paper with solution physics nsep-2012

7/30/2019 Paper With Solution Physics NSEP-2012

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INDIAN ASSOCIATION OF PHYSICS TEACHERS

NATIONAL STANDARD EXAMINATION IN PHYSICS 2012-2013

Date of Examination : 24th November 2012

Time 9.30 to 11.30 Hrs

Q. P. Code No.1

ONE

5 4

FIVE FOUR

Instruction to candidates

1. In addition to this question paper, you are given answer sheet for part A and an answer paper

for part B.

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

answer.

4. Part A has two parts. In part A1 (Q.No. 1 to 10) each question has Four alternatives, our of

which only one is correct. Choose the correct alternative and mark a cross (X) in the

corresponding box on the answer sheet.

For example, Q. b ca d

22

Part A2 (Q.Nos. 41 to 50) has four alternatives in each question, but any number of these (4, 3,

2, or 1) may be correct. You have to mark ALL correct alternatives and mark a cross (X) for

each, like

5. For Part A1, each correct answer gets 3 marks; wrong answer gets a penalty of 1 mark. For

Part A2 full marks are 6 for each question, you get them when ALL correct answer only are

marked.

6. Any rough work should be done only on the sheets provided with part B answer paper.

7. Use of nonprogrammable calculator is allowed.

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

9. After submitting your answer papers, read the instructions regarding evaluation given at the

end of the question paper.

PLEASE DO NOT MAKE ANY MARKS OTHER THAN (X) IN THE SPACE PROIDED ON

THE ANSWER SHEET OF PART A.Answer sheets are evaluated with the help of a machine. Hence, CHANGE OF ENTRY IS

NOT ALLOWED.

Scratching or overwriting may result in wrong score.


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DO NOT WRITE ANYTHING ON THE BACK SIDE OF PART A ANSWER SHEET.

INSTRUCTIONS TO CANDIDATES:

1. The answers / solutions to this question paper will be available on our website-

www.iapt.org.in by 3 rd Decembet 2012.

EVALUATION PROCEDURE (NSEP) :

2. Part 'A' Answers of ALL the candidates are examined.

3. Part B is evaluated of only those students who get marks above a certain' "cut oft" marks

in Part A. (e.g. NSEP Total marks for Part A are 180. Students getting (say) 100 or more

than 100 marks In Part A are identified and their Part B is evaluated. Thus "cut off" marks

are 100 in this example.)

4. PART B IS NOT EVALUATED OF ALL THE CANDIDATES.

CERTIFICATES & AWARDS

Following certificates are awarded by the I.A.P.T. to students successful in NSEP.

i) Certificate for Centre Top 10% students on the basis of marks in part A only.

ii) Merit certificates to statewise Top 1% students on the basis of (A+B) marks.

iii) Merit certificate and a prize in the form of a book to Nationwise Top 1% students based

on (A+B) marks.

5. Result sheets and the centre top 10% certificates of NSEP 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 2013 end.

6. TOP 300 (or so) 'students are called for the next examination -Indian National Physics

Olympiads (INPhO). Individual letters are sent to these students ONLY.

7. Gold medals will be awarded to TOP 35 students in this entire process.

8. No queries will be entertained in this regard.
http://www.iapt.org.in/


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NSEP-2012-13

INDIAN ASSOCIATION OF PHYSICS TEACHERS

NATIONAL STANDARD EXAMINATION IN PHYSICS 2012-2013

Total Time : 120 Minutes (A-1, A-2 & B)

PART A MARKS = 180

SUB-PART A-1 : ONLY ONE OUT OF FOUR OPTIONS IS CORRECT

N.B. : Physical constants are given at the end.

SUB-PART A1

1. The angle of refraction of a very thin prism is 1. A light ray is incident normally on one of the refracting

surfaces. The ray that ultimately emerges from the first surface, after suffering reflection from the second

surface. makes an angle of 3.32 with the normal. The deviation of the ray emerging from the second surfaceand the refractive index of the material of the prism respectively, are :

(A) 0.66, 1.66 (B) 1.66, 1.5 (C) 1.5, 1.66 (D) 0.66, 1.5

Sol.

n

32.3= 2

n = 1.66 Ans. (A)

2. A beam of light from a distant axial point source is incident on the plane surface of a thin planoconvex lens;

a real image is formed at a distance of 40 cm. Now if the curved surface is silvered, the real image is formed

at a distance of 7.5 cm. The radius of curvature of the curved surface of the lens and the refractive index of the

material of the lens respectively, are :

