MHT CET 2021 Physics Question Paper with Answer and Solution

(B) Given: Moment of inertia $I = 1.2 \; kg \cdot m^2$,initial angular velocity $\omega_0 = 0$,rotational kinetic energy $K_r = 1500 \; J$,and angular acceleration $\alpha = 25 \; rad/s^2$.
The formula for rotational kinetic energy is $K_r = \frac{1}{2} I \omega^2$.
Substituting the values: $1500 = \frac{1}{2} \times 1.2 \times \omega^2$.
$1500 = 0.6 \times \omega^2 \Rightarrow \omega^2 = \frac{1500}{0.6} = 2500$.
Thus,the final angular velocity $\omega = \sqrt{2500} = 50 \; rad/s$.
Using the kinematic equation for rotation: $\omega = \omega_0 + \alpha t$.
$50 = 0 + 25 \times t$.
$t = \frac{50}{25} = 2 \; s$.

(C) In an elastic collision,both linear momentum and kinetic energy are conserved.
Let the initial velocities be $v_{a}$ and $-v_{b}$ (since they move in opposite directions).
After the collision,the velocities are $-v_{b}$ and $v_{a}$ respectively.
According to the law of conservation of linear momentum:
$m_{a}v_{a} - m_{b}v_{b} = m_{a}(-v_{b}) + m_{b}v_{a}$
Rearranging the terms:
$m_{a}v_{a} + m_{a}v_{b} = m_{b}v_{a} + m_{b}v_{b}$
$m_{a}(v_{a} + v_{b}) = m_{b}(v_{a} + v_{b})$
Since $(v_{a} + v_{b}) \neq 0$,we can divide both sides by $(v_{a} + v_{b})$:
$m_{a} = m_{b}$
Therefore,the ratio $\frac{m_{a}}{m_{b}} = 1$.

(D) Let $u$ be the initial velocity of mass $m$ and $v_M$ be the velocity of mass $M$ after the collision. The mass $m$ comes to rest,so its final velocity is $0$.
By the law of conservation of linear momentum:
$mu + M(0) = m(0) + Mv_M$
$mu = Mv_M$
$v_M = \frac{m}{M}u$
The coefficient of restitution $e$ is defined as the ratio of the relative velocity of separation to the relative velocity of approach:
$e = \frac{v_M - 0}{u - 0} = \frac{v_M}{u}$
Substituting the value of $v_M$:
$e = \frac{(m/M)u}{u} = \frac{m}{M}$

(A) In an elastic collision with a stationary wall,the molecule rebounds with the same speed $v$ in the opposite direction.
Change in momentum of the molecule in one collision = $mv - (-mv) = 2mv$.
By the law of conservation of momentum,the change in momentum of the wall in one collision is equal and opposite to the change in momentum of the molecule,which is $2mv$.
Since there are $5$ collisions per second,the total change in momentum of the wall per second is $5 \times 2mv = 10mv$.

(B) Given: Mass of first block $M_1 = m$,initial velocity $u_1 = v$,final velocity $v_1 = 0$.
Mass of second block $M_2 = 4m$,initial velocity $u_2 = 0$,final velocity $v_2 = ?$.
According to the law of conservation of linear momentum:
$M_1 u_1 + M_2 u_2 = M_1 v_1 + M_2 v_2$
$m(v) + 4m(0) = m(0) + 4m(v_2)$
$mv = 4mv_2$
$v_2 = \frac{v}{4}$
The coefficient of restitution $e$ is defined as:
$e = \frac{v_2 - v_1}{u_1 - u_2}$
Substituting the values:
$e = \frac{\frac{v}{4} - 0}{v - 0} = \frac{v/4}{v} = \frac{1}{4} = 0.25$.

(B) The relationship between kinetic energy $K$ and momentum $p$ is given by $K = \frac{p^2}{2m}$,where $m$ is the mass of the body.
Rearranging this formula,we get $p = \sqrt{2mK}$.
Since the kinetic energy $K$ is the same for both bodies,the momentum $p$ is directly proportional to the square root of the mass,i.e.,$p \propto \sqrt{m}$.
Therefore,the body with the larger mass will have a greater momentum.
Thus,the heavy body has greater momentum.

(A) The acceleration due to gravity at a depth $d$ below the surface of the Earth is given by the formula:
$g^{\prime} = g \left(1 - \frac{d}{R}\right)$
Given that the acceleration due to gravity at depth $d$ is $\frac{g}{n}$,we substitute this into the equation:
$\frac{g}{n} = g \left(1 - \frac{d}{R}\right)$
Dividing both sides by $g$:
$\frac{1}{n} = 1 - \frac{d}{R}$
Rearranging the terms to solve for $d$:
$\frac{d}{R} = 1 - \frac{1}{n}$
$\frac{d}{R} = \frac{n-1}{n}$
$d = \frac{R(n-1)}{n}$

(A) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by the formula: $g' = g\left(1 - \frac{d}{R}\right)$.
According to the problem,the value of acceleration due to gravity at depth $d$ is $\frac{1}{n}$ times the value at the surface,so $g' = \frac{g}{n}$.
Substituting this into the formula: $\frac{g}{n} = g\left(1 - \frac{d}{R}\right)$.
Dividing both sides by $g$: $\frac{1}{n} = 1 - \frac{d}{R}$.
Rearranging the terms to solve for $d$: $\frac{d}{R} = 1 - \frac{1}{n} = \frac{n-1}{n}$.
Therefore,$d = R\left(\frac{n-1}{n}\right)$.

(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 $g' = \frac{g}{3}$,we substitute this into the equation:
$\frac{g}{3} = g \left( \frac{R}{R+h} \right)^2$.
Dividing both sides by '$g$',we get:
$\frac{1}{3} = \left( \frac{R}{R+h} \right)^2$.
Taking the square root on both sides:
$\frac{1}{\sqrt{3}} = \frac{R}{R+h}$.
Cross-multiplying gives:
$R + h = \sqrt{3} R$.
Rearranging for '$h$':
$h = \sqrt{3} R - R = (\sqrt{3} - 1) R$.

(C) The acceleration due to gravity at a distance $r$ from the center of the earth is given by $g' = \frac{GM}{r^2}$.
In the first case,the distance from the surface is $R$,so the distance from the center is $r_1 = R + R = 2R$. Given $g_1 = \frac{g}{4}$.
In the second case,the distance from the surface is $\frac{R}{2}$,so the distance from the center is $r_2 = R + \frac{R}{2} = \frac{3R}{2}$.
Using the ratio: $\frac{g_2}{g_1} = \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{2R}{3R/2}\right)^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9}$.
Therefore,$g_2 = \frac{16}{9} \times g_1 = \frac{16}{9} \times \frac{g}{4} = \frac{4g}{9}$.

(C) The acceleration due to gravity $g'$ 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 the height $h = R$,we substitute this into the formula:
$g' = g \left( \frac{R}{R+R} \right)^2$
$g' = g \left( \frac{R}{2R} \right)^2$
$g' = g \left( \frac{1}{2} \right)^2$
$g' = \frac{g}{4}$
Therefore,the gravitational acceleration at a height $R$ above the earth's surface is $\frac{g}{4}$.

(A) The time period of a seconds pendulum is $T = 2 \,s$. The formula for the time period is $T = 2 \pi \sqrt{\frac{\ell}{g}}$.
Since $T$ is constant $(2 \,s)$, we have $\ell \propto g$, which implies $\frac{\ell'}{\ell} = \frac{g'}{g}$.
The acceleration due to gravity is given by $g = \frac{GM}{R^2}$.
For the planet, $M' = 1.5M$ and $R' = 1.5R$.
Thus, $g' = \frac{G(1.5M)}{(1.5R)^2} = \frac{1.5}{2.25} \frac{GM}{R^2} = \frac{1}{1.5} g$.
Substituting this into the length ratio: $\ell' = \ell \times \frac{g'}{g} = 1 \,m \times \frac{1}{1.5} = \frac{1}{1.5} \,m \approx 0.67 \,m$.

(B) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by the formula:
$g' = g(1 - \frac{d}{R})$
where $g$ is the acceleration due to gravity at the surface and $R$ is the radius of the earth.
Given that $g' = 60\% \text{ of } g$,we have:
$g' = 0.6g$
Substituting this into the formula:
$0.6g = g(1 - \frac{d}{R})$
$0.6 = 1 - \frac{d}{R}$
Rearranging the terms to solve for $d$:
$\frac{d}{R} = 1 - 0.6$
$\frac{d}{R} = 0.4$
$\frac{d}{R} = \frac{4}{10} = \frac{2}{5}$
Therefore,$d = \frac{2}{5}R$.

