MHT CET 2024 Physics Question Paper with Answer and Solution

(D) The height of capillary rise is given by $h = \frac{2T \cos \theta}{r \rho g}$.
For a given value of $T$ and $r$,we have $h \propto \frac{\cos \theta}{\rho}$.
Since the liquids rise to the same height $(h_1 = h_2 = h_3)$,we have $\frac{\cos \theta_1}{\rho_1} = \frac{\cos \theta_2}{\rho_2} = \frac{\cos \theta_3}{\rho_3}$.
Given $\rho_1 > \rho_2 > \rho_3$,it follows that $\cos \theta_1 > \cos \theta_2 > \cos \theta_3$.
Since the cosine function is a decreasing function in the interval $[0, \frac{\pi}{2}]$,we have $\theta_1 < \theta_2 < \theta_3$.
Therefore,$0 \le \theta_1 < \theta_2 < \theta_3 < \frac{\pi}{2}$.

(D) diatomic gas molecule typically has $3$ translational degrees of freedom and $2$ rotational degrees of freedom.
It is given that the molecule has one additional vibrational mode.
Each vibrational mode contributes $2$ degrees of freedom (one for kinetic energy and one for potential energy).
Therefore, the total number of degrees of freedom $f = 3$ (translational) $+ 2$ (rotational) $+ 2$ (vibrational) $= 7$.
The molar specific heat at constant volume is given by the formula $C_V = \frac{fR}{2}$.
Substituting $f = 7$, we get $C_V = \frac{7R}{2}$.

(C) Let $x_1$ and $x_2$ be the initial distances of particles $m_1$ and $m_2$ from the centre of mass $C$ respectively. For the centre of mass to remain at $C$,the condition is $m_1 x_1 = m_2 x_2$ ... $(i)$
When particle $m_1$ is moved by distance $d$ towards the centre of mass,its new distance from $C$ becomes $(x_1 - d)$.
Let the second particle $m_2$ be moved by a distance $d'$ to keep the centre of mass unchanged. Its new distance from $C$ will be $(x_2 - d')$.
For the centre of mass to remain unchanged,the new condition is $m_1(x_1 - d) = m_2(x_2 - d')$ ... (ii)
Expanding equation (ii): $m_1 x_1 - m_1 d = m_2 x_2 - m_2 d'$
Since $m_1 x_1 = m_2 x_2$ from equation $(i)$,we can substitute this into the expanded equation:
$m_2 x_2 - m_1 d = m_2 x_2 - m_2 d'$
$-m_1 d = -m_2 d'$
$d' = \frac{m_1}{m_2} d$
Since the displacement $d'$ is positive in the direction towards the centre of mass,the second particle must also be moved towards the centre of mass.

(D) For the rod to be in rotational equilibrium,the sum of torques about any point must be zero.
Taking the torque about the knife-edge $B$:
$\sum \tau_B = 0$
$N_A \cdot r - W \cdot (r - x) = 0$
Where $N_A$ is the normal reaction at $A$.
$N_A \cdot r = W(r - x)$
$N_A = \frac{W(r - x)}{r}$

(B) The centre of mass of a system of two particles is defined by the position vector $\vec{R}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2}$.
Since the position vector of the centre of mass is a weighted average of the position vectors of the two particles,it must lie on the line segment connecting the two particles.
If the masses are equal $(m_1 = m_2)$,the centre of mass lies exactly at the midpoint.
If the masses are different,it lies closer to the heavier mass,but it always remains on the line joining the two particles.
Therefore,option $B$ is correct.

(B) Since the three balls are identical and have the same radius,their masses are equal. Let the mass of each ball be '$m$'.
Let the coordinates of the centres of the three balls be $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$.
The centre of mass $(X_{cm}, Y_{cm})$ of the system is given by:
$X_{cm} = \frac{m x_1 + m x_2 + m x_3}{m + m + m} = \frac{x_1 + x_2 + x_3}{3}$
$Y_{cm} = \frac{m y_1 + m y_2 + m y_3}{m + m + m} = \frac{y_1 + y_2 + y_3}{3}$
These coordinates represent the centroid of the triangle formed by the centres of the balls.
The centroid of a triangle is the point of intersection of its medians.

(A) Let $u$ be the initial velocity of mass $m$ and $0$ be the initial velocity of mass $M$.
Let $v_1$ be the final velocity of mass $m$ and $v_2$ be the final velocity of mass $M$.
Given that the particle $m$ stops after the collision,$v_1 = 0$.
By the law of conservation of linear momentum: $mu + M(0) = m(0) + Mv_2$.
This simplifies to $mu = Mv_2$,so $v_2 = \frac{mu}{M}$.
The coefficient of restitution $e$ is defined as $e = \frac{v_2 - v_1}{u_1 - u_2}$.
Substituting the known values: $e = \frac{\frac{mu}{M} - 0}{u - 0} = \frac{mu/M}{u} = \frac{m}{M}$.

(D) The external force acting on the system of two particles is the gravitational force: $F_{ext} = (m_1 + m_2)g$ (acting downwards).
According to Newton's second law,the rate of change of momentum is equal to the external force: $F_{ext} = \frac{\Delta P}{\Delta t}$.
Here,$\Delta P = (m_1 \vec{V}_1^{\prime} + m_2 \vec{V}_2^{\prime}) - (m_1 \vec{V}_1 + m_2 \vec{V}_2)$ and $\Delta t = 2t - 0 = 2t$.
Substituting these into the equation: $(m_1 + m_2)g = \frac{(m_1 \vec{V}_1^{\prime} + m_2 \vec{V}_2^{\prime}) - (m_1 \vec{V}_1 + m_2 \vec{V}_2)}{2t}$.
Therefore,the change in momentum is: $[(m_1 \vec{V}_1^{\prime} + m_2 \vec{V}_2^{\prime}) - (m_1 \vec{V}_1 + m_2 \vec{V}_2)] = 2(m_1 + m_2)gt$.

(A) The kinetic energy $(K.E.)$ of a body is related to its momentum $(p)$ by the formula: $K.E. = \frac{p^2}{2m}$.
When the momentum increases by $20 \%$,the new momentum $p'$ becomes:
$p' = p + 0.20p = 1.2p$.
The new kinetic energy $(K.E.')$ is:
$K.E.' = \frac{(p')^2}{2m} = \frac{(1.2p)^2}{2m} = \frac{1.44p^2}{2m}$.
Since $K.E. = \frac{p^2}{2m}$,we have $K.E.' = 1.44 \times K.E$.
The percentage increase in kinetic energy is given by:
$\Delta K.E. \% = \frac{K.E.' - K.E.}{K.E.} \times 100 \%$
$= \frac{1.44 K.E. - K.E.}{K.E.} \times 100 \%$
$= 0.44 \times 100 \% = 44 \%$.

(C) Let the initial velocity of mass $m_1$ be $v_1$ and the final velocity of mass $m_2$ be $v_2$.
According to the given condition,the final velocity of mass $m_1$ is $v_1' = \frac{v_1}{1.5} = \frac{2}{3} v_1$.
Assuming an elastic collision,the coefficient of restitution $e = 1$.
The formula for the coefficient of restitution is $e = \frac{v_2 - v_1'}{v_1 - 0}$.
Substituting the values: $1 = \frac{v_2 - \frac{2}{3}v_1}{v_1} \implies v_2 = v_1 + \frac{2}{3}v_1 = \frac{5}{3}v_1$.
By the law of conservation of linear momentum: $m_1 v_1 = m_1 v_1' + m_2 v_2$.
Substituting the velocities: $m_1 v_1 = m_1 (\frac{2}{3} v_1) + m_2 (\frac{5}{3} v_1)$.
Rearranging the terms: $m_1 v_1 - \frac{2}{3} m_1 v_1 = m_2 \frac{5}{3} v_1$.
$\frac{1}{3} m_1 v_1 = \frac{5}{3} m_2 v_1$.
Therefore,$\frac{m_1}{m_2} = \frac{5}{1}$.

(B) Given that:
Number of balls $N = 1000 = 10^3$
Mass of each ball $m = 1 \text{ g} = 10^{-3} \text{ kg}$
Area $A = 1 \text{ cm}^2 = 10^{-4} \text{ m}^2$
Velocity $v = 50 \text{ m/s}$
The change in momentum for each collision is $\Delta p = m[v - (-v)] = 2mv$.
Since there are $N$ collisions per second,the total force $F$ exerted on the surface is:
$F = N \times \Delta p = N \times 2mv$
$F = 10^3 \times 2 \times 10^{-3} \text{ kg} \times 50 \text{ m/s} = 100 \text{ N}$.
Pressure $P$ is defined as force per unit area:
$P = \frac{F}{A} = \frac{100 \text{ N}}{10^{-4} \text{ m}^2} = 10^6 \text{ N/m}^2$.

(D) The acceleration due to gravity at a height $h$ above the earth's surface is given by the formula: $g' = g \left( \frac{R}{R+h} \right)^2$.
Given that $g' = \frac{g}{n}$,we can write: $\frac{g}{n} = g \left( \frac{R}{R+h} \right)^2$.
Dividing both sides by $g$,we get: $\frac{1}{n} = \left( \frac{R}{R+h} \right)^2$.
Taking the square root of both sides: $\frac{1}{\sqrt{n}} = \frac{R}{R+h}$.
Rearranging the equation to solve for $h$: $R+h = R \sqrt{n}$.
Therefore,$h = R \sqrt{n} - R = R(\sqrt{n}-1)$.

(B) The weight of a body at height $h$ is given by $W_h = W \left( \frac{R}{R+h} \right)^2$,where $W$ is the weight on the surface of the earth.
Given that $W_h = \frac{1}{16} W$,we have:
$\frac{1}{16} W = W \left( \frac{R}{R+h} \right)^2$
$\frac{1}{16} = \left( \frac{R}{R+h} \right)^2$
Taking the square root on both sides:
$\frac{1}{4} = \frac{R}{R+h}$
$R + h = 4R$
$h = 3R$
Therefore,the height at which the weight becomes $\frac{1}{16}$ of its surface weight is $3R$.

