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GATE
2001 – 2017
PHYSICS-PH
First Edition : January 2018
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CONTENTS
PAPER WITH SOLUTION : 2001 TO 2017
PAPER
PAGES
GATE 2001 01-08
GATE 2002 09-16
GATE 2003 17-28
GATE 2004 29-40
GATE 2005 41-52
GATE 2006 53-63
GATE 2007 64-76
GATE 2008 77-92
GATE 2009 93-102
GATE 2010 103-111
GATE 2011 112-120
GATE 2012 121-130
GATE 2013 131-137
GATE 2014 138-145
GATE 2015 146-155
GATE 2016 156-162
GATE 2017 163-168
SOLUTION
GATE 2001 169-195
GATE 2002 196-214
GATE 2003 215-235
GATE 2004 236-256
GATE 2005 257-276
GATE 2006 277-297
GATE 2007 298-317
GATE 2008 318-339
GATE 2009 340-356
GATE 2010 357-369
GATE 2011 370-380
GATE 2012 381-394
GATE 2013 395-409
GATE 2014 410-424
GATE 2015 425-441
GATE 2016 442-456
GATE 2017 457-470
PAPER
2001 – 2017
FDFDFD
GATE-PH 2001 QUESTION PAPER 1
SECTION-A
1. This question consists of TWENTY FIVE sub-questions (1.1 to 1.25) of ONE mark each. For each
of these sub-questions, four possible answers (a, b, c and d) are given, out of which only one is
correct. [25×1 = 25]
1.1. If S is the closed surface enclosing a volume V and n̂ is the unit normal vector to the surface and r is
ˆ is
the position vector, then the value of the following integral r ndS
s
(a) V (b) 2V (c) 0 (d) 3V
+
1.2. For any operator A, i(A – A) is
(a) Hermitian (b) anti-Hermitian (c) unitary (d) orthogonal
1.3. The value of the integral z10 dz , where C is the unit circle with the origin as the centre is
C
(a) 0 (b) z11/11 (c) 2iz11 / 11 (d) 1/11
1 1 1
1.4. Consider the set of vectors (1,1, 0), (0,1,1) and (1, 0,1)
2 2 2
(a) The three vectors are orthonormal
(b) The three vectors are linearly independent
(c) The three vectors cannot form a basis in a three-dimensional real vector space
1 1 1
(d) (1,1,0) can be written as a linear combination of (0,1,1) and (1, 0,1)
2 2 2
1.5. The Lagrangian for the Kepler problem is given by
1 2 22
L
2
r
r r , ( 0),
where ( r , ) denote the polar coordinates and the mass of the particle is unity. Then
(a) p 2r 2 (b) pr 2r
(c) the angular momentum of the particle about the centre of attraction is a constant
(d) the total energy of the particle is time dependent
1.6 Which of the following equations is relativistically invariant? ( , , and are constants of suitable
dimensions)
( x, t ) 2 ( x, t ) 2 ( x, t ) 2( x, t )
(a) (b)
t x 2 t 2 x 2
2 ( x, t ) ( x, t ) ( x, t ) 3( x, t )
(c) 2
(d)
t x t x 3
1.7. The Lagrangian for a three particles system is given by:
1 2
L
2
1 22 32 a2 (12 22 32 13 ),
where a is real. Then one of the normal coordinates has a frequency given by
2 2
(a) 2 a 2 (b) 2 a 2 / 2 (c) 2 2a 2 (d) 2a
GATE-PH 2001 QUESTION PAPER 2
1.8. Two point charges Q1 = 1nC and Q2 = 2nC are kept in free space such that the distance between them
is 0.1 m.
(a) The force on Q2 is along the direction from Q2 to Q1
(b) The force on Q2 is the same in magnitude as that on Q1
(c) The force on Q1 is attractive
(d) A point charge Q3 = –3nC, placed at the midpoint between Q1 and Q2, experiences no net force
1.9. A current I flows in the anticlockwise direction through a square loop of side a lying in the xoy plane with
its centre at the origin. The magnetic induction at the centre of the square loop is
2 2 0 I 2 2 0 I 2 2 0 I 2 2 0 I
(a) eˆx (b) eˆz (c) eˆz (d) eˆx
a a a 2 a 2
1.10. A thin conducting wire is bent into a circular loop of radius r and placed in a time dependent magnetic field
of magnetic induction.
B(t ) B0e t eˆz , ( B0 0 and 0),
such that, the plane of the loop is perpendicular to B (t ) . Then the induced emf in the loop is
(a) r 2 B0 e t (b) r 2 B0 e t (c) r 2 B0 e t (d) r 2 B0 e t
1.11. Consider an electric field E existing in the interface between a conductor and free space. Then the electric
field E is
(a) external to the conductor and normal to the conductor’s surface
(b) internal to the conductor and normal to the conductor’s surface
(c) external to the conductor and tangential to the conductor’s surface
(d) both external and internal to the conductor and normal to the conductor’s surface
1.12. A spinless particle moves in a central potential V(r)
(a) The kinetic energy and the potential energy of the particle cannot simultaneously have sharp values
(b) The total energy and the potential energy of the particle can simultaneously have sharp values
(c) The total energy and the square of the orbital angular momentum about the origin cannot simultaneously
have sharp values.
(d) The total energy of the particle can have only discrete eigenvalues
1.13. Which of the following functions represents acceptable wave function of a particle in the range x .
(a) ( x) A tan x, A 0 (b) ( x) B cos x, B real
2
(c) ( x) C exp( D / x ), C 0, D 0 (d) ( x) E x exp( Fx 2 ), E , F 0
1.14. A quantum harmonic oscillator is in the energy eigenstate n . A time independent perturbation (at a)2 acts
on the particle, where is a constant of suitable dimensions and a and a† are lowering and raising
operators respectively. Then the first order energy shift is given by
2
(a) n (b) n (c) n 2 (d) ( n)2
1.15. Two particles are said to be distinguishable when
(a) the average distance between them is large compared to their de Broglie wavelengths
(b) the average distance between them is small compared to their de Broglie wavelengths
(c) they have overlapping wavepackets
(d) their total wave function is symmetric under particle exchange
GATE-PH 2001 QUESTION PAPER 3
1.16. For an energy state E of a photon gas, the density of states is proportional to
(a) E (b) E (c) E 3/2 (d) E 2
1.17. X-rays were produced using Cobalt (Z = 27) as target. It was observed that the X-ray spectrum contained
a strong K line of wavelength 0.1785 nm and a weak K line of wavelength 0.1930 nm. Then, the weak
K line is due to an impurity whose atomic number is
(a) 25 (b) 26 (c) 28 (d) 30
1.18. A sample of Silicon of thickness 200m is doped with 10 Phosphorous atoms per m3. If the sample is
23
kept in a magnetic field of 0.2 Wb/m2 and a current of 1 mA is passed through the sample, the Hall voltage
produced is
(a) 62.5 V (b) 6.25 V (c) 6.25 V (d) 62.5 V
1.19. The probability that a state which is 0.2 eV above the Fermi energy in a metal atom at 700K is
(a) 96.2% (b) 62.3% (c) 3.5% (d) 37.7%
1.20. The distance between the adjacent atomic planes in CaCO3 is 0.3 nM. The smallest angle of Bragg
scattering for 0.03 nm X-ray is
(a) 2.9º (b) 1.5º (c) 0.29º (d) 5.8º
1.21. Infrared absorption can be observed in which of the following molecules?
(a) N 2 (b) O 2 (c) HCl (d) C 2
1.22. The cross-sections of the reactions p K and p K – at a given energy are the
same due to
(a) baryon number conservation (b) time-reversal invariance
(c) charge conjugation (d) parity conservation
1.23. RAM and ROM are
(a) charge coupled devices used in computers (b) computer memories
(c) logic gates (d) binary counters used in computers
1.24. In an n-p-n transistor, the leakage current consists of
(a) electrons moving from the base to the emitter
(b) electrons moving from the collector to the base
(c) electrons moving from the collector to the emitter
(d) electrons moving from the base to the collector
1.25. A piece of semiconducting material is introduced into a circuit. If the temperature of the material is raised,
the circuit current will
(a) increase (b) remain the same (c) decrease (d) cease to flow
2. This question consists of TWENTY FIVE sub-questions (2.1 to 2.25) of ONE mark each. For each
of these sub-questions, four possible answers (a, b, c and d) are given, out of which only one is
correct. [25×2 = 50]
2.1. If A x eˆx y eˆy z eˆz , then 2 A equals
(a) 1 (b) 3 (c) 0 (d) –3
sin z
2.2. The value of the residue of is
z6
1 1 2i 2i
(a) (b) (c) (d)
5! 5! 5! 5!
GATE-PH 2001 QUESTION PAPER 4
1
f ( x )e ikx dx, then Fˆ [ f ( x)] is equal to
2
2.3. If F [ f ( x )]
2
(a) f ( x) (b) f ( x) (c) f ( x) (d) [ f ( x) f ( x )] / 2
2.4. A particle of mass m is constrained to move on the plane curve xy C (C 0) under gravity (y axis
vertical). The Lagrangian of the particle is given by
1 2 C 2 mgC 1 2 C 2 mgC
(a) m x 1 4 (b) m x 1 4
2 x x 2 x x
1 2 C mgC 1 2 C mgC
(c) m x 1 2 (d) m x 1 2
2 x x 2 x x
2.5. If poisson bracket [ q, f ( p )] f ( p ), where is a scalar then f p is equal to
p p
(a) ep (b) ep (c) e (d) e
2.6. x and p are two operators which satisfy [ x, p ] i . The operators X and P are defined as
X x cos p sin and Y x sin p cos , for real. Then [ X , Y ] equals
(a) 1 (b) –1 (c) i (d) –i
2.7. A quantum particle of mass m is confined to a square region in xoy-plane whose vertices are given by
(0, 0), (L, 0), (L, L) and (0, L). Which of the following represents an admissible wave function of the
particle (for l, m, n positive integers)?
2 nx my 2 l x ny
(a) sin cos (b) cos cos
L L L L L L
2 mx ny 2 nx l y
(c) sin sin (d) cos sin
L L L L L L
2.8. Let L ( Lx , Ly , Lz ) denote the orbital angular momentum operators of a particle and let L Lx iL y and
L– Lx – iLy . The particle is in an eigenstate of L2 and Lz eigenvalues 2 ( 1) and respectively. The
expectation value of L+ L– in this state is
(a) 2 (b) 2 2 (c) 0 (d)
2.9. A normalized state of a particle moving in a potential V(x) is given by ( x, t ) Cn n ( x)e Ent / h
n 0
where n ( x) ’s are the normalized eigenfunctions of the particle corresponding to the energies En’s. Then
(a) | C n |2 1
n 0
(b) The average energy of the particle in the state ( x, t ) is | C
n
n | En
(c) ( x, t ) is an eigenfunction of the Hamiltonian of the particle
(d) ( x, t ) is an eigenfunction of the momentum operator
GATE-PH 2001 QUESTION PAPER 5
2.10. A coaxial cable of uniform cross-section contains an insulating material of dielectric constant 3.5. The radius
of the central wire is 0.01 m and that of the sheath is 0.02 m. The capacitance per kilometer of a cable
is
(a) 280.5 nF (b) 28.05 nF (c) 56.10 nF (d) 2.805 nF
2.11. The xoy plane carries a uniform surface current of density K 50eˆz A/m. The magnetic field at the point
z 0.5m is
(a) 10 106 Wb (b) 1 106 Wb (c) 106 Wb (d) 10 106 Wb
2.12. Four point charges are placed at the corners of a square whose center is at the origin of a Cartesian
coordinate system. A point dipole p is placed at the centre of the square as shownq–in the figure.q–Then,
(a) there is no force acting on the dipole
(b) there is no torque about the centre of O on the dipole
x O
p
(c) the dipole has minimum energy if it is in eˆx direction
(d) the force on the dipole is increased if the medium is replaced q q
by another medium with larger dielectric constant z
2.13. The electric field E (r , t ) at a point r at time t in a metal due to the passage of electrons can be described
by the equation
2
1 2 E (r , t ) 2
E r ,t 2 E (r , t )
c t 2
where is a characteristic associated with the metal and c is the speed of light in vacuum. The dispersion
relation corresponding to the plane wave solutions of the form exp i (k r t ) is given by
(a) 2 c 2 k 2 2 (b) 2 c 2 k 2 2 (c) ck (d) ck
–6 2
2.14. A copper wire of uniform cross-sectional area 1.0 × 10 m carries a current of 1A. Assuming that each
copper atom contributes one electron to the electron gas, the drift velocity of the free electrons (density of
copper is 8.94 × 103 kg/m3 and its atomic mass is 1.05 × 10–25 kg) is
(a) 7.4 × 10–4 m/s (b) 74 × 10–4 m/s (c) 74 × 10–3 m/s (d) 7.4 × 10–5 m/s
2.15. The number of hyperfine components observed in the electronic transition 2 p1/ 2 2 S1/ 2 of an atom with
1
nuclear spin is
2
(a) 3 (b) 4 (c) 6 (d) 5
2.16. Which of the following functions describes the nature of interaction potential V(r) between two quarks inside
a nucleon? (r is the distance between the quarks and a and b positive constants of suitable dimensions)
a a a a
(a) V (r ) br (b) V (r ) br (c) V (r ) br (d) V (r ) br
r r r r
2.17. Which of the following reactions violates lepton number conservation?
–
(a) e e (b) e p n
(c) e n p (d) e
GATE-PH 2001 QUESTION PAPER 6
2.18. The Lande g-factor for the 3P1 level of an atom is
(a) 1/2 (b) 3/2 (c) 5/2 (d) 7/2
2.19. The pure rotational levels of a molecule in the far-infrared region follows the formula F(J) = BJ (J+1), where
F(J) is the energy of the rotational level with quantum number J and B is the rotational constant. The lowest
rotational energy gap in rotational Raman spectrum is
(a) 2B (b) 4B (c) 6B (d) 8B
2.20. The total number of Zeeman components observed in an electronic transition 2 D5/ 2 2 P3/ 2 of an atom in
a weak field is
(a) 4 (b) 6 (c) 12 (d) 10
2.21. A resistance of 600 is parallel to an inductance of reactance 600 applied voltage, then the total
impedance of the circuit is
(a) 628 (b) 268 (c) 424 (d) 300
2.22. An n-channel silicon (dielectric constant = 12) FET with a channel width a = 2 × 10–6 m is doped with
1021 electrons / m3. The pinch-off voltage is
(a) 0.86V (b) 0.68V (c) 8.6V (d) 6.8V
2.23. The solution of the system of differential equations
dy dz
y z and 4 y z
dx dx
is given by (for A and B are arbitrary constants)
3x x 3x x
(a) y ( x) Ae3 x Be x ; z ( x) 2 Ae3 x 2 Be x (b) y ( x) Ae Be ; z ( x) 2 Ae 2 Be
3x x 3x x
(c) y ( x) Ae3 x Be x ; z ( x) 2 Ae3 x 2 Be x (d) y ( x) Ae Be ; z ( x) 2 Ae 2 Be
2.24. If u ( x, y, z , t ) f ( x i y vt ) g ( x i y vt ), where f and g are arbitrary and twice differentiable
functions, is a solution of the wave function
2 u 2 u 1 2u
then is
x 2 y 2 c 2 t 2
1/ 2
v
1/ 2
v v2 v2
(a) 1 (b) 1 (c) 1 2 (d) 1 c 2
c c c
2.25. The rotational partition function for a diatomic molecule of moment of inertia I at a temperature T is given
by
Ik BT 2 Ik BT 3Ik BT Ik BT
(a) (b) (c) (d)
2 2 2 2 2
GATE-PH 2001 QUESTION PAPER 7
SECTION-B
This section consists of TWENTY questions of FIVE marks each. ANY FIFTEEN out of these
questions have to answered on the Answer Book provided. [75 Marks]
3. Given A y 2 eˆx 2 yxeˆy ( xye z sin x )eˆz , calculate the value of
S
A nˆ ds over the part of the
sphere x 2 y 2 z 2 1 above the xoy plane.
4. Find the matrix of the linear transformation T on V3(R) (i.e., three dimensional real vector space) defined
a a b 1 0 0
as T b b c , with respect to the basis B {eˆ1 , eˆ2 , eˆ3 }, where eˆ1 0 , eˆ2 1 and eˆ3 0 .
c c a 0 0 1
Also calculate the matrix representation of T–1.
2
5. Find the general solution of 4 x d y2 2 dy y 0, using the Frobenius power series method.
dx dx
2
q ln q
6. Consider the Lagrangian L q qf (q ), where f ( q ) is an arbitrary function and 0 .
2
(a) Write down the Euler-Lagrange equation of motion
(b) Does the Lagrangian transform covariantly under the transformation q q for a real constant?
(c) Calculate the Hamiltonian of the system
(d) Is total energy a constant of motion?
7. A square lamina OABC of side l and negligible thickness is lying in the XOY plane of a Cartesian coordinate
system such that O is at the origin and the sides OA and OC are along the positive X and Y directions
respectively. Calculate the moment of inertia tensor and the directions of the three principal moments. The
mass of the lamina is m.
8. A particle of mass m is subjected to a potential V ( x ) k x , k 0
(a) If H is the Hamiltonian of the particle, calculate [ H , V ( x)]
(b) Use the uncertainty principle in the form x p ~ to estimate the ground state energy of the particle.
2
9. A particle of mass m in the one-dimensional energy well
0 0 x L
V ( x)
elsewhere,
is in a state whose coordinate wave function is given by ( x) Cx( L x), where c is the normalization
constant.
(a) Determine the expectation value of the energy in the state ( x)
(b) Calculate the probability that on measurement of energy, the particle will be found in its ground state.
L L
2 2 L3 4 L3
Standard Integrals : dx x sin( x / L ) L / , dx x sin( x / L ) 3
0 0
GATE-PH 2001 QUESTION PAPER 8
10. Consider the harmonic oscillator in the form H ( p 2 x 2 ) / 2 (we have set m 1, 1 and 1 ). The
harmonic oscillator is in its nth energy eigenstate and subjected to a time-independent perturbation
( xp px), for real. Calculate the first-order energy shift and first-order correction to the wave function.
11. An ideal electron gas is confined to an area A in a two-dimensional plane at temperature T. Calculate
(a) the density of states
(b) N, the number of electrons
(c) EF, the Fermi energy as a function of N
12. Write down the partition function of a particle of mass m whose potential energy is given by
V ( x, y , z ) ax 2 b( y 2 z 2 )1/ 2 , where a and b are positive constants of suitable dimensions. Also calculate
the average energy of the particle.
2 2
Standard Integral : dx e x / 2 2
13. Given that the molecular weight of K Cl is 74.6 and its density is 1.99 × 103 kg/m3, calculate the following:
(a) the distance between the atomic planes
(b) the lattice constant
14. The reaction 3 H ( p, n)2 He has a Q value of –0.764 MeV. Calculate the threshold energy of incident
protons for which neutrons are emitted in the forward direction.
15. A circular conducting loop C1 of radius 2m is located in the XOY plane such that its centre is at
(0, 0, 0). Another circular conducting loop C2 of radius 2m is located at (0, 0, 4) such that the plane of
C2 is parallel to the XOY plane. A current of 5A is flowing in each of these loops such that the positive
Z-axis lies to the left of the directions of the currents. Find the magnetic induction B produced at
(0, 0, 0), neglecting the mutual induction of the loops.
16. Draw the electrical circuits for each of the following ************ source (battery), a detector (lamp),
and switch (es).
(a) AND (b) OR (c) NOT (d) NAND (e) NOR
17. The pinch-down voltage of a p-channel junction FET is VP = 5V and the drain-to-source saturation current
IDSS = –40mA. The value of drain-source voltage VDS is such that the transistor is operating in the saturated
region. The drain current is given as ID = –15 mA. Find the gate-source voltage VGS.
18. A narrow beam of electrons, accelerated under a potential difference, incident on a crystal whose grating
space is 0.3 nm. If the first diffraction ring is produced at an angle 5.8º from the incident beam, find the
momentum of the electrons and the potential difference applied.
19. The region z 0 of a Cartesian coordinate system contains a linear isotropic dielectric of dielectric constant
2.0. The region z 0 is the free space. A free space charge density of 5nC/m2 is at the interface z 0 .
If the displacement vector in the dielectric is D2 3eˆx 4eˆy 6eˆz nC/m2, find the corresponding displacement
vector D1 in the free space.
20. The series limit of the Balmer series for hydrogen atom is given as 360 nm. Calculate the atomic number
of the element that gives the lowest x-ray wavelength at 0.1 nm of the K-series.
21. The first few electronic energy states for neutral copper atom (Z = 29) are given as E1 < E2 < E3, where
E1 being the ground electronic state. The states E2 and E3 are doubly degenerate due to spin splitting. Write
the electron configuration of the states and arrange the spectral terms of the split levels following Hund’s
rules. Explain why E2 < E3.
22. The rotational lines of the CN band system at 3883.4Å is represented by a formula v = (25798 + 3.850m
+ 0.068 m2) cm–1, where m is a running number. Calculate the values of the rotational constants Bv and
Bv , the location of the band head and the degradation of the band.
GATE-PH 2002 QUESTION PAPER 9
SECTION-A
1. This question consists of TWENTY FIVE sub-questions (1.1 to 1.25) of ONE mark each. For each
of these sub-questions, four possible answers (a, b, c and d) are given, out of which only one is
correct. [25×1 = 25]
1.1. If two matrices A and B can be diagonalized simultaneously, which of the following is true?
(a) A2B = B2A (b) A2B2 = B2A (c) AB = BA (d) AB2AB = BABA2
1 1
1.2. Which one of the following matrices is the inverse of the matrix ?
0 1
1 1 1 0 1 1 1 1
(a) (b) (c) (d)
1 1 1 1 0 1 0 1
1.3. If a function f ( z ) u ( x, y ) iv ( x, y ) of the complex variable z x iy , where x, y , u and v are real,
is analytic in a domain D of z, then which of the following is true?
u v u v u v
(a) (b) and
x x x y y x
u v u v 2u 2v
(c) and (d)
x x y y xy xy
1.4. The homogeneity of time leads to the law of conservation of
(a) linear momentum (b) angular momentum
(c) energy (d) parity
1.5. Hamilton canonical equations of motion for a conservative system are
dqi H dp H dpi H dpi H
(a) and i (b) and
dt pi dt qi dt pi dt qi
dqi H dpi H dqi H dpi H
(c) and (d) and
dt pi dt qi dt pi dt qi
1.6. If R1 is the value of the Rydberg constant assuming mass of the nucleus to be infinitely large compared to
that of an electron, and if R2 is the Rydberg constant taking nuclear mass to be 7500 times the mass of
the electron, then the ratio R2/R1 is
(a) a little less than unity (b) a little more than unity
(c) infinitely small (d) infinitely large
1.7. Consider an infinitely long straight cylindrical conductor of radius R with its axis along the z-direction, which
carries a current of 1A uniformly distributed over its cross section. Which of the following statements is
correct?
(a) B 0 everywhere (b) B 0 2 zˆ everywhere,
R
(c) B 0 for r R , (d) B 0 2 zˆ for r R
R
where r is the radial distance from the axis of the cylinder.
GATE-PH 2002 QUESTION PAPER 10
1.8. Consider a set of two stationary point charges q1 and q2 as shown in the figure. Which of the following
statements is correct?
q2 q1
Surface S
Contour C P
(a) The electric field at P is independent of q2
(b) The electric flux crossing the closed surface S is independent of q2
(c) The line integral of the electric field E over the closed contour C depends on q1 and q2.
(d) E 0 everywhere
1.9. If the wave function of a particle trapped in space between x 0 and x L is given by
2x
( x) A sin , where A is a constant, for which value(s) of x will the probability of finding the
L
particle be the maximum?
L L L L L 3L
(a) (b) (c) and (d) and
4 2 6 3 4 4
1.10. In a Stern-Gerlach experiment, the magnetic field is in +z direction. A particle comes out of this experiment
in ẑ state. Which of the following statements is true?
(a) The particle has a definite value of the y-component of the spin angular momentum
(b) The particle has a definite value of the square of the spin angular momentum
(c) The particle has a definite value of the x-components of spin angular momentum
(d) The particle has definite values of x-and y-components of spin angular momentum
1.11. If is the total cross-section and f (), being the angle of scattering, is the scattering amplitude for a
quantum mechanical elastic scattering by a spherically symmetric potential, then which of the following is
true? Note that k is the magnitude of the wave vector along the ẑ direction.
4
(a) | f () |2 | f ( 0) |2
(b)
k
4 4
(c) Imaginary part of [ f ( 0)] (d) | f () |2
k k
1.12. In a classical micro-canonical ensemble for a system of N non-interacting particles, the fundamental volume
in phase space which is regarded as “equivalent to one micro-state” is
(a) h3N (b) h 2 N (c) h N (d) h
where h is the Planck’s constant
1.13. Which of the following conditions should be satisfied by the temperature T of a system of N non-interacting
particles occupying a volume V, for Bose-Einstein condensation to take place?
3/2 3/ 2
2
N h2 V
(a) T h (b) T
3 3
2mk B V 2 2mk B N 2
1/ 2 1/ 2
h2 N h 2 V
(c) T (d) T
2mk B V 32 2mk B N 32
where m is the mass of each particle of the system, kB is the Boltzmann constant, h is the Planck’s constant
and is the well known Zeta function.
GATE-PH 2002 QUESTION PAPER 11
1.14. A large circular coil of N turns and radius R carries a time varying current I I 0 sin(t ) . A small circular
coil of n turns and radius r ( r R ) is placed at the center of the large coil such that the coils are concentric
and coplanar. The induced emf in the small coil
(a) leads the current in the large coil by / 2 (b) lags the current in the large coil by
(c) is the phase with the current in the large coil (d) lags the current in the large coil by / 2
1.15. Four charges are placed at the four corners of a square of side a as shown in the figure. The electric dipole
moment of this configuration is y
(a) p qaiˆ qajˆ
q –2q
(b) p qaiˆ qajˆ
O x
(c) p qaiˆ qajˆ q a q
(d) p qaiˆ qajˆ
1.16. Which of the following statements is true?
(a) In a micro-canonical ensemble the total number of particles N and the energy E are constants while
in a canonical ensemble N and temperature T are constants
(b) In a micro-canonical ensemble the total number of particles N is a constant but the energy E is variable
while in a canonical ensemble N and T are constants
(c) In a micro-canonical ensemble N and E are constants while in a canonical ensemble N and T both vary
(d) In a micro-canonical ensemble N and E are constants while in a canonical ensemble N is a constant
but T varies
1.17. In a one-dimensional Kronig Penny model, the total number of possible wave functions is equal to
(a) twice the number of unit cells (b) number of unit cells
(c) half the number of unit cells (d) independent of the number of unit cells
1.18. The potential in a divalent solid at a particular temperature is represented by a one-dimensional periodic
model. The solid should behave electrically as
(a) a semiconductor (b) a conductor
(c) an insulator (d) a superconductor
1.19. In a cubic system with cell edge a, two phonons with wave vectors q1 and q2 collide and produce a third
phonon with a wave vector q3 such that
q1 q2 q3 R,
where R is a lattice vector. Such a collision process will lead to
(a) finite thermal resistance (b) zero thermal resistance
(c) an infinite thermal resistance (d) a finite thermal resistance for certain | R | only
1.20. The baryon number of proton, the lepton number of proton, the baryon number of electron and the lepton
number of electron are respectively
(a) zero, zero, one and zero (b) one, one, zero and one
(c) one, zero, zero and one (d) zero, one, one and zero
1.21. Typical energies released in a nuclear fission and a nuclear fusion reaction are respectively
(a) 50 MeV and 1000 MeV (b) 200 MeV and 1000 MeV
(c) 1000 MeV and 50 MeV (d) 200 MeV and 10 MeV
GATE-PH 2002 QUESTION PAPER 12
1.22. Nuclear forces are
(a) spin dependent and have no non-central part (b) spin dependent and have a non-central part
(c) spin independent and have no non-central part (d) spin independent and have a non-central part
1.23. The nuclear spins of 6C14 and 12Mg25 nuclei are respectively
(a) zero and half-integer (b) half-integer and zero
(c) an integer and half-integer (d) both half-integers
1.24. The asymmetry terms in the Weizsacker semi-empirical mass formula is because of
(a) non-spherical shape of the nucleus
(b) non-zero spin of nucleus
(c) unequal number of protons and neutrons inside the nucleus
(d) odd number of protons inside the nucleus
1.25. Which of the following options is true for a two input XOR gate?
Input Output
A B
(a) 0 1 1
(b) 1 0 0
(c) 0 0 1
(d) 1 1 1
2. This question consists of TWENTY FIVE sub-questions (2.1 to 2.25) of ONE mark each. For each
of these sub-questions, four possible answers (a, b, c and d) are given, out of which only one is
correct. [25×2 = 50]
2.1. Which of the following vectors is orthogonal to the vector ( aiˆ bjˆ ), where a and b ( a b) are constants,
and iˆ and ĵ are unit orthogonal vectors?
(a) biˆ ajˆ (b) aiˆ bjˆ (c) aiˆ bjˆ (d) biˆ ajˆ
2.2. Fourier transform of which of the following functions does not exist?
2 2 2
(a) e|x| (b) xe x (c) e x (d) e x
2.3. The unit vector normal to the surface 3 x 2 4 y z at the point (1, 1, 7) is
(a) ( 6iˆ 4 ˆj kˆ ) / 53 (b) (4iˆ 6 ˆj kˆ) / 53
(c) (6iˆ 4 ˆj kˆ) / 53 (d) (4iˆ 6 ˆj kˆ) / 53
2.4. The solution of the differential equation
d 2 y ( x) dy ( x)
(1 x ) 2
x y ( x ) 0 is
dx dx
x
(a) Ax 2 B (b) Ax Be (c) Ax Be x (d) Ax Bx 2
where A and B are constants
2.5. A particle of mass M moving in a straight line with speed v collides with a stationary particle of the same
mass. In the center of mass coordinate system, the first particle is deflected by 90º.The speed of the second
particle, after collision, in the laboratory system will be
(a) v / 2 (b) 2v (c) v (d) v/2
GATE-PH 2002 QUESTION PAPER 13
2.6. The scalar potential corresponding to the force field F iˆ( y z )
(a) is y 2 / 2 (b) is 1 (c) is zero (d) does not exist
2.7. Two particles of equal mass are connected by an inextensible string of length L. One of the masses is
constrained to move on the surface of a horizontal table. The string passes through a small hole in the table
and the other mass is hanging below the table. The only constraint is that the first mass moves on the surface
of the table. The number of degrees of freedom of the masses-string system is
(a) five (b) four (c) two (d) one
2.8. An electron is accelerated from rest by 10.2 million volts. The percent increase in its mass is
(a) 20,000 (b) 2,000 (c) 200 (d) 20
2.9. An infinitely long closely wound solenoid carries a sinusoidally varying current. The induced electric field is
(a) zero everywhere
(b) non-zero inside and zero outside the solenoid
(c) non-zero inside as well as outside the solenoid
(d) zero inside and non-zero outside the solenoid
2.10. A laser beam of wavelength 600 nm with a circular cross section having a radius of 10mm falls normally
on a lens of radius 20 mm and focal length 10 cm. The radius of the focussed spot is approximately
(a) 0.3 m (b) 0.6 m (c) 3.0 m (d) 6.0 m
2.11. A left circularly polarized light beam of wavelength 600 nm is incident on a crystal of thickness d and
propagates perpendicular to its optic axis. The ordinary and extraordinary refractive indices of the crystal
are n0 = 1.54 and ne = 1.55 respectively. The emergent light will be right circularly polarized if d is
(a) 120 m (b) 60 m (c) 30 m (d) 15 m
2.12. In a two beam interference pattern, the maximum and minimum intensity values are found to be 25I 0 and
9I 0 respectively, where I 0 is a constant. The intensities of the two interfering beams are
(a) 16 I 0 and I 0 (b) 5 I 0 and 3I 0 (c) 17 I 0 and 8 I 0 (d) 8 I 0 and 2 I 0
2.13. An electron propagating along the x-axis passes through a slit of width y 1nm . The uncertainty in the
y-component of its velocity after passing through the slit is
(a) 7.322 × 105 m/s (b) 1.166 × 105 m/s
(c) 3.436 × 105 m/s (d) 2.326 × 104 m/s
2.14.  and B̂ are two quantum mechanical operators. If [ Aˆ , Bˆ ] stands for the commutator of  and B̂ , then
[ Aˆ , Bˆ ],[ Bˆ , Aˆ ] is equal to
ˆ ˆ ˆ ˆ BABA
(a) ABAB ˆˆ ˆˆ (b) Aˆ ( AB
ˆ ˆ BA
ˆ ˆ ) Bˆ ( BA
ˆ ˆ AB
ˆ ˆ)
2
(c) zero
(d) [ Aˆ , Bˆ ]
3 / 2
2.15. An electron is in a state with spin wave function s in the Sz representation. What is the
1/ 2
probability of finding the z-component of its spin along the ẑ direction?
(a) 0.75 (b) 0.50 (c) 0.35 (d) 0.25
GATE-PH 2002 QUESTION PAPER 14
2.16. If the wavelength of the first line of the Balmer series in the hydrogen spectrum is , then the wavelength
of the first line of the Lyman series is
(a) (27 / 5) (b) (5 / 27) (c) (32 / 27) (d) (27 / 32)
2.17. A vector A (5 x 2 y )iˆ (3 y z ) ˆj (2 x az ) kˆ is solenoidal if the constant a has a value
(a) 4 (b) –4 (c) 8 (d) –8
2.18. The mean free path of the particles of a gas at a temperature T0 and pressure p0 has a value 0 . If the
pressure is increased to 1.5p0 and the temperature is reduced to 0.75 T0, the mean free path
(a) remains unchanged (b) is reduced to half
(c) is doubled (d) is equal to 1.125 0
2.19. Which of the following relations between the particle number density n and temperature T must hold good
for a gas consisting of non-interacting particles to be described by quantum statistics?
(a) n / T 1/ 2 1 (b) n / T 3/ 2 1
3/ 2 1/ 2 3/ 2
(c) n / T 1 (d) n / T and n / T can have any value
2.20. For a prefect free-electron gas in a metal, the magnitudes of phase velocity (vp) and group velocity (vg)
are such that
1
(a) v p vg (b) v p vg
2
(c) v p 2vg (d) v p 2vg
2.21. A metal has free-electron density n = 1029 m–3. Which of the following wavelengths will excite plasma
oscillations?
(a) 0.033 m (b) 0.330 m (c) 3.300 m (d) 33.000 m
2.22. For an NaCl crystal, the cell-edge a = 0.563 nm. The smallest angle at which Bragg reflection can occur
corresponds to a set of planes whose incides are
(a) 100 (b) 110 (c) 111 (d) 200
2.23. Silicon has diamond structure with unit-cell-edge a = 0.563 nm. The interatomic separation is
(a) 0.122 nm (b) 0.234 nm (c) 0.383 nm (d) 0.542 nm
9
2.24. The spin and parity of 4Be nucleus, as predicted by the shell model, are respectively.
(a) 3/2 and odd (b) 1/2 and odd (c) 3/2 and even (d) 1/2 and even
2.25. The feedback ratio of an amplifier, which on application of a negative feedback, changes the voltage gain
from –250 to –100 is
(a) –0.250 (b) –0.025 (c) –0.060 (d) –0.006
GATE-PH 2002 QUESTION PAPER 15
SECTION-B
This section consists of TWENTY questions of FIVE marks each. ANY FIFTEEN out of these
questions have to answered on the Answer Book provided. [75 Marks]
d 2 y( x) dy ( x )
3. Given the differential equation 2
2 5 y ( x ) 0 find its solution that satisfies the initial conditions
dx dx
dy
y 0 and x 0 and 1 at x 0
dx
0 1
4. Find the matrix that diagonalizes the matrix
1 0
dx
5. Using the residue theorem, compute the integral I 4
.
0
(1 x )
6. A particle of mass M is attached to two identical springs of unstretched length L0 and spring constant k.
The entire system is placed on a horizontal frictionless table as shown in the figure. The mass is slightly puled
along the surface of the table and perpendicular to the lengths of the springs and then let go. Using the
Lagrangian equation (s) of motion, show whether the mass will execute simple harmonic motion. If so, find
the time period.
L0 m L0
7. A uniform thin circular disc of mass M and radius R lies in the X-Y plane with its centre at the origin. Find
the moments of inertia tensor. What are the values of the principal moments of inertia? Find the principal
axes.
8. Two events, 10–7 s apart in time, take place at two points 50 m apart on the X-axis. Find the speed of
an observer moving along the X-axis who observes the two events simultaneously. What will be the spatial
separation between these two events as seen by this observer?
9. Consider a parallel plate air filled capacitor with a plate area of 10 cm2 separated by a distance of 2 mm.
The potential difference across the plates varies as
V = 360 sin (2 × 106 t) volts,
where t is measured in seconds. Neglecting fringe effects, calculate the displacement current flowing through
the capacitor.
10. The potential of a spherically symmetric charge distribution is given by
a r2
V (r ) 4 ; rR
3 R 2 for
aR
; for r R ,
r
a and R being constants. Find the corresponding charge distribution.
11. Consider a plane electromagnetic wave propagating in free space and having an electric field distribution
given by
3
E E0 ˆj 1 iˆ exp i t 3 x 1 y ,
2
2 2 2
where E0 , and a are constants. Calculate the corresponding magnetic field B .
GATE-PH 2002 QUESTION PAPER 16
12. A particle in the ground state of an infinitely deep one dimensional potential well of width a is subject to
a perturbation of the form
x
V V0 cos 2
a
where V0 is a constant. Find the shift in energy of the particle in the lowest order perturbation theory.
13. A quantum particle is in a state which is the superposition of the eigenstates of the momentum operator
px i . If the probability of finding the momentum k of the particle is 90%, compute its wave
x
function.
( x2 y2 z 2 )
14. The wave function of a free particle is given by (r ) Ce , where C is a constant. Compute the
momentum space probability density, normalize it to 1 and hence find the value of C.
15. Carbon monoxide has a bond length of 0.1132 nm. What will be the frequency of rotation of the molecule
for its lowest excited state?
16. 1 Kg of water at a temperature of 353 K is mixed adiabatically with an equal mass of water at 293 K.
Find the change in entropy of the universe assuming the specific heat of water to be constant equal to
238 J.kg–1. K–1.
17. A conductor having a free electron gas is maintained at a very low temperature (T 0K). Find the average
energy per electron in terms of the electrons density and the electron mass.
18. A small concentration of minority carries is injected into a homogeneous semiconductor held at 300K. An
electric field of 30 V/cm is applied across the width of the crystal. As a result, the minority carriers move
a distance of 1.5 cm in a time of 300 s. What is the diffusion coefficient of the minority carriers in the
semiconductor?
19. A phonon with wave vector q gets absorbed on collision with an electron of wave vector k . The electron
is considered free and its energy is much larger than that of the phonon. If the electron is scattered at an
angle , show that 2sin 1 (q / 2k ) .
20. In spherical coordinates, the wave function describing a state of a system is
3/ 2
1 1 r 4/ 2 a0
( r , , ) e sin e i
8 a0 a0
where a0 is a constant. Find the parity of the system in this state.
21. Calculate the minimum kinetic energy that the neutron should have in order to induce the reaction
O16(n1, He4)C13
in which C13 is left in an excited state of energy 1.79 MeV. Given:
Mass of O16 = 16.000000 amu
Mass of n1 = 1.008986 amu
Mass of He4 = 4.003874 amu
Mass of C13 = 13.007490 amu
22. Calculate the dc collector voltage (Vc) with respect to ground in the amplifier circuit shown in the figure.
The current gain for the transistor is 200.
+9V
4k
6k
VC
Vin 0.7V
~ 3k 2.3k
GATE-PH 2003 QUESTION PAPER 17
PHYSICS-PH
Q.1 – Q.30 : Carry ONE mark each.
1. The two vectors P i, q (i j ) / 2 are
(a) related by a rotation (b) related by a reflection through the xy-plane
(c) related by an inversion (d) not linearly independent
2. A 3 × 3 matrix has eigenvalues 0, 2 i and 2 i . Which one of the following statements is correct?
(a) The matrix is Hermitian (b) The matrix is unitary
(c) The inverse of the matrix exists (d) The determinant of the matrix is zero
2
3. The value of the integral dz / z , where z is a complex variable and C is the unit circle with the origin
C
as its centre, is
(a) 0 (b) 2i (c) 4i (d) 4i
4. A particle with an initial velocity v0 iˆ enters a region with an electric field E0 ˆj and a magnetic field B0 ˆj .
The trajectory of the particle will
(a) be an ellipse (b) be a cycloid
(c) be a helix with constant pitch (d) not be confined to any plane
5. An object of mass m rests on a surface with coefficient of static friction . Which of the following is NOT
correct?
(a) The force of friction is exactly mg
(b) The maximum force of friction is mg
(c) The force of friction is along the surface
(d) The force of friction opposes any effort to move the object
6. The Lagrangian of a particle of mass m moving in a plane is given by L = (1/2) m[vx2 v y2 ] a ( xv y yvx ),
where vx and vy are velocity components and a is a constant. The canonical momenta of the particle are
given by
(a) px mvx and p y mv y (b) px mvx ay and p y mv y ax
(c) px mvx ay and p y mv y ax (d) px mvx ay and p y mv y ax
7. Two events are separated by a distance of 6 × 105 km and the first event occurs 1sec before the second
event. The interval between the two events
(a) is time-like (b) is light-like (null)
(c) is space-like (d) cannot be determined from the information given
8. An electric charge, +Q is placed on the surface of a solid, conducting sphere of radius a. The distance
measured from the centre of the sphere is denoted as r. Then
(a) the charge gets distributed uniformly through the volume of the sphere
(b) the electrostatic potential has the same value for r a
(c) an equal and opposite charge gets induced in the bottom half of the sphere
(d) the electric field is given by 1 / (40 r 2 ) for r a
GATE-PH 2003 QUESTION PAPER 18
9. An electric field applied along the length of a long cylinder produces a polarization P. The depolarization
field produced in this configuration is
(a) 4P / 3 (b) 4P / 3 (c) 2P (d) 0
10. Which one of the following Maxwell’s equations implies the absence of magnetic monopoles?
(a) E / 0 (b) B 0
(c) E B / t (d) B (1 / c 2 )B / t 0 J
11. An electromagnetic wave is propagating in free space in the z-direction. If the electric field is given by
E cos(t kz )i, where t ck , then the magnetic field is given by
(a) B (1 / c) cos(t kz ) j (b) B (1/ c) sin(t kz ) j
(c) B (1/ c) cos(t kz )i (d) B (1 / c) cos(t kz ) j
12. Given a wave with the dispersion relation ck m for k 0 and m 0, which one of the following
is true?
(a) The group velocity is greater than the phase velocity
(b) The group velocity is less than the phase velocity
(c) The group velocity and the phase velocity are equal
(d) There is no definite relation between the group velocity and the phase velocity
13. Which of the following is a valid normalized wave function for a particle in a one dimensional infinite potential
well of width L centered at x 0 ?
(a) (2 / L)[cos(2x / L) sin(2 x / L)] (b) (2 / L)1/ 2 sin[nx / L] for odd n
(c) (2 / L)1/ 2 cos[nx / L] for odd n (d) (2 / L) cos(x / L)
14. The commutator [ x, P 2 ], where x and P are position and momentum operators respectively, is
(a) 2iP (b) iP (c) 2ixP (d) 2ixP
15. A spin half particle is in the state S z / 2 . The expectation values of S x , S x2 , S y , S y2 are given by
(b) 0, / 4, / 4,0
2 2
(a) 0, 0, 2 / 4, 2 / 4
(c) 0, 2 / 4, 0, 2 / 4 (d) 2 / 4, 2 / 4,0, 0
16. The spectral term for the atom with 70% filled subshell and only S = 3/2 is
3 4
(a) P0 (b) F9/ 2 (c) 3 F1/ 2 (d) 4 P1/ 2
17. The hyperfine splitting of the spectral lines of an atom is due to
(a) the coupling between the spins of two or more electrons
(b) the coupling between the spins and the orbital angular momenta of the electrons
(c) the coupling between the electron spins and the nuclear spin
(d) the effect of external electromagnetic fields
GATE-PH 2003 QUESTION PAPER 19
18. A piston containing an ideal gas is originally in the state X (see figure). The gas is taken through a thermal
cycle X Y X as shown. The work done by the gas is positive if the direction of the thermal cycle is
X
(a) clockwise
(b) counter-clockwise P
(c) neither clockwise nor counter-clockwise Y
(d) clockwise from X Y and counter-clockwise from Y X
V
19. A second order phase transition is one in which
(a) the plot of entropy as a function of temperature shows a discontinuity
(b) the plot of specific heat as a function of temperature shows a discontinuity
(c) the plot of volume as a function of pressure shows a discontinuity
(d) the plot of comprehensibility as a function of temperature is continuous
20. Consider the Fermi-Dirac distribution function f(E) at room temperature (300 K) where E refers to energy.
If EF is the Fermi energy, which of the following is true?
(a) f(E) is a step function
(b) f(EF) has a value of 1/2
(c) States with E < EF are filled completely
(d) f(E) is large and tends to infinity as E decreases much below EF
21. If the ionic radii of Mn and S are 0.80 and 0.184 nm respectively, the structure of MnS will be
(a) cubic closed packed (b) body centered cubic
(c) NaCl type (d) primitive cubic cell
22. A cubic cell consists of two atoms of masses m1 and m2 (m1 > m2) with m1 and m2 atoms situated on
alternate planes. Assuming only nearest neighbor interactions, the centre of mass of the two atoms
(a) moves with the atoms in the optical mode and remains fixed in the acoustic mode
(b) remains fixed in the optical mode and moves with the atoms in the acoustic mode
(c) remains fixed in both optical and acoustic modes
(d) moves with the atoms in both optical and acoustic modes
23. In simple metals the phonon contribution to the electrical resistivity at temperature T is
(a) directly proportional to T above Debye temperature and to T3 below it
(b) inversely proportional to T for all temperatures
(c) independent of T for all temperatures
(d) directly proportional to T above Debye temperature and to T5 below it
24. The effective mass of an electron in a semiconductor can be
(a) negative near the bottom of the band (b) a scalar quantity with a small magnitude
(c) zero at the center of the band (d) negative near the top of the band
25. The dielectric constant of water is 80. However its refractive index is 1.75 invalidating the expression
n 1/ 2 . This is because
(a) the water molecule has a permanent dipole moment
(b) the boiling point of water is 100ºC
(c) the two quantities are measured in different experiments
(d) water is transparent to visible light
GATE-PH 2003 QUESTION PAPER 20
26. The nucleus of the atom 9Be4 consists of
(a) 13 up quarks and 13 down quarks (b) 13 up quarks and 14 down quarks
(c) 14 up quarks and 13 down quarks (d) 14 up quarks and 14 down quarks
27. Which one of the following nuclear reactions is possible?
(a) 14
N 7 13C6 vc (b) 13 N 7 13C6 vc
(c) 13
N 7 13C6 (d) 13 N 7 13C7 vc
28. Suppose that a neutron at rest in free space decays into a proton and an electron. This process would
violate
(a) conservation of charge (b) conservation of energy
(c) conservation of linear momentum (d) conservation of angular momentum
29. Which one of the following is TRUE for a semiconductor pn junction with no external bias?
(a) The total charge in the junction is not conserved (b) The p side of the junction is positively charged
(c) The p side of the junction is negative charged (d) No charge develops anywhere in the junction
30. Which one of the set of values given below does NOT satisfy the Boolean relation R PQ (where Q
denotes NOT Q)?
(a) P 1, Q 1, R 0 (b) P 1, Q 1, R 1
(c) P 0, Q 0, R 0 (d) P 0, Q 1, R 1
Q.31 – Q.90 : Carry ONE mark each.
31. The curl of the vector A zi xj yk is given by
(a) i j k (b) i j k (c) i j k (d) i j k
32. Consider the differential equation d 2 x / dt 2 2dx / dt x 0 . At time t 0, it is given that x 1 and
dx / dt 0 . At t 1, the value of x is given by
(a) 1/ e (b) 2 / e (c) 1 (d) 3 / e
33. Sij and Aij represent a symmetric and an antisymmetric real-valued tensor respectively in three dimensions.
The number of independent components of Sij and Aij are
(a) 3 and 6 respectively (b) 6 and 3 respectively
(c) 6 and 6 respectively (d) 9 and 6 respectively
34. Consider the four statements given below about the function f ( x ) x 4 x 2 in the range x .
Which one of the following statements is correct?
P the plot of f ( x) versus x has two maxima and two minima
Q the plot of f(x) versus x cuts the x axis at four points
R the plot of f(x) versus x has three extrema
S no part of the plot f(x) versus x lies in the fourth quadrant
Pick the right combination of correct choices from those given below
(a) P and R (b) R only (c) R and S (d) P and Q
35. The Fourier transform of the function f ( x ) is F (k ) eikx f ( x )dx . The Fourier transform of df ( x) / dx
is
(a) dF ( k ) / dk (b) F (k ) / dk (c) ikF (k ) (d) ikF (k )
GATE-PH 2003 QUESTION PAPER 21
36. A particle of mass m is moving in a potential of the form V ( x, y , z ) (1/ 2)m2 (3 x 2 3 y 2 2 z 2 2 xy ) .
The oscillation frequencies of the three normal modes of the particle are given by
(a) , 3 and 3 (b) 2, 3 and 3
(c) 2, 2 and 2 (d) 2, 2 and 2
37. The speed of a particle whose kinetic energy is equal to its rest mass energy is given by (c is the speed
of light in vacuum)
3
(a) c / 3 (b) 2c / 3 (c) c / 2 (d) c
2
38. Electromagnetic waves are propagating along a hollow, metallic waveguide whose cross-section is a square
of side W. The minimum frequency of the electromagnetic waves is
(a) c / W (b) 2c / W (c) c / W (d) 2c / W
39. Consider the given statements about E(r, t) and B ( r , t ), the electric and magnetic vectors respectively in
a region of free space
P Both E and B are conservative vector fields
Q Both E and B are central force fields
R E and B are mutually perpendicular in the region
S Work done by B on a moving charge in the region is zero
Choose the right combination of correct statements from the following
(a) P and R (b) R and S (c) S only (d) P and Q
40. An infinite conducting sheet in the x-y plane carries a surface current density K along the y-axis. The
magnetic field B for z 0 is
(a) B 0 (b) B 0 Kkˆ / 2 (c) B 0 Kiˆ / 2 (d) B 0 Kjˆ / ( x 2 z 2 )0.5
41. A parallel beam of infrared radiation of wavelength of 1.01 × 10–6 m is incident normally on a screen with
two slits 5 × 10–6 m apart and the resulting interference pattern is observed on a distant screen. What is
the largest number of maxima that can be observed on the screen?
(a) 4 (b) 9 (c) 13 (d) infinitely many
42. A parallel beam of electrons of a given momentum pass through a screen S1 containing a slit and then
produces a diffraction pattern on a screen S2 placed behind it. The width of the central maximum observed
on the screen S2 can be increased by
(a) decreasing the distance between the screens S1 and S2
(b) increasing the width of the slit in screen S1
(c) decreasing the momentum of the electrons
(d) increasing the momentum of the electrons
43. An electron in a time independent potential is in a state which is the superposition of the ground state
(E0 = 11eV) and the first excited state (E1 = 1eV). The wave function of the electron will repeat itself with
a period of
(a) 3.1 × 10–18 s (b) 2.1 × 10–15 s
(c) 1.2 × 10–12 s (d) 1.0 × 10–9 s
GATE-PH 2003 QUESTION PAPER 22
44. A particle has the wave function ( x, t ) A exp(it ) cos( kx) . Which one of the following is correct?
(a) This is an eigenstate of both energy and momentum
(b) This is an eigenstate of momentum and not energy
(c) This is an eigenstate of energy and not momentum
(d) This is not an eigenstate of energy of momentum
45. A free particle with energy E whose wave-function is a plane wave with wavelength enters a region of
constant potential V 0 where the wavelength of the particle is 2 . The ratio (V/E) is
(a) 1/2 (b) 2/3 (c) 3/4 (d) 4/5
46. The vibrational spectrum of a molecule exhibits a strong line with P and R branches at a frequency v1 and
a weaker line at a frequency v2. The frequency v3 is not shown up. Its vibrational Raman spectrum shows
a strongly polarized line at frequency v3 and no feature at v1 and v2.
(a) the molecule could be linear
(b) the molecule lacks a center of inversion
(c) v1 arises from a symmetric stretching mode
(d) v3 arises from a bending mode
47. Three values of rotational energies of molecules are given below in different units
P 10 cm–1
Q 10–23 J
R 104 MHz
Choose the correct arrangement in the increasing order of energy
(a) P, Q, R (b) R, Q, P (c) R, P, Q (d) Q, R, P
48. The short wavelength cut off of the continuous X-ray spectrum from a nickel target is 0.0825 nm. The
voltage required to be applied to an X-ray tube is
(a) 0.15 KV (b) 1.5 KV (c) 15 KV (d) 150 KV
49. The spin-orbit coupling constant for the upper state of sodium atom which emits D lines of wave numbers
16956.2 and 16973.4 cm–1 is
(a) 15 cm–1 (b) 11.4 cm–1 (c) 12.5 cm–1 (d) 15.1 cm–1
50. Consider the following statements about molecular spectra
P CH4 does not give pure rotational Raman lines
Q SF6 could be studied by rotational Raman spectroscopy
R N2 shows infrared absorption spectrum
S CH3CH3 shows vibrational Raman and infrared absorption lines
T H2O2 shows pure rotational spectrum
Choose the right combination of correct statements
(a) P and Q (b) P, R and T (c) P, S and T (d) Q and R
51. The temperature of a cavity of fixed volume is doubled. Which of the following is true for the black-body
radiation inside the cavity?
(a) its energy and the number of photons both increase 8 times
(b) its energy increases 8 times and the number of photons increases 16 times
(c) its energy increases 16 times and the number of photons increases 8 times
(d) its energy and the number of photons both increase 16 times
GATE-PH 2003 QUESTION PAPER 23
52. A sample of ideal gas with initial pressure P and volume V is taken through an isothermal expansion proceed
during which the change in entropy is found to be S . The universal gas constant is R. Then the work done
by the gas is given by
(a) ( PV S ) / (nR ) (b) nRS (c) PV (d) ( PS ) / ( nRV )
53. Hydrogen molecules (mass m) are in thermal equilibrium at a temperature T. Assuming classical distribution
of velocity, the most probable speed at room temperature is
(a) (k BT ) / m (b) 2k BT / m (c) ( 2k BT / m ) (d) m / ( 2k BT )
54. Consider the energy E in the first Brillouin zone as a function of the magnitude of the wave vector k for
a crystal of lattice constant a. Then
(a) the slope of E versus k is proportional to the group velocity
(b) the slope of E versus k has its maximum value at | k | / a
(c) the plot of E versus k will be parabolic in the interval ( / a ) k ( / a )
(d) the slope of E versus k is non-zero for all k the interval ( / a) k ( / a)
55. An external magnetic field of magnitude H is applied to a Type-I superconductor at a temperature below
the transition point. Then which one of the following statements is NOT true for H less than the critical field
HC?
(a) the sample is diamagnetic
(b) it magnetization varies linearly with H
(c) the lines of magnetic induction are pushed out from the sample
(d) the sample exhibits mixed states of magnetization near HC
56. A ferromagnetic material has a Curie temperature 100K. Then
(a) its susceptibility is doubled when it is cooled from 300K to 200K
(b) all the atomic magnets in it get oriented in the same direction above 100K
(c) the plot of inverse susceptibility versus temperature is linear with a slope TC
(d) the plot of its susceptibility versus temperature is linear with an intercept TC
57. The point group symmetrics of the three molecules shown in Figs. 1–3 are respectively
H H Cl Cl H Cl
C C C C C C
H H H H Cl H
Fig. 1 Fig. 2 Fig. 3
[notation : C 2v = 2mm; C 2h = 2/m; D2h = mmm]
(a) C2h, C2v, C2h (b) C2v, C2h, C2h
(c) D2h, C2v, C2h (d) C2v, D2h, C2h
58. The energy density of states of an electron in a one dimensional potential well of infinitely high walls is (the
symbols have their usual meaning)
(a) L m / [ (2 E )] (b) Lm / ( E )
(c) Lm / [ (2 E )] (d) L m / (2E )
59. Which one of the following statements concerning the Compton effect is NOT correct?
(a) The wavelength of the scattered photon is greater than or equal to the wavelength of the incident photon
(b) The electron can acquire a kinetic energy equal to the energy of the incident photon
(c) The energy of the incident photon equals to the kinetic energy of the electron plus the energy of the
scattered photon
(d) The kinetic energy acquired by the electron in the largest when the incident and scattered photons move
in opposite directions
GATE-PH 2003 QUESTION PAPER 24
60. If the photon were to have a finite mass, then the Coulomb potential between two stationary charges
separated by a distance r would
(a) be strictly zero beyond some distance (b) fall off exponentially for large values of r
(c) fall off as 1 / r 3 for large values of r (d) fall off as 1/ r for large values of r
61. A stationary particle in free space is observed to spontaneously decay into two photons. This implies that
(a) the particle carries electric charge
(b) the spin of the particle must be greater than or equal to 2
(c) the particle is a boson
(d) the mass of the particle must be greater than or equal to the mass of the hydrogen atom
62. The masses of a hydrogen atom, neutron and 238U92 are given by 1.0078, 1.0087 and 238.0508 respectively.
The binding energy of 238U92 is therefore approximately equal to (taking 1 a.m.u. = 931.64 MeV)
(a) 120 MeV (b) 1500 MeV (c) 1600 MeV (d) 1800 MeV
63. A bistable multivibrator with a saturation voltage 5V is shown in the diagram. The positive and negative
threshold at the inverting terminal for which the multivibrator will switch to the other state are
2M
–
+ O
2M
0.01F
200k
(a) 5 / 11V (b) 10 / 11V (c) 5V (d) 11V
64. An avalanche effect is observed in a diode when
(a) the forward voltage is less than the breakdown voltage
(b) the forward voltage exceeds the breakdown voltage
(c) the reverse voltage exceeds the breakdown voltage
(d) the diode is heavily doped and forward biased
65. Which of the given relations between the Boolean variables P and Q is NOT correct? (In the notation used
here, P denotes NOT P and Q denotes NOT Q)
(a) PQ PQ P (b) ( PQ) P Q
(c) PQ ( P Q) (d) PQ Q P
Data for Q. No. 66 to 67
Consider the vector V r / r 3
66. The surface integral of this vector over the surface of a cube of size a and centered at the origin
(a) 0 (b) 2 (c) 2a 3 (d) 4
67. Which one of the following is NOT correct?
(a) Value of the line integral of this vector around any closed curve is zero
(b) This vector can be written as the gradient of some scalar function
(c) The line integral of this vector from point P to point Q is independent of the path taken
(d) This vector can represent the magnetic field of some current distribution
GATE-PH 2003 QUESTION PAPER 25
Data for Q. No. 68 to 69
Consider the motion of a particle in the potential V ( x) shown in the figure.
V1
V(x)
P Q R S T
x
68. Suppose the particle has a total energy E = V1 in the figure. Then the speed of the particle is zero when
it is at
(a) point P (b) point Q (c) point S (d) point T
69. Which one of the following statements is NOT correct about the particle?
(a) It experience no force when its position corresponds to the point Q on the curve
(b) It experience no force when its position corresponds to the point R on the curve
(c) Its speed is the largest when it is at S
(d) It will be in a closed orbit between P and R if E < V1
Data for Q. No. 70 to 71
A particle of mass m moving with speed v collides with a stationary particle of equal mass. After the
collision, both the particles move. Let be the angle between the two velocity vectors
70. If the collision is elastic, then
(a) is always less than 90º (b) is always equal to 90º
(c) is always greater than 90º (d) cannot be deduced from the given data
71. If the collision is inelastic, then
(a) is always less than 90º
(b) is always equal to 90º
(c) is always greater than 90º
(d) could assume any value in the range 0º to 180º
Data for Q. No. 72 to 73
Consider two conducting plates of infinite extent, one plate at z 0 and the other at z L, both parallel
to the xy plane. The vector and scalar potential in the region between the plates is given by
A(r , t ) A0iˆ cos(kz ) cos(kct );
( r , t ) 0
72. For this to represent a standing wave in the empty region between the plates
(a) k / L and 0 (b) k 2 / L and / 2
(c) k / (2 L) and / 2 (d) k / 2 L and 0
GATE-PH 2003 QUESTION PAPER 26
73. The energy density at z 0 and t 0 is
2 2 2
(a) 0 (b) 0 c k A0
(c) (1 / 2) 0 A0 2 k 2 (d) (1 / 2) 0 A0 2 k 2 (1 / 2) 0 c 2 k 2 A0 2
Data for Q. No. 74 to 75
A particle is located in a three dimensional cubic well of width L with impenetrable walls.
74. The sum of the energies of the third and the fourth levels is
(a) 102 2 / mL2 (b) 10 2 2 / 3mL2
(c) 11 2 2 / 2mL2 (d) 152 2 / 2mL2
75. The degeneracy of the fourth level is given by
(a) 1 (b) 2 (c) 3 (d) 4
Data for Q. No. 76 to 77
The normalized wave functions 1 and 2 correspond to the ground state and the first excited state of
a particle in a potential. You are given the information that the operator  acts on the wave functions as
Â1 2 and  2 1 .
76. The expectation value of A for the state (31 4 2 ) / 5 is
(a) –0.32 (b) 0.0 (c) 0.75 (d) 0.96
77. Which of the following are eigenfunctions of Â2 ?
(a) 1 and 2 (b) 2 and not 1
(c) 1 and not 2 (d) neither 1 nor 2
Data for Q. No. 78 to 79
In the presence of an inhomogeneous weak magnetic field, spectral lines due to transitions between two sets
of states were observed.
(1) 5l5 5 H 4 and (2) 2 D5/ 2 2 P3/ 2
78. The types of Zeeman effect observed in (1) and (2) respectively are
(a) normal, normal (b) anomalous, anomalous
(c) anomalous, normal (d) normal, anomalous
79. The number of levels into which each of the above four terms split into respectively is
(a) 6, 4, 10, 8 (b) 4, 6, 10, 12 (c) 11, 9, 6, 4 (d) 9, 5, 12, 10
Data for Q. No. 80 to 82
A system consists of three spin-half particles, the z components of whose spins S z (1), S z (2) and S z (3)
can take value +1/2 and –1/2. The total spin of the system is S z S z (1) S z (2) S z (3) .
80. The total number of possible micro-states of this system is
(a) 3 (b) 6 (c) 7 (d) 8
GATE-PH 2003 QUESTION PAPER 27
81. The total number of micro-states with S z 1/ 2 is
(a) 3 (b) 5 (c) 6 (d) 7
82. Consider an ensemble of systems where each microstate has equal probability. The ensemble average of
Sz is
(a) –1/2 (b) 0 (c) 1/2 (d) 3/2
Data for Q. No. 83 to 84
A gas of N particles is enclosed in a volume V at a temperature T. The logarithm of the partition function
is given by ln Z N ln{(V bN ) (k BT )3/ 2 } where b is a constant with appropriate dimensions
83. If P is the pressure of the gas, the equation of state is given by
(a) P(V bN ) Nk BT (b) P(V bN ) k BT
(c) P(V b) Nk BT (d) P(V b) Nk BT
84. The interval energy of the gas is given by
(a) U (1/ 2)k BT (b) U Nk BT
(c) U (3 / 2) Nk BT (d) U 2 Nk BT
Data for Q. No. 85 to 86
A crystal belongs to a face centered cubic lattice with four atoms in the unit cell. The size of the crystal
is 1 cm and its unit cell dimension is 1 nm. f is the scattering factor of the atom.
85. The number of atoms in the crystal is
(a) 2 × 1021 (b) 4 × 1021
(c) 2 × 1023 (d) 4 × 1024
86. The structure factors for (0 1 0) and (2 0 0) reflections respectively are
(a) 2f and zero (b) zero and 4f
(c) 2f and 2f (d) zero and zero
Data for Q. No. 87 to 88
An atomic bomb consisting of 235U explodes and releases an energy of 1014 J. It is known that each 235U
which undergoes fission releases 3 neutrons and about 200 MeV of energy. Further, only 20% of the 235U
atoms in the bomb undergo fission.
87. The total number of neutrons released is about
(a) 4.7 × 1024 (b) 9.7 × 1024
(c) 1.9 × 1025 (d) 3.7 × 1025
88. The mass of 235U in the bomb is about
(a) 1.5 kg (b) 3.0 kg (c) 6.1 kg (d) 12 kg
GATE-PH 2003 QUESTION PAPER 28
Data for Q. No. 89 to 90
The circuit below represents a non-inverting integrator
R
–
V1 + V0
R
C
89. For high frequencies ( ) the input impedance is
(a) 0 (b) R (c) R / (1 RC ) (d)
90. For low frequencies ( ) the input impedance is
(a) V0 (1/ RC ) V1dt
(b) The voltages at the inverting and non-inverting terminals of the op-amp are nearly
(c) The voltage at the non-inverting terminal of the op-amp and the current in the resistor attached to it
are / 2 out of phase
(d) The current in the two resistors are in phase
GATE-PH 2004 QUESTION PAPER 29
PHYSICS-PH
Q.1 – Q.30 : Carry ONE mark each.
1. For the function x 2 y xy , the value of at x y 1 is
(a) 5 (b) 5 (c) 13 (d) 13
2. The average of the function f ( x) sin x in the interval (0, ) is
1 2 1 4
(a) (b) (c) (d)
2
3. Identify the points of unstable equilibrium for the potential shown in the figure.
V(x)
r
s t u
q x
(a) p and s (b) q and t (c) r and u (d) r and s
4. Which one of the following remains invariant under Lorentz transformations ?
1 2 2 2 1 2
(a) 2 (b)
x y z c t x 2 y 2 z 2 c 2 t 2
2 2 2 1 2 2 2 2
(c) (d)
x 2 y 2 z 2 c 2 t 2 x 2 y 2 z 2
5. A charge q is kept at a distance of 2R from the centre of a grounded conducting sphere of radius R. The
image charge and its distance from the centre are, respectively
q R q R R q R
(a) and (b) and (c) q and (d) and
2 2 2 4 2 2 2
6. The state of polarization of light with the electric field vector E iˆ E0 cos (kz t ) ˆj E0 cos (kz t ) is
(a) linearly polarized along z-direction (b) linearly polarized at –45º to x-axis
(c) circularly polarized (d) ellliptically polarized with the major axis along x-axis
7. The resonance widths of , and particle resonances satisfy the relation p . Their life-times
satisfy the relation
(a) p (b) p (c) p (d) p
8. The time-independent Schrödinger equation of a system represents the conservation of the
(a) total binding energy of the system (b) total potential energy of the system
(c) total kinetic energy of the system (d) total energy of the system
9. In a hydrogen atom, the accidental or Coulomb degeneracy for the n 4 state is
(a) 4 (b) 16 (c) 18 (d) 32
GATE-PH 2004 QUESTION PAPER 30
p2
10. The Hamiltonian of a particle is given by H V r r L S , where S is angular momentum.
2m
The Hamiltonian does NOT commute with
(a) L S (b) S 2 (c) Lz (d) L2
11. The spectral terms for a certain electronic configuration are given by 3 D, 1D, 3P, 1P, 5S, 3S . The term with
the lowest energy is
(a) 5 (b) 3 (c) 3 (d) 3 S
S P D
12. The degeneracy of the spectral term 3 F is
(a) 7 (b) 9 (c) 15 (d) 21
13. The Lande g factor for the level 3 D3 is
2 3 3 4
(a) (b) (c) (d)
3 2 4 3
14. All vibrations producing a change in the electric dipole moment of a molecule yield
(a) Raman spectra (b) Infrared spectra (c) Ultra-violet spectra (d) X-ray spectra
15. For any process, the second law of thermodynamics requires that the change of entropy of the universe be
(a) positive only (b) positive or zero (c) zero only (d) negative or zero
16. The dimension of phase space of ten rigid diatomic molecules is
(a) 5 (b) 10 (c) 50 (d) 100
17. The specific heat of an ideal Fermi gas in 3-dimension at very low temperatures (T) varies as
(a) T (b) T3/2 (c) T2 (d) T3
18. Which one of the following is a first order phase transition ?
(a) Vaporization of a liquid at its boiling point (b) Ferromagnetic to paramagnetic
(c) Normal liquid He to superfluid He (d) Superconducting to normal state
19. The c/a ratio for an ideal hexagonal closed packed structure is
2 8
(a) (b) 8 (c) 5 (d)
3 3
20. The number of independent elastic constants in an isotropic cubic solid is
(a) 1 (b) 2 (c) 3 (d) 4
21. The effective mass of an electron in a semiconductor
(a) can never be positive (b) can never be negative
(c) can be positive or negative (d) depends on its spin
22. The critical magnetic field for a solid in superconducting state
(a) does not depend upon temperature (b) increases if the temperature increases
(c) increases if the temperature decreases (d) does not depend on the transition temperature
23. The volume of a nucleus in an atom is proportional to the
(a) mass number (b) proton number (c) neutron number (d) electron number
24. As one moves along the of stability from 56 Fe to 235 U nucleus, the nuclear binding energy per particle de-
creases from about 8.8 MeV to 7.6 MeV. This trend is mainly due to the
(a) short range nature of the nuclear forces (b) long range nature of the Coulomb forces
(c) tensor nature of the nuclear forces (d) spin dependence of the nuclear forces
GATE-PH 2004 QUESTION PAPER 31
235
25. A thermal neutron having speed v impinges on a U nucleus. The reaction cross-section is proportional to
(a) v 1 (b) v (c) v1/2 (d) v 1/2
26. Choose the particle with zero Baryon number from the list given below.
(a) pion (b) neutron (c) protron (d) +
27. A bipolar junction transistor with one junction forward biased and either the collector or emitter open, operates
in the
(a) cut-off region (b) saturation region (c) pinch-off region (d) active region
28. A field effect transistor is a
(a) unipolar device (b) special type of biopolar junction transistor
(c) unijunction device (d) device with low input impedance
29. The inverting input terminal of an operational amplifier (op-amp) is shorted with the output terminal apart from
being grounded. A voltage signal vi is applied to the non-inverting input terminal of the op-amp. Under this
configuration, the op-amp functions as
(a) an open loop inverter (b) a voltage to current converter
(c) a voltage follower (d) an oscillator
30. A half-adder is a digital circuit with
(a) three inputs and one output (b) three inputs and two outputs
(c) two inputs and one output (d) two inputs and two outputs
Q.31 – Q.90 : Carry TWO marks each.
31. A real traceless 4 × 4 unitary matrix has two eigenvalues –1 and +1. The other eigenvalues are
(a) zero and +2 (b) zero and +1 (c) zero and +2 (d) –1 and +1
1 i
32. The eigenvalues of the matrix are
i 1
(a) +1 and +1 (b) zero and +1 (c) zero and +2 (d) –1 and +1
3 4i
33. The inverse of the complex number is
3 4i
2 24 7 24 7 24 7 25
(a) i (b) i (c) i (d) i
25 25 25 25 25 25 25 25
dz
34. The value of ( z 2 a 2 ) , where C is a unit circle (anti-clockwise) centered at the origin in the complex z-
C
plane is
1 1
(a) for a 2 (b) zero for a (c) 4 for a 2 (d) for a
2 2 2
35. The Laplace transform f (t ) sin t is F ( s) , s 0 . Therefore, the Laplace transform of
( s 2 )
2
t sin t is
2 2s 2
(a) (b) (c) (d)
2 2
s (s ) 2 2 2
s (s ) 2 2 2
(s ) 2 2
( s 2 ) 2
2
GATE-PH 2004 QUESTION PAPER 32
36. A periodic function f ( x) x for x has the Fourier series representation
2
f ( x ) (1) n sin nx . Using this, one finds the sum n 2 to be
n 1 n n 1
2 2
(a) 2 ln 2 (b) (c) (d) ln 2
3 6
37. The Fourier transform F ( k ) of a function f ( x ) is defined as F (k ) dx f ( x) exp (ikx) . The F ( k ) for
f ( x) exp ( x 2 ) is [Given : exp ( x
2
) dx x ]
k 2 k 2
(a) exp ( k ) (b) exp (c) exp (d) 2 exp (k 2 )
4
2 2
38. The Lagrangian of a particle moving in a plane under the influence of a central potential is given by
1
L m (r 2 r 2 2 ) V (r )
2
The generalized momenta corresponding to r and are given by
(a) mr and mr 2 (b) mr and mr (c) mr2 and mr 2 (d) mr2 and mr 2 2
39. A particle of mass m is attached to a thin uniform rod of length a and mass 4m. The distance of the particle from
the center of mass of the rod is a/4. The moment of inertia of the combination about an axis passing through O
normal to the rod is O
64 2 91 2
(a) ma (b) ma a/2
48 48
a/4
27 51 2
(c) ma 2 (d) ma
48 48
40. A rigid frictionless rod rotates anticlockwise in a vertical plane with angular velocity . A bead of mass m
moves outward along the rod with constant velocity u0 . The bead will experience a coriolis force
(a) 2mu0ˆ (b) 2mu0ˆ (c) 4mu0ˆ (d) mu0ˆ
41. The Hamiltonian corresponding to the Langrangian L ax 2 by 2 kxy is
2 2 2
px2 p y px2 p y px2 p y px2 p 2y
(a) kxy (b) kxy (c) kxy (d) kxy
2a 2b 4a 4b 4a 4b 4ab
42. The value of the Poisson bracket a r , b p , where a and b are constant vectors, is
(a) a b (b) a b (c) a b (d) a b
GATE-PH 2004 QUESTION PAPER 33
43. A mass m is connected on either side with a spring each of spring constant k1 and k2 . The free ends of springs
are tied to rigid supports. The displacement of the mass is x from equilibrium position. Which one of the
following is TRUE ?
(a) The force acting on the mass is (k1 k2 )1/2 x
(b) The angular momentum of the mass is zero about the equilibrium point and its Lagrangian is
1 2 1
mx (k1 k2 ) x 2
2 2
1
m
(c) The total energy of the system is mx 2
2
m 2 1
(d) The angular momentum of the mass is mxx and the Lagrangian of the system is x ( k1 k 2 ) x 2
2 2
44. An electron gains energy so that its mass becomes 2m0 . Its speed is
3 3 3 3
(a) c (b) c (c) c (d) c
2 4 2 2
45. 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
46. A conducting sphere of radius R is placed in a uniform electric field E0 directed along z axis. The electric
R3
potential for outside points is given as Vout E0 1 r cos , where r is the distance from the centre
r 3
and is the polar angle. The charge density on the surface of the sphere is
0
(a) 3 0 E0 cos (b) 0 E0 cos (c) 3 0 E0 cos (d) E0 cos
3
47. A circular arc QTS is kept in an external magnetic field B0 as shown in figure. The arc carries a current I. The
magnetic field is directed normal and into the page. The force acting on the arc is
y
B0 x x x T x x x x
x x x x I x x x x
x x xQ x x x S x x
x x x x x x x x x
R R
60º
x
(a) 2I B0 Rjˆ (b) I B0 Rjˆ (c) 2I B0 Rjˆ (d) I B0 Rjˆ
48. A plane electromagnetic wave of frequency is incident on an air-dielectric interface. The dielectric is linear,,
isotropic, non-magnetic and its refractive index is n. The reflectance (R) and transmittance (T) from the inter-
face are
2
n 1 4n n 1 2
(a) R , T (b) R , T
n 1 (n 1)2 n 1 (n 1)2
3 n 12
n 1 4n3 4n2
(c) R , T (d) R , T
n 1 ( n 1)3 n 1 (n 1) 2
GATE-PH 2004 QUESTION PAPER 34
49. The electric field of a plane e.m. wave is E E0 exp i ( xk cos yk sin t ) . If xˆ , yˆ and zˆ are cartesian
unit vectors, the wave vector k̂ of the e.m. wave is
(a) ẑk ˆ sin yk
(b) xk ˆ cos
ˆ cos yk
(c) xk ˆ sin (d) zk
ˆ
50. The dispersion relation for a low density plasma is 2 20 c 2 k 2 , where 0 is the plasma frequency and c
is the speed of light in free space. The relationship between the group velocity (vg ) and phase velocity (v p )
is
(a) v p vg (b) v p v1/2
g (c) v p v g c 2 (d) vg v1/2
p
51. A Michelson interferometer is illuminated with monochromatic light. When one of the mirrors is moved through
a distance of 25.3 µm, 92 fringes pass through the cross-wire. The wavelength of the monochromatic light is
(a) 500 nm (b) 550 nm (c) 600 nm (d) 650 nm
52. A beam of mono-energetic particles having speed v is described by the wave function ( x) u ( x) exp (ikz ) ,
where u ( x) is a real function. This corresponds to a current density
2
(a) u 2 ( x)v (b) v (c) zero (d) u ( x )
53. The wave function of a spin-less particle of mass m in a one-dimensional potential V ( x) is
( x) A exp ( 2 x 2 ) corresponding to an eigenvalue E0 2 2 /m . The potential V ( x) is
(a) 2 E0 (1 2 x 2 ) (b) 2 E0 (1 2 x 2 ) (c) 2E0 2 x 2 (d) 2 E0 (1 2 2 x 2
54. Two spin S1 and S2 interact via a potential V (r ) S1 S2 V0 (r ) . The contribution of this potential in the
singlet and triplet states, respectively, are
3 1 1 3
(a) V0 (r ) and V0 ( r ) (b) V0 (r ) and V0 (r )
2 2 2 2
1 3 3 1
(c) V0 (r ) and V0 (r ) (d) V0 (r ) and V0 ( r )
4 4 4 4
2 x2
55. The wave function of a one-dimensonal harmonic oscillator is 0 A exp for the ground state
2
4
mw x
E0 , where 2 . In the presence of a perturbing potential of E0 , the first order change in
2 10
the ground state energy is
Given : (x 1) t x exp ( t ) dt
0
1 3
(a) E0 10 4 (b) (3E0 ) 10 4 (c) E0 10 4 (d) ( E0 ) 10 4
2 4
GATE-PH 2004 QUESTION PAPER 35
56. The L, S and J quantum numbers corresponding to the ground state electronic configuration of Boron ( Z 5)
are
1 3 1 1
(a) L 1, S , J (b) L 1, S , J
2 2 2 2
3 1 3 3
(c) L 1, S , J (d) L 1, S , J
2 2 2 2
57. The degeneracies of the J-states arising from the 3P term with spin-orbit interaction are
(a) 1, 3, 5 (b) 1, 2, 3 (c) 3, 5, 7 (d) 2, 6, 10
58. Assuming that the L-S coupling scheme is valid, the number of permitted transitions from 2P3/2 to 2S1/2 due to
a weak magnetic field is
(a) 2 (b) 4 (c) 6 (d) 10
59. Consider the pure rotational spectrum of a diatomic rigid rotor. The separation between two consecutive lines
( v ) in the spectrum
(a) is directly proportional to the moment of inertial of the rotor
(b) in inversely proportional to the moment of inertia of the rotor
(c) depends on the angular momentum
(d) is directly proportional to the square of the interatomic separation
60. Light of wavelength 1.5 µm incident on a material with a characteristic Raman frequency of 20 × 1012 Hz results
in a Stokes-shifted line of wavelength
[Given : c = 3 × 108 m.s–1]
(a) 1.47 µm (b) 1.57 µm (c) 1.67 µm (d) 1.77 µm
61. Consider black body radiation in a cavity maintained at 2000 K. If the volume of the cavity is reversibly and
adiabatically increased from 10 cm3 to 640 cm3, the temperature of the cavity changes to
(a) 800 K (b) 700 K (c) 600 K (d) 500 K
62. The equation of state of a dilute gas at very high temperature is described by pv 1 B (T ) , where v is the
k BT v
volume per particle and B (T ) is a negative quantity. One can conclude that this is a property of
(a) a van der Waals gas (b) an ideal Fermi gas
(c) an ideal Bose gas (d) an ideal inert gas
63. In the region of co-existence of a liquid and vapor phases of a material
1 V are both finite
(a) C p and Cv are both infinite (b) Cv and
V T P
1 V are both finite (d) C , and K are all infinite
(c) Cv and K T p
V T
64. A doped Germanium crystal of length 2 cm, breadth 1 cm and width 1 cm, carries a current of 1 mA along its
length parallel to +x-axis. A magnetic field of 0.5 T is applied along +z-axis. Hall voltage of 6 mV is measured
with negative polarity at y 0 plane. The sign and concentration of the majority charge carrier are, respec-
tively. [Given : e = 1.6 × 10–19 C]
y
O x
z
GATE-PH 2004 QUESTION PAPER 36
(a) positive and 5.2 1019 m 3 (b) negative and 5.2 1019 m 3
(c) positive and 10.4 1019 m 3 (d) negative and 10.4 1019 m 3
65. The temperature dependence of the electrical conductivity of two intrinsic semiconductors A and B is shown
in the figure. If E A and EB are the band gaps of A and B respectively, which one of the following is TRUE ?
ln A B
T –1
(a) E A EB (b) E A EB
(c) E A EB (d) E A and EB both depend on temperature
66. If the static dielectric constant of NaCl crystal is 5.6 and its optical refractive index is 1.5, the ratio of its electric
polarizability to its total polarizability is
(a) 0.5 (b) 0.7 (c) 0.8 (d) 0.9
67. Which one of the following statements is TRUE ?
(a) Magnetic tapes are made of Iron
(b) Permanent magnets are made from ferrites
(c) Ultrasonic transducers are made from quartz crystals
(d) Optoelectronic devices are made from soft ferrites
68. Which one of the following statements is NOT TRUE ?
(a) Entropy decreases markedly on cooling a superconductor below the critical temperature, Tc
(b) The electronic contribution to the heat capacity in the superconducting state has an exponential form with
an argument proportional to T –1, suggestive of an energy gap
(c) A type I superconductor is a perfect diamagnet
(d) Critical temperature of superconductors does not vary with the isotopic mass
r 3
69. The form factor F ( q ) exp i q ( r ) d r of Rutherford scattering is obtained by choosing a delta
function for the charge density ( r ) . The value of the form factor is
(a) unity (b) infinity (c) zero (d) undefined
70. Deuteron in its ground state has a total angular momentum J 1 and a positive parity. The corresponding
orbital angular momentum L and spin S combinations are
(a) L 0, S 1 and L 2, S 0 (b) L 0, S 1 and L 1, S 1
(c) L 0, S 1 and L 2, S 1 (d) L 1, S 1 and L 2, S 1
71. Which one of the following reaction is allowed ?
(a) p n e + (b) p e + ve (c) p + (d) p n 0
GATE-PH 2004 QUESTION PAPER 37
72. What should be the values of the components R and R2 such that the frequency of the Wien Bride oscillator is
300 Hz ?
[Given : C 0.01 F and R1 12 k ]
R1 R2
V0
C R
R C
(a) R 48 k and R2 12 k (b) R 26 k and R2 24 k
(c) R 530 and R2 1 M (d) R 53 k and R2 24 k
73. Figure shows a common emitter amplifier with 100 . What is the maximum peak to peak input signal (vs )
for which is distortion-free output may be obtained ?
[Assume VBE 0 and re 20 ]
+10 V
500 k 2 k
v0
vs
(a) 40 mV (b) 60 mV (c) 80 mV (d) 100 mV
74. Calculate the collector voltage (vc ) of the transistor circuit is shown in the figure.
[Given : 0.96, ICB 0 20 A, VBE 0.3V , RB 100 k , VCC 10V and RC 2.2 k ]
VCC
RC
RB
vc
(a) 3.78 V (b) 3.82 V (c) 4.72 V (d) 9.7 V
GATE-PH 2004 QUESTION PAPER 38
75. Figure shows a practical integrator with RS 30 M, RF 20 M and C F 0.1 F . If a step (dc) voltage
of +3 V is applied as input for 0 t 4 (t is in seconds), the output voltage is
RF
RS CF
v0
+3V
ROM
(a) a ramp function of –6 V (b) a step function of –12 V
(c) a ramp function of –15 V (d) a ramp function of –4 V
76. The Boolean expression Y ABCD ABCD ABCD ABCD reduces to
(a) AB (b) D (c) A (d) AD
Common Data for Q. 77 and Q. 78
Consider the differential equation y p ( x ) y q ( x) y ( x) 0 .
77. If xp( x) and x 2 q ( x ) have the Taylor series expansions
xp( x) 4 x x 2 ......
x 2 q ( x ) 2 3x 5 x 2 ......
then the roots of the incidicial equation are
(a) –1, 0 (b) –1, –2 (c) –1, 1 (d) –1, 2
78. If p ( x) 0 with the Wronskian at x 0 as W ( x 0) 1 and one of the solutions is x, then the other linearly
independent solution which vanishes at x 1/2 is
(a) 1 (b) 1 4x 2 (c) x (d) –1 + 2x
Common Data for Q. 79 and Q. 80
Consider a comet of mass m moving in a parabolic orbit around the Sun. The closest distance between the
comet and the Sun is b, the mass of the Sun is M and the universal gravitation constant is G.
79. The angular momentum of the comet is
(a) M Gmb (b) b GmM (c) G mMb (d) m 2GMb
80. Which one of the following is TRUE for the above system ?
(a) The acceleration of the comet is maximum when it is closest to the Sun
(b) The linear momentum of the comet is a constant
(c) The comet will return to the solar system after a specified period
(d) The kinetic energy of the comet is a constant
GATE-PH 2004 QUESTION PAPER 39
Common Data for Q. 81 and Q. 82
Let E xE
ˆ 0 exp i k r t , where k zˆ (k cos i k sin ), k 1 k and xˆ , yˆ and zˆ are cartesian unit
vectors, represent an electric field of plane electro magnetic wave of frequency .
81. Which one of the following statements is TRUE ?
(a) The magnitude of the electric field is attenuated as the wave propagates
(b) The energy of the e.m. wave flows along the x-direction
(c) The magnitude of the electric field of the wave is a constant
(d) The speed of the wave is the same as c (speed of light in free space)
82. The magnetic field B of the wave is
k
(a) yˆ E0 exp ( zk sin ) exp[i ( zk cos t )]
k
(b) yˆ E0 exp ( zk sin ) exp[i ( zk cos t )]
k
(c) yˆ E0 exp[i ( zk cos t )]
k
(d) yˆ E0 exp ( zk cos ) exp[i ( zk sin t )]
Common Data for Q. 83 and Q. 84
A particle is confined to the region 0 x L in one dimension
83. If the particle is in the first excited state, then the probability of finding the particle is maximum at
L L L L 3L
(a) x (b) x (c) x (d) x and
6 2 3 4 4
84. If the particle is in the lowest energy state, then the probability of finding the particle in the region 0 x L is
4
1 1 1 1 1 1
(a) (b) (c) (d)
4 (2) 4 4 (2) 2
Common Data for Q. 85 and Q. 86
The one-electron states for non-interacting electrons confined in a cubic box of side a are 0 1 2 3 4
etc.
85. The energy of the lowest state is
2 2 2 2 3 2 2
(a) zero (b) (c) (d)
2ma 2 ma 2 2ma 2
86. The degeneracy (including spin) of the level 3 is
(a) 2 (b) 4 (c) 6 (d) 8
GATE-PH 2004 QUESTION PAPER 40
Common Data for Q. 87 and Q. 88
An ensemble of N three level systems with energies 0 , 0, 0 is in thermal equilibrium at temperature
T. Let (k BT ) 1 .
87. If 0 2 , the probability of finding the system in the level 0 is
(cosh 2)
(a) (b) (cosh 2)1 (c) (2 cosh 2)1 (d) (1 2 cosh 2)1
2
88. The free energy of the system at high temperature (i.e., x 0 1) is approximately
(a) Nk BTx 2 (b) Nk BT [ln 2 x 2 /2]
(c) Nk BT [ln 3 x 2 /3] (d) Nk BT ln 3
Common Data for Q. 89 and Q. 90
The nucleus 41Ca can be described by the single particle shell model.
89. The single particle states occupied by the last proton and the last neutron, respectively, are given by
(a) d5/2 and f7/2 (b) d3/2 and f5/2 (c) d5/2 and f5/2 (d) d3/2 and f 7/2
90. The ground state angular momentum and parity of 41Ca are
7 3 5 5
(a) (b) (c) (d)
2 2 2 2
GATE-PH 2005 QUESTION PAPER 41
PHYSICS-PH
Q.1 – Q.30 : Carry ONE mark each.
