JEE Main 2021 Physics Question Paper with Answer and Solution

(C) Given that power $P$ is constant. We know that $P = Fv = mav = m \left( \frac{dv}{dt} \right) v$.
Integrating this expression: $\frac{P}{m} dt = v dv$.
Integrating both sides: $\int \frac{P}{m} dt = \int v dv \implies \frac{P}{m} t = \frac{v^2}{2}$.
Thus,$v^2 = \frac{2P}{m} t$,which gives $v = \sqrt{\frac{2P}{m}} t^{1/2}$.
Since $v = \frac{ds}{dt}$,we have $ds = \sqrt{\frac{2P}{m}} t^{1/2} dt$.
Integrating with respect to $t$: $s = \int \sqrt{\frac{2P}{m}} t^{1/2} dt = \sqrt{\frac{2P}{m}} \left( \frac{t^{3/2}}{3/2} \right) = \sqrt{\frac{2P}{m}} \left( \frac{2}{3} t^{3/2} \right)$.
Therefore,$s \propto t^{3/2}$.

(B) Let the radius of each small drop be $R$. The volume of two small drops is $V_{initial} = 2 \times (\frac{4}{3}\pi R^3) = \frac{8}{3}\pi R^3$.
Let the radius of the large drop be $R'$. Since the volume remains constant,$\frac{4}{3}\pi (R')^3 = \frac{8}{3}\pi R^3$,which gives $R' = 2^{1/3}R$.
The surface energy of a drop is given by $E = T \times A$,where $T$ is surface tension and $A$ is surface area.
Initial surface energy $E_i = 2 \times (T \times 4\pi R^2) = 8\pi R^2 T$.
Final surface energy $E_f = T \times 4\pi (R')^2 = 4\pi (2^{1/3}R)^2 T = 4\pi 2^{2/3} R^2 T$.
The ratio of initial to final surface energy is $\frac{E_i}{E_f} = \frac{8\pi R^2 T}{4\pi 2^{2/3} R^2 T} = \frac{2}{2^{2/3}} = 2^{1 - 2/3} = 2^{1/3}$.
Thus,the ratio is $2^{1/3}:1$.

(C) When two soap bubbles of radii $r_1$ and $r_2$ coalesce in a vacuum under isothermal conditions,the total number of moles of air remains constant. Since the temperature is constant,the product of pressure and volume $(PV)$ remains constant for the air inside the bubbles.
For a soap bubble,the excess pressure is $P_{ex} = \frac{4T}{r}$,where $T$ is the surface tension. The absolute pressure inside is $P = P_{atm} + \frac{4T}{r}$. In a vacuum,$P_{atm} = 0$,so $P = \frac{4T}{r}$.
The volume of a bubble is $V = \frac{4}{3}\pi r^3$.
For the first bubble: $P_1 V_1 = (\frac{4T}{r_1})(\frac{4}{3}\pi r_1^3) = \frac{16}{3}\pi T r_1^2$.
For the second bubble: $P_2 V_2 = (\frac{4T}{r_2})(\frac{4}{3}\pi r_2^3) = \frac{16}{3}\pi T r_2^2$.
For the resultant bubble of radius $R$: $P_R V_R = (\frac{4T}{R})(\frac{4}{3}\pi R^3) = \frac{16}{3}\pi T R^2$.
Since the total amount of air is conserved: $P_1 V_1 + P_2 V_2 = P_R V_R$.
$\frac{16}{3}\pi T r_1^2 + \frac{16}{3}\pi T r_2^2 = \frac{16}{3}\pi T R^2$.
Dividing by $\frac{16}{3}\pi T$,we get $R^2 = r_1^2 + r_2^2$,or $R = \sqrt{r_1^2 + r_2^2}$.

(B) The ratio of specific heats $\gamma$ is defined as the ratio of molar specific heat at constant pressure $C_p$ to molar specific heat at constant volume $C_v$,i.e.,$\gamma = \frac{C_p}{C_v}$.
According to the equipartition theorem,$C_v = \frac{f}{2}R$ and $C_p = C_v + R = (\frac{f}{2} + 1)R$.
Therefore,$\gamma = \frac{(\frac{f}{2} + 1)R}{\frac{f}{2}R} = \frac{\frac{f+2}{2}}{\frac{f}{2}} = 1 + \frac{2}{f}$.
Rearranging the equation: $\gamma - 1 = \frac{2}{f}$.
Solving for $f$,we get $f = \frac{2}{\gamma - 1}$.

(D) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $T V^{\gamma - 1} = \text{constant}$.
Thus,$T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$,which implies $\frac{T_1}{T_2} = (\frac{V_2}{V_1})^{\gamma - 1}$.
Since the gas is monoatomic,the adiabatic index $\gamma = 5/3$. Therefore,$\gamma - 1 = 5/3 - 1 = 2/3$.
The volume of the gas in a cylinder is $V = A \times L$,where $A$ is the cross-sectional area and $L$ is the length of the gas column.
Substituting $V_1 = A L_1$ and $V_2 = A L_2$,we get $\frac{V_2}{V_1} = \frac{A L_2}{A L_1} = \frac{L_2}{L_1}$.
Substituting these into the temperature ratio equation: $\frac{T_1}{T_2} = (\frac{L_2}{L_1})^{2/3}$.

(B) When two plates of equal thickness $l$ and cross-sectional area $A$ are connected in series,the equivalent thermal resistance $R_{eq}$ is the sum of individual thermal resistances $R_1$ and $R_2$.
$R_{eq} = R_1 + R_2$
Since thermal resistance $R = \frac{l}{KA}$,we have:
$\frac{2l}{K_{eq} A} = \frac{l}{K_1 A} + \frac{l}{K_2 A}$
Dividing both sides by $\frac{l}{A}$:
$\frac{2}{K_{eq}} = \frac{1}{K_1} + \frac{1}{K_2}$
$\frac{2}{K_{eq}} = \frac{K_2 + K_1}{K_1 K_2}$
$K_{eq} = \frac{2 K_1 K_2}{K_1 + K_2}$

(B) The total energy $E$ of a particle in simple harmonic motion is given by $E = \frac{1}{2} m \omega^2 a^2$.
The kinetic energy $K$ at a displacement $y$ is given by $K = \frac{1}{2} m \omega^2 (a^2 - y^2)$.
Taking the ratio of kinetic energy to total energy:
$\frac{K}{E} = \frac{\frac{1}{2} m \omega^2 (a^2 - y^2)}{\frac{1}{2} m \omega^2 a^2} = \frac{a^2 - y^2}{a^2} = 1 - \frac{y^2}{a^2}$.
Given that $K = 3E/4$,we have:
$\frac{3E/4}{E} = 1 - \frac{y^2}{a^2}$.
$\frac{3}{4} = 1 - \frac{y^2}{a^2}$.
$\frac{y^2}{a^2} = 1 - \frac{3}{4} = \frac{1}{4}$.
Taking the square root of both sides,we get $y = a/2$.

(B) The initial moment of inertia of the ring about its axis is $I = Mr^2$.
The initial angular momentum is $L = I\omega = Mr^2\omega$.
When two objects of mass $m$ are attached to the opposite ends of a diameter,the new moment of inertia $I'$ becomes the sum of the ring's moment of inertia and the moment of inertia of the two point masses: $I' = Mr^2 + m(r)^2 + m(r)^2 = (M + 2m)r^2$.
According to the principle of conservation of angular momentum,the external torque is zero,so $L_{initial} = L_{final}$.
$Mr^2\omega = (M + 2m)r^2\omega'$
Solving for the new angular velocity $\omega'$:
$\omega' = \frac{Mr^2\omega}{(M + 2m)r^2} = \frac{M\omega}{M + 2m}$.

(B) The velocity of the particle is given by $v = At + Bt^2$.
Since $v = \frac{ds}{dt}$,we have $ds = (At + Bt^2) dt$.
To find the distance travelled between $t = 1 \ s$ and $t = 2 \ s$,we integrate the velocity with respect to time:
$s = \int_{1}^{2} (At + Bt^2) dt$
$s = \left[ \frac{At^2}{2} + \frac{Bt^3}{3} \right]_{1}^{2}$
$s = \left( \frac{A(2)^2}{2} + \frac{B(2)^3}{3} \right) - \left( \frac{A(1)^2}{2} + \frac{B(1)^3}{3} \right)$
$s = \left( 2A + \frac{8B}{3} \right) - \left( \frac{A}{2} + \frac{B}{3} \right)$
$s = (2A - \frac{A}{2}) + (\frac{8B}{3} - \frac{B}{3})$
$s = \frac{3}{2}A + \frac{7}{3}B$
Since the velocity does not change sign in the interval $[1, 2]$,the distance is equal to the magnitude of displacement.

(B) Let $d$ be the distance of the escalator.
Velocity of Preeti on the stationary escalator is $v_1 = \frac{d}{t_1}$.
Velocity of the moving escalator is $v_2 = \frac{d}{t_2}$.
When Preeti walks on the moving escalator,her net velocity with respect to the ground is $v = v_1 + v_2$.
$v = \frac{d}{t_1} + \frac{d}{t_2} = d \left( \frac{t_1 + t_2}{t_1 t_2} \right)$.
The time $t$ taken to cover the distance $d$ with net velocity $v$ is:
$t = \frac{d}{v} = \frac{d}{d \left( \frac{t_1 + t_2}{t_1 t_2} \right)} = \frac{t_1 t_2}{t_1 + t_2}$.

(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature in Kelvin.
Given $\eta_1 = 1/6$,so $1/6 = 1 - \frac{T_2}{T_1}$,which implies $\frac{T_2}{T_1} = 5/6$ or $T_2 = \frac{5}{6}T_1$ $...(i)$.
When the sink temperature is reduced by $62^{\circ}C$ (which is equivalent to a change of $62 \ K$),the new efficiency $\eta_2 = 2 \times \eta_1 = 2 \times (1/6) = 1/3$.
The new sink temperature is $T_2' = T_2 - 62$.
Using the efficiency formula: $1/3 = 1 - \frac{T_2 - 62}{T_1}$.
Rearranging gives $\frac{T_2 - 62}{T_1} = 1 - 1/3 = 2/3$.
Substituting $T_2 = \frac{5}{6}T_1$ from equation $(i)$:
$\frac{\frac{5}{6}T_1 - 62}{T_1} = 2/3$.
$\frac{5}{6} - \frac{62}{T_1} = 2/3$.
$\frac{62}{T_1} = \frac{5}{6} - \frac{4}{6} = 1/6$.
$T_1 = 62 \times 6 = 372 \ K$.
To convert to Celsius: $T(^{\circ}C) = 372 - 273 = 99^{\circ}C$.

