WBJEE 2013 Physics Question Paper with Answer and Solution

(C) Let the velocity of the third fragment be $v'$.
According to the law of conservation of linear momentum,since no external force acts on the system,the total initial momentum is equal to the total final momentum.
Initial momentum $P_i = 5 M \times 0 = 0$.
Let the velocity of the first fragment $(M)$ be $\vec{v}_1 = 2v \hat{i}$ and the velocity of the second fragment $(2M)$ be $\vec{v}_2 = -v \hat{i}$ (since they move in opposite directions).
Final momentum $P_f = M(2v \hat{i}) + 2M(-v \hat{i}) + 2M(\vec{v}') = 0$.
$2Mv \hat{i} - 2Mv \hat{i} + 2M(\vec{v}') = 0$.
$0 + 2M(\vec{v}') = 0$.
Therefore,$\vec{v}' = 0$.
The third fragment will be at rest.

(A) The gravitational force $F$ between two masses $m_1$ and $m_2$ separated by a distance $r$ is given by $F = \frac{G m_1 m_2}{r^2}$.
Here,$m_1 = m$ and $m_2 = (M - m)$.
Substituting these values,we get $F = \frac{G m(M - m)}{r^2} = \frac{G}{r^2} (Mm - m^2)$.
For the force to be maximum,the derivative of $F$ with respect to $m$ must be zero,i.e.,$\frac{dF}{dm} = 0$.
$\frac{d}{dm} [\frac{G}{r^2} (Mm - m^2)] = 0$.
$\frac{G}{r^2} (M - 2m) = 0$.
Since $\frac{G}{r^2} \neq 0$,we have $M - 2m = 0$,which implies $m = \frac{M}{2}$.
Thus,the mass of the second piece is $(M - m) = M - \frac{M}{2} = \frac{M}{2}$.
The ratio of the masses is $\frac{m}{M-m} = \frac{M/2}{M/2} = \frac{1}{1}$.

(B) For a mono-atomic gas,the degree of freedom $f_1 = 3$. For a diatomic gas,the degree of freedom $f_2 = 5$.
The number of moles are $n_1 = 3$ and $n_2 = 1$.
The degree of freedom for the mixture is given by:
$f_{\text{mix}} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2} = \frac{3 \times 3 + 1 \times 5}{3 + 1} = \frac{9 + 5}{4} = \frac{14}{4} = 3.5$.
The adiabatic index $\gamma$ for the mixture is given by:
$\gamma_{\text{mix}} = 1 + \frac{2}{f_{\text{mix}}} = 1 + \frac{2}{3.5} = 1 + \frac{2}{7/2} = 1 + \frac{4}{7} = \frac{11}{7}$.

(D) The formula for the r.m.s. speed of gas molecules is $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$.
Given the initial temperature $T_1 = 100^{\circ} C = 373 \text{ K}$.
Thus,$v = \sqrt{\frac{3R(373)}{M}}$.
We want to find the temperature $T_2$ such that the new speed $v' = \sqrt{3}v$.
Since $v_{\text{rms}} \propto \sqrt{T}$,we have $\frac{v'}{v} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values,$\sqrt{3} = \sqrt{\frac{T_2}{373}}$.
Squaring both sides,$3 = \frac{T_2}{373}$.
$T_2 = 3 \times 373 = 1119 \text{ K}$.
Converting back to Celsius,$T_2 = 1119 - 273 = 846^{\circ} C$.

(B) Given: mass $m = 0.1 \ kg$,change in momentum $\Delta p = 2 \ kg \ m/s$,time $t = 1 \ s$,$g = 10 \ m/s^2$.
Since $\Delta p = F_{net} \times t$,we have $F_{net} = \frac{\Delta p}{t} = \frac{2}{1} = 2 \ N$.
For the block,the net force is $T - mg = F_{net}$,where $T$ is the tension in the string. Since the string is massless and the pulley is frictionless,$T = F$.
Thus,$F - mg = 2 \ N \Rightarrow F - (0.1 \times 10) = 2 \Rightarrow F - 1 = 2 \Rightarrow F = 3 \ N$.
So,the tension in the string is $T = F = 3 \ N$.
Acceleration $a = \frac{F_{net}}{m} = \frac{2}{0.1} = 20 \ m/s^2$.
Displacement $S = \frac{1}{2}at^2 = \frac{1}{2} \times 20 \times (1)^2 = 10 \ m$.
Work done by tension $W_T = T \times S = 3 \times 10 = 30 \ J$.
Work done against gravity $W_g = mg \times S = (0.1 \times 10) \times 10 = 10 \ J$.

(A) Let $V$ be the total volume of the cylindrical block and $\rho$ be its density.
In the first case,the block floats in a liquid of density $\rho_{1}$. According to the law of floatation,the weight of the block equals the buoyant force:
$V \rho g = V x_{1} \rho_{1} g$
$\Rightarrow \rho = x_{1} \rho_{1} \quad ... (1)$
In the second case,the block is fully immersed in two immiscible liquids of densities $\rho_{1}$ and $\rho_{2}$. The volume fraction $x_{2}$ is in liquid $\rho_{1}$ and the remaining fraction $(1 - x_{2})$ is in liquid $\rho_{2}$. The total buoyant force equals the weight of the block:
$V \rho g = V x_{2} \rho_{1} g + V (1 - x_{2}) \rho_{2} g$
$\Rightarrow \rho = x_{2} \rho_{1} + (1 - x_{2}) \rho_{2} \quad ... (2)$
Equating $(1)$ and $(2)$:
$x_{1} \rho_{1} = x_{2} \rho_{1} + (1 - x_{2}) \rho_{2}$
$x_{1} \rho_{1} - x_{2} \rho_{1} = (1 - x_{2}) \rho_{2}$
$\rho_{1} (x_{1} - x_{2}) = \rho_{2} (1 - x_{2})$
$\frac{\rho_{1}}{\rho_{2}} = \frac{1 - x_{2}}{x_{1} - x_{2}}$