1. The average value of the function f ( x) 4 x3 in the interval 1 to 3 is
(a) 15 (b) 20 (c) 40 (d) 80
2. The unit normal to the curve x3 y 2 xy 17 at the point (2, 0) is
(a)
iˆ ˆj (b) iˆ (c) ĵ (d) ĵ
2
dz
3. The value of the integral z 3 , where C is a circle (anticlockwise) with z 4 , is
C
(a) 0 (b) i (c) 2 i (d) 4 i
4. The determinant of a 3 × 3 real symmetric matrix is 36. If two of its eigen values are 2 and 3 then the third
eigenvalue is
(a) 4 (b) 6 (c) 8 (d) 9
5. For a particle moving in a central field
(a) the kinetic energy is a constant of motion (b) the potential energy is velocity dependent
(c) the motion is confined in a plane (d) the total energy is not conserved
6. A bead of mass m slides along a straight frictionless rigid wire rotating in a horizontal plane with a constant
angular speed . The axis of rotation is perpendicular to the wire and passes through one end of the wire. If r
is the distance of the mass from the axis of rotation and v is its speed then the magnitude of the Coriolis force
is
mv 2 2mv 2
(a) (b) (c) mv (d) 2mv
r r
7. If for a system of N particles of different masses m1, m2 ,...., mN with position vectors r1, r2 ,...., rN and
corresponding velocities v1 , v2 ,...., vN , respectively, such that vi 0 , then
i
(a) the total momentum MUST be zero
(b) the total angular momentum MUST be independent of the choice of the origin
(c) the total force on the system MUST be zero
(d) the total torque on the system MUST be zero
8. Although mass-energy equivalence of special relativity allows conversion of a photon to an electron-positron
pair, such a process cannot occur in free space because
(a) the mass is not conserved (b) the energy is not conserved
(c) the momentum is not conserved (d) the charge is not conserved
9. Three infinitely long wires are placed equally apart on the circumference of a circle of radius a, perpendicular
to its plane. Two of the wires carry current I each, in the same direction, while the third
carries current 2I along
the direction opposite to the other two. The magnitude of the magnetic induction B at a distance r from the
centre of the circle, for r a , is
2 0 I 2 0 I 20 I a
(a) 0 (b) (c) (d)
r r r2
GATE-PH 2005 QUESTION PAPER 42
10. A solid sphere of radius R carries a uniform volume charge density . The magnitude of electric field inside the
sphere at a distance r from the centre is
r R R 2 R3
(a) (b) (c) (d)
30 2 0 r 0 r 20
11. The electric field E ( r , t ) for a circularly polarized electromagnetic wave propagating along the position z-
direction is
(a) E0 ( xˆ yˆ ) exp[i (kz t )] (b) E0 ( xˆ i yˆ ) exp[i (kz t )]
(c) E0 ( xˆ i yˆ ) exp[i (kz t )] (d) E0 ( xˆ yˆ ) exp[i (kz t )]
12. The electric (E) and magnetic (B) field amplitudes associated with an electromagnetic radiation from a point
source behave at a distance r from the source as
1 1
(a) E = constant, B = constant (b) E , B
r r
1 1 1 1
(c) E 2
, B (d) E , B
r r2 r 3
r3
13. The parities of the wave functions
(i) cos ( kx), and
(ii) and tan h( kx) are
(a) (i) odd, (ii) odd (b) (i) even, (ii) even (c) (i) odd, (ii) even (d) (i) even, (ii) odd
14. The commutator, [ Lz , Ylm (, )] , where Lz is the z-component of the orbital angular momentum and Ylm (, )
is a spherical harmonic, is
(a) l (l 1) Ylm (, ) (b) m Ylm (, ) (c) m Ylm (, ) (d) l Ylm (, )
15. A system in a normalized state c1 1 c2 2 , with 1 and 2 representing two different
eigenstates of the system, requires that the constants c1 and c2 must satisfy the condition
(a) c1 c2 1 (b) c1 c2 1 (c) c1 c2 2 1 (d) c1 2 c2 2 1
16. A one dimensional harmonic oscillator carrying a charge q is placed in a uniform electric field E along the
positive x-axis. The corresponding Hamiltonian operator is
2 d 2 1 2 2 d 2 1 2
(a) kx qEx (b) kx qEx
2m dx 2 2 2m dx 2 2
2 d 2 1 2 2 d 2 1 2
(c) kx qEx (d) kx qEx
2m dx 2 2 2m dx 2 2
17. The L line of X-rays emitted from an atom with principal quantum numbers n 1, 2, 3,...., arises from the
transition
(a) n 4 n 2 (b) n 3 n 2 (c) n 5 n 2 (d) n 3 n 1
18. For an electron in hydrogen atom, the states are characterized by the usual quantum numbers n, l , ml . The
electric dipole transition between any two states requires that
(a) l 0, ml 0, 1 (b) l 1, ml 1, 2
(c) l 1, ml 0, 1 (d) l 1, ml 0, 2
GATE-PH 2005 QUESTION PAPER 43
1
19. If the equation of state for a gas with internal energy U is pV U , then the equation for an adiabatic process
3
is
1/3 2/3 4/3 3/5
(a) pV constant (b) pV constant (c) pV constant (d) pV constant
20. The total number of accessible states of N non-interacting particles of spin - 12 is
(a) 2 N (b) N 2 (c) 2 N /2 (d) N
21. The pressure for a non-interacting Fermi gas with internal energy U at temperature T is
3U 2U 3U 1 U
(a) p (b) p (c) p (d) p
2 V 3 V 5 V 2 V
22. A system of non-interacting Fermi particles with Fermi energy EF has the density of states proportional to
E , where E is the energy of a particle. The average energy per particle at temperature T 0 is
1 1 2 3
(a) EF (b) EF (c) EF (d) EF
6 5 5 5
23. In crystallographic notations the vector OP in the cubic cell shown in the figure is
z
a
a/2
P
O y
(a) [221] (b) [122] (c) [121] (d) [112]
24. Match the following and choose the correct combination
Group-I Group-II
P. Atomic configuration 1s22s22p63s23p6 1. Na
Q. Strongly electropositive 2. Si
R. Strongly electronegative 3. Ar
S. Covalent bonding 4. Cl
(a) P-1, Q-2, R-3, S-4 (b) P-3, Q-2, R-4, S-1
(c) P-3, Q-1, R-4, S-2 (d) P-3, Q-4, R-1, S-2
25. The evidence for the non-conservation of parity in - decay has been obtained from the observation that the
intensity
(a) antiparallel to the nuclear spin directions is same as that along the nuclear spin direction
(b) antiparallel to the nuclear spin direction is not the same as that along the nuclear spin direction
(c) shows a continuous distribution as a function of momentum
(d) is independent of the nuclear spin direction
GATE-PH 2005 QUESTION PAPER 44
26. Which of the following expressions for total binding energy B of a nucleus is correct (a1, a2 , a3 , a4 0) ?
Z ( Z 1) ( A 2 Z )2
(b) B a1 A a2 A2/3 a3 a4
A1/3 A
2/3 Z ( Z 1) ( A 2Z ) 2
(b) B a1 A a2 A a3 a4
A1/3 A
Z ( Z 1) ( A 2 Z )2
(c) B a1 A a2 A1/3 a3 a4
A1/3 A
( A 2Z ) 2
Z ( Z 1)
(d) B a1 A a2 A1/3 a3 a4
A1/3 A
27. Which of the following decay is forbidden ?
(a) e v vc (b) v
(c) e ve (d) e e e
28. With reference to nuclear forces which of the following statements is NOT true ?
The nuclear forces are
(a) short range (b) charge independent (c) velocity dependent (d) spin independent
29. A junction field effect transistor behaves as a
(a) voltage controlled current source (b) voltage controlled voltage source
(c) current controlled voltage source (d) current controlled current source
30. The circuit shown can be used as
Vin1
Vout
Vin2
(a) NOR gate (b) OR gate (c) NAND gate (d) AND gate
Q.31 – Q.80 : Carry TWO marks each.
31.
If a vector field F x iˆ 2 y ˆj 3z kˆ, then F is
(a) 0 (b) iˆ (c) 2 ˆj (d) 3kˆ
32. All solutions of the equation e z 3 are
(a) z i n ln 3, n 1, 2,.... (b) z ln 3 i (2n 1) , n 0, 1, 2,....
(c) z ln 3 i 2n, n 0, 1 2,.... (d) z i 3n, n 1 2,....
33. If f ( s ) is the Laplace transform of f (t ) the Laplace transform of f (at ) , where a is a constant, is
1 1
(a) f ( s) (b) f ( s /a ) (c) f ( s ) (d) f (s /a )
a a
GATE-PH 2005 QUESTION PAPER 45
34. Given the four vectors
1 3 2 3
u1 2 , u2 5 , u3 4 , u4 6
1 1 8 12
the linearly dependent pair is
(a) u1 u2 (b) u1 u3 (c) u1 u4 (d) u3 u4
sin z
35. Consider the following function: f z . Which of the following statemens is are TRUE?
z
(a) z = 0 is pole of order 1 (b) z = 0 is a removable singular point
(c) z = 0 is a pole order 3 (d) z = 0 is an essential singular point
36. Eigen values of the matrix
0 1 0 0
1
0 0 0
are
0 0 0 2i
0 0 2i 0
(a) –2, –1, 1, 2 (b) –1, 1, 0, 2 (c) 1, 0, 2, 3 (d) –1, 1, 0, 3
37. If a particle moves outward in a plane along a curved trajectory described by r a, t , where a and
are constants, then its
(a) kinetic energy is conserved (b) angular momentum is conserved
(c) total momentum is conserved (d) radial momentum is conserved
38. A circular hoop of mass M and radius a rolls without slipping with constant angular speed along the horizon-
tal x-axis in the xy-plane. When the centre of the hoop is at a distance d 2a from the origin, the magnitude
of the total angular momentum of the hoop about the origin is
(a) Ma 2 (b) 2Ma 2 (c) 2Ma 2 (d) 3Ma 2
39. Two solid spheres of radius R and mass M each are connected by a thin rigid rod of negligible mass. The
distance between the centres is 4R. The moment of inertia about an axis passing through the centre of symmetry
and perpendicular to the line joining the spheres is
11 22 44 88
(a) MR 2 (b) MR 2 (c) MR 2 (d) MR 2
5 5 5 5
40. A car is moving with constant linear acceleration a along horizontal x-axis. A solid sphere of mass M and radius
R is found rolling without slipping on the horizontal floor of the car in the same direction as seenfrom an inertial
frame outside the car. The acceleration of the sphere in the inertial frame is
a 2a 3a 5a
(a) (b) (c) (d)
7 7 7 7
41. A rod of length l0 makes an angle 0 with the y-axis in its rest frame, while the rest frame moves to the right
1/2
v2
along the x-axis with relativistic speed v with respect to the lab frame. If 1 2 , the angle in the
c
lab frame is
(a) tan 1 ( tan 0 ) (b) tan 1 ( cot 0 )
1 1
(c) tan 1 tan 0 (d) tan 1 cot 0
GATE-PH 2005 QUESTION PAPER 46
1 1
42. m2 x 2 mv 2 , where x is the position coordinate, v is
A particle of mass m moves in a potential V ( x )
2 2
the speed, and and are constants. The canonical (conjugate) momentum of the particle is
(a) p m(1 )v (b) p mv (c) p m v (d) p m(1 ) v
43. Consider the following three independent cases :
(i) Particle A of charge q moves in free space with a constant velocity v (v speed of light)
(ii) Particle B of charge q moves in free space in a circle of radius R with same speed v as in case (i)
(iii) Particle C having charge q moves as in case (ii)
If the power radiated by A, B and C are PA , PB and PC , respectively, then
(a) PA 0, PB PC (b) PA 0, PB PC (c) PA PB PC (d) PA PB PC
44. If the electrostatic potential were given by 0 ( x 2 y 2 z 2 ), where 0 is constant, then the charge den-
sity giving rise to the above potential would be
6
(a) 0 (b) 6 0 0 (c) 2 0 0 (d) 0
0
45. The work done in bringing a charge q from infinity in free space, to a position at a distance d in front of a
semi-infinite grounded metal surface is
q2 q2 q2 q2
(a) (b) (c) (d)
4 0 ( d ) 4 0 (2d ) 4 0 (4d ) 4 0 (6d )
46. A plane electromagnetic wave travelling in vaccum is incident normally on a non-magnetic, non-absorbing
medium of refractive index n. The incident ( Ei ) , reflected ( Er ) and transmitted ( Et ) electric fields are given
as, Ei E exp i (kz t ) , Er E0r exp i(kr z t ) , Et E0t exp i(kt z t ) . If E 2 V/m and
n 1.5 , then the application of appropriate boundary conditions leads to
3 7 1 8
(a) E0r V/m, E0t V/m (b) E0r V/m, E0t V/m
5 5 5 5
2 8 4 6
(c) E0r V/m, E0t V/m (d) E0r V/m, E0t V/m
5 5 5 5
Q
47. For a vector potential A, the divergence of A is A 0 2 , where Q is a constant of appropriate
4 r
dimension. The corresponding scalar potential ( r , t ) that makes A and Lorentz gauge invariant is
1 Q 1 Qt 1 Q 1 Qt
(a) (b) (c) (d)
4 0 r 4 0 r 4 0 r 2 4 0 r 2
48. An infinitely long wire carrying a current I (t ) I 0 cos (t ) is placed at a distance a from a square loop of side
a as shown in the figure. If the resistance of the loop is R, then the amplitude of the induced current in the loop
is
a
I (t)
0 aI 0 0 aI 0 20 aI 0 0 aI 0
(a) ln 2 (b) ln 2 (c) ln 2 (d)
2 R R R 2 R
GATE-PH 2005 QUESTION PAPER 47
49. The de Broglie wavelength for an electron of energy 150 eV is
(a) 10–8 m (b) 10–10 m (c) 10–12 m (d) 10–14 m
50. A particle is incident with a constant energy E on a one-dimensional potential barrier as shown in the figure. The
wavefunctions in regions I and II are respectively
(a) decaying, oscillatory (b) oscillatory, oscillatory
(c) oscillatory, decaying (d) decaying, decaying
51. The expectation value of the z-coordinate, (z) in the ground state of the hydrogen atom (wavefunction :
100 (r ) Ae r / a0 , where A is the normalization constant and a0 is the Bohr radius), is
Region II
Region I
V0
E
x
a0 a0
(a) a0 (b) (c) (d) 0
2 4
52. The degeneracy of the n 2 level for a three dimensional isotropic oscillator is
(a) 4 (b) 6 (c) 8 (d) 10
53. For a spin - 12 particle, the expectation value of sx s y sz , where sx , s y and sz are spin operators, is
i 3 i 3 i 3 i 3
(a) (b) (c) (d)
8 8 16 16
54. An atom emits a photon of wavelength 600 nm by transition from an excited state of lifetime 8 × 10–9 s. If
v represents the minimum uncertainty in the frequency of the photon, the fractional width v /v of the
spectral line is of the order of
(a) 10–4 (b) 10–6 (c) 10–8 (d) 10–10
55. The sodium doublet lines are due to transitions from 2 P3/2 and 2 P1/2 levels to 2 S1/2 level. On application of a
weak magnetic field, the total number of allowed transitions becomes
(a) 4 (b) 6 (c) 8 (d) 10
56. A three level system of atoms has N1 atoms in level E1, N 2 in level E2 , and N 3 in level E3 ( N 2 N1 N 3
and E1 E2 E3 ). Laser emission is possible between the levels
(a) E3 E1 (b) E2 E1 (c) E3 E2 (d) E2 E3
57. In an Raman scattering experiment, light of frequency v from a laser is scattered by diatomic molecules having
the moment of intertia I. The typical Raman shifted frequency depends on
(a) v and I (b) only v (c) only I (d) neither v nor I
58. For a diatomic molecule with the vibrational quantum number n and rotational quantum number J, the vibra-
tional level spacing En En En 1 and the rotational level spacing EJ EJ EJ 1 are approximately
(a) En = constant, EJ = constant (b) En = constant, EJ J
(c) En n, EJ J (d) En n, E J J 2
GATE-PH 2005 QUESTION PAPER 48
59. The typical wavelengths emitted by diatomic molecules in purely vibrational and purely rotational transitions are
respectively in the region of
(a) infrared and visible (b) visible and infrared
(c) infrared and microwave (d) microwave and infrared
60. In a two electron atomic system having orbital and spin angular momenta l1, l2 and s1, s2 respectively, the
coupling strengths are defined as l1 l2 , s1 s2 , l1 s1 , l2 s2 , l1 s2 and l2 s1 . For the jj coupling scheme to
be applicable, the coupling strengths MUST satisfy the condition
(a) l1 s2 , s1 s2 l1 s1 , l2 s2 (b) l1 s1 , l2 s2 l1 l2 , s1 s2
(c) l1 s2 , l2 s1 l1 l2 , s1 s2 (d) l1 s2 , l2 s1 l1 s1 , l2 s2
61. If the probability that x lies between x and x dx is p ( x ) dx ae ax dx , where 0 x , a 0 , then prob-
ability that x lies between x1 and x2 ( x2 x1 ) is
ax ax
(a) e 1 e 2
ax ax
(b) a e 1 e 2
ax
ax ax
(c) e 2 e 1 e 2 ax
ax ax
(d) e 1 e 1 e 2
kT
62. If the partition function of a harmonic oscillator with frequency at a temperature T is , then the free
h
energy of N such independent oscillator is
3
(a) NkT (b) kT ln (c) NkT ln (d) NkT ln
2 kT kT 2kT
63. The partition function of two Bose particles each of which can occupy any of the two energy levels 0 and is
(a) 1 e2/ kT 2e/ kT (b) 1 e2/ kT e/ kT
(c) 2 e2/ kT e/ kT (d) e 2 / kT e / kT
64. A one dimensional random walker takes steps to left or right with equal probability. The probability that the
random walker starting from origin is back to origin after N even number of steps is
N 2N N
N! 1 N! 1 1
(a) (b) (c) 2 N ! (d) N !
N N 2 N N 2 2
! ! ! !
2 2 2 2
65. The number of states for a system of N identical free particles in a three dimensional space having total energy
between E and E E (E E ) , is proportional to
3 N 1
(a) E 2 E (b) E N /2 E (c) NE1/2 E (d) E N E
66. The energy of a ferromagnet as a function of magnetization M is given by
F ( M ) F0 2(T Tc ) M 2 M 4 , F0 0.
The number of minima in the function F ( M ) for T Tc is
(a) 0 (b) 1 (c) 3 (d) 4
GATE-PH 2005 QUESTION PAPER 49
67. For a closed packed BCC structure of hard spheres, the lattice constant a is related to the sphere radius R as
4R
(a) a (b) a 4 R 3 (c) a 4 R 2 (d) a 2 R 2
3
68. An n-type semiconductor has an electron concentration of 3 1020 m3 . If the electron drift velocity is 100
ms 1 in an electric field of 200 Vm 1 , the conductivity (in 1 m 1 ) of this material is
(a) 24 (b) 36 (c) 48 (d) 96
69. Density of states of free electrons in a solid moving with an energy 0.1 eV is given by 2.15 1021 eV1 cm 3 .
The density of states (in eV1 cm3 ) for electrons moving with an energy of 0.4 eV will be
(a) 1.07 1021 (b) 1.52 10 21 (c) 3.04 1021 (d) 4.30 1021
70. The effective density of states at the conduction band edge of Ge is 1.04 1019 cm 3 at room temperature
(300 K). Ge has an optical bandgap of 0.66 eV. The intrinsic carrier concentration (in cm3 ) in Ge at room
temperature (300 K) is approximately
(a) 3 × 1010 (b) 3 × 1013 (c) 3 × 1016 (d) 3 × 1016
71. For a conventional superconductor, which of the following statements is NOT true ?
(a) Specific heat is discontinuous at transition temperature Tc
(b) The resistivity falls sharply at Tc
(c) It is diamagnetic below Tc
(d) It is paramagnetic below Tc
72. A nucleus having mass number 240 decays by emission to the ground state of its daughter nucleus. The Q
value of the process is 5.26 MeV. The energy (in MeV) of the particle is
(a) 5.26 (b) 5.17 (c) 5.13 (d) 5.09
73. The threshold temperature above which the thermonuclear reaction
3
2 He 32 He 24 He 211 H 12.86 MeV can occur is (use e 2/40 1.44 1015 MeVm)
(a) 1.28 1010 K (b) 1.28 109 K (c) 1.28 108 K (d) 1.28 107 K
74. According to the shell model, the ground state of 158 O nucleus is
3 1 3 1
(a) (b) (c) (d)
2 2 2 2
75. The plot of log A vs. time t, where A is activity, as shown in the figure, corresponds to decay
Log A
t
(a) from only one kind of radioactive nuclei having same half life
(b) from only neutron activated nuclei
(c) from a mixture of radioactive nuclei having different half lives
(d) which is unphysical
GATE-PH 2005 QUESTION PAPER 50
76. For the rectifier circuit shown in the figure, the sinusoidal voltage (V1 or V2 ) at the output of the transformer has
a maximum value of 10 V. The load resistance RL is k . If I ave is the average current through the resistor RL
the circuit corresponds
V1
RL
Vin Vout
V2
(a) full wave rectifier with I av 20/ mA (b) half wave rectifier with I av 20/ mA
(c) half wave rectifier with I av 10/ mA (d) full wave rectifier with I av 10/ mA
77. The Boolean expression : B ( A B ) A . ( B A) can be realized using minimum number of
(a) 1 AND gate (b) 2 AND gates (c) 1 OR gate (d) 2 OR gates
78. The output V0 of the ideal opamp circuit shown in the figure is
5 k
1 k
+2
Vo
+1
1 k
1 k
(a) –7 V (b) –5 V (c) 5 V (d) 7 V
79. The circuit shown in the figure can be used as a
Vin C Vout
(a) high pass filter or a differentiator (b) high pass filter or an integrator
(c) low pass filter or a differentiator (d) low pass filter or an integrator
80. In the circuit shown in the figure the Thevenin voltage VTh and Thevenin resistance RTh as seen by the load
resistance RL ( 1 k ) are respectively
2k 2k
20 V RL 10 V
(a) 15 V, 1 k (b) 30 V, 4 k (c) 20 V, 2 k (d) 10 V, 5 k
GATE-PH 2005 QUESTION PAPER 51
Linked Answer Questions : Q. 81a to Q. 85b carry two marks each.
Statement for Linked Answer Q.81a and Q.81b :
d2y dy
For the differential equation 2 y 0
dx 2 dx
81a. One of the solutions is
2 2
(a) e x (b) ln x (c) e x (d) e x
81b. The second linearly independent solution is
(a) e x (b) xe x (c) x 2e x (d) x 2e x
Statement for Linked Answer Q.82a and Q.82b :
The Langrangian of two coupled oscillators of mass m each is
1 1
L m x12 x22 m02 x12 x22 m02 x1 x2
2 2
82a. The equation of motion are
x1 02 x1 ,
(a) x2 02 x2 02 x2 x1 02 x2 02 µx1 0,
(b) x2 02 x2 02 x1
x1 02 x1,
(c) x2 02 x2 02 x2 x1 02 x1,
(d) x2 02 x2 02 x1
82b. The normal modes of the system are
(a) 0 2 1, 0 2 1 (b) 0 1 2 , 0 1 2
(c) 0 1, 0 1 (d) 0 1 , 0 1
Statement for Linked Answer Q.83a and Q.83b :
An infinitely long hollow cylinder of radius R carrying a surface density is rotated about its cylindrical axis
with a constant angular speed
83a. The magnitude of the surface current is
(a) R 2 (b) 2 R (c) R (d) 2 R
83b. The magnitude of vector potential inside the cylindrical at a distance are from its axis is
1 1
(a) 20Rr (b) 0Rr (c) 0Rr (d) 0Rr
2 4
Statement for Linked Answer Q.84a and Q.84b :
A particle is scattered by a spherically symmetric potential. In the centre of mass (CM) frame the wavefunction
of the incoming particle is Aeikz , where k is the wavevector and A is a constant.
84a. If f () is an angular function then in the asymtotic region the scattered wavefunction has the form
A f () eikr A f () e ikr A f () eikr A f () e ikr
(a) (b) (c) (d)
r r r2 r2
84b. The differential scattering cross section () in CM frame is
2
2 f ()
(a) () A 2
(b) () A 2 f () 2
r
2
(c) () f () (d) () A f ()
GATE-PH 2005 QUESTION PAPER 52
Statement for Linked Answer Q.85(a) and Q.85(b) :
Lead has atomic weight of 207.2 amu and density of 11.35 gm cm–3
85.a (A) Number of atoms per cm3 for lead is
(a) 1.1 × 1025 (b) 3.3 × 1022 (c) 1.1 × 1022 (d) 3.3 × 1025
85.b (B) If the energy of vacancy formation in lead is 0.55 eV/atom, the number of vacancies/cm3 at 500 K is
(a) 3.2 × 1016 (b) 3.2 × 1019 (c) 9.5 × 1019 (d) 9.5 × 1016
GATE-PH 2006 QUESTION PAPER 53
PHYSICS-PH
Q.1 – Q.20 : Carry ONE mark each.
1. The trace of a 3 × 3 matrix is 2. Two of its eigenvalues are 1 and 2. The third eigenvalue is
(a) –1 (b) 0 (c) 1 (d) 2
2. The value of A dl along a square loop of side L in a uniform field A is
c
(a) 0 (b) 2LA (c) 4LA (d) L2A
3. A particle of charge q, mass m and linear momentum p enters an electromagnetic field of vector potential A
and scalar potential . The Hamiltonian of the particle is
1 q
2
p2 A2
(a) q (b) p A q
2m 2m 2m c
1 q
2
p2
(c) p A p A (d) q p A
2m c 2m
4. A particle is moving in an inverse square force field. If the total energy of the particle is positive, then trajectory
of the particle is
(a) circular (b) elliptical (c) parabolic (d) hyperbolic
5. In an electromagnetic field, which one of the following remains invariant under Lorentz transformation ?
(a) E B (b) E 2 c 2 B 2 (c) B 2 (d) E 2
6. A sphere of radius R has uniform volume charge density. The electric potential at a point r(r < R) is
(a) due to the charge inside a sphere of radius r only
(b) due to the entire charge of the sphere
(c) due to charge in the spherical shell of inner and outer radii r and R, only
(d) independent of r
7. A free particle is moving in +x-direction with a linear momentum p. The wavefunction of the particle normalised
in a length L is
p p
1 p 1 p 1 i x 1 i x
(a) sin x (b) cos x (c) e (d) e
L L L L
8. Which one of the following relations is true for Pauli matrices x , y and z ?
(a) x y y x (b) x y z (c) x y i z (d) x y y x
9. The free energy of a photon gas enclosed in a volume V is given by F 1 aVT 4 , where a is a constant and
3
T is the temperature of the gas. The chemical potential of the photon gas is
4 1 4
(a) 0 (b) aVT 3 (c) aT (d) aVT 4
3 3
10. The wavefunctions of two identical particles in state n and s are given by n (r1 ) and s (r2 ) , respectively. The
particles obey Maxwell-Boltzmann statistics. The state of the combined two particle system is expressed as
1
(a) n (r1 ) s (r2 ) (b) n (r1 ) s (r2 ) n (r2 ) s (r1 )
2
1
(c) n (r1 ) s (r2 ) n (r2 ) s (r1 ) (d) n (r1 ) s (r2 )
2
GATE-PH 2006 QUESTION PAPER 54
11. The target of an X-ray tube is subjected to an excitation voltage V. The wavelength of the emitted X-rays is
proportional to
1 1
(a) (b) V (c) (d) V
V V
12. The principal series of spectral lines of lithium is obtained by transitions between
(a) nS and 2P, n 2 (b) nD and 2P, n 2 (c) nP and 2S, n 2 (d) nF and 3D, n 3
13. Which one of the following is NOT a correct statement about semiconductors ?
(a) The electrons and holes have different mobilities in a semiconductor
(b) In an n-type semiconductor, the Fermi level lies closer to the conduction band edge
(c) Silicon is a direct band gap semiconductor
(d) Silicon has diamond structure
14. Which one of the following axes of rotational symmetry is NOT permissible in single crystals ?
(a) two-fold axis (b) three-fold axis (c) four-fold axis (d) five-fold axis
15. Weak nuclear forces act on
(a) both hadrons and leptons (b) hadrons only
(c) all particles (d) all charged particle
16. Which one of the following disintegration series of the heavy elements will give 209Bi as a stable nucleus ?
(a) Thorium series (b) Neptunium series (c) Uranium series (d) Actinium series
17. The order of magnitude of the binding energy per nucleon in a nucleus is
(a) 10–5 MeV (b) 10–3 MeV (c) 0.1 MeV (d) 10 MeV
18. The interaction potential between two quarks, separated by a distance r inside a nucleon, can be described by
(a, b and are positive constants)
a a a
(a) aer (b) br (c) br (d)
r r r
19. The high input impedance of field effect transistor (FET) amplifier is due to
(a) the pinch-off voltage (b) its very low gate current
(c) the source and drain being far apart (d) the geometry of the FET
20. The circuit shown in the figure function as
+VCC
A B
(a) an OR gate (b) an AND gate (c) a NOR gate (d) a NAND gate
GATE-PH 2006 QUESTION PAPER 55
Q.21 – Q.85 : Carry TWO marks each.
x1
x x
21. A linear transformation T, defined as T x2 1 2 , transform a vector x for a 3-dimensional real
x x2 x3
3
space to a 2-dimensional real space. The transformation matrix T is
1 1 0 1 0 0 1 1 1 1 0 0
(a) (b) (c) (d)
0 1 1 0 1 0 1 1 1 0 0 1
r dS
22. The value of , where r is the position vector and S is a closed surface enclosing the origin, is
S r3
(a) 0 (b) (c) 4 (d) 8
e2 z
23. The value of ( z 1) 4 dz , where C is a circle defined by z 3 , is
C
8i 2 8i 1 8i 8i 2
(a) e (b) e (c) e (d) e
3 3 3 3
24. The kth Fourier component of f ( x ) ( x ) is
(a) 1 (b) 0 (c) (2)1/2 (d) (2)3/2
25.
An atom with net magnetic moment and net angular momentum L L is kept in a uniform magnetic
induction B B0 kˆ . The magnetic moment ( x ) is
d 2 x d 2 x dx
(a) B0 x 0 (b) B0 x 0
2 2 dt
dt dt
d 2 x d 2 x
(c) 2 B02 x 0 (d) 2 B0 x 0
dt 2 dt 2
26. A particle is moving in a spherically symmetric potential V(r ) r 2 , where is a positive constant. In a
stationary state, the expectation value of the kinetic energy T of the particle is
(a) T V (b) T 2 V (c) T 3 V (d) T 4 V
27. A particle of mass 2 kg is moving such that at time t, its position, in metre, is given by r (t ) 5iˆ 2t 2 ˆj . The
angular momentum of the particle at t 2 s about the origin, in kg m 2 s1 , is
(a) 40 kˆ (b) 80 kˆ (c) 80 kˆ (d) 40 kˆ
28. A system of four particles is in xy-plane. Of these, two particles each of mass m are located at (1, 1) and
(–1, –1). The remaining two particles each of mass 2m are located at (–1, 1) and (1, –1). The xy-
component of the moment of inertia tensor of this system of particles is
(a) 10m (b) –10m (c) 2m (d) –2m
GATE-PH 2006 QUESTION PAPER 56
29. For the given transformations (i) Q p, P q and (ii) Q p, P q , where p and q are canonically conju-
gate variables, which one of the following statement is true ?
(a) Both (i) and (ii) are canonical (b) Only (i) is canonical
(c) Only (ii) is canonical (d) Neither (i) nor (ii) is canonical
2m0
30. The mass m of a moving particle is , where m0 is its rest mass. The linear momentum of the particle is
3
2m0 m0c
(a) 2m0c (b) (c) m0c (d)
3 3
31. Three point charges q, q and 2q are located at (0, a, a), (0, a, a) and (0, 0, a) , respectively. The net
dipole moment of this charge distribution is
(a) 4qa kˆ (b) 2qa kˆ (c) 4qa iˆ (d) 2qa ˆj
32. A long cylindrical kept along z-axis carries a current density ˆj J 0rkˆ, where J 0 is a constant and r is the radial
distance form the axis of the cylinder. The magnetic induction B̂ inside the conductor at a distance d from the
axis of the cylinder is
0 J0 d ˆ 0 J0 d 2 ˆ 0 J0 d 3 ˆ
(a) 0 J 0ˆ (b) (c) (d)
2 2 4
33. The vector potential in a region is given as A ( x, y, z ) y iˆ 2 x ˆj . The associated magnetic induction is B
is
(a) iˆ kˆ (b) 3 kˆ (c) iˆ 2 ˆj (d) iˆ ˆj kˆ
34. At the interface between two linear dielectrics (with dielectric constants 1 and 2 ), the electric field lines
bend, as shown in the figure. Assume that there are no free charges at the interface. The ratio 1 /2 is
1
1
2
2
tan 1 cos 1 sin 1 cot 1
(a) tan (b) cos (c) sin (d) cot
2 2 2 2
35. Which one of the following sets of Maxwell’s equations for time-independent charge density and current
density Ĵ is correct ?
(a) E /0 (b) E /0 (c) E 0 (d) E /0
B 0 B 0 B 0 B 0 Jˆ
B
E E 0 E 0 E 0
t
E E
B 00 B 0Jˆ B 0Jˆ B 00
t t
GATE-PH 2006 QUESTION PAPER 57
36. A classical charged particle moving with frequency in a circular orbit of radius a, centred at the origin in the
xy-plane, electromagnetic radiation. At points (b, 0, 0) and (0, 0, b) , where b a , the electromagnetic waves
are
(a) circularly polarized and elliptically polarized, respectively
(b) plane polarized and elliptically polarized, respectively
(c) plane polarized and circularly polarized, respectively
(d) circularly polarized and plane polarized, respectively
37. A particle of mass m is represented by the wavefunction ( x ) Aeikx , where k is the wavevector and A is a
constant. The magnitude of the probability current density of the particle is
2 k 2 k (k )2 2 2 ( k )
2
(a) A (b) A (c) A (d) A
m 2m m 2m
1
38. A one-dimensional harmonic oscillator is in the state ( x ) 30 ( x) 21( x) 2 ( x) , where 0 ( x) ,
14
1 ( x ) and 2 ( x) are the ground, first excited and second excited states, respectively. The probability of
finding the oscillator in the ground state is
3 9
(a) 0 (b) (c) (d) 1
14 14
39. The wavefunction of a particle in a one-dimensional potential at time t 0 is
1
20 ( x) 1 ( x) ,
( x, t 0)
5
where 0 ( x) and 1 ( x ) are the ground and the first excited states of the particle with corresponding energies
E0 and E1 . The wavefunction of the particle at a time t is
i (E0 E1 ) t i E0 t
1 2 1
(a) e 20 ( x) 1 ( x) (b) e 2 0 ( x) 1 ( x)
5 5
E0 t E1 t
i E1 t 1
(c)
1
e 20 ( x) 1 ( x) (d) 5 2 0 ( x) e 1 ( x) e
5
40. The commutator Lx , y , where Lx is the x-component of the angular momentum operator and y is the y-
component of the position operator, is equal to
(a) 0 (b) i x (c) i y (d) i z
41. In hydro genic states, the probability of finding an electron at r 0 is
(a) zero in state 1s (r ) (b) non-zero in state 1s (r )
(c) zero in state 2 s (r ) (d) zero in state 2 p (r )
42. Each of the two isolated vessels, A and B of fixed volumes, contains N molecules of a perfect monatomic gas
at a pressure P. The temperatures of A and B are T1 and T2 , respectively. The two vessels are brought into
thermal contact. At equilibrium, the change in entropy is
3 T 2 T22 T
(a) N k B ln 1 (b) N k B ln 2
2 4 T1 T2 T1
3 (T1 T2 ) 2
(c) N k B ln (d) 2 N k B
2 4 T1 T2
GATE-PH 2006 QUESTION PAPER 58
3 a
43. The internal energy of n moles of a gas is given E nRT , where V is the volume of the gas at tempera-
2 V
ture T and a is a positive constant. One mole of the gas in state (T1 , V1 ) is allowed to expand adiabatically into
vacuum to a final state (T2 , V2 ) . The temperature T2 is
1 1 2 1 1
(a) T1 Ra (b) T1 Ra
V2 V1 3 V2 V1
2 1 1 1 1 1
(c) T1 Ra (d) T1 Ra
3 V2 V1 3 V2 V1
44. The mean internal of a one-dimensional classical harmonic oscillator in equilibrium with a heat bath of tempera-
ture T is
1 3
(a) k BT (b) k BT (c) k BT (d) 3k BT
2 2
45. A monatomic crystalline solid comprises of N atms, out of which n atoms are in interstitial positions. If the
available interstitial sites are N , then number of possible microstates is
(N n)! N! N!
(a) (b) n !(N n)! n !(N n)!
n !N!
N! N! N!
(c) n !(N n)! (d) n !(N n)! n !(N n)!
46. A system of N localized, non-interacting spin - 12 ions of magnetic moment µ each is kept in an external mag-
netic field H. If the system is in equilibrium at temperature T, then Helmholtz free energy of the system is
H H
(a) N k BT ln cosh (b) N k BT ln 2 cosh
k BT k BT
H H
(c) N k BT ln 2cosh (d) N k BT ln 2sinh
k BT k BT
47. The phase diagram of a free particle of mass m kinetic energy E, moving in one-dimensional box with perfectly
elastic walls at x 0 and x L , is given by
px px
2mE
x x
L L
(a) (b)
2mE
px px
2mE 2mE
x x
L –L L
(c) (d)
2mE
GATE-PH 2006 QUESTION PAPER 59
48. In the microwave spectrum of identical rigid diatomic molecules, the separation between the spectral lines is
recorded to be 0.7143 cm–1. The moment of inertia of the molecule, in kg m2, is
(a) 2.3 1036 (b) 2.3 1040 (c) 7.8 1042 (d) 7.8 1046
49. Which one of the following electronic transitions in Neon is NOT responsible for LASER action in a helium-
neon laser ?
(a) 6s 5 p (b) 5s 4 p (c) 5s 3 p (d) 4s 3 p
50. In the linear Stark effect, the application of an electric field
(a) completely lifts the degeneracy of n 2 level on hydrogen atom and splits n 2 level into four levels
(b) partially lifts the degeneracy of n 2 level on hydrogen atom and splits n 2 level into three levels
(c) partially lifts the degeneracy of n 2 level on hydrogen atom and splits n 2 level into two levels
(d) does not affect the n 2 levels
51. In hyperfine interaction, there is coupling between the electron angular momentum J and nuclear angular mo-
mentum I , forming resultant angular momentum F . The selection rules for the corresponding quantum number
F in hyperfine transitions are
(a) F 2 only (b) F 1 only (c) F 0, 1 (d) F 1, 2
52. A vibrational electronic spectrum of homonuclear binary molecules, involving electronic ground state and
excited , exhibits a continuum at v cm 1 . If the total energy of the dissociated atoms in the excited state
exceeds the total energy of the dissociated atoms in the ground state by Eex cm 1 , then dissociation energy of
the molecule in the ground state is
v Eex v Eex
(a)
2
(b)
2
(c) v Eex (d) v 2 Eex2
53. The NMR spectrum of ethanol (CH3 CH 2 OH) comprises of three bunches of spectral lines. The number of
spectral lines in the bunch corresponding to CH 2 group is
(a) 1 (b) 2 (c) 3 (d) 4
54. The energy E ( k ) of electrons of wavevector k in a solid is given by E (k ) Ak 2 Bk 4 , where A and B are
constants. The effective mass of the electron at k k0 is
2 2 2
(a) Ak02 (b) (c) (d)
2A 2 A 12 Bk02 Bk02
55. Which one of the following statements is NOT correct about the Brillouin zones (BZ) of a square lattices with
constant a ?
(a) The first BZ is a square of side 2 /a in k x k y plane
(b) The areas of the first BZ and third BZ are the same
(c) The k-points are equidistant in k x as well as in k y directions
(d) The area of the second BZ is twice that of the first BZ
GATE-PH 2006 QUESTION PAPER 60
56. In a crystal of N primitive cells, each cell contains two monovalent atoms. The highest occupied energy band of
the crystal is
(a) one-fourth filled (b) one-third filled (c) half filled (d) completely filled
57. If the number density of a free electron gas changes from 1028 to 1026 electrons/m3, the value of plasma
frequency (in Hz) changes from 5.7 × 1015 to
(a) 5.7 1013 (b) 5.7 1014 (c) 5.7 1016 (d) 5.7 1017
58. Which one of the following statements about superconductors is NOT true ?
(a) A type I superconductor is completely diamagnetic
(b) A type II superconductor exhibits Meissner effect upto the second critical magnetic field (H c2 )
(c) A type II superconductor exhibits zero resistance upto the second critical magnetic field
(d) Both type I and type II superconductors exhibits sharp fall in resistance at the superconducting transition
temperature
59. Two dielectric materials A and B exhibit both ionic and orientational polarizabilities. The variation of their
susceptibilities ( r , 1) with temperature T is shown in the figure, where r is the relative dielectric
constant. It can be inferred from the figure that
(r – 1)
1/T
(a) A is more polar and it has a smaller value of ionic polarizability than that of B
(b) A is more polar and it has a higher value of ionic polarizability than that of B
(c) B is more polar and it has a higher value of ionic polarizability than that of A
(d) B is more polar and it has a smaller value of ionic polarizability than that of A
60. The experimentally measured spin g factors of proton and a neutron indicate that
(a) Both proton and neutron are elementary point particles
(b) Both proton and neutron are not elementary point particles
(c) While proton is an elementary point particle, neutron is not
(d) While neutron is an elementary point particle, proton is not
61. By capturing an electron, 54
25 Mn 29 transform into
54
25 Cr30 releasing
(a) a neutrino (b) an antineutrino (c) an - particle (d) a positron
62. Which one of the following nuclear processes is forbidden ?
0
(a) v p n e (b) e ve
(c) p n K K (d) e ve v
63. To explain the observed magnetic moment of deuteron (0.8574 N ), its ground state wavefunction is taken to
be an admixture of S and D states. The expectation values of the z-component of the magnetic moment in pure
S and pure D states are 0.8797 N and 0.3101 N respectively. The contribution of the D state to the mixed
ground state is approximately
(a) 40% (b) 4% (c) 0.4% (d) 0.04%
GATE-PH 2006 QUESTION PAPER 61
64. A sinusoidal input voltage vin of frequency is fed to the circuit shown in the figure, where C1 C 2 . If vm
is the peak value of the input voltage, then output voltage (vout ) is
C1
vin C2 vout
vm
(a) 2vm (b) 2v0 sin t (c) 2vm sin t (d)
2
65. The low-pass active filter shown in the figure has a cut-off frequency of 2 kHz and a pass band gain of 1.5. The
values of the resistors are
15 k
R1
vout
vin R2 0.047µF
(a) R1 10 k ; R 2 1.3 (b) R1 30 k ; R 2 1.3
(c) R1 10 k ; R 2 1.7 (d) R1 30 k ; R 2 1.7
d 2v dv
66. In order to obtain a solution of the differential equation 2
2 v1 0 , involving voltages v(t ) and v1 ,
dt dt
an operational amplifier (Op-Amp) circuit would require at least
(a) two Op-Amp integrators and one Op-Amp adder
(b) two Op-Amp differentiators and one Op-Amp adder
(c) one Op-Amp integrator and one Op-Amp adder
(d) one Op-Amp integrator, one Op-Amp differentiator and one Op-Amp adder
67. In the given digital logic circuit, A and B form the input. The output Y is
A
B
Y
(a) Y A (b) Y AB (c) Y A B (d) Y B
68. The largest analog output voltage from a 6-bit digital to analog converter (DAC) which produces 1.0 V output
for a digital input of 010100, is
(a) 1.6 V (b) 2.9 V (c) 3.15 V (d) 5.0 V
69. A ripple counter designed with JK flip-flops provided with CLEAR (CL) input is shown in the figure. In order
that this circuit functions as a MOD-12 counter, the NAND gate input (X1 and X 2 ) should be
D J C J B J A J
CL K CL K CL K CL K
X1
X2
(a) A and C (b) A and D (c) B and D (d) C and D
GATE-PH 2006 QUESTION PAPER 62
70. The tank circuit of a Hartley oscillator is shown in the figure. If M is the mutual inductance between the
inductors, the oscillation frequency is
1 1 L1 L2
(a) (b)
2 (L1 L 2 2M)C 2 (L1 L 2 2M)C
1 1
(c) (d)
2 (L1 L 2 M)C 2 (L1 L 2 M)C C
Common Data for Q.71, Q.72 and Q.73 :
1 0
An unperturbed two-level system has energy eigenvalues E1 and E 2 , and eigen functions and . When
0 1
E1 A
perturbed, its Hamiltonian is represented by
A E 2
71. The first-order correct to E1 is
(a) 4A (b) 2A (c) A (d) 0
72. The second-order correction to E1 is
A2 A2
(a) 0 (b) A (c) (d)
E 2 E1 E1 E 2
1
73. The first-order correction to the eigenfunction is
0
0 0 A /(E1 E 2 ) 1
(a) (b) (c) (d)
A /(E1 E 2 ) 1 0 1
Common Data for Q.74 and Q.75 :
2 3 0
One of the eigenvalues of the matrix 3 2 0 is 5.
0 0 1
74. The other two eigenvalues are
(a) 0 and 0 (b) 1 and 1 (c) 1 and –1 (d) –1 and –1
75. The normalized eigenvector corresponding to the eigenvalue 5 is
0 1 1 1
1 1 1 1
(a) 1 (b) 1 (c) 0 (d) 1
2 2 2 2
1 0 1 0
Linked Answer Questions : Q. 76 to Q. 85 carry two marks each.
Statement for Linked Answer Q.76 and Q.77 :
The powder diffraction pattern of a body centred cubic crystal is recorded by using Cu K X-rays of wave-
length 1.54 Å.
76. If the (002) planes diffract at 60º, then lattice parameter is
(a) 2.67 Å (b) 3.08 Å (c) 3.56 Å (d) 5.34 Å
GATE-PH 2006 QUESTION PAPER 63
77. Assuming the atomic mass of the constituent atoms to be 50.94 amu, then density of the crystal in units of kg
m 3 is
(a) 3.75 × 103 (b) 4.45 × 103 (c) 5.79 × 103 (d) 8.89 × 103
Statement for Linked Answer Q.78 and Q.79 :
A particle of mass m is constrained to move in a vertical plane along a trajectory given by x A cos ,
y A sin , where A is a constant.
78. The Lagrangian of the particle is
1 1
(a) mA 2 2 mg A cos (b) mA 2 2 mg A sin
2 2
1 1
(c) mA 2 2 (d) mA 2 2 mg A cos
2 2
79. The equation of motion of the particle is
g g g
(a)
cos 0 (b)
sin 0 (c) 0 (d)
sin 0
A A A
Statement for Linked Answer Q.80 and Q. 81 :
A dielectric sphere of radius R carries polarization P kr 2 rˆ , where r is the distance from the centre and k is
a constant. In the spherical polar coordinate system, rˆ, ˆ and ˆ are the unit vectors.
80. The bound volume charge density inside the sphere at a distance r from the centre is
(a) 4kR (b) 4kr (c) 4kr 2 (d) 4kr 3
81. The electric field inside the sphere at a distance d from the centre is
kd 2 kR 2 kd 2 ˆ kR 2 ˆ
(a) rˆ (b) rˆ (c) (d)
0 0 0 0
Statement for Linked Answer Q. 82 and Q. 83 :
Consider Fermi theory of -decay..
82. The number of final states states of electrons corresponding to momenta between p and p dp is
(a) independent of p (b) proportional to pdp
2 3
(c) proportional to p dp (d) proportional to p dp
83. The number of emitted electrons with momentum p and energy E, in the allowed approximation, is proportional
to ( E0 is the total energy given up by the nucleus).
(a) (E0 E) (b) p(E 0 E) (c) p 2 (E 0 E) 2 (d) p (E 0 E) 2
Statement for Linked Answer Question 84 and 85 :
Consider a radiation cavity of volume V at temperature T.
84. The density of states at energy E of the quantized radiation (photons) is
8V 8V 8V 8V
(a) 3 3
E2 (b) 3 3
E3/2 (c) 3 3
E (d) 3 3
E1/2
hc hc hc hc
85. The average number of photons in equilibrium inside the cavity is proportional to
(a) T (b) T2 (c) T3 (d) T4
GATE-PH 2007 QUESTION PAPER 64
PHYSICS-PH
Q.1 – Q.20 : Carry ONE mark each.
1. The eigenvalues of a matrix are i, –2i and 3i. The matrix is
(a) unitary (b) anti-unitary
(c) Hermitian (d) anti-Hermitian
2. A space station moving in a circular orbit around the Earth goes into a new bound orbit by firing its engine
radially outwards. This orbit is
(a) a larger circle (b) a smaller circle
(c) an ellipse (d) a parabola
3. A power amplifier gives 150W output for an input of 1.5W. The gain, in dB, is
(a) 10 (b) 20 (c) 54 (d) 100
4. Four point charges are placed in a plane at the following positions: +Q at (1, 0), –Q at (–1, 0), +Q at
(0, 1) and –Q at (0, –1). At large distances the electrostatic potential due to this charge distribution will
be dominated by the
(a) monopole moment (b) dipole moment
(c) quadrupole moment (d) octopole moment
5. A charged capacitor (C) is connected in series with an inductor (L). When the displacement current reduces
to zero, the energy of the LC circuit is
(a) stored entirely in its magnetic field
(b) stored entirely in its electric field
(c) distributed equally among its electric and magnetic fields
(d) radiated out of the circuit
6. Match the following
P. Frank-Hertz experiment 1. electronic excitation of molecules
Q. Hartree-Fock method 2. wave function of atoms
R. Stern-Gerlach experiment 3. spin angular momentum of atoms
S. Franck-Condon principle 4. energy levels in atoms
(a) (b) (c) (d)
P-4 P-1 P-3 P-4
Q-2 Q-4 Q-2 Q-1
R-3 R-3 R-4 R-3
S-1 S-2 S-1 S-2
7. The wavefunction of a particle, moving in a one-dimensional time-independent potential V ( x ), is given by
( x) e iax b , where a and b are constants. This means that the potential V ( x) is of the form.
(a) V ( x ) x (b) V ( x) x 2 (c) V ( x) 0 (d) V ( x) e ax
8. The D1 and D2 lines of Na (32P1/2 32S1/2, 32P3/2 32S1/2) will split on the application of a weak
magnetic field into
(a) 4 and 6 lines respectively (b) 3 lines each
(c) 6 and 4 lines respectively (d) 6 lines each
GATE-PH 2007 QUESTION PAPER 65
9. In a He-Ne laser, the laser transition takes place in
(a) He only (b) Ne only
(c) Ne first, then in He (d) He first, then in Ne
10. The partition function of a single gas molecule is Z . The partition function of N such non-interacting gas
molecules is then given by
(Z ) N (Z ) N
(a) (b) ( Z ) N (c) N ( Z ) (d)
N! N
11. A solid superconductor is placed in an external magnetic field and then cooled below its critical temperature.
The superconductor
(a) retains its magnetic flux because the surface current supports it
(b) expels out its magnetic flux because it behaves like a paramagnetic materials
(c) expels out its magnetic flux because it behaves like an anti-ferromagnetic material
(d) expels out its magnetic flux because the surface current induces a field in the direction opposite to the
applied magnetic field
12. A particle with energy E is in a time-independent double well potential as shown in the figure.
V
Which of the following statements about the particle is NOT correct?
(a) The particle will always be in a bound state
(b) The probability of finding the particle in one well will be time-dependent
(c) The particle will be confined to any one of the wells
(d) The particle can tunnel from one well to the other, and back
13. It is necessary to apply quantum statistics to a system of particles if
(a) there is substantial overlap between the wavefunctions of the particles
(b) the mean free path of the particles is comparable to the inter-particle separation
(c) the particles have identical mass and charge
(d) the particles are interacting
14. When liquid oxygen is poured down close to a strong bar magnet, the oxygen stream is
(a) repelled towards the lower field because it is diamagnetic
(b) attracted towards the higher field because it is diamagnetic
(c) repelled towards the lower field because it is paramagnetic
(d) attracted towards the higher field because it is paramagnetic
15. Fission fragments are generally radioactive as
(a) they have excess of neutrons
(b) they have excess of protons
(c) they are products of radioactive nuclides
(d) their total kinetic energy is of the order of 200 MeV
GATE-PH 2007 QUESTION PAPER 66
16. In a typical npn transistor the doping concentrations in emitter, base and collector regions are CE, CB and
CC respectively. These satisfy the relation
(a) CE > CC > CB (b) CE > CB > CC
(c) CC > CB > CE (d) CE = CC > CB
2
17. The allowed states for He (2p ) configuration are
(a) 1S0, 3S1, 1P1, 3P0,1,2, 1D2 and 3D1,2,3 (b) 1S0, 3P0,1,2 and 1D2
(c) 1P1 and 3P0,1,2 (d) 1S0 and 1P1
18. The energy levels of a particle of mass m in a potential of the form
V ( x ) , x 0
1
m2 x 2 , x 0
2
are given, in terms of quantum number n 0,1, 2,3....., by
1 1
(a) n (b) 2n
2 2
3 3
(c) 2n (d) n
2 2
19. The electromagnetic field due to a point charge must be described by Lienard Weichert potentials when
(a) the point charge is highly accelerated
(b) the electric and magnetic fields are not perpendicular
(c) the point charge is moving with velocity close to that of light
(d) the calculation is done for the radiation zone, i.e. far away from the charge
20. The strangeness quantum number is conserved in
(a) strong, weak and electromagnetic interactions
(b) weak and electromagnetic interactions only
(c) strong and weak interactions only
(d) strong and electromagnetic interactions only
Q.21 – Q.75 : Carry TWO marks each.
5 4
21. The eigenvalues and eigenvectors of the matrix are
1 2
4 1 4 1
(a) 6, 1 and , 1 (b) 2, 5 and , 1
1 1
1 1 1 1
(c) 6, 1 and , 1 (d) 2, 5 and , 1
4 4
y2
22. A vector field is defined everywhere as F iˆ zkˆ . The net flux of F associated with a cube of side
L
L, with one vertex at the origin and sides along the positive X, Y, and Z axes, is
(a) 2L3 (b) 4L3 (c) 8L3 (d) 10L3
GATE-PH 2007 QUESTION PAPER 67
23. If r xiˆ yjˆ , then
(a) r 0 and | r | r (b) r 2 and | r | rˆ
rˆ rˆ
(c) r 2 and | r | (d) r 3 and | r |
r r
24. Consider a vector p 2iˆ 3 ˆj 2kˆ in the coordinate system (iˆ, ˆj, kˆ) . The axes are rotated anti-clockwise
about the Y axis by an angle of 60º. The vector p in the rotated coordinate system (iˆ, ˆj , kˆ) is
(a) (1 3)iˆ 3 ˆj (1 3) kˆ (b) (1 3)iˆ 3 ˆj (1 3) kˆ
(c) (1 3)iˆ (3 3) ˆj 2kˆ (d) (1 3)iˆ (3 3) ˆj 2kˆ
dz
25. The contour integral z
a4
is to be evaluated on a circle of radius 2a centered at the origin. It will
4
have contributions only from the points
1 i 1 i
(a) a and a (b) ia and ia
2 2
1 i 1 i 1 i
(c) ia, ia,
1 i
a and a (d) a, a, 1 i a and 1 i a
2 2 2 2 2 2
s 1
26. Inverse Laplace transform of is
s2 4
1 1
(a) cos 2 x sin 2 x (b) cos x sin x
2 2
1 1
(c) cosh x sinh x (d) cosh 2 x sinh 2 x
2 2
27. The points, where the series solution of the Legendre differential equation
2d2y dy 3 3
(1 x ) 2 2 x 1 y 0 will diverge, are located at
dx dx 2 2
3 5
(a) 0 and 1 (b) 0 and –1 (c) –1 and 1 (d) and
2 2
dy
28. Solution of the differential equation x y x 4 , with the boundary condition that y 1, at x 1, is
dx
x4 4x
(a) y 5 x 4 4 (b) y
5 5
4 x4 1 x4 4
(c) y (d) y
5 5x 5 5x
GATE-PH 2007 QUESTION PAPER 68
29. Match the following
P. rest mass 1. time like vector
Q. charge 2. Lorentz invariant
R. four-momentum 3. tensor of rank 2
S. electromagnetic field 4. conserved and Lorentz invariant
(a) (b) (c) (d)
P-2 P-4 P-2 P-4
Q-4 Q-2 Q-4 Q-2
R-3 R-1 R-1 R-3
S-1 S-3 S-3 S-1
30. The moment of inertia of a uniform sphere of radius r about an axis passing through its centre is given by
2 4 5 . A rigid sphere of uniform mass density and radius R has two smaller spheres of radius
r
5 3
R / 2 hollowed out of it, as shown in the figure. The moment of inertia of the resulting body about the Y
axis is Y
R5 5R5
(a) (b)
4 12
7R 5 3R5 X
(c) (d)
12 4
31. The Lagrangian of a particle of mass m is
2 2 2
m dx dy dz V 2
L ( x y 2 ) W sin t , where V, W and are constants.
2 dt dt dt 2
The conserved quantities are
(a) energy and z-component of linear momentum only
(b) energy and z-component of angular momentum only
(c) z-components of both linear and angular momenta only
(d) energy and z-components of both linear and angular momenta
32. Three particles of mass m each situated at x1 (t ), x2 (t ) and x3 (t ) respectively are connected by two springs
of spring constant k and un-stretched length l. The system is free to oscillate only in one dimension along
the straight line joining all the three particles. The Lagrangian of the system is
dx 2 dx
2
dx
2
(a) L m 1 2 3 k ( x1 x2 l ) 2 k ( x3 x2 l )2
2 dt dt dt 2 2
2 2 2
m dx1 dx2 dx3 k 2 k 2
(b) L ( x1 x3 l ) ( x3 x2 l )
2 dt dt dt 2 2
2 2 2
m dx1 dx2 dx3 k 2 k 2
(c) L ( x1 x2 l ) ( x3 x2 l )
2 dt dt dt 2 2
m dx1 2 dx2 2 dx3 2 k 2 k 2
(d) L ( x1 x2 l ) ( x3 x2 l )
2 dt dt dt 2 2
GATE-PH 2007 QUESTION PAPER 69
p2
33. The Hamiltonian of a particle is H pq, where q is the generalized coordinate and p is the
2m
corresponding canonical momentum. The Lagrangian is
2 2
m dq m dq
(a) q (b) q
2 dt 2 dt
2 2
m dq dq 2
m dq dq 2
(c) 2 dt q dt q (d) 2 dt q dt q
34. A toroidal coil has N closely-wound turns. Assume the current through the coil to be I and the toroid is
filled with a magnetic material of relative permittivity r . The magnitude of magnetic induction B inside the
toroid, at a radial distance r from the axis, is given by
r 0 NI
(a) r 0 NIr (b)
r
r 0 NI
(c) (d) 2 r 0 NIr
2r
35. An electromagnetic wave with E ( z, t ) E0 cos(t kz )iˆ is travelling in free space and crosses a disc of
radius 2m placed perpendicular to the z-axis. If E0 60Vm 1 , the average power, in Watt, crossing the
disc along the z-direction is
(a) 30 (b) 60 (c) 120 (d) 270
36. Can the following scalar and vector potentials describe an electromagnetic field?
( x , t ) 3xyz 4t
A( x , t ) (2 x t )iˆ ( y 2 z ) ˆj ( z 2 xeiax ) kˆ
where is a constant
(a) Yes, in the Coulomb gauge (b) Yes, in the Lorentz gauge
(c) Yes, provided 0 (d) No
1
37. For a particle of mass m in a one-dimensional harmonic oscillator potential of the form V ( x) m2 x 2 ,
2
2
the first excited energy eigenstate is ( x ) xe ax . The value of a is
(a) m / 4 (b) m / 3
(c) m / 2 (d) 2m / 3
If [ x, p ] i, the value of [ x , p] is
3
38.
(a) 2ix 2 (b) 2ix 2
(c) 3ix 2 (d) 3ix 2
GATE-PH 2007 QUESTION PAPER 70
39. There are only three bound states for a particle of mass m in a one-dimensional potential well of the form
shown in the figure. The depth V0 of the potential satisfies V
–a/2 +a/2
22 2 9 2 2 2 2 22 2 x
(a) V0 (b) V0
ma 2 2ma 2 ma 2 ma 2 –V0
22 2 82 2 22 2 502 2
(c) V0 (d) V0
ma 2 ma 2 ma 2 ma 2
40. An atomic state of hydrogen is represented by the following wavefunction:
3/2
1 1 r r /2 a0
(r , , ) 1 e cos
2 a0 2 a0
where a0 is a constant. The quantum numbers of the state are
(a) l 0, m 0, n 1 (b) l 1, m 1, n 2
(c) l 1, m 0, n 2 (d) l 2, m 0, n 3
41. Three operators X, Y and Z satisfy the commutation relations
[ X , Y ] iZ , [Y , Z ] iX and [ Z , X ] iY .
The set of all possible eigenvalues of the operator Z, in units of , is
1 3 5
(a) {0, 1, 2, 3,...} (b) ,1, , 2, ,...
2 2 2
1 3 5 1 1
(c) 0, , 1, , 2, ,... (d) ,
2 2 2 2 2
42. A heat pump working on the Carnot cycle maintains the inside temperature of a house at 22ºC by supplying
450 kJ s–1. If the outside temperature is 0ºC, the heat taken, in kJ s–1, from the outside air is approximately
(a) 487 (b) 470 (c) 467 (d) 417
43. The vapour pressure p (in mm of Hg) of a solid, at temperature T, is expressed by ln p 23 3863 / T
and that of its liquid phase by ln p 19 3063 / T . The triple point (in Kelvin) of the material is
(a) 185 (b) 190 (c) 195 (d) 200
a
44. The free energy for a photon gas is given by F VT 4 , where a is a constant. The entropy S and
3
the pressure P of the photon gas are
4 a 1 4 4a 3
(a) S aVT 3 , P T 4 (b) S aVT , P T
3 3 3 3
4 a 1 3 4a 4
(c) S aVT 4 , P T 3 (d) S aVT , P T
3 3 3 3
GATE-PH 2007 QUESTION PAPER 71
45. A system has energy levels E0 , 2 E0 , 3E0 ,..., where the excited states are triply degenerate. Four non-
interacting bosons are placed in this system. If the total energy of these bosons is 5E0 , the number of
microstates is
(a) 2 (b) 3 (c) 4 (d) 5
3
46. In accordance with the selection rules for electric dipole transitions, the 4 P1 state of helium can decay by
photon emission to the states
(a) 21S0, 21P1 and 31D2 (b) 31P1, 31D2 and 31S0
(c) 33P2, 33D3 and 33P0 (d) 23S1, 33D2 and 33D1
47. If an atom is in the 3D3 state, the angle between its orbital and spin angular momentum vectors ( L and
S ) is
1 2 1 1 1 3
(a) cos 1 (b) cos 1 (c) cos (d) cos
3 3 2 2
48. The hyperfine structure of Na(32P3/2) with nuclear spin I = 3/2 has
(a) 1 state (b) 2 states (c) 3 states (d) 4 states
49. The allowed rotational energy levels of a rigid hetero-nuclear diatomic molecule are expressed as
j BJ ( J 1), where B is the rotational constant and J is a rotational quantum number..
In a system of such diatomic molecules of reduced mass ; some of the atoms of one element are replaced
by a heavier isotope, such that the reduced mass is changed to 1.05 . In the rotational spectrum of the
system, the shift in the spectral line, corresponding to a transition J 4 J 5, is
(a) 0.475 B (b) 0.50 B (c) 0.95 B (d) 1.0 B
50. The number of fundamental vibrational modes of CO2 molecule is
(a) four : 2 are Raman active and 2 are infrared active
(b) four : 1 is Raman active and 3 are infrared active
(c) three : 1 is Raman active and 2 are infrared active
(d) three : 2 are Raman active and 1 is infrared active
51. A piece of paraffin is placed in a uniform magnetic field H0. The sample contains hydrogen nuclei of mass
mp, which interact only with external magnetic field. An additional oscillating magnetic field is applied to
observe resonance absorption. If gi is the g-factor of the hydrogen nucleus, the frequency, at which
resonance absorption takes place, is given by
3g i eH 0 3g i eH 0 gi eH 0 gi eH 0
(a) (b) (c) (d)
2m p 4m p 2m p 4m p
P Q
52. The solid phase of an element follows van der Waals bonding with inter-atomic potential V (r ) ,
r 6 r12
where P and Q are constants. The bond length can be expressed as
6 6 6 6
2Q Q P P
(a) (b) (c) (d)
P P 2Q Q
GATE-PH 2007 QUESTION PAPER 72
53. Consider the atomic packing factor (APF) of the following crystal structures:
P. Simple Cubic
Q. Body Centred Cubic
R. Face Centred Cubic
S. Diamond
T. Hexagonal Close Packed
Which two of the above structures have equal APF?
(a) P and Q (b) S and T (c) R and S (d) R and T
54. In a powder diffraction pattern recorded from a face-centred cubic sample using x-rays, the first peak
appears at 30º. The second peak will appear at
(a) 32.8º (b) 33.7º (c) 34.8º (d) 35.3º
55. Variation of electrical resistivity with temperature T of
three solids is sketched (on different scales) in the figure, as
curves P, Q and R.
Q
Resistivity
Which one of the following statements describes the P
variations most appropriately?
(a) P is for a superconductor, and R for a semiconductor
R
(b) Q is for a superconductor, and P for a conductor 0
Temperature
(c) Q is for a superconductor, and R for a conductor
(d) R is for a superconductor, and P for a conductor
56. An extrinsic semiconductor sample of cross-section A and length L is doped in such a way that the doping
x
concentration varies as N D ( x ) N 0 exp , where N0 is a constant. Assume that the mobility of the
L
majority carriers remains constant. The resistance R of the sample is given by
L L
(a) R [exp(1.0) 1] (b) R [exp(1.0) 1]
AeN 0 eN 0
L L
(c) R [exp(1.0) 1] (d) R
AeN 0 AeN 0
57. A ferromagnetic mixture of iron and copper having 75% atoms of Fe exhibits a saturation magnetization of
1.3 × 106 Am–1. Assume that the total number of atoms per unit volume is 8 × 1028 m–3. The magnetic
moment of an iron atom, in terms of the Bohr Magneton, is
(a) 1.7 (b) 2.3 (c) 2.9 (d) 3.8
8 3
58. Half life of a radio-isotope is 4 × 10 years. If there are 10 radioactive nuclei in a sample today, the number
of such nuclei in the sample 4 × 109 years ago were
(a) 128 × 103 (b) 256 × 103 (c) 512 × 103 (d) 1024 × 103
59. In the deuterium + tritium (d + t) fusion more energy is released as compared to deuterium + deuterium
(d + d) fusion because
(a) tritium is radioactive
(b) more nucleons participate in fusion
(c) the Coulomb barrier is lower for the d + t system than d + d system
(d) the reaction product 4He is more tightly bound
GATE-PH 2007 QUESTION PAPER 73
60. According to the shell model the ground state spin of the 17O nucleus is
3 5 3– 5–
(a) (b) (c) (d)
2 2 2 2
61. A relativistic particle travels a length of 3 × 10–3 m in air before decaying. The decay process of the particle
is dominated by
(a) strong interactions (b) electromagnetic interactions
(c) weak interactions (d) gravitational interactions
62. The strange baryon has the quark structure
(a) uds (b) uud (c) uus (d) us
63. A neutron scatters elastically from a heavy nucleus. The initial and final states of the neutron have the
(a) same energy (b) same energy and linear momentum
(c) same energy and angular momentum (d) same linear and angular momenta
64. The circuit shown is based on ideal operational amplifiers. It acts as a
R
R
R
(a) subtractor (b) buffer amplifier V1 – R
–
(c) adder (d) divider +
V2 +
65. Identify the function F generated by the logic network shown
Z
(a) F ( X Y )Z (b) F Z Y YX Y
(c) F ZY (Y X ) (d) F XYZ F
66. In the circuit shown, the ports Q1 and Q2 are in the state Q1 = 1, Q2 = 0. The circuit is now subjected
to two complete clock pulses. The state of these ports now becomes
Q1 Q2
(a) Q2 = 1, Q1 = 0 (b) Q2 = 0, Q1 = 1 1 J Q 1 J Q
(c) Q2 = 1, Q1 = 1 (d) Q2 = 0, Q1 = 0 CLK CLK
1 K Q 1 K Q
67. The registers QD, QC, QB and QA shown in the figure are initially in the state 1010 respectively. An input
sequence SI = 0101 is applied. After two clock pulses, the state of the shift registers (in the same sequence
QD QC QB QA) is
(a) 1001 (b) 0100 CLK SI 0101
QD QC QB QA
(c) 0110 (d) 1010
68. For the circuit shown, the potential difference (in Volts) across RL is
5 2
(a) 48 (b) 52
38V RL = 4 76V
(c) 56 (d) 65
GATE-PH 2007 QUESTION PAPER 74
69. In the circuit shown, the voltage at test point P is 12V and the voltage between gate and source is –2V.
The value of R (in k ) is VDD = 16V P
4k
(a) 42 (b) 48 2k
(c) 56 (d) 70
R 42k
70. When an input voltage Vi, of the form shown, is applied to the circuit given below, the output voltage V0
is of the form
+12V
R
V1 V0
3V
–12V
Si diode
12V
12V
(a) (b) 3V
0V 0V
12V
0V
(c) 2.3V (d)
0V –12V
Common Data Questions
Common Data for Questions 71, 72, 73:
A particle of mass m is confined in the ground state of a one-dimensional box, extending from x 2L
x
to x 2L . The wavefunction of the particle in this state is ( x) 0 cos , where 0 is a constant
4L
71. The normalization factor 0 of this wavefunction is
2 1 1 1
(a) (b) (c) (d)
L 4L 2L L
72. The energy eigenvalue corresponding to this state is
22 22 2 2 2 2
(a) (b) (c) (d)
2mL2 4mL2 16mL2 32mL2
73. The expectation value of p2 (p is the momentum operator) in this state is
2 2 2 2 2 2
(a) 0 (b) (c) (d)
32L2 16L2 8L2
Common Data for Questions 74, 75:
The Fresnel relations between the amplitudes of incident and reflected electromagnetic waves at an interface
between air and a dielectric of refractive index , are
cos r cos i incident cos r cos i incident
Ereflected E reflected
and E E
cos r cos i cos r cos i
The subscripts and refer to polarization, parallel and normal to the plane of incidence respectively..
Here, i and r are the angles of incidence and refraction respectively
GATE-PH 2007 QUESTION PAPER 75
74. The coordination for the reflected ray to be completely polarized is
(a) cos i cos r (b) cos i cos r
(c) cos i cos r (d) cos i cos r
75. For normal incidence at an air-glass interface with 1.5 the fraction of energy reflected is given by
(a) 0.40 (b) 0.20 (c) 0.16 (d) 0.04
Linked Answer Questions : Q. 76 to Q.85 carry two marks each.
Statement for Linked Answer Questions 76 & 77:
In the laboratory frame, a particle P of rest mass m0 is moving in the positive x direction with a speed of
5c 2c
. It approaches an identical particle Q, moving in the negative x direction with a speed of .
19 5
76. The speed of the particle P in the rest frame of the particle Q is
7c 13c 3c 63c
(a) (b) (c) (d)
95 85 5 95
77. The energy of the particle P in the rest frame of the particle Q is
1 5 19 11
(a) m0 c 2 (b) m0 c 2 (c) m0 c 2 (d) m0 c 2
2 4 13 9
Statement for Linked Answer Questions 78 & 79
The atomic density of a solid is 5.85 × 1028 m–3. Its electrical resistivity is 1.6 × 10–8 m. Assume that
electrical conduction is described by the Drude model (classical theory), and that each atom contributes one
conduction electron.
78. The drift mobility (in m2 V–1 s–1) of the conduction electrons is
(a) 6.67 × 10–3 (b) 6.67 × 10–6 (c) 7.63 × 10–3 (d) 7.63 × 10–6
79. The relaxation time (mean free time), in seconds, of the conduction electrons is
(a) 3.98 × 10–15 (b) 3.79 × 10–14 (c) 2.84 × 10–12 (d) 2.64 × 10–11
Statement for Linked Answer Questions 80 & 81:
A sphere of radius R carries a polarization P kr , where k is a constant and r is measured from the
centre of the sphere.
80. The bound surface and volume charge densities are given, respectively, by
(a) k | r | and 3k (b) k | r | and 3k
(c) k | r | and 4kR (d) k | r | and 4kR
81. The electric field E at a point r outside the sphere is given by
kR( R 2 r 2 )
(a) E 0 (b) E rˆ
0r 3
kR( R 2 r 2 ) 3k ( r R )
(c) E rˆ (d) E rˆ
0r 5 40 r 4
GATE-PH 2007 QUESTION PAPER 76
Statement for Linked Answer Questions 82 & 83:
An ensemble of quantum harmonic oscillators is kept at a finite temperature T 1/ k
1
82. The partition function of a single oscillator with energy levels ( n ) is given by
2
e /2 e/ 2
(a) Z (b) Z
1 e 1 e
1 1
(c) Z (d) Z
1 e 1 e
83. The average number of energy quanta of the oscillations is given by
1 e
(a) n (b) n
e 1 e 1
1 e
(c) n (d) n
e 1 e 1
Statement for Linked Answer Questions 84 & 85:
A 16 A beam of alpha particles, having cross-sectional area 10–4 m2, is incident on a rhodium target of
thickness 1 m. This produces neutrons through the reaction
100 Rh 101 Pd 3n
84. The number of alpha particles hitting the target per second is
(a) 0.5 × 1014 (b) 1.0 × 1014
(c) 2.0 × 1020 (d) 4.0 × 1020
85. The neutrons are observed at the rate of 1.806 × 108 s–1. If the density of rhodium is approximated as
104 kg m–3 the cross-section for the reaction (in barns) is
(a) 0.1 (b) 0.2 (c) 0.4 (d) 0.8
GATE-PH 2008 QUESTION PAPER 77
PHYSICS-PH
Q.1 – Q.20 : Carry ONE mark each.
1. For arbitrary matrices E , F , G and H, if EF FE 0 then Trace( EFGH ) is equal to
(a) Trace( HGFE ) (b) Trace( E ).Trace( F ).Trace(G ).Trace( H )
(c) Trace(GFEH ) (d) Trace( EGHF )
aei b
2. An unitary matrix i is given, where a, b, c, d , and are real. The inverse of the matrix is
ce d
aei cei aei cei ae i b ae i ce i
(a) (b) (c) i (d)
b d b d ce d b d
3. The curl of a vector field F is 2xˆ . Identify the appropriate vector field F from the choices given below
(a) F 2 zxˆ 3zyˆ 5 yzˆ (b) F 3 zyˆ 5 yzˆ
(c) F 3 xyˆ 5 yzˆ (d) F 2 xˆ 5 yzˆ
4. A rigid body is rotating about its centre of mass, fixed at the origin, with an angular velocity and angular
acceleration . If the torque acting on it is and its angular momentum is L , the rate of change of its
kinetic energy is
1 1 1 1
(a) . (b) L. (c) ( . L. ) (d) L.
2 2 2 2
5. A cylinder of mass M and radius R is rolling down without slipping on an inclined plane of angle of inclination
. The number of generalized coordinates required to describe the motion of this system is
(a) 1 (b) 2 (c) 4 (d) 6
6. A parallel plate capacitor is being discharged. What is the direction of the energy flow in terms of the
Poynting vector in the space between the plates?
z
+
– y
x r
(a) Along the wire in the positive z-axis (b) Radially inward ( r̂ )
(c) Radially outward ( r̂ ) (d) Circumferential ( )
7. Unpolarized light falls from air to a planar air-glass interface (refractive index of glass is 1.5) and the
reflected light is observed to be plane polarized. The polarization vector and the angle of incidence i are
(a) perpendicular to the plane of incidence and i 42º
(b) parallel to the plane of incidence and i 56º
(c) perpendicular to the plane of incidence and i 56º
(d) parallel to the plane of incidence and i 42º
GATE-PH 2008 QUESTION PAPER 78
8. A finite wave train, of an unspecified nature, propagates along the positive x-axis with a constant speed v
and without any change of shape. The differential equation among the four listed below.
2 1 2 2 1 2
(a) x 2 v 2 t 2 ( x, t ) 0
(b) v 2 t 2 (r , t ) 0
2 2 2 2
(c) 2m x 2 i ( x, t ) 0 (d) a (r , t ) 0
t t
9. Let 0 denote the ground state of the hydrogen atom. Choose the correct statement from those given
below:
(a) [ Lx , Ly ] 0 0 (b) J 2 0 0
(c) L.S 0 0 (d) [S x , S y ] 0 0
10. Thermodynamic variables of a system can be volume V, pressure P, temperature T, number of particles N,
internal energy E and chemical potential , etc. For a system to be specified by Microcanonical (MC),
Canonical (CE) and Grand Canonical (GC) ensembles, the parameters required for the respective ensembles
are:
(a) MC : ( N , V , T ); CE : ( E , V , N ); GC : (V , T , ) (b) MC : ( E , V , N ); CE : ( N ,V , T ); GC : (V , T , )
(c) MC : (V , T , ); CE : ( N , V , T ); GC : ( E, V , N ) (d) MC : ( E, V , N ); CE : (V , T , ); GC : ( N, V , T )
11. The pressure versus temperature diagram of a given system at certain low temperature range is found to
be parallel to the temperature axis in the liquid-to-solid transition region. The change in the specific volume
remains constant in the region. The conclusion one can get from the above is
(a) the entropy of solid is zero in this temperature region
(b) the entropy increases when the system goes from liquid to solid phase in this temperature region
(c) the entropy decreases when the system transforms from liquid to solid phase in this region of temperature
(d) the change in entropy is zero in the liquid-to-solid transition region
12. The radial wave function of the electrons in the state of n 1 and l 0 in a hydrogen atom is
2 r
R10 3/ 2
exp , a0 is the Bohr radius. The most probable value of r for an electron is
a0 a0
(a) a0 (b) 2a0 (c) 4a0 (d) 8a0
13. The last two terms of the electronic configuration of manganese (Mn) atom is 3d 5 4s 2 . The term factor of
Mn4+ ion is
(a) 4D1/2 (b) 4F3/2 (c) 3F9/2 (d) 3D7/2
14. The coherence length of laser light is
(a) directly proportional to the length of the active lasing medium
(b) directly proportional to the width of the spectral line
(c) inversely proportional to the width of the spectral line
(d) inversely proportional to the length of the active lasing medium
GATE-PH 2008 QUESTION PAPER 79
15. Metallic monovalent sodium crystallizes in body centered cubic structure. If the length of the unit cell is
4×10–8 cm, the concentration of conduction electrons in metallic sodium is
(a) 6.022 × 1023 cm–3 (b) 3.125 × 1022 cm–3
(c) 2.562 × 1021 cm–3 (d) 1.250 × 1020 cm–3
16. The plot of inverse magnetic susceptibility 1 / versus temperature T of an antiferromagnetic sample
corresponds to
1/
1/
(a) (b)
0 TC T TC 0 T
1/ 1/
(c) (d)
0 TC T 0 TC T
17. According to the quark model, the K+ meson is composed of the following quarks:
(a) u u d (b) u c (c) u s (d) s u
18. An O16 nucleus is spherical and has a charge radius R and a volume V 4 R 3 . According to the empirical
3
128
observations of the charge radii, the volume of the 54Xe nucleus, assumed to be spherical, is
(a) 8V (b) 2V (c) 6.75V (d) 1.89V
19. A common emitter transistor amplifier circuit is operated under a fixed bias. In this circuit, the operating
point.
(a) remains fixed with an increase in temperature
(b) moves towards cut-off region with an increase in temperature
(c) moves towards the saturation region with a decrease in temperature
(d) moves towards the saturation region with an increase in temperature
20. Under normal operating conditions, the gate terminal of an n-channel junction field effect transistor (JFET)
and an n-channel metal oxide semiconductor field effect transistor (MOSFET) are
(a) both biased with positive potentials
(b) both biased with negative potentials
(c) biased with positive and negative potentials, respectively
(d) biased with negative and positive potentials, respectively
GATE-PH 2008 QUESTION PAPER 80
Q.21 – Q.75 : Carry TWO marks each.
cos sin
21. The eigenvalues of the matrix are
sin cos
1 1
(a) ( 3 i ) when 45º (b) ( 3 i ) when 30º
2 2
1
(c) 1 since the matrix is unitary (d) (1 i ) when 30º
2
22. If the Fourier transform F [( x a)] exp( i 2va ), then F 1 (cos 2av ) will correspond to
(a) ( x a) ( x a ) (b) a constant
1 1
(c) [( x a ) i( x a )] (d) [( x a ) ( x a)]
2 2
23. If I dzLn( z ), where C is the unit circle taken anticlockwise and Ln( z ) is the principal branch of the
C
Logarithm function, which one of the following is correct?
(a) I 0 by residue theorem
(b) I is not defined since Ln( z ) has a branch cut
(c) I 0
dzLn( z
2
(d) ) 2I
C
i
24. The value of ( z 1)dz
i
is
(a) 0 (b) 2i (c) 2i (d) ( 1 2i )
d 2 y 1 dy
25. Consider the Bessel equation (v 0), y 0 . Which one of the following statements is correct?
dz 2 z dz
(a) Equation has regular singular points at z 0 and z
(b) Equation has 2 linearly independent solutions that are entire
(c) Equation has an entire solution and a second linearly independent solution singular at z 0
(d) Limit z , taken along x-axis, exists for both the linearly independent solutions
26. Under a certain rotation of coordinate axes, a rank-1 tensor va (a 1, 2, 3) transforms according to the
1 1
orthogonal transformation defined by the relations v1 (v1 v2 ); v2 (v1 v2 ); v3 v3 . Under
2 2
the same rotation a rank-2 tensor Ta ,b would transform such that
(a) T1,1 T1,1T1,2 (b) T1,1 T1,1
1
(c) T1,1 T1,1 2T2,2 T2,3 (d) T1,1 (T1,1 T2,2 T1,2 T2,1 )
2
GATE-PH 2008 QUESTION PAPER 81
1 2 1
27. The Lagrangian of a system is given by L q qq q 2 . It describes the motion of
2 2
(a) a harmonic oscillator (b) a damped harmonic oscillator
(c) an anharmonic oscillator (d) a system with unbounded motion
8 0 4
28. The moment of inertia tensor of a rigid body is given by I 0 4 0 . The magnitude of the moment
4 0 8
1 3
of inertia about an axis n , , 0 is
2 2
(a) 6 (b) 5 (c) 2 (d) 8/3
29. A hoop of radius R is pivoted at a point on the circumference. The period of small oscillations in the plane
of the hoop is
2R R R 9R
(a) 2 (b) 2 (c) 2 (d) 2
g 4g g 7g
30. A mass m is constrained to move on a horizontal frictionless surface. It is set in circular motion with radius
r0 and angular speed 0 by an applied force F communicated through an inextensible thread that passes
through a hole on the surface as shown in the figure. This force is then suddenly doubled. The magnitude
of the radial velocity of the mass
(a) increases till the mass falls into the hole
(b) decreases till the mass falls into the hole
(c) remains constant
(d) becomes zero at a radius r1 where 0 r1 r0 F
1 2 1 2 p iq
31. For a simple harmonic oscillator the Lagrangian is given by L q q . If A( p, q) and
2 2 2
H ( p, q) is the Hamiltonian of the system, the Poisson bracket { A( p, q), H ( p, q)} is given by
*
(a) iA( p, q ) (b) A* ( p, q ) (c) iA ( p, q) (d) iA( p, q )
32. A plane electromagnetic wave is given by E0 ( xˆ ei yˆ ) exp{i ( kz t )} . At a given location, the number
of times E vanishes in one second is
(a) An integer near when n and zero when n, n is integer
(b) An integer near and is independent of
(c) An integer near when n and zero when n, n is integer
2
(d) An integer near and is independent of
2
GATE-PH 2008 QUESTION PAPER 82
33. A dielectric sphere is placed in a uniform electric field directed along the positive y-axis. Which one of the
following represents the correct equipotential surfaces?
y y
x x
(a) (b)
y y
x x
(c) (d)
34. A rod of length L with uniform charge density per unit length is in the xy-plane and rotating about z-
axis passing through one of its edge with an angular velocity as shown in the figure below. ( rˆ, ˆ , zˆ ) refer
to the unit vectors at Q, A is the vector potential at a distance d from the origin O along z-axis for d L
and J is the current density due to the motion of the road. Which one of the following statements is
correct?
1
(a) J along rˆ; A along zˆ; | A |
d
z
1
(b) J along ˆ ; A along ˆ ; | A | 2 P
d
d
(c) J along rˆ; A along zˆ; | A | 12 O y
d L Q
1 x r
(d) J along ˆ ; A along ˆ ; | A |
d
0 r cos
35. A circular disc of radius a on the xy plane has a surface charge density . The electric dipole
a
moment of this charge distribution is
0 a 4 0 a 3 0 a 3 0 a 4
(a) xˆ (b) ˆ
x (c) ˆ
x (d) xˆ
4 4 4 4
36. At time t 0, a charge distribution ( r , 0) exists within an ideal homogeneous conductor of permittivity
and conductivity . At a later time ( r , t ) is given by
t (r , 0)
(a) (r , t ) (r , 0) exp (b) (r , t )
1 (t / ) 2
t 2 t
(c) ( r , t ) ( r , 0) exp (d) (r , t ) (r , 0) sin
t
GATE-PH 2008 QUESTION PAPER 83
37. A nonrelativistic charged particle moves along the positive x-axis with a constant positive acceleration axˆ .
The particle is at the origin at t 0 . Radiation is observed at t 0 at a distant point (0, d , 0) on the y-
axis. Which one of the following statements is correct?
(a) The radiation is unpolarized
(b) The radiation is plane polarized with polarization parallel to the x-axis
(c) The radiation is plane polarized with polarization parallel to the xy plane along a line inclined to the x-
axis
(d) The radiation is elliptically polarized
38. For a physical system, two observer O1 and O2 are known to be compatible. Choose the correct implication
from amongst those given below:
(a) Every eigenstate of O1 must necessarily be an eigenstate of O2
(b) Every non-degenerate eigenstate of O1 must necessarily be an eigenstate of O2
(c) When an observation of O1 is carried out on an arbitrary state of the physical system, a subsequent
observation of O2 leads to an unambiguous result
(d) Observation of O1 and O2, carried out on an arbitrary state of the physical system, lead to the
identical results irrespective of the order in which the observations are made.
39. An exact measurement of the position of a simple harmonic oscillator (SHO) is made with the result x x0 .
[The SHO has energy levels En (n = 0, 1, 2...) and associated normalized wave-functions 0 ]. Subsequently,,
an exact measurement of energy E is made. Using the general notation Pr( E E) denoting the probability
that a result E is obtained for this measurement, the following statements are written. Which one of the
following statements is correct?
(a) Pr( E E0 ) 0 (b) Pr( E En ) 1 for some value of n
(c) Pr( E En ) 0 ( x) (d) Pr( E E) 0 for any E
40. Consider the combined system of proton and electron in the hydrogen atom in its (electronic) ground state.
Let I denote the quantum number associated with the total angular momentum and let µ j denote the
magnitude of the expectation value of the net magnetic moment in the state. Which of the following pairs
represents a possible state of the system ( B is Bohr magneton)?
1
(a) I 0, µ j 0 (b) I , µ j 1 B
2
(c) I 1, µ j 1 B (d) I 0, µ j 2 B
41. A particle is placed in a one dimensional box of size L along the x-axis (0 x L) . Which of the following
is true?
(a) In the ground state, the probability of finding the particle in the interval ( L / 4,3L / 4) is half.
(b) In the first excited state, the probability of finding the particle in the interval ( L / 4,3L / 4) is half. This
also holds for states with n 4, 6,8,...
(c) For an arbitrary state , the probability of finding the particle in the left half of the well is half
(d) In the ground state, the particle has a definite momentum.
GATE-PH 2008 QUESTION PAPER 84
42. An inelastic ball of mass m has been thrown vertically upwards from the ground at z 0 . The initial kinetic
energy of the ball is E. The phase trajectory of the ball after successive bouncing on the ground is
pz pz
(a) z (b) O
z
O
pz pz
(c) z (d) z
O O
1
43. A system containing N non-interacting localized particles of spin and magnetic moment each is kept
2
in constant external magnetic field B and in thermal equilibrium at temperature T. The magnetization of the
system is,
B B
(a) N coth (b) N tanh
k BT k BT
B B
(c) N sinh (d) N cosh
k BT k BT
44. Two identical particles have to be distributed among three energy levels. Let rB , rF and rC represent the
ratios of probability of finding two particles to that of finding one particle in a given energy state. The
subscripts B, F and C correspond to whether the particles are bosons, fermions and classical particles,
respectively. Then, rB : rF : rC is equal to
1 1 1 1 1
(a) : 0 :1 (b) 1: :1 (c) 1: : (d) 1: 0 :
2 2 2 2 2
45. A photon gas is at thermal equilibrium at temperature T. The mean number of photons in an energy state
is
(a) exp 1 (b) exp 1
k BT k BT
1 1
(c) exp 1 (d) exp 1
k BT k BT
GATE-PH 2008 QUESTION PAPER 85
46. Consider a system of N atoms of an ideal gas of type A at temperature T and volume V. It is kept in diffusive
contact with another system of N atoms of another ideal gas of type B at the same temperature T and
volume V. Once the combined system reaches equilibrium.
(a) the total entropy of the final system is the same as the sum of the entropy of the individual system always
(b) the entropy of mixing is 2 Nk B ln 2
(c) the entropy of the final system is less than that of sum of the initial entropies of the two gases
(d) the entropy of mixing is non-zero when the atoms A and B are of the same type.
47. Consider a system of two non-interacting classical particles which can occupy any of the three energy levels
with energy values E 0, and and 2 having degeneracies g ( E ) 1, 2 and 4 respectively. The mean
energy of the system is
2 2
4exp k BT 8exp k BT 2exp k BT 8exp k BT
(a) (b)
1 2 exp 4exp 2 1 2 exp 4exp 2
k BT
k BT k BT k BT
2
2exp 2
k T 4 exp 2 k T exp k BT 2 exp k BT
(c) B B
(d)
1 2exp 4 exp 2 1 exp exp 2
k BT
k BT k BT k BT
48. Three consecutive absorption lines at 64.275 cm–1, 77.130 cm–1 and 89.985 cm–1 have been observed
in a microwave spectrum for a linear rigid diatomic molecule. The moments of inertia IA and IB are (IA is
with respect to the bond axis passing through the centre of mass and IB is with respect to an axis passing
through the centre of mass and perpendicular to bond axis).
2 2
(a) both equal to gm cm2 (b) zero and gm cm2
12.855hc 12.855hc
2 2
2
(c) both equal to gm cm (d) zero and gm cm2
6.427hc 6.427hc
49. A pure rotational Raman spectrum of a linear diatomic molecule is recorded using electromagnetic radiation
of frequency c . The frequency of two consecutive Stokes lines are
(a) c 10 B, c 14 B (b) c 2 B, c 4 B
(c) c 10 B, c 14 B (d) c 2 B, c 4 B
50. Which one of the following statement is INCORRECT in vibrational spectroscopy with anharmonicity?
(a) The selection rule for vibrational spectroscopy is v 1, 2,...
(b) Anharmonicity leads to multiple absorption lines
(c) The intensities of hot band lines are stronger than the fundamental absorption
(d) The frequencies of hot band lines are smaller than the fundamental absorption
51. The molecular spectra of two linear molecules O-C-O and O-C-S are recorded in the microwave region.
Which one of the following statement is correct?
(a) Both the molecules would show absorption lines
(b) Both the molecules would not show absorption lines
(c) O-C-O would show absorption lines, but not O-C-S
(d) O-C-S would show absorption lines, but not O-C-O
GATE-PH 2008 QUESTION PAPER 86
52. When the refractive index of the active medium changes by in a laser resonator of length L, the
change in the spectral spacing between the longitudinal modes of the laser is (c is the speed of light in free
space)
c c c
(a) (b) (c) (d) zero
2( ) L 2L 2 L ( )
a a
53. The primitive translation vectors of the body centered cubic lattice are a ( xˆ yˆ zˆ) , b ( xˆ yˆ zˆ)
2 2
a
and c ( xˆ yˆ zˆ) . The primitive translation vectors A, B and C of the reciprocal lattice are
2
2 2 2
(a) A ( xˆ yˆ ); B ( yˆ zˆ); C ( xˆ zˆ )
a a a
2 2 2
(b) A ( xˆ yˆ ); B ( yˆ zˆ); C ( xˆ zˆ )
a a a
2 2 2
(c) A ( xˆ yˆ ); B ( yˆ zˆ); C ( xˆ zˆ )
a a a
2 2 2
(d) A ( xˆ yˆ ); B ( yˆ zˆ); C ( xˆ zˆ)
a a a
54. The structure factor of a single cell of identical atoms of form factor f is given by
S hkl f exp( i 2( x j h y j k z j l )) where ( x j , y j , z j ) is the coordinate of an atom, and hkl are the
f
Miller indices. Which one of the following statement is correct for the diffraction peaks of body centered
cubic (BCC) and face centered cubic (FCC) lattices?