(B) In $SHM$,the velocity $V$ of a particle at a distance $x$ from the mean position is given by $V = \omega \sqrt{a^2 - x^2}$,where $a$ is the amplitude and $\omega$ is the angular frequency.
Squaring both sides,we get $V^2 = \omega^2(a^2 - x^2)$.
For distances $x_1$ and $x_2$,we have:
$V_1^2 = \omega^2(a^2 - x_1^2) \dots (i)$
$V_2^2 = \omega^2(a^2 - x_2^2) \dots (ii)$
Subtracting equation $(ii)$ from $(i)$:
$V_1^2 - V_2^2 = \omega^2(a^2 - x_1^2 - a^2 + x_2^2)$
$V_1^2 - V_2^2 = \omega^2(x_2^2 - x_1^2)$
$\omega^2 = \frac{V_1^2 - V_2^2}{x_2^2 - x_1^2}$
Since the time period $T = \frac{2\pi}{\omega}$,we have $\omega = \frac{2\pi}{T}$.
Substituting this into the equation for $\omega^2$:
$\left(\frac{2\pi}{T}\right)^2 = \frac{V_1^2 - V_2^2}{x_2^2 - x_1^2}$
$T^2 = 4\pi^2 \left(\frac{x_2^2 - x_1^2}{V_1^2 - V_2^2}\right)$
$T = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{V_1^2 - V_2^2}}$

(C) The potential energy $(PE)$ of a particle in $SHM$ is given by $PE = \frac{1}{2} m \omega^2 x^2$.
The kinetic energy $(KE)$ of a particle in $SHM$ is given by $KE = \frac{1}{2} m \omega^2 (A^2 - x^2)$.
Given that $PE = KE$,we have:
$\frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m \omega^2 (A^2 - x^2)$
$x^2 = A^2 - x^2$
$2x^2 = A^2$
$x^2 = \frac{A^2}{2}$
$x = \pm \frac{A}{\sqrt{2}}$
Thus,the position of the particle is $\frac{A}{\sqrt{2}}$.

(A) Let the magnitude of vectors be $|\vec A| = |\vec B| = A$.
Given that $|\vec A + \vec B| = n |\vec A - \vec B|$.
Squaring both sides,we get $|\vec A + \vec B|^2 = n^2 |\vec A - \vec B|^2$.
Using the vector identity $|\vec A \pm \vec B|^2 = A^2 + B^2 \pm 2AB \cos \theta$,we have:
$A^2 + A^2 + 2A^2 \cos \theta = n^2 (A^2 + A^2 - 2A^2 \cos \theta)$.
$2A^2 (1 + \cos \theta) = n^2 [2A^2 (1 - \cos \theta)]$.
Dividing both sides by $2A^2$:
$1 + \cos \theta = n^2 (1 - \cos \theta)$.
$1 + \cos \theta = n^2 - n^2 \cos \theta$.
$\cos \theta (1 + n^2) = n^2 - 1$.
$\cos \theta = \frac{n^2 - 1}{n^2 + 1}$.
Therefore,$\theta = \cos^{-1} \left[ \frac{n^2 - 1}{n^2 + 1} \right]$.

(B) Let the original length be $l$ and the cross-sectional area be $A$. The Young's modulus $Y$ is given by $Y = \frac{T}{A} \cdot \frac{l}{\Delta l}$,where $\Delta l$ is the change in length.
For tension $T_1$,the length is $l_1$,so $\Delta l_1 = l_1 - l$. Thus,$Y = \frac{T_1}{A} \cdot \frac{l}{l_1 - l}$.
For tension $T_2$,the length is $l_2$,so $\Delta l_2 = l_2 - l$. Thus,$Y = \frac{T_2}{A} \cdot \frac{l}{l_2 - l}$.
Since the material is the same,$Y$ is constant. Equating the two expressions:
$\frac{T_1}{A} \cdot \frac{l}{l_1 - l} = \frac{T_2}{A} \cdot \frac{l}{l_2 - l}$
$\frac{T_1}{l_1 - l} = \frac{T_2}{l_2 - l}$
$T_1(l_2 - l) = T_2(l_1 - l)$
$T_1 l_2 - T_1 l = T_2 l_1 - T_2 l$
$T_2 l - T_1 l = T_2 l_1 - T_1 l_2$
$l(T_2 - T_1) = T_2 l_1 - T_1 l_2$
$l = \frac{T_2 l_1 - T_1 l_2}{T_2 - T_1}$

(D) Given that $1 \text{ Main Scale Division (MSD)} = a \ cm$.
Let $1 \text{ Vernier Scale Division (VSD)} = a' \ cm$.
According to the problem,the $n^{\text{th}}$ division of the vernier scale coincides with the $(n-1)^{\text{th}}$ division of the main scale:
$n \times a' = (n-1) \times a$
$a' = \frac{(n-1)a}{n} \ cm$.
The least count ($L$.$C$.) is defined as the difference between one main scale division and one vernier scale division:
$L.C. = 1 \text{ MSD} - 1 \text{ VSD} = (a - a') \ cm$.
Substituting the value of $a'$:
$L.C. = a - \frac{(n-1)a}{n} = \frac{na - na + a}{n} = \frac{a}{n} \ cm$.
To convert the least count into $mm$,we multiply by $10$ (since $1 \ cm = 10 \ mm$):
$L.C. = \left(\frac{a}{n}\right) \times 10 \ mm = \frac{10a}{n} \ mm$.

(B) $1$. Resolve the applied force $F$ into horizontal and vertical components: $F_{x} = F \cos \theta$ and $F_{y} = F \sin \theta$.
$2$. The vertical forces acting on the block are the normal force $N$ (upwards),the vertical component of the applied force $F \sin \theta$ (upwards),and the weight $mg$ (downwards). Since there is no vertical motion,the net vertical force is zero: $N + F \sin \theta = mg$,which gives $N = mg - F \sin \theta$.
$3$. The kinetic friction force $f_{k}$ is given by $f_{k} = \mu_{K} N = \mu_{K}(mg - F \sin \theta)$.
$4$. The net horizontal force acting on the block is the horizontal component of the applied force minus the friction force: $F_{net} = F \cos \theta - f_{k} = F \cos \theta - \mu_{K}(mg - F \sin \theta)$.
$5$. Using Newton's second law,$F_{net} = ma$,we get $ma = F \cos \theta - \mu_{K}(mg - F \sin \theta)$.
$6$. Dividing by mass $m$,the acceleration is $a = \frac{F}{m} \cos \theta - \mu_{K}(g - \frac{F}{m} \sin \theta)$.

(A) The total pressure $P$ at a depth $d$ is given by $P = P_{0} + \rho gd$,where $P_{0}$ is the atmospheric pressure.
Given $P_{1} = 3 \times 10^{5} \; Pa$ and $P_{0} = 1 \times 10^{5} \; Pa$.
So,$\rho gd = P_{1} - P_{0} = 3 \times 10^{5} - 1 \times 10^{5} = 2 \times 10^{5} \; Pa$.
If the depth is doubled $(d' = 2d)$,the new pressure $P_{2}$ is $P_{2} = P_{0} + \rho g(2d) = P_{0} + 2(\rho gd)$.
Substituting the values: $P_{2} = 1 \times 10^{5} + 2(2 \times 10^{5}) = 1 \times 10^{5} + 4 \times 10^{5} = 5 \times 10^{5} \; Pa$.
The percentage increase in pressure is given by $\frac{P_{2} - P_{1}}{P_{1}} \times 100$.
Percentage increase $= \frac{5 \times 10^{5} - 3 \times 10^{5}}{3 \times 10^{5}} \times 100 = \frac{2 \times 10^{5}}{3 \times 10^{5}} \times 100 = \frac{200}{3} \%$.

(C) According to the law of conservation of angular momentum,the angular momentum of the comet remains constant at all points in its orbit.
Let $r_1$ and $v_1$ be the distance and speed at the nearest point (perihelion),and $r_2$ and $v_2$ be the distance and speed at the farthest point (aphelion).
Given:
$r_1 = 8.0 \times 10^{10} \ m$
$v_1 = 6 \times 10^{4} \ m/s$
$r_2 = 1.6 \times 10^{12} \ m$
Using the conservation of angular momentum:
$m v_1 r_1 = m v_2 r_2$
$v_2 = \frac{v_1 r_1}{r_2}$
Substituting the values:
$v_2 = \frac{(6 \times 10^{4}) \times (8.0 \times 10^{10})}{1.6 \times 10^{12}}$
$v_2 = \frac{48 \times 10^{14}}{1.6 \times 10^{12}}$
$v_2 = 30 \times 10^{2} \ m/s = 3 \times 10^{3} \ m/s$
Thus,the speed at the farthest point is $3 \times 10^{3} \ m/s$.

(C) The total pressure $P$ of a mixture of ideal gases is given by the ideal gas equation $PV = n_{total} RT$,where $n_{total} = n_1 + n_2 + n_3$.
First,calculate the number of moles for each gas:
$n_{O_2} = \frac{16 \, g}{32 \, g/mol} = 0.5 \, mol$
$n_{N_2} = \frac{28 \, g}{28 \, g/mol} = 1.0 \, mol$
$n_{CO_2} = \frac{44 \, g}{44 \, g/mol} = 1.0 \, mol$
Total moles $n_{total} = 0.5 + 1.0 + 1.0 = 2.5 \, mol = \frac{5}{2} \, mol$.
Substituting into the ideal gas equation:
$PV = (2.5) RT$
$P = \frac{5}{2} \frac{RT}{V}$.

(A) In thermodynamics,heat and work are classified as path functions.
According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Here,$\Delta U$ (internal energy) is a state function,meaning it depends only on the initial and final states of the system.
However,the work done $(\Delta W)$ depends on the specific process or path taken to change the state of the system.
Since $\Delta U$ is a state function and $\Delta W$ is a path function,the heat exchanged $(\Delta Q)$ must also depend on the path taken.
Therefore,both heat and work are path functions.

(C) The moment of inertia of a point mass is given by $I = m r^2$,where $r$ is the perpendicular distance from the axis of rotation.
Let the axis pass through $A$ and be parallel to the diagonal $DB$.
The perpendicular distances of the four masses from this axis are:
$1$. For mass at $A$: $r_A = 0$ (since the axis passes through $A$).
$2$. For mass at $B$: $r_B = l \sin(45^\circ) = l / \sqrt{2}$.
$3$. For mass at $D$: $r_D = l \sin(45^\circ) = l / \sqrt{2}$.
$4$. For mass at $C$: $r_C = l \sin(45^\circ) + l \sin(45^\circ) = l / \sqrt{2} + l / \sqrt{2} = 2l / \sqrt{2} = l \sqrt{2}$.
The total moment of inertia is $I = m(r_A^2 + r_B^2 + r_D^2 + r_C^2)$.
$I = m [0^2 + (l / \sqrt{2})^2 + (l / \sqrt{2})^2 + (l \sqrt{2})^2]$
$I = m [0 + l^2/2 + l^2/2 + 2l^2]$
$I = m [l^2 + 2l^2] = 3 m l^2$.

(D) When the lift is stationary,the time period of the simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g}}$.
When the lift moves upwards with an acceleration $a = g/2$,a pseudo force acts downwards on the pendulum bob.
The effective acceleration due to gravity becomes $g_{eff} = g + a = g + \frac{g}{2} = \frac{3g}{2}$.
The new time period $T'$ is given by $T' = 2 \pi \sqrt{\frac{L}{g_{eff}}}$.
Substituting $g_{eff} = \frac{3g}{2}$,we get $T' = 2 \pi \sqrt{\frac{L}{3g/2}} = 2 \pi \sqrt{\frac{2L}{3g}}$.
Comparing this with the original time period $T$,we have $T' = \sqrt{\frac{2}{3}} \times (2 \pi \sqrt{\frac{L}{g}}) = \sqrt{\frac{2}{3}} T$.

