JEE Main 2021 Physics Question Paper with Answer and Solution

(A) For the isothermal process $A-B$:
$W_{AB} = nRT \ln \left(\frac{V_B}{V_A}\right) = RT \ln \left(\frac{2V_1}{V_1}\right) = RT \ln 2$.
For the isochoric process $B-C$:
Since the volume is constant,$W_{BC} = 0$.
For the adiabatic process $C-A$:
The work done in an adiabatic process is $W = \frac{P_i V_i - P_f V_f}{\gamma - 1}$.
Here,$P_C = \frac{P_1}{4}$,$V_C = 2V_1$,$P_A = P_1$,$V_A = V_1$.
$W_{CA} = \frac{P_C V_C - P_A V_A}{\gamma - 1} = \frac{(\frac{P_1}{4})(2V_1) - P_1 V_1}{\gamma - 1} = \frac{\frac{P_1 V_1}{2} - P_1 V_1}{\gamma - 1} = \frac{-\frac{P_1 V_1}{2}}{\gamma - 1} = -\frac{RT}{2(\gamma - 1)}$.
The net work done is $W_{net} = W_{AB} + W_{BC} + W_{CA} = RT \ln 2 + 0 - \frac{RT}{2(\gamma - 1)} = RT \left(\ln 2 - \frac{1}{2(\gamma - 1)}\right)$.

(A) travelling wave is represented by a function of the form $y = f(x \pm vt)$.
For a wave to be a travelling wave,the variables $x$ and $t$ must appear in the combination $(x \pm vt)$ within the function.
In option $A$,$y = A \sin (15 x - 2 t)$,which can be rewritten as $y = A \sin [15(x - \frac{2}{15}t)]$. This matches the form $f(x - vt)$ where $v = \frac{2}{15}$.
Options $B$,$C$,and $D$ do not represent a travelling wave because they do not satisfy the condition of being a function of $(x \pm vt)$.

(C) Let $\sigma$ be the uniform surface mass density of the disc.
Let $M_1$ be the mass of the full disc of radius $a$ and $M_2$ be the mass of the removed circular hole of radius $\frac{a}{2}$.
$M_1 = \sigma \pi a^2$,with its center of mass at $x_1 = a$.
$M_2 = \sigma \pi \left(\frac{a}{2}\right)^2 = \frac{\sigma \pi a^2}{4}$,with its center of mass at $x_2 = \frac{3a}{2}$.
The center of mass of the remaining portion is given by:
$x_{COM} = \frac{M_1 x_1 - M_2 x_2}{M_1 - M_2}$
$x_{COM} = \frac{(\sigma \pi a^2)(a) - (\frac{\sigma \pi a^2}{4})(\frac{3a}{2})}{\sigma \pi a^2 - \frac{\sigma \pi a^2}{4}}$
$x_{COM} = \frac{a - \frac{3a}{8}}{1 - \frac{1}{4}} = \frac{\frac{5a}{8}}{\frac{3}{4}} = \frac{5a}{8} \times \frac{4}{3} = \frac{5a}{6}$

(A) Given the force $F = -\alpha x^{2}$.
According to Newton's second law,$ma = -\alpha x^{2}$,so $a = -\frac{\alpha}{m} x^{2}$.
We know that $a = v \frac{dv}{dx}$.
Substituting this,we get $v \frac{dv}{dx} = -\frac{\alpha}{m} x^{2}$.
Separating variables,we have $v dv = -\frac{\alpha}{m} x^{2} dx$.
Integrating both sides with limits from $v = v_{0}$ to $v = 0$ and $x = 0$ to $x = x_{f}$:
$\int_{v_{0}}^{0} v dv = -\frac{\alpha}{m} \int_{0}^{x_{f}} x^{2} dx$.
Evaluating the integrals:
$\left[ \frac{v^{2}}{2} \right]_{v_{0}}^{0} = -\frac{\alpha}{m} \left[ \frac{x^{3}}{3} \right]_{0}^{x_{f}}$.
$0 - \frac{v_{0}^{2}}{2} = -\frac{\alpha}{m} \frac{x_{f}^{3}}{3}$.
$\frac{v_{0}^{2}}{2} = \frac{\alpha x_{f}^{3}}{3m}$.
Solving for $x_{f}$:
$x_{f}^{3} = \frac{3 m v_{0}^{2}}{2 \alpha}$.
$x_{f} = \left( \frac{3 m v_{0}^{2}}{2 \alpha} \right)^{\frac{1}{3}}$.

(B) The weight of the body at the poles is given by $W_p = mg = 49 \, N$.
Since $g = 9.8 \, m/s^2$,the mass of the body is $m = \frac{49}{9.8} = 5 \, kg$.
At the equator,the effective acceleration due to gravity is $g_e = g - R\omega^2$,where $\omega$ is the angular velocity of the Earth.
The angular velocity $\omega = \frac{2\pi}{T}$,where $T = 24 \times 3600 \, s$.
$R\omega^2 = R \left(\frac{2\pi}{T}\right)^2 = 6.4 \times 10^6 \times \left(\frac{2 \times 3.14}{86400}\right)^2 \approx 0.0337 \, m/s^2$.
The weight at the equator is $W_e = m(g - R\omega^2) = 5 \times (9.8 - 0.0337) = 5 \times 9.7663 = 48.8315 \, N$.
Rounding to two decimal places,the weight is $48.83 \, N$.

(A) The formula for Young's modulus $Y$ is given by $Y = \frac{F \cdot \ell}{A \cdot \Delta \ell}$,where $A$ is the cross-sectional area.
Rearranging for elongation $\Delta \ell$,we get $\Delta \ell = \frac{F \cdot \ell}{Y \cdot A}$.
Since $A = \pi r^2$ (where $r$ is the radius,which is half the diameter $D$),we have $\Delta \ell = \frac{F \cdot \ell}{Y \cdot \pi r^2} = \frac{4 F \cdot \ell}{Y \cdot \pi D^2}$.
From this,we see that $\Delta \ell \propto \frac{\ell}{D^2}$.
Let the initial length be $\ell_1$ and initial diameter be $D_1$. Let the final length be $\ell_2 = 2\ell_1$ and final diameter be $D_2 = 2D_1$.
Then,$\frac{\Delta \ell_2}{\Delta \ell_1} = \left( \frac{\ell_2}{\ell_1} \right) \left( \frac{D_1}{D_2} \right)^2$.
Substituting the values: $\frac{\Delta \ell_2}{\Delta \ell_1} = (2) \left( \frac{1}{2} \right)^2 = 2 \cdot \frac{1}{4} = \frac{1}{2}$.
Given $\Delta \ell_1 = 0.04\, m$,we have $\Delta \ell_2 = \frac{0.04}{2} = 0.02\, m$.
Converting to centimeters,$\Delta \ell_2 = 2\, cm$.

(B) The total length of the bar is $L = 2.4 \,m$. Since it is bent into a hexagon with $6$ sides,the length of each side is $\ell = \frac{2.4}{6} = 0.4 \,m$.
The mass of the total bar is $M = 6 \,kg$,so the mass of each side is $m = \frac{6}{6} = 1 \,kg$.
For an equilateral hexagon,the distance $r$ from the center to the midpoint of a side is given by $r = \ell \sin 60^{\circ} = \ell \frac{\sqrt{3}}{2}$.
The moment of inertia of one side about an axis passing through its center and perpendicular to its length is $I_{cm} = \frac{m \ell^2}{12}$.
Using the parallel axis theorem,the moment of inertia of one side about the center of the hexagon is $I_1 = I_{cm} + m r^2 = \frac{m \ell^2}{12} + m \left(\frac{\ell \sqrt{3}}{2}\right)^2 = \frac{m \ell^2}{12} + \frac{3 m \ell^2}{4} = \frac{m \ell^2 + 9 m \ell^2}{12} = \frac{10 m \ell^2}{12} = \frac{5}{6} m \ell^2$.
Since there are $6$ such sides,the total moment of inertia is $I = 6 \times I_1 = 6 \times \frac{5}{6} m \ell^2 = 5 m \ell^2$.
Substituting the values $m = 1 \,kg$ and $\ell = 0.4 \,m$:
$I = 5 \times 1 \times (0.4)^2 = 5 \times 0.16 = 0.8 \,kg \cdot m^2$.
Thus,$I = 8 \times 10^{-1} \,kg \cdot m^2$.

(B) The kinetic energy $K$ of a body with mass $m$ and linear momentum $P$ is given by the formula $K = \frac{P^2}{2m}$.
Since both solids $A$ and $B$ have equal linear momentum,we have $P_A = P_B = P$.
Therefore,the kinetic energy is inversely proportional to the mass: $K \propto \frac{1}{m}$.
Taking the ratio of the kinetic energies of $A$ and $B$,we get $\frac{(K.E.)_A}{(K.E.)_B} = \frac{m_B}{m_A}$.
Given $m_A = 1\, kg$ and $m_B = 2\, kg$,we substitute these values: $\frac{(K.E.)_A}{(K.E.)_B} = \frac{2}{1}$.
Comparing this with the given ratio $\frac{A}{1}$,we find that $A = 2$.

(C) The root mean square speed $(v_{rms})$ of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $R$ and $M$ are constants for a given gas,$v_{rms} \propto \sqrt{T}$.
Note that $v_{rms}$ is independent of pressure.
Given $T_1 = 27^{\circ} C = 300\, K$ and $(v_{rms})_1 = 200\, ms^{-1}$.
Given $T_2 = 127^{\circ} C = 400\, K$.
Using the ratio: $\frac{(v_{rms})_2}{(v_{rms})_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{400}{300}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.
Therefore,$(v_{rms})_2 = \frac{2}{\sqrt{3}} \times (v_{rms})_1 = \frac{2}{\sqrt{3}} \times 200 = \frac{400}{\sqrt{3}}\, ms^{-1}$.
Comparing this with $\frac{x}{\sqrt{3}}$,we get $x = 400$.

(A) Given: Speed of each car $v_s = v_o = 7.2\, km/hr = 7.2 \times \frac{5}{18} = 2\, m/s$. Frequency of horn $f_0 = 676\, Hz$. Velocity of sound $v = 340\, m/s$.
Each driver hears two sounds: one directly from the other car and one reflected from the other car.
$1$. Frequency of sound heard directly from the other car $(f_d)$:
$f_d = f_0 \left( \frac{v + v_o}{v - v_s} \right) = 676 \left( \frac{340 + 2}{340 - 2} \right) = 676 \left( \frac{342}{338} \right) = 684\, Hz$.
$2$. Frequency of sound reflected from the other car $(f_r)$:
The sound from the other car travels to the driver's car,reflects,and returns. Effectively,the other car acts as a source moving towards the driver,and the driver acts as a source reflecting the sound back. The frequency heard after reflection is:
$f_r = f_0 \left( \frac{v + v_s}{v - v_s} \right) \left( \frac{v + v_o}{v - v_o} \right) = 676 \left( \frac{342}{338} \right) \left( \frac{342}{338} \right) = 676 \left( \frac{342}{338} \right)^2 \approx 692\, Hz$.
Beat frequency = $f_r - f_d = 692 - 684 = 8\, Hz$.

(D) For a polytropic process, the equation is $PV^{x} = \text{constant}$.
Given the process equation is $PV^{1/2} = \text{constant}$.
Using the ideal gas equation $PV = nRT$, we can write $P = \frac{nRT}{V}$.
Substituting this into the process equation:
$\left(\frac{nRT}{V}\right) V^{1/2} = \text{constant}$
$T V^{-1} V^{1/2} = \text{constant}$
$T V^{-1/2} = \text{constant}$
Thus, $T_{1} V_{1}^{-1/2} = T_{2} V_{2}^{-1/2}$.
Rearranging for the ratio of temperatures:
$\frac{T_{2}}{T_{1}} = \left(\frac{V_{2}}{V_{1}}\right)^{-1/2} = \left(\frac{V_{1}}{V_{2}}\right)^{1/2}$.
Given $V_{2} = 2V_{1}$, we have $\frac{V_{1}}{V_{2}} = \frac{1}{2}$.
Therefore, $\frac{T_{2}}{T_{1}} = \left(\frac{1}{2}\right)^{1/2} = \frac{1}{\sqrt{2}}$.
Hence, the correct option is $D$.

