Csir Ugc Net_physics_ Free Solved Paper

7/27/2019 Csir Ugc Net_physics_ Free Solved Paper

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C SIR NET, GATE, IIT-JAM, UGC NET , TIFR, IISc , JEST , JNU, BHU , ISM , IBPS, CSAT, SLET, NIMCET, CT ET

Phone: 0744-2429714 Mobile: 9001297111, 9829567114, 9001297243Website: www.vpmclasses.com E-Mail: vpm classe [email protected] / inf [email protected]

A ddress: 1-C-8, Shee la Chowdhary Road, SFS, TALWANDI, KOTA , RAJASTHA N, 324005 Page 1

For IIT-JAM, JNU, GATE, NET, NIMCET and Other Entrance Exams

Patte rn of questions : MCQs

This paper contains 55 Multiple Choice Questions

part A 15, part B 20 and part C 20

Each question in Part 'A' carries two marks

Part 'B' carries 3.5 marks

Part 'C' carries 5 marks respectively.

There will be negative marking @ 25% for each wrong answer.

CSIR NET - PHYSICAL SCIENCE

1-C-8, Sheela Chowdhary Road, Talwandi, Kota (Raj.) Tel No. 0744-2429714

Web Site www.vpmclasses.com [email protected]

Total marks : 200

Duration of test : 3 Hours

MOCK TEST PAPER


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PART A (1-15)

1. Tw enty four clerk can clear 180 f iles in 15 days. Number of clerk required to clear240 files in 12 days is

(1) 38

(2) 39

(3) 40

(4) 422. In the given f igure, RA = SA = 9cm and QA = 7cm. If PQ is the diameter, then radius is

AR S

Q

P

(1) 65 cm7

(2) 130 cm7

(3) 8 cm(4) None

3. If the circ les are draw n with radii R 1, R2, R3 w ith centre at the vertices of a triangle as show nin figure. Side of triangle is a, b, c respectively, then R 1 + R2 + R3 is equal to


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R 1

R3

(1) 3(a + b + c)

(2)1

(a b c)3

+ +

(3)1

(a b c)2

+ +

(4) 2(a + b + c)

4. Study the f ollow ing graph and ans w er the question given below it

10

15

20

2530

35

4045

50

24

6

8

1012

14

1618

20

1984 1985 1986 1987 1988 1989

.

Years

T o t a l v a

l u e o

f t o o

l s ( i n

R s c r o r e s )

Production in a Tool Factory

Number of Tools ----- Value What w as the value of each tool in 1985?


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(1) Rs1

53

thousand

(2) Rs 50 thousand

(3) Rs 5, 103

(4)5

59

5. The tota l adults in a city is 60000. The various sections of them ar e indicated below in thecircle

What percentage of the employed persons is self employed?

(1) 5519

(2) 1195

(3) 20

(4) 5

6. Look at this series: 14, 28, 20, 40, 32, 64, ... What number should come next?

(1) 52

(2) 56

(3) 96

108II

54 18

III

IV

V

I employees in the public sectorII employees in the private sectorIII employees in the corporate sectorIV self employed

unemployed

V


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(4) 128

7. A car ow ner buys petrol at Rs.7.50, Rs. 8 and Rs. 8.50 per l iter for three success ive years.

What approximately is the average cost per liter of petrol if he spends Rs. 400 0 each year?

(1) Rs. 7.98

(2) Rs. 8

(3) Rs. 8.50

(4) Rs. 9

8. In a certain store, the prof it is 320% of the cost. If the cost increases by 25% but the selling

price remains constant, approx imately w hat percentage of the selling price is the profit?

(1) 30%

(2) 70%

(3) 100%

(4) 250%

9. Today is Friday af ter 62 days, it w ill be :

(1) Thursday

(2) Friday

(3) Wednesday

(4) Tuesday


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10. A car travelling w ith of its actual speed covers 42 km in 1 hr 40 min 48 sec. Find the

actual speed of the car.

(1)6

17 km /hr7

(2) 25 km/hr

(3) 30 km/hr

(4) 35 km/hr

11. P is a w orking and Q is a sleeping partner. P puts in Rs. 3400 and Q puts Rs.6500. P

receives 20% of the prof its for managing. The rest is distributed in proportion to their

capitals. Out of a total profit of Rs.990, how much did P get ?

(1) 460

(2) 470

(3) 450

(4) 480

12. A lawn is the form of a rectangle having its s ide in the ratio 2:3 The area of the law n is 1/6hectares. Find the length and breadth of the lawn.

(1) 25m(2) 50m

(3) 75m

(4) 100 m

13. An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cov er thesame distance in 1 hours , it must travel at a speed of :

(1) 300 kmph


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(2) 360 kmph

(3) 600 kmph

(4) 720 kmph

14. Find out the missing nu mber of the given question:

2 7 45 2 3

1 ? 6

10 42 72(1) 2(2) 4

(3) 5(4) 3

15. All of the follow ing are the same in a manner. Fi nd out the one which is different amongthem:

(1) BFJQ

(2) RUZ G(3) GJOV(4) ILQX

PART B (16 -35)

16 . What is the solution of integral

0

[cos(3x) 2] (x )dx

+ (1) 0

(2) 2(3) 3

(4) 1

17 Solve the integral equation0

1 , 0 1f( ) cos d

0, 1

=

>


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(1) 22(1 cos )f( )

=

(2) 22cosf( )

=

(3)2

(1 cos )f( )

2

=

(4)2

(1 cos )f( )

=

18 . Find the fun ction whose laplace t ransform is2

2 2 2s(s a )+

(1) +1 [at sin at cos a t]2a

(2) 1 (a sin a t cos at )2a

+

(3)1

[a cos at sinat]2a

+ +

(4)1

[at cosat sin at]2a

+

19 . The Lagrange equation of m oti on of two rigid bodies of masses m and 2m are connected by a lightflexible spring of spring constan t K. what is the Lagrange equation o f motion.

