WBJEE 2018 Physics Question Paper with Answer and Solution

(C) The escape velocity $v_{e}$ from the surface of a planet is given by $v_{e} = \sqrt{\frac{2GM}{R}}$.
Since $g = \frac{GM}{R^{2}}$,we have $v_{e} = \sqrt{2gR}$.
The acceleration due to gravity is $g = \frac{GM}{R^{2}} = \frac{G}{R^{2}} \cdot \frac{4}{3}\pi R^{3}\rho = \frac{4}{3}G\pi R\rho$.
Thus,the radius $R$ is proportional to $\frac{g}{\rho}$,i.e.,$R \propto \frac{g}{\rho}$.
Substituting this into the expression for $v_{e}$:
$v_{e} = \sqrt{2g \cdot \left(\frac{3g}{4\pi G\rho}\right)} = \sqrt{\frac{3g^{2}}{2\pi G\rho}} \propto \frac{g}{\sqrt{\rho}}$.
Given $\frac{g_{1}}{g_{2}} = \frac{5}{2}$ and $\frac{\rho_{1}}{\rho_{2}} = \frac{2}{1}$,the ratio of escape velocities is:
$\frac{v_{1}}{v_{2}} = \frac{g_{1}}{g_{2}} \cdot \sqrt{\frac{\rho_{2}}{\rho_{1}}} = \frac{5}{2} \cdot \sqrt{\frac{1}{2}} = \frac{5}{2\sqrt{2}}$.
Therefore,the ratio is $5:2\sqrt{2}$.

(C) For a process,the molar heat capacity is given by $C = C_{V} + \frac{P}{n} \frac{dV}{dT}$.
For a monoatomic gas,$C_{V} = \frac{3}{2} R$.
The process is a straight line in the $V-T$ diagram passing through $(V_{0}, T_{0})$ and $(2V_{0}, 3T_{0})$.
The equation of the line is $V - V_{0} = \frac{2V_{0} - V_{0}}{3T_{0} - T_{0}} (T - T_{0}) = \frac{V_{0}}{2T_{0}} (T - T_{0})$.
Thus,$\frac{dV}{dT} = \frac{V_{0}}{2T_{0}}$.
Using the ideal gas law $PV = nRT$,we have $P = \frac{nRT}{V}$.
At the point $(V_{0}, T_{0})$,$P = \frac{nRT_{0}}{V_{0}}$.
Substituting these into the heat capacity formula:
$C = \frac{3}{2} R + \left( \frac{nRT_{0}}{V_{0}} \right) \frac{1}{n} \left( \frac{V_{0}}{2T_{0}} \right) = \frac{3}{2} R + \frac{R}{2} = 2R$.

(B) Let $N_{1}$ be the normal force between the two blocks and $N_{2}$ be the normal force between the lower block and the table.
For the upper block of mass $m_{1}$,the normal force is $N_{1} = m_{1}g$.
For the lower block of mass $m_{2}$,the total downward force is $N_{2} = m_{2}g + N_{1} = (m_{1} + m_{2})g$.
The horizontal force $F$ on the upper block creates a frictional force $f_{1}$ on the lower block,where $f_{1} \leq \mu_{1}N_{1} = \mu_{1}m_{1}g$.
For the lower block to never move,the maximum static frictional force from the table must be greater than or equal to the maximum possible frictional force exerted by the upper block on the lower block.
Thus,$\mu_{2}N_{2} \geq \mu_{1}N_{1}$.
Substituting the values,$\mu_{2}(m_{1} + m_{2})g \geq \mu_{1}m_{1}g$.
Dividing both sides by $\mu_{2}m_{1}g$,we get $\frac{m_{1} + m_{2}}{m_{1}} \geq \frac{\mu_{1}}{\mu_{2}}$.
Therefore,$\frac{\mu_{1}}{\mu_{2}} \leq 1 + \frac{m_{2}}{m_{1}}$.
The maximum possible value is $1 + \frac{m_{2}}{m_{1}}$.

(C) Given,$v = \frac{k}{\sqrt{x}} = k x^{-1/2}$,where $k$ is a constant.
Acceleration $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$.
First,find $\frac{dv}{dx}$:
$\frac{dv}{dx} = k \cdot (-\frac{1}{2}) x^{-3/2} = -\frac{k}{2x^{3/2}}$.
Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula:
$a = (k x^{-1/2}) \cdot (-\frac{k}{2x^{3/2}}) = -\frac{k^2}{2x^2}$.
According to Newton's second law,$F = ma$.
The magnitude of the force is $F = |ma| = m \cdot \frac{k^2}{2x^2}$.
Since $m$ and $k$ are constants,$F \propto \frac{1}{x^2}$.

