mock test paper

7/30/2019 Mock Test Paper

1/45

For CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, GATE, UGC NET, TIFR, IISc etc.

Phone: 0744-2429714, 0744-2427711 Mobile: 9001297111, 9829567114, 9829597114

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Page 1

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

Pattern 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.

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CSIR NET - PHYSICAL SCIENCE

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

Web Site www.vpmclasses.com [email protected]

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Total marks : 200

Duration of test : 3 Hours

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MOCK TEST PAPER


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PART A (1-15)

1. Twenty four clerk can clear 180 files in 15 days. Number of clerk required to clear

240 files in 12 days is

(1) 38

(2) 39

(3) 40

(4) 42

2. In the given figure, 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 circles are drawn with radii R1, R2, R3 with centre at the vertices of a triangle as shownin figure. Side of triangle is a, b, c respectively, then R1 + R2 + R3 is equal to

R1

R3

(1) 3(a + b + c)


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(2)1

(a b c)3

(3)

1

(a b c)2

(4) 2(a + b + c)

4. Study the following graph and answer the question given below it

10

15

20

25

30

35

40

45

50

2

4

6

8

10

12

14

16

18

20

1984 1985 1986 1987 1988 1989

No.ofTools(in'0

00)

Years

Totalvalueoftools(inRscrores)

Production in a Tool Factory

Number of Tools ----- Value

What was the value of each tool in 1985?

(1) Rs1

53

thousand

(2) Rs 50 thousand

(3) Rs 5, 103

(4)5

59

5. The total adults in a city is 60000. The various sections of them are indicated below in thecircle


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108II

54 18

III

IV

V

I employees in the public sectorII employees in the private sector

III employees in the corporate sector

IV self employed

unemployed

V

What percentage of the employed persons is self employed?

(1)5

519

(2)1

195

(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

(4) 128

7. A car owner buys petrol at Rs.7.50, Rs. 8 and Rs. 8.50 per liter for three successive years.

What approximately is the average cost per liter of petrol if he spends Rs. 4000 each year?

(1) Rs. 7.98

(2) Rs. 8

(3) Rs. 8.50


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(4) Rs. 9

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

price remains constant, approximately what percentage of the selling priceis the profit?

(1) 30%

(2) 70%

(3) 100%

(4) 250%

9. Today is Friday after 62 days, it will be :

(1) Thursday

(2) Friday

(3) Wednesday

(4) Tuesday

10. A car travelling with 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 /hr 7

(2) 25 km/hr

(3) 30 km/hr

(4) 35 km/hr


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

receives 20% of the profits 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 side in the ratio 2:3 The area of the lawn 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 cover thesame distance in 1 hours, it must travel at a speed of:

(1) 300 kmph

(2) 360 kmph(3) 600 kmph

(4) 720 kmph

14. Find out the missing number of the given question:

2 7 4

5 2 3

1 ? 6

10 42 72

(1) 2(2) 4

(3) 5

(4) 3


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15. All of the following are the same in a manner. Find out the one which is different amongthem:

(1) BFJQ(2) RUZG

(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 equation

0

1 , 0 1f( ) cos d

0, 1

(1)2

2(1 cos )f( )

(2) 22cos

f( )

(3)2

(1 cos )f( )

2

(4)2

(1 cos )f( )

18. Find the function whose laplace transform is2

2 2 2

s

(s a )

(1) 1

[at sinat cos at]2a

(2)1

(a sin at cos at)2a

(3)1

[a cosat sinat]2a


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(4)1

[at cosat sinat]2a

19. The Lagrange equation of motion of two rigid bodies of masses m and 2m are connected by a light

flexible spring of spring constant K. what is the Lagrange equation of motion.

