MHT CET 2023 Physics Question Paper with Answer and Solution

(B) According to Charles's Law,the volume $(V)$ of a fixed mass of an ideal gas is directly proportional to its absolute temperature $(T)$ when the pressure is held constant.
Mathematically,this is expressed as $V \propto T$ or $V = kT$,where $k$ is a constant.
This relationship represents a straight line passing through the origin on a $V$ versus $T$ graph.
Among the given options,graph $Q$ shows a straight line passing through the origin,which correctly represents the variation of volume with temperature at constant pressure.

(D) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$.
For oxygen gas $(O_2)$: $P_1 V_1 = \frac{m_1}{M_1} RT_1$ ... $(i)$
For hydrogen gas $(H_2)$: $P_2 V_2 = \frac{m_2}{M_2} RT_2$ ... (ii)
Given that $P_1 = P_2$,$V_1 = V_2$,and $m_1 = m_2$,dividing equation $(i)$ by (ii) gives:
$\frac{P_1 V_1}{P_2 V_2} = \frac{m_1 / M_1}{m_2 / M_2} \cdot \frac{T_1}{T_2}$
$1 = \frac{M_2}{M_1} \cdot \frac{T_1}{T_2}$
$\frac{T_1}{T_2} = \frac{M_1}{M_2}$
Substituting the molar masses $M_1 (O_2) = 32 \text{ g/mol}$ and $M_2 (H_2) = 2 \text{ g/mol}$:
$\frac{T_1}{T_2} = \frac{32}{2} = \frac{16}{1}$
Therefore,the ratio of their absolute temperatures is $16: 1$.

(A) According to the kinetic theory of gases,the absolute temperature $T$ of an ideal gas is directly proportional to the average translational kinetic energy of its molecules.
Mathematically,the average kinetic energy per molecule is given by $K_{avg} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant.
Therefore,temperature is a measure of the average kinetic energy of the gas molecules.

(A) The pressure $P$ exerted by an ideal gas on the walls of a container is given by the kinetic theory of gases as $P = \frac{1}{3} \rho v_{rms}^{2}$.
From the ideal gas equation,$PV = nRT$,we have $P = \frac{nRT}{V}$.
Since $n$,$R$,and $V$ are constant for a closed container,the pressure $P$ is directly proportional to the temperature $T$,i.e.,$P \propto T$.
The force $F$ exerted on the walls is given by $F = P \times A$,where $A$ is the surface area of the wall.
Since $A$ is constant for a closed container,$F \propto P$.
Therefore,$F \propto T$,which implies $F \propto T^{1}$.
Comparing this with $T^{x}$,we get $x = 1$.

(C) For an adiabatic compression,the relation is $PV^{\gamma} = \text{constant}$.
Given: $V_{\text{new}} = \frac{1}{4} V$ and $\gamma = \frac{3}{2}$.
Using the adiabatic process equation:
$P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$
$P \cdot V^{\gamma} = P_{\text{new}} \cdot \left(\frac{V}{4}\right)^{\gamma}$
$\frac{P_{\text{new}}}{P} = \left(\frac{V}{V/4}\right)^{\gamma} = (4)^{\gamma}$
$\frac{P_{\text{new}}}{P} = (4)^{3/2} = (2^2)^{3/2} = 2^3 = 8$
Therefore,$P_{\text{new}} = 8P$.

(D) The pressure $P$ and volume $V$ of an ideal gas are related to the number of molecules $N$ and the average kinetic energy per molecule $\langle K \rangle$ by the relation:
$PV = \frac{2}{3} N \langle K \rangle$
Given:
$V = 500 \text{ c.c.} = 500 \times 10^{-6} \text{ m}^3 = 5 \times 10^{-4} \text{ m}^3$
$P = 2 \times 10^5 \text{ N/m}^2$
$\langle K \rangle = 6 \times 10^{-21} \text{ J}$
Rearranging the formula to solve for $N$:
$N = \frac{3PV}{2\langle K \rangle}$
Substituting the values:
$N = \frac{3 \times (2 \times 10^5) \times (5 \times 10^{-4})}{2 \times (6 \times 10^{-21})}$
$N = \frac{3 \times 10^2}{12 \times 10^{-21}} = \frac{300}{12} \times 10^{21} = 25 \times 10^{21} = 2.5 \times 10^{22}$
Thus,the number of gas molecules is $2.5 \times 10^{22}$.

(B) The pressure $P$ of an ideal gas is given by the kinetic theory formula:
$P = \frac{1}{3} \frac{N m}{V} v_{rms}^2$
where $N$ is the number of molecules,$m$ is the mass of each molecule,$V$ is the volume,and $v_{rms}$ is the root-mean-square speed.
From this relation,we can see that $P \propto m \cdot v_{rms}^2$.
Let the initial mass be $m_1 = m$ and initial speed be $v_1 = v$. The initial pressure is $P_1 = P$.
Given the new conditions:
$m_2 = \frac{m}{2}$
$v_2 = 2v$
Now,calculating the new pressure $P_2$:
$\frac{P_2}{P_1} = \frac{m_2}{m_1} \times \left( \frac{v_2}{v_1} \right)^2$
$\frac{P_2}{P} = \left( \frac{m/2}{m} \right) \times \left( \frac{2v}{v} \right)^2$
$\frac{P_2}{P} = \left( \frac{1}{2} \right) \times (2)^2$
$\frac{P_2}{P} = \frac{1}{2} \times 4 = 2$
Therefore,$P_2 = 2P$.

(D) According to the ideal gas equation,$PV = Nk_B T$,where $N$ is the number of molecules and $k_B$ is the Boltzmann constant.
For jar $A$: $PV = N_A k_B T$ --- $(i)$
For jar $B$: $(2P) \times (V/4) = N_B k_B (2T)$
Simplifying the left side: $(1/2) PV = N_B k_B (2T)$
$PV = 4 N_B k_B T$ --- $(ii)$
Equating $(i)$ and $(ii)$: $N_A k_B T = 4 N_B k_B T$
$N_A = 4 N_B$
Therefore,the ratio $N_A / N_B = 4/1$ or $4: 1$.

(D) Conservation of volume: $\frac{4}{3} \pi R^3 = n \times \frac{4}{3} \pi r^3 \implies R^3 = n r^3$.
Energy released due to decrease in surface area: $\Delta U = T \times (n \times 4 \pi r^2 - 4 \pi R^2)$.
Since $n = \frac{R^3}{r^3}$, we have $\Delta U = 4 \pi T \left( \frac{R^3}{r} - R^2 \right) = 4 \pi T R^3 \left( \frac{1}{r} - \frac{1}{R} \right)$.
This energy is converted into kinetic energy $(K.E.)$ of the big drop: $K.E. = \frac{1}{2} M v^2$, where $M = \rho \times \frac{4}{3} \pi R^3$.
Equating energy: $\frac{1}{2} (\rho \times \frac{4}{3} \pi R^3) v^2 = 4 \pi T R^3 \left( \frac{1}{r} - \frac{1}{R} \right)$.
Simplifying: $\frac{2}{3} \rho \pi R^3 v^2 = 4 \pi T R^3 \left( \frac{1}{r} - \frac{1}{R} \right)$.
$v^2 = \frac{4 \pi T R^3 \times 3}{2 \pi \rho R^3} \left( \frac{1}{r} - \frac{1}{R} \right) = \frac{6 T}{\rho} \left( \frac{1}{r} - \frac{1}{R} \right)$.
Therefore, $v = \sqrt{\frac{6 T}{\rho} \left( \frac{1}{r} - \frac{1}{R} \right)}$.

(A) Step $1$: Isothermal expansion from $V$ to $2V$.
For an isothermal process,$P_1 V_1 = P_2 V_2$.
Given $P_1 = P$,$V_1 = V$,and $V_2 = 2V$.
$P(V) = P_2(2V) \implies P_2 = P/2$.
Step $2$: Adiabatic expansion from $2V$ to $16V$.
For an adiabatic process,$P_2 V_2^\gamma = P_3 V_3^\gamma$.
Given $P_2 = P/2$,$V_2 = 2V$,$V_3 = 16V$,and $\gamma = 5/3$.
$P_3 = P_2 (V_2 / V_3)^\gamma = (P/2) (2V / 16V)^{5/3}$.
$P_3 = (P/2) (1/8)^{5/3}$.
Since $8 = 2^3$,$(1/8)^{5/3} = (1/2^3)^{5/3} = (1/2)^5 = 1/32$.
$P_3 = (P/2) \times (1/32) = P/64$.

(A) From the ideal gas equation,$PV = nRT$,where $n = m/M$ (mass/molar mass).
Substituting $n$,we get $PV = (m/M)RT$.
Since density $\rho = m/V$,we can write $P = (\rho/M)RT$,which implies $\rho = (PM)/(RT)$.
Therefore,$\rho \propto P/T$.
Given: Initial state is $(\rho_0, P_0, T_0)$ and final state is $(\rho', 3P_0, 2T_0)$.
Using the proportionality $\rho' / \rho_0 = (P'/P_0) \times (T_0/T')$,we get:
$\rho' = \rho_0 \times (3P_0 / P_0) \times (T_0 / 2T_0)$.
$\rho' = \rho_0 \times 3 \times (1/2) = \frac{3}{2} \rho_0$.

(B) The pressure of an ideal gas is given by the kinetic theory formula: $P = \frac{1}{3} \frac{N m}{V} v_{rms}^2$.
From this expression,we can see that the pressure $P$ is directly proportional to the mass of the molecules $m$ and the square of their root-mean-square velocity $v_{rms}^2$,assuming the number of molecules $N$ and volume $V$ remain constant: $P \propto m v_{rms}^2$.
Let the initial state be $P_1 = P_0$,mass $m_1 = m$,and velocity $v_1 = v$.
In the final state,the mass is halved: $m_2 = \frac{m}{2}$,and the velocity is doubled: $v_2 = 2v$.
The ratio of the final pressure $P_2$ to the initial pressure $P_1$ is given by:
$\frac{P_2}{P_1} = \frac{m_2 v_2^2}{m_1 v_1^2} = \frac{(\frac{m}{2}) (2v)^2}{m v^2}$.
Simplifying the expression:
$\frac{P_2}{P_1} = \frac{(\frac{m}{2}) (4v^2)}{m v^2} = \frac{2 m v^2}{m v^2} = 2$.
Therefore,the final pressure is $P_2 = 2 P_1 = 2 P_0$.

