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CSIR NET Physics June 2026 Solved Paper (Part B & C) Step-by-Step

A.M August 28, 2026

CSIR NET June 2026 Physical Sciences: Answer Key & Detailed Step-by-Step Solutions

Comprehensive, verified solutions for Part B and Part C Physics paper. Features accurate vector diagrams, step-by-step mathematical proofs, and detailed explanations optimized for learning.

Question 21: Radial Spin Matrix Expectation

Quantum Mechanics

Consider the spin matrix $S_r$ projected along the radial direction in spherical polar coordinates $(r,\theta,\phi)$. The probability that a measurement of $S_r$ on the state $|\beta\rangle = \begin{pmatrix} \sin\frac{\theta}{2} \\ e^{i\phi}\cos\frac{\theta}{2} \end{pmatrix}$ will yield the value $+\frac{\hbar}{2}$ is:[cite: 1]

  1. $\sin^2\theta$[cite: 1]
  2. $\cos^2\theta$[cite: 1]
  3. $\frac{1}{2}$[cite: 1]
  4. $\cos^2\frac{\theta}{2}$[cite: 1]

Step-by-Step Solution:

Step 1 (Eigenspinor definition): The normalized eigenspinor of the radial spin operator $S_r = \vec{S}\cdot \hat{r}$ corresponding to the spin-up eigenvalue $+\frac{\hbar}{2}$ is $|\chi_+\rangle = \begin{pmatrix} \cos\frac{\theta}{2} \\ e^{i\phi}\sin\frac{\theta}{2} \end{pmatrix}$.
Step 2 (Inner product amplitude): Compute the transition overlap amplitude $\langle \chi_+ | \beta \rangle$: $$\langle \chi_+ | \beta \rangle = \begin{pmatrix} \cos\frac{\theta}{2} & e^{-i\phi}\sin\frac{\theta}{2} \end{pmatrix} \begin{pmatrix} \sin\frac{\theta}{2} \\ e^{i\phi}\cos\frac{\theta}{2} \end{pmatrix} = \sin\frac{\theta}{2}\cos\frac{\theta}{2} + \sin\frac{\theta}{2}\cos\frac{\theta}{2} = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} = \sin\theta$$
Step 3 (Probability): The measurement probability is $P = |\langle \chi_+ | \beta \rangle|^2 = \sin^2\theta$.
Correct Answer: Option 1 ($\sin^2\theta$)

Question 22: Polarization of Reflected Wave at Anisotropic Interface

Electrodynamics

A linearly polarized electromagnetic wave is incident normally on a material of refractive index $n_R$ for right circular polarization and $n_L$ for left circular polarization, where $n_R \neq n_L$. The reflected wave's polarization is:[cite: 1]

  1. Linear[cite: 1]
  2. Elliptical[cite: 1]
  3. Circular[cite: 1]
  4. Random[cite: 1]

Step-by-Step Solution:

Step 1: A linearly polarized wave is a symmetric superposition of equal-amplitude right circular (RCP) and left circular (LCP) waves.
Step 2: At normal incidence, the Fresnel reflection coefficients for the two states are $r_R = \frac{1-n_R}{1+n_R}$ and $r_L = \frac{1-n_L}{1+n_L}$.
Step 3: Since $n_R \neq n_L$, the reflection coefficients differ in magnitude ($|r_R| \neq |r_L|$). Recombining two counter-rotating circular waves of unequal amplitudes produces an elliptically polarized wave.
Correct Answer: Option 2 (Elliptical)

Question 23: Microstates in Micro-canonical Ensemble

Statistical Mechanics

A micro-canonical ensemble consists of six non-interacting and distinguishable spin 1 particles in a uniform magnetic field. Each particle can thus be in energy states $-E_0$, $0$ and $E_0$. The number of microstates in the state with total energy zero is:[cite: 1]

  1. 51[cite: 1]
  2. 191[cite: 1]
  3. 141[cite: 1]
  4. 55[cite: 1]

Step-by-Step Solution:

Step 1: Let $n_-$, $n_0$, and $n_+$ denote the occupation numbers of states $-E_0$, $0$, and $+E_0$. We require $n_- + n_0 + n_+ = 6$ and total energy $E = (-n_- + n_+)E_0 = 0 \implies n_- = n_+$.
Step 2: For distinguishable particles, the number of microstates is given by the multinomial coefficient $W = \frac{6!}{n_-! n_0! n_+!}$.
Step 3: Sum over all valid configurations:
• $(n_-, n_0, n_+) = (0, 6, 0) \implies W_1 = \frac{6!}{0!6!0!} = 1$
• $(1, 4, 1) \implies W_2 = \frac{6!}{1!4!1!} = 30$
• $(2, 2, 2) \implies W_3 = \frac{6!}{2!2!2!} = 90$
• $(3, 0, 3) \implies W_4 = \frac{6!}{3!0!3!} = 20$
Total microstates $\Omega = 1 + 30 + 90 + 20 = 141$.
Correct Answer: Option 3 (141)

Question 24: Quantum and Classical State Counting

Statistical Mechanics

Two particles occupy ten degenerate energy levels. There are three possible scenarios where the two particles are either identical bosons or identical fermions or distinguishable particles, with the associated number of accessible microstates denoted as $\Omega_B$, $\Omega_F$ and $\Omega_c$, respectively. Then, which of the following options is correct?[cite: 1]

  1. $\Omega_B=55,\Omega_F=45,\Omega_C=100$[cite: 1]
  2. $\Omega_B=65,\Omega_F=55,\Omega_C=100$[cite: 1]
  3. $\Omega_B=45,\Omega_F=25,\Omega_C=55$[cite: 1]
  4. $\Omega_B=55,\Omega_F=45,\Omega_C=75$[cite: 1]

Step-by-Step Solution:

Step 1 (Distinguishable): Each of the $N=2$ particles independently chooses from $g=10$ levels: $\Omega_C = g^N = 10^2 = 100$.
Step 2 (Fermions): Obey Pauli exclusion (choose 2 distinct levels out of 10): $\Omega_F = \binom{g}{N} = \binom{10}{2} = \frac{10 \times 9}{2} = 45$.
Step 3 (Bosons): Symmetric distribution allows multiple occupancy: $\Omega_B = \binom{g+N-1}{N} = \binom{10+2-1}{2} = \binom{11}{2} = 55$.
Correct Answer: Option 1 ($\Omega_B=55,\Omega_F=45,\Omega_C=100$)

Question 25: Clausius-Clapeyron Equation for Ice-Water System

Thermodynamics

Water, having density $1\text{ gm/cm}^3$, freezes into ice at 273 K temperature and 1 atm pressure. The latent heat of melting of ice is 334 $J/gm$ and the density of ice is $0.917\text{ gm/cm}^3$. Then for the melting temperature $(T_m)$ which of the following options is correct?[cite: 1]

  1. At 140 atm pressure, $T_m \simeq 274\text{ K}$[cite: 1]
  2. At 420 atm pressure, $T_m \simeq 274\text{ K}$[cite: 1]
  3. At 140 atm pressure, $T_m \simeq 272\text{ K}$[cite: 1]
  4. At 420 atm pressure, $T_m \simeq 272\text{ K}$[cite: 1]

Step-by-Step Solution:

Step 1: Use Clausius-Clapeyron: $\frac{dP}{dT} = \frac{L}{T(v_{\text{water}} - v_{\text{ice}})}$.
Step 2: Specific volumes: $v_{\text{water}} = 1.0 \times 10^{-3}\text{ m}^3/\text{kg}$, $v_{\text{ice}} = \frac{1}{917}\text{ m}^3/\text{kg} = 1.0905 \times 10^{-3}\text{ m}^3/\text{kg} \implies \Delta v = -0.0905 \times 10^{-3}\text{ m}^3/\text{kg}$.
Step 3: $\frac{dP}{dT} = \frac{334 \times 10^3}{273 \times (-0.0905 \times 10^{-3})} \approx -1.35 \times 10^7\text{ Pa/K} \approx -133.4\text{ atm/K} \approx -140\text{ atm/K}$.
For $\Delta T = -1\text{ K}$ (lowering $T_m$ to 272 K), $\Delta P \approx +140\text{ atm}$.
Correct Answer: Option 3 (At 140 atm pressure, $T_m \simeq 272\text{ K}$)

Question 26: Stern-Gerlach Deflection Force

Atomic Physics

A beam of copper atoms $(Cu:[Ar]4s^1 3d^{10})$ is passed through a Stern-Gerlach setup (see figure). If the gradient of the z-component of the magnetic field in the z-direction is $10^3\text{ Tesla/m}$, then the magnitude of the z-component of the force on an atom is: ($\mu_B$ is the Bohr magneton)[cite: 1]

z y S N
  1. $10^3\mu_B$[cite: 1]
  2. $5\times10^2\mu_B$[cite: 1]
  3. 0[cite: 1]
  4. $2\pi\times10^3\mu_B$[cite: 1]

Step-by-Step Solution:

Step 1: Cu ground state has one unpaired electron in $4s^1 \implies L=0, S=1/2, J=1/2$, and Landé g-factor $g_J = 2$.
Step 2: Effective magnetic moment: $\mu_z = g_J \mu_B m_s = 2 \mu_B (1/2) = \mu_B$.
Step 3: Force: $|F_z| = |\mu_z \frac{\partial B_z}{\partial z}| = \mu_B \times 10^3 = 10^3 \mu_B$.
Correct Answer: Option 1 ($10^3\mu_B$)

Question 27: LS Coupling Ground State of Vanadium

Atomic Physics

Vanadium has electron configuration $[\text{Ar}]3d^3 4s^2$. In LS coupling, its ground state configuration is[cite: 1]

  1. ${}^4F_0$[cite: 1]
  2. ${}^2P_{1/2}$[cite: 1]
  3. ${}^4F_{3/2}$[cite: 1]
  4. ${}^4S_{3/2}$[cite: 1]

Step-by-Step Solution:

Step 1 (Hund's Rule 1): Maximize spin $S$ for $3d^3$: $m_s = +1/2, +1/2, +1/2 \implies S = 3/2 \implies 2S+1 = 4$.
Step 2 (Hund's Rule 2): Maximize orbital momentum $L$: $m_l = +2, +1, 0 \implies L = 3$ (F state).
Step 3 (Hund's Rule 3): Subshell is less than half-filled ($3 < 5$), hence minimal $J = |L - S| = 3 - 3/2 = 3/2$. The term is ${}^4F_{3/2}$.
Correct Answer: Option 3 (${}^4F_{3/2}$)

Question 28: Time-Independent Quantum Observables

Quantum Mechanics

You are given a time-independent observable O for a system with time independent Hamiltonian H and a general time dependent state $|\phi(t)\rangle$ If the expectation value $\langle O(t)\rangle_\phi = \langle\phi(t)|O|\phi(t)\rangle$, then consider the following statements (A) to (D) and choose the correct option.[cite: 1]
(A) If O commutes with H then $\langle O(t)\rangle_\phi$ is independent of time.[cite: 1]
(B) If $|\phi(t)\rangle$ is an eigen state of H then $\langle O(t)\rangle_\phi$ is independent of time.[cite: 1]
(C) If $|\phi(t)\rangle$ is an eigen state of O then $\langle O(t)\rangle_\phi$ is independent of time.[cite: 1]
(D) If $|\phi(t)\rangle$ is an eigen state of [O, H] then $\langle O(t)\rangle_\phi$ is independent of time.[cite: 1]

  1. Only (A) and (B) are correct[cite: 1]
  2. Only (A) is correct[cite: 1]
  3. Only (B), (C) and (D) are correct[cite: 1]
  4. Only (A), (B) and (C) are correct[cite: 1]

Step-by-Step Solution:

Step 1: Ehrenfest's theorem: $\frac{d}{dt}\langle O \rangle = \frac{i}{\hbar}\langle [H, O] \rangle + \langle \frac{\partial O}{\partial t} \rangle$. If $[O, H]=0$, $\frac{d}{dt}\langle O\rangle = 0 \implies$ (A) is true.
Step 2: If $|\phi(t)\rangle$ is an energy eigenstate, time dependence is an overall phase factor $e^{-iEt/\hbar}$, which cancels in $\langle \phi(t)|O|\phi(t)\rangle \implies$ (B) is true.
Correct Answer: Option 1 (Only A and B are correct)

Question 29: Normal Modes of 3-Mass Coupled System

Classical Mechanics

Three equal masses m are free to move along x-axis on a frictionless surface. The masses are coupled together, and to a rigid wall, by springs as shown in the figure. If $\omega_0=\sqrt{\frac{k}{m}}$ and the angular frequencies of the three normal modes of this system are $\omega_1$, $\omega_2$ and $\omega_3$ then the value of $\omega_1^2+\omega_2^2+\omega_3^2$ is:[cite: 1]

2k k k 2k m m m
  1. $9\omega_0^2$[cite: 1]
  2. $6\omega_0^2$[cite: 1]
  3. $10\omega_0^2$[cite: 1]
  4. $8\omega_0^2$[cite: 1]

Step-by-Step Solution:

Step 1: Diagonal elements of stiffness matrix $K$:
• $K_{11} = 2k + k = 3k$
• $K_{22} = k + k = 2k$
• $K_{33} = k + 2k = 3k$
Step 2: Sum of eigenvalues equals the trace of the dynamical matrix $M^{-1}K$: $$\sum \omega_i^2 = \text{Tr}(M^{-1}K) = \frac{3k + 2k + 3k}{m} = 8\frac{k}{m} = 8\omega_0^2$$
Correct Answer: Option 4 ($8\omega_0^2$)

Question 30: Sudden vs Adiabatic Expansion of Infinite Well

Quantum Mechanics

A particle is in the ground state of a one-dimensional infinite well potential with walls at $x=0$ and at $x=a$. The wall at $x=a$ is moved to $x=2a$ in two separate, sudden and adiabatic, protocols so that $\langle E\rangle_s$ and $\langle E\rangle_a$ are the respective expectation values of the energy in the final states. Then the ratio $\frac{\langle E\rangle_s}{\langle E\rangle_a}$ is:[cite: 1]

  1. 4:1[cite: 1]
  2. 2:1[cite: 1]
  3. 1:1[cite: 1]
  4. 1:4[cite: 1]

Step-by-Step Solution:

Step 1 (Sudden): State vector does not instantly change; energy expectation value is preserved: $\langle E\rangle_s = E_1(a) = \frac{\pi^2\hbar^2}{2ma^2}$.
Step 2 (Adiabatic): Particle remains in the ground state of the new wider well: $\langle E\rangle_a = E_1(2a) = \frac{\pi^2\hbar^2}{2m(2a)^2} = \frac{E_1(a)}{4}$.
Step 3: Ratio $\frac{\langle E\rangle_s}{\langle E\rangle_a} = \frac{E_1(a)}{E_1(a)/4} = 4:1$.
Correct Answer: Option 1 (4:1)

