I

CSIR/NET/JRFPHYSICALSCIENCES

Previous Year’s Solved Papers

Publishing House

EXUDE TALENT PUBLISHING HOUSE

ii

Author:Sushil K. TomarM.Sc (Physics), IIT-D, Ph.D. (Pursuing)

Former Faculty Department of Physics,

MKM girls Degree College, Palwal.

Affiliated to MDU, Rohtak.

Kasturi Baluja(Retd.) Proffessor of Department of PhysicsUniversity of Delhi

Publishing House

EXUDE TALENT PUBLISHING HOUSEHead Office: Plot No. 63, Saubhagyapuram, Dasna-201015, Ghaziabad (U.P.)Mob.: 9891878971, 8273189173Email: exudetalentph@gmail.comWebsite: www.exudetalent.in

First Edition: August 2017Copyright © Publishers

ISBN No. 978-81-931770-4-4

Printed by : Akhilesh Tomar for exude talent publishing house

© All rights reserved by Publisher. No part of this book may be reproduced or distributed in any form or any means, electronic,

mechanical photocopying or otherwise without the prior permission of the Author. All Disputes are in the jurisdiction of District

Court of Ghaziabad.

Exude Talent has taken due care in writing the theory, questions (solutions), previous year questions and their solutions, before publishing book.

Inspite of this, if any inaccuracy or printing error occurs then, neither Author nor Exude Talent shall be responsible for any damage. Author is

providing the complete coverage of the syllabus, however necessary guidance from the expert is also needed. Author will be grateful if you could

point out any such error. Your suggestions will be appreciated for second edition.

III

CONTENTS___________________________________________

Syllabus iv

June-2017 M1-39

December-2016 L1-38

June-2016 K1-35

December-2015 J1-38

June-2015 I1-34

December-2014 H1-38

June-2014 G1-43

December-2013 F1-34

June-2013 E1-40

December-2012 D1-38

June-2012 C1-33

December-2011 B1-34

June-2011 A1-38

iv

CSIR-UGC National Eligibility Test (NET) for Junior Research Fellowshipand Lecturer-ship

PHYSICAL SCIENCESPART ‘A’ CORE

I. Mathematical Methods of Physics

Dimensional analysis. Vector algebra and vector calculus. Linear algebra, matrices, Cayley-Hamilton Theorem. Eigenvaluesand eigenvectors. Linear ordinary differential equations of first & second order, Special functions (Hermite, Bessel, Laguerreand Legendre functions). Fourier series, Fourier and Laplace transforms. Elements of complex analysis, analytic functions;Taylor & Laurent series; poles, residues and evaluation of integrals. Elementary probability theory, random variables,binomial, Poisson and normal distributions. Central limit theorem.

II. Classical Mechanics

Newton’s laws. Dynamical systems, Phase space dynamics, stability analysis. Central force motions. Two body Collisions- scattering in laboratory and Centre of mass frames. Rigid body dynamicsmoment of inertia tensor. Non-inertial framesand pseudoforces. Variational principle. Generalized coordinates. Lagrangian and Hamiltonian formalism and equationsof motion. Conservation laws and cyclic coordinates. Periodic motion: small oscillations, normal modes. Special theory ofrelativityLorentz transformations, relativistic kinematics and mass–energy equivalence.

III. Electromagnetic Theory

Electrostatics: Gauss’s law and its applications, Laplace and Poisson equations, boundary value problems. Magnetostatics:Biot-Savart law, Ampere’s theorem. Electromagnetic induction. Maxwell’s equations in free space and linear isotropicmedia; boundary conditions on the fields at interfaces. Scalar and vector potentials, gauge invariance. Electromagneticwaves in free space. Dielectrics and conductors. Reflection and refraction, polarization, Fresnel’s law, interference, coherence,and diffraction. Dynamics of charged particles in static and uniform electromagnetic fields.

IV. Quantum Mechanics

Wave-particle duality. Schrodinger equation (time-dependent and time-independent). Eigenvalue problems (particle in abox, harmonic oscillator, etc.). Tunneling through a barrier. Wave-function in coordinate and momentum representations.Commutators and Heisenberg uncertainty principle. Dirac notation for state vectors. Motion in a central potential: orbitalangular momentum, angular momentum algebra, spin, addition of angular momenta; Hydrogen atom. Stern-Gerlachexperiment. Timeindependent perturbation theory and applications. Variational method. Time dependent perturbation theoryand Fermi’s golden rule, selection rules. Identical particles, Pauli exclusion principle, spin-statistics connection.

V. Thermodynamic and Statistical Physics

Laws of thermodynamics and their consequences. Thermodynamic potentials, Maxwell relations, chemical potential, phaseequilibria. Phase space, micro- and macro-states. Micro-canonical, canonical and grand-canonical ensembles and partitionfunctions. Free energy and its connection with thermodynamic quantities. Classical and quantum statistics. Ideal Bose andFermi gases. Principle of detailed balance. Blackbody radiation and Planck’s distribution law.

VI. Electronics and Experimental Methods

Semiconductor devices (diodes, junctions, transistors, field effect devices, homo- and hetero-junction devices), devicestructure, device characteristics, frequency dependence and applications. Opto-electronic devices (solar cells, photo-detectors,

V

LEDs). Operational amplifiers and their applications. Digital techniques and applications (registers, counters, comparatorsand similar circuits). A/D and D/A converters. Microprocessor and microcontroller basics. Data interpretation and analysis.Precision and accuracy. Error analysis, propagation of errors. Least squares fitting,

PART ‘B’ ADVANCED

I. Mathematical Methods of Physics

Green’s function. Partial differential equations (Laplace, wave and heat equations in two and three dimensions). Elementsof computational techniques: root of functions, interpolation, extrapolation, integration by trapezoid and Simpson’s rule,Solution of first order differential equation using Runge Kutta method. Finite difference methods. Tensors. Introductorygroup theory: SU(2), O(3).

II. Classical Mechanics

Dynamical systems, Phase space dynamics, stability analysis. Poisson brackets and canonical transformations. Symmetry,invariance and Noether’s theorem. Hamilton-Jacobi theory.

III. Electromagnetic Theory

Dispersion relations in plasma. Lorentz invariance of Maxwell’s equation. Transmission lines and wave guides. Radiation-from moving charges and dipoles and retarded potentials.

IV. Quantum Mechanics

Spin-orbit coupling, fine structure. WKB approximation. Elementary theory of scattering: phase shifts, partial waves,Born approximation. Relativistic quantum mechanics: Klein-Gordon and Dirac equations. Semi-classical theory of radiation.

V. Thermodynamic and Statistical Physics

First- and second-order phase transitions. Diamagnetism, paramagnetism, and ferromagnetism. Ising model. Bose-Einsteincondensation. Diffusion equation. Random walk and Brownian motion. Introduction to nonequilibrium processes.

VI. Electronics and Experimental Methods

Linear and nonlinear curve fitting, chi-square test. Transducers (temperature, pressure/vacuum, magnetic fields, vibration,optical, and particle detectors). Measurement and control. Signal conditioning and recovery. Impedance matching,amplification (Op-amp based, instrumentation amp, feedback), filtering and noise reduction, shielding and grounding.Fourier transforms, lock-in detector, box-car integrator, modulation techniques. High frequency devices (including generatorsand detectors).

VII. Atomic & Molecular Physics

Quantum states of an electron in an atom. Electron spin. Spectrum of helium and alkali atom. Relativistic corrections forenergy levels of hydrogen atom, hyperfine structure and isotopic shift, width of spectrum lines, LS & JJ couplings.Zeeman, Paschen-Bach & Stark effects. Electron spin resonance. Nuclear magnetic resonance, chemical shift. Frank-Condon principle. Born-Oppenheimer approximation. Electronic, rotational, vibrational and Raman spectra of diatomicmolecules, selection rules. Lasers: spontaneous and stimulated emission, Einstein A & B coefficients. Optical pumping,population inversion, rate equation. Modes of resonators and coherence length.

vi

VIII. Condensed Matter PhysicsBravais lattices. Reciprocal lattice. Diffraction and the structure factor. Bonding of solids. Elastic properties, phonons,lattice specific heat. Free electron theory and electronic specific heat. Response and relaxation phenomena. Drude modelof electrical and thermal conductivity. Hall effect and thermoelectric power. Electron motion in a periodic potential, bandtheory of solids: metals, insulators and semiconductors. Superconductivity: type-I and type-II superconductors. Josephsonjunctions. Superfluidity. Defects and dislocations. Ordered phases of matter: translational and orientational order, kinds ofliquid crystalline order. Quasi crystals.

IX. Nuclear and Particle PhysicsBasic nuclear properties: size, shape and charge distribution, spin and parity. Binding energy, semiempirical mass formula,liquid drop model. Nature of the nuclear force, form of nucleon-nucleon potential, charge-independence and charge-symmetry of nuclear forces. Deuteron problem. Evidence of shell structure, single-particle shell model, its validity andlimitations. Rotational spectra. Elementary ideas of alpha, beta and gamma decays and their selection rules. Fission andfusion. Nuclear reactions, reaction mechanism, compound nuclei and direct reactions. Classification of fundamental forces.Elementary particles and their quantum numbers (charge, spin, parity, isospin, strangeness, etc.). Gellmann-Nishijimaformula. Quark model, baryons and mesons. C, P, and T invariance. Application of symmetry arguments to particlereactions. Parity non-conservation in weak interaction. Relativistic kinematics.

M - 5

Ans-aThe diagonal of the largest square fitting in the crcle is equal to the diagonal of the circle as shown in the fig.

