AIPMT 2015 Physics Question Paper with Answer and Solution

(A) The gravitational force exerted by the Earth on the satellite always acts towards the centre of the Earth. According to Newton's second law,$F = ma$,the acceleration of the satellite is always directed towards the centre of the Earth.
Since the gravitational force is a central force,the torque acting on the satellite about the centre of the Earth is zero. Therefore,the angular momentum $L$ of the satellite remains constant in both magnitude and direction.
According to the law of conservation of energy,in the absence of non-conservative forces,the total mechanical energy of the satellite remains constant throughout its orbit.
Since the distance $r$ of the satellite from the Earth varies in an elliptical orbit,the orbital velocity $v$ must change to conserve angular momentum $(L = mvr \sin \theta)$. Consequently,the linear momentum $p = mv$ is not constant.

(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 $\gamma = \frac{C_P}{C_V} = \frac{R(1 + f/2)}{(f/2)R} = \frac{1 + f/2}{f/2} = \frac{2}{f} + 1 = 1 + \frac{2}{f}$.

(C) Let the dimensional formula of surface tension $(S)$ be $S = E^x V^y T^z$.
The dimensions of surface tension are $[S] = [MT^{-2}]$.
The dimensions of the fundamental quantities are:
$[E] = [ML^2T^{-2}]$
$[V] = [LT^{-1}]$
$[T] = [T]$
Substituting these into the equation:
$[MT^{-2}] = [ML^2T^{-2}]^x [LT^{-1}]^y [T]^z$
$[MT^{-2}] = [M^x L^{2x+y} T^{-2x-y+z}]$
Comparing the powers of $M, L,$ and $T$ on both sides:
For $M$: $x = 1$
For $L$: $2x + y = 0 \Rightarrow 2(1) + y = 0 \Rightarrow y = -2$
For $T$: $-2x - y + z = -2 \Rightarrow -2(1) - (-2) + z = -2 \Rightarrow -2 + 2 + z = -2 \Rightarrow z = -2$
Thus,the dimensional formula is $[EV^{-2}T^{-2}]$.

(B) Given the dimensional relation: $[v_c] = [\eta^x \rho^y r^z]$ $(i)$
Writing the dimensions of the physical quantities:
$[v_c] = [M^0 L T^{-1}]$
$[\eta] = [M L^{-1} T^{-1}]$
$[\rho] = [M L^{-3} T^0]$
$[r] = [M^0 L T^0]$
Substituting these into equation $(i)$:
$[M^0 L T^{-1}] = [M L^{-1} T^{-1}]^x [M L^{-3} T^0]^y [M^0 L T^0]^z$
$[M^0 L T^{-1}] = [M^{x+y} L^{-x-3y+z} T^{-x}]$
Applying the principle of homogeneity of dimensions,we equate the powers of $M, L,$ and $T$ on both sides:
$x + y = 0$ $(ii)$
$-x - 3y + z = 1$ $(iii)$
$-x = -1$ $(iv)$
From $(iv)$,we get $x = 1$.
Substituting $x = 1$ into $(ii)$,we get $1 + y = 0$,so $y = -1$.
Substituting $x = 1$ and $y = -1$ into $(iii)$,we get $-1 - 3(-1) + z = 1 \implies -1 + 3 + z = 1 \implies 2 + z = 1 \implies z = -1$.
Thus,the values are $x = 1, y = -1, z = -1$.

(B) The velocity of the particle is given as $v(x) = \beta x^{-2n}$.
To find the acceleration $a$,we use the relation $a = v \frac{dv}{dx}$.
First,differentiate $v$ with respect to $x$:
$\frac{dv}{dx} = \frac{d}{dx} (\beta x^{-2n}) = \beta (-2n) x^{-2n-1} = -2n \beta x^{-2n-1}$.
Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula:
$a = (\beta x^{-2n}) \times (-2n \beta x^{-2n-1})$.
$a = -2n \beta^2 x^{-2n + (-2n) - 1}$.
$a = -2n \beta^2 x^{-4n-1}$.

(B) Let the initial position of ship $A$ be at the origin $(0, 0)$ moving along the negative $x$-axis. Its position at time $t$ is $\vec{r}_A = (-10t, 0)$.
Ship $B$ is initially at $(0, -100)$ moving along the positive $y$-axis. Its position at time $t$ is $\vec{r}_B = (0, -100 + 10t)$.
The relative position vector is $\vec{r}_{BA} = \vec{r}_B - \vec{r}_A = (10t, -100 + 10t)$.
The square of the distance $D^2 = (10t)^2 + (-100 + 10t)^2 = 100t^2 + 10000 - 2000t + 100t^2 = 200t^2 - 2000t + 10000$.
To find the shortest distance,differentiate $D^2$ with respect to $t$ and set it to zero:
$\frac{d(D^2)}{dt} = 400t - 2000 = 0$.
$400t = 2000 \Rightarrow t = 5 \, hr$.
Alternatively,using relative velocity: $\vec{v}_{BA} = \vec{v}_B - \vec{v}_A = (0, 10) - (-10, 0) = (10, 10) \, km \, h^{-1}$.
The magnitude of relative velocity is $|\vec{v}_{BA}| = \sqrt{10^2 + 10^2} = 10\sqrt{2} \, km \, h^{-1}$.
The shortest distance occurs when the relative position vector is perpendicular to the relative velocity vector. From the geometry,the time taken is $t = 5 \, hr$.

(D) Given position vector: $\vec{R} = 4\sin(2\pi t)\hat{i} + 4\cos(2\pi t)\hat{j}$.
Since $x = 4\sin(2\pi t)$ and $y = 4\cos(2\pi t)$,then $x^2 + y^2 = 16(\sin^2(2\pi t) + \cos^2(2\pi t)) = 16$. This represents a circle of radius $4 \ m$.
Velocity $\vec{v} = \frac{d\vec{R}}{dt} = 8\pi\cos(2\pi t)\hat{i} - 8\pi\sin(2\pi t)\hat{j}$.
Magnitude of velocity $|\vec{v}| = \sqrt{(8\pi\cos(2\pi t))^2 + (-8\pi\sin(2\pi t))^2} = 8\pi \ m/s$.
Acceleration $\vec{a} = \frac{d\vec{v}}{dt} = -16\pi^2\sin(2\pi t)\hat{i} - 16\pi^2\cos(2\pi t)\hat{j} = -4\pi^2\vec{R}$.
Since $\vec{a} = -4\pi^2\vec{R}$,the acceleration is directed towards the origin (along $-\vec{R}$).
For uniform circular motion,$a = \frac{V^2}{R}$. Here $V = 8\pi$ and $R = 4$,so $a = \frac{(8\pi)^2}{4} = 16\pi^2$. This matches the magnitude of $\vec{a}$.
Statement $(D)$ claims the magnitude of velocity is $8 \ m/s$,but it is $8\pi \ m/s$. Thus,$(D)$ is the wrong statement.

(D) Two vectors $\overrightarrow {A}$ and $\overrightarrow {B}$ are orthogonal to each other if their scalar product is zero,i.e.,$\overrightarrow {A} \cdot \overrightarrow {B} = 0$.
Given $\overrightarrow {A} = \cos \omega t \hat i + \sin \omega t \hat j$ and $\overrightarrow {B} = \cos \frac{\omega t}{2} \hat i + \sin \frac{\omega t}{2} \hat j$.
Calculating the dot product:
$\overrightarrow {A} \cdot \overrightarrow {B} = (\cos \omega t \hat i + \sin \omega t \hat j) \cdot (\cos \frac{\omega t}{2} \hat i + \sin \frac{\omega t}{2} \hat j)$
$= \cos \omega t \cos \frac{\omega t}{2} + \sin \omega t \sin \frac{\omega t}{2}$
Using the trigonometric identity $\cos(A - B) = \cos A \cos B + \sin A \sin B$:
$\overrightarrow {A} \cdot \overrightarrow {B} = \cos(\omega t - \frac{\omega t}{2}) = \cos(\frac{\omega t}{2})$
Since the vectors are orthogonal,$\overrightarrow {A} \cdot \overrightarrow {B} = 0$,so:
$\cos(\frac{\omega t}{2}) = 0$
We know $\cos(\frac{\pi}{2}) = 0$,therefore:
$\frac{\omega t}{2} = \frac{\pi}{2}$
$t = \frac{\pi}{\omega}$.

(B) Given: Masses $M_A = 4 \, kg, M_B = 2 \, kg, M_C = 1 \, kg$ and applied force $F = 14 \, N$.
First,calculate the acceleration $(a)$ of the entire system:
$a = \frac{F}{M_A + M_B + M_C} = \frac{14}{4 + 2 + 1} = \frac{14}{7} = 2 \, m/s^2$.
The contact force between $A$ and $B$ (let it be $F_{AB}$) is the force required to accelerate blocks $B$ and $C$ together.
$F_{AB} = (M_B + M_C) \times a$
$F_{AB} = (2 + 1) \times 2 = 3 \times 2 = 6 \, N$.
Alternatively,considering the free body diagram of block $A$:
$F - F_{AB} = M_A \times a$
$14 - F_{AB} = 4 \times 2$
$14 - F_{AB} = 8$
$F_{AB} = 14 - 8 = 6 \, N$.

