WBJEE 2017 Physics Question Paper with Answer and Solution

(D) The time taken for the first fall is $t_0 = \sqrt{\frac{2h}{g}}$.
After the first impact,the ball rises to height $h_1 = e^2 h$ and falls back,taking time $t_1 = 2 \sqrt{\frac{2h_1}{g}} = 2e \sqrt{\frac{2h}{g}}$.
Similarly,for subsequent bounces,the time taken is $t_n = 2e^n \sqrt{\frac{2h}{g}}$.
The total time $T$ is the sum of the initial fall and all subsequent bounces:
$T = \sqrt{\frac{2h}{g}} + 2e \sqrt{\frac{2h}{g}} + 2e^2 \sqrt{\frac{2h}{g}} + \dots = \sqrt{\frac{2h}{g}} \left( 1 + 2e + 2e^2 + \dots \right)$.
Using the sum of a geometric series for $e < 1$,$T = \sqrt{\frac{2h}{g}} \left( 1 + 2e \frac{1}{1-e} \right) = \sqrt{\frac{2h}{g}} \left( \frac{1+e}{1-e} \right)$.
Given $h = 0.4 \text{ m}$,$g = 10 \text{ m/s}^2$,and $T = 10 \text{ s}$:
$10 = \sqrt{\frac{2 \times 0.4}{10}} \left( \frac{1+e}{1-e} \right) = \sqrt{0.08} \left( \frac{1+e}{1-e} \right) = 0.2828 \left( \frac{1+e}{1-e} \right)$.
Actually,using $g = 9.8 \text{ m/s}^2$ or simplifying the expression: $10 = \sqrt{0.08} \frac{1+e}{1-e} \implies \frac{10}{0.2828} = \frac{1+e}{1-e} \approx 35.35$.
Solving for $e$ yields $e \approx 0.944$,which corresponds to $\frac{17}{18} \approx 0.944$.

(B) Let the mass of the bullet be $m = 4.2 \times 10^{-2} \text{ kg}$.
Mass of the block $M = 9m = 9 \times 4.2 \times 10^{-2} \text{ kg}$.
Initial velocity of the bullet $v = 300 \text{ m/s}$.
By the law of conservation of linear momentum, the momentum before collision equals the momentum after collision:
$mv = (m + M)V$
$m(300) = (m + 9m)V$
$300m = 10mV$
$V = 30 \text{ m/s}$.
Heat generated is equal to the loss in kinetic energy:
$\Delta K = K_i - K_f = \frac{1}{2}mv^2 - \frac{1}{2}(m + M)V^2$
$\Delta K = \frac{1}{2} \times (4.2 \times 10^{-2}) \times (300)^2 - \frac{1}{2} \times (10 \times 4.2 \times 10^{-2}) \times (30)^2$
$\Delta K = \frac{1}{2} \times 4.2 \times 10^{-2} \times (90000 - 10 \times 900)$
$\Delta K = 2.1 \times 10^{-2} \times (90000 - 9000) = 2.1 \times 10^{-2} \times 81000 = 1701 \text{ J}$.
Since $1 \text{ cal} = 4.2 \text{ J}$, the heat in calories is:
$\text{Heat} = \frac{1701}{4.2} = 405 \text{ cal}$.

(A) The centre of mass of a system moves according to the net external force acting on it,as given by the equation $F_{ext} = M a_{cm}$.
Since the particles $A$ and $B$ are moving under a mutual force of attraction,there is no external force acting on the system $(F_{ext} = 0)$.
Initially,both particles are at rest,so the initial velocity of the centre of mass is $v_{cm, initial} = 0$.
Since $F_{ext} = 0$,the acceleration of the centre of mass is $a_{cm} = 0$,which implies that the velocity of the centre of mass remains constant over time.
Therefore,the velocity of the centre of mass at any instant remains equal to its initial velocity,which is $0$.

(B) Let $v$ be the velocity at the highest point. The forces acting on the particle at the highest point are the gravitational force $Mg$ (downwards),the electrostatic repulsive force $F_e = \frac{1}{4 \pi \epsilon_0} \frac{Q^2}{R^2}$ (upwards),and the tension $T$ (downwards).
Given $Mg = \frac{Q^2}{4 \pi \epsilon_0 R^2}$,the net force towards the center is $F_{net} = Mg - F_e + T = \frac{M v^2}{R}$.
At the highest point,$T = 0$,so $Mg - \frac{Q^2}{4 \pi \epsilon_0 R^2} = \frac{M v^2}{R}$.
Since $Mg = \frac{Q^2}{4 \pi \epsilon_0 R^2}$,we have $0 = \frac{M v^2}{R}$,which implies $v = 0$.
Now,applying the law of conservation of energy between the lowest point (velocity $v_0$) and the highest point (velocity $v = 0$):
Total energy at lowest point = Total energy at highest point.
$\frac{1}{2} M v_0^2 + U_{lowest} = \frac{1}{2} M v^2 + U_{highest}$.
The potential energy includes both gravitational and electrostatic components.
$U_{grav} = Mgh$,$U_{elec} = \frac{1}{4 \pi \epsilon_0} \frac{Q^2}{r}$.
Change in energy: $\frac{1}{2} M v_0^2 = Mg(2R) + \Delta U_{elec}$.
Since the distance $R$ from the center is constant,the electrostatic potential energy remains constant throughout the circular path.
Thus,$\frac{1}{2} M v_0^2 = Mg(2R)$.
$v_0^2 = 4gR$.
$v_0 = 2 \sqrt{gR}$.

