WBJEE 2023 Physics Question Paper with Answer and Solution

(C) For a one-dimensional elastic collision between two masses $m_1$ and $m_2$,where $m_2$ is initially at rest,the velocity $v_2$ of the second mass after the collision is given by $v_2 = \frac{2m_1}{m_1 + m_2} v_1$.
1st collision: Mass $m_1 = m$ hits $m_2 = \frac{m}{2}$ with velocity $v_0$. The velocity of the second ball $v_1$ is:
$v_1 = \frac{2m}{m + \frac{m}{2}} v_0 = \frac{2m}{\frac{3m}{2}} v_0 = \frac{4}{3} v_0$.
2nd collision: Mass $m_2 = \frac{m}{2}$ hits $m_3 = \frac{m}{4}$ with velocity $v_1 = \frac{4}{3} v_0$. The velocity of the third ball $v_2$ is:
$v_2 = \frac{2(\frac{m}{2})}{\frac{m}{2} + \frac{m}{4}} v_1 = \frac{m}{\frac{3m}{4}} v_1 = \frac{4}{3} v_1 = \left(\frac{4}{3}\right)^2 v_0$.
Following this pattern,for the $n^{\text{th}}$ ball,the velocity $v_{n-1}$ after $(n-1)$ collisions will be:
$v_{n-1} = \left(\frac{4}{3}\right)^{n-1} v_0$.

(A) According to the principle of conservation of angular momentum,since no external torque acts on the system,the initial angular momentum equals the final angular momentum.
Initial angular momentum $L_i = I \omega_0$,where $\omega_0$ is the initial angular velocity.
When the mouse of mass $m$ lands on the edge at radius $R$,the new moment of inertia of the system becomes $I' = I + m R^2$.
Let the new angular velocity be $\omega$. Then,$L_f = (I + m R^2) \omega$.
Equating $L_i = L_f$,we get $I \omega_0 = (I + m R^2) \omega$.
Thus,$\omega = \frac{I \omega_0}{I + m R^2}$.
The fractional loss of angular velocity is given by $\frac{\omega_0 - \omega}{\omega_0} = 1 - \frac{\omega}{\omega_0}$.
Substituting the value of $\omega$,we get $1 - \frac{I}{I + m R^2} = \frac{I + m R^2 - I}{I + m R^2} = \frac{m R^2}{I + m R^2}$.

(A) The given equation is $V = -\frac{GM}{r} + \frac{A}{r^2}$.
According to the principle of homogeneity of dimensions,the dimensions of each term on both sides must be the same.
Therefore,the dimension of $\frac{A}{r^2}$ must be equal to the dimension of $\frac{GM}{r}$.
$[V] = [\frac{GM}{r}] = [\frac{A}{r^2}]$
From this,we get $[A] = [\frac{GM}{r}] \times [r^2] = [GM] \times [r]$.
We know that the gravitational potential $V$ has the dimensions of energy per unit mass,which is $[L^2 T^{-2}]$.
Also,$\frac{GM}{r}$ represents the gravitational potential,so $[\frac{GM}{r}] = [L^2 T^{-2}]$.
Since the velocity of light $c$ has dimensions $[L T^{-1}]$,then $c^2$ has dimensions $[L^2 T^{-2}]$.
Thus,$[\frac{GM}{r}] = [c^2]$.
Substituting $[r] = \frac{[GM]}{[c^2]}$ into the expression for $[A]$:
$[A] = [GM] \times \frac{[GM]}{[c^2]} = \frac{G^2 M^2}{c^2}$.

(C) For a satellite orbiting near the surface of a planet of radius $R$ and density $\rho$,the gravitational force provides the necessary centripetal force.
$m g = m \omega^2 R$
Since $g = \frac{G M}{R^2}$ and $M = \rho \cdot \frac{4}{3} \pi R^3$,we have $g = \frac{G \rho \frac{4}{3} \pi R^3}{R^2} = \frac{4}{3} \pi G \rho R$.
Substituting this into the force equation:
$m (\frac{4}{3} \pi G \rho R) = m \omega^2 R$
$\omega^2 = \frac{4}{3} \pi G \rho$
Since $\omega = \frac{2 \pi}{T}$,we get $(\frac{2 \pi}{T})^2 = \frac{4}{3} \pi G \rho$,which simplifies to $T^2 = \frac{3 \pi}{G \rho}$.
Thus,$T \propto \frac{1}{\sqrt{\rho}}$.
Since both satellites have the same time period $T$,their densities must be equal,i.e.,$\rho_m = \rho_e$.

(B) The average speed $\overline{V}$ is calculated as the arithmetic mean of the velocities: $\overline{V} = \frac{1 + 3 + 5 + 5 + 6 + 5}{6} = \frac{25}{6} \approx 4.16 \,m/s$.
The root mean square speed $V_{rms}$ is calculated as the square root of the mean of the squares of the velocities: $V_{rms} = \sqrt{\frac{1^2 + 3^2 + 5^2 + 5^2 + 6^2 + 5^2}{6}} = \sqrt{\frac{1 + 9 + 25 + 25 + 36 + 25}{6}} = \sqrt{\frac{121}{6}} = \frac{11}{\sqrt{6}} \approx 4.49 \,m/s$.
Comparing the two values,we find that $4.49 > 4.16$,therefore $V_{rms} > \overline{V}$.

