WBJEE 2022 Physics Question Paper with Answer and Solution

(D) According to the law of conservation of energy,the total energy at the surface of the Earth is equal to the total energy at infinity (or a very large distance where the potential energy is zero).
Total energy at the surface: $E_i = K_i + U_i = \frac{1}{2} m(\sqrt{3} v_e)^2 - \frac{GMm}{R}$.
We know that the escape velocity $v_e = \sqrt{\frac{2GM}{R}}$,so $v_e^2 = \frac{2GM}{R}$,which implies $\frac{GM}{R} = \frac{v_e^2}{2}$.
Substituting this into the energy equation: $E_i = \frac{1}{2} m(3 v_e^2) - m(\frac{v_e^2}{2}) = \frac{3}{2} m v_e^2 - \frac{1}{2} m v_e^2 = m v_e^2$.
At a very large distance,the potential energy is $0$,so the final energy is $E_f = \frac{1}{2} m v^2$.
Equating $E_i = E_f$: $m v_e^2 = \frac{1}{2} m v^2$.
$v^2 = 2 v_e^2 \implies v = \sqrt{2} v_e$.

(D) Let the actual length of the string be $L$ and the force constant be $k = \frac{AY}{L_0}$.
According to Hooke's Law,the extension is proportional to the applied force: $F = k \Delta L$.
For the first case: $F_1 = k(L_1 - L)$.
For the second case: $F_2 = k(L_2 - L)$.
Dividing the two equations: $\frac{F_1}{F_2} = \frac{L_1 - L}{L_2 - L}$.
Cross-multiplying gives: $F_1(L_2 - L) = F_2(L_1 - L)$.
$F_1 L_2 - F_1 L = F_2 L_1 - F_2 L$.
Rearranging to solve for $L$: $F_2 L - F_1 L = F_2 L_1 - F_1 L_2$.
$L(F_2 - F_1) = F_2 L_1 - F_1 L_2$.
Therefore,$L = \frac{F_2 L_1 - F_1 L_2}{F_2 - F_1}$.

(D) The impulse imparted to the ball is equal to the change in its linear momentum.
Impulse $= \int F \, dt = \text{Area under the } F-t \text{ graph}$.
Given that the force decreases linearly from $F$ to $0$ in time $t = 0.02 \text{ s}$, the area is a triangle with base $0.02 \text{ s}$ and height $F$.
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 0.02 \times F = 0.01F$.
The change in momentum is $\Delta p = m(v_f - v_i) = \frac{50}{1000} \text{ kg} \times (100 \text{ m/s} - 0) = 0.05 \times 100 = 5 \text{ kg m/s}$.
Equating impulse and change in momentum: $0.01F = 5$.
$F = \frac{5}{0.01} = 500 \text{ N}$.

(C) Let the radius of each small drop be $r$ and the radius of the big drop be $R$.
Since the volume remains conserved,$V_{big} = 27 \times V_{small}$.
$\frac{4}{3} \pi R^3 = 27 \times \frac{4}{3} \pi r^3 \Rightarrow R^3 = 27r^3 \Rightarrow R = 3r$.
The initial surface energy is $U_i = 27 \times (S \times 4 \pi r^2) = 108 \pi r^2 S$.
The final surface energy is $U_f = S \times 4 \pi R^2 = S \times 4 \pi (3r)^2 = 36 \pi r^2 S$.
The relative increase in surface energy is given by $\frac{\Delta U}{U_i} = \frac{U_f - U_i}{U_i} = \frac{36 \pi r^2 S - 108 \pi r^2 S}{108 \pi r^2 S}$.
$\frac{\Delta U}{U_i} = \frac{-72 \pi r^2 S}{108 \pi r^2 S} = -\frac{72}{108} = -\frac{2}{3}$.

(A, D) Since both wires are made of the same material,their Young's modulus $(Y)$ is identical.
Young's modulus is given by $Y = \frac{F L}{A x}$,where $F$ is the load,$L$ is the length,$A$ is the cross-sectional area,and $x$ is the elongation.
Rearranging for the load,we get $F = (\frac{Y A}{L}) x$.
Since $Y$ and $L$ are the same for both wires,the slope of the $F-x$ graph is proportional to the cross-sectional area $A$ (i.e.,$\text{slope} = \frac{Y A}{L} \propto A$).
From the graph,the slope of line $A$ is greater than the slope of line $B$,which implies that the cross-sectional area of $A$ is greater than that of $B$ $(A_A > A_B)$.
Therefore,statement $A$ is true and statement $D$ is true.

