the two ends is loaded at the centre. The depression at the centre is propotional to..............

(A) 5.

1 1 (D) Y Y2 A wire is loaded by 6 kg at its one end, the increase in length is 12 mm. If the radius of the wire is doubled and all other magnitudes are unchanged, then increase in length will be............. (A) 6 mm (B) 3 mm (C) 24 mm (D) 48 mm On increasing the length by 0.5 mm in a steel wire of length 2 m and area of cross-section 2 mm2 the force required is................

(A) 6.

7.

Y2

(B)

11 [Y for steel  2.2  10

(A) 8.

1.1  105 N

Y

(C)

1.1  10 4 N

(C)

N ] m

(B)

1.1  103 N

(D)

1.1  10 2 N

A stress of 3.18  108 Nm 2 is applied to steel rod of length 1 m along its length. Its young's 11 modulus is 2  10

(A)

3.18

N . Then what is the elongation produced in the rod in mm ? m2 (B) 6.36 (C) 5.18 (D) 1.59

207

9.

KP   Two springs P & Q of force constant KP & KQ  KQ   are stretched by applying force equal 2   magnitude. If the energy stored in Q is E. Then what is the energy stored in P ? E E (D) 2 4 A force F is needed to break a copper wire having radius R, The force needed to break a copper wire of radius 2R will be........

(A) 10.

E

(B)

2E

(C)

F F (B) 2F (C) 4F (D) 2 4 A rubber cord 10m long is suspended vertically. How much does it stretch under its own weight.

(A) 11.

( Density of rubber is 1500 (A) 12.

13.

(B)

yx2

(B)

(C)

12  10 4 m

(D)

25  10 4 m

2yx2

(C)

1 2 y x 2

(D)

1 yx 2 2

A steel wire of cross-sectional area 3  10 6 m 2 can with stand a maximum strain of 10-3 Young's 11 modulus of steel is 2  10

(A) 15.

7.5  104 m

The young's modulus of the material of a wire is equal to the ........... (A) stress required to increase its length four times (B) strtess required to prdouce unit strain (C) strain produced in it (D) stress acting on it If x, longitudinal strain is produced in a wire of young's modulus y then energy stroed in the material of the wire per unit volume is.......... (A)

14.

15  10 4 m

kg N m , Y  5  108 2 , g  10 2 ) 3 m m s

40 kg

(B)

m  N .  g  10 2  2 The maximum mass the wire can hold is............. s  m 

60 kg

(C)

80 kg

The young's modulus of a rubber string 8 cm long and density 1.5

(D)

100 kg

kg N 5  108 2 . What will 3 is m m

be the length increase due to its own weight ?

16.

17.

(A)

9.6  105 m

(B)

9.6  10 11 m

(C)

9.6  103 m

(D)

9.6 m

A and B are two wires. The radius of A is twice that of B. They are stretched by the same load. Then what is the stress on B ? (A) Equal to that on A (B) Four times that on A (C) Two times that on A (D) Half that on A If the length of wire is reduced to half then it can hold the .............load. (A) Half (B) Same (C) Double (D) One fourth 208

18.

There are two wires of same material and same length. While the diameter of second wire is 2 times, the diameter of first wire. Then what will be the ratio of extension produced in the wire by applying same load ? (A)

19.

1:1

(B)

2:1

(C)

1:2

(D)

4:1

When the length of a wire having cross-section area 10 6 m 2 is stretched by 0.1% then tension in

it is 100 N. Young's modulus of material of wire is........... (A) 20.

21.

N m2

(B)

10 2

N m2

(C)

1010

N m2

(D)

1011

N m2

Two wires of equal lengths are made of the same material wire A has a diameter that is twice as that of wire B. If identical weights are suspended from the ends of these wires the increase in length is............ (A)

Four times for wire A as for wire B.

(B)

Twice for wire A as for wire B.

(C)

Half for wire A as for wire B.

(D)

One-fourth for wire A as for wire B.

Steel and copper wires of same length are stretched by the same weight one after the other. Young's modulus of steel the ratio increase in length ? (A)

22.

1012

2 5

(B)

3 5

(C)

5 4

(D)

5 2

An area of a cross-section of rubber string is 2 cm3. Its length is doubled when stretched with a linear force of 2  105 dynes. What will be young's modulus of the rubber in dynes ? (A)

23.

24.

4  105

(B)

1  105

(C)

2  105

(D)

1  10 4

A substance breaks down by a stress of If the density of the material of the wire is then the length of wire of the substance which will break under its own weight when suspended vertically is............. (A) 66.6 m (B) 60.0 m (C) 33.3 m (D) 30.0 m The temperature of a wire of length 1 meter and area of cross-sectional section 1 cm2 is increased from to If the rod is not allowed to increase in length. What will be the force required ?  5 o 11 N     10 / C, Y  10  m2  

25.

(A) 103 N (B) 104 N (C) 105 N (D) 109 N If longitudinal strain for a wire is 0.03 and its poisson's ratio is 0.5, then what is its lateral strain ? (A) 0.003 (B) 0.0075 (C) 0.015 (D) 0.4

26.

9 An aluminium rod (Young's modulus  7  10

N ) has a breaking strain of 0.2 % what is the m2 minimum cross-sectional area of the rod in order to support a load of 104 Newtons ?

(A)

1  10 2 m 2

(B)

1.4  10 3 m 2

(C)

3.5  10 3 m 2

(D)

7.1  10 4 m 2

209

27.

28.

Two wires of copper having the length in the ratio 4 : 1 and their are as 1 : 4 are stretched by the same force. What will be the ratio of longitudinal strain in the two wires ? (A) 1 : 16 (B) 16 : 1 (C) 1 : 64 (D) 64 : 1 A wire elongates by 1 mm when a load W is hanged from it. If the wire goes ever a pulley and two weight W each are hang at the two ends. What will be the elongation of the wire ? (in mm)  (D)  2 200 kg weight hanged at a free edge of vertical wire of length 600.5 cm when removed the weight, length reduced by 0.5 cm and it gets original status then what is the young moduluse of wire ? (A) 2.35  1012 N/m2 (B) 1.35  1010 N/m2 (C) 13.5  1011 N/m2 (D) 23.5  109 N/m2 The ratio of diameter of two wires of same material is n : 1 the length of wires are 4 m each. On applying the same load. What will be the increase in length of their wire ? (A) n2 times (B) n times (C) 2n times (D) None of the above

(A) 29.

30.

31.

32.

33.

35.

(B)

zero

(C)

 N 4 N   Y  10  2 is applied on a wire what is the % increase in length ? m2  mm  (A) 0.002 (B) 0.001 (C) 0.003 (D) 0.01 A steel wire is stretched with a definate load of If the young's modulus of the wire is Y. For decreasing the value of Y............ (A) Radius is to be decreased (B) Radius is to be increased (C) Length is to be increased (D) None of the above

Longitudinal stress of 1

 11 N   is 0.1 cm2, what will be the force The area of cross-section of a steel wire is  Y  2  10 m2   required to double its length ?

(A) 34.

2

2  1012 N

(B)

2  1011 N

(C)

2  1010 N

(D)

2  106 N

Two wires A & B are of same materials. Their lengths in the ratio 1 : 2 and diameters are in the ratio 2 : 1 when stretched by force FA and FB respectively, they get equal increase in lengths Then the ratio

FA should be........... FB

(A)

1:2

(B)

1:1

(C)

2:1

(D)

8:1

The mean distance between the atoms of iron is 3  10 10 and interatomic force constant for iron is 7

N . What is the young's modulus of elasticity for iron ? m

(A)

2.33  10 5

N m2

(B)

23.3  1010

N m2

(C)

233  1010

N m2

(D)

2.33  1010

N m2

210

36.

A force of 200 N is applied at one end of a wire of length 2 m and having area of cross-section 10 2 cm 2 , the other end of the wire is rigidly fixed. If of linear expansion of the wire N and its temperature is increased by m2 50C then the increase in the tension of the wire will be.......... 11   8  10 6 / o C and young's modulus Y  2.2  10

(A) 37.

39.

F AY

2.4 N

(D)

(B)

2F AY

(C)

1 2

 F     AY 

(D)

2.0  1010

N m2

(B)

4.0  1010

N m2

(C)

2.0  1011

N m2

(D)

4.0  1011

N m2

8.8 N

3F AY

A rubber cord catapult has cross-sectional area 25 mm2 and initial length of cord is 10 cm. It is stretched to 5 cm and then released to project a missile of mass 5 gm. Taking

20

N velocity of projected missile is............ m2

m s

(B)

100

m s 1

(C)

250

m s 1

(D)

200

m s 1

A wire of cross-section 4 mm2 is stretched by 0.1 mm by a certain weight. How far (length) will be wire of same material and length but of area 8 mm2 stretched under the action of same force. (A)

42.

(C)

(A)

(A)

41.

4.4 N

The length of a wire is 1.0 m and the area of cross-section is 1.0  10 2 cm 2 . If the work done for increase in length by 0.2 cm is 0.4 joule. Then what is the young's modulus ? Of material of the wire ?

Yrubber  5  108

40.

(B)

A uniform plank of young's modulus Y is moved over a smooth horizontal surface by a constant horizontal force F, The area of cross-section of the plank is A. What is the compressive strain on its plank in the direction by the force ? (A)

38.

4.2 N

0.05 mm

(B)

0.10 mm

(C)

0.15 mm

(D)

0.20 mm

According to Hooke's law of elasticity stress is increased the ratio of stress toi strain......... (A)

increases

(B)

Decreases

(C)

becomes zero

(D)

Remains constant

N and area of cross section is 2 cm2 if force of 2  105 dyne is m2 applied along its length then its initial length L becomes .............

Young's modulus of rubber is 10

4

(A)

3L

(B)

4L

(C)

2L

(D)

None of the above

211

43.

A copper wire of length 4 m and area of cross-section 1.2 cm2 is stretched with a force of 4.8  103 N. 11

If young's modulus for copper is 1.2  10 (A) 44.

1.33 mm

(B)

N What will be the length increase of the wire ? m2

1.33 cm

(C)

2.66 mm

(D)

2.66 cm

N . the force constant = m2

If the interatomic spacing in a steel wire is 3A o & YSteel  20  1010 ............. (A)

45.

9 (B) 6  10

N A

(C)

6  10 5

N A

(D)

6  10 5

N A

0.8 cm

(B)

1.6 cm

(C)

2.4 cm

(D)

3.2 cm

A wire of length L and radius r is rigidly fixed at one end on stretching the other end of the wire a force F the increase in its lengths is L. If another wire of same material but of length 2L and radius 2r is stretched with a force of 2F, the increase in its length will be .......... (A)

47.

N A

A wire of length 2 m is made from 10cm3 of copper. A force F is applied so that its length increases by 2mm. Another wire of length 8 m is made from the same volume of copper. If the force F is applied to it, its length will increase by............ (A)

46.

6  10 2

(B)



(C)

2

 2

(D)

11 In steel the young's modulus and the strain at the breaking point are 2  10

 4 N and 0.15 m2

respectively the stress at the breaking point for steel is therefore ...........

48.

49.

(A)

1.33  1011

N m2

(B)

1.33  1012

(C)

7.5  10 13

N m2

(D)

3  1010

N m2

N m2

Which of the following statement is correct (A)

Hooke's law is applicable only within elastic limit.

(B)

The adiabatic and isothermal elastic constants of a gas area equal.

(C)

Young's modulus is dimensionsless.

(D)

Stress multiplied by strain is equal to stored energy.

The force required to stretch a steel wire of 1 cm 2 cross-section to 1.1 times its length would be  11 N   Y  2  10  m2  

(A)

2  10 6 N

(B)

2  103 N

(C) 212

2  10 6 N

(D)

2  10 7 N

50.

51.

52.

Which one of the following quantities does not have unit of force per unit area.......... (A)

stress

(B)

strain

(C)

Young's modulus of elasticity

(D)

pressure

A copper wire and a steel wire of same diameter and length are connected end to end a force is applied, which stretches their combined length by 1 cm, the two wires will have....... (A)

different stresses and strains

(B)

the same stress and strain

(C)

the same strain but different stresses

(D)

the same stress but different strains

A steel ring of radius r and cross-section area 'A' is fitted on to a wooden disc of radius R(R > r ) If young's modulus be E then what is force with which the steel ring is expanded ? (A)

53.

55.

56.

R r

(B)

R r AE    r 

(C)

E  R r    A  R 

(D)

Er AR

A wire of diameter 1 mm breaks under a tension of 100 N. Another wire of same material as that of the first one, but of diameter 2 mm breaks under a tension of......... (A)

54.

AE

500 N

(B)

1000 N

(C)

10,000 N

(D)

4000 N

A fixed volume of iron is drawn into a wire of length L. The extension x produced in this wire be a constant force F is propotional to.......... 1 1 (A) (B) (C) L2 (D) L 2 L L 8 N On applying a stress of 20  10 the length of a perfect elastic wire is doubled. What will be its m2 Young's modulus ? (A)

40  108

N m2

(B)

20  108

(C)

10  108

N m2

(D)

5  108

N m2

N m2

To keep constant time, watches are fitted with balance wheel made of........ (A)

invar

(B)

stainless steel

(C)

Tungsten

(D)

platinum

57.

A wire is stretched by 0.01 m by a certain force F. Another wire of same material whose diameter and length are double to the original wire is stretched by the same force ? Then what will be its elongation ? (A) 0.005 m (B) 0.01 m (C) 0.02 m (D) 0.002 m

58.

The Coefficient of linear expansion of brass & steel are 1 &  2 If we take a brass rod of length 1 & steel rod of length  2 at 0 o C , their difference in length  2  1  will remain the same at a temperature if................. (A)

1  2   2  1

(B)

 2 1   2  1

(C)

 22  1   22  2

(D)

1  1   2  2

213

59.

A rod is fixed between two points at 20 o C The Coefficient of linear expansion of material of rad is 11

1.1  10 5 / o C and Young's modulus is 1.2  10

N . Find the stress developed in the rod if m2

temperature of rod becomes 10 oC . (A) 60.

61.

1.32  10 7

N m2

15

(B) 1.10  10

N N N 1.32  108 2 (D) 1.10  10 6 2 2 (C) m m m

How much force is required to produce an increase of 0.2% in the length of a bross wire of diameter 11 0.6 mm (Young's modulus for brass  0.9  10

N ). m2

(A)

(C)

Nearly 17 N

(B)

Nearly 34 N

Nearly 51 N

(D)

Nearly 68 N

 10 N   of diameter 3mm supports a 40 kg mass. In order A 5m long aluminium wire  Y  7  10 m2   10 to have the same elongation in a copper wire Y  12  10

N of the same length under the same m2

weight, the diameter should now be in mm............ 62.

(A) 1.75 (B) 1.5 (C) 2.5 (D) 5.0 Two similar wires under the same load yield elongation of 0.1 mm and 0.05 mm respectively. If the area of Cross - section of the first wire is 4mm2. Then what is the area of cross - section of the second wire ? (A)

63.

64.

(B)

8 mm 2

(C)

10 mm 2

(D)

12 mm 2

An iron rod of length 2m and cross-section area of 50 mm 2 stretched by 0.5 mm, when a mass of 250 kg is hung from its lower end. What is young's modulus of the iron rod ? (A)

19.6  1010

N m2

(B)

19.6  1015

N m2

(C)

19.6  1018

N m2

(D)

19.6  10 20

N m2

A load W produces an extension of 1mm in a thread of radius r. Now if the load is made 4 w and radius is made 2r all other things remaining same the extension will becomes.............. (A)

65.

6 mm 2

4 mm

(B)

16 mm

(C)

1 mm

(D)

0.25 mm

A steel wire of 1m long and 1 mm 2 cross sectional area is hung from rigid end when weight of 1 kg  11 N   is hung from it then change in length will be............  Y  2  10 m2  

(A)

0.5 mm

(B)

0.25 mm

(C) 214

0.05 mm

(D)

5 mm

66.

Calculate the work done, if wire is loaded by 'M g' weight and the increase in length is 'l' ?

(A) 67.

69.

1:2

(B)

1:4

(C)

2:1

Zero

(B)

0.05 Joule

(C)

100 Joule

(D)

500 Joule

(D)

2mgl

(D)

1:1

If the force constant of a wire is k. What is the work done in increasing the length of the wire by  ? k 2

(B)

k 2 2

(C)

k

(D)

k 2

Wire A and B are made from the same material. A has twice the diameter and three times the length of B. If the elastic limits are not reached when each is stretched by the same tension, what is the ratio of energy stored in A to that in B ? 2:3

(B)

3:4

(C)

3:2

(D)

6:1

A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower and. The weight stretches the wire by 1 mm. Then what is the elastic energy stored in the wire ? (A)

72.

mg 

(C)

(A)

(A) 71.

Zero

A 5 meter long wire is fixed to the ceiling. A weight of 10 kg is hung at the lower end and is 1 meter above the four. The wire was elongated by 1 mm. What is the stored in the wire due to stretching ?

(A) 70.

(B)

Two wires of same diameter of the same material having the length  and 2  . If the force F is applied on each, what will be the ratio of the work done in the two wires ? (A)

68.

mg l

0.1 J

(B)

0.2 J

(C)

10 J

(D)

20 J

A brass rod of cross sectional area 1 cm2 and length 0.2 m is compressed length wise by a weight of 11

5 kg. If young's modulus of elasticity of brass is 1  10

N m g  10 2 Then what will be 2 and m s

increase in the energy of rod ? (A) 73.

(B)

2.5  10 5 J

(C)

5  10 5 J

(D)

2. 5  104 J

Young's modulus of the material of a wire is Y. On pulling the wire by a force F the increase in its length is x, what is the potential energy of the stretched wire ? (A)

74.

10 5 J

1 Fx 2

(B)

1 Yx 2

(C)

1 Fx 2 2

(D)

None of these

The work per unit volume to stretch the length by 1% of a wire with cross - sectional area 1 mm2 will  11 N   be............  Y  9  10 m2  

(A)

9  1011 J

(B)

4.5  107 J

(C) 215

9  10 7 J

(D)

4.5  1011 J

75.

A wire of length 50 cm and cross - sectional area of 1 mm2 is extended by 1mm what will be the



required work ? Y  2  1011 Nm 2 (A) 76.

2  10 2 J

(C)

4  10 2 J

(D)

1  10 2 J

T2 2x

(B)

T2 2k

(C)

2x T2

(D)

2T 2 k

(D)

FL 2

On stretching a wire what is the elastic energy stored per unit volume ? (A)

78.

(B)

If a spring extends by x cm loading then what is the energy stored by the spring ? (If T is tension in the spring & K is spring constant) (A)

77.

6  102 J



F 2AL

(B)

FA 2L

(C)

FL 2A

When a force is applied on a wire of uniform cross-sectional area 3  10 6 m 2 and length 4m, the  11 N   increase in length is 1 mm. what will be energy stored in it ?  Y  2  10 m2  

(A) 79.

62. 50 J

(B)

0.177 J

(C)

0.075 J

(D)

0.150 J

k is the force constant of a spring what will be the work done in increasing its extension form

1 to  2 be ? (A) 80.

 2 

(B)

k   2  1  2

(C)



2

k  2  1

2



(D)

k 2 2  2  1 2





4.900 Joule

(B)

2.450 Joule

(C)

0.495 Joule

(D)

0.245 Joule

Pressure

(D)

Specific heat

The isothermal elasticity of a gas is equal to.................... (A)

82.

 1

When a 4 kg mass is hung vertically on a light spring that obeys Hook's law, the spring stretches by 2 cm what will be the work required to be done by an external agent in stretching this spring by 5 cm ? (A)

81.

k

Density

(B)

Volume

(C)

If the volume of a block of aluminium is decreased, by the pressure (stress) on its surface is increased by.................(Bulk modulus of A  7.5  1010 Nm 2 ) (A)

83.

7.5  1010

N m2

(B)

7.5  108

N m2

(C)

7.5  106

N m2

(D)

7.5  10 4

N m2

The specific heat at constant pressure and at constant volume for an ideal gas are C p and C v and isothermal elasticities are E  and E  respectively. What is the ratio of E  and E  . (A)

Cv Cp

(B)

Cp

(C)

Cv 216

Cp C v

(D)

1 Cp C v

84.

What is the ratio of the adiabatic to isothermal elasticities of a triatomic gas ? (A)

85.

3 4

(B)

4 3

due to gravity  10 (A)

kg N  9  108 2 ,acceleration 3 Bulk modulus of rubber m m

9m

(B)

18 m

(C)

4  10 5 CC

(B)

4  10 5 CC

(C)

108

3

90 m

0.025 CC

(D)

0.004 CC

kg m N & g  10 2 . Then waht will be the volume elasticity in 2 ? 3 m s m

(B)

2  108

(C)

0.01

(B)

0.06

(C)

109

(D)

2  109

0.02

(D)

0.03

When a pressure of 100 atmosphere is applied on a spherical ball then its volume reduces to 0.01% What is the bulk moduls of the material of the rubber in (A)

10  1012

(B)

1  1012

(C)

dyne . cm 2

100  1012

(D)

20  1012

The pressure applied from all directions on a cube is p. How much its temperature should be raised to maintain the orginal volume ? The volume elasticity. of the cube is B and the coefficient of volume expansion is  . (A)

91.

(D)

   For a constant hydraulic stress on an object, the fractional change in the object volume   and    its bulk modulus (B) are related as...............

(A)

90.

180 m

If a rubber ball is taken at the depth of 200m in a pool, Its volume decreases by 0.1% . If the density

(A)

89.

5 3

m ) s2

of the water is 1  10

88.

(D)

The compressibility of water 4  10 5 per unit atmospheric pressure. The decrease in volume of 100 cubic centimeter of water under a pressure of 100 atmosphere will be................ (A)

87.

1

To what depth below the surface of sea should a rubber ball be taken as to decrease its volume by 0.1% (Take : density of sea water  1000

86.

(C)

P 

(B)

P 

(C)

p 

(D)

 p

A uniform cube is subjected to volume compression. If each side is decreased by 1% Then what is bulk strain ? (A)

0.01

(B)

0.06

(C)

217

0.02

(D)

0.03

92.

The ratio of two specific heats of gas

Cp Cv

for Argon is 1.6 and for hydrogen is 1.4. Adiabatic

elasticity of Argon at pressure p is E. Adiabatic elasticity of hydrogen will also be equal to E at the pressure. (A) 93.

p

(B)

7 P 8

(C)

8 P 7

(D)

1.4 P

What is the isothermal bulk modulus of a gas at atmospheric pressure ? (A) 1 mm of Hg (B) 13.6 mm of Hg (C)

1.013  105

N m2

(D)

2.026  105

N m2

94.

The bulk modulus of an ideal gas at constant temperature......... (A) is equal to its volume V (B) is equal to P/2 (C) is equal to its pressure P (D) cannot be determined

95.

A material has poisson's ratio 0.50. If uniform rod of it suffers a longitudinal strain of 2  10 3 . Then what is percentage change in volume ? (A) 0.6 (B) 0.4 (C) 0.2 (D) 0 There is no change in the volume of a wire due to change in its length on stretching. What is the possion's ratio of the material of the wire.... (A) +0.5 (B) -0.50 (C) 0.25 (D) -0.25 Which statement is true for a metal......

96.

97.

(A) 98.

99.

Y

(B)

Y

(C)

Y

(D)

Y

1 η

Which of the following relation is true (A)

3Y  k 1  6

(B)

k

gny yn

(C)

6  6k  n  Y

(D)

6

0.5 Y  n n

Two wires A & B of same length and of the same material have the respective radius r, & r2 their one end is fixed with a rigid support and at the other end equal twisting couple is applied. Then what will we be the ratio of the angle of twist at the end of A and the angle of twist at the end of B. 2

(A)

r1 r22

2

(B)

4

r2 r12

(C)

100. The poisson's ratio can not have the value........... (A) 0.7 (B) 0.2 (C) 218

4

r2 r14

(D)

r1 r24

0.1

(D)

0.5

101. The value of poisson's ratio lies between........ (A)

 1 to

1 2

(B)



3 1 to 4 2

(C)



1 to 1 2

(D)

1 to 2

102. If the young's modulus of the material is 3 times its modulus of rigidity. Then what will be its volume elasticity ? (A)

zero

(B)

infinity

(C)

2  1010

N m2

(D)

3  1010

N m2

103. For a given material the Young's modulus is 2.4 times that of rigidity modulus. What is its poisson's ratio ? (A) 2.4 (B) 1.2 (C) 0.4 (D) 0.2 104. The lower surface of a cube is fixed. On its upper surface is applied at an angle of 300 from its surface. What will be the change of the type ? (A) shape (B) size (C) none (D) shape & size 105. The upper end of a wire of radius 4 mm and length 100 cm is clamped and its other end is twisted through an angle of 300. Then what is the angle of shear ? (A) 120 (B) 0.120 (C) 1.20 (D) 0.0120 106. Mark the wrong statement. (A) Sliding of moleculawr layer is much easier than compression or expansion. (B) Receiprocal of bulk modulus is called compressibility. (C) Twist is difficult in big rod as compared to small rod. (D) Which is more strong out of hollow and solid cylinder having equal length and mass ? 107. A 2m long rod of radius 1 cm which is fixed from one end is gien a twist of 0.8 radians. What will be the shear strain developed ? (A) 0.002 (B) 0.004 (C) 0.008 (D) 0.016 108. Shearing stress causes change in (A) length (B) breadth (C) shape (D) volume 109. What is the relationship between Young's modulus Y, Bulk modulus k and modulus of rigidity  ? (A)

Y

9 k  3k

(B)

Y

9y k y 3k

(C)

Y

110. What is the possible value of posson's ratio ? (A) 1 (B) 0.9 (C) 0.8 111. The graph shown was obtained from experimental measurements of the period of oscillations T for different masses M placed in the scale pan on the lower end of the passing through the origin is that the...... (A) Spring did not obey Hooke's law (B) Amplitude of the oscillations was large (C) Clock used needed regulating (D) Mass of the pan was neglected 219

9 k 3 k

(D)

Y

(D)

0.4

3 k 9  k

112. A graph is shown between stress and strain for metals. The part in which Hooke's law holds good is (A) OA (B) AB (C) BC (D) CD

113. The strain-stress curves of three wires of different materials are shown in the figure. P, Q and R the elastic limits of the wires the figure shows that (A) Elasticity of wire P is maximum. (B) Elasticity of wire Q is maximum. (C) Tensile strength of R is maximum (D) none of above 114. The diagram shows a force extension graph for a Rubber band consider the following statements. (I) It will be easier to compress these rubber than expand it. (II) Rubber does not return to its original length after it is stretched. (III) The rubber band will get heated if it is stretched and relased. Which of these can be deduced from the graph. (A) III only (B) II and III (C) I and III 115. The stress versus strain graph for wires of two material A & B are as shown in the figure. If YA & YB are the young's modulus of the materials then (A) YB = 2YA (B) YA = YB (C) YB = 3YA (D) YA = 3YB 116. The load versus elongation graph for four wires of the same material is shown in the figure. The thickest wire is represented by the line. (A) OD (B) OC (C) OB (D) OA

220

(D)

I only

117. The adjacent graph shows the extension of a wire of length 1m suspended from the top of a roof at one end with the load W connected to the other end. If the cross sectional area of the wire is 10-6 m2, calculate the young's modulus of the material of the wire. (A)

2  1011

(C)

3  10 12

N m2 N m2

(B)

2  10 11

N m2

(D)

2  10 12

N m2

118. The graph is drawn between the applied force F and the strain (x) for a thin uniform wire the wire behaves as a liquid in the part. (A) ab (B) bc (C) cd (D) oa 119. The graph shows the behaviour of a length of wire in the region for which the substance obey's Hooke's law. P & Q represent. (A) p = applied force, Q = extension (B) p = extension, Q = applied force (C) p = extension, Q = stored elastic energy (D) p = stored elastic energy, Q = extension 120. The potential energy U between two nmolecules as a function of the distance x between them has been shown in the figure. The two molecules are. (A) Attracted when x lies between A & B and are repelled when x lies between B & C. (B) Attracted when x lies between B and C and are repelled when x lies between A and B. (C) Attracted when they reach B. (D) Repelled when they reach B. 121. The value of force constant between the applied elastic force F and displacement will be (A) (C)

3 1 2

(B)

1 3

(D)

3 2

221

122. The diagram shows stress v/s strain curve for the materials A and B. From the curves we infer that (A) A is brittle but B is ductile (B) A is ductile and B is brittle (C) Both A & B are ductile (D) Both A & B are brittle 123. Which are of the following is the young's modulus (in N/m2) for the wire having the stress strain curve in the figure. (A)

24  1011

(B)

8.0  1011

(C)

10  1011

(D)

2.0  1011

124. The diagram shows the change x in the length of a thin uniform wire caused by the application of stress F at two different temperature T1 & T2. The variation (A)

T1  T2

(B)

T1  T2

(C)

T1  T2

(D)

None of these

125. The point of maximum and minimum attraction in the curve between potential energy (U) and distawnce (r) of a diatomic molecules are respectively. (A) S and R (B) T and S (C) R and S (D) S and T

Assertion and Reason : Read the assertion and reason carefully to mark the correct option out of the option given below (A) If both assertion and reason are true and reason is the correct explanation of the assertion. (B) If both assertion and reason are true but reason is not the correct explanation of the assertion. (C) If assertion is true but reason is false. (D) If assertion and reason both are false.

222

126. Assertion : The stretching of a coil is determined by its shear modulus. Reason : Shear modulus change only shape of a body keeping its dimensions unchanged. (A) a (B) b (C) c (D) d 127. Assertion : The bridges are declared unsafe after a long use. Reason : Elastic strength of bridges decrease with (A) a (B) b (C) c (D) d 128. Assertion : Two identical solid balls, one of ivory and the other of wet clay are dropped from the same height on the floor. Both the balls will rise to same height after. Reason : Ivory and wet-clay have same elasticity. (A) a (B) b (C) c (D) d 129. Assertion : Young's modulus for a perfectly plastic body is zero. Reason : For a perfectly plastic body is zero. (A) a (B) b (C) c (D) d 130. Assertion : Identical spring of steel and copper are equally stretched more will be done on the steel spring. Reason : Steel is more elastic than copper. (A) a (B) b (C) c (D) d 131. A wire is stretched to double the length which of the following is false in this context ? (A) Its volume increased (B) Its longitudinal strain is I (C) Stress = Young's modlus (D) Stress = 2x Young's modulus 132. Which is the dimensional formula for modulus of rigidity ? (A)

M1 L1 T 2

(B)

M1 L1 T 2

(C)

M1 L2 T 1

(D)

M1 L2 T 2

133. When more than 20 kg mass is tied to the end of wire it breaks what is maximum mass that can be tied to the end of a wire of same material with half the radius ? (A) 20 kg (B) 5 kg (C) 80 kg (D) 160 kg 134. When 100 N tensile force is applied to a rod of 10-6 m2 cross-sectional area, its length increases by 1% so young's modulus of material is.......... (A) 1012 Pa (B) 1011 Pa (C) 1010 Pa (D) 102 Pa 135. A composite wire is made by joining ends of two wires of equal dimensions, one of copper and the other of steel. When a weight is sttached to its end the ratio of increase in their length is ......... YSteel 

20 YCopper 7

(A) 20.7 (B) 10.7 (C) 7:20 (D) 1:7 136. A rubber ball when taken to the bottom of a 100 m deep take decrease in volume by 1% Hence, the bulk modulus of rubber is............ m   g  10 2  s  

(A)

106 Pa

(B)

108 Pa

(C) 223

107 Pa

(D)

109 Pa

137. Young's modulus of a rigid body is (A) 0 (B) 1

(C)



(D)

0.5

138. Pressure on an object increases from 1.01  105 Pa to 1.165  10 5 Pa. He volume decrease by 10% at constant temperature. Bulk modulus of material is........ (A)

1.55  105 Pa

(B)

51.2  105 Pa

(C)

102.4  105 Pa

(D)

204.8  105 Pa

139. Cross-sectional area oif wire of length L is A. Young's modulus of material is Y. If this wire acts as a spring what is the value of force constant ? (A)

YA L

YA 2L

(B)

(C)

2YA L

(D)

YL A

Surface Tension 140. Writing on black board with a pieace of chalk is possible by the property of (A)

Adhesive force

(B)

Cohesive force

(C)

Surface force

(D)

Viscosity

141. When there is no external force, the shape of liquid drop is determined by (A)

Surface tension of liquid

(B)

Density of Liquid

(C)

Viscosity of liquid

(D)

Tempreture of air only

142. Soap helps in cleaning because (A)

chemicals of soap change

(B)

It increase the surface tension of the soluiton.