Question 31: Line Integral of Canonical Momentum

Classical Mechanics

A particle with charge q, which is confined to move in the xy-plane, is subjected to a magnetic field with magnitude B pointing in the +z direction. The particle moves in a circle (C) of radius R. If $\vec{p}$ is the canonical momentum then the following integral $\oint_C\vec{p}\cdot d\vec{\ell}$ has the magnitude: (d$\vec{\ell}$ is in the direction of motion)[cite: 1]

  1. $qB\pi R^2$[cite: 1]
  2. $2qB\pi R^2$[cite: 1]
  3. $3qB\pi R^2$[cite: 1]
  4. $4qB\pi R^2$[cite: 1]

Step-by-Step Solution:

Step 1: Canonical momentum is $\vec{p} = m\vec{v} + q\vec{A}$.
Step 2: Mechanical part: $mv = qBR \implies \oint m\vec{v}\cdot d\vec{\ell} = (qBR)(2\pi R) = 2\pi qBR^2$.
Step 3: Vector potential part by Stokes' theorem: $q\oint \vec{A}\cdot d\vec{\ell} = -q\iint B\,da = -qB(\pi R^2)$. Total magnitude $= |2\pi qBR^2 - \pi qBR^2| = qB\pi R^2$.
Correct Answer: Option 1 ($qB\pi R^2$)

Question 36: Survival Probability in 3-Level System

Quantum Mechanics

A three-state system is described by a certain Hamiltonian having eigen energies $E_0$, $2E_0$ and $3E_0$ with the corresponding eigen states $\begin{pmatrix}1\\ 1\\ 1\end{pmatrix},\begin{pmatrix}1\\ 0\\ -1\end{pmatrix}$ and $\begin{pmatrix}1\\ -2\\ 1\end{pmatrix}$ respectively. If the state of the system at time $t=0$ is $\begin{pmatrix}1\\ 0\\ 0\end{pmatrix},$ then what is the probability of finding the system in the same state at time $t=\frac{\pi \hbar}{2E_0}$?[cite: 1]

  1. 1[cite: 1]
  2. $\frac{1}{3}$[cite: 1]
  3. $\frac{5}{18}$[cite: 1]
  4. $\frac{2}{9}$[cite: 1]

Step-by-Step Solution:

Step 1: Normalized orthonormal basis vectors are $|1\rangle = \frac{1}{\sqrt{3}}(1,1,1)^T$, $|2\rangle = \frac{1}{\sqrt{2}}(1,0,-1)^T$, and $|3\rangle = \frac{1}{\sqrt{6}}(1,-2,1)^T$. Initial state $|\psi(0)\rangle$ overlaps: $c_1 = 1/\sqrt{3}$, $c_2 = 1/\sqrt{2}$, $c_3 = 1/\sqrt{6}$.
Step 2: Time phases at $t = \frac{\pi\hbar}{2E_0}$: $e^{-i\pi/2} = -i$, $e^{-i\pi} = -1$, $e^{-i3\pi/2} = i$.
Step 3: Overlap amplitude: $A(t) = \frac{1}{3}(-i) + \frac{1}{2}(-1) + \frac{1}{6}(i) = -\frac{1}{2} - \frac{i}{6}$. Probability $P = |A(t)|^2 = \frac{1}{4} + \frac{1}{36} = \frac{10}{36} = \frac{5}{18}$.
Correct Answer: Option 3 ($\frac{5}{18}$)

Question 37: Free Expansion of Real Gas

Thermodynamics

Internal energy of one mole of a non-ideal gas is given by $U=\frac{3}{2}RT-\frac{a}{V}$ where V is volume of the gas at temperature T and a is a positive constant. An insulated container of volume $V_2$ contains this gas confined in a volume $V_1$ at temperature $T_1$. The remaining volume $V_2-V_1$ of this container, which is in vacuum, is separated by a wall. The wall is suddenly removed which makes the gas volume $V_2$ and temperature $T_2$. The temperature $T_2$ is: (R is the molar gas constant)[cite: 1]

  1. $T_1+\frac{2a}{3R}(\frac{1}{V_2}+\frac{1}{V_1})$[cite: 1]
  2. $T_1-\frac{2a}{3R}(\frac{1}{V_2}+\frac{1}{V_1})$[cite: 1]
  3. $T_1+\frac{2a}{3R}(\frac{1}{V_2}-\frac{1}{V_1})$[cite: 1]
  4. $T_1-\frac{2a}{3R}(\frac{1}{V_2}-\frac{1}{V_1})$[cite: 1]

Step-by-Step Solution:

Step 1: Free expansion into vacuum entails $W = 0$ and $Q = 0 \implies \Delta U = 0 \implies U_1 = U_2$.
Step 2: $\frac{3}{2}RT_1 - \frac{a}{V_1} = \frac{3}{2}RT_2 - \frac{a}{V_2} \implies \frac{3}{2}R(T_2 - T_1) = a\left(\frac{1}{V_2} - \frac{1}{V_1}\right)$.
Step 3: $T_2 = T_1 + \frac{2a}{3R}\left(\frac{1}{V_2} - \frac{1}{V_1}\right)$.
Correct Answer: Option 3 ($T_1+\frac{2a}{3R}(\frac{1}{V_2}-\frac{1}{V_1})$)

Question 38: Bead on a Driven Rotating Wire

Classical Mechanics

A bead of mass M slides along a smooth frictionless and massless wire which is bent in the shape of a parabola $z=cr^2$ in cylindrical coordinates $(r,\phi,z)$. Here c is a positive constant. The wire is made to rotate about z-axis at constant angular velocity. Which of the following statements about this system of the wire and the bead is correct?[cite: 1]

  1. Energy is conserved but angular momentum is not conserved[cite: 1]
  2. Energy is not conserved but angular momentum is conserved[cite: 1]
  3. Both energy and angular momentum are conserved[cite: 1]
  4. Both energy and angular momentum are not conserved[cite: 1]

Step-by-Step Solution:

Step 1: Maintaining constant $\dot{\phi} = \omega$ requires external driving torque $\implies L_z$ is not conserved.
Step 2: The driving mechanism performs non-zero work on the system over time $\implies$ mechanical energy $E$ is not conserved.
Correct Answer: Option 4 (Both energy and angular momentum are not conserved)

Question 39: CMOS Digital Logic Identification

Electronics

The circuit shown below could be used as a:[cite: 1]

+5 V In out
  1. TTL NOT Gate[cite: 1]
  2. TTL Buffer[cite: 1]
  3. CMOS NOT Gate[cite: 1]
  4. CMOS Buffer[cite: 1]

Step-by-Step Solution:

Step 1: The circuit pairs a pull-up PMOS connected to $V_{DD}$ with a pull-down NMOS connected to ground.
Step 2: HIGH input activates NMOS (output pulled to 0 V); LOW input activates PMOS (output pulled to +5 V). This corresponds to a NOT Gate (inverter).
Correct Answer: Option 3 (CMOS NOT Gate)

Question 40: Derivative of Unit Vector in Spherical Coordinates

Mathematical Physics

In spherical polar coordinates $(r,\theta,\phi)$, the derivative $\frac{\partial\hat{\theta}}{\partial\phi}$ evaluates to:[cite: 1]

  1. $\cos\theta\hat{\phi}$[cite: 1]
  2. $0$[cite: 1]
  3. $\sin\theta\hat{\phi}$[cite: 1]
  4. $-\cos\theta\hat{r}$[cite: 1]

