WBJEE 2025 Physics Question Paper with Answer and Solution

(D) The force $F_{1}$ required to push the body up the incline is given by $F_{1} = mg(\sin \theta + \mu \cos \theta)$.
The force $F_{2}$ required to prevent the body from sliding down the incline is given by $F_{2} = mg(\sin \theta - \mu \cos \theta)$.
Taking the ratio of the two forces:
$\frac{F_{1}}{F_{2}} = \frac{mg(\sin \theta + \mu \cos \theta)}{mg(\sin \theta - \mu \cos \theta)} = \frac{\sin \theta + \mu \cos \theta}{\sin \theta - \mu \cos \theta}$.
Dividing the numerator and denominator by $\cos \theta$:
$\frac{F_{1}}{F_{2}} = \frac{\tan \theta + \mu}{\tan \theta - \mu}$.
Given that $\tan \theta = 2\mu$,substitute this value into the equation:
$\frac{F_{1}}{F_{2}} = \frac{2\mu + \mu}{2\mu - \mu} = \frac{3\mu}{\mu} = 3$.

(D) The density of the rod varies as $\rho = \rho_0 \frac{x^2}{L^2}$.
Consider a small elemental disc of thickness $dx$ at a distance $x$ from the end $x=0$.
The mass of this element is $dm = \rho \cdot dV = \rho \cdot (\alpha dx) = \left( \rho_0 \frac{x^2}{L^2} \right) \alpha dx$.
The position of the centre of mass $X_{cm}$ is given by the formula:
$X_{cm} = \frac{\int x dm}{\int dm}$
Substituting the values:
$X_{cm} = \frac{\int_0^L x \left( \rho_0 \frac{x^2}{L^2} \alpha dx \right)}{\int_0^L \left( \rho_0 \frac{x^2}{L^2} \alpha dx \right)}$
$X_{cm} = \frac{\frac{\rho_0 \alpha}{L^2} \int_0^L x^3 dx}{\frac{\rho_0 \alpha}{L^2} \int_0^L x^2 dx}$
$X_{cm} = \frac{[x^4/4]_0^L}{[x^3/3]_0^L} = \frac{L^4/4}{L^3/3} = \frac{3}{4} L$.

(A) The ball falls from height $h$ and hits the floor. The velocity just before the first impact is $v_0 = \sqrt{2gh}$.
After the first impact,the velocity becomes $v_1 = e v_0$. The ball rises to a height $h_1 = \frac{v_1^2}{2g} = e^2 h$.
It then falls from $h_1$ and hits the floor again. The distance covered in the first bounce (up and down) is $2h_1 = 2e^2 h$.
After the second impact,it rises to $h_2 = e^2 h_1 = e^4 h$. The distance covered in the second bounce is $2h_2 = 2e^4 h$.
The total distance $D$ is the sum of the initial fall and all subsequent bounces:
$D = h + 2h_1 + 2h_2 + 2h_3 + ...$
$D = h + 2e^2 h + 2e^4 h + 2e^6 h + ...$
$D = h + 2e^2 h (1 + e^2 + e^4 + ...)$
Using the sum of an infinite geometric series $S = \frac{a}{1-r}$ where $a=1$ and $r=e^2$:
$D = h + 2e^2 h \left( \frac{1}{1-e^2} \right)$
$D = h \left( 1 + \frac{2e^2}{1-e^2} \right) = h \left( \frac{1-e^2+2e^2}{1-e^2} \right) = h \left( \frac{1+e^2}{1-e^2} \right)$.

