MHT CET 2022 Physics Question Paper with Answer and Solution

(A) The thermal resistance $R$ of a rod is given by the formula $R = \frac{l}{KA}$,where $l$ is the length,$K$ is the thermal conductivity,and $A$ is the area of cross-section.
Given that the area of cross-section is the same for both rods,$A_A = A_B = A$.
Given the ratio of thermal conductivities is $\frac{K_A}{K_B} = \frac{3}{2}$.
Since the thermal resistances are equal,$R_A = R_B$.
Substituting the formula,we get $\frac{l_A}{K_A A} = \frac{l_B}{K_B A}$.
Simplifying this,we get $\frac{l_A}{l_B} = \frac{K_A}{K_B}$.
Substituting the given ratio,$\frac{l_A}{l_B} = \frac{3}{2}$.

(A) The rate of heat conduction is given by the formula: $\frac{Q}{t} = \frac{k A (T_1 - T_2)}{l}$.
Since the volume of the material remains constant during melting and reshaping,we have $A_1 l_1 = A_2 l_2$.
Given that the new length $l_2 = 4 l_1$,we substitute this into the volume equation: $A_1 l_1 = A_2 (4 l_1)$,which gives $A_2 = \frac{A_1}{4}$.
For the new rod,the heat conducted in time $t$ is $Q_2 = \frac{k A_2 (T_1 - T_2) t}{l_2}$.
Substituting $A_2 = \frac{A_1}{4}$ and $l_2 = 4 l_1$ into the equation for $Q_2$:
$Q_2 = \frac{k (A_1 / 4) (T_1 - T_2) t}{4 l_1} = \frac{1}{16} \frac{k A_1 (T_1 - T_2) t}{l_1}$.
Since $Q_1 = \frac{k A_1 (T_1 - T_2) t}{l_1}$,we get $Q_2 = \frac{1}{16} Q_1$.
Therefore,$\frac{Q_1}{Q_2} = 16$.

(B) The coefficient of thermal conductivity $(K)$ is an intrinsic property of the material of the rod.
It represents the ability of a material to conduct heat.
It does not depend on the physical dimensions such as the area of cross-section,length,or the mass of the rod.
Therefore,it only depends on the material of the rod.

(B) The rate of heat flow $\dot{Q}$ through a rod is given by the formula: $\dot{Q} = \frac{KA \Delta T}{l}$.
Here,$K$ is the thermal conductivity,$A$ is the cross-sectional area,$\Delta T$ is the temperature difference,and $l$ is the length of the rod.
Given that the length $l$ and temperature difference $\Delta T$ are the same for both rods $P$ and $Q$,the rate of heat flow is directly proportional to the product $KA$.
According to the problem,the rate of heat flow through rod $Q$ is three times that of rod $P$:
$(\dot{Q})_Q = 3 (\dot{Q})_P$
Substituting the formula:
$\frac{K_2 A_2 \Delta T}{l} = 3 \left( \frac{K_1 A_1 \Delta T}{l} \right)$
Since $l$ and $\Delta T$ are equal,they cancel out from both sides:
$K_2 A_2 = 3 K_1 A_1$ or $3 K_1 A_1 = K_2 A_2$.

(A) The heat current through the first slab is given by $\dot{Q}_1 = \frac{K_1 A (T_1 - T)}{d_1}$.
For the second slab,the heat current is $\dot{Q}_2 = \frac{K_2 A (T - T_2)}{d_2}$.
Since the slabs are connected in series,the same heat current flows through both,so $\dot{Q}_1 = \dot{Q}_2$.
Equating the two expressions: $\frac{K_1 A (T_1 - T)}{d_1} = \frac{K_2 A (T - T_2)}{d_2}$.
Canceling $A$ from both sides: $\frac{K_1 (T_1 - T)}{d_1} = \frac{K_2 (T - T_2)}{d_2}$.
Cross-multiplying gives: $K_1 d_2 (T_1 - T) = K_2 d_1 (T - T_2)$.
Expanding: $K_1 d_2 T_1 - K_1 d_2 T = K_2 d_1 T - K_2 d_1 T_2$.
Rearranging to solve for $T$: $K_1 d_2 T_1 + K_2 d_1 T_2 = T (K_1 d_2 + K_2 d_1)$.
Thus,$T = \frac{K_1 T_1 d_2 + K_2 T_2 d_1}{K_1 d_2 + K_2 d_1}$.

(D) The correct option is $D$.
Concept:
According to the Stefan-Boltzmann law,the emissive power $(E)$ of a body is given by $E = e \sigma T^4$,where $e$ is the emissivity,$\sigma$ is the Stefan-Boltzmann constant,and $T$ is the absolute temperature.
Given that the emissive power of both spheres is the same,we have $E_1 = E_2$.
Therefore,$e_1 \sigma T_1^4 = e_2 \sigma T_2^4$.
This implies $\frac{T_1^4}{T_2^4} = \frac{e_2}{e_1}$.
Given the ratio of emissivity $e_1 : e_2 = 1 : 4$,we have $\frac{e_2}{e_1} = \frac{4}{1} = 4$.
Thus,$\left(\frac{T_1}{T_2}\right)^4 = 4$.
Taking the fourth root on both sides,$\frac{T_1}{T_2} = (4)^{1/4} = (2^2)^{1/4} = 2^{1/2} = \sqrt{2}$.
So,the ratio $T_1 : T_2 = \sqrt{2} : 1$.

(D) According to Wien's displacement law,$\lambda_m T = b$,where $b$ is Wien's constant.
If $v_m$ is the frequency corresponding to the wavelength $\lambda_m$,then $\lambda_m = \frac{c}{v_m}$.
Substituting this into the displacement law,we get $\left(\frac{c}{v_m}\right) T = b$.
Rearranging the terms,we find $v_m = \left(\frac{c}{b}\right) T$.
Since $c$ (speed of light) and $b$ (Wien's constant) are constants,$v_m \propto T$.
This represents a linear relationship passing through the origin,which corresponds to the straight line labeled $B$ in the graph.
Therefore,option $D$ is the correct answer.

(D) According to Wien's displacement law,the product of the wavelength at which the radiation energy density is maximum $(\lambda_m)$ and the absolute temperature $(T)$ is a constant.
$\lambda_m T = \text{constant}$
Given that at temperature $T$,the maximum is at $\lambda_0$,we have:
$\lambda_0 T = \lambda' T'$
Here,$T' = 2T$. Substituting this into the equation:
$\lambda_0 T = \lambda' (2T)$
$\lambda' = \frac{\lambda_0 T}{2T} = \frac{\lambda_0}{2}$
Therefore,at temperature $2T$,the maximum energy density occurs at $\frac{\lambda_0}{2}$.

(C) According to Newton's law of cooling,the rate of cooling is given by: $\frac{dT}{dt} = -k(T_{avg} - T_0)$,where $T_{avg} = \frac{T_1 + T_2}{2}$ and $T_0$ is the surrounding temperature.
Case $(1)$: Cooling from $75^{\circ} C$ to $65^{\circ} C$ in $2 \text{ min}$.
$T_{avg} = \frac{75 + 65}{2} = 70^{\circ} C$.
Rate of cooling = $\frac{75 - 65}{2} = \frac{10}{2} = 5^{\circ} C/\text{min}$.
So,$5 = k(70 - 30) = k(40) \Rightarrow k = \frac{5}{40} = \frac{1}{8} \text{ min}^{-1}$.
Case $(2)$: Cooling from $55^{\circ} C$ to $45^{\circ} C$ in $t \text{ min}$.
$T_{avg} = \frac{55 + 45}{2} = 50^{\circ} C$.
Rate of cooling = $\frac{55 - 45}{t} = \frac{10}{t} ^{\circ} C/\text{min}$.
Using the law: $\frac{10}{t} = k(50 - 30) = k(20)$.
Substituting $k = \frac{1}{8}$:
$\frac{10}{t} = \frac{1}{8} \times 20 = 2.5$.
$t = \frac{10}{2.5} = 4 \text{ min}$.

(C) According to Wien's displacement law,the wavelength corresponding to maximum spectral intensity $\lambda_m$ is inversely proportional to the absolute temperature $T$:
$\lambda_m \propto \frac{1}{T}$
Since the frequency $f$ is related to wavelength $\lambda$ by $f = \frac{c}{\lambda}$,where $c$ is the speed of light,we have $\lambda \propto \frac{1}{f}$.
Substituting this into Wien's law:
$\frac{1}{f} \propto \frac{1}{T} \Rightarrow f \propto T$
Therefore,if the temperature $T$ is doubled,the frequency $f$ at which the spectral intensity becomes maximum will also be doubled.

(D) According to the Stefan-Boltzmann law,the rate of radiation $E$ from a black sphere of radius $R$ at temperature $T$ is given by $E = \sigma A T^4$,where $A = 4\pi R^2$ is the surface area of the sphere.
Thus,$E \propto R^2 T^4$.
Let the initial rate be $E_1 = E$ with radius $R_1 = R$ and temperature $T_1 = T$.
The new rate $E_2$ has radius $R_2 = \frac{R}{3}$ and temperature $T_2 = 3T$.
Taking the ratio: $\frac{E_2}{E_1} = \left(\frac{R_2}{R_1}\right)^2 \left(\frac{T_2}{T_1}\right)^4$.
Substituting the values: $\frac{E_2}{E} = \left(\frac{R/3}{R}\right)^2 \left(\frac{3T}{T}\right)^4 = \left(\frac{1}{3}\right)^2 (3)^4 = \frac{1}{9} \times 81 = 9$.
Therefore,$E_2 = 9E$.

