GATE PHYSICS 2005
Previous Year Question Paper with Solution.
1. The average value of the function f(x) = 4x3 in the interval 1 to 3 is
(a) 15
(b) 20
(c) 40
(d) 80
Ans. (c)
Sol. The average
2. The unit normal to the curve x3y2 = xy = 17 at the point (2, 0) is
(a)
(b)
(c)
(d)
Ans. (d)
Sol. Unit normal curve : = x3 y2 +xy – 17 at point (2, 0) is
3. The value of the integral , where C is a circle (anticlockwise) with |z| = 4, is
(a) 0
(b)
(c)
(d)
Ans. (c)
Sol.
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
Ans. (b)
Sol. Let be the eigenvaluesof the given matrix.
Determine of the matrix = Product of the eigenvalues
Given :
So, = 6
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
Ans. (c)
Sol. For central field potential is r dependent only i.e. V = V(r)
Therefore, energy and p0 (angular momentum) are conserved. Since angular momentum is conserved, motion must be confined to a plane.
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
(a)
(b)
(c)
(d)
Ans. (d)
Sol.
7. If for a system of N particles of different masses m1, m2, ...., mN with position vectors and corresponding velocities
, respectively, such that
, then
(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
Ans.
Sol.
So, force may not be zero
None of the option is correct.
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
Ans. (c)
Sol. Momentum is not conserved in such process
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 at a distance r from the centre of the circle, for r > a, is
(a) 0
(b)
(c)
(d)
Ans. (a)
Sol. According to Ampere's law,
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
(a)
(b)
(c)
(d)
Ans. (a)
Sol.
11. The electric field for a circularly polarized electromagnetic wave propagating along the position z-direction is
(a)
(b)
(c)
(d)
Ans. (b)
Sol. Superposition of two linear orthogonal polarized light of equal amplitude give circularly polarized light.
Wave propagating towards positive z-direction.
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
(a) E = constant, B = constant
(b)
(c)
(d)
Ans. (b)
Sol. For a point source particle both the electric k and magnetic field will be vary as
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
Ans. (d)
Sol.
14. The commutator, , where L2 is the z-component of the orbital angular momentum
and is a spherical harmonic, is
(a)
(b)
(c)
(d)
Ans. (c)
Sol.
15. A system in a normalized state 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
Ans. (d)
Sol.
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
(a)
(b)
(c)
(d)
Ans. (c)
Sol. Force on the charge – q, F = – qE
17. The line of X-rays emitted from an atom with principal quantum numbers n = 1, 2, 3, ..., arises from the transition
(a)
(b)
(c)
(d)
Ans. (b)
Sol. If vacancy is created in L shell, then electron from upper cells will have the transition to L shell (n = 2) givingL series. Transition n = 3
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)
(b)
(c)
(d)
Ans. (c)
Sol. Selection rule for electric dipole transition:
i.e., parity must change during transition and
19. If the equation of state for a gas with internal energy U is , then the equation for an adiabatic process is
(a) pV1/3 = constant
(b) pV2/3 = constant
(c) pV4/3 = constant
(d) pV3/5 = constant
Ans. (c)
Sol.
From first law of thermodynamic, dQ = dU + pdV, where dQ = 0 for an ideabatic process.
where c is a constant of integeration.
20. The total number of accessible states of N non-interacting particles of spin-½ is
(a) 2N
(b) N2
(c) 2N/2
(d) N
Ans. (a)
Sol. Number of accessible states for N non-interacting particles of spin s is = (2s + 1)N.
21. The pressure for a non-interacting Fermi gas with internal energy U at temperature T is
(a)
(b)
(c)
(d)
Ans. (b)
Sol. If u is the total internal energy density then pressure is given by
22. A system of non-interacting Fermi particles with Fermi energy EF has the density of states proportional to , where E is the energy of a particle. The average energy per particle at temperature T = 0 is
(a)
(b)
(c)
(d)
Ans. (a)
Sol. , where A is a constant of proportionality.
