AIEEE 2008 Physics Question Paper with Answer and Solution

(A) The pitch of the screw gauge is the distance moved in one full rotation. Since $2$ full turns cover $1 \ mm$,the pitch $= \frac{1 \ mm}{2} = 0.5 \ mm$.
The least count $(LC)$ is given by $\frac{\text{pitch}}{\text{total number of divisions}} = \frac{0.5 \ mm}{50} = 0.01 \ mm$.
The observed reading is calculated as: $\text{Main Scale Reading} + (\text{Circular Scale Division} \times LC) = 3 \ mm + (35 \times 0.01 \ mm) = 3.35 \ mm$.
The actual diameter is calculated by subtracting the zero error from the observed reading: $\text{Diameter} = \text{Observed Reading} - (\text{Zero Error}) = 3.35 \ mm - (-0.03 \ mm) = 3.35 \ mm + 0.03 \ mm = 3.38 \ mm$.

(C) For the body starting from rest at $x=0$ with constant acceleration $a$:
$x_{1} = \frac{1}{2}at^{2}$
For the body moving with constant speed $v$ starting from $x=0$:
$x_{2} = vt$
Let $f(t) = x_{1} - x_{2} = \frac{1}{2}at^{2} - vt$.
This is a quadratic equation in $t$ representing a parabola opening upwards.
At $t=0$,$f(0) = 0$.
The derivative is $f'(t) = at - v$.
Setting $f'(t) = 0$ gives $t = \frac{v}{a}$.
At $t = \frac{v}{a}$,the function reaches its minimum value: $f(\frac{v}{a}) = \frac{1}{2}a(\frac{v}{a})^{2} - v(\frac{v}{a}) = \frac{v^{2}}{2a} - \frac{v^{2}}{a} = -\frac{v^{2}}{2a}$.
Since the minimum value is negative and the parabola opens upwards,the graph starts at the origin,goes below the $t$-axis,reaches a minimum at $t = \frac{v}{a}$,and then increases,crossing the $t$-axis at $t = \frac{2v}{a}$.
This matches the graph shown in the solution image.

(B) The momentum $p$ is given by the product of mass $m$ and velocity $v$: $p = m \times v$.
Given $m = 3.513 \; kg$ (which has $4$ significant figures) and $v = 5.00 \; ms^{-1}$ (which has $3$ significant figures).
Calculating the product: $p = 3.513 \times 5.00 = 17.565 \; kg \; ms^{-1}$.
According to the rules of significant figures,the result of a multiplication should have the same number of significant figures as the measurement with the fewest significant figures.
Since $5.00$ has $3$ significant figures,the final result must be rounded to $3$ significant figures.
Rounding $17.565$ to $3$ significant figures gives $17.6 \; kg \; ms^{-1}$.

(A) The average speed of the athlete is $v = \frac{100 \ m}{10 \ s} = 10 \ m/s$.
The kinetic energy $(K.E.)$ is given by the formula $K.E. = \frac{1}{2}mv^2$.
Assuming the mass $(m)$ of an athlete typically ranges between $40 \ kg$ and $100 \ kg$:
For $m = 40 \ kg$: $K.E. = \frac{1}{2} \times 40 \times (10)^2 = 20 \times 100 = 2000 \ J$.
For $m = 100 \ kg$: $K.E. = \frac{1}{2} \times 100 \times (10)^2 = 50 \times 100 = 5000 \ J$.
Thus,the range of kinetic energy is $2000 \ J - 5000 \ J$.

(D) Given: $m_1 = 0.50 \ kg$,$u_1 = 2.00 \ m/s$,$m_2 = 1.00 \ kg$,$u_2 = 0 \ m/s$.
Since the bodies move together after the collision,it is a perfectly inelastic collision.
The formula for energy loss in a perfectly inelastic collision is given by:
$\Delta K = \frac{m_1 m_2}{2(m_1 + m_2)} (u_1 - u_2)^2$
Substituting the values:
$\Delta K = \frac{0.50 \times 1.00}{2(0.50 + 1.00)} (2.00 - 0)^2$
$\Delta K = \frac{0.50}{2(1.50)} (4)$
$\Delta K = \frac{0.50}{3.00} \times 4 = \frac{2}{3} \approx 0.67 \ J$.

