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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.

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



PART A (1-15)

1. Tw enty four clerk can clear 180 f iles 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 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 circles are draw n with radii R1, R2, R3 w ith centre at the vertices of a triangle as show n in f igure. Side of triangle is a, b, c respectively, then R1 + R2 + R3 is equal to



R1

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 follow ing graph and answ er 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.

of T

ools

(in

'00

0)

Years

Tota

l va

lue

of to

ols

(in

Rs

cror

es)

Production in a Tool Factory

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



(1) Rs 1

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 the circle

What percentage of the employed persons is self employed?

(1) 5519

(2) 1195

(3) 20

(4) 5

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

(1) 52

(2) 56

(3) 96

108°II

54° 18°

III

IV

V

I employees in the public sectorII employees in the private sector III employees in the corporate sectorIV self employed unemployed

→→→→→V



(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

(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, approximately w hat percentage of the selling price is the profit?

(1) 30%

(2) 70%

(3) 100%

(4) 250%

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

(1) Thursday

(2) Friday

(3) Wednesday

(4) Tuesday



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 side in the ratio 2:3 The area of the law n is 1/6 hectares. 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 the same 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

15. All of the follow ing are the same in a manner. Find out the one which is different among them:

(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∞

+ δ π∫

(1) 0

(2) 2

(3) 3

(4) 1

17 Solve the integral equation 0

1 – , 0 1f( ) cos d

0, 1

∞ α ≤ α ≤θ αθ θ =

α >∫



(1) 2

2(1 – cos )f( )

θθ =πθ

(2) 2

2cosf( )

θθ =πθ

(3) 2

(1 – cos )f( )

2

θθ =πθ

(4) 2

(1 – cos )f( )

θθ =πθ

18. Find the function whose laplace transform is 2

2 2 2

s(s a )+

(1) +1[at sin at cos at]

2a

(2) 1

(a sin at cos at )2a

+ −

(3) 1

[a cosat sinat]2a

+ +

(4) 1

[at cosat sin at]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

+ =ɺɺ

(2) k

x x 02m

+ =ɺɺ

(3) 3k

x x 02m

+ =ɺɺ

(4) 5k

x x 02m

+ =ɺɺ



20. In the fol lowing 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 is 2

2 2

1 r – 2rrF 1 –

r c

=

ɺ ɺɺ 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 rL – –

2 2 r rcθ= +ɺɺ ɺ

(2) 2 2 2 2

2 2

r r 1 2 rL

2 2 r c rθ

= + + +ɺɺ ɺ

(3) 2 2 2

2 2

r r 2 1 rL –

2 2 r c rθ

= + +ɺɺ ɺ

(4) 2 2

2 2

r 1 1 rL –

2 r c r= +

ɺ ɺ

22. Calculate the Fermi energy in electron volt for sodium assuming that i t has one free electron per atom. Given density 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



23. The paramagnetic contribution to the magnetic susceptibili ty per m3 of potassium, for which the Fermi energy is 2.1 eV is at wt. of potassium is 39.1 gm and density of potassium is 0.86 × 103 kgm–3.

(1) 420.5 × 10–8

(2) 450.2 ×10+8

(3) 420.5× 10–6

(4) 425.2×10+6

24. The figure shows the inverse magnetic susceptibili ty (1/χ) (dimensionless) a s a function of temperature for a paramagnetic material. Calculate the concentration of magnetic ions, i f they are assumed to be Co2+ with configuration 3d7.

(1) 5·4 × 1023

(2) 5·1 × 1026

(3) 6·9 × 1023

(4) 6·9 × 1026

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 will

be

(1) 0 0q qV8R 4 R

µ Ι ν µ Ι + π i

(2) 0 q 2 ˆ ˆ1 –8R

µ Ι ν + π i k

(3) 0 q ˆ–8

µ Ι νk

(4) Zero



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

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 molecule is 41·6 cm–1. Find the internuclear distance between the atoms forming the molecule. Given : h = 6·63 × 10–34 J s, c = 3·0 × 108 m s–1 and NA = 6·023 × 1023 mol–1.