(A) According to Bernoulli's principle for streamline flow in a horizontal tube, the sum of static pressure and dynamic pressure remains constant at all points along a streamline.
$p + \frac{1}{2} \rho v^{2} = p_{1} + \frac{1}{2} \rho v_{1}^{2}$
Given that at the second point, $p_{1} = p/2$.
Substituting this into the equation:
$p + \frac{1}{2} \rho v^{2} = \frac{p}{2} + \frac{1}{2} \rho v_{1}^{2}$
Rearranging the terms to solve for $v_{1}^{2}$:
$\frac{1}{2} \rho v_{1}^{2} = p - \frac{p}{2} + \frac{1}{2} \rho v^{2}$
$\frac{1}{2} \rho v_{1}^{2} = \frac{p}{2} + \frac{1}{2} \rho v^{2}$
Multiplying both sides by $2/\rho$:
$v_{1}^{2} = \frac{p}{\rho} + v^{2}$
$v_{1} = \sqrt{v^{2} + \frac{p}{\rho}}$

(D) The excess pressure 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.
For the smaller bubble of radius $r$,the pressure inside is $P_1 = P_{atm} + \frac{4T}{r}$.
For the bigger bubble of radius $2r$,the pressure inside is $P_2 = P_{atm} + \frac{4T}{2r} = P_{atm} + \frac{2T}{r}$.
Since $P_1 > P_2$,when the valve is opened,air flows from the region of higher pressure (smaller bubble) to the region of lower pressure (bigger bubble).
Consequently,the radius of the smaller bubble decreases and the radius of the bigger bubble increases.

(C) The terminal velocity $v$ of a sphere falling through a viscous liquid is given by the formula: $v = \frac{2r^2(\rho - \sigma)g}{9\eta}$
Where $r$ is the radius of the sphere,$\rho$ is the density of the sphere,$\sigma$ is the density of the liquid,$g$ is the acceleration due to gravity,and $\eta$ is the coefficient of viscosity.
Since the spheres are made of the same material and fall through the same liquid,the parameters $\rho, \sigma, g,$ and $\eta$ are constant.
Therefore,the terminal velocity is proportional to the square of the radius: $v \propto r^2$.
For radii $R$ and $3R$,the ratio of terminal velocities is: $\frac{v_1}{v_2} = \frac{R^2}{(3R)^2} = \frac{R^2}{9R^2} = \frac{1}{9}$.
Thus,the ratio is $1:9$.

(D) The elastic potential energy per unit volume $(u)$ of a stretched wire is given by the formula: $u = \frac{1}{2} \times Y \times (\text{strain})^2$,where $Y$ is Young's modulus and $\text{strain} = \frac{\Delta L}{L}$.
For the first wire: $\text{strain}_1 = \frac{l}{L}$.
Thus,$u_1 = \frac{1}{2} Y (\frac{l}{L})^2$.
For the second wire: $\text{strain}_2 = \frac{2l}{2L} = \frac{l}{L}$.
Thus,$u_2 = \frac{1}{2} Y (\frac{l}{L})^2$.
Since both wires are made of the same material,their Young's modulus $Y$ is the same.
The ratio of the stored elastic energy per unit volume is $\frac{u_1}{u_2} = \frac{\frac{1}{2} Y (l/L)^2}{\frac{1}{2} Y (l/L)^2} = 1:1$.

(A) Given,initial velocity vector $v = (3 \hat{i} + 10 \hat{j}) \text{ m/s}$.
Comparing with $v = v_x \hat{i} + v_y \hat{j}$,we get $v_x = 3 \text{ m/s}$ and $v_y = 10 \text{ m/s}$.
The maximum height $H$ attained by a projectile is given by the formula $H = \frac{v_y^2}{2g}$.
Substituting the values: $H = \frac{10^2}{2 \times 10} = \frac{100}{20} = 5 \text{ m}$.
The horizontal range $R$ is given by the formula $R = v_x \times T$,where $T = \frac{2v_y}{g}$ is the time of flight.
Calculating time of flight: $T = \frac{2 \times 10}{10} = 2 \text{ s}$.
Calculating range: $R = 3 \times 2 = 6 \text{ m}$.
Thus,the maximum height is $5 \text{ m}$ and the range is $6 \text{ m}$.

(A) Let the extension in the spring be $x$. The total length of the spring becomes $L+x$.
When the mass rotates in a horizontal circle,the centripetal force is provided by the horizontal component of the spring force.
The spring force is $F = Kx$,where $K$ is the spring constant.
The horizontal component of the spring force is $Kx \sin \theta = m \omega^{2} r$,where $r = (L+x) \sin \theta$ is the radius of the circular path.
Thus,$Kx \sin \theta = m \omega^{2} (L+x) \sin \theta$.
Canceling $\sin \theta$ from both sides,we get $Kx = m \omega^{2} (L+x)$.
We know that the angular frequency of vertical oscillations is $\omega_{0} = \sqrt{\frac{K}{m}}$,which implies $K = m \omega_{0}^{2}$.
Substituting $K$ in the equation: $m \omega_{0}^{2} x = m \omega^{2} (L+x)$.
Dividing by $m$: $\omega_{0}^{2} x = \omega^{2} L + \omega^{2} x$.
Rearranging terms to solve for $x$: $x(\omega_{0}^{2} - \omega^{2}) = \omega^{2} L$.
Therefore,$x = \frac{\omega^{2} L}{\omega_{0}^{2} - \omega^{2}}$.

(A) For the first $SHM$: $x_{1} = a \sin \omega t + a \cos \omega t = \sqrt{2} a \sin(\omega t + \frac{\pi}{4})$.
Amplitude $A_{1} = \sqrt{2} a$ and phase $\phi_{1} = \frac{\pi}{4}$.
For the second $SHM$: $x_{2} = a \sin \omega t + \frac{a}{\sqrt{3}} \cos \omega t$.
Amplitude $A_{2} = \sqrt{a^{2} + (\frac{a}{\sqrt{3}})^{2}} = \sqrt{a^{2} + \frac{a^{2}}{3}} = \sqrt{\frac{4a^{2}}{3}} = \frac{2a}{\sqrt{3}}$.
Phase $\phi_{2} = \tan^{-1}(\frac{a/\sqrt{3}}{a}) = \tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$.
Ratio of amplitudes $\frac{A_{1}}{A_{2}} = \frac{\sqrt{2} a}{2a/\sqrt{3}} = \frac{\sqrt{2} \cdot \sqrt{3}}{2} = \frac{\sqrt{6}}{2} = \sqrt{\frac{6}{4}} = \sqrt{\frac{3}{2}}$.
Phase difference $\Delta \phi = \phi_{1} - \phi_{2} = \frac{\pi}{4} - \frac{\pi}{6} = \frac{3\pi - 2\pi}{12} = \frac{\pi}{12}$.

