JEE Main 2020 Physics Question Paper with Answer and Solution

(A) The mean free path $\lambda$ is given by $\lambda = \frac{1}{\sqrt{2} \pi n_{v} d^{2}}$,where $n_{v}$ is the number density and $d$ is the diameter of the atom.
The mean free time $\tau$ is defined as $\tau = \frac{\lambda}{v_{rms}}$,where $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Substituting $\lambda$ and $v_{rms}$ into the expression for $\tau$:
$\tau = \frac{1}{\sqrt{2} \pi n_{v} d^{2}} \sqrt{\frac{M}{3RT}}$.
Since $n_{v}$,$R$,and $T$ are the same for both gases,the ratio of the mean free times $\tau_{Ar} / \tau_{Xe}$ is:
$\frac{\tau_{Ar}}{\tau_{Xe}} = \sqrt{\frac{M_{Ar}}{M_{Xe}}} \times \left( \frac{d_{Xe}}{d_{Ar}} \right)^{2}$.
Given $M_{Ar} = 40$,$M_{Xe} = 140$,$d_{Ar} = 2 \times 0.07 \; nm$,and $d_{Xe} = 2 \times 0.1 \; nm$:
$\frac{\tau_{Ar}}{\tau_{Xe}} = \sqrt{\frac{40}{140}} \times \left( \frac{0.1}{0.07} \right)^{2} = \sqrt{\frac{2}{7}} \times \left( \frac{10}{7} \right)^{2} \approx 0.5345 \times 2.0408 \approx 1.09$.

(C) Given: Initial velocity $\vec{u} = 3.0 \hat{i} \; m/s$,acceleration $\vec{a} = (6.0 \hat{i} + 4.0 \hat{j}) \; m/s^2$,and initial position $(0, 0)$ at $t=0$.
For motion in the $y$-direction:
$y = u_y t + \frac{1}{2} a_y t^2$
$32 = 0 \times t + \frac{1}{2} (4.0) t^2$
$32 = 2 t^2$
$t^2 = 16 \implies t = 4 \; s$.
For motion in the $x$-direction:
$x = u_x t + \frac{1}{2} a_x t^2$
$x = (3.0)(4) + \frac{1}{2} (6.0)(4)^2$
$x = 12 + 3.0 \times 16$
$x = 12 + 48 = 60 \; m$.
Thus,$D = 60$.

(C) $1$. At maximum height,the velocity of the first particle is $v_x = u \cos 60^{\circ} = \frac{u}{2}$ and $v_y = 0$.
$2$. The second particle has velocity $u \hat{i}$.
$3$. By conservation of linear momentum in the horizontal direction:
$m \left( \frac{u}{2} \right) + m(u) = (m + m) v^{\prime}$
$\frac{3mu}{2} = 2mv^{\prime} \implies v^{\prime} = \frac{3u}{4}$.
$4$. The maximum height $H$ is given by $H = \frac{u^2 \sin^2 60^{\circ}}{2g} = \frac{u^2 (3/4)}{2g} = \frac{3u^2}{8g}$.
$5$. The time taken to fall from height $H$ to the ground is $t = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2}{g} \cdot \frac{3u^2}{8g}} = \sqrt{\frac{3u^2}{4g^2}} = \frac{u \sqrt{3}}{2g}$.
$6$. The horizontal distance covered by the combined mass after collision is $d = v^{\prime} \cdot t = \left( \frac{3u}{4} \right) \left( \frac{u \sqrt{3}}{2g} \right) = \frac{3 \sqrt{3} u^2}{8g}$.

(B) By using the work-energy theorem,the work done by gravity equals the change in kinetic energy of the system.
$W_{g} = \Delta KE$
$(m_{1} - m_{2}) gh = \frac{1}{2} m_{1} v^{2} + \frac{1}{2} m_{2} v^{2} + \frac{1}{2} I \omega^{2}$
Since the string does not slip,$v = \omega R$. Substituting this into the equation:
$(m_{1} - m_{2}) gh = \frac{1}{2} (m_{1} + m_{2}) (\omega R)^{2} + \frac{1}{2} I \omega^{2}$
$(m_{1} - m_{2}) gh = \frac{\omega^{2}}{2} [(m_{1} + m_{2}) R^{2} + I]$
Solving for $\omega$:
$\omega = \sqrt{\frac{2(m_{1} - m_{2}) gh}{(m_{1} + m_{2}) R^{2} + I}}$

(A) The formula for escape velocity is $v_{e} = \sqrt{\frac{2GM}{R}}$.
For Planet $A$,the escape velocity is $v_{A} = \sqrt{\frac{2GM}{R}}$.
For Planet $B$,the mass is $M' = \frac{M}{2}$ and the radius is $R' = \frac{R}{2}$.
Thus,the escape velocity for Planet $B$ is $v_{B} = \sqrt{\frac{2G(M/2)}{R/2}} = \sqrt{\frac{2GM}{R}}$.
Comparing the two,we get $\frac{v_{A}}{v_{B}} = \frac{\sqrt{2GM/R}}{\sqrt{2GM/R}} = 1$.
Given that $\frac{v_{A}}{v_{B}} = \frac{n}{4}$,we have $1 = \frac{n}{4}$,which implies $n = 4$.

(D) The energy stored per unit volume $(u)$ is given by the formula: $u = \frac{1}{2} \times \text{Stress} \times \text{Strain} = \frac{1}{2} \frac{(\text{Stress})^2}{Y}$.
Since $\text{Stress} = \frac{F}{A} = \frac{F}{\pi (d/2)^2} = \frac{4F}{\pi d^2}$,we have $u = \frac{1}{2Y} \left( \frac{4F}{\pi d^2} \right)^2$.
Given that the load $(F)$,length,and Young's modulus $(Y)$ are the same for both wires,$u \propto \frac{1}{d^4}$.
Therefore,$\frac{u_1}{u_2} = \left( \frac{d_2}{d_1} \right)^4$.
Given $\frac{u_1}{u_2} = \frac{1}{4}$,we have $\frac{1}{4} = \left( \frac{d_2}{d_1} \right)^4$.
Taking the fourth root on both sides,$\frac{d_2}{d_1} = \left( \frac{1}{4} \right)^{1/4} = \left( \frac{1}{2^2} \right)^{1/4} = \frac{1}{2^{1/2}} = \frac{1}{\sqrt{2}}$.
Thus,$\frac{d_1}{d_2} = \sqrt{2} : 1$.

(D) To determine the correct order,we must perform the addition and round the result to the appropriate number of significant figures based on the least precise measurement (the one with the fewest decimal places).
$(i)$ $A_{1}+B_{1}+C_{1} = 24.36 + 0.0724 + 256.2 = 280.6324$. The least precise value is $256.2$ (one decimal place),so the sum is $280.6$.
$(ii)$ $A_{2}+B_{2}+C_{2} = 24.44 + 16.082 + 240.2 = 280.722$. The least precise value is $240.2$ (one decimal place),so the sum is $280.7$.
$(iii)$ $A_{3}+B_{3}+C_{3} = 25.2 + 19.2812 + 236.183 = 280.6642$. The least precise value is $25.2$ (one decimal place),so the sum is $280.7$.
$(iv)$ $A_{4}+B_{4}+C_{4} = 25 + 236.191 + 19.5 = 280.691$. The least precise value is $25$ (zero decimal places),so the sum is $281$.
Comparing the results: $281 > 280.7 = 280.7 > 280.6$. Thus,$A_{4}+B_{4}+C_{4} > A_{3}+B_{3}+C_{3} = A_{2}+B_{2}+C_{2} > A_{1}+B_{1}+C_{1}$.

(A) Step $1$: Adiabatic compression from $V_{1}$ to $V_{2} = V_{1}/16$.
For an adiabatic process,$T_{1}V_{1}^{\gamma-1} = T_{2}V_{2}^{\gamma-1}$.
Given $T_{1} = 300 \; K$,$\gamma = 1.4 = 7/5$,so $\gamma-1 = 0.4 = 2/5$.
$300 \times V_{1}^{2/5} = T_{2} \times (V_{1}/16)^{2/5}$.
$T_{2} = 300 \times (16)^{2/5} = 300 \times (2^{4})^{2/5} = 300 \times 2^{8/5}$.
$T_{2} = 300 \times 3.0314 = 909.42 \; K$.
Step $2$: Isobaric expansion from $V_{2}$ to $2V_{2}$.
For an isobaric process,$V/T = \text{constant}$,so $V_{2}/T_{2} = (2V_{2})/T_{f}$.
$T_{f} = 2 \times T_{2} = 2 \times 909.42 = 1818.84 \; K$.
Rounding to the nearest integer,the final temperature is $1819 \; K$. Given the options,$1818$ is the closest match.

(B) The work done by a variable force is given by the line integral $W = \int_{A}^{B} \overrightarrow{F} \cdot d\overrightarrow{r}$.
The equation of the line segment connecting $A(1, 0)$ and $B(0, 1)$ is $y = -x + 1$,or $x + y = 1$.
Thus,$d\overrightarrow{r} = dx \hat{i} + dy \hat{j}$.
The work done is $W = \int_{A}^{B} (-x \hat{i} + y \hat{j}) \cdot (dx \hat{i} + dy \hat{j}) = \int_{A}^{B} (-x dx + y dy)$.
Substituting the limits from $x=1$ to $x=0$ and $y=0$ to $y=1$:
$W = \int_{1}^{0} -x dx + \int_{0}^{1} y dy$
$W = \left[ -\frac{x^2}{2} \right]_{1}^{0} + \left[ \frac{y^2}{2} \right]_{0}^{1}$
$W = -\left( \frac{0^2}{2} - \frac{1^2}{2} \right) + \left( \frac{1^2}{2} - \frac{0^2}{2} \right)$
$W = -\left( 0 - 0.5 \right) + \left( 0.5 - 0 \right) = 0.5 + 0.5 = 1 \text{ J}$.

(C) The dimensions of the constants are:
$[h] = M^1 L^2 T^{-1}$
$[c] = L^1 T^{-1}$
$[G] = M^{-1} L^3 T^{-2}$
Substituting these into the expression for $f$:
$[f] = \sqrt{\frac{(M^1 L^2 T^{-1}) \times (L^1 T^{-1})^5}{M^{-1} L^3 T^{-2}}}$
$[f] = \sqrt{\frac{M^1 L^2 T^{-1} \times L^5 T^{-5}}{M^{-1} L^3 T^{-2}}}$
$[f] = \sqrt{\frac{M^1 L^7 T^{-6}}{M^{-1} L^3 T^{-2}}}$
$[f] = \sqrt{M^{1-(-1)} L^{7-3} T^{-6-(-2)}}$
$[f] = \sqrt{M^2 L^4 T^{-4}}$
$[f] = M^1 L^2 T^{-2}$
The dimension $M^1 L^2 T^{-2}$ corresponds to the dimension of Energy.

(A) Initially,the body of mass $m$ is moving in a circular orbit of radius $R$. So,it must be moving with orbital speed $v_{0} = \sqrt{\frac{GM}{R}}$.
After the collision,let the combined mass move with speed $v_{1}$.
Using the conservation of linear momentum:
$m v_{0} + \frac{m}{2} \left(\frac{v_{0}}{2}\right) = \left(m + \frac{m}{2}\right) v_{1}$
$m v_{0} + \frac{m v_{0}}{4} = \frac{3m}{2} v_{1}$
$\frac{5m v_{0}}{4} = \frac{3m}{2} v_{1}$
$v_{1} = \frac{5}{4} \times \frac{2}{3} v_{0} = \frac{5}{6} v_{0}$.
Since the speed after the collision $(v_{1} = \frac{5}{6} v_{0})$ is not equal to the orbital speed $(v_{0})$,the motion cannot be circular.
Since the velocity remains tangential,it cannot fall vertically towards the planet.
Also,the speed after the collision is less than the escape speed $(v_{e} = \sqrt{2} v_{0})$,so the body cannot escape the gravitational field.
Therefore,the combined body will move in an elliptical orbit around the planet.

(B) For a rigid diatomic molecule,the degrees of freedom $(f_A)$ are $5$ ($3$ translational + $2$ rotational).
For a diatomic molecule with an additional vibrational mode,the degrees of freedom $(f_B)$ are $7$ ($3$ translational + $2$ rotational + $2$ vibrational).
The molar specific heat at constant volume is given by $C_v = \frac{f}{2}R$.
Therefore,$C_v^A = \frac{5}{2}R$ and $C_v^B = \frac{7}{2}R$.
The ratio $\frac{C_v^A}{C_v^B} = \frac{\frac{5}{2}R}{\frac{7}{2}R} = \frac{5}{7}$.

(B) The pitch of the screw gauge is the distance moved by the screw in one complete rotation.
Given that $6$ rotations correspond to a movement of $3\; mm$ on the main scale.
Therefore,the pitch $= \frac{3\; mm}{6} = 0.5\; mm$.
The least count $(LC)$ of a screw gauge is defined as the ratio of the pitch to the total number of divisions on the circular scale.
$LC = \frac{\text{Pitch}}{\text{Number of circular scale divisions}}$.
Given the number of divisions $= 50$.
$LC = \frac{0.5\; mm}{50} = 0.01\; mm$.
Converting this to centimeters,$0.01\; mm = 0.001\; cm$.