(1)k

x x 0m

+ =

(2)k

x x 02m

+ =

(3)3k

x x 02m

+ =

(4) 5kx x 02m+ =


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20 . In the following indicator diagram, the net amount of work done will be

(1) Positive

(2) Negative

(3) Zero

(4) Infinity

21 . A particle moves in a plane under the influence of a for ce, acting towards a centre of force whose

magnitude is2

2 2

1 r 2rrF 1

r c

=

where r is the distance of the particle to the centre o f force, then

the Lagrangian for the motion in a plane i s

(1)2 2 2 2

2

r r 1 1 rL 2 2 r rc

= +

(2)2 2 2 2

2 2

r r 1 2 rL

2 2 r c r= + + +

(3)2 2 2

2 2

r r 2 1 rL

2 2 r c r= + +

(4)2 2

2 2

r 1 1 rL 2 r c r

= +

22 . Cal culate the Fermi energy in ele ctron vol t for sodium assum ing that it has one free electron peratom. Given densi ty of sodium = 0.97 g cm 3, atomic weight of sodium is 23.

(1) 3.541 eV

(2) 3.451 eV

(3) 5.135 eV

(4) 3.145 eV


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23 . The paramagnetic contribution to the magnetic susceptibility per m 3 of potassium, for which the Fermienergy is 2.1 eV is at wt. of potassium is 39.1 gm and density of potassium is 0.86 10 3 kgm 3 .

(1) 420.5 10 8

(2) 450.2 10 +8

(3) 420.5 10 6

(4) 425.210 +6

24 . The figure shows the inverse magnetic susceptibility (1/ ) (dimensionless) a s a function oftemperature for a paramagnetic material. Calculate the concentration of magnetic ions, if they areassumed to be Co 2+ with configuration 3d 7.

(1) 54 10 23

(2) 51 10 26

(3) 69 10 23

(4) 69 10 26

25 . A 3D structure of current carrying wire is as shown in the figure. The magnetic force experienced bycha rge particle of mass m and charge q, when it is crossing origin with velo city v along +ve Y-axis willbe

(1) 0 0q qV

8R 4 R +

i

(2) 0q 2 1

8R +

i k

(3) 0q

8

k

(4) Zero


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26 . If E f(0) and E f are Fermi levels of a metal at 0K and 30000 K, then what is the value off

f(0)

EE

if

E f(0) = 7 eV.

(1) 0.119

(2) 0 .88

(3) 1.113

(4) 1.188

27 . The small (rotational) Raman displacement for HCI mo lecule is 416 cm 1 . Find the internucleardi stance b etween the atoms form ing the m olecule. Given : h = 6 63 10 34 J s, c = 30 10 8 m s 1 and N A = 6023 10

23 mol 1 .

(1) 129 A

(2) 229 A

(3) 249 A(4) 064 A

28 . For the given circuit the the open loop gain is 12000 and R 1 = 120 k and R f = 600 k . V i = 1.2 V.Find the exact output vol tage fo r the inverting operational amplifie r.

(1) 5 V

(2) 6 V

(3) 599 V

(4) 499 V

29 . Oxygen has nuclear spin of 5/2. The NMR of oxygen gives

(1) 2 lines

(2) 3 lines

(3) 4 lines


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(4) 6 lines

30 . A metal strain gauge factor of two. ts nominal resistance i s 120 ohms. It undergoes strai n 10 5, thevalue of change of resistance in response to the strain is

(1) 240 ohms

(2) 2 10 5 ohm

(3) 2.4 10 3 ohm

(4) 1.2 10 3 ohm

31 . EvaluateV

Fd , where 2F xyz= over the prism placed at origin as sho wn in the Figure.

(1)13

(2)23

(3)1

9

(4)29

32 . A star initially has 10 40 deuterons. I t prod uces en ergy via the processes

1H2 + 1H2 1H3 + p and 1H2 + 1H3 2He 4 + nIf the average power radiated by the star is 10 16 W, the deuteron supply of the star is exhausted in atime of the order of(1) 10 6 sec(2) 10 8 sec(3) 10 12 sec(4) 10 16 sec

The masses of the nuclei are as follows:M (H2) = 2014 amu; M (p) = 1007 amu;

M (n) = 1008 amu; M (He 4) = 4001 amu.


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33 . A -ray photon produces an electron-positron pair, each moving with a K.E. of 001 MeV. The energyof the -ray photon is(1) 102 MeV(2) 104 MeV(3) 208 MeV(4) 103 MeV

34 . The temperature of the two outer surfaces of a composite slab, consisti ng of two materials havingcoefficients of thermal conductivity K and 2K and thickness x and 4x, respectively are T 2 and T 1 (T 2 >T 1).

2KT2 T1

x 4x

The rate of heat transferred through the slab, in a steady state is 2 1A(T T )K

f,x

with f equals to

(1) 1(2) 1/2(3) 2/3(4) 1/3

35 . In quark model what is the state of

(1) ud

(2) 1 (uu dd)2

(3) 1 (uu dd)2

+

(4) 1 (us su)2

PART C (36 -55)


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36 . The value of the counter integral6

3C

sin zdz

1z 6

, i f C i s the ci rcle z 1= .

(1) 20 i17

(2) 21 i16

(3)15

i7

(4)12

i13

37 . The matrix A, defined by

1 2 2A 2 a b

2 b a

=

Is orthogonal i f

(1) a = 1, b = 1

(2) a = 1 , b = 2

(3) a = 2, b = 1

(4) a = 2, b =

38 . A reversible engine works between three thermal reservoirs, A,B and C . The engine absorbs anequal amount of heat from the thermal reservoirs A and B kept at temperatures T A and T B respectively, and rejects heat to the thermal reservoir C kep t at temperature T C. The efficiency of theengine is times the efficiency of the reversible engine, which works between the two reservoirs Aan d C. which one of the following relation statement is correct ?