(C) The angular momentum $\vec{L}$ of a particle about the origin is given by $\vec{L} = \vec{r} \times \vec{p}$.
At the maximum height,the velocity of the particle is purely horizontal,given by $v_x = u \cos \theta$.
The position vector of the particle at maximum height is $\vec{r} = x \hat{i} + h_{\max} \hat{j}$,where $h_{\max} = \frac{u^2 \sin^2 \theta}{2g}$.
The momentum is $\vec{p} = m v_x \hat{i} = m u \cos \theta \hat{i}$.
The angular momentum is $\vec{L} = (x \hat{i} + h_{\max} \hat{j}) \times (m u \cos \theta \hat{i}) = -m u \cos \theta h_{\max} \hat{k}$.
The magnitude is $L = m u \cos \theta \left( \frac{u^2 \sin^2 \theta}{2g} \right) = \frac{m u^3 \sin^2 \theta \cos \theta}{2g}$.
Thus,$L \propto u^3$.

(A, B) Given position vector: $\vec{r} = b \cos \omega t \hat{i} + b \sin \omega t \hat{j}$.
Velocity vector: $\vec{v} = \frac{d\vec{r}}{dt} = -b \omega \sin \omega t \hat{i} + b \omega \cos \omega t \hat{j}$.
Magnitude of velocity: $v = |\vec{v}| = \sqrt{(-b \omega \sin \omega t)^2 + (b \omega \cos \omega t)^2} = b \omega$.
Since $v$ is constant,kinetic energy $E = \frac{1}{2}mv^2 = \frac{1}{2}mb^2\omega^2$ is constant. Thus,$\frac{E}{\omega} = \frac{1}{2}mb^2\omega$ is constant. Statement $(A)$ is true.
Trajectory: $x = b \cos \omega t$ and $y = b \sin \omega t$. Squaring and adding: $x^2 + y^2 = b^2(\cos^2 \omega t + \sin^2 \omega t) = b^2$. This is a circle. Statement $(B)$ is true.
Acceleration: $\vec{a} = \frac{d\vec{v}}{dt} = -b \omega^2 \cos \omega t \hat{i} - b \omega^2 \sin \omega t \hat{j} = -\omega^2 \vec{r}$.
Components: $a_x = -b \omega^2 \cos \omega t$ and $a_y = -b \omega^2 \sin \omega t$. Thus,$a_x^2 + a_y^2 = (b \omega^2)^2$,which is a circle in the $a_x-a_y$ plane. Statement $(C)$ is false.
Since $\vec{a} = -\omega^2 \vec{r}$ and $\vec{v} \perp \vec{r}$,$\vec{a}$ is not parallel to $\vec{v}$. Statement $(D)$ is false.

(B) The resultant amplitude $A_{\text{res}}$ of two simple harmonic motions with amplitudes $A_1$ and $A_2$ and phase difference $\delta$ is given by the formula: $A_{\text{res}}^2 = A_1^2 + A_2^2 + 2A_1A_2 \cos \delta$.
Given that $A_1 = A_2 = A$ and the resultant amplitude $A_{\text{res}} = A$,we substitute these values into the equation:
$A^2 = A^2 + A^2 + 2A^2 \cos \delta$.
$A^2 = 2A^2 + 2A^2 \cos \delta$.
$A^2 - 2A^2 = 2A^2 \cos \delta$.
$-A^2 = 2A^2 \cos \delta$.
$\cos \delta = -\frac{1}{2}$.
Since $\cos \delta = -\frac{1}{2}$,the phase difference $\delta = 120^{\circ}$ or $\delta = \frac{2\pi}{3}$ radians.

(A) The angular momentum $\vec{L}$ is defined as $\vec{L} = \vec{r} \times \vec{p}$. Since the particle moves in the $xy$-plane,$\vec{L}$ is directed along the $z$-axis (perpendicular to the plane of motion).
At point $A$,the particle is moving along the $y$-direction (tangent to the ellipse at the extreme right point),so the linear momentum $\vec{p}$ is along the $+y$ direction.
We need to find the direction of $\vec{\alpha} = \vec{p} \times \vec{L}$.
Using the right-hand rule for the cross product:
$\vec{p}$ is in the $+y$ direction $(\hat{j})$.
$\vec{L}$ is in the $+z$ direction $(\hat{k})$.
Therefore,$\vec{\alpha} = \vec{p} \times \vec{L} = (p\hat{j}) \times (L\hat{k}) = pL(\hat{j} \times \hat{k}) = pL\hat{i}$.
This corresponds to the $+x$ axis.