(C) Let the initial velocity of $S_2$ be $\vec{u}_2 = v \hat{i}$ and $S_1$ be $\vec{u}_1 = 0$. After the collision,$S_2$ moves with velocity $\vec{v}_2 = \frac{v}{2} \hat{j}$ (assuming it moves in the positive $y$-direction). Let the velocity of $S_1$ be $\vec{v}_1 = v_{1x} \hat{i} + v_{1y} \hat{j}$.
By conservation of linear momentum along the $x$-axis:
$m_2 v = m_1 v_{1x} + m_2(0) \implies v_{1x} = \frac{m_2}{m_1} v$
By conservation of linear momentum along the $y$-axis:
$0 = m_1 v_{1y} + m_2 \left(\frac{v}{2}\right) \implies v_{1y} = -\frac{m_2 v}{2m_1}$
The magnitude of velocity of $S_1$ is $v_1 = \sqrt{v_{1x}^2 + v_{1y}^2} = \sqrt{\left(\frac{m_2 v}{m_1}\right)^2 + \left(-\frac{m_2 v}{2m_1}\right)^2} = \frac{m_2 v}{m_1} \sqrt{1 + \frac{1}{4}} = \frac{m_2 v}{m_1} \frac{\sqrt{5}}{2}$.
The direction $\theta$ of $S_1$ with the $x$-axis is given by $\tan \theta = \frac{v_{1y}}{v_{1x}} = \frac{-m_2 v / 2m_1}{m_2 v / m_1} = -\frac{1}{2}$.
If $S_2$ had moved in the negative $y$-direction,$\tan \theta = \frac{1}{2}$.
Thus,$\theta = \tan^{-1}\left(\frac{1}{2}\right)$ or $\theta = \tan^{-1}\left(-\frac{1}{2}\right)$.

(C, D) The expressions for the speeds are:
$V_{\text{rms}} = \sqrt{\frac{3RT}{M}}$
$\overline{V} = \sqrt{\frac{8RT}{\pi M}}$
$V_{p} = \sqrt{\frac{2RT}{M}}$
Comparing these values,we find $V_{p} < \overline{V} < V_{\text{rms}}$. Thus,option $C$ is correct.
For the average kinetic energy of a molecule:
$K.E. = \frac{1}{2} m V_{\text{rms}}^{2} = \frac{1}{2} m \left( \frac{3RT}{M} \right)$.
Since $V_{p}^{2} = \frac{2RT}{M}$,we have $\frac{RT}{M} = \frac{V_{p}^{2}}{2}$.
Substituting this,$K.E. = \frac{1}{2} m \cdot 3 \cdot \left( \frac{V_{p}^{2}}{2} \right) = \frac{3}{4} m V_{p}^{2}$.
Thus,option $D$ is also correct.

(A) Given force $\vec{F} = a \hat{i} + b \hat{j} + c \hat{k}$.
Using Newton's second law,the acceleration $\vec{a}_{acc} = \frac{\vec{F}}{m} = \frac{a}{m} \hat{i} + \frac{b}{m} \hat{j} + \frac{c}{m} \hat{k}$.
Since the body starts from rest at the origin,the initial velocity $\vec{u} = 0$ and initial position $\vec{r}_0 = 0$.
Using the kinematic equation $\vec{s} = \vec{u}t + \frac{1}{2} \vec{a}_{acc} t^2$,we get $\vec{r} = \frac{1}{2} \vec{a}_{acc} t^2$.
Substituting the components:
$x = \frac{1}{2} (\frac{a}{m}) t^2 = \frac{at^2}{2m}$
$y = \frac{1}{2} (\frac{b}{m}) t^2 = \frac{bt^2}{2m}$
$z = \frac{1}{2} (\frac{c}{m}) t^2 = \frac{ct^2}{2m}$
Thus,the coordinates are $(\frac{at^2}{2m}, \frac{bt^2}{2m}, \frac{ct^2}{2m})$.

(A) Let $V_1$ be the volume of granite in water and $V_2$ be the volume of granite in mercury.
According to the principle of floatation,the total buoyant force acting on the granite must be equal to its weight.
The buoyant force from water is $F_{B1} = V_1 \rho_1 g$.
The buoyant force from mercury is $F_{B2} = V_2 \rho_2 g$.
The weight of the granite is $W = (V_1 + V_2) \rho g$.
Equating the forces: $V_1 \rho_1 g + V_2 \rho_2 g = (V_1 + V_2) \rho g$.
Dividing by $g$: $V_1 \rho_1 + V_2 \rho_2 = V_1 \rho + V_2 \rho$.
Rearranging the terms to group $V_1$ and $V_2$: $V_1 \rho_1 - V_1 \rho = V_2 \rho - V_2 \rho_2$.
$V_1(\rho_1 - \rho) = V_2(\rho - \rho_2)$.
Therefore,the ratio of the volume of granite in water $(V_1)$ to the volume of granite in mercury $(V_2)$ is:
$\frac{V_1}{V_2} = \frac{\rho - \rho_2}{\rho_1 - \rho} = \frac{\rho_2 - \rho}{\rho - \rho_1}$.