(B) The acceleration due to gravity $g$ on the surface of the Earth is given by the formula $g = \frac{GM}{R^2}$.
Substituting the mass of the Earth $M = \rho \cdot V = \rho \cdot \frac{4}{3} \pi R^3$,where $\rho$ is the average density and $R$ is the radius of the Earth:
$g = \frac{G}{R^2} \cdot \left( \frac{4}{3} \pi R^3 \rho \right) = \frac{4}{3} \pi G R \rho$.
Since $G$,$\pi$,and $R$ are constants for the Earth,we have $g \propto \rho$.
Therefore,the average density $\rho$ is directly proportional to the acceleration due to gravity $g$.

(C) The energy required to escape the earth's gravitational field is $E_{esc} = \frac{1}{2} m V_{e}^2$.
The initial kinetic energy given to the body is $K_i = \frac{1}{2} m (3 V_{e})^2 = \frac{9}{2} m V_{e}^2$.
According to the law of conservation of energy, the final kinetic energy $K_f$ after escaping the gravitational pull is the difference between the initial energy and the energy required to escape:
$K_f = K_i - E_{esc}$
$\frac{1}{2} m V^2 = \frac{9}{2} m V_{e}^2 - \frac{1}{2} m V_{e}^2$
$\frac{1}{2} m V^2 = 4 m V_{e}^2$
$V^2 = 8 V_{e}^2$
$V = \sqrt{8} V_{e} = 2 \sqrt{2} V_{e}$.

(A) To ensure the particle does not return,its total mechanical energy at infinity must be at least zero.
Let $v$ be the projection speed at height $h = 3R$ from the surface.
The distance from the center of the Earth is $r = R + 3R = 4R$.
Using the law of conservation of energy: $K_i + U_i = K_f + U_f$.
Here,$K_i = \frac{1}{2}mv^2$,$U_i = -\frac{GMm}{4R}$,$K_f = 0$,and $U_f = 0$.
$\frac{1}{2}mv^2 - \frac{GMm}{4R} = 0$.
$\frac{1}{2}v^2 = \frac{GM}{4R}$.
$v^2 = \frac{GM}{2R}$.
$v = [\frac{GM}{2R}]^{1/2}$.

(A) The formula for escape velocity is $V = \sqrt{\frac{2GM}{R}}$.
For Earth,the escape velocity is $V_e = \sqrt{\frac{2GM}{R}}$.
For the planet,the mass is $M_p = 6M$ and the radius is $R_p = 2R$.
Thus,the escape velocity from the planet is $V_p = \sqrt{\frac{2G(6M)}{2R}} = \sqrt{3 \times \frac{2GM}{R}}$.
Substituting $V_e$ into the equation,we get $V_p = \sqrt{3} V_e$.

(A) The escape velocity $v_e$ from the surface of a planet is given by the formula $v_e = \sqrt{2 g R}$,where $g$ is the acceleration due to gravity and $R$ is the radius of the planet.
Given the ratio of acceleration due to gravity $\frac{g_1}{g_2} = K_1$ and the ratio of radii $\frac{R_1}{R_2} = K_2$.
The ratio of escape velocities is $\frac{v_{e1}}{v_{e2}} = \sqrt{\frac{2 g_1 R_1}{2 g_2 R_2}}$.
Substituting the given ratios,we get $\frac{v_{e1}}{v_{e2}} = \sqrt{\frac{g_1}{g_2} \cdot \frac{R_1}{R_2}} = \sqrt{K_1 K_2}$.

(A) The acceleration due to gravity on Earth is given by $g = \frac{GM}{R^2}$.
For the planet, the mass is $M' = 4M$ and the radius is $R' = R$.
Therefore, the acceleration due to gravity on the planet is $g' = \frac{GM'}{R'^2} = \frac{G(4M)}{R^2} = 4g$.
Given $g = 10 \,ms^{-2}$, the acceleration due to gravity on the planet is $g' = 4 \times 10 = 40 \,ms^{-2}$.
The work done in lifting a body of mass $m = 5 \,kg$ through a height $h = 2 \,m$ is given by $W = m g' h$.
Substituting the values, $W = 5 \,kg \times 40 \,ms^{-2} \times 2 \,m = 400 \,J$.

(C) The formula for escape velocity is $v = \sqrt{\frac{2GM}{R}}$.
Since mass $M = \text{Volume} \times \text{density} = \frac{4}{3} \pi R^3 \rho$,we substitute this into the formula:
$v = \sqrt{\frac{2G}{R} \cdot \frac{4}{3} \pi R^3 \rho} = \sqrt{\frac{8}{3} \pi G \rho R^2} = R \sqrt{\frac{8 \pi G \rho}{3}}$.
Thus,$v \propto R \sqrt{\rho}$.
Given that the average mass densities $\rho$ are equal,the escape velocity is directly proportional to the radius: $v \propto R$.
Therefore,$\frac{V_P}{V_E} = \frac{R_P}{R_E}$.
Given $R_P = 2 R_E$,we get $\frac{V_P}{V_E} = 2$,which implies $V_P = 2 V_E$.

(C) According to Kepler's Third Law of Planetary Motion, the square of the period of revolution $T$ is proportional to the cube of the semi-major axis $R$ of the orbit: $T^2 \propto R^3$.
Given that the period of planet $A$ $(T_A)$ is $8$ times the period of planet $B$ $(T_B)$, we have $T_A = 8 T_B$.
Using the relation $\frac{T_A}{T_B} = \left(\frac{R_A}{R_B}\right)^{3/2}$, we can write:
$\left(\frac{T_A}{T_B}\right)^2 = \left(\frac{R_A}{R_B}\right)^3$
Substituting the given values:
$(8)^2 = \left(\frac{R_A}{R_B}\right)^3$
$64 = \left(\frac{R_A}{R_B}\right)^3$
Taking the cube root on both sides:
$\frac{R_A}{R_B} = (64)^{1/3} = 4$
Thus, the distance of planet $A$ from the sun is $4$ times the distance of planet $B$ from the sun.

(B) 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 or the orbital radius $(r)$:
$T^2 \propto r^3$
Given the initial time period $T_1 = 5 \text{ hours}$ and the initial radius $r_1 = r$.
The new radius is $r_2 = 4r$.
Using the ratio formula:
$\frac{T_2^2}{T_1^2} = \left(\frac{r_2}{r_1}\right)^3$
Substituting the values:
$\frac{T_2^2}{T_1^2} = \left(\frac{4r}{r}\right)^3 = (4)^3 = 64$
Taking the square root on both sides:
$\frac{T_2}{T_1} = \sqrt{64} = 8$
Therefore, the new time period is:
$T_2 = 8 \times T_1 = 8 \times 5 \text{ hours} = 40 \text{ hours}$.

(C) The orbital velocity of a satellite at a distance $r$ from the center of the earth is given by $v = \sqrt{\frac{GM}{r}}$.
For a satellite just above the earth's surface,the orbital radius is $r = R$,so the velocity is $V = \sqrt{\frac{GM}{R}}$.
For a satellite at an altitude $h = \frac{R}{2}$,the orbital radius is $r' = R + h = R + \frac{R}{2} = \frac{3R}{2}$.
The new orbital velocity $V'$ is given by $V' = \sqrt{\frac{GM}{r'}} = \sqrt{\frac{GM}{3R/2}} = \sqrt{\frac{2GM}{3R}}$.
Substituting $V = \sqrt{\frac{GM}{R}}$ into the expression,we get $V' = \sqrt{\frac{2}{3}} V$.

(D) The energy required to raise a satellite of mass '$m$' to a height '$h$' above the Earth's surface is equal to the change in its gravitational potential energy:
$W = U_f - U_i = -\frac{GMm}{R+h} - (-\frac{GMm}{R}) = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right) = \frac{GMmh}{R(R+h)}$.
The total energy of a satellite of mass '$m$' in a circular orbit at height '$h$' (distance $r = R+h$ from the center) is given by:
$E = \frac{GMm}{2(R+h)}$.
Now,the ratio of the energy required to raise the satellite to the energy required to put it into orbit is:
$\frac{W}{E} = \frac{\frac{GMmh}{R(R+h)}}{\frac{GMm}{2(R+h)}} = \frac{GMmh}{R(R+h)} \times \frac{2(R+h)}{GMm} = \frac{2h}{R}$.

(B) The kinetic energy $(K.E.)$ of a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$ from the center is given by $K.E. = \frac{GMm}{2r}$.
Here,$r$ is the distance from the center of the Earth,so $r = R_{earth} + h$.
For the first satellite at height $h_1 = R$,the orbital radius is $r_1 = R + R = 2R$.
For the second satellite at height $h_2 = 2R$,the orbital radius is $r_2 = R + 2R = 3R$.
Since both satellites have equal mass $m$,the kinetic energy is inversely proportional to the orbital radius: $K.E. \propto \frac{1}{r}$.
Therefore,the ratio of their kinetic energies is $\frac{(K.E.)_1}{(K.E.)_2} = \frac{r_2}{r_1} = \frac{3R}{2R} = \frac{3}{2}$.

