JEE Main 2025 Physics Question Paper with Answer and Solution

(C) Mass density $= \frac{\text{Mass}}{\text{Volume}} = \frac{M}{L^3} = [ML^{-3}T^0]$. Matches $(IV)$.
$(B)$ Impulse $= \text{Force} \times \text{Time} = [MLT^{-2}] \times [T] = [MLT^{-1}]$. Matches $(II)$.
$(C)$ Power $= \frac{\text{Work}}{\text{Time}} = \frac{[ML^2T^{-2}]}{[T]} = [ML^2T^{-3}]$. Matches $(I)$.
$(D)$ Moment of inertia $= \text{Mass} \times (\text{Distance})^2 = [M] \times [L^2] = [ML^2T^0]$. Matches $(III)$.
Therefore,the correct matching is $(A)-(IV), (B)-(II), (C)-(I), (D)-(III)$.

(A) The general equation of a wave is $y = A \sin(kx + \omega t + \phi)$.
Comparing this with the given equation $y = \sin(20 \pi x + 10 \pi t)$,we get the wave number $k = 20 \pi \ rad/m$.
The wavelength $\lambda$ is given by $\lambda = \frac{2 \pi}{k} = \frac{2 \pi}{20 \pi} = 0.1 \ m = 10 \ cm$.
Points having the same oscillating speed in a wave are separated by a distance of $\frac{\lambda}{2}$ (or multiples thereof).
Therefore,the minimum distance is $\frac{\lambda}{2} = \frac{10 \ cm}{2} = 5 \ cm$.

(C) Potential energy is defined only for conservative forces.
Conservative forces are those for which the work done in moving a particle between two points is independent of the path taken.
Examples of conservative forces include gravitational force,electrostatic (Coulomb's) force,and spring (restoring) force.
Frictional force is a non-conservative force because the work done against it depends on the path taken.
Therefore,potential energy cannot be defined for frictional force.

(C) Isothermal process: Temperature remains constant,so change in internal energy $\Delta U = 0$. Hence,$(A)-(IV)$.
$(B)$ Adiabatic process: No heat exchange with the surroundings,so $\Delta Q = 0$. Hence,$(B)-(II)$.
$(C)$ Isobaric process: Pressure remains constant. The internal energy changes as temperature changes,so $\Delta U \neq 0$. Hence,$(C)-(III)$.
$(D)$ Isochoric process: Volume remains constant,so work done $\Delta W = P \Delta V = 0$. Hence,$(D)-(I)$.

(D) Given:
Horizontal speed of helicopter,$u = 360 \ km/h = 360 \times \frac{5}{18} = 100 \ m/s$.
Altitude,$H = 2 \ km = 2000 \ m$.
Time taken to hit the ground,$t = 20 \ s$.
Step $1$: Calculate the horizontal displacement $(x)$ of the object.
$x = u \times t = 100 \ m/s \times 20 \ s = 2000 \ m = 2 \ km$.
Step $2$: The vertical displacement is the altitude $H = 2 \ km$.
Step $3$: Calculate the total displacement $(D)$ from the release point to point $O$.
$D = \sqrt{x^2 + H^2} = \sqrt{(2 \ km)^2 + (2 \ km)^2} = \sqrt{4 + 4} \ km = \sqrt{8} \ km = 2\sqrt{2} \ km$.
Thus,the correct option is $D$.

(D) Given mass $m = 500 \ g = 0.5 \ kg$.
The velocity is given by $v = 4 \sqrt{x}$.
Squaring both sides,we get $v^2 = 16x$.
Differentiating with respect to $x$,we get $2v \frac{dv}{dx} = 16$,which simplifies to $v \frac{dv}{dx} = 8$.
We know that acceleration $a = v \frac{dv}{dx}$. Therefore,$a = 8 \ m/s^2$.
Using Newton's second law,$F = ma = 0.5 \ kg \times 8 \ m/s^2 = 4 \ N$.

(C) The thermal resistance $R$ of a rod is given by $R = \frac{L}{KA}$,where $L$ is the length,$K$ is the thermal conductivity,and $A = \pi r^2$ is the cross-sectional area.
For the two rods in series,the rate of heat flow $\frac{dQ}{dt}$ is the same in steady state:
$\frac{dQ}{dt} = \frac{\Delta T}{R} = \frac{400 - T}{R_A} = \frac{T - 200}{R_B}$
$\frac{400 - T}{T - 200} = \frac{R_A}{R_B} = \left( \frac{L_A}{L_B} \right) \left( \frac{K_B}{K_A} \right) \left( \frac{r_B}{r_A} \right)^2$
Given $\frac{L_A}{L_B} = \frac{1}{2}$,$\frac{K_A}{K_B} = \frac{1}{2} \implies \frac{K_B}{K_A} = 2$,and $\frac{r_A}{r_B} = 2 \implies \frac{r_B}{r_A} = \frac{1}{2}$.
Substituting these values:
$\frac{R_A}{R_B} = \left( \frac{1}{2} \right) \times (2) \times \left( \frac{1}{2} \right)^2 = 1 \times \frac{1}{4} = \frac{1}{4}$
Now,$\frac{400 - T}{T - 200} = \frac{1}{4}$
$4(400 - T) = T - 200$
$1600 - 4T = T - 200$
$5T = 1800$
$T = 360 \ K$

(A) The moment of inertia of the complete disc of mass $M$ and radius $R$ about an axis passing through its centre $O$ and perpendicular to its plane is $I_1 = \frac{1}{2} MR^2$.
The mass of the removed disc of radius $r = R/3$ is $m = \frac{M}{\pi R^2} \times \pi (R/3)^2 = \frac{M}{9}$.
The moment of inertia of the removed disc about the same axis passing through $O$ is calculated using the parallel axis theorem: $I_2 = I_{cm} + md^2$,where $I_{cm} = \frac{1}{2} m r^2$ and $d = R - r = R - R/3 = 2R/3$.
$I_2 = \frac{1}{2} \left(\frac{M}{9}\right) \left(\frac{R}{3}\right)^2 + \left(\frac{M}{9}\right) \left(\frac{2R}{3}\right)^2 = \frac{MR^2}{162} + \frac{4MR^2}{81} = \frac{MR^2 + 8MR^2}{162} = \frac{9MR^2}{162} = \frac{MR^2}{18}$.
The moment of inertia of the remaining part is $I = I_1 - I_2 = \frac{1}{2} MR^2 - \frac{1}{18} MR^2 = \frac{9-1}{18} MR^2 = \frac{8}{18} MR^2 = \frac{4}{9} MR^2$.
Comparing this with $\frac{4}{x} MR^2$,we get $x = 9$.

(D) The length of the rod is $L$,and its linear mass density is $\lambda$. Therefore,the total mass of the rod is $M = \lambda L$.
When the rod is bent into a ring of radius $R$,the circumference of the ring is equal to the length of the rod: $2 \pi R = L$,which gives $R = \frac{L}{2 \pi}$.
The moment of inertia of a ring of mass $M$ and radius $R$ about its diameter is given by the formula $I = \frac{1}{2} M R^2$.
Substituting the values of $M$ and $R$ into the formula:
$I = \frac{1}{2} (\lambda L) \left( \frac{L}{2 \pi} \right)^2$
$I = \frac{1}{2} (\lambda L) \left( \frac{L^2}{4 \pi^2} \right)$
$I = \frac{\lambda L^3}{8 \pi^2}$.

(A) The formula for Young's modulus $(Y)$ is given by $Y = \frac{F \cdot L}{A \cdot \Delta L}$.
Here,$F = mg = 50 \times 3 \pi \ N$,$L = 3 \ m$,$r = 3 \times 10^{-3} \ m$,and $\Delta L = 0.1 \times 10^{-3} \ m$.
The area $A = \pi r^2 = \pi \times (3 \times 10^{-3})^2 = 9 \pi \times 10^{-6} \ m^2$.
Substituting these values into the formula:
$Y = \frac{(50 \times 3 \pi) \times 3}{(9 \pi \times 10^{-6}) \times (0.1 \times 10^{-3})}$
$Y = \frac{450 \pi}{0.9 \pi \times 10^{-9}} = \frac{450}{0.9} \times 10^9 = 500 \times 10^9 = 5 \times 10^{11} \ Nm^{-2}$.
Comparing this with $P \times 10^{11} \ Nm^{-2}$,we get $P = 5$.