(B) Initial momentum of the system: $\vec{P}_i = m(u\hat{i}) + m\left(\frac{u}{2}\hat{i} + \frac{u}{2}\hat{j}\right) = \frac{3}{2}mu\hat{i} + \frac{1}{2}mu\hat{j}$.
Since the collision is completely inelastic,the particles stick together and move with a common velocity $\vec{v}_f$. By conservation of linear momentum: $\vec{P}_i = (m+m)\vec{v}_f = 2m\vec{v}_f$.
$\vec{v}_f = \frac{1}{2m} \left(\frac{3}{2}mu\hat{i} + \frac{1}{2}mu\hat{j}\right) = \frac{3}{4}u\hat{i} + \frac{1}{4}u\hat{j}$.
Initial kinetic energy: $K_i = \frac{1}{2}mu^2 + \frac{1}{2}m\left|\frac{u}{2}\hat{i} + \frac{u}{2}\hat{j}\right|^2 = \frac{1}{2}mu^2 + \frac{1}{2}m\left(\frac{u^2}{4} + \frac{u^2}{4}\right) = \frac{1}{2}mu^2 + \frac{1}{4}mu^2 = \frac{3}{4}mu^2$.
Final kinetic energy: $K_f = \frac{1}{2}(2m)|\vec{v}_f|^2 = m\left(\left(\frac{3}{4}u\right)^2 + \left(\frac{1}{4}u\right)^2\right) = m\left(\frac{9}{16}u^2 + \frac{1}{16}u^2\right) = \frac{10}{16}mu^2 = \frac{5}{8}mu^2$.
Energy lost: $\Delta K = K_i - K_f = \frac{3}{4}mu^2 - \frac{5}{8}mu^2 = \frac{6-5}{8}mu^2 = \frac{1}{8}mu^2$.

(D) In the given $P-V$ diagram:
$1$. Process $1 \rightarrow 2$ is an adiabatic expansion (pressure decreases, volume increases).
$2$. Process $2 \rightarrow 3$ is an isobaric compression (pressure is constant, volume decreases).
$3$. Process $3 \rightarrow 1$ is an isochoric heating (volume is constant, pressure increases).
Now, let's analyze the $V-T$ diagrams:
- For process $1 \rightarrow 2$ (adiabatic): $TV^{\gamma-1} = \text{constant}$. Since volume increases, temperature must decrease.
- For process $2 \rightarrow 3$ (isobaric): $V \propto T$. Since volume decreases, temperature must decrease.
- For process $3 \rightarrow 1$ (isochoric): $P \propto T$. Since pressure increases, temperature must increase.
Comparing these with the given options, the $V-T$ diagram that shows these trends is represented in option $D$.

(A) Let the amplitude of each wave be $A$. Since intensity $I \propto A^2$,we have $I_0 = kA^2$,where $k$ is a constant.
The resultant wave equation is given by the sum of the three waves:
$y = A \sin(\omega t) + A \sin(\omega t - \frac{\pi}{4}) + A \sin(\omega t + \frac{\pi}{4})$
Using the trigonometric identity $\sin(x - y) + \sin(x + y) = 2 \sin x \cos y$:
$y = A \sin(\omega t) + A [2 \sin(\omega t) \cos(\frac{\pi}{4})]$
Since $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$:
$y = A \sin(\omega t) + A [2 \sin(\omega t) \cdot \frac{1}{\sqrt{2}}]$
$y = A \sin(\omega t) + \sqrt{2} A \sin(\omega t)$
$y = (1 + \sqrt{2}) A \sin(\omega t)$
The resultant amplitude $A_R = (1 + \sqrt{2}) A$.
The resultant intensity $I = k A_R^2 = k [(1 + \sqrt{2}) A]^2 = k A^2 (1 + \sqrt{2})^2$.
Since $I_0 = k A^2$:
$I = I_0 (1 + 2 + 2\sqrt{2}) = I_0 (3 + 2 \times 1.414) = I_0 (3 + 2.828) = 5.828 I_0$.
Thus,the resultant intensity is approximately $5.8 I_0$.

(A) The radius of each sphere is $r = d/2$. The distance from the centroid of the equilateral triangle to the center of each sphere is $R = \frac{d}{\sqrt{3}}$.
Using the parallel axis theorem for each sphere about the axis passing through the centroid $(I_0)$:
$I_0 = 3 \times [I_{cm} + mR^2] = 3 \times [\frac{2}{5}m(d/2)^2 + m(d/\sqrt{3})^2]$
$I_0 = 3 \times [\frac{1}{10}md^2 + \frac{1}{3}md^2] = 3 \times [\frac{3+10}{30}]md^2 = 3 \times \frac{13}{30}md^2 = \frac{13}{10}md^2$.
Now,for the axis passing through the center of one sphere $(A)$ and perpendicular to the plane,we use the parallel axis theorem between the centroid axis and the axis at $A$:
$I_A = I_0 + 3mR^2 = \frac{13}{10}md^2 + 3m(d/\sqrt{3})^2 = \frac{13}{10}md^2 + md^2 = \frac{23}{10}md^2$.
The ratio is $\frac{I_0}{I_A} = \frac{13/10}{23/10} = \frac{13}{23}$.

(C) The rate of flow of water is constant,so $A_{A} V_{A} = A_{B} V_{B}$.
Given $A_{A} = 40 \; cm^{2}$ and $A_{B} = 20 \; cm^{2}$,we have $40 V_{A} = 20 V_{B}$,which implies $V_{B} = 2 V_{A}$.
Using Bernoulli's theorem for a horizontal tube: $P_{A} + \frac{1}{2} \rho V_{A}^{2} = P_{B} + \frac{1}{2} \rho V_{B}^{2}$.
Rearranging gives $P_{A} - P_{B} = \frac{1}{2} \rho (V_{B}^{2} - V_{A}^{2})$.
Substituting the given values ($P_{A} - P_{B} = 700 \; Pa$,$\rho = 1000 \; kg/m^{3}$):
$700 = \frac{1}{2} \times 1000 \times ((2 V_{A})^{2} - V_{A}^{2})$
$700 = 500 \times (4 V_{A}^{2} - V_{A}^{2})$
$700 = 500 \times 3 V_{A}^{2}$
$V_{A}^{2} = \frac{700}{1500} = \frac{7}{15} \; m^{2}/s^{2}$
$V_{A} = \sqrt{\frac{7}{15}} \approx 0.683 \; m/s = 68.3 \; cm/s$.
The rate of flow $Q = A_{A} V_{A} = 40 \; cm^{2} \times 68.3 \; cm/s \approx 2732 \; cm^{3}/s$. Given the options,the closest value is $2720 \; cm^{3}/s$.

(C) Let the length of the bar be $\ell = 1\; m$ and its mass be $m$. The moment of inertia of the bar about the pivot point is $I = \frac{m\ell^2}{3}$.
Using the principle of conservation of mechanical energy,the initial potential energy of the center of mass of the bar is converted into rotational kinetic energy when it hits the table.
The initial height of the center of mass from the table is $h = \frac{\ell}{2} \sin 30^{\circ}$.
Initial potential energy $U_i = mgh = mg \left(\frac{\ell}{2}\right) \sin 30^{\circ}$.
Final potential energy $U_f = 0$ (at the table level).
Initial kinetic energy $K_i = 0$ (released from rest).
Final rotational kinetic energy $K_f = \frac{1}{2} I \omega^2$.
By conservation of energy: $U_i + K_i = U_f + K_f$
$mg \left(\frac{\ell}{2}\right) \sin 30^{\circ} = \frac{1}{2} \left(\frac{m\ell^2}{3}\right) \omega^2$
Substituting $\ell = 1\; m$ and $\sin 30^{\circ} = 0.5$:
$mg \left(\frac{1}{2}\right) (0.5) = \frac{1}{2} \left(\frac{m(1)^2}{3}\right) \omega^2$
$mg \left(\frac{1}{4}\right) = \frac{m}{6} \omega^2$
$\frac{g}{4} = \frac{\omega^2}{6} \Rightarrow \omega^2 = \frac{6g}{4} = 1.5g$.
Taking $g = 10\; m/s^2$,we get $\omega^2 = 1.5 \times 10 = 15$.
Thus,$\omega = \sqrt{15}\; s^{-1}$.
Comparing with $\sqrt{n}$,we get $n = 15$.

(D) Given $x^{2} = at^{2} + 2bt + c$.
Differentiating with respect to $t$:
$2x \frac{dx}{dt} = 2at + 2b$
$x v = at + b$,where $v = \frac{dx}{dt}$.
Differentiating again with respect to $t$:
$v \frac{dx}{dt} + x \frac{dv}{dt} = a$
$v^{2} + x a' = a$,where $a' = \frac{dv}{dt}$ is the acceleration.
From $x v = at + b$,we have $v = \frac{at+b}{x}$.
Substituting $v$ into the equation:
$x a' = a - v^{2} = a - \left(\frac{at+b}{x}\right)^{2}$
$x a' = \frac{ax^{2} - (at+b)^{2}}{x^{2}}$
Substitute $x^{2} = at^{2} + 2bt + c$:
$x a' = \frac{a(at^{2} + 2bt + c) - (a^{2}t^{2} + 2abt + b^{2})}{x^{2}}$
$x a' = \frac{a^{2}t^{2} + 2abt + ac - a^{2}t^{2} - 2abt - b^{2}}{x^{2}}$
$x a' = \frac{ac - b^{2}}{x^{2}}$
$a' = \frac{ac - b^{2}}{x^{3}}$
Thus,$a' \propto x^{-3}$. Comparing with $x^{-n}$,we get $n = 3$.

(D) The tension $T$ in the wire provides the necessary centripetal force for circular motion: $T = m \omega^{2} L$.
The breaking stress is defined as the maximum force per unit area: $\text{Breaking Stress} = \frac{T}{A}$.
Substituting the expression for tension: $\text{Breaking Stress} = \frac{m \omega^{2} L}{A}$.
Given: $m = 10 \; kg$,$L = 0.3 \; m$,$\text{Breaking Stress} = 4.8 \times 10^{7} \; N m^{-2}$,and $A = 10^{-2} \; cm^{2} = 10^{-2} \times 10^{-4} \; m^{2} = 10^{-6} \; m^{2}$.
Rearranging for $\omega^{2}$: $\omega^{2} = \frac{(\text{Breaking Stress}) \times A}{m \times L}$.
$\omega^{2} = \frac{4.8 \times 10^{7} \times 10^{-6}}{10 \times 0.3} = \frac{48}{3} = 16$.
Therefore,$\omega = \sqrt{16} = 4 \; rad \; s^{-1}$.

(D) From the graph,the displacement $x$ follows a cosine function starting from $x = A$ at $t = 0$,so $x(t) = A \cos(\omega t)$.
$(A)$ At $t = \frac{3T}{4}$,the displacement $x = A \cos(\omega \cdot \frac{3T}{4}) = A \cos(\frac{3\pi}{2}) = 0$. Since $F = -m\omega^2 x$,if $x = 0$,then $F = 0$. Thus,$(A)$ is true.
$(B)$ At $t = T$,$x = A \cos(\omega T) = A \cos(2\pi) = A$. The acceleration $a = -\omega^2 x = -\omega^2 A$. The magnitude of acceleration is maximum at the extreme positions $(x = \pm A)$. Thus,$(B)$ is true.
$(C)$ At $t = \frac{T}{4}$,$x = A \cos(\omega \cdot \frac{T}{4}) = A \cos(\frac{\pi}{2}) = 0$. The speed $v = \omega \sqrt{A^2 - x^2}$ is maximum when $x = 0$. Thus,$(C)$ is true.
$(D)$ At $t = \frac{T}{2}$,$x = A \cos(\omega \cdot \frac{T}{2}) = A \cos(\pi) = -A$. At $x = -A$,$P.E.$ is maximum and $K.E. = 0$. Therefore,$P.E. \neq K.E.$. Thus,$(D)$ is false.
Therefore,statements $(A), (B),$ and $(C)$ are true.

(C) The efficiency $\eta$ of a heat engine is defined as the ratio of the net work done to the total heat supplied.
$\eta = \frac{W}{Q_{\text{in}}} = \frac{Q_{\text{net}}}{Q_{\text{in}}}$
Here,the net heat exchanged in one cycle is $Q_{\text{net}} = 1915 - 40 + 125 - Q = 2000 - Q$.
The heat supplied $(Q_{\text{in}})$ is the sum of all positive heat exchanges: $Q_{\text{in}} = 1915 + 125 = 2040\, J$.
Given $\eta = 50.0\% = 0.5$,we have:
$0.5 = \frac{2000 - Q}{2040}$
$1020 = 2000 - Q$
$Q = 2000 - 1020 = 980\, J$.

(B) Let the dimensional formula for energy be $[E] = [P]^x [A]^y [T]^z$.
Substituting the dimensions of each quantity:
$[M L^2 T^{-2}] = [M L T^{-1}]^x [L^2]^y [T]^z$.
Equating the powers of $M, L,$ and $T$ on both sides:
$M^1 L^2 T^{-2} = M^x L^{x+2y} T^{-x+z}$.
Comparing the exponents:
For $M$: $x = 1$.
For $L$: $x + 2y = 2$. Substituting $x=1$,we get $1 + 2y = 2$,which implies $2y = 1$,so $y = 1/2$.
For $T$: $-x + z = -2$. Substituting $x=1$,we get $-1 + z = -2$,which implies $z = -1$.
Thus,the dimensional formula for energy is $[P^1 A^{1/2} T^{-1}]$ or $[P A^{1/2} T^{-1}]$.

