AP EAMCET 2022 Physics Question Paper with Answer and Solution

(B) The centre of mass $(COM)$ of a homogeneous semi-circular plate of radius $r$ is located on the axis of symmetry at a distance of $\frac{4r}{3\pi}$ from the centre of its straight edge $(O)$.
Therefore,the distance $OA = \frac{4r}{3\pi}$.

(A) Let the original disc have radius $R_1$ and mass $M_1 = \sigma \pi R_1^2$,where $\sigma$ is the surface mass density. Its center of mass is at the origin $(0,0)$.
Let the removed circular portion have radius $R_2$ and mass $M_2 = \sigma \pi R_2^2$. Its center is at $(R_1 - R_2, 0)$ as shown in the figure.
The center of mass of the remaining portion is given by:
$x_{CM} = \frac{M_1 x_1 - M_2 x_2}{M_1 - M_2}$
Substituting the values:
$x_{CM} = \frac{(\sigma \pi R_1^2)(0) - (\sigma \pi R_2^2)(R_1 - R_2)}{\sigma \pi R_1^2 - \sigma \pi R_2^2}$
$x_{CM} = \frac{-R_2^2(R_1 - R_2)}{R_1^2 - R_2^2}$
$x_{CM} = \frac{-R_2^2(R_1 - R_2)}{(R_1 - R_2)(R_1 + R_2)}$
$x_{CM} = -\frac{R_2^2}{R_1 + R_2}$

(A) The mass at position $x=N$ is given by $m_N = m \left(\frac{1}{3}\right)^N \frac{1}{N}$.
The centre of mass $X_{cm}$ along the $X$-axis is given by the formula:
$X_{cm} = \frac{\sum m_N x_N}{\sum m_N} = \frac{\sum_{N=2}^{\infty} \left[ m \left(\frac{1}{3}\right)^N \frac{1}{N} \right] \times N}{M}$
$X_{cm} = \frac{m}{M} \sum_{N=2}^{\infty} \left(\frac{1}{3}\right)^N$
This is an infinite geometric progression with the first term $a = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$ and common ratio $r = \frac{1}{3}$.
The sum of an infinite geometric series is $S = \frac{a}{1-r}$.
$X_{cm} = \frac{m}{M} \left[ \frac{1/9}{1 - 1/3} \right] = \frac{m}{M} \left[ \frac{1/9}{2/3} \right] = \frac{m}{M} \left[ \frac{1}{9} \times \frac{3}{2} \right] = \frac{m}{6M}$.

(A) The position of the centre of mass $X_{CM}$ is given by the formula $X_{CM} = \frac{\sum m_i x_i}{M}$.
Substituting the given values:
$X_{CM} = \frac{m(1) + (1/2)(m/2)(2) + (1/2)^2(m/3)(3) + \dots + (1/2)^{N-1}(m/N)(N) + \dots}{M}$
$X_{CM} = \frac{m + (1/2)m + (1/2)^2m + \dots + (1/2)^{N-1}m + \dots}{M}$
$X_{CM} = \frac{m}{M} [1 + 1/2 + (1/2)^2 + \dots + (1/2)^{N-1} + \dots]$
The term in the bracket is an infinite geometric progression with first term $a = 1$ and common ratio $r = 1/2$.
The sum of an infinite $G$.$P$. is $S = \frac{a}{1-r} = \frac{1}{1 - 1/2} = \frac{1}{1/2} = 2$.
Therefore,$X_{CM} = \frac{m}{M} \times 2 = \frac{2m}{M}$.
Since the masses are placed only along the $X$-axis,$Y_{CM} = 0$ and $Z_{CM} = 0$.
Thus,the centre of mass is $(\frac{2m}{M}, 0, 0)$.

(A) Applying the law of conservation of linear momentum before and after the collision:
$M(150) + 2M(-v) = M(0) + 2M(v')$ (Taking the direction of ball $A$ as positive)
$150 - 2v = 2v'$
$v' = 75 - v \dots (1)$
Now,the coefficient of restitution $e$ is given by:
$e = \frac{v_2 - v_1}{u_1 - u_2} = 1$
$1 = \frac{v' - 0}{150 - (-v)}$
$150 + v = v' \dots (2)$
Equating $(1)$ and $(2)$:
$75 - v = 150 + v$
$2v = -75$
Since we are looking for the speed (magnitude),we consider the relative velocities. Re-evaluating with standard convention: Let $A$ move at $+150$ and $B$ move at $-v$. After collision,$A$ is at $0$ and $B$ moves at $+v'$.
$150M - 2Mv = 2Mv' \Rightarrow 75 - v = v'$
$e = \frac{v' - 0}{150 - (-v)} = 1 \Rightarrow v' = 150 + v$
Solving $75 - v = 150 + v$ gives $v = -37.5$. The magnitude of speed is $37.5 \ m \ s^{-1}$.

(A) For ball $A$: $m_A = 1 \,kg$, $u_A = 4 \,ms^{-1}$. For ball $B$: $m_B = 3 \,kg$, $u_B = 0 \,ms^{-1}$.
Total initial momentum $p_i = m_A u_A + m_B u_B = (1 \times 4) + (3 \times 0) = 4 \,kg \cdot ms^{-1}$.
After collision, both bodies stick together and move with a common velocity $v$.
Total final momentum $p_f = (m_A + m_B)v = (1 + 3)v = 4v$.
By the law of conservation of linear momentum, $p_i = p_f$, so $4 = 4v$, which gives $v = 1 \,ms^{-1}$.
The force exerted on ball $B$ is the rate of change of momentum of ball $B$:
$F_B = \frac{\Delta p_B}{\Delta t} = \frac{m_B(v - u_B)}{\Delta t} = \frac{3(1 - 0)}{0.1} = \frac{3}{0.1} = 30 \,N$.

(C) The initial velocity of the ball is $u = 5 \,m \,s^{-1}$ at an angle of $30^{\circ}$ with the horizontal.
We resolve this velocity into horizontal and vertical components:
Horizontal component, $v_{1x} = u \cos 30^{\circ} = 5 \times \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2} \,m \,s^{-1}$.
Vertical component, $v_{1y} = u \sin 30^{\circ} = 5 \times \frac{1}{2} = 2.5 \,m \,s^{-1}$.
During the collision with the ground, the horizontal component of velocity remains unchanged because there is no impulse along the horizontal direction.
$v_{2x} = v_{1x} = \frac{5\sqrt{3}}{2} \,m \,s^{-1}$.
The vertical component changes according to the coefficient of restitution $e = 0.2$:
$v_{2y} = e \times v_{1y} = 0.2 \times 2.5 = 0.5 \,m \,s^{-1}$.
The final speed of the ball $v_f$ is the magnitude of the resultant velocity vector:
$v_f = \sqrt{v_{2x}^2 + v_{2y}^2} = \sqrt{\left(\frac{5\sqrt{3}}{2}\right)^2 + (0.5)^2}$
$v_f = \sqrt{\frac{25 \times 3}{4} + 0.25} = \sqrt{\frac{75}{4} + \frac{1}{4}} = \sqrt{\frac{76}{4}} = \sqrt{19} \,m \,s^{-1}$.

(A) Linear momentum is a vector quantity.
Let the direction towards the wall be positive.
Initial momentum $\vec{p}_i = m \times 10 \hat{i} = 0.5 \times 10 \hat{i} = 5 \hat{i} \ kg \ m \ s^{-1}$.
After rebounding,the ball moves in the opposite direction.
Final momentum $\vec{p}_f = -m \times v \hat{i} = -0.5 \times v \hat{i} \ kg \ m \ s^{-1}$.
The change in linear momentum is $\Delta \vec{p} = \vec{p}_f - \vec{p}_i$.
$\Delta \vec{p} = (-0.5v \hat{i}) - (5 \hat{i}) = -(0.5v + 5) \hat{i}$.
The magnitude of the change in linear momentum is given as $8.0 \ kg \ m \ s^{-1}$.
$|\Delta \vec{p}| = 0.5v + 5 = 8.0$.
$0.5v = 8.0 - 5 = 3.0$.
$v = \frac{3.0}{0.5} = 6.0 \ m \ s^{-1}$.

(A) Step $1$: The bullet of mass $m$ strikes $M_1$ and gets embedded in it. This is a perfectly inelastic collision. By the law of conservation of linear momentum:
$m v = (m + M_1) v_1$
where $v_1$ is the velocity of the $(m + M_1)$ system after the first collision.
$v_1 = \frac{m v}{m + M_1}$
Step $2$: The system $(m + M_1)$ now moves with velocity $v_1$ and collides elastically with $M_2$,which is initially at rest $(u_2 = 0)$.
For an elastic collision between two masses $m_A$ and $m_B$ where $m_B$ is initially at rest,the final velocity $v_B$ of mass $m_B$ is given by:
$v_B = \frac{2 m_A}{m_A + m_B} u_A$
Here,$m_A = (m + M_1)$,$m_B = M_2$,and $u_A = v_1$.
Substituting these values:
$v_2 = \frac{2(m + M_1)}{(m + M_1) + M_2} v_1$
$v_2 = \frac{2(m + M_1)}{m + M_1 + M_2} \cdot \frac{m v}{m + M_1}$
$v_2 = \frac{2 m v}{m + M_1 + M_2}$
Thus,the velocity of $M_2$ after the collision is $\frac{2 m v}{M_1 + M_2 + m}$.

(C) The gravitational acceleration $g$ at a distance $r$ from the centre of a uniform solid sphere of mass $M$ and radius $R$ (where $r \ge R$) is given by $g = \frac{GM}{r^2}$.
On the surface,$r = R$,so $a_o = \frac{GM}{R^2}$.
We want to find the distance $r$ where the acceleration becomes $a = \frac{a_o}{4}$.
Substituting the expressions,we get $\frac{GM}{r^2} = \frac{1}{4} \left( \frac{GM}{R^2} \right)$.
Canceling $GM$ from both sides,we get $\frac{1}{r^2} = \frac{1}{4R^2}$.
Taking the square root of both sides,$r^2 = 4R^2$,which gives $r = 2R$.
Thus,the distance from the centre of the sphere is $2R$.

(C) The acceleration due to gravity $g$ on the surface of a planet is given by $g = \frac{GM}{R^2}$.
Since density $\rho = \frac{M}{V} = \frac{M}{\frac{4}{3}\pi R^3}$,we can express the radius $R$ as $R = \left( \frac{3M}{4\pi \rho} \right)^{1/3}$.
Substituting $R$ into the formula for $g$:
$g = \frac{GM}{(\frac{3M}{4\pi \rho})^{2/3}} = G M^{1/3} (\frac{4\pi \rho}{3})^{2/3}$.
This shows that $g \propto M^{1/3} \rho^{2/3}$.
Given for the planet: $M^{\prime} = 8M$ and $\rho^{\prime} = 27\rho$.
Taking the ratio:
$\frac{g^{\prime}}{g} = \left( \frac{M^{\prime}}{M} \right)^{1/3} \left( \frac{\rho^{\prime}}{\rho} \right)^{2/3}$.
$\frac{g^{\prime}}{g} = (8)^{1/3} \times (27)^{2/3} = 2 \times (3^3)^{2/3} = 2 \times 3^2 = 2 \times 9 = 18$.
Therefore,$g^{\prime} = 18g$.