(A) 40 cm, 1.5 (B) 24 cm, 1.6 (C) 20 cm, 1.6 (D) 7.5 cm, 1.5

Sol. fm

= R/2

f= 40 m

F

1=

f

2

f

1

m

402

R2

5.71


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75

10

40

2

R

2

=15

2

20

1 =

1520

4015

R

2=

153

255

R = 24 cm. Ans. (B)

3. A convex lens forms the image of an axial point on a screen. A second lens with focal length f cm is placed

between the screen and the first lens at a distance of 10 cm from the screen. To view the image the screen

has to be shifted away from the lens by 5 cm. A third lens having focal length of the same magnitude f cm is

used to replace the second lens at the same position. But this time to view the image the screen has to be

shifted towards the lens by d cm. The values of f and d respectively, are :

(A) 30 cm, 2.5 cm (B) 30 cm, 5 cm (C) 7.5 cm, 2.5 cm (D) 7.5 cm, 5 cm

Sol.

In first case

u = +10

v = +15 cm

f

1

u

1

v

1

f1

101

151

f =5

1015

= 30 cm

For second case :

u = +10

v = ?

f = +30

30

1

10

1

v

1

30

4

30

1

10

1

v

1

v =2

15= 7.5 cm = (10 d)

d = 2.5 cm. Ans. (A)

4. Cerenkov effect: If the speed of an electron in a medium is greater than the speed of light in that medium then

the electron emits light. An electron beam in a medium is accelerated by a voltage V. The light that is emitted

just suffers total internal reflection at the boundary of the medium placed in air when the angle of incidence

is 45. The value of the voltage is :

(A) 63.91 kV (B) 255.64 kV (C) 200.34 kV (D) 127.82 kV


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Sol.

sinC =n1

n = 2

ve

= vlight

in that medium

=n

c=

2

103 8

3116

101.94

109

= 1.6 1019 v

46.1

1.99104

= V

127000V volt Ans. (D)

5. In an electrolytic process certain among of charge liberates 0.8 gm of oxygen. Then the amount of silver

liberated by the same amount of charge is :

(A) 10.8 gm (B) 1.08 gm (C) 0.9 gm (D) 9.0 gm

Sol. q =108

M1 = 2

16

8.0

M = 10.8 gm. Ans. (A)

6. A gas expands from i to f along the three paths indicated. The work done along the three paths denoted byW

1, W

2and W

3have the relationship :

(A) W1

< W2

< W3

(B) W2

< W1

= W3

(C) W2

< W1

< W3

(D) W1

> W2

> W3

Sol. Area under pv graph = work done by gas.

Ans. (A)

7. An ideal gas at 30C enclosed in cylinder with perfectly nonconducting side and a piston moving without

friction in it. The base of the cylinder is perfectly conducting. Cylinder is first placed on a heat source till the

gas is heated to 100C and the piston raised by 20 cm and the atmospheric pressure is 100 kPa. The piston

is then held in final position and cylinder is placed on the heat sink to cool the gas to 30C. Denoting Q1as

the heat supplied during heating and Q2

as the heat lost during the cooling, then [Q1

Q2

] would be

equal to :

(A) 436 J (B) 336 J (C) 236 J (D) 136 J


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Sol. Information is not sufficient

1= f

12nR(70) +

gas

2= 0 + f

12nR(70)

1

2=

gas.

8. Equal amounts liquid helium and water at their respective boiling points are boiled by supplying the heat

from identical heaters in time tHe

and tw

. The latent heats of vaporization of He and water are 2.09 104 J/kg

and 540 kcal / kg, then tHe

is :

(A) about 0.1 tw

(B) about 0.05 tw

(C) just greater than 0.01 tw

(D) just less than 0.01 tw

Sol. (D)

MLhe

= Qthc

MLW

= Qt2w

tHc

=w

He

L

Ltw

=2.410540

1009.23

4

tw

= 0.0092

the

just less than 0.01 tw

Ans. (D)

9. A 5 litre vessel contains 2 mole of oxygen gas at a pressure of 8 atm. The average tralslational kinetic energy

of an oxygen molecule under this condition is :

(A) 8.4 1014 J (B) 4.98 1021 J (C) 7.4 1016 J (D) 4.2 1021 J

Sol. (B)

pv = nRT (8 105) (5 103) = 2RT20 100 = RT ....(1)