(A) From the given velocity-displacement graph,for $0 \leq x \leq 200 \ m$,the graph is a straight line passing through $(0, 10)$ and $(200, 50)$.
The equation of the line is $v = mx + C$.
Slope $m = \frac{50 - 10}{200 - 0} = \frac{40}{200} = \frac{1}{5} \ m^{-1}s^{-1}$.
Intercept $C = 10 \ m/s$.
So,$v = \frac{1}{5}x + 10$.
Acceleration $a = v \frac{dv}{dx} = (\frac{1}{5}x + 10) \frac{d}{dx}(\frac{1}{5}x + 10) = (\frac{1}{5}x + 10)(\frac{1}{5}) = \frac{x}{25} + 2$.
At $x = 0$,$a = 2 \ m/s^2$.
At $x = 200$,$a = \frac{200}{25} + 2 = 8 + 2 = 10 \ m/s^2$.
For $x > 200 \ m$,$v = 50 \ m/s$ (constant).
Since $v$ is constant,$a = \frac{dv}{dt} = 0$.
Comparing this with the given options,option $A$ is the closest representation of the calculated values,noting that the peak value in the provided graph image is $18 \ m/s^2$ which corresponds to a different slope calculation if we assume the graph values are fixed. Based on the linear relationship $a = \frac{x}{25} + 2$,the acceleration increases linearly.

(D) The normal force $N$ provides the necessary centripetal force for circular motion.
$N = F_c = m \omega^2 R$
Since $\omega = \frac{2\pi}{T}$,we have $N = m \left( \frac{2\pi}{T} \right)^2 R = m \frac{4\pi^2}{T^2} R$.
Given values: $m = 200\, g = 0.2\, kg$,$R = 20\, cm = 0.2\, m$,and $T = 40\, s$.
Substituting these values into the formula:
$N = 0.2 \times \frac{4 \times (3.14159)^2}{(40)^2} \times 0.2$
$N = 0.2 \times \frac{4 \times 9.8696}{1600} \times 0.2$
$N = 0.04 \times \frac{39.4784}{1600}$
$N = 0.04 \times 0.024674 = 9.8696 \times 10^{-4}\, N$.
Rounding to the nearest option,we get $N \approx 9.859 \times 10^{-4}\, N$.

(C) Given: Mass $m = 20\, kg$,Radius $R = 0.2\, m$,Force $F = 20\, N$,Initial angular speed $\omega_0 = 0$,Final angular speed $\omega = 50\, rad/s$.
The torque $\tau$ applied by the force $F$ is $\tau = F \cdot R$.
The moment of inertia of the disk about its center is $I = \frac{1}{2} mR^2$.
Using $\tau = I\alpha$,the angular acceleration $\alpha$ is:
$\alpha = \frac{F \cdot R}{\frac{1}{2} mR^2} = \frac{2F}{mR} = \frac{2 \times 20}{20 \times 0.2} = \frac{2}{0.2} = 10\, rad/s^2$.
Using the kinematic equation $\omega^2 = \omega_0^2 + 2\alpha\Delta\theta$:
$(50)^2 = 0^2 + 2(10)\Delta\theta$
$2500 = 20\Delta\theta$
$\Delta\theta = 125\, rad$.
Since one revolution corresponds to $2\pi \approx 6.28\, rad$,the number of revolutions $n$ is:
$n = \frac{\Delta\theta}{2\pi} = \frac{125}{6.28} \approx 19.90$.
Rounding to the nearest integer,$n = 20$.

(D) The formula for resistance is $R = \frac{V}{I}$.
For division,the relative error is the sum of the relative errors of the individual quantities: $\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$.
To find the percentage error,multiply by $100$: $\frac{\Delta R}{R} \times 100 = \left( \frac{\Delta V}{V} \times 100 \right) + \left( \frac{\Delta I}{I} \times 100 \right)$.
Given $V = 50 \; V$,$\Delta V = 2 \; V$ and $I = 20 \; A$,$\Delta I = 0.2 \; A$.
Percentage error in $R = \left( \frac{2}{50} \times 100 \right) + \left( \frac{0.2}{20} \times 100 \right)$.
Percentage error in $R = 4\% + 1\% = 5\%$.
Thus,$x = 5$.

(A) Let the force $\overrightarrow{P}$ be applied at point $A$ in the downward vertical direction. We need to resolve this force along the directions of the rods $AB$ and $AC$.
Let $\overrightarrow{F_{AB}}$ and $\overrightarrow{F_{AC}}$ be the components of force $\overrightarrow{P}$ along $AB$ and $AC$ respectively.
According to the parallelogram law of vector addition,$\overrightarrow{P} = \overrightarrow{F_{AB}} + \overrightarrow{F_{AC}}$.
Using the Lami's theorem or the sine rule in the triangle formed by the force vectors:
The angle between the vertical force $\overrightarrow{P}$ and rod $AB$ is $35^{\circ}$ (as the angle between the vertical and the wall is $90^{\circ}$,and the angle between $AB$ and the wall is $70^{\circ}$,so the angle between $AB$ and the vertical is $90^{\circ} - 70^{\circ} = 20^{\circ}$ is incorrect based on the provided image geometry; let's use the provided diagram).
From the diagram,the angle between the vertical force $\overrightarrow{P}$ and rod $AC$ is $35^{\circ}$.
The component of force $\overrightarrow{P}$ along the direction of rod $AC$ is given by $P \cos(\theta)$,where $\theta$ is the angle between the force vector and the rod.
Thus,the magnitude of the component along $AC$ is $x = 100 \cos(35^{\circ}) \; N$.
Given $\cos(35^{\circ}) = 0.819$.
$x = 100 \times 0.819 = 81.9 \; N$.
Rounding to the nearest integer,we get $x = 82$.

(B) Initial momentum of the system along the $X$-axis is $P_i = m_1 v_1 = 10 \times 10 \sqrt{3} = 100 \sqrt{3} \, kg \cdot m/s$.
The second ball of mass $20 \, kg$ splits into two equal pieces of $10 \, kg$ each.
Let the velocity of the first piece be $v_y = 10 \, m/s$ along the $Y$-axis and the velocity of the second piece be $v_x = 20 \, m/s$ at an angle $\theta$ with the $X$-axis.
Applying the law of conservation of linear momentum along the $X$-axis:
$P_{ix} = P_{fx}$
$100 \sqrt{3} = (10 \times v_x \cos \theta) + (10 \times 0)$
$100 \sqrt{3} = 10 \times 20 \cos \theta$
$100 \sqrt{3} = 200 \cos \theta$
$\cos \theta = \frac{100 \sqrt{3}}{200} = \frac{\sqrt{3}}{2}$
$\theta = 30^{\circ}$.

(D) The coordinates of point $P$ are $(5, 5 \sqrt{3})$ cm. The coordinates of point $O$ are $(0, 0)$ and point $Q$ are $(10, 0)$.
Given force $\overrightarrow{F} = 4 \hat{i} - 3 \hat{j}$.
Position vector of $P$ with respect to $O$ is $\overrightarrow{r}_1 = 5 \hat{i} + 5 \sqrt{3} \hat{j}$.
Torque about $O$ is $\vec{\tau}_O = \overrightarrow{r}_1 \times \overrightarrow{F} = (5 \hat{i} + 5 \sqrt{3} \hat{j}) \times (4 \hat{i} - 3 \hat{j}) = (-15 \hat{k} - 20 \sqrt{3} \hat{k}) = (-15 - 20 \sqrt{3}) \hat{k}$.
Position vector of $P$ with respect to $Q$ is $\overrightarrow{r}_2 = (5-10) \hat{i} + 5 \sqrt{3} \hat{j} = -5 \hat{i} + 5 \sqrt{3} \hat{j}$.
Torque about $Q$ is $\vec{\tau}_Q = \overrightarrow{r}_2 \times \overrightarrow{F} = (-5 \hat{i} + 5 \sqrt{3} \hat{j}) \times (4 \hat{i} - 3 \hat{j}) = (15 \hat{k} + 20 \sqrt{3} \hat{k}) = (15 + 20 \sqrt{3}) \hat{k}$.
Thus,the values are $(-15 - 20 \sqrt{3})$ and $(15 + 20 \sqrt{3})$.

(A) Let $P_1$ and $P_2$ be the excess pressures inside the soap bubbles of radii $a$ and $b$ respectively.
The excess pressure inside a soap bubble is given by $P = \frac{4T}{r}$,where $T$ is the surface tension.
For the two bubbles,$P_1 = \frac{4T}{a}$ and $P_2 = \frac{4T}{b}$.
When they coalesce,they form a common interface with radius of curvature $r$.
The pressure difference across this common interface is $\Delta P = P_1 - P_2$ (since $a < b$,$P_1 > P_2$).
Thus,$\frac{4T}{r} = \frac{4T}{a} - \frac{4T}{b}$.
Dividing by $4T$,we get $\frac{1}{r} = \frac{1}{a} - \frac{1}{b}$.
$\frac{1}{r} = \frac{b-a}{ab}$.
Therefore,$r = \frac{ab}{b-a}$.

(A) For a polyatomic gas,the total degrees of freedom $f$ is given by the sum of translational,rotational,and vibrational degrees of freedom.
Translational degrees of freedom $= 3$.
Rotational degrees of freedom for a non-linear polyatomic molecule $= 3$.
Each vibrational mode contributes $2$ degrees of freedom (one for kinetic energy and one for potential energy).
Given $24$ vibrational modes,vibrational degrees of freedom $= 24 \times 2 = 48$.
Total degrees of freedom $f = 3 + 3 + 48 = 54$.
The adiabatic index $\gamma$ is given by $\gamma = 1 + \frac{2}{f}$.
Substituting the value of $f$: $\gamma = 1 + \frac{2}{54} = 1 + \frac{1}{27} = \frac{28}{27} \approx 1.037$.
Rounding to two decimal places,we get $\gamma = 1.03$.

(B) Given,positive zero error $= 0.2\, mm = 0.02\, cm$.
Main scale reading $(MSR) = 8.5\, cm$.
Vernier scale reading $(VSR) = \text{Vernier coincidence} \times \text{Least count} = 6 \times 0.01\, cm = 0.06\, cm$.
Observed reading $= MSR + VSR = 8.5\, cm + 0.06\, cm = 8.56\, cm$.
Correct reading $= \text{Observed reading} - \text{Zero error} = 8.56\, cm - 0.02\, cm = 8.54\, cm$.

(A) For a car moving in a circular path on a flat track,the centripetal force is provided by the static friction force $f_{s}$.
The maximum static friction force is given by $f_{s,max} = \mu_{s} N$,where $N$ is the normal force.
The centripetal force required is $F_{c} = \frac{mv^{2}}{R}$.
Thus,$\mu_{s} N = \frac{mv^{2}}{R}$,which gives $N = \frac{mv^{2}}{\mu_{s} R}$.
The normal force $N$ on the car is the sum of its weight $mg$ and the downward negative lift force $F_{L}$ (aerodynamic downforce).
Therefore,$N = mg + F_{L}$.
Substituting the expression for $N$,we get $\frac{mv^{2}}{\mu_{s} R} = mg + F_{L}$.
Solving for $F_{L}$,we get $F_{L} = \frac{mv^{2}}{\mu_{s} R} - mg = m \left(\frac{v^{2}}{\mu_{s} R} - g\right)$.

