JEE Main 2022 Physics Question Paper with Answer and Solution

(A) The ratio of specific heats is defined as $\gamma = \frac{C_P}{C_V}$.
According to the kinetic theory of gases,the molar specific heat at constant volume is $C_V = \frac{f}{2}R$.
The molar specific heat at constant pressure is $C_P = C_V + R = \frac{f}{2}R + R = R(1 + \frac{f}{2})$.
Therefore,the ratio $\gamma = \frac{C_P}{C_V} = \frac{R(1 + f/2)}{(f/2)R} = \frac{1 + f/2}{f/2} = \frac{2}{f} + 1 = 1 + \frac{2}{f}$.

(A) The vehicle moves down a frictionless inclined plane with an acceleration $a = g\sin \alpha$.
In the frame of the vehicle,a pseudo force $F_p = ma = mg\sin \alpha$ acts on the bob of the pendulum in the upward direction along the incline.
The effective acceleration $g_{eff}$ is the vector sum of the acceleration due to gravity $\vec{g}$ and the negative of the vehicle's acceleration $-\vec{a}$.
Resolving the components,the component of $g$ perpendicular to the incline is $g\cos \alpha$ and the component of $g$ parallel to the incline is $g\sin \alpha$. The pseudo force cancels the component $g\sin \alpha$.
Thus,the effective acceleration is $g_{eff} = g\cos \alpha$.
The time period of the simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}} = 2\pi \sqrt{\frac{L}{g\cos \alpha}}$.

(B) The position of car $P$ at any time $t$ is given by $x_P(t) = at + bt^2$.
The velocity of car $P$ is $v_P(t) = \frac{dx_P(t)}{dt} = a + 2bt$ ... $(i)$.
Similarly,for car $Q$,the position is $x_Q(t) = ft - t^2$.
The velocity of car $Q$ is $v_Q(t) = \frac{dx_Q(t)}{dt} = f - 2t$ ... $(ii)$.
Given that the cars have the same velocity,$v_P(t) = v_Q(t)$.
Therefore,$a + 2bt = f - 2t$.
Rearranging the terms to solve for $t$: $2bt + 2t = f - a$.
$2t(b + 1) = f - a$.
Thus,$t = \frac{f - a}{2(1 + b)}$.

(C) According to the ideal gas equation,
$PV = nRT$
Rearranging for $V$ in terms of $T$:
$V = \left( \frac{nR}{P} \right) T$
This represents a straight line passing through the origin,where the slope $m$ is given by:
$m = \frac{V}{T} = \frac{nR}{P}$
Since $n$ and $R$ are constants,the slope is inversely proportional to the pressure $P$ $(m \propto \frac{1}{P})$.
From the given figure,the angle $\theta_2 > \theta_1$,which implies that the slope of the line for $P_2$ is greater than the slope of the line for $P_1$:
$(\text{Slope})_2 > (\text{Slope})_1$
Since slope is inversely proportional to pressure,a higher slope corresponds to a lower pressure:
$P_2 < P_1$

(A) The efficiency of an ideal (Carnot) heat engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$.
Here,$T_1$ is the temperature of the source (boiling point of water) and $T_2$ is the temperature of the sink (freezing point of water).
Freezing point of water,$T_2 = 0^{\circ}C = 273\,K$.
Boiling point of water,$T_1 = 100^{\circ}C = (100 + 273)\,K = 373\,K$.
Substituting these values into the efficiency formula:
$\eta = 1 - \frac{273}{373} = \frac{373 - 273}{373} = \frac{100}{373}$.
To express this as a percentage:
$\% \eta = \left( \frac{100}{373} \right) \times 100 \approx 26.8\%$.

(D) In uniform circular motion,the acceleration is the centripetal acceleration $\vec{a}_c$,which is always directed towards the center of the circle.
For a point $P$ at an angle $\theta$ with the $x$-axis,the position vector makes an angle $\theta$ with the $x$-axis.
The centripetal acceleration vector $\vec{a}_c$ points from $P$ towards the origin $O$.
The magnitude of centripetal acceleration is $a_c = \frac{V^2}{R}$.
Resolving this vector into components:
The $x$-component is $a_x = -a_c \cos\theta = -\frac{V^2}{R} \cos\theta$.
The $y$-component is $a_y = -a_c \sin\theta = -\frac{V^2}{R} \sin\theta$.
Thus,the acceleration vector is $\vec{a} = -\frac{V^2}{R} \cos\theta \hat{i} - \frac{V^2}{R} \sin\theta \hat{j}$.

(D) The variation of acceleration due to gravity $g$ with distance $d$ from the center of the earth is given by:
$1$. Inside the earth $(d < R)$:
$g = \frac{GM}{R^3} d$
Since $G, M,$ and $R$ are constants,$g \propto d$. This represents a straight line passing through the origin.
$2$. At the surface of the earth $(d = R)$:
$g = \frac{GM}{R^2} = g_s$ (maximum value).
$3$. Outside the earth $(d > R)$:
$g = \frac{GM}{d^2}$
Here,$g \propto \frac{1}{d^2}$. This represents a rectangular hyperbola.
Combining these,the graph shows a linear increase from the center to the surface,followed by a hyperbolic decrease as distance increases beyond the surface. This matches the graph in option $D$.

(D) The total kinetic energy $(K_{total})$ of a rolling sphere is the sum of its linear kinetic energy $(K_{linear})$ and rotational kinetic energy $(K_{rotational})$.
$K_{total} = K_{linear} + K_{rotational} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
For a solid sphere,the moment of inertia is $I = \frac{2}{5}mr^2$ and the angular velocity is $\omega = \frac{v}{r}$.
Substituting these values into the rotational kinetic energy expression:
$K_{rotational} = \frac{1}{2} \left(\frac{2}{5}mr^2\right) \left(\frac{v}{r}\right)^2 = \frac{1}{5}mv^2$.
Now,calculating the total kinetic energy:
$K_{total} = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \left(\frac{5+2}{10}\right)mv^2 = \frac{7}{10}mv^2$.
The ratio of rotational kinetic energy to total kinetic energy is:
$\frac{K_{rotational}}{K_{total}} = \frac{\frac{1}{5}mv^2}{\frac{7}{10}mv^2} = \frac{1}{5} \times \frac{10}{7} = \frac{2}{7}$.
Thus,the ratio is $2:7$.

(A) The dimension of velocity gradient is $[T^{-1}]$.
The decay constant $\lambda$ is defined by the relation $N = N_0 e^{-\lambda t}$,which implies that $\lambda t$ is dimensionless. Thus,the dimension of the decay constant is $[T^{-1}]$.
Since both velocity gradient and decay constant have the same dimensions $[T^{-1}]$,option $A$ is correct.

(D) According to Kepler's third law of planetary motion,the square of the orbital period $T$ is directly proportional to the cube of the semi-major axis $R$ of its orbit: $T^2 \propto R^3$.
Given the initial distance is $R$ and the initial period $T = 1 \text{ year}$.
When the new distance $R' = 3R$,the new period $T'$ is given by the ratio:
$\frac{T'}{T} = \left(\frac{R'}{R}\right)^{3/2}$.
Substituting the values:
$\frac{T'}{1} = \left(\frac{3R}{R}\right)^{3/2} = (3)^{3/2}$.
$T' = 3^{1} \cdot 3^{1/2} = 3\sqrt{3} \text{ years}$.

(B) In a vertical circular motion with uniform speed $v$,the net centripetal force is provided by the tension $T$ and the component of gravity $mg \cos \theta$,where $\theta$ is the angle with the vertical.
At the highest point,the tension $T_{top}$ is given by $T_{top} + mg = \frac{mv^2}{r}$,so $T_{top} = \frac{mv^2}{r} - mg$.
At the lowest point,the tension $T_{bottom}$ is given by $T_{bottom} - mg = \frac{mv^2}{r}$,so $T_{bottom} = \frac{mv^2}{r} + mg$.
Since $T_{top} < T_{bottom}$,the tension is minimum at the highest position of the circular path.

(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_L}{T_H}$.
Given: $Q_H = 5000 \, kcal$,$T_H = 727 + 273 = 1000 \, K$,$T_L = 127 + 273 = 400 \, K$.
Efficiency $\eta = 1 - \frac{400}{1000} = 1 - 0.4 = 0.6$.
Work done $W = \eta \times Q_H = 0.6 \times 5000 \, kcal = 3000 \, kcal$.
Since $1 \, kcal = 4184 \, J$,then $W = 3000 \times 4184 \, J = 12,552,000 \, J$.
Rounding to the nearest value,$W \approx 12.6 \times 10^{6} \, J$.

(B) The maximum velocity of a simple harmonic oscillator is given by $V_{\max} = \omega A$,where $\omega = \sqrt{\frac{k}{m}}$ is the angular frequency and $A$ is the amplitude.
Given $V_{\max,1} = V_{\max,2}$,we have $\omega_1 A_1 = \omega_2 A_2$,which implies $\frac{A_1}{A_2} = \frac{\omega_2}{\omega_1}$.
Substituting $\omega = \sqrt{\frac{k}{m}}$,we get $\frac{A_1}{A_2} = \sqrt{\frac{k_2}{m_2}} \times \sqrt{\frac{m_1}{k_1}} = \sqrt{\frac{k_2}{k_1} \times \frac{m_1}{m_2}}$.
Given $k_1 = 2k$,$k_2 = 9k$,$m_1 = 50\,g$,and $m_2 = 100\,g$.
Thus,$\frac{A_1}{A_2} = \sqrt{\frac{9k}{2k} \times \frac{50}{100}} = \sqrt{\frac{9}{2} \times \frac{1}{2}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.

(C) The equilibrium position occurs where the force $F = -\frac{dU}{dr} = 0$.
Given $U = A r^{-10} - B r^{-5}$.
Differentiating with respect to $r$: $\frac{dU}{dr} = -10 A r^{-11} + 5 B r^{-6}$.
Setting $\frac{dU}{dr} = 0$ for equilibrium:
$-10 A r^{-11} + 5 B r^{-6} = 0$.
$5 B r^{-6} = 10 A r^{-11}$.
$r^5 = \frac{10A}{5B} = \frac{2A}{B}$.
Therefore,the equilibrium distance is $r = \left(\frac{2A}{B}\right)^{\frac{1}{5}}$.

(B) During ascent,both gravity and air resistance act downwards. The net retarding force is $F_{up} = mg + f = (5 \times 10) + 10 = 60 \, N$. The acceleration during ascent is $a_{up} = F_{up} / m = 60 / 5 = 12 \, ms^{-2}$.
During descent,gravity acts downwards and air resistance acts upwards. The net force is $F_{down} = mg - f = (5 \times 10) - 10 = 40 \, N$. The acceleration during descent is $a_{down} = F_{down} / m = 40 / 5 = 8 \, ms^{-2}$.
Let $h$ be the maximum height reached. For ascent,$h = \frac{1}{2} a_{up} t_1^2$,so $h = \frac{1}{2} \times 12 \times t_1^2 = 6 t_1^2$.
For descent,$h = \frac{1}{2} a_{down} t_2^2$,so $h = \frac{1}{2} \times 8 \times t_2^2 = 4 t_2^2$.
Equating the two expressions for $h$: $6 t_1^2 = 4 t_2^2$,which gives $\frac{t_1^2}{t_2^2} = \frac{4}{6} = \frac{2}{3}$.
Therefore,the ratio of time of ascent to the time of descent is $\frac{t_1}{t_2} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}$.