(D) planet revolves around the sun in an elliptical orbit under the influence of the gravitational force exerted by the sun.
Since the gravitational force $F$ always acts along the line joining the planet and the sun (the position vector $r$),the torque $\tau$ acting on the planet about the sun is given by $\tau = r \times F = rF \sin(180^{\circ}) = 0$.
According to the relation between torque and angular momentum,$\tau = \frac{dL}{dt}$.
Since $\tau = 0$,it follows that $\frac{dL}{dt} = 0$,which implies that the angular momentum $L$ is a constant with respect to time.
Therefore,the angular momentum of the planet remains constant.

(A) Let the rectangular frame have corners $A, B, C,$ and $D$ such that side $AD = BC = a$ and side $AB = CD = b$. The masses $m$ are placed at each corner.
The axis of rotation passes along the side of length $b$ (let this be side $CD$).
The moment of inertia $I$ of a system of particles is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th mass from the axis of rotation.
$1$. The masses at $C$ and $D$ lie on the axis of rotation,so their perpendicular distances are $r_C = 0$ and $r_D = 0$. Thus,their contribution to the moment of inertia is $m(0)^2 + m(0)^2 = 0$.
$2$. The masses at $A$ and $B$ are at a perpendicular distance $a$ from the axis $CD$. Thus,their contribution is $m(a)^2 + m(a)^2 = 2ma^2$.
Therefore,the total moment of inertia of the system about the axis $CD$ is $I = 0 + 2ma^2 = 2ma^2$.

(B) Initial velocity $u = 3.6 \ km/h = 3.6 \times \frac{5}{18} \ m/s = 1 \ m/s$.
Final velocity $v = 2 \times u = 2 \ m/s$.
Time $t = 10 \ s$.
Acceleration $a = \frac{v - u}{t} = \frac{2 - 1}{10} = 0.1 \ m/s^2$.
Distance covered $s = ut + \frac{1}{2}at^2 = (1)(10) + \frac{1}{2}(0.1)(10)^2 = 10 + 5 = 15 \ m$.
Circumference of the wheel $C = 2 \pi r = 2 \pi (0.25) = 0.5 \pi \ m$.
Number of revolutions $n = \frac{s}{C} = \frac{15}{0.5 \pi} = \frac{30}{\pi} \approx \frac{30}{3.14159} \approx 9.54$.
Wait,checking the calculation: $n = 9.54$ revolutions. Re-evaluating the question context: If the question implies angular displacement $\theta$ in radians,$\theta = \frac{s}{r} = \frac{15}{0.25} = 60 \ rad$. Number of revolutions $n = \frac{\theta}{2 \pi} = \frac{60}{2 \pi} = \frac{30}{\pi} \approx 9.55$. Given the options,there might be a unit or scale factor discrepancy in the problem statement,but based on standard kinematics,the value is approximately $9.5$. If the intended answer is $95$,it implies a factor of $10$ difference.

(A) The radius of the circular path of the mass is $R = L/2$. The linear velocity of the mass at the moment of detachment is $v = \omega R = \frac{\omega L}{2}$.
Since the mass is moving vertically upwards,the time of flight $T$ is given by $T = \frac{2v}{g} = \frac{\omega L}{g}$.
For the mass to reach the bar at the same point,the bar must complete an integer number of rotations $n$ in time $T$. The time for $n$ rotations is $T = n \cdot \frac{2\pi}{\omega}$.
Equating the two expressions for $T$: $\frac{\omega L}{g} = n \frac{2\pi}{\omega} \implies n = \frac{\omega^{2} L}{2\pi g}$. Thus,$n$ must be an integer.
For option $C$,the distance travelled by the mass in air is $2h = 2 \cdot \frac{v^2}{2g} = \frac{v^2}{g} = \frac{(\omega L/2)^2}{g} = \frac{\omega^2 L^2}{4g}$,which is proportional to $\omega^2$.

(C) The amount of heat $Q$ required to raise the temperature of a substance is given by $Q = \int m C \ dT$.
Given the specific heat $C = D T^{3}$.
Substituting this into the integral,we get $Q = \int_{20}^{30} m (D T^{3}) \ dT$.
Taking the constants $m$ and $D$ outside the integral: $Q = m D \int_{20}^{30} T^{3} \ dT$.
Evaluating the integral: $Q = m D \left[ \frac{T^{4}}{4} \right]_{20}^{30}$.
$Q = \frac{m D}{4} [ (30)^{4} - (20)^{4} ]$.
$Q = \frac{m D}{4} [ 810000 - 160000 ]$.
$Q = \frac{m D}{4} [ 650000 ]$.
$Q = \frac{650000}{4} m D = \frac{65}{4} \times 10^{4} m D$.

(B) The work done by a gas during a constant pressure (isobaric) process is given by the formula: $W = P \Delta V$.
Here, $P = 400 \text{ kPa} = 400 \times 10^3 \text{ Pa}$.
The change in volume $\Delta V$ is equal to the cross-sectional area $A$ multiplied by the displacement $h$.
Given $A = 0.3 \text{ m}^2$ and $h = 10 \text{ cm} = 0.1 \text{ m}$.
Therefore, $\Delta V = A \times h = 0.3 \text{ m}^2 \times 0.1 \text{ m} = 0.03 \text{ m}^3$.
Now, substitute the values into the work formula:
$W = (400 \times 10^3 \text{ Pa}) \times (0.03 \text{ m}^3)$
$W = 400,000 \times 0.03 = 12,000 \text{ J}$.
Converting to kilojoules: $W = 12 \text{ kJ}$.

(B) The heat supplied at constant pressure is given by $\Delta Q = n C_p \Delta T$.
The increase in internal energy is given by $\Delta U = n C_v \Delta T$.
The fraction of heat energy used to increase internal energy is $f = \frac{\Delta U}{\Delta Q} = \frac{n C_v \Delta T}{n C_p \Delta T} = \frac{C_v}{C_p}$.
We know that the ratio of specific heats is $\gamma = \frac{C_p}{C_v}$,so $\frac{C_v}{C_p} = \frac{1}{\gamma}$.
For a mono-atomic gas,the adiabatic index is $\gamma = 5/3$.
Therefore,the fraction is $f = \frac{1}{5/3} = 3/5$.