(C) Let $t _{1}$ be the time of acceleration and $t _{2}$ be the time of deceleration. The maximum velocity reached is $v _{0}$.
From the equations of motion:
$v _{0} = \alpha t _{1} \Rightarrow t _{1} = \frac{v _{0}}{\alpha}$
$0 = v _{0} - \beta t _{2} \Rightarrow t _{2} = \frac{v _{0}}{\beta}$
Given that the total time is $t = t _{1} + t _{2}$,we have:
$t = \frac{v _{0}}{\alpha} + \frac{v _{0}}{\beta} = v _{0} \left( \frac{\alpha + \beta}{\alpha \beta} \right)$
Thus,the maximum velocity is $v _{0} = \frac{\alpha \beta t}{\alpha + \beta}$.
The total distance travelled is the area under the $v-t$ graph,which is a triangle with base $t$ and height $v _{0}$:
Distance $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times t \times v _{0}$
Substituting $v _{0}$:
Distance $= \frac{1}{2} \times t \times \left( \frac{\alpha \beta t}{\alpha + \beta} \right) = \frac{\alpha \beta t ^{2}}{2(\alpha + \beta)}$

(D) The angular momentum is defined as $\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$.
For point $A$ (the center of the circular path): The position vector $\vec{r}_A$ is in the plane of the circle,and the velocity vector $\vec{v}$ is tangential to the circle. The cross product $\vec{r}_A \times \vec{v}$ always points in the positive $z$-direction (perpendicular to the plane of the circle). Since the speed $v$ and radius $r$ are constant,the magnitude $L_A = mvr$ is constant. Thus,$L_A$ is constant in both magnitude and direction.
For point $B$ (a point on the axis of rotation above $A$): The position vector $\vec{r}_B$ from $B$ to $M$ changes its direction as the mass $M$ rotates. Therefore,the angular momentum $\vec{L}_B = \vec{r}_B \times \vec{p}$ will have a constant magnitude,but its direction will change continuously as it precesses around the $z$-axis.

(D) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1 = 800\, K$ (source temperature) and $T_2 = 400\, K$ (sink temperature).
$\eta = 1 - \frac{400}{800} = 1 - 0.5 = 0.5$.
Also,efficiency is defined as $\eta = \frac{W}{Q_1}$,where $W$ is the work output and $Q_1$ is the heat supplied.
Given $W = 1200\, J$,we have $0.5 = \frac{1200}{Q_1}$.
Therefore,$Q_1 = \frac{1200}{0.5} = 2400\, J$.

(B) Given: Mass $m = 0.5\, kg$,Initial speed $u = 20\, ms^{-1}$.
Initial kinetic energy $K_i = \frac{1}{2} mu^2 = \frac{1}{2} \times 0.5 \times (20)^2 = 100\, J$.
After deflection,the final kinetic energy $K_f$ is $5\%$ of $K_i$.
$K_f = \frac{5}{100} \times 100 = 5\, J$.
Let the final speed be $v$. Then $K_f = \frac{1}{2} mv^2$.
$5 = \frac{1}{2} \times 0.5 \times v^2$.
$5 = 0.25 \times v^2$.
$v^2 = \frac{5}{0.25} = 20$.
$v = \sqrt{20} \approx 4.47\, ms^{-1}$.

(B) Let the final temperature of the mixture be $T$. Since there is no loss of energy,the total internal energy of the system is conserved.
The internal energy of a gas is given by $U = \frac{F}{2} n R T$,where $n$ is the number of moles. Since $n$ is proportional to the number of molecules $N$,we can write $U = \frac{F}{2} N k_{B} T$.
For the mixture,the total energy before mixing equals the total energy after mixing:
$U_{1} + U_{2} = U_{mix}$
$\frac{F_{1}}{2} n_{1} k_{B} T_{1} + \frac{F_{2}}{2} n_{2} k_{B} T_{2} = \frac{F_{1}}{2} n_{1} k_{B} T + \frac{F_{2}}{2} n_{2} k_{B} T$
Canceling the common terms $\frac{1}{2} k_{B}$ from both sides:
$F_{1} n_{1} T_{1} + F_{2} n_{2} T_{2} = T (F_{1} n_{1} + F_{2} n_{2})$
Solving for $T$:
$T = \frac{F_{1} n_{1} T_{1} + F_{2} n_{2} T_{2}}{F_{1} n_{1} + F_{2} n_{2}}$

(B) Given:
Initial angular speed,$\omega_1 = 900 \, rpm = \frac{900}{60} \, rev/s = 15 \, rev/s$.
Final angular speed,$\omega_2 = 2460 \, rpm = \frac{2460}{60} \, rev/s = 41 \, rev/s$.
Time interval,$t = 26 \, s$.
Since the angular acceleration is uniform,the average angular speed is $\omega_{avg} = \frac{\omega_1 + \omega_2}{2}$.
$\omega_{avg} = \frac{15 + 41}{2} = \frac{56}{2} = 28 \, rev/s$.
The total number of revolutions $N$ is given by the product of average angular speed and time:
$N = \omega_{avg} \times t = 28 \, rev/s \times 26 \, s = 728 \, revolutions$.

(A) The formula for escape velocity is $V_e = \sqrt{\frac{2GM}{R}}$.
Let the new radius be $R'$ such that the new escape velocity $V_e' = 10V_e$.
Thus,$10V_e = \sqrt{\frac{2GM}{R'}}$.
Dividing the two equations,we get $10 = \sqrt{\frac{R}{R'}}$.
Squaring both sides,$100 = \frac{R}{R'}$.
Therefore,$R' = \frac{R}{100}$.
Given $R = 6400 \ km$,we have $R' = \frac{6400}{100} = 64 \ km$.

(B) For Fig. $1$,the time period of oscillation for a single spring-mass system is given by:
$T_a = 2\pi \sqrt{\frac{M}{k}}$
For Fig. $2$,the two springs are connected in series. The equivalent spring constant $k_{eq}$ for two springs in series is given by:
$\frac{1}{k_{eq}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k} \Rightarrow k_{eq} = \frac{k}{2}$
The time period of oscillation for the series combination is:
$T_b = 2\pi \sqrt{\frac{M}{k_{eq}}} = 2\pi \sqrt{\frac{M}{k/2}} = 2\pi \sqrt{\frac{2M}{k}}$
Now,find the ratio $\frac{T_b}{T_a}$:
$\frac{T_b}{T_a} = \frac{2\pi \sqrt{\frac{2M}{k}}}{2\pi \sqrt{\frac{M}{k}}} = \sqrt{\frac{2M/k}{M/k}} = \sqrt{2}$
Given $\frac{T_b}{T_a} = \sqrt{x}$,we have $\sqrt{x} = \sqrt{2}$,which implies $x = 2$.

(C) For a body of mass $m$ and radius $R$ rolling down an inclined plane of angle $\theta$,the acceleration $a$ is given by $a = \frac{g \sin \theta}{1 + \frac{k^2}{R^2}}$,where $k$ is the radius of gyration.
The time taken to reach the bottom of an inclined plane of length $S$ is $t = \sqrt{\frac{2S}{a}} = \sqrt{\frac{2S}{g \sin \theta} \left(1 + \frac{k^2}{R^2}\right)}$.
For the time $t$ to be minimum,the ratio $\frac{k^2}{R^2}$ must be minimum.
Comparing the values of $\frac{k^2}{R^2}$ for the given bodies:
$(1)$ Ring: $k^2 = R^2 \Rightarrow \frac{k^2}{R^2} = 1$
$(2)$ Disc: $k^2 = \frac{R^2}{2} \Rightarrow \frac{k^2}{R^2} = 0.5$
$(3)$ Solid cylinder: $k^2 = \frac{R^2}{2} \Rightarrow \frac{k^2}{R^2} = 0.5$
$(4)$ Solid sphere: $k^2 = \frac{2R^2}{5} \Rightarrow \frac{k^2}{R^2} = 0.4$
Since the solid sphere has the minimum value of $\frac{k^2}{R^2}$,it will have the maximum acceleration and will reach the bottom of the inclined plane first.

(C) For the blocks to move together,the acceleration $a$ of the system must be such that the frictional force on the smaller block $m$ is sufficient to provide its acceleration.
The maximum frictional force available is $f_{\max} = \mu N = \mu mg$.
The maximum acceleration $a_{\max}$ that the block $m$ can have without slipping is $a_{\max} = \frac{f_{\max}}{m} = \mu g$.
Given $\mu = \frac{3}{7}$ and $g = 9.8\, m/s^2$,we have:
$a_{\max} = \frac{3}{7} \times 9.8 = 3 \times 1.4 = 4.2\, m/s^2$.
Now,for the entire system of mass $(M + m)$,the force $F$ is applied:
$F = (M + m) a_{\max}$.
Substituting the values:
$F = (4.5 + 0.5) \times 4.2 = 5 \times 4.2 = 21\, N$.
Thus,the maximum horizontal force is $21\, N$.

(B) The formula for the acceleration due to gravity is $g = \frac{4 \pi^2 \ell}{T^2}$.
Taking the relative error,we have $\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2 \frac{\Delta T}{T}$.
Given $\ell = 10 \ cm$ and $\Delta \ell = 1 \ mm = 0.1 \ cm$.
For $200$ oscillations,the total time $t = 100 \ s$ with resolution $\Delta t = 1 \ s$. The time period $T = \frac{t}{200} = \frac{100}{200} = 0.5 \ s$.
The error in time period is $\Delta T = \frac{\Delta t}{200} = \frac{1}{200} \ s$.
Substituting these values: $\frac{\Delta g}{g} = \frac{0.1}{10} + 2 \left( \frac{1/200}{0.5} \right) = 0.01 + 2 \left( \frac{1}{100} \right) = 0.01 + 0.02 = 0.03$.
Percentage error is $\frac{\Delta g}{g} \times 100 = 0.03 \times 100 = 3 \%$.

(A) According to the Law of Equipartition of Energy,for a system in thermal equilibrium at temperature $T$,the energy associated with each quadratic degree of freedom is $\frac{1}{2} k_{B} T$.
Therefore,the average energy per degree of freedom for an ideal gas is $\frac{1}{2} k_{B} T$.

(A) The adiabatic process occurs from $C$ to $D$.
The formula for work done in an adiabatic process is given by:
$W = \frac{P_D V_D - P_C V_C}{1 - \gamma}$
From the given $P-V$ diagram:
At point $C$: $P_C = 100 \, N/m^2$,$V_C = 4 \, m^3$
At point $D$: $P_D = 200 \, N/m^2$,$V_D = 3 \, m^3$
Substituting the values into the formula:
$W = \frac{(200 \times 3) - (100 \times 4)}{1 - 1.4}$
$W = \frac{600 - 400}{-0.4}$
$W = \frac{200}{-0.4}$
$W = -500 \, J$

(B) For a particle moving with constant acceleration $a$:
$1$. The acceleration-time graph is a horizontal line,as $a$ is constant.
$2$. The velocity-time relation is given by $v(t) = u + at$,which represents a straight line with a non-zero slope.
$3$. The position-time relation is given by $x(t) = x_0 + ut + \frac{1}{2}at^2$,which represents a parabola.
Comparing these characteristics with the given options,the second set of graphs (represented by image $981-$b614) correctly shows a constant acceleration,a linearly increasing velocity,and a parabolic position-time graph.