(C) Let $r$ be the radius of the capillary tube and $R$ be the radius of the meniscus.
From the geometry of the meniscus,the angle of contact $\theta$ is related to the angle between the tangents. Since the tangents make an angle of $60^{\circ}$ with each other,the angle between the tangent and the vertical wall is $30^{\circ}$. Thus,the contact angle $\theta = 30^{\circ}$.
From the geometry,$\cos \theta = \frac{r}{R}$,so $R = \frac{r}{\cos 30^{\circ}} = \frac{r}{\sqrt{3}/2} = \frac{2r}{\sqrt{3}}$.
Given $r = 0.15 \times 10^{-3} m$,we have $R = \frac{2 \times 0.15 \times 10^{-3}}{\sqrt{3}} = \frac{0.3 \times 10^{-3}}{\sqrt{3}} m$.
The height $h$ of the liquid column is given by $h = \frac{2T \cos \theta}{\rho g r}$.
Substituting the values: $h = \frac{2 \times 0.05 \times \cos 30^{\circ}}{667 \times 10 \times 0.15 \times 10^{-3}}$.
Alternatively,using $h = \frac{2T}{\rho g R}$:
$h = \frac{2 \times 0.05}{667 \times 10 \times (\frac{0.3 \times 10^{-3}}{\sqrt{3}})} = \frac{0.1 \times \sqrt{3}}{6670 \times 0.3 \times 10^{-3}} = \frac{0.1732}{2.001} \approx 0.0865\, m$.
Rounding to the nearest value,$h \approx 0.087\, m$.

(D) The mean free path $\lambda$ of molecules of an ideal gas is given by the formula:
$\lambda = \frac{V}{\sqrt{2} \pi d^2 N}$
where $V$ is the volume of the container,$d$ is the molecular diameter,and $N$ is the number of molecules.
Since the gas is in a closed container,the volume $V$ and the number of molecules $N$ remain constant as the temperature increases. Therefore,the mean free path $\lambda$ remains unchanged.
The mean collision time $\tau$ is defined as the ratio of the mean free path to the average speed of the molecules $(v_{av})$:
$\tau = \frac{\lambda}{v_{av}}$
Since the average speed $v_{av}$ of gas molecules is proportional to $\sqrt{T}$,we have:
$\tau \propto \frac{1}{\sqrt{T}}$
As the temperature $T$ increases,the average speed of the molecules increases,which causes the mean collision time $\tau$ to decrease.
Thus,statements $(B)$ and $(C)$ are correct.

(D) Since the discs are rotating in the same direction,the total angular momentum is conserved.
Initial angular momentum $L_{i} = I_{1}\omega_{1} + I_{2}\omega_{2}$
$L_{i} = (0.1 \times 10) + (0.2 \times 5) = 1 + 1 = 2 \; kg \cdot m^{2} \cdot s^{-1}$
Final angular momentum $L_{f} = (I_{1} + I_{2})\omega_{f}$
$2 = (0.1 + 0.2) \omega_{f} = 0.3 \omega_{f}$
$\omega_{f} = \frac{2}{0.3} = \frac{20}{3} \; rad \cdot s^{-1}$
Final kinetic energy $K_{f} = \frac{1}{2}(I_{1} + I_{2})\omega_{f}^{2}$
$K_{f} = \frac{1}{2}(0.3) \left(\frac{20}{3}\right)^{2} = 0.15 \times \frac{400}{9} = \frac{15}{100} \times \frac{400}{9} = \frac{60}{9} = 6.67 \; J$.

(B) Given that the fractional change in length is $\frac{\Delta L}{L} = 0.02 \% = 2 \times 10^{-4}$.
Since $\Delta L = L \alpha \Delta T$,we have $\alpha \Delta T = 2 \times 10^{-4}$.
The volume $V$ of the wire is $V = A \times L$,where $A$ is the cross-sectional area.
The density $\rho$ is given by $\rho = \frac{M}{V} = \frac{M}{AL}$.
Taking the logarithmic derivative,we get $\frac{\Delta \rho}{\rho} = - (\frac{\Delta A}{A} + \frac{\Delta L}{L})$.
Since $\frac{\Delta A}{A} = 2 \alpha \Delta T$ and $\frac{\Delta L}{L} = \alpha \Delta T$,the fractional change in density is $\frac{\Delta \rho}{\rho} = -(2 \alpha \Delta T + \alpha \Delta T) = -3 \alpha \Delta T$.
Substituting $\alpha \Delta T = 0.02 \%$,we get $\frac{\Delta \rho}{\rho} = -3 \times 0.02 \% = -0.06 \%$.
The magnitude of the percentage change is $0.06 \%$.

(C) Let the velocity of mass $m$ after collision be $v_1$ and the velocity of mass $10m$ after collision be $v_2$.
$1$. Conservation of momentum along the $y$-axis:
Since the initial momentum along the $y$-axis is zero,the final momentum components along the $y$-axis must cancel out:
$m v_1 \sin \theta_1 = 10m v_2 \sin \theta_2$
$v_1 \sin \theta_1 = 10 v_2 \sin \theta_2$ --- $(i)$
$2$. Kinetic energy condition:
The final kinetic energy of the particle of mass $m$ is half its initial kinetic energy:
$\frac{1}{2} m v_1^2 = \frac{1}{2} (\frac{1}{2} m u^2)$
$v_1^2 = \frac{u^2}{2} \Rightarrow v_1 = \frac{u}{\sqrt{2}}$ --- (ii)
$3$. Conservation of energy for elastic collision:
Initial kinetic energy = Final kinetic energy
$\frac{1}{2} m u^2 = \frac{1}{2} m v_1^2 + \frac{1}{2} (10m) v_2^2$
$u^2 = v_1^2 + 10 v_2^2$
Substitute $v_1^2 = \frac{u^2}{2}$:
$u^2 = \frac{u^2}{2} + 10 v_2^2$
$\frac{u^2}{2} = 10 v_2^2 \Rightarrow v_2^2 = \frac{u^2}{20} \Rightarrow v_2 = \frac{u}{\sqrt{20}}$ --- (iii)
$4$. Substituting (ii) and (iii) into $(i)$:
$(\frac{u}{\sqrt{2}}) \sin \theta_1 = 10 (\frac{u}{\sqrt{20}}) \sin \theta_2$
$\sin \theta_1 = 10 \cdot \frac{\sqrt{2}}{\sqrt{20}} \sin \theta_2$
$\sin \theta_1 = 10 \cdot \frac{1}{\sqrt{10}} \sin \theta_2$
$\sin \theta_1 = \sqrt{10} \sin \theta_2$
Comparing with $\sin \theta_1 = \sqrt{n} \sin \theta_2$,we get $n = 10$.

(D) Let $\sigma$ be the surface mass density of the disk material.
Mass of the complete disk,$M_1 = \sigma \pi a^2$,with its centre of mass at $O$ (coordinate $x_1 = 0$).
Mass of the removed square portion,$M_2 = \sigma l^2 = \sigma (\frac{a}{2})^2 = \sigma \frac{a^2}{4}$,with its centre of mass at distance $d = \frac{a}{2}$ from $O$ (coordinate $x_2 = \frac{a}{2}$).
The centre of mass of the remaining portion $X_{com}$ is given by:
$X_{com} = \frac{M_1 x_1 - M_2 x_2}{M_1 - M_2}$
$X_{com} = \frac{(\sigma \pi a^2)(0) - (\sigma \frac{a^2}{4})(\frac{a}{2})}{\sigma \pi a^2 - \sigma \frac{a^2}{4}}$
$X_{com} = \frac{-\frac{\sigma a^3}{8}}{\sigma a^2 (\pi - \frac{1}{4})} = \frac{-\frac{a}{8}}{\frac{4\pi - 1}{4}} = -\frac{a}{2(4\pi - 1)}$
Comparing this with $-\frac{a}{X}$,we get $X = 2(4\pi - 1) = 8\pi - 2$.
Using $\pi \approx 3.14159$,$X \approx 8(3.14159) - 2 = 25.1327 - 2 = 23.1327$.
The nearest integer value of $X$ is $23$.

(A) Given: Density $\rho = 9 \times 10^{-3} \,kg/cm^3 = 9 \times 10^{-3} / 10^{-6} \,kg/m^3 = 9000 \,kg/m^3$.
Length $L = 1 \,m$.
Strain $= 4.9 \times 10^{-4}$.
Young's modulus $Y = 9 \times 10^{10} \,N/m^2$.
For the lowest frequency (fundamental mode),$L = \lambda / 2$,so $\lambda = 2L = 2 \,m$.
The frequency $f$ is given by $f = v / \lambda = (1 / \lambda) \sqrt{T / \mu}$,where $T$ is tension and $\mu$ is mass per unit length.
Since $Y = (T/A) / \text{strain}$,we have $T = Y \cdot A \cdot \text{strain}$.
Also,$\mu = m/L = (\rho \cdot V) / L = (\rho \cdot A \cdot L) / L = \rho \cdot A$.
Substituting these into the frequency formula:
$f = (1 / 2L) \sqrt{(Y \cdot A \cdot \text{strain}) / (\rho \cdot A)} = (1 / 2L) \sqrt{(Y \cdot \text{strain}) / \rho}$.
Substituting the values:
$f = (1 / 2) \sqrt{(9 \times 10^{10} \times 4.9 \times 10^{-4}) / 9000} = (1 / 2) \sqrt{(9 \times 10^6 \times 4.9) / 9000} = (1 / 2) \sqrt{10^3 \times 4.9} = (1 / 2) \sqrt{4900} = 70 / 2 = 35 \,Hz$.

(D) From the law of conservation of linear momentum:
$\vec{P}_i = \vec{P}_f$
$m(u\hat{i}) + 3m(0) = m(v\hat{j}) + 3m\vec{v}_1$
$m(u\hat{i} - v\hat{j}) = 3m\vec{v}_1$
$\vec{v}_1 = \frac{u\hat{i} - v\hat{j}}{3}$
Taking the magnitude squared:
$v_1^2 = \frac{u^2 + v^2}{9} \quad \dots(1)$
Since the collision is perfectly elastic, kinetic energy is conserved:
$K_i = K_f$
$\frac{1}{2}mu^2 + 0 = \frac{1}{2}mv^2 + \frac{1}{2}(3m)v_1^2$
$u^2 = v^2 + 3v_1^2$
Substituting equation $(1)$ into the energy equation:
$u^2 = v^2 + 3\left(\frac{u^2 + v^2}{9}\right)$
$u^2 = v^2 + \frac{u^2 + v^2}{3}$
$3u^2 = 3v^2 + u^2 + v^2$
$2u^2 = 4v^2$
$v^2 = \frac{u^2}{2}$
$v = \frac{u}{\sqrt{2}}$

(D) $1$. Calculate the Least Count $(LC)$: $LC = 1\, MS D - 1\, VSD$. Given $10\, VSD = 9\, MSD$,so $1\, VSD = 0.9\, MSD = 0.9\, mm$. Thus,$LC = 1\, mm - 0.9\, mm = 0.1\, mm = 0.01\, cm$.
$2$. Calculate Zero Error: Since the zero of the vernier scale is to the right of the main scale zero,the error is positive. The $7^{th}$ division coincides,so $Zero\, Error = + (7 \times LC) = + (7 \times 0.01\, cm) = +0.07\, cm$.
$3$. Calculate Observed Reading: $Main\, Scale\, Reading (MSR) = 3.1\, cm$. $Vernier\, Scale\, Reading (VSR) = 4 \times LC = 4 \times 0.01\, cm = 0.04\, cm$. $Observed\, Reading = MSR + VSR = 3.1\, cm + 0.04\, cm = 3.14\, cm$.
$4$. Calculate Corrected Length: $Corrected\, Length = Observed\, Reading - Zero\, Error = 3.14\, cm - 0.07\, cm = 3.07\, cm$.

(C) The mass of the galaxy within a radius $R$ is calculated by integrating the density $\rho(r) = \frac{K}{r}$ over the volume.
$dm = \rho(r) \cdot 4\pi r^2 dr = \left(\frac{K}{r}\right) \cdot 4\pi r^2 dr = 4\pi K r dr$
$M(R) = \int_{0}^{R} 4\pi K r dr = 4\pi K \left[ \frac{r^2}{2} \right]_{0}^{R} = 2\pi K R^2$
For a star of mass $m$ in a circular orbit of radius $R$,the gravitational force provides the necessary centripetal force:
$\frac{G M(R) m}{R^2} = \frac{m v^2}{R}$
Substituting $M(R) = 2\pi K R^2$:
$\frac{G (2\pi K R^2) m}{R^2} = \frac{m v^2}{R} \Rightarrow 2\pi G K m = \frac{m v^2}{R} \Rightarrow v^2 = 2\pi G K R$
$v = \sqrt{2\pi G K R}$
The time period $T$ is given by $T = \frac{2\pi R}{v}$.
$T = \frac{2\pi R}{\sqrt{2\pi G K R}} = \sqrt{\frac{4\pi^2 R^2}{2\pi G K R}} = \sqrt{\frac{2\pi R}{G K}} \propto \sqrt{R}$
Squaring both sides,we get $T^2 \propto R$.

(D) For the rod to be in rotational equilibrium,the net torque about any point must be zero.
Let us calculate the torque about point $B$:
$\tau_{B} = 0$
Taking clockwise torques as positive and counter-clockwise as negative:
$(T_{A} \times 100) - (mg \times 50) - (2mg \times 25) = 0$
$100 T_{A} = 50mg + 50mg$
$100 T_{A} = 100mg$
$T_{A} = 1mg$
Thus,the tension in the string at $A$ is $1mg$.