(D) Let the axis of rotation be the $AX$ line,which passes through vertex $A$ and is perpendicular to $AB$.
The moment of inertia $I$ of the system is the sum of the moments of inertia of individual particles about the axis $AX$.
$I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th particle from the axis $AX$.
$1$. For the particle at vertex $A$: The distance $r_A = 0$,so $I_A = m(0)^2 = 0$.
$2$. For the particle at vertex $B$: The distance $r_B = a$,so $I_B = m(a)^2 = ma^2$.
$3$. For the particle at vertex $C$: The perpendicular distance from $AX$ is $r_C = a \cos 60^{\circ} = a \times \frac{1}{2} = \frac{a}{2}$.
So,$I_C = m(\frac{a}{2})^2 = \frac{ma^2}{4}$.
The total moment of inertia $I = I_A + I_B + I_C = 0 + ma^2 + \frac{ma^2}{4} = \frac{5ma^2}{4} \text{ } g-cm^2$.

(B) Given the internal energy $U = A P^2 V$.
For an adiabatic process, the first law of thermodynamics states $dQ = dU + dW = 0$, which implies $dU = -dW = -P dV$.
Thus, $dU = -P dV$.
Taking the differential of $U$ with respect to $V$: $dU = A P^2 dV + 2 A P V dP$.
Equating the two expressions for $dU$: $-P dV = A P^2 dV + 2 A P V dP$.
Dividing by $P$ (assuming $P \neq 0$): $-dV = A P dV + 2 A V dP$.
Rearranging the terms: $-dV - A P dV = 2 A V dP \Rightarrow -(1 + A P) dV = 2 A V dP$.
Separating the variables: $-\frac{dV}{V} = \frac{2 A dP}{1 + A P}$.
Integrating both sides: $-\int \frac{dV}{V} = \int \frac{2 A dP}{1 + A P}$.
$-\ln V = 2 \ln(1 + A P) + C$.
This simplifies to $\ln V + \ln(1 + A P)^2 = \text{constant}$.
Therefore, $V(1 + A P)^2 = \text{constant}$.

(A) The most probable speed of gas molecules is given by $v_p = \sqrt{\frac{2RT}{M}}$,which implies $v_p \propto \sqrt{T}$.
Isotherms represent lines of constant temperature. In a $P-V$ diagram,isotherms further from the origin correspond to higher temperatures.
From the figure,the isotherm $EF$ is the furthest from the origin,followed by $CD$,and then $AB$ is the closest.
Points $3$ and $4$ lie on the isotherm $EF$,so $T_3 = T_4$. Thus,$v_p$ at $3 = v_p$ at $4$.
Point $2$ lies on the isotherm $CD$,so $T_2$ is the temperature of $CD$.
Point $1$ lies on the isotherm $AB$,so $T_1$ is the temperature of $AB$.
Since the order of temperatures is $T_3 = T_4 > T_2 > T_1$,the order of most probable speeds is $v_p$ at $3 = v_p$ at $4 > v_p$ at $2 > v_p$ at $1$.

(B) For a diatomic gas,the total heat given is $\Delta Q_{total} = \Delta Q_{AB} + \Delta Q_{BC}$.
$1$. For process $AB$ (Isochoric process):
$\Delta Q_{AB} = n C_V \Delta T = n \left( \frac{R}{\gamma - 1} \right) \Delta T = \frac{P_f V_f - P_i V_i}{\gamma - 1}$
Given $n = 1$ mole,diatomic gas $\gamma = 1.4 = 7/5$.
$\Delta Q_{AB} = \frac{2 P_0 V_0 - P_0 V_0}{7/5 - 1} = \frac{P_0 V_0}{2/5} = 2.5 P_0 V_0$.
$2$. For process $BC$ (Isothermal process):
$\Delta Q_{BC} = W_{BC} = n R T \ln \left( \frac{V_f}{V_i} \right) = P_B V_B \ln \left( \frac{V_C}{V_B} \right)$
Since $P_B V_B = 2 P_0 V_0$ and $V_C = 2 V_0, V_B = V_0$:
$\Delta Q_{BC} = 2 P_0 V_0 \ln \left( \frac{2 V_0}{V_0} \right) = 2 P_0 V_0 \ln 2 = 2 P_0 V_0 (0.7) = 1.4 P_0 V_0$.
Total heat $\Delta Q_{total} = 2.5 P_0 V_0 + 1.4 P_0 V_0 = 3.9 P_0 V_0$.