(C) The initial temperature of the ideal gas is $T_{1} = 273 + 27 = 300 \ K$.
When the temperature is raised by $6^{\circ} C$,the final temperature is $T_{2} = 300 + 6 = 306 \ K$.
The root mean square (rms) velocity is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \sqrt{T}$.
Therefore,the ratio of the velocities is $\frac{v_{rms_{2}}}{v_{rms_{1}}} = \sqrt{\frac{T_{2}}{T_{1}}} = \sqrt{\frac{306}{300}} = \sqrt{1.02}$.
Using the binomial approximation $(1+x)^{n} \approx 1+nx$ for small $x$,we get $\sqrt{1.02} = (1 + 0.02)^{1/2} \approx 1 + \frac{1}{2}(0.02) = 1.01$.
Thus,$v_{rms_{2}} \approx 1.01 \ v_{rms_{1}}$,which represents an increase of $1 \%$.

(A) Given: Mass $m = 1 \ kg$, Force $F = kt$, where $k = 1 \ Ns^{-1}$.
According to Newton's second law, $F = ma$.
Since $a = \frac{dv}{dt}$, we have $m \frac{dv}{dt} = kt$.
Substituting $m = 1$ and $k = 1$, we get $\frac{dv}{dt} = t$.
Integrating with respect to time $t$ (starting from rest, $v=0$ at $t=0$):
$v = \int t \ dt = \frac{t^2}{2}$.
Since $v = \frac{dx}{dt}$, we have $\frac{dx}{dt} = \frac{t^2}{2}$.
Integrating again to find the displacement $x$ at $t = 6 \ s$:
$x = \int_{0}^{6} \frac{t^2}{2} \ dt = \left[ \frac{t^3}{6} \right]_{0}^{6}$.
$x = \frac{6^3}{6} = \frac{216}{6} = 36 \ m$.

(D) Let $A$ be the cross-sectional area of the capillary tube.
Initially,the tube is filled with air at pressure $p_{0}$ and volume $V = lA$.
When the tube is submerged by a length $x$,the air is compressed into a length $(l-x)$. Let the new pressure be $p^{\prime}$.
Using Boyle's Law: $p_{0}(lA) = p^{\prime}(l-x)A$,which gives $p^{\prime} = \frac{p_{0}l}{l-x}$.
Since the water levels inside and outside coincide,the pressure difference across the meniscus is given by the Young-Laplace equation: $p^{\prime} - p_{0} = \frac{2\gamma}{r}$.
Substituting $p^{\prime}$: $\frac{p_{0}l}{l-x} - p_{0} = \frac{2\gamma}{r}$.
$p_{0} \left( \frac{l}{l-x} - 1 \right) = \frac{2\gamma}{r} \implies p_{0} \left( \frac{l - l + x}{l-x} \right) = \frac{2\gamma}{r}$.
$\frac{p_{0}x}{l-x} = \frac{2\gamma}{r} \implies p_{0}xr = 2\gamma l - 2\gamma x$.
$x(p_{0}r + 2\gamma) = 2\gamma l$.
$x = \frac{2\gamma l}{p_{0}r + 2\gamma} = \frac{l}{\frac{p_{0}r}{2\gamma} + 1} = \frac{l}{1 + \frac{p_{0}r}{2\gamma}}$.

(A) The bulk modulus $k$ is defined as $k = -\frac{p}{\Delta V / V}$.
Since density $\rho = \frac{m}{V}$,we have $\frac{\Delta \rho}{\rho} = -\frac{\Delta V}{V}$.
Given that the density increases by $0.01 \%$,we have $\frac{\Delta \rho}{\rho} = 0.01 \% = \frac{0.01}{100} = 10^{-4}$.
Therefore,$-\frac{\Delta V}{V} = 10^{-4}$.
Substituting this into the bulk modulus formula: $p = k \times \left( -\frac{\Delta V}{V} \right)$.
$p = k \times 10^{-4} = \frac{k}{10000}$.

(B) The speed of a particle executing simple harmonic motion is given by $v = \omega \sqrt{a^2 - x^2}$,where $a$ is the amplitude,$\omega$ is the angular frequency,and $x$ is the displacement from the equilibrium position.
Squaring both sides,we get $v^2 = \omega^2 (a^2 - x^2)$.
For the two given cases:
$v_1^2 = \omega^2 (a^2 - x_1^2) \implies 13^2 = \omega^2 (a^2 - 3^2) \implies 169 = \omega^2 (a^2 - 9) \quad (1)$
$v_2^2 = \omega^2 (a^2 - x_2^2) \implies 12^2 = \omega^2 (a^2 - 5^2) \implies 144 = \omega^2 (a^2 - 25) \quad (2)$
Subtracting equation $(2)$ from $(1)$:
$169 - 144 = \omega^2 (a^2 - 9 - a^2 + 25)$
$25 = \omega^2 (16)$
$\omega^2 = \frac{25}{16} \implies \omega = \frac{5}{4} \ rad/s$.
The frequency $f$ is related to angular frequency by $\omega = 2 \pi f$.
Therefore,$f = \frac{\omega}{2 \pi} = \frac{5/4}{2 \pi} = \frac{5}{8 \pi} \ Hz$.

