WBJEE 2019 Physics Question Paper with Answer and Solution

(A) The total kinetic energy of the two blocks before the collision is $K.E. = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.
Since the collision is perfectly inelastic (the blocks melt),the maximum loss in kinetic energy is equal to the total initial kinetic energy,which is $mv^2$.
This energy is used to raise the temperature of the ice from $-8^{\circ}C$ to $0^{\circ}C$ and then to melt it.
The energy required for one block is $Q = ms\Delta\theta + mL$.
For two blocks,the total energy required is $2(ms\Delta\theta + mL)$.
Equating the energy loss to the heat required: $mv^2 = 2m(s\Delta\theta + L)$.
Canceling $m$ from both sides: $v^2 = 2(s\Delta\theta + L)$.
Given: $s = 2100 \ Jkg^{-1}K^{-1}$,$\Delta\theta = 8^{\circ}C$,$L = 3.36 \times 10^5 \ Jkg^{-1}$.
$v^2 = 2(2100 \times 8 + 3.36 \times 10^5) = 2(16800 + 336000) = 2(352800) = 705600$.
$v = \sqrt{705600} = 840 \ ms^{-1}$.

(A) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the orbital radius $r$:
$T^2 \propto r^3$
Since angular speed $\omega = \frac{2\pi}{T}$,we have $T = \frac{2\pi}{\omega}$. Substituting this into the law:
$(\frac{2\pi}{\omega})^2 \propto r^3 \Rightarrow \frac{1}{\omega^2} \propto r^3 \Rightarrow r^3 \omega^2 = \text{constant}$.
For the earth $(E)$ and the planet $(P)$:
$r_E^3 \omega_E^2 = r_P^3 \omega_P^2$
Given $r_E = R$ and $\omega_P = 2\omega_E$:
$R^3 \omega_E^2 = r_P^3 (2\omega_E)^2$
$R^3 \omega_E^2 = r_P^3 (4\omega_E^2)$
$R^3 = 4 r_P^3$
$r_P^3 = \frac{R^3}{4} = \frac{R^3}{2^2}$
Taking the cube root on both sides:
$r_P = \frac{R}{2^{2/3}} = 2^{-2/3} R$.

(B) For a gas at temperature $T$,the internal energy is given by:
$U = \frac{f}{2} \mu R T$
where $f$ is the degree of freedom.
The change in internal energy is:
$\Delta U = \frac{f}{2} \mu R \Delta T$
For a process at constant volume,the heat supplied is equal to the change in internal energy:
$\Delta Q_V = \mu C_V \Delta T = \Delta U$
Equating the two expressions for $\Delta U$:
$\mu C_V \Delta T = \frac{f}{2} \mu R \Delta T$
$C_V = \frac{f}{2} R$
For rigid diatomic molecules,the degree of freedom $f = 5$ ($3$ translational + $2$ rotational).
Substituting $f = 5$ into the formula:
$C_V = \frac{5}{2} R$

(D) Let $T$ be the tension in the string and $a$ be the magnitude of the acceleration of the masses.
Since $m_{2} > m_{1}$,the mass $m_{2}$ moves downwards with acceleration $a$,and the mass $m_{1}$ moves upwards with acceleration $a$.
For mass $m_{1}$ (moving upwards): $T - m_{1}g = m_{1}a$ ....$(i)$
For mass $m_{2}$ (moving downwards): $m_{2}g - T = m_{2}a$ ....(ii)
Adding equations $(i)$ and (ii),we get:
$(T - m_{1}g) + (m_{2}g - T) = m_{1}a + m_{2}a$
$(m_{2} - m_{1})g = (m_{1} + m_{2})a$
$a = \frac{m_{2} - m_{1}}{m_{1} + m_{2}} g$

(C) When a small spherical body falls through a viscous fluid,it experiences three forces: gravitational force (weight) acting downwards,buoyant force acting upwards,and viscous drag force acting upwards.
At terminal velocity $v$,the body moves with a constant velocity,which implies that the net acceleration of the body is zero.
According to Newton's second law of motion,$F_{\text{net}} = ma$. Since the acceleration $a = 0$,the net force $F_{\text{net}}$ acting on the body is $0$.
Therefore,the weight of the body is exactly balanced by the sum of the buoyant force and the viscous drag force.