(a) BCC : (200);(110); (222) (b) BCC : (210);(110); (222)
FCC : (111);(311); (400) FCC : (111);(311); (400)
(c) BCC : (200);(110); (222) (d) BCC : (200); (210);(222)
FCC : (111);(211);(400) FCC : (111);(211); (400)
55. The lattice specific heat C of a crystalline solid can be obtained using the Dulong Petit model, Einstein model
and Debye model. At low temperature k BT , which one of the following statements is true (a and
A are constants)
3
T
(a) Dulong Petit : C exp( a / T ); Einstein : C = constant; Debye: C
A
3
T
(b) Dulong Petit : C = constant; Einstein : C ; Debye: C exp( a / T )
A
3
e a /T T
(c) Dulong Petit : C = constant; Einstein : C 2
; Debye: C
T A
3
T e a /T
(d) Dulong Petit : C ; Einstein : C 2 ; Debye: C = constant
A T
GATE-PH 2008 QUESTION PAPER 87
56. A linear diatomic lattice of lattice constant a with masses M and m( M m) are coupled by a force
constant C. The dispersion relation is given by
2 1/ 2
M m 2 M m 4C
2
2 2 ka
C
2 C sin
Mm Mm Mm 2
Which one of the following statements is INCORRECT?
(a) The atoms vibrating in transverse mode correspond to the optical branch
(b) The maximum frequency of the acoustic branch depends on the mass of the lighter atom m
(c) The dispersion of frequency in the optical branch is smaller than that in the acoustic branch
(d) No normal modes exist in the acoustic branch for any frequency greater than the maximum frequency
at k / a
57. The kinetic energy of a free electron at a corner of the first Brillouin zone of a two dimensional square lattice
is larger than that of an electron at the mid-point of a side of the zone by a factor b. The value of b is
(a) b 2 (b) b 2 (c) b 4 (d) b 8
58. An intrinsic semiconductor with mass of a hole mh and mass of an electron me is at a finite temperature
T. If the top of the valence band energy is Ev and the bottom of the conduction band energy is Ec, the Fermi
energy of the semiconductor is
mh k T 3 m
(a) EF Ev Ec 3
4 k BT ln m (b) EF B ( Ev Ec ) ln h
2 e 2 4 me
E Ec 3 mh k T 3 m
(c) EF v 4 k BT ln m (d) EF B ( Ev Ec ) ln h
2 e 2 4 me
59. Choose the correct statement from the following
(a) The reaction K K pp can proceed irrespective of the kinetic energies of K+ and K–.
(b) The reaction K K pp is forbidden by the baryon number conservation
(c) The reaction K K 2 is forbidden by strangeness conservation
(d) The decay K 0 – proceeds via weak interactions
60. The following gives a list of pairs containing (i) a nucleus (ii) one of its properties. Find the pair which is
INAPPROPRIATE.
(a) (i) 10Ne20 nucleus; (ii) stable nucleus
(b) (i) A spheroidal nucleus; (ii) an electric quadrupole moment
(c) (i) 8O16 nucleus; (ii) nuclear spin J = 1/2
(d) (i) U238 nucleus; (ii) Binding energy = 1785 MeV (approximately)
61. The four possible configurations of neutrons in the ground state of 4Be9 nucleus, according to the shell
model, and the associated nuclear spin are listed below. Choose the correct one:
(a) (1s1/ 2 ) 2 (1 p3/ 2 ) 3 ; J 3 / 2 (b) (1s1/ 2 ) 2 (1 p1/ 2 ) 2 (1 p3/ 2 )1 ; J 3 / 2
(c) (1s1/ 2 )1 (1 p3/2 ) 4 ; J 1 / 2 (d) (1s1/ 2 ) 2 (1 p3/ 2 ) 2 (1 p1/ 2 )1 ; J 1 / 2
GATE-PH 2008 QUESTION PAPER 88
62. The mass difference between the pair of mirror nuclei 6C11 and 5B11 is given to be MeV / c 2 . According
to the semi-empirical mass formula, the mass difference between the pair of mirror nuclei 9F17 and 8O17
will approximately be (rest mass of proton mp = 938.27 MeV/c2 and rest mass of neutrons mn = 939.57
MeV/c2)
(a) 1.39 MeV / c 2 (b) (1.39 0.5)MeV / c 2
(c) 0.86MeV / c 2 (d) (1.6 0.78)MeV / c 2
63. A heavy nucleus is found to contain more neutrons than protons. This fact is related to which one of the
following statements
(a) The nuclear force between neutrons is stronger than that between protons
(b) The nuclear force between protons is of a shorter range than those between neutrons, so that a smaller
number of protons are held together by the nuclear force
(c) Protons are unstable, so their number in a nucleus diminishes
(d) It costs more energy to add a proton to a (heavy) nucleus than a neutron because of the Coulomb
repulsion between protons
64. A neutral pi meson (0 ) has a rest-mass of approximately 140 MeV/c2 and a lifetime of sec. A 0
produced in the laboratory is found to decay after 1.25 sec into two photons. Which of the following sets
represents a possible set of energies of the two photons as seen in the laboratory?
(a) 70 MeV and 70 MeV (b) 35 MeV and 100 MeV
(c) 75 MeV and 100 MeV (d) 25 MeV and 150 MeV
65. An a.c. voltage of 220 Vrms is applied to the primary of a 10:1 step-down transformer. The secondary of
the transformer is centre tapped and connected to a full wave rectifier with a load resistance. The d.c.
voltage appearing across the load is
22 31 62 44
(a) (b) (c) (d)
66. Let I1 and I2 represent mesh currents in the loop abcda and befcb respectively. The correct expression
describing Kirchoff’s voltage loop law in one of the following loops is
a 10 b 20 e
(a) 30 I1 15I 2 10 (b) 15I1 20 I 2 20
I1 I2 +
2A 5 15 20V
(c) 30 I1 15 I 2 10 (d) 15I1 20 I 2 20 –
c f
d
67. The simplest logic gate circuit corresponding to the Boolean expression, Y P PQ is
P
P
(a) Q Y (b) Y
P
Y
(c) QP Y (d)
Q
GATE-PH 2008 QUESTION PAPER 89
68. An analog voltage V is converted into 2-bit binary number. The minimum number of comparators required
and their reference voltage are
V V 3V V 2V
(a) 3, , , (b) 3, , ,V
4 2 4 3 3
V 2V 3V 4V V V 3V
(c) 4, , , , (d) 4, , , ,V
5 5 5 5 4 2 4
69. The following circuit (where RL R ) performs the operation of
R
V1
V0
R
V2 RL
V(1)
(a) OR gate for a negative logic system (b) NAND gate for a negative logic system
(c) AND gate for a positive logic system (d) AND gate for a negative logic system
70. In the T type master-slave JK flip flop is shown along with the clock and input waveforms. The Qn output
of flip flop was zero initially. Identify the correct output waveform.
Input
Clk J Q J Q
Clk > >
Input K Q K Q
(a) (b)
(c) (d)
Common Data Questions
Common Data for Questions 71, 72 and 73: A beam of identical particles of mass m and energy E is
incident from left on a potential barrier of width L (between 0 x L ) and height V0 as shown in the figure
( E V0 )
V(x)
V0
E
x
0
L
For x L , there is tunneling with a transmission coefficient T 0 . Let A0 , AB and AT denote the
amplitudes for the incident, reflected and the transmitted waves, respectively.
71. Throughout 0 x L, the wave-function
(a) can be chosen to be real (b) is exponentially decaying
(c) is generally complex (d) is zero
GATE-PH 2008 QUESTION PAPER 90
72. Let the probability current associated with the incident wave be S0. Let R be the reflection coefficient. Then
(a) the probability current vanishes in the classically forbidden region
(b) the probability current is TS0 for x L
(c) for, x 0 the probability current is S0 (1 R )
(d) for x L , the probability current is complex
73. The ratio of the reflected to the incident amplitude AB / A0 is
(a) 1 AT / A0 (b) (1 T ) in magnitude
2
A E
(c) a real negative number (d) 1 T
A0 V0 E
Common Data for Questions 74 and 75: Consider two concentric conducting spherical shells with inner
and outer radii a, b and c, d as shown in the figure. Both the shells are given Q amount of positive charges.
b
Q a c Q
74. The electric field in different regions are
Q Q
(a) E 0 for r a; E ˆ
r for a r b , E 0 for b r c; E rˆ for r d
40 r 2
4 0 r 2
Q Q Q
(b) E 2
ˆ
r for r a ; E 0 for a r b , E 2
ˆ
r for b r c; E rˆ for r d
40 r 40 r 4 0 r 2
Q 2Q
(c) E 2
ˆ
r for r a ; E 0 for a r b , E 0 for b r c; E rˆ for r d
40 r 4 0 r 2
Q 2Q
(d) E 0 for r a; E 0 for a r b , E 2
rˆ for b r c; E rˆ for r d
4 0 r 40 r 2
75. In order to have equal surface charge densities on the outer surfaces of both the shells, the following
conditions should be satisfied
(a) d 4b and c 2a (b) d 2b and c 2a
(c) d 2b and c a (d) d b and c 2a
GATE-PH 2008 QUESTION PAPER 91
Linked Answer Questions : Q.76 to Q.85 carry two marks each.
Statement for Linked Answer Questions 76 and 77:
Consider the -decay of a free neutron at rest in the laboratory..
76. Which of the following configurations of the decay products corresponds to the largest energy of the anti-
neutrino v ? (rest mass of electron me = 0.51 MeV/c2, rest mass of proton mp = 938.27 MeV/c2 and rest
mass of neutron mn = 939.57 MeV/c2)
(a) In the laboratory, proton is produced at rest
(b) In the laboratory, momenta of proton, electron and the anti-neutrino all have the same magnitude
(c) In the laboratory, proton and electron fly-off with (nearly) equal and opposite momenta
(d) In the laboratory, electron is produced at rest
77. Using the result of the above problem, answer the following. Which of the following represents approximately
the maximum allowed energy of the anti-neutrino v ?
(a) 1.3 MeV (b) 0.8 MeV (c) 0.5 MeV (d) 2.0 MeV
Statement for Linked Answer Questions 78 and 79:
Consider a two dimensional electron gas of N electrons of mass m each in a system of size L×L.
78. The density of states between energy and d is
4L2 m 4L2 m 1
(a) d (b) d
h2 h2
4L2 m 4L2 m
(c) d (d) d
h2 h2
79. The ground state energy E0 of the system in terms of the Fermi energy EF and the number of electrons N
is given by
1 1 2 3
(a) NEF (b) NEF (c) NEF (d) NEF
3 2 3 5
Statement for Linked Answer Questions 80 and 81:
The rate of a clock in a spaceship “Suryashakti” is observed from each to be 3/5 of the rate of the clocks
on earth.
80. The speed of the spaceship “Suryashakti” relative to earth is
4 3 9 2
(a) c (b) c (c) c (d) c
5 5 10 5
81. The rate of a clock in a spaceship “Aakashganga” is observed from earth to be 5/13 of the rate of the
clocks on earth. If both Aakashganga and Suryashakti are moving in the same direction relative to someone
on earth, then the speed of Aakashganga relative to Suryashakti is
12 4 8 5
(a) c (b) c (c) c (d) c
13 5 17 6
GATE-PH 2008 QUESTION PAPER 92
Statement for Linked Answer Questions 82 and 83:
The following circuit contains three operational amplifiers and resistors
R R
Va 3R
– R
3R – V01
Vb + +
Vc 3R
3R
R
R –
Va + V02
Vb R
Vc R
R
82. The output voltage at the end of second operational amplifier V01 is
1
(a) V01 3(Va Vb Vc ) (b) V01 (Va Vb Vc )
3
1 4
(c) V01 (Va Vb Vc ) (d) V01 (Va Vb Vc )
3 3
83. The output V02 (at the end of third op amp) of the above circuit is
(a) V02 2(Va Vb Vc ) (b) V02 3(Va Vb Vc )
1
(c) V02 (Va Vb Vc ) (d) Zero
2
Statement for Linked Answer Questions 84 and 85:
The set V of all polynomials of a real variable x of degree two or less and with real coefficients, constitutes
a real linear vector space V {c0 c1 x c2 x 2 : c0 , c1 , c2 R} .
2
84. For f ( x) a0 a1 x a2 x 2 V and g ( x ) b0 b1 x b2 x V , which one of the following constitutes an
acceptable scalar product?
(a) ( f , g ) a02b0 a12b1 a22b2 (b) ( f , g ) a02b02 a12 b12 a22b22
a1b1 a2b2
(c) ( f , g ) a0 b0 a1b1 a2b2 (d) ( f , g ) a0b0
2 3
85. Using the scalar product obtained in the above question, identify the subspace of V that is orthogonal to
(1 x) :
(a) { f ( x ) : b(1 x) cx 2 ; b, c R} (b) { f ( x) b(1 2 x) cx 2 ; b, c R}
(c) { f ( x) : b cx 2 ; b, c R} (d) { f ( x ) : bx cx 2 ; b, c R}
GATE-PH 2009 QUESTION PAPER 93
PHYSICS-PH
Q.1 – Q.20 : Carry ONE mark each.
1. The value of the contour integral, r d , for a circle C of radius r with center at the origin is
C
2
r
(a) 2 r (b) (c) r 2 (d) r
2
2. An electrostatic field
E exists in a given region R. Choose the WRONG statement.
(a) Circulation
of E is zero
(b) E can always be expressed as the gradient of a scalar field
(c) The potential difference between any two arbitrary points in the region R is zero
(d) The work done in a closed path lying entirely in R is zero
1
3. The Lagrangian of a free particle in spherical polar co-ordinates is given by L m r 2 r 2 2 r 22 sin 2 .
2
The quantity that is conserved is
L L L L
(a) (b) (c) (d) r
r
4. A conducting loop L of surface area S is moving with a velocity v in a magnetic field B r , t B0 t 2 , B0 is a
positive constant of suitable dimensions. The emf induced, Vemf , in the loop is given by
B
(a) S
t
dS (b) L v B dL
B B
(c) S
t
dS
L v
B dL (d) S t dS L v B
dL
0 i
5. The eigenvalues of the matrix A are
i 0
(a) real and distinct (b) complex and distinct
(c) complex and coinciding (d) real and coinciding
6. i (i 1, 2, 3) represent the Pauli spin matrices. Which one of the following is NOT true ?
(a) i j j i 2 i j (b) Tr ( i ) 0
(c) The eigenvalues of i are 1 (d) det ( i ) 1
d2
7. Which one of the functions given below represents the bound state eigenfunction of the operator in the
dx 2
region, 0 x , with the eigenvalue – 4 ?
(a) A0 e 2 x (b) A0 cosh 2 x (c) A0 e 2 x (d) A0 sinh 2 x
8. Pick the wrong statement
(a) the nuclear force is independent of electric charge
m
(b) the Yukawa potential is proportinoal to r 1 exp r . Where, r is the seperation between two
h
nucleons.
(c) The range of nuclear force is of the order of 1015 1014 m
(d) the nucleons interact among each other by the exchange of mesons.
GATE-PH 2009 QUESTION PAPER 94
9. If p and q are the position and momentum variables, which one of the following is NOT a canonical transfor-
mation ?
1
(a) Q q and P p, for 0
(b) Q q p and P q p for , real and 2 2 1
(c) Q p and P q
(d) Q p and P q
10. The Common Mode Rejection Ratio (CMRR) of a differential amplifier using an operational amplifier is 100
dB. The output voltage for a differential input of 200 µV is 2 V. The common mode gain is
(a) 10 (b) 0.1 (c) 30 dB (d) 10 dB
11. In an insulating solid which one of the following physical phenomena is a consequence of Pauli’s exclusion
principle ?
(a) Ionic conductivity (b) Ferromagnetism (c) Paramagnetism (d) Ferroelectricity
dx
12. Which one of the following curves gives the solution of the differential equation k1 k2 x k3 , where
dt
k1 , k 2 and k3 are positive constant with initial conditions x 0 at t 0 ?
x x
(a) (b)
t t
x x
(c) (d)
t t
13. Identify which one is a first order phase transition ?
(a) A liquid to gas transition at its critical temperature
(b) A liquid to gas transition close to its triple point
(c) A paramagnetic to ferromagnetic transition in the absence of a magnetic field
(d) A metal to superconductor transition in the absence of a magnetic field
14. Group- I lists some physical phenomena while Group-II gives some physical parameters. Match the phe-
nomena with the corresponding parameter.
Group I Group II
P. Doppler Broadening 1. Moment of inertia
Q. Natural Broadening 2. Refractive index
R. Rotational spectrum 3. Lifetime of the energy level
S. Total internal reflection 4. Pressure
(a) P-4, Q-3, R-1, S-2 (b) P-3, Q-2, R-1, S-4
(c) P-2, Q-3, R-4, S-1 (d) P-1, Q-4, r-2, S-3
GATE-PH 2009 QUESTION PAPER 95
15. The separation between the first stokes and corresponding anti-stokes lines of the rotational Raman spectrum
in terms of the rotational constant, B is :
(a) 2B (b) 4B (c) 6B (d) 12B
16. A superconducting ring is cooled in the presence of a magnetic field below its critical temperature (TC). The
total magnetic flux that passes through the ring is
h nh ne 2
(a) zero (b) n (c) (d)
2e 4 e hc
17. In a cubic crystal, atoms of mass M1 lie on one set of planes and atoms of mass M2 lie on planes interleaved
between those of the first set. If C is the force constant between nearest neighbour planes, the frequency of
lattice vibrations for the optical phonon branch with wavevector k = 0 is
1 1 1 1 1 1
(a) 2C (b) C (c) C (d) zero
M1 M 2 2M1 M 2 M1 2M 2
18. In the quark model which one of the following represents a proton?
(a) udd (b) uud (c) ub (d) cc
19. The circuit shown below:
VCC
R1
C1 T1
Vin
C2
Vo
R2 RE
(a) is a common-emitter amplifier (b) uses a pnp transistor
(c) is an oscillator (d) has a voltage gain less than one
20. Consider a nucleus with N neutrons and Z protons. If mp, mn and BE represents the mass of the proton,
the mass of the neutron and the binding energy of the nucleus respectively and c is the velocity of light in
free space, the mass of the nucleus is given by
BE BE
(a) Nm n Zm p (b) Nm p Zmn (c) Nm n Zm p 2 (d) Nm p Zmn 2
c c
Q.21 – Q.60 : Carry TWO marks each.
21. The magnetic field (in A m 1 ) inside a long solid cylindrical conductor of radius a 0.1 m , is
104 1 r ˆ
H
r 2 sin ( r ) cos ( r ) , where 2a . What is the total current (in A) in the conductor ?
800 400 300
(a) (b) (c) (d)
2a
22. Which one of the following current densities, J , can generate the magnetic vector potential A y iˆ x ˆj ?
2 2
2 ˆ ˆ 2 ˆ ˆ 2 ˆ ˆ 2 ˆ ˆ
(a) xi yj
(b) i j (c) i j
(d) xi yj
0 0 0 0
GATE-PH 2009 QUESTION PAPER 96
ez
23. The value of the integral C 2 dz , where the contour C is the circle z 3 is
z 3z 2 2
(a) 2 ie (b) ie (c) 2 ie (d) ie
24. In a non-conducting medium characterized by 0 , 0 and conductivity 0 , the electric field
(in V m 1 ) is given by E 20 sin [108 t kz ] ˆj . The magnetic field, H (in Am 1 ) , is given by
20k
(a) 20k cos 108 t kz iˆ (b) 8
sin 108 t kz ˆj
10 0
20k
(c) 8
sin 108 t kz iˆ (d) 20k cos 108 t kz ˆj
10 0
25. A cylindrical rod of length L and radius r, made of an inhomogeneous dielectric, is placed with its axis along the
z-direction with one end at the origin as shown below.
x
r
z
y L
If the rod carries a polarization, P (5 z 2 7) kˆ , the volume bound charge inside the dielectric is
(a) zero (b) 10 r 2 L (c) 5 r 2 L (d) 5 r 2 L2
26. Let Ti j ijk ak and k ijkTij , where ijk is the Levi-Civita density, defined to be zero if two of the
k i, j
indices coincide and +1 and –1 depending on whether ijk is even or odd permutation of 1, 2, 3. Then 3 is
equal to
(a) 2a3 (b) 2a3 (c) a3 (d) a3
27. The dependence of the magnetic susceptibility ( ) of a material with temperature (T) can be represented by
1
, where is the Curie-Weiss temperature. The plot of magnetic susceptibility versus temperature
T
is sketched in the figure, as curves P, Q and R with curve Q having 0 . Which one of the following state-
ments is correct ?
P
Q
R =0
T
(a) Curve R represents a paramagnet and Q a ferromagnet
(b) Curve Q represents a ferromagnet and P an antiferromagnet
(c) Curve R represents an antiferromagnet and Q a paramagnetic
(d) Curve R represents an antiferromagnet and Q a ferromagnet
GATE-PH 2009 QUESTION PAPER 97
28. The dielectric constant of a material at optical frequencies is mainly due to
(a) ionic polarizability (b) electronic polarizability
(c) dipolar polarizability (d) ionic and dipolar polarizability
29. An electron of wavevector ke , velocity ve and effective mass me is removed from a filled energy band. The
resulting hole has wavevector kh , velocity vh , and effective mass mh . Which one of the following statements
is correct ?
(a) kh ke ; vh ve ; mh me (b) kh ke ; vh ve ; mh me
(c) kh ke ; vh ve ; mh me (d) kh ke ; vh ve ; mh me
30. In a diatomic molecule, the internuclear separation of the ground and first excited electronic state are the same
as shown in the figure. If the molecule is initially in the lowest vibrational state of the ground state, then
the absorption spectrum will appear as
Energy
Energy
Energy
(a) (b)
–1
cm –1
cm r internuclear
distance
Energy
continuum
Energy
(c) (d)
–1 –1
cm cm
31. Five energy levels of a system including the ground state are shown below. Their lifetimes and the allowed
electric dipole transitions are also marked.
10–7S 4
–8
2×10 S 3
10–6S 2
10–8S 1
0(ground state)
Which one of the following transitions is the most suitable for a continuous wave (CW) laser?
(a) 1 0 (b) 2 0 (c) 4 2 (d) 4 3
32. Assuming the mean life time of a muon (in its rest frame) to be 2 10 6 s , its life time in the laboratory frame,
when it is moving with a velocity 0.95 c is
(a) 6.4 10 6 s (b) 0.62 10 6 s (c) 2.16 10 6 s (d) 0.19 10 6 s
33. Cesium has a nuclear spin of 7/2. The hyperfine spectrum of the D lines of the Cesium atom will consist of
(a) 10 lines (b) 4 lines (c) 6 lines (d) 14 lines
34. The probability that an energy level at a temperature T is unoccupied by a fermion of chemical potential
is given by
1 1 1 1
(a) ( )/ k BT (b) ( )/ k B T (c) ( )/ k BT (d) ( )/ k B T
e 1 e 1 e 1 e 1
GATE-PH 2009 QUESTION PAPER 98
35. Consider the following expression for the mass of a nucleus with Z protons and A nucleons.
1
M A, Z f A yZ zZ 2
c2
Here, f A is a function of A.
y 4a A
z ac A1/3 4a A A1
aA and ac are constants of suitable dimensions. For a fixed A, the expression of Z for the most stable nucleus
is
A/2 A/ 2
(a) Z (b) Z
ac 2/3 ac 2/3
1 A 1 A
aA 4a A
A/2 A
(c) Z (d) Z
ac 2/3 1 A2/3
1 A
4a A
36. The de-Broglie wavelength of particles of mass m with average momentum p at a temperature T in three
dimension is given by
h h h h
(a) (b) (c) (d)
2mk BT 3mk BT 2 k BT 3m
37. Assuming an ideal voltage source, Thevenin’s resistance and Thevenin’s voltage respectively for the below
circuit are
10 10
15 V 10 RL
(a) 15 and 7.5 V (b) 20 and 5 V (c) 10 and 10 V (d) 30 and 15 V
1 1 1 1
38. Let n and p denote the isospin state with I , I 3 and I , I 3 of a nuclear respectively..
2 2 2 2
Which one of the following two nuclear state has I 0, I3 0?
1 1
(a) nn pp (b) nn pp
2 2
1 1
(c) np pn (d) np pn
2 2
GATE-PH 2009 QUESTION PAPER 99
39. An amplifier of gain 1000 is made into a feedback amplifier by feeding 9.9 % of its output voltage in series
with the input opposing. If f L 20 Hz and f H 200 kHz for the amplifier without feedback, then due
to the feedback
(a) the gain decreases by 10 times (b) the output resistance increases by 10 times
(c) the f H increases by 100 times (d) the input resistance decreases by 100 times
40. Pick the correct statement based on the below circuit.
I RS= 1 k IL
Vin RL
VZ = 10 V
15-25 V
IZ
(a) The maximum zener current, I Z (max) , when RL 10 k is 15 mA
(b) The minimum zener current, I Z (min) , when RL 10 k is 5 mA
(c) With Vin = 20 V, I L I Z , when RL 2 k
(d) The power dissipated across the zener when RL 10 k and Vin = 20 V is 100 mW
41. The disintegration energy is defined to be the difference in the rest energy between the initial and final states.
Consider the following process;
94 Pu 240
92 U 236 2 He 4
The emitted -particle has a kinetic energy 5.17 MeV. The value of disintegration energy is
(a) 5.26 MeV (b) 5.17 MeV (c) 5.08 MeV (d) 2.59 MeV
42. A classical particle is moving in an external potential field V ( x, y, z ) which is invariant under the following
infinitesimal transformations
x x x x,
y y y y ,
x x x
y y RZ y ,
where RZ is the matrix corresponding to rotation about the z-axis. The conserved quantities are (the symbols
have their usual meaning)
(a) px , pz , Lz (b) px , p y , Lz , E (c) p y , Lz , E (d) p y , pz , Lx , E
1 0
43. The spin function of a free particle in the basis in which S z is diagonal, can be written as and with
0 1
eigenvalues and , respectively. In the given basis, the normalized eigenfunction of S y with eigenvalue
2 2
.
2
1 1 1 0 1 i 1 i
(a) (b) (c) (d)
2 i 2 i 2 0 2 1
GATE-PH 2009 QUESTION PAPER 100
44. Aˆ and Bˆ represent two physical characteristics of a quantum system. If  is Hermitian, then for the product
ˆ ˆ to be Hermitian, it is sufficient that
AB
(a) B̂ is Hermitian (b) B̂ is anti-Hermitian
(a) B̂ is Hermitian and Aˆ and Bˆ commute (d) B̂ is Hermitian and Aˆ and Bˆ anti-commute
45. Consider the set of vectors in three-dimensional real vector space 3 , S (1,1,1), (1, 1,1), (1,1, 1) . Which
one of the following statements is true ?
(a) S is not a linearly independent set (b) S is a basis for 3
(c) The vectors in S are orthogonal (d) An orthogonal set of vectors cannot be generated from S
46. For a Fermi gas of N particles in three dimensions at T = 0 K, the Fermi energy, EF is proportional to
(a) N 2/3 (b) N 3/ 2 (c) N 3 (d) N 2
m 2 k
47. The Lagrangian of a diatomic moleccule is given by L
2
x1 x22 x1 x2 , where m is the mass of each
2
of the atoms and x1 and x2 are the displacements of atoms measured from the equilibrium position and k 0 .
The normal frequencies are
1/ 2 1/ 4 1/4 1/2
k k k k
(a) (b) (c) (d)
m m 2m 2m
48. A particle is in the normalized state which is a superposition of the energy eigenstates E0 10 eV and
E1 30 eV . The average value of energy of the particle in the state is 20 eV..
1 3 1 2
(a) E0 10 eV E1 30 eV (b) E0 10 eV E1 30 eV
2 4 3 3
1 3 1 1
(c) E0 10 eV E1 30 eV (d) E0 10 eV E1 30 eV
2 4 2 2
mx 2 kx 2
49. The Lagrangian of a particle of mass m moving in one dimension is L exp ( t ) , where and k
2 2
are positive constant. The equation of motion of the particle is
k k k
(a)
x x 0 (b)
x x0 (c)
x x x 0 (d)
x x x 0
m m m
50. Two monochromatic waves having frequencies and and corresonding wavelength
and of same polarization, travelling along x-axis are superimposed on each other. The
phase velocity and group velocity of the resultant wave are respectively given by
2 2
(a) , (b) , (c) , (d) ,
2 2 2 2
GATE-PH 2009 QUESTION PAPER 101
Common Data for Questions 51 and 52 :
Consider a two level quantum system with energies 1 0 and 2 .
51. The Helmholtz free energy of the system is given by
(a) k BT ln 1 e / kBT (b) k BT ln 1 e / kBT
3
(c) k BT (d) k BT
2
52. The specific heat of the system is given by
e / k BT 2 e / k BT 2 e / k BT 2 e / k BT
(a) (b) (c) (d)
k BT 1 e / kBT 2 k BT 2 1 e / kBT 1 e / kBT
2
k BT 2 1 e / kBT 2
Common Data for Questions 53 and 54 :
A free particle of mass m moves along the x-direction. At t 0 , the normalized wave function of the particle is
1 x2
given by ( x, 0) 1/4
exp 2 ix , where is a real constant.
(2 ) 4
53. The expectation value of the momentum in this state is
(a) (b) (c) (d)
54. The expectation value of the particle energy is
2 1 2 2 2 4 2 1 2
(a) (b) (c) (d)
2m 2 3/ 2 2m 2m 4 3/ 2 8m 3/2
Common Data for Q.55 and Q.56:
Consider the Zeeman splitting of a single electron system for the 3d 3p electric dipole transition.
55. The Zeeman spectrum is :
(a) Randomly polarized (b) Only polarized
(c) Only polarized (d) Both and polarized
56. The fine structure line having the longest wavelength will split into
(a) 17 components (b) 10 components (c) 8 components (d) 4 components
Statement for Linked Answer Questions 57 and 58 :
The primitive translation vectors of the face centered cubic (fcc) lattice are
a ˆ ˆ a ˆ ˆ a ˆ ˆ
a1
2
j k , a2 2
i k , a3
2
i j
57. The primitive translation vectors of the fcc reciprocal lattice are
2 2 2
(a) b1
a
iˆ ˆj kˆ ; b2
a
iˆ ˆj kˆ ; b3
a
iˆ ˆj kˆ
(b) b1
a
iˆ ˆj kˆ ; b2
a
iˆ ˆj kˆ ; b3
a
iˆ ˆj kˆ
ˆ ˆ ˆ ˆ ˆ ˆ
(c) b1
2a
iˆ ˆj kˆ ; b2
2a
i j k ; b3 2a
i jk
3 3 3
(d) b1
a
iˆ ˆj kˆ ; b2
a
iˆ ˆj kˆ ; b3
a
iˆ ˆj kˆ
GATE-PH 2009 QUESTION PAPER 102
58. The volume of primitive cell of the fcc reciprocal lattice is
3 3 3 3
2 3
(a) 4 (b) 4 (c) 4 (d) 4
a a 2a a
Statement for Linked Answer Questions 59 and 60 :
The Karnaugh map of a logic circuit is shown below:
R R
PQ 1 1
PQ 1
PQ
PQ 1 1
59. The minimized logic expression for the above map is
(a) Y PR Q (b) Y Q PR (c) Y Q PR (d) Y Q PR
60. The corresponding logic implementation using gates is given as :
P P
(a) R Y (b) R Y
Q Q
P P
(c) R Y (d) R Y
Q Q
GATE-PH 2010 QUESTION PAPER 103
PHYSICS-PH
Q.1 – Q.25 : Carry ONE mark each.
1. Consider an anti-symmetric tensor P0 with the indices i and j running from 1 to 5. The number of independent
components of the tensor is
(a) 3 (b) 10 (c) 9 (d) 6
e z sin ( z )
2. The value of the integral dz , where the contour C is the unit circle z 2 1 , is
C z2
(a) 2 i (b) 4 i (c) i (d) 0
2 3 0
3. The eigenvalues of the matrix 3 2 0 are
0 0 1
(a) 5, 2, –2 (b) –5, –1, –1 (c) 5, 1, –1 (d) –5, 1, 1
0 for x 3
4. If f ( x) , then the Laplace transform of f ( x) is
x 3 for x 3
(a) s 2 e3 s (b) s 2 e3 s (c) s 2 (d) s 2 e 3 s
5. The valence electrons do not directly determine the following property of a metal.
(a) Electrical conductivity (b) Thermal conductivity
(c) Shear modulus (d) Metallic lustre
6. Consider X-ray diffraction from a crystal with a face-centered-cubic (fcc) lattice. The lattice plane for which
there is NO diffraction peak is
(a) (2, 1, 2) (b) (1, 1, 1) (c) (2, 0, 0) (d) (3, 1, 1)
7. The Hall coefficient, RH of sodium depends on
(b) The effective charge carrier mass and carrier density
(b) The charge carrier density and relaxation time
(c) The charge carrier density only
(d) The effective charge carrier mass
8. The Bloch theorem states that within a crystal, the wave function ( r ) , of an electron has the form
(a) ( r ) u ( r ) ei k r , where u ( r ) is an arbitrary function and k is an arbitrary vector..
(b) ( r ) u ( r ) ei Gr , where u ( r ) is an arbitrary function and G is a reciprocal lattice vector..
(c) (r ) u ( r ) ei Gr , where u ( r ) u ( r A), A is a lattice vector and G is a reciprocal lattice vector..
(d) ( r ) u ( r ) ei k r , where u ( r ) u ( r A), A is a lattice vector and k is an arbitrary vector..
9. In an experiment involving a ferromagnetic medium, the following observations were made. Which one of the
plots does NOT correctly represent the property of the medium ?
(TC is the Curie temperature)
Magnetization
Spontaneous
Magnetization
O Magnetic
(a) (b)
field
O 1/T 1/T
C
GATE-PH 2010 QUESTION PAPER 104
Specific heat
susceptibility
Magnetic
(c) (d)
O TC T O TC
10. The thermal conductivity of a given material reduces, when it undergoes a transition from its normalstate to the
superconducting state. The reason is
(a) The cooper pairs cannot transfer energy to the lattice.
(b) Upon the formation of Cooper pairs, the lattice becomes less efficient in heat transfer.
(c) The electrons in the normal state loose their ability to transfer heat because of their coupling to the Cooper
pairs.
(d) The heat capacity increases on transition to the superconducting state leading to a reduction in thermal
conductivity.
11. The basic process underlying the neutron -decay is
(a) d u e ve (b) d u e (c) s u e ve (d) u d e ve
12. In the nuclear shell model the spin parity of 15N is given by
1 1 3 3
(a) (b) (c) (d)
2 2 2 2
13. Match the reactions on the left with the associated interactions on the right
(1) µ vµ (i) Strong
0
(2) (ii) Electromagnetic
(3) 0 n p (iii) Weak
(a) 1-iii, 2-ii, 3-i (b) 1-i, 2-ii, 3-iii (c) 1-ii, 2-i, 3-iii (d) 1-iii, 2-i, 3-ii
14. To detect trace amounts of a gaseous species in a mixture of gases, the preferred probing tools is
(a) Ionization spectroscopy with X-rays (b) NMR spectroscopy
(c) ESR spectroscopy (d) Laser spectroscopy
15. A collection of N atoms is exposed to a strong resonant electromagnetic radiation with N g atoms in the
ground state and N e atoms in the excited state, such that N g N e N . This collection of two-level atoms
will have the following population distribution:
N N
(a) N g N e (b) N g N e (c) N g N e (d) N g N e
2 2
16. Two states of an atom have definite parities. An electric dipole transition between these states is
(a) Allowed if both the states have even parity
(b) Allowed if both the states have odd parity
(c) Allowed if both the states have opposite parities
(d) Not allowed unless a static electric field is applied
17. The spectrum of radiation emitted by a black body at a temperature 1000 K peaks in the
(a) Visible range of frequencies (b) Infrared range of frequencies
(c) Ultraviolet range of frequencies (d) Microwave range of frequencies
GATE-PH 2010 QUESTION PAPER 105
18. An insulating sphere of radius a carries a charge density ( r ) 0 ( a 2 r 2 ) cos ; r a . The leading order
term for the electric field at a distance d, far away from the charge distribution, is proportional to
(a) d 1 (b) d 2 (c) d 3 (d) d 4
19. The voltage resolution of a 12-bit digital to analog converter (DAC), whose output varies from –10 V to +10
V is, approximately
(a) 1 mV (b) 5 mV (c) 20 mV (d) 100 mV
20. In the of the following circuits, negative feedback does not operate for a negative input. Which one is it ? The
opamps are running from 15 V supplies.
5.1 V
(a) (b)
(c) (d)
21. A system of N non-interacting classical point particles is constrained to move on the two-dimensional surface
of a sphere. The internal energy of the system is
3 1 5
(a) Nk BT (b) Nk BT (c) Nk BT (d) Nk BT
2 2 2
22. Which of the following atoms cannot exhibit Bose-Einstein condensation, even in principle ?
40
(a) 1 H1 (b) 4
He2 (c) 23
Na11 (d) K19
23. For the set of all Lorentz transformation with velocities along the x-axis, consider the two statements given
below:
P : If L is a Lorentz transformation then, L–1 is also a Lorentz transformation
Q : If L1 and L2 are Lorentz transformation then, L1 L2 is necessarily a Lorentz transformation
Choose the correct option.
(a) P is true and Q is false (b) Both P and Q are true
(c) Both P and Q are false (d) P is false and Q is true
24. Which of the following is an allowed wavefunction for a particle in a bound state ? N is a constant and , 0.
e r
(a) N (b) N (1 e r )
r3
non-zero constant if r R
(d)
2
y2 z 2 )
(c) Ne x e ( x
0 if r R
–15
25. A particle is confined within a spherical region of radius one femtometer (10 m). Its momentum can be
expected to be about
keV keV MeV GeV
(a) 20 (b) 200 (c) 200 (d) 2
c c c c
GATE-PH 2010 QUESTION PAPER 106
Q.26 – Q.55 : Carry TWO marks each.
e z
e z
26. For the complex function, f ( z ) , which of the following statements is correct ?
sin z
(a) z 0 is a branch point (b) z 0 is a pole of order one
(c) z 0 is a removable singularity (d) z 0 is an essential singularity
d2y
27. The solution of the differential equation for y (t ) : y 2 cosh (t ) , subject to the initial conditions
dt 2
dy
y (0) 0 and 0 , is
dt t 0
1
(a) cosh (t ) t sinh (t ) (b) sinh (t ) t cosh (t )
2
(c) t cosh (t ) (d) t sinh (t )
28. Given the recurrence relation for the Legendre polynomials
(2n 1) x Pn ( x) (n 1) Pn 1 ( x ) n Pn 1 ( x),
which of the following integrals has a non-zero value ?
1 1
2
(a) x Pn ( x) Pn 1 ( x) dx (b) x P ( x) P
n n2 ( x) dx
1 1
1
1
2
(c)
2
x Pn ( x) dx (d) x
1
Pn ( x ) Pn 2 ( x) dx
1
29. For a two-dimensional free electron gas, the electronic density n, and the Fermi energy EF , are related by
(2mEF )3/2 mEF mEF 23/ 2 (mEF )1/ 2
(a) n (b) n (c) n (d) n
3 2 3 2 2 2
30. Far away from any of the resonance frequencies of a medium the real part of the dielectric permittivity is
(a) Always independent of frequency (b) Monotonically decreasing with frequency
(c) Monotonically increasing with frequency (d) A non-monotonic function of frequency
31. The ground state wavefunction of deuteron is in a superposition of s and d states. Which of the following is
NOT true as a consequence ?
(a) It has a non-zero quadrupole moment
(b) The neutron-proton potential is non-central
(c) The orbital wavefunction is not spherically symmetric
(d) The Hamiltonian does not conserve the total angular momentum
32. The first three energy levels of 228Th90 are shown below:
4 187 keV
2 57.5 keV
0 0 keV
The expected spin-parity and energy of the next level are given by
(a) (6 ; 400 keV) (b) (6 ; 300 keV) (c) (2 ; 400 keV) (d) (4 ; 300 keV)
GATE-PH 2010 QUESTION PAPER 107
33. The quark content of , K , and p is indicated:
uus ; K su ; ud ; p uud .
In the process, p K , considering strong interactions only, which of the following statements is
true ?
(a) The process is allowed because S 0
(b) The process is allowed because l3 0
(c) The process is not allowed because S 0 and l3 0
(d) The process is not allowed because the baryon number is violated
34. The three principal moments of inertia of a methanol (CH3OH) molecule have the property I x I y I and
I z I . The rotational energy eigenvalues are
2 2 ml2 1 1 2
(a) 2 I l ( l 1) (b) l (l 1)
2 Iz I 2I
2 m2 1 1 2 2 m2 1 1
(c) 2 I (d) 2 I l ( l 1)
z I 2 Iz I
35. A particle of mass m is confined in the potential
V(x)
1 2 2
m x for x0
V ( x) 2
for x0
x
0
1 2
Let the wavefunction of the particle be given by ( x) 0 1 , where 0 and 1 are the
5 5
eigenfunctions of the ground state and the first excited state respectively. The expectation value of the energy is
31 25 13 11
(a) (b) (c) (d)
10 10 10 10
36. Match the typical spectra of stable molecules with the corresponding wave-number range
1. Electronic spectra i. 106 cm–1 and above
2. Rotational spectra ii. 105 106 cm 1
3. Molecular dissociation iii. 100 10 2 cm 1
(a) 1-ii, 2-i, 3-iii (b) 1-ii, 2-iii, 3-i (c) 1-iii, 2-ii, 3-i (d) 1-i, 2-ii, 3-iii
37. Consider the operations P : r r (parity) and T : t t (time-reversal). For the electric and magnetic
fields E and B , which of the following set of transformations is correct ?
(a) P : E E , B B ; T : E E , B B
(b) P : E E , B B ; T : E E , B B
(c) P : E E , B B ; T : E E , B B
(d) P : E E , B B ; T : E E , B B
GATE-PH 2010 QUESTION PAPER 108
38. Two magnetic dipoles of magnitude m each are placed in a plane as shown.
m
45º 2
45º
m 1
The energy of interaction is given by
0 m 2 30 m 2 30 m 2
(a) zero (b) (c) (d)
4 d 3 2 d 3 8 d 3
39. Consider a conducting loop of radius a and total loop resistance R placed in a region with a magnetic field B
thereby enclosing a flux 0 . The loop is connected to an electronic circuit as shown, the capacitor being initially
uncharged.
C
B Vout
If the loop is pulled out of the region of the magnetic field at a constant speed u, the final output voltage Vout is
independent of
(a) 0 (b) u (c) R (d) C
40. The figure shows a constant current source charging a capacitor that is initially uncharged.
Vout
If the switch is closed at t = 0, which of the following plots depicts correctly the output voltage of the circuit as
a function of time ?
Vout Vout
(a) (b)
t t
Vout Vout
(c) (d)
t t
GATE-PH 2010 QUESTION PAPER 109
41. For any set of inputs A and B, the following circuits give the same output Q, except one. Which one is it ?
A
A
(a) B Q (b) Q
B
A
B Q
A
(c) (d) Q
B
42. CO2 molecule has the first few energy levels uniformly separated by approximately 2.5 meV. At a temperature
of 300 K, the ratio of the number of molecules in the 4th excited state to the number in the 2nd excited state is
about
(a) 0.5 (b) 0.6 (c) 0.8 (d) 0.9
43. Which among the following sets of Maxwell relations is correct ? (U-internal energy, H-enthalpy, A-Helmholtz
free energy and G-Gibb’s free energy)
U U H H
(a) T and P (b) V and T
V S S V P S S P
G G A A
(c) P and V (d) P and S
V T P S S T P V
2
44. For a spin-s particle, in the eigen basis of S 2 , S z , the expectation value s m S x s m is
2 s (s 1) m 2
(a) (b) 2 s ( s 1) 2m 2
2
(c) 2 s ( s 1) m 2 (d) 2 m 2
1 2 3
45. A particle is placed in a region with the potential V ( x) kx x , where k , 0 . Then,
2 3
k
(a) x 0 and x are points of stable equilibrium
k
(b) x 0 is a point of stable equilibrium and x is a point of unstable equilibrium
k
(c) x 0 and x are points of unstable equilibrium
(d) There are no points of stable or unstable equilibrium
46. A 0 meson at rest decays into two photons, which move along the x-axis. They are both detected simulta-
neously after a time, t 10 s . In an inertial frame moving with a velocity v 0.6 c in the direction of one of the
photons, the time interval between the two detections is
(a) 15 s (b) 0 s (c) 10 s (d) 20 s
47. A particle of mass m is confined in an infinite potential well :
0 if 0 x L
V ( x)
otherwise
GATE-PH 2010 QUESTION PAPER 110
2 x
It is subjected to a perturbing potential V p ( x ) V0 sin (1) (2)
within the well. Let E and E be the cor-
L
rections to the ground state energy in the first and second order in V0 , respectively. Which of the following are
true ?
V(x)
Vp(x)
0 L
(a) E (1) 0 ; E (2) 0 (b) E (1) 0 ; E (2) 0
(c) E (1) 0 ; E (2) depends on the sign of V0 (d) E (1) 0 ; E (2) 0
Common Data for Questions 48 and 49 :
In the presence of a weak magnetic field, atomic hydrogen undergoes the transition:
2
P1/2 1S1/2 by emission of radiation.
48. The number of distinct spectral lines that are observed in the resultant Zeeman spectrum is
(a) 2 (b) 3 (c) 4 (d) 6
1 1
49. The spectral line corresponding to the transition 2 P1/2 m j 1S1/2 m j is observed along the
2 2
direction of the applied magnetic field. The emitted electromagnetic field is
(a) Circularly polarized (b) Linearly polarized
(c) Unpolarized (d) Not emitted along the magnetic field direction
Common Data for Questions 50 and 51 :
The partition function for a gas of photons is given by
2 V ( k BT )3
ln Z
45 3C 3
50. The specific heat of the photon gas varies with temperature as
CV CV
(a) (b)
T T
CV CV
(c) (d)
T T
GATE-PH 2010 QUESTION PAPER 111
51. The pressure of the photon gas is
2 (k BT )3 2 ( k BT ) 4 2 ( k BT ) 4 (k BT )3/2
(a) (b) (c) (d)
15 3C 3 8 3C 3 45 3C 3 45 3C 3
Statement for Linked Answer Questions 52 and 53:
Consider the propagation of electromagnetic waves in a linear, homogeneous and isotropic material medium
with electric permittivity , and magnetic permittivity .
52. For a plane wave of angular frequency and propagation vector k̂ propagating in the medium Maxwell’ss
equations reduce to
(a) k E 0 ; k H 0 ; k E H ; k H E
(b) k E 0 ; k H 0 ; k E H ; k H E
(c) k E 0 ; k H 0 ; k E H ; k H E
(d) k E 0 ; k H 0 ; k E H ; k H E
53. If and assume negative values in a certain frequency range, then the directions of the propagation vector
k̂ and the Poynting vector S in that frequency range are related as
(a) k and S are parallel
(b) k and S are anti-parallel
(c) k and S are perpendicular to each other
(d) k and S make an angle that depends on the magnitude of and
Statement for Linked Answer Questions 54 and 55 :
The Lagrangian for a simple pendulum is given by :
1 2 2
L ml mgl (1 cos )
2
54. Hamilton’s equations are then given by
p p
(a) p mgl sin ; 2 (b) p mgl sin ; 2
ml ml
p g p
(c) p m ; (d) p ;
m l ml
55. The Poisson bracket between and is
1 1 g
(a) { , } 1 (b) { , } 2 (c) { , } (d) { , }
ml m l
GATE-PH 2011 QUESTION PAPER 112
PHYSICS-PH
Q.1 – Q.25 : Carry ONE mark each.
1. Two matrices A and B are said to be similar if B P 1 AP for some invertible matrix P. Which of the following
statements is NOT TRUE ?
(a) Det A = Det B (b) Trace of A = Trace of B
(c) A and B have the same eigenvectors (d) A and B have the same eigenvalues
2. If a force F is derivable from a potential function V ( r ) , where r is the distance from the origin of the
coordinate system, it follows that
(a) F 0 (b) F 0 (c) V 0 (d) 2V 0
3. The quantum mechanical operator for the momentum of a particle moving in one dimension is given by
d d 2 d 2
(a) i (b) i (c) i (d)
dx dx t 2m dx 2
4. A Carnot cycle operates on a working substance between two reservoirs at temperatures T1 and T2 , with
T1 T2 . During each cycle, an amout of heat Q1 is extracted from the reservoir at T1 and an amount Q2 is
delivered to the reservoir at T2 . Which of the following statements is INCORRECT ?
(a) work done in one cycle is Q1 Q2
Q1 Q2
(b) T T
1 2
(c) entropy of the hotter reservoir decreases
(d) entropy of the universe (consisting of the working substance and the two reservoirs) increases
5. In a first order phase transition, at the transition temperature, specfic heat of the system
(a) diverges and its entropy remains the same
(b) diverges and its entropy has finite discontinuity
(c) remains unchanged and its entropy has finite discontinuity
(d) has finite discontinuity and its entropy diverges
6. The semi-empirical mass formula for the binding energy of nucleus contains a surface correction term. This
term depends on the mass number A of the nucleus as
(a) A1/3 (b) A1/3 (c) A2/3 (d) A
7. The population inversion in a two level laser material cannot be achieved by optical pumping because
(a) the rate of upward transitions is equal to the rate of downward transitions.
(b) the upward transitions are forbidden but downward transitions are allowed.
(c) the upward transitions are allowed but downward transitions are forbidden.
(d) the spontaneous decay rate of the higher level is very low.
8. The temperature (T) dependence of magnetic susceptibility ( ) of a ferromagnetic substance with a Curie
temperature (Tc ) is given by
C C C C
(a) , for T Tc (b) , for T Tc (c) , for T Tc (d) , for all temperatures
T Tc T Tc T Tc T Tc
9. The order of magnitude of the energy gap of a typical superconductor is
(a) 1 MeV (b) 1 KeV (c) 1 eV (d) 1 meV
GATE-PH 2011 QUESTION PAPER 113
10. Which of the following statements is CORRECT for a common emitter amplifier circuit ?
(a) The output is taken from the emitter
(b) There is 180º phase shift between input and output voltages
(c) There is no phase shift between input and output voltages
(d) Both p-n junctions are forward biased
11. A 3 × 3 matrix has elements such that its trace is 11 and its determinant is 36. The eigenvalues of the matrix are
all known to be positive integers. The largest eigenvalue of the matrix is
(a) 18 (b) 12 (c) 9 (d) 6
12. A heavy symmetrical top is rotating about its own axis of symmetry (the z-axis). If I1 , I 2 and I 3 are the
principal moments of inertia along x, y and z respectively, then
(a) I 2 I 3 ; I1 I 2 (b) I1 I 3 ; I1 I 2 (c) I1 I 2 ; I1 I 3 (d) I1 I 2 I3
13. An electron with energy E is incident from left on a potential barrier, given by
0 for x 0
V ( x)
V0 for x 0
as shown in the figure.
V(x)
V0
E
x
0
For E V0 , the space part of the wavefunction for x 0 is of the form
(a) e x (b) e x (c) ei x (d) e i x
14. If Lx , Ly and Lz are respectively the x, y and z components of angular momentum operator L, the commu-
tator Lx Ly , Lz is equal to
(a) i ( L2x L2y ) (b) 2i Lz (c) i ( L2x L2y ) (d) 0
1 2 r/a
15. The normalized ground state wavefunction of a hydrogen atom is given by (r ) 3/2
e , where a is
4 a
the Bohr radius and r is the distance of the electron from the nucleus, located at the origin. The expectation
value 1/r 2 is
8 4 4 2
(a) (b) (c) (d)
a2 a2 a2 a2
GATE-PH 2011 QUESTION PAPER 114
16. Two charges q and 2q are placed along the x-axis in front of a grounded, infinite conducting plane, as shown
in the figure. They are located respectively at a distance of 0.5 m and 1.5 m from the plane. The force acting on
the charge q is
0.5 m q 2q
x
1.5 m
1 7q2 1 2 1 2 1 q2
(a) (b) 4 2q (c) 4 q (d)
4 0 2 0 0 4 0 2
17. A uniform surface current is flowing in the positive y-direction over an infinite sheet lying in xy-plane. The
direction of the magnetic field is
(a) along iˆ for z 0 and along iˆ for z 0 (b) along kˆ for z 0 and along kˆ for z 0
(c) along iˆ for z 0 and along iˆ for z 0 (d) along kˆ for z 0 and along kˆ for z 0
18. A magnetic dipole of dipole moment m is placed in a non-uniform magnetic field B . If the position vector of
the dipole is r , the torque acting on the dipole about the origin is
(a) r ( m B ) (b) r ( m B ) (c) m B (d) m B r (m B)
19. Which of the following expressions for a vector potential A DOES NOT represent a uniform magnetic field
of magnitude B0 along the z-direction ?
(a) A (0, B0 x, 0) (b) A ( B0 y, 0, 0)
Bx B y By Bx
(c) A 0 , 0 , 0 (d) A 0 , 0 , 0
2 2 2 2
20. A neutron passing through a detector is detected because of
(a) the ionization it produces
(b) the scintillation light it produces
(c) the electron-hole pair it produces
(d) the secondary particles produced in a nuclear reaction in the detector medium.
21. An atom with one outer electron having orbital angular momentum is placed in a weak magnetic field. The
number of energy levels into which the higher total angular momentum state splits, is
(a) 2 2 (b) 2 1 (c) 2 (d) 2 1
22. For a multi electron atom , L and S specify the one electron orbital angular momentum, total angular momentum
and total spin angular momentum respectively. The selection rules for electric dipole transition between the two
electronic energy levels, specified by , L and S are
(a) L 0, 1 ; S 0 ; 0, 1 (b) L 0, 1 ; S 0 ; 1
(c) L 0, 1 ; S 1 ; 0, 1 (d) L 0, 1 ; S 1 ; 1
GATE-PH 2011 QUESTION PAPER 115
23. For a three-dimensional crystal having N primitive unit cells with a basis of p atoms, the number of optical
branches is
(a) 3 (b) 3p (c) 3 p 3 (d) 3 N 3 p
24. For an intrinsic semiconductor, me and mh are respectively the effective masses of electrons and holes near
the corresponding band edge. At a finite temperature, the position of the Fermi level
(a) depends on me but not on mh (b) depends on mh but not on me
(c) depends on both me and mh (d) depends neither on me nor mh
25. In the following circuit, the voltage across and the current through the 2 k resistance are
500 1 k
20 V 10 V
30 V 2 k
(a) 20 V, 10 mA (b) 20 V, 5 mA (c) 10 V, 10 mA (d) 10 V, 5 mA
Q.26 – Q.55 : Carry TWO marks each.
26. The unit vector normal to the surface x 2 y 2 z 1 at the point P (1, 1, 1) is
iˆ ˆj kˆ 2iˆ ˆj kˆ iˆ 2 ˆj kˆ2iˆ 2 ˆj kˆ
(a) (b) (c) (d)
3 6 6 3
ˆ ˆ be the
27. Consider a cylinder of height h and a, closed at both ends, centered at the origin. Let r ix ˆjy kz
position vector and n̂ a unit vector normal to the surface. The surface integral r nˆ ds over the closed
s
surface of the cylinder is
z
O y
(a) 2 a 2 (a h) (b) 3 a 2 h (c) 2 a 2 h (d) zero
28. The solutions to the differential equation
dy x
dx y 1
are a family of
(a) circles with different radii (b) circles with different centers
(c) straight lines with different slopes (d) straight lines with different intercepts on the y-axis
29. A particle is moving under the action of a generalized potential
(1 q )
V (q, q )
q2
The magnitude of the generalized force is
GATE-PH 2011 QUESTION PAPER 116
2(1 q ) 2(1 q ) 2 q
(a) 3 (b) 3 (c) 3 (d) 3
q q q q
30. Two bodies of mass m and 2m are connected by a spring constant k. The frequency of the normal mode is
3k k 2k k
(a) (b) (c) (d)
2m m 3m 2m
31. Let ( p, q) and ( P , Q ) be two pairs of canonical variables. The transformation
Q q cos ( p )
P q sin ( p)
is canonical for
(a) 2, 1/2 (b) 2, 2 (c) 1, 1 (d) 1/2, 2
32. Two particles, each of rest mass m collide head-on and stick together. Before collision, the speed of each mass
was 0.6 times the speed of light in free space. The mass of the final entity is
5m 5m 25m
(a) (b) 2m (c) (d)
4 2 8
33. The normalized eigenstates of a particle in a one-dimensional potential well
0 ; if 0 x a
V ( x)
; otherwise
are given by
2 n x
n ( x) sin , where n 1, 2, 3,....
a a
The particle is subjected to a perturbation
x a
V ( x) V0 cos for 0 x
a 2
0 otherwise
The shift in the ground state energy due to the perturbation, in the first order perturbation theory, is
2V0 V0 V0 2V0
(a) (b) (c) (d)
3 3 3 3
34. If the isothermal compressibility of a solid is T 1010 (Pa) 1 , the pressure required to increase its density by
1 % is approximately
(a) 104 Pa (b) 106 Pa (c) 108 Pa (d) 1010 Pa
35. A system of N non-interacting and distinguishable particles of spin 1 is in thermodynamic equilibrium. The
entropy of the system is
(a) 2k B ln N (b) 3k B ln N (c) Nk B ln 2 (d) Nk B ln 3
36. A system has two energy levels with energies and 2 . The lower level is 4-fold degenerate while the upper
level is doubly degenerate. If there are N non-interacting classical particles in the system, which is thermody-
namic equilibrium at a temperature T, the fraction of particles in the upper level is
1 1 1 1
(a) (b) (c) (d)
1 e / k BT 1 2e / kBT 2e / k BT
4e 2 / k BT 2e / k BT
4e 2 / k B T
GATE-PH 2011 QUESTION PAPER 117
37. A spherical conductor of radius a is placed in a electric field E E0 kˆ . The potential at a point P ( r , ) for r a ,
E0 a 3
is given by (r , ) constant E0 r cos cos , where r is the distance of P from the center O of the
r2
sphere and is the angle OP makes with the z-axis.
P
r
O k
The charge density on the sphere at 30 is
(a) 3 3 0 E0 /2 (b) 3 0 E0 /2 (c) 3 0 E0 /2 (d) 0 E0 /2
38. According to the single particle nuclear shell model, the spin parity of the ground state of 17
8 O is
1 3 3 5
(a) (b) (c) (d)
2 2 2 2
39. In the - decay of neutron n p e e , the anti-neutrino e escapes detection. Its existence is inferred
from the measurement of
(a) energy distribution of electrons (b) angular distribution of electrons
(c) helicity distribution of electrons (d) forward-backward asymmetry of electrons
40. The isospin and the strangeness of baryon are
(a) 1, –3 (b) 0, –3 (c) 1, 3 (d) 0, 3
41. The life time of an atomic state is 1 nanosecond. The natural line width of the spectral line in the emission
spectrum of this state is of the order of
(a) 10 10 eV (b) 10 9 eV (c) 10 6 eV (d) 10 4 eV
42. The degeneracy of an excited state of nitrogen atom having electronic configuration 1s 2 2s 2 2p 2 3d1 is
(a) 6 (b) 10 (c) 15 (d) 150
43. The far infrared rotational absorption spectrum of a diatomic molecule shows equidistant lines with spacing 20
cm–1. The position of the first Stokes line in the rotational Raman spectrum of this molecule is
(a) 20 cm–1 (b) 40 cm–1 (c) 60 cm–1 (d) 120 cm–1
44. A metal with body centered cubic (bcc) structure shows the first (i.e. smallest angle) diffraction peak at a Bragg
angle of 30 . The wavelength of X-ray used is 2.1 Å. The volume of the PRIMITIVE unit cell of the metal
is
(a) 26.2 (Å)3 (b) 13.1 (Å)3 (c) 9.3 (Å)3 (d) 4.6 (Å)3
GATE-PH 2011 QUESTION PAPER 118
45. In the following circuit, Tr1 and Tr2 are identical transistors having VBE = 0.7 V. The current passing through the
transistor Tr2 is
+5V 100 Tr2
Tr1
(a) 57 mA (b) 50 mA (c) 48 mA (d) 43 mA
46. The following Boolean expression
Y A BC D A BC D A BC D A BC D A BC D A BC D
can be simplified to
(a) A B C A D (b) A B C A D (c) A B C A D (d) A B C A D
47. Consider the following circuit.
1 k 4 k
+10 V
V out
Vin
–10 V
Which of the following correctly represents the output Vout corresponding to the input Vin ?
+5V +5V
+2V +2V
Vin Vin
time time
–2V –2V
–5V –5V
+10V +10V
(a) (b)
Vout Vout
time time
–10V –10V
+5V +5V
+2V +2V
Vin Vin
time time
–2V –2V
–5V –5V
+10V +10V
(c) (d)
V out Vout
time time
–10V –10V
GATE-PH 2011 QUESTION PAPER 119
Common Data for Questions 48 and 49 :
z sin z
Consider a function f ( z ) of a complex variable z.
( z )2
48. Which of the following statements is TRUE for the function f ( z ) ?
(a) f ( z ) is analytic everywhere in the complex plane
(b) f ( z ) has a zero at z
(c) f ( z ) has a pole of order 2 at z
(d) f ( z ) has a simple pole at z
49. Consider a counterclockwise circular contour z 1 about the origin. The integral f ( z ) dz over this con-
tour is
(a) i (b) zero (c) i (d) 2i
Common Data for Questions 50 and 51 :
The tight binding energy dispersion (E-k) relation for electrons in a one-dimensional array of atoms having
lattice constant a and total length L is E E0 2 cos (ka) , where E0 , and are constant and k is the
wave-vector.
50. The density of states of electrons (including spin degeneracy) in the band is given by
L L L L
(a) a sin (ka ) (b) 2 a sin ( ka) (c) 2 a cos (ka) (d) a cos (ka)
51. The effective mass of electrons in the band is given by
2 2 2 2
(a) (b) (c) (d)
a 2 cos (ka ) 2 a 2 cos (ka ) a 2 sin (ka) 2 a 2 sin (ka)
Statement for Linked Answer Question 52 and 53 :
In a one-dimensional harmonic oscillator, 0 , 1 and 2 are respectively the ground, first and the second
excited states. These three states are normalized and are orthogonal to one another. 1 and 2 are two states
defined by
1 0 21 32
2 0 1 2
where is a constant.
52. The value of for which 2 is orthogonal to 1 is
(a) 2 (b) 1 (c) –1 (d) –2
53. For the value of determined in Q. 52, the expression value of energy of the oscillator in the state 2 is
3 9
(a) (b) (c) 3 (d)
2 2
GATE-PH 2011 QUESTION PAPER 120
Statement for Linked Answer Question 54 and 55 :
A plane electromagnetic wave has the magnetic field given by
k
B( x, y , z , t ) B0 sin ( x y ) t kˆ
2
where k is the wave number and iˆ, ˆj and kˆ are the Cartesian unit vectors in x, y and z direction, respectively..
54. The electric field E ( x, y, z , t ) corresponding to the above wave is given by
k (iˆ ˆj ) k (iˆ ˆj )
(a) cB0 sin ( x y ) t (b) cB0 sin ( x y ) t
2 2 2 2
k k
(c) cB0 sin ( x y ) t iˆ (d) cB0 sin ( x y ) t ˆj
2 2
55. The average Poynting vector is given by
cB02 (iˆ ˆj ) cB02 (iˆ ˆj ) cB02 (iˆ ˆj ) cB02 (iˆ ˆj )
(a) (b) (c) (d)
2 0 2 20 2 2 0 2 20 2
GATE-PH 2012 QUESTION PAPER 121
PHYSICS-PH
Q.1 – Q.25 : Carry ONE mark each.
1. Identify the CORRECT statement for the following vectors a 3iˆ 2ˆj and b ˆi 2ˆj
(a) The vectors a and b are linearly independent
(b) The vectors a and b are linearly dependent
(c) The vectors a and b are orthogonal
(d) The vectors a and b are normalized
2. Two uniform thin rods of equal length L, and masses M1 and M 2 are joined together along the length. The
moment of inertia of the combined rod of length 2L about an axis passing through the mid-point perpendicular
to the length of the rod is,
L2 L2 L2 L2
(a) M1 M 2 (b) M1 M 2 (c) M1 M 2 (d) M1 M 2
12 6 3 2
3. The space-time dependence of the electric field of a linearly polarized light in free space is given by
x̂E 0 cos t kz , where E 0 , and k are the amplitude, the angular frequency and the wavevector,,
respectively. The time averaged energy density associated with the electric field is
1 1
(a) 0 E 02 (b) 0 E 0
2
(c) 0 E 02 (d) 2 0 E 02
4 2
4. If the peak output voltage of a full wave rectifier is 10 V, its d.c. voltage is
(a) 10.0V (b) 7.07V (c) 6.36 V (d) 3.18V
5. A particle of mass m is confined in a two dimensional square well potential of dimension a. This potential
V x, y is given by
V x, y 0 for a x a and a y a
else where
The energy of the first excited state for this particle is given by,
2 2 2 2 2 52 2 4 2 2
(a) (b) (c) (d)
ma 2 ma 2 2ma 2 ma 2
6. The isothermal compressibility, of an ideal gas at temperature T0 and volume V0 , is given by
1 V 1 V P P
(a) (b) (c) V0 (d) V0
V0 P T0
V0 P T V T0 P T0
0
7. The ground state of sodium atom ( Na) is a 2S1/2 state. The difference in energy levels arising in the presence
11
of a weak external magnetic field B, given in terms of Bohr magneton, B is
(a) B B (b) 2 B B (c) 4 B B (d) 6 B B
8. For an ideal Fermi gas in three dimensions, the electron velocity VF at the Fermi surface is related to electron
concentration n as,
(a) VF n 2/3 (b) VF n (c) VF n1/2 (d) VF n1/3
9. Which one of the following sets corresponds to fundamental particles ?
(a) proton, electron and neutron (b) proton, electron and photon
(c) electron, photon and neutrino (d) quark, electron and meson
GATE-PH 2012 QUESTION PAPER 122
10. In case of a Geiger-Muller (GM) counter, which one of the following statements is CORRECT?
10
(a) Multiplication factor of the detector is of the order of 10 .
(b) Type of the particles detected can be identified.
(c) Energy of the particles detected can be distinguished.
(d) Operating voltage of the detector is few tens of Volts.
11. A plane electromagnetic wave traveling in free space is incident normally on a glass plate of refractive index 3/
2. If there is no absorption by the glass, Its reflectivity is
(a) 4% (b) 16 % (c) 20% (d) 50%
12. A Ge semiconductor is doped with acceptor impurity concentration of 1015 atoms /cm3. For the given hole
mobility of 1800 cm 2 / V s , the resistivity of this material is:
(a) 0.288 cm (b) 0.694 cm (c) 3.472 cm (d) 6.944 cm
13. A classical gas of molecules each of mass m, is in thermal equilibrium at the absolute temperature, T. The
velocity components of the molecules along the Cartesian axes are vx, vy and vz. The mean value of (vx + vy)2
is:
k T 3 k BT 1 k BT 2k B T
(a) B (b) 2 m (c) 2 m (d) m
m
14. In a central force field the trajectory of a particle of mass m and angular momentum L in plane polar coordinates
is given by
1 m
1 cos
r L2
Where, is the eccentricity of the particle’s motion. Which one of the following choices for gives rise to a
parabolic trajectory ?
(a) 0 (b) 1 (c) 0 1 (d) 1
15. Identify the CORRECT energy band diagram for silicon doped with Arsenic. Here CB, VB, ED and EF
conduction band, valence band, impurity level and Fermi level, respectively.
(a) (b) (c) (d)
16. The first stokes line of a rotational Raman spectrum is observed at 12.96 cm–1. Considering the rigid rotor
approximation, the rotational constant is given by
(a) 6.48 cm–1 (b) 3.24 cm–1 (c) 2.16 cm–1 (d) 1.62 cm–1
5
3
N
17. The total energy, E of an ideal non-relativistic Fermi gas in three dimensions is given by E 2
where N is
V 3
the number of particles and V is the volume of the gas. Identify the CORRECT equation of state ( P being the
pressure),
1 2 5
(a) PV E (b) PV E (c) PV E (d) PV E
3 3 3
18. Consider the wavefunction r1 , r2 s for a fermionic system consisting of two spin-half particles. The
spatial part of the wavefunction is given by
1
r1 , r2 1 r1 2 r2 2 r1 1 r2
2
GATE-PH 2012 QUESTION PAPER 123
Where 1 and 2 are single particle states. The spin part s of the wavefunction with spin states
1 1
and should be
2 2
1 1
(a) (b) (c) (d)
2 2
19. The electric and the magnetic fields E z, t and B z, t , respectively corresponding to the scalar poten-
tial z, t 0 and vector potential A z, t ˆitz are
(a) E ˆiz and B ˆjt (b) E ˆiz and B ˆjt (c) E ˆiz and B ˆjt (d) E ˆiz and B ˆjt
20. Consider the following OP-AMP circuit.
Which one of the following correctly represents the output Vout corresponding to the input Vin ?
(a) (b)
(c) (d)
GATE-PH 2012 QUESTION PAPER 124
21. Deuteron has only one bound state with spin parity 1+, isospin 0 and electric quadrupole moment 0.286 efm2.
These data suggest that the nuclear forces are having
(a) only spin and isospin dependence (b) no spin dependence and no tensor components
(c) spin dependence but no tensor components (d) spin dependence along with tensor components
2
22. A particle of unit mass moves along the x-axis under the influence of a potential, V x x x 2 . The
particle is found to be in stable equilibrium at the point x 2. The time period of oscillation of the particle is
3
(a) (b) (c) (d) 2
2 2
23. Which one of the following CANNOT be explained by considering a harmonic approximation for the lattice
vibrations in solids ?
(a) Debye’s T3 law (b) Dulong Petit’s law
(c) Optical branches in lattices (d) Thermal expansion
24. A particle is constrained to move in a truncated harmonic potential well (x > 0) as shown in the figure. Which
one of the following statements is CORRECT ?
(a) The parity of the first excited state is even (b) The parity of the ground state is even
1 7
(c) The ground state energy is (d) The first excited state energy is
2 2
25. The number of independent components of the symmetric tensor A ij with indices i, j = 1, 2, 3 is
(a) 1 (b) 3 (c) 6 (d) 9
Q.26 – Q.55 : Carry TWO marks each.
1 0
26. Consider a system in the unperturbed state described by the Hamiltonian, H 0 . The system is
0 1
subjected to a perturbation of the form H ' , where 1. The energy eigenvalues of the perturbed
system using the first order perturbation approximation are
(a) 1 and 1 2 (b) 1 and 1 (c) 1 2 and 1 2 (d) 1 and 1 2
27. Inverse susceptibility 1/ as a function of temperature, T for a material undergoing paramagnetic to ferro-
magnetic transition is given in the figure, where O is the origin. The values of the Curie constant, C and the
Weiss molecular field constant, , in CGS units, are
(a) C 5 105 , 3 10 2 (b) C 3 10 2 , 5 105
(c) C 3 102 , 2 104 (d) C 2 104 , 3 102
GATE-PH 2012 QUESTION PAPER 125
28. A plane polarized electromagnetic wave in free space at time t = 0 is given by E x, z 10ˆj exp i 6x 8z .
The magnetic field B x, z, t is given by
1
(a) B x, z, t
c
6kˆ 8iˆ exp i 6x 8z 10ct .
1
(b) B x, z, t
c
6kˆ 8iˆ exp i 6x 8z 10ct .
1
(c) B x, z, t
c
6kˆ 8iˆ exp i 6x 8z ct .
1
(d) B x, z, t
c
6kˆ 8iˆ exp i 6x 8z ct .
0 1 0
1 0 1
29. The eigenvalues of the matrix are
0 1 0
1 1
(a) 0,1,1 (b) 0, 2, 2 , ,0
(c) (d) 2, 2, 0
2 2
30. Match the typical spectroscopic regions specified in List-I with the corresponding type of transitions in List-II
and find the correct answer using the codes given below the list :
List-I List-II
P. Infrared region 1. Electronic transitions involving valence electrons
Q. Ultraviolet visible region 2. Nuclear transitions
R. X-ray region 3. Vibrational transitions of molecules
S. - rays region 4. Transitions involving inner shell electrons
Codes :
P Q R S
(a) 1 3 2 4
(b) 2 4 1 3
(c) 3 1 4 2
(d) 4 1 2 3
31. In the following circuit, for the output voltage to V0 V1 V2 / 2 the ratio R1 / R 2 is
(a) 1/2 (b) 1 (c) 2 (d) 3
GATE-PH 2012 QUESTION PAPER 126
32.
The terms Jˆ1 , Jˆ 2 arising from 2s 3d
j
1 1
electronic configuration in J J scheme are
1 3 1 5 1 1 1 3
(a) 2 , 2 and 2 , 2 (b) 2 , 2 and 2 , 2
2,1 3, 2 1, 0 2,1
1 1 1 5 3 1 1 5
(c) 2 , 2 and 2 , 2 (d) 2 , 2 and 2 , 2
1, 0 3, 2 2,1 3, 2
33. In the following circuit, the voltage drop across the ideal diode in forward bias condition is 0.7 V.
The current passing through the diode is
(a) 0.5 mA (b) 1.0 mA (c) 1.5 mA (d) 2.0 mA
34. Choose the CORRECT statement from the following
(a) Neutron interacts through electromagnetic interaction
(b) Electron does not interact through weak interaction
(c) Neutrino interacts through weak and electromagnetic interaction
(d) Quark interacts through strong interaction but not through weak interaction
35. A rod of proper length 0 oriented parallel to the x-axis moves with speed 2c/3 along the x-axis in the S-
frame, where c is the speed of the light in free space. The observer is also moving along the x-axis with speed
c/2 with respect to the S-frame. The length of the rod as measured by the observer is
(a) 0.35 0 (b) 0.48 0 (c) 0.87 0 (d) 0.97 0
36. A simple cubic crystal with lattice parameter ac undergoes transition into a tetragonal structure with lattice
parameter a t b t 2 a c and c t 2a c , below a certain temperature. The ratio of the interplanar spacings
of (101) planes for the cubic and the tetragonal structure is
1 1 3 3
(a) (b) (c) (d)
6 6 8 8
37. Consider the following circuit in which the current gain dc of the transistor is 100.
Which one of the following correctly represents the load line (collector current Ic with respect to collectr-
emitter voltage VCE ) and Q-point of this circuit?
GATE-PH 2012 QUESTION PAPER 127
(a) (b)
(c) (d)
38. Consider a system whose three energy levels are given by 0, and 2. The energy level is two-fold
1
degenerate and the other two are non-degenerate. The partition function of the system with is given
k BT
by
2
(a) 1 2e (b) 2e e 2 (c) 1 e (d) 1 e e 2
39. Two infinitely extended homogeneous isotropic dielectric media ( medium -1 and medium – 2 with dielectric
1
constants 2 and 2 5 , respectively) meet at the z = 0 plane as shown in the figure. A uniform electric
0 0
field exists everywhere. For z 0, the electric field is given by E1 2iˆ 3jˆ 5k.
ˆ The interface separating the
two media is charge free.
The electric displacement vector in the medium-2 is given by
(a) D2 0 10iˆ 15jˆ 10kˆ (b) D2 0 10iˆ 15jˆ 10kˆ
(c) D 2 0 4iˆ 6ˆj 10kˆ (d) D2 0 4iˆ 6ˆj 10kˆ
GATE-PH 2012 QUESTION PAPER 128
3/ 2
1 1
40. The ground state wavefunction for the hydrogen atom is given by 100 e r/a 0 , where a 0 is the
4 a 0
Bohr radius.
The plot of the radial probability density, P(r) for the hydrogen atom in the ground state is
(a) (b)
(c) (d)
41. Total binding energies of O15 , O16 and O17 are 111.96 MeV, 127.62 MeV and 131.76 MeV, respectively..
1 1
The energy gap between p 1 2 and d 5 2 neutron shells for the nuclei whose mass number is close to 16, is:
(a) 4.1 MeV (b) 11.5 MeV (c) 15.7 MeV (d) 19.8 MeV
42. A particle of mass m is attached to a fixed point ‘O’ by a weightless inextensible string of length a. It is rotating
under the gravity as shown in the figure.
The Lagrangian of the particle is
1
ma 2 2 sin 2 2 mga cos
L ,
2
where and are the polar angles.
The Hamiltonian of the particle is:
1 2 p 2 1 2 p 2
(a) H p
mga cos (b) H p
mga cos
2ma 2 sin 2 2ma 2 sin 2
1 1
(c) H
2ma 2
p2 p2 mga cos (d) H
2ma 2
p2 p2 mga cos
GATE-PH 2012 QUESTION PAPER 129
43.
Given F r B, where B B0 ˆi ˆj k is a constant vector and r is the position vector. The value of
F.dr , where C is a circle of unit radius centered at origin is,
C
(a) 0 (b) 2 B0 (c) 2 B0 (d) 1
The value of the integral e z dz , using the contour C of circle with unit radius |z| = 1 is:
1
44.
C
(a) 0 (b) 1 2i (c) 1 2i (d) 2i
45. A paramagnetic system consisting of N spin-half particles, is placed in an external magnetic field. It is found that
N/2 spins are aligned parallel and the remaining N/2 spins are aligned antiparallel to the magnetic field. The
statistical entropy of the system is,
N 3N
(a) 2Nk Bn 2 (b) 2 k B n 2 (c) k B n 2 (d) Nk B n 2
2 2
46. The equilibrium vibration frequency for a oscillator is observed at 2990 cm–1. The ratio of the frequencies
corresponding to the first and the fundamental spectral lines is 1.96. Considering the oscillator to be anharmonic,
the anharmonicity constant is
(a) 0.005 (b) 0.02 (c) 0.05 (d) 0.1
47. At a certain temperature T, the average speed of nitrogen molecules in air is found to be 400 m/s. The most
probable and the root mean square speeds of the molecules are, respectively,
(a) 355 m/s, 434 m/s (b) 820 m/s, 917 m/s
(c) 152 m/s, 301 m/s (d) 422 m/s, 600 m/s
Common data for Q.48 and Q.49
The wavefunction of a particle moving in free space is given by, eikx 2e ikx
48. The energy of the particle is
5 2 k 2 3 2 k 2 2 k 2 2 k 2
(a) (b) (c) (d)
2m 4m 2m m
49. The probability current density for the real part of the wavefunction is
k k
(a) 1 (b) (c) (d) 0
m 2m
Common data for Q.50 and Q. 51
The dispersion relation for a one dimensional monatomic crystal with lattice spacing a, which interacts via
nearest neighbour harmonic potential is given by
Ka
A sin ,
2
where A is a constant of a appropriate unit.
GATE-PH 2012 QUESTION PAPER 130
50. The group velocity at the boundary of the first Brillouin zone is
Aa 2 1 Aa 2
(a) 0 (b) 1 (c) (d)
2 2 2
51. The force constant between the nearest neighbour of the lattice is ( M is the mass of the atom)
MA 2 MA 2
(a) (b) (c) MA 2 (d) 2MA 2
4 2
Statement for Linked Answer Q.52 and Q.53
In a hydrogen atom, consider that the electronic charge is uniformly distributed in a spherical volume of radius
a 0.5 10 10 m around the proton. The atom is placed in a uniform electric field E 30 105 V/ m .
Assume that the spherical distribution of the negative charge remains undistorted under the electric field.
52. In the equilibrium condition, the separation between the positive and the negative charge centers is
(a) 8.66 10 16 m (b) 2.60 1015 m (c) 2.60 1016 m (d) 8.66 10 15 m
53. The polarizability of the hydrogen atom in unit of C2 m / N is
(a) 2.0 1040 (b) 1.4 1041 (c) 1.4 10 40 (d) 2.0 1039
Statement for Linked Answer Q.54 and Q.55
A particle of mass m slides under the gravity without friction along parabolic path y ax 2 as shown in the
figure. Here a is a constant.
54. The Lagrangian for this particle is given by,
1 1
(a) L mx 2 mgax 2 (b) L m 1 4a 2 x 2 x 2 mgax 2
2 2
1 1
mx 2 mgax 2 (d) L m 1 4a x x mgax
2 2 2 2
(c) L
2 2
55. The Lagrange’s equation of motion of the particle is
..
(a) x.. 2gax (b) m 1 4a 2 x 2 x 2mgax 4ma 2 x x 2
.. ..
(c) m 1 4a 2 x 2 x 2mgax 4ma 2 x x 2 (d) x 2gax
GATE-PH 2013 QUESTION PAPER 131
PHYSICS-PH
Q.1 – Q.25 : Carry ONE mark each.
1. f(x) is a symmetric periodic function of x i.e. f ( x ) f ( x ) . Then, in general, the Fourier series of the
function f ( x ) will be of the form
(a) f ( x) (an cos(nkx ) bn sin(nkx)) (b) f ( x) a0 (an cos(nkx ))
n 1 n 1
(c) f ( x ) (bn sin(nkx )) (d) f ( x) a0 (bn sin(nkx ))
n 1 n 1
2. In the most general case, which one of the following quantities is NOT a second order tensor?
(a) Stress (b) Strain (c) Moment of inertia (d) Pressure
3. An electron is moving with a velocity of 0.85c in the same direction as that of a moving photon. The relative
velocity of the electron with respect to photon is
(a) c (b) –c (c) 0.15c (d) –0.15c
4. If Planck’s constant were zero, then the total energy contained in a box filled with radiation of all frequencies
at temperature T would be (k is the Boltzmann constant and T is non zero)
3
(a) Zero (b) Infinite (c) kT (d) kT
2
5. Across a first order phase transition, the free energy is
(a) proportional to the temperature
(b) a discontinuous function of the temperature
(c) a continuous function of the temperature but its derivative is discontinuous
(d) such that the first derivative with respect to temperature is continuous
6. Two gases separated by an impermeable but movable partition are allowed to freely exchange energy. At
equilibrium, the two sides will have the same
(a) pressure and temperature (b) volume and temperature
(c) pressure and volume (d) volume and energy
7. The entropy function of a system is given by S ( E ) aE ( E0 E ) where a and E0 are positive constants.
The temperature of the system is
(a) negative for some energies (b) increases monotonically with energy
(c) decreases monotonically with energy (d) zero
8. Consider a linear collection of N independent spin 1/2 particles, each at a fixed location. The entropy of
this system is (k is the Boltzmann constant)
1
(a) Zero (b) Nk (c) Nk (d) Nk ln(2)
2
9. The decay process n p e ve violates
(a) baryon number (b) lepton number (c) isospin (d) strangeness
10. The isospin (I) and baryon number (B) of the up quark is
(a) I = 1, B = 1 (b) I = 1, B = 1/3 (c) I = 1/2, B = 1 (d) I = 1/2, B = 1/3
11. Consider the scattering of neutrons by protons at very low energy due to a nuclear potential of range r0.
Given that, cot( kr0 ) where is the phase shift, k the wave number and (– ) the logarithmic
k
derivative of the deuteron ground state wave function, the phase shift is
k
(a) kr0 (b) kr0 (c) kr0 (d) kr0
k 2 2
GATE-PH 2013 QUESTION PAPER 132
12. In the decay process, the transition 2 3 , is
(a) allowed both by Fermi and Gamow-Teller selection rule
(b) allowed by Fermi and but not by Gamow-Teller selection rule
(c) not allowed by Fermi but allowed by Gamow-Teller selection rule
(d) not allowed both by Fermi and Gamow-Teller selection rule
13. At a surface current, which one of the magnetostatic boundary condition is NOT CORRECT?
(a) Normal component of the magnetic field is continuous.
(b) Normal component of the magnetic vector potential is continuous.
(c) Tangential component of the magnetic vector potential is continuous.
(d) Tangential component of the magnetic vector potential is not continuous.
14. Interference fringes are seen at an observation plane z 0 , by the superposition of two plane waves
A1 exp[i (k1 r t )] and A2 exp[i (k2 r t )] , where A1 and A2 are real amplitudes. The condition for
interference maximum is
(a) (k1 k 2 ) r (2m 1) (b) (k1 k2 ) r 2m (c) (k1 k2 ) r (2m 1)
(d) (k1 k2 ) r 2m
15. For a scalar function satisfying the Laplace equation, has
(a) zero curl and non-zero divergence (b) non-zero curl and zero divergence
(c) zero curl and zero divergence (d) non-zero curl and non-zero divergence
16. A circularly polarized monochromatic plane wave is incident on a dielectric interface at Brewster angle.
Which one of the following statements is CORRECT?
(a) The reflected light is plane polarized in the plane of incidence and the transmitted light is circularly
polarized.
(b) The reflected light is plane polarized perpendicular to the plane of incidence and the transmitted light
is plane polarized in the plane of incidence.
(c) The reflected light is plane polarized perpendicular to the plane of incidence and the transmitted light
is elliptically polarized.
(d) There will be no reflected light and the transmitted light is circularly polarized.
17. Which one of the following commutation relations is NOT CORRECT? Here, symbols have their usual
meanings.
(a) [ L2 , Lz ] 0 (b) [ Lx , Ly ] iLz (c) [ Lz , L ] L (d) [ Lz , L– ] L–
18. The Lagrangian of a system with one degree of freedom q is given by L q 2 q 2 , where and
are non-zero constants. If pq denotes the canonical momentum conjugate to q then which one of the
following statements is CORRECT?
(a) pq 2q and it is a conserved quantity.. (b) pq 2q and it is not a conserved quantity..
(c) pq 2q and it is a conserved quantity.. (d) pq 2q and it is not a conserved quantity..
19. What should be the clock frequency of a 6-bit A/D converter so that its maximum conversion time is 32s ?
(a) 1 MHz (b) 2 MHz (c) 0.5 MHz (d) 4 MHz
20. A phosphorous doped silicon semiconductor (doping density: 1017 / cm3) is heated from 100ºC to 200ºC.
Which one of the following statements is CORRECT?
(a) Position of Fermi level moves towards conduction band
(b) Position of dopant level moves towards conduction band
(c) Position of Fermi level moves towards middle of energy gap
(d) Position of dopant level moves towards middle of energy gap
GATE-PH 2013 QUESTION PAPER 133
21. Considering the BCS theory of superconductors, which one of the following statements is NOT CORRECT?
(h is the Planck’s constant and e is the electronic charge)
(a) Presence of energy gap at temperatures below the critical temperature
(b) Different critical temperatures or isotopes
h
(c) Quantization of magnetic flux in superconducting ring in the unit of
e
(d) Presence of Meissner effect
22. Group I contains elementary excitations in solids. Group II gives the associated fields with these excitations.
MATCH the excitations with their associated field and select your answer as per codes given below.
Group-I Group-II
(P) phonon (i) photon + lattice vibration
(Q) plasmon (ii) electron + elastic deformation
(R) polaron (iii) collective electron oscillations
(S) polariton (iv) elastic wave
Codes:
(a) (P–iv), (Q-iii), (R-i), (S-ii) (b) (P–iv), (Q-iii), (R-ii), (S-i)
(c) (P–i), (Q-iii), (R-ii), (S-iv) (d) (P–iii), (Q-iv), (R-ii), (S-i)
23. The number of distinct ways of placing four indistinguishable balls into five distinguishable boxes is ______
24. A voltage regulator has ripple rejection of –50dB. If input ripple is 1mV, what is the output ripple voltage
in V? The answer should be up to two decimal places _______
25. The number of spectral lines allowed in the spectrum for the 3 2 D 3 2 P transition in sodium is ______
Q.26 – Q.55 : Carry TWO marks each.
26. Which of the following pairs of the given function F(t) and its Laplace transform f(s) is NOT CORRECT?
(a) F (t ) (t ), f ( s ) 1 , (Singularity at +0)
1
(b) F (t ) 1, f ( s ) (s > 0)
s
s
(c) F (t ) sin kt , f (s ) , ( s 0)
s2 k 2
kt 1
(d) F (t ) te , f (s) , ( s k , s 0)
(s k )2
27. If A and B are constant vectors, then ( A B r ) is
(a) A B (b) A B (c) r (d) zero
1
28. n is equal to [Given ( n 1) n(n) and (1 / 2) ]
2
n! 2n ! 2n ! n!
(a) n
(b) n
(c) 2n
(d)
2 n !2 n !2 22 n
29. The relativistic form of Newton’s second law of motion is
mcdv m c 2 v 2 dv mc 2 dv c 2 v 2 dv
(a) F (b) F (c) F 2 2 (d) F m
c 2 v 2 dt c dt c v dt c 2 dt
GATE-PH 2013 QUESTION PAPER 134
ia . p
30. Consider a gas of atoms obeying Maxwell-Boltzmann statistics. The average value of e over all the
momenta p of each of the particles (where a is a constant vector and a is its magnitude, m is the mass
of each atom, T is temperature and k is Boltzmann’s constant) is,
1 3
a 2m k T a 2m k T
(a) One (b) Zero (c) e 2 (d) e 2
q2
31. The electromagnetic form factor F(q2) of a nucleus is given by, F (q 2 ) exp
2Q 2
4 2 3
where Q is a constant. Given that F (q ) rdr (r ) sin qr
q 0 d r ( r ) 1
where ( r ) is the charge density, the root mean square radius of the nucleus is given by
(a) 1/Q (b) 2 / Q (c) 3 / Q (d) 6 / Q
32. A uniform circular disk of radius R and mass M is rotating with angular speed about an axis, passing
through its center and inclined at an angle 60 degrees with respect to its symmetry axis. The magnitude of
the angular momentum of the disk is,
3 3 7 7
(a) MR 2 (b) MR 2 (c) MR 2 (d) MR 2
4 8 8 4
33. Consider two small blocks, each of mass M, attached to two identical springs. One of the springs is
attached to the wall, as shown in the figure. The spring constant of each spring is k. The masses slide along
the surface and the friction is negligible. The frequency of one of the normal modes of the system is,
k k
M M
3 2 k 3 3 k 3 5 k 3 6 k
(a) (b) (c) (d)
2 M 2 M 2 M 2 M
34. A charge distribution has the charge density given by Q{( x x0 ) ( x x0 )} . For this charge distribution
the electric field at (2 x0 , 0, 0)
2Qxˆ Qxˆ Qxˆ Qxˆ
(a) (b) (c) (d)
9 0 x02 4 0 x03 4 0 x02 16 0 x02
35. A monochromatic plane wave at oblique incidence undergoes reflection at a dielectric interface. If kˆi , kˆr
and n̂ are the unit vectors in the directions of incident wave, reflected wave and the normal to the surface
respectively, which one of the following expression is correct?