(B) Since the container is insulated,$Q = 0$. Since the partition is removed without doing any work,$W = 0$.
According to the first law of thermodynamics,$\Delta U = Q + W = 0$.
This implies that the total internal energy remains constant: $U_{initial} = U_{final}$.
For an ideal gas,the internal energy is $U = n C_v T$.
Thus,$n_1 C_v T_1 + n_2 C_v T_2 = (n_1 + n_2) C_v T$.
Canceling $C_v$ from both sides,we get $T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$.
Using the ideal gas law $PV = nRT$,we have $n_1 = \frac{P_1 V_1}{R T_1}$ and $n_2 = \frac{P_2 V_2}{R T_2}$.
Substituting these into the expression for $T$:
$T = \frac{(\frac{P_1 V_1}{R T_1}) T_1 + (\frac{P_2 V_2}{R T_2}) T_2}{\frac{P_1 V_1}{R T_1} + \frac{P_2 V_2}{R T_2}} = \frac{P_1 V_1 + P_2 V_2}{\frac{P_1 V_1}{R T_1} + \frac{P_2 V_2}{R T_2}}$.
Simplifying the denominator:
$T = \frac{R(P_1 V_1 + P_2 V_2)}{\frac{P_1 V_1 T_2 + P_2 V_2 T_1}{T_1 T_2}} = \frac{T_1 T_2 (P_1 V_1 + P_2 V_2)}{P_1 V_1 T_2 + P_2 V_2 T_1}$.

(A) The center of mass $x_{CM}$ of a rod with linear mass density $\lambda(x)$ is given by:
$x_{CM} = \frac{\int_0^L x \lambda(x) dx}{\int_0^L \lambda(x) dx}$
Substituting $\lambda(x) = k(\frac{x}{L})^n$:
$x_{CM} = \frac{\int_0^L x \cdot k(\frac{x}{L})^n dx}{\int_0^L k(\frac{x}{L})^n dx} = \frac{\frac{k}{L^n} \int_0^L x^{n+1} dx}{\frac{k}{L^n} \int_0^L x^n dx} = \frac{[\frac{x^{n+2}}{n+2}]_0^L}{[\frac{x^{n+1}}{n+1}]_0^L} = \frac{L^{n+2}/(n+2)}{L^{n+1}/(n+1)} = L \frac{n+1}{n+2}$
Analyzing the function $f(n) = L \frac{n+1}{n+2} = L (1 - \frac{1}{n+2})$:
$1$. For $n = 0$,$x_{CM} = L(1 - 1/2) = L/2$.
$2$. As $n \to \infty$,$x_{CM} \to L$.
$3$. The derivative $\frac{dx_{CM}}{dn} = L \frac{(n+2) - (n+1)}{(n+2)^2} = \frac{L}{(n+2)^2} > 0$,so $x_{CM}$ is an increasing function.
$4$. The second derivative $\frac{d^2x_{CM}}{dn^2} = -\frac{2L}{(n+2)^3} < 0$,so the graph is concave down.
These properties match the graph in option $(A)$.

(A) The moment of inertia of a square plate of side '$a$' and mass '$m$' about an axis passing through its center of mass and perpendicular to its plane is given by:
$I_{cm} = \frac{1}{12}m(a^2 + a^2) = \frac{ma^2}{6}$
According to the parallel axis theorem,the moment of inertia about an axis passing through a corner and perpendicular to the plane is:
$I = I_{cm} + md^2$
Here,'$d$' is the distance from the center of the square to the corner. For a square of side '$a$',the diagonal is '$a\sqrt{2}$',so the distance from the center to a corner is:
$d = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$
Substituting the values:
$I = \frac{ma^2}{6} + m\left(\frac{a}{\sqrt{2}}\right)^2$
$I = \frac{ma^2}{6} + \frac{ma^2}{2}$
$I = \frac{ma^2 + 3ma^2}{6} = \frac{4ma^2}{6} = \frac{2}{3}ma^2$

(C) The formula for escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$.
Let $M_e$ and $R_e$ be the mass and radius of the Earth,and $M_p$ and $R_p$ be the mass and radius of the planet.
Given: $M_p = 10 M_e$ and $R_p = \frac{R_e}{10}$.
The ratio of escape velocities is given by:
$\frac{(v_e)_p}{(v_e)_e} = \sqrt{\frac{M_p}{M_e} \times \frac{R_e}{R_p}}$
Substituting the given values:
$\frac{(v_e)_p}{(v_e)_e} = \sqrt{\frac{10 M_e}{M_e} \times \frac{R_e}{R_e/10}} = \sqrt{10 \times 10} = \sqrt{100} = 10$.
Therefore,$(v_e)_p = 10 \times (v_e)_e = 10 \times 11 \ km/s = 110 \ km/s$.