(1)k

x x 0m

(2)k

x x 02m

(3)3k

x x 02m

(4)5k

x x 02m

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 force, acting towards a centre of force whose

magnitude is2

2 2

1 r 2rr F 1

r c

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

the Lagrangian for the motion in a plane is

(1)2 2 2 2

2

r r 1 1 r L

2 2 r r c


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(2)2 2 2 2

2 2

r r 1 2 r L

2 2 r c r

(3)

2 2 2

2 2

r r 2 1 r

L 2 2 r c r

(4)2 2

2 2

r 1 1 r L

2 r c r

22. Calculate the Fermi energy in electron volt for sodium assuming that it has one free electron per

atom. Given density of sodium = 0.97 g cm3

, atomic weight of sodium is 23.

(1) 3.541 eV

(2) 3.451 eV

(3) 5.135 eV

(4) 3.145 eV

23. The paramagnetic contribution to the magnetic susceptibility per m3 of potassium, for which the Fermienergy is 2.1 eV is at wt. of potassium is 39.1 gm and density of potassium is 0.86 103 kgm3.

(1) 420.5 108

(2) 450.2 10+8

(3) 420.5 106

(4) 425.210+6

24. The figure shows the inverse magnetic susceptibility (1/) (dimensionless) as a function oftemperature for a paramagnetic material. Calculate the concentration of magnetic ions, if they areassumed to be Co2+ with configuration 3d7.

(1) 54 1023

(2) 51 1026

(3) 69 1023

(4) 69 1026


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25. A 3D structure of current carrying wire is as shown in the figure. The magnetic force experienced by

charge particle of mass m and charge q, when it is crossing origin with velocity v

along +ve Y-axis willbe

(1) 0 0q qV8R 4 R

i

(2) 0q 2 1

8R

i k

(3) 0q

8

k

(4) Zero

26. If Ef(0)

and Ef

are Fermi levels of a metal at 0K and 30000 K, then what is the value of f

f (0)

E

Eif

Ef(0)

= 7 eV.

(1) 0.119

(2) 0.88

(3) 1.113

(4) 1.188

27. The small (rotational) Raman displacement for HCI molecule is 416 cm1. Find the internuclear

distance between the atoms forming the molecule. Given : h = 663 1034 J s, c = 30 108 m s1

and NA = 6023 1023 mol1.

(1) 129 A

(2) 229 A

(3) 249 A

(4) 064 A


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28. For the given circuit the the open loop gain is 12000 and R1

= 120 k and Rf= 600 k. V

i= 1.2 V.

Find the exact output voltage for the inverting operational amplifier.

(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

(4) 6 lines

30. A metal strain gauge factor of two. ts nominal resistance is 120 ohms. It undergoes strain 105, thevalue of change of resistance in response to the strain is

(1) 240 ohms

(2) 2 105 ohm

(3) 2.4 103 ohm

(4) 1.2 103 ohm

31. EvaluateV

Fd ,

where 2F xyz

over the prism placed at origin as shown in the Figure.

(1)1

3

(2)2

3

(3) 19

(4)2

9


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32. A star initially has 1040 deuterons. It produces energy via the processes

1H2 +

1H2

1H3 + p and

1H2 +

1H3

2He4 + n

If the average power radiated by the star is 1016 W, the deuteron supply of the star is exhausted in atime of the order of

(1) 106 sec

(2) 108 sec

(3) 1012 sec

(4) 1016 sec

The masses of the nuclei are as follows:

M (H2) = 2014 amu; M (p) = 1007 amu;

M (n) = 1008 amu; M (He4) = 4001 amu.

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, consisting of two materials havingcoefficients of thermal conductivity K and 2K and thickness x and 4x, respectively are T

2and T

1(T

2>

T1).

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


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(1) ud

(2)1

(uu dd)2

(3)1

(uu dd)2

(4)1

(us su)2

PART C (36-55)

36. The value of the counter integral

6

3C

sin z

dz1z

6

, if C is the circle z 1 .

(1)20

i17

(2)21

i16

(3)15

i

7

(4)12

i13

37. The matrix A, defined by

1 2 2

A 2 a b

2 b a

Is orthogonal if

(1) a = 1, b = 1

(2) a = 1 , b = 2


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(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 TA and TBrespectively, and rejects heat to the thermal reservoir C kept at temperature TC. The efficiency of the

engine is times the efficiency of the reversible engine, which works between the two reservoirs Aand C. which one of the following relation statement is correct ?