(C) Given: The fractional change in length is $\frac{\Delta l}{l} = -0.01$ (since it decreases by $1 \%$).
Poisson's ratio, $\sigma = 0.2$.
For a rod, the volume $V = A \times l$, where $A$ is the cross-sectional area.
The fractional change in volume is given by $\frac{\Delta V}{V} = \frac{\Delta A}{A} + \frac{\Delta l}{l}$.
Since $A = w \times t$ (width $\times$ thickness), $\frac{\Delta A}{A} = \frac{\Delta w}{w} + \frac{\Delta t}{t}$.
By definition of Poisson's ratio, $\sigma = -\frac{\Delta w / w}{\Delta l / l} = -\frac{\Delta t / t}{\Delta l / l}$.
Thus, $\frac{\Delta w}{w} = -\sigma \frac{\Delta l}{l}$ and $\frac{\Delta t}{t} = -\sigma \frac{\Delta l}{l}$.
Substituting these into the volume equation:
$\frac{\Delta V}{V} = -\sigma \frac{\Delta l}{l} - \sigma \frac{\Delta l}{l} + \frac{\Delta l}{l} = \frac{\Delta l}{l} (1 - 2\sigma)$.
Substituting the values: $\frac{\Delta V}{V} = -0.01 \times (1 - 2 \times 0.2) = -0.01 \times (1 - 0.4) = -0.01 \times 0.6 = -0.006$.
Therefore, the volume decreases by $0.6 \%$.

(A) Given,initial velocity $u = 10 \ m/s$.
Range $R = 5 \ m$.
Acceleration due to gravity $g = 10 \ m/s^2$.
The formula for the range of a projectile is $R = \frac{u^2 \sin(2\alpha)}{g}$.
Substituting the given values:
$5 = \frac{(10)^2 \sin(2\alpha)}{10}$
$5 = \frac{100 \sin(2\alpha)}{10}$
$5 = 10 \sin(2\alpha)$
$\sin(2\alpha) = \frac{5}{10} = 0.5$.
Since $\sin(30^{\circ}) = 0.5$,we have $2\alpha = 30^{\circ}$ or $2\alpha = 150^{\circ}$.
Therefore,$\alpha = 15^{\circ}$ or $\alpha = 75^{\circ}$.
Comparing with the given options,the correct value is $15^{\circ}$.

(A) Given,cross-sectional area of the nozzle $A = \frac{5}{\sqrt{21}} \times 10^{-3} \text{ m}^2$.
Volume flow rate $Q = \frac{1 \text{ m}^3}{10 \text{ s}} = 0.1 \text{ m}^3/\text{s}$.
The velocity of water $v$ is given by $v = \frac{Q}{A} = \frac{0.1}{\frac{5}{\sqrt{21}} \times 10^{-3}} = \frac{10^{-1} \times \sqrt{21}}{5 \times 10^{-3}} = 20\sqrt{21} \text{ m/s}$.
When the water hits a rigid wall,its kinetic energy is converted into heat energy. The maximum possible increase in temperature $\Delta T$ is given by the energy balance equation: $\frac{1}{2}mv^2 = ms\Delta T$.
Here,$s$ is the specific heat capacity of water $= 4.2 \times 10^3 \text{ J/(kg} \cdot ^{\circ}\text{C)}$.
Thus,$\Delta T = \frac{v^2}{2s} = \frac{(20\sqrt{21})^2}{2 \times 4.2 \times 10^3}$.
$\Delta T = \frac{400 \times 21}{8.4 \times 10^3} = \frac{8400}{8400} = 1^{\circ} \text{C}$.

(C) According to the Stefan-Boltzmann law,the thermal energy radiated per second $(Q)$ by a black body is given by $Q = \sigma A T^4$,where $\sigma$ is the Stefan-Boltzmann constant,$A$ is the surface area,and $T$ is the absolute temperature in Kelvin.
Given that the surface areas of both bodies $A$ and $B$ are equal $(A_A = A_B = A)$,the ratio of the energy radiated is:
$\frac{Q_A}{Q_B} = \frac{\sigma A T_A^4}{\sigma A T_B^4} = \left(\frac{T_A}{T_B}\right)^4$
Convert the temperatures from Celsius to Kelvin:
$T_A = 27^{\circ} C = 27 + 273 = 300 \ K$
$T_B = 177^{\circ} C = 177 + 273 = 450 \ K$
Substitute the values into the ratio:
$\frac{Q_A}{Q_B} = \left(\frac{300}{450}\right)^4 = \left(\frac{2}{3}\right)^4$
$\frac{Q_A}{Q_B} = \frac{16}{81}$
Thus,the ratio of the thermal energy radiated per second by $A$ to that by $B$ is $16: 81$.