(B) The formula for terminal velocity $v$ is given by $v = \frac{2}{9} r^{2} \frac{(\rho - \sigma)}{\eta} g$.
Since we neglect the buoyancy of air,we take $\sigma \approx 0$.
The formula simplifies to $v = \frac{2}{9} \frac{\rho}{\eta} r^{2} g$.
Given: diameter $d = 1.8 \times 10^{-3} \ m$,so radius $r = 0.9 \times 10^{-3} \ m$.
Density $\rho = 10^{3} \ kg \ m^{-3}$,viscosity $\eta = 1.8 \times 10^{-5} \ N \ s \ m^{-2}$,and $g = 9.8 \ m \ s^{-2}$.
Substituting the values: $v = \frac{2}{9} \times \frac{10^{3}}{1.8 \times 10^{-5}} \times (0.9 \times 10^{-3})^{2} \times 9.8$.
$v = \frac{2}{9} \times \frac{10^{3}}{1.8 \times 10^{-5}} \times (0.81 \times 10^{-6}) \times 9.8$.
$v = \frac{2}{9} \times \frac{10^{3} \times 0.81 \times 10^{-6} \times 9.8}{1.8 \times 10^{-5}} = \frac{2}{9} \times \frac{0.81 \times 10^{-3} \times 9.8}{1.8 \times 10^{-5}} = \frac{2}{9} \times 0.45 \times 10^{2} \times 9.8 = 0.1 \times 100 \times 9.8 = 98 \ m \ s^{-1}$.

(D) Given that stress $\sigma_{s} = \frac{1}{100} Y$,where $Y$ is the Young's modulus.
We know that $Y = \frac{\text{stress}}{\text{longitudinal strain}} = \frac{\sigma_{s}}{\Delta l / l}$.
Substituting the value of stress: $Y = \frac{Y / 100}{\Delta l / l} \implies \frac{\Delta l}{l} = \frac{1}{100} = 0.01$.
Poisson's ratio $\nu = -\frac{\Delta w / w}{\Delta l / l} = -\frac{\Delta t / t}{\Delta l / l} = 0.3$.
Thus,the lateral strain $\frac{\Delta w}{w} = \frac{\Delta t}{t} = -\nu \left( \frac{\Delta l}{l} \right) = -0.3 \times 0.01 = -0.003$.
The volume of a rectangular rod is $V = l \times w \times t$.
The fractional change in volume is $\frac{\Delta V}{V} \approx \frac{\Delta l}{l} + \frac{\Delta w}{w} + \frac{\Delta t}{t}$.
Substituting the values: $\frac{\Delta V}{V} = 0.01 + (-0.003) + (-0.003) = 0.01 - 0.006 = 0.004$.
Therefore,the percentage change in volume is $0.004 \times 100 \% = 0.4 \%$.

(C) For a simple harmonic motion,the displacement is given by $y = A \sin(\omega t)$ and velocity is $v = \frac{dy}{dt} = A \omega \cos(\omega t)$.
Rearranging these,we get $\frac{y}{A} = \sin(\omega t)$ and $\frac{v}{A \omega} = \cos(\omega t)$.
Squaring and adding,we get $\frac{v^2}{(A \omega)^2} + \frac{y^2}{A^2} = 1$,which represents an ellipse.
The major axis along the $X$-axis (velocity) is $2A \omega$ and the minor axis along the $Y$-axis (displacement) is $2A$.
The ratio of the major axis to the minor axis is $\frac{2A \omega}{2A} = \omega$.
Given that the ratio is $20 \pi$,we have $\omega = 20 \pi$.
Since $\omega = 2 \pi f$,we have $2 \pi f = 20 \pi$.
Therefore,$f = 10 \ Hz$.

(C) The electric field $E$ produced by an infinite plane sheet of charge with surface charge density $\sigma$ is given by $E = \frac{\sigma}{2 \varepsilon_{0}}$.
Since the bob carries a charge $q$,it experiences an upward electrostatic force $F_e = qE = \frac{q \sigma}{2 \varepsilon_{0}}$.
The effective acceleration due to gravity $g_{\text{eff}}$ acting on the bob is given by $m g_{\text{eff}} = mg - F_e = mg - \frac{q \sigma}{2 \varepsilon_{0}}$.
Thus,$g_{\text{eff}} = g - \frac{q \sigma}{2 \varepsilon_{0} m}$.
The time period $T$ of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g_{\text{eff}}}}$.
Substituting the value of $g_{\text{eff}}$,we get $T = 2 \pi \sqrt{\frac{L}{g - \frac{q \sigma}{2 \varepsilon_{0} m}}}$.