(A) For block $B$ (mass $m_2$) moving downwards with acceleration $a$: $m_2 g - T = m_2 a$ (Equation $1$)
For block $A$ (mass $m_1$) moving horizontally with acceleration $a$: $T - \mu_k m_1 g = m_1 a$ (Equation $2$)
Adding Equation $1$ and Equation $2$:
$(m_2 g - T) + (T - \mu_k m_1 g) = m_2 a + m_1 a$
$m_2 g - \mu_k m_1 g = (m_1 + m_2) a$
$a = \frac{(m_2 - \mu_k m_1) g}{m_1 + m_2}$
Now,substitute the value of $a$ into Equation $1$:
$T = m_2 g - m_2 a = m_2 (g - a)$
$T = m_2 \left( g - \frac{(m_2 - \mu_k m_1) g}{m_1 + m_2} \right)$
$T = m_2 g \left( \frac{m_1 + m_2 - m_2 + \mu_k m_1}{m_1 + m_2} \right)$
$T = \frac{m_1 m_2 (1 + \mu_k) g}{m_1 + m_2}$

(C) Let $\mu_s$ and $\mu_k$ be the coefficients of static and kinetic friction between the box and the plank,respectively.
When the angle of inclination $\theta$ reaches $30^o$,the block just starts to slide. Therefore,$\mu_s = \tan\theta = \tan 30^o = \frac{1}{\sqrt{3}} \approx 0.577 \approx 0.6$.
If $a$ is the acceleration produced in the block,then the equation of motion is:
$ma = mg\sin\theta - f_k$
$ma = mg\sin\theta - \mu_k N$
Since $N = mg\cos\theta$,we have:
$a = g(\sin\theta - \mu_k\cos\theta)$
Given $g = 10\, m/s^2$,$\theta = 30^o$,$s = 4.0\, m$,and $t = 4.0\, s$,we use the kinematic equation $s = ut + \frac{1}{2}at^2$ (with $u = 0$):
$4.0 = 0 + \frac{1}{2} a (4.0)^2$
$4.0 = 8a \implies a = 0.5\, m/s^2$.
Substituting $a$ into the acceleration equation:
$0.5 = 10(\sin 30^o - \mu_k \cos 30^o)$
$0.5 = 10(0.5 - \mu_k \frac{\sqrt{3}}{2})$
$0.05 = 0.5 - \mu_k \frac{\sqrt{3}}{2}$
$\mu_k \frac{\sqrt{3}}{2} = 0.45$
$\mu_k = \frac{0.9}{\sqrt{3}} = \frac{0.9}{1.732} \approx 0.519 \approx 0.5$.
Thus,$\mu_s \approx 0.6$ and $\mu_k \approx 0.5$.

(B) Let $v$ be the tangential speed of the heavier stone.
Given that the centripetal force experienced by both stones is the same.
The centripetal force for the lighter stone (mass $m$,radius $r$,speed $nv$) is given by:
$F_{lighter} = \frac{m(nv)^2}{r} = \frac{mn^2v^2}{r}$
The centripetal force for the heavier stone (mass $2m$,radius $r/2$,speed $v$) is given by:
$F_{heavier} = \frac{(2m)v^2}{r/2} = \frac{4mv^2}{r}$
Equating the two forces:
$\frac{mn^2v^2}{r} = \frac{4mv^2}{r}$
$n^2 = 4$
$n = 2$

(C) Case $(a)$: When stretched by the same amount $x$,the work done is $W = \frac{1}{2} K x^2$. Since $K_P > K_Q$ and $x$ is the same,$W_P = \frac{1}{2} K_P x^2$ and $W_Q = \frac{1}{2} K_Q x^2$. Therefore,$W_P > W_Q$.
Case $(b)$: When stretched by the same force $F$,the work done is $W = \frac{F^2}{2K}$. Since $K_P > K_Q$ and $F$ is the same,$W_P = \frac{F^2}{2K_P}$ and $W_Q = \frac{F^2}{2K_Q}$. Since $K_P > K_Q$,it follows that $\frac{1}{K_P} < \frac{1}{K_Q}$,therefore $W_P < W_Q$ or $W_Q > W_P$.

(A) Given: Mass $m = 10 \, kg$,initial velocity $v_i = 10 \, m/s$.
Initial kinetic energy $K_i = \frac{1}{2} m v_i^2 = \frac{1}{2} \times 10 \times (10)^2 = 500 \, J$.
The retarding force is $F = -0.1 \, x \, N$.
Work done by the retarding force $W = \int_{20}^{30} (-0.1 \, x) \, dx$.
$W = -0.1 \left[ \frac{x^2}{2} \right]_{20}^{30} = -0.1 \times \frac{1}{2} (30^2 - 20^2) = -0.05 \times (900 - 400) = -0.05 \times 500 = -25 \, J$.
According to the work-energy theorem,$W = K_f - K_i$.
$K_f = W + K_i = -25 \, J + 500 \, J = 475 \, J$.

(A) The power $P$ delivered to the particle is constant,so $P = k$.
Since $P = \frac{dW}{dt}$,we have $dW = k dt$.
Integrating both sides from $0$ to $t$,we get $W = kt$.
According to the work-energy theorem,the work done is equal to the change in kinetic energy:
$W = \Delta K = \frac{1}{2}mv^2 - 0$.
Equating the two expressions for $W$: $kt = \frac{1}{2}mv^2$,which gives $v = \sqrt{\frac{2kt}{m}}$.
The acceleration $a$ is the derivative of velocity with respect to time:
$a = \frac{dv}{dt} = \frac{d}{dt} \left( \sqrt{\frac{2k}{m}} t^{1/2} \right) = \sqrt{\frac{2k}{m}} \cdot \frac{1}{2} t^{-1/2} = \sqrt{\frac{k}{2mt}}$.
The force $F$ acting on the particle is $F = ma$:
$F = m \cdot \sqrt{\frac{k}{2mt}} = \sqrt{\frac{m^2 k}{2mt}} = \sqrt{\frac{mk}{2t}} = \sqrt{\frac{mk}{2}} t^{-1/2}$.

(B) The total initial kinetic energy of the two particles is given by $K_i = \frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2$.
During the collision,an amount of energy $\varepsilon$ is absorbed by one of the particles to reach an excited state.
Therefore,the total final kinetic energy of the particles $K_f = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$ must be less than the initial kinetic energy by the amount $\varepsilon$.
According to the law of conservation of energy: $K_i = K_f + \varepsilon$.
Substituting the expressions: $\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \varepsilon$.
Rearranging the terms,we get: $\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 - \varepsilon = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$.

(A) Let the particles $A$ and $B$ collide at time $t$. For their collision,the position vectors of both particles must be the same at time $t$,i.e.,
$\vec{r}_1 + \vec{v}_1 t = \vec{r}_2 + \vec{v}_2 t$
$\vec{r}_1 - \vec{r}_2 = (\vec{v}_2 - \vec{v}_1) t$ ... $(i)$
Taking the magnitude of both sides:
$|\vec{r}_1 - \vec{r}_2| = |\vec{v}_2 - \vec{v}_1| t$
$t = \frac{|\vec{r}_1 - \vec{r}_2|}{|\vec{v}_2 - \vec{v}_1|}$
Substituting this value of $t$ into equation $(i)$:
$\vec{r}_1 - \vec{r}_2 = (\vec{v}_2 - \vec{v}_1) \frac{|\vec{r}_1 - \vec{r}_2|}{|\vec{v}_2 - \vec{v}_1|}$
Rearranging the terms,we get:
$\frac{\vec{r}_1 - \vec{r}_2}{|\vec{r}_1 - \vec{r}_2|} = \frac{\vec{v}_2 - \vec{v}_1}{|\vec{v}_2 - \vec{v}_1|}$

(D) Let $m$ be the mass of the ball,$h = 20 \, m$ be the height,and $v$ be the velocity just before hitting the ground.
Using the law of conservation of energy for the downward motion:
$\frac{1}{2}mv^2 = \frac{1}{2}mv_0^2 + mgh$
$v^2 = v_0^2 + 2gh \quad ... (i)$
When the ball hits the ground,it loses $50\%$ of its kinetic energy. The remaining kinetic energy is $\frac{1}{2} \times (\frac{1}{2}mv^2) = \frac{1}{4}mv^2$.
This remaining energy allows the ball to rebound to the same height $h$:
$\frac{1}{4}mv^2 = mgh$
$v^2 = 4gh$
Substitute $v^2 = 4gh$ into equation $(i)$:
$4gh = v_0^2 + 2gh$
$v_0^2 = 2gh$
$v_0 = \sqrt{2gh}$
Given $g = 10 \, m/s^2$ and $h = 20 \, m$:
$v_0 = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20 \, m/s$.

(A) The collision is elastic,which means that both linear momentum and kinetic energy are conserved.
Let $v'$ be the speed of the second block after the collision.
According to the law of conservation of kinetic energy:
$\frac{1}{2}M V^2 + 0 = \frac{1}{2}M \left( \frac{V}{3} \right)^2 + \frac{1}{2}M (v')^2$
Dividing both sides by $\frac{1}{2}M$:
$V^2 = \frac{V^2}{9} + (v')^2$
$(v')^2 = V^2 - \frac{V^2}{9}$
$(v')^2 = \frac{8V^2}{9}$
$v' = \sqrt{\frac{8V^2}{9}} = \frac{2\sqrt{2}}{3}V$

(D) Given:
Volume of blood pumped, $V = 5 \, \text{litres} = 5 \times 10^{-3} \, \text{m}^3$.
Time, $t = 1 \, \text{min} = 60 \, \text{s}$.
Pressure, $P = 150 \, \text{mm of Hg} = 0.15 \, \text{m of Hg}$.
Density of mercury, $\rho = 13.6 \times 10^3 \, \text{kg/m}^3$.
Acceleration due to gravity, $g = 10 \, \text{m/s}^2$.
First, calculate the pressure in $\text{N/m}^2$ using $P = h \rho g$:
$P = (0.15 \, \text{m}) \times (13.6 \times 10^3 \, \text{kg/m}^3) \times (10 \, \text{m/s}^2)$
$P = 20.4 \times 10^3 \, \text{N/m}^2$.
Now, calculate the power using $Power = \frac{P \times V}{t}$:
$Power = \frac{(20.4 \times 10^3 \, \text{N/m}^2) \times (5 \times 10^{-3} \, \text{m}^3)}{60 \, \text{s}}$
$Power = \frac{20.4 \times 5}{60} \, \text{W}$
$Power = \frac{102}{60} \, \text{W} = 1.7 \, \text{W}$.

(A) According to $Wien's$ displacement law, the product of the wavelength corresponding to maximum intensity $(\lambda_m)$ and the absolute temperature $(T)$ is constant:
$\lambda_m T = \text{constant} \implies T \propto \frac{1}{\lambda_m}$
For star $P$, the intensity of violet colour is maximum. Since violet has the shortest wavelength among the three $(\lambda_V < \lambda_G < \lambda_R)$, star $P$ must have the highest temperature.
For star $R$, the intensity of green colour is maximum. Green light has a wavelength between violet and red $(\lambda_V < \lambda_G < \lambda_R)$.
For star $Q$, the intensity of red colour is maximum. Since red has the longest wavelength among the three, star $Q$ must have the lowest temperature.
Comparing the wavelengths: $\lambda_V < \lambda_G < \lambda_R$.
Therefore, the temperatures follow the inverse order: $T_P > T_R > T_Q$.

