AIPMT 1998 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: $16 = 25 - u^2$.
$u^2 = 25 - 16 = 9$.
$u = 3 \; km/hr$.

(A) The stopping distance $S$ of a vehicle is given by the formula $S = \frac{u^2}{2a}$,where $u$ is the initial velocity and $a$ is the magnitude of deceleration.
Since $a$ is constant for the same car and same braking force,we have $S \propto u^2$.
Given $u_1 = 40 \, km/h$ and $S_1 = 2 \, m$.
For $u_2 = 80 \, km/h$,we have the ratio $\frac{S_2}{S_1} = \left( \frac{u_2}{u_1} \right)^2$.
Substituting the values: $\frac{S_2}{2} = \left( \frac{80}{40} \right)^2 = (2)^2 = 4$.
Therefore,$S_2 = 2 \times 4 = 8 \, m$.

(A) The tension $T$ in the string provides the necessary centripetal force for the horizontal circular motion.
$T = \frac{mv^2}{r}$
Given: $m = 0.25 \, kg$,$r = 1.96 \, m$,and $T_{max} = 25 \, N$.
Substituting the values into the formula:
$25 = \frac{0.25 \times v^2}{1.96}$
$v^2 = \frac{25 \times 1.96}{0.25}$
$v^2 = 100 \times 1.96 = 196$
$v = \sqrt{196} = 14 \, m/s$.
Thus,the maximum speed is $14 \, m/s$.

(A) Given mass $m = 1\, kg$ and acceleration $a = 4.9\, m/s^2$. Taking $g = 9.8\, m/s^2$,we note that $a = g/2$.
Case $(i)$: When the mass is lifted up with acceleration $a$,the tension $T_1$ is given by $T_1 = m(g + a)$.
$T_1 = 1 \times (g + g/2) = 3g/2$.
Case $(ii)$: When the mass is lowered with acceleration $a$,the tension $T_2$ is given by $T_2 = m(g - a)$.
$T_2 = 1 \times (g - g/2) = g/2$.
The ratio of the tensions is $\frac{T_1}{T_2} = \frac{3g/2}{g/2} = \frac{3}{1}$.

(B) The thrust force $F$ required to lift the rocket with an upward acceleration $a$ is given by the equation: $F = v_{ex} \cdot \frac{dm}{dt} = m(g + a)$.
Here,$m = 5000\, kg$ is the mass of the rocket,$g = 10\, m/s^2$ is the acceleration due to gravity,$a = 20\, m/s^2$ is the required upward acceleration,and $v_{ex} = 800\, m/s$ is the exhaust speed.
Rearranging the formula to solve for the rate of mass ejection $\frac{dm}{dt}$:
$\frac{dm}{dt} = \frac{m(g + a)}{v_{ex}}$
Substituting the given values:
$\frac{dm}{dt} = \frac{5000 \times (10 + 20)}{800}$
$\frac{dm}{dt} = \frac{5000 \times 30}{800} = \frac{150000}{800} = 187.5\, kg/s$.
Thus,the amount of gas ejected per second is $187.5\, kg/s$.

(C) The force on the bullet is given by $F = 600 - 2 \times 10^5 t$.
When the bullet leaves the barrel,the force becomes zero,so $F = 0$.
$600 - 2 \times 10^5 t = 0$
$2 \times 10^5 t = 600$
$t = \frac{600}{2 \times 10^5} = 3 \times 10^{-3} \ s$.
Impulse $I$ is defined as the integral of force with respect to time: $I = \int_{0}^{t} F \ dt$.
$I = \int_{0}^{3 \times 10^{-3}} (600 - 2 \times 10^5 t) \ dt$
$I = [600t - 10^5 t^2]_{0}^{3 \times 10^{-3}}$
$I = 600(3 \times 10^{-3}) - 10^5(3 \times 10^{-3})^2$
$I = 1.8 - 10^5(9 \times 10^{-6})$
$I = 1.8 - 0.9 = 0.9 \ N-s$.

