solved PH'12

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DEC-2012 Physical Sciences

Booklet Series (A) Solved

Part-A

1. A granite block of 2m 5m 3 m size is cut into 5 cm thick slabs of 2m5m size. These slabs are laid over a2 m wide pavement. What is the length of the pavement that can be covered with these slabs?

(a) 100 m (b) 200 m (c) 300 m (d) 500 m

Soln. Let number of slabs is n.

Thus, 253 =5

n 2 5100

; n 60

Area covered by 60 slabs = 6025 m2.

Length of pavement =

260 2 5 m

2 m

= 300 m

Correct option is (c)

2. Which is the least among the following?

0.33 0.44 1/ 1/e0.33 , 0.44 , , e

(a) 0.330.33 (b) 0.440.44 (c)1/ (d) 1/ee

Soln. 1 4 1 13 9 e

1 1e0.33 0.441 4 1 10.33 ; 0.44 ; ; e

3 9 e

All terms are of the form: xy x

x1 dy dy

ny x nx nx 1 x nx 1y dx dx

For maxima or minima,dy 1

0 nx 1 0 xdx e

Since,

2

21

xe

d ypositive

dx

So, f(x) = x2has minima at1

xe

Correct option is (d)

3. What is the next number in this see and tell sequence?

1 11 21 1211 111221 _________

(a) 312211 (b) 1112221 (c) 1112222 (d) 1112131

Soln. Correct option is (a)

4. A vertical pole of length a stands at the centre of a horizontal regular hexagonal ground of side a. A rope

that is fixed taut in between a vertex on the ground and the tip of the pole has a length.

(a) a (b) 2 a (c) 3 a (d) 6 a

Soln. Using pythogorus theorem,

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F D

F C

A B

TB

a TB

A

a

2 2 2 2 2AT AB BT ; AT 2a ; AT 2 a Correct option is (b).

5. A peacock perched on the top of a 12 m high tree spots a snake moving towards its hole at the base of the tree

from a distance equal to thrice the height of the tree. The peacock flies towards the snake in a straight line and

they both move at the same speed. At what distance from the base of the tree will the peacock catch the snake?

(a) 16 m (b) 18 m (c) 14 m (d) 12 m

Soln. Distance travelled by peacock and snake must be same.

TC = BC

22 212 x 36 x

T

12m

A BCx 36x

2 2144 x 1296 72x x

72x 1152 ; x = 16Correct option is (a).

6. The cities of a country are connected by intercity roads. If a city is directly connected to an odd number of

other cities, it is called an odd city. If a city is directly connected to an even number of other cities, it is called

an even city. Then which of the following is impossible?

(a) There are an even number of odd cities (b) There are an odd number of odd cities.

(c) There are an even number of even cities (d) There are an odd number of even cities.

Soln. Correct option is (b).

7. In the figure ABC2

, AD = DE = EB

What is the ratio of the area of triangle ADC to that of triangle CDB?

A D E B

C

(a) 1:1 (b) 1:2 (c) 1:3 (d) 1:4

Soln. Area of ADC Pythagorus theorem area of ABC area of DBC

= 1 1

h 2x x hx2 2

C

BEDA x x x

h

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Area of 1

CDB 2x h xh2

Correct option is (b).

8. A rectangular sheet ABCD is folded in such a way that vertex A meets vertex C, thereby forming a line PQ.

Assuming AB = 3 and BC = 4, find PQ. NOte that AP = PC and AQ = QC.

D C

Q

P

A B

(a) 13/4 (b) 15/4 (c) 17/4 (d) 19/4

Soln. Correct option is (b).

9. A string of diameter 1 mm is kept on a table in the shape of a close flat spiral i.e. a spiral with no gap between

the turns. The area of the table occupied by the spiral is 1 m2. Then the length of the string is

(a) 10 m (b) 102m (c) 103m (d) 106m.

Soln. Let the length of string is lm.

Thus, 1

1 1000 m1000


10. 25% of 25% of a quantity is x% of the quantity where x is

(a) 6.25% (b) 12.5% (c) 25% (d) 50%

Soln. Let the given quantity is y.

25 25 Xx y y

100 100 100

;25

X X 6.254

Correct option is (a)

11. In sequence {an} every term is equal to the sum of all its previous terms.

If n 10n

n

aa 3, then lim is :

a

(a) 3 (b) 2 (c) 1 (d) e

Soln. a0= 3; a

1= a

0= 3; a

2= a

1+ a

0= 6; a

3= a

0+ a

1+ a

2= (a

0+a

1) + a

2= 12.

0 1 3 n 1n 1n n

n n

a a a a .......aalim lim

a a

n n n

n nn n

a a 2alim lim 2

a a


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12. In the figure below, angle ABC = /2. I, II, III are the areas of semicircles on the sides opposite angles B, A,and C respectively. Which of the following is always true?

A

B C

I

II

III

(a) II2+ III2= I2 (b) II + III = I (c) II2+ III2> I2 (d) II + III < I

Soln. Let length of AB is k and length of BC is h.