(B) Given:
Velocity of sound,$v = 340 \ m/s$
Velocity of listener (car),$v_L = 17 \ m/s$
Velocity of source (bus),$v_S = ?$
Frequency of horn emitted,$f = 640 \ Hz$
Apparent frequency,$f' = 680 \ Hz$
According to the Doppler effect,the apparent frequency $f'$ is given by:
$f' = f \left( \frac{v + v_L}{v - v_S} \right)$
Substituting the given values:
$680 = 640 \left( \frac{340 + 17}{340 - v_S} \right)$
Dividing both sides by $40$:
$17 = 16 \left( \frac{357}{340 - v_S} \right)$
$17(340 - v_S) = 16 \times 357$
$5780 - 17v_S = 5712$
$17v_S = 5780 - 5712$
$17v_S = 68$
$v_S = 4 \ m/s$

(B) Let the length of the closed pipe be $L_{1}$ and the length of the open pipe be $L_{2}$.
The fundamental frequency of a closed pipe is given by $v_{1} = \frac{v}{4 L_{1}}$.
The frequency of the second harmonic of an open pipe is given by $v_{2} = 2 \times \frac{v}{2 L_{2}} = \frac{v}{L_{2}}$.
According to the problem,the fundamental frequency of the closed pipe is equal to the frequency of the second harmonic of the open pipe,so $v_{1} = v_{2}$.
Substituting the expressions,we get $\frac{v}{4 L_{1}} = \frac{v}{L_{2}}$.
Rearranging the terms,we find $\frac{L_{1}}{L_{2}} = \frac{1}{4}$.
Thus,the ratio of their lengths is $1: 4$.

(C) The frequency of the wave is $f = 500 Hz$ and the velocity is $v = 300 ms^{-1}$.
First,we calculate the wavelength $\lambda$ using the relation $v = f \lambda$:
$\lambda = \frac{v}{f} = \frac{300}{500} = 0.6 m = 60 cm$.
The phase difference $\Delta \phi$ is given as $60^{\circ}$,which is $\frac{\pi}{3}$ radians.
The relationship between phase difference $\Delta \phi$ and path difference $\Delta x$ is given by $\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$.
Substituting the values: $\frac{\pi}{3} = \frac{2 \pi}{60 cm} \Delta x$.
Solving for $\Delta x$: $\Delta x = \frac{\pi}{3} \times \frac{60 cm}{2 \pi} = \frac{60}{6} cm = 10 cm$.

(D) Mass of bullet $= m$,Initial speed of bullet $= v$.
Mass of block $= M$,Initial speed of block $= 0$.
Let the common velocity of the system after collision be $V$.
According to the law of conservation of linear momentum:
$m v + M(0) = (m + M)V$
$V = \frac{m v}{m + M}$
Heat generated is equal to the loss in kinetic energy $(KE)$.
$\Delta KE = KE_{initial} - KE_{final}$
$\Delta KE = \frac{1}{2} m v^{2} - \frac{1}{2} (m + M) V^{2}$
Substituting the value of $V$:
$\Delta KE = \frac{1}{2} m v^{2} - \frac{1}{2} (m + M) \left( \frac{m v}{m + M} \right)^{2}$
$\Delta KE = \frac{1}{2} m v^{2} - \frac{1}{2} (m + M) \frac{m^{2} v^{2}}{(m + M)^{2}}$
$\Delta KE = \frac{1}{2} m v^{2} \left( 1 - \frac{m}{m + M} \right)$
$\Delta KE = \frac{1}{2} m v^{2} \left( \frac{m + M - m}{m + M} \right)$
$\Delta KE = \frac{1}{2} \frac{m M v^{2}}{m + M}$

(D) Power is defined as the rate of doing work,which is equal to the rate of change of kinetic energy $(KE)$ of the particle.
$P = \frac{dW}{dt} = \frac{d(KE)}{dt}$.
Since the power $(P)$ is given as a constant,the rate of change of kinetic energy,$\frac{d(KE)}{dt}$,must also be constant.
Therefore,the rate of change of kinetic energy remains constant.

(C) The given equation for alternating current is $I = I_0 \sin(\omega t + \phi)$,where $I_0 = 20 \ A$ and $\omega = 100 \pi \ rad/s$.
The root mean square (r.m.s.) value of the current is given by $I_{rms} = \frac{I_0}{\sqrt{2}}$.
Substituting the value of $I_0$: $I_{rms} = \frac{20}{\sqrt{2}} = 10 \sqrt{2} \ A$.
The angular frequency is $\omega = 2 \pi f = 100 \pi$.
Solving for frequency $f$: $f = \frac{100 \pi}{2 \pi} = 50 \ Hz$.
Therefore,the r.m.s. value is $10 \sqrt{2} \ A$ and the frequency is $50 \ Hz$.

(D) The total energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6 \ eV}{n^2}$.
For the $n=2$ state,the total energy is $E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \ eV$.
The relationship between potential energy $(PE)$ and total energy $(E)$ is $PE = 2E$.
Therefore,the potential energy in the $n=2$ state is $PE = 2 \times (-3.4 \ eV) = -6.8 \ eV$.

(D) Initially,the capacitor $C_{0}$ is charged to potential $V_{0}$. The total charge stored in the capacitor is $Q = C_{0}V_{0}$.
When the switch $S$ is closed,the two capacitors $C_{0}$ and $C$ are connected in parallel. The total charge $Q$ is now distributed across both capacitors,and they reach a common potential $V$.
According to the principle of conservation of charge,the total charge remains constant:
$Q = (C_{0} + C)V$
Substituting $Q = C_{0}V_{0}$:
$C_{0}V_{0} = (C_{0} + C)V$
$C_{0}V_{0} = C_{0}V + CV$
$CV = C_{0}V_{0} - C_{0}V$
$CV = C_{0}(V_{0} - V)$
$C = \frac{C_{0}(V_{0} - V)}{V}$

(A) The given arrangement consists of four plates. Let the plates connected to $P$ be the positive plates and the plates connected to $Q$ be the negative plates.
From the figure,we can see that there are two capacitors formed between the plates.
Each capacitor is formed by two adjacent plates separated by a distance $d$ with an area $a$.
The capacitance of a single parallel plate capacitor is given by $C = \frac{\varepsilon_{0} a}{d}$.
Since the two capacitors are connected in parallel,the equivalent capacitance $C_{eq}$ is the sum of the individual capacitances.
$C_{eq} = C_{1} + C_{2} = \frac{\varepsilon_{0} a}{d} + \frac{\varepsilon_{0} a}{d} = \frac{2 \varepsilon_{0} a}{d}$.