The average energy per particle at T = 0 K is
where f(E) = 1 at T = 0 K
23. In crystallographic notations the vector in the cubic cell shown in the figure is
(a) [221]
(b) [122]
(c) [121]
(d) [112]
Ans. (a)
Sol.
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
Ans. (c)
Sol. Since, Argonhaas 18 atomic number. Therfore, its atomic configuration 1s2 2s2 2p6 3s2 3p6. Na is strong electropositive and Cl is strong electronagive. Si is in semiconductor with covalent bonding.
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
Ans. (a)
Sol. The -intensity anti-prallel to the nuclear spin direction is same along the nuclear spin directions. This observation gives the evidence for the non-conservation of parity in
-decay.
26. Which of the following expressions for total binding energy B of a nucleus is correct (a1, a2, a3, a4 > 0) ?
(a)
(b)
(c)
(d)
Ans. (a)
Sol. Since, surface energy term also its sign is negative.
27. Which of the following decay is forbidden ?
(a)
(b)
(c)
(d)
Ans. (d)
Sol.
Therefore, this is forbidden.
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
Ans. (d)
Sol. Nuclear forces are short range, charge independent, velocity dependent and spin dependent.
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
Ans. (a)
Sol.
AC equivalent figure of FET
30. The circuit shown can be used as
(a) NOR gate
(b) OR gate
(c) NAND gate
(d) AND gate
Ans. (d)
Sol. Redraw this circuit
Case (1) : Both D1, D2 OFF Case 2 : Both D1, D2 ON
Case (3) : Either D1 or D2 is ON Truth Table
It is AND Gate.
31. If a vector field is
(a) 0
(b)
(c)
(d)
Ans. (a)
Sol.
32. All solutions of the equation ez = –3 are
(a)
(b)
(c)
(d)
Ans. (b)
Sol.
33. If is the Laplace transform of f(t) the Laplace transform of f(at), where a is a constant, is
(a)
(b)
(c)
(d)
Ans. (b)
Sol.
34. Given the four vectors
the linearly dependent pair is
(a) u1u2
(b) u1u3
(c) u1u4
(d) u3u4
Ans. (d)
Sol.
35. Which of the following functions of the complex variable z is NOT analytic everywhere ?
(a) ez
(b) sin z/z
(c) z3
(d) |z|3
Ans. (b)
Sol.
36. Eigen values of the matrix
are
(a) –2, –1, 1, 2
(b) –1, 1, 0, 2
(c) 1, 0, 2, 3
(d) –1, 1, 0, 3
Ans. (a)
Sol. Sum of the eigenvalues = Trace of the matrix = 0
Only option (a) satisfies this.
37. If a particle moves outward in a plane along a curved trajectory described by , 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
Ans. (d)
Sol.
It is time dependent therefore, not conserved, momentum cannot be conserved because force is not zero.
38. A circular hoop of mass M and radius a rolls without slipping with constant angular speed along the horizontal x-axis in the xy-plane. When the centre of the hoop is at a distance
from the origin, the magnitude of the total angular moment 4m of the hoop about the origin is
(a)
(b)
(c)
(d)
Ans. (c)
Sol.
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
(a)
(b)
(c)
(d)
Ans. (c)
Sol. Use parallel axes theorem
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 seen from an inertial frame outside the car. The acceleration of the sphere in the inertial frame is
(a)
(b)
(c)
(d)
Ans. (b)
Sol.
From equation (i)
Acceleration of sphere with respect to inertia frame.
41. A rod of length l0 makes an angle with the y-axis in its rest frame, while the rest frame moves to the right along the x-axis with relativistic speed v with respect to the lab frame. If
, the angle
in the lab frame is
(a)
(b)
(c)
(d)
Ans. (c)
Sol. Let be angle with x-axis as seen from s' frame
42. A particle of mass m moves in a potential , where x is the position coordinate, v is the speed, and
and µ are constants. The canonical (conjugate) momentum of the particle is
(a) p = m(1 + µ)v
(b) p = mv
(c) p = mµv
(d) p = m(1 – µ)v
Ans. (d)
Sol.