(D) Let the initial length be $L_{0}$ and initial breadth be $B_{0}$. The initial area is $S_{0} = L_{0} B_{0}$.
When the temperature increases by $\Delta t$,the new length $L_{t}$ and new breadth $B_{t}$ are given by:
$L_{t} = L_{0}(1 + \alpha_{1} \Delta t)$
$B_{t} = B_{0}(1 + \alpha_{2} \Delta t)$
The new area $S_{t}$ is:
$S_{t} = L_{t} \times B_{t} = L_{0}(1 + \alpha_{1} \Delta t) \times B_{0}(1 + \alpha_{2} \Delta t)$
$S_{t} = L_{0} B_{0} (1 + \alpha_{1} \Delta t + \alpha_{2} \Delta t + \alpha_{1} \alpha_{2} (\Delta t)^{2})$
Since $\alpha_{1} \Delta t \ll 1$ and $\alpha_{2} \Delta t \ll 1$,the term $\alpha_{1} \alpha_{2} (\Delta t)^{2}$ is negligible.
$S_{t} \approx S_{0} (1 + (\alpha_{1} + \alpha_{2}) \Delta t)$
Comparing this with the standard formula for surface expansion $S_{t} = S_{0} (1 + \beta \Delta t)$,where $\beta$ is the coefficient of surface expansion:
$\beta = \alpha_{1} + \alpha_{2}$

(C) For a monoatomic gas,the molar heat capacity at constant volume is $C_{v} = \frac{3}{2}R$. The change in internal energy is given by $\Delta U = n C_{v} \Delta T = n \left(\frac{3}{2}R\right) \Delta T$.
Using the ideal gas law $pV = nRT$,we have $\Delta T = \frac{p_f V_f - p_i V_i}{nR}$.
Substituting the values: $\Delta U = \frac{3}{2} (p_f V_f - p_i V_i) = \frac{3}{2} (4 p_{0} V_{0} - p_{0} V_{0}) = \frac{3}{2} (3 p_{0} V_{0}) = \frac{9}{2} p_{0} V_{0}$.
The work done $W$ is the area under the $p-V$ graph,which is a trapezoid: $W = \frac{1}{2} (p_i + p_f) (V_f - V_i) = \frac{1}{2} (p_{0} + 2 p_{0}) (2 V_{0} - V_{0}) = \frac{1}{2} (3 p_{0}) (V_{0}) = \frac{3}{2} p_{0} V_{0}$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + W$.
$\Delta Q = \frac{9}{2} p_{0} V_{0} + \frac{3}{2} p_{0} V_{0} = \frac{12}{2} p_{0} V_{0} = 6 p_{0} V_{0}$.

(C) The velocity of sound in an ideal gas is given by the formula $v = \sqrt{\frac{\gamma R T}{M}} = \sqrt{\frac{\gamma p}{\rho}}$.
$(i)$ From the relation $v = \sqrt{\frac{\gamma R T}{M}}$,it is clear that the velocity of sound is proportional to the square root of the absolute temperature,i.e.,$v \propto \sqrt{T}$.
(ii) From the relation $v = \sqrt{\frac{\gamma p}{\rho}}$,if the density $\rho$ is kept constant,the velocity of sound is proportional to the square root of the pressure $p$,i.e.,$v \propto \sqrt{p}$.
Comparing these with the given options,both $B$ and $C$ are physically correct statements based on the derived relations.

(A) The fundamental frequency of a string fixed at both ends is given by $v = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$,where $n$ is the harmonic number,$L$ is the length,$T$ is the tension,and $\mu$ is the linear mass density.
Linear mass density $\mu = \frac{M}{L}$.
Substituting $\mu$ into the formula:
$v = \frac{n}{2L} \sqrt{\frac{T}{M/L}} = \frac{n}{2L} \sqrt{\frac{TL}{M}}$.
Wait,let us re-evaluate: $v = \frac{n}{2L} \sqrt{\frac{T}{M/L}} = \frac{n}{2} \sqrt{\frac{T}{M/L \cdot L^2}} = \frac{n}{2} \sqrt{\frac{T}{ML}}$.
Thus,the correct formula is $v = \frac{n}{2} \sqrt{\frac{T}{ML}}$.

(C) The impedance $Z$ of a series $LCR$ circuit is given by the formula: $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$.
At very low frequencies $(\omega \to 0)$,the capacitive reactance $X_C = \frac{1}{\omega C}$ is very large,making the impedance $Z$ very high.
As the frequency $\omega$ increases,the term $(\omega L - \frac{1}{\omega C})^2$ decreases until it reaches zero at the resonance frequency $\omega_0 = \frac{1}{\sqrt{LC}}$. At this point,$Z = R$,which is the minimum value.
As the frequency increases further beyond $\omega_0$,the inductive reactance $X_L = \omega L$ becomes dominant,and the term $(\omega L - \frac{1}{\omega C})^2$ increases again,causing the impedance $Z$ to increase.
Therefore,the impedance first decreases and then increases.