(B) For an adiabatic process,the relationship between temperature and volume is given by $T V^{\gamma-1} = \text{constant}$.
Let the temperature after adiabatic compression be $T$. Then,$T_{0} V_{0}^{\gamma-1} = T \left(\frac{V_{0}}{2}\right)^{\gamma-1}$.
Solving for $T$,we get $T = T_{0} 2^{\gamma-1}$.
When the gas is allowed to come to thermal equilibrium with the surroundings at constant volume $\frac{V_{0}}{2}$,the heat released $\Delta Q$ is equal to the change in internal energy: $\Delta Q = n C_{V} \Delta T$.
Using $C_{V} = \frac{R}{\gamma-1}$ and the ideal gas law $n R T_{0} = p_{0} V_{0}$,we have:
$\Delta Q = n \left(\frac{R}{\gamma-1}\right) (T - T_{0}) = \frac{n R T_{0}}{\gamma-1} (2^{\gamma-1} - 1)$.
Substituting $n R T_{0} = p_{0} V_{0}$,the heat released is $\frac{p_{0} V_{0}}{\gamma-1} (2^{\gamma-1} - 1)$.

(A) This process is known as free expansion of an ideal gas.
Since the system is thermally isolated,there is no heat exchange with the surroundings,so $Q = 0$.
Since the gas expands into a vacuum,there is no external pressure against which the gas does work,so the work done $W = 0$.
According to the first law of thermodynamics,$\Delta U = Q - W$.
Since $Q = 0$ and $W = 0$,the change in internal energy $\Delta U = 0$.
For an ideal gas,the internal energy depends only on its temperature $(U \propto T)$.
Since $\Delta U = 0$,the temperature of the ideal gas remains constant.
Therefore,the final temperature attained at equilibrium will be $T$.

(C) Given,Power $(P) =$ constant.
Kinetic Energy $(KE) = \frac{1}{2} m v^{2}$.
We know that,$P = \frac{d(KE)}{dt} = \frac{d}{dt} (\frac{1}{2} m v^{2}) = m v \frac{dv}{dt}$.
Since $P$ is constant,$m v \frac{dv}{dt} = P$.
Integrating both sides with respect to time $t$:
$\int m v dv = \int P dt \Rightarrow \frac{1}{2} m v^{2} = P t$ (assuming initial velocity is $0$ at $t=0$).
$v^{2} = \frac{2 P}{m} t \Rightarrow v = \sqrt{\frac{2 P}{m}} t^{1/2}$.
Since $v = \frac{ds}{dt}$,we have $\frac{ds}{dt} = \sqrt{\frac{2 P}{m}} t^{1/2}$.
Integrating with respect to $t$:
$s = \int \sqrt{\frac{2 P}{m}} t^{1/2} dt = \sqrt{\frac{2 P}{m}} \cdot \frac{t^{3/2}}{3/2} = \frac{2}{3} \sqrt{\frac{2 P}{m}} t^{3/2}$.
Thus,$s \propto t^{3/2}$,which means $s = a t^{3/2}$ where $a$ is a constant.

(C) In a series $R-L-C$ circuit,the resonance frequency is given by $f_0 = \frac{1}{2\pi\sqrt{LC}}$.
Given: $R = 100 \Omega$,$L = 20 \text{ mH} = 20 \times 10^{-3} \text{ H}$,$f_0 = \frac{1250}{\pi} \text{ Hz}$.
Substituting the values into the resonance formula:
$\frac{1250}{\pi} = \frac{1}{2\pi\sqrt{20 \times 10^{-3} \times C}}$
$1250 = \frac{1}{2\sqrt{0.02 \times C}}$
$2500 = \frac{1}{\sqrt{0.02 \times C}}$
Squaring both sides:
$6.25 \times 10^6 = \frac{1}{0.02 \times C}$
$C = \frac{1}{0.02 \times 6.25 \times 10^6} = \frac{1}{0.125 \times 10^6} = 8 \times 10^{-6} \text{ F} = 8 \mu\text{F}$.
When charged by a $DC$ source of $V = 25 \text{ V}$,the charge $Q$ stored is:
$Q = C \times V = 8 \times 10^{-6} \text{ F} \times 25 \text{ V} = 200 \times 10^{-6} \text{ C} = 0.2 \times 10^{-3} \text{ C} = 0.2 \text{ mC}$.

(A) Given,inductance $L = 60 \text{ mH} = 60 \times 10^{-3} \text{ H}$.
Phase difference in $L-R$ circuit,$\theta_{1} = 60^{\circ}$.
Capacitance $C = 0.5 \text{ } \mu\text{F} = 0.5 \times 10^{-6} \text{ F}$.
Phase difference in $R-C$ circuit,$\theta_{2} = 30^{\circ}$.
For $L-R$ circuit,$\tan \theta_{1} = \frac{X_{L}}{R} = \frac{\omega L}{R} \quad \dots(i)$.
For $R-C$ circuit,$\tan \theta_{2} = \frac{X_{C}}{R} = \frac{1}{\omega CR} \quad \dots(ii)$.
Dividing $(i)$ by (ii): $\frac{\tan \theta_{1}}{\tan \theta_{2}} = \frac{\omega L / R}{1 / (\omega CR)} = \omega^{2} LC$.
Substituting values: $\frac{\tan 60^{\circ}}{\tan 30^{\circ}} = \frac{\sqrt{3}}{1/\sqrt{3}} = 3 = \omega^{2} LC$.
$\omega^{2} = \frac{3}{LC} = \frac{3}{60 \times 10^{-3} \times 0.5 \times 10^{-6}} = \frac{3}{30 \times 10^{-9}} = 10^{8}$.
$\omega = \sqrt{10^{8}} = 10^{4} \text{ rad/s}$.
Since $\omega = 2 \pi f$,then $f = \frac{\omega}{2 \pi} = \frac{10^{4}}{2 \pi} \text{ Hz}$.