(1) 1·29 A°

(2) 2·29 A°

(3) 2·49 A°

(4) 0·64 A°

28. For the given circuit the the open loop gain is 12000 and R1 = 120 kΩ and Rf = 600 kΩ . V i = 1.2 V. Find the exact output voltage for the inverting operational amplifier.

(1) – 5 V

(2) – 6 V

(3) – 5·99 V

(4) – 4·99 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. Ιts nominal resistance is 120 ohms. It undergoes strain 10–5, the value of change of resistance in response to the strain is

(1) 240 ohms

(2) 2 × 105 ohm

(3) 2.4 × 10–3 ohm

(4) 1.2 × 10–3 ohm

31. Evaluate V

Fd ,τ∫ ∫∫

where 2F xyz=

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

(1) 13

(2) 23

(3) 19

(4) 29

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 a time 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) = 2·014 amu; M (p) = 1·007 amu;

M (n) = 1·008 amu; M (He4) = 4·001 amu.



33. A γ-ray photon produces an electron-positron pair, each moving with a K.E. of 0·01 MeV. The energy of the γ-ray photon is

(1) 1·02 MeV

(2) 1·04 MeV

(3) 2·08 MeV

(4) 1·03 MeV

34. The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients 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 )Kf,

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 η°

(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 integral6

3C

sin zdz

1z –

6 π

∫ , i f C is the circle z 1= .

(1) 20

i17

π

(2) 21

i16

π

(3) 15

i7

π

(4) 12

i13

π

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 an equal amount of heat from the thermal reservoirs A and B kept at temperatures TA and TB respectively, and rejects heat to the thermal reservoir C kept at temperature TC. The efficiency of the engine is α times the efficiency of the reversible engine, which works between the two reservoirs A and C. which one of the fol lowing relation statement is correct ?

(1) ( ) ( )A A

B C

T T2 – 1 2 1 –

T T= α + α



(2) ( ) ( )A A

B C

T T2 – 1 2 – 1

T T= α + α

(3) ( ) ( ) CB

A A

TT2 – 1 2 1 –

T T= α + α

(4) ( ) ( )A A

C B

T T2 – 1 2 1–

T T= α + α

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 = π ℏ

2 2

2

9.

ma The 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 – 10°C 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 of this value, and taking the latent heat of fusion of ice at 0°C to be 335 J/g , the total entropy of the system 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 of the sphere from a fixed point on the plane, find the acceleration.

(1) 5

g sin7

α

(2) 5

g sin14

α

(3) mg sin α



(4) 7

g sin5

α

42. A symmetrical top with moments of inertia Ιn = Ιy and Ιz in the body axes frame is described by the Hamiltonian

( )2 2 2x y z

x

1 1H L L L

2I 2Ix= + +

Here moments of inertia are parameters and not operators. Lx Ly and Lz are the angular momentum operator 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

Ι Ι Ι

ℏℓ ℓ ℏ

(2) 2

2 2

x z x

1 1( + 1) + m

2 2 2

+ Ι Ι Ι

ℏℓ ℓ ℏ

(3) 2 2

2

x z x

1 1 m– ( 1)

2 2 2

+ + Ι Ι Ι

ℏℏ ℓ ℓ

(4) 2 2

2

x z x

1 1 m( 1)

2 2 2

+ + + Ι Ι Ι

ℏℏ ℓ ℓ

(ii) Expected value for a measurement of Lx + Ly + Lz for any state is

(1) Zero

(2) –ℏm

(3) ℏm

(4) m2ℏ

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

(1) 26

(2) 29

(3) 32

(4) 33



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

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

R = 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(εr1 = 8.5, µr1=1, σ1 = 0) to region 2, free space, with an angle of incidence of 15°. Given i

0E 1.0 V /m= µ , then r0E , is–

(1) 1.62 µV/m

(2) 0.623 µV/m

(3) 4.23 µV/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 elastic head-on coll ision with a stationary particle B of mass M. After the collision the particle A moves along the 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. 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 depth 1

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 a semiconductor 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), (i i ) and (ii i ) in the figure below can al l work as logic gates, where the input signals are either 0V or 5V and the output is V0.