(C) Consider the horizontal level at the bottom of the liquid with density $\rho_3$ in the left arm. The pressure at this level in both arms must be equal.
In the left arm,the pressure is due to the liquid of density $\rho_1$ (height $h$) and the liquid of density $\rho_3$ (height $h/2$).
$P_{left} = P_{atm} + \rho_1 gh + \rho_3 g(h/2)$
In the right arm,the pressure at the same horizontal level is due to the liquid of density $\rho_2$ (height $h$).
$P_{right} = P_{atm} + \rho_2 gh$
Equating the pressures: $P_{atm} + \rho_1 gh + \rho_3 g(h/2) = P_{atm} + \rho_2 gh$
Dividing by $gh$: $\rho_1 + \frac{\rho_3}{2} = \rho_2$
Rearranging for $\rho_3$: $\frac{\rho_3}{2} = \rho_2 - \rho_1 \implies \rho_3 = 2(\rho_2 - \rho_1)$

(B) The tension $T$ in the wire when the elevator moves upward with acceleration $a$ is given by $T = m(g + a)$.
Given $m = 10 \ kg$,$g = 10 \ m/s^2$,and $a = 2 \ m/s^2$,we have $T = 10(10 + 2) = 120 \ N$.
The longitudinal strain is defined as $\text{Strain} = \frac{\Delta \ell}{L} = \frac{\text{Stress}}{Y} = \frac{T}{AY}$.
Given cross-sectional area $A = 2 \ cm^2 = 2 \times 10^{-4} \ m^2$ and Young's modulus $Y = 2.0 \times 10^{11} \ N/m^2$.
Substituting the values: $\text{Strain} = \frac{120}{(2 \times 10^{-4}) \times (2.0 \times 10^{11})} = \frac{120}{4 \times 10^7} = 30 \times 10^{-7} = 3 \times 10^{-6}$.

(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g'}}$.
At a height $h = R$ (where $R$ is the radius of the Earth),the acceleration due to gravity $g'$ is given by $g' = g \left( \frac{R}{R+h} \right)^2 = g \left( \frac{R}{R+R} \right)^2 = \frac{g}{4}$.
Given $g = \pi^2 \ m/s^2$ at the surface,the effective gravity at height $h$ is $g' = \frac{\pi^2}{4} \ m/s^2$.
Substituting the values $\ell = 4.0 \ m$ and $g' = \frac{\pi^2}{4} \ m/s^2$ into the time period formula:
$T = 2\pi \sqrt{\frac{4}{\pi^2 / 4}} = 2\pi \sqrt{\frac{16}{\pi^2}} = 2\pi \times \frac{4}{\pi} = 8 \ s$.

(D) The given displacement equation for $SHM$ is $y = 2 \sin \left(\frac{\pi t}{2} + \phi\right) \text{ cm}$.
Comparing this with the standard $SHM$ equation $y = A \sin(\omega t + \phi)$,we get the amplitude $A = 2 \text{ cm}$ and angular frequency $\omega = \frac{\pi}{2} \text{ rad/s}$.
The formula for maximum acceleration in $SHM$ is $a_{\max} = \omega^2 A$.
Substituting the values,we get $a_{\max} = \left(\frac{\pi}{2}\right)^2 \times 2$.
$a_{\max} = \frac{\pi^2}{4} \times 2 = \frac{\pi^2}{2} \text{ cm/s}^2$.

(B) According to Newton's law of cooling,the rate of cooling $R$ is directly proportional to the temperature difference between the body and its surroundings,provided the difference is small.
Mathematically,$R = -\frac{d\theta}{dt} = k(\theta - \theta_0)$,where $k$ is a positive constant.
This equation is of the form $y = mx + c$,where $y = R$,$x = \theta$,$m = k$ (slope),and $c = -k\theta_0$ (y-intercept).
Since $k > 0$,the slope is positive.
When $\theta = \theta_0$,$R = 0$.
Therefore,the graph is a straight line passing through the point $(\theta_0, 0)$ with a positive slope,which matches the shape shown in option $B$.

