Mathematical Physics JEST-2016 · 2018-06-11 · fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Website:

fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES

Website: www.physicsbyfiziks.com Email: [email protected] 1

Head office fiziks, H.No. 40 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐16 Phone: 011‐26865455/+91‐9871145498

Branch office Anand Institute of Mathematics, 28‐B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi‐16

Mathematical Physics

JEST-2016

Q1. Given the condition 2 0φ∇ = , the solution of the equation 2 .kψ φ φ∇ = ∇ ∇ is given by

(a) 2

2kφψ = (b) 2kψ φ= (c) ln

2kφ φψ = (d) ln

2kφ φψ =

Ans: (a)

Solution: 2 0φ∇ = ( ). 0φ⇒∇ ∇ = ( ) ˆ ˆ ˆ. 0 x y zφ φ α β γ⇒∇ ∇ = ⇒∇ = + + x y zφ α β γ⇒ = + +

( )2 2 2.k kφ φ α β γ∇ ∇ = + +

If ( )2

2

2 2k k x y zφψ α β γ= = + +

( )2 2 2

2 2 2 22 2 2 k

x y zψ ψ ψψ α β γ∂ ∂ ∂

⇒∇ = + + = + +∂ ∂ ∂

2 .kψ φ φ⇒∇ = ∇ ∇

Q2. The mean value of random variable x with probability density

( ) ( )( )2

2

1 .exp2 2

x xp x

μ

σ π σ

⎡ ⎤+⎢ ⎥= −⎢ ⎥⎣ ⎦

is:

(a) 0 (b) 2μ (c)

2μ− (d) σ

Ans. : (a)

Solution: 2

2 21 exp exp 0

2 22x xx x dx x dxμσ σσ π

∞ ∞

−∞ −∞= − − =∫ ∫





Q3. Given a matrix2 11 2

M⎛ ⎞

= ⎜ ⎟⎝ ⎠

, which of the following represents cos6Mπ⎛ ⎞

⎜ ⎟⎝ ⎠

(a) 1 212 12⎛ ⎞⎜ ⎟⎝ ⎠

(b) 1 131 14

−⎛ ⎞⎜ ⎟−⎝ ⎠

(c) 1 131 14⎛ ⎞⎜ ⎟⎝ ⎠

(d) 1 31

2 3 1

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

Ans. : (b)

Solution: We have

2 10

1 2λ

λ−

=−

2 4 3 0 1λ λ λ⇒ − + = ⇒ = or 3λ =

For 1λ =

2 1 1 01 2 1 0

xy

−⎛ ⎞⎛ ⎞ ⎛ ⎞=⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠⎝ ⎠ ⎝ ⎠

gives

Thus 0x y x y+ = ⇒ = − . Taking 1x = , the eigenvector associated with 1λ = is

1

11

x ⎡ ⎤= ⎢ ⎥−⎣ ⎦

For 3λ =

1 1 0

1 1 0x

x yy

−⎛ ⎞⎛ ⎞ ⎛ ⎞= ⇒ =⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠⎝ ⎠ ⎝ ⎠

Taking 1x = , the eigenvectors associated with 3λ = is 2

11

x ⎡ ⎤= ⎢ ⎥⎣ ⎦

Thus

1 1 1 0 1/ 2 1/ 21 1 0 3 1/ 2 1/ 2

M−⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

01 1 1/ 2 1/ 261 1 1/ 2 1/ 26 0

6

ii M

i

ππ

π

⎡ ⎤⎢ ⎥ −⎡ ⎤ ⎡ ⎤

= ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎢ ⎥⎢ ⎥⎣ ⎦





66

6

1 1 0 1/ 2 1/ 21 1 1/ 2 1/ 2

0

ii M

i

ee

e

ππ

π

⎡ ⎤−⎡ ⎤ ⎡ ⎤⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎢ ⎥−⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

6 62 26 6

6 62 2 2 2

1 11 1 2 22 21 1 1 1

2 2 2 2

i ii ii i

i ii i i i

e e e ee e

e e e e e e

π ππ ππ π

π ππ π π π

⎡ ⎤⎡ ⎤ + +⎢ ⎥−−⎢ ⎥ ⎢ ⎥⎡ ⎤ ⎢ ⎥= = ⎢ ⎥⎢ ⎥− ⎢ ⎥⎣ ⎦ ⎢ ⎥+ +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎢ ⎥⎣ ⎦

3 3 34 4 4 4cos sin

6 6 3 3 34 4 4 4

i iM Mi

i iπ π

⎡ ⎤+ − +⎢ ⎥⎛ ⎞ ⎛ ⎞ ⎢ ⎥⇒ + =⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎝ ⎠

− + +⎢ ⎥⎣ ⎦

3 3 34 4 4 4cos sin

36 6 3 34 44 4

i iM Mi

i iπ π

⎡ ⎤ ⎡ ⎤−⎢ ⎥ ⎢ ⎥⎛ ⎞ ⎛ ⎞ ⎢ ⎥⇒ + = + ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

Thus

1 13cos1 16 4

Mπ −⎛ ⎞⎛ ⎞ = ⎜ ⎟⎜ ⎟ −⎝ ⎠ ⎝ ⎠

Q4. The sum of the infinite series 1 1 11 ...3 5 7

− + − + is

(a) 2π (b) π (c) 2π (d)

4π

Ans. : (d)

Solution: The series for 1tan x− for ,1x > is given by

3 5 7

1 1 1 1tan2 3 5 7

x

x x x xπ− = − + − + + ⋅⋅⋅⋅

Putting 1x = , we obtain

1 1 1 1tan 1 12 3 5 7π− ⎛ ⎞= − − + − + ⋅⋅⋅⎜ ⎟

⎝ ⎠1 1 11

4 2 3 5 7π π ⎛ ⎞⇒ = − − + −⎜ ⎟

⎝ ⎠⋅⋅⋅⋅





Q5. A semicircular piece of paper is folded to make a cone with the centre of the semicircle

as the apex. The half-angle of the resulting cone would be:

(a) o90 (b) o60 (c) o45 (d) o30

Ans. : (d)

Solution: When the semicircular piece of paper is folded to make a cone, the circumference of

base is equal to the circumference of the original semicircle. Let r be the radius of the

base of the core and R be the radius of the semicircle.

Hence, 22Rr R rπ π= ⇒ = .

The stay height of the come will also be R .

Hence, / 2 1sin2

RR

α = =

Thus, 030α =

JEST-2015

Q6. Given an analytic function ( ) ( ) ( ). ,f z x y i x y= +φ ψ , where ( ) yyxxyx 24, 22 +−+=φ .

If C is a constant, which of the following relations is true?

(a) ( ) 2, 4x y x y y Cψ = + + (b) ( ), 2 2x y xy x Cψ = − +

(c) ( ). 2 4 2x y xy y x Cψ = + − + (d) ( ) 2, 2x y x y x Cψ = − +

Ans. : (c)

Solution: ( ) 2 2, 4 2u Q x y x x y y= = + − +

From C.R. equation u vx y∂ ∂

=∂ ∂

u vy x∂ ∂

= −∂ ∂

2 4u xx∂

= +∂

R

r

α R





2 4v xy∂

= +∂

( )2 4v xy y f x= + + (i)

2 2u yy∂

= − +∂

2 2v yx∂

= + −∂

( )2 2v xy x f y= + +

( ) ( )2 4 2 2xy y f x xy x f y+ + = − + (ii)

( ) ( )2 , 4f x x f y y= =

2 4 2v xy y x c= + − +

Q7. If two ideal dice are rolled once, what is the probability of getting at least one ‘6’?

(a) 3611 (b)

361 (c)

3610 (d)

365

Ans: (a)

Solution: Number of point in sample space ( ) 11n S =

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )1,6 , 2,6 , 3,6 , 4,6 , 5,6 , 6,1 , 6,2 , 6,3 , 6,4 , 6,5 , 6,6⎡ ⎤⎣ ⎦

Number of point in population ( ) 26 36n P = =

Probability that at least one six on face of dice ( )( )

1136

n Sn P

= =





Q8. What is the maximum number of extrema of the function ( ) ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛+−

= 24

24 xx

k exPxf where

( )∞∞−∈ ,x and ( )xPk is an arbitrary polynomial of degree k ?

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

Ans: (c)

Solution: ( ) ( )4 2

4 2x x

xf x P x e⎛ ⎞

− +⎜ ⎟⎝ ⎠=

( ) ( ) ( ) ( )4 2

2 23x x

x xf x P x P x x x e⎛ ⎞

− +⎜ ⎟⎝ ⎠⎡ ⎤= + − +′ ′⎣ ⎦ ( ) 0f x′⇒ =

( ) ( ) ( )3 0xP x x x P x⎡ ⎤= + − =′⎣ ⎦ is polynomial if order 3k +

From the sign scheme maximum number of extrema 3k= +

Q9. The Bernoulli polynominals ( )sBn are defined by, ( )∑=− !1 nxsB

exe n

nx

xs

. Which one of the

following relations is true?

(a) ( )

( ) ( )∑ +=

−

−

!11

1

nxsB

exe n

nx

sx

(b) ( )

( ) ( ) ( )1

11 1 !

x s nn

nx

xe xB se n

−

= −− +∑

(c) ( )

( )( )∑ −−=−

−

!1

1

1

nxsB

exe n

nnx

sx

(d) ( )

( )( )∑ −=−

−

!1

1

1

nxsB

exe n

nnx

sx

Ans: (d)

Solution: ( )1

xS n

nx

xe xB Se n

=− ∑

Put ( )1S S= − , ( )

( )1

11

x S n

nx

xe xB Se n

−

= −− ∑

( ) ( ) ( )1 1 nnB S B S− = −

( )( ) ( )

1

11

x S nn

nx

xe xB Se n

−

= −− ∑





Q10. Consider the differential equation ( ) ( ) ( )xxkGxG δ=+′ ; where k is a constant. Which

following statements is true?

(a) Both ( )xG and ( )xG′ are continuous at 0=x .

(b) ( )xG is continuous at 0=x but ( )xG′ is not.

(c) ( )xG is discontinuous at 0=x .

(d) The continuity properties of ( )xG and ( )xG′ at 0=x depends on the value of k .

Ans: (c)

Q11. The sum ∑ = ++99

1 11

m mmis equal to

(a) 9 (b) 199 − (c) ( )1991−

(d) 11

Ans: (a)

Solution: 99

1

11m m m= + +

∑

( )99 99

1 1

1 11m m

m m m mm m= =

+ −= + −

+ −∑ ∑

2 1 3 2...... 100 99= − + − + −

100 1 10 1 9= − = − =

JEST-2014

Q12. What are the solutions to ( ) ( ) ( ) 02 =+′−′′ xfxfxf ?

(a) xec x /1 (b) xcxc /21 + (c) 21 cxec x + (d) xx xecec 21 +

Ans.: (d)

Solution: Auxilary equation 0122 =+− DD ( )21 0D⇒ − = 1, 1D⇒ = + +

∵ Roots are equal then ( ) ( )1 2xf x c c x e= + ( ) 1 2

x xf x c e c xe⇒ = +

Q13. The value of ∫2.2

2.0dxxex by using the one-segment trapezoidal rule is close to

(a) 11.672 (b) 11.807 (c) 20.099 (d) 24.119

Ans.: (c)

Solution: 22.02.2 =−=h ( ) ( )2.2 0.2 20.0992hI y y⎡ ⎤⇒ = + =⎣ ⎦ xy xe=∵





Q14. Given the fundamental constants (Planck’s constant), G (universal gravitation

constant) and c (speed of light), which of the following has dimension of length?

(a) 3cG (b) 5c

G (c) 3cG (d)

Gcπ8

Ans.: (a)

Solution: [ ][ ] 21

33

23112

⎥⎦

⎤⎢⎣

⎡−

−−−

TLTLMTML [ ] LL == 2

12

2 1ML T −⎡ ⎤= ⎣ ⎦ , [ ]2312

−−== TLMm

grG

Q15. The Laplace transformation of te t 4sin2− is

(a) 254

42 ++ ss

(b) 204

42 ++ ss

(c) 204

42 ++ ss

s (d) 2042

42 ++ ss

s

Ans.: (b)

Solution: sinatL e bt−⎡ ⎤⎣ ⎦∵( )2 2

bs a b

=+ +

2 sin 4tL e t−⎡ ⎤⇒ ⎣ ⎦ ( )2 2

42 4s

=+ + 2

44 20s s

=+ +

Q16. Let us write down the Lagrangian of a system as ( ) xcxkxxmxxxxL ++= 2,, . What is the

dimension of c ?

(a) 3−MLT (b) 2−MT (c) MT (d) 12 −TML

Ans.: (c)

Solution: According to dimension rule same dimension will be added or subtracted then

dimension of =xMx dimension of Cxx

[ ] [ ]2 2 3ML T C L LT− −⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦

[ ] [ ][ ] [ ]MT

MLTMLC == −

−

32

22





Q17. The Dirac delta function ( )xδ satisfies the relation ( ) ( ) ( )∫∞

∞−= 0fdxxxf δ for a well

behaved function ( )xf . If x has the dimension of momentum then

(a) ( )xδ has the dimension of momentum

(b) ( )xδ has the dimension of ( )2momentum

(c) ( )xδ is dimensionless

(d) ( )xδ has the dimension of ( ) 1momentum −

Ans.: (d)

Solution: ( ) ( ) ( )∫∞

∞−= 0fdxxxf δ

( ) ( ) ( ) ( )[ ] ( ) ( )[ ] ( ) [ ]100 −=⇒=⋅⇒= PxfPxxffdxxxf δδδ

Since, ( )[ ] ( )[ ]0fxf =

If ( ) βα += xxF is force [ ]2−TLM

( ) β=0F is also [ ]2TLM

Q18. The value of limit

11lim 6

10

++

→ zz

iz

is equal to

(a) 1 (b) 0 (c) -10/3 (d) 5/3

Ans.: (d)

Solution: 11lim 6

10

++

→ zz

iz 35

610

610lim

610lim

4

5

9

=⇒⇒⇒→→

zzz

iziz

Q19. The value of integral

∫ −=

cdz

zzIπ2

sin

with c a circle 2=z , is

(a) 0 (b) iπ2 (c) iπ (d) iπ−





Ans.: (c)

Solution: ∫ −=

C zzIπ2

sin pole 2

02 ππ =⇒=−⇒ zz

Residue at 2π

=z 2=z∵ so it will be lies within the contour

( ) 22

2

iz

emg C

eI R iz

ππ

= = ×⎛ ⎞−⎜ ⎟⎝ ⎠

∑∫

Res / 2

2

22 22

2

izi

z

z ee i

z

π

π

π

π=

⎛ ⎞−⎜ ⎟⎝ ⎠= = =

⎛ ⎞−⎜ ⎟⎝ ⎠

(taking imaginary part) ; Residue = 21

Now iiI ππ =×= 221

JEST-2013

Q20. A box contains 100 coins out of which 99 are fair coins and 1 is a double-headed coin.

Suppose you choose a coin at random and toss it 3 times. It turns out that the results of all

3 tosses are heads. What is the probability that the coin you have drawn is the double-

headed one?

(a) 0.99 (b) 0.925 (c) 0.75 (d) 0.01

Ans.: (c)

Q21. Compute ( ) ( )2

22

0

ImRelimz

zzz

+→

(a) The limit does not exist. (b) 1

(c) –i (d) -1

Ans.: (a)

Solution: ( ) ( )2 2 2 2 2 2

2 2 2 2 20 0 00

Re Im 2 2lim lim lim 12 2z z y

x

z z x y xy x y xyz x y ixy x y ixy→ → =

→

+ − + − += ⇒ =

− + − +





2 2

2 200

2lim 12x

y

x y xyx y ixy=

→

− +=

− + and

2 2

2 20

2lim2y x

x

x y xy ix y ixy=

→

− += −

− +

Q22. The vector field jyixz ˆˆ + in cylindrical polar coordinates is

(a) ( ) ( ) φρ φφρφφρ ezez ˆ1cossinˆsincos 22 −++

(b) ( ) ( ) φρ φφρφφρ ezez ˆ1cossinˆsincos 22 +++

(c) ( ) ( ) φρ φφρφφρ ezez ˆ1cossinˆcossin 22 +++

(d) ( ) ( ) φρ φφρφφρ ezez ˆ1cossinˆcossin 22 −++

Ans.: (a)

Solution: jyixzA ˆˆ += , , 0x y zA xz A y A⇒ = = =

( ) ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆˆx y zA A e A x e A y e A z eρ ρ ρ ρ ρ= ⋅ = ⋅ + ⋅ + ⋅

( ) ( )cos cos sin sin 0A zρ ρ φ φ ρ φ φ⇒ = + + ( )2 2 ˆcos sinA z eρ ρρ φ ρ φ⇒ = +

( ) ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆˆx y zA A e A x e A y e A z eφ φ φ φ φ= ⋅ = ⋅ + ⋅ + ⋅

( )cos sin sin cosA zφ ρ φ φ ρ φ φ⇒ = − + ⋅ ( ) ˆcos sin 1A z eφ φρ φ φ⇒ = ⋅ −

( ) ( )2 2ˆ ˆ ˆ ˆ ˆcos sin cos sin 1z zA A e A e A e z e z eρ ρ φ φ ρ φρ φ φ ρ φ φ= + + = + + −

Q23. There are on average 20 buses per hour at a point, but at random times. The probability

that there are no buses in five minutes is closest to

(a) 0.07 (b) 0.60 (c) 0.36 (d) 0.19

Ans.: (d)

Q24. Two drunks start out together at the origin, each having equal probability of making a

step simultaneously to the left or right along the x axis. The probability that they meet

after n steps is

(a) 2!!2

41

nn

n (b) 2!!2

21

nn

n (c) !221 nn (d) !

41 nn

Ans.: (a)





Solution: Into probability of taking ' 'r steps out of N steps

1 12 2r

r N r

CN−

⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

total steps 2N n n n= = + =

for taking probability of n steps out of N

( ) ( )

2 2

21 1 ! 1 1 2 ! 1 2 !2 2 ! ! 2 2 ! ! 2 ! 4n

n N n n n n n

C n

N n nP NN n n n n n

− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Q25. What is the value of the following series?

22

...!5

1!3

11....!4

1!2

11 ⎟⎠⎞

⎜⎝⎛ −+−+⎟

⎠⎞

⎜⎝⎛ −+−

(a) 0 (b) e (c) 2e (d) 1

Ans.: (d)

Solution: 2 4 3 5

cos 1 ..... , sin .....2! 4! 3! 5!θ θ θ θθ θ θ⇒ = − + = − +

11sin1cos!5

1!3

11...!4

1!2

11 2222

=+⇒⎟⎠⎞

⎜⎝⎛ +−+⎟

⎠⎞

⎜⎝⎛ +−⇒ 1cossin 22 =+ θθ∵

Q26. If the distribution function of x is ( ) λ/xxexf −= over the interval 0 < x < ∞, the mean

value of x is

(a) λ (b) 2λ (c) 2λ (d) 0

Ans.: (b)

Solution: ∵ it is distribution function so ( )( )

0

0

.x

x

x xe dxxf x dxx

f x dx xe dx

λ

λ

−∞∞

−∞∞

−∞

−∞

= = ∫∫∫ ∫

2

0

0

2

x

x

x e dx

xe dx

λ

λ

λ

−∞

−∞

⇒ =∫

∫





JEST-2012

Q27. The value of the integral ( )∫

∞

+0 22 1ln dx

xx is

(a) 0 (b) 4π− (c)

2π− (d)

2π

Ans. : (b)

Solution: ( ) ( )2 22 2

0 0

ln ln

1 1

x zdx dzx z

∞ ∞

=+ +

∫ ∫

Let us consider new function ( )2

2

ln1

zf zz

⎛ ⎞= ⎜ ⎟+⎝ ⎠, then

2

20

ln1

zI dzz

∞ ⎛ ⎞= ⎜ ⎟+⎝ ⎠∫

Pole at z i= ± is simple pole of second order.