(D) The rate of heat flow $(H)$ through a metal rod is given by the formula $H = \frac{dQ}{dt} = \frac{kA(T_2 - T_1)}{L}$,where $k$ is the thermal conductivity,$A$ is the cross-sectional area,$L$ is the length,and $(T_2 - T_1)$ is the temperature difference.
In the first case,the temperature difference is $\Delta T_1 = 110 ^\circ C - 100 ^\circ C = 10 ^\circ C$.
The rate of heat flow is $H_1 = 4.0 \ J/s$.
In the second case,the temperature difference is $\Delta T_2 = 210 ^\circ C - 200 ^\circ C = 10 ^\circ C$.
Since the rate of heat flow is directly proportional to the temperature difference $(H \propto \Delta T)$,and the temperature difference remains the same ($10 ^\circ C$ in both cases),the rate of heat flow will remain unchanged.
Therefore,$H_2 = H_1 = 4.0 \ J/s$.

(C) According to the ideal gas equation,the molecular weight $M$ of an ideal gas is given by:
$M = \frac{\rho R T}{P}$
where $P$ is the pressure,$T$ is the temperature,$\rho$ is the density,and $R$ is the universal gas constant.
For gases $A$ and $B$:
$M_A = \frac{\rho_A R T_A}{P_A}$ and $M_B = \frac{\rho_B R T_B}{P_B}$
The ratio of their molecular weights is:
$\frac{M_A}{M_B} = \left( \frac{\rho_A}{\rho_B} \right) \left( \frac{T_A}{T_B} \right) \left( \frac{P_B}{P_A} \right)$
Given:
$\frac{\rho_A}{\rho_B} = 1.5 = \frac{3}{2}$
$T_A = T_B \implies \frac{T_A}{T_B} = 1$
$P_A = 2 P_B \implies \frac{P_B}{P_A} = \frac{1}{2}$
Substituting these values:
$\frac{M_A}{M_B} = \left( \frac{3}{2} \right) \times (1) \times \left( \frac{1}{2} \right) = \frac{3}{4} = 0.75$

(C) Since the tension force acts towards the center of the circle,the torque about the center is zero. Therefore,the angular momentum of the mass $m$ is conserved.
Initial angular momentum $L_i = m v_0 R_0$.
Final angular momentum $L_f = m v R$,where $R = \frac{R_0}{2}$.
By conservation of angular momentum,$L_i = L_f$:
$m v_0 R_0 = m v \left( \frac{R_0}{2} \right)$
$v = 2 v_0$.
The initial kinetic energy is $K_i = \frac{1}{2} m v_0^2$.
The final kinetic energy is $K_f = \frac{1}{2} m v^2 = \frac{1}{2} m (2 v_0)^2 = \frac{1}{2} m (4 v_0^2) = 2 m v_0^2$.

(A) Let $N_1$ be the normal reaction on $A$ and $N_2$ be the normal reaction on $B$.
For vertical equilibrium,the sum of upward forces must equal the downward force:
$N_1 + N_2 = W$ --- $(i)$
For rotational equilibrium,we take the torque about the centre of mass of the rod (where the weight $W$ acts). The torque due to $W$ is zero.
Sum of torques about the centre of mass = $0$
$N_1 \cdot x = N_2 \cdot (d - x)$ --- $(ii)$
From equation $(i)$,$N_2 = W - N_1$.
Substitute this into equation $(ii)$:
$N_1 x = (W - N_1)(d - x)$
$N_1 x = W(d - x) - N_1(d - x)$
$N_1 x = W(d - x) - N_1 d + N_1 x$
$N_1 d = W(d - x)$
$N_1 = \frac{W(d - x)}{d}$

(A) The moment of inertia $I$ of the system about an axis passing through point $P$ at a distance $x$ from $m_1$ is given by:
$I = m_1 x^2 + m_2 (L - x)^2$
According to the work-energy theorem,the work done $W$ to set the rod rotating with angular velocity $\omega_0$ is equal to the rotational kinetic energy:
$W = \frac{1}{2} I \omega_0^2 = \frac{1}{2} [m_1 x^2 + m_2 (L - x)^2] \omega_0^2$
For the work $W$ to be minimum,we set the derivative with respect to $x$ to zero:
$\frac{dW}{dx} = \frac{1}{2} \omega_0^2 [2 m_1 x + 2 m_2 (L - x)(-1)] = 0$
$m_1 x - m_2 (L - x) = 0$
$m_1 x - m_2 L + m_2 x = 0$
$(m_1 + m_2) x = m_2 L$
$x = \frac{m_2 L}{m_1 + m_2}$

(B) For the conservation of angular momentum about the origin,the net torque $\overrightarrow{\tau}$ acting on the particle must be zero.
By definition,$\overrightarrow{\tau} = \overrightarrow{R} \times \overrightarrow{F}$.
Given $\overrightarrow{R} = 2\hat{i} - 6\hat{j} - 12\hat{k}$ and $\overrightarrow{F} = \alpha \hat{i} + 3\hat{j} + 6\hat{k}$.
Calculating the cross product:
$\overrightarrow{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -6 & -12 \\ \alpha & 3 & 6 \end{vmatrix}$
$\overrightarrow{\tau} = \hat{i}(-36 - (-36)) - \hat{j}(12 - (-12\alpha)) + \hat{k}(6 - (-6\alpha))$
$\overrightarrow{\tau} = \hat{i}(0) - \hat{j}(12 + 12\alpha) + \hat{k}(6 + 6\alpha)$
For $\overrightarrow{\tau} = 0$,both the $\hat{j}$ and $\hat{k}$ components must be zero:
$12 + 12\alpha = 0 \implies \alpha = -1$
$6 + 6\alpha = 0 \implies \alpha = -1$
Thus,the value of $\alpha$ is $-1$.

(B) Given:
Speed of the automobile,$v = 54 \,km h^{-1} = 54 \times \frac{5}{18} \,m s^{-1} = 15 \,m s^{-1}$.
Radius of the wheel,$R = 0.45 \,m$.
Moment of inertia of the wheel,$I = 3 \,kg m^2$.
Time taken to stop,$t = 15 \,s$.
The initial angular speed of the wheel is $\omega_i = \frac{v}{R} = \frac{15 \,m s^{-1}}{0.45 \,m} = \frac{1500}{45} \,rad s^{-1} = \frac{100}{3} \,rad s^{-1}$.
The final angular speed is $\omega_f = 0$ (as the vehicle comes to rest).
The angular retardation of the wheel is $\alpha = \frac{\omega_f - \omega_i}{t} = \frac{0 - \frac{100}{3}}{15} = -\frac{100}{45} \,rad s^{-2}$.
The magnitude of the average torque is $\tau = I |\alpha| = 3 \,kg m^2 \times \frac{100}{45} \,rad s^{-2} = \frac{300}{45} \,N m = \frac{20}{3} \,N m \approx 6.66 \,kg m^2 s^{-2}$.

(D) The moment of inertia of a thin spherical shell of mass $m$ and radius $r$ about its diameter is $I_{diam} = \frac{2}{3} mr^2$.
For the third shell,the axis $XX'$ passes through its diameter. Therefore,its moment of inertia is $I_3 = \frac{2}{3} mr^2$.
For the other two shells,the axis $XX'$ is tangent to them. According to the parallel axis theorem,the moment of inertia about a tangent is $I_{tangent} = I_{cm} + md^2$,where $d = r$. Thus,$I_1 = I_2 = \frac{2}{3} mr^2 + mr^2 = \frac{5}{3} mr^2$.
The total moment of inertia of the system about the $XX'$ axis is $I = I_1 + I_2 + I_3 = \frac{5}{3} mr^2 + \frac{5}{3} mr^2 + \frac{2}{3} mr^2 = \frac{12}{3} mr^2 = 4 mr^2$.

(B) The orbital speed of a satellite is given by the formula $v_0 = \sqrt{\frac{GM}{R+h}}$.
Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this into the orbital speed formula,we get $v_0 = \sqrt{\frac{gR^2}{R+h}} = R \sqrt{\frac{g}{R+h}}$.
Given values: $R = 6.38 \times 10^6 \, m$,$h = 0.25 \times 10^6 \, m$,and $g = 9.8 \, m s^{-2}$.
Total orbital radius $r = R + h = (6.38 + 0.25) \times 10^6 \, m = 6.63 \times 10^6 \, m$.
Now,$v_0 = \sqrt{\frac{9.8 \times (6.38 \times 10^6)^2}{6.63 \times 10^6}} = \sqrt{\frac{9.8 \times 40.7044 \times 10^{12}}{6.63 \times 10^6}} = \sqrt{60.16 \times 10^6} \approx 7.756 \times 10^3 \, m s^{-1}$.
Thus,$v_0 \approx 7.76 \, km s^{-1}$.

(B) The gravitational force of attraction between the sun and the planet provides the necessary centripetal force for the planet's orbit.
$\therefore \frac{GMm}{r^2} = \frac{mv^2}{r} \implies v^2 = \frac{GM}{r} \quad \dots(i)$
The time period $(T)$ of the planet is given by $T = \frac{2\pi r}{v}$.
Squaring both sides,we get $T^2 = \frac{4\pi^2 r^2}{v^2}$.
Substituting the value of $v^2$ from equation $(i)$:
$T^2 = \frac{4\pi^2 r^2}{(\frac{GM}{r})} = \frac{4\pi^2 r^3}{GM} \quad \dots(ii)$
According to the problem,Kepler's third law is given as:
$T^2 = Kr^3 \quad \dots(iii)$
Comparing equations $(ii)$ and $(iii)$,we get:
$K = \frac{4\pi^2}{GM} \implies GMK = 4\pi^2$.

(C) The initial distance between the centres is $12R.$ Collision occurs when the distance between the centres is equal to the sum of their radii,which is $R + 2R = 3R.$
Therefore,the total distance covered by both bodies before collision is $12R - 3R = 9R.$
Let $x_1$ and $x_2$ be the distances covered by the bodies of mass $M$ and $5M$ respectively. Since there is no external force,the centre of mass of the system remains stationary. Thus,$M x_1 = 5M x_2$,which gives $x_1 = 5x_2$.
Given $x_1 + x_2 = 9R$,we substitute $x_1 = 5x_2$ to get $5x_2 + x_2 = 9R$,so $6x_2 = 9R$,which means $x_2 = 1.5R$.
Then,$x_1 = 5(1.5R) = 7.5R$.
The distance covered by the smaller body (mass $M$) is $7.5R$.

