WBJEE 2024 Physics Question Paper with Answer and Solution

(D) Let the uniform plate be considered as a large rectangle of sides $2a$ and $2b$ with a smaller rectangle of sides $a$ and $b$ removed from it.
Let the mass of the removed rectangle be $m_1$ and the mass of the remaining shaded plate be $m_2$.
The area of the removed rectangle is $A_1 = a \times b$. Thus,$m_1 = \sigma ab$.
The area of the full rectangle is $A = 2a \times 2b = 4ab$. The area of the remaining plate is $A_2 = 4ab - ab = 3ab$. Thus,$m_2 = 3\sigma ab = 3m_1$.
The centre of mass of the full rectangle is at $(-a/2, -b/2)$ relative to the origin $(0,0)$.
The centre of mass of the removed rectangle is at $(a/2, b/2)$.
Using the principle of superposition for the centre of mass: $X_{cm} = \frac{m_2 x_2 + m_1 x_1}{m_2 + m_1} = 0$ (since the full rectangle's $CM$ is at the origin if we shift coordinates,but here we use the formula for a cavity).
For a cavity,the position of the centre of mass of the remaining part is given by $\vec{R} = \frac{M\vec{R}_{full} - m_1\vec{r}_1}{M - m_1}$.
Here,the full rectangle has its centre at $(-a/2, -b/2)$. Its mass is $M = 4m_1$.
The removed part has its centre at $(a/2, b/2)$.
$X_{cm} = \frac{4m_1(-a/2) - m_1(a/2)}{3m_1} = \frac{-2m_1a - 0.5m_1a}{3m_1} = \frac{-2.5a}{3} = -\frac{5a}{6}$.
Wait,re-evaluating based on the coordinate system provided: The origin $(0,0)$ is at the corner of the cavity.
The full rectangle (if it were complete) would span from $x = -a$ to $x = a$ and $y = -b$ to $y = b$. Its $CM$ is at $(0,0)$.
The removed rectangle spans $x = 0$ to $a$ and $y = 0$ to $b$. Its $CM$ is at $(a/2, b/2)$.
$X_{cm} = \frac{M(0) - m_1(a/2)}{3m_1} = -a/6$.
$Y_{cm} = \frac{M(0) - m_1(b/2)}{3m_1} = -b/6$.
Thus,the position is $(-\frac{a}{6}, -\frac{b}{6})$.

(B) The moment of inertia of a satellite of mass $m$ at a distance $R$ from the center of the Earth is $I = m R^2$.
The kinetic energy $(K)$ of the satellite in terms of angular momentum $(J)$ is given by $K = \frac{J^2}{2 I}$.
Substituting $I = m R^2$,we get $K = \frac{J^2}{2 m R^2}$.
For a satellite in a circular orbit,the total energy $(E)$ is related to the kinetic energy $(K)$ by the relation $E = -K$.
Therefore,$E = -\frac{J^2}{2 m R^2}$.
Thus,the correct option is $B$.

(B) The total area under the $P(v)$ versus $v$ curve represents the total number of particles $N$.
From the graph,the area is a triangle from $0$ to $v_0$ and a rectangle from $v_0$ to $2 v_0$.
Area $= \frac{1}{2} \times v_0 \times a + (2 v_0 - v_0) \times a = \frac{1}{2} v_0 a + v_0 a = \frac{3}{2} v_0 a$.
Since the total area equals $N$,we have $\frac{3}{2} v_0 a = N$,which implies $v_0 a = \frac{2}{3} N$.
We need to find the number of particles with speeds between $1.2 v_0$ and $1.8 v_0$. This corresponds to the area under the curve between these limits.
Since $P(v) = a$ for $v > v_0$,the area is a rectangle with width $(1.8 v_0 - 1.2 v_0) = 0.6 v_0$ and height $a$.
Number of particles $= 0.6 v_0 \times a = 0.6 (v_0 a) = 0.6 \times (\frac{2}{3} N) = 0.4 N$.