(C)

It absorbs the dirt.

(D)

It lowers the surface tension of the solution

143. A beaker of radius 15 cm is filled with liquid of surface tension 0.075 N/m. Force across an imaginary diameter on the surface ofliquid is (A)

0.075 N

(B)

1.5  102 N

(C)

0.225 N

(D)

2.25  10 2 N

144. A square frame of side L is dipped in a liquid on taking out a membrance is fomed if the surface tension of the liquid is T, the force acting on the frame will be. (A)

2 TL

(B)

4 TL

(C)

8 TL

(D)

10 TL

145. The force required to separate two glass plates of area 10-2 m2 with a film of water 0.05 mm thick between them is (surface tension of water is 70  10 (A)

28 N

(B)

14 N

3

(C)

224

N ) m

50 N

(D)

38 N

146. Surface tension of a liquid is found to be infuenced by (A)

It increases with the increase of temprature.

(B)

Nature of the liquid in contact.

(C)

Presense of soap that increaase it.

(D)

Its variation with the concentration of the liquid.

147. A thm metal disc of radius r floats on water surface B and bends the surface down wards along the perimeter making an angle Q with vertical edge of the disc. If the disc displaces a weight of water W and surface tension of water is T, then the weight at metal dis (A)

2rT  w

(B)

2rT cos   w

(C)

2rT cos   w

(D)

w  2rT cos 

148. A thin liquid film formed between a u-shaped wire and a light slider supports a weight of 1.5  102 N (see figure). The length of the slider is 30 cm and its weight negligible. The surface tension of the liquid film is. (A)

0.0125 N m 1

(B)

0.1 N m 1

(C)

0.05 N m 1

(D)

0.025 N m 1

Surface Energy 149. Radius of a soap bubble is 'r', surface tension of soap solution is T. Then without incresing the temprature how much energy will be needed to double its radius. (A)

4  r2 T

(B)

2  r2 T

(C)

12  r 2 T

(D)

24  r 2 T

150. The amount of work done in blowing a soap bubble such that its diameter increases from d to D is (T = Surface tension of solution) (A)

4 D 2  d 2 T





(B)

(C)

 D2  d 2 T

(D)





 2  D

 T

8 D 2  d 2 T 2

 d2

151. A soap bubble of radius r is blown up to form a bubble of radius 2r under isothermal conditions if the T is the surface tension of soap solution the energy spent in the slowing is. (A)

(B) 6Tr 2 (C) 12Tr 2 (D) 24Tr 2 3Tr 2 152. The surface tension of a liquid is 5 N/m. If a thin film of the area 0.02 m2 is formed on a loop, then its surface energy will be (A)

5  10 2 J

(B)

2.5  10 2 J

225

(C)

2  10 1 J

(D)

5  10 1 J

153. A frame made of a metalic wire enclosing O surface area A is covered with a soap film. If the area of the frame metalic wire is reduced by 50% the energy of the soap film will be changed by (A) 100% (B) 75% (C) 50% (D) 25% 154. Two small drops mercury, each of radius R, coaless the form a single large drop. The ratio of the total surface energies before and after the change is. (A)

1: 2

1 2

(B)

1 2

(C)

2 :1

2:1

(D)

1:2

155. The work done is blowing a soap bubble of 10 cm radius is (surface tension od the soap solution is

3 N/m) 100

(A)

75. 36  10 4 Joule

(B)

37. 68  10 4 Joule

(C)

150. 72  10 4 Joule

(D)

75. 36 Joule

156. The work done increasing the size of a soap film drom 10 cm  11 cm is 3  104 Joule . The surface tension of the film is (A)

1.5  10  2

N m

(B)

3.0  10  2

N m

(C)

6.0  10 2

N m

2 (D) 11.0  10

N m

157. A big drop of radius R is formed by 1000 small droplets of coater then the radius of small drop is (A)

R 2

(B)

R 5

(C)

R 6

(D)

R 10

158. 8000 identioal water drops are combined to form a bigdrop. Then the ration of the final surface energy to the intilial surface energy of all the drops together is (A) 1 : 10 (B) 1 : 15 (C) 1 : 20 (D) 1 : 25 159. The relation between surface tension T. Surface area A and surface energy E is given by. (A)

T

E A

(B)

(C)

T  EA

E

T A

Angle of Contact 160. If a coater drop is kept between two glasses plates then its shape is

(A)

(D)

(B)

(C)

None of these

226

(D)

T

A E

161. A liquid wets a solid completely. The menisions of the liquid in a surfficently long tube is (A) Flat (B) Concave (C) Convex (D) Cylindrical

Pressure Difference 162. When two soap bubbles of radius r1 and r2 (r2 > r1) coalesce, the radius of curvature of common surface is........... r2  r1 r1 r2 (A) r2  r1 (B) (C) (D) r2  r1 r1 r2 r2  r1 163. The excess of pressure inside a soap bubble than that of the other pressure is 2T 4T T T (B) (C) (D) r r 2r r 164. The radill of two soap bubbles are r1 and r2. In isothermal conditions two meet together is vacum Then the radius of the resultant bubble is given by

(A)

(A)

R  r1  r2  / 2

(B)

R  r1 r1  r2  r3 

(C)

R 2  r1  r2

2

(D)

R  r1  r2

2

165. A spherical drop of coater has radius 1 mm if surface tension of contex is 70  10 3 N / m difference of pressures between inside and outside of the spherical drop is N N N (B) 70 2 (C) 140 2 (D) zero 2 m m m 166. In capilley pressure below the curved surface at water will be (A) Equal to atomospheric (B) Equal to upper side pressure (C) More than upper side pressure (D) Lesser than upper side pressure 167. Two bubbles A and B (A>B) are joined through a narrow tube than (A) The size of a will increase (B) The size of B will increase (C) The size of B will increase untill thenpressure equals (D) None of these 168. If the excess pressure inside a soap bubble is balanced by oil column of height 2 mm then the surface tension of soap solution will be. Cr = 1 and density d = 0.8 gm/cc) N 2 N 3 N 1 N (A) 3.9 (B) 3.9  10 (C) 3.9  10 (D) 3.9  10 m m m m

(A)

35

Capillarity 169. A capillary tube at radius R is immersed in water and water rises in it to a height H. Mass of water in the capillary tube is M. If the radius of the tube is doubled. Mass of water that will rise in the capillary tube will now be (A) M (B) 2M (C) (D) 4M 227

170. A vesel whose bottom has round holes with diametre of 0.1 mm is filled with water. The maximum height to which the water can be filled without leakage is 75 dyne m g  1000 2 ) cm s (A) 100 cm (B) 75 cm (C) 50 cm (D) 30 cm The correct relation is 2T cos  T cos  hdg 2T dgh (A) r  (B) r  (C) r  (D) r  hdg 2hdg 2T cos  cos  In a capillary tube water rises by 1.2 mm. The height of water that will rise in another capillary tube having half the radius of the first is (A) 1.2 mm (B) 2.4 mm (C) 0.6 mm (D) 0.4 mm Water rises in a vertical capillary tube upto a beight of 2.0 cm. If tube is inclined at an angle of 600 with the vertical then the what length the water will rise in the tube. 4 cm (A) 2.0 cm (B) 4.0 cm (C) (D) 2 2 cm 3 The lower end of a glass capillary tube is dipped in water rises to a height of 8 cm the tube is then broken at a height of 6 cm. The height of water column and engled as contact will be

(S.T. of water =  171.

172.

173.

174.

(A)

6 cm . sin 1

3 4

(B)

6 cm . sin 1

4 5

3 1 (D) 6 cm . sin 1 4 2 175. A large number of water drops each of rdius r combine to have a drop of radius R. If the surface tension is T and the mechanical equllvalent at heat is J then the rise in tempreature will be 3J  1 1  2T  1 1  2T 3T       (A) (B) (C) (D) J r R  J r R  rJ RJ

(C)

6 cm . cos 1

Caraphical Question 176. The correct curve between the height or depression h of liquid in a capillary tube and its radius is

228

177. A soap bubble is blown with the help of a mechanical pump at the mouth-of a tube the pump produces a certion increase per minit in the volume of the bubble irrespective of its internal pressure the graph between the pressure inside the soap bubble and time t will be

178. Which graph present the variation of surface tension with temperature over small temperature ranges for coater.

Comprehension type question Passage - I When liquid medicine of desting S is to be put in the eye. It is done with the help of a dropper as the bulb on the top of the dropper is pressed a drop forms at the opening od the dropper we wish to estimate the size of the drop. We dirst assume that the drop formed at the opening is spherical because the requires a minimum increase in its surface energy. To determine the size we calculate the net vertical force due to surface tension T when the radius od the drop is R. When this force becomes smaller than the weight of the drop the drop gets detched from the dropper. 179. If the radius od the opening od the dropper is r ; the vertical force due to the surface tension on the drop of radius R. (cassuming r << R) is (A)

2rT

(B)

(C)

2R T

2 r2 T R

(D)

2R2 T r

180. If   5  10 4 m,   103 kg m 3  10 ms 2 T = 0.11 N m-1 the radius of the drop when it detaches from the dropper is approximately (A)

1.4  10 3 m

(B)

3.3  10 3 m

229

(C)

2.0  10 3 m

(D)

4.1  10 3 m

181. After the drop detaches its surface energy is (A)

1.4  10 6 J

(B)

2.7  10 6 J

(C)

5.4  10 6 J

(D)

8.1  10 6 J

Assertion & Reason type questions Read the assertion and reason carefully and mark the correct option given below. (a) If both assertion and reason are true and the reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not the correct explanation of the assertion. (c) If assertion is true but reason is false. (d) If the assertion and reason both are false. 182. Assertion : It is better to wash the clothes in cold soap solution. Reason : The surface tension of cold solution is more then the surface tension of hot solution. (A) a (B) b (C) c (D) d 183. Assertion : When height of a tube is less then liquid rise in the capillary tube the liquid does not overtow. Reason : Product of radius of meniscus and height of liquid incapilling tube always remains constant. (A) a (B) b (C) c (D) d 184. Assertion : The impurities always decrease the surface tension of a liquid. Reason : The change in surface tension of the liquid depends upon the degree of cont amination of the impurity. (A) a (B) b (C) c (D) d 185. Assertion : The concept of surface tension is help only for liquids. Reason : Surface tension does not hold for gases. (A) a (B) b (C) c (D) d 186. Assertion : The water rises higher in a capillary tube of small diametre than in the capillary tube of large diamtre. Reason : Height through which liquid rises in a capillay tube is inversely proportional to the diameter of the capillary tube. (A) a (B) b (C) c (D) d 187. Assertion : Tiny drops of liquid resist deforming forces better than bigger drops. Reason : Excess pressure inside a drop is directly proportional to surface tension. (A) a (B) b (C) c (D) d

Pressure and Density 188. When a large bubble rises drom the bottom of a lake to the surface. Its radius double S. It atmospheric pressure in euqal to that of colurnn of colurnn of water height H then the depth of Lake is (A) H (B) 2H (C) 7H (D) 8H 230

189. A trangular lamind of area A and, height h is immersed in a liquid of density S in a vertical plane with its base on the surface of the liquid. The thrust on lamina is (A)

1 Agh 2

(B)

1 Agh 3

(C)

1 Agh 6

(D)

2 Agh 3

190. The density S of coater of bulk modulus B at a depth y in the ocean is related to the density at surface so by the relation. (A)

 gy     0  1  0  B  

(B)

 gy     0  1  0  B  

(C)

 gyh     0  1  0 B  

(D)

 B    0  1   0gy  

191. By sucking through a straw, a student can reduce the pressure in his lungs to 750 mm of Hg (density  13.6

gm ) using the straw, he can drink water from a glass up to a maximum depth of cm 3

(A) 10 cm (B) 75 cm (C) 13.6 cm (D) 192. The pressure on a swimmer 20 m below the surface of coater at sea level is (A) 1.0 atm (B) 2.0 atm (C) 2.5 atm (D)

1.36 cm 3.0 atm

Pascal's Law and Archimedes Principal 193. A spherical solid ball of volume V is made of a material of density S. It is falling through a liquid of density S2 (S2 < S1). Assume that the liquid opplies a viscous force on the ball that is proportional to square of its speed V. i.e. Fviscous = - KV2 (K > 0) The terminal speed of the ball is (A)

vg K

vg K

(B)

(C)

vg  1  2  K

vg  1  2 

(D)

K

194. The fraction of floating object of volume VO and density do above the surface of a Liquid as density d will be (A)

d0 d

(B)

d d0 d  d0

(C)

d  d0 d

(D)

d d0 d  d0

195. A body floats in water with one-thired od its volume above the surface of water. It is placed in oil it floats with half of : Its volume above the surface of the oil. The specific gravity od the oil is. (A)

5 3

(B)

4 3

(C)

3 2

(D)

1

196. If there were no gravity which of the following will not be there for a fluid. (A)

Viscosity

(B)

Surface tension

(C)

pressure

(D)

Archime des's upward thrust

231

197. A piece of solid weighs 120 g in air, 80 g in water and 60 g in liquid the relative density of the solid and that of the solid and that of the liquid are respectively. (A)

3, 2

(B)

2,

3 4

(C)

3 ,2 4

(D)

3,

3 2

198. Ice pieces are floating in a beaker A containing watre and also in a beakre B containing miscible liquid of specific gravity 1.2 Ice melts the level of (A) water increases in A (B) water decreases in A (C) Liquid in B decrease B (D) Liquid in B increase

Fluid Flow 199. An engine pumps water continuously through a hose water leares the hose with a velocity V and m is the mass per unit length of the watre Jet what is the rate at which kinetic energy is imperted to water. (A)

1 mv 3 2

(B)

mv 3

(C)

1 mv 2 2

(D)

1 mv  v 2 2

200. The height of the dam in an hydro electric power station is 10 m. In order to generate 1 MW of electric power, the mass of water (in kg) that must full per second on the brades of turbine is (A) 106 (B) 105 (C) 103 (D) 104 201. Eight drops of a liquid of density 3 and each of radius a are fallingt through air with a constant velocity 3.75 cm S1 when the eight drops coalesce to form a single drop the terminal velocity of the new drop will be (A)

15  10 2 ms 1

(B)

2.4  10 2 m / s

(C)

0.75  10 2 ms 1 (D) 25  10 2 m / s

202. A cylinder of height 2.0 m is completely filled with water. The velocity of efflux of water cim m/s through a small hole on the side will of the cylinder mear its bottom is (A) 10 (B) 20 (C) 25.5 (D) 5 203. There is a hole in the bottom of tank having water. If total pressure at bottom 3 atm (1 atm ) then the velocity of water flowig drom hole is (A)

400

m s

(B)

60

m s

(C)

600

m s

(D)

None of these

204. Two drops of the same radius are falling through air with a steady velocity for 5 cm per sec. If the two drops coakesce the terminal velocity would be (A) 10 cm per sec (B) 2.5 cm per sec (C)

1

5  43 cm per sec

(D)

5 2 cm per sec

205. An application of Bernouli's equation for liquid flow is found in (A) Dynamic lift of an aeroplane (B) Viscocity metere (C) Capillary rise (D) tly dulic press

232

206. An L-Sp aped tube with a small office is held in a water stream as shwon in fig. The upper end of the tube is 10.6 cm above the surface of water. What will be the height of the set of water coming from the office velocity of water stream is 2.45 m/s. (A) zero (B) 20.0 cm (C) 10.6 cm (D) 40 cm 207. A tank is filled with water up to a height H. Water is allowed to come out of a hole P in one of the walls at a depth D below the surface of water express the horizontal distunce x in terms of H and D.

D H  D  2

(A)

x   D H  D 

(B)

x

(C)

x  2 D H  D 

(D)

x  4 D H  D 



208. An incomepressible fluid flows steadily through a cylinderical pipe which has radius 2r at point A and radius r at B further along the flow direction. It the velocity at point A is V, its velocity at point B. (A)

2v

(B)

v

v 2

(C)

(D)

4v

kg   209. If the terminal speed of a sphere of gold  density  19.5 3  is 2.0 m/s in a viscous liquid m   kg  kg     density  1.5 3  find the terminal speed of a sphere of siher  density  10.5 3  of the m  m    same size in the same liquid. m m m m (A) 0.133 (B) 0.2 (C) 0.1 (D) 0.4 s s s s 210. Two solid spheres of same metal but of mass M and 8M full simutineously on a viscous liquid and their terminal velocity are V and 'nv' then value of 'n' is (A) 16 (B) 8 (C) 4 (D) 2

211. Two metal Spheres are falling through a liquid of density 2 10 the meterical density at sphere 1 and sphere 2 are 8 10

3

3

kg with the same uniform speed m3

kg kg 11 103 3 respectively. The 3 and m m

ration of their radii is. (A)

11 8

(B)

11 8

(C)

233

3 2

(D)

3 2

212. Water is flowing continuously dram a temp having an internal diameter 8  103 m . The water velocity as it leaves the tap is 0.4 m/s. The diameter of the water stream at a distance 2  101 m below the tap is close to (A)

5.0  10 3 m

(B)

7.5  10 3 m

(C)

9.6  10 3 m

(D)

3.6  10 3 m

213. A large open tank has two holes in the wall one is a square hole of side L ata depth y froam the top and the other is a circular hole of radius R at a depth ay from the top. When the tank is completely filled with water the quantities of water following out per second from both the holes are the some then R is equal to (A)

(B)

2L

L 2

(C)

L

(D)

L 2

214. A block of ice floats on a liquid of density 1.2 in a beaker then level of liquid when ice completely melt. (A) Remains same (B) Rises (C) Lowers (D) (A) (B) or (C)

Assertion & Reason type questions Read the assertion and reason carefully to mark the correct option out of the options given below (a) If both assertion and reason are true and the reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not the correct explanation of the assertion. (c) If assertion is true but reason is false. (d) IF the assertion and reason are false. (e) If assertion is flase but reason is true. 215. Assertion : The blood pressure in humans is greater at the feet than at the brain. Reason : Pressure of liquid at any point is proportional to height clensity of liquid. (A) a (B) b (C) c (D) d 216. Assertion : To empty an oil tank two holes so it will made. Reason : Oil will come out of two holes so it will be emptied faster. (A) a (B) b (C) c (D) d 217. Assertion : A bubble comes from the bottom of a lake to the top. Reason : Its radius increases. (A) a (B) b (C) c (D) d

Ordinary Thinking Thermometary 218. Oxygen boils at 1830C. This temperature is approximately. (A) 2150 F (B) -2970 F (C) 3290 F 234

(D)

3610 F

219. The resistance of a rasistance thermometer has values 2.71 and 3.70 ohm at 100C and 1000C. The temprature at which the resistance is 3.26 ohm is (A) 400 C (B) 500 C (C) 600 C (D) 700 C 220. Maximum density of H2O is at the temprature. (A) 320 F (B) 39.20 F (C) 420 F (D) 40 F 221. At what temprature the centigrade (celsius) and Fahrenheit readings at the same. (A) -400 (B) +400 C (C) 36.60 (D) -370 C 222. Mercury thermometers can be used to measure tempratures up to (A) 1000 C (B) 2120 C (C) 3600 C (D) 5000 C 223. If temperature of an object is 1400 F then its temperature in centigratde is (A) 1050 C (B) 320 C (C) 1400 C (D) 600 C 224. When the room temprature becomes equal to the dew point the relative humidity of the room is (A) 100 % (B) 0 % (C) 70 % (D) 85 % 225. If the length of a cylinder on heating increases by 2% the area of its base will increase by. (A) 0.5 % (B) 2 % (C) 1 % (D) 4 % 0 0 226. Density of substance at 0 C is 10 gm/cc and at 100 Cits density is 9.7 gm/CC. The coefficient of linear expansion of the substance will be (A) 102 (B) 10-2 (C) 10-3 (D) 10-4 227. A beaker is completely filled with water at 40C It will overflow if (A) Heated above 40C (B) Cooled below 40C (C) Both heated and cooled above and below 40C respectively (D) None of these 228. An iron bar of length 10m is heated from 00C to 1000C. If the coefficient of linear thermal expansion 10 106 of iron is the increase in the length of bar is C (A) 0.5 cm (B) 1.0 cm (C) 1.5 cm (D) 2.0 cm

Calorimetry 229. Melting point of ice..... (A) Increases with increasing pressure (B) Decreases with increasing pressure (C) Is independent of pressure (D) is proportional of pressure 230. Amount of heat required to raise the temprature of a body through 1k is called it is (A) Water equivalent (B) Thermal capacity (C) Entropy (D) Specific heat 231. A vessel contains 110 g of water the heat capacity of the vessel is equal to 10 g of water The initical temprature of water in vessel is 100C If 220 g of hot water at 700C is poured in the vessel the Final temperature meglecting radiation loss will be (A) 700 C (B) 800 C (C) 600 C (D) 500 C 235

232. In a water fall the water falls from a height of 100 cm. If the entire K.E. of water is converted in to heat the rise in temperature of water will be (A) 0.230 C (B) 0.460 C (C) 2.30 C (D) 0.0230 C 233. The temprature at which the vapour pressure of a liquid becomes equals of the external pressure is its. (A) Melting point (B) sublimation point (C) Critical temprature (D) Boiling point 0 234. 10 g of ice at 0 C is mixed with 100 g of water at 500 C what is the resultant temprature of mixture. (A) 31.20 C (B) 32.80 C (C) 36.70 C (D) 38.20 C

Critical Thinking 235. Two metal strips that constituate a thermostant must necessarily in their (A) Mass (B) Length (C) Resistivity (D) Coefficient of liner expansion 0 236. 2 kg of Ice at -20 C is mixed with 5 kg of water at 200 C in an insulating vessel having a megligible heat capacity calculate the final mass of water remaining in the container. It is given that the specific heats of water and ice care 1 keal/kg per0C and 0.5 Keal/kg 10C while the latent heat of fusion of ice is 80 kcoil/kg. (A) 7 kg (B) 6 kg (C) 4 kg (D) 2 kg 0 237. A lead bullet at 27 C just melts when stopped by an obstancle Assuming that 25% of heat is obsorbed by the obstacle then the velocity of the bullet at the time of striking. [M.P. of lead = 3270 C, specific heat of lead = 0.03 cal/x latent heat of fusion of lead = 6 cal/g and J = 4.2 Jute / cal] (A) 410 m/s (B) 1230 m/s (C) 307.5 m/s (D) None of these 238. An electric kettle takes 4A current at 220V How much time will it take to boil 1 kg of water from temprature 20C ? The temprature of boiling water is 100 C. (A) 12.6 min (B) 4.2 min (C) 6.3 min (D) 8.4 min

Graphical options 0

239. Ablock of ice at -10 C is slowly heated and covered to steam at 1000 C which of the following curves represents the phenomenon qualitatively.

236

240. The variation of density of water with temprature is represented by the

Assertion & Reason

241.

242.

243.

244.

245.

246.

Read the assertion and reason carefully to mark the correct option out of the option given below. (a) If both asseration and reason are true and the reason is the correct explanation of the reason. (b) If both assertion and reason are true but reason is not the correct explanation of the assertion. (c) If assertion is true but reason is false. (d) If the assertion and reason both are false. (e) If assertion is false but reason is true. Assertion : The melting point of the ice decreases with increases pressure Reason : Ice contracts on melting (A) a (B) b (C) c (D) d Assertion : Fahrenheit is the smallest unit measuring temprature. Reason : Fahrenheit was the first temprature scale used for measuring temprature. (A) a (B) b (C) c (D) d Assertion : Melting of solid causes no change in internal energy. Reason :Latent heat is the heat required melt a unit mass of solid. (A) a (B) b (C) c (D) d Assertion : Specific heat capacity is the cause of formation of land and sea breeze. Reason : The specific heat of water is more then land. (A) a (B) b (C) c (D) d 0 0 Assertion : The moleculs of 0 C ice and 0 C water will have same potential energy. Reason : Potential energy depends only on temprature of the system. (A) a (B) b (C) c (D) d 0 Assertion : A beaker is completely filled with water at 4 C. It will overflow both where heated or cooled. Reason : There is expansion of water below and above 40C. (A) a (B) b (C) c (D) d 237

Answers Key 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 30 31 32 33 34 35 36 37 38 39 40 41

D B D B C B D D C C A B D B B B B D D D B B C B C D B D A A B D D D D D A C C A D

42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82

C A B D A D A A B D B D C B A A D A C C B A C C C A B C B A B A B C B A C D B C B

83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123

B B D A D B C A D B C C B A C D C A A B D D B C B C A D D A D A D A A B C B B B D

124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 238

A D A A D A A D B B C A B C A A A A D D C A D C D D D D C C B A B D C A C B C B C

165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205

C D A B B D A B B C C B A B C A B E A C B A B C B B C B D C B D D D A D A B A C A

206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246

B C D B C D D B B A C A B B B A C D A D D C B B B D A D D D B A C A A A C E A D A

Hint 1.

FL FV  V  2 V  A  L L   Ay A y  A



A2 y  F V

 F  A2

As cross sectional area of 2nd wire is 3 times therefore 9F force is required for same elongation. 2.

Stress  strain  stress 

3.

Y

F L L L   A  A d 2



L (As F and Y are constant) d2

The ratio of 4. 5.

6.

 L

L is maximum for case CD d2

Young modulus of wire does not vary with dimension of wire. It is the property of given material. Depression in beam.



wL2 4Y3bd3

L

1 r2

 

1 Y

If radius of the wire is doubled then increament in length will become 1/4 times, i.e. 7.

8.

9.

F

YA L

F Y  A Given stress 3.18  108 N2  m  kp Here KQ  2

 l 

According to Hooke's law

 Fp  k p x p

Fp  FQ (Given)



xp xQ



239

kQ kP

F/A Y

....1

12  3 mm 4

Energy stored in a spring is Up



10.

UQ



k p x 2P kp k 2Q 1    k q x 2 Q kQ k 2p 2

kp    kQ  2   

 UP 

UQ E  2 2

 

Breaking force  area of crossection of r 2 wire If radius of wire is doubled then breaking force will become four times. L2 2Y

11.



12.

Young's modulus of material 

13.

Linear stress Longitudin al strain

If longitudinal strain is equal to unity, then Y = Linear stress produced Energy stored per unit volume 

1  stress  strain 2

14.

stress Max . stress Y  Max strain  Strain V

15.



16.

force 1   stress  2 stress  Area r



mg Max strain  A Y



2

L2 dg 8  10 2  1.5  9.8   9.6  10 11 m 8 2Y 2  5  10 2

S  rA  2  S B   rB   2  SB  4 SA   A

17.

Breaking force  area of cross section of wire. i.e. load hold by the wire does not depend upon the length of the wire.

18.



19.

I   A A  10  6 m 2 Y           

20.



FL AY

 

1 (F, L & Y are constant) r2

FL 1    2 (F, Land Y are same) AY r

240

FL  Ya1  Steel  AY  a1 YSteel (F, Land Y are constant)

21.



22.

If length of the wire is doubled then strain = 1

23.

L

24.

F = force developed  YA  A

25.

 

26.

F F Y A A strain Yx strain

27.

Strain & stress 

28.

Elongation in the wire  

 Y  stress 

Force Area

P dg

lateral strain longitudinal strain

F A TL AY

 Elongation in wire tension in the wirein first case T1  wand in second case

T2 

2w w ww

As Tension in the wire in both the cases are equal.

 Elongation in the wire will be equal.

29.

Y

FA Al

30.



F 1    2 (F, Land Y are constant) 2 r y r

31.

Longitudinal strain =

32.

It is the specific property of a particular material at a given temperature which can be changed only by temperature variations.

33.

When the length of wire is doubled then   L and strain = 1  Y  strain 

34.

l stress = L Y

F A

 r2 FYA F (y and L are constant) L L 241

r 7  r0 3  10 10

35.

Y

36.

Increase in tension of wire  YA  Q

37.

F F Y  A  strain  strain AY

38.

W

39.

Potential energy stored in the rubber cord calapult will be covered into kinetic energy of mass

1 AL2 Y 2 L

1 1 YAL2 mv 2  2 2 L 40.



FL AY



1 (F, L and Y are constant) A

2 A 4 1  0.1  1    2  1   0.05mm 1 A 2 8 2 2 2

41.

Y

stress  cons tan t strain

42.



FL AY

43.



FL AY

44.

K  Yro

45.



It depends only on nature of material.

 Final length = initial lengh + increament = 2L

FL FL2 FL2   AY AL Y VY

   L2 If the volume of wire remains constant. FL FL F  2     2 (Y = constant) AY r Y r

46.



47. 48.

Breaking stress = Strain  young's modulus In accordance with Hook's law

49.

F  A  Y  Strain 242

50.

Because strain is a dimensionless and unitless quantity.

51.

stress 

52.

In the present case, force applied and area of cross-section of wires are same. Therefore stress has to be the same. Strain = stress /Y Since, the young's modulus of steel wire is greater than the copper wire, therefore strain in case of steel wire is less than that of in case of copper wire Knowledge base

53.

Breaking force  r 2

force area

If diameter becomes double then breaking force will become four times i.e.1000  4  4000 N 54.

FL FL2 FL2    AY AL Y AY If volume is fixed then   L2

55.

Young's modulus 

stress strain

As the length of wire get doubled therefore strain = 1  Y  strain  20  108

N m2

56.

Because dimension of invar does not vary with temperature.

57.



58.

 2   2 1   2 Q and L1   1  1  1  Q

FL r 2 y

 

L (Y & F are constant) r2

  2  1   l 2  l1   Q  2  2  1 1  now L 2  L1    2   1  So,  2 2  11  0 59.

Thermal stress = y   Q

60.

F

61.



YA L

FL 1  r 2  (F, Land l are constant) 2 r y Y

243

62.



FL r 2 y

 

1 (F, Land Y are constant) A

A2   1 A1 2

63.

Y

MgL A

64.



FL AY

65.

L

mgL YA

66.

Work done 

67.

w

 

F r2

1 Mg F  2 2 2

1 stress   Volume 2 Y

As F, A and Y are same - W  volume (area is same) w

(V = A1)

w1   1  1  1  w2  2 2 2

68.

w

1 F 2

69.

K

F 1 1 1 and w  F  k    k 2  2 2 2

70.

1 F2  L U  F  ; U  2 (F and Y are constant) 2 2AY r

71.

U

1 F 2

72.

U

1 stress    Volume 2 Y

73.

When a wire is stretched through a length then work has to be done. This work is stored in the wire in the form of elastic potential energy. Potential energy of stretched wire is

2

U

1  stress  strain 2

U

1 1  F  6  U  Fx 2 2

244

74.

U

1 2  YX strain  2

75.

w

YA 2 2L

76.

U

F2 T2  2K 2K

77.

Energy stored per unit volume 

78.

1 YA 2 U 2 L

79.

At extension l1 the stored energy 

1 F FL     2  A   L  2AL

1 2 k 1 2

 At extension l2 the stored energy 

Work done in increasing its extension from l1 to  2 

1 2 2 k  2  1 2

F X

80.

K

81.

Isothermal elasticity ki = P

82.

B

 PV V  B  V 1 P given  1%  Y V V 100

P

1.5  1010 N  1.5  108 2 100 m

83.

Ratio of adiabatic & isothermal elasticities  V P C  V P  P CV

84.

For triatomic gas V 

85.

k

86.

C

4 3

P  v v v P  0.4 CC v

245





1 2 k 1 2

P hsg  v v v v

87.

k

88.

3 If side of cube is L then V  L 

89.

k

90.