(D) The body experiences two components of acceleration: centripetal acceleration $(a_c)$ and tangential acceleration $(a_T)$.
Centripetal acceleration is given by $a_c = \frac{v^2}{r} = \frac{(2\sqrt{5})^2}{5} = \frac{20}{5} = 4 \,m/s^2$.
Tangential acceleration is given as $a_T = 3 \,m/s^2$.
The net acceleration is $a = \sqrt{a_c^2 + a_T^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \,m/s^2$.
The magnitude of the net force is $F = m \times a = 2 \,kg \times 5 \,m/s^2 = 10 \,N$.

(C) According to the equation of continuity, the volume flow rate is constant:
$AV = av \implies V = \frac{a}{A}v$
where $V$ is the velocity of the piston and $v$ is the velocity of the liquid emerging from the orifice.
Applying Bernoulli's principle between the inside of the cylinder (near the piston) and the orifice:
$P_{in} + \frac{1}{2} \rho V^2 = P_{out} + \frac{1}{2} \rho v^2$
The pressure inside the cylinder is $P_{in} = P_0 + \frac{F}{A}$, where $P_0$ is the atmospheric pressure. The pressure at the orifice is $P_{out} = P_0$.
Substituting these into the Bernoulli equation:
$(P_0 + \frac{F}{A}) + \frac{1}{2} \rho V^2 = P_0 + \frac{1}{2} \rho v^2$
$\frac{F}{A} = \frac{1}{2} \rho (v^2 - V^2)$
Substituting $V = \frac{a}{A}v$:
$\frac{F}{A} = \frac{1}{2} \rho (v^2 - (\frac{a}{A}v)^2) = \frac{1}{2} \rho v^2 (1 - \frac{a^2}{A^2})$
Since $A \gg a$, we have $\frac{a^2}{A^2} \approx 0$, so:
$\frac{F}{A} = \frac{1}{2} \rho v^2 \implies v = \sqrt{\frac{2F}{\rho A}}$

(D) When the $U$-tube is accelerated with acceleration '$f$' towards the right,a pseudo-force acts on the liquid in the opposite direction.
Let the angle of the liquid surface with the horizontal be $\theta$.
The effective acceleration acting on the liquid is the vector sum of the gravitational acceleration '$g$' (downwards) and the pseudo-acceleration '$f$' (towards the left).
The tangent of the angle $\theta$ is given by the ratio of the horizontal acceleration to the vertical acceleration:
$\tan \theta = \frac{f}{g}$
From the geometry of the $U$-tube,where '$h$' is the height difference and '$a$' is the horizontal distance between the two arms:
$\tan \theta = \frac{h}{a}$
Equating the two expressions for $\tan \theta$:
$\frac{h}{a} = \frac{f}{g}$
Therefore,the height difference is:
$h = \frac{f a}{g}$

(A) In simple harmonic motion,the acceleration $f$ is given by $f = -\omega^2 x$,where $\omega$ is the angular frequency and $x$ is the displacement.
Taking the magnitude,we have $|f| = \omega^2 |x|$.
The time period $T$ is given by $T = \frac{2\pi}{\omega}$.
Therefore,the product $|fT|$ is:
$|fT| = |f| \cdot T = (\omega^2 |x|) \cdot \left(\frac{2\pi}{\omega}\right) = 2\pi\omega |x|$.
Since $2\pi\omega$ is a constant,the relationship $|fT| = (2\pi\omega) |x|$ represents a linear equation of the form $y = mx$,where $y = |fT|$ and $x$ is the displacement.
Thus,the graph of $|fT|$ versus $x$ is a straight line passing through the origin.

(D) The angular momentum $L$ of a particle about the point of projection is given by $L = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$.
At the maximum height,the velocity of the particle is purely horizontal,given by $v_x = u \cos \theta$.
The position vector at maximum height has a vertical component equal to the maximum height $H = \frac{u^2 \sin^2 \theta}{2g}$.
The magnitude of angular momentum at maximum height is $L = m v_x H = m (u \cos \theta) \left( \frac{u^2 \sin^2 \theta}{2g} \right)$.
Simplifying this,we get $L = \frac{m u^3}{2g} \sin^2 \theta \cos \theta$.
For $\theta = 0^{\circ}$,$L = 0$. For $\theta = 90^{\circ}$ (or $\frac{\pi}{2}$),$L = 0$.
Between $0$ and $\frac{\pi}{2}$,the function $f(\theta) = \sin^2 \theta \cos \theta$ is positive and reaches a maximum value. This corresponds to the shape shown in graph $D$.