(1) A AB C

T T2 1 2 1

T T

(2) A AB C

T T2 1 2 1

T T

(3) CBA A

TT 2 1 2 1T T

(4) A AC B

T T2 1 2 1

T T

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

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

2

9.

maThe state is

(1) 4 fold degenerate

(2) 3-fold degenerate

(3) 2-fold degenerate

(4) Non-degenerate

40. Ten grammas of water at 20 C is converted into at 10C at constant atmospheric 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 JK1

(3) 15.63 JK1

(4) 15.63JK1


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

(1)5

g sin7

(2)5

g sin14

(3) mg sin

(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. Lx Ly and Lz 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) + m

2 2 2

(2)2

2 2

x z x

1 1( + 1) + m2 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 Lx + Ly + Lz for any state is

(1) Zero

(2) m

(3) m

(4) m2

43. If we take in the semi-empirical mass formula ac = 0.58 MeV and aa = 19.3 MeV. Then possibleatomic number of most stable nuclei of mass number 64.


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(1) 26

(2) 29

(3) 32

(4) 33

44. 36 g of water at 30C are converted into steam at 250C at constant atmospheric pressure. Thespecific heat of water is assumed constant at 4.2 J/g K and the latent heat of vaporization at 100C is

2260 J/g. For water vapour, assume pV = mRT where R = 0.4619 kJ/kg K, andpC

R= a + bT + cT,

where a = 3.634,

b = 1.195 103 K1 and c = 0.135 106 K2

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, free

space, with an angle of incidence of 15. Given i0E 1.0 V /m , thenr

0E , is

(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


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47. Calculate the amount of energy released if all the deuterium atoms in the water in the lake of area

about 105 sq. miles and of depth1

20the mile area used up in fusion.

(1) 2.18 1038 MeV

(2) 43 MeV

(3) 1.56 1039 MeV

(4) 6.9 1038 MeV

48. The maximum wave length of photons that can be detected by a photo diode made of asemiconductor of band gap 2 eV is about

(1) 620 nm

(2) 700 nm

(3) 740 nm

(4) 860 nm

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

0.

Identify the correct combination of logic gates (i), (ii), (iii) in the options given below.

(1) NOR, XOR, AND

(2) OR, NAND, NOR

(3) NAND, AND, XOR


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(4) AND, OR, NOR

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

Let f(z) = cos zsin z

z

for non-zero z C and f(0) = 0. Also, 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) Greater than 2

50(ii). Theng(z)

zf(z)has a pole at z = 0 of order

(1) 1

(2) 2

(3) 3

(4) Greater than 3

51. The Lagrangian of a system is given 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)


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

m x2

This particle has a charge q and is placed in a uniform electric field E parallel to the x axis, E = Ex .

52.(i). The Hamiltonian of the particle

(1)2

2 2P 1m x x2m 2

(2)2

2 2P 1m x x2m 2

(3)2

2 2P 1m x x2m 2

(4)2

2 2P 1m x x2m 2

52(ii). Perform a coordinate transformation y = ax+b (where a and b are constant / such that in the ycoordinate the Hamiltonian is similar to that of a one dimensional harmonic oscillator (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 = 1

52(iii). The energy eigenvalues of the system is

(1)2

2

1 1n

2 2 m

(2)2

2

1 3n

2 2 m

(3)2

2

1 1 ew n

2 2 2mw

(4)2

2

1 3n

2 2 2m


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53. The equation x3 x2 + 4x 4 = 0 is to be solved using the Newton-Rephson method. If x = 2 is takenas the initial approximation using this method will be

(1)2

3

(2)4

3

(3) 1

(4)3

2

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

(1) 1/ 22m (E mgz) dz

(2)1/ 2

2m (E mgz) dz (3) 2m(E mgz) dz

(4) 2m (E mgz) dz

55. At what temperature will the number of2 molecules in the = 1 level be one-tenth of that in the = 0level? Given : = 214.6 cm