(B) In the rotating frame of reference,the forces acting on the bead are the gravitational force $mg$ (downwards),the centrifugal force $m x \omega^2$ (outwards),and the normal reaction $N$ from the wire.
For the bead to remain stationary at point $P(a, b)$,the net force along the tangent to the parabola must be zero.
The slope of the parabola $y = 4Cx^2$ is given by $\frac{dy}{dx} = 8Cx$.
At point $P(a, b)$,the slope is $\tan \theta = 8Ca$,where $\theta$ is the angle the tangent makes with the horizontal.
Resolving forces along the tangent,we have $m x \omega^2 \cos \theta = mg \sin \theta$.
This simplifies to $x \omega^2 = g \tan \theta$.
Substituting $x = a$ and $\tan \theta = 8Ca$,we get $a \omega^2 = g(8Ca)$.
Thus,$\omega^2 = 8gC$,which gives $\omega = \sqrt{8gC} = 2\sqrt{2gC}$.

(A) The surface of a rotating liquid takes the shape of a paraboloid defined by the equation $z = \frac{\omega^2 r^2}{2g}$.
Here,the diameter of the vessel is $10 \, cm$,so the radius $R = 5 \, cm$.
The difference in height $h$ between the center $(r=0)$ and the side $(r=R)$ is given by:
$h = \frac{\omega^2 R^2}{2g}$
Substituting $R = 5 \, cm$:
$h = \frac{\omega^2 (5)^2}{2g} = \frac{25 \omega^2}{2g}$.

(D) To pull the cylinder over the step,the torque produced by the applied force $F$ about the edge of the step must be greater than or equal to the torque produced by the gravitational force $Mg$ about the same edge.
Let the edge of the step be point $P$. The perpendicular distance from $P$ to the line of action of force $F$ is $R$.
The perpendicular distance from $P$ to the line of action of the weight $Mg$ is $x$.
From the geometry of the cylinder,we have a right-angled triangle with hypotenuse $R$ and one side $(R-a)$. Thus,$x = \sqrt{R^2 - (R-a)^2}$.
For the cylinder to just start moving over the step,the torques must balance:
$F \times R = Mg \times x$
$F \times R = Mg \times \sqrt{R^2 - (R-a)^2}$
$F = \frac{Mg}{R} \sqrt{R^2 - (R-a)^2}$
$F = Mg \sqrt{\frac{R^2 - (R-a)^2}{R^2}}$
$F = Mg \sqrt{1 - \left(\frac{R-a}{R}\right)^2}$

(B) The internal energy $U$ of a gas mixture is given by the sum of the internal energies of its components: $U = U_1 + U_2$.
For an ideal gas,the internal energy is $U = \frac{f}{2} nRT$,where $f$ is the degrees of freedom.
Oxygen $(O_2)$ is a diatomic gas. Since the bond is rigid,its degrees of freedom $f_1 = 5$ ($3$ translational + $2$ rotational).
Argon $(Ar)$ is a monatomic gas,so its degrees of freedom $f_2 = 3$.
Given $n_1 = 3$ moles of oxygen and $n_2 = 5$ moles of argon.
Total internal energy $U = \frac{f_1}{2} n_1 RT + \frac{f_2}{2} n_2 RT$.
$U = \left( \frac{5}{2} \times 3 \times RT \right) + \left( \frac{3}{2} \times 5 \times RT \right)$.
$U = \frac{15}{2} RT + \frac{15}{2} RT = \frac{30}{2} RT = 15 RT$.
Thus,the total internal energy in units of $RT$ is $15$.

(A) Let the dimension of Young's modulus $Y$ be expressed as $Y = F^{x} A^{y} V^{z}$.
The dimensional formula for Young's modulus is $[M^{1} L^{-1} T^{-2}]$.
The dimensional formula for Force $F$ is $[M^{1} L^{1} T^{-2}]$.
The dimensional formula for Area $A$ is $[L^{2}]$.
The dimensional formula for Speed $V$ is $[L^{1} T^{-1}]$.
Substituting these into the equation:
$[M^{1} L^{-1} T^{-2}] = [M^{1} L^{1} T^{-2}]^{x} [L^{2}]^{y} [L^{1} T^{-1}]^{z}$
$[M^{1} L^{-1} T^{-2}] = [M]^{x} [L]^{x + 2y + z} [T]^{-2x - z}$
Comparing the powers of $M$,$L$,and $T$ on both sides:
For $M$: $x = 1$
For $T$: $-2x - z = -2 \Rightarrow -2(1) - z = -2 \Rightarrow z = 0$
For $L$: $x + 2y + z = -1 \Rightarrow 1 + 2y + 0 = -1 \Rightarrow 2y = -2 \Rightarrow y = -1$
Thus,the dimension of Young's modulus is $F^{1} A^{-1} V^{0}$.

(B) Let the direction of motion of train $A$ be positive $(+)$ and the direction of motion of train $B$ be negative $(-)$.
Velocity of train $A$ $(V_A)$ = $+36 \, km/h$.
Velocity of train $B$ $(V_B)$ = $-72 \, km/h$.
Velocity of the person with respect to train $A$ $(V_{m/A})$ = $-1.8 \, km/h$ (since the person is walking opposite to the motion of train $A$).
Velocity of the person with respect to the ground $(V_m)$ = $V_{m/A} + V_A = -1.8 + 36 = 34.2 \, km/h$.
Velocity of the person with respect to train $B$ $(V_{m/B})$ = $V_m - V_B = 34.2 - (-72) = 34.2 + 72 = 106.2 \, km/h$.
To convert $km/h$ to $m/s$,multiply by $\frac{5}{18}$:
$V_{m/B} = 106.2 \times \frac{5}{18} = 5.9 \times 5 = 29.5 \, m/s$.

(C) The fundamental frequency $f$ of a stretched string is given by the formula: $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ is the tension,and $\mu$ is the linear mass density.
Since the strings $X$ and $Z$ are identical and made of the same material,their length $L$ and linear mass density $\mu$ are the same.
Therefore,the frequency is directly proportional to the square root of the tension: $f \propto \sqrt{T}$.
Given $f_{x} = 450\, Hz$ and $f_{z} = 300\, Hz$,we have:
$\frac{f_{x}}{f_{z}} = \sqrt{\frac{T_{x}}{T_{z}}}$
Substituting the values:
$\frac{450}{300} = \sqrt{\frac{T_{x}}{T_{z}}}$
$1.5 = \sqrt{\frac{T_{x}}{T_{z}}}$
Squaring both sides:
$\frac{T_{x}}{T_{z}} = (1.5)^2 = 2.25$.

(A) For a diatomic gas with rigid molecules,the degrees of freedom $f = 5$ and the adiabatic index $\gamma = 1 + 2/f = 7/5 = 1.4$.
Initial temperature $T_1 = 20 + 273 = 293\,K$.
For an adiabatic process,$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Given $V_2 = V_1 / 10$,we have $T_2 = T_1 (V_1 / V_2)^{\gamma-1} = 293 \times (10)^{1.4-1} = 293 \times 10^{0.4}$.
Using $10^{0.4} \approx 2.5118$,$T_2 = 293 \times 2.5118 \approx 735.96\,K$.
The change in internal energy is $\Delta U = n C_v \Delta T = n (fR/2) (T_2 - T_1)$.
Using $n = 5$,$f = 5$,and $R = 8.314\,J/(mol\cdot K)$:
$\Delta U = 5 \times (5 \times 8.314 / 2) \times (735.96 - 293) = 12.5 \times 8.314 \times 442.96 \approx 46056\,J = 46.056\,kJ$.
Rounding to the nearest integer,$X \approx 46$.

(C) According to the work-energy theorem,the total work done by all forces on the block is equal to the change in its kinetic energy.
Since the block starts from rest at $B$ and comes to rest at $A$,the change in kinetic energy is $\Delta K = 0 - 0 = 0$.
The forces acting on the block are gravity and friction.
Work done by gravity $(W_g)$ = $mg \sin \theta \times (BC + AC)$.
Work done by friction $(W_f)$ = $-\mu mg \cos \theta \times AC$ (since friction only acts on the rough section $CA$).
Setting the total work to zero: $W_g + W_f = 0$.
$mg \sin \theta (BC + AC) - \mu mg \cos \theta (AC) = 0$.
Given $BC = 2AC$,substitute this into the equation:
$mg \sin \theta (2AC + AC) = \mu mg \cos \theta (AC)$.
$3mg \sin \theta (AC) = \mu mg \cos \theta (AC)$.
$3 \sin \theta = \mu \cos \theta$.
$\mu = 3 \tan \theta$.
Comparing this with $\mu = k \tan \theta$,we get $k = 3$.

(B) The gravitational field $E$ at a distance $r$ inside the planet is given by Gauss's Law for gravity: $E(4\pi r^{2}) = 4\pi G M(r)$,where $M(r)$ is the mass enclosed within radius $r$.
$M(r) = \int_{0}^{r} \rho(r) 4\pi r^{2} dr = 4\pi \rho_{0} \int_{0}^{r} \left(1 - \frac{r^{2}}{R^{2}}\right) r^{2} dr$.
$M(r) = 4\pi \rho_{0} \left[ \frac{r^{3}}{3} - \frac{r^{5}}{5R^{2}} \right]$.
Thus,$E = \frac{G M(r)}{r^{2}} = 4\pi G \rho_{0} \left( \frac{r}{3} - \frac{r^{3}}{5R^{2}} \right)$.
To find the maximum field,set $\frac{dE}{dr} = 0$:
$\frac{dE}{dr} = 4\pi G \rho_{0} \left( \frac{1}{3} - \frac{3r^{2}}{5R^{2}} \right) = 0$.
$\frac{1}{3} = \frac{3r^{2}}{5R^{2}} \Rightarrow r^{2} = \frac{5R^{2}}{9} \Rightarrow r = \sqrt{\frac{5}{9}} R$.

(C) The power $P$ supplied to the particle is constant. Since power is the rate of change of kinetic energy,we have $\frac{dK}{dt} = P$. Integrating this with respect to time,we get $K = Pt$ (assuming initial kinetic energy is zero).
Since $K = \frac{1}{2}mv^2$,we have $\frac{1}{2}mv^2 = Pt$,which implies $v = \sqrt{\frac{2P}{m}} t^{1/2}$.
Since velocity $v = \frac{ds}{dt}$,we have $\frac{ds}{dt} = \sqrt{\frac{2P}{m}} t^{1/2}$.
Integrating with respect to time,$s = \int \sqrt{\frac{2P}{m}} t^{1/2} dt = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3} t^{3/2}$.
Thus,$s \propto t^{3/2}$.
The graph of $s$ versus $t$ for $s \propto t^{3/2}$ is a curve that is concave upwards (increasing slope),which corresponds to Graph $C$.

(C) Given:
Heat at constant pressure: $Q_P = n C_P \Delta T_1 = 160 \text{ cal}$,where $\Delta T_1 = 50^{\circ} C$.
Heat at constant volume: $Q_V = n C_V \Delta T_2 = 240 \text{ cal}$,where $\Delta T_2 = 100^{\circ} C$.
From the first equation: $n C_P = \frac{160}{50} = 3.2$.
From the second equation: $n C_V = \frac{240}{100} = 2.4$.
Taking the ratio: $\frac{C_P}{C_V} = \gamma = \frac{3.2}{2.4} = \frac{4}{3}$.
The relation between degrees of freedom $f$ and adiabatic index $\gamma$ is $\gamma = 1 + \frac{2}{f}$.
Substituting $\gamma = \frac{4}{3}$: $\frac{4}{3} = 1 + \frac{2}{f} \Rightarrow \frac{1}{3} = \frac{2}{f} \Rightarrow f = 6$.

(B) At the equilibrium position,the velocity of the block is maximum,given by $V_0 = \omega_0 A = \sqrt{\frac{k}{m}} A$.
When half the mass breaks off at the equilibrium position,the velocity of the remaining mass $m/2$ remains the same as $V_0$ because there is no impulsive force acting on the system in the horizontal direction.
Let the new amplitude be $A'$. The new angular frequency is $\omega' = \sqrt{\frac{k}{m/2}} = \sqrt{\frac{2k}{m}} = \sqrt{2} \omega_0$.
Since the velocity at the equilibrium position is $V_0 = \omega' A'$,we have:
$\omega_0 A = \omega' A'$
$\omega_0 A = (\sqrt{2} \omega_0) A'$
$A' = \frac{A}{\sqrt{2}}$.
Thus,$f = \frac{1}{\sqrt{2}}$.

(A) $1$. Conservation of linear momentum during the perfectly inelastic collision:
$m_b u = (m_b + m_B) v$
$0.1 \times 20 = (0.1 + 1.9) v$
$2 = 2v \Rightarrow v = 1\, m/s$
$2$. Conservation of mechanical energy for the combined system falling from height $h = 1\, m$:
$KE_f = PE_i + KE_i$
$KE_f = mgh + \frac{1}{2} m v^2$
$KE_f = (2)(10)(1) + \frac{1}{2}(2)(1)^2$
$KE_f = 20 + 1 = 21\, J$

(A) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{dT}{dt} = -k(T - T_s)$,where $T_s$ is the surrounding temperature.
For the first interval: $\frac{50 - 40}{300} = k \left( \frac{50 + 40}{2} - 20 \right)$.
$\frac{10}{300} = k(45 - 20) \implies \frac{1}{30} = 25k \implies k = \frac{1}{750}$.
For the next $5$ minutes $(300 \, s)$,let the final temperature be $T$. Then: $\frac{40 - T}{300} = k \left( \frac{40 + T}{2} - 20 \right)$.
Substituting $k = \frac{1}{750}$: $\frac{40 - T}{300} = \frac{1}{750} \left( \frac{40 + T - 40}{2} \right)$.
$\frac{40 - T}{300} = \frac{T}{1500}$.
$5(40 - T) = T \implies 200 - 5T = T \implies 6T = 200$.
$T = \frac{200}{6} \approx 33.33^{\circ}C$. The closest integer value is $33^{\circ}C$.

