1.    In the nuclear reaction , the particle X is

    (a) an electron

    (b) an anti-electron

    (c) a muon

    (d) a pion

Ans.    (a)

Sol.    

    X should be electron (_–1e⁰)

    

    Therefore, charge Q will be conserved also electronic lepton number will be conserved.

2.    Two identical masses of 10 gin each are connected by a massless spring of spring constant 1 N/m. The non-zero angular eigenfrequency of the system is _______ rad/s. (up to two decimal places).

Ans.    14.14

Sol.    

    

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:

    (a)

    (b) 3k_B

    (c)

    (d) 6k_B

Ans.    (d)

Sol.    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 arc due to vibrational motion.

    According to law of equipartition of eneray. each translational and rotational degree of freedom contributes to the energy and each vibrational degree of frceddnTcontributes k_BT to the energy.

    Average energy

    Heat capacity,

4.    For the Hamiltonian where is a real vector, I is the identity matrix and are the Pauli matrices, the ground state energy is

    (a) |b|

    (b) 2a₀ – |b|

    (c) a₀ – |b|

    (d) a₀

Ans.    (c)

Sol.    

    If is the eigenvalue of Hamiltonian, then |H – I| = 0

    

    

    So, lowest energy = a₀ ± |b|

5.    The Poisson bracket [x, xp_y + yp_x] is equal to

    (a) – x

    (b) y

    (c) 2p_x

    (d) p_y

Ans.    (b)

Sol.    [x, xp_y + yp_x] = [x, xp_y] + [x, yp_x]

    = [x, x]p_y + x [x, p_y] + [x, y]p_x + y [x, p_x]

    = p_y × 0 + x × 0 + p_x × 0 + y × 1 = y

6.    The wavefunction of which orbital is spherically symmetric:

    (a) p_x

    (b) p_y

    (c) s

    (d) d_xy

Ans.    (c)

Sol.    For.s-orbitals,

     – 0, m –0, wavefunctions are only function of r. No angular dependence is there, hence s–orbitals are spherically symmetric.

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 ______ × 10^–12 Pa. (up to two decimal places) .

Ans.    8.85

Sol.    The pressure exerted on the mirror

    

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)

Ans.    (a)

Sol.    The energy of the nth state of hydrogen atom

    The highest possible energy state is

    

    E_highest = 0

9.    Consider a one-dimensional lattice with a weak periodic potential . The gap at the edge of the Brillouin zone is:

    (a) U₀

    (b)

    (c) 2U₀

    (d)

Ans.    (a)

Sol.    

    

    

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

    (a) 0, 0

    (b)

    (c)

    (d)

Ans.    (c)

Sol.    

    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 part circle which is present at point 3.

    So,

    And potential, V = V₁ + V₂ + V₃ + V₄ + V₅ =

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

    (a)

    (b)

    (c)

    (d)

Ans.    (b)

Sol.    The efficiency of a reversible Carnot engine is independent of any working substance and depends only on the temperature of the source (T₂) and of the sink (T₁). It is given by .

12.    The best resolution that a 7 bit A/D convertor with 5V full scale can achieve is _______ mV. (up to two decimal places).

Ans.    39.27

Sol.    Given 7 bit A to D converter

    Full scale voltage of 5 volt

    Voltage resolution =

    Resolution = 39.27 mV

13.    If the Lagrangian is modified to , which one of the following is 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

Ans.    (b)

Sol.    

    The given transformation satisfies, gague transformation of Lagrangian therefore equation of motion will not change. Since the term added contains q therefore generalised momentum will change, as seen below

    

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).

Ans.    11.29

Sol.    

    We have, Q =1 C, d = 10^–6 m, A = 1 m², = 8.854 × 10^–12

    

15.    The contour integral evaluated along a contour going from along the real axis and closed in the lower half-plane by a half circle is equal to ________. (up to two decimal places).

Ans.    3.14

Sol.    

    

    

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?

    (a)

    (b)

    (c)

    (d)

Ans.    (a)

Sol.    

    This circuit is a full wave rectifier. So, output wave form

    

    • Output wave form for

    

     Reason: Capacitor Charging and discharging

    

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).

Ans.    41.81

Sol.    

    

    

    

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

Ans.    (a)

Sol.    In electromagnetic interactions, all three C, CP and CPT should be conserved. Therefore, correct option is (a)

19.    The Compton wavelength of a proton is _____ fm. (up to two decimal places).

    (m_p = 1.67 × 10^–27 kg, h = 6.626 × 10^–34 Js, e = 1.602 × 10^–19 C, c = 3 × 10⁸ m s^–1)

Ans.    1.32

Sol.    

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

Ans.    (d)

Sol.    

    Charge is conserved, total lepton number and angular momentum is also conserved.