1,

exe = 0.6 cm1

, h = 6.63 1034

Js, c = 3.0 108ms

1and k = 1.38 10

23J/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 2

2 1 17 1 32 3 46 4

3 3 18 4 33 4 47 4

4 4 19 3 34 2 48 15 1 20 2 35 2 49 4

6 2 21 1 36 2 50(i) 3

7 1 22 4 37 2 50(ii) 3

8 2 23 1 38 1 51 3

9 4 24 2 39 2 52(i) 1

10 4 25 2 40 3 52(ii) 2

11 2 26 2 41 1 52(iii) 3

12 2 27 1 42(i) 1 53 2

13 4 28 2 42(ii) 3 54 1

14 4 29 4 43 2 55 3

15 1 30 3 44 3

Solutions

PART A (1-15)

1.(3) 1 1 2 2

1 2

m D m D

w w

2m 1224 15

180 24

m2 = 40

2.(1)RA SA 9 9

PA PAQA 7

Diameter = PA + AQ

81 1307

7 7

Radius =Diameter

2

65Radius

7

3.(3) R1 + R2 = a

R2 + R3 = b

R3 + R1 = c

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

1 2 3

a b cR R R

2


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4. (4) Value of each tool in 1985

=

7

3

10 10

18 10

[Since 1 crore = 107

]

=5

59

Thousand

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 multiplication and subtracting series: First, 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 6120

Rs. Rs.7.9876700 767

8.(2) Let C.P.= Rs. 100. Then, Profit = Rs. 320, S.P. = Rs. 420.

New C.P. = 125% of Rs. 100 = Rs. 125

New S.P. = Rs. 420.

Profit = Rs. (420 - 125) = Rs. 295.

Required percentage =%

295 1475100 % 70% (approximately)

420 21

A student multiplied a number by3

5instead of

5

3


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

So, after 63 days, it will be Friday. Hence after 63 days,

it will be Thursday.

Therefore the required day is Thursday.

10 .(4)4 51 126

40 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 profit = Rs. 990

Ration of their capitals = 34 : 65.

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

340.2 0.8

34 65

= 470

12.(2) Now area = (1/6 1000)sq m = 5000/3 sq m

2x 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

Speed = 1200/(5/3) km/hr. [We can write 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


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42? 3

14

15.(1) According to question,

Therefore, B F J Q is odd.

PART B (16-35)

16.(4) From the property of dirac delta 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

c0

2F f(x)cos sx dx

c0

22 (1 ), 0 1

F ( ) f( )cos d

0, 1

By inversion formula ofFc(), we get


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f() = c0

2F ( )xcos d

= 1

0 1

2 21 cos d 0 d

1

2

0

2 sin cosf( ) (1 ) ( 1)

=

2 2

2 cos 10 0

2

2(1 cos )f( )

18.(4) (ii) We have f1(s) = f

2=(s) =

2 2

sL{cosat}

s a

By convolution theorem

t

1

1 2 1 2

0L {f (s)f (s)} F (y)F (t y)dy

t2

1 1

2 2 2 2 2 2 2

0

s s sL L cos aycosa(t y)dy

(s a ) s a s a

=t

0

1[cosat cos(2ay at)]dy

2

0

1 1ycosat sin(2ay at)

2 2a

=1 1 1

tcosat sinat [at cos at sin at]2 a 2a

19.(3) Reduced mass of two body system is

M =m 2m 2m

m 2m 3

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


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Lagrangian equation is

d L L0

dt x x

d 2mx (kx) 0

dt 3

2mx kx 0

3

3kx x 0

2m

20.(2) Cycle 1 is clockwise so work done during cycle 1 is positive . 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.