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



Let f(z) = cos z sin 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). Then g(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= θ + φ θ φɺ ɺ

The equation of motion is

(1) 2 2d V(mr sin ) – 0

dt∂

θφ =∂φ

ɺ

(2) 2 2d V(mr sin ) – 0

dt∂

θφ =∂φ

ɺ

(3) 2 2d V(mr sin ) 0

dt∂

θφ + =∂φ

ɺ

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

dt∂θφ + =∂φ

ɺ

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

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

2ω



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

52.(i). The Hamiltonian of the particle

(1) 2

2 2P 1m x x

2m 2+ ω −ε

(2) 2

2 2P 1m x x

2m 2+ ω +ε

(3) 2

2 2P 1m x x

2m 2ω − ε

(4) 2

2 2P 1m x x

2m 2ω + ε

52(ii). Perform a coordinate transformation y = ax+b (where a and b are constant / such that in the y coordinate the Hamiltonian is similar to that of a one – dimensional harmonic oscil lator (with no charge) What are a and b

(1) a = 1 , b = ε / mω2

(2) a = 1 , b = – ε / mω2

(3) a = ε / mω2 , b = 1

(4) a =– ε / mω 2 , b = 1

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

(1) 2

2

1 1n

2 2 mε ω + − ω

ℏ

(2) 2

2

1 3n

2 2 mε ω + ω

ℏ

(3) 2

2

1 1 ew n

2 2 2mwℏ

+ −

(4) 2

2

1 3n

2 2 2mε ω + − ω

ℏ

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



(1) 23

(2) 43

(3) 1

(4) 32

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) 1/22m(E mgz) dz+∫

(3) 2m(E – mgz) dz∫

(4) 2m(E mgz) dz+∫

55. At what temperature will the number of Ι2 molecules in the ν = 1 level be one-tenth of that in the ν = 0 level? Given : ω ε = 214.6 cm–1,

ω exe = 0.6 cm–1, h = 6.63 × 10–34 Js, c = 3.0 × 108 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



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) 211 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 2

1 2

m D m D

w 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 = 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+ +

⇒ + + =

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 = ( )18

100360 –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 and then subtract 8.

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

litres7.50 8 8.50

+ +

consumed in 3 years 2 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, Prof 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.



Required percentage = %

295 1475100 % 70% (approximately)

420 21 × = =

A student mult iplied a number by 35

instead of 53

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

Rat ion of their capitals = 34 : 65.

Now, prof it amount got by P = 20% of total profit + P’s share in balance 80% prof it 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 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 “dirac delta function”

–

f (x ) (x – a)dx f(a)+∞

∞

δ =∫

so, 0

[cos(3x) 2] (x – )dx+∞

+ δ π∫

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

so, δ(x – π) = δ(0)

⇒ x = π

so, 0

[cos(3x) 2] (x – )dx∞

+ δ π∫ = cos(3π) + 2 = –1 + 2 = 1

17.(1) We know Fourier cos transform is–

42? 3

14= =



c 0

2F f(x)cos sx dx

∞=

π∫

c 0

22 (1– ), 0 1

F ( ) f( )cos d

0, 1

∞

⋅ ∝ ≤∝<∝ = θ ∝ θ θ = ππ ∝>

∫

By inversion formula of Fc(∝), we get

f(θ) = c0

2F ( )xcos d

∞θ θ ∝

π ∫ = ( )1

0 1

2 21 cos d 0 d

∞ − ∝ ∝ θ ∝ + ∝ π π

∫ ∫

1

20

2 sin cosf( ) (1 ) ( 1)

∝ θ ∝ θ θ = − ∝ − − − π θ θ =

2 2

2 cos 10 0

θ − − + π θ θ

2

2(1 cos )f( )