(D) Let $M$ be the mass of both spheres. For the solid sphere: $M = \rho_{1} \frac{4}{3} \pi R^{3}$.
For the hollow sphere with inner radius $r$: $M = \rho_{2} \frac{4}{3} \pi (R^{3} - r^{3})$.
Equating the masses: $\rho_{1} R^{3} = \rho_{2} (R^{3} - r^{3})$.
This gives $\frac{\rho_{1}}{\rho_{2}} = 1 - \frac{r^{3}}{R^{3}}$,so $\frac{r^{3}}{R^{3}} = 1 - \frac{\rho_{1}}{\rho_{2}}$,which implies $\frac{r}{R} = (1 - \frac{\rho_{1}}{\rho_{2}})^{1/3}$.
The moment of inertia of the solid sphere is $I_{S} = \frac{2}{5} M R^{2}$.
The moment of inertia of the hollow sphere is $I_{H} = \frac{2}{5} M \frac{R^{5} - r^{5}}{R^{3} - r^{3}}$.
Using $M = \rho_{2} \frac{4}{3} \pi (R^{3} - r^{3})$,we get $I_{H} = \frac{2}{5} (\rho_{2} \frac{4}{3} \pi (R^{3} - r^{3})) \frac{R^{5} - r^{5}}{R^{3} - r^{3}} = \frac{2}{5} \rho_{2} \frac{4}{3} \pi (R^{5} - r^{5})$.
Since $M = \rho_{1} \frac{4}{3} \pi R^{3}$,we have $I_{S} = \frac{2}{5} (\rho_{1} \frac{4}{3} \pi R^{3}) R^{2} = \frac{2}{5} \rho_{1} \frac{4}{3} \pi R^{5}$.
Therefore,$\frac{I_{H}}{I_{S}} = \frac{\rho_{2} (R^{5} - r^{5})}{\rho_{1} R^{5}} = \frac{\rho_{2}}{\rho_{1}} [1 - (\frac{r}{R})^{5}]$.
Substituting $\frac{r}{R} = (1 - \frac{\rho_{1}}{\rho_{2}})^{1/3}$,we get $\frac{I_{H}}{I_{S}} = \frac{\rho_{2}}{\rho_{1}} [1 - (1 - \frac{\rho_{1}}{\rho_{2}})^{5/3}]$.

(C) Let the initial mass of ice be $m \ g$.
According to the principle of calorimetry,heat lost by the system (calorimeter + water) = heat gained by the ice.
The system cools down to $0^{\circ} C$ because only half of the ice melts,implying the final equilibrium temperature is $0^{\circ} C$.
Heat lost by water and calorimeter: $Q_{lost} = (m_{water} + m_{cal}) \cdot s_{water} \cdot \Delta T = (50 + 10) \cdot 1.0 \cdot (15 - 0) = 60 \cdot 15 = 900 \ cal$.
Heat gained by ice: The ice warms from $-10^{\circ} C$ to $0^{\circ} C$ and then half of it melts.
$Q_{gained} = m \cdot s_{ice} \cdot \Delta T_{ice} + (m/2) \cdot L_f = m \cdot 0.5 \cdot (0 - (-10)) + (m/2) \cdot 80$.
$Q_{gained} = m \cdot 0.5 \cdot 10 + 40m = 5m + 40m = 45m$.
Equating heat lost and gained: $900 = 45m$.
$m = 900 / 45 = 20 \ g$.

(B) In a $p-V$ diagram, the slope of an adiabatic curve at any point is steeper than that of an isothermal curve at that same point.
Let the initial state be $A(p_{i}, V_{i})$.
During the reversible isothermal expansion to volume $V_{0}$, the gas moves along the isothermal curve to state $B(p_{2}, V_{0})$, where $p_{2} < p_{i}$.
During the subsequent reversible adiabatic compression from volume $V_{0}$ back to $V_{i}$, the gas moves along an adiabatic curve from state $B(p_{2}, V_{0})$ to state $C(p_{f}, V_{i})$.
Since the adiabatic curve is steeper than the isothermal curve, the path from $B$ to $C$ lies above the path from $A$ to $B$ in the $p-V$ plane.
Therefore, the final pressure $p_{f}$ at volume $V_{i}$ must be greater than the initial pressure $p_{i}$.
Thus, $p_{f} > p_{i}$.

(B, C) The internal energy $(U)$ of an ideal gas is a function of temperature $(T)$ only, given by $U = f(T)$.
In an isothermal process, the temperature of the gas remains constant $(\Delta T = 0)$.
Since the internal energy depends only on temperature, it remains constant in an isothermal process.
In an isobaric process, if the gas expands, the temperature increases (from $PV = nRT$, if $P$ is constant and $V$ increases, $T$ must increase), leading to an increase in internal energy.
Therefore, both statement $(B)$ and statement $(C)$ are true.