(A) According to the Work-Energy Theorem,the work done $W$ is equal to the change in kinetic energy: $W = \Delta K = \frac{1}{2} m(v_f^2 - v_i^2)$.
Given $x = 3t - 4t^2 + t^3$,the velocity $v$ is the derivative of position with respect to time: $v = \frac{dx}{dt} = 3 - 8t + 3t^2$.
At $t = 0 \,s$,$v_i = 3 - 8(0) + 3(0)^2 = 3 \,m/s$.
At $t = 4 \,s$,$v_f = 3 - 8(4) + 3(4)^2 = 3 - 32 + 48 = 19 \,m/s$.
The mass $m = 3 \,g = 0.003 \,kg$.
Substituting the values: $W = \frac{1}{2} \times 0.003 \times (19^2 - 3^2) = 0.0015 \times (361 - 9) = 0.0015 \times 352 = 0.528 \,J$.
Since $1 \,J = 1000 \,mJ$,the work done is $0.528 \times 1000 = 528 \,mJ$.

(D) In a one-dimensional elastic collision between two bodies of equal mass,the velocities of the bodies are interchanged after the collision.
Given: Initial velocity of mass $m_1$ is $u_1 = +3 \, m/s$ and initial velocity of mass $m_2$ is $u_2 = -5 \, m/s$.
Since $m_1 = m_2$,after the elastic collision,the final velocity of mass $m_1$ becomes $v_1 = u_2 = -5 \, m/s$ and the final velocity of mass $m_2$ becomes $v_2 = u_1 = +3 \, m/s$.

(B) Let the initial height be $h_1 = 5 \, m$ and the height after the bounce be $h_2 = 1.8 \, m$.
The velocity of the ball just before the impact is $v_1 = \sqrt{2gh_1}$ and the velocity just after the impact is $v_2 = \sqrt{2gh_2}$.
The coefficient of restitution $e$ is given by the ratio of velocities:
$e = \frac{v_2}{v_1} = \frac{\sqrt{2gh_2}}{\sqrt{2gh_1}} = \sqrt{\frac{h_2}{h_1}}$
Substituting the given values:
$e = \sqrt{\frac{1.8}{5}} = \sqrt{\frac{18}{50}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$
The velocity after the bounce is $v_2 = e v_1 = \frac{3}{5} v_1$.
The loss in velocity is $\Delta v = v_1 - v_2 = v_1 - \frac{3}{5} v_1 = \frac{2}{5} v_1$.
Therefore,the factor by which the ball loses its velocity is $\frac{\Delta v}{v_1} = \frac{2}{5}$.

(C) The change in internal energy $\Delta U$ for an ideal gas is given by the formula $\Delta U = n C_V \Delta T$.
Since $C_V = R/(\gamma - 1)$,we have $\Delta U = n (R/(\gamma - 1)) \Delta T$.
Using the ideal gas law $pV = nRT$,at constant pressure $p$,we have $p \Delta V = nR \Delta T$.
Substituting $n R \Delta T = p \Delta V$ into the equation,we get $\Delta U = p \Delta V / (\gamma - 1)$.
Given the volume changes from $V$ to $2V$,the change in volume is $\Delta V = 2V - V = V$.
Therefore,$\Delta U = pV / (\gamma - 1)$.

(A) According to the first law of thermodynamics,the heat supplied to the system $(\Delta Q)$ is equal to the sum of the change in internal energy $(\Delta U)$ and the work done by the system $(\Delta W)$:
$\Delta Q = \Delta U + \Delta W$
In an adiabatic process,there is no exchange of heat with the surroundings,so $\Delta Q = 0$.
Substituting $\Delta Q = 0$ into the equation,we get:
$0 = \Delta U + \Delta W$
Therefore,$\Delta U = - \Delta W$.

(C) According to the Stefan-Boltzmann law,the radiant energy emitted by a black body is proportional to the fourth power of its absolute temperature,i.e.,$E \propto T^4$.
Let $E_1$ be the initial radiant energy incident on the earth and $T_1$ be the initial temperature of the sun.
Given $E_1 = 20 \, kcal/(m^2 \cdot min)$.
Let $E_2$ be the new radiant energy when the temperature of the sun becomes $T_2 = 2T_1$.
Using the ratio formula:
$\frac{E_2}{E_1} = \left( \frac{T_2}{T_1} \right)^4$
Substituting the values:
$\frac{E_2}{20} = \left( \frac{2T_1}{T_1} \right)^4$
$\frac{E_2}{20} = (2)^4 = 16$
Therefore,$E_2 = 16 \times 20 = 320 \, kcal/(m^2 \cdot min)$.