(D) According to the Bohr's atomic model,the angular momentum $L$ is given by:
$L = mvr = \frac{nh}{2\pi} \dots (i)$
Since angular velocity $\omega = \frac{v}{r}$,we have $v = r\omega$.
Substituting $v = r\omega$ into equation $(i)$:
$m(r\omega)r = \frac{nh}{2\pi} \Rightarrow m\omega r^2 = \frac{nh}{2\pi} \Rightarrow \omega = \frac{nh}{2\pi mr^2} \dots (ii)$
The radius of the electron in the $n$-th orbit is given by:
$r = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2} \dots (iii)$
Substituting the expression for $r$ from equation $(iii)$ into equation $(ii)$:
$\omega = \frac{nh}{2\pi m} \left( \frac{\pi m Z e^2}{n^2 h^2 \epsilon_0} \right)^2$
$\omega = \frac{nh}{2\pi m} \cdot \frac{\pi^2 m^2 Z^2 e^4}{n^4 h^4 \epsilon_0^2}$
$\omega = \frac{\pi m Z^2 e^4}{2 h^3 \epsilon_0^2} \cdot \frac{1}{n^3}$
Therefore,$\omega \propto \frac{1}{n^3}$.

(A) When the capacitor is fully charged,the charge on it is $q = CE$.
The total work done by the $DC$ source (battery) is $W = qE = (CE)E = CE^2$.
The energy stored in the capacitor is $U = \frac{1}{2} CE^2$.
According to the law of conservation of energy,the work done by the battery is equal to the sum of the energy stored in the capacitor and the energy dissipated as heat in the resistor $R$.
Therefore,the energy dissipated in the resistance $R$ is given by:
$H = W - U$
$H = CE^2 - \frac{1}{2} CE^2$
$H = \frac{1}{2} CE^2$.

(D) The given circuit consists of two parallel branches connected across a $150 V$ battery.
Branch $1$ (top) contains two $20 \mu F$ capacitors in series. The equivalent capacitance $C_1$ is given by:
$\frac{1}{C_1} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10} \implies C_1 = 10 \mu F$.
Branch $2$ (bottom) also contains two $20 \mu F$ capacitors in series. The equivalent capacitance $C_2$ is:
$\frac{1}{C_2} = \frac{1}{20} + \frac{1}{20} = \frac{1}{10} \implies C_2 = 10 \mu F$.
Since these two branches are in parallel,the total equivalent capacitance $C_{eq}$ is:
$C_{eq} = C_1 + C_2 = 10 \mu F + 10 \mu F = 20 \mu F$.
The total charge $Q$ stored is given by $Q = C_{eq} V$:
$Q = 20 \times 10^{-6} F \times 150 V = 3000 \times 10^{-6} C = 3 \times 10^{-3} C$.

(C) In the given ladder network,let the current flowing through the first series resistor $R$ be $I_1$.
At each node,the current splits between the shunt resistor $2R$ and the next series resistor $R$.
By analyzing the circuit from right to left,we observe that the current halves at each stage due to the symmetry of the ladder network.
Specifically,if $I_1$ is the current entering the first series resistor,the current flowing through the subsequent shunt resistors will be $I_1/2, I_1/4, I_1/8$,and so on.
The current $I$ in the final branch is given by $I = I_1/8$.
Since the voltage across the first shunt resistor $2R$ is $V$,the current $I_1$ flowing through the first series resistor is $I_1 = V / (2R)$.
Substituting this value into the expression for $I$,we get:
$I = \frac{1}{8} \times \frac{V}{2R} = \frac{V}{16R}$.