(D) The electric field at the centre $O$ due to a charge $q'$ at distance $d$ is $E = \frac{1}{4 \pi \varepsilon_{0}} \frac{q'}{d^2}$.
Let the angles of $A, B, C$ with the positive $x$-axis be $\theta_A = 30^{\circ}$,$\theta_B = 150^{\circ}$,and $\theta_C = -30^{\circ}$.
The electric field components along the $x$-axis are:
$E_{Ax} = \frac{1}{4 \pi \varepsilon_{0}} \frac{|-4q|}{d^2} \cos(180^{\circ} + 30^{\circ}) = \frac{4q}{4 \pi \varepsilon_{0} d^2} (-\cos 30^{\circ}) = -\frac{4q \sqrt{3}}{8 \pi \varepsilon_{0} d^2} = -\frac{\sqrt{3} q}{2 \pi \varepsilon_{0} d^2}$.
$E_{Bx} = \frac{1}{4 \pi \varepsilon_{0}} \frac{2q}{d^2} \cos(150^{\circ}) = \frac{2q}{4 \pi \varepsilon_{0} d^2} (-\frac{\sqrt{3}}{2}) = -\frac{\sqrt{3} q}{4 \pi \varepsilon_{0} d^2}$.
$E_{Cx} = \frac{1}{4 \pi \varepsilon_{0}} \frac{|-2q|}{d^2} \cos(180^{\circ} - 30^{\circ}) = \frac{2q}{4 \pi \varepsilon_{0} d^2} (-\cos 30^{\circ}) = -\frac{\sqrt{3} q}{4 \pi \varepsilon_{0} d^2}$.
The total electric field along the $x$-direction is $E_x = E_{Ax} + E_{Bx} + E_{Cx} = -\frac{\sqrt{3} q}{2 \pi \varepsilon_{0} d^2} - \frac{\sqrt{3} q}{4 \pi \varepsilon_{0} d^2} - \frac{\sqrt{3} q}{4 \pi \varepsilon_{0} d^2} = -\frac{\sqrt{3} q}{\pi \varepsilon_{0} d^2}$.
The magnitude is $\frac{\sqrt{3} q}{\pi \varepsilon_{0} d^2}$.

(C) The area of the loop can be calculated by subtracting the area of the two triangular cutouts from the area of the rectangle.
The total area $A = (16 \; \text{cm} \times 4 \; \text{cm}) - 2 \times (\frac{1}{2} \times 2 \; \text{cm} \times 4 \; \text{cm}) = 64 \; \text{cm}^2 - 8 \; \text{cm}^2 = 56 \; \text{cm}^2 = 56 \times 10^{-4} \; \text{m}^2$.
The rate of change of the magnetic field is $\frac{dB}{dt} = \frac{B_f - B_i}{\Delta t} = \frac{500 - 1000}{5} \; \text{Gauss/s} = -100 \; \text{Gauss/s} = -100 \times 10^{-4} \; \text{T/s}$.
The magnitude of the induced $EMF$ is given by $\varepsilon = |\frac{d\Phi}{dt}| = |A \frac{dB}{dt}|$.
$\varepsilon = (56 \times 10^{-4} \; \text{m}^2) \times (100 \times 10^{-4} \; \text{T/s}) = 5600 \times 10^{-8} \; \text{V} = 56 \times 10^{-6} \; \text{V} = 56 \; \mu \text{V}$.

(C) For parallel combination,the effective capacitance is $C_{p} = C_{1} + C_{2} = 10\; \mu F$.
Energy stored in a capacitor is given by $U = \frac{1}{2}CV^{2}$.
Given that when connected to the same voltage $V$,$U_{2} = 4U_{1}$.
Substituting the formula,$\frac{1}{2}C_{2}V^{2} = 4 \times \frac{1}{2}C_{1}V^{2}$,which simplifies to $C_{2} = 4C_{1}$.
Substituting $C_{2} = 4C_{1}$ into the parallel equation: $C_{1} + 4C_{1} = 10\; \mu F \implies 5C_{1} = 10\; \mu F \implies C_{1} = 2\; \mu F$.
Then,$C_{2} = 4 \times 2 = 8\; \mu F$.
For series combination,the effective capacitance $C_{s}$ is given by $\frac{1}{C_{s}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = \frac{C_{1} + C_{2}}{C_{1}C_{2}}$.
$C_{s} = \frac{C_{1}C_{2}}{C_{1} + C_{2}} = \frac{2 \times 8}{2 + 8} = \frac{16}{10} = 1.6\; \mu F$.

(B) Let the resistance to be connected in parallel with the $10\, \Omega$ resistor be $R_p$.
For a balanced Wheatstone bridge, the ratio of resistances in adjacent arms must be equal: $\frac{P}{Q} = \frac{R}{S}$.
Here, the resistances are $15\, \Omega, 12\, \Omega, 4\, \Omega$ and the effective resistance of $10\, \Omega$ in parallel with $R_p$, which is $R_{eff} = \frac{10 R_p}{10 + R_p}$.
Setting the bridge balance condition: $15 \times 4 = 12 \times R_{eff}$.
$60 = 12 \times \frac{10 R_p}{10 + R_p}$.
$5 = \frac{10 R_p}{10 + R_p}$.
$5(10 + R_p) = 10 R_p$.
$50 + 5 R_p = 10 R_p$.
$5 R_p = 50$.
$R_p = 10\, \Omega$.

(C) The Lens-Maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a plano-convex lens,the radius of curvature of the curved surface is $R_1 = 30 \; cm$ and the flat surface has an infinite radius of curvature,so $R_2 = \infty$.
The refractive index of the lens material is $\mu = 1.5$.
Substituting these values into the formula:
$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{30} - \frac{1}{\infty} \right)$
$\frac{1}{f} = (0.5) \left( \frac{1}{30} - 0 \right)$
$\frac{1}{f} = \frac{0.5}{30} = \frac{1}{60}$
Therefore,the focal length $f = 60 \; cm$.

(B) The magnetic field inside a long solenoid is $B = \mu_{0} nI$. When an electron is projected radially with speed $v$,it experiences a Lorentz force $F = evB$ perpendicular to its velocity,causing it to move in a circular path of radius $r = \frac{mv}{eB}$.
For the electron not to hit the surface of the solenoid of radius $R$,the diameter of its circular path must be less than or equal to the radius of the solenoid. Thus,$2r \leq R$,which implies $r \leq \frac{R}{2}$.
Substituting $r = \frac{mv}{e(\mu_{0} nI)}$,we get $\frac{mv}{e\mu_{0} nI} \leq \frac{R}{2}$.
Therefore,the maximum speed $v_{\max}$ is $\frac{e \mu_{0} nIR}{2m}$.

(A) The direction of propagation is given by $\hat{n} = \frac{\hat{i}+\hat{j}}{\sqrt{2}}$.
The electric field polarization is given by $\hat{E} = \hat{k}$.
In an electromagnetic wave,the direction of propagation $\hat{n}$ is given by the cross product of the electric field direction $\hat{E}$ and the magnetic field direction $\hat{B}$,i.e.,$\hat{n} = \hat{E} \times \hat{B}$.
Substituting the known values: $\frac{\hat{i}+\hat{j}}{\sqrt{2}} = \hat{k} \times \hat{B}$.
We know that $\hat{k} \times \hat{i} = \hat{j}$ and $\hat{k} \times \hat{j} = -\hat{i}$.
Therefore,$\hat{k} \times \left( \frac{\hat{i}-\hat{j}}{\sqrt{2}} \right) = \frac{\hat{j} - (-\hat{i})}{\sqrt{2}} = \frac{\hat{i}+\hat{j}}{\sqrt{2}}$.
Thus,the direction of the magnetic field is $\hat{B} = \frac{\hat{i}-\hat{j}}{\sqrt{2}}$.
Hence,the magnetic field vector is $\vec{B} = B_{0} \frac{\hat{i}-\hat{j}}{\sqrt{2}} \cos \left(\omega t - \vec{k} \cdot \vec{r}\right)$.

(C) In the given circuit,the two diodes are connected in such a way that the top diode is forward-biased and the bottom diode is reverse-biased.
Therefore,the bottom branch containing the reverse-biased diode acts as an open circuit.
The circuit simplifies to a series combination of the $5 \ \Omega$ resistor (external),the $5 \ \Omega$ resistor (top middle),the $10 \ \Omega$ resistor (top right),and the $10 \ \Omega$ resistor (top left).
The total resistance $R_{eq} = 5 \ \Omega + 5 \ \Omega + 10 \ \Omega + 10 \ \Omega = 30 \ \Omega$.
The current $i$ is given by $i = \frac{V}{R_{eq}} = \frac{9 \ V}{30 \ \Omega} = 0.3 \ A$.

(B) The magnetic torque on the loop is given by $\vec{\tau} = \vec{M} \times \vec{B}$.
For a small angular displacement $\theta$,the magnitude of the torque is $\tau = MB \sin \theta \approx MB \theta$.
The restoring torque is $\tau = -MB \theta$.
Using Newton's second law for rotation,$\tau = I_{moment} \alpha$,where $I_{moment}$ is the moment of inertia of the circular loop about its diameter,$I_{moment} = \frac{ma^2}{2}$.
Thus,$\frac{ma^2}{2} \alpha = - (I \pi a^2) B \theta$.
$\alpha = - \frac{2 I \pi B}{m} \theta$.
Comparing this with the $SHM$ equation $\alpha = - \omega^2 \theta$,we get $\omega^2 = \frac{2 I \pi B}{m}$.
Therefore,$\omega = \sqrt{\frac{2 I \pi B}{m}}$.
The time period $T$ is given by $T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{m}{2 I \pi B}} = \sqrt{\frac{4 \pi^2 m}{2 I \pi B}} = \sqrt{\frac{2 \pi m}{IB}}$.

(A) Given: $L = 40 \times 10^{-3} \; H$,$C = 100 \times 10^{-6} \; F$,$V(t) = 10 \sin (314 t)$.
Here,$\omega = 314 \; rad/s$.
Inductive reactance $X_{L} = \omega L = 314 \times 40 \times 10^{-3} = 12.56 \; \Omega$.
Capacitive reactance $X_{C} = \frac{1}{\omega C} = \frac{1}{314 \times 100 \times 10^{-6}} = \frac{10^{4}}{314} \approx 31.85 \; \Omega$.
Since $X_{C} > X_{L}$,the circuit is capacitive.
The net reactance $X = X_{C} - X_{L} = 31.85 - 12.56 = 19.29 \; \Omega$.
The peak current $I_{m} = \frac{V_{m}}{X} = \frac{10}{19.29} \approx 0.52 \; A$.
In a capacitive circuit,the current leads the voltage by $\frac{\pi}{2}$.
Thus,$I(t) = I_{m} \sin (\omega t + \frac{\pi}{2}) = 0.52 \sin (314 t + \frac{\pi}{2}) = 0.52 \cos (314 t)$.

(A) The light source emits light in all directions,covering a total solid angle of $4\pi$ steradians.
Light emerges from the water surface only if the angle of incidence is less than or equal to the critical angle $\theta_c$.
Using Snell's law at the critical angle: $\mu \sin\theta_c = 1 \sin 90^{\circ}$,where $\mu = 4/3$.
$\sin\theta_c = 1 / (4/3) = 3/4$.
Then,$\cos\theta_c = \sqrt{1 - \sin^2\theta_c} = \sqrt{1 - 9/16} = \sqrt{7}/4$.
The solid angle $\Omega$ subtended by the cone of light that emerges is $\Omega = 2\pi(1 - \cos\theta_c)$.
Substituting the value of $\cos\theta_c$: $\Omega = 2\pi(1 - \sqrt{7}/4) = 2\pi(1 - 2.646/4) = 2\pi(1 - 0.6615) = 2\pi(0.3385) = 0.677\pi$.
The fraction of light emerging is $\frac{\Omega}{4\pi} = \frac{0.677\pi}{4\pi} \approx 0.169$.
Converting to percentage: $0.169 \times 100 \approx 17\%$.

(D) The ionization energy of a hydrogen-like ion in the ground state is given by $E = 13.6 Z^2 \text{ eV}$.
Given that the ionization energy is $9$ Rydbergs,and $1 \text{ Rydberg} = 13.6 \text{ eV}$,we have $13.6 Z^2 = 9 \times 13.6$.
Thus,$Z^2 = 9$,which implies $Z = 3$.
The energy of the emitted photon when the electron jumps from the second excited state $(n_2 = 3)$ to the ground state $(n_1 = 1)$ is given by the Rydberg formula:
$\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Substituting the values: $\frac{1}{\lambda} = R \times 3^2 \times \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R \times 9 \times \left( 1 - \frac{1}{9} \right) = R \times 9 \times \frac{8}{9} = 8R$.
Since $R \approx 1.097 \times 10^7 \text{ m}^{-1}$,we have $\frac{1}{\lambda} = 8 \times 1.097 \times 10^7 \text{ m}^{-1} = 8.776 \times 10^7 \text{ m}^{-1}$.
$\lambda = \frac{1}{8.776 \times 10^7} \text{ m} \approx 1.14 \times 10^{-8} \text{ m} = 11.4 \text{ nm}$.