(D) The phenomenon where the intensity of sound varies periodically in time is known as $beats$.
$Beats$ occur when two sound waves of slightly different frequencies,$f_1$ and $f_2$,interfere with each other.
The resulting intensity varies with a frequency equal to the difference of the two frequencies,$|f_1 - f_2|$.
Additionally,if the source intensity itself is modulated periodically,the observer will perceive a periodic change in intensity.
Comparing this with the given options,option $D$ describes the physical condition for $beats$,which is a standard cause for periodic intensity variation.

(D) In a series $L-C-R$ circuit,the current is the same at two different frequencies $f_1$ and $f_2$ if these frequencies are equidistant from the resonance frequency $f_0$ in a geometric sense.
The resonance frequency $f_0$ is given by the geometric mean of the two frequencies at which the current is equal:
$f_0 = \sqrt{f_1 \times f_2}$
Given:
$f_1 = 200 \ Hz$
$f_2 = 800 \ Hz$
Substituting the values:
$f_0 = \sqrt{200 \times 800}$
$f_0 = \sqrt{160000}$
$f_0 = 400 \ Hz$
Thus,the resonance frequency of the circuit is $400 \ Hz$.

(A) The linear velocity $v$ of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula:
$v = \frac{Z e^2}{2 \epsilon_0 n h}$
Where $Z$ is the atomic number,$e$ is the charge of the electron,$\epsilon_0$ is the permittivity of free space,$n$ is the principal quantum number,and $h$ is Planck's constant.
From this expression,it is clear that all terms except $n$ are constants for a given atom.
Therefore,the relationship is $v \propto \frac{1}{n}$.

(C) When the switch is turned on,the capacitor begins to charge. Initially,the charge on the capacitor is $0$,so the potential difference across it is $0 \ V$. Consequently,the entire battery voltage appears across the bulb,causing it to glow with maximum intensity.
As the capacitor charges,the potential difference across it $(V_c = q/C)$ increases over time. According to Kirchhoff's voltage law,the voltage across the bulb $(V_b)$ is given by $V_b = V_{battery} - V_c$. As $V_c$ increases,$V_b$ decreases.
Therefore,the intensity of the light,which depends on the power dissipated by the bulb $(P = V_b^2 / R)$,will be maximum at the start and will gradually decrease as the capacitor becomes fully charged.

(A) For an isolated charged parallel plate capacitor,the charge $q$ on the plates remains constant.
The electric field $E$ between the plates is given by $E = \frac{\sigma}{\varepsilon_{0}} = \frac{q}{A \varepsilon_{0}}$,which is constant since $q$,$A$,and $\varepsilon_{0}$ are constant.
The electrostatic force of attraction between the plates is $F = qE = \frac{q^2}{2A \varepsilon_{0}}$.
Since the force is constant,the acceleration $a$ of the plates is constant $(a = F/m)$.
The separation $d$ between the plates at time $t$ is given by $d(t) = d_{0} - \frac{1}{2} a t^2$,where $d_{0}$ is the initial separation.
The potential difference $V$ is given by $V = E \cdot d(t) = E (d_{0} - \frac{1}{2} a t^2)$.
This equation $V(t) = E d_{0} - (\frac{E a}{2}) t^2$ represents a downward-opening parabola,which corresponds to Graph $A$.

(D) To maximize the equivalent resistance across a diagonal,we must group the resistors such that the two parallel branches have the largest possible sum of resistances. Let the four resistors be $R_1, R_2, R_3, R_4$. When connected across a diagonal,they form two parallel branches,each containing two resistors in series. Let the branches be $(R_a + R_b)$ and $(R_c + R_d)$. The equivalent resistance is $R_{eq} = \frac{(R_a + R_b)(R_c + R_d)}{(R_a + R_b) + (R_c + R_d)}$. To maximize this,we should make the sums of the two branches as close to each other as possible. The total sum is $100 + 200 + 300 + 400 = 1000 \Omega$. Thus,we aim for each branch to be $500 \Omega$. We can pair them as $(400 + 100) = 500 \Omega$ and $(300 + 200) = 500 \Omega$. The equivalent resistance is then $\frac{500 \times 500}{500 + 500} = \frac{250000}{1000} = 250 \Omega$.

