WBJEE 2020 Physics Question Paper with Answer and Solution

(A) Let the velocity of the ball before collision be $v$. The velocity components are $v_x = v \sin \theta$ (parallel to the floor) and $v_y = v \cos \theta$ (perpendicular to the floor).
Since the floor is smooth,there is no impulsive force parallel to the floor,so the tangential component of velocity remains unchanged: $v'_x = v \sin \theta$.
For the normal component,the coefficient of restitution $\varepsilon$ is defined as the ratio of the velocity of separation to the velocity of approach along the normal: $\varepsilon = \frac{v'_y}{v_y}$.
Thus,$v'_y = \varepsilon v_y = \varepsilon v \cos \theta$.
Let $\theta'$ be the angle of reflection with the normal. Then,$\tan \theta' = \frac{v'_x}{v'_y} = \frac{v \sin \theta}{\varepsilon v \cos \theta} = \frac{\tan \theta}{\varepsilon}$.
Therefore,$\theta' = \tan ^{-1}\left(\frac{\tan \theta}{\varepsilon}\right)$.

(A) The work done against friction is converted into heat energy,which raises the temperature of the block.
Given: Mass $m = 20 \ kg = 20000 \ g$,velocity $v = 0.5 \ ms^{-1}$,time $t = 2.1 \ s$,coefficient of friction $\mu = 0.1$,specific heat $c = 0.1 \ cal \ g^{-1} \ ^{\circ}C^{-1} = 0.1 \times 4.2 \ J \ g^{-1} \ ^{\circ}C^{-1} = 0.42 \ J \ g^{-1} \ ^{\circ}C^{-1}$.
Frictional force $f = \mu mg = 0.1 \times 20 \times 10 = 20 \ N$.
Work done against friction $W = f \times d = f \times (v \times t) = 20 \times 0.5 \times 2.1 = 21 \ J$.
Heat energy $Q = mc \Delta T$.
Since $W = Q$,we have $21 = 20000 \times 0.42 \times \Delta T$.
$\Delta T = \frac{21}{20000 \times 0.42} = \frac{21}{8400} = 0.0025^{\circ} C$.

(A) Let a force $F$ be applied at an angle $\theta$ with the horizontal.
Resolving the forces,the vertical equilibrium gives: $N + F \sin \theta = mg$,so $N = mg - F \sin \theta$.
The horizontal force required to overcome friction is: $F \cos \theta = f_s = \mu N$.
Substituting $N$: $F \cos \theta = \mu (mg - F \sin \theta)$.
Rearranging for $F$: $F (\cos \theta + \mu \sin \theta) = \mu mg$,which gives $F = \frac{\mu mg}{\cos \theta + \mu \sin \theta}$.
To find the minimum force $F$,we maximize the denominator $D = \cos \theta + \mu \sin \theta$.
Setting the derivative with respect to $\theta$ to zero: $\frac{dD}{d\theta} = -\sin \theta + \mu \cos \theta = 0$.
This implies $\tan \theta = \mu$.
Using the identity $\cos \theta + \mu \sin \theta = \sqrt{1+\mu^2} \sin(\theta + \alpha)$ where $\tan \alpha = 1/\mu$,or simply substituting $\sin \theta = \frac{\mu}{\sqrt{1+\mu^2}}$ and $\cos \theta = \frac{1}{\sqrt{1+\mu^2}}$ into the expression for $F$:
$F_{\min} = \frac{\mu mg}{\frac{1}{\sqrt{1+\mu^2}} + \mu \frac{\mu}{\sqrt{1+\mu^2}}} = \frac{\mu mg}{\frac{1+\mu^2}{\sqrt{1+\mu^2}}} = \frac{\mu mg}{\sqrt{1+\mu^2}}$.

(C) The terminal velocity $v_{T}$ of a spherical body of radius $r$ and density $\rho_{s}$ falling through a liquid of density $\rho_{L}$ and viscosity $\eta$ is given by the formula:
$v_{T} = \frac{2r^{2}(\rho_{s} - \rho_{L})g}{9\eta}$
Assuming all other factors $(r, \rho_{s}, \rho_{L}, g)$ are constant,the relation between terminal velocity and viscosity is:
$v_{T} \propto \frac{1}{\eta}$
This represents a rectangular hyperbola,where $v_{T}$ decreases as $\eta$ increases. Therefore,the correct graph is the one showing an inverse relationship.

(A) Given:
Horizontal velocity of the plane,$u = 360 \text{ km/h} = 360 \times \frac{5}{18} \text{ m/s} = 100 \text{ m/s}$.
Altitude (vertical height),$h = 500 \text{ m}$.
Acceleration due to gravity,$g = 10 \text{ m/s}^2$.
When the bomb is dropped,it follows a projectile path. The time taken to reach the ground is given by $t = \sqrt{\frac{2h}{g}}$.
$t = \sqrt{\frac{2 \times 500}{10}} = \sqrt{100} = 10 \text{ s}$.
The horizontal distance covered by the bomb during this time is $R = u \times t$.
$R = 100 \text{ m/s} \times 10 \text{ s} = 1000 \text{ m}$.
Therefore,the bomb should be dropped at a distance of $1000 \text{ m}$ ahead of the target.