(A) The radius of the solid sphere is $R$ and its mass is $M$. The density of the sphere is $\rho = \frac{M}{\frac{4}{3} \pi R^3}$.
Each spherical cavity has a radius of $R/2$. The mass of each removed portion is $M' = \rho \times \text{Volume of cavity} = \frac{M}{\frac{4}{3} \pi R^3} \times \frac{4}{3} \pi (R/2)^3 = \frac{M}{8}$.
The center of the sphere is at the origin. Let the center of the left cavity be at $x = -R/2$ and the center of the right cavity be at $x = R/2$. The point mass $m$ is at $x = d$.
The gravitational force on $m$ is the force due to the full sphere minus the forces due to the two removed spherical masses.
$F = \frac{G M m}{d^2} - \frac{G M' m}{(d - R/2)^2} - \frac{G M' m}{(d + R/2)^2}$.
Substituting $M' = M/8$:
$F = \frac{G M m}{d^2} - \frac{G M m}{8(d - R/2)^2} - \frac{G M m}{8(d + R/2)^2}$.
Factoring out $\frac{G M m}{d^2}$:
$F = \frac{G M m}{d^2} \left[ 1 - \frac{1}{8(1 - R/2d)^2} - \frac{1}{8(1 + R/2d)^2} \right]$.

(A) The escape velocity is given by $v_E = \sqrt{\frac{2GM}{R}}$.
At the Earth's surface,the total mechanical energy is $E_1 = KE_1 + PE_1 = \frac{1}{2}mv^2 - \frac{GMm}{R}$.
Substituting $v = \alpha v_E = \alpha \sqrt{\frac{2GM}{R}}$,we get $KE_1 = \frac{1}{2}m(\alpha^2 \cdot \frac{2GM}{R}) = \frac{GMm\alpha^2}{R}$.
So,$E_1 = \frac{GMm\alpha^2}{R} - \frac{GMm}{R} = \frac{GMm}{R}(\alpha^2 - 1)$.
At the maximum height $h$,the velocity is $0$,so $KE_2 = 0$ and $PE_2 = -\frac{GMm}{R+h}$.
By the law of conservation of energy,$E_1 = E_2$:
$\frac{GMm}{R}(\alpha^2 - 1) = -\frac{GMm}{R+h}$.
$\alpha^2 - 1 = -\frac{R}{R+h} = -\frac{6400}{6400+800} = -\frac{6400}{7200} = -\frac{8}{9}$.
$\alpha^2 = 1 - \frac{8}{9} = \frac{1}{9}$.
Therefore,$\alpha = \frac{1}{3}$.

(C) Using the law of conservation of mechanical energy between the surface of the earth and the maximum height $h$:
Total energy at surface = Total energy at maximum height
$-\frac{G M m}{R_E} + \frac{1}{2} m v^2 = -\frac{G M m}{R_E + h} + 0$
Given $h = \frac{4}{5} R_E$,so $R_E + h = R_E + \frac{4}{5} R_E = \frac{9}{5} R_E$.
Substituting $GM = g R_E^2$:
$-\frac{g R_E^2 m}{R_E} + \frac{1}{2} m v^2 = -\frac{g R_E^2 m}{\frac{9}{5} R_E}$
$-g R_E + \frac{v^2}{2} = -\frac{5}{9} g R_E$
$\frac{v^2}{2} = g R_E - \frac{5}{9} g R_E = \frac{4}{9} g R_E$
$v^2 = \frac{8}{9} g R_E$
$v = \sqrt{\frac{8}{9} g R_E} = \frac{2}{3} \sqrt{2 g R_E}$
Since escape velocity $v_E = \sqrt{2 g R_E}$,we have $v = \frac{2}{3} v_E$.
Therefore,$\frac{v}{v_E} = \frac{2}{3}$.

(B) Let $V_e = V$ be the escape speed of the object on the surface of the earth. The formula for escape speed is $V = \sqrt{\frac{2GM}{R}}$.
By the law of conservation of mechanical energy,the total energy at the surface equals the total energy at infinity:
$K_i + U_i = K_f + U_f$
$\frac{1}{2}m(4V)^2 - \frac{GMm}{R} = \frac{1}{2}mV_0^2 + 0$
Since $V^2 = \frac{2GM}{R}$,we have $\frac{GM}{R} = \frac{V^2}{2}$.
Substituting this into the energy equation:
$\frac{1}{2}m(16V^2) - m(\frac{V^2}{2}) = \frac{1}{2}mV_0^2$
$8mV^2 - 0.5mV^2 = 0.5mV_0^2$
$7.5V^2 = 0.5V_0^2$
$V_0^2 = 15V^2$
$V_0 = \sqrt{15}V$

(B) 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 $d$ and radius $R$ as $M = d \times \frac{4}{3} \pi R^3$,we substitute this into the formula:
$v_e = \sqrt{\frac{2G}{R} \times \frac{4}{3} \pi R^3 d} = \sqrt{\frac{8}{3} G \pi R^2 d} = R \sqrt{\frac{8}{3} G \pi d}$
For the first planet $(A)$:
$v_1 = 16 \ km/s = R \sqrt{\frac{8}{3} G \pi d}$
For the second planet $(B)$ with radius $R' = 3R$ and density $d' = 2d$:
$v_2 = (3R) \sqrt{\frac{8}{3} G \pi (2d)} = 3R \sqrt{2} \sqrt{\frac{8}{3} G \pi d} = 3 \sqrt{2} \times v_1$
Substituting $v_1 = 16 \ km/s$:
$v_2 = 3 \sqrt{2} \times 16 = 48 \sqrt{2} \ km/s$
Given $v_2 = v \sqrt{2} \ km/s$,we find:
$v = 48$

(B) The time period $T$ of a satellite revolving around a planet is given by $T = 2\pi \sqrt{\frac{r^3}{GM}}$. Since both satellites are in the same circular orbit,their orbital radius $r$ is the same,meaning their time periods are identical. Thus,Statement $(A)$ is true.
The orbital velocity $v$ of a satellite is given by $v = \sqrt{\frac{GM}{r}}$. This shows that $v \propto \frac{1}{\sqrt{r}}$,meaning the orbital velocity is inversely proportional to the square root of the radius of the orbit. Thus,Statement $(B)$ is true.
The escape velocity $v_e$ from a point at distance $r$ from the center of a planet is given by $v_e = \sqrt{\frac{2GM}{r}}$. Since $r = R + h$ (where $R$ is the radius of the planet and $h$ is the altitude),the escape velocity depends on the altitude $h$. Thus,Statement $(C)$ is false.

(D) Given: $T_A = 8 T_B$.
According to Kepler's third law,$T^2 \propto R^3$,where $T$ is the time period and $R$ is the orbital radius.
Thus,$\frac{T_A^2}{T_B^2} = \frac{R_A^3}{R_B^3} \Rightarrow (8)^2 = \left(\frac{R_A}{R_B}\right)^3 \Rightarrow 64 = \left(\frac{R_A}{R_B}\right)^3$.
Taking the cube root,$\frac{R_A}{R_B} = (64)^{1/3} = 4$.
Orbital velocity $V$ is given by $V = \frac{2 \pi R}{T}$.
Therefore,the ratio of orbital velocities is $\frac{V_A}{V_B} = \frac{R_A}{R_B} \times \frac{T_B}{T_A} = 4 \times \frac{1}{8} = \frac{1}{2}$.
Hence,the ratio is $1: 2$.

(C) geostationary satellite is a satellite that orbits the Earth in the same direction as the Earth's rotation (from west to east).
Its orbital period is exactly equal to the Earth's rotational period, which is $24 \,h$ (or $1 \,day$).

(A) The total energy $E$ of a satellite of mass $m$ orbiting the Earth of mass $M_e$ at a distance $r$ is given by:
$E = -\frac{G M_e m}{2r}$
This shows that $E \propto \frac{1}{r}$.
Let the initial radius be $r_1$ and the initial energy be $E_1 = -\frac{G M_e m}{2r_1}$.
When the radius is reduced to half,$r_2 = \frac{r_1}{2}$.
The new energy is $E_2 = -\frac{G M_e m}{2r_2} = -\frac{G M_e m}{2(r_1/2)} = -\frac{G M_e m}{r_1} = 2 E_1$.
The change in total energy is $\Delta E = E_2 - E_1 = 2E_1 - E_1 = E_1$.
The percentage change is given by $\frac{|\Delta E|}{|E_1|} \times 100\% = \frac{|E_1|}{|E_1|} \times 100\% = 100\%$.

(A) The acceleration due to gravity on the surface of a planet is given by $g = \frac{GM}{R^2}$.
For Earth,$g_e = \frac{GM}{R^2}$.
For the new planet,the mass $M' = 100M$ and the radius $R' = 10R$.
Therefore,the acceleration due to gravity on the new planet is $g_p = \frac{G(100M)}{(10R)^2} = \frac{100GM}{100R^2} = \frac{GM}{R^2} = g_e$.
Since the height $h$ is the same and the acceleration due to gravity $g$ is the same,the time taken to reach the ground is given by $h = \frac{1}{2}gt^2$,which implies $t = \sqrt{\frac{2h}{g}}$.
Since $h$ and $g$ are identical for both,the time taken $t$ remains the same.

(B) In three-dimensional space,a molecule has a total of $3N$ degrees of freedom,where $N$ is the number of atoms. For a diatomic molecule,$N = 2$,so the total degrees of freedom are $3 \times 2 = 6$.
These $6$ degrees of freedom are categorized as follows:
$1$. Translational degrees of freedom: $3$ (along the $x, y,$ and $z$ axes).
$2$. Rotational degrees of freedom: $2$ (for a linear diatomic molecule).
$3$. Vibrational degrees of freedom: The remaining degrees of freedom are calculated as $6 - (3 + 2) = 1$.
Therefore,a diatomic molecule has $1$ vibrational degree of freedom.

(C) For a non-rigid diatomic molecule,the degrees of freedom $(f)$ are calculated as follows:
Translational degrees of freedom $= 3$
Rotational degrees of freedom $= 2$
Vibrational degrees of freedom $= 2$ (one for kinetic energy and one for potential energy).
Total degrees of freedom $(f) = 3 + 2 + 2 = 7$.
The ratio of specific heats is given by $\gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f}$.
Substituting $f = 7$,we get $\gamma = 1 + \frac{2}{7} = \frac{9}{7}$.
Therefore,$\frac{C_p}{C_v} = \frac{9}{7}$,which implies $7 C_p = 9 C_v$.
Squaring both sides,we get $49 C_p^2 = 81 C_v^2$.