− θθ =πθ

18.(4) (i i ) We have f1 (s) = f2 =(s) = 2 2

sLcosat

s a=

+

∴ By convolution theorem

t

11 2 1 2

0

L f (s)f (s) F (y)F (t y)dy− = −∫

∴t2

1 12 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+ −∫

t

0

1 1y cosat 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 2mm 2m 3

× =+



So, kinetic energy T = 21 2mx

2 3

ɺ

Potential energy of system V = 21kx

2

So, Lagrangian L = T–V

2 21 2m 1L x kx

2 3 2 = −

ɺ

Lagrangian equation is–

d L L0

dt x x∂ ∂ − = ∂ ∂ ɺ

d 2mx (–kx) 0

dt 3 − =

ɺ

2mx kx 0

3+ =ɺɺ

3kx x 0

2m+ =ɺɺ

20.(2) Cycle ‘1’ is clockwise so work done during cycle ‘1’ is positive . Similarly cycle ‘2’ is anticlockwise and work 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 Weber’s electrodynamics.

We have 2

r 2 2

1 r – 2rrF 1 –

r c

=

ɺ ɺɺ

Taking U = qφ – q(A ⋅⋅⋅⋅ v ) and Fr = 2

U d U–

r dt r∂ ∂ + ∂ ∂ ɺ

in usual notation,

2

r 2 2

1 r – 2r rF 1 –

r c

=

ɺ ɺɺ=

2 2 2 2

2 2 2 2

1 c – r 2r r 1 c – r 2rrr c c r

+ +=

ɺ ɺɺ ɺɺ

2p

V



= 2 2

2 2 2 2

1 c 1 2rr – rc r c r

⋅ +

ɺɺ ɺ =

2

2 2 2

1 1 2rr – r U d Ur dt rr c r

∂ ∂ + = + ∂ ∂

ɺɺ ɺ

ɺ

Yields,

2

U 1–

r r∂

=∂

which gives an integration U = 1r

= arbitrary constant say a function of r.

Assuming U = 2

2

1 1 r,

r rc

+

ɺ we get

2 2

2 2 2 2 2 2

U 1 r 1 1 r– – –

r r c r r c r∂ = + = ∂

ɺ ɺ and 2

U 1 2rr rc

∂= ⋅

∂ɺ

ɺ

so that 2

2 2 2

d U 2 d r 2 r – rdt r dt rc c r

∂ = = ∂

ɺ ɺɺ ɺ

Thus 2 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 ∂ ∂+ = + + ∂ ∂

ɺ ɺɺ ɺ ɺ ɺɺ (on simplification)

Justifies Fr = 2 2

1 r – 2rr1 –

r c

ɺ ɺɺ as given

As such the generalized potential U ≡ 2

2

1 1 rr rc

+ɺ

also T ≡ 2 2 21[r r ]

2+ θɺɺ

∴ Lagrangian L = T – U = 2 2 2 2

2

r r 1 1 r– –

2 2 r rcθ

+ɺɺ ɺ

22.(4) We know Fermi energy of electron

2 /32 2

fN

E 32m V

π=

ℏ

23N 1 6.06 10

V M× × × ρ=

= 231 6.06 10 0.97

23

−× × × = NV

= 2.55×1022/cm3 =2.55 × 1028 m–3

m = 9.1 ×10–31 kg

So, Ef = 34 2 2 /3

2 2831

(1.05 10 )3 (3.14) 2.55 10

2 9.1 10

−

−

× × × × ×



= 0.66× 10–37 [7542]2/3 (10)18

= 191/3

(754.2)0.66 10 Joule

(754.2)−×

Ef = 4.97×10–19 J

Ef = 19

19

4.97 10eV 3.10eV

1.6 10

−

−

× =×

23.(1) Paramagnetic susceptibili ty is given by

χp = 2

0 B

B F

nk T

µ µ

where µB is Bohr magneton = 9.3 × 10–24 J/tesla

µ0 is magnetic permeabili ty of free space = 4π × 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 is 23 3

3

6.02 10 0.86 103.91 10−

× × ××

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

3

6.02 10 0.86 1039.1 10−

× × ××

or n = 1.2 × 1028 m–3

EF = kBT F or TF = F

B

Ek

∴ TF = 19

423

2.1 1.6 102.43 10 K.

1.38 10

−

−

× × = ××

Substituting the values we have

7 28 24 2

8p 19

4 10 1.3 10 (9.3 10 )420.5 10 .