(A) For the block to just start moving,the horizontal component of the applied force $F$ must be equal to the limiting friction force.
$1$. Resolve the force $F$ into components:
Horizontal component: $F_{x} = F \cos 30^{\circ}$
Vertical component: $F_{y} = F \sin 30^{\circ}$
$2$. Determine the normal reaction $N$:
The vertical forces acting on the block are the weight $mg$ downwards,the normal reaction $N$ upwards,and the vertical component of the applied force $F \sin 30^{\circ}$ upwards.
$N + F \sin 30^{\circ} = mg$
$N = mg - F \sin 30^{\circ}$
Given $m = 10 \ kg$ and $g = 10 \ ms^{-2}$,so $mg = 100 \ N$.
$N = 100 - F \sin 30^{\circ} = 100 - 0.5F$
$3$. Apply the condition for motion:
The limiting friction is $f_{L} = \mu_{s} N$.
The block starts to move when $F \cos 30^{\circ} = \mu_{s} N$.
$F \cos 30^{\circ} = 0.25(100 - 0.5F)$
$F \frac{\sqrt{3}}{2} = 25 - 0.125F$
$F(0.866 + 0.125) = 25$
$F(0.991) = 25$
$F = \frac{25}{0.991} \approx 25.22 \ N$
Thus,the block will just start to move for $F \approx 25.2 \ N$.

(B) Let the height of the water level above the hole be $h = H - z$. The velocity of efflux is $v = \sqrt{2gh}$.
The time taken for the water to reach the ground is $t = \sqrt{\frac{2z}{g}}$.
The horizontal range $R$ is given by $R = v \cdot t = \sqrt{2g(H-z)} \cdot \sqrt{\frac{2z}{g}} = 2\sqrt{z(H-z)}$.
To maximize $R$,we maximize $R^2 = 4(zH - z^2)$.
Differentiating with respect to $z$ and setting to zero:
$\frac{d}{dz}(4zH - 4z^2) = 4H - 8z = 0$.
$8z = 4H \Rightarrow z = \frac{H}{2}$.
Thus,the range is maximum when the hole is at half the height of the vessel.

(D) Let $V$ be the total volume of the body,$d$ be its density,and $\sigma$ be the density of water.
For a floating body,the weight equals the buoyant force: $Vdg = V(1 - \frac{1}{n})\sigma g$.
Thus,$d = (\frac{n-1}{n})\sigma$.
When the body is submerged,the net upward force $F_{net}$ is $F_B - mg = V\sigma g - Vdg = V\sigma g - V(\frac{n-1}{n})\sigma g = V\sigma g (1 - \frac{n-1}{n}) = V\sigma g (\frac{1}{n})$.
The acceleration $a$ is given by $a = \frac{F_{net}}{m} = \frac{V\sigma g / n}{Vd} = \frac{\sigma g / n}{(\frac{n-1}{n})\sigma} = \frac{g}{n-1}$.
Using the kinematic equation $h = \frac{1}{2}at^2$,we have $h = \frac{1}{2} (\frac{g}{n-1}) t^2$.
Solving for $t$,we get $t = \sqrt{\frac{2h(n-1)}{g}}$.
Therefore,$t \propto \sqrt{n-1}$.

(B) The viscous force $F$ acting on the plate is given by Newton's law of viscosity: $F = \eta A \frac{dv}{dx}$.
Here,$\eta = 1.55 \,Ns\, m^{-2}$ is the coefficient of viscosity,$A = 10^{-2} \,m^2$ is the area of the plate,$v = 3 \times 10^{-2} \,ms^{-1}$ is the velocity,and $h = dx = 2 \times 10^{-3} \,m$ is the thickness of the oil layer.
Substituting the values: $F = 1.55 \times 10^{-2} \times \frac{3 \times 10^{-2}}{2 \times 10^{-3}}$.
$F = 1.55 \times 10^{-2} \times 1.5 \times 10^1$.
$F = 1.55 \times 1.5 \times 10^{-1} = 2.325 \times 10^{-1} = 0.2325 \,N$.

(B) The elastic potential energy stored in a strained body is defined as the work done in deforming the body.
For a material obeying Hooke's Law,the energy density (energy per unit volume) is given by $u = \frac{1}{2} \times \text{stress} \times \text{strain}$.
To find the total elastic potential energy $(U)$,we multiply the energy density by the total volume $(V)$ of the body.
Therefore,$U = u \times V = \frac{1}{2} \times \text{stress} \times \text{strain} \times V$.