2p

V

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

We have2

r 2 2

1 r 2rr F 1

r c

Taking U = q q(A v) and Fr=

2

U d U

r dt r

in usual notation,

2

r 2 2

1 r 2r r F 1

r c

=

2 2 2 2

2 2 2 2

1 c r 2r r 1 c r 2rr

r c c r

=2 2

2 2 2 2

1 c 1 2rr r

c r c r

=

2

2 2 2

1 1 2rr r U d U

r dt r r c r

Yields,

2

U 1

r r

which gives an integration U =

1

r= arbitrary constant say a function of r.

Assuming U =2

2

1 1 r,

r rc

we get


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2 2

2 2 2 2 2 2

U 1 r 1 1 r

r r c r r c r

and

2

U 1 2r

r rc

so that

2

2 2 2

d U 2 d r 2 r r

dt r dt r c 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 Fr=

2 2

1 r 2rr 1

r c

as given

As such the generalized potential U 2

2

1 1 r

r 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 r c

22.(4) We know Fermi energy of electron

2/ 32 2

f

NE 3

2m V

23N 1 6.06 10

V M

=231 6.06 10 0.97

23

=

N

V

= 2.551022

/cm3

=2.55 1028

m3

m = 9.1 1031

kg

So, Ef=34 2 2 / 3

2 28

31

(1.05 10 )3 (3.14) 2.55 10

2 9.1 10

= 0.66 1037

[7542]2/3

(10)18

= 191/ 3

(754.2)0.66 10 Joule

(754.2)

Ef= 4.971019

J

Ef=19

194.97 10 eV 3.10eV1.6 10

23.(1) Paramagnetic susceptibility is given by


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p

=2

0 B

B F

n

k T

where B

is Bohr magneton = 9.3 1024 J/tesla

0

is magnetic permeability of free space = 4 107 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

3

6.02 10 0.86 10

3.91 10

Hence no. of free electrons (per m3) n =23 3

3

6.02 10 0.86 10

39.1 10

or n = 1.2 1028 m3

EF

= kBT

For T

F= F

B

E

k

TF

=19

4

23

2.1 1.6 102.43 10 K.

1.38 10

Substituting the values we have

7 28 24 28

p 19

4 10 1.3 10 (9.3 10 )420.5 10 .

21. 1.6 10

24.(2) We know magnetic susceptibility

=2

20 BN g J(J 1)3KT

=

20N 4S(s 1)

3kT

For co+2

having 3d7

configuration,

ms=

s

1 1 1 3m s

2 2 2 2

From grap

Slopedy 1 4000 3000

dx T 20

=1000

5020

150

T

7 2

0 4 10 N/ A


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24 2

B 9.27 10 Am

K = 1.38 1023

J/K

J = S = 3/2

g = 2

So,2

0 B

3K( T)N

3 54

2 2

=23

7 24 2

3 1.38 10

4 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

= 0qv 2 + 1+

8R

k i

26.(2) We know Ef= E

f(0)

22

B

f(0)

K T1

12 E

Given Ef(0)

= 7 eV = 7 1.6 1019 J

T = 3 104 K

So, f

f (0)

E

E=

22 23 4

19

1.4 10 3 101

12 7 1.6 10

=

22 0.2 31

12 1.6

= 1

2

912 64

= 1 2(3.14) 3

4 64

= 1 .115


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f

f(0)

E0.88

E

27.(1) We know in Raman effect

=4B3

J2

So given 4B = 41.6 c.m1

So, rotation constant B = 141.6

10.4cm4

B = 1040 m1

We know B =2

h

8 C

So, I =2

h

8 BC

I =34

2 8

6.6 10

8 (3.14) (1040) (3 10 )

I = 2.7 1047

kgm2

So, r

reduce mass Of HCl molecule

So,H Cl

H Cl

M M

M M =

23 2

23

(1 35) /(6.023 10 )

(1 35)/(6.023 10 )

= 1.61 1024

gm

= 1.6 1027 kg.

So, r =47

27

2.7 10

1.6 10

= 1.29 1010

m

r 1.29A

28.(2) Given Rin

= R1

= 120 k

Rf= 600 k

Vi = 1.2 V

V0 = ?