Thus,2 2

AC h k

Area of I =

2 2 2 2h k h k

2 2 2 4

Area of II =

2 2h h

2 2 2 4

Area of III =

2 2k k

2 2 2 4

Area of II + area of III =

2 2 2 2h k h k

area of I2 4 4 2 4


13. What is the minimum number of days between one Friday the 13thand the next Friday the 13th? (Assume

that the year is a leap year).

(a) 28 (b) 56 (c) 91 (d) 84

Soln. Correct option is (c)

14. Suppose a person A is at the North-East corner of a square (see figure below). From that point he moves

along the diagonal and after covering 1/3rdportion of the diagonal, he goes to his left and after sometime he

stops, rotates 90 clockwise and moves straight. After a few minutes he stopes, rotates 180 anticlockwise.

Towards which direction he is facing now?

N

S

W E

A

(a) North-East (b) North-West (c) South-East (d) South-West

Soln. Correct option is (a)

15. Cucumber contains 99% water. Ramesh buys 100 kg of cucumbers. After 30 days of storing, the cucum-

bers lose some water. They now contain 98% water. What is the total weight of cucumbers now?

(a) 99 kg (b) 50 kg (c) 75 kg (d) 2 kg.

Soln. Water Cucumber Total weight

At the time of purchasing (99%) 99 kg (1%) 1 kg 100 kgAfter 30 days (98%) 49 kg (2%) 1 kg 50 kg

Correct option is (b)

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16. In a museum there were old coins with their respective years engraved on them, as follows

(a) 1837 AD (b) 1907 AD (c) 1947 AD (d) 200 BC

Soln. Option (d) is wrong because coins are not invented in 200 BC.

Correct option is (d).

17. A student observes the movement of four snails and plots the graphs of distance moved as a function of time

as given in figure (A), (B), (C) and (D).

Distance

Distance

Distance

Distance

(A) (B) (D)

Time Time Time Time

Which of the following is not correct?

(a) Graph (A) (b) Graph (B) (c) Graph (C) (d) Graph (D)

Soln. For a given time more than one distance as in graph C is not possibleCorrect option is (c).

18. Find the missing letter:

A

?

Q

U

B

EGK

OJF

C

P

V

R

D

(a) H (b) L (c) Z (d) Y

Soln. Letters are arranged as

E O

G J

K F

+1

+4

+3

? P

U R

Q V

+1

+4

+3

So, the required letter is (22 + 4) letter of english alphabet i.e. Z.

Correct option is (c).

19. Consider the following equation

2 2 2x 4y 9z 14x 28y 42z 147

where x, y and z are real numbers. Then the value of x 2y 3z is (a) 7 (b) 14 (c) 21 (d) Not unique.

Soln. 2 2 2x 4y 9z 14x 28y 42z 147 2 2 2

x 14x 4y 28y 9z 42z 147 0

2 2 2x 2.x.7 49 4y 2.2y.7 49 9z 2.3z.7 49 0

2 2 2

x 7 2y 7 3z 7 0

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Thus, each term must be separately zero.

7x 7 0 x 7; 2y 7 0 y 1; 3z 7 0 z

3

7 7x 2y 3z 7 2 3 21

2 3


20. The map given below shows a meandering river following a semi-circular path, along which two villages are

located at A and B, the distance between A and B along the east-west direction in the map is 7 cm. What is the

length of the river between A and B in the ground?

SCALE = 1:50000

A B

N

(a) 1.1 km (b) 3.5 km (c) 5.5 km (d) 11.0 km

Soln. Actual distance between A and B = (750000) cm = 3.5 km

Length of the river from A to B in the ground =3.5

km 5.5 km2


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Part-B

21. A 22 matrix A has eigenvalues i /5 i /6e and e . The smallest value of n such that AAn= 1 is:

(a) 20 (b) 30 (c) 60 (d) 120

Soln. A has eigenvalues i i5 6e and e

Anhas eigenvaluesn n

i i5 6e and e

So,nA product of eigenvalues =

11ni

30e

Since,

n n

i 11n /30 2m i

A I A 1

11n n 60e 1 e 2m

30 m 11

So, the smallest value of 'n' is 60.


22. The graph of the function f(x) as shown below is best described by

(a) The bessel function J0(x) (b) cos x

(b) excos x (d)1

cosx

xSoln. At 0x 0, Bessel function J x 1

J0(x) has zeroes at x = 2.4048, 5.5201, 8.6537 ................... etc.

0So, the graph is best described by J (x)

The other functions are following plots.

1

x

cos

cosx

1

x

e cosx

x

e cosx

x

cos x

x

1

00 0

cos x

x

Correct option is (a).

