GATE PHYSICS 2013
Previous Year Question Paper with Solution.
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)
(b)
(c)
(d)
Ans. (b)
Sol. 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.
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
Ans. (d)
Sol. Stress = Force/Area, these are two directions involved with it, i.e., one is directional of force and the other one is direction of area.
Strain = change in length/original length, there are two directions involved with it, ie., one is the direction of length and the direction along which length is changing.
Moment of inertia i.e., two direction involved one of angular momentum and other of angular velocity.
Pressure = Force/Area; But here direction of force direction of area ar same. Therefore, it is a first order tensor.
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
Ans. (b)
Sol.
velocity of electron w.r.t. photon is given as
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)
(a) Zero
(b) Infinite
(c)
(d) kT
Ans. (d)
Sol. Energy in case of blackbody radiation is given by
So using L, Hospital's rule, we have
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
Ans. (c)
Sol. The free energy is continuous but its derivative is discontinuous is first order phase transition.
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
Ans.
Sol. Once the gases are allowed to freely exchange of energy, both sides will attain thermal equilibrium maintaining the same temperature. Besides this, due to unbalanced pressure, both sides will attain mechanical equilibrium by balancing the pressure on both sides of the movable wall.
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
Ans. (a)
Sol. Given: S(E) = aE (E0 – E).
The entropy is defined as
For E0 < 2E, T is negative
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)
(a) Zero
(b) Nk
(c) Nk
(d) Nk ln(2)
Ans. (d)
Sol. The number of microstate for N independent spin s particles are
The entropy is
9. The decay process violates
(a) baryon number
(b) lepton number
(c) isospin
(d) strangeness
Ans. (c)
Sol.
If I3 is not conserved then isospin I will also be not conserved
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
Ans. (d)
Sol. Baryon number of upto quark is 1/3.
Isospin of up quark is 1/2.
11. Consider the scattering of neutrons by protons at very low energy due to a nuclear potential of range r0. Given that, where
is the phase shift, k the wave number and
the logarithmic derivative of the deuteron ground state wave function, the phase shift is
(a)
(b)
(c)
(d)
Ans. (a)
Sol. Consider the scattering of returns by protons at very low energy due to a nuclear potential of range r0.
12. In the decay process, the transition
, 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
Ans. (c)
Sol. According to Femi-Selection rule = 0
Gamow Teller rules
Therefore, the transition , parity does not change. The given transition is not allowed by Fermi but allowed by 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.
Ans. (d)
Sol. 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
14. Interference fringes are seen at an observation plane z = 0, by the superposition of two plane waves and
, where A1 and A2 are real amplitudes. The condition for interference maximum is
(a)
(b)
(c)
(d)
Ans. (b)
Sol. Equation for phase wave are given as
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
Ans. (c)
Sol. Consider a scalar function φ which is satisfying the Laplace equation,
Hence, has both zero divergence and zero curl.
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.
Ans. (c)
Sol. 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 components and refracted light has larger component in the plane of incidence than its component perpendicular to the plane. As a result the reflected light will be plane polarized light where as reflected light will be elliptically polarized light.
17. Which one of the following commutation relations is NOT CORRECT? Here, symbols have their usual meanings.
(a) [L2, Lz] = 0
(b)
(c)
(d)
Ans. (d)
Sol. We know that [L2, Lz] = 0, [Lx, Ly] = iHL
18. The Lagrangian of a system with one degree of freedom q is given by , 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) and it is a conserved quantity.
(b) and it is not a conserved quantity.
(c) and it is a conserved quantity.
(d) and it is not a conserved quantity.
Ans. (d)
Sol.
Canonical momentum is given as
Since q is not cyclic, pq is not conserved
19. What should be the clock frequency of a 6-bit A/D converter so that its maximum conversion time is 32µs?
(a) 1 MHz
(b) 2 MHz
(c) 0.5 MHz
(d) 4 MHz
Ans. (a, b)
Sol. Given 6-bit A/D converter analog to digital converter.
Maximum conversion time 32 µ sec
Note: Not mention which type of A to D converter is given.
Flash ADC > Successive approximation ADC > Counter ADC > Dual slope or integrating type
CT = 1 CT = nTC CT = 2n–1TC CT = 2nTC
CT = Converter time, n = number of bits, TC = CLK time period.
Assuming counter type ADC: CT = 2n–1TC TC = ?
Assuming Dual slope or integrating type ADCX : CT = 2nTC
Note: Option (a) or (b) is correct depending on the type of ADC.
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
Ans. (c)
Sol. With the increase in temperature, the concentration of minority carries starts increasing. Eventually, a temperature is reached called the critical temperature (85°C in case of germanium and 200 °C in case of silicon) when the number 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°C, silicon will have the Fermi-level in the middle of energy gap.