(D) The motion of a satellite depends on its horizontal velocity $(v)$ relative to the critical velocity $(v_c)$ and escape velocity $(v_e)$.
If $v < v_c$,the satellite will fall back to Earth.
If $v = v_c$,the satellite will revolve in a circular orbit.
If $v_c < v < v_e$,the satellite will revolve in an elliptical orbit with the Earth at one of the foci.
If $v = v_e$,the satellite will escape the gravitational field of the Earth.
Therefore,if the horizontal velocity is greater than the critical velocity but less than the escape velocity,the satellite will revolve in an elliptical orbit.

(A) The molar specific heat at constant volume,$C_{V}$,is given by the formula $C_{V} = f \times \frac{R}{2}$,where $f$ is the number of degrees of freedom.
Given $f = 6$,we substitute this into the formula:
$C_{V} = 6 \times \frac{R}{2} = 3R$.
Rearranging this equation to solve for $R$,we get:
$R = \frac{C_{V}}{3}$.

(C) For an ideal gas,the molar heat capacity at constant volume is given by $C_v = \frac{f}{2} R$,where $f$ is the degree of freedom.
Rearranging this formula,we get $R = \frac{2}{f} C_v$.
Comparing this with the given equation $R = \frac{2}{3} C_v$,we find that $f = 3$.
$A$ gas with $3$ degrees of freedom is monoatomic (e.g.,noble gases like Helium or Neon).

(A) For an ideal gas,the relationship between molar specific heats at constant pressure $(C_{P})$ and constant volume $(C_{V})$ is given by Mayer's relation: $C_{P} - C_{V} = R$.
From this,we can write $C_{P} = C_{V} + R$. Therefore,the expression $C_{V} = C_{P} + R$ given in option $(A)$ is incorrect.
Let us verify the other options:
$(B)$ $R = C_{P} - C_{V}$. Since $\gamma = \frac{C_{P}}{C_{V}}$,we have $C_{P} = \gamma C_{V}$. Substituting this,$R = \gamma C_{V} - C_{V} = C_{V}(\gamma - 1)$. This is correct.
$(C)$ $\frac{C_{V}}{C_{P}} = \frac{1}{\gamma}$. Since $\gamma = \frac{C_{P}}{C_{V}}$,this is correct.
$(D)$ $R = C_{P} - C_{V} = C_{P} - \frac{C_{P}}{\gamma} = C_{P}(1 - \frac{1}{\gamma}) = C_{P}(\frac{\gamma - 1}{\gamma})$. This is correct.

(B) The kinetic energy per mole of an ideal gas is given by the formula $E = \frac{3}{2} RT$,where $R$ is the universal gas constant and $T$ is the absolute temperature.
Rearranging the formula to solve for $R$:
$R = \frac{2 E}{3 T}$

(C) The pressure exerted by an ideal gas is given by the formula $P = \frac{1}{3} \rho v_{rms}^2$,where $\rho$ is the density and $v_{rms}$ is the root mean square velocity.
Since the pressures $P_1$ and $P_2$ are equal,we have $P_1 = P_2$.
Thus,$\frac{1}{3} \rho_1 v_{rms,1}^2 = \frac{1}{3} \rho_2 v_{rms,2}^2$.
This simplifies to $\rho_1 v_{rms,1}^2 = \rho_2 v_{rms,2}^2$.
Rearranging for the ratio of velocities,we get $\frac{v_{rms,1}^2}{v_{rms,2}^2} = \frac{\rho_2}{\rho_1}$.
Given the ratio of densities $\rho_1 : \rho_2 = 1 : 16$,we have $\frac{\rho_2}{\rho_1} = \frac{16}{1} = 16$.
Taking the square root on both sides,$\frac{v_{rms,1}}{v_{rms,2}} = \sqrt{16} = 4$.
Therefore,the ratio of their r.m.s. velocities is $4:1$.

(B) The translational kinetic energy of a gas molecule is given by the formula $E = \frac{3}{2}kT$,where $k$ is the Boltzmann constant and $T$ is the absolute temperature.
From this relation,it is clear that the kinetic energy $E$ is directly proportional to the absolute temperature $T$ $(E \propto T)$.
To double the kinetic energy $(E' = 2E)$,we must double the absolute temperature $(T' = 2T)$.
Therefore,the temperature must be increased to $2T$.

(A) The average translational kinetic energy of a gas molecule is given by $K.E. = \frac{3}{2} kT$.
Since $k = \frac{R}{N}$,we have $K.E. = \frac{3}{2} \frac{RT}{N}$.
The kinetic energy of an electron accelerated through a potential difference $V$ is $K.E. = eV$.
Equating the two energies:
$\frac{3}{2} \frac{RT}{N} = eV$
Solving for $T$:
$T = \frac{2 eVN}{3 R}$.

(B) The process is isothermal,which means the temperature $T$ of the gas remains constant throughout the process.
The $rms$ speed of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $R$ (universal gas constant),$M$ (molar mass),and $T$ (temperature) are constant,the $rms$ speed remains unchanged.

(B) The root mean square (r.m.s.) velocity of an ideal gas is given by the formula:
$V = \sqrt{\frac{3RT}{M_0}}$
Squaring both sides, we get:
$V^2 = \frac{3RT}{M_0}$
Rearranging the terms to isolate the variables $V$ and $T$:
$\frac{V^2}{T} = \frac{3R}{M_0}$
Since $R$ (universal gas constant) and $M_0$ (molar mass) are constants for a given gas, the ratio $\frac{3R}{M_0}$ is also a constant.
Therefore, $\frac{V^2}{T} = \text{constant}$.

(A) The root mean square speed $(v_{rms})$ of a gas molecule 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.
Since $M = m \times N_A$ (where $m$ is the mass of one molecule and $N_A$ is Avogadro's number),we can substitute this into the equation:
$v_{rms} = \sqrt{\frac{3RT}{mN_A}}$
Since $3$,$R$,and $N_A$ are constants,we have:
$v_{rms} \propto \sqrt{\frac{T}{m}}$
$v_{rms} \propto m^{-\frac{1}{2}} T^{\frac{1}{2}}$

(B) The pressure exerted by an ideal gas is given by the formula $P = \frac{1}{3} \rho c^2$,where $\rho$ is the density and $c$ is the root mean square (rms) speed.
Since the pressures are equal,we have $P_1 = P_2$.
Therefore,$\frac{1}{3} \rho_1 c_1^2 = \frac{1}{3} \rho_2 c_2^2$.
This simplifies to $\frac{c_1^2}{c_2^2} = \frac{\rho_2}{\rho_1}$.
Given the ratio of densities $\rho_1 : \rho_2 = 1 : 16$,we have $\frac{\rho_2}{\rho_1} = \frac{16}{1} = 16$.
Substituting this into the equation,we get $\frac{c_1^2}{c_2^2} = 16$.
Taking the square root on both sides,we find $\frac{c_1}{c_2} = \sqrt{16} = 4$.
Thus,the ratio of their rms speeds $(c_1: c_2)$ is $4: 1$.

(A) The root mean square (rms) speed of a gas molecule 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.
Since the rms speed depends only on the temperature $T$ and the molar mass $M$ of the gas,it is independent of the pressure $P$ of the gas.
Given that the temperature remains constant,the rms speed will remain unchanged even if the pressure is increased.
Therefore,the new rms speed will be $V$.

(D) The root mean square (r.m.s) speed of a gas molecule is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Here,$R$ is the universal gas constant,$T$ is the absolute temperature in Kelvin,and $M$ is the molar mass.
For helium $(He)$: $M_{He} = 4$,$T_{He} = 327^{\circ} C = 327 + 273 = 600 \ K$.
For oxygen $(O_2)$: $M_{O} = 32$,$T_{O} = 27^{\circ} C = 27 + 273 = 300 \ K$.
The ratio of the r.m.s speeds is:
$\frac{v_{He}}{v_{O}} = \sqrt{\frac{T_{He}}{T_{O}} \times \frac{M_{O}}{M_{He}}} = \sqrt{\frac{600}{300} \times \frac{32}{4}} = \sqrt{2 \times 8} = \sqrt{16} = 4$.
Thus,the ratio is $4:1$.

(D) Let $m$ be the mass of the man and $g$ be the acceleration due to gravity.
When the lift moves upwards with acceleration $a$, the apparent weight is $W_1 = m(g + a) = 620 \,N$ --- $(i)$
When the lift moves downwards with the same acceleration $a$, the apparent weight is $W_2 = m(g - a) = 340 \,N$ --- (ii)
Adding equations $(i)$ and (ii):
$m(g + a) + m(g - a) = 620 + 340$
$mg + ma + mg - ma = 960$
$2mg = 960$
$mg = 480 \,N$
Thus, the real weight of the man is $480 \,N$.

