2011 Paper

INDIAN ASSOCIATION OF PHYSICS TEACHERS NATIONAL STANDARD EXAMINATION IN ASTRONOMY 2011-2012 Date of Examination: 27th November 2011 Time 15.00 to 17.00 Hrs

Instruction to candidates 1.

In addition to this question paper, you are given a separate answer sheet.

2.

On the answer sheet fill up all the entries carefully in the space provided, ONLY In BLOCK CAPITALS. Incomplete / incorrect / carelessly filled information may disqualify your candidature.

3.

On the answer sheet, use only BLUE of BACK BALL PEN for making entries and marking answers.

4.

The question paper contains 80 multiple-choice questions. Each questions has 4 options, out of which only one is correct. Choose the correct answer and mark a cross in the corresponding box on the answer sheet as shown below:

5.

Any rough work should be done only on the sheet provided at the end of question paper.

6.

A correct answer carries 3 marks and 1 mark will be deducted for each wrong answer.

7.

Use of nonprogrammable calculator is allowed.

8.

No candidate should leave the examination hall before the completion of the examination.

9.

The answers/solution to this question paper will be available on our website - www.iapt.org.in by 3rd December 2011.

10.

Result sheets and the “centre top 10%” certificates of NSEA are dispatched to the Professor in charge of the centre. Thus you will get your marks from the Professor in charge of your centre by January 2012 end.

11.

TOP 300 (or so) students are called for the next examination-Indian National Astronomy Olympiads (INAO). Individual letters are sent to these students ONLY.

12.

No querries will be entertained in this regard. PLEASE DO NOT MAKE ANY MARK OTHER THAN (X) IN THE SPACE PROVIDED ON THE ANSWER SHEET. Answer sheets are evaluated with the help of a machine. Due to this, CHANGE OF ENTRY IS NOT ALLOWED. Scratching or overwriting may result in wrong score. DO NOT WRITE ANYTHING ON THE BACK SIDE OF ANSWER SHEET.

CERTIFICATES & AWARDS Following certificates are awarded by the I.A.P.T. to students successful in NSEA. i) Certificate for “Centre TOP 10%” students. ii) Merit certificates to statewise Top 1% students. iii) Merit certificate and a prize in the form of a book to Nationwise Top 1% students.

NSEA-2011-12

1.

Total solar eclipse can be observed from the moon. (a*) when total lunar eclipse is seen from the earth. (b) when total solar eclipse is seen from the earth. (c) on every full moon day (d) on every full new moon day

2.

A point mass is subjected to two simultaneous sinusoidal displacements in X direction, x1(t) = A sin (t) and 2    . Adding a third sinusoidal displacement x (t) = B sin(t + brings the mass to a x2 (t) = A sin  t  3 3  

complete rest. The values of B and  are (a) A 2 ,

Sol.

3 4

(b*) A,

4 3

(c) A 3 ,

5 6

(d) A,

 3

2    + Bsin(t + ) x = A sin(t) + Asin  t  3  

B =A&  =

4 to make the resultant 0. 3

3.

Consider the equation tan2  + 3 = 3 sec . Then, the given equation (a) is an identity (b) has no solution for 0   90º (c) has only one solution for 0   90º (d) has two solutions for 0   90º

Sol.

tan2 + 3 = 3sec  sec2 – 3sec + 2 = 0  (sec – 1) (sec –2) = 0  sec = 1 or sec = 2   = 2n or = 2n±

 , nI 3

Hence two solutions for 0    90º (d) is correct 4.

5.

The radiant of quadrantids meteor shower lies in (a) Pegasus (b) Carina (c) Orion

(d) Bootes.

      Vector p and q include an angle  between them. If (p  q) and (p  q) respectively include angles and  with p , then (tan + tan ) is

(a)

pq sin  p2  q2 cos2 

RESONANCE

(b)

2pq sin  p 2 – q2 cos 2 

p2 sin2  (c) 2 p  q2 cos2 

q2 sin2  (d) 2 p – q2 cos2 

PAGE - 2

NSEA-2011-12

Sol.

cos  =

p.q |p||q|

pq   .( p )  2  cos  =  pq p 2

p  q   . (p )  2  cos  = p  q = |p| 2

 tan  =

 tan  =

pq sin  p2  p . q

pq sin  p2  p.q

 p 2  p.q  p 2  p.q  2pq sin    = tan  + tan  = pq sin   2 2 2  ( p  p . q ) ( p  p . q ) p  q2 cos 2    (b) is correct 6.

Sol.

An alternative current of frequency  flows through a conductor with conductivity  . The material of the conductor has a magnetic permeability . The quantity ()–½ has the dimensions same as those of (a) charge (b) current (c) time (d*) length

1  F  T . 2 2   r E    Q   2 2  L  T 

1 / 2

1 / 2

1  2 L 

1 / 2

=L

7.

What is the position of the Red supergiant stars on the Hertzsprung-Russell diagram ? (a) upper left (b*) upper right (c) lower left. (d) lower right

8.

