MHT CET 2020 Physics Question Paper with Answer and Solution

(D) The rotational kinetic energy $(KE)$ is related to the angular momentum $(L)$ and moment of inertia $(I)$ by the formula: $KE = \frac{L^2}{2I}$.
Given that the rotational kinetic energies are equal $(KE_1 = KE_2)$,we have:
$\frac{L_1^2}{2I_1} = \frac{L_2^2}{2I_2}$
Substituting the given values $I_1 = I$ and $I_2 = 2I$:
$\frac{L_1^2}{2I} = \frac{L_2^2}{2(2I)}$
$\frac{L_1^2}{I} = \frac{L_2^2}{2I}$
$\frac{L_1^2}{L_2^2} = \frac{I}{2I} = \frac{1}{2}$
Taking the square root on both sides:
$\frac{L_1}{L_2} = \frac{1}{\sqrt{2}}$
Thus,the ratio of their angular momenta is $1:\sqrt{2}$.

(D) At the mean position,the potential energy is zero and the kinetic energy is maximum. Since no external horizontal force acts on the system during the placement of mass $m$,linear momentum is conserved.
Let $v_1$ be the velocity of mass $M$ at the mean position,and $v_2$ be the velocity of the combined mass $(M+m)$ immediately after placing $m$.
Conservation of momentum: $M v_1 = (M + m) v_2$.
The maximum velocity in $S.H.M.$ is given by $v_{max} = A \omega = A \sqrt{\frac{k}{m_{eff}}}$.
For the first case: $v_1 = A_1 \sqrt{\frac{k}{M}}$.
For the second case: $v_2 = A_2 \sqrt{\frac{k}{M+m}}$.
Substituting these into the momentum equation:
$M \left( A_1 \sqrt{\frac{k}{M}} \right) = (M + m) \left( A_2 \sqrt{\frac{k}{M+m}} \right)$.
$A_1 \sqrt{M k} = A_2 \sqrt{(M+m) k}$.
$A_1 \sqrt{M} = A_2 \sqrt{M+m}$.
Therefore,$\frac{A_1}{A_2} = \sqrt{\frac{M+m}{M}} = \left( \frac{M+m}{M} \right)^{\frac{1}{2}}$.

(D) Let $x_{1}$ and $x_{2}$ be the distances of masses $m_{1}$ and $m_{2}$ from the centre of mass,respectively.
By the definition of the centre of mass,$m_{1}x_{1} = m_{2}x_{2}$.
When the first particle is moved by a distance $d$ towards the centre of mass,its new distance becomes $(x_{1} - d)$.
Let the second particle be moved by a distance $D$ to keep the centre of mass unchanged.
For the centre of mass to remain at the same position,the new distances must satisfy: $m_{1}(x_{1} - d) = m_{2}(x_{2} - D)$.
Expanding this,we get $m_{1}x_{1} - m_{1}d = m_{2}x_{2} - m_{2}D$.
Since $m_{1}x_{1} = m_{2}x_{2}$,the equation simplifies to $-m_{1}d = -m_{2}D$.
Thus,$D = \frac{m_{1}}{m_{2}} d$.
Since the first particle moved towards the centre of mass,the second particle must move away from the centre of mass to maintain the balance.

$(A)$ According to the law of conservation of momentum, the total momentum before the collision equals the total momentum after the collision.
$m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$
Given $m_1 = 2 \,kg$, $u_1 = 10 \,ms^{-1}$, $m_2 = 3 \,kg$, and $u_2 = 0 \,ms^{-1}$.
$2 \times 10 + 3 \times 0 = (2 + 3)v$
$20 = 5v \implies v = 4 \,ms^{-1}$.
Initial kinetic energy $(K_i)$ = $\frac{1}{2} m_1 u_1^2 = \frac{1}{2} \times 2 \times (10)^2 = 100 \,J$.
Final kinetic energy $(K_f)$ = $\frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} \times (2 + 3) \times (4)^2 = \frac{1}{2} \times 5 \times 16 = 40 \,J$.
Loss in kinetic energy = $K_i - K_f = 100 \,J - 40 \,J = 60 \,J$.

(B) The coefficient of restitution $e$ is defined as the ratio of the relative velocity of separation to the relative velocity of approach.
For a perfectly elastic collision,the kinetic energy and momentum are conserved.
In such a collision,the relative velocity of separation is equal to the relative velocity of approach.
Therefore,$e = \frac{v_2 - v_1}{u_1 - u_2} = 1$.

(A) When a ball is released from height $h$,the time taken to reach the ground is given by $t = \sqrt{\frac{2h}{g}}$.
Since the collision is perfectly elastic,the ball rebounds with the same speed $v = \sqrt{2gh}$ and reaches the same height $h$.
The total time period $T$ of one complete oscillation (up and down) is $T = 2t = 2 \sqrt{\frac{2h}{g}}$.
The frequency $f$ is the reciprocal of the time period: $f = \frac{1}{T} = \frac{1}{2 \sqrt{\frac{2h}{g}}} = \frac{1}{2} \sqrt{\frac{g}{2h}}$.

(C) The force exerted on the wall is given by the rate of change of momentum.
Initial momentum of $N$ balls per second = $Nmu \hat{i}$.
Since the collision is elastic,the balls return with the same speed in the opposite direction.
Final momentum of $N$ balls per second = $-Nmu \hat{i}$.
Change in momentum per second = $\text{Final momentum} - \text{Initial momentum} = -Nmu \hat{i} - (Nmu \hat{i}) = -2Nmu \hat{i}$.
The magnitude of the change in momentum per second is $2Nmu$.
Since force is the rate of change of momentum,$F = \frac{\Delta p}{\Delta t} = \frac{2Nmu}{1} = 2Nmu \ N$.
Therefore,the correct option is $C$.

(A) According to the principle of conservation of linear momentum,the total momentum before the collision is equal to the total momentum after the collision.
Let the mass of each block be $m$.
The initial velocity of the first block is $3 \vec{V}$ and the second block is $0$.
After the collision,the two blocks stick together and move with a common velocity $V_{c}$.
Applying the conservation of momentum:
$m(3 \vec{V}) + m(0) = (m + m) V_{c}$
$3 m \vec{V} = 2 m V_{c}$
$V_{c} = \frac{3 m \vec{V}}{2 m} = \frac{3}{2} \vec{V}$

(A) Given: Height $h = 20 \,m$, acceleration due to gravity $g = 10 \,m/s^2$, and coefficient of restitution $e = 0.4$.
First, calculate the velocity $v$ of the ball just before it hits the ground using the equation of motion $v^2 = u^2 + 2gh$ (where initial velocity $u = 0$):
$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20 \,m/s$.
The coefficient of restitution $e$ is defined as the ratio of the velocity of separation (rebound velocity $v'$) to the velocity of approach (impact velocity $v$):
$e = \frac{v'}{v}$.
Therefore, the upward rebound velocity $v'$ is:
$v' = e \times v = 0.4 \times 20 = 8 \,m/s$.

(D) Initial mass $m_1 = m$,initial velocity $u_1 = 3 \ m/s$.
Initial mass $m_2 = 2m$,initial velocity $u_2 = 0 \ m/s$.
According to the law of conservation of linear momentum,the total initial momentum equals the total final momentum.
$P_i = m_1 u_1 + m_2 u_2 = m(3) + 2m(0) = 3m$.
After collision,the bodies coalesce,so the final mass is $M = m_1 + m_2 = m + 2m = 3m$.
Let the final velocity be $V$.
$P_f = (m_1 + m_2)V = 3mV$.
Equating $P_i = P_f$:
$3m = 3mV$.
$V = 1 \ m/s$.

(A) Given: Mass of the bullet $m = 20 \,g = 0.02 \,kg$. Initial velocity $u = 200 \,m/s$. Final velocity $v = 0 \,m/s$. Time taken $\Delta t = \frac{1}{50} \,s = 0.02 \,s$.
Impulse $(J)$ is equal to the change in momentum:
$J = \Delta p = m(v - u)$
$J = 0.02 \,kg \times (0 - 200) \,m/s = -4 \,kg \cdot m/s$.
The magnitude of impulse is $4 \,Ns$.
Average force $(F_{avg})$ is given by the rate of change of momentum:
$F_{avg} = \frac{J}{\Delta t} = \frac{4 \,Ns}{0.02 \,s} = 200 \,N$.
Thus,the impulse is $4 \,Ns$ and the average force is $200 \,N$.

(A) $1$. The impulse $P$ applied at end $B$ provides a linear momentum $P = mv_{cm}$,where $v_{cm}$ is the velocity of the center of mass. Thus,$v_{cm} = \frac{P}{m}$.
$2$. The impulse also provides an angular impulse about the center of mass: $J_{\theta} = P \cdot \frac{\ell}{2}$.
$3$. Since $J_{\theta} = I\omega$,where $I = \frac{m\ell^2}{12}$ is the moment of inertia about the center of mass,we have $\frac{P\ell}{2} = \frac{m\ell^2}{12} \omega$.
$4$. Solving for angular velocity $\omega$,we get $\omega = \frac{6P}{m\ell}$.
$5$. The time $t$ taken to rotate through an angle $\theta = \frac{\pi}{2}$ is given by $t = \frac{\theta}{\omega} = \frac{\pi/2}{6P/m\ell} = \frac{\pi m \ell}{12P}$.