(A) According to Gauss's Law for gravitation,the gravitational flux $\Phi_g$ through any closed surface is given by $\Phi_g = \oint \vec{E_g} \cdot d\vec{S} = -4\pi GM_{enclosed}$.
Statement-$1$ states the flux is $4\pi GM$,but the correct value is $-4\pi GM$ (due to the attractive nature of gravity). Thus,Statement-$1$ is false.
Statement-$2$ describes the general principle of Gauss's Law,which applies to any field following the inverse-square law (like gravity or electrostatics). This statement is true.
Therefore,Statement-$1$ is false and Statement-$2$ is true.

(A) From the figure,it is clear that liquid $1$ floats on liquid $2$.
Since the lighter liquid floats over the heavier liquid,we can conclude that $\rho_1 < \rho_2$.
The ball is in equilibrium at the interface of the two liquids.
For an object to float in a liquid,its density must be less than the density of the liquid.
Since the ball is partially submerged in liquid $1$ and partially in liquid $2$,its density $\rho_3$ must be greater than the density of the upper liquid $(\rho_1)$ and less than the density of the lower liquid $(\rho_2)$.
Therefore,the condition for equilibrium is $\rho_1 < \rho_3 < \rho_2$.

(B) At terminal speed $(v_t)$,the net force on the ball is zero.
Therefore,the downward weight of the ball is balanced by the upward buoyant force and the viscous force.
$Weight = \text{Buoyant force} + \text{Viscous force}$
$V\rho_1 g = V\rho_2 g + kv_t^2$
$kv_t^2 = Vg(\rho_1 - \rho_2)$
$v_t^2 = \frac{Vg(\rho_1 - \rho_2)}{k}$
$v_t = \sqrt{\frac{Vg(\rho_1 - \rho_2)}{k}}$

(D) The height of the liquid column in a capillary tube is given by the ascent formula: $h = \frac{2T \cos \theta}{r \rho g}$.
Here,$T$ is the surface tension,$\theta$ is the angle of contact,$r$ is the radius of the tube,$\rho$ is the density of the liquid,and $g$ is the acceleration due to gravity.
For water,the angle of contact $\theta$ is acute,resulting in a concave meniscus and a positive height $h$ (rise).
For a soap-water solution,the surface tension $T$ is significantly lower than that of pure water. Since $h \propto T$,the height of the soap solution column will be less than the height of the water column.
Both water and soap solution wet the glass,so both will form a concave meniscus (concave upwards) in the capillary tube.
Therefore,the correct representation shows both tubes with concave menisci,but with the liquid level in tube $(B)$ lower than in tube $(A)$.

(B) In a resonance column experiment,the frequency of the tuning fork $f$ remains constant.
For the first resonance in winter,the length is $\ell_1 = 18 \ cm$. The frequency is given by $f = \frac{v}{4 \ell_1} = \frac{v}{4 \times 18}$,where $v$ is the speed of sound in winter.
For the second resonance in summer,the length is $\ell_2 = x \ cm$. The frequency is given by $f = \frac{3v'}{4 \ell_2} = \frac{3v'}{4x}$,where $v'$ is the speed of sound in summer.
Equating the frequencies: $\frac{v}{4 \times 18} = \frac{3v'}{4x}$.
Solving for $x$: $x = 54 \times \frac{v'}{v} \ cm$.
Since the temperature in summer is higher than in winter,the speed of sound $v' > v$ (because $v \propto \sqrt{T}$).
Therefore,$x > 54 \ cm$.