Residue at z i= is

( ) ( )( ) ( )

22

2 2

ln zd z idz z i z i

= −− +

( )( )

2

2

ln zddz z i

=+

( ) ( ) ( ) ( )

( )

2 2

4

12 ln . ln .2z i z z z iz

z i

+ − +=

+

( ) ( ) ( )

( )

2

3

12ln ln .2z i z zz

z i

+ −=

+

( ) ( )

( )

2

3

12 2 ln ln 2

2

i i ii

i

× − ⋅=

2

4 22 2

8

i i

i

π π⎛ ⎞− ×⎜ ⎟⎝ ⎠=−

2

22

8

i

i

ππ +=

−

2

Res4 16z i

iπ π=

−⇒ = +

Similarly at z i= − ; 2

Res4 16z i

iπ π=−

−= −

2 2 22

20

ln 21 4 16 4 16

zI dz i i i iz

π π π ππ π∞ ⎛ ⎞−⎛ ⎞= = + − − = −⎜ ⎟⎜ ⎟+⎝ ⎠ ⎝ ⎠∫

( ) ( )2

R A B r A B

i f z dz f z dzπ⎛ ⎞ ⎛ ⎞

− = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ∫ ∫ ∫ ∫ ∫ ; vanish

A B∫ ∫

R

rA

B

•

•





Along path A; z x iε= − + and along path B; z x iε= − −

Thus ( ) ( )( )

( )( )

02

2 20

ln ln1 1A B

x i x ii f z dz dx dx

x i x iε ε

πε ε

∞

∞

⎡ ⎤ ⎡ ⎤⎛ ⎞ − + − −− = = − −⎢ ⎥ ⎢ ⎥⎜ ⎟

− + + − − +⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎣ ⎦ ⎣ ⎦∫ ∫ ∫ ∫

( )( )

( )( )

2 2

22 2

0 0

ln ln1 1

x i x ii dx dx

x i x iε ε

πε ε

∞ ∞⎡ ⎤ ⎡ ⎤− + − −⇒ − = −⎢ ⎥ ⎢ ⎥

− + + − − +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦∫ ∫

( ) ( )2 2

22 2

0 0

ln ln; 0

1 1x i x i

i dx dxx x

π ππ ε

∞ ∞+ −⎡ ⎤ ⎡ ⎤⇒ − = − →⎢ ⎥ ⎢ ⎥+ +⎣ ⎦ ⎣ ⎦

∫ ∫

( )( ) ( )( )( ) ( )

2 2

22 22 2

0 0

ln ln ln41 1

x i x i xi dx ix x

π ππ π

∞ ∞+ − −⇒ − = =

+ +∫ ∫

( )2

220

ln4 41

x iix

π ππ

∞ − −⇒ = =

+∫

Q28. If [x] denotes the greatest integer not exceeding x, then [ ]∫∞ −

0dxex x

(a) 1

1−e

(b) 1 (c) e

e 1− (d) 12 −e

e

Ans.: (a)

Solution: [ ]x

[ ] 010 ==<≤ xx , [ ] 121 ==<≤ xx , [ ] 232 ==<≤ xx

now [ ] [ ] [ ] [ ] [ ] dxexdxexdxexdxexdxex xxxxx −−−−−∞

∫∫∫∫∫ +++=4

3

3

2

2

1

1

00

dxedxedxe xxx ∫∫∫ −−− +++⇒4

3

3

2

2

1

.3.2.10

[ ] ( ) ( ) ....32 43

32

21 +−+−+−⇒ −−− xxx eee

+−+−+−+−⇒ −−−−−−−− 54433221 443322 eeeeeeee

∞++−++⇒ −−−− .....4321 eeee





11

1 1

1

−=

−⇒ −

−

eee

22 1 1

1

er e ee

−− + −

−

⎛ ⎞= = =⎜ ⎟

⎝ ⎠∵

Q29. As x → 1, the infinite series .......71

51

31 753 +−+− xxxx

(a) diverges (b) converges to unity

(c) converges to π / 4 (d) none of the above

Ans.: (c)

Solution: .......753

tan753

1 +−+−=− xxxxx4

1tan 1 π=⇒ −

Q30. What is the value of the following series? 22

.....!5

1!3

11....!4

1!2

11 ⎟⎠⎞

⎜⎝⎛ +++−⎟

⎠⎞

⎜⎝⎛ +++

(a) 0 (b) e (c) e2 (d) 1

Ans.: (d)

Solution: −++++=!3

1!2

11132

1e ,

.....!3

1!2

11132

1 −+−=−e

....!4

1!2

112

1cosh11

+++=+

=−ee

( ) ...!5

1!3

112

1sinh11

+++=−

=−ee

i.e 11sin1cos 22 =− hh

Q31. An unbiased die is cast twice. The probability that the positive difference (bigger -

smaller) between the two numbers is 2 is

(a) 1 / 9 (b) 2 / 9 (c) 1 / 6 (d) 1 / 3

Ans.: (a)

Solution: ( ) ( )( )SnEnp =2





The number of ways to come positive difference ( ) ( ) ( ) ( )[ ]4,6,3,5,2,4,1,3

( )91

3642 ==p

Q32. For an N x N matrix consisting of all ones,

(a) all eigenvalues = 1 (b) all eigenvalues = 0

(c) the eigenvalues are 1, 2, …., N (d) one eigenvalue = N, the others = 0

Ans.: (d)

Solution: 1 1

0, 21 1⎡ ⎤

=⎢ ⎥⎣ ⎦

3,0,0111111111

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

so far NN × matrix one eigen value is N and another’s eigen value is zero





Classical Mechanics

JEST-2016

Q1. A hoop of radius a rotates with constant angular velocity ω about

the vertical axis as shown in the figure. A bead of mass m can slide

on the hoop without friction. If 2g aω< at what angle θ apart from

0 and π is the bead stationary (i.e., 2

2 0d ddt dtθ θ= = )?

(a) 2tan ga

πθω

= (b) 2sin ga

θω

=

(c) 2cos ga

θω

= (d) 2tan ga

θπω

=

Ans: (c)

Solution: ( )2 2 2 21 sin cos2

L ma mgaθ θφ θ= + +

0d L Ldt θ θ

∂ ∂⎛ ⎞ ⎛ ⎞− =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠( )2 2 2sin cos sin 0ma ma mgaθ θ θφ θ⇒ − + =

( )2

2 22 0 sin cos sin 0d d ma mga

dt dtθ θ θ θφ θ= = ⇒ − + = ,

φ ω= and 2g aω< then 2cos ga

θω

=

a

θ





Q2. The central force which results in the orbit ( )1 cosr a θ= + for a particle is proportional

to:

(a) r (b) 2r (c) 2r− (d) None o the above

Ans: (c)

Solution: ( ) ( )1 11 cos

1 cosr a u

r aθ

θ= + ⇒ = = =

+

If J is angular momentum and m is mass of particle

( )

2 2

2 21 sin

1 cosJ d u duu fm u dd a

θθθ θ

⎛ ⎞ ⎛ ⎞− + = ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ +⎝ ⎠

( ) ( )

2 2

2 3 2sin cos2

1 cos 1 cosd ud a a

θ θθ θ θ

= − ++ +

( ) ( ) ( )2 2 2 2

2 3 2sin cos 1 12

1 cos1 cos 1 cosJ d u Ju fm m a ud a a

θ θθθ θ θ

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟− + = − − + + =⎜ ⎟ ⎜ ⎟⎜ ⎟+ ⎝ ⎠+ +⎝ ⎠ ⎝ ⎠

( ) ( ) ( )2 2

3 21 cos cos 1 12

1 cos1 cos 1 cosJ fm a ua a

θ θθθ θ

⎛ ⎞− ⎛ ⎞⎜ ⎟− − + + = ⎜ ⎟⎜ ⎟+ ⎝ ⎠+ +⎝ ⎠

Put ( )

11 cos

ua θ

=+

1,cos1au

θ =−

and solving

21f uu

⎛ ⎞ ∝⎜ ⎟⎝ ⎠

so ( ) 2f r r−∝





Q3. Light takes approximately 8 minutes to travel from the Sun to the Earth. Suppose in the

frame of the Sun an event occurs at 0t = at the Sun and another event occurs on Earth at

1t = minute. The velocity of the inertial frame in which both these events are

simultaneous is:

(a) 8c with the velocity vector pointing from Earth to Sun

(b) 8c with the velocity vector pointing from Sun to Earth : -

(c) The events can never be simultaneous - no such frame exists

(d) 211

8c ⎛ ⎞− ⎜ ⎟

⎝ ⎠ with velocity vector Pointing from to Earth

Ans: (a)

Solution: ' '2 1 8 60x x c− = × ×

' '2 1 60t t− =

( )' '

' '2 12 12 2 ' ' ' '

2 1 2 1 2 122 2

2 2

0 0 0

1 1

vx vxt t vc ct t t t x xcv v

c c

+ +− = ⇒ − = ⇒ − + − =

− −

( )' ' ' '2 1 2 12 260 8 60

8v v ct t x x c vc c

− + − ⇒ + × × ⇒ = −

Negative sign indicate frame is moving with the velocity 8c vector pointing from Earth to

Sun.





Q4. For the coupled system shown in the figure, the normal coordinates are 1 2x x+ and

1 2x x− corresponding to the normal frequencies 0ω and 03ω respectively.

At 0t = , the displacements are 1x A= , 2 0x = , and the velocities are 1 2 0v v= = . The

displacement of the second particle at time t is given by:

(a) ( ) ( ) ( )( )2 0 0cos cos 32Ax t t tω ω= + (b) ( ) ( ) ( )( )2 0 0cos cos 3

2Ax t t tω ω= −

(c) ( ) ( ) ( )( )2 0 0sin sin 32Ax t t tω ω= − (d) ( ) ( ) ( )2 0 0

1sin sin 32 3Ax t t tω ω⎛ ⎞

= −⎜ ⎟⎝ ⎠

Ans: (b)

Solution: Using boundary condition at 0t = 2 0x = and 2 0v =

Only ( ) ( ) ( )( )2 0 0cos cos 32Ax t t tω ω= − will satisfied

Q5. A cylindrical shell of mass in has an outer radius b and an inner radius a . The moment

of inertia of the shell about the axis of the cylinder is:

(a) ( )2 212

m b a− (b) ( )2 212

m b a+ (c) ( )2 2m b a+ (d) ( )2 2m b a−

Ans: (b)

( ) ( )2 2 2 22 2

22

b b

a a

m mx dm x xdx b ab a

ππ

= ⇒ +∫ ∫−

k 1xk

2xk

m m





JEST-2015

Q6. The distance of a star from the Earth is 25.4 light years, as measured from the Earth. A

space ship travels from Earth to the star at a constant velocity in 25.4 years, according to

the clock on the ship. The speed of the space ship in units of the speed of light is,

(a) 21 (b)

21 (c)

32 (d)

31

Ans: (b)

Solution: Proper life-time 04.25 4.25,t t

c vΔ = Δ =

0

2 2

121 /

tt v c

v c

ΔΔ = ⇒ =

−

Q7. A classical particle with total energy E moves under the influence of a potential

( ) 3223 223, yxyyxxyxV +++= . The average potential energy, calculated over a long

time is equal to,

(a)3

2E (b) 3E (c)

5E (d)

52E

Ans: None of the above is correct.

Solution: If one will use virial theorem then 2nT V= if nV r∝ according to problem 3n =

So E T V= + 23

E V V= + 25

V E=

But virial theorem is used only for conservative forces.

Force conservative 0F∇× = where F V= −∇

( ) 3 2 2 3, 3 2 2V x y x x y y x y= + + +∵ ( ) ( )2 2 2 2ˆ ˆ9 2 2 2 4 3V x xy y i x yx y j⇒∇ = + + + + +

0F⇒∇× ≠ i.e. non conservative in nature.

So we cannot use viral theorem. Therefore, none of the answer is correct





Q8, A chain of mass M and length L is suspended vertically with its lower end touching a

weighing scale. The chain is released and falls freely onto the scale. Neglecting the size

of the individual links, what is the reading of the scale when a length x of the chain has

fallen?

(a) L

Mgx (b) L

Mgx2 (c) L

Mgx3 (d) L

Mgx4

Ans: (c)

Solution: Reading of scale = impulse + actual weight ( )d mvdp Mgx Mgxdt L dt L

Δ= + = +

2 2 3M dx Mgx Mv Mgx Mgx Mgx MgxvL dt L L L L L L⎛ ⎞⇒ + = + = + =⎜ ⎟⎝ ⎠

2 2v gx=∵ and Mm dxL

Δ =

Q9. A bike stuntman rides inside a well of frictionless surface given by ( )22 yxaz += , under

the action of gravity acting in the negative z direction. zgg ˆ−= What speed should he

maintain to be able to ride at a constant height 0z without falling down?

(a) 0gz

(b) 03gz

(c) 02gz

(d) The biker will not be able to maintain a constant height, irrespective of speed.

Ans: (c)

Solution: ( )2 2z a x y= +

Using equation of constrain, we must solve the given system in cylindrical co-ordinate. 2z ar= 2z arr=

( )2 2 212

L m r r z mgz= + θ + −

( ) ( )( )2 2 2 2 2 2 2 2 2 2 21 14 1 42 2

L m r r a r r mgar m r a r r⇒ = + θ + − = + + θ





Equation of motion

0d L Ldt r r

∂ ∂⎛ ⎞ − =⎜ ⎟⎝ ⎠∂ ∂

( )2 2 2 21 4 2 0mr ra r mr arr mr mgar+ + − θ + =

At 0 0, 0,z z r r r= = =

So 20 02mr mgar− θ =

2 2 2ga gaθ = ⇒ θ = , 0

2v gar= , ( )2

0 0 02v ga r z ar= ⋅ =

1/ 20

02 2z

ga v gza

⎛ ⎞= ⋅ ⇒ =⎜ ⎟⎝ ⎠

Q10. The Lagrangian of a particle is given by qqqL −= 2 . Which of the following statements

is true?

(a) This is a free particle

(b) The particle is experiencing velocity dependent damping

(c) The particle is executing simple harmonic motion

(d) The particle is under constant acceleration.

Ans: (a)

Solution: 2L q qq= −∵ 2L q qq∂

⇒ = −∂

2d L q qdt q⎛ ⎞∂

⇒ = −⎜ ⎟∂⎝ ⎠

0d L Ldt q q⎛ ⎞∂ ∂

− =⎜ ⎟∂ ∂⎝ ⎠∵

2 0q q q⇒ − + = 2 0q⇒ =2

2 0d qdt

⇒ =dq Cdt

⇒ = q Ct α⇒ = +

Q11. How is your weight affected if the Earth suddenly doubles in radius, mass remaining the

same?

(a) Increases by a factor of 4 (b) Increases by a factor of 2

(c) Decreases by a factor of 4 (d) Decreases by a factor of 2

Ans: (a)





Solution: 2

GMW mR

= ⋅ and ( )22GMW m

R′ = ⋅

4WW ′⇒ =

Q12. A spring of force constant k is stretched by x . It takes twice as much work to stretch a

second spring by2x . The force constant of the second spring is,

(a) k (b) k2 (c) k4 (d) k8

Ans: (d)

Solution:

The relation between energy and maximum displacement is 21

12

E k A=

For A x= ; 21 2

12

E k x= and For 2xA = ;

22

1 2 21 12 2 8

xE k k x⎛ ⎞= =⎜ ⎟⎝ ⎠

2 12E E=∵ 2 22 1 2 1

1 12 88 2

k x k x k k∴ = × ⇒ = 2 8k k⇒ =

JEST-2014

Q13. A dynamical system with two generalized coordinates 1q and 2q has Lagrangian 22

21 qqL += . If 1p and 2p are the corresponding generalized momenta, the Hamiltonian

is given by

(a) ( ) 4/22

21 pp + (b) ( ) 4/2

221 qq + (c) ( ) 2/2

221 pp + (d) ( ) 4/2211 qpqp +

Ans.: (a)

Solution: i iH q p L= −∑ Lpqpq −+= 2211

2

2 1111

1

pqqpqL

=⇒==∂∂ and

22 2

2222

pqqpqL

=⇒==∂∂

4422

22

21

22

11 ppppppH −−⋅+⋅=

( )2 21 2

4

p pH

+⇒ =





Q14. In a certain inertial frame two light pulses are emitted, a distance 5 km apart and

separated by sμ5 . An observer who is traveling, parallel to the line joining the points

where the pulses are emitted, at a velocity V with respect to this frame notes that the

pulses are simultaneous. Therefore V is

(a) c7.0 (b) c8.0 (c) c3.0 (d) c9.0

Ans.: (c)

Solution: 3 62 1 2 15 10 , 5 10 secx x m t t −′ ′ ′ ′− = × − = ×

( ) ( )2 2 1 1 2 1 2 12 2 2

2 1 2 2 2

2 2 21 1 1

v v vt x t x t t x xc c ct t

v v vc c c

− −⎛ ⎞ ⎛ ⎞ ⎡ ⎤′ ′ ′ ′ ′ ′ ′ ′+ + − − −⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎣ ⎦− = − =

− − −

2 1t t=∵ 6 325 10 5 10 0v

c−⇒ × − × = 0.3v c⇒ =

Q15. A double pendulum consists of two equal masses m suspended by two strings of length

l . What is the Lagrangian of this system for oscillations in a plane? Assume the angles

21 , θθ made by the two strings are small (you can use 2/1cos 2θθ −= ).

Note: lg /0 =ω .

(a) ⎟⎠⎞

⎜⎝⎛ −−+≈ 2

220

21

20

22

21

2

21

21 θωθωθθmlL

(b) ⎟⎠⎞

⎜⎝⎛ −−++≈ 2

220

21

2021

22

21

2

21

21 θωθωθθθθmlL

(c) ⎟⎠⎞

⎜⎝⎛ −−−+≈ 2

220

21

2021

22

21

2

21


(d) ⎟⎠⎞

⎜⎝⎛ −−++≈ 2

220

21

2021

22

21

2

21






Ans.: (b)

Solution: 11 sinθlx = , 11 cosθly =

2 1 2 2 1 2sin cosx x l y y lθ θ= + = +

2 1 2 2 1 2sin sin , cos cosx l l y l lθ θ θ θ= + = +

2 1 1 2 2 2 1 1 2 2cos cos , sin sinx l l y l lθ θ θ θ θ θ θ θ= + = − −

21

21

22211

2222

22211

2222

22 sincoscos2coscos θθθθθθθθθθ llllyx +++=+

212122

222

2 sinsin2sin θθθθθθ ll ++

( )2 2 2 2 2 2 22 2 1 2 1 2 1 22 cosx y l l lθ θ θ θ θ θ⇒ + = + + − also 2 2 2 2

1 1 1x y l θ+ =

VTL −= ( ) 2122

22

21

212

1 mgymgyyxyxm −−+++=

( )( )2 2 2 2 2 2 21 1 2 1 2 1 2 1 2

1 2 cos 2 cos cos2

L m l l l l mgl mglθ θ θ θ θ θ θ θ θ⇒ = + + + − + +

2 22 2 2 1 2

1 2 1 21 2 11 12 2 2 2 2

g gL mll l

θ θθ θ θ θ⎡ ⎤⎡ ⎤ ⎡ ⎤

⇒ = + + + − + −⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

( )1 2cos 1θ θ− ≈∵

2 22 2 2 1 2

1 2 1 212 2 2 2 2

g g g gL mll l l l

θ θθ θ θ θ⎡ ⎤

⇒ = + + + − + −⎢ ⎥⎣ ⎦

comparing given options, option (b) is correct i.e.

2 2

2 2 2 20 11 2 1 2 0 2

1 12 2 4

L mlω θ

θ θ θ θ ω θ⎛ ⎞

= + + − −⎜ ⎟⎝ ⎠

Q16. A monochromatic wave propagates in a direction making an angle o60 with the x -axis

in the reference frame of source. The source moves at speed 54cv = towards the

observer. The direction of the (cosine of angle) wave as seen by the observer is

(a) 1413cos =′θ (b)

143cos =′θ (c)

613cos =′θ (d)

21cos =′θ

Ans.: (a)

Solution: 45cv = , ocos 60

2xcu c′ = = , o 3sin 60

2yu c c′ = =

1θ l

l

m

2θ





Now

2

42 5

412 5

x

c cu

c cc

+=

+ ⋅ 13

14c

=13cos14

θ⇒ =

Q17. The acceleration experienced by the bob of a simple pendulum is

(a) maximum at the extreme positions

(b) maximum at the lowest (central) positions

(c) maximum at a point between the above two positions

(d) same at all positions

Ans.: (a)

Solution: maT =θsin , mgT =θcos

θtanga = at o90=θ

a is maximum at extreme position.

Q18. Consider a Hamiltonian system with a potential energy function given by ( ) 42 xxxV −= .

Which of the following is correct?

(a) The system has one stable point (b) The system has two stable points

(c) The system has three stable points (d) The system has four stable points

Ans.: (a)

Solution: ( ) 42 xxxV −=

042 3 =−=∂∂ xx

xV [ ] 0212 2 =−⇒ xx

0,2

1±=x

2 22

2 212

12 12 2 12 4 02x

V Vxdx dx

=±

∂ ∂= − ⇒ = − × = − <

For stable point 0=∂∂

xV and 0

2

>∂∂

xV

020

2

2

>=∂∂

=xxV

θ l T θcosT

mgθsinT





Q19. Two point objects A and B have masses 1000 Kg and 3000 Kg respectively. They are

initially at rest with a separation equal to 1 m. Their mutual gravitational attraction then

draws them together. How far from A’s original position will they collide?

(a) 1/3 m (b) 1/2 m (c) 2/3 m (d) 3/4 m

Ans.: (d)

Solution: Since gravitational force is conservative, therefore they collide at their centre of mass

( ) 21 1 mxxm −=

( )4313 =⇒−= xxx

JEST-2013

Q20. In an observer’s rest frame, a particle is moving towards the observer with an energy E

and momentum P . If c denotes the velocity of light in vacuum, the energy of the

particle in another frame moving in the same direction as particle with a constant velocity

v is

(a) ( )( )2/1 cv

vpE

−

+ (b) ( )( )2/1 cv

vpE

−

− (c) ( )( )[ ]22/1 cv

vpE

−

+ (d) ( )( )[ ]22/1 cv

vpE

−

−

Ans.: (a)

Solution: 2 2

2 2 2

2 2 2

,1 1 1

vx x v vt x x xxc c c ct x x ct x ctcv v v

c c c

+ + +′′ ′ ′ ′= ⇒ = ⇒ = = =

− − −

∵

Now 2

2 2

2 2

, ,1 1

EE v E E Pvcx E x E E E mc E Pc P Ecv v

c c

+ +′ ′ ′ ′= = ⇒ = ⇒ = = ⇒ = ⇒ =

− −

Q21. The free fall time of a test mass on an object of mass M from a height 2R to R is

(a) ( )GMR3

12/ +π (b) GMR3

(c) ( )GMR3

2/π (d) GM

R32π

Ans.: (a)

1mA

x x−1 2mB

m1





Solution: Equation of motion AGMrA

dtrd

rGM

dtrd

rGMm

dtrmd

=−=⇒−=⇒−= ∵22

2

22

2

22

2

CrAv

rA

dtdv

dtd

dtdr

rA

dtdvv +=⇒⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⇒−=

22

22

2

when 0,2 == vRr

20 2 2 2 22 2 2 2 2 2 2

A A v A A A A dr A R rC C vR R r R r R dt rR

−= + ⇒ = − ⇒ = − ⇒ = − ⇒ =

∫ ∫−=−

R

R

tdt

RAdr

rRr

2 02

put ududrur 2,2 == when RuRuRrRr ==== ,2,,2

∫∫ ∫−

=−⇒−=×−

R

R

R

R

tdu

uRut

RAdt

RAudu

uRu

2 2

2

2 02 222

2

R

RRuRuRut

RA

2

12

2sin

222

22 ⎥

⎦

⎤⎢⎣

⎡+−−=−⇒ −

1 12 2 22 2 sin 2 2 sin2 2 22 2

A R R R R Rt R R R R RR R R

− −⎡ ⎤−⇒ − = − + + − −⎢ ⎥

⎣ ⎦

GMAGMRt

ARRtRRRt

RA

=⎟⎠⎞

⎜⎝⎛ +=⇒⎟

⎠⎞

⎜⎝⎛ +=⇒⎥⎦

⎤⎢⎣⎡ −+−

=−⇒ ∵3

12

12242

2 ππππ

Q22. Under a Galilean transformation, the coordinates and momenta of any particle or system

transform as: tt =' , tvrr +=' and vmpp +=' where v is the velocity of the boosted

frame with respect to the original frame. A unitary operator carrying out these

transformations for a system having total mass M , total momentum P and centre of

mass coordinate X is

(a) /./. PvtiXvMi ee (b) ( )2//./. 2 tvMiPvtiXvMi eee −−

(c) ( )2//./. 2 tvMiPvtiXvMi eee (d) ( )2//. 2 tvMiPvti ee −

Ans.: (b)





Q23. A spherical planet of radius R has a uniform density ρ and does not rotate. If the planet

is made up of some liquid, the pressure at point r from the center is

(a) ( )222

34 rRG

−πρ (b) ( )22

34 rRG

−πρ

(c) ( )222

32 rRG

−πρ (d) ( )22

2rRG

−ρ

Ans.: (c)

Solution: Pressure 2

32

2 4

4

4 rRrdrGMr

dpr

gdmdpA

gdmdpπ

πρ

π

⋅=⇒

⋅=⇒

⋅=

Grdrdpr

RrRdrGr

dp 22

332

34

43

44ρπ

π

πρπρ=⇒

⋅⋅⋅=⇒

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⇒⎟⎟

⎠

⎞⎜⎜⎝

⎛=⇒=∫ ∫ 223

423

43

4 222

222 rRGprGpGrdrdp

R

r

R

rρπρπρπ

( ) ( )222222

32

234 rRGprRGp −=⇒−=⇒ ρπρπ

Q24. A particle of mass m is thrown upward with velocity v and there is retarding air

resistance proportional to the square of the velocity with proportionality constant k . If

the particle attains a maximum height after time t , and g is the gravitational

acceleration, what is the velocity?