(A) The impulse imparted to an object is equal to the change in its linear momentum.
Let the initial velocity of the ball be $u = 6 \ m/s$ (towards the batsman).
After being hit,the ball moves towards the bowler with the same speed,so the final velocity is $v = -6 \ m/s$.
The mass of the ball is $m = 0.2 \ kg$.
Impulse $J = \Delta p = m(v - u)$.
$J = 0.2 \times (-6 - 6) = 0.2 \times (-12) = -2.4 \ Ns$.
The magnitude of the impulse imparted to the ball is $2.4 \ Ns$.

(A) According to the law of conservation of linear momentum,since the body is initially stationary,the total initial momentum is zero.
$P_{i} = 0$
After the explosion,the two parts move in opposite directions. Let the velocities be $v_{1}$ and $v_{2}$.
$P_{f} = M_{1}v_{1} - M_{2}v_{2} = 0$
$M_{1}v_{1} = M_{2}v_{2}$
$\frac{v_{1}}{v_{2}} = \frac{M_{2}}{M_{1}}$
The kinetic energy of a body is given by $E = \frac{p^{2}}{2M}$. Since the magnitudes of momenta are equal $(p_{1} = p_{2} = p)$,
$\frac{E_{1}}{E_{2}} = \frac{p^{2} / (2M_{1})}{p^{2} / (2M_{2})} = \frac{M_{2}}{M_{1}}$
Thus,the ratio of their kinetic energies is $\frac{M_{2}}{M_{1}}$.

(D) According to the law of conservation of linear momentum,if no external force acts on the system,the total momentum remains constant.
Initial momentum of the system = momentum of the bullet + momentum of the block = $mv + M(0) = mv$.
After the collision,the bullet is embedded in the block,so they move together as a single system of mass $(m+M)$ with a final velocity $V$.
Final momentum of the system = $(m+M)V$.
By the law of conservation of linear momentum:
$mv = (m+M)V$
$V = \frac{mv}{m+M}$.

(D) The motion of a rocket is based on the principle of conservation of linear momentum. As the fuel burns,the rocket ejects hot gases at high speed in the downward direction. According to Newton's third law of motion,the gases exert an equal and opposite force on the rocket,providing the necessary thrust. Since there is no external force acting on the system (rocket + fuel),the total linear momentum of the system remains conserved.

(A) Water vapour in the atmosphere is confined only to its lowest layer,i.e.,the troposphere.
This is the reason why all weather phenomena,such as cloud formation,rain,and storms,occur only in this layer.

(A) The four fundamental forces in nature are gravitational,electromagnetic,strong nuclear,and weak nuclear forces.
$1$. Gravitational force has an infinite range,making it the force with the maximum range.
$2$. Weak nuclear force has a very short range,approximately $10^{-16} \ m$,making it the force with the minimum range.
Therefore,the correct order is gravitational force (maximum) and weak nuclear force (minimum).

(C) We know that the acceleration due to gravity is given by $g = \frac{GM}{R^2}$.
Since the mass of a planet is $M = \rho \left( \frac{4}{3} \pi R^3 \right)$,we can substitute this into the gravity formula:
$g = \frac{G}{R^2} \left( \rho \frac{4}{3} \pi R^3 \right) = \frac{4}{3} \pi G \rho R$.
This shows that $g \propto \rho R$.
Given that $g_m = \frac{1}{6} g_e$ and the ratio of densities $\frac{\rho_e}{\rho_m} = \frac{5}{3}$,which implies $\frac{\rho_m}{\rho_e} = \frac{3}{5}$.
Using the proportionality $g \propto \rho R$,we have:
$\frac{g_m}{g_e} = \frac{\rho_m R_m}{\rho_e R_e}$
$\frac{1}{6} = \left( \frac{3}{5} \right) \left( \frac{R_m}{R_e} \right)$
$\frac{R_m}{R_e} = \frac{1}{6} \times \frac{5}{3} = \frac{5}{18}$
Therefore,$R_m = \frac{5}{18} R_e$.

(D) The acceleration due to gravity $g$ at a latitude $\phi$ is given by the formula: $g_{\phi} = g - \omega^2 R \cos^2 \phi$,where $g$ is the acceleration due to gravity at the poles,$\omega$ is the angular velocity of the earth,and $R$ is the radius of the earth.
At the equator,$\phi = 0^{\circ}$,so $\cos 0^{\circ} = 1$,which gives $g_{eq} = g - \omega^2 R$ (minimum value).
At the poles,$\phi = 90^{\circ}$,so $\cos 90^{\circ} = 0$,which gives $g_{pole} = g$ (maximum value).
Additionally,the earth is an oblate spheroid,meaning the radius at the equator is greater than the radius at the poles $(R_{eq} > R_{pole})$.
Since $g = \frac{GM}{R^2}$,a smaller radius at the poles results in a higher value of $g$.
Therefore,as we move from the equator to the poles,the value of acceleration due to gravity increases.

(A) The acceleration due to gravity at the surface of the earth is given by $g = \frac{GM}{R^2}$.
At a height '$h$' above the surface,the acceleration due to gravity '$g_h$' is given by $g_h = \frac{GM}{(R+h)^2}$.
Given that $g_h = \frac{g}{3}$,we substitute the expressions:
$\frac{g}{3} = \frac{GM}{(R+h)^2}$.
Substituting $g = \frac{GM}{R^2}$ into the equation:
$\frac{1}{3} \left( \frac{GM}{R^2} \right) = \frac{GM}{(R+h)^2}$.
$\frac{1}{3R^2} = \frac{1}{(R+h)^2}$.
Taking the square root on both sides:
$\frac{1}{\sqrt{3}R} = \frac{1}{R+h}$.
$R+h = \sqrt{3}R$.
$h = \sqrt{3}R - R$.
$h = R(\sqrt{3}-1)$.

(A) Let $M_E$ be the mass of the Earth and $M_M$ be the mass of the Moon. Given $M_E = 81 M_M$.
Let $x$ be the distance from the centre of the Earth where the net gravitational force on a test mass $m$ is zero.
At this point,the gravitational pull from the Earth must be equal in magnitude to the gravitational pull from the Moon.
$\frac{G M_E}{x^2} = \frac{G M_M}{(R - x)^2}$
Substituting $M_E = 81 M_M$:
$\frac{G (81 M_M)}{x^2} = \frac{G M_M}{(R - x)^2}$
$\frac{81}{x^2} = \frac{1}{(R - x)^2}$
Taking the square root on both sides:
$\frac{9}{x} = \frac{1}{R - x}$
$9(R - x) = x$
$9R - 9x = x$
$9R = 10x$
$x = \frac{9}{10} R$

(B) According to the law of conservation of energy,the total energy at the surface equals the total energy at infinity.
Initial energy at the surface: $E_i = K_i + U_i = \frac{1}{2} m(2 v_e)^2 - \frac{G M m}{R} = 2 m v_e^2 - m v_e^2 = m v_e^2$ (since $v_e^2 = \frac{2 G M}{R}$).
Final energy at infinity: $E_f = K_f + U_f = \frac{1}{2} m v^2 + 0$.
Equating $E_i = E_f$:
$m v_e^2 = \frac{1}{2} m v^2$
$v^2 = 2 v_e^2$
$v = \sqrt{2} v_e$ is incorrect based on the standard energy balance approach. Let's re-evaluate: The energy provided is $K_i = \frac{1}{2} m (2 v_e)^2 = 2 m v_e^2$. The energy required to escape is $K_{req} = \frac{1}{2} m v_e^2$. The remaining kinetic energy at infinity is $K_f = K_i - K_{req} = 2 m v_e^2 - 0.5 m v_e^2 = 1.5 m v_e^2$.
Thus,$\frac{1}{2} m v^2 = \frac{3}{2} m v_e^2$,which gives $v^2 = 3 v_e^2$,so $v = \sqrt{3} v_e$.

(B) The binding energy $(BE)$ of a satellite is defined as the negative of the total mechanical energy $(E)$.
$BE = -E = - (K.E. + P.E.) = - (\frac{GMm}{2r} - \frac{GMm}{r}) = \frac{GMm}{2r}$.
The potential energy $(P.E.)$ of a satellite is given by $P.E. = -\frac{GMm}{r}$.
Comparing the two expressions, we see that $P.E. = -2 \times BE$.
Given $BE = 3.5 \times 10^{8} \,J$.
Therefore, $P.E. = -2 \times (3.5 \times 10^{8} \,J) = -7.0 \times 10^{8} \,J$.

(A) The formula for escape velocity from the surface of a planet of mass $M$ and radius $R$ is given by $V_{e} = \sqrt{\frac{2GM}{R}}$.
For the Earth,$V_{e} = \sqrt{\frac{2GM}{R}}$.
For the new planet,the mass $M^{\prime} = 3M$ and the radius $R^{\prime} = 3R$.
The escape velocity $V_{e}^{\prime}$ for this planet is $V_{e}^{\prime} = \sqrt{\frac{2G(3M)}{3R}}$.
Simplifying the expression,we get $V_{e}^{\prime} = \sqrt{\frac{2GM}{R}} = V_{e}$.
Therefore,the escape velocity remains the same.