(A) Let $L$ and $A$ be the length and area of cross-section of each wire respectively.
In order to have the lower ends of the wires at the same level,the elongation produced in both wires must be equal.
Let $W_s$ and $W_b$ be the weights added to the steel and brass wires respectively.
By the definition of Young's modulus $Y = \frac{W/A}{\Delta L/L}$,the elongation produced in the steel wire is $\Delta L_s = \frac{W_s L}{Y_s A}$.
The elongation produced in the brass wire is $\Delta L_b = \frac{W_b L}{Y_b A}$.
Since $\Delta L_s = \Delta L_b$,we have $\frac{W_s L}{Y_s A} = \frac{W_b L}{Y_b A}$.
This simplifies to $\frac{W_s}{W_b} = \frac{Y_s}{Y_b}$.
Given that the Young's modulus of steel is twice that of brass,$Y_s = 2 Y_b$,so $\frac{Y_s}{Y_b} = 2$.
Therefore,$\frac{W_s}{W_b} = 2$,which means the ratio is $2:1$.

(C) Let $\rho_0$ and $\rho_T$ be the densities of glycerin at initial temperature and final temperature respectively.
The relationship between density and temperature is given by $\rho_T = \rho_0(1 - \gamma \Delta T)$,where $\gamma$ is the coefficient of volume expansion and $\Delta T$ is the change in temperature.
Rearranging the equation,we get $\frac{\rho_T}{\rho_0} = 1 - \gamma \Delta T$.
The fractional change in density is defined as $\frac{\rho_0 - \rho_T}{\rho_0}$.
From the equation,$\frac{\rho_0 - \rho_T}{\rho_0} = \gamma \Delta T$.
Given $\gamma = 5 \times 10^{-4} \ K^{-1}$ and $\Delta T = 40 \ K$.
Substituting the values,the fractional change in density = $(5 \times 10^{-4} \ K^{-1}) \times (40 \ K) = 200 \times 10^{-4} = 0.02$.

(C) According to the equation of continuity,the volume flow rate of the liquid remains constant throughout the tube.
The volume flow rate inside the cylindrical tube is given by $A_1 v_1 = \pi R^2 v$.
The total volume flow rate through the $n$ fine holes is given by $A_2 v_2 = n (\pi r^2) v_{ejection}$.
Equating the two flow rates:
$\pi R^2 v = n \pi r^2 v_{ejection}$.
Solving for the speed of ejection $(v_{ejection})$:
$v_{ejection} = \frac{\pi R^2 v}{n \pi r^2} = \frac{v R^2}{n r^2}$.

(C) The height to which water rises in a capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g}$.
If the actual length of the tube $L$ is less than the equilibrium height $h$,the water will rise to the top of the tube.
At the top,the radius of curvature $R$ of the meniscus will adjust such that $h' = \frac{2T \cos \theta'}{R \rho g} = L$,where $h' < h$.
Since the pressure at the top remains balanced by the surface tension,the water will not overflow but will stay at the top of the tube with a modified radius of curvature.

(C) Given:
Depth of ocean,$d = 2700 \, m$
Density of water,$\rho = 10^3 \, kg/m^3$
Compressibility of water,$K = 45.4 \times 10^{-11} \, Pa^{-1}$
Acceleration due to gravity,$g = 10 \, m/s^2$
The excess pressure at the bottom of the ocean is given by $\Delta P = \rho gd$.
$\Delta P = 10^3 \times 10 \times 2700 = 27 \times 10^6 \, Pa$.
Compressibility $K$ is defined as the reciprocal of the Bulk Modulus $B$,i.e.,$K = \frac{1}{B}$.
Since $B = \frac{\Delta P}{(\Delta V/V)}$,we have $\frac{\Delta V}{V} = K \times \Delta P$.
Substituting the values:
$\frac{\Delta V}{V} = (45.4 \times 10^{-11} \, Pa^{-1}) \times (27 \times 10^6 \, Pa)$
$\frac{\Delta V}{V} = 1225.8 \times 10^{-5} = 1.2258 \times 10^{-2} \approx 1.2 \times 10^{-2}$.

(C) Applying Bernoulli's theorem just above and just below the roof:
$P + \frac{1}{2}\rho v^2 = P_0 + 0$
Here,$P$ is the pressure just above the roof,$P_0$ is the atmospheric pressure inside the house,and $v = 40 \ m/s$ is the wind speed.
The pressure difference is $\Delta P = P_0 - P = \frac{1}{2}\rho v^2$.
The force exerted on the roof is $F = \Delta P \cdot A = \frac{1}{2}\rho A v^2$.
Substituting the given values:
$F = \frac{1}{2} \times 1.2 \ kg/m^3 \times 250 \ m^2 \times (40 \ m/s)^2$
$F = 0.6 \times 250 \times 1600 = 2.4 \times 10^5 \ N$.
Since the pressure inside the house $(P_0)$ is greater than the pressure just above the roof $(P)$,the net force acts in the upward direction.

(C) Since internal energy is a state function,the change in internal energy depends only on the initial and final states. Therefore,$\Delta U_{ABC} = \Delta U_{AC}$.
For process $AB$ (isochoric,$\Delta V = 0$):
$\Delta W_{AB} = 0$
$\Delta Q_{AB} = \Delta U_{AB} = 400 \, J$.
For process $BC$ (isobaric,$P = 6 \times 10^4 \, Pa$):
$\Delta W_{BC} = P \Delta V = 6 \times 10^4 \times (4 \times 10^{-3} - 2 \times 10^{-3}) = 6 \times 10^4 \times 2 \times 10^{-3} = 120 \, J$.
$\Delta Q_{BC} = \Delta U_{BC} + \Delta W_{BC} \implies 100 = \Delta U_{BC} + 120 \implies \Delta U_{BC} = -20 \, J$.
Total change in internal energy for path $ABC$ is:
$\Delta U_{AC} = \Delta U_{AB} + \Delta U_{BC} = 400 - 20 = 380 \, J$.
For process $AC$,the work done $\Delta W_{AC}$ is the area under the line $AC$ in the $PV$ diagram,which is a trapezoid:
$\Delta W_{AC} = \text{Area} = \frac{1}{2} \times (P_A + P_C) \times (V_C - V_A) = \frac{1}{2} \times (2 \times 10^4 + 6 \times 10^4) \times (4 \times 10^{-3} - 2 \times 10^{-3}) = \frac{1}{2} \times 8 \times 10^4 \times 2 \times 10^{-3} = 80 \, J$.
Using the first law of thermodynamics for process $AC$:
$\Delta Q_{AC} = \Delta U_{AC} + \Delta W_{AC} = 380 + 80 = 460 \, J$.

(B) We know that the change in internal energy for an ideal gas is given by $\Delta U = n C_v \Delta T$.
For a diatomic gas,the molar heat capacity at constant volume is $C_v = \frac{5R}{2}$.
Substituting this into the equation,we get $\Delta U = n \left( \frac{5R}{2} \right) (T_B - T_A)$.
Using the ideal gas law $PV = nRT$,we can write $T = \frac{PV}{nR}$,so $\Delta U = \frac{5}{2} (P_B V_B - P_A V_A)$.
From the given graph,at point $A$: $P_A = 5 \times 10^3 \, Pa$ and $V_A = 4 \, m^3$.
At point $B$: $P_B = 2 \times 10^3 \, Pa$ and $V_B = 6 \, m^3$.
Substituting these values:
$\Delta U = \frac{5}{2} [(2 \times 10^3 \times 6) - (5 \times 10^3 \times 4)]$
$\Delta U = \frac{5}{2} [12 \times 10^3 - 20 \times 10^3]$
$\Delta U = \frac{5}{2} [-8 \times 10^3]$
$\Delta U = -20 \times 10^3 \, J = -20 \, kJ$.

(C) The $P-V$ diagram of an ideal gas compressed from its initial volume $V_0$ to $\frac{V_0}{2}$ by several processes is shown in the figure.
Work done on the gas is equal to the area under the $P-V$ curve.
As the area under the $P-V$ curve is maximum for the adiabatic process,the work done on the gas is maximum for the adiabatic process.

(B) The coefficient of performance $(COP)$ of a refrigerator is given by the formula:
$\alpha = \frac{T_2}{T_1 - T_2}$
where $T_1$ is the temperature of the hot reservoir (surroundings) and $T_2$ is the temperature of the cold reservoir (freezer) in Kelvin.
Given: $\alpha = 5$,$T_2 = -20^{\circ}C = (-20 + 273) K = 253 K$.
Substituting the values into the formula:
$5 = \frac{253}{T_1 - 253}$
$5(T_1 - 253) = 253$
$5T_1 - 1265 = 253$
$5T_1 = 1518$
$T_1 = \frac{1518}{5} = 303.6 K$
Converting back to Celsius:
$T_1 = 303.6 - 273 = 30.6^{\circ}C \approx 31^{\circ}C$.

(A) Let $A$ be the amplitude and $\omega$ be the angular frequency of the simple harmonic motion.
The maximum acceleration is given by $\alpha = \omega^2 A$ $(i)$.
The maximum velocity is given by $\beta = \omega A$ $(ii)$.
Dividing equation $(i)$ by equation $(ii)$,we get:
$\frac{\alpha}{\beta} = \frac{\omega^2 A}{\omega A} = \omega$.
Therefore,the time period of vibration $T$ is given by:
$T = \frac{2\pi}{\omega} = \frac{2\pi}{(\alpha / \beta)} = \frac{2\pi \beta}{\alpha}$.

(C) Given the two displacements:
$y_1 = a \sin(\omega t)$
$y_2 = b \cos(\omega t) = b \sin(\omega t + \frac{\pi}{2})$
The resultant displacement $y = y_1 + y_2$ is given by:
$y = a \sin(\omega t) + b \cos(\omega t)$
To find the amplitude,we can write this in the form $y = A \sin(\omega t + \phi)$,where the resultant amplitude $A$ is:
$A = \sqrt{a^2 + b^2 + 2ab \cos(\frac{\pi}{2})}$
Since $\cos(\frac{\pi}{2}) = 0$,the amplitude is:
$A = \sqrt{a^2 + b^2}$
Thus,the resultant motion is simple harmonic with an amplitude of $\sqrt{a^2 + b^2}$.