21.    The coefficient of e^ikx in the Fourier expansion of u(x) = A sin² (ax) for k = is

    (a) A/4

    (b) –A/4

    (c) A/2

    (d) –A/2

Ans.    (b)

Sol.    Fourier series in complex form

    

    

22.    The phase space trajectory of an otherwise free particle bouncing between two hard walls elastically in one dimension is a

    (a) straight line

    (b) parabola

    (c) rectangle

    (d) circle

Ans.    (c)

Sol.    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

    where m is the mass of the particle.

    p = ± (2mE)^1/2

    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

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).

Ans.    2.53 × 10²² cm^–3

Sol.    The number of density of valence electrons

    

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

Ans.    (a)

Sol.    The number of degeneracy of nth energy level of 3-dimension isotropic harmonic oscillator is given by

    = (n + 1) (n + 2) = (3) × (4) (for third energy level n = 2)

25.    A one dimensional simple harmonic oscillator with Hamiltonian is subjected to a small perturbation, . The first order correction to the ground state energy is dependent on

    (a)

    (b)

    (c)

    (d)

Ans.    (d)

Sol.    The first order correction of the ground state energy

    

    

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 direction is ___ cm.

Ans.    0.28

Sol.    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 points is

    

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).

Ans.    0.28

Sol.    

    

28.    The energy density and pressure of a photon gas are given by u = aT⁴ and P = u/3, where T is the temperature and a is the radiation constant. The entropy per unit volume is given by aT³. The value of is _____. (up to two decimal places).

Ans.    1.33

Sol.    

    Using combined I and II law,

    TdS = dU + PdV, we have

    

    Comparing, we have

    

29.    A person weighs w_p at Earth's north pole and w_e at the equator. Treating the Earth as a perfect sphere of radius 6400 km, the value 100 × (w_p – w_e)/w_p is _____. (up to two decimal places).

    (Take g = 10 ms^–2).

Ans.    0.33

Sol.    We know due to rotation of earth, weight of person at latitude is given by

    

    

    

30.    The minimum number of NAND gates required to construct an OR gate is:

    (a) 2

    (b) 4

    (c) 5

    (d) 3

Ans.    (d)

Sol.    Minimum number of N AND gate required for OR-gale

    

    Applying bubble short concept

    

    Note: 3 number of NAND are needed.

31.    The total energy of an inert-gas crystal is given by (in eV), where R is the inter-atomic spacing in Angstroms. The equilibrium separation between the atoms is ________ Angstrons. (up to two decimal places).

Ans.    1

Sol.    

    

32.    The imaginary part of an analytic complex function is v(x, y) = 2xy + 3y. 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).

Ans.    3

Sol.    v(x, y) – 2xy + 3y

    According to Cauchy Reamann equation

    

    u(z = 1 + i) = u(x = 1,y = 1) = 3

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 ?

    (a)

    (b) N

    (c)

    (d)

Ans.    (a)

Sol.    The population of particles in the level with energy is

    

    In the high temperature limit,

    

34.    For the transistor amplifier circuit shown below with R₁ = 10 , R₂ = 10 , R₃ = 1 and = 99. Neglecting the emitter diode resistance, the input impedance of the amplifier looking into the base for small ac signal is _______ . (up to two decimal places).

Ans.    5

Sol.    

    • Its a CC-configuration

    

    Given: emitter diode resistance is zero; = 0

    

    If input impedance is calculated from input (looking from V_i)

    

    If input impedance is calculated from input (looking from V_i)

    R_1(overall) 10K || 10 K 100 = 5

35.    Which one of the following gases of diatomic molecules is Raman, infrared, and NMR active?

    (a) ¹H–¹H

    (b) ¹²C–¹⁶O

    (c) ¹H–³⁵Cl

    (d) ¹⁶O–¹⁶O

Ans.    (c)

Sol.    For the Raman active, diatomic molecules must have changing polarisability. For IR active moleculesmust have change in dipole moment. While for NMR, presence of non-zero nuclear spin is compulsory. Considering given option:

    (a) ¹H – ¹H : There is no change in dipole moment due to symmetry, therefore, it is not IR active.

    (b) ¹²C – ^l6O : Net nuclear spin zero, therefore not NMR active.

    (c) ¹H – ³⁵Cl : It has all the condition available for IR, NMR and Raman active.

    (d) ¹⁶O – ¹⁶O : There is no change in dipole moment, therefore it is not IR active.

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

Ans.    (c)

Sol.    Correct option is (c)

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).

Ans.    0.11

Sol.    The reflection coefficient is given by,

    

38.    An infinite solenoid carries a time varying current I(t) = At², with . the axis of the solenoid is along the direction. are the usual radial and polar directions in cylindrical polar coordinates. is the magnetic field at a point outside the solenoid. Which one of the following statements is true?

    (a)

    (b)

    (c)

    (d)

Ans.    (d)

Sol.    The magnetic field inside the solenoid will be along axis. So, B_z 0, B_r = B₀ = 0

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) ½

    (c) (i) 2/3 and (ii) ½

    (d) (i) 1 and (ii) 0

Ans.    (c)

Sol.    (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. Furthermore since level 1 already contains a particle, we have in total three ways to filled the levels.