(B) Using the law of conservation of energy,the total energy at the surface equals the total energy at the maximum height $h$.
Initial energy at surface: $E_i = \frac{1}{2}mV^2 - \frac{GMm}{R}$
Final energy at height $h$: $E_f = 0 - \frac{GMm}{R+h}$
Equating $E_i = E_f$:
$\frac{1}{2}mV^2 - \frac{GMm}{R} = - \frac{GMm}{R+h}$
$\frac{1}{2}V^2 = GM \left( \frac{1}{R} - \frac{1}{R+h} \right)$
Since $GM = gR^2$:
$\frac{1}{2}V^2 = gR^2 \left( \frac{R+h-R}{R(R+h)} \right) = gR \left( \frac{h}{R+h} \right)$
$\frac{V^2}{2gR} = \frac{h}{R+h}$
$V^2(R+h) = 2gRh$
$V^2R + V^2h = 2gRh$
$V^2R = h(2gR - V^2)$
$h = \frac{V^2R}{2gR - V^2}$

(A) The formula for escape velocity is $V_{e} = \sqrt{\frac{2GM}{R}}$.
For the Earth,$V_{e} = \sqrt{\frac{2GM}{R}}$.
For the given planet,the mass $M' = 6M$ and the radius $R' = 2R$.
The escape velocity $V_{e}'$ for this planet is:
$V_{e}' = \sqrt{\frac{2G(6M)}{2R}}$
$V_{e}' = \sqrt{3 \times \frac{2GM}{R}}$
$V_{e}' = \sqrt{3} V_{e}$.

(B) The gravitational potential energy $U$ of the body of mass $M$ at a distance $r/3$ from the Earth's centre is given by the sum of potentials due to Earth and Moon:
$U = -\frac{G M_{1} M}{r/3} - \frac{G M_{2} M}{2r/3} = -\frac{3 G M_{1} M}{r} - \frac{3 G M_{2} M}{2r} = -\frac{3 G M}{r} \left( M_{1} + \frac{M_{2}}{2} \right)$.
To escape to infinity,the total energy must be at least zero. Thus,the kinetic energy $K$ required is equal to the magnitude of the potential energy:
$K = \frac{1}{2} M V^{2} = |U| = \frac{3 G M}{r} \left( M_{1} + \frac{M_{2}}{2} \right)$.
Solving for $V$:
$V^{2} = \frac{6 G}{r} \left( M_{1} + \frac{M_{2}}{2} \right)$,
$V = \left[ \frac{6 G}{r} \left( M_{1} + \frac{M_{2}}{2} \right) \right]^{\frac{1}{2}}$.
Therefore,the correct option is $B$.

(C) The formula for escape velocity is $v_{e} = \sqrt{\frac{2GM}{R}}$.
Since the mass $M$ of a planet can be expressed in terms of its density $\rho$ and radius $R$ as $M = \text{Volume} \times \text{density} = \frac{4}{3} \pi R^{3} \rho$.
Substituting this into the escape velocity formula:
$v_{e} = \sqrt{\frac{2G}{R} \times \frac{4}{3} \pi R^{3} \rho} = \sqrt{\frac{8}{3} G \pi R^{2} \rho}$.
Simplifying the expression,we get $v_{e} = R \sqrt{\frac{8}{3} G \pi \rho}$.
Since $\frac{8}{3}$,$G$,and $\pi$ are constants,we find that $v_{e} \propto R \sqrt{\rho}$.

(B) The escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.
Given that the projection velocity $v = \frac{1}{3} v_e = \frac{1}{3} \sqrt{\frac{2GM}{R}}$.
Using the law of conservation of energy,the total energy at the surface equals the total energy at the maximum height $h$:
$TE_{\text{surface}} = TE_{\text{height } h}$
$-\frac{GMm}{R} + \frac{1}{2}mv^2 = -\frac{GMm}{R+h} + 0$
Substituting $v^2 = \frac{1}{9} \times \frac{2GM}{R}$:
$-\frac{GMm}{R} + \frac{1}{2}m \left( \frac{2GM}{9R} \right) = -\frac{GMm}{R+h}$
$-\frac{GMm}{R} + \frac{GMm}{9R} = -\frac{GMm}{R+h}$
$-\frac{8GMm}{9R} = -\frac{GMm}{R+h}$
$\frac{8}{9R} = \frac{1}{R+h}$
$8(R+h) = 9R$
$8R + 8h = 9R$
$8h = R$
$h = \frac{R}{8}$

(A) For a satellite to orbit the Earth,the gravitational force provides the necessary centripetal force.
$m r \omega^{2} = \frac{G M m}{r^{2}}$
Here,$r$ is the orbital radius,$M$ is the mass of the Earth,and $\omega$ is the angular velocity.
Simplifying the equation,we get $r^{3} = \frac{G M}{\omega^{2}}$.
We know that the acceleration due to gravity on the Earth's surface is $g = \frac{G M}{R^{2}}$,which implies $G M = g R^{2}$.
Substituting $G M$ in the expression for $r^{3}$,we get $r^{3} = \frac{g R^{2}}{\omega^{2}}$.
Therefore,the radius of the orbit is $r = \left(\frac{g R^{2}}{\omega^{2}}\right)^{1 / 3}$.

(B) The critical velocity (orbital velocity) of a satellite revolving around the Earth at a distance '$r$' from the center is given by the formula:
$v = \sqrt{\frac{GM}{r}}$
where '$G$' is the gravitational constant and '$M$' is the mass of the Earth.
For satellite '$A$' with radius '$R$':
$v_{A} = \sqrt{\frac{GM}{R}}$
For satellite '$B$' with radius '$2R$':
$v_{B} = \sqrt{\frac{GM}{2R}}$
Taking the ratio of the two velocities:
$\frac{v_{A}}{v_{B}} = \frac{\sqrt{\frac{GM}{R}}}{\sqrt{\frac{GM}{2R}}} = \sqrt{\frac{GM}{R} \times \frac{2R}{GM}} = \sqrt{2}$
Therefore,the ratio $\frac{v_{A}}{v_{B}}$ is $\sqrt{2}: 1$.

(C) The orbital velocity of a satellite is given by $v = \sqrt{\frac{GM}{r}}$.
The time period of revolution is $T = \frac{2\pi r}{v} = 2\pi \sqrt{\frac{r^3}{GM}}$.
The frequency of revolution is $f = \frac{1}{T} = \frac{1}{2\pi} \sqrt{\frac{GM}{r^3}}$.
As observed,the frequency '$f$' depends only on the mass of the Earth '$M$' and the radius of the orbit '$r$'.
It is independent of the mass of the satellite '$m$'.
Therefore,the ratio of their frequencies is $1:1$.

(A) The time period $T$ of a satellite revolving around a planet is given by the formula $T = 2 \pi \sqrt{\frac{r^3}{GM}}$,where $r$ is the orbital radius,$G$ is the gravitational constant,and $M$ is the mass of the planet.
From this expression,it is clear that the time period $T$ depends only on the mass of the planet $M$ and the orbital radius $r$.
It does not depend on the mass of the satellite $m$.
Therefore,the period of revolution is independent of the mass of the satellite.

(D) The energy required to raise the satellite to a height $h$ from the surface of the earth is the change in potential energy:
$U = -\frac{GMm}{R+h} - (-\frac{GMm}{R}) = GMm [\frac{1}{R} - \frac{1}{R+h}] = \frac{GMmh}{R(R+h)}$
The energy required to put the satellite into orbit at height $h$ is the kinetic energy required for circular motion:
$K = \frac{1}{2}mv_0^2 = \frac{1}{2}m(\frac{GM}{R+h}) = \frac{GMm}{2(R+h)}$
The ratio of the energy required to raise the satellite to the energy required to orbit is:
$\frac{U}{K} = \frac{GMmh}{R(R+h)} \times \frac{2(R+h)}{GMm} = \frac{2h}{R}$

(D) The kinetic energy $(K)$ of a satellite of mass $m$ revolving at a distance $r$ from the center of the Earth is given by $K = \frac{GMm}{2r}$.
Here,$r = R + h$,where $R$ is the radius of the Earth and $h$ is the height above the surface.
For satellite '$A$',$h_A = 2R$,so $r_A = R + 2R = 3R$.
For satellite '$B$',$h_B = 3R$,so $r_B = R + 3R = 4R$.
The ratio of kinetic energies is $\frac{K_A}{K_B} = \frac{r_B}{r_A} = \frac{4R}{3R} = \frac{4}{3}$.

(D) The orbital period $T$ of a satellite revolving around the Earth is given by the formula $T = 2 \pi \sqrt{\frac{r^3}{GM}}$,where $r$ is the orbital radius,$G$ is the gravitational constant,and $M$ is the mass of the Earth.
Since both satellites are revolving in the same orbit,their orbital radius $r$ is the same.
Because $T$ depends only on the orbital radius $r$ and the mass of the Earth $M$,and not on the mass of the satellite $m$,both satellites will have the same orbital period.
Therefore,the correct statement is that they have the same period.

(A) Let the vertices of the equilateral triangle be $A, B,$ and $C$. Let the mass $m_{2}$ be placed at point $D$,which is the midpoint of side $BC$.
The gravitational forces on $m_{2}$ due to the masses $m_{1}$ at $B$ and $C$ are equal in magnitude and opposite in direction,so they cancel each other out.
The distance of point $D$ from vertex $A$ is the altitude $h$ of the equilateral triangle.
In the right-angled triangle $ABD$,the hypotenuse $AB = \frac{L}{3}$ and $\angle BAD = 30^{\circ}$.
Thus,$h = AB \cos 30^{\circ} = \frac{L}{3} \times \frac{\sqrt{3}}{2} = \frac{L}{2\sqrt{3}}$.
The net force on $m_{2}$ is the force due to the mass $m_{1}$ at $A$:
$F = G \frac{m_{1} m_{2}}{h^{2}} = G \frac{m_{1} m_{2}}{\left(\frac{L}{2\sqrt{3}}\right)^{2}} = G \frac{m_{1} m_{2}}{\frac{L^{2}}{12}} = \frac{12 G m_{1} m_{2}}{L^{2}}$.