(A) The resistance $R$ is given by $R = \frac{\rho \ell}{A} = \frac{V}{I}$.
Thus,the resistivity $\rho$ is given by $\rho = \frac{AV}{I\ell} = \frac{\pi d^2 V}{4I\ell}$,where $A = \frac{\pi d^2}{4}$.
The relative error in resistivity is given by $\frac{\Delta \rho}{\rho} = 2\frac{\Delta d}{d} + \frac{\Delta V}{V} + \frac{\Delta I}{I} + \frac{\Delta \ell}{\ell}$.
Given values: $V = 5.0\, V, \Delta V = 0.1\, V, \ell = 10.0\, cm, \Delta \ell = 0.1\, cm, d = 5.00\, mm, \Delta d = 0.01\, mm, I = 2.00\, A, \Delta I = 0.01\, A$.
Substituting these values:
$\frac{\Delta \rho}{\rho} = 2\left(\frac{0.01}{5.00}\right) + \frac{0.1}{5.0} + \frac{0.01}{2.00} + \frac{0.1}{10.0}$
$\frac{\Delta \rho}{\rho} = 0.004 + 0.02 + 0.005 + 0.01 = 0.039$.
The percentage error is $\frac{\Delta \rho}{\rho} \times 100 = 0.039 \times 100 = 3.9\%$.

(B) According to Kepler's Third Law of Planetary Motion,the square of the time period $T$ is proportional to the cube of the orbital radius $R$,i.e.,$T^2 \propto R^3$.
Let the time period of the first satellite be $T_1 = T$ with radius $R_1 = R$.
Let the time period of the second satellite be $T_2$ with radius $R_2 = 9R$.
Using the ratio formula:
$\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{R_2}{R_1}\right)^3$
Substituting the given values:
$\left(\frac{T_2}{T}\right)^2 = \left(\frac{9R}{R}\right)^3$
$\left(\frac{T_2}{T}\right)^2 = (9)^3 = 729$
Taking the square root on both sides:
$\frac{T_2}{T} = \sqrt{729} = 27$
Therefore,$T_2 = 27T$.

(A) Let the two resistances be $R_1$ and $R_2$.
In series,the equivalent resistance is $S = R_1 + R_2$.
In parallel,the equivalent resistance is $P = \frac{R_1 R_2}{R_1 + R_2}$.
Given the condition $S = nP$,we substitute the expressions:
$R_1 + R_2 = n \left( \frac{R_1 R_2}{R_1 + R_2} \right)$.
Rearranging the terms,we get $(R_1 + R_2)^2 = n R_1 R_2$.
Using the algebraic identity $(R_1 + R_2)^2 = (R_1 - R_2)^2 + 4 R_1 R_2$,we have:
$(R_1 - R_2)^2 + 4 R_1 R_2 = n R_1 R_2$.
Dividing by $R_1 R_2$,we get $n = 4 + \frac{(R_1 - R_2)^2}{R_1 R_2}$.
Since the term $\frac{(R_1 - R_2)^2}{R_1 R_2}$ is always greater than or equal to $0$,the minimum value of $n$ occurs when $R_1 = R_2$,which gives $n = 4$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K_{\max} = hf - W_0$,where $W_0$ is the work function.
For the first photo-cathode: $hf_1 = W_0 + \frac{1}{2}mv_1^2$ ... $(i)$
For the second photo-cathode: $hf_2 = W_0 + \frac{1}{2}mv_2^2$ ... $(ii)$
Subtracting equation $(ii)$ from equation $(i)$:
$h(f_1 - f_2) = \frac{1}{2}m(v_1^2 - v_2^2)$
Rearranging the terms to solve for the velocity difference:
$v_1^2 - v_2^2 = \frac{2h}{m}(f_1 - f_2)$

(B) According to Einstein's photoelectric equation,the stopping potential $V_0$ is given by:
$\frac{hc}{e} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) = V_0$
For the first case:
$\frac{hc}{e} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) = 4.8 \quad ...(i)$
For the second case,where wavelength is $2\lambda$:
$\frac{hc}{e} \left( \frac{1}{2\lambda} - \frac{1}{\lambda_0} \right) = 1.6 \quad ...(ii)$
Dividing equation $(i)$ by $(ii)$:
$\frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{2\lambda} - \frac{1}{\lambda_0}} = \frac{4.8}{1.6} = 3$
$\frac{1}{\lambda} - \frac{1}{\lambda_0} = 3 \left( \frac{1}{2\lambda} - \frac{1}{\lambda_0} \right)$
$\frac{1}{\lambda} - \frac{1}{\lambda_0} = \frac{3}{2\lambda} - \frac{3}{\lambda_0}$
$\frac{3}{\lambda_0} - \frac{1}{\lambda_0} = \frac{3}{2\lambda} - \frac{1}{\lambda} = \frac{1}{2\lambda}$
$\frac{2}{\lambda_0} = \frac{1}{2\lambda}$
$\lambda_0 = 4\lambda$

(B) We know that the emitter current is the sum of the collector current and the base current:
$I_{E} = I_{C} + I_{B}$
Dividing both sides by $I_{C}$:
$\frac{I_{E}}{I_{C}} = 1 + \frac{I_{B}}{I_{C}}$
Since $\alpha = \frac{I_{C}}{I_{E}}$,it follows that $\frac{1}{\alpha} = \frac{I_{E}}{I_{C}}$.
Since $\beta = \frac{I_{C}}{I_{B}}$,it follows that $\frac{1}{\beta} = \frac{I_{B}}{I_{C}}$.
Substituting these into the equation:
$\frac{1}{\alpha} = 1 + \frac{1}{\beta}$
Rearranging to solve for $\frac{1}{\beta}$:
$\frac{1}{\beta} = \frac{1}{\alpha} - 1 = \frac{1 - \alpha}{\alpha}$
Therefore,$\beta = \frac{\alpha}{1 - \alpha}$.

(C) According to the law of reflection,the angle of incidence $i$ is equal to the angle of reflection $r$,so $i = r$.
Given that the reflected ray and the refracted ray are mutually perpendicular,the sum of the angle of reflection $r$,the angle between the reflected ray and the refracted ray $(90^{\circ})$,and the angle of refraction $r'$ is $180^{\circ}$.
Thus,$r + 90^{\circ} + r' = 180^{\circ}$,which implies $r' = 90^{\circ} - r$. Since $i = r$,we have $r' = 90^{\circ} - i$.
Applying Snell's law at the interface,the refractive index of the denser medium with respect to the rarer medium is given by $\mu = \frac{\sin r'}{\sin i}$.
We know that the critical angle $C$ is related to the refractive index by $\sin C = \frac{1}{\mu}$.
Therefore,$\sin C = \frac{\sin i}{\sin r'} = \frac{\sin i}{\sin(90^{\circ} - i)} = \frac{\sin i}{\cos i} = \tan i$.
Hence,the critical angle is $C = \sin^{-1}(\tan i)$.

(C) The refractive index $\mu$ of a prism is given by the formula: $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$.
Given that the angle of minimum deviation $\delta_m = A$,we substitute this into the formula:
$\mu = \frac{\sin(\frac{A + A}{2})}{\sin(\frac{A}{2})} = \frac{\sin(A)}{\sin(\frac{A}{2})}$.
Using the trigonometric identity $\sin(A) = 2\sin(\frac{A}{2})\cos(\frac{A}{2})$,we get:
$\mu = \frac{2\sin(\frac{A}{2})\cos(\frac{A}{2})}{\sin(\frac{A}{2})} = 2\cos(\frac{A}{2})$.
Rearranging for $A$:
$\cos(\frac{A}{2}) = \frac{\mu}{2} \Rightarrow \frac{A}{2} = \cos^{-1}(\frac{\mu}{2}) \Rightarrow A = 2\cos^{-1}(\frac{\mu}{2})$.

(C) Given: Diameter of lens $D = 6 \, cm$,so radius $r = 3 \, cm$. Thickness $y = 3 \, mm = 0.3 \, cm$. Speed of light in lens $v = 2 \times 10^8 \, m/s$. Speed of light in vacuum $c = 3 \times 10^8 \, m/s$.
$1$. Calculate the refractive index $\mu$:
$\mu = \frac{c}{v} = \frac{3 \times 10^8}{2 \times 10^8} = 1.5$.
$2$. Calculate the radius of curvature $R$ of the curved surface:
From the geometry of the lens,$R^2 = r^2 + (R - y)^2$.
$R^2 = r^2 + R^2 - 2Ry + y^2$.
$2Ry = r^2 + y^2$.
Since $y$ is very small,$y^2$ can be neglected.
$R = \frac{r^2}{2y} = \frac{3^2}{2 \times 0.3} = \frac{9}{0.6} = 15 \, cm$.
$3$. Calculate the focal length $f$ using the Lens Maker's Formula:
$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a plano-convex lens,$R_1 = R = 15 \, cm$ and $R_2 = \infty$.
$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{15} - \frac{1}{\infty} \right) = 0.5 \times \frac{1}{15} = \frac{1}{30}$.
Therefore,$f = 30 \, cm$.

(A) For a prism,the relationship between the angle of deviation $(\delta)$ and the angle of incidence $(i)$ is given by the formula $\delta = (i + e) - A$,where $e$ is the angle of emergence and $A$ is the angle of the prism.
As the angle of incidence $(i)$ increases,the angle of deviation $(\delta)$ initially decreases.
It reaches a minimum value known as the angle of minimum deviation $(\delta_m)$ at a specific angle of incidence.
After this point,as the angle of incidence $(i)$ continues to increase,the angle of deviation $(\delta)$ starts to increase again.
Therefore,the graph between $\delta$ and $i$ is a parabolic-like curve that shows a minimum,which corresponds to the graph shown in option $A$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max} = eV_s = \frac{hc}{\lambda} - \phi_0$,where $\phi_0$ is the work function.
Case $(i)$: For wavelength $\lambda$,stopping potential is $3V_0$:
$3eV_0 = \frac{hc}{\lambda} - \phi_0$ ......... $(1)$
Case $(ii)$: For wavelength $2\lambda$,stopping potential is $V_0$:
$eV_0 = \frac{hc}{2\lambda} - \phi_0$ ......... $(2)$
Multiply equation $(2)$ by $3$:
$3eV_0 = \frac{3hc}{2\lambda} - 3\phi_0$ ......... $(3)$
Equating $(1)$ and $(3)$:
$\frac{hc}{\lambda} - \phi_0 = \frac{3hc}{2\lambda} - 3\phi_0$
$2\phi_0 = \frac{3hc}{2\lambda} - \frac{hc}{\lambda} = \frac{hc}{2\lambda}$
$\phi_0 = \frac{hc}{4\lambda}$
Since the threshold wavelength $\lambda_0 = \frac{hc}{\phi_0}$,substituting $\phi_0$:
$\lambda_0 = \frac{hc}{hc / 4\lambda} = 4\lambda$.

(D) For an electron of energy $E$,the de-Broglie wavelength is given by $\lambda_{e} = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.
For a photon of energy $E$,the relation is $E = \frac{hc}{\lambda_{p}}$,which implies $\lambda_{p} = \frac{hc}{E}$.
Taking the ratio of the wavelengths:
$\frac{\lambda_{e}}{\lambda_{p}} = \frac{h}{\sqrt{2mE}} \times \frac{E}{hc} = \frac{1}{c} \times \frac{E^{1}}{\sqrt{E}} \times \frac{1}{\sqrt{2m}} = \frac{1}{c} \sqrt{\frac{E}{2m}} = \frac{1}{C}(\frac{E}{2m})^{1/2}$.