(D) The moment of inertia $(I)$ of a sphere about its diameter is given by $I = \frac{2}{5}MR^2$,where $M$ is the mass and $R$ is the radius.
Since the spheres have equal outer radii $(R_1 = R_2 = R)$ and equal moments of inertia $(I_1 = I_2)$,it follows that $\frac{2}{5}M_1R^2 = \frac{2}{5}M_2R^2$,which implies $M_1 = M_2$.
However,if the spheres are hollow or have different internal structures,they could have the same moment of inertia despite different masses if their mass distributions differ.
Given the problem states they have the same moment of inertia,the rotational kinetic energy $(K_r)$ is defined as $K_r = \frac{1}{2}I\omega^2$.
Since $I_1 = I_2$ and both are rotated with the same angular speed $(\omega_1 = \omega_2 = \omega)$,their rotational kinetic energies must be equal $(K_{r1} = K_{r2} = \frac{1}{2}I\omega^2)$.
Thus,option $(D)$ is the correct statement.

(A) For a simple pendulum of length $\ell$,the angular velocity $\omega_1$ at an angle $\theta$ below the horizontal is derived from energy conservation: $mg(\ell \sin \theta) = \frac{1}{2} m v^2 = \frac{1}{2} m (\omega_1 \ell)^2$. Thus,$\omega_1 = \sqrt{\frac{2g \sin \theta}{\ell}}$.
For a uniform bar of length $L$ pivoted at one end,the moment of inertia is $I = \frac{1}{3} m L^2$. The potential energy lost when it rotates by angle $\theta$ is $mg(\frac{L}{2} \sin \theta)$. By energy conservation: $mg(\frac{L}{2} \sin \theta) = \frac{1}{2} I \omega_2^2 = \frac{1}{2} (\frac{1}{3} m L^2) \omega_2^2$. Simplifying gives $\omega_2 = \sqrt{\frac{3g \sin \theta}{L}}$.
Since the motions are synchronous,$\omega_1 = \omega_2$,so $\frac{2}{\ell} = \frac{3}{L}$.
Therefore,$L = \frac{3}{2} \ell$.

(B) Let $Y_s$ and $Y_b$ be the Young's modulus of steel and brass respectively. Given $Y_s = 2 \times 10^{12} \,dynes/cm^{2}$ and $Y_b = 1 \times 10^{12} \,dynes/cm^{2}$.
Let $L = 50 \,cm$ be the length, $A = 0.005 \,cm^{2}$ be the cross-sectional area, and $\Delta L = 0.1 \,cm$ be the extension for both wires.
The tension $T$ in a wire is given by $T = \frac{Y A \Delta L}{L}$.
Since $A, \Delta L,$ and $L$ are the same for both wires, the tension $T$ is directly proportional to the Young's modulus $Y$ $(T \propto Y)$.
Let $T_s$ be the tension in the steel wire and $T_b$ be the tension in the brass wire.
$T_s = \frac{Y_s A \Delta L}{L}$ and $T_b = \frac{Y_b A \Delta L}{L}$.
Taking torque about the point where the load $W$ is applied at a distance $x$ from the steel wire:
$T_s \cdot x = T_b \cdot (15 - x)$.
Substituting the proportionality $T_s \propto Y_s$ and $T_b \propto Y_b$:
$Y_s \cdot x = Y_b \cdot (15 - x)$.
$(2 \times 10^{12}) \cdot x = (1 \times 10^{12}) \cdot (15 - x)$.
$2x = 15 - x$.
$3x = 15$.
$x = 5 \,cm$.

(D) Let $W$ be the water equivalent of the calorimeter in $g$.
Step $1$: Heat lost by hot water = Heat gained by cold water + Heat gained by calorimeter.
$100 \times 1 \times (100 - 20) = 300 \times 1 \times (20 - 10) + W \times 1 \times (20 - 10)$
$8000 = 3000 + 10W$
$10W = 5000 \implies W = 500 \ g$.
Step $2$: Heat lost by mixture = Heat gained by metallic block.
Mass of mixture = $100 \ g + 300 \ g = 400 \ g$.
Heat lost by mixture and calorimeter = $(400 \times 1 + 500) \times (20 - 19) = 900 \times 1 = 900 \ cal$.
Heat gained by metallic block = $m \times S_b \times \Delta T = 1000 \ g \times S_b \times (19 - 10) = 9000 \times S_b$.
Equating the two: $9000 \times S_b = 900$.
$S_b = 0.1 \ cal/g^{\circ} C$.

(C) The heat absorbed per cycle is $Q_H = 130 \text{ cal}$.
The heat rejected per cycle is $Q_C = 30 \text{ cal}$.
The net heat converted to work per cycle is $W_{\text{net}} = (Q_H - Q_C) \times 4.2 \text{ J/cal} = (130 - 30) \times 4.2 = 100 \times 4.2 = 420 \text{ J}$.
The energy consumed to overcome friction per cycle is $W_f = 2 \text{ J}$.
The useful work delivered to the load per cycle is $W_{\text{load}} = W_{\text{net}} - W_f = 420 \text{ J} - 2 \text{ J} = 418 \text{ J}$.
The engine operates at $90 \text{ cycles per minute}$, which is $90/60 = 1.5 \text{ cycles per second}$.
The power delivered to the load is $P = W_{\text{load}} \times \text{frequency} = 418 \text{ J} \times 1.5 \text{ s}^{-1} = 627 \text{ W}$.