(A, C) The equivalent resistance between points $A$ and $B$ remains unchanged when a resistance $R$ is connected between two points if the potential difference between those two points is zero. This is equivalent to the balanced condition of a Wheatstone bridge,where $\frac{P}{Q} = \frac{R}{S}$.
For option $(a)$: Connecting points $2$ and $6$ creates a bridge with arms of resistances $(2r, r)$ and $(4r, 2r)$. Since $\frac{2r}{4r} = \frac{r}{2r} = \frac{1}{2}$,the bridge is balanced.
For option $(b)$: Connecting points $3$ and $6$ creates a bridge with arms of resistances $(3r, r)$ and $(3r, 2r)$. Since $\frac{3r}{3r} \neq \frac{r}{2r}$,the bridge is not balanced.
For option $(c)$: Connecting points $4$ and $7$ creates a bridge with arms of resistances $(4r, 2r)$ and $(2r, r)$. Since $\frac{4r}{2r} = \frac{2r}{r} = 2$,the bridge is balanced.
For option $(d)$: Connecting points $4$ and $6$ creates a bridge with arms of resistances $(4r, r)$ and $(2r, 2r)$. Since $\frac{4r}{2r} \neq \frac{r}{2r}$,the bridge is not balanced.
Thus,the equivalent resistance remains unchanged for options $(a)$ and $(c)$.

(D) According to the balanced condition of a Wheatstone bridge,the ratio of resistances in opposite arms is equal: $\frac{P}{Q} = \frac{R}{S}$.
In the first case:
$\frac{P}{Q} = \frac{5}{S} \quad \dots(i)$
In the second case,$R$ is increased to $7 \Omega$ and $S$ is increased by $3 \Omega$ (i.e.,$S + 3 \Omega$):
$\frac{P}{Q} = \frac{7}{S + 3} \quad \dots(ii)$
Equating $(i)$ and $(ii)$:
$\frac{5}{S} = \frac{7}{S + 3}$
$5(S + 3) = 7S$
$5S + 15 = 7S$
$2S = 15$
$S = 7.5 \Omega$
Thus,the initial value of $S$ is $7.5 \Omega$.

(D) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m(KE)}}$.
Since the particles are accelerated by the same potential difference $V$,the kinetic energy $KE = qV$.
Thus,$\lambda = \frac{h}{\sqrt{2mqV}}$.
Since $h$,$q$,and $V$ are constants for both particles,we have $\lambda \propto \frac{1}{\sqrt{m}}$.
Therefore,$\frac{\lambda_{p}}{\lambda_{e}} = \sqrt{\frac{m_{e}}{m_{p}}}$.
Given $m_{p} = 2000 m_{e}$,we substitute this into the ratio:
$\frac{\lambda_{p}}{\lambda_{e}} = \sqrt{\frac{m_{e}}{2000 m_{e}}} = \sqrt{\frac{1}{2000}} = \frac{1}{\sqrt{400 \times 5}} = \frac{1}{20 \sqrt{5}}$.
Hence,$\lambda_{p} = \frac{\lambda_{e}}{20 \sqrt{5}}$.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $T$ of the emitted photoelectrons is given by $T = hv - \phi$,where $h$ is Planck's constant,$v$ is the frequency of incident light,and $\phi$ is the work function of the metal.
From this equation,it is clear that $T$ increases linearly with the frequency $v$ of the incident light.
Additionally,the number of photoelectrons emitted per second is directly proportional to the intensity $J$ of the incident light,provided the frequency $v$ is greater than the threshold frequency $v_{0}$.
Since both statements $(B)$ and $(D)$ are physically correct,in the context of standard multiple-choice questions where only one answer is expected,$(D)$ is often cited as a fundamental observation regarding intensity,while $(B)$ describes the energy dependence. Given the options,$(D)$ is a standard result of the photoelectric effect.

(B, D) According to Faraday's law of electromagnetic induction, an emf is induced in a loop when the magnetic flux $\Phi = B \cdot A \cos \theta$ linked with the loop changes with time.
$1$. If the loop is moved along the direction of $B$, the magnetic field $B$, area $A$, and the angle $\theta$ remain constant. Thus, the flux remains constant, and no emf is induced.
$2$. If the loop is squeezed to a smaller area, the area $A$ changes with time. Therefore, the magnetic flux $\Phi$ changes, and an emf is induced.
$3$. If the loop is rotated about its axis, the orientation of the area vector relative to the magnetic field remains constant $(\theta = 0^\circ)$. Thus, the flux remains constant, and no emf is induced.
$4$. If the loop is rotated about one of its diameters, the angle $\theta$ between the magnetic field $B$ and the area vector changes with time. Therefore, the magnetic flux $\Phi$ changes, and an emf is induced.
Since both $(b)$ and $(d)$ result in a change in magnetic flux, both conditions will induce an emf.