(a) ⎟⎟⎠

⎞⎜⎜⎝

⎛t

kg

gk tan (b) ⎟⎟

⎠

⎞⎜⎜⎝

⎛t

kggk tan

(c) ( )tgkkg tan (d) ( )tgkgk tan

Ans.: (c)

rR

( )dm massof elementary part

dr





Solution: Equation of motion 2 2

2

mdv dv k dvmg kv g v dtkdt dt m g vm

= + ⇒ = + ⇒ =+

t

kgmv

kgmk

mdtv

kgm

mk

dvdtv

mkg

dv=×⇒=

⎟⎠⎞

⎜⎝⎛ +

⇒=+

⇒ −∫∫∫∫ 1

22tan1

Q25. Consider a uniform distribution of particles with volume density n in a box. The particles

have an isotropic velocity distribution with constant magnitude v . The rate at which the

particles will be emitted from a hole of area A on one side of this box is

(a) nvA (b) 2Anv (c)

4Anv (d) none of the above

Ans.: (c)

Q26. If, in a Kepler potential, the pericentre distance of particle in a parabolic orbit is pr while

the radius of the circular orbit with the same angular momentum is cr , then

(a) rc = 2rp (b) rc = rp (c) 2rc = rp (d) pc rr 2=

Ans.: (a)

Solution: Ionic equation θcos1 erl

+= for parabola 1=e for circle, 0=e , 0θ =

1 , 1, 2 , 2p C p Cp C

l le l r l r r rr r= + = = = ⇒ =

Q27. A K meson (with a rest mass of 494 MeV) at rest decays into a muon (with a rest mass of

106 MeV) and a neutrino. The energy of the neutrino, which can be massless, is

approximately

(a) 120 MeV (b) 236 MeV (c) 300 MeV (d) 388 MeV

Ans.: (b)

Solution: ( )

22 2 2 2 2 2 2

2

494 494 106 106

, 4942 2

k

k

cm m c c c c ck Em

c

μνμ ν

⎛ ⎞× − ×⎜ ⎟− ⎝ ⎠→ + = ⇒×

2366275.235988

11236244036 ≈=−⇒ MeV





Q28. A light beam is propagating through a block of glass with index of refraction n . If the

glass is moving at constant velocity v in the same direction as the beam, the velocity of

the light in the glass block as measured by an observer in the laboratory is approximately

(a) ⎟⎠⎞

⎜⎝⎛ −+= 2

11n

vncu (b) ⎟

⎠⎞

⎜⎝⎛ −−= 2

11n

vncu

(c) ⎟⎠⎞

⎜⎝⎛ ++= 2

11n

vncu (d)

ncu =

Ans.: (a)

Solution: now 1

2

11

−

⎟⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛ +=

⋅⋅

+

+=

cnv

ncv

nccv

ncv

u2

2 21c v vvn cn c n

⎛ ⎞⎛ ⎞= + − +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

3

2

222

32

cncv

cnv

nc

ncv

cnvv +−++−⇒ 2

11cu vn n

⎛ ⎞⇒ = + −⎜ ⎟⎝ ⎠

Q29. The period of a simple pendulum inside a stationary lift is T . If the lift accelerates

downwards with an acceleration 4g , the period of the pendulum will be

(a) T (b) T / 4 (c) 3/2T (d) 5/2T

Ans.: (c)

Solution: ⇒=glT π2 lift accelerates down wards then

4

22gg

lTgg

lT−

=⇒′−

= ππgl

gl

322

342 ×⇒= ππ

32TT =′





Q30. The velocity of a particle at which the kinetic energy is equal to its rest energy is (in

terms of c , the speed of light in vacuum)

(a) 2/3c (b) 4/3c (c) c5/3 (d) 2/c

Ans.: (a)

Solution: 20

2. cmmcEK −= , rest mass energy 20cm=

=..EK rest mass energy 2

02

02 cmcmmc =−

2 202mc m c=

2

1

12

1 2

2

20

2

2

2

0 =

−

⇒=

−cv

cmc

cv

m114 2

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛−⇒

cv 2

2

34 32

v v cc

⇒ = ⇒ =

Q31. If the Poisson bracket , 1x p = − , then the Poisson bracket 2 ,x p p+ is ?

(a) 2x− (b) 2x (c) 1 (d) 1−

Ans.: (a)

Solution: pppxppx ,,, 22 +=+ 0,, ++⇒ xpxpxx ( ) ( )1 1 2x x x⇒ − + − ⇒ −

Q32. The coordinate transformation

yxyyxx 8.06.0,6.08.0 −=′+=′

represents

(a) a translation (b) a proper rotation

(c) a reflection (d) none of the above

Ans.: (b)





Q33. A small mass M hangs from a thin string and can swing like a pendulum. It is attached

above the window of a car. When the car is at rest, the string hangs vertically. The angle

made by the string with the vertical when the car has a constant acceleration 21.2 /a m s=

is approximately

(a) 01 (b) 07 (c) 015 (d) 090

Ans.: (b)

Solution: sinT maθ = , mgT =θcos , ga

=θtan 1 1 0 01.2tan tan 6.98 79.8

ag

θ − − ⎛ ⎞⇒ = = = ≈⎜ ⎟⎝ ⎠

JEST-2012

Q34. For small angular displacement (i.e., sinθ ≈ θ), a simple pendulum oscillates

harmonically. For larger displacements, the motion

(a) becomes a periodic

(b) remains periodic with the same period

(c) remains periodic with a higher period

(d) remains periodic with a lower period

Ans. : (c)

Q35. A planet orbits a massive star in a highly elliptical orbit, i.e., the total orbital energy E is

close to zero. The initial distance of closest approach is R0. Energy is dissipated through

tidal motions until the orbit is circularized with a final radius of Rf. Assume that orbital

angular momentum is conserved during the circularization process. then

(a) Rf = R0/2 (b) Rf = R0 (c) 02RR f = (d) Rf = 2R0

Ans. : (d)

Solution: For elliptically motion r

GMmmrJrmE −+= 2

22

221

0=E and closest approach is 0R at 00 =⇒ rR

020

2

200

RGMm

mRJ

−+=





020

2

2 RGMm

MRJ

=

022 2 RGMmJ =

from condition of circular orbit

( )rVrf

mRJ

f ∂∂

−==3

2

2

3 2f f

J GMmmR R

=

20

3 2

2

f f

GMm R GMmmR R

=

02fR R=

Q36. A binary system consists of two stars of equal mass m orbiting each other in a circular

orbit under the influence of gravitational forces. The period of the orbit is т. At t = 0, the

motion is stopped and the stars are allowed to fall towards each other. After what time t,

expressed in terms of т, do they collide?

∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

−−

ααα

α

xxxx

dxx 12

2

2

sin22

(a) τ2 (b) 2τ (c)

22τ (d)

24τ

Ans. : (d)

Solution: 22

2

xGMm

dtxdM −=

22

2

xGM

dtxd

−=

22

2

xA

dtxd

−=

dtdx

xA

dtdvv 2

−=





⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛xA

dtdv

dtd

2

2

CxAv+=

2

2

where Rx = 0=v

RA

xAv−=

2

2

RxAv 112 −=

xxR

RA

dtdx −

=2

dtRAdx

xRx t

R∫∫ =

− 0

0 2

put 2ux =

ududx 2= , 0,0 == ux

RuRx == ,

∫∫ =−

t

R

dtRAdu

uRu

0

0

2

2 22

tRA

RuRuRu

R

2sin22

20

12 =⎥⎦

⎤⎢⎣

⎡+−− −

tRA

RRRRRR 2sin

222 1 =⎥

⎦

⎤⎢⎣

⎡+−+ −

tRAR 21sin

22 1 =× −

2222 π

××=R

RAt





ARRt

22×=

π

GMRt

23

221 π

= (1)

and

2

2

RGMm

Rmv

=

RGMv =2

τπRv 2

=

RGMR

=2

224τπ

2324 τπ=

GMR

22

2323 τππτ =⇒=GMR

GMR

242221 ττ

==t

Q37. In a certain intertial frame two light pulses are emitted at point 5 km apart and separated

in time by 5 μs. An observer moving at a speed V along the line joining these points

notes that the pulses are simultaneous. Therefore V is

(a) 0.7c (b) 0.8c (c) 0.3c (d) 0.9c

Ans. : (c)

Solution: 0=Δt , stt μ512 =′−′ , kmxx 512 =′−′ Vv −=

2

2

121

2

2

222

12

11CV

xC

Vt

CV

xC

Vttt

−

′⎟⎠⎞

⎜⎝⎛ −+′

−

−

′⎟⎠⎞

⎜⎝⎛ −+′

=−





( ) ( )0

1 2

2

12212

=

−

⎥⎦⎤

⎢⎣⎡ ′−′−′−′

⇒

CV

xxCVtt

0105105 32

6 =××−×⇒ −

CV

93

6

2 10105105 −

−

=××

=⇒CV CCV 3.010103 98 =×××=⇒ −

Q38. A jet of gas consists of molecules of mass m, speed v and number density n all moving

co-linearly. This jet hits a wall at an angle θ to the normal. The pressure exerted on the

wall by the jet assuming elastic collision will be

(a) θ22 cos2mnvp = (b) θcos2 2mnvp =

(c) ( ) θ2cos2/3 mnvp = (d) p = mnv2

Ans.: (a)

Solution: change in momentum along −y direction will be cancelled out

∵change in momentum along −x direction

θcos2mvp =Δ

LApv

vLA

ptA

pAtp

.cos

cos.Area

ForcePressure θ

θ

Δ=

⋅

Δ=

ΔΔ

=ΔΔ

==

vNvmvp θθ coscos2Pressure ⋅

=′ , ( ), AreaNn V L A LV

⎛ ⎞= = × = ×⎜ ⎟⎝ ⎠

∵ ,

θ22 cos2mnvp =′

Q39. If the coordinate q and the momentum p from a canonical pair (q, p), which one of the

sets given below also forms a canonical?

(a) (q - p) (b) (q2 , p2) (c) (p, - q) (d) (q2, - p2)

Ans.: (c)

Solution: for canonical pair ( )qp −,

( ) ( ) ( )qq

pp

pq

qp

∂−∂

⋅∂∂

−∂−∂

⋅∂∂

= ( ) 110 =−−=





Q40. A girl measures the period of a simple pendulum inside a stationary lift and finds it to be

T seconds. If the lift accelerates upward with an acceleration g / 4, then the pendulum

will be

(a) T (b) T / 4 (c) 52T (d) 52T

Ans.: (c)

Solution: glT π2=

Since lift accelerated upward then

gglT

′+=′ π2

4

2gg

lT+

=′ π

45

2 ×=′glT π

522 ×=′

glT π

52TT =′⇒





Electromagnetic Theory

JEST-2016

Q1. The maximum relativistic kinetic energy of β particles from a radioactive nucleus is

equal to the rest mass energy of the particle. A magnetic field is applied perpendicular to

the beam of β particles, which bends it to a circle of radius R . The field is given by:

(a) 03m ceR

(b) 02m ceR

(c) 03m ceR

(d) 032

m ceR

Ans: (c)

Solution: 2 2 2max 0 0KE mc m c m c= − = 02m m⇒ =

0 002 2

2 2

322

1 1

m mm m v cv vc c

= ⇒ = ⇒ =

− −

∵

mvReB

=∵ 0 02 332

m m cmvB ceR eR eR

⇒ = = =

Q2. The strength of magnetic field at the center of a regular hexagon with sides of length a

carrying a steady current I is:

(a) 0

3Ia

μπ

(b) 06 Iaμ

π (c) 03 I

aμπ

(d) 03 Iaμ

π

Ans14: (d)

0 3cos302

d a a= =

( )02 1sin sin

4IBd

μ θ θπ

= −∵

0 00 0 01 2sin 30 2sin 30

4 3 2 342

I I IBd aa

μ μ μπ ππ

⇒ = = =

0 0 01

3 36 62 3 3

I I IB Baa a

μ μ μππ π

⇒ = = × = =

a

C

I

060I

a d

C





Q3. A spherical shell of radius R carries a constant surface charge density σ and is rotating

about one of its diameters with an angular velocityω . The magnitude of the magnetic

moment of the shell is:

(a) 44 Rπσω (b) 44

3Rπσω (c)

4415

Rπσω (d) 44

9Rπσω

Ans : (b)

Solution: The total charge on the shaded ring is

(2 sin )dq R Rdσ π θ θ=

Time for one revolution is 2dt πω

=

⇒Current in the ring 2 sindqI R ddt

σω θ θ= =

Area of the ring 2(R sin )= π θ , so the magnetic moment of the

ring is

2 2 2( sin ) sindm R d Rσω θ θ π θ= ×

4 3 40

4sin3

m R d Rπσω θ θ π σω= = ×∫ 44 ˆ3

m R zπ σω⇒ =

Q4. The electric field ( )0 0ˆ ˆsin 2 sin2

E E t kz x E t kz yπω ω⎛ ⎞= − + − +⎜ ⎟⎝ ⎠

represents:

(a) a linwearly polarized wave

(b) a right-hand circularly polarized wave

(c) a left-hand circularly polarized wave

(d) an elliptically polarized wave

Ans: (d)

zω θsinR

θRd

θdθ

R





Q5. Suppose yz plane forms the boundary between two linear dielectric media I and II

with dielectric constant 3I∈ = and 4II∈ = , respectively. If the electric field in region I at

the interface is given by ˆ ˆ ˆ4 3 5IE x y z= + + , then the electric field IIE at the interface in

region II is:

(a) ˆ ˆ ˆ4 3 5x y z+ + (b) ˆ ˆ ˆ4 0.75 1.25x y z+ −

(c) ˆ ˆ ˆ3 3 5x y z− + + (d) ˆ ˆ ˆ3 3 5x y z+ + Ans: (d)

Solution: ˆ ˆ3 5I II IIE E E y z= ⇒ = +∵ and 3 ˆ ˆ4 34

II I III I

I II II

E E E x xE

ε εε ε

⊥⊥ ⊥

⊥ = ⇒ = = =

ˆ ˆ ˆ3 3 5IIE x y z⇒ = + +

Q6. How much force does light from a 1.8 W laser exert when it is totally absorbed by an

object?

(a) 96.0 10 N−× (b) 90.6 10 N−× (c) 80.6 10 N−× (d) 94.8 10 N−×

Ans: (a)

Solution: Radiation Pressure F I P PFA c Ac c

= = ⇒ = 98

1.8 6.0 103 10

F N−⇒ = = ××

Q7. Self inductance per unit length of a long solenoid of radius R with n turns per unit

length is:

(a) 2 20 R nμ π (b) 2

02 R nμ π

(c) 2 202 R nμ π (d) 2

0 R nμ π Ans: (a)





JEST-2015

Q8. A circular loop of radius R , carries a uniform line charge densityλ . The electric field,

calculated at a distance z directly above the center of the loop, is maximum if z is equal

to,

(a) 3

R (b) 2

R (c)2R (d) R2

Ans: (b)

Solution: ( )( )3/ 22 2

0

214

R zE

R z

λ ππε

×=

+

For maximum( )

( )

3/ 22 2 2 2

32 20

2 3 / 2 2, 0 04

dE R R z z R z zEdz R z

λ ππε

⎡ ⎤× + − × + ×= ⇒ =⎢ ⎥

⎢ ⎥+⎣ ⎦

( )3/ 22 2 2 2 2 2 2 2 2 23 3 22

RR z z R z R z z R z z⇒ + = + ⇒ + = ⇒ = ⇒ =

Q9. Consider two point charges q and qλ located at the points, ax = and ax μ= ,

respectively. Assuming that the sum of the two charges is constant, what is the value of

λ for which the magnitude of the electrostatic force is maximum?

(a) μ (b) 1 (c) μ1 (d) μ+1

Ans: (b)

Solution: ( )( ) ( ) ( ) ( )

2 2

2 2 2 22 20 0 0

1 1 14 4 11 4 1

q q q cFa a a a

λ λ λπε πε λμ μ πε μ

×= = =

+− − − q q cλ+ =∵

For maximum( )

( ) ( )( )

2 2 2

2 420

1 1 2 1, 0 014 1

dF c cFdz a

λ λ λλπε μ

⎡ ⎤+ − × += ⇒ =⎢ ⎥

+− ⎢ ⎥⎣ ⎦

( ) ( )2 2 21 2 1 1 2 1c cλ λ λ λ λ λ⇒ + = × + ⇒ + = ⇒ =





Q10. A spherical shell of inner and outer radii a and b , respectively, is made of a dielectric

material with frozen polarization ( ) rrkrP ˆ= ,where k is a constant and r is the distance

from the its centre. The electric field in the region bra << is,

(a) 0

ˆkE rr

=∈

(b) 0

ˆkE rr

= −∈

(c) 0=E (d) 20

ˆkE rr

=∈

Ans: (b)

Solution: 2b 2 2

1 k k.P rr rr r

∂ −⎛ ⎞ρ = −∇ = − =⎜ ⎟∂ ⎝ ⎠ and b

kˆP.r (at r b)bˆP.n

kˆP.r (at r a)a

⎧ ⎫+ = =⎪ ⎪⎪ ⎪σ = = ⎨ ⎬−⎪ ⎪− = =⎪ ⎪⎩ ⎭

For a< r <b ; r2 2enc a 2

k kQ 4 a 4 r dra r

− −⎛ ⎞⎛ ⎞= × π + π⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ( )4 ka 4 k r a 4 kr= − π − π − = − π

enc20

Q1E4 r

=πε 0

k ˆE rr

−⇒ =

ε

Q11. The electrostatic potential due to a charge distribution is given by ( )r

eArVrλ−

= where

A and λ are constants The total charge enclosed within a sphere of radius λ1 , with its

origin at 0=r is given by,

(a) 08 Ae

π ∈ (b) 04 Ae

π ∈ (c) 0 Ae

π ∈ (d) 0

Ans: (a)

Solution: ( )reV r A

r

λ−

=∵

( ) ( )2 2ˆ ˆ1r r rre e AeE V A r r r

r r

λ λ λλλ

− − −⎡ ⎤× − −= −∇ = − = +⎢ ⎥

⎣ ⎦

( ) ( )2

20 0 02

0 0

ˆ ˆ. 1 . sin 4 1r

renc

AeQ E da r r r d d r Ae rr

π π λλε ε λ θ θ φ πε λ

−−= = + = +∫ ∫ ∫





Thus total charge enclosed within a sphere of radius 1rλ

= is

1

00

814 1encAQ Ae

eλ

λ πεπε λλ

− ⎛ ⎞= + =⎜ ⎟⎝ ⎠

Q12. The skin depth of a metal is dependent on the conductivity ( )σ of the metal and the

angular frequency ω of the incident field. For a metal of high conductivity, which of the

following relations is correct? (Assume that ωσ >>∈ , where ∈ is the electrical

permittivity of the medium.)

(a) ωσ

∝d (b) σω1

∝d

(c) σω∝d (d) σω

∝d

Ans: (b)

Solution: Skin depth 2dσμω

=

Q13. The wavelength of red helium-neon laser in air iso

6328 A . What happens to its frequency

in glass that has a refractive index of 50.1 ?

(a) Increases by a factor of 3

(b) Decreases by a factor of 5.1

(c) Remains the same

(d) Decreases by a factor of 0.5

Ans: (c)

Solution: Frequency of electromagnetic wave does not change when it enter in medium of any

refractive index.





Q14. The approximate force exerted on a perfectly reflecting mirror by an incident laser beam

of power 10mW at normal incidence is

(a) N1310− (b) N1110− (c) N910− (d) N1510−

Ans: (b)

Solution: When electromagnetic wave is reflected by mirror the momentum transferred to the

mirror per unit area per second is twice the momentum of the light striking the mirror per

unit area per second

i.e. 3

11 28

2 10 102 6.6 10 /3 10

dp Power kg m sdt c

−−

−

× ×= = × = ×

×

The force exerted on the reflecting mirror is 116.6 10dpF Ndt

−= = ×

Thus best suitable answer is option (b).