(B) The heat generated in a resistance $R$ connected to a cell of emf $E$ and internal resistance $r$ is given by $H = I^2 R t = \left( \frac{E}{R+r} \right)^2 R t$.
Given $H_1 = H$ for $R_1$ and $H_2 = 4H$ for $R_2$ in the same time $t$,we have:
$H = \frac{E^2 R_1}{(R_1+r)^2} t$ and $4H = \frac{E^2 R_2}{(R_2+r)^2} t$.
Dividing the two equations: $\frac{4H}{H} = \frac{R_2 (R_1+r)^2}{R_1 (R_2+r)^2} \Rightarrow 4 = \frac{R_2 (R_1+r)^2}{R_1 (R_2+r)^2}$.
Taking the square root on both sides: $2 = \frac{\sqrt{R_2} (R_1+r)}{\sqrt{R_1} (R_2+r)}$.
$2 \sqrt{R_1} (R_2+r) = \sqrt{R_2} (R_1+r) \Rightarrow 2 \sqrt{R_1} R_2 + 2 \sqrt{R_1} r = \sqrt{R_2} R_1 + \sqrt{R_2} r$.
Rearranging for $r$: $r (2 \sqrt{R_1} - \sqrt{R_2}) = \sqrt{R_2} R_1 - 2 \sqrt{R_1} R_2$.
$r = \frac{\sqrt{R_1 R_2} (\sqrt{R_1} - 2 \sqrt{R_2})}{2 \sqrt{R_1} - \sqrt{R_2}}$.
Multiplying numerator and denominator by $-1$: $r = \sqrt{R_1 R_2} \frac{2 \sqrt{R_2} - \sqrt{R_1}}{\sqrt{R_2} - 2 \sqrt{R_1}}$.

(B) Let the total current from the battery be $I$ and the current through the right loop be $I_1$. Applying Kirchhoff's voltage law to the right loop (loop $BCDEB$):
$2 I_1 + 4 I_1 + 2 I_1 - 8(I - I_1) = 0$
$8 I_1 - 8 I + 8 I_1 = 0$
$16 I_1 = 8 I \Rightarrow I_1 = 0.5 I$.
Now,applying Kirchhoff's voltage law to the left loop (loop $ABEF$):
$9 - 3 I - 8(I - I_1) - 2 I = 0$
$9 - 5 I - 8(I - 0.5 I) = 0$
$9 - 5 I - 8(0.5 I) = 0$
$9 - 5 I - 4 I = 0$
$9 I = 9 \Rightarrow I = 1 A$.
Therefore,the current through the $4 \Omega$ resistor is $I_1 = 0.5 \times 1 A = 0.5 A$.

(C) The drift velocity $v_{d}$ is given by the formula $v_{d} = \frac{I}{neA}$,where $I$ is the current,$n$ is the electron density,$e$ is the charge of an electron,and $A$ is the cross-sectional area.
Since $I = \frac{E}{R}$ and $R = \rho \frac{l}{A}$,we have $I = \frac{E A}{\rho l}$.
Substituting this into the drift velocity formula:
$v_{d} = \frac{E A}{\rho l n e A} = \frac{E}{\rho l n e}$.
From this expression,we see that $v_{d} \propto \frac{1}{l}$.
If the length is changed to $l' = 2l$,the new drift velocity $v_{d}'$ is:
$v_{d}' = \frac{E}{\rho (2l) n e} = \frac{1}{2} \left( \frac{E}{\rho l n e} \right) = \frac{v_{d}}{2}$.

(C) The given circuit can be simplified by observing the symmetry.
Let the points be labeled. The circuit consists of two loops in series.
Each loop has two resistors of resistance $r$ in series,which are in parallel with another branch of two resistors of resistance $r$ in series.
For the first part (left side),the two branches are in parallel,each having a resistance of $r + r = 2r$. The equivalent resistance of this part is $R_1 = \frac{2r \times 2r}{2r + 2r} = \frac{4r^2}{4r} = r$.
Similarly,for the second part (right side),the two branches are in parallel,each having a resistance of $r + r = 2r$. The equivalent resistance of this part is $R_2 = \frac{2r \times 2r}{2r + 2r} = r$.
Since these two parts are connected in series between points $a$ and $b$,the total equivalent resistance is $R_{eq} = R_1 + R_2 = r + r = 2r$.

(B) The de-Broglie wavelength is given by the formula: $\lambda = \frac{h}{\sqrt{2mK}}$.
Given values are:
$h = 6.6 \times 10^{-34} \ J \cdot s$
$m = 1 \times 10^{-30} \ kg$
$K = 200 \ eV = 200 \times 1.6 \times 10^{-19} \ J = 3.2 \times 10^{-17} \ J$.
Substituting these values into the formula:
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times (1 \times 10^{-30}) \times (3.2 \times 10^{-17})}}$
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{6.4 \times 10^{-47}}}$
$\lambda = \frac{6.6 \times 10^{-34}}{8 \times 10^{-23.5}}$
Using the calculation $\sqrt{64 \times 10^{-48}} = 8 \times 10^{-24}$,we get:
$\lambda = \frac{6.6 \times 10^{-34}}{8 \times 10^{-24}} = 0.825 \times 10^{-10} \ m = 8.25 \times 10^{-11} \ m$.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy of photoelectrons is given by $K_{max} = h v - \phi_{0}$,where $\phi_{0}$ is the work function of the metal.
Since $K_{max} = e V_{s}$,where $V_{s}$ is the stopping potential,we have $e V_{s} = h v - \phi_{0}$.
For the first case: $e V_{1} = h v_{1} - \phi_{0}$ ---$(i)$
For the second case: $e V_{2} = h v_{2} - \phi_{0}$ ---(ii)
Subtracting equation $(i)$ from equation (ii):
$e V_{2} - e V_{1} = (h v_{2} - \phi_{0}) - (h v_{1} - \phi_{0})$
$e(V_{2} - V_{1}) = h(v_{2} - v_{1})$
$v_{2} - v_{1} = \frac{e}{h}(V_{2} - V_{1})$
$v_{2} = v_{1} + \frac{e}{h}(V_{2} - V_{1})$

(B) Given,$B = 2t + 4t^{2}$.
At $t = 0 \ s$,$B_{1} = 2(0) + 4(0)^{2} = 0 \ T$.
At $t = 2 \ s$,$B_{2} = 2(2) + 4(2)^{2} = 4 + 16 = 20 \ T$.
The magnetic flux is $\phi = B \cdot A = B \cdot \pi r^{2}$.
The change in magnetic flux is $\Delta \phi = \phi_{2} - \phi_{1} = \pi r^{2} (B_{2} - B_{1})$.
Substituting the values,$\Delta \phi = \pi r^{2} (20 - 0) = 20 \pi r^{2} \ Wb$.
The induced charge $Q$ is given by $Q = \frac{\Delta \phi}{R}$.
Therefore,$Q = \frac{20 \pi r^{2}}{R} \ C$.