(B) Let the total height of the building be $H$. The first stone is dropped from the top $(u=0)$.
When it has fallen $5 \, m$,its velocity $v$ is given by $v^2 = u^2 + 2gh = 0 + 2 \times 10 \times 5 = 100$,so $v = 10 \, m/s$.
The time taken by the first stone to fall $5 \, m$ is $t_1 = v/g = 10/10 = 1 \, s$.
At this instant,the second stone is dropped from a point $25 \, m$ below the top. Let the total time taken by the first stone to reach the ground be $T$. Then the time taken by the second stone is $(T - 1) \, s$.
The distance covered by the first stone is $H = \frac{1}{2} g T^2$.
The distance covered by the second stone is $H - 25 = \frac{1}{2} g (T - 1)^2$.
Subtracting the two equations: $H - (H - 25) = \frac{1}{2} g [T^2 - (T - 1)^2] \Rightarrow 25 = 5 [T^2 - (T^2 - 2T + 1)] \Rightarrow 25 = 5(2T - 1) \Rightarrow 5 = 2T - 1 \Rightarrow 2T = 6 \Rightarrow T = 3 \, s$.
Substituting $T = 3 \, s$ into the first equation: $H = \frac{1}{2} \times 10 \times (3)^2 = 5 \times 9 = 45 \, m$.

(B) According to the equipartition theorem,the average energy associated with each degree of freedom is $\frac{1}{2} k_B T$.
For a diatomic molecule,the translational degrees of freedom are $3$,so the average translational kinetic energy is $E_{trans} = 3 \times (\frac{1}{2} k_B T) = \frac{3}{2} k_B T$.
The rotational degrees of freedom for a rigid diatomic molecule are $2$,so the average rotational kinetic energy is $E_{rot} = 2 \times (\frac{1}{2} k_B T) = k_B T$.
Statement $I$ is false because rotational energy levels are quantized and do not follow the continuous Maxwell-Boltzmann distribution in the same way as translational energy.
Statement $II$ is false because $E_{rot} = k_B T$ while $E_{trans} = \frac{3}{2} k_B T$,so they are not equal.

(D) When a block of mass $m$ is connected to two springs of spring constant $k_1$ and $k_2$ in parallel,the effective spring constant is $k_{eq} = k_1 + k_2$.
In this problem,both springs have a spring constant of $2k$.
Therefore,the equivalent spring constant is $k_{eq} = 2k + 2k = 4k$.
The time period $T$ of a simple harmonic oscillator is given by the formula $T = 2\pi \sqrt{\frac{m}{k_{eq}}}$.
Substituting the value of $k_{eq}$,we get $T = 2\pi \sqrt{\frac{m}{4k}}$.
Simplifying this,$T = 2\pi \frac{1}{2} \sqrt{\frac{m}{k}} = \pi \sqrt{\frac{m}{k}}$.

(A) The total initial energy of the sphere is $E_i = \frac{1}{2}mv_0^2 + \frac{1}{2}I\omega_0^2$.
Since it rolls without slipping,$\omega_0 = \frac{v_0}{a}$.
For a solid sphere,the moment of inertia is $I = \frac{2}{5}ma^2$.
Substituting these into the energy equation: $E_i = \frac{1}{2}mv_0^2 + \frac{1}{2}(\frac{2}{5}ma^2)(\frac{v_0}{a})^2 = \frac{1}{2}mv_0^2 + \frac{1}{5}mv_0^2 = \frac{7}{10}mv_0^2$.
At the maximum height $h$,the sphere momentarily comes to rest,so its final kinetic energy is zero. By the law of conservation of energy,$E_i = E_f = mgh$.
Therefore,$mgh = \frac{7}{10}mv_0^2$,which gives $h = \frac{7v_0^2}{10g}$.
The distance $d$ traveled along the incline is related to height $h$ by $h = d \sin \theta$.
Thus,$d \sin \theta = \frac{7v_0^2}{10g}$,which implies $d = \frac{7v_0^2}{10g \sin \theta}$.
This matches option $A$.

(C) The displacement equation is $Y = A \sin (\omega t + \phi_{0})$.
At $t = 0$,$Y = A \sin(\phi_{0}) = \frac{A}{2}$.
This implies $\sin(\phi_{0}) = \frac{1}{2}$.
Thus,$\phi_{0}$ can be $\frac{\pi}{6}$ or $\frac{5 \pi}{6}$.
The velocity of the particle is given by $v = \frac{dY}{dt} = A \omega \cos(\omega t + \phi_{0})$.
At $t = 0$,$v = A \omega \cos(\phi_{0})$.
Since the particle is moving in the negative direction,$v < 0$,which means $\cos(\phi_{0}) < 0$.
For $\phi_{0} = \frac{\pi}{6}$,$\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} > 0$.
For $\phi_{0} = \frac{5 \pi}{6}$,$\cos(\frac{5 \pi}{6}) = -\frac{\sqrt{3}}{2} < 0$.
Therefore,the initial phase angle is $\phi_{0} = \frac{5 \pi}{6}$.

(A) The Coulomb force between two charges $e$ separated by distance $r$ is given by $F = \frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{r^{2}}$.
From this,the dimension of $\frac{e^{2}}{4 \pi \varepsilon_{0}}$ is $[F r^{2}] = [M L T^{-2} \cdot L^{2}] = [M L^{3} T^{-2}]$.
The energy of a photon is $E = \frac{hc}{\lambda}$,so the dimension of $hc$ is $[E \lambda] = [M L^{2} T^{-2} \cdot L] = [M L^{3} T^{-2}]$.
Dividing these two quantities,the dimensions are $\frac{[M L^{3} T^{-2}]}{[M L^{3} T^{-2}]} = [M^{0} L^{0} T^{0}]$.
Thus,the quantity is dimensionless.

(D) Given,the point $A$ covers an angle $\theta = 30^{\circ} = \frac{\pi}{6} \text{ radians}$ in time $t = 0.1\, s$.
Since the speed is uniform,the angular velocity $\omega$ is given by $\omega = \frac{\theta}{t} = \frac{\pi/6}{0.1} = \frac{\pi}{0.6} = \frac{10\pi}{6} = \frac{5\pi}{3} \text{ rad/s}$.
The radius of the circle $R = 0.36\, m$,which represents the amplitude $A_{amp}$ of the simple harmonic motion.
The restoring force $F$ on a particle of mass $m$ executing simple harmonic motion is given by $F = -m\omega^2 x$,where $x$ is the displacement.
When $P$ touches $M$,the displacement $x$ is equal to the amplitude $R = 0.36\, m$.
The restoring force per unit mass is $\frac{F}{m} = \omega^2 R$.
Substituting the values: $\frac{F}{m} = \left(\frac{5\pi}{3}\right)^2 \times 0.36 = \frac{25\pi^2}{9} \times 0.36$.
Using $\pi^2 \approx 9.87$,we get $\frac{F}{m} = \frac{25 \times 9.87}{9} \times 0.36 = 25 \times 9.87 \times 0.04 = 25 \times 0.3948 = 9.87\, N/kg$.

(C) Using the law of conservation of energy between the surface of the earth and the maximum height $h = 10 R$:
Total energy at surface = Total energy at height $h$.
$-\frac{GMm}{R} + \frac{1}{2}mv_{i}^{2} = -\frac{GMm}{R+h} + 0$.
Since $h = 10R$,the total distance from the center is $R + 10R = 11R$.
$-\frac{GMm}{R} + \frac{1}{2}mv_{i}^{2} = -\frac{GMm}{11R}$.
$\frac{1}{2}mv_{i}^{2} = GMm \left( \frac{1}{R} - \frac{1}{11R} \right) = GMm \left( \frac{10}{11R} \right)$.
$v_{i}^{2} = \frac{20GM}{11R}$.
We know the escape velocity $v_{e} = \sqrt{\frac{2GM}{R}}$,so $v_{e}^{2} = \frac{2GM}{R}$.
Substituting this into the equation for $v_{i}^{2}$:
$v_{i}^{2} = \frac{10}{11} \times \left( \frac{2GM}{R} \right) = \frac{10}{11} v_{e}^{2}$.
Comparing this with $v_{i} = \sqrt{\frac{x}{y}} v_{e}$,we get $\frac{x}{y} = \frac{10}{11}$.
Thus,$x = 10$.

(B) The efficiency of a heat engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature.
Given $\eta = \frac{1}{4}$,we have $\frac{1}{4} = 1 - \frac{T_2}{T_1}$,which implies $\frac{T_2}{T_1} = \frac{3}{4}$,so $T_2 = 0.75 T_1$.
When the sink temperature is reduced by $52 \, K$,the new efficiency $\eta' = 2 \times \eta = 2 \times \frac{1}{4} = \frac{1}{2}$.
The new sink temperature is $T_2' = T_2 - 52$.
Using the efficiency formula for the new state: $\frac{1}{2} = 1 - \frac{T_2 - 52}{T_1}$.
This simplifies to $\frac{T_2 - 52}{T_1} = \frac{1}{2}$,or $T_2 - 52 = 0.5 T_1$.
Substituting $T_2 = 0.75 T_1$ into the equation: $0.75 T_1 - 52 = 0.5 T_1$.
$0.25 T_1 = 52$,which gives $T_1 = \frac{52}{0.25} = 208 \, K$.

(B) The speed of a transverse wave in a stretched string is given by the formula $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Taking the natural logarithm on both sides,we get $\ln v = \frac{1}{2} \ln T - \frac{1}{2} \ln \mu$.
Differentiating both sides,we obtain the relative error formula: $\frac{\Delta v}{v} = \frac{1}{2} \frac{\Delta T}{T}$.
Given that the percentage increase in tension is $\frac{\Delta T}{T} \times 100 = 4\, \%$.
Substituting this value into the error formula: $\frac{\Delta v}{v} \times 100 = \frac{1}{2} \times (4\, \%) = 2\, \%$.
Therefore,the percentage increase in the speed of the wave is $2\, \%$.

(C) The cross product of two vectors is anti-commutative,meaning $\overrightarrow{ P } \times \overrightarrow{ Q } = -(\overrightarrow{ Q } \times \overrightarrow{ P })$.
Given the condition $\overrightarrow{ P } \times \overrightarrow{ Q } = \overrightarrow{ Q } \times \overrightarrow{ P }$,we can substitute the anti-commutative property:
$-(\overrightarrow{ Q } \times \overrightarrow{ P }) = \overrightarrow{ Q } \times \overrightarrow{ P }$.
This implies $2(\overrightarrow{ Q } \times \overrightarrow{ P }) = 0$,so $\overrightarrow{ Q } \times \overrightarrow{ P } = 0$.
The magnitude of the cross product is given by $|\overrightarrow{ Q } \times \overrightarrow{ P }| = PQ \sin \theta = 0$.
Since $P$ and $Q$ are non-zero vectors,$\sin \theta = 0$.
In the range $0^{\circ} < \theta < 360^{\circ}$,$\sin \theta = 0$ at $\theta = 180^{\circ}$ (since $0^{\circ}$ and $360^{\circ}$ are excluded).
Therefore,the value of $\theta$ is $180^{\circ}$.

(A) The kinetic energy $K$ of a particle with mass $m$ and linear momentum $p$ is given by the formula $K = \frac{p^2}{2m}$.
Since the kinetic energies of the two particles are equal,we have $K_1 = K_2$.
Substituting the formula,we get $\frac{p_1^2}{2m_1} = \frac{p_2^2}{2m_2}$.
Given $m_1 = 4\, g$ and $m_2 = 16\, g$,we have $\frac{p_1^2}{2 \times 4} = \frac{p_2^2}{2 \times 16}$.
Simplifying this,$\frac{p_1^2}{4} = \frac{p_2^2}{16}$,which implies $\frac{p_1^2}{p_2^2} = \frac{4}{16} = \frac{1}{4}$.
Taking the square root of both sides,$\frac{p_1}{p_2} = \frac{1}{2}$.
We are given that the ratio of the magnitudes of their linear momentum is $n : 2$,so $\frac{p_1}{p_2} = \frac{n}{2}$.
Comparing $\frac{n}{2} = \frac{1}{2}$,we find $n = 1$.