For inverting O.P Amp voltage gain


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AV =f f

in 1

R R

R R =

600

120

AV = 5

AV =0

i

V

V

So, Vout = AV Vi

= 5 1.2

outV 6V

29.(4) Given P

=5

2

We know number of spectral levels (lines) in

NMR is = (2p + 1)

2 5

2+ 1

6 lines

30.(3) g = 2

5= 10

R = 120 ohm

So, New resistance R = Rg

= 120 2 105

= 240 105

3R' 2.4 10 ohm

31.(3) x runs from 0 to (1y)

2Fd xyz dx dy dz


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=1 2 2

2

0 0

(1 y)z y dy dz

2

=

1 2

2 2

0 0

1

z y(1 y) dy dz2

=1 2

2 3 2

0 0

1z (y y 2y )dy dz

2

=

21 2 4 32

0 0

1 y y 2yz dz

2 2 4 3

=1

2

0

1 4 16 2.8z dz

2 2 4 3

=

11 32

0 0

1 2 1 2 Z 1 1 1z dz2 3 2 3 3 3 3 9

32.(3)

2 2 3

1 1 1

2 2 4

1 1 2

2 4

1 2

H H H P

H H He n

3 H He P n

m = (3M(1H2) M(2He

4) MP Mn)

m = 3 2.014 4.001 1.007 1.008

m = 0.026 amu

So, energy E = m c2

= 0.026 931 1.6 10

13

JE = 38.72 10

13J

So, energy of total deuteron

W = ET = 1040

38.72 1013

J

P =W

t

t =W

P=

40 13

16

38.72 10 10

10

= 38.72 10

11

12t 3.87 10 sec.

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

h = E+ + E + 2m0c2

E+kinetic energy of positron


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E kinetic energy of electron

hincident photon energy

Given E+ + E = 0.01 MeV

Rest mass energy 2m0c2 = 2 9.1 1031 (3 108)2 J.

2m0c2

= 1.02 MeV

So, h = 0.01 MeV + 1.02 MeV

h 1.03MeV

34.(2) Photons are particles having spin 1 (integer) and pions are spin less particle 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)

Put f(z) = sin6

z, which is analytic within and on the circle |z| = 1, whose centre is z = 0 and radius 1,

also put n = 2 and a =1

,6

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

f

6

3C

1 2! sin z

dz.6 2 i 1z

6

.(1)

Now f(z) = sin6

z which gives f(z) = 6 sin5

z cos z

and f(z) = 6[5 sin4

z cos2

z sin6

z] = 6 sin4

z (5 cos2

z sin2

z)

f(/6) = 6 sin4 2 21 1 1

5 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 21i.

161

z 6

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


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Page 34

then

1 2 21

A 2 1 13

2 1 1

AT =

1 2 21

2 1 13

2 1 1

A square finite matrix A is said to orthogonal if

AAT

=

T

1 2 2 1 2 21 1

AA 2 1 1 2 1 13 3

2 1 1 2 1 1

1 4 4 2 2 2 2 2 21

2 2 2 4 1 1 4 1 19

2 2 2 4 1 1 4 1 1

9 2 2

1 2 6 29

2 2 6

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

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

Then A

1 2 21

2 1 23

2 2 1

AT =

1 2 21

2 1 2

3 2 2 1

AAT

=

1 2 2 1 2 21

2 1 2 2 1 29

2 2 1 2 2 1

=

9 0 01

0 9 09

0 0 9

=

1 0 01

9 0 1 09

0 0 1

=

1 0 0

0 1 0

0 0 1

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

38.(1) ofH.E. between A and C

CA

A

T1

T


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Page 35

of engine = C

A

T1

T

Here Q2 = 1 1C 3 CA B

Q QT , Q TT T

Total Heat rejection

(Q2 + Q3) = Q1TCA B

1 1

T T

Total Heat input = 2Q1

of engine =1 C

A B

1

1 1Q T

T T1

2Q

C C CA A B

T T T 1

T 2T 2T

Multiply both side by TA and divide by TC

or A A A

C C B

T T T1 1

T T 2 2 T

or

A A

B 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)

given E = 2 2

2

9

ma

E = 2 2

2

18

2ma...(2)

compere equation (1) and equation (2).