(A-D) According to Bohr's model for a hydrogen atom:
$v_{n} \propto \frac{1}{n}$
$E_{n} \propto \frac{1}{n^{2}}$
$r_{n} \propto n^{2}$
For option $A$: $\frac{E_{n} r_{n}}{E_{1} r_{1}} \propto \frac{(1/n^{2}) \cdot n^{2}}{1} = 1$. This is a constant,so the slope is $0$.
For option $B$: $\frac{r_{n} v_{n}}{r_{1} v_{1}} \propto \frac{n^{2} \cdot (1/n)}{1} = n$. This is a straight line with slope $1$.
For option $C$: $\frac{r_{n}}{r_{1}} = n^{2}$. Taking natural log on both sides: $\ln \left(\frac{r_{n}}{r_{1}}\right) = 2 \ln(n)$. This is a straight line with slope $2$.
For option $D$: $\frac{r_{n}}{E_{n}} \propto \frac{n^{2}}{1/n^{2}} = n^{4}$. Thus,$\frac{r_{n} E_{1}}{E_{n} r_{1}} = n^{4}$. Taking natural log: $\ln \left(\frac{r_{n} E_{1}}{E_{n} r_{1}}\right) = 4 \ln(n)$. This is a straight line with slope $4$.
All options $A, B, C,$ and $D$ are mathematically correct based on Bohr's model.

(C) When capacitors are connected in series,the equivalent capacitance $C_{eq}$ is given by:
$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$
$\frac{1}{C_{eq}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{5} = \frac{10 + 5 + 2}{10} = \frac{17}{10} \ \mu F^{-1}$
$C_{eq} = \frac{10}{17} \ \mu F$
The charge $Q$ on each capacitor in a series combination is the same:
$Q = C_{eq} \times V = \left( \frac{10}{17} \ \mu F \right) \times 10 \ V = \frac{100}{17} \ \mu C$
The potential difference $V_2$ across the $2.0 \ \mu F$ capacitor is:
$V_2 = \frac{Q}{C_2} = \frac{100/17 \ \mu C}{2 \ \mu F} = \frac{50}{17} \ V$

(D) tetrahedron has $4$ vertices and $6$ edges. Let the current enter at vertex $1$ and leave at vertex $2$.
There is a direct wire between $1$ and $2$ with resistance $r$.
There are two paths from $1$ to $2$ through the other two vertices ($3$ and $4$):
Path $1$: $1 \rightarrow 3 \rightarrow 2$ (two resistors in series,total $2r$)
Path $2$: $1 \rightarrow 4 \rightarrow 2$ (two resistors in series,total $2r$)
These two paths are in parallel with each other,so their equivalent resistance $R_p$ is given by $\frac{1}{R_p} = \frac{1}{2r} + \frac{1}{2r} = \frac{2}{2r} = \frac{1}{r}$,which means $R_p = r$.
Finally,this $R_p$ is in parallel with the direct wire between $1$ and $2$ (which also has resistance $r$).
Therefore,the total equivalent resistance $R_{eq}$ is $\frac{1}{R_{eq}} = \frac{1}{R_p} + \frac{1}{r} = \frac{1}{r} + \frac{1}{r} = \frac{2}{r}$.
Thus,$R_{eq} = \frac{r}{2}$.

(C) For the given circuit,the equivalent resistance $R^{\prime}$ of the parallel combination of $R$ and $X$ is given by:
$\frac{1}{R^{\prime}} = \frac{1}{R} + \frac{1}{X} = \frac{R+X}{RX} \implies R^{\prime} = \frac{RX}{R+X}$
The total current $i$ in the circuit is:
$i = \frac{E}{R + R^{\prime}} = \frac{E}{R + \frac{RX}{R+X}} = \frac{E(R+X)}{R^2 + RX + RX} = \frac{E(R+X)}{R^2 + 2RX}$
The voltage $V_X$ across the resistance $X$ is equal to the voltage across $R^{\prime}$:
$V_X = i R^{\prime} = \left( \frac{E(R+X)}{R^2 + 2RX} \right) \left( \frac{RX}{R+X} \right) = \frac{ERX}{R^2 + 2RX} = \frac{EX}{R + 2X}$
The power $P_X$ dissipated in resistance $X$ is:
$P_X = \frac{V_X^2}{X} = \frac{(EX)^2}{X(R+2X)^2} = \frac{E^2 X}{(R+2X)^2}$
For $P_X$ to be independent of small variations in $X$,we set $\frac{dP_X}{dX} = 0$:
$\frac{dP_X}{dX} = E^2 \left[ \frac{(R+2X)^2(1) - X(2)(R+2X)(2)}{(R+2X)^4} \right] = 0$
$(R+2X) - 4X = 0$
$R - 2X = 0 \implies X = \frac{R}{2}$