(B) In $SHM$,the velocity $V$ of a particle at a distance $x$ from the mean position is given by $V = \omega \sqrt{a^2 - x^2}$,where $a$ is the amplitude and $\omega$ is the angular frequency.
Squaring both sides,we get $V^2 = \omega^2(a^2 - x^2)$.
For distances $x_1$ and $x_2$,we have:
$V_1^2 = \omega^2(a^2 - x_1^2) \dots (i)$
$V_2^2 = \omega^2(a^2 - x_2^2) \dots (ii)$
Subtracting equation $(ii)$ from $(i)$:
$V_1^2 - V_2^2 = \omega^2(a^2 - x_1^2 - a^2 + x_2^2)$
$V_1^2 - V_2^2 = \omega^2(x_2^2 - x_1^2)$
$\omega^2 = \frac{V_1^2 - V_2^2}{x_2^2 - x_1^2}$
Since the time period $T = \frac{2\pi}{\omega}$,we have $\omega = \frac{2\pi}{T}$.
Substituting this into the equation for $\omega^2$:
$\left(\frac{2\pi}{T}\right)^2 = \frac{V_1^2 - V_2^2}{x_2^2 - x_1^2}$
$T^2 = 4\pi^2 \left(\frac{x_2^2 - x_1^2}{V_1^2 - V_2^2}\right)$
$T = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{V_1^2 - V_2^2}}$

(A) Given:
Frequency of source,$f_{0} = 100 \, Hz$
Velocity of source,$v_{s} = 19.4 \, m s^{-1}$
Velocity of sound in air,$v = 330 \, m s^{-1}$
The component of the source velocity along the line joining the source and the observer is $v_{s} \cos 60^{\circ}$. Since the source is moving away from the observer,the apparent frequency $f'$ is given by the Doppler effect formula:
$f' = f_{0} \left( \frac{v}{v + v_{s} \cos 60^{\circ}} \right)$
Substituting the values:
$f' = 100 \left( \frac{330}{330 + 19.4 \times \cos 60^{\circ}} \right)$
$f' = 100 \left( \frac{330}{330 + 19.4 \times 0.5} \right)$
$f' = 100 \left( \frac{330}{330 + 9.7} \right)$
$f' = 100 \left( \frac{330}{339.7} \right) \approx 97.14 \, Hz$
Rounding to the nearest integer,the apparent frequency is $97 \, Hz$.

(A) For a string fixed at both ends,the resonant frequencies are given by $v_n = n \cdot v_0$,where $v_0$ is the fundamental frequency (lowest resonant frequency) and $n = 1, 2, 3, \dots$.
The difference between any two consecutive resonant frequencies is $\Delta v = v_{n+1} - v_n = (n+1)v_0 - nv_0 = v_0$.
Given that $420\, Hz$ and $315\, Hz$ are resonant frequencies with no other resonant frequencies between them,they must be consecutive harmonics.
Therefore,the fundamental frequency $v_0$ is the difference between these two frequencies:
$v_0 = 420\, Hz - 315\, Hz = 105\, Hz$.

(B) Since $4.0 \, g$ of a gas occupies $22.4 \, L$ at $NTP$,the molar mass of the gas is $M = 4.0 \, g \, mol^{-1} = 4.0 \times 10^{-3} \, kg \, mol^{-1}$.
The speed of sound in an ideal gas is given by $v = \sqrt{\frac{\gamma R T}{M}}$.
Rearranging for $\gamma$,we get $\gamma = \frac{M v^2}{R T}$.
Given $v = 952 \, m s^{-1}$,$R = 8.3 \, J K^{-1} mol^{-1}$,and $T = 273 \, K$ at $NTP$:
$\gamma = \frac{(4.0 \times 10^{-3} \, kg \, mol^{-1}) \times (952 \, m s^{-1})^2}{(8.3 \, J K^{-1} mol^{-1}) \times (273 \, K)} \approx \frac{3625.216}{2265.9} \approx 1.6$.
We know that $\gamma = \frac{C_p}{C_v}$,so $C_p = \gamma C_v$.
Given $C_v = 5.0 \, J K^{-1} mol^{-1}$,we have $C_p = 1.6 \times 5.0 \, J K^{-1} mol^{-1} = 8.0 \, J K^{-1} mol^{-1}$.

(C) The efficiency $(\eta)$ of a Carnot engine and the coefficient of performance $(\beta)$ of a refrigerator are related as:
$\beta = \frac{1 - \eta}{\eta}$
Given $\eta = 1/10$,we calculate the coefficient of performance:
$\beta = \frac{1 - 1/10}{1/10} = \frac{9/10}{1/10} = 9$
The coefficient of performance $(\beta)$ is also defined as the ratio of heat absorbed from the cold reservoir $(Q_2)$ to the work done $(W)$ on the system:
$\beta = \frac{Q_2}{W}$
Given $W = 10 \ J$ and $\beta = 9$,we have:
$9 = \frac{Q_2}{10 \ J}$
$Q_2 = 9 \times 10 \ J = 90 \ J$
Therefore,the energy absorbed from the reservoir at a lower temperature is $90 \ J$.

(B) The fundamental frequency of a closed organ pipe is given by $f_c = \frac{v}{4\ell_c}$,where $\ell_c = 20\; cm$.
The frequencies of an open organ pipe are given by $f_n = \frac{nv}{2\ell_o}$,where $n = 1, 2, 3, \dots$.
The first overtone is $n=2$,and the second overtone is $n=3$.
Thus,the second overtone frequency of an open organ pipe is $f_{o,2} = \frac{3v}{2\ell_o}$.
According to the problem,the fundamental frequency of the closed pipe equals the second overtone of the open pipe:
$\frac{v}{4\ell_c} = \frac{3v}{2\ell_o}$
Canceling $v$ from both sides:
$\frac{1}{4\ell_c} = \frac{3}{2\ell_o}$
Rearranging to solve for $\ell_o$:
$\ell_o = \frac{3 \times 4\ell_c}{2} = 6\ell_c$
Substituting $\ell_c = 20\; cm$:
$\ell_o = 6 \times 20\; cm = 120\; cm$.

(C) Given the potential function: $V = 6xy - y + 2yz$.
The electric field $\vec{E}$ is related to the potential $V$ by the relation: $\vec{E} = -\nabla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right)$.
Calculating the partial derivatives:
$\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(6xy - y + 2yz) = 6y$
$\frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(6xy - y + 2yz) = 6x - 1 + 2z$
$\frac{\partial V}{\partial z} = \frac{\partial}{\partial z}(6xy - y + 2yz) = 2y$
Substituting these into the expression for $\vec{E}$:
$\vec{E} = -[ (6y)\hat{i} + (6x - 1 + 2z)\hat{j} + (2y)\hat{k} ]$.
Now,evaluating at the point $(1, 1, 0)$:
$\vec{E}_{(1, 1, 0)} = -[ (6 \times 1)\hat{i} + (6 \times 1 - 1 + 2 \times 0)\hat{j} + (2 \times 1)\hat{k} ]$
$\vec{E}_{(1, 1, 0)} = -(6\hat{i} + 5\hat{j} + 2\hat{k}) \ N/C$.

(D) Initially,the capacitor is charged to $q = CV$. When the cell is disconnected,the charge $q$ remains constant on the plates.
After inserting the dielectric slab,the new capacitance becomes $C' = KC$.
The potential difference between the plates becomes $V' = \frac{q}{C'} = \frac{CV}{KC} = \frac{V}{K}$. Thus,the potential difference decreases $K$ times.
The initial energy stored is $U_1 = \frac{q^2}{2C} = \frac{1}{2}CV^2$.
The final energy stored is $U_2 = \frac{q^2}{2C'} = \frac{q^2}{2KC} = \frac{U_1}{K}$. Thus,the energy decreases $K$ times.
The change in energy is $\Delta U = U_2 - U_1 = \frac{q^2}{2KC} - \frac{q^2}{2C} = \frac{1}{2}CV^2\left(\frac{1}{K} - 1\right)$.
Since the capacitor is disconnected from the cell,the charge $q$ on the plates remains constant (conserved). Therefore,the statement that the charge is not conserved is incorrect.

(C) The force of attraction $F$ between the plates of a parallel plate capacitor is given by the formula:
$F = \frac{Q^2}{2 \varepsilon_0 A}$
where $Q$ is the charge on the capacitor,$\varepsilon_0$ is the permittivity of free space,and $A$ is the area of each plate.
We know that the charge $Q$ is related to capacitance $C$ and potential difference $V$ by:
$Q = CV$
Also,the capacitance of a parallel plate capacitor is given by:
$C = \frac{\varepsilon_0 A}{d} \implies \varepsilon_0 A = Cd$
Substituting these expressions into the force formula:
$F = \frac{(CV)^2}{2(Cd)}$
$F = \frac{C^2 V^2}{2Cd}$
$F = \frac{CV^2}{2d}$

(C) According to Gauss's Law,the net electric flux $\phi$ through a closed surface is given by $\phi = \oint E \cdot dA = \frac{q_{en}}{\varepsilon_0}$.
Given the electric field $E = Ar$ is acting radially outward,for a spherical surface of radius $a$,the electric field at the surface is $E = Aa$.
The surface area of the sphere is $S = 4\pi a^2$.
Since the field is uniform over the surface of the sphere and directed radially outward,the flux is $\phi = E \times S = (Aa) \times (4\pi a^2) = 4\pi A a^3$.
Equating this to Gauss's Law: $4\pi A a^3 = \frac{q}{\varepsilon_0}$.
Therefore,the charge contained is $q = 4\pi \varepsilon_0 A a^3$.