(B) The acceleration due to gravity at a height $h$ above the earth's surface is given by the formula: $g_h = g \left( \frac{R}{R+h} \right)^2$.
Given that $g_h = \frac{g}{3}$,we substitute this into the equation:
$\frac{g}{3} = g \left( \frac{R}{R+h} \right)^2$.
Dividing both sides by $g$:
$\frac{1}{3} = \left( \frac{R}{R+h} \right)^2$.
Taking the square root of both sides:
$\frac{1}{\sqrt{3}} = \frac{R}{R+h}$.
Rearranging the terms to solve for $h$:
$R + h = \sqrt{3} R$.
$h = \sqrt{3} R - R$.
$h = (\sqrt{3} - 1) R$.

(B) The frequency of the pendulum at the surface is given by $f = \frac{1}{2 \pi} \sqrt{\frac{g}{l}}$.
At a depth $d$,the gravitational acceleration is given by $g_{eff} = g \left(1 - \frac{d}{R}\right)$.
For $d = \frac{R}{4}$,the effective acceleration is $g_{eff} = g \left(1 - \frac{R/4}{R}\right) = g \left(1 - \frac{1}{4}\right) = \frac{3}{4} g$.
The frequency at depth $d$ is $f_d = \frac{1}{2 \pi} \sqrt{\frac{g_{eff}}{l}} = \frac{1}{2 \pi} \sqrt{\frac{3g}{4l}}$.
Taking the ratio of the new frequency to the original frequency $n$:
$\frac{f_d}{n} = \frac{\frac{1}{2 \pi} \sqrt{\frac{3g}{4l}}}{\frac{1}{2 \pi} \sqrt{\frac{g}{l}}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.
Therefore,the new frequency is $f_d = \frac{\sqrt{3}}{2} n$.

(B) The acceleration due to gravity at the surface of the earth is $g = \frac{GM}{R^2}$.
At a height $h$,the acceleration due to gravity is $g_h = \frac{GM}{(R+h)^2}$.
Given that the value of $g$ is reduced by $64 \%$,the remaining value is $g_h = (100 \% - 64 \%) \text{ of } g = 36 \% \text{ of } g = 0.36g$.
Therefore,$\frac{g_h}{g} = 0.36 = \frac{36}{100}$.
Substituting the expressions for $g$ and $g_h$:
$\frac{R^2}{(R+h)^2} = \frac{36}{100}$.
Taking the square root on both sides:
$\frac{R}{R+h} = \frac{6}{10} = \frac{3}{5}$.
Cross-multiplying gives:
$5R = 3(R+h) = 3R + 3h$.
$2R = 3h$.
$h = \frac{2}{3} R$.

(B) The acceleration due to gravity at a depth $d$ is given by the formula: $g_d = g\left(1 - \frac{d}{R}\right)$.
Given that $g_d = \frac{g}{n-1}$,we substitute this into the equation:
$\frac{g}{n-1} = g\left(1 - \frac{d}{R}\right)$.
Dividing both sides by $g$:
$\frac{1}{n-1} = 1 - \frac{d}{R}$.
Rearranging the terms to solve for $d$:
$\frac{d}{R} = 1 - \frac{1}{n-1}$.
$\frac{d}{R} = \frac{n-1-1}{n-1} = \frac{n-2}{n-1}$.
Therefore,$d = R\left(\frac{n-2}{n-1}\right)$.

(B) The acceleration due to gravity $g$ at the surface of a planet is given by $g = \frac{GM}{R^2}$.
Since mass $M = \text{Volume} \times \text{Density} = \frac{4}{3} \pi R^3 \rho$,we can write $g = \frac{G}{R^2} \times \frac{4}{3} \pi R^3 \rho = \frac{4}{3} \pi R \rho G$.
Given that $g$ is the same for both the earth and the planet,we have $R_e \rho_e = R_p \rho_p$.
Here,$R_e = R$ and $\rho_p = 3 \rho_e$.
Substituting these values: $R \times \rho_e = R_p \times (3 \rho_e)$.
Therefore,$R_p = \frac{R}{3}$.

(D) The weight of an object is given by $W = mg$,where $m$ is the mass and $g$ is the acceleration due to gravity.
At the surface of the earth $(h=0)$,the acceleration due to gravity $g$ is maximum.
As we move into a mine of depth $h_1$,the acceleration due to gravity is $g_1 = g(1 - h_1/R)$,so $W_1 = mg(1 - h_1/R) < W_2$.
As we move to the top of a mountain of height $h_3$,the acceleration due to gravity is $g_3 = g(1 - 2h_3/R)$,so $W_3 = mg(1 - 2h_3/R) < W_2$.
Since $W_2$ is the weight at the surface,it is greater than both $W_1$ and $W_3$.
Therefore,the correct relation is $W_1 < W_2 > W_3$.

(B) The acceleration due to gravity $g$ at the surface of a planet is given by the formula: $g = \frac{4}{3} \pi \rho G R$,where $\rho$ is the density,$G$ is the gravitational constant,and $R$ is the radius of the planet.
Given that the density of the planet $\rho_p = 2 \rho_e$ and the acceleration due to gravity $g_p = g_e$.
Equating the expressions for both the planet and the earth:
$\frac{4}{3} \pi \rho_p G R_p = \frac{4}{3} \pi \rho_e G R_e$
$\rho_p R_p = \rho_e R_e$
Substituting $\rho_p = 2 \rho_e$ and $R_e = R$:
$(2 \rho_e) R_p = \rho_e R$
$2 R_p = R$
$R_p = \frac{R}{2}$

(A) The effective acceleration due to gravity $g'$ at a latitude $\theta$ is given by $g' = g - R\omega^2 \cos^2 \theta$.
At the equator,the latitude $\theta = 0^\circ$,so $\cos 0^\circ = 1$. Thus,$g' = g - R\omega^2$.
Given that the weight becomes $\frac{3}{5}$ of its present value,the effective gravity $g'$ must be $\frac{3}{5}g$.
Substituting this into the equation: $\frac{3}{5}g = g - R\omega^2$.
Rearranging the terms: $R\omega^2 = g - \frac{3}{5}g = \frac{2}{5}g$.
Solving for $\omega$: $\omega^2 = \frac{2g}{5R}$,which gives $\omega = \sqrt{\frac{2g}{5R}}$.

(A) The weight of an object at the surface of the earth is $W = mg = 72 \ N$.
At a height $h$ above the surface,the acceleration due to gravity $g_h$ is given by the formula: $g_h = g \left( \frac{R}{R+h} \right)^2$.
Given that the height $h = \frac{R}{2}$,we substitute this into the formula:
$g_h = g \left( \frac{R}{R + \frac{R}{2}} \right)^2 = g \left( \frac{R}{\frac{3R}{2}} \right)^2 = g \left( \frac{2}{3} \right)^2 = \frac{4}{9} g$.
The weight at height $h$ is $W_h = m g_h = m \left( \frac{4}{9} g \right) = \frac{4}{9} W$.
Substituting $W = 72 \ N$:
$W_h = \frac{4}{9} \times 72 = 4 \times 8 = 32 \ N$.

(C) The particle on the equator of the earth has linear speed $V = R \omega$,where $R$ is the radius of the earth and $\omega$ is the angular velocity of the earth's rotation.
At a latitude $\theta$,the particle moves in a circle of radius $r = R \cos \theta$.
The linear speed $V'$ of the particle at latitude $\theta$ is given by $V' = r \omega = (R \cos \theta) \omega$.
Substituting $\theta = 30^{\circ}$:
$V' = R \omega \cos 30^{\circ} = R \omega \left( \frac{\sqrt{3}}{2} \right)$.
Since $V = R \omega$,we have $V' = \frac{\sqrt{3}}{2} V$.

(C) For a point outside the sphere $(r > R)$,the gravitational field is given by $F = \frac{G M}{r^2}$.
Thus,for $r_1 > R$,$F_1 = \frac{G M}{r_1^2}$.
For a point inside the sphere $(r < R)$,the gravitational field is given by $F = \frac{G M r}{R^3}$.
Thus,for $r_2 < R$,$F_2 = \frac{G M r_2}{R^3}$.
Taking the ratio of $F_1$ to $F_2$:
$\frac{F_1}{F_2} = \frac{G M}{r_1^2} \times \frac{R^3}{G M r_2} = \frac{R^3}{r_1^2 r_2}$.

(C) The mass of the Earth $M$ can be expressed in terms of its volume $V$ and mean density $\rho$ as:
$M = V \rho = \frac{4}{3} \pi R^3 \rho$ ... $(i)$
The acceleration due to gravity $g$ on the surface of the Earth is given by:
$g = \frac{GM}{R^2}$ ... $(ii)$
Substituting the expression for $M$ from equation $(i)$ into equation $(ii)$:
$g = \frac{G}{R^2} \times \left( \frac{4}{3} \pi R^3 \rho \right)$
$g = \frac{4}{3} \pi R G \rho$
Rearranging the equation to solve for the mean density $\rho$:
$\rho = \frac{3 g}{4 \pi R G}$

(D) The gravitational potential energy $U$ of the body of mass $M$ at a distance $r/3$ from the Earth and $2r/3$ from the moon is given by:
$U = -\frac{G M_1 M}{r/3} - \frac{G M_2 M}{2r/3} = -\frac{3 G M_1 M}{r} - \frac{3 G M_2 M}{2r} = -\frac{3 G M}{r} \left( M_1 + \frac{M_2}{2} \right)$.
To escape to infinity,the total energy must be at least zero. If $v$ is the required escape speed,the kinetic energy is $\frac{1}{2} M v^2$.
By conservation of energy: $\frac{1}{2} M v^2 + U = 0$.
$\frac{1}{2} M v^2 = \frac{3 G M}{r} \left( M_1 + \frac{M_2}{2} \right)$.
Solving for $v$:
$v^2 = \frac{6 G}{r} \left( M_1 + \frac{M_2}{2} \right)$.
$v = \left[ \frac{6 G}{r} \left( M_1 + \frac{M_2}{2} \right) \right]^{1/2}$.