Average T.K.E. =

2

3

AN

RT

= 2

3

23106

2000

5 1021

J Ans. (B)

10. Six identical conducting rods are joined as shown. The ends A and D are maintained at 200C and 20 C

respectively. No heat is lost to surroundings. The temperature of the junction C will be :


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(A) 60C (B) 80C (C) 100C (D) 120C

Sol. i =R3

180=

R

60

T 20 = RR

60

T = 80C. Ans. (B)

11. Three corners of an equilateral triangle of side a are occupied by three charges of magnitude q. If the charges

are transfered to infinity, their kinetic energy will be04

1

times

(A)a

q2(B)

a

q3 2(C)

a3

q2(D)

a

q3

Sol.

a

kq3 2= (K.E.) Ans. (B)

12. A LDR (light dependent resistance) is connected to an appropriate voltage source and a current measuring

meter in series. (Assuming that the LDR current is proportional to the intensity of the incident light). The LDR

is illuminated with light from a distant metal filament bulb. The filament voltage V, the distance d of LDR from

the bulb and the LDR current I are noted. If both V and d are doubled, the LDR current is :

(A) 16 (B) 4 (C) (D) less than

Sol. LDR current i Intansity of light 2r

sourceofPower

If V is doubled, P will be four times

r is double

current will be same Ans. (C)

13. A point source is placed at a distance of 30 cm from a convex lens of focal length f on its axis and the image

is formed on a screen at a distance of 60 cm from the lens. Now the lens is split into two halves. One half is

moved perpendicular to the lens axis through a distance of 5 cm. It is found that the two halves of the lens

form two images on the screen and the images are separated by a distance d. The values of f and d

respectively, are :

(A) 20 cm, 15 cm (B) 20 cm, 10 cm (C) 30 cm, 10 cm (D) 30 cm, 5 cmSol. (A)


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60

1

30

1

=

f

1

f = 20 cm

m =u

v=

30

60

= 2

0

i

h

h

= 2

hi= 10 cm

d = 15 cm Ans. (A)

14. Two pendulums differ in lengths by 22 cm. They oscillate at the same place so that one of them makes

30 oscillations and the other makes 36 oscillations during the same time.The lengths (in cm) of the pendulum

are :

(A) 72 and 50 (B) 60 and 38 (C) 50 and 28 (D) 80 and 58

Sol. T1 = g2 1

T2

= g2 2

30T

t

1

6

5

T

T

1

2

2T

t= 36 6T

2= 5T

1

T1

2 =10

26

11

36

100

88 =5

26

5

26=

g2 1

25

236 = 4 10

101

Ans. (A)

15. Three waves of the same amplitude have frequencies (n 1), n and (n + 1)Hz. They superpose on one

another to produce beats. The number of beats produced per second is :

(A) n (B) 2 (C) 1 (D) 3nSol. f

1f2

f3

1hz 2hz 3hz

t = 0 0 0

t = 1 sec. 1/2 sec. 1/3 sec.

T = 1sec.

f = 1 hz. Ans. (C)

16. A spherical ball of mass m1collides head on with another ball of mass m

2at rest. The collision is elastic. The

fraction of kinetic energy lost by m1

is :

(A) 221

21

)mm(

mm4

(B) 21

1

mm

m

(C) 21

2

mm

m

(D) 221

21

)mm(

mm


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Sol. u = v1

+ 21

2 vm

m....(1)

v2 v

1= u ....(2)

1

11

i

if

k

kk =

2

1

u

v1

= 2

21

21

)mm(

mm4

Ans. (A)

17. Two equal masses are connected by a spring satisfying Hooke's law and are placed on a frictionless table.

The spring is elongated a little and allowed to go. Let the angular frequency of oscillations be . Now one ofthe masses is stopped. The square of the new angular frequency is :

(A) 2 (B)2

2(C)

3

2(D) 22

Sol.

k

= 21

21

mm

mm

=m

k2

2

=m

k

2

2 =m

k=

2

2. Ans. (B)

18. When a particle oscillates in simple harmonic motion, both in potential energy and kinetic energy vary

sinusoidally with time. If v be the frequency of the motion of the particle, the frequency associated with the

kinetic energy is :