(A) For a cyclic process, the change in internal energy $\Delta U$ over the complete cycle is zero.
According to the first law of thermodynamics, $\Delta Q = \Delta U + W$.
For the complete cycle $abca$, $\Delta U_{net} = 0$, so $\Delta Q_{net} = W_{net}$.
The net heat exchanged is $\Delta Q_{net} = \Delta Q_{ab} + \Delta Q_{bc} + \Delta Q_{ca}$.
Given:
$\Delta Q_{ab} = -50 \,J$ (heat rejected)
$\Delta Q_{bc} = 0 \,J$ (adiabatic process)
$\Delta Q_{ca} = 80 \,J$ (heat absorbed)
Thus, $\Delta Q_{net} = -50 + 0 + 80 = 30 \,J$.
Since $\Delta Q_{net} = W_{net}$, the net work done by the gas in the cycle is $30 \,J$.
The area enclosed by the $P-V$ diagram represents the net work done in the cycle.
Therefore, the area of the closed curve $abca$ is $30 \,J$.

(B) At equilibrium, the pressure $P_1$ and $P_2$ on both sides of the partition must be equal $(P_1 = P_2)$, and the temperature $T$ is the same for both gases.
Let $A$ be the cross-sectional area of the container. Let $x$ be the distance of the partition $P$ from the right wall $B$. Then the distance from the left wall $A$ is $(5 - x)$.
The volume of the left compartment is $V_1 = A(5 - x)$ and the volume of the right compartment is $V_2 = Ax$.
Using the ideal gas equation $PV = nRT = (m/M)RT$, where $m$ is the mass and $M$ is the molar mass:
For the left side: $P_1 V_1 = (m/32)RT$
For the right side: $P_2 V_2 = (m/18)RT$
Since $P_1 = P_2$ and $T$ is constant, we have $\frac{V_1}{V_2} = \frac{m/32}{m/18} = \frac{18}{32} = \frac{9}{16}$.
Substituting the volumes: $\frac{A(5 - x)}{Ax} = \frac{9}{16} \implies \frac{5 - x}{x} = \frac{9}{16}$.
$16(5 - x) = 9x \implies 80 - 16x = 9x \implies 25x = 80 \implies x = 3.2 \,m$.
The distance from the left wall $A$ is $5 - x = 5 - 3.2 = 1.8 \,m$.

(C) The time period of the pendulum is $T = 2 \,s$. The angular frequency is $\omega = \frac{2 \pi}{T} = \pi \,rad/s$.
At the lowest point,the displacement is $x = 0$,so the equation of motion is $x(t) = A \sin(\omega t)$,where $A$ is the amplitude.
The velocity at any time $t$ is $v(t) = \frac{dx}{dt} = A \omega \cos(\omega t)$.
At $t = 0$,$v(0) = v_{0} = A \omega$,so $A = \frac{v_{0}}{\omega}$.
We need the height $h$ at $t = 2.25 \,s$. Since $T = 2 \,s$,$t = 2.25 \,s = T + 0.25 \,s = T + \frac{T}{8}$.
At $t = \frac{T}{8} = 0.25 \,s$,the velocity is $v = v_{0} \cos(\omega \cdot \frac{T}{8}) = v_{0} \cos(\frac{2 \pi}{T} \cdot \frac{T}{8}) = v_{0} \cos(\frac{\pi}{4}) = \frac{v_{0}}{\sqrt{2}}$.
Using the principle of conservation of mechanical energy: $\frac{1}{2} m v_{0}^{2} = \frac{1}{2} m v^{2} + mgh$.
Substituting $v = \frac{v_{0}}{\sqrt{2}}$,we get $\frac{1}{2} v_{0}^{2} = \frac{1}{2} (\frac{v_{0}}{\sqrt{2}})^{2} + gh$.
$\frac{1}{2} v_{0}^{2} = \frac{1}{4} v_{0}^{2} + gh$.
$gh = \frac{1}{4} v_{0}^{2} \implies h = \frac{v_{0}^{2}}{4g}$.

(B) In a steady state,the capacitor acts as an open circuit,meaning no current flows through the branch containing the capacitor.
Let the current in the left loop be $I$. The circuit simplifies to a single loop consisting of the $2 \ V$ cell,the $2 \ V$ cell,the $4 \ \Omega$ resistor,the $2 \ \Omega$ resistor,and the $8 \ \Omega$ resistor in series.
Applying Kirchhoff's Voltage Law $(KVL)$ to the loop:
$2 \ V + 2 \ V - I(4 \ \Omega + 2 \ \Omega + 8 \ \Omega) = 0$
$4 \ V = I(14 \ \Omega)$
$I = \frac{4 \ V}{14 \ \Omega} = \frac{2}{7} \ A \approx 0.2857 \ A \approx 0.29 \ A$.
Thus,the current through the $2 \ \Omega$ resistor is approximately $0.29 \ A$.