(B) The kinetic energy $K$ is related to momentum $p$ and mass $m$ by the formula $K = \frac{p^2}{2m}$.
Rearranging this formula for momentum,we get $p^2 = 2mK$,which implies $p = \sqrt{2mK}$.
Since all three bodies have equal kinetic energy ($K$ is constant),the momentum is directly proportional to the square root of the mass: $p \propto \sqrt{m}$.
Comparing the masses: $m_P = m$,$m_Q = 2m$,and $m_R = 3m$.
Since $m_R > m_Q > m_P$,it follows that $p_R > p_Q > p_P$.
Therefore,the body with the greatest mass,which is $R$,will have the greatest momentum.

(A) The area under the $F-t$ graph represents the impulse, which is equal to the change in momentum of the body.
Since the body is initially at rest, the impulse equals the final momentum of the body after $1 \,s$.
$\text{Impulse} = \text{Area under } F-t \text{ graph}$
$\text{Impulse} = (10 \,N \times 0.5 \,s) + (20 \,N \times 0.5 \,s)$
$\text{Impulse} = 5 \,N-s + 10 \,N-s = 15 \,N-s$
Since $\text{Impulse} = \Delta p = m \times v - m \times u$ and $u = 0$, we have:
$15 = 2 \,kg \times v$
$v = \frac{15}{2} = 7.5 \,m/s$

(C) For a particle moving in a circular orbit,the centripetal force is provided by the central attractive force.
Given $F = \frac{k}{r}$,the magnitude of the centripetal force is $F_c = m r \omega^2$.
Equating the two: $m r \omega^2 = \frac{k}{r}$.
Rearranging for angular velocity $\omega$: $\omega^2 = \frac{k}{m r^2}$.
Taking the square root: $\omega = \sqrt{\frac{k}{m}} \cdot \frac{1}{r}$.
Since the periodic time $T = \frac{2\pi}{\omega}$,we have $T = 2\pi \sqrt{\frac{m}{k}} \cdot r$.
Therefore,$T \propto r$.

(D) The angle of banking $\theta$ is given by $\tan \theta = \frac{h}{x}$,where $h$ is the elevation of the outer rail and $x$ is the gauge of the railway line.
Given,the gauge $x = 1 \text{ m}$ and $\tan \theta = \frac{1}{20}$.
Substituting these values into the formula:
$\frac{1}{20} = \frac{h}{1 \text{ m}}$
$h = \frac{1}{20} \text{ m} = 0.05 \text{ m}$.
Converting this to centimeters:
$h = 0.05 \times 100 \text{ cm} = 5 \text{ cm}$.
Thus,the elevation of the outer rail above the inner rail is $5 \text{ cm}$.

(D) Let the length of the pendulum be $r$. When the bob is displaced by $90^{\circ}$ from the mean position,its height relative to the lowest point is $r$.
By the law of conservation of energy,the potential energy at the highest point is converted into kinetic energy at the lowest point:
$PE_{top} = KE_{bottom}$
$mgr = \frac{1}{2}mv^2$
$mv^2 = 2mgr$
At the lowest position,the forces acting on the bob are the tension $T$ (upwards) and weight $mg$ (downwards). The net force provides the necessary centripetal force:
$T - mg = \frac{mv^2}{r}$
$T = mg + \frac{mv^2}{r}$
Substituting $mv^2 = 2mgr$ into the equation:
$T = mg + \frac{2mgr}{r} = mg + 2mg = 3mg$.

(B) When the lift is stationary, the reading of the spring balance is equal to the weight of the bag: $W = mg = 49 \,N$.
Given $g = 9.8 \,m/s^2$, the mass of the bag is $m = \frac{49}{9.8} = 5 \,kg$.
When the lift moves downward with an acceleration $a = 5 \,m/s^2$, the apparent weight $T$ is given by the formula: $T = m(g - a)$.
Substituting the values: $T = 5(9.8 - 5)$.
$T = 5(4.8) = 24 \,N$.
Therefore, the reading of the spring balance will be $24 \,N$.

(B) The height $h$ to which water rises in a capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g}$,which implies $h \propto \frac{1}{r}$.
The mass of water $m$ in the capillary is given by $m = \text{Volume} \times \text{Density} = (\pi r^2 h) \rho$.
Substituting $h \propto \frac{1}{r}$ into the mass equation,we get $m \propto r^2 \times \frac{1}{r}$,which simplifies to $m \propto r$.
For a new radius $r' = \frac{r}{3}$,the new mass $m'$ will be $m' = m \times \frac{r'}{r} = m \times \frac{r/3}{r} = \frac{m}{3}$.

(C) The height of water rise in a capillary tube is given by the formula: $h = \frac{2 T \cos \theta}{r \rho g}$.
From this,we can see that $h \propto \frac{1}{r}$.
The area of cross-section is $A = \pi r^2$,which implies $r = \sqrt{\frac{A}{\pi}}$,so $r \propto \sqrt{A}$.
Substituting this into the proportionality for height,we get $h \propto \frac{1}{\sqrt{A}}$.
Given the initial area is $A_1 = A$ and the final area is $A_2 = \frac{A}{9}$,we have the ratio:
$\frac{h_2}{h_1} = \sqrt{\frac{A_1}{A_2}} = \sqrt{\frac{A}{A/9}} = \sqrt{9} = 3$.
Therefore,the new height $h_2 = 3 h$.

(B) The excess pressure inside a water drop is given by the formula: $P = \frac{2T}{R}$,where $T$ is the surface tension and $R$ is the radius of the drop.
Given:
Excess pressure $P = 72 \ Nm^{-2}$
Surface tension $T = 7.2 \times 10^{-2} \ Nm^{-1}$
Substituting the values into the formula:
$72 = \frac{2 \times 7.2 \times 10^{-2}}{R}$
$R = \frac{2 \times 7.2 \times 10^{-2}}{72}$
$R = \frac{14.4 \times 10^{-2}}{72}$
$R = 0.2 \times 10^{-2} \ m = 2 \times 10^{-3} \ m$
Since $1 \ mm = 10^{-3} \ m$,we have $R = 2 \ mm$.

(A) Let $R$ be the radius of the large drop and $r$ be the radius of each small droplet.
Since the volume remains constant,we have $\frac{4}{3} \pi R^3 = 512 \times \frac{4}{3} \pi r^3$.
This simplifies to $R^3 = 512 r^3$,which gives $R = 8r$ or $r = \frac{R}{8}$.
The initial surface energy is $U = 4 \pi R^2 T$,where $T$ is the surface tension.
The new surface energy $U'$ is the sum of the surface energies of $512$ droplets: $U' = 512 \times (4 \pi r^2 T)$.
Substituting $r = \frac{R}{8}$ into the equation: $U' = 512 \times 4 \pi (\frac{R}{8})^2 T$.
$U' = 512 \times 4 \pi \frac{R^2}{64} T = 8 \times (4 \pi R^2 T) = 8 U$.

(B) The phase difference $\phi$ between voltage and current in an $RL$ circuit is given by $\tan \phi = \frac{X_L}{R}$.
Given $\phi = 45^{\circ}$,$R = 100 \ \Omega$,and $f = 1000 \ Hz$.
Since $\tan 45^{\circ} = 1$,we have $X_L = R$.
Substituting $X_L = 2 \pi f L$,we get $2 \pi f L = R$.
Therefore,$L = \frac{R}{2 \pi f}$.
Substituting the values: $L = \frac{100}{2 \pi \times 1000} = \frac{100}{2000 \pi} = \frac{1}{20 \pi} = \frac{0.05}{\pi} \ H$.

(D) When the capacitance is removed,the circuit becomes an $LR$ circuit. The phase difference is given by $\tan \phi = \frac{X_L}{R}$.
Given $\phi = 60^{\circ}$,we have $\tan 60^{\circ} = \frac{X_L}{R} = \sqrt{3}$,so $X_L = R\sqrt{3}$.
When the inductance is removed,the circuit becomes an $RC$ circuit. The phase difference is given by $\tan \phi = \frac{X_C}{R}$.
Given $\phi = 60^{\circ}$,we have $\tan 60^{\circ} = \frac{X_C}{R} = \sqrt{3}$,so $X_C = R\sqrt{3}$.
Since $X_L = X_C$,the circuit is in resonance.
At resonance,the impedance $Z = R$.
Given $R = 500 \ \Omega$,the impedance $Z = 500 \ \Omega$.