(D) The rate of heat loss by radiation is given by Stefan-Boltzmann Law: $dQ/dt = e \sigma A (T^4 - T_0^4)$.
Since the spheres are at the same temperature $T$ and in the same environment $T_0$,the rate of heat loss is proportional to the surface area $A = 4 \pi R^2$.
Thus,$dQ/dt \propto R^2$.
The rate of cooling is defined as $dT/dt = (dQ/dt) / (mc)$,where $m$ is the mass and $c$ is the specific heat capacity.
Mass $m = \rho V = \rho (4/3 \pi R^3)$,where $\rho$ is the density.
Therefore,the rate of cooling $dT/dt \propto R^2 / R^3 = 1/R$.
The ratio of the initial rates of cooling is $(dT/dt)_1 / (dT/dt)_2 = R_2 / R_1$.

(A) According to Stefan-Boltzmann Law,the total emissive power $E$ of a black body is given by $E = \sigma T^4$,where $\sigma$ is the Stefan-Boltzmann constant and $T$ is the absolute temperature.
According to Wien's Displacement Law,the wavelength $\lambda_m$ corresponding to maximum intensity is inversely proportional to the absolute temperature $T$,i.e.,$T = \frac{b}{\lambda_m}$,where $b$ is Wien's constant.
Substituting $T$ in the expression for $E$,we get $E \propto (\frac{1}{\lambda_m})^4 = \frac{1}{\lambda_m^4}$.
Note that the emissive power $E$ of a black body depends only on its temperature,not on its surface area or radius.
Given $\lambda_x = 200 \ nm$,$\lambda_y = 300 \ nm$,and $\lambda_z = 400 \ nm$.
Since $E \propto \frac{1}{\lambda_m^4}$,we have:
$E_x \propto \frac{1}{200^4}$
$E_y \propto \frac{1}{300^4}$
$E_z \propto \frac{1}{400^4}$
Comparing the values,since $200 < 300 < 400$,it follows that $\frac{1}{200^4} > \frac{1}{300^4} > \frac{1}{400^4}$.
Therefore,$E_x > E_y > E_z$,which means $E_x$ is the maximum.

(C) According to Stefan-Boltzmann law,the emissive power $E$ of a black body is directly proportional to the fourth power of its absolute temperature $T$ (in Kelvin).
$E = \sigma T^4$
Given:
Initial temperature $T_1 = 273^{\circ} C = 273 + 273 = 546 \ K$.
Final temperature $T_2 = 0^{\circ} C = 0 + 273 = 273 \ K$.
Initial emissive power $E_1 = R$.
Using the ratio formula:
$\frac{E_2}{E_1} = \left( \frac{T_2}{T_1} \right)^4$
$\frac{E_2}{R} = \left( \frac{273}{546} \right)^4$
$\frac{E_2}{R} = \left( \frac{1}{2} \right)^4 = \frac{1}{16}$
$E_2 = \frac{R}{16}$.

(C) According to Wien's displacement law,the wavelength $\lambda_m$ corresponding to maximum energy emission is inversely proportional to the absolute temperature $T$:
$\lambda_m T = b$ (where $b$ is Wien's constant).
Therefore,$\lambda_{mA} T_A = \lambda_{mB} T_B$.
Given that $T_A = 3 T_B$,we substitute this into the equation:
$\lambda_{mA} (3 T_B) = \lambda_{mB} T_B \implies \lambda_{mB} = 3 \lambda_{mA}$.
We are also given the difference in wavelengths:
$\lambda_{mB} - \lambda_{mA} = 4 \mu m$.
Substituting $\lambda_{mB} = 3 \lambda_{mA}$ into the difference equation:
$3 \lambda_{mA} - \lambda_{mA} = 4 \mu m \implies 2 \lambda_{mA} = 4 \mu m \implies \lambda_{mA} = 2 \mu m$.
Now,calculate $\lambda_{mB}$:
$\lambda_{mB} = 3 \lambda_{mA} = 3(2 \mu m) = 6 \mu m$.

(C) Concept: According to the Stefan-Boltzmann law,the total radiant energy $E$ emitted by a black body of surface area $A$ at temperature $T$ is given by $E = \sigma A T^4$,where $\sigma$ is the Stefan-Boltzmann constant.
For a spherical star of radius $R$,the surface area is $A = 4 \pi R^2$.
Thus,$E = \sigma (4 \pi R^2) T^4$.
Given for star $(P)$: Radius $= R$,Temperature $= T$.
So,$E_P = \sigma (4 \pi R^2) T^4$.
Given for star $(Q)$: Radius $= 8R$,Temperature $= T/4$.
So,$E_Q = \sigma (4 \pi (8R)^2) (T/4)^4$.
Calculating the ratio:
$\frac{E_P}{E_Q} = \frac{\sigma (4 \pi R^2) T^4}{\sigma (4 \pi (64 R^2)) (T^4 / 256)} = \frac{T^4}{64 R^2 \cdot (T^4 / 256)} = \frac{256}{64} = 4$.
Wait,re-evaluating: $\frac{E_P}{E_Q} = \frac{R^2 T^4}{(8R)^2 (T/4)^4} = \frac{R^2 T^4}{64 R^2 (T^4 / 256)} = \frac{256}{64} = 4$.
The ratio of radiant energy emitted by $(P)$ to that by $(Q)$ is $4:1$.

(D) black body is an idealized physical body that absorbs all incident electromagnetic radiation.
According to Planck's law of black body radiation,the intensity of radiation emitted by a black body varies with wavelength.
The emission spectrum is continuous,meaning it emits radiation across all wavelengths.
However,the intensity is not constant for all wavelengths; it follows a specific distribution curve (Planck's distribution) that depends on the temperature of the body.
Therefore,the statement that intensity is the same for all wavelengths is incorrect.

(B) According to Wien's displacement law,the temperature $T$ is inversely proportional to the wavelength of maximum emission: $T \propto \frac{1}{\lambda_{\max }}$.
From Stefan-Boltzmann law,the power radiated $P$ by a blackbody is given by $P = \sigma A T^4$,where $A$ is the surface area and $\sigma$ is the Stefan-Boltzmann constant.
Since $A = \pi r^2$ for a disc (assuming radiation from both sides,$A = 2\pi r^2$),we have $P \propto r^2 T^4$.
Substituting $T \propto \frac{1}{\lambda_{\max }}$,we get $P \propto \frac{r^2}{\lambda_{\max }^4}$.
Given radii $r_x = 2 \ m, r_y = 2 \ m, r_z = 6 \ m$ and wavelengths $\lambda_x = 3 \ \mu m, \lambda_y = 4 \ \mu m, \lambda_z = 5 \ \mu m$:
$P_x \propto \frac{2^2}{3^4} = \frac{4}{81} \approx 0.049$
$P_y \propto \frac{2^2}{4^4} = \frac{4}{256} = 0.0156$
$P_z \propto \frac{6^2}{5^4} = \frac{36}{625} = 0.0576$
Comparing the values,$P_z > P_x > P_y$. Since the question asks for the maximum power,$P_z$ is the maximum.

(D) The correct option is $D$.
Concept: The length of a metallic rod at a temperature $T$ is given by $l = l_0(1 + \alpha \Delta T)$,where $l_0$ is the initial length,$\alpha$ is the coefficient of linear expansion,and $\Delta T$ is the change in temperature.
Let the lengths of the brass and steel rods at temperature $T$ be $l_b$ and $l_s$ respectively.
$l_b = l_1(1 + \alpha_1 \Delta T)$
$l_s = l_2(1 + \alpha_2 \Delta T)$
The difference in lengths is given by $l_s - l_b = l_2(1 + \alpha_2 \Delta T) - l_1(1 + \alpha_1 \Delta T)$.
$l_s - l_b = (l_2 - l_1) + (l_2 \alpha_2 - l_1 \alpha_1) \Delta T$.
Since the difference $(l_2 - l_1)$ is maintained constant at all temperatures,the term involving $\Delta T$ must be zero.
Therefore,$l_2 \alpha_2 - l_1 \alpha_1 = 0$,which implies $l_1 \alpha_1 = l_2 \alpha_2$.

(D) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.
Taking the derivative,the fractional change in time period is $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta l}{l}$.
Since $\frac{\Delta l}{l} = \alpha \Delta \theta$,we have $\frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta \theta$.
The change in temperature is $\Delta \theta = 20^{\circ} C - 15^{\circ} C = 5^{\circ} C$.
The total time in a day is $T = 24 \times 60 \times 60 = 86,400 \ s$.
The error in time per day is $\Delta T = \frac{1}{2} \alpha \Delta \theta \times T$.
Substituting the values: $\Delta T = \frac{1}{2} \times (1.2 \times 10^{-5}) \times 5 \times 86,400$.
$\Delta T = 0.6 \times 10^{-5} \times 5 \times 86,400 = 3 \times 10^{-5} \times 86,400 = 2.592 \ s \approx 2.6 \ s$.

(A) The change in volume $\Delta V$ is given by the formula: $\Delta V = V \times \gamma \times \Delta T$, where $\gamma$ is the coefficient of volume expansion.
Since $\gamma = 3\alpha$, the formula becomes: $\Delta V = V \times (3\alpha) \times \Delta T$.
Given: $V = 500 \,cm^3$, $\alpha = 12 \times 10^{-6} /^{\circ} C$, and $\Delta T = 100^{\circ} C - 0^{\circ} C = 100^{\circ} C$.
Substituting the values:
$\Delta V = 500 \times (3 \times 12 \times 10^{-6}) \times 100$
$\Delta V = 500 \times (36 \times 10^{-6}) \times 100$
$\Delta V = 500 \times 0.0036 = 1.8 \,cm^3$.