(B) The wavelength $\lambda$ of radiation emitted by a hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right]$.
For each spectral series,the range of wavelengths is determined by the transition from $n_i = n_f + 1$ to $n_i = \infty$.
$1$. Lyman series $(n_f = 1)$: $\lambda$ ranges from $\frac{1}{R}$ to $\frac{4}{3R}$.
$2$. Balmer series $(n_f = 2)$: $\lambda$ ranges from $\frac{4}{R}$ to $\frac{36}{5R}$.
$3$. Paschen series $(n_f = 3)$: $\lambda$ ranges from $\frac{9}{R}$ to $\frac{144}{7R}$.
Comparing the given options,the range $\frac{7}{5R}$ to $\frac{19}{5R}$ does not correspond to any of the spectral series of the hydrogen atom.

(D) Given: $R = 400 \Omega$,$L = 250 \text{ mH} = 0.25 \text{ H}$,$C = 2.5 \mu \text{F} = 2.5 \times 10^{-6} \text{ F}$,$V_0 = 5 \text{ V}$,$\omega = 2000 \text{ rad/s}$.
First,calculate the inductive reactance: $X_L = \omega L = 2000 \times 0.25 = 500 \Omega$.
Next,calculate the capacitive reactance: $X_C = \frac{1}{\omega C} = \frac{1}{2000 \times 2.5 \times 10^{-6}} = \frac{1}{0.005} = 200 \Omega$.
The impedance of the circuit is $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{400^2 + (500 - 200)^2} = \sqrt{400^2 + 300^2} = 500 \Omega$.
The peak current in the circuit is $I_0 = \frac{V_0}{Z} = \frac{5}{500} = 0.01 \text{ A}$.
The peak voltage across the capacitor is $(V_C)_0 = I_0 X_C = 0.01 \times 200 = 2 \text{ V}$.
The peak electrostatic energy stored in the capacitor is $(U_C)_{\max} = \frac{1}{2} C (V_C)_0^2 = \frac{1}{2} \times 2.5 \times 10^{-6} \times (2)^2 = 5 \times 10^{-6} \text{ J} = 5 \mu \text{J}$.

(D) Let the equivalent resistance of the infinite network be $x$. Since the network is infinite,adding one more section to the front will not change the equivalent resistance. Thus,the circuit can be represented as a resistor $2R$ in series with the parallel combination of $R$ and $x$.
The equivalent resistance $x$ is given by:
$x = 2R + \frac{R \cdot x}{R + x}$
Multiplying both sides by $(R + x)$:
$x(R + x) = 2R(R + x) + Rx$
$x^2 + Rx = 2R^2 + 2Rx + Rx$
$x^2 - 2Rx - 2R^2 = 0$
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{2R \pm \sqrt{(-2R)^2 - 4(1)(-2R^2)}}{2(1)}$
$x = \frac{2R \pm \sqrt{4R^2 + 8R^2}}{2}$
$x = \frac{2R \pm \sqrt{12R^2}}{2}$
$x = \frac{2R \pm 2R\sqrt{3}}{2}$
$x = R(1 \pm \sqrt{3})$
Since resistance cannot be negative,we take the positive root:
$x = (1 + \sqrt{3})R$

(A) Let $G$ be the resistance of the galvanometer and $I_{g}$ be the current at full-scale deflection.
For the voltmeter conversion:
$V_{0} = I_{g}(G + R_{1})$
$G + R_{1} = \frac{V_{0}}{I_{g}}$
$G = \frac{V_{0}}{I_{g}} - R_{1} \dots (1)$
For the ammeter conversion:
$I_{g} G = (I_{0} - I_{g}) R_{2}$
$G = \frac{(I_{0} - I_{g}) R_{2}}{I_{g}} \dots (2)$
Equating $(1)$ and $(2)$:
$\frac{V_{0}}{I_{g}} - R_{1} = \frac{I_{0} R_{2}}{I_{g}} - R_{2}$
$\frac{V_{0} - I_{0} R_{2}}{I_{g}} = R_{1} - R_{2}$
$I_{g} = \frac{V_{0} - I_{0} R_{2}}{R_{1} - R_{2}}$

(B) The intensity $I$ of light from a point source follows the inverse square law,$I \propto \frac{1}{d^2}$,where $d$ is the distance from the source.
When the distance $d$ is doubled,the intensity $I$ becomes $\frac{1}{4}$ of its original value.
The photoelectric current is directly proportional to the intensity of incident light.
Since the intensity decreases,the number of photons hitting the surface per unit time decreases,resulting in a decrease in the photoelectric current.
Stopping potential and maximum kinetic energy depend on the frequency of the incident light,not on its intensity.
Therefore,the correct result is that the photoelectric current will decrease.

(C) In an $LR$ circuit,when a $DC$ voltage $V$ is applied,the current $I$ at any time $t$ is given by the equation $I(t) = \frac{V}{R}(1 - e^{-Rt/L})$.
Initially,at $t = 0$,the current is $I = 0$ because the inductor opposes the change in current.
As time $t$ increases,the current gradually rises and approaches a steady-state value of $I_{max} = \frac{V}{R}$ as $t \to \infty$.
This behavior matches the description provided in the question.
Therefore,the circuit contains a resistor and an inductor in series.