(D) According to the kinetic theory of gases,the pressure exerted by a gas is due to the collision of gas molecules with the walls of the container,not due to collisions between the molecules themselves.
Collisions between molecules are considered to be negligible in terms of contributing to the macroscopic pressure.
Therefore,statement $D$ is incorrect.

(A) The root mean square (r.m.s.) speed of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
From this relation,we can see that $v_{rms} \propto \sqrt{T}$.
Given initial temperature $T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$.
Given final temperature $T_2 = 927^{\circ} C = 927 + 273 = 1200 \ K$.
Taking the ratio of the speeds:
$\frac{v_{rms_2}}{v_{rms_1}} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{1200}{300}} = \sqrt{4} = 2$.
Therefore,$v_{rms_2} = 2 \cdot v_{rms_1}$.
The r.m.s. speed becomes twice the initial speed.

(C) The root mean square (r.m.s.) speed of gas molecules is given by the formula: $V_{rms} = \sqrt{\frac{3RT}{M}}$.
Since the gas is the same,$V_{rms} \propto \sqrt{T}$.
First,convert the temperatures from Celsius to Kelvin:
$T_1 = 127^{\circ} C = 127 + 273 = 400 \ K$
$T_2 = 527^{\circ} C = 527 + 273 = 800 \ K$
The ratio of the r.m.s. speeds is:
$\frac{V_1}{V_2} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{400}{800}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Thus,the ratio is $1: \sqrt{2}$.

(C) The initial temperature $T_1 = -68^{\circ} C = -68 + 273 = 205 \ K$.
The root mean square (r.m.s.) velocity of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \sqrt{T}$.
Let the initial velocity be $(v_{rms})_1$ and the final velocity be $(v_{rms})_2 = 2(v_{rms})_1$.
Using the ratio: $\frac{(v_{rms})_2}{(v_{rms})_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $2 = \sqrt{\frac{T_2}{205}}$.
Squaring both sides: $4 = \frac{T_2}{205}$.
Therefore,$T_2 = 4 \times 205 = 820 \ K$.
Converting back to Celsius: $T_2 = 820 - 273 = 547^{\circ} C$.

(A) The $r.m.s.$ speed of a gas is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \frac{1}{\sqrt{M}}$.
Let the first gas have molecular mass $M_1 = 32$ and $r.m.s.$ speed $(v_{rms})_1$.
Let the second gas have molecular mass $M_2$ and $r.m.s.$ speed $(v_{rms})_2 = 4(v_{rms})_1$.
Using the relation $\frac{(v_{rms})_2}{(v_{rms})_1} = \sqrt{\frac{M_1}{M_2}}$,we substitute the given values:
$4 = \sqrt{\frac{32}{M_2}}$.
Squaring both sides,we get $16 = \frac{32}{M_2}$.
Therefore,$M_2 = \frac{32}{16} = 2$.

(A) In a conical pendulum,the forces acting on the bob are the tension $T$ in the string and the gravitational force $mg$.
Resolving the tension $T$ into components:
$T \cos \theta = mg$ (vertical equilibrium)
$T \sin \theta = F_c$ (where $F_c$ is the centripetal force)
From the geometry of the pendulum,$\sin \theta = \frac{r}{L}$.
Then,$\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{r^2}{L^2}} = \frac{\sqrt{L^2-r^2}}{L}$.
From the first equation,$T = \frac{mg}{\cos \theta} = \frac{mgL}{\sqrt{L^2-r^2}}$.
Substituting $T$ into the centripetal force equation:
$F_c = T \sin \theta = \left( \frac{mgL}{\sqrt{L^2-r^2}} \right) \times \left( \frac{r}{L} \right) = \frac{mgr}{\sqrt{L^2-r^2}}$.

(B) The maximum safe speed $V$ of a vehicle on a circular road of radius $r$ is given by $V = \sqrt{\mu rg}$,where $\mu$ is the coefficient of friction.
For the dry road: $V = \sqrt{\mu rg}$ $(i)$
For the wet road,let the new coefficient of friction be $\mu^{\prime}$. The new speed is $\frac{V}{2} = \sqrt{\mu^{\prime} rg}$ (ii)
Dividing equation $(i)$ by equation (ii):
$\frac{V}{V/2} = \frac{\sqrt{\mu rg}}{\sqrt{\mu^{\prime} rg}}$
$2 = \sqrt{\frac{\mu}{\mu^{\prime}}}$
Squaring both sides:
$4 = \frac{\mu}{\mu^{\prime}}$
$\therefore \mu^{\prime} = \frac{\mu}{4}$

(A) The moment of inertia of a solid sphere is $I = \frac{2}{5} MR^2$.
Since the sphere rolls without slipping,the condition $v = R\omega$ holds.
The rotational kinetic energy $E_{rot}$ is given by $E_{rot} = \frac{1}{2} I \omega^2 = \frac{1}{2} (\frac{2}{5} MR^2) \omega^2 = \frac{1}{5} MR^2 \omega^2 = \frac{1}{5} Mv^2$.
The total kinetic energy $E_{total}$ is the sum of translational and rotational kinetic energy: $E_{total} = \frac{1}{2} Mv^2 + \frac{1}{2} I \omega^2 = \frac{1}{2} Mv^2 + \frac{1}{5} Mv^2 = (\frac{5+2}{10}) Mv^2 = \frac{7}{10} Mv^2$.
The ratio of total kinetic energy to rotational kinetic energy is $\frac{E_{total}}{E_{rot}} = \frac{\frac{7}{10} Mv^2}{\frac{1}{5} Mv^2} = \frac{7}{10} \times 5 = \frac{7}{2}$.

(A) For a banked track,the angle of banking $\theta$ is given by $\tan \theta = \frac{v^2}{rg}$.
From the geometry of the track,where $h$ is the elevation and $d$ is the distance between the rails,we have $\tan \theta = \frac{h}{d}$.
Equating the two expressions for $\tan \theta$:
$\frac{v^2}{rg} = \frac{h}{d}$
Rearranging the equation to solve for the radius of curvature $r$:
$r = \frac{v^2 d}{gh}$

(D) For a solid sphere,the moment of inertia is $I_{S} = \frac{2}{5} MR^2$.
For a solid cylinder,the moment of inertia is $I_{C} = \frac{1}{2} MR^2$.
The acceleration of a body rolling down an inclined plane is given by $a = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$.
Therefore,the ratio of the acceleration of the cylinder '$a_{c}$' to that of the sphere '$a_{s}$' is:
$\frac{a_{c}}{a_{s}} = \frac{1 + \frac{I_{S}}{MR^2}}{1 + \frac{I_{C}}{MR^2}} = \frac{1 + \frac{2}{5}}{1 + \frac{1}{2}} = \frac{\frac{7}{5}}{\frac{3}{2}} = \frac{7}{5} \times \frac{2}{3} = \frac{14}{15}$.

(C) Let the natural length of the spring be $L$ and the extension be $x$.
The total radius of the circular path is $R = L + x$.
The restoring force provided by the spring acts as the centripetal force for the circular motion.
$F_{\text{restoring}} = F_{\text{centripetal}}$
$kx = m(L + x)\omega^2$
Given: $m = 200 \text{ g} = 0.2 \text{ kg}$,$k = 12.5 \text{ N/m}$,$\omega = 5 \text{ rad/s}$.
Substituting the values:
$12.5x = 0.2(L + x)(5)^2$
$12.5x = 0.2(L + x)(25)$
$12.5x = 5(L + x)$
$12.5x = 5L + 5x$
$7.5x = 5L$
$\frac{x}{L} = \frac{5}{7.5} = \frac{50}{75} = \frac{2}{3}$
Therefore,the ratio of extension to natural length is $2:3$.

(C) The electric force acting on the bob is $F_{\text{electric}} = qE$,which acts in the upward direction as per the figure.
The effective weight of the bob is given by $mg_{\text{eff}} = mg - F_{\text{electric}}$.
Therefore,the effective acceleration due to gravity is $g_{\text{eff}} = g - \frac{qE}{m}$.
The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g_{\text{eff}}}}$.
Substituting the value of $g_{\text{eff}}$,we get $T = 2 \pi \sqrt{\frac{L}{g - \frac{qE}{m}}} = 2 \pi \left[ \frac{L}{g - \frac{qE}{m}} \right]^{\frac{1}{2}}$.

(B) On the right side of the block,two springs with force constants $K$ and $2K$ are connected in parallel to the wall. Their equivalent spring constant is $K_R = K + 2K = 3K$.
On the left side of the block,two springs with force constants $2K$ and $2K$ are connected in series to the wall. Their equivalent spring constant $K_L$ is given by $\frac{1}{K_L} = \frac{1}{2K} + \frac{1}{2K} = \frac{2}{2K} = \frac{1}{K}$,so $K_L = K$.
Since the block is between these two sets,the effective spring constant of the system is $K_{eff} = K_R + K_L = 3K + K = 4K$.
The frequency of oscillation is given by $f = \frac{1}{2 \pi} \sqrt{\frac{K_{eff}}{M}} = \frac{1}{2 \pi} \sqrt{\frac{4K}{M}}$.

(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 force constant of a spring is inversely proportional to its length,i.e.,$k \propto \frac{1}{l}$,which implies $kl = \text{constant}$.
Let $k_1$ and $k_2$ be the force constants of the two springs of lengths $l_1$ and $l_2$ respectively.
Given $l = l_1 + l_2$ and $l_1 = n l_2$.
From the relation $kl = k_1 l_1 = k_2 l_2$,we have:
$k_1 = \frac{kl}{l_1}$ and $k_2 = \frac{kl}{l_2}$.
Since $l_1 = n l_2$,we have $l = n l_2 + l_2 = l_2(n+1)$.
Substituting $l$ in the expression for $k_1$:
$k_1 = \frac{k \cdot l_2(n+1)}{n l_2} = \frac{K(n+1)}{n}$.