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23. In a series of five cricket matches, one of the captains calls Heads every time when the toss is taken. The

probability that he will win 3 times and lose 2 times is

(a) 1/8 (b) 5/8 (c) 3/16 (d) 5/16

Soln. Number of trials i.e. tosses n = 5

Number of success i.e. getting heads r = 3

Probability of success i.e. getting head in one trial p = 1/2Probability of failure i.e. getting tail in one trial q = 1/2

2 2

53

1 1 5! 1 5.4 1 5Required probability: P C

3 2 3!2! 32 2 32 16


24. The unit normal vector at the pointa b c

, ,3 3 3

on the surface of the ellipsoid

2 2 2

2 2 2

x y z1, is

a b c

(a) 2 2 2

bci caj abk

a b c

(b) 2 2 2

ai bj ck

a b c

(c) 2 2 2

bi cj ak

a b c

(d) i j k

3

Soln. Given surface of ellipsoid:

2 2 2

2 2 2

x y z: 1

a b c

Normal to the surface at pointa b c

, ,3 3 3

:

a b c 2 2 2, ,3 3 3

2x 2y 2z 2 2 2 i j k i j k a b c 3 a 3 b 3 c

Unit normal vector to the surface ata b c, ,3 3 3

:

2 2 2 2 2 2

2 2 2

2 1 1 1 i j k bci caj abka b c3n

2 1 1 1 b c a c a b

a b c3

Option (a) is correct but it is wrongly printed in the question.

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25. A solid cylinder of height H, radius R and density , floats vertically on the surface of a liquid of density 0 .

The cylinder will be set into oscillatory motion when a small instantaneous downward force is applied. The

frequency of oscillation is

(a)

0

g

H

(b)0

g

H

(c)

0

g

H

(d)0g

H

Soln.

fig(a) fig(b)

H0

Hx

H0

y

(a) Cylinder is partially submerged and in equilibrium

(b) Cylinder is not in equilibrium since the force of bouyancy on cylinder is not equal to weight of the cylinder.

20 0AH g AH g for equilibrium A R For a displacement y, the equation of motion is

2

0 02

d yAH AH g A H y g

dt

2

0 0 0 0 02

H y g H g y gd yg g

H H Hdt

Since 0 0H H

2 20 0

2 2

g gd y d yg g y; y

H Hdt dt

Acceleration in proportional to displacment, motion is simple harmonic

2 0 0g g

H H


26. Three particles of equal mass m are connected by two identical massless springs of stiffness constant k as

shown in the figure.

k k

m m m

If x1, x

2and x

3denote the displacements of the masses from their respective equilibrium positions, the potential

energy of the system is:

(a) 2 2 21 2 31

k x x x2

(b) 2 2 21 2 3 2 1 31

k x x x x x x2

(c) 2 2 2

1 2 3 2 1 3

1

k x 2x x 2x x x2 (d) 2 2 2

1 2 3 2 1 3

1

k x 2x x 2x x x2

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Soln. Potential energy = 2

2 1

1k x x first spring

2

Potential energy = 2

3 2

1k x x second spring

2

Potential energy of the system,

22

2 1 3 2

1 1k x x k x x

2 2

2 2 2 21 2 1 2 2 3 2 3

1k x x 2x x x x 2x x

2

2 2 21 2 3 2 1 31

k x 2x x 2x x x2


27. Let v, p and E denotes the speed, the magnitude of the momentum, and the energy of a free particle of rest massm. Then

(a)dE

constantdp

(b) p mv (c)2 2 2

cpv

p m c

(d) 2E mc

Soln. The energy of a particle of rest mass m is given by 2 2 2 4E c p m c

The velocity of particle, v = vg= group velocity

=

22

2 2 2 4 2 2 2 2 2 2

12 c p

dE c p cp2dp c p m c c p m c p m c


28. A binary star system consists of two stars S1and S

2, with masses m and 2m respectively separated by a

distannce r. If both S1and S

2individually follow circular orbits around the centre of the mass with intantaneous

speeds v1and v

2respectively, the ratio of speeds v

1/v

2is:

(a) 2 (b) 1 (c) 1/2 (d) 2

Soln.C.M.

r

r1 r2

2mm

Let centre of mass C.M. lie at a distance r1from m and r

2from 2m, then by definition of C.M.

1 2mr 2mr

1 2 11 2

2

dr dr v 2mm 2m ; mv 2mv ; 2

dt dt v m ;

1

2

v2

v


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29. Three charges are located on the circumference of a circle of radius R as shown in the figure below. The two

charges Q subtend an angle 90 at the centre of the circle .The charge q is symmetrically placed with respect

to the charges Q. If the electric field at the centre of the circle is zero, what is the magnitude of Q?

QQ

q

(a) q/ 2 (b) 2q (c) 2q (d) 4q

Soln. What is the value of Q such that electricfield at centre is zero?

Let radius of circle is R.

Electric field due to charge q at centre

1 20

q

E 4 R

(up direction) ... (1)

Electric field due to both Q charges is

2 3 20

QE E

4 R

(aligned 45 from downward direction)

The resultant electric field of two Q charges will be in downward directions.