(C) The fractional change in volume is given by $\frac{\Delta V}{V} = \gamma \Delta T$,where $\gamma$ is the coefficient of volume expansion.
Given $\frac{\Delta V}{V} = 0.30 \% = 0.003$ and $\Delta T = 50^{\circ} C$.
Substituting these values: $0.003 = \gamma (50^{\circ} C) \Rightarrow \gamma = \frac{0.003}{50} = 6 \times 10^{-5} /^{\circ} C$.
We know that the coefficient of volume expansion $\gamma$ is related to the coefficient of linear expansion $\alpha$ by $\gamma = 3\alpha$.
Therefore,$\alpha = \frac{\gamma}{3} = \frac{6 \times 10^{-5}}{3} = 2 \times 10^{-5} /^{\circ} C$.

(D) For an ideal gas,the internal energy is a function of temperature only. When no work is done,the process is isochoric (constant volume),so $\Delta W = 0$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Since $\Delta W = 0$,the heat supplied is equal to the change in internal energy: $\Delta Q = \Delta U = n C_V \Delta T$.
For a monoatomic gas,the molar heat capacity at constant volume is $C_V = \frac{3}{2} R$.
Given $n = 2$ moles,$\Delta T = 373 \ K - 273 \ K = 100 \ K$.
Substituting the values: $\Delta Q = 2 \times \frac{3}{2} R \times 100 = 300 R$.

(D) The work done by a gas during expansion is given by the area under the $P-V$ curve.
For a given change in volume from $V_1$ to $V_2$,the pressure $P$ remains higher in an isobaric process compared to an isothermal or adiabatic process.
Since $W = \int_{V_1}^{V_2} P \, dV$,the area under the curve is largest for the isobaric process.
Therefore,the work done by the gas is maximum when the expansion is isobaric.

(C) For a monoatomic gas at constant pressure,the work done is given by $W = p \Delta V$.
Using the ideal gas equation $pV = nRT$,we have $W = nR \Delta T$.
At constant volume,the heat supplied is equal to the change in internal energy,given by $Q = n C_v \Delta T$.
For a monoatomic gas,the molar heat capacity at constant volume is $C_v = \frac{3}{2} R$.
Substituting this into the heat equation,we get $Q = n \left( \frac{3}{2} R \right) \Delta T$.
Since $W = nR \Delta T$,we can substitute $nR \Delta T$ with $W$.
Therefore,$Q = \frac{3}{2} W$.

(A) According to the first law of thermodynamics,the change in internal energy $\Delta U$ is given by the equation:
$\Delta U = \Delta Q - \Delta W$
Given:
Heat absorbed by the system,$\Delta Q = 68 \,J$
Work done by the system,$\Delta W = 30 \,J$
Substituting these values into the equation:
$\Delta U = 68 \,J - 30 \,J = 38 \,J$
Therefore,the change in internal energy of the system between $A$ and $C$ is $38 \,J$.

(A) For an adiabatic process,the relation between temperature and volume is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Therefore,the ratio of temperatures is $\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1}$.
For a monoatomic ideal gas,the adiabatic exponent is $\gamma = \frac{5}{3}$.
Thus,$\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$.
Since the gas is in a cylinder of constant cross-sectional area $A$,the volume is $V = A \times L$. Therefore,$V_1 = A L_1$ and $V_2 = A L_2$.
Substituting these into the temperature ratio equation:
$\frac{T_2}{T_1} = \left(\frac{A L_1}{A L_2}\right)^{2/3} = \left(\frac{L_1}{L_2}\right)^{2/3}$.

(B) The correct option is $B$.
Concept: In an adiabatic process,the system is thermally insulated from its surroundings,meaning no heat is transferred into or out of the system $(dQ = 0)$.
Reason: This process typically occurs rapidly,preventing sufficient time for heat exchange to take place between the system and the surroundings.

(C) The change in internal energy $(\Delta U)$ of an ideal gas depends only on the change in temperature and is given by the formula: $\Delta U = nC_v\Delta T$.
For an ideal gas, the molar heat capacity at constant volume is $C_v = \frac{R}{\gamma-1}$.
Substituting this into the internal energy formula, we get: $\Delta U = n \left(\frac{R}{\gamma-1}\right) (T_2 - T_1)$.
Therefore, the change in internal energy is $\frac{nR}{\gamma-1}(T_2 - T_1)$.

(D) For a sudden compression,the process is adiabatic.
For an adiabatic process,the relation between pressure and volume is $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$.
Given: $V_1 = V$,$V_2 = \frac{V}{4}$,$P_1 = P$,and $\gamma = \frac{5}{2}$.
Substituting these values into the adiabatic equation:
$P \cdot V^{\gamma} = P_2 \cdot \left(\frac{V}{4}\right)^{\gamma}$
$P_2 = P \cdot \left(\frac{V}{V/4}\right)^{\gamma}$
$P_2 = P \cdot (4)^{\gamma}$
$P_2 = P \cdot (4)^{5/2}$
$P_2 = P \cdot (2^2)^{5/2}$
$P_2 = P \cdot 2^5$
$P_2 = 32 P$.

(A) According to the first law of thermodynamics,the change in internal energy $\Delta U$ is given by $\Delta Q = \Delta U + W$,where $\Delta Q$ is the heat supplied to the system and $W$ is the work done by the system.
In an adiabatic process,there is no exchange of heat between the system and the surroundings,so $\Delta Q = 0$.
Substituting this into the first law equation: $0 = \Delta U + W$.
Therefore,$\Delta U = -W$.

(C) In an isochoric process,the volume of the gas remains constant.
According to Gay-Lussac's law,for a fixed mass of gas at constant volume,the pressure is directly proportional to its absolute temperature $(P \propto T)$.
Therefore,$\frac{P_1}{P_2} = \frac{T_1}{T_2}$.
First,convert the temperatures from Celsius to Kelvin:
$T_1 = 27 + 273 = 300 \ K$
$T_2 = 127 + 273 = 400 \ K$
Substituting these values into the ratio:
$\frac{P_1}{P_2} = \frac{300}{400} = \frac{3}{4}$.

(D) In an isothermal process,the temperature remains constant,so $\Delta T = 0$. However,heat exchange $Q$ is not necessarily zero; it is equal to the work done $(Q = W)$.
In an adiabatic process,the heat exchange $Q$ is zero.
Therefore,the statement 'In an isothermal process,$Q = 0$' is incorrect.

(A) i. Free expansions are adiabatic processes where there is no exchange of heat between the system and its environment,so $Q=0$.
ii. In a free expansion,the gas expands into a vacuum,meaning there is no external pressure to push against,so no work is done by the system,$W=0$.
iii. According to the first law of thermodynamics,$\Delta U = Q - W$. Since $Q=0$ and $W=0$,the change in internal energy $\Delta U = 0$.
iv. An example is the sudden rupture of a balloon or a tire,where gas rushes out into the surroundings without performing work on a piston or boundary.
v. $A$ free expansion is an uncontrolled,irreversible,and instantaneous change where the system is not in thermodynamic equilibrium during the process,and it cannot be represented as a continuous path on a $p-V$ diagram.

(B) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$.
Thus,$PV = \frac{m}{M}RT$,which implies $m = \frac{PVM}{RT}$.
For container $A$:
$m_A = \frac{PVM}{RT} \quad (1)$
For container $B$:
$m_B = \frac{(2P)(2V)M}{R(T/2)} = \frac{4PVM}{RT/2} = \frac{8PVM}{RT} \quad (2)$
Taking the ratio of $m_A$ to $m_B$:
$\frac{m_A}{m_B} = \frac{PVM/RT}{8PVM/RT} = \frac{1}{8}$.
Therefore,the ratio is $1: 8$.

(A) An isochoric process is a thermodynamic process that occurs at a constant volume.
Since the volume remains constant throughout the process,the change in volume is zero.
Therefore,the equation for an isochoric process is $\Delta V = 0$.

(B) For an adiabatic process,the relationship between pressure $p$ and volume $V$ is given by $p V^{\gamma} = \text{constant}$.
Let the initial pressure be $p$ and the initial volume be $V$.
The final volume is $V' = \frac{V}{8}$.
Using the adiabatic equation: $p V^{\gamma} = P' (V')^{\gamma}$.
Substituting the given values: $p V^{4/3} = P' \left(\frac{V}{8}\right)^{4/3}$.
$P' = p \left(\frac{V}{V/8}\right)^{4/3} = p (8)^{4/3}$.
Since $8 = 2^3$,we have $P' = p (2^3)^{4/3} = p (2^4) = 16p$.
Therefore,the new pressure is $16p$.

(B) For an adiabatic process,the relation between temperature and volume is $T V^{\gamma-1} = \text{constant}$.
For the adiabatic path $BC$,the points $B$ and $C$ lie on the adiabatic curve connecting the isothermals $T_1$ and $T_2$. Thus:
$T_1 V_b^{\gamma-1} = T_2 V_c^{\gamma-1}$
$\Rightarrow \left(\frac{V_b}{V_c}\right)^{\gamma-1} = \frac{T_2}{T_1} \quad ---(1)$
For the adiabatic path $AD$,the points $A$ and $D$ lie on the adiabatic curve connecting the isothermals $T_1$ and $T_2$. Thus:
$T_1 V_a^{\gamma-1} = T_2 V_d^{\gamma-1}$
$\Rightarrow \left(\frac{V_a}{V_d}\right)^{\gamma-1} = \frac{T_2}{T_1} \quad ---(2)$
Equating $(1)$ and $(2)$:
$\left(\frac{V_b}{V_c}\right)^{\gamma-1} = \left(\frac{V_a}{V_d}\right)^{\gamma-1}$
$\Rightarrow \frac{V_b}{V_c} = \frac{V_a}{V_d}$
Rearranging the terms,we get:
$\frac{V_b}{V_a} = \frac{V_c}{V_d}$
Therefore,option $(B)$ is correct.