(B) The time period of a mass-spring system is given by $T = 2 \pi \sqrt{\frac{m}{k}}$.
When an additional mass $M_1$ is added,the spring extends by $x$. According to Hooke's Law,the restoring force is $F = kx = M_1 g$.
Therefore,the spring constant is $k = \frac{M_1 g}{x}$.
The total mass oscillating on the spring is $m = M + M_1$.
Substituting these values into the time period formula:
$T = 2 \pi \sqrt{\frac{M + M_1}{k}} = 2 \pi \sqrt{\frac{M + M_1}{\frac{M_1 g}{x}}} = 2 \pi \sqrt{\frac{(M + M_1) x}{M_1 g}}$.

(C) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$,which implies $T \propto \sqrt{m}$.
For the first case,$T \propto \sqrt{m_1}$.
For the second case,the total mass is $(m_1 + m_2)$,so the new time period $T' \propto \sqrt{m_1 + m_2}$.
Given $T' = \frac{3}{2} T$,we have:
$\frac{T'}{T} = \frac{\sqrt{m_1 + m_2}}{\sqrt{m_1}} = \frac{3}{2}$.
Squaring both sides:
$\frac{m_1 + m_2}{m_1} = \frac{9}{4}$.
$1 + \frac{m_2}{m_1} = \frac{9}{4}$.
$\frac{m_2}{m_1} = \frac{9}{4} - 1 = \frac{5}{4}$.
Therefore,$\frac{m_1}{m_2} = \frac{4}{5}$.

(C) The two springs are connected in parallel. So,the effective spring constant is $k_{eff} = k + k = 2k$.
The time period $T$ of the spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k_{eff}}}$.
Substituting the values,we have $2 = 2\pi \sqrt{\frac{12}{2k}}$.
Dividing by $2$,we get $1 = \pi \sqrt{\frac{6}{k}}$.
Squaring both sides,$1 = \pi^2 \times \frac{6}{k}$.
Given $\pi^2 = 10$,we have $1 = 10 \times \frac{6}{k}$.
Therefore,$k = 60 \ Nm^{-1}$.

(A) According to the equation of continuity,$A_1 V_1 = A_2 V_2$.
Since the product $Av$ is constant,at the narrowest part where the cross-sectional area $A$ is minimum,the velocity $v$ must be maximum.
According to Bernoulli's principle for a horizontal pipe ($h$ is constant),$P + \frac{1}{2} \rho v^2 = \text{constant}$.
As the velocity $v$ increases,the pressure $P$ must decrease to keep the sum constant.
Therefore,at the narrowest part,the velocity is maximum and the pressure is minimum.

(B) According to the equation of continuity for an incompressible fluid,the product of the cross-sectional area and the velocity of the fluid remains constant at all points along the pipe.
$A_1 V_1 = A_2 V_2$
Here,$A_1$ is the area of the pipe,$V$ is the velocity in the pipe,$A_n$ is the area of the nozzle,and $V_1$ is the velocity at the nozzle.
$A_1 = \frac{\pi d^2}{4}$ and $A_n = \frac{\pi d_n^2}{4}$,where $d_n$ is the diameter of the nozzle.
Substituting these into the continuity equation:
$\frac{\pi d^2}{4} \times V = \frac{\pi d_n^2}{4} \times V_1$
$d^2 V = d_n^2 V_1$
$d_n^2 = d^2 \frac{V}{V_1}$
$d_n = d \sqrt{\frac{V}{V_1}}$

(A) Let $V$ be the volume of the body. When the body hits the surface of the lake,its velocity $v_0$ is given by $v_0^2 = 2gh$.
Inside the water,the forces acting on the body are the weight $W = \rho V g$ (downward) and the buoyant force $F_B = \sigma V g$ (upward).
The net upward force is $F_{net} = F_B - W = Vg(\sigma - \rho)$.
The acceleration $a$ of the body inside the water is $a = \frac{F_{net}}{m} = \frac{Vg(\sigma - \rho)}{\rho V} = \frac{g(\sigma - \rho)}{\rho}$.
Since the force is upward,the acceleration is directed upward (opposing the motion),so $a = -\frac{g(\sigma - \rho)}{\rho}$.
Let $H$ be the maximum depth reached. At this depth,the final velocity $v = 0$.
Using the equation of motion $v^2 - u^2 = 2aH$,we have:
$0 - (\sqrt{2gh})^2 = 2 \left( -\frac{g(\sigma - \rho)}{\rho} \right) H$
$-2gh = -\frac{2g(\sigma - \rho)}{\rho} H$
$h = \frac{(\sigma - \rho)}{\rho} H$
$H = \frac{h \rho}{(\sigma - \rho)}$.

(A) In a streamline flow,the velocity of a fluid at a specific point is always constant over time. This is because each particle of the fluid follows a well-defined path,and there is no mixing or crossing of paths between particles. Therefore,the statement that the velocity at a given point is 'never constant' is incorrect.

(C) Venturimeter is a device used to measure the rate of flow of an incompressible fluid flowing through a pipe. It operates on the principle of Bernoulli's theorem,where a constriction in the pipe causes a pressure drop that is proportional to the square of the flow rate.

(D) Molecules inside a liquid are surrounded by other molecules on all sides,resulting in a net attractive force of zero.
However,molecules on the surface of a liquid experience a net inward attractive force because there are no liquid molecules above them to balance the downward pull.
To bring a molecule from the interior to the surface,work must be done against this inward attractive force.
This work is stored as potential energy in the molecule.
Therefore,the potential energy of a molecule on the surface is higher than that of a molecule inside the liquid.

(C) According to the equation of continuity and Torricelli's law,the rate of flow of water is given by:
$A \left( -\frac{dh}{dt} \right) = a \sqrt{2gh}$
Rearranging the terms for integration:
$\int_{h}^{0} \frac{-dh}{\sqrt{h}} = \int_{0}^{t} \frac{a}{A} \sqrt{2g} dt$
Integrating both sides:
$[2\sqrt{h}]_{0}^{h} = \frac{a}{A} \sqrt{2g} t$
$2\sqrt{h} = \frac{a}{A} \sqrt{2g} t$
Thus,the time taken to empty the vessel is proportional to the square root of the height:
$t \propto \sqrt{h}$
Given that for height $h$,the time is $t_1 = t$,and for height $h_2 = 4h$,the time is $t_2$:
$\frac{t_2}{t_1} = \sqrt{\frac{h_2}{h_1}} = \sqrt{\frac{4h}{h}} = \sqrt{4} = 2$
Therefore,$t_2 = 2t_1 = 2t$.

(C) The formula for excess pressure inside a soap bubble is given by $P = \frac{4T}{R}$,where $T$ is the surface tension and $R$ is the radius of the bubble.
Given values are $P = 25.6 \ Nm^{-2}$ and $T = 3.2 \times 10^{-2} \ Nm^{-1}$.
Rearranging the formula to solve for the radius $R$:
$R = \frac{4T}{P}$
Substituting the values:
$R = \frac{4 \times 3.2 \times 10^{-2}}{25.6}$
$R = \frac{12.8 \times 10^{-2}}{25.6} = 0.5 \times 10^{-2} \ m = 0.5 \ cm$.
The diameter $D$ is twice the radius:
$D = 2R = 2 \times 0.5 \ cm = 1 \ cm$.

(D) The excess pressure $\Delta P$ inside a soap bubble of radius $r$ is given by $\Delta P = \frac{4T}{r}$,where $T$ is the surface tension of the soap solution.
From this,we see that $\Delta P \propto \frac{1}{r}$.
Let the excess pressure in the first bubble be $\Delta P_1$ and in the second bubble be $\Delta P_2$. Given $\Delta P_1 = 3 \Delta P_2$.
Therefore,$\frac{\Delta P_1}{\Delta P_2} = 3$.
Since $\frac{\Delta P_1}{\Delta P_2} = \frac{r_2}{r_1}$,we have $\frac{r_2}{r_1} = 3$,which implies $\frac{r_1}{r_2} = \frac{1}{3}$.
The volume $V$ of a spherical bubble is given by $V = \frac{4}{3} \pi r^3$,so $V \propto r^3$.
The ratio of the volumes is $\frac{V_1}{V_2} = \left( \frac{r_1}{r_2} \right)^3$.
Substituting the ratio of radii,$\frac{V_1}{V_2} = \left( \frac{1}{3} \right)^3 = \frac{1}{27}$.

(B) soap film has two surfaces. The change in surface area $\Delta A$ is given by $2 \times l \times \Delta x$, where $l = 10 \,cm = 0.1 \,m$ and $\Delta x = 1 \,mm = 10^{-3} \,m$.
$\Delta A = 2 \times 0.1 \,m \times 10^{-3} \,m = 2 \times 10^{-4} \,m^2$.
The work done $W$ is equal to the change in surface energy, given by $W = T \times \Delta A$.
Given $T = 65 \times 10^{-3} \,N/m$.
$W = (65 \times 10^{-3} \,N/m) \times (2 \times 10^{-4} \,m^2) = 130 \times 10^{-7} \,J = 1.3 \times 10^{-5} \,J$.
Wait, re-evaluating the provided surface tension value $65 \times 10^{-2} \,N/m$ from the prompt: $W = (65 \times 10^{-2}) \times (2 \times 10^{-4}) = 130 \times 10^{-6} = 1.3 \times 10^{-5} \,J$.
Given the options, if $T = 65 \times 10^{-3} \,N/m$ (standard value), $W = 1.3 \times 10^{-5} \,J$. If $T = 65 \times 10^{-2} \,N/m$, $W = 1.3 \times 10^{-4} \,J = 13.0 \times 10^{-5} \,J$. Thus, option $B$ is correct.