2 2 2x y zn n n 18

If nx = 4, ny = 1, nz = 1then 2 2 2x y zn n n 16 1 1 18

So, possible values (combinations) of nxn

yn

zare (n

x, n

y, n

z) = (4, 1,1) (1, 4, 1), (1, 1, 4)

So, the state is 3-fold degenerate.


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40.(3)273

p

2 1

293

mc dTS S

T

2730.01 4.2 In kJ /K

293

S2 S1 = 0.00297 kJ/K = 2.97 J/K

3 2

mLS S

T

=0.01 335 1000

273

S3 S2 = 12.271 J/K

268p

4 13

273T

mc dT 4.2 268S S 0.01 In kJ /K

T 2 273

= 0.3882 J/K

So, S4 S1 = S2 S1 + S3 S2 + S4 S3 =15.63 JK1

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

2 2

2

nE

2mL

So,

2

2

n'E

L

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

2 2 2 2 2 2

x y z z zx z x x z x

1 1 1 1 1 1

H (L L L ) L L L2 2 2 2 2 2

where L is the total angular momentum.

We know eigen values of L2

= ( + 1)2

and, L2 = m

22 2

m

x z x

1 1E ( 1) m

2 2 2

So the eigenstates of the Hamiltonian are those of L2 and Lz, i.e., the s [hetrical harmonic with the

eigen energies Em

..

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

z

L L L LY ( , ) | L | Y ( , )

2 2i

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

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

Eb

=avA = a

sA2/3 a

cz (z1) A1/3 a

a(A-2z)2A1 apA3/4


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

For maximum binding energy

Eb

=Ebmax

when bdE

0

dz

For maximum binding energy

Eb

= Ebmax

whendEb

0dz

ac(2z1)A1/3 + 4a

a(A 2z) A1 =0

1/ 3

a

1/ 3 1

c a

aA 4az

2a A 8a A

1/ 3

1/ 3

0.58(64) 4 19.3

8 19.32 0.58(64)

64

=0.145 77.2

28.620.29 2.4125

z 29

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

T1

= 30C = 303 K

T2

= 523 K

(S)water

= P373

mc n kJ/K303

= 0.03143 kJ/K


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(S)Vaporization

=2

mL

T

=0.036 2260

373

= 0.21812 kJ/K

(S)Vapour

=523

p

373

dTmc

T

=523

373

amR b CT dT

T

=

5232

373

CTmR a n T bT

2

=

5232

373

CTmR a n T bT2

= 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

= 0

r1

120129

8.5

and

2

= 0

= 120

and the angle of transmission is given by

0

sin15

sin 8.5

or

t= 48.99

Thenr

0 0 i 1 t

i

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 after collision is one-ninths of its initial kinetic 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 initial momentum of A and p is the momentum of B after the collision.

[The final momentum of particle A is negative since its direction is reversed].

Therefore, p = 4p/3


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The kinetic energy gained by particle B due to the collision is p2/2M where M is the mass of particle

B.

The kinetic energy lost by particle A due to the collision is (8/9)p2/2m.

[Note that the initial kinetic energy of particle A is p

2

/2m and its final kinetic 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

p2/2M = (8/9) p

2/2m

Substituting for p = 4p/3, we have

(16/9) (p2/2M) = (8/9) p

2/2m from which M = 2 m.

47.(4) The volume of water = 105 1

20= 5000 cubic miles.

Mass of water = 5000 (1.609 103)3 103 kg = 2.08 1016 kg.

or No. of molecules of water = 2.08 1016 6.02214 1026/18 = 6.97 1041 molecules.

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

= 6.97 1041 2 0.0156 102 = 2.18 1038.