(C) The frequency of a spring-mass system is given by the formula $n = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$,where $k$ is the spring constant and $m$ is the mass.
From this relation,we can see that $n \propto \frac{1}{\sqrt{m}}$.
Let the initial frequency be $n_1 = n$ for mass $m_1 = m$,and the new frequency be $n_2$ for mass $m_2 = 4m$.
Using the proportionality,we have $\frac{n_1}{n_2} = \sqrt{\frac{m_2}{m_1}}$.
Substituting the values,$\frac{n}{n_2} = \sqrt{\frac{4m}{m}} = \sqrt{4} = 2$.
Therefore,$n_2 = \frac{n}{2}$.

(B) For a damped driven oscillator,the amplitude $A$ is given by $A = \frac{F/m}{\sqrt{(\omega_0^2 - \omega^2)^2 + (b\omega/m)^2}}$.
Amplitude is maximum when the denominator is minimum,which occurs at $\omega_1 = \sqrt{\omega_0^2 - 2(b/2m)^2}$. Thus,$\omega_1 \neq \omega_0$.
The energy of the oscillator is proportional to the square of the amplitude and is also related to the power absorbed. The energy of the particle is maximum at the velocity resonance frequency,which occurs when $\omega_2 = \omega_0$.
Therefore,the correct relationship is $\omega_1 \neq \omega_0$ and $\omega_2 = \omega_0$.

(C) Let $T_S$ be the time period of the shorter pendulum and $T_L$ be the time period of the longer pendulum.
Given lengths are $l_S = 0.5\, m$ and $l_L = 2.0\, m$.
The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
So,$T_S = 2\pi \sqrt{\frac{0.5}{g}}$ and $T_L = 2\pi \sqrt{\frac{2.0}{g}}$.
Taking the ratio,$\frac{T_L}{T_S} = \sqrt{\frac{2.0}{0.5}} = \sqrt{4} = 2$.
Thus,$T_L = 2T_S$.
Let $N_S$ be the number of oscillations of the shorter pendulum and $N_L$ be the number of oscillations of the longer pendulum when they are in the same phase again.
At this time $t$,$t = N_S T_S = N_L T_L$.
Substituting $T_L = 2T_S$,we get $N_S T_S = N_L (2T_S)$.
Therefore,$N_S = 2N_L$.
For the first time they are in phase again,we take the smallest integer values,$N_L = 1$,which gives $N_S = 2$.

(A) In a sinusoidal wave,the motion from maximum displacement (amplitude) to zero displacement corresponds to one-quarter of the total time period $(T)$.
Therefore,the time taken is $t = \frac{T}{4}$.
Since the frequency ($n$ or $f$) is the reciprocal of the time period,$T = \frac{1}{n}$.
Substituting this into the equation,we get $t = \frac{1}{4n}$.
Rearranging for frequency,$n = \frac{1}{4t}$.
Given $t = 0.170\,s$,we calculate $n = \frac{1}{4 \times 0.170} = \frac{1}{0.680} \approx 1.47\,Hz$.

(D) The given wave equation is $y = y_0 \sin \frac{2\pi}{\lambda} (vt - x)$.
Comparing this with the standard wave equation $y = a \sin \frac{2\pi}{\lambda} (vt - x)$,we identify the amplitude $a = y_0$ and wave velocity $v_{wave} = v$.
The particle velocity $v_p$ is given by the derivative of displacement with respect to time: $v_p = \frac{\partial y}{\partial t} = y_0 \cdot \frac{2\pi v}{\lambda} \cos \frac{2\pi}{\lambda} (vt - x)$.
The maximum particle velocity is $(v_{max})_{particle} = y_0 \cdot \frac{2\pi v}{\lambda}$.
According to the problem,$(v_{max})_{particle} = 2 \cdot v_{wave}$.
Substituting the values: $\frac{y_0 \cdot 2\pi v}{\lambda} = 2v$.
Canceling $v$ from both sides: $\frac{2\pi y_0}{\lambda} = 2$.
Solving for $\lambda$: $\lambda = \pi y_0$.

(A) The distance between two consecutive nodes is $\frac{\lambda}{2}$.
In a standing wave with $3$ nodes and $2$ antinodes,there are $2$ loops.
The total length $L$ between the two extreme nodes is given by $L = 2 \times \frac{\lambda}{2} = \lambda$.
Given that the distance between the two atoms (which act as the extreme nodes) is $1.21 \; \mathring{A}$,we have $L = 1.21 \; \mathring{A}$.
Therefore,the wavelength $\lambda = 1.21 \; \mathring{A}$.

(B) The Doppler effect occurs when there is a relative velocity between the source and the observer along the line joining them.
In this problem,the vehicle is moving in a direction perpendicular to the line joining the observer and the vehicle.
Therefore,the component of the velocity of the source along the line joining the observer and the source is $v_s \cos(90^{\circ}) = 0$.
Since there is no relative motion along the line of sight,the frequency perceived by the observer remains the same as the source frequency.
Thus,the observed frequency is $n' = n$.
Given that the observed frequency is $n + n_1$,we have $n + n_1 = n$,which implies $n_1 = 0$.

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

(D) Let $L$ be the length of the rigid rod $AB$. Let $x$ be the distance of $A$ from the corner and $y$ be the distance of $B$ from the corner.
From the geometry,we have $L^2 = x^2 + y^2$.
Differentiating both sides with respect to time $t$,we get:
$0 = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}$
$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$
Given that the velocity of $A$ is $v_A = -\frac{dx}{dt} = 10\; m/s$ (since $x$ is decreasing).
So,$\frac{dx}{dt} = -10\; m/s$.
Substituting this into the equation:
$x(-10) + y \frac{dy}{dt} = 0$
$y \frac{dy}{dt} = 10x$
$\frac{dy}{dt} = 10 \left( \frac{x}{y} \right)$
From the triangle,$\frac{x}{y} = \cot(\alpha)$.
At $\alpha = 60^{\circ}$,$\frac{x}{y} = \cot(60^{\circ}) = \frac{1}{\sqrt{3}}$.
Therefore,the velocity of $B$ is $v_B = \frac{dy}{dt} = 10 \times \frac{1}{\sqrt{3}} = \frac{10}{1.732} \approx 5.77\; m/s$.
Rounding to one decimal place,we get $5.8\; m/s$.

(C) Let $x$ be the perpendicular distance from the center $O$ of the equilateral triangle to each of its sides.
Since the triangle is equilateral,the distance from the center to each side is the same.
The torque $\tau$ produced by a force $F$ about point $O$ is given by $\tau = F \times x$.
Looking at the directions of the forces $\vec F_1, \vec F_2$,and $\vec F_3$ along the sides of the triangle,we can see that $\vec F_1$ and $\vec F_2$ produce torques in the same rotational sense (e.g.,clockwise) about $O$,while $\vec F_3$ produces a torque in the opposite sense (e.g.,counter-clockwise).
For the total torque about $O$ to be zero,the sum of the torques must be zero:
$\tau_1 + \tau_2 - \tau_3 = 0$
$F_1 x + F_2 x - F_3 x = 0$
Dividing by $x$ (since $x \neq 0$):
$F_1 + F_2 - F_3 = 0$
Therefore,$F_3 = F_1 + F_2$.

(D) To keep the mass $M$ at rest relative to the wedge,we analyze the forces in the non-inertial frame of the wedge.
When the wedge accelerates to the left with acceleration $a$,a pseudo force $F_p = Ma$ acts on the mass $M$ towards the right.
The forces acting on the mass $M$ along the incline are:
$1$. The component of pseudo force $Ma \cos \theta$ acting up the incline.
$2$. The component of gravitational force $Mg \sin \theta$ acting down the incline.
For the mass to remain at rest relative to the wedge,these two forces must balance:
$Ma \cos \theta = Mg \sin \theta$
$a \cos \theta = g \sin \theta$
$a = g \frac{\sin \theta}{\cos \theta}$
$a = g \tan \theta$
Thus,an acceleration $a = g \tan \theta$ must be applied to the wedge towards the left.

(C) The force experienced by a charge $q$ in a uniform electric field $E$ is given by $F = qE$.
Since the particle starts from rest and moves a distance $y$ in the direction of the force,the work done by the electric field on the particle is $W = F \times y = (qE) \times y = qEy$.
According to the work-energy theorem,the work done by the electric field is equal to the change in kinetic energy of the particle.
Since the initial kinetic energy is $0$,the final kinetic energy attained by the particle is $K = qEy$.

(A) For a hollow conducting sphere,the charge resides entirely on its outer surface.
According to Gauss's Law,the electric field inside a closed conducting shell is always zero because there is no enclosed charge within any Gaussian surface drawn inside the sphere.
Therefore,the electric field at the centre of the sphere is $0\,N/C$.

(D) The electric field at a point on the equatorial line (perpendicular bisector) of an electric dipole is given by the formula:
$E_{equatorial} = \frac{1}{4\pi\epsilon_0} \frac{p}{r^3}$
where $p$ is the dipole moment and $r$ is the distance from the center of the dipole.
From this expression,it is clear that the electric field $E$ is directly proportional to the dipole moment $p$ $(E \propto p)$ and inversely proportional to the cube of the distance $r$ $(E \propto r^{-3})$.
Therefore,the electric field at $Q$ is proportional to $p$ and $r^{-3}$.

(B) Let $G = 9 \, \Omega$ be the resistance of the galvanometer and $S = 2 \, \Omega$ be the resistance of the shunt.
Let $I = 1 \, A$ be the total current.
Let $I_s$ be the current passing through the shunt and $I_g$ be the current passing through the galvanometer.
According to the principle of parallel circuits, the voltage across the galvanometer and the shunt is the same: $V = I_g G = I_s S$.
Since $I = I_g + I_s$, we have $I_g = I - I_s$.
Substituting this into the voltage equation: $(I - I_s) G = I_s S$.
Rearranging to solve for $I_s$: $I G - I_s G = I_s S \implies I G = I_s (S + G)$.
Therefore, $I_s = I \times \frac{G}{S + G}$.
Substituting the given values: $I_s = 1 \times \frac{9}{2 + 9} = \frac{9}{11} \approx 0.818 \, A$.
Rounding to the nearest provided option, the current through the shunt is approximately $0.8 \, A$.

(D) Let the resistance of each resistor be $R$ and the source $e.m.f.$ be $V$.
In series connection,the equivalent resistance is $R_S = R + R + R = 3R$.
The power dissipated in series is $P_S = \frac{V^2}{R_S} = \frac{V^2}{3R} = 10 \ W$.
From this,we get $\frac{V^2}{R} = 30 \ W$.
In parallel connection,the equivalent resistance is $R_P = \frac{R}{3}$.
The power dissipated in parallel is $P_P = \frac{V^2}{R_P} = \frac{V^2}{R/3} = 3 \left( \frac{V^2}{R} \right)$.
Substituting the value $\frac{V^2}{R} = 30 \ W$,we get $P_P = 3 \times 30 = 90 \ W$.

(C) The heat produced in a conductor due to the flow of current is given by Joule's law of heating: $H = i^2Rt$.
Since the rise in temperature $\Delta T$ is directly proportional to the heat produced $H$,we have $\Delta T \propto i^2$.
Given that the initial rise in temperature $\Delta T_1 = 5\,^{\circ}\text{C}$ for current $i_1 = i$.
When the current is doubled,$i_2 = 2i$.
The new rise in temperature $\Delta T_2$ is given by the ratio:
$\frac{\Delta T_2}{\Delta T_1} = \left(\frac{i_2}{i_1}\right)^2 = \left(\frac{2i}{i}\right)^2 = 4$.
Therefore,$\Delta T_2 = 4 \times \Delta T_1 = 4 \times 5\,^{\circ}\text{C} = 20\,^{\circ}\text{C}$.

(B) The force per unit length between two long parallel current-carrying wires is given by the formula: $\frac{F}{l} = \frac{\mu_0}{2\pi} \frac{i_1 i_2}{r}$.
Given: $i_1 = 1 \ A$,$i_2 = 1 \ A$,$r = 1 \ m$,and $\frac{\mu_0}{4\pi} = 10^{-7} \ T \cdot m/A$.
Substituting these values into the formula:
$\frac{F}{l} = 2 \times 10^{-7} \times \frac{1 \times 1}{1} = 2 \times 10^{-7} \ N/m$.

(B) Iron is a ferromagnetic substance,which has high magnetic permeability. When placed in an external magnetic field,the magnetic field lines prefer to pass through the iron rather than the interior space. This phenomenon is known as magnetic shielding. Therefore,to protect sensitive equipment from an external magnetic field,it should be placed inside an iron can.

(B) For an ideal transformer,the power input equals the power output,which implies the relationship between current and the number of turns is given by the inverse ratio: $\frac{I_p}{I_s} = \frac{N_s}{N_p}$.
Given:
Primary to secondary turns ratio $\frac{N_p}{N_s} = \frac{1}{25}$,so $\frac{N_s}{N_p} = 25$.
Secondary current $I_s = 2\, A$.
Using the formula: $I_p = I_s \times \frac{N_s}{N_p}$.
$I_p = 2\, A \times 25 = 50\, A$.
Therefore,the current in the primary coil is $50\, A$.

(B) The induced $e.m.f.$ in the second coil is given by $e = M \frac{di}{dt}$.
Given $M = 0.005 \, H$,$I = I_0 \sin(\omega t)$,$I_0 = 10 \, A$,and $\omega = 100\pi \, rad/s$.
Substituting the expression for $I$:
$e = M \frac{d}{dt} (I_0 \sin(\omega t)) = M I_0 \omega \cos(\omega t)$.
The maximum value of $e.m.f.$ $(e_{\max})$ occurs when $\cos(\omega t) = 1$.
$e_{\max} = M I_0 \omega$.
Substituting the values:
$e_{\max} = 0.005 \times 10 \times 100\pi = 5\pi \, V$.

(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Using the approximation $hc \approx 12400 \; eV \cdot \mathring A$,we get $E = \frac{12400}{5000} = 2.48 \; eV$.
According to Einstein's photoelectric equation,$E = W_0 + K_{\max}$,where $W_0$ is the work function and $K_{\max}$ is the maximum kinetic energy.
Substituting the values: $2.48 = 1.9 + K_{\max}$.
Therefore,$K_{\max} = 2.48 - 1.9 = 0.58 \; eV$.

(D) According to Einstein's photoelectric equation: $K_{\max} = \frac{hc}{\lambda} - W_0 = \frac{1}{2}mv^2$,where $W_0 = \frac{hc}{\lambda_0}$ is the work function.
Thus,$v = \sqrt{\frac{2hc}{m} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)} = \sqrt{\frac{2hc}{m} \left( \frac{\lambda_0 - \lambda}{\lambda \lambda_0} \right)} \dots (i)$
When the wavelength is changed to $\lambda' = \frac{3\lambda}{4}$,the new velocity $v'$ is:
$v' = \sqrt{\frac{2hc}{m} \left( \frac{1}{3\lambda/4} - \frac{1}{\lambda_0} \right)} = \sqrt{\frac{2hc}{m} \left( \frac{4\lambda_0 - 3\lambda}{3\lambda \lambda_0} \right)} \dots (ii)$
Dividing $(ii)$ by $(i)$:
$\frac{v'}{v} = \sqrt{\frac{4\lambda_0 - 3\lambda}{3\lambda \lambda_0} \cdot \frac{\lambda \lambda_0}{\lambda_0 - \lambda}} = \sqrt{\frac{4}{3} \cdot \frac{\lambda_0 - 0.75\lambda}{\lambda_0 - \lambda}}$
Since $\lambda_0 > \lambda$,it follows that $(\lambda_0 - 0.75\lambda) > (\lambda_0 - \lambda)$.
Therefore,$\frac{\lambda_0 - 0.75\lambda}{\lambda_0 - \lambda} > 1$.
This implies $\frac{v'}{v} > \sqrt{\frac{4}{3}}$,or $v' > v(4/3)^{1/2}$.

(C) The centripetal force required for the circular motion of the electron is provided by the electrostatic Coulomb force between the proton and the electron.
Equating the centripetal force to the Coulomb force:
$\frac{m{v^2}}{{a_0}} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{e^2}}}{{a_0^2}}$
Canceling $a_0$ from both sides:
$m{v^2} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{e^2}}}{{a_0}}$
Solving for $v^2$:
${v^2} = \frac{{{e^2}}}{{4\pi {\varepsilon _0}{a_0}m}}$
Taking the square root of both sides,we get the speed of the electron:
$v = \frac{e}{{\sqrt {4\pi {\varepsilon _0}{a_0}m} }}$
Thus,the correct option is $C$.

(A) Let the percentage of $_5B^{10}$ atoms be $x$. Then the percentage of $_5B^{11}$ atoms is $(100 - x)$.
The average atomic weight is given by the formula:
$\text{Average Atomic Weight} = \frac{(10 \times x) + (11 \times (100 - x))}{100} = 10.81$
Multiplying by $100$:
$10x + 1100 - 11x = 1081$
$-x = 1081 - 1100$
$-x = -19$
$x = 19$
So,the percentage of $_5B^{10}$ is $19\%$ and the percentage of $_5B^{11}$ is $81\%$.
Therefore,the ratio of $_5B^{10} : _5B^{11}$ is $19 : 81$.