(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:
$\frac{2l}{\sigma_{eff} A} = \frac{l}{A} \left( \frac{\sigma_1 + \sigma_2}{\sigma_1 \sigma_2} \right)$
$\frac{2}{\sigma_{eff}} = \frac{\sigma_1 + \sigma_2}{\sigma_1 \sigma_2}$
$\sigma_{eff} = \frac{2 \sigma_1 \sigma_2}{\sigma_1 + \sigma_2}$

(C) The current $I$ flowing through the potentiometer wire is given by the total e.m.f. divided by the total resistance of the primary circuit:
$I = \frac{E_0}{r + r_1}$
The potential difference $V$ across the entire length $L$ of the potentiometer wire is:
$V = I r = \frac{E_0 r}{r + r_1}$
The potential gradient $k$ along the wire is the potential difference per unit length:
$k = \frac{V}{L} = \frac{E_0 r}{(r + r_1) L}$
Since the unknown e.m.f. $E$ is balanced at a length $l$,the potential drop across length $l$ must equal $E$:
$E = k l = \left( \frac{E_0 r}{(r + r_1) L} \right) l = \frac{E_0 r l}{(r + r_1) L}$

(C) The circuit is shown in the figure.
First,calculate the effective resistance of the ammeter $(R_A)$. Since the coil resistance $(480\,\Omega)$ and the shunt resistance $(20\,\Omega)$ are in parallel,the equivalent resistance is:
$R_A = \frac{480 \times 20}{480 + 20} = \frac{9600}{500} = 19.2\,\Omega$
Since the ammeter is connected in series with the $40.8\,\Omega$ resistor,the total resistance $(R_{total})$ of the circuit is:
$R_{total} = 40.8\,\Omega + 19.2\,\Omega = 60\,\Omega$
Using Ohm's law $(I = V/R)$,the total current flowing through the circuit is:
$I = \frac{30\,V}{60\,\Omega} = 0.5\,A$
Thus,the reading in the ammeter will be $0.5\,A$.

(A) Required potential gradient $k = 1\, mV/cm = 10^{-3}\, V / 10^{-2}\, m = 0.1\, V/m$.
Length of the potentiometer wire $L = 4\, m$.
Potential difference across the wire $V_w = k \times L = 0.1 \times 4 = 0.4\, V$.
The circuit consists of an accumulator of $E = 2\, V$ in series with a resistor $R$ and the potentiometer wire of resistance $R_w = 8\, \Omega$.
The current in the circuit is $I = \frac{E}{R + R_w} = \frac{2}{R + 8}$.
The potential difference across the wire is $V_w = I \times R_w$.
Substituting the values: $0.4 = \left( \frac{2}{R + 8} \right) \times 8$.
$0.4 = \frac{16}{R + 8}$.
$R + 8 = \frac{16}{0.4} = 40$.
$R = 40 - 8 = 32\, \Omega$.

(A) Let the total current flowing through the circuit be $I$. This current $I$ passes through voltmeter $A$.
At the junction,the current $I$ splits into two parallel branches containing voltmeters $B$ and $C$. Let $I_B$ and $I_C$ be the currents through $B$ and $C$ respectively.
Since $B$ and $C$ are in parallel,the potential difference across them is the same:
$V_B = V_C \implies I_B \times (1.5R) = I_C \times (3R) \implies I_B = 2I_C$.
Also,$I_B + I_C = I$. Substituting $I_B = 2I_C$,we get $3I_C = I$,so $I_C = I/3$ and $I_B = 2I/3$.
The readings are:
$V_A = I \times R = IR$
$V_B = I_B \times 1.5R = (2I/3) \times (3R/2) = IR$
$V_C = I_C \times 3R = (I/3) \times 3R = IR$
Thus,$V_A = V_B = V_C$.

(B) For a metallic conductor, the total current $I$ flowing through any cross-section must be the same due to the principle of conservation of charge (steady state flow).
Since $I = nAev_d$ and $J = I/A = nev_d$, if the cross-sectional area $A$ varies, the current density $J$ and drift velocity $v_d$ must also vary to keep $I$ constant.
The electric field $E = J/\sigma$ also varies with the cross-sectional area.
Therefore, the current $I$ is the only quantity that remains constant along the conductor.

(C) The given situation is shown in the figure. The wire consists of three parts: two semi-infinite straight wires parallel to the $X$-axis and a semicircular arc in the $Y-Z$ plane.
$1$. For the two semi-infinite straight wires (labeled $1$ and $3$): The magnetic field at $O$ due to a semi-infinite wire is $B = \frac{\mu_0 I}{4 \pi R}$. Using the right-hand rule,both wires produce a magnetic field in the $-\hat{k}$ direction at point $O$. Thus,$\vec{B}_{1} = \vec{B}_{3} = -\frac{\mu_{0} I}{4 \pi R} \hat{k}$.
$2$. For the semicircular arc (labeled $2$): The magnetic field at the center of a semicircular arc of radius $R$ is $B = \frac{\mu_0 I}{4 R}$. Using the right-hand rule,the field is directed in the $-\hat{i}$ direction. Thus,$\vec{B}_{2} = -\frac{\mu_{0} I}{4 R} \hat{i}$.
$3$. The net magnetic field at point $O$ is the vector sum: $\vec{B} = \vec{B}_{1} + \vec{B}_{2} + \vec{B}_{3}$.
Substituting the values: $\vec{B} = -\frac{\mu_{0} I}{4 \pi R} \hat{k} - \frac{\mu_{0} I}{4 R} \hat{i} - \frac{\mu_{0} I}{4 \pi R} \hat{k}$.
Factoring out $-\frac{\mu_{0} I}{4 \pi R}$: $\vec{B} = -\frac{\mu_{0} I}{4 \pi R} (\pi \hat{i} + 2 \hat{k})$.

(D) Let the distance of the center of the square frame from the wire be $x$. The left side of the frame is at a distance $(x - a/2)$ from the wire,and the right side is at a distance $(x + a/2)$ from the wire.
The magnetic field due to the long wire at distance $r$ is $B = \frac{\mu_0 I}{2\pi r}$.
The motional $emf$ induced in the left side (length $a$) is $\varepsilon_1 = B_1 a V = \frac{\mu_0 I}{2\pi (x - a/2)} a V$.
The motional $emf$ induced in the right side (length $a$) is $\varepsilon_2 = B_2 a V = \frac{\mu_0 I}{2\pi (x + a/2)} a V$.
The net $emf$ induced in the frame is $\varepsilon = \varepsilon_1 - \varepsilon_2$.
$\varepsilon = \frac{\mu_0 I a V}{2\pi} \left[ \frac{1}{x - a/2} - \frac{1}{x + a/2} \right]$
$\varepsilon = \frac{\mu_0 I a V}{2\pi} \left[ \frac{2}{2x - a} - \frac{2}{2x + a} \right]$
$\varepsilon = \frac{\mu_0 I a V}{\pi} \left[ \frac{(2x + a) - (2x - a)}{(2x - a)(2x + a)} \right]$
$\varepsilon = \frac{\mu_0 I a V}{\pi} \left[ \frac{2a}{(2x - a)(2x + a)} \right]$
Thus,$\varepsilon \propto \frac{1}{(2x - a)(2x + a)}$.

(A) The radius of a circular orbit for a charged particle in a uniform magnetic field $B$ is given by $R = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$, where $K$ is the kinetic energy.
Rearranging for kinetic energy $K$, we get $K = \frac{q^2 B^2 R^2}{2m}$.
For the proton $(p)$ and alpha particle $(\alpha)$, since $B$ and $R$ are equal, the ratio of their kinetic energies is:
$\frac{K_{\alpha}}{K_{p}} = \left(\frac{q_{\alpha}}{q_{p}}\right)^2 \left(\frac{m_{p}}{m_{\alpha}}\right)$.
Given $q_{\alpha} = 2q_{p}$ and $m_{\alpha} = 4m_{p}$, we have:
$\frac{K_{\alpha}}{K_{p}} = (2)^2 \times \left(\frac{1}{4}\right) = 4 \times \frac{1}{4} = 1$.
Therefore, $K_{\alpha} = K_{p} = 1 \, MeV$.

(D) The current $I$ produced by an electron of charge $e$ moving in a circular orbit with frequency $n$ (rotations per second) is given by $I = q \times f = e \times n$.
The magnetic field $B$ at the center of a circular current-carrying loop of radius $r$ is given by the formula $B = \frac{\mu_0 I}{2r}$.
Substituting the value of $I = ne$ into the magnetic field formula,we get:
$B = \frac{\mu_0 (ne)}{2r} = \frac{\mu_0 ne}{2r}$.

(C) The torque acting on a current-carrying coil in a magnetic field is given by $\tau = N I A B \sin \theta$,where $\theta$ is the angle between the normal to the plane of the coil and the magnetic field direction.
Given: $N = 50$,$I = 2\, A$,$B = 0.2\, Wb/m^2$,and area $A = 0.12\, m \times 0.1\, m = 0.012\, m^2$.
The plane of the coil is inclined at $30^{\circ}$ to the magnetic field,so the angle between the normal to the coil and the magnetic field is $\theta = 90^{\circ} - 30^{\circ} = 60^{\circ}$.
Substituting the values: $\tau = 50 \times 2 \times 0.012 \times 0.2 \times \sin 60^{\circ}$.
$\tau = 100 \times 0.0024 \times \frac{\sqrt{3}}{2} = 0.24 \times 0.866 = 0.2078\, Nm \approx 0.20\, Nm$.

(A) As the beam of light is incident normally on the face $AB$ of the right-angled prism $ABC$,no refraction occurs at face $AB$. The light passes straight and strikes the face $AC$ at an angle of incidence $i = 45^{\circ}$.
For total internal reflection to take place at face $AC$,the condition is $i > i_c$,where $i_c$ is the critical angle.
We know that $\sin i_c = \frac{1}{\mu}$. Therefore,the condition for total internal reflection is $\sin i > \frac{1}{\mu}$,or $\mu > \frac{1}{\sin i}$.
Given $i = 45^{\circ}$,we have $\sin 45^{\circ} = \frac{1}{\sqrt{2}} \approx 0.707$. Thus,the condition becomes $\mu > \sqrt{2} \approx 1.414$.
Comparing the refractive indices:
For red: $\mu_{\text{red}} = 1.39 < 1.414$.
For green: $\mu_{\text{green}} = 1.44 > 1.414$.
For blue: $\mu_{\text{blue}} = 1.47 > 1.414$.
Since $\mu_{\text{red}} < 1.414$,the red light will be refracted out of the prism through face $AC$. Since $\mu_{\text{green}}$ and $\mu_{\text{blue}}$ are both greater than $1.414$,both green and blue light will undergo total internal reflection at face $AC$.
Therefore,the prism will separate the red colour from the green and blue colours.