(C) diatomic molecule consists of two atoms connected by a rigid bond.
It can rotate about two axes that are perpendicular to the line joining the two atoms.
Rotation about the axis passing through the two atoms is generally neglected in classical kinetic theory because the moment of inertia about this axis is negligible.
Therefore,a diatomic molecule has $2$ rotational degrees of freedom.

(B) The heat supplied at constant pressure is given by the formula: $\Delta Q = n C_P \Delta T$.
Since nitrogen $(N_2)$ is a diatomic gas, its degrees of freedom $f = 5$.
The molar heat capacity at constant pressure is $C_P = (\frac{f}{2} + 1) R = (\frac{5}{2} + 1) R = \frac{7}{2} R$.
The number of moles $n = \frac{M}{M_0}$, where $M$ is the mass of the gas and $M_0$ is the molar mass $(28 \,g/mol)$.
Substituting the values: $498 = (\frac{M}{28}) \times (\frac{7}{2}) \times 8.3 \times 40$.
Simplifying the equation: $498 = M \times (\frac{7}{56}) \times 8.3 \times 40$.
$498 = M \times 0.125 \times 332$.
$498 = M \times 41.5$.
$M = \frac{498}{41.5} = 12 \,g$.

(A) Given: Initial volume $V_i = 100 \ cc$.
Initial pressure $P_i = 760 \ mm$ of $Hg$.
Initial temperature $T_i = 30^{\circ} C$.
Final temperature $T_f = 30^{\circ} C$ (since the temperature remains constant).
Final pressure $P_f = 400 \ mm$ of $Hg$.
Since the temperature is constant,we apply Boyle's Law: $P_i V_i = P_f V_f$.
Substituting the values: $760 \times 100 = 400 \times V_f$.
Solving for $V_f$: $V_f = \frac{760 \times 100}{400}$.
$V_f = \frac{76000}{400} = 190 \ cc$.

(A) In an adiabatic compression,work is done on the gas,which increases its internal energy. Since the internal energy of an ideal gas is directly proportional to its absolute temperature $(U \propto T)$,the temperature of the gas increases.
According to the kinetic theory of gases,the average kinetic energy of gas molecules is given by $K_{avg} = \frac{3}{2} k_B T$. As the temperature $T$ increases,the average kinetic energy also increases.
This increase in kinetic energy occurs because,during compression,the molecules collide with the moving piston (wall). When a molecule hits a wall moving towards it,the molecule gains speed (and thus kinetic energy) due to the elastic collision,similar to a ball bouncing off a moving bat. Therefore,both the assertion and the reason are true,and the reason correctly explains the assertion.

(C) For a gas mixture,the adiabatic index $\gamma_{\text{mix}}$ is given by $\gamma_{\text{mix}} = \frac{C_{p(\text{mix})}}{C_{V(\text{mix})}}$.
Using the formula for mixture properties: $\gamma_{\text{mix}} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 C_{V1} + n_2 C_{V2}}$.
For oxygen $(O_2)$,which is diatomic: $n_1 = \frac{4 \ g}{32 \ g/mol} = \frac{1}{8} \ mol$. $C_{p1} = \frac{7}{2}R$,$C_{V1} = \frac{5}{2}R$.
For helium (He),which is monatomic: $n_2 = \frac{4 \ g}{4 \ g/mol} = 1 \ mol$. $C_{p2} = \frac{5}{2}R$,$C_{V2} = \frac{3}{2}R$.
Substituting these values into the formula:
$\gamma_{\text{mix}} = \frac{(\frac{1}{8} \times \frac{7}{2}R) + (1 \times \frac{5}{2}R)}{(\frac{1}{8} \times \frac{5}{2}R) + (1 \times \frac{3}{2}R)}$
$\gamma_{\text{mix}} = \frac{\frac{7}{16} + \frac{5}{2}}{\frac{5}{16} + \frac{3}{2}} = \frac{\frac{7+40}{16}}{\frac{5+24}{16}} = \frac{47}{29}$.

(B) The root mean square (rms) speed of a gas molecule is given by the formula $V_{rms} = \sqrt{\frac{3RT}{M_0}}$, where $R$ is the universal gas constant, $T$ is the absolute temperature, and $M_0$ is the molar mass of the gas.
Since $R$ and $T$ are constant for both gases, we have $V_{rms} \propto \frac{1}{\sqrt{M_0}}$.
Therefore, the ratio of the rms speeds of hydrogen $(H_2)$ and oxygen $(O_2)$ is given by:
$\frac{(V_{rms})_{H_2}}{(V_{rms})_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.
Given $(V_{rms})_{O_2} = 500 \,m/s$, we find:
$(V_{rms})_{H_2} = 4 \times 500 \,m/s = 2000 \,m/s$.

(C) The root mean square speed of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{m_1}}$.
The average speed of a gas is given by $v_{avg} = \sqrt{\frac{8RT}{\pi m_2}}$.
Given that $v_1 = 1.5 v_2$,we substitute the expressions:
$\sqrt{\frac{3RT}{m_1}} = 1.5 \times \sqrt{\frac{8RT}{\pi m_2}}$.
Squaring both sides,we get:
$\frac{3RT}{m_1} = (1.5)^2 \times \frac{8RT}{\pi m_2}$.
$\frac{3}{m_1} = 2.25 \times \frac{8}{\pi m_2}$.
Rearranging for the ratio $\frac{m_1}{m_2}$:
$\frac{m_1}{m_2} = \frac{3 \pi}{2.25 \times 8} = \frac{3 \times 3.14159}{18} = \frac{9.42477}{18} \approx 0.52$.

(B) The root mean square (r.m.s.) velocity of a gas is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$, which implies $v_{rms} \propto \sqrt{T}$, where $T$ is the absolute temperature in Kelvin.
Let $v_0$ be the r.m.s. velocity at $T_0 = 0^{\circ} C = 273 \,K$.
Let $v_T$ be the r.m.s. velocity at temperature $T$.
According to the problem, $v_T = 3v_0$.
Using the proportionality $v_{rms} \propto \sqrt{T}$, we have:
$\frac{v_T}{v_0} = \sqrt{\frac{T}{T_0}}$
$3 = \sqrt{\frac{T}{273}}$
Squaring both sides:
$9 = \frac{T}{273}$
$T = 9 \times 273 = 2457 \,K$.
To convert this temperature to Celsius:
$T(^{\circ} C) = 2457 - 273 = 2184^{\circ} C$.

(A) For the given system,the equation of motion is derived from the net force divided by the total mass.
$a = \frac{M_2 g \sin \theta}{M_1 + M_2}$
When the mass of $M_1$ is doubled $(M_1' = 2M_1)$ and the mass of $M_2$ is halved $(M_2' = M_2/2)$,the new acceleration $a'$ is:
$a' = \frac{M_2' g \sin \theta}{M_1' + M_2'} = \frac{(M_2/2) g \sin \theta}{2M_1 + M_2/2}$
Multiplying the numerator and denominator by $2$:
$a' = \frac{M_2 g \sin \theta}{4M_1 + M_2}$
Now,express $a'$ in terms of $a$:
$a' = \left( \frac{M_2 g \sin \theta}{M_1 + M_2} \right) \times \left( \frac{M_1 + M_2}{4M_1 + M_2} \right) = a \left( \frac{M_1 + M_2}{4M_1 + M_2} \right)$

(C) The block has a tendency to slide downwards,so the friction force $f$ acts in the upward direction along the incline.
For the box not to slide,the maximum static friction must balance the component of weight acting downwards along the incline.
$f_{max} \geq mg \sin \theta$
Since $f_{max} = \mu N$,where $N$ is the normal reaction force.
The force $F$ is applied perpendicular to the incline,so the normal reaction $N$ is:
$N = mg \cos \theta + F$
Substituting $N$ into the friction equation:
$\mu(mg \cos \theta + F) \geq mg \sin \theta$
$\mu mg \cos \theta + \mu F \geq mg \sin \theta$
$\mu F \geq mg \sin \theta - \mu mg \cos \theta$
$F \geq \frac{mg(\sin \theta - \mu \cos \theta)}{\mu}$
Given: $m = 2 \ kg$,$g = 10 \ ms^{-2}$,$\theta = 30^{\circ}$,$\mu = 0.2$.
$F \geq \frac{2 \times 10 \times (\sin 30^{\circ} - 0.2 \times \cos 30^{\circ})}{0.2}$
$F \geq \frac{20 \times (0.5 - 0.2 \times 0.866)}{0.2}$
$F \geq \frac{20 \times (0.5 - 0.1732)}{0.2}$
$F \geq \frac{20 \times 0.3268}{0.2}$
$F \geq 100 \times 0.3268 = 32.68 \ N$
Rounding to one decimal place,the minimum value of $F$ is $32.7 \ N$.

(B) Let $\theta$ be the angle that the string makes with the vertical in equilibrium.
At equilibrium,the forces acting on the block are balanced:
$1$. Horizontal force balance: $F = T \sin \theta$ ... $(i)$
$2$. Vertical force balance: $mg = T \cos \theta$ ... (ii)
Dividing equation $(i)$ by equation (ii),we get:
$\frac{F}{mg} = \frac{T \sin \theta}{T \cos \theta} = \tan \theta$
Given $F = 40 \ N$,$m = 8 \ kg$,and $g = 10 \ m \ s^{-2}$:
$\tan \theta = \frac{40}{8 \times 10} = \frac{40}{80} = \frac{1}{2}$
Therefore,$\theta = \tan ^{-1}\left(\frac{1}{2}\right)$.

(B) For an object moving at a constant speed on a frictionless inclined plane, the net force acting on it must be zero.
When the object is pulled up the plane, the applied force $F$ must balance the component of the gravitational force acting down the plane.
The component of the weight $W = mg$ acting down the inclined plane is $mg \sin \theta$.
Given:
Applied force $F = 500 \,N$
Angle of inclination $\theta = 30^{\circ}$
Since the object moves at a constant speed, the forces are in equilibrium:
$F = mg \sin 30^{\circ}$
$500 \,N = mg \cdot \frac{1}{2}$
$mg = 500 \,N \times 2$
$mg = 1000 \,N$
Therefore, the weight of the object is $1000 \,N$.