(A) The angular momentum $\vec{L}$ of a particle about the origin is given by $\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times (m\vec{v})$.
Given $\vec{r} = x(t) \hat{i} + b \hat{j}$ and $\vec{v} = v \hat{i}$.
$\vec{L} = (x(t) \hat{i} + b \hat{j}) \times (m v \hat{i}) = x(t) m v (\hat{i} \times \hat{i}) + b m v (\hat{j} \times \hat{i})$.
Since $\hat{i} \times \hat{i} = 0$ and $\hat{j} \times \hat{i} = -\hat{k}$,we get $\vec{L} = -b m v \hat{k}$.
The magnitude of the angular momentum is $|\vec{L}| = | -b m v | = b m v$.
Since $b$,$m$,and $v$ are all constants,the magnitude of the angular momentum $|\vec{L}|$ is constant and independent of the angle $\theta$.
Therefore,the graph of $|\vec{L}|$ versus $\theta$ is a horizontal straight line.

(C) The forces acting on the ball are the tension $T$ in the string and the gravitational force $mg$ acting downwards.
Let the centre of the horizontal circle be $O$. The position vector $\vec{r}$ of the ball relative to $O$ lies in the horizontal plane.
The gravitational force $\vec{F}_g = m\vec{g}$ acts vertically downwards through the ball.
The torque $\vec{\tau}$ about point $O$ is given by $\vec{\tau} = \vec{r} \times \vec{F}$.
Since the force of gravity $\vec{F}_g$ is parallel to the vertical axis passing through $O$,and the position vector $\vec{r}$ is horizontal,the torque due to gravity is $\vec{\tau}_g = \vec{r} \times (m\vec{g})$. The magnitude is $mgr \sin(90^\circ) = mgr$,but the direction is tangential to the circle.
However,the tension force $\vec{T}$ acts along the string towards the point of suspension. The line of action of the tension force passes through the point of suspension,but not through $O$.
Crucially,for a particle moving in a horizontal circle with constant angular velocity,the net torque about the centre $O$ must be zero because the angular momentum $\vec{L} = \vec{r} \times \vec{p}$ is constant in magnitude and direction (the vector $\vec{L}$ points along the vertical axis). Since $\frac{d\vec{L}}{dt} = \vec{\tau}_{net} = 0$,the net torque about $O$ is zero.

(A) Since the surface of the bowl is smooth,there is no friction to cause the sphere to roll. The sphere will simply slide down the surface.
By the law of conservation of mechanical energy,the loss in potential energy is equal to the gain in kinetic energy.
The change in height of the center of mass of the sphere is $h = R - r$.
Potential energy lost = $mgh = mg(R - r)$.
Since the sphere is sliding (not rolling),its total kinetic energy is purely translational: $K = \frac{1}{2} mv^2$.
Therefore,the total kinetic energy at the lowest point is $K = mg(R - r)$.

(B, C) Let the rod make an angle $\theta$ with the $OX$ axis. The velocity of end $A$ is $v_A = 3 \ m/s$ along $OX$ and the velocity of end $B$ is $v_B = 4 \ m/s$ along $OY$ (downwards).
Using the condition that the length of the rod is constant,the component of velocity along the rod must be the same for both ends:
$v_A \cos \theta = v_B \sin \theta \implies 3 \cos \theta = 4 \sin \theta \implies \tan \theta = \frac{3}{4}$.
Thus,$\sin \theta = \frac{3}{5}$ and $\cos \theta = \frac{4}{5}$.
The angular velocity $\omega$ is given by $\omega = \frac{v_A \sin \theta + v_B \cos \theta}{\ell} = \frac{3(3/5) + 4(4/5)}{1} = \frac{9/5 + 16/5}{1} = \frac{25}{5} = 5 \ rad/s$.
Since the rod is rotating such that $A$ moves right and $B$ moves down,the rotation is clockwise.
The rotational kinetic energy is $(KE)_{rot} = \frac{1}{2} I \omega^2 = \frac{1}{2} (\frac{1}{12} m \ell^2) \omega^2 = \frac{1}{2} \times \frac{1}{12} \times 4 \times (1)^2 \times (5)^2 = \frac{1}{24} \times 4 \times 25 = \frac{25}{6} \ J$.
Both statements $B$ and $C$ are correct.