(a) (kˆ kˆ ) nˆ 0
i r (b) (kˆ kˆ ) nˆ 0
i r (c) (kˆ nˆ ) kˆ 0 (d) (kˆ nˆ ) kˆ 0
i r i r
36. In a normal Zeeman effect experiment, spectral splitting of the line at the wavelength 643.8 nm corresponding
to the transition 5 1D2 5 1P1 of cadmium atom is to be observed. The spectrometer has a resolution of
0.01 nm. The minimum magnetic field needed to observe this is (me = 9.1 × 10–31 kg, e = 1.6×10–19 C,
c = 3 × 108 m/s)
(a) 0.26 T (b) 0.52 T (c) 2.6 T (d) 5.2 T
GATE-PH 2013 QUESTION PAPER 135
37. The spacing between vibrational energy levels in CO molecule is found to be 8.44l × 10–2 eV. Given that
the reduced mass of CO is 1.14 × 10–26 kg, Planck’s constant is 6.626 × 10–34 Js and
1eV = 1.6 × 10–19J. The force constant of the bond is CO molecule is
(a) 1.87 N/m (b) 18.7 N/m (c) 187 N/m (d) 1870 N/m
38. A lattice has the following primitive vectors (in A): = a 2( ˆj kˆ ), b 2( kˆ iˆ), c 2(iˆ ˆj ) . The reciprocal
lattice corresponding to the above lattice is
(a) BCC lattice with cube edge of A1 (b) BCC lattice with cube edge of 2 A1
2
1
(c) FCC lattice with cube edge of A1 (d) FCC lattice with cube edge of 2 A
2
e2 B
39. The total energy of an ionic solid is given by an expression E 9 where is Madelung
40 r r
constant, r is the distance between the nearest neighbours in the crystal and B is a constant. If r0 is the
equilibrium separation between the nearest neighbours then the value of B is
e 2r08 e 2 r08 2 e2 r010 e2 r010
(a) (b) (c) (d)
360 4 0 9 0 36 0
40. A proton is confined to a cubic box, whose sides have length 10–12 m. What is the minimum kinetic energy
of the proton? The mass of proton is 1.67 × 10–27 kg and Planck’s constant is 6.63 × 10–34 Js.
(a) 1.1 × 10–17 J (b) 3.3 × 10–17 J (c) 9.9 × 10–17 J (d) 6.6 × 10–17 J
16 z
41. For the function f ( z ) , the residue at the pole z = 1 is (your answer should be an integer)
( z 3) ( z 1) 2
________.
42. The degenerate eigenvalue of the matrix
4 1 1
1 4 1
is (your answer should be an integer) _______________
1 1 4
43. Consider the decay of a pion into a muon and an anti-neutrino in the pion rest frame.
2
m 139.6 MeV / c 2 , m 105.7 MeV / c , mv 0 . The energy (in MeV) of the emitted neutrino, to the
nearest integer is _______
44. In a constant magnetic field of 0.6 Tesla along the z direction, find the value of the path integral .dl
A
in the units of (Tesla m2) on a square loop of side length (1 / 2) meters. The normal to the loop makes
an angle of 60º to the z-axis, as shown in the figure. The answer should be up to two decimal places. ____
60º z
GATE-PH 2013 QUESTION PAPER 136
45. A spin-half particle is in a linear superposition 0.8 0.6 of its spin-up and spin-down states. If
and are the eigen states of z then what is the expectation value, up to one decimal place, of the
operator 10 z 5 x ? Here, symbols have their usual meanings _________
46. Consider the wave function Aeikr (r0 / r ) , where A is the normalization constant. For r 2r0 , the magnitude
2
of probability current density up to two decimal places, in units of ( A k / m) , is _______.
47. An n-channel junction field effect transistor has 5 mA source to drain current at shorted gate (IDSS) and
5V pinch off voltage (VP). Calculate the drain current in mA for a gate-source voltage (VGS) of –2.5V. The
answer should be up to two decimal places ________
Common Data Questions
Common Data for Questions 48 and 49: There are four energy levels E, 2E, 3E and 4E (where
E > 0). The canonical partition function of two particles is, if these particles are
48. Two identical fermions
(a) e2E e 4E e6E e 8E (b) e 3E e 4E 2e5E e6E e 7E
(c) (e E e 2E e 3E e4E )2 (d) e2E e 4E e 6E e8E
49. Two distinguishable particles
(a) e2E e 4E e6E e 8E (b) e 3E e 4E 2e5E e6E e 7E
(c) (e E e 2E e 3E e4E )2 (d) e2E e 4E e 6E e8E
Common Data for Questions 50 and 51: To the given unperturbed Hamiltonian
5 2 0
2 5 0
0 0 2
we add a small perturbation given by
1 1 1
1 1 1
1 1 1
where is a small quantity..
50. The ground state eigen vector of the unperturbed Hamiltonian is
(a) (1 / 2,1 / 2, 0) (b) (1 / 2, 1 / 2, 0) (c) (0, 0, 1) (d) (1, 0, 0)
51. A pair of eigen values of the perturbed Hamiltonian, using first order perturbation theory, is
(a) 3 2, 7 2 (b) 3 2, 2 (c) 3, 7 2 (d) 3, 2 2
Linked Answer Questions
Statement for Linked Answer Q.52 and Q.53: In the Schmidt model of nuclear magnetic moments, We
have,
e
( g l g s S )
2Mc
where the symbols have their usual meaning
GATE-PH 2013 QUESTION PAPER 137
52. For the case J l 1/ 2 , where J is the total angular momentum, the expectation value of S J in the
nuclear ground state is equal to,
(a) (J – 1)/2 (b) (J + 1)/2 (c) J/2 (d) –J/2
53. For the O17 nucleus (A = 17, Z = 8), the effective magnetic moment is given by,
e
eff gJ
2 Mc
where g is equal to, (gs = 5.59 for proton and –3.83 for neutron)
(a) 1.12 (b) –0.77 (c) –1.28 (d) 1.28
Statement for Linked Answer Questions 54 and 55: Consider the following circuit
V(in) –
V(out)
1000pF +
54. For this circuit the frequency above which the gain will decrease by 20 dB per decade is
(a) 15.9 kHz (b) 1.2 kHz (c) 5.6 kHz (d) 22.5 kHz
55. At 1.2 kHz the closed loop gain is
(a) 1 (b) 1.5 (c) 3 (d) 0.5
GATE-PH 2014 QUESTION PAPER 138
PHYSICS-PH
Q.1 – Q.25 : Carry ONE mark each.
1. The unit vector perpendicular to the surface x 2 y 2 z 2 3 at the point (1, 1, 1) is
xˆ yˆ zˆ xˆ yˆ zˆ xˆ yˆ zˆ xˆ yˆ zˆ
(a) (b) (c) (d)
3 3 3 3
2. Which one of the following quantities is invariant under Lorentz transformation?
(a) Charge density (b) Charge (c) Current (d) Electric field
3. The number of normal Zeeman splitting components of 1 P 1D transition is
(a) 3 (b) 4 (c) 8 (d) 9
4. If the half-life of an elementary particle moving with speed 0.9c in the laboratory frame is 5×10–8s, then the
proper half-life is _____×10–8 s. (c = 3×108 m/s)
5. An unpolarized light wave is incident from air on a glass surface at the Brewster angle. The angle between
the reflected and the refracted wave is
(a) 0º (b) 45º (c) 90º (d) 120º
6. Two masses m and 3m are attached to the two ends of a massless spring with force constant K. If m = 100g
and K = 0.3 N/m, then the natural angular frequency of oscillation is _______ Hz.
7. The electric field of a uniform plane wave propagating in a dielectric, non-conducting medium is given by,
E xˆ10cos 6 107 t 0.4 z V /m
The phase velocity of the wave is _________×108 m/s.
1 1 1 i
8. The matrix A is
3 1 i 1
(a) orthogonal (b) symmetric (c) anti-symmetric (d) unitary
9. The recoil momemtum of an atom is pA when it emits an infrared photon of wavelength 1500 nm, and it is pB
p
when it emits a photon of visible wavelength 500 nm. The ratio A is
pB
(a) 1 : 1 (b) 1: 3 (c) 1 : 3 (d) 3 : 2
10. For a gas under isothermal conditions, its pressure P varies with volume V as P V 5/3 . The bulk modulus
B is proportional to
(a) V 1/2 (b) V 2/3 (c) V 3/5 (d) V 5/3
11. Which one of the following high energy processes is allowed by conservation laws?
0 0 0
(a) p p (b) p n
(c) n p e ve (d) µ e
2 2
12.
2
dx2
The length element ds of an arc is given by, ds 2 dx1 3 dx1dx 2 . The metric tensor gij is
3 3
2 2 1 1
2 3 2 2
(a) (b) 3 (c) 3 3 (d) 3
3 1 1 2
2 2 2
2
GATE-PH 2014 QUESTION PAPER 139
13. The ground state and the first excited state wave functions of a one dimensional infinite potential well are 1
and 2 , respectively. When two spin-up electrons are placed in this potential, which one of the following, with
x1 and x2 denoting the position of the two electrons, correctly represents the space part of the ground state
wave function of the system?
1 1
(a) 1 x1 2 x1 1 x2 2 x2 (b) 1 x1 2 x2 1 x2 2 x1
2 2
1 1
(c) 1 x1 2 x1 1 x2 2 x2 (d) 1 x1 2 x2 1 x2 2 x1
2 2
14. If the vector potential,
A xxˆ 2 yyˆ 3 zzˆ
satisfies the Coulomb gauge, the value of the constant is _________________
15. At a given temperature, T , the average energy per particle of a non-interacting gas of two-dimensional classical
harmonic oscillator is _____________ k BT .
16. Which one of the following is a fermion?
(a) particle (b) 4 Be7 nucleus (c) hydrogen atom (d) deuteron
17. Which one of the following three-quark states (qqq), denoted by X, CANNOT be a possible baryon? The
corresponding electric charge is indicated in the superscript
(a) X (b) X (c) X – (d) X
18. The Hamilton’s canonical equations of motion in terms of Poisson Brackets are
(a) q q, H ; p p, H (b) q H , q ; p H , q
(c) q H , p ; p H , q (d) q p, H ; p q, H
1 1 1
19. The Miller indices of a plane passing through the three points having coordinates (0, 0, 1), (1, 0, 0), , ,
2 2 4
are
(a) (2 1 2) (b) (1 1 1) (c) (1 2 1) (d) (2 1 1)
20. The plot of specific heat versus temperature across the superconducting transition temperature TC is most
appropriately represented by
Cp Cp Cp Cp
(a) (b) (c) (d)
T T T T
TC TC TC TC
21. If L is the orbital angular momentum and S is the spin angular momentum, then L.S does NOT commute
with
2
(a) S z (b) L2 (c) S 2 (d) L S
GATE-PH 2014 QUESTION PAPER 140
22. The energy, k for band electrons as a function of the wave vector, k in the first Brillouin zone k
a a
of a one dimensional monatomic lattice is shown as (‘a’ is lattice constant)
k
k
-/a /a
vk vk
O
–/a k –/a k
(a) O /a (b) /a
vk vk
–/a k –/a k
(c) O /a (d) O /a
23. For a free electron gas in two dimensions, the variation of the density of states, N(E) as a function of energy E,
is the best represented by
N(E) N(E) N(E) N(E)
(a) (b) (c) (d)
E E E E
24. The input given to be an ideal OP-AMP integrator circuit is
V
V0
t
t0
The correct output of the integrator circuit is
GATE-PH 2014 QUESTION PAPER 141
V V
V0 V0
(a) (b)
t t
t0 t0
V V
V0 V0
(c) (d)
t t
t0 t0
25. The minimum number of flip-flops required to construct a mod-75 counter is _________
Q.26 – Q.55 : Carry TWO marks each.
26. A bead of mass ‘m’ can slide without friction along a massless rod kept at 45º with the vertical as shown in the
figure. The rod is rotating about the vertical axis with a constant angular speed . At any instant, r is the
distance of the bead from the origin. The momentum conjugate to ‘r’ is
^
Z
m
45º
r
^x
1 1
(a) mr (b) mr (c) mr (d) 2mr
2 2
27. An electron in the ground state of the hydrogen atom has the wave function
1
e 0
r/a
r
a03
2 2
where a0 is constant. The expectation value of the operator Qˆ z r , where z r cos is:
r n n n 1!
(Hint: 0 e r dr )
n1 n1
(a) a02 /2 (b) a02 (c) 3a02 /2 (d) 2a02
28. For Nickel, the number densit y is 8×10 23 atoms/cm3 and electronic configurat ion is
1s 2 2 s 2 2 p 6 3s 2 3 p 6 3d 8 4 s 2 . The value of the saturation magnetization of Nickel in its ferromagnetic states is
________ × 109 A/m.
21
(Given the value of Bohr magneton µB 9.2110 Am2 )
29. A particle of mass ‘m’ is in a potential given by
a ar02
V r 3
r 3r
GATE-PH 2014 QUESTION PAPER 142
where a and r0 are positive constants. When disturbed slightly from its stable equilibrium position, it undergoes
a simple harmonic oscillation. The time period of oscillation is
mr03 mr03 2mr03 mr03
(a) 2 (b) 2 (c) 2 (d) 4
2a a a a
30. The donor concentration in a sample of n-type silicon is increased by a factor of 100. The shift in the position
of the Fermi level at 300K, assuming the sample to be non degenerate is ________ meV.
kBT 25meV at 300K
31. A particle of mass m is subjected to a potential
1
m 2 x 2 y 2 , x , y
V x, y
2
The state with energy 4 is g-fold degenerate. The value of ‘g’ is ____________________
32. A hydrogen atom is in the state
8 3 4
200 310 321
21 7 21
where n, , m in nm denote the principal, orbital and magnetic quantum numbers, respectively. If L is the
angular momentum operator, the average value of L2 is ______________ 2 .
k
33. A planet of mass m moves in a circular orbit of radius r0 in the gravitational potential V r where k is
r
a positive constant. The orbital angular momentum of the planet is
(a) 2r0 km (b) 2r0 km (c) r0 km (d) r0 km
34. The moment of inertia of a rigid diatomic molecule A is 6 times that of another rigid diatomic molecule B. If the
rotational energies of the two molecules are equal, then the corresponding values of the rotational quantum
numbers JA and JB are
(a) J A 2, J B 1 (b) J A 3, J B 1 (c) J A 5, J B 0 (d) J A 6, J B 1
35. The value of the integral
z2
e z 1 dz
C
where C is the circle |z| = 4, is
(a) 2 i (b) 2 2i (c) 4 3i (d) 4 2i
36. A ray of light insid Region 1 in the xy-plane is incident at the semicircle boundary that carries no free charges.
The electric field at the point P r0 , /4 in plane polar coordinates is E1 7eˆr 3eˆ , where eˆr and ê are
the unit vectors. The emerging ray in Region 2 has the electric field E2 parallel to x-axis. If 1 and 2 are the
2
dielectric constants of Region 1 and Region 2 respectively, then is ______________
1
GATE-PH 2014 QUESTION PAPER 143
P r0 , /4
O
1 2 x
Region 1 Region 2
37. The solution of the differential equation
d2y
y0
dt 2
subject to the boundary conditions y 0 1 and y 0, is
(a) cos t sin t (b) cosh t sinh t (c) cos t sin t (d) cosh t sinh t
38. Given that the linear transformation of a generalized coordinate ‘q’ and the corresponding momentum p,
Q q 4ap
p q2p
is canonical, the value of the constant ‘a’ is ___________
39. The value of the magnetic field required to maintain non-relativistic protons of energy 1 MeV in a circular
orbit of radius 100 mm is ________ Tesla.
(Given: m p 1.67 1027 kg , e 1.6 1019 C )
40. For a system of two bosons, each of which can occupy any of the two energy levels 0 and , the mean energy
1
of the system at a temperature T with is given by
k BT
e 2 e 2 1
(a) (b)
1 2e e2 2e e 2
2 e e 2 e 2 e 2
(c) (d)
2 e e2 2 e e2
41. In an interference pattern formed by two coherent sources, the maximum and the minimum of the intensities are
9I0 and I0, respectively. The intensities of the individual waves are
(a) 3I0 and I0 (b) 4I0 and I0 (c) 5I0 and 4I0 (d) 9I0 and I0
1
42. 1 and 2 are two orthogonal states of a spin system. It is given that
2
1 1 2 0
1
3 0 3 1
1 0
where and represent the spin-up and spin-down states, respectively. When the system is in the state
0 1
2 , its probability to be in spin-up state is ___________
GATE-PH 2014 QUESTION PAPER 144
43. Neutrons moving with speed 103 m/s are used for the determination of crystal structure. If the Bragg
angle for the first order diffraction is 30º, the interplanar spacing of the crystal is ________Å.
(Given: mn 1.675 1027 kg , h 6.626 10 34 J .s )
p2 q 2
44. The Hamiltonian of a particle of mass ‘m’ is given by H . Which of the following figures describes
2m 2
the motion of the particle in phase space?
p p
q q
(a) (b)
p p
q q
(c) (d)
45. The intensity of a laser in free space is 150 mW /m2 . The corresponding amplitude of the electric field of the
laser is ______ V/m. 0 8.854 1012 C 2 /N.m2
46. The emission wavelength for the transition 1 D2 1F3 is 3122 Å. The ratio of populations of the final to the
initial states at a temperature 5000 K is (h 6.626 1034 J .s, c 3 108 m /s , k B 1.380 1023 J / K )
(a) 2.03 105 (b) 4.02 105 (c) 7.02 105 (d) 9.83 105
47. Consider a system of 3 fermions, which can occupy any of the 4 available energy states with equal probability.
The entropy of the system is
(a) k B ln 2 (b) 2k B ln 2 (c) 2k B ln 2 (d) 3k B ln 4
48. A particle is confined to a one dimensional potential box with potential
V x 0, 0xa
, otherwise
If the particle is subjected to a perturbation, within the box, W x , where is a small constant, the first
order correction to the ground state energy is
(a) 0 (b) a /4 (c) a /2 (d) a
GATE-PH 2014 QUESTION PAPER 145
49. Consider the process µ µ . The minimum kinetic energy of the muons (µ) in the centre of
mass frame required to produce the pion pairs at rest is _________ MeV. (Given :
mµ 105 MeV /c 2 , m 140 MeV /c 2 )
50. A one dimensional harmonic oscillator is in the superposition of number states, n , given by
1 3
2 3
2 2
The average energy of the oscillator in the given state is _____________ .
51. A nucleus X undergoes a first forbidden -decay to a nucleus Y. If the angular momentum (I) and parity (P),
P 7
denoted by I as for X, which of the following is a possible IP value for Y?
2
1 1 3 3
(a) (b) (c) (d)
2 2 2 2
52. The current gain of the transistor in the following circuit is dc 100 . The value of collector current IC is
______________ mA.
12V
3k
20µF
V0
150k
Vi
20µF
3k
53. In order to measure a maximum of 1V with a resolution of 1mV using a n-bit A/D converter, working under the
principle of ladder network, the minimum value of n is ___________________________
54. If L and L are the angular momentum ladder operators, then, the expectation value of L L L L , in
the state 1, m 1 of an atom is ___________ 2
1
55. A low pass filter is formed by a resistance R and a capacitance C. At the cut-off angular frequency c ,
RC
the voltage gain and the phase of the output voltage relative to the input voltage respectively, are
(a) 0.71 and 45º (b) 0.71 and –45º (c) 0.5 and –90º (d) 0.5 and 90º
GATE-PH 2015 QUESTION PAPER 146
PHYSICS-PH
Q.1 – Q.25 : Carry ONE mark each.
1. A satellite is moving in a circular orbit around the Earth. If T, V and E are its average kinetic, average potential
and total energies, respectively, then which one of the following options is correct ?
(a) V 2T ; E T (b) V T ; E 0
T T 3T T
(c) V ;E (d) V ;E
2 2 2 2
2. The lattice parameters a, b, c of an orthorhombic crystal are related by a 2b 3c . In units of a, the interplanar
separation between the (110) planes is_______ (upto three decimal places).
3. Consider w f ( z ) u ( x, y ) iv( x, y ) to be an analytic function in a domain D. Which one of the following
options is NOT correct ?
(a) u( x, y ) satisfies Laplace equation in D
(b) v( x, y ) satisfies Laplace equation in D
z2
(c) f ( z ) dz is dependent on the choice of the contour between z 1 and z2 in D
z1
(d) f ( z ) can be Taylor expanded in D
4. Let L and p be the angular and linear momentum operators, respectively, for a particle. The commutator
Lx , p y gives
(a) i pz (b) 0 (c) i p x (d) i pz
5. The dispersion relation for photons in a one dimensional monatomic Bravais lattice with lattice spacing a and
2C
consisting of ions of masses M is given by, (k ) 1 cos (ka ) , where is the frequency of oscilla-
M
tion, k is the wavevector and C is the spring constant. For the long wavelength modes ( a ) , the ratio of the
phase velocity to the group velocity is_______
6. For a black body radiation in a cavity, photons are created and annihilated freely as a result of emission and
absorption by the walls of the cavity. This is because
(a) the chemical potential of the photons is zero (b) photons obey Pauli exclusion principle
(c) photons are spin-1 particles (d) the entropy of the photons is very large
7. Four forces are given below in Cartesian and spherical polar coordinates.
r2
(i) F1 K exp 2 rˆ (ii) F2 K x 3 yˆ y 3 zˆ
R
ˆ
(iii) F3 K x 3 xˆ y 3 yˆ (iv) F4 K
r
where K is a constant. Identify the correct option.
(a) (iii) and (iv) are conservative but (i) and (ii) are not
(b) (i) and (ii) are conservative but (iii) and (iv) are not
(c) (ii) and (iii) are conservative but (i) and (iv) are not
(d) (i) and (iii) are conservative but (ii) and (iv) are not
GATE-PH 2015 QUESTION PAPER 147
3
8. The value of t 2 (3t 6) dt is_______ (upto one decimal place)
0
9. The mean kinetic energy of a nucleon in a nucleus of atomic weight A varies as An, where n is_______ (upto
two decimal places)
10. In Bose-Einstein condensates, the particles
(a) have strong interparticle attraction (b) condense in real space
(c) have overlapping wavefunctions (d) have large and positive chemical potential
11. A beam of X-ray of intensity I0 is incident normally on a metal sheet of thickness 2 mm. The intensity of the
transmitted beam is 0.025I0. The linear absorption coefficient of the metal sheet (in m-1) is _______________
(upto one decimal place)
12. In a Hall effect experiment, the Hall voltage for an intrinsic semiconductor is negative. This is because (symbols
carry usual meaning)
(a) n p (b) n > p (c) µe > µh (d) me* mn*
1
13. The Pauli matrices for three spin- particles are 1 , 2 and 3 , respectively. The dimension of the Hilbert
2
space required to define an operator Ô 1 2 3 is_______
14. The decay e is forbidden, because it violates
(a) momentum and lepton number conservations
(b) baryon and lepton number conservations
(c) angular momentum conservation
(d) lepton number conservation
15. The space between two plates of a capacitor carrying charges +Q and –Q is filled with two different dielectric
materials, as shown in the figure. Across the interface of the two dielectric materials, which one of the following
statements is correct ?
+Q –Q
(a) E and D are continuous (b) E is continuous and D is discontinuous
(c) D is continuous and E is discontinuous (d) E and D are discontinuous
GATE-PH 2015 QUESTION PAPER 148
R R
16. Given that magnetic flux through the closed loop PQRSP is . If A dl 1 along PQR, the value of A dl
P P
along PSR is
Q
R
P
(a) 1 (b) 1 (c) 1 (d) 1
17. A point charge is placed between two semi-infinite conducting plates which are inclined at an angle of 30° with
respect to each other. The number of image charges is_______
1
18. Consider a complex function f ( z ) . Which one of the following statements is correct ?
1
z z cos ( z )
2
1
(a) f ( z ) has simple poles at z 0 and z
2
1
(b) f ( z ) has a second order pole at z
2
(c) f ( z ) has infinite number of second order poles
(d) f ( z ) has all simple poles
19. The energy dependence of the density of states for a two dimensional non-relativistic electron gas is given by,
g E CE n , where C is constant. The value of n is _______________
20. In an inertial frame S, two events A and B take place at c t A 0, rA 0 and c tB 0, rB 2 yˆ , respectively..
The times at which these events take place in a frame S moving with a velocity 0.6 c yˆ with respect to S are
given by
(a) c t A 0; c tB 3/2 (b) c tA 0; c t B 0
(c) c tA 0; c tB 3/2 (d) c t A 0; c tB 1/2
21. In the given circuit, the voltage across the source resistor is 1 V. The drain voltage (in V) is _______
25 V
5 k
2 M 500
GATE-PH 2015 QUESTION PAPER 149
2 2
22. If f ( x ) e x and g ( x ) x e x , then
(a) f and g are differentiable everywhere
(b) f is differentiable everywhere but g is not
(c) g is differentiable everywhere but f is not
(d) g is discontinuous at x = 0
1
23. Consider a system of N non-interacting spin- particles, each having a magnetic moment , is in a magnetic
2
field B B zˆ . If E is the total energy of the system, the number of accessible microstates is given by
E
N! N B !
(a) (b)
1 E 1 E E
N ! N ! N B !
2 B 2 B
1 E 1 E N!
(c) N ! N ! (d)
2 B 2 B E
N B !
24. Which one of the following DOES NOT represent an exclusive OR operation for inputs A and B ?
(a) ( A B) AB (b) AB BA (c) ( A B) ( A B ) (d) ( A B ) AB
1 B
25. An operator for a spin- particle is given by Aˆ B , where B ( xˆ yˆ ), denotes Pauli matrices
2 2
and is a constant. The eigenvalue of  are
B
(a) (b) B (c) 0, B (d) 0, B
2
Q.26 – Q.55 : Carry TWO marks each.
26. Match the phrases in Group I and Group II and identify the correct option.
Group I Group II
(P) Electron spin resonance (ESR) (i) radio frequency
(Q) Nuclear magnetic resonance (NMR) (ii) visible range frequency
(R) Transition between vibrational states of a molecule (iii) microwave frequency
(S) Electronic transition (iv) far-infrared range
(a) (P-i), (Q-ii), (R-iii), (S-iv) (b) (P-ii), (Q-i), (R-iv), (S-iii)
(c) (P-iii), (Q-iv), (R-i), (S-ii) (d) (P-iii), (Q-i), (R-iv), (S-ii)
aVE 3/2
27. The entropy of a gas containing N particles enclosed in a volume V is given by S Nk B n 5/2
, where
N
E is the total energy, a is a constant and k B is the Boltzmann constant. The chemical potential of the system
at a temperature T is given by
aVE 3/2 5 aVE 3/2 3
(a) k BT n 5/2
(b) k BT n 5/2
N 2 N 2
aVE 3/2 5 aVE 3/2 3
(c) k BT n 3/2
(d) k BT n 3/2
N 2 N 2
GATE-PH 2015 QUESTION PAPER 150
28. The atomic masses of 152 152 1
63 Eu , 62 Sm, 1 H and neutron are 151.921749, 151.919756, 1.007825 and 1.008665
in atomic mass units (amu), respectively. Using the above information, the Q-value of the reaction
152 3
63 Eu n 152
62 Sm p is_______ 10 amu (upto three decimal places)
3
29. A particle with rest mass M is at rest and decays into two particles of equal rest masses M which move
10
along the z-axis. Their velocities are given by
(a) v1 v2 (0.8c ) zˆ (b) v1 v2 (0.8c ) zˆ
(c) v1 v2 (0.6c ) zˆ (d) v1 (0.6c) zˆ ; v2 ( 0.8c ) zˆ
30. The band gap of an intrinsic semiconductor is E g 0.72 eV and mh* 6me* . At 300K, the Fermi level with
respect to the edge of the valence band (in eV) is at ________________ (upto three decimal places)
kB 1.38 1023 JK 1 .
31. A charge –q is distributed uniformly over a sphere, with a positive charge q at its center in (i). Also in (ii), a
charge –q is distributed uniformly over an ellipsoid with a positive charge q at its center. With respect to the
origin of the coordinate system, which one of the following statements is correct ?
X X
Z Z
Y Y
(i) (ii)
(a) The dipole moment is zero in both (i) and (ii)
(b) The dipole moment is non-zero in (i) but zero in (ii)
(c) The dipole moment is zero in (i) but non-zero in (ii)
(d) The dipole moment is non-zero in both (i) and (ii)
32. The number of permitted transitions from 2 P3/2 2S1/2 in the presence of a weak magnetic field is______
33. A particle is confined in a box of length L as shown below.
V0
L/2
If the potential V0 is treated as a perturbation, including the first order correction, the ground state energy is
2 2 2 2 V0 2 2 V0 2 2 V0
(a) E V0 (b) E (c) E (d) E
2mL2 2mL2 2 2mL2 4 2mL2 2
GATE-PH 2015 QUESTION PAPER 151
34. In the given circuit, if the open loop gain A = 105, the feedback configuration and the closed loop gain Af are
V1 +
V0
–
9 k
1 k RL
(a) series-shunt, Af = 9 (b) series-series, Af = 10
(c) series-shunt, Af = 10 (d) shunt-shunt, Af = 10
35. A plane wave ( xˆ iyˆ ) E0 exp[i (kz t )] after passing through an optical element emerges as
( xˆ iyˆ ) E0 exp[i ( kz t )] , where k and are the wavevector and the angular frequency, respectively. The
optical element is
(a) quarter wave plate (b) half wave plate
(c) polarizer (d) Faraday rotator
36. A particle of mass 0.01 kg falls freely in the earth’s gravitational field with an initial velocity v (0) 10 ms 1 . If
the air exerts a functional force of the form, f kv , then for k 0.05 Nm 1s , the velocity (in ms 1 ) at
time t = 0.2 s is_______ (upto two decimal places)
(use g 10 ms 2 and e 2.72)
37. The Lagrangian for a particle of mass m at a position r moving with a velocity v is given by
m 2
L v Cr v V ( r ) , where V ( r) is a potential and C is a constant. If pc is the canonical momentum,
2
then its Hamiltonian is given by
1 2 1 2
(a) pc C r V r (b) pc C r V r
2m 2m
pc2 1 2
(c) V r (d) pc C 2 r 2 V r
2m 2m
38. A long solenoid is embedded in a conducting medium and is insulated from the medium. If the current through
the solenoid is increased at a constant rate, the induced current in the medium as a function of the radial
distance r from the axis of the solenoid is proportional to
1 1
(a) r2 inside the solenoid and outside (b) r inside the solenoid and 2 outside
r r
1 1
(c) r2 inside the solenoid and 2 outside (d) r inside the solenoid and outside
r r
1
39. In the nuclear shell model, the potential is modeled as V ( r ) m 2r 2 L S , 0 . The correct spin-
2
13
parity and isospin assignments for the ground state of C is
1 1 1 1 3 1 3 1
(a) ; (b) ; (c) ; (d) ;
2 2 2 2 2 2 2 2
GATE-PH 2015 QUESTION PAPER 152
40. Which one of the following represents the electron occupancy for a superconductor in its normal and super-
conducting states?
Normal Superconducting Normal Superconducting
state state state state
(a) f (E ) f (E ) (b) f (E ) f (E )
E E E E
Normal Superconducting Normal Superconducting
state state state state
(c) f(E) f(E) (d) f(E) f(E)
E E E E
41. In a rigid-rotator of mass M, if the energy of the first excited state is 1 meV, then the fourth excited state energy
(in meV) is_______
42. The binding energy per molecule of NaCl (lattice parameter is 0.563 nm) is 7.95 eV. The repulsive term of the
K
potential is of the form , where K is a constant. The value of the Madelung constant is ____________(upto
r9
three decimal places)
(Electron charge e 1.6 1019 C ; 0 8.854 1012 C 2 N 1m 2 )
43. The Hamiltonian for a system of two particles of masses m1 and m2 at r1 and r2 having velocities v1 and v2 is
1 1 C
given by H m1v12 m2v22 2 zˆ r1 r2 , where C is a constant. Which one of the following
2 2 r1 r2
statements is correct ?
(a) The total energy and total momentum are conserved
(b) Only the total energy is conserved
(c) The total energy and the z-component of the total angular momentum are conserved
(d) The total energy and total angular momentum are conserved
44. Given that the Fermi energy of gold is 5.54 eV, the number density of electrons is _______________×1028
m–3 (upto one decimal place).
(Mass of electron = 9.11×10–21 kg; h = 6.626×10–24J.s; 1 eV = 1.6×10–19 J)
1 i
45. Suppose a linear harmonic oscillator of frequency and mass m is in the state 0 e 1 at
2
2
t = 0 where 0 and 1 are the ground and the first excited states, respectively. The value of x in
the units of at t = 0 is_______
m
GATE-PH 2015 QUESTION PAPER 153
46. Consider the motion of the Sun with respect to the rotation of the Earth about its axis. If Fc and FCo denote the
centrifugal and the Coriolis forces, respectively, acting on the Sun, then
(a) Fc is radially outward and FCo Fc (b) Fc is radially inward and FCo 2 Fc
(c) Fc is radially outward and FCo 2 Fc (d) Fc is radially outward and FCo 2 Fc
1 m2
47. A function y ( z ) satisfies the ordinary differential equation y y 2 y 0 , where m 0,1, 2,3,....
z z
Consider the four statements P, Q, R, S as given below.
P : z m and z m are linearly independent solutions for all values of m
Q : z m and z m are linearly independent solutions for all values of m > 0
R : n z and 1 are linearly independent solutions for m = 0
S : z m and n z are linearly independent solutions for all values of m
The correct option for the combination of valid statement is
(a) P, R and S only (b) P and R only (c) Q and R only (d) R and S only
48. The average energy U of a one dimensional quantum oscillator of frequency and in contact with a heat bath
at temperature T is given by
1 1 1 1
(a) U coth (b) U sinh
2 2 2 2
1 1 1 1
(c) U tanh (d) U cosh
2 2 2 2
3
49. Consider a system of eight non-interacting, identical quantum particles of spin- in a one dimensional box of
2
22
length L. The minimum excitation energy of the system, in units of is_______
2mL2
50. In the simple current source shown in the figure, Q1 and Q2 are identical transistors with current gain
100 and VBE 0.7 V
Vcc = 30 V
5 k
I0
Q1 Q2
The current I0 (in mA) is_______(upto two decimal places)
GATE-PH 2015 QUESTION PAPER 154
1 for t 0
51. The Heaviside function is defined as H (t ) and its Fourier transform is given by –2i/. The
1 for t 0
1 1 1
Fourier transform of H t H t is
2 2 2
sin cos
2 2
(a) (b) (c) sin (d) 0
2
2 2
52. Consider the circuit shown in the figure, where RC = 1. For an input signal Vi shown below, choose the correct
V0 from the options:
R
Vi
C
Vi
Vo
1
t
1 2 3
Vo Vo
1 1
(a) t (b) t
1 2 3 1 2 3
–1 –1
Vo Vo
0.1 1
(c) t (d) t
1 2 3 1 2 3
–0.1
GATE-PH 2015 QUESTION PAPER 155
1
53. Let the Hamiltonian for two spin- particles of equal masses m, momenta p1 and p2 and positions r1 and r2
2
1 2 1 2 1
be H p1 p2 m 2 r12 r22 k 1 2 , where 1 and 2 denote the corresponding Pauli
2m 2m 2
matrices, 0.1 eV and k 0.2 eV . If the ground state has net spin zero, then the energy (in eV )
is_______
54. The excitation wavelength of laser in a Raman effect experiment is 546 nm. If the Stokes line is observed at
552 nm, then the wave number of the anti-Stokes line (in cm–1) is_______
55. A monochromatic plane wave (wavelength = 600 nm) E0 exp i ( kz t ) is incident normally on a diffraction
grating giving rise to a plane wave E1 exp i k1 r t in the first order of diffraction. Here
1 3
E1 E0 and k1 k1 xˆ zˆ . The period (in m) of the diffraction grating is_______(upto one deci-
2 2
mal place)
GATE-PH 2016 QUESTION PAPER 156
PHYSICS-PH
Q.1 – Q.25 : Carry ONE mark each.
dy
1. Consider the linear differential equation xy . If y 2 at x 0, then the value of y at x 2 is given
dx
by
(a) e 2 (b) 2e 2 (c) e2 (d) 2e 2
2. Which of the following magnetic vector potentials gives rise to a uniform magnetic field B0 kˆ ?
(a) B0 zkˆ (b) B0 xjˆ
B0 B0 ˆ ˆ
(c) ( yiˆ xjˆ) (d) ( yi xj )
2 2
3. The molecule 17O2 is
(a) Raman active but not NMR (nuclear magnetic resonance) active
(b) Infrared active and Raman active but not NMR active
(c) Raman active and NMR active
(d) Only NMR active
4. There are four electrons in the 3d shell of an isolated atom. The total magnetic moment of the atom in units
of Bohr magneton is ______________.
5. Which of the following transitions is NOT allowed in the case of an atom, according to the electric dipole
radiation selection rule?
(a) 2s 1s (b) 2 p 1s (c) 2 p 2 s (d) 3d 2 p
6. In the SU(3) quark model, the triplet of mesons , 0 , – has
(a) Isospin = 0, Strangeness = 0 (b) Isospin = 1, Strangeness = 0
(c) Isospin = 1/2, Strangeness = +1 (d) Isospin = 1/2, Strangeness = –1
7. The magnitude of the magnetic dipole moment associated with a square shaped loop carrying a steady
current I is m. If this loop is changed to a circular shape with the same current I passing through it, the
pm
magnetic dipole moment becomes . The value of p is _____________
8. The total power emitted by a spherical black body of radius R at a temperature T is P1. Let P2 be the
total power emitted by another spherical black body of radius R/2 kept at temperature 2T. The ratio, P1 / P2
is _________. (Give your answer upto two decimal places).
9. The entropy S of a system of N spins, which may align either in the upward or in the downward direction,
is given by S k B N [ p ln p (1 p) ln(1 p)] . Here k B is the Boltzmann constant. The probability of
alignment in the upward direction is p. The value of p, at which the entropy is maximum, is _____. (Give
your answer upto one decimal place).
10. For a system at constant temperature and volume, which of the following statements is correct at equilibrium?
(a) The Helmholtz free energy attains a local minimum
(b) The Helmholtz free energy attains a local maximum
(c) The Gibbs free energy attains a local minimum
(d) The Gibbs free energy attains a local maximum
GATE-PH 2016 QUESTION PAPER 157
11. N atoms of an ideal gas are enclosed in a container of volume V. The volume of the container is changed
to 4V, while keeping the total energy constant. The change in the entropy of the gas, in units of Nk B ln 2,
is _______, where k B is the Boltzmann constant.
12. Which of the following is an analytic function of z everywhere in the complex plane?
* 2
(a) z 2 (b) ( z ) (c) | z |2 (d) z
13. In a Young’s double slit experiment using light, the apparatus has two slits of unequal widths. When only
slit-1 is open, the maximum observed intensity on the screen is 4I0. When only slit-2 is open, the maximum
observed intensity is I0. When both the slits are open, an interference pattern appears on the screen. The
ratio of the intensity of the principal maximum to that of the nearest minimum is _____________.
14. Consider a metal which obeys the Sommerfeld model exactly. If EF is the Fermi energy of the metal at
T = 0K and RH is its Hall coefficient, which of the following statements is correct?
(a) RH EF3/ 2 (b) RH EF2/3
(c) RH EF3/ 2 (d) RH is independent of EF
15. A one-dimensional linear chain of atoms contains two types of atoms of masses m1 and m2 (where m2>m1),
arranged alternately. The distance between successive atoms is the same. Assume that the harmonic
approximation is valid. At the first Brillouin zone boundary, which of the following statements is correct?
(a) The atoms of mass m2 are at rest in the optical mode, while they vibrate in the acoustical mode.
(b) The atoms of mass m1 are at rest in the optical mode, while they vibrate in the acoustical mode.
(c) Both types of atoms vibrate with equal amplitudes in the optical as well as in the acoustical modes.
(d) Both types of atoms vibrate, but with unequal, non-zero amplitudes in the optical as well as in the
acoustical modes.
16. Which of the following operators is Hermitian?
d d2 d2 d3
(a) (b) (c) i (d)
dx dx 2 dx 2 dx 3
17. The kinetic energy of a particle of rest mass m0 is equal to its rest mass energy. Its momentum in units of
m0 c, where c is the speed of light in vacuum, is _________. (Give your answer upto two decimal places)
18. The number density of electrons in the conduction band of a semiconductor at a given temperature is
2×1019 m–3. Upon lightly doping this semiconductor with donor impurities, the number density of conduction
electrons at the same temperature becomes 4×1020 m–3. The ratio of majority to minority charge carrier
concentration is _______________.
19. Two blocks are connected by a spring of spring constant k. One block has mass m and the other block
has mass 2m. If the ratio k/m = 4s–2, the angular frequency of vibration of the two block spring system
in s–1 is ___________. (Give your answer upto two decimal places).
20. A particle moving under the influence of a central force F ( r ) kr (where r is the position vector of the
particle and k is a positive constant) has non-zero angular momentum. Which of the following curves is a
possible orbit for this particle?
(a) A straight line segment passing through the origin (b) An ellipse with its center at the origin
(c) An ellipse with one of the foci at the origin (d) A parabola with its vertex at the origin
21. Consider the reaction 54
25 Mn e 54
24 Cr X . The particle X is
(a) (b) ve (c) n (d) 0
GATE-PH 2016 QUESTION PAPER 158
22. The scattering of particles by a potential can be analyzed by Born approximation. In particular, if the
scattered wave is replaced by an appropriate plane wave, the corresponding Born approximation is known
as the first Born approximation. Such an approximation is valid for
(a) large incident energies and weak scattering potentials
(b) large incident energies and strong scattering potentials
(c) small incident energies and weak scattering potentials
(d) small incident energies and strong scattering potentials
23. Consider an elastic scattering of particles in l 0 states. If the corresponding phase shift 0 is 90º and the
magnitude of the incident wave vector is equal to 2 fm–1 then the total scattering cross section in units
of fm2 is ___________.
24. A hydrogen atom is in its ground state. In the presence of a uniform electric field E E0 zˆ, the leading order
change in its energy is proportional to ( E0 ) n . The value of the exponent n is ___________
25. A solid material is found to have a temperature independent magnetic susceptibility, C . Which of the
following statements is correct?
(a) If C is positive, the material is a diamagnet
(b) If C is positive, the material is a ferromagnet
(c) If C is negative, the material could be a type I superconductor
(d) If C is positive, the material could be a type I superconductor
Q.26 – Q.55 : Carry TWO marks each.
26. An infinite, conducting slab kept in a horizontal plane carries a uniform charge density . Another infinite
slab of thickness t, made of a linear dielectric material of dielectric constant k, is kept above the conducting
slab. The bound charge density on the upper surface of the dielectric slab is
( k 2) ( k 1)
(a) (b) (c) (d)
2k k 2k k
27. The number of spectroscopic terms resulting from the L.S. coupling of a 3p electron and a 3d electron is
__________
28. Which of the following statements is NOT correct?
(a) A deuteron can be disintegrated by irradiating it with gamma rays of energy 4 MeV.
(b) A deuteron has no excited states
(c) A deuteron has no electric quadrupole moment
(d) The 1S0 state of deuteron cannot be formed
29. If s1 and s2 are the spin operators of the two electrons of a He atom, the value of s1 s2 for the ground
state is
3 2 3 2 1 2
(a) (b) (c) 0 (d)
2 4 4
30. A two-dimensional square rigid box of side L contains six non-interacting electrons at T 0K . The mass
2 2
of the electron is m. The ground state energy of the system of electrons, in units of is _______
2mL2
GATE-PH 2016 QUESTION PAPER 159
31. An alpha particle is accelerated in a cyclotron. It leaves the cyclotron with a kinetic energy of 16 MeV.
The potential difference between the D electrodes is 50 kilovolts. The number of revolutions the alpha
particle makes in its spiral path before it leaves the cyclotron is _______
32. Let Vi be the ith component of a vector field V , which has zero divergence. If j / x j , the expression
for ijk lmk j lVm is equal to
2 2
(a) j kVi (b) j kVi (c) jVi (d) jVi
1 2 1 2
33. The direction of f for a scalar field f ( x, y, z ) x xy z at the point P (1, 1, 2) is
2 2
( ˆj 2kˆ) ( ˆj 2kˆ) ( ˆj 2kˆ) ( ˆj 2kˆ)
(a) (b) (c) (d)
5 5 5 5
34. x , y and z are the Pauli matrices. The expression 2 x y y x is equal to
(a) 3i z (b) i z (c) i z (d) 3i z
35. A particle of mass m = 0.1 kg is initially at rest at origin. It starts moving with a uniform acceleration
a 10iˆ ms–2 at t 0 . The action S of the particle, in units of J-s, at t 2s is _________. (Give your
answer upto two decimal places).
36. A periodic function f(x) of period 2 is defined in the interval ( x ) as:
1, x 0
f ( x)
1, 0 x
The appropriate Fourier series expansion for f ( x ) is
(a) f ( x) (4 / )[sin x (sin 3 x) / 3 (sin 5 x) / 5 ...]
(b) f ( x) (4 / )[sin x (sin 3 x) / 3 (sin 5 x) / 5 ...]
(c) f ( x) (4 / )[cos x (cos 3 x) / 3 (cos 5 x ) / 5 ...]
(d) f ( x ) (4 / )[cos x (cos 3x ) / 3 (cos 5 x ) / 5 ...]
37. Atoms, which can be assumed to be hard spheres of radius R, are arranged in an fcc lattice with lattice
constant a, such that each atom touches its nearest neighbours. Take the center of one of the atoms as the
origin. Another atom of radius r (assumed to be hard sphere) is to be accommodated at a position
(0, a / 2, 0) without distorting the lattice. The maximum value of r / R is ____________. (Give your
answer upto two decimal places).
38. In an inertial frame of reference S, an observer finds two events occurring at the same time at co-ordinates
x1 0 and x2 d . A different inertial frame S moves with velocity v with respect to S along the positive
x-axis. An observer in S also notices these two events and finds them to occur at times t1 and t2 and
1
at positions x1 and x2 , respectively. If t t2 t1, x x2 x1 and , which of the following
v2
1 2
c
statements is true?
(a) t 0, x d (b) t 0, x d /
(c) t vd / c 2 , x d (d) t vd / c 2 , x d /
GATE-PH 2016 QUESTION PAPER 160
39. The energy vs. wave vector ( E k ) relationship near the bottom of a band for a solid can be approximated
as E A(ka )2 B(ka) 4 , where the lattice constant a 2.1Å. The values of A and B are 6.3 × 10–19J
and 3.2 × 10–20 J, respectively. At the bottom of the conduction band, the ratio of the effective mass of
the electron to the mass of free electron is ________. (Give your answer upto two decimal places)
(Take = 1.05 × 10–34 J-s, mass of free electron = 9.1 × 10–31kg)
40. The electric field component of a plane electromagnetic wave travelling in vacuum is given by
E ( z, t ) E0 cos(kz t )iˆ . The Poynting vector for the wave is
(a) (c0 / 2) E02 cos2 (kz t ) ˆj (b) (c0 / 2) E02 cos2 (kz t )kˆ
(c) c0 E02 cos 2 (kz t ) ˆj (d) c0 E02 cos 2 (kz t )kˆ
41. Consider a system having three energy levels with energies 0, 2 and 3 , with respective degeneracies of
2, 2 and 3. Four bosons of spin zero have to be accommodated in these levels such that the total energy
of the system is 10 . The number of ways in which it can be done is ___________.
42. The Lagrangian of a system is given by
1 2 2
L ml sin 2 2 mgl cos , where m, l and g are constants.
2
Which of the following is conserved?
(a) sin 2 (b) sin (c) (d)
sin sin 2
43. Protons and -particles of equal initial momenta are scattered off a gold foil in a Rutherford scattering
experiment. The scattering cross sections for proton on gold and -particle on gold are p and
respectively. The ratio / p is ___________.
44. For the digital circuit given below, the output X is
A
X
B
C
(a) A B C (b) A ( B C ) (c) A ( B C ) (d) A ( B C )
45. The Fermi energies of two metals X and Y are 5eV and 7eV and their Debye temperatures are 170 K
and 340 K, respectively. The molar specific heats of these metals at constant volume at low temperatures
can be written as (CV ) X X T AX T 3 and (CV )Y Y T AY T 3 , where and A are constants. Assuming
that the thermal effective mass of the electrons in the two metals are same, which of the following is correct?
X 7 AX X 7 AX 1
(a) 5 , A 8 (b) 5 , A 8
Y Y Y Y
X 5 AX 1 X 5 AX
(c) 7 , A 8 (d) 7 , A 8
Y Y Y Y
GATE-PH 2016 QUESTION PAPER 161
46. A two-level system has energies zero and E. The level with zero energy is non-degenerate, while the level
with energy E is triply degenerate. The mean energy of a classical particle in this system at a temperature
T is
Ee E / kBT Ee E / kBT 3Ee E / kBT 3Ee E / kBT
(a) (b) (c) (d)
1 3e E / kBT 1 e E / kBT 1 e E / kBT 1 3e E / kBT
47. A particle of rest mass M is moving along the positive x-direction. It decays into two photons 1 and 2
as shown in the figure. The energy of 1 is 1GeV and the energy of 2 is 0.82 GeV. The value of M (in
units of GeV/c2) is _____________. (Give your answer upto two decimal places)
1
M 45º
60º
2
48. If x and p are the x components of the position and the momentum operators of a particle respectively,
the commutator [ x 2 , p 2 ] is
(a) i( xp px) (b) 2i( xp px)
(c) i( xp px) (d) 2i( xp px )
49. The x-y plane is the boundary between free space and a magnetic material with relative permeability r .
The magnetic field in the free space is Bx iˆ Bz kˆ . The magnetic field in the magnetic material is
(a) Bx iˆ Bz kˆ (b) Bx iˆ r Bz kˆ
1 ˆ
(c) Bx i Bz kˆ (d) r Bxiˆ Bz kˆ
r
50. Let l , m be the simultaneous eigenstates of L2 and Lz. Here L is the angular momentum operator with
Cartesian components ( Lx , Ly , Lz ), l is the angular momentum quantum number and m is the azimuthal
quantum number. The value of 1, 0 | ( Lx iLy ) |1, 1 is
(a) 0 (b) (c) 2 (d) 3
51. For the parity operator P, which of the following statements is NOT true?
(a) P† P (b) P 2 P (c) P 2 I (d) P † P 1
52. For the transistor shown in the figure, assume VBE 0.7V and dc 100. If Vin 5V,, Vout (in Volts) is
_________. (Give your answer upto one decimal place).
GATE-PH 2016 QUESTION PAPER 162
10V
3k
Vin Vout
200k
1k
53. The state of a system is given by
1 2 2 3 3
where 1 , 2 and 3 form an orthonormal set. The probability of finding the system in the state 2
is __________. (Give your answer upto two decimal places)
54. According to the nuclear shell model, the respective ground state spin-parity values of 15 17
8 O and 8 O nuclei
are
– – – – –
1 1 1 5 3 5 3 1
(a) , (b) , (c) , (d) ,
2 2 2 2 2 2 2 2
55. A particle of mass m and energy E, moving in the positive x direction, is incident on a step potential at
x 0, as indicated in the figure. The height of the potential is V0, where V0 E . At x x0 , where x0 0,
2m(V0 E )
the probability of finding the electron is 1/ e times the probability of finding it at x 0. If ,
2
the value of x0 is
V0
E
x=0 x = x0
2 1 1 1
(a) (b) (c) (d)
2 4
GATE-PH 2017 QUESTION PAPER 163
PHYSICS-PH
Q.1 – Q.25 : Carry ONE mark each.
1. In the nuclear reaction 13C6 ve 13 N 7 X , the particle X is
(a) an electron (b) an anti-electron (c) a muon (d) a pion
2. Two identical masses of 10 gm each are connected by a massless spring of spring constant 1 N/m. The non-
zero angular eigen frequency of the system is _______ rad/s. (up to two decimal places).
3. Consider a triatomic molecule of the shape shown in the figure below in three dimensions. The heat capacity of
this molecule at high temperature (temperature much higher than the vibrational and rotational energy scales of
the molecule but lower than its bond dissociation energies) is:
3 9
(a) kB (b) 3k B (c) kB (d) 6k B
2 2
4. For the Hamiltonian H a0 I b . where a0 R, b is a real vector,, I is the 2 2 identity matrix and are
the Pauli matrices, the ground state energy is
(a) b (b) 2a0 b (c) a0 b (d) a0
5. The Poisson bracket x, xp y ypx is equal to
(a) x (b) y (c) 2 px (d) p y
6. The wavefunction of which orbital is spherically symmetric:
(a) px (b) p y (c) s (d) d xy
7. A monochromatic plane wave in free space with electric field amplitude of 1 V/m is normally incident on a fully
reflecting mirror. The pressure exerted on the mirror is ______ 1012 Pa. (up to two decimal places)
( 0 8.854 10 12 F/m) .
8. The electronic ground state energy of the Hydrogen atom is –13.6 eV. The highest possible electronic energy
eigenstate has an energy equal to
(a) 0 (b) 1 eV (c) +13.6 eV (d)
2x
9. Consider a one-dimensional lattice with a weak periodic potential U ( x) U 0 cos . The gap at the edge
a
of the Brillouin zone k is:
a
U0 U0
(a) U 0 (b) (c) 2U 0 (d)
2 4
10. Identical charge q are placed at five vertices of a regular hexagon of side a. The magnitude of the electric field
and the electrostatic potential at the centre of the hexagon are respectively
q q q 5q 5q 5q
(a) 0, 0 (b) 4 a 2 , 4 a (c) 4 a 2 , 4 a (d) 2
,
0 0 0 0 40 a 4 0 a
GATE-PH 2017 QUESTION PAPER 164
11. A reversible Carnot engine is operated between temperatures T1 and T2 T2 T1 with a photon gas as the
working substance. The efficiency of the engine is
3/4 4/3
3T1 T1 T1 T1
(a) 1 4T (b) 1 T (c) 1 (d) 1
2 2 T2 T2
12. The best resolution that a 7 bit A/D converter with 5V full scale can achieve is _______ mV. (up to two
decimal places).
2
1 dq 1 dq
13. If the Lagrangian L0 m m2 q 2 is modified to L L0 aq , which one of the following is
2 dt 2 dt
TRUE?
(a) Both the canonical momentum and equation of motion do not change
(b) Canonical momentum changes, equation of motion does not change
(c) Canonical momentum does not change, equation of motion changes
(d) Both the canonical momentum and equation of motion change
14. A parallel plate capacitor with square plates of side 1 m separated by 1 micro meter is filled with a medium of
dielectric constant of 10. If the charges on the two plates are 1 C and -1 C, the voltage across the capacitor is
______kV. (up to two decimal places). ( 0 8.854 10 12 F/m)
dz
15. The contour integral evaluated along a contour going from to along the real axis and closed
1 z 2
in the lower half-plane by a half circle is equal to ________. (up to two decimal places).
16. In the figure given below, the input to the primary of the transformer is a voltage varying sinusoidally with time.
The resistor R is connected to the centre tap of the secondary. Which one of the following plots represents the
voltage across the resistor R as a function of time?
V V
0 0
(a) (b)
t t
V V
0 0
(c) (d)
t t
17. Light is incident from a medium of refractive index n 1.5 onto vacuum. The smallest angle of incidence for
which the light is not transmitted into vacuum is ________ degrees. (up to two decimal places).
18. Electromagnetic interactions are :
(a) C conserving (b) C non-conserving but CP conserving
(c) CP non-conserving but CPT conserving (d) CPT non-conserving
GATE-PH 2017 QUESTION PAPER 165
19. The Compton wavelength of a proton is _____ fm. (up to two decimal places).
(m p 1.67 1027 kg, h 6.626 1034 Js, e 1.602 1019 C, c 3 108 ms 1 )
20. Which one of the following conservation laws is violated in the decay
(a) Angular momentum (b) Total Lepton number
(c) Electric charge (d) Tau number
21. The coefficient of eikx in the Fourier expansion of u x A sin 2 ax for k 2a is
(a) A/4 (b) A/4 (c) A/2 (d) A/2
22. The phase space trajectory of a free particle bouncing between two hard walls elastically in one dimension is
a
(a) straight line (b) parabola (c) rectangle (d) circle
23. The atomic mass and mass density of Sodium are 23 and 0.968 g cm 3 , respectively. The number density of
valence electrons is ______ 1022 cm 3 . (Up to two decimal places.)
(Avogadro number, N A 6.022 1023 ).
24. The degeneracy of the third energy level of a 3-dimensional isotropic quantum harmonic oscillator is
(a) 6 (b) 12 (c) 8 (d) 10
p2 1 2
25. A one dimensional simple harmonic oscillator with Hamiltonian H 0 kx is subjected to a small per-
2m 2
turbation, H1 x x 3 x 4 . The first order correction to the ground state energy is dependent on
(a) only (b) and (c) and (d) only
Q.26 – Q.55 : Carry TWO marks each.
26. Three charges (2 C, –1 C, –1 C) are placed at the vertices of an equilateral triangle of side 1m as shown in the
figure. The component of the electric dipole moment about the marked origin along the ŷ direction is ___ cm.
y
2C
1m
-1C -1C
0 1.5m x
27. An object travels along the x-direction with velocity c/2 in a frame O. An observer in a frame O sees the same
object travelling with velocity c / 4 . The relative velocity of O with respect to O in units of c is _______. (up
to two decimal places).
28. The energy density and pressure of a photon gas are given by u aT 4 and P u /3 , where T is the tempera-
ture and a is the radiation constant. The entropy per unit volume is given by aT 3 . The value of is _____.
(up to two decimal places).
29. A person weighs w p at Earth’s north pole and we at the equator. Treating the Earth as a perfect sphere of
radius 6400 km, the value 100 ( w p we )/wp is _____. (up to two decimal places).
2
(Take g 10 ms ).
GATE-PH 2017 QUESTION PAPER 166
30. The minimum number of NAND gates required to construct an OR gate is:
(a) 2 (b) 4 (c) 5 (d) 3
0.5 1
31. The total energy of an inert-gas crystal is given by E ( R )
(in eV), where R is the inter-atomic
R12 R 6
spacing in Angstroms. The equilibrium separation between the atoms is ________ Angstroms. (up to two
decimal places).
32. The imaginary part of an analytic complex function is v( x, y ) 2 xy 3 y . The real part of the function is zero
at the origin. The value of the real part of the function at 1+i is ______. (up to two decimal places).
33. Consider N non-interacting, distinguishable particles in a two-level system at temperature T. The energies of
the levels are 0 and , where 0 . In the high temperature limit (k BT ) , what is the population of
particles in the level with energy ?
N N 3N
(a) (b) N (c) (d)
2 4 4
34. For the transistor amplifier circuit shown below with R1 10 k, R 2 10 k, R 3 1 k, and 99 . Ne-
glecting the emitter diode resistance, the input impedance of the amplifier looking into the base for small ac
signal is _______ k . (up to two decimal places).
VCC
R1
C
B
Vin
E Vout
R2 R3
35. Which one of the following gases of diatomic molecules is Raman, infrared, and NMR active?
(a) 1 H 1H (b) 12 C 16 O (c) 1 H 35 Cl (d) 16 O 16 O
36. Let X be a column vector of dimension n 1 with at least one non-zero entry. The number of non-zero
eigenvalues of the matrix M XX T is
(a) 0 (b) n (c) 1 (d) n 1
37. A free electron of energy 1 eV is incident upon a one-dimensional finite potential step of height 0.75 eV. The
probability of its reflection from the barrier is _______ (up to two decimal places).
38. An infinite solenoid carries a time varying current I (t ) At 2 , with A 0 . The axis of the solenoid is along the
ẑ direction. rˆ and ˆ are the usual radial and polar directions in cylindrical polar coordinates.
B Br rˆ Bˆ Bz zˆ is the magnetic field at a point outside the solenoid. Which one of the following state-
ments is true?
(a) Br 0, B 0, Bz 0 (b) Br 0, B 0, Bz 0
(c) Br 0, B 0, Bz 0 (d) Br 0, B 0, Bz 0
GATE-PH 2017 QUESTION PAPER 167
39. Consider two particles and two non-degenerate quantum levels 1 and 2. Level 1 always contains a particle.
Hence, what is the probability that level 2 also contains a particle for each of the two cases:
(i) when the two particles are distinguishable and (ii) when the two particles are bosons?
(a) (i) 1/2 and (ii) 1/3 (b) (i) 1/2 and (ii) 1/2 (c) (i) 2/3 and (ii) 1/2 (d) (i) 1 and (ii) 0
a
40.
The real space primitive lattice vectors are a1 axˆ and a1 xˆ 3 yˆ . The reciprocal space unit vectors
2
b1 and b2 for this lattice are, respectively
2 yˆ 4 2 yˆ 4
(a) xˆ and yˆ (b) xˆ and yˆ
a 3 a 3 a 3 a 3
2 4 xˆ 2 4 xˆ
(c) xˆ and yˆ (d) xˆ and yˆ
a 3 a 3 a 3 a 3
41. The geometric cross-section of two colliding protons at large energies is very well estimated by the product of
the effective sizes of each particle. This is closest to
(a) 10 b (b) 10 mb (c) 10 µb (d) 10 pb
42. A uniform volume charge density is placed inside a conductor (with resistivity 102 m ). The charge density
becomes 1/(2.718) of its original value after time ______ femto seconds. (up to two decimal places)
(0 8.854 10 12 F/m)
43. Water freezes at 0ºC at atmospheric pressure (1.01 105 Pa) . The densities of water and ice at this tempera-
ture and pressure are 1000 kg/m3 and 934 kg/m3 respectively. The latent heat of fusion is 3.34 105 J/kg . The
pressure required for depressing the melting temperature of ice by 10ºC is ____ GPa. (up to two decimal
places)
2
44. The integral x 2 e x dx is equal to _______. (up to two decimal places).
0
45. J P for the ground state of the 13
C6 nucleus is
3 3 1
(a) 1 (b) (c) (d)
2 2 2
46. A uniform solid cylinder is released on a horizontal surface with speed 5 m/s without any rotation (slipping
without rolling). The cylinder eventually starts rolling without slipping. If the mass and radius of the cylinder are
10 gm and 1 cm respectively, the final linear velocity of the cylinder is _______ m/s. (up to two decimal places)
47. Consider a one-dimensional potential well of width 3 nm. Using the uncertainty principle x.p /2 , an
estimate of the minimum depth of the well such that it has at least one bound state for an electron is
( me 9.31 10 31 kg , h 6.626 10 34 J s, e 1.602 10 19 C ) :
(a) 1 µeV (b) 1 meV (c) 1 eV (d) 1 MeV
48. Consider an ideal operational amplifier as shown in the figure below with R1 5 k, R 2 1 k , R L 100 k .
For an applied input voltage V 10 mV , the current passing through R 2 is _______ µA. (up to two decimal
places).
GATE-PH 2017 QUESTION PAPER 168
R1
V
RL
R2
49. The decays at rest to and . Assuming the neutrino to be massless, the momentum of the neutrino is
________ MeV/c. (up to two decimal places)
m x 139MeV/c 2 , m 105 MeV/c 2 .
50. Consider the differential equation dy / dx y tan( x) cos( x ) . If y (0) 0, y ( / 3) is ______. (up to two
decimal places).
51. Consider a metal with free electron density of 6 1022 cm 3 . The lowest frequency electromagnetic radiation
to which this metal is transparent is 1.38 1016 Hz . If this metal had a free electron density of 1.8 1023 cm 3
instead, the lowest frequency electromagnetic radiation to which it would be transparent is _______ 1016 Hz .
(up to two decimal places).
52. Using Hund’s rule, the total angular momentum quantum number J for the electronic ground state of the nitro-
gen atom is
(a) 1/2 (b) 3/2 (c) 0 (d) 1
53. Consider a 2-dimensional electron gas with a density of 1019 m 2 . The Fermi energy of the system is _____eV..
(up to two decimal places).
me 9.311031 kg, h 6.626 1034 Js, e 1.602 1019 C
54. Which one of the following operators is Hermitian?
(a) i
p x x
2
x 2 px
(b) i
p x
x
2
x 2 px
(c) eipx a (d) e ip x a
2 2
55. Positronium is an atom made of an electron and a positron. Given that the Bohr radius for the ground state of
the Hydrogen atom to be 0.53 Angstroms, the Bohr radius for the ground state of positronium is ______
Angstroms. (up to two decimal places).
SOLUTION
2001 – 2017
FDFDFD
GATE PH-2001 SOLUTION 169
OBJECTIVE QUESTION
1.1. r ˆ
ndS
r dV (using Divergence Theorem)
S V
3 dV as r xiˆ yjˆ zkˆ 3
V
3V
Correct option is (d)
1.2. We know that,
If A is Hermitian then it is equal to its complex conjugate transpose i.e. M † M
If A is anti-Hermitian then it is equal to its complex conjugate transpose with negative sign i.e. M † M
If A is unitary matrix then the conjugate transpose is equal to its inverse i.e. M † M 1
If A is orthogonal then the transpose of the matrix is equal to its inverse i.e. M T M 1
†
i A† A i † A A† i A† A
Correct option is (a)
1.3. f z z10 is analytic everywhere in the complex argand plane.
So, according to Cauchy integral theorem,
10
z dz 0
C
Correct option is (a)
1 1 1
1.4. A 1,1, 0 ; B 0,1,1 ; C 1, 0,1
2 2 2
1 1
0
A1 B1 C1 2 2
1 1 1 1 1 1 1 1
A2 B2 C2 0 2 0 0 2 0 2 0 0
Now, A B C A
3
B3 C3
2 2 2 2 2
1 1
0
2 2
So, A, B, C are linearly independent.
Correct option is (b)
1.5. We have Lagrangian,
1 2 2 2 µ
L
2
r r
r
L
Canonical momentum corresponding to coordinates , P r 2
L
Linear momentum, Pr r
r
GATE PH-2001 SOLUTION 170
d L L
Equation of motion, 0 P r 2 constant
dt
Since, L does not depends on time explicitly. So, total energy of the system is conserved.
Correct option is (c)
1.6. Lagrangian transform,
x vt vx 1
x' x vt y ' y, z ' z , t ' t 2 ,
2
1 v / c 2 c 1 v2 / c2
x ' t ' v
x x ' x t ' x x ' c 2 t '
2 x ' t '
2
x x ' x x t ' x x
2 v 2 2 v 2 2 v 2 2 2
2
x '2 c 2 x ' t ' c 2 t ' x ' c 4 t '2
t '
And, v = v
t x ' t ' t x ' t ' x ' t '
2 x ' t '
2
t x ' t t t ' t t
2 2
2
2
2
2
2
v 2 2 v v
x '2 xt ' t ' x ' t '2
2 1 2 2 v 2 2 2 2 2
4
x 2 c 2 t 2 x '2 c t '2 c 2 t '2
2 1 2
x '2 c 2 t '2
2 2 2 2
x 2 t 2 x '2 t '2
Correct option is (b)
1.7. We have Lagrangian,
1 2 a2 2
L
2
1 22 32
2
1 22 32 13
1 2
2
Kinetic energy = 1 2 3
2 2
a2 2
Potential energy =
2
1 22 32 13
Corresponding matrices,
1 0 0 1 0 1
Tˆ 0 1 0 , Vˆ a 0 1 0
2
0 0 1 1 0 1
GATE PH-2001 SOLUTION 171
For frequency of normal modes
det Vˆ 2Tˆ 0
a2 2 0 a 2
0 a2 2 0 0
a2 0 a2 2
2 2 2
a 2 a 2 a 2 2 0 a a a 0
2 2 2 4
a2 2 a 2 2
2
a 2
2 0 and a 2 2 a4
2 a2 a 2 2 a 2
2 a2 2 0 and 2 2a 2
1.8. According to Coulomb law, the force on Q2 due to Q1 is
1 Q1Q2
F21 rˆ21
4 0 r21
And the force on Q1 due to Q2 is
1 Q1Q2
F12 rˆ12
4 0 r12
F12 F21
Correct answer is (b)
a
1.9. Magnetic field at distance from finite length wire is given by
2
µ0 I
B1 sin 1 sin 2 outward direction to the page
4 a /2
1 2
a
Since, 1 2 45º
µI µI 1
B1 0 sin 45º zˆ 0 zˆ
a a 2
Therefore, the total magnetic field at the centre of square is given by
4µ I 1 2 2 µ0 I
B B1 B2 B3 B4 4 B1 0 zˆ zˆ
a 2 a
Correct option is (b)
1.10. The flux through the circular disc is
r 2 B t r 2 B0e t
d
Therefore, the induce emf r 2 B0 e t
dt
Correct option is (a)
GATE PH-2001 SOLUTION 172
1.11. We know that when we introduce some charge on the conducting the charges is uniformly distributed on
the surface and produce a equipotential surface. So, accoding to Gauss law we can say the electric field
inside the conductor will be zero.
Now, if we say the electric field out side the conductor is maing a angle with surface at the point A.
So, the tangent component of the electric field is E cos . So, the potential difference betweeen A and B
points is E cos AB .
But the conducting surface is equipotential surface that says that
E cos AB 0
A B
cos 0 E and AB can ' t be zero
2
So, the electric field out side the conducting surface perpendicular to the conducting surface.
Correct option is (a)
1.12. Kinetic energy of the particle is a function momentum and potential energy of the particle is a function of
position. According to Heisenberg’s uncertainty principle, position and momentum of the particle cannot be
measured simultaneously accurately. Therefore, kinetic energy and potential energy of the particle cannot si-
multaneously have sharp values.
For a spinless particle of mass m moving in central potential V r , the Hamiltonian of the system can be
written as
ˆ 2 2 L2
H r V r
2mr 2 r r 2mr 2
Therefore, Hˆ , Lˆ 0 i.e. Energy of the particle and square of the orbital angular momentum can simulta-
2
neously have sharp values.
Correct option is (a)
1.13. For a particle moving in the range , , the acceptable wave function should be finite, single valued and
continuous everywhere in space and the wave function should be square integrable in nature.
3
x A tan x is not finite at x , ,..... Therefore, the function is not acceptable.
2 2
x B cos x is not square integrable as
2 2 B2
2
x dx B cos x dx
2 1 cos 2 x dx Finite Quantity
D
x C exp 2 is not finite at x = 0 (since D < 0)
x
x E x exp Fx 2 is finite, single valued and continuous everywhere in space. The function is also square
integrable in nature as
2
x dx E
2 2
x exp 2 Fx 2 dx Finite Quantity
Correct option is (d)
GATE PH-2001 SOLUTION 173
1.14. First order energy shift due to perturbation =
2 2
En n | H p | n n | aˆ †aˆ
1
| n n | aˆ † aˆ | n n2 n | n n2
Correct option is (c)
1.15. If the average distance between two particles is large compared to their de Broglie wavelengths, then the wave
function of those particles will not overlap with each other i.e. two particles will be distinguishable.
Correct option is (a)
1.16. The density of state in 3-D is
V 4 p 2 dp
g ( p ) dp
h3
dE
For photon gas, E pc dp
c
V 2 4 E 2 dE
Therefore, g ( E ) dE
h3c 2 c
Photons have two polarization states (One for each dimension transverse to the direction of propagation). So,
g(E) should be multiplied by a factor of 2.
g ( E) E 2
Correct option is (d)
1.17. According to Moseley’s law :
( Z ) 2 ... (i)
Here, v characteristic frequency emitted by an element
Z atomic number of element
Here, for strong K line
1
(27 1) 2 ... (ii)
0.1785 nm
Here, we have used 1 as screening factor
and for weak K line
1
(Z 1)2 ... (iii)
0.1930 nm
Now, from equations (ii) and (iii), we have
1
0.1785 nm 2
(26)
1 ( Z 1) 2
0.1930 nm
0.1785 262
(Z 1) 2 625 z 1 25 26
0.1930
Hence, correct option is (b).
GATE PH-2001 SOLUTION 174
1.18. Hall voltage is defined as :
BI
VH
tne
Where, B is magnetic field , I is current, t is the thickness, n is charge density and e is electron charge.
0.2 1 103
200 10 6 10 23 1.6 1019
0.2 101
0.625 10 4 62.5 µV.
200 1.6
Here, in this question, the silicon is doped with phosphorus (P) which is pentavalent. Therefore, the Hall
voltage will be produced by electrons and will be of negative type.
Hence, correct option is (d).
1.19. The probability in Fermi Dirac statistics is
1
p( E ) E EF , where E EF 0.2 eV and T 700 K .
k BT
e 1
1 1
So, p( E ) 0.035 3.5 %.
0.2 1.6 1019 2 1600
exp 23 1 exp 1
1.38 10 700 138 7
Correct option is (c).
1.20. Here, interatomic distance, d 0.3 nm
Wavelength of X-ray is, 0.03 nm
From Bragg’s law : n 2 d sin
sin ( smallest scattering is considered hence we have taken n = 1)
2d
0.03 nm 0.03 1
sin
2 0.3 nm 0.6 20
1
sin 1 sin 1 (0.05)
20
2.86 2.9
Hence, correct option is (a).
1.21. Since, we know that the infra-red absorption can be observed only on those molecules which have permanent
dipole moment. In the above options, only HCl has permanent dipole moment.
Hence, correct option is (c).
1.22. Charge conjugation because both of these nuclear reactions are outcome of strong interaction. Also, K meson
is present as a product of these reactions.
Correct option is (c)
GATE PH-2001 SOLUTION 175
1.23. RAM (Random Access memory)
ROM (Read Only Memory)
Both are the computer memories.
Correct option is (b)
1.24. For dc-biasing we apply input loop in forwards biased and output in reverse biased. When input loop is open
still these have current in output loop. Which is represent by ICBO it is called leakage current also it is known as
reverse saturation current of a reverse-biased diode.
n p n
IE IC
ICBO
IB
So, current will be flow from collector to base due to motion of electron from based to collectors.
Correct option is (d)
1.25. As the temperature increase the number of free electron at the conduction band increase. So, conductivity of
the semiconductor increase. So, in the circuit current will increase as the temperature of the semiconductor is
raised.
Correct option is (a)
2.1. A xeˆx yeˆy zeˆz
2 2 2
2 A 2 xeˆx yeˆy zeˆz 2 xeˆx yeˆy zeˆz 2 xeˆx yeˆy zeˆz = 0
x y z
Correct option is (c)
2.2. Laurent series expansion of f z about z = 0, will be
sin z 1 z3 z5 z 7
f z 6 z ........
z6 z 3! 5! 7!
1 1 1 11
......
z 5 3! z 3 5! z
1 1
Residue of f z = coefficient of
z 5!
Correct option is (b)
1 ikx
2.3. F f x g k f x e dx
2
1 ikx
F 2 f x F g k g k e dk
2
1 1
g k e ik x
dk f x F 1 g k g k e ikx
dx f x
2 2
GATE PH-2001 SOLUTION 176
2.4. xy C
xy
xy 0 y
yx
x y
The kinetic energy of the particle is
1 1 x 2 y 2 1 C2
T
2
m x 2 y 2 m x 2 2 mx 2 1 4
2 x 2 x
x
mgC
Potential energy, V mgy
x
Therefore, the Lagrangian of the particle is given by
1 2 C 2 mgC
L T V mx 1 4
2 x x
Correct option is (b)
2.5. q, f p qp f p
f
0 f p
p
f p
f p nf p p
p
f p e p
Correct option is (a)
2.6. X , Y x cos p sin , x sin p cos
cos sin x, x cos 2 x, p sin 2 p, x cos sin p, p
0 cos 2 i sin 2 i 0 i cos2 sin 2 i
Correct option is (c)
2.7. For a quantum particle of mass m confined to a square region in x-y plane, the wave function of the particle
should be zero at the boundary of the square for the sake of continuous nature of the wave function. Thus, the
admissible wave function of the particle should satisfy the following condition:
x 0, y x L, y x, y 0 x, y L 0
2 n x n y
Only x, y sin sin satisfy the above mentioned condition.
L L L
Correct option is (c)
2.8. L L Lx iLy Lx iLy L2x L2y i Ly Lx Lx Ly L2x L2y i Ly , Lx
L2 L2z i iLz L2 L2z Lz
Therefore, L L | L Lz Lz | l l 1 l l 2l
2 2 2 2 2 2
GATE PH-2001 SOLUTION 177
Correct option is (b)
2.9. For the given state of the particle to be normalized, the following condition should be satisfied:
x, t | x, t 1
C0 0 eiE0t / C1 1 eiE1t / ....... Cn n eiEnt / .......
C0 0 e iE0t / C1 1 e iE1t / ....... Cn n e iEnt / ....... 1
2 2 2
C0 0 | 0 C1 1 | 1 ............. Cn n | n ...... 1
2 2 2 2
C0 C1 ............. Cn ...... 1 C
n 0
n 1
Average energy of the particle in the given state will be
2 2 2 2
E | Hˆ | C0 E0 C1 E1 ............. Cn En ...... C n En
n 0
x, t will be an Eigen function of the Hamiltonian of the particle only if 0 x , 1 x ,...... n x ..... all have
same energy eigenvalues, i.e., the states are degenerate.
x, t is not an Eigen function of the momentum operator as
pˆ x x, t i x, t x, t
x
Correct option is (a)
2.10. Capacitance of coaxial cable is per unit length
2 r 0 2 3.5 8.85 1012
C 280.63 1012 F /m 280.63 nF / km
n b /a 0.02
n
0.01
Correct option is (a)
2.11. We know that magnetic field due to sheet carrying uniform current is given by
0
B K nˆ
2
0 K 4107 50
So, | B | 10106 Wb
2 2
Correct option is (d)
2.12. From figure we can say that there will be a net electric field along – ẑ direction and field is not uniform.
–q –q
So, some force will act on the dipole.
x O
If dipole is placed along z direction then torque
on it about the point O. P E PE sin180º 0 . +q +q
If the dipole is placed along x direction then potential energy. z
GATE PH-2001 SOLUTION 178
u P E PE cos 90º 0 , which is not minimum. Because minimum potential energy of dipole is –pE.
Correct option is (b)
Let E r , t E0 e
i k r t
2.13.
1 E r, t
2
2
We have, E r, t 2 2
'2 E r , t
c t
1
k 2 E 2 '2 E
2 (since, E0ei k r t 0 ]
c
k 2 c 2 2 '2
2 c 2 k 2 '2
Correct option is (b)
2.14. Free electron density of the copper wire is
8.94 103
n 1/ m3 8.51 1028 / m3
1.05 1025
We know that current density
I 1
J 6
106 A / m 2
A 1 10
Also, J nev , so, nev 106
106
v
8.51 1028 1.6 1019 7.4 10 m / sec
5
Correct option is (d)
2.15. For 2 p1/ 2 term ; J 1 , I 1
2 2
The allowed values of hyperfine structure quantum number are
F J I , J I 1 ,....., J I
1 1 1 1 1 1
, 1 ,....., 1, 0
2 2 2 2 2 2
1 1
Similarly, for 2 s1/2 term, J , I
2 2
Thus, F 1, 0
Now, selection rule F 0, 1 (0 0) will gives transitions as shown in figure.
F=1
2
p1/2
F=0
F=1
2
s1/2
F=0
Hence, total transitions are 3.
Hence, correct option is (a).
GATE PH-2001 SOLUTION 179
2.16. Intereaction potential can be given as
0
a
V (r ) br V(r) r
r
a
where, because nuclear forces are short range and decreases with decreasing distnace
r
br because Nuclear forces will be of repulsion type at long range.
Correct option is (b)
2.17. For (a), L = (–1 + 1) – (1 – 1) = (0) – (0) = 0
For (b), L = (1) – (1) = 0
For (c), L = (–1) – (1) = –2 0 violates lepton number conservation
For (d), L = (1) – (1) = 0
Correct option is (c)
2.18. For 3 p1 term, l 1, j 1, 2s 1 3 s 1
Therefore, Lande g-factor
j ( j 1) l (l 1) s ( s 1) 1(1 1) 1(1 1) 1(1 1) 2 2 2 1 3
g 1 1 1 1 .
2 j ( j 1) 2 1 (1 1) 4 2 2
Hence, correct option is (b).
2.19. Selection rules of rotational Raman spectrum,
J 0, 2
J 0 , corresponds to the Rayleigh line. However, J 2 , gives anti-Stokes line (+) and stokes line(–)
respectively.
The wave number of the rotation Raman lines
vRaman v F J 2 F J v B J 2 J 3 BJ J 1 F J BJ J 1
3
v 4B J
2
The rotation Raman shift
3
vRaman 4 B J
2
For J = 0, 1, 2, 3, ..., we get rotational Raman spectrum at distance 6B, 10B, 14B, 18B, ... with equidistant
separation 4B.
Correct option is (b)
GATE PH-2001 SOLUTION 180
2.20. The electronic transitions of 2 d5/ 2 2 p3/ 2 is shown in figure (below):
mj
5/2
3/2
2 1/2
d5/2
–1/2
–3/2
–5/2
3/2
2 1/2
p3/2
–1/2
–3/2
mj = +1 mj = 0 mj = –1
() () ()
The transitions with selection rules m j 0, 1 gives 12 Zeeman components in a weak field.
Hence, correct option is (c).
2.21. Total impedance of the circuit
1 1 1
z R jX L
Given : R 600 , X L 600
600 600 j 600
z z 424.26
600 j 600 2
Correct option is (c)
2.22. The pinch of voltage is given by
q N D a 2
Vp
2
q 1.6 1019 Columb ; N D Donor ions concentration 1021
a 2 106 m (Depletion width)
0 r rsi 11.7 to 12
0 8.854 10 12
2
1.6 1019 1021 2 10 6
Vp
2 8.854 1012 12
V p 3.01 Volt
dy dz
2.23. y z, 4 y z
dx dx
Equation can be expressed as
d
D 1 y z 0 ... (1) [where, D ]
dx
GATE PH-2001 SOLUTION 181
D 1 z 4 y 0 ... (2)
2
(1) × (D–1) D 1 y D 1 z 0
(2) × 1 4 y D 1 z 0
Subtracting we get,
________________________________________
D 12 y 4 y 0
D 2 y 2 Dy 3 y 0
d 2 y 2dy
3y 0
dx 2 dx
mx
Let the trial solution y e
m2 2m 3 0
m2 3m m 3 0
m 3 m 1 0
m 3, 1
y A e3x Be x
dy
So, z y z Ae3 x Be x 3 Ae3 x Be x
dx
2 Ae3 x 2 Be x
Correct option is (a)
2.24. u x, y , z f x i y vt g x i y vt
Let S x i y vt and V x i y vt
u f S g V f g
x S x V x S V
2u f g S f g V
x 2 S S V x V s V x
2 f 2 g 2 f 2 g
2
S S V V S V 2
u f S g V f g
i i
y S y V y S V
2 u f g S f g V
2
i i i i
y S S V y V S V y
2 f 2
2 g
2
2 f
2
2 g
2
S 2 S V V S V 2
GATE PH-2001 SOLUTION 182
u f S g V f g
v v
t S t V t S V
2u f g S f g g
2
v v v v
t S S V t V S V t
2 f
2
2
2 g
2
2 f
2
2 g
v v v v
S 2 S V V S V 2
2u 2u 1 2 u
0
x 2 y 2 c 2 t 2
2 f 2 g 2 f 2 g 2
2 f
2
2 g
2
2 f
2
2 y
S 2 S V V S V 2 S 2 S V V S V 2
v 2 2 f v2 2 g v 2 2 f v2 2 g
0
c 2 S 2 c 2 S V c 2 V S c 2 V 2
2 f
Coefficient should be zero
V 2
1/2
v2 2 v2 v2
1 2 2 0 1 2 1 2
c c c
Correct option is (c)
2.25. The rotational energy of a diatomic molecule of moment of inertia I is
2
E ( 1) ; 0, 1, 2,.... , where is the quantum number associated to the rotation.
2I
Ei 2
( 1)
k BT 2 I k BT
The partition function is Z gi e (2 1) e ; 0, 1, 2,....
i 0
For large value of I and T, the energy separation between the levels is small and we can replace summation by
2 2
2 I k BT
integration. So the approximate rotational partition function is Z 2 e d .
0
2
2
Letting a . Then, Z 2 e a d
2 I k BT
0
Now, taking 2 t . Then 2 d dt
at e at 1 2 I k BT
Therefore, Z e dt .
0
a
0
a 2
Correct option is (b).
GATE PH-2001 SOLUTION 183
DESCRIPTION
3. A nˆ dS
S
Using Divergence theorem
A dV 0
A 0
V
Correct answer is (0)
a 1 0 0 a
X b 0 1 0 b aeˆ1 beˆ2 ceˆ3
4.
c 0 0 1 c
TX aT eˆ1 bT eˆ2 cT eˆ3
a 1 1 0 a a b
T b a 0 b 1 c 1 T b b c
c 1 0 1 c c a
a 1 1 0 a
T b 0 1 1b
c 1 0 1c
1 1 0
T 0 1 1
1 0 1
T
1 1 1 1 1 1
1 Adj T 1 1
So, T 1 1 1 1 1 1
det T 2 2
1 1 1 1 1 1
d2y dy
5. 4x 2 2 y 0
dx dx
Since, x 0 is a regular singular point, we assume the solution,
dy
y ak x m k m k ak x m k 1
k 0 dx k 0
d2y
2
ak m k m k 1 x m k 2
dx k 0
Putting these value in the above equation, we get,
GATE PH-2001 SOLUTION 184
4 ak m k m k 1 x m k 1 2 ak m k x m k 1 ak x m k 0
2 ak m k 2m 2k 1 x m k 1 ak x m k 0
Now, eqating the coefficient of the lowest power of x i.e. xm–1to be equal to zero
1
2a0 m 2m 1 0 m 0 or m as a0 0
2
Now, equating the coefficient of the x m k to be equal to zero
ak
2ak 1 m k 1 2m 2k 1 ak 0 ak 1
2 m k 1 2m 2k 1
ak
The first solution of y for m = 0, ak 1
2 k 1 2k 1
a0
For k = 0, a1
2
a1 a
For k = 1, a2 0
2 23 24
a2 a0
For k = 2, a3
2 3 5 24 30
y1 x ak x m k ak x k
k 0 k 0
a0 a1 x a2 x 2 a3 x 3.......
x a a0
a0 a0 0 x 2 x3 ............
2 24 24 30
x x 2 x3
a0 1 ..........
2 24 720
1
The second solution for m
2
ak ak
ak 1
3 2 2k 3 k 1
2 k 2k 2
2
a0 a1 a a2 a
For k =0, a1 , For k = 1, a2 0 , For k = 2, a3
6 2 5 2 120 273 120 42
1
k
y 2 x ak x 2 a0 x1/2 a1 x 3/2 a2 x5/2 ..........
a0 3/2 a0 5/2
a0 x1/2 x x ................
6 120
x 3/2 x5/2
a0 x1/2 .........
6 120
x x2 x3 x 3/2 x5/2
y x Ay1 x By2 x A 1 .......... B x1/2 ..........
2 24 720 6 120
GATE PH-2001 SOLUTION 185
6. (a) L
q qf q
q ln q 2
2
The Euler-Lagrange equation,
d L L
0
dt q q
L
q qf q
q ln q 2
2
dL q qq q
2 2q 2
dq 2q q q
dL ln q
2
f q q
df
dq 2 dq
d L L
0
dt q q
d q ln q
2
f a q
f
0
dt q 2 q
qq ln q
qq
2
f q q
f
0
q 2
2 q
qq ln q
1 2
2
f q q
f
0
q 2 q
(b) Since, L is function of f , arbitrary function. So, lagrangian does not covariant under the transform
q q
q
(c) Hamiltonian, H pq L q
q ln q q qf a q ln q q 1 qf q
2 2
q 2 2
(d) Since, L does not depend on time explicitly. So, total energy of the motion is conserved.
7. Since, the lamina is very thin
x3 0
L L
1
I11 x22 x32 dm x22 dx2 dx1 m 2
3
0 0
1 2
Similarly, I 22 m
3
2 2
I 33 I11 I 22 m
3
GATE PH-2001 SOLUTION 186
L L
1
And I 21 I12 x1 x2 dx1dx2 x1dx1 x2 dx2 m 2 x2
0 0
4
And, I31 I13 I32 I 23 0 C B
Therefore, the moment of inertial tensor
O x1
A
1 2 1
m m2 0
I11 I12 I13 3 4
x3
1 1 2
I I 21 I 22 I 23 m2 m 0
4 3
I I 32 I33
31 2 2
0 0 m
3
Let I1, I2 and I3 are three principal moment of inertia.