(B) Let $\gamma_L$ be the real coefficient of volume expansion of the liquid.
Let $\gamma_C$ and $\gamma_S$ be the coefficients of volume expansion of copper and silver,respectively.
We know that the coefficient of volume expansion $\gamma = 3 \times \text{linear coefficient of expansion } \alpha$.
Given $\gamma_C = 3A$.
The apparent expansion coefficient is given by $\gamma_{app} = \gamma_L - \gamma_{vessel}$.
For copper: $C = \gamma_L - 3A \implies \gamma_L = C + 3A$.
For silver: $S = \gamma_L - \gamma_S \implies \gamma_S = \gamma_L - S$.
Substituting $\gamma_L = C + 3A$ into the equation for silver: $\gamma_S = C + 3A - S$.
Since $\gamma_S = 3 \alpha_S$,where $\alpha_S$ is the linear coefficient of expansion of silver:
$3 \alpha_S = C + 3A - S \implies \alpha_S = \frac{C + 3A - S}{3}$.

(D) In the given $P-T$ diagram:
$AB$: The pressure $P$ is constant,and temperature $T$ decreases. Since $PV = nRT$,if $P$ is constant and $T$ decreases,volume $V$ must decrease. Thus,$AB$ is an isobaric compression.
$BC$: The process is a vertical line,meaning temperature $T$ is constant. Since $P$ decreases,$V$ must increase to keep $PV$ constant. Thus,$BC$ is an isothermal expansion.
$CA$: The process is a line passing through the origin in a $P-T$ diagram,meaning $P \propto T$. Since $PV = nRT$,$P/T = nR/V$,so $V$ must be constant. Thus,$CA$ is an isochoric process.
Comparing these characteristics with the given options,the correct $P-V$ diagram is represented by option $D$.

(B) The standard equation for a stationary wave is $y = A \sin(kx) \cos(\omega t)$.
Comparing this with the given equation $y = 5 \sin \left( \frac{\pi x}{3} \right) \cos(40 \pi t)$,we identify the wave number $k = \frac{\pi}{3} \ cm^{-1}$.
The relationship between the wave number $k$ and the wavelength $\lambda$ is $k = \frac{2\pi}{\lambda}$.
Substituting the value of $k$: $\frac{\pi}{3} = \frac{2\pi}{\lambda}$,which gives $\lambda = 6 \ cm$.
The separation between two adjacent nodes in a stationary wave is given by $\frac{\lambda}{2}$.
Therefore,the separation $= \frac{6 \ cm}{2} = 3 \ cm$.

(A, B, D) The given equation is $y(x, t) = 0.02 \cos(50 \pi t + \frac{\pi}{2}) \cos(10 \pi x)$.
Nodes occur where the spatial part $\cos(10 \pi x) = 0$.
$10 \pi x = (n + \frac{1}{2}) \pi \implies x = \frac{n + 0.5}{10} = 0.05, 0.15, 0.25, \dots \ m$. Thus,a node occurs at $x = 0.15 \ m$.
Antinodes occur where $|\cos(10 \pi x)| = 1$.
$10 \pi x = n \pi \implies x = \frac{n}{10} = 0, 0.1, 0.2, 0.3, \dots \ m$. Thus,an antinode occurs at $x = 0.3 \ m$.
Comparing with the standard stationary wave equation $y = A \cos(\omega t + \phi) \cos(kx)$,we get $k = 10 \pi$ and $\omega = 50 \pi$.
Wavelength $\lambda = \frac{2 \pi}{k} = \frac{2 \pi}{10 \pi} = 0.2 \ m$.
Wave speed $v = \frac{\omega}{k} = \frac{50 \pi}{10 \pi} = 5 \ m/s$.