Q15. Which of the following expressions represents an electric field due to a time varying

magnetic field?

(a) ( )zzyyxxK ˆˆˆ ++ (b) ( )zzyyxxK ˆˆˆ −+

(c) ( )yyxxK ˆˆ − (d) ( )zzyxyyK ˆ2ˆˆ +−

Ans: (d)

Solution: For time varying fields 0E∇× ≠

(a)ˆ ˆ ˆ/ / /x y z

E K x y zx y z

∇× = ∂ ∂ ∂ ∂ ∂ ∂ ˆ ˆ ˆ 0z y x z y xx y zy z z x x y

⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞= − + − + − =⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠

(b)ˆ ˆ ˆ/ / /x y z

E K x y zx y z

∇× = ∂ ∂ ∂ ∂ ∂ ∂−

ˆ ˆ ˆ 0z y x z y xx y zy z z x x y

⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞= − − + + + − =⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠

(c)ˆ ˆ ˆ/ / /

0

x y zE K x y z

x y∇× = ∂ ∂ ∂ ∂ ∂ ∂

−

ˆ ˆ ˆ0 0 0y x y xx y zz z x y

⎛ ⎞∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞= + + − + − − =⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠





(d)ˆ ˆ ˆ/ / /

2

x y zE K x y z

y x z∇× = ∂ ∂ ∂ ∂ ∂ ∂

−

( ) ( )2 2ˆ ˆ ˆ

z zx x y yx y zy z z x x y

∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂= + + − − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠

ˆ 0z= − ≠

Q16. A charged particle is released at time 0=t , from the origin in the presence of uniform

static electric and magnetic fields given by yEE ˆ0= and zBB ˆ0= respectively. Which of

the following statements is true for 0>t ?

(a)The particle moves along the x -axis.

(b) The particle moves in a circular orbit.

(c) The particle moves in the ( )yx, plane.

(d) Particle moves in the ( )zy, plane

Ans: (c)

Solution: In a cycloid charged particle will always confine in a plane perpendicular to B.

JEST-2014

Q17. For an optical fiber with core and cladding index of 45.11 =n and 44.12 =n , respectively,

what is the approximate cut-off angle of incidence? Cut-off angle of incidence is defined

as the incidence angle below which light will be guided.

(a) o7 (b) o22 (c) o5 (d) o0

Ans.: (a)

Solution:

2/12

1

21 1sin⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−= −

nn

θ where 45.1,44.1 12 == nn

2/1

1

45.145.144.144.11sin ⎟

⎠⎞

⎜⎝⎛

××

−= −θ ( )11sin 0.11726θ −⇒ = 0 06.67 7θ⇒ = ≈





Q18. Two large nonconducting sheets one with a fixed uniform positive charge and another

with a fixed uniform negative charge are placed at a distance of 1 meter from each other.

The magnitude of the surface charge densities are 2/8.6 mCμσ =+ for the positively

charged sheet and 2/3.4 mCμσ =− for the negatively charged sheet. What is the electric

field in the region between the sheets?

(a) CN /1030.6 5× (b) CN /1084.3 5×

(c) CN /1040.1 5× (d) CN /1016.1 5×

Ans.: (a)

Solution: Electric field between the sheet is 6 6

0 0 0 0

6.8 10 4.3 102 2 2 2σ σ − −

+ − × ×= + = +

∈ ∈ ∈ ∈

66 5

12

11.2 10 0.626 10 6.3 10 /2 8.86 10

N C−

−

×⇒ = × ⇒ ×

× ×

Q19. A system of two circular co-axial coils carrying equal currents I along same direction

having equal radius R and separated by a distance R (as shown in the figure below).

The magnitude of magnetic field at the midpoint P is given by

(a) RI

220μ

(b) RI

554 0μ

(c) RI

558 0μ

(d) 0

Ans.: (c)

Solution: ( )

20

32 2 22

IRBR d

μ=

+∵

20

1 32 2

224

IRB

RR

μ⇒ =

⎛ ⎞+⎜ ⎟

⎝ ⎠

, 2

02 3

2 222

4

IRB

RR

μ=

⎛ ⎞+⎜ ⎟

⎝ ⎠

2Rd =∵

21 BBB += 23

0

452

2

⎟⎠⎞

⎜⎝⎛

×=

R

Iμ32

0 032

4 85 55

I IB

RR

μ μ⇒ = =

P

I I

X

R

Y

R





Q20. Find the resonance frequency (rad/sec) of the circuit shown in the figure below

(a) 1.41 (b) 1.0 (c) 2.0 (d) 1.73

Ans.: (b)

Solution: 2

2

1 1.0RLC L

ω = − = (where 2 , 2 , 0.25R L H C F= Ω = = )

Q21. An electron is executing simple harmonic motion along the y-axis in right handed

coordinate system. Which of the following statements is true for emitted radiation?

(a) The radiation will be most intense in xz plane

(b) The radiation will be most intense in xy plane

(c) The radiation will violate causality

(d) The electron’s rest mass energy will reduce due to radiation loss

Ans.: (a)

Solution: Oscillating electron does not emit radiation in the direction of oscillation.

In the perpendicular direction of oscillation intensity is maximum.

So in this case the intensity will be maximum along x and z - axis or xz - plane

(intensity is also en xy -plane but less).

Q22. A conducting sphere of radius r has charge Q on its surface. If the charge on the sphere

is doubled and its radius is halved, the energy associated with the electric field will

(a) increase four times (b) increase eight times

(c) remain the same (d) decrease four times

Ans.: (b)

Solution: 2

2 2 2 20 01 22 0

0 0

ˆ 4 44 2 2 8

R

R

Q QE r W E r dr E r dr Wr R

π ππ π

∞∈ ∈= = + ⇒ =

∈ ∈∫ ∫

( )2 2

00

2 8 888

2

Q QW WR Rππ′⇒ = = =

∈∈

V F25.0Ω3

Ω2

H2





JEST-2013

Q23. At equilibrium, there can not be any free charge inside a metal. However, if you forcibly

put charge in the interior then it takes some finite time to ‘disappear’ i.e. move to the

surface. If the conductivity σ of a metal is ( ) 1610 m −Ω and the dielectric constant

120 1085.8 −×=ε Farad/m, this time will be approximately:

(a) 510 sec− (b) 1110 sec− (c) 910 sec− (d) 1710 sec−

Ans.: (d)

Solution: Characteristic time: 186

12

1085.810

1085.8 −−

×=×

=∈

=σ

τ

Q24. The electric fields outside ( )r R> and inside ( )r R< a solid sphere with a uniform

volume charge density are given by rrqE Rr ˆ

41

20πε

=> and rrRqE Rr ˆ

41

30πε

=<

respectively, while the electric field outside a spherical shell with a uniform surface

charge density is given by 20

1 ˆ ,4r R

qE r qrπε< = being the total charge. The correct ratio

of the electrostatic energies for the second case to the first case is

(a) 1: 3 (b) 9 :16 (c) 3 : 8 (d) 5 : 6

Ans.: (d)

Solution: Electrostatic energy in spherical shell drrEdrrEwR

R

sp ∫∫∞∈

+∈

= 22

20

0

22

10 4

24

2ππ

( ) Rq

rqdrr

rq

RR

18

18

442 0

2

0

22

420

20

∈=⎟

⎠⎞

⎜⎝⎛−

∈=

∈

∈⇒

∞∞

∫ πππ

π

Electrostatic energy in solid sphere drrEdrrEwR

R

s ∫∫∞∈

+∈

= 222

0

0

221

0 42

42

ππ

∞

⎥⎦⎤

⎢⎣⎡−

∈+⎥

⎦

⎤⎢⎣

⎡×

∈⇒

R

R

rqr

Rq 1

851

8 0

2

0

5

60

2

ππ

Rq

Rq

Rqws

0

2

0

2

0

2

406

81

85 ∈=

∈+⋅

∈×=

πππ





Now

2

02

0

8 56 6

40

spherical

sphere

qW

qWR

π

π

∈= =

∈

Q25. A thin uniform ring carrying charge Q and mass M rotates about its axis. What is the

gyromagnetic ratio (defined as ratio of magnetic dipole moment to the angular

momentum) of this ring?

(a) ( )MQ π2/ (b) Q/M (c) Q/(2M) (d) ( )MQ π/

Ans.: (c)

Solution: Magnetic dipole moment 2

2 222 2

Q Q Q rM IA r rT T

ωπ π ππ

′ = = ⇒ × × =

Angular momentum 2

2M QJ MrJ M

ω′

= ⇒ =

Q26. The electric and magnetic field caused by an accelerated charged particle are found to

scale as nrE −∝ and mrB −∝ at large distances. What are the value of n and m ?

(a) 1, 2n m= = (b) 2, 1n m= = (c) 1, 1n m= = (d) 2, 2n m= =

Ans.: (c)

Solution: For large distance r

Br

Er

qaBr

qaF 1,1sin,sin∝∝⇒==

θθ

So 1== nm

Q27. If kxzjyzixyE ˆ3ˆ2ˆ1 ++= and ( ) kyzjzxyiyE ˆ2ˆ2ˆ 22

2 +++= then

(a) Both are impossible electrostatic fields

(b) Both are possible electrostatic fields

(c) Only 1E is a possible electrostatic field

(d) Only 2E is a possible electrostatic field

Ans.: (d)

Solution: For electrostatic field 0=×∇ E





2

2 2

ˆˆ ˆ

2 2

i j k

Ex y z

y xy z yz

∂ ∂ ∂∇× =

∂ ∂ ∂+

( ) ( ) 0ˆ220ˆ22 =−++− zyyizz

( )1

ˆˆ ˆ

ˆ ˆ0 2 0 0

i j k

E y i xjx y z

xy yz yxz

∂ ∂ ∂∇× = = − + + ≠

∂ ∂ ∂∂

Q28. A charge q is placed at the centre of an otherwise neutral dielectric sphere of radius a and

relative permittivity rε . We denote the expression 204/ rq πε by ( )E r . Which of the

following statements is false?

(a) The electric field inside the sphere, r a< , is given by ( ) rrE ε/

(b) The field outside the sphere, r a> , is given by ( )E r

(c) The total charge inside a sphere of radius r a> is given by q .

(d) The total charge inside a sphere of radius r a< is given by q .

Ans.: (d)

Q29. An electromagnetic wave of frequency ω travels in the x - direction through vacuum. It

is polarized in the y - direction and the amplitude of the electric field is 0E . With kcω

=

where c is the speed of light in vacuum, the electric and the magnetic fields are then

conventionally given by

(a) ( )xtkyEE ˆcos0 ω−= and ( )ztkycEB ˆcos0 ω−=

(b) ( )ytkxEE ˆcos0 ω−= and ( )ztkxc

EB ˆcos0 ω−=

(c) ( )ztkxEE ˆcos0 ω−= and ( ) ytkyc

EB ˆcos0 ω−=

(d) ( )xtkxEE ˆcos0 ω−= and ( ) ytkyc

EB ˆcos0 ω−=





Ans.: (b)

Solution: ( )0 ˆcosE E kx t yω= −

( ) ( )01 1ˆ ˆ ˆcosB k E B x E kx t yc c

ω⎡ ⎤= × ⇒ = × −⎣ ⎦

( )( ) ( )( )0 0ˆ ˆ ˆcos cosE E

B kx t x y B kx t zc c

ω ω⇒ = − × ⇒ = −

JEST-2012

Q30. A magnetic field ( )kjiBB ˆ4ˆ2ˆ0 −+= exists at point. If a test charge moving with a

velocity, ( )kjivv ˆ2ˆˆ30 +−= experiences no force at a certain point, the electric field at

that point in SI units is

(a) ( )kjiBvE ˆ4ˆ2ˆ300 −−−= (b) ( )kjiBvE ˆ7ˆˆ00 ++−=

(c) ( )kjBvE ˆ7ˆ1400 += (d) ( )kjBvE ˆ7ˆ1400 +−=

Ans. : (d)

Solution: [ ] 0=×+= BvEqF ( )BvE ×−=⇒

( )kjBv ˆ7ˆ1400 +−=

Q31. An observer in an inertial frame finds that at a point P the electric field vanishes but the

magnetic field does not. This implies that in any other inertial frame the electric field E

and the magnetic field B satisfy

(a) 22

BE = (b) 0=⋅ BE (c) 0=× BE (d) 0=E

Ans.: (b)

Q32. A circular conducting ring of radius R rotates with constant angular velocity ω about its

diameter placed along the x-axis. A uniform magnetic field B is applied along the y-axis.

If at time t = 0 the ring is entirely in the xy-plane, the emf induced in the ring at time

t > 0 is

(a) tRB 22πω (b) ( )2 tanB R tωπ ω

(c) ( )2 sinB R tωπ ω (d) ( )2 cosB R tωπ ω

( ) ( ) ( ) 0 0ˆˆ ˆ4 4 2 12 6 1E v B i j k⇒ = − − + + + +





Ans. : (c)

Solution: tBABAABm ωθφ coscos ==⋅=

( ) [ ]cosmd d dB A BA tdt dt dtφε ω= − = − ⋅ = − ( )ωω tBA sin−−=

2 sinB R tε π ω ω⇒ = 2 sinB R tε ωπ ω⇒ =

Q33. An electric field in a region is given by ( ) kbyjcziaxzyxE ˆ6ˆˆ,, ++= . For which values of

a, b, c does this represent an electrostatic field?

(a) 13, 1, 12 (b) 17, 6, 1 (c) 13, 1, 6 (d) 45, 6, 1

Ans.: (c)

Solution: For electrostatic field 0=×∇ E

ˆˆ ˆ

0

6

i j k

Ex y z

ax cz by

⎡ ⎤⎢ ⎥

∂ ∂ ∂⎢ ⎥∇× = =⎢ ⎥∂ ∂ ∂⎢ ⎥⎢ ⎥⎣ ⎦

( ) [ ] [ ]ˆˆ ˆ6 0 0 0 0E b c i j k⇒ ∇× = − + − + =

( ) ˆ6 0b c i⇒ − = 6c b⇒ =

Q34. A capacitor C is connected to a battery 0V through three equal resistors R and a switch S

as shown below:

The capacitor is initially uncharged. At time 0t = , the switch S is closed. The voltage

across the capacitor as a function of time t for 0t > is given by

(a) ( ) ( )( )0 / 2 1 exp / 2V t RC− − (b) ( ) ( )( )0 / 3 1 exp / 3V t RC− −

(c) ( ) ( )( )0 / 3 1 exp 3 / 2V t RC− − (d) ( ) ( )( )0 / 2 1 exp 2 / 3V t RC− −

Ans.: (d)

R R

CR0V

S





Solution:

Apply KVL in loop 1: ( ) ( )0 1 1 2 0V i t R i i R− + + − = ( ) ( )1 2 02i t R i t R V⇒ − = ….. (i)

Apply KVL in loop 2:

( ) 0121

022 =−−+ ∫ Riidti

CRi

t0 2

2 20

1 2 02

t V i Ri dt i RC

+⎛ ⎞⇒ + − =⎜ ⎟⎝ ⎠∫

02 2

0

1 32 2

t Vi dt i RC

−⇒ + =∫ ……………(ii)

22

1 3 02

dii RC dt

⇒ + = 22

3 12

diR idt C

⇒ = − 22

23

di idt RC

−⇒ = ( )

23

2

tRCi t Ke−

⇒ =

Initial Conditions

( )R

VRR

V

RRRRR

Vi 0001 3

2

2

0 =+

=

+×

+=+ and ( ) 0 0

2203 3V VRi

R R R R+ = × =

+, ( )0 0Cv + =

( ) 02 0

3ViR

+ =∵ 0

3VKR

⇒ = ( )2

0 32 3

tRCVi t e

R

−

⇒ =

( )2

0 32

0 0

13

t t tRC

CVv t i dt e dt

C RC

−

= =∫ ∫2

30

0

233

tt

RCV eRC RC

−⎡ ⎤⎢ ⎥

= ⎢ ⎥−⎢ ⎥⎣ ⎦

( )2

0 33 13 2

tRC

CV RCv t eRC

−⎡ ⎤−⇒ = × −⎢ ⎥

⎣ ⎦( )

20 31

2

tRC

CVv t e

−⎡ ⎤⇒ = −⎢ ⎥

⎣ ⎦

R R

CR0V

S

1i 2i

1 2i i−

+ − + −

+

−





Q35. A small magnet is dropped down a long vertical copper tube in a uniform gravitational

field. After a long time, the magnet

(a) attains a constant velocity (b) moves with a constant acceleration

(c) moves with a constant deceleration (d) executes simple harmonic motion

Ans. : (a)

Q36. Consider a particle of electric charge e and mass m moving under the influence of a

constant horizontal electric field E and constant vertical gravitational field described by

acceleration due to gravity g. If the particle starts from rest, what will be its trajectory?

(a) parabolic (b) elliptic (c) straight line (d) circular

Ans.: (c)

Solution: Equation of motion 112

2

ctdtdx

dtxmdqE +=⇒= α

at ,0,0 == vtmqE

=1α2

11 2

tdx t xdt

αα= ⇒ =

similarly, 2

2

md ymgdt

=

gxyty ==⇒= 21

22

2 ,2

αα

αα

Q37. A point charge +q is placed at (0, 0, d) above a grounded infinite conducting plane

defined by z = 0. There are no charges present anywhere else. What is the magnitude of

electric field at (0, 0, - d)?

(a) ( )208/ dq ∈π (b) - ∞ (c) 0 (d) ( )2

016/ dq ∈π

Ans.: (d)

Solution: Electric field at Q

( )( )z

dqE ˆ24 2

0∈−

=π 2

0

ˆ16

q zdπ

−=

∈

2016 d

qE∈

=π





Q38. A time-dependent magnetic field ( )tB is produced in a circular region of space, infinitely

long and of radius R. The magnetic field is given as ztBB ˆ0= and is zero for r > R,

where B0 is a positive constant. The electric field at point 2r R= is

(a) 0 ˆ2

B R r (b) 0 ˆ4

B R θ− (c) 0 ˆ2

B R r− (d) 0 ˆ4

B R θ

Ans.: (b)

Solution: Solution: line

BE dl dat

⎛ ⎞∂⋅ = − ⋅⎜ ⎟∂⎝ ⎠

∫ ∫ 202E r B Rπ π⇒ × = −

2

0 2RE B

r⇒ = − θ

2

20

rRB

E −=⇒

The electric field at point 2r R= is 0 ˆ4

B RE θ= −

Q39. When unpolarised light is incident on a glass plate at a particular angle, it is observed that

the reflected beam is linearly polarized. What is the angle of the refracted beam with

respect to the surface normal?

(a) 56.7°

(b) 33.4°

(c) 23.3°

(d) The light is completely reflected and there is no refracted beam.

Ans.: (b)

Solution: Since 1 1n = , 52.12 =n

Brewster angle 01

1

21 7.56152.1tantan =⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

nn

Bθ

Now 0180 90 56.7 33.4Rθ = − − =

Q40. A cube has a constant electric potential V on its surface. If there are no charges inside the

cube, the potential at the center of the cube is

(a) V (b) V / 8 (c) 0 (d) V / 6

Ans.: (a)





Quantum Mechanics

JEST-2016

Q1. The wavefunction of a hydrogen atom is given by the following superposition of energy

eigen functions ( )nlm rυ/ ( , ,n l m are the usual quantum numbers):

( ) ( ) ( ) ( )100 210 3222 3 17 14 14

r r r rυ υ υ υ= − +/ / / /

The ratio of expectation value of the energy to the ground state energy and the

expectation value of 2L are, respectively:

(a) 229504

and 212

7 (b) 101

504 and

2127

(c) 101504

and 2 (d) 229504

and 2

Ans: (a)

Solution: 0 0 00

2 9 1 2297 1 14 4 14 9 504

E E EE E= × + × + × =

2 2 2 2 2 22 9 1 24 120 2 67 14 14 14 7

L = × + × + × = =

Q2. A spin- 12

particle in a uniform external magnetic field has energy eigenstates 1 and

2 . The system is prepared in ket-state ( )1 2

2

+at time t = 0. It evolves to the state

described by the ket ( )1 2

2

− in time T . The minimum energy difference between two

levels is:

(a) 6hT

(b) 4hT

(c) 2hT

(d) hT

Ans: (c)





Solution: ( ) ( ) ( )1 21 2 exp1 2

02 2

E t E ti it t tψ ψ

⎛ ⎞⎛ ⎞ ⎛ ⎞− + −⎜ ⎟ ⎜ ⎟⎜ ⎟+ ⎝ ⎠ ⎝ ⎠⎝ ⎠= = ⇒ = =

( )

( )2 1

1

1 2 exp

2

E E ti

E tt t iψ

⎛ ⎞⎛ ⎞−+ −⎜ ⎟⎜ ⎟⎜ ⎟⎛ ⎞ ⎝ ⎠⎝ ⎠= = −⎜ ⎟

⎝ ⎠

( )2 1exp 1E E t

i⎛ ⎞−− = −⎜ ⎟⎝ ⎠

( ) ( )2 12 1 2

E E T hE ET Tππ

−= ⇒ − = =

Q3. The energy of a particle is given by E p q= + where p and q are the generalized

momentum and coordinate, respectively. All the states with 0E E≤ are equally probable

and states with 0E E> are inaccessible. The probability density of finding the particle at

coordinate q , with 0q > is:

(a) ( )020

E qE+

(b) 20

qE

(c) ( )020

E qE−

(d) 0

1E

Ans: (c)

Solution: For condition E p q= + total no of accessible state upto energy 0E for 0q > is area

under the curve 2 20 0

1 22

E E× × =

The probability density of finding the particle at coordinate q , with 0q >

( )02 2 20 0 0

E q dqdpdq pdqE E E

−= ⇒

For probability at point point q dq is insignificant so ( )( ) ( )020

E qp q

E−

=





Q4. Consider a quantum particle of mass m in one dimension in an infinite potential well,

i.e., ( ) 0V x = for 2 2a ax−< < and ( )V x = ∞ for

2ax ≥ . A small perturbation,

( ) 2 xV x

a∈

′ = is added. The change in the ground state energy to ( )O ∈ is:

(a) ( )22 4

2π

π∈

− (b) ( )22 4

2π

π∈

+

(c) ( )2

2 42π π∈

+ (d) ( )2

2 42π π∈

−

Ans: (a)

Solution: ( )2 21 * ' 2

1 1 1

2 2

2 2 cos

a a

a a

xE V x dx x dxa a a

πφ φ− −

∈= ⇒∫ ∫

2 2 222 2

0 0 0

2 2 4 1 2 2 2.2. cos cos 1 cos 12

a a a

x x xx dx x dx x dxa a a a aa a

π π π∈ ∈ ∈⎛ ⎞ ⎛ ⎞= ⇒ + ⇒ + ⇒∫ ∫ ∫⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

20

2 2cos 1

a

xx dxaaπ∈ ⎛ ⎞+ =∫ ⎜ ⎟

⎝ ⎠( )2

2 42

ππ∈

−

Q5. If ( )22 2, 212xyY Y Y −= − where ,l mY are spherical harmonics then which of the following is

true?

