JEE Main 2026 Physics Question Paper with Answer and Solution

(C) The force constant $K$ of a spring is inversely proportional to its length $\ell$,i.e.,$K \ell = \text{constant}$.
Let the original length be $\ell$ and the original force constant be $K = 15 \ N/m$.
The spring is cut into two pieces with lengths in the ratio $1 : 3$. Thus,the lengths of the pieces are $\ell_1 = \frac{\ell}{4}$ and $\ell_2 = \frac{3\ell}{4}$.
The smaller piece has length $\ell_1 = \frac{\ell}{4}$.
Using the relation $K \ell = K^{\prime} \ell_1$,we get:
$15 \times \ell = K^{\prime} \times (\frac{\ell}{4})$
$K^{\prime} = 15 \times 4 = 60 \ N/m$.

(D) The heat supplied to the water is given by $Q = m S \Delta T$.
Given $m = 4 \ kg$,$S = 4200 \ J/kg \cdot K$,and $\Delta T = 20 - 4 = 16 \ K$.
$Q = 4 \times 4200 \times 16 = 268800 \ J$.
The work done by the water during expansion is $W = P \Delta V$.
The change in volume is $\Delta V = m(\frac{1}{\rho_f} - \frac{1}{\rho_i}) = 4(\frac{1}{998} - \frac{1}{1000}) = 4(\frac{1000 - 998}{998000}) = 4(\frac{2}{998000}) = \frac{8}{998000} \ m^3$.
Given $P = 10^5 \ Pa$,the work done is $W = 10^5 \times \frac{8}{998000} = \frac{800000}{998000} \approx 0.8016 \ J$.
According to the first law of thermodynamics,$\Delta U = Q - W$.
$\Delta U = 268800 - 0.8016 = 268799.1984 \ J \approx 268799.2 \ J$.

(A) At equilibrium,the weight of the block is balanced by the buoyant force:
$\rho A h g = M g$
where $h$ is the submerged depth.
When the block is depressed by a small distance $x$,the additional buoyant force acting upwards is $F_b = \rho A x g$.
Since this force acts in the direction opposite to the displacement,the restoring force is $F = -\rho A x g$.
Using Newton's second law,$M a = -\rho A x g$,which gives the acceleration $a = -\left(\frac{\rho A g}{M}\right) x$.
Comparing this with the standard $SHM$ equation $a = -\omega^2 x$,we get $\omega^2 = \frac{\rho A g}{M}$,so $\omega = \sqrt{\frac{\rho A g}{M}}$.
The period of oscillation $T$ is given by $T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{M}{\rho A g}}$.

(B) For an isobaric process,the work done is given by $W = P \Delta V = nR \Delta T = 100 \ J$.
For a diatomic gas,the degree of freedom $f = 5$.
The molar heat capacity at constant pressure is $C_p = (\frac{f}{2} + 1)R = (\frac{5}{2} + 1)R = \frac{7}{2}R$.
The heat supplied to the gas is $Q = nC_p \Delta T = n(\frac{7}{2}R) \Delta T$.
Substituting $nR \Delta T = 100 \ J$,we get $Q = \frac{7}{2} \times 100 = 350 \ J$.

(B) The terminal velocity $v_T$ of a spherical body falling through a viscous fluid is given by the formula: $v_T = \frac{2}{9} \frac{r^2 g (\rho - \sigma)}{\eta}$.
Here,$r$ is the radius of the ball,$\rho$ is the density of the material of the ball,$\sigma$ is the density of the fluid,$\eta$ is the coefficient of viscosity,and $g$ is the acceleration due to gravity.
Since the material of the ball and the fluid are the same,$v_T \propto r^2$.
Given: $r_1 = 6 \ mm$,$v_{T1} = 20 \ cm/s$,and $r_2 = 3 \ mm$.
Using the proportionality: $\frac{v_{T2}}{v_{T1}} = (\frac{r_2}{r_1})^2$.
Substituting the values: $\frac{v_{T2}}{20} = (\frac{3}{6})^2 = (\frac{1}{2})^2 = \frac{1}{4}$.
Therefore,$v_{T2} = \frac{20}{4} = 5 \ cm/s$.

(C) Given: $m = 2 \ kg$,$x(t) = t^2 + t + 1 \ m$.
The velocity is $v(t) = \frac{dx}{dt} = 2t + 1 \ m/s$.
The acceleration is $a(t) = \frac{dv}{dt} = 2 \ m/s^2$.
Since the acceleration is constant,the force acting on the body is $F = m \times a = 2 \ kg \times 2 \ m/s^2 = 4 \ N$.
The displacement $s$ during the interval $t = 2 \ s$ to $t = 3 \ s$ is $s = x(3) - x(2)$.
$x(3) = (3)^2 + 3 + 1 = 9 + 3 + 1 = 13 \ m$.
$x(2) = (2)^2 + 2 + 1 = 4 + 2 + 1 = 7 \ m$.
$s = 13 - 7 = 6 \ m$.
The work done is $W = F \times s = 4 \ N \times 6 \ m = 24 \ J$.

(D) For the body of mass $M$ to remain stuck to the inner wall of the rotating drum,the normal force $N$ exerted by the wall provides the necessary centripetal force:
$N = M \omega^2 R$
The frictional force $f$ acts vertically upwards to balance the gravitational force $Mg$ acting downwards:
$f = Mg$
Since the body is on the verge of slipping,the frictional force is at its maximum value,given by $f = \mu N$:
$Mg = \mu N$
Substituting the expression for $N$ into the friction equation:
$Mg = \mu (M \omega^2 R)$
Dividing both sides by $M$ and solving for $\omega$:
$g = \mu \omega^2 R$
$\omega^2 = \frac{g}{\mu R}$
$\omega = \sqrt{\frac{g}{\mu R}}$

(C) Given: Width of the river $d = 200 \ m$,velocity of river $v_R = 18 \ km/h = 18 \times \frac{5}{18} = 5 \ m/s$,velocity of boat in still water $v_{BR} = 36 \ km/h = 36 \times \frac{5}{18} = 10 \ m/s$.
To cross the river in minimum time,the boat must head perpendicular to the river flow.
The time taken to cross the river once is $t = \frac{d}{v_{BR}} = \frac{200}{10} = 20 \ s$.
For a round trip (going and coming back),the total time taken is $T = 20 + 20 = 40 \ s$.
During this time,the boat is carried downstream by the river flow. The velocity of the boat relative to the ground along the river bank is $v_R = 5 \ m/s$.
The displacement along the river bank is $x = v_R \times T = 5 \times 40 = 200 \ m$.

(A) The formula for capillary rise is given by $h = \frac{2T \cos \theta}{\rho g r}$.
Assuming the contact angle $\theta$ is the same for both liquids and the densities $\rho$ are equal,we have $h \propto \frac{1}{r}$.
Given $r_1 > r_2$,it follows that $\frac{1}{r_1} < \frac{1}{r_2}$.
Therefore,$h_1 < h_2$.
Since the problem specifies $T_1 = T_2$,the correct relation is $h_1 < h_2$ and $T_1 = T_2$.

(B) The terminal velocity $v_0$ of a spherical body is given by Stokes' Law: $v_0 = \frac{2}{9} \frac{r^2 g}{\eta} (\sigma - \rho)$.
Rearranging the formula to solve for viscosity $\eta$: $\eta = \frac{2}{9} \frac{r^2 g}{v_0} (\sigma - \rho)$.
Since $\sigma, \rho$,and $g$ are constants with negligible errors,the relative error in $\eta$ is determined by the variables $r$ and $v_0$.
Using the rules of error propagation for multiplication and division,the maximum relative error is: $\frac{\Delta \eta}{\eta} = 2 \frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}$.

(D) To find the sum of physical quantities,we add the given values: $52.01 \ m + 153.2 \ m + 0.123 \ m = 205.333 \ m$.
According to the rules of significant figures for addition,the final result should be reported to the same number of decimal places as the measurement with the fewest decimal places.
The values are $52.01$ (two decimal places),$153.2$ (one decimal place),and $0.123$ (three decimal places).
The measurement with the fewest decimal places is $153.2 \ m$,which has one decimal place.
Therefore,we round the sum $205.333 \ m$ to one decimal place,which gives $205.3 \ m$.

(A) The angular frequency of the kinetic energy oscillation is given as $\omega_k = 176 \ rad/s$.
For a simple harmonic oscillator,the kinetic energy $K = \frac{1}{2} m v^2 = \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t + \phi)$.
Using the identity $\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$,we get $K = \frac{1}{4} m \omega^2 A^2 (1 - \cos(2\omega t + 2\phi))$.
This shows that the kinetic energy oscillates with an angular frequency $\omega_k = 2\omega$,where $\omega$ is the angular frequency of the oscillator.
Given $\omega_k = 176 \ rad/s$,we have $2\omega = 176 \ rad/s$,which implies $\omega = 88 \ rad/s$.
The frequency of the oscillator is $f = \frac{\omega}{2\pi} = \frac{88}{2 \times (22/7)} = \frac{88 \times 7}{44} = 2 \times 7 = 14 \ Hz$.

(C) Let the radius of the rim be $r$. The moment of inertia $I$ of the pulley consists of the rim and two rods.
$I = I_{\text{rim}} + 2 \times I_{\text{rod}}$
$I = Mr^2 + 2 \times \left( \frac{M(2r)^2}{12} \right) = Mr^2 + 2 \times \left( \frac{4Mr^2}{12} \right) = Mr^2 + \frac{2}{3}Mr^2 = \frac{5}{3}Mr^2$.
Let $a$ be the acceleration of the blocks and $T_1, T_2$ be the tensions in the string.
The equations of motion are:
$Mg - T_2 = Ma$ ... $(1)$
$T_1 - mg = ma$ ... $(2)$
$(T_2 - T_1)r = I \alpha = I \left( \frac{a}{r} \right) \implies T_2 - T_1 = \frac{I}{r^2} a = \frac{5}{3}Ma$ ... $(3)$
Adding $(1)$,$(2)$,and $(3)$:
$(M - m)g = (M + m + \frac{5}{3}M)a$
$(M - m)g = (\frac{8}{3}M + m)a$
$a = \frac{(M - m)g}{\frac{8}{3}M + m}$.

(B) The angular momentum of a particle relative to another particle is given by $L = m \cdot v_{\text{rel}} \cdot r_{\perp}$.
Here,the mass of car $A$ is $m = 10^3 \text{ kg}$.
The relative velocity of car $A$ with respect to car $B$ is $v_{\text{rel}} = v_A - v_B = 72 \text{ km/h} - 36 \text{ km/h} = 36 \text{ km/h}$.
Converting the relative velocity to $SI$ units: $v_{\text{rel}} = 36 \times \frac{5}{18} \text{ m/s} = 10 \text{ m/s}$.
The perpendicular distance between the tracks is $r_{\perp} = 10 \text{ m}$.
Substituting these values into the formula: $L = 1000 \times 10 \times 10 = 10^5 \text{ kg} \cdot \text{m}^2/\text{s}$ (or $\text{J} \cdot \text{s}$).

(C) The formula for r.m.s. speed is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$.
Given that $V_{rms, O_2} = V_{rms, H_2}$.
Temperature of oxygen $T_{O_2} = 273 + 47 = 320 \ K$.
Equating the speeds: $\sqrt{\frac{3RT_{O_2}}{M_{O_2}}} = \sqrt{\frac{3RT_{H_2}}{M_{H_2}}}$.
Squaring both sides and simplifying: $\frac{T_{O_2}}{M_{O_2}} = \frac{T_{H_2}}{M_{H_2}}$.
Substituting the values: $\frac{320}{32} = \frac{T_{H_2}}{2}$.
$10 = \frac{T_{H_2}}{2} \implies T_{H_2} = 20 \ K$.
Converting to Celsius: $T(^{\circ}C) = 20 - 273 = -253^{\circ}C$.