(C) Assuming Hooke's law is valid,the tension $T$ is proportional to the extension $\Delta \ell = \ell - \ell_{0}$,where $\ell_{0}$ is the original length.
$T = k(\ell - \ell_{0})$
For the two given states:
$T_{1} = k(\ell_{1} - \ell_{0})$ --- $(1)$
$T_{2} = k(\ell_{2} - \ell_{0})$ --- $(2)$
Dividing equation $(1)$ by $(2)$:
$\frac{T_{1}}{T_{2}} = \frac{\ell_{1} - \ell_{0}}{\ell_{2} - \ell_{0}}$
Cross-multiplying:
$T_{1}(\ell_{2} - \ell_{0}) = T_{2}(\ell_{1} - \ell_{0})$
$T_{1}\ell_{2} - T_{1}\ell_{0} = T_{2}\ell_{1} - T_{2}\ell_{0}$
Rearranging to solve for $\ell_{0}$:
$T_{2}\ell_{0} - T_{1}\ell_{0} = T_{2}\ell_{1} - T_{1}\ell_{2}$
$\ell_{0}(T_{2} - T_{1}) = T_{2}\ell_{1} - T_{1}\ell_{2}$
$\ell_{0} = \frac{T_{2}\ell_{1} - T_{1}\ell_{2}}{T_{2} - T_{1}}$

(C) Let the frequency of tuning fork $A$ be $f_A$. The frequency of the known fork is $f = 340 \, Hz$.
The initial beat frequency is $|f_A - 340| = 5 \, Hz$,which implies $f_A = 345 \, Hz$ or $f_A = 335 \, Hz$.
When a tuning fork is filed,its frequency increases.
Case $1$: If $f_A = 345 \, Hz$,filing increases the frequency to $f_A' > 345 \, Hz$. The new beat frequency would be $|f_A' - 340| > 5 \, Hz$. This contradicts the given information that the beat frequency decreases to $2 \, Hz$.
Case $2$: If $f_A = 335 \, Hz$,filing increases the frequency to $f_A' = 335 + \Delta f$. The new beat frequency is $|340 - (335 + \Delta f)| = |5 - \Delta f| = 2 \, Hz$.
This gives $5 - \Delta f = 2$,so $\Delta f = 3 \, Hz$,which is possible.
Therefore,the original frequency of fork $A$ is $335 \, Hz$.

(A) The given equation of the trajectory is $y = \alpha x - \beta x^2$.
Comparing this with the standard equation of a projectile trajectory: $y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$.
By comparing the coefficients of $x$,we get $\tan \theta = \alpha$,which implies $\theta = \tan^{-1} \alpha$.
By comparing the coefficients of $x^2$,we get $\beta = \frac{g}{2u^2 \cos^2 \theta}$.
Rearranging for $u^2$,we find $u^2 = \frac{g}{2\beta \cos^2 \theta}$.
The formula for maximum height is $H = \frac{u^2 \sin^2 \theta}{2g}$.
Substituting $u^2$ into the formula: $H = \frac{g}{2\beta \cos^2 \theta} \cdot \frac{\sin^2 \theta}{2g} = \frac{\tan^2 \theta}{4\beta}$.
Since $\tan \theta = \alpha$,we have $H = \frac{\alpha^2}{4\beta}$.

(B) According to the law of conservation of energy,the loss in potential energy of the weight is equal to the gain in the total kinetic energy of the system (rotational kinetic energy of the wheel + translational kinetic energy of the weight).
Loss in potential energy = $mgh$
Gain in kinetic energy = $\frac{1}{2} I \omega^2 + \frac{1}{2} mv^2$
Since the cord does not slip,the linear velocity $v$ of the weight is related to the angular velocity $\omega$ of the wheel by $v = \omega r$.
Substituting $v = \omega r$ into the energy equation:
$mgh = \frac{1}{2} I \omega^2 + \frac{1}{2} m(\omega r)^2$
$mgh = \frac{1}{2} (I + mr^2) \omega^2$
Solving for $\omega^2$:
$\omega^2 = \frac{2 mgh}{I + mr^2}$

(B) For an ideal gas,the internal energy is given by $U = \frac{f}{2} PV$,where $f$ is the degree of freedom.
Given the relation $U = 3PV + 4$,we equate the two expressions:
$\frac{f}{2} PV = 3PV + 4$
Dividing both sides by $PV$,we get:
$\frac{f}{2} = 3 + \frac{4}{PV}$
$f = 6 + \frac{8}{PV}$
Since the term $\frac{8}{PV}$ is positive for a gas,$f > 6$.
$A$ gas with degrees of freedom $f > 6$ is classified as polyatomic.

(D) The acceleration of the center of mass of the system is given by $a_{cm} = \frac{ F_{ext} }{ M_{total} }$.
Here,the external force is $F$ and the total mass is $M + M = 2M$.
So,$a_{cm} = \frac{ F }{ 2M }$.
Also,$a_{cm} = \frac{ m_A a_A + m_B a_B }{ m_A + m_B }$,where $a_A = a$ (acceleration of mass $A$) and $a_B$ is the acceleration of mass $B$.
Substituting the values: $\frac{ F }{ 2M } = \frac{ M(a) + M(a_B) }{ 2M }$.
Canceling $2M$ from both sides,we get $F = Ma + Ma_B$.
Rearranging for $a_B$,we get $Ma_B = F - Ma$.
Therefore,$a_B = \frac{ F - Ma }{ M }$.

(B) Let $v_{\max}$ be the maximum velocity attained by the scooter.
During the acceleration phase,the final velocity is $v_{\max} = 0 + a_{1}t_{1}$,so $t_{1} = \frac{v_{\max}}{a_{1}}$.
During the retardation phase,the final velocity is $0 = v_{\max} - a_{2}t_{2}$,so $t_{2} = \frac{v_{\max}}{a_{2}}$.
Taking the ratio of the two times:
$\frac{t_{1}}{t_{2}} = \frac{v_{\max} / a_{1}}{v_{\max} / a_{2}} = \frac{a_{2}}{a_{1}}$.
Alternatively,using the velocity-time graph,the slope of the acceleration part is $a_{1} = \tan \theta_{1} = \frac{v_{\max}}{t_{1}}$ and the magnitude of the slope of the retardation part is $a_{2} = \tan \theta_{2} = \frac{v_{\max}}{t_{2}}$.
Thus,$\frac{t_{1}}{t_{2}} = \frac{a_{2}}{a_{1}}$.

(B) second's pendulum is defined as a pendulum whose time period of oscillation is $2 \, s$.
Therefore,Statement $I$ is false.
The time taken to move from one extreme position to the other extreme position is equal to half of the time period $(T/2)$.
Since $T = 2 \, s$,the time taken is $2 \, s / 2 = 1 \, s$.
Therefore,Statement $II$ is true.

(B) The work done by a gas is given by $W = \int P\,dV$.
From the ideal gas law,$P = \frac{nRT}{V}$.
Substituting this into the work formula: $W = \int \frac{nRT}{V}\,dV$.
Given the relation $V = KT^{2/3}$,we differentiate with respect to $T$: $dV = K \cdot \frac{2}{3} T^{-1/3}\,dT$.
Substituting $V$ and $dV$ into the integral: $W = \int_{T_1}^{T_2} \frac{nRT}{KT^{2/3}} \cdot \left( \frac{2}{3} K T^{-1/3} \right) dT$.
Simplifying the expression: $W = \int_{T_1}^{T_2} \frac{2}{3} nR \cdot \frac{T \cdot T^{-1/3}}{T^{2/3}} \,dT = \int_{T_1}^{T_2} \frac{2}{3} nR \,dT$.
Integrating,we get $W = \frac{2}{3} nR (T_2 - T_1)$.
Given $\Delta T = T_2 - T_1 = 90\,K$ and assuming $n = 1$ mole: $W = \frac{2}{3} \cdot 1 \cdot R \cdot 90 = 60R$.
Thus,$x = 60$.

(D) The velocity of a particle in $S.H.M.$ at displacement $y$ is given by $v = \omega \sqrt{a^2 - y^2}$,where $a$ is the amplitude and $\omega$ is the angular frequency.
Maximum speed is $v_{\max} = a\omega$.
Given that the speed $v = \frac{v_{\max}}{2}$,we have $\frac{a\omega}{2} = \omega \sqrt{a^2 - y^2}$.
Squaring both sides: $\frac{a^2}{4} = a^2 - y^2$.
Rearranging for displacement $y$: $y^2 = a^2 - \frac{a^2}{4} = \frac{3a^2}{4}$.
Thus,$y = \frac{\sqrt{3} a}{2}$.
Comparing this with the given expression $\frac{\sqrt{x} a}{2}$,we find $x = 3$.

(A) Starting from the mean position,the particle moves to the extreme position ($1/4$ oscillation),back to the mean position ($1/2$ oscillation),and then to the other side.
$5/8$ of an oscillation is equivalent to $1/2 + 1/8$ of an oscillation.
In terms of phase,$1/2$ oscillation corresponds to a phase change of $\pi$ radians.
The remaining $1/8$ oscillation corresponds to a phase change of $\frac{1}{8} \times 2\pi = \frac{\pi}{4}$ radians.
However,the question asks for the time taken to reach the position corresponding to $5/8$ of the cycle. Using the reference circle method,the displacement $y = A \sin(\omega t)$.
For $5/8$ of the cycle,the phase angle $\phi = \frac{5}{8} \times 2\pi = \frac{5\pi}{4}$.
Since the particle starts from the mean position,we consider the phase relative to the mean position.
The time $t$ is given by $\omega t = \phi$,where $\omega = \frac{2\pi}{T}$.
Thus,$t = \frac{\phi}{\omega} = \frac{5\pi/4}{2\pi/T} = \frac{5}{8} T$.
Wait,checking the provided solution logic: The particle reaches the extreme at $T/4$,returns to mean at $T/2$. To complete $5/8$,it travels an additional $1/8$ of the cycle from the mean position in the negative direction. The displacement is $y = A \sin(\omega t)$. At $t = 5T/8$,$y = A \sin(2\pi/T \times 5T/8) = A \sin(5\pi/4) = -A/\sqrt{2}$.
If the question implies the time to reach the displacement corresponding to $5/8$ of the cycle,then $\alpha = 5$.

(C) Let $R$ be the radius of the Earth. Point $A$ is at a distance $r$ from the center $O$ inside the Earth. Point $C$ is at a height $h = 3200 \text{ km} = R/2$ above the surface $B$.
The acceleration due to gravity at point $A$ (inside the Earth) is given by $g_A = \frac{G M r}{R^3}$.
The acceleration due to gravity at point $C$ (outside the Earth) is given by $g_C = \frac{G M}{(R + h)^2} = \frac{G M}{(R + R/2)^2} = \frac{G M}{(3R/2)^2} = \frac{4 G M}{9 R^2}$.
Given that $g_A = g_C$,we equate the expressions:
$\frac{G M r}{R^3} = \frac{4 G M}{9 R^2}$
Solving for $r$:
$r = \frac{4 R}{9}$.
Thus,$OA = r = \frac{4 R}{9}$.
The distance $AB = R - r = R - \frac{4 R}{9} = \frac{5 R}{9}$.
Therefore,the ratio $OA : AB = \frac{4 R}{9} : \frac{5 R}{9} = 4 : 5$.
Since the ratio is $x : y = 4 : 5$,the value of $x$ is $4$.