(a) xyY is an eigenfunction of both 2L and zL

(b) xyY is an eigenfunction of 2L but not zL

(c) xyY is an eigenfunction both of zL but not 2L

(d) xyY is not an eigenfunction of either 2L and zL

Ans: (b)

Solution: The ( )2 21xy xyL Y l l Y= + where 2l = and z xy xyL Y mY≠

So xyY is an eigenfunction of 2L but not zL





Q6. A spin-1 particle is in a state υ/ described by the colunm matrix 1 2, 2, 210

i⎛ ⎞⎜ ⎟⎝ ⎠

in the zS basis. What is the probability that a measurement of operator zS will yield the

result h for the state xS υ/ ?

(a) 12

(b) 13

(c) 14

(d) 16

Ans: (c)

Solution: 0 1 01 0 1

2 0 1 0xS

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

2

22i

ψ

⎛ ⎞⎜ ⎟

= ⎜ ⎟⎜ ⎟⎝ ⎠

22 2

2xS iψ

⎛ ⎞⎜ ⎟

= +⎜ ⎟⎜ ⎟⎝ ⎠

1 0 00 0 00 0 1

zS⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟−⎝ ⎠

the eigen state for eigen value of zs is 100

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

( ) 2 2 12 4 2 8 4

p = = =+ +

Q7. The Hamiltonian of a quantum particle of mass in confined to a ring of unit radius is:

22

2H i

mα

θ∂⎛ ⎞= − −⎜ ⎟∂⎝ ⎠

where θ is the angular coordinate, α is a constant. The energy eigenvalues and

eigenfunctions of the particle are ( n is an integer):

(a) ( )2

in

ne θ

υ θπ

=/ and ( )2

2

2nE nm

α= − (b) ( ) ( )sin2n

nθυ θ

π=/ and ( )

22

2nE nm

α= −

(c) ( ) ( )cos2n

nθυ θ

π=/ and ( )

22

2nE nm

α= − (d) ( )2

in

ne θ

υ θπ

=/ and ( )2

2

2nE nm

α= +

Ans: (a)





Solution: 22 2 2

22 2

2 2H i i E

m mψ ψα α α ψ ψ

θ θ θ⎡ ⎤∂ ∂ ∂⎛ ⎞= − − ⇒ − + + =⎜ ⎟ ⎢ ⎥∂ ∂ ∂⎝ ⎠ ⎣ ⎦

By inspection ( )2

in

ne θ

ψ θπ

= wich will also satisfied boundary condition

( ) ( )2n nψ θ π ψ θ+ = will satisfied the Hamiltonian and value of ( )22

2n

Emα−

=

Q8. The adjoint of a differential operator ddx

acting on a wavefunction ( )xυ/ for a quantum

mechanical system is:

(a) ddx

(b) didx

− (c) ddx

− (d) didx

Ans: (c)

Q9. For a quantum mechanical harmonic oscillator with energies, 12nE n ω⎛ ⎞= +⎜ ⎟

⎝ ⎠, where

0,1,2...n = , the partition function is:

(a) 2 1

B

B

k T

k T

e

e

ω

ω−

(b) 2 1Bk Teω

− (c) 2 1Bk Teω

+ (d) 2

1

B

B

k T

k T

e

e

ω

ω−

Ans: (d)

Solution: 3 5 7exp exp exp exp ......2 2 2 2

zkT kT kT kTω ω ω ω

= − + − + − + −

2exp 1 exp exp .....2

zkT kT kTω ω ω⎛ ⎞= − + − + −⎜ ⎟

⎝ ⎠

exp exp12 2

1 exp exp exp exp 12 2

kT kTz

kT kT kT kT

ω ω

ω ω ω ω

−= ⇒ ⇒

− − − −





Q10. In the ground state of hydrogen atom, the most probable distance of the electron from the

nucleus, in units of Bohr radius 0a is:

(a) 12

(b) 1 (c) 2 (d) 32

Ans: (d)

Solution: 032

r a= most probable distance is 0pr a=

32p

rr

=

Q11. For operators P and Q , the commutator 1,P Q−⎡ ⎤⎣ ⎦ is

(a) [ ]1 1,Q P Q Q− − (b) [ ]1 1,Q P Q Q− −− (c) [ ]1 ,Q P Q Q− (d) [ ] 1,Q P Q Q−−

Ans (b) 1 1 1,P Q PQ Q P− − −⎡ ⎤ = −⎣ ⎦

[ ] [ ]1 1 1 1 1 1 1 1 1 1, ,Q P Q Q Q PQ QP Q Q PQQ QPQ Q P PQ P Q− − − − − − − − − −⎡ ⎤ ⎡ ⎤− ⇒ − − = − − = − + =⎣ ⎦ ⎣ ⎦

Q12. A spin 12

particle is in a state ( )

2

↑ + ↓ where ↑ and ↓ are the eigenstates of zS

operator. The expectation value of the spin angular momentum measured along x

direction is:

(a) (b) − (c) 0 (d) 2

Ans: (d)

Solution: ( )

12

122

⎛ ⎞⎜ ⎟↑ + ↓⎜ ⎟⇒⎜ ⎟⎜ ⎟⎝ ⎠

0 11 02xs ⎛ ⎞

= ⎜ ⎟⎝ ⎠





10 11 1 21 0 12 22 2

2

xs

⎛ ⎞⎜ ⎟⎛ ⎞⎛ ⎞ ⎜ ⎟= =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎜ ⎟⎝ ⎠

JEST-2015

Q13. Consider a harmonic oscillator in the state2

2 0ae eα

αψ+−

= , where 0 is the ground

state, +a is the raising operator and α is a complex number. What is the probability that

the harmonic oscillator is in the n -th eigenstate n ?

(a) !

22

ne

nαα− (b) !2

2

naa n

e−

(c) !

2

ne

nαα− (d) !

2

2

2

ne

nαα−

Ans: (a)

Solution: ( )2 2

2 20 0n

a

n

ae e e

n

α αα

αψ

++

− −= = ∑ and

( ) ( )0 0n

nan a n n

n

++= ⇒ =

( )2

2

n

n

ne n

n

α αψ

−= ∑

( )( )

2*

2

nn

e n nn

α α αψ ψ −⇒ = ∑

2 2 2

1n

n

e e en

α α αα− −= = =∑

Probability that ψ is in n state 2

2nn

ψψ

ψ ψ=

( )2

2

n

n

ne n

n

α αψ

−= ∑

2

2 1n

ne n

n

α

α−

= ∑

22

22 1n n

n

en e n nn n

αα

ψ α α−

−⇒ = =∑

22

2n

n en

α αψ −⇒ =





Q14. For non-interacting Fermions in −d dimensions, the density of states ( )ED varies as

⎟⎠⎞

⎜⎝⎛ −1

2d

E . The Fermi energy FE of an N particle system in −− 2,3 and −1 dimensions

will scale respectively as,

(a) NNN ,, 3/22 (b) 23/2 ,, NNN

(c) 3/22 ,, NNN (d) 23/2 ,, NNN

Ans: (d)

Q15. A particle of mass m moves in −1 dimensional potential ( )xV , which vanishes at infinity.

The exact ground state eigenfunction is ( ) ( )secx A h xυ λ=/ where A and λ are

constants. The ground state energy eigenvalue of this system is,

(a) m

E22λ

= (b) m

E22λ

−=

(c) m

E2

22λ−= (d)

mE

2

22λ=

Ans: (d)

Solution: ( ) ( )secx A h xψ λ=∵ ( ) ( )sec tanhd A h x xdxψ λ λ λ⇒ = −

( ) ( ) ( ) ( )2

2 22 sec tan sec secd A h x h x h x h x

dxψ λ λ λ λ λ λ λ⎡ ⎤⇒ = − − +⎣ ⎦

( ) ( ) ( )2 2 2sec tan secA h x h x h xλ λ λ λ⎡ ⎤⎡ ⎤= − − +⎣ ⎦⎣ ⎦

( ) ( ) ( )2 2 2sec sec tanA h x h x h xλ λ λ λ⎡ ⎤⎡ ⎤= − −⎣ ⎦⎣ ⎦

( ) ( ) ( )2 2 2sec sec 1 secA h x h x h xλ λ λ λ⎡ ⎤⎡ ⎤⎡ ⎤= − − −⎣ ⎦⎣ ⎦⎣ ⎦

( ) ( )2 2tan 1 sech x h xλ λ= −∵

( ) ( ) ( )2 2 2sec sec 1 secA h x h x h xλ λ λ λ⎡ ⎤⎡ ⎤= − − +⎣ ⎦⎣ ⎦

( ) ( )2

2 32 2sec secd A h x h x

dxψ λ λ λ⎡ ⎤⇒ = − −⎣ ⎦





Now put the value 2

2

ddxψ in equation ( ) ( ) ( )

2 2

22d V x x E x

m dxψ ψ ψ− + =

( ) ( ) ( ) ( ) ( )2

2 32sec sec sec sec2

A h x h x V x A h x EA h xmλ λ λ λ λ⎡ ⎤− − + =⎣ ⎦

( ) 0V x as x→ →∞∵

( ) ( ) ( )2 2 2

2 3sec 2 sec sec2 2

A h x A h x EA h xm m

λλ λ λ λ⇒ + − =

Now we have to do approximation i.e. ( )3sec h xλ dacays very fastly as x →∞ so second

term

( )2 2

32 sec 02

A h xmλ λ = . Thus ( ) ( )

2 2

sec sec2

A h x EA h xmλ λ λ=

2 2

2E

mλ

⇒ =

Q16. Consider a spin21

− particle characterized by the Hamiltonian zSH ω= . Under a

perturbation xgSH =′ , the second order correction to the ground state energy is given by,

(a) ω4

2g− (b)

ω4

2g (c) ω2

2g− (d)

ω2

2g

Ans: (a)

Solution: 1 00 12z zH s and sω⎡ ⎤

= = ⎢ ⎥−⎣ ⎦∵

1 00 12

H ω ⎛ ⎞⇒ = ⎜ ⎟−⎝ ⎠

and 1 00 12x

gH gs⎛ ⎞′ = = ⎜ ⎟−⎝ ⎠

Ground state energy is 2ω

− with eigenvector 1

01

φ⎛ ⎞

= ⎜ ⎟⎝ ⎠

and first excited state energy is 2ω with eigenvector 2

10

φ⎛ ⎞

= ⎜ ⎟⎝ ⎠

Second order correction in ground state 2 2

1 122 0 0

1 1

2 2

m m

m m

H HE

E Eφ φ φ φ

ω ω≠

′ ′= =

− − −∑





( )2

2 222

0 1 01 0

1 0 124

2

gE ω

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⇒ =−

2 2 2

4 4g gω ω

= − = −

Q17. Given that 1υ/ and 2υ/ are eigenstates of a Hamiltonian with eigenvalues 21 EandE

respectively, what is the energy uncertainty in the state ( )21 υυ /+/ ?

(a) 21EE− (b) 2121 EE −

(c) ( )2121 EE + (d) 122

1 EE −

Ans: (b)

Solution: ( )2 2

1 22 2 21 2

1 12 2 2

E EE E E

+= + = and 1 2

1 12 2

E E E= +

( ) ( )2 2 2 2 2 2

22 1 22 1 2 1 2 1 21 2

2 2 212 4 4

E E E E E E E EE E E E E+ + − − −

Δ = − = − + =∵

2 21 2 1 2

1 22 1

4 2E E E EE E E+ −

⇒ Δ = = −

Q18. A particle moving under the influence of a potential ( )2

2krrV = has a wavefunction

( )tr,υ/ . If the wavefunction changes to ( )tr,αυ/ , the ratio of the average final kinetic

energy to the initial kinetic energy will be,

(a) 2

1α

(b) α (c) α1 (d) 2α

Ans: (c)

Solution: For ( ),r tψ the average kinetic energy ( ) ( )2

* 2 2

0,

2T r t r dr

mψ ψ

∞ ⎛ ⎞= − ∇⎜ ⎟

⎝ ⎠∫ 2∇ is

written in spherical polar coordinate, which is dimension of ( ) 2length −

For wave function ( ),r tψ α





( ) ( )( )2

* 2 2

0, ,

2T r t r t r dr

mα ψ α ψ α∞ ⎛ ⎞

= − ∇⎜ ⎟⎝ ⎠

∫

Put r rα ′= or r drr drα α′ ′

= ⇒ = and 2 2 2r rα∇ = ∇

( ) ( )2 2

* 2 23 0

, ,2

T r t r t r drmα

α ψ ψα

∞ ⎛ ⎞′ ′ ′ ′= − ∇⎜ ⎟

⎝ ⎠∫ ( ) ( )

2* 2 2

0

1 , ,2

r t r t r drm

ψ ψα

∞ ⎛ ⎞′ ′ ′ ′= − ∇⎜ ⎟

⎝ ⎠∫

TTα α

⇒ =1T

Tα

α⇒ =

Q19. If a Hamiltonian H is given as ( )01101100 −+−= iH , where 0 and 1 are

orthonormal states, the eigenvalues of H are

(a) 1± (b) i± (c) 2± (d) 2i±

Ans: (c)

Solution: ( )0 0 1 1 0 1 1 0H i= − + −

0 0 1H i= − and 1 1 0H i= − +

The matrix representation of H is 0 0 0 1

1 0 1 1

H H

H H1

1i

i⎛ ⎞

= ⎜ ⎟− −⎝ ⎠

Eigenvalue of H 1

01

iiλ

λ−⎛ ⎞

=⎜ ⎟− − −⎝ ⎠( )21 1 0λ⇒ − − − = − 2λ⇒ = ±

JEST-2014

Q20. Suppose a spin 2/1 particle is in the state

⎥⎦

⎤⎢⎣

⎡ +=/ 2

16

1 iυ

If xS ( x component of the spin angular momentum operator) is measured what is the

probability of getting 2/+ ?

(a) 3/1 (b) 3/2 (c) 6/5 (d) 6/1

Ans.: (c)





Solution: 0 11 02xS⎡ ⎤

= ⎢ ⎥⎣ ⎦

with eigenvalues 2

± and eigenvector corresponding to 2

is 1112⎛ ⎞⎜ ⎟⎝ ⎠

Now probability getting 2

+

[ ]

[ ]

2

211 1 11 1 1 222 6 512

112 61 61 2 626

ii

pi

i

φ ψψ ψ

+⎡ ⎤⋅ ⎢ ⎥ + +

⎣ ⎦⎛ ⎞ = = = =⎜ ⎟ +⎡ ⎤⎝ ⎠ ×− ⎢ ⎥⎣ ⎦

Q21. The Hamiltonian operator for a two-state system is given by

( )12212211 ++−= αH ,

where α is a positive number with the dimension of energy. The energy eigenstates

corresponding to the larger and smaller eigenvalues respectively are:

(a) ( ) ( )2121,2121 −++−

(b) ( ) ( )2121,2121 +−−+

(c) ( ) ( ) 2112,2121 −+−+

(d) ( ) ( ) 2112,2121 +−+−

Ans.: (b)

Solution: ( )12212211 ++−= αH ( )1 1 2H α⇒ = + , ( )212 −= αH

Lets check for option (b): ( ) ( )1 2 1 2 , 1 2 1 2+ − − +

Now ψαψ =H ( )1 2 1 2H ⎡ ⎤⇒ + −⎣ ⎦ ( )1 2 1 2H H= + +

( )1 2 1 2H ⎡ ⎤+ −⎣ ⎦ ( ) ( )1 2 1 2H H⇒ + − ( ) ( ) ( )1 2 2 1 1 2α α⇒ + + − −

( )1 2 1 1 1 2 1 2α α ⎡ ⎤⎡ ⎤⇒ + − + − −⎣ ⎦ ⎣ ⎦ ( )22212 −+⇒ αα

( )[ ]21212 −+⇒α

Now ( )2121 +−H ( )[ ]2121 +−⇒ H ( )2121 +−⇒ HH





( ) ( )( )1 2 2 1 1 2α α ⎡ ⎤⇒ + − + −⎣ ⎦ ( ) ( )1 2 1 1 1 2 1 2α α⇒ − − + + +

( ) 22212 αα ++−⇒ ( )[ ]22112 +−−⇒ α

Q22. Consider an eigenstate of 2L and zL operator denoted by ml, . Let LnA ⋅= ˆ denote an

operator, where n is a unit vector parametrized in terms of two angles

as ( ) ( )θφφθ cos,sin,cossin, =zyx nnn . The width AΔ in ml, state is:

(a) ( ) θcos2

1 2mll −+ (b) ( ) θsin2

1 2mll −+

(c) ( ) θsin1 2mll −+ (d) ( ) θcos1 2mll −+

Ans.: (b)

Solution: ˆA n L= ⋅ x y zx y zA L L Lr r r

⇒ = ⋅ + ⋅ + ⋅

sin sinsin cos cosx y z

rr rA L L Lr r r

θ φθ φ θ⇒ = ⋅ + ⋅ + ⋅

sin cos sin sin cosx y zA L L Lθ φ θ φ θ⇒ = + ⋅ +

Now 22 AAA −=Δ

sin cos sin sin cosx y zA L L Lθ φ θ φ θ= + +

( )cosA m θ= 0, 0x yL L= =∵

θφθφθ 222222222 cossinsincossin zyx LLLA ++=

( ) ( )2 22 2 2 2 2 2 2 2 2 2

1 1sin cos sin sin cos

2 2

l l m l l mA mθ φ θ φ θ

⎡ ⎤+ − ⎛ ⎞+ −⎣ ⎦⇒ = + +⎜ ⎟⎜ ⎟⎝ ⎠

( ) 22 2 2 2 2 2 2 2

1sin sin cos cos

2

l l mA mθ φ φ θ

⎡ ⎤+ −⎣ ⎦ ⎡ ⎤⇒ = + +⎣ ⎦

( ) 22 2 2 2 2 2

1sin cos

2

l l mA mθ θ

⎡ ⎤+ −⎣ ⎦⇒ = +





( )( )222 2 2 2 2 2 2 2 2

1sin cos cos

2

l l mA A A m mθ θ θ

+ −Δ = − = + −

( ) 21sin

2

l l mA θ

⎡ ⎤+ −⎣ ⎦Δ =

Q23. Consider a three-state system with energies EE, and gE 3− (where g is a constant) and

respective eigenstates

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=/01

1

21

1υ ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=/

211

61

2υ ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=/

111

31

3υ

If the system is initially (at 0=t ), in state ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=/

001

iυ

what is the probability that at a later time t system will be in state ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=/

100

fυ

(a) 0 (b) ⎟⎠⎞

⎜⎝⎛

23sin

94 2 gt

(c) ⎟⎠⎞

⎜⎝⎛

23cos

94 2 gt (d) ⎟

⎠⎞

⎜⎝⎛ −

23sin

94 2 gtE

Ans.: (b)

Q24. A hydrogen atom in its ground state is collided with an electron of kinetic energy 13.377

eV. The maximum factor by which the radius of the atom would increase is

(a) 7 (b) 8 (c) 49 (d) 64

Ans.: (c)

Solution: 2

13.6nE eV

n−

=

1 13.6E eV⇒ = − , 2 3.4E eV= − , 3 1.5E eV= , 4 0.85E eV= , 5 0.54E eV=

6 0.3777E eV= , 7 0.2775E eV=





Since Electron have kinetic energy 13.377 13.6 0.2775 7eV eV n= − + ⇒ = 2

0nr a n=∵ 049nr a⇒ =

Q25. The lowest quantum mechanical energy of a particle confined in a one-dimensional box

of size L is 2 eV. The energy of the quantum mechanical ground state for a system of

three non-interacting spin 21 particles is

(a) 6 eV (b) 10 eV (c) 12 eV (d) 16 eV

Ans.: (c)

Solution: 2 2

1 2 22

E eVml

π= = , 2 14 8E E eV= =

∵ spin is 21

then degeneracy 12 1 2 1 22

S + = × + =

⇒ground state 2 2 1 8 12eV eV eV× + × =

Q26. A ball bounces off earth. You are asked to solve this quantum mechanically assuming the

earth is an infinitely hard sphere. Consider surface of earth as the origin implying

( ) =∝0V and a linear potential elsewhere (i.e. ( ) mgxxV −= for 0>x ). Which of the

following wave functions is physically admissible for this problem (with 0>k ):

(a) xe kx /−=/υ (b) 2kxxe−=/υ (c) kxAxe−=/υ (d)

2kxAe−=/υ

Ans.: (b)

Solution: 2kxxe−=ψ

For given potential, at 0,x = and x = ∞ wave function must vanish.