(C) The height of the liquid rise in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.
Given that $h, T, r$,and $g$ are constant for all three liquids.
Therefore,$\frac{\cos \theta}{\rho} = \text{constant}$.
This implies: $\frac{\cos \theta_1}{\rho_1} = \frac{\cos \theta_2}{\rho_2} = \frac{\cos \theta_3}{\rho_3}$.
Since $\rho_1 > \rho_2 > \rho_3$,it follows that $\cos \theta_1 > \cos \theta_2 > \cos \theta_3$.
Because the cosine function is a decreasing function in the range $[0, \pi/2]$,a larger cosine value corresponds to a smaller angle.
Therefore,$\theta_1 < \theta_2 < \theta_3$.

(A) The surface tension of a liquid arises due to the cohesive forces between molecules in the bulk and the surface.
As the temperature of a liquid increases,the kinetic energy of the molecules increases,which weakens the cohesive forces.
At the critical temperature $(T_c)$,the distinction between the liquid phase and the vapor phase disappears,meaning the density of the liquid becomes equal to the density of its saturated vapor.
Since the cohesive forces effectively vanish at this point,the surface tension of the liquid becomes $0$.

(C) Work done,$W = S \times \Delta A$,where $\Delta A$ is the change in surface area.
Initial surface area of the big drop,$A_{\text{initial}} = 4 \pi R^2$.
Let the radius of each small drop be $r$. The total volume remains constant:
$\frac{4}{3} \pi R^3 = 8 \times \frac{4}{3} \pi r^3 \implies R^3 = 8r^3 \implies R = 2r$ or $r = R/2$.
Final surface area of $8$ small drops,$A_{\text{final}} = 8 \times 4 \pi r^2 = 32 \pi (R/2)^2 = 32 \pi (R^2/4) = 8 \pi R^2$.
Change in surface area,$\Delta A = A_{\text{final}} - A_{\text{initial}} = 8 \pi R^2 - 4 \pi R^2 = 4 \pi R^2$.
Work done,$W = S \times \Delta A = S \times 4 \pi R^2 = 4 \pi R^2 S$.

(D) The height of liquid rise in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.
Given that the diameter $(2r)$,angle of contact $(\theta)$,and acceleration due to gravity $(g)$ are constant,the height is proportional to the ratio of surface tension to density: $h \propto \frac{T}{\rho}$.
Therefore,the ratio of the heights is given by: $\frac{h_1}{h_2} = \frac{T_1}{\rho_1} \times \frac{\rho_2}{T_2} = \left(\frac{T_1}{T_2}\right) \times \left(\frac{\rho_2}{\rho_1}\right)$.
Given $\frac{T_1}{T_2} = \frac{6}{5}$ and $\frac{\rho_1}{\rho_2} = \frac{4}{3}$,we have $\frac{\rho_2}{\rho_1} = \frac{3}{4}$.
Substituting these values: $\frac{h_1}{h_2} = \frac{6}{5} \times \frac{3}{4} = \frac{18}{20} = 0.9$.

(B) The volume of the original drop is equal to the sum of the volumes of the $n$ smaller droplets.
Let $r$ be the radius of each small droplet.
$n \cdot \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$
$r^3 = \frac{R^3}{n} \implies r = \frac{R}{n^{1/3}}$
Work done $W$ is equal to the change in surface area multiplied by the surface tension $T$.
$W = (A_{\text{final}} - A_{\text{initial}}) T$
$W = (n \cdot 4 \pi r^2 - 4 \pi R^2) T$
Substitute $r = R \cdot n^{-1/3}$:
$W = (n \cdot 4 \pi (R \cdot n^{-1/3})^2 - 4 \pi R^2) T$
$W = (4 \pi R^2 \cdot n \cdot n^{-2/3} - 4 \pi R^2) T$
$W = 4 \pi R^2 (n^{1/3} - 1) T$

(D) The excess pressure inside a spherical drop of radius $r$ and surface tension $T$ is given by $P = \frac{2T}{r}$.
Let $r_1$ and $r_2$ be the radii of the first and second drops,respectively.
Given that the excess pressure in the first drop is three times that of the second drop: $P_1 = 3P_2$.
Substituting the formula: $\frac{2T}{r_1} = 3 \left( \frac{2T}{r_2} \right)$.
This simplifies to $\frac{1}{r_1} = \frac{3}{r_2}$,which means $\frac{r_1}{r_2} = \frac{1}{3}$.
The mass of a drop is given by $m = V \rho = \frac{4}{3} \pi r^3 \rho$,where $\rho$ is the density of water.
Since both drops are of water,the density $\rho$ is the same for both.
Therefore,the ratio of the masses is $\frac{m_1}{m_2} = \frac{\frac{4}{3} \pi r_1^3 \rho}{\frac{4}{3} \pi r_2^3 \rho} = \left( \frac{r_1}{r_2} \right)^3$.
Substituting the ratio of radii: $\frac{m_1}{m_2} = \left( \frac{1}{3} \right)^3 = \frac{1}{27}$.

(D) Let $R$ be the radius of the original large drop and $r$ be the radius of each small drop.
Since the volume remains constant,the volume of the large drop equals the sum of the volumes of the $1000$ small drops:
$\frac{4}{3} \pi R^3 = 1000 \times \frac{4}{3} \pi r^3$
$R^3 = 1000 r^3 \implies R = 10r$.
The surface energy $E$ of a spherical drop is given by $E = T \times A$,where $T$ is the surface tension and $A$ is the surface area $(4\pi r^2)$.
$E_1 = T(4\pi R^2)$
$E_2 = 1000 \times T(4\pi r^2)$
Taking the ratio:
$\frac{E_1}{E_2} = \frac{T(4\pi R^2)}{1000 \times T(4\pi r^2)} = \frac{R^2}{1000 r^2}$
Substituting $R = 10r$:
$\frac{E_1}{E_2} = \frac{(10r)^2}{1000 r^2} = \frac{100 r^2}{1000 r^2} = \frac{1}{10}$
Given $\frac{E_1}{E_2} = \frac{x}{10}$,comparing both sides,we get $x = 1$.

(D) soap film has two surfaces, so the total change in area is $2 \times \Delta A$.
Initial area $A_1 = 3 \times 3 \,cm^2 = 9 \times 10^{-4} \,m^2$.
Final area $A_2 = 5 \times 5 \,cm^2 = 25 \times 10^{-4} \,m^2$.
Change in area $\Delta A = A_2 - A_1 = (25 - 9) \times 10^{-4} \,m^2 = 16 \times 10^{-4} \,m^2$.
Surface tension $T = 2.5 \times 10^{-2} \,N/m$.
Work done $W = T \times (2 \Delta A) = 2.5 \times 10^{-2} \times 2 \times 16 \times 10^{-4} \,J$.
$W = 5 \times 10^{-2} \times 16 \times 10^{-4} \,J = 80 \times 10^{-6} \,J$.

(A) Let $R$ be the radius of the large coalesced drop.
Since the total volume remains constant:
$\frac{4}{3} \pi R^3 = 8 \times \frac{4}{3} \pi r^3$
$R^3 = 8r^3$
$R = 2r$
Surface energy $E$ is given by $E = T \times A$,where $T$ is surface tension and $A$ is the surface area.
Initial surface energy $E_1 = 8 \times (T \times 4 \pi r^2) = 32 \pi r^2 T$.
Final surface energy $E_2 = T \times 4 \pi R^2 = T \times 4 \pi (2r)^2 = 16 \pi r^2 T$.
The ratio of total surface energy before and after the change is:
$\frac{E_1}{E_2} = \frac{32 \pi r^2 T}{16 \pi r^2 T} = \frac{2}{1}$.
Thus,the ratio is $2: 1$.

(D) Let the radius of each small droplet be $r = 0.1 \,mm = 10^{-4} \,m$ and the number of droplets be $n = 27$.
When $n$ droplets merge to form a single large drop of radius $R$, the volume remains constant: $\frac{4}{3} \pi R^3 = n \times \frac{4}{3} \pi r^3$.
Thus, $R = n^{1/3} r = (27)^{1/3} \times r = 3r$.
The energy released is equal to the decrease in surface area multiplied by the surface tension $T$:
$\Delta E = T \times (A_{initial} - A_{final}) = T \times (n \times 4 \pi r^2 - 4 \pi R^2)$.
Substituting $R = 3r$:
$\Delta E = 4 \pi T r^2 (n - 9)$.
Given $n = 27$, $\Delta E = 4 \pi T r^2 (27 - 9) = 4 \pi T r^2 (18) = 72 \pi T r^2$.
Substituting values: $\Delta E = 72 \times 3.14 \times 0.072 \times (10^{-4})^2$.
$\Delta E \approx 1.627 \times 10^{-7} \,J$.

(C) The Rydberg formula for the hydrogen spectrum is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For the Lyman series,$n_1 = 1$. The longest wavelength corresponds to the smallest energy transition,which occurs at $n_2 = 2$.
$\frac{1}{\lambda_{\max(L)}} = R \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = R \left[ 1 - \frac{1}{4} \right] = \frac{3R}{4} \implies \lambda_{\max(L)} = \frac{4}{3R}$.
For the Balmer series,$n_1 = 2$. The longest wavelength corresponds to the smallest energy transition,which occurs at $n_2 = 3$.
$\frac{1}{\lambda_{\max(B)}} = R \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] = R \left[ \frac{1}{4} - \frac{1}{9} \right] = R \left[ \frac{9-4}{36} \right] = \frac{5R}{36} \implies \lambda_{\max(B)} = \frac{36}{5R}$.
The ratio of the longest wavelengths is $\frac{\lambda_{\max(L)}}{\lambda_{\max(B)}} = \frac{4}{3R} \times \frac{5R}{36} = \frac{4 \times 5}{3 \times 36} = \frac{20}{108} = \frac{5}{27}$.

(D) According to de Broglie's hypothesis,the wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.
According to Bohr's postulate for the quantization of angular momentum,$L = mvr_n = \frac{nh}{2\pi}$.
Rearranging this equation to solve for the momentum $mv$,we get $mv = \frac{nh}{2\pi r_n}$.
Substituting this expression for $mv$ into the de Broglie wavelength formula:
$\lambda = \frac{h}{mv} = \frac{h}{(nh / 2\pi r_n)} = \frac{2\pi r_n}{n}$.
Thus,the de Broglie wavelength of an electron in the $n^{\text{th}}$ orbit is $\frac{2\pi r}{n}$.