(C) In a steady state,a capacitor acts as an open circuit,meaning no direct current flows through the branches containing capacitors.
Therefore,we can neglect the branches with the $1 \mu F$ and $2 \mu F$ capacitors.
The circuit simplifies to a single series loop containing the $6 V$ battery,the $200 \Omega$ resistor,and the $400 \Omega$ resistor.
The total resistance of the circuit is $R_{\text{net}} = 200 \Omega + 400 \Omega = 600 \Omega$.
Using Ohm's law,the current $I$ in the circuit is:
$I = \frac{V}{R_{\text{net}}} = \frac{6 V}{600 \Omega} = 0.01 A$.
Converting to milliamperes,$I = 0.01 \times 1000 mA = 10 mA$.

(B, D) Let the battery voltage be $V = 10 \ V$. When the key $K$ is open,the circuit is a series-parallel combination. The equivalent resistance is $R_{eq} = (200 + 300) \parallel (100 + G) = \frac{500(100+G)}{600+G}$. The current through the galvanometer branch is $I_{G, open} = \frac{V}{100+G}$.
When the key $K$ is closed,the circuit becomes a Wheatstone bridge. The condition that the current through the galvanometer does not change means the bridge is balanced,i.e.,$\frac{200}{100} = \frac{300}{G}$.
Solving for $G$: $G = \frac{300 \times 100}{200} = 150 \ \Omega$. Thus,option $D$ is correct.
With $G = 150 \ \Omega$,the current through the galvanometer when the key is closed is $I_{G, closed} = \frac{V}{R_{eq}'}$. Since the bridge is balanced,the potential at the two ends of the galvanometer is the same,and the current is determined by the potential divider: $I_G = \frac{10}{100+150} = \frac{10}{250} = 0.04 \ A = 40 \ mA$. Thus,option $B$ is correct.
Since the bridge is balanced,the ratio of resistances in the arms is equal,so the currents in the upper and lower branches are independent of the galvanometer connection,but the current through $200 \ \Omega$ is not necessarily equal to the current through $300 \ \Omega$ unless the resistances are equal. Here $200 \ \Omega \neq 300 \ \Omega$,so option $C$ is incorrect.

(D) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2 m_e e V}}$.
From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.
Therefore,$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}}$.
Given that $\lambda_1 = \lambda$ and $\lambda_2 = 2\lambda$,we have $\frac{\lambda}{2\lambda} = \sqrt{\frac{V_2}{10000}}$.
Squaring both sides,we get $\frac{1}{4} = \frac{V_2}{10000}$.
Thus,$V_2 = \frac{10000}{4} = 2500 \ V$.

(A) The magnetic field $B$ at a distance $y$ from a long straight wire carrying current $I$ is given by $B = \frac{\mu_{0} I}{2 \pi y}$.
Let the loop have length $l$ and breadth $b$. The loop moves with velocity $v$ away from the wire. Let $y$ be the distance of the inner edge of the loop from the wire at $t=0$. At time $t$,the distance is $y(t) = y_0 + vt$.
Since the breadth $b$ is very small $(b \ll y)$,the magnetic field across the loop can be approximated as uniform,and the induced emf $E$ is given by the difference in motional emf across the two sides of length $l$:
$E = E_1 - E_2 = v l (B_1 - B_2) = v l \left( \frac{\mu_{0} I}{2 \pi y} - \frac{\mu_{0} I}{2 \pi (y+b)} \right)$
$E = \frac{\mu_{0} I v l}{2 \pi} \left( \frac{y+b-y}{y(y+b)} \right) = \frac{\mu_{0} I v l b}{2 \pi y(y+b)}$
Since $b \ll y$,we have $y+b \approx y$,so $E \approx \frac{\mu_{0} I v l b}{2 \pi y^2}$.
Since $y = y_0 + vt$,for large $t$,$y \approx vt$. Thus,$E \propto \frac{1}{(vt)^2} \propto \frac{1}{t^2}$.

(D) The centripetal force required for circular motion is provided by the electrostatic force of attraction:
$F_{\text{centripetal}} = m r \omega^{2} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1} q_{2}}{r^{2}}$
From this,we find the angular velocity $\omega$:
$\omega^{2} = \frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} m r^{3}} \implies \omega \propto \frac{1}{r^{3/2}}$
The moving charge $q_{1}$ constitutes a current $i$ given by:
$i = \frac{q_{1}}{T} = \frac{q_{1} \omega}{2 \pi}$
The magnetic field $B$ at the center of a circular current loop is:
$B = \frac{\mu_{0} i}{2 r} = \frac{\mu_{0}}{2 r} \left( \frac{q_{1} \omega}{2 \pi} \right) = \frac{\mu_{0} q_{1} \omega}{4 \pi r}$
Since $\omega \propto r^{-3/2}$,we have:
$B \propto \frac{\omega}{r} \propto \frac{r^{-3/2}}{r} = r^{-5/2}$
Thus,$B \propto \frac{1}{r^{5/2}}$.