(C) Equilibrium points occur where the force $F = -\frac{dV}{dx} = 0$,which corresponds to the points where the slope of the $V(x)$ versus $x$ graph is zero (i.e.,the peaks and valleys of the curve).
Looking at the graph,there are three such points: two local maxima (peaks) and one local minimum (valley).
$1$. At a local minimum,the potential energy is at a minimum,so $\frac{d^2V}{dx^2} > 0$. This represents a stable equilibrium point.
$2$. At a local maximum,the potential energy is at a maximum,so $\frac{d^2V}{dx^2} < 0$. This represents an unstable equilibrium point.
Since there are two peaks (unstable) and one valley (stable),there are three equilibrium points in total: one stable and two unstable.

(D) The applied voltage is given by $V = 50 \sqrt{2} \sin \omega t$. The peak voltage is $V_0 = 50 \sqrt{2} \ V$,so the $rms$ voltage is $V_{rms} = \frac{V_0}{\sqrt{2}} = 50 \ V$.
In a series $LCR$ circuit,the relationship between the $rms$ voltages is given by $V_{rms}^2 = V_R^2 + (V_L - V_C)^2$.
Given: $V_R = 30 \ V$,$V_C = 90 \ V$,and $V_{rms} = 50 \ V$.
Substituting the values: $50^2 = 30^2 + (V_L - 90)^2$.
$2500 = 900 + (V_L - 90)^2$.
$(V_L - 90)^2 = 1600$.
$V_L - 90 = \pm 40$.
Case $1$: $V_L = 90 + 40 = 130 \ V$.
Case $2$: $V_L = 90 - 40 = 50 \ V$.
Assuming the standard case where the inductor voltage is the peak value asked,we look for the peak voltage across the inductor. The $rms$ voltage $V_L$ is $50 \ V$ or $130 \ V$. The peak voltage is $(V_L)_{peak} = V_L \sqrt{2}$.
For $V_L = 50 \ V$,$(V_L)_{peak} = 50 \sqrt{2} \ V$.

(C, D) In a series $\text{LCR}$ circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
At resonance (point $C$),$X_L = X_C$,so $Z$ is minimum.
For frequencies lower than the resonant frequency (portion $AC$),$X_C > X_L$,which means the circuit is capacitive.
For frequencies higher than the resonant frequency (portion $BC$),$X_L > X_C$,which means the circuit is inductive.
Therefore,the impedance $Z$ is capacitive in the portion $AC$ and inductive in the portion $BC$.

(D) In the given circuit,the two capacitors on the left are connected in series. Let their equivalent capacitance be $C_s$.
$C_s = \frac{C \times C}{C + C} = \frac{C^2}{2C} = \frac{C}{2}$.
Now,this equivalent capacitor $C_s$ is in parallel with the third capacitor $C$ connected between the middle junction and point $N$.
Therefore,the total equivalent capacitance $C_{eq}$ between points $P$ and $N$ is:
$C_{eq} = C_s + C = \frac{C}{2} + C = \frac{3C}{2}$.

(C) According to the maximum power transfer theorem,the power delivered to an external circuit is maximum when the external resistance $R_{\text{ext}}$ is equal to the internal resistance $r$ of the cell,i.e.,$R_{\text{ext}} = r$.
First,we determine the equivalent external resistance $R_{\text{ext}}$ between terminals $A$ and $B$. The circuit consists of three resistors of resistance $R$ connected in parallel.
For three resistors $R$ in parallel,the equivalent resistance $R_{\text{ext}}$ is given by:
$\frac{1}{R_{\text{ext}}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R}$
Therefore,$R_{\text{ext}} = \frac{R}{3}$.
Setting $R_{\text{ext}} = r$ for maximum power transfer:
$\frac{R}{3} = r$
$R = 3r$
Thus,the value of $R$ for which the power is maximum is $3r$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K$ of the emitted electrons is given by:
$K = \frac{hc}{\lambda} - \phi$
When an electron of charge $q$ and mass $m$ moves perpendicular to a magnetic field $B$,it follows a circular path of radius $R$ given by:
$R = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$
Squaring both sides,we get:
$R^2 = \frac{2mK}{q^2 B^2}$
Rearranging for $B^2$:
$B^2 = \frac{2mK}{q^2 R^2}$
Substituting the expression for $K$:
$B^2 = \frac{2m}{q^2 R^2} \left( \frac{hc}{\lambda} - \phi \right)$
$B^2 = \left( \frac{2mhc}{q^2 R^2} \right) \frac{1}{\lambda} - \left( \frac{2m\phi}{q^2 R^2} \right)$
This is an equation of a straight line of the form $y = mx + c$,where $y = B^2$,$x = \frac{1}{\lambda}$,slope $m = \frac{2mhc}{q^2 R^2}$ (positive),and intercept $c = -\frac{2m\phi}{q^2 R^2}$ (negative).
Thus,the graph is a straight line with a positive slope and a negative y-intercept. This corresponds to Graph $C$.