(C) According to the First Law of Thermodynamics,$\Delta Q = \Delta U + \Delta W$,where $\Delta U$ is the change in internal energy.
Since internal energy is a state function,$\Delta U$ is the same for any path between the same two states $A$ and $C$.
For the direct path $A \rightarrow C$:
$\Delta W_{AC} = 200 \ J$
$\Delta Q_{AC} = 280 \ J$
$\Delta U = \Delta Q_{AC} - \Delta W_{AC} = 280 \ J - 200 \ J = 80 \ J$.
For the path $A \rightarrow B \rightarrow C$:
$\Delta W_{ABC} = 80 \ J$
Since $\Delta U$ is a state function,$\Delta U = 80 \ J$ for this path as well.
Therefore,the heat absorbed $\Delta Q_{ABC} = \Delta U + \Delta W_{ABC} = 80 \ J + 80 \ J = 160 \ J$.

(A) The heat supplied at a constant rate is given by $\Delta H = mC \Delta \theta$.
Since the rate of heat supply $\frac{dH}{dt}$ is constant,we can write:
$\frac{dH}{dt} = mC \frac{d\theta}{dt}$.
This implies $\frac{d\theta}{dt} = \frac{1}{mC} \left( \frac{dH}{dt} \right)$.
Since $m$ and $\frac{dH}{dt}$ are constant,the slope of the $\theta-t$ graph is inversely proportional to the specific heat $C$ (i.e.,$\text{slope} \propto \frac{1}{C}$).
From the figure,the slope of line $B$ is greater than the slope of line $A$ (i.e.,$\text{slope}_B > \text{slope}_A$).
Therefore,$C_B < C_A$,which means the specific heat of $A$ is greater than that of $B$.

(A, B, C, D) In a $p-v$ diagram:
$1$. The process $1 \rightarrow 2$ is a horizontal line,meaning pressure $p$ is constant. Thus,it is an isobaric process.
$2$. The process $3 \rightarrow 1$ is a vertical line,meaning volume $v$ is constant. Thus,it is an isochoric process.
$3$. The process $2 \rightarrow 3$ is a curve,which represents an adiabatic process in this cycle.
$4$. For any complete cyclic process,the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics,$Q = \Delta U + W$. Since $\Delta U = 0$,$Q = W$. The work done $W$ in a cyclic process is equal to the area enclosed by the cycle,which is non-zero. Therefore,both work done and heat absorbed are non-zero.
Thus,statements $A$,$B$,$C$,and $D$ are all correct.

(A) The general equation for a wave travelling in the positive $x$-direction is $y = A \sin(\omega t - kx + \phi)$.
At $t=0$,the equation is $y = 3 \sin(-2\pi x / 3)$,which implies $k = 2\pi / 3$.
Given the wave velocity $v = 4 \ m/s$,we use the relation $v = \omega / k$.
Thus,$\omega = v \cdot k = 4 \times (2\pi / 3) = 8\pi / 3$.
The general wave equation is $y = 3 \sin \left(\frac{8\pi}{3}t - \frac{2\pi}{3}x\right)$.
At $t=4 \ s$,substituting the value of $t$:
$y = 3 \sin \left(\frac{8\pi}{3}(4) - \frac{2\pi}{3}x\right)$
$y = 3 \sin \left(\frac{32\pi}{3} - \frac{2\pi}{3}x\right)$
Factoring out $2\pi$:
$y = 3 \sin 2\pi \left(\frac{16}{3} - \frac{x}{3}\right) = 3 \sin 2\pi \left(-\frac{x-16}{3}\right)$.

(D) The mass per unit length of the rope is $\lambda = \frac{M}{L} = \frac{0.4 \,kg}{4 \,m} = 0.1 \,kg/m$.
The length of the hanging part is $l = 0.6 \,m$.
The mass of the hanging part is $m = \lambda \times l = 0.1 \,kg/m \times 0.6 \,m = 0.06 \,kg$.
The center of mass of the hanging part is at a distance $h = \frac{l}{2} = \frac{0.6 \,m}{2} = 0.3 \,m$ below the table edge.
The work done to pull the rope onto the table is equal to the change in potential energy of the hanging part,which is $W = mgh$.
Substituting the values: $W = 0.06 \,kg \times 10 \,m/s^2 \times 0.3 \,m = 0.18 \,J$.

(A, B, C) $1$. With respect to the girl, the initial velocity of the ball is $0$ and final velocity is $v$. The kinetic energy is $E_k = \frac{1}{2}mv^2$. Thus, the work done by the girl is $W = \Delta E_k = \frac{1}{2}mv^2$. So, option $A$ and $B$ are correct.
$2$. With respect to the ground, the initial velocity of the ball is $u$ and the final velocity is $v+u$. The work done by the train is the change in kinetic energy of the ball as seen from the ground minus the work done by the girl: $W_{\text{train}} = \Delta E_{k, \text{ground}} - W_{\text{girl}} = [\frac{1}{2}m(v+u)^2 - \frac{1}{2}mu^2] - \frac{1}{2}mv^2 = \frac{1}{2}m(v^2 + u^2 + 2vu - u^2) - \frac{1}{2}mv^2 = mvu$. So, option $C$ is correct.
$3$. The gain in kinetic energy as measured by a person on the ground is $\Delta E_{k, \text{ground}} = \frac{1}{2}m(v+u)^2 - \frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mvu$. Thus, option $D$ is incorrect.