21. 1.6 10

− −−

−π× × × × ×χ = = ×

× ×

24.(2) We know magnetic susceptibili ty

χ=2

20 BNg J(J 1)

3KTµ µ + =

20N 4S(s 1)

3kTθµ µ +

For co+2 having 3d7 configuration,

ms=



s

1 1 1 3m s

2 2 2 2= = + + =∑

From grap

Slope dy 1 4000 3000dx T 20

−= =χ

= 100050

20=

150

T=

χ

7 20 4 10 N/ A−µ = π ×

24 2B 9.27 10 Am−µ = ×

K = 1.38 ×10–23J/K

J = S = 3/2

g = 2

So, 2

0 B

3K( T)N

3 54

2 2

χ= µ µ

= 2 3

7 2 4 2

3 1.38 104 10 (9.27 10 ) 15 50

−

− −× ×

π × × × ×

26N 5.1 10= ×

25.(2) The magnetic field at O is

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

µ Ι µ Ι µ Ι µ Ι π π

B k k i i k

0 2ˆ ˆ ˆ ˆ= q(v × ) = q ×8R

µ Ι ν + + π

F B j i 1 k k

= 0 qv 2ˆ ˆ– + 1 +8R

µ Ι π

k i

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

2B

f( 0)

K T1 –

12 E

π

Given Ef(0) = 7 eV = 7 × 1.6 × 10–19 J



T = 3 × 104 K

So, f

f(0 )

EE

= 22 –23 4

–1 9

1.4 10 3 101 –

12 7 1.6 10 π × × × × ×

= 22 0.2 3

1 –12 1.6π ×

= 1 – 2 9

12 64π ×

×

= 1 – 2(3.14) 3

4 64×

× = 1 – .115

f

f(0 )

E0.88

E=

27.(1) We know in Raman effect

∆υ =4B3

J2

+

So given 4B = 41.6 c.m–1

So, rotation constant B = – 141.610.4cm

4=

B = 1040 m–1

We know B = 2

h8 Cπ Ι

So, I = 2

h8 BCπ

I = 34

2 8

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

−×× × × ×

I = 2.7 × 10–47 kgm2

So, rΙ=µ

µ → reduce mass Of HCl molecule

So, H Cl

H Cl

M MM M

µ =+

= 23 2

23

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

× ×+ ×



= 1.61 × 1024 gm

µ = 1.6× 10–27 kg.

So, r = 47

27

2.7 101.6 10

−

−××

= 1.29× 10–10m

r 1.29A= °

28.(2) Given Rin = R1 = 120 kΩ

Rf = 600 kΩ

V i = 1.2 V

V0 = ?

∵ For inverting O.P Amp voltage gain

AV = f f

in 1

–R –RR R

= = –600120

AV = –5

∵ AV = 0

i

VV

So, Vout = AV × Vi

= –5 × 1.2

outV – 6V=

29.(4) Given ΙP = 52

We know number of spectral levels (lines) in

NMR is = (2Ιp + 1)

⇒ 2 × 52

+ 1

⇒ 6 lines

30.(3) g = 2

–5=10∆ℓ

ℓ



R = 120 ohm

So, New resistance R’ = Rg ∆ℓ

ℓ

= 120 × 2 × 10–5

= 240 × 10–5

–3R' 2.4 10 ohm= ×

31.(3) x runs from 0 to (1–y)

2Fd xyz dx dy dzτ =∫ ∫∫ ∫ ∫∫

= 1 2 2

2

0 0

(1 y)z y dy dz

2

−

∫ ∫

= 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 – dz

2 2 4 3

+

∫

= 1

2

0

1 4 16 2.8z dz

2 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 31 1 1

2 2 41 1 2

2 41 2

H H H P

H H He n

3 H He P n

+ → +

+ → +→ + +

∆m = (3M(1H2) – M(2He4) – 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 J

E = 38.72 × 10–13 J

So, 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 × 1011

12t 3.87 10 sec.= ×

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

hν = E+ + E– + 2m0c2

E+ → kinetic energy of positron

E– → kinetic energy of electron

hν → incident photon energy

Given E+ + E– = 0.01 MeV

Rest mass energy 2m0c2 = 2 × 9.1 × 10–31 × (3 × 108)2 J.