(D) The potential energy of the water at height $h$ is converted into heat energy when it hits the pool.
$mgh = ms \Delta T$
Here,$m$ is the mass of water,$g = 10 \ m/s^2$ is the acceleration due to gravity,$h = 200 \ m$ is the height,and $s = 4200 \ J/(kg \ K)$ is the specific heat capacity of water.
Canceling $m$ from both sides,we get:
$gh = s \Delta T$
$\Delta T = \frac{gh}{s}$
Substituting the values:
$\Delta T = \frac{10 \times 200}{4200} = \frac{2000}{4200} = \frac{20}{42} = \frac{10}{21} \ K$
$\Delta T \approx 0.476 \ K \approx 0.48 \ K$

(C) The quantity is given by $Q = X^{-2} Y^{\frac{3}{2}} Z^{-\frac{2}{5}}$.
Using the rule for propagation of errors,the maximum fractional error in $Q$ is given by:
$\frac{\Delta Q}{Q} = |-2| \frac{\Delta X}{X} + |\frac{3}{2}| \frac{\Delta Y}{Y} + |-\frac{2}{5}| \frac{\Delta Z}{Z}$
Given fractional errors are $\frac{\Delta X}{X} = 0.1$,$\frac{\Delta Y}{Y} = 0.2$,and $\frac{\Delta Z}{Z} = 0.5$.
Substituting these values:
$\frac{\Delta Q}{Q} = 2(0.1) + \frac{3}{2}(0.2) + \frac{2}{5}(0.5)$
$\frac{\Delta Q}{Q} = 0.2 + 0.3 + 0.2$
$\frac{\Delta Q}{Q} = 0.7$

(C) Since the gas is in a thermally insulated container and is compressed suddenly,the process is adiabatic.
For an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $P_i V_i^\gamma = P_f V_f^\gamma$.
Here,$V_f = \frac{1}{8} V_i$,which implies $\frac{V_i}{V_f} = 8$.
The ratio of final pressure to initial pressure is $\frac{P_f}{P_i} = \left(\frac{V_i}{V_f}\right)^\gamma$.
Substituting the values,$\frac{P_f}{P_i} = (8)^{5/3}$.
Since $8 = 2^3$,we have $(2^3)^{5/3} = 2^5 = 32$.
Thus,the ratio of final pressure to initial pressure is $32$.

(B) The velocity of a transverse wave in a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Linear mass density $\mu = \rho A = \rho (\pi R^2)$,where $\rho$ is the density of the material and $R$ is the radius of the cross section.
Since both strings are made of the same material,$\rho$ is constant. Given that the tension $T$ is the same for both strings,we have $v \propto \frac{1}{\sqrt{R^2}} \propto \frac{1}{R}$.
Therefore,$\frac{v_2}{v_1} = \frac{R_1}{R_2}$.
Given $R_1 = R$ and $R_2 = R/2$,we get $\frac{v_2}{v_1} = \frac{R}{R/2} = 2$.

(B) Given: Mass $m = 2 \ kg$, Initial velocity $\overrightarrow{u} = 3 \hat{i} + 4 \hat{j} \ ms^{-1}$, Force $\overrightarrow{F} = 6 \hat{k} \ N$, Time $t = \frac{5}{3} \ s$.
Using Newton's second law, the acceleration is $\overrightarrow{a} = \frac{\overrightarrow{F}}{m} = \frac{6 \hat{k}}{2} = 3 \hat{k} \ ms^{-2}$.
Using the first equation of motion, $\overrightarrow{v} = \overrightarrow{u} + \overrightarrow{a}t$.
Substituting the values: $\overrightarrow{v} = (3 \hat{i} + 4 \hat{j}) + (3 \hat{k}) \times \left(\frac{5}{3}\right)$.
$\overrightarrow{v} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \ ms^{-1}$.

(A) The maximum height of a projectile is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
Given that $(H_{\max})_1 = 8 \times (H_{\max})_2$,we have:
$\frac{u^2 \sin^2 \theta_1}{2g} = 8 \times \frac{u^2 \sin^2 \theta_2}{2g}$.
Simplifying this,we get $\sin^2 \theta_1 = 8 \sin^2 \theta_2$,which implies $\sin \theta_1 = \sqrt{8} \sin \theta_2 = 2\sqrt{2} \sin \theta_2$.
The time of flight is given by $T = \frac{2u \sin \theta}{g}$.
Therefore,the ratio of the times of flight is $\frac{T_1}{T_2} = \frac{2u \sin \theta_1 / g}{2u \sin \theta_2 / g} = \frac{\sin \theta_1}{\sin \theta_2}$.
Substituting the value of $\sin \theta_1$,we get $\frac{T_1}{T_2} = \frac{2\sqrt{2} \sin \theta_2}{\sin \theta_2} = 2\sqrt{2}$.
Thus,the ratio is $2\sqrt{2} : 1$.

(D) The resultant wave is given by $y = y_1 + y_2 = A \sin(kx - \omega t + \phi)$.
Using the phasor addition method for two waves with amplitudes $A_1 = 4$ and $A_2 = 2$ and a phase difference $\Delta\phi = \frac{2\pi}{3} = 120^{\circ}$,the resultant amplitude $A$ is:
$A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\Delta\phi)}$
$A = \sqrt{4^2 + 2^2 + 2(4)(2) \cos(120^{\circ})}$
$A = \sqrt{16 + 4 + 16(-0.5)} = \sqrt{20 - 8} = \sqrt{12} = 2\sqrt{3}$.
The phase angle $\phi$ of the resultant wave is given by:
$\tan \phi = \frac{A_2 \sin(\Delta\phi)}{A_1 + A_2 \cos(\Delta\phi)}$
$\tan \phi = \frac{2 \sin(120^{\circ})}{4 + 2 \cos(120^{\circ})} = \frac{2(\frac{\sqrt{3}}{2})}{4 + 2(-0.5)} = \frac{\sqrt{3}}{4 - 1} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$.
Thus,$\phi = \frac{\pi}{6}$.

(B) Given:
Natural length of spring $L = 2 \ m$
Mass of block $m = 2 \ kg$
Spring constant $k = 200 \ N/m$
Initial compression $x_i = L - 1 = 2 - 1 = 1 \ m$
Final compression at distance $x$ from the wall is $x_f = L - x = (2 - x) \ m$
Using the principle of conservation of mechanical energy:
$K_i + U_i = K_f + U_f$
Since the block is released from rest,$K_i = 0$.
$0 + \frac{1}{2} k x_i^2 = \frac{1}{2} m v^2 + \frac{1}{2} k x_f^2$
$\frac{1}{2} m v^2 = \frac{1}{2} k (x_i^2 - x_f^2)$
$m v^2 = k (x_i^2 - x_f^2)$
Substituting the values:
$2 \times v^2 = 200 \times [1^2 - (2 - x)^2]$
$v^2 = 100 \times [1 - (2 - x)^2]$
$v = 10 \sqrt{1 - (2 - x)^2} \ m/s$
$v = 10 [1 - (2 - x)^2]^{1/2} \ m/s$

(B) Given: Initial pressure $(P_i) = 1 \ atm$,Final pressure $(P_f) = 5 \ atm$.
Change in pressure $(dP) = P_f - P_i = 4 \ atm = 4 \times 10^5 \ Pa$.
Change in volume $(dV) = -0.8 \ cm^3 = -0.8 \times 10^{-6} \ m^3$.
Bulk modulus $(B) = 2 \ GPa = 2 \times 10^9 \ Pa$.
The formula for bulk modulus is $B = -\frac{dP}{dV/V}$,which implies $V = -\frac{B \cdot dV}{dP}$.
Substituting the values: $V = -\frac{2 \times 10^9 \times (-0.8 \times 10^{-6})}{4 \times 10^5}$.
$V = \frac{1.6 \times 10^3}{4 \times 10^5} = 0.4 \times 10^{-2} \ m^3 = 4 \times 10^{-3} \ m^3$.
Since $1 \ m^3 = 1000 \ litres$,$V = 4 \times 10^{-3} \times 1000 = 4 \ litres$.