(A) According to the Bohr quantization condition,the angular momentum is given by $mvr = \frac{nh}{2 \pi}$.
Therefore,$vr = \frac{nh}{2 \pi m} \dots (i)$.
For a particle moving in a circular path under a magnetic field,the magnetic force provides the centripetal force: $qvB = \frac{mv^2}{r}$.
This simplifies to $mv = qBr$,or $v = \frac{qBr}{m} \dots (ii)$.
Substituting $v$ from $(ii)$ into $(i)$,we get $(\frac{qBr}{m})r = \frac{nh}{2 \pi m}$,which implies $r^2 = \frac{nh}{2 \pi qB}$.
The kinetic energy $E$ is given by $E = \frac{1}{2}mv^2$. Since $mv = qBr$,we have $E = \frac{1}{2}m(\frac{qBr}{m})^2 = \frac{q^2 B^2 r^2}{2m}$.
Substituting $r^2$ into the energy expression: $E = \frac{q^2 B^2}{2m} \times \frac{nh}{2 \pi qB} = \frac{nhqB}{4 \pi m}$.

(D) According to Bohr's theory of the hydrogen atom,the total energy $E_n$ of an electron in the $n^{th}$ stationary orbit is given by the formula:
$E_n = -\frac{m Z^2 e^4}{8 \varepsilon_0^2 h^2 n^2}$
Where $m$ is the mass of the electron,$Z$ is the atomic number,$e$ is the charge of the electron,$\varepsilon_0$ is the permittivity of free space,$h$ is Planck's constant,and $n$ is the principal quantum number.
From this expression,it is clear that the total energy $E_n$ is inversely proportional to the square of the principal quantum number $n$.
Therefore,$E_n \propto \frac{1}{n^2}$.

(B) The centripetal force $F$ acting on the electron is given by $F = \frac{mv^2}{r}$.
According to Bohr's theory,the velocity $v$ of the electron in the $n^{th}$ orbit is proportional to $\frac{1}{n}$ $(v \propto \frac{1}{n})$.
The radius $r$ of the $n^{th}$ orbit is proportional to $n^2$ $(r \propto n^2)$.
Substituting these proportionalities into the force equation:
$F \propto \frac{v^2}{r}$
$F \propto \frac{(1/n)^2}{n^2}$
$F \propto \frac{1/n^2}{n^2}$
$F \propto \frac{1}{n^4}$ or $F \propto n^{-4}$.

(B) The velocity of an electron in the $n^{th}$ Bohr orbit is given by the formula $v_n = \frac{Ze^2}{2 \varepsilon_0 nh}$.
From this expression,we can see that the velocity is inversely proportional to the principal quantum number $n$,i.e.,$v_n \propto \frac{1}{n}$.
For the first orbit,$n_1 = 1$,and for the second orbit,$n_2 = 2$.
Therefore,the ratio of the velocities is $\frac{v_1}{v_2} = \frac{n_2}{n_1} = \frac{2}{1}$.
Thus,the ratio is $2:1$.

(C) The moment of inertia $I$ of an electron of mass $m$ moving in an orbit of radius $r$ is given by $I = mr^2$.
The radius of the $n^{\text{th}}$ Bohr orbit is given by the formula:
$r_n = \frac{\varepsilon_0 h^2 n^2}{\pi m e^2}$
For the second Bohr orbit,$n = 2$. Substituting $n = 2$ into the formula:
$r_2 = \frac{\varepsilon_0 h^2 (2)^2}{\pi m e^2} = \frac{4 \varepsilon_0 h^2}{\pi m e^2}$
Now,calculate the moment of inertia $I$ for the second orbit:
$I = m \times (r_2)^2$
$I = m \times \left( \frac{4 \varepsilon_0 h^2}{\pi m e^2} \right)^2$
$I = m \times \frac{16 \varepsilon_0^2 h^4}{\pi^2 m^2 e^4}$
$I = \frac{16 \varepsilon_0^2 h^4}{\pi^2 m e^4}$

(C) According to Bohr's postulate,the angular momentum $L$ of an electron in the $n^{\text{th}}$ orbit is given by $L = \frac{nh}{2\pi}$.
Change in angular momentum $\Delta L = L_2 - L_1$.
Here,$n_1 = 4$ and $n_2 = 5$.
$\Delta L = \frac{n_2 h}{2\pi} - \frac{n_1 h}{2\pi} = \frac{h}{2\pi} (n_2 - n_1)$.
Substituting the values: $\Delta L = \frac{6.63 \times 10^{-34}}{2 \times 3.14} (5 - 4)$.
$\Delta L = \frac{6.63 \times 10^{-34}}{6.28} \approx 1.055 \times 10^{-34} \text{ J s}$.
Rounding to the nearest given option,the change is approximately $1 \times 10^{-34} \text{ J s}$.

(B) Given the initial capacitance $C = \frac{\varepsilon_0 A}{d} = 2.5 \mu F$.
When the capacitor is half-filled with a dielectric as shown,it acts as two capacitors in parallel,each with plate area $A/2$ and separation $d$.
The first part is filled with air (dielectric constant $K_1 = 1$):
$C_1 = \frac{\varepsilon_0 (A/2)}{d} = \frac{1}{2} \left( \frac{\varepsilon_0 A}{d} \right) = \frac{2.5 \mu F}{2} = 1.25 \mu F$.
The second part is filled with a dielectric (dielectric constant $K_2 = K$):
$C_2 = \frac{K \varepsilon_0 (A/2)}{d} = \frac{K}{2} \left( \frac{\varepsilon_0 A}{d} \right) = K \times 1.25 \mu F$.
Since they are in parallel,the equivalent capacitance $C_{eq} = C_1 + C_2$.
Given $C_{eq} = 5 \mu F$:
$5 = 1.25 + 1.25 K$
$3.75 = 1.25 K$
$K = \frac{3.75}{1.25} = 3$.
Therefore,the dielectric constant is $3$.

(B) When capacitors are connected in series,the charge $Q$ on each capacitor is the same.
Given: $C_1 = 3 \mu F$,$C_2 = 2 \mu F$.
The potential difference across a capacitor is given by $V = \frac{Q}{C}$.
Since $Q$ is constant for both capacitors in series,$V \propto \frac{1}{C}$.
Therefore,the ratio of potential across $C_2$ to $C_1$ is:
$\frac{V_2}{V_1} = \frac{Q/C_2}{Q/C_1} = \frac{C_1}{C_2}$.
Substituting the values:
$\frac{V_2}{V_1} = \frac{3 \mu F}{2 \mu F} = \frac{3}{2}$.
Thus,the ratio is $3:2$.

(B) The potential difference $V$ between the plates is given by the difference of the potentials on the two plates.
$V = V_1 - V_2 = 20 \,V - (-20 \,V) = 40 \,V$.
The charge on the capacitor plate is given as $Q = 40 \,C$.
The capacitance $C$ of a capacitor is defined by the formula $C = \frac{Q}{V}$.
Substituting the values,we get $C = \frac{40 \,C}{40 \,V} = 1 \,F$.
Therefore,the capacitance of the capacitor is $1 \,F$.

(D) Let the four capacitors be $C_1, C_2, C_3, C_4$ in a row from left to right.
Terminal $A$ is connected to the left plate of $C_1$.
The right plate of $C_1$,left plate of $C_2$,and right plate of $C_4$ are connected together.
The right plate of $C_2$ and left plate of $C_3$ are connected.
The right plate of $C_3$ and left plate of $C_4$ are connected to terminal $B$.
By analyzing the circuit,we see that $C_1, C_2, C_3$ are in series,and this combination is in parallel with $C_4$.
However,looking at the diagram,the first three capacitors are in series between $A$ and $B$,and the fourth capacitor is connected between $A$ and $B$ as well.
Actually,the circuit shows three capacitors in series between $A$ and $B$,and one capacitor in parallel with the series combination.
Equivalent capacitance of three capacitors in series: $\frac{1}{C_s} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C} \implies C_s = \frac{C}{3}$.
Now,this $C_s$ is in parallel with the fourth capacitor $C$.
$C_{eq} = C_s + C = \frac{C}{3} + C = \frac{4C}{3}$.

(A) The potential energy of a charged capacitor is given by $U_0 = \frac{Q^2}{2C}$,where $Q$ is the charge on the plates and $C$ is the initial capacitance.
When a dielectric slab of constant $K$ is inserted between the plates,the new capacitance becomes $C' = KC$.
Assuming the capacitor is isolated (charge $Q$ remains constant),the new potential energy $U'$ is given by:
$U' = \frac{Q^2}{2C'} = \frac{Q^2}{2(KC)}$
$U' = \frac{1}{K} \left( \frac{Q^2}{2C} \right)$
$U' = \frac{U_0}{K}$

(D) The capacitor can be modeled as two capacitors in series,each with plate area $A$ and dielectric thicknesses $d_1 = \frac{d}{4}$ and $d_2 = \frac{3d}{4}$.
Capacity of the $1^{\text{st}}$ capacitor:
$C_1 = \frac{K_1 \varepsilon_0 A}{d/4} = \frac{4 K_1 \varepsilon_0 A}{d}$
Capacity of the $2^{\text{nd}}$ capacitor:
$C_2 = \frac{K_2 \varepsilon_0 A}{3d/4} = \frac{4 K_2 \varepsilon_0 A}{3d}$
Since they are in series,the equivalent capacitance $C$ is given by:
$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$
$\frac{1}{C} = \frac{d}{4 K_1 \varepsilon_0 A} + \frac{3d}{4 K_2 \varepsilon_0 A}$
$\frac{1}{C} = \frac{d}{4 \varepsilon_0 A} \left[\frac{1}{K_1} + \frac{3}{K_2}\right]$
$\frac{1}{C} = \frac{d}{4 \varepsilon_0 A} \left[\frac{K_2 + 3 K_1}{K_1 K_2}\right]$
Therefore,$C = \frac{4 \varepsilon_0 A}{d} \left[\frac{K_1 K_2}{3 K_1 + K_2}\right]$.