Therefore, the characteristic determinant,
1 2 1
m I m2 0
3 4
1 1 2
m2 m I 0 0
4 3
2 2
0 0 m I
3
1 2 7 2 2 2
m I m I m I 0
12 12 3
1 7 2
I1 m 2 , I 2 m 2 , I 3 m 2
12 12 3
1
Corresponding to I1 m 2
12
I11 I1 I12 I13 1
I 21 I 22 I1 I 23 2 0
I I 33 I1
31 I32 3
1 2 1 2 1 2
m m 1 m 2 0 0
3 12 4
1 1 1 2 1
m 21 m 2 m2 2 0 0 0 0 m2 m2 3 0
4 3 2 3 12
1 2 and 3 0
xˆ11 xˆ22
The principal axis is along xˆ1 xˆ2
Similarly, the principal axis corresponding to I2 is along xˆ2 xˆ1 or xˆ1 xˆ2
And the principal axis corresponding to I3 is along x̂3
GATE PH-2001 SOLUTION 187
8. (a) V x k x , k 0
px2
Therefore, Hamiltonian, H V x
2m
p2 k 2k
H , V x x V x , V x px2 , x
2 m 2 m 2m px px , x
kp x ik
m
px , x
m
px for x 0
kp ik
x px , x px for x 0
m m
(b) x 2 x
p
2 x 4 x
p 2 2
V(x)
E k x k 2x
2m 32 x 2 m
For minimum energy, –kx kx
dE 2 2 x=0
x
2k 0
dx 32m x3
1/3
32 1 2
x x
64km 4 2km
1/3 1/3
2 2k 2 2k 2
So, EG.S .
m
1 2
2/3 4 km
32
16 km
9. V x 0 0 xL
otherwise
x cx L x
According to normalized condition,
L
2
x dx 1
0
L L
2 2 2
x L 2Lx x dx 1
2 2 2 2
c x L x dx 1 c
0 0
L L
2 L2 x3 2 Lx 4 x5 2
2 2 3 4
c L x 2 Lx x dx 1 c 1
0 3 4 5 0
GATE PH-2001 SOLUTION 188
2 L5 L5 L5
c2 1
3 2 5
5
2 10 15 6 5 2 L 30
c L 1 c 1 c 5
30 30 L
L L
2 d 2 30 2
E * x
2m dx 2 5 2 dx
x dx x L x
0 L 0 2m
L L
2 30 30 2 x 2 x3 30 2 L3 L3 30 2 L3 5 2
5 x L x dx L
m L 0 mL5 2 3
0
mL5 2 3 mL
5 6 mL2
(b) The ground state of the particle in a box
2
0 sin x
L L
Therefore, the probability of finding the particle in the ground state is
L 2 L 2
* 2 30
0 dx sin x 5 x L x dx
0 0
L L L
2 L
2 30 2
5 xL sin L x x sin x dx
L L 0
L
2 2
60 L2 L3 4 L3 60 4 L3 60 16 960
6 L 3 3 6
0.9985
L L 961.32
10. We have, H ' xp px
The first order energy correction,
E n H ' n n xp px n n xp px 2 px n n | x, p | n 2 n px n
I
2 †
n i n 2 n px n i 2 n i
4
a a a a† n
i 2
i 2 n a† a a 2 a † aa† n
2
2 x
2m
a a†
i i n a †a a 2 a† aa† n
m †
p i
2
a a
i i n a †a n n a 2 n n a †2 n n aa † n
i i n 0 0 n 1 i i 0
i i n n 1 i i 0
The first order correction in wave function,
GATE PH-2001 SOLUTION 189
m H'n
n1 m
m n En0 Em0
mH'n m xp px n m xp px 2 px n
En0 Em0
m En0 Em0
m En0 Em0
m
mn m n m n
m xi p n 2 m px n
En0 Em0
m
mn
i †
m i n 2 m
2
a a a a† n
En0 Em0
m
mn
2
i m a† a aa † a 2 a † n
0 En0 Em0
m
mn
2
i m a †a aa† a 2 a† n
En0 Em0
m
mn
m a† a n m a †a n m a2 n m a†2 n
m m m m
E 0 Em0 En0 Em0 En0 Em0 En0 Em0
m n n mn m n m n
m a†2 n m a2 n
0 0 m m
0 0
En0 Em0
m n En Em mn
n 1 n 2 n n 1
n2 n2
n 1 n 5 1 3
n n
2 2 2 2
n 1 n 2 n 2 n n 1 n 2
2
2 k 2 2k
11. Density of states, d 2 k 2 kdk and E dE dk
2m m
The probability f k that the quantum state of wave-vector k is occupied by an electron is given by the Fermi
dirac-distribution function
1
f k
E k EF
1 exp
kT
GATE PH-2001 SOLUTION 190
d 2k
Number of electrons, N 2 A
2
2
f k
Factor of 2 comes from two spin of electron
d 2k kdk m
2 A A A 2 dE
2
2
0
0
And we know,
d 2k 1 1
N 2 A 2
dEg 2 D E mA
2 1 exp E EF 0 E EF g 2 D E 2
kT 1 exp kT
At T = 0K,
EF
Am
N dEg 2 D E f E E F dEg E E
2D 2 F
0 0
N m Am
n 2 EF N EF
A 2
1 1 m m
At T 0 K , n dEg 2 D E kT log 1 e EF /kT and n 2 EF
A 0 E EF
kT
2
1 e
EkTF
So, E F T kT log e 1 Exact relations
12. The total energy is
p2 p2 2 2 2 2
E V ( x, y , z ) ax 2 b( y 2 z 2 )1/2 , where p p x p y p z
2m 2m
1
The partition function is Z 3 e E / kBT d 3 p d 3 x
h
1
p x2 /2 m k BT p 2y /2 m k BT 2
3 e dp x e dp y e pz /2m kBT dpz
h
2
b ( y 2 z 2 )1/2 / k BT
e ax / k BT
dx e dy dz
x2
Now using, 2 2
dx e 2 , we have
1 k BT b ( y 2 z 2 )1/2 / k BT
Z 3 2m k BT 2mk BT 2mk BT e dy dz
h a
1/2 2 br
1 3/2 k BT k BT
2mkBT re dr d
h3 a 0 0
where we have subsitute y r cos and z r sin
GATE PH-2001 SOLUTION 191
1/2 2
2 3/2 k BT k BT
2mkBT
h3 a b
The average energy of the particle is
E k BT 2 ln Z 4k BT
T
13. (a) The interplaner distance of atomic plane (1 0 0)
a
d100 a 6.2911Å
h2 k 2 2
The interplanar distance of atomic plane (1 1 0)
a
d110 4.448 Å
2
The interplanar distance of atomic plane (1 1 1)
a
d111 3.6321Å
3
(b) KCL has fcc structure, so the number of atom in the unit cell is n = 4
The lattice constant of the KCL crystal,
1/3 1/3
nM 4 74.6 1/3 1/3
a
NA
26
6.02 10 1990
0.02490 1026
249 1030 6.2911Å
m
14. We know the the threshold energy, Eth Q 1 x
mX
Given : Q = – 0.764 MeV
mx 2 p 2 1.00814 2.01628 a.m.u.
MX M H 3.0169982 a.m.u.
3
Therefore, the threshold energy 0.764 1 2.01628 1.02745 MeV
3.0169982
15. The magnetic field at z = 0 due current flow in the loop C2 is
2m
I = 5A
C2
z = 4m
y
2m I = 5A
C1
x
GATE PH-2001 SOLUTION 192
µ0 I a2 zˆ µ0 I
4 2 µ0 I
3/2 3/2 3/2
B2 zˆ zˆ
2 z 2 a2 2 20 20
The magnetic field at zero due to current flow in the loop C1 is
µI µI µI
B1 0 0 0 zˆ
2a 2 2 4
So, total magnetic field at z = 0
µ I 2 µ0 I
B 0 zˆ
4 20 3/2
1 2
B µ0 I 3/2
zˆ µ0 I 0.272 zˆ
4 20
4 107 5 0.272 zˆ 15.00 10 7 Tesla 1.743 µ Tesla
16. (a) AND
a b
V F = ab, Bulb will glow when a and b both switch is closed.
(b) OR
b f ab
V Bulb will glow when any input switch is closed
(c) NOT
V f c
c
Bulb will glow when any input switch is closed
(d) NAND
a
V f a b
b
Bulb will glow when any switch is open
(e) NOR
GATE PH-2001 SOLUTION 193
V a b f ab
Bulb will glow when both switch is open
17. Given pinchoff voltage, VP 5V
Drain to source saturation current, IDSS 40 mA
Drain current, ID 15 mA
2
VGS
We know that, I D I DSS 1
VP
2 2
V 3 V
15 40 1 GS 1 GS
5 8 5
VGS
0.612 1
5
VGS
0.387
5
VGS 1.94 Volt
VGS 1.94 Volt
18. 2d sin n given : 2 5.8º, 2.9º
2d sin 2.90
2d sin 2.9 2 0.3 109 0.05 0.03 nm or 0.3 Å
h h .62 1034
We know, p 2.2 1023 kgms 1
p 0.03 109
150.4 150.4 150.4
Also, Å V 2 1671 V 1.67 kV
V 0.3 0.3
19. Let us take z < 0 as region1 and z > 0 as region 2
According to boundary condition,
D2 n D1n free and E2 E1
D2 D1
2 1
1D2 2 D1
D2 n 6eˆz and D2 3eˆx 4eˆ y
0 3eˆx 4eˆ y 2 0 D1
And D2 n D1n free = 5
D1n D2n 5 1
GATE PH-2001 SOLUTION 194
1
D1 3eˆx 4eˆ y eˆz
2
D1 1.5eˆx 2eˆ y eˆz
1 1 1
20. Balmer series for Hydrogen is R 22 n 2
2
If series limit n2 is 360 nm
1 1 1 R 4 4
R 2 R
2 4 360 nm
By Moseley’s law:
1 2 1 1
R z 1 2 2
1 n
For minimum of this series occur when n ,
1 2 360 nm
R Z 1 Z 1
2
900
4 0.1 nm
Z 1 30
Z 31
21. Cu(Z = 29) = [ Ar] 3d104s1 = Ground state electronic configuration
First excited state of Cu = [Ar]3d10 4s0 4p1
Second excited state of Cu = [Ar] 3d10 4s0 4p0 5s0 5p1
First and second states are written keeping in mind the selection rule 1
Ground state energy = E1
First excited state = E2
Second excited state = E3
2
P1/2
1 1
E3 5 p1 s , L
2 2 2
P3/2
2
P1/2
1
E2 4 p1 s , L 1
2 2
P3/2
1
E1 4s1 s , L 0 2
s1/2
2
State E2 and E3 are doubly degenerate (because of splitting) E3 corresponds to principal quantum num-
ber, n = 5, while E2 corresponds to n = 4. So, E3 has higher n value. That’s why energy of E3 will be more
than energy of E2 E2 E3
GATE PH-2001 SOLUTION 195
22. The given rotational line
v 28798 3.85m 0.068m 2
We know that the lines of P and R branches of a band are represented by a single formula
v v0 B 'v B "v m B 'v B "v m 2
Comparing it with the given equation, we have
v0 28798 cm 1
B 'v B "v 3.85 cm1
B 'v B "v 0.068 cm1
B 'v 1.959 cm 1 and B "v 1.891 cm 1
Now, the m value corresponding to the vertex of the Fortral Parabola (band-head) is given by
B 'v B "v 3.850
mvertex 28.30 28
2 B 'v B "v 2 0.068
2 2
v0
B 'v B "v 25798 3.850 25744 cm 1
Therefore, the band head vhead
4 B 'v B "v 4 0.068
The band head (25744 cm–1) lies toward the lower frequency side of the band origin (25798 cm–1). This
says that the band is degrarded toward the low-frequency side, that is toward the violet-degraded band
28
vhead
m=0
v0 v
–28 Violet degraded band
GATE-PH 2002 SOLUTION 196
OBJECTIVE QUESTION
1.1. A and B matrices can be diagonalized simultaneously i.e. both A and B can be diagonalized using same matrix.
DA P 1 AP, DB P 1BP
The diagonal matrices DA and DB will commute with each other.
DA DB DB DA
P 1 APP 1 BP P 1BPP 1 AP
P 1 ABP P 1 BAP AB BA
Correct option is (c)
a b d b
1.2. For a 2×2 matrix A , Adj A c a
c d
adjA 1 1 1 1 1
A 1
A 1 0 1 0 1
Correct option is (c)
1.3. A complex function f z u x, y iv x, y will be analytic in nature if its real part and imaginary part
satisfies Cauchy-Reamann equation i.e.
u v u v
and
x y y x
Correct option is (b)
1.4. Homogeneity of time means time is not explicitly present in Lagrangian. Therefore energy is conserved.
Hence correct answer is (c)
1.5. Hamilton’s equations (canonical equations) are
dqi H dpi H
,
dt pi dt qi
Hence correct answer is (d)
1.6. Rydberg constant for an atom with infinitely large nucleus is given by
me 4
R ... (i)
8 02 ch3
where m is mass of electron.
and Rydberg constant for an atom with nucleus of finite mass M is given by
e 4 ... (ii)
RM
8 02 ch3
mM
where, µ , is reduced mass
(m M )
mM e4 M me
4
RM (8 2 ch3 ) m M
8 2 ch3
mM 0 0
GATE-PH 2002 SOLUTION 197
1
RM R ... (iii)
m
1
M
m 1
According to given question R R1 and RM R2 and
M (7500)
Thus, equation (iii), give
1
R2 R1 R2 7500 R2
1 i.e., 1
1 R1 7501 R1
7500
Hence, correct option is (a).
1.7. We can write from Ampere law,
µI
B 0 ˆ rR
2 r
µ Ir
0 2 ˆ r R
2 R
1 µ0 I
B 0 r R Here, is in cylindrical coordinates
r r 2
1 µ0 Ir 2
B zˆ
r r 2 R
µ0 I
zˆ 0 rR
R2
Correct option is (c)
1.8. (i) The electric field at p depends on both charge
(ii) The electric flux through the closed surface S.
q
E dS 10
Therefore, the electric flux crossing the closed surface S is independent of q2.
Correct option is (b)
1.9. Probability density corresponding to the given wave function of the particle is
2 2 x
P x x A2 sin 2
L
2 x L 3L
Therefore, Probability density will have a maximum value when sin 2 1 i.e. at x 4 and 4
L
Correct option is (d)
1.10. In a Stern-Gerlach experiment, a particle comes out in z; state i.e. spin up sate and it can mathematically
expressed as
1 1 1
z ; ,
2 2 0
GATE-PH 2002 SOLUTION 198
1 1
We know that, , is a simultaneous Eigen state of S 2 and S z corresponding to eigenvalues s s 1 2 and
2 2
ms respectively. Thus, the particle will have definite values of S 2 and S z .
Correct option is (b)
1.11. According to Partial wave analysis, for a spherically symmetric potential the scattering amplitude and total
cross section can be written as
1
f 2l 1 eil sin l Pl cos
k l 0
4
k2
2l 1 sin 2 l
l 0
where l is the phase shift of the individual partial waves due to scattering. Now for 0 , the scattering
amplitude will be
1 i l 1
f 0 2 l 1 e sin l 2l 1 cos l sin l i sin 2 l
k l 0 k l 0
4
Therefore, Imaginary part of f 0
k
Correct option is (c)
1.12. For one particle, V dx dpx dy dp y dz dpz h3
For N particles, V (h3 ) N h3 N
Correct option is (a).
1.13. For Bose-Einstein condensation to take place, the temperature of the Boson gas should be less than TB ,
3/2
h2 N
where TB is given by TB
2mk B V (3/2)
Correct option is (a).
1.14. We have, I I 0 sin t
The magnetic field at the centre of large circular surface is
µ0 IN µ0 NI 0
B sin t
2R 2R r
Therefore, the magnetic flux passing through the smaller circular surface. R
µ0 NI 0 2
r sin t
2R
d µ NI
emf 0 0 r 2 cos t
dt 2R
µ0 NIr 2
emf cos t
2R
GATE-PH 2002 SOLUTION 199
2 t
I I 0 sin
T
t
µ0 NI 0 r 2
cos t
emf 2R
t
Therefore, induce emf in the small coil lags behind the current in the large coil by
2
Correct option is (d)
1.15. The electric dipole moment of the given configuratino with respect to origin,
P qi ri q 0 qaiˆ qajˆ 2aq iˆ ˆj aqiˆ aqjˆ
i
Correct option is (c)
1.16. Micro-canonical ensemble : It is an ensemble in which all element have same macrostate represented by same
number of particles, same volume and same energy.
Canonical ensemble : It is an ensemble in which all element have same macrostate represented by same
number of particles, same volume and same temperature.
Correct option is (a).
1.17. For one dimensional Kronig Penney model, total number of posible states or wave function is equal to the
number of unit cells.
Correct option is (b)
1.18. Divalent solids are Alkaline earth metals. They behave like a conductor
Correct option is (b)
1.19. In q1 q2 q3 R , the phonon q3 destroys momentum R . So conservation of momentum does not hold.
Therefore, a finite value of thermal resistance comes out.
Hence, correct option is (a).
1.20. Since, proton belong to Baryon family so its Baryon number is one.
Since, proton do not belong to lepton family. Therefore, Lepton number of proton is zero.
Since, electron is a lepton, its lepton number is equal to one and its Baryon number is zero
Therefore, correct results are one, zero, zero and one.
Correct option is (c)
1.21. 200 MeV and 10 MeV
Correct option is (d)
1.22. Nuclear force have the following properties
(1) These forces are short range forces
(2) These forces are change independent
(3) These forces are strongest known forces
(4) these forces are spin dependent
(5) These forces are velocity dependent
Correct option is (b)
GATE-PH 2002 SOLUTION 200
1.23. Since, 6C14 Z = 6, N = 8, Nuclear spin = 0
( 6C14 have even number of protons and neutrons)
And 12Mg25 Z = 12 (even number), N = 13 (odd number)
5
13N (1s1/2)2 (1p3/2)4 (1p1/2)2 (1d5/2)5 J =
2
Therefore, spin = half integer
Correct option is (a)
1.24. The Easymmetry term originates from the asymmetry between the number of protons and the number of neutrons
in the nucleus. Nuclear data for stable nuclei indicate that for lighter nuclei, the number of protons is almost
equal to that of neutrons, i.e. N = Z. As mass number ‘A’ increases, the symmetry of proton and neutron
number is lost and as the number of neutrons exceeds that of protons to maintain the nuclear stability. This
excess of neutrons over protons i.e. (N – Z) is the measure of the asymmetry and it decreases the stability or
the binding energy of medium or heavy nuclei.
Easym. (N – Z)
Easym. (N – Z)/A
( N Z )2 ( A 2Z )2
Easym. = aasym. aasym.
A A
arises due to unequal number of protons and neutrons
Correct option is (c)
1.25. XOR Exclusive OR gate
A
Y A B AB AB
B
Truth Table A B A AB B AB Y
0 0 1 0 1 0 0
0 1 1 1 0 0 1
1 0 0 0 1 1 1
1 1 0 0 0 0 0
XOR operator .
2.1. aiˆ bjˆ . biˆ ajˆ ab ab 0
[ two vectors are said to be orthogonal if their dot product is zero]
Correct option is (a)
2.2. For the existance of fourier transform of a function f x , f x should be either finite or zero at x . But
2 x2
f x e x is infinite at x , therefore, Fourier transform of e will not exist.
Correct option is (c)
2.3. Unit vector normal to the surface : 3 x 2 4 y z 0 at the ponit (1, 1, 7) is
nˆ
6 x iˆ 4 ˆj kˆ
6 iˆ 4 ˆj kˆ
36 x 2 16 1 53
1,1,7
Correct option is (c)
GATE-PH 2002 SOLUTION 201
dy x d2y
2.4. y Ax Be x A Be 2 Be x
dx dx
d2y dy
1 x 2
x y
dx dx
1 x Be x x A Be x Ax Be x
Be x xBe x Ax xBe x Ax Be x 0
Correct option is (b)
2.5. c 90º
angle of scattering in lab frame
sin c sin 900
tan 1 1
1 cos c 1 cos 90º
1 45º v
m
sin c m m º
tan 2 1 v
1 cos c rest º
v
m
2 45º
Using conservation of momentum we can conclude that the two masses move with equal speed. Using conser-
vation of kinetic energy
v
v 2 v2 v2 v
2
Hence correct answer is (a)
2.6. F iˆ y z
iˆ ˆj kˆ
F iˆ(0 0) ˆj 0 1 kˆ 0 1 ˆj kˆ 0
x y z
yz 0 0
Force is not conservative so we cannot define potential.
Hence correct answer is (d)
2.7.
A
B
Particle A moves on a surface therefore it has 2 degree of freedom. Particle B moves in space so it has three
degree of freedm. However between the particles there is a constraint that length of string is constant. There-
fore total degree of freedom for A + B is 2 + 3 –1 = 4.
Hence correct answer is (b)
GATE-PH 2002 SOLUTION 202
2.8. Kinetic energy = qV
mc 2 m0 c 2 qV
m m0 qV 10.2 106 eV
m0 m0 c 2 0.51 106 eV
m m0
% increase in mass 100
m0
10.2
100 2000%
0.51
Hence correct answer is (b)
B
2.9. According Maxwell equation, E
t
E ds t B ds
E dl
t
0 I 0 sin t n r 2 for r a
n I cos t r 2 ˆ
E 0 0
2 r
n 0 I 0 cos t ˆ
r
2
And for r a
n I cos t a 2 ˆ
E 0 0
2 r
n 0 I 0a 2 cos t ˆ
2r
Correct option is (c)
2.10. The radius of the focal spot
f
b
a a
b
We have 600 109 m
f 10 cm 101 m f
a 10 103 m
600 109 10 1
b 6 m
10 103
Correct option is (b)
GATE-PH 2002 SOLUTION 203
y
2.11. Two components of left circular light as
Ex E0 cos t kz x
E y E0 sin t kz
Let after travelling ‘d’ distance the left circular polarized light become right circular polarized.
Ex E0 cos t kz
E y E0 sin t kz
Therefore, the phase introduce between two component y
2
ne n0 d
x
600
d 30 µm
2 ne n0 2 0.01
Correct option is (c)
I max 25I 0 25
2.12.
I min 9I0 9
2
a b 25 ab 5
ab 9 ab 3
Taking positive value
a 4
3a 3b 5a 5b 2a 8b b 1
2
I1 a 16 I 0
I 2 b I0
Correct option is (a)
2.13. Uncertainty in the y component of position of the electron is y 1 nm . According to Heisenberg’s uncertainty
principle, the uncertainty in the y-component of velocity of the electron after passing through the slit is
v y
1.06 10 34
m / sec 1.164 105 m / sec
m y 9.1 10
31
10 9
Correct option is (b).
2.14. Aˆ , Bˆ , Bˆ , Aˆ Aˆ , Bˆ , Aˆ , Bˆ 0 (since, Bˆ , Aˆ Aˆ , Bˆ )
Correct option is (c)
2.15. Given state of the electron:
3 / 2 3 1 1 0 3 1
s zˆ , zˆ , C zˆ, C zˆ ,
1 / 2 2 0 2 1 2 2
2 1
Probability of finding the z-component of spin along direction ẑ = C 0.25 .
4
Correct option is (d)
GATE-PH 2002 SOLUTION 204
2.16. For Balmer series, the first line is given by
1 1 1
R 2 2
2 3
1 1 1 5R
R
4 9 36
36 ... (i)
R
5
Now, the first line of the Lyman series is given by
1 1 1 1 3R 3 36 27
R 2 2 R 1
Lyman 1 2 4 4 4 5 5
1 5
.
Lyman 27
Hence, correct option is (b).
2.17. A will solenoidal if A 0 5 3 a 0 a 8
Correct option is (c)
k BT
2.18. The mean free path is , where d is the diameter of the molecule.
2 d 2 P
Given : when P p0 , T T0 , 0
k B T0 ... (i)
0 2
2 d p0
Now, if P 1.5 p0 and T 0.75 T0 , we have
k B (0.75) T0 ... (ii)
new 2
2 d (1.5 p0 )
new 0.75 1 0
Dividing (ii) by (i), new
0 1.50 2 2
Correct option is (b).
2.19. The quantum statistics applies where the interparticle distance between the particles is comparable to the de-
Broglie wavelength of them.
If n is the number density of the particle then,
N N
n
V 4 r 3 , where N is the total number of particles
3
r n 1/3 , where r is the interparticle distance.
Also, the de-Broglie wavelength is
h h h
p 2mE 2mk BT
So for quantum statistics, r n1/3 1
GATE-PH 2002 SOLUTION 205
hn1/3 n1/3 (2mk B )1/2 n (2m k B )3/2 22 n
1 3/2 10 1
(2m k BT )1/2 T 1/2 h T h3 T 3/2
Correct option is (c).
2.20. For a perfect free electron gas in a metal, the angular frequency and wavenumber are related by the following
k 2
relation:
2m
k d k
So, Phase velocity v p and Group velocity vg .
k 2m dk m
1
Therefore, relation between phase velocity and group velocity is v p vg .
2
Correct option is (b)
2.21. The wavelength of plasma oscillation.
2 c 2 c 108
p n 1028 /m3
p 18 n0 3 n0 0
0.33m
Correct option is (b)
2.22. For cubic system, the relation between interatomic distance (d) and cell-edge (a) is given by
a
d where, (hkl) = miller indices ... (i)
h k2 l2
2
From Bragg’s law : 2d sin n ... (ii)
n
sin
2d
For minimum , n 1, d must be maximum.
For maximum d , ( h 2 k 2 l 2 ) will be minimum. According to given options only “(a) (100)” shows mini-
mum (h 2 k 2 l 2 ) , which gives smallest .
Hence, correct option is (a).
2.23. Unit-cell-edge ( a) 0.542 nm
The interatomic separation ( d ) ?
Since diamond has fcc structure : 4r a 2
a 2 a 1.414
2r 0.707 0.542
2 2
d 2r 0.383 nm
Hence, correct option is (c).
GATE-PH 2002 SOLUTION 206
2.24. 4
Be9 has Z = 4, N = 9 – 4 = 5
5N (1 s1/ 2 ) 2 (1 p3/2 )3
Therefore, J 3 , l = 1 (for p) and parity = (1)l (1)1 1 (odd)
2
3
Therefore, spin and parity = and odd.
2
Correct option is (a)
AOL
2.25. Closed loop gain for negative feed-back ACL 1 A
OL
where, AOL (open loop gain) , ACL (closed loop gain), feedback ratio
250
100 1 250 2.5
1 250
250 1.5 6 103 0.006
Correct option is (d)
DESCRIPTION
d2y dy
3. 2
2 5y 0
dx dx
Let, y C e mx be the trial solution
2 4 20
m 2 2m 5 0 m 1 2i
2
y A e
1 2 i x
B e
1 2i x
So,
Given : x 0, y 0 A B 0
dy i
x 0, 1 A 1 2i 1 2i B 1 2i A B 1 A B
dx 2
i i
So, A , B
4 4
i 1 2 i x 1 2i x i i 1
Thus, y x e e e x e 2 ix e 2ix e x 2i sin 2 x e x sin 2 x
4 4 4 2
0 1
4. A
1 0
The eigen value equation,
1
A I 0 0 2 1 0, 1
1
Let, the eigenvector corresponding, 1 is X1
GATE-PH 2002 SOLUTION 207
AX 1 X1
0 1 x1 x1
1 0 x2 x2
x2 x1
Let x1 k x2 k
k
X1
k
1
Applying normalization condition i.e. X 1† X 1 1 k
2
1 1
So, X1
2 1
1 1
Similarly, the eigenvector corresponding to the eigen value 1 is X 2
2 1
Diagonalizing matrix,
1 1 1
P X1 X 2
2 1 1
dx
5. I 4
0 1 x
Lets t 1 x
dt dx
dt t 41 1 1
I 4 0 1 = 0.33
1
t 4 1 1 3 3
l0 l0
6. Kinetic energy T 1 mx 2
2 x
2
potential energy V 1 k 20 x 2 0 20 x 2
2
Gravitational potential energy term is not introduced because motion is taking place in horizontal plane.
2
1 1
Lagrangian L T V
2
mx 2 k
2
20 x 2 0
Lagrange’s equation
d L L
0
dt x x
x
mx 2k 20 x 2 0 20 x 2
0
GATE-PH 2002 SOLUTION 208
0
mx 2kx 1 0
20 x 2
x 2 1/ 2
for x 0 mx 2kx 1 1 2 0
0
x2 kx 3
mx 2kx 1 1 2 0 mx 2 0
x x3
2 0 0
Motion is not simple harmonic.
7. x, y , z axes are symmetric therefore these are Principal axes
Thus I xy I yx I yz I zy I xz I zx 0
y
MR 2
I xx I diameter I yy
4
MR 2 z
x
I zz I xx I yy
2
1 0 0
MR 2
inertia tensor II 0 1 0
4 0 0 2
Here I xx , I yy , I zz are principal moment of inertia.
8. t 107 sec x 50 meter
In observer’s frame t 0
xv 50.v
t t 2 ; 0 107 2 ; v
c 2 108 3
c
c c 5 5
Spatial separation between these events in observer’s frame
3 8
107
x x v t
x v t 50 5 3 10
1 v 2 /c 2 3
2
1
5
50 18 32 5
40 meters
4/5 4
9. We know that displacement current,
A V
Id 0
d t
A A
d t d
0 360 sin 2 106 t 0 2 106 360 cos 2 106 t
10 104
2 10 3
8.85 1012 2 360 106 cos 2 106 t 0.01cos 2 106 t
GATE-PH 2002 SOLUTION 209
a r2
10. We have, V r 4 rR
3 R3
aR
rR
r
V a 2r
E rˆ rˆ rR
r 3 R3
aR
2 rˆ rR
r
1 d 2 2a 2a
E 2 r r rR
r dr 3R 3 R 3
1 d 2 aR
r 0 rR
r 2 dr r 2
According to Poisson’s equation,
2a
E 0 3
0 R
R
2a 0 4 R 3 8 a 0
2
Therefore, total charge, Q 4 r dr 3 .
0
R 3 3
3 1
3 i t x y
11. We have, E E0 ˆj 1 iˆ e 2 2
2 2
3 1
k iˆ ˆj
2 2
For plane progressive wave
K E B
3 1
1 1 3 ˆ 1 ˆ 3 ˆ 1 ˆ i t x y
B
k E
2
i j
2 2
j i E0 e
2
2 2
3 1
E 3 1 i t 2
x y
2
0 zˆ zˆ e
4 4
3 1
i t x y
E 2 2
0 zˆ e
12. The ground state wave function of infinitely deep potential will is given
2 x
sin
a a
x
We have, H ' V0 cos 2
a
GATE-PH 2002 SOLUTION 210
Therefore, the first order energy correction,
a a
I 2V x 2 x
2 V 2 2 x
E H ' 0 cos a sin a dx 2a0 sin a dx
a 0 0
a
4 x
a sin
V0 4 x V0 a V0 V0
1 cos dx x 4 a 0
4a 0 a 4a 4a 4
a 0
13. i px
x
ip x
d ipx x
e
dx
The wave function is for a free particle and it can carry momentum px k or px k
A eikx B e ikx
Applying normalization condition,
2 2 2 3
A B 1 Given: A 0.9 A
10
1
0.9 B 1 B
2
10
3 ikx 1 ikx
So, e e
10 10
14. We have, r Ce
x2 y2 z2 x y z
x y z
2
x x e x
ikx 2 k2
1 x 2 ik x x 1 x 2 ik x x 1 1 4x
px e e dx e dx e 4 e
2 2 2 1 2
Therefore, p px p y pz
2
1 p 2 p y p z2
x
p
C
1
e 4
k x2 k y2 k z2 C
e
4 2 2 2
2 23 2 23
Therefore, the probability density in the momentum space
C 2 2 px p y pz
1 2 2 2
2
P p p 3 e 2
8
Therefore, according normalization condition,
2
c 2 2 p 2y /2 2 p 2z /2 2
P p dp 1 3 e px /2 dpx e dp y e 1
8
GATE-PH 2002 SOLUTION 211
2
C
3
2 2 2 2 2 2 1
8
2
C 3/2
2 1
8
1/2
8
C 3/2
2
15. The quantum-mechanical energy of a rotating diatomic molecule is given by
h2
EJ J J 1
8 2 I
where, I is the moment of inertia of the molecule about the axis of rotation.
The lowest rotational energy level corresponds to J = 1, and for this level in CO
h2 2 h2
EJ 1
8 2 I 4 2 I
Substituting the value of h in M.K.S. units and the given value of I, we have
2
E J 1
6.62 10 J sec
34
23
2 7.6110 J
4 3.14 1.46 10 kg.m
46 2
The reduced mass of CO:
MC MO
CO
MC MO
2
12 16 / 6.023 1023
12 16 / 6.023 1023
12 16
1.14 1023 g
28 6.023 10
23
co 1.14 10 23 g 1.14 10 26 kg
Moment of inertia
I r 2 ; where r is bond length
2
1.14 1026 kg 0.1132 109 m
1.46 1046 kg m 2
The angular frequency of the co molecular
2E
10.2 1011 rad/ sec
I
Frequency
10.2 1011
2 1.623 1011 Hz
2 2
GATE-PH 2002 SOLUTION 212
16. Given : 1 kg water 1 kg water
(353 K) (293 K)
m1 = 1 kg m2 = 1 kg
Let the final temperature after mixing waters be T f . Then,
By the principle of calorimetry, m1 s1 (T f T1 ) m2 s2 (T2 T f ) , where s1 s2 specific heat of water..
T1 T2 353 293
Tf m1 m2
2 2
T f 323 K
Now change in entropy is
Tf Tf
dQ dQ dT dT
S (S )1 (S ) 2 1 2 m1 s1 m2 s2 dQ msdT
T T T
T T
T
1 2
Tf Tf 323 323
m1 s1 n m2 s2 n 238 n n 2.06 J K 1.
T1 T1 353 293
17. Electrons are fermions. So they follow Fermi-Dirac statistics. Their distribution is
1 1 for E EF
f E f E 0 for E E F when T 0 K
e( E EF )/ kBT 1
The density of state is
4V
g ( p)dp 3 p 2 dp,
h
m
where p 2mE dp dE
2E
4V m
g ( E )dE 3
2mE
dE
h 2E
Since, electron spin degeneracy is 2 so we should multiply the density of state by the factor of two
8V m
g E 3
2mE
h 2E
EF EF
8V 3/2
The total number of particles are N f E g ( E ) dE 3
2m E1/2 dE
0 h 0
8V 2 E 3/2
N 2m3/2 F ... (i)
h3 3
N 8 2
and electrons density, n 3 2m3/2 EF3/2 ... (ii)
V h 3
Also, the total energy is
E E f ( E ) g ( E ) dE
0
GATE-PH 2002 SOLUTION 213
EF
8V
E 3
2m3/2 E1/2 dE
0 h
5/2
8V 3/2 2 EF
2m ... (iii)
h3 5
The average energy per electron
E 2
5
EF5/2 3
EF ... (iv)
N 2
3
EF3/2 5
8 2 3/2
Now, using equation (ii), n 3
2m3/2 EF
h 3
2/3 2/3
3nh3 3nh3 1
EF
16 2m3/2 16 2 m
2/3
E 3 3nh3 1
Therefore, average energy per electron
N 5 16 2 m
18. The Einstein relationship between mobility diffusion constant.
D k BT
µ e
k BT k T d 1 1.38 1019 300 1.5 102 102
D µ B 19
6
4.315 104 m 2 / sec
e e t E 1.6 10 300 10 30
19. According to momentum conservation,
k ' q k
where, q is the wave vector of phonon , k & k ' are the wave vector of electron before and after collision,
respectively.
q k ' k
q 2 k '2 k 2 2kk 'cos
Since, energy of the electron much much larger than the phonon. So, we can take k ' k
q 2 4k 2 sin 2
2
q
q 2k sin 2 sin 1
2 2k
3/2 r
1 r 2a0
1
20. r , , e sin e i
8 a
0 a0
Operate parity operator on :
3/2 r
1 1 r 2a0
pˆ r , , r , , e sin e i
8 a0 a0
GATE-PH 2002 SOLUTION 214
3/2 r
1 1 r 2 a0
e sin e i r , ,
8 a0 a0
So, the parity of the function is odd.
21. The Q value of the reaction is given by
Q M 0 M n M He M C Eexcitation
2.215 1.79 MeV 0.4251 MeV
M 1.008986
Eth Q 1 n 0.4251 1 0.4519 MeV
M0 16
22. To calculate Vc ?
200 ; VBE 0.7 Volt
1
Applying DC - Analysis capacitors are open circuited because f = 0 for DC source so, X c
2 fc
9V
6k 4k
VC
+
VE
+
3k
– –
Reducing circuit “Thevenin equivalent”
9V
4k
RTh= 6||3k VC = ?
2k + I
93 0.7– E
VTh 3V
63 2.3k
“KVL”
input
Input KVL, 3 2 IB 0.7 2.3 I E
IE 201 IB ; IE 1mA
Output KVL, 9 4 IC VC
200
IE IC 1 mA is very large
So, VC 5 Volt
GATE-PH 2003 SOLUTION 215
OBJECTIVE QUESTION
1. P iˆ cos 0º iˆ cos 90º ˆj
Q iˆ ˆj / 2 cos 45º iˆ sin 45º ˆj
y y
Q
Rotation
by 45º
45º
x x
P
Correct option is (a)
2. Determinant of matrix = product of the eigenvalues = 0
Correct option is (d)
1 y
3. f z has a pole of order 2 at z = 0
z2
dz x
Therefore, z 2
2 i [Residue at z = 0]
C
z=0
1 d 2 1
= 2 i z 2 0 C : |z| = 1
2 1 ! dz z z 0
Correct option is (a)
4. Total force acting on particle
F qE q( v B )
F qE0 ˆj qv0 B0 kˆ
So, particle will not be confined to any plane.
5. Static friction is given as f s N f s mg
Correct option is (a)
1
6. L m vx2 v 2y a xv y yvx
2
L
px mvx ay
vx
L
And py mv y ax
v y
Hence correct answer is (c)
7. The distance between two event is
r 6 105 km 6 108 m
The time difference between two event is
t 1sec
c t 3 108 1 3 108 m
GATE-PH 2003 SOLUTION 216
r ct
Therefore, interval between the events is space like
Hence correct answer is (c)
8. As soon as we put same charge on the sphere it will distribute uniformly over the whole surface. So, field
inside the sphere will be zero. As a result potential will be constant.
Correct option is (b)
9. Let is the surface charge density on the cross-section area appear due to polarization. So, the depolarization
field
L
Ep 1 2 L R
2 0 2
L R
Thus, Ep 0
Correct option is (d)
10. We know that, B 0
( B ) dV 0
B dS 0
So, magnetic flux through a close surface is zero.
This say that monopole magnetic does exist.
Correct option is (b)
11. We know in free space for progressive wave.
B k E
k
B ( zˆ iˆ) cos(t kz )
1
B cos(t kz ) ˆj
c
Correct option is (a)
m d
12. Phase velocity v p c and Group velocity vg c vg v p m 0 and k 0
k k dk
Correct option is (b)
13. Normalized wave function for a particle in a one dimensional infinite potential well of width L centered at x = 0,
is
2 n x
cos for n 1,3,5,....
L L
x
2 n x
sin for n 2, 4, 6,....
L L
Correct option is (c)
14. x, P 2 2 P x, P 2iP (Since, Aˆ , Bˆ n nBˆ n1 Aˆ , Bˆ )
GATE-PH 2003 SOLUTION 217
1
15. State of the spin half particle is S z ;
2 0
0 1 1
Therefore, S x | S x | 1 0 0
2 1 0 0
0 i 1
S y | S y | 1 0 0
2 i 0 0
2 1 0 1 2
S x2 | S x2 | 1 0 0 1 0 4
4
2 1 0 1 2
S y2 | S y2 | 1 0 0 1 0 4
4
Correct option is (c)
3
16. For s , means either 3 electron or 3 unpaired subshell. As the shell is 70% filled, therefore it can not have
2
3 9
only three electrons. Thus we have 7 d electrons. Thus J L s 3
2 2
According to the Hund’s rule for more than half filled shell, we take highest value of J for ground state.
Hence, correct option is (b).
17. The hyperfine splitting of the spectral lines of an atom is due to the coupling between the electron spin and the
nuclear spin.
Hence, correct option is (c).
18. Work done = Area between P-V curve and volume axis.
So, WXY Area XYABX and WYX Area YXBAY
Total work done, W = WXY + WYX
Since, WXY is positive as volume is increased and WYX is negative as volume is decreased.
So, W = Area XYABX – Area YXBAY
Since Area XYABX Area YXBAY , W is positive.
P
Y
B V A
So work done by the gas is positive if the direction of the thermal cycle is clockwise.
Correct option is (a).
19. The second derivative of Gibbs free energy with respect to temperature and pressure are discontinuous in
second order phase transition.
G V G S Cv
i.e. V and
T P T P T P P T P T T T T
GATE-PH 2003 SOLUTION 218
So, isothermal compressibility ( ) and specific heat at constant volume (Cv ) are discontinuous in second
order phase transition.
Correct option is (b).
1
20. f E f(E)
E EF T = 0K
exp 1
kT 1
So, At T = 300 K and E = EF T > 0K
0
EF E
1 1
f EF
E EF 2
exp F 1
kT
Correct option is (b)
21. Ionic solid MnS will have NaCl type crystal structure.
Hence, correct option is (c).
A m
22. For optical brands : 1 . This indicates that the two atoms move in opposite directions and their
B m2
amplitudes are inversely proportional to their masses so that the center of mass of the unit cell remains unchanged.
A
For acoustical branch : 1 . This means that the two atoms of different masses move in the same
B
direction with the same amplitude and there is a movement of their centers of masses as well.
Hence, correct option is (b).
23. The phonon contribution to the electrical resistivity at temperature T varies as:
T (at temperatures higher than Debye temperature i.e. T D ).
and T 5 (at temperatures lower than Debye temperature i.e. T D ) .
Hence, correct option is (d).
24. The effective mass (m*) can be represented as
1
2k 2 d 2 2 d 2E
k 2 m 2
* 2
2m dk m dk
m*
–/a +/a
k
k
k1 0 k1 ;
a a –m*
From this figure, it is clear that m* is negative near the top of the band.
Hence, correct option is (a).
GATE-PH 2003 SOLUTION 219
25. Water is a polar molecule or dielectric material. Its molecule has a permanent dipole moment that’s why its
dielectric constant and refractive index do not validate the expression n 1/ 2 .
Hence, correct option is (a).
26. 4
Be9 number of protons (Z) = 4 and number of neutrons (N) = 9 – 4 = 5
For one proton, there will be uud (quark structure) and for one neutron, there will be udd (quark
structure).
For 4 proton, (4u, 4u, 4d) and for 5 neutrons (5u, 5d, 5d).
Therefore, total up quarks for 4Be9 = 13 and total down quarks for 4Be9 = 14
Correct option is (b)
27. The basic equations of -disintegrations is
17
N7 has 7 protons and 6 neutrons, 13C6 has 6 protons and 7 neutrons. So, one proton is changed into
neutron.
p n e ve
Out of these options, correct option is (b)
28. In the process of neutron decay into proton and electron, the conservation of angular momentum will be
violated (according to Fermi theory)
Correct option is (d)
29. At the pn-junction of a semiconductor, when no external bias, due to diffusion there will be no majority
charge carries. But negative minority carrier on the p-side and positive minority carrier on the n-side.
Correct option is (c)
30. By putting the value of given option check which is satisfy this relation R PQ
(a) P 1, Q 1, R 0
Put the value in the given expression
0 1. 1
1 0
0 1
So, satisfying the condition,
(b) P 1, Q 1, R 1
1 1. 1 Not satisfying the condition
(c) P 0, Q 0, R 0
0 0.0 Satisfying the condition
(d) P 0, Q 1, R 1
1 0. 1 Not satisfying the condition
So, (b) and (d) is not satisfying.
iˆ ˆj kˆ
31. A
x y z
iˆ ˆj kˆ
z x y
Correct option is (a)
d 2x dx
32. 2
2 x 0
dt dt
Let, x C .e mt be the trial solution.
GATE-PH 2003 SOLUTION 220
So, m 2 2m 1 C.e mt 0 m 1, 1
Therefore, solution will be x C1 C2t e t
At t = 0, x = 1 C1 = 1 x 1 C2t e t
dx
Now, C2 e t 1 C2 t e t
dt
dx
At t 0, 0 C2 1 0 C2 1
dt
2
Therefore, x 1 t e t and x t 1
e
Correct option is (b)
1
33. Number of independent components of a symmetric tensor of rank 2 in 3-D will be 3(3 1) 6 .
2
1
Number of independent components of an anti-symmetric tensor of rank 2 in 3-D will be 3(3 1) 3 .
2
Correct option is (b)
34. f x x4 x2
f ' x 4 x3 2 x
For maximum or minimum of f x : f ' x 0
4 x3 2 x 0
1 1
x 0, ,
2 2
Now, f " x 12 x 2 2
f " x 0 2 0 . So, at x 0 f x has a maximum value
1 1
f " x 4 0 . So, at x f x has a minimum value
2
2
So, f x has three extremum points i.e. statement R is correct.
When plot of f x vs x, cuts x-axis, then f x 0
x4 x 2 0
x2 x2 1 0
x 0, 0, 1 i.e. plot will cut x-axis at three points.
For 0 x 1, x 4 x 2 and f x 0
So, the corresponding part of the plot lies in the 4th quadrant.
Correct option is (b)
GATE-PH 2003 SOLUTION 221
df df ikx
35. F e dx
dx dx
eikx f x ikeikx f x dx
0 ik f x eikx dx ikF k
Correct option is (c)
1
36. V x m2 3x 2 3 y 2 2 z 2 2 xy
2
3 1 0 1 0 0
Vˆ m2 1 3 0 , Tˆ m 0 1 0
0 0 2 0 0 1
If be frequency of normal modes
det Vˆ 2Tˆ 0
3m2 m2 m2 0
det m2 3m2 m2 0 0
2 2
0 0 2m m
2 2
2m 2
m2 3m2 m2 m2 0
22 2 2
2
3 2
2 2 0
32 2 2 2 , 2
frequencies of normal modes are
2 , 2 , 2 .
Hence correct answer is (c)
37. Kinetic energy m0 c 2
mc 2 m0 c 2 m0 c 2 m 2m0
m0 v2 1 3
2m0 1 v c
2 2
1 v /c c2 4 2
38. The cut-off frequency of elecromagnetic wave inside a waveguide is given by
2 2
m n
c
a b
For square cross-section, a b W
2
c
Therefore, the minimum is c (take m 1, n 0 )
W W
Correct option is (c)
GATE-PH 2003 SOLUTION 222
39. Since, the displacement of the charge particle is always perpendicular to the magnetic force. So, the work done
by the magnetic field is zero.
Correct option is (c)
0
40. B K nˆ
2
from z 0 , nˆ zˆ and K kyˆ
0 k n
B ( yˆ zˆ) 0 xˆ
2 2
Correct option is (c)
41. For nth order maxima
d sin n (for largest maximum = 90º)
d n
d 5 106 500
n 4 (since, n is an integer)
1.01106 101
Correct option is (a)
42. Condition for the first order minima is a sin , where a is the slit width and is the angle of diffraction for
first order minima. Width of the central maximum is given by,
2 2
W D D
a ap
whereD is the distance between the slit and screen &p is momentum of the electrons.
Thus the width of the central maxima can be increased by one of the following ways: (i) by increasing D, (ii) by
decreasing p, (iii) decreasing a.
Correct option is (c).
43. Given wave function of the particle: x, t 0 C0 0 C1 1
E0 E1
i t i t
Wave function of the particle at time ‘t’ will be x, t C0 0e C1 1e
Wave function of the particle will repeat itself after time
t
2
h
6.6 1034
0.411015 sec
E0 E1 E0 E1 10 1.6 1019
Correct option is (b)
44. pˆ x x, t i Aeit cos kx iAeit k sin kx x, t
x
Hˆ x, t i Aeit cos kx iAeit i cos kx x, t x, t
t
Correct option is (b)
GATE-PH 2003 SOLUTION 223
2 2
45. Wavelength of the particle in Region-I:
kI 2mE
2 2
Wavelength of the particle in Region-II: 2
k II 2m E V
E V 1 V 1 V 3
Thus, 1
E 2 E 4 E 4
Correct option is (c)
46. Since, the frequencies do not coincide in Raman and vibrational spectra thus by mutual exclusion principle
molecule should have centre of symmetry. This could be satisfied by linear molecule e.g. CO2(O = C = O).
Correct option is (a)
47. Energy , E hc 1 hc vhc
1
where v ; wave number
For P : Given wave number v 10 m 1 10 102 cm 1
E vhc 10 102 m 1 6.63 10 34 J S 3 108 m /s 19.89 1023 J
For Q : E 1023 J
For R : Given frequency v 104 MHz 104 106 Hz
Energy E hv 6.63 1034 J .S 104 106 s 1 6.63 1024 J
Thus, increasing order of energy is
19.89 1023 J 1023 J 6.63 1024 J
RQ P
Hence correct option is (b)
hc
48. For X-ray, E eV
hc 6.6 1034 3 108
V 19 9
150 102 V 15 103 V 15 kV.
e 1.6 10 0.0825 10
Hence, correct option is (c).
49. The spin-orbit interaction energy is given by
a
ES , L j j 1 1 s s 1
2
where j s (for one electron atom)
where a is coupling constant.
1 1
Ground state configuration of sodium spectrum (that emit D line) is 2S. So 0, s j s
2 2
2
Ground state energy level for S1/ 2 S J
2 s 1
is
GATE-PH 2003 SOLUTION 224
a 1 1 1 1
ESo 2 S1/2 1 0 1 0
2 2 2 2 2
1 1 3 1
Excited state term is 2P, i.e. 1, s , j 1 ,
2 2 2 2
This level will split into two energy level 2 P1/2 , 2 P3/2 .
For 2 P1/ 2 : Let a be the split-coupling constant for 2P level.
a 1 1 1 1
Es0 2 P1/ 2 1 11 1 1 a
2 2 2 2 2
a 3 3 1 1 a
Es0 2 P3/2 1 11 1 1
2 2 2 2 2 2
Therefore energy level is
2
p3/2
a
Es0
2P 2
Es0 a
2
P1/2
E0
2
2S S1/2
For the first lines transition 2
p3/2 2 S1/2
a
D1 16973.4 cm 1 E0 , where E0 is central energy..
2
For the second lines transition 2
p1/2 2 S1/2
D2 16956.2 cm 1 E0 a
3
a 16973.4 16956.2 cm 1
2
2
a 17.2 11.4 cm 1
3
Hence correct option is (b)
50. (P) for pure rotational Raman spectrum molecule should have permanent electric dipolemoment but CH4 does
not have permanent dipole. So, it will not show pure rotational Raman spectrum.
(Q) SF6 also has not permanent dipolemoment. So, it will not show pure rotational spectrum.
(R) IR-absorption occurs from the stretching and bending at the covalent bonds in molecules. Since, N2 has
symmetric bonds and bond streching does not change the dipole moment of the molecule. So, it will not show
IR spectrum.
(S) For vibrational Raman and infrared absorption lines observe if the polarisability of the molecule change as
molecule vibrates. C2H6 follow the above condition.
(T) Since, H2O2 has premanent dipole moment so it will shows the pure rotational spectrum
Hence, correct option is (c).
GATE-PH 2003 SOLUTION 225
1
51. In case of photons, n and density of states is
k BT
e 1
2 2
2V d V d
g () d 4 8
h3 c c 2 c 3
V 2
So the average energy in cavity, E E n( E ) g ( E ) dE n g d / k BT
2 3 d
0 0 0 e 1 c
Letting x , then
k BT
( k BT ) 4 V x3 dx
E ex 1
2 3c 3 0
E T 4 ... (i)
Also, the number of photons are given by
3
V
2d V k T x2
N n( E ) g ( E ) dE e/ k BT 2 3 B e x 1 dx
0 2 c 3 0 1 c 0
N T 3 ... (ii)
From equations (i) and (ii), if temperature is doubled, then E 16 E and N 8 N
Correct option is (c).
52. For isothermal process, dT 0
dU 0 for an ideal gas
From first law of thermodynamics, Q U PV
T S PV W
PV
W S for an ideal gas PV nRT
nR
Correct option is (a).
53. The Maxwell-Boltzmann distribution is given by
3/2 mv 2
m
f (v ) 4 v 2e 2 k BT
2 k BT
df (v)
For most probable speed, 0
dv
mv 2 mv 2
2 k BT 2mv 2 2 kBT
2ve v e 0
2 k BT
GATE-PH 2003 SOLUTION 226
3
mv2
mv 2 k BT
2v
k BT
0 e 0
as this means v is infinite
2 k BT
v v 0
m
Correct option is (c).
2 k 2
54. E-k diagram is represented in the figure. Also it is expressed as E ... (i)
2m
E
k
k1 0 k1
a a
It is clear from equation (i) that the plot of E versus k will be parabolic in the interval k .
a a
Hence, correct option is (c).
55. Below HC, the sample behaves as a diamagnetic. Its magnetization varies linearly with H (slope = –1) and
Meissner effect will be followed by the superconductor. The statement “the sample exhibits mixed states of
magnetization near HC” is not true for type-I superconductor. It is true for type-II superconductors.
–M
O HC H
Hence, correct option is (d).
56. The magnetic susceptibility of a ferromagnetic material is given from Curie-Weiss equations
C
... (i)
(T TC )
Above TC, atomic magnets will be randomly oriented and the material will become paramagnetic.
1/
O TC T O TC T
GATE-PH 2003 SOLUTION 227
C C
At T = 200 K and TC = 100 K; 200 K (200 100) 100 ... (ii)
C C
At T = 300 K ; 300 K ... (iii)
(300 100) 200
From equations (ii) and (iii), we get 200 K 2 300 K .
Hence, correct option is (a).
H H
57. (a) C2 (axis) 180º
(1) 3C2
H H (2) 2v
C2 (axis) (3) 1h
C2 (axis)
C 2 2C 2 2 v 1 h D 2h
Cl Cl
(1) 1C2
(2) 2v
(b)
H H
1C 2 2 v C2v
H Cl
(1) 1C2
(c) (2) 1h
Cl H
1C 2 1 h C 2h
Correct option is (c)
58. Energy eigenvalues of an electron in a one dimensional potential well of infinitely high walls is
n 2 2 2 L
E n 2mE
2mL 2
dn L 1 L m
Energy density of states is 2m
dE 2 E 2E
Correct option is (a).
GATE-PH 2003 SOLUTION 228
59. In Compton effect experiment, the relation between the wavelength of the scattered photon ( ' ) and wavelength
of the incident photon ( ) are related as
h
' 1 cos i.e. '
m0 c
Energy conservation Law: h m0c 2 h ' m0c 2 Ek ,electron h h ' Ek ,electron
h 2 2
1 cos
m0c 2
Kinetic energy acquired by the electron is Ek ,electron .
h
1 1 cos
m0c 2
0
Thus, kinetic energy will be largest if 180 i.e. incident and scattered photons move in the opposite direction.
If the electron acquires a kinetic energy equal to energy of the incident photon, then energy and momentum
conservation cannot be satisfied simultaneously.
Correct option is (b)
60. Form of Coulamb potential will be replaced by Yukawa potential
e mr
V in natural units
4 r
Correct option is (b)
61. Since, photons are bosons having integral spins. Therefore, the particle will be a boson.
Correct option is (c)
62. Binding energy is given as
BE mass of constituents mass of atom c2
238
B.E. of 92U is
92 1.0078 238 92 1.0087 238.0508 c 2
1804.59 MeV use 1 amu 931 MeV
Correct option is (d)
63. Given circuit is positive feedback “Schmitt trigger”
2M
Vo = 5V
0.01µF
+ 2M = 2000k
200k Vf
–
Output of positive feedback circuit is V0 VSAT
GATE-PH 2003 SOLUTION 229
V0 200
Vf Put V0 5 (apply voltage division rule)
200 2000
5
Vf
11
5
Vf Put V0 5
11
5
V f (feed back voltage) Volt
11
Correct option is (a)
Forward
Current
Forward
Bias
Region
64. Reverse Bias
Forward Bias
Avalanche
Breakdown Reverse
Current Current
Correct option is (c)
65. (a) PQ PQ P
P Q Q P Correct.
(b) PQ P Q Correct by De-morgan’s law..
(c) PQ P Q By De-morgan’s law x y x y
P Q PQ correct x x Axioms.
(d) PQ Q P By Distributive law x xy x x x y
P Q Q Q P Q P incorrect
Correct option is (d)
66. V ˆ
nds V dv
using Divergence theorem
S v
r
3 dv 4 3 r dv
v r v
4 x y z dv 4
v
Correct option is (d)
GATE-PH 2003 SOLUTION 230
r
67. 3 0
r
r
Therefore, V 3 is a conservative field.
r
Q
So, V dr 0 and V and V dr independent of path
P
C
But, magnetic field B is not conservative.
Correct option is (d)
68. Energy line intersects potential at P, Therefore at P speed is zero.
Hence correct answer is (a)
69. At S potential energy is not minimum therefore speed cannot be maximum at this point.
Hence correct answer is (c)
70. v1
m
m 1
v
rest 2
v2
m
using conservation of momentum we get
v v1 cos v1 v2 cos 2 ... (i)
0 v1 sin 1 v2 sin 2 ... (ii)
square and add to get
v 2 v12 v22 2v1v2 cos 1 2 ... (iii)
If collision is elastic
1 2 1 2 1 2
mv mv1 mv2
2 2 2
v 2 v12 v22
from equation (iii) we get
v12 v22 v12 v22 2v1 v2 cos 1 2
cos 1 2 0
1 2 90º i.e. two particles move at 90º to each other..
Hence correct answer is (b)
71. If collision is inelastic
v 2 v12 v22
v 2 v12 v22 0
from equation (iii) 2v1v2 cos 1 2 0
cos 1 2 0
1 2 90º
Hence correct answer is (a)
GATE-PH 2003 SOLUTION 231
72. We have, A r , t A0iˆ cos kz cos kct , r , t 0
A A
E 0 A0 kc cos kz sin kct
t t
For standing wave the boundary condition is
E 0, t E L, t 0
cos 0;
2
And cos kL 0 z= 0 z=L
kL 2m 1 m 0,1, 2,3, ....
2
5 2
kL 2 k
2 2 2 L
2
k and
L 2
Correct option is (b)
1 2 1 2 2
73. The energy density, u 0 E 0 Akc sin kct cos kz
2 2
At t = 0 and z = 0, u = 0
Correct option is (a)
74. Energy eigenvalues of a particle located in three dimensional cubic well of width L with impenetrable walls, are
22 2 2 2
Enx ny nz
2mL2
nx n y nz where nx , n y , nz 1, 2,3,.....
Sum of third level energy and fourth level energy = 2nd excited state energy + 3rd excited state energy =
9 2 2 11 2 2 10 2 2
E122 E113
2mL2 2mL2 mL2
Correct option is (a)
75. Degeneracy of the fourth level i.e. third excited state is 3 and corresponding combinations of nx , n y , nz are
(1,1,3), (1,3,1) and (3,1,1).
Correct option is (c).
3 4 3 4 3 4 3 4
76. A | Aˆ | 1 2 Aˆ 1 2 1 2 2 1
5 5 5 5 5 5 5 5
12 12 24
0.96
25 25 25
Correct option is (d)
GATE-PH 2003 SOLUTION 232
77. Aˆ 2 1 AA
ˆ ˆ Aˆ i.e. Â2 is an Eigen function of .
1 2 1 1
Aˆ 2 2 AA
ˆ ˆ Aˆ i.e. Â2 is an Eigen function of .
2 1 2 2
Correct option is (a).
78. Both transitions are between multiplets. Therefore, both are anomalous Zeeman effect.
Correct option is (b)
5 3
79. In anomolous Zeeman effect, each J level splits in 2J + 1 levels. Here, J values are 5, 4, and . Therefore,
2 2
2J + 1 values are equal to 11, 9, 6 and 4.
Correct option is (c)
80. Number of possible micro-states for a collection of N particles having spin quantum numbers
(2S 1) N
3
1 1
2 1 8
2 s 2
Correct option is (d).
1
81. For S z , only the following combinations of S z 1 , S z 2 , S z 3 are possible.
2
1 1 1 1 1 1 1 1 1
, , , , , , , ,
2 2 2 2 2 2 2 2 2
Sz S z 1 , S z 2 , S z 3 P
3 1 1 1 1
, ,
2 2 2 2 8
1 1 1 1 1 1 1 1 1 1 3
, , , , , , , ,
2 2 2 2 2 2 2 2 2 2 8
1 1 1 1 1 1 1 1 1 1 3
, , , , , , , ,
2 2 2 2 2 2 2 2 2 2 8
3 1 1 1 1
, ,
2 2 2 2 8
1 3 3 1 3 1 1 3
Sz 0
8 2 8 2 8 2 8 2
So the total number of possible microstates are 3.
Correct option is (a).
GATE-PH 2003 SOLUTION 233
82. The following microstates are possible:
S z 1 S z 2 S z 3 Sz
½ ½ ½ 32
½ ½ ½ ½
½ ½ ½ ½
½ ½ ½ ½
½ ½ ½ ½
½ ½ ½ ½
½ ½ ½ ½
½ ½ ½ 32
Since each state is equally probable, only above configurations are possible. Hence,
ensemble average
32 3 21 3 12 23 0.
4
Correct option is (b).
83. Given : n Z N n (V bN ) (k BT )3/2
The Helmhotz free energy is
F k BT n Z
k BT N n (V bN ) (k BT )3/2
The pressure is given by
3/2
F ( k BT )
P Nk BT
V T , N (V bN ) (k BT )3/2
P(V bN ) N k BT
Correct option is (a).
84. The internal energy is
1
U n Z k BT 2 n Z
T k BT
k BT 2
T
N n (V bN ) (k BT )3/2
3 k B3/2T 1/2
N k BT 2
2(k BT )3/2
3
N k BT
2
Correct option is (c).
GATE-PH 2003 SOLUTION 234
85. Unit cell dimension (a) = 1 nm = 1 × 10–9 m and size of the crystal = 1 cm.
1 cm 1 cm 1 cm 102 102 102 m 3 106
Number of unit cells in the crystal 1021.
109 m 109 m 109 m 109 109 109 m 3 1027
Number of atoms in the crystal = (number of atoms/unit cell) × number of unit cells 4 1021 atoms.
Hence, correct option is (b).
1 1 1 1
86. The amplitude of diffracted X-ray beam for fcc lattice having four atoms at (0, 0, 0), , 0, , , , 0
2 2 2 2
1 1
and 0, , positions is given by
2 2
F f 1 e i ( h l ) e i ( h k ) e i ( k l ) ... (i)
It is obvious that the amplitude of diffracted X-ray beam is non-zero only if h, k and l are all even or all odd
and has a value equal to 4f. The amplitude of diffracted X-ray beam vanishes for all other odd-even combina-
tions of h, k and l .
Hence, for (010) :
F f 1 e i (0 0) e i (0 1) e i (1 ) f 1 e0 e i e i f 1 1 1 1 0
And for (200) : F f 1 e i (2 0) e i (0 0) e i (0 2) f 1 e 2 i e 0 i e2 i f 1 1 1 1 4 f .
Hence, correct option is (b).
87. Since, each nuclei of 235U releases 3 neutrons but only 20% of the 235U atoms in bomb undergoes fission,
hence total number of neutrons released are
6.1103 20
6.023 10 23 3 26
0.0938 10 9.38 10
24
235 100
Correct option is (b)
88. First we will solve (ii),
Let the total mass of 235U is m kg.
m
Therefore, the number of nuclei in m kg mass are 6.023 1023
235
According to question, each 235U gives 200 MeV but 20% of 235U nuclei in the bomb undergoes fission.
Hence, the total energy produced after the fission of m mass atomic bomb
m 20
6.023 1023 200 MeV
235 100
m
1014 joule 6.023 10 23 40 1.6 10 13 joule
235
235 1014
m 23 13
0.6096 1014 2313 0.6096 10 4 g 6.096 kg 6.1 kg
6.023 10 40 1.6 10
Correct option is (c)
GATE-PH 2003 SOLUTION 235
89. C
R (a)
R (b) Vo
V1
i1
C
An ideal op-amp has infinite input resistance. So, there will be no current pass through op-amp. So, Va and Vb
will be virtual short i.e. Va = Vb.
V1 i1 z
V1 1
z R
i1 CS
1
z j R S j
jC
z j R
Correct option is (b)
90. Applying voltage divider rule at node ‘b’
1
Vi ( s ) Vi ( s )
Vb CS Vb Va Virtual ground concept
1 1 RCS
R
CS
By applying KCL at node ‘a’
Va Va Vo s
0
R 1
CS
1 1 RCS
V0( s ) CS Va CS V0 s CS Va
R R
Put value of Va
Vi s 1 RCS
Vo s CS
1 RCS R
V0 s Vi ( s )
? V0( s )
Vi s RCS
V0( s ) 1 1
V0 (t ) Vi (t ) dt By taking inverse Laplace, this is given in option (a)
Vi ( s ) RCS RC
(b) Va Vb (Virtual ground concept) perfectly equal.
(c) Vi and i1 (through R) are in equal phase
(d) Current will be in opposite direction in both Resistor (180º) out of phase
Correct option is (a)
GATE-PH 2004 QUESTION PAPER 236
OBJECTIVE QUESTION
Q.1 – Q.30 : Carry ONE mark each.
1. x 2 y xy 2 xy y iˆ x 2 x ˆj
3iˆ 2 ˆj 13
1,1,1 1,1,1
Correct option is (d)
2. f x sin x , 0 x
1 1 1 2
Average value of f x = f x dx sin xdx cos x 0
0
0
Correct option is (b)
dV
3. At equilibrium points 0
dr
and at unstable equilibrium there is local maxima
Hence correct answer is (c)
2 2 2 1 2
4. quantity 2
x y 2 z 2 c 2 t 2
is invariant under Lorentz transformation as shown below.
According to Lorentz transformation
x x vt
1
y y
1 v 2 /c 2
z z
t t x v /c 2
x t v
x x x t x x c 2 t
2 x t
x 2 x x x x x t x x
2 2 v 2 v 2 2 v 2 2
2 –
x2 c 2 xt c 2 t x c 4 t 2
x t
Similarly,
t x t t t
v
t x t
2 x t
t 2 t t x t t t t t
2 2
2 2
2
2
v2 2 v v
x2 xt xt t 2
GATE-PH 2004 QUESTION PAPER 237
since y y, z z
2 2
2
2
, 2
2
y y z z
2 2 1 2 2 2v 2 2v 2 2v2 2 2 1
2
2
2
2 2
2
2
2
4 2
2
2 2 v2 2 2
x y z c t x c xt c xt c t y z c x
v 2 2 v 2 2 2 2
c 2 xt c 2 xt c 2 t 2
v2 2 2 2 2 v2 2
2 1 2 2 2 2 2 1 2 2
c x y z c c t
2 2 2 1 2
x2 y2 z2 c 2 t 2
Correct option is (c)
5. The image charge,
qa qR q
q'
d 2R 2
And distance from the centre
a2 R2 R
d'
d 2R 2
Correct option is (a)
6. We have,
ˆ 0 cos kz t ˆjE0 cos kz t
E iE y
So, Ex E0 cos kz t
x
E y E0 cos kz t 45º
Ey
1
Ex
E y Ex
Therefore, it is a linear plarized light at –45ºto x-axis.
Correct option is (b)
7. E t where E
t
Life time, i.e. inversely proportional to
Correct option is (c)
8. Correct option is (d)
2
9. Coulomb degeneracy of nth state of hydrogen atom is g n n . For n = 4 state, g4 = 16
Correct option is (b)
GATE-PH 2004 QUESTION PAPER 238
p2 p2
10. H V r r L S V r r Lx S x L y S y Lz S z
2m 2m
p2 1
2m 2
V r r J 2 L2 S 2
H , S 2 0 H , L S 0
H , L2 0
H , Lz 0
Correct option is (a)
11. Term with largest multiplicity (2S + 1) will be ground state term. 5S term has largest multiplicity equal to 5.
Correct option is (a)
12. For 3F term ; 2s 1 3 s 1 and l 3
Therefore, total angular momentum J (l s) to l s (3 1) to (3 1) 4, 3, 2.
4
Now, total number of m j (2 J 1) (2 4 1) (2 3 1) (2 2 1) 9 7 5 21.
J2
Therefore, degeneracy of spectral term 3F is 21.
Hence, correct option is (d).
13. For 3 D3 , 2 s 1 3 s 1 and l 2 and J 3
J ( J 1) s ( s 1) l (l 1)
Therefore, g-factor (g ) 1
2 J ( J 1)
3(3 1) 1(1 1) 2(2 1) 12 2 6 1 4
1 1 1 .
2 3 (3 1) 24 3 3
Hence, correct option is (d).
14. Infrared spectra will be observed when there is a change in the electric dipole moment of a molecule.
Hence, correct option is (b).
15. The second law of thermodynamics states that the entropy of the universe always increases in a irreversible
process and remains constant in a reversible process, i.e., dS 0 .
Correct option is (b).
16. The degree of freedom of each rigid diatomic molecules, f = (3N–C) = 3×2–1 = 5
So, the phase space dimension of each diatomic molecules = 2f = 10
So, the phase space dimension of 10 diatomic molecules = 100
Correct option is (d)
17. The specific heat of Fermi gas is given by
2 T
Cv Nk . So, C T
2 TF v
Correct option is (a)
GATE-PH 2004 QUESTION PAPER 239
18. The first order phase transition involves the concept of latent heat. Moreover, the vaporization needs latent heat
of vaporization and first derivative of Gibbs free energy is discontinious
Correct option is (a).
19. Let a is interatomic distance within a layer and c is height of the unit cell
Unit cell of hexagonal closed packed structure can be made as:
A conventional
unit cell
B C a
c
A
a
2r
A
c 8
From this figure, . a/2 a/2
a 3
F G
a /2 a /2 a a/2 30° D a/2
In figure : BD
cos 30 30°
3 2 3 B a/2 E a/2 C
In the center of spheres in contact, then
2 2
c a
2 2 a2 c2 2a 2 c 2 c 8
a a 1.633.
2 3 3 4 3 4 a 3
Hence, correct option is (d).
20. Youngs Modulus, Bulk Modulus and Poissions Ratio.
Correct option is (c)
21. An electron moving in a crystal has an effective mass m*,
E
2
m*
d 2E
2 k
dk
O m*
k
m * can be positive in the lower half of the energy band O
and negative in the upper half.
Hence, correct option is (c).
GATE-PH 2004 QUESTION PAPER 240
T2
22. H C (T ) H C (0) 1 2 , where H C (0) is critical field at 0 K. Now, if T increases, T 2 will also in-
TC
crease.
HC
TC
T
Thus, H C (T ) will decrease and vice-versa
Hence, correct option is (c).
23. Since, R A1/3 R0 A1/3
3 3
R A R0 A ... (1)
4 3 4
Therefore, volume of nucleus (V) = R ( R03 A) V A
3 3
Correct option is (a)
24. The decrease of binding energy from 56Fe to 235U nuclear is due to the long range nature of Coulomb forces.
Correct option is (b)
25. The reaction cross-section is given by
R
nvN
where, R is the reaction rate density
n is the number density of the particles in the neutron beam.
v is the velocity of neutron
N is the number of density of the nuclei in the tangent.
Correct option is (a)
26. Since, neutron, proton and are member of Baryon family so they will have B = 1
But pion belong to Hadrons family (meson). Therefore, it will have B = 0
Correct option is (a)
27. Correct option is (c)
28. FET is unipolar because it has only majority carriers.
FET is high impedance device
FET is having low gain
FET “VCCS” voltage control current source.
Correct option is (a)
GATE-PH 2004 QUESTION PAPER 241
29. Drawing circuit as per statement
Vi
V0
V0 = Vi
Voltage follower
Note: Used for impedance matching.
Correct option is (c)
30. x Sum S x y x
er S
add y
lf
y Ha Carry C xy
C
Hence, it has 2 input and 2 output (Sum, Carry)
Note : Similarly for full addar it has 3 input and 2 output.
Correct option is (d)
Q.31 – Q.85 : Carry TWO marks each.
31. Sum of the eigenvalues = Trace of the matrix
1 2 3 4 0 1 1 3 4 0 3 4 0
So, 3 1, 4 1
Correct option is (d)
1 i 2
32. Eigenvalue equation : 0 1 i 2 0 2 2 0 0, 2
i 1
Correct option is (c)
1 2
3 4i 3 4i 3 4i 3 4i 3 4i 7 24i 7 24
33. 2
i
3 4i 3 4i 3 4i 3 4i 9 16i 25 25 25
Correct option is (d)
1
34. f z has simple poles at z ia
z a2
2
For a = 2, poles at z 2i lies outside the circle |z| = 1. y
dz
So, z 2
a2
0 C
C x
1 i
For a , poles at z lies within the circle
2 2
C : |z| = 1
dz i i
C
2
z a 2
2 i Res. f
z
Res.
2
f z
2
1
1
2 i 0
2 z z i /2 2 z z
i
2
Correct option is (b)
GATE-PH 2004 QUESTION PAPER 242
n
n n d
35. Property : L f x f s then
L x f n f s
x 1
ds
1 d d 2 s
L t sin t 1 L sin t 2 2
ds ds s s 2 2 2
Correct option is (c)
2 n
36. f x x 1 sin nx
n 1 n
2 2 2
x 2 sin x sin 2 x sin 3 x sin 4 x
2 3 4
Using, Perseval’s formula, we get
4 4 4
x dx 4 2
2
2
2
2 ........
3 4
2 3 1 1 1 1
4 2 2 2 2 ..............
3 1 2 3 4
2 2
n
6 n 1
Correct option is (c)
2 2 2 k2
b2 / 4 a )
37. F e x x ikx
e e dx e
x ikx
dx exp (Using ax 2
bx
dx
4 e
a
e
Correct option is (b)
1
38. L m r 2 r 2 2 V r
2
L
pr mr
r
L
p mr 2
Hence correct answer is (a)
39. I 0 I rod I particle
2
1 3a
4m a 2 m 3a
3 4 4
a/4
4ma 2 9ma 2 91 2 m
ma
3 16 48
Hence correct answer is (b)
GATE-PH 2004 QUESTION PAPER 243
z
40. Fcoriolis 2 m v
2mu0 rˆ zˆ u0
2mu0 ˆ
Hence correct answer is (b)
41. L ax 2 by 2 kxy
L p
px 2ax x x
x 2a
L p
py 2by y y
y 2b
H pi qi L px x p y y ax 2 by 2 kxy
i
2 2
px2 p y px2 p y
H kxy
2a 2b 4a 4b
2
px2 p y
H kxy
4a 4b
Hence correct answer is (c)
42.
a r , b p ai ri , b j p j
ai b j ri , p j ai b j ij
ai bi a b
Hence correct answer is (d)
1
43. force x k1 k 2 , U F dx k1 k 2 x 2
2
1 2 1
L T U mx k1 k2 x 2
2 2
k1 x k2
at any instant r and p are along same line. Therefore
Lrp0
1 1
total energy E m x 2 k1 k2 x 2
2 2
Hence correct answer is (b)
m0 m0
44. m ; 2m0
2 2
1 v /c 1 v 2 /c 2
v2 1 3
1 2
v c
c 4 2
Hence correct answer is (a)
GATE-PH 2004 QUESTION PAPER 244
Q2
45. Energy u
R
u1 Q2 / R
u2 4Q / R / 2 u2 8u1
Correct option is (b)
Vout
46. 0
r rR
2 R3
0 E0 cos 3 cos 3 0 E0 cos
r rR
Correct option is (c)
Fr
y
dl=Rd^
47. d
x
The force on the elementary length
dF IRdˆ B0 kˆ IRB0 d rˆ I R B0 (cos ˆj sin iˆ) d
/6
So, F I R B0 (cos ˆj sin iˆ) d IRB0 ˆj
/6
Correct option is (b)
48. For normal incidence, 1 2 0
2 2
n n n 1
R R R| | 1 2
n1 n2 n 1
4n1n2 4n
T T T| | 2
n1 n2 n 12
Correct option is (a)
49. We have, E E0 exp i ( xk cos yk sin t )
Comparing this with plane wave equation
E E0ei kr t
ˆ cos ˆjk sin
We can write, k ik
Correct option is (c)
GATE-PH 2004 QUESTION PAPER 245
2 2
50. 2 02 k 2 c 2 2 d c 2 2kdk d c k vg c v p vg c 2
dk vp
Correct option is (c)
51. For the mth order bright fring
2d m
Given: 2 d d m 92
2d 92
2 25.3 10 6
550 nm
92
52. u x eikz
* ikz
u x e
i
So, probability current density, J * *
2m
i u ikz u ikz
e iku 2 x * e iku 2 x
2m x x
i k 2
2m
2iku 2 x
m
u x vu 2 x
Correct option is (a)
53. According to Schrodinger equation,
2 d 2
V x x E x
2m dx 2
2 d
2m dx 2 2
A2 2 xe x V x x E0 x
2
2m 2 2 2 2
2 A 2e x 4 A 4 x 2e x V x x E0 x
2
2m
2 2 4 4 x 2 V x E0
2 2 2
V x E0
2m
2 2 4 4 x 2 E0
m
E0 E0 2 E0 2 x 2 2 E0 2 x 2
Correct option is (c)
54. V (r ) S1 S 2 V0 (r )
1 ˆ2 ˆ2 ˆ2
2
S S1 S2 V0 (r )
GATE-PH 2004 QUESTION PAPER 246
1
Ep s( s 1) s1 ( s1 1) s2 ( s2 1) V0 (r )
2
1 1
For singlet state s 0, s1 , s2
2 2
1 1 3 1 3 3
Ep V0 (r ) V0 (r )
2 2 2 2 2 4
1 1
For triplet state s 1, s1 , s2
2 2
1 3 1 3 1
Ep 2 2 2 V0 (r )
2 4 2 2 4
Correct option is (d)
2 x2
55. 0 A exp
2
Applying normalization condition,
*
0 0 dx 1
2 2 x 2 ax 2 bx b2 / 4 a
A e dx 1 [using e
dx
a
e ]
2 2
A 1 A
2
1/ 2
2 2
So, 0 e x /2
So, first order correction to energy,
4
x E0 4 4 2 x 2
E0 0 E0
I
0 xe dx
10 104
E0 5 2 x2
4
2 x 4e dx
10 0
2 2 2 z 1 dz
[Put x z x 2 2 xdx 2
dz dx ]
2 2 x
E0 5 3/2 z dz E0 z 3/2 E0 5
z e e z dz
10 4
0
5 104 0 10 4
2
3 1 E0
3
E0 104
4
10 2 2 4
Correct option is (c)
GATE-PH 2004 QUESTION PAPER 247
56. Atomic number of boron, z = 5.
The electronic configuration of boron is given by 1s 2 , 2 s 2 , 2 p1 i.e., there is one unpaired electron in 2p subshell.
1
Thus, l 1, s
2
According to L-S coupling, J will have values from (l s) to l s .
1 1 3 1
Thus, J 1 to 1 , .
2 2 2 2
For less than half filled orbital, lowest J value corresponds to ground state.
1 1
Thus, the ground state is characterized by L 1, S , J
2 2
Hence, correct option is (b).
57. For the spectroscopic term 3p we have S = 1
Multiplicity 2s 1 3 l 1
Now, for spin-orbit interaction, total angular momentum can have values from (l s) to l s .
Thus, J (1 1) to 1 1 2, 1, 0.
So, the term is 3 p0,1, 2 i.e., 3 p0 , 3 p1 and 3 p2
Now, degeneracy of individual terms are given by
For J 0 (2 J 1) 1
For J 1 (2 J 1) 3
For J 2 (2 J 1) 5
Hence, correct option is (a).
mj
3/2
1/2
2
58. p3/2
–1/2
–3/2
1/2
2
s1/2
–1/2
m j 1, m j 0, m j 1
() ( ) ( )
According to selection rule, m j 0, 1 (0 0) , total number of permitted transitions from 2 p3/ 2 to 2 s1/ 2
due to a weak magnetic field is 6.
Hence, correct option is (c).
59. For pure rotational spectrum of a diatomic rigid rotor, the separation between two consecutive lines in the
spectrum is given by
2h
2 B
8 2 Ic
GATE-PH 2004 QUESTION PAPER 248
where, I is moment of inertia.
Thus, the separation is inversely proportional to the moment of inertia of the rotor..
Hence, correct option is (b).
60. The frequency of the light of wavelength 1.5 µm incident on material is
c 3 108
6
2 1014 Hz
1.5 10
This will be the exciting line i.e. Rayleigh line and the frequency of Raman line is Raman 20 1012 Hz
Therefore, the Raman shift
Raman 2 1014 20 1012 Hz
180 1012 Hz
In terms of wavelength, the Raman shift is given by
3 108
12
1.67 106 m
180 10
1.67 m
Hence, correct option is (c).
61. According to Wien’s displacement law of blackbody radiation, TV 1/3 constant.
T1V11/3 T2V21/3
where T1 2000 K, V1 10 cm 3 , T2 ?, V2 640 cm 3
1/3 1/3
V 10
Therefore, T2 T1 1 2000 500 K
V2 640
Correct option is (d).
p B (T )
62. Given equation : 1 ... (i)
k BT
a
For a vander Waal’s gas, p 2 ( b) k BT
k BT a
p 2
b
p a
k B T b k BT
1
b a
1
k BT
b a b
1 considering 1
k BT v
a 1
1 b
k BT
GATE-PH 2004 QUESTION PAPER 249
a
So, B (T ) b
k BT
pv f5/2 ( z ) z 2 z3
For an ideal Fermi gas, k T N f ( z ) , where f ( z ) z
B 3/2 2 3
pv
N
g5/2 ( z ) z 2 z3
For an ideal Bose gas, k T g ( z ) z
B g3/2 ( z ) , where 2 3
pv
For an ideal gas, k T N
B
Clearly, B (T ) can be negative for Vander waals’ gas only, i.e., B (T ) 0
a a
b 0 or T
k BT b kB
Correct option is (a).
63. The region of co-existance of a liquid and a vapour is a region of first order phase transition and the specific
heats C p and CV both diverges in it.
Correct option is (a).
y
IB
64. Hall voltage, VH
enb
IB 1 103 0.5 x
n 19 2 3
5.2 1019 m 3
ebVH 1.6 10 10 6 10
z
and this is n-type
Correct option is (b)
E
65. Slope of the n versus T 1 graph is , also from the given figure, slope of A < slope of B.
2k B
(E)A (E )B
E A EB
2k B 2k B
Hence, correct option is (a).
1 N
66. Clausius-Mossotti equation ; ... (i)
2 3 0
where, is total polarizability.
N 5.6 1 4.6
For 5.6, 0.605 ... (ii)
3 0 5.6 2 7.6
Now, since refractive index, n 1.5
r 1.5 r 2.25
GATE-PH 2004 QUESTION PAPER 250
r 1 N e N e 2.25 1
Again from equation (i), 0.2941.
r 2 3 0 3 0 2.25 2
N e
Electric polarizability e 3 0 0.294
0.486 0.5.
Total polarizability N 0.605
3 0
Hence, correct option is (a).
67. Parmanent magnets are made from hard ferromagnetic materials.
Correct option is (b)
68. From option (a) entropy always decreases on cooling a superconductor below its critical temperature. This
indicates that superconducting state is more ordered than normal state.
From option (b) electronic contribution to heat capacity also has exponential form.
From option (c) A type-I superconductor is also perfect diamagnet
From option (d) Critical temperature of superconductor vary with isotopic mass and this is known as isotopic
effect.
Correct option is (d)
r
69. F q exp iq r d 3r
Now, given r r f x x dx f 0
r
exp iq r d 3 r 1
Correct option is (a)
70. In ground state deutron remains 96% of time in 3S state for which (L = 0, S = 1) and 4% of time in 3D state for
which (L = 2, S = 1)
Correct option is (c)
71. (a) p n e ; here Lepton number and angular momentum is not consreved. Hence reaction is not allowed.
(b) p e ve ; Baryon number and angular momentum is not conserved. Hence, reaction is not allowed.
(c) p ; this violates the Baryon number conservation. Hence, this is not allowed reaction.
(d) p n 0
Charge (Q): 1 0 1 0 Q 0
Baryon number (B): 1 1 0 0 B 0
Strangeness (S): 0 0 0 0 S 0
This reaction is allowed.
Correct option is (d).
GATE-PH 2004 QUESTION PAPER 251
72. Given circuit is standard Wein Bridge Oscillator. The condition for sustaining oscillation in the Wein bridge
oscillator is Gain A 3
R2
Therefore, Gain 3 1
R1
1
• Frequency of oscillation f 0
2RC
frequency can be calculated by finding feedback factor “” and equating it’s imaginary part to zero
R2
3 1 R1 12 k
R1
R2
2 R2 24 k
12
1 1
f 300
2 RC 2 R 0.01F
R 53 k
Correct option is (d)
RC
73. The gain of the transistor is equal to g m RC
re
RC 2 103
Av 100
rE 20
Av 100
Vout t AvVin t
Now, the input should be such that the transistor does not go in to saturation region
V0 max 10 V
10 10 1000
Vin 100 mV
100 100 1000
Thus theoretically on input signal of 100 mV can be applied.
Correct option is (d)
74. 10V
100k 2.2k
Vc = 0.96
+
0.3V –
, put value of
1
24
GATE-PH 2004 QUESTION PAPER 252
Given I CBO 20 A
We know I C I B 1 ICBO
By putting value I C 24 I B 25 20 µA
IC 24 I B 500 µA ... (1)
Apply KVL at collector-base junction
10 100 I B 0.3
I B 0.097 mA
Put the value of IB in equation (1)
IC 24 0.097 mA 0.5 mA
IC 2.328 0.5 mA 2.828 mA
Apply KVL at output
10 IC 2.2 VC
VC 3.778 Volt
Correct option is (a)
75. 3V is applied for 0 t 4sec.
1 RF
RF
CF S
1 RS CF
RF
V0 ( s ) z ( s ) CF S Vo
Vi ( s ) RS RS 3V
RF ROM
V0 ( s) 1 RF CF S
Vi ( s ) RS
V0 RF 1
( s)
Vi RS 1 RF CF S
Put the given value
V0 20 1 V0 2 1
( s) s 1
Vi 30 1 2S Vi 3
2 S
2
Note: Vi 3 Vi ( s ) 3 By Laplas transform
S
1 1 3 1
V0 (s ) V ( s )
3 S 0.5 S
0
S S 0.5
2 2 1 1 4 sec
V0 ( s ) V0 (s ) 2 t
S S 0.5 S S 0.5
By applying Inverse Laplace Transforms –2
V0 (t ) 2 1 e 0.5 t V0 (t ) 1.73 Volt
It is Ramp function.
Nearest option is (d)
GATE-PH 2004 QUESTION PAPER 253
76. Y ABCD ABCD ABCD ABCD
ABCD ABCD AB C C D
ABCD ABCD ABD
ABCD ACD BCD ABD
A BD CD D CD BD BCD
A D D CD BCD
A D BCD
AD BCD
77. Consider the given differential equation,
y " p x y ' q x y x 0
We can apply Frobenius power series method
y Cn x k n
n 0
Indicial equation will be
k 2 p 1 k q 0
By Taylor series expansion, p = 4, q = 2
k 2 4 1 k 2 0
k 2 3k 2 0
k 1, 2
Correct option is (b)
x1 x2
78. W 0
x '1 x '2
For solution to be linearly independent
x y
1
dy W W 0 condition n x 0 1, y 0, at x 2
1
dx
dy
x y W
dx
1 1 dy dy
At x , y 0 W 2W
2 2 dx dx
y 2Wx c
GATE-PH 2004 QUESTION PAPER 254
1
At x , y 0
2
1
0 2W c
2
c W y 2 x 1 W
Correct option is (d)
79. Energy of comet in parabolic orbit is zero.
E 0
near the sun
1 2 GMm 2GM b
mv 0 v Sun
2 b b v
angular momentum
2GM
L mvb m b L m 2GMb
b
Hence correct answer is (d)
80. Nearest to the Sun, force is maximum, therefore acceleration is maximum.
Hence correct answer is (a)
i k r t
81. ˆ
Given : E xE0 e
k zˆ k cos ik sin
E xˆ E0 ei zˆ k cos ik sin r t
ˆ 0 e k sin z ei k cos z t
xE
This says that the amplitude decays as the wave propagates.
Correct option is (a)
82. We know,
k E B
k
B
kˆ E
We have,
ˆ 0 e
i k r t
E xE ˆ 0 e kz sin ei kz cos t and kˆ zˆ
xE
k
B E0 e kz sin e
i kz cos t
zˆ xˆ
k
E0 e kz sin e
i kz cos t
yˆ
Correct option is (b)
GATE-PH 2004 QUESTION PAPER 255
n
83. The wave function of nth state is n 2/L sin x
L
2
Therefore, 2 2/L sin x
L
The probability density in the first excited state, is
2 2 2
P 2 2 sin x
L L
2
Probability of finding the particle, will be maximum if, sin x 1
L
2 L L 3L
x (2n 1) x (2n 1) x ,
L 2 4 4 4
Correct option is (d)
84. The wave function of the lowest energy state, is
2 x
1 sin
L L
L
The probability of finding the particle in the region 0 x is
4
L /4 L /4
2 x 1 2x 1 L L 1 1
P sin 2 dx 1 cos dx
L 0
L L 0
L L 4 2 4 2
Correct option is (a)
85. The energy of the electron in nth state of the cubic box is given by
22 2 2 2
Enx n y nz
2ma 2
nx n y nz
The energy of the lowest state nx 1, n y 1, nz 1
3 2 2
E111 0
2ma 2
Correct option is (d)
11 2 2
86. The energy of the third excited state is 3 2
i.e. nx2 n y2 nz2 11
2ma
Possible combination of nx , n y , nz are (3, 1, 1), (1, 3, 1), (1, 1, 3)
So, the degeneracy (including spin) of the level 3 = 3×2 = 6
Correct option is (c)
87. The partition function is Z ei e 0 e0 e0
i
1 2 cosh (0 )
1 2 cosh (2) 0 2
GATE-PH 2004 QUESTION PAPER 256
e 0 1 1
Required probability 1 2 cosh (2)
Z 1 2 cosh (2)
Correct option is (d).
88. The free energy is
F N k BT n Z
N k BT n 1 e 0 e 0
N k BT n 1 e x e x x 0
At high temperature, x 1 e x 1 x
So, F N k BT n 1 1 x 1 x
N k BT n 3
Correct option is (d).
1 2 1 4 1 2 1 6 2 2 1 4
89. Ca Z = 20 ( s1/2) ( p3/2) ( p1/2) ( d5/2) ( s1/2) ( d3/2)
41
20
N = 21 (1s1/2)2 (1p3/2)4 (1p1/2)2 (1d5/2)6 (2s1/2)2 (1d3/2)4 (1f7/2)1
Last proton (1d3/2)4 and last neutron (1f7/2)1
Correct option is (d)
90. N = 21, last neutron (1f7/2)1
7
J= , l = 3 (for f)
2
Parity = (–1)3 = (–1)3 = –1 (odd)
7
Angular momentum and parity =
2
Correct option is (a)
GATE-PH 2005 SOLUTION 257
OBJECTIVE QUESTION
1. f x 4 x3 1 x 3
3 3 3
1 3 1 3 1 4x4 1 4
The average value of f x 4 x dx 4 x dx 3 1 40
3 1 1 21 2 4 1 2
Correct option is (c)
2. Unit normal vector to the curve : x 3 y 2 xy 17 at point (2, 0) is
nˆ /
3 x 2 y 2 y iˆ 2 x 3 y x ˆj
2 ˆj ˆ
j
2,0 2
2,0
Correct option is (d)
1
3. f z has a simple pole at z = –3
z 3
dz 1
C z 3 2 i
Res. f z 3
2 i z 3 2 i
z 3
Correct option is (c)
4. Lets 1, 2 and 3 are the eigenvalue of the matrix
Determinant of the matrix = Product of the eigenvalues 36 12 3
Given : 1 2, 2 3
So, 3 6
Correct option is (b)
5. For central field potential is r dependent only i.e. V V r
1
L m r 2 r 2 2 V r
2
Therefore, energy and p (angular momentum are conserved) since angular momentum is conserved, motion
must be confined in a plane.