A

C D

B

Let ‘s’ is the side of the suqare. The diagonal of the suqare is given by,2 2 2 2 2AD AC CD r s s r (SJ17.01)

The radius of the smaller circle is

2 22

rEF r s r r (SJ17.02)

13. In how many ways can you place N coins on a board with N rows and N columns such that every row and every column containsexactly one coin?

a. N b. 1 2 ...2 1N N N

c. N2 d. NN

Ans-b

There are N ways to distribute in rows, 1N ways for next 1N rows, 2N ways for 2N rays and so on. Thus, thetotal number of ways so that there shall be exactly one coin in each cell is

1 2 ...3.2.1N N N

14. A 100 m long train crosses a bridge 200m long and 20m wide bridge in 20 seconds What is the speed of the train in km/hr?a. 45 b. 36

c. 54 d. 57.6

Ans-cWe have

100 , 200 , 20secT SL m L m T (SJ17.01)The speed of the train in km/hour is

200 100 1815 / 15 / 54 /20 5

T SL Lv m s km hr km hrT (SJ17.02)

15. My brthday is in January. What would be a sufficient number of questions with ‘Yes/No’answers that will enable one to findmy birth date?a. 6 b. 3

c. 5 d. 2

Ans-cIn order to determine the date of brith in January, oner must ask 5 question, Let DOB is 7 Jan 2017.1. Is the DOB an odd number?Ans. Yes [if NO than date is even number]

M - 15

35. If the root-mean squared momentum of a particle in the ground state of a one-dimensional simple harmonic potential is P0then its root-mean-squared momentum in the first excited state is

a. 0 2p b. 0 3p

c. 0 2 / 3p d. 0 3 / 2p

Ans-bThe expectation value of p2 for nth excited state is

2 2 12

mp n (SJ17.01)

For ground state, expectation value of squared of momentum operator and average moment operator are

20 02.0 1 ; 0

2 2g gm mp p (SJ17.02)

The rms momentum of the ground state is,

220 0 0 02rms g g

mp p p p (SJ17.03)

Similarly for first excited state, the expectation value of squared of momentum operator is

21 1

32.1 1 ; 0

2 2g gm mp p (SJ17.04)

The rms momentum for first excited state is

221 1 1 03 3

2rms g gmp p p p (SJ17.05)

36. Consider a potential barier A of height V0 and wdth b, and another potential barrier B of height 2V0 and the same width b. Theratio A BT / T of tunneling probabilities A BT & T through barriers A and B respectively, for a particle of energy 0V / 100, isbest approximated by

a. 2 20exp 1.99 0.99 8 /mV b b. 2 2

0exp 1.98 0.98 8 /mV b

c. 2 20exp 2.99 0.99 8 /mV b d. 2 2

0exp 2.98 0.98 8 /mV b

Ans-aThe transmission probability through rectangular barrier of length b is given by

2 bT Ae (SJ17.01)

where A is constant, b width and is defined as

2

2m V E (SJ17.02)

we have

00 0, 2 ,

100a bVV V V V E (SJ17.03)

The ratio of transmission probability is2

2 2 202 1.99 0.99 8 /

ab a

b

bbA

bB

T Ae e Exp mV bT Ae

(SJ17.04)

M - 27

c. 1 00 1 : ,0 0 1

ab a b R d.

1 01 0 : ,

0 0 1

ab a b R

Ans-cGroup of matrix satisfies four property(a) The product of two matrix .AB G(b) Associativity (AB)C A(BC)(c) Identity matrix(d) The matrix must be invertible i.e. A exist.The upper triangle matrix in general form the group of matrixLet us define the matrices A and B as,

1 0 1 00 1 , 0 10 0 1 0 0 1

a rA b B s (SJ17.01)

The product of the two matrices is given by,

1 0 1 0 1 00 1 0 1 0 10 0 1 0 0 1 0 0 1

a r a rAB b s b s (SJ17.02)

The product also .AB G Similarly the two matrices are defined as,

1 0 1 00 1 , 0 10 0 1 0 0 1

a rA b B s (SJ17.03)

The product of the two matrices is given by,

1 0 1 0 10 1 , 0 1 0 10 0 1 0 0 1 0 0 1

a r a r sAB b B s b s (SJ17.04)

The product AB does not belong to the group. Rest of the properties are satisfied by both matrices. Hence the correct optionis c.

55. The Lagrangian of a free relativistic particle(in one dimension) of mass is given by 2L = -m 1- x where x = dx / dt. If sucha particle is acted upon by a constant force in the direction of the motion, the phase space trajectories obtained from thecorresponding Hamiltonian area. ellipses b. cycloids

c. hyperbolas d. parabolas

Ans-cWe have

21L m x V x (SJ17.01)

The Hamiltonian for the system is defined asH xp L (SJ17.02)

The conjugate momenta is given by,

2 2 21

L mx pp xx x m p

(SJ17.03)

M - 39

Ans-cWe have

19 2 1 121 2.1 10 , 3000ÅB m s (SJ17.01)

The ratio of Einstein coefficient is3 3

2121 213 3

21

8 8A hv hvA BB c c (SJ17.02)

Inserting eq. (SJ17.01) in eq. (SJ17.02), we obtain34 19

621 310

8 3.14 6.6 10 2.1 10 8 3.14 6.6 2.110

273000 10A (SJ17.03)

The life time of the excited state is approximately

6

21

1 2710 80

8 3.14 6.6 2.1ns

A (SJ17.04)

75. If the binding energies of the electron in the K and L shells of silver atom are 25.4 keV and 3.34 keV, respectively, then thekinetic energy of the Auger electron will be approximatelya. 22 keV b. 9.3keV

c. 10.5 keV d. 18.7 keV

Ans-dThe kinetic energy of an ejected Augar electron is equal to the energy h of the charecterstic X-ray minus the binding energyof the ejected electron in the respective shell

. auger eK E hv BE (SJ17.01)

Silver atom is bombarded with gak radiation from Tungsten (energy 59.1 KeV) (z 44) with

254 , 3.34k LE keV E keV (SJ17.02)

24.9 , 22.1k kE Ag keV E Ag keV (SJ17.03)

The kinetic energy of the Augar electron ejected from the L shell by k x-rays is

. 24.9 3.34 21.56k E keV (SJ17.04)by k x-rays is

. 22.1 3.34 18.76k E keV (SJ17.05)

L - 1

December-2016Part-A

1. Find out the missing pattern.

+ 18 25 37 6 9

2 ÷ 7?

a. 2

714? b. 14

72?

c. 14

72 ? d.

1472

?Ans-bOne can see that the missing bracket must contain sign of multiplication and the product of the two numbers are multiplied togive the third term in lower triangle of the suqare i.e.,

27

14?2. Seeds when soaked in water gain about 20% by weight and 10% by volume. By what factor does the density increase?

a. 1.20 b.1.10

c. 1.11 d.1.09

Ans-dWe have,

20% 1.2 ; 10% 1.1M m of m m V v of v v (SD16.01)Let ‘m’ and ‘v’ are the initial mass and volume, respectively. The initial density is defined as,

/M V (SD16.02)The new density of the seeds is

1.2 1.09 1.091.1

M m mV v v (SD16.03)

3. Retarding frictional force f, on a moving ball, is proportional to its velocity V . Two identical balls roll down identicalslopes (A & B) from different heights. Compare the retarding forces and the velocities of the balls at the bases of theslopes.

a. ;A B A Bf f V V b. ;A B B Af f V V

c. ;B A B Af f V V d. ;B A A Bf f V VAns-a

L-12

C

+RR

Residue of the function f(z) at the pole 2z i is,

2 2

2Re lim 2

2 2 2 2

iik k

z i

e esidue z i f zi

(SD16.04)

Applying the cauchy’s integral formulae, we obtain2 2

2

22 Re2 2 2 2

ikz k ke dz ie ei si F zz i

(SD16.05)

and also,2

2 2 2 2 22 2 2 2 2 2

ikz ikx ikz ikx ikz kR

RC

e dz e dx e dz e dx e dz eIz x z x z (SD16.06)

When ,R we obtain required integral and the second integral along the curve vanishes, as

2

10F z dz

z(SD16.07)

30. A screen has two slits, each of width w with their centres at a distance 2w apart. It is illuminated by a monochromaticplane wave travelling along the x-axis.

The intensity of the interference pattern, measured on a distant screen, at an angle to the x-axis isa. zero for n 1,2,3... b. maximum for n 1,2,3...

c. maximum for 1 3 5

, , ...2 2 2

n d. zero for only n 0 only

Ans-aA screen has two slits, each of width w with their centres at a distance 2w apart. It is illuminated by a monochromatic planewave travelling along the x-axis.

L - 19Assuming that the interference takes place only between light reflected by the bottom surface of the top plate and thetop surface of bottom plate, the distance d is closest toa. 12 m b. 24 m

c. 60 m d. 120 m

Ans-dThe constructive interference for such a system are obtained, if it satisfies the condition,

2 1/ 2d n (SD16.01)where, ‘n’ are the integer number. The distance between two plates are

4952 120

2 4 4d d m m (SD16.02)

42. The I-V characteristics of a device can be expressed as SaVI = I exp - 1 ,T

where T is the temperature and a and IS

are constants independent of T and V. Which one of the following plots is correct for a fixed applied voltage V?

a.

3

3

2

2

1

10

0aV/T

b.

aV/T

2

2-3

3

-2

1

1

-1

0

0

c.

aV/T

2

2-3

3

-2

1

1

-1

0

0

d.

321

0

0aV/T

2

-3

3

-2

1

-1

4

Ans-dThe I-V characteristics of a device can be expressed as

exp 1 ,SaVI IT

(SD16.01)

where T is the temperature and a and IS are constants independent of T and V. The eq. (SD16.01) is re-expressed as

exp 1 exp 1 ln ln ln exp 1 ln lns s s savI I I I x I I x I I xT (SD16.02)

where, we have used, / .x aV T The eq. (SD16.02),for a fixed applied voltage V, is correctly shown in the option 4.

43. The active medium in a blue LED (light emitting diode) is a GaxIn1–xN alloy. The band gaps of GaN and InN are 3.5eVand 1.5eV respectively. If the band gap of GaxIn1–xN varies approximately linearly with x, the value of x required forthe emission of blue light of wavelength 400 nm is (take hc 1200 eV-nm)a. 0.95 b. 0.75

c. 0.50 d. 0.33

Ans-bWe have,

3.5 , 1.5GaN InNE eV E eV (SD16.01)

L-38

V a

a

B in z direction

bB

a

y

x

V

V

b

B

Let , &dv B d are the drift velocity of electron, applied magnetic field and d is the width along y direction, respectively. The

drift velocity of the electron is inversely proportional to the width along x-direction 1/ .dv x The hall voltage across theplates is defined as,

,H dV V Bd (SD16.01)The ratio of the hall voltage developed across the plate in two different orientation is

2 2

1 1

22

22

2 22 : 2 :11 1

H d

H d

V v Bd a a a bV v Bd bb (SD16.02)

Where, we have used,

1 21 2, ; ;d dk kv d b v d aa b (SD16.03)

K - 1

June-2016Part-A

1. “My friend Raju has more than 1000 books”, said Ram. “Oh no, he has less than 1000 books”, said Shyam. “Well, Rajucertainly has at least one book”, said Geeta. If only one of these statements is true, how many books does Raju have?a. 1 b. 1000

c. 999 d. 1001

Ans-b Let Ram has books 1000,x shyam has books, 1000,y and Raju has books 1.z The possible distribution of books areshown below.

1000 1001 999F T FF F TT T T

Since, only one statement is true, hence raju has 1000 books.

2. Of the following, which is the odd one out?a. Cone b. Torus

c. Sphere d. Ellipsoid

Ans- a or bCone in the odd one, because it one side in plane.

3. An infinite number of identical circular discs each of radius 1/2are tightly packed such that the centres of the discs are atinteger values of coordinates x and y. The ratio of the area of the uncovered patches to the total area isa. 1 / 4 b. / 4

c. 1 d.