(D) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Using $hc \approx 12400 \ \text{eV} \cdot \mathring{A}$,we get $E = \frac{12400}{4770} \ \text{eV} \approx 2.6 \ \text{eV}$.
For photoelectric emission to occur,the energy of the incident photon must be greater than or equal to the work function $(\phi)$ of the metal $(E \ge \phi)$.
Comparing $E = 2.6 \ \text{eV}$ with the work functions:
For metal $A$: $2.6 \ \text{eV} < 4.2 \ \text{eV}$ (No emission)
For metal $B$: $2.6 \ \text{eV} < 3.7 \ \text{eV}$ (No emission)
For metal $C$: $2.6 \ \text{eV} < 3.2 \ \text{eV}$ (No emission)
For metal $D$: $2.6 \ \text{eV} > 2.3 \ \text{eV}$ (Emission occurs)
Therefore,electrons will be emitted only from metal $D$.

(C) Let the distance of the bar from the vertex at time $t$ be $x = vt$.
In the isosceles triangle formed by the plates and the bar,the length of the bar $\ell$ can be calculated using trigonometry.
Considering the right-angled triangle formed by the bisector of the angle $\theta$,we have:
$\frac{\ell/2}{x} = \tan \frac{\theta}{2}$
$\ell = 2x \tan \frac{\theta}{2} = 2vt \tan \frac{\theta}{2}$.
The motional emf induced in a conductor of length $\ell$ moving with velocity $v$ in a magnetic field $B$ is given by $\varepsilon = B \ell v$.
Substituting the value of $\ell$ at $t = 1 \text{ s}$:
$\ell = 2v(1) \tan \frac{\theta}{2} = 2v \tan \frac{\theta}{2}$.
Therefore,the induced emf is:
$\varepsilon = B(2v \tan \frac{\theta}{2})v = 2Bv^2 \tan \frac{\theta}{2}$.

(D) Longitudinal waves cannot be polarized. Polarization is a property that applies only to transverse waves,where the oscillations occur perpendicular to the direction of wave propagation. In longitudinal waves,the oscillations occur parallel to the direction of wave propagation. Since there is no asymmetry in the direction of oscillation relative to the direction of propagation,longitudinal waves cannot be restricted to a single plane of vibration.

(B) The given electric field is $\vec{E} = E_0 \hat{n} e^{i k_0[(x+y+z) - ct]}$.
Comparing this with the standard form $\vec{E} = E_0 \hat{n} e^{i(\vec{k} \cdot \vec{r} - \omega t)}$,we identify the wave vector $\vec{k} = k_0(\hat{i} + \hat{j} + \hat{k})$.
The magnitude of the wave vector is $k = |k_0(\hat{i} + \hat{j} + \hat{k})| = k_0 \sqrt{1^2 + 1^2 + 1^2} = k_0 \sqrt{3}$.
The speed of the wave in the medium is $v = \frac{\omega}{k} = \frac{k_0 c}{k_0 \sqrt{3}} = \frac{c}{\sqrt{3}}$.
The refractive index $n = \frac{c}{v} = \sqrt{3}$.
Since the wave is transverse,$\vec{E} \cdot \vec{k} = 0$,which implies $\hat{n} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$.
Given that $\vec{E}$ is polarized in the $x-z$ plane,$\hat{n}$ must be in the $x-z$ plane,so $\hat{n} = a\hat{i} + b\hat{k}$.
From $\hat{n} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$,we get $a + b = 0$,so $a = -b$.
Normalizing $\hat{n}$,we get $\hat{n} = \frac{\hat{i} - \hat{k}}{\sqrt{2}}$.
Thus,both options $B$ and $C$ are correct.