(B) The minimum kinetic energy required to escape from the gravitational influence of the earth is given by $K_e = \frac{GMm}{R}$.
Given that the kinetic energy of projection is $K = \frac{1}{2} K_e = \frac{GMm}{2R}$.
At the highest point,the velocity of the body becomes zero,so its kinetic energy is zero.
By the law of conservation of energy,the loss in kinetic energy equals the gain in potential energy:
$K = U_f - U_i$
$\frac{GMm}{2R} = \left( -\frac{GMm}{R+h} \right) - \left( -\frac{GMm}{R} \right)$
$\frac{GMm}{2R} = \frac{GMm}{R} - \frac{GMm}{R+h}$
Dividing both sides by $GMm$:
$\frac{1}{2R} = \frac{1}{R} - \frac{1}{R+h}$
$\frac{1}{R+h} = \frac{1}{R} - \frac{1}{2R} = \frac{1}{2R}$
$R+h = 2R$
$h = R$.

(A) From the first law of thermodynamics,$Q = \Delta U + W$.
Given that $W = \frac{Q}{3}$,the change in internal energy is $\Delta U = Q - W = Q - \frac{Q}{3} = \frac{2}{3} Q$.
For a diatomic gas,the internal energy change is given by $\Delta U = n C_v \Delta T$. Since $n = 1$,$\Delta U = C_v \Delta T = \frac{5}{2} R \Delta T$.
Equating the two expressions for $\Delta U$: $\frac{2}{3} Q = \frac{5}{2} R \Delta T$,which implies $Q = \frac{15}{4} R \Delta T$.
The molar heat capacity $C$ is defined as $C = \frac{Q}{n \Delta T}$.
Substituting $n = 1$ and $Q = \frac{15}{4} R \Delta T$,we get $C = \frac{\frac{15}{4} R \Delta T}{1 \cdot \Delta T} = \frac{15 R}{4}$.

(D) The molar heat capacity at constant volume is given by $C_{v} = \frac{n R}{2}$.
The molar heat capacity at constant pressure is given by $C_{p} = C_{v} + R$.
Substituting the value of $C_{v}$,we get $C_{p} = \frac{n R}{2} + R = R \left(1 + \frac{n}{2}\right)$.
The ratio $\gamma = \frac{C_{p}}{C_{v}}$ is calculated as:
$\gamma = \frac{R \left(1 + \frac{n}{2}\right)}{\frac{n R}{2}} = \frac{\frac{2+n}{2}}{\frac{n}{2}} = \frac{2+n}{n} = 1 + \frac{2}{n}$.

(C) For a monoatomic gas,the number of degrees of freedom is $f_m = 3$. Thus,$\gamma_m = 1 + \frac{2}{3} = \frac{5}{3}$.
For a rigid diatomic gas,the number of degrees of freedom is $f_d = 5$. Thus,$\gamma_d = 1 + \frac{2}{5} = \frac{7}{5}$.
The ratio of $\gamma_d$ to $\gamma_m$ is $\frac{\gamma_d}{\gamma_m} = \frac{7/5}{5/3} = \frac{7}{5} \times \frac{3}{5} = \frac{21}{25}$.

(B) Given,$\frac{R}{C_{V}} = 0.4$.
From Mayer's relation,$C_{P} - C_{V} = R$,so $C_{P} = C_{V} + R$.
Substituting $R = 0.4 C_{V}$,we get $C_{P} = C_{V} + 0.4 C_{V} = 1.4 C_{V}$.
The adiabatic index $\gamma$ is defined as $\gamma = \frac{C_{P}}{C_{V}}$.
Therefore,$\gamma = \frac{1.4 C_{V}}{C_{V}} = 1.4$.
For a rigid diatomic gas,the degrees of freedom $f = 5$.
The adiabatic index is $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{5} = 1 + 0.4 = 1.4$.
Thus,the gas is made up of rigid diatomic molecules.

(C) Given: $\frac{R}{C_{v}} = 0.67$.
We know that the gas constant $R = C_{p} - C_{v}$.
Substituting this into the given equation: $\frac{C_{p} - C_{v}}{C_{v}} = 0.67$.
$\frac{C_{p}}{C_{v}} - 1 = 0.67$.
Since the adiabatic index $\gamma = \frac{C_{p}}{C_{v}}$,we have $\gamma - 1 = 0.67$,which implies $\gamma = 1.67$.
For a monoatomic gas,the degrees of freedom $f = 3$,so $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = 1 + 0.666... \approx 1.67$.
Therefore,the gas is monoatomic.

(D) From the ideal gas equation,we have $PV = nRT$.
Here,$n$ is the number of moles,$R$ is the universal gas constant,and $T$ is the temperature.
We know that the number of moles $n = \frac{N}{N_A}$,where $N$ is the total number of molecules and $N_A$ is Avogadro's number.
Substituting this into the ideal gas equation: $PV = \left(\frac{N}{N_A}\right) RT$.
Rearranging the terms,we get $PV = N \left(\frac{R}{N_A}\right) T$.
Since the Boltzmann constant $k = \frac{R}{N_A}$,the equation becomes $PV = NkT$.
Therefore,$\frac{PV}{kT} = N$,which represents the total number of molecules of the gas.

(C) Initially,each gas has pressure $P$ and volume $V$. When they are mixed,the total volume of the mixture is $V_{1} = V + V = 2V$.
Since the temperature is constant,we use Boyle's Law: $P_{1}V_{1} = P_{2}V_{2}$.
The initial pressure of the mixture $P_{1}$ is $P$ (as the pressure of each gas is $P$ and they are at the same temperature and volume).
Given the final volume $V_{2} = \frac{V}{2}$.
Substituting the values: $P \times (2V) = P_{2} \times (\frac{V}{2})$.
Solving for $P_{2}$: $P_{2} = \frac{P \times 2V}{V/2} = 4P$.

(B) The ideal gas equation is given by $PV = nRT$,where $n$ is the number of moles. Since the number of molecules $N$ is proportional to the number of moles $(N = nN_A)$,the ratio of the number of molecules is equal to the ratio of the number of moles.
For the first sample: $P_1 = P, V_1 = V, T_1 = T$. Thus,$n_1 = \frac{PV}{RT}$.
For the second sample: $P_2 = 2P, V_2 = V/4, T_2 = 2T$. Thus,$n_2 = \frac{(2P)(V/4)}{R(2T)} = \frac{PV/2}{2RT} = \frac{PV}{4RT}$.
The ratio of the number of molecules is $\frac{N_1}{N_2} = \frac{n_1}{n_2} = \frac{PV/RT}{PV/4RT} = \frac{1}{1/4} = 4:1$.

(C) The root mean square (r.m.s.) velocity of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.
In an isothermal process,the temperature $T$ of the gas remains constant.
Since $v_{rms}$ depends only on the temperature $T$ (assuming the gas composition $M$ remains constant),if $T$ is constant,then $v_{rms}$ must also remain constant.
Therefore,when a gas is compressed isothermally,the r.m.s. velocity of its molecules remains the same.

(B) The root mean square ($R$.$M$.$S$.) velocity of a gas molecule is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Given that the $R$.$M$.$S$. velocities of hydrogen $(H_2)$ and oxygen $(O_2)$ are equal,we have $v_{H} = v_{O}$.
Substituting the formula: $\sqrt{\frac{3RT_H}{M_H}} = \sqrt{\frac{3RT_O}{M_O}}$.
Squaring both sides and simplifying: $\frac{T_H}{M_H} = \frac{T_O}{M_O}$.
Given $T_O = 47^{\circ} C = 47 + 273 = 320 \ K$,$M_H = 2$,and $M_O = 32$.
Rearranging for $T_H$: $T_H = T_O \times \frac{M_H}{M_O}$.
Substituting the values: $T_H = 320 \times \frac{2}{32} = 320 \times \frac{1}{16} = 20 \ K$.

(A) The r.m.s. velocity of a gas molecule is given by the formula $C = \sqrt{\frac{3RT}{M}}$,which implies $C \propto \sqrt{T}$.
Given initial temperature $T_{1} = 27^{\circ} C = 27 + 273 = 300 \ K$.
Given final temperature $T_{2} = 327^{\circ} C = 327 + 273 = 600 \ K$.
The ratio of the velocities is $\frac{C_{2}}{C_{1}} = \sqrt{\frac{T_{2}}{T_{1}}}$.
Substituting the values,we get $\frac{C_{2}}{C_{1}} = \sqrt{\frac{600}{300}} = \sqrt{2}$.

(D) According to the Rydberg formula:
$\frac{1}{\lambda} = R \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right]$
In the first case,the electron jumps from the third excited state $(n_i = 4)$ to the second excited state $(n_f = 3)$:
$\frac{1}{\lambda_1} = R \left[ \frac{1}{3^2} - \frac{1}{4^2} \right] = R \left[ \frac{1}{9} - \frac{1}{16} \right] = R \left[ \frac{16 - 9}{144} \right] = \frac{7}{144} R$ .... $(i)$
In the second case,the electron jumps from the second excited state $(n_i = 3)$ to the first excited state $(n_f = 2)$:
$\frac{1}{\lambda_2} = R \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] = R \left[ \frac{1}{4} - \frac{1}{9} \right] = R \left[ \frac{9 - 4}{36} \right] = \frac{5}{36} R$ .... $(ii)$
To find the ratio $\frac{\lambda_1}{\lambda_2}$,we divide the expression for $\frac{1}{\lambda_1}$ by $\frac{1}{\lambda_2}$:
$\frac{\lambda_2}{\lambda_1} = \frac{\frac{5}{36} R}{\frac{7}{144} R} = \frac{5}{36} \times \frac{144}{7} = \frac{5 \times 4}{7} = \frac{20}{7}$
Therefore,the ratio $\frac{\lambda_1}{\lambda_2} = \frac{20}{7}$.