(B) The given wave equation is $y(x, t) = 0.005 \cos(\alpha x - \beta t)$.
Comparing this with the standard wave equation $y(x, t) = a \cos(kx - \omega t)$,we identify the wave number $k = \alpha$ and the angular frequency $\omega = \beta$.
The wave number is defined as $k = \frac{2\pi}{\lambda}$. Given $\lambda = 0.08 \ m$,we have $\alpha = \frac{2\pi}{0.08} = 25\pi \ rad/m$.
The angular frequency is defined as $\omega = \frac{2\pi}{T}$. Given $T = 2.0 \ s$,we have $\beta = \frac{2\pi}{2.0} = \pi \ rad/s$.
Thus,$\alpha = 25\pi$ and $\beta = \pi$.

(B) travelling microscope is a precision instrument used to measure small distances.
It consists of a main scale and a vernier scale attached to the microscope assembly.
When measuring the refractive index of a glass slab,the microscope is focused on a mark on the table,then on the mark through the glass slab,and finally on a dust particle on the top surface of the slab.
The vertical displacement is measured accurately using the vernier scale provided on the microscope.

(B) For oxygen $(O_2)$:
Molar mass,$M_1 = 32 \,g/mol$.
Heat capacity ratio,$\gamma_1 = C_p / C_V = 7/5$ (for diatomic gas).
Speed of sound,$v_1 = 460 \,ms^{-1}$.
For helium $(He)$:
Molar mass,$M_2 = 4 \,g/mol$.
Heat capacity ratio,$\gamma_2 = C_p / C_V = 5/3$ (for monoatomic gas).
The speed of sound in an ideal gas is given by $v = \sqrt{\frac{\gamma RT}{M}}$.
Taking the ratio of speeds: $\frac{v_1}{v_2} = \sqrt{\frac{\gamma_1 / M_1}{\gamma_2 / M_2}} = \sqrt{\frac{\gamma_1}{\gamma_2} \cdot \frac{M_2}{M_1}}$.
Substituting the values: $\frac{460}{v_2} = \sqrt{\frac{7/5}{5/3} \cdot \frac{4}{32}} = \sqrt{\frac{21}{25} \cdot \frac{1}{8}} = \sqrt{\frac{21}{200}} \approx 0.324$.
Wait,re-calculating: $\frac{460}{v_2} = \sqrt{\frac{7}{5} \cdot \frac{3}{5} \cdot \frac{4}{32}} = \sqrt{\frac{21}{25} \cdot \frac{1}{8}} = \sqrt{\frac{21}{200}} \approx 0.324$.
Actually,$\frac{v_2}{v_1} = \sqrt{\frac{\gamma_2}{\gamma_1} \cdot \frac{M_1}{M_2}} = \sqrt{\frac{5/3}{7/5} \cdot \frac{32}{4}} = \sqrt{\frac{25}{21} \cdot 8} = \sqrt{\frac{200}{21}} \approx \sqrt{9.52} \approx 3.085$.
$v_2 = 460 \times 3.085 \approx 1419.1 \,ms^{-1} \approx 1420 \,ms^{-1}$.

(C) The magnetic force $F$ on a charge $q$ moving with velocity $v$ in a magnetic field $B$ is given by the Lorentz force formula: $F = qvB \sin \theta$.
Considering dimensions,we have $[F] = [q][v][B]$.
The dimensions are: $[F] = MLT^{-2}$,$[q] = C$,and $[v] = LT^{-1}$.
Rearranging for $B$,we get: $[B] = \frac{[F]}{[q][v]} = \frac{MLT^{-2}}{C \cdot LT^{-1}}$.
Simplifying the expression: $[B] = M \cdot L^1 L^{-1} \cdot T^{-2} T^1 \cdot C^{-1} = MT^{-1}C^{-1}$.

(B) According to Gauss's Law,for a thin spherical shell of radius $R$ with a uniformly distributed charge $Q$:
$1$. Inside the shell $(r < R)$: The electric field $E$ is zero because there is no enclosed charge.
$2$. Outside the shell $(r \ge R)$: The shell acts as a point charge located at its center,so the electric field is given by $E = k\frac{Q}{r^2}$,where $k = \frac{1}{4\pi\epsilon_0}$.
Thus,the graph of $E(r)$ versus $r$ should show $E = 0$ for $0 \le r < R$ and a $1/r^2$ decay for $r \ge R$. This corresponds to the graph shown in option $B$.