(C) Using the ideal gas equation,$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
Given:
$P_1 = 2\,atm$
$V_1 = 60\,cm^{3}$
$T_1 = 27^{\circ}C = 27 + 273 = 300\,K$
$V_2 = 20\,cm^{3}$
$T_2 = 77^{\circ}C = 77 + 273 = 350\,K$
Substituting the values into the equation:
$\frac{2 \times 60}{300} = \frac{P_2 \times 20}{350}$
$\frac{120}{300} = \frac{P_2 \times 20}{350}$
$0.4 = P_2 \times \frac{20}{350}$
$P_2 = 0.4 \times \frac{350}{20} = 0.4 \times 17.5 = 7\,atm$.

(B) Using the principle of conservation of mechanical energy,the loss in potential energy of the $2m$ mass equals the gain in kinetic energy of the system.
Loss in potential energy of $2m$ mass = $(2m)gh$.
Gain in potential energy of $m$ mass = $mgh$.
Net loss in potential energy = $(2m)gh - mgh = mgh$.
This energy is converted into the kinetic energy of the two masses and the rotational kinetic energy of the pulley.
Total kinetic energy = $K_{m} + K_{2m} + K_{pulley} = \frac{1}{2}mv^2 + \frac{1}{2}(2m)v^2 + \frac{1}{2}I\omega^2$.
Since $I = \frac{1}{2}Mr^2 = \frac{1}{2}(30m)r^2 = 15mr^2$ and $\omega = \frac{v}{r}$,we have $K_{pulley} = \frac{1}{2}(15mr^2)(\frac{v^2}{r^2}) = 7.5mv^2$.
Equating energy: $mgh = \frac{1}{2}mv^2 + mv^2 + 7.5mv^2 = 9mv^2$.
$v^2 = \frac{gh}{9} = \frac{10 \times 3.6}{9} = 4$.
$v = 2 \ m/s$.

(A) Statement $I$ is true by definition: the total kinetic energy of a system of particles is the scalar sum of the kinetic energies of individual particles,$K = \sum \frac{1}{2} m_i v_i^2$.
Statement $II$ is also true based on the theorem of kinetic energy for a system of particles. If $\vec{v}_i$ is the velocity of the $i$-th particle relative to the origin,$\vec{V}_{cm}$ is the velocity of the center of mass,and $\vec{v}_i'$ is the velocity of the $i$-th particle relative to the center of mass,then $\vec{v}_i = \vec{V}_{cm} + \vec{v}_i'$.
The total kinetic energy is $K = \sum \frac{1}{2} m_i v_i^2 = \sum \frac{1}{2} m_i (\vec{V}_{cm} + \vec{v}_i')^2 = \frac{1}{2} M V_{cm}^2 + \sum \frac{1}{2} m_i (v_i')^2$,where $M = \sum m_i$.
Thus,the total kinetic energy is the sum of the kinetic energy of the center of mass and the kinetic energy relative to the center of mass.

(D) Statement-$I$: Incorrect.
The work done by a force $\vec{F}$ is defined as $W = \int_{r_{1}}^{r_{2}} \vec{F} \cdot d\vec{r}$. The potential energy change $\Delta U$ is defined as $-\int_{r_{1}}^{r_{2}} \vec{F} \cdot d\vec{r}$. Thus,the given expression for work is incorrect.
Statement-$II$: Incorrect.
$A$ conservative force is defined as a force for which the work done is independent of the path taken between two points. Therefore,the work done does not change with the path followed.

(B) The time period of a satellite orbiting close to the Earth's surface is given by $T = 2\pi\sqrt{\frac{R_e^3}{GM}}$.
Since the mass of the Earth $M = \rho \cdot \frac{4}{3}\pi R_e^3$,where $\rho$ is the density of the Earth,we can substitute $M$ into the formula:
$T = 2\pi\sqrt{\frac{R_e^3}{G(\rho \cdot \frac{4}{3}\pi R_e^3)}} = 2\pi\sqrt{\frac{3}{4\pi G\rho}}$.
This shows that $T$ depends on the density $\rho$ of the Earth. Thus,Statement $I$ is true.
For a satellite very close to the surface,the gravitational force provides the centripetal force,leading to $g = \frac{GM}{R_e^2}$.
Substituting $GM = gR_e^2$ into the general time period formula $T = 2\pi\sqrt{\frac{R_e^3}{GM}}$,we get $T = 2\pi\sqrt{\frac{R_e^3}{gR_e^2}} = 2\pi\sqrt{\frac{R_e}{g}}$.
Thus,Statement $II$ is true.

(D) The height $h$ of a liquid in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{\rho gr}$.
Assuming the angle of contact $\theta$ and acceleration due to gravity $g$ remain constant, the relative change is given by: $\frac{\Delta h}{h} = \frac{\Delta T}{T} - \frac{\Delta \rho}{\rho} - \frac{\Delta r}{r}$.
Given that $T$, $\rho$, and $r$ each decrease by $1\%$, we have $\frac{\Delta T}{T} = -0.01$, $\frac{\Delta \rho}{\rho} = -0.01$, and $\frac{\Delta r}{r} = -0.01$.
Substituting these values: $\frac{\Delta h}{h} = (-0.01) - (-0.01) - (-0.01) = -0.01 + 0.01 + 0.01 = +0.01$.
Therefore, the height will change by $+1\%$.

(B) For an open organ pipe,the frequency of the $n^{\text{th}}$ harmonic is given by $f_n = n \left( \frac{v}{2L} \right)$,where $v$ is the speed of sound and $L$ is the length of the pipe.
Given $v_3 = 3 \left( \frac{v}{2L} \right)$ and $v_6 = 6 \left( \frac{v}{2L} \right)$.
According to the problem,$v_6 - v_3 = 2200 \text{ Hz}$.
Substituting the expressions: $6 \left( \frac{v}{2L} \right) - 3 \left( \frac{v}{2L} \right) = 2200$.
$3 \left( \frac{v}{2L} \right) = 2200$.
Given $v = 330 \text{ m/s}$,we have $3 \left( \frac{330}{2L} \right) = 2200$.
$990 / (2L) = 2200$.
$2L = 990 / 2200 = 0.45 \text{ m}$.
$L = 0.225 \text{ m} = 225 \text{ mm}$.

(D) Let the radius of each small bubble be $r$ and the charge on each be $q$. The potential of each small bubble is $V_i = \frac{kq}{r}$.
When three such bubbles coalesce,the volume is conserved. Let the radius of the bigger bubble be $R$.
$3 \times (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3 \implies R^3 = 3r^3 \implies R = 3^{1/3}r$.
The total charge on the bigger bubble is $Q = 3q$. The potential of the bigger bubble is $V_f = \frac{kQ}{R} = \frac{k(3q)}{3^{1/3}r}$.
The ratio of the potentials is $\frac{V_i}{V_f} = \frac{kq/r}{3kq / (3^{1/3}r)} = \frac{1}{3 / 3^{1/3}} = \frac{3^{1/3}}{3} = \frac{1}{3^{1 - 1/3}} = \frac{1}{3^{2/3}}$.

(A) Let the center of mass of the rod be the origin. The moment of inertia of the rod about its center is $I_{rod} = \frac{(20m)(12)^2}{12} = 240m \text{ cm}^2$.
Using the conservation of angular momentum about the center of mass of the rod:
$L_i = L_f$
Initial angular momentum $L_i = (2m)(v)(2) + (m)(v)(4) = 4mv + 4mv = 8mv$.
The final moment of inertia of the system is $I_f = I_{rod} + I_{mass1} + I_{mass2} = 240m + (2m)(2)^2 + (m)(4)^2 = 240m + 8m + 16m = 264m \text{ cm}^2$.
Since $L_f = I_f \omega$,we have $8mv = 264m \omega$.
Therefore,$\frac{v}{\omega} = \frac{264}{8} = 33$.

(B) The collision frequency $(Z)$ is given by the formula: $Z = \sqrt{2} \pi d^2 N \bar{v}$,where $\bar{v} = \sqrt{\frac{8RT}{\pi M}}$.
Since temperature $(T)$ and number density $(N)$ are the same for both gases,we have $Z \propto d^2 \sqrt{\frac{1}{M}}$.
Given: $d_A = \frac{d_B}{2}$ and $M_A = 4M_B$.
Therefore,the ratio of collision frequencies is:
$\frac{Z_A}{Z_B} = \left( \frac{d_A}{d_B} \right)^2 \sqrt{\frac{M_B}{M_A}}$
$\frac{Z_A}{Z_B} = \left( \frac{1}{2} \right)^2 \sqrt{\frac{M_B}{4M_B}} = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$.
Given $Z_B = 32 \times 10^{18} /s$,we get:
$Z_A = \frac{32 \times 10^{18}}{8} = 4 \times 10^{18} /s$.

(B) The time period $T$ of a simple pendulum is given by the formula: $T = 2\pi \sqrt{\frac{L}{g}}$.
Squaring both sides,we get: $T^2 = \frac{4\pi^2 L}{g}$.
Taking the reciprocal of both sides,we get: $\frac{1}{T^2} = \frac{g}{4\pi^2 L}$.
This equation is of the form $y = \frac{k}{x}$,where $y = \frac{1}{T^2}$,$x = L$,and $k = \frac{g}{4\pi^2}$ is a constant.
This represents a rectangular hyperbola,which corresponds to the plot shown in option $B$.

(C) The expression is $\frac{\epsilon_0 E}{t}$.
We know that $\epsilon_0 E$ represents the electric displacement field $D$,which has the same dimensions as surface charge density $\sigma = \frac{q}{A}$.
Thus,the dimensions of $\epsilon_0 E$ are $[I T L^{-2}]$.
Dividing by time $t$ (dimension $[T]$),we get:
$\frac{[I T L^{-2}]}{[T]} = [I L^{-2}]$.
Since the dimension of current $I$ is $A$ (Ampere) and length $L$ is $m$ (meter),the unit is $A / m^2$.

(C) The zero of the circular scale lies $3$ divisions above the pitch line,which indicates a negative zero error.
Zero error $e = -3 \times LC = -3 \times 0.01 \ mm = -0.03 \ mm$.
The observed reading is given by: $\text{Pitch scale reading} + (\text{Circular scale reading} \times LC)$.
Observed reading $= 1 \ mm + 51 \times 0.01 \ mm = 1 \ mm + 0.51 \ mm = 1.51 \ mm$.
The correct reading is calculated as: $\text{Correct reading} = \text{Observed reading} - \text{Zero error}$.
Correct reading $= 1.51 \ mm - (-0.03 \ mm) = 1.51 \ mm + 0.03 \ mm = 1.54 \ mm$.

(B) The loss in kinetic energy during a perfectly inelastic collision is given by: $\Delta K = \frac{1}{2} \left( \frac{m_1 m_2}{m_1 + m_2} \right) (v_1 + v_2)^2$.
Substituting the given values: $m_1 = 15 \ kg$,$m_2 = 25 \ kg$,$v_1 = 10 \ m/s$,$v_2 = 30 \ m/s$.
$\Delta K = \frac{1}{2} \left( \frac{15 \times 25}{15 + 25} \right) (10 + 30)^2 = \frac{1}{2} \left( \frac{375}{40} \right) (40)^2 = \frac{1}{2} \times 375 \times 40 = 7500 \ J$.
This energy is converted into heat: $Q = (m_1 + m_2) S \Delta T$.
Given $S = 31 \ cal/kg \cdot ^{\circ}C = 31 \times 4.2 \ J/kg \cdot ^{\circ}C = 130.2 \ J/kg \cdot ^{\circ}C$.
$7500 = (15 + 25) \times 130.2 \times \Delta T$.
$7500 = 40 \times 130.2 \times \Delta T$.
$\Delta T = \frac{7500}{5208} \approx 1.44^{\circ}C$.