(A) Let the resistance of each resistor be $R = 1 k\Omega = 1000 \Omega$. The circuit is a ladder network. Let the current in the rightmost vertical resistor $X$ be $I_4 = 1 mA$. The current in the horizontal resistor connected to it is also $1 mA$.
Applying Kirchhoff's Current Law $(KCL)$ at the node above $X$,the current in the vertical resistor to the left of $X$ is $I_v = 1 mA + 1 mA = 2 mA$.
Moving left,the current in the next horizontal resistor is $I_h = 1 mA + 2 mA = 3 mA$.
The current in the next vertical resistor is $I_v = 3 mA + 2 mA = 5 mA$.
Continuing this process:
- Current in the next horizontal resistor: $I_h = 3 mA + 5 mA = 8 mA$.
- Current in the next vertical resistor: $I_v = 8 mA + 5 mA = 13 mA$.
- Current in the first horizontal resistor: $I_h = 8 mA + 13 mA = 21 mA$.
- Current in the first vertical resistor: $I_v = 21 mA + 13 mA = 34 mA$.
The potential difference between $A$ and $B$ is the voltage across the first vertical resistor: $V_{AB} = I_{v1} \times R = 34 mA \times 1 k\Omega = 34 V$.

(B) The torque $\tau$ experienced by a current-carrying coil in a magnetic field is given by $\tau = N I A B \sin \theta$,where $N$ is the number of turns,$I$ is the current,$A$ is the area of the coil,$B$ is the magnetic field,and $\theta$ is the angle between the normal to the coil and the magnetic field.
Since the current $I$,magnetic field $B$,and number of turns $N$ are the same for both galvanometers,the ratio of the torques depends only on the ratio of the areas of the coils.
For the first coil (square of side $a$): $A_1 = a^2$.
For the second coil (circular of radius $r = \frac{a}{\sqrt{\pi}}$): $A_2 = \pi r^2 = \pi \left( \frac{a}{\sqrt{\pi}} \right)^2 = \pi \left( \frac{a^2}{\pi} \right) = a^2$.
Since $A_1 = A_2$,the torque experienced by both coils is the same.
Therefore,the ratio of the torque experienced by the first coil to that experienced by the second one is $1: 1$.

(B) $1$. First,calculate the total resistance of the circuit. The resistors $1 \text{ k}\Omega$ and $2 \text{ k}\Omega$ are in series with the $3 \text{ k}\Omega$ resistor. Total resistance $R_{eq} = 1 \text{ k}\Omega + 2 \text{ k}\Omega + 3 \text{ k}\Omega = 6 \text{ k}\Omega = 6000 \ \Omega$.
$2$. The current $I$ flowing through the circuit is $I = \frac{E}{R_{eq}} = \frac{3 \text{ V}}{6000 \ \Omega} = 0.5 \times 10^{-3} \text{ A} = 0.5 \text{ mA}$.
$3$. The potential at point $B$ relative to $A$ is $V_{AB} = I \times R_{1k} = 0.5 \text{ mA} \times 1 \text{ k}\Omega = 0.5 \text{ V}$. Thus,$V_A - V_B = 0.5 \text{ V}$.
$4$. The potential at point $C$ relative to $A$ is determined by the voltage divider rule for the capacitors. Since the capacitors are in series and no $DC$ current flows through them,the potential $V_C$ is determined by the ratio of their capacitances. $V_A - V_C = V_{AD} \times \frac{C_2}{C_1 + C_2}$,where $V_{AD} = I \times (1 \text{ k}\Omega + 2 \text{ k}\Omega) = 0.5 \text{ mA} \times 3 \text{ k}\Omega = 1.5 \text{ V}$.
$5$. $V_A - V_C = 1.5 \text{ V} \times \frac{2 \mu\text{F}}{1 \mu\text{F} + 2 \mu\text{F}} = 1.5 \times \frac{2}{3} = 1.0 \text{ V}$.
$6$. Now,$V_{BC} = V_B - V_C = (V_A - V_C) - (V_A - V_B) = 1.0 \text{ V} - 0.5 \text{ V} = 0.5 \text{ V}$.

(D) The de-Broglie wavelength $\lambda$ is related to kinetic energy $E$ by the formula: $\lambda = \frac{h}{\sqrt{2mE}}$.
From this,we can see that $\lambda \propto \frac{1}{\sqrt{E}}$,which implies $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{E_2}{E_1}}$.
Given:
$\lambda_1 = 0.4 \times 10^{-10} \ m$
$E_1 = 1.0 \ keV$
$\lambda_2 = 1.0 \times 10^{-10} \ m$
Substituting these values into the ratio:
$\frac{0.4 \times 10^{-10}}{1.0 \times 10^{-10}} = \sqrt{\frac{E_2}{1.0 \ keV}}$
$0.4 = \sqrt{E_2}$
Squaring both sides:
$E_2 = (0.4)^2 = 0.16 \ keV$.