(C) In a regular hexagon,the distance from the centre $O$ to any vertex is equal to the side length $a$.
To maximize the electric field at the centre,we place the four charges at vertices $A, B, C,$ and $F$ as shown in the figure.
The electric field due to charges at $F$ and $C$ are equal in magnitude and opposite in direction,so they cancel each other out $(E_F + E_C = 0)$.
The net electric field at the centre $O$ is due to the charges at $A$ and $B$.
The magnitude of the electric field due to each charge is $E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{a^2}$.
The angle between the electric field vectors $E_A$ and $E_B$ is $60^{\circ}$.
The resultant electric field $E_{\text{net}}$ is given by:
$E_{\text{net}} = \sqrt{E_A^2 + E_B^2 + 2 E_A E_B \cos 60^{\circ}}$
Since $E_A = E_B = E$,we have:
$E_{\text{net}} = \sqrt{E^2 + E^2 + 2 E^2 (1/2)} = \sqrt{3E^2} = E\sqrt{3}$
Substituting the value of $E$:
$E_{\text{net}} = \frac{\sqrt{3}Q}{4 \pi \epsilon_0 a^2}$

(C) $1$. The charge $+Q$ at the center $O$ contributes a total flux of $\frac{Q}{\varepsilon_{0}}$ through the entire cube. By symmetry,the flux through each of the $6$ faces is $\frac{Q}{6\varepsilon_{0}}$.
$2$. The charge $+Q$ at point $P$ is located at a distance $a/2$ from the center of the face $ABCD$. This charge creates an electric field. The flux due to charge $+Q$ at $P$ through the face $ABCD$ is $\frac{Q}{2\varepsilon_{0}}$ (since the charge is at a distance $a/2$ from the center of the face,it subtends a solid angle of $2\pi$ steradians at the face,or by considering an identical cube placed adjacent to the first one).
$3$. The net flux through face $ABCD$ due to both charges is $\phi_{ABCD} = \phi_{O, ABCD} + \phi_{P, ABCD} = \frac{Q}{6\varepsilon_{0}} - \frac{Q}{2\varepsilon_{0}} = -\frac{Q}{3\varepsilon_{0}}$ (the flux from $P$ enters the cube,hence negative).
$4$. Total flux through the cube due to both charges is $\phi_{total} = \frac{Q_{enclosed}}{\varepsilon_{0}} = \frac{Q}{\varepsilon_{0}}$.
$5$. The flux through the remaining $5$ faces is $\phi_{5} = \phi_{total} - \phi_{ABCD} = \frac{Q}{\varepsilon_{0}} - (-\frac{Q}{3\varepsilon_{0}}) = \frac{4Q}{3\varepsilon_{0}} = \frac{8Q}{6\varepsilon_{0}}$.

(B) The potential $V$ at a distance $x$ from the center of a ring of radius $r$ carrying charge $q$ is given by $V = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{\sqrt{x^{2} + r^{2}}}$.
For point $A$ at distance $x_A = \frac{4}{3}r$,the distance from the ring circumference is $d_A = \sqrt{(\frac{4}{3}r)^2 + r^2} = \sqrt{\frac{16}{9}r^2 + r^2} = \sqrt{\frac{25}{9}r^2} = \frac{5}{3}r$.
Thus,$V_A = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{5r/3} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{3q}{5r}$.
For point $B$ at distance $x_B = \frac{3}{4}r$ on the opposite side,the distance from the ring circumference is $d_B = \sqrt{(\frac{3}{4}r)^2 + r^2} = \sqrt{\frac{9}{16}r^2 + r^2} = \sqrt{\frac{25}{16}r^2} = \frac{5}{4}r$.
Thus,$V_B = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{5r/4} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{4q}{5r}$.
The work done $W$ in moving a charge $-q$ from $A$ to $B$ is $W = (-q)(V_B - V_A)$.
$W = -q \left( \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{4q}{5r} - \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{3q}{5r} \right) = -q \left( \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{5r} \right) = -\frac{1}{5} \cdot \frac{q^2}{4 \pi \varepsilon_{0} r}$.

(A) The total electric potential $V$ at a point $P$ between the charges is given by $V = \frac{1}{4\pi\varepsilon_0} \left( \frac{Q}{r} + \frac{4Q}{d-r} \right)$,where $r$ is the distance of $P$ from $A$.
For the potential to be minimum,we set $\frac{dV}{dr} = 0$.
$\frac{dV}{dr} = \frac{1}{4\pi\varepsilon_0} \left( -\frac{Q}{r^2} + \frac{4Q}{(d-r)^2} \right) = 0$.
This implies $\frac{Q}{r^2} = \frac{4Q}{(d-r)^2}$,which is equivalent to the condition where the net electric field $E = 0$.
Taking the square root on both sides: $\frac{1}{r} = \frac{2}{d-r}$.
Solving for $r$: $d - r = 2r \Rightarrow 3r = d \Rightarrow r = \frac{d}{3}$.
Thus,the potential is minimum at a distance of $\frac{d}{3}$ units to the right of $A$.