(A) The angular momentum $L$ is given by $L = I\omega$,where $I$ is the moment of inertia and $\omega$ is the angular velocity. Thus,$\omega = \frac{L}{I}$.
First,locate the centre of mass $(CM)$ of the system. Let the mass $m$ be at $x = 0$ and $2m$ be at $x = d$. The position of the $CM$ is $x_{cm} = \frac{m(0) + 2m(d)}{m + 2m} = \frac{2md}{3m} = \frac{2d}{3}$.
The distance of mass $m$ from the $CM$ is $r_1 = \frac{2d}{3}$,and the distance of mass $2m$ from the $CM$ is $r_2 = d - \frac{2d}{3} = \frac{d}{3}$.
The moment of inertia $I$ about the axis passing through the $CM$ is $I = m(r_1)^2 + 2m(r_2)^2 = m(\frac{2d}{3})^2 + 2m(\frac{d}{3})^2$.
$I = m(\frac{4d^2}{9}) + 2m(\frac{d^2}{9}) = \frac{4md^2 + 2md^2}{9} = \frac{6md^2}{9} = \frac{2}{3}md^2$.
Substituting $I$ into the formula for angular velocity: $\omega = \frac{L}{\frac{2}{3}md^2} = \frac{3L}{2md^2}$.

(D) At point $A$, the kinetic energy is $(KE)_A = K = \frac{1}{2}mu^2$.
At the highest point $B$, the vertical component of velocity is zero, so the velocity is $v_B = u \cos 60^{\circ} = \frac{u}{2}$.
The kinetic energy at point $B$ is $(KE)_B = \frac{1}{2}m(\frac{u}{2})^2 = \frac{1}{4}(\frac{1}{2}mu^2) = \frac{K}{4}$.
Since points $A$ and $C$ are at the same horizontal level, the speed at $C$ is equal to the speed at $A$. Thus, $(KE)_C = (KE)_A = K$.
The difference in kinetic energies at points $B$ and $C$ is $|(KE)_C - (KE)_B| = |K - \frac{K}{4}| = \frac{3K}{4}$.
The ratio of this difference to the kinetic energy at point $A$ is $\frac{3K/4}{K} = \frac{3}{4}$.

(A) Young's modulus $(Y)$ is defined as the ratio of stress to strain: $Y = \frac{\text{Stress}}{\text{Strain}}$.
In the given graph,strain is plotted on the $y$-axis and stress is plotted on the $x$-axis.
Therefore,the slope of the graph is $\text{Slope} = \frac{\text{Strain}}{\text{Stress}} = \frac{1}{Y}$.
To have the largest Young's modulus $(Y)$,the value of $\frac{1}{Y}$ must be the smallest.
This means the slope of the strain-stress graph should be the minimum.
Looking at the figure,material $C$ has the smallest slope.
Thus,material $C$ has the largest Young's modulus.

(D) Let the mass of each cylinder be $M' = M/4$.
Consider the two cylinders perpendicular to the axis (labeled $I_1$ in the diagram). The axis passes through their centers perpendicular to their length. The moment of inertia of one such cylinder is $I_1 = \frac{M'R^2}{4} + \frac{M'L^2}{12}$.
Consider the two cylinders parallel to the axis (labeled $I_2$ in the diagram). The axis is at a distance $L/2$ from their centers. The moment of inertia of one such cylinder about its own center (axis along length) is $\frac{M'R^2}{2}$. By the parallel axis theorem,$I_2 = \frac{M'R^2}{2} + M'(L/2)^2 = \frac{M'R^2}{2} + \frac{M'L^2}{4}$.
The total moment of inertia is $I_{net} = 2I_1 + 2I_2 = 2(\frac{M'R^2}{4} + \frac{M'L^2}{12}) + 2(\frac{M'R^2}{2} + \frac{M'L^2}{4})$.
$I_{net} = \frac{M'R^2}{2} + \frac{M'L^2}{6} + M'R^2 + \frac{M'L^2}{2} = \frac{3}{2}M'R^2 + \frac{2}{3}M'L^2$.
Substituting $M' = M/4$: $I_{net} = \frac{3}{2}(M/4)R^2 + \frac{2}{3}(M/4)L^2 = \frac{3}{8}MR^2 + \frac{1}{6}ML^2$.

(C) For a spring-mass system in equilibrium,the spring force balances the gravitational force: $kx = mg$,where $x$ is the extension of the spring.
Thus,the extension is $x = \frac{mg}{k}$.
The elastic potential energy $U$ stored in the spring is given by $U = \frac{1}{2}kx^2$.
Substituting the value of $x$,we get $U = \frac{1}{2}k\left(\frac{mg}{k}\right)^2 = \frac{m^2g^2}{2k}$.
From this expression,we see that $U \propto \frac{m^2}{k}$.
Given $m_{A} = 100 \text{ g}$,$m_{B} = 200 \text{ g}$,and $4k_{A} = 3k_{B}$,we can write the ratio of energies as:
$\frac{U_{A}}{U_{B}} = \left(\frac{m_{A}}{m_{B}}\right)^2 \cdot \left(\frac{k_{B}}{k_{A}}\right)$.
Substituting the values: $\frac{E}{U_{B}} = \left(\frac{100}{200}\right)^2 \cdot \left(\frac{4}{3}\right) = \left(\frac{1}{2}\right)^2 \cdot \left(\frac{4}{3}\right) = \frac{1}{4} \cdot \frac{4}{3} = \frac{1}{3}$.
Therefore,$U_{B} = 3E$.

(C) The time period $T$ of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$,which implies $T \propto \sqrt{\ell}$.
Given that the time duration $t$ is constant,the number of oscillations $n$ is inversely proportional to the time period $T$ $(n = \frac{t}{T})$.
Therefore,$n \propto \frac{1}{\sqrt{\ell}}$,which means $n^2 \propto \frac{1}{\ell}$ or $\ell \propto \frac{1}{n^2}$.
Initially,$n_1 = 20$ and $\ell_1 = 30 \ cm$.
Finally,$n_2 = 40$.
Using the ratio: $\frac{\ell_2}{\ell_1} = \left( \frac{n_1}{n_2} \right)^2$.
$\ell_2 = 30 \times \left( \frac{20}{40} \right)^2 = 30 \times \left( \frac{1}{2} \right)^2 = 30 \times \frac{1}{4} = 7.5 \ cm$.

(C) Step $1$: Calculate the mean length of the rod:
$\ell_{\text{mean}} = \frac{20.00 + 19.75 + 17.01 + 18.25}{4} = \frac{75.01}{4} = 18.7525 \ cm \approx 18.75 \ cm$.
Step $2$: Calculate the absolute errors for each measurement:
$|\Delta \ell_1| = |20.00 - 18.75| = 1.25 \ cm$
$|\Delta \ell_2| = |19.75 - 18.75| = 1.00 \ cm$
$|\Delta \ell_3| = |17.01 - 18.75| = 1.74 \ cm$
$|\Delta \ell_4| = |18.25 - 18.75| = 0.50 \ cm$
Step $3$: Calculate the mean absolute error:
$\Delta \ell_{\text{mean}} = \frac{1.25 + 1.00 + 1.74 + 0.50}{4} = \frac{4.49}{4} = 1.1225 \ cm \approx 1.12 \ cm$.
Step $4$: Calculate the relative error:
$\text{Relative error} = \frac{\Delta \ell_{\text{mean}}}{\ell_{\text{mean}}} = \frac{1.12}{18.75} \approx 0.06$.

(B) $1$. Let $l = 1 \ m$ be the length of the rod. The velocity $V_A$ of bob $A$ at the lowest point is given by the conservation of mechanical energy:
$mgl(1 - \cos \theta) = \frac{1}{2} m V_A^2$
$V_A = \sqrt{2gl(1 - \cos 60^{\circ})} = \sqrt{2 \times 10 \times 1 \times (1 - 0.5)} = \sqrt{10} \ m/s$.
$2$. Since the collision between identical bobs $A$ and $B$ is elastic,the velocities are exchanged. Thus,after the collision,bob $B$ moves with velocity $V_B = V_A = \sqrt{10} \ m/s$.
$3$. For bob $B$ to just complete a vertical circular path of radius $R$,the minimum velocity at the bottom of the circular track must be $V_{min} = \sqrt{5gR}$.
$4$. Equating the velocities: $\sqrt{10} = \sqrt{5gR} \implies 10 = 5 \times 10 \times R \implies 10 = 50R \implies R = \frac{10}{50} = \frac{1}{5} \ m$.

(C) The moment of inertia of a circular disc about its central axis is given by $I = \frac{1}{2} M R^2$.
Since the discs are made of the same material,their density $\rho$ is the same.
The mass of a disc is $M = \text{Volume} \times \text{density} = (\pi R^2 T) \rho$.
Therefore,the moment of inertia is $I = \frac{1}{2} (\pi R^2 T \rho) R^2 = \frac{1}{2} \pi \rho R^4 T$.
Given that $I_1 = I_2$,we have:
$\frac{1}{2} \pi \rho R_1^4 T_1 = \frac{1}{2} \pi \rho R_2^4 T_2$
$R_1^4 T_1 = R_2^4 T_2$
$\frac{T_1}{T_2} = \frac{R_2^4}{R_1^4} = \left( \frac{R_2}{R_1} \right)^4$.
Given $\frac{R_1}{R_2} = 2$,so $\frac{R_2}{R_1} = \frac{1}{2}$.
$\frac{T_1}{T_2} = \left( \frac{1}{2} \right)^4 = \frac{1}{16}$.
Comparing this with $\frac{T_1}{T_2} = \frac{1}{\alpha}$,we get $\alpha = 16$.

(B) Let the distance of point $A$ from the line joining the speakers be $D = 40 \ m$. The speakers $L_1$ and $L_2$ are separated by $10 \ m$,so their coordinates relative to the midpoint of the line joining them are $(0, 5)$ and $(0, -5)$. Point $A$ is at $(40, 0)$.
When the recorder moves to point $B$ at a distance of $25 \ m$ from $A$,its coordinates are $(40, 25)$.
The distances from the speakers to point $B$ are:
$L_1B = \sqrt{40^2 + (25-5)^2} = \sqrt{40^2 + 20^2} = \sqrt{1600 + 400} = \sqrt{2000} = 20\sqrt{5} \ m$.
Given $\sqrt{5} = 2.23$,$L_1B = 20 \times 2.23 = 44.6 \ m$.
$L_2B = \sqrt{40^2 + (25+5)^2} = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \ m$.
The path difference at point $B$ is $\Delta x = L_2B - L_1B = 50 - 44.6 = 5.4 \ m$.
Since the recorder undergoes $10$ cycles of minima and maxima,point $B$ corresponds to the $10^{\text{th}}$ maxima,so $\Delta x = n\lambda$,where $n = 10$.
$5.4 = 10 \times \lambda \implies \lambda = 0.54 \ m$.
The frequency $f$ is given by $f = \frac{v}{\lambda} = \frac{324}{0.54} = 600 \ Hz$.

$(D)$. Spring constant $(k)$: From $F = kx$, we have $[k] = [F]/[x] = [MLT^{-2}]/[L] = [ML^0 T^{-2}]$. Thus, $A-II$.
$B$. Thermal conductivity $(k)$: From $dQ/dt = kA(\Delta T/l)$, we have $[k] = [ML^2 T^{-3}][L]/([L^2][K]) = [MLT^{-3} K^{-1}]$. Thus, $B-IV$.
$C$. Boltzmann constant $(k_B)$: From $E = (3/2)k_B T$, we have $[k_B] = [E]/[T] = [ML^2 T^{-2}]/[K] = [ML^2 T^{-2} K^{-1}]$. Thus, $C-I$.
$D$. Inductive reactance $(X_L)$: $X_L = \omega L$. The dimension is the same as resistance $(R = V/I)$. $[R] = [ML^2 T^{-3} A^{-2}]$. Thus, $D-III$.
Therefore, the correct matching is $A-II, B-IV, C-I, D-III$.