QQ

q

45 45

E2

E3

E1

E sin452 E sin453

(E + E )cos452 3Hence, 2 20 0

2Q 2Q 1sin45

4 R 4 R 2

... (2)

From (1) and (2), electric fields must be equal

2 20 0

q 2Q 1

4 R 4 R 2

qQ

2


30. Consider a hollow charged shell of inner radius a and outer radius b. The volume charge density is

2k

r

r

(where k is a constant) in the region a < r < b. The magnitude of the electric field produced at

distance r > a is:

(a)

20

k b afor r a

r

(b)

2 20 0

k b a kbfor a r b and for r b

r r

(c)

2 20 0

k r a k b afor a r b and for r b

r r

(d)

2 20 0

k r a k b afor a r b and for r b

a a

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Soln. 2k

a r br

Electric field for a < r < b

Charge enclosed by Gaussian surface

r2

enc 2

a

kq .4 r dr

r = 4 k r a

From Gauss law, 2enc

S0 0

4 k r aqE.dS E.4 r

2

0

k r aE r for a

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32. Four charges (two +q and two q) are kept fixed at the four vertices of a square of side a as shown

P

q

q

q

q

aR

At the point P which is at a distance R from the centre (R > > a), the potential is proportional to

(a)1

R(b) 2

1

R(c) 3

1

R(d) 4

1

R

Soln.

-q +q

-q+q

R P

p1

p2

p4

p3

Monopole moment of this charge configuration = +qq+qq = 0

Dipole moment = 1 2 3 4p p p p 0

But quadrupole moment 0

hence, V for (R >> a), 31V

R


33. A point charge q of mass m is kept at a distance d below a grounded infinite conducting sheet which lies

in the xy-plane. What is the value of d for which the charge remains stationary?

(a) 0q/4 mg (b) 0q/ mg

(c) There is no finite value of d (d) 0mg /q

Soln. Force on a point charge q due to infinite grounding conducting plane is given by

2

20

1 qupward direction

4 4d... (1)

While its weight = mg (downward direction) ... (2)

V=0

d

+qUpward coulomb force in equation (1) balances the

downward weight in equation (2) hence by (1) and (2), we get

2

20 0

1 q qmg d

4 4d 4 mg

Correct option is (a).

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34. The wave function of a state of the hydrogen atom is given by

200 211 210 21 12 3 2

where n m denotes the normalized eigen function of the state with quantum numbers n,land m in the usual

notation. The expectation value of Lzin the state is:

(a) 1516

(b) 1116

(c) 38 (d)

8

Soln. The given wavefunction is200 211 210 21 12 3 2

The expectation value of Lzis given by

2

z

z 2

C mL

L| C

= 1 0 4 2 9 0 2 1 4 2 2

1 4 9 2 16 16 8


35. The energy eigenvalues of a particle in the potential 2 21V x m x ax are2

(a)

2

n 2

1 aE n

2 2m

(b)

2

n 2

1 aE n

2 2m

(c)

2

n 2

1 aE n

2 m

(d) n

1E n

2

Soln. The potential 2 21

V x m x ax2

The Schrodinger equation, 2 2

2

dV x x E x

2m dx

2

2 2

2 2

d x 2m 1For S.E., E m x 0 ... 1

dx 2

1The energy is given by E n

2

2 2

2 22

d 1m x ax x E x2m dx 2

2

2 2

02 2

d 2m 1E m x ax

dx 2

2 2 2

2 2

2 2 2 2

d 2m a 1 aE m x ax x 0

dx 2m 2 2m

2 2 2

2 2

2 2 2 2 4 2

d 2m a 1 a 2axE m x x 0

dx 2m 2 m m

Replace,

2

2 2

a aE E ' , x x ' & dx dx '

2m m

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2 2

2 2

d 2m 1E ' m x ' x ' 0

dx ' 2

... (2)

On comparing equation (2) by equation (1), the energy corresponding to equation (2) are given by

1E ' n

2

2

2

a 1E n

2m 2

2

2

1 aE n

2 2m


36. If a particle is represented by the normalized wave function

2 2

52

15 a xfor a x a

x4a

0 otherwise

the uncertainty p in its momentum is

(a)2

5a

(b)

5

2a

(c)

10

a

(d)

5

2 a

Soln. The given normalized wavefunction is

52

2 215 a x

for a x ax 4a

0 otherwise

* d

p i dxdx

5 52 2

a a

2 2 2 2

a a

d 15 d 15 d0 i 0 dx a x i a x dx 0 i 0 dx

dx dx dx4a 4a

a

2 2

5

a

15i a x 2x dx 0

16a

[As the function given in the integral is odd]

2

2 * 22dp dxdx

a a2 2 2

2 2 2 2 2 2 2

2 5 2 2

a a

d 15 d d0. .0 dx a x a x dx 0. 0 dx

dx 16 a dx dx

a

2 2 2

5

a

15a x 2 dx

16 a

aa2 2 3 2 3 3

2 2 2 3 3

5 5 5

aa

15 15 x 15 a aa x dx a x a a

8a 8 a 5 8 a 3 3

=

2 3 2

5 2

15 4 a 5

8 a 3 2 a

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222 5 5p p p

2 a a2


37. Given the usual canonical commutation relations, the commutator y xA, B of A i xp yp and

z yB yp zp is :

(a) z xxp p z (b) z xxp p z (c) z xxp p z (d) z xxp p z

Soln. x z yA i xp yp & B yp zp

y x z yA, B i xp yp , yp zp By using the distributive properties,

AB, C A B,C A,C B & A, BC B A,C A,B C

And associative properties A, B C A,B A,C We can simplify

x z y x yA, B i xp yp , yp i xp yp , zp

z x z y y x yi xp , yp i yp , yp i xp , zp i yp , zp [* out of 4 commutation relations 2ndand 3rdwill be zero as there will be no canonically conjugate pair term.