43. Consider the following three independent cases :
(i) Particle A of charge +q moves in free space with a constant velocity 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
Ans. (b)
Sol. Radiation power from accelerated charge particle depends on charge and acceleration as
where, q is the charge of the particle and a is the acceleration of particle.
In first case, particle moving with constant speed. So, PA = 0.
In second and third cases, particle are moving with same (centripetal) acceleration. So, PB = PC.
44. If the electrostatic potential were given by where
is constant, then the charge density giving rise to the above potential would be
(a) 0
(b)
(c)
(d)
Ans. (b)
Sol. We know Lapase equation,
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
(a)
(b)
(c)
(d)
Ans. (c)
Sol. The work done in bringing a charge particle from infinite to a position at a distance d in front of a semi-infinite grounded metal surface is
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, . If E = 2 V/m and n = 1.5, then the application of appropriate boundary conditions leads to
(a)
(b)
(c)
(d)
Ans. (c)
Sol.
47. For a vector potential the divergence of
is
is a constant of appropriate dimension. The corresponding scalar potential
that makes
and
Lorentz gauge invariant is
(a)
(b)
(c)
(d)
Ans. (d)
Sol. Lorentz Gauge condition,
48. An infinitely long wire carrying a current 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)
(b)
(c)
(d)
Ans. (a)
Sol.
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
Ans. (a)
Sol. The de-Broglie wavelength,
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
Ans. (c)
Sol. One dimensional time independent Schrodinger equation,
For the existence of bound state,
51. The expectation value of the z-coordinate, (z) in the ground state of the hydrogen atom (wavefunction : , where A is the normalization constant and a0 is the Bohr radius), is
(a) a0
(b)
(c)
(d) 0
Ans. (d)
Sol.
52. The degeneracy of the n = 2 level for a three dimensional isotropic oscillator is
(a) 4
(b) 6
(c) 8
(d) 10
Ans. (b)
Sol. The degeneracy of nth state of three dimensional isotropic harmonic oscillator is
53. For a spin-½ particle, the expectation value of sxsysz, where sx, sy and sz are spin operators, is
(a)
(b)
(c)
(d)
Ans. (a)
Sol.
54. An atom emits a photon of wavelength by transition from an excited state of lifetime 8 × 10–9 s. If
represents the minimum uncertainty in the frequency of the photon, the fractional width
of the spectral line is of the order of
(a) 10–4
(b) 10–6
(c) 10–8
(d) 10–10
Ans. (c)
Sol. The time-energy uncertainity principle
55. The sodium doublet lines are due to transitions from 2P3/2 and 2P1/2 levels to 2S1/2 level. On application of a weak magnetic field, the total number of allowed transitions becomes
(a) 4
(b) 6
(c) 8
(d) 10
Ans. (d)
Sol.
For sodium doublet lines the total number of transitions from 2p3/2 and 2p1/2 levels to 2s1/2 level is 10.
56. A three level system of atoms has N1 atoms in level E1, N2 in level E2, and N3 in level E3(N2 > N1 > N3 and E1 < E2 < E3). Laser emission is possible between the levels
(a)
(b)
(c)
(d)
Ans. (b)
Sol. Here, in the given three levels system of atoms, the level with energy E2 has largest atoms N2 amongst all three levels, which confirms that is a metastable state. Similarly, E1 and E2 respersents the ground and exicited state respectively. Thus, the laser emission will be possible only between E2 to E1 levels.
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
Ans. (c)
Sol. Raman shift (or displacement) is property of the substance or sample to be studied and is independent of incident radiation. Thus, it is confirmed that the typical Raman shift depends only on I.