(C) From Einstein's photoelectric equation,$h v = W + e V$,where $W$ is the work function,$v$ is the frequency of incident light,and $V$ is the stopping potential.
For frequency $v_{1}$,we have: $h v_{1} = W + e V_{1}$ --- $(i)$
For frequency $v_{2}$,we have: $h v_{2} = W + e V_{2}$ --- (ii)
Subtracting equation $(i)$ from equation (ii):
$h(v_{2} - v_{1}) = e(V_{2} - V_{1})$
$e = \frac{h(v_{2} - v_{1})}{V_{2} - V_{1}}$
However,we need to express $e$ in terms of $W$. From $(i)$,$h = \frac{W + e V_{1}}{v_{1}}$. Substituting this into (ii):
$(\frac{W + e V_{1}}{v_{1}}) v_{2} = W + e V_{2}$
$W v_{2} + e V_{1} v_{2} = W v_{1} + e V_{2} v_{1}$
$e(V_{1} v_{2} - V_{2} v_{1}) = W v_{1} - W v_{2}$
$e(V_{1} v_{2} - V_{2} v_{1}) = -W(v_{2} - v_{1})$
$e = \frac{W(v_{2} - v_{1})}{V_{2} v_{1} - V_{1} v_{2}}$
This matches option $C$.

(A) Let the total charge be $Q = 0.8 \text{ C}$.
Let $Q_{1} = q$,then $Q_{2} = Q - q = (0.8 - q)$.
The electrostatic force between the two charges is given by Coulomb's Law:
$F = \frac{k Q_{1} Q_{2}}{r^{2}} = \frac{k q (0.8 - q)}{r^{2}}$
For the force $F$ to be maximum,the derivative of $F$ with respect to $q$ must be zero:
$\frac{dF}{dq} = \frac{k}{r^{2}} \frac{d}{dq} (0.8q - q^{2}) = 0$
$0.8 - 2q = 0$
$2q = 0.8$
$q = 0.4 \text{ C}$
Thus,$Q_{1} = 0.4 \text{ C}$ and $Q_{2} = 0.8 - 0.4 = 0.4 \text{ C}$.
Therefore,the force is maximum when $Q_{1} = Q_{2} = 0.4 \text{ C}$.

(D) The particle moves with constant velocity $u$ along the $X$-axis, so $x = ut$, which implies $t = x/u$.
Along the $Y$-axis, the force on the particle is $F = eE$, so the acceleration is $a_y = \frac{eE}{m}$.
The displacement along the $Y$-axis is $y = \frac{1}{2} a_y t^2 = \frac{1}{2} \left( \frac{eE}{m} \right) \left( \frac{x}{u} \right)^2 = \left( \frac{eE}{2mu^2} \right) x^2$.
This is the equation of a parabola of the form $x^2 = 4ay$.
Rearranging the equation: $x^2 = \left( \frac{2mu^2}{eE} \right) y$.
Comparing this with the standard form $x^2 = 4ay$, we get $4a = \frac{2mu^2}{eE}$.
Therefore, the focal length $a = \frac{2mu^2}{4eE} = \frac{mu^2}{2eE}$.

(A) According to Gauss's law,the total electric flux $\phi_{total}$ through any closed surface is given by $\phi_{total} = \frac{Q}{\varepsilon_{0}}$,where $Q$ is the net charge enclosed by the surface.
Since the charge $Q$ is placed at the centre of the cube,the electric flux is symmetrically distributed through all $6$ faces of the cube.
Therefore,the electric flux $\phi_{face}$ through any one face of the cube is $\phi_{face} = \frac{\phi_{total}}{6} = \frac{Q}{6 \varepsilon_{0}}$.

(B) For a finite wire segment of length $L$ carrying current $I$,the magnetic field $B$ at a distance $r$ on its perpendicular bisector is given by the Biot-Savart law integration:
$B = \frac{\mu_{0} I}{4 \pi r} (\sin \theta_{1} + \sin \theta_{2})$
For a segment of length $L$,$\sin \theta_{1} = \sin \theta_{2} = \frac{L/2}{\sqrt{r^{2} + (L/2)^{2}}}$.
Substituting this,we get $B = \frac{\mu_{0} I}{4 \pi r} \cdot \frac{L}{\sqrt{r^{2} + (L/2)^{2}}}$.
Since $r >> L$,the denominator $\sqrt{r^{2} + (L/2)^{2}} \approx r$.
Thus,$B \approx \frac{\mu_{0} I L}{4 \pi r^{2}}$.
Therefore,the magnetic field decreases as $\frac{1}{r^{2}}$.

(C) For the third wire to experience no net magnetic force,the magnetic fields produced by the first and second wires at the position of the third wire must be equal in magnitude and opposite in direction.
The magnetic field $B$ due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.
Since the currents in the first and second wires are in opposite directions,the magnetic fields produced by them will be in the same direction in the region between them and in opposite directions in the regions outside them.
Let the third wire be placed at a distance $x$ from the first wire. If it is placed outside the region between the two wires (on the side of the smaller current,i.e.,the $1 \ A$ wire),the fields will be in opposite directions.
Equating the magnitudes of the magnetic fields:
$B_1 = B_2$
$\frac{\mu_0 I_1}{2 \pi x} = \frac{\mu_0 I_2}{2 \pi (0.1 + x)}$
Substituting the given values $I_1 = 1 \ A$ and $I_2 = 2 \ A$:
$\frac{1}{x} = \frac{2}{0.1 + x}$
$0.1 + x = 2x$
$x = 0.1 \ m$
Thus,the third wire is placed at a distance of $0.1 \ m$ from the first wire,away from the second wire.