(B) For a polytropic process $PV^n = \text{constant}$, the molar heat capacity is $C = C_V + \frac{R}{1-n}$.
Given $n = 3$ and $C_V = \frac{3R}{2}$ for a monoatomic gas.
Thus, $C = \frac{3R}{2} + \frac{R}{1-3} = \frac{3R}{2} - \frac{R}{2} = R$.
From the ideal gas law $PV = RT$ (for $1$ mole), we have $T = \frac{PV}{R}$.
Initial state: $T_1 = \frac{P_1 V_1}{R}$.
Final state: Since $P_1 V_1^3 = P_2 V_2^3$, we have $P_2 = P_1 \left( \frac{V_1}{V_2} \right)^3$.
$T_2 = \frac{P_2 V_2}{R} = \frac{P_1 V_1^3}{R V_2^2}$.
The heat absorbed is $Q = n C \Delta T = 1 \cdot R \cdot (T_2 - T_1)$.
$Q = R \left( \frac{P_1 V_1^3}{R V_2^2} - \frac{P_1 V_1}{R} \right) = P_1 V_1 \left( \frac{V_1^2}{V_2^2} - 1 \right)$.

(D) According to the work-energy theorem,the work done by the net force is equal to the change in kinetic energy: $W = \Delta E_k$.
For an infinitesimal displacement $dx$,the work done is $dW = F \cdot dx$.
Therefore,$F = \frac{dE_k}{dx}$.
This means the force acting on the particle is equal to the slope of the $E_k$ versus $X$ graph.
We need to find the force at $X = 10 \ m$. This point lies on the line segment between $X = 8 \ m$ and $X = 12 \ m$.
The coordinates of the endpoints of this segment are $(8, 40)$ and $(12, 20)$.
The slope of this line is $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{20 - 40}{12 - 8} = \frac{-20}{4} = -5 \ N$.
Since the slope is constant between $X = 8 \ m$ and $X = 12 \ m$,the force at $X = 10 \ m$ is $-5 \hat{i} \ N$.

(B) According to Faraday's Law of electromagnetic induction,the induced electromotive force (Emf) is given by: $\epsilon = -\frac{d\phi}{dt}$.
Since the circuit has a resistance $R$,the induced current $i$ is: $i = \frac{\epsilon}{R} = -\frac{1}{R} \frac{d\phi}{dt}$.
We know that current $i = \frac{dq}{dt}$,where $dq$ is the small amount of charge flowing in time $dt$.
Substituting this,we get: $\frac{dq}{dt} = -\frac{1}{R} \frac{d\phi}{dt}$.
Integrating both sides,we find the total charge $q$ that flows:
$q = \int dq = -\frac{1}{R} \int d\phi = \frac{\Delta\phi}{R}$.
Given $\Delta\phi = 5 \text{ Wb}$ and $R = 100 \Omega$,the charge $q$ is:
$q = \frac{5}{100} = 0.05 \text{ C}$.

(D) From the figure,the current $I$ leads the emf $E$ by a phase angle $\phi = \frac{\pi}{4}$. This indicates that the circuit is a capacitive circuit ($RC$ series circuit).
In an $RC$ series circuit,the phase angle $\phi$ is given by $\tan \phi = \frac{X_C}{R}$.
Given $\phi = \frac{\pi}{4}$,we have $\tan(\frac{\pi}{4}) = 1$,so $X_C = R$.
We know $X_C = \frac{1}{\omega C}$.
Given $\omega = 100 \text{ rad/s}$ and $R = 1 \text{ k}\Omega = 1000 \text{ }\Omega$.
Substituting these values: $1000 = \frac{1}{100 \times C}$.
$C = \frac{1}{100 \times 1000} = \frac{1}{10^5} = 10 \times 10^{-6} \text{ F} = 10 \mu\text{F}$.
Thus,$R = 1 \text{ k}\Omega$ and $C = 10 \mu\text{F}$.

(D) In the given hypothetical world,the angular momentum is $L = 2n' \frac{h}{2\pi} = n' \frac{h}{\pi}$,where $n' = 1, 2, 3, \dots$.
Comparing this with the standard Bohr quantization $L = n \frac{h}{2\pi}$,we see that the allowed orbits correspond to $n = 2n'$.
The energy of an orbit is $E_n = -\frac{13.6}{n^2} \text{ eV}$. Substituting $n = 2n'$,we get $E_{n'} = -\frac{13.6}{(2n')^2} = -\frac{13.6}{4n'^2} = -\frac{3.4}{n'^2} \text{ eV}$.
For the visible range,the transition must end at the first excited state of this system. The ground state is $n'=1$ $(n=2)$,and the first excited state is $n'=2$ $(n=4)$.
The transition for the largest wavelength (smallest energy) in the visible range is from $n'=2$ to $n'=1$.
The energy difference is $\Delta E = E_2 - E_1 = -\frac{3.4}{2^2} - (-\frac{3.4}{1^2}) = -0.85 + 3.4 = 2.55 \text{ eV}$.
The wavelength is $\lambda = \frac{hc}{\Delta E} = \frac{1242 \text{ eV-nm}}{2.55 \text{ eV}} \approx 487 \text{ nm}$.