If the coefficient of volume expansion is  and raise in temprature is  then

dv 3dL  v L

100 N dynes  106 atm 1011 2  1012 0.01 m cm 2 100

v  v  

v     volume elasticity   P  P v v   2 v If side of the cube is L then V = L 

91.



dv 3dL  v L

% change in voulume  3  (% change in length)  3  1%  3% v  0.03 v Adiabatic elasticity   vp  Bulk modulus 

92.

For Argon EAr = 1.6 p For hydrogon EH2 = 1.4 pl As elasticity of hydrogen & Argon are equal  1.6 P  1.4 Pl

 P' 

8 P 7

93.

5 Isothermal elasticity P  Ki  1 atm  1.013  1.013  10

94.

Isothermal bulk modulus = pressure of gas

95.

dv dL  1  2  v L

96.

We know that

dv

3 3  v  2  2  10  4  10

dv dL  1  2   v L

If  

i.e. there is no change in volume 246

N m2

1    6  0.5 2   

1 dv then 0 2 v

97.

Y = 2N ( 1 +  )

98.

Y  2n 1      

99.

 n r4  Twisting coulple C  2

0.5 y  n n

If material and length of the wires A and B equal twisting coulple are applied then 1  4 r

r    1   2  2  r1 

4

100. Value of possion's ratio die in range of 1 to 1/2. 101. y  3k 1  2; y  2n 1   For Y=0 we get 1  2   0 also 1    0

  lies between 102. y  2n 1    3n  2n 1  



3 1 1 2 2

Now, substituting the value of  in the following expression

k

y  3 1  2 

103. Y  2n  1   104. There will be both shear stress and normal stress. 105. Angle of shear  

r 4  10 1   0.12' L 100  30'

106. For twisting, angle of shear  

1 i.e. if L is more then  will be small L

107. r   L  10 2  0.8  2    0.004 108. Y  3K 1  2  and Y  2n 1  

9nk Eliminating  we got Y  n  3k

109. Poisson's ratio varies between - 1 and 0.5 110. T  2 

n  T2  M k

If we draw a graph of between T2 & M then it will be straight line and for M  0 ; T 2  0 i.e. graph should pass through the origin but from the graph it is not reflected it means the mass of pan was neglected. 247

111. In the region OA, stress  strain . i.e.Hooke's las hold good. 112. As stress is shown on X axis and strain on Y axis so we can say that Y  cot  

1 1  tan  slope

So elasticity of wire P is minimum and of wire R is maximum. 113. Area of hysterisis loop gives the energy loss in the process of stretching and understretching of rubber band and this loss will appear in the form of heating.

YA tan  A  YB tan  B 115. i.e.for the same load thickest wire will show minimum elongation so graph D present the thickest wire. 114.

116. From the graph Y 

  10  4 M, F  20 N

A  10  6 M 2 ,   1 M

FL 20  1 N 10  6  2  1011 2  4  20  10 Al 10  10 m

117. At point b yeilding of material starts. 118. Graph between applied force and extension will be straight line because in elastic range. Applied force extension. but the graph between extension and stored elastic energy will be parabolic in nature. As U 

1 kx 2 or U  x 2 2

 dv  119. F      dx 

In the region BC slope of the graph is positive.

 F  negative i.e. force is attractive in nature.

In the region AB slope of the graph is negative.

 F  positive i.e. force is repulsive in nature.

120. Force constant K = tan 300 

1 3

121. In ductile materials, yeild point exist which in brittle material failure would occur without yeilding. 122. Young's modulus is defined

Y

stress strain

123. Elasticity of wire decreases at high temperatare i.e. at higher temprature slope of graph will be less, So T1 > T2. 124. Attraction will be minimum when the distance b/w the molecule is maximum. Altraction will be maximum at that point where the positive slope is maximum b'sc F 

du dx

248

125. B'se stretching of coil simply chnages its shape without any change in the length of the wire used in coil Due to which shear of elasticity is prevolved. 126. A bridge during its use undergoes alternating strains for a large number of times each day depending upon the movement of vehicle on it. When a bridge is used for long time. It losses its strength Due to which the amount of strain in the bridge for a given stress will become large and ultimately the bridge may collapse. This may not happen if the fridges are declared unsafe. 127. Ivory is more elastic than wet-clay. Hence, the ball of ivory will rise to a greater height. Infact the ball of wet day will not rise at all it will be same, what flattended permanently. 128. Young's modulus of a material. Y  Here, stress force 

stress strain

Re staring force Area

As restoring force is zero.  Y  0

1 1 2  stress  strain   Y  strain  Since, elasticity of steel is more than copper,, 2 2 hence, more work has to be done in order to stretch the steel.

129. Work done 

Ft 130.  A x y W 133. Y  A A  139. Y 

F L A L

249

Hint 143. Soap helps to lower the surface tension of solution thus soap get stick to the dist partcles and grease and these are removed by action of water. F

145. Force required to separate the plates

2TA t

147.

 weight of displaced coater  T cos   2 r   w  2rT cos 

148. 2TL  mg

T



149. w  8T R 2 2  R 12



150. w  T  8 r2 2  r12

mg 2L



 8 T 2r   r 





 D2 d2   T  8    4 4  

2

2



 24  r 2 T





 2 D 2  d 2 T

151. Energy spent = T  increase in surface ared



 T  2 4 2r   4r 2 2







 24  T r 2 Joule

152. w  T  A 153. Surface energy = Surface tension  surface ared E  T  2A

A New surface energy F1  T  2   2

% decrase in surface energy 

E  Ei  100 E 1

1

154. The ration of the total surface energies before and after the change  n 2 : 1  2 3 : 1 155. w  8 R 2 T 156. w  T  A

157.

T

w A

4  R3 3

250

158. As volume remains constant R 3  8000r 3 R  20 r Surface energy of one big drop 4R 2T  Surface energy of 8000 small drop 8000 4 r 2  1 159. Tension 

surface energy Area

or T 

E A

160. Angle of contact is a 00 165. p 

2T R

167. rA  rB and p 

1 So PA  PB r

So air will flow from B to A 168.

4T  hdg R

T 

i.e. size of A will increase

Rhdg 4

 3.9  10  2

N m

169. Mass of liquid in capillary tube 1 M  R2    R M  R If radius become double then mass will become twice.

M  R2  

172. h  173.  

2T rdg

h1

 r1 h1  r2 h 2

r

h    4.0 cm cos  cos 60

174. When a capillary tube is broken at a height of 6 cm the height of water column will be 6 cm. As h 

25 cos  h or  cons tan t rg cos 

8 6 6 cos 0o 3  or cos    cos 0o cos  8 4

1  3     cos  

4

175. Rise in tempreture  

176. h 

3T  1 1     Jsd  r R 

2T cos  rdg

p 

4T r

 

3T  1  1    J r R 

(For water S = 1 and d = 1)

h 

1 r

so the graph between h and r will be rectangular hyperbold

p 

1 r

As radius of soap bubble increases with time  P  251

1 t

178. TC  To I  t  i.e. Surface tension decreases with increase in temperature 179. Due to surface tension vertical force on drop

Fv  T2  r Sm   T2  r

r 2 r2 T R R

2  r2 4 T    R3 . 9 180. R 3



3 181. U  T  A  0.11  4  1.4  10

183. h 

184.

185.

186.

187.

2T 2T  hR  Rdg dg



2

 hR  constant

Hence when the tube is of insufficient length radius of curvature of the liquid meniscus increasses so as to maintain the product hR a finite constant. i.e. as h decreases R increases and the liquid meniscus becomes more and more flat but the liquid does not overflow. The presence of impurities either on the liquid surface or dissolved in it considerably affect the force of surface tension depending upon the degree of surface tension depending upon the degree of contamination. A highly soluable substance like sodium chioride when dissolved in water increase the surface tension. But the sparing soluable or substance like phenol when dissolved in water reduces the surface tension of water. We know that the intermolecular distance between the gas moleculas is Large as compaired to that of liquid Due to it the forces of cohesion in the gas moleculas are very small and these are quite Large for liquids. Therefor the concept of surface tension is applicable to Liquid but not to gasses. The height of capillary rise is inversly propostional to radius (or diametre) of capillary tube 1 i.e. h  so, for smaller r the value of his higher r When a drop of Liquid is poured on a glass, plate, the shape of the drop also is governed the force of gravity for every small drops the potential energy due to gravity is insignification. Compared to that due to surface tention. Hence, in this case the shape of the drop is determined by sufrace tension alone and drope becomes spherical.

188. p1 v1  p 2 v 2

p  hg  4 3

 r3  p0 

4 2r 3 3

252

189. Thrust on lamina = pressure at centroid  area 

hg 1 A  Agh 3 3

190. Bulk modulas B   V0

p    v  v0  1  B  

p p  v  v 0 v B

p   Density   0  1  B  

1

p    0  1  B  

191. = 760 - 750 = mnat Hg = 4g 192. Here, h = 20 m kg m3

3 Density of water   10

Atmospheric pressure Pa  1.01  105 pa p  pa  gh 193. Weight of the ball = Buyoant force + viscous force 194. For the floatation Vo dog = Vin dg Vin  V0

do d

Vout  V0  Vin  V0  V0

195. Weight of body Weight of water displaced Specitic grvity of oil 

do d

Weight of oil displaced

0 4  w 3

196. Aschemedies principal explains buoyant force and bouant force depends on acceleration due to gravity 204. If two drops of same radius r coalesce then radius of new drop is given by R 4 4 4  R 3   r 3   r 3  R 3  2r 3 3 3 3 1

R  2 3 r Is drop of radius r is falling in viscous medium then it acquire a critical velocity V and V r 2

 13  2 r v2  R      v1  r   r   

2

2

2 3

2 3

v 2  2 v1

2 5 253

1

 5  4 3

m s

206. According to Bernouli's theoram 2

 2.45  v2   0.30 cm h  h   9  2  10   30.0 cm

207. Time taken by outer to reach the bottom = 2  H  D

t 

g

and velocity of water coming out of hole

 v  2gD Horizontal distance covered

x  vt 209. Terminal speed v 

2 r 2g    9 n

  Density of the substance

Where,

  Density of the liquid

If n and r are constant then

v     210. Mass = volume  Density  M

4 3 r   3

As the density remains constant M  r3 211. The terminal velocity of the spherical body of radius R density S falling through a liquid of density  is given by Vt 

2 2 R     g 9 n

where n is the coefficient of vescosity of the liquid. 2

VT1 

2 R1  1    9n

2

g

and Vt 2 

2 R2  2    9n

According to the given problem

254

g

212. Diameter  8  10 3 m V1  0.4

m s

v 2  v12  2gh A 1v 1 = A 2v 2 M 215. The volume of liquid displaced by floating ice VD   

Volume of water formed by melting ice, VF 

M w

M M  i.e VD  VF L w

-

Height of the blood column in the human body is more at feet than at the brain as P = hpg there face the blood exeists more pressure at the feet than at the brain. When to holes are made in the Hn air keeps on externing through the other hole Due to this the pressure inside the tin does not become less than at mosphere pleassure which happen only when hole is made.

218.

C F  32 183 F  32     F   297 o F 5 9 5 9

219. Change in resistonce 3.70  2.71  0.99  to interval of temprature 900 C so change in resistance 3.26  2.71  0.55  corresponds to change in temprature.



90  0.55  50 o C 0.99

220. Maxmimum density of water is at 40 C also C F  32  5 9

221.

C F  32  5 9

222. The boiling point of mercury is 4000 C. Therefore the mercury thermeter can be used to measure the range upto 3600 C. 223.

C F  32  5 9

2 225. A  L 

A L  2. A L

255

226. Coffcient of volume expansion

r

   2  1  . T  

Hence, cofficent of linear expansion 227. Water has maximum density at 40C so if the water is heated above 40C or Density cooled below 40C density decreases. In other words it expands so it overflows in both cases. 228. Increase in length L  Lo  

Calorimetry 229. Melting point of ice decreases with increase in pressuire. 230.   m. c ; if   1k then   mc  Thermal capacity 231. Let final temprature of water bc  heat taken = Heat given

110  1  10  10  10  220  1 70      48. 8o C  50o C 232.   0.0023h  0.0023  100  0.23o C 233. At boiling point vapour pressure becomes equal to the external pressure. miLi cw mi  mw

mw w  234.  mix 

Critical Thinking 235. Thermostat is used in electric opporatas like refrigerator iron etc for automatic cut off Therefore for metallic strips to bend on heating their coefficient of linear expansion should be different. 236. Initially ice will absorb heat to rise its temprature to 00C then its melting takes place. If mi = Initial mass of ice mi-1 = Mass of ice that melts and mw = Initial mass of water By low of mixture Heat gained by ice = Heat lost by water  mi  c  20   mi  L  mwC w 20  2  0.5 20  mi  80  5 1  20  mi 7  1 kg So final mass water Initial mass of water + mass of ice that melts = 5 + 1 = 6 kg 237. If mass of the bullet is m gm then total heat required for bullet to just melt down.

1  mc  mL Now when bullet is stopped by the obstacle the loss in its mechanical energy  (As mg  m  10 3 kg ) As 25% of this energy is absorbed by the 256

1 cm  10 3 ) v 2 9 2

75 1 3 2 3 2 3 2  100  2 mv  10  8 mv  10 J

Now the bullet melt if  2  1 238. P  t  mc

t

mc 4200 m 4200  m     p p VI

  1  oC   water  4200 kg    t

4200  1  100  20   381sec  6.3 min 220  4

Graphical Optaions 239. Initially on heating temprature rises from -100 C to 00 C Then ice melt and temprature does not rise After the whole ice has melted temprature begins to rise until reaches 1000 C Then it becomes constant as at the boiling point will not rise. 240. Density of water is maximum at 40 C and is less on either side of this temprature.

Assertion & Reason 241. With rise in pressure melting point of ice decreases also ice contracts on melting. 242. Celsius scale was the first temprature scale and fahrenheit is the smallest unit measuring temprature. 243. Melting is associated with increasing of internal energy without change in temprature in view of the reason being correct the amount of heat absorbed or given out during change of state is expressed where m is the mass of the substances and L is the latent heat of the substance. 244. If bath assertion and reason are true and the reason is the correct explanation of the assertion. 245. The potential energy of water molecules is more. The heat given to melt the ice at 00C used up in increasing the potentical energy of water molecules formed at 00 C 246. Water has maximum density at 40 C on heating above 40 C or cooling below 40C density of water decreases and its volume increases. Therefore water overflows in both the cases.

257

SUMMARY    

Thermal equilibrium, teroth law of thermodynamics, Concept of temperature. Heat, work and internal energy. First law of thermodynamics. Second law of thermodynamics, reversible and irrevessible processes. Isothermal and adiabatic process. Carnot engine and its efficency Refrigerators.



C  0 F  32 C F    0 0 0 100 180 100 1800



w  P  V (Isothesmal process)  RT  n

v2 v1 (Isothesmal process)

PV = Const (Isothesmal process) P v = Const (adiabasic process) P1- T = Const TV-1 = Constant r=

cp cv

Monoatomic gases C v = Diatomic gases C v =

5 7 R, Cp = R, γ = 1.4 2 2

Polgatomic gases C v =

W=

= 

3 5 R, = Cp = R,  = 1.67 2 2

7R 9R  , Cp = , = 1.4 2 2

1 (P V  P2 V2 ) (adiabafic Process) γ 1 1 1

mR (T1  T2 ) γ 1

 Q   u   W (First law of thermodynamics) Q  U Cvt (V = Const) Q Cpt ( P = Const)

u  0 (isothesmal process cyelic process)

259



Coefricent of Performance of refrigesatos

n  QW1 Q1 Q 2



Q1





T1  T2 T1

Coefcient of performance of refrigetors. 

Q2 w



Q2 Q1  Q 2



T2 T1  T2



I deal gas Cp  Cv  R

1.

A difference of temperature of 250 C is equivalent to a difference of (A) 720 F (B) 450 F (C) 320 F (D) 250 F What is the value of absolute temperature on the Celsius Scale ? (A) -273.150 C (B) 1000 C (C) -320 C (D) 00 C The temperature of a substance increases by 270C What is the value of this increase of Kelvin scale? (A) 300K (B) 2-46K (C) 7 K (D) 27 K At Which temperature the density of water is maximum? (A) 40 F (B) 420 F (C) 320 F (D) 39.20 F The graph AB Shown in figure is a plot of a temperature of a body in degree Fahrenhit than slope of line AB is

2. 3.

4. 5.

(A)

5 9

(B)

9 5

(C)

260

1 9

(D)

3 9

6.

7.

8.

9.

10.

11.

The temperature on celsius scale is 250 C What is the corresponding temperature on the Fahrenheit Scale? (A) 400 F (B) 450 F (C) 500 F (D) 770 F The temperature of a body on Kelvin Scale is found to be x.K.when it is measured by Fahrenhit thesmometes. it is found to be x0F, then the value of x is . (A) 313 (B) 301.24 (C) 574-25 (D) 40 A Centigrade and a Fahrenhit thesmometes are dipped in boiling wates-The wates temperature is lowered until the Farenhit thesmometes registered 1400 what is the fall in thrmometers (A) 800 (B) 600 (C) 400 (D) 300 A uniform metal rod is used as a bas pendulum. If the room temperature rises by 100C and the efficient of line as expansion of the metal of the rod is, 2 106 0C1 what will have percentage increase in the period of the pendulum ? (A) -2  10-3 (B) 1  10-3 (C) -1  10-3 (D) 2  10-3 A gas expands from 1 litre to 3 litre at atmospheric pressure. The work done by the gas is about (A) 200 J (B) 2 J (C) 300 J (D) 2  105 J Each molecule of a gas has f degrees of freedom. The radio

(A) 1  12.

f 2

1 f

(C) 1 

2 f

(D) 1 

f

 1 3

Is the cyclic Process Shown on the V  P diagram, the magnitude of the work done is

 P2  P1   (A)    2 

(C) 13.

(B) 1 

CP = γ for the gas is CV



2

 V2  V1   (B)    2 

 P2V2  PV 1 1

(D)

2

 P2  P1 V2  V1  4

If the ratio of specific heat of a gas at Consgant pressure to that at constant volume is , the Change in internal energy of the mass of gas, when the volume changes from V to 2V at Constant Pressure p, is

PV (A) γ 1

(B)

R γ 1

(C) PV 261

(D)

γ PV γ 1

14. 15.

16.

17.

The change in internal energy, when a gas is cooled from 9270C ³ to 270C (A) 200% (B) 100% (C) 300% (D) 400% For hydrogen gas Cp - Cv = a and for oxygen gas Cp - Cv = b, The relation between a and b is given by (A) a = 4b (B) a = b (C) a = 16b (D) a = 8b In a thermodynamic process, pressure of a fixed mass of a gas is changed in such a manner that the gas release 20J of heat and 8J of work has done on the gas- If the inifial internal energy of the gas was 30j, then the final internal energy will be (A) 58 J (B) 2 J (C) 42 J (D) 18 J If for a gas

cp = 1.67, this gas is made up to molecules which are cv

(A) diatomic (C) monoatomic 18.

An ideal monoatomic gas is taken around the cycle ABCDA as Shown in the P  V diagram. The work done during the cycle is given by

(A) PV 19.

21.

(B)

1 PV 2

(C) 2 PV

(D) 4 PV

A given mass of a gas expands from state A to B by three different paths 1, 2 and 3 as shown in the figure. If W1,W2 and W3 respectively be the work done by the gas along the three paths, then

(A) W1>W2>W3 20.

(B) Polytomic (D) mixnese of diatomic and polytomic molecules

(B) W1<W2<W3

 

One mole of a monoatomic gas  γ 

(C) W1=W2=W3 (D) W1<W2, W1<W3

5 7   is mixed with one mole of A diatomic gas  γ   what 3 5 

will be value of  for mixture ? (A) 1-454 (B) 1-4 (C) 1-54 (D) 1-5 If du represents the increase in internal energy of a thesmodynamic system and dw the work done by the system, which of the following statement is true ? (A) du = dw in isothermal process (C) du = - dw in an aidabadic process (B) du = dw in aidabadic process (D) da = - dw in an isothermal process

262

22.

23.

24.

25.

One mole of a monoatomic ideal gas is mixed with one mole of a diatomic ideal gas The molas specific heat of the micture at constant volume is .......... (A) 4 R (B) 3 R (C) R (D) 2R One mole of a monoatomic gas is heate at a constant pressure of 1 atmosphere from 0k to 100 k. If the gas constant R = 8.32 J/mol k the change in internal energy of the gas is approximate ? (A) 23 J (B) 1.25  103 J (C) 8.67  103 J (D) 46 J A gas mixture consists of 2 mde of oxygen and 4 mole of argon at tempressure T.Neglecting all vibrational modes, the total internal energy of the system is (A) 11 RT (B) 9 RT (C) 15 RT (D) 4 RT A monoatomic ideal gas, initially at temperature T1 is enclosed in a cylindes fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston suddenly T1 If L1 and L2 the lengths of the gas colum be fore and afters expansion respectively, then then T is 2

given by 2

L  (A)  1   L2 

26.

27.

28.

3

2

L  (B)  2   L1 

3

L1

(C) L 2

L2

(D) L 1

Starting with the same intial Conditions, an ideal gas expands from Volume V1 to V2 in three different ways. The Work done by the gas is W1 if the process is purely isothermal, W2 if purely isobasic and W3 if purely adiabatic Then

(A) W2>W1>W3 (B) W2>W3>W1 (C) W1>W2>W3 (D)W1>W3>W2 In a given process on an ideal gas dw = 0 and dQ < 0. Then for the gas (A) the volume will increase (B) the pressure will semain constant (C) the temperature will decrease (D) the temperature will increase Wafer of volume 2 filter in a containes is heated with a coil of 1kw at 270C. The lid of the containes J is open and energy dissipates at the late of 160 . In how much time tempreture will rise from S

270 C to 770C. Specific heat of wafers is 4.2 (A) 7 min

KJ Kg

(B) 6 min 2s

(C) 14 min 263

(D) 8 min 20 S

29.

30.

70 calorie of heat are required to raise the temperature of 2 mole of an ideal gas at constant pressure from 300C to 350C The amount of heat required to raise the temperature of the same gas through the same range at constant volume is .................. calorie. (A) 50 (B) 30 (C) 70 (D) 90 When an ideal diatomic gas is heated at constant pressure, the Section of the heat energy supplied which increases the infernal energy of the gas is.. (A)

31.

32.

3 7

34. 35. 36. 37.

38. 39.

3 5

(C)

2 5

(D)

5 7

Two cylinders A and B fitted with piston contain equal amounts of an ideal diatomic gas at 300 k. The piston of A is free to move, While that of B is held fixed. The same amount of heat is given to the gas in each cylindes. If the rise in temperature of the gas in A is 30K, then the rise in temperature of the gas in B is. (A) 30 K (B) 42 K (C) 18 K (D) 50 K An insulated containes containing monoatomic gas of molas mass Mo is moving with a velocity, V.If the container is suddenly stopped, find the change in temperature.

Mov2 (A) 5R 33.

(B)

Mov2 (B) 4R

Mov2 (C) 3R

Mov2 (D) 2R

A Small spherical body of radius r is falling under gravity in a viscous medium. Due to friction the medium gets heated. How does the late of heating depend on radius of body when it attains terminal velocity! (A) r2 (B) r3 (C) r4 (D) r5 The first law of thermodynamics is concerned with the conservation of (A) momentum (B) energy (C) mass (D) temperature If heat given to a system is 6 k cal and work done is 6kj. The change in internal energy is .......... KJ. (A) 12.4 (B) 25 (C) 19.1 (D) 0 The internal energy change in a system that has absorbed 2 Kcal of heat and done 500J of work is (A) 7900 J (B) 4400 J (C) 6400 J (D ) 8900 J Which of the following is not a thermodynamical function. (A) Enthalpy (B) Work done (C) Gibb's energy (D) Internal energy Which of the following is not a thermodynamic co-ordinate. (A) R (B) P (C) T (D) V The work of 62-25 KJ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by 50C The gas is _______ R = 8-3 (A) triatomic (B) diatomic

(C) monoatomic (D) a mixture of monoatomic and diatomic

264

J molk

40.

Cp and Cv denote the specific heat of oxygen per unit mass at constant Pressure and volume respectively, then (A) cp - cv =

R 16

(C) Cp - Cv = 32 R 41.

(B) Cp - Cv = R (D) Cp - Cv =

R 32

When a System is taken from State i to State f along the path iaf, it is found that Q = 70 cal and w = 30 cal, along the path ibf. Q=52cal. W atoug the path ibf is

(A) 6 cal

(B) 12 cal

(C) 24 cal

(D) 8 cal

N 5kg what is the 2 The density of the gas is m m3

42.

5 One kg of adiatomic gas is at a pressure of 5 10

43.

energy of the gas due to its thermal motion ? (A) 2.5  105 J (B) 3.5  105 J (C) 4.5  105 J (D) 1.5  105 J 200g of water is heated from 250 C0 450C Ignoring the slight expansion of the water the change in its internal energy is (Specific heat of wafer 1

44.

(A) 33.4 KJ (B) 11.33 KJ (C) 5.57 KJ (D) 16.7 KJ During an adiabatic process, the pressure of a gas ifound to be propostional to the fifth power of its absolute temperature. The radio (A)

45.

cal ) 90 C

4 5

cp for the gas is cv

(B)

3 4

(C)

(D) 4

One mole of oxygen is heated at constant pressure stasting at 00 C. How much heat energy in cal must be added to the gas to double its volume ? Take R = 2

46.

5 4 cal molk

(A) 1938 (B) 1920 (C) 1911 (D) 1957  moles of a gas filled in a containes at temperature T is in equilibrium inidially - If the gas is compressed slowly and is thesmally to half its initial volume the work done by the atmosphere on the piston is (A) -

µRT 2

(B)

RT 2

1 (C) RT ln (2 - ) 2

265

(D) –RTln 2

47.

48.

49.

50.

51.

52.

53.

54.

Heat capacity of a body depends on the .......... as well as on .......... (A) material of the body, its mass (C) mass of the body, itd temperature (B) material of the body, its temperature (D) Volume of the body, its mass In thesmodynamics, the work done by the system is considered ......... and the work done on the system is Considered ........... (A) Positive, zero (B) nagative, Positive (C) zero, negative (D) Positive, negative A thesmodynamic system goes from States (i) P,V to 2P, V (ii) P,V to P, 2V. Then what is work done in the two Cases. (A) Zero, PV (B) Zero, Zero (C) PV, Zero (D) PV,PV For free expansion of the gas which of the following is true ? (A) Q = 0, W > 0 and  Eint = -W (B) W = 0, Q >0 and  Eint = Q (C) W > 0, Q < 0 and  Eint = 0 (D) Q = W = 0 and  Eint = 0 For an adiabatic process involving an ideal gas (A) P  – 1 = T  – 1 = constant (B) P 1 –  = T = constant (C) PT  – 1 = constant (D) P  – 1 T = constant Figure shows four P  V diagrams. which of these curves represent. isothermal and adiabatic processes ?

(A) A and C (B) A and B (C) C and D (D) B and D An ideal gas is taken through cyclic process as shown in the figure. The net work done by the gas is

(A) PV (B) 2 PV (C) 3 PV (D) zero  moles of gas expands from volume V1 to V2 at constant temperature T. The work done by the gas is

 V2  (A) RT  V   1

V  (B) RT lnn 2   V1 

 V2  (C) RT  V 1  1  266

 V2  (D) RT lnn V 1  1 

55.

A Cyclic Process ABCD is Shown in the P  V diagam. which of the following curves represent the same Process ?

(A)

56.

(B)

(C)

(D)

A Cyclic process is Shown in the P  T diagram. Which of the curve show the same process on a V  T diagram ?

(A)

(B)

(C)

57.

(D)

One mole of an ideal gas

Cp Cv

 γ at absolute temperature T1 is adiabatically compressed from an

initial pressure P1 to a final pressure P2 The resulting temperature T2 of the gas is given by. γ  p 2  γ 1

(A) T2 = T1 

  p1 

 p2  (C) T2 = T1    p1 

γ 1  γ

p (B) T2 = T1  2   p1 

γ

p  (D) T2 = T1  2   p1  267

γ 1

58.

59.

An ideal gas is taken through the cycle A  B  C  A as shown in the figure. If the net heat supplied to the gas in the cycle is 5J, the work done by the gas in the process C  A is

(A) - 5 J (B) -10 J (C) -15 J (D) - 20 J 0 In anisothermal reversible expansion, if the volume of 96J of oxygen at 27 C is increased from 70 liter to 140 liter, then the work done by the gas will be (A) 300 R loge(2)

60.

For an iso thermal expansion of a Perfect gas, the value of (A)

61.

1 - γ2

γ

Δr v

(B) 

1 γ

2 (C) -γ

v v

ΔV V

(D) γ 

ΔV V

P is equal to P

2 (C) -γ

(B) -1

v2 v1

Δv v

(D) - γ

Δv v

(C) 

(D)  + 1

v2 (B) RT log10 v 1

v1 (C) RT log10 v

2

v1 (D) RT log e v

2

The isothermal Bulk modulus of an ideal gas at pressure P is (A) vP

65.

ΔV V

P is equal to P

Work done permol in an isothermal ? change is (A) RT log e

64.

(B)  γ

(D) 100 R log10(2)

If r denotes the ratio of adiabatic of two specific heats of a gas. Then what is the ratio of slope of an adiabatic and isothermal P  V curves at their point of intersection ? (A)

63.

ΔV V

For an aidabatic expansion of a perfect gas, the value of (A) -

62.

(C) 200 R og10 2

(B) 81 R loge(2)

(B) P

(C)

p 2

(D)

p v

The isothermal bulk modulus of a perfect gas at a normal Pressure is N (A) 1.013  106 m2

(B) 1.013  10

11

N m2

268

(C) 1.013  10

5

N m2

(D) 1.013  10

11

N m2

66.

An adiabatic Bulk modulus of an ideal gas at Pressure P is (A)  P

67.

68.

(B)

p γ

(C) P

What is an adiabatic Bulk mudulus of hydrogen gas at NTP? r = 1.4 N N N (A) 1.4 2 (B) 1.4  105 2 (C) 1  10-8 M 2 M M

(D)

P 2

(D) 1  105

N M2

If a quantity of heat 1163.4 J is supplied to one mole of nitrogen gas, at room temprature at constant pressure, then the rise intemperature is R = 8.31

J m.l.k

(A) 28 K

(C) 54 K

(B) 65 K

(D) 40 K

69.

One mole of O2 gas having a Volume equal to 22.4 liter at Oc and 1 atmosiheric Pressure is Compressed isothermally so that its volume reduces to 11.2 lites. The work done in this Process is (A) 1672.4 J (B) -1728J (C) 1728J (D) -1572.4J

70.

The Specific heat of a gas in an isothermal Process is (A) zero

71.

73.

(C) Infinite

(B) Coppes

(C) glass

75.

(D) Cloth

A thermodynamic Process in which temprature T of the system remains constant through out Variable P and V may Change is called (A) Isothermal Process (B) Isochoric Process (C) Isobasic Process (D) None of this 5 N When 1g of wates Oc and 10 m 2 Pressure is Converted into ice of Volume 1.091 cm3 the external

work done be .......... J. (A) 0.0182 (B) -0.0091 74.

(D) Remairs

A Containes that suits the occurrence of an isothermal process should be made of (A) Wood

72.

(B) Negative

(C) -0.0182

(D) 0.0091

J If the work done in the Process of expansion g of 1g is 168J. then increase in internal energy is .......... J (A) 2072 (B) 2408 (C) 2240 (D) 1904 The Volume of an ideal gas is 1 liter column and its Pressure is equal to 72 cm of Hg. The Volume of gas is made 900 cm3 by compressing it isothermally. The stress of the gas will be ............. Hg column. (A) 4 cm (B) 6 cm (C) 7 cm (D) 8 cm The letent heat of Vaporisation of water is 2240

269

76. 77.

In adiabatic expansion (A)  u=0

79.

(C) Ju = Nagative

(D)  w = 0

1 mm3 Of a gas is compressed at 1 atmospheric pressure and temperature 27 C to 627 C What is the final pressure under adiabatic condition. r = 1.5 5 (A) 80  10

78.

