GATE PHYSICS 2011
Previous Year Question Paper with Solution.
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
Ans. (c)
Sol. Two similar matrices A and B have same eigenvalues, same determinant and same trace.
2. If a force is derivable from a potential function V(r), where r is the distance from the origin of the coordinate system, it follows that
(a)
(b)
(c)
(d)
Ans. (a)
Sol. Correct option is (a)
3. The quantum mechanical operator for the momentum of a particle moving in one dimension is given by
(a)
(b)
(c)
(d)
Ans. (b)
Sol. Quantum mechanical operator for the momentum of a particle moving in one dimension is
4. A Carnot cycle operates on a working substance between two reservoirs at temperatures T1 and T2, with T1 > T2. During each cycle, an amount 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
(b)
(c) entropy of the hotter reservoir decreases
(d) entropy of the universe (consisting of the working substance and the two reservoirs) increases
Ans. (d)
Sol. The Carnot cycle is reversible and hence the entropy change of the universe is zero.
5. In a first order phase transition, at the transition temperature, specific 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
Ans. (b)
Sol. The key point are
(i) The specific heat divergence for both I and II order phase transition.
(ii) The entropy has a finite discontinuity for I order phase transition whereas it is continous for II order phase transition.
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) A–1/3
(b) A1/3
(c) A2/3
(d) A
Ans. (c)
Sol.
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.
Ans.
Sol. In case of two level, the rate of upward transitions is equal to the rate of downward transitions which gives N2 = N1.
Thus, there will be no population inversion.
8. The temperature (T) dependence of magnetic susceptibility of a ferromagnetic substance with a Curie temperature (Tc) is given by
(a)
(b)
(c)
(d)
Ans. (b)
Sol.
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
Ans. (d)
Sol. The order of magnitude of the energy gap of a typical superconductor is ~1 meV.
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
Ans. (b)
Sol. Common emittor configuration/amplifier
Output is taken from collector
It has 180° 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.
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
Ans. (d)
Sol.
12. A heavy symmetrical top is rotating about its own axis of symmetry (the z-axis). If I1, I2 and I3 are the principal moments of inertia along x, y and z respectively, then
(a)
(b)
(c)
(d)
Ans. (c)
Sol.
Linear, I1 = I2, I33 = 0
13. An electron with energy E is incident from left on a potential barrier, given by
as shown in the figure.
For E < V0, the space part of the wavefunction for x > 0 is of the form
(a)
(b)
(c)
(d)
Ans. (b)
Sol. For the region, x > 0, the 1-D time independent Schordinger equation will be
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 the form = Be–ax.
14. If Lx, Ly and Lz are respectively the x, y and z components of angular momentum operator L, the commutator [LxLy, Lz] is equal to
(a)
(b)
(c)
(d) 0
Ans. (c)
Sol.
15. The normalized ground state wavefunction of a hydrogen atom is given by , where a is the Bohr radius and r is the distance of the electron from the nucleus, located at the origin. The expectation value is
(a)
(b)
(c)
(d)
Ans. (d)
Sol.
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
(a)
(b)
(c)
(d)
Ans. (a)
Sol.
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)
(b)
(c)
(d)
Ans. (a)
Sol.
18. A magnetic dipole of dipole moment is placed in a non-uniform magnetic field . If the position vector of the dipole is , the torque acting on the dipole about the origin is
(a)
(b)
(c)
(d)
Ans. (d)
Sol. If we placed a magnetic dipole in non-uniform magnetic field.
The net torque anad on it is given by
19. Which of the following expressions for a vector potential does not represent a uniform magnetic field of magnitude B0 along the z-direction ?
(a)
(b)
(c)
(d)
Ans. (c)
Sol.
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.
Ans. (d)
Sol. 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.
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)
(b)
(c)
(d)
Ans. (a)
Sol. In weak field, the higher total angular momentum state splits into (2J+1) states. Possible J values for one electron (1s1),
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)
(b)
(c)
(d)
Ans. (b)
Sol. The selection rules for orbital angular momentum:
For one electron, most the transition take place when value changes by one unit, i.e., because, according to Laporte rule, the parity should change in a electric dopole transition.