(D) The initial capacitance of the air-filled capacitor is $C_0 = \frac{\epsilon_0 A}{d} = 9 \ pF$.
When the space is filled with two dielectrics of thicknesses $d_1 = d/3$ and $d_2 = 2d/3$,the system acts as two capacitors connected in series.
The capacitance of the first part is $C_1 = \frac{k_1 \epsilon_0 A}{d_1} = \frac{3 \epsilon_0 A}{d/3} = 9 \frac{\epsilon_0 A}{d} = 9 C_0 = 9 \times 9 = 81 \ pF$.
The capacitance of the second part is $C_2 = \frac{k_2 \epsilon_0 A}{d_2} = \frac{6 \epsilon_0 A}{2d/3} = 9 \frac{\epsilon_0 A}{d} = 9 C_0 = 9 \times 9 = 81 \ pF$.
Since they are in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$.
$\frac{1}{C_{eq}} = \frac{1}{81} + \frac{1}{81} = \frac{2}{81}$.
Therefore,$C_{eq} = \frac{81}{2} = 40.5 \ pF$.

(B) Let $V_{P1} = 0\,V$. Then the potential at $P_2$ is $V$.
Using nodal analysis at node $P_2$:
$\frac{V - 5}{2} + \frac{V}{10} + \frac{V - 2}{1} = 0$
Multiply by $10$:
$5(V - 5) + V + 10(V - 2) = 0$
$5V - 25 + V + 10V - 20 = 0$
$16V = 45$
$V = \frac{45}{16} = 2.8125\,V$
The current through the $10\,\Omega$ resistor is $I = \frac{V}{R} = \frac{2.8125}{10} = 0.28125\,A \approx 0.28\,A$.
Given the options,the closest value is $0.27\,A$. Since $V_{P2} > V_{P1}$,the current flows from $P_2$ to $P_1$.

(C) In a meter bridge,the balancing condition is given by the formula: $\frac{P}{Q} = \frac{l_1}{l_2}$,where $P$ and $Q$ are the resistances in the two gaps,and $l_1$ and $l_2$ are the corresponding lengths of the wire.
Given,$P = 55 \, \Omega$ and $l_1 = 20 \, \text{cm}$.
The total length of the meter bridge wire is $100 \, \text{cm}$,so $l_2 = 100 - 20 = 80 \, \text{cm}$.
Substituting these values into the balancing condition:
$\frac{55}{R} = \frac{20}{80}$
$\frac{55}{R} = \frac{1}{4}$
$R = 55 \times 4 = 220 \, \Omega$.

(B) Let $j$ be the current density.
Since the current $I$ spreads over a hemispherical surface,$j \times 2 \pi r^2 = I$,which gives $j = \frac{I}{2 \pi r^2}$.
Using Ohm's law,the electric field is $E = \rho j = \frac{\rho I}{2 \pi r^2}$.
The potential difference due to current $I$ entering at $A$ between points $B$ and $C$ is $\Delta V_{BC, A} = V_B - V_C = \int_{r_B}^{r_C} E dr = \int_{a}^{a+b} \frac{\rho I}{2 \pi r^2} dr = \frac{\rho I}{2 \pi} \left[ -\frac{1}{r} \right]_{a}^{a+b} = \frac{\rho I}{2 \pi} \left( \frac{1}{a} - \frac{1}{a+b} \right)$.
Similarly,for current $I$ leaving at $D$,the potential difference $\Delta V_{BC, D}$ is also $\frac{\rho I}{2 \pi} \left( \frac{1}{a} - \frac{1}{a+b} \right)$.
By the superposition principle,the total potential difference is $\Delta V = \Delta V_{BC, A} + \Delta V_{BC, D} = 2 \times \frac{\rho I}{2 \pi} \left( \frac{1}{a} - \frac{1}{a+b} \right) = \frac{\rho I}{\pi a} - \frac{\rho I}{\pi(a+b)}$.

(D) Let $j$ be the current density.
Since the current $I$ spreads over a hemispherical surface of area $2 \pi r^2$,the current density is $j = \frac{I}{2 \pi r^2}$.
Using Ohm's law in the form $E = \rho j$,we get:
$E = \rho \left( \frac{I}{2 \pi r^2} \right) = \frac{\rho I}{2 \pi r^2}$.