(A) The electric field at point $P$ at distance $x$ due to the three charges is:
$E = \frac{1}{4 \pi \epsilon_0} \left[ \frac{q}{(x-a)^2} - \frac{2q}{x^2} + \frac{q}{(x+a)^2} \right]$
$E = \frac{q}{4 \pi \epsilon_0} \left[ \frac{(x+a)^2 - 2(x^2-a^2) + (x-a)^2}{x^2(x^2-a^2)} \right]$
$E = \frac{q}{4 \pi \epsilon_0} \left[ \frac{x^2 + 2ax + a^2 - 2x^2 + 2a^2 + x^2 - 2ax + a^2}{x^2(x^2-a^2)} \right]$
$E = \frac{q}{4 \pi \epsilon_0} \left[ \frac{4a^2}{x^2(x^2-a^2)} \right]$
Since $x \gg a$,we can approximate $x^2 - a^2 \approx x^2$:
$E = \frac{q}{4 \pi \epsilon_0} \left[ \frac{4a^2}{x^2(x^2)} \right] = \frac{4qa^2}{4 \pi \epsilon_0 x^4}$
Given $qa^2 = Q$,we have $E = \frac{4Q}{4 \pi \epsilon_0 x^4}$.
Comparing this with $E = \frac{\alpha Q}{4 \pi \epsilon_0 x^\beta}$,we get $\alpha = 4$ and $\beta = 4$.
Therefore,$\alpha = \beta$.

(C) According to Gauss's Law,the total electric flux $\phi$ through any closed surface is equal to the net charge $Q_{\text{enclosed}}$ inside the surface divided by the permittivity of free space $\epsilon_0$.
Mathematically,$\phi = \frac{Q_{\text{enclosed}}}{\epsilon_0}$.
Since the charge $Q$ is placed at the centre of the cube,the entire charge is enclosed by the cube.
Therefore,the total flux through the six surfaces of the cube is $\phi = \frac{Q}{\epsilon_0}$.

(A, B, D) $1$. Gauss's Law is derived from Coulomb's Law and the superposition principle. Therefore,it inherently contains both of these fundamental principles of electrostatics.
$2$. The area element $d\vec{s}$ (or $d\vec{A}$) is defined as a vector pointing outward,normal to the surface. Since it is defined as the cross product of two displacement vectors (e.g.,$d\vec{l}_1 \times d\vec{l}_2$),it behaves as a pseudo-vector (or axial vector) because its direction depends on the orientation of the surface coordinate system,not just the position.

(B) The magnetic force is given by $\vec{F} = q(\vec{v} \times \vec{B})$.
Since $\vec{F}$ is perpendicular to $\vec{v}$,their dot product must be zero: $\vec{F} \cdot \vec{v} = 0$.
$(F_1 \hat{i} + F_2 \hat{j}) \cdot (v_1 \hat{i} + v_2 \hat{j}) = F_1 v_1 + F_2 v_2 = 0 \Rightarrow \frac{F_1}{F_2} = -\frac{v_2}{v_1} \quad (I)$.
Also,$\vec{F}$ is perpendicular to $\vec{B}$,so $\vec{F} \cdot \vec{B} = 0$.
Let $\vec{B} = B_1 \hat{i} + B_2 \hat{j} + B_3 \hat{k}$. Then $(F_1 \hat{i} + F_2 \hat{j}) \cdot (B_1 \hat{i} + B_2 \hat{j} + B_3 \hat{k}) = F_1 B_1 + F_2 B_2 = 0 \Rightarrow \frac{F_1}{F_2} = -\frac{B_2}{B_1} \quad (II)$.
Comparing $(I)$ and $(II)$,we get $\frac{v_2}{v_1} = \frac{B_2}{B_1}$,which implies $\frac{v_1}{v_2} = \frac{B_1}{B_2}$.
Thus,$\vec{B} = B_1 \hat{i} + B_2 \hat{j} + B_3 \hat{k}$ satisfies the condition $\frac{v_1}{v_2} = \frac{B_1}{B_2}$.