2m0c2 = 1.02 MeV

So, hν = 0.01 MeV + 1.02 MeV

h 1.03 MeVν =

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

35.(2) η° is meson having zero charge.

So, quark structure is uu or dd

So, normalized quark structure is

( )1uu – dd

2ψ =

PART C (36-55)

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

n! f(z )dz,

2 i (z – a) +π ∫



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

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

f” 6

3C

1 2! sin zdz.

6 2 i 1z –

6

π = π π

∫ ….(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”( π/6) = 6 sin4 2 21 1 15 cos – sin

6 6 6 π π π

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

∴ From (1) we have 6

3C

sin z dz 21i.

161z –

6

= π π

∫

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

then

1 2 21

A 2 1 –13

–2 1 –1

=

∴ AT = 1 2 –2

12 1 1

32 –1 –1

A square finite matrix A is said to orthogonal if

AAT = Ι

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– 2

12 2 – 2 4 1 1 –4 1 1

9–2 2 – 2 –4 1 1 4 1 1

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

9 2 –2

12 6 –2

9–2 –2 6

= ≠ Ι

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

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



Then A

1 2 21

2 1 –23

–2 2 –1

=

∴ AT = 1 2 –2

12 1 2

32 –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 0

19 0 1 0

90 0 1

×

= 1 0 00 1 0

0 0 1

= Ι

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

38.(1) η of H.E. between A and C

CA

A

T1 –

T

η =

η of engine = C

A

T1 –

T

α

Here Q2 = 1 1C 3 C

A B

Q QT , Q T

T T× = ×

∴ Total Heat rejection

(Q2 + Q3) = Q1TC A B

1 1T T

+

Total Heat input = 2Q1

η of engine = 1 C

A B

1

1 1Q T

T T1 –

2Q

+



∴ C C C

A A B

T T T– 1 – –

T 2T 2Tαα =

Multiply both side by TA and divide by TC

or A A A

C C B

T T T1 1– – –

T T 2 2 Tα α =

or A A

B C

T T(2 – 1) 2(1 – )

T T= α + α

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

π

=ℏ

2 2

2E2ma

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

given E = π ℏ

2 2

2

9ma

E = π ℏ

2 2

2

182ma

...(2)

compere equation (1) and equation (2). + + =2 2 2

x y zn n n 18

If nx = 4, ny = 1, nz = 1

then + + = + + =2 2 2x y zn n n 16 1 1 18

So, possible values (combinations) of nx ny nz are (nx, ny, nz) = (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= × ×

S2 – S1 = – 0.00297 kJ/K = 2.97 J/K

3 2mL

S ST

−− =

= 0.01 335 1000273

− × ×

S3 – S2 = – 12.271 J/K 268

p4 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 JK–1

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

2 2

2

nE

2mLπ∆ = ℏ So,

2

2

n 'E

L∆ ∝

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

= + + + − = + − Ι Ι Ι Ι Ι Ι

where L is the total angular momentum.