(B) From the first law of thermodynamics,$Q = \Delta U + W.$
Given that $W = Q/5,$ we have $Q = \Delta U + Q/5.$
Therefore,$\Delta U = Q - Q/5 = 4Q/5.$
For a rigid diatomic gas,the internal energy change is $\Delta U = n C_v \Delta T,$ where $C_v = \frac{5}{2}R.$
Also,$Q = n C \Delta T,$ where $C$ is the molar heat capacity.
Substituting these into the equation $\Delta U = \frac{4}{5}Q,$ we get $n C_v \Delta T = \frac{4}{5} (n C \Delta T).$
$C_v = \frac{4}{5} C \implies C = \frac{5}{4} C_v.$
Substituting $C_v = \frac{5}{2}R,$ we get $C = \frac{5}{4} \times \frac{5}{2} R = \frac{25}{8} R.$
Comparing this with $\frac{xR}{8},$ we find $x = 25.$

(A) Let the magnitude of each vector be $\lambda$.
$\overrightarrow{ OA } = \lambda (\cos 30^{\circ} \hat{i} + \sin 30^{\circ} \hat{j}) = \lambda (\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j})$
$\overrightarrow{ OB } = \lambda (\cos 60^{\circ} \hat{i} - \sin 60^{\circ} \hat{j}) = \lambda (\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j})$
$\overrightarrow{ OC } = \lambda (\cos 45^{\circ} (-\hat{i}) + \sin 45^{\circ} \hat{j}) = \lambda (-\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j})$
Now,calculate $\overrightarrow{ R } = \overrightarrow{ OA } + \overrightarrow{ OB } - \overrightarrow{ OC }$.
$\overrightarrow{ R } = \lambda [(\frac{\sqrt{3}}{2} + \frac{1}{2} - (-\frac{1}{\sqrt{2}})) \hat{i} + (\frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}) \hat{j}]$
$\overrightarrow{ R } = \lambda [(\frac{\sqrt{3}+1+\sqrt{2}}{2}) \hat{i} + (\frac{1-\sqrt{3}-\sqrt{2}}{2}) \hat{j}]$
The angle $\theta$ with the $x$-axis is given by $\tan \theta = \frac{R_y}{R_x}$.
$\tan \theta = \frac{\frac{1-\sqrt{3}-\sqrt{2}}{2}}{\frac{\sqrt{3}+1+\sqrt{2}}{2}} = \frac{1-\sqrt{3}-\sqrt{2}}{1+\sqrt{3}+\sqrt{2}}$
Therefore,$\theta = \tan ^{-1} \frac{1-\sqrt{3}-\sqrt{2}}{1+\sqrt{3}+\sqrt{2}}$.

(A) For a uniform spherical shell of mass $M$ and radius $R$,the gravitational field $E$ at any point inside $(r < R)$ is given by $E = 0$.
Since the gravitational field $E = -dV/dr$,if $E = 0$,then the potential $V$ must be constant throughout the interior.
The value of this constant potential is $V = -GM/R$,which is not zero.
Therefore,statements $(a)$,$(c)$,and $(d)$ are correct,while $(b)$ is incorrect.
Thus,the correct option is $(a), (c)$ and $(d)$ only.

(B) The pressure at points $A$ and $B$ at the same horizontal level must be equal,so $P_A = P_B$.
The pressure just below the meniscus in the two limbs is given by $P_{atm} - \frac{2T}{r_1}$ and $P_{atm} - \frac{2T}{r_2}$ respectively.
Equating the pressures at the same horizontal level:
$P_{atm} - \frac{2T}{r_1} + \rho g(x + \Delta h) = P_{atm} - \frac{2T}{r_2} + \rho g x$
$\rho g \Delta h = 2T \left( \frac{1}{r_1} - \frac{1}{r_2} \right)$
Given $r_1 = 2.5 \, mm = 2.5 \times 10^{-3} \, m$ and $r_2 = 4.0 \, mm = 4.0 \times 10^{-3} \, m$.
$\Delta h = \frac{2T}{\rho g} \left( \frac{1}{r_1} - \frac{1}{r_2} \right)$
$\Delta h = \frac{2 \times 7.3 \times 10^{-2}}{1.0 \times 10^3 \times 10} \left( \frac{1}{2.5 \times 10^{-3}} - \frac{1}{4.0 \times 10^{-3}} \right)$
$\Delta h = \frac{14.6 \times 10^{-2}}{10^4} \times 10^3 \left( \frac{1}{2.5} - \frac{1}{4.0} \right)$
$\Delta h = 14.6 \times 10^{-3} \times (0.4 - 0.25) = 14.6 \times 10^{-3} \times 0.15 = 2.19 \times 10^{-3} \, m = 2.19 \, mm$.

(A) According to the First Law of Thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Given:
Heat supply rate $\frac{\Delta Q}{\Delta t} = \frac{6000 \, J}{60 \, sec} = 100 \, W$.
Power delivered by the system $\frac{\Delta W}{\Delta t} = 90 \, W$.
Change in internal energy $\Delta U = 2.5 \times 10^{3} \, J = 2500 \, J$.
Using the power equation: $\frac{\Delta Q}{\Delta t} = \frac{\Delta U}{\Delta t} + \frac{\Delta W}{\Delta t}$.
$100 = \frac{2500}{\Delta t} + 90$.
$100 - 90 = \frac{2500}{\Delta t}$.
$10 = \frac{2500}{\Delta t}$.
$\Delta t = \frac{2500}{10} = 250 \, sec$.

(A) The formula for the $rms$ speed of gas molecules is $V_{RMS} = \sqrt{\frac{3RT}{M_{W}}}$,where $R$ is the universal gas constant,$T$ is the absolute temperature,and $M_{W}$ is the molar mass of the gas.
Since the temperature $T$ is the same for all gases,we have $V_{RMS} \propto \frac{1}{\sqrt{M_{W}}}$.
The molar masses are: $M_{H} = 2 \ g/mol$,$M_{O} = 32 \ g/mol$,and $M_{C} = 44 \ g/mol$.
Since $M_{H} < M_{O} < M_{C}$,it follows that $\frac{1}{\sqrt{M_{H}}} > \frac{1}{\sqrt{M_{O}}} > \frac{1}{\sqrt{M_{C}}}$.
Therefore,$V_{H} > V_{O} > V_{C}$.

(C) $1$. Calculate the Least Count $(LC)$: The pitch is $0.5 \, mm$ and the number of circular scale divisions is $50$. Thus,$LC = \frac{0.5 \, mm}{50} = 0.01 \, mm$.
$2$. Determine the Zero Error: The $5^{th}$ division coincides with the reference line when the ratchet is closed. Since it is positive (implied by standard convention for this type of problem),Zero Error $= +5 \times LC = 5 \times 0.01 = 0.05 \, mm$.
$3$. Calculate the Observed Reading: Observed Reading = Main Scale Reading + (Circular Scale Division $\times LC$) = $5 \, mm + (20 \times 0.01 \, mm) = 5.20 \, mm$.
$4$. Calculate the True Reading: True Reading = Observed Reading - Zero Error = $5.20 \, mm - 0.05 \, mm = 5.15 \, mm$.

(D) The thrust force exerted on the rocket is given by $F_{\text{thrust}} = \left(\frac{dm}{dt}\right) V_{\text{rel}}$.
The net force acting on the rocket is $F_{\text{net}} = F_{\text{thrust}} - mg = ma$.
Substituting the given values: $m = 1000 \, \text{kg}$,$a = 20 \, \text{m s}^{-2}$,$V_{\text{rel}} = 500 \, \text{m s}^{-1}$,and $g = 10 \, \text{m s}^{-2}$.
$\left(\frac{dm}{dt}\right) \times 500 - (1000 \times 10) = 1000 \times 20$
$\left(\frac{dm}{dt}\right) \times 500 - 10000 = 20000$
$\left(\frac{dm}{dt}\right) \times 500 = 30000$
$\frac{dm}{dt} = \frac{30000}{500} = 60 \, \text{kg s}^{-1}$.

(D) The dimensions of the given quantities are:
$E = [ML^2 T^{-2}]$
$L = [ML^2 T^{-1}]$
$m = [M]$
$G = [M^{-1} L^3 T^{-2}]$
Given the formula $P = E L^2 m^{-5} G^{-2}$,we substitute the dimensions:
$[P] = [ML^2 T^{-2}] \cdot [ML^2 T^{-1}]^2 \cdot [M]^{-5} \cdot [M^{-1} L^3 T^{-2}]^{-2}$
$[P] = [ML^2 T^{-2}] \cdot [M^2 L^4 T^{-2}] \cdot [M^{-5}] \cdot [M^2 L^{-6} T^4]$
$[P] = [M^{1+2-5+2} L^{2+4-6} T^{-2-2+4}]$
$[P] = [M^0 L^0 T^0]$
Thus,the dimensions of $P$ are dimensionless.

(A) Let the total length of the chain be $L = 3\, m$ and total mass be $M = 3\, kg$. The linear mass density is $\lambda = M/L = 3/3 = 1\, kg/m$.
Initially,$2\, m$ of the chain is on the table and $1\, m$ is hanging. The center of mass of the hanging part is at a distance of $0.5\, m$ below the table level. Taking the table surface as the reference level $(U = 0)$,the initial potential energy is $U_i = -(m_{hanging})g(h_{cm}) = -(1\, kg)(10\, m/s^2)(0.5\, m) = -5\, J$.
The initial kinetic energy is $K_i = 0$.
Finally,when the chain completely slips off,the entire chain of length $3\, m$ hangs vertically. Its center of mass is at $1.5\, m$ below the table level. The final potential energy is $U_f = -(M)g(h_{cm}) = -(3\, kg)(10\, m/s^2)(1.5\, m) = -45\, J$.
By the law of conservation of mechanical energy,$K_i + U_i = K_f + U_f$.
$0 + (-5\, J) = K_f + (-45\, J)$.
$K_f = 45 - 5 = 40\, J$.
Thus,the value of $k$ is $40$.

(C) Given:
Speed of source $V_S = 20 \, m/s$
Speed of observer/detector $V_O = 20 \, m/s$
Speed of sound $V = 340 \, m/s$
Observed frequency $f' = 1800 \, Hz$
According to the Doppler effect,when the source and observer move away from each other,the observed frequency is given by:
$f' = f \left( \frac{V - V_O}{V + V_S} \right)$
Substituting the values:
$1800 = f \left( \frac{340 - 20}{340 + 20} \right)$
$1800 = f \left( \frac{320}{360} \right)$
$1800 = f \left( \frac{8}{9} \right)$
$f = \frac{1800 \times 9}{8}$
$f = 225 \times 9 = 2025 \, Hz$
Therefore,the original frequency of the source is $2025 \, Hz$.

(D) Let the first ball be thrown at $t = 0$ and the second ball at $t = 3 \, \text{s}$.
Let $t$ be the time elapsed since the first ball was thrown.
The position of the first ball is $y_1 = u t - \frac{1}{2} g t^2 = 35 t - 5 t^2$.
The position of the second ball is $y_2 = u(t - 3) - \frac{1}{2} g(t - 3)^2 = 35(t - 3) - 5(t - 3)^2$.
At the point of collision, $y_1 = y_2$.
$35 t - 5 t^2 = 35(t - 3) - 5(t^2 - 6 t + 9)$
$35 t - 5 t^2 = 35 t - 105 - 5 t^2 + 30 t - 45$
$0 = 30 t - 150$
$30 t = 150 \implies t = 5 \, \text{s}$.
Substituting $t = 5 \, \text{s}$ into the equation for $y_1$:
$h = 35(5) - 5(5)^2 = 175 - 125 = 50 \, \text{m}$.
Thus, the balls collide at a height of $50 \, \text{m}$.

(C) The excess pressure inside a soap bubble of radius $r$ is given by $\Delta P = \frac{4S}{r}$,where $S$ is the surface tension.
For a soap bubble of radius $r_2 = 3 \, cm$ formed inside a soap bubble of radius $r_1 = 6 \, cm$,the pressure inside the smaller bubble relative to the atmospheric pressure is the sum of the excess pressure due to the smaller bubble and the excess pressure due to the larger bubble.
$\Delta P_{total} = \frac{4S}{r_2} + \frac{4S}{r_1}$
We want to find the radius $R_{eq}$ of an equivalent single soap bubble such that its excess pressure equals $\Delta P_{total}$:
$\frac{4S}{R_{eq}} = \frac{4S}{r_2} + \frac{4S}{r_1}$
Dividing both sides by $4S$:
$\frac{1}{R_{eq}} = \frac{1}{r_2} + \frac{1}{r_1}$
Substituting the given values $r_1 = 6 \, cm$ and $r_2 = 3 \, cm$:
$\frac{1}{R_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2}$
Therefore,$R_{eq} = 2 \, cm$.

(B) For a node,the amplitude of the standing wave must be zero.
Given the equation ${y} = 1.0 \, \text{mm} \cos(1.57 \, \text{cm}^{-1} x) \sin(78.5 \, \text{s}^{-1} t)$,the spatial part is $\cos(1.57 \, \text{cm}^{-1} x)$.
Setting the spatial part to zero: $\cos(1.57 \, \text{cm}^{-1} x) = 0$.
This occurs when the argument is an odd multiple of $\frac{\pi}{2}$. For the node closest to the origin $(x > 0)$,we take the first positive value:
$1.57 \, \text{cm}^{-1} x = \frac{\pi}{2}$.
Using $\pi \approx 3.14$,we have $1.57 \approx \frac{\pi}{2}$.
Therefore,$x = \frac{\pi}{2 \times 1.57} \, \text{cm} = \frac{1.57}{1.57} \, \text{cm} = 1 \, \text{cm}$.