Ans- a

4. It takes 5 days for a steamboat to travel from A to B along a river. It takes 7 days to return from B to A. How many days willit take for a raft to drift from A to B (all speeds stay constant)?a. 13 b. 35

c. 6 d. 12

Anc-bLet u and v are the speed of boat and speed of water current, respectively. Let D is the distance between the pointsbetween A and B. The time taken by the boat in travelling A to B in upstream is

77up

D DT days u vu v (SJ16.01)

Similarly, the time taken by the boat in travelling B to A in downstream is

55down

D DT days u vu v (SJ16.02)

Adding the eq. (SJ16.01) and eq. (SJ16.02), and solving for velocity of water current, one obtains

K - 12

2 3 2L 5 L 5 L

If its energy is measured, the possible outcomes and the average value of energy

are, respectively

a. 2 2

2 2

2,2

h hmL mL

and2

2

7350

hmL

b. 2 2

2 2

2,8

h hmL mL

and 2

2

1940

hmL

c. 2 2

2 2

2,2

h hmL mL

and2

2

1910

hmL

d. 2 2

2 2

2,8

h hmL mL

and2

2

73200

hmL

Ans-aThe state of a particle of mass in a one-dimensional rigid box in the interval 0 to L is given by the normalised wavefunction,

2 42 3 2 4 4 3 4sin sin

5 5 5 5x xx

L L L (SJ16.01)

If a measurement is carried out on the system, we would obtain2 2 2

22nnE

mawith a corresponding probability

2.n n nP E (SJ16.02)

Since the initial wavefunction contain only two eigenstates of 2 4ˆ , & ,H x x the results of the energy measurement

along with the corresponding probabilities are, for the state 2 ,x

2 2 2 2

2 2 2 2 2 22 2

4 9ˆ ;252 2

hE H P EmL mL

(SJ16.03)

and for state 4 ,x

2 2 2 2

4 4 4 4 4 42 2

16 2 16ˆ ;252

hE H P Ema mL

(SJ16.04)

The average energy is2 2

2 4 2 2 2

9 16 9 64 7325 25 5050 50 2n n

n

h hE P E E EmL mL mL (SJ16.05)

31. A magnetic field B is in the region x 0 and zero elsewhere. A rectangular loop, in the xy-plane, of sides l (along the x-direction) and h (along the y-direction) is inserted into the x > 0 region from the x < 0 region at a constant velocity ˆWhich of the following values of l and h will generate the largest EMF?a. l 8, h 3 b. l 4, h 6

c. l 6, h 4 d. l 12, h 2

Ans-bThe induced emf associated with the motion of rectangular loop in the presence of magnetic field is given by

d d dA dxBA B Bh Bhvdt dt dt dt (SJ16.01)

Where, B is the magnetic field, A is the area of rectangular loop, h is the width of rectangular loop and v is the velocity of loop.

The induced emf will be maximum for higher value of width h. Since h is maximum in option (2), so max

is maximum in this case.

32. The x- and z-components of a static magnetic field in a region are 2 2x 0B = B x - y and zB = 0, respectively. Which of the

following solutions for its y-component is consistent with the Maxwell equations?

K - 21

Ans-bWe have,

1

n

nn

df x x xdx (SJ16.01)

The fourier transform of

1ƒ

nikx ikx ikx

nn

df k dxe x dxe x dxe x

dx (SJ16.02)

Let us compute the value of first and second term in the eq. (SJ16.02), we get

0 1, 1n

n nikx k ikxn

ddxe x e dxe x ikdx

(SJ16.03)

Inserting eq. (SJ16.03) in eq. (SJ16.02), one obtains

2 3 4

11 1 1 ..n n

nF k ik ik ik ik ik

2 1 11 1 .. 11 1 1

ikik ik ik f kik ik ik (SJ16.04)

49. The integral equation-ik x-x +i

32 2 2 2

dx, t = dx dt x , tÎ2

is equivalent to the differential equation

a. 2 2

2 32 2

1, ,6

m i x t x tt x b.

2 22 2

2 2 , ,m i x t x tt x

c. 2 2

2 22 2 , 3 ,m i x t x t

t x d. 2 2

2 32 2 , ,m i x t x t

t x

Ans-d

50. A canonical transformation q,p Q,P is made through the generating function 2F q,p = q P on the Hamiltonian

24

2

pH q,p = + q42

where and are constants. The equations of motion for (Q, P) are

a. /Q P and P Q b. 4 /Q P and / 2P Q

c. /Q P and 22PP Q

Q d. 2 /Q P and P Q

Ans-bWe have,

24

2,42

pH q p qq (SJ16.01)

The hamiltonian in the system (q,p) is transformed through canonical transformation to a new system (Q, P) by using thelegendre transformation, genrating function,

22 ,F q P q P (SJ16.02)

The position and momenta co-ordinates in the new system are obtained from Hamiltonian equation of motion i.e.,

22 22 ;F F

p qP Q qq P (SJ16.03)

K - 35

74. Let ES denote the contribution of the surface energy per nucleon in the liquid drop model. The ratio 27 64S 13 S 30E AI : E Zn is

a. 2:3 b. 4:3

c. 5:3 d. 3:2

Ans-bLet ES denote the contribution of the surface energy per nucleon in the liquid drop model and is equal to

2 2/34S SE ER kA

A A(SJ16.01)

where, we have used,1/3

0R R A (SJ16.02)

The energy of 27 6413 30&S SE AI E Zn are

2/327 2/3 6427 ; 64S SE EA k Zn kA A (SJ16.03a, 03b)

Dividing eq. (SJ16.03a) by eq. (SJ6.03b), we get27 2/3

64

27 64 9 464 27 16 3

S

S n

E A A ZincA AE Z (SJ16.04)

75. In the large hadron collider (LHC), two equal energy proton beams traverse in opposite directions along a circular path oflength 27 km. If the total centre of mass energy of a proton-proton pair is 14 TeV, which of the following is the bestapproximation for the proper time taken by a proton to traverse the entire path?a. 12 ns b. 1.2 sc. 1.2 ns d. 0.12 s

Ans-aWe have,

827 , 27 , 3 10 /S km E TeV c m s (SJ16.01)The time taken to complete one round trip is,

3

8

27 10 903 10

S mT sc s

(SJ16.02)

The value of relative exponent is given by,12

32 2

7 10 7.48 10938

pp

p p p

Ep cp eVp m cm c MeVm c m c (SJ16.03)

The proper time taken by a proton to traverse the entire path

0 090 127.48

T nsT T T ns (SJ16.04)

J- 1

December-2015Part-A

1. A shopkeeper purchases a product for Rs. 100 and sells it making a profit of 10%. The coustomer resells it to the sameshopkeeper incurring a loss of 10%. In these dealings the shopkeeper makesa. no profit, no loss b. Rs. 11

c. Re. 1 d. Rs. 20

Ans-bThe cost price of the object for the shopkeeper is Rs 100, the shopkeeper sells the object at profit of 10%, so the selling priceof the object is

1010% 100 100 110100

SP CP of CP Rs (SD15.01)

The customer resale that object at loss of 10%, so the selling price for the customer is

11010% 110 100 99100

SP CP of CP Rs (SD15.02)

The profit of the shopkeeper is

10 1 11profit selling reselling Rs (SD15.03)

2. A vessel is partially filled with water. More water is added to it at a rate directly proportional to time dv

i.e., t .dt Which

of the following graphs depicts correctly the variation of total volume V of water with time t?

a.

V

0 t b.

V

0 t

c.

V

0 t

d.

V

0 t

Ans-bWe have,

2

2dv tt v a cdt

(SD15.01)

Applying the boundary condition 00 ,v t v on the eq. (SD15.01), we get

0 00 0v t v c c v (SD15.02)Inserting eq. (SD15.02) in eq. (SD15.01), we obtain

20 0 0; / 2v v a t a a (SD15.03)

The eq. (SD15.03) is correctly represented by the curve shown in the figure given below.

V

0 t

quadratic

J- 13

b. 1

n1

yy

d. 1

n1

yy

Ans-dWe have,

2

2

1 1tanh 1

x x x

x x x

Cosh x e e eyx Sinh x e e e (SD15.01)

Simplifying the eq. (SD15.01), and solving for x, one obtains

2 2 2 1 1 1 111 2 1 1

x x x y y ye y y e e x n x ny y y (SD15.02)

30. A Hermitian operator O has two normalised eigenvalues 1 and 2 with eigenvalues 1 and 2, respectively. The two states

u = cos and are such that ˆv O v = 7 / 4 and u v = 0. Which of the following are

possible values of and ?

a. 6

and 3

b. 6

and 3

c. 4

and 4

d. 3

and 6

Ans-aWe have,

cos 1 sin 2 ; cos 1 sin 2u v (SD15.01)Now applying the orthonormality relation,we get

cos 1 sin 2 cos 1 sin 2 cos cos 1 1 sin sin 2 2 0u v

cos cos sin sin 0 cos 0 cos 0 / 2 , (SD15.02)This condition is satisfied by option (1) and (4) :

1 : ; 2 :3 6 2 4 4 2

option option (SD15.03)

Applying the operator O on the wavefunction ,v we get

ˆ cos 1 sin 2 cos 1 2sin 2O (SD15.04)

The expectation value of O is

2 2 7ˆ ˆ| | cos 1 sin 2 cos 1 2sin 2 cos 2sin4

O v O v (SD15.05)

Inserting the value of / 3 in eq. (SD15.06), we obtain,2

2 2 2 3 3 7cos 2sin 1 sin 1 13 3 3 2 4 4 (SD15.06)

31. A beam of unpolarized light in a medium with dielectirc constant 1 is reflected from a plane interface formed with another

medium of dielectric constant 2 13 . The two media have identical magnetic permeability. If the angle of incidence is 60°,then the reflected lighta. is plane polarized perpendicular to the plane of incidence

J- 25

Straight linewith shape

51. The Hermite polynomial Hn(x) satisfies the differential equation 2

n nn2

d H dH- 2x + 2nH x = 0.

dxdx The corresponding

generating function nn=0 n

1G t, x = H x t

n! satisfies the equation

a. 2

2 2 2 0G G Gx tx tx

b. 2

22 2 2 0G G Gx t

x tx

c. 2

2 2 2 0G G Gxx tx

d. 2 2

2 2 2 0G G Gx tx x tx

Ans-aThe Hermite polynomial Hn(x) satisfies the differential equation

2

2 2 2 0.n nn

d H dHx nH x

dxdx(SD15.01)

The generating function

01,!

nn nG t x H x t

n (SD15.02)

satisfies the following differential equation i.e.,22

2 22 2 0 2 2 0!

nn n

nn

d H dHG G G tx t x nH xx t n dxx dx (SD15.03)

yielding,2

2 2 2 0.n nn

d H dHx nH xdxdx

(SD15.04)

where, we have used,22

2 2

1 1 2; ; 2! ! !

n nn n nn

H x H xG G G nt t t t H xn x n x t nx x

(SD15.05)

52. A dipole of moment p, oscillating at frequency radiates spherical waves. The vector potential at large distance is

ikr0 eA r = i

4To order (1/r) the magnetic field B at a point ˆr = rn is

a. 2

0 ˆ ˆ4

ikren p nc r

b. 2

0 ˆ4

ikren p nc r

c. 20 ˆ ˆ4

ikrek n p pr

d. 2

0 ˆ4

ikrepc r

Ans-bThe vector potential associated with a dipole of moment ,p oscillating with frequency , at large distance is

J-38Which of the following statements is true?a. Process (i) obeys all conversation laws Processb. Process (ii) conserveses baryon number, but violates energy-momentum conservationc. Process (iii) is not allowed by strong interactions, but is allowed by weak interactionsd. Process (iv) conserves baryon number, but violates lepton number conservation

Ans-bIn the process given below:

: 1 1 0P n Baryon no (SD15.01)

The Baryon Number is conserved. In the centre of mass frame of reference, the total momentum of P n is zero, but then has to be created at rest to conserve momentum which is not possible. Thus, the reaction violates the conservation ofmomentum.