(B) According to the first law of thermodynamics,$\Delta U = Q - W$.
$1$. In an adiabatic process,$Q = 0$ by definition,so it does not satisfy the condition.
$2$. In an isothermal process,the temperature remains constant,which implies $\Delta U = 0$ for an ideal gas,so it does not satisfy the condition.
$3$. In an isochoric process,the volume remains constant,which implies $W = P \Delta V = 0$,so it does not satisfy the condition.
$4$. In an isobaric process,the pressure remains constant. During expansion,the volume changes $(W \neq 0)$,the temperature changes $(\Delta U \neq 0)$,and heat is exchanged with the surroundings $(Q \neq 0)$. Therefore,all three quantities are non-zero.

(C) According to Boyle's Law, at constant temperature, $PV = \text{constant}$.
Let the initial pressure be $P_1 = P$ and the initial volume be $V_1 = V$.
If the pressure is decreased by $20 \%$, the new pressure $P_2$ is $P - 0.20P = 0.8P$.
Using the relation $P_1 V_1 = P_2 V_2$:
$P \cdot V = (0.8P) \cdot V_2$
$V_2 = \frac{PV}{0.8P} = \frac{V}{0.8} = 1.25V$.
The change in volume is $\Delta V = V_2 - V_1 = 1.25V - V = 0.25V$.
The percentage change in volume is $\frac{\Delta V}{V_1} \times 100 = \frac{0.25V}{V} \times 100 = 25 \%$.
Since the value is positive, the volume increases by $25 \%$.

(B) The work done by or on a gas is given by the formula $\Delta W = P \Delta V$,where $P$ is the pressure and $\Delta V$ is the change in volume.
For the work done to be zero,the change in volume $\Delta V$ must be zero.
This implies that the volume $V$ remains constant throughout the process.
$A$ thermodynamic process in which the volume remains constant is known as an isochoric process.
Therefore,in an isochoric process,the work done is zero.

(C) Since the compression is sudden,it is an adiabatic process.
For an adiabatic process,the relationship between temperature and volume is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Given: $T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$,$V_1 = V$,$V_2 = V/8$,and $\gamma = 5/3$.
Substituting the values:
$T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma-1}$
$T_2 = 300 \times \left( \frac{V}{V/8} \right)^{(5/3) - 1}$
$T_2 = 300 \times (8)^{2/3}$
$T_2 = 300 \times (2^3)^{2/3} = 300 \times 2^2 = 300 \times 4 = 1200 \ K$.

(A) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by the equation $T V^{\gamma-1} = \text{constant}$.
Therefore,for two different states $(P_1, V_1, T_1)$ and $(P_2, V_2, T_2)$,the relation holds as $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Thus,the correct option is $A$.

(D) For an ideal gas,the molecules are assumed to be point masses with no intermolecular forces of attraction or repulsion.
Since there are no intermolecular forces,there is no potential energy associated with the configuration of the molecules.
Therefore,the total internal energy of an ideal gas consists solely of the kinetic energy of its molecules due to their random motion.
Thus,the internal energy is only kinetic energy.

(A) The dimension of pressure $P$ is $[M^1 L^{-1} T^{-2}]$.
The dimension of volume $V$ is $[L^3]$.
Therefore,the dimension of the product $PV$ is $[M^1 L^{-1} T^{-2}] \times [L^3] = [M^1 L^2 T^{-2}]$.
Since the dimension of energy (work) is $[M^1 L^2 T^{-2}]$,the dimension of $PV$ is the same as energy.

(B) The given wave equation is $Y = Y_0 \sin(2 \pi n t - \frac{2 \pi x}{\lambda})$.
Comparing this with the standard form $Y = Y_0 \sin(\omega t - kx)$,we get angular frequency $\omega = 2 \pi n$.
The maximum particle velocity $v_{p, \text{max}} = Y_0 \omega = Y_0 (2 \pi n) = 2 \pi n Y_0$.
The wave velocity $v = n \lambda$.
According to the problem,the wave velocity is $(1/8)^{\text{th}}$ of the maximum particle velocity:
$v = \frac{1}{8} v_{p, \text{max}}$
$n \lambda = \frac{1}{8} (2 \pi n Y_0)$
$n \lambda = \frac{\pi n Y_0}{4}$
$\lambda = \frac{\pi Y_0}{4}$.

(B) For a string fixed at both ends,the $n^{th}$ overtone corresponds to the $(n+1)^{th}$ harmonic.
Here,the fifth overtone is the sixth harmonic $(n=6)$.
The length of the string $L = 2.4 \ m$.
The condition for the $n^{th}$ harmonic is $L = n \frac{\lambda}{2}$.
Substituting the values: $2.4 = 6 \times \frac{\lambda}{2}$.
This gives $\frac{\lambda}{2} = \frac{2.4}{6} = 0.4 \ m$.
Thus,$\lambda = 0.8 \ m$.
The distance between a node and the successive antinode is always $\frac{\lambda}{4}$.
Therefore,distance $= \frac{0.8 \ m}{4} = 0.2 \ m$.

(A) The general equation for a wave is $y = A \sin(\omega t)$,where $\omega = 2 \pi f$.
Given $y_1 = 0.35 \sin(316 t)$,the angular frequency is $\omega_1 = 316 \text{ rad/s}$. Thus,the frequency $f_1 = \frac{\omega_1}{2 \pi} = \frac{316}{2 \pi} \text{ Hz}$.
Given $y_2 = 0.35 \sin(310 t)$,the angular frequency is $\omega_2 = 310 \text{ rad/s}$. Thus,the frequency $f_2 = \frac{\omega_2}{2 \pi} = \frac{310}{2 \pi} \text{ Hz}$.
The number of beats produced per second is the beat frequency $f_b = |f_1 - f_2|$.
$f_b = \frac{316}{2 \pi} - \frac{310}{2 \pi} = \frac{6}{2 \pi} = \frac{3}{\pi} \text{ beats per second}$.

(D) Let the original frequency be $f$. The apparent frequency $f^{\prime}$ drops by $20 \%$,so $f^{\prime} = f - 0.2f = 0.8f = \frac{4}{5}f$.
Since the frequency decreases,the source is moving away from the stationary observer.
Using the Doppler effect formula for a source moving away from a stationary observer:
$f^{\prime} = f \left[ \frac{V}{V + V_s} \right]$
where $V = 350 \ m/s$ is the speed of sound and $V_s$ is the speed of the engine.
Substituting the values:
$\frac{4}{5}f = f \left[ \frac{350}{350 + V_s} \right]$
$\frac{4}{5} = \frac{350}{350 + V_s}$
$4(350 + V_s) = 5(350)$
$1400 + 4V_s = 1750$
$4V_s = 350$
$V_s = 87.5 \ m/s$.

(B) According to the Doppler effect,the frequency heard by an observer when the source is approaching is $n_{\text{before}} = \left(\frac{V}{V - v_c}\right) n$.
When the source is receding,the frequency heard is $n_{\text{after}} = \left(\frac{V}{V + v_c}\right) n$.
Given the ratio $\frac{n_{\text{before}}}{n_{\text{after}}} = \frac{11}{9}$.
Substituting the expressions: $\frac{\frac{V}{V - v_c} n}{\frac{V}{V + v_c} n} = \frac{V + v_c}{V - v_c} = \frac{11}{9}$.
Cross-multiplying: $9(V + v_c) = 11(V - v_c)$.
$9V + 9v_c = 11V - 11v_c$.
$20v_c = 2V$.
$v_c = \frac{2V}{20} = \frac{V}{10}$.

(C) Light is an electromagnetic wave in nature.
Electromagnetic waves do not require any material medium for their propagation.
Therefore,the statement that 'Light requires a material medium' is incorrect and is not a property of light.

(D) The absolute permittivity of a medium $(\varepsilon)$ is related to the permittivity of free space $(\varepsilon_0)$ and the relative permittivity $(\varepsilon_r)$ by the formula: $\varepsilon = \varepsilon_r \varepsilon_0$.
From this relation,the relative permittivity $(\varepsilon_r)$ is given by: $\varepsilon_r = \frac{\varepsilon}{\varepsilon_0}$.
Therefore,the correct option is $D$.

(A) The initial force between the two charges is given by Coulomb's Law: $F = k \frac{q_1 q_2}{l^2}$.
When one charge is doubled $(q_1' = 2q_1)$ and the distance is halved $(l' = l/2)$,the new force $F'$ is:
$F' = k \frac{(2q_1) q_2}{(l/2)^2}$
$F' = k \frac{2q_1 q_2}{l^2 / 4}$
$F' = 8 \left( k \frac{q_1 q_2}{l^2} \right)$
$F' = 8F$.
Thus,the magnitude of the force becomes $8$ times the original force,so $n = 8$.

(B) The initial electrostatic force between two identical balls with charge $Q$ at distance $r$ is given by Coulomb's Law: $F = \frac{kQ^2}{r^2}$.
When an identical uncharged ball touches one of the charged balls,the charge $Q$ is shared equally between them. Thus,the touched ball now has charge $Q/2$ and the third ball also acquires a charge $Q/2$.
The third ball is placed at the midpoint ($r/2$ from each). Let the charges be $q_1 = Q/2$ (touched ball),$q_2 = Q$ (untouched ball),and $q_3 = Q/2$ (third ball).
The force on the third ball due to the first ball is $F_1 = \frac{k(Q/2)(Q/2)}{(r/2)^2} = \frac{kQ^2/4}{r^2/4} = \frac{kQ^2}{r^2} = F$ (directed away from the first ball).
The force on the third ball due to the second ball is $F_2 = \frac{k(Q)(Q/2)}{(r/2)^2} = \frac{kQ^2/2}{r^2/4} = \frac{2kQ^2}{r^2} = 2F$ (directed away from the second ball).
Since the forces are in opposite directions,the net force is $F_{\text{net}} = |F_2 - F_1| = |2F - F| = F$.

