WBJEE 2010 Physics Question Paper with Answer and Solution

(B) The acceleration due to gravity at a height $h$ above the Earth's surface is given by the formula: $g' = g \left(1 + \frac{h}{R}\right)^{-2}$.
Given that the acceleration due to gravity at height $h$ is $1 \%$ of its value at the surface,we have $g' = \frac{1}{100} g$.
Substituting this into the formula:
$\frac{g}{100} = g \left(1 + \frac{h}{R}\right)^{-2}$
$\frac{1}{100} = \left(1 + \frac{h}{R}\right)^{-2}$
Taking the square root on both sides:
$\frac{1}{10} = \left(1 + \frac{h}{R}\right)^{-1}$
$1 + \frac{h}{R} = 10$
$\frac{h}{R} = 9$
$h = 9 R$.
Therefore,the height is $9 R$.

(A) The gravitational potential energy $U$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.
At the surface of the Earth,$r = R$,so $U_i = -\frac{GMm}{R}$.
At a height $h = nR$ above the surface,the distance from the center is $r = R + nR = R(1+n)$.
So,$U_f = -\frac{GMm}{R(1+n)}$.
The change in gravitational potential energy is $\Delta U = U_f - U_i = -\frac{GMm}{R(1+n)} - (-\frac{GMm}{R})$.
$\Delta U = \frac{GMm}{R} \left(1 - \frac{1}{1+n}\right) = \frac{GMm}{R} \left(\frac{1+n-1}{1+n}\right) = \frac{GMm}{R} \left(\frac{n}{1+n}\right)$.
Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this,$\Delta U = \frac{gR^2 m}{R} \left(\frac{n}{n+1}\right) = \left(\frac{n}{n+1}\right) mgR$.

(B) The root mean square speed $(v_{rms})$ of an ideal gas is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.
From this relation,we can see that $v_{rms} \propto \sqrt{T}$.
Let $v_1$ be the speed at temperature $T_1 = 120 \,K$ and $v_2$ be the speed at temperature $T_2 = 480 \,K$.
Given $v_1 = v$.
Using the proportionality,we have: $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $\frac{v_2}{v} = \sqrt{\frac{480}{120}}$.
$\frac{v_2}{v} = \sqrt{4} = 2$.
Therefore,$v_2 = 2v$.

(C) The boy is climbing the pole at a constant speed, which means his acceleration is $0$. Therefore, the net force acting on him in the vertical direction is $0$.
The forces acting on the boy in the vertical direction are his weight $(mg)$ acting downwards and the frictional force $(f)$ acting upwards.
For equilibrium, $f = mg$.
Given that the frictional force $f = \mu N$, where $\mu$ is the coefficient of friction and $N$ is the normal force (the horizontal force applied by the boy on the pole).
So, $\mu N = mg$.
Substituting the given values: $0.8 \times N = 40 \,kg \times 10 \,m/s^2$.
$0.8 \times N = 400 \,N$.
$N = \frac{400}{0.8} = 500 \,N$.
Thus, the horizontal force applied by the boy is $500 \,N$.

(B) Step $1$: Calculate the common acceleration of the system.
The total mass of the system is $M = 2 \,kg + 1 \,kg = 3 \,kg$.
The applied force is $F = 3 \,N$.
Using Newton's second law, $F = Ma$, we get:
$a = \frac{F}{M} = \frac{3 \,N}{3 \,kg} = 1 \,m/s^2$.
Step $2$: Calculate the contact force between the blocks.
Consider the $1 \,kg$ block. The only horizontal force acting on it is the contact force $N_1$ exerted by the $2 \,kg$ block.
Applying Newton's second law to the $1 \,kg$ block:
$N_1 = m_2 \times a = 1 \,kg \times 1 \,m/s^2 = 1 \,N$.
Therefore, the contact force between the two blocks is $1 \,N$.

(B) The force constant $k$ of a spring is inversely proportional to its natural length $l$,i.e.,$k \propto \frac{1}{l}$.
When a spring of length $L$ and force constant $k$ is cut into $n$ equal parts,the length of each part becomes $l' = \frac{L}{n}$.
Since $k' l' = k L$,we have $k' = k \frac{L}{l'} = k \frac{L}{L/n} = n k$.
In this problem,the spring is cut into $n = 3$ equal parts.
Therefore,the force constant of each part is $k' = 3 k$.

(C) For a body floating in a liquid, the weight of the body equals the weight of the displaced liquid. Let $V$ be the total volume of the body and $d$ be its density.
Case $1$: Floating in water (density $\rho_w = 1 \text{ g/cm}^3$)
Volume immersed $V_{in} = V - 0.4V = 0.6V$.
Weight of body = Weight of displaced water
$V \cdot d \cdot g = (0.6V) \cdot \rho_w \cdot g$
$d = 0.6 \cdot 1 = 0.6 \text{ g/cm}^3$.
Case $2$: Floating in oil (density $\rho_{oil}$)
Volume immersed $V_{in} = V - 0.6V = 0.4V$.
Weight of body = Weight of displaced oil
$V \cdot d \cdot g = (0.4V) \cdot \rho_{oil} \cdot g$
$0.6 = 0.4 \cdot \rho_{oil}$
$\rho_{oil} = \frac{0.6}{0.4} = 1.5$.
Thus, the relative density of the oil is $1.5$.