(D) For a reversible adiabatic process, the relation between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = \text{constant}$.
Let the initial state be $(T_1, V_1)$ and the final state be $(T_2, V_2)$.
Given: $V_2 = 8V_1$ and $T_2 = T_1/4$.
Using the adiabatic relation: $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Substituting the values: $T_1 V_1^{\gamma-1} = (T_1/4) (8V_1)^{\gamma-1}$.
Dividing both sides by $T_1 V_1^{\gamma-1}$: $1 = (1/4) \cdot 8^{\gamma-1}$.
$4 = 8^{\gamma-1}$.
Expressing in base $2$: $2^2 = (2^3)^{\gamma-1} = 2^{3\gamma-3}$.
Equating the exponents: $2 = 3\gamma - 3$.
$3\gamma = 5$, which gives $\gamma = 5/3$.
A gas with adiabatic index $\gamma = 5/3$ is a monoatomic gas.
Among the given options, Helium (He) is a monoatomic gas. Therefore, the correct option is $D$.

(A) Let the thermal resistance of rod $x$ be $R = \frac{\ell}{k_x A}$. Since the thermal conductivity of $x$ is $3$ times that of $y$ $(k_x = 3k_y)$,the thermal resistance of rod $y$ is $3R$.
From the figure,the network between $B$ and $E$ consists of two parallel branches: one with rods $y$ and $y$ (total resistance $3R + 3R = 6R$) and one with rods $x$ and $x$ (total resistance $R + R = 2R$).
The equivalent resistance $R_{BE}$ between $B$ and $E$ is $\frac{1}{R_{BE}} = \frac{1}{6R} + \frac{1}{2R} = \frac{1+3}{6R} = \frac{4}{6R} = \frac{2}{3R}$,so $R_{BE} = 1.5R$.
The total resistance of the circuit is $R_{total} = R_{AB} + R_{BE} + R_{EF} = R + 1.5R + 3R = 5.5R = \frac{11R}{2}$.
The total heat current $H = \frac{100 - 40}{5.5R} = \frac{60}{5.5R} = \frac{120}{11R}$.
Now,$T_B = 100 - H \cdot R = 100 - \frac{120}{11R} \cdot R = 100 - 10.91 \approx 89.09^{\circ}C$.
And $T_E = 40 + H \cdot (3R) = 40 + \frac{120}{11R} \cdot 3R = 40 + 32.73 \approx 72.73^{\circ}C$.
Thus,$T_B \approx 89^{\circ}C$ and $T_E \approx 73^{\circ}C$.

(A) Consider a small element of gas of length $dx$ at a distance $x$ from the axis of rotation $A$. Let $A$ be the cross-sectional area of the tube.
The net force on this element providing the centripetal acceleration is $(P+dP)A - PA = (dm) \omega^2 x$.
$AdP = (dm) \omega^2 x$.
Since $dm = \rho A dx$,we have $dP = \rho \omega^2 x dx$.
Using the ideal gas law $PM = \rho RT$,we have $\rho = \frac{PM}{RT}$.
Substituting $\rho$,we get $dP = \left(\frac{PM}{RT}\right) \omega^2 x dx$.
Rearranging the terms,$\frac{dP}{P} = \frac{M \omega^2}{RT} x dx$.
Integrating from $x=0$ to $x=l$ and $P=P_A$ to $P=P_B$:
$\int_{P_A}^{P_B} \frac{dP}{P} = \frac{M \omega^2}{RT} \int_0^l x dx$.
$\ln\left(\frac{P_B}{P_A}\right) = \frac{M \omega^2}{RT} \left[\frac{x^2}{2}\right]_0^l = \frac{M \omega^2 l^2}{2RT}$.
Therefore,$P_B = P_A \exp\left(\frac{M \omega^2 l^2}{2RT}\right)$.

(A) Let the distance from the centre to each corner be $r$. Let a test mass $m_{0}$ be placed at the centre.
In the initial configuration,the forces exerted by the masses at the corners on $m_{0}$ are directed towards the corners. Let $k = \frac{Gm_{0}}{r^{2}}$.
The forces are $F_{M} = kM$,$F_{2M} = 2kM$,$F_{3M} = 3kM$,and $F_{4M} = 4kM$.
Opposite pairs are $(M, 3M)$ and $(2M, 4M)$.
The net force along the diagonal with $M$ and $3M$ is $F_{diag1} = (3M - M)k = 2kM$ (towards $3M$).
The net force along the diagonal with $2M$ and $4M$ is $F_{diag2} = (4M - 2M)k = 2kM$ (towards $4M$).
Since the diagonals are perpendicular,$F_{1} = \sqrt{(2kM)^{2} + (2kM)^{2}} = 2\sqrt{2}kM$.
In the new configuration,$3M$ and $4M$ are interchanged. The corners now have $M, 2M, 4M,$ and $3M$ in order.
Opposite pairs are $(M, 4M)$ and $(2M, 3M)$.
The net force along the diagonal with $M$ and $4M$ is $F'_{diag1} = (4M - M)k = 3kM$ (towards $4M$).
The net force along the diagonal with $2M$ and $3M$ is $F'_{diag2} = (3M - 2M)k = kM$ (towards $3M$).
$F_{2} = \sqrt{(3kM)^{2} + (kM)^{2}} = \sqrt{9+1}kM = \sqrt{10}kM$.
Ratio $\frac{F_{1}}{F_{2}} = \frac{2\sqrt{2}kM}{\sqrt{10}kM} = \frac{2\sqrt{2}}{\sqrt{2}\sqrt{5}} = \frac{2}{\sqrt{5}}$.
Comparing with $\frac{\alpha}{\sqrt{5}}$,we get $\alpha = 2$.

(D) According to Pascal's law,the pressure at any point in a fluid at rest is the same in all directions.
Fluid pressure exists at every point within the liquid,not just at the boundaries or on solid surfaces in contact. Therefore,Statement $I$ is false.
Regarding Statement $II$,molecules at the surface of a liquid have fewer neighbors than those in the interior,leading to a net inward force. This results in higher potential energy for surface molecules compared to interior molecules,which is the physical origin of surface tension. Therefore,Statement $II$ is true.

(D) In projectile motion,the horizontal component of velocity remains constant throughout the flight.
Let the initial speed be $u$. The horizontal component of the initial velocity is $u_x = u \cos 60^{\circ}$.
Given that at a certain point,the speed is $v = 20 \text{ m/s}$ at an angle of $45^{\circ}$ with the horizontal,the horizontal component of this velocity is $v_x = v \cos 45^{\circ} = 20 \cos 45^{\circ} = 20 \times \frac{1}{\sqrt{2}} = \frac{20}{\sqrt{2}} \text{ m/s}$.
Since $u_x = v_x$,we have:
$u \cos 60^{\circ} = \frac{20}{\sqrt{2}}$
$u \times \frac{1}{2} = \frac{20}{\sqrt{2}}$
$u = \frac{40}{\sqrt{2}} = 20\sqrt{2} \text{ m/s}$.

(C) The formula for escape velocity is $V_{e} = \sqrt{\frac{2GM}{R}}$. Substituting mass $M = \rho \times \frac{4}{3}\pi R^{3}$,we get $V_{e} = \sqrt{\frac{2G \times \rho \times 4\pi R^{3}}{3R}} = R \sqrt{\frac{8\pi G \rho}{3}}$.
Thus,$V_{e} \propto R\sqrt{\rho}$.
Given that for planet $B$,$\rho_{B} = 0.1 \rho_{A}$ and $R_{B} = 0.1 R_{A}$.
The ratio of escape velocities is $\frac{(V_{e})_{B}}{(V_{e})_{A}} = \frac{R_{B}}{R_{A}} \times \sqrt{\frac{\rho_{B}}{\rho_{A}}} = (0.1) \times \sqrt{0.1} = \frac{1}{10} \times \frac{1}{\sqrt{10}} = \frac{1}{10\sqrt{10}}$.
Given $(V_{e})_{A} = 10 \ km/s = 10000 \ m/s$.
Therefore,$(V_{e})_{B} = 10000 \times \frac{1}{10\sqrt{10}} = \frac{1000}{\sqrt{10}} = 100\sqrt{10} \ m/s$.

(D) The moment of inertia of a solid sphere about its tangent is given by the parallel axis theorem: $I = I_{cm} + MR^2 = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$.
For the system of two spheres,the total moment of inertia about the common tangent passing through the point of contact is the sum of the moments of inertia of each sphere about that same axis.
$I_{total} = I_1 + I_2 = \frac{7}{5}m_1R_1^2 + \frac{7}{5}m_2R_2^2 = \frac{7}{5}[m_1R_1^2 + m_2R_2^2]$.
Given: $m_1 = 5 \ kg$,$R_1 = 0.1 \ m$; $m_2 = 10 \ kg$,$R_2 = 0.2 \ m$.
$I = \frac{7}{5} [5 \times (0.1)^2 + 10 \times (0.2)^2]$.
$I = \frac{7}{5} [5 \times 0.01 + 10 \times 0.04] = \frac{7}{5} [0.05 + 0.4] = \frac{7}{5} [0.45]$.
$I = 7 \times 0.09 = 0.63 \ kg \cdot m^{2}$.

(C) The change in internal energy $\Delta U$ for an ideal gas is given by the formula: $\Delta U = n C_v \Delta T$.
Given that the process occurs at constant volume,$C_v = C_p - R$.
Substituting the given values: $C_v = 7 - 2 = 5 \text{ cal/mol}^{\circ} C$.
Number of moles $n = 10 \text{ mol}$.
Change in temperature $\Delta T = 40^{\circ} C - 30^{\circ} C = 10^{\circ} C$.
Therefore,$\Delta U = 10 \times 5 \times 10 = 500 \text{ cal}$.

(B) Let the vertical rod be $1$ and the horizontal rod be $2$.
For rod $1$,the moment of inertia about an axis passing through $P$ (its end) and perpendicular to its length is $I_1 = \frac{ML^2}{3}$.
For rod $2$,the moment of inertia about its center of mass is $I_{cm} = \frac{ML^2}{12}$. Using the parallel axis theorem,the moment of inertia about an axis passing through $P$ and perpendicular to the plane is $I_2 = I_{cm} + Md^2$,where $d = L$ is the distance from $P$ to the center of rod $2$.
$I_2 = \frac{ML^2}{12} + M(L)^2 = \frac{13ML^2}{12}$.
The total moment of inertia is $I = I_1 + I_2 = \frac{ML^2}{3} + \frac{13ML^2}{12} = \frac{4ML^2 + 13ML^2}{12} = \frac{17ML^2}{12}$.
Comparing this with $\frac{x}{12} ML^2$,we get $x = 17$.

(B) In steady state,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor.
The circuit simplifies to a $10 \ V$ battery connected to a $1 \ \Omega$ resistor in series with a parallel combination of two $8 \ \Omega$ resistors.
The equivalent resistance of the two $8 \ \Omega$ resistors in parallel is $R_p = (8 \ \Omega \times 8 \ \Omega) / (8 \ \Omega + 8 \ \Omega) = 4 \ \Omega$.
The total resistance of the circuit is $R_{eq} = 1 \ \Omega + 4 \ \Omega = 5 \ \Omega$.
The total current drawn from the battery is $I = V / R_{eq} = 10 \ V / 5 \ \Omega = 2 \ A$.
The ammeter is placed in the return path of the circuit. Since the capacitor branch is open,the current $I = 2 \ A$ flows through the $1 \ \Omega$ resistor and then splits equally into the two $8 \ \Omega$ resistors. The ammeter measures the current flowing through the right-hand side of the circuit,which is the current through the rightmost $8 \ \Omega$ resistor. Since the current $I = 2 \ A$ splits equally into the two $8 \ \Omega$ resistors,the current through each is $I' = I / 2 = 2 \ A / 2 = 1 \ A$.
Therefore,the reading of the ammeter is $1 \ A$.

(C) The Zener diode is used for voltage regulation. The power dissipation $P_D$ of the Zener diode is given by $P_D = V_Z \times i$,where $V_Z$ is the breakdown voltage and $i$ is the current through the diode.
Given $P_D = 0.4 \ W$ and $V_Z = 10 \ V$,we have:
$0.4 = 10 \times i$
$i = \frac{0.4}{10} = 0.04 \ A$
The protective resistance $R$ is connected in series with the Zener diode. The voltage drop across the resistance $R$ is $V_R = V_{source} - V_Z = 15 \ V - 10 \ V = 5 \ V$.
Using Ohm's law,$V_R = i \times R$,we get:
$R = \frac{V_R}{i} = \frac{5 \ V}{0.04 \ A} = 125 \ \Omega$.