(B) $1$. For the circular loop: The circumference is $2 \pi R = L$, so $R = \frac{L}{2 \pi}$.
$2$. For the square loop: The perimeter is $4s = a$, where $s$ is the side length of the square. Thus, $s = \frac{a}{4}$.
$3$. The magnetic field $B$ at the centre of the circular loop due to a current $I$ flowing through it is $B = \frac{\mu_{0} I}{2R}$.
$4$. Since $a \ll R$, the magnetic field $B$ is approximately uniform over the area of the square loop. The magnetic flux $\phi$ linked with the square loop is $\phi = B \times \text{Area of square} = B \times s^2$.
$5$. Substituting the values: $\phi = \left( \frac{\mu_{0} I}{2R} \right) \times s^2 = \left( \frac{\mu_{0} I}{2(L / 2 \pi)} \right) \times \left( \frac{a}{4} \right)^2 = \left( \frac{\mu_{0} I \pi}{L} \right) \times \frac{a^2}{16} = \frac{\mu_{0} \pi a^2}{16 L} I$.
$6$. The mutual inductance $M$ is given by $M = \frac{\phi}{I} = \frac{\mu_{0} \pi a^2}{16 L}$.

(A, C, D) The net force on a charged particle moving with velocity $\vec{v}$ in the presence of electric field $\vec{E}$ and magnetic field $\vec{B}$ is given by the Lorentz force: $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$.
For the velocity to remain constant,the net force must be zero $(\vec{F} = 0)$.
Case $(i)$: If $\vec{E} = 0$ and $\vec{B} = 0$,then $\vec{F} = 0$. The particle moves with constant velocity.
Case (ii): If $\vec{E} \neq 0$ and $\vec{B} \neq 0$,the forces can cancel each other if $\vec{E} = -(\vec{v} \times \vec{B})$. This is possible.
Case (iii): If $\vec{E} = 0$ and $\vec{B} \neq 0$,the force is $\vec{F} = q(\vec{v} \times \vec{B})$. If $\vec{v}$ is parallel or anti-parallel to $\vec{B}$,then $\vec{v} \times \vec{B} = 0$,so $\vec{F} = 0$. The particle moves with constant velocity.
Case (iv): If $\vec{E} \neq 0$ and $\vec{B} = 0$,the force is $\vec{F} = q\vec{E}$. For $\vec{F} = 0$,$\vec{E}$ must be zero,which contradicts the assumption. Thus,this case is not possible.
Therefore,cases $(A)$,$(C)$,and $(D)$ are possible.

(A) The electric field $E_{x}$ due to two positive point charges $q_{1}$ and $q_{2}$ on the $x$-axis is given by the superposition principle: $E_{x} = \frac{k q_{1}}{x^{2}} + \frac{k q_{2}}{(x-10)^{2}}$ for $x < 0$,$E_{x} = \frac{k q_{1}}{x^{2}} - \frac{k q_{2}}{(x-10)^{2}}$ for $0 < x < 10$,and $E_{x} = -\frac{k q_{1}}{x^{2}} - \frac{k q_{2}}{(x-10)^{2}}$ for $x > 10$.
$1$. For $x < 0$: Both charges contribute to a negative electric field in the $x$-direction,so $E_{x} < 0$. As $x \to 0^{-}$,$E_{x} \to -\infty$.
$2$. For $0 < x < 10$: The field due to $q_{1}$ is positive and the field due to $q_{2}$ is negative. There is a point between the charges where $E_{x} = 0$. As $x \to 0^{+}$,$E_{x} \to +\infty$. As $x \to 10^{-}$,$E_{x} \to -\infty$.
$3$. For $x > 10$: Both charges contribute to a positive electric field in the $x$-direction,so $E_{x} > 0$. As $x \to 10^{+}$,$E_{x} \to +\infty$. As $x \to \infty$,$E_{x} \to 0$.
Comparing these characteristics with the given options,the graph in option $A$ correctly represents these behaviors.

(B) The electric field between the plates is $E = \frac{V}{d}$.
The force on an electron of charge $e$ is $F = eE = \frac{eV}{d}$.
The acceleration of the electron in the transverse direction ($y$-axis) is $a_y = \frac{F}{m} = \frac{eV}{md}$.
The time taken to travel an axial distance $L$ with constant horizontal velocity $v_0$ is $t = \frac{L}{v_0}$.
The transverse velocity $v_y$ acquired after time $t$ is $v_y = a_y t = \left(\frac{eV}{md}\right) \left(\frac{L}{v_0}\right) = \frac{eVL}{mdv_0}$.
The horizontal velocity remains constant at $v_x = v_0$.
The angle $\theta$ that the beam makes with the plates is given by $\tan \theta = \frac{v_y}{v_x}$.
Substituting the values,$\tan \theta = \frac{eVL / mdv_0}{v_0} = \frac{eVL}{mdv_0^2}$.
Therefore,$\theta = \tan ^{-1}\left(\frac{eVL}{mdv_0^2}\right)$.