‘The dirty ice ball’ theory of comets was formulated by (a) Encke (b*) Fred Whipple (c) Halley

RESONANCE

(d) Brian Marsden PAGE - 3

NSEA-2011-12

9.

The value of the expression

(a) 2

Sol.

7 cos 70º 3 cos 55 º cos ec 35 º + is 2 sin 20º 2 tan 25º tan 45º tan 65º

(b*) 5

(c)

7 2

(d)

3 2

7 cos 70 º 3 cos 55º cos ec 35º + 2 sin 20 º 2 tan 25º tan 45º tan 65º 7 cos 70 º 3 cos 55º 7 3 = 2 cos 70 º + = + =5 2 sin 35º ( tan 25º tan 65º ).1 2 2 (b) is correct

10.

The quantity

o  o has the units same as those of

(a) frequency Sol.

F=

F=

(b) speed

(c*) resistance

(d) conductance

 0i1i2  2r

q1q2 4c 0r 2

F 0  0

i2  q2

F 2r 2 i 2  q2

Fr 2

=

Fr w = iq iq

v = iR =

R=

11.

12.

W q

w iq

The spectrum of the sun belongs to spectral class (a) A (b) K (c*) G

(d) B

Two charges q1 and q2 are placed 30 cm apart as shown in the figure. A third charge q3 is moved along the arc

 q3  of a circle of radius 40 cm from C to D. The change in potential energy of the system is  4  k , where k 0   is :

RESONANCE

PAGE - 4

NSEA-2011-12

Sol.

(a*) 8 q2 U = U2 – U1 = UD – UC

(b) 6 q2

(c) 8 q1

(d) 6 q1

 Kq Kq2   Kq1 Kq 2   = q3  1    40 10   40 50   = Kq2q3 ×

4 50

q3 2 1  2 = 4 q0 × 25 10 0 13.

14.

Sol.

15.

16.

= 8q2

The maximum number of solar eclipses in a year is (a) three (b) four (c*) five

(d) seven

If A = {2, 4, 6, 8, .....400}, so that n(A) = 200 and B = {3, 6, 9, 12, ...... 750}, so that n(B) = 250, then (a) n (A  B) = 125 and n(A  B) = 325 (b) n (A  B) = 124 and n(A  B) = 326 (c) n(A  B) = 66 and n(A  B) = 384 (d) n (A  B) = 65 and n(A  B) = 385 A = { 2,4,6,8 ....... 400} B = { 3,6,9,12 ....... 750} A  B = { 6 ,12, 18 , ..... , 396}  n (A B) = 66 and n (AB) = n(A) +n(B) – n(AB) = 200 + 250 – 66 = 384 (c) is correct The type of Andromeda galaxy is (a*) spiral (b) barred spiral

(c) elliptical

(d) irregular

A rod PQ of length l is resting on two mutually perpendicular axes as shown in the figure. If the tip P begins to slide down with a constant speed of

(a)

3 m/s

RESONANCE

(b*) 1 m/s

3 m/s, the speed of tip Q when  = 30º will be

(c) 3 m/s

(d) 2 m/s

PAGE - 5

NSEA-2011-12

Sol.

3 sin  = v2 cos  v2 =

3 tan 30º = 1

(b) is correct 17.

The figure shows beams of light incident on a thin film and subsequently emerging from it. Here each beam is represented by a single ray. Let r be the ratio of amplitude of the reflected beam to that of the incident beam, Amplitude of the incident beam a0 is assumed to be unity. Neglecting absorption in the film, amplitude of beam a will be

(a*) 1 – r2 Sol.

(b) (1 – r)2

Energy incident =

(c) r2

(d) 1 – r 2

1 2 kq0 2

Energy refracted after first reflection =

1 2 kq0 (1  r 2 ) 2

Energy refracted after second reflection =

1 2 kq0 (1  r 2 )(1  r 2 ) 2

1 2 K kA t = q02 (1  r 2 )2 2 2 q0 = 1 At2 = (1 – r2)2 At = (1 – r2)

18.

Two plane mirrors are inclined at an angle  as shown. The number of images formed of an object O is n. Then, the angle 

(a) must be

2 

(c) lies between

(b*) must be

2 2 and n n 1

RESONANCE

(d) must be

2 n 1 2 n–1 PAGE - 6

NSEA-2011-12

Sol.

360 1 

n=

(n + 1) =

==

Sol.

2 n 1

or n =

2 

2 n

=

19.

360 

The sum of the third and the seventh terms of an A.P. is 6 and their product is 8. The sum of the first 16 terms of this A.P. will be (a) 20 only (b) 76 only (c) 20 or 76 (d) 8.5 or – 2.5 Let a is first term and d is common difference a3 + a7 = 6  (a + 2d) + (a + 6d) = 6 .........(1) a3a7 = 8  (a + 2d) (a + 6d) = 8 .........(2) by solving (1) and (2) a + 2d = 2 and a + 6d = 4 or a + 2d = 4 and a + 6d = 2

1 a  2d  2   d = 2 and a = 1 a  6d  4 

Case I If

S16=

16 2

1  2(1)  (15 ). 2   

 19  = 8   = 76 2

Case II If

a  2d  4  1  d =–  a = 5 a  6d  2  2

S16=

16   1  2(5)  (15 )   2   2 

15   = 8 10   = 80 – 60 = 20 2 

(c) is correct. 20.