(A) For the body to be in equilibrium,the net force must be zero,i.e.,$\Sigma F_x = 0$ and $\Sigma F_y = 0$.
Resolving the forces into $x$ and $y$ components:
$F_{1x} = F_1 \cos 30^{\circ} = F_1 \frac{\sqrt{3}}{2}$,$F_{1y} = F_1 \sin 30^{\circ} = F_1 \frac{1}{2}$
$F_{2x} = -F_2 \cos 60^{\circ} = -F_2 \frac{1}{2}$,$F_{2y} = F_2 \sin 60^{\circ} = F_2 \frac{\sqrt{3}}{2}$
$F_{3x} = F_3 \sin 45^{\circ} = F_3 \frac{1}{\sqrt{2}}$,$F_{3y} = -F_3 \cos 45^{\circ} = -F_3 \frac{1}{\sqrt{2}}$
For $\Sigma F_x = 0$:
$F_1 \frac{\sqrt{3}}{2} - F_2 \frac{1}{2} + F_3 \frac{1}{\sqrt{2}} = 0 \Rightarrow \sqrt{3} F_1 - F_2 + \sqrt{2} F_3 = 0$ ... $(i)$
For $\Sigma F_y = 0$:
$F_1 \frac{1}{2} + F_2 \frac{\sqrt{3}}{2} - F_3 \frac{1}{\sqrt{2}} = 0 \Rightarrow F_1 + \sqrt{3} F_2 - \sqrt{2} F_3 = 0$ ... (ii)
Adding $(i)$ and (ii):
$(\sqrt{3}+1) F_1 + (\sqrt{3}-1) F_2 = 0 \Rightarrow F_2 = -F_1 \frac{\sqrt{3}+1}{\sqrt{3}-1} = -F_1 \frac{(\sqrt{3}+1)^2}{3-1} = -F_1 \frac{4+2\sqrt{3}}{2} = -F_1(2+\sqrt{3})$
Substituting $F_2$ in $(i)$:
$\sqrt{3} F_1 - [-F_1(2+\sqrt{3})] + \sqrt{2} F_3 = 0$
$\sqrt{3} F_1 + 2 F_1 + \sqrt{3} F_1 + \sqrt{2} F_3 = 0$
$\sqrt{2} F_3 = -F_1(2+2\sqrt{3}) \Rightarrow F_3 = -F_1 \frac{2(1+\sqrt{3})}{\sqrt{2}} = -\sqrt{2} F_1(1+\sqrt{3}) = -F_1(\sqrt{2}+\sqrt{6})$
Note: The option $C$ is $F_3 = -F_1(\sqrt{2}+\sqrt{6})$,which is equivalent to $F_3 = \frac{-4 F_1}{\sqrt{6}-\sqrt{2}}$.

(B) Initial speed of the body on the horizontal rough surface,$u = 10 \,ms^{-1}$.
Final velocity,$v = 4 \,ms^{-1}$ and time,$t = 2 \,s$.
If $a$ is the retardation due to friction,then by using the equation of motion:
$v = u - at$
$4 = 10 - a \times 2$
$2a = 6 \Rightarrow a = 3 \,ms^{-2}$.
According to Newton's second law,the frictional force is given by $f_k = ma$.
Since $f_k = \mu_k N = \mu_k mg$,we have:
$\mu_k mg = ma$
$\mu_k = \frac{a}{g} = \frac{3}{10} = 0.3$.

(B) The weight of the object is $W = mg = 150 \ N$.
To move the object on a horizontal surface,the applied force $F$ must be at least equal to the limiting friction force $f_l$.
The limiting friction force is given by $f_l = \mu N$,where $\mu$ is the coefficient of friction and $N$ is the normal reaction.
For a horizontal surface,the normal reaction $N$ is equal to the weight of the object,so $N = mg = 150 \ N$.
Given that the force required to move the object is $F = 75 \ N$,we have:
$F = \mu N$
$75 = \mu \times 150$
$\mu = \frac{75}{150} = 0.5$.
Thus,the coefficient of friction is $0.5$.

(A) The block stops on the inclined plane when its initial kinetic energy is used to do work against friction and is partly converted into potential energy.
Given the angle of inclination is $\theta = 45^{\circ}$,the distance covered on the incline $(s)$ is related to the vertical height $(h)$ as:
$\sin 45^{\circ} = \frac{h}{s} \implies s = \frac{h}{\sin 45^{\circ}} = h \sqrt{2}$.
According to the work-energy theorem,the initial kinetic energy $(K.E.)$ is equal to the work done against friction plus the gain in potential energy:
$K.E. = W_{\text{friction}} + \Delta U$
$K.E. = (\mu mg \cos \theta) \cdot s + mgh$
Substituting $s = h \sqrt{2}$ and $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$:
$100 = (0.5 \times 5 \times 10 \times \frac{1}{\sqrt{2}}) \times (h \sqrt{2}) + (5 \times 10 \times h)$
$100 = (0.5 \times 5 \times 10 \times h) + 50h$
$100 = 25h + 50h$
$100 = 75h$
$h = \frac{100}{75} = \frac{4}{3} \text{ m}$.
Now,the distance covered $s$ is:
$s = h \sqrt{2} = \frac{4}{3} \sqrt{2} \text{ m}$.

(B) The graph of friction versus applied force shows that the maximum value of static friction (limiting friction) is greater than the kinetic friction acting during motion.
Since the limiting friction is given by $f_{s,max} = \mu_s N$ and kinetic friction is $f_k = \mu_k N$,where $N$ is the normal force.
From the graph,$f_{s,max} > f_k$.
Therefore,$\mu_s N > \mu_k N$,which implies $\mu_s > \mu_k$.

(D) When the bus accelerates forward with acceleration $a$,a pseudo force $F_p = ma$ acts on the block in the backward direction relative to the bus.
For the block to remain stationary relative to the bus,the static friction force $f_s$ must balance this pseudo force.
The maximum static friction force is given by $f_{s,max} = \mu N = \mu mg$.
For the block to remain stationary,the pseudo force must be less than or equal to the maximum static friction force:
$ma \leq \mu mg$
$a \leq \mu g$
Substituting the given values:
$a \leq 0.2 \times 10 \ m/s^2$
$a \leq 2 \ m/s^2$
Therefore,the maximum acceleration of the bus such that the block remains stationary is $2 \ m/s^2$.

(B) The total mass of the system is $M = 40 \ kg + 60 \ kg = 100 \ kg$.
Applying Newton's second law to the entire system,the acceleration $a$ is given by:
$a = \frac{F}{M} = \frac{1000 \ N}{100 \ kg} = 10 \ m/s^2$.
Now,consider the $40 \ kg$ block. The only horizontal force acting on it is the tension $T$ in the string.
Applying Newton's second law to the $40 \ kg$ block:
$T = m_1 \times a$
$T = 40 \ kg \times 10 \ m/s^2 = 400 \ N$.

(B) Given:
Mass $m = 50 \ g = 50 \times 10^{-3} \ kg = 0.05 \ kg$
Initial velocity $u = 50 \ cm \ s^{-1} = 50 \times 10^{-2} \ m \ s^{-1} = 0.5 \ m \ s^{-1}$
Final velocity $v = 0 \ m \ s^{-1}$
Time $t = 0.5 \ s$
Initial momentum $p_i = m \times u = 0.05 \ kg \times 0.5 \ m \ s^{-1} = 0.025 \ kg \ m \ s^{-1}$
Final momentum $p_f = m \times v = 0.05 \ kg \times 0 \ m \ s^{-1} = 0 \ kg \ m \ s^{-1}$
According to Newton's second law,the force $F$ is the rate of change of momentum:
$F = \frac{\Delta p}{t} = \frac{p_f - p_i}{t}$
$F = \frac{0 - 0.025 \ kg \ m \ s^{-1}}{0.5 \ s}$
$F = -0.05 \ N$
The magnitude of the force applied to stop the ball is $0.05 \ N$.

(A) The component of the force responsible for horizontal motion is $F \cos \theta$.
So,the acceleration of the bar is $a = \frac{F \cos \theta}{m}$.
Since acceleration $a = v \frac{d v}{d x}$,we have:
$v \frac{d v}{d x} = \frac{F \cos \theta}{m}$
Given $\theta = k x$,we have $d \theta = k d x$,or $d x = \frac{d \theta}{k}$.
Substituting these into the equation:
$v d v = \frac{F \cos \theta}{m} \cdot \frac{d \theta}{k}$
Integrating both sides with initial conditions $v = 0$ at $\theta = 0$:
$\int_{0}^{v} v d v = \frac{F}{m k} \int_{0}^{\theta} \cos \theta d \theta$
$\frac{v^2}{2} = \frac{F}{m k} [\sin \theta]_{0}^{\theta}$
$\frac{v^2}{2} = \frac{F \sin \theta}{m k}$
$v^2 = \frac{2 F \sin \theta}{m k}$
$v = \sqrt{\frac{2 F \sin \theta}{m k}}$

(C) An object begins to slide down an inclined plane when the angle of inclination (also called the angle of repose) is equal to the angle of friction. Let $\theta$ be the angle at which a mass begins to slide down an incline.
At the verge of sliding,the force of static friction $f_s$ is equal to the downward component of gravity.
$f_s = mg \sin \theta$
Since $f_s = \mu N$ and the normal force $N = mg \cos \theta$,we have:
$\mu (mg \cos \theta) = mg \sin \theta$
$\mu = \frac{\sin \theta}{\cos \theta} = \tan \theta$
$\theta = \tan^{-1} \mu$
Since the angle of repose $\theta$ depends only on the coefficient of static friction $\mu$ and is independent of the mass $m$ of the body,all three masses will begin to slide at the same instant at the same inclination angle $\theta$.

(D) Consider a wire of negligible mass suspended from a ceiling. $A$ force $F$ is applied at the lower end of the wire.
Since the wire is massless,the net force on any segment of the wire must be zero for it to be in equilibrium.
If we take any cross-section of the wire and consider the lower part of the wire below that section,the only forces acting on this part are the tension $T$ acting upwards at the cross-section and the weight $F$ acting downwards at the end.
Applying Newton's second law,$T - F = ma$. Since the mass $m$ of the wire is zero,$T - F = 0$,which gives $T = F$.
Therefore,the tension at any cross-section of the wire is $F$.

(B) Given: Mass $m = 0.6 \, kg$, Radius $r = 1 \, m$, Frequency $f = \frac{900}{\pi} \, \text{rpm}$.
First, convert the frequency to revolutions per second (Hz):
$f = \frac{900}{\pi} \times \frac{1}{60} = \frac{15}{\pi} \, \text{Hz}$.
The angular velocity $\omega = 2 \pi f = 2 \pi \times \frac{15}{\pi} = 30 \, \text{rad/s}$.
The linear velocity $v = \omega r = 30 \times 1 = 30 \, \text{m/s}$.
Kinetic energy $K = \frac{1}{2} m v^2$.
$K = \frac{1}{2} \times 0.6 \times (30)^2$.
$K = 0.3 \times 900 = 270 \, \text{J}$.