Q27. The operator A and B share all the eigenstates. Then the least possible value of the

product of uncertainties BAΔΔ is

(a) (b) 0 (c) 2/ (d) Determinant (AB)

Ans.: (b)





Solution: [ ]2

ABBA ≥Δ⋅Δ

0≥Δ⋅Δ BA A∵ and B have share their eigen values

so [ ] 0=AB

Q28. Consider a square well of depth 0V− and width a with aV0 fixed. Let ∞→0V and

0→a . This potential well has

(a) No bound states (b) 1 bound state

(c) 2 bound states (d) Infinitely many bound states

Ans.: (b)

Solution: It forms delta potential so it has only one bound state.





JEST-2013

Q29. A particle of mass m is contained in a one-dimensional infinite well extending from

2Lx = − to

2Lx = . The particle is in its ground state given by ( ) ( )LxLx /cos/20 πϕ = .

The walls of the box are moved suddenly to form a box extending from x L= − to x L= .

what is the probability that the particle will be in the ground state after this sudden

expansion?

(a) ( )23/8 π (b) 0 (c) ( )23/16 π (d) ( )23/4 π

Ans.: (a)

Solution: Probability Lx

LLx

L 2cos

22,cos2, 10

210

πφπφφφ =

Since the wall of box are moved suddenly then

Probability 2

2/

2/

22/

2/ 2coscos2

212

2coscos12 dx

Lx

Lx

Ldx

Lx

Lx

LLL

L

L

L ∫∫ −−⋅=⋅⋅=

ππππ

22 / 2/ 2

/ 2/ 2

2 1 3 2 1 2 3 2cos cos sin sin2 2 2 2 3 2 2

LL

LL

x x L x L xdxL L L L L L

π π π ππ π−

−

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎡ ⎤⇒ ⋅ + ⇒ ⋅ +⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎣ ⎦⎣ ⎦∫

22 1 2 3 3 2sin sin sin sin

2 3 4 4 4 4L L

Lπ π π π

π π⎡ ⎤⎛ ⎞ ⎛ ⎞⇒ ⋅ + + +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

22

382

32

πππ=+⇒

Q30. A quantum mechanical particle in a harmonic oscillator potential has the initial wave

function ( ) ( ),10 xx ψψ /+/ where 0ψ/ and 1ψ/ are the real wavefunctions in the ground and

first excited state of the harmonic oscillator Hamiltonian. For convenience we take

1=== ωm for the oscillator. What is the probability density of finding the particle at

x at time π=t ?

(a) ( ) ( )( )201 xx ψψ /−/ (b) ( )( ) ( )( )2

02

1 xx ψψ /−

(c) ( ) ( )( )201 xx ψψ /+/ (d) ( )( ) ( )( )2

02

1 xx ψψ /+/





Ans.: (a)

Solution: ( ) ( ) ( )xxx 10 ψψψ +=

( ) ( ) ( )0 10 1, i iE t E tx t x e x eψ ψ ψ− −= +

Now probability density at time t

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2* *0 1 0 1 1 0, , , 2Re cos tx t x t x t x x x x E Eψ ψ ψ ψ ψ ψ ψ= = + + −

putting π=t

( ) ( ) ( ) ( ) ( )2 2 2 *0 1 0 1 1 0, 2 Re cos 1x t x x x x E Eψ ψ ψ ψ ψ π ω= + + − = =∵

( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 2*0 1 0 1 1 0, 2Rex t x x x x x xψ ψ ψ ψ ψ ψ ψ⎡ ⎤= + − = −⎣ ⎦

Q31. If ,x yJ J and zJ are angular momentum operators, the eigenvalues of the operator

( )x yJ j+ are:

(a) real and discrete with rational spacing

(b) real and discrete with irrational spacing

(c) real and continuous

(d) not all real

Ans.: (b)

Solution: ( ) ( )0 1 0 01 , ,0 0 1 02 2x y

iJ J J J J J J J+ − − + + −

⎡ ⎤ ⎡ ⎤= + = − ⇒ = =⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦

0 1 0 1 0 11,1 0 1 0 1 02 2 2

x yx y

J J iiJ Ji

+− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = ⇒ =⎢ ⎥ ⎢ ⎥ ⎢ ⎥+⎣ ⎦ ⎣ ⎦ ⎣ ⎦

eigen value 2021

121 2 ±=⇒=−⇒⎟⎟

⎠

⎞⎜⎜⎝

⎛−+−−

λλλ

λi

i





Q32. A simple model of a helium-like atom with electron-electron interaction is replaced by

Hooke’s law force is described by Hamiltonian

( ) ( )42

12

22

21

222

21

2 λω −++∇+∇− rrm

m 2

212 rrm −ω .

What is the exact ground state energy?

(a) ( )λω ++= 1123E (b) ( )λω += 1

23E

(c) λω −= 123E (d) ( )λω −+= 11

23E

Ans.: (b)

Q33. Consider the state ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

2/12/12/1

corresponding to the angular momentum 1l = in the zL basis

of states with 1, 0, 1m = + − . If 2zL is measured in this state yielding a result 1, what is the

state after the measurement?

(a) ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

001

(b) ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

3/20

3/1 (c)

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

100

(d) ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

2/10

2/1

Ans.: (d)

Solution: ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=

100000001

,100

000001

2zz LL , eigenvector

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

100

,010

,001

Corresponding eigenvalue 1,0,1

Now state after measurement yielding ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=+⇒

101

21

101

1 31 φφ





Q34. What are the eigenvalues of the operator aH ⋅= σ , where σ are the three Pauli matrices

and a is a vector?

(a) zaandyx aa + (b) yzx iaaa ±+ (c) ( )zyx aaa ++± (d) a±

Ans.: (d)

Solution: ( )zzyyxx aaaaH ... σσσσ ++=⋅=

0 1 0 1 01 0 0 0 1x y z

ia a a

i−⎛ ⎞ ⎛ ⎞ ⎛ ⎞

= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠

( )( )

( ) ( )( ) ( )⎟⎟⎠

⎞⎜⎜⎝

⎛+−+

−−⇒⎟⎟

⎠

⎞⎜⎜⎝

⎛−+−

⇒λ

λ

zyx

yxz

zyx

yxz

aiaaiaaa

aiaaiaaa

( )( ) ( )( )yxyxzz iaaiaaaa +−−+−−⇒ λλ

2 2 2 2 0z x ya a aλ− + − − =

2 2 2 2x y za a aλ = + +

a±=⇒ λ

Q35. The hermitian conjugate of the operator ⎟⎠⎞

⎜⎝⎛∂∂−x

is

(a) x∂∂ (b)

x∂∂

− (c) x

i∂∂ (d)

xi∂∂

−

Ans.: (a)

Solution: ( ) ( ) ( ) ( )† *

* xx x x

x xψ

ψ ψ ψ⎛ ⎞−∂∂⎛ ⎞⇒ − = ⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

( ) ( ) ( ) ( ) ( ) ( )*

* * xx x dx x x x dx

x xψ

ψ ψ ψ ψ ψ∞ ∞∞

−∞−∞ −∞

∂∂⎡ ⎤⇒ − − − −⎢ ⎥∂ ∂⎣ ⎦∫ ∫

( ) ( )* x

x dxx

ψψ

∞

−∞

∂⇒

∂∫





Q36. If the expectation value of the momentum is p for the wavefunction ( )xψ , then the

expectation value of momentum for the wavefunction ( )/i k xe xψ is

(a) k (b) kp − (c) kp + (d) p

Ans.: (c)

Solution: ( ) ( )* x i x dx px

ψ ψ∞

−∞

∂⎛ ⎞− =⎜ ⎟∂⎝ ⎠∫

Now

( ) ( )*ikx ikx

e x i e x dxx

ψ ψ∞ −

−∞

∂⎛ ⎞−⎜ ⎟∂⎝ ⎠∫ ( )( ) ( ) ( )*ikx ikx ikxike x i e x e x

xψ ψ ψ

−∞

−∞

⎡ ⎤∂⇒ − +⎢ ⎥∂⎣ ⎦∫

( ) ( ) ( ) ( )* *.ikx ikx ikxike x i x e i e x x dx

xψ ψ ψ ψ

−∞ − ∞

−∞ −∞

∂⎛ ⎞⇒ − + −⎜ ⎟∂⎝ ⎠∫ ∫

( ) ( ) ( ) ( )* *x i x k x xx

ψ ψ ψ ψ∞ ∞

−∞ −∞

∂⎡ ⎤⇒ − +⎢ ⎥∂⎣ ⎦∫ ∫ KP +⇒

Q37. Two electrons are confined in a one dimensional box of length L . The one-electron states

are given by ( ) 2 sinnn xx

L Lπψ ⎛ ⎞= ⎜ ⎟

⎝ ⎠. What would be the ground state wave function

( )1 2,x xψ if both electrons are arranged to have the same spin state?

(a) ( ) ⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=/

Lx

Lx

LLx

Lx

Lxx 2121

21 sin2sin22sinsin22

1, ππππυ

(b) ( ) ⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=/

Lx

Lx

LLx

Lx

Lxx 2121

21 sin2

sin22sinsin2

21,

ππππυ

(c) ( ) ⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=/

Lx

Lx

Lxx 21

212sinsin2, ππ

υ

(d) ( ) ⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=/

Lx

Lx

Lxx 21

21 sin2sin2, ππυ

Ans.: (b)





Solution: Electrons are Fermions of spin 21 and it wave functions are anti symmetric

Spin part is symmetric and space part will be anti symmetric (since total wave function is

anti symmetric)

Then

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=

Lx

Lx

LLx

Lx

L2121 sin.

2sin22

sin.sin22

1 ππππ

Q38. The operator

⎟⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛ − x

dxdx

dxd

is equivalent to

(a) 22

2

xdxd

− (b) 122

2

+− xdxd

(c) 122

2

+− xdxdx

dxd (d) 2

2

2

2 xdxdx

dxd

−−

Ans.: (b)

Solution: ( )xfxdxdx

dxd

⎟⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛ −⇒ ( ) ( )⎥⎦

⎤⎢⎣⎡ +⎟⎠⎞

⎜⎝⎛ −⇒ xxfxf

dxdx

dxd

( ) ( ) ( ) ( )xfxxfdxdxxxfxf

dxd

dxd 2−−⎥⎦

⎤⎢⎣⎡ +⇒

( ) ( ) ( ) ( ) ( )2

22

df xd df x f x x x f x x f xdx dx dx

⇒ + + − −

( ) ( ) ( ) ( )xfxdxdxfxfxxf

dxd

⎟⎟⎠

⎞⎜⎜⎝

⎛+−=+−⇒ 12

2

22

2

2





Q39. If a proton were ten times, the ground state energy of the electron in a hydrogen atom

would be

(a) less

(b) more

(c) the same

(d) less, more or equal depending on the electron mass

Ans.: (b)

Solution: 2

0.9999513.6 13.59932 0.99995en e

e

mE mn m

μ−= × ⇒ − =∵

JEST-2012

Q40. The ground state (apart from normalization) of a particle of unit mass moving in a one-

dimensional potential V(x) is ( ) ( )xx 2cosh2/exp 2− . The potential V(x), in suitable

units so that h = 1, is (up to an addiative constant.)

(a) π2/2 (b) ( )xx 2tanh22/2 −π

(c) ( )xx 2tan22/2 −π (d) ( )xx 2coth22/2 −π

Ans. : (b)

Q41. Consider the Bohr model of the hydrogen atom. If α is the fine-structure constant, the

velocity of the electron in its lowest orbit is

(a) α+1

c (b) ( )corc αα

−+

11 2 (c) α2 c (d) α c

Ans. : (d)

Solution: mvr n=

2

2

0

2

41

rze

rmv

∈=

π

2

20

14

zermrπ

=∈





2

20

14

zemv nmvπ

⋅ =∈

2

04zev

nπ=

∈ and fine structure constant

2

04e

cα

π=

∈

2

04zev

nπ=

∈

for lowest orbit 2

04zevπ

=∈

2

04ze cv

cπ⇒ =

∈

cv α=

Q42. Define ( ),ffx += +σ and ( ),ffiy −−= +σ where the σ’ are Pauli spin matrices and f,f†

obey anticommutation relations 1,,0, † == ffff . Then σ2 is given by

(a) ,f† f-1 (b) 2f† f-1 (c) 2f† f + 1 (d) f† f

Ans. : (c)

Solution: zyx iσσσ =

yxzi σσσ =

( )( )ffffii

i yxz −+−

== ++σσσ 1

( )[ ]22 ffffff −+−−= +++

( )[ ]ffff .1 ++ −+−−=

[ ]ff +−−= 21

12 −= + ff

Q43. Consider a system of two spin-1/2 particles with total spin S = s1 +s2, where s1 and s2 are

in terms of Pauli matrices σi. The spin triplet projection operator is

(a) 2141 ss ⋅+ (b) 214

3 ss ⋅− (c) 2143 ss ⋅+ (d) 214

1 ss ⋅−

Ans. : (c)





Solution: 21 SSS +=⇒ 2122

21

2 2 SSSSS ⋅++=

2 21 2

3 3 2.4 4

S S S⎛ ⎞= + + ⋅⎜ ⎟⎝ ⎠

1,0=S∵

2 21 2

324

S S S⎡ ⎤= + ⋅⎢ ⎥⎣ ⎦ for Triplet projection operator

( ) 2 21 2

31 24

s s S S⎡ ⎤+ = + ⋅⎢ ⎥⎣ ⎦ 1=S

( ) ⎟⎠⎞

⎜⎝⎛ ⋅+=+ 214

32111 SS ISS =⋅+⇒ 2143

Q44. Consider a spin-1/2 particle in the homogeneous magnetic field of magnitude B along z-

axis which is prepared initially in a state ( )↓+↑=2

1ψ at time t = 0. at what time t

will the particles be in the state ψ− (μB is Bohr magneton)?

(a) B

tBμπ

= (b) B

tBμπ2

= (c) B

tBμ

π2

= (d) Never

Ans.: (a)

Solution: zBE B ˆ⋅= μ ⎟⎟⎠

⎞⎜⎜⎝

⎛=

11

21ψ

( )iEt

etx−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

11

21,ψ ( ) ψψ −=⇒ tx,1

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

11

21

11

21 Bti B

eμ

1−=− Bti B

eμ

πμ coscos =⎟

⎠⎞

⎜⎝⎛ tBB

BttB

B

B

μππ

μ=⇒=





Q45. The ground state energy of 5 identical spin-1/2 particles which are subject to a one-

dimensional simple harmonic oscillator potential of frequency ωis

(a) (15/2)ћω (b) (13/2)ћω (c) (1/2)ћω (d) 5ћω

Ans. : (b)

Solution: ⇒ degeneracy 2121212 =+×=+s

1 3 52 2 12 2 2groundE ω ω ω= × + × + × ω

213

=

Q46. The spatial part of a two-electron state is symmetric under exchange. If ↑ and ↓

represent the spin-up and spin-down states respectively of each particle, the spin-part of

the two-particle state is

(a) ↓↑ (b) ↑↓

(c) ( ) 2/↓↑−↑↓ (d) ( ) 2/↓↑+↑↓

Ans. : (c)

Solution: Since electron are Fermion and Fermions have antisymmetric wave function

∵ spatial part is symmetric then its spin part is antisymmetric to maintain antisymmtric

wave function

( ) ( )↓↑−↑↓=2

1xψ

Q47. The wave function of a free particle in one dimension is given by

( ) xBxAx 3sinsin +=/υ . Then ( )xυ/ is an eigenstate of

(a) the position operator (b) the Hamiltonian

(c) the momentum operator (d) the parity operator

Ans. : (d) ( ) ( )xx ψψ =−

( )xψ−= parity (even and odd)

( ) ( ) ( )xBxAx 3sinsin −+−=−ψ [ ]xBxA 3sinsin +−=

( ) ( )⇒−=− xx ψψ parity i.e. parity operator





Q48. The quantum state ( ) ,cosexpsin ↓+↑ xix φ where 0=↓↑ and x, ф are,real, is

orthogonal to:

(a) ↑xsin (b) ( ) ↓+↑ xix sinexpcos φ

(c) ( ) ↓−↑− xix sinexpcos φ (d) ( ) ↓+↑−− xxi sincosexp φ

Ans.: (d)

Solution: 0=↓↑ , ( ) ↓+↑= xix cosexpsin φψ

( ) ( ) ( ) ( )2exp cos sin exp exp cos sin exp cos sini x x i i x x i x xψ ψ φ φ φ φ′ = − ↑ ↑ − ↓ ↑ + ↓ ↑ + ↓ ↓

( ) ( ) 0sincosexpsincosexp =+−= xxixxi φφ

Q49. The binding energy of the hydrogen atom (electron bound to proton) is 13.6 eV. The

binding energy of peritoneum (electron bound to positron) is

(a) 13.6 / 2 eV (b) 13.6 / 1810 eV

(c) 13.6 × 1810 eV (d) 13.6 × 2 eV

Ans.: (a)

Solution: men

Enμ

2

6.13−=′

2m

memememe

=+⋅

=μ

meme

nEn 2

6.132 ⋅−=′

2

12

6.13n

En ×−=′

Thus binding energy will be eV2

6.13





Thermodynamics & Statistical Mechanics

JEST-2016

Q1. An ideal gas with adiabatic exponent γ undergoes a process in which its pressure P is

related to its volume V by the relation 0P P Vα= − , where 0P and α are positive

constants. The volume starts from being very close to zero and increases monotonically

to 0Pα

. At what value of the volume during the process does the gas have maximum

entropy?

(a) ( )

0

1P

α γ+ (b)

( )0

1Pγ

α γ− (c)

( )0

1Pγ

α γ+ (d)

( )0

1P

α γ−

Ans: (c)

Solution: 1V

nRdTTdS nC dT PdV TdS PdVγ

= + ⇒ = +−

For maximum entropy 0dS =

For Ideal gas PV nRT PdV VdP nRdT= ⇒ + =

1 1 1VPdV VdP pV VdPTdS nC dT PdV TdS PdV dS PdV

nRγ

γ γ γ+

= + ⇒ = + ⇒ = +− − −

dP dVα= −

( ) ( )1 1 1 1PV VdV dS nRP nRdS PdV VnR dV PV PV

γ α γ αγ γ γ γ

= − ⇒ = −− − − −

For maximum entropy ( )00 0dS P V P V VdV

γ α γ α α= ⇒ − = ⇒ − =

( )0

1PV γ

α γ=

+





Q2. A point charge q of mass m is released from rest at a distance d from an infinite

grounded conducting plane (ignore gravity). How long does it take for the charge to hit

the plane?

(a) 3 3

02 mdq

π ε (b)

302 md

qπ ε

(c) 3 3

0mdq

π ε (d)

30md

qπ ε

Ans: (a)

Solution: 2 2

2 20

14 4

d x qF ma mdt dπε

= = = − 2

2 2

d x Adt x

⇒ = − where2

016qAmπ ε

= .

( )22 2

12

dv A dv dv A dx d d Av v vdt x dt dt x dt dt dt x

⎛ ⎞⇒ = − ⇒ = − ⇒ = ⎜ ⎟⎝ ⎠

2

2v A C

x⇒ = + at , 0 Ax d v C

d⇒ = = ⇒ = −

1 12v Ax d

⎛ ⎞⇒ = −⎜ ⎟⎝ ⎠

.