(A) In circuit $A$,the diode is reverse-biased. Therefore,no current flows through the circuit. The capacitor $A$ remains fully charged. Thus,$Q_{A} = CV$.
In circuit $B$,the diode is forward-biased. The capacitor discharges through the resistor $R$. The charge on the capacitor at any time $t$ is given by $q(t) = Q_{0} e^{-\frac{t}{RC}}$,where $Q_{0} = CV$.
At time $t = CR$,the charge $Q_{B}$ is:
$Q_{B} = CV e^{-\frac{CR}{RC}} = CV e^{-1} = \frac{CV}{e}$.
Therefore,$Q_{A} = CV$ and $Q_{B} = \frac{CV}{e}$.

(A) The acceleration of the electron is $a = \frac{|e|E}{m}$.
Since the electron starts from rest $(u = 0)$,its velocity at time $t$ is $v = at = \frac{|e|E}{m}t$.
The de-Broglie wavelength is given by $\lambda = \frac{h}{mv} = \frac{h}{m(\frac{|e|E}{m}t)} = \frac{h}{|e|Et}$.
To find the rate of change of wavelength,we differentiate $\lambda$ with respect to $t$:
$\frac{d\lambda}{dt} = \frac{d}{dt} (\frac{h}{|e|Et}) = \frac{h}{|e|E} \cdot \frac{d}{dt}(t^{-1}) = \frac{h}{|e|E} (-t^{-2}) = -\frac{h}{|e|Et^2}$.

(D) For a meter bridge,the balancing condition is given by $\frac{R}{S} = \frac{l}{100 - l}$.
Given $l = 25 \; cm$,we have $\frac{R}{S} = \frac{25}{100 - 25} = \frac{25}{75} = \frac{1}{3}$.
Thus,$S = 3R$.
The resistance of a wire is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi (d/2)^2} = \frac{4 \rho l}{\pi d^2}$.
When the length is halved $(l^{\prime} = l/2)$ and the diameter is halved $(d^{\prime} = d/2)$,the new resistance $R^{\prime}$ becomes:
$R^{\prime} = \frac{4 \rho (l/2)}{\pi (d/2)^2} = \frac{4 \rho l / 2}{\pi d^2 / 4} = 2 \left( \frac{4 \rho l}{\pi d^2} \right) = 2R$.
Now,for the new balancing length $l^{\prime}$,the condition is $\frac{R^{\prime}}{S} = \frac{l^{\prime}}{100 - l^{\prime}}$.
Substituting $R^{\prime} = 2R$ and $S = 3R$:
$\frac{2R}{3R} = \frac{l^{\prime}}{100 - l^{\prime}}$
$\frac{2}{3} = \frac{l^{\prime}}{100 - l^{\prime}}$
$200 - 2l^{\prime} = 3l^{\prime}$
$5l^{\prime} = 200$
$l^{\prime} = 40 \; cm$.

(D) The total resistance in the circuit is $R = 200 \; \Omega + 200 \; \Omega = 400 \; \Omega$.
The voltage drop across the resistors is $V_R = V_{in} - V_{out} = 12 \; V - 8 \; V = 4 \; V$.
The current flowing through the circuit is $I = \frac{V_R}{R} = \frac{4 \; V}{400 \; \Omega} = 0.01 \; A = 10 \; mA$.
Since the two identical Zener diodes are connected in series, the voltage across each diode is $V_d = \frac{8 \; V}{2} = 4 \; V$.
The power dissipated in each diode is $P = V_d \times I = 4 \; V \times 10 \; mA = 40 \; mW$.

(A) The width of the screen portion is constant for both cases.
Let the width of the screen portion be $L$.
The fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$.
The number of fringes $n$ observed in a length $L$ is given by $L = n \times \beta = n \times \frac{\lambda D}{d}$.
For the first case: $L = 15 \times \frac{500 \; nm \times D}{d}$.
For the second case: $L = 10 \times \frac{\lambda \times D}{d}$.
Equating the two expressions for $L$:
$15 \times 500 = 10 \times \lambda$.
$\lambda = \frac{15 \times 500}{10} = 15 \times 50 = 750 \; nm$.

(D) The electric field is given by $\overrightarrow{E} = 4x \hat{i} - (y^2 + 1) \hat{j}$.
Surface $ABCD$ lies in the $xy$-plane at $z = 2$. The area vector for this surface is $\overrightarrow{S}_I = S \hat{k}$. Since $\overrightarrow{E} \cdot \hat{k} = 0$,the flux $\phi_I = 0$.
Surface $BCGF$ lies in the $yz$-plane at $x = 3$. The area vector for this surface is $\overrightarrow{S}_{II} = 4 \hat{i}$ (area = $2 \times 2 = 4$).
At $x = 3$,the electric field is $\overrightarrow{E} = 4(3) \hat{i} - (y^2 + 1) \hat{j} = 12 \hat{i} - (y^2 + 1) \hat{j}$.
The flux $\phi_{II} = \int \overrightarrow{E} \cdot d\overrightarrow{S} = \int_{0}^{2} \int_{0}^{2} (12 \hat{i} - (y^2 + 1) \hat{j}) \cdot (dy dz \hat{i}) = \int_{0}^{2} \int_{0}^{2} 12 \, dy dz = 12 \times 4 = 48 \text{ Nm}^2/C$.
Thus,$\phi_I - \phi_{II} = 0 - 48 = -48 \text{ Nm}^2/C$.

(C) The electric fields are $\overrightarrow{E}_{1}=E_{0} \hat{j} \cos (\omega t-kx)$ and $\overrightarrow{E}_{2}=E_{0} \hat{k} \cos (\omega t-ky)$.
At $t=0$ and origin $(0,0,0)$,$\overrightarrow{E}_{1} = E_{0} \hat{j}$ and $\overrightarrow{E}_{2} = E_{0} \hat{k}$.
The corresponding magnetic fields are given by $\overrightarrow{B} = \frac{1}{c} (\hat{n} \times \overrightarrow{E})$,where $\hat{n}$ is the direction of propagation.
For $\overrightarrow{E}_{1}$,$\hat{n} = \hat{i}$,so $\overrightarrow{B}_{1} = \frac{1}{c} (\hat{i} \times E_{0} \hat{j}) = \frac{E_{0}}{c} \hat{k}$.
For $\overrightarrow{E}_{2}$,$\hat{n} = \hat{j}$,so $\overrightarrow{B}_{2} = \frac{1}{c} (\hat{j} \times E_{0} \hat{k}) = \frac{E_{0}}{c} \hat{i}$.
The Lorentz force is $\overrightarrow{F} = q(\overrightarrow{E}_{1} + \overrightarrow{E}_{2}) + q(\overrightarrow{v} \times (\overrightarrow{B}_{1} + \overrightarrow{B}_{2}))$.
Substituting values: $\overrightarrow{F} = q(E_{0} \hat{j} + E_{0} \hat{k}) + q(0.8 c \hat{j} \times (\frac{E_{0}}{c} \hat{k} + \frac{E_{0}}{c} \hat{i}))$.
$\overrightarrow{F} = q E_{0} \hat{j} + q E_{0} \hat{k} + 0.8 q E_{0} (\hat{j} \times \hat{k}) + 0.8 q E_{0} (\hat{j} \times \hat{i})$.
Using cross products $\hat{j} \times \hat{k} = \hat{i}$ and $\hat{j} \times \hat{i} = -\hat{k}$:
$\overrightarrow{F} = q E_{0} \hat{j} + q E_{0} \hat{k} + 0.8 q E_{0} \hat{i} - 0.8 q E_{0} \hat{k} = q E_{0} (0.8 \hat{i} + \hat{j} + 0.2 \hat{k})$.

(D) To find the electric field due to the remaining portion,we use the principle of superposition by considering the sphere as a combination of a full sphere of charge density $+\rho$ and a smaller sphere of charge density $-\rho$.
At point $A$ (which is at the center of the carved-out sphere and at a distance $R/2$ from the center of the large sphere):
$|\overrightarrow{E}_{A}| = |\overrightarrow{E}_{large} + \overrightarrow{E}_{small}| = |\frac{\rho (R/2)}{3\epsilon_0} + 0| = \frac{\rho R}{6\epsilon_0}$.
At point $B$ (which is at the bottom of the large sphere,distance $R$ from its center and $3R/2$ from the center of the carved-out sphere):
$|\overrightarrow{E}_{B}| = |\frac{\rho R}{3\epsilon_0} - \frac{\rho (R/2)^3}{3\epsilon_0 (3R/2)^2}| = \frac{\rho R}{3\epsilon_0} - \frac{\rho R^3/8}{3\epsilon_0 (9R^2/4)} = \frac{\rho R}{3\epsilon_0} (1 - \frac{1}{18}) = \frac{\rho R}{3\epsilon_0} (\frac{17}{18}) = \frac{17\rho R}{54\epsilon_0}$.
Taking the ratio:
$\frac{|\overrightarrow{E}_{A}|}{|\overrightarrow{E}_{B}|} = \frac{\rho R / 6\epsilon_0}{17\rho R / 54\epsilon_0} = \frac{1}{6} \times \frac{54}{17} = \frac{9}{17} = \frac{18}{34}$.

(A) Let the total current be $I$ and current density be $J = \frac{I}{\pi a^2}$.
For a point inside the wire at distance $r < a$,using Ampere's law: $\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{enclosed}$.
$B(2\pi r) = \mu_0 (J \cdot \pi r^2) \Rightarrow B = \frac{\mu_0 J r}{2}$.
At $r = \frac{a}{3}$,$B_A = \frac{\mu_0 J (a/3)}{2} = \frac{\mu_0 J a}{6}$.
For a point outside the wire at distance $r > a$,the wire acts as a long straight wire carrying current $I$ at its axis.
$B(2\pi r) = \mu_0 I = \mu_0 (J \pi a^2) \Rightarrow B = \frac{\mu_0 J a^2}{2r}$.
At $r = 2a$,$B_B = \frac{\mu_0 J a^2}{2(2a)} = \frac{\mu_0 J a}{4}$.
The ratio is $\frac{B_A}{B_B} = \frac{\mu_0 J a / 6}{\mu_0 J a / 4} = \frac{4}{6} = \frac{2}{3}$.

(C) The de Broglie wavelength $\lambda$ of a particle with kinetic energy $E$ is given by $\lambda = \frac{h}{\sqrt{2mE}}$.
When the energy is increased by $\Delta E$,the new kinetic energy becomes $E' = E + \Delta E$,and the new wavelength is $\lambda' = \frac{\lambda}{2}$.
Using the formula for the new wavelength: $\lambda' = \frac{h}{\sqrt{2m(E + \Delta E)}}$.
Since $\lambda' = \frac{\lambda}{2}$,we have $\frac{h}{\sqrt{2m(E + \Delta E)}} = \frac{1}{2} \cdot \frac{h}{\sqrt{2mE}}$.
Squaring both sides: $\frac{1}{2m(E + \Delta E)} = \frac{1}{4} \cdot \frac{1}{2mE}$.
Simplifying the equation: $\frac{1}{E + \Delta E} = \frac{1}{4E}$.
Therefore,$4E = E + \Delta E$,which gives $\Delta E = 3E$.

(B) The apparent depth $(d_{app})$ of an object seen through multiple layers of liquids is given by the sum of the apparent depths of each layer: $d_{app} = \sum \frac{h_i}{\mu_i}$.
Here, the total depth is $2h$. The lower half has depth $h$ and refractive index $\mu_1 = 2\sqrt{2}$. The upper half has depth $h$ and refractive index $\mu_2 = \sqrt{2}$.
The apparent depth of the bottom surface as seen from the top is:
$d_{app} = \frac{h_1}{\mu_1} + \frac{h_2}{\mu_2}$
$d_{app} = \frac{h}{2\sqrt{2}} + \frac{h}{\sqrt{2}}$
$d_{app} = \frac{h + 2h}{2\sqrt{2}} = \frac{3h}{2\sqrt{2}}$
Rationalizing the denominator:
$d_{app} = \frac{3h \times \sqrt{2}}{2\sqrt{2} \times \sqrt{2}} = \frac{3\sqrt{2}h}{4} = \frac{3}{4}h\sqrt{2}$.

(A) Let the work function be $\phi$.
According to Einstein's photoelectric equation,the maximum kinetic energy $KE_{\max}$ is given by:
$KE_{\max} = \frac{hc}{\lambda} - \phi$.
When an electron with charge $q$ and mass $m$ enters a magnetic field $B$ perpendicular to its velocity,it follows a circular path of radius $R = \frac{\sqrt{2m KE_{\max}}}{qB}$.
Squaring both sides,we get $R^2 = \frac{2m KE_{\max}}{q^2 B^2}$,which implies $KE_{\max} = \frac{R^2 q^2 B^2}{2m}$.
Substituting the values: $h = 6.63 \times 10^{-34} \; J \cdot s$,$c = 3 \times 10^8 \; m/s$,$\lambda = 6561 \times 10^{-10} \; m$,$R = 10 \times 10^{-3} \; m$,$q = 1.6 \times 10^{-19} \; C$,$m = 9.1 \times 10^{-31} \; kg$,$B = 3 \times 10^{-4} \; T$.
Energy of incident photon $E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{6561 \times 10^{-10}} \approx 3.03 \times 10^{-19} \; J = 1.89 \; eV$.
Kinetic energy $KE_{\max} = \frac{(10^{-2})^2 \times (1.6 \times 10^{-19})^2 \times (3 \times 10^{-4})^2}{2 \times 9.1 \times 10^{-31}} \approx 1.26 \times 10^{-20} \; J = 0.079 \; eV$.
Thus,$\phi = E - KE_{\max} = 1.89 \; eV - 0.079 \; eV \approx 1.81 \; eV$.
Wait,re-evaluating the calculation: $E = 1.89 \; eV$. $KE_{\max} = 0.079 \; eV$. $\phi = 1.81 \; eV$.
Given the options,$1.8 \; eV$ is the closest value.