(B) Given that the flywheel starts from rest,the initial angular velocity $\omega_0 = 0$.
Using the equation of rotational motion $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$,where $\alpha$ is the angular acceleration.
For the first second $(t = 1 \, s)$,the angle rotated is $\theta_1 = 5 \, rad$:
$5 = 0(1) + \frac{1}{2} \alpha (1)^2 \implies \alpha = 10 \, rad/s^2$.
Now,we find the total angle rotated in the first two seconds $(t = 2 \, s)$:
$\theta_2 = 0(2) + \frac{1}{2} (10) (2)^2 = 5 \times 4 = 20 \, rad$.
The angle rotated in the next second (from $t = 1$ to $t = 2$) is the difference between the total angle in $2 \, s$ and the angle in $1 \, s$:
$\Delta \theta = \theta_2 - \theta_1 = 20 - 5 = 15 \, rad$.

(C) The kinetic energy of the hammer is given by $K.E. = \frac{1}{2} mv^2 = \frac{1}{2} \times 1.5 \times (60)^2 = 0.75 \times 3600 = 2700 \, J$.
According to the problem,one-fourth of this energy is used to heat the iron nail.
Heat energy absorbed by the nail $Q = \frac{1}{4} \times 2700 = 675 \, J$.
The formula for heat absorbed is $Q = mc\Delta T$,where $m = 100 \, g$,$c = 0.42 \, Jg^{-1} {}^{\circ}C^{-1}$,and $\Delta T$ is the rise in temperature.
Substituting the values: $675 = 100 \times 0.42 \times \Delta T$.
$675 = 42 \times \Delta T$.
$\Delta T = \frac{675}{42} \approx 16.07 \, ^{\circ}C$.

(A) Let the initial velocity be $u$. The horizontal and vertical components of velocity at any time $t$ are given by:
$v_x = u \cos 45^{\circ} = \frac{u}{\sqrt{2}}$
$v_y = u \sin 45^{\circ} - gt = \frac{u}{\sqrt{2}} - 10(2) = \frac{u}{\sqrt{2}} - 20$
Given that the resultant velocity at $t = 2 \, s$ is $20 \, m/s$,we have:
$v^2 = v_x^2 + v_y^2$
$20^2 = \left(\frac{u}{\sqrt{2}}\right)^2 + \left(\frac{u}{\sqrt{2}} - 20\right)^2$
$400 = \frac{u^2}{2} + \frac{u^2}{2} - 20\sqrt{2}u + 400$
$0 = u^2 - 20\sqrt{2}u$
Since $u \neq 0$,we get $u = 20\sqrt{2} \, m/s$.
The maximum height $H$ is given by:
$H = \frac{u^2 \sin^2 45^{\circ}}{2g} = \frac{(20\sqrt{2})^2 \times (1/2)}{2 \times 10} = \frac{800 \times 0.5}{20} = \frac{400}{20} = 20 \, m$.

(D) The equation of a stationary wave is given by $y = A_{res} \sin \omega t$,where $A_{res}$ is the amplitude of the particle at position $x$.
Comparing the given equation $y = (10 \cos \pi x) \sin \frac{2 \pi t}{T}$ with the standard form,the amplitude at any position $x$ is $A(x) = |10 \cos \pi x|$.
Given $x = \frac{4}{3} \, cm$,we substitute this value into the expression:
$A = |10 \cos(\pi \cdot \frac{4}{3})|$
$A = |10 \cos(\frac{4 \pi}{3})|$
Since $\cos(\frac{4 \pi}{3}) = \cos(\pi + \frac{\pi}{3}) = -\cos(\frac{\pi}{3}) = -\frac{1}{2}$,
$A = |10 \cdot (-0.5)| = |-5| = 5 \, cm$.
Thus,the amplitude is $5 \, cm$.

(C) According to the first law of thermodynamics,$\Delta Q = \Delta U + W$.
Given,$W = \frac{Q}{4}$.
Therefore,$\Delta U = \Delta Q - W = Q - \frac{Q}{4} = \frac{3Q}{4}$.
For a monoatomic gas,the change in internal energy is $\Delta U = n C_v \Delta T = n (\frac{3}{2}R) \Delta T$.
So,$n (\frac{3}{2}R) \Delta T = \frac{3Q}{4} \Rightarrow n \Delta T = \frac{Q}{2R}$.
The molar heat capacity $C$ is defined as $C = \frac{Q}{n \Delta T}$.
Substituting $n \Delta T$,we get $C = \frac{Q}{Q / (2R)} = 2R$.
Thus,the value of $x$ is $2$.

(B) According to Newton's law of cooling,the temperature difference $\Delta T$ at time $t$ is given by $\Delta T = \Delta T_{0} e^{-\lambda t}$,where $\Delta T_{0}$ is the initial temperature difference.
From the graph,at $t = 0$,$\Delta T_{0} = 60^{\circ}C$.
At $t = 6 \text{ min}$,$\Delta T = 40^{\circ}C$. Substituting these values: $40 = 60 e^{-6\lambda} \Rightarrow e^{-6\lambda} = \frac{40}{60} = \frac{2}{3} \Rightarrow 6\lambda = \ln(1.5)$.
At $t = t_{2}$,$\Delta T = 20^{\circ}C$. Substituting these values: $20 = 60 e^{-t_{2}\lambda} \Rightarrow e^{-t_{2}\lambda} = \frac{20}{60} = \frac{1}{3} \Rightarrow t_{2}\lambda = \ln(3)$.
Dividing the two equations: $\frac{t_{2}\lambda}{6\lambda} = \frac{\ln(3)}{\ln(1.5)} \Rightarrow \frac{t_{2}}{6} = \frac{1.0986}{0.4055} \approx 2.709$.
Therefore,$t_{2} = 6 \times 2.709 \approx 16.25 \text{ min}$.
Rounding to the nearest integer,$t_{2} \approx 16 \text{ min}$.

(C) The bulk modulus $B$ is defined as $B = -\frac{\Delta P}{\Delta V / V}$.
Given,$B = 3 \times 10^{10} \ Nm^{-2}$.
The fractional change in volume is $\frac{\Delta V}{V} = -2\% = -0.02$ (negative sign indicates reduction in volume).
Rearranging the formula for pressure change $\Delta P$: $\Delta P = -B \times \left(\frac{\Delta V}{V}\right)$.
Substituting the values: $\Delta P = -(3 \times 10^{10}) \times (-0.02)$.
$\Delta P = 3 \times 10^{10} \times 0.02 = 0.06 \times 10^{10} = 6 \times 10^{8} \ Nm^{-2}$.
Thus,the required pressure is $6 \times 10^{8} \ Nm^{-2}$.

(D) $1$. Wave number $(k = 1/\lambda)$ and Rydberg's constant $(R)$ both have dimensions of $[L^{-1}]$.
$2$. Stress and Coefficient of elasticity (Young's modulus,Bulk modulus,etc.) both have dimensions of $[M L^{-1} T^{-2}]$.
$3$. Coercivity $(H)$ and Magnetisation $(M)$ have different dimensions. Coercivity is the magnetic field strength $(H)$ with dimensions $[A L^{-1}]$,while Magnetisation is magnetic moment per unit volume with dimensions $[A L^{-1}]$. Wait,both are $[A L^{-1}]$. Let us re-evaluate.
$4$. Specific heat capacity $(S)$ has dimensions $[L^2 T^{-2} K^{-1}]$ (from $S = Q / (m \Delta T)$),while Latent heat $(L)$ has dimensions $[L^2 T^{-2}]$ (from $L = Q / m$).
Since $[L^2 T^{-2} K^{-1}] \neq [L^2 T^{-2}]$,the pair with different dimensions is Specific heat capacity and Latent heat.

(D) The horizontal range $R$ is given by $R = \frac{V^2 \sin(2\theta)}{g} = \frac{2V^2 \sin\theta \cos\theta}{g}$.
The time taken to reach the highest point (where inclination is zero) is $t = \frac{V \sin\theta}{g}$.
From this,$V \sin\theta = gt$,so $V = \frac{gt}{\sin\theta}$.
Substitute $V$ into the range formula: $R = \frac{2(gt/\sin\theta)^2 \sin\theta \cos\theta}{g} = \frac{2g^2 t^2 \sin\theta \cos\theta}{g \sin^2\theta} = \frac{2gt^2 \cos\theta}{\sin\theta} = \frac{2gt^2}{\tan\theta}$.
Given $g = 10 \, m/s^2$,we have $R = \frac{2(10)t^2}{\tan\theta} = \frac{20t^2}{\tan\theta}$.
Rearranging for $\theta$,we get $\tan\theta = \frac{20t^2}{R}$,which implies $\cot\theta = \frac{R}{20t^2}$.
Therefore,$\theta = \cot^{-1}\left(\frac{R}{20t^2}\right)$.

(B) The frictional force acting on the block is $f = \mu N = \mu mg$.
Using Newton's second law,the retardation $a$ is given by $ma = -\mu mg$,so $a = -\mu g$.
Given $\mu = 0.5$ and $g = 9.8\, m/s^2$,the acceleration is $a = -0.5 \times 9.8 = -4.9\, m/s^2$.
Using the kinematic equation $v^2 = u^2 + 2as$,where final velocity $v = 0$ and initial velocity $u = 9.8\, m/s$:
$0 = (9.8)^2 + 2(-4.9)s$
$9.8s = 9.8 \times 9.8$
$s = 9.8\, m$.

(C) The centripetal force required for circular motion is provided by the tension in the string: $T = M \omega^{2} R$.
Given: $T = 80 \,N$,$M = 100 \,g = 0.1 \,kg$,$R = 2 \,m$.
Substituting the values: $80 = 0.1 \times \omega^{2} \times 2$.
$80 = 0.2 \omega^{2} \implies \omega^{2} = 400 \implies \omega = 20 \,rad/s$.
Since $\omega = 2 \pi f$,where $f$ is the frequency in $rev/s$: $2 \pi f = 20 \implies f = \frac{10}{\pi} \,rev/s$.
To convert frequency to $rev/min$,multiply by $60$: $f = \frac{10}{\pi} \times 60 = \frac{600}{\pi} \,rev/min$.
Comparing this with $\frac{K}{\pi} \,rev/min$,we get $K = 600$.

(C) According to the Work-Energy Theorem,the change in kinetic energy $(\Delta KE)$ is equal to the work done $(W)$ by the force on the particle.
The work done is given by the line integral: $W = \int \overrightarrow{F} \cdot d\overrightarrow{r} = \int (4x \hat{i} + 3y^2 \hat{j}) \cdot (dx \hat{i} + dy \hat{j})$.
This simplifies to: $W = \int_{1}^{2} 4x \, dx + \int_{2}^{3} 3y^2 \, dy$.
Evaluating the integrals:
$W = [2x^2]_{1}^{2} + [y^3]_{2}^{3}$.
$W = (2(2)^2 - 2(1)^2) + (3^3 - 2^3)$.
$W = (8 - 2) + (27 - 8)$.
$W = 6 + 19 = 25 \, J$.
Therefore,the kinetic energy changes by $25 \, J$.