(1) ( ) ( )A AB C

T T2 1 2 1 T T

= +


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(2) ( ) ( )A AB C

T T2 1 2 1

T T= +

(3) ( ) ( ) CBA A

TT 2 1 2 1 T T

= +

(4) ( ) ( )A AC B

T T2 1 2 1 T T

= +

39 . A quantum mechanical particle of mass m is confined in three-di mensional infinite square well

po tential of side a. The eigen-energy of particle is given as E = 2 2

2

9.ma The state i s

(1) 4 fold degenerate

(2) 3-fold degenerate

(3) 2-fold degenerate

(4) Non-degenerate

40 . Ten grammas of water at 20 C i s con verted into at 10C at constant atmosph eric pressure.Assuming the specific heat of liquid water to remain constant at 4.2 J/gK and that of ice to be half ofthis value, and taking the latent heat of fusion of ice at 0C to be 335 J/g , the total entropy of thesystem is .

(1) Zero

(2) 16.02 JK 1

(3) 15.63 JK 1

(4) 15.63JK 1

41 . A sphere rolls down a rough included plane; i f x be the distance of the point of contact o f the spherefrom a fixed point on the plane, find the acceleration.

(1) 5 g sin7

(2)5

g sin14

(3) mg sin


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(4) 7 g sin5

42 . A symmetrical top with moments of inertia n = y and z in the body axes frame is described by theHamiltonian

( )2 2 2x y zx

1 1H L L L

2I 2Ix= + +

Here moments of inertia are parameters and not operators. L x Ly and L z are the angular momentumoperator in the body axes frame.

(i) The eigenvalues of the Hamiltonian

(1)2

2 2

x z x

1 1( + 1) + m2 2 2

(2)2

2 2

x z x

1 1( + 1) + m

2 2 2

+

(3)2 2

2

x z x

1 1 m ( 1)2 2 2

+ +

(4)

2 2

2

x z x

1 1 m( 1)2 2 2

+ + +

(ii) Expected value for a measurement of L x + Ly + Lz for a ny state is

(1) Zero

(2) m

(3) m

(4) m2

43 . If we take in the semi-empirical mass formula a c = 0.58 MeV and a a = 19.3 MeV. Then possibleatomic number of most stable nuclei of m ass number 64.(1) 26

(2) 29(3) 32

(4) 33


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44 . 36 g of water at 30C are converted into steam at 250C at constant atmospheric pressure. Thespe cific heat of water is assumed constant at 4.2 J/g K and the latent heat of vaporization at 10 0C is

2260 J/g. For wate r vapour, assum e pV = mRT where R = 0.4619 kJ/ kg K , and pCR

= a + bT + cT ,

where a = 3.634,

b = 1.195 10 3 K 1 and c = 0.135 10 6 K 2

Calculate the entropy change of the system.

(1) 0.2181 kJ/K(2) 0.0235 kJ/K

(3) 273.1 J/K(4) 314.3 J/K

45 . A perpendicularly polarized wave propagates from region 1(r1 = 8.5, r1=1, 1 = 0) to region 2, freespa ce, with an angle of incidence of 15. Given i0E 1.0 V / m= , then r0E , i s

(1) 1.62 V/m

(2) 0.623 V/m

(3) 4.23 V/m

(4) 7.75 mV/m

46 . A particle A of mass m moving along the positive x-direction with kinetic energy K suffers an elastichead-on collision with a stationary particle B of mass M. After the collision the particle A moves alongthe negative x-direction with kinetic energy K/9. What is the mass of particle B?

(1) 9 m

(2) 6 m

(3) 3 m

(4) 2 m

47 . Cal culate the amount of energy released i f all the deuteri um atoms in the water in the lake of area

about 10 5 sq. miles and o f depth1

20the mile area used up in fusio n.

(1) 2.18 10 38 MeV


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(2) 43 MeV

(3) 1.56 10 39 MeV

(4) 6.9 10 38 MeV

48 . The maximum wa ve length of photons that can be detected by a photo diode made of asem iconductor of band gap 2 eV is about

(1) 620 nm

(2) 700 nm

(3) 740 nm

(4) 860 nm

49 . The three electroni c circuits marked (i), (ii ) and (iii ) in the figure below can al l work as logic gates,where the input signals are either 0V or 5V and the output is V 0.

Identi fy the correct combina tion of l ogic gates (i ), (i i ), (iii ) in the options gi ven below.

(1) NOR, XOR, AND

(2) OR, NAND, NOR

(3) NAND, AND, XOR

(4) AND, OR, NOR

Statement for Linked Answer Question 50(i) and 50 (ii)


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Let f(z) = cos zsin z

z

for non-zero z C and f(0) = 0. Al so, let g (z) = sinh z for z C.

50(i). Then f(z) has a zero at z = 0 of order

(1) 0

(2) 1

(3) 2

(4) Greate r than 2

50(ii). Theng(z)zf(z)

has a pole at z = 0 of order

(1) 1

(2) 2

(3) 3

(4) Greate r than 3

51 . The Lagrangian of a system is gi ven by

2 2 2 21L mr ( sin ) V(r, , )2

= +

The equation of motion is

(1) 2 2d V

(mr sin ) 0dt

=

(2) 2 2d V

(mr sin ) 0dt

=

(3)2 2d V

(mr sin ) 0dt

+ =

(4) 2 2d V(mr cos ) 0dt

+ =

Linked question 52(i), 52(ii), 52(iii)

A one-dimensional harmonic oscillator of a particle with m ass an and potential energy v(x) = 2 21 m x2


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This pa rticle ha s a charge q and is placed in a uniform electric field E pa ralle l to the x axis, E = Ex .