A window is 50 cm in height. A stone falling from a height of 40 cm above the window crosses the window in time t equal to (a*)

1 s 7

RESONANCE

(b)

3 s 7

(c)

5 s 7

(d)

4 s 7

PAGE - 7

NSEA-2011-12

Sol.

vavt = 0.5  2g(0.4)  2g(0.9)   t  1   2 2  

1 t=

0.2g (2  3) 1

=

=

21. Sol.

0 .2  9 . 8 ( 5 )

1 7

The number of points on the circle 2x2 + 2y2 – 3x = 0 having distance 2 units from the point (–2, 1) is (a) zero (b) 1 (c) 2 (d) 4 2 2 2x + 2y – 3x = 0

3 x=0 2

 x2 + y2 –

S(–2, 1) = 4 + 1 + 3 = 8 > 0 Hence (–2, 1) lies out side the circle 3  distance of center  , 0  from point (–2, 1) is 4  2

=

3    2   (0  1) 2 = 4 

2

 11    1 = 4

137 11 > 16 4

Hence there are no point on the circle (a) is correct. 22.

23.

Outer planet becomes retrograde (a) before conjunction (b) before opposition

(d*) after opposition

A conical vessel of radius 6 cm and height 8 cm is completely filled with water. A metal shpere is now lowered into the water. The size of the sphere is such that when it touches the inner surface, it just gets immersed. The fraction of water that overflows from the conical vessel is (a*)

Sol.

(c) after conjuction

3 8

(b)

5 8

(c)

7 8

(d)

5 16

R=6,H=8 V=

1 ( 6)2 (8) 3

sin  =

R 3 3 = (since tan  = ) 8 R 5 4

5R = 24 – 3R R=3 V=

3 8

RESONANCE

PAGE - 8

NSEA-2011-12

24.

To determine the absolute magnitude of a star, it is considered to be at a distance of (a) 10 AU (b*) 10 parsec (c) 1 parsec (d) 1 light year

25.

Which one of the following never moves in retrograde motion ? (a) Mars (b) Venus (c*) the moon

(d) the moon’s nodes

x

26.

 x 3 lim   is equal to x   x  2  (a) e5

Sol.

(b) e

lim  x  3  x   x 2

=

e

(d) e–5

x

3   1  x  = xlim 2    1  x 

=

(c) e–1

x

1 form

 x 3  lim  1 . x x2 

x  

 –5 x  lim    x2  x   e

= e–5

Hence (d) is correct.

27.

28.

The zero shadow days CANNOT be observed in the city of (a*) Amritsar (b) Nagpur (c) Chennai

If a2 + b2 + c2 + d2 = 45, out of the following statements the correct one is (a) ab + bc + cd + da  45 (b) ab + bc + cd + da  45 (c) ab + bc + cd + da 

Sol.

(d) Sriharikota

45 2

(d) ab + bc + cd + da 

45 2

AM  GM Now a2 + b2  2ab b2 + c2  2bc c2 + d2  2cd d2 + a2  2da On addition 2(a2 + b2 + c2 + d2)  2(ab + bc + cd + da)  a2 + b2 + c2 + d2  ab + bc + cd + da  ab + bc + cd + da  45 (a) is correct.

29.

An ellipse has foci (3, 1) and (1, 1). It passes through the point P(1, 3). Therefore, the eccentricity of the ellipse is (a)

2 –1

RESONANCE

(b)

3 –1

(c)

2 1 2

(d)

3 1 2 PAGE - 9

NSEA-2011-12

Sol.

foci (3, 1) and (1, 1)  major axes y = 1, center (2, 1) minor axies x = 2 Let equation of ellipse ( x  2) 2 a2

( y  1)2

+

b2

=1

it passes through P(1, 3) so that 1 a

2



4 b2

=1

....(2)

distance between foci is = 2  2ae = 2 Now e2 = 1 –

b2 a2

 a2e2 = a2 – b2  1 = a2 – b2 Put in (2)

1 1 b

2

+

4 b2

= 1  b4 – 4b2 = 4

 b2 = 2 + 2 2  a2 = 3 + 2 2  e2 = 1 –

22 2 32 2

1 =

32 2

1

e=

( 2  1)

= ( 2  1)

(a) is correct. 30.

Apart from the twelve constellations, the sun passes through the constellation of (a) Orion (b) Ursa Major (c*) Opiuchus (d) Perseus

31.

For the quadratic equation 3x2 + px + 3 = 0, where p > 0, one of the roots is square of the other. Hence, the value of p is (a)

Sol.