(B) The weight of a body at height $h$ is given by $W' = m g'$,where $g' = g \left( \frac{R}{R+h} \right)^2$.
Given that the weight at height $h$ is $\frac{1}{3}$ of the weight on the surface,we have $m g' = \frac{1}{3} m g$,which implies $g' = \frac{g}{3}$.
Substituting the expression for $g'$,we get $g \left( \frac{R}{R+h} \right)^2 = \frac{g}{3}$.
Taking the square root on both sides,we get $\frac{R}{R+h} = \frac{1}{\sqrt{3}}$.
Rearranging the equation,$R+h = R\sqrt{3}$,so $h = R(\sqrt{3} - 1)$.
Substituting the values $R = 6400 \, km$ and $\sqrt{3} = 1.732$,we get $h = 6400 \times (1.732 - 1) = 6400 \times 0.732$.
$h = 4684.8 \, km \approx 4685 \, km$.

(A) Given amplitudes are $A_{1} = 5 \, cm$ and $A_{2} = 3 \, cm$.
The phase difference $\Delta \phi$ between the two waves is determined by the difference in their arguments:
$\Delta \phi = 2 \pi (x - vt + 1.5) - 2 \pi (x - vt) = 2 \pi (1.5) = 3 \pi$.
The resultant amplitude $A_{net}$ is given by the formula:
$A_{net} = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos(\Delta \phi)}$.
Substituting the values:
$A_{net} = \sqrt{5^{2} + 3^{2} + 2(5)(3) \cos(3 \pi)}$.
Since $\cos(3 \pi) = -1$:
$A_{net} = \sqrt{25 + 9 + 30(-1)} = \sqrt{34 - 30} = \sqrt{4} = 2 \, cm$.

(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_L}{T_H}$,where $T_L$ is the sink temperature and $T_H$ is the source temperature.
Given $T_L = 27\,^{\circ} C = 300\, K$ and $\eta = 25\% = 0.25$.
Substituting the values: $0.25 = 1 - \frac{300}{T_H} \implies \frac{300}{T_H} = 0.75 \implies T_H = \frac{300}{0.75} = 400\, K$.
If the efficiency is increased by $100\%$ of its original value,the new efficiency $\eta' = \eta + 100\% \text{ of } \eta = 0.25 + 0.25 = 0.50$ or $50\%$.
Let the new source temperature be $T_H'$.
$0.50 = 1 - \frac{300}{T_H'} \implies \frac{300}{T_H'} = 0.50 \implies T_H' = \frac{300}{0.50} = 600\, K$.
The change in temperature is $\Delta T = T_H' - T_H = 600\, K - 400\, K = 200\, K$.
Since a change in temperature of $1\, K$ is equivalent to a change of $1\,^{\circ} C$,the temperature should be increased by $200\,^{\circ} C$.

(B) In steady-state heat conduction,the rate of heat flow $H$ through both blocks is the same.
$H = \frac{K_{1} A \Delta T_{1}}{\ell_{1}} = \frac{K_{2} A \Delta T_{2}}{\ell_{2}}$
Given: $\ell_{1} = 16 \text{ cm}$,$\ell_{2} = 8 \text{ cm}$,$K_{2} = K$,$K_{1} = xK$.
Temperature difference across $M_{1}$ is $\Delta T_{1} = 100^{\circ}C - 80^{\circ}C = 20^{\circ}C$.
Temperature difference across $M_{2}$ is $\Delta T_{2} = 80^{\circ}C - 0^{\circ}C = 80^{\circ}C$.
Since the area $A$ is the same,we have:
$\frac{K_{1} \Delta T_{1}}{\ell_{1}} = \frac{K_{2} \Delta T_{2}}{\ell_{2}}$
$\frac{(xK) \times 20}{16} = \frac{K \times 80}{8}$
$x \times \frac{20}{16} = 10$
$x = 10 \times \frac{16}{20} = 8$.
Thus,the thermal conductivity of $M_{1}$ is $8K$.

(A) Given mass of $N_{2} = 0.056 \, kg = 56 \, g$.
Molar mass of $N_{2} = 28 \, g/mol$.
Number of moles $n = \frac{56}{28} = 2 \, moles$.
Initial temperature $T_{1} = 127 + 273 = 400 \, K$.
Since the root mean square speed $v_{rms} \propto \sqrt{T}$,to double the speed,the temperature must increase by a factor of $2^{2} = 4$.
So,$T_{2} = 4 \times T_{1} = 4 \times 400 = 1600 \, K$.
Change in temperature $\Delta T = T_{2} - T_{1} = 1600 - 400 = 1200 \, K$.
Nitrogen is a diatomic gas,so the degree of freedom $f = 5$.
The heat required is given by $Q = \frac{f}{2} n R \Delta T$.
Substituting the values: $Q = \frac{5}{2} \times 2 \times 2 \times 1200 = 5 \times 2400 = 12000 \, cal = 12 \, kcal$.

(A) Let the height of the tower be $h$ and the speed of projection in the first two cases be $u$. Taking the downward direction as positive.
For case-$I$ (thrown upward): The displacement is $h$. The initial velocity is $-u$. The acceleration is $g$.
Using the equation $s = ut + \frac{1}{2}at^2$:
$h = -u(6) + \frac{1}{2}g(6)^2$
$h = -6u + 18g$ ... $(i)$
For case-$II$ (thrown downward): The displacement is $h$. The initial velocity is $u$. The acceleration is $g$.
$h = u(1.5) + \frac{1}{2}g(1.5)^2$
$h = 1.5u + 1.125g$ ... $(ii)$
To eliminate $u$,multiply equation $(ii)$ by $4$:
$4h = 6u + 4.5g$ ... $(iii)$
Adding equation $(i)$ and equation $(iii)$:
$h + 4h = (-6u + 18g) + (6u + 4.5g)$
$5h = 22.5g$
$h = 4.5g$ ... $(iv)$
For case-$III$ (released from rest): The initial velocity is $0$. The acceleration is $g$.
$h = 0(t) + \frac{1}{2}gt^2$
$h = \frac{1}{2}gt^2$ ... $(v)$
Substituting equation $(iv)$ into equation $(v)$:
$4.5g = \frac{1}{2}gt^2$
$t^2 = 9$
$t = 3 \, s$.

(D) Let the mass of the ball be $m = 100 \, g = 0.1 \, kg$. The height from which it is dropped is $h = 10 \, cm = 0.1 \, m$. The compression in the spring is $x = \frac{h}{2} = 5 \, cm = 0.05 \, m$.
By the principle of conservation of mechanical energy,the loss in gravitational potential energy of the ball equals the gain in elastic potential energy of the spring.
The total vertical displacement of the ball is $h + x = h + \frac{h}{2} = \frac{3h}{2}$.
Therefore,$mg \left( h + \frac{h}{2} \right) = \frac{1}{2} kx^2$.
Substituting the values: $0.1 \times 10 \times \left( 0.1 + 0.05 \right) = \frac{1}{2} \times k \times (0.05)^2$.
$1 \times 0.15 = \frac{1}{2} \times k \times 0.0025$.
$0.15 = k \times 0.00125$.
$k = \frac{0.15}{0.00125} = \frac{150000}{1250} = 120 \, N \, m^{-1}$.

(B) Let the mass of the metre scale be $m$. The centre of mass of the uniform metre scale is at the $50.0\, cm$ mark.
When the scale is balanced at the $40.0\, cm$ mark,the torque due to the coins must balance the torque due to the weight of the scale.
The mass of the two coins is $2 \times 10\, g = 20\, g = 0.02\, kg$.
The distance of the coins from the knife edge is $|40.0\, cm - 10.0\, cm| = 30.0\, cm = 0.3\, m$.
The distance of the centre of mass of the scale from the knife edge is $|50.0\, cm - 40.0\, cm| = 10.0\, cm = 0.1\, m$.
Applying the principle of moments (balancing torques about the knife edge):
$(0.02\, kg) \times g \times (0.3\, m) = m \times g \times (0.1\, m)$
$0.006 = 0.1m$
$m = 0.06\, kg = 6 \times 10^{-2}\, kg$.
Comparing this with $x \times 10^{-2}\, kg$,we get $x = 6$.

(A) For the same range,the angles of projection must satisfy $\theta_{1} + \theta_{2} = 90^{\circ}$,which means $\theta_{2} = 90^{\circ} - \theta_{1}$.
The maximum heights are given by $h_{1} = \frac{u^{2} \sin^{2} \theta_{1}}{2g}$ and $h_{2} = \frac{u^{2} \sin^{2} \theta_{2}}{2g} = \frac{u^{2} \cos^{2} \theta_{1}}{2g}$.
Multiplying these heights,we get $h_{1} h_{2} = \left(\frac{u^{2} \sin^{2} \theta_{1}}{2g}\right) \cdot \left(\frac{u^{2} \cos^{2} \theta_{1}}{2g}\right)$.
This can be rewritten as $h_{1} h_{2} = \frac{u^{4} \sin^{2} \theta_{1} \cos^{2} \theta_{1}}{4g^{2}} = \left(\frac{u^{2} \cdot 2 \sin \theta_{1} \cos \theta_{1}}{4g}\right)^{2} = \left(\frac{u^{2} \sin(2\theta_{1})}{4g}\right)^{2}$.
Since the range $R = \frac{u^{2} \sin(2\theta_{1})}{g}$,we have $h_{1} h_{2} = \left(\frac{R}{4}\right)^{2} = \frac{R^{2}}{16}$.
Therefore,$R^{2} = 16 h_{1} h_{2}$,which implies $R = 4 \sqrt{h_{1} h_{2}}$.
Both Assertion $A$ and Reason $R$ are true,and $R$ provides the correct mathematical derivation for $A$.

(B) For the beaker to revolve with the disc without slipping,the necessary centripetal force must be provided by the static friction force.
The required centripetal force is $F_{c} = m \omega^{2} R$,where $m$ is the mass of the beaker.
The maximum available static friction force is $f_{s,max} = \mu N = \mu mg$,where $N = mg$ is the normal force.
For the beaker to move in a circular path with the disc,the static friction must be greater than or equal to the required centripetal force:
$f_{s} \leq f_{s,max}$
Substituting the expressions:
$m \omega^{2} R \leq \mu mg$
Dividing both sides by $m \omega^{2}$:
$R \leq \frac{\mu g}{\omega^{2}}$

(B) The increase in volume is given by $\Delta V = \gamma V_{0} \Delta T$.
Since $\gamma = 3\alpha$,we have $\Delta V = (3\alpha) V_{0} \Delta T$.
The total surface area of a cube is $6a^{2}$,where $a$ is the side length.
Given $6a^{2} = 24 \; m^{2}$,so $a^{2} = 4 \; m^{2}$,which means $a = 2 \; m$.
The initial volume $V_{0} = a^{3} = (2)^{3} = 8 \; m^{3}$.
Substituting the values: $\Delta V = (3 \times 5.0 \times 10^{-4} \; ^{\circ}C^{-1}) \times (8 \; m^{3}) \times (10 \; ^{\circ}C)$.
$\Delta V = 15 \times 10^{-4} \times 80 = 1200 \times 10^{-4} = 0.12 \; m^{3}$.
Since $1 \; m^{3} = 10^{6} \; cm^{3}$,then $\Delta V = 0.12 \times 10^{6} \; cm^{3} = 1.2 \times 10^{5} \; cm^{3}$.