(C) From the given velocity-time graph,the initial velocity at $t = 0 \ s$ is $u = 50 \ m/s$. The velocity becomes $0 \ m/s$ at $t = 10 \ s$.
The acceleration $a$ is the slope of the graph: $a = \frac{v_f - v_i}{t_f - t_i} = \frac{0 - 50}{10 - 0} = -5 \ m/s^2$.
The velocity at $t = 2 \ s$ is given by $v = u + at = 50 + (-5)(2) = 40 \ m/s$.
According to the work-energy theorem,the work done $W$ is equal to the change in kinetic energy:
$W = \Delta K = K_f - K_i = \frac{1}{2} m v^2 - \frac{1}{2} m u^2$
$W = \frac{1}{2} \times 10 \ kg \times ((40 \ m/s)^2 - (50 \ m/s)^2)$
$W = 5 \times (1600 - 2500) = 5 \times (-900) = -4500 \ J$.

(A) The root mean square voltage is given as $V_{rms} = 220 \ V$.
The peak voltage $V_0$ is related to $V_{rms}$ by the formula $V_0 = V_{rms} \sqrt{2}$.
Therefore, $V_0 = 220 \sqrt{2} \ V$.
The angular frequency $\omega$ is given by $\omega = 2 \pi f$, where $f = 50 \ \text{Hz}$.
Thus, $\omega = 2 \pi \times 50 = 100 \pi \ \text{rad/s}$.
The instantaneous potential difference is given by $V(t) = V_0 \cos(\omega t)$.
Substituting the values, we get $V(t) = 220 \sqrt{2} \cos(100 \pi t)$.

(A) For the Lyman series,the wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right)$,where $n = 2, 3, 4, \dots$
$1$. The minimum wavelength $(\lambda_{\min})$ occurs for the transition from $n = \infty$ to $n = 1$:
$\frac{1}{\lambda_{\min}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R$
Given $\lambda_{\min} = P$,we have $P = \frac{1}{R}$.
$2$. The maximum wavelength $(\lambda_{\max})$ occurs for the transition from $n = 2$ to $n = 1$:
$\frac{1}{\lambda_{\max}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right)$
$3$. Substituting $R = \frac{1}{P}$ into the equation:
$\frac{1}{\lambda_{\max}} = \frac{1}{P} \cdot \frac{3}{4}$
$\lambda_{\max} = \frac{4 P}{3}$.

(C) The energy of the photon emitted during the transition from $n=2$ to $n=1$ is given by:
$E = E_2 - E_1 = -\frac{13.6}{2^2} - (-\frac{13.6}{1^2}) = -3.4 + 13.6 = 10.2 \ eV$.
According to Einstein's photoelectric equation:
$E = \phi + K_{max}$,where $\phi$ is the work function and $K_{max} = eV_s$ is the maximum kinetic energy.
$10.2 \ eV = 4.2 \ eV + eV_s$.
$eV_s = 10.2 \ eV - 4.2 \ eV = 6.0 \ eV$.
Therefore,the stopping potential $V_s = 6 \ V$.

(B) In a steady state, the capacitors act as open circuits, so no current flows through the branches containing the capacitors.
Therefore, the current $(I)$ flows only through the $1 \text{ k}\Omega$ resistor and the $2 \text{ k}\Omega$ resistor in series.
The total resistance of the circuit is $R_{eq} = 1 \text{ k}\Omega + 2 \text{ k}\Omega = 3 \text{ k}\Omega = 3000 \Omega$.
The current in the circuit is $I = \frac{V}{R_{eq}} = \frac{6 \text{ V}}{3000 \Omega} = 2 \times 10^{-3} \text{ A} = 2 \text{ mA}$.
The potential difference across the $2 \text{ k}\Omega$ resistor is $V_{AB} = I \times R = (2 \times 10^{-3} \text{ A}) \times (2000 \Omega) = 4 \text{ V}$.
Since the capacitors are in parallel with the $2 \text{ k}\Omega$ resistor, the potential difference across both capacitors is $4 \text{ V}$.
The charge on the $1 \mu F$ capacitor is $q_1 = C_1 \times V = (1 \mu F) \times (4 \text{ V}) = 4 \mu C$.
The charge on the $2 \mu F$ capacitor is $q_2 = C_2 \times V = (2 \mu F) \times (4 \text{ V}) = 8 \mu C$.