(C) Using the lens maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For the two identical plano-convex lenses ($f_1$ and $f_2$):
$\frac{1}{f_1} = (1.5 - 1) \left( \frac{1}{20} - \frac{1}{\infty} \right) = 0.5 \times \frac{1}{20} = \frac{1}{40} \, cm^{-1}$.
Similarly,$\frac{1}{f_2} = \frac{1}{40} \, cm^{-1}$.
The oil lens formed in the middle is a biconcave lens with radii of curvature $R_1 = -20 \, cm$ and $R_2 = 20 \, cm$ (or vice versa,resulting in the same focal length).
For the oil lens $(f_3)$:
$\frac{1}{f_3} = (1.7 - 1) \left( \frac{1}{-20} - \frac{1}{20} \right) = 0.7 \times \left( -\frac{2}{20} \right) = 0.7 \times \left( -\frac{1}{10} \right) = -\frac{0.7}{10} = -\frac{7}{100} \, cm^{-1}$.
The focal length of the combination is given by $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}$.
$\frac{1}{f} = \frac{1}{40} + \frac{1}{40} - \frac{7}{100} = \frac{2}{40} - \frac{7}{100} = \frac{1}{20} - \frac{7}{100}$.
$\frac{1}{f} = \frac{5 - 7}{100} = -\frac{2}{100} = -\frac{1}{50}$.
Therefore,$f = -50 \, cm$.

(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:
$90^o - A/2 = (A+\delta)/2$
$180^o - A = A + \delta$
$\delta = 180^o - 2A$

(D) According to Einstein's photoelectric equation,the maximum kinetic energy of the emitted electron is $K_{\max} = \frac{hc}{\lambda} - \phi_0$,where $\lambda$ is the wavelength of incident light and $\phi_0$ is the work function.
Given: $\lambda = 500\, nm$,$hc = 1240\, eV\, nm$,and $\phi_0 = 2.28\, eV$.
$K_{\max} = \frac{1240\, eV\, nm}{500\, nm} - 2.28\, eV = 2.48\, eV - 2.28\, eV = 0.2\, eV$.
The de Broglie wavelength $\lambda_e$ is related to kinetic energy $K$ by $\lambda_e = \frac{h}{\sqrt{2mK}}$. Since $K \le K_{\max}$,the minimum wavelength corresponds to the maximum kinetic energy: $\lambda_{\min} = \frac{h}{\sqrt{2mK_{\max}}}$.
Using $h = 6.6 \times 10^{-34}\, J\, s$,$m = 9.1 \times 10^{-31}\, kg$,and $K_{\max} = 0.2 \times 1.6 \times 10^{-19}\, J$:
$\lambda_{\min} = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 0.32 \times 10^{-19}}} \approx 2.8 \times 10^{-9}\, m$.
Since $K \le K_{\max}$,it follows that $\lambda_e \ge \lambda_{\min}$. Thus,$\lambda_e \ge 2.8 \times 10^{-9}\, m$.

(B) Let $\phi_{0}$ be the work function of the surface of the material. According to Einstein's photoelectric equation,the maximum kinetic energy of the emitted photoelectrons in the first case is $K_{\max 1} = \frac{hc}{\lambda} - \phi_{0}$.
In the second case,the wavelength is $\lambda/2$,so the maximum kinetic energy is $K_{\max 2} = \frac{hc}{\lambda/2} - \phi_{0} = \frac{2hc}{\lambda} - \phi_{0}$.
Given that $K_{\max 2} = 3 K_{\max 1}$,we substitute the expressions:
$\frac{2hc}{\lambda} - \phi_{0} = 3 \left( \frac{hc}{\lambda} - \phi_{0} \right)$.
Expanding the equation: $\frac{2hc}{\lambda} - \phi_{0} = \frac{3hc}{\lambda} - 3\phi_{0}$.
Rearranging the terms: $3\phi_{0} - \phi_{0} = \frac{3hc}{\lambda} - \frac{2hc}{\lambda}$.
$2\phi_{0} = \frac{hc}{\lambda}$.
Therefore,the work function is $\phi_{0} = \frac{hc}{2\lambda}$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max} = eV_s = \frac{hc}{\lambda} - \phi_0$,where $\phi_0$ is the work function.
Case $(i)$: For wavelength $\lambda$,stopping potential is $3V_0$:
$3eV_0 = \frac{hc}{\lambda} - \phi_0$ ......... $(1)$
Case $(ii)$: For wavelength $2\lambda$,stopping potential is $V_0$:
$eV_0 = \frac{hc}{2\lambda} - \phi_0$ ......... $(2)$
Multiply equation $(2)$ by $3$:
$3eV_0 = \frac{3hc}{2\lambda} - 3\phi_0$ ......... $(3)$
Equating $(1)$ and $(3)$:
$\frac{hc}{\lambda} - \phi_0 = \frac{3hc}{2\lambda} - 3\phi_0$
$2\phi_0 = \frac{3hc}{2\lambda} - \frac{hc}{\lambda} = \frac{hc}{2\lambda}$
$\phi_0 = \frac{hc}{4\lambda}$
Since the threshold wavelength $\lambda_0 = \frac{hc}{\phi_0}$,substituting $\phi_0$:
$\lambda_0 = \frac{hc}{hc / 4\lambda} = 4\lambda$.

(D) The de-Broglie wavelength $\lambda$ is related to the momentum $p$ of a particle by the equation $\lambda = \frac{h}{p}$, where $h$ is Planck's constant.
This equation can be rewritten as $p \lambda = h$.
Since $h$ is a constant, the product of $p$ and $\lambda$ is constant $(p \lambda = \text{constant})$.
This relationship represents a rectangular hyperbola in the $p-\lambda$ plane, where $p$ decreases as $\lambda$ increases.
Therefore, the graph shown in Figure $D$ correctly represents this inverse relationship.

(D) As the electron moves from $X$ to $Y$,it creates a magnetic field. According to the right-hand rule,the magnetic field lines pass through the coil $abcd$ in a direction perpendicular to the plane of the coil (into the page).
As the electron approaches the coil,the magnetic flux linked with the coil increases. According to Lenz's Law,the induced current will oppose this increase by creating a magnetic field in the opposite direction (out of the page),which corresponds to an anticlockwise direction $(adcb)$.
As the electron moves away from the coil,the magnetic flux linked with the coil decreases. The induced current will now oppose this decrease by creating a magnetic field in the same direction as the original field (into the page),which corresponds to a clockwise direction $(abcd)$.
Therefore,the current will reverse its direction as the electron goes past the coil.

(C) The current $i$ in a series $R-C$ circuit is given by:
$i = \frac{V_0}{Z} = \frac{V_0}{\sqrt{R^2 + X_C^2}} = \frac{V_0}{\sqrt{R^2 + (1/\omega C)^2}}$
The voltage across the capacitor is $V = i X_C = i \cdot \frac{1}{\omega C}$.
Substituting the expression for $i$:
$V = \frac{V_0}{\sqrt{R^2 + (1/\omega C)^2}} \cdot \frac{1}{\omega C} = \frac{V_0}{\sqrt{R^2 \omega^2 C^2 + 1}}$
When the capacitor is filled with mica (dielectric constant $K > 1$),its capacitance increases,so $C_b > C_a$.
$1$. For current: As $C$ increases,$X_C = 1/\omega C$ decreases,so the impedance $Z = \sqrt{R^2 + X_C^2}$ decreases. Thus,$i_b > i_a$.
$2$. For voltage across the capacitor: From $V = \frac{V_0}{\sqrt{R^2 \omega^2 C^2 + 1}}$,as $C$ increases,the denominator increases,which means $V$ decreases. Therefore,$V_b < V_a$,or $V_a > V_b$.

(A) Case $I$: The power consumed by a purely resistive circuit is given by $P = V_{\text{rms}} I_{\text{rms}}$.
Since $I_{\text{rms}} = \frac{V_{\text{rms}}}{R}$,we have $P = \frac{V_{\text{rms}}^2}{R}$,which implies $V_{\text{rms}}^2 = P R$ ... $(i)$.
Case $II$: When an inductance $L$ is connected in series with the resistance $R$,the circuit becomes an $LR$ circuit with impedance $Z$.
The power consumed in an $AC$ circuit is given by $P' = V_{\text{rms}} I_{\text{rms}} \cos \phi$,where $\cos \phi = \frac{R}{Z}$ is the power factor.
Here,$I_{\text{rms}} = \frac{V_{\text{rms}}}{Z}$.
Substituting these values,we get $P' = V_{\text{rms}} \times \left( \frac{V_{\text{rms}}}{Z} \right) \times \left( \frac{R}{Z} \right) = V_{\text{rms}}^2 \frac{R}{Z^2}$.
Using equation $(i)$,$V_{\text{rms}}^2 = P R$,so $P' = (P R) \frac{R}{Z^2} = P \left( \frac{R}{Z} \right)^2$.

(B) The energy of a photon is given by $E = pc$,where $p$ is the momentum of the photon and $C$ is the speed of light.
Therefore,the initial momentum of the radiation incident on the surface is $p_i = E/C$.
Since the surface is perfectly reflecting,the radiation reflects back with the same magnitude of momentum but in the opposite direction.
Thus,the final momentum of the radiation is $p_f = -E/C$.
The momentum transferred to the surface is the change in momentum of the radiation,given by $\Delta p = p_i - p_f$.
Substituting the values,we get $\Delta p = E/C - (-E/C) = E/C + E/C = 2E/C$.

(B) The energy $E$ of a photon is related to its wavelength $\lambda$ by the formula $\lambda = \frac{hc}{E}$.
Given,$E = 15\, keV = 15 \times 10^3\, eV$.
The value of $hc$ is approximately $1240\, eV\, nm$.
Substituting these values,we get $\lambda = \frac{1240\, eV\, nm}{15 \times 10^3\, eV} \approx 0.083\, nm$.
The wavelength range for $X$-rays typically spans from $1\, nm$ to $10^{-3}\, nm$.
Since $0.083\, nm$ falls within this range,the electromagnetic waves belong to the $X$-ray part of the spectrum.