(B) Let the initial angle between the strings be $2\theta$. The angle each string makes with the vertical is $\theta$.
For a pith ball of mass $m$ and charge $q$,the forces acting are tension $T$,gravitational force $mg$,and electrostatic force $F_e = \frac{kq^2}{(2L \sin \theta)^2}$.
Equating forces in equilibrium:
$T \sin \theta = F_e = \frac{kq^2}{4L^2 \sin^2 \theta}$
$T \cos \theta = mg$
Dividing the two equations:
$\tan \theta = \frac{kq^2}{4mgL^2 \sin^2 \theta} \implies \tan \theta \sin^2 \theta = \frac{kq^2}{4mgL^2} = C$ (constant).
When charge becomes $3q$,the angle between strings becomes $2(2\theta) = 4\theta$,so the angle with vertical becomes $2\theta$.
Thus,$\tan \theta \sin^2 \theta = \tan(2\theta) \sin^2(2\theta)$.
Using $\sin(2\theta) = 2 \sin \theta \cos \theta$ and $\tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta}$:
$\tan \theta \sin^2 \theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \cdot (2 \sin \theta \cos \theta)^2$
$\tan \theta \sin^2 \theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \cdot 4 \sin^2 \theta \cos^2 \theta$
$1 = \frac{8 \cos^2 \theta}{1 - \tan^2 \theta} = \frac{8 \cos^2 \theta}{1 - \frac{\sin^2 \theta}{\cos^2 \theta}} = \frac{8 \cos^4 \theta}{\cos^2 \theta - \sin^2 \theta} = \frac{8 \cos^4 \theta}{\cos(2\theta)}$.
Alternatively,using the ratio method:
$\frac{\tan 2\theta}{\tan \theta} = \frac{q'^2}{q^2} \cdot \frac{\sin^2 \theta}{\sin^2 2\theta} = 9 \cdot \frac{\sin^2 \theta}{4 \sin^2 \theta \cos^2 \theta} = \frac{9}{4 \cos^2 \theta} = \frac{9}{4} \sec^2 \theta$.
$\frac{2 \tan \theta}{1 - \tan^2 \theta} \cdot \frac{1}{\tan \theta} = \frac{9}{4} (1 + \tan^2 \theta) \implies \frac{2}{1 - \tan^2 \theta} = \frac{9}{4} (1 + \tan^2 \theta)$.
Let $x = \tan^2 \theta$: $\frac{2}{1-x} = \frac{9}{4}(1+x) \implies 8 = 9(1-x^2) \implies 9x^2 = 1 \implies x = 1/3$.
$\tan^2 \theta = 1/3 \implies \tan \theta = 1/\sqrt{3} \implies \theta = 30^{\circ}$.
The initial angle between the strings is $2\theta = 60^{\circ}$.

(B) To find the electric field inside a charged solid cylinder,we use Gauss's Law.
Consider a Gaussian surface in the form of a cylinder of radius '$x$' and length '$\ell$' coaxial with the given cylinder.
The total charge enclosed by this Gaussian surface is $q_{enc} = \rho \times V = \rho (\pi x^{2} \ell)$.
According to Gauss's Law,the electric flux through the Gaussian surface is $\oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{k \varepsilon_{0}}$.
Since the electric field is radial and uniform on the curved surface of the Gaussian cylinder,the flux is $E(2 \pi x \ell)$.
Equating the two expressions: $E(2 \pi x \ell) = \frac{\rho \pi x^{2} \ell}{k \varepsilon_{0}}$.
Solving for $E$,we get $E = \frac{\rho x}{2 k \varepsilon_{0}}$.

(C) According to the principle of conservation of energy,the initial total energy of the system must equal the final total energy.
Since the particles are released from rest,the initial kinetic energy $KE_i = 0$.
The initial potential energy $PE_i$ is the sum of the electrostatic potential energies of all pairs of charges.
In a square of side $a$,there are $4$ pairs of charges at distance $a$ (sides) and $2$ pairs of charges at distance $\sqrt{2}a$ (diagonals).
$PE_i = 4 \times \frac{kq^2}{a} + 2 \times \frac{kq^2}{\sqrt{2}a} = \frac{kq^2}{a} (4 + \sqrt{2})$.
When the particles are infinitely far apart,the final potential energy $PE_f = 0$.
Thus,$KE_f = PE_i = \frac{kq^2}{a} (4 + \sqrt{2})$.

(A) The magnetic force on a current-carrying conductor in a uniform magnetic field is given by $\overrightarrow{F} = I(\overrightarrow{L} \times \overrightarrow{B})$,where $\overrightarrow{L}$ is the displacement vector from the starting point to the end point of the conductor.
For any closed loop,the starting point and the end point are the same,which means the total displacement vector $\overrightarrow{L}$ is zero.
Therefore,the net magnetic force $\overrightarrow{F}$ on any closed current loop placed in a uniform magnetic field is always zero.