z x yA, B i xp , yp i yp , zp

z z y x y y xix p , yp i x, yp p iy p , zp i y, zp p [2ndand 3rdwill be also zero, reason will be same as above]

z y xA, B ix p , yp i y, zp p

z y z y x y x||||00

ixy p ,p ix p , y p iz y, p p i y, z p p

z x

z x x x

z x

A, B ix i p iz i p

xp zp , as zp p z

xp p z


38. The entropy of a system, S, is related to the accessible phase space volume by BS k n E, N, V where E, N and V are the energy, number of particles and volume respectively. From this one can conclude

that (a) does not change during evolution to equilibrium

(b) Oscillates during evolution to equilibrium

(c) Is a maximum in equilibrium

(d) Is a minimum in equilibrium

Soln. In the micrononical ensemble, the phase volume is maximized to obtain the characteristic thermodynamics

quantity which is entropy


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39. Let W be the work done in a quasistatic reversible thermodynamics process. Which of the following state-

ments about W is correct?(a) W is a perfect differential if the process is isothermal(b) W is a perfect differential if the process is adiabatic(c) W is always a perfect differential

(d) W cannot be a perfect differential.

Soln. According to the first law of thermodynamics

dQ = dU + dW

where dQ is the amount of heat absorbed by a system, dU is the change in the internal energy of the system

which depends only on the temperature, and dW is the amount of work done. If dQ 0 , then dW is not aperfect differential as it depends upon the path chosen. However for an adiabatic process dQ = 0,

dU dW Since dU is a perfect differential, therefore dW is also a perfect differential.


40. A binary star system consists of two stars S1and S2, with masses m and 2m respectively separated by adistannce r. If both S

1and S

2individually follow circular orbits around the centre of the mass with intantaneous

speeds v1and v

2respectively, the ratio of speeds v

1/v

2is:

(a) 2 (b) 1 (c) 1/2 (d) 2

Soln. The energy of the system is given by

1 2 2 3 3 1E J S S S S S S , where J is positive constant and each among S1, S2and S3can take +1 and 1values.

Suppose S1, S

2, S

3all have value +1, then the energy of the system is

E J 1 1 1 1 1 1 3J Suppose S1, S

2, S

3all have value 1, then the energy of the system is

E J 1 1 1 1 1 1 3J If S

1, S

2, S

3have different values then energy will be greater than 3J.


41. The minimum energy of a collection of 6 non-interacting electrons of spin1

2 placed in a one dimensional

infinite square well potential of width L is

(a) 2 2 214 / mL (b) 2 2 291 / mL (c) 2 2 27 / mL (d) 2 2 23 / mL

Soln. The energy of a particle of mass m in a one-dimension square well potential of width L is given by2 2 2

2

nE , n 1,2,3,4.................

2mL

Only two electron with spin1

2can fill in the same state.

n = 3

n = 2

n = 1(corresponding to minimum energy)

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The minimum energy of the system is

2 2 2 2 2 2 2 2

2 2 2 2

4 9 142

2mL 2mL 2mL mL


42. A live music broadcast consists of a radio-wave of frequency 7 MHz, amplitude-modulated by a microphoneoutput consisting of signals with a maximum frequency of 10 KHz. The spectrum of modulated output will be

zero outside the frequency band

(a) 7.00 MHz to 7.01 MHz (b) 6.99 MHz to 7.01 MHz

(c) 6.99 MHz to 7.00 MHz (d) 6.995 MHz to 7.005 MHz

Soln. Carrier frequency fc= 7 MHz

Modulation frequency fm=10 KHz = 0.01 MHz

Modulated signal will have frequency components (fc+ f

m) = 7.01 MHz and (f

c f

m) = 6.99 MHz


43. In the op-amp circuit shown in the figure, Viis a sinusoidal input signal of frequency 10 Hz and V

0is the output

signal.

+

1K

10K

V0

Vi

0.01 F

The magnitude of the gain and the phase shift, respectively, are close to the values

(a) 5 2 and2

(b) 5 2 and

2

(c) 10 and zero (d) 10 and

Soln. 0 f

in 1

V Rsin ce inverting amplifier

V R

fR 10k || 0.01 f =

10k0.01

1

10k 0.01

= 3 6

10k

1 10 2 10 10 0.01 10

3 6

10k

1 10 2 10 10 0.01 10

f3

10kR

1 6.28 10

0 f

33in 1

V R 10k 1 10

V R 1k 1 6.28 101 6.28 10

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0

in

V10

V

Phase difference is 180 = Correct options is (d)

44. The logic circuit shown in the figure below.

A

HIGH

B

y

implements the Boolean expression

(a) y A.B (b) y A.B (c) y A.B (d) y A B

Soln.

A

HIGH

B

y

High means 1.