1 3

(b) 1

(c) 3

(d)

2 3

Let Roots are  & 2   + 2 = –

p & .2 = 1 3

( + 2)3 = –

p3 27

3 + 6 + 3.2( + 2) = –

p3 27

 p p3 1 + 1 + 3    = –  3 27

RESONANCE

PAGE - 10

NSEA-2011-12

32.

54 – 27p = –p3 p3 – 27p + 54 = 0 (p – 3) (p2 + 3p – 18) = 0 p = 3 or p2 + 3p – 18 = 0 p = 3 or p2 + 3p – 18 = 0 p = 3 or (p + 6) (p – 3) = 0  p = 3 or p = –6 but p > 0 p=3 (c) is correct.       Two vectors ( a  2b ) and (5 a  4b ), where a and b are unit vectors, are perpendicular to each other..   therefore, the angle between a and b is (a) 45º

Sol.

(b*) 60º

 1 (c) cos–1   3

 1 (d) cos–1   6

  a b cos  = ab     (a  2b).(5a – 4b) = 0,   5a 2 – 4a.b  10a.b – 8b 2  0

cos  =

cos  =

3 6 1 2

 = 60º (b) is correct 33.

34.

The north galactic pole lies in the constellation of (a*) Ursa Minor (b) Ursa Major (c) Coma Berenices The transits of Mercury occur in the month of (a) January and July. (c*) May and November

(d) Draco

(b) March and September (d) June and December

Sol. 35.

If the sun is rising in Gemini, the zodiacal constellation on the meridian will be (a) Virgo (b) Sagittarius (c) Pisces (d) Libra

Sol.

36.

1 1 1   1 1 1 If A =  , out of the following statements the correct one is 1 1 1

(a) A3 = 27 A

RESONANCE

(b) A3 = 3A

(c) A3 = A

(d) A + A + A = A2

PAGE - 11

NSEA-2011-12

Sol.

 1 1 1   1 1 1 A=   1 1 1 

A – I =

1– 

1

1

1

1– 

1

1

1

1– 

=0

 (1 – ) ( 2 – 2) – 1 (1 –  – 1) + 1 (1 – 1 + ) = 0 = (1 – ) ( 2 – 2) +  +  = 0 =  2 – 2 –  3 + 2 2 + 2= 0 = –  3 + 3 2 = 0  3 = 3 2 A3 = 3A2 (d) is correct 37.

In the spectrum of a quasar, the observed wavelength of an element (having rest wavelength of 1216 Å) is found to be 2432 Å. Therefore, the value of red shift (z) of that quasar is (a) –1 (b) – 0.5 (c) 0.5 (d*) 1

Sol.

Z=

 obs   emit  emit

Z=

2432  1216 =1 1216

Z > 0 is known as red shift. 38.

Which of the following is the cube root of 64 ? (a) 2 3 i  2

Sol.

(b) 2 3 i – 2

(c) 2 – 3 2 i

(d) 3 – 2 3 i

Cube root of 64 is x = (64)1/3 (1)1/3 x = 4, 4, 42

  1 3 i      4   1 3 i  x = 4, 4  ,    2 2     x = 4, – 2 + 2 3 i, –2 – 2 3 i so x = 2 3 i – 2 (b) is correct 39.

Sol.

The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. Therefore, the wavelength of the second spectral line in the Balmer series of singly ionized helium atom is (a*) 1215 Å (b) 1640 Å (c) 2430 Å (d) 4687 Å n1 = 2 n1 = 2 n2 = 3 n2 = 4 Z=1 Z=2  1 = 6561 2 = ? 1  1 1  R   1 4 9 1 1 1   Rx 2 2    2  4 16 

2 5  16  1 4  9  4  3

 2 = 243 × 5  2 = 1215 Aº

RESONANCE

PAGE - 12

NSEA-2011-12

40.

If x  0, x  1 and x x

x

(a) 0.444... Sol.

xx

x

xx

x

x

 

– x x

= 0, the value of x is

(b) 2.25

(c) 4

(d) 9

– ( x x )x = 0 = ( x x )x

x x log x = x log x x x 3/2 log x =

3x log x 2

 3 / 2 3x  –  =0 logx  x 2  

logx = 0, x 3/2 –

3x =0 2

3  x = 1, x x –  = 0 2 



x = 0 and x 

9 = 2.25 4

(b) is correct 41.

An alpha particle of 10 MeV energy collides head on with a copper nucleus (Z = 29) and is deflecteds back. The distance of closest approach of the alpha particle is about (a) 4.2 × 10–15 m (b*) 8.3 × 10–15 (c) 1.6 × 10–14 m (d) 2.1 × 10–14 m

Sol.

10 MeV =

Kq1q2 r

10 × 1.6 × 10–19 + 6 =

9  10 9 38  2  2.56  29 1.6  10 12

= 18 × 1.6 ×29 × 1021 – 38 18 × 1.6 × 2y × 10–17 = 8.3 × 10–15 m 42.