(B)  u = Positive

N m2

5 (B) 36  10

N m2

5 (C) 56  10

N m2

5 (D) 27  10

N m2

5 1 is suddenly Compressed to of its original volume adiabatically 3 8 then the final Pressure of gas is ............. times its intial Pressure. 24 40 (A) 8 (B) 32 (C) (D) 5 3 1 1 The Pressure and density of a diatomic gas  = 7 5 Change adiabatically from (P,d) to (P ,d ) If

A monoatomic gas for it  =

d' p' =32 then Should be d p

1 (C) 32 (D) None of this 128 8 80. An ideal gas at 27 C is Compressed adiabatically, to of its original Volume. If v = 5 3 , then the 27 rise in temperaure is (A) 225 k (B) 450 K (C) 375 K (D) 405 K (A) 128

81.

(B)

A diatomic gas intially at 18 C is Compressed adiabatically to one eight of its original volume. The temperature after Compression will be (A) 10 C

82.

Cal mol o K

(A) 600

(C) 60

(D) 54

(B) 2.5 erg

(C) 250 W

(D) 250 N

P1 In an isochoric Process T1  27 C and T2  127 C then P will be equal to 2

(A) 85.

(B) 546

N A gas expands 0.25m 3 at Constant Pressure 103 m2 the work done is

(A) 250 J 84.

(D) 144 C

Work done by 0.1 mole of a gas at 27 C to double its volume at constant Pressure is Cal. R = 2

83.

(C) 887 C

(B) 668 K

9 59

(B)

2 3

(C)

4 3

(D)

3 4

If the temperature of 1 mole of ideal gas is changed from OC to 100 C at constant pressure, then work done in the process is ............ J. R = 8-3 (A) 8-3  10-3

(B) 8-3  102

J molk

(C) 8-3  10-2 270

(D) 8-3  103

86.

88.

A mono atomic gas is supplied the heat Q very slowly keeping the pressure constant The work done by the gas. 2 2 1 3 (A) Q (B) Q (C) Q (D) Q 5 3 5 5 The Volume of air increases by 5% in an adiabatic expansion. The percentange lecrease in its Pressure will br. (A) 5% (B) 6% (C) 7% (D) 8% In P  V diagram given below, the isochoric, isothermal and isobaric path respectively are

89.

(A) BA,AD,DC (B) DC,CB,BA (C) AB,BC,CD In the following indicatos diagram, the net amount of work done will be

(D) Infinity

90.

(A) Negative (B) zero (C) Positive Work done in the given P  V diagram in the cyclic process is

(A) 4 PV

(D)

87.

91.

(B) 3 PV

(D) CA,DA,AB

(C) 2 PV

PV 2

In the Cyclic Process shown is the figure, the work done by the gas in one cycle

(A) 40 P1V1

(B) 20 P1V1

271

(C) 10 P1V1

(D) 5 P1V1

92.

An ideal gas is taken V path ACBA as Shown in figure, The net work done in the whole cycle is

(A)  3 P1V1 93.

94.

(B) Zero

(C) 5 P1V1

(D) 3 P1V1

A thermodynamic system is taken from state A to B along ACB and is brought back to A along BDA as Shown in the P  V diagram. The net work done during the complete cycle is given by the area.

(A) P1 ACB P2 P1 (B) ACBDA (C) ACBB1A1A (D) ADBB1A 1A Two identical samples of a gas are allowed to expand (i) isothermally (ii) adiabatically work done is

(A) more inanisothermal process (C) equal in both process.

(B) more is an adiabatic process (D) neithes of them

95.

What is the relationship Pressure and temperature for an ideal gas under going adiabatic Change.  (A) PT = Const (B) PT-1 +  = Const (C) P1 -  T = Const (D) P  - 1T  = Const

96.

For adiabatic Process which relation is true mentioned below ? γ =

97.

98.

(A) pV = Const (B) TV = Const (C) TV  =Const (D) TV–1 = Const For adiabatic Process which one is wrong statement? (A) dQ = 0 (B) entropy is not constant (C) du = - dw (D) Q = constant Air is filled in a motor tube at 27 C and at a Pressure of a atmosphere. The tube suddenly bursts. Then what is the temperature of air. given r of air = 1.5 (D) 27.5 C If v is the radio of Specigic heats and R is the universal gas constant, then the molar Specific heat at constant Volume Cv is given by ............ (A) 150 K

99.

Cp Cv

(A)

vR v 1

(B) 150 C

(C) 75 K

(B) vR

272

(C)

R v 1

(D)

 v  1 R r

100. A Car not engine operating between temperature T1 and T2 has efficiency 0.4, when T2 lowered by 50K, its efficiency uncreases to 0.5. Then T1 and T2 are respectively. (A) 300 K and 100 K (B) 400 K and 200 K (C) 600 K and 400 K (D) 400 K and 300 K 101. A monoatomic gas is used in a car not engine as the working substance, If during the adiabatic expansion part of the cycle the volume of the gas increases from V to 8V1 the efficiency of the engine is .. (A) 60% (B) 50% (C) 75% (D) 25% 102. A System under goes a Cyclic Process in which it absorbs Q1 heat and gives out Q2heat. The efficiency of the Process is n and the work done is W. Which formula is wrong ? (A) W  Q1  Q2

Q2 (B) n  Q 1

W (C) n  Q 1

Q2 (D) n  1  Q 1

103. A car not's engine whose sink is at a temperature of 300K has an efficiency of 40% By space should the temperature of the source be increase the efficiency to 60% (A) 275 K (B) 325 K (C) 300 K (D) 250 K 104. An ideal gas heat engine is operating between 227 C and 127 C . It absorks 104 J Of heat at the higher temperature. The amount of heat Converted into. work is .......... J. (A) 2000 (B) 4000 (C) 5600 (D) 8000 105. Efficiency of a car not engine is 50%, when temperature of outlet is 500K. in order to increase efficiency up to 60% keeping temperature of intake the same what is temperature of out let. (A) 200 K (B) 400 K (C) 600 K (D) 800 K 106. A car not engine takes 3 106 cal of heat from a reservoir at 627 C , and gives to a sink at 27 C . The work done by the engine is (A) 4.2 106 J

(B) 16.8 106 J

(C) 8.4 106 J

(D) Zero

107. For which combination of working temperatures the efficiency of Car not's engine is highest. (A) 80 K, 60 K (B) 100 K, 80 K (C) 60 K, 40 K (D) 40 K, 20 K 108. An ideal heat engine working between temperature T1 and T2 has an efficiency n. The new efficiency if both the source and sink temperature are doubled, will be (A) n

(B) 2n

(C) 3n

(D)

n 2

109. An ideal refrigerator has a freetes at a temperature of 13 C , The coefficent of perfomance of the engine is 5. The temperature of the air to which heat is rejected will be. (A) 325 C

(B) 39 C

(C) 325 K

(D) 320 C

110. An engine is supposed to operate between two reservoirs at temperature 727 C and 227 C . The maximum possible efficiency of such an engine is (A)

3 4

(B)

1 4

(C)

273

1 2

(D) 1

111. A car not engine Convertsm one sixth of the heat input into work. When the temperature of the sink is reduces by 62 C the efficiency of the engine is doubled. The temperature of the source and sink are (A) 800 C, 370 C (B) 950 C, 280 C (C) 900 C, 370 C (D) 990 C, 370 C 112. Car not engine working between 300 K and 600 K has work output of 800J per cycle. What is amount of heat energy supplied to the engine from source per cycle (A) 1600

J cycle

(C) 1 0 0 0 c ycJ le

J (B) 2000 cycle

113. What is the value of sink temperature when efficiency of engine is 100% (A) 300 K (B) 273 K (C) 0 K

(D) 1 8 0 0 c ycJ le

(D) 400 K

1 as heat engine is used as a refrigerators. if the work 10 done on the system is 10J. What is the amount of energy absorbed from the reservoir at lowes temperature ! (A) 1 J (B) 90 J (C) 99 J (D) 100 J

114. A car not engine having a efficiency of n =

115. The temperature of sink of car not engine is 27 C . Efficiency of engine is 25% Then find the temperature of source. (A) 2270 C (B) 3270 C (C) 270 C (D) 1270 C 116. The efficiency of car not's engine operating between reservoirs, maintained at temperature 27 C

117.

118.

119.

120.

and 123 C is ............ (A) 0.5 (B) 0.4 (C) 0.6 (D) 0.25 If a heat engine absorbs 50KJ heat from a heat source and has efficiency of 40%, then the heat released by it in heat sink is ........... (A) 40 KJ (B) 30 KJ (C) 20 J (D) 20 KJ The efficiency of heat engine is 30% If it gives 30KJ heat to the heat sink, than it should have absorbed ........... KJ heat from heat source. (A) 42.8 (B) 39 (C) 29 (D) 9 If a heat engine absorbs 2KJ heat from a heat source and release 1.5KJ heat into cold reservoir, then its efficiency is ........... (A) 0.5% (B) 75% (C) 25% (D) 50% If the doors of a refrigerators is kept open, then which of the following is true? (A) Room is cooled (B) Room is eithers cooled or heated (C) Room is neither cooled nor heated (D) Room is heated

274

Assertion-Reason Instructions :Read the assertion and reason carefully to mask the correct option out of the options given below. (A) If both assertion and reason are true and the reason is the correct explanation of the assertion. (B) If both assertion and reason are true but reason is not be correct explanation of assertion. (C) If assertion is true but reason is false. (D) If the assertion and reason both are false. 121. Assertion : The melting point of ice decreases with increase of Pressure Reason : Ice contracts on melting. (A) C (B) B (C) A (D) D 122. Assertion : Fahrenhit is the smallest unit measuring temperature. Reason : Fahrenhit was the first temperature scale used for measuring temperature. (A) A (B) C (C) B (D) D 123. Assertion : A beakes is completely, filled with water at 4 C . It will overlow, both when heated or cooled. Reason : These is expansion of water below 40 C (A) A (B) B (C) C (D) D 124. Assertion : The total translation kinetic energy of all the molecules of a given mass of an ideal gas is 1-5 times the product of its Pressure and its volume. Reason : The molecules of a gas collide with each other and velocities of the molecules change due to the collision (A) D (B) C (C) A (D) B 125. Assertion : The car not is useful in understanding the perfomance of heat engine Reason : The car not cycle provides a way of determining the maximum possible efficiency achivable with reservoirs of given temperatures. (A) A (B) B (C) C (D) D

275

Match column 126. Heat given to process is positive, match the following column I with the corresponding option of column I1

Colum-i (A) JK

Colum-ii (p)  W >0

(B) KL

(q)  Q <0 (r)  W <0

(C) LM

(D) MJ (s)  Q >0 (A) A-p, B-q, C-r, D-s (C) A-r, B-s, C-p, D-q (B) A-q, B-p, C-s, D-r (D) A-s, B-r, C-q, D-p 127. In Column I different Process is given match corresponding option of column I1 Column - I Column - II (A) adiabatic process (p)  p = 0 (B) Isobaric process (C) Isochroic process (D) Isothermal process (A) A-p, B-s, C-r, D-q (B) A-s, B-q, C-p, D-r

(a)  u = 0 (r)  Q = 0 (s)  W = 0 (C) A-r, B-p, C-s, D-q (D) A-q, B-r, C-q, D-p

Comprehehsion Type In a containes of negligible heat capacity, 200g ice at 0 C and 100g steam at 100 C are added to 200g of water that has temperature 55 C .Assume no heat is lost to the surroundings and the pressure in the container is constant 1 atm. 128. What is the final temperature the System ? (A) 72 C (B) 48 C (C) 100 C 129. At the final temperature, mass of the toal water present in the system is (A) 493.6g (B) 483.3g (C) 472.6 g 130. Amount of the Sm left in the system, is equal to (A) 16-7 g (B) 8-4 g (C) 12 g

276

(D) 94 C (D) 500 g (D) 0 g

KEY NOTE 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

B A D D A D C C B A C D A C B D C A B D C D D A B

26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

A C D A D B C D B C A B A C D B A D C C B A D A D

51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75

B C A B C D B A C D D C A B C A B D D C B A D A D

277

76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

C D B A C B C A D B A C D C A B D B A C D B A C D

101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130

C B D A B C D A B C D A C B D A B A C D C B A D A B C C B A

HINT 1.

C F 25 F    0 0 100 180 100 180

2.

T = 273.15 +  0 C

3. 4.

0 = 273.15 +  0 C equal 40 C

c F  32  5 9 5.

c F  32  5 9

y = m x + c with comparision m = 6.

c f  32  5 9

7.

F  32 K  273  9 5

8.

ΔTc ΔTF = 100 180

5 9

ΔTc TC  400 C 100 9.

T 1 1  2 0   2 106  10  105 T 2 2 % in increase =

T 100  10  5  100  1  10 3 % T

10.

W = pΔV

11.

f  f  Cp    1 R, Cv  R 2 2 

Cp Cv

 f  f 1   1 R f 2 2 2   2 =    1 f f f f R 2 2

278

12.

W = area inside the closed curve 1 1   (p 2  p1 )  V2  V1  [Treat Circle as an cllipse] 2 2

P(V2 -V1) PV = γ-1 γ-1

13.

Δu =

14.

uT

15.

For all gases Cp - Cv = R

16.

 Q  u  w

17. 18.

Cp Cv

u1 T u u T T  1  1 2  1 2 u2 T2 u2 T2

= +

2 f

W = area enclose by P  V graph  AB X BC  (2P  P)  (2V  V)  PV

19.

20.

W = area under the P  V curve Smallest for curve 1, largest for curve 3 3 5 monoatomic gas cv  R , diatomic gas cv  R , 2 2 1 3 5  one mole of each gas is mixed Cv (mix) =  R  R   2R 2 2 2 

Cp (mix) = R + Cv = R + 2R = 3R r=

C p mix C v mix

=

3R = 1.5 2R

21.

dQ = du + dw

22.

Cv mix =

23.

ΔU = γ C v ΔT

24.

u =

n1 Cv1  n 2Cv2 n1  n 2

nfRt 2

n = number of mole of the gas f = number of degree of freedom

U Total = U 02  U AB 279

25.

TV = [an adiabatic process]

T1V1 = T2 T2 V  T1 = 2 T2  V1  26.

W = area under the curve and volume axis on the P  V diagram.

27.

dQ = du + dw

28.

energy required in heating water = ms  from coil,

energy avilable  Power of coil  Power lost S

For 840 J time required = 1S  4.2  104 J = ?

t

29.

4.2  105 500 J  8.33min  8 min 205 840

Cp =

 Q  P r T

cv  cp  R

cal = 7-2 = 5 mol K

30.

u 1 5   Q  7  u = n Cv  T  Q = n Cp  T

C u 1 5  v   Q Cp r 7 31.

For A (  Q) P = r Cp (  T) A (Isobaric process) For B (  Q) V = Cv (  T) B (Isochroic process) (  Q )P = (  Q)V

1 mv 2 = γ C v ΔT 2

32.

Decrease in K.E = increase in I.E

33.

Rate of heat produced = (Viscous force F)  (Velocity V)

dQ   nrv 2 dt

 2 (ρ  ρo ) r 2 g    n  g 

2

280

34. 36. 39.

Q = U +W Q = U + W Q = U +W  u =  u - 62.25 (adiabatic process) u = 62.25 x 103 J

u n t

Cv =

3

3

J

for monoatomic gas Cv = 2 R  2  8  3  12 .45 mol K 40.

If CP1 and CV1 is a molar specific heat CP1 - CV1 = R ------------ (i) Cp 

Cp1 Cv , Cv  1 [molar mass of O2 = 32] 32 32

From equation (i) 32Cp - 32Cv = R Cp - Cv = 41.

R 32

(i) Q = w + u (ii) Q  W  u (u equal)

42.

5 RT 2 PV  RT [diatomic gas] 5 u PV 2 u 

V

mass 1 3  m density 5

43.

u  mcT

44.

given P  T 5 an adiabatic process P  T

45.



 1

  5  1

281

46.

W = Patm V = Patm

V (i) 2

Intially Patm V = rRT  V  49.

 RT pat

(i) Volume Constant W  PV  0 2V

(ii) Pressure Constant W  P



dv  P V

V

50.

For free expansion Q = W = O T = const

51.



 Eint  o

PV v = Const (an adiabatic process) PV = m Rt  V

µ RT P

52.

For both solpe is negative & Slope of adiabatic curve is more

53.

W  area of P  V diagarm

= 54.

57.

1 P( 2V )  PV 2

AB  Constant P, increasing V, increasing T BC  Constant T, increasing V, decreasing P CD  Constant V, decreasing P, decreasing T DA  Constant T, decreasing V, increasing P Also BC is at highes temperature than AD PV = const [adiabatic process] ideal gas PV = RT (m = 1) V  

 RT   P  = const  P  58.

RT P

Tr = const r -1  P

Q = WAB +WBC +WCA 5 = 10+0+WCA

59.

W = µRT oge

V2 V2  M   2-3   RT og10 V1 V1  Mo 

282

60.

PV = constant (isothermal Process)

PΔV - VΔP = 0  61.

ΔP -ΔV = P V

PV = Constant (an adiabatic Process) Pr v-1  V + V  P = 0

62.

 PP adiabatic  PP isothermal

63.

V2 V2 W = mRT ln V = R T loge V 1 1

64.

isothermal bulk modulus B = P

64.

N isothermal bulk modulus B = P = 1.013  105 m 2

66.

an adiabatic bulk modulus B = vP

67.

an adiabatic bulk modulus B = 1.4  1  105 = 1.4  105

68. 69.

dQ =  Cp dT W = -  RT log e (isothermally compressed)

70.

Ciso =

71.

For isothermal process base is conductory

73. 74. 74. 76.

W = P  V (isothemal process) u = Q - w P1V2 = P2V2 (isothermal process)  Q =  u+  W

-γ =

-

V V V = g V

Q =  [  T = 0, isothermal process] mT

0 = u +W 77.

T = Constant [adiabatic change] P -1 3

1

2 T   P2      2   P1   T1 

1

2

 P2  2  900   5  =  300     10 

3

2

283

N m2

78.

PV = Constant (adiabatic compressed) 8

V  P2 = 1 P1  V2  79.

PV = Constant (adiabatic process

80.

T2  V1  =  T1  V2 

81.

TV  - 1 = Constant

 -1

 V1  T2 = T1    V2  82.

V1 T1  V2 T2

 -1



V 300  2V T2

T2  600K

W = PV 83. 84.

W = P  V = 103  0.25 = 250 J PT (constant volume) P1 T1  P2 T2

85.

W = PV

86.

W  (Q) P  u  (Q) P  (Q) V  Q V   (Q) P  1    Q P   CV   (Q) P 1    CP 

87.

dp dv × 100 = -  × 100 p V

89.

cyclic process

90. 91.

(adiabatic expansion)

1 nagative work 2 net positive W  area of closed P  V diagram = (3V-V) (3P-P) = 4PV W = area under curve 284

92. 94. 95. 96.

W = area under curve  (Area)iso >(Area)adi Wiso > wad: PV = Con PV = Con

97.

Q = O, Q = const, du = - dw (aidabatic process)  1  P2  

98.

T2 =   T1  P1 

99.

CP = CV

 CP =  CV

but CP - CV = R 100. n = 1-

T2 100. T1

 -1 = T2 V2 -1 (qdiabafic process) 101. T1V1

V  T1 =T2  2   V1 

 -1

T2 103. n  1  T 1

104. n = 1 -

T2 1-400 1 = = T1 500 5

W = n Q1 105. n  1 

T2 T2 , T1 T1

106. (i) n  1 

T2 T1

(ii ) n1  1 

109.  

Should be minimum

2T2 T  1 2  n 2T1 T1

T2 T1  T2

285

T2 500 1 110. n  1  T  1  1000  2 1

111. (i)

n=1-

(ii ) n1  1 

 1

T2 W 1 1  =  n = - (1) T1 Q1 6 6 T2  62 T1

T2 62  T1 T1

 n

62  ( 2) T1

Now, n1 = 2n 112. n = 1 -

113. n  1 

T2 W = T1 Q1

 T  Q=  1  T1 -T2

 w 

T2 T1

T2 114. n  1  T1

T  W = Q2  1 -1  T2  115. n = 1 -

T2 T1

116. n = 1 -

T2 T1

117. n = 1 -

Q2 Q1

118. n = 1 -

Q2 Q1

286

119. n = 1 -

Q2 Q1

121. with rise in pressure melting point of ice decreases. Also ice contracts on melting. 122. celcius scale was the first temperature scale and Fahrenhit is the smallest unit measuring. 123. Water has maximum density at 4 C on heating above 4 C or cooling below 4 C density of water decreases and its volume increases, therefore, water overflows in the both cases. 124.

1 3 m (υ2 ) = RT 2 2

125. car not cycle has maximum efficiency 126. (a) isochoric process P T

w  0

 Q  u

P decrease, T also decrease  u negative  Q  0

(b) isobasic process, volume increase  W  0 (c) isochoric process W  0 Q  u P  T, P increase  T increase  Q  0 (d) Volumedecrease W  0 127. adiabatic process Q  0 Isobasic process P = const

 P  0

Isochroic process V = const  W  0 Isothermal process T = const  u  0 128 to 130. head rewuired by ice and water to go up to 1000 C = m1L + m1sw  T + mw sw  T = 200  80+200  1  100+200  1  45 = 16,000+20,000+9,000 = 45,000 cal = give by ms mass of steam = ms L 45,000  83.3 g convert into waters of 1000 C 540 Total water = 200 + 200 + 83.3 = 483.3 g steam left = 100 - 83.3 = 16.79

ms =

287

SUMMARY * 1.

Gas Laws: Boyle's law: For a given mass of an ideal gas at constant temperature, the volume of a gas is inversely proportional to its pressure, i.e. Vα

1 P

or PV = constant  P1V1 = P2V2

P1 P2 m P (i) PV  P  = constant  = constant or     1 2  m Where density,   , and m = constant V

(ii) As number of molecules per unit volume n 

N V

 V 

 N (iii) PV  P  = n

constant 

N also N = const. n

P1 P2 P = constant or n  n n 1 2

(iv) Graphical representation: (If m and T are constant)

2.

Charle's law: At constant pressure, the volume of the given mass of a gas is directly proportional to its absolute temperature. i.e. V  T  (i)

V1 V2 V = constant = T  T 1 2 T

V m  m   = constant  V   T T ρ   or T = constant = 1T1  2T2 289

(ii)

3.

Graphical representation: (If m and P are constant)

Gay-Lussac`s Law or pressure law: According to it for a given mass of an ideal gas at constant volume, pressure of a gas is directly proportional to its absolute temperature. i.e. P  T or (i)

*

P = constant T



P1 P2  T1 T2

Graphical representation: (If m and V are constant)

Avogadro's law: Equal Volume of all the gases under similar conditions of temperature and pressure contain equal number of molecules. i.e. N1 = N2 Avogadro Number: The number of particles (atoms or molecules) in one mole of substance (gas) is called Avogadro number (NA) which has a magnitude NA = 6.023  1023 mol-1 290

*

Equation of State OR Ideal Gas Equation The equation which relates the pressure (P), volame (V) and temperature (T) of the given state of an of an ideal gas is known as ideal gas equation or equation of state. Ideal gas equation is PV = µRT

*

where µ = number of mole R = universal gas constant = 8.314 J.mol-1 K-1 Different forms of Ideal gas-state-equation

 N    R PV  µRT    kB   RT  Nk BT   NA   NA 

(i)

where kB = Boltzmann`s constant = 1.38  10-23 JK-1 (ii)

P

N k BT  nk BT V

where n 

(iii)

PV 

N V

=

number density of gas

=

number of molecules per unit volume

 M M  RT  µ   Mo Mo   where Mo = P

*

molar mass of the gas M RT ρRT m = ( ρ  density of the gas)  V Mo Mo V

The work done during the change in volume of the gas: Vf

It can be obtained from the graph of P – V W   P dν Vi

*

Assumption of Ideal gases (or kinetic theory of gases) Kinetic theory of gases relates the macroscopic properties of gases (such as pressure, temperature etc.) to the microscopic properties of the gas molecules (such as momentum, speed, kinetic enengy of molecule etc.) Assumptions: 1. Gas is made up of tiny particles, These particles are called molecules of the gas. 2. The molecules of a gas are identical, spherical, rigid and perfectly elastic point masses. 3. The molecules of a gas perform incessant random motion. 4. The molecules of a gas follow Newton's laws of motion. 5. The number of molecules in a gas is very large. 6. The volume of molecules is negligible in comparision with the volume of gas. 291

7. 8.

*

Inter molecular forces act only when two molecules come close to each other or collide. The time spent in a collision between two moleules is negligible in comparision to time between two successive collisions. The collisions between the molecules and between a molecule and the wall of a container are elastic. Pressure of an Ideal gas P

*

1 1 ρ  ν2   ρ ν 2 rms 3 3

where  = density of the gas v2rms = <2> = mean squre velocity of molecule Relation between pressure and kinetic energy P

1 ρ ν 2 rms        (i) 3

K.E. per unit volume is E =

1 M 2 1  rms =  2rms        (ii) 2 V 2

From equation (i) and (ii) P

*

2 E 3

Root mean square speed (vrms) : It is defined as the square root of mean of squares of the speed of the speed of different molecules. i.e. rms =

ν12  ν22  ν32  ........  νN2 N

From the expression of pressure 1  2rms  rms = 3

P =

=

3RT = M

where ρ  m

3P = ρ

3PV M

3k BT m

M = density of the gas V

M  mass of each molecule NA

-

νrms  T

-

with increase in molecular weight rms speed of gas molecule decreases as vrms  292

1 M

*

rms speed of gas molecules does not depend on the pressure of gas (If temperature remains constant) At T= 0 K , vrms= 0, i.e. the rms speed of molecules of a gas is zero at O K. This temperature is called absolute zero. Kinetic interpretation of temperature Kinetic energy of of 1 mole ideal gas E

1 1  3RT  3 M ν 2 rms  M    RT 2 2  M  2

-

For 1 molecule E 

3 kBT , kB = Boltzmann`s constant 2

-

For N molecule E 

3 N kBT 2

-

Kinetic energy per nolecule of gas does not depend upon the mass of the molecule but only depends on the temperature. Mean free path:The distance travelled by a gas molecule between two successive collisions is known as free path. The average of such free paths travelled by a molecule is called mean free path. mean free path,

*

Total distance travelled by a gas molecule between two successive collisions  = Total number of collisions 1 2 nd 2

-

 

-



-

Collision frequency

 1 k BT P    P  nk BT  n   2 2  k BT  2 nπd 2 π Pd  = number of collisions per second. 

νrms 



*

-

Degrees of Freedom : The term degrees of freedom of a molecule or gas are the number of independent motions that a molecule or gas can have. The independent motion of a system can be translational, rotational or vibrational or any combination of these. Degress of freedom, f = 3A - B; where A = Number of independent particles, B = Number of independent restrictions monoatomic gas  3 degrees of freedom ( All translational) 293

-

Diatomic gas

-

triatomic gas

 5 degrees of freedom ( 3 translational + 2 rotational) 

(Non-linear)

6 degrees of freedom (at room temperature) ( 3 translational + 3 rotational)



8 degrees of freedom (at very high temperature) (3 translational + 3 rotational + 2 vibrational)

*

Law of equipartition of energy (Boltzmann law) According to this law, for any system in equilibrium, the total energy is equally distributed among its various degrees of freedom and each degrees of freedom is associated with energy

-

1 k B T where kB = Boltzmann`s constant. 2 At a given temperataue T, all ideal gas molecules no matter what their mass have the same average 3 k BT 2 At same temperature gases with different degrees of freedom (i.e. H2 and He) will have different

translational kinetic energy = -

average energy 

f k BT 2

( f = degress of freedom different for different gases.) f k BT 2

-

The total energy associated with each modlec 

*

Specific heat of a gas :

-

Specific heat at constant volume (Cv)

-

The amount of heat required to change the temperature of l mole of gas by 1 K, keeping its volume constant, is called specific heat of the gas at constant volume.

-

Specific heat at constant volume (Cp) The amount of heat required to change the temperature of l mole of gas by 1 K, keeping its pressure constant, is called specific heat of the gas at constant pressure. Molar specific heat:

-

The quantity of heat required to change the temperature of 1 mole of gas by 1 K (or 1o C) is called molar specific heat of the gas. Ratio of CP and CV is . f   1 R  C 2 2   P   1 f CV f R 2

 1  f    CV  f R , C P    1 R  2 2   

294

MCQ Choose the correct alternative from given options. 1.

Volume, pressure and temperature of an ideal gas are V, P and T respectively. If mass of molecule is m, then its density is [ kB = Boltzmann`s constant] P (A) k T B

2.

3.

4.

5.

6.

9.

10.

(D) mk B T

(B)

p 8

(C) 2 P

(D) P

The equation of state for 5 g of oxygen at a pressure P and temperature T, when occupying a volume V, will be  5  (A) PV    RT  32 

8.

.P (C) k TV B

The temperature of an ideal gas at atmospheric pressure is 300 K and volume 1 m3. If temperature and volume become double, then pressure will be (A) 4  105 Nm–2 (B) 2  105 Nm–2 (C) 1  105 Nm–2 (D) 0.5  105 Nm–2 At 100 K and 0.1 atmospheric pressure, the volume of helium gas is 10 litres. If volume and pressure are doubled, its temperature will change to (A) 127 K (B) 400 K (C) 25 K (D) 200 K What is the mass of 2 litres of nitrogen at 22.4 atmospheric pressure and 273 K. (R = 8.314 Jmol k-1) (A) 14  22.4 g (B) 56 g (C) 28 g (D) None of these. An electron tube was sealed off during manufacture at a pressure of 1.2  10-7 mm of mercury at 270C. Its volume is 100 cm3. The number of molecules that remain in the tube is _______ (density of mercury is 13.6 gcm–3) (A) 3.9  1011 (B) 3  1016 (C) 2  1014 (D) 7  1011 A vessel contains 1 mole of O2 gas (relative molar mass 32) at a temperature T. The pressure of the gas is P. An identical vessel containing 1 mole of He gas (relative molar mass 4) at a temperature 2T has pressure of .......... (A) 8 P

7.

Pm (B) k T B

5 (B) PV    RT  16 

5 (C) PV    RT 2

(D) PV = 5 RT

A gas at 1 atmosphere and having volume 100 ml is mixed with another gas of equal moles at 0.5 atm and having volume 50 ml in flask of one litre, what is the final pressure? (A) 0.125 atm (B) 0.75 atm (C) 1 atm (D) 0.5 atm PV The quantity k T represents B

(A) mass of gas (B) number of moles of gas (C) number of molecules in gas (D) K. E. of gas Equation of gas in terms of pressure (P), absolute temperature (T) and density () is P1T1 P2T2 (A) ρ  ρ 1 2

P1 P2 (B) T ρ  T ρ 1 1 2 2

(C) 295

P1ρ 2 P2ρ1  T1 T2

P1ρ1 P2ρ 2 (D) T  T 1 2

11.

02 gas is filled in a vessel. If pressure is doubled, temperature becomes four times, how many times

its density will become. (A) 4 12.

13.

(B)

1 4

15.

1 2

(A) Varies inversely as the square of its mass

(B) Varies inversely as its mass

(C) is independent of its mass

(D) Varies linearly as its mass

If pressure of a gas contained in a closed vessel is increased by 0.4% when heated by 10 C the initial temperature must be. (B) 250 C

(C) 250 K

(D) 2500 K

To decrease the volume of a gas by 5% at constant temperature the pressure should be (A) Incseased by 5.26%

(B) Decreased by 5.26%

(C) Decreased by 11%

(D) Increased by 11%

A gas at the temperature 250 K is contained in a closed vessel. If the gas is heated through 1 K, then the percentage increase in its pressurse will be (A) 0.4%

16.

(D)

At a given volume and temperature the pressure of a gas

(A) 2500 C 14.

(C) 2

(B) 0.1 %

(C) 0.8%

(D) 0.2%

The product of the pressure and volume of an ideal gas is (A) A constant (B) Directly proportional to its temperature. (C) Inversely proportional to its temperature. (D) Approx. equal to the universal gas constant.