(B) The potential energy $U$ of an electron in a hydrogen atom at a state $n$ is given by $U_n = -27.2 / n^2 \ eV$.
For the first excited state $(n = 2)$,the potential energy is $U_2 = -27.2 / 2^2 = -6.8 \ eV$.
According to the problem,we assume $U_2 = 0$. This implies we are adding a constant $C = 6.8 \ eV$ to the potential energy scale.
The total energy $E_n$ of a hydrogen atom is given by $E_n = -13.6 / n^2 \ eV$.
For $n = \infty$,the standard total energy is $E_{\infty} = 0 \ eV$.
Since we shifted the potential energy scale by adding $6.8 \ eV$,the new total energy $E'_{\infty}$ will be $E_{\infty} + 6.8 \ eV = 0 + 6.8 \ eV = 6.8 \ eV$.

(A) Let the charges on the two plates be $Q_1 = 12 \mu C$ and $Q_2 = 6 \mu C$.
When two large conducting plates are placed parallel to each other,the charge on the outer surfaces is equal to $\frac{Q_1 + Q_2}{2}$ and the charge on the inner surfaces is equal to $\frac{Q_1 - Q_2}{2}$ and $\frac{Q_2 - Q_1}{2}$ respectively.
For surface $A$ (outer surface of first plate): $q_A = \frac{Q_1 + Q_2}{2} = \frac{12 + 6}{2} = \frac{18}{2} = 9 \mu C$.
For surface $B$ (inner surface of first plate): $q_B = \frac{Q_1 - Q_2}{2} = \frac{12 - 6}{2} = \frac{6}{2} = 3 \mu C$.
For surface $C$ (inner surface of second plate): $q_C = \frac{Q_2 - Q_1}{2} = \frac{6 - 12}{2} = \frac{-6}{2} = -3 \mu C$.
For surface $D$ (outer surface of second plate): $q_D = \frac{Q_1 + Q_2}{2} = \frac{12 + 6}{2} = \frac{18}{2} = 9 \mu C$.
Thus,the charge distribution on surfaces $A, B, C$ and $D$ is $9 \mu C, 3 \mu C, -3 \mu C$ and $9 \mu C$ respectively.

(A, C, D) Initially,when the switch $K$ is closed,both capacitors $A$ and $B$ are in parallel with the battery of voltage $V$. The total energy stored is $U = \frac{1}{2}CV^2 + \frac{1}{2}CV^2 = CV^2$.
When the switch $K$ is opened,capacitor $A$ remains connected to the battery,while capacitor $B$ is isolated. The charge on $B$ remains constant at $q_B = CV$.
After inserting the dielectric $(K=3)$:
For capacitor $A$: The new capacitance is $C_A' = KC = 3C$. The voltage remains $V$. The energy stored is $U_A = \frac{1}{2}C_A'V^2 = \frac{1}{2}(3C)V^2 = \frac{3}{2}CV^2$.
For capacitor $B$: The charge $q_B = CV$ remains constant. The new capacitance is $C_B' = KC = 3C$. The energy stored is $U_B = \frac{q_B^2}{2C_B'} = \frac{(CV)^2}{2(3C)} = \frac{C^2V^2}{6C} = \frac{1}{6}CV^2$.
The total energy stored is $U_{total} = U_A + U_B = \frac{3}{2}CV^2 + \frac{1}{6}CV^2 = \frac{9+1}{6}CV^2 = \frac{10}{6}CV^2 = \frac{5}{3}CV^2$.
Thus,statements $A$,$C$,and $D$ are correct.

(C) Let the resistance of the voltmeter be $R_v$. The voltmeter is connected in parallel with the $160 \Omega$ resistor. The equivalent resistance of this parallel combination is $R_p = \frac{160 R_v}{160 + R_v}$.
The total resistance of the circuit is $R_{eq} = R_p + 20 = \frac{160 R_v}{160 + R_v} + 20$.
The total current in the circuit is $I = \frac{V}{R_{eq}} = \frac{10}{\frac{160 R_v}{160 + R_v} + 20}$.
The voltage across the parallel combination (voltmeter reading) is $V_p = I \times R_p = 8 \,V$.
Substituting $I$ and $R_p$: $\frac{10}{\frac{160 R_v}{160 + R_v} + 20} \times \frac{160 R_v}{160 + R_v} = 8$.
Let $x = \frac{160 R_v}{160 + R_v}$. Then $\frac{10x}{x + 20} = 8 \implies 10x = 8x + 160 \implies 2x = 160 \implies x = 80 \Omega$.
Now, $\frac{160 R_v}{160 + R_v} = 80 \implies 160 R_v = 80(160 + R_v) \implies 160 R_v = 12800 + 80 R_v \implies 80 R_v = 12800 \implies R_v = 160 \Omega$.