A non-conducting ring charged uniformly is rotating about its axis (passing through its centre and perpendicular to its plane) with a constant angular velocity. Let E, V and B be respectively the electrostatic field, electrostatic potential and magnetic field produced at the centre of the ring. Then, (a) E = 0, V  0, B = 0 (b*) E = 0, V  0, B  0 (c) E  0, V = 0, B  0 (d) E  0, V = 0, B = 0

Sol.

E=0 V0 B0

RESONANCE

PAGE - 13

NSEA-2011-12

43.

In the circuit given below the voltage across the inductor (a) is 90 volt. (b) is zero volt. (c) cannot be greater that 150 volt. (d*) is 180 volt.

Sol.

irms =

v 120 = C 45 60

vC = 90 1502 = 1202 + (vL – 90)2

(150 2  120 2 ) = |(vL – 90)| |VL – 90| = 10 × 9 vL = 180 or vL = 0 44.

Sol.

(not possible)

The successive resonance frequencies of a pipe are 425 Hz, 595 Hz and 765 Hz respectively. If the speed of sound in air is 340 m/s, the length of the pipe is (a) 0.85 m. (b*) 1.0 m. (c) 0.5 m (d) 1.7 m. 425 : 595 : 765 85 : 119 : 153 2v 4

595 – 425 = =1m 45.

If m = 6 + (a)

Sol.

35 and n = 6 – (b)

14

m=6+ n=6–

35 , the value of 21

 m – n  is (c)

18

(d)

10

35 35

m · n = 36 – 35 = 1 Now

2

 m – n

= m + n – 2 mn

= 12 – 2.1 = 10 

m– n =

10

(d) is correct 46.

Mercury stands at a height h in a barometer. A small hole is made at a height h' (less that h). The mercury comes out of the hole with a speed v equal to (a)

2gh'

RESONANCE

(b)

2g(h – h' )

(c)

2gh

(d*) zero

PAGE - 14

NSEA-2011-12

Sol.

(h – h´) g =

(h – h´)  g =

1 2 v  p 0 2 1  v2 + h  g 2

v2 = [(h – h´)g – h] × 2 v=0

47.

1     Let  and  be the cube roots of unity. If A =   and B =   , then the product AB is     1 

1 1 (a)   1 2

Sol.

 – 1 – 1 (b)    – 1 – 2

 1 – 1 (c)   – 1 2 

 – 1 – 1 (d)   – 1 2 

      2   1       AB =     =  2    2      1  =  = 2 1 +  + 2 = 0 2 = 1  – 1 – 1 So AB =   – 1 2 

(d) is correct 48.

Sol.

A uniform wire of length  and mass m is bent in the form of a rectangle PQRS such that PQ : QR = 2 : 1. The moment of inertia of this rectangle about side QR is (a)

11 m 2 243

=

(2a )(2a)2  2  (a )a 2  4 3

(b)

7 m 2 216

16  14  m 3 28 3  = m  a  = a  4  = 3  9 6a 3  6

=

(c)

5 m 2 136

(d*)

7 m 2 162

2

14 m 2 9  36

=

7 m 2 162

RESONANCE

PAGE - 15

NSEA-2011-12

49.

A planoconvex lens form a lens combination with a planconcave lens such that their curved surfaces have equal radii of curvature and are in contact with each other. If µ1 and µ2 are the refractive indices of the materials of the two lenses, the focal length of the combination is proportional to (a) | µ1 – µ2 |

Sol.

2 (c*) | µ  µ | 1 2

2 (b) µ  µ 1 2

(d) None of the above

1  1  1  (1  1)   ( 2  1)   f R    R

=

1 ( 1  1  1   2 ) R

R f =   1 2 50.

Refer to the cicuit. The current through 10 ohm resistance is

(a) 371 mA from B to A (c*) 200 mA from B to A

Sol.

(b) 371 mA from A to B (d) 200 mA from A to B

20  6 10 6  40 20 40 EqL = = 3 1 1  40 20 40 14 2 3 i= 40 = 10 = 0.2 A 10  3 i = 200 mA BA

51.

Given that 1 + x 2 = 2x cos , the value of (a) 2 cos 3

Sol.

(b) 3 cos 2

x6  1 x3

is

(c) 2 cos 2

(d) 3 cos 3

1 + x = 2 cos  x 3

1    x  = 8 cos 3  x  x3 +

1 x

x6  1 x

3

3

1  + 3   x  = 8 cos 3 x 

+ 3 (2cos ) = 8 cos 3 

RESONANCE

PAGE - 16

NSEA-2011-12

x6  1 x3 x6  1 x3 Alter : 

+ 3 (2 cos ) = 8 cos 3 

= 2(4cos 3  – 3cos ) = 2 cos 3

1 + x = 2 cos  x x = cos  + i sin  x6  1

Now

x

3

= x3 +

1 x3

= (cos + isin)2 + (cos  – i sin )3 = 2 cos3 (a) is correct 52.

Sol.

53.