(D) Given: Mass $m = 1 \ kg$.
Initial angular velocity $\omega_{i} = 1800 \ rpm = 1800 \times \frac{2 \pi}{60} = 60 \pi \ rad/s$.
Final angular velocity $\omega_{f} = 2100 \ rpm = 2100 \times \frac{2 \pi}{60} = 70 \pi \ rad/s$.
External torque $\tau_{ext} = 25 \pi \ Nm$.
Time $t = 40 \ s$.
Using the equation of rotational motion,$\omega_{f} = \omega_{i} + \alpha t$:
$70 \pi = 60 \pi + \alpha(40) \implies 10 \pi = 40 \alpha \implies \alpha = \frac{\pi}{4} \ rad/s^2$.
For a disk rotating about its diameter,the moment of inertia is $I = \frac{mR^2}{4}$.
Using $\tau = I \alpha$:
$25 \pi = \left( \frac{1 \times R^2}{4} \right) \times \frac{\pi}{4}$.
$25 = \frac{R^2}{16} \implies R^2 = 400 \implies R = 20 \ m$.
The diameter of the disk is $D = 2R = 2 \times 20 = 40 \ m$.

(A) Given: Side of the cube $a = 10 \ cm = 0.1 \ m$. Volume of the cube $V = a^3 = (0.1)^3 = 10^{-3} \ m^3$.
Let the mass of the cube be $m$ and its density be $\rho$. Thus,$m = \rho V = \rho \times 10^{-3} \ kg$.
The distance of the unknown mass from the wedge is $2 \ cm = 0.02 \ m$,and the distance of the $200 \ g$ $(0.2 \ kg)$ mass from the wedge is $25 \ cm = 0.25 \ m$.
When half the volume of the cube is submerged,the buoyant force $F_B$ acting on it is given by $F_B = \rho_{water} \times V_{submerged} \times g = 1000 \ kg/m^3 \times (0.5 \times 10^{-3} \ m^3) \times 10 \ m/s^2 = 5 \ N$.
For the rod to be in rotational equilibrium about the wedge point $O$,the net torque must be zero:
$\tau_{net} = (mg - F_B) \times 0.02 \ m - (0.2 \ kg \times 10 \ m/s^2) \times 0.25 \ m = 0$
$(m \times 10 - 5) \times 0.02 = 2 \times 0.25$
$(10m - 5) \times 0.02 = 0.5$
$10m - 5 = 0.5 / 0.02 = 25$
$10m = 30$
$m = 3 \ kg$.

(C) Let the voltage of the mains supply be $V$. The resistance of each bulb is $R = V^2/P$.
When $n$ identical bulbs are connected in series,the total resistance of the combination is $R_{eq} = nR$.
The total power $P_s$ drawn by the series combination is given by $P_s = V^2 / R_{eq}$.
Substituting $R_{eq} = nR$,we get $P_s = V^2 / (nR)$.
Since $R = V^2/P$,we substitute this into the equation: $P_s = V^2 / (n(V^2/P))$.
Simplifying this,we get $P_s = P/n$.

(C) Statement $(I):$ The energy of a photon is given by $E = hf$. The dimensions of Planck's constant $h$ are $[h] = [E]/[f] = [ML^2T^{-2}]/[T^{-1}] = [ML^2T^{-1}]$. The dimensions of angular momentum $L = mvr$ are $[L] = [M][LT^{-1}][L] = [ML^2T^{-1}]$. Thus,Statement $(I)$ is correct.
Statement $(II):$ According to Bohr's postulate,the angular momentum $L$ of an electron in a stable orbit is an integral multiple of $h/(2\pi)$,i.e.,$L = nh/(2\pi)$,where $n$ is an integer. The statement claims it is an integral multiple of $h$,which is incorrect because the factor $1/(2\pi)$ is missing. Thus,Statement $(II)$ is incorrect.

(A) To obtain an equivalent resistance of $6 \ \Omega$,we analyze the circuit in image $A$:
In the upper branch,$R_1$ and $R_2$ are in series,so $R_{up} = R_1 + R_2 = 5 \ \Omega + 5 \ \Omega = 10 \ \Omega$.
In the lower branch,$R_3$ and $R_4$ are in series,so $R_{low} = R_3 + R_4 = 5 \ \Omega + 10 \ \Omega = 15 \ \Omega$.
These two branches are in parallel,so the equivalent resistance $R_P$ is given by:
$\frac{1}{R_P} = \frac{1}{R_{up}} + \frac{1}{R_{low}} = \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6} \ \Omega^{-1}$.
Therefore,$R_P = 6 \ \Omega$.
Thus,the circuit in image $A$ provides the required equivalent resistance.

(B) The electric field due to a charged arc at its center is directed along the angle bisector. For arc $AB$ (a quarter circle),the electric field $E$ at $O$ is directed at $45^\circ$ from the radii $OA$ and $OB$.
This field $E$ can be resolved into two components: $E_x = E \cos(45^\circ) = E / \sqrt{2}$ along the negative $x$-axis (towards $D$) and $E_y = E \sin(45^\circ) = E / \sqrt{2}$ along the negative $y$-axis (towards $C$).
Arc $ABC$ consists of two identical quarter-circle arcs: $AB$ and $BC$.
For arc $AB$,the field is $\vec{E}_{AB} = (-E/\sqrt{2}) \hat{i} + (-E/\sqrt{2}) \hat{j}$.
For arc $BC$,the field is $\vec{E}_{BC} = (E/\sqrt{2}) \hat{i} + (-E/\sqrt{2}) \hat{j}$.
Adding these vectorially,the $x$-components cancel out,and the $y$-components add up:
$\vec{E}_{ABC} = \vec{E}_{AB} + \vec{E}_{BC} = 0 \hat{i} + (-2E/\sqrt{2}) \hat{j} = -\sqrt{2} E \hat{j}$.
The magnitude is $\sqrt{2} E$.

(C) Initially,all capacitors have capacitance $C = 5 \mu F$.
When a dielectric of constant $K = 4$ is inserted into $C_1$,its new capacitance becomes $C_1' = K \times C = 4 \times 5 \mu F = 20 \mu F$.
The capacitors $C_2$ and $C_3$ remain unchanged,so $C_2 = 5 \mu F$ and $C_3 = 5 \mu F$.
From the circuit diagram,$C_1'$ and $C_2$ are connected in series. Let their equivalent capacitance be $C_{12}$.
$C_{12} = \frac{C_1' \times C_2}{C_1' + C_2} = \frac{20 \times 5}{20 + 5} = \frac{100}{25} = 4 \mu F$.
This combination $C_{12}$ is connected in parallel with $C_3$.
Therefore,the effective capacitance $C_{eq} = C_{12} + C_3 = 4 \mu F + 5 \mu F = 9 \mu F$.

(A) Let the intensity of light incident on $P_1$ be $I_0$. Since $P_1$ and $P_2$ are crossed,the angle between them is $90^{\circ}$ or $\frac{\pi}{2}$.
Let the angle between $P_1$ and $P_3$ be $\theta$. Then the angle between $P_3$ and $P_2$ is $(\frac{\pi}{2} - \theta)$.
According to Malus' Law,the intensity after passing through $P_3$ is $I_1 = I_0 \cos^2 \theta$.
The intensity after passing through $P_2$ is $I_{\text{net}} = I_1 \cos^2(\frac{\pi}{2} - \theta) = I_0 \cos^2 \theta \sin^2 \theta$.
$I_{\text{net}} = I_0 (\sin \theta \cos \theta)^2 = I_0 (\frac{\sin 2\theta}{2})^2 = \frac{I_0}{4} \sin^2(2\theta)$.
For maximum intensity,$\sin^2(2\theta)$ must be $1$,so $2\theta = 90^{\circ}$ or $\theta = 45^{\circ}$ or $\frac{\pi}{4}$.
The angle between $P_3$ and $P_2$ is $\frac{\pi}{2} - \theta = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.

(A) In an $n$-type semiconductor,electrons are majority carriers and holes are minority carriers. This statement is correct.
$(B)$ To create an $n$-type semiconductor,a pentavalent atom (donor) is added. This statement is correct.
$(C)$ According to the law of mass action,the product of electron and hole concentrations is constant at a given temperature,$n_e n_h = n_i^2$. Therefore,the statement $n_e n_h \neq n_i^2$ is incorrect.
$(D)$ Since $n_e n_h = n_i^2$,the statement $n_e n_h \geq n_i^2$ is technically true (as equality holds),but in the context of multiple-choice questions regarding semiconductor properties,$(A), (B),$ and $(E)$ are the standard defining characteristics.
$(E)$ In an $n$-type semiconductor,holes are generated primarily due to thermal excitation,not by the donor atoms. This statement is correct.
Thus,the correct statements are $(A), (B),$ and $(E)$.