(A) The escape velocity $v_e$ is given by the formula: $v_e = \sqrt{\frac{2GM}{R}}$.
Since the mass $M$ of a planet can be expressed in terms of its density $\rho$ and radius $R$ as $M = \frac{4}{3} \pi R^3 \rho$, we substitute this into the formula:
$v_e = \sqrt{\frac{2G}{R} \cdot \frac{4}{3} \pi R^3 \rho} = \sqrt{\frac{8}{3} G \pi \rho R^2} = R \sqrt{\frac{8}{3} G \pi \rho}$.
Given that the density $\rho$ is the same for both the Earth and the planet, we have $v_e \propto R$.
Let $v_e$ be the escape velocity of Earth and $v_e'$ be the escape velocity of the planet.
Given $R' = 2R$, we have:
$\frac{v_e'}{v_e} = \frac{R'}{R} = \frac{2R}{R} = 2$.
Therefore, $v_e' = 2 \times v_e = 2 \times 11 \,km/s = 22 \,km/s$.

(C) The escape velocity $v_e$ of a planet is given by the formula: $v_e = \sqrt{\frac{2GM}{R}}$.
Since the mass $M$ of a planet can be expressed in terms of its density $\rho$ and radius $R$ as $M = \frac{4}{3}\pi R^3 \rho$,we substitute this into the escape velocity formula:
$v_e = \sqrt{\frac{2G}{R} \cdot \frac{4}{3}\pi R^3 \rho} = \sqrt{\frac{8}{3}G\pi\rho R^2} = R \sqrt{\frac{8}{3}G\pi\rho}$.
Given that the densities $\rho$ are the same for both the planet and the earth,the escape velocity is directly proportional to the radius: $v_e \propto R$.
Therefore,the ratio of the escape velocities is: $\frac{V_p}{V_E} = \frac{R_p}{R_E}$.
Given that the radius of the planet is double that of the earth $(R_p = 2R_E)$,we have:
$\frac{V_p}{V_E} = \frac{2R_E}{R_E} = 2$.
Thus,the value of $x$ is $2$.

(B) The escape velocity $v_e$ is given by the formula $v_e = \sqrt{\frac{2GM}{R}}$.
Substituting $M = \frac{4}{3} \pi R^3 \rho$,we get $v_e = \sqrt{\frac{2G}{R} \cdot \frac{4}{3} \pi R^3 \rho} = R \sqrt{\frac{8}{3} \pi G \rho}$.
This implies $v_e \propto R \sqrt{\rho}$.
Given $R_p = 4R_E$ and $\rho_p = 4\rho_E$,the ratio of escape velocity on Earth $(v_E)$ to that on the planet $(v_p)$ is:
$\frac{v_E}{v_p} = \frac{R_E \sqrt{\rho_E}}{R_p \sqrt{\rho_p}} = \frac{R_E \sqrt{\rho_E}}{(4R_E) \sqrt{4\rho_E}} = \frac{1}{4 \times 2} = \frac{1}{8}$.
Thus,the ratio is $1: 8$.

(B) According to the law of conservation of energy,the total mechanical energy at the initial position is equal to the total mechanical energy at the surface of the earth.
Initial energy $E_i = K_i + U_i = 0 - \frac{GMm}{R_0}$.
Final energy $E_f = K_f + U_f = \frac{1}{2}mv^2 - \frac{GMm}{R}$.
Equating $E_i = E_f$:
$-\frac{GMm}{R_0} = \frac{1}{2}mv^2 - \frac{GMm}{R}$.
Rearranging the terms:
$\frac{1}{2}mv^2 = GMm\left(\frac{1}{R} - \frac{1}{R_0}\right)$.
$v^2 = 2GM\left(\frac{1}{R} - \frac{1}{R_0}\right)$.
Therefore,the velocity $v$ is:
$v = \sqrt{2GM\left(\frac{1}{R} - \frac{1}{R_0}\right)}$.

(A) According to the law of conservation of energy:
$(K.E + P.E)_{\text{surface}} = (K.E + P.E)_{\text{max height}}$
Since the velocity at maximum height is $0$,we have:
$-\frac{GMm}{R} + \frac{1}{2} m(n V_e)^2 = -\frac{GMm}{R + h} + 0$
We know that the escape velocity $V_e = \sqrt{\frac{2GM}{R}}$,so $V_e^2 = \frac{2GM}{R}$.
Substituting this into the equation:
$\frac{1}{2} m n^2 \left( \frac{2GM}{R} \right) = GMm \left( \frac{1}{R} - \frac{1}{R + h} \right)$
$\frac{n^2 GMm}{R} = GMm \left( \frac{R + h - R}{R(R + h)} \right)$
$\frac{n^2}{R} = \frac{h}{R(R + h)}$
$n^2 = \frac{h}{R + h}$
$n^2(R + h) = h$
$n^2 R + n^2 h = h$
$n^2 R = h(1 - n^2)$
$h = \frac{n^2 R}{1 - n^2}$

(A) The orbital speed of a planet at a distance $r$ from the sun is given by the formula: $v_{\text{orb}} = \sqrt{\frac{GM}{r}}$.
For planet $A$,the orbital speed is $v_A = v = \sqrt{\frac{GM}{r_A}}$.
For planet $B$,the orbital speed is $v_B = \sqrt{\frac{GM}{r_B}}$.
Substituting $r_B = 100 r_A$ into the expression for $v_B$:
$v_B = \sqrt{\frac{GM}{100 r_A}} = \frac{1}{\sqrt{100}} \sqrt{\frac{GM}{r_A}} = \frac{1}{10} \sqrt{\frac{GM}{r_A}}$.
Since $v = \sqrt{\frac{GM}{r_A}}$,we have $v_B = \frac{v}{10}$.

(A) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is directly proportional to the cube of the semi-major axis $(r)$ of the orbit: $T^2 \propto r^3$.
This implies that $\frac{T_p^2}{T_e^2} = \frac{r_p^3}{r_e^3}$,where $T_p$ and $r_p$ are the time period and orbital radius of the planet,and $T_e$ and $r_e$ are those of the earth.
Given that $T_p = 8 T_e$,we substitute this into the ratio:
$\frac{(8 T_e)^2}{T_e^2} = \frac{r_p^3}{r_e^3}$
$64 = \frac{r_p^3}{r_e^3}$
Taking the cube root on both sides:
$\frac{r_p}{r_e} = (64)^{1/3} = 4$.
Therefore,the ratio of the radius of the planet's orbit to the radius of the earth's orbit is $4$.

(C) From Kepler's third law,the time period $T$ is given by $T = 2 \pi \sqrt{\frac{r^3}{GM}}$.
Since the satellite is revolving close to the surface of the planet,we take the orbital radius $r = R$.
Thus,$T = 2 \pi \sqrt{\frac{R^3}{GM}} \dots (i)$.
The mass $M$ of the planet can be expressed in terms of its density $\rho$ and radius $R$ as $M = \text{Volume} \times \text{Density} = \frac{4}{3} \pi R^3 \rho \dots (ii)$.
Substituting the value of $M$ from equation $(ii)$ into equation $(i)$:
$T = 2 \pi \sqrt{\frac{R^3}{G \times \frac{4}{3} \pi R^3 \rho}}$
$T = 2 \pi \sqrt{\frac{3}{4 \pi G \rho}}$
$T = 2 \pi \times \frac{1}{2} \sqrt{\frac{3}{\pi G \rho}}$
$T = \sqrt{\frac{4 \pi^2 \times 3}{4 \pi G \rho}} = \sqrt{\frac{3 \pi}{\rho G}}$.

(D) For a planet to orbit a star in a circular path,the gravitational force provides the necessary centripetal force.
Given that the gravitational force $F_G \propto R^{-5/2}$.
The centripetal force is given by $F_c = m \omega^2 R$,where $\omega = \frac{2\pi}{T}$.
Thus,$F_c = m \left(\frac{4\pi^2}{T^2}\right) R$.
Equating the forces: $m \left(\frac{4\pi^2}{T^2}\right) R \propto R^{-5/2}$.
Since $m$ and $4\pi^2$ are constants,we have $\frac{R}{T^2} \propto R^{-5/2}$.
Rearranging for $T^2$: $T^2 \propto R \cdot R^{5/2} = R^{7/2}$.
Taking the square root on both sides: $T \propto R^{7/4}$.

(C) The gravitational force provides the necessary centripetal force for the satellite to revolve in a circular orbit:
$\frac{mv^2}{r} = \frac{GMm}{r^2}$
Solving for the orbital velocity $v$:
$v^2 = \frac{GM}{r} \implies v = \sqrt{\frac{GM}{r}}$
The angular momentum $L$ of the satellite is defined as $L = mvr$.
Substituting the expression for $v$:
$L = m \left( \sqrt{\frac{GM}{r}} \right) r$
$L = m \sqrt{GM} \cdot \sqrt{r} = m(GMr)^{1/2}$

(D) The energy required to raise a satellite to a height $h$ is the change in potential energy: $E_1 = U_f - U_i = -\frac{GMm}{R+h} - (-\frac{GMm}{R}) = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right) = \frac{GMmh}{R(R+h)}$.
Using $g = \frac{GM}{R^2}$,we get $GM = gR^2$. Substituting this,$E_1 = \frac{gR^2mh}{R(R+h)} = \frac{mgh}{1 + h/R}$.
The energy required to put the satellite into orbit at height $h$ is the kinetic energy $E_2 = \frac{1}{2}mv_0^2$. Since $v_0 = \sqrt{\frac{GM}{R+h}}$,$E_2 = \frac{1}{2}m \left( \frac{GM}{R+h} \right) = \frac{GMm}{2(R+h)}$.
Substituting $GM = gR^2$,$E_2 = \frac{gR^2m}{2(R+h)} = \frac{mgR}{2(1 + h/R)}$.
Taking the ratio: $\frac{E_1}{E_2} = \frac{mgh / (1 + h/R)}{mgR / (2(1 + h/R))} = \frac{h}{R/2} = \frac{2h}{R}$.