Note : The combined rest masses of P n is about 940 2 1880MeV and that of is about 140 MeV, so lot of rest mass

1880 140 1740MeV MeV is available to create large number of pions. But creating of only pion is not possible.

I - 1

June-2015Part-A

1. Each of the following pairs of words hides a number, based on which you can arrange them in ascending order. Pick thecorrect answer:I. Cloth reelJ Silent wonderK Good toneL. Bronz rod

a. L, K, J, I b. I, J, K, L

c. K, L, J, I d. K, J, I, L

Ans- aThe hidden word in these word are:

I. Cloth reel three 3J. Silent Wonder two 2K. Good tone one 1L. Bronze rod zero 0

These numbers are arranged in the following manner:0 1 2 3 , , ,L K J I (SJ15.01)

2. Which of the following values is same as 2222 ?

a. 26 b. 28

c. 216 d. 2222

Ans-c

The number 2222 is simplified as:

22 42 2 162 2 2 (SJ15.01)

3. A 12m 4m rectangular roof is resting on four 4 m tall thin poles. Sunlight falls on the roof at an angle of 45° from the east,creating a shadow on the ground. What will be the area of the shadow?a. 24 m2 b. 36 m2

c. 48 m2 d. 60 m2

Ans- cSunlight travels in a straight line, so the shadow will be of rectangular shape with an area 212 4 48 .m m m The dimension 4mis due to 45° inclination.

4. If

22

684

8 6

ab

c

d

I-14

The action of an operator T on x is given by

ˆ ,T x x a (SJ15.02)where a is a constant. Applying the operator T on eq. (SJ15.01), is

/ /ˆ ipx ipxT p T x e dx x a e dx

/ / / /ip x a ipa ipx ipax e d x a e x e dx e p (SJ15.03)

36. If Li are the components of the angular momentum operator L, then the operator i=1, 2, 3 i iL, L , L equals

a. L b. 2L

c. 3L d. L

Ans-bLet us compute the value of commutator in the following manner:

3

1 1 2 2 3 31

, , , , , , , ,i ii

L L L L L L L L L L L L (SJ15.01)

where, the angular momentum is defined as

1 2 3ˆˆ ˆL iL jL kL (SJ15.02)

Let us evaluate the value of first commutator, we get

1 1 3 2 1 3 1 2 1 2 3ˆ ˆ ˆˆ ˆ ˆ, , , , ,L L L jL kL L j L L k L L jL kL (SJ15.03)

Where, we have used,

1 1 2 3ˆˆ ˆ, , iL L iL jL kL L

1 1 2 1 3 1 3 2ˆ ˆˆ ˆ ˆ, , ,i L L j L L k L L j L k L (SJ15.04)

Similarly, the second and third commutator brakets also give

2 2 1 3 3 3 1 2ˆˆ ˆ ˆ, , ; , ,L L L iL kL L L L iL jL (SJ15.05)

Inserting the eq. (SJ15.03) and eq.(SJ15.05) in eq. (SJ15.01), we get3

1 2 31

ˆˆ ˆ, , 2 2 2 2i ii

L L L iL jL kL L (SJ15.06)

37. A particle moves in one dimension in the potential 21V = k t x ,2

where k(t) is a time dependent parameter. Then d v ,dt

the

rate of change of the expectation value v of the potential energy, is

a. 212 2

dk kx xp pxdt m b. 2 21 1

2 2dk

x pdt m

c. 2k xp pxm d. 21

2dk xdt

Ans-aA particle moves in one dimension in the potential

21 ,2

V k t x (SJ15.01)

where k(t) is a time dependent parameter. The rate of change of an operator A is given by,,

1 ,Ad A A H

dt i t (SJ15.02)

I - 27

Where, we have used,

1ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ, , ,2

a H a a a a a a a t (SJ15.03)

59. A particle of energy E scatters off a repulsive spherical potential

0V for r < aV r =

0 for r awhere V0 and a are positive constants. In the low energy limit, the total scattering cross-section is

22 1

ka

where 22

02m

k = V - E > 0. In the limit 0V the ratio of to the classical scattering cross-section off a sphere of

radius a isa. 4 b. 3

c. 1 d. 1/2

Ans-aThe total scattering cross-section of a particle of energy E scatters off a repulsive spherical potential, is

22

02

1 24 tanh 1 , 0ma ka k V Eka (SJ15.01)

In the limit 0 ,V it behaves like a hard sphere. The total scattering cross-section becomes in

4T a (SJ15.02)The geometric cross-section area of the particle is

cs a (SJ15.03)The ratio of to the classical scattering cross-section off a sphere of radius a is

2

4 4T

cs

aa (SJ15.04)

60. Two different sets of orthogonal basis vectors 1 0 1 11 1, and ,0 1 1 -12 2

are given for a two-dimensional real

vector space. The matrix representation of a linear operator A in these bases are related by a unitary transformation. Theunitary matrix may be chosen to be

a. 0 11 0 b.

0 11 0

c. 1 111 12 d.

1 011 12

Ans-cThis problem is solved through options. Check all the options, but only option (3) work as follows:

1 1 1 1 1 1 0 11 1 1 1;1 1 0 1 1 1 1 12 2 2 2 (SJ15.01)

61. A large number N of Brownian particles in one dimension start their diffusive motion from the origin at time t = 0. Thediffusion cofficient is D. The number of particles crossing a point at a distance L from the origin, per unit time, depends onL and time t as

I-34

Magic Numbers are ( k = 0, 1, 2 --- etc) : 2, 8, 20, 40, 70, 112. Thus, the atoms 4 162 8&He O have 2, 8 magic numbers.

74. The charm quark is assigned a charm quantum number C = 1. How should the Gellmann-Nishijima formula for electriccharge be modified for four flavours of quarkS?

a. 31

I2

B S C b. 31

I2

B S C

c. 31

I2

B S C d. 31

I2

B S C

Ans-dThe properties of the different quarks are given below:

Quark Charge B S C I3Up +2/3 1/3 0 0 1/2Down 1/3 1/3 0 0 1/2Charm +2/3 1/3 0 1 0Strange 1/3 1/3 1 0 0

The formulae, which can correctly, obtains these properties is given below.

312

Q B S C (SJ15.01)

75. The reaction 2 2 4 01 1 2D + D He + cannot proceed via strong interactions because it violates the conversation of

a. angular momentum b. electric charge

c. baryon number d. isospin

Ans-dWe have,

2 2 4 01 1 2D D He (SJ15.01)

Applying the conservation laws, one obtainsElectric charge 1 1 2 0 (SJ15.01)

(Baryon No.)B 2 2 4 0 (SJ15.01)

vector additon: : 1 1 0,1, 2, 0 0Angular momentum (SJ15.01)

Isospin: 0 0 0 1 (SJ15.01)Since the reaction violates the Isospin conservation, hence it cannot proceed via strong interactions.

H - 1

December-2014Part-A

1. Lunch-dinner pattern of a person of m days is given below. He has a choice of a VEG or a NON-VEG meal for his lunch/dinner(a) if he takes a NON-VEG lunch he will have only VEG dinner

(b) He takes NON-VEG dinner for exactlly 9 days

(c) He takes VEG lunch for exactly 15 days

(d) He takes a total of 14 NON-VEG meals What is m?

a. 18 b. 24

c. 20 d. 38

Ans-cLet V denotes the vegetarian meal, and N is the non vegetarian meal. The sequence of lunch and dinner for 20 days arementioned below,

sequence for lunch : (15V)(5N) Total 20sequence for dinner : (9N)(11V) Total 20

2. Two locomotives are running towards each other with speed of 60 and 40 Km/h. An object keeps on flying to and fro from thefront tip of one locomotive to the front tip of other with a speed of 70 km/h. After 30 minutes, the two locomotives collide andthe object cover before being crushed?a. 50 km b. 15 km

c. 35 km d. 10 km

Ans-cThe distance covered by flying object in 30 min is

170 352

Distance speed time km (SD14.01)

3. A sphere is made up of very thin concentric shells of increasing radii (leaving no gaps). The mass of an arbitrarily choosenshell isa. Equal to the mass of the preceding shell b. proportional to its volume

c. proportional to its radius d. proportional to its surface area

Ans-dConsider a sphere made up of large number thin spherical shells. Let us consider a thin spherical shell of radius ‘r’ and thikness‘dr’, as shown in the figure given below.

shell of radius r

Mass of shell dm

rdr

The mass of the thin spherical shell is2 24 4dm dV r dr dm surface area r (SD14.01)

K - 21

Ans-bWe have,

1

n

nn

df x x xdx (SJ16.01)

The fourier transform of

1ƒ

nikx ikx ikx

nn

df k dxe x dxe x dxe x

dx (SJ16.02)

Let us compute the value of first and second term in the eq. (SJ16.02), we get

0 1, 1n

n nikx k ikxn

ddxe x e dxe x ikdx

(SJ16.03)

Inserting eq. (SJ16.03) in eq. (SJ16.02), one obtains

2 3 4

11 1 1 ..n n

nF k ik ik ik ik ik

2 1 11 1 .. 11 1 1

ikik ik ik f kik ik ik (SJ16.04)

49. The integral equation-ik x-x +i

32 2 2 2

dx, t = dx dt x , tÎ2

is equivalent to the differential equation

a. 2 2

2 32 2

1, ,6

m i x t x tt x b.