58. For a diatomic molecule with the vibrational quantum number n and rotational quantum number J, the vibrational level spacing and the rotational level spacing
are approximately
(a)
(b)
(c)
(d)
Ans. (b)
Sol. The energy of potential level is given by
EJ = BJ (J + 1)
= EJ – EJ–1 = BJ (J + 1) – B (J – 1) (J – 1 + 1) = B [J (J + 1)] – J (J – 1)] = BJ (J + 1 – J + 1) = 2BJ
Vibrational spacing,
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
Ans. (c)
Sol. The typical wavelengths emitted by diatomic moleculs in purely vibrational transitions lie in infrared region(1 µm – 102 µm). In purley rotational transitions, the typical wavelength lie in microwave region (103 µm – 104 µm)
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 . For the jj coupling scheme to be applicable, the coupling strengths MUST satisfy the condition
(a)
(b)
(c)
(d)
Ans. (b)
Sol. In J-J coupling spin orbit interaction dominates over spin-spin and orbit-orbit interaction.
61. If the probability that x lies between x and x + dx is p(x)dx = ae–ax dx, where , then probability that x lies between x1 and x2(x2 > x1) is
(a)
(b)
(c)
(d)
Ans. (a)
Sol. Given : px(dx) = ae–ax dx,
Probability that x lies between x1 and x2 (x2 > x1) is
62. If the partition function of a harmonic oscillator with frequency at a temperature
, then the free energy of N such independent oscillator is
(a)
(b)
(c)
(d)
Ans. (c)
Sol.
63. The partition function of two Bose particles each of which can occupy any of the two energy levels 0 and is
(a)
(b)
(c)
(d)
Ans. (b)
Sol. The possible configuration are
Hence, the partition function is
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
(a)
(b)
(c)
(d)
Ans. (a)
Sol. Since the random walker takes steps left or right with equal probability,
Probability of taking step to right .
The random walker comes back to origin after N even steps if he takes steps to the right and
step to the left.
65. The number of states for a system of N identical free particles in a three dimensional space having total energy between , is proportional to
(a)
(b)
(c)
(d)
Ans. (c)
Sol. The density of state in three dimensional space is
Since electron spin degeneracy is 2 sow we should multiply g(E) by 2.
For N-identical particles
66. The energy of a ferromagnet as a function of magnetization M is given by
F(M) = F0 + 2(T – Tc)M2 + M4, F0 > 0.
The number of minima in the function F(M) for T > Tc is
(a) 0
(b) 1
(c) 3
(d) 4
Ans. (b)
Sol. F(M) = F0 + 2 (T – TC) M2 + M4 …(i)
For, M = 0,
Therefore, F(M) have a minima at M = 0 for T > TC
= – 8T + 8TC < 0 (for T > TC)
67. For a closed packed BCC structure of hard spheres, the lattice constant a is related to the sphere radius R as
(a)
(b)
(c)
(d)
Ans. (a)
Sol. For body cntered cubic (bcc) stucture, the atoms will lie at corners as well as body centers of cube.
68. An n-type semiconductor has an electron concentration of 3 × 1020 m–3. If the electron drift velocity is 100 ms–1 in an electric field of 200 Vm–1, the conductivity of this material is
(a) 24
(b) 36
(c) 48
(d) 96
Ans.
Sol.
69. Density of states of free electrons in a solid moving with an energy 0.1 eV is given by 2.15 × 1021 eV–1 cm–3. The density of states (in eV–1 cm–3) for electrons moving with an energy of 0.4 eV will be
(a) 1.07 × 1021
(b) 1.52 × 1021
(c) 3.04 × 1021
(d) 4.30 × 1021
Ans.
Sol.
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 cm–3) in Ge at room temperature (300 K) is approximately
(a) 3 × 1010
(b) 3 × 1013
(c) 3 × 1016
(d) 3 × 1016
Ans. (d)
Sol. The concentration of electron in the conduction band
By multiplying equation (1) and (2)
The intrinsic carrier concentration
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
Ans.
Sol.
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
Ans. (b)
Sol.