(D) For the inductor coil at $50 \, Hz$:
$X_L = \frac{V}{I} = \frac{100}{8} = 12.5 \, \Omega$.
For the resistor:
$R = \frac{V}{I} = \frac{100}{10} = 10 \, \Omega$.
At the new frequency $f' = 40 \, Hz$, the new inductive reactance $X_L'$ is:
$X_L' = X_L \times \frac{f'}{f} = 12.5 \times \frac{40}{50} = 10 \, \Omega$.
When connected in series, the impedance $Z$ is:
$Z = \sqrt{R^2 + (X_L')^2} = \sqrt{10^2 + 10^2} = 10\sqrt{2} \, \Omega$.
The current $I$ in the series circuit is:
$I = \frac{V}{Z} = \frac{100}{10\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \, A$.

(A) The initial impedance is $Z = \sqrt{R^2 + X_c^2}$,where $X_c = \frac{1}{\omega C}$.
The initial current is $I = \frac{V}{Z}$.
When the frequency changes to $\omega' = \frac{\omega}{3}$,the new capacitive reactance is $X_c' = \frac{1}{\omega' C} = \frac{1}{(\omega/3) C} = 3X_c$.
The new impedance is $Z' = \sqrt{R^2 + (X_c')^2} = \sqrt{R^2 + (3X_c)^2}$.
Given that the new current $I' = \frac{I}{2}$,we have $\frac{V}{Z'} = \frac{1}{2} \frac{V}{Z}$,which implies $Z' = 2Z$.
Squaring both sides,$(Z')^2 = 4Z^2$,so $R^2 + 9X_c^2 = 4(R^2 + X_c^2)$.
$R^2 + 9X_c^2 = 4R^2 + 4X_c^2$.
$5X_c^2 = 3R^2$.
$\frac{X_c^2}{R^2} = \frac{3}{5} = 0.6$.
Therefore,the ratio $\frac{X_c}{R} = \sqrt{0.6}$.

(C) In an a.c. circuit,when the electromotive force $(e)$ and current $(i)$ reach their zero,minimum,and maximum values at the same time,it indicates that there is no phase difference between them.
This means the phase angle $\phi = 0$.
In a purely resistive circuit,the voltage and current are always in phase.
Therefore,the circuit element connected to the source must be a pure resistor.

(D) In an $LCR$ series resonant circuit,the voltage across the inductor $(V_L)$ leads the current by $90^{\circ}$,while the voltage across the capacitor $(V_C)$ lags the current by $90^{\circ}$.
Therefore,the phase difference between $V_L$ and $V_C$ is $90^{\circ} - (-90^{\circ}) = 180^{\circ}$.
Since they are $180^{\circ}$ out of phase,they act in opposite directions.
At resonance,the inductive reactance equals the capacitive reactance $(X_L = X_C)$,which implies $V_L = V_C$.
Because they are equal in magnitude and $180^{\circ}$ out of phase,they cancel each other out.

(D) The initial impedance is $Z = \sqrt{R^2 + X_c^2}$,where $X_c = \frac{1}{\omega C}$. The initial current is $I = \frac{V}{Z}$.
When the frequency changes to $\omega' = \frac{\omega}{3}$,the new reactance is $X_c' = \frac{1}{\omega' C} = \frac{1}{(\omega/3)C} = 3X_c$.
The new impedance is $Z' = \sqrt{R^2 + (3X_c)^2} = \sqrt{R^2 + 9X_c^2}$.
The new current is $I' = \frac{V}{Z'} = \frac{I}{2}$,which implies $Z' = 2Z$.
Squaring both sides,$Z'^2 = 4Z^2$,so $R^2 + 9X_c^2 = 4(R^2 + X_c^2)$.
$R^2 + 9X_c^2 = 4R^2 + 4X_c^2$.
$5X_c^2 = 3R^2$.
$\frac{X_c^2}{R^2} = \frac{3}{5}$.
Therefore,the ratio $\frac{X_c}{R} = \sqrt{\frac{3}{5}}$.

(B) In a series $RL$ circuit,the phase angle $\phi$ between voltage and current is given by the formula: $\tan \phi = \frac{X_L}{R}$.
Given that the phase angle $\phi = 45^{\circ}$ and the resistance $R = 40 \ \Omega$.
Substituting these values into the formula:
$\tan 45^{\circ} = \frac{X_L}{40}$.
Since $\tan 45^{\circ} = 1$,we have:
$1 = \frac{X_L}{40}$.
Therefore,the inductive reactance $X_L = 40 \ \Omega$.

(C) The energy stored in the capacitor is $U_E = \frac{1}{2}CV^2$.
When the capacitor is connected to the inductor,the energy oscillates between the electric field of the capacitor and the magnetic field of the inductor.
At the moment of maximum current $i_0$,all the energy is stored in the inductor as magnetic energy $U_B = \frac{1}{2}Li_0^2$.
By the law of conservation of energy: $\frac{1}{2}CV^2 = \frac{1}{2}Li_0^2$.
Solving for $i_0$: $i_0^2 = \frac{CV^2}{L} \implies i_0 = V \sqrt{\frac{C}{L}}$.

(D) In a series $LCR$ circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + (2\pi fL - \frac{1}{2\pi fC})^2}$.
The r.m.s. current is given by $I = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + (2\pi fL - \frac{1}{2\pi fC})^2}}$.
At resonance,the inductive reactance $X_L$ equals the capacitive reactance $X_C$,which makes the impedance $Z$ minimum $(Z = R)$.
Consequently,the current $I$ becomes maximum at the resonant frequency $f_0 = \frac{1}{2\pi\sqrt{LC}}$.
As the frequency $f$ increases from zero,the current first increases,reaches a maximum at resonance,and then decreases. This behavior is correctly represented by graph $S$.

(D) For the given part of the circuit containing an inductor $L$ and a resistor $R$ in series,the potential difference between points $A$ and $B$ is given by the equation:
$V_A - V_B = L \frac{dI}{dt} + IR$
Given values are $V_A - V_B = 0.5 \ V$,$I = 0.5 \ A$,$R = 0.2 \ \Omega$,and $\frac{dI}{dt} = 8 \ A/s$.
Substituting these values into the equation:
$0.5 = L(8) + (0.5)(0.2)$
$0.5 = 8L + 0.1$
$8L = 0.5 - 0.1$
$8L = 0.4$
$L = \frac{0.4}{8} = 0.05 \ H$
Therefore,the inductance of the coil is $0.05 \ H$.

(C) Given voltage $V = 20 \cos (2000 t)$,comparing with $V = V_0 \cos (\omega t)$,we get peak voltage $V_0 = 20 \text{ V}$ and angular frequency $\omega = 2000 \text{ rad/s}$.
From the circuit,total resistance $R = 10 \Omega + 5 \Omega = 15 \Omega$,inductance $L = 5 \text{ mH} = 5 \times 10^{-3} \text{ H}$,and capacitance $C = 50 \mu\text{F} = 50 \times 10^{-6} \text{ F}$.
Inductive reactance $X_L = \omega L = 2000 \times 5 \times 10^{-3} = 10 \Omega$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{2000 \times 50 \times 10^{-6}} = \frac{1}{0.1} = 10 \Omega$.
Since $X_L = X_C$,the circuit is in resonance,and the impedance $Z = R = 15 \Omega$.
The peak current $I_0 = \frac{V_0}{Z} = \frac{20}{15} = \frac{4}{3} \text{ A}$.
The r.m.s. current $I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{4/3}{\sqrt{2}} = \frac{2 \sqrt{2}}{3} \text{ A}$.

(C) In an $LCR$ series circuit,the voltage across the inductor $V_L$ and the voltage across the capacitor $V_C$ are $180^{\circ}$ out of phase with each other.
Given that $V_L = 60 \,V$ and $V_C = 60 \,V$.
The resultant voltage across the $LC$ combination is given by $V_{LC} = |V_L - V_C|$.
Substituting the values,we get $V_{LC} = |60 \,V - 60 \,V| = 0 \,V$.
Therefore,the voltage across the $LC$ combination is $0 \,V$.

(A) In a series $LCR$ circuit,the current $I$ is the same through all components.
$1$. The potential difference across the resistor $(V_R)$ is in phase with the current $I$.
$2$. The potential difference across the capacitor $(V_C)$ lags behind the current $I$ by $\frac{\pi}{2}$.
$3$. The potential difference across the inductor $(V_L)$ leads the current $I$ by $\frac{\pi}{2}$.
Evaluating the options:
$(A)$ The potential difference across $R$ $(V_R)$ is in phase with $I$,and the potential difference across the capacitor $(V_C)$ lags behind $I$ by $\frac{\pi}{2}$. Therefore,the phase difference between $V_R$ and $V_C$ is $\frac{\pi}{2}$. This is $CORRECT$.
$(B)$ The applied e.m.f. $(V)$ leads or lags the current $I$ by a phase angle $\phi$,while $V_R$ is in phase with $I$. Thus,$V$ and $V_R$ are not in the same phase. This is $INCORRECT$.
$(C)$ The phase difference between the applied e.m.f. $(V)$ and $V_L$ is $(\frac{\pi}{2} - \phi)$. This is $INCORRECT$.
$(D)$ $V_L$ leads $I$ by $\frac{\pi}{2}$ and $V_C$ lags $I$ by $\frac{\pi}{2}$. Thus,the phase difference between $V_L$ and $V_C$ is $\pi$. This is $INCORRECT$.