The weight of a metal sphere when immersed in water at 0°C is W 1 and that at 50°C is W 2. If the coefficient of cubical expansion of the metal is less than that of water, then (a) W 1 = W 2 (b*) W 1 < W 2 (c) W 1 > W 2 (d) nothing can be said as the data are insufficient. 3s < vL W 1 = W 0 – 1 =vg  < 0  W  W2 > W1 An organ pipe of length  (open at both ends) is sounded together with another organ pipe of length ( + x) where x << , to produce their fundamental notes. If speed of sound in air is v, the beat frequency is (a)

vx

(b*)

2



vx 2

2

vx (c)

2 x

(d)

vx 4 2

Sol.

  2

  (  x ) 2

V 2  f = f1 – f2 f1 

f2 

=

1  V 1 –  2     x 

=

Vx x << l 2(  x )

f =

Vx 2 2

RESONANCE

V 2(  x )

(b)

PAGE - 17

NSEA-2011-12

54.

A circuit connects a pure inductor (L) to a dc souce through a rheostat as shown. The sliding contact of the rheostat is pulled to the right so as to increase the resistance. At an instant when the resistanc in 20 ohm, the current in the circuit is

(a) 0.5 A (b) less than 0.5 A (c*) more than 0.5 A. (d) Less than or more than 0.5 A depending on the value of L. Sol.

55.

  and . The longest side of this triangle has length 1 3 . Therefore, 6 4



Two angles of a triangle are



the area of the triangle is

1 (a)

1 (b)

3 –1

(c)

3 1

3 +1

(d)

3 –1

Sol.

Angle are

  7 , , 4 6 12

a = BC = 1 + 3 a b c   sin A sin B sin C b 1 3 = sin  / 6 sin 105



b=

1 (1  3 ) = 2  3  1    2 2   

Area of ABC

2

=

1 ab sin C 2

=

1 1 3 2

=

1 3 2



  2   sin 4 

1 =

RESONANCE

3 1

(a) is correct

PAGE - 18

NSEA-2011-12

56.

A given mass of an ideal gas expands to 2.5 times its initial valume (V) at a constant pressure (p). If  is the ratio of specific heats,the change in internal energy of the gas is (a)

Sol.

5pV 2(  – 1)

(b*)

3pV 2(  – 1)

(c)

pV 2(  – 1)

(d)

pV (  – 1)

u = n CV T =

nR ( T )  –1

=

P(2.5 – 1)V  –1

=

3 PV 2 (  – 1)

Correct option is (b)

57.

Sol.

Let the energy, the magnitude of linear momentum and the angular frequency of an electron revolving in an orbit with quantum number n be E, p and  respectively then E (a)   varies at n2  

 Ep   is independent of n. (b*)    

(c) (p) varies as

(d) (Ep) is independent of n.

n

Energy E Liner momentum p Angular frequency w Quntam no. n

Ep indendent of n w (b) option is correct. 58.

Which of the following corresponds to (sin–1 x + sin–1 y) ? 2 2  (a) sin–1  x 1 – y – y 1 – x   

2 2   (b) sin–1  x 1 – y  y 1 – x   

2 2  (c) sin–1  x 1 – x – y 1 – y   

2 2   (d) sin-1  x 1 – x  y 1 – y   

Sol.

(b) Standard Result

59.

     Given A  B = C . If B is reversed, the resulatant is D . Then, we have

Sol.

(a) C2 – D2 = A2 – B2 (c) C2 + D2 = A2 + B2    A +B = C    and A– B = D Then C2 + D2 = 2(A2 + B2) (d) is correct

RESONANCE

(b) C2 – D2 = 2(A2 – B2) (d) C2 + D2 = 2(A2 + B2)

PAGE - 19

NSEA-2011-12

60.

Sol.

61.

        1  cot  – sec     1  cot   sec     equals 2   2     (a) cot  (b) 2 cot  (1 + cot  + cosec  ) ( 1 + cot  – cosec) = (1 + cot )2 – (cosec2) = 1 + cot2 + 2 cot – cosec2 = 2 cot  (b) is correct

(c) 3 cot 

(d) 4 cot 

Which of the following is INCORRECT ? (a) electric field exerts a force on a stationary charge. (b) Electric field exerts a force on a moving charge. (c*) Magnetic field exerts a force on a stationary charge. (d) Magnetic field exerts a force on a moving charge.

Sol. 62.

Sol.

Consider the two circles x 2 + y2 = 4 and x 2 + y2 – 6x – 8y = 24. The number of common tangents to these circles is (a*) 1 (b) 2 (c) 3 (d) 4 C1 (0, 0), C2 = (3, 4) r1 = 2,

r2 =

9  16  24 = 7

C1C2 = 5, r1 + r2 = 7 |r1 – r2| = C1C2 < r1 + r2 Circles are internally touching so one tangent can be drawn. 63.

A body of mass m suspended from a spring of length  and force constant k oscillates with periodic time T. Now, the spring is cut into two pieces of lengths in the ratio 2 : 3. The same mass is attached to the two pieces as shown and set into oscillations on a smooth surface. The periodic time of oscillations T is

T (a) Sol.