(C) The heat lost by the copper block as it cools from $500^{\circ} C$ to $0^{\circ} C$ is given by the formula $\Delta Q = mc\Delta T$.
Mass of copper $m = 5.0 \, kg = 5000 \, g$.
Specific heat of copper $c = 0.39 \, J g^{-1} {}^{\circ} C^{-1}$.
Temperature change $\Delta T = 500^{\circ} C - 0^{\circ} C = 500^{\circ} C$.
Heat released $\Delta Q_1 = 5000 \times 0.39 \times 500 = 975,000 \, J$.
Let $m_{ice}$ be the mass of ice that melts. The heat absorbed by the ice to melt at $0^{\circ} C$ is $\Delta Q_2 = m_{ice} \times L_f$,where $L_f = 335 \, J g^{-1}$.
Equating the heat lost and heat gained: $\Delta Q_1 = \Delta Q_2$.
$975,000 = m_{ice} \times 335$.
$m_{ice} = \frac{975,000}{335} \approx 2910.45 \, g$.
Converting to $kg$,$m_{ice} \approx 2.91 \, kg$.

(A) Given the equation: $z = a^{2} x^{3} y^{1/2}$.
Since $a$ is a constant,its relative error $\frac{\Delta a}{a} = 0$.
The relative error in $z$ is given by: $\frac{\Delta z}{z} = 3 \left( \frac{\Delta x}{x} \right) + \frac{1}{2} \left( \frac{\Delta y}{y} \right)$.
To find the percentage error,multiply by $100$:
$\frac{\Delta z}{z} \times 100 = 3 \left( \frac{\Delta x}{x} \times 100 \right) + \frac{1}{2} \left( \frac{\Delta y}{y} \times 100 \right)$.
Given $\frac{\Delta x}{x} \times 100 = 4\%$ and $\frac{\Delta y}{y} \times 100 = 12\%$.
Substituting the values: $\frac{\Delta z}{z} \times 100 = 3(4\%) + \frac{1}{2}(12\%) = 12\% + 6\% = 18\%$.

(A) The maximum speed $v$ of a car on a level curved road is given by the formula $v = \sqrt{\mu Rg}$,where $\mu$ is the coefficient of friction,$R$ is the radius of the curve,and $g$ is the acceleration due to gravity.
Since $\mu$ and $g$ are constant,we have $v \propto \sqrt{R}$.
Therefore,the ratio of the speeds is $\frac{v_2}{v_1} = \sqrt{\frac{R_2}{R_1}}$.
Given $v_1 = 30 \, m/s$,$R_1 = 75 \, m$,and $R_2 = 48 \, m$.
Substituting these values: $\frac{v_2}{30} = \sqrt{\frac{48}{75}}$.
Simplifying the fraction inside the square root: $\frac{48}{75} = \frac{16}{25}$.
So,$\frac{v_2}{30} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
$v_2 = 30 \times \frac{4}{5} = 6 \times 4 = 24 \, m/s$.

(A) The mass of the block is $m = 200\, g = 0.2\, kg$. The weight of the block is $W = mg = 0.2 \times 10 = 2\, N$ (taking $g = 10\, m/s^2$).
For the block to remain stationary on the smooth inclined plane,the forces acting along the plane must balance.
The component of the weight acting down the plane is $mg \sin 60^{\circ}$.
The component of the horizontal force $F$ acting up the plane is $F \cos 60^{\circ}$.
Equating these forces for equilibrium: $F \cos 60^{\circ} = mg \sin 60^{\circ}$.
Substituting the values: $F \times \frac{1}{2} = 2 \times \frac{\sqrt{3}}{2}$.
$F = 2\sqrt{3} = \sqrt{4 \times 3} = \sqrt{12}$.
Given $F = \sqrt{x}$,we have $\sqrt{x} = \sqrt{12}$,therefore $x = 12$.

(C) The radius of all bodies is $2R$ and mass is $M$.
$I_{1} = \frac{2}{5} M (2R)^{2} = \frac{8}{5} MR^{2}$
$I_{2} = \frac{1}{2} M (2R)^{2} = 2 MR^{2}$
$I_{3} = \frac{M (2R)^{2}}{4} = MR^{2}$
$I_{4} = \frac{M (2R)^{2}}{2} = 2 MR^{2}$
Given equation: $2(I_{2} + I_{3}) + I_{4} = x I_{1}$
Substitute the values: $2(2 MR^{2} + MR^{2}) + 2 MR^{2} = x (\frac{8}{5} MR^{2})$
$2(3 MR^{2}) + 2 MR^{2} = x (\frac{8}{5} MR^{2})$
$6 MR^{2} + 2 MR^{2} = x (\frac{8}{5} MR^{2})$
$8 MR^{2} = x (\frac{8}{5} MR^{2})$
$x = 5$

(A) The orbital speed $V$ of a satellite revolving around a planet of mass $M$ at a distance $r$ is given by the formula $V = \sqrt{\frac{GM}{r}}$.
Given the radii of the orbits are $R_{1} = 3200 \, km$ and $R_{2} = 800 \, km$.
The ratio of the speeds is $\frac{V_{1}}{V_{2}} = \frac{\sqrt{\frac{GM}{R_{1}}}}{\sqrt{\frac{GM}{R_{2}}}} = \sqrt{\frac{R_{2}}{R_{1}}}$.
Substituting the values: $\frac{V_{1}}{V_{2}} = \sqrt{\frac{800}{3200}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Comparing this with $\frac{1}{x}$,we get $x = 2$.

(C) For a gas in a closed vessel,the volume $V$ remains constant.
According to the ideal gas law,$pV = nRT$.
Since $V$,$n$,and $R$ are constant,we have $p \propto T$,which implies $\frac{\Delta p}{p} = \frac{\Delta T}{T}$.
Given that the pressure increases by $0.4 \%$,we have $\frac{\Delta p}{p} = \frac{0.4}{100}$.
The change in temperature is $\Delta T = 1^{\circ} C = 1 \ K$.
Substituting these values into the equation: $\frac{0.4}{100} = \frac{1}{T}$.
Solving for $T$,we get $T = \frac{100}{0.4} = 250 \ K$.

(C) Given the expression $Z = \frac{A^{2} B^{3}}{C^{4}}$.
According to the rules of propagation of errors,for a quantity $Z = \frac{A^p B^q}{C^r}$,the relative error is given by $\frac{\Delta Z}{Z} = p \frac{\Delta A}{A} + q \frac{\Delta B}{B} + r \frac{\Delta C}{C}$.
Applying this rule to the given expression,where $p=2$,$q=3$,and $r=4$:
$\frac{\Delta Z}{Z} = 2 \frac{\Delta A}{A} + 3 \frac{\Delta B}{B} + 4 \frac{\Delta C}{C}$.
Therefore,the relative error in $Z$ is $\frac{2 \Delta A}{A} + \frac{3 \Delta B}{B} + \frac{4 \Delta C}{C}$.

(C) Given that $|\vec{A}| = c$,where $c$ is a nonzero constant.
The cross product of any vector with itself is defined as $\vec{A} \times \vec{A} = |\vec{A}| |\vec{A}| \sin(\theta) \hat{n}$,where $\theta$ is the angle between the vector and itself.
Since the angle between a vector and itself is $0^{\circ}$,we have $\theta = 0^{\circ}$.
Substituting this into the formula: $\vec{A} \times \vec{A} = |\vec{A}|^2 \sin(0^{\circ}) \hat{n}$.
Since $\sin(0^{\circ}) = 0$,the expression becomes $\vec{A} \times \vec{A} = 0$.

(B) For two unit vectors $\hat{A}$ and $\hat{B}$ with angle $\theta$ between them:
$|\hat{A}+\hat{B}| = \sqrt{|\hat{A}|^2 + |\hat{B}|^2 + 2|\hat{A}||\hat{B}| \cos \theta} = \sqrt{1+1+2 \cos \theta} = \sqrt{2(1+\cos \theta)} = \sqrt{2(2 \cos^2 \frac{\theta}{2})} = 2 \cos \frac{\theta}{2}$.
$|\hat{A}-\hat{B}| = \sqrt{|\hat{A}|^2 + |\hat{B}|^2 - 2|\hat{A}||\hat{B}| \cos \theta} = \sqrt{1+1-2 \cos \theta} = \sqrt{2(1-\cos \theta)} = \sqrt{2(2 \sin^2 \frac{\theta}{2})} = 2 \sin \frac{\theta}{2}$.
Dividing the two magnitudes:
$\frac{|\hat{A}-\hat{B}|}{|\hat{A}+\hat{B}|} = \frac{2 \sin \frac{\theta}{2}}{2 \cos \frac{\theta}{2}} = \tan \frac{\theta}{2}$.
Therefore,$|\hat{A}-\hat{B}| = |\hat{A}+\hat{B}| \tan \frac{\theta}{2}$.

(B) The torque $\vec{\tau}$ is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$:
$\vec{\tau} = \vec{r} \times \vec{F}$
Using the determinant form for the cross product:
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 2 \\ 3 & 4 & -2 \end{vmatrix}$
Expanding the determinant:
$\vec{\tau} = \hat{i} [(1)(-2) - (2)(4)] - \hat{j} [(2)(-2) - (2)(3)] + \hat{k} [(2)(4) - (1)(3)]$
$\vec{\tau} = \hat{i} [-2 - 8] - \hat{j} [-4 - 6] + \hat{k} [8 - 3]$
$\vec{\tau} = -10 \hat{i} + 10 \hat{j} + 5 \hat{k}$

(D) Let $R$ be the radius of the Earth. The diameter of the Earth is $2R$.
Given that the height $h$ of point $P$ above the surface is equal to the diameter of the Earth,so $h = 2R$.
The distance of point $P$ from the center of the Earth is $r = R + h = R + 2R = 3R$.
The acceleration due to gravity at the surface of the Earth is $g = \frac{GM}{R^2}$.
The acceleration due to gravity at point $P$ at a distance $r$ from the center is $g' = \frac{GM}{r^2}$.
Substituting $r = 3R$ into the equation,we get:
$g' = \frac{GM}{(3R)^2} = \frac{GM}{9R^2}$.
Since $g = \frac{GM}{R^2}$,we can write:
$g' = \frac{1}{9} \left( \frac{GM}{R^2} \right) = \frac{g}{9}$.

(A) The magnetic field $B$ inside a long solenoid is given by the formula $B = \mu_0 n i$,where $n$ is the number of turns per unit length and $i$ is the current.
Let the initial magnetic field be $B = \mu_0 n i$.
When the current is doubled,the new current $i' = 2i$.
When the number of turns per cm is halved,the new number of turns per unit length $n' = n/2$.
The new magnetic field $B'$ is given by $B' = \mu_0 n' i' = \mu_0 (n/2) (2i) = \mu_0 n i$.
Therefore,$B' = B$.

(C) The total $e.m.f.$ of the series combination is $E_{eq} = E + E = 2E$.
The total resistance of the circuit is $R_{eq} = R + r_1 + r_2$.
The current $I$ flowing through the circuit is given by $I = \frac{2E}{R + r_1 + r_2}$.
The potential difference $V_1$ across the first cell (with internal resistance $r_1$) is given by $V_1 = E - I r_1$.
Given that $V_1 = 0$,we have $E - I r_1 = 0$,which implies $E = I r_1$.
Substituting the expression for $I$:
$E = \left( \frac{2E}{R + r_1 + r_2} \right) r_1$.
Dividing both sides by $E$ (assuming $E \neq 0$):
$1 = \frac{2r_1}{R + r_1 + r_2}$.
$R + r_1 + r_2 = 2r_1$.
$R = 2r_1 - r_1 - r_2 = r_1 - r_2$.
Thus,the correct option is $C$.