We know eigen values of L2 = ℓ(ℓ + 1) ℏ2 and, L2 = mℏ

22 2

mx z x

1 1E ( 1) m

2 2 2 = + + − Ι Ι Ι

ℓ

ℏℓ ℓ ℏ

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

(ii)(3) m mx y zY ( , ) | (L L L ) | Y ( , )⟨ θ φ + + θ φ >

ℓ ℓ

m mz

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

2 2i+ − + −+ − < θ φ + + θ φ ⟩

ℓ ℓ

= m mzY ( , ) | L | Y ( , )< θ φ θ φ >ℓ ℓ = mℏ

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

Eb =avA = asA2/3 – acz (z–1) A–1/3 –aa(A-2z)2A–1 – apA–3/4

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

∴ For maximum binding energy

Eb =Ebmax when bdE0

dz=

∴ For maximum binding energy

Eb = Ebmax whendEb

0dz

=

∴ –ac(2z–1)A–1/3 + 4aa (A– 2z) A–1 =0

∴ 1/ 3

a1/ 3 1

c a

aA 4az

2a A 8a A

−

− −+=+



∴ 1/3

1/3

0.58(64) 4 19.38 19.3

2 0.58(64)64

−

−

+ ××× +

= 0.145 77.228.62

0.29 2.4125+ =

+

z 29≈

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

T 1 = 30°C = 303 K

T 2 = 523 K

(∆S)water

= P373

mc n kJ/K303

ℓ

= 0.03143 kJ/K

(∆S)Vaporization = 2

mLT

= 0.036 2260

373×

= 0.21812 kJ/K

(∆S)Vapour = 523

p373

dTmc

T∫

= 523

373

amR b CT dT

T + + ∫



= 5 232

3 73

CTmR a n T bT

2

+ +

ℓ

= 5 232

3 73

CTmR a n T bT

2

+ +

ℓ

= 2 2523 CmR a n b (523 – 373) (523 – 373 )

373 2 + × +

ℓ

= 0.023556 kJ/K

(∆S)System= (∆S)water + (∆S)vaporization + (∆S)vapour = 273.1 J/K.

45.(2) The intrinsic impedances are

η1 = 0

r1

120129

8.5

η= = Ω

ε and η2 = η0 = 120π Ω

and the angle of transmission is given by

t 0

sin15sin 8.5

° ε=θ ε

or θt = 48.99°

Then r0 0 i 1 ti

2 i 1 t0

E cos coscos cosE

η θ − η θ=η θ + η θ

= 0.623 or r0E = 0.623 µV/m

46.(4) Since the kinetic energy of A after coll ision is one-ninths of i ts initial kinetic energy, the momentum of A after collision is one-third of i ts 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 coll ision.

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

Therefore, p’ = 4p/3

The kinetic energy gained by particle B due to the coll ision is p ’2/2M 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 p2/2m and its final kinetic energy is (1/9) p2/2m].

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

p ’2/2M = (8/9) p2/2m

Substituting for p’ = 4p/3, we have

(16/9) (p2/2M) = (8/9) p2/2m 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 × 10–2 = 2.18 × 1038.

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

= 1.56 × 1039 MeV.

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

λE = 12400, very nearly.

Therefore, λ = 12400/E

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

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

Photons with wave length greater than 640 nm wil l have energy less than 2 eV so that they will be unable 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 ⋅⋅⋅⋅ B

Circuit (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+

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

–z

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

⇒ cos zsin z

–z

= 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!

+ + =

⇒ z2 1 1– ........ 0

3! 2! + =

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

50(ii).(3) g(z) sinhzzf (z ) zcosz – sinz

=

To find poles z cos z – sin z = 0

⇒ 3 5z z

z – – .......2! 4!

+

–

3zz – ....... 0

3!

+ =

⇒ 3 51 1 1 1z – 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 21mr ( sin ) – V(r, , ).

2θ + θφ θ φɺ ɺ ..(1)

In this case the only two generalized co-ordinates are θ and φ, therefore there will be only two Lagrangian equations, one in θ and the other in φ.

The Lagrangian equation in coordinate θ is given by

d L L

– 0.dt

∂ ∂ = ∂θ∂θ ɺ ..(2)

From equation (1), we have

2 2 2L L Vmr and mr sin cos –

∂ ∂ ∂= θ = θ θφ

∂θ ∂θ∂θɺ ɺ

ɺ

With these substitutions, equation (2) becomes

2 2 2d V(mr ) – mr sin cos 0

dt∂θ θ θφ + =∂θ

ɺ ɺ …(3)



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

d L L

– 0dt ∂ ∂

= ∂φ∂φ ɺ

Again equation (1) gives

2 2L V L– and mr sin

∂ ∂ ∂= = θφ∂φ ∂φ ∂φ

ɺɺ

…(4)

With these substitutions equation (4) becomes

2 2d V(mr sin ) 0

dt∂

θφ + =∂φ

ɺ

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

E = – ∇φ …(1)