(C) The angular resolution of a telescope is given by the formula $\Delta \theta = \frac{1.22 \lambda}{D}$, where $\lambda$ is the wavelength of light and $D$ is the diameter of the aperture.
Given: $\lambda = 5500 \; \mathring{A} = 5500 \times 10^{-10} \; m$, $D = 5 \; m$, and distance $d = 4 \times 10^{5} \; km = 4 \times 10^{8} \; m$.
The linear separation $x$ between two objects on the moon that can be just resolved is given by $x = d \cdot \Delta \theta$.
Substituting the values:
$x = d \times \frac{1.22 \lambda}{D} = \frac{4 \times 10^{8} \times 1.22 \times 5500 \times 10^{-10}}{5}$.
$x = \frac{4 \times 1.22 \times 5.5 \times 10^{-2}}{5} \times 10^{8} = 0.8 \times 1.22 \times 5.5 \times 10^{-2} \times 10^{8} = 53.68 \; m$.
Rounding this value to the nearest provided option, we get $60 \; m$.

(B) Let the potential at point $A$ be $V_A = 20 \text{ V}$ and at point $C$ be $V_C = 0 \text{ V}$.
First,find the potentials at $B$ and $D$ using the voltage divider rule or nodal analysis.
The branch $AB$ and $BC$ are in series,and $AD$ and $DC$ are in series.
For the upper branch $ABC$: The total resistance is $1 \Omega + 2 \Omega = 3 \Omega$. The current is $I_1 = 20 \text{ V} / 3 \Omega = 6.67 \text{ A}$. The potential at $B$ is $V_B = V_A - I_1 \times 1 \Omega = 20 - 6.67 = 13.33 \text{ V}$.
For the lower branch $ADC$: The total resistance is $4 \Omega + 3 \Omega = 7 \Omega$. The current is $I_2 = 20 \text{ V} / 7 \Omega = 2.86 \text{ A}$. The potential at $D$ is $V_D = V_A - I_2 \times 4 \Omega = 20 - 11.44 = 8.56 \text{ V}$.
Since $V_B > V_D$,current flows from $B$ to $D$.
Alternatively,using Kirchhoff's Current Law $(KCL)$ at nodes $B$ and $D$ or nodal analysis,we find the current in the wire $BD$ is $2 \text{ A}$ as shown in the provided solution image.

(B) Using the work-energy theorem: $W_{net} = \Delta K$
$qE(2a) = \frac{1}{2}m(2v)^2 - \frac{1}{2}mv^2 = \frac{3}{2}mv^2$
$E = \frac{3mv^2}{4qa}$. Thus,$(A)$ is correct.
Rate of work done by electric field $P_E = \vec{F}_E \cdot \vec{v} = (qE\hat{i}) \cdot (v\hat{i}) = qEv = q\left(\frac{3mv^2}{4qa}\right)v = \frac{3mv^3}{4a}$. Thus,$(B)$ is correct.
At $Q$,velocity is $-2v\hat{j}$. Electric force is $qE\hat{i}$. Since $\vec{F}_E \perp \vec{v}$,power by electric field is $0$. Magnetic force is always perpendicular to velocity,so power by magnetic field is $0$. Thus,$(C)$ is correct.
Angular momentum $L = \vec{r} \times \vec{p}$. At $P$,$\vec{r}_P = a\hat{j}$,$\vec{p}_P = mv\hat{i}$,so $L_P = (a\hat{j}) \times (mv\hat{i}) = -mav\hat{k}$. Magnitude $|L_P| = mav$.
At $Q$,$\vec{r}_Q = 2a\hat{i}$,$\vec{p}_Q = -2mv\hat{j}$,so $L_Q = (2a\hat{i}) \times (-2mv\hat{j}) = -4mav\hat{k}$. Magnitude $|L_Q| = 4mav$.
Difference $= 4mav - mav = 3mav$. Thus,$(D)$ is incorrect.

(C) Given the dipole moment $\overrightarrow{p} = (-\hat{i} - 3\hat{j} + 2\hat{k}) \times 10^{-29} \; C \cdot m$ and position vector $\overrightarrow{r} = \hat{i} + 3\hat{j} + 5\hat{k}$.
Since $\overrightarrow{r} \cdot \overrightarrow{p} = (1)(-1) + (3)(-3) + (5)(2) = -1 - 9 + 10 = 0$,the point lies on the equatorial plane of the dipole.
For a point on the equatorial plane,the electric field $\overrightarrow{E}$ is given by $\overrightarrow{E} = -\frac{1}{4\pi\epsilon_0} \frac{\overrightarrow{p}}{r^3}$.
This implies that the electric field $\overrightarrow{E}$ is antiparallel to the dipole moment $\overrightarrow{p}$.
Therefore,$\overrightarrow{E} \parallel -\overrightarrow{p}$.
Since $\overrightarrow{p} = (-\hat{i} - 3\hat{j} + 2\hat{k}) \times 10^{-29}$,then $-\overrightarrow{p} = (\hat{i} + 3\hat{j} - 2\hat{k}) \times 10^{-29}$.
Thus,the electric field is parallel to $(\hat{i} + 3\hat{j} - 2\hat{k})$.

(B) The induced electromotive force $(EMF)$ in an inductor is given by the formula: $V = L \left| \frac{di}{dt} \right|$.
Here,the change in current $\Delta i = 0.25 \; A - 0 \; A = 0.25 \; A$.
The time interval $\Delta t = 0.025 \; ms = 0.025 \times 10^{-3} \; s$.
The induced voltage $V = 100 \; V$.
Rearranging the formula to solve for self-inductance $L$: $L = \frac{V}{|\Delta i / \Delta t|} = \frac{V \cdot \Delta t}{\Delta i}$.
Substituting the values: $L = \frac{100 \times 0.025 \times 10^{-3}}{0.25}$.
$L = \frac{100 \times 0.025}{0.25} \times 10^{-3} = 100 \times 0.1 \times 10^{-3} = 10 \times 10^{-3} \; H$.
Since $1 \; H = 1000 \; mH$,we have $L = 10 \; mH$.

(C) In the given circuit,diode $D_{1}$ is connected to $12.7\; V$ and diode $D_{2}$ is connected to $4\; V$.
Since the potential at the anode of $D_{1}$ $(12.7\; V)$ is higher than the potential at the anode of $D_{2}$ $(4\; V)$,diode $D_{1}$ will be forward biased and diode $D_{2}$ will be reverse biased.
For a forward-biased diode with a built-in potential of $0.7\; V$,the output voltage at point $A$ is given by $V_{A} = V_{in} - V_{barrier}$.
Substituting the values,we get $V_{A} = 12.7\; V - 0.7\; V = 12\; V$.
Thus,the voltage at point $A$ is $12\; V$.

(C) The time period of a charged particle in a uniform magnetic field is given by $T = \frac{2 \pi m}{qB}$.
The proton completes $10$ revolutions,so the total time spent in the magnetic field is $t = 10T = 10 \times \frac{2 \pi m}{qB}$.
The velocity component parallel to the magnetic field is $v_{\parallel} = v \cos(60^{\circ}) = v \times \frac{1}{2} = \frac{v}{2}$.
Since the velocity component parallel to the magnetic field remains constant,the distance $l$ covered in the direction of the field is $l = v_{\parallel} \times t = \frac{v}{2} \times 10 \times \frac{2 \pi m}{qB} = \frac{10 \pi m v}{qB}$.
Substituting the given values:
$l = \frac{10 \times 3.14 \times 1.67 \times 10^{-27} \times 4 \times 10^{5}}{1.6 \times 10^{-19} \times 0.3}$
$l = \frac{20.942 \times 10^{-21}}{0.48 \times 10^{-19}} = \frac{20.942}{48} \approx 0.436 \, m$.
Rounding to the nearest option,$l \approx 0.44 \, m$.

(A) The magnetic moment $\vec{M}$ of a current loop is given by $\vec{M} = I\vec{A}$,where $\vec{A}$ is the area vector.
For the loop $A, B, C, D, E, F, A$,we can consider it as two loops $ABCD$ and $DEFA$ joined together.
The magnetic moment of loop $ABCD$ (in the $XY$-plane) is $\vec{M}_1 = I(ab)\hat{k} = abI\hat{k}$.
The magnetic moment of loop $DEFA$ (in the $XZ$-plane) is $\vec{M}_2 = I(ab)\hat{j} = abI\hat{j}$.
The total magnetic moment is $\vec{M} = \vec{M}_1 + \vec{M}_2 = abI(\hat{j} + \hat{k})$.
The magnitude is $|\vec{M}| = abI\sqrt{1^2 + 1^2} = \sqrt{2}abI$.
The direction is given by the unit vector $\frac{\vec{M}}{|\vec{M}|} = \frac{abI(\hat{j} + \hat{k})}{\sqrt{2}abI} = \left(\frac{\hat{j}}{\sqrt{2}} + \frac{\hat{k}}{\sqrt{2}}\right)$.
Thus,the magnetic moment is $\sqrt{2}abI$ along $\left(\frac{\hat{j}}{\sqrt{2}} + \frac{\hat{k}}{\sqrt{2}}\right)$.

(A) In a potentiometer,the balancing length $l$ is always measured from the high potential end $P$.
Given the total length of the wire $PQ = 100\,cm$.
The null point is obtained at $49\,cm$ from $Q$.
Therefore,the balancing length from $P$ is $l = 100\,cm - 49\,cm = 51\,cm$.
The emf of the cell $E_{2}$ is balanced by the potential drop across the length $l$ of the potentiometer wire.
The formula for the emf is $E_{2} = \phi \times l$,where $\phi$ is the potential gradient.
Substituting the given values: $1.02\,V = \phi \times 51\,cm$.
Solving for $\phi$: $\phi = \frac{1.02}{51}\,V/cm = 0.02\,V/cm$.
Thus,the potential gradient in the potentiometer wire is $0.02\,V/cm$.

(A) Let the length of the segment be $\ell$.
Let $N$ be the number of fringes in the segment $\ell$,and $w$ be the fringe width.
The relationship between the number of fringes and fringe width is given by $N w = \ell$.
Since the fringe width $w = \frac{\lambda D}{d}$,we can write $N \left( \frac{\lambda D}{d} \right) = \ell$.
For a fixed segment length $\ell$,the product $N \lambda$ remains constant because $D$ and $d$ are constant.
Therefore,$N_1 \lambda_1 = N_2 \lambda_2$.
Given $N_1 = 16$,$\lambda_1 = 700 \,nm$,and $\lambda_2 = 400 \,nm$.
Substituting the values: $16 \times 700 = N_2 \times 400$.
$N_2 = \frac{16 \times 700}{400} = \frac{16 \times 7}{4} = 4 \times 7 = 28$.
Thus,the number of fringes observed is $28$.

(B) In a hydrogen atom,the energy of the $n^{\text{th}}$ level is given by $E_n = -\frac{E_0}{n^2}$,where $E_0$ is the ionization energy of hydrogen.
For a transition from $(n+1)$ to $n$,the energy of the emitted radiation $\Delta E$ is the difference between the energy levels:
$\Delta E = E_{n+1} - E_n = E_0 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right)$
Using the relation $\Delta E = h\nu$,we have:
$h\nu = E_0 \left( \frac{(n+1)^2 - n^2}{n^2(n+1)^2} \right) = E_0 \left( \frac{2n+1}{n^2(n+1)^2} \right)$
For $n >> 1$,we can approximate $(n+1) \approx n$ and $(2n+1) \approx 2n$:
$h\nu \approx E_0 \left( \frac{2n}{n^2 \cdot n^2} \right) = E_0 \left( \frac{2n}{n^4} \right) = \frac{2E_0}{n^3}$
Thus,the frequency $\nu$ is proportional to $\frac{1}{n^3}$.

(C) Let the output of the $NAND$ gate be $P = \overline{AB}$.
Let the output of the $OR$ gate be $Q = A+B$.
The $AND$ gate takes $P$ and $Q$ as inputs,so its output is $R = P \cdot Q = (\overline{AB}) \cdot (A+B)$.
Using Boolean algebra: $R = (\bar{A} + \bar{B}) \cdot (A+B) = \bar{A}A + \bar{A}B + \bar{B}A + \bar{B}B = 0 + \bar{A}B + A\bar{B} + 0 = A \oplus B$ ($XOR$ operation).
The final $NOR$ gate takes $P$ and $R$ as inputs,so $Z = \overline{P+R} = \overline{(\overline{AB}) + (A \oplus B)}$.
Let's evaluate for each input $(A, B)$:
$1$. $(1, 0): P = \overline{1 \cdot 0} = 1, Q = 1+0 = 1, R = 1 \cdot 1 = 1. Z = \overline{1+1} = 0$.
$2$. $(0, 0): P = \overline{0 \cdot 0} = 1, Q = 0+0 = 0, R = 1 \cdot 0 = 0. Z = \overline{1+0} = 0$.
$3$. $(1, 1): P = \overline{1 \cdot 1} = 0, Q = 1+1 = 1, R = 0 \cdot 1 = 0. Z = \overline{0+0} = 1$.
$4$. $(0, 1): P = \overline{0 \cdot 1} = 1, Q = 0+1 = 1, R = 1 \cdot 1 = 1. Z = \overline{1+1} = 0$.
The outputs are $0, 0, 1, 0$.