(D) The total energy $E$ of a satellite of mass $m$ revolving in an orbit of radius $r$ around a planet of mass $M$ is given by $E = -\frac{GMm}{2r}$.
From this expression,we can see that $E \propto \frac{m}{r}$.
Given the mass ratio $\frac{m_A}{m_B} = \frac{3}{1}$ and the radius ratio $\frac{r_A}{r_B} = \frac{r}{4r} = \frac{1}{4}$.
Therefore,the ratio of the total energies is $\frac{E_A}{E_B} = \frac{m_A}{m_B} \times \frac{r_B}{r_A}$.
Substituting the given values: $\frac{E_A}{E_B} = \frac{3}{1} \times \frac{4r}{r} = \frac{12}{1}$.
Thus,the ratio is $12: 1$.

(B) Let $M_e$ be the mass of the earth,$m$ be the mass of the satellite,and $r = R_e + h$ be the orbital radius.
The potential energy $U$ of the satellite is given by $U = -\frac{GM_e m}{r}$.
The orbital velocity $v$ of the satellite is given by $v = \sqrt{\frac{GM_e}{r}}$.
The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2 = \frac{1}{2}m \left(\frac{GM_e}{r}\right) = \frac{GM_e m}{2r}$.
Comparing the magnitudes,we have $K = \frac{1}{2} |U|$.
Therefore,the ratio of kinetic energy to the magnitude of potential energy is $K : |U| = 1 : 2$.

(A) The total energy of a satellite in a circular orbit at a distance $r = R + h$ from the center of the planet is given by $E_0 = -\frac{GMm}{2r}$.
Given the altitude $h = 2R$,the orbital radius is $r = R + 2R = 3R$.
Thus,the orbital energy is $E_0 = -\frac{GMm}{2(3R)} = -\frac{GMm}{6R}$.
The potential energy of the satellite on the surface of the planet is $E_i = -\frac{GMm}{R}$.
The minimum energy required to launch the satellite is the difference between the final orbital energy and the initial surface energy:
$\Delta E = E_0 - E_i = -\frac{GMm}{6R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{6R} = \frac{5GMm}{6R}$.

(C) The time period $T$ of a satellite orbiting just above the surface of a planet of radius $R$ is given by $T = 2 \pi \sqrt{\frac{R}{g}}$.
Squaring both sides,we get $T^2 = 4 \pi^2 \frac{R}{g}$.
We know that the acceleration due to gravity $g$ at the surface of a planet is given by $g = \frac{GM}{R^2}$.
Since the mass $M$ of the planet is $M = \text{Volume} \times \text{Density} = \frac{4}{3} \pi R^3 \rho$,we substitute $M$ into the expression for $g$:
$g = \frac{G}{R^2} \left( \frac{4}{3} \pi R^3 \rho \right) = \frac{4}{3} \pi \rho G R$.
Now,substitute this expression for $g$ into the equation for $T^2$:
$T^2 = 4 \pi^2 \frac{R}{\frac{4}{3} \pi \rho G R} = 4 \pi^2 \times \frac{3}{4 \pi \rho G} = \frac{3 \pi}{\rho G}$.
Rearranging the terms,we get $T^2 \rho = \frac{3 \pi}{G}$.

(B) The gravitational force between two spheres of mass $M$ separated by a distance $d$ is given by $F = \frac{GM^2}{d^2}$.
Since the spheres are in contact,the distance between their centers is $d = 2R$.
Thus,$F = \frac{GM^2}{(2R)^2} = \frac{GM^2}{4R^2}$.
The mass $M$ of each sphere is given by $M = \text{density} \times \text{volume} = \rho \times \frac{4}{3} \pi R^3$.
Substituting the value of $M$ into the force equation:
$F = \frac{G}{4R^2} \left( \rho \cdot \frac{4}{3} \pi R^3 \right)^2$
$F = \frac{G}{4R^2} \cdot \rho^2 \cdot \frac{16}{9} \pi^2 R^6$
$F = \frac{4}{9} G \pi^2 R^4 \rho^2$
Therefore,$F \propto R^4 \rho^2$.

(D) We know that angular momentum $L = m v R$ ... $(i)$
For a planet revolving around the sun,the gravitational force provides the necessary centripetal force:
$\frac{m v^2}{R} = \frac{G M m}{R^2}$
$v^2 = \frac{G M}{R}$
$v = \sqrt{\frac{G M}{R}}$ ... (ii)
Substituting equation (ii) into $(i)$:
$L = m \times \sqrt{\frac{G M}{R}} \times R$
$L = m \sqrt{G M R}$
Since $m$,$G$,and $M$ are constants,we have:
$L \propto \sqrt{R}$
$L \propto R^{1/2}$
Comparing this with $L \propto R^n$,we get $n = 0.5$.

(A) We know that for an ideal gas,the Mayer's relation is $C_{p} - C_{v} = R$.
Given that $\gamma = \frac{C_{p}}{C_{v}}$,we can write $C_{v} = \frac{C_{p}}{\gamma}$.
Substituting this into the Mayer's relation:
$C_{p} - \frac{C_{p}}{\gamma} = R$
$C_{p} \left(1 - \frac{1}{\gamma}\right) = R$
$C_{p} \left(\frac{\gamma - 1}{\gamma}\right) = R$
Therefore,$C_{p} = \frac{\gamma R}{\gamma - 1}$.

(B) The kinetic energy of the gas inside the container is given by $K.E. = n \left( \frac{1}{2} m V^2 \right)$,where $n$ is the number of moles and $m$ is the molar mass.
When the container is stopped suddenly,this kinetic energy is converted into the internal energy of the gas.
The change in internal energy is given by $\Delta U = n C_v \Delta T$.
For a diatomic gas,the molar heat capacity at constant volume is $C_v = \frac{5}{2} R$.
Since the energy is conserved,$\Delta U = K.E.$
Substituting the values: $n \left( \frac{5}{2} R \right) \Delta T = n \left( \frac{1}{2} m V^2 \right)$.
Simplifying the equation: $\frac{5}{2} R \Delta T = \frac{1}{2} m V^2$.
Therefore,the change in temperature is $\Delta T = \frac{mV^2}{5R}$.

(B) $C_p = C_v + R$
$\therefore C_p$ for hydrogen,$C_{p_1} = \frac{5}{2} R + R = \frac{7}{2} R$
$C_p$ for helium,$C_{p_2} = \frac{3}{2} R + R = \frac{5}{2} R$
$C_p$ for water vapour,$C_{p_3} = 3 R + R = 4 R$
Given: $n_1 = 4, n_2 = 2, n_3 = 1$
$C_p \text{ of mixture} = \frac{n_1 C_{p_1} + n_2 C_{p_2} + n_3 C_{p_3}}{n_1 + n_2 + n_3}$
$= \frac{4 \times \frac{7}{2} R + 2 \times \frac{5}{2} R + 1 \times 4 R}{4 + 2 + 1}$
$= \frac{14 R + 5 R + 4 R}{7} = \frac{23 R}{7}$

(C) We know that the molar specific heat at constant volume is given by $C_{v} = \frac{n R}{2}$.
From Mayer's relation,$C_{p} - C_{v} = R$,we can write $C_{p} = C_{v} + R$.
Substituting the value of $C_{v}$,we get $C_{p} = \frac{n R}{2} + R = R \left( \frac{n}{2} + 1 \right)$.
Now,the ratio $\gamma = \frac{C_{p}}{C_{v}}$ is calculated as:
$\gamma = \frac{R \left( \frac{n}{2} + 1 \right)}{\frac{n R}{2}} = \frac{\frac{n+2}{2}}{\frac{n}{2}} = \frac{n+2}{n} = 1 + \frac{2}{n}$.

(D) For a monoatomic gas,the degrees of freedom $f = 3$.
At constant pressure,the molar heat capacity is $C_P = \frac{5}{2}R$ and the molar heat capacity at constant volume is $C_V = \frac{3}{2}R$.
The total heat supplied is $\Delta Q = n C_P \Delta T$.
The change in internal energy is $\Delta U = n C_V \Delta T$.
The fraction of heat used to change internal energy is $\frac{\Delta U}{\Delta Q} = \frac{n C_V \Delta T}{n C_P \Delta T} = \frac{C_V}{C_P}$.
Substituting the values,$\frac{\Delta U}{\Delta Q} = \frac{\frac{3}{2}R}{\frac{5}{2}R} = \frac{3}{5} = 0.6$.
Therefore,the percentage of total heat used is $0.6 \times 100 = 60 \%$.

(B) real gas behaves as an ideal gas under conditions of low pressure and high temperature.
At high temperatures,the kinetic energy of the gas molecules is very high,which makes the intermolecular forces of attraction negligible.
At low pressures,the volume occupied by the gas molecules is negligible compared to the total volume of the container,effectively making the gas particles behave like point masses.
Therefore,the assumptions of the kinetic theory of gases are satisfied,and the gas follows the ideal gas equation $PV = nRT$.
Thus,option $B$ is correct.

(D) The average translational kinetic energy of a gas molecule is given by the formula: $E = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
From this relation,it is clear that the translational kinetic energy depends only on the temperature $T$ and is independent of the molar mass of the gas.
Given for nitrogen $(N_2)$: $E_1 = 0.042 \ eV$ at temperature $T_1 = T$.
For oxygen $(O_2)$: The temperature is doubled,so $T_2 = 2T$.
Since $E \propto T$,we have:
$\frac{E_2}{E_1} = \frac{T_2}{T_1} = \frac{2T}{T} = 2$
Therefore,$E_2 = 2 \times E_1 = 2 \times 0.042 \ eV = 0.084 \ eV$.