(A) 4 v (B) 2 v (C) v (D)2

v

Sol. x = A sin wt

K.E. =2

1KA2w2 cos2wt

PE =2

1mA2w2 sin2 wt

frequancy of kinetic energy is 2V Ans. (B)

19. The energy state of doubly ionized lithium having the same energy as that of the first excited state of

hydrogen is :

(A) 4 (B) 6 (C) 3 (D) 2

Sol. (13.6) 2Li

2Li

n

z= (13.6)

4

12

2Lin = 36 nLi = 6 Ans. (B)


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20. The logic circuit shown below is equivalent to :

(A)

(B)

(C)

(D)

Ans. (D)

21. In the circuit shown below, the potential of A with respect to B of the capacitor C is :

(A) 2.00 volt (B) 2.00 volt (C) 1.50 volt (D) +1.50 volt

Sol.

i =2010

15.2

=30

5.1=

20

1A

vBA = 2.5

20

1

10

0.5 = 1.5 V

VAB

= 1.5 V Ans. (C)


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22. Two thermally insulated compartments 1 and 2 are filled with a perfect gas and are connected by a short tube

having a valve which is closed. The pressures, volumes and absolute temperatures of the two compartments

are respectively (p1, V

1, T

1) and (p

2, V

2, T

2). After opening the value, the temperature and the pressure of both

the compartments respectively are :

(A)21

2211

222111

221121

VV

VpVp,

)TVpTVp(

)VpVp(TT

(B) ,TT 2121

2211

VV

VpVp

(C)2211

222111

222111

221121

TVTVTVpTVp,

)TVpTVp()VpVp(TT

(D) 21221121

VVVpVp,

2TT

Sol. n1

=1

11

RT

VPn

2=

2

22

RT

VP

n = n1

+ n2

(number of moles are conserved)

Finally pressure in both parts & temperature of the both the gases will become equal.

RT

)VV(P 21 =

1

11

RT

VP+

2

22

RT

VP

From energy conservation

2

fn

1RT

1+

2

fn

2RT

2=

2

fn

1RT +

2

fn

2RT

T =122211

212211

TVPTVP)TT()VPVP(

P =21

2211

VV

VPVP

Ans. Most approprite is (A)Bonus

23. The voltage drop across a forward biased diode is 0.7 volt. In the following circuit, the voltages across the 10

ohm resistance in series with the diode and 20 ohm resistance are :

(A) 0.70 V, 4.28 V (B) 3.58 V, 4.28 V (C) 5.35 V, 2.14 V (D) 3.58 V, 9.3 V


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Sol.

20

10x +

10

0x + = 10

x 10 + 2x + 2x 18.6 = 0

5x = 28.6

x =5

6.28= 5.72 V

V20

= 9.3 5.72 = 3.58 Ans.

V20

= 10 5.72 = 4.28 Ans. (B)

24. The magnetic flux through a stationary loop of wire having a resistance R varies with time as = at2 + bt (aand b are positive constants). The average emf and the total charge flowing in the loop in the time interval

t = 0 to t = respectively are :

(A)R

ba,ba

2 (B)

R2

ba,ba

2

(C)

R

ba,

2

ba 2 (D) 2(a + b),

R2

ba 2

Sol. = at2 + bt

e = dt

d= at + b

=

dt

edt

=t

btat2 = at + b

q = idt =R

dt =R

=

R

ba 2 Ans. (A)

25. An inductance coil is connected to a an ac source through a 60 ohm resistance in series. The source

voltage, voltage across the coil and voltage across the resistance are found to be 33 V, 27 V and 12 V

respectively. Therefore, the resistance of the coil is :

(A) 30 ohm (B) 45 ohm (C) 105 ohm (D) 75 ohm


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Sol.

272 = VL2 + V

r2 ....(1)

332 = VL2 + (V

R+ V

r)2 ....(2)

332 272 = VR

2 + Vr2 + 2V

RV

r V

r2

332 272 = 122 + 2 12 Vr

Vr=

122

122733 222

= 9

Vr= ir r = 9 5 = 45 Ans. (B)

26. An ideal inductance coil is connected to a parallel plate capacitor. Electrical oscillations with energy W areset up in this circuit. The capacitor plates are slowly drawn apart till the frequency of oscillations is doubled.