(A) When a battery remains connected to a parallel plate capacitor,the potential difference $V$ between the plates remains constant.
The electric field $E$ between the plates is given by the formula $E = V/d$,where $d$ is the distance between the plates.
Since the battery is connected,$V$ is constant. When the dielectric slab is inserted,the capacitance increases,but the potential difference $V$ across the plates remains equal to the battery voltage.
When the dielectric slab is taken out,the system returns to its original state because the battery maintains the potential difference $V$ at all times.
Therefore,the electric field $E$ remains the same as it was before the insertion of the slab.

(C) The initial capacitance of the air-filled parallel plate capacitor is given by $C_1 = \frac{A \varepsilon_0}{d}$.
When the space is filled with a dielectric of constant $K = 8$ and the distance is changed to $d'$,the new capacitance is $C_2 = \frac{K A \varepsilon_0}{d'} = \frac{8 A \varepsilon_0}{d'}$.
Given that the capacity increases $16$ times,we have $C_2 = 16 C_1$.
Substituting the expressions,we get $\frac{8 A \varepsilon_0}{d'} = 16 \left( \frac{A \varepsilon_0}{d} \right)$.
Simplifying the equation: $\frac{8}{d'} = \frac{16}{d}$.
Solving for $d'$,we get $d' = \frac{8d}{16} = \frac{d}{2}$.

(C) The initial capacitance of the air-filled parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d} = 10 \mu F$.
When the area is divided into two equal halves and filled with dielectrics $K_1$ and $K_2$,the two parts act as two capacitors connected in parallel.
The area of each part is $A' = \frac{A}{2}$ and the distance between the plates remains $d$.
The capacitance of the first part is $C_1 = \frac{K_1 \varepsilon_0 A'}{d} = \frac{K_1 \varepsilon_0 A}{2d} = \frac{K_1}{2} C$.
The capacitance of the second part is $C_2 = \frac{K_2 \varepsilon_0 A'}{d} = \frac{K_2 \varepsilon_0 A}{2d} = \frac{K_2}{2} C$.
Since they are in parallel,the equivalent capacitance is $C_{\text{eq}} = C_1 + C_2 = \frac{C}{2} (K_1 + K_2)$.
Substituting the given values: $C_{\text{eq}} = \frac{10 \mu F}{2} (2 + 4) = 5 \mu F \times 6 = 30 \mu F$.

(D) Initial energy of the system,$U_i = \frac{1}{2} CV_1^2 + \frac{1}{2} CV_2^2 = \frac{1}{2} C(V_1^2 + V_2^2)$.
When the capacitors are connected in parallel,the common potential is $V = \frac{CV_1 + CV_2}{C + C} = \frac{V_1 + V_2}{2}$.
Final energy of the system,$U_f = \frac{1}{2}(2C)V^2 = C \left(\frac{V_1 + V_2}{2}\right)^2 = \frac{1}{4} C(V_1 + V_2)^2$.
Decrease in energy,$\Delta U = U_i - U_f = \frac{1}{2} C(V_1^2 + V_2^2) - \frac{1}{4} C(V_1 + V_2)^2$.
$\Delta U = \frac{1}{4} C [2V_1^2 + 2V_2^2 - (V_1^2 + V_2^2 + 2V_1V_2)]$.
$\Delta U = \frac{1}{4} C(V_1^2 + V_2^2 - 2V_1V_2) = \frac{1}{4} C(V_1 - V_2)^2$.

(A) For a parallel plate capacitor,the energy density $u = \frac{1}{2} \varepsilon_0 E^2$.
Since the electric field $E$ between the plates is given by $E = \frac{\sigma}{\varepsilon_0}$,where $\sigma = \frac{q}{A}$ is the surface charge density.
Substituting the value of $E$ in the energy density formula:
$u = \frac{1}{2} \varepsilon_0 \left( \frac{\sigma}{\varepsilon_0} \right)^2 = \frac{\sigma^2}{2 \varepsilon_0}$.
Now,substituting $\sigma = \frac{q}{A}$:
$u = \frac{(q/A)^2}{2 \varepsilon_0} = \frac{q^2}{2 \varepsilon_0 A^2}$.

(D) Initial capacitance $C_0 = \frac{\varepsilon_0 A}{d}$.
Initial charge $Q = C_0 V_0$.
Since the battery is disconnected,the charge $Q$ remains constant.
Final separation $d' = 3d$.
Final capacitance $C' = \frac{\varepsilon_0 A}{3d} = \frac{C_0}{3}$.
Initial potential energy $U_i = \frac{Q^2}{2C_0} = \frac{1}{2} C_0 V_0^2$.
Final potential energy $U_f = \frac{Q^2}{2C'} = \frac{Q^2}{2(C_0/3)} = \frac{3Q^2}{2C_0} = \frac{3}{2} C_0 V_0^2$.
Work done $W = U_f - U_i = \frac{3}{2} C_0 V_0^2 - \frac{1}{2} C_0 V_0^2 = C_0 V_0^2$.
Substituting $C_0 = \frac{\varepsilon_0 A}{d}$,we get $W = \frac{\varepsilon_0 A V_0^2}{d}$.

(C) We know that the capacitance of a spherical conductor is $C = 4 \pi \epsilon_0 r$. Since $C \propto r$,the ratio of capacitances is equal to the ratio of radii: $\frac{C_1}{C_2} = \frac{r_1}{r_2} = \frac{3}{2}$.
Given that both spheres are at the same potential $V$,the charges are $Q_1 = C_1 V$ and $Q_2 = C_2 V$.
The surface charge density is given by $\sigma = \frac{Q}{A} = \frac{Q}{4 \pi r^2}$.
Substituting $Q = CV$,we get $\sigma = \frac{CV}{4 \pi r^2}$.
Since $C = 4 \pi \epsilon_0 r$,we have $\sigma = \frac{(4 \pi \epsilon_0 r) V}{4 \pi r^2} = \frac{\epsilon_0 V}{r}$.
Thus,$\sigma \propto \frac{1}{r}$.
Therefore,$\frac{\sigma_1}{\sigma_2} = \frac{r_2}{r_1} = \frac{2 \ cm}{3 \ cm} = \frac{2}{3}$.

(B) The energy stored in a capacitor is given by $U = \frac{Q^2}{2C}$.
Since $C$ is constant,$U \propto Q^2$.
Let the initial charge be $Q_1$ and the initial energy be $U_1$.
When the charge is increased by $3 \ C$,the new charge is $Q_2 = Q_1 + 3$.
The new energy $U_2$ is $U_1 + 44\% \text{ of } U_1 = 1.44 \ U_1$.
Using the proportionality $U \propto Q^2$,we have:
$\frac{U_2}{U_1} = \left( \frac{Q_2}{Q_1} \right)^2$
$1.44 = \left( \frac{Q_1 + 3}{Q_1} \right)^2$
Taking the square root on both sides:
$1.2 = \frac{Q_1 + 3}{Q_1}$
$1.2 \ Q_1 = Q_1 + 3$
$0.2 \ Q_1 = 3$
$Q_1 = \frac{3}{0.2} = 15 \ C$.

(B) Given: $C_1: C_2 = 1: 2$. Therefore,$C_2 = 2C_1$.
Equivalent capacitance in parallel: $C_P = C_1 + C_2 = C_1 + 2C_1 = 3C_1$.
Equivalent capacitance in series: $C_S = \frac{C_1 C_2}{C_1 + C_2} = \frac{C_1(2C_1)}{3C_1} = \frac{2}{3}C_1$.
Let $V_P$ and $V_S$ be the potential differences applied across the parallel and series combinations respectively.
Since the energy stored $E = \frac{1}{2}CV^2$ is the same in both cases:
$\frac{1}{2} C_P V_P^2 = \frac{1}{2} C_S V_S^2$
$\frac{V_P^2}{V_S^2} = \frac{C_S}{C_P} = \frac{\frac{2}{3}C_1}{3C_1} = \frac{2}{9}$
Taking the square root on both sides:
$\frac{V_P}{V_S} = \sqrt{\frac{2}{9}} = \frac{\sqrt{2}}{3}$.

(C) In a parallel combination,the potential difference $V$ across each capacitor is the same.
Given $V = 5 \text{ V}$,$C_1 = 2C$,and $C_2 = C$.
Charge stored in a capacitor is given by $Q = CV$.
$Q_1 = C_1 V = (2C)(5) = 10C \text{ C}$.
$Q_2 = C_2 V = (C)(5) = 5C \text{ C}$.
Energy stored in a capacitor is given by $E = \frac{1}{2}CV^2$.
$E_1 = \frac{1}{2} C_1 V^2 = \frac{1}{2} (2C) (5)^2 = 25C \text{ J}$.
$E_2 = \frac{1}{2} C_2 V^2 = \frac{1}{2} (C) (5)^2 = 12.5C \text{ J}$.
Now,calculate the ratio $\frac{E_1-E_2}{Q_1-Q_2}$:
$\frac{E_1-E_2}{Q_1-Q_2} = \frac{25C - 12.5C}{10C - 5C} = \frac{12.5C}{5C} = \frac{12.5}{5} = 2.5 = \frac{5}{2}$.

(C) The equivalent capacitance of three capacitors connected in series is given by:
$\frac{1}{C_{s}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C}$
Therefore,$C_{s} = \frac{C}{3}$.
This combination is connected in parallel with another identical capacitor of capacitance $C$.
The total equivalent capacitance $C_{eq}$ is:
$C_{eq} = C_{s} + C = \frac{C}{3} + C = \frac{4C}{3}$.