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
(c) Quantization of magnetic flux in superconducting ring in the unit of
(d) Presence of Meissner effect
Ans. (c)
Sol. Considering the BCS theory of superconductors statement "(c) Quantization of magnetic flux in superconducting ring in the unit of (h/e)" is not correct. It correctly predict the Meissner effect, isotope effect and bond gap in the superconductor, It is quantised in unit of .
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)
Ans. (b)
Sol. Group-I Group-II
P. Phonon iv. elastic wave
Q. Plasmon iii. collective oscillations
R. Polaron ii. electron + elastic formation
S. Polariton i. photon + lattice vibration
23. The number of distinct ways of placing four indistinguishable balls into five distinguishable boxes is ______
Ans. 70
Sol. 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 µV? The answer should be up to two decimal places _______
Ans. 3.16
Sol. Ripple Rejection of voltage regulator = – 50 dB
input ripple is 1 mV
output ripple voltage in µV
Ripple Rejcection =
–50 dB = 20 log10 (Ripple Rejection)
Ripple rejection = 3.162 × 10–3
O/P ripple voltage = 3.162 × 10–6 Volt
Correct answer is (3.16)
25. The number of spectral lines allowed in the spectrum for the transition in sodium is ______
Ans. 3
Sol.
Selection rules for sodium (alkali spectra) are = 0, + 1. So, only three spectra lines are allowed.
26. Which of the following pairs of the given function F(t) and its Laplace transform f(s) is NOT CORRECT?
(a)
(b)
(c)
(d)
Ans. (c)
Sol.
27. If and
are constant vectors, then
is
(a)
(b)
(c)
(d) zero
Ans. (b)
Sol.
If are constant vectors, then
is also constant vector.
For constant vectors
Therefore,
28.
(a)
(b)
(c)
(d)
Ans. (c)
Sol.
29. The relativistic form of Newton's second law of motion is
(a)
(b)
(c)
(d)
Ans. (c)
Sol. For one dimensional motion, Newton's equation of motion is written as
30. Consider a gas of atoms obeying Maxwell-Boltzman statistics. The average value of over all the momenta
of each of the particles (where
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,
(a) One
(b) Zero
(c)
(d)
Ans. (c)
Sol.
Here, we have used Maxwell-Boltzmann distribution function f(vx, vy, vz) and it is equal to
31. The electromagnetic form factor F(q2) of a nucleus is given by,
where Q is a constant. Given that
where is the charge density, the root mean square radius of the nucleus is given by
(a) 1/Q
(b)
(c)
(d)
Ans. (c)
Sol.
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,
(a)
(b)
(c)
(d)
Ans. (c)
Sol. Angular momentum is disc is
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,
(a)
(b)
(c)
(d)
Ans. (a)
Sol.
Let x1 and x2 be displacement of masses from their respective equilibrium positions. Therefore,
Corresponding matrices are
34. A charge distribution has the charge density given by . For this charge distribution the electric field at (2x0, 0, 0)
(a)
(b)
(c)
(d)
Ans. (a)
Sol. We have one dimension charge density
Therefore, the total charge,
Therefore, the field at (2x0, 0, 0) point is
35. A monochromatic plane wave at oblique incidence undergoes reflection at a dielectric interface. If , and
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)
(b)
(c)
(d)
Ans. (c)
Sol. We know that all are in the same plane.
The vector has its direction perpendicular to the plane containing the vector
,
36. In a normal Zeeman effect experiment, spectral splitting of the line at the wavelength 643.8 nm corresponding to the transition 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
Ans. (b)
Sol.
Given: wavelenth, = 643.8 nm
wavelength is related to wave-number by
In normal Zeeman effect experiment. Wave number separation between Zeeman level are equal and is given by
where, B is the applied magnetic field.
Putting, e = 1.6 × 10–19 C, me = 9.1 × 10–31 kg, c = 3 × 108 m/s, we get
As the spectrometer has resolution to observe a minimum spread in wavelength = 0.01 nm, therefore from equation (1) and equation (2).
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
Ans. (c)
Sol. Energy level of harmonic oscillator
where v = vibration quantum number
vosci is classical frequency given by
; k = force constant, µ = reduced mass
38. A lattice has the following primitive vectors (in A): = . The reciprocal lattice corresponding to the above lattice is
(a) BCC lattice with cube edge of
(b) BCC lattice with cube edge of
(c) FCC lattice with cube edge of
(d) FCC lattice with cube edge of
Ans. (a)
Sol. Primitive vectors …(i)
Volume of unit cell, = 0 [0 – 4] – 2 [0 – 4] + 2 [4 – 0] = 8 + 8 = 16 Å3
Now, translation vectors of reciprocal lattice
Therefore, volume of the reciprocal lattice,
which is the volume of primitive unit cell of bcc with cube edge .