The work done in this process will be :

(A) W (B) 2W (C) 3W (D) 4W

Sol. f =LC2

1

if f' = 2f C' =4

cU = W =

C2

Q2

Work done Uf U

i = 'C2

Q2

C2

Q2

= 3W Ans. (C)

27. In the circuit shown below, the switch is in position 1 for a long time. At some moment after that the switch

is through in position 2. The quantity of heat generated in the resistance of 375 ohm after the switch is

changed to position 2 is :

(A) 0.15 J (B) 0.25 J (C) 0.50 J (D) 0.10 J


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Sol.

U =2

18 106 (250)2 = 4 25 25 104

= 0.25 J

U375

=250375

37525.0

= 0.15 J Ans. (A)

28. A conducting square frame of side a and a long straight wire carrying current I are located in the same plane

as shown in the figure. The frame moves to the right with a constant velocity v. The emf induced in the frame

will be proportional to :

(A) 2x

1(B) 2)ax2(

1

(C) 2)ax2(

1

(D)

)ax2()ax2(

1

Sol.

BAB

=

2

ax2

i0


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BCD

=

2

ax2

i0

e = VBAB

a VBCD

a

=

2

2iVa 0

ax2

1

ax2

1

=

iVa 0

)ax2()ax2(

ax2ax2

=

iVa2 02

)ax2()ax2(1

Ans. (D)

29. In the circuit shown below, the switch S is closed at the moment t = 0. As a result the voltage across the

capacitor C will change with time as :

(A) (B)

(C) (D)

Sol. = C.2

R, q

max.= 50 C


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NSEP-2012-13

Ans. (C)

30. The ratio of the rotational kinetic energy to the total kinetic energy of one mole of a gas of rigid diatomic

molecules is :

(A)3

2(B)

5

2(C)

5

3(D)

2

5

Sol. Rotational degree of freedom = 2

Translational degree of freedom = 3

Total

Rotational

.E.K

.E.K=

TK2

15

K2

12

B

BT

=

5

2. Ans. (B)

31. When a metal surface is illuminated with light of wavelength , the stopping potential is V0. When the same

surface is illuminated with light of wavelength 2, the stopping potential is4

V0. If the velocity of light in air is

c, the threshold frequency of photoelectric emission is :

(A) 6c

(B) 3c

(C) 3c2

(D) 3c4

Sol.

hc= eV

0+

0........(i)

2hc

= 00

4

ev ........(ii)

By (i) & (ii),

21

1hc

=

4

11ev0

0ev4

3

2

hc

ev0=

hc

3

2

Put value of ev0in equation (i)

hc

= 0hc

3

2

0

=3

hc

hv0=

3hc


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NSEP-2012-13

v0=

3c

Ans. (B)

32. Two elastic waves move along the same direction in the same medium. The pressure amplitudes of both the

waves are equal, but the wavelength of the first wave is three times that of the second. If the average power

transmitted through unit area by the first wave is W1

and that by the second is W2, then.

(A) W1

= W2

(B) W1

= 3W2

(C) W2= 3W

1(D) W

1= 9W

2

Sol. I =Area

Power=

v2

P20

I = I1

= I2

w1

= w2

Ans. (A)

33. A metal cylinder of length L is subjected to a uniform compressive force F as shown in the figure. The

material of the cylinder has Young's modulus Y and Poisson's ratio . The change in volume of the cylinderis :

(A)Y

FL(B)

Y

FL)1( (C)

Y

FL)21( (D)

Y

FL)21(

Sol. Valumetric change

V = L3V + V = (L + L) (LL1)2

V

V=

L

L1 1

L

L1

21

V

V= 1

L

L1

21

V

V~ (1 2)

L

L................(1)

Y =

V

LA

F

L =

AY

FL

From eq. (1)

V

V~ (1 2 )

LAY

FL

V ~ (1 2 )Y

FLAns. (D)


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34. Three persons A, B, and C note the time taken by their train to cover the distance between two successivestations by observing the digital clocks on the platforms of two stations. The clocks display time in hours andminutes. The three persons find the intervals 3. 5 and 4 minutes respectively. Assume the maximum discrep-ancy of 2 seconds in actual starting and stopping of the train and the observations by A, B and C. Then,(A) All A, B and C can be correct(B) Only A and B or B and C can be correct(C) Only one of A, B and C can be correct(D) C is correct since it is equal to the average of the three observations.