52.(i). The Hamiltonian of the particle

(1)2

2 2P 1 m x x2m 2

+

(2)2

2 2P 1 m x x2m 2

+ +

(3)

22 2P 1

m x x2m 2

(4)2

2 2P 1 m x x2m 2

+

52(ii). Perform a coordinate transformation y = ax+b (wh ere a and b are constant / such that in the ycoo rdinate the Hamiltonian i s similar to that of a one dimensional harmonic oscilla tor (with nocharge) What are a and b

(1) a = 1 , b = / m2

(2) a = 1 , b = / m2

(3) a = / m2 , b = 1

(4) a = / m2 , b = 152(iii). The energy eigenvalues of the system is

(1)2

2

1 1n

2 2 m +

(2)2

2

1 3n2 2 m

+

(3) 22

1 1 ew n2 2 2mw +

(4) 2

2

1 3n2 2 2m

+

53 . The equation x 3 x 2 + 4 x 4 = 0 is to be solved using the Newton-Rephson method. If x = 2 is takenas the ini tial approximation using this metho d will be


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(1) 23

(2)43

(3) 1

(4) 32

54 . A mass m is released from rest at height h. Find the Hamilton character isti c function of the system

(1) 1 /22m (E mgz) dz (2) 1/22m(E mgz) dz+ (3) 2m(E mgz) dz

(4) 2m (E mgz) dz+ 55 . At what temperatu re will the number of 2 molecules in the = 1 level be one-tenth of that i n the = 0

le vel? Given : = 214.6 cm 1 ,

e xe = 0.6 cm 1, h = 6.63 10 34 Js, c = 3.0 10 8 ms 1 and k = 1 .38 10 23 J/K.

(1) 155.3 K(2) 135.5K

(3) 133.5 K

(4) 127.5 K

Answer key


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Que. Ans. Que. Ans. Que. Ans. Que. Ans.1 3 16 4 31 3 45 22 1 17 1 32 3 46 43 3 18 4 33 4 47 44 4 19 3 34 2 48 15 1 20 2 35 2 49 46 2 21 1 36 2 50(i) 37 1 22 4 37 2 50(ii) 38 2 23 1 38 1 51 39 4 24 2 39 2 52 i 110 4 25 2 40 3 52(ii) 2

11 2 26 2 41 1 52(iii) 312 2 27 1 42(i) 1 53 213 4 28 2 42 ii 3 54 114 4 29 4 43 2 55 315 1 30 3 44 3

Solutions

PART A (1-15)

1.(3) 1 1 2 21 2

m D m Dw w

=

2m 1224 15180 24

=

m2 = 40

2.(1) RA SA 9 9

PA PAQA 7 = =

Diameter = PA + AQ

81 13077 7

+ =

Radius = Diameter2

65Radius7

=

3.(3) R1 + R2 = aR2 + R3 = b

R3 + R1 = cR1 + R2 + R2 + R3 + R3 + R1 = a + b + c


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1 2 3

a b cR R R

2+ +

+ + =

4. (4) Value of each tool in 1985

=7

3

10 1018 10

[Since 1 crore = 10 7]

=5

59

T housand

5.(1) The required percentage =( )

18100

360 18

(since total employed = 360 unemployed)

=18 5

100 5 %342 19

=

6.(2) This is an alternating m ultiplication and subtracting series: Firs t, multiply by 2 andthen subtract 8.

7.(1) Total quantity of petrol =4000 4000 4000

litres7.50 8 8.50 + +

consumed in 3 years2 1 2

4000 liters15 8 17 + +

=76700

litres51

Total amount spent = Rs. (3 x 4000) = Rs. 12000.

Average cost = 12000 51 6120Rs. Rs.7.9876700 767 = =

8.(2) Let C.P.= Rs. 100. Then, Pr of it = Rs. 320, S.P. = Rs. 420.New C.P. = 125% of Rs. 100 = Rs. 125

New S.P. = Rs. 420.

Prof it = Rs. (420 - 125) = Rs. 295.


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Required percentage =%

295 1475100 % 70% (approximately)

420 21 = =

A student multiplied a number by35

instead of53

9.(4) Each day of the w eek is repeated after 7 days .

So, af ter 63 days, it w ill be Friday . Hence after 63 days,it w ill be Thursday.

Therefore the required day is Thursday .

10 .(4) 4 51 12640 min 1 hrs hrs.5 75 75

= =

Time taken = 1 hr 40 min 48 sec = 1 hr

Let the actual speed be x km/hr.

Then, 5 126

x 427 75

=

x =42 7 75

35km/ hr.5 126 =

11.(2) Given, Total prof it = Rs. 990

Ration of their capit als = 34 : 65.

Now , prof it amount got by P = 20% of total profit + Ps share in balance 80%profit for his capital

340.2 0.834 65

+ + = 470

12.(2) Now area = (1/6 1000)sq m = 5000/3 sq m2x 3x = 5000/3 =>x x = 2500 / 9

x = 50/3

length = 2x = 100/3 m and breadth = 3x = 3 (50/3) = 50m

13. (4) Distance = (240 x 5) = 1200 km.

Speed = Distance/Time


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Speed = 1200/(5/3) km/hr. [We can w rite 1 hours as 5/3 hours]

Required speed = 1200 x 3 km/hr = 720 km/hr.

14.(4) As, 2 5 1 = 20

and 4 3 6 = 72

Similarly , 7 2 ? = 42

15.(1) According to question,

Therefore, B F J Q is odd.

PART B (16 -35)16.(4) From the property of d irac del ta function

f (x ) (x a)dx f(a)+

=

so,0

[cos(3x) 2] (x )dx+

+ Here f(x) = cos(3x) + 2

so, (x ) = (0) x =

so,0

[cos(3x) 2] (x )dx

+ = cos(3 ) + 2 = 1 + 2 = 1 17.(1) We know Fourier cos transform is

42? 3

14= =


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c 0

2F f(x)cos sx dx

=

c 0

22 (1 ), 0 1F ( ) f( )cos d

0, 1

< = =

>

By inversion formula of Fc (), we get

f() = c02 F ( )xcos d

= ( )1

0 1

2 2 1 cos d 0 d

+

1

20

2 sin cosf( ) (1 ) ( 1) =

= 2 2

2 cos 10 0 +

2

2(1 cos )f( )