(A) Given:
Velocity of proton $v = 10^{6} \ m/s$ along the $Y$-axis.
Electric field $E = 2 \times 10^{4} \ V/m$ along the negative $X$-axis.
Specific charge $\frac{e}{m} = 10^{8} \ C/kg$.
Since the proton moves with a uniform velocity,the net force on it is zero. Thus,the magnetic force balances the electric force:
$qE = qvB$
$B = \frac{E}{v} = \frac{2 \times 10^{4}}{10^{6}} = 2 \times 10^{-2} \ T$
When the electric field is switched off,the proton moves in a circular path due to the magnetic force acting as the centripetal force:
$R = \frac{mv}{qB} = \frac{v}{(e/m)B}$
$R = \frac{10^{6}}{10^{8} \times 2 \times 10^{-2}}$
$R = \frac{10^{6}}{2 \times 10^{6}} = 0.5 \ m$
Therefore,the radius of the circle is $0.5 \ m$.

(B) The relationship between magnetic susceptibility $\chi$ and relative permeability $\mu_r$ is given by $\chi = \mu_r - 1$.
Also,the absolute permeability $\mu$ is related to relative permeability by $\mu = \mu_r \mu_0$.
For a paramagnetic substance,$\chi$ is small and positive $(\chi > 0)$,which implies $\mu_r > 1$,and therefore $\mu > \mu_0$.
For a diamagnetic substance,$\chi$ is small and negative $(\chi < 0)$,which implies $\mu_r < 1$,and therefore $\mu < \mu_0$.
For a ferromagnetic substance,$\chi$ is very large and positive $(\chi \gg 1)$,which implies $\mu_r \gg 1$,and therefore $\mu \gg \mu_0$.
Comparing these with the given options,both $(b)$ and $(d)$ are scientifically correct statements.

(A) The half-life of Radon-$222$ is $T_{1/2} = 3.8 \text{ days}$.
Total time elapsed is $t = 19 \text{ days}$.
The number of half-lives $n$ is given by $n = \frac{t}{T_{1/2}} = \frac{19}{3.8} = 5$.
The remaining quantity $N$ is given by the formula $N = N_0 \left(\frac{1}{2}\right)^n$,where $N_0$ is the initial amount.
Given $N_0 = 0.064 \ kg$ and $n = 5$,we have:
$N = 0.064 \times \left(\frac{1}{2}\right)^5$
$N = 0.064 \times \frac{1}{32}$
$N = 0.002 \ kg$.
Thus,the quantity of Radon-$222$ left after $19$ days is $0.002 \ kg$.

(C) Let the unit negative charge be at a distance $x$ from the midpoint along the perpendicular bisector. The distance of this charge from each fixed charge $+Q$ is $r = \sqrt{x^2 + a^2}$.
The electrostatic force exerted by each $+Q$ charge on the unit negative charge is $F = \frac{1}{4 \pi \varepsilon_0} \frac{Q \cdot 1}{r^2} = \frac{Q}{4 \pi \varepsilon_0 (x^2 + a^2)}$.
The components of these forces perpendicular to the line connecting the charges cancel out,while the components along the perpendicular bisector add up.
The net restoring force is $F_{\text{net}} = -2F \cos \theta$,where $\cos \theta = \frac{x}{r} = \frac{x}{\sqrt{x^2 + a^2}}$.
$F_{\text{net}} = -2 \left( \frac{Q}{4 \pi \varepsilon_0 (x^2 + a^2)} \right) \left( \frac{x}{\sqrt{x^2 + a^2}} \right) = -\frac{2Qx}{4 \pi \varepsilon_0 (x^2 + a^2)^{3/2}}$.
Since $x \ll a$,we can approximate $(x^2 + a^2)^{3/2} \approx (a^2)^{3/2} = a^3$.
Thus,$F_{\text{net}} \approx -\left( \frac{2Q}{4 \pi \varepsilon_0 a^3} \right) x = -\left( \frac{Q}{2 \pi \varepsilon_0 a^3} \right) x$.
This is the equation of simple harmonic motion $F = -kx_{eff}$,where $k_{eff} = \frac{Q}{2 \pi \varepsilon_0 a^3}$.
The frequency of oscillation is $f = \frac{1}{2 \pi} \sqrt{\frac{k_{eff}}{M}} = \frac{1}{2 \pi} \sqrt{\frac{Q}{2 \pi \varepsilon_0 M a^3}}$.
Comparing this with the given options,none of the options match the calculated frequency.

(B) The point object is placed at the focus of the equiconvex lens.
When light rays from an object at the focus pass through the lens,they become parallel to the principal axis.
These parallel rays then strike the plane mirror placed horizontally below the lens.
The plane mirror reflects these rays back along the same path.
These reflected rays,being parallel,pass through the lens again and converge at the focal point of the lens.
Therefore,the final image is formed at the same position as the object,which is $0.1 \ m$ above the center of the lens.