(D) The maximum charge $Q_{\text{max}}$ that a spherical conductor of radius $R$ can hold in an external electric field $E$ before the air around it breaks down is given by the condition where the electric field at the surface due to the charge equals the breakdown field strength. However,in this context,the formula for the induced charge capacity is $Q_{\text{max}} = 4 \pi \varepsilon_0 R^2 E$.
Given:
Diameter $d = 6 \ mm$,so radius $R = 3 \ mm = 3 \times 10^{-3} \ m$.
Electric field $E = 2 \times 10^7 \ N/C$.
Constant $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \ N \cdot m^2/C^2$.
Substituting the values:
$Q_{\text{max}} = \frac{1}{9 \times 10^9} \times (3 \times 10^{-3})^2 \times (2 \times 10^7)$
$Q_{\text{max}} = \frac{1}{9 \times 10^9} \times (9 \times 10^{-6}) \times (2 \times 10^7)$
$Q_{\text{max}} = 10^{-15} \times 2 \times 10^7 = 2 \times 10^{-8} \ C$
$Q_{\text{max}} = 0.02 \times 10^{-6} \ C = 0.02 \ \mu C$.

(A) Concept: When a conductor is charged,the charge resides entirely on its outer surface due to the mutual repulsion between like charges.
Since both spheres have the same radius,their outer surface areas are identical.
Therefore,the maximum amount of charge that can be stored on the surface before dielectric breakdown of the surrounding medium occurs is the same for both.
Thus,both metallic spheres (solid or hollow) can be charged equally.

(B) According to Coulomb's law, the force between two point charges is given by $F = \frac{k Q q}{r^2}$.
Initially, $F_1 = \frac{k Q q}{r^2} = 100 \,N$.
When $Q$ is increased by $10 \%$, the new charge is $Q' = Q + 0.1 Q = 1.1 Q$.
When $q$ is decreased by $10 \%$, the new charge is $q' = q - 0.1 q = 0.9 q$.
The new force $F_2$ is given by $F_2 = \frac{k (1.1 Q) (0.9 q)}{r^2} = (1.1 \times 0.9) \frac{k Q q}{r^2}$.
$F_2 = 0.99 \times F_1 = 0.99 \times 100 \,N = 99 \,N$.
The change in force is $\Delta F = F_1 - F_2 = 100 \,N - 99 \,N = 1 \,N$.
Therefore, the force decreases by $1 \,N$.

(C) Let the side of the square be $a$. The distance between adjacent corners is $a$,and the distance between opposite corners (diagonal) is $\sqrt{2}a$.
According to Coulomb's Law,the force between two charges is $F = \frac{1}{4 \pi \varepsilon_0} \frac{q_i q_j}{r^2}$.
For charges $q_1$ and $q_2$ at adjacent corners,the distance is $r = a$. Thus,$F_{12} = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{a^2}$.
For charges $q_1$ and $q_3$ at opposite corners,the distance is $r = \sqrt{2}a$. Thus,$F_{13} = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_3}{(\sqrt{2}a)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_3}{2a^2}$.
Given $q_1 = q_2 = q_3 = q$,we have:
$\frac{F_{12}}{F_{13}} = \frac{\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{a^2}}{\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{2a^2}} = \frac{1}{1/2} = 2$.

(C) Let the charges be placed at positions $x=0$,$x=\frac{d}{2}$,and $x=d$ respectively.
To make the net force on the charge $q$ at $x=0$ equal to zero,the force exerted by $Q$ and the force exerted by $+4q$ must be equal in magnitude and opposite in direction.
Using Coulomb's law,the force exerted by $Q$ on $q$ is $F_Q = \frac{k q Q}{(d/2)^2}$ and the force exerted by $+4q$ on $q$ is $F_{+4q} = \frac{k q (4q)}{d^2}$.
For the net force to be zero,the sum of these forces must be zero:
$\frac{k q Q}{(d/2)^2} + \frac{k q (4q)}{d^2} = 0$
$\frac{k q Q}{d^2/4} + \frac{4 k q^2}{d^2} = 0$
$\frac{4 k q Q}{d^2} + \frac{4 k q^2}{d^2} = 0$
Dividing by $\frac{4 k q}{d^2}$ (assuming $q \neq 0$):
$Q + q = 0$
$Q = -q$

(A) The potential energy $U$ of an electric dipole in an external electric field is given by $U = -\overrightarrow{p} \cdot \overrightarrow{E} = -pE \cos \theta$.
Initially,the dipole is lying along the electric field,so $\theta_1 = 0^{\circ}$.
Initial potential energy $U_1 = -pE \cos 0^{\circ} = -pE(1) = -pE$.
Finally,the dipole is rotated by $90^{\circ}$,so $\theta_2 = 90^{\circ}$.
Final potential energy $U_2 = -pE \cos 90^{\circ} = -pE(0) = 0$.
The work done $W$ is equal to the change in potential energy: $W = U_2 - U_1$.
$W = 0 - (-pE) = pE$.

(D) The electric potential due to a point charge $q$ at a distance $r$ is given by $V = \frac{1}{4 \pi \epsilon_0} \frac{q}{r}$.
For the given dipole,the distance of point $P$ from charge $-q$ is $(x+a)$ and from charge $+q$ is $(x-a)$.
The potential due to $-q$ is $V_{-q} = \frac{1}{4 \pi \epsilon_0} \frac{(-q)}{(x+a)}$.
The potential due to $+q$ is $V_{+q} = \frac{1}{4 \pi \epsilon_0} \frac{(+q)}{(x-a)}$.
The total potential $V_P$ at point $P$ is the algebraic sum of potentials due to individual charges:
$V_P = V_{-q} + V_{+q} = \frac{q}{4 \pi \epsilon_0} \left[ \frac{1}{x-a} - \frac{1}{x+a} \right]$.
$V_P = \frac{q}{4 \pi \epsilon_0} \left[ \frac{(x+a) - (x-a)}{(x-a)(x+a)} \right] = \frac{q}{4 \pi \epsilon_0} \left[ \frac{2a}{x^2-a^2} \right]$.
$V_P = \frac{2aq}{4 \pi \epsilon_0(x^2-a^2)} = \frac{aq}{2 \pi \epsilon_0(x^2-a^2)}$.

(B) The electrostatic force acting on a charged particle in a uniform electric field $E$ is given by $F = qE$.
For an electron,the magnitude of charge is $e$,so the force is $F_e = eE$. The acceleration is $a_e = \frac{F_e}{m_e} = \frac{eE}{m_e}$.
For a proton,the magnitude of charge is also $e$,so the force is $F_p = eE$. The acceleration is $a_p = \frac{F_p}{m_p} = \frac{eE}{m_p}$.
Taking the ratio of the acceleration of the electron to the acceleration of the proton:
$\frac{a_e}{a_p} = \frac{eE / m_e}{eE / m_p} = \frac{m_p}{m_e}$.

(D) Electric field lines are imaginary lines representing the electric field.
$1$. They never intersect each other because at the point of intersection,there would be two directions for the electric field,which is impossible.
$2$. They do not pass through the interior of a conductor in electrostatic equilibrium because the electric field inside a conductor is zero.
$3$. They originate from positive charges and terminate on negative charges.
$4$. Electric field lines $CAN$ pass through insulators (dielectrics),as insulators do not have free charges to cancel the field.
Therefore,the statement that they do not pass through an insulator is incorrect.

(D) The electric field $E$ between two parallel plates separated by a distance $d$ with a potential difference $V$ is given by $E = \frac{V}{d}$.
The force $F$ acting on a particle of charge $q$ in this electric field is $F = qE = \frac{qV}{d}$.
According to Newton's second law of motion,the acceleration $a$ is given by $a = \frac{F}{m}$.
Substituting the value of $F$,we get $a = \frac{qV}{md}$.

(D) According to Gauss's law,the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_0}$.
Here,$q_{enclosed}$ is the total charge enclosed by the surface.
When the balloon is blown up,its size increases,but the total charge $q$ on the surface of the balloon remains constant.
Since the charge enclosed by the surface does not change,the total electric flux $\phi$ passing through the surface remains unchanged.

(C) According to Gauss's Law,the total electric flux $\phi_{T}$ through a closed surface is given by $\phi_{T} = \frac{q}{\epsilon_0}$,where $q$ is the net charge enclosed by the surface.
For the given hollow cylinder,the total flux is the sum of the flux through the two plane surfaces ($A$ and $C$) and the curved surface $(B)$.
Let $\phi_A$,$\phi_B$,and $\phi_C$ be the fluxes through surfaces $A$,$B$,and $C$ respectively.
Given $\phi_B = \phi$. Due to the symmetry of the cylinder,the flux through the two plane ends must be equal,i.e.,$\phi_A = \phi_C$.
Therefore,$\phi_A + \phi_B + \phi_C = \frac{q}{\epsilon_0}$.
Substituting the values,we get $2\phi_A + \phi = \frac{q}{\epsilon_0}$.
Solving for $\phi_A$,we get $2\phi_A = \frac{q}{\epsilon_0} - \phi$,which implies $\phi_A = \frac{1}{2}\left(\frac{q}{\epsilon_0} - \phi\right)$.