=

18.(4) (ii ) We ha ve f 1 (s) = f 2 =(s) = 2 2s L{cosat}

s a=

+

By con volution theorem

t1

1 2 1 20

L {f (s)f (s)} F (y)F (t y)dy =

t21 1

2 2 2 2 2 2 20

s s sL L cos aycos a(t y)dy

(s a ) s a s a = = + + +

=t

0

1 [co s at co s( 2a y at) ]d y2

+ t

0

1 1y cosat sin(2ay at)2 2a

= +

= 1 1 1t cos at sinat [at cos at sin at]2 a 2a

+ = +

19.(3) Red uced m ass of two bo dy system is

M = m 2m 2mm 2m 3

=+


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So, kinetic energy T = 21 2m x2 3

Potential energy of system V = 21 kx2

So, Lagrangian L = TV

2 21 2m 1L x kx2 3 2

=

Lagrangian equation is

d L L 0dt x x

=

d 2mx (kx) 0

dt 3 =

2m x kx 03

+ =

3kx x 0

2m+ =

20.(2) Cycle 1 is clockwise so work done during cycle 1 is po si tive . Similarly cycle 2 is anticlockwise andwork done during cycle 2 becomes negative.

But area of cycle 2 is greater than area of cycle 1. So resultant work is negative.

21.(1) Here the expression for F represents the force between two charges in Webers electrodynamics.

We have2

r 2 2

1 r 2rrF 1

r c

=

Taking U = q q( A v ) and F r = 2U d U

r dt r

+ in usual notation,

2

r 2 2

1 r 2r rF 1

r c

=

=2 2 2 2

2 2 2 2

1 c r 2r r 1 c r 2rrr c c r

+ +=

2p

V


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=2 2

2 2 2 2

1 c 1 2rr rc r c r

+

=2

2 2 2

1 1 2rr r U d Ur dt rr c r

+ = +

Yields,

2

U 1

r r = whi ch gives an i nteg ration U =

1r

= arbitrary constant say a function of r.

Assuming U =2

2

1 1 r,

r rc

+

we g et

2 2

2 2 2 2 2 2

U 1 r 1 1 r

r r c r r c r

= + =

and 2U 1 2r

r rc

=

so that2

2 2 2

d U 2 d r 2 r rdt r dt rc c r

= =

Thus2 2 2

2 2 2 2 2 2 2 2 2

U d U 1 r 2rr 2r 1 r 2rr 1

r dt r r c r c r c r r c + = + +

(on simplification)

Justifies F r = 2 21 r 2rr

1 r c

as given

As such the generalized potential U 2

2

1 1 rr rc

+

also T 2 2 21 [r r ]2

+

Lagrangian L = T U = 2 2 2 2

2

r r 1 1 r

2 2 r rc+

22.(4) We know Fermi energy of electron2/ 32 2

fN

E 32m V

=

23N 1 6.06 10V M

=

= 231 6.06 10 0.97

23

= NV

= 2 .5510 22 /cm 3 =2.55 10 28 m 3

m = 9.1 10 31 kg

So, E f =34 2 2/ 3

2 2831

(1.05 10 )3 (3.14) 2.55 10

2 9.1 10


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= 0.66 10 37 [7542] 2/3 (10)18

= 19 1/3(754.2)

0.66 10 Joule(754.2)

E f = 4.9710 19 J

E f = 19

19

4.97 10eV 3.10eV

1.6 10

=

23.(1) Paramagnetic suscepti bility is gi ven by

p =2

0 B

B F

nk T

where B is Bohr magneton = 9.3 10 24 J/tesla

0 is magnetic permeability of free space = 4 10 7 hnry/ m.n is no. of free electrons per unit volume.

Assuming one free electron per atom the number of atoms per cubic meter of potassium is23 3

36.02 10 0.86 10

3.91 1 0

Hence no. of free electrons (per m 3) n =23 3

36.02 10 0.86 10

3 9.1 10

or n = 1.2 10 28 m 3

E F = kBT F or TF = FB

Ek

TF =19

423

2.1 1.6 102.43 10 K.

1.38 10

=

Substi tuting the values we have7 28 24 2

8p 19

4 10 1.3 10 (9.3 10 ) 420.5 10 .21. 1.6 10

= =

24.(2) We kno w magnetic suscep ti bility

=2

20 BN g J(J 1)3KT + =

20N 4S(s 1)

3kT +

For co +2 having 3d 7 configuration ,

m s=


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s

1 1 1 3m s

2 2 2 2= = + + =

From grap

Slopedy 1 4000 3000dx T 20

= =

= 1000 5020

=

150

T=

7 20 4 10 N/ A

= 24 2

B 9.27 10 Am =

K = 1.38 10 23 J/K

J = S = 3/2g = 2

So ,2

0 B

3K( T)N3 542 2

=

=

2 3

7 2 4 2

3 1.38 104 10 (9.27 10 ) 15 50

26N 5.1 10=

25.(2) The magnetic field at O is

0 0 0 0 2 = + + = + 1 +4 R 8R 8R 8R

B k k i i k

0 2 = q(v ) = q 8R

+ + F B j i 1 k k

= 0 qv 2 + 1 +8R k i

26.(2) We know E f = E f(0) 2

2B

f(0)

K T1 12 E

Given E f(0) = 7 eV = 7 1.6 10 19 J


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T = 3 10 4 K

So, ff(0)

EE

=22 23 4

19

1.4 10 3 101

12 7 1.6 10

=22 0.2 3

1 12 1.6

= 1 2 9

12 64

= 1 2(3.14) 3

4 64

= 1 .115

f

f(0)

E 0.88E

=

27.(1) We know in Raman effect

=4B 3J2

+

So given 4B = 41.6 c.m 1

So , rotation constant B = 141.6 10.4cm4

=

B = 1040 m 1

We kno w B = 2h

8 C

So, I = 2h

8 BC

I = 34

2 8

6.6 108 (3.14) (1040) (3 10 )

I = 2.7 10 47 kgm 2

So, r=

reduce mass Of HCl molecule

So , H ClH Cl

M MM M

=+

= 23 2

23

(1 35) /(6.023 10 )(1 35)/(6.023 10 )

+


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= 1.61 10 24 gm

= 1.6 10 27 kg.