(B) $1$. Capacitors $C_3$ and $C_4$ are in series. Their equivalent capacitance $C_S$ is given by:
$\frac{1}{C_S} = \frac{1}{C_3} + \frac{1}{C_4} = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4} \implies C_S = 4 \mu F$.
$2$. Capacitors $C_5$ and $C_6$ are in parallel. Their equivalent capacitance $C_P$ is:
$C_P = C_5 + C_6 = 4 \mu F + 4 \mu F = 8 \mu F$.
$3$. Now,$C_2$ and $C_P$ are in series. Their equivalent capacitance $C_{2P}$ is:
$C_{2P} = \frac{C_2 \cdot C_P}{C_2 + C_P} = \frac{8 \cdot 8}{8 + 8} = \frac{64}{16} = 4 \mu F$.
$4$. This $C_{2P}$ is in parallel with $C_S$. Their equivalent capacitance $C_{total}'$ is:
$C_{total}' = C_{2P} + C_S = 4 \mu F + 4 \mu F = 8 \mu F$.
$5$. Finally,$C_1$ and $C_{total}'$ are in series. The resultant capacitance $C_{AB}$ is:
$C_{AB} = \frac{C_1 \cdot C_{total}'}{C_1 + C_{total}'} = \frac{8 \cdot 8}{8 + 8} = \frac{64}{16} = 4 \mu F$.

(A) The circuit consists of two main parts connected in series.
$1$. The first part (left side) consists of three capacitors of capacitance $C$ connected in parallel between the input node and the intermediate node. The equivalent capacitance is $C_1 = C + C + C = 3C$.
$2$. The second part (right side) consists of three capacitors of capacitance $C$ connected in parallel between the intermediate node and point $B$. The equivalent capacitance is $C_2 = C + C + C = 3C$.
$3$. Now,these two equivalent capacitors $C_1$ and $C_2$ are in series.
$4$. The total equivalent capacitance $C_{AB}$ is given by $\frac{1}{C_{AB}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{3C} + \frac{1}{3C} = \frac{2}{3C}$.
$5$. Therefore,$C_{AB} = \frac{3C}{2} = 1.5C$.

(A) Let $m$ capacitors be connected in parallel and $n$ capacitors be connected in series,such that the total number of capacitors is $m + n = 7$.
For $m$ capacitors in parallel,the equivalent capacitance is $C_p = mC$.
For $n$ capacitors in series,the equivalent capacitance is $C_s = C/n$.
When these two groups are connected in series,the total equivalent capacitance $C_{\text{net}}$ is given by:
$\frac{1}{C_{\text{net}}} = \frac{1}{C_p} + \frac{1}{C_s} = \frac{1}{mC} + \frac{n}{C} = \frac{1 + mn}{mC}$.
Given $C_{\text{net}} = \frac{10}{11} \mu F$ and $C = 2 \mu F$:
$\frac{11}{10} = \frac{1 + mn}{2m} \implies \frac{11}{5} = \frac{1 + mn}{m} = \frac{1}{m} + n$.
Testing the options where $m + n = 7$:
If $m = 5$ and $n = 2$,then $\frac{1}{5} + 2 = 0.2 + 2 = 2.2 = \frac{11}{5}$.
This matches the required value. Thus,$5$ capacitors in parallel and $2$ in series is the correct combination.

(B) The two capacitors are connected in parallel to the battery.
In a parallel combination,the equivalent capacitance $C_{eq}$ is the sum of individual capacitances:
$C_{eq} = C + \frac{C}{2} = \frac{3C}{2}$
The work done $W$ by the battery to charge the capacitors is equal to the energy stored in the equivalent capacitor:
$W = \frac{1}{2} C_{eq} V^2$
Substituting the value of $C_{eq}$:
$W = \frac{1}{2} \left( \frac{3C}{2} \right) V^2$
$W = \frac{3}{4} CV^2$

(B) When an electron is accelerated through a potential difference $V$,its kinetic energy is given by $K = eV$.
Since the kinetic energy is $K = \frac{1}{2}mv^2$,we have:
$\frac{1}{2}mv^2 = eV$
Rearranging for $V$:
$V = \frac{mv^2}{2e}$
Substituting the given values:
$m = 9 \times 10^{-31} \ kg$,$v = 1.6 \times 10^7 \ m/s$,$e = 1.6 \times 10^{-19} \ C$
$V = \frac{(9 \times 10^{-31}) \times (1.6 \times 10^7)^2}{2 \times 1.6 \times 10^{-19}}$
$V = \frac{9 \times 10^{-31} \times 2.56 \times 10^{14}}{3.2 \times 10^{-19}}$
$V = \frac{23.04 \times 10^{-17}}{3.2 \times 10^{-19}}$
$V = 7.2 \times 10^2 \ V$

(B) The formula for the equivalent e.m.f. of the parallel combination of batteries is given by $\varepsilon_{eq} = r_{eq} \left( \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2} \right)$.
First,we calculate the equivalent internal resistance $r_{eq}$:
$\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} = \frac{1}{2} + \frac{1}{1} = \frac{3}{2} \Omega^{-1}$.
Therefore,$r_{eq} = \frac{2}{3} \Omega$.
Now,substitute the values into the formula for equivalent e.m.f.:
$\varepsilon_{eq} = \frac{2}{3} \left( \frac{12}{2} + \frac{6}{1} \right) = \frac{2}{3} (6 + 6) = \frac{2}{3} \times 12 = 8 \text{ V}$.
The voltmeter measures the potential difference across the parallel combination,which is equal to the equivalent e.m.f. of the circuit.
Thus,the reading of the voltmeter is $8 \text{ V}$.

(C) The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by distance $r$ is given by $F = \frac{\mu_0}{4\pi} \frac{2 I_1 I_2}{r} \cdot l$.
Given the initial force $F = 1 \times 10^{-3} \ N$.
When the current in both wires is doubled,the new currents become $I_1' = 2I_1$ and $I_2' = 2I_2$.
The new force $F'$ is given by $F' = \frac{\mu_0}{4\pi} \frac{2(2I_1)(2I_2)}{r} \cdot l = 4 \times \left( \frac{\mu_0}{4\pi} \frac{2 I_1 I_2}{r} \cdot l \right)$.
Substituting the initial force value: $F' = 4 \times (1 \times 10^{-3} \ N) = 4 \times 10^{-3} \ N$.

(B) We know that the force on the electron is $F = qE$ and from Newton's second law,$F = ma$.
Equating these,we get $ma = qE$,which implies the acceleration $a = \frac{qE}{m}$.
Using the third equation of motion,$v^2 - u^2 = 2aS$,where $u = 0$ (starts from rest) and $S = L$:
$v^2 - 0^2 = 2 \left( \frac{qE}{m} \right) L$
$v^2 = \frac{2qEL}{m}$
$v = \sqrt{\frac{2qEL}{m}}$

(A) To find the potential difference $(V_A - V_B)$, we apply Kirchhoff's Voltage Law $(KVL)$ starting from point $A$ to point $B$ along the circuit path.
Starting from $A$, we move through the resistor $(12 \, \Omega)$, the inductor $(6 \, H)$, and the battery $(12 \, V)$ to reach $B$.
The current $I = 2 \, A$ flows from $A$ to $B$.
The potential drop across the resistor is $V_R = I \cdot R = 2 \, A \times 12 \, \Omega = 24 \, V$.
The potential drop across the inductor is $V_L = L \frac{dI}{dt} = 6 \, H \times (-1 \, A/s) = -6 \, V$.
The potential drop across the battery is $12 \, V$ (moving from positive to negative terminal).
Applying $KVL$: $V_A - I \cdot R - L \frac{dI}{dt} - 12 = V_B$.
Rearranging gives: $V_A - V_B = I \cdot R + L \frac{dI}{dt} + 12$.
Substituting the values: $V_A - V_B = (2)(12) + 6(-1) + 12$.
$V_A - V_B = 24 - 6 + 12 = 30 \, V$.

(C) When two conductors are connected by a thin conducting wire,charge flows until they reach the same potential $V$.
Since $V_1 = V_2$,we have $\frac{k q_1}{R} = \frac{k q_2}{r}$,which implies $\frac{q_1}{q_2} = \frac{R}{r}$.
The surface charge density $\sigma$ is defined as $\sigma = \frac{q}{4 \pi R^2}$.
Therefore,the ratio of the surface charge densities is:
$\frac{\sigma_1}{\sigma_2} = \frac{q_1 / (4 \pi R^2)}{q_2 / (4 \pi r^2)} = \left( \frac{q_1}{q_2} \right) \left( \frac{r^2}{R^2} \right)$.
Substituting $\frac{q_1}{q_2} = \frac{R}{r}$,we get:
$\frac{\sigma_1}{\sigma_2} = \left( \frac{R}{r} \right) \left( \frac{r^2}{R^2} \right) = \frac{r}{R}$.
Thus,the ratio is $r: R$.

(B) According to Kirchhoff's first law (junction rule), the algebraic sum of currents meeting at a junction is zero. Let the current $I$ in conductor $PQ$ be directed away from point $P$.
Sum of incoming currents = Sum of outgoing currents
$5 \,A + 4 \,A = 5 \,A + 3 \,A + I$
$9 \,A = 8 \,A + I$
$I = 9 \,A - 8 \,A = 1 \,A$
Since the result is positive, the assumed direction (away from $P$) is correct. Therefore, $1 \,A$ current flows from $P$ to $Q$.

(B) Let the total current in the circuit be $I$.
Given that the current through the galvanometer $(I_g)$ is $4 \%$ of $I$,so $I_g = 0.04 I$.
The current through the shunt resistance $(I_s)$ is $I - I_g = I - 0.04 I = 0.96 I$.
Since the galvanometer and the shunt resistance are in parallel,the potential difference across them is the same: $I_g G = I_s S$.
Substituting the values: $(0.04 I) G = (0.96 I) S$.
Solving for $S$: $S = \frac{0.04 I G}{0.96 I} = \frac{4}{96} G = \frac{G}{24}$.
Therefore,the shunt resistance is $\frac{G}{24}$.

(A) Let the total current be $I$ and the current through the galvanometer be $I_g$.
The resistance of the galvanometer is $G$ and the shunt resistance is $S = 0.1 G$.
According to the current divider rule,the current through the galvanometer is given by:
$I_g = I \left( \frac{S}{S + G} \right)$
Substituting the value of $S$:
$I_g = I \left( \frac{0.1 G}{0.1 G + G} \right)$
$I_g = I \left( \frac{0.1 G}{1.1 G} \right)$
$I_g = I \left( \frac{1}{11} \right)$
Therefore,the part of the total current that flows through the galvanometer is $\frac{1}{11} I$.