(B) The necessary centripetal force for a car on a curved unbanked road is provided by the static friction force between the tyres and the road.
For safe turning,the centripetal force must be less than or equal to the maximum static friction force:
$\frac{mV^2}{r} \leq \mu mg$
To find the maximum radius $r$ for a given speed $V$,we use the limiting case:
$\frac{mV^2}{r} = \mu mg$
$r = \frac{V^2}{\mu g}$
Given values are $V = 10 \,ms^{-1}$,$\mu = 0.4$,and $g = 10 \,ms^{-2}$.
Substituting these values into the equation:
$r = \frac{10^2}{0.4 \times 10}$
$r = \frac{100}{4}$
$r = 25 \,m$
Thus,the maximum radius of curvature is $25 \,m$.

(D) For springs connected in series,the effective spring constant $K_{\text{eff}}$ is given by:
$\frac{1}{K_{\text{eff}}} = \frac{1}{K_1} + \frac{1}{K_2}$
Given $K_1 = K_2 = 10 \text{ N/m}$,we have:
$\frac{1}{K_{\text{eff}}} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5}$
Thus,$K_{\text{eff}} = 5 \text{ N/m}$.
To convert $K_{\text{eff}}$ to $\text{dyne/cm}$:
$1 \text{ N} = 10^5 \text{ dyne}$ and $1 \text{ m} = 100 \text{ cm}$.
$K_{\text{eff}} = 5 \times \frac{10^5 \text{ dyne}}{100 \text{ cm}} = 5 \times 10^3 \text{ dyne/cm}$.
The work done $W$ to stretch the system by $x = 1 \text{ cm}$ is:
$W = \frac{1}{2} K_{\text{eff}} x^2 = \frac{1}{2} \times (5 \times 10^3) \times (1)^2 = 2500 \text{ erg}$.

(A) Given: Initial position $x_i = 3 \ cm$,final position $x_f = -5 \ cm$,initial velocity $\vec{u} = 12 \hat{i} \ cm \ s^{-1}$,and time $t = 2 \ s$.
Displacement $\vec{s} = x_f - x_i = (-5 - 3) \hat{i} = -8 \hat{i} \ cm$.
Using the equation of motion $\vec{s} = \vec{u}t + \frac{1}{2} \vec{a}t^2$:
$-8 \hat{i} = (12 \hat{i})(2) + \frac{1}{2} \vec{a} (2)^2$
$-8 \hat{i} = 24 \hat{i} + 2 \vec{a}$
$2 \vec{a} = -8 \hat{i} - 24 \hat{i} = -32 \hat{i}$
$\vec{a} = -16 \ cm \ s^{-2} \hat{i}$.

(A) The capacitive reactance $X_C$ of a capacitor is given by the formula: $X_C = \frac{1}{2 \pi f C}$.
Here,$f$ is the frequency of the $AC$ source and $C$ is the capacitance.
From the formula,it is clear that $X_C \propto \frac{1}{f}$.
Given,initial capacitive reactance $X_{C1} = 3 \ k\Omega$ at frequency $f_1 = f$.
When the frequency is doubled,$f_2 = 2f$.
The new capacitive reactance $X_{C2}$ will be: $X_{C2} = \frac{1}{2 \pi (2f) C} = \frac{1}{2} \times \left( \frac{1}{2 \pi f C} \right) = \frac{X_{C1}}{2}$.
Substituting the value: $X_{C2} = \frac{3 \ k\Omega}{2} = 1.5 \ k\Omega$.

(B) Given: Inductance $L = 70 mH = 70 \times 10^{-3} H$,Voltage $V_{rms} = 220 V$,Frequency $f = 50 Hz$.
Inductive reactance is given by $\chi_{L} = \omega L = 2 \pi f L$.
Substituting the values: $\chi_{L} = 2 \times \pi \times 50 \times 70 \times 10^{-3} = 7 \pi \Omega$.
The rms current is $i_{rms} = \frac{V_{rms}}{\chi_{L}} = \frac{220}{7 \pi} \approx \frac{10}{\pi} A$ (Note: Based on standard textbook values,if $L = 70 mH$,the result is $\frac{220}{7 \pi} A$. However,checking the options,if $L = 70 mH$ and $V = 220 V$,the calculation yields $10 A$ only if $\pi \approx 3.14$ and $L$ was adjusted. Re-evaluating: $i_{rms} = \frac{220}{2 \pi \times 50 \times 70 \times 10^{-3}} = \frac{220}{7 \pi} \approx 10 A$ is not exact. Given the options,$10 A$ is the intended answer assuming approximation or specific values).

(D) The impedance $Z$ of an $LR$ circuit is given by the formula: $Z = \sqrt{R^2 + X_L^2}$,where $X_L = \omega L = 2 \pi f L$.
Given: $R = 8 \Omega$,$L = \frac{60}{\pi} \text{ mH} = \frac{60}{\pi} \times 10^{-3} \text{ H}$,and $f = 50 \text{ Hz}$.
First,calculate the inductive reactance $X_L$:
$X_L = 2 \pi \times 50 \times \left( \frac{60}{\pi} \times 10^{-3} \right) = 100 \pi \times \frac{60}{\pi} \times 10^{-3} = 6000 \times 10^{-3} = 6 \Omega$.
Now,calculate the impedance $Z$:
$Z = \sqrt{R^2 + X_L^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \Omega$.

(B) The resonance frequency of a series $L-C-R$ circuit is given by the formula:
$f_0 = \frac{1}{2 \pi \sqrt{L C}} \quad \dots (i)$
When the inductance is reduced to $L^{\prime} = \frac{L}{4}$ and the capacitance is increased to $C^{\prime} = 16 C$,the new resonance frequency $f_0^{\prime}$ is:
$f_0^{\prime} = \frac{1}{2 \pi \sqrt{L^{\prime} C^{\prime}}}$
Substituting the new values:
$f_0^{\prime} = \frac{1}{2 \pi \sqrt{(\frac{L}{4}) \times (16 C)}}$
$f_0^{\prime} = \frac{1}{2 \pi \sqrt{4 L C}}$
$f_0^{\prime} = \frac{1}{2} \times \frac{1}{2 \pi \sqrt{L C}}$
Using equation $(i)$,we get:
$f_0^{\prime} = \frac{f_0}{2}$

(A) Given: Capacitance $C = \frac{50}{\pi} \mu F = \frac{50}{\pi} \times 10^{-6} \, F$.
Source voltage $V_{rms} = 250 \, V$ and frequency $f = 50 \, Hz$.
The capacitive reactance $X_C$ is given by the formula $X_C = \frac{1}{2 \pi f C}$.
Substituting the values: $X_C = \frac{1}{2 \pi \times 50 \times (\frac{50}{\pi} \times 10^{-6})} = \frac{1}{100 \times 50 \times 10^{-6}} = \frac{1}{5000 \times 10^{-6}} = \frac{1}{0.005} = 200 \, \Omega$.
The rms current $I_{rms}$ is given by $I_{rms} = \frac{V_{rms}}{X_C}$.
$I_{rms} = \frac{250}{200} = 1.25 \, A$.

(A) Let us assume the current distribution in the given circuit is as shown.
Now,we apply Kirchhoff's loop rule in loop $1$ and loop $2$ to get the following equations:
In loop $1$,
$-i R - (i + i_1) R - 2 E + E = 0$
$-2 i R - i_1 R = E$
$i_1 R + 2 i R = -E$ ... $(i)$
In loop $2$,
$-3 E + i_1 R + \frac{i_1}{2} R + i_1 R + 2 E + (i + i_1) R = 0$
$\frac{7}{2} i_1 R + i R = E$
$7 i_1 R + 2 i R = 2 E$ ... (ii)
Now,$7 \times$ eq. $(i)$ - eq. (ii) gives:
$7(i_1 R + 2 i R) - (7 i_1 R + 2 i R) = 7(-E) - 2 E$
$7 i_1 R + 14 i R - 7 i_1 R - 2 i R = -9 E$
$12 i R = -9 E$
$i = -\frac{9 E}{12 R} = -\frac{3 E}{4 R}$

(A) Given: $V_{rms} = 100 \text{ V}$, $f = 50 \text{ Hz}$, $C = 100 \mu F = 100 \times 10^{-6} \text{ F}$.
First, calculate the capacitive reactance $X_C$ using the formula $X_C = \frac{1}{2 \pi f C}$.
$X_C = \frac{1}{2 \times \pi \times 50 \times 100 \times 10^{-6}} = \frac{1}{100 \pi \times 10^{-4}} = \frac{1}{0.01 \pi} = \frac{100}{\pi} \Omega$.
The rms value of the current $I_{rms}$ is given by $I_{rms} = \frac{V_{rms}}{X_C}$.
$I_{rms} = \frac{100}{100 / \pi} = \pi \text{ A}$.
Since $\pi \approx 3.14$, the current is $3.14 \text{ A}$.

(D) The inductive reactance is given by $X_L = \omega L = 2 \pi f L$.
Substituting the given values: $X_L = 2 \times \pi \times 350 \times 0.1 = 70 \pi \approx 70 \times 3.14159 = 219.9 \Omega$.
In an $RL$ series circuit,the phase difference $\phi$ between the voltage and the current is given by $\tan \phi = \frac{X_L}{R}$.
Substituting the values: $\tan \phi = \frac{219.9}{110} \approx 1.999 \approx 2$.
Therefore,the phase difference is $\phi = \tan ^{-1}(2)$.

(D) In an $R-L-C$ series circuit,the given values are:
$R = 150 \Omega$
$C = 20 \mu F = 20 \times 10^{-6} F = 2 \times 10^{-5} F$
$L = 500 mH = 0.5 H$
$\omega = 400 rad s^{-1}$
First,calculate the inductive reactance $X_L$:
$X_L = \omega L = 400 \times 0.5 = 200 \Omega$
Next,calculate the capacitive reactance $X_C$:
$X_C = \frac{1}{\omega C} = \frac{1}{400 \times 2 \times 10^{-5}} = \frac{1}{800 \times 10^{-5}} = \frac{10^5}{800} = \frac{1000}{8} = 125 \Omega$
The phase angle $\phi$ is given by the formula:
$\tan \phi = \frac{X_L - X_C}{R}$
Substituting the values:
$\tan \phi = \frac{200 - 125}{150} = \frac{75}{150} = 0.5$
Therefore,the phase angle is $\phi = \tan^{-1}(0.5)$.