17.

o

o

At O C the density of a fixed mass of a gas divided by pressure is x. At 100 C, the ratio will be  273  x (B)   373 

(A) x 18.

(B) 206 kPa

(D) 212 kPa

(B) 17.26 ml

(C) 19.27 ml

(D) 192.7 ml

o

2g of O2 gas is taken at 27 C and pressure 76 cm Hg. Find out volume of gas (ln litre) (A) 3.08

21.

(C) 200 kPa

The volume of a gas at 20 C is 200 ml. If the temperature is reduced to –20o C at constant pressure, its volume will be. (A) 172.6 ml

20.

 100  x (D)   273 

Air is pumped into an automobile tube upto a pressure of 200 kPa in the morning when the air o o temperature is 22 C. During the day, temperature rises to 42 C and the tube expands by 2% The pressure of the air in the tube at this temperature will be approximately. (A) 209 kPa

19.

 373  x (C)   273 

(B) 44.2

(C) 1.53

(D) 2.44

1 mole of gas occupies a volume of 100 ml at 50 mm pressure. What is the volume occupied by two moles of gas at 100 mm pressure and at same temperature (A) 50 ml

(B) 200 ml

(C) 100 ml 296

(D) 500 ml

22.

23.

A partition divides a container, having insulated walls, into two compartments, I and II. The same gas is filled the compartments. The ratio of number of molecules in compartments I and II is. (A) 6:1 (B) 1:6 (C) 4:1

P, V, T

2P, 2V, T

I

II

(D) 1:4

A cylinder contains 10 kg of gas at pressure of 107 N/m2. The quantity of gas taken out of the cylinder, if final pressure is 2.5  10 6 Nm 2 , will be ______ (temperature of gas is constant) (A) 5.2 kg

24.

(C) 7.5 kg

(D) 1 kg o

The volume of a gas at pressure 21  104 Nm–2 and temperature 27 C is 83 Litres. If R = 8.3 J mol–1K–1. Then the quantity of gas in g-mole will be (A) 42

25.

(B) 3.7 kg

(B) 7

(C) 14

(D) 15 o

The pressure and temperature of an ideal gas in a closed vessel are 720 kPa and 40 C respectively. If

1 4

th

of the gas is released from the vessel and the temperature of the remaning gas is raised o

to 353 C, final pressure of the gas is (A) 1440 kPa 26.

27.

(B) 2 P

o

o

o

(C) 851 C

o

o

(B) 182 C

(C) 646 C

o

o

(B) 600

(D)None of these o

(D) 546 C

(C) 327

o

(D) 270

At constant temperature on incerasing the pressure of a gas 5% its volume will decrease by (A) 5%

(B) 5.26%

(C) 4.76%

(D) 4.26%

o

o

At on 0 C pressure measured by barometer is 760 mm. what will be pressure at 100 C (A) 780 mm

33.

o

(B) 651 C

To double the volume of a given mass at an ideal gas at 270 C keeping the pressure constant one must raise the temperature in degree centigrade (A) 54

32.

(D) 6 P

At what temperature volume of an ideal gas becomes triple (A) 819 C

31.

(C) 4 P

Air is filled in a bottle at atmospheric pressure and it is corked at 35 C, If the cork can come out at 3 atmospheric pressure then upto what temperature should the bottle be heated in order to remove the cork. o

30.

(D) 720 kPa

o

(A) 325.5 C 29.

(C) 1080 kPa

Suppose ideal gas equation follows VP 3 = constant, Initial temperature and volume of the gas are T and V respectively. If gas expand to 27 V, then temperature will become (A) 9 T (B) 27 T (C) T 9 (D) T The temperature of a gas at pressure P and volume V is 270 C Keeping its volume constant if its temperature is raised to 9270 C, then its pressure will be (A) 3 P

28.

(B) 540 kPa

(B) 760 mm

(C) 730 mm

(D) None of these.

Hydrogen gas is filled in a ballon at 200 C. If temperature is made 400 C, pressure remaining the same what fraction of haydrogen will come out (A) 0.75

(B) 0.07

(C) 0.25 297

(D) 0.5

34.

35.

36.

37.

38.

When the pressure on 1200 ml of a gas is increased from 70 cm to 120 cm of mercury at constant temperature, the new volume of the gas will be (A) 400 ml (B) 600 ml (C) 700 ml (D) 500 ml 0 A gas at 27 C temperature and 30 atmospheric pressure is allowed to expand to the atmospheric pressure if the volume becomes two times its initial volume, then the final temperature becomes (A) 2730 C (B) -1730 C (C) 1730 C (D) 1000 C A gas at 270 C has a volume V and pressure P. On heating its pressure is doubled and volume becomes three times. The resulting temperature of the gas will be (A) 15270 C (B) 6000 C (C) 1620 C (D) 18000 C o A perfect gas at 270 C is heated at constant pressure to 3270 C. If original volume of gas at 27 C o is V then volume at 327 C is (A) 2 V (B) V (C) V 2 (D) 3 V A vessel contains 1 mole of O2 gas (molar mass 32) at a temperature T. The pressure of the gas is P. An identical vessel containing one mole of He gas (molar mass 4) at temperature 2T has a pressure of P 8 The pressure and temperature of two different gases P and T having the volumes V for each. They are mixed keeping the same volume and temperature, the pressure of the mixture will be,

(A) 2 P 39.

(B) P

41.

42.

43.

(B)

2 E 3

3 E 2

E 2 The root mean square speed of hydrogen molecules of an ideal hydrogen kept in a gas chamber at 00C is 3180 ms–1. The pressure on the hydrogen gas is (Density of hydrogen gas is

(A) P  44.

(D)

P (C) 4 P (D) 2 P 2 Air is filled at 600 C in a vessel of open mouth. The vessel is heated to a temperature T so that th 1 part of air escapes. Assuming the volume of the vessel remaining constant the value of T is. 4 (A) 3330 C (B) 1710 C (C) 4440 C (D) 800 C A gas is filled in a cylinder, its temperature is incresecd by 20% on kelvin scale and volume is reduced by 10%. How much percentage of the gas will leak out (A) 15% (B) 25% (C) 40% (D) 30% The pressure is exerted by the gas on the walls of the container because (A) It sticks with the walls (B) It is accelerated towards the walls (C) It loses kinetic energy (D) On collision with the walls there is a change in momentum The relation between the gas pressure P and average kinetic energy per unit volume E is

(A) P 40.

(C) 8 P

(B) P 

(C) P  E

(D) P =

8.99  10 2 kg m 3 , 1 atm  1.01 105 Nm 2 ) (A) 1.0 atm

(B) 3.0 atm

(C) 2.0 atm 298

(D) 1.5 atm

45.

Gas at a pressure Po is contained in a vessel. If the masses of all the molecules are halved and

their speeds are doubled, the resulting pressure will be equal to (A) 2 P0 46.

(B) 4 P0

(C)

P0 2

(D) P0

A cylinder of capacity 20 litres is filled with H 2 gas. The total average kinetic energy of translatory motion of its molecules is 1.5  105 J. The pressure of hydrogen in the cylinder is (A) 4  106 Nm–2 (B) 3  106 Nm–2 (C) 5  106 Nm–2 (D) 2  106 Nm–2

47.

The average kinetic energy per molecule of a gas at -230 C and 75 cm pressure is 5  10 14 erg for H 2 . The mean kinetic energy per molecule of the O2 at 227 C and 150 cm pressure will be

48. 49.

50.

51.

52. 53. 54.

55.

(A) 80  10 14 erg

(B) 10  10 14 erg

(C) 20  10 14 erg

(D) 40  10 14 erg

The ratio of mean kinetic energy of hydrogen and oxygen at a given temperature is (A) 1:8 (B) 1:4 (C) 1:16 (D) 1:1 The ratio of mean kinetic energy of hydrogen and nitrogen at temperature 300 K and 450 K respectively is (A) 2:3 (B) 3:2 (C) 4:9 (D) 2:2 Pressure of an ideal gas is increased by keeping temperature constant what is the effect on kinetic energy of molecules. (A) Decrease (B) Increase (C) No change (D) Can`t be determined A sealed container with negligible co-efficient of volumetric expansion contains helium (a monoatomic gas) when it is heated from 200 K to 600 K, the averagy K.E. of helium atom is (A) Halved (B) Doubled (C) Unchanged (D) Increased by factor 2 The mean kinetic energy of a gas at 300 K is 100J. mean energy of the gas at 450 K is equal to (A) 100 J (B) 150 J (C) 3000 J (D) 450 J At what temperature is the kinetic energy of a gas molecule double that of its value at 27 C (A) 540 C (B) 1080 C (C) 3270 C (D) 3000 C The average kinetic energy of a gas molecule at 270 C is 6.21  10–21 J. Its average kinetic energy at 2270 C will be (A) 5.22  10 21 J (B) 11.35 10 21 J (C) 52.2  10 21 J (D) 12.42  1021 J The average translational energy and rms speed of molecules in sample of oxygen gas at 300 K are 6.21 1021 J and 484 m s respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour) (A) 6.21 1021 J , 968 m s (B) 12.42  1021 J , 684 m s (C) 12.42  1021 J , 968 m s (D) 8.78  1021 J , 684 m s

299

56.

57. 58.

The average translational kinetic energy of O2 (molar mass 32) molecules at a particular temperature is 0.068 eV. The translational kinetic energy of N2 (molar mass 28) molecules in eV at the same temperature is (A) 0.003 eV (B) 0.068 eV (C) 0.056 eV (D) 0.678 eV At O K which of the follwing properties of a gas will be zero (A) Kinetic energy (B) Density (C) Potential energy (D) Vibrational energy The kinetic energy of one mole gas at 300 K temperatue is E. At 400 K temperature kinetic enrgy is E'. The value of

16 4 (C) 1.33 (D) 9 3 The average kinetic energy of hydrogen molecules at 300 K is E. At the same temperature the average kinetic energy of oxygen molecules will be

(A) 2 59.

60.

(B)

(A) E 16 (B) E (C) 4 E (D) E 4 The temperature at which the average translational kinetic energy of a molecule is equal to the energy gained by an electron accelerating from rest through a potential differencc of 1 volt is (A) 4.6 103 K

61.

62.

63.

64.

66.

(B) 7.7 103 K

(C) 11.6 103 K

(D) 23.2 103 K

At a given temperature the rms velocity of molecules of the gas is (A) Proportional to molecular weight (B) Inversely proportional to molecular weight (C) Inversely proportional to square root of molecular weight (D) Proportional to square of molecular weight According to the kinetic theroy of gases the r.m.s velocity of gas molecules is directly proportional to 1 (A) T 2 (B) T (C) T (D) T The speeds of 5 molecules of a gas (in arbitrary units) are as follws: 2, 3, 4, 5, 6, The root mean squre speed for these molecules is (A) 4.24 (B) 2.91 (C) 4.0 (D) 3.52 o To what temperature should the hydrogen at room temperature (27 C) be heated at constant pressuse so that the rms velocity of its molecule becomes double of its previous value (A) 927 C

65.

E' is E

(B) 600 C

(C) 108 C (D) 1200 C Root mean square velocity of a molecule is  at pressure P. If pressure is increased two times, then the rms velocity becomes (A) 3 (B) 2 The rms speed of gas molecules is given by (A) 2.5

Mo RT

RT

(C) 0.5

(D) 

RT

(B) 2.5 M o

(C) 1.73 M o

300

(D) 1.73

Mo RT

o

67.

A sample of gas is at O C. To what temperature it must be raised in order to double the rms speed of molecule. o o o o (A) 270 C (B) 819 C (C) 100 C (D) 1090 C

68.

If the ratio of vapour density for hydrogen and oxygen is ratio of their rms velocities will be (A) 4:1 (B) 1:16

69.

1 , then under constant pressure the 16

(C) 16:1

(D) 1:4

The molecules of a given mass of a gas have a rms velocity of 200 m s at 27 C and 1.0  105 Nm 2 pressure when the temperature is 127 C and pressure is 0.5  10 5 Nm 2 , the rms velocity in m s will be (A) 100 2

70.

(B)

100 2 3

(C)

400 3

(D) None of these

If the molecular weight of two gases are M1 and M 2 , then at a given temperature the ratio of root mean square velocity 1 and  2 will be

71.

72.

(A)

M1 M2

(B)

M2 M1

(C)

M1  M 2 M1  M 2

(D)

M1  M 2 M1  M 2

To what temperature should the hydrogen at 327 C cooled at constant pressure, so that the root mean square velocity of its molcules become half of its previous value (A) 100 C (B) 123 C (C) 0 C (D) 123 C At what temperature is the root mean square velocity of gaseous hydrogen molecules equal to that of oxygen molecules at 47 C ? (A) -73 K

73.

(D) 3 K th

(B) 0 C

(C) O K

(D) 100 C

At room temperature ( 27 C ), the rms speed of the molecules of certain diatomic gas is found to be 1930 m/s. The gas is (A) O2

75.

(C) 20 K

The root mean square velocity of the molecules in a sample of helium is 57 that of the molecules in a sample of hydrogen. If the temperature of hydorgen sample is 00C, then the temperature of the helium sample is about (A) 273 C

74.

(B) 80 K

(B) C 2

(C) H2

(D) F2

If three molecules have velocities 0.5, 1 and 2 the ratio of rms speed and average speed is (The velocities are in km/s) (A) 0.134 (B) 1.34 (C) 1.134 (D) 13.4 301

76.

77.

78.

79.

80.

81.

At what temperature pressure remaining constant will the rms speed of a gas molecules increase by 10% of the rms speed at NTP? (A) 57.3 K (B) 57.30 C (C) 557.3 K (D) -57.30 C When temperature of an ideal gas is increased from 27o C to 227o C, its rms speed changed from 400 ms–1 to Vs. The Vs is (A) 516 ms–1 (B) 746 ms–1 (C) 310 ms–1 (D) 450 ms–1 At what temperature the molecules of nitrogen will have the same rms. velocity as the molecules of Oxygen at 127oC. (A) 2730 C (B) 3500 C (C) 770 C (D) 4570 C The temperature of an ideal gas is increased from 270 C to 1270 C, then percentage increase in rms is (A) 33% (B) 11% (C) 15.5% (D) 37% Let A and B the two gases and given : TA  4 M A . Where T is the temperature and M is TB MB A molecular mass. If A and B are the r.m.s speed, then the ratio  will be equal to __________ B (A) 2 (B) 4 (C) 0.5 (D) 1 -1 The rms. speed of the molecules of a gas in a vessel is 400 ms . If half of the gas leaks out, at constant temperature, the r.m.s speed of the remaining molecules will be (A) 800 ms-1

82.

83.

84.

85.

86.

87.

(B) 200 ms-1

(C) 400 2 ms-1 (D) 400 ms-1 At which temperature the velocity of 02 molecules will be equal to the rms velocity of N2 molecules at 00 C (A) 930 C (B) 400 C (C) 390 C (D) can not be calculated. The rms speed of the molecules of a gas at a pressure 105 Pa and temperature 00 C is 0.5 km/s. If the pressure is kept constant but temperature is raised to 8190 C, the rms speed becomes (A) 1.5 kms-1 (B) 2 kms-1 (C) 1 kms-1 (D) 5 kms-1 The root mean square velocity of a gas molecule of mass m at a given temperature is proportional to (A) m0 (B) m-1/2 (C) m1/2 (D) m The ratio of the vapour densities of two gases at a given temperature is 9:8, The ratio of the rms velocities of their molecule is (A) 3 : 2 2 (B) 2 2 : 3 (C) 9:8 (D) 8:9 At what temperature, pressure remaining unchanged, will the rms velocity of a gas be half its value at OoC ? (A) 204.75 K (B) 204.750 C (C) -204.75 K (D) -204.750 C The rms velocity of gas molecules is 300 ms-1. The rms velocity of molecules of gas with twice the molecular weight and half the absolute temperature is (A) 300 ms-1 (B) 150 ms-1 (C) 600 ms-1 (D) 75 ms-1 302

88.

89.

90.

Calculate the temperature at which rms velocity of S02 molecules is the same as that of O2 molecules at 270 C. Molecular weights of Oxygen and SO2 are 32 g and 64 g respectively (A) 3270 C (B) 327 K (C) 1270 C (D) 2270 C For a gas, the rms speed at 800 K is (A) Four times the value at 200 K (B) Twice the value at 200 K (C) Half the value at 200 K (D) same as at 200 K A mixture of 2 moles of helium gas (atomic mass = 4 amu), and 1 mole of argon gas (atomic mass = 40 amu) is kept at 300 K in a container. The ratio of the rms speeds

91.

92.

93.

94.

95.

(A) 0.45 (B) 2.24 (C) 3.16 (D) 0.32 0 0 The temperature of an ideal gas is increased from 27 C to 927 C. The root mean square speed of its molecules becomes (A) Four times (B) One-fourth (C) Half (D) Twice At a given temperature the root mean square velocities of Oxygen and hydrogen molecules are in the ratio (A) 1:4 (B) 1:16 (C) 16:1 (D) 4:1 If mass of He atom is 4 times that of hydrogen atom, then rms speed of the is (A) Two times of H  rms speed. (B) Four times of H  rms speed. (C) Same as of H  rms speed. (D) half of H  rms speed. At temperature T, the rms speed of helium molecules is the same as rms speed of hydrogen mdecules at normal temperature and pressure. The value of T is (A) 5460 C (B) 00 C (C) 2730 C (D) 136.50 C The root mean square speed of hydrogen molecules at 300 K is 1930 m/s. Then the root mean square speed of Oxygen molecules at 900 K will be (A) 836 m/s

96.

97.

98.

vrms (helium) is vrms (argon)

(B) 643 m/s

(C) 1930 3 m/s

(D)

1930 m/s 3

If rms speed of a gas is rms = 1840 m/s and its density  = 8.99  10-2 kg/m3 , the pressure of the gas will be (A) 1.01  103 Nm–2 (B) 1.01  105 Nm–2 (C) 1.01  107 Nm–2 (D) 1.01 Nm–2 When the temperature of a gas is raised from 270 C to 900 C , the percentage increase in the rms velocity of the molecules will be (A) 15% (B) 17.5% (C) 10% (D) 20% The rms speed of a gas at a certain temperature is 2 times than that of the Oxygen molecule at that temperature, the gas is_____ (A) SO2 (B) CH4 (C) H2 (D) He 303

99.

The temperature at which the rms speed of hydrogen molecules is equal to escape velocity on earth surface will be (A) 5030 K (B) 10063 K (C) 1060 K (D) 8270 K 100. What is the meanfree path and collision frequency of a nitrogen molecule in a cylinder containing 

nitrogen at 2 atm and temperature 17oC ? Take the radius of nitrogen molecule to be 1A . M olecular mass of nitrogen = 28 , k B = 1.38  10–23 JK–1, 1 atm = 1.013  105 Nm–2 (A) 2.2 10 7 m, 2.58  10 9

(B) 1.110 7 m, 4.58  108

(C) 1.110 7 m, 4.58 10 9

(D) 2.2 10 7 m, 3.58  109 0

101. The radius of a molecule of Argon gas is 1.78 A . Find the mean free path of molecules of Argon at 00 C temperature and 1 atm pressure.

k B  1.38  10 23 JK 1 (A) 6.65  10 8 m

(B) 3.65  10 7 m

(C) 6.65  10 7 m

(D) 3.65  10 8 m

102. A monoatomic gas molecule has (A) Three degrees of freedom (B) Five degrees of freedom (C) Six degrees of freedom (D) Four degrees of freedom 103. A diatomic molecule has how many degrees of freedom (For rigid rotator) (A) 4 (B) 3 (C) 6 (D) 5 104. The degrees of freedom for triatomic gas is ______ (At room temperature) (A) 8 (B) 6 (C) 4 (D) 2 105. If the degrees of freedom of a gas are f, then the ratio of two specific heats

Cp Cv

is given by

2 2 1 3 1 (B) 1  (C) (D) 1  f f f f 106. A diatomic gas molecule has translational, rotational and vibrational degrees of freedom. The (A) 1 

Cp Cv

is

(A) 1.29 (B) 1.33 (C) 1.4 107. The value of Cv for one mole of neon gas is (A)

3 R 2

(B)

7 R 2

(C)

1 R 2

108. The relation between two specific heats of a gas is (A) CV  CP  R J

(B) CP  CV  J

(C) CP  C V  R J

(D) CV  CP  J

304

(D) 1.67

(D)

5 R 2

109. The molar specific heat at constant pressure for a monoatomic gas is (A)

3 R 2

110. For a gas

(B)

5 R 2

(C) 4 R

(D)

7 R 2

R  0.67 . This gas is made up of molecules which are CV

(A) Diatomic (B) monoatomic (C) polyatomic (D) mixture of diatomic and polyatomic molecules 111. The specific heat of an ideal gas is (A) Proportional to T2 (B) Proportional to T3 (C) Proportional to T (D) Independent of T 112. The specific heats at constant pressure is greater than that of the same gas at constant volume because (A) At constant volume work is done in expanding the gas. (B) At constant pressure work is done in expanding the gas. (C) The molecular vibration increases more at constant pressure. (D) The molecular attraction increases more at constant pressure.  7 is mixed with one mole of diatomic gas     . 3  What is  for the mixture?  denotes the ratio of specific heat at constant pressure to that at constant volume.

113. One mole of ideal monoatomic gas

(A) 114.

35 23

23 15

(C)

3 2

(D)

4 3

For a gas if ratio of specific heats at constant pressure and volume is  , then value of degrees of freedom is 2

(A)   1 115.

(B)

5     3   

(B)

25   1 2

3  1

(C) 2   1

The molar specific heat at constant pressure of an ideal gas is

(D)

7 R . The ratio of specific heat 2

at constant pressure to that ratio at constant volume is (A) 116.

5 7

For a gas  

(B)

9 7

(C)

8 7

(D)

7 5

7 , the gas may probably be 5

(A) Neon (B) Argon (C) Helium (D) Hydrogen 117. From the following P - T graph, what inference can be drawn (A ) V 2< V1 (B) V2= V1 (C) V 2>V1 (D) none of these

305

9   1 2

118. The figure shows the volume V versus temperature T graphs for a certain mass of a perfect gas at two constant pressure of P1 and P2. What inference can you draw from the graphs. (A) P1< P2 (B) P1>P2 (C) P1= P2 (D) No inference can be drawn due to insufficient information. 119. Which one the following graphs represents the behaviour of an ideal gas

(A)

(B)

(C)

120. Under constant temperature, graph between p and 1 V is (A) Hyperbola (B) Circle (C) Parabola

(D)

(D) Straight line

Direction:(a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. (b) If both Assertion and Reason are true, but Reason is not correct explanation of the Assertion. (c) If Assertion is true; the Reason is false. (d) If Assertion is false, but the reason is true. 121. Assertion : 300 cc of a gas at 270 C is cooled at -30 C at constant pressure. The final volume of the gas would be 270 cc Reason

V2 T2 : This is as per charle's law V  T 1 1

(A) a (B) b (C) c (D) d -8 122. Assertion : The time of collision of molecules is of the order of 10 s, which is very very small compared to the time between two successive collisions. Reason : This is an experimental fact. (A) a (B) b (C) c (D) d 123. Assertion : Mean free path of gas varies inversly as density of the gas. Reason : Mean free path varies inversely as pressure of the gas. (A) a (B) b (C) c (D) d

306

KEY NOTE 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

1 B 2 C 3 B C 4 A 5 C 6 A 7 A 8 C B 9 D 10 D 11 C 12A A 13 B 14 B 15A 16A 17 C C 18 D 19 C 20 B 21 C 22 23 24 25

B C B C A C A A C B D D C A A B B A A C C D C B C

26 26 27 27 28 28 29 29 30 30 31 31 32 33 32 34 33 35 34 36 35 37 38 36 39 37 40 38 41 39 42 40 43 44 41 45 42 46 43 47 44 48 45 49 50 46 47 48 49 50

A C B D C C D B C B A A A D B B D A B C B D A A C

A51 C52 B53 D 54 C C55 D56 B57 C 58 B A59 A60 A61 D62 B 63 B D64 A65 B66 A67 C 68 B D69 A70 C71

72 73 74 75

B 51 B 52 C 53 54 D 55 B 56 B 57 A 58 C 59 60 B 61 B 62 C 63 B 64 65 A 66 A 67 D 68 C 69 B 70 71 A 72 C 73 B 74 D 75 C B C C

76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

307

BB AB CC D C B AB DA CC CB B BC BB DA BA D A C BB CA DC AB D D C CB AC BC C B B C

101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123

A 76 77 A 78 D 79 B 80 D 81 B 82 83 A 84 C 85 B 86 B 87 88 D 89 90 C 91 A 92 D 93 94 D 95 C 96 B 97 C 98 D 99 100 A A B

B A C C A D C C B B D B A B C D A D C A B C B B C

101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123

A A D B C B A C B B D B C A D D C B C D A A B

Hint / Solution 1.

M

PV = µRT  PV  M

(B) 

PM o M   RT V

 

2.

RT

O

(C)

PM o P  m  N A Pm Pm    RT RT  R  k BT  T  NA 

PV  RT  P

T V

If T and V both doubled, then pressure remains same 3.

(B)

PV  μRT  PV α T

If P and V doubled, then T becomes four times, 4.

(C)

PV  µRT  µ 

PV RT

 Mass of  litre nitrogen = µ  M o

5.

(A)

Gas equation for N molecules, PV = NkBT

 N 

6.

(C)

PV k BT

PV  µRT  P 

µRT V

For same µ, R and V, P  T 7.

(A)

molecular weight of oxygen = 32 (g) number of moles in 5 g of oxygen =



Equation of state is PV = µRT  PV 

8.

(A)

5 32

5 RT 32

Total number of moles is conserved, 

PV PV PV 1 1  2 2  RT RT RT

1 100 0 .5  5 0 P 1000 + = ( 1 L = 100 m  ) RT RT RT  P  0.125 atm 308 

9.

(C)

The ideal gas equation is PV  µRT

PV  RT  µ  1   R  PV  k B N AT   kB  NA   PV   N A  Avogadro's number.. k BT 10.

(B).

PV  µRT

 M   PV    RT  Mo 

 M  RT     V  Mo  RT Mo 

11.

(D)

 M       V 

P R  = const. T M o

PV  µRT M  P    µRT   

M  M  P  P   RT    T     Mo  12.

(D)

PV  µRT

M   PV    RT  P  M  Mo  13.

(C). PV  µRT  P T



14.

( V, T  Constant )

( Closed vessel i.e. volume is constant)

P1 T1  P2 T2

(A). PV  µRT = Constant ( temp. is const.)

 P1V1  P2V2  95   PV 1 1  P2  V1  100 

 V2  95% V1  309

 P2  1.0526 P1  P1  0.0526 P1  P1  5.26% P1 15.

 Pressure 5.26% increases. (A) Closed Vessel . i.e. volume remains constant.

From, PV  µRT P α T 



P2 T2  P1 T1

P2  P1 T2  T1  P1 T1

16.

(B)

 PV  T (  µ,R  Constant)

17.

(B)

PV  µRT

 M     Mo 

RT

M Mo  PV RT density Mo  = P RT 

 density     P At 0 C

 density   P    At 100 C

M  x...........................(i) R  273

=

=

M .........................(ii)  R  373

 density   273   x   P    At 100 C  373  18.

(A)

PV =µR = constant  P1V1  P2 V2 T T1 T2

19.

(A)

PV=µRT

since P is const. V  T 

20.

(C)

V1 T1  V2 T2

PV= µRT  M  =  RT  Mo 

V=

MRT M oP

310

21. (C)

PV= µRT



22.

(D)

P1V1 µ  1 P2 V2 µ2

PV  Nk B T

Now, N '  23.

(C)

( T const.)

 N

PV k BT

 2P  2V 

 PV   4 4N  k BT 

k BT



N 1  N' 4

PV=µRT

 M   PV =   RT  P  M  Mo  

( V, R, T - Constant)

P1 M1 107 10     M2 = 2.5 kg 6 P2 M 2 2.5  10 M2

 Hence mass of the gas taken out of the cylinder = 10 – 2.5 = 7.5 kg

PV RT

24.

(B)

PV = µRT   =

25.

(C)

 M  PV = µRT  PV =   RT  P  MT  Mo 

26.

(A)

VP3 = constant = k  P 3 

1 1  P 1 V V 3

k

 P

V k

PV = µRT 

V k V

 V 2

 V1  Hence    V2  

1 T  9 T2

3

2

3

2

3



1

1

3

V  µRT 3

 µRT

µRT k 2

T  1 T2

 V     27 V 

3



T T2

 T2  9 T

311



P2  M 2   T2     P1  M1   T1 

P1 T1 Using Gay-lussac's law P  T 2 2

27.

(C)

28.

(B) At constant volume P1 T1

29.



P2 T2

 P  T2   2   P1 



T1

(D) At constant Pressure V  T



V2 T2  V1 T1

30.

(C)

V  T

31.

(C)

P

V  T2   2  V1

  T1 

V1 T1  V2 T2

1 V P 100 100  2 1   V2  V1  0.9524 V1 V V1 P2 105 105

 V2  1  0.0476  V1  V1  0.0476 V1  V1  4.76% V1 32.

(D)

PT 

P2 T2 T   P2  P1 2 P1 T1 T1

33.

(B)

V T 

V2 T2 V  V1 T2  T1   2  V1 T1 V1 T1

 34.

(C)

313  293 V  273  40    273  20    0.07  293 V 273  20

At constant Pressure PV = constant  P1V1  P2 V2 

P1 V2  P2 V1

PV   T2   2 2  T1  P1V1 

35.

(B)

P1V1 P2 V2  T1 T2

36.

(A)

PV P1V1 PV  2 2  T2  2 2 T1 Pl V1 T1 T2

37.

(A)

VT 

V1 T1  V2 T2

312

38.

(A)

PV = µRT  P=µT ( V and R = constant )

P2 µ 2 T2  P1 µ1T1



39.

(D)

PV= µRT  µ1 

 P ' 40.

(B)

 µ1  µ 2  RT



V

PV PV and µ 2  RT RT

2 PV RT   2P RT V

For open mouth vessel, pressure is constant. volume is also given constant. Hence from  PV  µRT PV 

 

41.

(B)

M 1 RT  T  Mo M



T1 M 2  T2 M1

1 th part escapes, so remaining mass in the vessel is 4 3 M 3 273  60 4 1 M 2  M1   T  444 K  1710 C  4 T M1 Let initial conditions = V, T final conditions = V', T'

By Charle's law, V  T ( P remains constant ) V V' V V'     V '  1.2 V T T' T 1.2 T

But as per question, volume is reduced by 10% means  V' = 0.9 V so percentage of volume leaked out 

1.2  0.9  V 1.2 V

42.

(D)

Pressure P 

43.

(A)

P 

44.

(B)

 100 %  25%

F 1 P ( P = change in momentum)  A A t

2 ( Energy per unit volume ) 3 2 3

 rms 

E V



2 E 3

3P 2   P  rms  3

313

3P 3PV   M

45.

(A)

 rms 

46.

(C)

E

47.

(B)

The average kinetic energy E

  rms 

P M

1  2



P1 M 2  P2 M1

3 2E PV  P  2 3V 3 E T k BT  1  1 2 E 2 T2

48.

(D)

Kinetic energy is a function of temperature.

49.

(A)

ET

50.

(C)

51.

(B)

Kinetic energy of ideal gas depends only on its temperature. Hence, it remains constant whether pressure is increased or decreased. Kinetic energy is directly proportional to temperature. Hence if temperature is doubled, kinetic energy will also be doubled.

52.

(B)

53.

(C)

ET 

E1 T1  E 2 T2

54.

(D)

ET 

E1 T1  E 2 T2

55.

(B)

Average translational K.E. of a molecule is =



E1 T1  E 2 T2

Average kinetic energy  Temperature 

E1 T1  E 2 T2

At 300 K, average K.E.

= 6.21  10 21 J

At 600 K average K.E.

= 2  6.21  10 21

3 k BT 2

= 12.42  1021 J We know that rms =

3k BT m

At 300 K, rms = 484 ms–1 At 600 K, rms = 56.