(D) The expression is $X = \varepsilon_0 L \frac{\Delta V}{\Delta t}$.
We know that the electric flux $\phi_E = E \cdot A$,where $E = \frac{V}{L}$ and $A = L^2$.
Thus,$\phi_E = \frac{V}{L} \cdot L^2 = V \cdot L$.
Substituting this into the expression for $X$,we get $X = \varepsilon_0 \frac{\Delta \phi_E}{\Delta t}$.
According to the definition of displacement current $i_d = \varepsilon_0 \frac{d\phi_E}{dt}$,the quantity $X$ represents the displacement current.
Therefore,the dimension of $X$ is the same as that of current.

(A) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.
Initially,let the mass be $m$,speed be $v$,and kinetic energy be $E = \frac{1}{2}mv^2$. Thus,$\lambda = \frac{h}{\sqrt{2mE}}$.
After some passengers leave,let the new mass be $m'$,new speed be $v' = 2v$,and new kinetic energy be $E' = 2E$.
Using the kinetic energy formula: $E' = \frac{1}{2}m'(v')^2$.
Substituting the values: $2E = \frac{1}{2}m'(2v)^2 = \frac{1}{2}m'(4v^2) = 2m'v^2$.
Since $E = \frac{1}{2}mv^2$,we have $2(\frac{1}{2}mv^2) = 2m'v^2$,which implies $m' = \frac{m}{2}$.
Now,the new de-Broglie wavelength $\lambda'$ is:
$\lambda' = \frac{h}{\sqrt{2m'E'}} = \frac{h}{\sqrt{2(\frac{m}{2})(2E)}} = \frac{h}{\sqrt{2mE}} = \lambda$.
Therefore,the new de-Broglie wavelength remains $\lambda$.

(A) The power $P$ of the lamp is given by the total energy emitted per second,which is the product of the number of photons emitted per second $N$ and the energy of a single photon $E = \frac{hc}{\lambda}$.
Given: $N = 10^{20} \ s^{-1}$,$\lambda = 660 \ nm = 660 \times 10^{-9} \ m$,$h = 6.6 \times 10^{-34} \ J \cdot s$,and $c = 3 \times 10^8 \ m/s$.
$P = N \times \frac{hc}{\lambda}$
$P = 10^{20} \times \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{660 \times 10^{-9}}$
$P = 10^{20} \times \frac{19.8 \times 10^{-26}}{660 \times 10^{-9}}$
$P = 10^{20} \times 0.03 \times 10^{-17}$
$P = 30 \ W$.

(B) The force on the particle is $F = qE = qE_0 \sin(\omega t)$.
Using Newton's second law,$ma = qE_0 \sin(\omega t)$,so $a = \frac{qE_0}{m} \sin(\omega t)$.
The velocity $v(t)$ is the integral of acceleration: $v(t) = \int a \, dt = \int \frac{qE_0}{m} \sin(\omega t) \, dt = -\frac{qE_0}{m\omega} \cos(\omega t) + C$.
Since the particle starts from rest at $t = 0$,$v(0) = 0$,which gives $C = \frac{qE_0}{m\omega}$.
Thus,$v(t) = \frac{qE_0}{m\omega} (1 - \cos(\omega t))$.
The maximum speed occurs when $\cos(\omega t) = -1$,so $v_{\max} = \frac{2qE_0}{m\omega}$.
Substituting the values: $q = 1 \text{ C}$,$E_0 = 2 \text{ N/C}$,$m = 1 \text{ g} = 10^{-3} \text{ kg}$,and $\omega = 1000 \text{ rad/s}$.
$v_{\max} = \frac{2 \times 1 \times 2}{10^{-3} \times 1000} = \frac{4}{1} = 4 \text{ m/s}$.

(A) The work done in moving a charge $Q$ in an electrostatic field is path-independent and depends only on the initial and final positions.
Work done $W = Q(V_{final} - V_{initial})$.
Here, the initial point is $C$ and the final point is $D$.
Distance $AC = L$, $CB = L$, $BD = L$.
Potential at $C$ $(V_C)$: $V_C = \frac{1}{4 \pi \epsilon_0} [\frac{q}{AC} + \frac{-q}{CB}] = \frac{1}{4 \pi \epsilon_0} [\frac{q}{L} - \frac{q}{L}] = 0$.
Potential at $D$ $(V_D)$: $V_D = \frac{1}{4 \pi \epsilon_0} [\frac{q}{AD} + \frac{-q}{BD}] = \frac{1}{4 \pi \epsilon_0} [\frac{q}{3L} - \frac{q}{L}] = \frac{1}{4 \pi \epsilon_0} [\frac{q - 3q}{3L}] = \frac{-2q}{12 \pi \epsilon_0 L} = \frac{-q}{6 \pi \epsilon_0 L}$.
Since the electrostatic force is conservative, the work done along any path from $C$ to $D$ is the same:
$W_1 = W_2 = Q(V_D - V_C) = Q(\frac{-q}{6 \pi \epsilon_0 L} - 0) = \frac{-Qq}{6 \pi \epsilon_0 L}$.
Thus, $W_1 = W_2 = \frac{-Qq}{6 \pi \epsilon_0 L}$.

(C) The magnetic moment $M$ of a particle of charge $q$ moving in a circular orbit is given by $M = IA$,where $I$ is the current and $A$ is the area of the orbit.
The current $I = \frac{q}{T} = \frac{q\omega}{2\pi}$.
The area $A = \pi r^2$.
Thus,$M = \left(\frac{q\omega}{2\pi}\right)(\pi r^2) = \frac{q\omega r^2}{2}$.
The angular momentum $L$ of the particle is $L = mvr = m(\omega r)r = m\omega r^2$.
The ratio of the magnetic moment to the angular momentum is $\frac{M}{L} = \frac{q\omega r^2 / 2}{m\omega r^2} = \frac{q}{2m}$.
This ratio is known as the gyromagnetic ratio and depends only on the charge '$q$' and mass '$m$' of the particle.

(A, C) The radius of a nucleus is given by the formula $r = R_0 A^{1/3}$,where $R_0$ is a constant and $A$ is the mass number.
Since $A_A = 64$ and $A_B = 125$,we have $r_A = R_0 (64)^{1/3} = 4R_0$ and $r_B = R_0 (125)^{1/3} = 5R_0$.
Thus,$r_A < r_B$.
Regarding binding energy per nucleon $(E_{bn})$,the value of $E_{bn}$ increases with mass number for light nuclei and reaches a maximum near $A = 56$ (Iron). For nuclei with $A > 60$,$E_{bn}$ gradually decreases.
Since $A_A = 64$ is closer to $56$ than $A_B = 125$,the binding energy per nucleon for nucleus $A$ is greater than that for nucleus $B$,i.e.,$E_{bnA} > E_{bnB}$.
Therefore,both options $A$ and $C$ are correct.

(A) The law of radioactive decay is given by $N = N_0 e^{-\lambda t}$.
Activity $A$ is defined as the rate of decay of nuclei,given by $A = -\frac{dN}{dt}$.
Differentiating $N$ with respect to $t$,we get:
$\frac{dN}{dt} = -\lambda N_0 e^{-\lambda t} = -\lambda N$.
Therefore,the magnitude of activity is $A = |-\frac{dN}{dt}| = \lambda N$.
This equation $A = \lambda N$ represents a linear relationship between $N$ and $A$,where $\lambda$ is the decay constant.
Since $A$ is directly proportional to $N$ $(A \propto N)$,the graph of $N$ versus $A$ will be a straight line passing through the origin. Thus,option $A$ is correct.

(A) Let the initial nucleus be represented as $_Z X^A$.
$1$. An $\alpha$-decay reduces the mass number by $4$ and atomic number by $2$.
$2$. $A$ $\beta$-decay increases the atomic number by $1$ and keeps the mass number unchanged.
$3$. $A$ $\gamma$-decay does not change the mass number or atomic number.
Given the decay chain:
$X(Z, A) \xrightarrow{\alpha} X_1(Z-2, A-4) \xrightarrow{\beta} X_2(Z-1, A-4) \xrightarrow{\alpha} X_3(Z-3, A-8) \xrightarrow{\gamma} X_4(Z-3, A-8)$
Given $X_4$ has atomic number $69$ and mass number $172$:
$Z - 3 = 69 \implies Z = 72$
$A - 8 = 172 \implies A = 180$
Thus,the atomic number and mass number of $X$ are $72$ and $180$ respectively.