(A) According to Gauss's Law,the electric flux $\phi$ linked with a closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_0}$.
From the figure,the charges enclosed by the closed surface are $2.35 \ C$,$5 \ C$,$2 \ C$,and $-0.5 \ C$.
The total charge $q_{enclosed} = (2.35 + 5 + 2 - 0.5) \ C = 8.85 \ C$.
Given $\varepsilon_0 = 8.85 \times 10^{-12} \ C^2 N^{-1} m^{-2}$.
Substituting the values,we get $\phi = \frac{8.85 \ C}{8.85 \times 10^{-12} \ C^2 N^{-1} m^{-2}} = 10^{12} \ N m^2 C^{-1}$.

(A) According to Gauss's Law,the total electric flux $\phi_{total}$ through a closed surface enclosing a charge $Q$ is given by $\phi_{total} = \frac{Q}{\varepsilon_0}$.
Since the charge is placed at the center of the cube,the flux is distributed equally among all $6$ faces due to symmetry.
Therefore,the flux through one face is $\phi_{one} = \frac{\phi_{total}}{6} = \frac{Q}{6 \varepsilon_0}$.
The flux through two opposite faces is the sum of the flux through each of these two faces,which is $\phi_{two} = 2 \times \phi_{one} = 2 \times \frac{Q}{6 \varepsilon_0} = \frac{Q}{3 \varepsilon_0}$.

(C) According to Gauss's law,the electric flux $\phi$ through any closed surface is given by the formula:
$\phi = \frac{Q_{\text{enclosed}}}{\varepsilon_0}$
Here,$Q_{\text{enclosed}}$ is the total charge enclosed by the Gaussian surface and $\varepsilon_0$ is the permittivity of free space.
From this expression,it is clear that the electric flux depends only on the magnitude of the charge enclosed within the surface.
It does not depend on the shape or the size (radius $R$) of the Gaussian surface.
Therefore,if the radius of the Gaussian surface is doubled,the enclosed charge $Q$ remains the same,and consequently,the outward electric flux will remain the same.

(A) According to Gauss's Law,the electric flux $\Phi_E$ through a closed surface is given by $\Phi_E = \frac{q_{enclosed}}{\epsilon_0}$.
Here,$q_{enclosed}$ is the total charge enclosed by the Gaussian surface and $\epsilon_0$ is the permittivity of free space.
Since the point charge $q$ remains the same regardless of the radius of the spherical Gaussian surface,the enclosed charge $q_{enclosed} = q$ remains constant.
Therefore,the electric flux $\Phi_E = \frac{q}{\epsilon_0}$ remains unchanged when the radius of the surface is increased.

(B) The energy density $U$ in an electric field is given by the formula $U = \frac{1}{2} \varepsilon_0 E^2$,where $E$ is the electric field intensity.
For an isolated conducting sphere of radius $r$ carrying charge $Q$,the electric field just outside its surface is $E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2}$.
Substituting this into the energy density formula: $U = \frac{1}{2} \varepsilon_0 \left( \frac{Q}{4 \pi \varepsilon_0 r^2} \right)^2 = \frac{Q^2}{32 \pi^2 \varepsilon_0 r^4}$.
Since $Q$ is constant,we have $U \propto \frac{1}{r^4}$.
Given the radii $R_A = R$,$R_B = 2R$,and $R_C = 3R$,we find:
$U_A \propto \frac{1}{R^4}$,$U_B \propto \frac{1}{(2R)^4} = \frac{1}{16R^4}$,and $U_C \propto \frac{1}{(3R)^4} = \frac{1}{81R^4}$.
Comparing these values,it is clear that $U_A > U_B > U_C$.

(B) Statement $A$ is incomplete as stated in the original prompt,but generally,for a point outside a charged spherical shell $(r > R)$,the electric field $E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}$,which is inversely proportional to the square of the distance from the center. However,inside the shell $(r < R)$,the electric field is zero. Since the statement does not specify the region,it is technically ambiguous or incorrect in a general context.
Statement $B$ is correct. The electric potential $V$ due to a point charge $q$ at a distance $r$ is given by $V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$,which shows $V \propto \frac{1}{r}$.
Therefore,only statement $B$ is universally correct as stated.

(C) Let the corners of the square be $A, B, C, D$ in order,with charges $+q$ at $A$ and $B$,and $-q$ at $C$ and $D$. The side length is $2r$. Point $P$ is the midpoint of side $CD$.
Thus,the distances are: $DP = PC = r$.
The distances from $A$ and $B$ to $P$ are: $AP = BP = \sqrt{AD^2 + DP^2} = \sqrt{(2r)^2 + r^2} = \sqrt{5r^2} = r\sqrt{5}$.
The electric potential $V_P$ at point $P$ is the sum of potentials due to all four charges:
$V_P = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q}{AP} + \frac{q}{BP} + \frac{-q}{CP} + \frac{-q}{DP} \right]$
$V_P = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q}{r\sqrt{5}} + \frac{q}{r\sqrt{5}} - \frac{q}{r} - \frac{q}{r} \right]$
$V_P = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{2q}{r\sqrt{5}} - \frac{2q}{r} \right]$
$V_P = \frac{1}{4 \pi \varepsilon_0} \frac{2q}{r} \left[ \frac{1}{\sqrt{5}} - 1 \right]$

(C) The relationship between electric field $\vec{E}$ and electric potential $V$ is given by the equation: $dV = -\vec{E} \cdot d\vec{l}$.
Given $\vec{E} = 20x^2 \hat{i}$ and $d\vec{l} = dx \hat{i}$.
Substituting these into the equation,we get: $dV = -(20x^2 \hat{i}) \cdot (dx \hat{i}) = -20x^2 dx$.
To find the potential difference $V_A - V_0$,we integrate from $x=0$ to $x=3 \ m$:
$V_A - V_0 = \int_{V_0}^{V_A} dV = \int_{0}^{3} -20x^2 dx$.
$V_A - V_0 = -20 \left[ \frac{x^3}{3} \right]_{0}^{3}$.
$V_A - V_0 = -20 \left( \frac{3^3}{3} - 0 \right) = -20 \left( \frac{27}{3} \right) = -20 \times 9 = -180 \ V$.

(D) The total electrostatic potential energy $U$ of a system of point charges is the sum of the potential energies of all distinct pairs of charges.
For the given configuration,the pairs are $(Q, +q)$,$(+q, +q)$,and $(Q, +q)$ with distances $l$,$l$,and $\sqrt{2}l$ respectively.
The total potential energy is given by:
$U = \frac{kQq}{l} + \frac{kq^2}{l} + \frac{kQq}{\sqrt{2}l} = 0$
Dividing by $k/l$ (assuming $k \neq 0$ and $l \neq 0$):
$Qq + q^2 + \frac{Qq}{\sqrt{2}} = 0$
$Qq(1 + \frac{1}{\sqrt{2}}) = -q^2$
$Qq(\frac{\sqrt{2}+1}{\sqrt{2}}) = -q^2$
$Q = -q^2 \cdot \frac{\sqrt{2}}{q(\sqrt{2}+1)}$
$Q = -\frac{\sqrt{2}q}{\sqrt{2}+1}$

(B) The force per unit length between two long parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $d$ is given by the formula: $f = \frac{F}{l} = \frac{\mu_0 I_1 I_2}{2 \pi d}$.
Initially,the force is $F = \frac{\mu_0 I_1 I_2 l}{2 \pi d}$.
According to the problem,the new current $I_1' = 3 I_1$,the new distance $d' = 2 d$,and the current $I_2$ remains the same (the direction change only affects the nature of the force,not its magnitude).
The new force $F'$ is given by: $F' = \frac{\mu_0 (3 I_1) I_2 l}{2 \pi (2 d)}$.
Simplifying this expression,we get: $F' = \frac{3}{2} \left( \frac{\mu_0 I_1 I_2 l}{2 \pi d} \right)$.
Substituting the initial force $F$ into the equation,we get: $F' = \frac{3}{2} F$.

(A) The torque $\tau$ on a current-carrying loop in a magnetic field is given by $\tau = N i A B \sin \theta$,where $N$ is the number of turns,$i$ is the current,$A$ is the area,$B$ is the magnetic field,and $\theta$ is the angle between the normal to the loop and the magnetic field.
For a wire of length $L$ formed into a circular coil of $N$ turns with radius $r$,the circumference is $L = N(2 \pi r)$,so $r = \frac{L}{2 \pi N}$.
The area of the coil is $A = \pi r^2 = \pi \left( \frac{L}{2 \pi N} \right)^2 = \frac{L^2}{4 \pi N^2}$.
Substituting $A$ into the torque equation: $\tau = N i \left( \frac{L^2}{4 \pi N^2} \right) B \sin \theta = \frac{i L^2 B \sin \theta}{4 \pi N}$.
To maximize the torque,we set $\sin \theta = 1$ and choose the minimum number of turns,$N = 1$.
Therefore,the maximum torque is $\tau_{\max} = \frac{i L^2 B}{4 \pi}$.