(B) Since both metal wires are of identical dimensions,their length and area of cross-section are the same. Let them be $l$ and $A$ respectively.
The resistance of the first wire is $R_1 = \frac{l}{\sigma_1 A}$ ...$(i)$
The resistance of the second wire is $R_2 = \frac{l}{\sigma_2 A}$ ...$(ii)$
Since they are connected in series,their effective resistance is $R_s = R_1 + R_2$.
$R_s = \frac{l}{\sigma_1 A} + \frac{l}{\sigma_2 A} = \frac{l}{A} \left( \frac{1}{\sigma_1} + \frac{1}{\sigma_2} \right)$ ...$(iii)$
If $\sigma_{eff}$ is the effective conductivity of the combination,then the total length is $2l$ and the total resistance is $R_s = \frac{2l}{\sigma_{eff} A}$ ...$(iv)$
Equating equations $(iii)$ and $(iv)$,we get:
$\frac{2l}{\sigma_{eff} A} = \frac{l}{A} \left( \frac{\sigma_1 + \sigma_2}{\sigma_1 \sigma_2} \right)$
$\frac{2}{\sigma_{eff}} = \frac{\sigma_1 + \sigma_2}{\sigma_1 \sigma_2}$
$\sigma_{eff} = \frac{2 \sigma_1 \sigma_2}{\sigma_1 + \sigma_2}$

(B) The formula for the refractive index $\mu$ of a prism in terms of the refracting angle $A$ and the angle of minimum deviation $\delta$ is given by:
$\mu = \frac{\sin((A+\delta)/2)}{\sin(A/2)}$
Given $\mu = \cot(A/2) = \frac{\cos(A/2)}{\sin(A/2)},$ we substitute this into the formula:
$\frac{\cos(A/2)}{\sin(A/2)} = \frac{\sin((A+\delta)/2)}{\sin(A/2)}$
Canceling $\sin(A/2)$ from both sides,we get:
$\cos(A/2) = \sin((A+\delta)/2)$
Using the trigonometric identity $\cos(\theta) = \sin(90^o - \theta)$:
$\sin(90^o - A/2) = \sin((A+\delta)/2)$
Equating the angles:
$90^o - A/2 = (A+\delta)/2$
$180^o - A = A + \delta$
$\delta = 180^o - 2A$

(A) The de Broglie wavelength of an electron with kinetic energy $E$ is given by:
$\lambda_e = \frac{h}{\sqrt{2 m_e E}}$ .... $(i)$
The wavelength of a photon with energy $E$ is given by:
$\lambda_p = \frac{hc}{E}$ or $E = \frac{hc}{\lambda_p}$ .... $(ii)$
From equation $(i)$,squaring both sides:
$\lambda_e^2 = \frac{h^2}{2 m_e E}$ or $E = \frac{h^2}{2 m_e \lambda_e^2}$ .... $(iii)$
Equating the expressions for $E$ from $(ii)$ and $(iii)$:
$\frac{hc}{\lambda_p} = \frac{h^2}{2 m_e \lambda_e^2}$
Rearranging for $\lambda_p$:
$\lambda_p = \left( \frac{2 m_e c}{h} \right) \lambda_e^2$
Since $\frac{2 m_e c}{h}$ is a constant,we have:
$\lambda_p \propto \lambda_e^2$

(A) The force acting on the electron due to the electric field is $\vec{F} = q\vec{E} = (-e)(-E_0 \hat{i}) = eE_0 \hat{i}$.
The acceleration produced in the electron is $\vec{a} = \frac{\vec{F}}{m} = \frac{eE_0}{m} \hat{i}$.
The velocity of the electron at time $t$ is given by $\vec{v}_t = \vec{v} + \vec{a}t = \left(V_0 + \frac{eE_0}{m}t\right) \hat{i}$.
The magnitude of the velocity is $v_t = V_0 + \frac{eE_0}{m}t = V_0 \left(1 + \frac{eE_0}{mV_0}t\right)$.
The de-Broglie wavelength at time $t$ is $\lambda_t = \frac{h}{mv_t}$.
Substituting $v_t$,we get $\lambda_t = \frac{h}{m V_0 \left(1 + \frac{eE_0}{mV_0}t\right)}$.
Since the initial de-Broglie wavelength is $\lambda_0 = \frac{h}{mV_0}$,we have $\lambda_t = \frac{\lambda_0}{\left(1 + \frac{eE_0}{mV_0}t\right)}$.

(A) For the particles to be in limiting equilibrium,the electrostatic repulsive force must be equal to the maximum static frictional force.
Electrostatic force $F_e = \frac{kq^2}{L^2}$,where $k = 9 \times 10^9 \, N \cdot m^2/C^2$.
Frictional force $F_f = \mu mg$,where $\mu = 0.25$,$m = 10 \times 10^{-3} \, kg$,and $g = 10 \, m/s^2$.
Equating the two: $\frac{kq^2}{L^2} = \mu mg$.
$L^2 = \frac{kq^2}{\mu mg} = \frac{(9 \times 10^9) \times (2.0 \times 10^{-7})^2}{0.25 \times (10 \times 10^{-3}) \times 10}$.
$L^2 = \frac{9 \times 10^9 \times 4 \times 10^{-14}}{0.25 \times 0.1} = \frac{36 \times 10^{-5}}{0.025} = \frac{36 \times 10^{-5}}{25 \times 10^{-3}} = 1.44 \times 10^{-2} \, m^2$.
$L = \sqrt{1.44 \times 10^{-2}} = 1.2 \times 10^{-1} \, m = 0.12 \, m = 12 \, cm$.

(B) To find the equivalent resistance of $\frac{22}{3} \, \Omega$,we test the given options.
For option $A$: Parallel combination of $A$ and $C$ is $R_{p} = \frac{A \times C}{A + C} = \frac{2 \times 6}{2 + 6} = \frac{12}{8} = 1.5 \, \Omega$.
Adding $B$ in series: $R_{eq} = R_{p} + B = 1.5 + 4 = 5.5 \, \Omega$.
For option $B$: Parallel combination of $A$ and $B$ is $R_{p} = \frac{A \times B}{A + B} = \frac{2 \times 4}{2 + 4} = \frac{8}{6} = \frac{4}{3} \, \Omega$.
Adding $C$ in series: $R_{eq} = R_{p} + C = \frac{4}{3} + 6 = \frac{4 + 18}{3} = \frac{22}{3} \, \Omega$.
Thus,the correct combination is the parallel combination of $A$ and $B$ connected in series with $C$.

(C) Magnetic retentivity determines the magnetism left in the material after the magnetizing field has been switched off.
Electromagnets are operated in conditions requiring fast reversal of polarity,so high retentivity is undesirable.
Permeability is directly related to susceptibility; high permeability means the material is easily magnetized.
Therefore,electromagnets must be made of materials having high permeability and low retentivity.
Soft iron is such a material,making it ideal for electromagnets.

(D) The radius $R$ of a charged particle moving in a uniform magnetic field is given by $R = \frac{mv}{qB}$.
Since kinetic energy $K = \frac{1}{2}mv^2$, we have $v = \sqrt{\frac{2K}{m}}$.
Substituting $v$ into the radius formula: $R = \frac{m}{qB} \sqrt{\frac{2K}{m}} = \frac{\sqrt{2mK}}{qB}$.
Since $K$ and $B$ are constant for all particles, $R \propto \frac{\sqrt{m}}{q}$.
For a proton $(p)$: mass $= m$, charge $= e$. So, $R_p \propto \frac{\sqrt{m}}{e}$.
For a deuteron $(d)$: mass $= 2m$, charge $= e$. So, $R_d \propto \frac{\sqrt{2m}}{e}$.
For an $\alpha$-particle $(\alpha)$: mass $= 4m$, charge $= 2e$. So, $R_{\alpha} \propto \frac{\sqrt{4m}}{2e} = \frac{2\sqrt{m}}{2e} = \frac{\sqrt{m}}{e}$.
The ratio $R_p : R_d : R_{\alpha} = \frac{\sqrt{m}}{e} : \frac{\sqrt{2m}}{e} : \frac{\sqrt{m}}{e} = 1 : \sqrt{2} : 1$.

(C) Statement-$I$ is true. The net reactance $X = X_L - X_C$. If $X_L = X_C$, then $X = 0$. This occurs at resonance in an $LCR$ or $LC$ circuit, which contains both an inductor and a capacitor.
Statement-$II$ is false. The average power in an $ac$ circuit is given by $P_{avg} = V_{rms} I_{rms} \cos \phi$. If the circuit is purely inductive or purely capacitive, the phase angle $\phi = 90^{\circ}$, so $\cos 90^{\circ} = 0$. Thus, the average power delivered by the source becomes zero in such cases.

(A) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$,where $q$ is the charge and $C$ is the capacitance.
Since $C$ is constant,$U \propto q^2$.
Let the initial charge be $q$ and the initial energy be $U_i = \frac{q^2}{2C}$.
When the charge is increased by $2 \ C$,the new charge is $q' = q + 2$.
The new energy is $U_f = \frac{(q+2)^2}{2C}$.
Given that the energy increases by $44 \%$,the new energy is $U_f = U_i + 0.44 \, U_i = 1.44 \, U_i$.
Substituting the expressions for $U_f$ and $U_i$:
$\frac{(q+2)^2}{2C} = 1.44 \times \frac{q^2}{2C}$.
$(q+2)^2 = 1.44 \, q^2$.
Taking the square root on both sides:
$q + 2 = \sqrt{1.44} \, q$.
$q + 2 = 1.2 \, q$.
$0.2 \, q = 2$.
$q = \frac{2}{0.2} = 10 \ C$.

(A) For a point outside the cylinder at a distance $r$ from the axis,we use Gauss's Law: $\oint E \cdot dA = \frac{q_{enclosed}}{\varepsilon_{0}}$.
Considering a Gaussian surface of radius $r$ and length $\ell$,the enclosed charge is $q_{enc} = \rho \cdot \pi R^{2} \ell$.
Thus,$E(2 \pi r \ell) = \frac{\rho \pi R^{2} \ell}{\varepsilon_{0}}$,which gives the electric field $E = \frac{\rho R^{2}}{2 \varepsilon_{0} r}$.
The electrostatic force provides the necessary centripetal force for circular motion: $qE = \frac{mv^{2}}{r}$.
Substituting $E$: $q \left( \frac{\rho R^{2}}{2 \varepsilon_{0} r} \right) = \frac{mv^{2}}{r}$.
Simplifying,we get $mv^{2} = \frac{q \rho R^{2}}{2 \varepsilon_{0}}$.
The kinetic energy $K = \frac{1}{2} mv^{2} = \frac{q \rho R^{2}}{4 \varepsilon_{0}}$.