(C) For a thin hollow cylinder of radius $R$ carrying a uniform current $i$ along its length:
$1$. Inside the cylinder $(d < R)$,the magnetic field $B$ is zero because the enclosed current is zero.
$2$. Outside the cylinder $(d \geq R)$,the magnetic field $B$ is given by Ampere's Law as $B = \frac{\mu_{0} i}{2 \pi d}$. This shows that $B \propto \frac{1}{d}$.
Therefore,the graph of $B$ versus $d$ will show $B = 0$ for $d < R$ and a hyperbolic decay for $d \geq R$. This corresponds to the plot shown in figure $C$.

(B) The magnetic field at the centre of a circular loop of radius $r$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2r}$.
Case $(I)$: For a loop of radius $r$,the length is $l_1 = 2\pi r$. The resistance is $R_1 = \rho \frac{l_1}{A} = \rho \frac{2\pi r}{A}$. The current is $I_1 = \frac{V}{R_1}$. The magnetic field is $B_1 = \frac{\mu_0 I_1}{2r}$.
Case $(II)$: For a loop of radius $2r$,the length is $l_2 = 2\pi(2r) = 2l_1$. The resistance is $R_2 = \rho \frac{l_2}{A} = 2R_1$. The current is $I_2 = \frac{V}{R_2} = \frac{V}{2R_1} = \frac{I_1}{2}$.
The new magnetic field is $B_2 = \frac{\mu_0 I_2}{2(2r)} = \frac{\mu_0 (I_1/2)}{4r} = \frac{1}{4} \left( \frac{\mu_0 I_1}{2r} \right) = \frac{B_1}{4}$.

(B) The time period $T$ of a charged particle moving in a circular path in a uniform magnetic field is given by the formula: $T = \frac{2\pi m}{qB}$.
This formula shows that the time period $T$ depends only on the mass $m$,the charge $q$,and the magnetic field strength $B$.
Since $m$,$q$,and $B$ remain constant,the time period $T$ is independent of the speed $v$ of the proton.
Therefore,even if the speed is increased to $\sqrt{2}v$,the time period will remain $T$.

(D) Given,half-life $t_{1/2} = 3$ days.
Number of active nuclei remaining after time $t$ is given by $N(t) = N_0 (1/2)^{t/t_{1/2}}$.
At the start of the third day ($t = 2$ days),the number of nuclei remaining is $N_1 = N_0 (1/2)^{2/3} = N_0 / (2^{2/3}) = N_0 / (4^{1/3})$.
Given $\sqrt[3]{0.25} \approx 0.63$,we have $N_1 = N_0 \times 0.63$.
At the end of the third day ($t = 3$ days),the number of nuclei remaining is $N_2 = N_0 (1/2)^{3/3} = 0.5 N_0$.
The number of nuclei that decay during the third day is the difference between the nuclei present at the start and end of the day:
$\Delta N = N_1 - N_2 = 0.63 N_0 - 0.5 N_0 = 0.13 N_0$.
Thus,the fraction of the initial nuclei that decay on the third day is $0.13$.

(A, C) The critical angle $\theta_{C} = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45^{\circ}$.
Applying Snell's law at the incident surface $S$ $(n_{1} \sin \theta_{1} = n_{2} \sin \theta_{2})$:
$1 \cdot \sin 45^{\circ} = \sqrt{2} \cdot \sin \theta \implies \sin \theta = \frac{1}{2} \implies \theta = 30^{\circ}$.
At point $P$,the angle of incidence is $90^{\circ} - 15^{\circ} = 75^{\circ}$. Since $75^{\circ} > 45^{\circ}$,total internal reflection $(TIR)$ occurs at $P$.
At point $Q$,the angle of incidence is $15^{\circ}$. Since $15^{\circ} < 45^{\circ}$,partial reflection and refraction occur at $Q$.
The ray emerging at $R$ is parallel to the incident ray at $S$ because the geometry of the prism and the path of the ray result in a net deviation of $0^{\circ}$ (or $360^{\circ}$),meaning they are parallel. Thus,statement $A$ and $C$ are correct.