(C) The potential at any point on a conducting spherical shell is the sum of the potentials due to all the charges present on the shells.
For a point at distance $r$ from the center,the potential due to a shell of radius $R$ and charge $q$ is $\frac{kq}{R}$ if $r \le R$ and $\frac{kq}{r}$ if $r > R$.
For sphere $A$ (radius $a$): It is inside $B$ and $C$,so the potential is $V_A = \frac{kq_1}{a} + \frac{kq_2}{b} + \frac{kq_3}{c} = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{a} + \frac{q_2}{b} + \frac{q_3}{c} \right)$.
For sphere $B$ (radius $b$): It is on the surface of $B$,inside $C$,and outside $A$. So,$V_B = \frac{kq_1}{b} + \frac{kq_2}{b} + \frac{kq_3}{c} = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1+q_2}{b} + \frac{q_3}{c} \right)$.
For sphere $C$ (radius $c$): It is on the surface of $C$,and outside $A$ and $B$. So,$V_C = \frac{kq_1}{c} + \frac{kq_2}{c} + \frac{kq_3}{c} = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1+q_2+q_3}{c} \right)$.

(A) The electrostatic potential is given by $V = ar^3 + b$.
The electric field $E$ is related to the potential by $E = -\frac{dV}{dr}$.
Differentiating $V$ with respect to $r$,we get $E = -\frac{d}{dr}(ar^3 + b) = -3ar^2$.
According to Gauss's Law,the total charge $q_{enc}$ enclosed in a sphere of radius $r=1$ is given by $q_{enc} = \epsilon_0 \oint E \cdot dA$.
For a spherical surface of radius $r=1$,the area $A = 4\pi r^2 = 4\pi(1)^2 = 4\pi$.
Substituting the values at $r=1$,$E = -3a(1)^2 = -3a$.
Thus,$q_{enc} = \epsilon_0 \times (-3a) \times 4\pi = -12\pi a \epsilon_0$.
Comparing this with the given expression $\alpha \times \pi a \epsilon_0$,we find $\alpha = -12$.

(D) Statement $I$ is false because the binding energy per nucleon is not a monotonically increasing function of the mass number. It increases initially,reaches a maximum for iron $(Fe)$,and then decreases for heavier nuclei.
Statement $II$ is true because nuclei with lower binding energy per nucleon are less stable. Through processes like nuclear fission (for heavy nuclei) or nuclear fusion (for light nuclei),they tend to transform into more stable nuclei with higher binding energy per nucleon to achieve a lower energy state.

(B) Initially,for a meter bridge,the balance condition is given by $\frac{P}{Q} = \frac{l}{100-l}$.
Given $P = 2 \Omega$ and $Q = 3 \Omega$,we have $\frac{2}{3} = \frac{l}{100-l}$.
$2(100-l) = 3l \Rightarrow 200 - 2l = 3l \Rightarrow 5l = 200 \Rightarrow l = 40 \text{ cm}$.
When an unknown resistor $R$ is connected in parallel with $3 \Omega$,the new resistance $Q'$ becomes $\frac{3R}{3+R}$.
The null point shifts by $22.5 \text{ cm}$ toward $Y$,so the new length $l' = 40 + 22.5 = 62.5 \text{ cm}$.
The new balance condition is $\frac{2}{Q'} = \frac{62.5}{100-62.5} = \frac{62.5}{37.5} = \frac{5}{3}$.
Substituting $Q'$,we get $\frac{2}{\frac{3R}{3+R}} = \frac{5}{3} \Rightarrow \frac{2(3+R)}{3R} = \frac{5}{3}$.
$6(3+R) = 15R \Rightarrow 18 + 6R = 15R \Rightarrow 9R = 18 \Rightarrow R = 2 \Omega$.

(C) Radio waves are produced by the rapid acceleration and deceleration of electrons in aerials.
Microwaves are produced by special vacuum tubes like klystrons, magnetrons, and Gunn diodes.
Infrared waves are produced by hot bodies and molecules due to changes in their vibrational and rotational modes.
$X$-rays are produced when high-energy electrons are suddenly stopped or when inner-shell electrons transition from a higher energy level to a lower energy level.
Therefore, the correct matching is: $A - IV, B - I, C - II, D - III$.

(C) The power factor of an $LCR$ series circuit is given by $\cos \phi = \frac{R}{Z}$.
First,we calculate the impedance $Z$ of the circuit:
$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$
Given $R = 60 \ \Omega$,$X_{L} = 70 \ \Omega$,and $X_{C} = 150 \ \Omega$.
$Z = \sqrt{60^{2} + (70 - 150)^{2}}$
$Z = \sqrt{60^{2} + (-80)^{2}}$
$Z = \sqrt{3600 + 6400} = \sqrt{10000} = 100 \ \Omega$.
Now,the power factor is:
$\cos \phi = \frac{R}{Z} = \frac{60}{100} = \frac{6}{10}$.
Comparing this with the given power factor $\frac{\alpha}{10}$,we get $\alpha = 6$.

(A) For a compound microscope in normal adjustment,the objective lens forms the image at the focal point of the eyepiece.
The tube length $L$ is the distance between the second focal point of the objective and the first focal point of the eyepiece.
The magnification $M$ of a compound microscope in normal adjustment is given by the formula: $M = \left( \frac{L}{f_o} \right) \times \left( \frac{D}{f_e} \right)$.
Given: $L = 10 \ cm$,$f_o = 2 \ cm$,$f_e = 5 \ cm$,and the least distance of distinct vision $D = 25 \ cm$.
Substituting the values: $M = \left( \frac{10}{2} \right) \times \left( \frac{25}{5} \right) = 5 \times 5 = 25$.
We are given $M = (5)^k$.
Therefore,$25 = (5)^k \implies 5^2 = 5^k$.
Comparing the exponents,we get $k = 2$.

(A) For a hydrogen-like atom,the velocity of an electron in the $n^{th}$ orbit is given by $v \propto \frac{Z}{n}$.
The radius of the $n^{th}$ orbit is given by $r \propto \frac{n^2}{Z}$.
From these two relations,we can write $r \propto \frac{n}{v}$ (since $Z \propto \frac{n}{v} \cdot n = \frac{n^2}{r}$,substituting $Z$ into the velocity relation).
Given that the radii are nearly the same $(r_1 \approx r_2)$,we have $\frac{n_1}{v_1} = \frac{n_2}{v_2}$.
Therefore,$\frac{n_1}{n_2} = \frac{v_1}{v_2} = \frac{3 \times 10^5}{2.5 \times 10^5} = \frac{3}{2.5} = \frac{6}{5}$.
Thus,the possible order of energy states ($n_1$ and $n_2$) is $6$ and $5$.

(A) For the condition of minimum deviation,the angle of incidence $i$ is equal to the angle of emergence $e$,and the angle of refraction $r$ is given by $r = \frac{A}{2}$,where $A$ is the prism angle.
At the exit surface,the light ray strikes the interface between the prism (refractive index $n$) and the coating (refractive index $n/2$).
For the condition of the critical angle $\theta_{c}$,the angle of refraction $r$ must be equal to $\theta_{c}$.
According to Snell's law at the exit surface: $n \sin(r) = (n/2) \sin(90^{\circ})$.
Since $\sin(90^{\circ}) = 1$,we have $n \sin(r) = n/2$.
Dividing both sides by $n$,we get $\sin(r) = 1/2$.
Since $r = A/2$,we have $\sin(A/2) = 1/2$.
Therefore,$A/2 = 30^{\circ}$,which gives $A = 60^{\circ}$.

(C) Let the two resistors be $R_1 = 100 \Omega$ and $R_2 = 100 \Omega$. The voltmeter with resistance $R_v = 400 \Omega$ is connected in parallel with $R_2$.
The equivalent resistance of the parallel combination of $R_2$ and $R_v$ is:
$R_p = \frac{R_2 \times R_v}{R_2 + R_v} = \frac{100 \times 400}{100 + 400} = \frac{40000}{500} = 80 \Omega$
The total resistance of the circuit is $R_{eq} = R_1 + R_p = 100 + 80 = 180 \Omega$.
The total current in the circuit is $I = \frac{V}{R_{eq}} = \frac{9}{180} = \frac{1}{20} \text{ A} = 0.05 \text{ A}$.
The voltage drop across the parallel combination (which is the voltmeter reading) is $V_v = I \times R_p = 0.05 \times 80 = 4 \text{ V}$.

(C) The dipole moment $\vec{p}$ of a system of charges is given by $\vec{p} = \sum q_i \vec{r}_i$.
Given charges are $q_1 = +2q$ at $\vec{r}_1 = (0, -3a) = -3a\hat{j}$,$q_2 = +3q$ at $\vec{r}_2 = (2a, 0) = 2a\hat{i}$,and $q_3 = -4q$ at $\vec{r}_3 = (-2a, 0) = -2a\hat{i}$.
Substituting these values into the formula:
$\vec{p} = (2q)(-3a\hat{j}) + (3q)(2a\hat{i}) + (-4q)(-2a\hat{i})$
$\vec{p} = -6qa\hat{j} + 6qa\hat{i} + 8qa\hat{i}$
$\vec{p} = (6qa + 8qa)\hat{i} - 6qa\hat{j}$
$\vec{p} = 14qa\hat{i} - 6qa\hat{j}$
$\vec{p} = 2qa(7\hat{i} - 3\hat{j})$

(D) $1$. Magnetic induction $(B)$: Using $F = qvB$, we have $[B] = [F] / ([q][v]) = [MLT^{-2}] / ([AT][LT^{-1}]) = [MT^{-2}A^{-1}]$. This matches $III$.
$2$. Magnetic flux $(\phi)$: $\phi = B \cdot A$. Thus, $[\phi] = [MT^{-2}A^{-1}] \cdot [L^2] = [ML^2T^{-2}A^{-1}]$. This matches $IV$.
$3$. Magnetic permeability $(\mu)$: Using $B = \mu_0 H$ or $F = \frac{\mu_0 I_1 I_2 L}{2\pi r}$, we find $[\mu] = [MLT^{-2}A^{-2}]$. This matches $I$.
$4$. Self inductance $(L)$: Using $U = \frac{1}{2}LI^2$, we have $[L] = [U] / [I^2] = [ML^2T^{-2}] / [A^2] = [ML^2T^{-2}A^{-2}]$. This matches $II$.
Therefore, the correct matching is $A-III, B-IV, C-I, D-II$.

(C) The shift in the central bright fringe in a Young's double slit experiment when a thin transparent sheet is introduced is given by the formula: $y_{\text{shift}} = \frac{(\mu - 1) t D}{d}$.
Given values are: $d = 0.4 \ mm = 0.4 \times 10^{-3} \ m$,$D = 1 \ m$,$t = 20 \ \mu m = 20 \times 10^{-6} \ m$,and $y_{\text{shift}} = 20 \ mm = 20 \times 10^{-3} \ m$.
Substituting these values into the formula:
$20 \times 10^{-3} = \frac{(\mu - 1) \times 20 \times 10^{-6} \times 1}{0.4 \times 10^{-3}}$
$20 \times 10^{-3} = \frac{(\mu - 1) \times 20 \times 10^{-6}}{0.4 \times 10^{-3}}$
$20 \times 10^{-3} = (\mu - 1) \times 50 \times 10^{-3}$
$\mu - 1 = \frac{20}{50} = 0.4$
$\mu = 1.4$
Since the refractive index is given as $\frac{\alpha}{10} = 1.4$,we get $\alpha = 14$.