(C) For the convex mirror,the object distance $u = -30 \ cm$ and focal length $f = +30 \ cm$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} + \frac{1}{-30} = \frac{1}{30}$
$\frac{1}{v} = \frac{1}{30} + \frac{1}{30} = \frac{2}{30} = \frac{1}{15}$
So,$v = +15 \ cm$. This image is formed $15 \ cm$ behind the convex mirror.
The distance of this image from the object is $30 \ cm + 15 \ cm = 45 \ cm$.
For the plane mirror to produce an image that coincides with the image from the convex mirror,the plane mirror must be placed such that the object distance from the plane mirror equals the image distance from the plane mirror.
Let the plane mirror be at a distance $x$ from the object. The image formed by the plane mirror is at a distance $x$ behind the plane mirror.
The distance of this image from the convex mirror is $(30 - x) + x = 30 \ cm$ (relative to the plane mirror position).
For the images to coincide,the plane mirror must be placed such that its image is at the same position as the convex mirror's image ($15 \ cm$ behind the convex mirror).
Let $d$ be the distance between the mirrors. The object is $30 \ cm$ from the convex mirror. Let the plane mirror be at distance $d$ from the convex mirror. The object is at distance $(30 - d)$ from the plane mirror.
The image formed by the plane mirror is at distance $(30 - d)$ behind the plane mirror,which is at a distance $d + (30 - d) = 30 \ cm$ from the convex mirror.
Wait,the condition is that the images coincide. The convex mirror forms an image at $15 \ cm$ behind its pole. The plane mirror must form an image at the same location.
If the plane mirror is at distance $d$ from the convex mirror,the object is at distance $(30 - d)$ from the plane mirror. The image is at distance $(30 - d)$ behind the plane mirror.
Distance of this image from the convex mirror pole $= d + (30 - d) = 30 \ cm$ (in front of the convex mirror).
This is not correct. Let the plane mirror be at distance $d$ from the convex mirror. The object is at distance $(30-d)$ from the plane mirror. The image is at distance $(30-d)$ behind the plane mirror. The distance of this image from the convex mirror is $d + (30-d) = 30 \ cm$ behind the plane mirror,which is $30-d$ from the convex mirror.
Equating the image positions: $15 = -(30 - 2d) \Rightarrow 15 = -30 + 2d \Rightarrow 2d = 45 \Rightarrow d = 22.5 \ cm$.

(A) Given:
Charge $q = 1.6 \ \mu C = 1.6 \times 10^{-6} \ C$
Mass $m = 16 \ \mu g = 16 \times 10^{-9} \ kg$
Magnetic field $B = 6.28 \ T$
Since the particle is fired perpendicular to the magnetic field,it will perform uniform circular motion.
The time period $T$ for one complete revolution is given by the formula:
$T = \frac{2 \pi m}{qB}$
Substituting the values:
$T = \frac{2 \times 3.14 \times 16 \times 10^{-9}}{1.6 \times 10^{-6} \times 6.28}$
$T = \frac{6.28 \times 16 \times 10^{-9}}{1.6 \times 10^{-6} \times 6.28}$
$T = \frac{16 \times 10^{-9}}{1.6 \times 10^{-6}}$
$T = 10 \times 10^{-3} \ s = 0.01 \ s$
Thus,the time required to return to the original location is $0.01 \ s$.

(C) The velocity of light in a medium is given by $v = \frac{1}{\sqrt{\mu \epsilon}}$.
Since $\mu = \mu_0 \mu_r$ and $\epsilon = \epsilon_0 \epsilon_r$,we have $v = \frac{1}{\sqrt{\mu_0 \mu_r \epsilon_0 \epsilon_r}} = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \cdot \frac{1}{\sqrt{\mu_r \epsilon_r}}$.
Given that the velocity of light in vacuum is $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$,we can write $v = \frac{c}{\sqrt{\mu_r \epsilon_r}}$.
Given $\mu_r = \frac{10}{\pi}$ and $\epsilon_r = \frac{1}{0.0885 \times 10^{-10}}$ (Note: $8.85 \times 10^{-12} = 0.0885 \times 10^{-10}$),let's calculate the product $\mu_r \epsilon_r$.
Actually,using the provided values: $\mu_r \epsilon_r = \frac{10}{\pi} \times \frac{1}{0.0885} \approx 3.183 \times 11.299 \approx 36$.
Thus,$v = \frac{c}{\sqrt{36}} = \frac{c}{6}$.
Therefore,$c = 6v$. The velocity of light in vacuum is $6$ times greater than the velocity in the medium.

(D) The width of the central maximum of a single slit diffraction pattern is given by $w = \frac{2 \lambda D}{a}$,where $a$ is the slit width.
The distance between adjacent maxima in a Young's double slit experiment is the fringe width $\beta = \frac{\lambda D}{d}$.
The total width occupied by $20$ maxima (which corresponds to $20$ fringe widths) is $20 \times \beta = \frac{20 \lambda D}{d}$.
Given that these $20$ maxima are contained within the central maximum of the diffraction pattern:
$\frac{20 \lambda D}{d} = \frac{2 \lambda D}{a}$
$\frac{10}{d} = \frac{1}{a}$
$a = \frac{d}{10}$
Given $d = 1.5 \ mm = 0.15 \ cm = 1.5 \times 10^{-1} \ cm$.
$a = \frac{1.5 \times 10^{-1}}{10} \ cm = 15 \times 10^{-3} \ cm$.
Comparing this with $x \times 10^{-3} \ cm$,we get $x = 15$.

(A) Given: Self-inductance $L = 1 \ H$,Resistance $R = 100 \pi \ \Omega$,Voltage $V_{rms} = 100 \pi \ V$,Frequency $f = 50 \ Hz$.
First,calculate the inductive reactance $X_L = \omega L = 2 \pi f L = 2 \pi \times 50 \times 1 = 100 \pi \ \Omega$.
The impedance of the $LR$ series circuit is $Z = \sqrt{R^2 + X_L^2} = \sqrt{(100 \pi)^2 + (100 \pi)^2} = \sqrt{2 \times (100 \pi)^2} = 100 \pi \sqrt{2} \ \Omega$.
The $RMS$ current is $I_{rms} = \frac{V_{rms}}{Z} = \frac{100 \pi}{100 \pi \sqrt{2}} = \frac{1}{\sqrt{2}} \ A$.
The maximum current $I_{max}$ is given by $I_{max} = I_{rms} \sqrt{2} = \frac{1}{\sqrt{2}} \times \sqrt{2} = 1 \ A$.

(C) The resultant amplitude $A$ of two waves with amplitudes $A_1$ and $A_2$ and phase difference $\phi$ is given by the formula: $A = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2 \cos \phi}$.
Here,$A_1 = E_0$,$A_2 = E_0$,and $\phi = \frac{\pi}{3}$.
Substituting these values into the formula:
$A = \sqrt{E_0^2 + E_0^2 + 2(E_0)(E_0) \cos(\frac{\pi}{3})}$
Since $\cos(\frac{\pi}{3}) = 0.5$,we have:
$A = \sqrt{E_0^2 + E_0^2 + 2E_0^2(0.5)}$
$A = \sqrt{2E_0^2 + E_0^2} = \sqrt{3E_0^2}$
$A = \sqrt{3} E_0 \approx 1.732 E_0$.

(D) Let the resistance of each of the $6$ segments be $r$. Since the total resistance of the wire is $R$,we have $6r = R$,which implies $r = R / 6$.
Looking at the circuit,we can identify that the structure is equivalent to a Wheatstone bridge configuration. The two resistors connected to the top node and the two resistors connected to the central node form a balanced bridge between points $A$ and $B$.
Specifically,the path from $A$ to $B$ consists of two parallel branches:
$1$. $A$ branch with two resistors in series: $r + r = 2r$.
$2$. Another branch with two resistors in series: $r + r = 2r$.
$3$. $A$ direct resistor between $A$ and $B$ with resistance $r$.
Thus,the equivalent resistance $R_{AB}$ is given by:
$\frac{1}{R_{AB}} = \frac{1}{2r} + \frac{1}{2r} + \frac{1}{r} = \frac{1}{r} + \frac{1}{r} = \frac{2}{r}$.
Substituting $r = R / 6$:
$R_{AB} = \frac{r}{2} = \frac{R / 6}{2} = \frac{R}{12}$.
Comparing this with $R / n$,we get $n = 12$.