(A) If there were $12$ equal charges $+Q$ at all hour positions,the electric field at the centre would be zero due to symmetry,as each charge would be cancelled by an equal and opposite charge at the diametrically opposite position.
Let $\vec{E}_{10}$ be the electric field at the centre due to a charge $+Q$ at the $10$ o'clock position. The net electric field $\vec{E}_{net}$ with the charge at $10$ missing is given by:
$\vec{E}_{total} = \vec{E}_{net} + \vec{E}_{10} = 0$
Therefore,$\vec{E}_{net} = -\vec{E}_{10}$.
The magnitude of the electric field due to a single charge $+Q$ at distance $r$ is $E = \frac{Q}{4 \pi \varepsilon_{0} r^{2}}$.
The direction of $\vec{E}_{10}$ is from the $10$ o'clock position towards the centre.
Thus,$-\vec{E}_{10}$ is a vector of magnitude $\frac{Q}{4 \pi \varepsilon_{0} r^{2}}$ directed from the centre towards the $10$ o'clock position.

(A) Let the two positive charges $Q$ be located at $(0, d)$ and $(0, -d)$. The negative charge $-q$ is displaced to $(x, 0)$.
The distance $r$ between each positive charge $Q$ and the negative charge $-q$ is $r = \sqrt{x^2 + d^2}$.
The magnitude of the electrostatic force exerted by each positive charge on the negative charge is $F_e = \frac{kQq}{r^2}$.
The components of these forces along the $y$-axis cancel out due to symmetry. The components along the $x$-axis add up:
$F = -2 F_e \cos \theta$,where $\cos \theta = \frac{x}{r}$.
Substituting the expressions:
$F = -2 \left( \frac{kQq}{r^2} \right) \left( \frac{x}{r} \right) = -\frac{2kQqx}{r^3}$.
Since $r = (x^2 + d^2)^{1/2}$,we have:
$F = -\frac{2kQqx}{(x^2 + d^2)^{3/2}}$.
Given the condition $x \ll d$,we can approximate $x^2 + d^2 \approx d^2$ in the denominator:
$F \approx -\frac{2kQqx}{(d^2)^{3/2}} = -\frac{2kQqx}{d^3}$.
Thus,the magnitude of the force $F$ is directly proportional to the displacement $x$,i.e.,$F \propto x$.

(B) The magnetic field $B$ at a distance $x$ from an infinitely long wire carrying current $I$ is given by $B = \frac{\mu_0 I}{2 \pi x}$.
As the current $I$ in the wire is suddenly halved, the magnetic flux $\phi$ linked with the square loop decreases.
According to Lenz's law, the induced current in the loop will flow in such a direction as to oppose this decrease in magnetic flux. This means the induced current will create a magnetic field in the same direction as the original magnetic field.
For this to happen, the induced current in the loop must flow in the clockwise direction.
Now, consider the forces on the sides of the loop:
$1$. The side of the loop closer to the wire (let's call it $CD$) carries current in the same direction as the main wire. Thus, it experiences an attractive force $F_{CD}$ towards the wire.
$2$. The side of the loop farther from the wire (let's call it $AB$) carries current in the opposite direction to the main wire. Thus, it experiences a repulsive force $F_{AB}$ away from the wire.
Since the magnetic field is stronger closer to the wire, the attractive force on the closer side is greater than the repulsive force on the farther side $(F_{CD} > F_{AB})$.
Therefore, the net force on the loop is directed towards the wire, and the loop will move towards the wire.

(D) The current $I$ enters at one point and leaves at the diametrically opposite point of the circular wire. This divides the wire into two semicircular arcs,each carrying a current of $I/2$.
For each semicircular arc,the magnetic field produced at the center of the circle can be calculated using the Biot-Savart law. The magnetic field due to a semicircular arc at its center is $B = \frac{\mu_{0} I_{arc}}{4r}$.
Since the two semicircular arcs carry current in opposite directions relative to the center,the magnetic fields produced by them at the center are equal in magnitude but opposite in direction.
Specifically,if one arc produces a magnetic field $B$ pointing into the page,the other arc produces a magnetic field $B$ pointing out of the page.
Therefore,the net magnetic field $B_{net}$ at the center of the circle is $B - B = 0$.
The magnetic force $F$ on a charge $q$ moving with velocity $v$ in a magnetic field $B$ is given by $F = q(v \times B)$.
Since the net magnetic field $B_{net}$ at the center is $0$,the magnetic force experienced by the particle is $F = q(v \times 0) = 0$.