(D) For a series $RLC$ circuit,the resonant frequency is given by $\omega_r = \frac{1}{\sqrt{LC}}$.
Given that both circuits have the same resonant frequency $f_r$,we have:
$\omega_r = \frac{1}{\sqrt{L_1 C_1}} = \frac{1}{\sqrt{L_2 C_2}} \implies L_1 C_1 = L_2 C_2$.
When the two circuits are connected in series,the equivalent inductance is $L_{eq} = L_1 + L_2$ and the equivalent capacitance is $C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$.
The new resonant frequency $\omega'$ is given by:
$\omega' = \frac{1}{\sqrt{L_{eq} C_{eq}}} = \frac{1}{\sqrt{(L_1 + L_2) \cdot \frac{C_1 C_2}{C_1 + C_2}}}$
Substituting $L_1 = \frac{L_2 C_2}{C_1}$:
$\omega' = \frac{1}{\sqrt{(\frac{L_2 C_2}{C_1} + L_2) \cdot \frac{C_1 C_2}{C_1 + C_2}}} = \frac{1}{\sqrt{L_2 C_2 (\frac{C_2 + C_1}{C_1}) \cdot \frac{C_1 C_2}{C_1 + C_2}}} = \frac{1}{\sqrt{L_2 C_2}} = \omega_r$.
Thus,the new resonant frequency remains $f_r$.

(A) At resonance,the impedance of the circuit is equal to the resistance,$Z = R$.
Since the circuit is in series,the current $I$ is the same through all components.
$I = \frac{V_R}{R} = \frac{60 \ V}{120 \ \Omega} = 0.5 \ A$.
At resonance,the inductive reactance $X_L$ is given by $X_L = \frac{V_L}{I}$.
$X_L = \frac{40 \ V}{0.5 \ A} = 80 \ \Omega$.
We know that $X_L = \omega L$,where $\omega = 4 \times 10^5 \ rad \ s^{-1}$.
$L = \frac{X_L}{\omega} = \frac{80}{4 \times 10^5} = 20 \times 10^{-5} \ H$.
$L = 2 \times 10^{-4} \ H = 0.2 \times 10^{-3} \ H = 0.2 \ mH$.

(A) In a series $LCR$ circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + (\omega L - 1/\omega C)^2}$.
At resonance,the inductive reactance equals the capacitive reactance,i.e.,$\omega L = 1/\omega C$.
Therefore,the net reactance $(\omega L - 1/\omega C)$ becomes zero.
Substituting this into the impedance formula,we get $Z = \sqrt{R^2 + 0^2} = R$.
The peak current $I_0$ is given by $I_0 = E_0 / Z$.
Substituting $Z = R$,we get $I_0 = E_0 / R$.

(D) When a d.c. voltage is applied, the coil acts as a pure resistor because the inductive reactance $X_L = 2 \pi f L = 0$ for $f = 0$.
$R = \frac{V}{I_{dc}} = \frac{200}{1} = 200 \, \Omega$
When an a.c. voltage is applied, the impedance $Z$ of the $LR$ circuit is given by:
$Z = \frac{V}{I_{ac}} = \frac{200}{0.5} = 400 \, \Omega$
The impedance is related to resistance and inductive reactance by $Z^2 = R^2 + X_L^2$.
$(400)^2 = (200)^2 + X_L^2$
$160000 = 40000 + X_L^2$
$X_L^2 = 120000$
$X_L = \sqrt{120000} = 200 \sqrt{3} \, \Omega$
Since $X_L = 2 \pi f L$, we have:
$2 \pi f \left( \frac{2 \sqrt{3}}{\pi} \right) = 200 \sqrt{3}$
$4 \sqrt{3} f = 200 \sqrt{3}$
$f = \frac{200}{4} = 50 \, Hz$

(B) The given equation is $V = 80 \sin(100 \pi t) \cos(100 \pi t)$.
Using the trigonometric identity $\sin(2\theta) = 2 \sin \theta \cos \theta$,we can rewrite the equation as:
$V = 40 \times (2 \sin(100 \pi t) \cos(100 \pi t))$
$V = 40 \sin(200 \pi t)$.
Comparing this with the standard form of alternating voltage $V = V_0 \sin(\omega t)$,where $V_0$ is the peak voltage,
we get $V_0 = 40 \text{ V}$.

(D) The given equation for the instantaneous current is $I = 50 \sin(100 \pi t)$.
We need to find the time $t$ when $I = 25 \ A$.
Substituting the values into the equation:
$25 = 50 \sin(100 \pi t)$
$\frac{25}{50} = \sin(100 \pi t)$
$0.5 = \sin(100 \pi t)$
Since $\sin 30^{\circ} = 0.5$ and $30^{\circ} = \frac{\pi}{6}$ radians,we have:
$100 \pi t = \frac{\pi}{6}$
Dividing both sides by $100 \pi$:
$t = \frac{\pi}{6 \times 100 \pi}$
$t = \frac{1}{600} \ s$.

(D) The formula for inductive reactance $(X_L)$ is given by $X_L = 2 \pi f L$,where $f$ is the frequency and $L$ is the inductance.
Given that the initial inductive reactance is $R = 2 \pi f L$.
If the new inductance $L' = 2L$ and the new frequency $f' = 2f$,the new inductive reactance $X_L'$ will be:
$X_L' = 2 \pi f' L' = 2 \pi (2f) (2L) = 4 (2 \pi f L) = 4R$.
Therefore,the new inductive reactance is $4R$.

(A) The r.m.s. current $I_{rms}$ in a capacitive circuit is given by $I_{rms} = \frac{V_{rms}}{X_C}$,where $X_C = \frac{1}{\omega C}$ is the capacitive reactance.
Substituting $X_C$,we get $I_{rms} = V_{rms} \cdot \omega \cdot C$.
Given values are $V_{rms} = 230 \text{ V}$,$\omega = 300 \text{ rad/s}$,and $C = 100 \text{ pF} = 100 \times 10^{-12} \text{ F}$.
$I_{rms} = 230 \times 300 \times 100 \times 10^{-12} \text{ A}$.
$I_{rms} = 230 \times 3 \times 10^4 \times 10^{-12} \text{ A}$.
$I_{rms} = 690 \times 10^{-8} \text{ A} = 6.9 \times 10^{-6} \text{ A}$.

(D) The given equation for alternating emf is $e = e_0 \cos \omega t$.
Given peak value $e_0 = 10 \ V$ and frequency $f = 50 \ Hz$.
The angular frequency is $\omega = 2 \pi f = 2 \pi \times 50 = 100 \pi \ rad/s$.
Substituting the values into the equation at $t = \frac{1}{600} \ s$:
$e = 10 \cos(100 \pi \times \frac{1}{600})$
$e = 10 \cos(\frac{\pi}{6})$
Since $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$,we get:
$e = 10 \times \frac{\sqrt{3}}{2} = 5 \sqrt{3} \ V$.

(B) The inductive reactance $X_{L}$ of an inductor is given by the formula $X_{L} = 2 \pi f L$,where $f$ is the frequency of the $a.c.$ supply and $L$ is the self-inductance.
From this relation,it is clear that $X_{L} \propto f$.
Therefore,if the frequency $f$ increases,the inductive reactance $X_{L}$ also increases linearly.

(A) Given the alternating e.m.f. equation: $e = e_0 \sin \omega t$.
We want to find the time $t$ when $e = \frac{e_0}{2}$.
Substituting the value into the equation: $\frac{e_0}{2} = e_0 \sin \omega t$.
This simplifies to: $\sin \omega t = \frac{1}{2}$.
Since $\sin 30^{\circ} = \frac{1}{2}$,we have $\omega t = 30^{\circ} = \frac{\pi}{6} \text{ radians}$.
Substituting $\omega = \frac{2\pi}{T}$: $\left( \frac{2\pi}{T} \right) t = \frac{\pi}{6}$.
Solving for $t$: $t = \frac{T}{12}$.