Hence correct answer is (c)
6. Coriolis force 2m v
v
2mv r
Hence correct answer is (d)
7. None of the given options are correct.
8. Energy is not conserved in such process
Hence correct answer is (b)
9. B d µ0 I1 I 2 I 3 µ0 I I 2 I 0
B=0
Correct option is (a)
GATE-PH 2005 SOLUTION 258
4 r3
10. E dS 3 0
4 r 3 r
E 4 r 2 E
3 0 3 0
Correct option is (a)
11. Super position of two linear orthogonal polarized light of equal amplitude give circular polarized light.
Wave propagating towards positive z-direction.
Correct option is (b)
12. For a point source charge particle both the electric and magnetic field will be vary as
1 1
E and B
r r
Correct option is (b)
13. pˆ cos ( kx) cos ( kx) cos kx (even parity)
e kx e kx e kx e kx
pˆ tanh (kx) pˆ kx
kx xk xk
tanh (kx) (odd parity)
e e e e
Correct option is (d)
14. Lz , Ym , ,
i Ym , , iYm ,
Y
i Ym , i m , iYm ,
i im Ym , , Ym , eim
Lz , Ym , i im Ym , mYm ,
Correct option is (c)
15. c1 1 c2 2
According to normalization condition, | 1
2 2
c1 c2 1 ( n | m m n )
Correct option is (d)
16. Force on the charge q qE
dV
qE V qEx
dx
Hˆ kinetic energy potential energy
GATE-PH 2005 SOLUTION 259
2 d 2 1 2
kx qEx
2m dx 2 2
Correct option is (c)
17. If vacancy is created in L shell, then electron from upper shells will have the transition to L shell (n = 2) giving
L series. Transition n = 3 n = 2 L
n=5
n=4
L
n=3
L L
n=2
n=1
n = 4 n = 2 L etc.
Hence, correct option is (a).
18. Selection rule for electric dipole transition : l 1
i.e., parity must change during transition and ml 0, 1, ml 0 ml 0 if l 0
Hence, correct option is (c).
1
19. Given : pV U
3
From first law of thermodynamics, dQ dU pdV , where dQ 0 for an adiabatic process.
dU pdV 0
4dV 3dp
d (3 pV ) pdV 0 4 pdV 3V dp 0 0
V p
4
nV n p n c
3
where c is a constant of integration.
pV 4/3 constant.
Correct option is (c).
20. Number of accessible states for N non-interacting particles of spin s is (2 s 1) N .
So for spin 12 N-particles, 2 N .
Correct option is (a).
21. If u is the total internal energy density then pressure is given by
2 2U
p u (U = uV, the total internal energy)
3 3V
Correct option is (b)
22. Given : g ( E ) E g ( E ) A E , where A is a constant of proportionality..
The average energy per particle at T 0 K is
GATE-PH 2005 SOLUTION 260
EF
E f ( E ) g ( E ) dE At T 0
0
E EF 1 E EF
f E
f ( E ) g ( E ) dE 0 E EF
0
where f ( E ) 1 at T 0 K
EF EF
A E E dE E 3/2 dE
0 0 3
E EF
EF
EF
5
A E dE E1/2 dE
0 0
Correct option is (d).
a 1
23. OP a, a, 11 221.
2 2
a
Here, a, a, are the intercepts along x, y and z-axis.
2
Hence, correct option is (a).
24. Since, Argon has 18 atomic number. Therefore, its atomic configuration is 1s 2 2s 2 2 p 6 3s 2 3 p 6 . Na is strong
electropositive and Cl is strong electronegative. Si is a semiconductor with covalent bonding.
Hence, correct option is (c).
25. The -intensity anti-parallel to the nuclear spin directions is same as that along the nuclear spin directions. This
observation gives the evidence for the non-conservation of parity in -decay..
Correct option is (a)
26. Since, surface energy term Es R 2 A2/3 also its sign is negative.
Correct option is (a)
27. (a) µ
e vµ ve
Charge (Q) : 1 1 0 0 Q 0
Electron Lepton No : 0 1 1 0 1 Le 0
Muonie Lepton No : 1 0 1 0 L
Baryon number (B) : 0 0 0 0 B 0 (Allowed)
(b)
µ vµ
Charge (Q) : 1 1 0 Q 0
Munonic Lepton number (Lµ) : 0 1 1 Lµ 0
Baryon number (B) : 0 0 0 B 0 (Allowed)
(c)
e ve
GATE-PH 2005 SOLUTION 261
Charge (Q) : 1 1 0 Q 0
Electronic Lepton number (Le) : 0 1 1 Le 0
Baryon number (B) : 0 0 0 B 0 (Allowed)
(d) µ
e e e
Charge (Q) : 1 1 1 1 Q 0
Electronic Lepton No : 0 1 1 1 Le 0
Muonic Lepton No: 1 0 0 0 Lµ 0
Lepton number (L) : 1 1 1 1 L 0
Baryon number (B) : 0 0 0 0 B 0
Therefore, this is forbidden.
Correct option is (d)
28. Nuclear forces are short reange, charge independent, velocity dependent and spin dependent.
Correct option is (d)
29. Drain
rd RL
gmVGS
Voltage control current source
AC equivalent figure of FET
Correct option is (a)
30. Redraw this circuit
VCC
D1
V1
Pull up resistor is R
V0
V2
D2
Case (1)
D1 , D2 OFF
VCC
R
V1
V0
V2
Case - (2) Case - (3)
D1 ,D2 ON either D1 ,D2 is ON
GATE-PH 2005 SOLUTION 262
VCC VCC
R R
V0 V0 = 0
low logic V2
Truth Table
V1 V2 D1 D2 V0
0 0 ON ON 0
0 1 ON OFF 0
1 0 OFF ON 0
1 1 OFF OFF 1
Logic AND Gate
Correct option is (d)
iˆ ˆj kˆ
31. F xiˆ 2 yjˆ 3 zkˆ F iˆ 0 ˆj 0 kˆ 0 0
x y z
x 2y 3z
F 0 F 0
Correct option is (a)
e z 3 e z e
i 2 n 1
32. 3
z ln 3 i 2n 1 n 0, 1, 2, .......
Correct option is (b)
33. L f t fˆ s e st f t dt
0
z
s dz
L f at e st f at dt e a
f z
0 0 a (putting, at z a dt dz ]
1 s / a z 1
= a e f z dz fˆ s /a
0
a
Correct option is (b)
3
34. u4 u3 i.e. u3 , u4 are linearly dependent
2
Correct option is (d)
sin z
35. z = 0 is a removable singular point of f z as
z
GATE-PH 2005 SOLUTION 263
sin z 1 z3 z5 z 7 z2 z4
z ..... 1 .... (no negative powers of z)
z z 3! 5! 7! 3! 5!
Correct option is (b)
36. Sum of the eigenvalues = Trace of the matrix = 0
Only option (a) satisfies this.
Correct option is (a)
37. r a, t r at
radial acceleration ar r r 2 0 at 2 a2 t constant
Transverse acceleration a 2r r
2a 0 2a2
a 0 , force is not radial, therefore angular momentum is not conserved.
1 2 1 1 1
Kinetic energy mv m r 2 r 2 2 m a 2 2 a 2 2t 2 2 ma2 1 2t 2
2 2 2 2
it is time dependent therefore not conserved momentum cannot be conserved because force is not zero.
Radial momentum pr mvr mr ma it is conserved.
Hence correct answer is (d)
38. L0 Mva I c , v a v
a
M a 2 Ma 2 c v
2
a
2M a 0
Hence correct answer is (c)
39. Use parallel axes theorem
I 2 I c Md 2 I Ic
2 2
2 MR 2 M 2 R
5 4R
2 44 d
2 4 MR 2 MR 2
5 5
Hence correct answer is (c)
40. a
floor of
car a
let us observe from frame of car equation of motion for translation
ma f r ma
f
a a r ... (i)
m pseudo
force
equation of motion for rotation ma a
fr R
2
f r R I MR 2
5
5 fr
... (ii)
2 MR
GATE-PH 2005 SOLUTION 264
for rolling a R
f 5 fr
a r
m 2 m
7 fr 2ma
a fr
2m 7
from equation (i)
2ma 5a
a a
7m 7
acceleration of sphere with respect to inertial frame
5a 2 a
a a a
7 7
Hence correct answer is (b)
41. L et be angle with x axis as seen from s frame
tan 0 s y s
tan tan 0
1 v 2 /c 2 0
v
If be angle with y axis as seen from s then 0
x x
rest frame another frame
90 or 90 , and 0 90 0
cot cot 0
1
tan tan 0
1
tan 1 tan 0
Hence correct answer is (c)
1 1
42. V x m2 x 2 mv 2
2 2
1 2 1 1
Lagrangian L T V mv m2 x 2 m v 2
2 2 2
L
p mv m v mv 1
v
Hence correct answer is (d)
43. Radiation power from accelerated charge particle dependency on charge and acceleration as
P q 2a 2
where, q is the charge of the particle and a is the acceleration of particle.
In case one particle is moving with constant speed. So, pa 0
In case of two and three particles are moving with same (centripital) acceleration. So, pb pc
Correct option is (b)
GATE-PH 2005 SOLUTION 265
44. We know Laplace equation,
2 2
0 0 60 0
Correct option is (b)
45. The work will be done to bringing a charge particle from infinite to a position at a distance ‘d’ in front of
a semiinfinite grounded metal surface is
1 q2 -q q
W
4 0 4d
2d
Er n1 n2 1 1.5 0.5 1 2
46. Er
Ei n1 n2 2.5 2.5 5 5
Et 2n1 2 4 8
Eot
Ei n1 n2 2.5 5 5
Correct option is (c)
47. Lorent Gauge condition,
A µ 0 0
t
µ0 Q
2
µ0 0 0
4 r t
Qt
4 0 r 2
Correct option is (d)
48. d B adx
dx a
µI
d 0 adx
2 x
2a x a
aµ0 I 0 cos t dx a I cos t dx
d 0 0
2 x 2 a
x
a0 I 0 cos t 2a aµ0 I 0 cos t
ln ln 2
2 a 2
d aµ0 I 0 sin t
emf ln 2
dt R
emf µ aI sin t
So, current, I 0 0 ln 2
R 2 R
Correct option is (b)
GATE-PH 2005 SOLUTION 266
49. The de-Broglie wavelength,
h h 6.6 1034
108
p 2mE 31
2 9.1 10 150 1.6 10 19
Correct option is (a)
50. 1-D, time independent, Schrodinger equation,
2 d 2
V x x E x
2m dx 2
In region-I, V x 0
d 2 2mE
2 0
dx 2
2mE
1 A sin k1 x B cos k1 x oscillating nature where, k1
2
In region-II, V x V0 E
d 2 2m
V0 E 0
dx 2 2
2m V0 E
II A e k2 x Be k2 x where k2 2
For the existance of bound state,
At x , II 0 0 B 0
So, II A e kx decaying nature
Correct option is (c)
2 2r
2 a0 2
A r cos e r sin d d dr
0 0 0
z r cos 2r
51. 2 2
2 a0
A e r 2 sin drd d
0 0 0
2r 2
3 a0
r e dr sin cos d d
0 0 0
2r
0
2
2 a0
r e dr sin d d
0 0 0
Correct option is (d)
GATE-PH 2005 SOLUTION 267
1
52. The degeneracy of nth state of 3D harmonic oscillator is n 1 n 2
2
1
For n = 2 level, g 2 3 4 6
2
Correct option is (b)
3 0 1 0 i 1 0 i3
53. Sx S y Sz I
8 1 0 i 0 0 1 8
i3 i3
S x S y Sz S x S y S z I
8 8
Correct option is (a)
54. The time-energy uncertainty principle
E t hvt
1
vt
2
1
v
2t
v 600 109
9 8
108
v 2tc 8 10 3 10 2
Correct option is (c)
mj
3/2
2 1/2
55. p3/2
–1/2
–3/2
2
1/2
p1/2
–1/2
D1 D2
2
1/2
s1/2
–1/2
mj = 1 mj = 0 mj = –1 mj = 1 mj = 0 mj = –1
() () () () () ()
For sodium doublet lines, the total number of transitions from 2 p3/2 and 2 p1/ 2 levels to 2 s1/2 level is 10.
Hence, correct option is (d).
56. Here, in the given three level system of atoms, the level with energy E2 has largest atoms N 2 amongest all three
levels, which confirms that this is a metastable state. Similarly, E1 and E 2 represents the ground and excited
state repectively. Thus, the laser emission will be possible only between E2 to E1 levels.
Hence, correct option is (b).
GATE-PH 2005 SOLUTION 268
3
57. The rotational Raman shift is given by Raman 4B J 6B, 10B, 14B,.... where J 0, 1, 2,.... and
2
h
B . Thus, it is confirmed that the typical Raman shift depends only on I.
82 Ic
Hence, correct option is (c).
58. The energy of rotational level is given by
EJ BJ J 1
E E J E J 1 BJ J 1 B J 1 J 1 1 B J J 1 J J 1 BJ J 1 J 1 2 BJ
E J
1
Also, vibrational energy is En n
2
Vibrational spacing,
E En E n 1 i.e. constant (independent of n)
Correct option is (a)
59. The typical wavelengths emitted by diatomic molecules in purely vibrational transitions lie in infrared region
(1 m 102 m) . In purely rotational transitions, the typical wavelength lie in microwave region
(103 m 104 m) .
Hence, correct option is (c).
60. Out of above options, only (b) 1 s2 2 s2 1 2 s1 s2 is valid.
Hence, correct option is (b).
ax
61. Given : p( x) dx ae dx ,
Probability that x lies between x1 and x2 ( x2 x1 ) is
x2 x2
p x dx / p x dx a e
ax
dx / a e ax dx e ax1 e ax2
x1 0 x1 0
Correct option is (a).
kT
62. Given : Z1
N
N kT
The partition for N such independent oscillator is Z N Z1
kT
The free energy is F kT n Z N NkT n NkT n .
kT
Correct option is (c).
GATE-PH 2005 SOLUTION 269
63. The possible configurations are
0
••
••
• •
The partition function hence is
z e 0/ kT e 2 / kT e / kT = 1 e 2 / kT e / kT
Correct option is (b).
64. Since the random walker takes steps to left or rightwith equal probability,
1
Probability of taking step to right =
2
1
And probability of taking step to left =
2
N N
The random walker comes back to origin after N even step if he takes steps to the right and steps to the
2 2
left.
N /2 N /2 N
1N 1 N! 1
Therefore, required probability = C N /2
2 2 N N 2
!
!
22
Correct option is (a)
65. The density of state in 3-D is
4p 2
g ( p )p p ,
h3
where p 2mE for free particle
2m
p E
2 E
4 2m E 1 2m 4 2 3/2 1/2
g ( E ) E E m E E
h3 2 E h3
Since electron spin degeneracy is 2 so we should multiply g(E) by 2.
8 2m E 1 2m
Therefore, g ( E ) E E
h3 2 E
For N-identical particles,
8 2
g ( E ) E 3
m3/2 NE1/2E
h
g ( E ) E N E E
Correct option is (c).
GATE-PH 2005 SOLUTION 270
66. F ( M ) F0 2 (T TC ) M 2 M 4 ... (i)
F (M )
For minima, 0 2 (T TC ) 2 M 4 M 3 0
M
M (T TC ) M 2 0
M 0 or M 2 (TC T ) M (TC T ) ... (ii)
2 F (M )
and 0 4 (T TC ) 12 M 2 4(T TC ) 12 M 2
M 2
For, M = 0,
2 F (M ) 2 F (M )
4(T TC ) 0 for T TC 0
M 2 M 2
Therefore, F ( M ) have a minima at M 0 for T > TC
For M (TC T ),
2 F (M )
4(T TC ) 12 M 2 4(T TC ) 12(TC T )
M 2
8T 8TC 0 (for T TC )
F ( M ) have a maxima at M (T TC ) for T > TC
Function F ( M ) has only one minima for T TC at M 0 .
Hence, correct option is (b).
67. For body centered cubic (bcc) structure, the atoms will lie at corners as well as body centers of cube.
R
R
a 3
R
R
R
R
a 3 4R
From the figure, 4 R a 3 R a .
4 3
Hence, correct option is (a).
J nevd 3 1020 1.6 1019 100
68. Conductivity, 2.4 10 24 1 m 1.
E E 200
Hence, correct option is (a).
69. The density of state of free electrons moving in a solid is proportional to square root of its energy,
dn E ... (i)
GATE-PH 2005 SOLUTION 271
For E1 0.1 eV, dn1 2.15 10 21 eV 1cm 3
For E2 0.4 eV, dn2 ?
dn1 E1
From equation (i), we have
dn2 E2
2.15 10 21 0.1 1
dn2 2 2.15 10 21 eV 1cm 3
dn2 0.4 2
dn2 4.3 10 21 eV 1cm 3 .
Hence, correct option is (d).
70. The concentration of electron in the conduction band
n Nc e c F B
E E /k T
... (1)
E E / k BT
Similarly, for hole p N v e F V ... (2)
By multiplying equation (1) and (2)
np N c N v e c v B
E E /k T
np ni2 by mass action law..
The intrinsic carrier concentration
E
ni ( N c N v ) exp g
2 k BT
0.66 1.6 10 19
1.04 1019 exp 23 3 1013 cm 3 .
2 1.38 10 300
Hence, correct option is (b).
71. The statement “It is paramagnetic below TC” is not correct as B 0 0 ( H M ) 0 H M
1 , which implies that superconductor is diamagnetic below TC.
Hence, correct option is (d).
A 240 236
72. Q k 5.26 k k 5.26 5.17 MeV
A 4 236 240
Correct option is (b)
73. If T be the threshold temperature of the reaction then,
3 4 e2 e2 15 13
kT 4 1.44 10 1.6 10 Jm
2 4 0 r 0
r 2r A1/3 2 1.2 1015 31/3 m
2 4 1.44 10 15 1.6 10 13 0
T k 1.38 1023 kgm 2 sec 2 K 1
3 1.38 10 23 2.4 1.44 10 5
1.28 1010 K
Correct option is (a)
GATE-PH 2005 SOLUTION 272
74. 8
O15 has Z = 8, N = 7 (1s1/2)2 (1p3/2)4 (1p1/2)1
1
J=
2
Parity = (–1)1 = –1 (odd)
1
Angular momentum and parity =
2
Correct option is (d)
75. The activity A of a sample decays exponentially with time, At A0 e t
log At log A0 – t (2.303)
It will be a straight line (figure (a)) between log At and t with a negative slope.
log At log At
t t
Figure (a) Figure (b)
But for complex decay i.e. for mixture of two half lives, we get the graph between log At vs. t as shown
in figure (b).
Correct option is (c)
76. Given circuit consist of two diode its a full wave rectifier
Number of Diode Rectifier
1 HWR half wave rectifier
2 FWR full wave rectifier
4 BWR Bridge wave
IL
Im
RL 1k
V2max 10 V
I Lmax 10 mA
2 I max 20
I DC I average mA is average value
Correct option is (a)
GATE-PH 2005 SOLUTION 273
77. f B A B A B A
f AB B AB A
f A B
A
f
B
1 OR- gate needed
Note: 1 A 1 1 B 1 Axioms
Correct option is (c)
5k
78.
1k
2V a
1k Vo
1V
1k
Vx
1 1 1
Vx Volt 0.5 Volt
11 2
As ideal op-amp has infinite input resistance. So, there will be no current pass through op-amp i.e. Va=Vx due
to virtual short.
KCL at (a)
0.5 2 0.5 V0
0
1 5
3 V0 0.5
V0 7 Volt
2 5
Correct option is (a)
79. R
+
Vin C Vout
–
By applying voltage divider rule.
1
Vi s
V0 s CS
1
R
CS
V0 s 1
its a low pass filter OR integrator..
Vi s 1 RCS
Note: If R and C are interchanged circuit will high pass filter OR differenciator.
GATE-PH 2005 SOLUTION 274
+ C +
Vi R V0
– –
V0 s RCS
Vi s 1 RCS
Correct option is (d)
80. Calculation of Thevenin votlage across RL 1k
Thevenin voltage is also referred as open circuit voltage
Open the load resistance of the circuit for the calculation of VTh and RTh.
2k 2k
+
20V VTH 10V
i –
20 10
i 2.5 mA
4
Apply KVL
20 2 i VTH
vTH 15 Volt
For RTH (Thevenin resistance) calculation
Voltage source is short circuited
Condition
Current source is open circuited
2k 2k 2k 2k 22
RTH
RTH 1 k
22
Option (a) is correct
81. (a, b). Here option (a) is correct
d2y dy
2
2 y 0
dx dx
mx
Let the trial solution be y C .e
e m 2m 1 0
mx 2
2
m 1 0 since, e mx
0
m 1, 1
The solution will be y c1 c2 x e mx c1 c2 x e x c1 e x c2 x e x
Correct option is (a) and (b)
GATE-PH 2005 SOLUTION 275
82. (a) We have the Lagrangian,
1 1
L m x12 x22 . m02 x12 x22 m02 µx1 x2
2 2
The equation of motion are
d L L
0
dt x1 x1
mx1 m02 x1 m02 µx2 0
x1 02 x1 02 x2 0
d L L
and 0
dt x2 x2
mx2 m02 x2 m02 µx1 0
x2 02 x2 02 x1 0
Correct option is ( )
m 0 ˆ m02 m02 µ
82. (b) Tˆ , V
2 2
0 m m0 µ m0
Vˆ 2T 0
m02 m 2 m02 µ
0
m02 µ m02 2 m
2
m 2
0 2 m m 204 µ2 0
m 2
0 2 m m02 µ
2 02 02 µ
0 1 µ , 0 1 µ
Correct option is (b)
83. (a) Surface current density = v R
(b) We know, B A
B dS A dS
µ0 I r 2 A d A 2 r
1 1
Aµ0 Ir µ0 R r
2 2
Correct option is (c)
GATE-PH 2005 SOLUTION 276
84. (a) The scattered wave is not spherically symmetric, its amplitude depends on the direction (, ) along which
is scattered. So, scattered wave is given by
ei k r
sca (r ) Af (, )
r
Correct option is (a)
2
(b) The differential scattering cross-section () in C.M. frame is given by () f () .
Correct option is (c)
85. (a) We have, 11.35 g/cm3 and atomic weight M 207.2 amu
Now, the volume of the lead can be calculated by
mass ( M ) 207.2 amu
Volume (V )
density ( ) 11.35 g/cm 3
207.2 amu
There are 6.023 × 1023 atoms present in 207.2 amu of lead and the volume is
11.35 g/cm3
6.023 1023 11.35
Therefore, number of atoms per cm3 = 0.3299 1023 atom/cm 3
207.2
3.3 1022 atom/cm3
Hence, correct option is (b).
(b) Number of atoms per cm3 (n) 3.3 1022 atoms/cm3
Energy of vancancy formation ( EV ) 0.55 eV and temperature (T ) 500 K
E 0.55
The number of vacancies per cm3 = nV n exp V 3.3 1022 exp 5
k BT 8.62 10 500
3.3 1022 exp ( 12.75) 9.577 1016 vacancies/cm3
Hence, correct option is (d).
GATE-PH 2006 SOLUTION 277
OBJECTIVE QUESTION
1. Let 1, 2 and 3 are eigen values of 3×3 matrix.
Trace of the matrix = sum of eigen values of the matrix
Therefore, 1 2 3 2
1 2 3 2 3 1
Correct option is (a)
2. According to Stokes Theorem,
A
C
d
S
A dS
Since, A is uniform. So, A 0
Therefore, A d 0
C
Correct option is (a)
q
2
p c A qA
3. Hamiltonian H q is correct p ' p for electromagnetic fields
2m c
Hence correct answer is (b)
4. For positive energy eccentricity becomes greater than one. Therefore path is hyperbolic
Hence correct answer is (d)
5. E x E 'x Bx B ' x
E y E ' y vB 'z By B ' y vE 'z / c 2
Ez E ' z vB ' y Bz B 'z vE ' y / c 2
Now, E 2 c2B2
E x2 E y2 E z2 c 2 Bx2 B y2 Bz2
E '2x c 2 B '2x 2 E '2y E '2z B '2y c 2 B '2z c 2
E '2x c 2 B '2x 2 [ E '2y 2 E ' y B 'z v v 2 B '2z E '2z 2 E 'z vB ' y v 2 B '2y
v 2 E '2z v 2 E '2y
c 2 B '2y 2 B ' y E 'z v B ' 2 2
z c 2 B ' z E ' y v ]
c2 c2
E '2x c 2 B '2x 2 1 v 2 /c 2 E '2y 2 1 v 2 /c 2 E '2z 2 v 2 c 2 B '2z 2 v 2 b 2 B '2y
E '2x E '2y E '2z c 2 B '2x B '2y B '2z
E '2 c 2 B '2
Correct option is (b)
GATE-PH 2006 SOLUTION 278
R
6. We know that V (r ) E dr Ein dr Eout dr
r r R
Eout depends on total charge. Therefore, V ( r ) also depends on total charge.
Correct option is (b)
7. Since, particle is moving along +x direction. The normalized wave function wtihin length L is
p
1 ix
e
L
p
[Since, wave function is just a plane wave e ik x , where k normalised over length L]
Correct option is (d)
0 1 0 i i 0 1 0
8. x y i i z
1 0 i 0 0 i 0 1
Correct option is (c)
a
9. Given : F VT 4
3
F
The chemical potential is 0
N V , T
Correct option is (a).
10. The particles are distinguishable in Maxwell-Boltzmann statistics and hence their wave functions do not over-
lap. So, the state of the combined two particle system, will be
ns (r1 , r2 ) n (r1 ) s (r2 )
Correct option is (d).
11. For radiation (X-ray) E eV hc hc 1 .
eV V
Hence, correct option is (c).
12. The principal series of spectral lines of lithium is obtained by transition between nP to 2S transitions (for
n 2 ).
Hence, correct option is (c).
13. Silicon is an indirect band gap semiconductor.
Hence, correct option is (c).
N
14. cos n , where N is an integer
2
B' C'
360º
And the number of possible fold axis, n
n n n
A B C D
Since, 1 cos n 1
GATE-PH 2006 SOLUTION 279
So, possible n are 360º, 180º, 120º, 90º, 60º
Corresponding the number of fold axis are 1, 2, 3, 4, 6
So, five-fold axis is not possible.
Correct option is (d)
15. Correct option is (c)
16. Since, Thorium series is 4n series.
Uranium series is (4n + 2) series.
Actinium series is (4n + 3) series and Neptunium series is (4n + 1) series.
209
Bi have mass number (A) = 209. 209 4 52 1
Therefore, it belongs to Neptunium series.
Correct option is (b)
17. Since average binding energy per nucleon i.e. B/A for all the nuclei 8 MeV. (i.e. of order 10 MeV)
Correct option is (d)
18. Interaction potential can be given as
0
a
V (r ) br V(r) r
r
a
where, because nuclear forces are short range and decreases with decreasing distance,
r
br because nuclear forces will be of repulsion type at long range.
Correct option is (c)
VDD
RD
19.
RG IG
+
VGS –
VGG RS
input junction
Ri
VGS
Input resistance Ri
IG
The larger input resistance of JFET is due to
(1) Reverse biasing gate to source junction
(2) Neglegible gate current IG(nA)
GATE-PH 2006 SOLUTION 280
RiFET
• Input junction of FET is always reverse bias
• I g nA practically
RiPractically 106 108
So, correct option is (b)
20. BJT as Digital switch.
Vcc Vcc
Vcc Vcc
Final
Rc Rc A B A B CKT
BJT
R
a a y y as switch
equivalent R R
figure.
Truth Table
A B Y
0 0 0
0 1 1
1 0 1 Then, Y = A + B. So, given circuit is OR gate.
1 1 1
Note: Switch closed current will flow y will have output
y 0 , when both switches are open.
Correct option is (a)
x1
1 1 0 x1 x2
21. 0 1 1 x2 x x
2 3
x3
Correct option is (a)
r
r ds
22. s r 3 V r 3 dV (using divergence theorem)
Since the closed surface encloses the origin i.e. r 0
r 3
So, 3 4 r
r
r 3
So, r 3 dV 4 r dV 4
V V
Correct option is (c)
e2 z
23. f z 4 has a pole of order 4 at z = –1.
z 1
Residue of f z at z = –1
GATE-PH 2006 SOLUTION 281
1 d3 4 e2 z C : |z| = 3
z 1 4
4 1! dz 3 z 1 z 1
y
1 4
8e2 z e 2 x
3! z 1 3 z = –1
4 8 i 2
So, I 2 i e 2 e
3 3
Correct option is (a)
1 ikx 1
24. k Fourier component of x
th
x e dx
2 2
Correct option is (c)
25. We know that, µ B
iˆ ˆj kˆ
µx µy µz iˆ µ y B0 ˆj µx B0 0kˆ
0 0 B0
x µ y B0 , y µx B0 , z 0
We know, torque = rate of change of angular momentum,
d
dt
Lxiˆ L y ˆj Lz kˆ x iˆ y ˆj z kˆ
d d
dt
Lx µy B0 ;
dt
Ly µx B0
µ
µ L; L
d µx
µy B0 ;
d µy µ B
x 0
dt dt
d 2 µx d 2 µx 2 2 d 2 µx
B0 µx B0 2
B µ
0 x 2
2 B02 µx 0
dt dt dt
Correct option is (c)
26. V r r 2 r n n 2
We know from Virial theorem
n
If V r n then T V
2
2
T
V T V
2
Hence correct answer is (a)
GATE-PH 2006 SOLUTION 282
27. r 5 iˆ 2t 2 ˆj
dr
v 4t ˆj
dt
p m v 2 4t ˆj 8t ˆj
L r p 5 iˆ 2t ˆj 8t ˆj
2
40t kˆ 40 2 kˆ 80 kˆ
Hence correct answer is (b)
28. I xy mi xi yi
m 11 m 1 1 2 m 11 2 m 1 1
2m 2m
Hence correct answer is (c)
29. For canonical transformation
Poisson Bracket Q, P q , p 1
If Q p, P q
Q P Q P
Q, P q, p 0 1 1 1
q p p q
It Q p, P q
Q P Q P
Q, P q, p 0 1 1 1
q p p q
only (i) is canonical transformation
Hence correct answer is (b)
2m0
30. m
3
m0 2m0 3
1 v 2 /c 2
1 v 2 /c 2 3 2
c
v
2
2m0 c m0c
momentum p mv
3 2 3
Hence correct answer is (d)
31. The total charge , Q qi q q 2q 0
So, the dipole moment of this charge distribution does not depend on choice of origin
Therefore, net dipole moment,
GATE-PH 2006 SOLUTION 283
p qi ri q ajˆ akˆ q ajˆ akˆ 2q akˆ
qajˆ qakˆ qajˆ qakˆ 2aqkˆ = 4aqkˆ
Correct option is (a)
32. According to Ampere’s law,
B d µ0 ds
J
d 2
B 2 d µ0 J 0 rdrd
0 0
d3
B 2 d µ0 J 0 2
3
µ0 J 0 d 2
B ˆ
3
Correct option is (c)
33. A yiˆ 2 xjˆ
iˆ ˆj kˆ
B A 2 1 kˆ 3kˆ
x y z
y 2x 0
Correct option is (b)
34. We know the boundary condition, D2 nˆ D1 nˆ
D2 nˆ D1 nˆ (since surface is charge free)
Or, D2 cos 2 D1 cos 1 2 E2 cos 2 1 E1 cos 1 ... (i)
And E1 E2
E2 sin 2 E1 sin 1 ... (ii)
Therefore, (ii) divided by (i), we get
tan 1 1
tan 2 2
Correct option is (a)
35. The Maxwell’s equations are
E ; B 0 ; E B
0 t
E
B µ0 J 0 µ0
t
E
For steady state current flow 0
t
GATE-PH 2006 SOLUTION 284
B
B µ0 J and since J is time independent 0
t
Therefore, the Maxwell’s equation for time independent charge density and current density are
.E / 0
.B 0
E 0
B Jˆ 0
Correct option is (b)
36. Observer will see linear polarized light when he will look from x-direction. But if observer looks from z-axis
then he will see electric field vary as a circular polarization light with time.
Correct option is (c)
37. Probability current density,
i i 2 k
J
2m
* *
2m
2
2
A ik ik A A
m
Correct option is (a)
3 2 1
38. x 0 x 1 x 2 x C0 0 x C1 1 x C2 2 x
14 14 14
The probability of finding the oscillator in ground state, is
2 2 9
P0 0 | C0
14
Correct option is (c)
2 1
39. x, t 0 0 x 1 x
5 5
Therefore, wave function at time t is
iE0t iE1t
2
1
x, t 0 x e e
5 5
Correct option is (d)
40. Lx , y ypz zp y , y y, y pz y pz , y z, y p y z p y , y
0 0 0 z i i z
Lx , y iz
Correct option is (d)
41. The probability of finding the electron at r = 0 is zero for 1s, 2s, 2p
Hence, (a) , (c) and (d) are correct.
A B A B
N, P N, P
42.
T1 T2
GATE-PH 2006 SOLUTION 285
T1 T2
The final temperature would be T f as both the vessels have all same parameters.
2
Now change in entropy is
Tf Tf
dT dT T f2
S S1 S2 Cv Cv = Cv n
T T T1T2
T1 T2
3
where Cv N k B (for monoatomic gas).
2
3 T f2
S N k B n
2 T1 T2
3 (T T )2
N k B n 1 2
2 4 T1 T2
Correct option is (c).
43. Since the gas is expanded adiabatically into vaccum, it is a case of free expansion. Hence, internal energy does
not change
3 a 3 a
E1 E2 2
nRT1 nRT2
V1 2 V2
3 3 1 1 2a 1 1
nRT2 nRT1 a T2 T1
2 2 V2 V2 3nR V2 V2
2a 1 1
So for one mole, T2 T1 3R V V
2 2
Correct option is (c).
44. The Hamiltonian for one-dimension harmonic oscillator is
px2 1 2
H kx
2m 2
Since the Hamiltonian has two quadratic terms and as per as law of equipartition of energy, each contributes
1
k BT .
2
1
So, mean internal energy = 2 k BT k BT
2
Correct option is (b).
45. Number of ways of choosing n-atoms out of N-atoms N Cn
and number of ways for putting n-atoms in N -sites N
Cn .
N N! N !
Therefore, total number of microstates, N Cn Cn
n !( N n)! n !( N n)!
Correct option is (d).
GATE-PH 2006 SOLUTION 286
46. The energy of a ion is
H for spin 2 ion
1
E H
1
H for spin 2 ion
H H
H
k BT k BT
So the partition of an ion is Z1 e e 2 cosh
k BT
N
H
N
The partition for N such non-interacting ions is ZN (Z1) 2cosh
kBT
The Helmholtz free energy is
H
F k BT n Z N N k BT n 2 cosh
k BT
Correct option is (b).
47. px = +
– 2mE
x
x=0 x=L
There is no force on the particle except due to collision.
Therefore momentum of particle remains constant and changes its direction due to collision.
Hence correct answer is (a)
48. Microwave region 104 m 10 2 m of the electromagnetic spectrum corresponds to the pure rotational spectra.
Energy of the level (in terms of wave number)
E J B J J 1 , J 0, 1, 2, ...
Selection rules
J 1 ,
Therefore, wave number of emitted or absorbed radiation is
2 B J 1
Putting J = 0, 1, 2, 3
2 B, 4 B, 6 B is the position of spectral lines.
Separation between spectral lines
4 B 2 B 2 B = constant
Given :
2 B 0.7143 cm 1 0.7143 10 2 m 1
0.7143
B 102 m 1 0.3571 10 2 m 1
2
h
Since, B 2
8 Ic
h 6.63 1034
moment of inertia I 2 2 2 8
7.8 1046
8 BC 8 3.14 0.357110 3 10
Hence correct option is (d)
GATE-PH 2006 SOLUTION 287
49. For He-Ne laser the laser transition takes place in Ne only.
Helium 5s
H3 21s E6
Neon
H2 23s 4p
E5
4s
E4
3p
E3
3p
E2
1s2 2s2 2p6
H1
Figure : (Relevant energy level of the He-Ne laser)
According to above diagram, the electronic transition 6s 5p is not responsible for LASER action in He-Ne
laser.
Hence, correct option is (a).
50. For n 2 level, the allowed values of n, n1 , n 2 and m are given in following table :
n n2 n1 (n 2 n1 ) 2 (n 2 n1 ) m [n n 2 n1 1]
1 1 0 0 1
1 0 1 2 0
2
0 1 1 2 0
0 0 0 0 1
The Stark shift of hydrogen n 2 energy level is shown below :
n 2 n1 n (n 2 n1 ) m
1 2 0
n2
0 0 1, 1
1 2 0
From above figure, it is clear that two levels corresponding to m 1 and m 1 falls together i.e. application
of electric field partially lifts the degeneracy of n 2 level and splits n 2 into three levels.
Hence, correct option is (b).
51. The selection rule for the quantum number F in hyperfine transitions : F 0, 1 but F 0 F 0 .
Hence, correct option is (c).
GATE-PH 2006 SOLUTION 288
D Eex
Energy
cm
–1
52.
D Dg
Illustration of dissociation
From the above figure vcm 1 Dg E ex cm 1 Dg v E ex cm 1
Therefore, dissociation energy of the homonuclear molecule will be ( E ex. ) .
Hence, correct option is (c).
53. In the NMR spectroscopy of CH3CH2OH three bunches are observed
Hc Hb
Hc C C OHa
Hc Hb
Three type of proton exist in this molecule so multiplicity (2NI + 1) for every proton
For–CH3 : 3H c 2NI 1
where, N is number of near by proton (H)
I = spin of proton (H)
1
2 2 1 3 triplet
2
1
For –CH2 group : 2H b (2NI 1) 2 3 1 4 (quartet)
2
For –OH group : a H alcoholic hydrogen broad signal
So, Due to –CH2 group quartet are obtain four time are observed
Hence, correct option is (d).
2
54. Effective mass (m*) ... (i)
d 2E
2
dk
Here, E ( k ) Ak 2 Bk 4
dE 3 d 2E
2 Ak 4 Bk ; 2
2 A 12 Bk 2
dk dk
d 2E
At k k0 , 2 A 12 Bk02 ... (ii)
dk 2 k k0
GATE-PH 2006 SOLUTION 289
2
From equation (ii), equation (i) implies m * .
(2 A 12 Bk02 )
Hence, correct option is (c).
2 4
55. The first BZ is a square of side and second BZ is a square of side a .
a
2
2 4 2
Area of first BZ ... (i)
a a2
1 2 4 2
and area of second BZ = 4 ... (ii)
2 a a a2
Therefore, area of second BZ = area of first BZ.
Hence, correct option is (d).
56. The highest occupied energy band of the crystal is half filled.
Hence, correct option is (c).
57. The plasma frequency (p) is directly proportional to density of free electrons
i.e., p N ... (i)
Initially the electron density, N1 1028 electrons/m3
Finally the electron density, N 2 10 26 electrons/m3 and initial plasma frequency ( p )1 5.7 1015 Hz.
Let the final plasma frequency ( p ) 2 x
( p )1 N1 1028 5.7 1015 1028
( p )2 N2 1026 x 1026
1026 1
x 5.7 1015 28
5.7 1015 x 5.7 1014 Hz
10 10
Hence, correct option is (b).
58. Type-I superconductors exhibit complete Meissner effect and are completely diamagnetic. Type-II supercon-
ductors exhibit complete Meissner effect upto H C1 only. Above H C1 , the flux begins to penetrate the speciman
and for H H C2 , the complete penetration occurs and material becomes completely normal conductor. In the
region between the field H C1 and H C2 , the Meissner effect is not strictly followed.
–M –M I II III
H H
Type-I superconductor Type-II superconductors
Therefore, the statement in option (b) is not true.
Hence, correct option is (b).
GATE-PH 2006 SOLUTION 290
N Np 2
59. The susceptibility ( ) is given by ( r 1) ( e i )
0 3 0 k BT
A Np2
30kB = slope
(= r – 1)
B
N
Intercept = ( + i)
0 e
1/T
Since, slope of B > slope of A, therefore, B is more polar.
Since intercept of A is larger than B, therefore ionic polarizability of B is lesser in comparison to A.
Hence, correct option is (d).
60. The experimentally measured spin g-factors of proton and neutron indicate that both proton and neutron are
not elementary point particles.
Correct option is (b)
61. 1 e0 25 Mn29
54
24 Cr3054 .............
For the conservation of lepton number in the above reaction (equation), we need to add a lepton member
on the right side. i.e. either electron or electron neutrino or muon or muon neutrino.
1 e0 25 Mn29
54
24 Cr3054 e
Since, on left side there is an electron so we need a neutrino on right side.
Correct option is (a)
62. For (a) p n e
Charge Q : 0 1 0 1, Q 0
L : 1 0 0 1, L 0
B : 0 1 1 0, B 0 (allowed)
0
For (b) e e
Charge Q : 1 1 0 0 Q 0
Electronic Lepton No : 0 1 0 1 Le 0 (forbidden)
For (c) p n K K
Charge Q : 1 (1) 0 1 1, Q 0
L : 0 0 0 0 0 L 0
B : 0 1 1 0 0 B 0 (allowed)
For (d) e e
Charge Q : 1 1 0 0 Q 0
GATE-PH 2006 SOLUTION 291
Electronic Lepton No : 1 0 0 1 Le 0
Muonic Lepton No : 1 0 0 1 Lµ 0
B : 0 0 0 0 B 0 (allowed)
Correct option is (b)
63. Deutron is a single two nucleon (one p and one n) bound system which is found in nature. The important
experimental determinal properties about deutron are given below:
(i) The B.E. of deutron/nucleon is very small compare to other nuclei i.e. it is a weakly system.
(ii) The ground state spin of deutron Id 1 (iii) The ground state parity of deutron = even (+)
(iv) The quadrupole momento of deutron Qd 0
(v) The magnetic moment of deutron is slightly different from the sum of intrinsic mag. moments of neutron
+ proton i.e. n p d 0
3 3
These factor represents that the ground state of Deutron is a mixture of L S0 & L D2 states in which 3 S
contribution is 96% and 3 D contribution is only 4%.
Correct option is (b)
64. Considering positive haf cycle
D1 ON and D2 OFF
VC1
+ –
Vm + C1 + +
+
– Vin C2 Vout Vm L
KV
–Vm – – –
Vm VC1 0
VC1 Vm
Considering (–)ve half cycle.
D1 OFF , D2 ON
VC1
+ –
– +
+
Vm KVL VC 2 Vout VC2 Vout
–
+ –
Vm VC1 VC2 0
Vm Vm VC2 0
VC2 Vout 2Vm
Vout 2Vm
Note : This circuit is also called voltage doubler circuit
Consider magnitude. So, change Vout (Polarity)
Correct option is (a)
GATE-PH 2006 SOLUTION 292
65. Given fcutoff 2 KHz
f H 2 KHz LPF
Gain = 1.5
R1 15k
R2 Vout
Vin(s) C = 0.047 µF
1
Vi s
V x s CS
1
R2
CS
Vi ( s )
Vx ( s )
1 R2CS
Virtual ground concept at inverting input
15
15 Vi ( s ) 1
Vout 1 Vout ( s ) R1 Gain
;
R1 1 R2CS Vi ( s ) 1 R2CS 1 R2CS
15
Gain 1.5 1 R1 30 k
R1
Gain Gain A0 1
equating ; fH f H ; upper-3dB frequency
1 R2C j 1 j 2fR2C 1 j f 2R2 C
fH
1
2 103
2R2 0.047 106
R2 1.7 k ; R1 30k and R2 1.7 k
Correct option is (d)
d 2v dv
66. Consider the differential equation, 2
2 v1 0
dt dt
This is a second order differential equation.
If we can integrate it two times, we can easily get the solution of the above equation.
(Since, because it is involving second order derivatives)
So, two op-amp integrator must be there.
Since, there will be two solution of differential equation v t and v1 . So, we also need an adder to add the
outputs.
Correct option is (a)
67. Output Y = ?
A A
A+B
B
B
B Y
A A+B
GATE-PH 2006 SOLUTION 293
Y A B A B y AB AB B
Y B 1 A A Y B
Note: 1 a 1 1 a a 1 (by boolean rule)
Correct option is (d)
68. 1 volt analog output for 010100
digital input
20 in decimal
1
for 000001 Volt step size/least count.
20
63 1
for 111 111 Volt
20
26 1 63
(max) input 111 111 output is 3.15 Volt
Correct option is (c)
69. CLOCK signal is applied to Flip-Flop “A”, so it becomes “LSB” of Flip-Flop.
(i) Here clock is applied at negative edge trigger and output is taken at Q so it is up-counter.
(ii) It is MOD 12 counter and here QA is LSB so output must be 1100 (DCBA) to reset the counter.
Truth table of given counter :
D C B A
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
Counter will
0 1 1 0 count till 11 and
0 1 1 1 12 output will be
1 0 0 0 discarded.
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
CLR
X1
Hence X1 X2 are C and D
X2
It is clear from truth table, that as soon as 1100 appears, input to the NAND gate becomes 11 which reset the
flip-flop with clear = 1
So, X1 and X2 should be connected QC = 1 and QD = 1 i.e. C and D.
Correct option is (d)
GATE-PH 2006 SOLUTION 294
L1 L2
70. Hartely oscillator,
M
Lequivalent L1 L2 2 M (Same behaviour)
1
f0 Leq Lequivalent
2 Leq .C
Put value
1
f0 Hz
2 L1 L2 2M C
71. Given : perturbed Hamiltonian
E1 A E1 0 0 A
H * * H0 H '
A E2 0 E2 A 0
Therefore, the first order correction in E1 is
0 A 1 0
1 H ' 1 1 0 * 1 0 * 0
A 0 0 A
Correct option is (d)
72. The second order correction in E1, is
2 2
0 A 1 0
2 H ' 1
2 0 1 * 0 1 * 2
A 0 0 A A
E1 E2 E1 E2 E1 E2 E1 E2
Correct option is (d)
1
73. The first order correction to the eigenfunction is
0
0 A 1 0
0 1 * 0 1 * *
0
2 H ' 1 A 0 0 0 A 0 A 0 *
2 1 1 1 A
E1 E2 E1 E2 E1 E2 E1 E2
E1 E2
Correct option is (a)
74. Sum of the eigenvalues = Trace of the matrix = 5
1 2 3 5
5 2 3 5 2 3 0
So, 2 1 and 2 1
Correct option is (c)
75. Eigenvalue equation : A I X 0
2 3 0 x1
3
2 0 x2 0
0 0 1 x3
GATE-PH 2006 SOLUTION 295
Putting 5 , we get
3 3 0 x1
3 3 0 x 0
2
0 0 4 x3
3 x1 3x2 0 and 4 x3 0 x1 x2 and x3 0
Assuming, x1 x2 k , x3 0
k
X k ,
0
putting normalization condition X † X 1 k 1
2
1
1
So, X 1
2
0
Correct option is (d)
76. (hkl ) (002), 60, 1.54 Å
a a a a
From equation, d hkl
2
h k l 2 2 2
0 0 2 2 2
4 2
From Bragg’s law, 2d sin n
a
2 sin 6 0º 1 1.54 Å (Taking first order reflections, n 1 )
2
1 1.54 2
a 1.78 Å (No option matches the result)
3
2 1.54 2
By taking second order reflections i.e., n 2 : a 3.556 Å 3.56 Å
3
Hence, correct option is (c).
nM 2 50.94
77. Density of the crystal ( ) kg/m 3
Na 3
6.023 1026 (3.56 1010 )3
0.3749 104 kg/m3 3.75 103 kg/m3
Hence, correct option is (a).
78. L T V x
1 y m
L m x 2 y 2 mgx
2 x
y
x A cos , y A sin
x A sin , y A cos
GATE-PH 2006 SOLUTION 296
1
L m A2 sin 2 2 A2 cos 2 2 mg A cos
2
1
L m A2 2 mg A cos
2
Hence correct answer is (a)
79. Equation of motion
d L L
0
dt
d
mA2 mg A sin 0
dt
m A2
mg A sin 0
g
sin 0
A
Hence correct answer is (d)
1 2 2
80. P 2
r r
r kr 4kr
Correct option is (b)
dV
81. E dS 0
2
d
4kr 2 kd 2
E 4 d r sin d d dr E rˆ
0
0 0
Correct option is (a)
82. The number of final states of electrons corresponding to momenta between p and (p + dp) is proportional to
p 2 dp .
Correct option is (c)
83. The number of emitted electrons with momentum (p) and energy (E) in the allowed approximation is
proportional to p E0 E .
Correct option is (b)
84. The density of state is
4p 2 dpV
g ( p ) dp ,
h3
dE
where E pc for photons dp
c
4V E 3 dE 4V 2
g ( E ) dE g p dp 3 3 E dE
h3 c 2 c hc
4V
g( E) 3 3
E2
hc
GATE-PH 2006 SOLUTION 297
Since photons have two planes of polarization, we should multiply the density of state by the factor of two.
8V
Therefore, g ( E ) 3 3
E2
hc
Correct option is (a).
85. The energy density is
8V
g ( E ) dE 3 3
E 2dE , where E for photons
hc
V 2 h
g () d g E dE 2 3
c
d 2
1
Also, f ()
Also, for bosons (e.g., photons)
k BT
e 1
The average number of photons are
V 2 d
N f ( E ) g ( E ) dE f () g () d 1
2 c 3 0 k BT
0 0 e
3
V k BT x2
ex 1 dx x
2 c 3 0 k BT
Clearly, N T 3
Correct option is (c).
GATE-PH 2007 SOLUTION 298
OBJECTIVE QUESTION
Q.1 – Q.20 : Carry ONE mark each.
1. Eigenvalues of an anti-hermitian matrix are either zero or purely imaginary
Correct option is (d)
2. When fire engine radially outward, its velocity is not perpendicular to radial direction. Therefore new bound
orbit cannot be circle. So new bound orbit must be ellipse.
Hence correct answer is (c)
3. P0 150 W ; Pi 1.5W
150
Power Gain 100
1.5
Gain in dB 10 log10 100 20 dB
Note: For power we take 10 log P
4. Since, total charge is zero. So, monopole moment is zero.
Also, quadrupole moment, ij 3 xik x jk rij ij qk 0
So, electrostatic potential will be due to charge distribution of dipole moment.
5. When displacement current reduces to zero this says that the current in the circuit is will be zero. So, total
energy will stored entirely in its electric field.
6. P : Frank-Hertz experiment was performed to identify energy levels in atoms.
Q : Hartree-Fock method is used for the wave functions of atoms.
R : Stern-Gerlach experiment was performed to check the space quantisation of “spin angular momentum of
atoms”.
S : Frank-Condon principle was given for the electronic excitation of molecules.
Hence, correct option is (a).
7. x e iax b eb e iax A e iax
This is wavefunction of free partile. So, V x 0
Correct option is (c)
8. In weak magnetic field (Anomalous Zeeman effect ), selection rules are
M J 0, 1, M J 0 0 , if J 0 . Thus, D1 and D2 will give 4 and 6 transitions.
M M
J J
½ 3/2
½
2 2
P1/2 P1/2
–3/2
–½ –½
½ ½
2 2
S1/2 S1/2
–½ –½
4 transitions 6 transitions
Correct option is (a)
GATE-PH 2007 SOLUTION 299
9. In a He-Ne laser (Four level laser), the population inversion is achieved by electric discharge He atoms help in
achieving a population inversion in the Ne atoms. Thus, the laser transition takes place in Ne only.
Hence, correct option is (b).
10. If we take classical approach, the gas molecules are taken to be distinguishable. Hence,
N
Z N Z
But if take semi-classical approach, the gas molecules are taken to be indistinguishable.
N
Z
Therefore, Z N
N!
Correct option is (a) or (b).
11. According to Meissner effect, the superconductor expels out its magnetic flux because the surface current
induces a field in the direction opposite to the applied magnetic field and it behaves like a perfect diamagnet
Hence, correct option is (d).
12. Since, potential is time independent, therefore the probability of finding the particle in one well will be time
independent.
Correct option is (b)
13. The quantum statistics applies when the de-Broglie wavelength of the particles (or associated with the par-
ticles) is comparable to the size of the particles. Under this condition, their wavefunctions start overlapping.
Correct option is (a).
14. Electron configuration of oxygen, O8 :1s 2 2 s 2 2 p 4
So, it has two free electron in the outer shell.
So, it will be attracted towards the higher field because it is paramagnetic.
Correct option is (d)
15. Since number of neutron increases in comparision to number of protons with atomic number i.e. (N–Z) in-
creases with Z. For the radioactive elements like 92U235 and 92U238, (N-Z) values are 51 and 54 respectively.
When a uranium nucleus, undergoes fission fragments (Kr and Ba) generally contain many neutrons.
Correct option is (a)
16. N E NC N B
As per manufacturing of BJT doping of emitter > collector > base.
Correct option is (a)
17. For He (2p2) configuration, we have two equivalent electrons in 2p2 state.
1
Thus, For one electron, l1 1, s1
2
1
For other electron, l2 1, s2
2
Thus, possible values of s and l are :
1 1 1 1
s s1 s2 , s1 s2 1,...., ( s1 s2 ) , 0, 1
2 2 2 2
GATE-PH 2007 SOLUTION 300
Multiplicity (2s 1) 1, 3 and
l l1 l2 , l1 l2 1,...., (l1 l2 ) 1 1 , 1 1 1,...., (1 1) 0, 1, 2 (S, P, D, states)
Thus, in total we have six terms, three singlet and three triplet terms.
All these terms are even because the configuration 2p2 is even (l 1 1 2) . We can write these terms as
1
S, 1P, 1D, 3S, 3 P, 3 D
To take into account spin-orbit interaction, let us combine L and S to form J.
Now, J L S ,..... (L S).
For single terms, we have For triplet terms, we have
S 0 S 1
; J 0 ; 1S0 ; J 1 ; 3S1
L 0 L 0
S 0 S 1
; J 1 ; 1P1 ; J 2, 1, 0 ; 3 P0,1 2
L 1 L 1
S 0 S 1
; J 2 ; 1D 2 ; J 3, 2, 1 ; 3 D1, 2 3
L 2 L 2
For the equivalent electrons (according Hund’s rule), terms 1 P1 , 3S1 , 3D1, 2, 3 will not be present and terms
1
S0 , 1D 2 , 3P0,1, 2 will be present.
Hence, correct option is (b).
18. For the given potential,
x 0 for x > 0 and x 0 for x < 0. For the sake of continuity at x = 0
x should be zero at x 0
The normalized wavefunction of a particle moving under a linear harmonic oscillator potential is given by
2 x 2 m
(where, 2 )
n Ae 2 H n x
n x 0 will be non-zero for n = 0, 2, 4, 6 ............. and will be zero 1, 3, 5, 7........ So, possible value of
‘n’ is odd.
The energy eigenvalues of the particle will be
1 3
En 2n 1 2n
2 2
Correct option is (c)
19. Correct option is (c)
20. Strangeness is conserved in strong and electromagnetic both interactions.
Correct option is (d)
Q.21 – Q.75 : Carry TWO marks each.
5 4
21. A ; eigenvalue equation : A I 0
1 2
GATE-PH 2007 SOLUTION 301
5 4
0 2 5 4 0 2 7 6 0 6, 1
1 2
Now, A I X 0
5 4 x1
0
1 2 x2
1 4 x1
Putting 6 0 x1 4 x2 0 x1 4 x2
1 4 x2
x1 4
So, X
x2 1
4 4 x1
Putting 1 0 4 x1 4 x2 0 x1 x2
1 1 x2
x1 1
So, X
x2 1
Correct option is (a)
y2
22. F ˆj zkˆ
L
The net flux of F associated with the cube
F ds
F dV (using Divergence theorem)
L L L L L L
2y
dxdydz dxdydz L3 L3 2 L3
0 0 0
L 0 0 0
Correct option is (a)
23. r xiˆ yjˆ r 2
r
r r 1 rˆ
r
Correct option is (c)
24. The rotation matrix corresponding to the anti-clockwise rotation about y-axis by an angle will be
cos 0 sin
Ry 0 1 0
sin 0 cos
x ' cos 60º 0 sin 60º 2 1 3
y ' 0 1 0 3 3
z ' sin 60º 0 cos 60º 2 1 3
Correct option is (a)
GATE-PH 2007 SOLUTION 302
1
25. f z has simple poles at z ia
z a2
2
Correct option is (b)
s 1 s 1 1 2 1
26. L1 2 L1 2 2
L 2 2
cosh 2 x sinh 2 x
s 4 s 2 2 s 2 2
Correct option is (b)
27. The solution of Legendre differential equation will be
y A Pn x BQn x
where Pn x is Legendre Polynomial of order ‘n’ and Qn x is Legendre function of second kind and
given by
1 1 x
Q0 x n
2 1 x
x 1 x
Q1 x n 1
2 1 x
This shows that, Qn x will diverge at x 1 .
Correct option is (c)
dy dy y
28. x y x4 x3
dx dx x
I .F . e
dx / x
e ln x x
3 x5
Solution will be xy x xdx xy c
5
4
Putting the condition, at x = 1, y = 1, we get c
5
x5 4 x4 4
So, xy y
5 5 5 5x
Correct option is (d)
29. (P) rest mass (2) Lorentz invariant
(Q) charge (4) conserved and Lorentz invariant
(R) Fourmomentum (1) time like vector
(S) Electromagnetic field (3) tensor of rank 2
Hence correct answer is (c)
4 5
30. M r
3
5
2 4 5 7 4 r
I yy r 2
5 3 5 3 2
4 5 7 4 5 25 5
r 2 r r5
3 5 16 15 16 12
Hence correct answer is (b)
GATE-PH 2007 SOLUTION 303
31. L explicitly depends on t so energy is not conserved
z is cyclic coordinate so pz is conserved
L is invariant under rotation about z axis so Lz is conserved
Hence correct answer is (c)
32. Kinetic energy
1
T m x12 x22 x32
2
Potential energy k k
x1
1 2 1 x2
V k x2 x1 k x3 x2 x3
2 2
p2
33. H pq
2m
H p
Hamilton’s equation q q
p m
p m q q
Lagrangian L pq H
p2 p2
pq
pq p q q
2m 2m
Putting the value of p from above
2
2m q q
L m q q
2
1 2
L m q q
2
Hence correct answer is (b)
µr µ0 IN
34. d µr µ0 IN B 2 r
B
Correct option is (c)
35. We have, I 150 Wm 2
1 2
We know, I u c 0 E0 c
2
2 1 12 2 2 8
Therefore, the average power I R 8.85 10 60 2 3 10 120
2
Correct option is (c)
36. A 3 zy iˆ 3 xyjˆ 3 xykˆ 0 does not follow coulomb gauge
1 4
A 2
4 0 does not follow Lorentz gauge
C t C
Correct option is (d)
GATE-PH 2007 SOLUTION 304
37. Schrodinger equation,
2 d 2
V x x E x
2m dx 2
2 2 3
2m
6a 4a 2 x 2 x e ax V x x e ax x e ax
2
6 2 a 2 a 2 2 x 2 1 3
m 2 x 2
2m m 2 2
2
Comparing the coefficient of x from both sides,
m
a
2
Correct option is (c)
38. We have x, p i
x3 , p
x , p nx
n n 1
x, p
3 x 2 x, p 3 x 2i 3ix 2
Correct option is (c)
3
39. There will be only three bound state if R
2
9 2
2 R 2
4
mV0 a 2 92
2
2 2 4
2 2 2 9 2 2
V0
m a02 2m a02
Correct option is (a)
3/2
1 1 r r /2a0
40. r , , 1 e cos
2 a0 2a0
r
r
2 a0
Compare the e with
. We can say n = 2
na0
e
Since, there are only one cosine term so 1
Since, there are no term containing , so m = 0
So, the state is 2P.
Correct option is (c)
41. X , Y iZ , Y , Z iX and Z , X iY
Since, component of total angular moment follow the above relation
GATE-PH 2007 SOLUTION 305
J x , J y iJ z , J y , J z i J x J z , J x i J z
1 3
So, z operator is behaving like Jz and it has eigen values mJ where mJ 0, , 1, .......
2 2
Correct option is (c)
42. Given : T1 0 ºC 273 K, Q1 ? and T2 22 ºC 295 K, Q2 450 kJs 1 T2
Q2
Q Q
Using 1 2 , we have W
T1 T2
T1 273
Q1 Q2 450 417 kJs 1 Q1
T2 295 T1
Correct option is (d).
43. At triple point, vapour pressure of a solid phase gets equal to that of liquid phase.
3863 3063 800
23 19 4 T 200 K
T T T
Correct option is (d).
a
44. Given : F VT 4
3
F 4a
The entropy is S VT 3
T V 3
F a 4
and the pressure is P T
V T 3
Correct option is (a).
45. Given : Every excited state is 3-fold degenerate.
2E0 3E0
i.e., E0 and so on.
The possible states where the total energy is 5E0 are as follows :
2E0 2E0 2E0
E0 E0 E0
So number of possible states are three.
Correct option is (b).
46. For Helium : 4 3P1
For the electric dipole transition, 1 i.e. the parity must change. Since the parities are determined
by (–1), the transitions corresponding to 0, 2, 4,.... are not allowed, because in these cases the
parities of the initial and the final eigen function would be the same. The transitions allowed by parity are
1, 3, 5,....
The selection rule m 0, 1 disallowes the transitions 3, 5,.... because then m may be
greater than 1. Hence, the selection rule for is 1 . Out of these four options, only option (d)
corresponds to 1 .
Hence, correct option is (d).
GATE-PH 2007 SOLUTION 306
47. For 3 D3 state, L 2, J 3, 2S 1 3 S 1
The angle between the orbital and spin angular momentum can be calculated using formula,
J J 1 L L 1 S S 1
cos cos (L S)
2 L(L 1) S(S 1)
3(3 1) 2(2 1) 1(1 1)
2 2(2 1) 1(1 1)
12 (6 2) 4 1
2 6 2 22 3 3
1
cos 1 (L S) cos1
3
Hence, correct option is (a).
3 3 1
48. For Na 3 2 P3/ 2 with nuclear spin I , we have, J , 2S 1 2 S and L 1
2 2 2
For hyperfine structure,
Total angular momentum F = total angular momentum of electron + total nuclear spin
J I , J I 1,...., J I
3 3 3 3
F J I to 3, 2, 1, 0
2 2 2 2
3
Thus the hyperfine structure of Na 3 2 P1/ 2 with nuclear spin I has 4 states as shown below..
2
F
3
2
3 2P3/2
1
0
Correct option is (d)
49. The allowed rotational energy
E J BJ (J 1) ... (i)
The wave number () of the emitted or absorbed radiation under allowed transition is given by
2B (J 1) ... (ii)
h
where J is rotational quantum number and B is rotational constant.
8 2 Ic
For the reduced mass µ the wave number corresponding to the transition J 4 5 is given by
2B (4 1) 10B 10 2 10 h
h
... (iii)
8 Ic 8 2 (r 2 )c
[We take lower value of J i.e. J = 4, here]
GATE-PH 2007 SOLUTION 307
For the reduced mass 1.05 and J 4 5 transition the wave number is given by
h
2B (4 1) 10B 10 2
8 Ic
10 h 10 h
2 ... (iv)
8 (r )c 8 (1.05 r 2 )c
2 2
From equations (iii) and (iv), we get
10 h 1
22 1 10 B 1 0.9524
8 (r )c 1.05
10 B 0.0476 0.476 B
Hence, correct option is (a).
50. The number of fundamental vibrational modes of CO2 molecule is three. Out of these, two are infrared active
and one is Raman active.
eh
51. The magnetic moment of the nucleus µ g I
2 mp
The energy absorbed at resonance frequency
gehH 0 1
µ Hˆ 0 mI
2 m p M I 2
If v0 is the resonance frequency, then
gehH 0 1 1 geH 0
h 0 0
2 m p 2 2 2 m p
Correct option is (c)
P Q
52. V (r ) ,
r 6 r 12
d V (r )
At r r0 , potential energy will be minimum i.e., 0
dr at r r0
P ( 6r0 7 ) Q ( 12 r0 12 1 ) 0
6P 12
7
Q 13 0
r0 r0
1/6
12Q 2Q
6P r0
r06 P
1/6
2Q
Bond length ( r0 )
P
Hence, correct option is (a).
GATE-PH 2007 SOLUTION 308
53.
Crystal structure APF
P : Simple cubic 52%
Q : Body-centered cubic 68%
R : Face-centered cubic 74%
S : Diamond 34%
T : Hexagonal close packed 74%
From above table, it is clear that, the APF of face-centered cubic (R) and hexagonal close packed (T) are
same.
Hence, correct option is (d).
54. For face-centered cubic sample, 1 30, 2 ?
For first peak, (hkl) = (111)
For second peak, (hkl) = (200)
Therefore, from Bragg’s law: 2d sin
2a
sin ... (i)
h2 k 2 l 2
2a 2a sin
For first peak ; sin 1
h2 k 2 l 2 12 12 12
2a sin 30 a ... (ii)
3 3
2a 2a sin 2
For second peak ; sin 2
2 2 2
2 0 0 4
a sin 2 ... (iii)
a 1
Therefore, from equations (ii) and (iii), we get, a sin 2 sin 2 0.577
3 1.732
2 sin 1 (0.5773) 35.26º
Hence, correct option is (d).
Q
55. For superconductors, T graph is
O T
P
For conductors, T graph is
O T
GATE-PH 2007 SOLUTION 309
For semiconductors, T graph is
R
O T
Hence, correct option is (b).
56. Conductivity, eNµ0 eN 0 µe x / L
Consider an elementary length dx.
Therefore, the resistance of the elementary length
x
dx 1 dx 1
dR L
e dx
A A eN 0 µA
L x L
1 L Lx L
R e L
dx e e1 1
eN0 µA 0 eN0 µA 0 eN 0 µA
Correct option is (a)
57. Ferromagnetic mixture of Iron (Fe) and Copper (Cu)
Amount (%) of Fe = 75%; % of Cu = 25%
N
Total number of atoms, n 8 1028 m 3
V
75% of n contributes for saturation magnetisation
So, neff 0.75 8 1028 6 1028 m 3
µ M
M Nµ tanh
k BT
1.0 tanh(m/t)
Graphical solution of this equations is
M k BT
m and t
Nµ Nµ2
m
m
m tanh 1 when it gets saturated
t
M 1.3 106
µ; µ 28
0.216 1022
N 6 10
0.216 1022
µ 2.3 Bohr magneton
9.27 1024
Correct option is (b)
GATE-PH 2007 SOLUTION 310
58. Given that half life T1/2 = 4 × 108 years
Total time = 4 × 109 years.
4 109
Therefore, number of half life time = 10
4 108
Therefore, remaining number of particles after 10 half life time
10
1 N
10 = N N 0 0 N 0 1024 103 radioactive nuclei.
3
2 1024
Correct option is (d)
59. Since, tritium is radioactive, therefore more energy releases in (deuterium + tritium) fusion.
Correct option is (a).
60. For 8O17, Z = 8, N = 9
9N = (1s1/2)2 (1p3/2)4 (1p1/2)2 (1d5/2)2
5
J , l = 2 (for d)
2
Therefore, parity = (–1)2 = +1 (even)
5
Therefore, spin-parity =
2
Correct option is (b)
61. Travelling distance = 3 × 10–3 m
The time taken by the particle to travel the given distance
3 103
t sec = 10–11 sec, life time = 10–11 sec
3 108
Therefore, interaction is weak.
Correct option is (c)
62. For baryon, quark structure is ‘uus’
u u s
2 2 1
Since, charge Q 1 1
3 3 3
1 1 1
Baryon number B 1 1
3 3 3
Strangeness S 1 0 0 1 1
Correct option is (c)
63. Since there is no loss of energy in elastic scattering. Therefore, the initial and final states of the neutron have
the same energy.
Correct option is (a)
GATE-PH 2007 SOLUTION 311
64. R
R
R
V1 R
A
V2 B V3
Standard Standard
inverting inverting
Opamp Opamp
R R
V2 V1 V3 V2
R R
Ideal op-amp has infinite input resistance so there will no current pass through op-amp and VA and VB equal
to zero due to virtual ground concept.
V3 V1
• Output same as input. So, it is work as Buffer amplifier
Correct option is (b)
Z YZ
65.
Y
–
F = (X + Y) YZ
X –
X+Y
Output expression F X Y YZ
XYZ YYZ
Hence, F XYZ YY 0
Correct option is (d)
66. Given Q1 1 , Q2 = 0
CLK J1 K1 J2 K2 Q1 Q2
0 1 1 1 1 1 0
1 1 1 1 1 0 0
2 1 1 1 1 1 1
For J = K = 1 output is complemented (Toggle), and Q2 will be toggle when Q1 become from 0 1 (as it is
clock pulse of Q2)
So, after two clock pulse output will be 11
Correct option is (c)
GATE-PH 2007 SOLUTION 312
67. Serial input is SI 0101
msb lsb
Parallel output is QD QC QB QA
CLK SIPO 0101
QD QC QB QA
Clock 1 0 1 0 0101 = SI
1 0 1 0 0 Serial input “0” first CLK.
2 1 0 0 1
3 0 0 1 0
4 0 1 0 1
Parallel output of 2nd clock = 1 0 0 1
Correct option is (a)
68. To find potential difference (in volts) across RL.
2 Va 2
38V 76V
RL = 4
Nodal-Analysis at (a)
Va 38 Va Va 76
0
5 4 2
Va 48 Volt
Correct option is (a)
69. 16V VP=12V
2k S 4k
–
ID
+G 1
I1
R 42k
16 12
ID 2 mA
2
VGS 2Volt
KVL (1)
VGS ID 4 42 I1
2 I D 4 42 I1
Pul ID 2 mA
1
6 42 I1 ; I1
mA
7
KVL as per dashed line shown
GATE-PH 2007 SOLUTION 313
1
16 I1 R 2 8 Put I1 mA
7
16 I1 R 6
10 I1 R Put I1 value
R 70 k
Correct option is (d)
70. Vy Si = 0.3V
Equivalent circuit.
R
R
12V + + 12V
2.3V + +
Vin 0.7V Vout Vin Vout
2.3V 2.3V
3V – –
–12V – –
–12V
Case - (1) Case - (2)
0 Vin 2.3V 2.3V Vi 12V
Diode - ON Diode - OFF
R R
+ + + +
Vi Vo=2.3V Vi Vo Vo = Vi
2.3V 2.3V
– – – –
Case - (3)
Vin 0
Diode - ON
output 12V
wave
2.3V
Note: When the diode is upward direction the signal will transmitted above the reference voltage minus cut in
voltage of diode (Vr = 0.7) (Plus minus depend on polarity of reference voltage).
Correct option is (c)
x
71. x 0 cos 2 L x 2 L
4L
Normalization condition:
2L
2
dx 1
2 L
2L 2 2L
2 x 2 0 x
0 cos 4 L dx 1 2 1 cos 2 L dx 1
2 L 2 L
GATE-PH 2007 SOLUTION 314
2L
x
sin 2 L
2
0 2 1 1
4L 1 0 0
2L 2L
2 / 2L
2 L
Correct option is (c)
72. The energy of the ground state will be
2 2 2 2
E1 2
2m 4 L 32mL2
Correct option is (d)
73. The expectation value of
2 2 2 2
P 2 E 2m 2m
32mL2 16 L2
Correct option is (c)
74. When the incident angle i is become i p Brewster angle then the reflected light is completely polarized.
E||reflected 0 i
n1 = 1
cos r µ cos i 0
90º
µ cos i cos r n2 = 1.5
r
Correct option is (a)
75. We know that
2
n n
R 1 2 for normal incidence
n1 n2
2 2
1 1.5 0.5 1
0.04
1 1.5 2.5 25
Correct option is (d)
76. Speed of P in rest frame of Q 2c/5
5c
5c 2c P
19
Q
VP VQ
63 c 3 c
VPQ 19 5
VP VQ 5 2 21 5 5
1 1
c2 19 5
Hence correct answer is (c)
77. Energy of P in rest frame of Q is
m0 c 2 m0 c 2
5m0 c 2
E
1 vQ2 /c 2 3
2 4
1
5
Hence correct answer is (b)
GATE-PH 2007 SOLUTION 315
78. We know that conductivity,
n eµ
1 1
µ 8 28 19
6.67 103 m 2V 1 s 1
ne ne 1.6 10 5.85 10 1.6 10
Correct option is (a)
me 6.67 103 9.1 10 31
79. Relaxation time ( ) 19
37.93 1015 3.79 1014 sec.
e 1.6 10
Hence, correct option is (b).
80. Bound surface charge density
b P nˆ P rˆ kr k r
Bound volume charge density,
1 2
P 2
r r
r kr 3k
Correct option is (b)
81. The electric field at the out side of the sphere
3 4 3
Qb 4 kr 3 r 3k
E dS 0 0
0
E 0 for r R
Correct option is (a)
1
82. The energy is En n ; n 0,1, 2,...
2
The partition function of a single oscillator is
En
Z e
k BT
e
2 kBT
n 1
n n
n
2k BT k BT
e e
n
2
2 k BT k BT
e 1 e e kBT
1
2 k BT
e
k BT
1 e
e /2 1
, where
1 e k BT
Correct option is (a).
GATE-PH 2007 SOLUTION 316
83. The average number of energy quanta are
2
n 1
n eE ne n e n 0 e 2e 2 3e 3
n n0
n 0
eE e
2
n 1
e n 1 e e 2
n 0
n 0
1 2e 3e 2
n e
1 e e 2
Letting e x , then
x 1 2 x 3 x 2 x (1 x )2
n
2
1 x x (1 x)1
x e 1
1 x 1 e e 1
Correct option is (a).
84. Given that, I 16µA 16 106 A
Charge q 2e 2 1.6 1019 C
q n 2e
I for t 1 sec
t t
I 16 106
nq I n
q 2 1.6 1019
n 0.5 1014
Correct option is (a)
85. For a reaction, X x y Y or X x, y Y
where x, y are incident and emitted particles.
The cross-section x, y of the reaction is given by
Iy
x, y
I x number of nuclei x per unit area
where I x is number of particles incident per second
I y is number of particles outgoing per second
100 101
In reaction, Rh
Pd 3n
x X Y y
I x 0.5 1014 particles/sec as calculated in Q.84
GATE-PH 2007 SOLUTION 317
The above formula is for 1 to 1 corresponding i.e. for 1x particle 1y should be formed. In this equation
3n are formed.
At which number of -particles hitting the target
I x 0.5 1014 / sec
At which neutrons are produced = 1.806 108 / sec
Since, 3n are produced
1
I y 1.806 108 / sec 0.602 108 / sec
3
Now, to calculate number of Rh nuclei per unit area
100 gm of Rh possesss = 6.022 1023 nuclei Avagadro No.
Since, 1gm of Rh possess = 6.022 1021 nuclei Avagadro No.
Since, density = 104 kg/m3 = 107 gm/m3
Therefore, 107 gm of Rh possess = 6.022×1028 nuclei/m3
Now, 1 area × thickness possess = 6.022×1028 nuclei/m3
1 area × 1µm possess = 6.022×1028 × 10–6 nuclei/area
Therefore, number of Rh nuclei per unit area = 6.0222×1022
Iy 0.602 108
x, y 14 22
0.2 1028 m 2
I x number of Rh nuclei per unit area 0.5 10 6.022 10
0.2 barn 1 barn 1028 m2
Correct option is (b)
GATE-PH 2008 SOLUTION 318
OBJECTIVE QUESTION
Q.1 – Q.20 : Carry ONE mark each.
1. Tr EFGH Tr HEFG Tr HFEG
Correct option is (a)
1 a e i
† c e i
2. For unitary matrix, U U
b d
Correct option is (d)
3. F 3 zyˆ 5 yzˆ
iˆ ˆj kˆ
F 2iˆ
x y z
0 3z 5y
Correct option is (b)
4. Rotational Kinetic energy is given as
1
TR I
2
dTR 1 d d
I I
dt 2 dt dt
1 1
2 2
I I L
where, L I angular momentum
I torque
Correct option is (c)
5. We need two generalised coordinates, one for rotation and the other for translation. However, after imposing
the condition for rolling, one coordinate gets eliminated.
Hence correct answer is (b)
6. The electric field inside the two parallel plate is along ẑ and from right handed rule we can say the magnetic
field along ˆ .
Therefore, poynting vector, s E H
Therefpre. direction of poynting s zˆ ˆ rˆ
Correct option is (b) p
n1
7. We know that at p 90º
n2
n2 1.5
tan p
n1 1
p tan 1 1.5 56º
Correct option is (c)
GATE-PH 2008 SOLUTION 319
8. Since, the wave train is propagating along the positive x-axis with constant speed v.
i kx t
So, wave equation will be e x, t
2 2
2
k 2 and 2
i i 2
x t
Speed = v
k
2 1 2 k2
2 2 2
x , t k 2
2
2 0
x v t
2 1 2
2 2 2
x, t 0
x v t
Correct option is (a)
9. Since, the ground state of hydrogen atom does not depends on the
Lx , Ly 0 iLˆz 0 2 0 0
Correct option is (a)
10. The parameters required are
MC : ( E , V , N )
CE : ( N , V , T )
GC : (V , T , )
Correct option is (b).
11. P-T curve is parallel to the temperature axis
dP
0
dT
From Clausius-Clapeyron equation,
dP L S
dT T V V
S 0
Correct option is (d).
12. Radial probability density of finding the electron
2 4
P r R10 r 2 e 2r / a0 r 2
a03
dP 4 2r 2
For most probable value of r , 0 3 e 2r / a0 2r 0
dr a0 a
0
2r 2
2r 0 r a0
a0
Correct option is (a)
GATE-PH 2008 SOLUTION 320
13. For Mn4+ electronic configuration is 3d3. To fill 3 electrons
ms ½ ½ ½
ml 2 1 0 –1 –2
3
S ms ; L m 3
2
3 3 3 5 7 9
J 3 to 3 , , ,
2 2 2 2 2 2
3
For less than half filled ground state is for J
2
Therefore, term is 4 F3/ 2
Correct option is (b)
14. Length ‘L’ of the active medium is related with the wavelength of laser light by
n ... (i)
L
2
c nc nc
Using c ; L n
2 2L
c 1 2L
Frequency width n 1 n
2L c
Coherence length is given by c, where is coherent time
1 2L
But c L
c
i.e., coherent length is proportional to length of the using medium.
Correct option is (a)
15. The number of atom in the unit cell of B.C.C. = 2. Each atom will give one free electrons. So, the concentration
of conduction electrons.
2
24
3.125 10 22 cm 3
64 10
Correct option is (b)
c 1 1
16. Susceptibility, T T T Tc / c at T Tc , 0
c
Correct option is (b)
17. K+ has charge (Q) = +1, Strangeness (S) = 1, Baryon number (B) = 0
Since, the Baryon number is zero. Therefore, we need two quarks. One with positive sign and other with
negative sign. Also strangeness is 1, we can take one strange quark with negative sign i.e. s .
Now in order to give it charge (Q) = +1, we should add one up quark u so that its charge become
2 1
1 .
3 3
GATE-PH 2008 SOLUTION 321
Therefore, quark structure of K+ meson is u s .
Correct option is (c)
4 3
18. Since, radius R A1/3 and V R
3
For 8O16 , RO 161/3 ... (i)
and for 54 Xe128 , RXe 1281/3 ... (ii)
equation (i) RO 161/3 RO3 16 1
1/3
3
equation (ii) RXe 128 RXe 128 8
4 3 4
3
RXe RO3 8 RXe 8 RO3
3 3
VXe 8 VO
Therefore, VXe 8 V ( VO V )
Correct option is (a)
19. As temperature increase the of the transistor increase and as a result IC also start to increase. So, Q
point will be shift towards saturated region
Correct option is (d)
Q.21 – Q.75 : Carry TWO marks each.
20. n-FET
Drain
D
NMOS
n
Gate n
+ + VDD n channel
P n P G
n VDS
+
P-substrate
VGG Source S
VGS – VGS (–)ve
VGS (–)ve for n FET
Hence, both biased with negative potential.
Correct option is (b)
21. Let the is the eigenvalue of the matrix. Therefore, the eigenvalue equation,
A I 0
cos sin
0
sin cos
22
cos sin cos i sin
1
cos i sin
2
3 i when 30º
Correct option is (b)
GATE-PH 2008 SOLUTION 322
22. F x a exp i 2 va
F 1 exp i 2 va x a
1 1
F 1 cos 2 va F exp i 2 va exp i 2 va
2
1
x a x a
2
Correct option is (d)
23. z = 0 is branch point of f z and will behave like a simple pole.
1 1 ln t
Res f z 0 Lt z 0 ln z Lt ln Lt 0
z 0 t t t t t
Correct option is (a)
i i
z2 i2 i2
24. i z 1 dz z i i 2 i
2 i 2 2
Correct option is (b)
25. Consider the equation,
d2y 1 dy
2
y0
dz z dz
d2y
2 dy
z 2
z z2 y 0
dz dz
This is a Bessel differential equation with v = 0
dy
coefficient of
Here, a1 z dz 1
a2 z d2y z
coefficient of 2
dz
a0 z coefficient of y
1
a2 z d2y
coefficient of 2
dz
a1 z
Since, a z is not defined at z = 0, therefore z = 0 is a regular singular point of the differential equation.
2
The given Bessel differential equation has two linearly independent solutions i.e.
(i) J0(z) which is defined everywhere and it is entire solution
(ii) Y0(z) which is not defined at z = 0 and it is singular (not analytic) at z = 0
Correct option is (c)
1 1
26. v '1 v1 v2 , v '2 v1 v2 , v '3 v3
2 2
GATE-PH 2008 SOLUTION 323
1 1
Consider, a11 , a12 , a13 0
2 2
1 1
a21 , a22 , a23 0
2 2
a31 1, a32 0, a33 0
Rotation in rank-2 tensor matrix is given as
3 3
Tij ai a jmTm and keeping the same rotation we can write
1 m1
T 'ij ai1a jmT1m ai 2 a jmT2m ai 3a jmT3m ai1 a j1T11 ai1 a j 2T12 ...........
m 1
a11a11T11 a11a12T12 a11a13T13 a12 a11T21 a12 a12T22
T '11
a12 a13T23 a13a11T31 a13a12T32 a13a13T33
1 1 1 1
T '11 T11 T12 T22 T21
2 2 2 2
1
T '11 T11 T22 T12 T21
2
Correct option is (d)
1 2 1
27. L q q q q 2
2 2
Equation of motion
d L L d
dt q q
0; q q q q 0
dt
q q q q 0
q q 0
This is equation of motion of a harmonic oscillator
Hence correct answer is (a)
28. Moment of inertia about given axis is
8 0 4 12
1 3
nT I n 0 0 4 0 23
2 2 4 0 8 0
4
1 3
0 2 3 2 3 0 5
2 2 2
Hence correct answer is (b)
GATE-PH 2008 SOLUTION 324
29. Time period of oscillation is
I
given as t 2
mgd
I = moment of inertia about axis of rotation
d = distance between point of suspension and center of mass.
Here I 2 MR 2 , d R
2MR 2 2R
t 2 2
Mg R g
Hence correct answer is (a)
30. Since the force is radial, angular momentum is conserved. If mass fall into the hole, its angular momentum will
become zero. Which will violate conservation of angular momentum. Therefore it will not fall into the hole.
Hence correct answer is (d)
p iq
31. A
2
1 2 1 2
L q q
2 2
p2 1 2
Hamiltonian is H q
2 2
A H A H
Poisson bracket A, H
q p p q
p 1 1 i
p q ip q p iq iA
2 2 2 2
Hence correct answer is (a)
32. We have, E E0 ( xˆ ei yˆ ) exp{i ( kz t )}
So, Ex E0 , E y E0ei
For n, n 0,1, 2..... it will be plane wave.
So, electric field will be zero times in one second.
But if n, n 0,1, 2 ,
It will not be linear polarized light. E will not be zero.
Correct option is (a)
33. Since electronic permeability of dielectric material is greater than the 0 . So, the field intensity inside the
material will be increase.
Correct option is (d)
GATE-PH 2008 SOLUTION 325
34. Since rod is rotated in the xy-plane the direction of J and A will be along ˆ
Since, d L , so we can consider rod as magnetic dipole, the vector potential due to magnetic dipole as
µ0 M d
A
4 d 3
1
A 2
d
Correct option is (b)
35. The elementary area on the disc.
ds rdrd y
Total charge on the disc is given by
rdrd
2 a
0 2
Q ds cos d r dr 0 x
a a 0
So, dipole moment independent of choice of origin.
So, the position of any point on the disc is given by
r r (cos xˆ sin yˆ )
The dipole moment,
2 a
2 3 0 a 3
p r dq 0 0 cos d 0 r drxˆ 4 xˆ
a
Correct option is (b)
36. We know that, from equation of continuity
d
J 0
t
d
J
dt
d
E
dt
J E
d d
dt
t
ln tc
t
r, t r , 0 e
t
r, t r, 0 e
Correct option is (a)
37. Correct option is (b)
GATE-PH 2008 SOLUTION 326
38. Since, Q1 and Q2 are compatible variable, therefore, they will commute with each other.
So, Q1Q2 Q2Q1
Correct option is (d)
39. Wrong question
1
40. For ground state of hydrogen atom, 0 s
2
1
Therefore, j s
2
q qg
Magnetic moment, J g J J µJ J J x iˆ J y ˆj J z kˆ
2m 2m
e
J g J J z kˆ (for ground state of hydrogen atom J x J y 0 )
2m
e e ˆ
2 m j kˆ k µB kˆ
2m 2m
Therefore, µJ µB
Correct option is (b)
2 n x
41. n sin 0 x L
L L
L
The probability of finding the particle between 0 to is
2
L /2 L /2
2 2 n x 1 2n
P sin dx 1 cos x dx
L 0
L L 0
L
L /2
2n x
sin
1 L/2 L
x 0 = 1
L 2 n / L 2
0
Correct option is (c)
42. At z = 0, momentum will be maximum and when z will be maximum then momentum will be zero. Again when
z will start decrease momentum will be start to increase but direction will be opposite. And since in the inelastic
collision. So, the momentum of the ball will be less than the initial value.
Correct option is (c)
43. The energy of a particle with magnetic moment µ in an external magnetic field B is
B for s 2 particle
1
E B 1
B for s 2 particle
The partition function of a particle is
Z1 eB / kBT eB / kBT 2cosh
B
k BT
The partition function for N non-interacting such particles is
GATE-PH 2008 SOLUTION 327
N
N B
Z N Z1 2 cosh
k BT
The Helmholtz free energy is F k BT n Z N
B
F N k BT n 2 cosh
k BT
The magnetization is
F
m N k T
2sinh
B
k BT
B
B T 2 cosh kT
B
k BT
B
B
N tanh
k BT
Correct option is (b).
44. The distributions are as follows :
Bosons Fermions Classical particles
1 2 3 1 2 3 1 2 3
• • • • A B
• • • • B A
• • • • A B
•• P(2) 0 B A
rF 0
•• P(1) 2/3 A B
•• B A
P(2) 1/6 1 AB
rB
P(1) 2/6 2 AB
AB
P(2) 1/6 1
rC
P(1) 4/6 4
Here we have choosen 1-state to be the given energy state.
1 1 1
So the ratio is rB : rF : rC : 0 : 1: 0 :
2 4 2
Correct option is (d).
45. The photons are bosons and hence follow Bose-Einstein statistics.
The mean number of photons are
1
n , where 0 (for photons) and (given)
exp 1
k BT
1
1
n
exp 1
k BT
e k BT 1
Correct option is (d).
GATE-PH 2008 SOLUTION 328
46. A B AB
N atoms N atoms 2 N atoms
T,V T,V T , 2V
Since the gases are ideal in nature, then internal energy depends only on the temperature. Furthermore, tem-
perature of the mixture remains T as TA TB T .
i.e., (T )mixture 0
U 0
From first law of thermodynamics,
dQ dU pdV
pdV dU 0
The entropy change of the system is
S (S )1 (S )2
dQ1 dQ
2
T T
pdV pdV
T T
2V 2V
dV dV
N kB N kB pV N kBT
V
V V
V
2 N k B n 2
Correct option is (b)
47. The mean energy is
Ei g ( Ei ) e Ei / k BT
i
E
g ( Ei ) e Ei / k BT
i
0.1 e 0/ kBT .2.e / k BT 2.4.e 2/ kBT
1 e 0/ kBT 2e / kBT 4e 2 / kBT
2 e / kBT 8 e 2 / k BT
1 2e / kBT 4e2 / kBT
2e / k BT 8e2 / k BT
/ k T 2 / k BT
1 2e B 4e
Correct option is (b).
48. The wave number of three consecutive lines are 64.275 cm–1, 77.130 cm–1 and 89.985 cm–1.
.......... 89.985 cm 1 2 B
.......... 77.130 cm 1
2B
.......... 64.275 cm 1
GATE-PH 2008 SOLUTION 329
Therefore, the difference in wave number
2B 89.985 77.130
h
2B 12.855 2 12.855
8 2 I B c
2h 2
IB gm cm 2 and
8 2 c (12.855) 12.855 hc
IA 0 ( r 0 , from the bond axis passing through the center of mass).
Hence, correct option is (b).
49. The wave number of the rotational Raman lines are given by
3
v ve 4 B J ; J 0,1, 2,3,....
2
ve 6 B, 10 B, 14 B, 18 B,....
Stoke’s lines is corresponding of (–) values
i.e., vstokes ve 10 B, ve 14 B, ve 18B,...
Correct option is (a)
50. For vibrational spectroscopy with anharmonicity the selection rule is 1, 2, 3,.... The anharmonicity
leads to multiple absorption lines. The intensity of hot band lines are stronger than the fundamental absorption
is incorrect statement as fundamental lines are most intense.
Hence, correct option is (c).
51. The CO2 molecule do not have permanent dipole moment while COS have permanent dipole moment. Thus
CO2 molecule will not show absorption lines.