(B) The torque on a current-carrying coil in a magnetic field is given by $\tau = NIAB \sin \theta$. For a radial magnetic field,the plane of the coil is always parallel to the magnetic field,so $\theta = 90^{\circ}$ and $\sin 90^{\circ} = 1$.
Thus,the magnetic torque is $\tau_m = NIAB$.
The restoring torque provided by the suspension fiber is $\tau_r = K \phi$,where $K$ is the torsional constant and $\phi$ is the angle of rotation.
Equating the two torques for equilibrium: $NIAB = K \phi$.
Given: $A = 0.05 \ m^2$,$B = 0.01 \ Wb/m^2$,$K = 5 \times 10^{-9} \ Nm/\text{degree}$,$I = 300 \ \mu A = 300 \times 10^{-6} \ A$,and assuming $N = 1$.
Substituting the values: $\phi = \frac{NIAB}{K} = \frac{1 \times 300 \times 10^{-6} \times 0.01 \times 0.05}{5 \times 10^{-9}}$.
$\phi = \frac{300 \times 10^{-6} \times 5 \times 10^{-4}}{5 \times 10^{-9}} = \frac{300 \times 10^{-10}}{5 \times 10^{-9}} = 60 \times 10^{-1} = 6^{\circ}$.
Wait,re-calculating: $\phi = \frac{300 \times 10^{-6} \times 0.0005}{5 \times 10^{-9}} = \frac{150000 \times 10^{-12}}{5 \times 10^{-9}} = 30000 \times 10^{-3} = 30^{\circ}$.
Therefore,the angle of rotation is $30^{\circ}$.

(C) The force per unit length between two long parallel current-carrying conductors is given by the formula $F = \frac{\mu_0 I_1 I_2}{2\pi r}$.
Here,$r$ is the separation distance,and $I_1$ and $I_2$ are the currents in the conductors.
According to the right-hand rule,when two parallel wires carry current in the same direction,they exert an attractive force on each other.
Given that $I_1 = I_2 = I$,the magnitude of the force per unit length is $F = \frac{\mu_0 I^2}{2\pi r}$.
Therefore,the wires attract each other with a force of $\frac{\mu_0 I^2}{2\pi r}$ per unit length.

(A) The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $d$ is given by the formula: $F = \frac{\mu_0 I_1 I_2}{2 \pi d}$.
Since the current in both wires remains the same,the force is inversely proportional to the distance between them: $F \propto \frac{1}{d}$.
If the new distance $d' = \frac{d}{3}$,the new force $F'$ will be: $F' = \frac{\mu_0 I_1 I_2}{2 \pi (d/3)} = 3 \times \left( \frac{\mu_0 I_1 I_2}{2 \pi d} \right) = 3F$.
Therefore,the magnitude of the force becomes $3$ times the original value.

(C) The correct option is $C$.
Concept: For a wire to experience no force, the net magnetic field at its position due to the other two wires must be zero.
The magnetic field $B$ due to a long straight wire carrying current $I$ at a distance $r$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.
Let $x$ be the distance of wire $C$ from wire $A$. The distance of wire $C$ from wire $B$ is $(15 - x) \,cm$.
For the net magnetic field at $C$ to be zero, the magnetic fields produced by wire $A$ and wire $B$ must be equal in magnitude and opposite in direction.
$\frac{\mu_0 I_A}{2 \pi x} = \frac{\mu_0 I_B}{2 \pi (15 - x)}$
$\frac{I_A}{x} = \frac{I_B}{15 - x}$
Given $I_A = 15 \,A$ and $I_B = 10 \,A$:
$\frac{15}{x} = \frac{10}{15 - x}$
$15(15 - x) = 10x$
$225 - 15x = 10x$
$25x = 225$
$x = 9 \,cm$.

(B) Using the Biot-Savart law,the magnetic field at a point on the axis of a straight current-carrying conductor is zero. Therefore,the magnetic fields due to the straight segments $AO$,$OC$,$DE$,and $FG$ are zero at point $O$.
The magnetic field at the center of a full circular loop is $B = \frac{\mu_0 I}{2R}$.
For a circular arc subtending an angle $\theta$ at the center,the magnetic field is $B = \frac{\mu_0 I \theta}{4\pi R}$.
Here,both arcs $CD$ and $EF$ subtend an angle of $90^\circ$ or $\frac{\pi}{2}$ radians at point $O$.
For arc $CD$ with radius $R_1$:
$B_{CD} = \frac{\mu_0 I (\pi/2)}{4\pi R_1} = \frac{\mu_0 I}{8 R_1}$ (directed into the plane of the paper).
For arc $EF$ with radius $R_2$:
$B_{EF} = \frac{\mu_0 I (\pi/2)}{4\pi R_2} = \frac{\mu_0 I}{8 R_2}$ (directed into the plane of the paper).
Since both fields are in the same direction,the total magnetic field at $O$ is:
$B = B_{CD} + B_{EF} = \frac{\mu_0 I}{8 R_1} + \frac{\mu_0 I}{8 R_2} = \frac{\mu_0 I}{8} \left( \frac{1}{R_1} + \frac{1}{R_2} \right) = \frac{\mu_0 I}{8} \left( \frac{R_1 + R_2}{R_1 R_2} \right)$.

(D) Let the distance between the wires be $2r$. The distance from each wire to the midpoint is $r$. The magnetic field due to a long wire at distance $r$ is $B = \frac{\mu_0 I}{2 \pi r}$.
When currents are in the same direction,the fields at the midpoint are in opposite directions. The net field is $B_1 = \frac{\mu_0}{2 \pi r} (I_1 - I_2) = 6 \times 10^{-6} \ T$.
When the direction of $I_2$ is reversed,the fields at the midpoint are in the same direction. The net field is $B_2 = \frac{\mu_0}{2 \pi r} (I_1 + I_2) = 3 \times 10^{-5} \ T$.
Dividing the two equations: $\frac{I_1 - I_2}{I_1 + I_2} = \frac{6 \times 10^{-6}}{3 \times 10^{-5}} = \frac{6}{30} = \frac{1}{5}$.
$5(I_1 - I_2) = I_1 + I_2 \implies 5I_1 - 5I_2 = I_1 + I_2 \implies 4I_1 = 6I_2$.
Therefore,$\frac{I_1}{I_2} = \frac{6}{4} = \frac{3}{2}$.

(B) The magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 i}{2r}$.
Given that $B_1 = B_2$,we have $\frac{i_1}{r_1} = \frac{i_2}{r_2}$.
Since $r_1 = 2r_2$,we get $\frac{i_1}{2r_2} = \frac{i_2}{r_2}$,which implies $i_1 = 2i_2$.
The resistance of a wire is $R = \rho \frac{L}{A}$. Since both coils are made of the same wire,$\rho$ and $A$ are constant. The length $L = 2\pi r$.
Thus,$R_1 = 2\pi r_1$ and $R_2 = 2\pi r_2$. Since $r_1 = 2r_2$,$R_1 = 2R_2$.
The potential difference is $V = iR$. Therefore,$\frac{V_1}{V_2} = \frac{i_1 R_1}{i_2 R_2}$.
Substituting the values: $\frac{V_1}{V_2} = \frac{(2i_2)(2R_2)}{i_2 R_2} = 4$.
Thus,the ratio is $4: 1$.

(B) To find the magnetic field inside the toroid,we use Ampere's Circuital Law.
Consider an Amperian loop of radius $r$ inside the toroid,where $r = \frac{r_1 + r_2}{2}$.
According to Ampere's Law: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$.
Since the magnetic field $B$ is uniform along the circular path and parallel to the length element $d\vec{l}$,the angle between them is $0^\circ$.
Thus,$B \oint dl = \mu_0 N I$.
The circumference of the loop is $2 \pi r = 2 \pi \left( \frac{r_1 + r_2}{2} \right) = \pi(r_1 + r_2)$.
Substituting this into the equation: $B \cdot \pi(r_1 + r_2) = \mu_0 N I$.
Therefore,the magnetic field is $B = \frac{\mu_0 N I}{\pi(r_1 + r_2)}$.

(A) The magnetic field $B$ at the centre of a circular coil with $N$ turns,radius $r$,and current $I$ is given by $B = \frac{\mu_0 N I}{2r}$.
Let the total length of the wire be $L$.
For a single turn $(N_1 = 1)$,the circumference is $2 \pi r_1 = L$,so $r_1 = \frac{L}{2 \pi}$.
The magnetic field is $B_1 = \frac{\mu_0 (1) I}{2 (L / 2 \pi)} = \frac{\mu_0 I \pi}{L}$.
For three turns $(N_2 = 3)$,the total length is $3(2 \pi r_2) = L$,so $r_2 = \frac{L}{6 \pi}$.
The magnetic field is $B_2 = \frac{\mu_0 (3) I}{2 (L / 6 \pi)} = \frac{9 \mu_0 I \pi}{L}$.
The ratio of the magnetic inductions is $\frac{B_1}{B_2} = \frac{\mu_0 I \pi / L}{9 \mu_0 I \pi / L} = \frac{1}{9}$.

(C) The magnetic field $B$ produced by a long straight current-carrying wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2\pi r}$.
Since the currents $I_1$ and $I_2$ are in opposite directions,by the Right-Hand Thumb Rule,the magnetic fields produced by both wires at the midpoint will point in the same direction.
Therefore,the resultant magnetic field is the sum of the individual fields: $B_{net} = B_1 + B_2$.
The distance of the midpoint from each wire is $r = \frac{d}{2}$.
Substituting the values: $B_{net} = \frac{\mu_0 I_1}{2\pi(d/2)} + \frac{\mu_0 I_2}{2\pi(d/2)}$.
Simplifying the expression: $B_{net} = \frac{\mu_0 I_1}{\pi d} + \frac{\mu_0 I_2}{\pi d} = \frac{\mu_0(I_1+I_2)}{\pi d}$.