(B) Let $r$ be the distance from each vertex to the center of the equilateral triangle.
The electric potential $V$ at the center is the algebraic sum of potentials due to individual charges:
$V = \frac{1}{4 \pi \varepsilon_{0}} \left( \frac{2q}{r} + \frac{-q}{r} + \frac{-q}{r} \right)$
$V = \frac{1}{4 \pi \varepsilon_{0} r} (2q - q - q) = 0$
The electric field $\vec{E}$ at the center is the vector sum of fields due to individual charges. Since the charges are not equal in magnitude and are not symmetrically arranged in a way that their field vectors cancel out,the net electric field $\vec{E}$ is non-zero.
Specifically,the field due to $2q$ points away from it,while the fields due to the two $-q$ charges point towards them. These vectors do not sum to zero.
Therefore,the field is non-zero but the potential is zero.

(C) When $S_1$ is closed,the capacitor $C$ is charged to a potential difference $V$. The energy stored in the capacitor is $U_E = \frac{1}{2}CV^2$.
When $S_1$ is opened and $S_2$ is closed,the capacitor $C$ and inductor $L$ form an $LC$ oscillating circuit.
The energy oscillates between the electric field of the capacitor and the magnetic field of the inductor.
The maximum current $I_{max}$ in the circuit occurs when all the electric energy is converted into magnetic energy.
By the law of conservation of energy:
$\frac{1}{2}CV^2 = \frac{1}{2}LI_{max}^2$
$I_{max}^2 = \frac{C}{L}V^2$
$I_{max} = V\sqrt{\frac{C}{L}}$
Thus,the instantaneous current in the circuit can reach this maximum value.

(B) The impedance of an $LCR$ series circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
At resonance,the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$,i.e.,$X_L = X_C$.
Substituting this into the impedance formula:
$Z = \sqrt{R^2 + (0)^2} = R$.
Given $R = 12 \text{ } \Omega$,the impedance at resonance is $Z = 12 \text{ } \Omega$.

(D) The phase angle $\phi$ between voltage and current in an $RL$ circuit is given by $\tan \phi = \frac{X_L}{R}$.
Given $\phi = 45^{\circ}$,$R = 80 \Omega$,and $f = 480 \text{ Hz}$.
Since $\tan 45^{\circ} = 1$,we have $1 = \frac{X_L}{R}$,which implies $X_L = R$.
Substituting $X_L = 2 \pi f L$,we get $2 \pi f L = R$.
Solving for $L$: $L = \frac{R}{2 \pi f}$.
Substituting the values: $L = \frac{80}{2 \pi \times 480} = \frac{80}{960 \pi} = \frac{1}{12 \pi} \text{ H}$.

(B) The given alternating voltage is $v = 200 \sqrt{2} \sin(100 t)$.
Comparing this with the standard form $v = V_0 \sin(\omega t)$, we get the peak voltage $V_0 = 200 \sqrt{2} \text{ V}$ and angular frequency $\omega = 100 \text{ rad/s}$.
The capacitive reactance $X_C$ is given by $X_C = \frac{1}{\omega C}$.
Given $C = 1 \mu F = 10^{-6} \text{ F}$, we have $X_C = \frac{1}{100 \times 10^{-6}} = 10^4 \Omega$.
The peak current $I_0$ is $I_0 = \frac{V_0}{X_C} = \frac{200 \sqrt{2}}{10^4} = 2 \sqrt{2} \times 10^{-2} \text{ A}$.
The $A.C.$ ammeter measures the root mean square $(RMS)$ value of the current, $I_{rms}$.
$I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{2 \sqrt{2} \times 10^{-2}}{\sqrt{2}} = 2 \times 10^{-2} \text{ A} = 20 \text{ mA}$.

(D) At resonance,the impedance of the circuit is $Z = R$.
Voltage across the resistance is given by $V_R = I \times R$ --- $(i)$
Voltage across the inductor is given by $V_L = I \times X_L = I \times \omega L$ --- (ii)
Dividing equation (ii) by equation $(i)$:
$\frac{V_L}{V_R} = \frac{I \times \omega L}{I \times R}$
$\frac{V_L}{V_R} = \frac{\omega L}{R}$
Rearranging for $L$:
$L = \frac{V_L R}{V_R \omega}$

(A) When $80 \ V$ $d.c.$ is applied,the solenoid acts as a pure resistor because the frequency of $d.c.$ is zero,so inductive reactance $X_L = 2 \pi fL = 0$.
Resistance $R = \frac{V}{I_{dc}} = \frac{80 \ V}{0.8 \ A} = 100 \ \Omega$.
When $80 \ V$ $a.c.$ is applied,the impedance $Z$ is given by $Z = \frac{V}{I_{ac}} = \frac{80 \ V}{0.4 \ A} = 200 \ \Omega$.
The impedance of an $RL$ circuit is $Z = \sqrt{R^2 + X_L^2}$.
Substituting the values: $200 = \sqrt{100^2 + X_L^2}$.
$40000 = 10000 + X_L^2 \Rightarrow X_L^2 = 30000$.
$X_L = \sqrt{30000} \approx 173.2 \ \Omega$.
Since $X_L = 2 \pi fL$,we have $L = \frac{X_L}{2 \pi f} = \frac{173.2}{2 \times 3.14 \times 50} = \frac{173.2}{314} \approx 0.55 \ H$.
Thus,the impedance is $200 \ \Omega$ and the inductance is $0.55 \ H$.

(C) For maximum current to be drawn,the circuit must be in resonance condition.
At resonance,the inductive reactance equals the capacitive reactance: $X_L = X_C$.
This implies $2 \pi f L = \frac{1}{2 \pi f C}$.
Rearranging for capacitance $C$,we get $C = \frac{1}{4 \pi^2 f^2 L}$.
Given $L = 2 \mu H = 2 \times 10^{-6} \text{ H}$,$f = 5 \text{ kHz} = 5 \times 10^3 \text{ Hz}$,and $\pi^2 = 10$.
Substituting the values: $C = \frac{1}{4 \times 10 \times (5 \times 10^3)^2 \times 2 \times 10^{-6}}$.
$C = \frac{1}{40 \times 25 \times 10^6 \times 2 \times 10^{-6}} = \frac{1}{40 \times 25 \times 2} = \frac{1}{2000} \text{ F}$.
Comparing this with $\frac{1}{x} \text{ F}$,we find $x = 2000$.

(D) The impedance $Z$ of an $LCR$ series circuit is given by the formula: $Z = \sqrt{R^2 + (X_L - X_C)^2}$,where $X_L = 2\pi fL$ and $X_C = \frac{1}{2\pi fC}$.
As the frequency $f$ increases,$X_L$ increases and $X_C$ decreases.
At low frequencies,$X_C$ dominates,so $Z$ decreases as $f$ increases.
At the resonant frequency $f_0 = \frac{1}{2\pi\sqrt{LC}}$,$X_L = X_C$,and the impedance $Z$ reaches its minimum value,which is equal to the resistance $R$.
As the frequency increases further beyond $f_0$,$X_L$ dominates,causing the impedance $Z$ to increase.
Therefore,the impedance first decreases,reaches a minimum at resonance,and then increases.

(A) In circuit $(A)$,the impedance is $Z_A = R = 40 \ \Omega$.
In circuit $(B)$,the impedance is $Z_B = \sqrt{R^2 + (X_L - X_C)^2}$.
Since $Z_B = \sqrt{R^2 + (X_L - X_C)^2} \ge R$,it follows that $Z_B \ge Z_A$.
The r.m.s. current is given by $I = V/Z$. Since the voltage $V$ is the same $(220 \ V)$ for both circuits,$I_B = V/Z_B$ and $I_A = V/Z_A$.
Since $Z_B \ge Z_A$,it implies $I_B \le I_A$. Thus,the r.m.s. current in circuit $(B)$ can never be greater than that in circuit $(A)$.

(A) The resonant frequency $f_r$ of a series $LCR$ circuit is given by the formula:
$f_r = \frac{1}{2\pi \sqrt{LC}}$
For the resonant frequency to remain the same,the product $LC$ must remain constant:
$L_1 C_1 = L_2 C_2$
Given that $L_1 = L$,$C_1 = C$,and $C_2 = 3C$,we substitute these values into the equation:
$L \cdot C = L_2 \cdot (3C)$
$L_2 = \frac{L \cdot C}{3C}$
$L_2 = \frac{L}{3}$
Therefore,the inductance should be changed to $\frac{L}{3}$.

(C) In the steady state,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor and the $4 \Omega$ resistor. The circuit effectively consists of three resistors of $1 \Omega, 2 \Omega,$ and $3 \Omega$ connected in parallel across the $6 \text{ V}$ battery.
The equivalent resistance $R_{eq}$ of the parallel combination is given by:
$\frac{1}{R_{eq}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{6 + 3 + 2}{6} = \frac{11}{6} \Omega$
Therefore,the current $I$ through the battery is:
$I = \frac{V}{R_{eq}} = \frac{6}{11/6} = \frac{36}{11} \text{ A} \approx 3.27 \text{ A}$.
Wait,let's re-evaluate the calculation: $I = \frac{V}{R_{eq}} = 6 \times \frac{6}{11} = \frac{36}{11} \text{ A}$.
Looking at the options,there seems to be a discrepancy in the provided solution's calculation. Let's re-calculate: $I_1 = 6/1 = 6 \text{ A}$,$I_2 = 6/2 = 3 \text{ A}$,$I_3 = 6/3 = 2 \text{ A}$. Total current $I = 6 + 3 + 2 = 11 \text{ A}$.
The charge on the capacitor is $Q = C \times V = 0.5 \mu\text{F} \times 6 \text{ V} = 3 \mu\text{C}$.
Thus,the current is $11 \text{ A}$ and the charge is $3 \mu\text{C}$.

(B) The resonant frequency of an $LC$ circuit is given by $f = \frac{1}{2 \pi \sqrt{LC}}$.
This implies $f \propto \frac{1}{\sqrt{LC}}$.
Given the initial values $L_1 = L$ and $C_1 = C$,the initial frequency is $f = \frac{1}{2 \pi \sqrt{LC}}$.
The new values are $L_2 = 3L$ and $C_2 = C + 3C = 4C$.
The new resonant frequency $f'$ is given by $f' = \frac{1}{2 \pi \sqrt{L_2 C_2}} = \frac{1}{2 \pi \sqrt{(3L)(4C)}} = \frac{1}{2 \pi \sqrt{12LC}}$.
Comparing the two frequencies: $\frac{f'}{f} = \frac{\frac{1}{2 \pi \sqrt{12LC}}}{\frac{1}{2 \pi \sqrt{LC}}} = \sqrt{\frac{LC}{12LC}} = \frac{1}{\sqrt{12}} = \frac{1}{2 \sqrt{3}}$.
Therefore,$f' = \frac{f}{2 \sqrt{3}}$.