(B) According to Gauss's Law,the electric flux through a spherical surface of radius $R$ is given by $\oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\varepsilon_{0}}$.
For a sphere of radius $R$,the electric field $E$ at the surface is uniform,so $E(4 \pi R^{2}) = \frac{q_{enclosed}}{\varepsilon_{0}}$.
The total charge $q_{enclosed}$ is calculated by integrating the volume charge density $\rho = k r$ over the volume of the sphere:
$q_{enclosed} = \int_{0}^{R} \rho (4 \pi r^{2}) dr = \int_{0}^{R} (k r) (4 \pi r^{2}) dr = 4 \pi k \int_{0}^{R} r^{3} dr = 4 \pi k \left[ \frac{r^{4}}{4} \right]_{0}^{R} = \pi k R^{4}$.
Substituting this into Gauss's Law:
$E(4 \pi R^{2}) = \frac{\pi k R^{4}}{\varepsilon_{0}}$.
Solving for $E$:
$E = \frac{\pi k R^{4}}{4 \pi R^{2} \varepsilon_{0}} = \frac{k R^{2}}{4 \varepsilon_{0}}$.

(C) The work done in moving a charge in an electrostatic field is equal to the change in the electrostatic potential energy of the system.
$W = U_{f} - U_{i}$
$U = \frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r}$
Initial potential energy $U_{i}$ when the charge $+5 \text{ C}$ is at $A(2, 0) \text{ m}$:
$U_{i} = \frac{1}{4 \pi \varepsilon_{0}} \frac{(2)(5)}{2} = \frac{1}{4 \pi \varepsilon_{0}} (5)$
Final potential energy $U_{f}$ when the charge $+5 \text{ C}$ is at $B(0, 2) \text{ m}$:
$U_{f} = \frac{1}{4 \pi \varepsilon_{0}} \frac{(2)(5)}{2} = \frac{1}{4 \pi \varepsilon_{0}} (5)$
Since the distance $r$ from the origin $O$ to point $A$ is $2 \text{ m}$ and from the origin $O$ to point $B$ is also $2 \text{ m}$,the potential energy remains the same.
$W = U_{f} - U_{i} = 0 \text{ J}$.

(C) The force $\vec{F}$ on a current-carrying wire is given by $\vec{F} = I(\vec{L} \times \vec{B})$.
Here,$I = 1 \ A$,$\vec{L} = 0.5 \hat{i} \ m$,and $\vec{B} = (2\hat{i} + 4\hat{j}) \ T$.
Substituting the values:
$\vec{F} = 1 \times (0.5 \hat{i} \times (2\hat{i} + 4\hat{j}))$
$\vec{F} = 0.5 \times 2(\hat{i} \times \hat{i}) + 0.5 \times 4(\hat{i} \times \hat{j})$
Since $\hat{i} \times \hat{i} = 0$ and $\hat{i} \times \hat{j} = \hat{k}$:
$\vec{F} = 0 + 2\hat{k} = 2\hat{k} \ N$.
The magnitude of the force is $|\vec{F}| = 2 \ N$,and the direction is along the positive $z$-axis.

(A) Given,magnetic moment $M = 200 \text{ A m}^2$.
Magnetic field $B = 0.30 \text{ N A}^{-1} \text{ m}^{-1}$.
Angle $\theta = 30^{\circ}$.
We know that the torque $\tau$ acting on a magnetic dipole in a magnetic field is given by the formula:
$\tau = M B \sin \theta$.
Substituting the given values:
$\tau = 200 \times 0.30 \times \sin(30^{\circ})$.
Since $\sin(30^{\circ}) = \frac{1}{2}$,
$\tau = 200 \times 0.30 \times \frac{1}{2} = 100 \times 0.30 = 30 \text{ N m}$.
Thus,the torque required is $30 \text{ N m}$.

(D) The current $I$ enters at $A$ and splits into two paths: one through $AB$ and the other through $AC$. Since the wires are uniform, the resistance of path $AB$ is $R$ and the resistance of path $AC$ is $R$.
At the midpoint of $BC$ (let's call it $M$), the current from $B$ to $M$ and $C$ to $M$ recombines to exit.
Due to the symmetry of the equilateral triangle and the current distribution, the magnetic field produced by the current in segment $AB$ and $BM$ at the centroid $O$ is equal and opposite to the magnetic field produced by the current in segment $AC$ and $CM$ at the centroid $O$.
Specifically, the magnetic field at the centroid $O$ due to the current flowing through the left half of the triangle $(A-B-M)$ is equal in magnitude but opposite in direction to the magnetic field due to the current flowing through the right half of the triangle $(A-C-M)$.
Therefore, the net magnetic field at the centroid $O$ is $B_{net} = B_{left} + B_{right} = 0$.

(A) The force on a charged particle moving with velocity $\vec{v}$ in an electric field $\vec{E}$ and magnetic field $\vec{B}$ is given by the Lorentz force equation: $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$.
For the particle to move with a uniform velocity,the net force must be zero,i.e.,$\vec{F} = 0$.
Case $1$: If $E=0$ and $B=0$,the particle moves with uniform velocity (Newton's first law).
Case $2$: If $E=0$ and $B \neq 0$,the particle moves with uniform velocity if it moves parallel or anti-parallel to the magnetic field $(\vec{v} \times \vec{B} = 0)$.
Case $3$: If $E \neq 0$ and $B \neq 0$,the particle moves with uniform velocity if the electric force and magnetic force cancel each other,which occurs when $\vec{E} = -(\vec{v} \times \vec{B})$. In terms of magnitudes,this implies $E = vB$ (where $\vec{v} \perp \vec{B}$ and $\vec{E} \perp \vec{v}$).
Thus,the condition $E=vB$ is a valid possibility for uniform velocity.