(C) The de Broglie wavelength $\lambda$ of a particle accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mqV}}$.
Given: $h = 6.6 \times 10^{-34} \text{ J} \cdot \text{s}$,$q = 3 \times 10^{-19} \text{ C}$,$m = 6 \times 10^{-27} \text{ kg}$,and $V = 1.21 \text{ V}$.
Substituting the values into the formula:
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times (6 \times 10^{-27}) \times (3 \times 10^{-19}) \times 1.21}}$
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{36 \times 10^{-46} \times 1.21}}$
$\lambda = \frac{6.6 \times 10^{-34}}{6 \times 10^{-23} \times 1.1}$
$\lambda = \frac{6.6 \times 10^{-34}}{6.6 \times 10^{-23}}$
$\lambda = 10^{-11} \text{ m} = 10 \times 10^{-12} \text{ m}$.
Comparing this with $\alpha \times 10^{-12} \text{ m}$,we get $\alpha = 10$.

(A) The conducting current density is given by $j_c = \sigma E$.
For an electromagnetic wave,$E = E_0 \sin(\omega t - kx)$,so the maximum conducting current density is $(j_c)_{\max} = \sigma E_0$.
The displacement current density is given by $j_d = \epsilon_0 \frac{\partial E}{\partial t}$.
Substituting $E$,we get $j_d = \epsilon_0 E_0 \omega \cos(\omega t - kx)$,so the maximum displacement current density is $(j_d)_{\max} = \epsilon_0 E_0 \omega$.
The ratio is $\frac{(j_c)_{\max}}{(j_d)_{\max}} = \frac{\sigma E_0}{\epsilon_0 \omega E_0} = \frac{\sigma}{\epsilon_0 \omega}$.
Given $\sigma = 10 \text{ mho/m}$,$f = 100 \times 10^6 \text{ Hz}$,and $\omega = 2 \pi f = 2 \pi \times 10^8 \text{ rad/s}$.
Also,$\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9$,so $\frac{1}{\epsilon_0} = 4 \pi \times 9 \times 10^9 = 36 \pi \times 10^9$.
Ratio $= \frac{10 \times 36 \pi \times 10^9}{2 \pi \times 10^8} = \frac{360 \pi \times 10^9}{2 \pi \times 10^8} = 180 \times 10 = 1800$.

(A) Let the prism be $PDC$ with base $DC$. The incident ray is parallel to the base $DC$.
For grazing emergence at the second surface,the angle of refraction $r_2$ is equal to the critical angle $C$.
Given $\mu = \sqrt{2}$,we have $\sin r_2 = \frac{1}{\mu} = \frac{1}{\sqrt{2}}$,which gives $r_2 = 45^{\circ}$.
Since the incident ray is parallel to the base,the angle of incidence $i$ at the first surface is equal to the base angle $45^{\circ}$.
Applying Snell's Law at the first surface: $1 \times \sin(45^{\circ}) = \sqrt{2} \times \sin(r_1)$.
$\frac{1}{\sqrt{2}} = \sqrt{2} \sin(r_1) \implies \sin(r_1) = \frac{1}{2} \implies r_1 = 30^{\circ}$.
The angle of the prism $A$ at the top vertex is $A = r_1 + r_2 = 30^{\circ} + 45^{\circ} = 75^{\circ}$.
In the triangle formed by the prism,the sum of angles is $180^{\circ}$. Thus,$45^{\circ} + \theta + A = 180^{\circ}$.
$45^{\circ} + \theta + 75^{\circ} = 180^{\circ} \implies \theta + 120^{\circ} = 180^{\circ} \implies \theta = 60^{\circ}$.

(B) The magnetic field at the centre $O$ is the vector sum of the fields due to the straight segments $AB$,$DE$ and the circular arc $BCD$.
Using the Biot-Savart law,the field due to a semi-infinite wire at a distance $r$ from its end is $\vec{B} = \frac{\mu_0 I}{4 \pi r} \hat{k}$.
For segment $AB$,the field at $O$ is $\vec{B}_{AB} = \frac{\mu_0 I}{4 \pi r} \hat{k}$.
For segment $DE$,the field at $O$ is $\vec{B}_{DE} = \frac{\mu_0 I}{4 \pi r} \hat{k}$.
For the circular arc $BCD$,the angle subtended is $2\pi - 0 = 2\pi$ (a full circle minus the gap). However,looking at the geometry,the arc is $2\pi - \theta$ where $\theta$ is the angle of the gap. Assuming the gap is negligible,the arc is a full circle. The field is $\vec{B}_{BCD} = -\frac{\mu_0 I}{2 r} \hat{k}$.
Summing these: $\vec{B}_O = \vec{B}_{AB} + \vec{B}_{DE} + \vec{B}_{BCD} = \frac{\mu_0 I}{4 \pi r} \hat{k} + \frac{\mu_0 I}{4 \pi r} \hat{k} - \frac{\mu_0 I}{2 r} \hat{k} = \frac{\mu_0 I}{2 \pi r} \hat{k} - \frac{\mu_0 I}{2 r} \hat{k} = \frac{\mu_0 I}{2 \pi r} (1 - \pi) \hat{k} = -\frac{\mu_0 I}{2 \pi r} (\pi - 1) \hat{k}$.
Note: The direction is along the $z$-axis $(\hat{k})$,not $\hat{i}$ as given in the options. Assuming the question intended $\hat{k}$ for the $z$-axis direction.

(C) According to Bohr's postulate,the angular momentum $L$ of an electron in an orbit is given by $L = \frac{nh}{2\pi}$,where $n$ is the principal quantum number.
Rearranging this,we get $\frac{2\pi L}{h} = n$.
The energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{E_0}{n^2}$,where $E_0$ is the magnitude of the ground state energy.
Given that $E_n = -0.04 E_0$,we have $-\frac{E_0}{n^2} = -0.04 E_0$.
Dividing both sides by $-E_0$,we get $\frac{1}{n^2} = 0.04$.
$n^2 = \frac{1}{0.04} = \frac{100}{4} = 25$.
Therefore,$n = 5$.
Since $\frac{2\pi L}{h} = n$,the value is $5$.

(A) From the circuit diagram,the resistance $2R$ and $X$ are in series. Their equivalent resistance is $R_{s} = 2R + X$.
This combination is in parallel with the resistance $R$. The equivalent resistance $R_{eq}$ between points $A$ and $B$ is given by:
$R_{eq} = \frac{R \cdot (2R + X)}{R + (2R + X)}$
Given that $R_{eq} = X$,we have:
$X = \frac{R(2R + X)}{3R + X}$
$X(3R + X) = 2R^{2} + RX$
$3RX + X^{2} = 2R^{2} + RX$
$X^{2} + 2RX - 2R^{2} = 0$
Using the quadratic formula $X = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$:
$X = \frac{-2R \pm \sqrt{(2R)^{2} - 4(1)(-2R^{2})}}{2(1)}$
$X = \frac{-2R \pm \sqrt{4R^{2} + 8R^{2}}}{2}$
$X = \frac{-2R \pm \sqrt{12R^{2}}}{2} = \frac{-2R \pm 2R\sqrt{3}}{2}$
$X = -R \pm R\sqrt{3}$
Since resistance must be positive,we take the positive root:
$X = (\sqrt{3} - 1)R$

(D) The current $I$ in the circuit is given by $I = \frac{E}{R+r}$.
The power $P$ consumed by the external resistance $R$ is $P = I^2 R = \left(\frac{E}{R+r}\right)^2 R$.
To find the condition for maximum power,we differentiate $P$ with respect to $R$ and set it to zero: $\frac{dP}{dR} = 0$.
$\frac{dP}{dR} = E^2 \left[ \frac{(R+r)^2(1) - R(2)(R+r)}{(R+r)^4} \right] = 0$.
This implies $(R+r)^2 - 2R(R+r) = 0$.
Dividing by $(R+r)$,we get $(R+r) - 2R = 0$,which simplifies to $r - R = 0$ or $R = r$.
Thus,the power consumption is maximum when the external resistance equals the internal resistance of the battery.

(C) The angular fringe width $\theta$ in a Young's double slit experiment is given by the formula $\theta = \frac{\lambda}{d}$,where $\lambda$ is the wavelength of the light used and $d$ is the distance between the two slits.
Statement $I$: The angular fringe width depends only on the wavelength $\lambda$ and the slit separation $d$. It is independent of the distance $D$ between the screen and the slits. Therefore,moving the screen away does not change the angular separation. Thus,Statement $I$ is false.
Statement $II$: Since $\theta = \frac{\lambda}{d}$,the angular fringe width is directly proportional to the wavelength $\lambda$. If the source is replaced by one with a higher wavelength,$\theta$ will increase. Thus,Statement $II$ is true.
Conclusion: Statement $I$ is false and Statement $II$ is true.

(B) The given circuit represents a Wheatstone bridge configuration where the galvanometer shows zero deflection at point $P$.
This implies that the bridge is balanced.
For a balanced Wheatstone bridge,the ratio of resistances in the arms must be equal:
$\frac{R_1}{R_2} = \frac{R_{AP}}{R_{PB}}$
Given $R_1 = 6 \ \Omega$ and $R_2 = 4 \ \Omega$.
Let $L$ be the total length of the wire $AB = 50 \text{ cm}$.
Let $\ell_{AP} = x$ and $\ell_{PB} = 50 - x$.
Since the resistance of the wire is proportional to its length $(R = \rho \frac{\ell}{A})$,we have:
$\frac{6}{4} = \frac{x}{50 - x}$
$\frac{3}{2} = \frac{x}{50 - x}$
$3(50 - x) = 2x$
$150 - 3x = 2x$
$5x = 150$
$x = 30 \text{ cm}$.
Therefore,the length $AP$ is $30 \text{ cm}$.

(B) In the steady state,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor.
Looking at the circuit,the diode in the branch with $3 \Omega$ is reverse-biased,so no current flows through that branch.
The diode in the branch with $4 \Omega$ is forward-biased,so current flows through the branch containing the $2.5 \text{ V}$ battery,the $1 \Omega$ resistor,and the $4 \Omega$ resistor.
The total resistance in this path is $R_{eq} = 1 \Omega + 4 \Omega = 5 \Omega$.
The current in the circuit is $i = \frac{V}{R_{eq}} = \frac{2.5 \text{ V}}{5 \Omega} = 0.5 \text{ A}$.
The voltage across the capacitor $V_C$ is equal to the voltage across the $4 \Omega$ resistor because they are in parallel.
$V_C = i \times 4 \Omega = 0.5 \text{ A} \times 4 \Omega = 2 \text{ V}$.
The charge stored by the capacitor is $Q = C \times V_C = 5 \mu\text{F} \times 2 \text{ V} = 10 \mu\text{C}$.

(A) Let the initial separation be $r_i = 4R$ and the final separation when they just touch be $r_f = 2R$.
By the principle of conservation of energy,the total initial energy equals the total final energy.
Initial Energy: $E_i = 2 \times (\frac{1}{2} m u^2) - \frac{G m^2}{4R} + \frac{k Q^2}{4R} = m u^2 - \frac{G m^2}{4R} + \frac{k Q^2}{4R}$.
Final Energy (at the moment of touching,speed is zero): $E_f = 0 - \frac{G m^2}{2R} + \frac{k Q^2}{2R}$.
Equating $E_i = E_f$:
$m u^2 - \frac{G m^2}{4R} + \frac{k Q^2}{4R} = - \frac{G m^2}{2R} + \frac{k Q^2}{2R}$.
$m u^2 = \frac{k Q^2}{2R} - \frac{k Q^2}{4R} - \frac{G m^2}{2R} + \frac{G m^2}{4R}$.
$m u^2 = \frac{k Q^2}{4R} - \frac{G m^2}{4R} = \frac{1}{4R} (k Q^2 - G m^2)$.
$u^2 = \frac{1}{4 m R} (k Q^2 - G m^2) = \frac{k Q^2}{4 m R} (1 - \frac{G m^2}{k Q^2})$.
$u = \sqrt{\frac{k Q^2}{4 m R} (1 - \frac{G m^2}{k Q^2})}$.