Output of first EXOR gate is A

Output of second EXOR gate is B

Output of OR gate is A B

As per demorgans theorem A.B A B Correct options is (a)

45. A diode D as shown in the circuit as an iv relation which can be proximated by

2

D D DD

D

v 2v , for v 0i

0, for v 0

iD

DD10V

1

+

The value of vDin the circuit is:

(a) 1 11 V (b) 8V (c) 5V (d) 2V

Soln.

iD

DD10V

1

+

Apply KVL

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D D10 i 1 v

2D D Di v 2v given in questions

2D D D10 1 v 2v v

2

Dv 3D 10 0 v 5, 2 neglect 5 Correct option is (d)

Part-C

46. The Taylor expansion of the function n cosh x , where x is real, about the point x = 0 starts with the

following terms:

(a)2 41 1

x x ..........2 12

(b)2 41 1

x x ..........2 12

(c)2 41 1x x ..........

2 6

(d) 2 41 1

x x ..........

2 6

Soln. Taylor series expansion of f(x) about x = x0is:

20 0

0 0 0 0 0

f " x f "' xf x f x f ' x x x x x x x ...............

2! 3!

Now, in question 0x 0, f x n cosh x

1

f ' x sinh x tanh xcosh x

2f " x sec h x ;

2

f "' x 2sech x sech x. tanh x 2sech x.tanh x 2 2f "" x 2 2sech x sech x.tanh x tanh x 2sech x.sech x = 2 2 44sech x. tanh x 2sech x

Required Taylor series expansion: 2 41 1

n cosh x 0 0 .1.x 0 2 x .................2 4!

2 41 1n cosh x x x ....................

2 12


47. Given a 22 unitary matrix U satisfying U'U UU' 1 with deti

U e

, one can construct a unitary matrix V V 'V VV ' 1 with det V = 1 from it by

(a) Multiplying U by i /2e (b) Multiplying any single element of U by ie

(c) Multiplying any row or column of U by i /2e (d) Multiplying U byie .

Soln. Leti

a bU and det U ad bc e

c d

(Given)

Let,i /2 i /2

i /2

i /2 i /2

e a e bV e U V

e c e d

i i i i idet V e ad e bc e ad bc e e 1

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i /2We have to multiply U with e to get V with determinant 1


48. The value of the integral

3

2

C

z dz

z 5z 6 , where C is a closed contour defined by the equation 2 z 5 0 ,

traversed in the anti-clockwise direction, is:

(a) 16 i (b) 16 i (c) 8 i (d) 2 i

Soln. 3

2

zf z

z 5z 6

Poles:2z 5z 6 0 Given closed contour:

5c : 2 z 5 0 z

2

z 3 z 2 0

So, z = 3 point (3, 0)

centre (0, 0), radius 5/2 units

z = 2 point (2, 0)

Only z = 2 i.e. point (2, 0) is within c

3

z 2

z 2

zRes.f z z 2 8

z 3 z 2

Given integral: I 2 i sum of the residues 16 i


49. A function f x obeys the differential equation

2

2

d f3 2i f 0

dx and satisfies the conditions f 0 1 and

f x 0 as x . The value of f is :

(a)2e (b) 2e (c) 2e (d) 2 ie

Soln. Given differential equation: 2

2

d f3 4i f 0

dx

(where 2 is wrongly printed in place of 4 in the question)

Let, mx

f c.ebe trial solution.

2mx 2 mx

2

df d f c.m e c.m e

dx dx

Substituting in the given equation, we get

2 mx

2 2 2

m 3 4i 0 e 0

m 3 4i 4 1 4i 2 i 2.2.i 2 i 2 i

Therefore, 2 i x 2 i xf x A e B e

Since, f 0 1 A B 1 Since, f x 0 when x A 0, then B 1

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2 i x 2 i 2 if x e f e e e 2f e


50. A planet of mass m moves in the gravitational field of the Sun (mass M). If the semi-major and semi-minor

axes of the orbit are a and b respectively, the angular momentum of the planet is:

(a) 22GMm a b (b) 22GMm a b

(c)

22GMm ab

a b (d)

22GMm ab

a b

Soln. Consider a circular orbit (a = b =r) for which it is easy to calculate angular momentum L

L = mvr;2

2

mv MmG

r r ;

GMv

r

2GM

L mr GMm r r

Only option (d) reduces to this form when a = b = r


51. The Hamiltonian of a simple pendulum consisting of a mass m attached to a massless string of length l is

2

2

pH mg 1 cos

2m

. If L denotes the Lagrangian, the value ofdL

is :dt

(a)2g

p sin

(b)gp sin 2

(c)

gp cos

(d)

2p cos

Soln. 2

2

pH mg 1 cos

2m

; i ii

L p q H p H

2

pH H; p mg sin

p m

2 2

2 2 2

p p pL p mg 1 cos ; L mg 1 cos

m 2m 2m

dL L LL L , p , p

dt p

2 2

p pL L, p mg sin , mg sin ,

p m m

2 2

p pdL 2gmg sin mg sin p sin

dt m m


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52. Which of the following set of phase-space trajectories which one is not possible for a particle obeying Hamiltons

equations of motion (for a time-independent Hamiltonian)?

(a)

x

(b)

x

(c)

x

(d)

x

Soln. According to Louvilles theorem, phase space trajectories cannot cross.