(B) The intensity $I$ of the electromagnetic wave at a distance $r$ from a point source is given by $I = \frac{\eta P}{4 \pi r^2}$,where $\eta = 0.035$ is the efficiency and $P = 200 \, W$ is the power.
Also,the intensity is related to the peak magnetic field $B_0$ by $I = \frac{c B_0^2}{2 \mu_0}$.
Equating the two expressions: $\frac{\eta P}{4 \pi r^2} = \frac{c B_0^2}{2 \mu_0}$.
Rearranging for $B_0$: $B_0 = \sqrt{\frac{\mu_0 \eta P}{2 \pi c r^2}} = \frac{1}{r} \sqrt{\frac{\mu_0 \eta P}{2 \pi c}}$.
Given $\frac{\mu_0}{4 \pi} = 10^{-7} \, T \cdot m/A$,$c = 3 \times 10^8 \, m/s$,$r = 4 \, m$,$\eta = 0.035$,and $P = 200 \, W$.
$B_0 = \frac{1}{4} \sqrt{\frac{2 \times 10^{-7} \times 0.035 \times 200}{3 \times 10^8}} = \frac{1}{4} \sqrt{\frac{14 \times 10^{-7}}{3 \times 10^8}} = \frac{1}{4} \sqrt{4.66 \times 10^{-16}} = \frac{2.16 \times 10^{-8}}{4} \approx 0.54 \times 10^{-8} \, T$.
Re-evaluating the provided formula in the prompt: $B_0 = \frac{1}{r} \sqrt{\frac{\mu_0 \eta P}{2 \pi c}} = \frac{1}{4} \sqrt{\frac{2 \times 10^{-7} \times 0.035 \times 200}{3 \times 10^8}} = 1.71 \times 10^{-8} \, T$ is obtained if the formula used is $B_0 = \sqrt{\frac{\mu_0 \eta P}{2 \pi c r^2}}$. The calculation yields $1.71 \times 10^{-8} \, T$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of emitted electrons is given by $K_{max} = E - \Phi$,where $E$ is the photon energy and $\Phi$ is the work function.
For the first photon,$E_1 = 3.8 \, eV$ and $\Phi = 0.6 \, eV$. Thus,$K_{max1} = 3.8 - 0.6 = 3.2 \, eV$.
For the second photon,$E_2 = 1.4 \, eV$ and $\Phi = 0.6 \, eV$. Thus,$K_{max2} = 1.4 - 0.6 = 0.8 \, eV$.
Since $K_{max} = \frac{1}{2} m v_{max}^2$,the ratio of the maximum speeds is $\frac{v_1}{v_2} = \sqrt{\frac{K_{max1}}{K_{max2}}}$.
Substituting the values,$\frac{v_1}{v_2} = \sqrt{\frac{3.2}{0.8}} = \sqrt{4} = 2$.
Therefore,the ratio is $2: 1$.

(D) Given the ratio of intensities $I_1 : I_2 = 9 : 4$.
Let $I_1 = 9k$ and $I_2 = 4k$.
The amplitudes are proportional to the square root of intensities,so $a_1 = \sqrt{I_1} = 3\sqrt{k}$ and $a_2 = \sqrt{I_2} = 2\sqrt{k}$.
The ratio of maximum intensity to minimum intensity is given by the formula:
$\frac{I_{max}}{I_{min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2$.
Substituting the values:
$\frac{I_{max}}{I_{min}} = \left( \frac{3\sqrt{k} + 2\sqrt{k}}{3\sqrt{k} - 2\sqrt{k}} \right)^2 = \left( \frac{5\sqrt{k}}{1\sqrt{k}} \right)^2 = 5^2 = 25$.
Thus,the ratio is $25 : 1$.

(B) In Bohr's model for a hydrogen atom,the energy levels are given by $E_n = -\frac{13.6 \text{ eV}}{n^2}$. As the electron transitions to a higher level ($n$ increases),$E$ increases (becomes less negative).
For a hydrogen-like atom,the kinetic energy is $K = -E = \frac{13.6 \text{ eV}}{n^2}$. As $n$ increases,$K$ decreases.
The potential energy is $P = 2E = -\frac{27.2 \text{ eV}}{n^2}$. As $n$ increases,the magnitude of $P$ decreases,meaning $P$ becomes less negative,so $P$ increases.
Therefore,as the electron moves to a higher energy level,$K$ decreases,while $P$ and $E$ increase.

(B) The speed of light in a dielectric medium is given by $v = \frac{c}{\sqrt{\mu_{r} \varepsilon_{r}}}$.
Given $\mu_{r} = 1$ and $\varepsilon_{r} = 6.25$,the speed $v = \frac{3 \times 10^{8}}{\sqrt{6.25}} = \frac{3 \times 10^{8}}{2.5} = 1.2 \times 10^{8} \, m/s$.
For an antenna to radiate effectively,its length $L$ should be at least $\frac{\lambda}{4}$.
Thus,$\lambda = 4L = 4 \times 5.0 \times 10^{-3} \, m = 20 \times 10^{-3} \, m = 0.02 \, m$.
The frequency $f$ is given by $f = \frac{v}{\lambda} = \frac{1.2 \times 10^{8}}{0.02} = 60 \times 10^{8} \, Hz = 6 \, GHz$.

(C) The total resistance of the primary circuit is $R_{total} = R_{wire} + R_{external} = 20 \,\Omega + 30 \,\Omega = 50 \,\Omega$.
The current in the primary circuit is $I = \frac{V}{R_{total}} = \frac{25 \,V}{50 \,\Omega} = 0.5 \,A$.
The potential drop across the entire potentiometer wire is $V_{wire} = I \times R_{wire} = 0.5 \,A \times 20 \,\Omega = 10 \,V$.
The potential gradient along the wire is $k = \frac{V_{wire}}{L} = \frac{10 \,V}{10 \,m} = 1 \,V/m$.
The balancing length is $l = 250 \,cm = 2.5 \,m$.
The emf of the cell is $E = k \times l = 1 \,V/m \times 2.5 \,m = 2.5 \,V$.
Given $E = \frac{x}{10}$,we have $2.5 = \frac{x}{10}$,which implies $x = 25$.

(A) The Zener diode is connected in parallel with the load resistor $R_{L}$.
Since the Zener diode is in the breakdown region,the voltage across the load resistor $R_{L}$ is equal to the Zener breakdown voltage,which is $V_{Z} = 5 \, V$.
The load current $I_{L}$ is given by Ohm's law:
$I_{L} = \frac{V_{Z}}{R_{L}}$
Given $V_{Z} = 5 \, V$ and $R_{L} = 1 \, k\Omega = 1000 \, \Omega$.
$I_{L} = \frac{5 \, V}{1000 \, \Omega} = 0.005 \, A = 5 \, mA$.

(B) The number of atoms $N$ in a mass $m$ with atomic weight $M$ is given by $N = \frac{m}{M} N_A$,where $N_A$ is Avogadro's number.
Initial number of atoms: $N_{A,0} = \frac{m}{M_A} N_A$ and $N_{B,0} = \frac{m}{M_B} N_A$.
Given $m_A = m_B = 10^{-2} \ kg$ and $\frac{M_A}{M_B} = \frac{1}{2}$,so $M_B = 2M_A$.
Thus,$\frac{N_{A,0}}{N_{B,0}} = \frac{M_B}{M_A} = 2$.
After time $t = 16 \ s$,the number of remaining atoms is $N(t) = N_0 (0.5)^{\frac{t}{T_{1/2}}}$.
For substance $A$: $N_A(16) = N_{A,0} (0.5)^{\frac{16}{4}} = N_{A,0} (0.5)^4 = \frac{N_{A,0}}{16}$.
For substance $B$: $N_B(16) = N_{B,0} (0.5)^{\frac{16}{8}} = N_{B,0} (0.5)^2 = \frac{N_{B,0}}{4}$.
The ratio is $\frac{N_A(16)}{N_B(16)} = \frac{N_{A,0}}{16} \times \frac{4}{N_{B,0}} = \frac{1}{4} \times \frac{N_{A,0}}{N_{B,0}} = \frac{1}{4} \times 2 = \frac{1}{2} = \frac{50}{100}$.
Therefore,$x = 50$.

(B) The formula for lateral shift $d$ is given by: $d = t \frac{\sin(i-r)}{\cos r}$,where $t$ is the thickness,$i$ is the angle of incidence,and $r$ is the angle of refraction.
Using Snell's Law: $\sin i = \mu \sin r \Rightarrow \sin 60^{\circ} = \sqrt{3} \sin r$.
$\frac{\sqrt{3}}{2} = \sqrt{3} \sin r \Rightarrow \sin r = \frac{1}{2} \Rightarrow r = 30^{\circ}$.
Now,substitute the values into the lateral shift formula:
$4\sqrt{3} = t \frac{\sin(60^{\circ}-30^{\circ})}{\cos 30^{\circ}}$.
$4\sqrt{3} = t \frac{\sin 30^{\circ}}{\cos 30^{\circ}} = t \tan 30^{\circ}$.
$4\sqrt{3} = t \left(\frac{1}{\sqrt{3}}\right)$.
$t = 4\sqrt{3} \times \sqrt{3} = 4 \times 3 = 12 \, cm$.

(D) The maximum induced electromotive force $(EMF)$ in a rotating coil is given by the formula: $\varepsilon_{\max} = BAN\omega$.
Given:
$B = 0.07 \, T$ (Magnetic field)
$A = 1 \, m^{2}$ (Area of the coil)
$N = 1000$ (Number of turns)
$f = 1 \, \text{rev/s}$ (Frequency)
Angular velocity $\omega = 2\pi f = 2\pi(1) = 2\pi \, \text{rad/s}$.
Substituting the values:
$\varepsilon_{\max} = 0.07 \times 1 \times 1000 \times 2\pi$
$\varepsilon_{\max} = 70 \times 2\pi = 140\pi$.
Using $\pi \approx 3.14159$:
$\varepsilon_{\max} = 140 \times 3.14159 \approx 439.82 \, V$.
Rounding to the nearest integer, we get $440 \, V$.

(A) The magnetic force on a moving charged particle is given by $\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B})$.
Since the force $\overrightarrow{F}$ is the cross product of velocity $\overrightarrow{v}$ and magnetic field $\overrightarrow{B}$,the force is always perpendicular to the velocity $(\overrightarrow{F} \perp \overrightarrow{v})$.
Work done by the magnetic force is $W = \int \overrightarrow{F} \cdot d\overrightarrow{s} = \int \overrightarrow{F} \cdot \overrightarrow{v} dt$.
Since $\overrightarrow{F} \perp \overrightarrow{v}$,the dot product $\overrightarrow{F} \cdot \overrightarrow{v} = 0$,so the work done is $0$.
According to the work-energy theorem,the change in kinetic energy is equal to the work done. Since work done is $0$,the kinetic energy remains constant.
Since kinetic energy $K = \frac{1}{2}mv^2$ is constant,the speed $v$ of the particle also remains constant.
Thus,both Assertion $(A)$ and Reason $(R)$ are true,and $(R)$ is the correct explanation of $(A)$.

(C) The two $20 \,\Omega$ resistors are connected in parallel,so the equivalent external resistance $R$ is given by $\frac{1}{R} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10}$,which means $R = 10 \,\Omega$.
The two identical cells of emf $E = 1.5 \,V$ and internal resistance $r$ are connected in parallel. The equivalent emf $E_{eq} = E = 1.5 \,V$ and equivalent internal resistance $r_{eq} = \frac{r}{2}$.
The terminal voltage $V$ across the external resistance is given by $V = E_{eq} - I r_{eq}$,where $I$ is the total current.
Given $V = 1.2 \,V$,we have $1.2 = 1.5 - I \left(\frac{r}{2}\right)$,which implies $I \left(\frac{r}{2}\right) = 0.3$.
Also,the total current $I$ is given by $I = \frac{E_{eq}}{R + r_{eq}} = \frac{1.5}{10 + \frac{r}{2}}$.
Substituting $I$ into the equation $I \left(\frac{r}{2}\right) = 0.3$:
$\left(\frac{1.5}{10 + \frac{r}{2}}\right) \left(\frac{r}{2}\right) = 0.3$
$1.5 \left(\frac{r}{2}\right) = 0.3 \left(10 + \frac{r}{2}\right)$
$0.75r = 3 + 0.15r$
$0.6r = 3$
$r = \frac{3}{0.6} = 5 \,\Omega$.