(C) Given: Inductance $L = \frac{1}{6 \pi} \text{ H}$,Resistance $R = 15 \text{ } \Omega$,Voltage $V = 100 \text{ V}$,Frequency $f = 60 \text{ Hz}$.
First,calculate the inductive reactance $X_L$:
$X_L = L \omega = L(2 \pi f) = \left(\frac{1}{6 \pi}\right) \times 2 \pi \times 60 = 20 \text{ } \Omega$.
Next,calculate the impedance $Z$ of the $LR$ series circuit:
$Z = \sqrt{R^2 + X_L^2} = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25 \text{ } \Omega$.
The current $I$ in the circuit is:
$I = \frac{V}{Z} = \frac{100}{25} = 4 \text{ A}$.
The phase difference $\phi$ between voltage and current is given by:
$\tan \phi = \frac{X_L}{R} = \frac{20}{15} = \frac{4}{3}$.
Therefore,$\phi = \tan^{-1}\left(\frac{4}{3}\right)$.

(B) The capacitive reactance $X_C$ of a capacitor is given by the formula $X_C = \frac{1}{2 \pi f C}$,where $f$ is the frequency of the $AC$ source and $C$ is the capacitance.
From this relation,we can see that $X_C \propto \frac{1}{f}$.
Let the initial frequency be $f_1$ and the initial reactance be $X_{C1} = 6 \ k\Omega$.
Let the new frequency be $f_2 = 2f_1$.
The new reactance $X_{C2}$ is given by the ratio:
$\frac{X_{C2}}{X_{C1}} = \frac{f_1}{f_2} = \frac{f_1}{2f_1} = \frac{1}{2}$.
Therefore,$X_{C2} = \frac{X_{C1}}{2} = \frac{6 \ k\Omega}{2} = 3 \ k\Omega$.

(B) The average power dissipated in an $LR$ series circuit is given by the formula:
$P_{avg} = I_{rms}^2 R = \frac{V_{rms}^2 R}{Z^2}$
Here,$Z$ is the impedance of the circuit,given by $Z = \sqrt{X_L^2 + R^2}$.
Given values are $X_L = 1 \ \Omega$,$R = 3 \ \Omega$,and $V_{rms} = 10 \ V$.
First,calculate the impedance $Z$:
$Z = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10} \ \Omega$.
Now,calculate the average power $P_{avg}$:
$P_{avg} = \frac{(10)^2 \times 3}{(\sqrt{10})^2} = \frac{100 \times 3}{10} = 30 \ W$.
Thus,the power dissipated in the circuit is $30 \ W$.

(A) The energy of a photon emitted during a transition is given by $E = 13.6 Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ eV}$.
For hydrogen $(Z=1)$,the transition from $n=4$ to $n=2$ is:
$E_H = 13.6 \times 1^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{16} \right) = 13.6 \left( \frac{3}{16} \right) = 2.55 \text{ eV}$.
Given that the frequency (and thus energy) of the $Li^{2+}$ transition $(Z=3)$ is such that $E_H = \frac{3}{7} E_{Li}$,we have $E_{Li} = \frac{7}{3} E_H = \frac{7}{3} \times 2.55 = 5.95 \text{ eV}$.
For $Li^{2+}$,$E_{Li} = 13.6 \times 3^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = 122.4 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = 5.95 \text{ eV}$.
$\left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{5.95}{122.4} \approx 0.0486$.
Testing transitions:
For $n_2=4$ to $n_1=3$: $\frac{1}{9} - \frac{1}{16} = 0.111 - 0.0625 = 0.0486$.
Thus,the transition is $4$ to $3$.

(C) The wavelength $\lambda$ for the hydrogen spectrum is given by the Rydberg formula: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Paschen series,the transition occurs to the $n_1 = 3$ energy level.
The shortest wavelength corresponds to the transition from the highest energy level,$n_2 = \infty$.
Substituting these values into the formula:
$\frac{1}{\lambda_S} = R_H \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) = R_H \left( \frac{1}{9} - 0 \right) = \frac{R_H}{9}$.
Therefore,$\lambda_S = \frac{9}{R_H}$.
Given $R_H = 1.097 \times 10^7 \ m^{-1}$,we have:
$\lambda_S = \frac{9}{1.097 \times 10^7} \ m \approx 8.204 \times 10^{-7} \ m$.
Converting to nanometers $(1 \ nm = 10^{-9} \ m)$:
$\lambda_S = 820.4 \ nm$.

(D) In the Paschen series,an electron transitions from a higher energy state to the $n=3$ state. The energy of a photon in the Paschen series is given by $E = 13.6 \left( \frac{1}{3^2} - \frac{1}{n^2} \right) \text{ eV}$,where $n = 4, 5, 6, \dots$
The minimum energy corresponds to the transition from $n=4$ to $n=3$:
$E_{\min} = 13.6 \left( \frac{1}{9} - \frac{1}{16} \right) = 13.6 \left( \frac{16-9}{144} \right) = 13.6 \left( \frac{7}{144} \right) \approx 0.66 \text{ eV}$.
The maximum energy corresponds to the transition from $n=\infty$ to $n=3$:
$E_{\max} = 13.6 \left( \frac{1}{9} - 0 \right) = \frac{13.6}{9} \approx 1.51 \text{ eV}$.
For a photon to eject a photoelectron,its energy must be greater than or equal to the work function $\phi_0$ of the metal. Thus,$\phi_0 \leq E_{\text{photon}}$. Since the maximum energy of a Paschen photon is $1.51 \text{ eV}$,the work function must satisfy $\phi_0 \leq 1.51 \text{ eV}$.
Among the given options,$1.1 \text{ eV}$ is the only value that satisfies this condition.

(D) According to Bohr's postulate,the orbital angular momentum $L$ of an electron in a hydrogen atom or a hydrogen-like ion is given by the formula:
$L = \frac{n h}{2 \pi}$
where $n$ is the principal quantum number,$h$ is Planck's constant,and $\pi$ is a mathematical constant.
For the ground state,the principal quantum number is $n = 1$ for all hydrogen-like atoms and ions.
Since $h$ and $\pi$ are universal constants,the value of $L$ for $n = 1$ is $L = \frac{h}{2 \pi}$,which is independent of the atomic number $Z$.
Therefore,the orbital angular momentum of the electron is the same for all hydrogen-like atoms and ions in their ground state.

(A) According to Bohr's postulate,the angular momentum $L$ of an electron in the $n^{th}$ orbit is given by $L = \frac{nh}{2\pi}$.
Given that the electron excites from the $n_1 = 2$ orbit to the $n_2 = 4$ orbit.
The change in angular momentum $\Delta L$ is given by:
$\Delta L = L_2 - L_1 = \frac{n_2 h}{2\pi} - \frac{n_1 h}{2\pi} = \frac{h}{2\pi} (n_2 - n_1)$.
Substituting the values:
$\Delta L = \frac{6.64 \times 10^{-34}}{2 \times 3.14} (4 - 2)$.
$\Delta L = \frac{6.64 \times 10^{-34}}{6.28} \times 2$.
$\Delta L = 1.0576 \times 10^{-34} \times 2 \approx 2.11 \times 10^{-34} \ J \cdot s$.

(B) For a hydrogen atom,the energy of the ground state $(n=1)$ is $E_1 = -\frac{13.6}{1^2} = -13.6 \text{ eV}$.
For the second excited state,$n=3$. The energy is $E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 \text{ eV}$.
The energy required to excite the electron from the ground state to the second excited state is $\Delta E = E_3 - E_1 = -1.51 - (-13.6) = 12.09 \text{ eV} \approx 12 \text{ eV}$.
$(b)$ The ionization energy is the energy required to remove the electron from the ground state $(n=1)$ to infinity $(n=\infty)$.
$E_{\infty} = 0 \text{ eV}$.
Energy required = $E_{\infty} - E_1 = 0 - (-13.6) = 13.6 \text{ eV}$.

(C) The total energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} eV$.
For the excited state $n=4$,the total energy is $E_4 = -\frac{13.6}{4^2} = -\frac{13.6}{16} = -0.85 eV$.
The potential energy $(PE)$ of an electron in a Bohr orbit is related to the total energy $(E)$ by the relation $PE = 2E$.
Therefore,$PE = 2 \times (-0.85 eV) = -1.7 eV$.

(A) The radius of the $n^{\text{th}}$ Bohr orbit is given by the formula $r_n = a_0 n^2$,where $a_0$ is the Bohr radius and $n$ is the principal quantum number.
Therefore,the ratio of the radii of the $3^{\text{rd}}$ and $6^{\text{th}}$ orbits is given by $\frac{r_3}{r_6} = \frac{n_3^2}{n_6^2}$.
Substituting the values $n_3 = 3$ and $n_6 = 6$,we get $\frac{r_3}{r_6} = \frac{3^2}{6^2} = \frac{9}{36}$.
Simplifying this fraction,we obtain $\frac{r_3}{r_6} = \frac{1}{4} = 0.25$.

(C) The energy of the $n^{th}$ level in a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$.
The energy difference between consecutive levels is $\Delta E = E_{n+1} - E_n = 13.6 \left[ \frac{1}{n^2} - \frac{1}{(n+1)^2} \right]$.
Simplifying this expression,we get $\Delta E = 13.6 \left[ \frac{(n+1)^2 - n^2}{n^2(n+1)^2} \right] = 13.6 \left[ \frac{2n+1}{n^2(n+1)^2} \right]$.
For large values of $n$,$\Delta E \approx 13.6 \left[ \frac{2n}{n^4} \right] = \frac{27.2}{n^3}$.
Since $\Delta E \propto \frac{1}{n^3}$,as the quantum number $n$ increases,the energy difference $\Delta E$ decreases.

(B) In a hydrogen atom,the potential energy $(PE)$ of an electron in the $n$th orbit is given by $PE = -\frac{kZe^2}{r}$.
For higher values of $n$,the radius $r$ increases because $r \propto n^2$.
As $r$ increases,the magnitude of the negative potential energy decreases,meaning the value of the potential energy increases (becomes less negative).
Thus,statement $(A)$ is correct.
In Thomson's model,an atom is described as a spherical cloud of positive charge with electrons embedded within it,similar to a plum pudding. Thus,statement $(B)$ is correct.
According to Bohr's model,an electron in an atom is located at a definite distance from the nucleus and revolves with a definite velocity. However,Heisenberg's uncertainty principle states that it is impossible to determine the exact position and momentum of a moving electron simultaneously.
Therefore,Bohr's model contradicts the uncertainty principle,making statement $(C)$ incorrect.
Hence,option $(b)$ is the correct answer.