53. Two bodies of equal mass m are connected by a massless rigid rod of length l lying in the xy-plane with the

centre of the rod at the origin. If this system is rotating about the z-axis with a frequency , its angularmomentum is

(a)2m /4 (b) 2m /2 (c) 2m (d) 22m

Soln.m m

O

l/2 l/2

z-axis

L = angular momentum = I, where I is moment of inertia of the rod about an axis z with centre at O.2 2 4 2 2m m m

I m m2 2 4 4 2

2mL I

2


54. An infinite solenoid with its axis of symmetry along the z-direction carries a steady current I.

R

z

The vector potential A

at a distance R from the axis.

(a) Is constant inside and varies as R outside the solenoid.

(b) Varies as R inside and is constant outside the solenoid.

(c) Varies as 1/R inside and as R outside the solenoid.

(d) Varies as R inside and as 1/R outside the solenoid.

Soln. Infinite solenoid carrying a current I. The vector potential A

at a distance R from the axis, then

We know that, S S

A.d A .dS B.dS

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Let radius of solenoid is r.

then inside,SB.dS

=

2B. R & A.d A.2 R

Hence,2A2 R B R

BRA

2

but 0B nI

Hence,0nIRA2

inA R

Outside, 2A.2 R B. r 2

0out

nIrA

2R

out

1A

R

z

R


55. Consider an infinite conducting sheet in the xy-plane with a time dependent current density Kti , where K is a

constant. The vector potential at (x, y, z) is given by

20K A ct z i

4c

The magnetic field B

is:

(a)0Kt j2

(b)

0Kzj2c

(c) 0

K ct z i2c

(d) 0

K ct z j2c

Soln. Vector potential 20k A ct z i

4c

We know that, magnetic field

x

i j k

B Ax y z

A 0 0

x x i 0 0 j 0 A k 0 Az y

20 0k k j ct z ct z 1 j

z 4c 2c

= 0

k ct z j2c


56. When a charged particle emits electromagnetic radiation, the electric field E

and the Poynting vector

0

1S E B

at a large distance r from the emitter vary as n m

1 1and

r rrespectively. Which of the following

choices for n and m are correct?

(a) n 1 and m 1 (b) n 2 and m 2 (c) n 1 and m 2 (d) n 2 and m 4

Soln. For electromagnetic radiation

1 1E and Br r

. Hence, poynting vector 20

E B 1Sr

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Hence, n 1 m 2


57. The energies in the ground state and first excited state of a particle of mass1

m2

in a potential V(x) are 4

and 1, respectively, (in units in which1

). If the corresponding wavefunctions are related by

1 0x x sinh x , then the ground state eigenfunction is

(a) 0 x sec hx (b) 0 x sechx

(c) 20 x sec h x (d) 3

0 x sec h x

Soln. The time independent schrodinger equation 2 2

2

dV x x E x

2m dx

On taking

1

m , 12

2

2

dV x x E x

dx

For ground state, 2

0 02

dV x x 4 x

dx

2

00 02

dV x x 4 x

dx

... (1)

For first excited state, 2

1 12

d V x x xdx

Since, 1 0x x sinh x

2

0 02

dV x x sinh x x sinh x

dx

2

0 0 02

dx sinh x V x x sinh x x sinh x

dx

0

0 0 0

dd

x cosh x sinh x V x sinh x x sinh xdx dx

2

0 0 00 0 02

d d dx sinh x cosh x cosh x sinh x V x x sinh x x sinh x

dx dx dx

2

0 002

d d2cosh x sinh x V x x sinh x 0

dx dx

2

0 002

d d2cosh xV x x

dx sinh x dx

... (2)

Substituting the value of 0V x x from equation (1) into equation (2)

2 2

0 0 002 2

d d d2cosh x4 x

dx sinh x dx dx

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0 0d sinh x

2 xdx cosh x

;

0

0

d 2sinh xdx

x cosh x

0 21

n x 2 n cosh x ncosh x

20n x n sec h x

20 x sec h x Correct option is (c).

58. The perturbation

b a x a x aH '

0 otherwise

acts on a particle of mass m confined in an infinite square well potential

0 a x a

V xotherwise

The first order correction to the ground state energy of the particle is

(a)ba

2(b)

ba

2(c) 2ba (d) ba

Soln. The ground state wavefunction for the given potential 2 x 1 x

x cos cos2a 2a 2aa

The perturbation, b a x a x a

H '0 otherwise

The first order correction in energy,*E ' H dx

a

a

1 x 1 xcos b a x cos dx

2a 2aa a

=

a

2

a

b xcos a x dx

a 2a

a

a

x1 cos

b aa x dx

a 2

=

a a a

a a a||0

b x xa x dx a cos dx x cos dx

2a a a

as integral is odd

a

22

a

xasin

b x baax 2a 0 ab

2a 2 2a

a


59. Let 0 and 1 denote the normalized eigenstates corresponding to the ground and the first excited states of

a one-dimensional harmonic oscillator. The uncertainty x in the state

1

0 1 is :2

(a) x / 2m (b) x / m (c) x 2 / m (d) x / 4m

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22 x x xm 2m 2m


60. What would be the ground state energy of the Hamiltonian

2 22dH x

2m dx

if vibrational principle is used to estimate it with the trial wavefunction 2

xx A e with b as the variational

parameter?