1 12dx Adt x d

⎛ ⎞⇒ − = −⎜ ⎟⎝ ⎠

0

0

2t

d

xd dx A dtd x

⇒ = −−∫ ∫

Put 2sin 2 sin cosx d dx x d dθ θ θ θ= ⇒ = =

( )20

2/ 2

sin2 sin cos 2

cos

d dd d At

dπ

θθ θ θ

θ⇒ = −∫

0 03/ 2 2

/ 2 / 2

sin2 2 sin cos 2 sincos

At d d d d dπ π

θ θ θ θ θ θθ

⇒ − = =∫ ∫

( ) 003/ 2 3/ 2 3/ 2

/ 2/ 2

1 cos 2 sin 22 22 2 2

At d d d dππ

θ θ πθ θ− ⎡ ⎤⇒ − = = − = −⎢ ⎥⎣ ⎦∫

3/ 222

At d π⇒ − = −

23/ 2

0

216 2

q t dm

ππ ε

⇒ − × = −

3 303/ 2 0

2

282

mdmt dq q

π επ επ⇒ = × =

P

d

q+

0

x

q−

d





Q3. A two dimensional box in a uniform magnetic field B contains 2N localised spin- 1

2

particles with magnetic moment μ , and 2N free spinless particles which do not interact

with each other. The average energy of the system at a temperature T is:

(a) 13 sinh2 B

BNkT N Bk Tμμ

⎛ ⎞− ⎜ ⎟

⎝ ⎠ (b) 1 tanh

2 B

BNkT N Bk Tμμ

⎛ ⎞− ⎜ ⎟

⎝ ⎠

(c) 1 1 tanh2 2 B

BNkT N Bk Tμμ

⎛ ⎞− ⎜ ⎟

⎝ ⎠ (d) 3 1 cosh

2 2 B

BNkT N Bk Tμμ

⎛ ⎞+ ⎜ ⎟

⎝ ⎠

Ans: (c)

Solution: For 2N free particle in two dimension energy is

2N kT , for

2N localized spin- 1

2

particle the energy is 1 tanh2 B

BN Bk Tμμ

⎛ ⎞− ⎜ ⎟

⎝ ⎠

1 tanh2 2 B

NkT BN Bk Tμμ

⎛ ⎞− ⎜ ⎟

⎝ ⎠

Q4. An ideal gas has a specific heat ratio 2P

V

CC

= . Starting at a temperature 1T the gas under

goes an isothermal compression to increase its density by a factor of two. After this an

adiabatic compression increases its pressure by a factor of two. The temperature of the

gas at the end of the second process would be:

(a) 1

2T (b) 12T (c) 12T (d) 1

2T

Ans (b)

During the isothermal process 1T T= is constant

Let us assume the adiabatic process started at point A ( )1 1,P T and at point B the

coordinate is ( )2 2,P T it is given

1 1 221 1 1 1

1 1 2 2 2 1 2 12 12

P PP T P T T T T TP P

γγγ γ γ γ

− −

− − ⎛ ⎞ ⎛ ⎞= ⇒ = ⇒ =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

2 12T T=





Q5. A gas of N molecules of mass m is confined in a cube of volume 3V L= at temperature

T . The box is in a uniform gravitational field ˆgz− . Assume that the potential energy of a

molecule is U mgz= where [ ]0,z L∈ is the vertical coordinate inside the box. The

pressure ( )P z at height z is:

(a) ( )

2exp

2sinh

2

B

B

Lmg z

k TN mgLP zV mgL

k T

⎛ ⎞⎛ ⎞−⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟−⎜ ⎟⎜ ⎟⎝ ⎠=

⎛ ⎞⎜ ⎟⎝ ⎠

(b) ( )

2exp

2cosh

2

B

B

Lmg z

k TN mgLP zV mgL

k T

⎛ ⎞⎛ ⎞−⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟−⎜ ⎟⎜ ⎟⎝ ⎠=

⎛ ⎞⎜ ⎟⎝ ⎠

(c) ( ) Bk TNP zV

= (d) ( ) NP z mgzV

=

Ans: (c)

The partition function of a system is given by 32

22 1 exp

N NNB B

NB

mk T k TV mglZmgL k Th

π ⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞= − −⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

Helmohtz free energy is given by lnB NF k T Z= −

Pressure is given by ,

B

T N

k TNFPV V∂⎛ ⎞= − =⎜ ⎟∂⎝ ⎠





JEST-2015

Q6. For a system in thermal equilibrium with a heat bath at temperature T , which one of the

following equalities is correct? 1

Bk T⎛ ⎞β =⎜ ⎟⎝ ⎠

(a) 22 EEE −=∂∂β

(b) 22 EEE −=∂∂β

(c) 22 EEE +=∂∂β

(d) ( )22 EEE +−=∂∂β

Ans: (a)

Solution:

i

i

Ei

iE

i

E eE

e

β

β

−

−=∑∑

∵

22 2 2 2

2 2

i i i i i

i i

i i

E E E E Ei i i i

i i i iE E

E Ei i

i i

E e E e e E e E eEe e

e e

β β β β β

β ββ β

β

− − − − −

− −− −

⋅∂

= − + = − +∂ ⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ ∑ ∑ ∑∑ ∑∑ ∑

2 2E

E Eβ

∂⇒ = −

∂

Q7. An ideal gas is compressed adiabatically from an initial volume V to a final volume Vα

and a work W is done on the system in doing so. The final pressure of the gas will be

⎟⎟⎠

⎞⎜⎜⎝

⎛=

V

P

CC

γ

(a) γγ ααγ

−−1

VW (b) γγ αα

γ−−1

VW

(c) γααγ

−−1

VW (d) γαα

γ−−1

VW

Ans: (c)

Solution: Work done in adiabatic process

2 2 1 1

1PV PVW

γ−

=−

2 2 1 1PV PVγ γ=





21 2

1

VP P

V

γ⎛ ⎞

= ⎜ ⎟⎝ ⎠

( )1 2P P γα⇒ =

( )2 2

1P V P V

Wγα α

γ−

=−

( )( )2

1WPV γ

γα α−

=−

Q8. What is the area of the irreducible Brillouin zone of the crystal structure as given in the

figure?

(a) 2

2

32

Aπ

(b) 2

2

23Aπ

(c) 2

22Aπ

(d) 2

2

3Aπ

Ans: (a)

Solution: Area of the Brillouin zone can be related to the area of normal cell as

2 2

Area of B.Z.Area of cell A B

π π= =

×

( )2 0 23sin sin 602

A B A B A Aθ× = = =

2

2

2Area of Brillouin zone3Aπ

∴ =

o60 A

BABA ==

A

B060

A B A= =





Q9. A particle in thermal equilibrium has only 3 possible states with energies , 0,−∈ ∈ . If

the system is maintained at a temperatureB

Tk∈

>> , then the average energy of the particle

can be approximated to,

(a) 22

3 Bk T∈ (b)

223 Bk T− ∈

(c) 2

Bk T−∈ (d) 0

Ans: (b)

Solution: 0

1

kT kT

kT kT

e eEe e

ε ε

ε ε

ε ε+ −

−

− + +=

+ + 1

kT kT

kT kT

e e

e e

ε ε

ε εε−

−

⎛ ⎞−⎜ ⎟= ⎜ ⎟⎜ ⎟+ +⎝ ⎠

1 1

1 1 1

kT kTE

kT kT

ε ε

ε ε

⎡ ⎤⎛ ⎞ ⎛ ⎞− − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦⇒ =⎛ ⎞ ⎛ ⎞+ − + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

223kTε−

=

Q10. The blackbody at a temperature of 6000 K emits a radiation whose intensity spectrum

peaks at nm600 . If the temperature is reduced to K300 , the spectrum will peak at,

(a) mμ120 (b) 12 mμ (c) 12mm (d)120mm

Ans: (b)

Solution: 1 1 2 2T Tλ λ= 1 12

2

TTλλ⇒ =

600 20 12000 12300

nm mμ×= = =





Q11. The entropy-temperature diagram of two Carnot engines, A and B , are shown in the

figure 4. The efficiencies of the engines are Aη and Bη respectively. Which one of the

following equalities is correct?

(a) 2B

Aη

η =

(b) BA ηη =

(c) BA ηη 3=

(d) BA ηη 2=

Ans: (d)

Solution: 1

WQ

η Δ= where WΔ = area under the curve , 1Q = area under high temperature

( )2 12 2 2A

T T TT T

η−

= = = and ( ) ( )

( )4 3 4

4 4B

T T S ST S S

η− −

=−

14 4TT

= =

1/ 2 21/ 4

A

B

ηη

⇒ = = 2A Bη η⇒ =

Q12. Electrons of mass m in a thin, long wire at a temperature T follow a one-dimensional

Maxwellian velocity distribution. The most probable speed of these electrons is,

(a) ⎟⎠⎞

⎜⎝⎛

mkTπ2

(b) ⎟⎠⎞

⎜⎝⎛

mkT2 (c) 0 (d) ⎟

⎠⎞

⎜⎝⎛

mkTπ8 .

Ans: (c)

Solution: ( )21/ 2

2 ;2

xmvkT

x x xmf v e dv vkTπ

−⎛ ⎞= −∞ < < ∞⎜ ⎟⎝ ⎠

Most probable speed 0xv =

S

A B

T

( )xf v

xv





JEST-2014

Q13. A monoatomic gas consists of atoms with two internal energy levels, ground state 00 =E

and an excited state EE =1 . The specific heat of the gas is given by

(a) k23 (b)

( )2/2

/2

1 kTE

kTE

ekTeE+

(c) ( )2/2

/2

123

kTE

kTE

ekTeEk+

+ (d) ( )2/2

/2

123

kTE

kTE

ekTeEk+

−

Ans.: (c)

Solution: 0 10,E E E= =

iEz e β−= ∑ 0 Ez e eβ β− × −⇒ = +

( )11lnln Eez β−+=

( ) ( ) ( )1ln ln 11

E EE

U E z e E ee

β βββ β

− −−

−∂ ∂= = = − + = − −

∂ ∂ +

1

E

BE

EeU k Te

β

β β−

−= =+

∵

2 2

2

1 . .

1

B B B B

B

E E E Ek T k T k T k T

B BV E

vk T

E Ee E e Ee ek T k TU C

Te

− − − −

−

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ − ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠∂⎛ ⎞ ⎝ ⎠= =⎜ ⎟∂⎝ ⎠ ⎛ ⎞

+⎜ ⎟⎜ ⎟⎝ ⎠

2 22 2 2

2 2 2 2 2

2 2 2

2 21 1 1

B B B

B B

B B B

E E EE Ek T k T k T

k T k TB B B

V E E Ek T k T k T

B B

E E Ee e ek T k T k T E e E eC

e k T e k T e

− − −−

− −

+ −= = =

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

If gas will classically allowed then 32V BC k=





and due to quantum mechanically 2

2

2 1

B

B

Ek T

V Ek T

B

E eC

k T e

=⎛ ⎞+⎜ ⎟⎜ ⎟

⎝ ⎠

( )

2 /

22 /

32 1

E kT

V B E kT

E eC kkT e

∴ = ++

Q14. The temperature of a thin bulb filament (assuming that the resistance of the filament is

nearly constant) of radius r and length L is proportional to

(a) 2/14/1 −Lr (b) rL2 (c) 14/1 −rL (d) 12 −Lr

Ans.: (a)

Q15. Ice of density 1ρ melts at pressure P and absolute temperature T to form water of

density 2ρ . The latent heat of melting of 1 gram of ice is L . What is the change in the

internal energy UΔ resulting from the melting of 1 gram of ice?

(a) ⎟⎟⎠

⎞⎜⎜⎝

⎛−+

12

11ρρ

PL (b) ⎟⎟⎠

⎞⎜⎜⎝

⎛−−

12

11ρρ

PL

(c) ⎟⎟⎠

⎞⎜⎜⎝

⎛−−

21

11ρρ

PL (d) ⎟⎟⎠

⎞⎜⎜⎝

⎛−+

21

11ρρ

PL

Ans.: (c)

Solution: dU dQ W dQ pdVδ= − = −

dU mL pdV= −2

1

2

1dU L P dρ

ρ

ρρ

⎛ ⎞⇒ = − −⎜ ⎟

⎝ ⎠∫

1 2

1 1L Pρ ρ⎡ ⎤

= − −⎢ ⎥⎣ ⎦

2

1 1V dV dρρ ρ

= ⇒ = −∵

Q16. What is the contribution of the conduction electrons in the molar entropy of a metal with

electronic coefficient of specific heat?

(a) Tγ (b) 2Tγ (c) 3Tγ (d) 4Tγ

Ans.: (a)

Solution: 3VC BT AT= +





Q17. Consider a system of 2N non-interacting spin 2/1 particles each fixed in position and

carrying a magnetic momentμ . The system is immersed in a uniform magnetic field B.

The number of spin up particle for which the entropy of the system will be maximum is

(a) 0 (b) N (c) N2 (d) 2/N

Ans.: (b)

Solution: Let us consider n number of spin out of N2 particle have spin up remaining nN −2 is

down.

Number of ways nCN2=ω for spin 21 (up)

22

N nCNω−

= for spin 21 down

entropy ωlnkS = 2 22ln lnN N

N n nS k C k C−⇒ = +

( )( ) ( )( )2 ! 2 !ln ln

! 2 ! ! 2 !N NS K

n N n n N n

⎧ ⎫⎡ ⎤ ⎡ ⎤⎪ ⎪= +⎢ ⎥ ⎢ ⎥⎨ ⎬− −⎢ ⎥ ⎢ ⎥⎪ ⎪⎣ ⎦ ⎣ ⎦⎩ ⎭

( )( )[ ]!2ln!ln!2ln2 nNnNkS −−−=

( ) ( ) ( ) 2 2 ln 2 2 ln 2 ln 2 2S K N N N n n n N n N n N n⎡ ⎤= − − + − − − − −⎣ ⎦!ln!ln NNNN −=∵

( ) ( ) ( )2 2 ln 2 2 ln 2 ln 2 ln 2 2S K N N N n n n N N n n N n N n⎡ ⎤= − − + − − + − + −⎣ ⎦

( ) ( )2 2 ln 2 ln 2 ln 2 ln 2S K N N n n N N n n N n⎡ ⎤= − − − + −⎣ ⎦

now for entropy maximum at equilibrium for spin 21 up particle

0=dndS

( ) ( ) ( )22 1 ln 1 1 ln 22 2

n N nK n N nn N n N n

⎡ ⎤= − ⋅ − − − + − + −⎢ ⎥− −⎣ ⎦

( )⎥⎦⎤

⎢⎣⎡ −+

−−

−+−−= nN

nNn

nNNnK 2ln

222ln12





( ) ⎥⎦⎤

⎢⎣⎡ −−+

−−

+−= nnNnNnNK ln2ln

2212

( ) 02ln112 =⎥⎦⎤

⎢⎣⎡ −

++−=n

nNk

02 ≠k∵

02ln =⎟⎠⎞

⎜⎝⎛ −

∴n

nN 12=

−⇒

nnN nN 22 =⇒ n N⇒ =

Q18. For which gas the ratio of specific heats ( )vp CC / will be the largest?

(a) mono-atomic (b) di-atomic (c) tri-atomic (d) hexa-atomic

Ans.: (a)

Solution: 21P

V

CC f

γ⎛ ⎞

= = +⎜ ⎟⎝ ⎠

where f is degree of freedom.

For monoatomic: 3f = , For diatomic: 6f = , For Triatomic: 9f =

For hexaatomic: 18f =

JEST-2013

Q19. Consider a system of two particles A and B . Each particle can occupy one of three

possible quantum states 2,1 and 3 . The ratio of the probability that the two particles

are in the same state to the probability that the two particles are in different states is

calculated for bosons and classical (Maxwell-Boltzmann) particles. They are respectively

(a) 1, 0 (b) 1 , 12

(c) 11,2

(d) 10,2

Ans.: (c)

Solution: For two particle in same state:

3 3 3 3 3 3AB AB

2 2 2 2 2 2AB AB

1 1 1 1 1 1AB AB

Boson Boltzman)-(Maxwell Classical





Probability ratio: 1/ 3 11/ 3

=

For two particle in different states

Probability ratio: 21

3/23/1=

Q20. For a diatomic ideal gas near room temperature, what fraction of the heat supplied is

available for external work if the gas is expanded at constant pressure?

(a) 17

(b) 57

(c) 34

(d) 27

Ans.: (d)

Solution: It is isobaric process (constant pressure)

Then pnC T W nR Tδθ = Δ ⇒ Δ = Δ

In this process δθ is heat exchange during process.

Function of heat supplied

1 11

1p

W nR T RQ nC T R

δ γγ γ γ

γ

Δ −= = = = = −Δ Δ

−

11121

pp

V

C RCC

f

γγγ

⇒ − = ⇒ =−⎛ ⎞

+⎜ ⎟⎝ ⎠

21

+−⇒

ff =f degree of freedom, for diatomic molecule 5f =

72

2551 ⇒+

−⇒

Boson Classical (Maxwell-Boltzmann)

3 3 3 3 3 3B B B 3 3 3BA A

2 2 2 2 2 2B A B 2 2 2A BA

1 1 1 1 1 1 1 1 1BA A A B A





Q21. Consider the differential equation

( ) ( ) ( )xxkGdx

xdG δ=+ ,

where k is a constant. Which of the following statements is true?

(a) Both ( )G x and ( )G x′ are continuous at 0x =

(b) ( )G x is continuous at 0x = but ( )G x′ is not.

(c) ( )G x is discontinuous at 0x =

(d) The continuity properties of ( )G x and ( )G x′ at 0x = depend on the value of k .

Ans.: (c)

Q22. A metal bullet comes to rest after hitting its target with a velocity of 80 m/s. If 50% of the

heat generated remains in the bullet, what is the increase in its temperature? (The specific

heat of the bullet 0160 / /Joule kg C= )

(a) 014 C (b) 012.5 C (c) 010 C (d) 08.2 C

Ans.: (c)

Solution: Conservation of momentum 21 150 % 80 80 1602 2

mv mc T T× = Δ ⇒ × = Δ

080 80 1 104 160

T C×⇒ Δ = × =

Q23. Consider a particle with three possible spin states: 0s = and 1± . There is a magnetic

field h present and the energy for a spin state s is hs− . The system is at a

temperatureT . Which of the following statements is true about the entropy ( )S T ?

(a) ( ) ln 3 at 0, and 3 at high S T T T= = (b) ( ) ln 3 at 0, and 0 at high S T T T= =

(c) ( ) 0 at 0, and 3 at high S T T T= = (d) ( ) 0 at 0, and ln 3 at high S T T T= =

Ans.: (d)

Solution: ωlnkS = where =ω number of microstates

3lnkS = at high T 3=ω

and at 0=T it is perfect ordered i.e. 0=S





Q24. Consider three situations of 4 particles in one dimensional box of width L with hard

walls. In case (i), the particles are fermions, in case (ii) they are bosons, and in case (iii)

they are classical. If the total ground state energy of the four particles in these three cases

are ,F BE E and clE respectively, which of the following is true?

(a) F B clE E E= = (b) F B clE E E> =

(c) F B clE E E< < (d) F B clE E E> >

Ans.: (b)

Solution: For fermions 02

22

2=∈

mlπ

0 0 0 0 01 1 4 1 9 1 16 30×∈ + × ∈ + × ∈ + × ∈ = ∈

0 04 , 4For Boson For Maxwell= ×∈ = ×∈

F B clE E E> =

JEST-2012

Q25. A monoatomic ideal gas at 170C is adiabatically compressed to 1/8 of its original

volume. The temperature after compression is

(a) 2.1 oC (b) 17oC (c) -200.5oC (d) 887oC

Ans. : (d)

Solution: costantPV γ⇒ = , RTPV =

costantTVV

γ

=

1 costantTV γ −⇒ =

1 11 1 2 2TV T Vγ γ− −⇒ =

1

12 1

2

rV

T TV

−⎛ ⎞

⇒ = ⎜ ⎟⎝ ⎠

( ) 64.154140.3443443 6. =×=>= θ

CT 02 12627364.1541 θ=−=

⎟⎠⎞

⎜⎝⎛ +=

121r∵ 66.1

321 =+=





Q26. Consider a system of particles in three dimensions with momentum p and energy

cpcE ,= being a constant. The system is maintained at inverse temperature β, volume

V and chemical potential μ. What is the grand partition function of the system?

(a) ( )[ ]3/8exp chVe βπβμ (b) ( )2/6 chVe βπβμ

(c) ( )[ ]3/6exp chVe βπβμ (d) ( )2/8 chVe βπβμ

Ans. : (a)

Solution: canonical partition function

3

1 HN x y zz e dp dp dp dxdydz

hβ−= ∫ pcE =

dpephVz E

Nβπ −

∞

∫=0

23 4

( ) ( )3330

23

8344hc

Vch

VdpephV pc

βπ

βππ β =⋅== −

∞

∫

grand canonical partition function ⎥⎦

⎤⎢⎣

⎡= N

kTu zez

μ

exp( ) ⎥

⎦

⎤⎢⎣

⎡⋅= 3

8exphcVe kT

βπμ

( ) ⎥⎦

⎤⎢⎣

⎡⋅⇒ 3

8exphcVe

βπβμ

Q27. Consider a system maintained at temperature T, with two available energy states E1 and

E2 each with degeneracies g1 and g2. If p1 and p2 are probabilities of occupancy of the two

energy states, what is the entropy of the system?

(a) ( ) ( )[ ]222111 /ln/ln gppgppkS B +−=

(b) ( ) ( )[ ]222111 lnln gppgppkS B +−=

(c) ( ) ( )[ ]212211 lnln gg

B ppppkS +−=

(d) ( ) ( ) ( ) ( )[ ]222111 /ln/1/ln/1 gppgppkS B +−=

Ans. : (a)

Solution: iE

ii

g epz

β−Σ= where z is partition function

zEgp iii lnlnln −−= β





zEgp iii lnlnln −−=− β zkTF ln−=∵

kTFE

gp

ii

i +−= βln

FEgp

ii

i ββ +−=ln

FEgp

ii

i ββ +−=ln

[ ]ln i

i

p F Ug

β= − F U TS= −∵

TSgp

i

i ×−= βln kT1

=β

1 21 2

1 2

ln ln ln lni ii

i i

p p p pS k k p k p pg g g g

⎛ ⎞ ⎡ ⎤= − = − = − +⎜ ⎟ ⎢ ⎥

⎣ ⎦⎝ ⎠∑

Q28. Consider an ideal gas of mass m at temperature 1T which is mixed isobarically (i.e. at

constant pressure) with an equal mass of same gas at temperature 2T in a thermally

insulated container. What is the change of entropy of the universe?

(a) ⎟⎟⎠

⎞⎜⎜⎝

⎛ +

21

21

2ln2

TTTTmC p (b) ⎟

⎟⎠

⎞⎜⎜⎝

⎛ −

21

21

2ln2

TTTTmC p

(c) ⎟⎟⎠

⎞⎜⎜⎝

⎛ +

21

21

2ln2

TTTT

mC p (d) ⎟⎟⎠

⎞⎜⎜⎝

⎛ −

21

21

2ln2

TTTTmC p

Ans. : (a)

Solution: Let us consider final temperature will be T

( ) ( )21 TTmcTTmc −=−

221 TTT +

=

TTmcS p

Δ=Δ 1





now 21 SSS Δ+Δ=Δ ∫∫ +=Δ⇒T

Tp

T

Tp T

dTmcTdTmcS

21

1 2

ln lnp pT TS mc mcT T⎛ ⎞ ⎛ ⎞

Δ = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

1 2

1 2 1 2

2 ln ln2p pT TTS mc mc

TT TT

⎛ ⎞+Δ = = ⎜ ⎟⎜ ⎟

⎝ ⎠

1 2

1 2

2 ln2pT TS mc

TT

⎛ ⎞+⇒ Δ = ⎜ ⎟⎜ ⎟

⎝ ⎠

Q29. A collection of N two-level systems with energies 0 and E > 0 is in thermal equilibrium

at temperature T. For T → ∞, the specific heat approaches

(a) 0 (b) NkB (c) 3NkB/2 (d) ∞

Ans.: (a)

Solution: ∑ −×−− +== ii EE eeeZ βββ 0 EeZ β−+=⇒ 1 ( )Eez β−+=⇒ 1lnln

( ) ( )1ln ln 1 ,2 1 1

EE E

E E

EeU E z e e Ee e

ββ β

β ββ β

−− −

− −

∂ ∂= = − = − + = − × − =

∂ + +

now 1

EkT

V EV kT

U EeCT T

e

−

−

⎛ ⎞∂ ∂⎛ ⎞ ⎜ ⎟= =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎜ ⎟+⎝ ⎠

2

2

2

22

2

2

2

2

1 ⎟⎟⎠

⎞⎜⎜⎝

⎛+

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

=−

−−−

kTE

kTE

kTE

kTE

V

e

ekTEe

kTEe

kTE

C 2

2

2

1 ⎟⎟⎠

⎞⎜⎜⎝

⎛+

=⇒−

−

kTE

kTE

V

e

ekTE

C 0V TC

→∞⇒ =

Q30. Efficiency of a perfectly reversible (Carnot) heat engine operating between absolute

temperature T and zero is equal to

(a) 0 (b) 0.5 (c) 0.75 (d) 1

Ans. : (d)

Solution: 10

111

2 =−=−=T

TT

η





Q31. A thermally insulated ideal gas of volume V1 and temperature T expands to another

enclosure of volume V2 through a prous plug. What is the change in the temperature of

the gas?

