AIPMT 2000 Physics Question Paper with Answer and Solution

(B) The shortest path to cross a river is the straight line perpendicular to the river flow.
Let $v_b = 5 \; km/hr$ be the velocity of the boat and $u$ be the velocity of the river stream.
The resultant velocity $v_r$ of the boat relative to the ground,when crossing the river along the shortest path,is given by $v_r = \sqrt{v_b^2 - u^2}$.
Given width $d = 1 \; km$ and time $t = 15 \; min = 0.25 \; hr = \frac{1}{4} \; hr$.
The resultant velocity is $v_r = \frac{d}{t} = \frac{1}{1/4} = 4 \; km/hr$.
Substituting the values: $4 = \sqrt{5^2 - u^2}$.
Squaring both sides: $16 = 25 - u^2$.
$u^2 = 25 - 16 = 9$.
$u = 3 \; km/hr$.

(D) The displacement is given by $s = 3t^3 + 7t^2 + 14t + 8$.
To find the velocity $v$,we differentiate $s$ with respect to time $t$:
$v = \frac{ds}{dt} = \frac{d}{dt}(3t^3 + 7t^2 + 14t + 8) = 9t^2 + 14t + 14$.
To find the acceleration $a$,we differentiate the velocity $v$ with respect to time $t$:
$a = \frac{dv}{dt} = \frac{d}{dt}(9t^2 + 14t + 14) = 18t + 14$.
At time $t = 1 \ s$,the acceleration is:
$a = 18(1) + 14 = 18 + 14 = 32 \ m/s^2$.

(A) The horizontal range $R$ of a projectile is given by the formula $R = \frac{u^2 \sin(2\theta)}{g}$.
Given that the initial velocity $u$ is the same for both projectiles.
For the first projectile,$\theta_1 = 60^o$,so $R_1 = \frac{u^2 \sin(120^o)}{g} = \frac{u^2 \sin(60^o)}{g}$.
For the second projectile,$\theta_2 = 30^o$,so $R_2 = \frac{u^2 \sin(60^o)}{g}$.
Since $\sin(120^o) = \sin(60^o)$,the horizontal range $R$ is the same for both angles.
Angles $\theta$ and $(90^o - \theta)$ are complementary angles,and for any pair of complementary angles,the horizontal range is identical.

(C) Let the masses be $m_1 = 5 \ kg$ and $m_2 = 10 \ kg$.
The acceleration $a$ of the system is given by the formula:
$a = \frac{m_2 - m_1}{m_1 + m_2} g$
Substituting the given values:
$a = \frac{10 - 5}{10 + 5} g$
$a = \frac{5}{15} g$
$a = \frac{g}{3}$

(A) The weight of a body at the surface of the earth is given by $W = mg = 72 \ N$.
At a height $h$ above the surface of the earth,the acceleration due to gravity $g'$ is given by the formula $g' = g \left( \frac{R}{R + h} \right)^2$,where $R$ is the radius of the earth.
Given $h = \frac{R}{2}$,we substitute this into the formula:
$g' = g \left( \frac{R}{R + \frac{R}{2}} \right)^2 = g \left( \frac{R}{\frac{3R}{2}} \right)^2 = g \left( \frac{2}{3} \right)^2 = \frac{4}{9}g$.
The weight of the body at height $h$ is $W' = mg' = m \left( \frac{4}{9}g \right) = \frac{4}{9} W$.
Substituting the value of $W = 72 \ N$:
$W' = \frac{4}{9} \times 72 = 4 \times 8 = 32 \ N$.

(C) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.
Given that the mass of the planet $M_p = M_e$ (mass of Earth) and the radius $R_p = \frac{R_e}{4}$ (where $R_e$ is the radius of Earth).
The escape velocity for the planet is $v_p = \sqrt{\frac{2GM_p}{R_p}} = \sqrt{\frac{2GM_e}{R_e/4}}$.
This simplifies to $v_p = \sqrt{4 \times \frac{2GM_e}{R_e}} = 2 \times \sqrt{\frac{2GM_e}{R_e}}$.
Since the escape velocity of Earth is $v_e = 11.2 \ km/s$,the escape velocity for the planet is $v_p = 2 \times 11.2 \ km/s = 22.4 \ km/s$.

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

(D) The efficiency of a reversible engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature in Kelvin.
Initially,$\eta = \frac{1}{6}$,so $1 - \frac{T_2}{T_1} = \frac{1}{6} \implies \frac{T_2}{T_1} = \frac{5}{6} \implies T_2 = \frac{5}{6}T_1$ ...$(i)$
When the sink temperature is reduced by $62^\circ C$ (which is equivalent to $62 \ K$),the new efficiency $\eta' = 2\eta = 2 \times \frac{1}{6} = \frac{1}{3}$.
The new sink temperature is $T_2' = T_2 - 62$.
Thus,$\eta' = 1 - \frac{T_2 - 62}{T_1} = \frac{1}{3}$.
Substituting $T_2 = \frac{5}{6}T_1$ into the equation: $1 - \frac{\frac{5}{6}T_1 - 62}{T_1} = \frac{1}{3}$.
$1 - (\frac{5}{6} - \frac{62}{T_1}) = \frac{1}{3} \implies 1 - \frac{5}{6} + \frac{62}{T_1} = \frac{1}{3}$.
$\frac{1}{6} + \frac{62}{T_1} = \frac{1}{3} \implies \frac{62}{T_1} = \frac{1}{3} - \frac{1}{6} = \frac{1}{6}$.
$T_1 = 62 \times 6 = 372 \ K = (372 - 273)^\circ C = 99^\circ C$.
Now,$T_2 = \frac{5}{6} \times 372 = 310 \ K = (310 - 273)^\circ C = 37^\circ C$.

(A) Natural convection occurs due to the difference in density between two regions of a fluid. When a fluid is heated,it expands and its density decreases,causing it to rise,while cooler,denser fluid sinks to take its place. This movement is driven by the gravitational force acting on the density gradient. Therefore,gravitational force is essential for natural convection.

(C) When the bob is displaced by an angle $\theta$,it rises to a vertical height $h$ above its mean position.
By the law of conservation of mechanical energy,the potential energy at the extreme position is converted into kinetic energy at the mean (lowest) position.
$mgh = \frac{1}{2}mv_{\max}^2$
$v_{\max} = \sqrt{2gh}$
From the geometry of the pendulum,the vertical distance from the point of suspension to the bob at the extreme position is $l \cos \theta$.
Therefore,the height $h$ is given by:
$h = l - l \cos \theta = l(1 - \cos \theta)$
Substituting the value of $h$ into the velocity equation:
$v_{\max} = \sqrt{2gl(1 - \cos \theta)}$

(B) The fundamental frequency of a stretched wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $l$ is the length,$T$ is the tension,and $\mu$ is the mass per unit length.
Since the wires are similar,the mass per unit length $\mu$ and tension $T$ are the same for all segments.
Thus,$n \propto \frac{1}{l}$,which implies $nl = k$ (a constant).
For the three wires,we have $n_1 l_1 = n_2 l_2 = n_3 l_3 = k$.
This gives $l_1 = \frac{k}{n_1}$,$l_2 = \frac{k}{n_2}$,and $l_3 = \frac{k}{n_3}$.
The total length of the combined wire is $l = l_1 + l_2 + l_3$.
Substituting the values,we get $\frac{k}{n} = \frac{k}{n_1} + \frac{k}{n_2} + \frac{k}{n_3}$.
Dividing by $k$,we obtain $\frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3}$.

(A) The time $t$ taken by a body rolling down an inclined plane of length $l$ and inclination $\theta$ is given by $t = \sqrt{\frac{2l(1 + K^2/R^2)}{g \sin \theta}}$,where $K$ is the radius of gyration and $R$ is the radius of the body.
For a solid cylinder,the moment of inertia $I = \frac{1}{2}MR^2$,so $K^2 = \frac{1}{2}R^2$,which means $K^2/R^2 = 0.5$.
For a hollow cylinder,the moment of inertia $I = MR^2$,so $K^2 = R^2$,which means $K^2/R^2 = 1$.
Since the time $t$ is directly proportional to $\sqrt{1 + K^2/R^2}$,the body with the smaller $K^2/R^2$ ratio will take less time to reach the bottom.
Comparing the two,the solid cylinder has a smaller $K^2/R^2$ ratio $(0.5 < 1)$,therefore,the solid cylinder will reach the bottom first.

(A) Given equations are $x=a \cos (\omega t+\delta)$ and $y=a \cos (\omega t+\alpha)$.
Substituting $\delta=\alpha+\frac{\pi}{2}$ into the equation for $x$:
$x=a \cos (\omega t+\alpha+\frac{\pi}{2}) = -a \sin (\omega t+\alpha)$.
Now,squaring and adding both equations:
$x^2+y^2 = (-a \sin (\omega t+\alpha))^2 + (a \cos (\omega t+\alpha))^2 = a^2 (\sin^2 (\omega t+\alpha) + \cos^2 (\omega t+\alpha)) = a^2$.
This is the equation of a circle with radius $a$.
To determine the direction,at $t=0$,$x = -a \sin \alpha$ and $y = a \cos \alpha$. As $t$ increases,the point moves in the clockwise $(c.w)$ direction.

(C) Let $v$ be the speed of sound and $n$ be the frequency of the sources. Since $\lambda = v/n$,we have $n = v/\lambda$.
For the $1^{st}$ source,the observer is moving away with velocity $u$. The apparent frequency is $n_1 = n \left( \frac{v - u}{v} \right) = n \left( 1 - \frac{u}{v} \right)$.
For the $2^{nd}$ source,the observer is moving towards it with velocity $u$. The apparent frequency is $n_2 = n \left( \frac{v + u}{v} \right) = n \left( 1 + \frac{u}{v} \right)$.
The beat frequency is the difference between the two apparent frequencies:
$|n_2 - n_1| = n \left( 1 + \frac{u}{v} \right) - n \left( 1 - \frac{u}{v} \right) = n \left( \frac{2u}{v} \right)$.
Substituting $n = v/\lambda$,we get:
Beat frequency $= \left( \frac{v}{\lambda} \right) \left( \frac{2u}{v} \right) = \frac{2u}{\lambda}$.

(C) The moment of inertia $(I)$ of a body depends on the distribution of its mass relative to the axis of rotation. The farther the mass is from the axis,the greater the moment of inertia.
In the given right-angled triangle $ABC$ with sides $AB=4$,$BC=3$,and $AC=5$,the moments of inertia about the axes passing through the sides are:
$I_{1}$ is the moment of inertia about side $AB$.
$I_{3}$ is the moment of inertia about side $BC$.
$I_{2}$ is the moment of inertia about side $AC$.
Comparing the distances of the mass distribution from these axes,we find that $I_{1} < I_{3} < I_{2}$.
Therefore,the relation $I_{1} > I_{2}$ is incorrect.

(B) The tension $T$ at any point in a vertical circular motion is given by:
$T = \frac{mv^2}{l} + mg \cos \theta$
where $\theta$ is the angular displacement from the lowest point,$l$ is the length of the string,and $m$ is the mass of the particle.
At the lowest point $(B)$,$\theta = 0^\circ$,so $\cos \theta = 1$,and the tension is $T_B = \frac{mv_B^2}{l} + mg$.
At the highest point $(A)$,$\theta = 180^\circ$,so $\cos \theta = -1$,and the tension is $T_A = \frac{mv_A^2}{l} - mg$.
It is clear that the tension is maximum at the lowest point $(B)$ because the centripetal force and the component of gravity both contribute to the tension. If the average velocity of the particle is increased,the tension at the lowest point will increase the most. Therefore,the string has the maximum probability of breaking at point $B$.

(A) The instantaneous power $P$ is given by the dot product of the force vector $\overrightarrow{F}$ and the velocity vector $\overrightarrow{V}$.
$P = \overrightarrow{F} \cdot \overrightarrow{V}$
Substituting the given vectors:
$P = (60 \hat{i} + 15 \hat{j} - 3 \hat{k}) \cdot (2 \hat{i} - 4 \hat{j} + 5 \hat{k})$
Using the property of the dot product ($\hat{i} \cdot \hat{i} = 1, \hat{j} \cdot \hat{j} = 1, \hat{k} \cdot \hat{k} = 1$ and cross terms are $0$):
$P = (60 \times 2) + (15 \times -4) + (-3 \times 5)$
$P = 120 - 60 - 15$
$P = 45 \; W$

(A) We analyze the dimensions of each pair:
$1$. Force: $[MLT^{-2}]$; Impulse: $[MLT^{-1}]$. These are not equal.
$2$. Angular momentum: $[ML^2T^{-1}]$; Planck constant: $[ML^2T^{-1}]$. These are equal.
$3$. Energy: $[ML^2T^{-2}]$; Torque: $[ML^2T^{-2}]$. These are equal.
$4$. Elastic modulus: $[ML^{-1}T^{-2}]$; Pressure: $[ML^{-1}T^{-2}]$. These are equal.
Therefore,the pair that does not have equal dimensions is Force and Impulse.

(C) According to the law of conservation of mechanical energy,the total mechanical energy remains constant in the absence of non-conservative forces like friction.
For both the man and the bag,the initial potential energy is $PE_i = mgh$ and the initial kinetic energy is $KE_i = 0$.
At the bottom,the potential energy is $PE_f = 0$ and the kinetic energy is $KE_f = \frac{1}{2}mv^2$.
Equating the energies: $mgh = \frac{1}{2}mv^2$.
Solving for velocity: $v = \sqrt{2gh}$.
Since both the man and the bag fall through the same vertical height $h$ and are subject to the same gravitational acceleration $g$,their final velocities at the bottom will be equal,regardless of the path taken (as long as the plane is frictionless).

(C) Initial velocity $u = 100 \; m/s$. Acceleration $g = -10 \; m/s^2$.
Velocity of the mass at $t = 5 \; s$ is $v = u + at = 100 - 10 \times 5 = 50 \; m/s$ (upward).
Let the total mass be $M = 1 \; kg$. The mass splits into $m_1 = 0.4 \; kg$ and $m_2 = 0.6 \; kg$.
According to the law of conservation of momentum,the momentum before the explosion equals the momentum after the explosion.
$Mv = m_1 v_1 + m_2 v_2$
Taking upward direction as positive:
$1 \times 50 = 0.4 \times (-25) + 0.6 \times v_2$
$50 = -10 + 0.6 \times v_2$
$60 = 0.6 \times v_2$
$v_2 = \frac{60}{0.6} = 100 \; m/s$.
Since the result is positive,the velocity is $100 \; m/s$ upward.

(B) The frequency of a simple pendulum is given by $f = \frac{1}{2 \pi} \sqrt{\frac{g}{l}}$.
Given the frequency relation $f_{A}=2 f_{B}$.
Substituting the formula for frequency:
$\frac{1}{2 \pi} \sqrt{\frac{g}{l_{A}}} = 2 \times \frac{1}{2 \pi} \sqrt{\frac{g}{l_{B}}}$
Squaring both sides:
$\frac{g}{l_{A}} = 4 \times \frac{g}{l_{B}}$
$\frac{1}{l_{A}} = \frac{4}{l_{B}}$
Therefore,$l_{A} = \frac{l_{B}}{4}$.
Since the frequency of a simple pendulum depends only on the length $l$ and acceleration due to gravity $g$,it does not depend on the mass $M$ of the bob.

(C) The change in momentum occurs only in the direction perpendicular to the wall.
Let the velocity be $v = 10 \, m/s$ and mass $m = 3 \, kg$.
The component of velocity perpendicular to the wall is $v_{\perp} = v \sin(60^\circ)$.
Initial momentum perpendicular to the wall: $p_i = mv \sin(60^\circ)$.
Final momentum perpendicular to the wall (after reflection): $p_f = -mv \sin(60^\circ)$.
Change in momentum: $\Delta p = p_f - p_i = -mv \sin(60^\circ) - mv \sin(60^\circ) = -2mv \sin(60^\circ)$.
The magnitude of the change in momentum is $|\Delta p| = 2mv \sin(60^\circ)$.
Force exerted on the wall $F = \frac{|\Delta p|}{\Delta t} = \frac{2mv \sin(60^\circ)}{\Delta t}$.
Substituting the values: $F = \frac{2 \times 3 \times 10 \times \sin(60^\circ)}{0.2} = \frac{60 \times (\sqrt{3}/2)}{0.2} = \frac{30\sqrt{3}}{0.2} = 150\sqrt{3} \, N$.

(A) The circuit is a Wheatstone bridge. Let the resistors be $P = 3 \, \Omega$,$Q = 4 \, \Omega$,$R = 6 \, \Omega$,and $S = 8 \, \Omega$. The central resistor is $G = 7 \, \Omega$.
Check the ratio of the arms: $\frac{P}{R} = \frac{3}{6} = 0.5$ and $\frac{Q}{S} = \frac{4}{8} = 0.5$.
Since $\frac{P}{R} = \frac{Q}{S}$,the Wheatstone bridge is balanced.
In a balanced Wheatstone bridge,no current flows through the central resistor $G$. Therefore,it can be removed from the circuit.
After removing the $7 \, \Omega$ resistor,the circuit consists of two parallel branches:
Branch $1$ (upper): $3 \, \Omega$ and $4 \, \Omega$ in series,so $R_1 = 3 + 4 = 7 \, \Omega$.
Branch $2$ (lower): $6 \, \Omega$ and $8 \, \Omega$ in series,so $R_2 = 6 + 8 = 14 \, \Omega$.
The equivalent resistance $R_{eq}$ between $A$ and $B$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{7} + \frac{1}{14}$.
$\frac{1}{R_{eq}} = \frac{2 + 1}{14} = \frac{3}{14}$.
Thus,$R_{eq} = \frac{14}{3} \, \Omega$.

(C) When a battery is being charged,the terminal voltage $V$ is given by the formula $V = E + Ir$,where $E$ is the $e.m.f.$,$I$ is the charging current,and $r$ is the internal resistance.
Given:
$E = 12\,V$
$I = 60\,A$
$r = 5 \times 10^{-2}\,\Omega = 0.05\,\Omega$
Substituting the values:
$V = 12 + (60 \times 0.05)$
$V = 12 + 3$
$V = 15\,V$.

(C) The power rating of an electrical appliance is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.
For the first bulb $(40 \; W, 200 \; V)$:
$R_{40} = \frac{V^2}{P_1} = \frac{(200)^2}{40} = \frac{40000}{40} = 1000 \; \Omega$.
For the second bulb $(100 \; W, 200 \; V)$:
$R_{100} = \frac{V^2}{P_2} = \frac{(200)^2}{100} = \frac{40000}{100} = 400 \; \Omega$.
Comparing the two values,we find that $1000 \; \Omega > 400 \; \Omega$,which implies $R_{40} > R_{100}$.

(D) The magnetic field due to a long straight wire is given by $B = \frac{\mu_0 I}{2 \pi r}$.
For the wire carrying $5 \text{ A}$ current at a distance of $2.5 \text{ m}$ from $P$,the magnetic field is:
$B_1 = \frac{\mu_0 \times 5}{2 \pi \times 2.5} = \frac{2 \mu_0}{2 \pi} \otimes$ (directed into the page).
For the wire carrying $2.5 \text{ A}$ current,the distance from $P$ is $5 \text{ m} - 2.5 \text{ m} = 2.5 \text{ m}$. The magnetic field is:
$B_2 = \frac{\mu_0 \times 2.5}{2 \pi \times 2.5} = \frac{\mu_0}{2 \pi} \odot$ (directed out of the page).
The net magnetic field at $P$ is:
$B_{net} = B_1 - B_2 = \frac{2 \mu_0}{2 \pi} - \frac{\mu_0}{2 \pi} = \frac{\mu_0}{2 \pi} \otimes$.

(A) The quality factor $(Q)$ of an $LCR$ circuit is defined as the ratio of the voltage across the inductor $(V_L)$ or capacitor $(V_C)$ to the voltage across the resistor $(V_R)$ at resonance.
Mathematically,$Q = \frac{V_L}{V_R} = \frac{I \cdot X_L}{I \cdot R} = \frac{X_L}{R}$.
Since the inductive reactance is given by $X_L = \omega L$,we substitute this into the expression.
Therefore,the quality factor is $Q = \frac{\omega L}{R}$.

(A) Einstein explained the photoelectric effect by proposing that light consists of discrete packets of energy called photons. The energy of each photon is given by the equation $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency of the radiation. This equation forms the basis of Einstein's photoelectric equation,$K_{max} = h\nu - \Phi_0$. Therefore,the correct option is $A$.

(D) In $\beta^+$ decay (positron emission), a proton inside the nucleus is converted into a neutron, a positron $(\beta^+)$, and a neutrino $(\nu)$.
The reaction is given by: $_6C^{11} \to _5B^{11} + _1e^0 + \nu$.
According to the law of conservation of lepton number, the lepton number must be conserved. Since the positron $(\beta^+)$ has a lepton number of $-1$, a neutrino $(\nu)$ with a lepton number of $+1$ must be emitted to balance the equation.
Therefore, the particle $X$ is a neutrino.

(B) The radioactive decay law is given by $N(t) = N_0 e^{-\lambda t}$.
By definition,at half-life $T_{1/2}$,the number of radioactive nuclei remaining is half of the initial amount,i.e.,$N(T_{1/2}) = \frac{N_0}{2}$.
Substituting this into the decay law: $\frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}}$.
$\frac{1}{2} = e^{-\lambda T_{1/2}}$.
Taking the natural logarithm on both sides: $\ln(1/2) = -\lambda T_{1/2}$.
$-\ln 2 = -\lambda T_{1/2}$.
Therefore,$T_{1/2} = \frac{\ln 2}{\lambda}$.

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

(D) The Boolean expression for an $AND$ gate is $R = P \cdot Q$.
Checking the values from the truth table:
For $P=1, Q=1$,$R = 1 \cdot 1 = 1$.
For $P=1, Q=0$,$R = 1 \cdot 0 = 0$.
For $P=0, Q=1$,$R = 0 \cdot 1 = 0$.
For $P=0, Q=0$,$R = 0 \cdot 0 = 0$.
Since the calculated values match the given truth table,the gate is an $AND$ gate.

(B) The focal length of a silvered lens is given by the formula $\frac{1}{F} = \frac{2}{f_l} + \frac{1}{f_m}$,where $f_l$ is the focal length of the lens and $f_m$ is the focal length of the mirror.
For a plano-convex lens,the focal length $f_l$ is given by $\frac{1}{f_l} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.
Here,$R_1 = 10 \; cm$ and $R_2 = \infty$,so $\frac{1}{f_l} = (1.5 - 1)(\frac{1}{10} - 0) = \frac{0.5}{10} = \frac{1}{20}$. Thus,$f_l = 20 \; cm$.
The silvered plane surface acts as a plane mirror,so its focal length $f_m = \infty$.
Therefore,$\frac{1}{F} = \frac{2}{20} + \frac{1}{\infty} = \frac{1}{10} + 0 = \frac{1}{10}$.
Hence,$F = 10 \; cm$.

(C) The formation of a rainbow is a complex optical phenomenon involving multiple processes.
$1$. Sunlight enters a water droplet and undergoes refraction and dispersion,splitting into its constituent colors.
$2$. The light then undergoes total internal reflection $(TIR)$ at the back surface of the droplet.
$3$. Finally,the light refracts again as it exits the droplet.
Therefore,the primary processes involved are refraction,dispersion,and total internal reflection.

(A) The electromagnetic spectrum is ordered by increasing wavelength or decreasing frequency. The order of frequency for these electromagnetic waves is given by: $\nu_{\gamma-rays} > \nu_{X-rays} > \nu_{UV-rays}$.
Given that $\gamma-$ rays are $(b)$,$X-$ rays are $(a)$,and $UV-$ rays are $(c)$,the frequency order is $b > a > c$.

(A) The energy of a photon emitted during an electronic transition is given by $\Delta E = E_{initial} - E_{final} = h\nu$,where $\nu$ is the frequency.
For emission to occur,the electron must transition from a higher energy level to a lower energy level. Thus,options $(b)$ and $(d)$ are excluded as they represent absorption.
Comparing the energy differences for the remaining transitions:
For $n = 2$ to $n = 1$: $\Delta E_1 = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \left( 1 - 0.25 \right) = 13.6 \times 0.75 = 10.2 \text{ eV}$.
For $n = 6$ to $n = 2$: $\Delta E_2 = 13.6 \left( \frac{1}{2^2} - \frac{1}{6^2} \right) = 13.6 \left( 0.25 - 0.0277 \right) = 13.6 \times 0.2223 \approx 3.02 \text{ eV}$.
Since $\nu = \frac{\Delta E}{h}$,the transition with the largest energy difference will emit the photon with the highest frequency.
Comparing $\Delta E_1$ and $\Delta E_2$,we see that $\Delta E_1 > \Delta E_2$.
Therefore,the transition from $n = 2$ to $n = 1$ emits the photon of the highest frequency.

(B) diode is in a forward-biased condition when the potential at the $p$-terminal (anode) is higher than the potential at the $n$-terminal (cathode).
Let $V_A$ be the potential at the anode and $V_K$ be the potential at the cathode.
For forward bias,the condition is $V_A > V_K$.
Checking the options:
$(A)$ $V_A = 0 \ V$,$V_K = 2 \ V$. Here $V_A < V_K$,so it is reverse-biased.
$(B)$ $V_A = 0 \ V$,$V_K = -2 \ V$. Here $V_A > V_K$ $(0 > -2)$,so it is forward-biased.
$(C)$ $V_A = -5 \ V$,$V_K = -2 \ V$. Here $V_A < V_K$ $(-5 < -2)$,so it is reverse-biased.
$(D)$ $V_A = 5 \ V$,$V_K = 12 \ V$. Here $V_A < V_K$,so it is reverse-biased.
Therefore,option $(B)$ is the correct answer.

(C) Let the thickness of the glass slab be $x$.
When viewed from one side,the apparent depth is $d_1 = 5 \ cm$.
When viewed from the other side,the apparent depth is $d_2 = 2 \ cm$.
The relationship between real depth and apparent depth is given by $\mu = \frac{\text{real depth}}{\text{apparent depth}}$.
Let $x_1$ be the real distance of the bubble from the first side and $x_2$ be the real distance from the second side,such that $x_1 + x_2 = x$.
From the formula,$x_1 = \mu d_1 = 1.5 \times 5 = 7.5 \ cm$.
Similarly,$x_2 = \mu d_2 = 1.5 \times 2 = 3.0 \ cm$.
Therefore,the total thickness of the slab is $x = x_1 + x_2 = 7.5 + 3.0 = 10.5 \ cm$.

(D) potentiometer is considered the best device for measuring the potential difference $(p.d.)$ or electromotive force $(EMF)$ of a cell because it operates on the principle of a null deflection method.
In this method,when the potentiometer is balanced,no current flows through the galvanometer connected to the cell being measured.
Since no current is drawn from the source,the terminal potential difference measured is equal to the actual $EMF$ of the cell.
Therefore,it measures the $p.d.$ in an open circuit condition,which avoids the error caused by the internal resistance of the source.

(C) The spectral series of a hydrogen atom is determined by the final energy level $(n_f)$ of the electron transition.
$1$. Transitions to $n_f = 1$ result in the Lyman series.
$2$. Transitions to $n_f = 2$ result in the Balmer series.
$3$. Transitions to $n_f = 3$ result in the Paschen series.
In this case,the electron transitions from $n_i = 4$ to $n_f = 2$. Since the final state is $n_f = 2$,the emission belongs to the Balmer series.
The lines of the Balmer series are defined by transitions from $n_i = 3, 4, 5, ...$ to $n_f = 2$:
- The first line corresponds to $n_i = 3$ to $n_f = 2$.
- The second line corresponds to $n_i = 4$ to $n_f = 2$.
Therefore,the transition from $n=4$ to $n=2$ represents the second line of the Balmer series.

(C) To see the full image of an object of height $H$ in a plane mirror,the minimum length of the mirror required is $H/2$.
Given the height of the man is $H = 6$ feet.
Therefore,the minimum length of the mirror required is $6 / 2 = 3$ feet.

(B) The radius of the helical path of a charged particle moving in a uniform magnetic field is given by the formula:
$r = \frac{mv_{\perp}}{qB} = \frac{mv \sin \theta}{qB}$
Given values:
$q/m = 10^8 \, C/kg \implies m/q = 10^{-8} \, kg/C$
$v = 3 \times 10^5 \, m/s$
$B = 0.3 \, T$
$\theta = 30^{\circ}$
Substituting the values into the formula:
$r = \left( \frac{m}{q} \right) \frac{v \sin 30^{\circ}}{B}$
$r = (10^{-8}) \times \frac{3 \times 10^5 \times 0.5}{0.3}$
$r = 10^{-8} \times \frac{1.5 \times 10^5}{0.3}$
$r = 10^{-8} \times 5 \times 10^5 = 5 \times 10^{-3} \, m$
Converting meters to centimeters:
$r = 5 \times 10^{-3} \times 10^2 \, cm = 0.5 \, cm$.

(B) $J$.$J$. Thomson determined the specific charge $(e/m)$ of an electron using his cathode ray tube experiment. Later,Robert $A$. Millikan determined the charge $(e)$ of an electron using his oil-drop experiment. By combining these two values,the mass $(m)$ of the electron was evaluated indirectly as $m = e / (e/m)$.

(B) Initially,the capacitor of capacitance $C$ is charged to a potential $V$ by a battery. The energy stored is $U = \frac{1}{2} CV^2 = \frac{Q^2}{2C}$,where $Q$ is the initial charge.
When the battery is disconnected and an identical capacitor $C$ is connected in parallel,the total charge $Q$ is redistributed between the two capacitors.
Since the capacitors are in parallel,they will share the charge equally,so each capacitor now has a charge $Q' = \frac{Q}{2}$.
The new potential difference across each capacitor is $V' = \frac{Q'}{C} = \frac{Q}{2C} = \frac{V}{2}$.
The energy stored in each capacitor is $U' = \frac{1}{2} C(V')^2 = \frac{1}{2} C \left(\frac{V}{2}\right)^2 = \frac{1}{2} C \frac{V^2}{4} = \frac{1}{4} \left(\frac{1}{2} CV^2\right) = \frac{U}{4}$.

(C) According to Gauss's Law,the total electric flux through a closed surface is $\frac{1}{\varepsilon_0}$ times the net charge enclosed by the surface.
When a charge $Q$ is placed at the corner of a cube,it is shared by $8$ such identical cubes to form a larger symmetric closed surface (a Gaussian surface).
Therefore,the total flux through the entire Gaussian surface is $\frac{Q}{\varepsilon_0}$.
Since the charge is symmetrically distributed among $8$ cubes,the flux passing through one cube is $\frac{1}{8}$ of the total flux.
Thus,the electric flux through the cube is $\Phi = \frac{Q}{8\varepsilon_0}$.

(A) Consider a small element of length $dl = a d\theta$ at an angle $\theta$ with the axis of symmetry $PO$. The charge on this element is $dq = \lambda dl = \lambda a d\theta$.
The electric field $dE$ at the centre $O$ due to this element is $dE = \frac{1}{4\pi \varepsilon_0} \frac{dq}{a^2} = \frac{1}{4\pi \varepsilon_0} \frac{\lambda a d\theta}{a^2} = \frac{\lambda d\theta}{4\pi \varepsilon_0 a}$.
By symmetry,the components of the electric field perpendicular to the axis of symmetry $PO$ cancel out,while the components along $PO$ add up.
The component of $dE$ along $PO$ is $dE_{\parallel} = dE \cos \theta = \frac{\lambda}{4\pi \varepsilon_0 a} \cos \theta d\theta$.
Integrating from $\theta = -\pi/2$ to $\pi/2$:
$E = \int_{-\pi/2}^{\pi/2} \frac{\lambda}{4\pi \varepsilon_0 a} \cos \theta d\theta = \frac{\lambda}{4\pi \varepsilon_0 a} [\sin \theta]_{-\pi/2}^{\pi/2} = \frac{\lambda}{4\pi \varepsilon_0 a} (1 - (-1)) = \frac{2\lambda}{4\pi \varepsilon_0 a} = \frac{\lambda}{2\pi \varepsilon_0 a}$.