(B) In an $LR$ circuit,the impedance is given by $Z = \sqrt{R^2 + X_L^2}$,where $X_L = \omega L$ is the inductive reactance.
When a capacitor is connected in series,the circuit becomes an $LCR$ circuit.
The new impedance of the $LCR$ circuit is given by $Z' = \sqrt{R^2 + (X_L - X_C)^2}$,where $X_C = \frac{1}{\omega C}$ is the capacitive reactance.
If the circuit was initially inductive ($X_L > X_C$ or $X_L$ is present),adding a capacitor introduces $X_C$,which partially cancels the effect of $X_L$ if $X_L > X_C$,thereby reducing the total impedance $Z$.
Since the current $I = \frac{V}{Z}$,a decrease in impedance $Z$ leads to an increase in the current $I$ flowing through the circuit.

(A) At resonance frequency $f$,the inductive reactance equals the capacitive reactance: $X_L = X_C \implies 2\pi f L = \frac{1}{2\pi f C}$.
When the frequency is doubled to $f' = 2f$,the new inductive reactance is $X_L' = 2\pi(2f)L = 2(2\pi f L) = 2X_L$.
The new capacitive reactance is $X_C' = \frac{1}{2\pi(2f)C} = \frac{1}{2} \left( \frac{1}{2\pi f C} \right) = \frac{1}{2} X_C$.
Since $X_L = X_C$,we can write $X_C = X_L$.
Substituting this into the expression for $X_C'$,we get $X_C' = \frac{1}{2} X_L$.
Since $X_L = \frac{1}{2} X_L'$,we have $X_C' = \frac{1}{2} (\frac{1}{2} X_L') = \frac{1}{4} X_L'$.

(C) The $r.m.s.$ current $I$ in the circuit is given by $I = \frac{V}{Z}$.
Given $V = 220 \ V$ and $Z = 33 \ \Omega$,we have $I = \frac{220}{33} = \frac{20}{3} \ A$.
The true power $P$ consumed in an $a.c.$ circuit is given by $P = I^2 R$.
Substituting the values,$P = \left(\frac{20}{3}\right)^2 \times 18$.
$P = \frac{400}{9} \times 18 = 400 \times 2 = 800 \ W$.

(C) The resonant frequency of an $LCR$ circuit is given by the formula: $\omega = \frac{1}{\sqrt{LC}}$.
Since the resonant frequency remains the same,we have: $\omega_1 = \omega_2$.
Therefore,$\frac{1}{\sqrt{L_1 C_1}} = \frac{1}{\sqrt{L_2 C_2}}$,which implies $L_1 C_1 = L_2 C_2$.
Given $L_1 = L$,$C_1 = C$,and $L_2 = 9 L$,we substitute these values into the equation:
$L \times C = 9 L \times C_2$.
Solving for $C_2$: $C_2 = \frac{L \times C}{9 L} = \frac{C}{9}$.

(B) Given: $R=200 \Omega$, $L=663 \text{ mH}$, $C=26.5 \mu F$, $V_{0}=50 \text{ V}$, $X_{L}=250 \Omega$, $X_{C}=100 \Omega$.
The impedance $Z$ of a series $LCR$ circuit is given by $Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$.
Substituting the values: $Z = \sqrt{200^{2} + (250 - 100)^{2}} = \sqrt{40000 + 150^{2}}$.
$Z = \sqrt{40000 + 22500} = \sqrt{62500} = 250 \Omega$.
The peak current $i_{0}$ is given by $i_{0} = \frac{V_{0}}{Z}$.
$i_{0} = \frac{50}{250} = 0.2 \text{ A}$.

(B) At resonance, the inductive reactance equals the capacitive reactance, i.e., $X_{L} = X_{C}$.
In an $LCR$ series circuit at resonance, the impedance $Z$ is equal to the resistance $R$.
Therefore, the current $I$ in the circuit is given by $I = \frac{V}{R} = \frac{0.2 \, V}{4 \, \Omega} = 0.05 \, A$.
The inductive reactance is $X_{L} = \omega L = \frac{1}{\sqrt{LC}} \times L = \sqrt{\frac{L}{C}}$.
Substituting the given values: $X_{L} = \sqrt{\frac{200 \times 10^{-3} \, H}{80 \times 10^{-6} \, F}} = \sqrt{\frac{200000}{80}} = \sqrt{2500} = 50 \, \Omega$.
The voltage drop across the inductor $V_{L}$ is $V_{L} = I \times X_{L}$.
$V_{L} = 0.05 \, A \times 50 \, \Omega = 2.5 \, V$.

(A) The instantaneous e.m.f. is given by $E = E_{0} \sin \omega t$.
The average power dissipated in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$,where $\cos \phi$ is the power factor.
Here,$V_{rms} = \frac{E_{0}}{\sqrt{2}}$ and $I_{rms} = \frac{V_{rms}}{Z} = \frac{E_{0}}{\sqrt{2} Z}$.
The power factor is $\cos \phi = \frac{R}{Z}$.
Thus,$P = \left( \frac{E_{0}}{\sqrt{2}} \right) \left( \frac{E_{0}}{\sqrt{2} Z} \right) \left( \frac{R}{Z} \right) = \frac{E_{0}^{2} R}{2 Z^{2}}$.
For an $LR$ series circuit,the impedance $Z$ is given by $Z = \sqrt{R^{2} + X_{L}^{2}}$.
Given $X_{L} = R$,we have $Z = \sqrt{R^{2} + R^{2}} = \sqrt{2 R^{2}} = R \sqrt{2}$.
Therefore,$Z^{2} = 2 R^{2}$.
Substituting $Z^{2}$ into the power equation: $P = \frac{E_{0}^{2} R}{2 (2 R^{2})} = \frac{E_{0}^{2} R}{4 R^{2}} = \frac{E_{0}^{2}}{4 R}$.

(B) The capacitive reactance $X_{C}$ is given by the formula: $X_{C} = \frac{1}{2 \pi f C}$.
From this formula,it is clear that the capacitive reactance is inversely proportional to the frequency,i.e.,$X_{C} \propto \frac{1}{f}$.
Let the initial frequency be $f$ and the initial reactance be $X_{C}$.
Let the new frequency be $f' = 4f$ and the new reactance be $X_{C}'$.
Using the proportionality $X_{C} \cdot f = \text{constant}$,we get:
$X_{C}' \cdot f' = X_{C} \cdot f$
$X_{C}' \cdot (4f) = X_{C} \cdot f$
$X_{C}' = \frac{X_{C}}{4}$.

(B) Given: Peak current $I_{0} = \frac{2}{\pi} \ A$,frequency $f = 50 \ Hz$,and mutual inductance $M = 1 \ H$.
The angular frequency is $\omega = 2 \pi f = 2 \pi (50) = 100 \pi \ rad/s$.
The current in the primary coil is $I = I_{0} \sin(\omega t)$.
The induced e.m.f. in the secondary coil is $\mathcal{E} = M \frac{dI}{dt}$.
Differentiating the current with respect to time,$\frac{dI}{dt} = I_{0} \omega \cos(\omega t)$.
The peak value of the induced e.m.f. is $\mathcal{E}_{0} = M \cdot I_{0} \cdot \omega$.
Substituting the values: $\mathcal{E}_{0} = 1 \times (\frac{2}{\pi}) \times (100 \pi) = 200 \ V$.

(A) The given equation for the alternating e.m.f. is $e = e_{0} \sin(\omega t)$.
We know that $\omega = \frac{2\pi}{T}$,so the equation becomes $e = e_{0} \sin\left(\frac{2\pi t}{T}\right)$.
We want to find the time $t$ when $e = \frac{e_{0}}{2}$.
Substituting this into the equation: $\frac{e_{0}}{2} = e_{0} \sin\left(\frac{2\pi t}{T}\right)$.
$\frac{1}{2} = \sin\left(\frac{2\pi t}{T}\right)$.
Since $\sin(30^{\circ}) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$,we have $\frac{2\pi t}{T} = \frac{\pi}{6}$.
Solving for $t$: $t = \frac{\pi}{6} \times \frac{T}{2\pi} = \frac{T}{12}$.

(D) The circular coil is divided into two parts by points $P$ and $Q$ with resistances $R_1 = 30 \Omega$ and $R_2 = 10 \Omega$. These two parts are connected in parallel between points $P$ and $Q$.
The equivalent resistance $R_{PQ}$ of the parallel combination is given by:
$\frac{1}{R_{PQ}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{30} + \frac{1}{10} = \frac{1+3}{30} = \frac{4}{30}$
$R_{PQ} = \frac{30}{4} = 7.5 \Omega$
The total resistance of the circuit including the internal resistance $r = 0.5 \Omega$ is:
$R_{total} = R_{PQ} + r = 7.5 \Omega + 0.5 \Omega = 8 \Omega$
The current $I$ flowing through the circuit is given by Ohm's law:
$I = \frac{V}{R_{total}} = \frac{16 \text{ V}}{8 \Omega} = 2 \text{ A}$

(C) Given frequency $f = 50 \,Hz$. The time period $T$ is given by $T = \frac{1}{f} = \frac{1}{50} = 0.02 \,s$.
The time taken to reach from zero to the maximum value (peak) is one-quarter of the time period:
$t = \frac{T}{4} = \frac{0.02}{4} = 0.005 \,s$.
The peak current $I_0 = 14.14 \,A$. The r.m.s. value of the current $I_{rms}$ is given by:
$I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{14.14}{1.414} = 10 \,A$.
Thus,the time taken is $0.005 \,s$ and the r.m.s. value is $10 \,A$.