(D) The magnetic field at the center $O$ due to the circular loop of radius $r$ is $B_{loop} = \frac{\mu_{0} I}{2r}$ (directed outwards).
The magnetic field at the center $O$ due to the square loop of side $a$ is $B_{square} = 4 \times \frac{\mu_{0} I}{4 \pi (a/2)} (\sin 45^{\circ} + \sin 45^{\circ}) = \frac{4 \mu_{0} I}{\pi a} \times \frac{2}{\sqrt{2}} = \frac{4 \sqrt{2} \mu_{0} I}{\pi a}$ (directed inwards).
Given $a/r = 8/\pi$,so $r = \frac{\pi a}{8}$.
Substituting $r$ in $B_{loop}$,we get $B_{loop} = \frac{\mu_{0} I}{2(\pi a / 8)} = \frac{4 \mu_{0} I}{\pi a}$.
The net magnetic field $B_{net} = B_{square} - B_{loop} = \frac{4 \sqrt{2} \mu_{0} I}{\pi a} - \frac{4 \mu_{0} I}{\pi a} = \frac{4 \mu_{0} I}{\pi a} (\sqrt{2} - 1)$.
This can be rewritten as $\frac{2 \mu_{0} I}{\pi a} \times 2(\sqrt{2} - 1)$,which matches option $D$.

(C) The magnetic flux $\phi$ through the loop is given by $\phi = B \cdot A = 20 \times 10^{-2} \times 2 \sin(50 \pi t) = 0.4 \sin(50 \pi t) \,Wb$.
The induced current is $I = \frac{e}{R} = -\frac{1}{R} \frac{d\phi}{dt}$.
The charge flowing through the loop is $\Delta q = \int_{0}^{t} I \,dt = -\frac{1}{R} \int_{\phi_1}^{\phi_2} d\phi = -\frac{\phi_2 - \phi_1}{R}$.
At $t = 0$, $\phi_1 = 0.4 \sin(0) = 0 \,Wb$.
At $t = 20 \,ms = 0.02 \,s$, $\phi_2 = 0.4 \sin(50 \pi \times 0.02) = 0.4 \sin(\pi) = 0 \,Wb$.
Thus, the net charge $\Delta q = -\frac{0 - 0}{20} = 0 \,C$.

(A) The binding energy per nucleon $(BE/A)$ for a nucleus of mass number $A = 238$ is $7.6 \text{ MeV}$, and for a nucleus of mass number $A = 119$, it is $8.6 \text{ MeV}$.
Total binding energy of the parent nucleus $(A = 238)$ is $E_1 = 238 \times 7.6 \text{ MeV} = 1808.8 \text{ MeV}$.
When the nucleus splits into two fragments of mass number $119$ each, the total binding energy of the product nuclei is $E_2 = 2 \times (119 \times 8.6 \text{ MeV}) = 238 \times 8.6 \text{ MeV} = 2046.8 \text{ MeV}$.
The energy released in the fission process is $\Delta E = E_2 - E_1 = 2046.8 \text{ MeV} - 1808.8 \text{ MeV} = 238 \text{ MeV}$.
Among the given options, $214 \text{ MeV}$ is the closest approximate value.

(C) The nuclear reaction for beta-minus decay is given by: $_{Z}X^{A} \rightarrow _{Z+1}Y^{A} + _{-1}e^{0} + \bar{\nu} + Q$.
The $Q$-value of the reaction is the energy released,which is given by the mass defect multiplied by $c^{2}$.
The atomic mass $M_{x}$ of nucleus $X$ is related to its nuclear mass $m_{x}$ by $M_{x} = m_{x} + Zm_{e}$. Thus,$m_{x} = M_{x} - Zm_{e}$.
The atomic mass $M_{y}$ of nucleus $Y$ is related to its nuclear mass $m_{y}$ by $M_{y} = m_{y} + (Z+1)m_{e}$. Thus,$m_{y} = M_{y} - (Z+1)m_{e}$.
The energy released $Q$ is given by $Q = (m_{x} - m_{y} - m_{e})c^{2}$.
Substituting the expressions for nuclear masses:
$Q = [(M_{x} - Zm_{e}) - (M_{y} - (Z+1)m_{e}) - m_{e}]c^{2}$
$Q = [M_{x} - Zm_{e} - M_{y} + Zm_{e} + m_{e} - m_{e}]c^{2}$
$Q = (M_{x} - M_{y})c^{2}$.
Therefore,the maximum energy of the emitted beta particle is $(M_{x} - M_{y})c^{2}$.

(A) For the convex mirror,the object distance $u = -60 \ cm$ and focal length $f = +30 \ cm$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} + \frac{1}{-60} = \frac{1}{30}$
$\frac{1}{v} = \frac{1}{30} + \frac{1}{60} = \frac{2+1}{60} = \frac{3}{60} = \frac{1}{20}$
So,$v = +20 \ cm$ (behind the mirror).
The image formed by the convex mirror is at a distance of $20 \ cm$ behind the mirror.
The total distance of this image from the object is $60 \ cm + 20 \ cm = 80 \ cm$.
Let the plane mirror be placed at a distance $x$ from the object. The image formed by the plane mirror will be at a distance $x$ behind the plane mirror.
For no parallax,the image formed by the plane mirror must coincide with the image formed by the convex mirror.
The distance of the plane mirror's image from the object is $2x$.
Equating the distances: $2x = 80 \ cm$,which gives $x = 40 \ cm$.