(D) Let the handle be a rod of length $L = 6r$ and the ring be a circular hoop of radius $R = r$. The mass of both is $M$.
$1$. Moment of inertia of the handle about an axis through its center of mass (perpendicular to it) is $I_{cm, rod} = \frac{M(6r)^2}{12} = 3Mr^2$.
The distance of this axis from the rotation axis is $d_1 = (6r/2) - r/2 = 5r/2$.
Using the parallel axis theorem,$I_{rod} = I_{cm, rod} + M(d_1)^2 = 3Mr^2 + M(5r/2)^2 = 3Mr^2 + 6.25Mr^2 = 9.25Mr^2$.
$2$. Moment of inertia of the ring about its diameter is $I_{cm, ring} = \frac{MR^2}{2} = \frac{Mr^2}{2} = 0.5Mr^2$.
The distance of the center of the ring from the rotation axis is $d_2 = 6r - r/2 + r = 6.5r = 13r/2$.
Using the parallel axis theorem,$I_{ring} = I_{cm, ring} + M(d_2)^2 = 0.5Mr^2 + M(13r/2)^2 = 0.5Mr^2 + 42.25Mr^2 = 42.75Mr^2$.
Total moment of inertia $I = I_{rod} + I_{ring} = 9.25Mr^2 + 42.75Mr^2 = 52Mr^2$.

(D) The source of microwave frequency is a magnetron.
$(b)$ The source of infrared frequency is the vibration of atoms and molecules.
$(c)$ The source of Gamma rays is the radioactive decay of the nucleus.
$(d)$ The source of $X$-rays is the transition of inner shell electrons.
Therefore,the correct matching is $(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)$.

(A) Given:
Conductivity $\sigma = 5 \times 10^{7} \, S/m$
Radius $r = 0.5 \, mm = 5 \times 10^{-4} \, m$
Electric field $E = 10 \, mV/m = 10 \times 10^{-3} \, V/m = 10^{-2} \, V/m$
Using Ohm's law in microscopic form,current density $J = \sigma E$:
$J = (5 \times 10^{7}) \times (10^{-2}) = 5 \times 10^{5} \, A/m^{2}$
Current $i = J \times A = J \times (\pi r^{2})$:
$i = (5 \times 10^{5}) \times \pi \times (5 \times 10^{-4})^{2}$
$i = 5 \times 10^{5} \times \pi \times 25 \times 10^{-8}$
$i = 125 \times 10^{-3} \, \pi \, A$
$i = 125 \, \pi \, mA$
Given that $i = x^{3} \, \pi \, mA$,we have:
$x^{3} = 125$
$x = \sqrt[3]{125} = 5$

(C) The charge is placed at a distance $a/2 = 6 \,cm$ from the center of a square of side $a = 12 \,cm$.
By considering the square as one face of a cube of side $a = 12 \,cm$,the charge is at the center of this cube.
According to Gauss's Law,the total electric flux through the entire cube is $\phi_{total} = \frac{q}{\varepsilon_{0}}$.
Since the cube has $6$ identical faces,the flux through one face (the square) is $\phi = \frac{1}{6} \frac{q}{\varepsilon_{0}}$.
Given $q = 12 \times 10^{-6} \,C$ and $\varepsilon_{0} = 8.854 \times 10^{-12} \,C^{2}/Nm^{2}$.
$\phi = \frac{12 \times 10^{-6}}{6 \times 8.854 \times 10^{-12}} = \frac{2 \times 10^{-6}}{8.854 \times 10^{-12}} \approx 0.2259 \times 10^{6} \,Nm^{2}/C$.
$\phi \approx 226 \times 10^{3} \,Nm^{2}/C$.

(D) The total attenuation in the cable is given by: $\text{Total Attenuation} = (\text{Attenuation per km}) \times (\text{Length}) = -5 \, dB/km \times 20 \, km = -100 \, dB$.
The formula for gain in $dB$ is: $\text{Gain} = 10 \log_{10}(\frac{P_0}{P_i})$.
Here,$\text{Gain} = -100 \, dB$,$P_i = 0.1 \, kW = 100 \, W = 10^2 \, W$.
Substituting the values: $-100 = 10 \log_{10}(\frac{P_0}{10^2})$.
$-10 = \log_{10}(\frac{P_0}{10^2})$.
$10^{-10} = \frac{P_0}{10^2}$.
$P_0 = 10^{-10} \times 10^2 = 10^{-8} \, W$.
Comparing $P_0 = 10^{-8} \, W$ with $10^{-x} \, W$,we get $x = 8$.

(D) At resonance,the inductive reactance $X_{L}$ and capacitive reactance $X_{C}$ cancel each other out,meaning the total impedance of the circuit is equal to the resistance $R$,i.e.,$Z = R$.
The power consumed in an $LCR$ circuit at resonance is given by the formula:
$P = \frac{V^{2}}{R}$
Given values are:
$P = 16 \, W$
$V = 120 \, V$
Rearranging the formula to solve for $R$:
$R = \frac{V^{2}}{P}$
Substituting the values:
$R = \frac{(120)^{2}}{16}$
$R = \frac{14400}{16}$
$R = 900 \, \Omega$
Thus,the value of resistance $R$ is $900 \, \Omega$.

(A) The frequency of the wave is $f = 3 \, GHz = 3 \times 10^9 \, Hz$.
In vacuum,the wavelength is $\lambda_0 = \frac{c}{f} = \frac{3 \times 10^8 \, m/s}{3 \times 10^9 \, Hz} = 0.1 \, m = 10 \, cm$.
The refractive index of the medium is given by $n = \sqrt{\epsilon_r \mu_r}$. Assuming the medium is non-magnetic,$\mu_r = 1$.
Thus,$n = \sqrt{2.25} = 1.5$.
The wavelength in the medium is $\lambda_m = \frac{\lambda_0}{n} = \frac{10 \, cm}{1.5} = \frac{100}{15} \, cm = 6.666... \, cm \approx 6.67 \, cm$.
Expressing this in the form $x \times 10^{-2} \, cm$,we get $6.67 \, cm = 667 \times 10^{-2} \, cm$.

(B) In an $n$-type semiconductor,the Fermi level $E_F$ lies below the conduction band $E_C$. As the doping concentration $N_D$ increases,the Fermi level moves closer to the conduction band,which means it shifts upward.
In a $p$-type semiconductor,the Fermi level $E_F$ lies above the valence band $E_V$. As the doping concentration $N_A$ increases,the Fermi level moves closer to the valence band,which means it shifts downward.
Therefore,the Fermi level of $p$-type semiconductors moves downward and the Fermi level of $n$-type semiconductors moves upward.

(D) In a ferromagnetic material,below the Curie temperature,the atoms spontaneously align themselves in a common direction over a macroscopic volume. This region is called a magnetic domain. Within each domain,the magnetic dipoles are aligned in the same direction,resulting in saturation magnetization for that specific region. Therefore,a domain is a macroscopic region with saturation magnetization.

(D) When a message signal is amplitude modulated with a carrier signal,the resulting modulated wave contains the carrier frequency $f_{c}$ and the sideband frequencies $(f_{c} + f_{m})$ and $(f_{c} - f_{m})$.
However,the carrier signal is the primary component that is radiated through the antenna to carry the information.
The wavelength $\lambda$ of the radiated signal is determined by the carrier frequency $f_{c}$ as $\lambda = \frac{c}{f_{c}}$,where $c$ is the speed of light in air.

(B) According to Gauss's Law,the total electric flux through a closed surface is $\frac{q}{\varepsilon_{0}}$.
Since the charge $q$ is placed at one corner of the cube,it is shared by $8$ such identical cubes to enclose the charge completely.
Therefore,the total flux through the cube is $\phi_{\text{cube}} = \frac{q}{8\varepsilon_{0}}$.
The flux through the three faces of the cube that meet at the corner where the charge is placed is zero because the electric field lines are parallel to these surfaces.
The remaining three faces are symmetric with respect to the charge,so the flux through each of these three faces is equal.
Let the shaded area be one of these faces. The flux through the shaded area is $\phi_{\text{shaded}} = \frac{1}{3} \times \phi_{\text{cube}} = \frac{1}{3} \times \frac{q}{8\varepsilon_{0}} = \frac{q}{24\varepsilon_{0}}$.

(D) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$, where $R \approx 1.097 \times 10^7 \, m^{-1}$ is the Rydberg constant.
For a transition from $n_i = 2$ to $n_f = 1$:
$\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{2^2} \right)$
$\frac{1}{\lambda} = 1.097 \times 10^7 \left( 1 - 0.25 \right) = 1.097 \times 10^7 \times 0.75$
$\frac{1}{\lambda} = 0.82275 \times 10^7 \, m^{-1}$
$\lambda = \frac{1}{0.82275 \times 10^7} \approx 1.215 \times 10^{-7} \, m = 121.5 \, nm$.
Using the precise value of $R$ and energy levels, the standard value is approximately $121.8 \, nm$.

(D) Given: $R = 110 \, \Omega$,$V = 220 \, V$,$\omega = 300 \, rad/s$.
When capacitance is removed,the circuit is an $LR$ circuit. The phase angle is given by $\tan \phi = \frac{X_L}{R}$. Since current lags by $45^{\circ}$,$\tan 45^{\circ} = \frac{X_L}{R} = 1$,so $X_L = R$.
When the inductor is removed,the circuit is an $RC$ circuit. The phase angle is given by $\tan \phi = \frac{X_C}{R}$. Since current leads by $45^{\circ}$,$\tan 45^{\circ} = \frac{X_C}{R} = 1$,so $X_C = R$.
Since $X_L = R$ and $X_C = R$,it implies $X_L = X_C$.
In an $LCR$ circuit,when $X_L = X_C$,the circuit is in resonance.
At resonance,the impedance $Z = R$.
The rms current $I = \frac{V}{Z} = \frac{V}{R} = \frac{220}{110} = 2 \, A$.

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass,and $v$ is the speed.
Given that both particles move with the same speed $v$,the ratio of their wavelengths is:
$\frac{\lambda_{e}}{\lambda_{p}} = \frac{\frac{h}{m_{e}v}}{\frac{h}{m_{p}v}} = \frac{m_{p}}{m_{e}}$.
Substituting the given relation $m_{p} = 1836 m_{e}$:
$\frac{\lambda_{e}}{\lambda_{p}} = \frac{1836 m_{e}}{m_{e}} = 1836$.

(B) When an electron enters the electric field between parallel plates,the electric force acts only in the direction perpendicular to the plates (vertical direction).
Therefore,there is no force acting in the direction parallel to the plates (horizontal direction).
As a result,the component of velocity parallel to the plates remains constant throughout the motion.
Let $v_{1}$ be the initial velocity and $v_{2}$ be the final velocity.
The horizontal component of velocity at entry is $v_{1} \cos \alpha$.
The horizontal component of velocity at exit is $v_{2} \cos \beta$.
Since the horizontal component is constant,we have: $v_{1} \cos \alpha = v_{2} \cos \beta$.
This implies: $\frac{v_{1}}{v_{2}} = \frac{\cos \beta}{\cos \alpha}$.
The kinetic energy $K$ is given by $K = \frac{1}{2}mv^{2}$.
Therefore,the ratio of kinetic energies is: $\frac{K_{1}}{K_{2}} = \frac{\frac{1}{2}mv_{1}^{2}}{\frac{1}{2}mv_{2}^{2}} = \left(\frac{v_{1}}{v_{2}}\right)^{2}$.
Substituting the velocity ratio: $\frac{K_{1}}{K_{2}} = \left(\frac{\cos \beta}{\cos \alpha}\right)^{2} = \frac{\cos^{2} \beta}{\cos^{2} \alpha}$.

(B) The given circuit consists of two $AND$ gates,one $OR$ gate,and one $NOT$ gate (forming a $NOR$ gate at the output). The inputs to the first $AND$ gate are $A$ and $\bar{B}$. The inputs to the second $AND$ gate are $\bar{A}$ and $B$.
The output of the $OR$ gate is $(A \cdot \bar{B} + \bar{A} \cdot B)$.
The final output $Y$ is the inversion of this,so $Y = \overline{(A \cdot \bar{B} + \bar{A} \cdot B)}$.
Using De Morgan's theorem,$Y = \overline{(A \cdot \bar{B})} \cdot \overline{(\bar{A} \cdot B)} = (\bar{A} + B) \cdot (A + \bar{B})$.
Expanding this,$Y = \bar{A} \cdot A + \bar{A} \cdot \bar{B} + B \cdot A + B \cdot \bar{B} = 0 + \bar{A} \cdot \bar{B} + A \cdot B + 0 = A \cdot B + \bar{A} \cdot \bar{B}$.
This is the Boolean expression for an $XNOR$ gate.