2 22 2

2 2 , ,m i x t x tt x

c. 2 2

2 22 2 , 3 ,m i x t x t

t x d. 2 2

2 32 2 , ,m i x t x t

t x

Ans-d

50. A canonical transformation q,p Q,P is made through the generating function 2F q,p = q P on the Hamiltonian

24

2

pH q,p = + q42

where and are constants. The equations of motion for (Q, P) are

a. /Q P and P Q b. 4 /Q P and / 2P Q

c. /Q P and 22PP Q

Q d. 2 /Q P and P Q

Ans-bWe have,

24

2,42

pH q p qq (SJ16.01)

The hamiltonian in the system (q,p) is transformed through canonical transformation to a new system (Q, P) by using thelegendre transformation, genrating function,

22 ,F q P q P (SJ16.02)

The position and momenta co-ordinates in the new system are obtained from Hamiltonian equation of motion i.e.,

22 22 ;F F

p qP Q qq P (SJ16.03)

K - 35

74. Let ES denote the contribution of the surface energy per nucleon in the liquid drop model. The ratio 27 64S 13 S 30E AI : E Zn is

a. 2:3 b. 4:3

c. 5:3 d. 3:2

Ans-bLet ES denote the contribution of the surface energy per nucleon in the liquid drop model and is equal to

2 2/34S SE ER kA

A A(SJ16.01)

where, we have used,1/3

0R R A (SJ16.02)

The energy of 27 6413 30&S SE AI E Zn are

2/327 2/3 6427 ; 64S SE EA k Zn kA A (SJ16.03a, 03b)

Dividing eq. (SJ16.03a) by eq. (SJ6.03b), we get27 2/3

64

27 64 9 464 27 16 3

S

S n

E A A ZincA AE Z (SJ16.04)

75. In the large hadron collider (LHC), two equal energy proton beams traverse in opposite directions along a circular path oflength 27 km. If the total centre of mass energy of a proton-proton pair is 14 TeV, which of the following is the bestapproximation for the proper time taken by a proton to traverse the entire path?a. 12 ns b. 1.2 sc. 1.2 ns d. 0.12 s

Ans-aWe have,

827 , 27 , 3 10 /S km E TeV c m s (SJ16.01)The time taken to complete one round trip is,

3

8

27 10 903 10

S mT sc s

(SJ16.02)

The value of relative exponent is given by,12

32 2

7 10 7.48 10938

pp

p p p

Ep cp eVp m cm c MeVm c m c (SJ16.03)

The proper time taken by a proton to traverse the entire path

0 090 127.48

T nsT T T ns (SJ16.04)

H - 38

The binding energy B of a nucleus (mass number A and charge Z) is2 2

2/31/3

2 cv s sym

A Z a ZB a A a A aA A

(SD14.02)

The question has wrongly mentioned 22 ,Z A instead of 22 ,A Z as it gives negative answer and from physics point of

view syma term is representing excess of neutron over protons.Thus, we have proceeded with correct energy expression.Differentiating eq. (SD14.02) w.r.t z,and solving for Zmin, one obtains

min 2/30

2 / 2c sym

dB Azdz a a A (SD14.03)

Inserting eq. (SD14.01) in eq. (SD14.03), we obtain the most stable isobar for a nucleus is

min 2/3

216 84.3 842 0.75 / 48 216

z (SD14.04)

G - 1

June-2014Part-A

1. Find the missing letterA B C DF I L OK P U ZP W D ?

a. P b. K

c. J d. L

Ans-cOne can easily see that each coulmn contains three constant and a vowel, so in the last coulmn, the last element must beconstant as it has already contains vowel ‘O’. This constant is J as P, K and L are already prtesent in first three coulmn. Thecorrected last coulmn is mentioned below.

DOZJ

2. Consider a right-angle tringle ABC where AB = AC = 3. A rectangle APOQ is drawn inside it, as shown such that the heightof the rectangle is twice its width. The rectangle is moved horizontlly by a distance 0.2 as shown schematically in the diagram(not to sale).

C

Q SO

A P B

T

What is the ratio of ? of

Area ABCArea OST

a. 625 b. 400

c. 225 d. 125

Ans-cWe have,

3, 2 , 0.2, 0.2AB AC AQ AP OS RT OR ST (SJ14.01)The modified diagram after shifting the rectangular is shown below.

One can see that the OST is congruent to ,ORT and hence these two triangles have the same area. The area of the ORT is

1 0.2 0.2 0.022

Area ORT (SJ14.02)

G-16

On equating the eq. (SJ14.01) with eq. (SJ14.02), we get

1 2

2 2 2 2 2 21 2 2 2

, 1 22 2

5 52 2n n

n nE n n

ma ma(SJ14.03)

and solving for n1 and n2, one obtains

1 2 1 21, 2 2, 1n n or n n (SJ14.04)The wave function corresponding to these values, for the fermions, must be antisymmetric i.e.,

1,22 2 2sin sin sin sinx x x xa a a a a (SJ14.05)

32. A particle of mass m in the potential 2 2 21V x, y = m2

is in eigenstate of energy 5

E =2

The corresponding

un-normalized eigenfunction is

a. 2 2exp 22my x y b. 2 2exp 2

2m

x x y

c. 2 2exp2my x y d. 2 2exp

2mxy x y

Ans-aWe have,

2 2 2,

1 5, 4 ,

2 2x yn nV x y m x y E (SJ14.01)

The Hamiltonian corresponding to this potential is,2 2 2 2

2 2 2 22 2

1 1ˆ ˆ ˆ22 2 2 2 x yn nH m x m y H H

m mx y (SJ14.02)

The wave function corresponding to this hamiltonian is

,x yn nx y x y (SJ14.03)

where,

21/ 2

2 /2 ;2 !x y

m xn n nn

mx H x e xn

21/ 2

/2

2 !m y

nn

mH y en

(SJ14.04)

Inserting eq. (SJ14.03) in eq. (SJ14.02), yielding

, ,

ˆ ,ˆ , ,,x y x yn n n n

H x yH x y E x y E

x y

,

ˆˆyx

x y x y

nnn n n n

H yH xE E E

x y (SJ14.05)

where,

1 12 ,2 2x yn x n yE n E n (SJ14.06)

G-28

A

B

A B CB C

C

52. A spectral line due to a transition from an electronic state p to an s state splits into three Zeeman lines in the presence of astrong magnetic filed. At intermediate filed strengths the number of spectral lines isa. 10 b. 3

c. 6 d. 9

Ans-dFor p state, the possible values of total angular momentum are

1 1/ 2 , 1 1/ 2 1/ 2,3 / 2J S to S (SJ14.01)Where, we have used

1, 1/ 2S (SJ14.02)

The number of different energy level corresponding to angular momentum 1/ 2,3/ 2J are

1/ 2, 1 / 2,1 / 2 ; 3 / 2, 3 / 2, 1 / 2,1 / 2,3 / 2j jFor J m For J m (SJ14.03)Thus, there are 6 different energy level corresponding to p state. Similarly, for s state, the possible values of total angularmomentum are

0 1/ 2 , 0 1/ 2 1/ 2J S to S (SJ14.04)Where, we have used

0, 1/ 2S (SJ14.01) (SJ14.05)

The number of different energy level corresponding to angular momentum 1/ 2J are1/ 2, 1 / 2,1 / 2jFor J m (SJ14.06)

Thus, there are 2 different energy level corresponding to s state. Thus, the total number of possible transition from P to S stateare 12. However, applying the dipole transition relation,

0, 1jm (SJ14.07)

The transition corresponding to 2jm are not allowed i.e.,

3 / 2 1/ 2 , 3 / 2 1/ 2j j j jm m m m (SJ14.08)Hence, the total number of observed spectral line, as shown in the figure, in the presence of strong magnetic field are

12 2 10spectral lines (SJ14.09)

53. A particle in the infinite square well

0 0 < x < aV x =

otherwise

is prepared in a state with the wavefuction

3Asin 0 < x < aa

0 otherwise

The expectation value of the energy of the particle is

G - 43

Ans-bWe have,

17 6

0

110 / , 10 , 1%

2Irad s mI (SJ14.01)

The intensity of an em wave inside the conducting medium is

0xI I e (SJ14.02)

Where, is an attenuation constant and it is defined as

2 2 22 /skin depth (SJ14.03)

Inserting eq. (SJ14.01) in eq. (SJ14.03), we get

67 7 3 110

2 2 4 10 10 2 102

m (SJ14.04)

The penetration distance of the em wave inside the conducting medium is

00 0

1x I II I e n x x nI I

(SJ14.05)

Inserting eq. (SJ14.04) and eq. (SJ14.01) in eq. (SJ14.05), we obtain the penetration distance as

23

1 2 10 2 2.310 2.32 10

n mx n x mm (SJ14.06)

Note: If attennuatian of only E or B field is asked then

1skin depth , (SJ14.07)

and your answer will be 4.6mm but since intensity square of amplitude of E (or B), one must use the skin depth relation as

2skin depth

(SJ14.08)

F - 1

December-2013Part-A

1. Three fishermen caught fishes and went to sleep. One of them woke up, took away one fish and 1/3rd of the remainder as hisshare, without other’s knowledge. Later, the three of them divided the remained equally. Howmany fishes were caught?a. 58 b. 19

c. 76 d. 88

Ans -bWe solve this problem using option:Let the total fish is 19, the first person takes a fish and 1/3 his share and the number of remaining fishes

Re 19 1 18 / 3 12maining fishes (SD13.01)These share of each person in the remaining fishes are

Re 12 / 3 4Share of each person maining fish (SD13.02)Note: In other option, the remaining fishes are not divisible by 3.

2. What is the arithmetic mean of 1 1 1 1 1, , , , ..........., ?1× 2 2×3 3× 4 4× 5 100×101

a. 0.01 b. 1

101

c. 0.00111 d. 1 1

49 50 50 512

Ans -bWe have

1 1 1 1 1, , , ,...,1 2 2 3 3 4 4 5 100 101 (SD13.01)

Rearranging eq. (SJ13.01), we get

1 1 1 1 1 1 1 1 100........ 11 2 2 3 3 4 101 101 101 (SD13.02)

The number of terms in the sequence is 100. So, the arthemetic mean is of all the terms 100 /101 1

total number of terms 100 101sumArthemetic mean (SD13.03)

3. Every time a ball falls to ground, it bounces back to half the height it fell from. A ball is dropped from a height of 1024 cm. Themaximum height from the ground to which it can rise after the tenth bounce isa. 102.4 cm b. 1.24 cm

c. 1 cm d. 2 cm

Ans - cLet the initial height be ‘x’, the height after first drop is

1 / 2h x (SD13.01)Similarly, the height after second drop is

22 / 2h x (SD13.02)

Similarly, the height after 10th drop is10

10 / 2 1024 /1024 1h x cm (SD13.03)

F - 12

a. C B AT T T b. A C BT T T

c. B C AT T T d. B A CT T T

Ans-bLet us consider the entropy vs energy curve as shown in the question. Entropy for a system is defined as

1dQ dSdS m slopeT T dQ (SD13.01)

One can see that the slope of phase in the region A is smaller than the slope of phase in region C i.e.,

1 31

A CT T T as m mm (SD13.02)

Since the slope in the region B is zero, this does not means that temperature becomes infinity in this region. However, thisshows that temperature in this region must be lower than that of in region A and C as entropy remains constant with theincrease in energy. Thus the temperature dependence in the three region is given by

A C BT T T . (SD13.03)

31. The physical phenomenon that cannot be used for memory storage applications isa. Large variation in magneto resistance as a function of applied magnetic fieldb. Variation in magnetization of a ferromagnet as a function of applied magnetic fieldc. Variation in polarization of a ferroelectric as a function of applied electric fieldd. Variation in resistance of a metal as a function of applied electric field

Ans-dThe variation in resistance of a metal as a function of applied electric field can not be used for memory storage function. Whilein other three cases memory can be stored as they are related to the orientation of spins and change in the resistance ofmaterial in the presence of magnetic field.

32. Two identical Zener diodes are placed back to back in series and are connected to a variable DC power supply. The bestrepresentation of the I - V characteristics of the circuit is

a.

I

Vb.

I

V

c.

I

Vd.

I

V

Ans-cThe V-I characteristics for first diode, the reverse current increases abruptly for the voltage greater than breakdown voltage.The output of the first zener diode is applied as the input to the second zener diode, thus, the second diode reverse thedirection of current i.e. the forward current increases abruptly for the voltage greater than the breakdown voltage. Thesecharacterstic are correctly shown in the option (c).