(B) For a prism,the angle of deviation $\delta$ is given by the formula: $\delta = i + e - A$,where $i$ is the angle of incidence,$e$ is the angle of emergence,and $A$ is the prism angle.
From the graph,we observe that for a deviation $\delta = 30^{\circ}$,there are two possible angles of incidence: $i_1 = 15^{\circ}$ and $i_2 = 60^{\circ}$.
By the principle of reversibility of light,if $i = 15^{\circ}$,then $e = 60^{\circ}$,and if $i = 60^{\circ}$,then $e = 15^{\circ}$.
Substituting these values into the formula: $A = i + e - \delta$.
$A = 15^{\circ} + 60^{\circ} - 30^{\circ}$.
$A = 75^{\circ} - 30^{\circ} = 45^{\circ}$.
Therefore,the prism angle is $45^{\circ}$.

(A) In the circuit,the diode is in parallel with the resistance $R$.
For $V < 0$ (reverse bias),the diode acts as an open circuit (assuming an ideal diode),so the current flows only through the resistance $R$. According to Ohm's law,$I = V/R$,which is a linear relationship passing through the origin with a negative slope in the third quadrant.
For $V > 0$ (forward bias),the diode conducts after the knee voltage. Before the knee voltage,the current flows through the resistor. After the knee voltage,the diode offers very low resistance,so the total current increases rapidly.
Combining these,the graph shows a linear relationship for $V < 0$ and a non-linear,rapidly increasing current for $V > 0$. Option $A$ correctly represents this behavior.

(A) Given:
Zener voltage $V_z = 5.6 \, V$
Maximum power dissipation $P_{z \max} = \frac{1}{4} \, W = 0.25 \, W$
Input voltage $V_{in} = 10 \, V$
The maximum current $I_z$ that the zener diode can handle is given by:
$P_{z \max} = V_z \times I_z$
$0.25 = 5.6 \times I_z$
$I_z = \frac{0.25}{5.6} \, A$
In the circuit, the current through the series resistance $R_s$ is $I_s = I_z$. The voltage drop across the resistance $R_s$ is:
$V_{R_s} = V_{in} - V_z = 10 \, V - 5.6 \, V = 4.4 \, V$
Using Ohm's law for the resistance $R_s$:
$R_s = \frac{V_{R_s}}{I_s} = \frac{4.4}{I_z} = \frac{4.4}{(0.25 / 5.6)}$
$R_s = \frac{4.4 \times 5.6}{0.25} = 4.4 \times 5.6 \times 4 = 98.56 \, \Omega$
Thus, the minimum value of the resistance $R_s$ is $98.56 \, \Omega$.

(B) The given circuit consists of two $NOT$ gates connected to the inputs $A$ and $B$,followed by a $NOR$ gate.
$1$. The outputs of the two $NOT$ gates are $\overline{A}$ and $\overline{B}$.
$2$. These are fed as inputs to a $NOR$ gate.
$3$. The output $Y$ of the $NOR$ gate is given by $Y = \overline{\overline{A} + \overline{B}}$.
$4$. Using De Morgan's theorem,$\overline{\overline{A} + \overline{B}} = \overline{\overline{A}} \cdot \overline{\overline{B}} = A \cdot B$.
$5$. The expression $A \cdot B$ represents the operation of an $AND$ gate.
Therefore,the combination represents an $AND$ gate.

(B) The angular width of the central maximum in a single slit diffraction pattern is given by $\theta = \frac{2\lambda}{a}$,where $\lambda$ is the wavelength of light and $a$ is the slit width.
Since the wavelength of blue light $(\lambda_{blue})$ is smaller than the wavelength of red light $(\lambda_{red})$,the angular width of the diffraction fringes decreases.
Therefore,the fringes become narrower and more crowded together.