The selection rules for total orbital angulara momentum:
For total angular momentum, the selection rule for electric dipole transition is = 0.
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) 3p – 3
(d) 3N – 3p
Ans. (c)
Sol. 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.
24. For an intrinsic semiconductor, 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)
(b)
(c)
(d)
Ans. (c)
Sol. The fermi level of an instrinsic semiconductor is given by
25. In the following circuit, the voltage across and the current through the 2 resistance are
(a) 20 V, 10 mA
(b) 20 V, 5 mA
(c) 10 V, 10 mA
(d) 10 V, 5 mA
Ans. (d)
Sol.
For given circuit to check zener in breakdown region.
For give circuit to check zener in breakdown region.
Zener-ON
Final circiut
26. The unit vector normal to the surface x2 + y2 – z = 1 at the point P(1, 1, 1) is
(a)
(b)
(c)
(d)
Ans. (b)
Sol.
27. Consider a cyclinder of height h and a, closed at both ends, centered at the origin. Let be the position vector and a unit vector normal to the surface. The surface integral over the closed surface of the cylinder is
(a)
(b)
(c)
(d) zero
Ans. (b)
Sol.
28. The solutions to the differential equation
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
Ans. (a)
Sol.
This represents a family of circles having centre at (0, –1) and radii of units.
29. A particle is moving under the action of a generalized potential
The magnitude of the generalized force is
(a)
(b)
(c)
(d)
Ans. (c)
Sol.
Generalized force is given as
30. Two bodies of mass m and 2m are connected by a spring constant k. The frequency of the normal mode is
(a)
(b)
(c)
(d)
Ans. (a)
Sol. Frequency of normal mode is given as
31. Let (p, q) and (P, Q) be two pairs of canonical variables. The transformation
is canonical for
(a)
(b)
(c)
(d)
Ans. (d)
Sol.
For canonical transformation
Poisson Bracket {Q, p}q, p = 1
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
(a)
(b) 2m
(c)
(d)
Ans. (c)
Sol.
Initial momentum of two particles is zero. Therefore final momentum also be zero. So final entity will be at rest.
final :
Apply conservation of energy.
Einitial = Efinal
33. The normalized eigenstates of a particle in a one-dimensional potential well
are given by
The particle is subjected to a perturbation
The shift in the ground state energy due to the perturbation, in the first order perturbation theory, is
(a)
(b)
(c)
(d)
Ans. (a)
Sol. First order correction to ground state energy is
34. If the isothermal compressibility of a solid is = 10–10 (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
Ans. (c)
Sol.
35. A system of N non-interacting and distinguishable particles of spin 1 is in thermodynamic equilibrium. The entropy of the system is
(a) 2kB ln N
(b) 3kB ln N
(c) NkB ln 2
(d) NkB ln 3
Ans. (d)
Sol. Number of microstates for N non-interaacting particles with spin s is
= (2s + 1)N = 3N [ for s = 1]
The entropy of the system is S = kBln = N kB ln 3
36. A system has two energy levels with energies . The lower level is 4-fold degenerate while the upper level is doubly degenerate. If there are N are non-interacting classical particles in the system, which is thermodynamic equilibrium at a temperature T, the fraction of particles in the upper level is
(a)
(b)
(c)
(d)
Ans. (a)
Sol.
37. A spherical conductor of radius a is placed in a electric field . The potential at a point , is given by , where r is the distance of P from the center O of the sphere and is the angle OP makes with the z-axis.
The charge density on the sphere at is
(a)
(b)
(c)
(d)
Ans. (a)
Sol.