Hence, correct option is (d).
n
52. Length of the resonator must satisfy relation L , where is wavelength of laser light, n is number of
2
c c
mode. Velocity of light in a medium of refractive index µ is v , where is frequency of
light
n nc nc c
L L . For n 1 frequency
2 2 2L 2L
If length L is kept constant but refractive index µ is changed from , frequency becomes
c
2 ( ) L
Change in frequencies
c c c 1 1 c u
2L 2( ) L 2 L µ µ µ 2 L µ µ µ
c
2 L ( )
Correct option is (c)
GATE-PH 2008 SOLUTION 330
53. Volume of the bcc lattice, V a b c
a a a
xˆ yˆ zˆ xˆ yˆ zˆ xˆ yˆ zˆ
2 2 2
1 1 1
a3 a3
1 1 1 1(1 1) 1(1 1) 1(1 1)
8 1 1 1 8
a3 a3
2 2
8 2
Primitive translation vectors of reciprocal lattice are
a a
2 ( xˆ yˆ zˆ) xˆ yˆ zˆ
2 b c
A
2 2
a b c a
3
2
2 a 2
xˆ xˆ xˆ yˆ xˆ zˆ yˆ xˆ yˆ yˆ yˆ zˆ zˆ xˆ zˆ yˆ zˆ zˆ
a3 4
2
2 2
zˆ yˆ zˆ xˆ yˆ xˆ 2 xˆ yˆ xˆ yˆ
2a a a
2 2
Similarly, B yˆ zˆ and C zˆ xˆ
a a
Hence, correct option is (d).
1 1 1
54. For bcc structure: (0, 0, 0) and , , are atomic positions
2 2 2
S hkl f 1 ei ( h k l ) ... (i)
For odd value of ( h k l ) above term will be zero. For, even value of ( h k l ) , S hkl 2 f .
Therefore, for bcc structure, reflections like (100), (111), (210) etc. are missing whereas the diffraction lines
corresponding to (110), (200), (222) etc. are present.
For fcc structure: An fcc unit cell has four identical atoms. One of the atoms is contributed by corners and
may arbitrarily be assigned coordinates (0, 0, 0), whereas other three are contributed by face-centers and have
1 1 1 1 1 1
the coordinates , 0, , , , 0 and 0, , .
2 2 2 2 2 2
S hkl f 1 e i ( h l ) e i ( h k ) e i ( k l ) ... (ii)
From above equation (ii), it is clear that structure factor is non-zero only if h, k and l are all even or all odd and
has a value equal to 4. Hence reflections of the type (111), (200), (222) etc. are present, whereas those of the
type (100), (110), (211) etc. are absent for an fcc crystal.
bcc : (200); (110); (222)
From the given options : (a) is correct.
fcc : (111); (311); (400)
Hence, correct option is (a).
GATE-PH 2008 SOLUTION 331
55. According to Dulong-Petit law : CV 3R constant ... (i)
2
e a /T
According to Einstein model : CV 3 Nk B E e E /T 2 ... (ii)
T T
3 3
12 T T
According to Debye model : CV 4 Nk B ... (iii)
5 D A
Hence, correct option is (c).
2C
56. Maximum frequency of aquostic branch =
M
It depends on mass of heavier atoms, not on the mass of lighter atoms
1 1
2C Optical Branch
m M
2C /m 2C /m
Forbidden frequency gap
2C /M 2C /M
m<M Acoustic
Branch
2a
O 2a
First Brillouin zone
Dispersion curves for linear diatomic lattice showing acoustical and optical modes.
From above figure, it is clear that optical is larger than acoustical .
Hence, correct option is (b).
57. The kinetic energy of a free electron at a corner and center of the first Brillouin zone in two-dimensional square
lattice is given by
2 2
2 2 2 2 2
Ecorner
2m
k x k y 2m a a 2m a 2
2 2
2
2 2 2 2 2
and Ecenter
2m
k x 2m a 2m a 2
The ratio of kinetic energies = b
2 2 2
Ecorner 2m a 2
b b
Ecenter 2 2
2m a 2
b2
Hence, correct option is (b).
GATE-PH 2008 SOLUTION 332
58. We know that the electron density in the conduction band
3/ 2
1 2m k T
n e 2B e F C B
E E / k T
4
And the density of hole at the valence band
3/2
1 2m k T
p h 2B e V F B
E E /k T
4
For an intrinsic semiconductor the numbers of electrons in the conduction band is equal to the number of holes
in the valence band only when an electron make a transition to the condition band.
n p
3/ 2 3/ 2
1 2me k BT EF EC / kB T 1 2m k T
e h 2B e Ev EF / kBT
4 2 4
3/ 2 E E 2 EF
v C
me k BT
e
mh
Taking log both side we get
3 me Ev EC 2 EF Ev Ec 3 m
n EF k B T n h
2 mh k BT 2 4 me
Correct option is (c)
59. For (a) the reaction K K
pp can proceed irrespective of the kinetic energies of K+ and K–, this
statement is not correct.
For (b) K K
pp
Baryon number (B): 0 0 1 1 B 0 (allowed). This statement is not correct.
For (c) K K
2 ;
Strangeness (S) : 1 1 0 0 S 0 (allowed). This statement is not correct.
; proceeds via weak interaction. This statement is correct.
(d) The decay K 0
Correct option is (d)
60. For (a), 10Ne20 nucleus is stable ( A = 20 is a magic number).
For (b), a spheroidal nucleus shows an electric quadrupole moment.
For (c), 8O16
For Z = N = even i.e. even-even nuclei, nuclear spin J = 0 and not ½.
Correct option is (c)
61. Since, 4Be9 has Z = 4, N = 9 – 4 = 5
Therefore, for N = 5, (1s1/2)2 (1p3/2)3
3
and nuclear spin J
2
Correct option is (a)
GATE-PH 2008 SOLUTION 333
62. For mirror nucleus 6C11 and 5B11
Mass difference between mirror nuclei is
M A, Z M A, Z 1 M p M n a3 A2/3
MeV
For 6C11 and 5B11 mass difference is
c2
2/3 2/3
M p M n a3 11 1.3 a3 11
2/3
a3 1.3 / 11
Now for 17 and 8 O17
9F
2/3
Mass difference = M p M n a3 17
Putting value of a3.
2/3
17 MeV
Therefore, mass difference 1.3 1.3 1.3 1.336 1.73 1.336 0.437
11 c2
Correct option is (b)
63. Since, proton is charged particle. Therefore, it costs more energy to add extra proton in nucleus in order
to minimize the Coulomb repulsion.
Correct option is (d)
64.
65.
+ +
V1 V2
– –
10:1
V1 N1
We know transfermer ratio V N
2 2
V1 10 220
V2rms
V2 1 10
V2rms V2rms 22
For full wave rectifier output will be
VL
2Vm 2 22 44
DC voltage, VDC
Correct option is (d)
GATE-PH 2008 SOLUTION 334
10 20
66.
I1 I2 +
2A 5 15 20V
–
(1) (2)
KVL
Applying KVL at Loop (2)
20 I 2 20 15 I 2 I1 0
15I1 35 I 2 20 15 I1 35 I 2 20
Applying KVL at Loop (1)
10 I1 15 I1 I 2 5 I1 2 0
30 I1 15I 2 10
I
R
Note:-
+ V – V = IR
Correct option is (a)
67. Given Y P PQ
Y P P P Q
Y PQ {A + BC = (A + B)(A+C), by distribution theorem}
{ A A 1 by boolean theorem}
Check the output of given option i.e. Y = P + Q
P Y = PQ
(a) Q
P P
(b) Y Y PPQ
P̄
Q Y 1 Q
P̄ + Q
Y 1
Note:- x x 1
P P 1 axioms/property
P
(c) Q Y Y PQ
(d) P P
P̄ Y= P·P̄Q
Q Y=0
Q P̄Q
Correct option is (c)
GATE-PH 2008 SOLUTION 335
68. 2-bit flash “ADC”
VR Vanalog
Comparator
R
(a)
VR ×3R
4R R b1
4×2
b0
(b)
VR ×2R Priority
4R R encoder
VR ×R
4R R
• Number of comparator 2n 1 = 22–1 = 3
Note : (i) Number of resistor 2 n
(ii) Reference voltage are mentioned at point (a), (b) and (c) (using voltage division rule)
Correct option is (a)
69. [Analysing CKT for Positive Logic]
R D1
V1
V0
V2
R D2
RL
Case - (1) Case - (2)
V1 High V1 low
V2 Low V2 low
R R
V1 V1
V0 = High V0 = Low
V2 V2
R R
RL RL
I=0
Case - (3) :
V1 and V2 High
R
V1
V0 = High
V2
R
RL
Truth Table
V1 V2 D1 D2 V0
0 0 OFF OFF 0
0 1 OFF ON 1
1 0 ON OFF 1
1 1 ON ON 1
GATE-PH 2008 SOLUTION 336
Note: CKT is (+)ve logic-OR operation or , (–)ve logic AND operation.
Duality convert (+)ve logic to (–)ve logic and vice versa.
Correct option is (d)
70. Clock
Input
–
Q
Output
71. The schrodinger equation between 0 x L
2 d 2
V0 E
2m dx 2
d 2 2m d 2 2m
V0 E 0 2 k 2 0 , where k 2 V0 E
dx 2
dx
A e kx B e kx
Correct option is (a)
transmitted current density transmitted current density
72. Transmission coefficient, T T
Incident current density S0
transmitted current = TS0
Correct option is (b)
73. According to energy conservation
R+T=1
2 2 2
AR AT AB AB
2
2
1 1 T 1 T
A0 A0 A0 A0
Correct option is (b)
74. Since, charge remain outer surface of a conductor.
So, E = 0 for 0 < r < a (since there are no charge for r < a)
Also, E = 0 for a < r < b (since there are no charge for a < r < b)
1 Q
E
4 0 r 2 for b r c (due to charge on the outer surface of the inner sphere)
1 2Q
And E for r > d (Q charge for outer sphere and another Q induce charge will be appear at the
4 0 r 2
outer surface)
Correct option is (d)
GATE-PH 2008 SOLUTION 337
75. Since, both surface have equal surface charge density
Q 2Q
2
d 2 2b 2 d 2 b and c a
4 b 4 d 2
Correct option is (c)
76. -de cay of a free neutron
Z XA Z 1Y
A
1e
0
ve
which is written in terms of n and p as
n Pe
For anti-neutrino to have maximum energy, electron should produce at rest, as P or nucleus Y will have almost
negligible energy as they are heavier. Energy released Q is always distributed among electron and ve as their
kinetic energies. For maximum anti-neutrino energy electron should be produces at rest.
Correct option is (d)
77. n Pe
2
Q value = mn m p me mve c 939.57 938.27 0.51 0 c
2
0.79 MeV 0.8 MeV
This whole energy goes to ve -for maximum energy..
Correct option is (b)
78. The density of state for 2-D electron gas is
2p Adp
g ( p ) dp 2
, where A L2 and p 2mE
h
2L2 2m
g ( E ) dE 2mE dE
h2 2 E
2mL2
dE
h2
Since electron spin degeneracy is 2 so we should multiply density of states by a factor of 2.
4mL2
g ( E ) dE dE
h2
Correct option is (a).
79. Total number of particles is
1, E EF
N f ( E ) g ( E ) dE , where f ( E ) 0, E EF
at 0 K
0
EF
4L2 m 4L2 m
dE EF
0 h2 h2
h2 N
EF
4L2 m
Also the ground state energy is
GATE-PH 2008 SOLUTION 338
EF EF
4 L2 m
E0 E f ( E ) g ( E ) dE E dE
0 0 h2
4 L2 m EF2 N EF2 N
EF
2 2 EF 2 2
h
Correct option is (b).
80. According to question
3 5
Rate spaceship Rate earth
i.e. time spaceship time earth
5 3
using time dilation formula we get
time earth 5
time earth
1 v /c 2 2 3
2
2 3 2 4
1 v /c v c
5 5
Hence correct answer is (a)
13
81. time Aakashganga time earth
5
time earth 13
time earth
12
1 v /c 2 5
1 13 12
v c
v12 5 13
1
c2
Therefore, Speed of Aakashganga relative to Suryashakti is
12 4
c c
v v 13 5 8c
Vas
v v 48 17
1 2 1
c 65
Hence correct answer is (c)
R Standard inverting Opamp
3R R
82. Va
3R R
Vb V0 1 3R
3R (A)
Vc Vx R (B)
R V0 2
Va
R R
V01 Vx Vb Y Vy
R R
Vc
V01 Vx
R
GATE-PH 2008 SOLUTION 339
As ideal op-amp has infinite input resistance, so there will no current pass through the op-amp and VA will be
virtual ground i.e. VA = 0
(i) Applying KCL at node (A)
0 Va 0 Vb 0 Vc 0 Vx
0
3R 3R 3R R
1
Vx Va Vb Vc
3
Va Vb Vc
V01 Vx
3
Correct option is (c)
83. (ii) Applying KCL at node (B)
(iii) As ideal op-amp has infinite input resistance. So, VB & Vy will be virtual short i.e. VB = Vy
Vy V01 Vy V02
0 For calculation of Vy, applying KCL at node (Y)
R 3R
1 1 V0 V0 Vy Vy Va Vy Vb Vy Vc
Vy 1 2 0
R 3R R 3R R R R R
4 V0 V0 Va Vb Vc
Vy 1 2 4Vy
3R R 3R 1
Put value of V y and V01
Va Vb Vc Va Vb Vc V02
3 3 3
V02 0
Correct option is (d)
84. We have, f x a0 a1 x a2 x 2 and q x b0 b1 x b2 x 2
f x g x a0b0 a1b1 a2 b2
85. Let R x 1 x 1 x 0 x 2
If f x is the subspace of V that is orthogonal to R x then R x f x 0
Only option (a) follow the above condition
Correct option is (a)
GATE-PH 2009 SOLUTION 340
OBJECTIVE QUESTION
Q.1 – Q.20 : Carry ONE mark each. y
2
1. r d rd sin 90º r
C C
0
d 2 r
x
r d
Correct option is (a)
2. Let any two points on the region A and B
We know that,
VA A
Potential difference dV E dr V AB E AB 0
VB B
Correct option is (c)
1
3. L m r 2 r 2 2 r 2 2 sin 2
2
L
coordinate is cyclic, threfore canonical momentum p is conserved.
Hence correct answer is (c)
4. Let consider an elementary surface is
Therefore, the magnetic flux pass through the elementary area dS is
d B dS
B
Induced emf dS
dt t
Correct option is (a)
i
5. Eigenvalue equation : 0 2 i 2 0 i
i
Correct option is (b)
6. det i 1
Correct option is (d)
7. For bound state 0 as x
d2 d
A e2 x 2 A0 e2 x 4 A0 e 2 x
2 0
dx dx
Correct option is (c)
8. Nuclear force is charge independent.
Yukawa potential is given by
e ar
V r V0
r
where, V0 z , is fine structure constant
z is atomic number, a is screening parameter
The range of the nuclear force is of the order of two fermi
According to Yukawa Mesons Theorem, mesons are exchanged during nucleons interaction.
Wrong statement is (b)
GATE-PH 2009 SOLUTION 341
9. For canonical transforma PB Q , Pq , p 1
Q P Q P
i.e. 1
q p p q
Q P Q P 1
(a) 0 1
q p p q
Q P Q P
(b) 2 2 1
q p p q
Q P Q P
(c) 0 1.1 1
q p p q
Q P Q P
(d) 0 1 1 1
q p p q
Therefore, transformation given in option (c) is not a canonical transformation.
Hence correct answer is (c)
10. Given CMRR 100 dB
CMRR
2V
20log10 100 dB dB V0 for differential input 200 V
Ad (differential gain) = 10000
Ad dB 20 log10 10000
Ad
Ac
Ad = differential gain, Ac = Common mode gain
dB Ad dB AcdB
100 80 AcdB 20 dB AcdB
20 20 log10 Ac Ac 0.1
Correct option is (b)
11. For ferromagnetic substance, the specific heat is maximum at Tc. Thus correct graph of specific heat versus T
M s 0 h M s
is expressed as : CV 2 M s , where ; M s N tan h
T k BT
Specific heat
O T
T = TC
Thus, option (d) is not correctly shown.
Hence, correct option is (d).
GATE-PH 2009 SOLUTION 342
dx dx k2 k
12. k1 k2 x k3 x 3
dt dt k1 k1
k2
k1 dt
e
k 2 / k1 t
I .F . e
Therefore, the solution will be
k3 k
x e k2 / k1 t exp 2 t dt
k1 k1
k3 k
x e k2 / k1 t exp 2 t c
k2 k1
k3 k
x c exp 2 t
k2 k1
k3
Putting the condition, at t = 0, x = 0, we get C =
k2
k3 k2
Therefore, x k 1 exp k t
2 1
Correct option is (a)
13. The first order phase transition involves the concept of latent heat that is needed when a liquid transform into a
gas at its critical temperature.
Correct option is (a).
14. P : Doppler broadening gives pressure.
Q : Natural broadening gives life time of the energy level
R : Rotational spectrum gives moment of inertia
S : Total internal reflection gives refractive index
Hence, correct option is (a).
15.
4B 4B 6B 6B 4B 4B
e
Stokes lines Anti-stokes lines
The separation between the first stokes and corresponding anti-stokes lines of the rotational Raman spectrum
in terms of the rotational constant is 12B.
Hence, correct option is (d).
h
16. Flux gets quantised in the units of
2e
Hence, correct option is (b)
GATE-PH 2009 SOLUTION 343
1 1
17. For optical phonon branch, frequency of lattice vibration is given by 2C .
M1 M 2
Hence, correct option is (a).
18. Proton is represented as ‘uud’.
u u d
Since, it has charge Q 2 2 1 3 1
3 3 3 3
1 1 1
and it has Baryon Number B 1
3 3 3
and it has strangeness Number S 0 0 0 0
1 1 1 1 1
I3 =
2 2 2 2 2
1 1 1
Y=1 1
3 3 3
Correct option is (b)
19. From figure
• Input is base (B) (E)
• Output is emitter
npn-BJT
Av ; voltage gain of common collector is 1
CC-config.
Av ideally 1
• CC configurationb is also referred as emitter follower.
Correct option is (d)
20. Since, BE = mc 2
= (mass of total protons + mass of total neutrons – mass of nucleus) × c2
BE ( Zm p Nmn A) c 2
BE BE
2
Zm p Nmn A A Zm p Nmn 2
c c
Correct option is (c)
Q.21 – Q.60 : Carry TWO marks each.
21. J H
1 4 1 r
10 2 sin r cos r kˆ
r r
GATE-PH 2009 SOLUTION 344
10 4 1 1 ˆ
r cos r cos r r sin r k
104 sin r kˆ
I J ds
a a
10 4 sin r 2 rdr 104 2 sin r rdr
0 0
a
cos r r sin r cos a sin a
2 10 4 2 10 4 a
2
2
0
cos sin 8a 2
2 a 2 800
2 104 104
2
4a 2
Correct option is (b)
22. We know that.
B µ0 J
A B
1 1
J
µ
A
0
µ
0
2 x 2 y kˆ
1 2 ˆ ˆ
µ0
2 iˆ ˆj
µ0
ij
Correct option is (b)
ez
23. f z has simple poles at z = 1 and z = 2
z 2 3z 2 C:|Z| = 3/2
Only z = 1 lies within the circle.
I 2 i Residue of f z at z 1
z=1 z=2
ez
2 lim z 1
z 1 z 2 2 i e 2 ie
z 1
Correct option is (c)
24. We know the relation between E , k , H is
k E µ0 H
k
H
µ0
k E
We have,
E 20 sin 108 t kz ˆj
kˆ zˆ , 108
GATE-PH 2009 SOLUTION 345
20k 20k
H
µ0
zˆ ˆj sin 108 t kz 8 sin 108 t kz iˆ
10 µ0
Correct option is (c)
25.
Given : Polarization, P 5 z 2 7 kˆ
The volume charge density, b P 10 z
Therefore, volume bound change inside the dielectric
2
Qb r 2 dz 10 z 5 z 0 r 2 5 r 2 L
Correct option is (c)
26. 3 ij 3Tij 113T11 223T22 333T33 123T12 213T21 233T23 323T32 133T13 313T31
i, j
123T12 213T21 T12 T21
3 12 k ak 21k ak 123 a3 213 a3 2a3
k k
Correct option is (a)
27. From the given figure, it is clear that curve P, Q and R represents versus T graphs for ferromagnetic, para-
magnetic and antiferromagnetic materials respectively.
Hence, correct option is (c).
28. Optical frequency range 390 THz to 770 THz
In this range dielectric constant of a material due to electronic polarizability.
UHF to
micro
waves IR UV
Total polarizabilities
d
(real part)
i
T
e
Frequency
Correct option is (b)
29. For hole, kh ke , vh ve and mh me
For velocity part h kh e ke Vh Ve
These relations are obtained by using mass and momentum conservation laws.
Hence, correct option is (d).
30. For diatomic molecule, the internuclear separation of the ground and first excited state are the same. The
absorption spectrum will appear for the transition from lowest vibrational state of the ground state will appear
as shown in figure :
Intensity
0 cm–1
Hence, correct option is (a).
GATE-PH 2009 SOLUTION 346
31. In all the transitions given above, the transition from 2 0 is the most suitable for a continuous wave (cw)
laser because the state 2 is a metastable state with highest life time of 10–6 sec.
Hence, correct option is (b).
time in rest frame
32. Time in lab frame
1 v 2 /c 2
2 106
6.4 10 6 sec
2 .
1 0.95
Hence correct answer is (a)
33. The nuclear spin of Cesium (I) = 7/2
For D-lines of the Cesium atom, the transitions will be from 2 P3/ 2 2S1/2 and 2 P1/ 2 2S1/ 2
3 7
For 2 P3/ 2 term, we have J , I
2 2
Since, F = I J to I J F = 5, 4, 3, 2 ... (i)
1 7
For 2 P1/ 2 term, we have J , I .
2 2
Therefore, F = 4, 3 ... (ii)
1 7
Also for 2 S1/ 2 term, we have J , I
2 2
Therefore, F = 4, 3 ... (iii)
Now, for hyperfine spectrum, the lines are shown with selection rule F 0, 1 (0 0)
5
2 4
P3/2
3
2
2
P1/2 4
3
2
S1/2 4
3
Thus, these are total 10 lines in the hyperfine spectrum.
Hence, correct option is (a).
34. The probability of occupied state for fermions is
1
P
e( )/ kBT 1
So probability of unoccupied state is
e( )/ kBT 1
P 1 P ( )/ k BT
( )/ k BT
e 1 1 e
Correct option is (c).
GATE-PH 2009 SOLUTION 347
1
35. Since, M ( A, Z ) 2
[ f ( A) yZ zZ 2 ]
c
dM ( A, Z )
For stable nucleus, 0
dZ
d 1 2
dZ c 2 f ( A) yZ zZ 0
y z (2 Z ) 0 y 2 zZ 0
4a A 2( ac A1/3 4a A A1 ) Z 0
4a A 2( ac A1/3 4a A A1 ) Z 0
4a A 4a A
Z
2[ac A 4a A A1 ]
1/3
1
2 4 a A A1 ac A1/3
4
(a A / 2) aA / 2 (a A/ 2 ) A/ 2
aA 1 1/3 aA a aA a a
ac A 1 c A2/3 1 c A2/3 1 c A2/3
A 4 A 4a A A 4aA 4a A
Correct option is (c)
36. The energy in 3-D is
3 p2
E k BT p 3m k BT
2 2m
The de-Broglie wavelength is
h h
p 3m k BT
Correct option is (b).
37. For calculation of Thevenin voltage (VTH) or open circuit voltage (VOC) Remove RL
Find VTH
10 10
+ +
15V 10
– –
15 10
By applying voltage divider, VTH Volt = VOC
10 10
VTH 7.5 Volt
For the calculation of thevenin resistance
Condition Voltage-source is short-circuit and current-source is open-circuit }
GATE-PH 2009 SOLUTION 348
• Equivalent circuit
10 10 10 10
10 RTH 10 10 RTH 5 RTH=15
10 10
5
10 10
Correct option is (a)
1 1 1 1
38. For n ; I , I 3 for p , I , I 3
2 2 2 2
1
For option (a), I nn pp 1 12 12 12 12 0
2 2
1 1 1 1 1 1 1 1 1
I3 2 2 2 2 0
2 2 4 4 2 2
1
For option (b), I nn pp 1 12 12 12 12 1 0
2 2 2 2
1 1 1 1 1 1 1 1 1
I3 2 2 2 2 0
2 2 4 4 2 2
1
For option (c), I np pn 1 12 12 12 12 1 14 14 0
2 2 4
1 1 1 1 1 1 1 1
I3 0
2 2 2 2 2 2 4 4
1
For option (d), I np pn 1 12 12 12 12 1 12 1 0
2 2 2 2 2
1 1 1 1 1 1 1 1 1 1 1
I3 2 2 2 2 0
2 2 4 4 2 2 2 2
Therefore, correct option is (c)
39. Given : Gain of Amplifier = 1000
“feedback” = 9.9% of output
f L 20 Hz “lower 3-dB frequency”
f H 200 KHz “upper 3-dB frequency”
A0
A0 / 2
fL fH
GATE-PH 2009 SOLUTION 349
fL
f Lnew f H new f H 1 A
1 A
9.9
f H new f H 1 1000
100
f H new 100 times f H
Note: (1) Feedback decreases f L
(2) Feedback increases f H
(3) Feedback decreases gain.
Correct option is (c)
40. Check the value of given option
(a) I zmax 15 mA, RL 10 k
25 10
The given circuit zener diode will conduct in reverse bias if VOC > VBR i.e VOC 25.72 V
10 1
10
Then, VR Vz 10 V . Hence, I L = 1 mA
L 10 k
IS=15mA
RS=1k IZ IL=1mA
+
25V VZ=10V 10k
–
25 10
IS IZ IL I L 1 mA, I S 15
1
So, I Z 14 mA
Hence false I Z 15 mA given
(b) RL 10 k
5mA = IS
1k
+ Iz
10V 10k
15V –
4mA 1mA = IL
I Z 4 mA
15 10
VOC 13.63 . So, VOC VBR . Again, zener diode will conduct in reverse bias .
10 1
10 15 10
So, VR VZ 10 . So, I L 1 mA , I S 1 k 5mA
10 k
Here, I S I Z I L
So, IZ = 5–1 = 4 mA
GATE-PH 2009 SOLUTION 350
Hence invalid I Z min 5 mA (given)
(c) RL 2 k
Firstly check diode is conducting or not.
20 2 40
VOC 13.33V
1 2 3
Since, VOC VBR
Then, zener diode will conduct i.e. VL VZ 10V
10mA
1k
+ Iz 5mA
10V
20V – 2K
5mA
20 10 V 10
I S I Z I L (Here, I S 10 mA , I L L 5 mA )
1 k 2 2
10 = IZ + 5 IZ = 5
Hence, IZ = IL
Correct option is (c)
M He
41. Since, Q T 1 M ... (i)
U
where, M U A of uranium and M He A of He = 4
where T kinetic energy of -particle = 5.17 MeV and Q = disintegration energy = ?
Therefore, equation (i)
4
Q 5.17 1 5.257 MeV
236
Correct option is (a)
42. x x x x
y y y y
are translation, invariance of lagrangian under translation leads to conservation of linear momentum in that
direction therefore, px and py are conserved.
Invariance of Lagrangian under rotation leads to conservation of angular momentum along that axis.
Therefore, Lz is also conserved.
Since potential doesn’t explicitly depend on time and is also independent of velocity therefore energy is con-
served.
Hence, correct answer is (b)
0 i
43. Sy
2i 0
Eigenvalue equation of Sy corresponding to
2
GATE-PH 2009 SOLUTION 351
0 i x1 x1
2 i 0 x2 2 x2
ix2 x1 x1 ix2
Assuming x2 1, x1 i
1 i
Therefore, the normalized eigen function of Sy corresponding to eigenvalue is
2 2 1
Correct option is (d)
44. AB † B† A† B A If A and B are hermitian
AB If A and B commutes
Correct option is (c)
45. Consider the set of vectors in 3-dimensional real vector space 3
S 1,1,1 , 1, 1,1 , 1,1, 1
Volume spanned by these vectors
1 1 1
v 1 1 1 1 0 1 2 11 1 2 2 4 0
1 1 1
Therefore, S is a set of linearly independent vector and they will form a basis in 3
Correct option is (b)
46. The density of state in 3-D is
4V
g ( p ) dp 3
p 2 dp ,
h
m
where p 2mE dp dE
2E
4V m
g ( E ) dE g p dp 3
2mE dE
h 2E
The number of particles is
1, E EF
N f ( E ) g ( E ) dE , where f ( E ) at 0 K
0 0, E E F
EF
8 V m
N 3
2mE dE
0 h 2E
EF
4 2V 3/2
3 m
3/2 1/2
E dE 4 2V m3/2 2 EF
h 0 h3 3
GATE-PH 2009 SOLUTION 352
E F N 2/3
Correct option is (a).
1 1
47. L m x12 x22 kx1 x2
2 2
1 1 xx x x
T m x12 x22 , V k 1 2 2 1
2 2 2 2
Corresponding matrices are
1 0 ˆ 0 1/2
Tˆ m , V k
0 1 1/2 0
for frequency of normal modes
m2 k /2
det 2Tˆ Vˆ 0 0
k /2 m2
2
m2 k2 0
2
k k
m2
2 2m
Hence correct answer is (d)
48. Let the state is A 0 B 1
Therefore, the average energy of the particle
2 2
E PE
i i A E0 B E1 20
2 2
A 10 30 B 20
2 2
A 3 B 2
2 2 2 2
According to normalized condition A B 1 A 1 B
Putting this value in above equation
2 2 2
1 B 3 B 2 1 2 B 2
1 1
B A
2 2
1 1
Therefore, 0 1
2 2
1 1
E0 10 eV E1 30 eV .
2 2
Correct option is (d)
1 1
49. L exp t mx 2 kx 2
2 2
equation of motion
GATE-PH 2009 SOLUTION 353
d L L
0
dt x x
d t
dt
e mx e t kx 0
et mx me t
x e t kx 0
k
x x
x0
m
Hence correct answer is (d)
50. The phase velocity by p
(2/) 2
2
The group velocity is given by g
k 2 2
Hence, correct option is (a).
51. The partition function is
Z e 0/ kBT e/ kBT 1 e / kBT
The Helmholtz free energy is
F k BT n Z k BT n (1 e / kBT )
Correct option is (a).
52. The internal energy is
1
U n Z k BT 2 n Z
T k BT
k BT 2
T
n 1 e / k BT
e / k BT
k BT 2
1 e / k BT k BT 2
e / kBT
1 e / kBT
The specific heat is
U
1 e / k BT
e / k BT
k BT 2
e
/ k BT
k T2 e
B
/ k BT
Cv 2
T V
1 e / kBT
2 e / k BT
k BT 2 1 e / k BT 2
Correct option is (d).
GATE-PH 2009 SOLUTION 354
53. Dimension of wave function in 1-D is (length)–1/2. But in the given question, dimension of the wave function is
(length)–1/4. Hence question is not correct.
54. Dimension of wave function in 1-D is (length)–1/2. But in the given question, dimension of the wave function is
(length)–1/4. Hence question is not correct.
1 1 1 3 5
55. For 3d : L = 2, S J 2 , 2 , Terms are 2 D5/ 2 and 2 D
2 2 2 2 2 3/2
1 1 1 1 3
For 3p : L 1, S J 1 , 1 , Terms are 2 P and 2 P
2 2 2 2 2 3/ 2 1/ 2
2
D5/2 Since, transitions is between doubles, thus it is
3d anomalous Zeeman effect. Here, each J will split
2 into MJ values. Selection rules are
D3/2
2
P3/2 M J 0, 1, M J 0 0, if J=0
3p Also M J 0 for transitions and M J 1 for
2
P1/2 transitions. Here, both type of transitions will be seen
M
J
M
5/2 J
3/2 3/2
1/2 1/2
2 J = 5/2 2 J = 3/2
D5/2 P3/2
–1/2 –1/2
–3/2 –3/2
–5/2
3/2
1/2 1/2
2 J = 3/2 2 J = 1/2
D3/2 P1/2
–1/2 –1/2
–3/2
Hence, correct option is (d).
2
D5/2
3d
56. 2
D3/2
2
P3/2
3p
2
P1/2
GATE-PH 2009 SOLUTION 355
Longest wavelength means smallest energy gap which is for 2 D3/ 2 2 P3/ 2
Total transitions = 4 J1 2 J 2 1 10 where J1 J 2
3
Here, J1 J 2
2
Correct option is (b)
57. Volume of primitive unit cell of fcc lattice,
a ˆ ˆ a ˆ ˆ a ˆ ˆ
V a1 a2 a3
2
jk i k i j
2 2
0 a /2 a /2
a2 a a2 a a2
a /2 0 a /2 0 0 0 0
4 2 4 2 4
a /2 a /2 0
a3 a3 a3
8 8 4
Therefore, primitive translation vectors of reciprocal lattice are
2
a ˆ ˆ a ˆ ˆ 2 a
2 a a 2 2 i k
2
i j
4 iˆ iˆ iˆ ˆj kˆ iˆ kˆ ˆj
b1 2 3
a1 a2 a3 a3 a3
4 4
2 2
a
0 kˆ ˆj iˆ
a
iˆ ˆj kˆ
2
a ˆ ˆ a ˆ ˆ 2 a
2 a a 2 2 i j
2
j k
4 iˆ ˆj iˆ kˆ ˆj ˆj ˆj kˆ
b2 3 1
a1 a2 a3 a3 a3
4 4
2 ˆ ˆ ˆ 2 ˆ ˆ ˆ
a
k ji
a
i jk
2
Similarly, b3
a
iˆ ˆj kˆ
Hence, correct option is (a).
2 2 ˆ ˆ ˆ 2 ˆ ˆ ˆ
58.
V b1 b2 b3
a
iˆ ˆj kˆ
a
i jk
a
i jk
3 1 1 1 3 3 3
2 2 2 2
1 1 1 1 (1 1) 1 (1 1) 1 (1 1) (2 2) 4
a a a a
1 1 1
Hence, correct option is (a).
GATE-PH 2009 SOLUTION 356
R
PQ 0 1
59. Q̄
00 1 1
01 1 P̄ R̄
11 1
10 1 1
Note:- Form group of 1’s in K-map
f PR Q simplified expression.
Correct option is (a)
60. (a) P P̄ P R PR
R
R̄
Q Y
Q̄
Y PR Q
Y PR Q
Note:- Apply Demorgan’s law.
P̄ P̄ R̄
P
R
(b) R̄
Q Y
Q̄
Y PR Q
Hence K-map expression is present in option (b)
Correct option is (b)
GATE-PH 2010 SOLUTION 357
OBJECTIVE QUESTION
Q.1 – Q.25 : Carry ONE mark each.
1 2
1. The number of independent component of a antisymmetric tensor Pij of rank ‘2’, is given by
2
N N
Here, the indices i and j are running from 1 to 5. So, N = 5
Therefore, number of independent components
1 2 1
2
N N 25 5 10
2
Correct option is (b)
e z sin z y C:|Z–2| = 1
2. f z has pole of order ‘2’ at z = 0
z2
Since, z = 0 lies outside the circle x
z
z=0 z=2
e sin z
So,
C
z2
dz 0
Correct option is (d)
3. Sum of the eigenvalues = Trace of the matrix i.e. 1 2 3 5
Product of the eigenvalues = Determinant of the matrix i.e. 12 3 5
Only option (c) satisfy both relations.
Correct option is (c)
sx e sx e sx e sx e 3 s
4. L f x e x 3 dx x 3 1 dx 2
3 s 3
3 s s 3
s2
Correct option is (d)
5. Electrical conductivity, thermal conductivity and metallic lustre properties are valence electron dependent while
shear modulus is independent of valence electrons.
Hence, correct option is (c).
6. For fcc lattice, the amplitude of diffracted X-ray beam is given by
F f 1 e i ( h k ) e i ( k l ) e i ( l h ) ... (i)
For (212), F f 1 e i (2 1) e i (1 2) e i (2 2) f (1 1 1 1) 0 (i.e., no diffraction)
For (111), F f 1 e i (1 1) e i (1 1) e i (1 1) f (1 1 1 1) 4 f
For (200), F f 1 e i (2 0) e i (0 0) e i (0 2) f (1 1 1 1) 4 f
For (311), F f 1 e i (3 1) e i (1 1) e i (1 3) f (1 1 1 1) 4 f
Hence, correct option is (a).
1
7. Hall coefficient, RH
ne
Correct option is (c)
8. Option (d) is the correct statement of Bloch theorem.
Hence, correct option is (d).
GATE-PH 2010 SOLUTION 358
9. For ferromagnetic substance, the specific heat is maximum at Tc. Thus correct graph of specific heat versus T
M s 0 h M s
is expressed as : CV 2 M s
, where ; M s N tan h
T k BT
Specific heat
O T
T = TC
Thus, option (d) is not correctly shown.
Hence, correct option is (d).
10. The thermal conductivity reduces in the transition from the normal state to the superconducting state because
Cooper pairs have very less energy ( 103 eV) . They can not transfer energy to the lattice.
Hence, correct option is (a).
11. In -decay process one neutron decay into one proton and one electron
n p e ve
Quark structure of u is udd and p is uud
Thus, reaction is written as udd uud e ve or d u e ve
Correct option is (a)
15
12. Since, 7 N has Z = 7, N = 15 – 7 = 8
Therefore, 7Z, (1s1/2)2 (1p3/2)4 (1p1/2)1
1
J , l 1 (for p) and parity = (–1)1 = –1(odd)
2
Thus, spin-parity of 15
7 N = 1
2
Correct option is (a)
13. (1) Lepton are produced by weak interaction. I3 is not conserved.
(2) -ray produced through electromagnetic interaction.
(3) Baryon are produced by strong interaction. Isospin I and I3 are conserved.
Correct option is (a)
14. For a gaseous species in a mixture of gases, the preferred probing tools is ESR spectroscopy.
Correct option is (c)
15. Since, to establish equilibirium
Correct option is (c)
16. Electric dipole transition between two states will be allowed if both the states have opposite parities. Since,
selection rules are 1 and m 0, 1
Hence, correct option is (c).
GATE-PH 2010 SOLUTION 359
17. Using Wein’s displacement law,
2.897 103 2.897 103
max m-K 2.897 106 m 2897 Å
T 1000
This wavelength lies in infrared region
Correct option is (b)
18. From the figure we can say that total charge is zero.
z
y
x
But dipole moment is not zero. So, electric field will be due to dipole moment.
Correct option is (d)
19. 12-bit DAC
Range of voltage –10 to 10 = 20V.
VPP
Resolution of DAC (R) =
2n 1
VH H L 10 10
Voltage Resolution 5mV
2n 1 212 1
Correct option is (b)
R2
R1
Vi
V0
20. (a)
V0 R
2
input output Vi R1
180º
Vz
– +
(b) Vi
R V0
input (–)ve will make zener - ON
Since, Zener diode operates in reverse break down voltage
(c) At the time of negative input of the signal diode become off and output V0 = 0
GATE-PH 2010 SOLUTION 360
R
Vi
V0
Diode in reverse bias for negative input, So V0 = 0
(d) D2
D1
Vi
V0
D2 conduct for positive input. D1 conduct for negative input.
Correct option is (c)
21. We need two independent coordinates for each particles to know its position on the surface of the sphere.
Number of degree of freedom for N particles = 2N
and energy per degree of freedom 12 k BT .
So internal energy of the system 12 k BT 2 N N k BT .
Correct option is (c).
22. To exhibit Bose-Einstein condensation, a particle must be Boson, i.e., it must have an integral spin. Since all
protons, electrons and neutrons have half integral spin, only those atoms whose n p ne nn is even will be a
boson.
Particle n p ne nn n p ne nn
1
H1 1 1 0 2 (even)
4
He 2 2 2 2 6 (even)
23
Na11 11 11 12 34 (even)
40
K19 19 19 21 59 (odd)
Since n p ne nn for 40
K19 is odd, it cannot exhibit Bose-Einstein condensation.
Correct option is (d).
23. Lorentz transformation satisfies group properties.
Therefore both statements P and Q are correct.
Hence correct answer is (B)
24. Any allowed wave function of the particle in bound state should satisfy the following conditions:
(1) x should be singled valued, finite, continuous everywhere in space,
(2) x should become zero at x , r .
e r
N is not defined a r = 0
r3
N 1 e r is not zero at r
Ne x e
x2 y 2 z 2
is zero at x
GATE-PH 2010 SOLUTION 361
The function given in option (d) is discontinuous at r = R. Therefore, the function given is option (c) will be an
allowed wave function for a particle in a bound state.
Correct option is (c)
25. Momentum of the particle is,
1.06 1034 3 108 eV MeV
p pmin 15 19
200
xmax 10 1.6 10 c c
Correct option is (c)
Q.26 – Q.55 : Carry TWO marks each.
26. f z
e z
e z
2
sinh z is not analytic at z = 0
sin z sin z
1
cosh z
lim 2
z
sinh
2 lim 2 z 2
z 0
sin z cos z
1
z0
2 z
So, z = 0 is a removable singular point.
Correct option is (c)
d2y
27. y 2 cosh t
dt 2
C.F . Aet Be t
1 1
P.I . 2 2 cosh t et e t
D 1 D 1
2
1 t t
t
2D
e e t et e t t sinh t
2
Total solution, y Aet Be t t sinh t
Putting the condition, y t 0 0 A B 0
dy
Putting the condition, 0 A B 0
dt t 0
Solving above two equations, A 0, B 0
Therefore, y t sinh t
Correct option is (d)
1 1
28. x 2 Pn x Pn 2 x dx xPn x xPn 2 x dx
1 1
1 n 1 n n 3 n 2 P x dx
Pn 1 x Pn 1 x Pn 3 x
1
2n 1 2n 1 2n 5 2n 5 n1
1 n 1 n 2 P x P x dx
n1
1 2 n 1 2 n 5
n1
GATE-PH 2010 SOLUTION 362
1 2
[Using the orthonormal property, Pm x Pn x dx m ]
1 2n 1
n 1 n 2 2
0
2n 1 2n 5 2n 3
Correct option is (d)
29. The density of state is 2-D is
2A
g ( p ) dp p dp ,
h2
m
where p 2mE or dp dE
2E
m 2A 2A
g ( E ) dE 2
dE 2 m dE
2mE
h 2E h
Since, we can have an electron with spin up or spin down in each state, we need to multiply density of state by
the factor of two.
4A
g ( E)
m
h2
The number of particles is
1, E EF
N f ( E ) g ( E ) dE , where f ( E )
0, E E F
0
4A N 4EF m
N 2
mEF n
h A h2
N
where n electronic density
A
4mEF 4mEF h
n
h 2
( 2) 2 2
mEF
n
h2
Correct option is (b)
30. Correct option is (d).
31. Correct option is (d)
2
32. E I I 1 where I is spin and is moment of inertia
2
Now, for I = 2, E = 57.5 KeV
2 2 57.5
57.5 23
2 2 6
Next level is 6 +.
Correct option is (a)
GATE-PH 2010 SOLUTION 363
33. Given nuclear reaction, p K
1 1 1 1
I 3 1 1 1 0
2 2 2 2
S (0 0) (1 1) 2 0
B (0 1) (0 1) 0
Correct option is (c)
34. CH3OH is example of symmetric top molecules, where I x I y I and I z I
Energy level of symmetric top molecule is
EJ BJ J 1 A B mJ2 hc
where, mJ 0, 1, 2, ............ J and J 0,1, 2,3,......
h h
Also, B 2 and A 2
8 I B c 8 I Ac
where, IB and IA is the moment of inertia about orthogonal and parallel to the molecules axis i.e.
I B I x I y I and I A I z .
Here, J is same as , therefore, energy levels can be written as
2 2 I I
E 1 m2
2I 2 Iz IB
2 2 1 1
Thus, E 1 m2 IB I
2I 2 Iz I
Correct option is (a)
35. For the given Half harmonic oscillator potential, energy eigenvalues of the particle are
3
En 2n
2
Expectation value of the energy will be
1 3 4 7 31
E P0 E0 P1E1
5 2 5 2 10
Correct option is (a)
36. Here the molecular has been compared with their corresponding wave number or we can say that energy of
these spectra has been normalized in the units of cm–1. So, in terms of energy rotational spectra corresponds to
microwave radiation which has the lowest energy. Then comes vibrational spectra (IR region) slightly greater
than both rotational and vibration.
Molecular dissociation corresponds to very high energy.
So, Erotational < Evibrational < Eelectronic < Emolecular dissociation
Rotational 10º – 102 cm–1
Electronics 105–106 cm–1
Molecular 106 cm–1 and above
Correct option is (b)
GATE-PH 2010 SOLUTION 364
Quantity T P
E E E
37.
B B B
Correct option is (a)
38. The interaction energy between two magnetic dipole moment is given by
µ0 m1 r12 r12 m1
w12 3 3 m2
4 r125 r12
µ0 m1 r12 m2 r12 m1 m2
3
4 r125 r123
µ0 3m 2 cos2 45º
0 m1 m2
4 d3
2 2
µ 3m 1 µ 3m
0 3 0 3
4 d 2 8 d
Correct option is (d)
39. When the loop will be pulled out of the region of the magnetic field, then amount of flux passing through
the loop will gets changed. This change in flux will induced an emf in the circuit
d
Vemf
dt
At t = 0, loop was completly inside the magnetic field
So, t 0 0 constant say
Given circuit is op-amp integrator circuit
1
Vout Vemf dt Vemf will acts as an input voltage for op-amp
RC
1 d
dt
RC dt
1
[This will be depending on u also]
RC
Vout is independent of 0
Correct option is (a)
40. Since current flow in the circuit is constant I 0.
So, initially capacitor will start to charging
dq
I q It
dt
v0 vc qC CIt (linear with time)
After certain time capacitor will be fully charged.
So, Voltage will not change with time.
Correct option is (d)
GATE-PH 2010 SOLUTION 365
41. (a) a ab + ab
b Q
b
Q ab ab b Q ab
Note: x x 0 0 x x
(b) a Q = ab
b
(c) a ab
b Q ab 0 ab
ab
0
b ab b 0
Q ab 0 Q ab
(d) ab Q a b
Option (d) is different except other.
42. Number of molecules in a given energy states is given by Boltzmann law.
E2 E1
k BT
N 2 N1 e
where, N2 = number of molecules in energy state E2
n=4
N1 = number of molecules in energy state E1 2×2.5 = 5 meV
n=3
T = temperature of the system n=2
2.5 meV
kB = Boltzmann constant. n=1
Ground state
E4 E2
Thus, N4 N2 e k BT
Ratio of molecules,
E4 E2
N4 k BT
e
N2
As energy is equally spaced, therefore,
E E4 E2 2 2.5 meV 5 meV
5103 eV
N4 8.62105 eVk 1300 K
e
e 0.193 0.826 0.8
N2
Correct option is (c)
43. The enthalpy is defined as
H U PV
dH dU PdV VdP where dU TdS PdV
GATE-PH 2010 SOLUTION 366
dH TdS VdP
So we have T
H H
and V
S P P S
Correct option is (b).
44. 1
S x2 s, m | S x2 | s, m
4
s, m | S2 S2 S S S S | s , m
1 1
s , m | S S | s, m s, m | S S | s, m
4 4
1 1 1
s m s m 1 2 s m s m 1 2 s s 1 m 2 2
4 4 2
Correct option is (a)
1 2 3
45. V ( x) kx x , k, 0
2 3
dV kx x 2 0
for equilibrium 0
dx k
x 0,
d 2V
k 2x
dx 2
d 2V
at x 0, k 0
dx 2
Therefore x = 0 stable equilibrium point.
k d 2V
at x k 0
dx 2
k
Therefore, x is unstable equilibrium point.
Hence correct answer is (b)
S S
v
46.
c
–c
The position of two photon in the s-frame when they are detected at
x1 10c m , x2 10c m
So, the time internal between two detection in s ' frame is
1
v x2 x1 / c 2 0.6c 20c
t ' t '2 t '1 c 2 15 s
1 v2 / c 2 1 0.6
2
Correct option is (a)
GATE-PH 2010 SOLUTION 367
47. First order correction to ground state energy is,
L L
2V x 2 x V 2 x 2 x
E 1 | V p | 1 0 sin 2
1
sin dx 0 1 cos sin dx 0
L 0 L L L 0 L L
Second order correction to the ground state energy is always less than equal to zero i.e. E 2 0
Correct option is (a)
1 1
48. Consider the transition, 2 P½ 2 S½ . These terms are correspond to J1 and J 2 . In a weak
2 2
magnetic field it breaks into 2 J1 1 2 J 2 1 2 components
mJ
½
2
P½
–½
2
½
S½
–½
The selection rule of transition
mJ 0, 1
Thus, in total four spectral lines are observed in transition from 2 P1/2 1S1/2
1 1
Also, by formula, total number of transitions = 4 J1 2 J 2 1 4 2 1 4
2 2
Correct option is (c)
1 1
49. Given transition, mJ mJ is corresponds to mJ 1
2 2
Transition mJ 1 corresponds to components. Electric field of the emitted radiation oscillates
perpendicular to the applied magnetic field. Classically, this situation corresponds to two parallel dipoles
oscillating with phase difference 90º. Superposition of two dipoles produces circulating current. Thus,
in direction of the magnetic field, circularly polarized light is emitted.
For mJ 1 , it is clockwise and
mJ 1 , it is anti-clockwise circular
Correct option is (a)
50. The partition function of the photon gas is
2 V (kBT )3
n Z
45 3c3
The integral energy is
2 2 V ( k BT ) 3
U k BT n Z k BT 2 3 3
T T 45 c
GATE-PH 2010 SOLUTION 368
2
2 V 2 3 V k 4 T 4
k BT 2 k 3
B 3T B
45 3c 3 45 3c3
U 3 2V 4
Now, Cv k B 4T 3 Cv T 3
3 3
T V 45 c
Correct option is (b).
51. The free energy is
2 V ( k BT ) 3
F k B T n Z k BT 3 3
45 c
F 2 ( k B T ) 3 2 ( k BT ) 4
The pressure is given by P k BT
3 3 3 3
V N , T 45 c 45 c
Correct option is (c).
52. Let E E0 ei k r t
We know that Maxwell’s equations,
E 0
B 0
B
E
t
E
B 0 µ0
t
Replacing by ik and by i , we get
t
ik E 0 k E 0
ik B 0 k B 0
ik E i µH k E µ H
ik H i E k H E
Correct option is (d)
53.
S E H * E k E* µ1
1 * 1 2 k 2
µ
k E.E k E E *
µ
k E
µ
E kˆ
Since µ is negative.
k 2
S E kˆ
µ
Therefore, k and S are anti-parallel.
Correct option is (b)
GATE-PH 2010 SOLUTION 369
1 22
54. L m mg 1 cos
2
Corresponding Hamittonian of system is
p2
H mg 1 cos
2m2
Hamilton’s equations are
H H
and p
p
Therefore
p
2 , p mg sin
m
Hence correct answer is (a)
55. Poisson Bracket ,
p 1 1 1
2
, 2 , p 1
m m m m 2
2
Hence correct answer is (b)
GATE-PH 2011 SOLUTION 370
OBJECTIVE QUESTION
Q.1 – Q.25 : Carry ONE mark each.
1. Two similar matrices A and B have same eigenvalues, same determinant and same trace.
Correct option is (c)
2. Correct option is (a)
3. Quantum mechanical operator for the momentum of a particle moving in one dimension is
pˆ x i
x
Correct option is (b)
4. The Carnot cycle is a reversible cycle and hence the entropy change of the universe is zero.
i.e., (S )universe 0
Correct option is (d).
5. The key points are :
(i) The specific heat diverges for both I and II order phase transition.
(ii) The entropy has a finite discontinuity for I order phase transition whereas it is continuous for II order phase
transition.
Correct option is (b).
6. Surface energy Es as A2/3 Es A2/3
Correct option is (c)
7. In case of two level, the rate of upward transitions is equal to the rate of downward transitions which gives
N 2 N1.
Thus, there will be no population inversion.
Correct option is (a).
C
8. For ferromagnetic materials, , for T TC .
(T TC )
O TC T
Correct option is (b).
9. The order of magnitude of the energy gap of a typical superconductor is 1 meV .
Correct option is (d). (Reference C. Kittel)
GATE-PH 2011 SOLUTION 371
10. Common emitter configuration / amplifier
Circuit
VCC
RC
IB IC
B C RB V0 180º phase between
Input Output input and output
E E Vi
• Output is taken from collector
• It has 180º phase shift between input and output.
• Amplifier is an application in active region. So, one junction is forward bias and other junction is reversed
bias.
Correct option is (b)
11. If 1 , 2 , 3 are the eigenvalues of matrix, then
Trace = 1 2 3 11
Determinant = 1 2 3 36
So, the possible values of 1 , 2 and 3 are (2, 3, 6)
Correct option is (d)
12. Linear, I1 I 2 , I 3 0 z
Symmetric top, I1 I 2 I3
y
Spherical top, I1 I 2 I 3
Asymmetric top, I1 I 2 I3 x
Correct answer is (c)
13. For the region x > 0, the 1-D time independent Schrodinger equation will be
d 2 2m d 2 2m
E V0 0 2 0 where 2 2 V0 E
dx 2
2
dx 2
Solution will of the form: Ae x Be x
Since, the wave function should be zero at x , therefore A = 0 and the space part of the wave function for
the region x > 0 will be of the form Be x .
Correct option is (b)
14. Lx Ly , Lz Lx , Lz L y Lx L y , Lz i Ly L y Lx i Lx i L2x L2y
Correct option is (c)
GATE-PH 2011 SOLUTION 372
1 1 1 1
15. 2
* r 2 r dV 3 r 2
e 2 r / a r 2 sin drd d
r r a
2
1 2 r / a 2
3
a e
r 0
dr sin d d a
0 0
2
Correct option is (d)
1 2q 2 q 2 2q 2
F
16. 4 0 4 1 1
0.5m 0.5m
1 7q 2 –2q –q q 2q
4 0 2 1.5m 1.5m
Correct option is (a)
0
17. B K nˆ , where K is the surface current density
2
0 K ˆ
i for z 0
2
0 K ˆ
i for z 0
2
Correct option is (a)
18. Correct option is (c)
19. We know that, A B
B0 xJˆ B0 kˆ is uniform field.
B0 yJˆ B0 kˆ is uniform field.
B x By
0 iˆ 0 ˆj 0
2 2
By Bx B
0 iˆ 0 ˆj 0 kˆ is uniform field.
2 2 2
Correct option is (c)
20. Neutrons cannot produce ionization because they are electrically neutral. They can be detected because of the
secondary particles produced in a nuclear reaction in the detector medium.
Correct option is (d)
21. In weak field, the higher total angular momentum state splits into (2J + 1) states. Possible J values for one
1 1 1
electron 1s1 , s , J ,
2 2 2
1 1
Higher J value i.e. will split into 2 J 1 2 1 2 2
2 2
GATE-PH 2011 SOLUTION 373
Correct option is (a).
22. The selection rules for orbital angular momentum :
For one electron, most of the transitions take place when value changes by one unit, i.e., 1 because,
according to Laporte rule, the parity should change in a electric dipole transition.
The selection rules for total orbital angular momentum:
For total angular momentum, the selection rule for the electric dipole transition is L 0, 1.
The selection rules for total spin angular momentum :
For total spin angular momentum, the selection rule for electric dipole transition is S 0.
Correct option is (b).
23. If there are p atoms in a primitive cell, then total number of branches are 3p. Out of 3p branches, 3 will
be acoustical branches and remaining (3p-3) will be optical branches. The number of branches does not
depend on the number of primitive cells.
Correct option is (c)
24. The fermi level of an intrinsic semiconductor is given by
Ec Ev 3 mh*
EF k BT ln *
2 4 me
Correct option is (c)
25. Voltage across 2k
+ +
Va Vz
– – if Va Vz means zener ON
0.5k 1k
+ + +
30V V a1 Va2 2k
– – –
For given circuit to check zener in breakdown region.
30 3
Va1 Va1 25.71
0.5 3
Vz1 20 Volt
Va1 Vz1 so, Zener - ON
0.5k 1k
+
30V 20V Va 2 2k
–
equivalent circuit
20 2
Va2 Va2 13.33 Volt
3
Va2 Vz2 So, Va2 10 V
zener - ON
GATE-PH 2011 SOLUTION 374
Final circuit
0.5k 1k
+
30V 20V 10V 2k V2k = 10V
–
10
I 2k 5 mA
2
Correct option is (d)
Q.26 – Q.55 : Carry TWO marks each.
2iˆ 2 ˆj kˆ
26. Unit vector normal to the surface nˆ
3
(1,1,1)
Correct option is (d)
27. r nˆ dS ( r ) dV (Using divergence theorem)
S V
3 dV 3V 3 a 2h
V
Correct option is (b)
dy x
28. ( y 1) dy x dx
dx y 1
y2 x2
y c
2 2
x 2 y 2 2 y 2c
x 2 ( y 1)2 2c 1
This represents a family of circles having center at (0, –1) and radii of 2c 1 units.
Correct option is (a)
1 q
29. V q, q
q2
Generalised force is given as
V d V
Q
q dt q
2 1 q d 1 2 1 q 2q 2
2 3 3
q3 dt q q3 q q
Correct answer is (c)
30. Frequency of normal mode is given as
k mm
, 1 2 reduced mass
m1 m2
GATE-PH 2011 SOLUTION 375
m 2m 2 m
m 2m 3
k k 3k
2m /3 2m
Hence correct answer is (a)
31. Q q cos p
P q sin p
For canonical transformation
Poisson Bracket Q , Pq , p 1
Q P Q P
or, 1
q p p q
or, q 1 cos p q cos p q sin p q 1 sin p 1
or, q 2 1 1 q 0
2 1 0 1/2
1
1 1 2
2
Correct answer is (d)
32. Initial : 0.6c m
m 0.6c
Initial momentum of two particles is zero. Therefore final momentum will also be zero. So final entity will be at
rest.
final : M rest
Apply conservation of energy.
Einitial E final
mc 2
2 Mc 2
2
1 0.6
2m 5
M m
0.8 2
Correct answer is (c)
33. First order correction to ground state energy is
a /2 a/2
1 2V x x 2V
E x V x 1 x dx 0
*
1
'
sin
2
cos dx 0
0 a 0 a a 3
x
[Assume sin p to solve the integration]
a
Correct option is (a)
GATE-PH 2011 SOLUTION 376
34. Given : T 1010 (Pa) 1
1 dV m
By definition, T , where V ; is the density
V dP
m2 d
dV d
V m /
d 1 (d /)
T dP
dP T
d
For 1% 0.01 and T 1010 , we have
0.01
dP 10
108 Pa
10
Correct option is (c).
35. Number of microstates for N non-interacting particles with spin s is
(2 s 1) N 3 N for s 1
The entropy of the system is S k B ln N k B ln 3
Correct option is (d).
36. The partition function is Z g (i ) ei / kBT 4e / k BT 2e2 / kBT
i
2 e 2 / k BT 1
The fraction of particles in the upper level / k BT 2 / k BT
/ k BT
4e 2e 2e 1
Correct option is (b).
37. 0
r r a
0 2 E0 a3 3 3 3 3
0 E0 cos cos 0 E0 2 E0 0 E0
r 3
ra 2 2 2
Correct option is (a)
38. We have nucleus 17
8 O,
Number of protons (Z) = 8 (even) and number of neutrons (N) = 9 (odd)
9N = ( 1s1/2)2 (1p3/2)4 (1p1/2)2 (1d5/2)1
5
J , l 2 (for d)
2
Parity = (–1)2 = +1 (positive)
5
Spin parity of ground state =
2
Correct option is (d)
GATE-PH 2011 SOLUTION 377
39. Correct option is (a)
40. For baryon , strangeness S 3 and isospin (I3)
Correct option is (b)
41. According to Heisenberg uncertainty relation
h h
E t (suppose) t 1 ns 1 109 s
2 2
h 6.6 1034
E Joule
2 t 2 3.14 109
h 6.6 1034
E eV = 0.65 × 10–6 eV 10–6 eV
2 t 2 3.14 10 9 1.6 1019
Correct option is (c)
42. Total degeneracy, D D1 D2 , where D1 is degeneracy of 2 p 2 and D1 is degeneracy of 3d1.
Now, D1
41 2 ! 4 1 2 6! 6 5 15
41 2 N ! N ! 4 1 2 2 !2! 4!2! 2
And D2
4 2 2 ! 10! 10 D D D 15 10 150
1 2
4 2 2 1!1! 9!
Correct option is (d).
43. The line spacing in the infrared rotational absorption spectrum 2B = 20 cm–1 = constant B 10 cm 1
Since, the distance of first Stokes lines from Rayleigh line is 6B. Therefore, the position of first Stokes line in the
rotational Raman spectrum = 6 × 10 cm–1 = 60 cm–1.
Correct option is (c).
44. Given that : 2.1 Å, 30 , bcc structure
2.1 Å 2.1 Å
From Bragg’s law : 2d sin d 2.1 Å.
2 sin 2 sin 30 2 1
2
For first diffraction peak ( hkl ) (100)
a
d ad h 2 k 2 l 2 a 2.1 Å 12 02 02 a 2.1 Å
2 2 2
h k l
Therefore, volume of the unit cell a3 (2.1 Å)3 9.261 Å 3 9.3 Å 3
Correct option is (c).
45. Given VBE 0.7 V (identical transistor)
Ic1
100 Ic2
VL
K
+ Tr1 Tr2
5V 0.7 –
KVL
GATE-PH 2011 SOLUTION 378
Here, Ic2 Ic1 because it is current mirror circuit.
By applying KVL through dotted loop
5 Ic1 100 0.7
Ic1 43 mA Ic2
Correct option is (d)
10
00
0
0 01
0
011
46. YABCDA BCDABCDABCDA BCDABCD
01 01 0 111 10 01
It is a 4 variable boolean function which can be minimized by using K-map
CD
AB 00 01 11 10
00 1 1
01 AD
1 1
11
10 1 1
ABC
f ABC AD
Correct option is (c)
47. Lets redraw the circuit.
+10V
Vin
V0 10V
Feedback –10V
voltage + 4k
Vf 1k
–
V0
10 1 10 1 V f Vin Vsat
Vf Vf
1 4 1 4 V f Vin Vsat
V f VUT 2 Volt V f 2 Volt
VUT 2V VLT 2V
Upper threshold Lower threshold
O/P of this ve feedback circuit is a square wave [Schmitt Trigger]
Note: (1) When input wave crosses VUT , V0 makes transition from VSAT to VSAT
(2) When input wave drops below VLT , V0 makes transition from VSAT to VSAT .
Correct option is (a)
GATE-PH 2011 SOLUTION 379
z sin z
48. f (z) putting z t
( z )2
(t ) sin (t ) (t ) [ sin t ]
t2 t2
sin t sin t 1 t 3 t 5 1 t 3 t5
f ( z) 2 t ...... 2 t ......
t t t 3! 5! t 3! 5!
Highest negative power in the Laurrent series expansion is 1/t
So, f ( z ) has a simple pole at z
Correct option is (d)
49. Since, z lies outside the circle.
y
C : |z| = 1
x
z=
z sin z
Therefore,
( z )2
C
dz 0
Correct option is (b)
L L L
50. Density of states .
dE d 2 a sin ka
E0 2 cos (ka)
dk dk
Hence, correct option is (b).
2 2 2
51. Effective mass (m*)
d 2E d d d 2 a sin (ka )
2 dk dk 0
E 2 cos ( ka ) dk
dk
2
m*
2 a 2 cos (ka)
Hence, correct option is (b).
52. 2 will be orthogonal to 1 if
1 | 2 0 0 2 1 3 2 0 1 2 0 1 2 3 0 1
Correct option is (c)
1 1 1
53. Normalized form of 2 : 2 0 1 2
3 3 3
1 1 1 3 1 5 3
Therefore, E P0 E0 P1 E1 P2 E2
3 2 3 2 3 2 2
Correct option is (d)
GATE-PH 2011 SOLUTION 380
k k ˆ ˆ
54. B B0 sin x y t kˆ , wavevector K
2
ij
2
We know the solution,
1 1
E K B c2 K B
µ 0
0
k 1 ˆ
cB0 sin x y
2
t iˆ ˆj
k
2
k 1
cB0 sin x y
2
t ˆj iˆ
2
cB0
ˆj iˆ sin x y k
t
2 2
Correct option is (a)
cB02
T
2 k ˆ ˆ
i j
55. S E H* Tµ0 0
sin x y
2
t 2
dt
cB02 1 T 2 k 1
iˆ ˆj sin x y t dt
2 2 µ0 T 0 2 2
Correct option is (c)
GATE-PH 2012 SOLUTION 381
OBJECTIVE QUESTION
Q.1 – Q.25 : Carry ONE mark each.
1. Given vectors are a 3iˆ 2 ˆj, b iˆ 2 ˆj
If a and b are linearly dependent then we must have
a mb 0, 0, m 0
3iˆ 2 ˆj m iˆ 2 ˆj 0
3 m 0 and 2 2m 0
These equations do not have non-zero solution i.e., 0, m 0
Therefore, a and b are linearly independent.
Further, a.b 0 , not orthogonal
a 1, b 1 not normalized.
Correct option is (a)
2. Moment of inertia of a thin rod about an axis passing through its one end and perpendicular to its length is
mL2
, where m = mass, L length of rod
3
Therefore, for given system
M 1L2 M 2 L2 L2
Moment of inertia M1 M 2
3 3 3
Correct option is (c)
1 1
3. E E0 cos t kz u E 0 E 2 uE 0 E02 cos 2 t kz
2 2
1 1 1
cos t kz 2
2
u E 0 E02 cos 2 t kz 0 E02
2 4
Correct option is (a)
4. VDC for full wave rectifier
1 2V Vm
VDC Vm sin t dt m
0
0 2 3 4
2 Vm
VDC , putting the value of Vm , we get
2 10
VDC 6.36 Volt
DC value is also referred as average value.
Correct optino is (c)
GATE-PH 2012 SOLUTION 382
5. Energy eigenvalues of the particle confined in a two dimensional square well potential of width 2a is
22 22 2
2 x
Enx n y n ny
2 2
2 x
n n y2
2 m 2a 8ma
2 2 2 2 5 2 2
Energy of the first excited state of the particle is E21 or E12
8ma 2
1 2 8ma2
Note : The question in the original paper is wrong.
1 V
6. The isothermal compressibility is defined as T .
V0 P T0
It actually explains how does the volume of a substance change when pressure on it changes at constant
temperature and hence explains how easily a material can be compressed.
Correct option is (a).
1 1
7. For 2S1/2 state : 0, 2s 1 2 s , J
2 2
For weak field, this state will split into two 2 J 1 2 energy levels.
The energy of splitted levels is given by E g B J B gµB BJ cos gµB Bm j
1 3 1 3 3 3
0 1
J (J 1) ( 1) s (s 1) 2 2 2 2 4 4
2
For S1/2 state : g 1 1 1 1 1 2.
2J (J 1) 1 3 6
2
2 2 4
1
mj
2
2
S1/2
B0 1
mj
2
1 1
Therefore, the energy of upper level, m j (2) ( B ) B B B
2 2
1 1
The energy of lower level, m j (2) ( B ) B B B
2 2
Therefore, the difference in energy levels B B B B 2 B B .
Hence, correct option is (b).
8. The Fermi energy of a Fermi gas can be given as
2 2/3
EF
2m
3 2 n EF n 2/3 ...(i)
1
Now, this Fermi energy is related to velocity vF at the Fermi surface as EF mvF2
2
1 2 2/3
2
mvF2
2m
3 2 n vF2 n 2/3 vF n1/3 .
Hence, correct option is (d).
GATE-PH 2012 SOLUTION 383
9. Electron, photon and neutrino do not have substructures and hence are the fundamental particles.
Correct option is (c)
10. Correct option is (a)
2
3
n n 1
2
1 1
11. Reflectivity is given as R 1 2 2 100 4%
n1 n2 1 3 25 25
2
Correct option is (a)
1 1
12. ne 15 3.472 cm
ne 10 1.6 10 19 1800
Correct option is (c)
2
13. v x v y v x2 v y2 2vx . v y
2
v x vy vx2 v y2 2 vx . v y
From equipartition theorem,
1 1 k T
m vx2 k BT vx2 B
2 2 m
k T
Similarly, v 2y B
m
vx v y vx v y 0 as vx 0 and v y 0
2 k BT k BT 2k T
v x vy
m
m
0 B
m
Correct option is (d)
14. From mathematical concepts of conic section it is known that for parabola 1
Correct option is (b)
15. Arsenic is penta-valent.
Note : In N-type semi conductor as doping increases Fermi level moves towards the conduction band.
Hence, it is lies nearly donar energy level.
Correct option is (b)
16. The position of first stokes line of a rotational Raman spectrum 6B = 12.96 cm–1 B 2.16 cm 1
Hence, the rotational constant B 2.16 cm 1
Hence, correct option is (c).
5 5
3 3 5
kN E 2 N 2 k N 3 2 E
17. E 2
P k P . . 2 .
V 3 V S , N 3 V 3 V V 3 3 V
2
PV E
3
Correct option is (b)
GATE-PH 2012 SOLUTION 384
18. For a system of identical Fermions, total wave function should be anti-symmetric in nature. Since, the given
spatial part of the wave function is symmetric under particle exchange, therefore the spin part of the wave
function should be anti-symmetric under particle exchange.
Correct option is (b)
A
19. E E izˆ
t
iˆ ˆj kˆ iˆ ˆj kˆ
B A ˆj 0 t ˆjt
x y z x y z
Ax Ay Az tz 0 0
Correct option is (d)
+10V
20.
Vin +
4k a Vout
By applying voltage +5V –
divider rule
1k –10V
51
Va
4 1
Va 1 volt
Input condition V0
V V VSAT
V V VSAT
• When Vin is greater than Va ; output V0 VSAT
Correct option is (a)
21. The nuclear forces are same for neutron-neutron, neutron-proton and proton-proton inside nucleus. There-
fore, it suggests that these forces are not isospin dependent. The spin and quadrupole moment is non-zero,
therefore, they have spin dependence along with tensor components.
Correct option is (d)
2
22. Given : V x x x 2 , m = 1
m
Time period 2 k 0
k
d 2V d 2V d 2
k
x 2 2 x x 2 x 2
dx 2 at stable point
dx 2 x2
dx
2 x 2 2 x 2 2 x x 2 4
m 1
time period 2 2
k 4
Correct option is (b)
GATE-PH 2012 SOLUTION 385
23. Expression for potential energy of thermal expansion is given by U ( x) (cx 2 gx 3 fx 4 ) . This is a non-
harmonic approximation in the lattice vibration in solids. Here x is the displacement of their equilibirium posi-
tion.
Hence, correct option is (d).
24. Since the potential under which the particle is moving i.e. truncated harmonic oscillator potential , is not sym-
metric about x = 0, therefore the wave function for the different states of the particle should not have a definite
parity.
Energy eigenvalues of a particle moving under a truncated harmonic oscillator potential are
3
E n 2 n
2
3 7
Therefore, ground state energy and first excited state energy will be and respectively.
y.
2 2
Correct option is (d)
25. For symmetric tensor, Aij A ji
Therefore, if 3 of the off-diagonal elements are known then the other 3 off-diagonal elements are also known.
Hence, number of independent components = 3 diagonal + 3 off-diagonal = 6
Correct option is (c)
Q.26 – Q.55 : Carry TWO marks each.
1 0
26. Energy eigenvalues corresponding to unperturbed Hamiltonian H 0 is E0 1,1 .
0 1
1 0
Since the unperturbed Hamiltonian H 0 and perturbing Hamiltonian H ' commutes with
0 1
each other, therefore the eigenvalues of the perturbing Hamiltonian H ' will give the first order cor--
rection to energy.
Eigenvalue equation of H ' :
H ' I 0 0 2 , 0
Energy eigenvalues of the perturbed system using the first order perturbation approximation, will be 1 2 and
1 respectively.
Correct option is (a)
27. The magnetic susceptibilty of a ferromagnetic material is given by
C
...(i)
T C
C C 1
At T = 0 K; T 0 K 0 C C
1
2 104 2 10 4
T 0 K
GATE-PH 2012 SOLUTION 386
TC 600
And also TC C C 4
300 104 3 102
2 10
C 3 102
Hence, correct option is (c).
28. E 10 ˆj exp i 6 x 8 z E x, z , t 10 ˆj exp i 6 x 8 z 10ct ck 10c
k 6 xˆ 8 zˆ k 10
1 1
c c
B kˆ E x, z , t 6kˆ 8iˆ exp i 6 x 8 z 10ct
Correct option is (a)
29. Trace of the matrix = 1 2 3 0
Only option (b) satisfies this.
Correct option is (b)
30. P. Infrared region 3. Vibrational transitions of molecules
Q. Ultraviolet visible region 1. Electronic transitions involving valence electrons
R. X-ray region 4. Transitions involving inner shell electrons
S. - rays region 2. Nuclear transitions
Hence, correct option is (c).
R
31.
R (a)
V1
R1 (b) V0
V2
R2
By applying voltage division rule
V2 R2
Vb= Va (Since, Va = Vb due to virtual short)
R1 R2
(Since, Va = Vb due to virtual short)
V2 V V V2 R2
Given V0 V1 ; a b
R1 R2
2
Applying KCL at node - (a)
Va V1 Va V0
0
R R
2Va V1 V0 ; Put Va value
2V2 R2
V1 V0
R1 R2
Compare this equation with given V0 value
1 2 R2 1 R2 R1 R2
4
2 R1 R2 4 R1 R2 R2
R1 R1
4 1 3
R2 R2
Correct option is (d)
GATE-PH 2012 SOLUTION 387
32. The given electronic configuration 2s1 3d1
1
For 2s1 electron : 0, s
2
1
For 3d1 electron : 2, s
2
Thus, the total angular momentum (J1 ) for 2s1 electron
1 1 1
J1 s to s 0 to 0 .
2 2 2
The total angular momentum (J 2 ) for 3d1 electron
1 1 5 3
J 2 s to s 2 to 2 , .
2 2 2 2
Thus, total angular momentum J in j–j , coupling is given by
1 3 1 3 1 3
For J1 and J 2 : J J1 J 2 to J1 J 2 to 2, 1.
2 2 2 2 2 2
1 5 1 5 1 5
For J1 and J 2 : J to 2, 3.
2 2 2 2 2 2
1 3 1 5
Thus, we can write it as , and ,
2 2 2,1 2 2 3, 2
Hence, correct option is (a).
33. Given V si 0.7 volt
12k
(2)
0.7V
V0.7 24V i1 6k i2
(equivalent figure) (1) 3.3k
Solve this circuit i2 1 mA
Applying KVL rule,
Loop-(1) : 24 12 i1 6 i1 i2 ... (i)
Loop-(2) : 0.7 3.3 i2 6 i2 i1 ... (ii)
Solving (i) and (ii), we get i2 1 mA
Correct option is (b)
34. Charged particles interact through electromagnetic interactions. In weak all particles interact and neutrino
interact through weak.
Correct option is (a)
GATE-PH 2012 SOLUTION 388
35. s rod 2c
c/2
3
observer
velocity of rod with respect to observer
2c c
vr v0 3 2 c
vro
v v 2c c 4
1 r 2 0
c 1 3 2 2
c
length of rod in observer’s frame is
2 2
v c /4 0 15
0 1 r0 0 1 0.97 0
c c 4
Correct option is (d)
36. Relation between lattice parameters : a, b and c ; and interplanar spacings of (hkl) plane is given by
1 ... (i)
d hkl
2 2 2
h k l
2 2 2
a b c
For simple cubic crystal, a b c ac , d (101) ?
1 1 a
d hkl (d101 )c c ... (ii)
2 2 2
1 0 1 2 2
2
2 2 2
ac ac ac ac
For tetragonal crystal, at bt 2 ac , ct 2 ac
1
d hkl (d101 )t
12 02 12
2
2
2
2 ac 2 ac 2 ac
1 1 2a ... (iii)
c
1 1 3 3
2 ac 4 ac2
2 2
4 ac
ac
(d101 )c 3 ac 3
2 .
(d101 )t 2 a c 2 2 ac 8
3
Hence, correct option is (c).
GATE-PH 2012 SOLUTION 389
37. dc 100
15V
100k 900
+
+ VCE
VBE – –
100
• For drawing DC load line High
Applying KVL at emitter collector loop.
15 900 IC VCE 100 IC ; Load line equation
Put VCE 0 I C 15 mA
Put IC = 0 VCE 15 V
IC
15mA
Q-point(VCE ,IC)
(2V, 13mA)
VCE
15V
Now KVL at Base-Emitter loop
15 100 I B 0.7 0.1 I E
I E I B ; I E 1 I B ; I E 101 I B
I B 0.13 mA
I C I B 100 0.13
I C 13 mA
Now KVL at o/p (output)
15 0.9 I C VCE 0.1 I E
Put value of I C and I E
VCE 2 V
Correct option is (a)
38. The partition function is
Z g (i ) e i 1 e 0 2 e 1e 2
i
1 2 e e 2 (1 e ) 2
Correct option is (c).
GATE-PH 2012 SOLUTION 390
39. Since surface is charge free
D2 nˆ D1 nˆ
5 0 E2 n 2 0 5
E2 n 2kˆ
And
E2 E1 ; E2 2iˆ 3 ˆj
E2 2iˆ 3 ˆj 2kˆ
D2 5 E 10iˆ 15 ˆj 10kˆ
0 2 0
Correct option is (b)
40. Radial probability density for the ground state of Hydrogen atom is
2 1 2 2 r / a0
P r 100 r 2 r e
4 a03
For small ‘r’, the term ‘ r 2 ’ will dominate and for large ‘r’, the term ‘ e 2 r / a 0 ’ will dominate.
Correct option is (d)
41. For 1 p1/2 , O15 0n1 O16
Therefore, binding energy of n in 1 p1/2
B.E . O16 B.E. O15 as n has zero B.E.
= 127.62 – 111.96 = 15.66 MeV
For 1d5/2 , O16 0n1 O17
Therefore, binding energy of n in 1 p1/2
B.E . O17 B.E. O16 131.76 127.62 4.14 MeV
Therefore, energy difference = 15.66–4.14 = 11.52 MeV
Correct option is (b)
1
42. L , ma 2 2 sin 2 2 mga cos
2
L p
p ma 2 2
ma
L p
p ma 2 sin 2
ma sin 2
2
Hamiltonian is defined as
H pi qi L
GATE-PH 2012 SOLUTION 391
1
H p p ma 2 2 sin 2 2 mga cos
2
1 2 p2
put and to get H p
mga cos
2ma 2 sin 2
Correct option is (b)
43. F r B
F dr F dS r B dS
C S
B r r B r B B r dS
S
B 0 0 3B dS 2 B dS 2 B0 2 B0
Correct option is (c)
1 1 1 y
44. f ( z ) e1/ z 1 2
......
z 2! z C : |z| = 1
f ( z ) has an essential singular point at z = 0.
x
z=0
1/ z
e dz 2 i Residue of f ( z ) at z 0
C
2 i co-efficients of 1/z 2 i
Correct option is (d)
45. The number of microstates when N particles are distributed into N1 and N 2 particles are :
N!
, where N1 N 2 N
N1 ! N 2 !
N N!
Given : N1 N 2 2
, N! N!
2 2
The entropy is S kB n k B n( N )! 2n
N2 !
Using Stirling approximation, n N ! N n N N (for large N )
N N N
S k B ( N n N N ) 2 n
2 2 2
N
k B N n N N n N k B n 2
2
Correct option is (d).
46. The vibrational terms (energies in wave number unit m–1 or cm–1) of diatomic molecules are given by
2
1 1
G v e v e xe v
2 2
where, v is called vibrational quantum number
GATE-PH 2012 SOLUTION 392
e is the wave-number spacing of energy levels that would occur if the potential were a parabola.
e xe is called an harmonicity constant that arise if potential curve is not a perfect parabola.
1 x
Thus, G 0 e e e ... (1)
2 4
3 9
G 1 e e xe ... (2)
2 4
5 25
G 2 e e xe ... (3)
2 4
Fundamental band corresponds to the transition, v 0 v 1 , therefore, wave number
v0 G 1 G 0 e 2e xe
First spectral lines (or first overtone band) corresonds to
v 0 v 2 , thus v1 G 2 G 0 2e 6e xe
Ratio of frequencies,
v1 2 e 3e xe
1.96 Given
v0 e 2e xe
xe 0.01923 xe 0.02
Correct option is (b)
8 k BT
47. The average speed, vav. ... (i)
m
2k BT
The most probable speed, v p ... (ii)
m
3k BT
and the r.m.s. speed, vr.m.s. ... (iii)
m
vp 2
From equations (i) and (ii),
vav. 8/ 2
vp vav. 400 354.49 m/s 355 m /s
2 2
vr.m.s. 3 3 3
Now from equations (i) and (iii), vr.m.s. vav. 400 434.16 m/s 434 m /s
vav. 8/ 8 8
Correct option is (a).
2 2 ikx 2 2 k 2 ikx 2 k 2
48. Hˆ e 2e ikx
ik ik e ikx
ik ik 2e ikx
e 2e ikx
2m .
2m x 2 2m 2m
2 k 2
Energy of the particle is E .
2m
Correct option is (c)
GATE-PH 2012 SOLUTION 393
49. Real part of the wave function is Re cos kx 2 cos kx .
Probability current density for the real part of the wave function is
i *
* 0 Since for real wavefunction
*
J
2 m x x
Correct option is (d)
Ka
50. A sin ... (i)
2
At the boundary of first Brillouin zone, i.e. at K , equation (i) gives
a
a
A sin A sin A ... (ii)
a 2 2
d d Ka Aa Ka
It can also be solved as : Group velocity vg A sin cos ... (iii)
dK dK 2 2 2
At the boundary of first Brillouin zone i.e. for K , equation (ii) gives
a
Aa
vg cos vg 0.
2 2
Standing waves produced at Brillouin zone boundary.
Hence, correct option is (a).
51. The dispersion relation ( versus K ) for one-dimensional monatomic lattice is given by
4C Ka
sin ... (iv)
M 2
Comparing equation (iv) and equation (i), we get (where, C is force constant)
4C 4C MA2
A A2 C .
M M 4
Correct option is (a).
1 p
52. Since, E 3
4 0 a
p qd
a3
4 0 E 4 0 E
4 0 a 3 E 4 3.14 8.854 (0.5 1010 )3 30 105 1012
d 19
260.63 1018 m
q 1.6 10
d 2.60 1016 m.
Correct option is (c).
GATE-PH 2012 SOLUTION 394
53. Since, dipole moment ( p ) q d E
qd 1.6 1019 2.6 1016
0.139 10 40
E 30 105
1.4 10 41 .
Correct option is (b).
54. Given y ax 2 y 2axx
L=T–V
1
m x 2 y 2 mgy
2
1
m x 2 4a 2 x 2 x 2 mgax 2
2
1
L m 1 4a 2 x 2 x 2 mgax 2
2
Correct answer is (b)
55. Equation of motion is
d L L
dt x x 0
d
m 1 4a 2 x 2 x 4ma 2 x 2 x 2mgax 0
dt
m 1 4a x
x 8ma 2 xx 2 4ma 2 x 2 x 2mgax 0
2 2
m 1 4a x
x 4ma 2 x x 2 2mgax 0
2 2
or 1 4a x x 4a x x
2 2 2 2
2 gax 0
Correct answer is (b)
GATE-PH 2013 SOLUTION 395
OBJECTIVE QUESTION
Q.1 – Q.25 : Carry ONE mark each.
1. For a symmetric periodic function i.e., f ( x ) f ( x ) , fourier co-efficient bn 0
The Fourier series contains constant term and cosine terms.
f ( x ) a0 an cos (n k x )
n 1
Correct option is (b)
2. Stress = Force/Area, there are two directions involved with it, i.e., one is direction of force and the other one
is direction of area.
Strain = change in length/original length, there are two directions involved with it, i.e., one is the direction of
length and the direction along which length is changing.
Moment of inertia Angular momentum i.e., two directions involved one of angular momentum and other
Angular velocity
of angular velocity
Pressure = Force/Area ; But here direction of force and direction of area are same. Therefore, it is a first order
tensor.
Correct option is (d)
3. 0.85c c
electron photon
velocity of electron w.r.t. photon is given as
ve v p 0.85c c c 0.85 1
vep c
ve v p 1 0.85 1 0.85
1 2
c
vep c
Correct option is (b)
4. Energy in case of blackbody radiation is given by
E / k BT
e 1
0
If 0 , we have form.
0
So using L. Hospital’s rule, we have
k BT
E lim / k B T
lim k BT
0 e 1 0 1
e / k BT .
k BT
Correct option is (d)
5. The free energy is continuous but its derivative is discontinuous in first order phase transition.
Correct option is (c)
GATE-PH 2013 SOLUTION 396
6. Once the gases are allowed to freely exchange of energy, both sides will attain thermal equilibriumby maintain-
ing the same temperature. Besides this, due to unbalanced pressure, both sides will attain mechanical equilib-
rium by balancing the pressure on both sides of the movable wall.
Correct option is (a)
7. Given: S E aE E0 E .
The entropy is defined as
dE
dS
T
1 dS
aE0 2aE
T dE
1
T
a E0 2 E
For E0 2 E, T is negative.
Correct option is (a)
8. The number of microstates for N independent spin s particles are
N
2 s 1 2 N s 1/2
The entropy is
S k ln Nk ln 2
Correct option is (d)
9. p e 1
n ve
B :1 1 0 0 B 0
L:0 0 1 1 L 0
I 3 : ½ ½ 0 0 I 1 0
S :0 0 0 0 S 0
Correct option is (c)
10. Baryon number of up quark is 1/3.
Isospin number of up quark is 1/2.
Correct option is (d)
11. Consider the scattering of neutrons by protons at very low energy due to a nuclear potential of range r0.
Given that cot kr0 here, phase shift
k
k k
tan kr0 kr0
k
kr0
Correct option is (a)
GATE-PH 2013 SOLUTION 397
12. According to Fermi-Selection rules I 0
Gamow Teller rules I 1, 0 except 0 0
Therefore, the transition 2 3 , I 1 , parity does not change. The given transition is not allowed
by Fermi but allowed by Gamow Teller selection rule.
Correct option is (c)
13. At a surface current (using boundary condition), normal component of magnetic field is continuous and
tangential component of magnetic field is discontinuous.
For vector potential : Both the normal component and tangential component are continuous
Correct option is (d)
14. Equation for plane waves are given as
1 r , t A1 ei k1r t ; 2 r , t A2 ei k2r t
Phase difference = k1 k2 path difference
Condition for interference maxima k1 k2 r 2m
Correct option is (b)
15. Consider a scalar function which is satisfying the Laplace equation, 2 0
0
Also, 0
Hence, has both zero divergence and zero curl.
Correct option is (c)
16. Circularly polarized light can be decomposed into two perpendicular components. One in the plane of
incidence and the other perpendicular to the plane. At Brewster angle the reflected light contains only
perpendicular component and refracted light has larger component in the plane of incidence than its com-
ponent perpendicular to the plane. As a result the reflected light will be plane polarized light where as
refracted light will be elliptically polarized light.
Correct option is (c)
17. We know that, L2 , Lz 0, Lx , Ly iLz
Lz , L = Lz , Lx iLy Lz , Lx i Lz , Ly iLy i iLx
i Ly iLx L
Lz , L Lz , Lx iLy iL incorrect
Correct option is (d)
18. L q 2 q 2
L
canonical momentum is given as pq 2q
q
Since q is not cyclic, pq is not conserved
Correct answer is (d)
GATE-PH 2013 SOLUTION 398
19. Given 6-bit A/D converter analog to digital convertor.
Maximum convertion time = 32 µ sec
Note: Not mention which type of A to D convertor is given.
Flash ADC > Successive approximation ADC > Counter type ADC > Dual slope or integrating type
CT 1 CT nTC CT 2n 1TC CT 2n TC
Assuming counter type ADC: CT 2n 1 TC TC ?
6 6 1
32 10 2 TC
TC 1 106 fC 1 MHz
Assuming Dual slope or integrating type ADC : CT 2n TC
32 106 26 TC TC 0.5 106 fC 2 MHz
Note: Option (a) or (b) is correct depending on the type of ADC.
20. With the increase in temperature, the concentration of minority carriers starts increasing. Eventually, a
temperature is reached called the critical temperature (85 ºC in case of germanium and 200 ºC in case of
silicon) when the number of covalent bonds that are broken is very large and the number of holes is
approximately equal to the number of electrons. The extrinsic semiconductor now behaves essentially like
an intrinsic semiconductor. Therefore, at 200 ºC, silicon will have the Fermi-level in the middle of energy
gap.
Hence, correct option is (c).
21. Considering the BCS theory of superconductors the statement “(c) Quantization of magnetic flux in supercon-
ducting ring in the unit of (h/e)” is not correct. It correctly predicts the Meissner effect, isotope effect and band
h
gap in the superconductor. It is quantised in units of
2e
Hence, correct option is (c).
22. Group-I Group-II
P. Phonon iv. elastic wave
Q. Plasmon iii. collective electron oscillations
R. Polaron ii. electron + elastic deformation
S. Polariton i. photon + lattice vibration
Hence, correct option is (b).
23. The number of distinct ways of placing four indistinguishable balls into five distinguishable boxes is
Ni g 1
i C Ni , N i 4, g i 5 degeneracy
4 51 8! 8 7 6 5 4!
C4 8 C4 70
4! 4! 24 4!
Correct answer is (70)
24. Ripple Rejection of voltage regulator = –50 dB
• input ripple is 1 mV
• output ripple voltage in µV
output ripple voltage
Ripple Rejection
input ripple voltage
GATE-PH 2013 SOLUTION 399
50 dB 20 log10 Ripple Rejection
Ripple rejection 3.162 103
O/P ripple voltage
3.162 103
1 103
O/P ripple voltage 3.162 106 Volt 3.16 µV
Correct answer is (3.16)
2
D5/2
25. 2
D
2
D3/2
2
P3/2
2
P
2
P1/2
Selection rules for sodium (alkali spectra) are L 1, S 0, J 0, 1 (but J 0 J 0 ). So, only
three spectral lines are allowed.
Correct answer is (3)
Q.26 – Q.55 : Carry TWO marks each.
26. L [ (t )] (t ) e st dt 1
0
e st
st 1
L [1] 1 e dt
0 s 0 s
k
L [sin kt ] sin kt e st dt
0
s k2
2
1
L t t e st dt
0
s2
1 1
L [te kt ] L [t ]s s k
s2 ss k (s k )2
Correct option is (c)
27.
A B r r A B A B r
If A and B are constant vectors, then A B is also constant vector..