(C) Given,radius $R=0.05 \ m = 5 \times 10^{-2} \ m$.
Refractive index $n=1.5$.
The critical angle $c$ is given by $\sin c = \frac{1}{n} = \frac{1}{1.5} = \frac{2}{3}$.
From the geometry of the problem,the ray that strikes the curved surface at the critical angle $c$ will emerge tangent to the surface and hit the table at a distance $x$ from the edge of the cylinder.
In the right-angled triangle formed by the radius $R$ and the distance $(R+x)$,the angle at the center is $c$.
Thus,$\cos c = \frac{R}{R+x}$.
Since $\sin c = \frac{2}{3}$,we have $\cos c = \sqrt{1 - \sin^2 c} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.
Equating the two expressions for $\cos c$:
$\frac{\sqrt{5}}{3} = \frac{R}{R+x}$.
$\sqrt{5}(R+x) = 3R$.
$\sqrt{5}x = 3R - \sqrt{5}R = R(3 - \sqrt{5})$.
$x = R \frac{3 - \sqrt{5}}{\sqrt{5}} = R \left( \frac{3}{\sqrt{5}} - 1 \right) = R \left( \frac{3\sqrt{5}}{5} - 1 \right)$.
Wait,re-evaluating the geometry: The distance from the center to the point where the ray hits the table is $d = \frac{R}{\cos c}$.
So,$R+x = \frac{R}{\cos c} = \frac{R}{\sqrt{5}/3} = \frac{3R}{\sqrt{5}} = \frac{3\sqrt{5}R}{5}$.
$x = \frac{3\sqrt{5}R}{5} - R = R \left( \frac{3\sqrt{5}-5}{5} \right)$.
Given $R = 5 \times 10^{-2} \ m$,we get $x = 5 \times 10^{-2} \left( \frac{3\sqrt{5}-5}{5} \right) = (3\sqrt{5}-5) \times 10^{-2} \ m$.

(D) The angle $\theta$ subtended by the sun at the mirror is given by the ratio of the diameter of the sun $(D)$ to the distance between the earth and the sun $(d_{SE})$.
Since the sun is at a very large distance,its image is formed at the focus of the concave mirror.
The angular size $\theta$ is given by $\theta = \frac{D}{d_{SE}} = 0.009 \ rad$.
The diameter of the image $(d)$ formed at the focal plane is given by $d = f \times \theta$,where $f$ is the focal length of the mirror.
The focal length $f$ is half of the radius of curvature $R = 0.4 \ m$,so $f = \frac{R}{2} = \frac{0.4}{2} = 0.2 \ m$.
Substituting the values,we get $d = 0.2 \ m \times 0.009 = 0.0018 \ m$.
Therefore,$d = 1.8 \times 10^{-3} \ m$.

(D) $p-n$ junction diode acts as a rectifier,which allows current to flow in only one direction.
When the $p$-region is connected to the positive terminal of the battery and the $n$-region to the negative terminal,the diode is in forward bias,and current flows through the circuit.
When the polarity of the battery is reversed,the $p$-region is connected to the negative terminal and the $n$-region to the positive terminal. This is reverse bias,where the depletion region widens,and practically no current flows through the circuit.

(B) The circuit consists of two $AND$ gates, two $NOT$ gates, and one $NOR$ gate. The Boolean expression for the output $Y$ is $Y = \overline{(A \cdot B) + (\overline{A} \cdot \overline{B})}$.
Case $1$: When $A = 1$ and $B = 1$, the output of the top $AND$ gate is $1 \cdot 1 = 1$. The output of the bottom $AND$ gate is $\overline{1} \cdot \overline{1} = 0 \cdot 0 = 0$. The $NOR$ gate receives inputs $1$ and $0$, so $Y = \overline{1 + 0} = \overline{1} = 0$.
Case $2$: When $A = 0$ and $B = 0$, the output of the top $AND$ gate is $0 \cdot 0 = 0$. The output of the bottom $AND$ gate is $\overline{0} \cdot \overline{0} = 1 \cdot 1 = 1$. The $NOR$ gate receives inputs $0$ and $1$, so $Y = \overline{0 + 1} = \overline{1} = 0$.
Thus, the outputs are $0$ and $0$.

(A) The resultant intensity $I$ of two superposed coherent light beams is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$,where $\phi$ is the phase difference between the beams.
For maximum intensity,$\cos \phi = 1$ (constructive interference):
$I_{\max} = I_1 + I_2 + 2\sqrt{I_1 I_2} = (\sqrt{I_1} + \sqrt{I_2})^2$
Given $I_1 = L$ and $I_2 = \frac{L}{4}$:
$I_{\max} = (\sqrt{L} + \sqrt{\frac{L}{4}})^2 = (\sqrt{L} + \frac{\sqrt{L}}{2})^2 = (\frac{3\sqrt{L}}{2})^2 = \frac{9L}{4}$
For minimum intensity,$\cos \phi = -1$ (destructive interference):
$I_{\min} = I_1 + I_2 - 2\sqrt{I_1 I_2} = (\sqrt{I_1} - \sqrt{I_2})^2$
$I_{\min} = (\sqrt{L} - \frac{\sqrt{L}}{2})^2 = (\frac{\sqrt{L}}{2})^2 = \frac{L}{4}$
Thus,the maximum and minimum intensities are $\frac{9L}{4}$ and $\frac{L}{4}$ respectively.