57.

2  484  684 ms 1

3 k BT 2

(B)

Average translational K.E. of a molecules 

(A)

(Where, kB = Boltzmann's constant ) This is same, for all gases at same temperature. At 0 K Kinetic energy is zero. 314

3 E' T' RT  E  T   2 E T

58.

(C)

E

59.

(B)

E T

60.

(B)

2 eV 2 1.6  10 19 3    7.7  103 K k B T  1 eV  T  23 3 kB 3 1.38  10 2

61.

(C)

rms =

3RT M o  rms 

62.

(B)

rms 

T 2

2

2

1 Mo

2

2

63.

(C)

rms =

1   2   3   4   5 5

64.

(A)

rms 

  rms 2  T    rms  1

65.

(D)

rms velocity does not depend on pressure.

66.

(C)

rms =

3RT Mo

67.

(B)

rms 

T , To double the rms speed temperature should be made four times i.e.



3

= 4.24

T2 T1

RT Mo

 1.73

RT Mo

 T2  4T1 3P   1   2

2  1

16  4 :1 1

68.

(A)

vrms 

69.

(C)

rms velocity doesn't depend on pressure, it depends upon temperature only.

 rms =

3RT   rms  T  T   2 rms Mo



1 T1  2 T2

70.

(B)

 rms =

3RT 1 1   1   1  and  2  Mo M1 M2 2

71.

(D)

 rms =

3RT   rms  T  T   2rms Mo

72.

(C)

 rms 

3RT Mo

 T  M o (  rms , R  constant)

315

M2 M1

73.

(B)

3RT T   rms  Mo Mo

 rms 



THe  M H 2  He 5    H2 7 TH 2 M He

 THe 

25 4   273 49 2

 273 K  00 C 74.

(C)

3 RT Mo

 rms 

 Mo 

3 RT 3  8.3  300   2  103 kg  2 g 2  2 rms 1920 

Gas is hydrogen. 75.

(C)

rms speed, rms =

1   2   3 3

average speed,  

76.

(B)

As

 1.1 

  rms t   rms O

(A)

 rms  T

78.

(C)

 rms 

79.

TN2 TO 2

(C)



T and rms speed increases by 10% T0

273  t 273  t 2  1.1  1.21 or 273 273

77.





12   22  32 3

2 T2  1 T1

3RT  T  Mo Mo

 M O N  M O O

 rms 



 t  273 1.21  1  57.30 C

(  rms , R  constant)

2

2

3RT Mo

% increase in  rms 

3RT2 3 RT1  Mo Mo 3RT1 Mo

 100% 

316

20  17.32 100%  15.5% 17.32

80.

(A)

TA T 4 B MA MB 



TA MA

 2

3 RTA 3 RTB  2 MA MB

  A  2 B 

A 2 B

81.

(D)

Since temperature is constant. so vrms remains same.

82.

(C)

 rms =



3 RT  T  Mo Mo

TO2



TN2

 M o O  Mo N

(  rms , R  Const. )

2

2

  rms 1   rms 2



T1 T2

83.

(C)

 rms  T 

84.

(B)

 rms 

85.

(B)

At a given temperature  rms 

86.

(D)

 rms  T

  rms 2   rms 1





87.

TB MB

3k BT -1   rms  m 2 m

and   rms  2 

T 1  T0 2





1 

1   rms 1 2

273  t 273  0



1 2

273  t 1 273  t  273 = 68.25-273 = -204.750 C 273 4 4

(B)

 rms 

3 RT Mo

  rms 1 

3 RT ; Mo

  rms 2 

3R  T2  3RT  4M o 2M o

317



1 3RT 2 4M o



 vrms 1 2

= 150 ms–1

88.

(A)

3 RT Mo

 rms 

 3RT  rms = =  Mo 

89.

(B)

28×10-3

 

  rms 1    rms 2

here 

 3 8.314 290  508.24 ms1 

TO 2

 M o O

 2

TSO2

 Mo SO 

 rms  T

90.

(C)

  rms He   rms Ar

91.

(D)

 rms  T 

92.

(A)

 rms 



1 Mo

300 TSO2 0   Tso 2  600 K  327 C 32 64

 2

v1 T  1 v2 T2 3RT  Mo He 3RT  M o Ar

40  10 4



2 T2  1 T1

So

 rms O   rms H

2

=

2

 M o H  M o O

2

2

93.

(D)

 rms 

1  mH  He  m H m He

94.

(C)

rms =

 M o He T 3RT  T  M o  He = TH  M o H M0

95.

(A)

rms =

3 RT Mo

96.

(B)

 rms =

3P 

 3.16



 rms O   rms H

2

2

=

TO2 ×  M o H

2

TH2 ×  M o  O

2

2 or P =   rms 3

318

97.

(C)

3RT Mo

 rms =



2 = 1

T2 = T1

273+90 273+27

=1.1

 2   % increase =   1 ×100% = 0.1×100% =10%  1 

98.

(B)

 rms 

1 Mo



1 = 2

 M o 2  M o 1

 M o 2

1 = 2



  M o  2 = 16

32

 Hence the gas is CH 4

99.

(B)

Escape velocity from the earth's surface is 11.2 kms–1 2

So, rms = Vescape

100.

(C)  =

k BT 2πPd 2

 escape   Mo 3RT  T Mo 3R

1.38×10   290 =  2  3.14   2.026×10  2×10  -23

-5

-10

Collision Frequency = no. of collision per second 

   rms =   101. (A)  

1 = 2πnd 2

2

= 1.11 10-7 m

508.24  rms = = 4.58×109 -7 1.11×10 

3RT = Mo

 38.314 290   508.24 ms1  28×10-3

 

k BT  6.65  108 m 2 2 πPd

102. (A) A monoatomic gas molecule has only three translational degrees of freedom. 103. (D) A diatomic molecule has three translational and two rotational degrees of freedom. Hence total degrees of freedom, f = 3+2 = 5 104. (B) For a triatomic gas f = 6 ( 3 translation + 3 rotational ) 105.

(C)

Cp 2    1 Cv f

106. (B) Degrees of freedom = 3 ( translatory ) + 2 ( rotatory ) + 1 ( vibratory ) = 6 Cp 2 2 1    1 = 1 = 1 Cv f 6 3



4  1.33 3

107. (A) Neon gas is mono atomic and for mono atomic gases C v  3 R 2 108.

(C) When Cp and Cv are given with caloric and R with Joule then Cp – Cv = 319

R J

109. (B) C p  C v  R

110.

(B) C v =

 Cp  R  Cv  R 

f 3 R = R+ R  f = 3 2 2 5  R 2

R 3 = 1.5 R = R 0.67 2

This is the value for mono atomic gases 111. According to the equilibrium theorem, the molar heat capacities should be independent of temperature How ever, variations in Cv and Cp are observed as the temperature changes. At very high temperatures, vibrations are also inportant and that affects the values of Cv and Cp for diatomic and poly atomic gases. Here in the question according to given information (D) may be correct answer. 112. (B) 1 53 1 75 µ11 µ 2  2    53  1 75  1  3  1.5 1  1  2  1    113. (C) mix µ1 µ 1 1 2  2  7 5 1  1  2 1  1  1 3  5    

2 2 f 1 2   1    f  f f 2  1  1

114.

(A)

115.

(D) molar specific heat at constant pressure, Cp  Since C p  C v  R  C v  Cp  R 

116. 117.

118.

(D)  

7 R 2

7 5 R R  R 2 2

7 for a diatomic gas. 5

T T (C) As 2  1  tan 2 > tan 1        P  2  P 1 T Also from PV  µRT ,  V  V2  V1 P V V (B) As 1  2  tan 1 < tan 2        T 1  T 2

1 1 V 1 Hence  P    P   P1  P2   1  2 T P 119. (C) For an ideal gas PV = constant i.e PV doesn't vary with V 1 120. (D) At constant temperature, PV = constant  P  V from PV = µRT ;

320

Unit - 10 Ocsillations And Waves

321

SUMMARY 1.

Waves : The motion of the disturbance in the medium (or in free space) is called wave pulse or generally a wave.

2.

Amplitude of a wave : Amplitude of oscillation of particles of the medium is called the amplitude of a wave.

3.

Wavelength and frequency : The linear distance between any two points or particles having phase difference of 2  rad is called the wavelength (λ) of the wave. Frequency of wave is just the frequency of oscillation of particles of the medium. Relation between wavelength and frequency :

v = f,  =

 where, v is the speed of wave in the medium. k

4.

Mechanical waves : The waves which require elastic medium for their transmission are called mechanical waves, e.g. sound waves.

5.

Transverse and longitudinal waves : Waves in which the oscillations are in a direction perpendicular to the direction of wave propagation are called the transverse wave. Waves in which the oscillations of the particles of medium are a!cng the direction of wave propagation are called longitudinal waves.

6. Wave Equation : The equation which describe the displacement for any particle of medium at a required time is called wave equation. Various forms of wave equations are as follows :

 t x –  T 

(ii) y  A sin 

(i) y  A sin (t - kx)

 

(iii) y  A sin 2  t –

x  v

(iv) y  A sin

2 

 vt  x

The above equations are for the wave travelling in the direction of increasing value of x. If the wave is travelling in the direction of decreasing value of x then put '+' instead of '—' in above equations. 7.

The elasticity and inertia of the medium are necessary for the propagation of the mechanical waves.

8.

The speed of the transverse waves in a medium like string kept under tension, v  where, T = Tension in the string and (I = mass per unit length of the string -- y

9.

Speed of sound waves in elastic medium, v 

E 

where, E = Elastic constant of a medium,  = Density of the medium. Speed of longitudinal waves in a fluid, v 

B  

322

P 

T 

Cp

where, B — Bulk modulus of a medium y 

CV

= 1.41 (for air)

Speed of longitudinal waves in a linear medium like a rod, v 

 

where,  = Young modulus,  = Density of a medium At constant pressure and constant humidity, speed of sound waves in gas is directly proportional to the square root of its absolute temperature.

v

RT  v M

T

The speed of sound in a gas does not depend on the pressure variation. 10.

Principle of Superposition : When a particle of medium comes under the influence of two or more waves simultaneously, its net displacement is the vector sum of displacement that could occur under the influence of the individual waves.

11.

Stationary Waves : When two waves having same amplitude and frequency and travelling in mutually opposite directions are superposed the resultant wave formed loses the property of propagation. Such a wave is .called a stationary wave. Equation of stationary wave : y = – 2 A sinkx cos t Amplitude of stationary wave : 2 A sin kx Position of nodes in stationary wave xn =

n 2

where, n = 1,2, 3.....At all these points the amplitude is zero. Position of antinodes in stationary wave's, xn = (2n – 1)

 where , n — 1, 2, 3,.... 4

The amplitude of all these points is 2A. 12.

Frequencies corresponding to different normal modes of vibration in a stretched string of length L fixed at both the ends are given by,

fn  13.

nv n T  where n — 1, 2, 3...... 2L 2L 

In a closed pipe the values of possible wavelengths required for stationary wave pattern are given by.

n 

4L v  (2n – 1) f1 and possible frequencies, fn  (2n – 1) (2n - 1) 4L

where, n = 1, 2, 3,..... and L = length of pipe. In a closed pipe only odd harmonics f1, 3f1, 5f1 .... are possible.

323

14.

In an open pipe the values of possible wavelength required for stationary waves are given by,

n 

2L nv  nf1 where, n — 1, 2, 3,......and and possible frequencies, fn  n 2L

L - length of pipe. In open pipe of the harmonics like f1, 2f1, 3f1 ..... are possible. 15.

Beat: The phenomenon of the loudness of sound becoming maximum periodically due to superposition of two sound waves of equal amplitude and slightly different frequencies is called the 'beats'. Number of beats produced in unit time = f1 – f2.

16.

Doppler Effect : Whenever there is a relative motion between a source of sound and a listener with respect to the medium in which the waves are propagating the frequency of sound experienced by the listener is different from that which is emitted by the source. This phenomenon is called Doppler effect. Frequency listened by the listener, f L 

v  vL fS v  vS

Where, v = velocity of sound, vL = velocity of a listener, vS = velocity of a source, fS = frequency of sound emitted by the source. 17. If a body repeats its motion along a certain path, about a fixed point, at a definite interval of time, it is said to have periodic motion. 18.

If a body moves to and fro, back and forth, or up and down about a fixed point in a fixed interval of time, such a motion is called an oscillatory motion.

19.

When a body moves to and fro repeatedly about an equilibrium position under a restoring force, which is always directed towards equilibrium position and whose magnitude at any instant is directly proportional to the displacement of the body from the equilibrium position of that instant then such a motion is known as simple harmonic motion.

20.

The maximum displacement of the oscillator on cither side of mean position is called amplitude of the oscillator.

21.

The time taken by the oscillator to complete one oscillation is known as periodic time or time period or period (T) of the oscillator.

22.

The number of oscillation completed by the simple harmonic oscillator in one second is known as its frcquency(f).

23.

2  times the frequency of oscillator is the angular frequency CO of that oscillator..

24.

T

25.

For simple harmonic motion, the displacement y(t} of a particle from its equilibrium position is represented by sine, cosine or its linear combination like

1 2 1 2  or f  or  = f  T T

y( t ) = A sin (t +  ) y(t) = B cos (t +  ) y( t )  A' sin t + B' cost whereA'  Acos  and B' = Bsin

324

A2 – y2

26.

The velocity of SHO is given by v   

27.

The acceleration of SHO is given by a = -2y

28.

A particle of mass m oscillating under the influence of Hook's Law exhibits simple harmonic motion with



k ; m

T  2

m k

d2y + 2 y = 0 29. Differential equation for SHM is dth 30. For scries combination of n spring of spring constants k1, k2, k3,..., kn, the equivalent spring constant is

1 1 1 1 = =  ... the periodic time T  2  m k k1 k2 kn k 31.

For parallel combination of n springs of spring constants k constant is k = k1 + k2 + k3 + .... + kn and period T  2 

ky k

...., kn, the equivalent spring

m k

1 m2 (A2 - y2) 2

32.

The kinetic energy of the SHO is K =

33.

The potential energy of the SHO is U =

34.

The total mechanical energy of SHO is E = K + U =

35.

For SHO, at y — 0, the potential energy is minimum (U = 0) and the kinetic energy is maximum

(K 

36.

1 2 ky 2 1 2 1 m2 A2 = kA 2 2

1 kA2  E ) 2

For SHO, at y =  A, the potential energy is maximum (U 

1 kA2  E ) and the kinetic energy gy 2

is minimum (K = 0) 37.

Simple harmonic motion is the projection of uniform circular motion on a diameter of the reference circle.

325

38.

For simple pendulum, for small angular displacement

1 and g

T = 2

  2f =

2  T

g l

39.

For simple pendulum, T is independent of the mass of the bob as well as the amplitude of the oscillaions.

40.

The differential equaiton for damped harmonic oscillation is with the displacement

m

d2y dt

 b

2

dy = + ky = 0 dt

k b2 – m 4m2

and angular frequency ' 

1 kA2 e –etlm gives the mechanical energy of damped oscillation at time t. 2

41.

E( t ) 

42.

A system oscillates under the influence of external periodic force are forced oscillations.

43.

The differential equation for forced oscillations is

d2y dt 44.

2

b dy k F + y = 0 sint m dt m m



The amplitude for forced oscillation is

F0

A 2

2

2

2

2

[m ( 0 –  ) + b  ]

1 2

326

MCQ For the answer of the following questions choose the correct alternative from among the given ones. SECTION – I: 1.

If the equation for a particle performing S.H.M. is given by y = Sin2t + 3 Cos2t, its periodic time will be ………..s. (A) 21

2.

(C) 2p

(D) 4p.

The distance travelled by a particle performing S.H.M. during time interval equal to its periodic time is …….. (A) A

3.

(B)p

(B) 2A

(C) 4A

(D) Zero.

A person standing in a stationary lift measures the periodic time of a simple pendulum inside the lift to be equal to T. Now, if the lift moves along the vertically upward direction with an acceleration of g , then the periodic time of the lift will now be ……… 3

(A) 4.

3T

T 3

(D)

T 3

(C)  

(B) è

 2

(D)  

 . 2

When a body having mass m is suspended from the free end of two springs suspended from a rigid support, as shown in figure, its periodic time of oscillation is T. If only one of the two springs are used, then the periodic time would be ……… (A) (C)

6.

(C)

If the equation for displacement of two particles executing S.H.M. is given by y1 = 2Sin(10t+è) and y2 = 3Cos10t respectively, then the phase difference between the velocity of two particles will be ……….. (A) – è

5.

3 T 2

(B)

T 2

2T

(B)

T 2

(D) 2T

If the maximum velocity of two springs ( both has same mass ) executing S.H.M. and having force constants k1 and k2 respectively are same, then the ratio of their amplitudes will be ………. k1 (A) k 2

k2 (B) k 1

(C)

327

k1 k2

(D)

k2 k1

7.

8.

As shown in figure, two masses of 3.0 kg and 1.0 kg are attached at the two ends of a spring having force constant 300 N m- 1 . The natural frequency of oscillation for the system will be …………hz. ( Ignore friction ) (A) ¼

(B) 1/3

(C) 4

(D) 3

3kg

1kg

The bob of a simple pendulum having length ‘ l’ is displaced from its equilibrium position by an angle

of è and released. If the velocity of the bob, while passing through its equilibrium position is v, then v = ………..

9.

(A)

2 gl (1  Cos )

(B)

2 gl (1  Sin )

(C)

2 gl (1  Sin )

(D)

2 gl (1  Cos )

If

1 of a spring having length l is cutoff, then what will be the spring constant of remaining part? 4

(A) k 10.

(B) 4k

(C)

4k 3

(D)

3k 4

The amplitude for a S.H.M. given by the equation x = 3Sin3pt + 4Cos3pt is ………m. (A) 5

(B) 7

(C) 4

(D) 3.

11.

When an elastic spring is given a displacement of 10mm, it gains an potential energy equal to U. If this spring is given an additional displacement of 10 mm, then its potential energy will be …….. (A) U (B) 2U (C) 4U (D) U/4.

12.

The increase in periodic time of a simple pendulum executing S.H.M. is ………….when its length is increased by 21%. (A) 42 %

13.

(B) 10%

(D) 21%.

A particle executing S.H.M. has an amplitude A and periodic time T. The minimum time required by the particle to get displaced by (A) T

14.

(C) 11%

A 2

from its equilibrium position is …….. s.

(B) T/4`

(C) T/8

(D) T/16.

If a body having mass M is suspended from the free ends of two springs A and B, their periodic time are found to be T1 and T2 respectively. If both these springs are now connected in series and if the same mass is suspended from the free end, then the periodic time is found to be T. Therefore ………….. (A) T = T1 + T2

1 1 1 (B) T  T  T 1 2

(C) T2 = T12 + T22

328

1 1 1 (D) T 2  2  2 . T1 T2

15.

The displacement of a S.H.O. is given by the equation x = A Cos ( ùt +

 ). At what time will it attain 8

maximum velocity? (A) 16.

3 8

18.

(C)

A 2

(B) 

A

A

(C) 

2

(A) 1.5

(B) 3.0

(C) 4.5

(D) 6.0.

(D)

 . 16

3

(D)  3 A .

For particles A and B executing S.H.M., the equation for displacement is given by y1 = 0.1Sin(100t+p/3) and y2 = 0.1Cospt respectively. The phase difference between velocity of particle A with respect to that of B is …………

 3

(B)

 6

(C) 

 6

(D)

 3

The periodic time of a simple pendulum is T1. Now if the point of suspension of this pendulum starts moving along the vertical direction according to the equation y = kt 2, the periodic time of the pendulum becomes T 2. Therefore, (A) 6/5

20.

3 16

Three identical springs are shown in figure. When a 4 kg mass is suspended from spring A, its length increases by 1cm. Now if a 6 kg mass is suspended from the free end of spring C, then increase in its length is ………cm.

(A)  19.

8 3

At what position will the potential energy of a S.H.O. become equal to one third its kinetic energy? (A) 

17.

(B)

(B) 5/6

T1

2

T2

2

(C) 4/5

= …( k = 1 m/s 2 & g= 10 m/s 2 ) (D) 1

A hollow sphere is filled with water. There is a hole at the bottom of this sphere. This sphere is suspended with a string from a rigid support and given an oscillation. During oscillation, the hole is opened up and the periodic time of this oscillating system is measured. The periodic time of the system…………. (A) will remain constant (B) Will increase upto a certain time (C) Increases initially and then decreases to attain its initial periodic time (D) Initially decreases and then will attain the initial periodic time value. 329

21.

The periodic time of a S.H.O. oscillating about a fixed point is 2 s. After what time will the kinetic energy of the oscillator become 25% of its total energy? (A) 1/12 s

22.

(C) ¼ s

(D) 1/3 s.

A body having mass 5g is executing S.H.M. with an amplitude of 0.3 m. If the periodic time of the system is

 s, then the maximum force acting on body is ………. 10

(A) 0.6 N 23.

(B) 1/6 s

(B) 0.3 N

(C) 6 N

(D) 3 N

As shown in figure, a body having mass m is attached with two springs having spring constants k1 and k2. The frequency of oscillation is f. Now, if the springs constants of both the springs are increased 4 times, then the frequency of oscillation will be equal to ………….

24.

(A) 2f

(B) f/2

(C) f/4

(D) 4f

The figure shows a graph of displacement versus time for a particle executing S.H.M. The acceleration of the S.H.O. at the end of time t =

(A)

3 2  32

2 (C) 32 25.

(B) 

2 32

(D) 

3 2  32

4 second is ………..cm.s – 2 3

As shown in figure, the object having mass M is executing S.H.M. with an amplitude A. The amplitude of point P shown in figure will be ……. (A)

k1 A k2

k2 A (B) k 1

(C)

k1 A k1  k 2

(D)

k2 A k1  k 2

26.

A particle is executing S.H.M. between x = - A and x = +A. If the time taken by the particle to travel from x = 0 to A/2 is T1 and that taken to travel from x = A/2 to x = A is T2, then ………. (A) T1 < T2 ( B ) T1 > T2 (C) T1 = 2T2 (D) T1 = T2

27.

For a particle executing S.H.M., when the potential energy of the oscillator becomes 1/8 the maximum potential energy, the displacement of the oscillator in terms of amplitude A will be ………. (A)

A 2

(B)

A

(C)

2 2

330

A 2

(D)

A 3 2

.

28.

The average values of potential energy and kinetic energy over a cycle for a S.H.O. will be ……………….. respectively. 1 m 2 A 2 2

(A) 0 ,

(C) 29.

1

(B)

5

(D)

1 1 m 2 A 2 , m 2 A2 . 4 4

1 5

(C)

5 1

(D)

5 1

When a mass M is suspended from the free end of a spring, its periodic time is found to be T. Now, if the spring is divided into two equal parts and the same mass M is suspended and oscillated, the periodic time of oscillation is found to be T’. Then ……….. (A) T < T’

31.

1 m 2 A 2 , 0 2

The ratio of force constants of two springs is 1:5. The equal mass suspended at the free ends of both springs are performing S.H.M. If the maximum acceleration for both springs are equal, the ratio of amplitudes for both springs is ……… (A)

30.

1 1 m 2 A 2 , m 2 A 2 2 2

(B)

(B) T = T’

(C) T > T’

The periodic time of two oscillators are T and

(D) Nothing can be said.

5T respectively. Both oscillators starts their oscillation 4

simultaneously from the mid point of their path of motion. When the oscillator having periodic time T completes one oscillation, the phase difference between the two oscillators will be ……… (A) 900 32.

(C) 720

(D) 450

A rectangular block having mass m and cross sectional area A is floating in a liquid having density r. If this block in its equilibrium position is given a small vertical displacement, its starts oscillating with periodic time T. Then in this case….. (A) T

33.

(B) 1120

1 m

(C) T

(B) T 

1 A

(D) T

As shown in figure, a spring attached to the ground vertically has a horizontal massless plate with a 2 kg mass in it. When the spring ( massless ) is pressed slightly and released, the 2 kg mass, starts executing S.H.M. The force constant of the spring is 200 N m- 1. For what minimum value of amplitude, will the mass loose contact with the plate? ( Take g = 10 ms – 2 ) (A) 10.0 cm

(B) 8.0 cm

(C) 4.0 cm

(D) For any value less than 12.0 cm. 331

1 

34.

Which of the equation given below represents a S.H.M.? (A) acceleration = - k ( x + a )

(B) acceleration = k ( x + a )

(C) acceleration = kx

(D) acceleration = - k0x + k1x2

{ Here k, k0 and k1 are force constants and units of x and a is meter } 35.

The displacement for a particle performing S.H.M. is given by x = A Cos( ùt + Ô). If the initial position of the particle is 1 cm and its initial velocity is p cm s- 1 , then what will be its initial phase? The angular frequency of the particle is p s-1. (A)

36.

2 4

(B)

(B) 12

3 4

(D) 20

(C) 0.25 s- 1

(D) 2 s- 1

(B) 288p2

(C) 188p2

(D) None of these.

A simple pendulum having length l is given a small angular displacement at time t = 0 and released. After time t, the linear displacement of the bob of the pendulum is given by ……………… l t g

(B) x = aCos2p

g g t (C) x = aSin t l l

(D) x = aCos

g t l

Two masses m1 and m2 are attached to the two ends of a massless spring having force constant k. When the system is in equilibrium, if the mass m1 is detached, then the angular frequency of mass m2 will be ………….

(A) 41.

(D)

A particle having mass 1 kg is executing S.H.M. with an amplitude of 0.01 m and a frequency of 60 hz. The maximum force acting on this particle is ……….. N

(A) x = aSin2p 40.

(C) 21

(B) 3.14 s- 1

(A) 144p2 39.

5 4

The maximum velocity and maximum acceleration of a particle executing S.H.M. are 1 m/s and 3.14 m/s2 respectively. The frequency of oscillation for this particle is …….. (A) 0.5 s- 1

38.

(C)

Two simple pendulums having lengths 144 cm and 121 cm starts executing oscillations. At some time, both bobs of the pendulum are at the equilibrium positions and in same phase. After how many oscillations of the shorter pendulum will both the bob’s pass through the equilibrium position and will have same phase? (A) 11

37.

7 4

k m1

(B)

k m2

(C)

k  m1 m2

(D)

k m1  m2

When the displacement of a S.H.O. is equal to A/2, what fraction of total energy will be equal to kinetic energy? { A is amplitude } (A) 2/7

(B) ¾

(C) 2/9

332

(D) 5/7

42.

The speed of a particle executing motion changes with time according to the equation y = aSinùt + bCosùt, then …….. (A) Motion is periodic but not a S.H.M. (B) It is a S.H.M. with amplitude equal to a+b (C) It is a S.H.M. with amplitude equal to a2 + b2 (D) Motion is a S.H.M. with amplitude equal to a 2  b 2 .

43.

44.

A body is placed on a horizontal plank executing S.H.M. along vertical direction. Its amplitude of oscillation is 3.92 x 10 – 3m. What should be the minimum periodic time so that the body does not loose contact with the plank? (A) 0.1256 s (B) 0.1356 s (C) 0.1456 s (D) 0.1556 s 2 If the kinetic energy of a particle executing S.H.M. is given by K = K0 Cos ùt, then the displacement of the particle is given by ……….

 K0   (A)  2  m  45.

46.

1/ 2

Sint

Sint

1 n   (B) k   n 

 2 2 (C)   mK 0

  

1/ 2

Sint

 2K 0   (D)   m 

1/ 2

Sint

(C) k

(D)

k (n  1)

When a mass m is suspended from the free end of a massless spring having force constant k, its oscillates with frequency f. Now if the spring is divided into two equal parts and a mass 2m is suspended from the end of anyone of them, it will oscillate with a frequency equal to …………. (A) f

48.

1/ 2

The equation for displacement of two identical particles performing S.H.M. is given by x1 =4Sin(20t+p/6) and x2 =10Sinùt. For what value of ù will both particles have same energy? (A) 4 units (B) 8 units (C) 16 units (D) 20 units A spring having length l and spring constant k is divided into two parts having lengths l1 and l2. If l1 = nl2, the force constant of the spring having length l2 is ………. (A) k(1+n)

47.

 2K 0   (B)  2  m 

(B) 2f

(C)

f 2

(D)

2f

A mass m on an inclined smooth surface is attached to two springs as shown in figure. The other ends of both springs are attached to rigid surface. If the force constant of both spring is k, then the periodic time of oscillation for the system is ……… 1/ 2

M  (A) 2    2k 

 MgSin   (C) 2   2k 

 2M   (B) 2   k  1/ 2

1/ 2

1/ 2

 2Mg   (D) 2   k 

333

49.

50.

A body of mass 1 kg suspended from the free end of a spring having force constant 400 Nm-1 is executing S.H.M. When the total energy of the system is 2 joule, the maximum acceleration is ………ms – 2 . (A) 8 ms – 2 (B) 10 ms – 2 (C) 40 ms – 2 (D) 40 cms – 2 When a block of mass m is suspended from the free end of a massless spring having force constant k, its length increases by y. Now when the block is slightly pulled downwards and released, it starts executing S.H.M with amplitude A and angular frequency ù. The total energy of the system comprising of the block and spring is ………. 1 1 1 2 1 2 1 1 2 2 2 2 2 m 2 A 2 (B) m A  ky (C) ky (D) m A  ky . 2 2 2 2 2 2 A spring is attached to the center of a frictionless horizontal turn table and at the other end a body of mass 2 kg is attached. The length of the spring is 35 cm. Now when the turn table is rotated with an angular speed of 10 rad s – 1 , the length of the spring becomes 40 cm then the force constant of the spring is ……….. N/m. (A) 1.2 x 103 (B) 1.6 x 103 (C) 2.2 x 103 (D) 2.6 x 10 3 As shown in figure (a) and (b), a body of mass m is attached at the ends of the spring system. All springs have the same spring constant k. Now when both systems oscillates along vertical direction, the ratio of their periodic time is …….. (A) ¼ (B) ½ (C) 2 (D)4 A simple pendulum is executing S.H.M. around point O between the end points B and C with a periodic time of 6 s. If the distance between B and C is 20 cm then in what time will the bob move from C to D? Point D is at the mid-point of C and O. (A) 1 s (B) 2 s (C) 3 s (D) 4 s A small spherical steel ball is placed at a distance slightly away from the center of a concave mirror having radius of curvature 250 cm. If the ball is released, it will now move on the curved surface. What will be the periodic time of this motion? Ignore frictional force and take g = 10 m/s2.

(A) 51.

52.

53.

54.

  s (B) p s (c) s (D) 2p s 4 2 Two identical springs are attached at the opposite ends of a rod having length l and mass m. The rod could rotate about its mid-point O as shown in figure. Now, if the point A of the rod is pressed slightly and released, the rod starts executing oscillatory motion. The periodic time of this motion is ……… (A)

55.

(A) 2

(C) 

m 2k 2m 3k

(B) 2

(D) 

2m k

k

k

3m 2k 334

l 2

l 2

56.

A simple pendulum having length l is suspended at the roof of a train moving with constant acceleration ‘a’ along horizontal direction. The periodic time of this pendulum is ……… (A) T  2

57.

l (B) T  2 g

l l (D) T  2 . 2 g  a2 g a

l (C) T  2 ga

A trolley is sliding down a frictionless slope having inclination è. If a simple pendulum is suspended l on top of this trolley, its periodic time is given by T  2 g , where gefff = ……….. eff

(A) g 58.

(C) g cos θ

(D) g tan θ

One end of a massless spring having force constant k and length 50 cm is attached at the upper end of a plane inclined at an angle è = 300.When a body of mass m = 1.5 kg is attached at the lower end of the spring, the length of the spring increases by 2.5 cm. Now, if the mass is displaced by a small amount and released, t he amplitude of the resultant oscillation is ……….. (A)

59.

(B) g sin θ

 7

(B)

2 7

(C)

 5

Two blocks A and B are attached to the two ends of a spring having length L and force constant k on a horizontal surface. Initially the system is in equilibrium. Now a third block having same mass m, moving with velocity v collides with block A. In this situation………

(D)

2 5

v

(A) During maximum contraction of the spring, the kinetic energy of the system A-B will be zero. (B) During maximum contraction of the spring, the kinetic energy of the system A-B will be mv2 / 4

60.