(D) Since the velocity $\vec{v}$ is perpendicular to the magnetic field $\vec{B}$,the charged particle moves in a circular path with radius $R = \frac{mv}{qB}$.
In region $2$,the particle completes a semi-circle of radius $R_2 = \frac{mv}{qB_2}$ and returns to the interface at point $D$. The distance $AD = 2R_2 = \frac{2mv}{qB_2}$.
Then,the particle enters region $1$ and completes a semi-circle of radius $R_1 = \frac{mv}{qB_1}$ and returns to the interface at point $C$. The distance $AC = 2R_1 = \frac{2mv}{qB_1}$.
The net displacement along the interface is the distance between the starting point $A$ and the final point $C$. Since the particle moves in opposite directions in the two regions relative to the interface,the displacement is $AC = AD - CD$ is incorrect based on the geometry. Looking at the path,the particle moves from $A$ to $D$ (in region $2$) and then from $D$ to $C$ (in region $1$).
The total displacement along the interface is $AC = AD - CD$ is not correct; rather,the particle moves from $A$ to $D$ and then from $D$ to $C$. The distance $AC = AD - CD$ is not the correct interpretation. The distance $AC = 2R_1 - 2R_2 = \frac{2mv}{qB_1} - \frac{2mv}{qB_2} = \frac{2mv}{q} \left(\frac{1}{B_1} - \frac{1}{B_2}\right) = \frac{2mv}{qB_1} \left(1 - \frac{B_1}{B_2}\right)$.

(D) According to Maxwell's modification of Ampere's Law,the displacement current $i_d$ is defined as:
$i_d = \varepsilon_0 \frac{d \phi_{E}}{dt}$
Since $i_d$ represents a current,its dimensions are the same as those of electric current.
Therefore,the dimensions of $\left(\varepsilon_0 \frac{d \phi_{E}}{dt}\right)$ are equivalent to the dimensions of electric current.

(A) The magnetic field inside an air-cored solenoid is $B_0 = \mu_0 n I$.
When the core is filled with a material of relative permeability $\mu_r$,the new magnetic field is $B = \mu n I = \mu_0 \mu_r n I$.
The change in magnetic field is $\Delta B = B - B_0 = \mu_0 (\mu_r - 1) n I$.
The percentage increase is given by $\frac{\Delta B}{B_0} \times 100 \% = (\mu_r - 1) \times 100 \%$.
Since magnetic susceptibility $\chi = \mu_r - 1$,the percentage increase is $\chi \times 100 \%$.
Given $\chi_{mg} = 1.2 \times 10^{-5}$,the percentage increase is $(1.2 \times 10^{-5}) \times 100 \% = 1.2 \times 10^{-3} \%$.

(B) Using the lens maker's formula: $\frac{1}{f} = (\frac{\mu_L}{\mu_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.
For air,$\mu_m = 1$ and $f = 12 \ cm$:
$\frac{1}{12} = (1.6 - 1)(\frac{1}{R_1} - \frac{1}{R_2})$
$\frac{1}{12} = 0.6(\frac{1}{R_1} - \frac{1}{R_2})$
$(\frac{1}{R_1} - \frac{1}{R_2}) = \frac{1}{12 \times 0.6} = \frac{1}{7.2} \ cm^{-1}$.
For water,$\mu_m = 1.28$:
$\frac{1}{f_w} = (\frac{1.6}{1.28} - 1)(\frac{1}{7.2})$
$\frac{1}{f_w} = (1.25 - 1)(\frac{1}{7.2}) = 0.25 \times \frac{1}{7.2} = \frac{1}{4 \times 7.2} = \frac{1}{28.8} \ cm^{-1}$.
Thus,$f_w = 28.8 \ cm = 288 \ mm$.

(C) The given current is $i = 5 \sqrt{2} + 10 \cos \left(650 \pi t + \frac{\pi}{6}\right)$.
To find the $r.m.s$ value,we use the formula $I_{rms} = \sqrt{\langle i^2 \rangle}$.
First,calculate $i^2 = \left(5 \sqrt{2} + 10 \cos \left(650 \pi t + \frac{\pi}{6}\right)\right)^2$.
$i^2 = (5 \sqrt{2})^2 + (10 \cos \left(650 \pi t + \frac{\pi}{6}\right))^2 + 2(5 \sqrt{2})(10 \cos \left(650 \pi t + \frac{\pi}{6}\right))$.
$i^2 = 50 + 100 \cos^2 \left(650 \pi t + \frac{\pi}{6}\right) + 100 \sqrt{2} \cos \left(650 \pi t + \frac{\pi}{6}\right)$.
Taking the average over a full cycle,$\langle \cos \theta \rangle = 0$ and $\langle \cos^2 \theta \rangle = \frac{1}{2}$.
$\langle i^2 \rangle = 50 + 100 \left(\frac{1}{2}\right) + 0 = 50 + 50 = 100$.
Therefore,$I_{rms} = \sqrt{100} = 10 \text{ A}$.

(D) Given focal lengths are $f_1 = 30 \ cm = 0.3 \ m$ and $f_2 = 10 \ cm = 0.1 \ m$.
The distance between the lenses is $d = 1 \ cm = 0.01 \ m$.
The equivalent focal length $f_{eq}$ of two lenses separated by a distance $d$ is given by the formula:
$\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$
Substituting the values in meters:
$\frac{1}{f_{eq}} = \frac{1}{0.3} + \frac{1}{0.1} - \frac{0.01}{0.3 \times 0.1}$
$\frac{1}{f_{eq}} = 3.33 + 10 - \frac{0.01}{0.03}$
$\frac{1}{f_{eq}} = 3.33 + 10 - 0.33 = 13 \ D$
Wait,re-calculating: $\frac{1}{0.3} = 3.33$,$\frac{1}{0.1} = 10$,$\frac{0.01}{0.03} = 0.333$.
$P = 3.33 + 10 - 3.33 = 10 \ D$.
Thus,the power of the combination is $10 \ D$.

(C) First,we check if the Zener diode is in breakdown by calculating the voltage across the parallel branch without the Zener diode:
$V_{open} = \frac{400 \ \Omega}{100 \ \Omega + 400 \ \Omega} \times 12 \ V = \frac{400}{500} \times 12 \ V = 0.8 \times 12 \ V = 9.6 \ V$.
Since the calculated voltage $9.6 \ V$ is greater than the Zener breakdown voltage $V_z = 4 \ V$,the Zener diode operates in the breakdown region.
Therefore,the voltage across the $400 \ \Omega$ resistor is fixed at the Zener voltage,$V = 4 \ V$.
The current $I$ through the ammeter is given by Ohm's law:
$I = \frac{V}{R} = \frac{4 \ V}{400 \ \Omega} = 0.01 \ A = 10 \ mA$.

(D) The energy of an electron in a hydrogen-like ion is given by $E_n = -13.6 \frac{Z^2}{n^2} \ eV$.
For the ground state,$n_1 = 1$. For the $2^{\text{nd}}$ excitation state,$n_2 = 3$ (since $1^{\text{st}}$ excitation is $n=2$ and $2^{\text{nd}}$ excitation is $n=3$).
The energy difference is given by $\Delta E = E_3 - E_1 = 13.6 Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Substituting the given values: $108.8 = 13.6 Z^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right)$.
$108.8 = 13.6 Z^2 \left( 1 - \frac{1}{9} \right) = 13.6 Z^2 \left( \frac{8}{9} \right)$.
$Z^2 = \frac{108.8 \times 9}{13.6 \times 8} = 8 \times 9 = 9$.
$Z = 3$.

(B) The wavelength $\lambda$ for a transition in a hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$,where $R$ is the Rydberg constant.
For the Lyman series,the largest wavelength corresponds to the transition from $n_i = 2$ to $n_f = 1$:
$\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4} \implies \lambda_L = \frac{4}{3R}$.
For the Balmer series,the largest wavelength corresponds to the transition from $n_i = 3$ to $n_f = 2$:
$\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5R}{36} \implies \lambda_B = \frac{36}{5R}$.
The ratio of the largest wavelength of the Lyman series to that of the Balmer series is:
$\frac{\lambda_L}{\lambda_B} = \frac{4/3R}{36/5R} = \frac{4}{3} \times \frac{5}{36} = \frac{5}{27}$.
Thus,the ratio is $5: 27$.

(D) The magnetic force provides the necessary centripetal force for circular motion:
$\frac{mv^2}{r} = qvB$
$mv = qBr$
$r = \frac{mv}{qB}$
Since kinetic energy $E = \frac{1}{2}mv^2$,we have $v = \sqrt{\frac{2E}{m}}$.
Substituting $v$ in the expression for $r$:
$r = \frac{m}{qB} \sqrt{\frac{2E}{m}} = \frac{\sqrt{2mE}}{qB}$
Thus,$r = \left( \frac{\sqrt{2m}}{qB} \right) \sqrt{E}$
This implies $r \propto \sqrt{E}$.
The graph of $r$ versus $E$ is a parabola opening towards the $E$-axis,which corresponds to Graph $D$.