F - 24

60. The fourier transform of the derivative of the Dirac function, namely is proportional toa. 0 b. 1

c. sin k d. i k

Ans-dWe have,

'f x x (SD13.01)

The fourier transform of the derivative of the function is,

12

ikxF x e x dx (SD13.02)

Integrating the eq. (SD13.01) using by parts method, we get

1 '2

ikx ikxF x e x dx ik e x dx

0

2 2ikik ike x dx (SD13.03)

Where, we have used

( ) ( ) (0)f x x dx f and ( ) 0x dx (SD13.04)

61. If ˆ ˆ ˆA = iyz + jxz + kxy, then the integral A.dl (where C is along the perimeter of a rectangular area bounded by x = 0, x = aand y = 0, y = b) is

a. 3 312

a b b. 2 2ab a b

c. 3 3a b d. 0

Ans-dWe have

ˆˆ ˆA i yz jxz kxy (SD13.01)Applying the Stoke’s theorem, we get

ˆs

A dl A nds (SD13.02)

The curl of eq. (SD13.01) is given

ˆˆ ˆ

( ) ( ) ( ) ( ) ( ) ( )ˆˆ ˆ 0

i j kxy xz xy yz xz yzA i j k

x y y y z x z x yyz xz xy

(SD13.03)

Inserting eq. (SD13.03) in eq. (SD13.02), we obtain

ˆ0 0s

C

A dl nds (SD13.04)

62. Consider an n ×n n > 1 matrix A, in which Aij is the product of the indicies i and j (namely Aij = ij). The matrix AAa. has one degenerate eigenvalue with degeneracy (n 1)b. has two degenerate eigenvalues with degeneracies 2 and (n 2)c. has one degenerate eigenvalue with degeneracy nd. Does not have any degeenrate eigenvalue

F - 34

where N is the number of particle, q is the position, p is the momentum and E is the energy of the system. The energy of theparticle is

221

2 2pE krm

(SD13.03)

Inserting eq. (SD13.03) in eq. (SD13.02), we get2 2

3 32 21 3

1 p krmZ e d p e d r

h2 22 2 3/ 2

2 22 23 3 30 0

4 42

P krm mp e dp r e dr

kh h (SD13.04)

The integral in eq. (SD13.) are solved in following manner: Let us consider the first integral2 3/2

2 3/ 2 120 0

1 22

ptm mI p e dp t e dt

3/ 2 3/ 21 2 23 / 22 4

m m(SD13.05)

where, we have used,2

1/ 222

p mt p t dpm

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

2 2m mt dt p dp t dt (SD13.06)

Similarly, the second integral is evaluated as2 3/2

2 3/ 2 122 0 0

1 22

krtI r e dr t e dt

k3/2 3/ 2

1 2 23 / 22 4k k (SD13.07)

where, we have used,2

1/ 222kr t r t dr

k3/ 2

1/ 2 2 1/21 2 1 22 2

t dt r dr t dtk k (SD13.08)

The internal energy of the system is defined as

13ln ln ln 3ln 3N

NNU Z Z N A NKT (SD13.09)

Where, we have used2 23/ 2 3/ 2

31 13 3

4 4ln ln 3ln ;

2 2m mZ Z A Ak kh h

(SD13.10)

75. A child makes a random walk on a square lattice of lattice constant a taking a step in the north, east, south, or west directionswith probabilities 0.255, 0.255, 0.245 and 0.245, respectively. After a large number of steps, N, the expected position of thechild with respect to the starting point is at a distance

a. 22 10 Na in the north - east direction b. 22 10N a in the north - east direction

c. 22 2 10 Na in the south - west direction d. 0

Ans-a

E - 1

June 2013Part-A

1. There is an equilateral triangle in the XY-plane with its centre at the origin. The distance of its sides from the origin is 3.5cm. The area of its circumcircle in cm2 isa. 38.5 b. 49

c. 63.65 d.154

Ans-dThe radius of the circle is

2 3.5 7OS r cm (SJ13.01)The area of the circle is

2 2 2227 7 7 1547

A r cm (SJ13.02)

The figure shows the triangle inside the triangle is given below.

DP

Q

R

S

T

UE

B G

F

O

O

3.5

2. A sphere of iron of radius R/2 fixed to one end of a string was lowered into water in a cylindrical container of base radius Rto keep exactly half the sphere dipped. The rise in the level of water in the container will be

a. R/3 b. 49

c. 63.65 d. 154

Ans-dWhen a sphere of iron is dipped inside the cylindrical container such that it is exactly half dipped in the water, the water in thecontainer will rise equal to the half volume displaced by the sphere i.e.,

3 32 21 41/ 2

2 3 2 12 12Cylind sphR R RV V R h R h h (SJ13.01)

E - 13

a. 910 b. 810

c. 710 d. 610

Ans-bWe have,

Let i is the total current supplied by the battery, now applying the KVL in the resistance and capacitor circuit i.e.,

0c C CCdV dV dVC X C X i R X Rdt dt dt

C

dV dtiR VV R X (SJ13.01)

On integrating the eq. (SJ13.01), we get

15

150

1 15n ln 1150C C

tV tC R X C R X

2 2 2

1000 10001 21

j tj RC RCR C (SJ13.02)

The value of the resistance is

8 86

1000 1000 1.414 10 102 5 10 2

RC

(SJ13.03)

30. The approximation cos is valid up to 3 decimal placed as long as is less than: (take 180° / )a. 1.28° b. 1.81°

c. 3.28° d. 4.01°

Ans-cWe have,

cos 1 (SJ13.01)Applying the taylor’s expansion of cos and considering the first 3 terms only in eq.(SJ13.01), we obtain

2 2 4 22 21 ... 1 0 12

2! 4! 4! 2!

0 0, 12 3.46 (SJ13.02)

E - 28

Let us evaluate the value of each of these inner product separately. So the value of first term of inner product is2

2 2 2 22| 2 4 2 | 4d b b x x b b x

dx(SJ13.06)

or

2 21/22

2 2 22

2| 2 4 b xd bb b x e dxdx

1/22 2

3/2

2 1 12 4 2 2 44 42

bb b b b bbb

(SJ13.07)

Similarly, the value second term of inner product is

21/ 2

22 bxbx x x x e dx

21/ 2 1/ 2

2

0

2 2bx

x

b be (SJ13.08)

substituting eq. (SJ13.07) and eq. (SJ13.08) in eq. (SJ13.05), we get1/ 22 2

2b bE b

m(SJ13.09)

The value of variation parameterb is obtained by differentiating E(b) given in eq. (SJ13.09) w.r.t b i.e1/ 22 2 4

1/24

2 1 202 2

dE b mb bdb m (SJ13.10)

Substituting eq. (SJ13.10) in eq. (SJ13.09), we have2

min 2

mE (SJ13.11)

54. Consider two diferent systems each with three identical non - interacting particles. Both have single particle states with

energies 0 0 and 0 05 One system is populated by spin 1/2 fermions and the other by bosons. What is the value

of FE - E , where EF and EB are the ground state energies of the fermionic and bosonic systems respectively?

a. 06 b. 02

c. 04 d. 0

Ans-bThe ground state of the fermion and boson particle are shown in the figure given below:

Boson

05

03

0

Fermion

03

0

05

E - 40

74. Consider the laser resonator cavity shown in the figure. If I1 is the intensity of the radiation at mirror M1 and is the gaincoefficient of the medium between the mirrors, then the energy densityof photons in the plane P at a distance x from M1 is

R1=1 R2=R

l

X

PM1 M2

a. 1 / i xI c e b. 1 / xI c e

c. 1 / x xI c e e d. 1 / xI c e

Ans-c

75. A spin 1/2 particle A undergoes the decay, A B + C + D; where it is known that B and C are also spin 1/2 particles. Thecomplete set of allowed values of the spin of the particle D is

a. 1 3 5,1, ,2, ,3,....2 2 2

b. 0, 1

c. 1/2 only d. 1 3 5 7

, , , ,.....2 2 2 2

Ans-dWe have,

A B C D (SJ13.01)also the spins of the particles A, B and C is 1/2 respectively. The possible values of D is given by

1 1 1 112 2 2 2

A B C D D D (SJ13.02)

In order to satisfy the eq. (SJ13.02), the particle D can not have any integer values, so the possible values of D are

1 3 5 7, , , ,.....2 2 2 2

D (SJ13.03)

D-1

December - 2012Part-A

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

AI

II

III

B C

a. II2 III2 I2 b. II III I

c. II2 III2 I2 d. II III I

Ans-bWe have,

AI

II

III

B C

Applying Pythagoras theoram in ,ABC we obtain2 2 2AC AB BC (SD12.01)

The area of the semicircle for the region I is2 2/ 2

2 8AC ACI (SD12.02)

The area of the semicircle for the region II is2 2/ 2

2 8BC BCII (SD12.03)

The area of the semicircle for the region III is2 2/ 2

2 8AB ABIII (SD12.04)

Now adding eq. (SD12.03) and eq. (SD12.04), we get2 2

2 2 2

8 8 8 8BC ABII III AB BC AC I II III I (SD12.05)

2. A peacock perched on the top of a12 m high tree spots a snake moving towards its hole at the base of the tree from adistance equal to thrice the height of the tree. The peacock files towards the snake in a straight line and they both movethe same speed. At what distance the base of the tree will the peacock catch the snake?a. 16m b. 18 m

c. 14 m d. 12 m

D-12

28. A binary star system consists of two stars S1 and S2, with masses m and 2m respectively separated by a distance r. Ifboth S1 and S2 individually follow circular orbits around the centre of mass with instantaneous speeds v1 and v2 respec-tively, the speeds ratio v1/v2 is

a. 2 b. 1

c. 1/2 d. 2

Ans-dLet the m and 2m are the masses of the two stars moving with velocities v1 & v2 respectively. The angular momentum for thestars, moving in the circular orbit, is always conserved i.e.,

1 2 1 22L L mv r mv r (SD12.01)From the eq. (SD12.01), one, therefore, obtains the ratio of velocities for the two stars as

1 2: 2 :1v v (SD12.02)

29. Three charges are located on the circumference of radius R as shown in the figure below. The two charges Q subtendan angle 90° at the centre of the circle. The charge q is symmetrically placed with respect to the charges Q. If theelectric field at the centre of the circle is zero, what is the magnitude of Q?

a.2

qb. 2q

c. 2q d. 4q

Ans-aThe three charges are placed at the circumference of the radius R. The two charges Q are placed such that they subtends anangle 90° at the centre and the third charge q is placed symmetrically w.r.t Q as shown in figure given below.