(C) Given: Resistance $R = 12 \ \Omega$ and inductive reactance $X_L = 5 \ \Omega$.
In an $L-R$ series circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + X_L^2}$.
Substituting the values: $Z = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \ \Omega$.
The phase angle $\phi$ between the current and the potential difference is given by $\tan \phi = \frac{X_L}{R}$.
Therefore,$\tan \phi = \frac{5}{12}$,which implies $\phi = \tan^{-1}\left(\frac{5}{12}\right)$.
Alternatively,using $\cos \phi = \frac{R}{Z} = \frac{12}{13}$,we get $\phi = \cos^{-1}\left(\frac{12}{13}\right)$.
Since the provided options in the original question were inconsistent with standard calculations,the correct expression is $\tan^{-1}\left(\frac{5}{12}\right)$ or $\cos^{-1}\left(\frac{12}{13}\right)$. Based on the provided choices,option $C$ is the most accurate representation.

(A) In a parallel resonant circuit,the capacitive reactance $(X_{C})$ and inductive reactance $(X_{L})$ are equal $(X_{L} = X_{C})$.
Since the inductor and capacitor are in parallel,the voltage across them is the same.
Therefore,the magnitudes of the currents are equal: $i_{L} = i_{C} = \frac{e}{X_{L}} = \frac{e}{X_{C}}$.
However,the current through the inductor lags the voltage by $90^{\circ}$,and the current through the capacitor leads the voltage by $90^{\circ}$.
Thus,the currents $i_{L}$ and $i_{C}$ are $180^{\circ}$ out of phase with each other.
The net current $i$ in the circuit is the phasor sum of $i_{L}$ and $i_{C}$,which is $i = |i_{L} - i_{C}| = 0$.
Hence,the condition is $i = 0$ and $i_{L} = i_{C} \neq 0$.

(A) Given that,number of turns in primary winding,$N_{p} = 300$.
Number of turns in secondary winding,$N_{s} = 450$.
Primary voltage,$V_{p} = 150 \ V$.
Primary current,$I_{p} = 9 \ A$.
For a transformer,the relationship between voltage and turns is given by $\frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}}$.
Substituting the values: $\frac{V_{s}}{150} = \frac{450}{300}$.
$\Rightarrow V_{s} = \frac{450}{300} \times 150 = 1.5 \times 150 = 225 \ V$.
Assuming an ideal transformer,the power input equals the power output: $V_{p} I_{p} = V_{s} I_{s}$.
Substituting the values: $150 \times 9 = 225 \times I_{s}$.
$\Rightarrow I_{s} = \frac{1350}{225} = 6.0 \ A$.
Thus,the secondary current is $6.0 \ A$ and the secondary voltage is $225 \ V$.

(D) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
For the transition from the $3^{rd}$ orbit to the $2^{nd}$ orbit:
$\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9-4}{36} \right) = \frac{5R}{36}$.
For the transition from the $4^{th}$ orbit to the $3^{rd}$ orbit,let the wavelength be $\lambda'$:
$\frac{1}{\lambda'} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) = R \left( \frac{16-9}{144} \right) = \frac{7R}{144}$.
Now,taking the ratio of the two equations:
$\frac{\lambda'}{\lambda} = \frac{5R/36}{7R/144} = \frac{5}{36} \times \frac{144}{7} = \frac{5 \times 4}{7} = \frac{20}{7}$.
Therefore,$\lambda' = \frac{20}{7} \lambda$.

(A) The magnetic field at the center of a circular current loop is given by $B = \frac{\mu_{0} I}{2r}$.
For an electron revolving in an orbit,the equivalent current is $I = ef = e \left( \frac{v}{2 \pi r} \right) = \frac{ev}{2 \pi r}$.
Substituting this into the magnetic field formula: $B = \frac{\mu_{0} ev}{4 \pi r^{2}}$.
For the ground state of a hydrogen atom,the velocity $v = \frac{e^{2}}{2 \epsilon_{0} h}$ and the radius $r = \frac{h^{2} \epsilon_{0}}{\pi m e^{2}}$.
Substituting $v$ and $r$ into the expression for $B$:
$B = \frac{\mu_{0} e}{4 \pi r^{2}} \left( \frac{e^{2}}{2 \epsilon_{0} h} \right) = \frac{\mu_{0} e^{3}}{8 \pi \epsilon_{0} h r^{2}}$.
Now,substituting $r = \frac{h^{2} \epsilon_{0}}{\pi m e^{2}}$:
$B = \frac{\mu_{0} e^{3}}{8 \pi \epsilon_{0} h} \left( \frac{\pi m e^{2}}{h^{2} \epsilon_{0}} \right)^{2} = \frac{\mu_{0} e^{3}}{8 \pi \epsilon_{0} h} \cdot \frac{\pi^{2} m^{2} e^{4}}{h^{4} \epsilon_{0}^{2}} = \frac{\mu_{0} \pi m^{2} e^{7}}{8 \epsilon_{0}^{3} h^{5}}$.

(B) The Lyman series of the hydrogen atom corresponds to transitions ending at the ground state $(n_f = 1)$,which results in photon emissions in the ultraviolet region.
The Balmer series lies in the visible region.
The Paschen,Brackett,and Pfund series lie in the infrared region.

(B) The Rydberg formula is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)$.
For the Balmer series,$n_{1} = 2$. The series limit occurs at $n_{2} = \infty$. Thus,$\frac{1}{\lambda_{1}} = R \left( \frac{1}{2^{2}} - \frac{1}{\infty^{2}} \right) = \frac{R}{4}$,which implies $\lambda_{1} = \frac{4}{R}$.
For the Brackett series,$n_{1} = 4$. The longest wavelength occurs at $n_{2} = 5$. Thus,$\frac{1}{\lambda_{2}} = R \left( \frac{1}{4^{2}} - \frac{1}{5^{2}} \right) = R \left( \frac{1}{16} - \frac{1}{25} \right) = R \left( \frac{25 - 16}{400} \right) = \frac{9R}{400}$.
This implies $\lambda_{2} = \frac{400}{9R}$.
Now,taking the ratio: $\frac{\lambda_{2}}{\lambda_{1}} = \frac{400}{9R} \times \frac{R}{4} = \frac{100}{9} = 11.11$ (This seems to be a calculation check,let's re-evaluate).
Wait,$\frac{\lambda_{2}}{\lambda_{1}} = \frac{400}{9R} / \frac{4}{R} = \frac{400}{36} = 11.11$. Let's re-check the ratio $\frac{\lambda_{1}}{\lambda_{2}} = \frac{4}{R} \times \frac{9R}{400} = \frac{36}{400} = 0.09$.
Therefore,$\lambda_{1} = 0.09 \lambda_{2}$.

(A) The energy of a photon emitted during a transition is given by $E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the wavelength.
From this relation,we have $E \propto \frac{1}{\lambda}$.
This means that for a transition to emit radiation of maximum wavelength,the energy difference $(E)$ must be minimum.
For a transition to emit radiation of minimum wavelength,the energy difference $(E)$ must be maximum.
Let's calculate the energy differences for the given transitions:
Transition $A$: $\Delta E = 0 - (-2) = 2 \text{ eV}$ (Minimum energy difference)
Transition $B$: $\Delta E = 0 - (-4.5) = 4.5 \text{ eV}$
Transition $C$: $\Delta E = -2 - (-4.5) = 2.5 \text{ eV}$
Transition $D$: $\Delta E = -2 - (-10) = 8 \text{ eV}$ (Maximum energy difference)
Since transition $A$ has the minimum energy difference,it corresponds to the maximum wavelength.
Since transition $D$ has the maximum energy difference,it corresponds to the minimum wavelength.
Therefore,the transitions for maximum and minimum wavelength are $A$ and $D$ respectively.

(B) The Rydberg formula for the hydrogen spectrum is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Comparing this with the given equation $\frac{1}{\lambda} = R \left( \frac{1}{4^2} - \frac{1}{n^2} \right)$,we find that $n_1 = 4$.
The spectral series corresponding to $n_1 = 4$ is the Brackett series.
The Brackett series transitions occur between higher energy levels $(n_2 = 5, 6, 7, \ldots)$ and the $n_1 = 4$ level.
This series lies in the infrared region of the electromagnetic spectrum.

(B) The centripetal acceleration $a$ is given by $a = \frac{v^2}{r}$.
In Bohr's model,the velocity $v \propto \frac{1}{n}$ and the radius $r \propto n^2$.
Substituting these into the expression for acceleration: $a \propto \frac{(1/n)^2}{n^2} = \frac{1}{n^4}$.
Therefore,the ratio of centripetal acceleration for the $3^{\text{rd}}$ orbit $(n_1 = 3)$ to the $5^{\text{th}}$ orbit $(n_2 = 5)$ is:
$\frac{a_3}{a_5} = \frac{n_2^4}{n_1^4} = \frac{5^4}{3^4} = \frac{625}{81}$.