(A) The horizontal component of Earth's magnetic field is given by $H = B \cos \delta$,where $B$ is the total magnetic field and $\delta$ is the angle of dip.
Assuming the total magnetic field $B$ is the same at both locations:
$H_1 = B \cos 30^{\circ}$ and $H_2 = B \cos 45^{\circ}$.
Taking the ratio:
$\frac{H_1}{H_2} = \frac{B \cos 30^{\circ}}{B \cos 45^{\circ}} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \frac{\sqrt{3}}{2} \times \sqrt{2} = \frac{\sqrt{3}}{\sqrt{2}}$.
Thus,the ratio is $\sqrt{3} : \sqrt{2}$.

(A) The alpha particle $\left({ }^{4} He\right)$ consists of $2$ protons and $2$ neutrons.
The mass of the constituents is $m_{c} = 2(m_{p} + m_{n}) = 2(1.00783 + 1.00867) = 2(2.01650) = 4.03300 \ amu$.
The mass defect $\Delta m$ is given by $\Delta m = m_{c} - m_{He} = 4.03300 - 4.00300 = 0.0300 \ amu$.
The binding energy $E$ is calculated using the conversion factor $1 \ amu = 931 \ MeV$.
$E = \Delta m \times 931 \ MeV/amu = 0.0300 \times 931 = 27.93 \ MeV$.
Thus,the binding energy is approximately $27.9 \ MeV$.

(A) The radioactive decay law is given by $N(t) = N_{0} e^{-\lambda t}$,where $\lambda = \frac{\ln 2}{T}$.
Let $t_{1}$ be the time when the number of atoms is $N_{0}/2$ and $t_{2}$ be the time when the number of atoms is $N_{0}/10$.
For $N(t_{1}) = N_{0}/2$,we have $\frac{N_{0}}{2} = N_{0} e^{-\lambda t_{1}} \Rightarrow e^{\lambda t_{1}} = 2 \Rightarrow \lambda t_{1} = \ln 2$.
For $N(t_{2}) = N_{0}/10$,we have $\frac{N_{0}}{10} = N_{0} e^{-\lambda t_{2}} \Rightarrow e^{\lambda t_{2}} = 10 \Rightarrow \lambda t_{2} = \ln 10$.
The time interval required is $\Delta t = t_{2} - t_{1} = \frac{\ln 10 - \ln 2}{\lambda} = \frac{\ln(10/2)}{\lambda} = \frac{\ln 5}{\lambda}$.
Since $\lambda = \frac{\ln 2}{T}$,we substitute $\lambda$ to get $\Delta t = \frac{\ln 5}{(\ln 2 / T)} = T \frac{\ln 5}{\ln 2} = T \log_{2} 5$.

(C) Let the two fixed charges $Q$ be located at positions $-d$ and $+d$.
When the charge $q$ is displaced by a distance $x$ from the origin,the net force acting on it is:
$F = F_{left} - F_{right} = \frac{1}{4\pi\varepsilon_{0}} \frac{Qq}{(d-x)^{2}} - \frac{1}{4\pi\varepsilon_{0}} \frac{Qq}{(d+x)^{2}}$
$F = \frac{Qq}{4\pi\varepsilon_{0}} \left[ \frac{(d+x)^{2} - (d-x)^{2}}{(d^{2}-x^{2})^{2}} \right] = \frac{Qq}{4\pi\varepsilon_{0}} \left[ \frac{4dx}{(d^{2}-x^{2})^{2}} \right]$
Since $x \ll d$,we can approximate $(d^{2}-x^{2})^{2} \approx d^{4}$:
$F \approx \frac{Qq}{4\pi\varepsilon_{0}} \cdot \frac{4dx}{d^{4}} = \frac{Qqx}{\pi\varepsilon_{0}d^{3}}$
Since the force is directed towards the mean position,$F = -kx$,where $k = \frac{Qq}{\pi\varepsilon_{0}d^{3}}$.
The time period $T$ is given by:
$T = 2\pi \sqrt{\frac{M}{k}} = 2\pi \sqrt{\frac{M \pi \varepsilon_{0} d^{3}}{Qq}} = 2 \sqrt{\frac{\pi^{3} M \varepsilon_{0} d^{3}}{Qq}}$

(B) For a long-sighted person (hypermetropia),the person cannot see objects clearly at the normal near point $(u = -25 \ cm)$. However,the problem states the person's current near point is $v = -60 \ cm$. We want to bring the near point to $u = -12 \ cm$ using a lens.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
Here,$v = -60 \ cm$ (image formed at the person's actual near point) and $u = -12 \ cm$ (object placed at the desired near point).
$\frac{1}{f} = \frac{1}{-60} - \frac{1}{-12} = \frac{-1 + 5}{60} = \frac{4}{60} = \frac{1}{15} \ cm^{-1}$.
Since $f$ is in $cm$,$f = 15 \ cm = 0.15 \ m$.
The power of the lens $P = \frac{1}{f(m)} = \frac{1}{0.15} = \frac{100}{15} = +\frac{20}{3} \ D$.

(A) For a convex lens,the magnification $m$ is given by $m = \frac{f}{f+u}$,where $u$ is the object distance (using sign convention,$u$ is negative).
Given that the magnification $m$ is the same for two different object positions,this is only possible if one image is real and the other is virtual.
For the first case,$u_1 = -16 \ cm$. The magnification is $m = \frac{f}{f-16}$. Since $m > 1$,this must be a virtual image,so $m = \frac{f}{f-16}$.
When the object is moved towards the lens by $8 \ cm$,the new object distance is $u_2 = -(16 - 8) = -8 \ cm$. The magnification is $m' = \frac{f}{f-8}$. Since the magnitude of magnification is the same,we have $|\frac{f}{f-16}| = |\frac{f}{f-8}|$.
Since $m > 1$,one must be positive and one negative: $\frac{f}{f-16} = -\frac{f}{f-8}$.
Dividing by $f$ (as $f \neq 0$): $\frac{1}{f-16} = -\frac{1}{f-8}$.
$\Rightarrow f - 8 = -(f - 16) = -f + 16$.
$\Rightarrow 2f = 24$.
$\Rightarrow f = 12 \ cm$.