(A) Step $1$: Find the focal length of the lens using the empty vessel condition.
For an empty vessel, the object distance $u = -80 \,cm$ and the image distance $v = +20 \,cm$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
$\frac{1}{f} = \frac{1}{20} - \frac{1}{-80} = \frac{1}{20} + \frac{1}{80} = \frac{4+1}{80} = \frac{5}{80} = \frac{1}{16} \,cm^{-1}$.
So, the focal length $f = 16 \,cm$.
Step $2$: Find the new object distance after pouring water.
When water of height $h = 64 \,cm$ is poured, the apparent depth of the coin is $d' = \frac{h}{\mu} = \frac{64}{4/3} = 64 \times \frac{3}{4} = 48 \,cm$.
The distance of the coin from the lens is $80 \,cm$. The distance of the apparent image from the bottom is $64 - 48 = 16 \,cm$. Thus, the new object distance $u'$ from the lens is $80 - 16 = 64 \,cm$ (or $16 \,cm$ air + $48 \,cm$ apparent depth).
So, $u' = -64 \,cm$.
Step $3$: Find the new image position $v'$.
Using the lens formula $\frac{1}{f} = \frac{1}{v'} - \frac{1}{u'}$:
$\frac{1}{16} = \frac{1}{v'} - \frac{1}{-64} \implies \frac{1}{v'} = \frac{1}{16} - \frac{1}{64} = \frac{4-1}{64} = \frac{3}{64}$.
$v' = \frac{64}{3} \approx 21.33 \,cm$.
Therefore, the image is formed $21.33 \,cm$ above the lens.

(A) Given:
$\beta = 48$
$I_{B} = 200 \mu A = 200 \times 10^{-6} \ A$
$R_{C} = 500 \ \Omega$
$V_{CC} = 5 \ V$
Step $1$: Calculate the collector current $I_{C}$.
$I_{C} = \beta I_{B} = 48 \times 200 \times 10^{-6} \ A = 9600 \times 10^{-6} \ A = 9.6 \times 10^{-3} \ A = 9.6 \ mA$.
Step $2$: Apply Kirchhoff's voltage law to the output loop.
$V_{CC} = I_{C} R_{C} + V_{CE}$
$V_{Y} = V_{CE} = V_{CC} - I_{C} R_{C}$
$V_{Y} = 5 \ V - (9.6 \times 10^{-3} \ A) \times (500 \ \Omega)$
$V_{Y} = 5 \ V - 4.8 \ V = 0.2 \ V$.
Thus,the voltage at terminal $Y$ is $0.2 \ V$.

(D) The voltage gain $A_v$ of a common emitter amplifier is defined as the ratio of the change in output voltage to the change in input voltage.
$A_v = \frac{\Delta V_{out}}{\Delta V_{in}} = \frac{\Delta I_C \times R_L}{\Delta V_{BE}}$
Given:
Load resistance $R_L = 6 k\Omega = 6000 \Omega$
Input signal voltage $\Delta V_{BE} = 15 mV = 15 \times 10^{-3} V$
Change in collector current $\Delta I_C = 1.8 mA = 1.8 \times 10^{-3} A$
Substituting the values into the formula:
$A_v = \frac{1.8 \times 10^{-3} A \times 6000 \Omega}{15 \times 10^{-3} V}$
$A_v = \frac{1.8 \times 6000}{15} = \frac{10800}{15} = 720$
Thus, the voltage gain of the amplifier is $720$.

(D) The condition for the $n^{th}$ order minima in a single slit diffraction is given by $a \sin \theta = n \lambda$. For small angles,$\sin \theta \approx \theta = \frac{y}{D}$.
Thus,the position of the $n^{th}$ minima is $y_n = \frac{n \lambda D}{a}$.
The linear separation between the first order minima on either side of the central maxima is the width of the central maxima,which is given by $w = y_1 - (-y_1) = 2y_1 = \frac{2 \lambda D}{a}$.
Given: $a = 0.5 ~mm = 0.5 \times 10^{-3} ~m$,$\lambda = 600 ~nm = 600 \times 10^{-9} ~m$,and $D = 50 ~cm = 0.5 ~m$.
Substituting the values: $w = \frac{2 \times 600 \times 10^{-9} \times 0.5}{0.5 \times 10^{-3}} = 1200 \times 10^{-6} ~m = 1.2 ~mm$.

(B) Let the intensities of light from the two slits be $I_1$ and $I_2$. Given that $I_1 = 1.5 I_2$,we have the ratio $\frac{I_1}{I_2} = 1.5 = \frac{3}{2}$.
The ratio of maximum intensity to minimum intensity in an interference pattern is given by the formula:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right)^2$.
Dividing the numerator and denominator by $\sqrt{I_2}$,we get:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{I_1/I_2} + 1}{\sqrt{I_1/I_2} - 1} \right)^2$.
Substituting $\frac{I_1}{I_2} = \frac{3}{2} = 1.5$:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{1.5} + 1}{\sqrt{1.5} - 1} \right)^2 \approx \left( \frac{1.225 + 1}{1.225 - 1} \right)^2 = \left( \frac{2.225}{0.225} \right)^2 \approx (9.88)^2 \approx 97.7 \approx 98$.