(B) The given combination consists of three lenses in contact: two identical equiconvex glass lenses ($L_1$ and $L_3$) and one water lens $(L_2)$ formed in the space between them.
Let $f_1$ be the focal length of each glass lens and $f_2$ be the focal length of the water lens.
The combination of lenses is in contact,so the equivalent focal length $F$ is given by:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}$
Since $L_1$ and $L_3$ are identical,$f_1 = f_3 = f$. The water lens $L_2$ is a biconcave lens,so its focal length $f_2$ is negative.
Thus,$\frac{1}{F} = \frac{1}{f} + \frac{1}{f_2} + \frac{1}{f} = \frac{2}{f} + \frac{1}{f_2}$.
Let $R$ be the radius of curvature. For the glass lenses,$\frac{1}{f} = (\mu_g - 1)(\frac{2}{R})$. For the water lens,$\frac{1}{f_2} = -(\mu_w - 1)(\frac{2}{R})$.
Since $\mu_g > \mu_w$,the magnitude of the power of the glass lens is greater than that of the water lens,meaning $|\frac{1}{f}| > |\frac{1}{f_2}|$.
Therefore,$\frac{1}{F} = \frac{2}{f} - |\frac{1}{f_2}|$. Since $|\frac{1}{f_2}| > 0$,$\frac{1}{F} < \frac{2}{f}$,which implies $F > \frac{f}{2}$.
Also,since $\frac{1}{F} = \frac{2}{f} - |\frac{1}{f_2}|$,it is clear that $\frac{1}{F} > \frac{1}{f}$ (because we are subtracting a positive value from $\frac{2}{f}$ but the net power is still positive and less than $\frac{2}{f}$),thus $F < f$.
Combining these,we get $\frac{f}{2} < F < f$.

(D) The air bubble is located at the centre $O$ of the solid glass sphere.
When light rays originating from the air bubble travel towards the surface of the sphere,they move along the radius of the sphere.
Since the radius is always perpendicular to the surface of the sphere,the light rays strike the surface at an angle of incidence $i = 0^\circ$.
According to Snell's law,when light is incident normally on a surface,it does not undergo any refraction or deviation.
Therefore,the light rays continue to travel in a straight line without bending.
As a result,the observer looking from outside sees the image of the bubble at the same position as the object.
Thus,the apparent distance of the bubble from the centre of the sphere is zero.

(A) Let the source be at $S(0,1)$ and the point on the mirror be $M$. The reflected ray passes through $P(3,3)$.
Since the mirror is along the $X$-axis,the image of the source $S(0,1)$ formed by the mirror is $I(0,-1)$.
The path length of the reflected ray is $SM + MP$.
By the law of reflection,$SM = IM$.
Therefore,the total path length is $IM + MP = IP$.
The distance $IP$ is the distance between points $I(0,-1)$ and $P(3,3)$.
Using the distance formula: $IP = \sqrt{(3-0)^2 + (3 - (-1))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

(A) In the given circuit,the Zener diode is connected in parallel with the $4 \text{ k}\Omega$ resistor. Since the Zener diode is operating in its breakdown region,the voltage across it is constant at $6 \text{ V}$.
Because the Zener diode and the $4 \text{ k}\Omega$ resistor are in parallel,the voltage across the $4 \text{ k}\Omega$ resistor is also $6 \text{ V}$.
The total voltage in the circuit is $10 \text{ V}$. The voltage drop across the $6 \text{ k}\Omega$ series resistor is the difference between the source voltage and the Zener breakdown voltage:
$V_{6\text{k}\Omega} = 10 \text{ V} - 6 \text{ V} = 4 \text{ V}$.
The current flowing through the $6 \text{ k}\Omega$ resistor is given by Ohm's law:
$I = \frac{V_{6\text{k}\Omega}}{R} = \frac{4 \text{ V}}{6 \text{ k}\Omega} = \frac{4}{6} \text{ mA} = \frac{2}{3} \text{ mA}$.

(C) The circuit consists of two cross-coupled $NAND$ gates, which form an $S-R$ latch.
Given $A=1$ and $B=0$.
The output $Y$ of the bottom $NAND$ gate is $Y = \overline{B \cdot X} = \overline{0 \cdot X} = \overline{0} = 1$.
Now, using this value of $Y$ in the top $NAND$ gate, the output $X$ is $X = \overline{A \cdot Y} = \overline{1 \cdot 1} = \overline{1} = 0$.
Thus, the stable state is $X=0$ and $Y=1$.

(C) In Young's double slit experiment,white light consists of wavelengths ranging from $4000 \ Å$ to $7000 \ Å$.
At the central position on the screen,the path difference for all wavelengths is zero.
Since the path difference is zero,all wavelengths interfere constructively at the center,resulting in a white fringe.
As we move away from the center,the path difference increases,causing different wavelengths to interfere constructively at different positions,resulting in coloured fringes.