As the fusion of 6 deuterium atoms gives an energy release of 43 MeV, hence the total energyreleased = 2.18 1038 (43/6).

= 1.56 1039 MeV.

48.(1) The wave length (in Angstrom unit) of a photon of energy E (in electron volt) is given by

E = 12400, very nearly.

Therefore, = 12400/E

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

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

Photons with wave length greater than 640 nm will have energy less than 2 eV so that they will beunable to produce electron hole pairs in the semiconductor of band gap 2 eV. So the correct option is(1).

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

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

Y = A B

Similarly circuit (iii) is shown the logic circuit of NOR GATE and output is

Y = A B


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50(i).(3)Since f(z) = coszsinz

z

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

cos z sinzz

= 0

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!

z21 1

........ 0

3! 2!

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

50(ii).(3)g(z) sinhz

zf (z) zcoszs inz

To find poles z cos z sin z = 0

3 5z z

z .......2! 4!

3zz ....... 0

3!

3 51 1 1 1

z z 0

3! 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 in 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


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With these substitutions, equation (2) becomes

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

(3)

The Lagrangian equation for conservative system in the variable is given by

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 ) 0dt

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. Intergrating (1) we obtain (x) = - x + c, where c is aconstant of integration. Let us choose c = 0; then

x x (2)

The total Hamiltonian is

22 2p 1H m x x

2m 2 (3)

The first term on the right-hand side of (3) is the standard kinetic term, the second term is theharmonic oscillator potential energy, and the third term is the electrical potential 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 H0 is a constant and y = ax + b. Consider the kinetic term. We see that py = px, so a = 1. Now wecan substitute y = x + b into (4) and obtain

Hy = 2 2

22 2 2 2x x0 0

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

2m 2 2m 2 2 (5)

From Eq. (3) of part (i) we see that Hx = Hy only if b = /m2and H0 =

2/ 2m2 .


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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

n

1 1E w n

2 2

..(6)

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

eigenvalues are now.

E n =2

2

1 1 1n

2 2 2 m

53.(2) f(x) = x3 x2 + 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

= 8

2 2 2 /3 4 /312

54.(1) Taking earth as reference level for zero potential energy, we have

V(q) = mgz ..(1)

The Hamilton-Jacobi equation for Hamiltons principal function is

2

1 S SV(q) 0

2m q t

..(2)

The first term in bracket is function of q only, while the second term is function of t only, thereforeeach term must be equal to the same constant with opposite signs.

2

1 Si.e. V(q)

2m q

Sand

t

...(3)

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

W being a constant of integration.


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Ass W

,q q

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

the form

2

1 W V(q)2m q

...(5)

This gives W

2m V(q)q

...(6)

Integrating above expression, we get

W = 1/2

2m V(q) dq ...(7)

Now 1/ 2S W

p 2m V(q)q q

...(8)

and

S W

t

= 1/ 2

2m V(q) dq t

i.e.1/ 2

m dqt

2 [ V(q)

...(9)

The Hamiltonian of system is

H =2p

mgz E2m

...(10)

From (9) 1/ 21/2m dz m 2

t (E mgz)2 2 mg[E mgz]

= 1/ 21 2

(E mgz)g m

...(11)

Thus 22

1 2( t)

mg (E mgz)

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 mgt u sin g (12)z

...(14)

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

2

p mgt

...(15)

These are required equations with S = W Et

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

55.(3) From Maxwell-Boltzmann distribution law, the number of molecules in the th state relative to that inthe = 0 (lowest) state at T Kelvin temperature is given by

0G ( )hc / kT

0

Ne

N

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

where G() =2

e e e

1 1 x

2 2

2

e e e e e( n x n x n)hc /k Tn

0

Ne

N

For the n = 1 level, we have

e e e( 2 x )hc /kT1

0

Ne

N

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 e

10

or

1 1(213.4 cm )hc /kT (21340 m )hc /kT

10 e e Taking natural logarithm:

Loge 10 = (21340 m1

) hc/kT.


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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.

mock test paper

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