Ans. (A)

35. When two drops of water coalesce (I) Total surface area decreases. (II) There is some rise in temperature.Which of the following is correct ?(A) Both (I) and (II) are wrong statements(B) Statement (I) is true but (II) is not true.(C) Both (I) and (II) are true and the two statements are independent of each other.(D) Both (I) and (II) are true and (I) is the cause of (II)

Sol. Temp. will increaseS.Ai = 8r

2

S.Af = 4r2/22/3

Final Area < Initial Area. Ans. (D)

36. Two capacitors 0.5 F and 1.0 F in series are connected to a dc source of 30 V. The voltages across thecapacitors respectively are :(A) 10 V, 20 volt (B) 15 V, 15 V (C) 20 V, 10 V (D) 30 V, 30 V

Sol. Q = CVQ = 1/3 30= 10 C

1/2

10V1 = 20 volt

V2 = 10 volt Ans. (C)

37. The 23290Th atom has successive alpha and beta decays to the end product20882Pb . The numbers of alpha

and beta particles emitted in the process respectively are :(A) 4, 6 (B) 4, 4 (C) 6, 2 (D) 6, 4

Sol. No. of particle =4

208232 = 6

particle = 4. Ans. (D)

38. If the breakdown field of air is 2.0 106 V/ m, the maximum charge that can be given to a sphere of diameter10 cm is :(A) 2.0 104 C (B) 5.6 107 C (C) 5.6 105 C (D) 2.0 102 C

Sol. 2 106 = 2

9

)2/1.0(

Q109 Q =

49

102 5

=510

18

1 = 5.6 107 Ans. (B)

39. Density of ocean water varies with depth. This is due to :

(A) elasticity (B) viscosity (C) surface tension (D) all the threeAns. (A)

40. A spring of certain length and having spring constant k is cut into two pieces of lengths in a ratio 1 : 2. Thespring constants of the two pieces are in a ratio :(A) 1 : 1 (B) 1 : 4 (C) 1 : 2 (D) None of these

Sol.

2

1

k

k=

2/k3

k3=

1

2Ans. (D)


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SUB- PART A-2

In question 41 to 50 any number of options (1 or 2 or 3 or all 4) may be correct. You are to identify

all of them correctly to get 6 marks. Even if one answer identified is incorrect or one correct

answer is missed, you get zero score.

41. A cube floats both in water and in a liquid of specific gravity 0.8. Therefore,

(A) apparent weight of the cube is the same in water and in the liquid.

(B) the cube has displaced equal volume of water and the liquid while floating(C) the cube has displaced equal weight of water and the liquid while floating.

(D) if some weights are placed on the top surface of the cube to make it just sink, the load is case of water

will be 0.8 times of that to be used in case of the liquid.

Sol. = density of cube = 0.4 (say)

= density of liquid = 0.8

w

= density of water = 1.0

For liquid

mg = ()

2

(2) g

m =2

3 =

2

)8.0( 3

= (0.4) 3

For water

m'g = w (0.6) 2 gm' = (1) (0.6) 3

m' = (0.6) 3

m

'm=

4

6=

2

3 m' = 1.5 m Ans. (A,C)

42. On the basis of the kinetic theory of gases one compares 1 gram of hydrogen with 1 gram of argon both at

0C . Then,

(A) the same temperature implies that the average kinetic energy of the molecules is the same in both the

cases.

(B) the same temperature implies that the average potential energy of the molecules is the same in both the

case.

(C) internal energies in both the cases are equal.

(D) when both the samples are heated through 1C, the total energy added to both of them is not the same.


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NSEP-2012-13

Sol. Average K.E. of a molecule

=2

RT3(as molecule is taken as spherical in nature)

Internal energy depends upon number of moles. Ans. (AD)

43. While explaining the action of heat engine, one can say that

(A) heat cannot be fully converted into mechanical work.

(B) the first law of thermodynamics is necessary but not sufficient(C) heat under no circumstances can flow from lower to higher temperature.

(D) A body can not be cooled to absolute zero.

Sol. (ABD)

44. The rate of change of angular momentum of a system of particles about the centre of mass is equal to the

sum of external torques about the centre of mass when the centre of mass is :

(A) Fixed with respect to an inertial frame.

(B) in linear acceleration

(C) in rotational motion.

(D) is in a translational motion.