(A) Given: Length $l = 2 \ m$,Area $A = 0.2 \ mm^{2} = 0.2 \times 10^{-6} \ m^{2}$,Current $I = 1.6 \ A$,Voltage $V = 2 \ V$,Concentration $n = 5 \times 10^{28} \ m^{-3}$,Charge $e = 1.6 \times 10^{-19} \ C$.
The drift velocity $V_{d}$ is given by $V_{d} = \frac{I}{neA}$.
The mobility $\mu$ is defined as $\mu = \frac{V_{d}}{E}$,where $E = \frac{V}{l}$ is the electric field.
Substituting $V_{d}$ and $E$,we get $\mu = \frac{I}{neA} \times \frac{l}{V}$.
Substituting the values: $\mu = \frac{1.6 \times 2}{5 \times 10^{28} \times 1.6 \times 10^{-19} \times 0.2 \times 10^{-6} \times 2}$.
$\mu = \frac{3.2}{5 \times 10^{28} \times 1.6 \times 10^{-19} \times 0.2 \times 10^{-6} \times 2} = \frac{3.2}{3.2 \times 10^{3}} = 1 \times 10^{-3} \ m^{2}/V \cdot s$.
Comparing with $\alpha \times 10^{-3} \ m^{2}/V \cdot s$,we get $\alpha = 1$.

(D) Initial charge on capacitor $P$ is $Q_1 = C_1 V_1 = (10 \times 10^{-6} \text{ F}) \times (6.0 \text{ V}) = 60 \times 10^{-6} \text{ C} = 6 \times 10^{-5} \text{ C}$.
When the two capacitors are connected in parallel,the charge is redistributed until they reach a common potential $V$.
The common potential $V$ is given by $V = \frac{Q_{total}}{C_{total}} = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$.
Given $C_1 = 10 \times 10^{-6} \text{ F}$,$V_1 = 6.0 \text{ V}$,$C_2 = 20 \times 10^{-6} \text{ F}$,and $V_2 = 0 \text{ V}$.
$V = \frac{(10 \times 10^{-6} \times 6) + (20 \times 10^{-6} \times 0)}{10 \times 10^{-6} + 20 \times 10^{-6}} = \frac{60 \times 10^{-6}}{30 \times 10^{-6}} = 2 \text{ V}$.
The charge on capacitor $Q$ at equilibrium is $Q_2 = C_2 V = (20 \times 10^{-6} \text{ F}) \times (2 \text{ V}) = 40 \times 10^{-6} \text{ C} = 4 \times 10^{-5} \text{ C}$.
Comparing this with $\alpha \times 10^{-5} \text{ C}$,we get $\alpha = 4$.

(B) The induced $EMF$ in a rotating loop is given by $E = B A \omega \sin(\theta)$,where $\theta = \omega t$ is the angle between the area vector and the magnetic field.
Given: $B = 0.5 \ T$,$\omega = 100 \ rad/s$,$E = 15.4 \ mV = 15.4 \times 10^{-3} \ V$,and $\theta = 30^{\circ}$.
The area of the circular loop is $A = \pi r^2$.
Substituting the values into the formula: $15.4 \times 10^{-3} = 0.5 \times (\pi r^2) \times 100 \times \sin(30^{\circ})$.
Since $\sin(30^{\circ}) = 0.5$,we have: $15.4 \times 10^{-3} = 0.5 \times \pi r^2 \times 100 \times 0.5$.
$15.4 \times 10^{-3} = 25 \times \pi r^2$.
Using $\pi = 22/7$: $15.4 \times 10^{-3} = 25 \times (22/7) \times r^2$.
$r^2 = \frac{15.4 \times 10^{-3} \times 7}{25 \times 22} = \frac{107.8 \times 10^{-3}}{550} = 0.196 \times 10^{-3} = 196 \times 10^{-6} \ m^2$.
$r = \sqrt{196 \times 10^{-6}} = 14 \times 10^{-3} \ m = 14 \ mm$.

(A) According to Einstein's photoelectric equation,the energy of the incident photon is equal to the work function when the kinetic energy of the emitted photoelectrons is zero.
$ E = h\nu = \phi $
Where $ \phi = 110 \times 10^{-20} \ J $ is the work function and $ h = 6.63 \times 10^{-34} \ J \cdot s $ is Planck's constant.
The frequency $ \nu $ is given by $ \nu = \frac{\phi}{h} $.
The angular frequency $ \omega $ is related to frequency by $ \omega = 2\pi\nu $.
Substituting the values:
$ \omega = 2\pi \left( \frac{\phi}{h} \right) = \frac{2 \times 3.14 \times 110 \times 10^{-20}}{6.63 \times 10^{-34}} $
$ \omega \approx 1.04 \times 10^{16} \ rad/s $.

(D) The intensity $I$ of an electromagnetic wave is given by $I = \frac{1}{2} \epsilon_0 E_0^2 c$.
From this,the amplitude of the electric field $E_0$ is $E_0 = \sqrt{\frac{2I}{\epsilon_0 c}}$.
We know that the relation between the amplitudes of the electric and magnetic fields is $E_0 = B_0 c$,which implies $B_0 = \frac{E_0}{c}$.
Substituting $E_0$,we get $B_0 = \frac{1}{c} \sqrt{\frac{2I}{\epsilon_0 c}} = \sqrt{\frac{2I}{\epsilon_0 c^3}}$.
Given $I = 4.0 \times 10^{14} \ W/m^2$,$\epsilon_0 = 8.85 \times 10^{-12} \ C^2/Nm^2$,and $c = 3 \times 10^8 \ m/s$:
$B_0 = \sqrt{\frac{2 \times 4.0 \times 10^{14}}{8.85 \times 10^{-12} \times (3 \times 10^8)^3}} = \sqrt{\frac{8.0 \times 10^{14}}{8.85 \times 10^{-12} \times 27 \times 10^{24}}} = \sqrt{\frac{8.0 \times 10^{14}}{238.95 \times 10^{12}}} = \sqrt{\frac{800}{238.95}} \approx \sqrt{3.348} \approx 1.83 \ T$.

(A) $1$. Electric potential $(V)$ is a scalar quantity. The potential at the center $O$ due to a single charge $q$ at a distance $r$ is $V_i = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r}$. Since there are five such charges,the total potential is the algebraic sum: $V = 5 \times \frac{1}{4 \pi \varepsilon_0} \frac{q}{r} = \frac{5 q}{4 \pi \varepsilon_0 r}$.
$2$. Electric field $(\vec{E})$ is a vector quantity. For a regular polygon with identical charges at all vertices,the electric field vectors due to individual charges cancel each other out due to symmetry. Therefore,the net electric field at the center $O$ is $\vec{E} = 0$.

(A) The angular width of the central maximum in single slit diffraction is given by $\theta = \frac{2\lambda}{a}$, and the linear width is $\beta_{cm} = \frac{2\lambda D}{a}$, where $\lambda$ is the wavelength, $D$ is the distance to the screen, and $a$ is the slit width.
$(A)$ Since $\beta_{cm} \propto \lambda$, the width of the central maximum increases as the wavelength increases. Thus, $(A)$ is correct.
$(B)$ Since $\beta_{cm} \propto \lambda$, the width decreases as wavelength decreases. Thus, $(B)$ is incorrect.
$(C)$ Since $\beta_{cm} \propto \frac{1}{a}$, the width increases as the slit width $a$ decreases. Thus, $(C)$ is correct.
$(D)$ Since $\beta_{cm} \propto \frac{1}{a}$, the width decreases as the slit width $a$ increases. Thus, $(D)$ is incorrect.
$(E)$ The intensity of the central maximum is proportional to the square of the slit width and inversely proportional to the square of the wavelength $(\text{Intensity} \propto \frac{a^2}{\lambda^2})$. Therefore, as $\lambda$ decreases, the intensity (brightness) of the central maximum increases. Thus, $(E)$ is correct.
Conclusion: Statements $(A)$, $(C)$, and $(E)$ are correct.

(A) At the moment the switch $K$ is turned $ON$ $(t = 0)$,the current through the inductors cannot change instantaneously. Therefore,the inductors act as open circuits (infinite resistance).
In the given circuit,the branch containing the first inductor and the branch containing the second inductor will have zero current.
Only the middle branch,which contains a single resistor of $9 \ \Omega$,will conduct current.
The equivalent resistance of the circuit at $t = 0$ is $R_{eq} = 9 \ \Omega$.
The voltage of the battery is $V = 9 \ V$.
Using Ohm's law,the current $I$ through the ammeter is:
$I = \frac{V}{R_{eq}} = \frac{9 \ V}{9 \ \Omega} = 1 \ A$.

(C) The frequency $f$ of light remains constant when it travels from one medium to another.
Since $v = f \lambda$ and $v = \frac{c}{\mu}$,we have $\frac{c}{\mu} = f \lambda$,which implies $\mu \lambda = \frac{c}{f} = \text{constant}$.
Therefore,$\mu_1 \lambda_1 = \mu_2 \lambda_2$.
Given: $\mu_1 = \frac{4}{3}$,$\lambda_1 = 540 \ nm$,and $\mu_2 = \frac{3}{2}$.
Substituting the values: $\left(\frac{4}{3}\right) \times 540 = \left(\frac{3}{2}\right) \times \lambda_2$.
$\lambda_2 = 540 \times \frac{4}{3} \times \frac{2}{3}$.
$\lambda_2 = 540 \times \frac{8}{9} = 60 \times 8 = 480 \ nm$.

(A) Given: Power delivered $P_{out} = 1 \ kW = 1000 \ W$,Voltage $V = 250 \ V$,Resistance $R = 2 \ \Omega$.
The current $I$ flowing through the line is $I = \frac{P_{out}}{V} = \frac{1000}{250} = 4 \ A$.
The power loss in the transmission line is $P_{loss} = I^2 R = (4)^2 \times 2 = 16 \times 2 = 32 \ W$.
The total power generated at the source is $P_{in} = P_{out} + P_{loss} = 1000 + 32 = 1032 \ W$.
The efficiency $\eta$ is given by $\eta = \left( \frac{P_{out}}{P_{in}} \right) \times 100$.
$\eta = \left( \frac{1000}{1032} \right) \times 100 \approx 96.898 \% \approx 96.9 \%$.

(B) The logic circuit consists of an $AND$ gate $(A, B)$,followed by a $NOT$ gate,an $OR$ gate $(C, D)$,an $AND$ gate,and a final $NOT$ gate.
Let the output of the first $AND$ gate be $A \cdot B$. After the $NOT$ gate,it becomes $\overline{A \cdot B}$.
The output of the $OR$ gate is $C + D$.
These two signals are fed into an $AND$ gate,giving $(\overline{A \cdot B}) \cdot (C + D)$.
The final $NOT$ gate gives the output $Y = \overline{(\overline{A \cdot B}) \cdot (C + D)}$.
Using De Morgan's Law,$Y = \overline{(\overline{A \cdot B})} + \overline{(C + D)} = (A \cdot B) + (\overline{C + D})$.
Evaluating for the given inputs:
$1$. $A=1, B=1, C=0, D=1$: $Y = (1 \cdot 1) + \overline{(0+1)} = 1 + 0 = 1$.
$2$. $A=0, B=0, C=1, D=1$: $Y = (0 \cdot 0) + \overline{(1+1)} = 0 + 0 = 0$.
$3$. $A=1, B=0, C=1, D=0$: $Y = (1 \cdot 0) + \overline{(1+0)} = 0 + 0 = 0$.
$4$. $A=1, B=1, C=1, D=1$: $Y = (1 \cdot 1) + \overline{(1+1)} = 1 + 0 = 1$.
Comparing with the options,option $B$ is correct.