(D) The energy stored in an $LR$ circuit is given by the formula:
$E = \frac{1}{2} LI^2$ ...$(i)$
Given that the current is reduced to half,the new current $I^{\prime}$ is:
$I^{\prime} = \frac{I}{2}$
Substituting this into the energy formula,the new energy $E^{\prime}$ is:
$E^{\prime} = \frac{1}{2} L (I^{\prime})^2$
$E^{\prime} = \frac{1}{2} L \left(\frac{I}{2}\right)^2$
$E^{\prime} = \frac{1}{2} L \left(\frac{I^2}{4}\right)$
$E^{\prime} = \frac{1}{4} \left(\frac{1}{2} LI^2\right)$
Using equation $(i)$,we get:
$E^{\prime} = \frac{1}{4} E$
Therefore,the energy stored becomes $(1/4)^{\text{th}}$ of the original energy.

(D) The inductive reactance of a coil is given by the formula $X_{L} = 2 \pi f L$,where $f$ is the frequency and $L$ is the inductance.
Given that the initial inductive reactance is $X_{L1} = R = 2 \pi f L$.
If the inductance is tripled $(L' = 3L)$ and the frequency is tripled $(f' = 3f)$,the new inductive reactance $X_{L2}$ is:
$X_{L2} = 2 \pi f' L' = 2 \pi (3f) (3L) = 9 (2 \pi f L)$.
Substituting the initial value $R = 2 \pi f L$,we get:
$X_{L2} = 9R$.

(D) The resonant frequency of a series $LC$ circuit is given by the formula:
$f = \frac{1}{2 \pi \sqrt{LC}}$
When the inductance is changed to $L' = 3L$ and the capacitance is changed to $C' = 6C$,the new resonant frequency $f'$ is:
$f' = \frac{1}{2 \pi \sqrt{L' C'}} = \frac{1}{2 \pi \sqrt{(3L)(6C)}}$
$f' = \frac{1}{2 \pi \sqrt{18 LC}} = \frac{1}{2 \pi \sqrt{9 \cdot 2 \cdot LC}}$
$f' = \frac{1}{2 \pi \cdot 3 \sqrt{2} \sqrt{LC}}$
Since $f = \frac{1}{2 \pi \sqrt{LC}}$,we can substitute this into the equation:
$f' = \frac{f}{3 \sqrt{2}}$

(B) The impedance $Z$ of a series $LCR$ circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Given $R = \frac{X_L}{2}$,we have $X_L = 2R$.
Given $R = 2X_C$,we have $X_C = \frac{R}{2}$.
Substituting these values into the impedance formula:
$Z = \sqrt{R^2 + (2R - \frac{R}{2})^2} = \sqrt{R^2 + (\frac{3R}{2})^2} = \sqrt{R^2 + \frac{9R^2}{4}} = \sqrt{\frac{13R^2}{4}} = \frac{\sqrt{13}}{2} R$.
The phase difference $\phi$ is given by $\tan \phi = \frac{X_L - X_C}{R}$.
Substituting the values: $\tan \phi = \frac{2R - R/2}{R} = \frac{3R/2}{R} = \frac{3}{2}$.
Therefore,$\phi = \tan^{-1}(\frac{3}{2})$.

(D) In a series $LCR$ circuit,the current $I$ is given by $I = \frac{E}{Z} = \frac{E}{\sqrt{R^2 + (X_L - X_C)^2}}$.
Here,$X_L = 2 \pi f L$ and $X_C = \frac{1}{2 \pi f C}$.
To keep the current $I$ constant for a given voltage $E$ and resistance $R$,the impedance $Z$ must remain constant.
This requires the reactances $X_L$ and $X_C$ to remain unchanged.
If the frequency $f$ is doubled $(f' = 2f)$,then for $X_L$ to remain constant: $2 \pi (2f) L' = 2 \pi f L \implies L' = \frac{L}{2}$.
Similarly,for $X_C$ to remain constant: $\frac{1}{2 \pi (2f) C'} = \frac{1}{2 \pi f C} \implies C' = \frac{C}{2}$.
Therefore,both $L$ and $C$ should be halved simultaneously.

(B) Given: $R = X_C = \frac{1}{\omega C}$.
Initial impedance $Z = \sqrt{R^2 + X_C^2} = \sqrt{R^2 + R^2} = \sqrt{2} R$.
Initial peak current $I_0 = \frac{V_0}{Z} = \frac{V_0}{\sqrt{2} R}$,which implies $\frac{V_0}{R} = \sqrt{2} I_0$.
When angular frequency $\omega' = \frac{\omega}{\sqrt{3}}$,the new capacitive reactance $X_C' = \frac{1}{\omega' C} = \frac{1}{(\omega / \sqrt{3}) C} = \sqrt{3} \left(\frac{1}{\omega C}\right) = \sqrt{3} R$.
New impedance $Z' = \sqrt{R^2 + (X_C')^2} = \sqrt{R^2 + (\sqrt{3} R)^2} = \sqrt{R^2 + 3R^2} = \sqrt{4R^2} = 2R$.
New peak current $I_0' = \frac{V_0}{Z'} = \frac{V_0}{2R} = \frac{1}{2} \left(\frac{V_0}{R}\right) = \frac{1}{2} (\sqrt{2} I_0) = \frac{I_0}{\sqrt{2}}$.

(D) The impedance $Z$ of an $LCR$ series circuit is given by the formula: $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
We know that inductive reactance $X_L = L\omega = 2\pi fL$ and capacitive reactance $X_C = \frac{1}{C\omega} = \frac{1}{2\pi fC}$.
As the frequency $f$ increases,$X_L$ increases linearly while $X_C$ decreases.
At low frequencies,$X_C$ dominates,so $Z$ decreases as $f$ increases.
At the resonant frequency $f_0 = \frac{1}{2\pi\sqrt{LC}}$,$X_L = X_C$,making the impedance $Z = R$,which is the minimum value.
As the frequency increases beyond $f_0$,$X_L$ dominates,causing the impedance $Z$ to increase.
Therefore,the impedance first decreases,becomes minimum at resonance,and then increases.

(A) The phase difference $\phi$ between the voltage and current in an $LCR$ series circuit is given by: $\tan \phi = \frac{X_L - X_C}{R}$.
Given $\phi = 45^{\circ}$,so $\tan 45^{\circ} = 1$.
Thus,$\frac{\omega L - \frac{1}{\omega C}}{R} = 1$.
$\omega L - \frac{1}{\omega C} = R$.
$\omega L = R + \frac{1}{\omega C} = \frac{R \omega C + 1}{\omega C}$.
Since $\omega = 2 \pi f$,we have $L = \frac{R \omega C + 1}{\omega^2 C} = \frac{R(2 \pi f)C + 1}{(2 \pi f)^2 C}$.
$L = \frac{1 + 2 \pi fCR}{4 \pi^2 f^2 C}$.

(C) The phase angle $\phi$ in an $LCR$ series circuit is given by $\tan \phi = \frac{X_L - X_C}{R}$.
Given: $L = \frac{300}{\pi} \text{ mH} = \frac{0.3}{\pi} \text{ H}$,$C = \frac{1}{\pi} \text{ mF} = \frac{1}{\pi} \times 10^{-3} \text{ F}$,$R = 20 \ \Omega$,$f = 50 \text{ Hz}$.
Angular frequency $\omega = 2 \pi f = 2 \times \pi \times 50 = 100 \pi \text{ rad/s}$.
Inductive reactance $X_L = \omega L = 100 \pi \times \frac{0.3}{\pi} = 30 \ \Omega$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100 \pi \times \frac{1}{\pi} \times 10^{-3}} = \frac{1}{0.1} = 10 \ \Omega$.
Substituting these values into the formula:
$\tan \phi = \frac{30 - 10}{20} = \frac{20}{20} = 1$.
Therefore,$\phi = \tan^{-1}(1)$.

(A) For $d.c.$ supply,the inductor acts as a pure resistor:
$R = \frac{V}{I} = \frac{100 \ V}{1 \ A} = 100 \ \Omega$.
For $a.c.$ supply,the impedance $Z$ is given by:
$Z = \frac{V}{I} = \frac{100 \ V}{0.5 \ A} = 200 \ \Omega$.
The impedance of an $LR$ circuit is $Z = \sqrt{R^2 + X_L^2}$,where $X_L = 2 \pi f L$.
Substituting the values: $200 = \sqrt{100^2 + X_L^2}$.
Squaring both sides: $40000 = 10000 + X_L^2 \Rightarrow X_L^2 = 30000$.
$X_L = \sqrt{30000} = 100\sqrt{3} \ \Omega$.
Since $X_L = 2 \pi f L$,we have $100\sqrt{3} = 2 \pi (50) L$.
$100\sqrt{3} = 100 \pi L \Rightarrow L = \frac{\sqrt{3}}{\pi} \ H$.

(C) At resonance,the angular frequency is given by $\omega = \frac{1}{\sqrt{LC}}$,which implies $C = \frac{1}{\omega^2 L}$.
In a series $L-C-R$ circuit,the current $I$ is the same through all components.
The voltage across the resistor is $V_R = IR$,so $I = \frac{V_R}{R}$.
The voltage across the inductor is $V_L = I X_L = I \omega L$.
Substituting $I$ into the expression for $V_L$,we get $V_L = \left( \frac{V_R}{R} \right) \omega L$.
Solving for $L$,we find $L = \frac{V_L R}{V_R \omega}$.
Now,substitute $L$ into the expression for $C$:
$C = \frac{1}{\omega^2 \left( \frac{V_L R}{V_R \omega} \right)} = \frac{V_R \omega}{V_L R \omega^2} = \frac{V_R}{V_L R \omega}$.