    

    Therefore, P (level 2 also contains a particle) =

    (ii) When particles A and Bare 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.

    

    Therefore, P(level 2 also contains a particle) =

40.    The real space primitive lattice vectors are . The reciprocal space unit vectors for this lattice are, respectively

    (a)

    (b)

    (c)

    (d)

Ans.    (a)

Sol.    

    The reciprocal space vector of this lattice are

    

    

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

Ans.    (b)

Sol.    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 r² – 10^–30 m = 10^–2 barn    ( 1 barn = 10^–28 m²)

42.    A uniform volume charge density is placed inside a conductor (with resistivity 10^–2). The charge density becomes 1/(2.718) of its original value after time ______ femto seconds. (up to two decimal places) ( = 8.854 × 10^–12 F/m)

Ans.    88.50

Sol.    

    

43.    Water freezes at 0ºC at atmospheric pressure (1.01 × 10⁵ Pa). The densities of water and ice at this temperature 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)

Ans.    0.173

Sol.    The charge in the temperature of the melting point of a solid dT due to a change of pressure dP is given ba Clapeyron's equtaion

    

    where Z, is the latent heat offusion ofthe solid and (V₂ – V₁) the change in volume, T is melting point

    Specific volume of ice

    Specific volume of water

    

    dT = 10°C = 10K, L = 3.34 × 10⁵ J/kg, T = 273 K

    

    = 0.173 × 10⁹ Pa = 0.173 × GPa

    Correct answer is (0.173)

44.    The integral is equal to _______. (up to two decimal places).

Ans.    0.44

Sol.    

    

 

45.    J^P for the ground state of the ¹³C₆ nucleus is

    (a) 1⁺

    (b)

    (c)

    (d)

Ans.    (d)

Sol.    ¹³C₆

    Proton number P = 6 and Neutron number N = 7

    

    The total angular moment is due to momentum oflast unpaired neutron.

    

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)

Ans.    3.57

Sol.    

    

    

47.    Consider a one-dimensional potential well of width 3 nm. Using the uncertainty principle , an estimate of the minimum depth of the well such that it has at least one bound state for an electron is (m_e = 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

Ans.    (b)

Sol.    

    

    So, the minimum depth of the well should be 1 me V.

48.    Consider an ideal operational amplifier as shown in the figure below with R₁ = 5 , R₂ = 1 , R_L = 100 . For an applied input voltage V = 10 mV, the current passing through R₂ is _______ µA. (up to two decimal places).

Ans.    1

Sol.    Given R₁ = 5

    R₂ = 1

    R_L = 100

    V = 10 mV

    Current through R₂__ µA (micro ampere)

    As ideal op-amp has infinite input resistance. So, there will be no current pass through op-amp.

    So, V₊ = V_– (Virtual ground concept)

    = 10mV

    

49.    The decays at rest to µ⁺ and V_µ. Assuming the neutrino to be momentum of the neutrino is ________ MeV/c. (up to two decimal places)

    (m_x = 139 MeV/c², m_µ = 105 MeV/c²).

Ans.    29.84

Sol.    

    By conservation of momentum, P_µ = P_v                         ...(1)

    By conservation of energy,

     = E_µ + E_v                                        ...(2)

    If m_x, m_µ and 0 are rest masses of , v and v respectively. Then equation (2) becomes

    

50.    Consider the differential equation dy/dx + y tan(x) = cos(x). If y(0) = 0, y(/3) is ______. (up to two decimal places).

Ans.    0.5233

Sol.    Given differential equation is dy

     st order linear differential equation)

        

    Therefore, solution will be

    

    

    

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).

Ans.    2.39

Sol.    We know that the lowest frequency of electromagnetic wave that can propagate through the metal is proportional to square root of density of free electrons.

    

    

52.    Using Hund's rule, the total angular momentum quantum number J for the electronic ground state of the nitrogen atom is

    (a) 1/2

    (b) 3/2

    (c) 0

    (d) 1

Ans.    (b)

Sol.    Electronic configuration of the ₇N = 1s² 2s² sp³ i.e., it has three electrons outside closed shells.

    Chart of nitrogen statem

        

    

    i.e., total spin S = . multiplicity

    Total 1 = 0, i.e. s-orbital

    Total angular momentum

    Therefore, ground state term is ⁴S_3/2.

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).

    (m_e = 9.31 × 10^–31 kg, h = 6.626 × 10^–34 J s, e = 1.602 × 10^–19 C)

Ans.    2.34

Sol.    

    = 0.3756 × 10^–18 J = 0.2345 × 10 eV = 2.34 eV

54.    Which one of the following operators is Hermitian?

    (a)

    (b)

    (c)

    (d)

Ans.    (a)

Sol.    

55.    Positronium is an atom made of an electron and a positron. Given 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).

Ans.    1.06

Sol.    Radius ofthe Bohr orbit is ; reduced mass of the hydrogen like system

    For hydrogen atom: m_p = 1836 m_e