$A$	$B$	$Y$
$0$	$0$	$1$
$0$	$1$	$0$
$1$	$0$	$0$
$1$	$1$	$1$

(C) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = \phi + eV_s$,where $\phi$ is the work function and $V_s$ is the stopping potential.
For the first case: $\frac{1240}{491} = \phi + 0.71 \implies 2.525 = \phi + 0.71 \implies \phi = 1.815\, eV$.
For the second case: $\frac{1240}{\lambda} = \phi + 1.43$.
Substituting $\phi = 1.815\, eV$: $\frac{1240}{\lambda} = 1.815 + 1.43 = 3.245\, eV$.
Therefore,$\lambda = \frac{1240}{3.245} \approx 382\, nm$.

(A) Rectifier: $A$ device used to convert alternating current $(a.c.)$ into direct current $(d.c.)$. Thus,$(a)-(ii)$.
$(b)$ Stabilizer: $A$ device used to maintain a constant output voltage even when the input voltage or load current changes. Thus,$(b)-(iv)$.
$(c)$ Transformer: $A$ device used to step up or step down the $a.c.$ voltage. Thus,$(c)-(i)$.
$(d)$ Filter: $A$ circuit used to remove any ripples (fluctuations) in the rectified output voltage to provide a smoother $d.c.$ output. Thus,$(d)-(iii)$.
Therefore,the correct matching is $(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)$.

(D) The angular width of the central diffraction maximum for a circular aperture is given by $\theta \approx \frac{1.22 \lambda}{D}$,where $D$ is the diameter of the pinhole and $\lambda$ is the wavelength of light.
As the diameter $D$ of the pinhole increases,the angular width $\theta$ decreases,which means the size of the diffraction pattern decreases.
Since the total amount of light passing through the pinhole increases as the area of the pinhole increases (area $\propto D^2$),and this light is now concentrated into a smaller area,the intensity of the diffraction pattern increases.
Therefore,the size decreases and the intensity increases.

(D) The power of the bulb is $P = 80\, W$. The efficiency is $10\, \%$,so the radiated power is $P_{rad} = 80 \times 0.10 = 8\, W$.
Since it is a point source,the intensity $I$ at a distance $r = 10\, m$ is given by $I = \frac{P_{rad}}{4 \pi r^{2}} = \frac{8}{4 \pi (10)^{2}} = \frac{8}{400 \pi} = \frac{1}{50 \pi} \, W/m^{2}$.
The intensity of an electromagnetic wave is related to the peak electric field $E_{0}$ by $I = \frac{1}{2} c \epsilon_{0} E_{0}^{2}$.
Using $\epsilon_{0} = \frac{1}{\mu_{0} c^{2}}$,we get $I = \frac{1}{2} c \left( \frac{1}{\mu_{0} c^{2}} \right) E_{0}^{2} = \frac{E_{0}^{2}}{2 \mu_{0} c}$.
Equating the two expressions for $I$: $\frac{E_{0}^{2}}{2 \mu_{0} c} = \frac{1}{50 \pi}$.
$E_{0}^{2} = \frac{2 \mu_{0} c}{50 \pi} = \frac{\mu_{0} c}{25 \pi}$.
$E_{0} = \sqrt{\frac{\mu_{0} c}{25 \pi}} = \frac{1}{5} \sqrt{\frac{\mu_{0} c}{\pi}} = \frac{2}{10} \sqrt{\frac{\mu_{0} c}{\pi}}$.
Comparing this with $\frac{x}{10} \sqrt{\frac{\mu_{0} c}{\pi}}$,we find $x = 2$.

(D) Let $m = 10 \, mg = 10 \times 10^{-6} \, kg$,$L = 0.5 \, m$,$r = 0.2 \, m$,and $g = 10 \, ms^{-2}$.
At equilibrium,the forces acting on each sphere are tension $T$,weight $mg$,and electrostatic force $F_e = \frac{kq^2}{r^2}$.
From the geometry,$\sin \theta = \frac{r/2}{L} = \frac{0.1}{0.5} = 0.2$. Thus,$\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - 0.04} = \sqrt{0.96}$.
Resolving forces: $T \cos \theta = mg$ and $T \sin \theta = F_e$.
Dividing the two equations: $\tan \theta = \frac{F_e}{mg} = \frac{kq^2}{r^2 mg}$.
$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{0.2}{\sqrt{0.96}} = \frac{0.2}{0.9798} \approx 0.204$.
$q^2 = \frac{r^2 mg \tan \theta}{k} = \frac{(0.2)^2 \times (10^{-5}) \times (0.204)}{9 \times 10^9} = \frac{0.04 \times 10^{-5} \times 0.204}{9 \times 10^9} \approx 9.06 \times 10^{-17} \, C^2$.
$q \approx 9.52 \times 10^{-9} \, C = 0.952 \times 10^{-8} \, C$.
Given $q = \frac{a}{21} \times 10^{-8} \, C$,so $\frac{a}{21} = 0.952 \implies a = 0.952 \times 21 \approx 20$.

(C) The energy of an $X$-ray photon is given by $E = \frac{hc}{\lambda}$.
According to the problem,the mass $m$ of a fictitious particle having the same energy is given by $E = mc^2$.
Equating the two expressions: $\frac{hc}{\lambda} = mc^2$.
Solving for $m$: $m = \frac{h}{c\lambda}$.
Given $\lambda = 10 \, \mathring{A} = 10 \times 10^{-10} \, \text{m} = 10^{-9} \, \text{m}$.
Substituting the value of $c = 3 \times 10^8 \, \text{m/s}$:
$m = \frac{h}{(3 \times 10^8) \times 10^{-9}} = \frac{h}{3 \times 10^{-1}} = \frac{h}{0.3} = \frac{10h}{3}$.
Comparing this with the given expression $\frac{x}{3} h$,we get $x = 10$.

(C) When two identical conducting spheres are brought into contact,the total charge is redistributed equally between them.
Total charge $Q = 2.1 \, nC + (-0.1 \, nC) = 2.0 \, nC$.
Charge on each sphere after contact $q = \frac{Q}{2} = \frac{2.0 \, nC}{2} = 1.0 \, nC = 1.0 \times 10^{-9} \, C$.
The distance between the spheres is $r = 0.5 \, m$.
The electrostatic force is given by Coulomb's Law: $F = k \frac{q_{1}q_{2}}{r^{2}}$.
Substituting the values: $F = (9 \times 10^{9}) \times \frac{(1.0 \times 10^{-9}) \times (1.0 \times 10^{-9})}{(0.5)^{2}}$.
$F = \frac{9 \times 10^{9} \times 10^{-18}}{0.25} = \frac{9 \times 10^{-9}}{0.25} = 36 \times 10^{-9} \, N$.
Thus,the force is $36 \times 10^{-9} \, N$.

(A) The circuit consists of two parallel branches connected between points $P$ and $R$.
Branch $1$ consists of the wire $PQ$ and $QR$ in series. The resistance of this branch is $R_1 = 2 \,\Omega + 2 \,\Omega = 4 \,\Omega$.
Branch $2$ consists of the wire $PR$ directly,with resistance $R_2 = 2 \,\Omega$.
The total current $I = 6 \, A$ divides into these two parallel branches.
Using the current divider rule,the current $i_1$ flowing through the branch $PQ$ (and then $QR$) is given by:
$i_1 = I \times \frac{R_2}{R_1 + R_2}$
$i_1 = 6 \times \frac{2}{4 + 2} = 6 \times \frac{2}{6} = 2 \, A$.
Thus,the current $i_1$ is $2 \, A$.

(D) Given $\lambda = \frac{C}{V}$.
Since $C = \frac{Q}{V}$,we have $\lambda = \frac{Q}{V^{2}}$.
We know that $V = \frac{W}{Q}$,where $W$ is work and $Q$ is charge.
Substituting $V$ in the expression for $\lambda$:
$\lambda = \frac{Q}{(W/Q)^{2}} = \frac{Q^{3}}{W^{2}}$.
Using dimensional formulas: $[Q] = [IT]$,$[W] = [ML^{2}T^{-2}]$.
$\lambda = \frac{[IT]^{3}}{[ML^{2}T^{-2}]^{2}} = \frac{[I^{3}T^{3}]}{[M^{2}L^{4}T^{-4}]}$.
$\lambda = [M^{-2} L^{-4} I^{3} T^{3 - (-4)}] = [M^{-2} L^{-4} I^{3} T^{7}]$.

(A) The motional $EMF$ induced in the wings is given by $\epsilon = B_v \cdot v \cdot l$,where $B_v$ is the vertical component of the Earth's magnetic field.
Given: $l = 10 \, m$,$v = 180 \, km/h = 180 \times \frac{5}{18} = 50 \, m/s$,$B = 2.5 \times 10^{-4} \, T$,and angle of dip $\delta = 60^{\circ}$.
The vertical component $B_v = B \sin \delta = 2.5 \times 10^{-4} \times \sin 60^{\circ}$.
Substituting the values: $\epsilon = (2.5 \times 10^{-4} \times \sin 60^{\circ}) \times 50 \times 10$.
$\epsilon = 2.5 \times 10^{-4} \times \frac{\sqrt{3}}{2} \times 500$.
$\epsilon = 2.5 \times 10^{-4} \times 0.866 \times 500 = 0.10825 \, V$.
Converting to $mV$: $\epsilon = 0.10825 \times 1000 = 108.25 \, mV$.

(C) Assertion $A$ is true. In a simple microscope,the virtual image formed by the convex lens subtends the same angle at the eye as the object itself,because the rays from the object appear to originate from the image position.
Reason $R$ is also true. $A$ simple microscope (convex lens) allows an object to be placed at a distance $u_0$ which is less than the least distance of distinct vision $(D = 25\, cm)$.
Since the angular magnification $m = \frac{\theta^{\prime}}{\theta} = \frac{D}{u_0}$,and $u_0 < D$,we get $m > 1$. The reason explains why we get magnification,which is the purpose of the microscope,and it correctly justifies the assertion regarding the angular size.
Therefore,both $A$ and $R$ are true and $R$ is the correct explanation of $A$.

(B) Analysis of Statement $I$: According to Gauss's Law,the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{in}}{\varepsilon_0}$. For an electric dipole,the net charge $q_{in} = +q + (-q) = 0$. Thus,the flux $\phi = 0$. However,the electric field $\vec{E}$ due to the dipole is non-zero at every point inside the sphere. Therefore,Statement $I$ is true.
Analysis of Statement $II$: For a solid metallic sphere of radius $R$ carrying charge $Q$,the charge resides entirely on the outer surface. For any Gaussian surface of radius $r < R$,the enclosed charge $q_{in} = 0$. By Gauss's Law,the electric flux $\phi = \frac{q_{in}}{\varepsilon_0} = 0$. Also,the electric field $\vec{E}$ inside a conductor is zero. The statement claims the flux is not zero,which is incorrect. Therefore,Statement $II$ is false.
Conclusion: Statement $I$ is true and Statement $II$ is false.

(A) The energy released when an electron transitions from $n = 5$ to $n = 1$ is given by:
$\Delta E = E_5 - E_1 = -0.54 \, eV - (-13.6 \, eV) = 13.06 \, eV$.
By the law of conservation of linear momentum,the initial momentum of the atom is zero,so the final momentum of the atom must be equal and opposite to the momentum of the emitted photon:
$P_{\text{atom}} = P_{\text{photon}} = \frac{h}{\lambda}$.
Since the energy of the photon is $\Delta E = \frac{hc}{\lambda}$,we have $\frac{1}{\lambda} = \frac{\Delta E}{hc}$.
Substituting this into the momentum equation,the recoil momentum of the atom is $Mv = \frac{\Delta E}{c}$,where $M$ is the mass of the hydrogen atom $(M \approx 1.67 \times 10^{-27} \, kg)$ and $c$ is the speed of light $(3 \times 10^8 \, m/s)$.
The recoil speed $v$ is given by:
$v = \frac{\Delta E}{Mc} = \frac{13.06 \times 1.6 \times 10^{-19} \, J}{1.67 \times 10^{-27} \, kg \times 3 \times 10^8 \, m/s} \approx 4.17 \, m/s$.