(C) For a $d.c.$ source,the inductor acts as a simple resistor because the frequency is zero. Thus,the resistance $R$ is given by $R = \frac{V}{I} = \frac{12 \ V}{4 \ A} = 3 \ \Omega$.
For an $a.c.$ source,the impedance $Z$ is given by $Z = \frac{V}{I} = \frac{12 \ V}{2.4 \ A} = 5 \ \Omega$.
The impedance of an $RL$ circuit is $Z = \sqrt{R^2 + X_L^2}$,where $X_L$ is the inductive reactance.
Squaring both sides,$Z^2 = R^2 + X_L^2$,so $X_L^2 = Z^2 - R^2 = 5^2 - 3^2 = 25 - 9 = 16$.
Thus,$X_L = 4 \ \Omega$.
Since $X_L = 2 \pi f L$,we have $L = \frac{X_L}{2 \pi f} = \frac{4}{2 \pi \times 50} = \frac{4}{100 \pi} = \frac{4}{\pi} \times 10^{-2} \ H$.

(B) The impedance $Z$ of a series $LCR$ circuit is given by the formula: $Z = \sqrt{R^2 + (2 \pi v L - \frac{1}{2 \pi v C})^2}$.
At very low frequencies $(v \to 0)$,the capacitive reactance $X_C = \frac{1}{2 \pi v C}$ approaches infinity,so $Z \to \infty$.
At the resonant frequency $v_r = \frac{1}{2 \pi \sqrt{LC}}$,the inductive reactance $X_L$ equals the capacitive reactance $X_C$,making the net reactance zero. Thus,$Z = R$,which is the minimum value of impedance.
For frequencies higher than the resonant frequency $(v > v_r)$,the inductive reactance $X_L = 2 \pi v L$ dominates,and $Z$ increases as $v$ increases.
Therefore,the graph of $Z$ versus $v$ starts from a high value,decreases to a minimum value $R$ at $v_r$,and then increases again. This behavior is correctly represented by graph $(B)$.

(B) The capacitive reactance of a capacitor is given by the formula $X_{C} = \frac{1}{2 \pi f C}$.
Here,$f$ is the frequency of the $A.C.$ source and $C$ is the capacitance.
When the capacitance $C$ is reduced,the capacitive reactance $X_{C}$ increases.
Similarly,when the frequency $f$ is reduced,the capacitive reactance $X_{C}$ increases.
Since the bulb is in series with the capacitor,the total impedance $Z$ of the circuit is given by $Z = \sqrt{R^2 + X_{C}^2}$.
As $X_{C}$ increases in both cases,the total impedance $Z$ of the circuit increases.
According to Ohm's law for $A.C.$ circuits,the current $I = \frac{V}{Z}$.
Since the impedance $Z$ increases,the current $I$ flowing through the bulb decreases.
The brightness of the bulb depends on the power dissipated,$P = I^2 R$.
As the current $I$ decreases,the power dissipated and the brightness of the lamp decrease in both cases.

(C) In an $LCR$ series circuit,the current $I$ is the same through all components. The voltage across the resistor $V_R$ is in phase with the current. The voltage across the inductor $V_L$ leads the current by $90^\circ$,and the voltage across the capacitor $V_C$ lags behind the current by $90^\circ$.
Thus,$V_L$ and $V_C$ are in opposite directions. The net reactive voltage is $(V_L - V_C)$.
Using the phasor diagram,the resultant voltage $V$ is the vector sum of $V_R$ and $(V_L - V_C)$,which are perpendicular to each other.
Applying the Pythagorean theorem: $V^2 = V_R^2 + (V_L - V_C)^2$.

(D) Given:
Inductance $L = 2 \text{ mH} = 2 \times 10^{-3} \text{ H}$
Capacitance $C = 20 \mu\text{F} = 20 \times 10^{-6} \text{ F}$
Resistance $R = 50 \Omega$
Condition for resonance: Inductive reactance $X_L$ equals capacitive reactance $X_C$.
$X_L = X_C = \omega L = \frac{1}{\omega C}$
From this,$\omega^2 = \frac{1}{LC}$.
$\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{2 \times 10^{-3} \times 20 \times 10^{-6}}} = \frac{1}{\sqrt{40 \times 10^{-9}}} = \frac{1}{\sqrt{4 \times 10^{-8}}} = \frac{1}{2 \times 10^{-4}} = 0.5 \times 10^4 = 5000 \text{ rad/s}$.
Now,calculate the reactance $X_L = \omega L$:
$X_L = 5000 \times 2 \times 10^{-3} = 10 \Omega$.
Therefore,the reactance of either component is $10 \Omega$.

(C) Given equations are $V = 100 \sin(100t) \text{ V}$ and $I = 100 \sin(100t + \frac{\pi}{3}) \text{ mA}$.
Comparing with standard forms $V = V_0 \sin(\omega t)$ and $I = I_0 \sin(\omega t + \phi)$,we get:
$V_0 = 100 \text{ V}$
$I_0 = 100 \text{ mA} = 100 \times 10^{-3} \text{ A} = 0.1 \text{ A}$
Phase difference $\phi = \frac{\pi}{3}$.
The average power dissipated in an $A$.$C$. circuit is given by $P = V_{\text{rms}} I_{\text{rms}} \cos \phi$.
$P = \left(\frac{V_0}{\sqrt{2}}\right) \left(\frac{I_0}{\sqrt{2}}\right) \cos \phi = \frac{V_0 I_0}{2} \cos \phi$.
Substituting the values:
$P = \frac{100 \times 0.1}{2} \times \cos(\frac{\pi}{3})$
$P = \frac{10}{2} \times \frac{1}{2} = 5 \times 0.5 = 2.5 \text{ W}$.

(D) The given equation for the alternating current is $I = 100 \sin(200 \pi t)$.
Comparing this with the standard form $I = I_0 \sin(\omega t)$,the peak value $I_0$ is $100 \ A$.
The current reaches its peak value when $\sin(200 \pi t) = 1$.
This occurs when the argument of the sine function is equal to $\frac{\pi}{2}$.
So,$200 \pi t = \frac{\pi}{2}$.
Dividing both sides by $\pi$,we get $200 t = \frac{1}{2}$.
Therefore,$t = \frac{1}{2 \times 200} = \frac{1}{400} \ s$.

(C) The power consumed in an $A$.$C$. circuit containing a resistor is given by the formula:
$P = I_{rms}^2 R$
Given:
$P = 108 \text{ W}$
$I_{rms} = 3 \text{ A}$
Substituting the values into the formula:
$108 = (3)^2 \times R$
$108 = 9 \times R$
$R = \frac{108}{9} = 12 \ \Omega$
Therefore,the resistance in the circuit is $12 \ \Omega$.

(D) Given equations are $E = 200 \sin(50t) \text{ V}$ and $I = 100 \sin(50t + \frac{\pi}{3}) \text{ mA}$.
Comparing these with standard forms $E = E_0 \sin(\omega t)$ and $I = I_0 \sin(\omega t + \phi)$,we get:
$E_0 = 200 \text{ V}$
$I_0 = 100 \text{ mA} = 100 \times 10^{-3} \text{ A} = 0.1 \text{ A}$
Phase difference $\phi = \frac{\pi}{3} = 60^{\circ}$.
The average power dissipated in an $AC$ circuit is given by $P = E_{\text{rms}} I_{\text{rms}} \cos \phi$.
We know that $E_{\text{rms}} = \frac{E_0}{\sqrt{2}}$ and $I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$.
Substituting these values:
$P = \left( \frac{E_0}{\sqrt{2}} \right) \left( \frac{I_0}{\sqrt{2}} \right) \cos \phi = \frac{E_0 I_0}{2} \cos 60^{\circ}$.
$P = \frac{200 \times 0.1}{2} \times 0.5 = \frac{20}{2} \times 0.5 = 10 \times 0.5 = 5 \text{ W}$.

(A) Given: Inductance $L = 42 \, mH = 42 \times 10^{-3} \, H$, Voltage $V_{rms} = 200 \, V$, Frequency $f = 50 \, Hz$.
First, calculate the inductive reactance $X_L$ using the formula $X_L = \omega L = 2 \pi f L$.
Substituting the values: $X_L = 2 \times \frac{22}{7} \times 50 \times 42 \times 10^{-3} \, \Omega$.
$X_L = 2 \times 22 \times 50 \times 6 \times 10^{-3} \, \Omega = 4400 \times 6 \times 10^{-3} \, \Omega = 13.2 \, \Omega$.
The r.m.s. current $I_{rms}$ is given by $I_{rms} = \frac{V_{rms}}{X_L}$.
$I_{rms} = \frac{200}{13.2} \approx 15.15 \, A$.

(A) The given alternating current equation is $I = I_0 \sin(\omega t)$.
Comparing this with the given equation $I = 100 \sin(50 \pi t)$, we get angular frequency $\omega = 50 \pi \text{ rad/s}$.
We know that $\omega = 2 \pi f$, where $f$ is the frequency.
So, $2 \pi f = 50 \pi$, which gives $f = 25 \text{ Hz}$.
This means the current completes $25$ cycles in one second.
In one complete cycle of a sine wave, the current becomes zero twice (at the start/end and at the half-cycle point).
Therefore, in $25$ cycles, the current will become zero $25 \times 2 = 50$ times in one second.

(C) $1$. Ideal Inductor:
- The voltage and current in an ideal inductor are out of phase by $90^{\circ}$,with the current lagging the voltage by $90^{\circ}$.
- The instantaneous power is given by $p(t) = v(t) \times i(t)$.
- Over a complete cycle,the positive and negative power values cancel out,resulting in an average power of zero.
$2$. Ideal Capacitor:
- In an ideal capacitor,the voltage and current are also out of phase by $90^{\circ}$,but the current leads the voltage by $90^{\circ}$.
- Similar to the inductor case,the instantaneous power alternates between positive and negative,resulting in an average power of zero over a complete cycle.