(C) The distances from the wires to point $P$ are $r_1 = 3 \ m$ and $r_2 = 4 \ m$. The distance between the wires is $5 \ m$. Since $3^2 + 4^2 = 5^2$,the triangle formed by the two wires and point $P$ is a right-angled triangle with the right angle at $P$.
The magnetic field due to the first wire at $P$ is $B_1 = \frac{\mu_0 I}{2 \pi r_1} = \frac{\mu_0 I}{2 \pi (3)} = \frac{\mu_0 I}{6 \pi}$.
The magnetic field due to the second wire at $P$ is $B_2 = \frac{\mu_0 I}{2 \pi r_2} = \frac{\mu_0 I}{2 \pi (4)} = \frac{\mu_0 I}{8 \pi}$.
Since the currents are in opposite directions,the magnetic field vectors $\vec{B}_1$ and $\vec{B}_2$ are perpendicular to each other at point $P$ because the triangle is right-angled at $P$. Thus,the net magnetic field is $B = \sqrt{B_1^2 + B_2^2}$.
$B = \sqrt{\left( \frac{\mu_0 I}{6 \pi} \right)^2 + \left( \frac{\mu_0 I}{8 \pi} \right)^2} = \frac{\mu_0 I}{\pi} \sqrt{\frac{1}{36} + \frac{1}{64}} = \frac{\mu_0 I}{\pi} \sqrt{\frac{16 + 9}{576}} = \frac{\mu_0 I}{\pi} \sqrt{\frac{25}{576}} = \frac{5}{24} \left( \frac{\mu_0 I}{\pi} \right)$.

(D) When a charged particle moves with velocity $v$ perpendicular to a uniform magnetic field $B$ (i.e.,$\theta = 90^{\circ}$),it experiences a magnetic Lorentz force $F = qvB$.
This force acts perpendicular to the velocity,providing the necessary centripetal force for circular motion:
$qvB = \frac{mv^2}{r}$
Solving for the radius $r$:
$r = \frac{mv}{qB}$
Since the momentum of the particle is $p = mv$,we can write:
$r = \frac{p}{qB}$
From this expression,it is clear that the radius $r$ is directly proportional to the momentum $p$ of the particle.
Therefore,option $(D)$ is correct.

(A) According to the question,the charge on the particle is $q$ and its velocity is $v$.
Due to the uniform electric field,the electric force on the particle is $F_{\text{electric}} = qE$.
Due to the uniform magnetic field,the magnetic force on the particle is given by $F_{\text{magnetic}} = q(v \times B)$.
For the particle to move undeflected,the net Lorentz force must be zero,which means the electric force and magnetic force must be equal in magnitude and opposite in direction: $F_{\text{electric}} = F_{\text{magnetic}}$.
Substituting the magnitudes,we get $qE = qvB$.
Solving for velocity,we find $v = \frac{E}{B}$.
Thus,the condition for the particle to move undeflected is $v = \frac{E}{B}$.

(B) The decay process is $X \xrightarrow{\alpha} Y \xrightarrow{\beta} Z$.
Let $N_X$ and $N_Y$ be the number of nuclei of $X$ and $Y$ present at any time $t$.
The rate of change of the number of daughter nuclei $Y$ is given by: $\frac{dN_Y}{dt} = \lambda_X N_X - \lambda_Y N_Y$.
Since the rate of emission of $\beta$-particles $(\lambda_Y N_Y)$ is constant over several months,we have $\frac{dN_Y}{dt} \approx 0$.
This implies $\lambda_X N_X = \lambda_Y N_Y$.
The rate of emission of $\alpha$-particles is $\lambda_X N_X$,and the rate of emission of $\beta$-particles is $\lambda_Y N_Y$.
Given that the rate of $\beta$-emission is $10^{7} / \text{h}$,the rate of $\alpha$-emission must also be $10^{7} / \text{h}$.
Therefore,the number of $\alpha$-particles emitted in one hour is $10^{7}$.

(D) Given: Focal length $f = 0.05 \ m$,object distance $u = -0.2 \ m$.
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Substituting the values: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{0.05} - \frac{1}{0.2} = 20 - 5 = 15 \ m^{-1}$.
Thus,$v = \frac{1}{15} \ m$.
Differentiating the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ with respect to time,we get: $-\frac{dv}{v^2} + \frac{du}{u^2} = 0$.
This implies $dv = \left( \frac{v^2}{u^2} \right) du$.
The amplitude of the image oscillation $A_{image} = |dv|$ is given by $A_{image} = \left( \frac{v}{u} \right)^2 A_{object}$.
Given $A_{object} = A \ cm = A \times 10^{-2} \ m$.
$A_{image} = \left( \frac{1/15}{0.2} \right)^2 \times A \times 10^{-2} \ m = \left( \frac{1}{15 \times 0.2} \right)^2 \times A \times 10^{-2} \ m = \left( \frac{1}{3} \right)^2 \times A \times 10^{-2} \ m = \frac{A}{9} \times 10^{-2} \ m$.