Sol.dt

Ld C

= C

Ans. (A,B,C,D)

45. Light is traveling in vacuum along the Z axis. The sets of possible electric and magnetic fields could be :

(A) )kztsin(BjB),kztsin(EiE 00

(B) )kztcos(BjB),kztsin(EiE 00

(C) )kztsin(BiB),kztsin(EjE 00

(D) )kztsin(BjB),kztsin(EiE 00

Sol. LBE

Ans. (ACD)

46. In case of photoelectric effect,

(A) since photons are absorbed as a single unit, there is no significant time delay in the emission of photo-

electrons.

(B) Einstein's analysis gives a critical frequency v0

=h

e, where is the work function and the light of this

frequency ejects electrons with maximum kinetic energy.

(C) only a small fraction of the incident photons succeed in ejecting photoelectrons while most of them are

absorbed by the system as a whole and generate thermal energy.(D) the maximum kinetic energy of the electrons is dependent on the intensity of radiation.


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NSEP-2012-13

Sol. hv = + Kmax

v = v0 K

max= 0

hv0

= v0

=h

Ans. (A,C)

47. A parallel combination of an inductor coil and a resistance of 60 ohm is connected to an ac source. The

current in the coil, current in the resistance and the source current are respectively 3A, 2.5 A and 4.5 A

respectively. Therefore,(A) Kirchhoff's current law is NOT applicable to ac circuits.

(B) impedance of the coil is 50 ohm

(C) electric power dissipated in the coil is 150 watt.

(D) impedance of the circuit is 33.3 ohm.

Sol. ZLis impedence of inductor

3ZL= 2.5 60

ZL= 50

(4.5)2 = (2.5)2 + 32 + 2(2.5) (3) cos

cos =1

3

L

X

8= R

L

ZL2 = 9R

L2

502 = 9RL2

RL=

50

3

Power dissipaled in coil

=2 503 150 watt

3

2 2

total

1 1 1 1 12 cos

Z 60 5060 50

Z = 33.3 Ans. (B,C,D)

48. The nuclear forces

(A) are stronger being roughly hundred times that of electromagnetic forces

(B) have a short range dominant over a distance of about a few fermi

(C) are central forces independent of the spin of the nucleons.

(D) are independent of the nuclear charge.

Sol. (ABD)

49. Consider a mole of a sample of hydrogen gas at NTP.

(A) The volume of the gas is exactly 2.24 102 m3.

(B) The volume of the gas is approximately 2.24 102 m3.

(C) The gas will be in thermal equilibrium with 1 mole of oxygen gas at NTP

(D) The gas will be in thermodynamic equilibrium with 1 mole of oxygen at NTP.

Ans. (BCD)

50. A particle moves in one dimension in a conservation force field. The potential energy is depicted in the graph

below.


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NSEP-2012-13

If the particle starts to move from rest from the point A, then

(A) the speed is zero at the point A and E.

(B) the acceleration vanished at the points A, B, C, D, E

(C) the acceleration vanished at the points B, C, D.(D) the speed is maximum at the point D.

Sol. VA

+ KA

= VE

+ KE

VA

= VE

& KA

= 0 KE

= 0

F =dx

dV= 0,

Slope = 0 at points B, C & D

Ans. (AC)

PART B MARKS : 60

All questions are compulsory. All questions carry equal marks.

1. (a) A conductor having resistance R (independent of temperature) and thermal capacity C is initially at

temperature T0

same as that of the surrounding. At time t = 0 it is connected to a source with constant

voltage V. The thermal power dissipated by the conductor to the surrounding varies as q = k (T T0).

Determine the temperature T of the conductor at any time t and at the time t =k

C.

(b) A particle moves rectilinearly in an electric field E = E0 ax where a is a positive constant and x is the

distance from the point where the particle is initially at rest. Let the particle have a specific chargem

q. Find

(I) the distance covered by the particle till the moment at which it once again comes to rest, and (II) accelera-

tion of the particle at this moment.

Sol.

(a) t = 0, T = T0

= TS

Rate of heat loss q = k(T T0)

Power delivered bysource

R

v2= )TT(k

dt

dTC 0


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R

v)TT(k

dt

dTC

2

0

T

T

20

0R/v)TT(k(

dT=

t

0C

dt

R/v

)TT(kR/vn2

0

2

=ck t

0

tc

k2

Te1Rk

vT

At t =k

c

T =Rk

v2(1 e) + T

0

(b) Particle is performing SHM having mean position at x =a

E0an amplitude

a

E0

E = E0 ax

So, maximum distance travelled =a

E2 0

and Required acceleration =m

qE0 .