(D) The magnetic field $B$ due to a long straight current-carrying wire at a distance $r$ is given by the formula:
$B = \frac{\mu_0 I}{2 \pi r}$
Given:
$I = 100 \, A$
$r = 4 \, m$
$\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$
Substituting the values:
$B = \frac{(4\pi \times 10^{-7}) \times 100}{2 \pi \times 4}$
$B = \frac{2 \times 10^{-7} \times 100}{4}$
$B = \frac{2 \times 10^{-5}}{4} = 0.5 \times 10^{-5} = 5 \times 10^{-6} \, T$
According to the Right-Hand Thumb Rule,if the thumb points in the direction of the current (West),the curled fingers at the point below the wire will point towards the South.

(C) For a diamagnetic material,the relative permeability $\mu_r$ is always less than $1$ (i.e.,$\mu_r < 1$).
For any dielectric material,the relative permittivity $\varepsilon_r$ is always greater than $1$ (i.e.,$\varepsilon_r > 1$).
Comparing these conditions with the given options,we find that option $C$ satisfies both conditions: $\varepsilon_r = 1.5 > 1$ and $\mu_r = 0.5 < 1$.

(C) The lens equation is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Using sign convention for a convex lens,the object distance is negative,so we take $u = -x$ (where $x > 0$).
The equation becomes $\frac{1}{v} - \frac{1}{-x} = \frac{1}{f}$,which simplifies to $\frac{1}{v} + \frac{1}{x} = \frac{1}{f}$.
Rearranging for $v$,we get $\frac{1}{v} = \frac{1}{f} - \frac{1}{x} = \frac{x - f}{xf}$,so $v = \frac{xf}{x - f}$.
As $x$ increases from $f$ to $\infty$,$v$ decreases from $\infty$ to $f$. This represents a hyperbolic curve in the first quadrant of the $v-u$ plane (considering magnitudes of $u$ and $v$),which corresponds to the shape shown in option $C$.

(A) The formula for the mutual inductance $M$ of two coaxial solenoids is given by $M = \frac{\mu_0 N_1 N_2 A}{\ell}$.
Given values are:
$A = 10 \ cm^2 = 10 \times 10^{-4} \ m^2 = 10^{-3} \ m^2$
$\ell = 20 \ cm = 0.2 \ m$
$N_1 = 300$
$N_2 = 400$
$\mu_0 = 4\pi \times 10^{-7} \ T \ m \ A^{-1}$
Substituting these values into the formula:
$M = \frac{(4\pi \times 10^{-7}) \times 300 \times 400 \times 10^{-3}}{0.2}$
$M = \frac{4\pi \times 10^{-7} \times 120000 \times 10^{-3}}{0.2}$
$M = \frac{4\pi \times 10^{-7} \times 120}{0.2}$
$M = 4\pi \times 10^{-7} \times 600$
$M = 2400\pi \times 10^{-7} \ H = 2.4\pi \times 10^{-4} \ H$.

(A) When electrons pass through a narrow slit of width '$d$' comparable to their de Broglie wavelength, they exhibit wave-like behavior and undergo diffraction.
According to the single-slit diffraction pattern, the intensity (or number of electrons '$N$') is maximum at the center $(y = 0)$.
The first minima occur at positions given by the condition $d \sin \theta = \lambda$, where $\theta \approx y/D$.
Thus, $y = \pm \lambda D / d$.
Since the wavelength $\lambda$ is comparable to '$d$', the width of the central maximum is significant, and the diffraction pattern spreads out.
Graph $A$ represents the standard intensity distribution for single-slit diffraction, where the central maximum is at $y = 0$ and the intensity drops to zero at $y = \pm \lambda D / d$.

(C) The force acting on the electron is $F = \frac{k}{r}$.
For circular motion,this force provides the necessary centripetal force: $\frac{k}{r} = \frac{mv^2}{r}$.
This implies $mv^2 = k$,which means the kinetic energy $K_n = \frac{1}{2}mv^2 = \frac{k}{2}$ is constant and independent of $n$.
According to the Bohr quantization condition,$mvr = \frac{nh}{2\pi}$.
Since $v = \sqrt{\frac{k}{m}}$ is constant,we have $m \sqrt{\frac{k}{m}} r_n = \frac{nh}{2\pi}$.
Therefore,$r_n \propto n$.