(B) The magnetic field $B$ at the center of a circular current loop is given by $B = \frac{\mu_{0} I}{2r}$.
Here, the current $I$ is due to the revolving electron: $I = \frac{e}{T} = \frac{e}{2 \pi r / v} = \frac{ev}{2 \pi r}$.
Substituting $I$ into the magnetic field formula: $B = \frac{\mu_{0}}{2r} \left( \frac{ev}{2 \pi r} \right) = \frac{\mu_{0} ev}{4 \pi r^{2}}$.
The angular momentum $L$ of the electron is $L = mvr$, which implies $v = \frac{L}{mr}$.
Substituting $v$ into the expression for $B$: $B = \frac{\mu_{0} e}{4 \pi r^{2}} \left( \frac{L}{mr} \right) = \frac{\mu_{0} eL}{4 \pi m r^{3}}$.

(D) According to Bohr's theory,the angular momentum of an electron in the ground state $(n=1)$ is given by $L = mvr = \frac{h}{2 \pi}$.
From this,the velocity $v$ is $v = \frac{h}{2 \pi mR}$.
The time period $T$ of the electron revolving in the orbit is $T = \frac{2 \pi R}{v} = \frac{2 \pi R}{h / (2 \pi mR)} = \frac{4 \pi^2 mR^2}{h}$.
The equivalent current $I$ due to the revolving electron is $I = \frac{e}{T} = \frac{e}{4 \pi^2 mR^2 / h} = \frac{eh}{4 \pi^2 mR^2}$.
The orbital magnetic moment $M$ is given by $M = I \times A$,where $A = \pi R^2$ is the area of the orbit.
Substituting the values,$M = \left( \frac{eh}{4 \pi^2 mR^2} \right) \times (\pi R^2) = \frac{eh}{4 \pi m}$.
Thus,the correct option is $D$.

(A) According to the de-Broglie hypothesis,the wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.
For an electron in the $n^{th}$ Bohr orbit,the quantization condition is $mvr = \frac{nh}{2\pi}$,which implies $mv = \frac{nh}{2\pi r}$.
Substituting this into the wavelength formula: $\lambda = \frac{h}{nh / (2\pi r)} = \frac{2\pi r}{n}$.
For a hydrogen atom,the radius of the $n^{th}$ orbit is $r_n = a_0 n^2$,where $a_0$ is the Bohr radius.
Thus,$\lambda_n = \frac{2\pi (a_0 n^2)}{n} = 2\pi a_0 n$.
Therefore,the ratio of wavelengths for the first $(n=1)$ and second $(n=2)$ orbits is $\frac{\lambda_1}{\lambda_2} = \frac{2\pi a_0 (1)}{2\pi a_0 (2)} = \frac{1}{2}$.

(D) For an electron revolving in a Bohr orbit of a hydrogen atom at a distance $r$ from the nucleus:
Kinetic energy $(K.E.)$ is given by $K = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{e^{2}}{2r} = \frac{e^{2}}{8 \pi \epsilon_{0} r}$.
Potential energy $(P.E.)$ is given by $P = -\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{e^{2}}{r}$.
Taking the ratio of $K.E.$ to $P.E.$:
$\frac{K}{P} = \frac{\frac{e^{2}}{8 \pi \epsilon_{0} r}}{-\frac{e^{2}}{4 \pi \epsilon_{0} r}} = -\frac{4 \pi \epsilon_{0} r}{8 \pi \epsilon_{0} r} = -\frac{1}{2}$.

(C) For a hydrogen-like atom,the energy levels are given by:
$T.E. = -13.6 \frac{Z^2}{n^2} \text{ eV}$
$P.E. = 2 \times T.E. = -27.2 \frac{Z^2}{n^2} \text{ eV}$
$K.E. = -T.E. = 13.6 \frac{Z^2}{n^2} \text{ eV}$
When an electron transitions from an excited state to the ground state,the principal quantum number $n$ decreases.
As $n$ decreases:
$1$. $K.E. = 13.6 \frac{Z^2}{n^2}$ increases.
$2$. $T.E. = -13.6 \frac{Z^2}{n^2}$ becomes more negative,so it decreases.
$3$. $P.E. = -27.2 \frac{Z^2}{n^2}$ becomes more negative,so it decreases.
Therefore,$K.E.$ increases,while $P.E.$ and $T.E.$ decrease.

(D) Given: Radius $r = 0.53 \text{ Å} = 0.53 \times 10^{-10} \text{ m}$.
Time period $T = 1.571 \times 10^{-16} \text{ s}$.
The velocity $v$ of the electron is given by the formula $v = \frac{\text{Distance}}{\text{Time}} = \frac{2 \pi r}{T}$.
Substituting the values: $v = \frac{2 \times 3.142 \times 0.53 \times 10^{-10}}{1.571 \times 10^{-16}}$.
Since $2 \times 3.142 = 6.284$,we have $v = \frac{6.284 \times 0.53 \times 10^{-10}}{1.571 \times 10^{-16}}$.
Note that $\frac{6.284}{1.571} = 4$.
So,$v = 4 \times 0.53 \times 10^{6} \text{ m/s}$.
$v = 2.12 \times 10^{6} \text{ m/s}$.

(C) According to Bohr's model,the energy of an electron in the $n^{th}$ orbit is given by $E_{n} = -\frac{13.6}{n^{2}} \text{ eV}$.
Thus,$E_{n} \propto \frac{1}{n^{2}}$.
The angular momentum of an electron in the $n^{th}$ orbit is given by $L_{n} = \frac{nh}{2\pi}$.
Thus,$L_{n} \propto n$,which implies $n \propto L_{n}$.
Substituting $n \propto L_{n}$ into the energy relation,we get $E_{n} \propto \frac{1}{L_{n}^{2}}$.
Therefore,the correct relation is $E_{n} \propto \frac{1}{L_{n}^{2}}$.

(A) According to Bohr's quantization condition,the angular momentum $L$ is given by $L = \frac{nh}{2\pi}$.
For the second orbit,$n = 2$,so $L = \frac{2h}{2\pi} = \frac{h}{\pi}$.
The rotational kinetic energy $E$ is given by $E = \frac{L^{2}}{2I}$.
Substituting the value of $L$,we get $E = \frac{(h/\pi)^{2}}{2I} = \frac{h^{2}}{2I\pi^{2}}$.

(C) According to Bohr's theory,the centripetal force acting on the electron is given by $F = \frac{mv^2}{r}$.
From Bohr's model,the velocity of the electron is $v \propto \frac{1}{n}$ and the radius of the orbit is $r \propto n^2$.
Substituting these proportionalities into the force equation:
$F \propto \frac{(1/n)^2}{n^2} = \frac{1/n^2}{n^2} = \frac{1}{n^4}$.
Therefore,the force is related to the principal quantum number as $F \propto n^{-4}$.

(C) According to Bohr's theory,the velocity $(v)$ of an electron in the $n^{th}$ orbit is given by $v = \frac{e^{2}}{2 \epsilon_{0} n h}$.
For the first orbit,$n = 1$,so the velocity is $v = \frac{e^{2}}{2 \epsilon_{0} h}$.
The ratio of the velocity of the electron $(v)$ to the velocity of light $(c)$ is given by $\frac{v}{c} = \frac{e^{2}}{2 \epsilon_{0} h c}$.

(D) The energy of a photon emitted during an electronic transition is given by $E = Rhc \left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)$.
Here,$R$ is the Rydberg constant,$h$ is Planck's constant,and $c$ is the speed of light.
$(i)$ For the transition from the second $(n_2 = 2)$ to the first $(n_1 = 1)$ energy level:
$E_{1} = Rhc \left( \frac{1}{1^{2}} - \frac{1}{2^{2}} \right) = Rhc \left( 1 - \frac{1}{4} \right) = \frac{3}{4} Rhc$.
$(ii)$ For the transition from the highest energy level $(n_2 = \infty)$ to the second $(n_1 = 2)$ energy level:
$E_{2} = Rhc \left( \frac{1}{2^{2}} - \frac{1}{\infty^{2}} \right) = Rhc \left( \frac{1}{4} - 0 \right) = \frac{1}{4} Rhc$.
The ratio of the energies is $\frac{E_{1}}{E_{2}} = \frac{\frac{3}{4} Rhc}{\frac{1}{4} Rhc} = \frac{3}{1}$.
Therefore,the ratio is $3: 1$.

(A) For the hydrogen atom,the energy of the $n^{th}$ orbit is given by $E_n = \frac{-13.6}{n^2} \ eV$.
For the second excited state,$n = 3$.
Therefore,the total energy $E_3 = \frac{-13.6}{3^2} = \frac{-13.6}{9} \approx -1.51 \ eV$.
In any Bohr orbit,the kinetic energy $(K.E.)$ is equal to the negative of the total energy $(E)$: $K.E. = -E = -(-1.51 \ eV) = +1.51 \ eV$.
The potential energy $(P.E.)$ is equal to twice the total energy $(E)$: $P.E. = 2E = 2 \times (-1.51 \ eV) = -3.02 \ eV$.
Thus,the kinetic energy is $+1.51 \ eV$ and the potential energy is $-3.02 \ eV$.