(C) Let $K$ be the kinetic energy of the electron.
The de-Broglie wavelength of the electron is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$.
Squaring both sides,we get $\lambda^2 = \frac{h^2}{2mK}$,which implies $K = \frac{h^2}{2m\lambda^2}$.
The cut-off wavelength $\lambda_0$ for continuous $X$-rays is determined by the maximum energy of the photon,which equals the kinetic energy of the incident electron: $E = \frac{hc}{\lambda_0} = K$.
Substituting the value of $K$,we get $\frac{hc}{\lambda_0} = \frac{h^2}{2m\lambda^2}$.
Solving for $\lambda_0$,we find $\lambda_0 = \frac{hc \cdot 2m\lambda^2}{h^2} = \frac{2mc\lambda^2}{h}$.
Therefore,option $(C)$ is correct.

(C) The original capacitance of the parallel plate capacitor is $C_{0} = \frac{\epsilon_{0} A}{d}$.
When a dielectric slab of thickness $t = \frac{3}{4}d$ is inserted,the system can be modeled as two capacitors $C_{1}$ and $C_{2}$ in series.
$C_{1}$ is the capacitance of the part filled with dielectric: $C_{1} = \frac{K \epsilon_{0} A}{3d/4} = \frac{4 K \epsilon_{0} A}{3d}$.
$C_{2}$ is the capacitance of the air gap: $C_{2} = \frac{\epsilon_{0} A}{d - 3d/4} = \frac{\epsilon_{0} A}{d/4} = \frac{4 \epsilon_{0} A}{d}$.
Since they are in series,the equivalent capacitance $C'$ is given by $\frac{1}{C'} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$.
Substituting the values:
$\frac{1}{C'} = \frac{3d}{4 K \epsilon_{0} A} + \frac{d}{4 \epsilon_{0} A}$.
$\frac{1}{C'} = \frac{d}{4 \epsilon_{0} A} \left( \frac{3}{K} + 1 \right) = \frac{d}{4 \epsilon_{0} A} \left( \frac{3+K}{K} \right)$.
Therefore,$C' = \frac{4 K \epsilon_{0} A}{d(3+K)} = \frac{4K}{3+K} C_{0}$.

$(A)$ For a prism, the angle of deviation $\delta$ is given by $\delta = i + e - A$, where $i$ is the angle of incidence, $e$ is the angle of emergence, and $A$ is the prism angle.
At the angle of minimum deviation:
$1$. The angle of incidence is equal to the angle of emergence, i.e., $i = e$. This implies that the incident ray and emergent ray are symmetric with respect to the prism, making statement $(A)$ and $(C)$ true.
$2$. The refracted ray inside the prism becomes parallel to its base, which implies $r_1 = r_2$. This makes statement $(B)$ true.
Since statements $(A)$, $(B)$, and $(C)$ are all correct conditions for minimum deviation, the correct option is $(A)$.

(A) The relationship between the electric field $\overrightarrow{E}$ and the magnetic field $\overrightarrow{B}$ in an electromagnetic wave is given by $\overrightarrow{E} = c(\overrightarrow{B} \times \hat{n})$,where $\hat{n}$ is the direction of propagation.
Given,the wave travels along the $y$-direction,so $\hat{n} = \hat{y}$.
The magnetic field is $\overrightarrow{B} = 8.0 \times 10^{-8} \hat{z}\, T$.
Substituting the values:
$\overrightarrow{E} = (3 \times 10^{8}\, m/s) \times (8.0 \times 10^{-8} \hat{z} \times \hat{y})$
Since $\hat{z} \times \hat{y} = -\hat{x}$,we get:
$\overrightarrow{E} = (3 \times 8.0) \times (-\hat{x}) = -24 \hat{x}\, V/m$.

(C) The length of the magnet $2l = 14 \, cm$,so $l = 7 \, cm = 0.07 \, m$.
The distance of the neutral point from the center is $d = 18 \, cm = 0.18 \, m$.
The distance $r$ from each pole to the neutral point is $r = \sqrt{d^2 + l^2} = \sqrt{18^2 + 7^2} = \sqrt{324 + 49} = \sqrt{373} \, cm$.
The magnetic field due to one pole at the neutral point is $B_0 = \frac{\mu_0}{4\pi} \frac{m}{r^2}$.
At the neutral point,the horizontal component of the Earth's magnetic field $B_H$ is balanced by the resultant magnetic field of the two poles: $B_H = 2 B_0 \sin \theta$,where $\sin \theta = \frac{l}{r}$.
Substituting the values: $B_H = 2 \left( \frac{\mu_0}{4\pi} \frac{m}{r^2} \right) \left( \frac{l}{r} \right) = \frac{\mu_0}{4\pi} \frac{2ml}{r^3} = \frac{\mu_0}{4\pi} \frac{M}{r^3}$,where $M = m(2l)$ is the magnetic moment.
$0.4 \times 10^{-4} = 10^{-7} \times \frac{M}{(r \times 10^{-2})^3} = 10^{-7} \times \frac{M}{(373 \times 10^{-4})^{3/2}}$.
$M = \frac{0.4 \times 10^{-4} \times (373 \times 10^{-4})^{3/2}}{10^{-7}} = 0.4 \times 10^3 \times (373)^{3/2} \times 10^{-6} = 0.4 \times 10^{-3} \times 7203.82 \approx 2.88 \, J \, T^{-1}$.

(A) The initial resistance of the wire is $R = \frac{\rho l}{A}$.
According to the question,the new length $l' = 2l$ and the new area of cross-section $A' = \frac{A}{2}$.
The new resistance $R'$ is given by:
$R' = \rho \frac{l'}{A'} = \rho \frac{2l}{A/2} = \frac{4 \rho l}{A} = 4R$.
The new current $I'$ is given by Ohm's law:
$I' = \frac{V}{R'} = \frac{V}{4R} = \frac{V}{4(\rho l / A)} = \frac{1}{4} \frac{VA}{\rho l}$.

(D) For an antenna to be effective,its length $L$ should be a fraction of the wavelength $\lambda$ of the signal it transmits.
Typically,the minimum length of an antenna is given by $L = \frac{\lambda}{4}$.
Given the length of the antenna $L = 25\, m$.
Substituting the value into the formula: $25 = \frac{\lambda}{4}$.
Solving for $\lambda$: $\lambda = 25 \times 4 = 100\, m$.
Thus,the wavelength of the signal is $100\, m$.

(A) In an electromagnetic wave $(EMW)$,the average energy density associated with the electric field is given by $U_{e} = \frac{1}{4} \epsilon_{0} E_{0}^{2}$.
The average energy density associated with the magnetic field is given by $U_{m} = \frac{1}{4} \frac{B_{0}^{2}}{\mu_{0}}$.
Since $E_{0} = c B_{0}$ and $c = \frac{1}{\sqrt{\mu_{0} \epsilon_{0}}}$,we have $E_{0}^{2} = c^{2} B_{0}^{2} = \frac{B_{0}^{2}}{\mu_{0} \epsilon_{0}}$.
Substituting this into the expression for $U_{e}$,we get $U_{e} = \frac{1}{4} \epsilon_{0} \left( \frac{B_{0}^{2}}{\mu_{0} \epsilon_{0}} \right) = \frac{1}{4} \frac{B_{0}^{2}}{\mu_{0}} = U_{m}$.
Thus,the average energy densities due to electric and magnetic fields are equal.

(D) The circuit shown is a series $RC$ circuit where the output voltage is measured across the capacitor $C$.
When the input square wave is high (from $t_1$ to $t_2$),the capacitor charges through the resistor $R$. The voltage across the capacitor follows the exponential charging curve: $V_C(t) = V_0(1 - e^{-t/RC})$.
When the input square wave is low (from $t_2$ to $t_3$),the capacitor discharges through the resistor $R$. The voltage across the capacitor follows the exponential discharging curve: $V_C(t) = V_0 e^{-t/RC}$.
Combining these two processes,the output waveform across the capacitor will show an exponential rise followed by an exponential decay,which matches the pattern shown in Option $D$.

(D) According to Einstein's photoelectric equation,$K_{max} = h\nu - \phi_0$,where $K_{max} = eV_s$.
Thus,$eV_s = h\nu - \phi_0$,which implies $V_s = \frac{h}{e}\nu - \frac{\phi_0}{e}$.
Here,$V_s$ is the stopping potential,$h$ is Planck's constant,$\nu$ is the frequency of incident radiation,and $\phi_0$ is the work function.
Since $V_s$ is a linear function of $\nu$,the stopping potential depends on the frequency of the incident electromagnetic radiation.

(C) When a conducting bar of length $L$ moves with velocity $v$ in a magnetic field $B$ directed into the page,an induced electromotive force $(EMF)$ $\varepsilon = BvL$ is generated across the bar.
According to Fleming's Right-Hand Rule,the direction of the induced current in the moving bar is from bottom to top.
This moving bar acts as a battery with its positive terminal at the top and negative terminal at the bottom.
For the left loop containing $R_{1}$,the current $I_{1}$ flows from the top of the bar,through $R_{1}$,and back to the bottom of the bar,which corresponds to a clockwise direction.
For the right loop containing $R_{2}$,the current $I_{2}$ flows from the top of the bar,through $R_{2}$,and back to the bottom of the bar,which corresponds to an anticlockwise direction.
Therefore,$I_{1}$ is in the clockwise direction and $I_{2}$ is in the anticlockwise direction.

(A) First,calculate the equivalent resistance of the three $3\, k\Omega$ resistors connected in parallel:
$\frac{1}{R_p} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} = 1 \implies R_p = 1\, k\Omega$.
Now,the circuit consists of the $5\, k\Omega$ resistor,the equivalent resistance $R_p = 1\, k\Omega$,and the internal resistance of the battery $r = 1\, k\Omega$ all in series.
The total resistance $R_{eq} = 5\, k\Omega + 1\, k\Omega + 1\, k\Omega = 7\, k\Omega$.
The total current $I$ flowing through the circuit is given by Ohm's law:
$I = \frac{V}{R_{eq}} = \frac{21\, V}{7\, k\Omega} = 3\, mA$.
Since the $5\, k\Omega$ resistor is in series with the rest of the circuit,the same current $I = 3\, mA$ flows through it.

(B) The formula for fringe width in Young's Double Slit Experiment is given by $\beta = \frac{\lambda D}{d}$.
Here,$\beta = 6 \ mm = 6 \times 10^{-3} \ m$,$d = 1 \ mm = 1 \times 10^{-3} \ m$,and $D = 10 \ m$.
Rearranging the formula to solve for wavelength $\lambda$:
$\lambda = \frac{\beta d}{D} = \frac{(6 \times 10^{-3} \ m) \times (1 \times 10^{-3} \ m)}{10 \ m}$.
$\lambda = \frac{6 \times 10^{-6}}{10} \ m = 6 \times 10^{-7} \ m$.
To convert this to nanometers $(nm)$,we multiply by $10^9$:
$\lambda = 6 \times 10^{-7} \times 10^9 \ nm = 600 \ nm$.
Thus,the value of $x$ is $600$.