(A) The total charge $Q$ passing through the coil is equal to the area under the current-time graph. Since the current decreases uniformly from $I_0$ to $0$ in time $T$,the graph is a triangle with base $T$ and height $I_0$.
$Q = \frac{1}{2} I_0 T \Rightarrow I_0 = \frac{2 Q}{T}$
The current as a function of time is given by $I(t) = I_0 \left(1 - \frac{t}{T}\right) = \frac{2 Q}{T} \left(1 - \frac{t}{T}\right)$.
The heat generated $H$ in the coil is given by $H = \int_0^T I^2 R \, dt$.
$H = R \int_0^T \left[ \frac{2 Q}{T} \left(1 - \frac{t}{T}\right) \right]^2 \, dt = \frac{4 Q^2 R}{T^2} \int_0^T \left(1 - \frac{t}{T}\right)^2 \, dt$.
Let $u = 1 - \frac{t}{T}$,then $du = -\frac{1}{T} dt$,so $dt = -T du$.
When $t=0, u=1$; when $t=T, u=0$.
$H = \frac{4 Q^2 R}{T^2} \int_1^0 u^2 (-T \, du) = \frac{4 Q^2 R}{T} \int_0^1 u^2 \, du$.
$H = \frac{4 Q^2 R}{T} \left[ \frac{u^3}{3} \right]_0^1 = \frac{4 Q^2 R}{3 T}$.

(D) According to the right-hand thumb rule,if the current in the wire flows upward,the magnetic field lines in the region of the coil are directed into the plane of the paper.
If the current in the wire flows downward,the magnetic field lines in the region of the coil are directed out of the plane of the paper.
For a clockwise induced current in the loop,the induced magnetic field must be directed into the plane of the paper (by the right-hand grip rule).
According to Lenz's law,the induced current opposes the change in magnetic flux.
Case $1$: If the current is flowing upward and increasing,the magnetic field into the plane increases,so the induced current will be counter-clockwise to oppose it. This does not match.
Case $2$: If the current is flowing upward and decreasing,the magnetic field into the plane decreases,so the induced current will be clockwise to support it.
Case $3$: If the current is flowing downward and increasing,the magnetic field out of the plane increases,so the induced current will be counter-clockwise to oppose it.
Case $4$: If the current is flowing downward and decreasing,the magnetic field out of the plane decreases,so the induced current will be clockwise to support it.
Since the current must be time-dependent to induce a current,and both decreasing upward current and decreasing downward current can produce a clockwise induced current,the most appropriate general description is that the current is time-dependent.

(A) When the bar magnet falls through the conducting ring,the magnetic flux linked with the ring changes. According to Lenz's law,an induced current is produced in the ring that opposes the motion of the magnet.
As the magnet approaches the ring,the induced current creates a magnetic field that repels the magnet,causing its acceleration to be less than $g$ (acceleration due to gravity).
As the magnet passes through the center of the ring,the magnetic flux linked with the ring starts decreasing. The induced current now creates a magnetic field that attracts the magnet,again opposing its downward motion.
Throughout the process,the net force on the magnet is $F_{net} = mg - F_{mag}$,where $F_{mag}$ is the magnetic force. The speed of the magnet increases continuously,but its acceleration decreases when it is near the ring due to the opposing magnetic force. The graph in option $A$ correctly shows this behavior,where the slope (acceleration) decreases as the magnet passes through the ring and then increases again.

(B) For an electromagnetic wave,the direction of propagation is given by the vector $\vec{k}_{dir} = \hat{k}$ (since the phase is $kz - \omega t$).
The electric field vector is $\vec{E} = E_0(\hat{i} + \hat{j})$.
The magnetic field $\vec{B}$ is perpendicular to both the direction of propagation and the electric field $\vec{E}$.
The direction of $\vec{B}$ is given by the cross product $\vec{k}_{dir} \times \vec{E}$.
$\vec{B}_{dir} = \hat{k} \times (\hat{i} + \hat{j}) = (\hat{k} \times \hat{i}) + (\hat{k} \times \hat{j}) = \hat{j} - \hat{i} = -\hat{i} + \hat{j}$.
Thus,the direction of the magnetic field is $-\hat{i} + \hat{j}$.

(B) The electric field $E$ due to an infinite charged cylinder at a distance $r$ from its axis is given by $E \propto \frac{1}{r}$.
Since the $+Q$ charge is closer to the cylinder than the $-Q$ charge,the electric field at the position of $+Q$ $(E_1)$ is stronger than the electric field at the position of $-Q$ $(E_2)$,i.e.,$E_1 > E_2$.
The force on the $+Q$ charge is $F_1 = Q E_1$ (directed away from the cylinder,i.e.,to the right).
The force on the $-Q$ charge is $F_2 = Q E_2$ (directed towards the cylinder,i.e.,to the left).
Since $E_1 > E_2$,the net force $F_{\text{net}} = F_1 - F_2$ is directed to the right.
Regarding torque,the force $F_1$ acts at a greater distance from the center of the rod and is larger,while $F_2$ acts at a smaller distance. The combination of these forces creates a clockwise torque about the center of the rod.