(C) Maximum contraction of the spring is v

m k

(D) Maximum contraction of the spring is v

2m k

The displacement of a particle executing S.H.M. is given by y = 4Cos2(t/2)Sin1000t. This displacement is due to superposition of ………… S.H.M.’s. (A) 2

61.

(B) 3

(C) 4

The displacement of a particle is given by x = A Cost. Which of the following graph represents variation in potential energy as a function of time t and displacement x. (A) I, III

(B) II, IV

(C) II, III

(D) I, IV 335

(D) 5 PE

O

t

62.

A system is executing S.H.M. The potential energy of the system for displacement x is E1 and for a displacement of y, the potential energy of the system is E2. The potential energy for a displacement of (x+y) is ……… (A) E1 + E2

63.

4

(B)

5 5

(C) E1 + E2 + 2

E1 E 2 (D)

E1 E 2

5

(C)

4 5

8 4 5

(D)

8 5 5

(B) 2.5

(C) 3.33

(D) 2.33

A block is placed on a horizontal table. The table executes S.H.M. along the horizontal plane with a period T. The coefficient of static friction between the table and block is µ. The maximum amplitude of oscillation should be ………so that the block does not slide off the table.

(A) 66.

2

The periodic time of a simple pendulum is 3.3 s. Now if the point of support of the pendulum starts moving along the vertically upward direction with a velocity v = kt ( where k = 2.1 m/s2 ), then the new periodic time is …………… s. { Take g = 10 m/s2 } (A) 3

65.

2

E1  E 2

A system is executing S.H.M. with a periodic time of 4/5 s under the influence of force F1. When a force F2 is applied, the periodic time is (2/5) s. Now if F1 and F2 are applied simultaneously along the same direction, the periodic time will be ………

(A) 64.

(B)

gT 5

(B)

gT 2 4 2

(C)

gT 2

(D) µgT

As shown in figure, a block A having mass M is attached to one end of a massless spring. The block is on a frictionless horizontal surface and the free end of the spring is attached to a wall. Another block B having mass ‘m’ is placed on top of block A. Now on displacing this system horizontally and released, it executes S.H.M. What should be the maximum amplitude of oscillation so that B does not slide off A? Coefficient of static friction between the surfaces of the block’s is µ. (A) Amax =

mg k

(B) A max =

 (m  M ) g k

( C) Amax =

 ( M  m) g k

(D) Amax =

2 ( M  m) g k

336

67.

A particle is executing S.H.M. about the origin at x = 0. Which of the following graph shows variation in potential energy with displacement?

U(x)

U(x) (A)

(B)

U(x)

U(x) (C)

68.

(D)

A horizontal plank is executing SHM along the vertical direction with angular frequency ù. A coin is placed on top of this plank. If the amplitude of oscillation is increased gradually, for what maximum amplitude will the coin be on the verge of loosing contact with the plank? (A) When is plank is at its maximum height

(B) When the plank is at the midpoint.

g (C) When the amplitude is 2 

g2 (D) When the amplitude is 2  SECTION : II

Assertion – Reason type questions : Note: For the following questions, statement as well as the reason(s) are given. Each questions has four options. Select the correct option. (a)

Statement – 1 is true, statement- 2 is true; statement-2 is the correct explanation of statement – 1 .

(b)

Statement – 1 is true, statement- 2 is true but statement-2 is not the correct explanation of statement –1.

(c)

Statement – 1 is true, statement- 2 is false

(d)

Statement – 1 is false, statement- 2 is true (A) a

69.

(B) b

(C) c

(D) d

Statement – 1 : If a spring having spring constant k is divided into equal parts, then the spring constant of each part will be 2k.

337

Statement – 2 : When the length of the elastic spring is increased ( stretched ) by x, then the amount of work required to be done is (A) a 70.

(B) b

(C) c

(D) d

Statement – 1 : The periodic time of a S.H.O. depends on its amplitude and force constant. Statement – 2 : The elasticity and inertia decides the frequency of S.H.O. (A) a

71.

1 2 kx 2

(B) b

(C) c

(D) d

Statement – 1 : For small amplitude, the motion of a simple pendulum is a S.H.M. with periodic time T  2

l . For large amplitudes, periodic time is greater than 2 g

l . g

Statement – 2 : For large amplitude, the speed of the bob is more when it passes through the mid-point ( equilibrium point ). (A) a 72.

(D) d

(B) b

(C) c

(D) d

Statement – 1: The periodic time of a simple pendulum increases on the surface of moon. Statement – 2 : Moon is very small as compared to Earth. (A) a

74.

(C) c

Statement – 1 : Periodic time of a simple pendulum is independent of the mass of the bob. Statement – 2 : The restoring force does not depend on the mass of the bob. (A) a

73.

(B) b

(B) b

(C) c

(D) d

Statement – 1: If the length of a simple pendulum is increased by 3%, then the periodic time changes by 1.5%. Statement – 2 : Periodic time of a simple pendulum is proportional to its length. (A) a

75.

(B) b

(C) c

(D) d

Statement – 1: For a particle executing S.H.M. with an amplitude of 0.01 m and frequency 30 hz, the maximum acceleration is 36p2 m/s2. Statement – 2 : The maximum acceleration for the above particle is + ù2A, where A is amplitude. (A) a

76.

(B) b

(C) c

(D) d

Statement – 1 : The periodic time of a stiff spring is less than that of a soft spring. Statement – 2 : The periodic time of a spring depends on its force constant value and for a stiff spring, it is more. (A) a

(B) b

(C) c

338

(D) d

77.

St at ement – 1 : The amplitude of an oscillator decreases with time.

Statement – 2 : The frequency of an oscillator decreases with time. (A) a 78.

(B) b

(C) c

(D) d

Statement – 1 : For a particle executing SHM, the amplitude and phase is decided by its initial position and initial velocity. Statement – 2 : In a SHM, the amplitude and phase is dependent on the restoring force. (A) a

(B) b

(C) c

(D) d

SECTION – III COMPREHENSION BASED QUESTIONS NOTE: Questions 79 to 81 are based on the following passage. Passage – 1: As shown in figure, two light springs having force constants k1 = 1.8 N m – 1 and k2 = 3.2 N m – 1 and a block having mass m = 200 g are placed on a frictionless horizontal surface. One end of both springs are attached to rigid supports. The distance between the free ends of the spring is 60 cm and the block is moving in this gap with a speed v = 120 cm s – 1 .

v

79.

When the block is moving towards spring k2, what will be the time taken for the spring to get maximum compressed from point D? (A) p s

80.

(C) (p/3) s

(D) (p/4) s

When the block is moving towards k1, what will be the time taken for it to get maximum compressed from point C? (A) p s

81.

(B) (p/2) s

(B) (2/3) s

(C) (p/3) s

(D) (p/4) s

What will be the periodic time of the block, between the two springs? (A) 1+ (5p/6) s

(B) 1+ (7p/6) s

(C) 1+ (5p/12) s

(D) 1+ (7p/12) s

NOTE: Questions 82 to 84 are based on the following passage. Passage – 2 : A block having mass M is placed on a horizontal frictionless surface. This mass is attached to one end of a spring having force constant k. The other end of the spring is attached to a rigid wall. This system consisting of spring and mass M is executing SHM with amplitude A and frequency f. When the block is passing through the mid-point of its path of motion, a body of mass m is placed on top of it, as a result of which its amplitude and frequency changes to A’ and f’. 339

82.

f' The ratio of frequencies = ……….. f

(A)

83.

84.

 M    mM 

(B)

 m    mM 

(C)

 MA   '   mA 

 ( M  m) A '   (D)  mA  

If the velocity before putting the mass and after putting it is v and v1 respectively, then  M   (A)  mM 

M m  (B)   M 

The ratio of amplitudes

A1 = ……. A

(A)

 M  m    m 

(B)

1  M  m A   (C) M m A

 m     M  m

(C)

 M     M  m

v1 = ………. v

1 M m A   (D) .  M  m A

(D)

 M  m    M 

NOTE: Questions 85 to 90 are based on the following passage. Passage – 3: The equation for displacement of a particle at time t is given by the equation y = 3Cos2t + 4Sin2t. . 85.

The motion of the particle is ……… (A) Damped motion

86.

(D) 2p s

(B) 3

(C) 5

(D) 7

(B) 12

(C) 20

(D) 28

If the mass of the particle is 5 gm, then the total energy of the particle is ……erg. (A) 250

90.

(C) (/2) s

The maximum acceleration of the particle is ……..cm / s2. (A) 4

89.

(B) p s

The amplitude of oscillation is ……cm (A) 1

88.

(C) Rotational motion (D) S.H.M.

The periodic time of oscillation is ……… (A) 2 s

87.

(B) Periodic motion

(B) 125

(C) 500

(D) 375

(C) (1/2p)

(D) (p/2)

The frequency of the particle is ………s- 1 . (A) (1/p)

(B) p 340

Waves SECTION – I : 91.

Equation for a harmonic progressive wave is given by y = A sin ( 15pt + 10px + p/3) where x is in meter and t is in seconds. This wave is ………. (A) Travelling along the positive x direction with a speed of 1.5 ms – 1 . (B) Travelling along the negative x direction with a speed of 1.5 ms – 1 . (C) Has a wavelength of 1.5 m along the – x direction. (D) Has a wavelength of 1.5 m along the positive x – direction.

92.

If the velocity of sound wave in humid air is vm and that in dry air is vd, then…… (A) vm > vd

93.

( D ) vm >> vd

(B) 5:2

(C) 3:5

(D) 5:3

If the maximum frequency of a sound wave at room temperature is 20,000 hz then its minimum wavelength will be approximately ……… ( v = 340 ms – 1 ) (A) 0.2 A 0

95.

( C ) vm = vd

The ratio of frequencies of two waves travelling through the same medium is 2:5. The ratio of their wavelengths will be ……… (A) 2:5

94.

( B ) vm < vd

( B ) 5 A0

( C ) 5 cm to 2 m

(D) 20 mm

If the equation of a wave in a string having linear mass density 0.04 kg m – 1 is given by y = 0.02   t x    , then the tension in the string is …………..N. ( All values are in mks ) Sin  2    0.04 0.50  (A) 6.25

96.

(B) 4.0

(C) 12.5

(D) 0.5

 t x If the equation for a transverse wave is y = A Sin2p    , then for what wavelength will the T  

maximum velocity of the particle be double the wave velocity? (A) 97.

A 4

(B)

A 2

(C) pA

(D) 2pA

Consider two points lying at a distance of 10 m and 15 m from an oscillating source. If the periodic time of oscillation is 0.05 s and the velocity of wave produced is 300 m/s, then what will be the phase difference the two points? (A) p

(B) p/6

(C) p/3

341

(D) 2p/3

98.

A string is divided into three parts having lengths l1, l2 and l3 each. If the fundamental frequency of these parts are f1, f2 and f3 respectively, then the fundamental frequency of the original string f = ………. (A)

f 

f1 

f2 

f

(B) f = f1 + f2 + f3

3

1 1   (D) f f1

1 1 1 1 (C) f  f  f  f 1 2 3

99.

1 f

 2

1 f

3

Waves produced by two tuning forks are given by y1 = 4Sin500pt and y2 = 2Sin506pt. The number of beats produced per minute is ……. (A) 360

(B) 180

(c) 60

(D) 3

100. Equation for a progressive harmonic wave is given by y = 8Sin2p( 0.1x – 2t), where x and y are in cm and t is in seconds. What will be the phase difference between two particles of this wave separated by a distance of 2 cm? (A) 180

(B) 360

(C) 720

(D) 540

101. As shown in figure, two pulses in a string having center to center distance of 8 cm are travelling along mutually opposite direction. If the speed of both the pulse is 2 cm/s, then after 2 s, the energy of these pulses will be ……… (A) zero (B) totally kinetic energy (C) totally potential energy (D) Partially potential energy and partially kinetic energy. 102. Two waves are represented by y1 = ASinùt and y2 = aCosùt. The phase of the first wave, w.r.t. to the second wave is ………. (A) more by radian

(B) less by p radian

(C) more by p/2

(D) less by p/2

103. If the resultant of two waves having amplitude b is b, then the phase difference between the two waves is ……. (A) 1200

(B) 600

(C) 900

(D) 1800

104. If two antinodes and three nodes are formed in a distance of 1.21 A0, then the wavelength of the stationary wave is ………. ( A ) 2.42 A0

(B) 6.05 A0

(C) 3.63 A0

(D) 1.21 A0

105. The function Sin2(ùt ) represents…… (A) A SHM with periodic time p/ù.

(B) A SHM with a periodic time 2p/ù.

(C) A periodic motion with periodic time p/ù.

(D) A periodic motion with period 2p/ù.

342

106. If two almost identical waves having frequencies n1 and n2, produced one after the other superposes then the time interval to obtain a beat of maximum intensity is …….. 1 (A) n  n 1 2

1 1 (B) n  n 1 2

1 1 (C) n  n 1 2

1 (D) n  n 1 2

107. When two sound waves having amplitude A , angular frequency ù and a phase difference of p/2 superposes, the maximum amplitude and angular frequency of the resultant wave is ………

108.

109.

110.

111.

112.

A   (C) ,ù (D) 2 A, 2 2 2 2 The amplitude of a wave in a string is 2 cm. This wave is propagating along the x-direction with a speed of 128 m/s. Five such waves are accommodated in 4 m length of the string. The equation for this wave is ……. (A) y = 0.02Sin(15.7x – 2010t ) m (B) y = 0.02Sin(15.7x + 2010t ) m (C) y = 0.02Sin(7.85x – 1005t ) m (D) y = 0.02Sin(7.85x + 1005t ) m A string of length 70 cm is stretched between two rigid supports. The resonant frequency for this string is found to be 420 hz and 315 hz. If there are no resonant frequencies between these two values, then what would be the minimum resonant frequency of this string? (A) 10.5 hz (B) 1.05 hz (C) 105 hz (D) 1050 hz Sound waves propagates with a speed of 350 m/s through air and with a speed of 3500 m/s through brass. If a sound wave having frequency 700 hz passes from air to brass, then its wavelength …………. (A) decreases by a fraction of 10 (B) increases 20 times (C) increases 10 times (D) decreases by a fraction of 20 A transverse wave is represented by y = ASin (ùt-kx). For what value of its wavelength will the wave velocity be equal to the maximum velocity of the particle taking part in the wave propagation? (A) 2pA (B) A (C) pA (D) pA/2 Two monoatomic ideal gases 1 and 2 has molecular weights m1 and m2. Both are kept in two different containers at the same temperature. The ratio of velocity of sound wave in gas 1 and 2 is ………. (A)

2A, ù

(B)

(A)

m2 m1

(B)

A

,

m1 m2

(C)

m1 m2

(D)

m2 m1

113. A wire having length L is kept under tension between x = 0 and x = L. In one experiment, the  x  equation of the wave and energy is given by y1 = ASin   Sinùt and E1 respectively. In another  L  2x   Sin2ùt and E2. Then …… experiment, it is y2 = ASin   L 

(A) E2 = E1

(B) E2 = 2E1

(C) E2 = 4E1 343

(D) E2 = 16E1

114. Twenty four tuning forks are arranged in such a way that each fork produces 6 beats/s with the preceding fork. If the frequency of the last tuning fork is double than the first fork, then the frequency of the second tuning fork is ……… (A) 132

(B) 138

(C) 276

(D) 144

115. If two SHM’s are given by the equation y1 = 0.1 Sin(100pt + p/3) and y2 = 0.1 Cospt, then the phase difference between the velocity of particle 1 and 2 is ……… (A) p/6

(B) - p/3

(C) p/3

(D) - p/6 –1

116. The wave number for a wave having wavelength 0.005 m is …….. m . (A) 5

(B) 50

(C) 100

(D) 200

117. An listener is moving towards a stationary source of sound with a speed ¼ times the speed of sound. What will be the percentage increase in the frequency of sound heard by the listener? (A) 20%

(B) 25%

(C) 2.5%

(D) 5%

118. When the resonance tube experiment, to measure speed of sound is performed in winter, the first harmonic is obtained for 16 cm length of air column. If the same experiment is performed in summer, the second harmonic is obtained for x length of air column. Then …. (A) 32 > x > 16

(B) 16 > x

(C) x > 48

(D) 48 > x > 32

119. What should be the speed of a source of sound moving towards a stationary listener, so that the frequency of sound heard by the listener is double the frequency of sound produced by the source? { Speed of sound wave is v } (A) v

( B ) 2v

(C) v/2

(D) v /4

120. A metal wire having linear mass density 10 g/m is passed over two supports separated by a distance of 1 m. The wire is kept in tension by suspending a 10 kg mass. The mid point of the wire passes through a magnetic field provided by magnets and an a.c. supply having frequency n is passed through the wire. If the wire starts vibrating with its resonant frequency, what is the frequency ofa.c. supply? (A) 50 hz

(B) 100 hz

(C) 200 hz

(D) 25 hz

121. If the listener and the source of sound moves along the same direction with the same speed, then…….. (A)

fL <1 fs

fL (B) f =0 s

fL (C) f = 1 s

fL (D) f >1 s

122. A wire of length 10 m and mass 3 kg is suspended from a rigid support. The wire has uniform cross sectional area. Now a block of mass 1 kg is suspended at the free end of the wire and a wave having wavelength 0.05 m is produced at the lower end of the wire. What will be the wavelength of this wave when it reached the upper end of the wire? (A) 0.12 m

(B) 0.18 m

(C) 0.14 m

(D) 0.10 m

123. If the mass of 1 mole of air is 29 x 10 – 3 kg, then the speed of sound in it at STP is …….. ( ã=7/5). { T = 273 K, P = 1.01 x 105 Pa } (A) 270 m/s

(B) 290 m/s

(C) 330 m/s

344

(D) 350 m/s

124. A wave travelling along a string is described by y = 0.005Sin(40x – 2t) in SI units. The wavelength and frequency of the wave are……… (A) (p/5) m ; 0.12 hz

(B) (p/10) m ; 0.24 hz (C) (p/40) m ; 0.48 hz (D) (p/20) m ; 0.32 hz

125. Two sitar strings A and B playing the note “Dha” are slightly out of time and produce beats of frequency 5 hz. The tension of the string B is slightly increased and the beat frequency is found to decrease to 3 hz. What is the original frequency of B if the frequency of A is 427 hz? (A) 432

(B) 422

(C) 437

(D) 417

126. A rocket is moving at a speed of 130 m/s towards a stationary target. While moving, it emits a wave of frequency 800 hz. Calculate the frequency of the sound as detected by the target. ( Speed of wave = 330 m/s) (A) 1320 hz

(B) 2540 hz

(C) 1270 hz

(D) 660 hz

127. Length of a steel wire is 11 m and its mass is 2.2 kg. What should be the tension in the wire so that the speed of a transverse wave in it is equal to the speed of sound in dry air at 200 C temperature? (A) 2.31 x 104 N

(B) 2.25 x 104 N

(C) 2.06 x 104 N

(D) 2.56 x 104 N

128. A wire stretched between two rigid supports vibrates with a frequency of 45 hz. If the mass of the wire is 3.5 x 10 – 2 kg and its linear mass density is 4.0 x 10 - 2 kg/m, what will be the tension in the wire ? (A) 212 N

(B) 236 N

(C) 248 N

(D) 254 N

129. Tube A has both ends open while tube B has one end closed, otherwise they are identical. The ratio of fundamental frequency of tube A and B is …….. (A) 1:2

(B) 1:4

(C) 2:1

(D) 4:1

130. A tuning fork arrangement produces 4 beats/second with one fork of frequency 288 hz. A little wax is applied on the unknown fork and it then produces 2 beats/s. The frequency of the unknown fork is ……….hz. (A) 286

(B) 292

(C) 294

(D) 288

131. A wave y = aSin(ùt – kx ) on a string meets with another wave producing a node at the equation of the unknown wave is ……… (A) y = a Sin(ùt + kx ) (B) y = - aSin( ùt+kx) (C) y = aSin(ùt – kx )

x = 0. Then

(D) y = - aSin(ùt – kx)

132. When temperature increases, the frequency of a tuning fork……. (A) Increases

(B) Decreases

(C) remains same

(D) Increases or decreases depending on the material.

133. A tuning fork of known frequency 256 hz makes 5 beats per second with the vibrating string of a piano. The beats frequency decreases to 2 beats/s when the tension in the piano string is slightly increased. The frequency of the piano string before increase in the tension was ……..hz. (A) 256 + 2

(B) 256 – 2

(C) 256 – 5

345

(D) 256 + 5.

134. An observer moves towards a stationary source of sound with a velocity one – fifth the velocity of sound. What is the percentage increase in the apparent frequency? (A) 5%

(B) 20%

(C) zero

(D) 0.5%

135. The speed of sound in Oxygen ( O2) at a certain temperature is 460 m/s. The speed of sound in helium at the same temperature will be …….…….ms- 1 . (Assume both gases to be ideal) (A) 330 (B) 460 (C) 5002 (D) None of these 136. In a longitudinal wave, pressure variation and displacement variation are……… (A) In phase (B) 900 out of phase (C) 450 out of phase (D) 1800 out of phase 137. A tuning fork of frequency 480 hz produces 10 beats/s when sounded with a vibrating sonometer string. What must have been the frequency of the string if a slight increase in tension produces fewer beats per second than before? (A) 480 (B) 490 hz (C) 460 hz (D) 470 hz 138. Which of the following functions represents a wave? (A) ( x – vt)2

(B) ln( x + vt )

(C) e  ( x  vt ) 2

1 x  vt

(D)

139. Two sound waves are represented by y = aSin(ùt-kx) and y = aCos(ùt-kx). The phase difference between the waves in water is …….. (A)

 2

(B)

 4

(C) 

(D)

3 4

140. A string of linear density 0.2 kg/m is stretched with a force of 500 N. A transverse wave of length 4.0 m and amplitude 1/l meter is travelling along the string. The speed of the wave is …….. m/s. (A) 50 (B) 62.5 (C) 2500 (D) 12.5 141. Two wires made up of same material are of equal lengths but their radii are in the ratio 1:2. On stretching each of these two strings by the same tension, the ratio between their fundamental frequency is …….. (A) 1:2 (B) 2:1 (C) 1:4 (D) 4:1 142. The tension in a wire is decreased by 19%, then the percentage decrease in frequency will be ……… (A) 19% (B) 10% (C) 0.19% (D) None of these 143. An open organ pipe has fundamental frequency 100 hz. What frequency will be produced if its one end is closed? (A) 100, 200, 300, …. (B) 50, 150, 250…. (C) 50, 100, 200, 300… (D) 50, 100, 150, 200,….. 144. A closed organ pipe has fundamental frequency 100 hz. What frequencies will be produced if its other end is also opened? (A) 200, 400, 600, 800,…..

(B) 200, 300, 400, 500, …..

(C) 100, 300, 500, 700,……

(D) 100, 200, 300, 400, …….

346

145. A column of air of length 50 cm resonates with a stretched string of length 40 cm. The length of the same air column which will resonate with 60 cm of the same string at the same tension is …….. (A) 100 cm

(B) 75 cm

(C) 50 cm

(D) 25 cm

146. Two forks A and B when sounded together produce 4 beats/s. The fork A is in unison with 30 cm length of a sonometer wire and B is in unison with 25 cm length of the same wire at the same tension. The frequencies of the fork are ……. (A) 24 hz, 28 hz

(B) 20 hz, 24 hz

(C) 16 hz , 20 hz

(D) 26 hz, 30 hz

147. A tuning fork of frequency 200 hz is in unison with a sonometer wire. The number of beats heard per second when the tension is increased by 1 % is ……. (A) 1

(B) 2

(C) 4

(D) 0.5

148. A bus is moving with a velocity of 5 m/s towards a huge wall. The driver sounds a horn of frequency 165 hz. If the speed of sound in air is 335 m/s, the number of beats heard per second by the passengers in the bus will be ……. (A) 3

(B) 4

(C) 5

(D) 6

149. A vehicle with a horn of frequency n is moving with a velocity of 30 m/s in a direction perpendicular to the straight line joining the observer and the vehicle. The observer perceives the sound to have a frequency ( n + n1 ). If the sound velocity in air is 300 m/s, then…… (A) n1 = 10n

(B) n1 = 0

(C) n1 = 0.1 n

(D) n1 = - 0.1n

150. In a sine wave, position of different particles at time t = 0 is shown in figure. The equation for this wave travelling along the positive x – direction can be ……. y

(A) y = Asin ( ω t – kx ) (B) y = Acos (kx – ω t)

0

x

(C) y = Acos( ω t – kx ) (D) y = Asin(kx – ω t) 151. Which of the following changes at an antinode in a stationary wave? (A) Density only

(B) Pressure only

(C) Both pressure and density

(D) Neither pressure nor density

152. A sonometer wire supports a 4 kg load and vibrates in fundamental mode with a tuning fork of frequency 416 hz. The length of the wire between the bridges is now doubled. In order to maintain fundamental mode, the load should be changed to ……. (A) 1 kg

(B) 2 kg

(C) 8 kg

(D) 16 kg

153. In brass, the velocity of a longitudinal wave is 100 times the velocity of a transverse wave. If Y = 1 x 1011 N/m2, then stress in the wire is ………… (A) 1 x 1013 N/m2

(B) 1 x 109 N/m2

(C) 1 x 1011 N/m2 347

(D) 1 x 107 N/m2.

154. The frequency of tuning fork A is 2% more than the frequency of a standard fork. Frequency of tuning fork B is 3% less than the frequency of the standard fork. If 6 beats per second are heard when the two forks A and B are excited, then frequency of A is …… hz. (A) 120

(B) 122.4

(C) 116.4

(D) 130

155. Fundamental frequency of a sonometer wire is n. If the length and diameter of the wire are doubled keeping the tension same, the new fundamental frequency is ……… (A)

2n 2

(B)

n

(C)

2 2

2n

(D)

n 4

156. A car blowing its horn at 480 hz moves towards a high wall at a speed of 20 m/s. If the speed of sound is 340 m/s, the frequency of the reflected sound heard by the driver sitting in the car will be closest to ……..hz. (A) 540

(B) 524

(C) 568

(D) 480

157. A cylindrical tube open at both ends has a fundamental frequency f in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now……… (A) f/2 (B) f (C) 3f/4 (D) 2f 158. Three sound waves of equal amplitudes have frequencies (v-1 ), v, ( v+1). They superpose to give beats. The number of beats produced per second will be ……. (A) 3

(B) 2

(C) 1

(D) 4

159. A wave travelling along the x-axis is described by the equation y(x,t) = 0.005Cos(áx - ât). If the wavelength and the time period of the wave are 0.08 m and 2.0 s respectively, then á and â in appropriate units are ……… (A) á = 12.50p , â= p/2.0

(B) á = 25p , â= p

(C) á = 0.08/p , â= 2.0/p

(D) á = 0.04/p , â= 1.0/p

160. A wave travelling along a string is described by the equation y = ASin(ùt-kx). The maximum particle velocity is ……….. (A) Aù

(B) ù/k

(C) dù/dk

(D) x

161. A string is stretched between fixed points seperated by 75 cm. It is observed to have a resonant frequencies of 420 hz and 315 hz. There are other resonant frequencies between these two. Then the lowest frequency for this string is ……..hz. (A) 1.05

(B) 1050

(C) 10.5

(D) 105

162. Two tuning forks P and Q when set vibrating gives 4 beats/second. If the prong of fork P is filed, the beats are reduced to 2/s. What is the frequency of P, if that of Q is 250 hz.? (A) 246 hz

(B) 250 hz

(C) 254 hz

(D) 252 hz

163. The length of a string tied across two rigid supports is 40 cm. The maximum wavelength of a stationary wave that can be produced in it is ………. cm. (A) 20

(B) 40

(C) 80 348

(D) 120

164. A stationary wave of frequency 200 hz are formed in air. If the velocity of the wave is 360 m/s, the shortest distance between two antinodes is …….m (A) 1.8

(B) 3.6

(C) 0.9

(D) 0.45

165. A tuning fork produces 8 beats per second with both 80 cm and 70 cm of stretched wire of a sonometer. Frequency of the fork is …….hz. (A) 120

(B) 128

(C) 112

(D) 240

166. An open pipe is in resonance in 2nd harmonic with frequency f1. Now one end of the tube is closed and frequency is increased to f2, such that the resonance again occurs in the nth harmonic. Choose the correct option. (A) n = 3, f2 = (3/4)f1 (B) n = 3, f2 = (5/4)f1

(C) n = 5, f2 = (5/4)f1

(D) n = 5, f2 = (3/4)f1

SECTION : II Assertion – Reason type questions : Note: For the following questions, statement as well as the reason(s) are given. Each questions has four options. Select the correct option. (a)

Statement – 1 is true, statement- 2 is true; statement-2 is the correct explanation of statement – 1 .

(b)

Statement – 1 is true, statement- 2 is true but statement-2 is not the correct explanation of statement –1.

(c)

Statement – 1 is true, statement- 2 is false

(d)

Statement – 1 is false, statement- 2 is true

167. Statement – 1: Two waves moving in a uniform string having uniform tension cannot have different velocities. Statement – 2 : Elastic and inertial properties of string are same for all waves in same string. Moreover speed of wave in a string depends on its elastic and inertial properties only. (A) a

(B) b

(C) c

(D) d

168. Statement – 1: When a sound source moves towards observer, then frequency of sound increases. Statement – 2 : Wavelength of sound in a medium moving towards the observer decreases. (A) a

(B) b

(C) c

(D) d

169. Statement – 1: Newton’s equation for speed of sound was found wrong because he assumed the process to be isothermal. Statement – 2 : When sound propagates, the compressions and rarefactions happen so rapidly that there is not enough time for heat to be distributed. (A) a

(B) b

(C) c

349

(D) d

170. Statement – 1 : When pressure in a gas changes, velocity of sound in gas may change. Statement – 2 : Velocity of sound is directly proportional to square root of pressure. (A) a

(B) b

(C) c

(D) d

171. Statement – 1 : If wave enters from one medium to another medium then sum of amplitudes of reflected wave and transmitted wave is equal to the amplitude of incident wave. Statement – 2 : If wave enters from one medium to another medium some part of energy is transmitted and rest of the energy is reflected back. (A) a

(B) b

(C) c

(D) d

SECTION – III COMPREHENSION BASED QUESTIONS NOTE: Questions 172 to 174 are based on the following passage. Passage – 1 A string 25 cm long and having a mass of 2.5 g is under tension. A pipe closed at one end is 40 cm long. When the string is set vibrating in its first overtone and the air in the pipe in its fundamental frequency, 8 beats per second is heard. It is observed that decreasing the tension in the string decreases the beat frequency. The speed of sound in air is 320 ms – 1 . 172. The frequency of the fundamental mode of the closed pipe is ….. hz (A) 100

(B) 200

(C) 300

(D) 400

173. The frequency of the string vibrating in its 1st overtone is …… hz (A) 92

(B) 108

(C) 192

(D) 208.