(D) For the $1^{\text{st}}$ line of the Balmer series $(n_1 = 2, n_2 = 3)$:
$\frac{1}{\lambda_{1}} = R \left(\frac{1}{2^{2}} - \frac{1}{3^{2}}\right) = R \left(\frac{1}{4} - \frac{1}{9}\right) = R \left(\frac{5}{36}\right) \implies \lambda_{1} = \frac{36}{5R}$.
For the $3^{\text{rd}}$ line of the Balmer series $(n_1 = 2, n_2 = 5)$:
$\frac{1}{\lambda_{3}} = R \left(\frac{1}{2^{2}} - \frac{1}{5^{2}}\right) = R \left(\frac{1}{4} - \frac{1}{25}\right) = R \left(\frac{21}{100}\right) \implies \lambda_{3} = \frac{100}{21R}$.
Calculating the ratio $\frac{\lambda_{1}}{\lambda_{3}}$:
$\frac{\lambda_{1}}{\lambda_{3}} = \left(\frac{36}{5R}\right) \times \left(\frac{21R}{100}\right) = \frac{36 \times 21}{500} = \frac{756}{500} = 1.512$.
Expressing as $x \times 10^{-1}$:
$1.512 = 15.12 \times 10^{-1}$.
Rounding to the nearest integer,$x \approx 15$.

(A) $1$. The voltage across the series resistor $R_{S}$ is $V_{R_{S}} = V_{in} - V_{Z} = 22 \text{ V} - 15 \text{ V} = 7 \text{ V}$.
$2$. The total current flowing through the series resistor $R_{S}$ is $I = \frac{V_{R_{S}}}{R_{S}} = \frac{7 \text{ V}}{35 \Omega} = 0.2 \text{ A} = \frac{1}{5} \text{ A}$.
$3$. The current flowing through the load resistor $R_{L} = 90 \Omega$ is $I_{L} = \frac{V_{Z}}{R_{L}} = \frac{15 \text{ V}}{90 \Omega} = \frac{1}{6} \text{ A}$.
$4$. The current flowing through the Zener diode is $I_{Z} = I - I_{L} = \frac{1}{5} \text{ A} - \frac{1}{6} \text{ A} = \frac{6 - 5}{30} \text{ A} = \frac{1}{30} \text{ A}$.
$5$. The power dissipated across the Zener diode is $P = V_{Z} \times I_{Z} = 15 \text{ V} \times \frac{1}{30} \text{ A} = 0.5 \text{ W}$.
$6$. Expressing this in the form $x \times 10^{-1} \text{ W}$, we get $0.5 \text{ W} = 5 \times 10^{-1} \text{ W}$.
$7$. Therefore, the value of $x$ is $5$.

(B) At resonance,the impedance of the $LCR$ circuit is purely resistive,i.e.,$Z = R = 8\, \Omega$.
The peak voltage is $V_0 = 250\, V$. The root mean square $(RMS)$ voltage is given by $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{250}{\sqrt{2}}\, V$.
The power dissipated at resonance is given by the formula $P = \frac{(V_{rms})^2}{R}$.
Substituting the values: $P = \frac{(250 / \sqrt{2})^2}{8} = \frac{62500 / 2}{8} = \frac{31250}{8} = 3906.25\, W$.
Converting to $kW$: $P = 3.90625\, kW$.
Rounding to the nearest integer,we get $x = 4$.

(A) Given inputs are $A = 0$ and $B = 1$.
$1$. The input $A=0$ goes to the $NOR$ gate and also to a $NOT$ gate. The output of this $NOT$ gate is $1$.
$2$. The input $B=1$ goes to a $NOT$ gate,so its output is $0$. This $0$ is fed to the $NAND$ gate and also to another $NOT$ gate,which outputs $1$.
$3$. The $NOR$ gate receives $A=0$ and the output of the second $NOT$ gate,which is $1$. Thus,the $NOR$ gate output is $\overline{0+1} = \overline{1} = 0$.
$4$. The $NAND$ gate receives the output of the first $NOT$ gate $(1)$ and the output of the $NOT$ gate connected to $B$ $(0)$. Thus,the $NAND$ gate output is $\overline{1 \cdot 0} = \overline{0} = 1$.
$5$. Finally,the $AND$ gate receives the outputs of the $NOR$ gate $(0)$ and the $NAND$ gate $(1)$. The output $Y$ is $0 \cdot 1 = 0$.
Therefore,$x = 0$.

(B) According to Bohr's theory of the hydrogen atom,the velocity $(v_n)$ of an electron in the $n^{\text{th}}$ orbit is given by the formula:
$v_n = \frac{2 \pi k Z e^2}{n h}$
Where:
$k$ is Coulomb's constant,
$Z$ is the atomic number (for hydrogen,$Z=1$),
$e$ is the charge of the electron,
$h$ is Planck's constant,
$n$ is the principal quantum number (orbit number).
Since $2, \pi, k, Z, e^2,$ and $h$ are constants,the velocity is inversely proportional to the orbit number $n$.
Therefore,$v_n \propto \frac{1}{n}$.

(B) The given current is $I = I_{1} \sin \omega t + I_{2} \cos \omega t$.
This can be rewritten in the form $I = I_{0} \sin(\omega t + \phi)$,where $I_{0}$ is the peak current.
The amplitude $I_{0}$ is given by $\sqrt{I_{1}^{2} + I_{2}^{2}}$.
$A$ hot wire ammeter measures the root mean square $(RMS)$ value of the current.
The $RMS$ value is defined as $I_{rms} = \frac{I_{0}}{\sqrt{2}}$.
Substituting the value of $I_{0}$,we get $I_{rms} = \sqrt{\frac{I_{1}^{2} + I_{2}^{2}}{2}}$.

(A) The magnetic flux density $B$ inside a solenoid is given by the formula $B = \mu n I$,where $\mu = \mu_0 \mu_r$.
Given:
Number of turns per unit length $n = 1000 \, m^{-1} = 10^3 \, m^{-1}$.
Relative permeability $\mu_r = 500$.
Current $I = 5 \, A$.
Permeability of free space $\mu_0 = 4 \pi \times 10^{-7} \, H/m$.
Substituting these values into the formula:
$B = (4 \pi \times 10^{-7}) \times 500 \times 1000 \times 5$
$B = 4 \pi \times 10^{-7} \times 500 \times 10^3 \times 5$
$B = 4 \pi \times 10^{-7} \times 2.5 \times 10^6$
$B = 10 \pi \times 10^{-1} = \pi \, T$.
Therefore,the magnetic flux density is $\pi \, T$.

(B) $1$. The first two gates are $NOT$ gates (since inputs $A$ and $B$ are connected to both terminals of the $NOR$ gates). Thus,the outputs are $\bar{A}$ and $\bar{B}$.
$2$. These outputs are fed into a $NOR$ gate. The output of this gate is $\overline{\bar{A} + \bar{B}}$.
$3$. By De Morgan's theorem,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.
$4$. This result $(A \cdot B)$ is then fed into a final $NOT$ gate (a $NOR$ gate with shorted inputs). The final output $Y$ is $\overline{A \cdot B}$.
$5$. The expression $\overline{A \cdot B}$ represents a $NAND$ gate.

(B) The energy of an electron in a hydrogen-like atom is given by the formula $E_n = -13.6 \frac{Z^2}{n^2} \text{ eV}$.
For the hydrogen atom,the atomic number $Z = 1$ and for the ground state,$n = 1$.
Thus,the ground state energy of hydrogen is $E_H = -13.6 \times \frac{1^2}{1^2} = -13.6 \text{ eV}$.
For a singly ionized carbon atom $(C^+)$,the atomic number $Z = 6$.
We want to find the energy level $n$ such that the energy $E_C$ equals the ground state energy of hydrogen:
$-13.6 \frac{6^2}{n^2} = -13.6$.
Dividing both sides by $-13.6$,we get $\frac{36}{n^2} = 1$.
Therefore,$n^2 = 36$,which gives $n = 6$.

(B) Given: Current $i = 10\, A$,Area $A = 5\, mm^{2} = 5 \times 10^{-6}\, m^{2}$,Drift velocity $v_{d} = 2 \times 10^{-3}\, m/s$,and charge of an electron $e = 1.6 \times 10^{-19}\, C$.
We know the relation between current and drift velocity is $i = neAv_{d}$.
Rearranging for the number density $n$,we get $n = \frac{i}{eAv_{d}}$.
Substituting the values: $n = \frac{10}{1.6 \times 10^{-19} \times 5 \times 10^{-6} \times 2 \times 10^{-3}}$.
$n = \frac{10}{16 \times 10^{-28}} = 0.625 \times 10^{28} = 625 \times 10^{25}\, m^{-3}$.

(A) The distance to the horizon $d$ for an antenna of height $h$ is given by the formula $d = \sqrt{2Rh}$,where $R$ is the radius of the earth.
The service area $A$ covered by the antenna is the area of a circle with radius $d$,given by $A = \pi d^{2}$.
Substituting the expression for $d$,we get $A = \pi (\sqrt{2Rh})^{2} = 2\pi Rh$.
Given $R = 6400\, km = 6400 \times 10^{3}\, m$ and $h = 30\, m$.
Converting $R$ to $km$ to match the required unit for area: $R = 6400\, km$ and $h = 30\, m = 0.03\, km$.
$A = 3.14 \times 2 \times 6400 \times 0.03$.
$A = 3.14 \times 384 = 1205.76\, km^{2}$.
Rounding off to the nearest integer,we get $A \simeq 1206\, km^{2}$.

(D) Let the capacitance of each pair of adjacent plates be $C_{0} = \frac{\varepsilon_{0} A}{d}$,where $A = l \times b = 2 \times \frac{3}{2} = 3 \, cm^{2}$.
From the figure,plates $B$ and $D$ are connected together.
The capacitor is formed between plates $A$ and $B$,$B$ and $C$,and $C$ and $D$.
Since $B$ and $D$ are connected,the plates $B$ and $D$ are at the same potential.
The circuit consists of a capacitor between $A$ and $B$ (let this be $C_{1}$),and two capacitors in parallel between $B$ and $C$ (one between $B$ and $C$,and one between $C$ and $D$).
Thus,$C_{eq} = C_{AB} \text{ in series with } (C_{BC} \parallel C_{CD})$.
$C_{eq} = \frac{C_{0} \times (C_{0} + C_{0})}{C_{0} + (C_{0} + C_{0})} = \frac{C_{0} \times 2C_{0}}{3C_{0}} = \frac{2}{3} C_{0}$.
Substituting $C_{0} = \frac{\varepsilon_{0} A}{d} = \frac{3 \varepsilon_{0}}{d}$,we get $C_{eq} = \frac{2}{3} \times \frac{3 \varepsilon_{0}}{d} = \frac{2 \varepsilon_{0}}{d}$.
Comparing this with $\frac{x \varepsilon_{0}}{d}$,we find $x = 2$.

(D) Initial energy stored in the capacitor is given by $U_i = \frac{1}{2} C V^2$.
Substituting the values: $U_i = \frac{1}{2} \times 14 \, pF \times (12 \, V)^2 = \frac{1}{2} \times 14 \times 144 = 1008 \, pJ$.
When the battery is disconnected,the charge $Q$ on the plates remains constant.
After inserting the dielectric slab of constant $k = 7$,the new capacitance becomes $C' = kC = 7 \times 14 \, pF = 98 \, pF$.
The new energy stored in the capacitor is $U_f = \frac{Q^2}{2C'} = \frac{Q^2}{2(kC)} = \frac{U_i}{k}$.
$U_f = \frac{1008 \, pJ}{7} = 144 \, pJ$.
The loss in electrical potential energy is converted into the mechanical energy of the slab.
Mechanical energy $= U_i - U_f = 1008 \, pJ - 144 \, pJ = 864 \, pJ$.