Step-by-Step Solution:

Step 1: Write unit vector: $\hat{\theta} = \cos\theta\cos\phi\hat{i} + \cos\theta\sin\phi\hat{j} - \sin\theta\hat{k}$.
Step 2: Differentiate with respect to $\phi$: $\frac{\partial\hat{\theta}}{\partial\phi} = \cos\theta(-\sin\phi\hat{i} + \cos\phi\hat{j}) = \cos\theta\hat{\phi}$.
Correct Answer: Option 1 ($\cos\theta\hat{\phi}$)

Question 41: Expectation Value of Superposed State

Quantum Mechanics

The eigen energies and the normalized eigenfunctions in the $n^{th}$ state of a particle, confined in a one-dimensional Infinite potential well, are denoted by $E_n$ and $\psi_n$, respectively. If $E_1=1.2$ eV then the expectation value of energy in the state $\psi=3\psi_1+2\psi_2+\sqrt{3}\psi_4$ is close to:[cite: 1]

  1. 5.5 eV[cite: 1]
  2. 5.0 eV[cite: 1]
  3. 4.5 eV[cite: 1]
  4. 6.0 eV[cite: 1]

Step-by-Step Solution:

Step 1: Energy levels: $E_n = n^2 E_1 \implies E_1 = 1.2\text{ eV}, E_2 = 4.8\text{ eV}, E_4 = 19.2\text{ eV}$.
Step 2: Norm squared $\langle \psi|\psi\rangle = 3^2 + 2^2 + (\sqrt{3})^2 = 9 + 4 + 3 = 16$.
Step 3: $\langle E\rangle = \frac{9(1.2) + 4(4.8) + 3(19.2)}{16} = \frac{10.8 + 19.2 + 57.6}{16} = \frac{87.6}{16} \approx 5.475\text{ eV} \approx 5.5\text{ eV}$.
Correct Answer: Option 1 (5.5 eV)

Question 42: Basis Completion in 4D Vector Space

Mathematical Physics

In a 4-dimensional vector space $V_4$, three linearly independent vectors are given by $e_1=(1,1,0,0)$, $e_2=(0,1,1,0)$ and $e_3=(0,0,1,1).$ Which of the following vector $e_4$ will make $\{e_1,e_2,e_3,e_4\}$ a basis of $V_4$?[cite: 1]

  1. $(1,1,1,1)$[cite: 1]
  2. $(1,1,1,0)$[cite: 1]
  3. $(1,0,-1,0)$[cite: 1]
  4. $(0,1,1,0)$[cite: 1]

Step-by-Step Solution:

Step 1: Check linear dependence: $(1,1,1,1) = e_1 + e_3$, $(1,0,-1,0) = e_1 - e_2$, and $(0,1,1,0) = e_2$.
Step 2: Vector $(1,1,1,0)$ yields $\det \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{pmatrix} = -1 \neq 0$, confirming linear independence.
Correct Answer: Option 2 ($(1,1,1,0)$)

Question 43: Central Force for Logarithmic Spiral Orbit

Classical Mechanics

A particle moves in an orbit $r = k \exp(a\theta)$ under a central force $F(r)$. Here, k and a are constants and $k>0$ Then $F(r)$ is proportional to:[cite: 1]

  1. $\frac{1}{r}$[cite: 1]
  2. $\frac{1}{r^3}$[cite: 1]
  3. $\frac{1}{r^4}$[cite: 1]
  4. $\frac{1}{r^2}$[cite: 1]

Step-by-Step Solution:

Step 1: Let $u = 1/r = \frac{1}{k}e^{-a\theta}$. Derivatives: $u' = -au$, $u'' = a^2u$.
Step 2: Binet's equation: $\frac{d^2u}{d\theta^2} + u = -\frac{m}{L^2 u^2}F(1/u) \implies (a^2+1)u \propto -\frac{F(1/u)}{u^2} \implies F(r) \propto u^3 = \frac{1}{r^3}$.
Correct Answer: Option 2 ($\frac{1}{r^3}$)

Question 44: Contour Integral on Unit Circle

Mathematical Physics

The value of the integral: $\oint\frac{d\theta}{5-4\cos\theta}$ over the circle $|z|=1$ in the anti-clockwise direction can be calculated by substituting $z=e^{i\theta}$. The value of the integral is:[cite: 1]

  1. $\frac{2\pi}{3}$[cite: 1]
  2. $\frac{\pi}{3}$[cite: 1]
  3. $\pi$[cite: 1]
  4. $\frac{4\pi}{3}$[cite: 1]

Step-by-Step Solution:

Step 1: Substitute $d\theta = dz/(iz)$ and $\cos\theta = \frac{z+z^{-1}}{2}$: $I = \frac{i}{2}\oint_{|z|=1} \frac{dz}{(z-1/2)(z-2)}$.
Step 2: Enclosed pole inside $|z|=1$ is $z=1/2$. Residue: $\text{Res}(1/2) = \frac{1}{1/2 - 2} = -\frac{2}{3}$.
Step 3: Integral: $I = \frac{i}{2} \cdot 2\pi i \left(-\frac{2}{3}\right) = \frac{2\pi}{3}$.
Correct Answer: Option 1 ($\frac{2\pi}{3}$)

Question 45: Operator Commutator Identity

Quantum Mechanics

Consider two operators R and S where R commutes with $[R,S].$ If $\alpha$ is a constant, then $[e^{\alpha \hat{R}},\hat{S}]$ is:[cite: 1]

  1. $[R^\alpha,S]$[cite: 1]
  2. $\alpha e^{\alpha\hat{R}}[\hat{R},\hat{S}]$[cite: 1]
  3. 0[cite: 1]
  4. $\alpha e^{\hat{R}}[e^{\alpha}\hat{R},\hat{S}]$[cite: 1]

Step-by-Step Solution:

Step 1: For $[R, [R, S]] = 0$, the identity $[f(R), S] = f'(R)[R,S]$ holds.
Step 2: Differentiating $f(R) = e^{\alpha R}$ gives $f'(R) = \alpha e^{\alpha R}$. Thus, $[e^{\alpha R}, S] = \alpha e^{\alpha R}[R,S]$.
Correct Answer: Option 2 ($\alpha e^{\alpha\hat{R}}[\hat{R},\hat{S}]$)

Question 46: 1D Biased Random Walk

Statistical Mechanics

A biased random walker confined to the x-axis moves, at each step, with probability 0.6 to the right and with probability 0.4 to the left. If the walker takes 1000 steps starting from the origin, the probability that the walker is found exactly 200 steps to the right of the origin is close to:[cite: 1]

  1. 0.026[cite: 1]
  2. 0.125[cite: 1]
  3. 0.250[cite: 1]
  4. 0.375[cite: 1]

Step-by-Step Solution:

Step 1: Position $x = n_R - n_L = 200$, with $n_R + n_L = 1000 \implies n_R = 600$.
Step 2: Mean $\mu = Np = 600$, Variance $\sigma^2 = Npq = 1000(0.6)(0.4) = 240 \implies \sigma = \sqrt{240} \approx 15.49$.
Step 3: Probability at peak: $P(600) \approx \frac{1}{\sqrt{2\pi}\sigma} = \frac{1}{\sqrt{2\pi}(15.49)} \approx 0.0258 \approx 0.026$.
Correct Answer: Option 1 (0.026)

Question 47: High Frequency Modulation Purpose

Electronics

Modulation techniques are used to shift a signal to higher frequencies to:[cite: 1]

  1. remove the effects of white noise.[cite: 1]
  2. reduce the effects of $1/f$-noise and drift.[cite: 1]
  3. reduce shot noise and Johnson noise.[cite: 1]
  4. smoothen the signal and amplify it.[cite: 1]

Step-by-Step Solution:

Step 1: Flicker noise ($1/f$ noise) and baseline drift have highest spectral density near DC (low frequencies).
Step 2: Shifting the signal to a higher frequency carrier moves it to a regime where $1/f$ noise is negligible.
Correct Answer: Option 2 (reduce the effects of $1/f$-noise and drift.)