(D) The magnetic field at the center of a circular coil of $N$ turns,radius $r$,and current $I$ is given by $B = \frac{N \mu_0 I}{2r}$.
Initially,for one turn $(N=1)$ and radius $r$,the magnetic field is $B = \frac{\mu_0 I}{2r}$.
When the same wire of length $L = 2 \pi r$ is bent into $n$ turns,the new radius $r^{\prime}$ is given by $L = n(2 \pi r^{\prime})$,which implies $r^{\prime} = \frac{r}{n}$.
The new magnetic field $B^{\prime}$ is $B^{\prime} = \frac{n \mu_0 I}{2r^{\prime}}$.
Substituting $r^{\prime} = \frac{r}{n}$ into the expression for $B^{\prime}$:
$B^{\prime} = \frac{n \mu_0 I}{2(r/n)} = n^2 \left( \frac{\mu_0 I}{2r} \right) = n^2 B$.

(B) The magnetic field $B$ on the axis of a circular coil of radius $R$ carrying current $I$ at a distance $x$ from its center is given by: $B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.
At the center of the coil $(x = 0)$,the magnetic field $B_0$ is: $B_0 = \frac{\mu_0 I}{2R}$.
According to the problem,the magnetic field at point $P$ is $\frac{1}{8}$ of the field at the center: $B_P = \frac{1}{8} B_0$.
Substituting the expressions: $\frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} = \frac{1}{8} \left( \frac{\mu_0 I}{2R} \right)$.
Simplifying the equation: $\frac{R^2}{(R^2 + x^2)^{3/2}} = \frac{1}{8R} \Rightarrow \frac{R^3}{(R^2 + x^2)^{3/2}} = \frac{1}{8}$.
Taking the cube root of both sides: $\frac{R}{(R^2 + x^2)^{1/2}} = \frac{1}{2}$.
Squaring both sides: $\frac{R^2}{R^2 + x^2} = \frac{1}{4}$.
Cross-multiplying: $4R^2 = R^2 + x^2 \Rightarrow x^2 = 3R^2 \Rightarrow x = \sqrt{3}R$.

(A) Let $r$ be the radius of the circular loop.
Since the area $A = \pi r^2$,we have $r = \sqrt{\frac{A}{\pi}}$.
The magnetic field at the centre of a circular loop is given by $B = \frac{\mu_0 I}{2r}$.
Substituting the value of $r$,we get $B = \frac{\mu_0 I}{2 \sqrt{A/\pi}}$.
Solving for current $I$,we get $I = \frac{2B}{\mu_0} \sqrt{\frac{A}{\pi}}$.
The magnetic moment $M$ of the loop is defined as $M = I A$.
Substituting the expression for $I$,we get $M = \left( \frac{2B}{\mu_0} \sqrt{\frac{A}{\pi}} \right) A = \frac{2 B A^{3/2}}{\mu_0 \sqrt{\pi}}$.
Thus,the correct option is $A$.

(A) The gyromagnetic ratio is defined as the ratio of the magnetic moment $(M_0)$ to the angular momentum $(L_0)$ of a particle.
Mathematically,it is expressed as:
$\text{Gyromagnetic ratio} = \frac{M_0}{L_0}$
For an electron revolving in an orbit,the magnetic moment $M_0 = \frac{e}{2m} L_0$,where $e$ is the charge and $m$ is the mass of the electron.
Thus,the ratio $\frac{M_0}{L_0} = \frac{e}{2m}$,which is a constant.

(D) magnetic wire of length $l$ has a magnetic moment $M = m l$,where $m$ is the pole strength of each end.
When the wire is bent into an $L$-shape,the two segments of length $l/2$ are perpendicular to each other.
The effective length (displacement between the poles) becomes the hypotenuse of a right-angled triangle with sides $l/2$ and $l/2$.
Effective length $l' = \sqrt{(l/2)^2 + (l/2)^2} = \sqrt{l^2/4 + l^2/4} = \sqrt{l^2/2} = \frac{l}{\sqrt{2}}$.
The new magnetic moment $M'$ is given by $M' = m l' = m \left( \frac{l}{\sqrt{2}} \right)$.
Since $M = m l$,we get $M' = \frac{M}{\sqrt{2}}$.

(A) Let the length of the wire be $l$. For a square loop,the side of the square is $a = l/4$.
Hence,its magnetic moment is $M_{sq} = i A = i (l/4)^2 = i l^2 / 16$.
For a circular loop,the circumference is $2 \pi r = l$,so the radius is $r = l / (2 \pi)$.
Hence,its magnetic moment is $M_{cir} = i A = i \pi r^2 = i \pi (l / (2 \pi))^2 = i \pi (l^2 / 4 \pi^2) = i l^2 / (4 \pi)$.
The ratio of the magnetic moment of the circle to that of the square is:
$\frac{M_{cir}}{M_{sq}} = \frac{i l^2 / (4 \pi)}{i l^2 / 16} = \frac{16}{4 \pi} = \frac{4}{\pi}$.

(D) When a charged particle moves in a perpendicular magnetic field, it undergoes uniform circular motion.
The magnetic force provides the necessary centripetal force: $\frac{mv^2}{R} = qvB$.
This simplifies to $v = \frac{qBR}{m}$.
The kinetic energy $E$ is given by $E = \frac{1}{2}mv^2 = \frac{1}{2}m \left(\frac{qBR}{m}\right)^2 = \frac{q^2B^2R^2}{2m}$.
Since $q$, $m$, and $R$ are constant, $E \propto B^2$.
If the magnetic field $B$ is increased by $4$ times $(B' = 4B)$, the new kinetic energy $E'$ will be:
$E' \propto (4B)^2 = 16B^2$.
Therefore, $E' = 16E$. The kinetic energy increases by $16$ times.

(C) The kinetic energy of the electrons is given by $\frac{1}{2}mv^2 = eV$,which implies the velocity $v = \sqrt{\frac{2eV}{m}}$.
The magnetic force experienced by the electrons is $F = evB = eB\sqrt{\frac{2eV}{m}}$.
This shows that $F \propto \sqrt{V}$.
Given that the new force $F^{\prime} = 2F$ when the potential is $V^{\prime}$,we have $\frac{F^{\prime}}{F} = \sqrt{\frac{V^{\prime}}{V}}$.
Substituting the values,$2 = \sqrt{\frac{V^{\prime}}{V}}$.
Squaring both sides,$4 = \frac{V^{\prime}}{V}$,which means $V^{\prime} = 4V$.
Therefore,the ratio $\frac{V}{V^{\prime}} = \frac{V}{4V} = \frac{1}{4}$.

(D) stationary electric charge produces an electric field in the space around it.
When an electric charge moves with a uniform velocity,it constitutes an electric current.
According to the Oersted experiment,an electric current produces a magnetic field in the surrounding space.
Since the charge is moving,it continues to possess its inherent electric field while simultaneously generating a magnetic field due to its motion.
Therefore,a charge moving with uniform velocity produces both electric and magnetic fields.

(B) The magnetic force acting on a charged particle is given by $\vec{F} = q(\vec{v} \times \vec{B})$.
Since the force is always perpendicular to the velocity vector $\vec{v}$,the work done by the magnetic force is $W = \int \vec{F} \cdot d\vec{r} = \int \vec{F} \cdot \vec{v} dt = 0$.
According to the work-energy theorem,the change in kinetic energy is equal to the work done,so the kinetic energy remains constant.
However,because the direction of the velocity vector changes continuously during the motion,the momentum $\vec{p} = m\vec{v}$ changes.
Therefore,the momentum changes while the kinetic energy remains constant.

(D) When a charged particle of mass $m$ and charge $q$ moves in a circular path perpendicular to a uniform magnetic field $B$,the magnetic Lorentz force provides the necessary centripetal force.
$qvB = \frac{mv^2}{r}$
From this,the radius of the path is $r = \frac{mv}{qB}$.
The time period $T$ is the time taken to complete one full circle,which is the circumference divided by the velocity:
$T = \frac{2 \pi r}{v} = \frac{2 \pi (mv/qB)}{v} = \frac{2 \pi m}{qB}$.
As seen from the formula $T = \frac{2 \pi m}{qB}$,the time period $T$ depends only on the mass $m$,charge $q$,and magnetic field $B$.
It is independent of the velocity $v$ of the charge.

(D) The magnetic force $F$ on a charged particle of mass $m$ and charge $q$ moving with velocity $v$ in a uniform magnetic field $B$ is given by $F = Bqv$.
When an electron is accelerated through a potential difference $V$,its kinetic energy is given by $\frac{1}{2}mv^2 = eV$.
From this,the velocity $v$ is $v = \sqrt{\frac{2eV}{m}}$.
Substituting this into the force equation,we get $F = B e \sqrt{\frac{2eV}{m}} = B e \sqrt{\frac{2e}{m}} \sqrt{V}$.
This shows that $F \propto \sqrt{V}$.
If the potential difference is increased to $2V$,the new force $F'$ will be $F' \propto \sqrt{2V}$.
Therefore,$\frac{F'}{F} = \frac{\sqrt{2V}}{\sqrt{V}} = \sqrt{2}$.
Thus,$F' = \sqrt{2}F$.

(B) The radius of curvature $r$ of a charged particle of mass $m$ and charge $q$ moving in a region with a uniform perpendicular magnetic field $B$ is given by the formula: $r = \frac{mv}{Bq} = \frac{p}{Bq}$.
Here,$p$ is the momentum of the charged particle.
Since the momenta $p$ are the same for both the electron and the proton,and the magnetic field $B$ is also the same,the radius of the circular path depends only on the charge $q$ of the particle.
However,the magnitude of the charge for both an electron and a proton is equal $(|q_e| = |q_p| = e)$.
Therefore,$r_e = \frac{p}{Be}$ and $r_p = \frac{p}{Be}$,which implies $r_e = r_p$.
Thus,both particles will follow a curved path of the same radius,neglecting the sense of revolution (direction of curvature).