Q Q

q

o

R RE'

E

45°

r

Esin Esin

The electric field produced by the two charges at centre ‘O’ is evaluated as follows:The electric field produced by the chargeQ in component form is written as: the horizontal component of electric field at point ‘O’ is

°2 2

0 0

1 1cos 454 4 2H

Q QER R (SD12.01)

Similarly, the perpendicular component of electric field at point ‘O’ is

°2 2

0 0

1 1sin 454 4 2

Q QER R (SD12.02)

The perpendicular components of two charges are equal and opposite, and hence the cancel out. The resultant electric fielddue to charges ‘Q’ at point ‘O’ is equal to the sum of the horizontal component of electric field i.e.,

20

224 2r H

QE ER (SD12.03)

D-28

a. 0 ˆ2Kt

j b. 0 ˆ2Kz jc

c. 0 ˆ2

K ct z ic

d. 0 ˆ2

K ct z jc

Ans-dWe have,

20 ˆ4

kA ct z ic

(SD12.01)

The magnetic field in terms of vector potential is expressed as

B A (SD12.02)Inserting eq. (SD12.02) in eq. (SD12.01), the magnetic field is

0

20

ˆ2

0 04

x y z

i j k i j kkB A ct z j

x y z x y z cA A A k ct z

c

(SD12.03)

56. When a charged particle emits electromagnetic radiation, the electricfield E and the Poynting vector 0

1S = E×B at

a large distance r from the emitter vary as 1/rn and 1/rm respectively. Which of the following choices for n and m arecorrect?a. n = 1 and m = 1 b. n = 2 and m = 2

c. n = 1 and m = 2 d. n = 2 and m = 4

Ans-cThe variation of the electric field and poynting vector with distance ‘r’ is

1 1;n mE Sr r

(SD12.01)In radiation zone, the electric field, magnetic field and poynting vector are given by

20

1 1 1 1; ;E B S E B

r r r (SD12.02)

On equating eq. (SD12.01) with eq. (SD12.02), we obtain the value of constant n and m as1, 2n m (SD12.03)

57. The energies in the ground states and first excited state of a particle of mass m 1/2 in a potential V(x) are -4 and -1,respectively, (in units in which = 1 ). If the correspondig wavefunctions are related by 1 0 sin hx, then theground state eigenfunction is

a. 0 ( ) sech xx b. 0 ( ) secx hx

c. 20 ( ) secx h x d. 3

0 ( ) secx h x

Ans-c

The ground state wavefunction and its corresponding energy, in the unit 1 , in the potential V(x) for a particle of mass(m 1/2)is

0 0 0, 4,x E (SD12.01)

D-38

where, the concentration of free eletrons in fcc structure is

3 3

. . 4 1 4volume of unit cell

no of atoms no freeena a

(SD12.04)

73. The muon has mass 105 MeV/c2 and mean lifetime 2.2 in its rest frame. The mean distance traversed by a muon ofenergy 315 MeV/c2 before decaying is approximatelya. 3 105 km b. 2.2 cm

c. 6.6 m d. 1.98 km

Ans-dWe have,

2 20 105 / , 2.2 , 315 /m MeV c s m MeV c (SD12.01)

The lorentz factor is obtained from the expression of the relativistic mass of the particle i.e.,

002 2, 315 105 3

1 /

mm m MeV MeV

v c (SD12.02)

Now, the dilated time t taken by the particle before decaying is3 2.2 6.6t s (SD12.03)

The distance traversed by the muon before decaying is,5 63 10 6.6 10 1.98d c t km km (SD12.04)

74. Consider the following particles; the proton p, the neutron n, the neutral pion 0 and the delta resonance + Whenordered in terms of decreasing lifetime, the correct arrangement is as follows:a. 0 , , ,n p b. 0, , ,p n

c. 0, , ,p n d. 0, , ,n p

Ans-cThe correct specification of the particles are given below.

Particles Rest Mass 2/MeV c Life timeP 93.83 stable

n 939.3 925135.0 178 10- -

So, the correct arrangement in decreasing order of lifetime is, , , , .p n

75. The single particle energy difference between the p-orbitals (i.e, p3/2 and p1/2) of the nucleus 11450 Sn is 3MeV. The energy

difference between the states in its 1 f orbital isa. 7MeV b. 7 MeV

c. 5MeV d. 5MeV

Ans-b

C-1

June - 2012Part-A

1. In still air, fragrance of a burning incense stick will be smelt by an observer quickest when the experiment is carriedout ata. Low altitude and high air temperature b. high altitude and low air temperature

c. low altitude and low air temperature d. hig altitude and high air temperature

Ans-d

Diffusion of air accelerates at high altitude and high temperature of air.

2. How many squares are there in this figure?

a. 9 b. 14

c. 15 d. 17

Ans-c

3. A mountain road has 3 sections of different slopes as shown. What is the average slope m of the entire climb?

h

h

h

60°

45°

30°

a. 1 b. 1/ 3 1/ 2m

c. 1 3m d. 1/ 3 1m

Ans-dThe average slpoes of the entire climb, as shown in the figure, is

Average slope 1 2 3

3m m mm (SJ12.01)

One can divide the climb into 3 triangles and determine the slopes of each triangle seprately. The slope of the ABC is

11tan 303

m (SJ12.02)

C-13

36. A particle of mass m is in a cubic box of size a. The potential inside the box 0 x a,0 y a,0 z a is zero and

infinite outside. If the particle is in an eigenstate of energy 2 2

2

14E = ,2ma

it wavefunction is

a.3/22 3 5 6sin sin sinx y z

a a a a b.

3/ 22 7 4 3sin sin siny za x a a

c.3/22 4 8 2sin sin sinx y z

a a a a d.

3/ 22 2 3sin sin sinx y za a a a

Ans-dA energy of the particle of mass m confined in a cubical box of size is

2 22 2 2

22 x y zE n n nma

(SJ12.01)

and its corresponding wave function are3/22 sin sin sinyx z

n yn x n ya a a a (SJ12.02)

The particle of an eigenstate of energy is given by,2 2

2

14 ;2

Ema

(SJ12.03)

where a is the size of cubic box. On comapring eq. (SJ12.01) with eq. (SJ12.02), we get

1, 2,3 , 1,3,2 , 3,1, 2 , 2,1,3 , 3, 2,1 , 2,3,1 (SJ12.04)The wavefunction corresponding to the given energy eigenvalue can take anyone of the degenrate state i.e

3/ 22 2 3sin sin sinx y za a a a (SJ12.05)

37. Let nlm denote the eigenfunction of Hamiltonian for a spherically symmetric potential V(r). The wavefunction

210 21-1 21314

is an eigenfunction only of

a. H, L2 and Lz b. H and Lz

c. H and L2 d. L2 and Lz

Ans-cWe have

210 21 1 2111 5 104

(SJ12.01)

Any function is said to be an eigenfunction of an operator A if it satisfies the conditionA a (SJ12.02)

Now applying the Hamiltonian operator H on the wave function is

210 21 1 2111 5 104

H H H H

2 210 2 21 1 2 2111 5 104

E E E

2210 21 1 211 2

ˆ 5 104

EH E (SJ12.03)

C-25

Ans-cThe Charges Q, Q and -2Q are placed on the vertices of an equilateral triangle ABC of sides of length as shown in the givenfigure. The co-ordinates of these charges are shown in the figure.

-2Q C

QQD

a a

i

J

A(0,0) B(a,0)

3,2 2

a a

The dipole moment for a system of discrete charges are obtained by the relation

ˆ ii

p dipole moment q with sign r positionof charge at that point (SJ12.01)

Since the total charge for this system of charges is zero, the dipole moment for this system is independent of the co-ordinatesystem or the origin chosen at any charge. Let us now consider the origin at co-ordinate A, the dipole moment of this systemis

ˆ ˆ ˆˆ 0 2 / 2 3 / 2a a b b c cp q r q r q r Q Q ai Q a i j

ˆ3aQ j (SJ12.02)

59. The vector potential A due to a magnetic moment m at a point r is given by 3

m× rA = .r

If m is directed along the

positive z-axis, the x-component of the magnetic field, at the point r, is

a. 5

3myzr

b. 5

3mxyr

c. 5

3mxzr

d. 2

5

3m z xy

r

Ans-c

The vector potential A due to a magnetic moment m at a point r is given by

3

m rAr

(SJ12.01)

If m is directed along the positive z-axis, the magnetic field, at the point r , is

3 33 .

m rB A r m r r m r

r(SJ12.02)

Or

35

32B r m r r m m r rr

3 35

32 3 .r m r m m r rr

(SJ12.03)

Or

3 5

3 .mB m r rr r

(SJ12.04)

C-33

2 2; 1, 2,3,......& , 1,2,3,......n mk n g nL L

(SJ12.06)

73. The ground state of 20782 Pb nucleus has spin parity P 1J = ,

2 while the first excited state has

-P 5J = .

2 The electromag-

netic radiation emitted when the nucleus makes a transition from the first excited state to the ground state area. E2 and E3 b. M2 and E3

c. E2 and M3 d. M2 and M3

Ans-c

The ground state of 20782 Pb nucleus has spin parity 1/ 2pJ , while the first excited state has 5 / 2 .pJ The condition for

E type transistion, there is no change in parity i.e parity before and after transition remains same. Similarly, for M typetransition, there occurs change in parity i.e., even parity changes to odd parity and vice versa. The change in angular momen-tum is

1 2 1 2| |,| | 3, 2J J J J J (SJ12.01)The condition of parity is given by

1 jP (SJ12.02)

One can see that for 2,J there is no change in parity, thus it leads to E2. For 3,J there is change in parity M type, thusit leads to M3.

74. The dominant interactions underlying the following process

P. K p

Q. K K

R. 0p are

a. P: strong, Q : electromagnetic and R : weakb.P : strong, Q : weak and R : weakc. P: weak, Q : electromagnetic and R : strongd.P: weak, Q: electromagnetic and R : weak

Ans-c

75. If a Higgs boson of mass mH with a speed decays into a pair photons, then the invariant mass of the photonpair is[Note: The invariant mass of a system of two particles, with four - momenta p1 and p2 is (p1 p2)

2]a. Hm b. mH

c. 21Hm d. 2/ 1Hm

Ans-bThe mass of Higgs boson is equal to invariant mass of the photon is i.e. mH.

B - 1

December - 2011Part-A

1. The most abudnant element by mass in the human body isa. carbon b. hydrogen

c. calcium d. oxygen

Ans-dOxygen is the most abundant element by mass in the human body.

2. A number system consists of digits 0, 1, 2, 3, 4 and 5. What is the decimal equivalent of 15 in this number system?a. 15 b. 13

c. 11 d. 12

Ans-c

The decimal equivalent of 15 in the given number system is 1 06

15 1 6 5 6 11.

3. A segment of a circle (slightly greater than a semicircle, whose centre is O) is given below. Identify the correct statementregarding the three angles A, B and C.a. A is equal to B but not equal to Cb. A, B and C are equal and have a value of 85°c. A, B and C are unequald. A, B and C are each equal to 95°

A

O

B C

170°

Ans-bThe angle subtended by an arc at the circumference is half of the angle subtended by an arc at the centre.

2 2 2POQ PAQ PBQ PCQ (SD11.01) So, the angle at the point A, B and C are

1 1 170 852 2

PAQ POQ (SD11.02)

Thus, A, B, O and C are equal and have a value at 85°.