(A) The impedance $Z$ of an $RL$ circuit is given by $Z = \sqrt{R^2 + X_L^2} = 100\, \Omega$.
Given that the voltage leads the current by $\phi = 45^{\circ}$,we have $\tan \phi = \frac{X_L}{R}$.
Since $\tan 45^{\circ} = 1$,it follows that $X_L = R$.
Substituting $X_L = R$ into the impedance equation: $\sqrt{X_L^2 + X_L^2} = 100 \Rightarrow \sqrt{2} X_L = 100$.
Thus,$X_L = \frac{100}{\sqrt{2}} = 50\sqrt{2}\, \Omega$.
Since $X_L = 2\pi f L$,we have $L = \frac{X_L}{2\pi f} = \frac{50\sqrt{2}}{2 \times \pi \times 1000}$.
$L = \frac{25\sqrt{2}}{1000\pi} \approx \frac{25 \times 1.414}{3141.59} \approx 0.01125\, H = 1.125 \times 10^{-2}\, H$.

(C) The particle is subjected to two constant forces: the gravitational force $mg$ acting downwards (along the $y$-axis) and the electric force $qE$ acting horizontally (along the $x$-axis).
Since the initial velocity is zero, the net force $F_{net} = \sqrt{(mg)^2 + (qE)^2}$ is constant in both magnitude and direction.
According to Newton's second law, the acceleration $a = F_{net}/m$ is also constant.
$A$ particle starting from rest and moving under the influence of a constant net force will travel in a straight line path along the direction of the net force.
Therefore, the trajectory is a straight line.

(B) Initially, the charge on the $10\,\mu F$ capacitor is:
$Q = C_1 V_1 = (10\,\mu F)(50\, V) = 500\,\mu C$
When this capacitor is connected in parallel to an uncharged capacitor of capacitance $C_2$, the total charge remains conserved.
Let the final potential difference be $V = 20\, V$.
The total charge $Q$ is distributed between the two capacitors:
$Q = (C_1 + C_2)V$
$500\,\mu C = (10\,\mu F + C_2)(20\, V)$
Dividing both sides by $20\, V$:
$25\,\mu F = 10\,\mu F + C_2$
$C_2 = 25\,\mu F - 10\,\mu F = 15\,\mu F$

(A) The direction of propagation of an electromagnetic wave is given by the direction of the Poynting vector,which is $\hat{E} \times \hat{B}$.
Given,the direction of the electric field is $\hat{E} = \hat{k}$.
The direction of the magnetic field is given by the vector $\vec{B} = 2\hat{i} - 2\hat{j}$.
The unit vector for the magnetic field is $\hat{B} = \frac{\vec{B}}{|B|} = \frac{2\hat{i} - 2\hat{j}}{\sqrt{2^2 + (-2)^2}} = \frac{2\hat{i} - 2\hat{j}}{\sqrt{8}} = \frac{2\hat{i} - 2\hat{j}}{2\sqrt{2}} = \frac{1}{\sqrt{2}}(\hat{i} - \hat{j})$.
The direction of propagation $\hat{C}$ is $\hat{E} \times \hat{B} = \hat{k} \times [\frac{1}{\sqrt{2}}(\hat{i} - \hat{j})]$.
Using the cross product rules $\hat{k} \times \hat{i} = \hat{j}$ and $\hat{k} \times \hat{j} = -\hat{i}$,we get:
$\hat{C} = \frac{1}{\sqrt{2}}(\hat{k} \times \hat{i} - \hat{k} \times \hat{j}) = \frac{1}{\sqrt{2}}(\hat{j} - (-\hat{i})) = \frac{1}{\sqrt{2}}(\hat{i} + \hat{j})$.

(D) Let the mass of the particle be $m$ and the mass of the electron be $m_e = 9.1 \times 10^{-31} \ kg$.
Let the speed of the electron be $v_e = V$.
Then,the speed of the particle is $v_p = 5V$.
The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
For the particle: $\lambda_p = \frac{h}{m(5V)}$.
For the electron: $\lambda_e = \frac{h}{m_e V}$.
The ratio of the wavelengths is $\frac{\lambda_p}{\lambda_e} = \frac{h}{5mV} \times \frac{m_e V}{h} = \frac{m_e}{5m}$.
Given $\frac{\lambda_p}{\lambda_e} = 1.878 \times 10^{-4}$,we have $\frac{m_e}{5m} = 1.878 \times 10^{-4}$.
Solving for $m$: $m = \frac{m_e}{5 \times 1.878 \times 10^{-4}} = \frac{9.1 \times 10^{-31}}{9.39 \times 10^{-4}} \approx 9.7 \times 10^{-28} \ kg$.

(A) Given that all resistances $R_1$ to $R_5$ are $2\, \Omega$. When the capacitor is fully charged,no current flows through the branch containing the capacitor.
Let the nodes be marked as shown in the solution figure. The branch $ADB$ contains the capacitor,so no current flows through it.
The circuit simplifies to a combination of resistors where $R_1$ and $R_2$ are in series,and this combination is in parallel with $R_3$. This whole block is in series with $R_4$.
However,looking at the circuit,the current $i$ from the $10\, V$ source splits. The effective resistance $R_{eq}$ is calculated as follows:
The branch $AEB$ is not carrying current through the capacitor. The current flows through $R_1$ and $R_2$ in series $(2+2=4\, \Omega)$,which is in parallel with $R_3$ $(2\, \Omega)$.
Equivalent resistance of this part $= (4 \times 2) / (4 + 2) = 8 / 6 = 4/3\, \Omega$.
Adding $R_4$ in series,$R_{eq} = 4/3 + 2 = 10/3\, \Omega$.
Total current $i = V / R_{eq} = 10 / (10/3) = 3\, A$.
The potential difference across the capacitor is the potential difference between points $A$ and $B$,which is $V_{AB} = V_{AE} + V_{EB}$.
$V_{AE} = I_{R2} \times R_2 = 1\, A \times 2\, \Omega = 2\, V$.
$V_{EB} = I_{R4} \times R_4 = 3\, A \times 2\, \Omega = 6\, V$.
Thus,$V_{AB} = 2\, V + 6\, V = 8\, V$.

(B) Let the point of incidence be $A$ and the point of reflection/refraction be $B$. The center of the sphere is $O$.
By Snell's law at point $A$:
$1 \times \sin 60^{\circ} = \sqrt{3} \times \sin \theta$
$\frac{\sqrt{3}}{2} = \sqrt{3} \sin \theta$
$\sin \theta = \frac{1}{2} \Rightarrow \theta = 30^{\circ}$.
In the triangle $\triangle OAB$,$OA = OB = R$ (radius of the sphere). Thus,$\triangle OAB$ is an isosceles triangle.
The angle of incidence at point $B$ is also $\theta = 30^{\circ}$.
The angle of reflection at $B$ is $r' = 30^{\circ}$ (angle with the normal $OB$).
By Snell's law at $B$ for refraction:
$\sqrt{3} \times \sin 30^{\circ} = 1 \times \sin r''$
$\sqrt{3} \times \frac{1}{2} = \sin r'' \Rightarrow \sin r'' = \frac{\sqrt{3}}{2} \Rightarrow r'' = 60^{\circ}$.
The angle between the normal $OB$ and the reflected ray is $30^{\circ}$.
The angle between the normal $OB$ and the refracted ray is $60^{\circ}$.
The angle between the reflected and refracted rays is the sum of these angles with the normal: $30^{\circ} + 60^{\circ} = 90^{\circ}$.

(A) From the figure,the radius of curvature $R = 16 \ cm$. Since the inner surface is reflecting,it is a concave mirror.
Focal length $f = \frac{R}{2} = \frac{-16}{2} = -8 \ cm$.
The object is placed at $u = -10 \ cm$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} + \frac{1}{-10} = \frac{1}{-8}$
$\frac{1}{v} = \frac{1}{10} - \frac{1}{8} = \frac{4 - 5}{40} = \frac{-1}{40}$
$v = -40 \ cm$.
Magnification $m = \frac{-v}{u} = \frac{-(-40)}{-10} = -4$.
Since $m$ is negative,the image is real and inverted. Since $|m| = 4 > 1$,the image is magnified.

(B) The pitch of a helical path is given by the formula: $P = v \cos \theta \times T = v \cos \theta \times \frac{2 \pi m}{qB}$.
Given values:
Speed $v = 4 \times 10^{5} \ m/s$
Magnetic field $B = 0.3 \ T$
Angle $\theta = 60^{\circ}$
Mass $m = 1.67 \times 10^{-27} \ kg$
Charge $q = 1.6 \times 10^{-19} \ C$
Substituting the values:
$P = \frac{2 \times 3.14 \times 1.67 \times 10^{-27} \times 4 \times 10^{5} \times \cos 60^{\circ}}{1.6 \times 10^{-19} \times 0.3}$
Since $\cos 60^{\circ} = 0.5$:
$P = \frac{2 \times 3.14 \times 1.67 \times 10^{-27} \times 4 \times 10^{5} \times 0.5}{1.6 \times 10^{-19} \times 0.3}$
$P = \frac{10.4872 \times 10^{-22}}{0.48 \times 10^{-19}}$
$P \approx 0.0437 \ m \approx 4.37 \ cm$.
The closest option is $4 \ cm$.

(D) The resistivity values at room temperature $(20^{\circ}C)$ for the given materials are approximately:
Copper $(\rho_{C})$: $1.72 \times 10^{-8} \ \Omega \cdot m$
Aluminium $(\rho_{A})$: $2.82 \times 10^{-8} \ \Omega \cdot m$
Tungsten $(\rho_{T})$: $5.60 \times 10^{-8} \ \Omega \cdot m$
Mercury $(\rho_{M})$: $98.0 \times 10^{-8} \ \Omega \cdot m$
Comparing these values,we get $\rho_{M} > \rho_{T} > \rho_{A} > \rho_{C}$.
Among the given options,the correct relationship is $\rho_{M} > \rho_{A} > \rho_{C}$.

(A) For a permanent magnet $(P)$,the material must have high retentivity so that it remains magnetized,and high coercivity so that it is not easily demagnetized by external magnetic fields or temperature fluctuations.
For a transformer core $(T)$,the material must have high retentivity but very low coercivity to minimize energy loss due to hysteresis during the rapid magnetization and demagnetization cycles of alternating current.
Therefore,the correct property is that $T$ requires large retentivity and small coercivity.

(A) Given:
Slit separation $d = 1 \, mm = 10^{-3} \, m$
Wavelength $\lambda = 632.8 \, nm = 632.8 \times 10^{-9} \, m$
Distance $D = 100 \, cm = 1 \, m$
Position of bright fringe $y = 1.27 \, mm = 1.27 \times 10^{-3} \, m$
The condition for a bright fringe is $y = \frac{n D \lambda}{d}$,where $n$ is the order of the fringe.
Calculating $n$:
$n = \frac{y d}{D \lambda} = \frac{1.27 \times 10^{-3} \times 10^{-3}}{1 \times 632.8 \times 10^{-9}} = \frac{1.27 \times 10^{-6}}{632.8 \times 10^{-9}} = \frac{1270}{632.8} \approx 2$
The path difference $\Delta x$ for a bright fringe is given by $\Delta x = n \lambda$.
Substituting the values:
$\Delta x = 2 \times 632.8 \, nm = 1265.6 \, nm = 1.2656 \, \mu m \approx 1.27 \, \mu m$.

(B) The total energy density $u$ of an electromagnetic wave is given by $u = \frac{B_0^2}{2 \mu_0}$.
Given $u = 1.02 \times 10^{-8} \ J/m^3$ and $\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$.
Rearranging the formula for the magnetic field amplitude $B_0$:
$B_0^2 = 2 \mu_0 u$
$B_0^2 = 2 \times (4 \pi \times 10^{-7}) \times (1.02 \times 10^{-8})$
$B_0^2 = 8 \pi \times 1.02 \times 10^{-15}$
$B_0^2 \approx 25.13 \times 1.02 \times 10^{-15} \approx 25.63 \times 10^{-15} \approx 256.3 \times 10^{-16}$
Taking the square root:
$B_0 \approx 16 \times 10^{-8} \ T$
Since $1 \ nT = 10^{-9} \ T$,we have $B_0 = 160 \times 10^{-9} \ T = 160 \ nT$.

(B) $1$. Calculate the number of uranium atoms in $2 \, kg$ of ${ }_{92} U ^{235}$:
$n = \frac{\text{mass}}{\text{molar mass}} \times N_A = \frac{2 \, kg}{235 \, kg/kmol} \times 6.023 \times 10^{26} \, \text{atoms/kmol} \approx 5.126 \times 10^{24} \, \text{atoms}$.
$2$. Calculate the total energy released $(E)$:
$E = n \times 200 \, MeV = 5.126 \times 10^{24} \times 200 \times 10^6 \times 1.6 \times 10^{-19} \, J \approx 1.64 \times 10^{14} \, J$.
$3$. Calculate the power output $(P)$:
$P = \frac{E}{t}$, where $t = 30 \, \text{days} = 30 \times 24 \times 3600 \, s = 2.592 \times 10^6 \, s$.
$P = \frac{1.64 \times 10^{14}}{2.592 \times 10^6} \approx 6.326 \times 10^7 \, W = 63.26 \, MW$.
The value is closest to $60 \, MW$.