(A) The frequency of an $LC$ oscillator is given by the formula:
$F = \frac{1}{2 \pi \sqrt{LC}}$
From this relation,we can see that the frequency $F$ is inversely proportional to the square root of the capacitance $C$ (i.e.,$F \propto \frac{1}{\sqrt{C}}$).
Let $F_1 = F$ and $C_1 = 0.1 \ \mu F$.
Let $F_2$ be the new frequency and $C_2 = 0.2 \ \mu F$.
Using the ratio method:
$\frac{F_2}{F_1} = \sqrt{\frac{C_1}{C_2}}$
$\frac{F_2}{F} = \sqrt{\frac{0.1}{0.2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$
Therefore,$F_2 = \frac{F}{\sqrt{2}} \ Hz$.

(A) The applied voltage is $e = e_0 \sin \omega t$.
In an inductor,the current lags the voltage by $\frac{\pi}{2}$. The inductive reactance is $X_L = \omega L$. Thus,the instantaneous current $i_L$ is:
$i_L = \frac{e_0}{\omega L} \sin \left(\omega t - \frac{\pi}{2}\right) = -\frac{e_0}{\omega L} \cos \omega t$
In a capacitor,the current leads the voltage by $\frac{\pi}{2}$. The capacitive reactance is $X_C = \frac{1}{\omega C}$. Thus,the instantaneous current $i_C$ is:
$i_C = \frac{e_0}{X_C} \sin \left(\omega t + \frac{\pi}{2}\right) = e_0 \omega C \cos \omega t$
Therefore,the currents are $-\frac{e_0}{\omega L} \cos \omega t$ and $e_0 \omega C \cos \omega t$.

(A) rejector circuit,also known as a parallel resonant circuit,is a circuit in which an inductor $(L)$ and a capacitor $(C)$ are connected in parallel. At the resonant frequency,the impedance of this parallel combination becomes maximum,which effectively rejects or blocks the current at that specific frequency. Therefore,$L-C-R$ components are connected in parallel to form a rejector circuit.

(C) For an ideal transformer, the input power equals the output power, meaning $P_{in} = P_{out}$.
Since $P = V \times I$, we have $V_{p}I_{p} = V_{s}I_{s}$, which implies $\frac{V_{s}}{V_{p}} = \frac{I_{p}}{I_{s}}$.
In a step-down transformer, the secondary voltage $V_{s}$ is less than the primary voltage $V_{p}$ $(V_{s} < V_{p})$.
According to the relation $\frac{V_{s}}{V_{p}} = \frac{I_{p}}{I_{s}}$, if $V_{s} < V_{p}$, then $I_{s} > I_{p}$.
Therefore, the current increases in the secondary coil.

(C) step-up transformer increases the voltage,meaning $V_s > V_p$.
Since the power in an ideal transformer is conserved $(P_p = P_s)$,we have $V_p I_p = V_s I_s$.
Because $V_s > V_p$,it follows that $I_p > I_s$.
Therefore,the current in the primary coil is greater than the current in the secondary coil.

(B) The input power to the transformer is given by $P_{\text{in}} = V_1 I_1$.
The output power from the transformer is given by $P_{\text{out}} = V_2 I_2$.
The ratio of input power to output power is $\frac{P_{\text{in}}}{P_{\text{out}}} = \frac{V_1 I_1}{V_2 I_2}$.
Assuming an ideal transformer,the input power equals the output power,so the ratio is $1:1$. However,based on the provided options and the general definition of power,the ratio is $\frac{V_1 I_1}{V_2 I_2}$. Since this specific expression is not explicitly listed in the options,we re-evaluate the question context. In many textbook problems,if the transformer is assumed to be ideal,the ratio is $1$. Given the options provided,option $B$ represents the ratio of input power to output power.

(B) Given, turns ratio $\frac{N_p}{N_s} = \frac{20}{1}$.
Secondary voltage $V_s = 8 \, V$ and secondary resistance $R_s = 0.4 \, \Omega$.
The secondary current $I_s$ is calculated using Ohm's law: $I_s = \frac{V_s}{R_s} = \frac{8}{0.4} = 20 \, A$.
For an ideal transformer, the relation between currents and turns ratio is $\frac{I_p}{I_s} = \frac{N_s}{N_p}$.
Substituting the values: $\frac{I_p}{20} = \frac{1}{20}$.
Therefore, the primary current $I_p = 1 \, A$.

(C) For an ideal transformer, the power input equals the power output, which implies $V_p I_p = V_s I_s$.
Also, the transformation ratio is given by $\frac{N_s}{N_p} = \frac{V_s}{V_p} = \frac{I_p}{I_s}$.
Given the ratio of primary to secondary windings is $\frac{N_p}{N_s} = 1:20$, we have $\frac{N_s}{N_p} = 20$.
Using the relation $\frac{I_p}{I_s} = \frac{N_s}{N_p}$, we get $I_p = I_s \times \left(\frac{N_s}{N_p}\right)$.
Given $I_s = 2 \,A$ and $\frac{N_s}{N_p} = 20$, we calculate $I_p = 2 \,A \times 20 = 40 \,A$.

(D) The electron is initially in the third excited state,which corresponds to the principal quantum number $n = 4$ (since the ground state is $n = 1$,first excited state is $n = 2$,second excited state is $n = 3$,and third excited state is $n = 4$).
When the electron transitions from an excited state $n$ to the ground state,the maximum number of spectral lines emitted is given by the formula:
$N = \frac{n(n - 1)}{2}$
Substituting $n = 4$ into the formula:
$N = \frac{4(4 - 1)}{2} = \frac{4 \times 3}{2} = 6$
Therefore,the maximum number of spectral lines emitted is $6$.

(A) The series limit of the Lyman series corresponds to the transition from $n = \infty$ to $n = 1$. Using the Rydberg formula $\frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2})$:
$\frac{1}{\lambda_1} = R(\frac{1}{1^2} - \frac{1}{\infty^2}) = R$
The series limit of the Balmer series corresponds to the transition from $n = \infty$ to $n = 2$:
$\frac{1}{\lambda_3} = R(\frac{1}{2^2} - \frac{1}{\infty^2}) = \frac{R}{4}$
The first line of the Lyman series corresponds to the transition from $n = 2$ to $n = 1$:
$\frac{1}{\lambda_2} = R(\frac{1}{1^2} - \frac{1}{2^2}) = R(1 - \frac{1}{4}) = R - \frac{R}{4}$
Substituting the values of $R$ and $\frac{R}{4}$ from the previous equations:
$\frac{1}{\lambda_2} = \frac{1}{\lambda_1} - \frac{1}{\lambda_3}$
Rearranging the terms,we get:
$\frac{1}{\lambda_1} - \frac{1}{\lambda_2} = \frac{1}{\lambda_3}$

(D) The wave number $\bar{\nu}$ for a spectral line is given by the Rydberg formula: $\bar{\nu} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Balmer series,$n_1 = 2$.
The last line of the Balmer series corresponds to the transition from $n_2 = \infty$ to $n_1 = 2$.
Substituting these values: $\bar{\nu} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4}$.
Given $R = 10^7 \ m^{-1}$,we have $\bar{\nu} = \frac{10^7}{4} = 0.25 \times 10^7 \ m^{-1} = 25 \times 10^5 \ m^{-1}$.

(C) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Balmer series,$n_1 = 2$ and $n_2 = 3, 4, 5, \dots$.
Maximum wavelength $(\lambda_{max})$ occurs for the transition $n_2 = 3$ to $n_1 = 2$:
$\frac{1}{\lambda_{max}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right) \implies \lambda_{max} = \frac{36}{5R}$.
Minimum wavelength $(\lambda_{min})$ occurs for the transition $n_2 = \infty$ to $n_1 = 2$:
$\frac{1}{\lambda_{min}} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4} \implies \lambda_{min} = \frac{4}{R}$.
The ratio of maximum to minimum wavelength is:
$\frac{\lambda_{max}}{\lambda_{min}} = \frac{36/5R}{4/R} = \frac{36}{5R} \times \frac{R}{4} = \frac{9}{5}$.

(C) For a hydrogen-like atom,the energies are given by:
Potential Energy $(P.E.)$ = $-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Ze^2}{r}$
Kinetic Energy $(K.E.)$ = $-\frac{1}{2} (P.E.) = \frac{1}{8 \pi \varepsilon_0} \cdot \frac{Ze^2}{r}$
Total Energy $(T.E.)$ = $\frac{1}{2} (P.E.) = -\frac{1}{8 \pi \varepsilon_0} \cdot \frac{Ze^2}{r}$
As the electron transitions from an excited state to the ground state,the radius $(r)$ of the orbit decreases.
Since $K.E. = \frac{k}{r}$,as $r$ decreases,$K.E.$ increases.
Since $P.E. = -\frac{k'}{r}$,as $r$ decreases,$P.E.$ becomes more negative,meaning it decreases.
Since $T.E. = -\frac{k''}{r}$,as $r$ decreases,$T.E.$ becomes more negative,meaning it decreases.
Therefore,$K.E.$ increases,while $P.E.$ and $T.E.$ decrease.