(A, B) The energy required for a transition in $He^{+}$ is given by $E = 13.6 \times Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$. For $He^{+}$,$Z=2$,so $E = 13.6 \times 4 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = 54.4 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) eV$.
For a transition from $n=2$ to $n=4$,$E = 54.4 \left( \frac{1}{4} - \frac{1}{16} \right) = 54.4 \left( \frac{3}{16} \right) = 10.2 eV$.
Since the incident photon energy is $10.2 eV$,the $He^{+}$ electron can be excited from $n=2$ to $n=4$.
Once in the $n=4$ state,the number of spectral lines emitted is given by $\frac{n(n-1)}{2} = \frac{4(4-1)}{2} = 6$.
Therefore,statements $A$ and $B$ are correct.

(B) By analyzing the circuit diagram,we can see that all three capacitors are connected in parallel between points $A$ and $B$.
In a parallel combination,the equivalent capacitance $C_{eq}$ is given by the sum of individual capacitances:
$C_{eq} = C_1 + C_2 + C_3$
Since all capacitors have capacitance $C$,we have:
$C_{eq} = C + C + C = 3C$
Therefore,the equivalent capacitance between $A$ and $B$ is $3C$.

(A) The terminal voltage $V_R$ across the external resistance $R$ is given by the formula:
$V_R = \frac{E}{(R + r)} \times R$
Dividing the numerator and denominator by $R$,we get:
$V_R = \frac{E}{(1 + r/R)}$
For the battery to act as a constant voltage source,the terminal voltage $V_R$ should be approximately equal to the emf $E$,regardless of the variations in the external load $R$.
This condition is satisfied when $r/R$ is very small,which implies $r << R$.

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant,$m$ is the mass of the particle,and $K$ is the kinetic energy.
Since the de-Broglie wavelength $\lambda$ is the same for all three particles,we have $\sqrt{2mK} = \text{constant}$,which implies $mK = \text{constant}$ or $K \propto \frac{1}{m}$.
The masses of the particles are $m_e$ (electron),$m_p$ (proton),and $m_{\alpha}$ (alpha particle). We know that $m_e < m_p < m_{\alpha}$.
Since $K$ is inversely proportional to mass,we have $\varepsilon_1 > \varepsilon_3 > \varepsilon_2$ (where $\varepsilon_1$ corresponds to the electron,$\varepsilon_3$ to the proton,and $\varepsilon_2$ to the alpha particle).

(B) Let the charge $q$ be displaced by a small distance $y$ along the $y$-axis. The position of the charge becomes $(0, y)$.
The distance $r$ of this charge from each of the charges $-q$ located at $(-a, 0)$ and $(a, 0)$ is $r = \sqrt{a^2 + y^2}$.
The force exerted by each charge $-q$ on $q$ is $F' = \frac{kq^2}{r^2} = \frac{kq^2}{a^2 + y^2}$.
The components of these forces along the $x$-axis cancel out due to symmetry.
The components along the $y$-axis are both directed towards the origin (negative $y$-direction).
The net force is $F = -2 F' \sin \theta$,where $\sin \theta = \frac{y}{r} = \frac{y}{\sqrt{a^2 + y^2}}$.
$F = -2 \left( \frac{kq^2}{a^2 + y^2} \right) \left( \frac{y}{\sqrt{a^2 + y^2}} \right) = -\frac{2kq^2 y}{(a^2 + y^2)^{3/2}}$.
Since the displacement $y$ is very small $(y \ll a)$,we can approximate $(a^2 + y^2)^{3/2} \approx (a^2)^{3/2} = a^3$.
Thus,$F \approx -\frac{2kq^2}{a^3} y$.
Therefore,$F \propto -y$.

(C) When a charge $q_a$ is placed inside a cavity of a conducting sphere,it induces a charge $-q_a$ on the inner surface of the cavity and a charge $+q_a$ on the outer surface of the conducting sphere.
Similarly,placing $q_b$ in the other cavity induces $-q_b$ on its inner surface and $+q_b$ on the outer surface of the conducting sphere.
The total charge on the outer surface of the conducting sphere becomes $q_a + q_b$. Since the sphere is conducting,the electric field inside the bulk of the conductor is zero.
The charge $q_a$ experiences a force due to the field produced by the induced charge $-q_b$ on the inner surface of the second cavity and the field produced by the charge distribution on the outer surface of the sphere.
However,due to the electrostatic shielding property of the conductor,the electric field produced by the charges outside the cavity (including the outer surface charge and the other cavity's induced charge) is zero inside the cavity containing $q_a$.
Therefore,the charge $q_a$ only experiences the field produced by the charge $-q_b$ induced on the inner surface of its own cavity,which is zero at its center due to symmetry.
Thus,the net force on $q_a$ due to $q_b$ and the induced charges is zero.