(B) The impedance $Z$ of an $LR$ series circuit is given by $Z = \sqrt{R^2 + X_L^2}$.
Given $X_L = 2R$,we have $Z^2 = R^2 + (2R)^2 = R^2 + 4R^2 = 5R^2$.
The average power dissipated in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$,where $\cos \phi = \frac{R}{Z}$ is the power factor.
Since $V_{rms} = \frac{E_0}{\sqrt{2}}$ and $I_{rms} = \frac{V_{rms}}{Z} = \frac{E_0}{\sqrt{2}Z}$,the power is $P = \frac{E_0}{\sqrt{2}} \times \frac{E_0}{\sqrt{2}Z} \times \frac{R}{Z} = \frac{E_0^2 R}{2 Z^2}$.
Substituting $Z^2 = 5R^2$,we get $P = \frac{E_0^2 R}{2(5R^2)} = \frac{E_0^2 R}{10 R^2} = \frac{E_0^2}{10 R}$.

(C) The average power consumed in an $AC$ circuit is given by $P = E_{rms} I_{rms} \cos \phi$.
Here,the power factor is $\cos \phi = \frac{R}{Z}$.
The $rms$ current is $I_{rms} = \frac{E_{rms}}{Z} = \frac{E_0}{\sqrt{2} Z}$.
Substituting these into the power formula:
$P = \left( \frac{E_0}{\sqrt{2}} \right) \left( \frac{E_0}{\sqrt{2} Z} \right) \left( \frac{R}{Z} \right) = \frac{E_0^2 R}{2 Z^2} \dots (i)$.
Given that the inductive reactance $X_L = R$,the impedance $Z$ of the $L-R$ circuit is:
$Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + R^2} = \sqrt{2} R$.
Substituting $Z^2 = 2 R^2$ into equation $(i)$:
$P = \frac{E_0^2 R}{2 (2 R^2)} = \frac{E_0^2 R}{4 R^2} = \frac{E_0^2}{4 R}$.

(D) In an $LCR$ series circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
At resonance,the inductive reactance $X_L$ equals the capacitive reactance $X_C$,i.e.,$X_L = X_C$.
Therefore,the net reactance $X = X_L - X_C = 0$.
The phase angle $\phi$ is given by $\tan \phi = \frac{X_L - X_C}{R} = \frac{0}{R} = 0$.
Thus,$\phi = 0$,which means the current and voltage are in the same phase at resonance.

(D) In an $A.C.$ circuit containing a coil (inductor with resistance),the phase difference $\phi$ between the voltage and the current is given by the formula: $\tan \phi = \frac{X_L}{R}$.
Given that the reactance $X_L = \sqrt{3} R$.
Substituting this value into the formula:
$\tan \phi = \frac{\sqrt{3} R}{R} = \sqrt{3}$.
Therefore,the phase difference is $\phi = \tan^{-1}(\sqrt{3})$.

(D) For the voltage to be maximum,the sine function must equal $1$.
$\sin \left(\omega t+\frac{\pi}{3}\right) = 1$
Since $\sin \left(\frac{\pi}{2}\right) = 1$,we have:
$\omega t + \frac{\pi}{3} = \frac{\pi}{2}$
$\omega t = \frac{\pi}{2} - \frac{\pi}{3}$
$\omega t = \frac{\pi}{6}$
Substituting $\omega = \frac{2\pi}{T}$:
$\left(\frac{2\pi}{T}\right) t = \frac{\pi}{6}$
$t = \frac{\pi}{6} \times \frac{T}{2\pi}$
$t = \frac{T}{12}$

(A) The capacitive reactance is given by $X_{C} = \frac{1}{2 \pi f C}$.
If the capacitance $(C)$ is increased,the capacitive reactance $(X_{C})$ will decrease.
The total impedance $(Z)$ of the circuit is given by $Z = \sqrt{R^2 + X_{C}^2}$.
As $X_{C}$ decreases,the total impedance $(Z)$ of the circuit decreases.
According to Ohm's law for $A.C.$ circuits,the current $(I = V/Z)$ will increase.
Since the brightness of the lamp depends on the power dissipated $(P = I^2 R)$,an increase in current leads to an increase in the brightness of the lamp.

(C) Given equations are $e = 160 \sin(100t) \text{ V}$ and $i = 250 \sin(100t + \frac{\pi}{3}) \text{ mA}$.
Comparing these with standard forms $e = e_0 \sin(\omega t)$ and $i = i_0 \sin(\omega t + \phi)$,we get:
Peak voltage $e_0 = 160 \text{ V}$.
Peak current $i_0 = 250 \text{ mA} = 250 \times 10^{-3} \text{ A} = 0.25 \text{ A}$.
Phase difference $\phi = \frac{\pi}{3}$.
The average power dissipated in an $AC$ circuit is given by $P_{avg} = V_{rms} I_{rms} \cos(\phi) = \frac{e_0}{\sqrt{2}} \cdot \frac{i_0}{\sqrt{2}} \cos(\phi) = \frac{e_0 i_0}{2} \cos(\phi)$.
Substituting the values:
$P_{avg} = \frac{160 \times 0.25}{2} \times \cos(\frac{\pi}{3})$
$P_{avg} = \frac{40}{2} \times \frac{1}{2} = 20 \times 0.5 = 10 \text{ W}$.

(D) $1$. According to Faraday's law of induction,eddy currents are loops of electrical current induced within conductors by a changing magnetic field.
$2$. When a solid metal core is used in a transformer,the changing magnetic flux induces large eddy currents,which lead to significant energy loss in the form of heat.
$3$. By laminating the core (using thin,insulated sheets of metal),the path for these eddy currents is restricted,significantly reducing their magnitude.
$4$. Therefore,the core is laminated to reduce eddy current losses and improve the overall efficiency of the device.

(B) Given: Efficiency $\eta = 90 \% = 0.9$, Input Voltage $V_p = 200 \ V$, Input Power $P_{in} = 3 \ kW = 3000 \ W$, Secondary Current $I_s = 6 \ A$.
First, calculate the primary current $I_p$ using the input power formula: $P_{in} = V_p \times I_p$.
$3000 \ W = 200 \ V \times I_p \implies I_p = \frac{3000}{200} = 15 \ A$.
Next, calculate the output power $P_{out}$ using efficiency: $P_{out} = \eta \times P_{in}$.
$P_{out} = 0.9 \times 3000 \ W = 2700 \ W$.
Finally, calculate the secondary voltage $V_s$ using the output power formula: $P_{out} = V_s \times I_s$.
$2700 \ W = V_s \times 6 \ A \implies V_s = \frac{2700}{6} = 450 \ V$.
Thus, the secondary voltage is $450 \ V$ and the primary current is $15 \ A$.

(C) For an ideal transformer,the power input at the primary coil is equal to the power output at the secondary coil.
Given power $P = 6.6 \ kW = 6.6 \times 10^3 \ W$.
Secondary voltage $V_s = 4.4 \ kV = 4.4 \times 10^3 \ V$.
Since $P = V_s \times I_s$ for an ideal transformer,
$I_s = \frac{P}{V_s} = \frac{6.6 \times 10^3}{4.4 \times 10^3} \ A$.
$I_s = \frac{6.6}{4.4} \ A = 1.5 \ A$.
Thus,the current rating of the secondary coil is $1.5 \ A$.

(C) Given: Primary turns $N_p = 120$, Primary current $I_p = 5 \text{ A}$, Input power $P_{in} = 1 \text{ kW} = 1000 \text{ W}$, Secondary voltage $V_s = 560 \text{ V}$.
First, calculate the primary voltage $V_p$ using the power formula $P = V_p I_p$:
$V_p = \frac{P_{in}}{I_p} = \frac{1000 \text{ W}}{5 \text{ A}} = 200 \text{ V}$.
Using the transformer turns ratio formula $\frac{N_s}{N_p} = \frac{V_s}{V_p}$:
$\frac{N_s}{120} = \frac{560}{200}$.
Solving for $N_s$:
$N_s = \frac{120 \times 560}{200} = \frac{120 \times 56}{20} = 6 \times 56 = 336$.
Therefore, the number of turns in the secondary coil is $336$.

(C) In an ideal transformer,the induced electromotive force $(EMF)$ per turn is the same for both the primary and secondary coils.
Given:
Number of turns in primary,$N_p = 1000$
Number of turns in secondary,$N_s = 3000$
Voltage applied to primary,$V_p = 80 \ V$
The potential difference per turn is given by the ratio $\frac{V}{N}$.
For the primary coil,the potential difference per turn is $\frac{V_p}{N_p} = \frac{80 \ V}{1000} = 0.08 \ V$.
Since the flux linked per turn is the same for both coils in an ideal transformer,the potential difference per turn in the secondary coil is equal to the potential difference per turn in the primary coil.
Therefore,the potential difference per turn of the secondary coil is $0.08 \ V$.

(D) In a transformer, energy is lost due to various factors, one of which is the formation of eddy currents in the iron core.
When the magnetic flux through the core changes, it induces circulating currents (eddy currents) within the bulk of the core material, leading to significant heat dissipation ($I^2R$ loss).
By laminating the core—using thin, insulated sheets of metal instead of a solid block—the path for these eddy currents is restricted.
This increase in resistance significantly reduces the magnitude of the eddy currents, thereby minimizing the energy loss in the form of heat.

(D) The impedance matching condition for a transformer is given by the relation: $\frac{Z_p}{Z_s} = \left(\frac{N_p}{N_s}\right)^2$.
Given: Primary impedance $Z_p = 8000 \ \Omega$ and secondary impedance $Z_s = 8 \ \Omega$.
Substituting the values: $\frac{8000}{8} = \left(\frac{N_p}{N_s}\right)^2$.
$1000 = \left(\frac{N_p}{N_s}\right)^2$.
Taking the square root on both sides: $\frac{N_p}{N_s} = \sqrt{1000} = 10\sqrt{10}$.
Since $\sqrt{10} \approx 3.16$,we get $\frac{N_p}{N_s} \approx 10 \times 3.16 = 31.6 \approx 32$.
Therefore,the ratio of primary to secondary turns is $32: 1$.