(B) Given,$C_1 = C_2 = 8 \mu F = 8 \times 10^{-6} \ F$ and $V = 10 \ V$.
Since $C_1$ and $C_2$ are connected in parallel,the initial equivalent capacitance is $C_{eq} = C_1 + C_2 = 16 \mu F = 1.6 \times 10^{-5} \ F$.
The initial total charge is $q_i = C_{eq} \cdot V = 1.6 \times 10^{-5} \times 10 = 1.6 \times 10^{-4} \ C = 160 \mu C$.
We know that capacitance $C = \frac{\varepsilon_0 A}{d}$,so $C \propto \frac{1}{d}$.
If the plate separation $d$ is reduced to $40 \%$ of its initial value,the new separation is $d' = 0.4d$. Thus,the new capacitance $C'$ becomes $C' = \frac{C}{0.4} = \frac{C}{2/5} = 2.5C = \frac{5}{2}C$.
Wait,the problem states the separation is reduced $TO$ $40 \%$,meaning $d' = 0.4d$. So $C' = \frac{C}{0.4} = 2.5 \times 8 \mu F = 20 \mu F$.
New total capacitance $C_{eq}' = C' + C_2 = 20 \mu F + 8 \mu F = 28 \mu F$.
New charge $q_f = C_{eq}' \cdot V = 28 \mu F \times 10 \ V = 280 \mu C$.
Increase in charge $\Delta q = q_f - q_i = 280 \mu C - 160 \mu C = 120 \mu C$.

(D) The initial capacitance of the capacitor is $C_0 = \frac{\varepsilon_0 A}{d}$.
When two dielectric slabs of thickness $d/2$ are placed between the plates,the arrangement acts as two capacitors in series,each with plate separation $d/2$.
The capacitance of the first slab is $C_1 = \frac{K_1 \varepsilon_0 A}{d/2} = \frac{2 K_1 \varepsilon_0 A}{d} = 2 K_1 C_0$.
The capacitance of the second slab is $C_2 = \frac{K_2 \varepsilon_0 A}{d/2} = \frac{2 K_2 \varepsilon_0 A}{d} = 2 K_2 C_0$.
Since they are in series,the equivalent capacitance $C_{eq}$ is given by:
$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{2 K_1 C_0} + \frac{1}{2 K_2 C_0} = \frac{1}{2 C_0} \left( \frac{1}{K_1} + \frac{1}{K_2} \right) = \frac{1}{2 C_0} \left( \frac{K_1 + K_2}{K_1 K_2} \right)$.
Therefore,$C_{eq} = 2 C_0 \left( \frac{K_1 K_2}{K_1 + K_2} \right)$.

(C) Given, $C_1 = C_2 = C_3 = C_4 = C_5 = 4 \mu F$.
Looking at the circuit, we can identify that the capacitors $C_1, C_2, C_3, C_4, C_5$ form a bridge network.
Let the nodes be $X$ (input), $Y$ (output), and intermediate nodes $A, B, C$.
By analyzing the potential distribution, we can see that the circuit is equivalent to a parallel combination of two branches.
Specifically, the branch containing $C_1$ and $C_2$ in series is in parallel with the branch containing $C_4$ and $C_5$ in series, and so on.
However, a simpler way to view this is to recognize that the circuit simplifies to two parallel branches, each consisting of two capacitors in series.
$C_{eq} = (C_1 \text{ in series with } C_2) + (C_4 \text{ in series with } C_5)$.
Since all $C = 4 \mu F$, the series combination of two $4 \mu F$ capacitors is $\frac{4 \times 4}{4 + 4} = 2 \mu F$.
Thus, $C_{eq} = 2 \mu F + 2 \mu F = 4 \mu F$.

(A) The capacitance of a parallel plate capacitor is given by $C = \frac{K A \varepsilon_0}{d}$.
Initially,for air,$K_1 = 1$ and $d_1 = d$,so $C_1 = \frac{A \varepsilon_0}{d} = 12 \mu F$.
When the distance is doubled,$d_2 = 2d$,and the dielectric constant is $K_2 = 4$.
The new capacitance $C_2$ is given by $C_2 = \frac{K_2 A \varepsilon_0}{d_2}$.
Substituting the values,we get $C_2 = \frac{4 A \varepsilon_0}{2d} = 2 \left( \frac{A \varepsilon_0}{d} \right) = 2 \times C_1$.
Therefore,$C_2 = 2 \times 12 \mu F = 24 \mu F$.

(A) Given:
$C_1 = 10 \mu F = 10^{-5} \text{ F}$
$V_1 = 9 \text{ V}$
Initial charge on capacitor $C_1$:
$q_1 = C_1 V_1 = 10^{-5} \times 9 = 9 \times 10^{-5} \text{ C}$
When the uncharged capacitor $C_2 = 20 \mu F = 2 \times 10^{-5} \text{ F}$ is connected in parallel to the charged capacitor $C_1$,charge flows from $C_1$ to $C_2$ until both capacitors reach a common potential $V$.
The common potential $V$ is given by:
$V = \frac{\text{Total Charge}}{\text{Total Capacitance}} = \frac{q_1}{C_1 + C_2}$
$V = \frac{9 \times 10^{-5}}{10^{-5} + 2 \times 10^{-5}} = \frac{9 \times 10^{-5}}{3 \times 10^{-5}} = 3 \text{ V}$
The charge on capacitor $C_2$ at equilibrium is:
$q_2 = C_2 V = (2 \times 10^{-5} \text{ F}) \times (3 \text{ V}) = 6 \times 10^{-5} \text{ C}$

(A) Initial stored energy $U_i = \frac{1}{2} C_1 V_1^2$
$U_i = \frac{1}{2} \times 500 \times 10^{-12} \times (100)^2 = 2.5 \ \mu J$
When the capacitors are connected,the charge is redistributed until the potential becomes common.
Common potential $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} = \frac{500 \times 10^{-12} \times 100 + 0}{500 \times 10^{-12} + 500 \times 10^{-12}} = 50 \ V$
Final stored energy $U_f = \frac{1}{2} (C_1 + C_2) V^2$
$U_f = \frac{1}{2} \times (1000 \times 10^{-12}) \times (50)^2 = 1.25 \ \mu J$
Energy lost $\Delta U = U_i - U_f = 2.5 \ \mu J - 1.25 \ \mu J = 1.25 \ \mu J$

(A) Initially,when $C_1$ is connected to a $9 \ V$ battery,the charge on it is $q_0 = C_1 V = 1 \ \mu F \times 9 \ V = 9 \ \mu C$.
When $C_1$ is connected to $C_2$ and $C_3$ in parallel,the total charge $q_0$ is redistributed among the three capacitors such that they all have the same potential difference $V'$.
By the law of conservation of charge,$q_1 + q_2 + q_3 = q_0 = 9 \ \mu C$.
Since they are in parallel,$V' = \frac{q_1}{C_1} = \frac{q_2}{C_2} = \frac{q_3}{C_3}$.
Substituting the values,$\frac{q_1}{1} = \frac{q_2}{2} = \frac{q_3}{3} = V'$.
Thus,$q_1 = V'$,$q_2 = 2V'$,and $q_3 = 3V'$.
Substituting these into the conservation equation: $V' + 2V' + 3V' = 9 \ \mu C$.
$6V' = 9 \ \mu C \Rightarrow V' = 1.5 \ V$.
The charge on $C_3$ is $q_3 = C_3 V' = 3 \ \mu F \times 1.5 \ V = 4.5 \ \mu C = 4.5 \times 10^{-6} \ C$.

(B) The given circuit is a Wheatstone bridge configuration. Let the capacitors be $C_1 = 1 \mu F$,$C_2 = 3 \mu F$,$C_3 = 2 \mu F$,$C_4 = 6 \mu F$,and the central capacitor $C_5 = 5 \mu F$.
Check the ratio of capacitors in the arms: $\frac{C_1}{C_3} = \frac{1}{2}$ and $\frac{C_2}{C_4} = \frac{3}{6} = \frac{1}{2}$.
Since $\frac{C_1}{C_3} = \frac{C_2}{C_4}$,the Wheatstone bridge is balanced.
In a balanced Wheatstone bridge,no charge flows through the central capacitor $C_5$,so it can be removed from the circuit.
Now,the circuit consists of two parallel branches: one with $C_1$ and $C_2$ in series,and the other with $C_3$ and $C_4$ in series.
Equivalent capacitance of the upper branch: $C_{up} = \frac{C_1 C_2}{C_1 + C_2} = \frac{1 \times 3}{1 + 3} = \frac{3}{4} \mu F$.
Equivalent capacitance of the lower branch: $C_{low} = \frac{C_3 C_4}{C_3 + C_4} = \frac{2 \times 6}{2 + 6} = \frac{12}{8} = \frac{3}{2} \mu F$.
Since these two branches are in parallel,the total equivalent capacitance $C_{AB} = C_{up} + C_{low} = \frac{3}{4} + \frac{3}{2} = \frac{3 + 6}{4} = \frac{9}{4} \mu F$.

(B) The magnitude of the charge transferred between the plates is given by $Q = N \cdot e$.
Given $N = 10^{12}$ electrons and $e = 1.6 \times 10^{-19} \,C$.
So,$Q = 10^{12} \times 1.6 \times 10^{-19} = 1.6 \times 10^{-7} \,C$.
The potential difference developed is $V = 10 \,V$.
The capacitance $C$ is defined as $C = \frac{Q}{V}$.
Substituting the values,$C = \frac{1.6 \times 10^{-7}}{10} = 1.6 \times 10^{-8} \,F$.

(D) In a series combination of $n$ identical capacitors each of capacitance $C$,the equivalent capacitance $C_{eq}$ is given by:
$C_{eq} = \frac{C}{n}$
Given $n = 25$ and $C = 2 \mu F = 2 \times 10^{-6} F$.
$C_{eq} = \frac{2 \times 10^{-6}}{25} F = 0.08 \times 10^{-6} F = 8 \times 10^{-8} F$.
The potential difference applied is $V = 100 \ V$.
The total charge $Q$ stored on the series combination is the same as the charge on each capacitor,given by:
$Q = C_{eq} \times V$
$Q = (8 \times 10^{-8} F) \times (100 \ V)$
$Q = 8 \times 10^{-6} C$.

(B) coaxial cable,also known as $TV$ aerial cable or coax,is primarily used to carry video and data signals from an antenna to a device such as a satellite dish or television.
This is due to the well-insulated conductor wire,which prevents frequency interference.
Coaxial cables can transmit a high range of frequencies due to low power losses.
The maximum number of $TV$ signals that can be transmitted simultaneously along a single coaxial cable is $125$.

(A) The $UHF$ (Ultra High Frequency) range typically spans from $300 \ MHz$ to $3000 \ MHz$.
Due to their high frequency,these waves cannot be reflected by the ionosphere (sky waves) and are heavily attenuated by the ground (ground waves).
Therefore,they propagate primarily through line-of-sight communication,which is achieved by means of space waves.

(D) Microwaves have a short wavelength and travel in straight lines. They do not bend significantly around corners of obstacles in their path. Due to this property,they are highly effective for detection and tracking,which is why they are extensively used in radar systems for remote sensing and navigation.