(A) The potential at point $C$ due to charges $q_1$ and $q_2$ is:
$V_C = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{AC} + \frac{q_2}{BC} \right)$
Given $AC = 40\ cm = 0.4\ m$. Since $AB = 30\ cm$ and $AC = 40\ cm$,by Pythagoras theorem,$BC = \sqrt{0.3^2 + 0.4^2} = 0.5\ m$.
So,$V_C = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{0.4} + \frac{q_2}{0.5} \right)$.
The potential at point $D$ due to charges $q_1$ and $q_2$ is:
$V_D = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{AD} + \frac{q_2}{BD} \right)$
Given $AD = 40\ cm = 0.4\ m$ (radius of the circular path) and $BD = AD - AB = 40\ cm - 30\ cm = 10\ cm = 0.1\ m$.
So,$V_D = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{0.4} + \frac{q_2}{0.1} \right)$.
The change in potential energy $\Delta U$ is given by:
$\Delta U = q_3 (V_D - V_C) = q_3 \left[ \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{0.4} + \frac{q_2}{0.1} \right) - \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{0.4} + \frac{q_2}{0.5} \right) \right]$
$\Delta U = \frac{q_3}{4 \pi \epsilon_0} \left( \frac{q_2}{0.1} - \frac{q_2}{0.5} \right) = \frac{q_3}{4 \pi \epsilon_0} (10 q_2 - 2 q_2) = \frac{8 q_2 q_3}{4 \pi \epsilon_0}$.
Comparing this with $\frac{q_3 K}{4 \pi \epsilon_0}$,we get $K = 8 q_2$.

(B) Let the input voltage be $V_{\text{in}}$. The circuit consists of two potential dividers in parallel.
For the left branch,the voltage at point $A$ with respect to the bottom node is $V_A = V_{\text{in}} \cdot \frac{-jX_C}{R - jX_C}$.
For the right branch,the voltage at point $B$ with respect to the bottom node is $V_B = V_{\text{in}} \cdot \frac{R}{R - jX_C}$.
Thus,the potential difference $(V_B - V_A)$ is given by:
$V_B - V_A = V_{\text{in}} \cdot \frac{R + jX_C}{R - jX_C}$.
Let $Z = R - jX_C$. Then $V_B - V_A = V_{\text{in}} \cdot \frac{R + jX_C}{R - jX_C}$.
The phase of $(V_B - V_A)$ relative to $V_{\text{in}}$ is the phase of $\frac{R + jX_C}{R - jX_C}$.
Let $\tan \theta = \frac{X_C}{R}$. Then the phase of $(R + jX_C)$ is $\theta$ and the phase of $(R - jX_C)$ is $-\theta$.
The phase of the ratio is $\theta - (-\theta) = 2\theta$.
Given the phase difference is $90^{\circ}$,we have $2\theta = 90^{\circ}$,so $\theta = 45^{\circ}$.
Therefore,$\tan 45^{\circ} = \frac{X_C}{R} \implies 1 = \frac{1}{\omega RC} \implies \omega = \frac{1}{RC}$.
Substituting the values: $R = 10^5 \ \Omega$,$C = 100 \times 10^{-12} \ F = 10^{-10} \ F$.
$\omega = \frac{1}{10^5 \times 10^{-10}} = \frac{1}{10^{-5}} = 10^5 \ rad/sec$.
Comparing with $10^x$,we get $x = 5$.

(C) The apparent depth $d'$ of the bottom of the container when viewed from above through multiple layers of liquids is given by the formula:
$d' = \frac{h_1}{\mu_1} + \frac{h_2}{\mu_2}$
Here,$h_1 = 60 \ cm$,$\mu_1 = 1.2$,$h_2 = H$,and $\mu_2 = 1.6$.
So,$d' = \frac{60}{1.2} + \frac{H}{1.6} = 50 + \frac{H}{1.6}$.
The actual depth of the bottom is $d = 60 + H$.
The apparent shift is given by the difference between the actual depth and the apparent depth:
$\text{Shift} = d - d'$
$40 = (60 + H) - (50 + \frac{H}{1.6})$
$40 = 10 + H - \frac{H}{1.6}$
$30 = H(1 - \frac{1}{1.6})$
$30 = H(\frac{1.6 - 1}{1.6}) = H(\frac{0.6}{1.6})$
$30 = H(\frac{6}{16}) = H(\frac{3}{8})$
$H = 30 \times \frac{8}{3} = 80 \ cm$.

(A) The phenomenon described is known as electrostatic shielding. According to the properties of conductors in electrostatic equilibrium,the electric field inside a closed metallic cavity is always zero,regardless of the external electric field or charge distribution. When lightning strikes an aircraft,the metal body acts as a Faraday cage,ensuring that the electric field inside remains zero and the occupants remain safe. Therefore,Assertion $(A)$ is correct because it describes the protective effect,and Reason $(R)$ is the correct scientific explanation for this phenomenon.

(B) The density of a nucleus is given by $\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{A \cdot m_n}{\frac{4}{3} \pi R^3}$,where $A$ is the mass number and $m_n$ is the average mass of a nucleon.
Since the radius of a nucleus is given by $R = R_0 A^{1/3}$,substituting this into the density formula gives $\rho = \frac{A \cdot m_n}{\frac{4}{3} \pi (R_0 A^{1/3})^3} = \frac{A \cdot m_n}{\frac{4}{3} \pi R_0^3 A} = \frac{m_n}{\frac{4}{3} \pi R_0^3}$.
This shows that the nuclear density $\rho$ is independent of the mass number $A$.
Therefore,the density of all nuclei is approximately constant and does not depend on the element.
Thus,Assertion $(A)$ is incorrect because the densities are equal,while Reason $(R)$ is a correct statement regarding the relationship between nuclear radius and mass number.

(B) The intensity $I$ of an electromagnetic wave is given by the relation $I = \frac{1}{2} \varepsilon_0 E_0^2 c$,where $E_0$ is the amplitude of the electric field.
Rearranging the formula for $E_0$,we get $E_0^2 = \frac{2 I}{\varepsilon_0 c}$,which implies $E_0 = \sqrt{\frac{2 I}{\varepsilon_0 c}}$.
Since $E_0$ represents the electric field,its $SI$ unit is the same as the electric field,which is $\text{Volt per meter}$ $(V/m)$ or $\text{Newton per Coulomb}$ $(N/C)$.
Comparing this with the given options,the correct unit is $N/C$.

(D) The speed of light in vacuum is given by $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$.
Also,the characteristic impedance of free space is given by $Z_0 = \sqrt{\frac{\mu_0}{\epsilon_0}}$.
The unit of impedance is $\text{ohm} (\Omega)$,which is the same as the unit of resistance.
Therefore,the dimension of $\sqrt{\frac{\mu_0}{\epsilon_0}}$ is the same as the dimension of resistance.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E.)$ of the emitted photoelectrons is given by: $K.E. = E - W$,where $E = \frac{hc}{\lambda_i}$ is the energy of the incident photon and $W = \frac{hc}{\lambda_0}$ is the work function of the substance.
Substituting these,we get: $K.E. = \frac{hc}{\lambda_i} - \frac{hc}{\lambda_0} = hc \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)$.
The de-Broglie wavelength $\lambda_c$ of an electron with kinetic energy $K.E.$ is given by $\lambda_c = \frac{h}{\sqrt{2m(K.E.)}}$.
Squaring both sides,we get $\lambda_c^2 = \frac{h^2}{2m(K.E.)}$,which implies $K.E. = \frac{h^2}{2m\lambda_c^2}$.
Equating the two expressions for $K.E.$: $\frac{h^2}{2m\lambda_c^2} = hc \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)$.
Solving for $\lambda_c$: $\lambda_c^2 = \frac{h^2}{2mhc \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)} = \frac{h}{2mc \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)}$.
Therefore,$\lambda_c = \sqrt{\frac{h}{2mc \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)}}$.