(B) Given: Resistance of galvanometer $G = 20 \Omega$.
Full scale deflection current $I_g$: Since $5$ divisions correspond to $1 \text{ mA}$,$50$ divisions correspond to $I_g = (1 \text{ mA} / 5) \times 50 = 10 \text{ mA} = 0.01 \text{ A}$.
Required voltage range $V = 25 \text{ V}$.
To convert a galvanometer into a voltmeter,a high resistance $R$ must be connected in series with it.
The formula is $V = I_g(R + G)$.
Rearranging for $R$: $R = (V / I_g) - G$.
Substituting the values: $R = (25 / 0.01) - 20 = 2500 - 20 = 2480 \Omega$.
Thus,we must connect $2480 \Omega$ in series.

(D) Let the total current be $I$ and the current through the galvanometer be $I_g$.
According to the problem,$I_g = 1\% \text{ of } I = \frac{I}{100}$.
The current through the shunt resistance $S$ is $I_s = I - I_g = I - \frac{I}{100} = \frac{99I}{100}$.
Since the galvanometer and the shunt are in parallel,the potential difference across them is equal:
$I_g \times G = I_s \times S$.
Substituting the values:
$\frac{I}{100} \times G = \frac{99I}{100} \times S$.
Solving for $S$:
$S = \frac{G}{99}$.

(A) The galvanometer has a resistance $G$ and a full-scale deflection voltage $V_g$. The full-scale current $I_g$ is given by $I_g = \frac{V_g}{G}$.
To convert the galvanometer into a voltmeter of range $V$,a series resistance $R$ must be connected in series with the galvanometer.
The total resistance of the circuit becomes $R + G$.
According to Ohm's law,for the new range $V$,the current $I_g$ remains the same:
$V = I_g(R + G)$
Substituting $I_g = \frac{V_g}{G}$ into the equation:
$V = \left(\frac{V_g}{G}\right)(R + G)$
$\frac{V}{V_g} = \frac{R+G}{G}$
$\frac{V}{V_g} = \frac{R}{G} + 1$
$\frac{R}{G} = \frac{V}{V_g} - 1$
$R = G\left(\frac{V}{V_g} - 1\right)$

(A) In a potentiometer,the balancing length $l$ is directly proportional to the e.m.f. of the cell,i.e.,$E \propto l$ or $E = kl$,where $k$ is the potential gradient.
When cells are connected in series with the same polarity,the effective e.m.f. is $E_1 + E_2 = k l_1$.
Given $l_1 = 64 \ cm$,so $E_1 + E_2 = 64k$ --- (Equation $1$).
When the polarity of $E_2$ is reversed,the effective e.m.f. is $E_1 - E_2 = k l_2$.
Given $l_2 = 32 \ cm$,so $E_1 - E_2 = 32k$ --- (Equation $2$).
Dividing Equation $1$ by Equation $2$:
$\frac{E_1 + E_2}{E_1 - E_2} = \frac{64k}{32k} = \frac{2}{1}$.
Applying componendo and dividendo:
$\frac{(E_1 + E_2) + (E_1 - E_2)}{(E_1 + E_2) - (E_1 - E_2)} = \frac{2 + 1}{2 - 1}$.
$\frac{2E_1}{2E_2} = \frac{3}{1}$.
Therefore,$\frac{E_1}{E_2} = 3: 1$.

(A) The potential gradient $K$ of the potentiometer wire is given by:
$K = \frac{V_{wire}}{L} = \frac{E_{driving} \times R}{(R + r) L}$
Given $E_{driving} = 5 \, V$, $r = 4 \, \Omega$, $L = 5 \, m$, and $R = 16 \, \Omega$:
$K = \frac{5 \times 16}{(16 + 4) \times 5} = \frac{80}{100} = 0.8 \, V/m$
Case $1$: Cells assist each other $(E_{net} = E_1 + E_2)$
$E_1 + E_2 = K l_1$
$1.3 + 1.1 = 0.8 \times l_1$
$2.4 = 0.8 \times l_1 \implies l_1 = 3 \, m$
Case $2$: Cells oppose each other $(E_{net} = E_1 - E_2)$
$E_1 - E_2 = K l_2$
$1.3 - 1.1 = 0.8 \times l_2$
$0.2 = 0.8 \times l_2 \implies l_2 = 0.25 \, m$
Thus, the balancing lengths are $3 \, m$ and $0.25 \, m$.

(C) Given,the length of the wire at which the cell balances is $l = 300 \ cm = 3 \ m$.
Resistance per unit length is $2 \ \Omega/m$.
Total resistance of the wire segment $R = 3 \ m \times 2 \ \Omega/m = 6 \ \Omega$.
Since the potentiometer is balanced,the potential difference across the wire segment is equal to the e.m.f. of the cell,which is $1.5 \ V$.
Using Ohm's law,$V = IR$,we have:
$I = V / R = 1.5 \ V / 6 \ \Omega = 0.25 \ A$.
Converting to milliamperes,$I = 0.25 \times 1000 \ mA = 250 \ mA$.

(B) The total resistance of the circuit is $R_{total} = R_{wire} + R_{series} + r = 5 \, \Omega + 992 \, \Omega + 3 \, \Omega = 1000 \, \Omega$.
The current flowing through the potentiometer wire is $I = \frac{E}{R_{total}} = \frac{4 \, V}{1000 \, \Omega} = 0.004 \, A$.
The potential drop across the entire $4 \, m$ wire is $V_{wire} = I \times R_{wire} = 0.004 \, A \times 5 \, \Omega = 0.02 \, V$.
The potential gradient (potential drop per unit length) is $k = \frac{V_{wire}}{L} = \frac{0.02 \, V}{4 \, m} = 0.005 \, V/m$.
The e.m.f. balanced by a length of $0.75 \, m$ is $E' = k \times l = 0.005 \, V/m \times 0.75 \, m = 0.00375 \, V$.
Converting to millivolts, $E' = 0.00375 \times 1000 \, mV = 3.75 \, mV$.

(D) Let $E_1$ and $E_2$ be the e.m.f.s of the two cells.
When cells are in series,the total e.m.f. is $E_1 + E_2$,and the balancing length is $l_1 = 8 \ m$. So,$E_1 + E_2 = k l_1 = 8k$,where $k$ is the potential gradient.
When cells are in opposition,the total e.m.f. is $E_1 - E_2$,and the balancing length is $l_2 = 4 \ m$. So,$E_1 - E_2 = k l_2 = 4k$.
Dividing the two equations: $\frac{E_1 + E_2}{E_1 - E_2} = \frac{8k}{4k} = 2$.
Applying componendo and dividendo: $\frac{(E_1 + E_2) + (E_1 - E_2)}{(E_1 + E_2) - (E_1 - E_2)} = \frac{2 + 1}{2 - 1}$.
This simplifies to $\frac{2E_1}{2E_2} = \frac{3}{1}$.
Therefore,$\frac{E_1}{E_2} = 3:1$.

(C) Let $R_{G}$ be the resistance of the galvanometer and $I_{g}$ be the full-scale deflection current.
The original voltage range of the galvanometer is $V_{g} = I_{g}R_{G}$.
When a series resistance $R_{s} = nR_{G}$ is added to convert it into a voltmeter,the new voltage range $V$ is given by:
$V = I_{g}(R_{s} + R_{G})$
Substitute $R_{s} = nR_{G}$ into the equation:
$V = I_{g}(nR_{G} + R_{G})$
$V = I_{g}R_{G}(n + 1)$
Since $V_{g} = I_{g}R_{G}$,we have:
$V = V_{g}(n + 1)$
Therefore,the voltmeter is now capable of measuring $(n + 1)$ times the original voltage range of the $MCG$.

(D) According to Kirchhoff's Current Law $(KCL)$,the sum of currents entering a junction is equal to the sum of currents leaving the junction.
Let the current in conductor '$OP$' be $x$ and assume it is flowing away from point '$O$' (from $O$ to $P$).
Sum of currents entering the junction '$O$' = $10 \ A + 2.5 \ A + 5 \ A = 17.5 \ A$.
Sum of currents leaving the junction '$O$' = $6 \ A + x$.
Applying $KCL$: $17.5 \ A = 6 \ A + x$.
$x = 17.5 \ A - 6 \ A = 11.5 \ A$.
Since the result is positive,our assumption that the current flows from $O$ to $P$ is correct.
Therefore,the magnitude is $11.5 \ A$ and the direction is from $O$ to $P$.

(D) The e.m.f. of the cell is $E = 3 \ V$.
The terminal voltage across the voltmeter is $V = 2.5 \ V$.
The resistance of the voltmeter is $R = 150 \ \Omega$.
The internal resistance $r$ of the cell is given by the formula:
$r = \left[ \frac{E}{V} - 1 \right] R$
Substituting the given values:
$r = \left[ \frac{3}{2.5} - 1 \right] 150$
$r = [1.2 - 1] \times 150$
$r = 0.2 \times 150$
$r = 30 \ \Omega$
Therefore,the internal resistance of the cell is $30 \ \Omega$.

(C) The wire of resistance $R_w = 20 \, \Omega$ and the external resistor $R = 10 \, \Omega$ are connected in series.
Total resistance of the circuit is $R_{net} = R_w + R = 20 \, \Omega + 10 \, \Omega = 30 \, \Omega$.
The current $I$ flowing through the circuit is given by $I = \frac{E}{R_{net}} = \frac{3 \, V}{30 \, \Omega} = 0.1 \, A$.
The potential drop across the wire is $V_w = I \times R_w = 0.1 \, A \times 20 \, \Omega = 2 \, V$.
The potential gradient is defined as the potential drop per unit length of the wire: $\text{Potential gradient} = \frac{V_w}{L} = \frac{2 \, V}{10 \, m} = 0.2 \, V/m$.