(D) The electric potential $V$ due to an electric dipole at a point $(r, \theta)$ is given by $V = \frac{p \cos \theta}{4 \pi \varepsilon_0 r^2}$.
Point $A(a, 0, 0)$ lies on the equatorial plane of the dipole (which is along the $z$-axis),so the angle $\theta_A = 90^{\circ}$. Thus,the potential at $A$ is $V_A = \frac{p \cos 90^{\circ}}{4 \pi \varepsilon_0 a^2} = 0$.
Point $B(0, 0, a)$ lies on the axial line of the dipole,so the angle $\theta_B = 0^{\circ}$. Thus,the potential at $B$ is $V_B = \frac{p \cos 0^{\circ}}{4 \pi \varepsilon_0 a^2} = \frac{p}{4 \pi \varepsilon_0 a^2}$.
The work done $W$ to move a charge $q$ from $A$ to $B$ is $W = q(V_B - V_A)$.
Substituting the values,$W = q \left( \frac{p}{4 \pi \varepsilon_0 a^2} - 0 \right) = \frac{p q}{4 \pi \varepsilon_0 a^2}$.

(D) The linear charge density is given by $\lambda = \lambda_0 \sin \theta$. Consider a small element of the rod subtending an angle $d\theta$ at the centre. The charge on this element is $dq = \lambda (R d\theta) = \lambda_0 R \sin \theta d\theta$.
The electric field $dE$ at the centre due to this element is $dE = \frac{k dq}{R^2} = \frac{1}{4 \pi \varepsilon_0} \frac{\lambda_0 R \sin \theta d\theta}{R^2} = \frac{\lambda_0}{4 \pi \varepsilon_0 R} \sin \theta d\theta$.
Due to symmetry,the $x$-components of the electric field cancel out. The $y$-component is $dE_y = -dE \sin \theta = -\frac{\lambda_0}{4 \pi \varepsilon_0 R} \sin^2 \theta d\theta$.
Integrating from $\theta = 0$ to $\pi$:
$E_y = -\frac{\lambda_0}{4 \pi \varepsilon_0 R} \int_0^{\pi} \sin^2 \theta d\theta = -\frac{\lambda_0}{4 \pi \varepsilon_0 R} \int_0^{\pi} \frac{1 - \cos 2\theta}{2} d\theta = -\frac{\lambda_0}{8 \pi \varepsilon_0 R} [\theta - \frac{\sin 2\theta}{2}]_0^{\pi} = -\frac{\lambda_0}{8 \pi \varepsilon_0 R} (\pi) = -\frac{\lambda_0}{8 \varepsilon_0 R}$.
Thus,$\vec{E} = -\frac{\lambda_0}{8 \varepsilon_0 R} \hat{j}$. Since this is not among the options,the correct answer is $D$.

(B) The electric field component $E_{x}$ is given by the negative gradient of the potential with respect to $x$,i.e.,$E_{x} = -\frac{dV}{dx}$.
From the graph,the potential difference $\Delta V = V_{2} - V_{1} = 4 \ V - 2 \ V = 2 \ V$.
The distance along the $x$-axis between these two equipotential lines is $\Delta x = 4 \ cm - 2 \ cm = 2 \ cm = 0.02 \ m$.
Therefore,the magnitude of the $x$-component of the electric field is $E_{x} = -\frac{\Delta V}{\Delta x} = -\frac{2 \ V}{0.02 \ m} = -100 \ V/m$.
Thus,the correct option is $B$.

(C) The magnetic force on a current-carrying wire in a uniform magnetic field is given by $\vec{F} = I (\vec{L}_{eff} \times \vec{B})$,where $\vec{L}_{eff}$ is the effective displacement vector from the starting point to the end point of the wire.
For the given curve $y = A \sin \left(\frac{2 \pi}{\lambda} x\right)$,the starting point is at $x=0$,which gives $y=0$. The end point is at $x=\lambda$,which gives $y = A \sin(2\pi) = 0$.
Thus,the effective displacement vector $\vec{L}_{eff}$ is the vector from $(0, 0)$ to $(\lambda, 0)$,which is $\vec{L}_{eff} = \lambda \hat{i}$.
The magnetic field is in the $z$-direction,so $\vec{B} = B \hat{k}$.
The magnetic force is $\vec{F} = I (\lambda \hat{i} \times B \hat{k}) = I \lambda B (\hat{i} \times \hat{k}) = -I \lambda B \hat{j}$.
The magnitude of the magnetic force is $|\vec{F}| = I \lambda B$.

(B) The magnetic field is directed along the $z$-axis $(B = B_0 \hat{k})$.
The velocity of the particle is $v = 3 \hat{i} + 4 \hat{k} \text{ m/s}$.
The velocity component perpendicular to the magnetic field is $v_{\perp} = 3 \hat{i} \text{ m/s}$,which causes circular motion in the $x-y$ plane.
The velocity component parallel to the magnetic field is $v_{\parallel} = 4 \hat{k} \text{ m/s}$,which remains constant as there is no force acting in this direction.
Since the particle has both perpendicular and parallel velocity components,the trajectory is a helical path.
The distance to be covered along the $z$-axis is $s = 2 \text{ m}$.
Using the formula $s = v_{\parallel} \times t$,we get $t = \frac{s}{v_{\parallel}} = \frac{2}{4} = \frac{1}{2} \text{ s}$.