2

2 3T 5

(c*)

6T 5

T (d)

5

k,  T  2

64.

(b)

m  1 1 5k    2 3

The polar equation (a) a parabola

RESONANCE

= 2

6m 25k

=

6 T 5

8 = cos  – 1 represents r (b) an ellipse

(c) a hyperbola

(d) a pair of straight lines

PAGE - 20

NSEA-2011-12

Sol.

8 = cos  – 1 r 8 = r cos  – r 8=x–

x2  y2

 (x – 8)2 = x 2 + y2 x 2 – 16x + 64 = x 2 + y2  y2 + 16x – 64 = 0 (Parabola) (a) is correct 65.

A non-uniform rod of length  is placed along positive X axis with one of its ends at (0, 0). Linear mass density of the rod varies as ( – x). Its centre of mass is at   (a)  , 0  6 

  (b*)  , 0  3 

 2  (c)  , 0   3 

 5  (d)  , 0   6 

Sol.

 = (l – x)

 xdm  dm 

2 3 – 3  0  2 =  2 2  – ( – x )dx 2

 ( – x)x dx





3 6   .0 3 2 2

0

66.

A satellite revolves round the earth along an elliptical orbit of eccentricity e. It T is the periodic time of revolution, the time taken by the satellite to travel from one extreme of the minor axis to the other through perigee is (a*)

T e  1   2 2 

(b)

T   1   2 2e 

(c)

T e 1    2

(d)

T 1 1   2 e

Sol.

RESONANCE

PAGE - 21

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a2(1 – e)a 1 – ae(1 – e) a = Area of shaded part 2 2 e  2 a a (1 – e)   –  = Area 2  dA L  dt 2m

 2m   a (1 – e) a × 2 T=   L 

e  2m    a2 (1 – e)   –  t=  L 2    T 2  e t   –  2 

67.

t

T  e  –  2  2

t

e  T 1 –  2  2

Consider the points A(a, b + c), B(b, c + a) and C(c, a + b). The area of ABC is (a) 2(a2 + b2 + c2)

(b)

a2  b2  c 2 6

(c) 2(ab + bc + ca)

(d) none of the above

a bc 1 Sol.

1 Area of ABC = 2

b ca 1 c ab 1

C1  C1 + C2 1 bc 1

1 1 ca 1 (a + b + c) =0 2 1 ab 1

=

(d) is correct 68.

In a certain village of Maharashtra everybody knows Marathi language. However, 52% people can understand only Marathi, 35% are able to understand Hindi while 23% can understand English. The number of people who can understand Marathi, Hindi as well as English is 1500. Hence the population of this village is (a) 13750 (b) 15000 (c) 16250 (d) 18125

Sol.

n H  E = 52%





n(H) = 35% n(E) = 23%

...(1)



100% – n(H  E) = 52%

   

n (H  E) = 48% n(H  E) = n(H) + n(E) – n(H  E) 48% = 35% + 23% – n(H  E) n(H  E) = 10%

So total = 1500 ×

100 = 15000 10

(b) is correct

RESONANCE

PAGE - 22

NSEA-2011-12

69.

The value of [1.5 + 0.5 + 0.25 + 0.15 + 0.1 +......+

15 (a) 1 17 Sol.

Tn =

(b) 2

(c) 2

16 17

(d) 3

1 17

3 n2  n

3 1 = 2 850 n n

Sn =

8 17

1 ] is 850

n

3 2

n



=

n = 50

 (n  1) – (n)   n(n  1) 

 3

1  1 Sn = 3  –   n n  1

 1 1   1 1   1 1  1   1 S50 = 3  1 – 2    2 – 3    3 – 4   .... 50 – 51          1  = 3 1 –  51  

50 16 = 2 17 17 (c) is correct =

70.

Sol.

If (12!)–1 – (14!)–1 = x(15!)–1, the value of x is (a) 2565 (b) 2645

(c) 2715

(d) 2875

1 1 x – = 12! 14 ! 15 !

1 12 !



  1 x 1 –  = ( 14 ) ( 13 ) (15) (14) (13) 12!  

181 x = (14) (13) 15(14) (13)

 x = 181 × 15 x = 2715 (c) is correct 71.

Sol.

Consider the curve given by y = x4 + 8x3 + 26x2 + 32x – 73. The number of points on this curve at which tangent to the curve is parallel to X axis is (a) 1 (b) 2 (c) 3 (d) 4 dy = 4x3 + 24x2 + 52x + 32 = 0 (for tangent parallel to x-axis) dx

x = –1 satisfy (x + 1) (4x2 + 20x + 32) = 0  4(x + 1) (x2 + 5x + 8) = 0 x = –1 only one point (1, –6) (a) is correct

RESONANCE

PAGE - 23

NSEA-2011-12

72.

If 53x–4 + 54 – 3x = 2, the value of x is (a)

Sol.

4 3

Sol.

4 3

(c)

3 4

(d) log(2)

AM  GM. equality holds only if 53x–4 = 54–3x 

73.