(C) Given magnification $m = \frac{1}{4}$. Since the image is smaller than the object and formed by a mirror,we assume a concave mirror forming a real image.
For a real image,$m = -\frac{v}{u} = -\frac{1}{4}$,so $u = 4v$.
The distance between the object and the image is $|u - v| = 40 \ cm$.
Since $u$ and $v$ are on the same side for a real image,$u - v = 40 \ cm$.
Substituting $u = 4v$,we get $4v - v = 40 \implies 3v = 40 \implies v = \frac{40}{3} \ cm$.
Then $u = 4 \times \frac{40}{3} = \frac{160}{3} \ cm$.
Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ with sign convention ($u = -\frac{160}{3} \ cm$,$v = -\frac{40}{3} \ cm$):
$\frac{1}{f} = -\frac{3}{40} - \frac{3}{160} = \frac{-12 - 3}{160} = -\frac{15}{160} = -\frac{3}{32}$.
$f = -\frac{32}{3} \approx -10.7 \ cm$.
The magnitude of the focal length is $10.7 \ cm$.

(A) The electric field $E$ between the plates of the capacitor is given by $E = \frac{V}{d}$,where $V = 5 \ V$ and $d = 0.5 \ mm = 5 \times 10^{-4} \ m$.
$E = \frac{5}{5 \times 10^{-4}} = 10^4 \ V/m$.
The dipole moment $p$ is given by $p = q \times a$,where $q = 2 \ \mu C = 2 \times 10^{-6} \ C$ and $a = 0.5 \ \mu m = 5 \times 10^{-7} \ m$.
$p = 2 \times 10^{-6} \times 5 \times 10^{-7} = 1 \times 10^{-12} \ C \cdot m$.
The torque $\tau$ acting on the dipole is given by $\tau = pE \sin \theta$,where $\theta = 30^{\circ}$.
$\tau = (1 \times 10^{-12} \ C \cdot m) \times (10^4 \ V/m) \times \sin(30^{\circ})$.
$\tau = 10^{-8} \times 0.5 = 5 \times 10^{-9} \ N \cdot m$.

(A) From the circuit diagram,the output of the $AND$ gate is $Y_1 = A \cdot B$.
The inputs to the $OR$ gate are $B$ and $\overline{A}$,so its output is $Y_2 = \overline{A} + B$.
These two outputs $Y_1$ and $Y_2$ are fed into a $NAND$ gate,so the final output is $Y = \overline{Y_1 \cdot Y_2} = \overline{A \cdot B} \cdot \overline{(\overline{A} + B)}$ is incorrect; the correct expression is $Y = \overline{Y_1 \cdot Y_2} = \overline{(A \cdot B) \cdot (\overline{A} + B)}$.
Simplifying the expression: $Y = \overline{A \cdot B \cdot \overline{A} + A \cdot B \cdot B} = \overline{0 + A \cdot B} = \overline{A \cdot B}$.
For $Y = 0$,we need $\overline{A \cdot B} = 0$,which implies $A \cdot B = 1$.
This occurs only when $A = 1$ and $B = 1$.

(C) The refractive index of a medium is a measure of how much the speed of light is reduced within that medium compared to a vacuum. Glass is optically denser than air,so its refractive index is higher. This makes Assertion $(A)$ correct.
The optical density of a medium is not directly related to its mass density. For example,turpentine has a higher mass density than water but is optically less dense. Therefore,the statement that optical density is directly proportionate to mass density is false. This makes Reason $(R)$ incorrect.
Thus,$(A)$ is correct but $(R)$ is not correct.

(C) Assertion $(A)$ is correct because, according to Gauss's Law for magnetism, the net magnetic flux through any closed surface is zero $( \oint \vec{B} \cdot d\vec{A} = 0)$. This implies that there are no isolated magnetic charges (monopoles).
Reason $(R)$ is also correct because magnetic field lines always form continuous closed loops, starting from the North pole and ending at the South pole outside the magnet, and continuing from South to North inside the magnet.
Since magnetic field lines form closed loops, they do not have a starting or ending point (source or sink), which directly explains why magnetic monopoles cannot exist. Therefore, $(R)$ is the correct explanation of $(A)$.

(C) For total internal reflection to occur at the top surface,the angle of incidence at that surface must be at least the critical angle $\theta_C$.
From the geometry of the triangle formed inside the block,the angle of refraction $r$ at the first surface and the angle of incidence $\theta_C$ at the top surface satisfy $r + \theta_C = 90^{\circ}$,so $r = 90^{\circ} - \theta_C$.
Applying Snell's Law at the first surface: $\mu_1 \sin \theta = \mu_2 \sin r$.
Substituting $r = 90^{\circ} - \theta_C$: $\sin \theta = \frac{\mu_2}{\mu_1} \sin(90^{\circ} - \theta_C) = \frac{\mu_2}{\mu_1} \cos \theta_C$.
For total internal reflection,$\sin \theta_C = \frac{\mu_1}{\mu_2} = \frac{1.0}{1.25} = \frac{1}{5/4} = \frac{4}{5}$.
Thus,$\cos \theta_C = \sqrt{1 - \sin^2 \theta_C} = \sqrt{1 - (4/5)^2} = \sqrt{1 - 16/25} = \sqrt{9/25} = 3/5$.
Substituting this into the equation for $\sin \theta$: $\sin \theta = \frac{1.25}{1.0} \times \frac{3}{5} = \frac{5}{4} \times \frac{3}{5} = \frac{3}{4}$.
Therefore,$\theta = \sin^{-1}(3/4)$.

(A) The induced charge $Q_{in}$ on a dielectric slab placed in an electric field is given by the formula: $Q_{in} = Q \left(1 - \frac{1}{K}\right)$, where $Q$ is the charge on the capacitor plates and $K$ is the dielectric constant.
Given: $Q = 5 \times 10^{-6} \ C$ and $Q_{in} = 4 \times 10^{-6} \ C$.
Substituting the values into the formula:
$4 \times 10^{-6} = 5 \times 10^{-6} \left(1 - \frac{1}{K}\right)$
$0.8 = 1 - \frac{1}{K}$
$\frac{1}{K} = 1 - 0.8 = 0.2$
$K = \frac{1}{0.2} = 5$.
Therefore, the dielectric constant of the slab is $5$.

(B) The impedance $Z$ of an $LCR$ series circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Given $R = 50 \ \Omega$, $X_L = 100 \ \Omega$, and $X_C = 50 \ \Omega$.
$Z = \sqrt{50^2 + (100 - 50)^2} = \sqrt{50^2 + 50^2} = \sqrt{2500 + 2500} = \sqrt{5000} = 50\sqrt{2} \ \Omega$.
The average power dissipated in an $AC$ circuit is $P = V_{rms} I_{rms} \cos \phi$, where $\cos \phi = \frac{R}{Z}$ is the power factor.
$P = V_{rms} \times \left( \frac{V_{rms}}{Z} \right) \times \left( \frac{R}{Z} \right) = \frac{V_{rms}^2 R}{Z^2}$.
Given $V_{rms} = 10 \ V$.
$P = \frac{10^2 \times 50}{(50\sqrt{2})^2} = \frac{100 \times 50}{2500 \times 2} = \frac{5000}{5000} = 1 \ W$.

(D) The electric field is given by $\vec{E} = (2 \hat{i} + 4 \hat{j} + 6 \hat{k}) \times 10^3 \ N/C$.
Since the surface is parallel to the $x-z$ plane,its area vector $\vec{A}$ must be perpendicular to the $x-z$ plane,which means it is directed along the $y$-axis. Thus,$\vec{A} = A \hat{j}$.
The electric flux $\phi$ is given by the dot product $\phi = \vec{E} \cdot \vec{A}$.
Substituting the values: $\phi = (2 \hat{i} + 4 \hat{j} + 6 \hat{k}) \times 10^3 \cdot (A \hat{j}) = 4 \times 10^3 \times A$.
Given $\phi = 6.0 \ N m^2 C^{-1}$,we have $6.0 = 4 \times 10^3 \times A$.
Solving for $A$: $A = \frac{6.0}{4 \times 10^3} = 1.5 \times 10^{-3} \ m^2$.
Converting to $cm^2$: $A = 1.5 \times 10^{-3} \times (10^2 \ cm)^2 = 1.5 \times 10^{-3} \times 10^4 \ cm^2 = 15 \ cm^2$.

(B) $1$. For a uniformly charged spherical shell of radius $R$ and charge $Q$,the electric field inside the shell is zero $(E = 0)$.
$2$. Since the electric field is zero inside,the potential $V$ is constant throughout the interior and is equal to the potential at the surface,given by $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$.
$3$. The work done $W$ in moving a test charge $q$ between two points $1$ and $2$ is given by $W = q(V_2 - V_1)$.
$4$. Since $V_1 = V_2$ at any two points inside the shell,$W = q(0) = 0$.
$5$. Thus,Assertion $A$ is true because the potential is constant,and Reason $R$ is true and provides the correct explanation for why the work done is zero.