(A) Statement-$1$ is true. Energy is released in nuclear fission of heavy nuclei and nuclear fusion of light nuclei because the binding energy per nucleon of the product nuclei is higher than that of the reactant nuclei,leading to a more stable configuration.
Statement-$2$ is false. According to the binding energy per nucleon curve,for heavy nuclei (mass number $A > 170$),the binding energy per nucleon decreases as $Z$ increases. For light nuclei (mass number $A < 30$),the binding energy per nucleon generally increases as $Z$ increases. The statement provided in Statement-$2$ is the exact opposite of this physical reality.

(B) transistor consists of two $pn$ junctions.
$1$. No conduction between $P$ and $Q$ implies that $P$ and $Q$ are the collector and emitter terminals (or vice versa),as there is no direct $pn$ junction between them.
$2$. When the negative terminal of the multimeter is connected to $R$ and the positive terminal is connected to $P$ or $Q$,conduction (resistance) is observed.
$3$. In a multimeter,the common (negative) terminal is connected to the internal battery's negative pole,and the positive terminal is connected to the positive pole.
$4$. Conduction occurs when the $pn$ junction is forward-biased.
$5$. Since the negative terminal is at $R$ and the positive terminal is at $P$ or $Q$,$R$ must be the $p$-type material (base) and $P, Q$ must be $n$-type materials.
$6$. This configuration corresponds to an $npn$ transistor where $R$ is the base.

(A) The given circuit consists of two diodes connected in parallel with a common resistor connected to the ground.
Let us analyze the truth table for this circuit:

$A$	$B$	$C$
$0$	$0$	$0$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$1$

$1$. When both inputs $A$ and $B$ are at low potential $(0)$,both diodes are reverse-biased (or non-conducting),so the output $C$ is at ground potential $(0)$.
$2$. When either $A$ or $B$ is at high potential $(1)$,the corresponding diode becomes forward-biased and conducts,pulling the output $C$ to a high potential $(1)$.
$3$. This behavior corresponds to the Boolean expression $C = A + B$,which is the characteristic operation of an $OR$ gate.

(B) The condition for constructive interference in crystal diffraction (Bragg's Law) is given by $2d \sin \theta = n \lambda$, where $\theta$ is the glancing angle.
From the figure, the angle of incidence $i$ is measured with respect to the normal. The glancing angle $\theta$ is the angle between the incident ray and the crystal plane, so $\theta = 90^{\circ} - i$.
Given $i = 30^{\circ}$, we have $\theta = 90^{\circ} - 30^{\circ} = 60^{\circ}$.
For first-order diffraction, $n = 1$. Substituting the values:
$2 \times (1 \times 10^{-10}\; m) \times \sin(60^{\circ}) = 1 \times \lambda$
$\lambda = 2 \times 10^{-10} \times \frac{\sqrt{3}}{2} = \sqrt{3} \times 10^{-10}\; m \approx 1.732 \times 10^{-10}\; m$.
The de-Broglie wavelength of an electron accelerated by potential $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}}$.
Squaring both sides:
$\lambda^2 = \frac{h^2}{2meV} \Rightarrow V = \frac{h^2}{2me\lambda^2}$.
Substituting the values:
$V = \frac{(6.6 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times (\sqrt{3} \times 10^{-10})^2}$
$V = \frac{43.56 \times 10^{-68}}{29.12 \times 10^{-50} \times 3} = \frac{43.56 \times 10^{-68}}{87.36 \times 10^{-50}} \approx 0.4986 \times 10^{18} \times 10^{-68} \times 10^{50} \approx 49.86\; V$.
Thus, $V$ should be about $50\; V$.

(D) For constructive interference,the path difference between waves reflected from successive atomic planes must be an integral multiple of the wavelength,which is given by Bragg's Law: $2d \sin \theta = n \lambda_{dB}$,where $\theta$ is the glancing angle (the angle between the incident ray and the crystal plane).
From the provided figure,the angle $i$ is the angle between the incident ray and the normal to the crystal plane. Therefore,the glancing angle $\theta$ is given by $\theta = 90^{\circ} - i$.
Substituting this into Bragg's Law:
$2d \sin(90^{\circ} - i) = n \lambda_{dB}$
Since $\sin(90^{\circ} - i) = \cos i$,we get:
$2d \cos i = n \lambda_{dB}$