(a) 0 (b) T ln(V1 / V2) (c) T ln(V2 / V1) (d) T ln(V2 – V1) / V2)

Ans. : ()

Solution: VdPTdSdH += , for porous plug Joul Thomshon 0=dH and 0=TdS since it is

thermally insulated ideal gas

0=VdP

VnRTdVnRdTpdVnRdTVdP =⇒=⇒= 0∵

1

2ln2

1 VVTdT

VdVTdT

VdVTdT

V

V=⇒=⇒= ∫





Electronics

JEST-2016

Q1. It is found that when the resistance R indicated in the figure below is changed from

1 kΩ to 10 kΩ the current flowing through the resistance R′ does not change. What is

the value of the resistor R′?

(a) 5 kΩ (b) 100 kΩ (c) 10 kΩ (d) 1 kΩ

Ans: (b)

Solution: Apply Wheatstone bridge condition

31

2 4

RRR R

=1 10

1R⇒ =

′

Q2. A transistor in common base configuration has ratio of collector current to emitter current

β and ratio of collector to base currentα . Which of the following is true?

(a) ( )1αβ

α=

+ (b) ( )1α

βα+

=

(c) ( )1αβ

α=

− (d) ( )1α

βα−

=

Ans: (a)

Solution: 1 11 1E BE C B

C C

I II I II I β α

= + ⇒ = + ⇒ = +∵1αβα

⇒ =+

5V

10kΩ

1kΩ

10kΩ

R

R′

1kΩ

5V

1kΩ

10kΩ

R

R′

1kΩ

10kΩ

A• •B

1R =

2R =

3R =

4R =





JEST-2015

Q3. What is the voltage at the output of the following operational amplifier circuit. [See in the

figure]?

(a) 1V

(b) 1mV

(c) 1 Vμ

(d)1nV

Ans: (b)

Solution: Output of first Op-Amp ( )( )3 9 501 10 10 1 10 10v volt− −= − × × = −

Output of first Op-Amp 5 3991 10 10 11outv volts mV− −⎛ ⎞= + × = =⎜ ⎟

⎝ ⎠

Q4. The reference voltage of an analog to digital converter is V1 . The smallest voltage step

that the converter can record using a 12 -bit converter is,

(a) V24.0 (b) mV24.0 (c) Vμ24.0 (d) nV24.0

Ans: (b)

Smallest voltage step 12

1 0.242 1

mV= ≈−

Q5. In Millikan’s oil drop experiment the electronic charge e could be written as 5.1ηk ,where

κ is a function of all experimental parameters with negligible error. If the viscosity of air

η is taken to be %4.0 lower than the actual value, what would be the error in the

calculated value of e ?

(a) %5.1 (b) %7.0 (c) %6.0 (d) %4.0

Ans: (d)

Solution: Electronic charge is proportional to the viscosity i.e. 1.5 3/ 2e K Kη η= =

Now error in the measurement of charge is 2

2 2e

eησ σ

η⎛ ⎞∂

= ⎜ ⎟∂⎝ ⎠

−

+−

+

LRoutV

Ωk99Ωk1

An1

Ωk10





ee

ησ ση

⎛ ⎞∂⇒ = ⎜ ⎟∂⎝ ⎠

where 1/ 232

e Kηη∂

=∂

1/ 2 3 / 23 3 32 2 2e K K eη η

η

σ σσ η σ η

η η⎛ ⎞∴ = = =⎜ ⎟⎝ ⎠

32

e

eησση

⇒ =

Given 0.4%ηση

=

3 0.4% 0.6%2

e

eσ

∴ = × = . Thus correct answer is option (c).

Q6. For the logic circuit shown in figure 5, the required input condition ( )CBA ,, to

make the output ( ) 1=X is,

(a) 1,0,1

(b) 1,0,0

(c) 1,1,1

(d) 1,1,0

Ans: (d)

Solution: XOR is inequality comparator and XNOR is equality comparator. In AND gate output

will be high when all the input is 1.

AB

C

X

AND

XOR

XNOR

1U

2U

3U





JEST-2014

Q7. Which of the following circuits will act like a 4-input NAND gate?

(a) (b)

(c) (d)

Ans.: (d)

Solution:

ABC

D

AB ABABC

ABCD

ABC

D

ABABC

ABCD

ABCD

AB

CD

ABCD

ABCD

AB

CD

AB

CD

ABCD





Q8. The formula for normal strain in a longitudinal bar is given by ,AEF

=ε where F is

normal force applied, A is cross-sectional area of the bar and E is Young’s modulus. If 2002.02.0,5.050 mANF ±=±= and 99 10110210 ×±×=E Pa, the maximum error in

the measurement of strain is

(a) 12100.1 −× (b) 111095.2 −× (c) 91022.1 −× (d) 91019.1 −×

Ans.: (b)

Solution: AEF

∈=

9

9

10210101

2.0002.

505.0

××

++=Δ

+Δ

+Δ

=∈∈Δ

EE

AA

FF

.02476 0.2476Δ∈= ⇒ Δ∈= ×∈

∈11

9

0.2476 50 2.95 100.2 210 10

−×= = ×

× ×

Q9. A 100 ohms resistor carrying current of 1 Amp is maintained at a constant temperature of

Co30 by a heat bath. What is the rate of entropy increase of the resistor?

(a) 3.3 Joules/K/sec (b) 6.6 Joules/K/sec

(c) 0.33 Joules/K/sec (d) None of the above

Ans.: (c)

Solution: .qVω = W i t R⇒ = ⋅ ⋅ 2W i Rt=

now2 1 100 0.33

30 273W i RtT T

∂ ×= = =

∂ +

JEST-2012

Q10. The ratio of maximum to minimum resistance that can be obtained with N 1-Ω resistors is

(a) N (b) N2 (c) 1 (d) ∞

Ans.: (b)

Solution: resistance in series is maximum and minimum in parallel

NNRs =++++= .....1111





NNRp

=++++=1.....

11

11

11

11

111

2NNNRpRs

=×=N

Rp 1=⇒

Q11. The net charge of an n-type semiconductor is

(a) positive (b) zero (c) negative (d) dependent

Ans.: (b)





Solid State Physics

JEST-2016

Q1. If k is the wavevector of incident light ( 2k πλ

= , λ is the wavelength of light) and G is

a reciprocal lattice vector, then the Bragg’s law can be written as:

(a) 0k G+ = (b) 22 . 0k G G+ =

(c) 22 . 0k G k+ = (d) . 0k G = Ans. : (b)

Solution: By means of Eward construction we can write the Bragg’s law in

vector form

,G OB K AO′= =

For diffraction it is necessary that vector K G′ + , that is vector AB

be equal in magnitude to the vector K or

( )2 2 22 0K G K K G G+ = ⇒ ⋅ + =

Q2. The number of different Bravais lattices possible in two dimensions is:

(a) 2 (b) 3 (c) 5 (d) 6

Ans. : (c)

Solution: Five Bravais lattices in 2D are:

(i) square lattice

(ii) Rectangular ( )P lattice

(iii) Rectangular ( )C lattice

(iv) Hexagonal lattice

(v) Oblique lattice

AK

G

K

B

O





JEST-2015

Q3. For a 2 - dimensional honeycomb lattice as shown in the figure 3, the first Bragg spot

occurs for the grazing angle 1θ while sweeping the angle from o0 . The next Bragg spot is

obtained at 2θ given by

(a) ( )11 sin3sin θ− (b) ⎟

⎠⎞

⎜⎝⎛−

11 sin

23sin θ

(c) ⎟⎟⎠

⎞⎜⎜⎝

⎛−1

1 sin23sin θ (d) ( )1

1 sin3sin θ−

Ans: (c)

Solution: According to Bragg’s law, the condition for first Bragg spot and second spot is

1 12 sind nθ λ= and 2 22 sind nθ λ=

1 11 1 2 2 2 1

2

2 sin 2 sin sin sindd dd

θ θ θ θ− ⎛ ⎞∴ = ⇒ = ⎜ ⎟

⎝ ⎠

For 2 - dimensional honeycomb lattice, the lattice constant ‘ a ’ and interplanar spacing

‘ d ’ is linked as 2 2

2 2 21 1

32 4 2a ad a d a a⎛ ⎞= − ⇒ = − =⎜ ⎟

⎝ ⎠ and 2d a=

12 1

3sin sin2

θ θ− ⎛ ⎞∴ = ⎜ ⎟⎜ ⎟

⎝ ⎠

Q4. A particle of mass m is confined in a potential well given by ( ) 0V x = for 22LxL

<<−

L/2 and ( ) ∞=xV elsewhere. A perturbing potential ( ) axxH =′ has been applied to the

system. Let the first and second order corrections to the ground state be ( )10E and ( )2

0E ,

respectively. Which one of the following statements is correct?

(a) ( ) ( ) 0and0 20

10 >< EE (b) ( ) ( ) 0and0 2

01

0 >= EE

(c) ( ) ( ) 0Eand0 20

10 <>E (d) ( ) ( )1 2

0 00 and E 0E = < Ans: (d)

o120

o120o120aA

B

1d060

a

a





Solution: ( )0 / 2 / 2L x L

V xelsewhere

− < < +⎧= ⎨∞⎩

and ( )H x xα′ =

For ground state 02 cos xL L

πφ =

( ) /21 0 0 20 /2

0 0

2 cos 0L

L

H xE xL L

φ φ παφ φ −

′= = =∫

( ) ( )2

02 2 0 00 0 00 0

0 0

0mm

m m

HE E E E

E Eφ φ

≠

′= ⇒ < <

−∑ ∵

Q5. Given the tight binding dispersion relation ( ) ⎟⎠⎞

⎜⎝⎛+=

2sin 2

0kaAEkE , where 0E and A are

constants and a is the lattice parameter. What is the group velocity of an electron at the

second Brillouin zone boundary?

(a) 0 (b) ha (c)

ha2 (d)

ha2

Ans: (a)

Solution: Group velocity is defined as 1g

dEvdk

=

since 20 sin

2kaE E A ⎛ ⎞= + ⎜ ⎟⎝ ⎠

sin cos sin2 2 2

dE ka ka aAaA kadk

⎛ ⎞ ⎛ ⎞⇒ = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

In one dimension, the Brillouin zone boundary is

The 1st Brillouin zone boundaries lie at aπ

±

The 2nd Brillouin zone boundaries lie at 2aπ

±

Thus, the group velocity at the second Brillouin zone boundary is

22sin sin 2

2 2ga

aA aAv aaππ

π±

⎛ ⎞= × =⎜ ⎟⎝ ⎠0gv⇒ =

2aπ− a

π− aπ 2

aπ0

K





Q6. The total number of +Na and −Cl ions per unit cell of NaCl is,

(a) 2 (b) 4 (c) 6 (d) 8

Ans: (d)

Solution: Total number of Na+ and Cl− ions per unit ( )d is

1 18 2Cl c fN n n− = + , 1 1

4Na e iN n n+ = + ×

where cn = number of ions at corner fn = number of ions at face en = number of ions at edges in = number of ions inside

1 1 18 6 12 1 1 1 3 3 1 88 2 4Cl NaN N N− += + = × + × + × + × = + + + =

JEST-2014

Q7. Circular discs of radius 1 m each are placed on a plane so as to form a closely packed

triangular lattice. The number of discs per unit area is approximately equal to

(a) 286.0 −m (b) 243.0 −m (c) 229.0 −m (d) 214.0 −m

Ans.: (c)

Solution: For closely packed hexagonal

,2ra = 1=r 1 1 16 2eff C f ln n n n= × + × + ×

1 13 0 1 06 2effn⇒ = × + × + × 0.5effn⇒ =

Occupancy ( )2

2effn ra r

Aπ×

= =∵

( )

2

2

0.53 2

4

r

r

π×⇒

×

0.5 0.90643π×

= =

Now number of disc per unit area will be 29.0302.3

9064.0≈=

Closely packed hexagonal

Cl −Na+





Q8. An ideal gas of non-relativistic fermions in 3-dimensions is at 0K. When both the number

density and mass of the particles are doubled, then the energy per particle is multiplied by

a factor

(a) 2/12 (b) 1 (c) 3/12 (d) 3/12−

Ans.: (d)

Solution: 32

2

43

2⎟⎠⎞

⎜⎝⎛=πn

mhEF at 0T K=

2n n′ =∵ and 2m m′ =

2 222 23 333 3 12 2

4 4 2 4 2Fh h nE nm mπ π⎛ ⎞ ⎛ ⎞′⇒ = × = × ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

212 3

33 22 4h nm π

−⎛ ⎞= ×⎜ ⎟⎝ ⎠

Q9. When two different solids are brought in contact with each other, which one of the

following is true?

(a) Their Fermi energies become equal

(b) Their band gaps become equal

(c) Their chemical potentials become equal

(d) Their work functions become equal

Ans.: (c)

JEST-2013

Q10. A flat surface is covered with non-overlapping disks of same size. What is the largest

fraction of the area that can be covered?

(a) π3 (b)

65π (c)

76 (d)

32π

Ans.: (d)

Solution: 1 1 11 6 1 33 2 3eff C f in n n n= + + × = × + = ra 2=

Now largest fraction of area i.e. packing fraction ( ) 322

436

3

436 2

2

2

ππ=

××

×=

××

×=

r

r

a

Aneff





Q11. A metal suffers a structural phase transition from face-centered cubic ( )FCC to the

simple cubic ( )SC structure. It is observed that this phase transition does not involve any

change of volume. The nearest neighbor distances fcd and SCd for the FCC and the SC

structures respectively are in the ratio fc

SC

dd

⎛ ⎞⎜ ⎟⎝ ⎠

[Given 132 1.26= ]

(a) 1.029 (b) 1.122 (c) 1.374 (d) 1.130

Ans. 1: ()

Solution: Nearest neighbour in 6. =→→ NCaSC

Nearest neighbour in 12.2

=→→ NCaFCC

707.0414.11

212 ====

a

a

dSCdFCC

JEST-2012

Q12. A beam of X-rays is incident on a BCC crystal. If the difference between the incident and

scatered wavevectors is zlykxK ˆˆˆ ++= where zyx ˆ,ˆ,ˆ are the unit vectors of the

associated cubic lattice, the necessary condition for the scattered beam to give a Laue

maximum is

(a) h + k + l = even (b) h + k + l = even

(c) h, k, l are all distinct (d) h + k + l = odd

Ans.: (a)

Solution: In BCC basis ( ) ⎟⎠⎞

⎜⎝⎛

21,

21,

21,0,0,0

Crystal structure factor F

( ) ( )2

1

effn n n

ni hu kV l

nF f S S e π ω+ +

=

= ⇒ =∑

( ) [ ]lkheefF ii ++⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+=

210 22 ππ





( )1 i h k lF f eπ + +⎡ ⎤= +⎣ ⎦

now for plane ( )011

( ) 2110 42 fIfF =⇒=

0111 =F 2

200 2F f I f= ⇒ =

i.e h k l+ + = even then plane will be present and if =++ lkh odd then plane will be

absent.

Q13. The second order maximum in the diffraction of X-rays of 0.20 nanometer wavelength

from a simple cubic crystal is found to occur at an angle of thirty degrees to the crystal

plane. The distance between the lattice planes is

(a) 1 Angstrom (b) 2 Angstrom (c) 4 Angstrom (d) 8 Angstrom

Ans.: (c)

Solution: λθ nd =sin2

λθ 2sin2 =d

md 9o 102.0230sin2 −××=××

md 9102.02 −××= o109 4104104.0 Ammd =×⇒×= −−

Q14. The Dulong –Petit law fails near room temperature (300 K) for many light elements (such

as boron and beryllium) because their Debye temperature is

(a) >> 300 K (b) ~ 300 K (c) << 300 K (d) 0 K

Ans.: (a)





Nuclear & Particle Physics

JEST-2016

Q1. The half-life of a radioactive nuclear source is 9 days. The fraction of nuclei which are

left under cayed after 3 days is:

(a) 78

(b) 13

(c) 56

(d) 13

1

2

Ans: (d)

Solution: 3/9

0 0 1/30

1 1 12 2 2

n NN N NN

⎛ ⎞ ⎛ ⎞= = ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

JEST-2015

Q2. The stable nucleus that has 31 the radius of Os189 nucleus is,

(a) Li (b) O16 (c) He4 (d) N14

Ans: (a)

Solution: ( ) ( )1/3 1/30 0

1 1 189 73 3OsR R R A R A= ⇒ = ⇒ =

Q3. The reaction γ→+ −+ ee is forbidden because,

(a) lepton number is not conserved

(b) linear momentum is not conserved

(c) angular momentum is not conserved

(d) charge is not conserved

Ans: (b)

Solution: In order to conserve linear momentum two photons are required that move in opposite

direction.





JEST-2014

Q4. In the mixture of isotopes normally found on the earth at the present time, U23892 has an

abundance of 99.3% and U23592 has an abundance of 0.7%. The measured lifetimes of

these isotopes are 91052.6 × years and 91002.1 × years, respectively. Assuming that they

were equally abundant when the earth was formed, the estimated age of the earth, in

years is

(a) 9100.6 × (b) 9100.1 × (c) 8100.6 × (d) 8100.1 ×

Ans.: (a)

Solution: If the number of 92 238U nuclei originally formed is N , the number present now is

/ / 6.52238

t T tN Ne Ne− −= =

where t is elapsed time in units of 910 year and T is life time of U . Since the number of 92 235U nuclei originally formed is. The number now present is

/1.02235

tN Ne−= The present abundance of 92 235U is

/1.02

3 0.827235 235/ 6.52 3

238 235 238

1 4.967 10 143 6.07 10 0.827

tt

t

N N Ne e tN N N Ne

−−

− −× = ≈ = = ≈ = = = =+ ×

That is, the elapsed time is 96.0 10t = × yr.

JEST-2013

Q5. 238U decays with a half life of 194.51 10× years, the decay series eventually ending at 206 Pb , which is stable. A rock sample analysis shows that the ratio of the numbers of

atoms of 206 Pb to 238U is 0.0058 . Assuming that all the 206 Pb has been produced by the

decay of 238U and that all other half-lives in the chain are negligible, the age of the rock

sample is

(a) 638 10× years (b) 648 10× years (c) 738 10× years (d) 748 10× years

Ans.: (a)

Solution: 1 ln pb u

u u

N Nt

Nλ+⎛ ⎞

= ⎜ ⎟⎝ ⎠





Q6. The binding energy of the k -shell electron in a Uranium atom ( )92, 238Z A= = will be

modified due to (i) screening caused by other electrons and (ii) the finite extent of the

nucleus as follows:

(a) increases due to (i), remains unchanged due to (ii)

(b) decreases due to (i), decreases due to (ii)

(c) increases due to (i), increases due to (ii)

(d) decreases due to (i), remains unchanged due to (ii)

Ans.: (b)





Optics

JEST 2015

Q1. Let λ be the wavelength of incident light. The diffraction pattern of a circular aperture of

dimension 0r will transform from Fresnel to Fraunhofer regime if the screen distance z

is,

(a) λ

20rz >> (b)

0

2

rz λ>> (c)

0

2

rz λ<< (d)

λ

20rz <<

Ans: (a)

Solution: Fraunhofer made an approximation on the quadratic phase function:

( )2 2 20 0 0

2 2 1k x y kri i

z ze e+

= ≈

If 2 2

0 0

2kr rz z π

λ>> ⇒ >>

20rzλ

⇒ >>

For this reason Fraunhofer diffraction is also called Far-field diffraction, whereas for

Fresnel diffraction, the condition is

z λ>> called near-field diffraction.

X

Y

0r = aperture





JEST 2014

Q2. A spherical air bubble is embedded in a glass slab. It will behave like a

(a) Cylindrical lens (b) Achromatic lens (c) Converging lens (d) Diverging lens

Ans.: (c)

Q3. The resolving power of a grating spectrograph can be improved by

(a) recording the spectrum in the lowest order

(b) using a grating with longer grating period

(c) masking a part of the grating surface

(d) illuminating the grating to the maximum possible extent

Ans.: (d)

Solution: R P nNλλΔ

⇒ ⋅ = = where N - Number of slit and n - order of diffraction.

Q4. Three sinusoidal waves have the same frequency with amplitude 2/, AA and 3/A while

their phase angles are 2/,0 π and π respectively. The amplitude of the resultant wave is

(a) 6

11A (b) 3

2A (c) 6

5A (d) 6

7A

Ans.: (c)

Solution: ( ) ( )πωπωω +=⎟⎠⎞

⎜⎝⎛ +=+= tAytAytAy sin

3,

2sin

2,0sin 321

tAtAtAyyyy ωωω sin3

cos2

sin321 −+=++=

tAtA ωω cos2

sin3

2+⇒

6

536

2549

423

2 22222 AAAAAAA ==+=⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=′





JEST 2013 Q5. The equation describing the shape of curved mirror with the property that the light from a

point source at the origin will be reflected in a beam of rays parallel to the x -axis is (with

a as some constant)

(a) y2 = ax + a2 (b) 2y = x2 + a2 (c) y2 = 2ax + a2 (d) y2 = ax3 + 2a2

Ans.: (c)

Documents

Mathematical Physics JEST-2016 · 2018-06-11 · fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Website: