JEST 2014 Question n Solution

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JEST PHYSICS-2014

PART-A: 3 Mark Questions

Q1. 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) ( ) 4/2221 pp + (b) ( ) 4/2221 qq + (c) ( ) 2/2221 pp + (d) ( ) 4/2211 qpqp + Ans.: (a)

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

22 1111

1

pqqpqL === and 22

2222

2

pqqpqL ===

4422

22

21

22

11 ppppppH += ( )2 21 2

4

p pH

+ = Q2. In a certain inertial frame two light pulses are emitted, a distance 5 km apart and separated

by s5 . 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 tv v vc c c

+ + = =

2 1t t= 6 325 10 5 10 0vc = 0.3v c =

Q3. Suppose a spin 2/1 particle is in the state

+=/ 21

61 i





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 02x

S = = with eigenvalues

2= and eigenvector corresponding to

2= is 11

12

Now probability getting 2=+

[ ]

[ ]

2

211 1 11 1 1 222 6 512

112 61 61 2 626

ii

pi

i

+ + + = = = = +

=

Q4. 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 where 45.1,44.1 12 == nn

2/1

1

45.145.144.144.11sin

= ( )11sin 0.11726 = 0 06.67 7 =

Q5. 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 , made by the two strings are small (you can use 2/1cos 2 = ). Note: lg /0 = .

(a)

+ 2220212022212 21

21 mlL





(b)

++ 222021202122212 21

21 mlL

(c)

+ 222021202122212 21

21 mlL

(d)

++ 222021202122212 21

21 mlL

Ans.: (b)

Solution: 11 sinlx = , 11 cosly = 2 1 2 2 1 2sin cosx x l y y l = + = + 2 1 2 2 1 2sin sin , cos cosx l l y l l = + = + 2 1 1 2 2 2 1 1 2 2cos cos , sin sinx l l y l l = + =

21

21

22211

2222

22211

2222

22 sincoscos2coscos llllyx +++=+

212122

222

2 sinsin2sin ll ++ ( )2 2 2 2 2 2 22 2 1 2 1 2 1 22 cosx y l l l + = + + also 2 2 2 21 1 1x y l + =

VTL = ( ) 212222212121 mgymgyyxyxm +++= ( )( )2 2 2 2 2 2 21 1 2 1 2 1 2 1 21 2 cos 2 cos cos2L m l l l l mgl mgl = + + + + +

2 22 2 2 1 2

1 2 1 21 2 11 12 2 2 2 2

g gL mll l

= + + + + ( )1 2cos 1

2 22 2 2 1 2

1 2 1 212 2 2 2 2

g g g gL mll l l l

= + + + +

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 ml = + +

1 l

l

m

2





Q6. 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 l

n n n n= + +

1 13 0 1 06 2eff

n = + + 0.5effn =

Occupancy ( )2 2effn r a rA= =

( )2

2

0.53 2

4

r

r

0.5 0.90643= =

Now number of disc per unit area will be 29.0302.3

9064.0 =

Q7. 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 2 xf x c c x e= + ( ) 1 2x xf x c e c xe = +

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

= n

mhEF at 0T K=

Closely packed hexagonal





2n n = and 2m m =2 2

22 23 333 3 12 2

4 4 2 4 2Fh h nE nm m = =

212 3

33 22 4h nm

=

Q9. The value of 2.22.0 dxxex 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=

Q10. The Hamiltonian operator for a two-state system is given by

( )12212211 ++= H , where 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 ++= H ( )1 1 2H = + , ( )212 = H Lets check for option (b): ( ) ( )1 2 1 2 , 1 2 1 2+ + Now =H ( )1 2 1 2H + ( )1 2 1 2H H= + +

( )1 2 1 2H + ( ) ( )1 2 1 2H H + ( ) ( ) ( )1 2 2 1 1 2 + + ( )1 2 1 1 1 2 1 2 + + ( )22212 +

( )[ ]21212 + Now ( )2121 +H ( )[ ]2121 + H ( )2121 + HH





( ) ( )( )1 2 2 1 1 2 + + ( ) ( )1 2 1 1 1 2 1 2 + + + ( ) 22212 ++ ( )[ ]22112 +

Q11. Given the fundamental constants = (Plancks 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)

Gc8=

Ans.: (a)

Solution: [ ][ ] 21

33

23112

TLTLMTML [ ] LL == 212

2 1ML T = = , [ ]2312 == TLMmgrG Q12. Consider an eigenstate of 2L

G and zL operator denoted by ml, . Let LnA

G= denote an operator, where n is a unit vector parametrized in terms of two angles

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

21 2 =mll + (b) ( ) sin

21 2 =mll +

(c) ( ) sin1 2=mll + (d) ( ) cos1 2=mll + Ans.: (b)

Solution: A n L= G x y zx y zA L L Lr r r = + +

sin sinsin cos cosx y z

rr rA L L Lr r r

= + + sin cos sin sin cosx y zA L L L = + +

Now 22 AAA = sin cos sin sin cosx y zA L L L = + +





( )cosA m = = 0 , 0x yL L= = 222222222 cossinsincossin zyx LLLA ++=

( ) ( )2 22 2 2 2 2 2 2 2 2 21 1sin cos sin sin cos2 2

l l m l l mA m + + = + +

= = =

( ) 22 2 2 2 2 2 2 21 sin sin cos cos2

l l mA m + = + + = =

( ) 22 2 2 2 2 21 sin cos2

l l mA m + = += =

( )( )222 2 2 2 2 2 2 2 21 sin cos cos2

l l mA A A m m + = = + = = =

( ) 21sin

2

l l mA + = =

Q13. 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: sina tL e bt ( )2 2b

S a b= + +

2 sin 4tL e t ( )2 24

2 4S= + + 204

42 ++= SS

Q14. 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

/ / 6.52238t 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

/1.02235tN 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. Q15. Which of the following circuits will act like a 4-input NAND gate?

(a) (b)

(c) (d)

Ans.: (d)

Q16. Consider a three-state system with energies EE, and gE 3 (where g is a constant) and respective eigenstates

=/01

1

21

1

=/

211

61

2

=/

111

31

3

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

=/

001

i





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

=/

100

f

(a) 0 (b)

=2

3sin94 2 gt

(c)

=2

3cos94 2 gt (d)

=23sin

94 2 gtE

Ans.: (b)

Q17. 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: 213.6

nE eVn=

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 =

Q18. The formula for normal strain in a longitudinal bar is given by ,AEF= where F is normal

force applied, A is cross-sectional area of the bar and E is Youngs 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

++=++=

EE

AA

FF

.02476 0.2476 = = 11

9

0.2476 50 2.95 100.2 210 10

= =

Q19. 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 kTEkTE

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 = 0 Ez e e = +

( )11lnln Eez += ( ) ( ) ( )

1ln ln 11

E EE

U E z e E ee

= = = + = +

1

E

BEEeU k Te

= =+

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 Tk T k T

B B BV E E E

k T k T k TB 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 B

C 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

= ++

Q20. 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 mC =+ for the positively charged sheet and 2/3.4 mC = 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 2 + = + = +

66 5

12

11.2 10 0.626 10 6.3 10 /2 8.86 10

N C

=

Q21. 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 = (b)

143cos = (c)

613cos = (d)

21cos =

Ans.: (a)





Solution: 45cv = , ocos 60

2xcu c = = , o 3sin 60

2yu c c = =

Now

2

42 5

412 5

x

c cu

c cc

+=

+ 13

14c= 13cos

14 =

Q22. 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

220 (b)

RI

554 0

(c) RI

558 0 (d) 0

Ans.: (c)

Solution: ( )2

03

2 2 22

IRBR d

=+

2

01 3

2 222

4

IRB

RR

= +

, 2

02 3

2 222

4

IRB

RR

= +

2Rd =

21 BBB += 23

0

452

2

=

R

I32

0 032

4 85 55

I IB

RR

= =

Q23. 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

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

V F25.03

2

H2

RP

i i

X

R

Y





Q24. 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)

Q25. Ice of density 1 melts at pressure P and absolute temperature T to form water of density 2 . The latent heat of melting of 1 gram of ice is L . What is the change in the internal energy U resulting from the melting of 1 gram of ice?

(a)

+

12

11PL (b)

12

11PL

(c)

21

11PL (d)

+

21

11PL

Ans.: (c)

Solution: dU dQ W dQ pdV= =

dU mL pdV= 2

1

2

1dU L P d

= 1 21 1L P

=

2

1 1V dV d = =

PART B: 1 MARK QUESTIONS

Q26. 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)

Q27. 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 =sin , mgT =cos tanga = at o90=

a is maximum at extreme position.

Q28. 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: .qV = W i t R = 2W i Rt=

now2 1 100 0.33

30 273W i RtT T

= = = +

Q29. 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 ==

xxxV [ ] 0212 2 = xx

0,2

1=x 2 2

22 2

12

12 12 2 12 4 02x

V Vxdx dx =

= = = <

For stable point 0=xV and 0

2

>xV

l T cosT

mgsinT





020

2

2

>=

=xxV

Q30. 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 22E eV

ml= == , 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 + =

Q31. The value of elastic constant for copper is about 1100 Nm and the atomic spacing is nm256.0 .

What is the amplitude of the vibration of the Cu atoms at 300 K as a percentage of the

equilibrium separation?

(a) 4.55 % (b) 3.55 % (c) 2.55 % (d) 1.55 %

Ans.: (b)

Solution: we know that

212

E A= ma 910256.0 = for one dimension

1.2 B

K E K T= and KTEP21. =

E kT= now 212B

K T A= 2 Bk TA =





TkA

2=23

232 1.30 10 300 8.28 10100

A

= = 129.09 10 0.0090 nm= =

Now x of 0.256 0.009nm nm= 0.00909 0.03551 3.5%0.256

nmxnm

= = = Q32. What is the contribution of the conduction electrons in the molar entropy of a metal with

electronic coefficient of specific heat?