(C) The shortest wavelength in the Lyman series occurs for the transition from $n = \infty$ to $n = 1$:
$\frac{1}{\lambda_{L}} = R \left[ \frac{1}{1^2} - \frac{1}{\infty^2} \right] = R$
$\therefore \lambda_{L} = \frac{1}{R} = 912 \ \text{Å}$
The longest wavelength in the Paschen series occurs for the transition from $n = 4$ to $n = 3$:
$\frac{1}{\lambda_{P}} = R \left[ \frac{1}{3^2} - \frac{1}{4^2} \right] = R \left[ \frac{1}{9} - \frac{1}{16} \right] = R \left[ \frac{16 - 9}{144} \right] = \frac{7R}{144}$
$\therefore \lambda_{P} = \frac{144}{7R}$
Substituting $\frac{1}{R} = 912 \ \text{Å}$:
$\lambda_{P} = \frac{144}{7} \times 912 \ \text{Å} \approx 18760 \ \text{Å}$

(B) The gyromagnetic ratio is defined as the ratio of the magnetic moment $\mu$ to the angular momentum $L$ of the electron.
$\text{Gyromagnetic ratio} = \frac{\mu}{L} = \frac{iA}{mvr}$
Using the Bohr model,the current $i = \frac{e}{T} = \frac{e \omega}{2 \pi}$ and the area $A = \pi r^2$.
The angular momentum $L = mvr = mr(r \omega) = mr^2 \omega$.
Substituting these values:
$\frac{\mu}{L} = \frac{(\frac{e \omega}{2 \pi})(\pi r^2)}{mr^2 \omega} = \frac{e}{2m}$
Since $e$ and $m$ are constants,the gyromagnetic ratio is independent of the orbit $n$ of the electron.

(A) The potential energy $U$ of an electron in a hydrogen atom at a distance $r$ from the nucleus is given by the formula:
$U = -\frac{1}{4 \pi \varepsilon_0} \cdot \frac{e^2}{r}$
As the electron moves from the ground state $(n=1)$ to the $5^{\text{th}}$ orbit $(n=5)$,the radius $r$ of the orbit increases $(r \propto n^2)$.
Since $U$ is inversely proportional to $r$ and carries a negative sign,as $r$ increases,the value of $U$ becomes less negative,which means it increases.
Therefore,the potential energy of the electron increases.

(B) The centripetal acceleration is given by $a = \frac{v^2}{r}$.
For a hydrogen atom,the velocity of an electron in the $n^{\text{th}}$ orbit is $v \propto \frac{1}{n}$ and the radius is $r \propto n^2$.
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 $3^{\text{rd}}$ and $5^{\text{th}}$ orbits is:
$\frac{a_3}{a_5} = \left(\frac{5}{3}\right)^4 = \frac{625}{81}$.

(B) In the Bohr model of the atom,the kinetic energy $(K)$ of an electron in the $n^{\text{th}}$ orbit is given by $K = \frac{kZe^2}{2r_n}$.
The total energy $(E)$ of the electron in the $n^{\text{th}}$ orbit is given by $E = -\frac{kZe^2}{2r_n}$.
Comparing these two expressions,we find that $K = -E$.
Therefore,the ratio of the kinetic energy to the total energy is $\frac{K}{E} = \frac{-E}{E} = -1$.
This can be expressed as the ratio $1:-1$.

(C) In the Bohr model of the hydrogen atom,the total energy $E$ of an electron in the $n^{\text{th}}$ orbit is given by $E \propto \frac{1}{n^2}$.
According to Bohr's quantization postulate,the angular momentum $L$ of an electron in the $n^{\text{th}}$ orbit is given by $L = \frac{nh}{2\pi}$,which implies $L \propto n$.
Substituting $n \propto L$ into the energy expression,we get $E \propto \frac{1}{L^2}$.
Therefore,$E \propto L^{-2}$.

(C) The energy of an electron in a hydrogen-like atom is given by the formula: $E_n = -13.6 \frac{Z^2}{n^2} \text{ eV}$.
For the second orbit of a hydrogen atom $(Z=1, n=2)$: $E = -13.6 \frac{1^2}{2^2} = -\frac{13.6}{4} \text{ eV}$.
Therefore,$13.6 \text{ eV} = 4E$.
For the third orbit of a helium ion $(Z=2, n=3)$: $E_3 = -13.6 \frac{2^2}{3^2} = -13.6 \frac{4}{9} \text{ eV}$.
Substituting $13.6 = -4E$ (considering magnitude or relative energy levels): $E_3 = -(-4E) \frac{4}{9} = \frac{16E}{9}$.

(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6 \ eV}{n^2}$.
For the first transition (ii to i): $n_i = 2$ to $n_f = 1$.
The energy of the emitted photon is $\Delta E_1 = E_2 - E_1 = -\frac{13.6}{2^2} - (-\frac{13.6}{1^2}) = 13.6(1 - \frac{1}{4}) = 13.6 \times \frac{3}{4} \ eV$.
For the second transition (highest to ii): $n_i = \infty$ to $n_f = 2$.
The energy of the emitted photon is $\Delta E_2 = E_{\infty} - E_2 = 0 - (-\frac{13.6}{2^2}) = \frac{13.6}{4} \ eV$.
The ratio of the energies is $\frac{\Delta E_1}{\Delta E_2} = \frac{13.6 \times \frac{3}{4}}{\frac{13.6}{4}} = \frac{3}{1} = 3:1$.

(C) The capacitance of a parallel plate capacitor is given by the formula $C = \frac{k A \varepsilon_0}{d}$, where $k$ is the dielectric constant, $A$ is the area of the plates, $\varepsilon_0$ is the permittivity of free space, and $d$ is the separation between the plates.
Initially, the capacitance is $C_1 = \frac{k A \varepsilon_0}{d}$.
When the plates are moved closer such that the new separation $d' = \frac{d}{2}$, the new capacitance $C_2$ becomes:
$C_2 = \frac{k A \varepsilon_0}{d'} = \frac{k A \varepsilon_0}{d/2} = 2 \left( \frac{k A \varepsilon_0}{d} \right)$.
Substituting $C_1$ into the equation, we get $C_2 = 2C_1$.

(B) Let $d$ be the initial distance between the plates. The initial capacitance is $C = \frac{A \epsilon_0}{d}$.
When a dielectric plate of thickness $t = 2 \,mm$ is inserted, the new capacitance $C_1$ is given by $C_1 = \frac{A \epsilon_0}{d - t + \frac{t}{k}}$.
When the distance between the plates is increased by $x = 1.6 \,mm$ to maintain the same potential difference (which implies the same capacitance $C_2 = C$), the new distance becomes $d' = d + x$.
The new capacitance is $C_2 = \frac{A \epsilon_0}{d + x - t + \frac{t}{k}}$.
Since $C_2 = C$, we have $d + x - t + \frac{t}{k} = d$.
Simplifying this, we get $x - t + \frac{t}{k} = 0$, or $t - x = \frac{t}{k}$.
Substituting the values $t = 2 \,mm$ and $x = 1.6 \,mm$:
$2 - 1.6 = \frac{2}{k}$
$0.4 = \frac{2}{k}$
$k = \frac{2}{0.4} = 5$.

(D) The capacitance of a parallel plate capacitor filled with a dielectric medium is given by $C = \frac{K \varepsilon_0 A}{d}$.
Given that the dielectric constant $K = 3$,the initial capacitance is $C = \frac{3 \varepsilon_0 A}{d}$.
When the oil (dielectric) is removed,the medium between the plates becomes air (or vacuum),for which the dielectric constant $K' = 1$.
The new capacitance $C'$ is given by $C' = \frac{\varepsilon_0 A}{d}$.
Comparing the two expressions,we find that $C' = \frac{C}{3}$.

(B) The capacitance of a parallel plate capacitor is given by $C = \frac{k A \varepsilon_0}{d}$.
Initially,for an air-filled capacitor,$C_1 = \frac{A \varepsilon_0}{d_1} = 2 \ pF$,where $k_1 = 1$.
Finally,the separation is doubled $(d_2 = 2d_1)$ and the space is filled with a dielectric of constant $k_2 = k$. The new capacitance is $C_2 = \frac{k A \varepsilon_0}{d_2} = 6 \ pF$.
Taking the ratio: $\frac{C_2}{C_1} = \frac{k \cdot A \varepsilon_0 / (2d_1)}{A \varepsilon_0 / d_1} = \frac{k}{2}$.
Given $\frac{C_2}{C_1} = \frac{6}{2} = 3$.
Therefore,$\frac{k}{2} = 3$,which gives $k = 6$.