(b) –

4 – 3x = 0



x=

4 3

The function f(x) = sin x cos x is increasing in the interval   (a)  0,   2

  3   (b)  , 2 2 

1 sin 2x 2 f '(x) = cos2x > 0

(for increasing)

  3   (c)  , 2 4 

 3  (d)  ,    4 

f(x) =

   3  x   0,    ,    4  4 

(d) is correct 74.

Sol.

The last three terms in the expansion of (a + b)n are 54880, 48020 and 16807 respectively. Therefore, the values of a, b and n respectively are (a) 5, 7, 4 (b) 2, 7, 7 (c) 4, 7, 5 (d) 7, 4, 5 n (a + b) T n+1 = nCn a°bn = bn = 16807 ...(1) T n = nCn–1 a1bn–1 = nabn–1 = 48020 ...(2) T n–1 = nCn–2 a2bn–2 = by (2) / (1)

by (3)/(2)

n(n – 1) 2 n–2 a b = 54880 2

na 48020 = b 16807

54880 (n – 1) a = 2 b 48020

by (4) & (5)

...(3)

...(4)

...(5)

54880 16807 n–1  = 2n 48020 48020



1844736320 n–1 = n 2305920400



1 1844736320 =1– n 2305920400

1 461184080 = n 2305920400  n=5 by (1) b5 = 16807  b = 7 and a = 4 (c) is correct 

RESONANCE

PAGE - 24

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75.

Two soap bubbles A and B respectively of radii 2 mm and 5 mm stick to each other to form a double bubble. Then, (a) The radius of curvature of common surface is

10 mm and concave on the side of bubble A. 3

(b) The radius of curvature of common surface is

7 mm and convex on the side of bubble A. 2

(c*) the radius of curvature of common surface is

10 mm and convex on the side of bubble A. 3

(d) the radius of curvature of common surface is

Sol.

7 mm and concave on the side of bubble A. 2

1 1 1  – r 2 5 r

5  2 10  mm 3 3

(c) Concave towards A 76.

Sol.

A body of mass 50g performs simple harmonic motion with amplitude 20 cm. If the total energy of the body is 2 J, the periodic time is (a) 1.4 ms (b) 1.4 s (c) 14 ms (d*) 140 ms

1 kA2 = 2 2 1 k (.2)2 = 2 2 k=

4 = 100 .04

T = 2

m k

.05 100

= 2

2 5 100

T

= 14.04 × 10–2 = 140 × 10–3 = 140 ms

 6

77.

The integral

cos 4 (3 x )

 sin (3x)  cos 4

0

(a)

Sol.

=

 12

(b)

/6

cos 4 (3 x )



cos 4 3 x  sin 4 (3 x )

0

 0

RESONANCE

(3 x )

dx equals

 6

dx

a

using properties

4

(c)

 3

(d)

 2

...(1) a

f ( x ) dx =

 f (a  x ) dx 0

PAGE - 25

NSEA-2011-12 /6

sin 4 (3 x )



cos 4 (3 x )  sin 4 (3 x )

=

0

dx

...(2)

adding /6

 1. dx

2 =

0

2 =

 6

 12 (a) is correct =

78.

Refer to the arrangement of logic gates given below. The outputs Y for A = 0, B = 0 and A = 1, B = 1 respectively are

(a) 0 and 1

(b*) 1 and 0

(c) 0 and 0

(d) 1 and 1

Sol.

Y = ( A  B ). B Y = (1 + 0) . 1 = 1 Y = (0 + 1) . 0 = 0

79.

An electron in the second excited state of a hydrogen atom has an associated de Broglie wavelength of about (a) 0.32 nm (b) 0.66 nm (c) 3.2 nm (d*) 1.0 nm

Sol.

d =

h = mV

h Z  m. 2.18  10 6  n 

h3

=

m 2.18  10 6



= 1 nm

80.

 Ans. (d)

x 4  log 5(5 x 1 )



5x  x5

dx gives

(a) 5x – x5

Sol.

(b) (5x – x5)–1

(c) log(5x + x5)

(d)

1 log( 5 x  x 5 ) 5

x 4  log 5(5 x 1 )

dx 5x  x5 Let 5x + x5 = t (5x log 5 + 5x4) dx = dt

=



=

1 1 dt = log t  c 5 5 t



1 log( 5 x  x 5 )  c 5 (d) is correct

=

RESONANCE

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NSEA-2011-12

Physical constants you many need : (1) Charge on electron e = 1.6 × 10-19 C (2) Mass of an electron me = 9.1 × 10–31 kg (3) Universal gravitational constant G = 6.67 × 10–11 N m2 / kg2 (4) Permittivity of free space 0 = 8.85 × 10–12 C2 / Nm2 (5) Gas constant R = 8.31 J / K mol (6) Planck constant h = 6.62 × 10–34 J s (7) Stefan constant  = 5.67 × 10–8 W/m2 k4 (8) Boltzmann constant k = 1.38 × 10–23 J/K

RESONANCE

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