73. The threshold temperature above which the thermonuclear reaction
can occur is
(a) 1.28 × 1010 K
(b) 1.28 × 109 K
(c) 1.28 × 108 K
(d) 1.28 × 107 K
Ans. (a)
Sol. If T be the threshold temperature of the reaction then,
74. According to the shell model, the ground state of nucleus is
(a)
(b)
(c)
(d)
Ans. (d)
Sol.
75. The plot of log A vs. time t, where A is activity, as shown in the figure, corresponds to decay
(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
Ans. (c)
Sol. The activity A of a sample decays exponentially with time, At =
It will be a straight line (figure (a)) between log At and t with a negative slope.
But for complex decay i.e.for mixture of wo half lves,we get the graph between log At vs. t as shown in figure (b).
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 . If Iave is the average current through the resistor RL the circuit corresponds
(a)
(b)
(c)
(d)
Ans. (a)
Sol. Given circuit consist of two diodis, its a full wave rectifier.
77. The Boolean expression : 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
Ans. (c)
Sol.
1 OR-gate needed
78. The output V0 of the ideal opamp circuit shown in the figure is
(a) –7 V
(b) –5 V
(c) 5 V
(d) 7 V
Ans. (a)
Sol.
79. The circuit shown in the figure can be used as a
(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
Ans. (d)
Sol. By applying voltage divider rule.
Note: If R and C are interchanged circuit with high pass filter OR differenciator.
80. In the circuit shown in the figure the Thevenin voltage VTh and Thevenin resistance RTh as seen by the load resistance are respectively
(a)
(b)
(c)
(d)
Ans. (a)
Sol. Calculation of Thevenin voltage across RL =
Thevenin voltage is also referred as open circuit voltage
Open the load resistance of the circuit for the calculation of VTH and RTH.
Statement for Linked Answer Q.81a and Q.81b :
For the differential equation
81a. One of the solutions is
(a) ex
(b) ln x
(c)
(d)
Ans. (a)
Sol. Correct option is (a)
81b. The second linearly independent solution is
(a) e–x
(b) xex
(c) x2ex
(d) x2e–x
Ans. (b)
Sol. Correct option is (b)
Statement for Linked Answer Q.82a and Q.82b :
The Langrangian of two coupled oscillators of mass m each is
82a. The equation of motion are
(a)
(b)
(c)
(d)
Ans. (b)
Sol. We have the Lagrangian,
The equation of motion are
82b. The normal modes of the system are
(a)
(b)
(c)
(d)
Ans. (d)
Sol.
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 magnetic of the surface current is
(a)
(b)
(c)
(d)
Ans. (a)
Sol. Surface current density k =
83b. The magnitude of vector potential inside the cylindrical at a distance from its axis is
(a)
(b)
(c)
(d)
Ans. (a)
Sol.
The rotating cylinder is equivalent to long solenoid therefore magnetic field inside is
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 , where k is the wavevector and A is a constant.
84a. If is an angular function then in the asymtotic region the scattered wavefunction has the form
(a)
(b)
(c)
(d)
Ans.
Sol.
84b. The differential scattering cross section in CM frame is
(a)
(b)
(c)
(d)
Ans.
Sol.
Statement for Linked Answer Q.85a and Q.85b :
Lead has atomic weight of 207.2 amu and density of 11.35 gm cm–3
85a. Number of atoms per cm3 for lead is
(a) 1.1 × 1025
(b) 3.3 × 1022
(c) 1.1 × 1022
(d) 3.3 × 1025
Ans. (b)
Sol. We have, = 11.35 g/cm3 and atomic weight M = 207.2 amu
Now, the volume of the lead can be calculated by
There are 6.023 × 1023 atoms present in 207.2 amu for lead and the volume is
Therefore, number of atoms per
= 3.3 × 1022 atom/cm3
85b. 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
Ans. (d)
Sol. Number of atoms per cm3 (n) = 3.3 × 1022atoms/cm3
Energy of vacancy formation (EV) = 0.55 eV and temperature (T) = 500 K
The number of vacancies per cm3 = nV = n exp
= 3.3 × 1022 exp (–12.75) = 9.577 × 1016 vacancies/cm3