(A) Given: $L = 100 \, mH = 0.1 \, H$,$C = 100 \, \mu F = 10^{-4} \, F$,$R = 120 \, \Omega$,$V = 30 \sin(100t) \, V$.
From the voltage equation,peak voltage $V_0 = 30 \, V$ and angular frequency $\omega = 100 \, rad/s$.
Inductive reactance $X_L = \omega L = 100 \times 0.1 = 10 \, \Omega$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100 \times 10^{-4}} = \frac{1}{0.01} = 100 \, \Omega$.
Impedance $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{120^2 + (10 - 100)^2} = \sqrt{120^2 + (-90)^2} = \sqrt{14400 + 8100} = \sqrt{22500} = 150 \, \Omega$.
Peak current $I_0 = \frac{V_0}{Z} = \frac{30}{150} = 0.2 \, A$.
Resonant frequency $f_r = \frac{1}{2\pi \sqrt{LC}} = \frac{1}{2\pi \sqrt{0.1 \times 10^{-4}}} = \frac{1}{2\pi \sqrt{10^{-5}}} = \frac{1}{2\pi \times 10^{-2.5}} \approx \frac{1}{2 \times 3.14 \times 0.00316} \approx 50.3 \, Hz \approx 50 \, Hz$.

(D) The forces acting on the body are gravity $(mg)$,normal force $(N)$,electric force $(F_e = qE)$,and friction $(f = \mu N)$.
$F_e = (5 \times 10^{-3} \, C) \times (200 \, N/C) = 1 \, N$.
Resolving forces perpendicular to the incline:
$N = mg \cos 30^{\circ} + F_e \sin 30^{\circ} = (1 \times 9.8 \times 0.866) + (1 \times 0.5) = 8.487 + 0.5 = 8.987 \, N \approx 9 \, N$.
Resolving forces parallel to the incline:
$F_{net} = mg \sin 30^{\circ} + F_e \cos 30^{\circ} - \mu N = (1 \times 9.8 \times 0.5) + (1 \times 0.866) - (0.2 \times 9) = 4.9 + 0.866 - 1.8 = 3.966 \, N$.
Acceleration $a = F_{net} / m = 3.966 / 1 = 3.966 \, m/s^2$.
The length of the incline $L = h / \sin 30^{\circ} = 1 / 0.5 = 2 \, m$.
Using $S = ut + \frac{1}{2}at^2$ with $u = 0$:
$2 = 0 + \frac{1}{2} \times 3.966 \times t^2 \implies t^2 = 4 / 3.966 \approx 1.008 \implies t \approx 1 \, s$.
Given the options provided,the closest value is $1.3 \, s$.

(D) The given circuit consists of a $NAND$ gate,an $OR$ gate,and a $NOT$ gate. Let the inputs be $A$ and $B$. The output of the $NAND$ gate is $\overline{A \cdot A} = \overline{A}$.
This output $\overline{A}$ and input $B$ are fed into an $OR$ gate,giving an intermediate output $Z = \overline{A} + B$.
Finally,this is passed through a $NOT$ gate to give the final output $Y = \overline{Z} = \overline{\overline{A} + B}$.
Using De Morgan's law,$\overline{\overline{A} + B} = \overline{\overline{A}} \cdot \overline{B} = A \cdot \overline{B}$.
Now,we analyze the truth table for $Y = A \cdot \overline{B}$:
- For $t = 0$ to $1$ s: $A=1, B=0 \implies Y = 1 \cdot \overline{0} = 1 \cdot 1 = 1$.
- For $t = 1$ to $2$ s: $A=1, B=1 \implies Y = 1 \cdot \overline{1} = 1 \cdot 0 = 0$.
- For $t = 2$ to $3$ s: $A=0, B=0 \implies Y = 0 \cdot \overline{0} = 0 \cdot 1 = 0$.
- For $t = 3$ to $4$ s: $A=1, B=1 \implies Y = 1 \cdot \overline{1} = 1 \cdot 0 = 0$.
- For $t = 4$ to $5$ s: $A=1, B=0 \implies Y = 1 \cdot \overline{0} = 1 \cdot 1 = 1$.
Thus,the output $Y$ is high $(1)$ from $t = 0$ to $1$ s and from $t = 4$ to $5$ s,and low $(0)$ otherwise. This matches the signal shown in option $D$.

(C) The activity of a radioactive sample at time $t$ is given by $A(t) = A_{0} e^{-\lambda t}$.
At time $t_{1}$,$A = A_{0} e^{-\lambda t_{1}}$ ... $(i)$
At time $t_{2}$,$\frac{A}{5} = A_{0} e^{-\lambda t_{2}}$ ... $(ii)$
Dividing equation $(i)$ by $(ii)$:
$\frac{A}{A/5} = \frac{A_{0} e^{-\lambda t_{1}}}{A_{0} e^{-\lambda t_{2}}}$
$5 = e^{\lambda(t_{2}-t_{1})}$
Taking the natural logarithm on both sides:
$\ln 5 = \lambda(t_{2}-t_{1})$
$\lambda = \frac{\ln 5}{t_{2}-t_{1}}$
The average life time $\tau$ is the reciprocal of the decay constant $\lambda$:
$\tau = \frac{1}{\lambda} = \frac{t_{2}-t_{1}}{\ln 5}$

(A) Initial resistance $R_{0} = 1\, \Omega$ and initial length $\ell_{0} = 1\, m$.
When the wire is stretched,its volume remains constant. Let the new length be $\ell_{1}$.
Given that the length increases by $25\, \%$,so $\ell_{1} = \ell_{0} + 0.25\ell_{0} = 1.25\ell_{0} = 1.25\, m$.
Since volume $V = A \ell$ is constant,$A_{0}\ell_{0} = A_{1}\ell_{1}$,which implies $A_{1} = A_{0}(\ell_{0} / \ell_{1})$.
The resistance $R$ is given by $R = \rho \frac{\ell}{A}$.
Therefore,the new resistance $R_{1} = \rho \frac{\ell_{1}}{A_{1}} = \rho \frac{\ell_{1}}{A_{0}(\ell_{0} / \ell_{1})} = \rho \frac{\ell_{1}^{2}}{A_{0}\ell_{0}} = R_{0} \left( \frac{\ell_{1}}{\ell_{0}} \right)^{2}$.
Substituting the values: $R_{1} = 1 \times (1.25)^{2} = 1.5625\, \Omega$.
The percentage change in resistance is $\frac{R_{1} - R_{0}}{R_{0}} \times 100\, \% = \frac{1.5625 - 1}{1} \times 100\, \% = 56.25\, \%$.
Rounding to the nearest integer,the percentage change is $56\, \%$.

(C) According to the law of reflection,the angle of incidence equals the angle of reflection,and the incident ray,reflected ray,and normal lie in the same plane.
Let the normal vector be $\overrightarrow{ c }$. The component of the incident ray $\overrightarrow{ a }$ along the normal is $(\overrightarrow{ a } \cdot \overrightarrow{ c }) \overrightarrow{ c }$.
The component of $\overrightarrow{ a }$ perpendicular to the normal is $\overrightarrow{ a } - (\overrightarrow{ a } \cdot \overrightarrow{ c }) \overrightarrow{ c }$.
Since the reflected ray $\overrightarrow{ b }$ has the same component perpendicular to the normal but the opposite component along the normal,we have:
$\overrightarrow{ b } = (\overrightarrow{ a } - (\overrightarrow{ a } \cdot \overrightarrow{ c }) \overrightarrow{ c }) - (\overrightarrow{ a } \cdot \overrightarrow{ c }) \overrightarrow{ c }$
$\overrightarrow{ b } = \overrightarrow{ a } - 2(\overrightarrow{ a } \cdot \overrightarrow{ c }) \overrightarrow{ c }$

(B) The bandwidth required for a single $AM$ broadcast station is given by twice the maximum modulating frequency.
Bandwidth per station $= 2 \times f_m = 2 \times 5\, kHz = 10\, kHz$.
To find the number of stations that can be accommodated in a total bandwidth of $90\, kHz$,we divide the total bandwidth by the bandwidth per station.
Number of stations $= \frac{\text{Total Bandwidth}}{\text{Bandwidth per station}} = \frac{90\, kHz}{10\, kHz} = 9$.
Therefore,$9$ stations can be accommodated.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $KE_{\max} = h\nu - \phi$,where $\phi$ is the work function.
Since $KE_{\max} = \frac{1}{2}mv^2$,we have $v = \sqrt{\frac{2(h\nu - \phi)}{m}}$.
For the first case,$h\nu_1 = 2\phi$. Thus,$v_1 = \sqrt{\frac{2(2\phi - \phi)}{m}} = \sqrt{\frac{2\phi}{m}}$.
For the second case,$h\nu_2 = 10\phi$. Thus,$v_2 = \sqrt{\frac{2(10\phi - \phi)}{m}} = \sqrt{\frac{18\phi}{m}}$.
The ratio of the maximum velocities is $\frac{v_1}{v_2} = \sqrt{\frac{2\phi}{18\phi}} = \sqrt{\frac{1}{9}} = \frac{1}{3}$.
Given the ratio is $x:y = 1:3$,the value of $x$ is $1$.

(C) Let the mirror be placed along the $y$-axis with its centre at the origin $(0,0)$. The mirror extends from $y = -25 \,cm$ to $y = +25 \,cm$.
The light source $S$ is at $(60 \,cm, 0)$.
The image $S'$ of the source $S$ in the plane mirror is formed at $(-60 \,cm, 0)$.
The man walks along the line $x = 120 \,cm - 60 \,cm = 60 \,cm$ relative to the mirror plane,or more simply,the man is at a distance of $1.2 \,m = 120 \,cm$ from the mirror.
The rays from the image $S'$ that define the field of view pass through the edges of the mirror at $y = 25 \,cm$ and $y = -25 \,cm$.
Using similar triangles,the ratio of the height to the distance from the image $S'$ is constant:
$\frac{y_{edge}}{distance_{S'}} = \frac{x}{distance_{S'} + distance_{man}}$
Here,the distance of the image $S'$ from the mirror is $60 \,cm$. The distance of the man from the mirror is $120 \,cm$.
So,the total distance of the man from the image $S'$ is $60 \,cm + 120 \,cm = 180 \,cm$.
Using the property of similar triangles:
$\frac{25 \,cm}{60 \,cm} = \frac{x}{180 \,cm}$
$x = \frac{25 \times 180}{60} = 25 \times 3 = 75 \,cm$.
This $x$ is the distance from the central axis to the extreme point on one side.
The total distance between the two extreme points is $2x = 2 \times 75 \,cm = 150 \,cm$.

(D) Let $r$ be the radius of each small drop and $R$ be the radius of the big drop.
Since the volume is conserved,$27 \times (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$.
This gives $R^3 = 27r^3$,so $R = 3r$.
Let $q$ be the charge on each small drop. The total charge on the big drop is $Q = 27q$.
The electrostatic potential energy of a charged spherical drop of radius $r$ and charge $q$ is given by $U_1 = \frac{3}{5} \frac{kq^2}{r}$.
The potential energy of the bigger drop is $U = \frac{3}{5} \frac{kQ^2}{R}$.
Substituting $Q = 27q$ and $R = 3r$ into the equation for $U$:
$U = \frac{3}{5} \frac{k(27q)^2}{3r} = \frac{3}{5} \frac{k \cdot 729q^2}{3r} = \frac{729}{3} \left( \frac{3}{5} \frac{kq^2}{r} \right)$.
$U = 243 U_1$.
Therefore,the potential energy of the bigger drop is $243$ times that of a smaller drop.