2. One mole of an ideal gas ( = 1.4) with initial pressure of 2 atm and temperature of 57C is taken to twice itsvolume through different processes that include isothermal, isobaric and adiabatic processes. Determine the

process where maximum work is done and the amount of work in this case. By what percentage is this work

larger than the work done in a process in which it is the least ?

Sol.

p0v

0= p

0v

0

W2= P

0v

0

W1= p

0v

0 n2

W3

=1

vpvp

1

)TT(nRU 00ffif

=

1

)21(nRT 1

=4.0

)330(R1

5/22

11 Ans.Ans.


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In isobaric process work done is maximum = P0v

0

In adiabatic process work done is minimum =)1(

)2/21(vp 00

% diff = %1001

=

1

)21( 1

100% =

5/22

11

4.0

1 100% = 65.2% Ans.

3. A railway carriage of mass Mcfilled with sand of mass M

smoves along the rails. The carriage is given an

impulse and it starts with a velocity v0. At the same time it is observed that the sand starts leaking through

a hole at the bottom of the carriage at a constant mass rate . Find the distance at which the carriagebecomes empty and the velocity attained by the carriage at that time. (Neglect the friction along the rails.)

Sol.

At t,

F = mdt

dv+ (v v

0)

dt

dm

0 = mdt

dv+ (v v

0)

dt

dm

dt

dv= 0

v = constant v = v

0Ans.

Also S = v0t

S = V0

sM

Ans.

4. Show that, for any angle of incidence on a prism

)A(2

1sin

)A(2

1sin

= )ei(

2

1cos

)rr(2

1cos 21

(symbols have usual meanings)

and that the right-hand side reduces to at minimum deviation

Sol. i + e A = i + e = A +

2

Asin

2

A

sin

=

2

eicos

2

rr

cos21


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NSEP-2012-13

2

rrsin

2

eisin

21

=

2

eicos

2

rrcos 21

(sin i sinr1) = ( sin r

2 sin e)

0 = 0.

5. (a) A small amount of solution containing Na24 nuclides with activity 20500 disintegrations per second was

injected in the blood stream of a person. The activity of 1 ml of blood sample taken after 5 hours later, was

found to be 20 disintegration per minute. The half life of the radioactive nuclides is 15 hours. Find the total

volume of the blood of this person.

(b) The wire loop shown in the figure lies in uniform magnetic induction B = B0cos t perpendicular to its

plane. (Given r1= 10 cm and r

2= 20 cm, B

0= 20 mT and = 100 ). Find the amplitude of the current induced

in the loop if its resistance is 0.1 /m.

Physical constants you may need ......

(1) Charge on electron e = 1.6 1019C

(2) Mass of electron me

= 9.1 031 kg

(3) Universal gravitational constant G = 6.67 10

11 Nm2/kg2(4) Permittivity of free space

0= 8.85 1012 C2 Nm2

(5) Gas constant R = 8.31 J/K mol

(6) Planck constant h = 6.62 1034 Js

(7) Stefan constant = 5.67 108 W/m2 K4

(8) Boltzman constant = 1.38 1023 J/K

Sol. Let volume of blood be V ml then activity per ml =

V

20500A0

activity after 5 hr = At= A

0et

60

20

dps =

5

15

2n

ev

20500

=2n

3

1

ev

20500

=

3/1

2

1n

ev

20500

3

1=

3/1

2

1.

v

20500

So, v = m)2(

2050033/1

= m

)2(

615003/1 Ans.


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NSEP-2012-13

(b)

B = B0cost

Net flux

= BA2 BA

1

So, induced emf

e = (A2 A

1)

dt

dB

So, i =dt

dB.

R

)AA( 12 = tsinB.

R

)rr.(0

21

22

So, i =

10010201030102

10)100400.( 321

4

= ampere. Ans.

Physical constants you may need

1. Charge on electron e = 1.6 1019 C

2. Mass of an electron me

= 9.1 1031kg

3. Universal gravitational constant G = 6.67 1011 N m2/kg2

4. Permittivity of free space 0

= 8.85 1012 C2 / N m2

5. Gas constant R = 8.31 J/K mol

6. Planck constant h = 6.62 1034 Js

7. Stefan constant = 5.67 108 W/m2 k4

8. Boltzmann constant k = 1.38 1023

J/K

paper with solution physics nsep-2012

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