[Hint: 12 n2n 2bx2

1x e dx 2b n

2

]

(a) 2 2m /2 (b) 2 22m / (c) 2 2m / (d) 2 2m /

Soln. The Hamiltonian, 2 2

2dH x

2m dx

The trial wave function 2

xx A e with b variational parameter..

2 2

2 2

2 2* bx bx

2

* bx bx

dA e x A e dx

H 2m dxH

|A e A e dx

2 2 2

2

2 22 bx bx 2bx

2

2 2bx

dA e e dx e x dx

2m dx

A e dx

2 2

2

2bx 2 2 bx

2bx

e 4b x 2b e dx2m

e dx

by using f x x a dx f a

2 2

2

2 22 2bx 2 2bx

2bx

4b e x dx 2b e dx2m 2m

H

e dx

2 2

2

2 2 22bx 2 2bx dx

0 0

2bx

0

4b 2be x dx e

2m 2m 2

e dx

3 122

12

2 2 24b 1 / 2 2b

2m 2 2m 22 2b2b

2 2b

12

12

2 2 2 2bb 2b b2b

2m 2m 2m

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H0

b

1

2

2 2 10

2m 2b

2

2m2 b

2 4

22 b 4m

2 2

4

2 mb

The ground state energy,

122 2 2 2 2

4 4

2 m 2 2 mE

2m

2 2 2

2 2 2

m 2 m m


61. The free energy difference between the superconducting and the normal states of a material is given by

2 4

S NF F F2

, where is an order parameter and and are constants such that 0

in the normal and 0 in the superconducting state, while 0 always. The minimum value of F in thesuperconducting state is

(a) 2 / (b) 2 / 2 (c) 23 / 2 (d) 25 / 2

Soln.2 4

F2

Let x ; 2 4F x x2

3F 2 x 2 x 0x

; 22x x 0

x 0 trivial solution

2 2x 0 x

2 4min

2x

F x x2

=

2

22

=

2 2 21

2 2


62. A given quantity of gas is taken from the state A C reversibly, by two paths, A C directly and

A B C as shown in the figure below..

A

B C

V

P

During the A C the work done by the gas is 100 J and the heat absorbed is 150 J. If during the process

A B C the work done by the gas is 30 J, the heat absorbed is:(a) 20J (b) 80J (c) 220 J (d) 280 J

Soln. 1dQ dU dW , 1dW 100J

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For path AC: 150 J = dU + 100 J

dU = 150 J 100J = 50 J

For path ABC: 2dQ = dU + dW = 50J + 30J = 80J


63. Consider a one-dimensional Ising model with N spins, at very low temperatures when almost all the spins arealigned parallel to each other. There will be a few spin flips with each flip costing an energy 2J. In a configura-

tion with r spin flips, the energy of the system is E = NJ + 2rJ and the number of configuration isNCr; r varies

from 0 to N. The partition function is

(a)

N

B

J

k T

(b) NJ/k TBe (c)

N

B

Jsinh

k T

(d)

N

B

Jcosh

k T

Soln. Let us consider only three energy levels, Er= 2J + 2rJ

E0= 2J, E

1= 0, E

2= 2J

2E E E2 2 2 20 1 2

2 0 1 2 r

r 0

Q C e C e C e / C

2 22J 0 2J J J J Je 2e e e e e e

4 4 2

2

2Q cosh J N

N

N

JQ cosh J cosh

kT


64. A magnetic field sensor based on the Hall effect is to be fabricated by implanting. As into a Si film of thickness

1 m. The specifications require a magnetic field sensitivity of 500 mV/Tesla at an excitation current of 1 mA.The implanation dose is to be adjusted such that the average carrier density, after activation, is

(a) 26 31.25 10 m (b) 22 31.25 10 m (c) 21 34.1 10 m (d) 20 34.1 10 m

Soln. Hall voltage to magnetic field ratio3HV 500 10 V/T

B

3

6

Current, I 1mA 10 Amp

Thickness of sample, t 1 10 m

We know that Hall voltage , H IBVtne

3

6 19 3H

IB 10concentration of dopants, n

t e V 10 1.6 10 500 10

2121 22 3100 10 100 10 1.25 10 / m

5 1.6 8


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65. Band-pass and band-reject filters can be implemented by combining a low pass and a high pass filter in series

and in parallel, respectively. If the cut-off frequencies of the low pass and high pass filters are LP HP0 0and ,

respectively, the condition required to implement the band-pass and band-reject filters are respectively.

(a) HP LP HP LP0 0 0 0and (b)

HP LP HP LP0 0 0 0and

(c)HP LP HP LP0 0 0 0and (d)

HP LP HP LP0 0 0 0and

Soln. Low pass cut off frequency LP0

High pass out off frequency HP0

Low pass response. High pass reponses

0LP 0

HP

For band pass filter responses is: For band reject filter response is

0

HP 0

LP

0HP 0

LP

Documents

solved PH'12