A B C

E

O

F

4. After bubbling air through pure water (pH 7.0), its pH decreased. Which of the following is responsible for the pHchange?a. Nitrogen b. Carbon dioxide

c. Oxygen d. Helium

B-10

The second integral in eq. (SD11.04) is evaluated as follows:

/ 2 / 02 0 0

11 4 1 4 1 0R rr R r RI e dV e r dV e

r (SD11.05)

where, we have used

2 1 4 rr (SD11.06)

The first integral in eq. (SD11.04) is evaluated as follows:

/ /1 3 30 0

11

R Rr R r Rr rI e dV e r dVRr r (SD11.07)

/ 2 / 13 30 0

1 4. 4 4 1

R Rr R r Rr rI e r dr e dr eR Rr r

(SD11.08)

Substituting eq. (SD11.05) and eq. (SD11.08) in eq. (SD11.04), we obtain1 1

0 0 00. 4 1 0 4 1

RQ E dV e e (SD11.09)

29. A counter consist of four flip - flops connected as shown in the figure.

If the counter is initialized as 0 1 2 3A A A A = 0110, the state after the next clock pulse isa. 1000 b. 0001

c. 0011 d. 1100

Ans-bSince the counter, given in question, is synchronous, all the flip flop are clocked simultaneously. The state of counter isstablized at the state mentioned below

0 1 2 3

0 1 2 3

0 1 1 0

0 1 1 0

A A A A

Q Q Q Q (SD11.01)

The input applied at each J-K flip flop, see the truth table, is

0 3 1 0 2 1 3 2J Q J Q J Q J Q

0 3 1 0 2 1 3 2K Q K Q K Q K Q (SD11.02)The truth table of J-K flip flop is

( )0 0 (Q )0 1 01 0 11 1

n

n

n

J K O utput QN o chnag e

Q

Thus, the output corresponding to each flip flops are

0 1 2 30 0 0 1J J J J

0 1 2 31 1 1 0K K K K (SD11.03)

B - 27

c.31 .....

4 3! 4x x d.

2 31 1 .....2! 312x xx

Ans-bWe have

sinf x x (SD11.01)The taylor series expansion of the function f(x) at x a is

2

( ) ( ) ....1! 2!

x a x af x f a f a f a (SD11.02)

Inserting eq. (SD11.01) in eq. (SD11.02) and taking / 4,a we get

21 1 1 ....4 42 2 2! 2

f x x x

21 11 ....4 2! 42

x x (SD11.03)

64. A one dimensional chain consists of a set of N each of length a. When stretched by a load, each rod can align either parallelor perpendicular to the length of the chain. The energy of a rod is - when aligend parallel to the length of the chain and is+ when perpendicular to it. When the chain is in thermal equilibrium at temeprature T, its average length is:a. Na/2 b. Na

c. 2 // 1 Bk TNa e d. 2 /1 Bk TNa e

Ans-cA one dimensional chain consists of a set of N each of length a. When stretched by a load, each rod has energy when alignparallel to the length of the chain and has when align perpendicular to the length of the chain. The partition function forsuch a system is

,E

rz e e e (SD11.01)

The system will be in equilibrium whenever its energy becomes minimum, which happens only when the rod are alignedparallel to the chain of the length. The probabilty of finding the chain in that states is

e ePz e e

(SD11.02)

The mean length of the segment is,

2 / ;1 Bk T

Nae NaP Nae e e

(SD11.03)

65. If the hyperfine interaction in an atom is given by e pH = aS ×S , where eS and pS denote the electron and proton spins,

respectively, the splitting between the 31S and 1

0S state is

a. 2 / 2a b. 2a

c. 2 / 2a d. 22aAns-bThe hyperfine interaction in an atom is

e pH aS S (SD11.01)

B-34

The electric field at a distance r ct from the infinite straight long wire is

2 2 20 1 ˆln2

IAE ct c t r kt r

02 2 2

ˆ.2

Ic kc t r

(SD11.02)

75. Monochromatic light of wavelength 660 nm and intensity 2 2100mW / cm mW / cm falls on a solar cell of area 30 3cm2. Theconversion efficiency of the solar cell is 10%. If each converted photon results in an electron hole pair, what is themaximum

circuit current supplied by the solar cell? (Take -34 8h = 6.6×10 Js,c = 3×10 m / s and -19e = 1.6×10 C).a. 160 mA b. 320 mA

c. 1600 mA d. 3200 mA

Ans-aWe have

2. 10%, 30%, 100 / , 660 .Q E A I mW cm nm (SD11.01)The quantum efficiency of the solar cell is defined as

1240 1240.( ) ( ) ( )

op

in

IQ E R

nm nm P W (SD11.02)

Where responsivity is defined as

;( )op

in

IR

P W (SD11.03)

Inserting the eq. (SD11.01) in eq. (SD11.02), the maximum current supplied by the solar cell is

1240 0.1 660 30.1 160660 3 1240

opop

II mA (SD11.04)

A-1

June - 2011Part-A

1. A physiological disorder X always leads to the disorder Y. However, disorder Y may occur by itself. A population shows4% incidence of disorder Y. Which of the following inferences is valid?a. 4% of the population suffers from both X and Yb.Less than 4% of the population suffers from Xc. At least 4% of the population suffers from X.d.There is no incidence of X in the given population

Ans-bIt is given that disorder Y occurs either through disorder X or by itself. Since, incidence of the disorder Y in the population is4%, one can inferred that there are less than 4% people of population who are suffering from disorder X.

2. Exposing an organism to a certain chemical can change nucleotide bases in a gene, causing mutation. In one suchmutated organism if a protein had only 70% of the primary amino acid sequence, which of the following is likely?a. Mutation broke the protein.b.The organism could not make amino acidsc. Mutation created a terminator codond.The gene was not transcribed.

Ans-bTerminator codon is produced through mutation.

3. The speed of a car increaes every minute as shown in the following table:

Time (Minutes) 1 2 3 24 25Speed

1.5 3.0 4.5 36.0 37.5(Meter/Second)

The speed at the end of the 19th minute would bea. 26.5 b. 28.0

c. 27.0 d. 28.5

Ans-dThe velocity of the car is in the arithmetic progression with respect to time. In the AP series, the n-th term is given by

1 ,nthV a n d (SJ11.01)Where d represents the common difference between two term and a is the first term of the A.P series. According to the question,a = 1.5 m/s, d = 3.0 - 1.5 = 1.5 and n = 19. Substituting these value in above equation, the velocity of the car at 19 min is givenby

19 1.5 19 1 1.5 28.5 /V m s (SJ11.02)

4. If Vinput is applied to the circuit shown, the output would be

Vin

Voutputt

A-13

The expectation value of eq. (SJ11.01) is

0( )u x u x (SJ11.02)According to equi partition theorem, we have

BV k T (SJ11.03)

On equating eq. (SJ11.02) with eq. (SJ11.03), we get the value of x as

0

Bk Tu x V x

u (SJ11.04)

31. Circularly polarized light with intensity I0 is incident normally on a glass prism as shown in the figure. The index ofrefraction of glass is 1.5. The intensity I of light emerging from the prism is

45°

90° 45°

I

I0

a. I0 b. 0.96 I0

c. 0.92I0 d. 0.881I0

Ans-bWe have

1 21.5, 1.0n n (SJ11.01)The transmission coefficient is defined as

1 22

1 2

4n nT

n n (SJ11.02)

where n1 & n2 are the refractive indices in two medium. Inserting eq. (SJ11.01) in eq. (SJ11.02), we obtain

2 2

4 1.5 6 6.00 0.96.6.251 1.5 2.5

T (SJ11.03)

So, the intensity of light emerging from the prism is

0 00.96I TI I (SJ11.04)

32. The acceleration due to gravity (g) on the surface of Earth is approximately 2.6 times that on the surface of Mars.Given that the radius of Mars is about one half the radius of Earth, the ratio of the escape velocity on Earth to that onMars is approximatelya. 1.1 b. 1.3

c. 2.3 d. 5.2

Ans-cThe escape velocity is defined as

2 2eGMv grr

(SJ11.01)

A-28

a. 750 W b. 675 W

c. 250 W d. 225 W

Q. The generated photo - current for a quantum efficiency of unity will bea. 360 A b. 400 A

c. 133 A d. 120 AP. Ans-aWe have

1 1000 ; 1%inc ref incP mW W P P

010 ; ln 4 ; 990W t m P W (SJ11.01)The power transmitted in the semiconductor is

0t

transP P e (SJ11.02)Where, P0 is the power transmission by the semiconductor in unexcited state, t is the thickness and is the absorption co-efficient. Substituting eq. (SJ11.01) in eq. (SJ11.02), we obtain

4 410 ln 4 100

990990 247.54

ttP P e e m (SJ11.03)

So, the power absorbed by the semiconductor is1000 247.5 10 750 .i r t a a i t rP P P P P P P P mW (SJ11.04)

Q. Ans-bWe have

660 ; . 1, 750transnm Q E P mW (SJ11.01)The quantum efficiency of the semiconductor is defined as

12401240. ;( )

OP OP

abs in

I A I ARQ E Rnm P W nm P W (SJ11.02)

Substituting eq. (SJ11.01) in eq. (SJ11.02), we obtain

750 660 4001240OPI A (SJ11.03)

54. The magnetic field of the TE11 mode of a rectangular waveguide of dimensions a b as shown in the figure is given by

z 0H = H cos 0.3 where x and y are in cm.

a

b

y

z

x

P. The dimensions of the waveguide area. 3.33 , 2.50a cm b cm b. 0.40 , 0.30a cm b cm

c. 0.80 , 0.60a cm b cm d. a 1.66 cm, b 1.25 cm

A-38

The number of particles N presents in all excited state in BE condensation is3

. 3 3/0 0

4 11

ss

B E s

V dN D F dh sa e

(SJ11.04)

One can see that from eq. (SJ11.04) that 0 .s N Further, the number of particle becomes zero when 0, for

1& 3.s s Therefore, the range of s for which the system undergoes a BE condensation at non zero temperature is givenby1 3.s

65. Two gravitating bodies A and B with masses mA and mB, respectively, are moving in circular orbit. Assume that B Am >> mand let the radius of the orbit of body A be RA. If the body A is losing mass adibatically. Its orbital radius RA isproportional toa. 1/mA b. 21/ Am

c. mA d. 2Am

Ans-bSince the body is loosing mass through adiabatic process, the angular momentum of the body remains conserved i.e.

tanA B A A A B B BL L cons t m V R m V R k (SJ11.01)Where LA & LB are the angular momentum of the bodies A and B. The velocity of the body A, from eq. (SJ11.01) is

/ ;A A AV k m R (SJ11.02) where, k is constant. The gravitational force acting between two bodies A and B moving in a circular orbit is

2 ,B A

A

Gm mF

R (SJ11.03)

The centripital force required for the circular motion of the body is2

A Acp

A

m vFR (SJ11.04)

On equating the eq. (SJ11.03) with eq. (SJ11.04), we obtain the variation of the orbital radius RA as22

2 2

1A A B A BA

A A A AA A

m v Gm m Gmk RR m R RR m

(SJ11.05)