So, r = 47

27

2.7 101.6 10

= 1.29 10 10 m

r 1.29A=

28 .(2) Given R in = R1 = 120 k

R f = 600 k Vi = 1.2 VV0 = ?

For inverting O.P Amp voltage gain

AV =f f

in 1

R RR R

= = 600120

AV = 5

AV =0

i

VV

So, V out = AV Vi = 5 1.2

outV 6V=

29.(4) Given P =52

We know number of spectral levels (lines) in

NMR is = (2 p + 1)

2 52 + 1

6 lines

30.(3) g = 2

5= 10


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R = 120 ohm

So, New resistance R = Rg

= 120 2 10 5

= 240 10 5

3R' 2.4 10 ohm=

31.(3) x runs from 0 to (1y)2Fd xyz dx dy dz =

=1 2 2

2

0 0

(1 y)z y dy dz2

=1 2

2 2

0 0

1z y(1 y) dy dz

2

=1 2

2 3 2

0 0

1z (y y 2y )dy dz

2

+

=21 2 4 3

2

0 0

1 y y 2yz dz2 2 4 3

+

=1

2

0

1 4 16 2.8z dz2 2 4 3

+

=11 3

2

0 0

1 2 1 2 Z 1 1 1z dz

2 3 2 3 3 3 3 9

= = =

32.(3) 2 2 3

1 1 1

2 2 41 1 2

2 41 2

H H H P

H H He n3 H He P n

+ ++ + + +

m = (3M( 1H2) M( 2He 4) MP Mn)


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m = 3 2.014 4.001 1.007 1 .008m = 0.026 amuSo, energy E = m c 2 = 0.026 931 1.6 10 13 J

E = 38.72 10 13 JSo, energy of total deuteron

W = ET = 1040 38.72 10 13 J

P = Wt

t =WP =

40 1 3

16

38.72 10 1010

= 38.72 10

11

12t 3.87 10 sec .=

33.(4) In pair production process, electron-positron pair is produce d. So,

h = E + + E + 2m 0c2 E + kine tic energy of positronE kinetic energy of electronh incident photon energyGiven E + + E = 0.01 MeV

Rest mass energy 2m 0c2 = 2 9.1 10 31 (3 10 8)2 J.

2m 0c2 = 1.02 MeV

So, h = 0.01 MeV + 1.02 MeV

h 1.03 MeV =

34.(2) Photons are pa rticles ha ving spin 1 (intege r) and pio ns are spin less parti cle so, they are Bosons.

35.(2) is meson having zero charge.

So, quark structure is uu or dd

So, normalized quark structure is

( )1 uu dd2

=

PART C (36 -55)

36.(2) We know f (n) (1) =n 1C

n! f(z )dz ,2 i (z a) +


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Put f(z) = sin 6 z, which is analytic within and on the circle |z| = 1, whose centre is z = 0 and radius 1,

al so put n = 2 and a = 1 ,6

which lies within the given circle C, in the above formula then we get

f6

3C

1 2! sin z dz.6 2 i 1z

6

=

.(1)

Now f(z) = sin 6 z which gi ves f(z) = 6 sin 5 z cos z

and f(z) = 6[5 sin 4 z cos 2 z sin 6 z] = 6 sin 4 z (5 cos 2 z sin 2 z)

f( /6) = 6 sin 4 2 21 1 15 cos sin6 6 6

= 6(1/2) 4 [5(3/2) 2 (1/2) 2] = (3/32) [5(3) 1]

From (1) we have6

3C

sin z dz 21 i.161z

6

=

37.(2) (i) If we take a = 1, b = 1

then1 2 2

1A 2 1 1

3 2 1 1

=

AT =1 2 2

12 1 1

32 1 1

A square finite matrix A is said to orthogonal if

AAT =

T

1 2 2 1 2 21 1AA 2 1 1 2 1 13 3 2 1 1 2 1 1

=

1 4 4 2 2 2 2 2 2

1 2 2 2 4 1 1 4 1 19 2 2 2 4 1 1 4 1 1

+ + + + = + + + + + + + + + +

9 2 21 2 6 29

2 2 6

=

So, this is not correct value of a, b.

(ii) If we take a = 1, b = 2


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Then A1 2 2

1 2 1 23 2 2 1

=

AT =1 2 2

1 2 1 23

2 2 1

AAT =1 2 2 1 2 2

12 1 2 2 1 2

9 2 2 1 2 2 1

=9 0 0

10 9 0

9 0 0 9

=1 0 0

19 0 1 0

90 0 1

=1 0 00 1 00 0 1

=

So, the matrix is orthogonal if a = 1, b = 2.

38.(1) of H.E. between A and C

CA

A

T1

T

=

of engine = CA

T1 T

Here Q 2 =1 1

C 3 CA B

Q QT , Q T

T T =

Total Heat rejection

(Q 2 + Q 3) = Q 1TC A B

1 1T T +

Total Heat input = 2Q 1

of engine =1 C

A B

1

1 1Q TT T

1 2Q

+


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C C C

A A B

T T T 1

T 2T 2T =

Multiply bo th side by T A and divide by T C

or A A AC C B

T T T1 1 T T 2 2 T

=

or A AB C

T T(2 1) 2(1 )

T T= +

39.(2) Energy eigen value of three dimension well is=

2 2

2E 2ma+ +2 2 2x y z(n n n ) ...(1)

gi ven E = 2 2

2

9ma

E = 2 2

2

182ma

...(2)

compere equation (1) and equation (2).+ + =2 2 2x y zn n n 18

If nx = 4, n y = 1, n z = 1then + + = + + =2 2 2x y zn n n 16 1 1 18

So, possible values (combinations) of n x n y n z are (n x, ny, n z) = (4, 1,1) (1, 4, 1), (1, 1, 4)So, the state is 3-fold degenerate.