(C) For a charged conductor,the surface charge density $\sigma$ is inversely proportional to the radius of curvature $(ROC)$ at that point,i.e.,$\sigma \propto \frac{1}{ROC}$.
From the figure,the radius of curvature at the sharpest point (top) is the smallest,and it increases as we move towards the flatter sides.
Comparing the points:
$1$. The point at $\sigma_1$ has the smallest radius of curvature.
$2$. The point at $\sigma_3$ has a larger radius of curvature than $\sigma_1$ but smaller than the sides.
$3$. The points at $\sigma_2$ and $\sigma_4$ are on the flatter sides,where the radius of curvature is the largest and equal.
Thus,the order of radius of curvature is $(ROC)_1 < (ROC)_3 < (ROC)_2 = (ROC)_4$.
Since $\sigma \propto \frac{1}{ROC}$,the order of surface charge densities is $\sigma_1 > \sigma_3 > \sigma_2 = \sigma_4$.
Therefore,the correct option is $(C)$.

(D) The system consists of a liquid lens (plano-concave) and a glass lens (concave-convex).
For the liquid lens: $\mu_l = 1.3$,$R_1 = \infty$,$R_2 = -30 \ cm$.
Using the lens maker's formula: $\frac{1}{f_l} = (\mu_l - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = (1.3 - 1) \left( \frac{1}{\infty} - \frac{1}{-30} \right) = 0.3 \times \frac{1}{30} = \frac{0.3}{30} = \frac{1}{100} \ cm^{-1}$.
For the glass lens: $\mu_g = 1.5$,$R_1 = -30 \ cm$,$R_2 = -20 \ cm$.
Using the lens maker's formula: $\frac{1}{f_g} = (\mu_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = (1.5 - 1) \left( \frac{1}{-30} - \frac{1}{-20} \right) = 0.5 \left( \frac{-2 + 3}{60} \right) = 0.5 \times \frac{1}{60} = \frac{1}{120} \ cm^{-1}$.
The total focal length $f$ is given by $\frac{1}{f} = \frac{1}{f_l} + \frac{1}{f_g} = \frac{1}{100} + \frac{1}{120} = \frac{6 + 5}{600} = \frac{11}{600} \ cm^{-1}$.
Therefore,$f = \frac{600}{11} \ cm$.

(D) According to Gauss's Law, the net electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enc}}{\varepsilon_0}$.
The wire passes through two maximally displaced corners of the cube, which means it passes along the body diagonal of the cube.
The length of the body diagonal of a cube with side length $a$ is $L = \sqrt{3}a$.
Given $a = \sqrt{3} \ cm = \sqrt{3} \times 10^{-2} \ m$, the length of the wire inside the cube is $L = \sqrt{3} \times (\sqrt{3} \times 10^{-2} \ m) = 3 \times 10^{-2} \ m$.
The charge enclosed by the cube is $q_{enc} = \lambda \cdot L = (2 \times 10^{-9} \ C/m) \times (3 \times 10^{-2} \ m) = 6 \times 10^{-11} \ C$.
Using $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9$, we have $\frac{1}{\varepsilon_0} = 36 \pi \times 10^9$.
Therefore, the flux $\phi = q_{enc} \cdot \frac{1}{\varepsilon_0} = (6 \times 10^{-11}) \times (36 \pi \times 10^9) = 216 \pi \times 10^{-2} = 2.16 \pi \ Nm^2 C^{-1}$.
Thus, $x = 2.16 \pi$.

(B) In the given circuit,diode $D_1$ has its anode connected to $+5 \ V$ and its cathode connected to $V_{\text{out}}$. Diode $D_2$ has its anode connected to $V_{\text{out}}$ and its cathode connected to ground $(0 \ V)$.
Assuming $D_2$ is forward biased,$V_{\text{out}}$ will be clamped to $0 \ V$ (since the diode is ideal).
If $V_{\text{out}} = 0 \ V$,then for $D_1$,the anode is at $+5 \ V$ and the cathode is at $0 \ V$. Thus,$D_1$ is forward biased.
However,if $D_1$ is forward biased,$V_{\text{out}}$ would be $+5 \ V$. If $V_{\text{out}} = +5 \ V$,then $D_2$ has its anode at $+5 \ V$ and cathode at $0 \ V$,making $D_2$ forward biased.
Since $D_2$ is forward biased,it pulls the output voltage to the potential of its cathode,which is $0 \ V$.
Therefore,the output voltage $V_{\text{out}} = 0 \ V$.

(D) The potential $V$ of a conducting sphere of radius $r$ with surface charge density $\sigma$ is given by $V = \frac{\sigma r}{\varepsilon_0}$.
When two conducting spheres are brought into contact,charge flows between them until their potentials become equal,i.e.,$V_1 = V_2$.
Substituting the expression for potential,we get $\frac{\sigma_1 r_1}{\varepsilon_0} = \frac{\sigma_2 r_2}{\varepsilon_0}$.
This simplifies to $\sigma_1 r_1 = \sigma_2 r_2$.
Given $r_1 = R$ and $r_2 = 3R$,we have $\sigma_1 R = \sigma_2 (3R)$.
Therefore,the ratio $\frac{\sigma_1}{\sigma_2} = \frac{3R}{R} = 3$.

(D) The focal length of the convex lens is $f_1 = +30 \ cm$ and the focal length of the concave lens is $f_2 = -20 \ cm$.
For a combination of two thin lenses in contact,the equivalent focal length $f$ is given by:
$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$
Substituting the values:
$\frac{1}{f} = \frac{1}{30} + \frac{1}{-20} = \frac{2 - 3}{60} = -\frac{1}{60}$
Thus,$f = -60 \ cm$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,where the object distance $u = -20 \ cm$:
$\frac{1}{v} - \frac{1}{-20} = \frac{1}{-60}$
$\frac{1}{v} + \frac{1}{20} = -\frac{1}{60}$
$\frac{1}{v} = -\frac{1}{60} - \frac{1}{20} = \frac{-1 - 3}{60} = -\frac{4}{60} = -\frac{1}{15}$
Therefore,$v = -15 \ cm$.
The distance of the image from the lens is $15 \ cm$.

(C) The total current $I$ divides into two parallel paths $\text{ABC}$ and $\text{ADC}$.
Since the resistance of $\text{ABC}$ is $r$ and $\text{ADC}$ is $2r$,the currents $i_1$ and $i_2$ are inversely proportional to the resistances:
$i_1 = I \cdot \frac{2r}{r + 2r} = \frac{2I}{3}$ (through $\text{ABC}$)
$i_2 = I \cdot \frac{r}{r + 2r} = \frac{I}{3}$ (through $\text{ADC}$)
The magnetic field at the centre due to a finite wire of length $a$ at distance $d = a/2$ is $B = \frac{\mu_0 i}{4 \pi d} (\sin \theta_1 + \sin \theta_2)$. For a square,$\theta_1 = \theta_2 = 45^\circ$,so $B = \frac{\mu_0 i}{4 \pi (a/2)} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 i}{\sqrt{2} \pi a}$.
The field due to $\text{ABC}$ (carrying $2I/3$) is $B_1 = 2 \times \frac{\mu_0 (2I/3)}{\sqrt{2} \pi a} = \frac{2\sqrt{2} \mu_0 I}{3 \pi a}$ (into the page).
The field due to $\text{ADC}$ (carrying $I/3$) is $B_2 = 2 \times \frac{\mu_0 (I/3)}{\sqrt{2} \pi a} = \frac{\sqrt{2} \mu_0 I}{3 \pi a}$ (out of the page).
The resultant field is $B_{net} = B_1 - B_2 = \frac{2\sqrt{2} \mu_0 I}{3 \pi a} - \frac{\sqrt{2} \mu_0 I}{3 \pi a} = \frac{\sqrt{2} \mu_0 I}{3 \pi a}$.

(B) For an interference pattern to be observed,the two sources must be coherent.
Coherent sources must emit light of the same frequency and wavelength.
$A$ red filter only allows red light (wavelength $\lambda_R \approx 650 \ nm$) to pass,while a green filter only allows green light (wavelength $\lambda_G \approx 550 \ nm$) to pass.
Since the two slits now emit light of different frequencies and wavelengths,the phase difference between the waves at any point on the screen will change rapidly with time.
Therefore,the sources are incoherent,and no stable interference pattern will be formed on the screen.