38. According to the single particle nuclear shell model, the spin parity of the ground state of is
(a)
(b)
(c)
(d)
Ans. (d)
Sol. We have nuclues ,
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
39. In the decay of neutron , the anti-neutrino 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
Ans. (b)
Sol. Correct option is (b)
40. The isospin and the strangeness of baryon are
(a) 1, –3
(b) 0, –3
(c) 1, 3
(d) 0, 3
Ans. (b)
Sol. For baryon , strangeness (S) = – 3 and isospin (1)
41. The life time of an atomic state is 1 ns. The neutral line width of the spectral line is 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
Ans. (c)
Sol. According to Heisenber uncertainity relation
42. The degeneracy of an excited state of nitrogen atom having electronic configuration 1s22s22p23d1 is
(a) 6
(b) 10
(c) 15
(d) 150
Ans. (d)
Sol. Total degeneracy, D = D1D2, where D1 is degeneracy 2p2 and D1 is degeneracy of 3d1.
Here, N is number of electrons in the given orbital.
43. The far infrared rotational absorption spectrum of a diatomic molecule shows equivalent 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
Ans. (c)
Sol. The line spacing in the infrared rotational absroption spectrum 2B = 10 cm–1 = constant B =10 cm–1
Since, the distanec of first Stocsk line in the rotational Raman spectrum = 6 × 10 cm–1 = 60 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 . 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
Ans. (b)
Sol.
For the diffraction peak in BCC (hkl) = (1, 1, 0)
45. In the following circuit, Tr1 and Tr2 are identical transistors having VBE = 0.7 V. The current passing through the transistor Tr2 is
(a) 57 mA
(b) 50 mA
(c) 48 mA
(d) 43 mA
Ans. (d)
Sol. Given BVE = 0.70 V (identical transistor)
Here, Ic2 Ic1 because it is current mirror circuit.
By applying KVL through dotted loop
5 = Ic1 × 100 + 0.7
Ic1 = 43 mA Ic2
46. The following Boolean expression
can be simplified to
(a)
(b)
(c)
(d)
Ans.
Sol.
It is a4 variable boolean function which can be minimized by using K-map
47. Consider the following circuit.
Which of the following correctly represents the output Vout corresponding to the input Vin?
(a)
(b)
(c)
(d)
Ans. (a)
Sol. Lets redraw the circuit.
Vf = VUT =2 Volt Vf = – 2 Volt
VUT = 2V VI.T = – 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 VLT, V0 makes transition from –VSAT to +VSAT.
Common Data for Questions 48 and 49 :
Consider a function of a complex variable z.
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
(c) f(z) has a pole of order 2 at
(d) f(z) has a simple pole at
Ans. (d)
Sol.
Highest negative power in the Laurrent series expansion is 1/t
So, f(z) has a simple pole at z =
49. Consider a counterclockwise circular contour |z| = 1 about the origin. The integral over this contour is
(a)
(b) zero
(c)
(d)
Ans. (b)
Sol. Since, z = lies outside the circle.
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 , where are constant and k is the wave-vector.
50. The density of states of electrons (including spin degeneracy) in the band is given by
(a)
(b)
(c)
(d)
Ans. (b)
Sol.
51. The effective mass of electrons in the band is given by
(a)
(b)
(c)
(d)
Ans. (b)
Sol.
Statement for Linked Answer Question 52 and 53 :
In a one-dimensional harmonic oscillator, are respectively the ground, first and the second excited states. These three states are normalized and are orthogonal to one another. are two states defined by
where is a constant.
52. The value of for which is orthogonal to is
(a) 2
(b) 1
(c) –1
(d) –2
Ans. (c)
Sol.
53. For the value of determined in Q. 52, the expression value of energy of the oscillator in the state is
(a)
(b)
(c)
(d)
Ans. (d)
Sol.
Statement for Linked Answer Question 54 and 55 :
A plane electromagnetic wave has the magnetic field given by
where k is the wave number and are the Cartesian unit vectors in x, y and z direction, respectively.
54. The electric field corresponding to the above wave is given by
(a)
(b)
(c)
(d)
Ans. (a)
Sol.
We know that solution,
55. The average Poynting vector is given by
(a)
(b)
(c)
(d)
Ans. (d)
Sol.