(A) In satellite communication,the uplink frequency is the frequency at which the ground station transmits signals to the satellite.
This frequency is kept high (in the microwave range) to ensure that the signals can penetrate the ionosphere without being reflected back to Earth.
The standard frequency band used for uplink satellite communication is $5.9-6.4 \text{ GHz}$.

(A) The modulation index $m$ is defined as the ratio of the peak voltage of the modulating signal $(E_m)$ to the peak voltage of the carrier signal $(E_c)$: $m = \frac{E_m}{E_c}$.
When the modulating signal is increased by $1 \%$,the new peak voltage is $E_m' = E_m(1 + 0.01) = 1.01 E_m$.
When the carrier signal is increased by $3 \%$,the new peak voltage is $E_c' = E_c(1 + 0.03) = 1.03 E_c$.
The new modulation index $m'$ is given by $m' = \frac{E_m'}{E_c'} = \frac{1.01 E_m}{1.03 E_c} \approx 0.9806 m$.
The percentage change in the modulation index is $\frac{m' - m}{m} \times 100 = (0.9806 - 1) \times 100 \approx -1.94 \% \approx -2 \%$.
Thus,the modulation index changes by approximately $-2 \%$.

(D) The modulation index $\mu$ is defined as the ratio of the peak voltage of the message signal $(A_m)$ to the peak voltage of the carrier signal $(A_c)$.
Given:
$A_m = 12 \text{ V}$
$A_c = 20 \text{ V}$
Formula:
$\mu = \frac{A_m}{A_c}$
Calculation:
$\mu = \frac{12}{20} = \frac{3}{5} = 0.6$
Therefore,the modulation index is $0.6$.

(B) Maximum amplitude of $AM$ wave, $A_{\max} = 20 \,V$.
Minimum amplitude of $AM$ wave, $A_{\min} = 4 \,V$.
The modulation index $\mu$ is given by the formula:
$\mu = \frac{A_{\max} - A_{\min}}{A_{\max} + A_{\min}}$
Substituting the values:
$\mu = \frac{20 - 4}{20 + 4} = \frac{16}{24} = \frac{2}{3} \approx 0.67$.

(B) The minimum size of a transmitting antenna is given by $l = \frac{\lambda}{4}$.
Given frequency $f = 1500 \text{ MHz} = 1500 \times 10^6 \text{ Hz}$.
The speed of light $c = 3 \times 10^8 \text{ m/s}$.
The wavelength $\lambda = \frac{c}{f} = \frac{3 \times 10^8}{1500 \times 10^6} = \frac{3}{15} = 0.2 \text{ m}$.
Therefore,the minimum length $l = \frac{0.2}{4} = 0.05 \text{ m}$.
Converting to centimeters,$l = 0.05 \times 100 = 5 \text{ cm}$.

(B) The modulation index $m$ is defined as the ratio of the message signal amplitude $A_m$ to the carrier signal amplitude $A_c$.
$m = \frac{A_m}{A_c} = \frac{5 \,V}{25 \,V} = \frac{1}{5} = 0.2$.
In amplitude modulation, the amplitude of each sideband (upper and lower) is given by the formula $\frac{m A_c}{2}$.
Substituting the values, we get:
Amplitude of side band $= \frac{0.2 \times 25 \,V}{2} = \frac{5 \,V}{2} = 2.5 \,V$.

(B) The modulation index $\mu$ is defined as the ratio of the amplitude of the modulating signal $(A_M)$ to the amplitude of the carrier wave $(A_C)$.
$\mu = \frac{A_M}{A_C}$
Given,$\mu = 90 \% = 0.9$ and $A_C = 60 \,V$.
Substituting the values into the formula:
$0.9 = \frac{A_M}{60}$
$A_M = 0.9 \times 60$
$A_M = 54 \,V$
Thus,the peak voltage of the modulating signal is $54 \,V$.

(B) Sky wave propagation,also known as skip propagation,involves the reflection of radio waves off the ionosphere to reach locations beyond the horizon.
This mode of communication is typically effective for frequencies in the range of $3 MHz$ to $30 MHz$,which is known as the High Frequency $(HF)$ band.
Among the given options,$10 MHz$ falls within this range,making it the most suitable frequency for sky wave communication.

(D) Let the electromotive force $(EMF)$ of the cell be $E$ and its internal resistance be $r$.
According to Ohm's law,the current $I$ in a circuit with external resistance $R$ is given by $I = \frac{E}{R + r}$.
For the first case: $1 = \frac{E}{2.5 + r} \implies E = 2.5 + r$ (Equation $1$).
For the second case: $0.5 = \frac{E}{10 + r} \implies E = 0.5(10 + r) = 5 + 0.5r$ (Equation $2$).
Equating the two expressions for $E$:
$2.5 + r = 5 + 0.5r$
$r - 0.5r = 5 - 2.5$
$0.5r = 2.5$
$r = \frac{2.5}{0.5} = 5 \ \Omega$.
Therefore,the internal resistance of the cell is $5 \ \Omega$.

(B) Given:
Electromotive force (emf) $\varepsilon = 12 \, V$
Total charge capacity $q = 80 \, A \cdot h$
Power output $P = 120 \, W$
The total energy $E$ stored in the battery is given by $E = q \cdot \varepsilon$.
Substituting the values: $E = 80 \, A \cdot h \times 12 \, V = 960 \, W \cdot h$.
Since power $P$ is the rate of energy delivery, $P = E / \Delta t$, where $\Delta t$ is the time.
Therefore, $\Delta t = E / P = 960 \, W \cdot h / 120 \, W = 8 \, h$.
Thus, the battery can deliver energy at the rate of $120 \, W$ for $8 \, h$.

(C) The terminal voltage $V$ of a battery is given by the relation $V = E - Ir$,where $E$ is the emf,$I$ is the current,and $r$ is the internal resistance.
Given:
External resistance $R = 8 \ \Omega$
Internal resistance $r = 0.2 \ \Omega$
Terminal voltage $V = 10 \ V$
The current $I$ flowing through the circuit is given by Ohm's law applied to the external resistor:
$I = \frac{V}{R} = \frac{10 \ V}{8 \ \Omega} = 1.25 \ A$
Now,substituting the values into the terminal voltage equation:
$V = E - Ir$
$10 = E - (1.25 \ A \times 0.2 \ \Omega)$
$10 = E - 0.25 \ V$
$E = 10 + 0.25 = 10.25 \ V$
Therefore,the emf of the battery is $10.25 \ V$.

(A) The current density $J$ is given as $J(r) = \beta(r + r_0)^2$.
Consider a small elemental area $dA$ at a radial distance $r$ with thickness $dr$ and angular width $d\theta$. The area element is $dA = r dr d\theta$.
The current $di$ through this element is $di = J(r) dA = \beta(r + r_0)^2 r dr d\theta$.
The shaded region consists of two sectors,each with an angular width of $\pi/6$. The total angular width is $\Delta \theta = \pi/6 + \pi/6 = \pi/3$.
The total current $I$ through the shaded region is obtained by integrating $di$ over $r$ from $0$ to $R$ and $\theta$ over the total angular width $\Delta \theta$:
$I = \int_0^{\pi/3} d\theta \int_0^R \beta(r^2 + 2rr_0 + r_0^2) r dr$
$I = \frac{\pi}{3} \beta \int_0^R (r^3 + 2r_0r^2 + r_0^2r) dr$
$I = \frac{\pi \beta}{3} \left[ \frac{r^4}{4} + \frac{2r_0r^3}{3} + \frac{r_0^2r^2}{2} \right]_0^R$
$I = \frac{\pi \beta}{3} \left[ \frac{R^4}{4} + \frac{2r_0R^3}{3} + \frac{r_0^2R^2}{2} \right]$
$I = \pi \beta \left[ \frac{R^4}{12} + \frac{2r_0R^3}{9} + \frac{r_0^2R^2}{6} \right]$

(D) The drift velocity $v_d$ is given by the ratio of length $L$ to the time $t$ taken to drift across the wire:
$v_d = \frac{L}{t} = \frac{2 \ m}{40 \times 10^3 \ s} = 0.5 \times 10^{-4} \ m/s = 5 \times 10^{-5} \ m/s$.
The current $I$ in a conductor is related to the number density $n$ of free electrons by the formula:
$I = n e A v_d$
where $e = 1.6 \times 10^{-19} \ C$ is the charge of an electron and $A$ is the cross-sectional area.
Given $A = 4 \ mm^2 = 4 \times 10^{-6} \ m^2$ and $I = 1.6 \ A$,we rearrange for $n$:
$n = \frac{I}{e A v_d}$
$n = \frac{1.6}{(1.6 \times 10^{-19}) \times (4 \times 10^{-6}) \times (5 \times 10^{-5})}$
$n = \frac{1.6}{1.6 \times 10^{-19} \times 20 \times 10^{-11}}$
$n = \frac{1}{20 \times 10^{-30}} = \frac{1}{2} \times 10^{29} = 5 \times 10^{28} \ m^{-3}$.

(C) Given,resistance of galvanometer $R_g = 100 \Omega$.
Full scale deflection current $I_g = 50 \mu A = 5 \times 10^{-5} \text{ A}$.
Required range of ammeter $I = 10 \text{ mA} = 10^{-2} \text{ A}$.
To convert a galvanometer into an ammeter,a shunt resistance $R_s$ is connected in parallel.
The formula for shunt resistance is $R_s = \frac{I_g}{I - I_g} \times R_g$.
Substituting the values:
$R_s = \frac{5 \times 10^{-5}}{10^{-2} - 5 \times 10^{-5}} \times 100$
$R_s = \frac{5 \times 10^{-5}}{100 \times 10^{-4} - 0.5 \times 10^{-4}} \times 100$
$R_s = \frac{5 \times 10^{-5}}{995 \times 10^{-5}} \times 100$
$R_s = \frac{500}{995} \approx 0.5025 \Omega$.
Thus,the resistance to be added is approximately $0.5 \Omega$.

(A) In a potentiometer experiment,the emf $E$ of a cell is directly proportional to its balancing length $l$,given by the relation $E \propto l$.
Therefore,for two cells $E_1$ and $E_2$ with balancing lengths $l_1$ and $l_2$,we have the ratio: $\frac{E_1}{E_2} = \frac{l_1}{l_2}$.
Rearranging for $E_2$,we get: $E_2 = E_1 \cdot \frac{l_2}{l_1}$.
Given values are $E_1 = 2.4 \ V$,$l_1 = 360 \ cm$,and $l_2 = 420 \ cm$.
Substituting these values into the formula: $E_2 = 2.4 \times \frac{420}{360}$.
Simplifying the fraction: $E_2 = 2.4 \times \frac{7}{6} = 0.4 \times 7 = 2.8 \ V$.
Thus,the emf of cell $B$ is $2.8 \ V$.