(B) The total voltage supplied is $V_{in} = 90\, V$ and the series resistance is $R_{s} = 4\, k\Omega = 4000\, \Omega$.
The Zener diode maintains a constant voltage of $V_{z} = 30\, V$ across it.
The voltage drop across the series resistor $R_{s}$ is $V_{R} = V_{in} - V_{z} = 90\, V - 30\, V = 60\, V$.
The total current $i$ flowing from the source through the series resistor is $i = \frac{V_{R}}{R_{s}} = \frac{60\, V}{4000\, \Omega} = 0.015\, A = 15\, mA$.
The load resistor is $R_{L} = 5\, k\Omega = 5000\, \Omega$. The current $i_{1}$ flowing through the load resistor is $i_{1} = \frac{V_{z}}{R_{L}} = \frac{30\, V}{5000\, \Omega} = 0.006\, A = 6\, mA$.
Applying Kirchhoff's Current Law at the junction,the current through the Zener diode $i_{z}$ is given by $i_{z} = i - i_{1} = 15\, mA - 6\, mA = 9\, mA$.

(D) The magnetic field on the axis of a current-carrying coil of radius $a$ at a distance $r$ from the centre is given by $B_{\text{axis}} = \frac{\mu_{0} i a^{2}}{2(a^{2} + r^{2})^{3/2}}$.
The magnetic field at the centre of the coil is $B_{\text{centre}} = \frac{\mu_{0} i}{2a}$.
The fractional change in magnetic field is defined as $\frac{B_{\text{centre}} - B_{\text{axis}}}{B_{\text{centre}}} = 1 - \frac{B_{\text{axis}}}{B_{\text{centre}}}$.
Substituting the expressions: $1 - \frac{\frac{\mu_{0} i a^{2}}{2(a^{2} + r^{2})^{3/2}}}{\frac{\mu_{0} i}{2a}} = 1 - \frac{a^{3}}{(a^{2} + r^{2})^{3/2}} = 1 - \left(1 + \frac{r^{2}}{a^{2}}\right)^{-3/2}$.
Using the binomial approximation $(1 + x)^{n} \approx 1 + nx$ for $x << 1$,where $x = \frac{r^{2}}{a^{2}}$ and $n = -3/2$:
$1 - (1 - \frac{3}{2} \frac{r^{2}}{a^{2}}) = \frac{3}{2} \frac{r^{2}}{a^{2}}$.

(D) The mirror used in a car as a rear-view mirror is a convex mirror.
For a spherical mirror,the velocity of the image $V_I$ with respect to the mirror is given by $V_{I/m} = -m^2 V_{O/m}$,where $m$ is the magnification and $V_{O/m}$ is the velocity of the object with respect to the mirror.
Given:
Relative speed of car $B$ with respect to car $A$ is $V_{O/m} = 40 \, ms^{-1}$.
Focal length of the convex mirror $f = +10 \, cm = +0.1 \, m$.
Object distance $u = -1.9 \, m$.
The magnification $m$ is given by $m = \frac{f}{f - u}$.
Substituting the values: $m = \frac{0.1}{0.1 - (-1.9)} = \frac{0.1}{2.0} = \frac{1}{20}$.
Now,calculate the velocity of the image:
$V_{I/m} = -m^2 V_{O/m} = -(\frac{1}{20})^2 \times 40 = -\frac{1}{400} \times 40 = -0.1 \, ms^{-1}$.
The negative sign indicates that the image moves in the direction opposite to the object's motion relative to the mirror. The speed of the image is $0.1 \, ms^{-1}$.

(D) The energy stored in an inductor is given by $U = \frac{1}{2} L i^2$. Given $U = 64 \, J$ and $i = 8 \, A$,we have $64 = \frac{1}{2} \times L \times (8)^2$. This simplifies to $64 = 32L$,so $L = 2 \, H$.
The rate of energy dissipation (power) is given by $P = i^2 R$. Given $P = 640 \, W$ and $i = 8 \, A$,we have $640 = (8)^2 \times R$. This simplifies to $640 = 64R$,so $R = 10 \, \Omega$.
The time constant $\tau$ of an $LR$ circuit is given by $\tau = \frac{L}{R}$. Substituting the values,$\tau = \frac{2}{10} = 0.2 \, s$.

(A) For maximum average power in an $LCR$ circuit, the circuit must be in resonance.
At resonance, the inductive reactance equals the capacitive reactance, i.e., $X_{L} = X_{C}$.
Given $X_{L} = 250 \pi \, \Omega$ and frequency $f = 50 \, Hz$.
The formula for capacitive reactance is $X_{C} = \frac{1}{2 \pi f C}$.
Equating the two: $250 \pi = \frac{1}{2 \pi (50) C}$.
$250 \pi = \frac{1}{100 \pi C}$.
$C = \frac{1}{250 \pi \times 100 \pi} = \frac{1}{25000 \pi^{2}}$.
Given $\pi^{2} = 10$, we have $C = \frac{1}{25000 \times 10} = \frac{1}{250000} \, F$.
$C = 4 \times 10^{-6} \, F = 4 \, \mu F$.

(C) The given circuit consists of two $NAND$ gates whose inputs are tied together,acting as $NOT$ gates,followed by an $AND$ gate.
Let the inputs be $A$ and $B$.
The outputs of the two $NOT$ gates are $X = \overline{A}$ and $Y = \overline{B}$.
The final output $Z$ is the $AND$ operation of $X$ and $Y$:
$Z = X \cdot Y = \overline{A} \cdot \overline{B}$.
Using De Morgan's theorem,$\overline{A} \cdot \overline{B} = \overline{A + B}$.
This is the Boolean expression for a $NOR$ gate.
Truth Table:

$A, B$	$X, Y$	$Z$
$0, 0$	$1, 1$	$1$
$0, 1$	$1, 0$	$0$
$1, 0$	$0, 1$	$0$
$1, 1$	$0, 0$	$0$

(D) The frequency of emitted radiation in a hydrogen-like ion is given by the Rydberg formula: $f = R c Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$,where $k = R c Z^2$ is a constant for a specific ion.
For the transition from $n=3$ to $n=1$: $f_1 = k \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = k \left( 1 - \frac{1}{9} \right) = k \left( \frac{8}{9} \right) = 2.92 \times 10^{15} \text{ Hz}$.
For the transition from $n=2$ to $n=1$: $f_2 = k \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = k \left( 1 - \frac{1}{4} \right) = k \left( \frac{3}{4} \right)$.
Taking the ratio: $\frac{f_1}{f_2} = \frac{k(8/9)}{k(3/4)} = \frac{8}{9} \times \frac{4}{3} = \frac{32}{27}$.
Therefore,$f_2 = f_1 \times \frac{27}{32} = 2.92 \times 10^{15} \times \frac{27}{32} \approx 2.46 \times 10^{15} \text{ Hz}$.

(A) The Einstein's photoelectric equation is given by $KE_{\max} = eV_S = \frac{hc}{\lambda} - \phi$.
For $\lambda_1 = 280 \, nm$,the energy of the incident photon is $E_1 = \frac{1240 \, eV \cdot nm}{280 \, nm} \approx 4.43 \, eV$.
The stopping potential $V_{S1}$ is $V_{S1} = \frac{E_1 - \phi}{e} = 4.43 - 2.5 = 1.93 \, V$.
For $\lambda_2 = 400 \, nm$,the energy of the incident photon is $E_2 = \frac{1240 \, eV \cdot nm}{400 \, nm} = 3.10 \, eV$.
The stopping potential $V_{S2}$ is $V_{S2} = \frac{E_2 - \phi}{e} = 3.10 - 2.5 = 0.60 \, V$.
The change in stopping potential is $\Delta V = V_{S1} - V_{S2} = 1.93 - 0.60 = 1.33 \, V \approx 1.3 \, V$.

(C) From the circuit diagram,the four resistors are connected in parallel between points $A$ and $B$. The values of the resistors are $4 \, \Omega, 8 \, \Omega, 12 \, \Omega$,and $6 \, \Omega$.
The equivalent resistance $R_{eq}$ of the parallel combination is given by:
$\frac{1}{R_{eq}} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{6}$
$\frac{1}{R_{eq}} = \frac{6 + 3 + 2 + 4}{24} = \frac{15}{24} = \frac{5}{8} \, \Omega^{-1}$
$R_{eq} = \frac{8}{5} = 1.6 \, \Omega$
The total resistance of the circuit $R_T$ including the internal resistance $r = 0.6 \, \Omega$ is:
$R_T = R_{eq} + r = 1.6 + 0.6 = 2.2 \, \Omega$
The power dissipated in the whole circuit is given by $P = \frac{E^2}{R_T}$,where $E = 2.2 \, V$ is the $emf$ of the cell:
$P = \frac{(2.2)^2}{2.2} = 2.2 \, W$
Thus,the power dissipated in the whole circuit is $2.2 \, W$.

(A) For a solid metal sphere of radius $R$ with charge $q$ placed inside a conducting spherical shell of inner radius $a$ and outer radius $b$:
$1$. For $r < R$: The electric field inside a conductor is $E = 0$.
$2$. For $R \leq r < a$: The electric field is due to the central sphere,so $E = \frac{k q}{r^2}$.
$3$. For $a \leq r < b$: This region is inside the material of the conducting shell. The induced charge $-q$ on the inner surface $a$ cancels the field of the central charge $q$,so $E = 0$.
$4$. For $r \geq b$: The total charge enclosed is $q + 0 = q$ (assuming the shell is neutral),so $E = \frac{k q}{r^2}$.
Thus,the electric field is zero for $r < R$,follows $1/r^2$ for $R \leq r < a$,is zero for $a \leq r < b$,and follows $1/r^2$ for $r \geq b$.

(B) The resistance $R$ of a wire is given by $R = \rho \frac{L}{A}$, where $\rho$ is resistivity, $L$ is length, and $A$ is the cross-sectional area.
Given diameter $d = 2 \; mm = 2 \times 10^{-3} \; m$, so radius $r = 1 \times 10^{-3} \; m$.
Area $A = \pi r^2 = \pi \times (10^{-3})^2 = \pi \times 10^{-6} \; m^2$.
Resistivity of iron $\rho_1 = 12 \; \mu\Omega \cdot cm = 12 \times 10^{-6} \times 10^{-2} \; \Omega \cdot m = 1.2 \times 10^{-7} \; \Omega \cdot m$.
Resistivity of alloy $\rho_2 = 51 \; \mu\Omega \cdot cm = 51 \times 10^{-6} \times 10^{-2} \; \Omega \cdot m = 5.1 \times 10^{-7} \; \Omega \cdot m$.
For parallel connection, $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \Rightarrow R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = 3 \; \Omega$.
$R_1 = \frac{\rho_1 L}{A}$ and $R_2 = \frac{\rho_2 L}{A}$.
$R_{eq} = \frac{(\rho_1 L / A) \cdot (\rho_2 L / A)}{(\rho_1 L / A) + (\rho_2 L / A)} = \frac{\rho_1 \rho_2 L}{A(\rho_1 + \rho_2)} = 3$.
$L = \frac{3 A (\rho_1 + \rho_2)}{\rho_1 \rho_2} = \frac{3 \times \pi \times 10^{-6} \times (1.2 + 5.1) \times 10^{-7}}{1.2 \times 10^{-7} \times 5.1 \times 10^{-7}} = \frac{3 \times \pi \times 6.3 \times 10^{-13}}{6.12 \times 10^{-14}} \approx 97 \; m$.

(C) Given: Resistivity $\rho = 200 \, \Omega \, m$,Capacitance $C = 2 \times 10^{-12} \, F$,Potential difference $V = 40 \, V$,Relative permittivity $K = 50$.
The resistance $R$ of the dielectric material between the plates is given by $R = \frac{\rho d}{A}$,and the capacitance is $C = \frac{K \varepsilon_0 A}{d}$.
Multiplying these,we get the time constant $\tau = RC = \left( \frac{\rho d}{A} \right) \left( \frac{K \varepsilon_0 A}{d} \right) = \rho K \varepsilon_0$.
The leakage current $I$ is given by Ohm's Law: $I = \frac{V}{R}$.
Substituting $R = \frac{\rho d}{A}$ and $C = \frac{K \varepsilon_0 A}{d} \implies \frac{A}{d} = \frac{C}{K \varepsilon_0}$,we get $R = \frac{\rho K \varepsilon_0}{C}$.
Thus,$I = \frac{V}{R} = \frac{VC}{\rho K \varepsilon_0}$.
Substituting the values: $I = \frac{40 \times 2 \times 10^{-12}}{200 \times 50 \times 8.854 \times 10^{-12}}$.
$I = \frac{80 \times 10^{-12}}{10000 \times 8.854 \times 10^{-12}} = \frac{80}{88540} \approx 0.000903 \, A = 0.903 \, mA$.
Rounding to the nearest option,the leakage current is $0.9 \, mA$.