39. The total energy of an ionic solid is given by an expression where
is Madelung 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
(a)
(b)
(c)
(d)
Ans. (a)
Sol.
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
Ans. (c)
Sol. Sides of box = 10–12 m
Charge of proton = 1.6 × 10–19 coulomb
Mass of proton (mp) = 1.67 × 10–27 kg
Planck's constant h = 6.63 × 10–34 J-s
41. For the function , the residue at the pole z = 1 is (your answer should be an integer) ________.
Ans. 3
Sol. has a pole of order 2 at z = 1
Residue of f(z) at z = 1
42. The degenerate eigenvalue of the matrix
is (your answer should be an integer) _______________
Ans. 5
Sol.
43. Consider the decay of a pion into a muon and an anti-neutrino in the pion rest frame.
. The energy (in MeV) of the emitted neutrino, to the nearest integer is _______
Ans. 30
Sol. According to energy conservation,
According to momentum conservation,
Using (i) and (ii),
Using (ii) and (iii),
44. In a constant magnetic field of 0.6 Tesla along the z direction, find the value of the path integral in the units of (Tesla m2) on a square loop of side length
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. ____
Ans. 0.15
Sol.
45. A spin-half particle is in a linear superposition of its spin-up and spin-down states. If
and
are the eigen states of
then what is the expectation value, up to one decimal place, of the operator
? Here, symbols have their usual meanings _________
Ans. 7.6
Sol.
46. Consider the wave function Aeikr (r0/r), where A is the normalization constant. For r = 2r0, the magnitude of probability current density up to two decimal places, in units of , is _______.
Ans. 0.25
Sol.
47. An n-channel junction field effect transistor has 5mA 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 ________
Ans. 1.25
Sol. IDSS = 5 mA
VP = – 5V
VGS = – 2.5V
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)
(b)
(c)
(d)
Ans. (b)
Sol. The fermions are distributed as
So the partition function is
49. Two distinguishable particles
(a)
(b)
(c)
(d)
Ans. (c)
Sol. The partition function of one distinguish particle is
Common Data for Questions 50 and 51: To the given unperturbed Hamiltonian
we add a small perturbation given by
where is a small quantity.
50. The ground state eigen vector of the unperturbed Hamiltonian is
(a)
(b)
(c) (0, 0, 1)
(d) (1, 0, 0)
Ans. (c)
Sol.
(3x1 + 3x2) = 0 …(1)
(2x1 + 3x2) = 0 …(2)
Solving equation (1) and (2),
x1 = 0, x2 = 0 and x3 is arbitrary.
So, eigenvector = (0, 0, 1)
51. A pair of eigen values of the perturbed Hamiltonian, using first order perturbation theory, is
(a)
(b)
(c)
(d)
Ans. (c)
Sol. Perturbed Hamiltonian,
Eigenvectors corresponding to eigenvalue, of the unperturbed Hamiltonian, E = 2, 3, 7 will be
First order correction to energy, will be
Statement for Linked Answer Questions 52 and 53: In the Schmidt model of nuclear magnetic moments, we have,
where the symbols have their usual meaning
52. For the case J = l + 1/2, where J is the total angular momentum, the expectation value of in the nuclear ground state is equal to,
(a) (J – 1)/2
(b) (J + 1)/2
(c) J/2
(d) –J/2
Ans. (b)
Sol.
Taking self dot product both sides
53. For the O17 nucleus (A = 17, Z = 8), the effective magnetic moment is given by,
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
Ans. (b)
Sol.
Statement for Linked Answer Questions 54 and 55: Consider the following circuit
54. For this circuit the frequency above which the gain will decrease by 20dB per decade is
(a) 15.9 kHz
(b) 1.2 kHz
(c) 5.6 kHz
(d) 22.5 kHz
Ans. (a)
Sol.
20 dB/decade frequency means cut off frequency.
From equation (a) and (b)
fH = 15.91 KHz
This is cut off frequency for LPF.
Circuit Analysis:
Capacitor is open circuit
At law frequency we are getting output.
Case-(2) at high frequency,
Capacitor is short circuit.
at high frequency no output
55. At 1.2 kHz the closed loop gain is
(a) 1
(b) 1.5
(c) 3
(d) 0.5
Ans. (b)
Sol.
Closed loop again
C = 1000 × 10–12 F
f = 1.2 × 103 Hz
Closed loop gain (T(if)) ~ 1.5