(B) Statement $(i)$ is false because the mass of a nucleus is always less than the sum of the masses of its individual nucleons. This mass difference,known as the mass defect,corresponds to the binding energy released when the nucleus is formed. Therefore,the mass energy of a nucleus is less than the total mass energy of its individual protons and neutrons.
Statement $(ii)$ is true. To separate a nucleus into its constituent nucleons,work must be done against the strong nuclear force,which requires an energy input equal to the binding energy.
Statement $(iii)$ is true. Binding energy per nucleon is a standard measure of nuclear stability; higher values indicate that nucleons are more tightly bound.
Statement $(iv)$ is true. In nuclear fission,a heavy nucleus splits into lighter nuclei with higher binding energy per nucleon,resulting in the release of energy.
Thus,statements $(ii)$,$(iii)$,and $(iv)$ are correct.

(D) For the first lens $L_1$,the object distance $u_1 = -4f$ and focal length $f_1 = f$. Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$:
$\frac{1}{v_1} - \frac{1}{-4f} = \frac{1}{f} \Rightarrow \frac{1}{v_1} = \frac{1}{f} - \frac{1}{4f} = \frac{3}{4f} \Rightarrow v_1 = \frac{4f}{3}$.
This image acts as an object for the second lens $L_2$. The distance between the lenses is $d = \frac{f}{2}$.
The object distance for the second lens is $u_2 = v_1 - d = \frac{4f}{3} - \frac{f}{2} = \frac{8f - 3f}{6} = \frac{5f}{6}$.
Since the object is to the left of $L_2$,$u_2 = -\frac{5f}{6}$.
Using the lens formula for $L_2$ with $f_2 = f$:
$\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2} \Rightarrow \frac{1}{v_2} - \frac{1}{-5f/6} = \frac{1}{f} \Rightarrow \frac{1}{v_2} + \frac{6}{5f} = \frac{1}{f}$.
$\frac{1}{v_2} = \frac{1}{f} - \frac{6}{5f} = \frac{5 - 6}{5f} = -\frac{1}{5f} \Rightarrow v_2 = -5f$.
The negative sign indicates that the final image is formed at a distance $5f$ to the left of $L_2$.

(D) Initially,with the empty tank,the object distance $u = -45 \ cm$ and the image distance $v = +36 \ cm$. Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
$\frac{1}{f} = \frac{1}{36} - \frac{1}{-45} = \frac{1}{36} + \frac{1}{45} = \frac{5+4}{180} = \frac{9}{180} = \frac{1}{20} \Rightarrow f = 20 \ cm$.
When the liquid is poured to a depth of $40 \ cm$,the apparent depth of the mark is $d' = \frac{40}{\mu}$. The remaining distance from the lens to the liquid surface is $45 - 40 = 5 \ cm$. Thus,the new object distance is $u' = -(\frac{40}{\mu} + 5) \ cm$. The new image distance is $v' = +48 \ cm$.
Using the lens formula again: $\frac{1}{f} = \frac{1}{v'} - \frac{1}{u'}$
$\frac{1}{20} = \frac{1}{48} - \frac{1}{-(\frac{40}{\mu} + 5)} = \frac{1}{48} + \frac{1}{\frac{40}{\mu} + 5}$
$\frac{1}{\frac{40}{\mu} + 5} = \frac{1}{20} - \frac{1}{48} = \frac{12 - 5}{240} = \frac{7}{240}$
$\frac{40}{\mu} + 5 = \frac{240}{7} \approx 34.286$
$\frac{40}{\mu} = 34.286 - 5 = 29.286$
$\mu = \frac{40}{29.286} \approx 1.366$.