(A, C) According to Gauss's Law,the total electric flux $\phi_{\text{total}}$ through any closed surface in a region with no enclosed charge is zero,i.e.,$\phi_{\text{total}} = 0$.
For the hemisphere,$\phi_{\text{total}} = \phi_{\text{curved}} + \phi_{\text{flat}} = 0$.
The flux through the flat circular surface is $\phi_{\text{flat}} = -E \cdot A = -E(\pi R^2)$ (since the field lines enter the surface).
Therefore,$\phi_{\text{curved}} = -\phi_{\text{flat}} = E \pi R^2$.
Thus,statement $(A)$ is correct and $(C)$ is correct.
Regarding the work done,for a uniform electric field,the potential difference $\Delta V = -\vec{E} \cdot \Delta \vec{r}$.
Since points $A$ and $B$ lie on the same equipotential plane perpendicular to the electric field $\vec{E}$,the potential difference between $A$ and $B$ is $\Delta V = 0$.
Therefore,the work done $W = q \Delta V = 0$,which is independent of the path and the radius $R$.

(A) Let $q$ be the charge appearing on the outer shell of radius $2R$ after earthing.
Since the outer shell is grounded,its potential must be zero.
The potential at the outer shell is the sum of potentials due to the inner charge $Q$ and the outer charge $q$:
$V_{\text{outer}} = \frac{KQ}{2R} + \frac{Kq}{2R} = 0$
Solving for $q$,we get $q = -Q$.
Now,we calculate the potential at a distance $r = \frac{3R}{2}$ from the center.
Since $R < r < 2R$,the point lies outside the inner sphere and inside the outer sphere.
The potential at this point is the sum of the potential due to the inner sphere (as a point charge) and the potential due to the outer shell (constant inside):
$V(r) = \frac{KQ}{r} + \frac{Kq}{2R}$
Substituting $r = \frac{3R}{2}$ and $q = -Q$:
$V = \frac{KQ}{3R/2} + \frac{K(-Q)}{2R} = \frac{2KQ}{3R} - \frac{KQ}{2R} = \frac{4KQ - 3KQ}{6R} = \frac{KQ}{6R} = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{6R}$

(A) The relation between electric field $E$ and electric potential $V$ is given by $E = -\frac{dV}{dx}$.
This means the electric field is the negative of the slope of the $V-X$ graph.
$1$. For the interval $X \in [-4, -2]$: The slope is $\frac{10 - 0}{-2 - (-4)} = \frac{10}{2} = 5$. Thus,$E = -5$.
$2$. For the interval $X \in [-2, 2]$: The potential is constant $(V = 10)$,so the slope is $0$. Thus,$E = 0$.
$3$. For the interval $X \in [2, 7]$: The slope is $\frac{0 - 10}{7 - 2} = \frac{-10}{5} = -2$. Thus,$E = -(-2) = 2$.
Comparing these values with the given options,option $A$ represents $E = -5$ for $X \in [-4, -2]$,$E = 0$ for $X \in [-2, 2]$,and $E = 2$ for $X \in [2, 7]$.

(D) The potential at the surface of shell $B$ is the sum of potentials due to all three shells $A, B$ and $C$.
$V_B = V_{A,B} + V_{B,B} + V_{C,B}$
Since the potential inside a shell is constant and equal to the potential at its surface,we have:
$V_{A,B} = \frac{k Q_A}{b} = \frac{1}{4\pi\varepsilon_0} \frac{\sigma(4\pi a^2)}{b} = \frac{\sigma a^2}{\varepsilon_0 b}$
$V_{B,B} = \frac{k Q_B}{b} = \frac{1}{4\pi\varepsilon_0} \frac{-\sigma(4\pi b^2)}{b} = -\frac{\sigma b}{\varepsilon_0}$
$V_{C,B} = \frac{k Q_C}{c} = \frac{1}{4\pi\varepsilon_0} \frac{\sigma(4\pi c^2)}{c} = \frac{\sigma c}{\varepsilon_0}$
Adding these,we get:
$V_B = \frac{\sigma a^2}{\varepsilon_0 b} - \frac{\sigma b}{\varepsilon_0} + \frac{\sigma c}{\varepsilon_0} = \frac{\sigma}{\varepsilon_0} \left( \frac{a^2}{b} - b + c \right)$