(A) In a nuclear fission reaction,both the mass number and the atomic number must be conserved on both sides of the equation.
Given reaction: $_{92}U^{235} + _0n^1 \to _{38}Sr^{90} + X$
Let the unknown product $X$ be $_{Z}A^{A}$.
Conservation of mass number: $235 + 1 = 90 + A \implies 236 = 90 + A \implies A = 146$.
However,the standard fission of $U^{235}$ producing $Sr^{90}$ releases neutrons. Checking the options,option $A$ provides $_{54}Xe^{143} + 3_0n^1$.
Sum of mass numbers: $90 + 143 + 3(1) = 90 + 143 + 3 = 236$.
Sum of atomic numbers: $38 + 54 + 3(0) = 92$.
Since both are conserved,the correct completion is $_{54}Xe^{143} + 3_0n^1$.

(C) The number of half-lives $n$ is given by $n = \frac{t}{T_{1/2}}$.
For substance $A$,$n_A = \frac{80}{20} = 4$.
For substance $B$,$n_B = \frac{80}{40} = 2$.
The number of remaining nuclei is given by $N = N_0 \left( \frac{1}{2} \right)^n$.
Since initial numbers $N_0$ are equal,the ratio is $\frac{N_A}{N_B} = \frac{N_0 (1/2)^{n_A}}{N_0 (1/2)^{n_B}} = \frac{(1/2)^4}{(1/2)^2} = \frac{2^2}{2^4} = \frac{4}{16} = \frac{1}{4}$.
Thus,the ratio is $1 : 4$.

(D) The initial nucleus is $_n{X^m}$.
When it emits one $\alpha$ particle $(_{2}He^{4})$,the atomic number decreases by $2$ and the mass number decreases by $4$. The nucleus becomes $_{n-2}X^{m-4}$.
When this nucleus emits one $\beta$ particle $(_{-1}e^{0})$,the atomic number increases by $1$ while the mass number remains unchanged. The new atomic number is $(n-2) + 1 = n-1$.
Thus,the resulting nucleus is $_{n-1}Z^{m-4}$.

(C) $PN-$junction diode allows current to flow when it is forward biased.
When the battery polarity is reversed,the $PN-$junction becomes reverse biased.
In reverse bias,the depletion layer width increases,offering very high resistance to the flow of current.
Consequently,the current drops almost to zero.
Therefore,the device is a $PN-$junction.

(D) Given: Transfer ratio (current gain) $\beta = 50$,Input resistance $R_i = 1 \; k\Omega = 1000 \; \Omega$,and Peak input voltage $V_i = 0.01 \; V$.
First,calculate the peak input current $i_b$ using Ohm's law: $i_b = \frac{V_i}{R_i} = \frac{0.01 \; V}{1000 \; \Omega} = 10^{-5} \; A$.
The relationship between collector current $i_c$ and base current $i_b$ is given by $\beta = \frac{i_c}{i_b}$.
Therefore,the peak collector current is $i_c = \beta \times i_b = 50 \times 10^{-5} \; A = 500 \times 10^{-6} \; A = 500 \; \mu A$.

(C) Let us analyze the output for each gate with the given inputs:
$(A)$ For an $AND$ gate,the output $Y = A \cdot B$. With inputs $1$ and $0$,$Y = 1 \cdot 0 = 0$.
$(B)$ For a $NOR$ gate,the output $Y = \overline{A + B}$. With inputs $0$ and $1$,$Y = \overline{0 + 1} = \overline{1} = 0$.
$(C)$ For a $NAND$ gate,the output $Y = \overline{A \cdot B}$. With inputs $0$ and $1$,$Y = \overline{0 \cdot 1} = \overline{0} = 1$.
$(D)$ For an $XNOR$ gate,the output $Y = A \odot B$ (or $\overline{A \oplus B}$). With inputs $0$ and $1$,$Y = \overline{0 \oplus 1} = \overline{1} = 0$.
Thus,the $NAND$ gate produces an output of $1$.

(D) For a $NAND$ gate,the output $C$ is given by the Boolean expression $C = \overline{A \cdot B}$.
Checking the values:
$1$. For $A = 0, B = 0$: $C = \overline{0 \cdot 0} = \overline{0} = 1$.
$2$. For $A = 0, B = 1$: $C = \overline{0 \cdot 1} = \overline{0} = 1$.
$3$. For $A = 1, B = 0$: $C = \overline{1 \cdot 0} = \overline{0} = 1$.
$4$. For $A = 1, B = 1$: $C = \overline{1 \cdot 1} = \overline{1} = 0$.
Comparing these results with the given table,the output matches the $NAND$ gate logic.