174. The tension in the string is very nearly equal to …… (A) 25 N

(B) 27 N

(C) 28 N

(D) 30 N

NOTE: Questions 175 to 178 are based on the following passage. Passage – 2 Standing waves are produced by the superposition of two waves y1 = 0.05Sin(3pt – 2x) and 0.05Sin(3pt + 2x) where x and y are in meters and t is in seconds. 175. The speed ( in ms – 1 ) of each wave is …… (A) 1.5

(B) 3.0

(C) 3p/2

(D) 3p

176. The distance ( in meters ) between two consecutive nodes is ……. (A) p/2

(B) p

(C) 0.5 350

(D) 1.0

y2 =

177. The amplitude of a particle at x = 0.5 m is …… (A) 1.08 x 10 – 1 m

(B) 5.4 x 10 –2 m

(C) (p/2) x 10 – 1 m

(D) p x 10 – 1 m

178. The velocity ( in ms – 1 ) of a particle at x = 0.25 m at t = 0.5 s is …… (A) 0.1p

(B) 0.3p

(C) zero

(D) 0.3

NOTE: Questions 179 to 181 are based on the following passage. Passage – 3 When two sound waves travel in the same direction in a medium, the displacement of a particle located at x at time t is given by y1 = 0.05Cos(0.50px - 100pt ) & y2 = 0.05Cos ( 0.46px - 92pt ), where y1, y2 and x are in meter and t is in seconds. 179. What is the speed of sound in the medium? (A) 332 m/s

(B) 100 m/s

(C) 92 m/s

(D) 200 m/s

180. How many times per second does an observer hear the sound of maximum intensity? (A) 4

(B) 8

(C) 12

(D) 16

181. At x = 0, how many times between t = 0 and t = 1 s does the resultant displacement become zero? (A) 46

(B) 50

(C) 92

(D) 100

NOTE: Questions 182 to 183 are based on the following passage. Passage – 4 The equation y = 10Sin

x Cos10t represents a stationary wave where x and y are in centimeter and t is 4

in seconds. 182. The amplitude of each component wave is ……. (A) 5 cm

(B) 10 cm

(C) 20 cm

(D) between 5 cm and 10 cm.

183. The separation between two consecutive nodes is ……… (A) 2 cm

(B) 4 cm

(C) 5 cm

351

(D) 8 cm

KEY NOTE 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 30 31 32 33 34 35 36 37 38 39 40

(B) (C) (B) (C) (A) (D) (D) (A) (C) (A) (C) (B) (C) (C) (A) (A) (B) (C) (A) (C) (D) (A) (A) (D) (D) (A) (B) (D) (C) (C) (C) (C) (A) (A) (D) (B) (A) (A) (D) (B)

41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

(B) (D) (A) (B) (B) (A) (A) (A) (C) (B) (B) (B) (A) (B) (C) (D) (C) (A) (B) (B) (A) (C) (A) (A) (B) (B) (D) (C) (B) (D) (B) (C) (B) (C) (B) (A) (C) (C) (D) (C)

81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120

352

(D) (A) (A) (C) (D) (B) (C) (C) (A) (A) (B) (A) (B) (D) (A) (C) (D) (C) (B) (C) (B) (D) (A) (D) (A) (A) (A) (C) (C) (C) (A) (A) (C) (B) (B) (D) (B) (C) (C) (A)

121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160

(C) (D) (C) (D) (B) (A) (A) (C) (C) (B) (B) (B) (C) (B) (D) (D) (D) (C) (A) (A) (B) (B) (B) (A) (B) (B) (A) (C) (B) (D) (D) (D) (D) (A) (D) (A) (B) (C) (B) (A)

161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183

(D) (A) (C) (C) (A) (C) (D) (A) (A) (B) (D) (B) (D) (B) (C) (A) (B) (C) (D) (A) (D) (A) (B)

HINT 1.

y  sin 2t  3 cos 2t

3  1   y  2  sin 2t  cos 2t  2  2 

 2 cos  sin 2t  sin  cos 2t  2sin(2t  ) Compaling than with y  A sin(wt  ), we get w 2

2  2 t   s T

2.

3.

T  2

 g When lift moves up with accleratim 3 the effective graritatianl acclenations g

in g1  g  g 3  4g 3 l'  new peliodic time T '  2  g '

4.

1 

dy1  2 10 cos(10t  ) dt

 2  3 10sin t  30 cos(10t   2 )





   Phase diffdence = 10t     10t  2   2 5.

K1k 2 k For series combination, K S  K  k  2 ( k1  k 2  K) 1 2

T'   T k

now T  1 6.

ks = 2  T' = 2 T k

For maximum velocity, A11  A 2 2



A1  2   A 2 1

k2 / m2  k1 / m1

k2 ( m1  m 2 ) k1

353

7.

Reduced mass of system  

m1m 2  0.75 kg m1  m 2

1 k 20   3Hz 2 m  KE at o = PE at A

 freq of oscillation f 

8.

1  m 2  mgh  mg 1  cos  2

  2g(1  cos )

9.

K11  K 2  2  k  3   K1    K 2     k  4 4   force constant of spring having lenght

k2 

3 l in 4

4 k 3

10. Amplitnde of SHM given by x = a sin t+b cos t in 1

A

2 2 2 a + b = (3  4 )  5m 2

2

2

11.

y  u u  y  2   2   u 2  4u u1  y1  2

12. T  l 1 because 2 and g are constants



T2  T1

l2 1.21l1   1.1 l1 l1

 %increase 

T2  T1  100  10% T1

13. y  A sin(t  )

A 1  A sin t   0   sin t   4 2 2 

2  .t  t  T 8 T 4

354

14. T1  2 

m1 4  2 .M 42 M  k1  and k  2 2 2 k1 T1 T2

for Series connection; T  2

 T  2

M kk where k  1 2 k k1  k 2

M1 2 2 (T1  T2 ) 2 4 M

 T  T12  T2 2  T 2  T12  T22 15.

dx  x  A cos( t   )  v   – A sin (t  ) 8 dt 2

If sin(t   8 )  1, then velocity will be maximam  t 

  3 3   t  t 8 2 8 8

16. 3u = k 1 1 A  3  ky 2  k(A 2  y 2 )  y   2 2 2

17. For spring A b restaning face F=kx  displacement x  F k

figure (b) if the resultant spring contant k 1k 2 1 m  k 1 , then k  k  k (Series) 1 2

k2  in figure (b) if on applying a force F`, if

displacement in x`, then x '  F ' k ' 

x ' F' k 6    2  3 x k' F 4

 x '  3x  3  1cm  3cm

18. 1 

dy1 dt

and v2 

dy2 dt

355

19. Here y  kt 2



dy d2y  2kt  2  2k  2 ms2 dt dt

m 2  the point of support in moving upwards with an accelaration of 2 s n  effective acc l . g '  g  a  12 m / s 2

Now T1  2

20.

T  2

  and T2  2 g g'

' as water leaks, the center of gravity moves down and hence “  ” increases. g

 T increases initially When all the water has leaked, the center of gravity moves up and hence “  ” decreases and hence T decreases Finally the centre of gravity steady at the center of sphde and so T will remain constant. 21. Kinetic energy = 25 % E

K 

22. Fmax  ma max

1 E 4

4 2  mA  mA 2  0.6 N. T 2

23. For parallel combination k p  k 1  k 2  freqrency f 

1 k 1 kl  k2   (1) 2 m 2 m

Now when k1 and k2 in increased 4 times,

f1 

1 4(K1  K 2 )  2f .(from(1)) 2 m

24. From graph A=1 cm  T=8s  y  A sin t  A sin

a  w2 y 

2 3 ty cm T 2

-42 31 31 2 cm .    2 T2 2 32 s

356

25. Now k 

1  k  contant 

 k11  k 2  2   2  k11  k +k   k2  Now, A = l1  1 2   l1 =  A  k2   k2  k2  26.   w A 2  x 2 , the velocity for moving form x=o to x  A 2 will ge more them for x  A 2 tox  A  T1  T2. 27. U 

1 u max 8

1 2 11 2  A2 2  ky   kA   y  2 8 2 8  28. In the expression for both Kinetic and potential energy, We have the square of the halmonic functions (sine or cisine). The average of which over a cycle is 1 2  u 

E 1  K  m2 A 2 2 4

29. Angnlar freqvency  

k m

Since ‘m’ is constant,   K

a max A

2 Now, a max  A   



1

a max a  k  max  A A K

A 

1 k

30. For a spring, T  2 

m k

 T 

1 k

357

( m is comtant)

2 31. Phase of 1st oscillater 1  1 t    T t   1 2

t+ For 2nd oscillater,  2  2 t    T2

phase diff θ1 -θ 2 32. Restoring force F = – Ayg = – (Ag)y = – ky  k  Ag  T  2

m 1  T k A1

33. To loose comtact, the condition in ; m2 A  mg A  g



2



mg ( k  mw 2 ) k

34. In SHM, accelelation and displacement are opposite in direction Also a  y.. 35. Here t = 0, x = 1 cm and    cm s 1 , w   s 1 Now, x  A cos(t  ) ---- (1) dx   A sin(t  ) ----- (2) dt Solved the equation (1) and (2)

Velocity  

36. T  2 144 and T  2 121 1 2 g g

 T1  T2

 When the shorter pendulum completes n oscillations, the longer one completes (n-1) oscilla tions (when in same phase).

 nT2  (n  1)T1 37.   

3.14 2 r  0.5s 1  3.14  2f  3.14  f  2 r

38. Maximum force  m2 A  m4 2f 2 A 39. Periodic time T  2

l 2 g and    g T l

Linear displacement x = a cos  t

358

40.  

k k '  or removing m, angular frequency m1  m 2 m2

41. Kinetic energy K 

1 m2 (A 2  y 2 ) 2

Now total energy E 

1 m 2 A 2 2

42. y = a sin  t + b cos  t, Taking a = Acosθ and b = Asinθ, y  A cos  sin t  A sin   cos t  A sin( t  )

Now a 2  b 2  A 2  A  a 2  b 2. 43. The body will not loose contact with the surface, 2 if mg  m r 

m42  r {where r is amplitude} T2

 T  2

1  2K 0  2 2 44. Maximum kinetic energy K 0  m A  A   2  2  m   Equation for displacement is ; 1

 2K o  2 y  A sin t   sin t 2   m  45. E 

1 m2 A 2  E2 A 2 2

 E(A) 2  (1A1 ) 2  (2 A 2 ) 2 46. k 2 be the spring constant of the spring having lenght  2. Now, 1   2.   n 2   2.   47. f 

1 k 1 2k and f '  {k'  2k} 2 m 2 2m f '  f

359

1

2

r g

48. Here both springs are in parallel. The restoring force on the system in only due to spring and not due to gravitational force  We can ignore the slope. Equivalent spring cantant =k+k=2k 1  Periodic time T  2 M 2k

49. Enelgy stoved =Work done E 

1 kA 2 2

Now maximum accelaration a max  2 A

50. Potential energy gainad by the spring on suspending mass “m” is

1 2 ky . 2

When system executes SHM, the energy gained by the system   total final energy of the system 

1 1 m2 A 2  ky 2 . 2 2

51. Radius of the rotational motion r =0.4 m When the turn table rotates, the restoring force developed in the spring = centrifugal force  Frestore  m2 r  2(10) 2  0.4  80N

Now increase in lenght of spring = 40-35 = 5 cm  Force constant k 

F 80   1.6  103 N / m. x 0.05

52. In case-I, springs are connected in parallel.  equivalent force constant k p  k1  k 2  2k.

m m  Peliodic time Tp  2 kp  2 2k

In case-II, spring are connected in series. k1k 2 k  Equivalent force constant k s  k  k  2 1 2 m

 periodic time Ts  2  k  2  s

2m k



360

Tp Ts



1 . 2

1 m 2 A 2 2

53. Here T=6 s 1 BC  10cm 2  OD =5 cm Now driplacement x  A sin(wt  )  (1)

Amplitude OB  OC 

2  where A=10 gm,   T  3 rad

B O

Now if at t=0, oscillator is at C. i.e. at t=0, x=A  A  A sin(  0  ) (OR) A=A sin   sin   1

   2 Putting this in eqn. (1) x  A sin(t   2 )  A cos t = 10 cos  t  for x = 5 cm,

5=10cos  t  cos t  1 2 t   3

t  1 S 54. Force responsible for oscillation in F  mg sin   mg  {  is small} x R Comparing this with F= - kx ;  mg 

mg R 55. Let the rod be pressed down by “x” at point A and released.  both spring gets displaced by “x”  Restoring torgue produced k

   = kx   2   kx   2      =kx 

Now tan  

x

 2



2x 

361

D

C

If  in small; tan     2x  x 

 2

k     torque   k      2  2 Now moment of inertia of rod with reference to O is if I, then 2

 k 2  Id 2     dt 2  2  Comparing with 



 k 2  d 2     dt 2  2I 

d 2  2 ; dt 2

k2  2I

2 m2 where   and I  T 12

2m 3k 56. Here 2 acceleration vectors g. and a are acting along mutually prependicular direction . T  

 effective acceleratioin l n g eff  g 2  a 2

 T  2

 g eff

57. g 2 off  a x 2  (g  ay) 2 here, ax gsinθ cosθ, ay = gsin 2 θ

 a x 2  g 2  a y 2  2ga y = a 2 sin 2  cos2   g 2  g 2 sin 2   2g 2 sin 2   g 2 (1  sin 2 )  g 2 cos 2   g eff  g cos  58. When the length of spring increases by x=2.5 cm force F  mg sin   force constant k 

 T  2

F mg sin   x x

m x   2  s. k g sin  7

362

59. Since the spring is massless, when C collides with A, both A and B will gain equal momentum. Also, since A and B have equal mass, both will have same velocity. Let this velocity be u.  Acc. to the law of conservation of momentum,

m  mu  mu  2mu u 

v 2

Now if the compression produced in the spring is x, then acc. to law of conservation of energy, 1 1 1 1 m 2  mu 2  mu 2  kx 2 2 2 2 2

 v 2  kx 2 kx 2  v  2u   2   m  4 m 2

2

kx 2 v 2 m    xv m 2 2k

4

     (1)

Now block A and B will have equal kinetic energy. 1 1 1  kx 2  mu 2  mu 2  mu 2 2 2 2  During max imum contraction,

kinetic energy of the system A-B is mu2 =

m .v 2 4

60. Displacement t y = 4 cos 2   sin100t 2

61. Here x = A cos  t 1 2 2 Now potential energy  m x {taking P.E. as a function of x} 2  when x=0, potential energy=0 graph (b)  III

Also, potential energy  1 m2 (A cos t) 2 {taking P. E. as a function of time} 2 1 m 2 A 2 2  graph -I

At t=0 potential energy 

363

1 1 2 2 2 62. E1  m x  E1  x m      (1) 2 2 1 1 E 2  m2 y 2  E2  y m2      (2) 2 2 E 

1 1 m2 (x  y)2  E  (x  y) m2      (3) 2 2 From (1), (2) {(3),

E  E1  E 2

or E  E1  E 2  2 E1E 2 2 63. 1 

k kx F   1    (1) m mx mx

2 Similarly, 2 

F2    (2) mx

If F1 and F2 acts simaltaneously, then angular frequency w2 

F1  F2    (3) mx

From (1), (2) and (3);  2  12   2 2

64. Initial periodic time T1  2

Now, use equ.  =

2 T

  (1) g

When pendulum moves along vertical direction, effictive acceleration g eff  g  a where ‘a’ in accleration of pendulum. Now, a 

d d (kt)   k  2.1 m.s 2 dt dt

  New periodic time T2  2 g  (2) eff T g  2  T1 g eff

1

364

65. Block will not slide if mg  ma

 g  a To prevent the block from sliding the maximum acceleration of table must be a max  g Now maximum accleration a max  2 A  2 A max  g

A max 

g gT 2  2 4 2

66. Angular frequency of system 1

 k      (M  m 

2

     (1)

Now to prevent B from sliding off A, the maximum force acting on B should not be more than the frictional force mg .  f max  ma max  m2 A max  (2)

From (1) & (2)  k   f max  m   A max mM

To prevent block from sliding,  f max  mg 

mkA max  mg mM

67. Restoring force F=- kx Now, F   x

 U(x)   k.x.dx  0

du dx

 kx  

du dx

 du  k.x.dx

kx 2 C 2

Where C in contant of integration. Now in a SHM, potential energy at the equilibrium position is zero.  u(x  0)  0  C  0

u(x) 

1 2 kx in an equation for a parabola. 2

365

68.

At the upper most end, when mg  R  m2 A, coin will loose contact. Taking R=0 m2 A  mg A g

2

69. Force required to increase the lenght by x in F=kx-----(1) After spring is divided into 2 equal parts, F  k 'x ' where x '  x 2  k'

x    (2) 2

from (1) & (2); k '  2k 70. Frequency of SHM depends on elasticity & inertia. 71. Restoring force F  mg sin  OR F   mg e where g e  g sin 

If  is small, sinθ  θ  Effective value of g is g e θ

For large oscillation , g sin   g  ( sin   ) 1  T  2 

g

72. Restoring force F=- mg sin  which depends on “m”

73. “g” in less on moon 1  form the equation T  2 g ,

T will increase As compared to earth, moon in small

366

74. Periodic time T 

T 



75.

1 2 

  { On differentiation}

T  1.5% T

a max  2 A  4 2 f 2 A  4 2 (30) 2  0.01

 362 m

s2

Since the oscillator moves between + A & -A, maximum acceleration  362 77. 78.

Energy dissipates & so amplitnde decreases. Statement -2 in false. Statement -1 in true. statement -2 in false. In a SHM, amplitnde & phase does not depend on restornig force.

79. Time taken by the spring k 2 to get maximum compressed from point D= half period of oscillation of the block. (if block in attached at the frce end of spring)

T2 1 m 1 0.2  i.e. t 2  2  2 2 k  2 2 0.3  4 s 2 T1   s 2 3 81. Time period of Block T=Time taken by the block to move from C to D and D to C

80. Similarly t1 

82. f 

1 k 2 M

and

f'

1 k 2 m  M

83. According to the law of conservation of momentum, MV  (M  m) ' 84. According to the law of conservation of energy, Kinetic Energy at mid point = potential Energy at the end points

1 1  Mv 2  kA 2 2 2

And

2 2 1 1 ( M  m ) v1  kA1 2 2

367

85. y = 3 cos 2t + 4 sin 2t  A sin   3 And A cos   4  y  A sin  cos 2t  A cos  sin 2t  y  A sin (t  )

Which shows that the motion is simple harmonic motion 86. T 

2 2  s  2

87. Amplitude A  32  42  5 cm 88. Maximum Accelaration of a particle  A2  5(2)2  20 cm s 2 89. Mechanical Energy 

1 m2 A 2  250 erg 2

1 1 1 90. frequency of the particle f   S T   91. On comparing y = A sin (15t  10x  ) 3

with y = A sin (t  kx  ) 92. At constant pressure density of water vapour is less than dry air.  with increase in humidity according to the equation v 

p the 

velocity of sound increases. 93.

f  1 

f1   2 f2 1

94. From the equation v = fλ, λ min = f

v

 17 mm which is nearer to 20 mm

max

95. On comparing with the wave equation  t x y = A sin 2     we get , T  

25 1 ms T = 0.04 s,  =0.5 m  v  2

 T  v 2   6.25 N

368

96. Maximum velocity of particle = A  wave velocity = f λ  Maximum velocity of particle = 2 x wave velocity  A = 2f     A

97. Putting values in   vT if phase diff.. = in the interval x is  then  

2 2 2 x   (15  10)   15 3

98. Freq. of a wave in a string f

 1 23

99.

1 

1 1 1 1     f f1 f2 f3

On comparing y1  4 sin 500 t with y1  A sin 1 t we get 1  2f 1 = 500  f1  250 Hz Similarly y 2  2sin 506t  2  2f 2  506  f 2  253 Hz  Freq. of beats  f 2 – f1  3  No.of beats heard per minute  3  60  180

100.

y  8sin 2 (0.1x  2t)

 t x  y  8sin 2 (2t  0.1x) comparing with y  A sin    T 

We get

1 =0.1    10cm 

Now path diffrence between 2 particles  

 

101.

2 .x  kx 

2  180  2 = 72o 10

Distance covered by the pulse = speed x time = 4 cm in 2 seconds both will cover 4 cm & the centre of both will superpose & potential energy will be zero.  Total energy will be in the from of kinetic energy..

369

102.

y1  a sin t & y 2  a cos t  a sin (t  π 2 )  1st wave is ilagging behind in phase by π 2

103.

Here A is the amplitude of resultant wave formed by 2 waves having amplitude A1 and A2 respectively. 2 2 A 2  A1  A 2  2A1 A 2 cos  Also  in the phase A1 , A 2

Now puttng A1  A 2  b & A  b , We get

b 2  2b 2 1  cos    cos    1 2

   1200

104.

As seen from fig., distance between 3 nodes in λ

105.

y  sin 2 t   υ=

1  cos 2 t 1 1   cos 2 t ____________  1 2 2 2

1 2 sin  2 t    sin 2 t 2

 a  22 cos 2t 1  = 2× 22  -Y  {From eng. (1)} 2  1  = –42   y  2   a α -y { SHM}

Now, 106.

2π π  2  T  T 

No.of beats produced per second =

 n1  n 2

 Time interval between 2 consecutive beats 

107.

1 n1  n 2

Since the phase difference between the 2 waves in perpendicalal direction.  Resultant ampltude =

A2  A2 

2 A

 Angular freq. will remain same.

370

π they are oscillating along mutually 2

108.

A  0.02 m, υ =

5λ = 4   =

  128 ms 1 k

4 2π m, k   2.5 π  7.85 5 λ

   128  k  128  7.85  1005

y  A sin  kx  t  109.

 y = 0.02 sin (7.85x - 1005 t) Let the number of loops obtained for 315Hz and 420Hz n and (n+1) respectively.

 f n  nf1  315

 f n 1   n  1 f1  420  f n 1  f n  f1  105 Hz 110.

When, sound waves travel fromone medium to another, its frequency does not change. f 



υa υ  b λa λb

λb 

111.

υ = consant λ

υb λ a  10λ a υa

Wave velocity = max. velo.of particle  1   A  A =  k k 2  λ = 2πA

112.



113.

RT m

Speed of sound in an ideal gas υ 

υ1 m2  υ2 m1

E

1      m 

1 1 m 2 A 2  m 2 2

4π 2 f 2 A 2 2

 E αf

2

2

E f   f  1  1  1     E 2  f 2   2f  4

 E 2  4E1

371

114.

Let hte freq. of 1st fork be f1 

frequency of 2nd fork = f1  6  f1  6  2  1



freq. of th 24th fork

= f1  6  24  1  f1  138

Now, freq. of 24th fork = 2 x freq. of 1st fork (given)  f1  138  2 f1

115.

 f1  138 Hz

π  Differentabing y1  0.1 sin  100 t +  w.r.t time, 3 

π  υ1  (0.1) (100) cos  100 π t +  3 

Similarly differentiating y 2  0.1 cos t w.r.t. time, π  υ 2  0.1 sin  π t   2   phase difference between the 2 velocities is

π  δ  πt+  3 

π π   π t    rad 2 6  1 1   200 m 1 λ 0.005

116.

Wave number 

117.

Frequency heard by the listener  υ + υL fL    υ

  fs 

f υ + υ2  L  fs υ  % increase 

118.

From υ =

  υs = 0  υ+ 

υ

υ 4



5 4

f L  fS  54 100    100  25% fS  4 

RT , υα T M

In summer, velocity increases & hence decreases and so L increases. The length of 2nd halmonics x = 3L1  3  16  48 cm In summer, velocity being more, x  3L1

 x  48

372

119.

 υ + υL  In f L    fs υ  υ s   putting υ L  0, f L  2fs ,

120.

1 2L

T µ

f L υ + υL  f S υ  υS

or

but, υ L  υS  122.

υs   υs

 υ  υ 2fs    υs   fs 2υ  υ  s 2  υ  υs  For resonance, the frequency of a.c. supply should be same as fundamnetal freq. of wire. f 

121.

υ  υ,

= 50 Hz f L υ  υL  fS υ - υS fL 1 fS

Since the rope is heavy, the tension at the lower end & top end of the rope will be different. Mass of rope m 2  3kg Mass of block m1  1 kg  tension at the lower end T1  m1 g  1 g N &

at the upper end in T2   m1  m 2  g  4 g N Now speed of wave in rope υ=

T T f λ = µ µ

 λ= T

( f ,μ are constants)

 Wave length at lower and λ1 = T1 & at

the upper end λ 2 = T2



123.

λ2 T2 T2   λ2  λ1 T1 T1

Speed of sound  ρ

P ρ

mass of 1 mole air Volume of 1 mole air

 speed 

= λ1 = λ 2  0.1m

7 1.01 105  5 1.3



29 103 kg  1.3 22.4  103 m3

 330 ms 1

373

124

From the phase angle (40-2t), we get k = 40 OR

2   40     20

1 and  = 2 OR 2f  2  f   Hz

125

Increase in tension of string increases its frequency. If the original frequency of B(fB) were greater than that of A(fA), further the increase in fB should have resulted in increase in the beat frequency. But the beat frequency is found to decrease. This shows that fA- fB = 5 Hz and fA=427 Hz, we get fB = 422 Hz

126

 υ  υL   330  0  fL    fS     800  1320 Hz  υ  υs   330  130 

127

υ

128

f 

4 T M 2 2.2  T  μυ 2  υ   (340) 2  T  2.3110 N µ L 11

1 2l

T  T  f 2 4l 2 µ µ

2

M 4f 2 M 2  T  4f   µ   248 N µ  µ  2

129

In tube A,  A  2l In tube B,  B  4l  υA 

130

υ υ υ υ υ 2   υB    A  λ A 2l λ B 4l υB 1

The was decreases the frequency of unknown fork. The possible unknown frequencies are, (288+4) Hz and (288-4) Hz. Wax reduces 284 Hz and so beats should increases. It is not given in the question. This frequency is ruled out. Wax reduces 292 Hz and so beats should decrease. It is given that the beats decrease from 2 to 4. Hence the unknown fork has frequency 292 Hz. consider option (a)

131

Stationary wave : Y = a sin (wt-kx) + a sin(wt+kx) When x = 0, Y  0. The option is not acceptable consider option (b) stationary wave : Y = a sin (wt-kx) - a sin(wt+kx) At x = 0, Y = 0. This option holds good. Option (c) gives Y = 2a sin(wt - kx) At x = 0, Y  0 Option (d) gives Y = 0. Hence option (b) holds good.

374

132

When temperature increases, l increases. Hence frequency decreases.

133

The possible frequency of piano are (256 + 5)Hz and (256 - 5)Hz. For a piano string υ 

1 2l

T When tension T increases v increases. µ

(i) If 261 Hz increases, beats / second increase. This is not given. (ii) If 251 Hz increases due to tension, beats / second decrease. This is given. 134

fL  υ  υL By Doppler’s effect, f   υ  υ S S   Fractional increase 

f L - fS f L 6 1  1  1  fS fS 5 5

 percentage increase 

135





P  

υ2  υ1

υυ  5 6   f L  υ  υ L  fS υ υ 5 

100  20% 5

RT M

γ He 32  4 γo2

Hence option (d) is correct. 136

In a longitudinal wave, pressure is maximum where displacement is minimum. Therefore pressure and displacement variations are 180 out of phase

137

Frequency of tuning fork f 1 = 480 Hz. Number of beats s-1, n = 10

Frequency of string f2 = (480 + 10)Hz. A slight increase in tension increase f2 f2 = 480 - 10 = 470 Hz. 138

(c) is the correct choice because its value is finite at all times.

139

As sin( 90  )  cos   The phase difference between the two waves is  2

140

υ

141

r1 1 Here, ρ1  ρ 2 , r  2 , T1  T2 2

T 500   50ms 1 µ 0.2

f1 

1 2l r1

T1 1 f2  , ρ1 2l r2

f r 2 T2  1  1  f 2 r2 1 ρ 2

375

142

143

f f f2 T 81 9  1 2  100  10%  2   f1 f1 T1 100 10

When one end is closed f1 

100  50H z 2

f 2 =3f1 =150Hz,f3 =5f1 =250Hz and so on... 144

When other end of pipe is opened, its fundamental frequency becomes 200Hz. The overtone have frequencies 400, 600, 800 Hz..

145

As

146

f1  2 25 5    f 2 1 30 6

2  2 ' 60  2 '      2 '  75cm 2  1 ' 40 50

f2 -f1 =4. on solving, weget f2 =24 Hz f1  20Hz 147

f2 101  1    1   f1 100  100   f 2  f1 

148

1

2

 1

1 200

f1 f  numbers of be ab s 1  f 2  f1  1  1 200 200

f L υ+υ L 1 1 = , f S υ+υ S Here  L  5 ms , S  5 ms , f S  165 Hz

 f L =170 Hz Number of be ab s 1  170  165  5 149

As the source is moving perpendicular to straight line joining the observer and source, (as if moving along a circle), apparent frequency is not affected n1 = 0

150

As is clear from figure, at t = 0, x = 0, displacement-y = 0 Therefore option (A) OR (D) may be correct. In case of (D) y = A sin (kx - wt)

dy dy dy dt dy dy  = -ω A cos(kx-t) = kA cos(kx-ωt)  dy  – v  dt  - v dx dt dx dx i.e. particle velocity = - (wave speed) xslope and slope at x = 0 and t = 0 is positive, in figure Therefore, particle velocity is in negative y - direction 151

At a displacement antinode, a pressure node is present. Since pressure does not change at its node, nor does density.

376

152

For a sonometer fundamental f 

1 T 22 

To maintatin the fundamental mode, in doubling the length, tension must be quadrupled. 153

velocity of transverse waves T 

velocity of longitudinal waves L 



154

Y 

L Y Y   2 T T / r stress

Let the frequency of standard fork = x f A 

102 97 97 x, f B  x , fB  x 100 100 100

Now f A  f B  155

T T  m r 2

102 x 97  x  x  120 Hz 100 100

If the length of the wire between the two bridges is  , then the frequency of vibration is

n

1 T 1 T  2l m 2l πr 2d

If the length and diameter of the wire are doubled keeping the tension same, then new fundamental frequency will be n/4 156

f L υ  υL  f S υ  υS using this equation the frequency of reflected sound heard by the girl, fL 

157

υ  υL fS υ – υS

υ f open  2  o pen

fclosed 



158

υ 4 closed



   As  closed  open   4 open /2  2  υ

υ  f open 2  open

i.e. frequency remains unchanged. If we assume that all the three waves are in same phase at t = 0, we shall hear only 1 beat s-1 377

159

y (x, t) = 0.005 cos (x   t ) compare it with standard equation 2   2 2 2   t    and  = y (x, t) = A cos (kx - wt) = A cos   T  T  

160

Given that the displacement of particle is y = A sin (t - kx) ..........(i) The particle velocity vp 

dy ................(ii)  dt

Now, on diffrentiating eqn.1 with respect to t

dy  A  cos(t  kx) dt

From eqn.(2 mental mode of the colsed pipe is f1 

173

υ 320   200 Hz 4L 4  0.40

Since the beat frequency is 8, the frequncy of the string vibrating in its first Overtone is 192 Hz or 208 Hz. Where for 1st Overtone frequency f1 ' 

1 T ........(1)  m

It is given that the beat frequency decreases if the tension in the string is decreased.

 f1 '  f1 Hence f1 ' = 208Hz and not 192Hz 174

substituting the values of .m and f1 ' in equation 1 we get T = 27.04 N

175

2π 2λ m λ 2πf 3λ  3π  υ  ms1 λ 2    m 2 2

176

Distance between two consecutive nodes =

177

The resultant displacement is given by, y = 0.1 cos 2x sin3  t Or y = A sin 3  t Where A is the Amplitude of standing waves given by 0.1 cos 2x  π   At x = 0.5m, cos 2x = cos (1rad) = cos    cos57.3  0.054 m  3.14 

Amplitude A at (x = 0.5m) = 0.1  0.54 = 0.54m 178

Particle velocity υ 

dy d  (0.1 Cos 2x sin 3πt) = 0.1 x 3 cos 2x sin 3t dt dt

at x = 0.25m and t = 0.5 s, v = 0 378

179

The two displacements can be written as

y1  A cos (k1x 1t) and y2  A cos (k 2x 2t) compare this equation with given equation and get solution. 180 181

1 2 – 2 2 The resultant displacement is given by

Beat frequncy = f1  f 2 =

y  y1  y2  A cos ( k1x 1t)  A cos (k2x 2t) For x = 0 we have y=A cos ω 1 t+A cos ω 2 t  y  0.10 cos (96t) cos (4t) Between t = 0 and t = 1 s, Cos 96 π t becomes zero 96times and cos 4 π t becomes zero 4 times Hence the resultant displacement Y at x = 0 becomes zero 100 times between t = 0 and t = 15.

182

y1 = A sin(kx+ t) yr  A sin (kx t)  y  yi  yr  y  2A sin kx cos t Here 2A=10  A  5

379

NOTES

380