(C) The magnetic force on a small current element $I d\vec{l}$ is given by $d\vec{F} = I (d\vec{l} \times \vec{B})$.
Here,the current element is $d\vec{l} = dx \hat{i}$ and the magnetic field is $\vec{B} = B_0 \left( 2 - \frac{x}{a} \right) \hat{k}$.
Thus,$d\vec{F} = I (dx \hat{i}) \times B_0 \left( 2 - \frac{x}{a} \right) \hat{k} = I B_0 \left( 2 - \frac{x}{a} \right) dx (\hat{i} \times \hat{k}) = I B_0 \left( 2 - \frac{x}{a} \right) dx (-\hat{j})$.
To find the total force,we integrate from $x = a$ to $x = 2a$:
$\vec{F} = \int_{a}^{2a} I B_0 \left( 2 - \frac{x}{a} \right) dx (-\hat{j})$
$\vec{F} = -I B_0 \hat{j} \int_{a}^{2a} \left( 2 - \frac{x}{a} \right) dx$
$\vec{F} = -I B_0 \hat{j} \left[ 2x - \frac{x^2}{2a} \right]_{a}^{2a}$
$\vec{F} = -I B_0 \hat{j} \left[ \left( 2(2a) - \frac{(2a)^2}{2a} \right) - \left( 2(a) - \frac{a^2}{2a} \right) \right]$
$\vec{F} = -I B_0 \hat{j} \left[ (4a - 2a) - (2a - 0.5a) \right]$
$\vec{F} = -I B_0 \hat{j} [ 2a - 1.5a ] = -I B_0 \left( \frac{a}{2} \right) \hat{j}$.
Comparing this with the given force $\vec{F} = I B_0 \left( \frac{ka}{2} \right) \hat{j}$,we get $k = -1$.

(C) The magnetic force $F_B$ acting on a current-carrying wire in a magnetic field is given by $F_B = I \vec{L}_{eff} \times \vec{B}$.
For a semi-circular wire of radius $r$,the effective length $\vec{L}_{eff}$ is the straight-line distance between the two ends,which is the diameter $2r$.
Therefore,the magnitude of the magnetic force is $F_B = I(2r)B = 2IrB$.
This total upward magnetic force is shared equally by the two identical springs $X$ and $Y$ attached to the ends of the wire.
Let $F$ be the force acting on each spring. Then,$2F = F_B$.
Substituting the value of $F_B$,we get $2F = 2IrB$.
Thus,the force on each spring is $F = I r B$.

(A, B, D) The motional electromotive force $(EMF)$ induced in the loop is $\varepsilon = B \times b \times v$,acting only across the segment $PS$ which is moving through the magnetic field.
The total resistance of the loop is $R_{\text{total}} = \lambda(2b + 2a)$.
The induced current is $I = \frac{\varepsilon}{R_{\text{total}}} = \frac{Bbv}{\lambda(2b + 2a)}$.
Using Lenz's Law,the magnetic flux into the page is decreasing as the loop moves out,so the induced current will be in the clockwise direction to oppose this change.
For part $SR$,the length vector $\vec{\ell}$ is parallel to the velocity vector $\vec{v}$,so the induced $EMF$ $\varepsilon = \vec{v} \times \vec{B} \cdot \vec{\ell} = 0$. Thus,no induction occurs in part $SR$.
Therefore,statements $A$,$B$,and $D$ are correct.

(C) Consider a segment of length $\ell$ on both line charges with linear charge densities $\lambda_1$ and $\lambda_2$.
The electric force $F_E$ per unit length between the two line charges is given by Coulomb's law for infinite lines: $F_E = \frac{2 k \lambda_1 \lambda_2 \ell}{d} = \frac{2 \lambda_1 \lambda_2 \ell}{4 \pi \varepsilon_0 d} = \frac{\lambda_1 \lambda_2 \ell}{2 \pi \varepsilon_0 d}$.
The moving charges constitute currents $I_1 = \lambda_1 v$ and $I_2 = \lambda_2 v$.
The magnetic force $F_B$ per unit length between two parallel current-carrying wires is given by: $F_B = \frac{\mu_0 I_1 I_2 \ell}{2 \pi d} = \frac{\mu_0 (\lambda_1 v) (\lambda_2 v) \ell}{2 \pi d} = \frac{\mu_0 \lambda_1 \lambda_2 v^2 \ell}{2 \pi d}$.
For the magnetic attraction to balance the electric repulsion,we set $F_E = F_B$:
$\frac{\lambda_1 \lambda_2 \ell}{2 \pi \varepsilon_0 d} = \frac{\mu_0 \lambda_1 \lambda_2 v^2 \ell}{2 \pi d}$.
Simplifying the equation,we get: $\frac{1}{\varepsilon_0} = \mu_0 v^2$.
Since the speed of light $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$,it follows that $c^2 = \frac{1}{\mu_0 \varepsilon_0}$,which implies $v^2 = c^2$ or $v = c$.