For constant vector a , a r a
Therefore, A B r A B
Correct option is (b)
GATE-PH 2013 SOLUTION 400
1
28. For n 0,
2
3 1 1 1
For n 1,
2 2 2 2
5 3 3 3 1
For n 2,
2 2 2 2 2
All these satisfies the form given in option (c)
Correct option is (c)
29. For one dimensional motion, Newton’s equation of motion is written as
dv
F 3 m0
dt
m0 dv
F
2 3/ 2
1 v 2
/c dt
m dv m0
F 2 2
, m
1 v /c dt 1 v 2 /c 2
Correct option is (c)
eia p e
i ax px a y p y az pz
30.
where, a ax2 a y2 az2 & p px2 p 2y pz2
ia y p y
eia p eiax px e eiaz pz
3/2 mv2x mv2y mvz2
m mia x vx
2 kT mia y v y
2 kT ima z v z
2 kT
e e dvx e e dv y e e dvz
2 kT
Here, we have used Maxwell-Boltzmann distribution function f vx , v y , vz and it is equal to
3/2
m
f vx , v y , vz e
m 2 2 2
2 kTvx v y vz
2 kT
mvx2 m 2 2 2 m m
mia x vx ax k T vx iax kT 2 2 kT 2 a2x kT
2 kT 2 kT 2 kT
Now, e dv x e e dvx e
m
3/2 3/ 2
e
m
ia p 2 kT
e
m 2 2 2
2
a x a y a z kT m
a 2 kT
e 2
2 kT m
Correct option is (c)
q2
31. The electromagnetic form factor F q 2 exp 2
2Q
And we know that, F q eiqr r dr
GATE-PH 2013 SOLUTION 401
q2 2
F q 1 r .............. q r 1
6
2 F q2 1
r 6 2 6 exp 2 2
q q 0 2Q 2Q q 0
3 3
r2
Q2 Q
Correct option is (c)
32. Angular momentum of disc is
2 2 2
L I xx x I yy y I zz z z
MR 2
Here, I xx I yy 60º
4
MR 2 x
I zz
2
3
x sin 60º
2
y 0, z cos 60º /2
2 2
MR 2 3 MR 2
L 0
4 2
2 2
MR 2 3 7 MR 2
1
4 4 8
Correct option is (c)
33.
x1 x2
k k
M M
Let x1 and x2 be displacement of masses from their respective equilibrium positions. Therefore,
Kinetic energy T 1 M x12 1 M x 22
2 2
1 2 1 2
Potential energy V k x1 0 k x2 x1
2 2
1
k 2 x12 x22 2 x1 x2
2
Corresponding matrices are
1 0 ˆ 2 1
Tˆ M ,V k
0 1 1 1
For frequency () of normal modes
det 2 Tˆ Vˆ 0
GATE-PH 2013 SOLUTION 402
2k M 2 k
0
k k M 2
or 2k M k M k
2 2 2
0
2k 2 3Mk 2 M 2 4 k 2 0
M 2 4 3Mk 2 k 2 0
3Mk 9M 2 k 2 4 M 2 k 2
2
2M 2
k k
2
2M
3 5
2M
3 5
Correct option is (c)
34. We have one dimension charge density
Q x x0 x x0
Therefore, the total charge,
Q dx Q x x0 dx Q x x0 dx
Q at x x0 Q at x x0
Therefore, the field at 2 x0 , 0, 0 point is
1 Q Q
E 2 2
4 0 x0 3x0 2x0 2x0
Q 9 1 2Q -Q Q
2 2
xˆ
4 0 9 x
0 9 x
0 0
Correct option is (a)
35. We know that ki , kr , nˆ all are in the same plane.
The vector kˆi nˆ has its direction perpendicular
to the plane containing the vectors kˆi , nˆ, kˆr
kˆ nˆ kˆ
i r
0 because kˆi nˆ kˆr k^i
n^
k^r
Correct option is (c)
36. 5 1D2 5 1P1
Given : wavelength, 643.8 nm
Wavelength is related to wave-number by
1
v v 2 ... (1)
GATE-PH 2013 SOLUTION 403
2 v
In normal Zeeman effect experiment. Wave number separation between consecutive Zeeman level are equal
and is given by
eB
v
4 me c
where, B is the applied magnetic field.
Putting, e 1.6 1019 C, me 9.1 1031 kg, c 3 108 m/s , we get
v 46.7 Bm 1 where, B is in Tesla ... (2)
As the spectrometer has resolution to observe a minimum spread in wavelength min 0.01 nm, therefore
from equation (1) and equation (2).
2 2
0.01 nm 643.8 nm v 643.8 nm 46.7 B m1
Thus, minimum magnetic field,
0.01 nm 0.01 109
Bmin 2
2
5.166 1010 109 0.52 T
643.8 nm 46.7 643.8 46.7
Correct option is (b)
37. Energy level of harmonic oscillator
1
Ev hvosci v where, v = vibrational quantum number
2
vosci is classical frequency given by
1 k
vosc ; k = force constant, µ = reduced mass
2 µ
Selection rule, v 1, gives
Ev hvosci equispaced energy level
Given : Ev 8.44 102 eV 8.44 102 1.6 1019 J 13.504 1021 J
2
1 k 4 2 µ Ev
Also, Ev h k
2 µ h2
2
k
2
4 3.14 1.14 10 26 kg 13.504 10 21 186.74 N /m
34 2
6.626 10
Correct option is (c)
38.
Primitive vectors, a 2 ˆj kˆ Å, b 2 kˆ iˆ Å, c 2 iˆ ˆj Å ... (i)
0 2 2
Volume of unit cell, V a b c 2 0 2 0 0 4 2 0 4 2 4 0 8 8 16 Å 3
2 2 0
Now, translation vectors of reciprocal lattice
2 b c
a*
2 2 kˆ iˆ 2 iˆ ˆj Å 2 8
kˆ iˆ kˆ ˆj iˆ iˆ iˆ ˆj Å 1
a b c 16 Å 3
16
GATE-PH 2013 SOLUTION 404
ˆ ˆ ˆ 1 ˆ ˆ ˆ 1
j i k Å i j k Å
2 2
Similarly, b * iˆ ˆj kˆ Å 1 ... (ii)
2
and c * iˆ ˆj kˆ Å 1 ... (iii)
2
Therefore, volume of the reciprocal lattice,
V * a * b * c *
2
iˆ ˆj kˆ 2 iˆ ˆj kˆ 2 iˆ ˆj kˆ
3 1 1 1
3 3 3
1 1 1 1(1 1) 1(1 1) 1(1 1) 4
2 8 8 2
1 1 1
which is the volume of primitive unit cell of bcc with cube edge 3 /2 .
Correct option is (a).
e2 B
39. Total energy, E 9 ... (i)
4 0 r r
dE
For equilibrium, 0, (r r0 (equilibrium separation))
dr r r0
e2 1 9B e2 9B
2 10 0 2
10
4 0 r0 r0 4 0 r0 r0
(9 B ) (4 0 )
8
e2 r08
r 0
B .
e2 36 0
Correct option is (a).
40. Sides of box = 10 12 m
Charge of proton = 1.6 10 19 coulomb
Mass of proton (m p ) 1.67 1027 kg
Planck’s constant h 6.63 1034 J-s
2 2 2 h2 3
Minimum K.E. ( EK ) 2 x n n 2
y nz
2
8m L2
(nx ny nz 1 for minimum energy)
2m L
(6.63 1034 )2 3
27 12 2
3 3.29 1017 9.9 1017 J.
8 1.67 10 (10 )
Correct option is (c).
16 z
41. f ( z) has a pole of order 2 at z = 1
( z 3) ( z 1) 2
GATE-PH 2013 SOLUTION 405
Residue of f ( z ) at z = 1
1 d 16 z ( z 3) 16 16 z 1 48
( z 1)2 2
2
3
(2 1)! dz ( z 3) ( z 1) z 1 ( z 3) z 1
( z 3)2 z 1
Correct answer is (3)
42. Trace of the matrix = 1 2 3 12
Determinant of the matrix = 1 2 3 50
The combination 1 5, 2 5, 3 2 satisfies the above two relations.
Correct answer is (5)
43. According to energy conservation,
Ev Eµ m c 2 ... (i)
According to momentum conservation,
0 Pv Pµ Pµ Pv
2
Ev Pv2c 2 mv c 4 Pv c mv 0 ... (ii)
Using (i) and (ii),
m c 2 Pµ2 c 2 mµ2 c 4 Pv c
2 m 2
mµ2 c 2
m c
2
Pv c P c m c
µ
2 2 2 4
µ
Pc
v
2m
... (iii)
Using (ii) and (iii), Ev
m 2
mµ2 c 2 139.6 2 105.7 2 2
c 29.7 MeV 30 MeV
2m 2 139.6
Correct answer is (30)
1 1 1
44. C A d
S A
dS S dS BA cos 0.6 2 cos 60º 0.6 2 2 0.15
B
Correct answer is (0.15)
1 0 0.8
45. 0.8 0.6 0.8 0.6
0 1 0.6
1 0 0 1 10 5
10 z 5 x 10 5
0 1 1 0 5 10
10 5 0.8 11
10 z 5 x 0.8 0.6 0.8 0.6 8.8 1.2 7.6
5 10 0.6 2
Correct answer is (7.6)
i ikr r
46. J
2m
* * , A e 0
r
GATE-PH 2013 SOLUTION 406
*
i ikr r0 * ikr r0 ikr r0 ikr r0
A e A e A e A e
2m r r r r r r
i 2 ikr r0 r r r
A e ik e ikr 0 e ikr 0 ikr eikr 0
2m r r r r
2 2 2 2
i 2 r0 r0 k 2 r k A
A ik ik A 2 0 0.25
2m r r 2m r m
Correct answer is (0.25)
47. I DSS 5 mA VP 5V VGS 2.5V
2 2 2
VGS 2.5 1 5
i
Drain current, d I DSS 1 5 1 5 1.25
VP 5 2 4
Correct answer is (1.25)
48. The fermions are distributed as
E 2 E 3E 4E
So the partition function is
E 2 E E 3 E E 4 E 2 E 3 E 2 E 4 E 3 E 4 E
Z e Ei e e e e e e
i
3 E
e e 4E 2e5E e 6E e 7 E
Correct option is (b)
49. The partition function of one distinguishable particle is
Z1 eE e 2E e3E e4E
Since distinguishable particles are non-interacting, the partition function of two such particles is
2 2
Z Z1 eE e 2E e3E e4E
Correct option is (c)
5 2 0
50. Unperturbed hamiltonian, H 0 2 5 0
0 0 2
GATE-PH 2013 SOLUTION 407
5 2 0
Eigenvalues of H 0 H 0 I 0 2 5 0 0
0 0 2
5 5 2 2 2 2 0
2
2 5 4 0 2 25 10 4 0
2
2
2 10 21 0 2,3, 7
Eigenvector for 2
3 2 0 x1 0
2 3 0 x 0
2
0 0 0 x3 0
2 3 x1 2 x2 0 x3 0 ... (1)
3 2 x1 3 x2 0 x3 0 ... (2)
2 - 1 5 x2 0, x2 0, x1 0, x3 arbitrary
So, eigenvector = (0, 0, 1)
Correct option is (c)
1 1 1
51. Perturbed Hamiltonian, H ' 1 1 1
1 1 1
Eigenvectors corresponding to eigenvalue, of the unperturbed Hamiltonain, E = 2, 3, 7 will be
0 1 1
1 1
1 0 , 2 1 , 3 2 1
1 2 0 0
First order correction to energy, will be
0
E '1 1 H ' 1 0 0 1 H ' 0 0
1
1
1 1
E '2 2 H ' 2 1 1 0 H ' 1 0
2 2
0
E '3 3 H ' 3 2
E1 2 , E2 3 0 , E3 7 2
Pair = 3, 7 2
Correct option is (c)
GATE-PH 2013 SOLUTION 408
52. J S J S
Taking self dot product both sides
J S J S
2 2 1
J S 2S J Given : J
2
2 2
J S 1 1 3 1 1 J 1
S J = J J 1 J J =
2 2 2 2 2 2 2
Correct option is (b)
e
53. ( g l g s S ) = gJµN
2Mc
2 4 2 1
For 8 O , N 9, s1/ 2 , p3/ 2 , p1/ 2 , d5/2
17 1 1 1 1
5
J ; magnetic moment = –1.91 µN
2
5 2 1.91 3.82
gµN 1.91 µN g 0.77
2 5 5
Correct option is (b)
1
Vi s
54. Vx s CS 10k
1 Vx(s)
R R
Vi
CS V0
Vi ( s ) C 1k
Vx s ... (i)
Vx(s)
1 RCS
1000 pF 2k
2
Vx s V0 ( s ) 2 1 ... (ii)
Using (i) and (ii)
V0 ( s ) 1.5 A0
Vi ( s ) 1 RCS 1 RCS
A0
S j ; T j
1 j 2 fRC
A0 gain of this circuit
• 20 dB/decade frequency means cut off frequency.
A0
T s T jf
1 jf /f H
From equation (a) and (b)
1
fH , Putting the values of R and C
2RC
1
fH
2 10 10 1000 10 12
3
GATE-PH 2013 SOLUTION 409
f H 15.91 KHz
This is cut off frequency for LPF.
Circuit Analysis:
Case-(1) at low frequency f 0 0 X C
Capacitor is open circuit
10k
Vi
V0
1k
At low frequency we are getting output.
2k
Case-(2) at high frequency, f X C 0
Capacitor is short circuit.
10k
Vi
V0
1k
at high frequency no output
2k
Correct option is (a)
1.5
55. T jf
1 jRC
1.5
T jf ( given f 1.2 103 Hz )
1 j 2fRC
Closed loop gain
1.5
T jf
2
1 2fRC
from circuit R 10 103
C 1000 10 12 F
f 1.2 103 Hz
Closed loop gain T jf 1.5
Correct option is (b)
GATE-PH 2014 SOLUTION 410
OBJECTIVE QUESTION
Q.1 – Q.25 : Carry ONE mark each.
1. Unit vector perpendicular to the surface
2 x iˆ 2 y ˆj 2 z kˆ iˆ ˆj kˆ
nˆ
2 x2 y2 z2 (1,1,1)
3
(1,1,1)
Correct option is (d)
2. Charge is conserved for isolated system so it is relativistically invariant.
Correct answer is (b)
1
3. D and 1P is singlet state, because multiplicity 2s +1 = 1 s = 0
Therefore, total angular momentum J = L + S is equal to the totat angular momentum L.
Each energy level splits into 2L+ 1 levels as shown in figure below.
m
+2
+1
1
D 0
–1
–2
+1
1
P 0
–1
Selection rule : m 0, 1
Energy of each level is equal E m µB B . Because of the uniform splitting of the levels. There are only three
different transition energies
eB eB
E0 , E0 , E0 , corresponding to m 1, 0, 1 respectively
2me 2me
The energy separation between two neighbouring levels in P and D states are equal.
Thus, these will be only 3 normal Zeeman splitting components of 1 P 1D transition.
Hence, correct option is (a).
proper
4. We know that Lab
1 v 2 /c 2
2
proper Lab 1 v 2 / c 2 5 108 1 0.9 2.18 108
Hence correct answer is (2.18)
5. When light is incident at Brewster angle, then the reflected light and the refracted light make 90º angle with
each other.
Correct option is (c)
K m m m 3m 3m
6. , 1 2
m1 m2 m 3m 4
GATE-PH 2014 SOLUTION 411
K 4K 4 0.3
2 rad/s
3m /4 3m 3 0.1
2 1
v 0.32 Hz
2 2
Hence correct answer is (0.32)
V
7. We have, E xˆ 10 cos 6 107 t 0.4 z
m
V
ˆ 0 cos t kz
Compared with electric field E xE
m
We get, 6 107 , k 0.4
6
Phase velocity, Vp 107 1.5 108 m / sec
k 0.4
Correct answer is (1.5)
1 1 1 i 1 1 1 i 1 3 0 1 0
8. AA† I
3 1 i 1 3 1 i 1 3 0 3 0 1
So, A is an unitary matrix
Correct option is (d)
h
9. p A 1 p A p photon
1
h
p B 2 pB p photon
2
p A 2 500 1
1: 3
pB 1 1500 3
Hence correct answer is (c)
10. Given : P V 5/3 P CV 5/3
where C is a constant of proportionality.
Bulk modulus is defined as
P 5 5
B V CV 8/3V CV 5/3
V T 3 3
B V 5/3
Correct option is (d)
GATE-PH 2014 SOLUTION 412
11. p p
0 0
Q : 1 1 0 0 Q 0
B :1 1 1 1 B 2 0
L:0 0 0 0 L 0
Therefore, it is not allowed reaction because Baryon number is not conserved.
p
0 n
Q : 1 1 0 0 Q 0
B:0 1 0 1 B 0
L:0 0 0 0 L 0
1 1
I 3 : 1 0 I 3 0
2 2
Therefore, it is an allowed reaction.
n
p e ve
Q:0 1 1 0 Q 0
Le : 0 0 1 1 Le 0
Therefore, it is not allowed reaction because electronic lepton number is not conserved
µ
e
Le : 0 1 0 Le 0
L : 1 0 0 L 0
It is not allowed reaction because electronic and muonic lepton numbers are not conserved.
Correct option is (b)
12. (ds)2 2(dx1 ) 2 (dx 2 )2 3(dx1 ) (dx 2 )
General form of metric tensor gij in 2-D will be (ds)2 g11 (dx1 )2 g22 (dx 2 )2 2 g12 (dx1 ) ( dx 2 )
(Here g12 g 21 )
3
2
So, metric tensor gij 2
3
1
2
Correct option is (b)
13. Two spin up fermion cannot occupy the same quantum state and the spin wave function is symmetric. So, the
space part wave function must be anti-symmetric, since the total wave function of a fermion is anti-symmetric.
1
1 x1 2 x2 1 x2 2 x1
2
Correct option is (d)
GATE-PH 2014 SOLUTION 413
14. The Coulomb gauge condition is
A= 0
x 2 y 3z 0
x y z
23 0
1
Correct answer is (1)
15. The Hamiltonian of a classical harmonic oscillator is given by
px2 1 2 2
p 2y 1
H m x m2 y 2
2m 2 2m 2
Since we have four quadratic terms in the Hamiltonian and according to the law of equipartition of energy each
1
would contribute k BT to the average energy..
2
So average energy per particle
1
4 k BT 2 k BT
2
Correct answer is (2)
16. Fermions have half integral spin. Proton, electron, neutron are fermions and have 1/2-spin.
particle 42 He nucleus 2 p 2n integer spin BOSON
7
4 Be nucleus 4 p 3n half-integer spin FERMION
1
H atom 1 H atom 1 p 1e integer spin BOSON
Deutron 12 H nucleus 1 p 1n integer spin BOSON
Correct option is (b)
17. Baryons have charges +2, +1, –1 but donot have –2 charge state.
Therefore, X does not represent a Baryon.
Correct option is (d)
18. According to Poisson’s equation
dA A
A, H
dt t
A
If A q, 0
t
dq
q, H q q, H
dt
A
If A p, 0
t
dp
p, H p p, H
dt
Hence correct answer is (a)
GATE-PH 2014 SOLUTION 414
x y z
19. Let the plane is 1 , where OA, OB and OC are intercepts on x, y and z-axis respectively..
OA OB OC
a a a
Now, for ( hkl ) plane, OA , OB , OC
h k l
xh yk zl
Thus above equation implies that 1
a a a
xh yk zl a
For (1, 0, 0), ah a h 1
For (0, 0, 1), al a l 1
1 1 1 a a a a a a 1
For , , , h k l a 1 k 1 a k
2 2 4 2 2 4 2 2 4 2
1
Thus, ( hkl ) 1 1 (2 1 2)
2
Hence, correct option is (a).
20. The specific heat C Cel. Clattice
Since, (Cel. )superconductor a exp , where a is a constant and is energy gap.
k
B T
i.e., the specific heat of a superconductor is discontinuous at transition temperature.
Therefore, the electronic specific heat of a metal in the superconducting state varies with temperature in an
exponential manner.
Hence, correct option is (a).
21. L S Lx S x Ly S y Lz S z
2
So, S z , L , S 2 will commute with L S but L S J 2 will not commute with L S
2
Correct option is (d)
22. The velocity v can be given by
k p
v ... (i)
m m
This shows that, for a free electron, v is proportional to k. However, in the band theory, Ek is generally not
proportional to k. The variation of E with k based on the band theory is shown in figure (below).
1 dE ... (ii)
Here, v
k dk Ek
k
O
a a
GATE-PH 2014 SOLUTION 415
This graph shows that the slope dE /dk of the E (k ) curve is not constant but changes with k. Using this curve
and employing equation (ii), one can obtain v versus k as shown in figure (below).
vk
a –k0
k
O k0
a
This curve indicates that the velocity of the electron is zero for k 0 and , where the slope dE /dk is zero.
a
i.e. at the top and bottom of the energy band (first Brillouin zone). For k k0 , where k0 corresponds to the
inflection point of E (k ) curve, the absolute value of the velocity attains a maximum value equal to free electron
velocity. Beyond this point, the velocity decreases with increase in energy.
Hence, correct option is (b).
23. Since the density of states, N(E) is independent of energy (E)
for a free electron gas in two dimensions. N (E )
Therefore,N(E) versus E graph can be represented as
mA O
Where N E constant E
2
Hence, correct option is (c).
24. V V0 V0
rat as
eg P
t0
or
int PAM
V0 t t
t0
O
t
t0 1
Vi t dt
RC Considering- Non integrator
Opamp as integrator
Most appropriate option is (a)
25. Given Modulus = 75
M 2n n is number of flip-flops
75 2 n for n 6 Hence (n = 6 not possible)
n=7
Correct answer is (7)
GATE-PH 2014 SOLUTION 416
Q.26 – Q.55 : Carry TWO marks each.
26. In spherical polar coordinate the Lagrangian of system is
1 z^
L m r 2 r 2 2 r 2 sin 2 2
2
mgr cos
Here 45º , 45º r
momentum conjugate to r is ^
x
L
pr mr
r
Hence correct answer is (a)
1
e 0
r /a
27. r
a03
2r
1 a0
Qˆ z 2 r 2 r 2 cos 2 r 2 3
a0
e 2 r 2 drd sin
0 0
2r
2
3 r 4e a0 dr sin 3 d
a0 0 0
2 4! 1
3 5
3sin sin 3 d
a0 2/a0 4 0
2 4 3 2 5 1 16
3
a0 2a02
a0 32 4 3
Correct option is (d)
28. For Nickel, number density, n 8 1023 atoms/cm3
And its electronic configuration 1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 8 4s 2
µB 9.21 1021 Am 2
n 8 1029 / m3
Magnetic moment of Ni = 0.6 µB (can be calculated using concept of atomic physics)
Saturation magnetization = N µ = 8×1029×0.6×9.2 ×10–21
44.16 108 4.4 109 Am 1
Correct answer is (4.4)
a a r2
29. V (r ) 03
r 3r
Time period of oscillation is given as
m d 2V
T 2 where k 2 at stable position
k dr
At equilibrium position r r
GATE-PH 2014 SOLUTION 417
dV a ar02
0 0
dr r r r 2 r 4
putting r r0
d 2V 2a 4ar02 2a 4ar02 2a
3 5 3 5 3 k
dr 2 r r
r r r0 r0 r0
m m m r03
T 2 2 2
k 2a / r03 2a
Hence correct answer is (a)
30. At T 300 K, k BT 25 meV
Let initial donor concentration N D and final donor concentration 100 N D
N
Initial Fermi energy ( EF )initial EC k BT n C
ND
NC
and final Fermi energy ( EF )final EC k BT n
100 N D
NC N
( EF )final ( EF )initial EC k BT n
EC k BT n C
100 N D ND
N C 100 N D
k B T n k BT n 100 2 (k BT ) n 10
ND NC
2 2.303 25 meV 115 meV
Hence, correct answer is 115 meV.
31. The energy of the two dimensional harmonic oscillator,
E nx n y 1
Given : E 4
nx n y 1 4
nx n y 3
nx ny
1 2 3
2 1 3 4-fold degeneracy, therefore g = 4
0 3 3
3 0 3
Correct answer is (4)
GATE-PH 2014 SOLUTION 418
8 3 4
32. 200 310 321
21 7 21
L2 , m 2 1 , m
8 9 4
L2
21
0. 2 2 2 6 2 22
21 21
Correct answer is (2)
k
33. V (r )
r
2 k 2
Effective potential Veff V r V
eff
2mr 2 r 2mr 2
d Veff k 2
for circular orbit 0 2 0, mkr0
dr r0 mr03
r r0
Hence correct answer is (d)
34. Rotational energy of the rigid diatomic molecules is given by
h2
E 2 J J 1
8 I
In terms of wave number unit,
E h
v 2 J J 1 B J J 1 cm 1
hc 8 Ic
where, J is rotational quantum number.
IA and IB is moment of inertia of rigid rotator A and B respectively.
h2 h2
Also, EA J J 1 and E J B J B 1
8 2 I B
A A B
8 2 I A
E A EB and I A 6 I B
J A J A 1 J B J B 1
J A J A 1 6 J B J B 1
6I B IB
This equation is satisfied for
J A 3 and J B 1
Correct option is (b) y
2 C : |z| = 4
z
35. f ( z) z z = i
e 1
4
Condition of singularity e z 1 0 z= 0
x
z = –i
e z 1 e ( 2 n 1) i
z i (2n 1) i , i 3 , i 5 .........
Only z i lies within the circle and it is a simple pole.
I 2 i [Residue of f ( z ) at z i + Residue of f ( z ) at z i ]
z2 z2
2 i z 2 i z 4 3i
e z i e z i
Correct option is (c)
GATE-PH 2014 SOLUTION 419
36. Given : Electric field in region 1 is
E1 7 eˆr 3 eˆ
According to boundary condition,
E1 E2
3 E2 E2 sin
4
E2 y
3
2 E2
E2 3 2
x
And D1n D2 n As boundary is charge free
1 7 2 E2 n 2 E2 cos
4
2 7 2 7
2.33
1 E2 3
d2y
37. y0
dt 2
mt
Assume, the trial solution to be y c e
m 2 1 0 m 1
So, y Aet B e t
Putting y (t 0) 1 , we get A B 1
Putting y(t ) 0 , we get A 0, B 1
Therefore, y (t ) e t cosh t sinh t
Correct option is (d)
38. For canonical transformation
Poisson Bracket Q , Pq , p 1
Q P Q P
1
q p p q
1 2 4a 1 1
2 4a 1
1
4a 1 a 0.25
4
Hence correct answer is (0.25)
39. Kinetic energy = 1 MeV
1 2
mv 1 106 1.6 1019 J
2
3.2 1013 3.2 1013
v 27
1.91677 1014 1.384474 10 7 m/sec
m 1.67 10
GATE-PH 2014 SOLUTION 420
mv 2
Now, for circular orbit, qvB
r
mv 1.67 1027 1.384474 107
B
qr 1.6 1019 101
1.67 1.384474
1.4450 Tesla
1.6
Correct answer is (1.4450)
40. Two bosons particles can be distributed as
0
So the partition function is
0
Z e0 e2 e 1 e e 2
The mean energy is
E
ln Z ln 1 e e 2
e 2e2 e 2e2
1 e e 2 1 e e 2
None of the options are correct.
2
I max a b 9 I 0 9 ab 3
41.
I min a b I0 1 ab 1
Taking positive value, we get
a b 3a 3b
a 2
2a 4b
b 1
2
I1 a 4 4 I 0
I 2 b 1 I 0
Correct option is (b)
1 1 2 0
42. 1
3 0 3 1
1 0
Let 2 C1 C 2
0 1
According to Normalization condition,
2 2
C1 C2 1
Given 1 and 2 are orthogonal i.e. 2 | 1 0
C1 2
C2 0 C1 2 C2
3 3
GATE-PH 2014 SOLUTION 421
2 1 1 0
2
3 0 3 1
2
So, the probability of finding the spin up in 2 is C12 0.66
3
34
43. Wavelength of neutrons can be given by h h 6.626 10 m 3.96 1010 m
p mv 1.675 10 27 103
3.96 Å
Here, n 1, 30, 3.96 Å, d ?
From Bragg’s law : 2d sin n
2 d sin 30 1 3.96 1010
1
2 d 3.96 1010 d 3.96 1010 m d 3.96 Å.
2
Hence, correct answer is (3.96).
44. For positive p arrow should point towards positive q and for negative p arrow should point towards negative
q therefore correct answer is (d).
Hence correct answer is (d)
1
45. We know that, I u c 0 E02 c
2
2I 2 150 103
E0 10.62 V/m
c 0 3 108 8.85 1012
Correct answer is (10.62)
46. [Note: According to the option given in the question, the ratio of population of the initial to the final state has to
be calculated].
The ratio of the molecule population of the initial states to final states is given by
Ei
Ei E f E
N Ji 2 J i 1 e k BT
2 J i 1 e k BT 2 Ji 1 e k T
B
N Jf
2J f 1
Ef
2J f 1
2J f 1
e kBT
Here, difference in the energy states
hc
E E J 2 E J 3
where, is the wavelength for the transition from state J = 2 to J = 3.
Given : 3122 Å
hc
N Ji 2 J 1 k T
i e B
N Jf 2J f 1
6.6261034 3108
N Ji 2
5
312210 10 1.381023 5000
5
e e 9.227 7.02 10 5
N Jf 3 7 7
Correct option is (c)
GATE-PH 2014 SOLUTION 422
47. The distribution of fermions is
1 2 3 4
• • •
• • •
• • •
• • •
So total number of microstates are 4, i.e., 4 .
The entropy is S k B ln 2 k B ln 2
Correct option is (b)
48. For one dimensional potential box, the wavefunction is given by
2 n x
n sin
a a
2 x
Ground state n 1 : 1 sin
a a
Perturbation, W x
First order correction to the ground state energy is
a
2 x
E1 1 W 1 x sin 2
1
dx
a0 a
a a
2 x 2 x
x 1 cos dx x x cos dx
a0 a a 0 a
2 a
x 2 xa 2 x a 2 x
sin cos
a 2 2 a 2 a 0
a2 2 a
a
0 1 1
a 2 2 2
Correct option is (c)
Correct option is (a)
49. According to the law of conservation of energy,
(K.E. + rest mass energy) of reactants = (K.E. + rest mass energy) of products
k m0 c 2 2 0 m0 c 2 2 ... (1)
From equation (1),
k m0 m0 c 2 140 105 35 MeV
Correct answer is (35)
1 3
50. 2 3
2 2
1 5 3 7 26
E Pi Ei P2 E2 P3 E3 3.25
4 2 4 2 8
Correct answer is (3.25)
GATE-PH 2014 SOLUTION 423
51. -decay
X Y
7
So, if angular momentum (I) and Parity (P) of ‘X’ is denoted by then,
2
For a first forbidden decay : I 0, 1, 2, 1 Parity change
7 7 7 5 7 3
Possible values of I for Y 0 or 1 or 2
2 2 2 2 2 2
7 5 3
Parity changes, so possible IP values for Y , ,
2 2 2
Correct option is (c)
52. dc 100 12V
IC ? 3k
VBE 0.7 V IC
150k
Apply DC analysis, it means capacitor is open circuit IB
KVL at input +
12 IC I B 3 150 I B 0.7 3 I E 0.7 – IE
3k
I C 100 I B
I E 101 I B
12 101 I B 3 150 I B 0.7 3 101 I B
I B 0.015 mA
I C I B 100 0.015 mA
I C 1.5 mA
Correct answer is (1.5)
53. Given : Resolution = 1 mV
To measure 1V
VPP
Resolution
2n
Number of distinct values to measure 1V using least count of 1 mV
1V
2n 1000
1 103 Volt
number of bits required to code 1000 different amplitudes
2n 1000 n = 10
Correct answer is (10)
54. 11 L L L L 11 2 11 L2x L2y 11 2 11 L2 L2z 11
2 L2 2 L2z 2 2 2 2 2 2 2
Correct answer is (2)
GATE-PH 2014 SOLUTION 424
55. LPF
R
+
Vi C +V
– 0
–
V0 ( s ) 1
T ( s)
Vi ( s ) 1 RCS
1
T ( j)
1 jRC
1 1 1
T j 0.71, magnitude at
RC 2 RC
1 1 1
T j tan RC put
RC RC
1 1
T j tan 1 45º
RC
1
Phase at
RC
Correct option is (b)
GATE-PH 2015 SOLUTION 425
OBJECTIVE QUESTION
Q.1 – Q.25 : Carry ONE mark each.
1. According to Virial theorem,
n
T V when,V r r n
2
The potential energy of the satellite V r r 1
1 1
T V T V
2 2
V 2T
Total average energy, E T V
E T 2T T
Correct option is (a)
a a
2. Lattice parameters a, b, c have relation a 2b 3c a a, b , c
2 3
and the miller indices of the plane (hkl) = (110) i.e., h 1, k 1, l 0
Thereore, interplanar separation
1 1 1 1 a
d 0.447 a
h2 k 2 l 2 12 12 02 1 4 0 5 5
a 2 b2 c 2 a 2 a 2 a 2 a 2 a2 a 2 a2
2 3
Hence, correct answer is 0.477 a.
3. f z u x, y iv x, y
The real part u x, y and imaginary part v x, y of a complex analytic function f z are harmonic functions
i.e. they will satisfy Laplace’s equation.
z2
Since, f z is analytic in the domain D, then f z can be Taylor expanded and the value of z f z dz will
1
be independent of the choice of the contour.
Correct option is (c)
4. We know that, Li , p j i ijk pk
Lx , p y ipz
Correct option is (d)
5. The k relation for monatomic lattice is given by
2C
(k ) 1 cos (ka) ... (i)
M
2C 2 ka ... (ii)
(k ) 1 1 2 sin
M 2
GATE-PH 2015 SOLUTION 426
2C 2 ka
(k ) 1 1 2 sin 2
M
4C ka ... (iii)
(k ) sin
M 2
ka
Here, we have taken modulus of sin because ( k ) cannot be negative.
2
2 ka ka
For long wavelength modes ( a ), i.e., k 0 sin
2 2
4C ka
equation (iii) (k ) ... (iv)
M 2
4C a
Phase velocity, (v p ) ... (v)
k M 2
and group velocity, (vg ) d 4C a ... (vi)
dk M 2
4C a
vp M 2 1
Ratio of phase velocity to the group velocity
vg 4C a
M 2
Hence, correct answer is (1).
6. Since photon has zero chemical potential, the number of the photons cannot be conserved. Thus, photons can
be created and annihilated.
Correct option is (a)
rˆ rˆ r sin ˆ
1
7. F1 2
0
r sin r
k .exp r 2 /R 2 0 0
iˆ ˆj kˆ
F2 3kx 2 kˆ
x y z
0 kx 3 kz 3
iˆ ˆj kˆ rˆ rˆ r sin ˆ
1 1 k k
F3 0 ; F4 2 2
rˆ 2 3 ˆ
x y z r sin r r sin r r sin
kx3 ky 3 0 k
0 0
r
Correct option is (d)
GATE-PH 2015 SOLUTION 427
3 3
2 1 2 1
8. t 3t 6 dt
30
t t 2 dt Using c t t0 t t0
c
0
1
4 1.67
3
9. Total kinetic energy (K.E.) of N-neutrons and Z-protons is given by
5/3 5/3
4 3 h2 4 3 h2
K .E. 2/3
(Z 5/3 N 5/3 ) 2/3
( Z 5/3 N 5/3 )
5m 8 V 5m 8 (CA)
[Volume of nucleus V Atomic weight of nucleus A. Hence, V = CA, where, C is a constant of
proportionality]
K.E. A–2/3
2
n 0.67
3
Correct answer is (–0.67)
10. Since the interparticle distance between the bosons is comparable to their de-Broglie wavelengths, their
wavefuntions overlap.
Correct option is (c)
11. If is the linear absorption coefficient of the metal sheet then the transmitted intensity is given by
I I 0 exp L
0.025 I 0 I 0 exp 2
exp 2 0.025
2 3.688 1.8444 mm 1 1844.4 m 1
Correct answer is (1844.4)
12. The hall coefficient of intrinsic semiconductor is given by
1 nh µh2 ne µe2
RH
q nh µh ne µe 2
1 IB
The hall voltage is given by VH
RH t
Given hall voltage is negative,
RH 0 nh µh2 ne µe2 0
µh µe in ideal intrinsic semiconductor, n e nh 1 1
*
* me* mh*
mh me
Correct option is (d)
13. The required number of dimensions
(2s 1) n , where s is the spin of the particle and n is the number of particle.
3
1
2 1 23 8.
2
Correct answer is (8)
GATE-PH 2015 SOLUTION 428
14. e
B: 000 B 0, conserved
L : 1 0 0 L 0, not conserved
Le : 0 1 0 Le 0, not conserved
Electronic and muonic lepton number should be conserved separately. Here both are not conserved.
Correct option is (d)
15. Suppose the line AB is the interface between two dielectrics.
We can use the boundary condition for E and D
Across interface, parallel component of E is continuous while perpendicular component suffers a discontinuity
of where is the permitivity of medium.
A
E1|| E2|| 2
1
Or 1E1 2 E2 D1|| D2|| +Q –Q
1E1 2 E2
D1 D2 if there is no charge Q Q 0 B
D1 D2 0 0
D is continuous in both cases parallel and perpendicular. While E is continuous only for parallel, not for per--
pendicular.
Correct option is (c)
16.
Magnetic flux through the loop i.e. B dS A dS , A d
S S PQRSP
A d A d
PQR RSP
1 A d A d 1 A d 1
RSP PSR PSR
Correct option is (b)
17. The number of image charges
360
n 1 where, is the angle between two conductor plates
360
1 11
30
Correct answer is (11)
GATE-PH 2015 SOLUTION 429
1 1
18. f z , putting z t
1 2
z z cos z
2
1 1
1 1 1
t 2 t cos t 2 t 2 t sin t
1 1 1 1 1 1
2 1 2t
1 3 3
2t 3
1 2t t t t .......... t
t 1
......
2 3! 3!
2
1 1 2t 2 2t 2
2
2 1 2t 4t ...... 1 ..... ..... ......
t t 3! 3!
1
Highest negative power term in the above Laurrent series expansion is
t2
1
Therefore, f z has a pole of order 2 at z
2
Correct option is (b)
19. The density of states in 2-D is given by
(p + dp)
2 Ap
g p dp dp
h2
p
where, p 2 2mE pdp mdE O
2A
g E dE mdE
h2
2A
g E m
h2
Since electron spin degeneracy is 2 so we should multiply density of states by a factor of 2.
4A
Therefore, g E m
h2
On comparing with g E CE n , we have n 0
Correct answer is (0)
20. According to Lorentz transform
t A vx A / c 2
t 'A 00 0
1 v 2 / c2
0.6 2
0
t B vyB / c 2 c 1.2 3 ct ' 3
t 'B B
1 v2 / c 2 0.64 0.8c 2c 2
Correct option is (a)
GATE-PH 2015 SOLUTION 430
21. Given; Source voltage 25V
Vs 1V ID
5k
To find Drain voltage VD n - FET
(D)
Applying KVL at input (G)
VS = 1V
VGS 1 0 ID
(S)
VGS 1 Volt 2M 0.5k VS = ID × 0.5
KVL at output 1 = ID × 0.5
ID = 2 mA
25 5 I D VD input KVL
VD 25 5 I D put I D 2 mA
VD 10 Volt
Correct answer is (10)
22. This question can be solved easily by plotting f x and g x
2
f x e x
f(x)
x=0 x
2
xe x , x0
g x x2
xe , x0
g(x)
x=0 x
From graph, it is clear, g ( x) is not differentiable at x 0 (due to presence of sharp dip at x = 0).
Correct option is (b)
1
23. Let N Number of particles with spin (or up)
2
1
and N Number of particles with spin (or down).
2
So, we have
N N N ... (i)
Also, the total energy is
E
E B N N N N ... (ii)
B
Solving (i) and (ii), we have
1 E 1 E
N N and N N
2 B 2 B
GATE-PH 2015 SOLUTION 431
Total number of microstates when N particles are distributed in two states N and N is
N! N!
N ! N ! 1 E 1 E
N ! N !
2 B 2 B
Correct option is (a)
24. (a) A B AB
A B A B (By apply De-Morgan’s law)
AB AB XOR-logic-gate
(b) AB BA XOR-logic-gate
(c) A B A B
A A AB AB A B
0
0
AB AB XOR logic-gate
Note: xx 0
(d) A B AB
AAB ABB AB AB AB
AND logic - gate
Note: A A A
Correct option is (d)
B y B 0 1 0 i B B 0 1 i
25. Â B x B
2 2 2 1 0 i 0 2 2 1 i 0
Eigenvalue equation:
A I 0
B
1 i
2 2 2B2
0 2 0 2 2 B 2 B
B 2
1 i
2
Correct option is (b)
Q.26 – Q.55 : Carry TWO marks each.
26. P-(iii): ESR (electron spin resonance) lies in GHz [8 – 10 GHz] frequency range. Thus, it lies in microwave
range.
Q-(i): NMR (nuclear magnetic resonance) lies in [100 – 1000] MHz frequency range. Thus, it is in
radiofrequency range.
R-(iv): Transition between vibrational states of a molecule lies in frequency range (1012 to 1014 Hz) . Thus, it
lies in far-infrared range.
S-(ii): Electronic transition lies in frequency range 1014 to 1015 Hz. Thus, it lies in visible range.
Hence, correct option is (d).
GATE-PH 2015 SOLUTION 432
aVE 3/2
27. Given: S Nk B ln 5/2
N
From first law of thermodynamics,
TdS dE PdV dN
At constant E and V,
S aVE 3/ 2 N 5/ 2 5 aVE 3/ 2
T T k ln
B 5/ 2
Nk B 3/ 2
7/ 2
N E ,V N aVE 2 N
aVE 3/ 2 5
k BT ln 5/ 2
N 2
Correct option is (a)
28. Q M Eu M n M sm M p
151.921749 1.008665 151.919756 1.007825 2.833 103 a.m.u.
Correct answer is (2.833)
29. According to energy conservation
3Mc 2 3Mc 2
Mc 2
10 1 v12 / c 2 10 1 v22 / c 2
Since, initial velocity was zero and two parts have same mass.
So, v1 v2 for momentum to be conserved
2 3 9
1 1 v12 /c 2
2
10 1 v / c
1
2 25
16 4
v1 c c 0.8c
25 5
v1 v2 0.8c zˆ
Correct option is (b)
30. E g 0.72 eV and mh* 6 me*
E EV k BT NC
T = 300K, EF C ln
2 2 NV
E EV k BT NC 0.72 eV 1.38 1023 300 3 me
EF EV C ln ln
2 2 NV 2 2 2 mn
3
0.36 1.38 10 23 150 ln 6 eV 0.395 eV
2
31. Total charge in the given figures (i) and (ii) are zero.
Therefore, value of dipole moment p is independent of origin.
GATE-PH 2015 SOLUTION 433
The charge distribution in (i) and (ii) are both symmetric about origin.
Dipole moment p due to symmetric charge distribution is zero.
Therefore, dipole moment is zero in both (i) and (ii).
Correct option is (a)
32. For 2 P3/ 2 2S1/ 2 :
1 1
In presence of an external weak field, the state 2S1/2 will split into two states m j , and 2 P3/2 state will
2 2
3 1 1 3
split into four state m j , , , .
2 2 2 2
mj
3/2
1/2
1
P3/2
–1/2
–3/2
1/2
1
S1/2
–1/2
Selection rule ; m j 0, 1, 0 0
Thus, there will be total six transitions permitted in 2 P3/ 2 2S1/ 2 .
Hence, correct answer is (6).
2 x
33. sin
L L
First order energy correction
L /2 L /2
I 2V0 x V 2 x V0 L V0
E0 sin 2 dx 0 1 cos dx
0
L L L 0
L L 2 2
2 n2 V0
The ground state energy is E
2mL2 2
Correct option is (d)
34. AOL 105 V1
Vo
open loop gain
AOL
ACL + 9k RL
1 AOL Vf 1k
closed loop gain –
V0 1 Vf 1
Vf feedback ratio
1 9 V0 10
GATE-PH 2015 SOLUTION 434
105 AOL
ACL 5
10 using
10 1 1 AOL
1
10
Non inverting amplifier is an example of voltage series topology
Voltage series OR series-shunt
Correct option is (c)
i kz t
35. Einitial xˆ iyˆ ei kz t xe
ˆ i kz t yˆ e 2
3
i kz t
E final xˆ iyˆ ei kz t xe
ˆ i kz t yˆ e 2
So, phase introduce by the optical system =
Correct option is (b)
dv dv
36. Equation of motion, m mg kv dt
dt k
g v
m
m k
ln g v t c
k m
At t 0, v 10 ms 1
m k
c ln g 10
k m
m k m k
ln g v t ln g 10
k m k m
k k
g v g v kt
m m m
k 10k ktm
ln t e m g v g
k g 10k 10k e
g m m
m m
m 10k ktm 0.01
0.05
0.2
v g g e 10 10 50 0.01 4.9
e
k m 0.05
m 2 L
37. L v C r v V r ; pc mv Cr
2 v
1
H pC v L mv Cr v mv 2 C r v V r
2
1
mv 2 C r v C r v V r
2
1 2
pC C r V r
2m
Correct option is (b)
GATE-PH 2015 SOLUTION 435
38. Consider a long solenoid which is embeded into a conducting medium and is insulated from the medium.
Suppose n is the number of turns per unit length of solenoid. So, total number of turns N = nL where L is the
length of solenoid.
µ0 NI
Magnetic field inside the solenoid B µ0 nI
2 r
where, r is the radius of solenoid.
B B
flux passing through the surface, B dS B 2 r
2
µ0 nI 2 r 2
d
µ0 n 2
d Ir 2
dt dt
dI N dI
2 µ0 nr 2 2 µ0 r 2
dt 2 r dt
Induced current will be proportional to the rate of change of flux.
d
So, Iinduced r Inside
dt
Outside the solenoid,
B dS B 2 r 2 P
Magnetic field at point due to this circular ring comes out to be
1
B well known
r3
1
So, outside 3
r2
r
d 1
I induced outside
dt r
Correct option is (d)
39. We have nucleus, 13 C6
Z = 6 (even)
N = 7(odd) 1s1/2 2 1 p3/4 2 1 p1/1 2
1 1
J , 1 for p , parity 1 1
2
1
Spin parity of ground state is
2
M 1 2 1 1
Isospin, I (M is the multiplicity 13 C , 12 C )
2 2 2
1 1
Therefore, Isospin I3 of 13 C is ( I 3 is assigned to proton)
2 2
Correct option is (a)
GATE-PH 2015 SOLUTION 436
40. The superconductor behaves like a normal conductor above transition temperature (TC ) and below TC, it
exhibits superconductivity.
In normal state the electrons will have 12 -integral spin and will follow Fermi-Dirac distribution function
Normal state
1
f (E )
e ( E E F ) k BT
1 f (E )
Therefore, f ( E ) versus E graph (T 0 K) will be E
In superconducting state, electrons form cooper pairs with spin = 0.
Hence these electrons in superconducting state will follow Bose-Einstein distribution function.
1 Superconducting
f (E ) state
e ( E EF ) k B T
1
Therefore, f ( E ) versus E graph (T 0 K) will be f (E )
Hence, correct option is (b).
E
41. Energy of a rigid rotator is
2
E J (J 1) ... (i)
2I
where J 0,1, 2,...
Energy of first excited state
2 2
(E1 ) (1) (1 1) 1 meV 1 meV ... (ii)
2I I
Energy of fourth excited state
2 20 2 2
E4 4(4 1) 10 10 1 meV 10 meV
2I 2 I I
Hence, correct answer is (10)
42. The cohesive energy of NaCl can be given by :
e2 1
U r r0
1 n ... (i) For NaCl, a 2 r0
4 0 r0
r0 r0
1
(1.6 1019 )2 1
7.95 1.6 1019 n
a Cl– Na+ Cl–
4 3.14 8.854 1012
2 a
1
(1.6 1019 ) 1
7.95 9 (Here n 9)
0.563
4 3.14 8.854 1012 10
9
2
9 7.95 4 3.14 8.854 0.563 10 21
174.98 102 1.7498 .
8 2 1.6 1019
Correct answer is ( –1.750).
GATE-PH 2015 SOLUTION 437
43. Since, the given Hamiltonian does not depends on time explicitly. So, total energy of the system will be con-
served.
Correct option is (b)
2/3
h2 3N
44. The Fermi energy ( EF ) is given by EF
2m 8 V
2/3
h2 3 N
EF n Let V n electron density
2m 8
2/3 3/2
3n 2m E F 3n 2m EF
2
2
8 h 8 h
3/2 3/2
31 19
8 2m EF 8 3.14 2 9.11 10 5.54 1.6 10
n
3 h 2 3 (6.626 1034 )2
3/2
3.6785 1050
8.373 8.373 7.055 10 27
1068
n 59.071515 10 27 per m 3
n 5.9 1028 per m3 .
Hence, correct answer is (5.9).
i
1 2
45. We have, 0 e 1
2
We know that,
xˆ
2m
aˆ aˆ †
x
i i
2m
0 e 2 1 a a
†
0 e 2 1
i i i
i i
2 2 2 2 2 e 2
0 e 1 1 e 2 e 0 e 0
2m
2m 2 m
Correct answer is (0)
46. Fc m 2 r is radially outward
FCO 2m v 2m 2 r 2 Fc
Correct option is (d)
GATE-PH 2015 SOLUTION 438
1 m2
47. y " y ' 2 y 0 ... (i)
z z
2d2y dy
z 2
z m 2 y 0 Cauchy Euler equation.
dz dz
... (ii)
dy dy d2y d 2 y dy
Putting z e x and replacing z by and z 2 2 by 2 in equation (ii)
dz dx dz dx dx
d 2 y dy dy
We get, 2
m2 y 0 ... (iii)
dx dx dx
Trial solution y C e kx , putting inequation (iii)
k 2
m2 0 k m
So, y C1e mx C2 e mx C1 z m C1 z m
Therefore, z m and z m are linearly independent solutions for m > 0.
For m 0, k 0 y C1 C2 x e0. x C1 C2 nz
Therefore, 1 and lnz are linearly independent solutions for m = 0
Correct option is (c)
48. The energy of a one-dimensional quantum oscillator with frequency is
1
En n , n 0, 1, 2,
2
The partition function is
1
n
n
Z eEn e 2
e 2
e
n n n
n
Now e 1 e e 2
n
Letting e x . Then,
e n
1 x x 2 1 1
n 1 x 1 e
2
e 1 1
Z
1 e
e 2
e 2 2sinh
2
e x e x
where we have used sinh x
2
The average energy U is
h
2 cosh
h 2 1 coth 1
U ln Z ln 2sinh
2
h 2 2 2
2 sinh
2
Correct option is (a)
GATE-PH 2015 SOLUTION 439
3
49. The number of degeneration including spin is 2s 1 2 1 4
2
22
Therefore, the minimum exitation energy of the system = 4 1 3 4 1 9 4 1 4 4
2mL2
22 5 2 2
25 20
2mL2 2mL2
E2 E3
E2
E1 E1
ground state first excited state
configuration configuration
Correct answer is (5)
50. • Given figure is a current mirror circuit. In this circuit output current is forced to be equal to input current.
• Output current is mirror image of input current
• It is basically voltage to current converter
30V
Iref
5k
2I0 I0
I0 IB IB
1 2
Q1 Q2
I0/ I0/
Given :
• 100
• VBE 0.7
• Here, Q1 and Q2 are identical transistor
So, 1 2 100, VBE1 VBE2 VBE 0.7 (Q1 and Q2 are in active region)
VCC VBE 30 0.7 29.3
I ref 5.86 mA
Rref 5K 5K
I ref IC1 I B1 I B2 (By applying KCL) ... (1)
1 2 . So, I B1 I B2 and I C1 I C2 I 0
We know that, I C I B
IC
So, I B
Put the value of IC1 , I B1 , I B2 in equation (1)
2IC 2I
I ref I 0 2 I B I 0 I0 0
GATE-PH 2015 SOLUTION 440
I ref
I0
1 2/
Put the value of I ref and
5.86
I0 5.74 mA
2
1
100
Correct answer is (5.74)
51. Shifting property:
If F f t f , then F f t a e i a f
2i 1 1 1 1
F H t ; F H t H t
2 2 2 2
1 i / 2 2i 1 i /2 2i i i / 2 i / 2 sin
e i 2
e e e 2i sin
2 2 2 / 2
Correct option is (a)
52. It is a differentiator circuit
I R
C IC
Vi
=0 V0
A
=0
As ideal op-amp has infinite input resistance. So, there will be no current passing through op-amp.
So, by applying KCL at node A.
dVi dVi
IC I ; I C C and VR RIC ; V0 VR ; V0 RC
dt dt
So, output signal can be obtained by differentiating the input signal.
Vi
1
t
0 1 2 3 4
V0 Given RC = 1
1
–1
Note: The circuit given is an inverting differentiator. (Just differentiate the given input signal)
Correct option is (b)
GATE-PH 2015 SOLUTION 441
1 2 1 2 1
53. H
2m
p1
2m
p2
2m
m2 r12 r22 k 1 2
1 2 1 1 2 1
p1 m2 r12 p2 m2 r22 k 1 2
2m 2 2m 2
4 2k
H1 H 2 k
s s H1 H 2 2 s 2 s12 s22
2 1 2
Energy eigenvalues are
3 3
E n1 n2 2k s s 1 s1 s1 1 s2 s2 1
2 2
For ground state, (n1 = 0, n2 = 0, s = 0)
3 3 3
EG .S . 2k 3 3k 0.3 eV
2 2 2
Correct answer is (–0.2)
54. The wave number of anti-stokes line is given by
anti Stokes Raman Raman Stokes
anti Stokes 2 Raman Stokes ... (1)
1 1 1
Given : vRaman
Raman 546 10 m 546 107 cm
9
1 1
vStokes
Stokes 552 107 cm
Substituting values in equation(1),
1 1
vanti Stokes 2
(546 10 ) (552 107 )
7
Stokes Raman anti Stokes
0.003663 107 0.0018116 107
36630 18116 per cm 18514 per cm
Correct answer is (18514)
55. The condition for nth order maxima is d sin n
For first order diffraction n = 1
d
sin
We have, k k 1 xˆ 3 zˆ
2 2
1/ 2 1
tan
3/2 3
1
tan 1 30º 600 109
;d 1.2 µm
3 sin 0.5
Correct answer is (1.2)
GATE-PH 2016 SOLUTION 442
OBJECTIVE QUESTION
Q.1 – Q.25 : Carry ONE mark each.
dy dy x2
1. We have, xy xdx ln y c
dx y 2
x2 y x
2
Given, x 0, y 2, c ln 2 ln y ln 2 ln
2 2 2
Putting x = 2, we get
y
ln 2
2
y 2e 2
Correct option is (d)
iˆ ˆj kˆ
2. We known, B A , A B0kˆ
x y z
B0 y B0 x
0
2 2
We have, B B0 kˆ , only option (c) satisfy the above condition
Correct option is (c)
3. For Raman active, there must be change in polarizability. Since 17O2 is homo diatomic molecule therefore, it will
have change in polarizability, thus it will be Raman active. Net dipole moment in 17O2 is zero, so IR inactive
17
O2 has nuclear spin I = 5/2, therefore, it is NMR active.
Correct option is (c)
4. For the 4, 3-d electrons
ml = –2 –1 0 1 2
4
3d =
s= ½ ½ ½ ½
L m 2 1 0 1 2
1 1 1 1
S s 2
2 2 2 2
Since shell is more than half filled total angular momentum
J L S 2 2 0
The resultant magnetic momentum is given by
g B J 0 J 0
Correct answer is (0)
5. For electric dipole radiation, selection rule is 1 .
Here, in option for transition 2s 1s, we have 0 , which is forbidden.
Thus, 2s 1s is not allowed.
Hence, correct option is (a).
GATE-PH 2016 SOLUTION 443
6. For triplet of mesons, 2I + 1 = 3 I = 1
Mesons are not strange particles. Therefore, S = 0
Correct option is (b)
7. We know magnetic moment, m Ia 2 nˆ
where a is the side of the square and n̂ is unit vector normal to the surface
a
If the square is converted into circle then the radius of the circle will be r
2
2 I a2 m
So, magnetic moment, mc I r nˆ nˆ
4 4
2
So, P 2.46
4
Correct answer is (2.46)
8. At R, T
Total power emitted = P1
R
At , 2T
2
Total power emitted = P2
Energy
Power
Time
Energy
Total power in space 4 R 2
Time
Energy T 4 Stefan's Boltzman law
T 4
P1 4 R 2 , t time
t
2
4 R
P2 2T 4
2
P1 T4 R2 1
4
2
4 0.25
P2 16T R 4
Correct option is (0.25)
9. Given : S k B N p ln p 1 p ln 1 p
For maxima and minima,
dS p 1 p
0 kB N ln p 1 ln 1 p 0
dp p 1 p
k B N 1 ln p 1 ln 1 p 0
p 1
ln 0 p 0.5
1 p 2
Correct answer is (0.5)
GATE-PH 2016 SOLUTION 444
10. The Helmholtz free energy (F) is defined as
F U TS
dF dU TdS SdT
TdS PdV TdS SdT dU TdS PdV
PdV SdT
At constant temperature and volume,
dF 0
So F is either maximum or minimum for a system at constant volume and temperature. Moreover the second
law of thermodynamics sets a limit that in any process at constant volume and temperature, the Helmholtz free
energy decreases.
i.e., dF 0 .
where equality holds for reversible and inequality for irreversible processes.
So, Helmholtz free energy decreases and hence minimum at equilibrium.
Correct option is (a)
11. From first law of thermodynamics,
dQ dU PdV
It is given that the total energy is constant.
dU 0 dQ PdV
dQ PdV
By definition, S
T T
4V
Nk B
dV for ideal gas, PV Nk BT
V
V
4V
Nk B ln V V Nk B ln 4 2 Nk B ln 2
Correct answer is (2)
12. Let z x iy
f z z 2 x 2 y 2 2 xyi u iv
Here, u x 2 y 2 , v 2 xy
u u
Now, 2 x, 2 y
x y
v v
2 y, 2x
x y
Therefore, Cauchy Reamann equation is satisfied i.e.
u v u v
&
x y y x
Correct option is (a)
13. When slit-1 open when slit-2 open
a2 4I0 b 2 I0
a 2 I0 b I0
GATE-PH 2016 SOLUTION 445
So, when both are open
2
I max a b
2
3 I0 9 :1
I min a b 2 2
I0
Correct answer is (9:1)
2/3
2 3 2 N
14. Fermi energy can be given by EF ... (i)
2 V
N
where, n electron concentration per unit volume.
V
EF n 2/3 ... (ii)
1 1
and Hall coefficient can be given by RH
ne n
1 1
RH 3/ 2 RH EF3/ 2 .
n EF
Hence, correct option is (d)
2K
15. acoustic at boundary
m2
2K
optical at boundary
m1
Correct option is (a).
* 2 * *
16. x 2 dx x x x dx
2 * 2 *
*
2
dx dx
x x
x 2
†
2 2
2
x x 2
Correct option is (b)
17. Given k m 0 c 2 here, k = kinetic energy
we know that;
1
p k k 2 m0 c 2
c
1
p m0 c 2 m0 c 2 2m0 c 2 3 m0c 1.732 m0 c
c
Hence correct answer is (1.73)
GATE-PH 2016 SOLUTION 446
18. In intrinsic semiconductor, ni 2 1019 m 3
In doped semiconductor, n 4 1020 m 3
By doping with donor impurities, the majority carriers will be electrons (n) and minority carriers will be holes
(p).
From equation ni2 np ... (i)
ni2 (2 1019 ) 2 4 1038
p 20
per m3 p 20
1 1018 m 3
n 4 10 4 10
Therefore, ratio of majority to minority charge carriers
n 4 1020
4 10 2 400
p 1 1018
Hence, correct answer is (400).
k
19. Given: m1 m, m2 2m, 4 s 2
m
Angular frequency for system shown in figure is given as
k m m
, 1 2 reduced mass
m1 k m2
m1 m2
m 2m 2 m
m 2m 3
k 3 k 3
4 6 2.45 rad /s
2m / 3 2 m 2
Hence correct answer is (2.45)
20. Given, F kr k xiˆ y ˆj , and angular momentum is non zero.
Particle is moving under two perpendicular restoring forces therefore its dynamics can be resolved into two
perpendicular simple harmonic motions. We know that superposition of two perpendicular simple harmonic
motions leads to three types of path; Straight line through origin, circle about origin, ellipse about origin. Since
angular momentum is non zero path cannot be straight line through origin.
Hence correct answer is (b)
21. Here, in this reaction, we have electron on left hand side, which belong to Lepton family. For the conservation
of Lepton number in the above equation, we need to add a Lepton member on the right hand side also i.e.
either an electron, or electron-neutrino or muon or muon-neutrino.
Since, we have electron at left side, we need to add electron neutrino on right side.
Correct option is (b)
22. First Born approximation state that the energy of the incident particle is large and scattering potential is
weak.
Correct option is (a)
GATE-PH 2016 SOLUTION 447
23. Total scattering cross-section
4
4 4
k2
2 1 sin 2
k2
2 0 1 sin 2 90º 2
2
0
2
24. In case of hydrogen atom, a uniform electric field is applied. So, this electric field will be treated as a
perturbation.
V p qE0 z eE0 r cos
First order correction to the energy, in the ground state
r
1 a0
EG1 .S . 100 eE0 r cos 100 0 100 e
3
a0
In presence of electric field, there is no correction to the first order,
Second order correction to the energy in the ground state
2
nm eE0 r cos 100
E02
n 1 E10 Em0
This is also well known result in every literature that presence of electric field lifts the degeneracy of energy
levels in second order correction which comes out to be proportional to E02
Hence, the value of exponent n is (2)
Correct answer is (2)
25. Magnetic susceptibility, C (independent of T) [Property of diamagnets]
For type-I superconductor, the magnetic susceptibility 1 , which is independent of temperature.
Hence, correct option is (c).
Q.26 – Q.55 : Carry TWO marks each.
26. The electric field at the upper surface will be E
k 0
0 k 1 k 1 ˆ
So, polarization, P 0 E nˆ n
k 0 k
k 1
So, surface bound charge, s
k
Correct option is (d)
27. For 3p 3d electron, we have
1 1
s1 , s2 , 1 1, 2 2
2 2
The possible values of S and L are:
S s1 s2 , s1 s2 1, , s1 s2
0,1; multiplicity 2 s 1 1,3
GATE-PH 2016 SOLUTION 448
and L 1 2 , 1 2 1, , 1 2
1, 2, 3 P, D, Fstates
Thus, we have in all six terms, three singlet terms and three triplet terms. All these terms are odd because the
configuration 3p3d is odd 1 2 3 .
We can write these terms as
1
P, 1D, 1F, 3 P, 3 D, 3 F
To take into account spin-orbit interaction, let us combine L and S to form J. Now
J L S , L S
Fo singlet terms, we have
S 0
; J 1; 1P1
L 1
S 0
; J 2; 1D 2
L 2
S 0
; J 3; 1F3
L 3
For triplet terms, we have
S 1
; J = 0,1, 2; 3 P0 , 3 P1 , 3 P2
L 1
S 1
; J = 1, 2,3; 3 D1 , 3 D 2 , 3 D3
L 2
S 1
; J = 2,3, 4; 3 F2 , 3 F3 , 3 F4
L 3
Thus a single degenerate level of configuration 3P 3d is splitted into 12 levels.
Hence, correct answer is (12).
28. Deuteron has binding energy 2.23 MeV. So, can be disintegrated by 4 MeV gamma rays. Deuteron has no
excited states. Deuteron ground state is triplet so it cannot have 1 S0 state. Its quadrupole moment is positive
i.e. prolate.
Correct option is (c)
1
29. s1 s2 s 2 s12 s22
2
2
s s 1 s1 s1 1 s2 s2 1
2
2
1 1 1 1
0 2 2 1 2 2 1 for ground state of He, S = 0
2
3 2
4
Correct option is (b)
GATE-PH 2016 SOLUTION 449
30.
(1, 2) (2, 1)
(1, 1)
2 2 2 2
So, the ground state energy of the system Enx n y
2mL2
nx n y
2 2 2 5 2 2 22
EG .S . 2 4 24
2mL2 2mL2 2mL2
Correct answer is (24)
31. Maximum kinetic energy, K .Emax 2 NqV
where N is number of revolutions
V 50 103 V
q 2e 2 1.6 1019 C of -particle
K .Emax 16 106 1.6 1019
N 80
2qV 2 2 1.6 10 19 50 103
Correct answer is (80)
32. ijk mk j Vm
kij k m j Vm
i jm im j j Vm
j iV j j jVi i jV j 2jVi 2jVi
2Vi Given: jV j 0
2jVi
Correct option is (d)
1 2 1
33. We have, f x, zy, z x xy z 2
2 2
ˆ ˆ 1 2 1
f iˆ j k x xy z 2 x y iˆ x ˆj zkˆ
x y z 2 2
f
1,1,2
ˆj 2kˆ
ˆj 2kˆ
Therefore, direction of f is
5
Correct option is (b)
34. 2 x y y x 2i z i z i z
Correct option is (c)
GATE-PH 2016 SOLUTION 450
t2
35. Action is defined as S Ldt where L T V is Lagrangian.
t1
1 2 1 2 1 2
T mv 0.1 u at 0.1 0 10t 5t 2
2 2 2
1 1
V Fdx ma dx 0.1 10 dx x ut at 2 0 10 t 2 5t 2
2 2
2 2 2
t 3 80
S T V dt 5t 5t 2 2
dt 10
3 0 3
26.67
0 0
S 26.67 J-s
Correct answer is (26.67)
1 x 0 a0
36. f x ; f x an cos nx bn sin nx
1 0 x 2 n n
Given f x is an odd function :
Therefore, a0 0, an 0
0
1 1
bn f x sin nxdx sin nxdx sin nxdx
0
0
1 cos nx cos n 1 1
n
n 0 n
n
n
1 1 1 1
0 for even n
2 1 n
n
1 1
bn 4
n for odd n
4 1 4 1 1
f x bn sin x sin nx sin x sin 3x sin 5 x ....... n is odd
n n1 n 3 5
Correct option is (a)
a
37. Along the Y-axis there are three atoms [two at corners and one at 0, , 0 ]. If a is the lattice parameters.
2
Y
a
0, 2 , 0
Radius R
Radius r
Z
Then a 2 R 2 r ... (i)
Diagonal of each face will have 4 atoms (two at corners and one at center).
2a 4 R ... (ii)
GATE-PH 2016 SOLUTION 451
2a
R
4
2a
Now putting value of R from equation (ii) to equation (i), we get a 2 2r
4
2 2a 4 2 2
2r a a
4 4
42 2
r a ... (iii)
8
42 2
a
r
8
4 4 2 2 2 2 2 2 2 2 1 2 1 1.414 1 0.414.
R 2a 8 2 2 2 2 2
4
Hence, correct answer is (0.41).
38. Given x x2 x1 d
t t2 t1 0
From Lorentz transformation
x x vt d
x v dv
t t 2 2
c c
Hence correct answer is (c)
39. The ratio of effective mass to the free electron mass
2 2
2
d E 2
d
me* dk dk
2 2ka A 4 Bk a 2a 2 A 12 Bk 2 a 4
2 3 4
me me me me
2
2a 2 A
{at the bottom of conduction band k 0}
me
(1.05 1034 )2
2 (2.1 10 10 ) 2 (6.3 1019 ) 1.05 1.05 10 68 1 102
(9.1 1031 ) 2 2.1 2.1 1020 6.3 10 19 9.1 10 31 8 6.3 9.1
10 2
0.218 0.22
458.64
Hence, correct answer is (0.22).
40. We have, E E0 cos kz t iˆ in free space.
We know, k E µ0 H
GATE-PH 2016 SOLUTION 452
E E
cµ0
cµ0
H 0 cos kz t kˆ iˆ 0 cos kz t ˆj
E2 1
And S E H 0 cos 2 kz t kˆ As, c
cµ0 µ0 0
c 0 E02 cos 2 kz t kˆ
Correct option is (d)
41. Let n1 , n2 and n3 be the number of particles with energy 0, 2 and 3 respectively. Then.
According to given,
n1 n2 n3 4 ... (i)
and n1 0 n2 2 n3 3 10
or 2n2 3n3 10 ... (ii)
Using equation (ii),
n2 n3
0 10/3
1 8/3
2 2
3 4/3
4 2/3
Since n1 , n2 and n3 all are integers, only possible values of n2 and n3 are 2 and 2 respectively..
Now from equation (i), n1 0
So possible values of n1 0, n2 2 and n3 2
Also, given that g1 2, g 2 2 and g3 3 .
The number of microstates for bosons is
W
3
ni g i 1! n1 g1 1 ! n2 g 2 1! n3 g3 1!
i 1 ni ! gi 1 ! n1 ! g1 1 ! n2 ! g 2 1! n3 ! g 3 1!
1! 3! 4!
1 3 6 18
0!1! 2!1! 2!2!
Correct answer is (18)
1 2 2
42. L
2
m sin 2 2 mg cos
is cyclic, therefore p is conserved
L
p Conserved
m 2 sin 2 is conserved
Therefore sin 2 is conserved.
Correct answer is (a)
GATE-PH 2016 SOLUTION 453
43. Rutherford scattering cross-section
NnLZ 2 k 2e 2
(since, momentum is equal)
2
4r 2 K .E sin 4
2
4
Z2 4
p 1
Correct answer is (4)
44. To find output X
A A
X
B
C B+C
X A B C X A BC X A B C
(By applying De-Morgan’s theorem)
A B A B
Correct option is (b)
45. The molar specific heat can be given by
2 N Ak B2 12 4 N Ak B 3
CV T 3 T ... (i)
2 EF 5 TD
CV T AT 3 ... (ii)
2 N A k B2 1
where, ... (iii)
2 EF EF
12 4 N A k B 1
and A 3
A 3 ... (iv)
5 TD TD
For X and Y :
3
1 1
3
3 3
X ( EF ) X
( EF )Y 7 eV 7 and AX (TD ) X (TD )Y (TD )Y 340 (2)3 8
3
Y 1 ( EF ) X 5 eV 5 AY 1 (TD )3X (TD ) X 170
( E F )Y
(TD )Y
Correct option is (a).
46. The mean energy is
i Ei gi e Ei / kBT
E
gi e Ei / kBT where gi is the degeneracy of the Ei th level.
i
0 1 e 0/ kBT E 3 e E / kBT 3 E e E / k BT
E
1 e 0/ kBT 3 e E / kBT 1 3 e E / k BT
Correct option is (d)
GATE-PH 2016 SOLUTION 454
47. Apply conservation of momentum in x direction to get
Mv
p1 cos 45º p2 cos 60º
v2
1 2
c
E
We have energy momentum relation p for photon
c
Mv 1 E E
1 2 ... (i)
2
1 v /c c 2 2
2
Apply conservation of energy; given E1 1GeV , E2 0.82 GeV
Mc 2
E1 E2 ... (ii)
1 v 2 /c 2
Divide (i) by (ii) to get
E1 E2 E E 1 0.82
c 1 2 c
v 1 2 2 ; 2 2 2 2
v
c 2 c E1 E2 E1 E2 1 0.82
v 0.614 c
From equation (ii) we get
2
E E2 1 0.82 1 0.614
M 1 1 v /c 2 2
GeV 1.44 GeV /c 2
c2 c2
Hence correct answer is (1.44).
48. x 2 , p 2 x 2 , p p x 2 , p p p x 2 , p i2 xp pi 2 x 2i xp px
Correct option is (d)
49. According to boundary condition,
1 B1t
B2 n B1n ; H 2t H1t ; µ B2t µ
2 1
µ2 µ2
We have, B1n Bz , B1t Bx , µr
µ1 µ0
B2n Bz , B2t µr Bx
So, magnetic field in the magnetic material will be
B B B µ B iˆ B kˆ
2n 2t r x z
Correct option is (d)
50. 1, 0 Lx iLy 1, 1 1, 0 L 1, 1 2 1 1, 0 |1, 0 2
L , m m m 1 , m 1
Correct option is (c)
GATE-PH 2016 SOLUTION 455
51. We know that, parity operator is hermitian i.e. P† P
Parity operator is also unitary operator i.e. P † P 1
P 1 P2 I
P2 I
Therefore, P 2 P not correct
Correct option is (b)
10V
52. Given Vin 5V IC
3k
VBE 0.7 V V0
200k
dc 100 5V
IB +
0.7 – IE
Applying KVL at Base-Emitter Junction 1k
5 200 I B 0.7 I E
I E 1 I B
So, 5 200 I B 0.7 101 I B
I B 0.0142 mA
I C I B 1.42 mA
Now, applying KVL at output
10 3IC V0 (Put value of I C )
V0 5.714 Volt
Correct answer is (5.7)
53. 1 2 2 3 3
A 1 2 2 3 3 , where A is a normalization constant
Using the normalization condition, | 1
2 1 1 2 3
A 1 4 9 1 A 1 2 3
14 14 14 14
2 4 2
The probability of finding the system in the state 2 is 2 | 0.2857
14 7
Correct answer is (0.29)
54. For 15
8 O , Z 8 even
N = 15–8 = 7
2 4 1
1
s1/ 2 1
p3/2 1
p1/ 2
1
Therefore, J , 1 for p and parity = 1 1
1
2
GATE-PH 2016 SOLUTION 456
1
Thus, spin-parity =
2
17
For 8 O , Z = 8 (even)
N = 17–8 = 9
2 4 2 1
1
s1/ 2 1
p3/ 2 1
p1/ 2 d
1
5/ 2
5
J , 2 for d
2
5
2
Therefore, parity = (–1) = +1 and spin-parity =
2
Correct option is (b)
55. Since, E V0 , so according to principle of quantum tunneling there will be a fiinite probability for particle to
cross the barrier.
Wevefunction of the particle in the region x 0 is
x D e x
So, probability of finding the particle in the region x 0 is
V0
2 2 2 x
Given, P x x D e E
1
P x x0 P x 0 x=0 x=x0
e
e2 x0 e 1
1
2 x0 1 x0
2
Correct option is (c)
GATE-PH 2017 SOLUTION 457
OBJECTIVE QUESTION
13
1. C6 ve 13 N 7 X
X should be electron 1 e 0
13
C6 ve 13 N 7 1e 0
Therefore, charge Q will be conserved also electronic lepton number will be conserved.
Correct option is (a)
2. m m
k
k mm m
, reduced mass
mm 2
k 2k 2 1
200 10 2 rad/s 10 1.414 14.14 rad/sec .
m /2 m 10 103
Hence correct answer is (14.14)
3. The degree of freedom is
f 3N C
where N is the number of atoms in a molecule and C is the number
of constraints.
Here, N 3 and C 3 . So, f = 6.
Out of these six degree of freedom, three are due to translational and rest three are due to rotational motion.
At high temperature, all constraints get zero. So total degree of freedom becomes f = 3N at high temperature
and hence, nine for the given molecule.
Another three degree of freedom are due to vibrational motion.
According to law of equipartition of energy, each translational and rotational degree of freedom contributes
1
k BT to the energy and each vibrational degree of freedom contributes kBT to the energy..
2
1
Average energy, U 6 k BT 3 k BT 6k BT
2
dU
Heat capacity, C 6k B
dT
Correct option is (d)
1 0 0 1 0 i 1 0 a0 bz bx iby
4. H a0 I b a0 bx by bz
0 1 1 0 i 0 0 1 bx iby a0 bz
If is the eigenvalue of Hamiltonian, then H I 0
a0 bz bx iby 2 2 2
bx iby a0 bz
0 a0 2 2
bz bx by 0 a0 bx2 by2 bz2
a0 b
So, lowest enregy = a0 b
Correct option is (c)
GATE-PH 2017 SOLUTION 458
5. x, x p y y p x x, x p y x, y p x
p y x, x x x, p y p x x, y y x, p x
p y 0 x 0 px 0 y 1 y
Hence correct answer is (b)
6. For s-orbitals,
0, m 0 , wavefunctions are only function of r. No angular dependence is there, hence s-orbitals are
spherically symmetric
Correct option is (c)
7. The pressure exerted on the mirror
1
P 2 u 2 0 E02 0 E02 8.85 10 12 Pa
2
Correct answer is (8.85)
8. The energy of the nth state of hydrogen atom
1
En
n2
The highest possible energy state is n
Ehighest 0
Correct option is (a)
2 x
9. U x U 0 cos 1 dimensional lattice
a
Magnitude of energy gap at the edge of Brillouin-zone k
a
a
2 2
Eg dx U x
0
2 x 2 x
where cos and sin are the normalized wavefunctions.
a a a a
a
2 2 x 2 x x
Eg dxU 0 cos cos sin 2
a0 a a a
4 x
a
1 cos
a a
2U 0 2 2 x 2U 0 a U0 4 x a
a 0 a 0
dx cos dx x sin a 4 U 0
a 2 a 0
Correct option is (a)
GATE-PH 2017 SOLUTION 459
q q
2 3
q q
10. 1 4
6 q5
At centre : Field due to charge at 1 will be cancelled by field of charge 4. Field due to 2 and 5 are also cancelled
at centre.
The resultant electric field will be for only one charge particle which is present at point 3.
1 q
So, E
4 0 a 2
5 q
And potential, V V1 V2 V3 V4 V5
4 0 a
Correct option is (c)
11. The efficiency of a reversible Carnot engine is independent of any working substance and depends only on the
T1
temperature of the source T2 and of the sink T1 . It is given by 1 .
T2
Correct option is (b)
12. Given 7 bit A to D converter
Full scale voltage of 5 volt
Full-Scale Voltage 5
Voltage resolution n
7
2 1 2 1
Resolution = 39.27 mV
1 1
13. L0 mq 2 m2 q 2
2 2
qdq d 1
L L0 L0 q 2
dt dt 2
The given transformation satisfies gauge transformation of Lagrangian, therefore equation of motion will not
change. Since the term added contains q therefore generalised momentum will change, as seen below
L0 L L0
pinitial mq ; pfinal qq ; p mq q
q q q q final
pfinal pinitial
Hence, correct option is (b)
Q Q Qd
14. We know, C V
V C A 0 k
We have, Q 1 C , d 106 m, A 1 m 2 , 0 8.854 1012
106
V 12
0.1129 105 11.29 kV
8.854 10 10
Correct answer is (11.29)
GATE-PH 2017 SOLUTION 460
dz dz
15. 1 z 2
z i z i
Therefore, poles are at z i, i . Only z i will lie inside the given contour
y
Therefore, Residue at z i
1 1 x
lim z i –R R
z i
z i z i 2i –i
dz
1 z 2
2 i Residual at z i
1
2 i 3.14
2i
D1
16.
D2
This circuit is a full wave rectifier. So, output wave form
o/p
t
• Output wave form for
D1
Reason: capacitor
charging and
discharging
D2
Capacitor charge
and discharge C R
Correct option is (a)
17. We know Snell’s law, n1 sin 1 n2 sin 2
At critical angle, n sin c 1
1
sin c c
1.5
c 41.81
Correct answer is (41.81)
GATE-PH 2017 SOLUTION 461
18. In electromagnetic interactions, all three C, CP and CPT should be conserved.
Therefore, correct option is (a)
h 6.626 1034
19. Compton wavelength, c 27 8
1.322 1015 m 1.322 fm
m p c 1.67 10 3 10
Correct answer is (1.32)
20. ; Le : 1 0 0 0
L 0 i.e. Tau number is not conserved
Charge is conserved, total lepton number and angular momentum is also conserved.
Correct option is (d)
21. Fourier series in complex form
f x C0 Ck eikx C k e ikx
k 1 k 1
2
eiax e iax A 2 iax 2iax
e e 2
2
u x A sin ax A
2i 4
A
Hence co-efficient of eikx when k 2a is
4
Correct option is (b)
22. The phase-space trajectory of a particle under any potential is the momentum-space diagram.
If the free particle has fixed amount of energy E, then
p2
E
2m
where m is the mass of the particle.
1/2
p 2mE
So we have two fixed values of p.
Moreover, the ball is bouncing between two hards walls. So position of it is also restricted, lets say
between, x = –a and x a . So the phase space is
p
p 2mE
x= –a x=a
x
p 2 mE
Correct option is (c)
N atomic
23. The number of density of valence electrons
M
6.022 1023
0.968 cm 3 2.53 1022 cm3
23
GATE-PH 2017 SOLUTION 462
24. The number of degeneracy of nth energy level of 3-dimension isotropic harmonic oscillator is given by
1 1
n 1 n 2 4 5 for third energy level n 3
2 2
10
Correct option is (d)
25. The first order correction of the ground state energy
E0 0 H ' 0 0 x 0 0 x3 0 0 x 4 0 0 0 0 x 4 0
I
Correct option is (d)
26. Total charge of the system is 2 – 1 – 1 = 0
So, the dipole moment does not depend on the origin
So, the dipole moment w.r.t. the point A is
1 3
P qi ri 1 2 iˆ 2 ˆj 3 ˆj 1.73 ˆj
i 2 2
Correct answer is (1.73)
27. v velocity of O w.r.t O
c
vx
2 v
v v c
vx x vx
v v 4
1 x 2 O x O x
c
c
c 2 v c v c 7v c 2c
4 1 v 4 8 2 v 8 4 v 7 0.28 c
2c
Hence correct answer is (0.28)
U
28. Given : (1) u aT 4 or U aT 4V dU a 4T 3VdT T 4 dV
V
u U
(2) P
3 3V
S
(3) aT 3 or S aT 3V dS a 3T 2VdT T 3dV
V
Using combined I and II law,
TdS = dU + PdV, we have
U
aT 3T 2VdT T 3dV a 4T 3VdT T 4 dV dV
3V
4
3aT 3VdT aT 4 dV 4aT 3VdT aT 3dV
3
Comparing, we have
4
1.33
3
Correct answer is (1.33)
GATE-PH 2017 SOLUTION 463
29. We know due to rotation of earth, weight of person at latitude is given by
mg ' mg m 2 R cos 2
w p mg (at north pole)
And we mg mw2 R
we wp mw2 R
w p we w2 R w2 R
wp g g
2 3
w p we 2 6400 10
100 100 0.003337 100 0.337
wp 24 3600 10
Correct answer is (0.33)
30. Minimum number of NAND gate required for OR-gate
a
y = a + b OR-gate
b
a
a
a b a b NAND-gate
b
b
a a+b a a+b
b b
Bubbled OR gate = NAND gate
Applying bubble short concept.
a+b
b
a
a + b Bubbled OR gate = NAND gate
b
Note: 3 number of NAND are needed.
Correct option is (d)
0.5 1 dE 0.5 12 6
31. E R 12
6 7
R R dR R13 R
dE
At the equilibrium dR 0
R R0
6 6
Therefore, R13 R 7 0 R0 1
0 0
Correct answer is (1.00)
GATE-PH 2017 SOLUTION 464
32. v x, y 2 xy 3 y
According to Cauchy Reamann equation
u v u v
2 x 3 u x2 3x f y ; 2 y u y 2 g x
x y x y
So, u x, y x 2 3x y 2
u z 1 i u x 1, y 1 3
Correct answer is (3)
33. The population of particles in the level with energy is
N e / kBT N e / kBT N
N 0/ k BT / k BT
/ k BT
/ k BT
e e 1 e e 1
In the high temperature limit,
k BT 1
k BT
N N
e / k BT 1 N
11 2
Correct option is (a)
34. Vcc
R1=10k
Vin
R2=10k Vout
input R3=1k output
• Its a CC-configuration
(a)
hie hfe Ib
(E) (E)
hie = r (emitter diode )
Given: emitter diode resistance is zero; r 0
Ib r= 0
+ (1 + hfe)Ib
Vin hfe.Ib R3 = 1k
–
h-parameter for CC model
GATE-PH 2017 SOLUTION 465
Vin r I b 1 h fe I b R3 r 0
Vin
Ri 1 h fe R3 1 99 1 100 k
Ib
Correct answer is (100)
35. For the Raman active, diatomic molecules must have changing polarisability. For IR active molecules must have
change in dipole moment. While for NMR, presence of non-zero nuclear spin is compulsory.
Considering given option :
(a) 1 H 1H : There is no change in dipole moment due to symmetry, therefore, it is not IR active.
12
(b) C 16O : Net nuclear spin zero, therefore not NMR active.
(c) 1 H 35Cl : It has all the condition available for IR, NMR and Raman active.
(d) 16 O 16O : There is no change in dipole moment, therefore it is not IR active.
Correct option is (c)
36. Correct option is (c)
37. The reflection coefficient,
2 2
E E V0 1 0.25 1 0.5 2 0.5 2 1
R 0.11
E E V
0 1 0.25 1 0.5 1.5 9
38. The magnetic field inside the solenoid will be along ẑ axis. So, Bz 0, Br B 0
Correct option is (d)
39. (i) When particles A and B are distinguishable.
Since particles are distinguishable, there is no limit on the filling of the particles in energy levels. Further-
more since level 1 already contains a particle, we have in total three ways to filled the levels.
1 2
AB
A B
B A
2
Therefore, P(level 2 also contains a particle)
3
(ii) When particles A and B are bosons.
Since bosons are indistinguishable particles and there is no restriction on the filling of them in different
energy levels, we have in total two ways to fill them.
1 2
1
Therefore, P(level 2 also contains a particle)
2
Correct option is (c)
a
40. a1 axˆ; a2
2
xˆ 3 yˆ
Let a3 zˆ
The reciprocal space vector of this lattice are
GATE-PH 2017 SOLUTION 466
a a 2 3 a 2 1
b1 a1* 2 2 3 axˆ yˆ xˆ yˆ
a1 a2 a3 3 2 2 2 a 3
a
2
* a3 a1 2 ayˆ 4 yˆ
b2 a2 2
a1 a2 a3 3 2 3a
a
2
Correct option is (a)
41. Consider two colliding protons at very large energies. The effective size of colliding protons is approximately
105 times smaller (in radius) than an atom.
Size of an atom is generally considered in Angstroms (Å)
1Å = 10–10 m
Effective size of proton will be 10–15 m
Geometric cross-section r 2 1030 m 102 barn 1 barn 10 28
m2
–2
= 10 b = 10 mb
Correct option is (b)
t 0
0
42. We know, 0e 0 t 1
e t 0
e 0
0
t 8.85 1012 102 8.85 1014 88.50 10 15 sec 88.50 Femto seconds
Correct answer is (88.50)
43. The charge in the temperature of the melting point of a solid dT due to a change of pressure dP is given by
Clapeyron’s equtaion
dp L
dT T v2 v1
where L is the latent heat of fusion of the solid and V2 V1 the change in volume, T is melting point
1
Specific volume of ice vice
ice
1
Specific volume of of water vwater
water
1 1
v2 vice , v1 vwater
ice water
dT 10º C 10 K , L 3.34 105 J / kg , T 273 K
LdT 3.34 105 10 3.34 934 109
dp
1 1 1 1 273 1000 934
273 273
ice water 934 1000
0.173 109 Pa 0.173 GPa
Correct answer is (0.173)
GATE-PH 2017 SOLUTION 467
2 x2 2
44. x e dx , [let x t 2xdx dt ]
0
2 dt
x x e x dx t1/ 2e t
0 0 2
1 32 1 t 1
t e dt 3/2 1 1 1/2 1 1 0.44
20 2 2 2 2 2 4
Correct answer is (0.44)
13
45. C6
Proton number P = 6 and Neutron number N = 7
P 6 :1S½2 1P3/42
N 7 :1S½2 1P3/42 1P½1
The total angular moment is due to momentum of last unpaired neutron.
1 1
J , the parity 1 1 1
2
1
1
JP
2
Correct option is (d)
46.
u v
R R
Angular momentum about point of contact is conserved
mu R mvR I c
2 v
mu R mvR mR 2
5 R
7
u v
5
5 5 5 25
v u 3.57 m /s
7 7 7
Hence correct answer is (3.57)
47. x max 3 nm
109
px min
2 3 109 6
2
6.626 1025
2
2
109
/ 6 2 36
p x
Emin 9.83 104 10 104 eV 1 meV
2m 2 9.311031 2 9.31 1031
So, the minimum depth of the well should be 1 meV.
Correct option is (b)
GATE-PH 2017 SOLUTION 468
48. Given R1 5 k
R2 1 k
RL 100 k
V 10 mV
Current through R2 ____ A (micro ampere)
V+
V–
V 10mV R1 RL
10mV R2 IR2
10mV 1k
As ideal op-amp has infinite input resistance. So, there will be no current pass through op-amp.
So, V V (Virtual ground concept)
= 10 mV
10 mV 10 103
I R2
R2 1103
I R2 1106 Ampere
I R2 1 A
Correct answer is (1)
49.
v
Rest
By conservation of momentum, P Pv ... (1)
By conservation of energy,
E E Ev ... (2)
If m , m and 0 are rest masses of , and v respectively. Then equation (2) becomes
2 2 2 2
2 2 2 4
Pc
m c m c
m c P m c Pv c
v
2m c 2
2 2
Pc
139 105 29.84 MeV Pv
29.84
MeV
v
2 139 c
Correct answer is (29.84)
50. Given differential equation is
dy
y tan x cos x (1st or order linear differential equation)
dx
I.F.= e
tan x dx
elogsec x sec x
Therefore, solution will be
y sec x cos x sec x dx c
GATE-PH 2017 SOLUTION 469
y sec x x c
y 0 0,
0 0c c 0
x
So, y sec x x y x cos x
sec x
1
y /3 cos = 0.5233
3 3 3 2 6
Correct answer is (0.5233)
51. We know that the lowest frequency of electromagnetic wave that can propagate through the metal is propor-
tional to square root of density of free electrons.
f n
f1 n
1
f2 n2
n2 1.8 1023
f 2 f1 1.38 1016 Hz 1.38 1016 3 2.39 1016 Hz
n1 6 1022
Correct answer is (2.39)
52. Electronic configuration of the nitrogen 7 N 1 s 2 2 s 2 2 p3
i.e., it has three electrons outside closed shells.
Chart of nitrogen state
me e
1
0
–1
S ms 32
L ml 0
3 3
i.e., total spin S , multiplicity 2 1 4
2 2
Total L 0 , i.e. s-orbital
3
Total angular momentum J L S .... to L S
2
Therefore, ground state term is 4 S3/2
Correct option is (b)
GATE-PH 2017 SOLUTION 470
i px x 2 x 2 p x
i x 2 px p x x 2 i p x
x
2
x 2 px
54. 2 2 2
Correct option is (a)
4 0 2 n 2 1 m m
55. Radius of the Bohr orbit is rn , i.e., rn , where 1 2 ; reduced mass of the hydro-
Z e 2 m1 m2
gen like system.
For hydrogen atom : m p 1836 me
me m p 1836 me
H me e–
me m p 1837 me
e+
For positronium atom : m m
e e
m m me
e e
p
m m 2
e e
RP H 1
Ratio of ground state orbit
RH P rn
m
RP H RH e 0.53 Å 2 0.53 Å 1.060Å 1.06Å
P me /2
Correct answer is (1.06)
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