(B) The incident ray vector is $\hat{e}_0$ and the reflected ray vector is $\hat{e}$. The normal vector is $\hat{n}$.
From the law of reflection,the angle of incidence equals the angle of reflection,and all three vectors lie in the same plane.
Let $\theta$ be the angle between the incident ray and the normal. Since $\hat{e}_0$ is directed towards the mirror,the angle between $\hat{e}_0$ and $\hat{n}$ is $(180^\circ - \theta)$.
Thus,$\hat{e}_0 \cdot \hat{n} = |\hat{e}_0| |\hat{n}| \cos(180^\circ - \theta) = -\cos \theta$.
We can resolve the incident ray $\hat{e}_0$ into components parallel and perpendicular to the normal $\hat{n}$:
$\hat{e}_0 = \hat{e}_{0\perp} + \hat{e}_{0\parallel} = (\hat{e}_0 \cdot \hat{n}) \hat{n} + (\hat{e}_0 - (\hat{e}_0 \cdot \hat{n}) \hat{n})$.
The reflected ray $\hat{e}$ has the same component parallel to the mirror surface but the component along the normal is reversed:
$\hat{e} = -(\hat{e}_0 \cdot \hat{n}) \hat{n} + (\hat{e}_0 - (\hat{e}_0 \cdot \hat{n}) \hat{n})$.
$\hat{e} = \hat{e}_0 - 2(\hat{e}_0 \cdot \hat{n}) \hat{n}$.
Therefore,the correct option is $(B)$.

(D) To determine if the Zener diode is in breakdown, we first calculate the voltage across the parallel combination of the $1 k\Omega$ resistor and the Zener diode without the Zener diode connected.
Using the voltage divider rule, the voltage $V_{AB}$ across the $1 k\Omega$ resistor is:
$V_{AB} = V_s \times \frac{R_{parallel}}{R_s + R_{parallel}} = 10 V \times \frac{1 k\Omega}{1 k\Omega + 1 k\Omega} = 10 V \times \frac{1}{2} = 5 V$.
Since the voltage across the terminals $A$ and $B$ is $5 V$, which is less than the Zener breakdown voltage of $6 V$, the Zener diode does not conduct.
Therefore, the current through the Zener diode is $0 A$ (Zero).

(B) According to De Morgan's first law,the complement of the sum of two variables is equal to the product of their individual complements.
Mathematically,this is expressed as: $\overline{A+B} = \bar{A} \cdot \bar{B}$.
Therefore,the expression $\bar{A} \cdot \bar{B}$ is equivalent to $\overline{A+B}$.

(A) The formula for fringe width in Young's double slit experiment is $\beta = \frac{\lambda D}{d}$.
Taking the logarithmic differentiation,we get the relative error formula: $\frac{\Delta \beta}{\beta} = \frac{\Delta D}{D} - \frac{\Delta d}{d}$.
Given that $D$ is increased by $0.5 \%$,so $\frac{\Delta D}{D} \times 100 = 0.5 \%$.
Given that $d$ is decreased by $0.3 \%$,so $\frac{\Delta d}{d} \times 100 = -0.3 \%$.
Substituting these values into the error formula:
$\frac{\Delta \beta}{\beta} \times 100 = 0.5 \% - (-0.3 \%) = 0.5 \% + 0.3 \% = 0.8 \%$.
Therefore,the fringe width increases by $0.8 \%$.

(C) The angular width of the central maximum in a single slit diffraction pattern is given by $\theta = \frac{2\lambda}{d}$,where $\lambda$ is the wavelength and $d$ is the slit width.
The change in angular width is $\Delta\theta = \frac{2\Delta\lambda}{d}$,where $\Delta\lambda = |\lambda_1 - \lambda_2|$.
Given frequencies $f_1 = 4 \times 10^{14} \ s^{-1}$ and $f_2 = 5 \times 10^{14} \ s^{-1}$.
Using $\lambda = \frac{c}{f}$,where $c = 3 \times 10^8 \ m/s$:
$\lambda_1 = \frac{3 \times 10^8}{4 \times 10^{14}} = 0.75 \times 10^{-6} \ m = 7.5 \times 10^{-7} \ m$.
$\lambda_2 = \frac{3 \times 10^8}{5 \times 10^{14}} = 0.60 \times 10^{-6} \ m = 6.0 \times 10^{-7} \ m$.
$\Delta\lambda = |7.5 \times 10^{-7} - 6.0 \times 10^{-7}| = 1.5 \times 10^{-7} \ m$.
Given $\Delta\theta = 0.6 \ \text{radian}$.
From $d = \frac{2\Delta\lambda}{\Delta\theta}$:
$d = \frac{2 \times 1.5 \times 10^{-7}}{0.6} = \frac{3.0 \times 10^{-7}}{0.6} = 5 \times 10^{-7} \ m$.