(C) The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} \times \vec{B})$.
Since the magnetic field $\vec{B}$ is perpendicular to the plane of the orbit,the force $\vec{F}$ acts towards the center of the circular path (centripetal force).
The torque $\vec{\tau}$ about the nucleus is given by $\vec{\tau} = \vec{r} \times \vec{F}$.
Since the force $\vec{F}$ is directed towards the center,the position vector $\vec{r}$ and the force vector $\vec{F}$ are collinear (anti-parallel).
Therefore,$\vec{\tau} = \vec{r} \times \vec{F} = 0$.
Since the torque is zero,the dot product of the torque vector with any vector,including the angular momentum vector $\vec{L}$,must be zero: $\vec{\tau} \cdot \vec{L} = 0$.

(C) The angular resolution $\theta$ is given by the ratio of the distance between two dots $x$ to the distance $d$ from the eye, i.e., $\theta = \frac{x}{d}$.
Given that the printer prints $300 \text{ dots per inch}$, the distance between two adjacent dots $x$ is $x = \frac{1 \text{ inch}}{300} = \frac{2.54 \text{ cm}}{300}$.
Substituting the values into the formula: $5.8 \times 10^{-4} = \frac{2.54 \text{ cm} / 300}{d}$.
Rearranging for $d$: $d = \frac{2.54}{300 \times 5.8 \times 10^{-4}} \text{ cm}$.
$d = \frac{2.54}{0.174} \text{ cm} \approx 14.597 \text{ cm}$.
Thus, the minimal distance is approximately $14.59 \text{ cm}$.

(B) The voltage across the series resistor $R_s$ is given by $V_{R_s} = V_{in} - V_z = 10 \, V - 6 \, V = 4 \, V$.
For the Zener diode to act as a voltage regulator, it must maintain a minimum current $I_{z,min} = 10 \, mA$ even when the load current $I_L$ is zero.
The total current $I$ flowing through $R_s$ is $I = I_z + I_L$.
To find the maximum value of $R_s$, we consider the condition where the load current $I_L$ is minimum (i.e., $I_L = 0$).
Thus, $I = I_{z,min} = 10 \, mA$.
Using Ohm's law, $R_s = \frac{V_{R_s}}{I} = \frac{4 \, V}{10 \, mA} = \frac{4 \, V}{10 \times 10^{-3} \, A} = 400 \, \Omega$.
Therefore, the maximum value of $R_s$ is $400 \, \Omega$.

(C) The intensity of light in an interference pattern is given by $I_{res} = 4I_0 \cos^2(\frac{\phi}{2})$,where $I_0$ is the intensity of each individual wave and $\phi$ is the phase difference.
At a path difference of $\Delta x = \lambda$,the phase difference is $\phi = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi$.
The intensity is $I = 4I_0 \cos^2(\frac{2\pi}{2}) = 4I_0 \cos^2(\pi) = 4I_0(1)^2 = 4I_0$.
Now,for a path difference of $\Delta x = \frac{\lambda}{4}$,the phase difference is $\phi' = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2}$.
The new intensity $I'$ is $4I_0 \cos^2(\frac{\pi/2}{2}) = 4I_0 \cos^2(\frac{\pi}{4})$.
Since $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$,we have $I' = 4I_0 (\frac{1}{\sqrt{2}})^2 = 4I_0 \cdot \frac{1}{2} = 2I_0$.
Since $I = 4I_0$,then $I_0 = \frac{I}{4}$.
Substituting this into the expression for $I'$,we get $I' = 2(\frac{I}{4}) = 0.5I$.

(A) The maximum intensity $I_{max}$ and minimum intensity $I_{min}$ in an interference pattern are given by $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
Given that $I_{max} = 4 I_{min}$,we have $(\sqrt{I_1} + \sqrt{I_2})^2 = 4(\sqrt{I_1} - \sqrt{I_2})^2$.
Taking the square root on both sides: $\sqrt{I_1} + \sqrt{I_2} = 2(\sqrt{I_1} - \sqrt{I_2})$.
Rearranging the terms: $\sqrt{I_1} + \sqrt{I_2} = 2\sqrt{I_1} - 2\sqrt{I_2}$.
This simplifies to $3\sqrt{I_2} = \sqrt{I_1}$.
Squaring both sides gives $9I_2 = I_1$,so the ratio $\frac{I_1}{I_2} = 9/1$.
The ratio of the intensities of the two waves is $I_2/I_1 = 1/9$.