Since B = 0, we seek a gauge in which A = 0. Intergrating (1) we obtain φ(x) = - εx + c, where c is a constant of integration. Let us choose c = 0; then

( )x xφ = −ε …(2)

The total Hamiltonian is

2

2 2p 1H m x x

2m 2= + ω − ε …(3)

The first term on the right-hand side of (3) is the standard kinetic term, the second term is the harmonic osci llator 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+ ω + …(4)

Where H0 is a constant and y = ax + b. Consider the kinetic term. We see that py = px, so a = 1. Now we can 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+ ω + + = + ω + ω + …(5)

From Eq. (3) of part (i) we see that Hx = Hy only i f b = –ε/mω2 and H0 = – ε2 / 2mω 2 .

52(iii).(3) To conclude, i f we perform the coordinate transformation y = x –ε/mω2 , we get a one-dimensional harmonic osci llator with no charge, and the energy eigenvalues of a one-dimensional harmonic osci llator are



n

1 1E w n

2 2ℏ

= +

..(6)

Corresponding to the eigenstate n| ψ . We have a shifted harmonic oscil lator; thus, the energy eigenvalues are now.

E n = 2

2

1 1 1n

2 2 2 mε ω + − ω

ℏ

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 = ( )82 – 2 – 2/3 4 /3

12= =

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

V(q) = mgz ..(1)

The Hamilton-Jacobi equation for Hamilton’s 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, therefore each term must be equal to the same constant with opposite signs.

21 S

i.e. V(q)2m q

Sand –

t

∂+ = α ∂

∂ = α ∂

...(3)

Then we have S = W(q, α)–αt, ...(4)

W being a constant of integration.



As s W

,q q

∂ ∂=∂ ∂

therefore the Hamilton-Jacobi equation for Hamilton’s characteristic function W takes

the form

2

1 WV(q)

2m q ∂ + = α ∂

...(5)

This gives W2m – V(q)

q∂ = α ∂

...(6)

Integrating above expression, we get

W = 1 /22m – V(q) dq α ∫ ...(7)

Now 1 /2S Wp 2m – V(q)

q q∂ ∂

= = = α ∂ ∂ ...(8)

and S W

– t∂ ∂

β = =∂α ∂α

= 1/22m – V(q) dq – t

∂ α ∂α ∫

i .e. 1/ 2

m dqt

2 [ – V(q)β + =

α∫ ...(9)

∴ The Hamiltonian of system is

H = 2p

mgz E2m

+ = = α ...(10)

From (9) 1/21/2

m dz m 2t (E – mgz)

2 2 mg[E – mgz] β + = = ⋅ ∫

= 1/21 2(E – mgz)

g m ...(11)

Thus 22

1 2( t)

mgβ + = ⋅ (E – mgz)

Solving for z, we get

z = 21 E– g( t)

2 mgβ + + ...(12)

Initial conditions are

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

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



E = mgh, h = 21 E– g

2 mgβ +

Solving these equations, we get β = 0

Now p = W

mz mgt usin g (12)z

∂= =

∂ɺ ...(14)

Thus we have from (12) and (14) 21

z h – gt2

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 νth state relative to that in the ν = 0 (lowest) state at T Kelvin temperature is given by

0–G ( )h c/ kT

0

Ne

Nνν =

= –G( ) – G(0 )hc /kTe ,ν

where G(ν) = 2

e e e

1 1– x

2 2 ω ν + ω ν +

∴ 2

e e e e e–( n – x n – x n)hc /kTn

0

Ne

Nω ω ω=

For the n = 1 level, we have

e e e–( – 2 x )hc / kT1

0

Ne

Nω ω=

Here –1 –11e e e

0

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

N 10= ω = ω =

∴ –1 – 1–( 214.6 cm – 1 .2 cm )hc /kT1

e10

=

or –1 –1(213 .4 cm )hc /kT (2134 0m )hc / kT10 e e= =

Taking natural logarithm:

Loge 10 = (21340 m–1) hc/kT.



∴ T = – 121340 m hc

2.303 k

= –3 4 8 –1

–1– 23 –1

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

1.38 10 JK× ×

× = 133.5 K.

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