(D) For a non-reflecting surface (complete absorption) with normal incidence,the force $F$ exerted by light is given by $F = \frac{P}{c}$,where $P$ is the power and $c$ is the speed of light.
Since power $P = I \times A$,where $I$ is the intensity (energy flux) and $A$ is the area,the force equation becomes $F = \frac{I A}{c}$.
Rearranging for intensity $I$,we get $I = \frac{F c}{A}$.
Given values: $F = 2.5 \times 10^{-6} \ N$,$A = 30 \ cm^{2}$,and $c = 3 \times 10^{8} \ m/s$.
Substituting these values: $I = \frac{2.5 \times 10^{-6} \times 3 \times 10^{8}}{30} = \frac{7.5 \times 10^{2}}{30} = \frac{750}{30} = 25 \ W/cm^{2}$.
The energy flux is $25 \ W/cm^{2}$.

(B) For the oil drop to be stationary,the electric force must balance the gravitational force: $qE = Mg$.
Here,$q = ne$,where $n$ is the number of excess electrons and $e = 1.6 \times 10^{-19} \, C$.
The mass $M = \text{density} (\rho) \times \text{volume} (V) = \rho \times \frac{4}{3} \pi r^3$.
Given: $\rho = 3 \, g \, cm^{-3} = 3000 \, kg \, m^{-3}$,$r = 2 \, mm = 2 \times 10^{-3} \, m$,$E = 3.55 \times 10^{5} \, V \, m^{-1}$,$g = 9.81 \, m \, s^{-2}$.
Substituting the values: $n \times (1.6 \times 10^{-19}) \times (3.55 \times 10^{5}) = 3000 \times \frac{4}{3} \times \pi \times (2 \times 10^{-3})^3 \times 9.81$.
$n \times 5.68 \times 10^{-14} = 4000 \times 3.14159 \times 8 \times 10^{-9} \times 9.81$.
$n \times 5.68 \times 10^{-14} = 9.833 \times 10^{-4}$.
$n = \frac{9.833 \times 10^{-4}}{5.68 \times 10^{-14}} \approx 1.73 \times 10^{10}$.

(A) The atmosphere is stratified based on altitude as follows:
$1$. Troposphere: $0$ to $12\, km$. Thus,$(a) \rightarrow (iv)$.
$2$. Stratosphere: $12$ to $50\, km$. Thus,$(b) \rightarrow (iii)$.
$3$. Mesosphere: $50$ to $85\, km$. Thus,$(c) \rightarrow (ii)$.
$4$. Thermosphere: Above $85\, km$. Thus,$(d) \rightarrow (i)$.
Therefore,the correct matching is $(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)$.

(B) The energy levels of a hydrogen-like atom are given by $E_n = -\frac{m e^4}{8 \epsilon_0^2 n^2 h^2}$,where $m$ is the mass of the orbiting particle.
Since $E \propto m$,the ionization potential $IP$ is directly proportional to the mass of the particle.
For a standard hydrogen atom,$IP_e = 13.6 \ eV$.
When the electron is replaced by a muon with mass $m_{\mu} = 207 m_e$,the new ionization potential $IP_{\mu}$ is given by:
$IP_{\mu} = IP_e \times \frac{m_{\mu}}{m_e}$
$IP_{\mu} = 13.6 \ eV \times 207$
$IP_{\mu} = 2815.2 \ eV$.

(C) The magnitude of the electric field $E$ is related to the magnetic field $B$ by the relation $E = B \cdot c$,where $c$ is the speed of light in vacuum $(c = 3 \times 10^8 \, m/s)$.
Given $B = 2.0 \times 10^{-8} \, T$,we have $E = (2.0 \times 10^{-8} \, T) \times (3 \times 10^8 \, m/s) = 6.0 \, V/m$.
The direction of propagation of an electromagnetic wave is given by the direction of the vector $\overrightarrow{E} \times \overrightarrow{B}$.
Here,the wave travels along the $x$-direction $(\hat{i})$,and $\overrightarrow{B}$ is along the $z$-direction $(\hat{k})$.
Since $\hat{i} = \hat{j} \times \hat{k}$,the electric field $\overrightarrow{E}$ must be along the $y$-direction $(\hat{j})$.
Therefore,$\overrightarrow{E} = 6.0 \hat{j} \, V/m$.

(D) The magnetic field $B$ at the center of a long solenoid is given by $B = \mu_0 n i$,where $n$ is the number of turns per unit length and $i$ is the current flowing through it.
Since the solenoids are identical,$n$ is the same for all.
Thus,$B \propto i$.
Let the total current entering the circuit be $I$. This current $I$ flows through solenoid $A$.
At the junction,the current $I$ splits into three identical parallel branches containing solenoids $B, C,$ and $D$.
Since the solenoids are identical,their resistances are equal,and the current $I$ divides equally among the three branches.
Therefore,the current through solenoid $C$ is $i_C = \frac{I}{3}$.
Given that the magnetic field at the center of $A$ is $B_A = 3\, T$,we have $B_A \propto I$,so $3\, T \propto I$.
The magnetic field at the center of $C$ is $B_C \propto i_C = \frac{I}{3}$.
Therefore,$B_C = \frac{B_A}{3} = \frac{3\, T}{3} = 1\, T$.

(C) For a radioactive sample undergoing two independent decay processes with decay constants $\lambda_1$ and $\lambda_2$,the total decay constant is $\lambda_{eq} = \lambda_1 + \lambda_2$.
Since the decay constant $\lambda$ is related to the half-life $T_{1/2}$ by the formula $\lambda = \frac{\ln 2}{T_{1/2}}$,we can write:
$\frac{\ln 2}{T_{1/2}} = \frac{\ln 2}{T_{1/2}^{(1)}} + \frac{\ln 2}{T_{1/2}^{(2)}}$.
Dividing both sides by $\ln 2$,we get:
$\frac{1}{T_{1/2}} = \frac{1}{T_{1/2}^{(1)}} + \frac{1}{T_{1/2}^{(2)}}$.
Solving for $T_{1/2}$,we find:
$T_{1/2} = \frac{T_{1/2}^{(1)} T_{1/2}^{(2)}}{T_{1/2}^{(1)} + T_{1/2}^{(2)}}$.

(B) The fringe width $\beta$ is given by the formula $\beta = \frac{\lambda D}{d}$.
Given: $\lambda = 5890 \, \text{Å} = 5890 \times 10^{-10} \, m$, $D = 0.5 \, m$, and $d = 0.5 \, mm = 0.5 \times 10^{-3} \, m$.
Substituting the values: $\beta = \frac{5890 \times 10^{-10} \times 0.5}{0.5 \times 10^{-3}} = 5890 \times 10^{-7} \, m = 589 \times 10^{-6} \, m$.
The distance between the $n^{th}$ and $m^{th}$ bright fringe is $(n-m) \beta$.
For the first and third bright fringe, the distance is $(3-1) \beta = 2 \beta$.
Distance $= 2 \times 589 \times 10^{-6} \, m = 1178 \times 10^{-6} \, m$.

(D) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.
Given the ratio of the de-Broglie wavelength of the particle $(p)$ to the electron $(e)$ is $\frac{\lambda_p}{\lambda_e} = 2:1$.
Also,the velocity of the particle is $v_p = 4v_e$.
Using the formula $\frac{\lambda_p}{\lambda_e} = \frac{m_e v_e}{m_p v_p}$,we substitute the known values:
$2 = \frac{m_e v_e}{m_p (4v_e)}$
Simplifying the equation:
$2 = \frac{m_e}{4m_p}$
Solving for $m_p$:
$m_p = \frac{m_e}{4 \times 2} = \frac{m_e}{8}$.
Therefore,the mass of the particle is $\frac{1}{8}$ times the mass of the electron.

(A) The resonance frequency of an $LCR$ circuit is given by $\omega_0 = 1 / \sqrt{LC}$. Since it depends only on $L$ and $C$,it remains constant when resistance $R$ is changed.
The bandwidth of an $LCR$ circuit is defined as $\Delta \omega = R / L$.
Since the bandwidth $\Delta \omega$ is directly proportional to the resistance $R$ $(\Delta \omega \propto R)$,increasing the resistance $R$ will increase the bandwidth.
The quality factor $Q$ is given by $Q = \omega_0 L / R = 1 / R \sqrt{L/C}$. Since $Q \propto 1/R$,increasing the resistance $R$ will decrease the quality factor.
Therefore,the correct statement is that the bandwidth of the resonance circuit will increase.

(A) The instantaneous current in an $AC$ circuit is given by $i = i_{0} \cos(\omega t)$,where $i_{0}$ is the peak current.
At $t = 0$,the current is at its maximum value,$i = i_{0}$.
The $rms$ value of the current is $i_{rms} = \frac{i_{0}}{\sqrt{2}}$.
We need to find the time $t$ when $i = \frac{i_{0}}{\sqrt{2}}$.
Setting $\cos(\omega t) = \frac{1}{\sqrt{2}}$,we get $\omega t = \frac{\pi}{4}$.
Substituting $\omega = 2\pi f$,we have $2\pi f t = \frac{\pi}{4}$.
Solving for $t$,$t = \frac{1}{8f}$.
Given $f = 50\, Hz$,$t = \frac{1}{8 \times 50} = \frac{1}{400}\, s$.
$t = 0.0025\, s = 2.5\, ms$.

(B) The inability to see distant objects clearly is a symptom of $Myopia$ (nearsightedness).
However,the specific observation that a uniform mesh appears distorted or non-uniform is the characteristic symptom of $Astigmatism$.
$Astigmatism$ occurs due to the irregular curvature of the cornea or lens,causing light to focus at different points on the retina rather than a single point.
Since the question describes both the inability to see distant objects clearly and the distortion of the image,the diagnosis is $Myopia$ with $Astigmatism$.

(A) When a flexible current-carrying loop is placed in an external magnetic field,every infinitesimal segment $(d\ell)$ of the wire experiences a magnetic force given by $dF = i(d\ell \times B)$.
This force acts perpendicular to both the current direction and the magnetic field.
Due to the symmetry of the magnetic force acting radially outward (or inward depending on the current direction),the wire experiences a uniform tension that forces it to expand into a circular shape to maximize the area enclosed by the loop,with its plane oriented normal to the magnetic field lines.

(B) The potential difference across the capacitor at $t = 0$ is given by $V = \frac{q}{C}$.
Given $q = 30\, \mu C$ and $C = 3\, \mu F$,we have $V = \frac{30\, \mu C}{3\, \mu F} = 10\, V$.
When the key is closed,the capacitor discharges through the resistor $R = 5\, M\Omega = 5 \times 10^6\, \Omega$.
The initial current $i_0$ flowing through the circuit at $t = 0$ is given by $i_0 = \frac{V}{R}$.
Substituting the values,$i_0 = \frac{10\, V}{5 \times 10^6\, \Omega} = 2 \times 10^{-6}\, A$.
Since $1\, \mu A = 10^{-6}\, A$,the current $i_0 = 2\, \mu A$.
Thus,the value of $'x'$ is $2$.