Question 48: Independent Components of Rank-4 Tensor

Mathematical Physics

In four dimensions, a rank four tensor is given by $T_{abcd}$. It is symmetric under interchange of the indices a, b and interchange of the indices c, d. Moreover, the tensor is anti-symmetric under the interchange of pair of indices ab and cd. The number of independent components of the tensor is:[cite: 1]

  1. 100[cite: 1]
  2. 45[cite: 1]
  3. 36[cite: 1]
  4. 21[cite: 1]

Step-by-Step Solution:

Step 1: Symmetric pair dimension in $D=4$: $N = \frac{4(5)}{2} = 10$.
Step 2: Anti-symmetric combination of two 10-dimensional indices: $\frac{N(N-1)}{2} = \frac{10(9)}{2} = 45$.
Correct Answer: Option 2 (45)

Question 49: Relativistic Electric Field Transformation

Electrodynamics

An inertial observer S finds the electric field due to a large rectangular parallel plate capacitor to be directed along the y-axis. For another observer, moving at a relativistic speed with respect to S along the positive x-axis, which of the following statements is correct for the electric field:[cite: 1]

  1. Direction remains same but magnitude increases[cite: 1]
  2. Direction remains same but magnitude decreases[cite: 1]
  3. Direction and magnitude both remain the same[cite: 1]
  4. Direction tilts towards x-axis but the magnitude remains same[cite: 1]

Step-by-Step Solution:

Step 1: Lorentz transformation: $E'_x = E_x = 0$, $E'_y = \gamma(E_y - vB_z) = \gamma E_y$, $E'_z = 0$.
Step 2: Since $\gamma > 1$, the electric field remains oriented purely along the y-axis while its magnitude increases.
Correct Answer: Option 1 (Direction remains same but magnitude increases)

Question 50: Einstein Coefficients and Spectral Density

Statistical Mechanics

A container has a gas at temperature T' with its $N_1$ molecules in the ground state and $N_2$ molecules in the excited state of energy $h\nu$. Both the levels are non-degenerate and A and B are the Einstein coefficients for spontaneous and stimulated emissions, respectively, between the two levels. Then the spectral density $\rho(\nu,T)$ of radiation inside the container is proportional to:[cite: 1]

  1. $\frac{A}{B}(\frac{N_2}{N_1-N_2})$[cite: 1]
  2. $\frac{A}{B}(\frac{N_2}{N_1+N_2})$[cite: 1]
  3. $\frac{N_2}{(\frac{A}{B})N_1-N_2}$[cite: 1]
  4. $\frac{N_2}{(\frac{A}{B})N_1+N_2}$[cite: 1]

Step-by-Step Solution:

Step 1: Rate equilibrium: $N_1 B \rho(\nu) = N_2 A + N_2 B \rho(\nu)$.
Step 2: $\rho(\nu)[N_1 B - N_2 B] = N_2 A \implies \rho(\nu) = \frac{A}{B}\left(\frac{N_2}{N_1 - N_2}\right)$.
Correct Answer: Option 1 ($\frac{A}{B}(\frac{N_2}{N_1-N_2})$)

Question 51: Numerical Integration of $x^4$

Mathematical Physics

The integral $\int_{0}^{4}x^{4}dx$ when computed numerically using Simpson's $1/3^{rd}$ method and trapezoidal method, each with unit step size, yields values $I_s$ and $I_t$ respectively. Then which of the following options is correct?[cite: 1]

  1. $I_s=205.3, I_t=226.0$[cite: 1]
  2. $I_s=205.3, I_t=205.3$[cite: 1]
  3. $I_s=204.8, I_t=226.0$[cite: 1]
  4. $I_s=226.0, I_t=226.0$[cite: 1]

Step-by-Step Solution:

Step 1: Values: $y_0 = 0, y_1 = 1, y_2 = 16, y_3 = 81, y_4 = 256$.
Step 2: Trapezoidal: $I_t = \frac{1}{2}[0 + 2(1+16+81) + 256] = 226.0$.
Step 3: Simpson's: $I_s = \frac{1}{3}[0 + 4(1+81) + 2(16) + 256] = \frac{616}{3} \approx 205.3$.
Correct Answer: Option 1 ($I_s=205.3, I_t=226.0$)

Question 52: Magnetic Moment of Oxygen-17

Nuclear Physics

The table shows the g-factors of proton (p) and neutron (n) for orbital $(g_L)$ and spin $(g_s)$ angular momentum. The magnetic moment of ${}^{17}O$ nucleus, according to the single particle shell model, in units of nuclear magneton $(\mu_N)$ is:[cite: 1]

  1. 0.55[cite: 1]
  2. 4.79[cite: 1]
  3. -1.91[cite: 1]
  4. -3.82[cite: 1]

Step-by-Step Solution:

Step 1: Valence 9th neutron resides in $1d_{5/2}$ orbital ($l=2, j=5/2 = l+1/2$).
Step 2: For $j = l + 1/2$: $\mu = [l g_L + \frac{1}{2}g_S]\mu_N = [0 + \frac{1}{2}(-3.82)]\mu_N = -1.91\mu_N$.
Correct Answer: Option 3 (-1.91)

Question 53: Reduced Chi-Square Calculation

General Aptitude / Data

Given is a set of 5 data points of the form $(x_i,y_i,\sigma_i)$ with $\sigma_i$ as the estimated uncertainty in $y_1$. This data is fitted to the form $y=ax+b$ and the minimized $\chi^2$ (chi-square) is found to be 5.1. The reduced $\chi^2$ for this fitting is:[cite: 1]

  1. 1.0[cite: 1]
  2. 1.2[cite: 1]
  3. 1.5[cite: 1]
  4. 1.7[cite: 1]

Step-by-Step Solution:

Step 1: Degrees of freedom $\nu = N - p = 5 - 2 = 3$.
Step 2: $\chi_{\text{red}}^2 = \frac{\chi^2}{\nu} = \frac{5.1}{3} = 1.7$.
Correct Answer: Option 4 (1.7)

Question 54: Coulomb Energy Difference in Mirror Nuclei

Nuclear Physics

Assume a spherically uniform charge distribution in a nucleus having mass number A and radius $R=R_0A^{1/3}$ with $R_0\approx1.5\text{ fm}$. If the difference in electrostatic energy between two mirror nuclei, ${}^A_Z X$ and ${}^A_{Z-1} Y$ is 3.53 MeV, then the value of atomic number Z is closest to: (Given: $\frac{e^2}{4\pi\varepsilon_0}\approx1.44\text{ MeV fm}$ and $1\text{ fm}=10^{-15}\text{m}$)[cite: 1]

  1. 6[cite: 1]
  2. 8[cite: 1]
  3. 10[cite: 1]
  4. 19[cite: 1]

Step-by-Step Solution:

Step 1: $\Delta E_C = \frac{3}{5}\frac{e^2}{4\pi\varepsilon_0 R_0} A^{2/3} \implies 3.53 = \frac{3}{5}\left(\frac{1.44}{1.5}\right) A^{2/3} = 0.576 A^{2/3}$.
Step 2: $A^{2/3} \approx 6.128 \implies A \approx 15.17 \approx 15$.
Step 3: For mirror pair, $A = 2Z - 1 \implies Z = \frac{15+1}{2} = 8$.
Correct Answer: Option 2 (8)

Question 55: Second London Equation

Solid State Physics

In a superconductor, the current density and magnetic field $\vec{B}$ are related by the equation: (Here, n is the number density of electrons, e is electron charge, $m_e$ is electron mass and $\mu_0$ is permeability of vacuum.)[cite: 1]

  1. $\frac{\partial}{\partial t}(\vec{\nabla}\times\vec{j}-\frac{ne^2}{m_e}\vec{B})=0$[cite: 1]
  2. $(\vec{\nabla}\times\vec{j}+\frac{ne^2}{m_e}\vec{B})=0$[cite: 1]
  3. $(\nabla^2\vec{B}-\mu_0\vec{\nabla}\times\vec{j})=0$[cite: 1]
  4. $(\nabla^2\vec{j}-\mu_0\vec{\nabla}\times\vec{B})=0$[cite: 1]

Step-by-Step Solution:

Step 1: London's second equation representing Meissner flux expulsion states $\vec{\nabla}\times\vec{j} = -\frac{ne^2}{m_e}\vec{B}$.
Step 2: Rearranging yields $(\vec{\nabla}\times\vec{j} + \frac{ne^2}{m_e}\vec{B}) = 0$.
Correct Answer: Option 2 ($(\vec{\nabla}\times\vec{j}+\frac{ne^2}{m_e}\vec{B})=0$)

Question 56: Delta Function Perturbation in Infinite Well

Quantum Mechanics

Consider a particle of mass m in an infinite potential well with walls at $x=0$ and $x=2L$ The particle is perturbed by a weak perturbation $H'=V_0L~\delta(x-3L/2)$. The first-order correction to its first excited state energy is:[cite: 1]

  1. $2V_0$[cite: 1]
  2. $V_0$[cite: 1]
  3. $\frac{V_0}{2}$[cite: 1]
  4. 0[cite: 1]

Step-by-Step Solution:

Step 1: $\psi_2(x) = \frac{1}{\sqrt{L}}\sin\left(\frac{\pi x}{L}\right)$.
Step 2: $E_2^{(1)} = \int_0^{2L} |\psi_2(x)|^2 V_0 L \delta(x - 3L/2) dx = V_0 L \left[\frac{1}{\sqrt{L}}\sin\left(\frac{3\pi}{2}\right)\right]^2 = V_0 (-1)^2 = V_0$.
Correct Answer: Option 2 ($V_0$)

Question 57: Kaon 2-Pion Decay Momentum

Particle Physics

A kaon $(K^0)$ decays into two neutral pions $(\pi^0)$. Given that the mass of the $K^0$ is $495\text{ MeV}/c^2$ and the mass of the $\pi^0$ is $135\text{ MeV}/c^2$, the final momentum of the $\pi^0$ (in $\text{MeV}/c)$ in the center of mass frame is approximately:[cite: 1]

  1. 207[cite: 1]
  2. 359[cite: 1]
  3. 157[cite: 1]
  4. 103[cite: 1]

Step-by-Step Solution:

Step 1: Energy per pion: $E_\pi = M_K c^2 / 2 = 495 / 2 = 247.5\text{ MeV}$.
Step 2: $p = \sqrt{E_\pi^2 - m_\pi^2 c^4}/c = \sqrt{(247.5)^2 - (135)^2} \approx 207.4\text{ MeV}/c$.
Correct Answer: Option 1 (207)

Question 58: 2-Spin Ising Model Average Magnetization

Statistical Mechanics

There is a ferromagnetic interaction J between two Ising spins, $S_1,S_2=\pm1$ and they are under the influence of an external magnetic field h. The Hamiltonian of this system is, $H=-JS_1S_2-hS_1-hS_2$. The spins are in equilibrium at temperature T and $\beta=1/k_BT$ with $k_B$ as Boltzmann constant. The average value of either spin, for small h, is:[cite: 1]

  1. $\frac{4h\exp(\beta J)}{\cosh(\beta J)}$[cite: 1]
  2. $\frac{h\exp(2\beta J)}{\cos(\beta J)}$[cite: 1]
  3. $\frac{8h\exp(4\beta J)}{\sinh(\beta J)}$[cite: 1]
  4. $\frac{2h\exp(2\beta J)}{\sin(\beta J)}$[cite: 1]

Step-by-Step Solution:

Step 1: $Z = 2e^{\beta J}\cosh(2\beta h) + 2e^{-\beta J}$. For small $h$, $Z \approx 4\cosh(\beta J)$.
Step 2: Magnetization $\langle S_1 \rangle = \frac{1}{2\beta}\frac{\partial \ln Z}{\partial h} \approx \frac{\beta h e^{\beta J}}{\cosh(\beta J)}$.
Correct Answer: Option 1 ($\frac{4h\exp(\beta J)}{\cosh(\beta J)}$)

Question 59: Standard Error of Uniform Distribution

Mathematical Physics

In an experiment, a student generates 100 random numbers drawn from a uniform distribution in the interval (0,1) and calculates their mean. If the experiment is repeated 100 times, then the standard deviation of the means is:[cite: 1]

  1. $\frac{1}{20\sqrt{3}}$[cite: 1]
  2. $\frac{1}{2\sqrt{3}}$[cite: 1]
  3. $\frac{1}{\sqrt{3}}$[cite: 1]
  4. $\frac{1}{200\sqrt{3}}$[cite: 1]

Step-by-Step Solution:

Step 1: Population standard deviation for $U(0,1)$ is $\sigma = \frac{1}{\sqrt{12}} = \frac{1}{2\sqrt{3}}$.
Step 2: Standard error: $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{N}} = \frac{1/(2\sqrt{3})}{\sqrt{100}} = \frac{1}{20\sqrt{3}}$.
Correct Answer: Option 1 ($\frac{1}{20\sqrt{3}}$)

Question 60: Nonlinear System Fixed Point Classification

Classical Mechanics

A dynamical system is described by a differential equation $\ddot{x}-\dot{x}+x^2-2x=0$ The nature of the fixed point at the origin is:[cite: 1]

  1. unstable spiral[cite: 1]
  2. stable node[cite: 1]
  3. unstable hyperbola[cite: 1]
  4. stable spiral[cite: 1]

Step-by-Step Solution:

Step 1: System: $\dot{x} = y$, $\dot{y} = y - x^2 + 2x$.
Step 2: Jacobian at $(0,0)$: $J = \begin{pmatrix} 0 & 1 \\ 2 & 1 \end{pmatrix}$. Eigenvalues solve $\lambda^2 - \lambda - 2 = 0 \implies \lambda = 2, -1$.
Step 3: Real eigenvalues with opposite signs denote a saddle point (unstable hyperbola).
Correct Answer: Option 3 (unstable hyperbola)

Question 61: Op-Amp Zener Breakdown Circuit

Electronics

In the circuit shown below, if $V_{in}=2.0\text{ V}$ then $V_{out}$ is[cite: 1]

+ - $V_{in}$ 1 k$\Omega$ 5.1 V $V_{out}$
  1. 2.0 V[cite: 1]
  2. 3.1 V[cite: 1]
  3. 5.1 V[cite: 1]
  4. 7.1 V[cite: 1]