(A) Let $\alpha$ be the angle of incidence at the end face of the rod and $r$ be the angle of refraction. By Snell's Law,$1 \cdot \sin \alpha = n \cdot \sin r$,so $\sin r = \frac{\sin \alpha}{n}$.
At the lateral surface,the angle of incidence is $i = 90^\circ - r$. For total internal reflection to occur,the angle of incidence $i$ must be greater than the critical angle $C$,where $\sin C = \frac{1}{n}$.
Thus,$i > C \implies \sin i > \sin C$.
Substituting $i = 90^\circ - r$,we get $\sin(90^\circ - r) > \sin C$,which simplifies to $\cos r > \frac{1}{n}$.
Squaring both sides,$\cos^2 r > \frac{1}{n^2} \implies 1 - \sin^2 r > \frac{1}{n^2}$.
Substituting $\sin r = \frac{\sin \alpha}{n}$,we have $1 - \frac{\sin^2 \alpha}{n^2} > \frac{1}{n^2}$.
Rearranging gives $1 > \frac{1 + \sin^2 \alpha}{n^2}$,or $n^2 > 1 + \sin^2 \alpha$.
Since this must hold for any angle of incidence $\alpha$,we consider the maximum value of $\sin^2 \alpha$,which is $1$ (when $\alpha = 90^\circ$).
Therefore,$n^2 > 1 + 1 = 2$,which means $n > \sqrt{2}$.

(C) For the convex lens,the object distance $u = -30 \,cm$ and focal length $f = +20 \,cm$. Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
$\frac{1}{20} = \frac{1}{v} - \frac{1}{-30}$
$\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3-2}{60} = \frac{1}{60}$
So,$v = 60 \,cm$. This means the lens forms a real image at $60 \,cm$ from the lens.
For the final image to coincide with the object,the light rays must fall normally on the convex mirror so that they retrace their path. This happens if the rays are directed towards the center of curvature of the mirror.
The radius of curvature of the mirror is $R = 10 \,cm$. Thus,the mirror must be placed such that its center of curvature coincides with the image formed by the lens.
The distance of the mirror from the lens is $d = v - R = 60 \,cm - 10 \,cm = 50 \,cm$.

(D) The energy $E$ of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Given:
$h = 6.63 \times 10^{-34}\,J\cdot s$
$c = 3 \times 10^8\,m/s$
$\lambda = 21\,cm = 0.21\,m = 21 \times 10^{-2}\,m$
Substituting these values:
$E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{21 \times 10^{-2}}$
$E = \frac{19.89 \times 10^{-26}}{21 \times 10^{-2}}$
$E \approx 0.947 \times 10^{-24}\,J$
$E \approx 10^{-24}\,J$.

(B) The magnetic field $B$ at the center of a circular coil of $N$ turns and radius $r$ carrying current $I$ is given by $B = \frac{\mu_0 N I}{2r}$.
Let $L$ be the total length of the wire.
For the first coil $(N_1 = 1)$: The circumference $2\pi r_1 = L$,so $r_1 = \frac{L}{2\pi}$. The magnetic field is $B_1 = \frac{\mu_0 (1) I}{2(L/2\pi)} = \frac{\mu_0 \pi I}{L}$.
For the second coil $(N_2 = 2)$: The total length $2(2\pi r_2) = L$,so $r_2 = \frac{L}{4\pi}$. The magnetic field is $B_2 = \frac{\mu_0 (2) I}{2(L/4\pi)} = \frac{4\mu_0 \pi I}{L}$.
Taking the ratio: $\frac{B_1}{B_2} = \frac{\mu_0 \pi I / L}{4\mu_0 \pi I / L} = \frac{1}{4}$.
Thus,the ratio is $1 : 4$.

(D) During the formation of a $p-n$ junction,holes from the $p-$region diffuse into the $n-$region and electrons from the $n-$region diffuse into the $p-$region.
When an electron meets a hole,they recombine and neutralize each other,creating a thin layer at the junction that is devoid of free charge carriers. This is known as the depletion layer.
Due to the diffusion process,immobile ionized atoms are left behind: negative ions on the $p-$side and positive ions on the $n-$side.
This accumulation of immobile charges creates an electric field and a potential difference across the junction,which is known as the potential barrier.