(B) For the water droplet to remain suspended,the upward electric force must balance the downward gravitational force.
$F_{e} = F_{g}$
$qE = mg$
Given:
Mass $m = 0.1 \, g = 0.1 \times 10^{-3} \, kg = 10^{-4} \, kg$
Electric field $E = 4.9 \times 10^{5} \, N/C$
Acceleration due to gravity $g = 9.8 \, m/s^{2}$
Substituting the values:
$q(4.9 \times 10^{5}) = (10^{-4})(9.8)$
$q = \frac{9.8 \times 10^{-4}}{4.9 \times 10^{5}}$
$q = 2 \times 10^{-9} \, C$
Thus,the charge on the droplet is $2.0 \times 10^{-9} \, C$.

(A) The instantaneous current in an $AC$ circuit is given by $I = I_0 \sin(\omega t)$.
At maximum value,the phase $\omega t_1 = \frac{\pi}{2}$.
At $rms$ value,the current is $I = \frac{I_0}{\sqrt{2}}$,so $\sin(\omega t_2) = \frac{1}{\sqrt{2}}$. The first instance after the peak is $\omega t_2 = \frac{3\pi}{4}$.
The phase difference is $\Delta \phi = \omega t_2 - \omega t_1 = \frac{3\pi}{4} - \frac{\pi}{2} = \frac{\pi}{4}$.
Given frequency $f = 50 \,Hz$,the angular frequency $\omega = 2\pi f = 2\pi \times 50 = 100\pi \,rad/s$.
The time taken is $t = \frac{\Delta \phi}{\omega} = \frac{\pi / 4}{100\pi} = \frac{1}{400} \,s$.
Converting to milliseconds: $t = 0.0025 \,s = 2.5 \,ms$.

(C) The speed of an electromagnetic wave in a medium is given by $v = \frac{c}{\sqrt{\mu_{r} \epsilon_{r}}}$.
Given $\mu_{r} = 1.61$ and $\epsilon_{r} = 6.44$,the speed $v = \frac{3 \times 10^{8}}{\sqrt{1.61 \times 6.44}} = \frac{3 \times 10^{8}}{\sqrt{10.3684}} = \frac{3 \times 10^{8}}{3.22} \approx 9.317 \times 10^{7} \; m s^{-1}$.
The relationship between electric field $E$ and magnetic field $B$ is $E = vB$.
Since $B = \mu_{0} \mu_{r} H$,we have $E = v \mu_{0} \mu_{r} H$.
$E = (9.317 \times 10^{7}) \times (4 \pi \times 10^{-7}) \times (1.61) \times (4.5 \times 10^{-2})$.
$E = 9.317 \times 4 \times 3.1416 \times 1.61 \times 4.5 \times 10^{-2} \approx 8.48 \; V m^{-1}$.

(C) According to Rutherford's model,the electron revolves around the nucleus in a circular orbit.
Since the electron is moving in a circular path,it undergoes centripetal acceleration.
According to classical electromagnetic theory,an accelerating charged particle must emit electromagnetic $(EM)$ radiation.
As the electron loses energy through radiation,its orbital radius decreases,and it should eventually spiral into the nucleus.
Therefore,a classical atom based on Rutherford's model is unstable and is doomed to collapse.

(D) The energy released in a nuclear fission reaction is given by the difference between the total binding energy of the products and the total binding energy of the reactant.
Total binding energy of the reactant (Nucleus $A$):
$BE_A = 220 \times 5.6 \, MeV = 1232 \, MeV$.
Total binding energy of the products (Nuclei $B$ and $C$):
$BE_{B+C} = (105 \times 6.4) + (115 \times 6.4) \, MeV = (105 + 115) \times 6.4 \, MeV = 220 \times 6.4 \, MeV = 1408 \, MeV$.
The energy $Q$ released is:
$Q = BE_{products} - BE_{reactant}$
$Q = 1408 \, MeV - 1232 \, MeV = 176 \, MeV$.

(C) The carrier frequency is $f_{c} = 3.5\, GHz = 3.5 \times 10^{9}\, Hz$.
The wavelength $\lambda$ of the carrier signal is given by $\lambda = \frac{c}{f_{c}}$,where $c = 3 \times 10^{8}\, m/s$ is the speed of light.
$\lambda = \frac{3 \times 10^{8}}{3.5 \times 10^{9}} = \frac{3}{35}\, m \approx 0.0857\, m = 85.7\, mm$.
The minimum length of the antenna required for efficient transmission is $\frac{\lambda}{4}$.
Minimum length $= \frac{85.7\, mm}{4} = 21.425\, mm \approx 21.4\, mm$.

(D) The maximum electric field $E$ that the dielectric can withstand is its dielectric strength,$E = 3.6 \times 10^{7} \, Vm^{-1}$.
The capacitance of a parallel plate capacitor with a dielectric is $C = \frac{K \varepsilon_{0} A}{d}$.
The maximum charge $q$ is given by $q = CV = C(Ed) = \left( \frac{K \varepsilon_{0} A}{d} \right) Ed = K \varepsilon_{0} A E$.
Rearranging for the dielectric constant $K$,we get $K = \frac{q}{\varepsilon_{0} A E}$.
Given $q = 7 \times 10^{-6} \, C$,$A = 30 \pi \times 10^{-4} \, m^{2}$,$E = 3.6 \times 10^{7} \, Vm^{-1}$,and $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \, Nm^{2}C^{-2}$,which implies $\varepsilon_{0} = \frac{1}{36 \pi \times 10^{9}}$.
Substituting these values: $K = \frac{7 \times 10^{-6}}{\left( \frac{1}{36 \pi \times 10^{9}} \right) \times (30 \pi \times 10^{-4}) \times (3.6 \times 10^{7})}$.
$K = \frac{7 \times 10^{-6} \times 36 \pi \times 10^{9}}{30 \pi \times 10^{-4} \times 3.6 \times 10^{7}} = \frac{7 \times 36 \times 10^{3}}{30 \times 3.6 \times 10^{3}} = \frac{252}{108} = 2.33$.

(C) The magnetic field at the centre of a circular coil is given by $B_{C} = \frac{\mu_{0} I}{2 r}$.
The magnetic field at a point on the axis at a distance $x$ from the centre is given by $B_{a} = \frac{\mu_{0} I r^{2}}{2(x^{2} + r^{2})^{3/2}}$.
Given $x = \frac{r}{2}$,we substitute this into the formula:
$B_{a} = \frac{\mu_{0} I r^{2}}{2((\frac{r}{2})^{2} + r^{2})^{3/2}}$
$B_{a} = \frac{\mu_{0} I r^{2}}{2(\frac{r^{2}}{4} + r^{2})^{3/2}} = \frac{\mu_{0} I r^{2}}{2(\frac{5r^{2}}{4})^{3/2}}$
$B_{a} = \frac{\mu_{0} I r^{2}}{2 \cdot r^{3} \cdot (\frac{5}{4})^{3/2}} = \frac{\mu_{0} I}{2 r} \cdot (\frac{4}{5})^{3/2}$
Since $B = \frac{\mu_{0} I}{2 r}$,we have:
$B_{a} = B \cdot (\frac{2}{\sqrt{5}})^{3}$.

(A) For a biconvex lens,the lens maker's formula is $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Given $f = 15 \, cm$ and $\mu = 1.5$,and for a symmetric biconvex lens $R_1 = R$ and $R_2 = -R$,we have:
$\frac{1}{15} = (1.5 - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = 0.5 \left( \frac{2}{R} \right) = \frac{1}{R}$.
Thus,$R = 15 \, cm$.
The combination consists of two biconvex lenses and a liquid lens in between. The liquid lens has refractive index $\mu_l = 1.25$ and its surfaces have radii of curvature $-R$ and $R$ (concave shape).
The focal length of the liquid lens $f_l$ is given by:
$\frac{1}{f_l} = (\mu_l - 1) \left( \frac{1}{-R} - \frac{1}{R} \right) = (1.25 - 1) \left( -\frac{2}{R} \right) = 0.25 \left( -\frac{2}{15} \right) = -\frac{0.5}{15} = -\frac{1}{30}$.
The equivalent focal length $f_{eq}$ of the combination is:
$\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_l} + \frac{1}{f_2} = \frac{1}{15} - \frac{1}{30} + \frac{1}{15} = \frac{2 - 1 + 2}{30} = \frac{3}{30} = \frac{1}{10}$.
Therefore,$f_{eq} = 10 \, cm$.

(B) The input resistance $r_i$ is calculated as the ratio of the change in base-emitter voltage to the change in base current:
$r_i = \frac{\Delta V_{BE}}{\Delta I_B} = \frac{10 \, mV}{10 \, \mu A} = \frac{10 \times 10^{-3} \, V}{10 \times 10^{-6} \, A} = 1000 \, \Omega = 1 \, k\Omega$.
The current gain $\beta$ is the ratio of the change in collector current to the change in base current:
$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{1.5 \, mA}{10 \, \mu A} = \frac{1.5 \times 10^{-3} \, A}{10 \times 10^{-6} \, A} = 150$.
The voltage gain $A_V$ for a common-emitter amplifier is given by the product of the current gain and the ratio of load resistance $R_L$ to input resistance $r_i$:
$A_V = \beta \times \frac{R_L}{r_i} = 150 \times \frac{5000 \, \Omega}{1000 \, \Omega} = 150 \times 5 = 750$.

(B) Given: Inductance $L = 200 \, mH = 0.2 \, H$,$V_{rms} = 220 \, V$,$f = 50 \, Hz$.
The inductive reactance $X_L$ is given by:
$X_L = 2 \pi f L = 2 \pi \times 50 \times 0.2 = 20 \pi \, \Omega$.
The peak voltage $V_0$ is:
$V_0 = V_{rms} \sqrt{2} = 220 \sqrt{2} \, V$.
The peak current $i_0$ is:
$i_0 = \frac{V_0}{X_L} = \frac{220 \sqrt{2}}{20 \pi} = \frac{11 \sqrt{2}}{\pi} \, A$.
We can rewrite this as:
$i_0 = \frac{\sqrt{11^2 \times 2}}{\pi} = \frac{\sqrt{121 \times 2}}{\pi} = \frac{\sqrt{242}}{\pi} \, A$.
Comparing this with the given expression $\frac{\sqrt{a}}{\pi} \, A$,we get:
$a = 242$.