(C) The Rydberg formula is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Lyman series,the smallest wavelength corresponds to $n_1 = 1$ and $n_2 = \infty$.
$\frac{1}{\lambda_{L,min}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R = \frac{1}{91} \ nm^{-1}$.
For the Balmer series,the largest wavelength corresponds to $n_1 = 2$ and $n_2 = 3$.
$\frac{1}{\lambda_{B,max}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = \frac{1}{91} \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{1}{91} \left( \frac{5}{36} \right)$.
$\lambda_{B,max} = \frac{91 \times 36}{5} = 655.2 \ nm$.
For the Paschen series,the largest wavelength corresponds to $n_1 = 3$ and $n_2 = 4$.
$\frac{1}{\lambda_{P,max}} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = \frac{1}{91} \left( \frac{1}{9} - \frac{1}{16} \right) = \frac{1}{91} \left( \frac{7}{144} \right)$.
$\lambda_{P,max} = \frac{91 \times 144}{7} = 1872 \ nm$.
The difference is $\Delta\lambda = \lambda_{P,max} - \lambda_{B,max} = 1872 - 655.2 = 1216.8 \ nm \approx 1217 \ nm$.

(B) In the parallax method,we need to form a real image of the object to observe the parallax between the image and the needle.
For a concave mirror,a real image is formed only when the object is placed beyond the focus $(F)$.
If the object is placed between the pole $(P)$ and the focus $(F)$,the image formed is virtual,erect,and magnified,which cannot be used for the parallax method as it cannot be captured on a screen or observed as a real point of coincidence.
Therefore,the object must be placed at any point beyond the focus $(F)$.

\(C _1=\frac{5 \in_0 A / 2}{d / 2}=\frac{5 \in_0 A}{ d }=5 C\)
\(C _2=\frac{2 \in_0 A / 2}{d / 2}=\frac{2 \in_0 A}{ d }=2 C\)
\(C _1 \& C _2\) in series.
\(C^{\prime}=\frac{C_1 C_2}{C_1+C_2}=\frac{(5 C)(2 C)}{7 C}=\frac{10}{7} C\)
\(C _3=\frac{3 \in_0 A / 2}{d / 2}=3 C\)
\(C _4=\frac{2 \epsilon_0 A / 2}{d / 2}=2 C\)
\(C _4 \& C _3\) in series; \(C ^{\prime \prime}=\frac{(2 C )(3 C )}{5 C }=\frac{6}{5} C\)
\(C ^{\prime} \& C ^{\prime \prime}\) in parallel;
So, \(C _{ eq }= C \left(\frac{6}{5}+\frac{10}{7}\right)= C \left(\frac{42+50}{35}\right)=\left(\frac{92}{35}\right) C\)
\(\frac{92}{35} C =\frac{ nC }{3}\)
\(n =\frac{92 \times 3}{35}=7.9 \Rightarrow n \simeq 8\)

(C) The fringe width $\beta$ in a Young's double-slit experiment is given by $\beta = \frac{D\lambda}{d}$.
Given that the fringe widths are equal,we have $\beta_1 = \beta_2$.
Therefore,$\frac{D_1 \lambda_1}{d_1} = \frac{D_2 \lambda_2}{d_2}$.
Rearranging the terms to find the ratio $\frac{D_1}{D_2}$,we get $\frac{D_1}{D_2} = \frac{\lambda_2}{\lambda_1} \times \frac{d_1}{d_2}$.
Given the ratios of slit separations $\frac{d_1}{d_2} = 2$ and the ratio of wavelengths $\frac{\lambda_1}{\lambda_2} = \frac{1}{2}$,which implies $\frac{\lambda_2}{\lambda_1} = 2$.
Substituting these values,we get $\frac{D_1}{D_2} = 2 \times 2 = 4$.

(B) In an $LCR$ circuit,the current is maximum at resonance.
At resonance,the angular frequency $\omega$ is given by $\omega = \frac{1}{\sqrt{LC}}$.
From the given voltage equation $V = 5 \sin(100t)$,we have $\omega = 100 \text{ rad/s}$.
Given $L = 2 \text{ H}$,we can substitute these values into the resonance formula:
$100 = \frac{1}{\sqrt{2 \times C}}$
Squaring both sides:
$10000 = \frac{1}{2C}$
$C = \frac{1}{2 \times 10000} = \frac{1}{20000} \text{ F}$
$C = 0.5 \times 10^{-4} \text{ F} = 50 \times 10^{-6} \text{ F} = 50 \mu\text{F}$.

(B) The given electric field equation is $E = E_0 \sin(\omega t - kx)$,where $E_0 = \sqrt{377} \text{ V/m}$.
The intensity (average power per unit area) of an electromagnetic wave is given by $I = \frac{1}{2} c \varepsilon_0 E_0^2$.
Since $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$,we can write $I = \frac{1}{2} \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \varepsilon_0 E_0^2 = \frac{1}{2} \sqrt{\frac{\varepsilon_0}{\mu_0}} E_0^2$.
Given $\sqrt{\frac{\mu_0}{\varepsilon_0}} = 377$,then $\sqrt{\frac{\varepsilon_0}{\mu_0}} = \frac{1}{377}$.
Substituting the values: $I = \frac{1}{2} \times \frac{1}{377} \times (\sqrt{377})^2 = \frac{1}{2} \times \frac{1}{377} \times 377 = \frac{1}{2} \text{ W/m}^2$.
Comparing this with $\frac{1}{\alpha} \text{ W/m}^2$,we get $\alpha = 2$.

(B) The induced $EMF$ in a rotating rod of length $\ell$ with angular velocity $\omega$ in a magnetic field $B$ is given by $\varepsilon = \frac{B \omega \ell^2}{2}$.
To find the maximum angular velocity $\omega_{\max}$,we use the principle of conservation of energy.
The potential energy at the release angle $\theta = 60^{\circ}$ is converted into kinetic energy at the lowest point.
$mg\ell(1 - \cos 60^{\circ}) = \frac{1}{2} I \omega_{\max}^2$,where $I = m\ell^2$.
$mg\ell(1 - 0.5) = \frac{1}{2} m\ell^2 \omega_{\max}^2$.
$g(0.5) = \frac{1}{2} \ell \omega_{\max}^2$.
$\omega_{\max} = \sqrt{\frac{g}{\ell}} = \sqrt{\frac{10}{0.1}} = \sqrt{100} = 10 \ rad/s$.
Now,substitute the values into the $EMF$ formula:
$\varepsilon_{\max} = \frac{2 \times 10 \times (0.1)^2}{2} = 10 \times 0.01 = 0.1 \ V$.
Converting to $mV$: $0.1 \ V = 100 \ mV$.

(C) Assertion $(A)$ is true because ferromagnetic materials consist of small regions called domains where atomic magnetic dipoles are aligned due to strong exchange interactions.
Reason $(R)$ is also true because thermal agitation at high temperatures disrupts the alignment of these magnetic dipoles,causing the domain structure to break down.
When the temperature exceeds the Curie temperature $(T_c)$,the material transitions from ferromagnetic to paramagnetic behavior,meaning the spontaneous magnetization disappears.
Since the disintegration of the domain structure at high temperatures is the fundamental reason why the ferromagnetic property (and thus spontaneous magnetization) is lost,Reason $(R)$ correctly explains Assertion $(A)$.

(B) The induced $EMF$ $(\varepsilon)$ in a conductor moving through a magnetic field is given by $\varepsilon = Bv\ell$.
First, calculate the velocity $(v)$ of the wire after falling a distance $(h = 200 \ m)$ under gravity using the equation $v^2 = u^2 + 2gh$. Since the initial velocity $(u = 0)$, $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 200} = \sqrt{4000} = 20\sqrt{10} \ m/s$.
The magnetic field $(B)$ is $0.5 \ Gauss = 0.5 \times 10^{-4} \ T$.
The length of the wire $(\ell)$ is $20 \ m$.
Substituting these values into the $EMF$ formula: $\varepsilon = (0.5 \times 10^{-4} \ T) \times (20\sqrt{10} \ m/s) \times (20 \ m)$.
$\varepsilon = 20\sqrt{10} \times 10^{-4} \times 10 = 20\sqrt{10} \times 10^{-3} \ V$.
Since $1 \ V = 1000 \ mV$, the induced $EMF$ is $20\sqrt{10} \ mV$.

(D) According to Gauss's Law,the total electric flux $\phi$ through a closed surface is $\phi = \frac{Q_{\text{in}}}{\varepsilon_0}$,where $Q_{\text{in}}$ is the net charge enclosed by the surface.
$1$. The charge $2q$ is placed at vertex $A$. $A$ vertex of a cube is shared by $8$ adjacent cubes. Therefore,the fraction of charge $2q$ enclosed within the cube is $\frac{2q}{8} = \frac{q}{4}$.
$2$. The charge $q$ is placed at the centre of the face $CDEF$. $A$ face of a cube is shared by $2$ adjacent cubes. Therefore,the fraction of charge $q$ enclosed within the cube is $\frac{q}{2}$.
$3$. The total enclosed charge $Q_{\text{in}}$ is the sum of these contributions: $Q_{\text{in}} = \frac{q}{4} + \frac{q}{2} = \frac{q + 2q}{4} = \frac{3q}{4}$.
$4$. Thus,the total electric flux passing through the cube is $\phi = \frac{Q_{\text{in}}}{\varepsilon_0} = \frac{3q}{4\varepsilon_0}$.

(B) The wavelength $\lambda$ for a transition is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
For the Lyman series $(n_f = 1)$,the maximum wavelength occurs at $n_i = 2$: $\frac{1}{\lambda_{max}} = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4} \implies \lambda_{max} = \frac{4}{3R}$. Thus,statement $A$ is correct.
The Balmer series $(n_f = 2)$ corresponds to the visible region. Thus,statement $B$ is correct.
For the Paschen series $(n_f = 3)$,the minimum wavelength occurs at $n_i = \infty$: $\frac{1}{\lambda_{min}} = R \left( \frac{1}{3^2} - 0 \right) = \frac{R}{9} \implies \lambda_{min} = \frac{9}{R}$. Thus,statement $C$ is correct.
For the Lyman series $(n_f = 1)$,the minimum wavelength occurs at $n_i = \infty$: $\frac{1}{\lambda_{min}} = R \left( 1 - 0 \right) = R \implies \lambda_{min} = \frac{1}{R}$. Thus,statement $D$ is incorrect.
Therefore,statements $A, B,$ and $C$ are correct.

(D) Given: Zener voltage $V_Z = 5 \text{ V}$,load current $I_L = 5 \text{ mA}$.
Maximum Zener current $I_Z = 4 \times I_L = 4 \times 5 \text{ mA} = 20 \text{ mA}$.
Total current through the series resistor $R_S$ is $I = I_L + I_Z = 5 \text{ mA} + 20 \text{ mA} = 25 \text{ mA} = 25 \times 10^{-3} \text{ A}$.
Maximum input voltage $V_{\text{in}} = 25 \text{ V}$.
The voltage drop across the series resistor $R_S$ is $V_{R_S} = V_{\text{in}} - V_Z = 25 \text{ V} - 5 \text{ V} = 20 \text{ V}$.
Using Ohm's law,$R_S = \frac{V_{R_S}}{I} = \frac{20 \text{ V}}{25 \times 10^{-3} \text{ A}} = 800 \Omega$.

(C) For total internal reflection to occur,light must travel from a denser medium to a rarer medium. Here,the speed of light in medium $A$ $(v_A = 2.4 \times 10^8 \ m/s)$ is less than in medium $B$ $(v_B = 2.7 \times 10^8 \ m/s)$,so medium $A$ is denser.
At the critical angle $c$,the angle of refraction is $90^\circ$. According to Snell's Law:
$\mu_A \sin c = \mu_B \sin 90^\circ$
Since $\mu = \frac{c_{light}}{v}$,we have $\frac{c_{light}}{v_A} \sin c = \frac{c_{light}}{v_B} \times 1$
$\sin c = \frac{v_A}{v_B} = \frac{2.4 \times 10^8}{2.7 \times 10^8} = \frac{24}{27} = \frac{8}{9}$
Now,using the trigonometric identity $\tan c = \frac{\sin c}{\sqrt{1 - \sin^2 c}}$:
$\tan c = \frac{8/9}{\sqrt{1 - (8/9)^2}} = \frac{8/9}{\sqrt{1 - 64/81}} = \frac{8/9}{\sqrt{17/81}} = \frac{8}{\sqrt{17}}$
Therefore,$c = \tan^{-1}(\frac{8}{\sqrt{17}})$.