(A) Given:
Distance between slits $d = 1\, mm = 1 \times 10^{-3}\, m$
Distance of screen $D = 1\, m$
Wavelength $\lambda = 500\, nm = 500 \times 10^{-9}\, m$
The width of the central maximum in a single-slit diffraction pattern is given by $w = \frac{2\lambda D}{a}$,where $a$ is the slit width.
The width of one fringe in a double-slit interference pattern is $\beta = \frac{\lambda D}{d}$.
According to the problem,the width of the central maximum of the single-slit pattern contains $10$ maxima of the double-slit pattern.
Therefore,$\frac{2\lambda D}{a} = 10 \times \frac{\lambda D}{d}$.
Simplifying the equation: $\frac{2}{a} = \frac{10}{d}$.
$a = \frac{2d}{10} = \frac{d}{5}$.
Substituting $d = 1\, mm$:
$a = \frac{1\, mm}{5} = 0.2\, mm$.

(B) The intensity $I$ of light is directly proportional to the width $W$ of the slit,and also to the square of the amplitude $A$ of the wave.
$\therefore \frac{I_1}{I_2} = \frac{W_1}{W_2} = \frac{A_1^2}{A_2^2}$
Given the ratio of widths $\frac{W_1}{W_2} = \frac{1}{25}$,we have:
$\frac{A_1^2}{A_2^2} = \frac{1}{25} \implies \frac{A_1}{A_2} = \sqrt{\frac{1}{25}} = \frac{1}{5}$
The ratio of maximum intensity to minimum intensity is given by:
$\frac{I_{max}}{I_{min}} = \frac{(A_1 + A_2)^2}{(A_1 - A_2)^2} = \left( \frac{\frac{A_1}{A_2} + 1}{\frac{A_1}{A_2} - 1} \right)^2$
Substituting $\frac{A_1}{A_2} = \frac{1}{5}$:
$\frac{I_{max}}{I_{min}} = \left( \frac{\frac{1}{5} + 1}{\frac{1}{5} - 1} \right)^2 = \left( \frac{\frac{6}{5}}{-\frac{4}{5}} \right)^2 = \left( -\frac{6}{4} \right)^2 = \left( -\frac{3}{2} \right)^2 = \frac{9}{4}$

(A) In single slit diffraction,the condition for the first minima on either side of the central maxima is given by $a \sin \theta = \pm \lambda$.
For small angles,$\sin \theta \approx \theta = \frac{\lambda}{a}$.
The distance of the first minima from the center of the screen is $y = D \tan \theta \approx D \theta = \frac{D \lambda}{a}$.
The central maxima lies between the first minima on both sides,so its total width is $2y = \frac{2D \lambda}{a}$.

(D) The situation is shown in the figure. In the figure,$A$ and $B$ represent the edges of the slit $AB$ of width $a$,and $C$ represents the midpoint of the slit. For the first minimum at $P$,the condition is given by:
$a \sin \theta = \lambda$ ...... $(i)$
where $\lambda$ is the wavelength of light.
The path difference $\Delta x$ between the wavelets from the edge $A$ and the midpoint $C$ is:
$\Delta x = \frac{a}{2} \sin \theta = \frac{1}{2}(a \sin \theta) = \frac{\lambda}{2}$ (using equation $(i)$).
The corresponding phase difference $\Delta \phi$ is given by the relation:
$\Delta \phi = \frac{2 \pi}{\lambda} \Delta x = \frac{2 \pi}{\lambda} \times \frac{\lambda}{2} = \pi \text{ rad}$.

(A) The wavelength of a spectral line in the Lyman series is given by $\frac{1}{\lambda_{L}} = R\left(\frac{1}{1^{2}} - \frac{1}{n^{2}}\right)$,where $n = 2, 3, 4, \dots$.
For the longest wavelength in the Lyman series,we take $n = 2$:
$\frac{1}{\lambda_{L}} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4} \implies \lambda_{L} = \frac{4}{3R}$.
The wavelength of a spectral line in the Balmer series is given by $\frac{1}{\lambda_{B}} = R\left(\frac{1}{2^{2}} - \frac{1}{n^{2}}\right)$,where $n = 3, 4, 5, \dots$.
For the longest wavelength in the Balmer series,we take $n = 3$:
$\frac{1}{\lambda_{B}} = R\left(\frac{1}{4} - \frac{1}{9}\right) = R\left(\frac{9-4}{36}\right) = \frac{5R}{36} \implies \lambda_{B} = \frac{36}{5R}$.
The ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is:
$\frac{\lambda_{L}}{\lambda_{B}} = \frac{4/3R}{36/5R} = \frac{4}{3R} \times \frac{5R}{36} = \frac{5}{27}$.

(B) According to Bohr's theory,the velocity of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula:
$v_n = v_0 \times \frac{Z}{n}$
where $v_0 = 2.18 \times 10^6 \; m/s$ is the velocity of an electron in the first orbit of the hydrogen atom.
For $He^{+}$,$Z = 2$ and for the $3^{rd}$ orbit,$n = 3$.
Substituting these values:
$v_3 = (2.18 \times 10^6) \times \frac{2}{3} \; m/s$
$v_3 = 2.18 \times 10^6 \times 0.666... \; m/s$
$v_3 \approx 1.453 \times 10^6 \; m/s \approx 1.46 \times 10^6 \; m/s$.

(D) If $\vec{p}_{Th}$ and $\vec{p}_{He}$ are the momenta of thorium and helium nuclei respectively,then according to the law of conservation of linear momentum:
$0 = \vec{p}_{Th} + \vec{p}_{He}$ or $\vec{p}_{Th} = -\vec{p}_{He}$.
The negative sign shows that both are moving in opposite directions. However,in magnitude,$p_{Th} = p_{He}$.
If $m_{Th}$ and $m_{He}$ are the masses of thorium and helium nuclei respectively,then the kinetic energy of the thorium nucleus is $K_{Th} = \frac{p_{Th}^2}{2m_{Th}}$ and that of the helium nucleus is $K_{He} = \frac{p_{He}^2}{2m_{He}}$.
Therefore,$\frac{K_{Th}}{K_{He}} = \left(\frac{p_{Th}}{p_{He}}\right)^2 \left(\frac{m_{He}}{m_{Th}}\right)$.
Since $p_{Th} = p_{He}$ and $m_{He} < m_{Th}$,it follows that $K_{Th} < K_{He}$ or $K_{He} > K_{Th}$.
Thus,the helium nucleus has more kinetic energy than the thorium nucleus.

(B) The radius of a nucleus is given by the formula $R = R_0 A^{1/3}$,where $A$ is the mass number and $R_0$ is a constant.
For the $^{27}_{13}Al$ nucleus,the mass number $A_{Al} = 27$. Thus,$R_{Al} = R_0 (27)^{1/3} = 3 R_0$.
For the $^{125}_{53}Te$ nucleus,the mass number $A_{Te} = 125$. Thus,$R_{Te} = R_0 (125)^{1/3} = 5 R_0$.
Taking the ratio of the two radii:
$\frac{R_{Te}}{R_{Al}} = \frac{5 R_0}{3 R_0} = \frac{5}{3}$.
Therefore,$R_{Te} = \frac{5}{3} R_{Al}$.

(A) Given, input signal $V_{i} = 2 \cos(15t + \frac{\pi}{3})$ and voltage gain $A_{v} = 150$.
The voltage gain is defined as $A_{v} = \frac{V_{o}}{V_{i}}$, so the output signal magnitude is $V_{o} = A_{v} \times V_{i}$.
A common emitter $(CE)$ amplifier introduces a phase shift of $\pi$ $(180^{\circ})$ between the input and output signals.
Therefore, the output signal is $V_{o} = 150 \times 2 \cos(15t + \frac{\pi}{3} + \pi)$.
Simplifying the phase, $\frac{\pi}{3} + \pi = \frac{4\pi}{3}$.
Thus, $V_{o} = 300 \cos(15t + \frac{4\pi}{3}) \text{ V}$.

(B) The potential difference across the resistance $R$ is given by subtracting the barrier potential of the diode from the total $e.m.f.$ of the circuit:
$V_R = E - V_{\text{barrier}} = 3.5 \,V - 0.5 \,V = 3.0 \,V$
According to Ohm's law,the current $I$ in the circuit is:
$I = \frac{V_R}{R} = \frac{3.0 \,V}{100 \,\Omega} = 0.03 \,A$
To convert the current into milliamperes $(mA)$:
$I = 0.03 \,A \times 1000 \,mA/A = 30 \,mA$
Therefore,the current in the circuit is $30 \,mA$.

(A) In normal adjustment,the distance between the objective lens and the eyepiece is $f_0 + f_e$.
Since the line of length $L$ is on the objective lens,it acts as an object for the eyepiece.
The distance of this object from the eyepiece is $u = -(f_0 + f_e)$.
The magnification of the eyepiece is given by $m_e = f_e / (f_e + u)$.
Substituting the value of $u$:
$m_e = f_e / (f_e - (f_0 + f_e)) = f_e / (-f_0) = -f_e / f_0$.
The magnitude of magnification is $|m_e| = l / L = f_e / f_0$.
Since the magnification of the telescope is $M = f_0 / f_e$,we have $M = L / l$.

(C) The inputs to the $NOR$ gate are $\bar{A}$ and $\bar{B}$ because they pass through $NOT$ gates. The output $Y$ of the $NOR$ gate is given by the Boolean expression: $Y = \overline{\bar{A} + \bar{B}}$.
According to De Morgan's theorem,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.
This is the Boolean expression for an $AND$ gate.
Truth table:

$A$	$B$	$\bar{A}$	$\bar{B}$	$\bar{A} + \bar{B}$	$Y = \overline{\bar{A} + \bar{B}}$
$0$	$0$	$1$	$1$	$1$	$0$
$0$	$1$	$1$	$0$	$1$	$0$
$1$	$0$	$0$	$1$	$1$	$0$
$1$	$1$	$0$	$0$	$0$	$1$

Thus,the combination is equivalent to an $AND$ gate.

(A) The circuit consists of a $p-n$ junction diode in series with a load resistor $R_L$.
When the input voltage is $+5\, V$,the diode is forward-biased. Assuming an ideal diode,it acts as a short circuit,and the entire input voltage of $+5\, V$ appears across the load resistor $R_L$.
When the input voltage is $-5\, V$,the diode is reverse-biased. It acts as an open circuit,and no current flows through the resistor $R_L$. Therefore,the output voltage across $R_L$ is $0\, V$.
Thus,the output signal is a square wave that varies between $+5\, V$ and $0\, V$.