(A, B, C, D) For $r < R$,using Faraday's law of induction: $\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_B}{dt}$.
$E(2\pi r) = \frac{d}{dt}[(4t^2 - 2t + 6) \pi r^2] = (8t - 2) \pi r^2$.
Thus,$E = \frac{(8t - 2)r}{2} = (4t - 1)r$. Since $E \propto r$,the electric field varies linearly with $r$ for $r < R$.
For $r > R$,the magnetic flux is confined to the region $r \le R$: $\oint \vec{E} \cdot d\vec{l} = -\frac{d}{dt}[B \cdot \pi R^2]$.
$E(2\pi r) = (8t - 2) \pi R^2 \implies E = \frac{(8t - 2)R^2}{2r}$.
At $r = 2R$ and $t = 2 \text{ s}$:
$E = \frac{(8(2) - 2)R^2}{2(2R)} = \frac{14R^2}{4R} = \frac{7R}{2}$.
The acceleration $a = \frac{Eq}{m} = \frac{7qR}{2m}$.
Induced electric fields are non-conservative and form closed loops. By Lenz's law,since $\vec{B}$ is increasing for $t > 0.25 \text{ s}$,the induced field opposes the change,resulting in a clockwise direction for a positive charge.

(A, B, C, D) For a charged particle in a magnetic field,the radius of the circular path is given by $r = \frac{mv}{qB}$.
Applying Bohr's quantization condition for angular momentum: $mvr = \frac{nh}{2\pi} = n\hbar$.
Substituting $v = \frac{qBr}{m}$ into the quantization condition: $m \left( \frac{qBr}{m} \right) r = n\hbar \implies qBr^2 = n\hbar \implies r_n = \sqrt{\frac{n\hbar}{qB}}$. Thus,$r_n \propto \sqrt{n}$. (Statement $A$ is true).
Now,$v_n = \frac{qBr_n}{m} = \frac{qB}{m} \sqrt{\frac{n\hbar}{qB}} = \sqrt{\frac{n q B \hbar}{m^2}}$. (Statement $B$ is true).
The energy of the $n^{\text{th}}$ level is $E_n = \frac{1}{2}mv_n^2 = \frac{1}{2}m \left( \frac{n q B \hbar}{m^2} \right) = n \left( \frac{q B \hbar}{2m} \right)$. Thus,$E_n \propto n$. (Statement $C$ is true).
The transition frequency $\omega$ is given by $\Delta E = \hbar \omega$. For successive levels,$\Delta E = E_{n+1} - E_n = \frac{q B \hbar}{2m}$. Therefore,$\omega = \frac{\Delta E}{\hbar} = \frac{qB}{2m}$,which is independent of $n$. (Statement $D$ is true).

(A) For the first lens:
$u_1 = -f/2$,$f_1 = f$
Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$:
$\frac{1}{v_1} - \frac{1}{-f/2} = \frac{1}{f} \Rightarrow \frac{1}{v_1} + \frac{2}{f} = \frac{1}{f} \Rightarrow \frac{1}{v_1} = -\frac{1}{f} \Rightarrow v_1 = -f$.
Magnification $m_1 = \frac{v_1}{u_1} = \frac{-f}{-f/2} = 2$.
For the second lens:
The image from the first lens acts as an object for the second lens. The distance between the lenses is $f$. So,$u_2 = -(f + |v_1|) = -(f + f) = -2f$.
Using the lens formula $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$:
$\frac{1}{v_2} - \frac{1}{-2f} = \frac{1}{f} \Rightarrow \frac{1}{v_2} + \frac{1}{2f} = \frac{1}{f} \Rightarrow \frac{1}{v_2} = \frac{1}{2f} \Rightarrow v_2 = 2f$.
Magnification $m_2 = \frac{v_2}{u_2} = \frac{2f}{-2f} = -1$.
For the third lens:
The image from the second lens acts as an object for the third lens. The distance between the second and third lens is $f$. The image $v_2 = 2f$ is formed behind the second lens. The object distance for the third lens is $u_3 = -(f - v_2) = -(f - 2f) = f$.
Using the lens formula $\frac{1}{v_3} - \frac{1}{u_3} = \frac{1}{f_3}$:
$\frac{1}{v_3} - \frac{1}{f} = \frac{1}{f} \Rightarrow \frac{1}{v_3} = \frac{2}{f} \Rightarrow v_3 = f/2$.
Magnification $m_3 = \frac{v_3}{u_3} = \frac{f/2}{f} = 1/2$.
Total magnification $M = m_1 \times m_2 \times m_3 = 2 \times (-1) \times (1/2) = -1$.
The final image is at $f/2$ behind the rightmost lens.