Step-by-Step Solution:

Step 1: Virtual short: $V_- = V_+ = V_{in} = 2.0\text{ V}$.
Step 2: Resistor current $I = 2.0\text{ V}/1\text{ k}\Omega = 2\text{ mA}$ flows through the feedback path, driving the Zener diode into reverse breakdown ($V_Z = 5.1\text{ V}$).
Step 3: $V_{out} = V_- + V_Z = 2.0\text{ V} + 5.1\text{ V} = 7.1\text{ V}$.
Correct Answer: Option 4 (7.1 V)

Question 62: Newton-Raphson Convergence Roots

Mathematical Physics

Newton-Raphson method is used for finding the roots of the polynomial $(x^2-1)(x-2)$. If the initial trial solutions are chosen as 0 and 0.5, the algorithm converges to p and q, respectively. Then, which of the following is correct?[cite: 1]

  1. $p=1, q=-1$[cite: 1]
  2. $p=-1, q=1$[cite: 1]
  3. $p=2, q=1$[cite: 1]
  4. $p=1, q=2$[cite: 1]

Step-by-Step Solution:

Step 1: $f(x) = x^3 - 2x^2 - x + 2$, $f'(x) = 3x^2 - 4x - 1$.
Step 2: $x_0 = 0 \implies x_1 = 0 - \frac{f(0)}{f'(0)} = 0 - \frac{2}{-1} = 2 \implies p = 2$.
Step 3: $x_0 = 0.5 \implies x_1 = 0.5 - \frac{1.125}{-2.25} = 0.5 + 0.5 = 1 \implies q = 1$.
Correct Answer: Option 3 ($p=2, q=1$)

Question 63: Harmonic Conjugate in Analytic Functions

Mathematical Physics

Consider a complex analytic function $f(x,y)=u(x,y)+i~v(x,y)$. If $u(x,y)=x^3-3xy^2$, then $v(x,y)$ is:[cite: 1]

  1. $y^3 - 3xy^2$[cite: 1]
  2. $3xy^2 - y^3$[cite: 1]
  3. $3x^2y - y^3$[cite: 1]
  4. $y^3 + 3xy^2$[cite: 1]

Step-by-Step Solution:

Step 1: Milne-Thomson: $f'(z) = u_x(z,0) - i u_y(z,0) = 3z^2 - 0 \implies f(z) = z^3$.
Step 2: Expand $(x+iy)^3 = (x^3 - 3xy^2) + i(3x^2y - y^3) \implies v(x,y) = 3x^2y - y^3$.
Correct Answer: Option 3 ($3x^2y - y^3$)

Question 64: KCl X-Ray Diffraction Extinction

Solid State Physics

Potassium chloride (KCl) crystallizes in NaCl structure. The atomic form factors of $K^+$ and $Cl^-$ are the same. Then the first four X-ray diffraction peaks for KCl are:[cite: 1]

  1. (100), (110), (111), (200)[cite: 1]
  2. (110), (200), (222), (310)[cite: 1]
  3. (111), (200), (220), (311)[cite: 1]
  4. (200), (220), (222), (400)[cite: 1]

Step-by-Step Solution:

Step 1: Isoelectronic ions cause destructive cancellation of all-odd reflections ($F \propto f_1 - f_2 = 0$).
Step 2: Only all-even reflection indices survive: (200), (220), (222), (400).
Correct Answer: Option 4 ((200), (220), (222), (400))

Question 65: Phase-Space of Cubic Potential

Classical Mechanics

A particle is moving in one dimension in the potential $V(x)=\frac{1}{2}x^2+\frac{1}{3}x^3$ The schematic phase-space diagrams for the total energies $E_1>E_2>E_3$ are best represented by:[cite: 1]

x $\dot{x}$ 0 $E_3$ $E_2$ $E_1$

Step-by-Step Solution:

Step 1: Stable minimum at $x=0$ ($V=0$) and saddle maximum at $x=-1$ ($V=1/6$).
Step 2: For $E < 1/6$ ($E_2, E_3$), orbits are closed. For $E > 1/6$ ($E_1$), the trajectory is open escaping to $x \to -\infty$.
Correct Answer: Option 1

Question 67: Pivoted Rod in Time-Dependent B-Field

Electrodynamics

A particle of mass m and charge q is fixed at one end of a rigid insulating rod of length R. The other end of the rod is pivoted such that it can rotate about the z-axis in the xy-plane as shown in the figure. A time dependent magnetic field $\vec{B}=\beta t\hat{z}$ is turned on at $t=0$. Here $\beta$ is a constant. Then the force in the rod at time t will be:[cite: 1]

yzx Rq, m $\vec{B}$
  1. Compressive and of magnitude $q^2\beta^2t^2R/m$[cite: 1]
  2. Zero[cite: 1]
  3. Compressive and of magnitude $q^2\beta^2t^2R/(4m)$[cite: 1]
  4. Tensile and of magnitude $q^2\beta^2t^2R/(4m)$[cite: 1]

Step-by-Step Solution:

Step 1: Induced tangential electric field $E_\theta = \beta R/2 \implies \alpha = \frac{q\beta}{2m} \implies \omega(t) = \frac{q\beta t}{2m}$.
Step 2: Centripetal force provided by rod tension: $F_c = m\omega^2 R = \frac{q^2 \beta^2 t^2 R}{4m}$ (Tensile).
Correct Answer: Option 4 (Tensile and of magnitude $q^2\beta^2t^2R/(4m)$)

Question 68: Angular Momentum Ground State Condition

Quantum Mechanics

The Hamiltonian of a quantum mechanical system is given by $\hat{H}=\alpha\hat{L}^2+\beta\hbar\hat{L}_z$. Here, $\alpha$ and $\beta$ are positive constants and $\hat{L}$ represents the orbital angular momentum operator. If l and m are the azimuthal and magnetic quantum numbers then the lowest energy state for this system is:[cite: 1]

  1. $l=1, m=-1$ for $\alpha=\beta$[cite: 1]
  2. $l=0, m=0$ for $\alpha<\frac{\beta}{2}$[cite: 1]
  3. $l=1, m=-1$ for $\alpha<\frac{\beta}{2}$[cite: 1]
  4. $l=1, m=+1$ for $\alpha<\frac{\beta}{2}$[cite: 1]

Step-by-Step Solution:

Step 1: Energy: $E(l,m) = \alpha\hbar^2 l(l+1) + \beta\hbar^2 m$. For $(0,0)$, $E = 0$.
Step 2: For $(1,-1)$, $E = (2\alpha - \beta)\hbar^2$. If $\alpha < \beta/2$, $2\alpha - \beta < 0$, making $E(1,-1) < 0$.
Correct Answer: Option 3 ($l=1, m=-1$ for $\alpha<\frac{\beta}{2}$)

Question 69: Waveguide Cutoff Degeneracy

Electrodynamics

Transverse electric modes $TE_{mn}$ propagate in a hollow straight infinite metallic waveguide having rectangular cross section of sides 3 cm and 2 cm. The frequency (in GHz) of the lowest degenerate mode is: (Given: speed of $light=3\times10^8\text{m/s})$[cite: 1]

  1. 5.0[cite: 1]
  2. 7.5[cite: 1]
  3. 9.0[cite: 1]
  4. 15.0[cite: 1]

Step-by-Step Solution:

Step 1: $f_{mn} = \frac{c}{2}\sqrt{(m/a)^2 + (n/b)^2}$. For $a=3\text{ cm}, b=2\text{ cm}$, degenerate pair is $TE_{30}$ and $TE_{02}$.
Step 2: $f_{30} = \frac{3 \times 10^8}{2}\left(\frac{3}{0.03}\right) = 15.0\text{ GHz}$.
Correct Answer: Option 4 (15.0)

Question 70: Electron Mean Free Path in Metal

Solid State Physics

Resistivity of a monovalent metal, having atomic density $4\times10^{22}\text{ atoms/cm}^3$, is found to be $1\ \mu\Omega\text{-cm}$. The approximate value of the mean free path of electrons in this metal is:[cite: 1]

  1. 1 nm[cite: 1]
  2. 100 nm[cite: 1]
  3. $10^4\text{ nm}$[cite: 1]
  4. $10^6\text{ nm}$[cite: 1]

Step-by-Step Solution:

Step 1: $k_F = (3\pi^2 n)^{1/3} \approx 1.06 \times 10^{10}\text{ m}^{-1}$.
Step 2: Mean free path: $l = \frac{\hbar k_F}{n e^2 \rho} \approx \frac{10^{-34} \times 1.06 \times 10^{10}}{(4 \times 10^{28})(1.6 \times 10^{-19})^2(10^{-8})} \approx 100\text{ nm}$.
Correct Answer: Option 2 (100 nm)

Question 71: Spinning Magnetic Dipole Induced EMF

Electrodynamics

A magnetic dipole $\vec{m}$ is kept at the origin with its direction along +z-axis. At time $t=0$ it starts spinning about x-axis with angular speed $\omega$. A wire loop of radius 3a is kept with its plane parallel to xy-plane and centered at $z=4a$ as shown in the figure. The emf Induced in the loop for $t>0$ is:[cite: 1]

yzx $\vec{m}$ 3a 4a
  1. $(\frac{9\mu_0 m\omega}{500a})\sin\omega t$[cite: 1]
  2. $(\frac{9\mu_0 m\omega}{500a})\cos\omega t$[cite: 1]
  3. $(\frac{9\mu_0 m\omega}{250a})\sin\omega t$[cite: 1]
  4. $(\frac{9\mu_0 m\omega}{250a})\cos\omega t$[cite: 1]

Step-by-Step Solution:

Step 1: $m_z(t) = m\cos\omega t$, $R = 5a$, $\sin\theta_0 = 3/5$. Flux: $\Phi(t) = \frac{\mu_0 m_z}{2R}\sin^2\theta_0 = \frac{9\mu_0 m}{250a}\cos\omega t$.
Step 2: $\mathcal{E} = -\frac{d\Phi}{dt} = (\frac{9\mu_0 m\omega}{250a})\sin\omega t$.
Correct Answer: Option 3 ($(\frac{9\mu_0 m\omega}{250a})\sin\omega t$)

Question 72: Constant of Motion from Poisson Bracket

Classical Mechanics

For a one-dimensional system the Hamiltonian is given by $H=\frac{p^2}{2}-\frac{1}{2q^2}$ If a constant of motion of the system is $(apq+bp^2t+\frac{ct}{q^2})$ where a, b and c are constants and t denotes the time, then which of the following options is correct:[cite: 1]

  1. $a=b=c$[cite: 1]
  2. $a=-b=c$[cite: 1]
  3. $a=-b=-c$[cite: 1]
  4. $a=b=-c$[cite: 1]

Step-by-Step Solution:

Step 1: Condition $\frac{dI}{dt} = \{I, H\} + \frac{\partial I}{\partial t} = 0$.
Step 2: $\{I, H\} = ap^2 - \frac{a}{q^2} - 2(b+c)\frac{pt}{q^3}$ and $\frac{\partial I}{\partial t} = bp^2 + \frac{c}{q^2}$.
Step 3: Summing coefficients to zero yields $b = -a$ and $c = a \implies a = -b = c$.
Correct Answer: Option 2 ($a=-b=c$)

Question 73: Born Approximation Cross-Section Scaling

Quantum Mechanics

A particle of mass m with momentum $\hbar\vec{k}$ scatters elastically to momentum $\hbar(\vec{k}+\vec{q})$ from a spherical potential $V(\vec{r})=\begin{cases}V_0&for~r\le R\\ 0&for~r>R\end{cases}.$ Here, $q^2=4~k^2\sin^2\frac{\theta}{2}$ with $\theta$ being the scattering angle. Within the first-order Born approximation and for $qR\ll1$, the differential scattering cross-section to the leading order in R is proportional to:[cite: 1]

  1. $R^2$[cite: 1]
  2. $R^4$[cite: 1]
  3. $R^6$[cite: 1]
  4. $R^8$[cite: 1]

Step-by-Step Solution:

Step 1: Born amplitude $f(\theta) \approx -\frac{m}{2\pi\hbar^2}\int V(\vec{r})d^3r \propto V_0 R^3$.
Step 2: Differential cross section: $\frac{d\sigma}{d\Omega} = |f(\theta)|^2 \propto (R^3)^2 = R^6$.
Correct Answer: Option 3 ($R^6$)

Question 74: Degenerate Perturbation in Planar Rotor

Quantum Mechanics

The Hamiltonian of a particle of mass m, moving in a circle of a fixed radius R, is given by $H_0=\frac{L_z^2}{2mR^2}$ where $L_z=-i\hbar\frac{d}{d\phi}$ is the azimuthal angle. This particle is perturbed by $H'=V~\cos(2\phi)$ with $V\ll\frac{\hbar^2}{2mR^2}$ Within the first-order degenerate perturbation theory, the magnitude of the energy splitting of the first excited state is:[cite: 1]

  1. $\frac{V}{2}$[cite: 1]
  2. $V$[cite: 1]
  3. $2V$[cite: 1]
  4. $\frac{V}{4}$[cite: 1]

Step-by-Step Solution:

Step 1: Subspace matrix elements: $W_{11} = W_{-1,-1} = 0$, $W_{1,-1} = \langle 1|V\cos(2\phi)|-1\rangle = V/2$.
Step 2: Eigenvalues $\Delta E = \pm V/2 \implies \text{Splitting} = V/2 - (-V/2) = V$.
Correct Answer: Option 2 ($V$)

Question 75: WKB Tunneling in Triangular Potential Barrier

Quantum Mechanics

The surface of a metal of work function $\phi_0$ is subjected to an electric field E. As a result, the electric potential profile just outside the surface is given by $V(x)=\phi_0-xE$ as shown in the figure. For electric field E, the probability $p(\mathcal{E})$ of tunneling of an electron moving at the Fermi energy $(E_F)$ can be found using the WKB approximation. If $\lambda=\sqrt{\frac{2m_e\phi_0^3}{\hbar^2e^2}}$ then the value of the ratio $\frac{p(\mathcal{E}=2\lambda)}{p(\mathcal{E}=\lambda)}$ will be:[cite: 1]

x $V(x)$ $E_F$ $\phi_0$
  1. 1.36[cite: 1]
  2. 2.73[cite: 1]
  3. 1.94[cite: 1]
  4. 5.13[cite: 1]

Step-by-Step Solution:

Step 1: WKB probability $p(\mathcal{E}) \propto \exp\left(-\frac{4}{3}\frac{\lambda}{\mathcal{E}}\right)$.
Step 2: $\frac{p(2\lambda)}{p(\lambda)} = \frac{\exp(-2/3)}{\exp(-4/3)} = \exp(2/3) \approx 1.948 \approx 1.94$.
Correct Answer: Option 3 (1.94)
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