(B) The orbital period $T$ is given by $T = \frac{2 \pi r}{v}$.
According to Bohr's model,the radius of the $n^{\text{th}}$ orbit is $r = \frac{n^{2} h^{2} \epsilon_{0}}{\pi m e^{2}}$.
The velocity of the electron in the $n^{\text{th}}$ orbit is $v = \frac{e^{2}}{2 \epsilon_{0} n h}$.
Substituting these expressions into the formula for $T$:
$T = \frac{2 \pi \left( \frac{n^{2} h^{2} \epsilon_{0}}{\pi m e^{2}} \right)}{\left( \frac{e^{2}}{2 \epsilon_{0} n h} \right)}$
$T = \frac{2 n^{2} h^{2} \epsilon_{0}}{m e^{2}} \times \frac{2 \epsilon_{0} n h}{e^{2}}$
$T = \frac{4 \epsilon_{0}^{2} n^{3} h^{3}}{m e^{4}}$.

(B) According to Bohr's model, the radius of the $n^{th}$ orbit of a hydrogen atom is given by the formula $r_n = a_0 n^2$, where $a_0$ is the Bohr radius and $n$ is the principal quantum number.
This implies that the radius $r$ is directly proportional to the square of the principal quantum number, i.e., $r \propto n^2$.
For the first four orbits $(n = 1, 2, 3, 4)$, the radii are proportional to $1^2, 2^2, 3^2, 4^2$.
Calculating these values, we get $1: 4: 9: 16$.
Therefore, the correct option is $B$.

(C) The de-Broglie wavelength $\lambda$ associated with an electron in an orbit of radius $r$ is given by the relation $2\pi r = n\lambda$,where $n$ is the principal quantum number.
For a hydrogen atom,the radius of the $n^{th}$ orbit is $r_n \propto n^2$.
Substituting this into the de-Broglie relation: $2\pi (k n^2) = n\lambda$,which simplifies to $\lambda \propto n$.
The ground state corresponds to $n = 1$.
The third excited state corresponds to $n = 1 + 3 = 4$.
Since the principal quantum number $n$ increases from $1$ to $4$,the de-Broglie wavelength $\lambda$ will increase.

(D) The initial charge stored in the capacitor is given by $Q = CV$.
When the potential is reduced by $V^{\prime}$,the new potential becomes $(V - V^{\prime})$.
The new charge stored is $Q^{\prime} = C(V - V^{\prime})$.
Expanding this,we get $Q^{\prime} = CV - CV^{\prime}$.
Since $Q = CV$,we can substitute $Q$ into the equation: $Q^{\prime} = Q - CV^{\prime}$.
Rearranging the terms to solve for $C$: $CV^{\prime} = Q - Q^{\prime}$.
Therefore,$C = \frac{Q - Q^{\prime}}{V^{\prime}}$.

(A) Assuming the Earth is a solid sphere of radius $R$.
The volume of the sphere is $V = \frac{4}{3} \pi R^{3}$.
The surface area of the sphere is $A = 4 \pi R^{2}$.
Dividing the volume by the surface area,we get:
$\frac{V}{A} = \frac{\frac{4}{3} \pi R^{3}}{4 \pi R^{2}} = \frac{R}{3}$.
From this,we find the radius $R = 3 \frac{V}{A}$.
The capacitance $C$ of an isolated spherical conductor is given by $C = 4 \pi \epsilon_{0} R$.
Substituting the value of $R$,we get:
$C = 4 \pi \epsilon_{0} \left( 3 \frac{V}{A} \right) = 12 \pi \epsilon_{0} \frac{V}{A}$.

(A) When a dielectric material is inserted between the plates of a capacitor,it gets polarized. This polarization creates an internal electric field that opposes the external electric field produced by the charges on the plates. As a result,the net electric field $E$ between the plates decreases. Since the potential difference $V$ is related to the electric field by $V = E \cdot d$ (where $d$ is the distance between the plates),a decrease in the electric field leads to a decrease in the potential difference between the plates for a given charge $Q$. Consequently,the capacitance $C = Q/V$ increases.

(D) The capacitance $C$ of a parallel plate capacitor is given by the formula $C = \frac{\epsilon_0 A}{d}$,where $\epsilon_0$ is the permittivity of free space,$A$ is the area of the plates,and $d$ is the distance between the plates.
From this formula,it is clear that the capacitance $C$ is inversely proportional to the distance $d$ between the plates $(C \propto \frac{1}{d})$.
Therefore,by decreasing the distance $d$ between the plates,the capacitance $C$ of the capacitor increases.

(B) Let the initial capacitance of each capacitor be $C$. When one capacitor is filled with a dielectric of constant $K$,its new capacitance becomes $C' = KC$.
The two capacitors are connected in series to a battery of e.m.f. $V$.
Let $C_1 = KC$ be the capacitance of the dielectric-filled capacitor and $C_2 = C$ be the capacitance of the air-filled capacitor.
The potential difference across the air-filled capacitor $(C_2)$ in a series combination is given by the voltage divider rule:
$V_2 = V \times \frac{C_1}{C_1 + C_2}$
Substituting the values:
$V_2 = V \times \frac{KC}{KC + C}$
$V_2 = V \times \frac{KC}{C(K + 1)}$
$V_2 = \frac{KV}{K + 1}$

(C) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon A}{d}$.
For air as the medium,the capacitance is $C_{0} = \frac{\varepsilon_{0} A}{d} = 3 \mu F$.
When a dielectric medium is introduced,the new capacitance is $C = \frac{\varepsilon A}{d} = 15 \mu F$.
Taking the ratio of the two capacitances:
$\frac{C}{C_{0}} = \frac{\varepsilon}{\varepsilon_{0}} = K$ (where $K$ is the dielectric constant).
$\frac{15}{3} = \frac{\varepsilon}{\varepsilon_{0}} \implies 5 = \frac{\varepsilon}{\varepsilon_{0}}$.
Therefore,the permittivity of the medium is $\varepsilon = 5 \times \varepsilon_{0}$.
$\varepsilon = 5 \times 8.85 \times 10^{-12} = 44.25 \times 10^{-12} \text{ F/m}$.
Converting to scientific notation: $\varepsilon = 0.4425 \times 10^{-10} \text{ F/m}$.

(A) The energy stored in a capacitor is given by $U = \frac{Q^2}{2C}$.
When the battery is removed,the charge $Q$ on the capacitor plates remains constant.
The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$,where $d$ is the distance between the plates.
If the distance $d$ is increased to $d' = 4d$,the new capacitance becomes $C' = \frac{\epsilon_0 A}{4d} = \frac{C}{4}$.
The new energy stored $U'$ is given by $U' = \frac{Q^2}{2C'} = \frac{Q^2}{2(C/4)} = 4 \times \frac{Q^2}{2C} = 4U$.
Therefore,the energy stored becomes $4U$.

(B) The energy density $u$ of an electric field $E$ is given by $u = \frac{1}{2} \epsilon_{0} E^{2}$.
The volume $V$ of the space between the plates of a parallel plate capacitor is the product of the area $A$ and the distance $d$,so $V = Ad$.
The total energy $U$ stored in the capacitor is the product of the energy density and the volume.
Therefore,$U = u \times V = (\frac{1}{2} \epsilon_{0} E^{2}) \times (Ad) = \frac{1}{2} \epsilon_{0} E^{2} Ad$.

(C) Let $C$ be the capacitance of each capacitor. The equivalent capacitance of three capacitors in parallel combination is $C_{p} = 3C$ and in series combination is $C_{s} = C/3$.
Let $V_{p}$ be the potential difference in the parallel combination and $V_{s}$ be the potential difference in the series combination. Given that the energy stored in both cases is the same:
$\frac{1}{2} C_{p} V_{p}^{2} = \frac{1}{2} C_{s} V_{s}^{2}$
$\frac{V_{p}^{2}}{V_{s}^{2}} = \frac{C_{s}}{C_{p}} = \frac{C/3}{3C} = \frac{1}{9}$
Taking the square root on both sides:
$\frac{V_{p}}{V_{s}} = \frac{1}{3}$

(B) Let the capacity of each capacitor be $C$.
When two capacitors are joined in series,the equivalent capacity $C_s$ is given by:
$\frac{1}{C_s} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C} \implies C_s = \frac{C}{2}$.
When two capacitors are joined in parallel,the equivalent capacity $C_p$ is given by:
$C_p = C + C = 2C$.
The ratio of the resultant capacity in series to that in parallel is:
$\frac{C_s}{C_p} = \frac{C/2}{2C} = \frac{1}{4}$.
Thus,the ratio is $1: 4$.

(C) Let the capacitance of each capacitor be $C$.
When connected in series,the equivalent capacitance $C_{1}$ is given by $\frac{1}{C_{1}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{4}{C}$,so $C_{1} = \frac{C}{4}$.
When connected in parallel,the equivalent capacitance $C_{2}$ is given by $C_{2} = C + C + C + C = 4C$.
Now,we calculate the ratio $\frac{C_{2}}{C_{1}}$:
$\frac{C_{2}}{C_{1}} = \frac{4C}{C/4} = 4 \times 4 = 16$.

(D) When capacitors are connected in series,the charge $Q$ remains the same on all the capacitors.
Given the relation $Q = C V$,the potential difference across each capacitor is $V_i = \frac{Q}{C_i}$.
Since $Q$ is constant for all capacitors in series,the potential difference $V_i$ is inversely proportional to the capacitance $C_i$.
Therefore,the ratio of potentials $V_1 : V_2 : V_3$ is given by:
$V_1 : V_2 : V_3 = \frac{Q}{C_1} : \frac{Q}{C_2} : \frac{Q}{C_3}$
$V_1 : V_2 : V_3 = \frac{1}{C_1} : \frac{1}{C_2} : \frac{1}{C_3}$