(a) T (b) 2T (c) 3T (d) 4T Ans.: (a)

Solution: 3VC BT AT= + Q33. Consider a system of 2N non-interacting spin 2/1 particles each fixed in position and carrying a

magnetic moment . 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= for spin 2

1 (up)

22

N nCN = for spin 2

1 down

entropy lnkS = 2 22ln lnN NN 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

= + +

( )

++= nNnNn

nNNnK 2ln

222ln12

( )

++= nnNnNnNK ln2ln

2212

( ) 02ln112 =

++=n

nNk

02 k 02ln =

nnN 12 =

nnN nN 22 = n N =

Q34. 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)

Q35. 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: 21PV

CC f

= = + where f is degree of freedom.





For monoatomic: 3f = , For diatomic: 6f = , For Triatomic: 9f = For hexaatomic: 18f =

Q36. 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 /=/ (b) 2kxxe=/ (c) kxAxe=/ (d) 2kxAe=/

Ans.: (b)

Solution: 2kxxe=

For given potential, at 0,x = and x = wave function must vanish. Q37. Which functional form of potential best describes the interaction between a neutral atom and an

ion at large distances (i.e. much larger than their diameters)

(a) 2/1 rV (b) rV /1 (c) reV ar // (d) 3/1 rV Ans.: (a)

Q38. 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 electrons 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).





Q39. 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 As 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

Q40. If a proton were ten times lighter, then the ground state energy of the electron in a hydrogen

atom would have been

(a) Less (b) More

(c) The same (d) Depends on the electron mass

Ans.: (b)

Solution: 526.1300545.16.136.13

22 ===e

e

en m

mnmn

E

1.0054510

p e pp e

e p

M m mM m

m m = = =+

Q41. Let us write down the Lagrangian of a system as ( ) xcxkxxmxxxxL ++= 2,, . What is the dimension of c ?

(a) 3MLT (b) 2MT (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 = [ ] [ ][ ] [ ]MTML TMLC ==

32

22

1mA

x x1 2mB

m1





Q42. The Dirac delta function ( )x satisfies the relation ( ) ( ) ( ) = 0fdxxxf for a well behaved function ( )xf . If x has the dimension of momentum then (a) ( )x has the dimension of momentum (b) ( )x has the dimension of ( )2momentum (c) ( )x is dimensionless (d) ( )x has the dimension of ( ) 1momentum

Ans.: (d)

Solution: ( ) ( ) ( ) = 0fdxxxf ( ) ( ) ( ) ( )[ ] ( ) ( )[ ] ( ) [ ]100 === PxfPxxffdxxxf Since, ( )[ ] ( )[ ]0fxf = If ( ) += xxF is force [ ]2TLM ( ) =0F is also [ ]2TLM

Q43. The operator A and B share all the eigenstates. Then the least possible value of the product of

uncertainties BA is (a) = (b) 0 (c) 2/= (d) Determinant (AB)

Ans.: (b)

Solution: [ ]2ABBA

0 BA A and B have share their eigen values so [ ] 0=AB

Q44. 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 nN = = where N - Number of slit and n - order of diffraction.

Q45. 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

= z

zz

iziz

Q46. 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

4 44 2 2 8

R

R

Q QE rW E r dr E r dr Wr R

= = + =

( )2 2

00

2 8 888

2

Q QW WR R = = =

Q47. Three sinusoidal waves have the same frequency with amplitude 2/, AA and 3/A while their

phase angles are 2/,0 and respectively. The amplitude of the resultant wave is (a)

611A (b)

32A (c)

65A (d)

67A

Ans.: (c)

Solution: ( ) ( ) +=

+=+= tAytAytAy sin3

,2

sin2

,0sin 321





tAtAtAyyyy sin3

cos2

sin321 +=++=

tAtA cos2

sin3

2 +

6

536

2549

423

2 22222 AAAAAAA ==+=

+

=

Q48. The value of integral

= c dzz zI 2sin with c a circle 2=z , is (a) 0 (b) i2 (c) i (d) i

Ans.: (c)

Solution: = C z zI 2sin pole 202 == zz Residue at

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

( ) =

=

C

iz

emg iRz

eI 22

2

Res 22

22

2 2/

2

ie

z

ez iiz

z

==

=

(taking imaginary part)

Residue = 21

Now iiI == 221

Q49. Consider a square well of depth 0V and width a with aV0 fixed. Let 0V and 0a . 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.

Q50. If hydrogen atom is bombarded by energetic electrons, it will emit

(a) K X - rays (b) -rays (c) Neutrons (d) none of the above

Ans.: (d)

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JEST 2014 Question n Solution