40.(3) 273

p2 1

293

mc dTS S

T =

2730.01 4.2 In kJ / K

293=

S 2 S 1 = 0.00297 kJ/K = 2 .97 J/K

3 2mL

S ST

=

= 0.01 335 1000273

S 3 S 2 = 12.271 J/K268

p4 13

2 73T

mc dT 4.2 268S S 0.01 In kJ / K

T 2 273 = =


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= 0.3882 J/K

So, S 4 S 1 = S 2 S 1 + S 3 S 2 + S 4 S 3 = 15.63 JK 1

41.(1) For one-dimensional free electron gas, energy level separation2 2

2

nE2mL

= So , 2

2

n 'EL

42.(i)(1) We begin by writing the Hamiltonian as

2 2 2 2 2 2x y z z z

x z x x z x

1 1 1 1 1 1H (L L L ) L L L

2 2 2 2 2 2

= + + + = +

where L is the total angular momentum.

We know eigen values of L 2 = ( + 1) 2 and, L 2 = m

22 2

mx z x

1 1E ( 1) m2 2 2

= + +

So the eigenstates of the Hamiltonian are those of L 2 and L z, i.e., the s [hetrical harmonic with theeigen energies E m..

(ii)(3) m mx y zY ( , ) | (L L L ) | Y ( , ) + + > m mzL L L L

Y ( , ) | L | Y ( , )2 2i

+ + + < + +

= m mzY ( , ) | L | Y ( , )< > = m

43.(2) Binding energy according to semimpirical mass formula

E b =a vA = a sA2/3 a cz (z1 ) A 1/3 a a(A-2z) 2A 1 apA 3/4

A nuclie will be most stable isobar which has maximum binding energy.

For maximum binding energy

E b =E bmax whenbdE 0

dz=

For maximum binding energy

E b = E bmax when dEb 0dz =

a c(2z1)A 1/3 + 4a a (A 2z) A 1 =0

1/ 3a

1/ 3 1c a

aA 4az2a A 8a A

+=+


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1/3

1/3

0.58(64) 4 19.38 19.3

2 0.58(64)64

+ +

= 0.145 77.2 28.620 .29 2.4125

+ =+

z 29

44.(3) m = 36 g = 0.036 kg

T 1 = 30C = 303 K

T 2 = 523 K

(S) water

= P37 3

mc n kJ /K303

= 0.03143 kJ/K

(S) Vaporization =2

mLT

= 0.036 2260373

= 0.21812 kJ/K

(S) Vapour =523

p373

dTmcT

=523

373

amR b CT dT

T + +


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=5 232

3 73

CTmR a n T bT2

+ +

=5 232

3 73

CTmR a n T bT

2

+ +

= 2 2523 C

mR a n b (523 373) (523 373 )373 2

+ +

= 0.023556 kJ/K

(S)System

= (S)water

+ (S)vaporization

+ (S)vapour

= 273.1 J/K.

45.(2) The intrinsic impedances are

1 = 0r1

120129

8.5

= =

an d 2 = 0 = 120

and the angle of transmission is given by

t 0

sin15sin 8.5

=

or t = 48.99

Thenr0 0 i 1 ti

2 i 1 t0

E cos cos

cos cosE

= +

= 0.623 or r0E = 0.623 V/m

46.(4) Since the kinetic energy of A af ter collision i s one-ninths of i ts initial kine tic energy, the momentum ofA after collision is one-third of its initial momentum.Since the momentum is to be conserved, we have

p = p p/3 where p is ini tial momentum of A and p is the momentum of B af ter the colli sion.[The final m omentum of particle A is neg ative since its di rection is reversed ].

Therefore, p = 4p/3The kin etic energy gained by particle B due to the coll ision is p 2 /2M where M is the m ass of parti cleB.

The kinetic energy lost by particle A due to the collision is (8/9)p 2 /2m.[Note that the initial kine tic energy of particle A is p 2 /2m and its final kine tic energy is (1/9) p 2 /2m].

Since the kinetic energy too is conserved in elastic collisions, the kinetic energy gained by particle Bis equal to the kinetic energy lost by particle A. Therefore, we have

p 2 /2M = (8/9) p 2 /2m

Substituting for p = 4p/3, we have

(16/9) (p 2 /2M) = (8/9) p 2 /2m from which M = 2 m.


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47.(4) The volume of water = 10 5 1

20= 5000 cubic miles.

Mass of water = 5000 (1.609 10 3)3 10 3 kg = 2.08 10 16 kg.

or No. of molecules of water = 2.08 10 16 6.0221 4 10 26 /18 = 6.97 10 41 molecu les.

As the abundance of deuterium is 0.0156% so that the total number of deuterium atoms

= 6.97 10 41 2 0.0156 10 2 = 2.18 10 38.As the fusion of 6 deu terium atoms give s an energy release of 43 MeV , hence the total energyreleased = 2.18 10 38 (43/6).

= 1.56 10 39 MeV.

48.(1) The wave length (in Angstrom unit) of a photon o f energy E (in electron volt) is given byE = 12400, very nearly.Therefore, = 12400/E

[The above expression can be easily obtained by remembering that a photon of energy 1 eV ha swave length 12400 and the energy is inversel y proportional to the wave length].

Since E = 2 eV we have = 12400/2 = 6200 = 620 nanometre.

Photons with wave length grea ter than 640 nm will have energy less than 2 eV so that th ey will beunable to produce electron hole pai rs in the semi conductor of band gap 2 eV. So the correct option i s

(1).

49.(4) Circuit (i) is shown the logic circuit of AND GATE and here output

Y = A BCircuit (ii) is shown the logic circui t of OR GATE and output

Y = A +BSimilarly circuit (iii) is shown the logic circuit of NOR GATE and output is

Y = A B+

50(i).(3) Since f(z) = cosz sinz z

to find zeros of f(z) put f(z) = 0

cos z sinz z

= 0


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3 5

2 4z z

z .....z z 3! 5!1 ......... 02! 4! z

+

+ =

2 4 2 4z z z z

1 ........ 1 ........ 02! 4! 3! 5!

+ + =

z 2 1 1 ........ 03! 2! + =

so z = 0 is a zero of f(z) of order 2

50(ii).(3) g(z) sinhzzf (z ) z cos z s in z

=

To find poles z cos z sin z = 0

3 5z zz .......

2! 4!

+

3zz ....... 03!

+ =

3 51 1 1 1z z 03! 2! 5! 4!

+ + =

g(z)zf(z)

have pole at z = 0 of order 3

51.(3) The Lagrangian of a is given by

L = T V = 2 2 2 21 mr ( sin ) V(r, , ).2

+ ..(1)

In this case the only two generalized co-ordinates are and , therefore there will be only twoLagrangian equations, one i n and the other in .The Lagrangian equation in coordinate is given by

d L L 0.dt

= ..(2)

From equation (1), we have

2 2 2L L Vmr and mr sin cos

= =

With these substitutions, equation (2) becomes

2 2 2d V(mr ) mr sin cos 0dt

+ =

(3)


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The Lagrangian equation for conservative system in the variable is given b y

d L L 0

dt =

Again equation (1) gives

2 2L V L and mr sin = =

(4)

With these substitutions equation (4) becomes

2 2d V(mr sin ) 0

dt

+ =

52(i).(1) We have E = Ex and we seek ( )x,t such that

E = (1) Since B = 0, we seek a gauge in which A = 0. Inte rgrating (1) we obtain (x) = - x + c, where c is acon stant of integration. Let us choose c = 0; then

( )x x = (2)

The total Hamiltonian is2

2 2p 1H m x x2m 2

= + (3)

The first term on the right-hand side of (3) is the standard kinetic term, the second term is theharmonic osci llator potential energy, and the thi rd term is the electrical po ten tial energy.

52(ii).(2) We will now write part (i) eq.(3) in the following form:

Hy =2y 2 2

y 0

p 1m y H

2m 2+ + (4)

Where H 0 is a constant and y = ax + b. Con sider the kinetic term. We see that p y = p x, so a = 1. Now wecan substitute y = x + b into (4) and obtain

Hy

=( )

2 222 2 2 2x x

0 0

p p1 1 1m x b H m bx m b H2m 2 2m 2 2

+ + + = + + +

(5)

From Eq. (3) of part (i) we see that H x = Hy only if b = /m2 and H 0 =

2 / 2m 2.

52(iii).(3) To conclude, if we perform the coordinate transformation y = x /m2 , we get a one-dimensionalharmonic oscillator with no charge, and the energy eigenvalues of a one-dimensional harmonicoscillator are


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n

1 1E w n2 2

= + ..(6)

Corresponding to the eigenstate n| . We have a shifted harmonic oscillator; thus, the energyeigenvalues are now.

E n =2

2

1 1 1n2 2 2 m

+

53.(2) f(x) = x 3 x 2 + 4x 4

f(x) = 3x2 2x + 4

f(2) = 8 4 + 8 4 = 8

f(2) = 12 4 + 4 = 12

nn 1 n

n

f(x )x x

f '(x )+ =

xn + 1 = ( )82 2 2/ 3 4 / 3

12= =

54.(1) Taking earth as refe rence l evel f or zero potential energy, we have

V(q) = mgz ..(1)

The Hamilton-Jacobi equation for Hamiltons principal function i s2

1 S SV(q) 02m q t

+ + = ..(2)

The fi rst term in bracket is function of q only, while the second te rm is fu nction of t only, thereforeeach term must be equal to the same constant with opposite signs.

21 S

i.e. V(q)2m qS

and t

+ =

=

...(3)

Then we have S = W(q, ) t,...(4)

W being a constant of integration.


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As s W ,q q

=

therefore the Hamilton-Jacobi equation for Hamiltons characteristic function W takes

the form2

1 W V(q)2m q

+ = ...(5)

This gives { }W 2m V(q)q

= ...(6)

Integrating above expression, we get

W ={ }

1 /22m V(q) dq

...(7)

Now { }1 /2S W

p 2m V(q)q q

= = = ...(8)

an dS W

t = =

= { }1/2

2m V(q) dq t

i.e. 1/ 2m dqt2 [ V(q)

+ = ...(9)

The Hamiltonian o f system is

H =2p mgz E

2m+ = = ...(10)

From (9) 1/21/2

m dz m 2t (E mgz)

2 2 mg[E mgz] + = =

= 1/21 2

(E mgz)g m

...(11)

Thus 22

1 2( t)mg

+ = (E m gz)

Solving for z, we get

z = 21 E g( t)2 mg

+ + ...(12)

Initial conditions are

At t = 0, z = h, v = z 0= and so p = mz 0= ...(13)

Using (12), equations (10) and (12) yield


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E = mgh, h = 21 E

g2 mg

+

Solving these equations, we get = 0

Now p =W

mz m gt usin g (12)z

= = ...(14)

Thus we have from (12) and (14) 21z h gt

2p mgt

=

=...(15)

These are requi red equ ations with S = W Et

where 1 /2W (2m) (E mgz ) dz=

55.(3) From M axwell-Boltzmann di stribution law, the number of m ole cules in the th state relative to that inthe = 0 (lowest) state at T Kelvin temperature is given by

0 G ( )h c/ kT

0

Ne

N =

= G( ) G( 0 )hc /kTe ,

where G( ) =2

e e e

1 1 x2 2

+ +

2

e e e e e ( n x n x n)hc /kTn

0

N eN

=

For the n = 1 level, we have

e e e ( 2 x )hc / kT1

0

N eN

=

Here 1 11 e e e0

N 1, 214.6cm and x 0.6 cm (given)

N 10= = =

1 1

( 214.6 cm 1 .2 cm )hc /kT1 e10

=

or 1 1

(213 .4 c m )hc /kT (2134 0m )hc / kT10 e e= =

Taking na tural logarithm:Lo ge 10 = (21340 m

1 ) hc/kT.


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Phone: 0744 2429714 Mobile: 9001297111 9829567114 9001297243

T = 121340 m hc

2.303 k

= 34 8 1

1 23 1

(6.63 10 Js)(3.0 10 ms )(9266 m )1.38 10 JK

= 133.5 K.

Csir Ugc Net_physics_ Free Solved Paper

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