(A) Let $\hat{t}$ be the unit vector along the normal to the surface.
Let $\hat{n}_1$ be the unit vector along the incident ray and $\hat{n}_2$ be the unit vector along the reflected ray.
From the geometry of reflection,the angle of incidence equals the angle of reflection,$\theta$.
We can resolve $\hat{n}_1$ and $\hat{n}_2$ into components parallel and perpendicular to the normal $\hat{t}$.
$\hat{n}_1 = -\cos \theta \hat{t} + \sin \theta \hat{u}$,where $\hat{u}$ is a unit vector parallel to the surface.
$\hat{n}_2 = \cos \theta \hat{t} + \sin \theta \hat{u}$.
Subtracting the two equations: $\hat{n}_2 - \hat{n}_1 = 2 \cos \theta \hat{t}$.
Since $\hat{n}_1 \cdot \hat{t} = \cos(180^\circ - \theta) = -\cos \theta$,we have $\cos \theta = -(\hat{n}_1 \cdot \hat{t})$.
Substituting this into the subtraction equation: $\hat{n}_2 - \hat{n}_1 = -2(\hat{n}_1 \cdot \hat{t}) \hat{t}$.
Therefore,$\hat{n}_2 = \hat{n}_1 - 2(\hat{n}_1 \cdot \hat{t}) \hat{t}$.

(B) In the given circuit,the top diode is in forward bias because its p-side is connected to the positive terminal of the battery. The bottom diode is in reverse bias because its n-side is connected to the positive terminal of the battery.
For the diode in forward bias,the resistance is $0 \ \Omega$. Thus,the current flows only through the top branch containing the $10 \ \Omega$ resistor.
For the diode in reverse bias,the resistance is $\infty$,so no current flows through the bottom branch.
Using Ohm's law,the current $I$ supplied by the cell is:
$I = \frac{V}{R} = \frac{2 \ V}{10 \ \Omega} = 0.2 \ A$.

(D) Let us trace the output of the logic gates step by step.
Step $1$: The first $AND$ gate takes inputs $X_0$ and $X_1$,giving output $Y_1 = X_0 X_1$.
Step $2$: The first $OR$ gate takes $Y_1$ and $X_2$,giving output $Y_2 = X_0 X_1 + X_2$.
Step $3$: The next $AND$ gate takes $Y_2$ and $X_3$,giving output $Y_3 = (X_0 X_1 + X_2) X_3 = X_0 X_1 X_3 + X_2 X_3$.
Step $4$: The next $OR$ gate takes $Y_3$ and $X_4$,giving output $Y_4 = X_0 X_1 X_3 + X_2 X_3 + X_4$.
Continuing this pattern,for an even $n$,the final output $Q$ will be of the form: $Q = X_0 X_1 X_3 X_5 \ldots X_{n-1} + X_2 X_3 X_5 \ldots X_{n-1} + X_4 X_5 \ldots X_{n-1} + \ldots + X_{n-2} X_{n-1} + X_n$.

(C) The condition for the $n^{th}$ order diffraction minimum in a single-slit experiment is given by $a \sin \theta = n \lambda$,where $a$ is the slit width and $\theta$ is the angle of diffraction.
For the $2^{nd}$ order minimum of $\lambda_1$,we have $a \sin \theta_1 = 2 \lambda_1$.
For the $3^{rd}$ order minimum of $\lambda_2$,we have $a \sin \theta_2 = 3 \lambda_2$.
Since the minima coincide,the angles of diffraction are equal,i.e.,$\theta_1 = \theta_2 = \theta$.
Therefore,$2 \lambda_1 = 3 \lambda_2$.
Rearranging the terms,we get $\frac{\lambda_1}{\lambda_2} = \frac{3}{2}$.

(D) For reflected light,the condition for destructive interference (dark fringe) is given by:
$2 \mu t \cos r = n \lambda$
where $\mu$ is the refractive index,$t$ is the thickness,$r$ is the angle of refraction,and $n$ is an integer $(n = 1, 2, 3, ...)$.
For the smallest thickness,we take $n = 1$:
$2 \mu t \cos r = \lambda$
Given values:
$\lambda = 6000 \text{ Å} = 6000 \times 10^{-10} \text{ m} = 6 \times 10^{-7} \text{ m}$
$\mu = 1.5$
$r = 60^{\circ} \Rightarrow \cos 60^{\circ} = 0.5$
Substituting these values into the equation:
$2 \times 1.5 \times t \times 0.5 = 6 \times 10^{-7}$
$1.5 \times t = 6 \times 10^{-7}$
$t = \frac{6 \times 10^{-7}}{1.5} = 4 \times 10^{-7} \text{ m}$