(C) For a single slit diffraction,the condition for the $n^{th}$ maxima is given by $a \sin \theta = (n + \frac{1}{2}) \lambda$.
For the first maxima,$n = 1$,so $a \sin \theta = \frac{3 \lambda}{2}$.
Since $\theta$ is very small,$\sin \theta \approx \theta = \frac{y}{L}$,where $y$ is the distance from the central maxima and $L$ is the distance to the screen.
Thus,$y = \frac{3 \lambda L}{2 a}$.
The separation between the first maxima for two wavelengths $\lambda_1 = 650\, nm$ and $\lambda_2 = 655\, nm$ is $\Delta y = y_2 - y_1 = \frac{3 L}{2 a} (\lambda_2 - \lambda_1)$.
Given $L = 2.0\, m$,$a = 0.5\, mm = 0.5 \times 10^{-3}\, m$,$\lambda_1 = 650 \times 10^{-9}\, m$,and $\lambda_2 = 655 \times 10^{-9}\, m$.
$\Delta y = \frac{3 \times 2.0}{2 \times 0.5 \times 10^{-3}} \times (655 - 650) \times 10^{-9}$.
$\Delta y = \frac{6}{10^{-3}} \times 5 \times 10^{-9} = 6 \times 5 \times 10^{-6} = 30 \times 10^{-6} = 3 \times 10^{-5}\, m$.
Comparing with $x \times 10^{-5}\, m$,we get $x = 3$.

(A) According to Einstein's photoelectric equation: $h\nu = h\nu_{0} + K_{\text{max}}$,where $K_{\text{max}} = \frac{1}{2}mv^{2}$.
Case $1$: When $\nu = 2\nu_{0}$:
$h(2\nu_{0}) = h\nu_{0} + \frac{1}{2}mv_{1}^{2}$
$h\nu_{0} = \frac{1}{2}mv_{1}^{2} \dots(1)$
Case $2$: When $\nu = 5\nu_{0}$:
$h(5\nu_{0}) = h\nu_{0} + \frac{1}{2}mv_{2}^{2}$
$4h\nu_{0} = \frac{1}{2}mv_{2}^{2} \dots(2)$
Dividing equation $(2)$ by $(1)$:
$\frac{4h\nu_{0}}{h\nu_{0}} = \frac{\frac{1}{2}mv_{2}^{2}}{\frac{1}{2}mv_{1}^{2}}$
$4 = \left(\frac{v_{2}}{v_{1}}\right)^{2}$
$\frac{v_{2}}{v_{1}} = \sqrt{4} = 2$
$v_{2} = 2v_{1}$
Given $v_{2} = xv_{1}$,therefore $x = 2$.

(A) In a potentiometer,the balancing length $\ell$ is directly proportional to the emf $\varepsilon$ of the cell,given by $\varepsilon \propto \ell$ or $\frac{\varepsilon_1}{\varepsilon_2} = \frac{\ell_1}{\ell_2}$.
Given $\varepsilon_1 : \varepsilon_2 = 3 : 2$ and $\ell_1 = 75 \, cm$.
Substituting the values: $\frac{3}{2} = \frac{75}{\ell_2}$.
Solving for $\ell_2$: $\ell_2 = \frac{75 \times 2}{3} = 50 \, cm$.
The difference in the balancing lengths is $\Delta \ell = |\ell_1 - \ell_2| = |75 - 50| = 25 \, cm$.

(A) The original capacitance of the parallel plate capacitor is given by $C_1 = \frac{\epsilon_0 A}{d}$.
When a metal sheet of thickness $t = \frac{d}{2}$ is introduced between the plates,the effective distance between the plates decreases.
The new capacitance $C_2$ is given by the formula $C_2 = \frac{\epsilon_0 A}{d - t}$.
Substituting $t = \frac{d}{2}$,we get $C_2 = \frac{\epsilon_0 A}{d - \frac{d}{2}} = \frac{\epsilon_0 A}{d/2} = \frac{2 \epsilon_0 A}{d}$.
Therefore,the ratio of the new capacitance to the original capacitance is $\frac{C_2}{C_1} = \frac{2 \epsilon_0 A / d}{\epsilon_0 A / d} = \frac{2}{1}$ or $2:1$.

(A) The total emf of the circuit is $2E$ and the total resistance is $R + r_{1} + r_{2}$.
The current $I$ in the circuit is given by:
$I = \frac{2E}{R + r_{1} + r_{2}} \quad \dots (i)$
The potential difference across the second cell (with emf $E$ and internal resistance $r_{2}$) is given by $V = E - Ir_{2}$.
Given that the potential difference across the second cell is zero:
$E - Ir_{2} = 0 \Rightarrow I = \frac{E}{r_{2}} \quad \dots (ii)$
Comparing the values of $I$ from equations $(i)$ and $(ii)$:
$\frac{E}{r_{2}} = \frac{2E}{R + r_{1} + r_{2}}$
$R + r_{1} + r_{2} = 2r_{2}$
$R = 2r_{2} - r_{2} - r_{1}$
$R = r_{2} - r_{1}$

$(A)$ According to Curie's law, the magnetic susceptibility $\chi$ of paramagnetic substances is inversely proportional to the absolute temperature $T$, given by $\chi = \frac{C}{T}$.
For ferromagnetic substances, the susceptibility follows the Curie-Weiss law: $\chi = \frac{C}{T - T_C}$, where $T_C$ is the Curie temperature. As $T$ decreases towards $T_C$, the susceptibility increases. Thus, Statement-$I$ is true.
Diamagnetism arises due to the orbital motion of electrons. When an external magnetic field is applied, it induces a change in the orbital motion, creating a magnetic moment that opposes the applied field (Lenz's law at the atomic level). Thus, Statement-$II$ is true.
Therefore, both statements are true. The correct option is $(A)$.

(A) Given: $V(t) = 210 \sin(3000t) \text{ V}$,$L = 10 \text{ mH} = 10^{-2} \text{ H}$,$C = 25 \mu\text{F} = 25 \times 10^{-6} \text{ F}$,$R = 100 \Omega$.
Angular frequency $\omega = 3000 \text{ rad/s}$.
Inductive reactance $X_L = \omega L = 3000 \times 10^{-2} = 30 \Omega$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{3000 \times 25 \times 10^{-6}} = \frac{1}{0.075} = \frac{1000}{75} = \frac{40}{3} \Omega \approx 13.33 \Omega$.
The phase difference $\Phi$ is given by $\tan \Phi = \frac{X_L - X_C}{R}$.
$\tan \Phi = \frac{30 - 13.33}{100} = \frac{16.67}{100} = 0.1667 \approx 0.17$.
Therefore,$\Phi = \tan^{-1}(0.17)$.

(A) The speed of electromagnetic waves in a medium is given by $v = \frac{1}{\sqrt{\mu \epsilon}}$.
Since $\mu = \mu_0 \mu_r$ and $\epsilon = \epsilon_0 \epsilon_r$,we have $v = \frac{1}{\sqrt{\mu_0 \mu_r \epsilon_0 \epsilon_r}} = \frac{c}{\sqrt{\mu_r \epsilon_r}}$,where $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} = 3.0 \times 10^8 \ m/s$.
Given $v = 2.0 \times 10^8 \ m/s$ and $\mu_r = 1.0$.
Substituting these values: $\frac{c}{v} = \sqrt{\mu_r \epsilon_r}$.
$\frac{3.0 \times 10^8}{2.0 \times 10^8} = \sqrt{1.0 \times \epsilon_r}$.
$1.5 = \sqrt{\epsilon_r}$.
Squaring both sides: $\epsilon_r = (1.5)^2 = 2.25$.

(D) Given the intensity ratio $\frac{I_1}{I_2} = 4$,we can write $I_1 = 4I_2$.
The ratio of maximum to minimum intensity in an interference pattern is given by:
$\frac{I_{\max}}{I_{\min}} = \left[ \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right]^2$
Substituting $I_1 = 4I_2$:
$\frac{I_{\max}}{I_{\min}} = \left[ \frac{\sqrt{4I_2} + \sqrt{I_2}}{\sqrt{4I_2} - \sqrt{I_2}} \right]^2 = \left[ \frac{2\sqrt{I_2} + \sqrt{I_2}}{2\sqrt{I_2} - \sqrt{I_2}} \right]^2 = \left( \frac{3}{1} \right)^2 = 9$.
Now,we need to calculate $\frac{I_{\max} + I_{\min}}{I_{\max} - I_{\min}}$:
$\frac{I_{\max} + I_{\min}}{I_{\max} - I_{\min}} = \frac{9 + 1}{9 - 1} = \frac{10}{8} = \frac{5}{4}$.
Comparing this with the given expression $\frac{5}{x}$,we get:
$\frac{5}{x} = \frac{5}{4} \implies x = 4$.

(D) According to Brewster's law,when unpolarized light is incident at Brewster's angle,the reflected light is completely plane-polarized with its electric field vector perpendicular to the plane of incidence.
If the incident light is already polarized such that its electric field vectors are completely removed in the plane of incidence (i.e.,the light is polarized perpendicular to the plane of incidence),then at Brewster's angle,there is no component of the electric field that can be reflected.
Consequently,there is no reflection,and the light is entirely transmitted into the prism.
This is a special case of total transmission as discussed in $NCERT$ Physics Part-$2$,Chapter $10$ (Wave Optics).
Therefore,option $(D)$ is the correct choice.

(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2Em}}$,where $h$ is Planck's constant,$E$ is the kinetic energy,and $m$ is the mass of the particle.
Since the energy $E$ is the same for all particles,we have $\lambda \propto \frac{1}{\sqrt{m}}$.
The masses of the particles are related as $m_{e} < m_{p} \approx m_{n} < m_{\alpha}$.
Specifically,$m_{p} \approx m_{n}$ and $m_{\alpha} \approx 4m_{p}$.
Since $\lambda$ is inversely proportional to the square root of mass,the particle with the smallest mass will have the largest wavelength.
Therefore,$\lambda_{e} > \lambda_{p} \approx \lambda_{n} > \lambda_{\alpha}$.

(B) The radius of a nucleus is given by the relation: $R = R_{0} A^{1/3}$,where $R_{0}$ is a constant and $A$ is the mass number.
Taking the natural logarithm on both sides:
$\ln \left(\frac{R}{R_{0}}\right) = \ln (A^{1/3})$
Using the logarithmic property $\ln(x^n) = n \ln(x)$,we get:
$\ln \left(\frac{R}{R_{0}}\right) = \frac{1}{3} \ln A$
This equation is of the form $y = mx$,which represents a straight line passing through the origin with a slope of $1/3$.
Therefore,the graph of $\ln \left(\frac{R}{R_{0}}\right)$ versus $\ln A$ is a straight line.

(A) The circuit consists of two $NOR$ gates whose inputs are shorted,followed by a $NOR$ gate.
$1$. The first two gates are $NOR$ gates with inputs $A$ and $A$,and $B$ and $B$ respectively. Their outputs are $\overline{A+A} = \overline{A}$ and $\overline{B+B} = \overline{B}$.
$2$. These outputs are fed into a final $NOR$ gate.
$3$. The output $Y$ is given by $Y = \overline{\overline{A} + \overline{B}}$.
$4$. Using De Morgan's Law,$\overline{\overline{A} + \overline{B}} = \overline{\overline{A}} \cdot \overline{\overline{B}} = A \cdot B$.
$5$. The expression $Y = A \cdot B$ represents an $AND$ gate.

(B) The correct matching is as follows:
$A.$ Facsimile is a method used to transmit a $Static\, Document\, Image$.
$B.$ Guided media channel refers to physical paths like $Optical\, Fiber$ through which signals are transmitted.
$C.$ Frequency Modulation is commonly used in $Local\, Broadcast\, Radio$.
$D.$ Digital Signal is represented by a $Rectangular\, wave$.
Therefore,the correct sequence is $A-I, B-IV, C-II, D-III$.