(B) From Snell's law,we have $\mu_X \sin i = \mu_Y \sin r$.
Since $\mu = \frac{c}{V}$,we can write $\frac{c}{V_X} \sin i = \frac{c}{V_Y} \sin r$,which simplifies to $\frac{\sin i}{\sin r} = \frac{V_X}{V_Y}$.
From the given graph,the slope of the line is $\tan 30^{\circ} = \frac{\sin r}{\sin i} = \frac{1}{\sqrt{3}}$.
Therefore,$\frac{\sin i}{\sin r} = \sqrt{3}$.
Substituting this into the velocity ratio,we get $\frac{V_X}{V_Y} = \sqrt{3}$,which implies $V_X = \sqrt{3} V_Y$.
Since the light travels from a rarer medium $(X)$ to a denser medium $(Y)$ (as $r < i$),total internal reflection cannot occur when light is incident from medium $X$ to $Y$. Frequency remains constant during refraction.

(B) To determine if the Zener diode is conducting, we first calculate the voltage across the load resistance $R_L$ assuming the diode is open-circuited.
Using the voltage divider rule, the voltage $V_{ab}$ across the load resistance $R_L = 100 \, \Omega$ is given by:
$V_{ab} = V_{source} \times \frac{R_L}{R + R_L}$
$V_{ab} = 9 \, V \times \frac{100 \, \Omega}{200 \, \Omega + 100 \, \Omega} = 9 \, V \times \frac{100}{300} = 3 \, V$
Since the calculated voltage $V_{ab} = 3 \, V$ is less than the Zener breakdown voltage $V_Z = 5 \, V$, the Zener diode does not enter the breakdown region and remains in the non-conducting $(OFF)$ state.
Therefore, the circuit behaves as a simple series circuit, and the voltage drop across the load resistance $R_L$ is $V_L = 3 \, V$.

(C) The output $Y$ of the given logic circuit is determined as follows:
$1$. The input $A$ passes through a $NOT$ gate to become $\overline{A}$.
$2$. The inputs $\overline{A}$ and $B$ are fed into a $NAND$ gate,producing $\overline{\overline{A} \cdot B}$.
$3$. The input $C$ passes through a $NOT$ gate to become $\overline{C}$.
$4$. The outputs $\overline{\overline{A} \cdot B}$ and $\overline{C}$ are fed into a $NOR$ gate,resulting in $Y = \overline{(\overline{\overline{A} \cdot B}) + \overline{C}}$.
$5$. Using De Morgan's theorem,$Y = \overline{(\overline{\overline{A} \cdot B})} \cdot \overline{(\overline{C})} = (\overline{A} \cdot B) \cdot C = \overline{A} \cdot B \cdot C$.
$6$. The output $Y=1$ only when $\overline{A}=1, B=1, C=1$,which means $A=0, B=1, C=1$.
$7$. There are $2^3 = 8$ total possible combinations for inputs $A, B, C$.
$8$. Since $Y=1$ for only $1$ combination,the number of combinations for which $Y=0$ is $8 - 1 = 7$.

(A) Given the intensity ratio $\frac{I_1}{I_2} = n$,so $I_1 = nI_2$.
$I_{\text{Max}} = (\sqrt{I_1} + \sqrt{I_2})^2 = (\sqrt{nI_2} + \sqrt{I_2})^2 = (\sqrt{n} + 1)^2 I_2$.
$I_{\text{Min}} = (\sqrt{I_1} - \sqrt{I_2})^2 = (\sqrt{nI_2} - \sqrt{I_2})^2 = (\sqrt{n} - 1)^2 I_2$.
Now,the ratio is:
$\frac{I_{\text{Max}} - I_{\text{Min}}}{I_{\text{Max}} + I_{\text{Min}}} = \frac{(\sqrt{n} + 1)^2 I_2 - (\sqrt{n} - 1)^2 I_2}{(\sqrt{n} + 1)^2 I_2 + (\sqrt{n} - 1)^2 I_2} = \frac{(\sqrt{n} + 1)^2 - (\sqrt{n} - 1)^2}{(\sqrt{n} + 1)^2 + (\sqrt{n} - 1)^2}$.
Expanding the terms:
$= \frac{(n + 1 + 2\sqrt{n}) - (n + 1 - 2\sqrt{n})}{(n + 1 + 2\sqrt{n}) + (n + 1 - 2\sqrt{n})} = \frac{4\sqrt{n}}{2(n + 1)} = \frac{2\sqrt{n}}{n + 1}$.
Let $f(n) = \frac{2\sqrt{n}}{n + 1}$. To find the maximum,we differentiate with respect to $n$ or observe that for $n=1$,$f(1) = \frac{2(1)}{1+1} = 1$,which is the maximum possible value for this expression since $(\sqrt{n}-1)^2 \ge 0$ implies $n+1 \ge 2\sqrt{n}$.

(D) When $X$-rays are incident on parallel atomic planes with spacing $d$ at a glancing angle $\theta$,the path difference between the rays reflected from successive planes is given by $\Delta x = 2 d \sin \theta$.
For constructive interference (maxima),the path difference must be an integer multiple of the wavelength $\lambda$.
Therefore,the condition for maxima is $2 d \sin \theta = n \lambda$,where $n = 1, 2, 3, \dots$ is the order of the interference fringe. This is known as Bragg's Law.