(A) According to the Lens Maker's Formula,the focal length $f'$ of a lens in a medium is given by $\frac{1}{f'} = (\frac{n_{1}}{n_{2}} - 1)(\frac{1}{R_{1}} - \frac{1}{R_{2}})$.
For a biconvex lens,$R_{1} = R$ and $R_{2} = -R$,so $(\frac{1}{R_{1}} - \frac{1}{R_{2}}) = \frac{2}{R}$.
Thus,$\frac{1}{f'} = (\frac{n_{1}}{n_{2}} - 1)(\frac{2}{R})$.
If $n_{1} > n_{2}$,then $(\frac{n_{1}}{n_{2}} - 1) > 0$,so $f' > 0$,and the lens behaves as a convex (converging) lens.
If $n_{1} < n_{2}$,then $(\frac{n_{1}}{n_{2}} - 1) < 0$,so $f' < 0$,and the lens behaves as a concave (diverging) lens.
Therefore,the behavior of the lens depends on the relative values of $n_{1}$ and $n_{2}$.

(D) For dispersion without deviation,the net deviation produced by the combination of two prisms must be zero.
Let the refractive indices of prisms $P_{1}$ and $P_{2}$ be $\mu$ and $\mu^{\prime}$ and their refracting angles be $A$ and $A^{\prime}$ respectively.
The condition for no deviation is given by $\delta + \delta^{\prime} = 0$,which implies $(\mu - 1)A = (\mu^{\prime} - 1)A^{\prime}$.
Given values are $\mu = 1.54$,$A = 4^{\circ}$,and $A^{\prime} = 3^{\circ}$.
Substituting these values into the equation: $(1.54 - 1) \times 4^{\circ} = (\mu^{\prime} - 1) \times 3^{\circ}$.
$0.54 \times 4 = (\mu^{\prime} - 1) \times 3$.
$2.16 = (\mu^{\prime} - 1) \times 3$.
$\mu^{\prime} - 1 = \frac{2.16}{3} = 0.72$.
$\mu^{\prime} = 0.72 + 1 = 1.72$.

(C) The current $I$ is given by $I = \frac{e}{T} = \frac{e \omega}{2 \pi}$. This is independent of $r$.
The magnetic moment $\mu = I A = \left( \frac{e \omega}{2 \pi} \right) (\pi r^2) = \frac{e \omega r^2}{2}$.
The angular momentum $L = m v r = m (\omega r) r = m \omega r^2$.
Comparing $\mu$ and $L$,we get $\mu = \frac{e}{2m} L$.
Thus,the magnetic moment is proportional to the angular momentum,and the ratio $\mu / L = e / (2m)$ is known as the gyromagnetic ratio. Statement $C$ is correct as it relates $\mu$ and $L$ via the factor $e / (2m)$.

(A) In an $n-p-n$ or $p-n-p$ transistor,the emitter is heavily doped to provide a large number of majority charge carriers. The base is very thin and lightly doped to minimize recombination. The collector is moderately doped and is physically larger than the emitter to handle the power dissipation. Therefore,the emitter has a higher degree of doping compared to the collector.

(D) Let the output of the $NOR$ gate be $Y$. The $NOR$ gate performs the operation $Y = \overline{A+B}$. The $NAND$ gate then takes $Y$ and $C$ as inputs to produce output $D = \overline{Y \cdot C}$.
Case $1$: $A=0, B=0, C=0$
$Y = \overline{0+0} = \overline{0} = 1$
$D = \overline{Y \cdot C} = \overline{1 \cdot 0} = \overline{0} = 1$
Case $2$: $A=1, B=0, C=1$
$Y = \overline{1+0} = \overline{1} = 0$
$D = \overline{Y \cdot C} = \overline{0 \cdot 1} = \overline{0} = 1$
Thus,the outputs are $1$ and $1$ respectively.

(B) For constructive interference,the path difference $\Delta x$ between the two waves must be an integral multiple of the wavelength,i.e.,$\Delta x = n\lambda$,where $n = 0, 1, 2, \dots$.
Let the point be $P(x, 0)$ on the $x$-axis.
The distance from $S_{1}(0,0)$ to $P(x,0)$ is $r_{1} = \sqrt{(x-0)^{2} + (0-0)^{2}} = x$.
The distance from $S_{2}(0,3\lambda)$ to $P(x,0)$ is $r_{2} = \sqrt{(x-0)^{2} + (0-3\lambda)^{2}} = \sqrt{x^{2} + 9\lambda^{2}}$.
The path difference is $\Delta x = |r_{2} - r_{1}| = |\sqrt{x^{2} + 9\lambda^{2}} - x|$.
Testing option $(B)$ $(4\lambda, 0)$:
$r_{1} = 4\lambda$.
$r_{2} = \sqrt{(4\lambda)^{2} + (3\lambda)^{2}} = \sqrt{16\lambda^{2} + 9\lambda^{2}} = \sqrt{25\lambda^{2}} = 5\lambda$.
Path difference $\Delta x = |5\lambda - 4\lambda| = \lambda$.
Since $\Delta x = 1\lambda$,which is an integral multiple of $\lambda$,constructive interference occurs at $(4\lambda, 0)$.

(A) Given: Mass of the particle $= M$,Charge $= q$,Electric field $= E$,Initial velocity $u = 0$.
Acceleration $a = F/M = qE/M$.
Distance travelled $S = ut + (1/2)at^2 = (1/2)(qE/M)t^2$.
Final velocity $v = u + at = (qE/M)t$.
Kinetic energy $T = (1/2)Mv^2 = (1/2)M(qEt/M)^2 = (1/2)(q^2E^2t^2/M)$.
Now,the ratio $T/S = [(1/2)(q^2E^2t^2/M)] / [(1/2)(qE/M)t^2] = qE$.
Since $q$ and $E$ are constants,the ratio $T/S = qE$ is independent of time $t$ and remains constant.

(B) The particle is accelerated by a uniform electric field $E$ over a distance $D$. The work done by the electric field is equal to the kinetic energy $(KE)$ gained by the particle.
$KE = W = q E D$
At the point of closest approach $(r_0)$,the entire kinetic energy of the particle is converted into electrostatic potential energy $(PE)$ due to the interaction with the fixed charge $Q$.
$PE = \frac{1}{4 \pi \varepsilon_0} \frac{q Q}{r_0}$
Equating $KE$ and $PE$ at the point of closest approach:
$q E D = \frac{1}{4 \pi \varepsilon_0} \frac{q Q}{r_0}$
Solving for $r_0$:
$r_0 = \frac{Q}{4 \pi \varepsilon_0 E D}$