(A) Let the vertical line passing through the center be the reference. The interface between the two liquids is at an angle $\theta$ from the vertical.
Since the tube is in a vertical plane,the pressure at the lowest point must be the same from both sides.
Let the density of the liquid on the left be $\delta$ and on the right be $\rho$.
The vertical height of the center of mass of the liquid column of density $\delta$ from the lowest point is $h_1 = R(1 - \cos \theta)$.
The vertical height of the center of mass of the liquid column of density $\rho$ from the lowest point is $h_2 = R(1 - \cos \theta)$.
For equilibrium,the pressure at the lowest point of the tube must be equal from both sides.
The pressure exerted by a liquid column in a circular tube is proportional to the vertical depth of its center of mass.
The condition for equilibrium is given by:
$\delta g R(\cos \theta + \sin \theta) = \rho g R(\cos \theta - \sin \theta)$
Dividing both sides by $gR$:
$\delta(\cos \theta + \sin \theta) = \rho(\cos \theta - \sin \theta)$
$\delta \cos \theta + \delta \sin \theta = \rho \cos \theta - \rho \sin \theta$
$\sin \theta(\rho + \delta) = \cos \theta(\rho - \delta)$
$\tan \theta = \frac{\rho - \delta}{\rho + \delta}$
$\theta = \tan ^{-1}\left(\frac{\rho - \delta}{\rho + \delta}\right)$

(C) The terminal velocity $v_t$ of a sphere of radius $r$ falling in a viscous liquid is given by $v_t = \frac{2r^2g(\rho - \sigma)}{9\eta}$,where $\rho$ is the density of the sphere and $\sigma$ is the density of the liquid.
Since the spheres are of the same metal,$\rho$ is constant,so $v_t \propto r^2$.
The mass $M$ of a sphere is given by $M = \frac{4}{3}\pi r^3 \rho$,which implies $M \propto r^3$,or $r \propto M^{1/3}$.
Substituting this into the proportionality for terminal velocity,we get $v_t \propto (M^{1/3})^2 = M^{2/3}$.
Given the masses $M_1 = M$ and $M_2 = 8M$,the ratio of their terminal velocities is $\frac{v_1}{v_2} = \left(\frac{M_1}{M_2}\right)^{2/3}$.
Substituting the values,$\frac{v}{nv} = \left(\frac{M}{8M}\right)^{2/3} = \left(\frac{1}{8}\right)^{2/3} = \left(\left(\frac{1}{2}\right)^3\right)^{2/3} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
Thus,$\frac{1}{n} = \frac{1}{4}$,which gives $n = 4$.

(A) The initial kinetic energy of the body is $K = \frac{1}{2} m u^2$.
At the highest point of the trajectory,the vertical component of velocity is zero,and the horizontal component remains $u \cos \beta$.
Thus,the kinetic energy at the highest point is $K' = \frac{1}{2} m (u \cos \beta)^2 = K \cos^2 \beta$.
Given that $K' = \frac{3}{4} K$,we have $K \cos^2 \beta = \frac{3}{4} K$.
This simplifies to $\cos^2 \beta = \frac{3}{4}$,which means $\cos \beta = \frac{\sqrt{3}}{2}$.
Therefore,$\beta = 30^{\circ}$.

(B) The motion of the ball is a horizontal projectile motion.
For vertical motion, the initial vertical velocity $u_y = 0 \text{ m/s}$.
The vertical displacement is $H = 19.6 \text{ m}$.
The acceleration due to gravity is $g = 9.8 \text{ m/s}^2$.
Using the equation of motion $H = u_y t + \frac{1}{2} g t^2$, we get:
$19.6 = 0 \times t + \frac{1}{2} \times 9.8 \times t^2$
$19.6 = 4.9 \times t^2$
$t^2 = \frac{19.6}{4.9} = 4$
$t = \sqrt{4} = 2 \text{ s}$.
Thus, the ball will take $2 \text{ s}$ to hit the ground.

(C) The total energy $E$ of a particle in simple harmonic motion is given by $E = \frac{1}{2} m \omega^2 A^2$.
Potential energy $U$ at displacement $x$ is given by $U = \frac{1}{2} m \omega^2 x^2$.
According to the problem,the potential energy is half of the total energy,so $U = \frac{E}{2}$.
Substituting the expressions,we get $\frac{1}{2} m \omega^2 x^2 = \frac{1}{2} \left( \frac{1}{2} m \omega^2 A^2 \right)$.
Simplifying the equation,we get $x^2 = \frac{A^2}{2}$.
Taking the square root on both sides,we find $x = \pm \frac{A}{\sqrt{2}}$.

(B) Let the particle be displaced vertically downwards by a small distance $x$.
The vertical spring stretches by $x$,providing a restoring force $F_1 = kx$ upwards.
The two inclined springs are at an angle of $135^\circ$ with the vertical. When the particle moves down by $x$,the change in length of each inclined spring is $\Delta l = x \cos(135^\circ - 90^\circ) = x \cos(45^\circ) = \frac{x}{\sqrt{2}}$.
The restoring force component along the vertical for each inclined spring is $F_2 = k \Delta l \cos(45^\circ) = k (\frac{x}{\sqrt{2}}) (\frac{1}{\sqrt{2}}) = \frac{kx}{2}$.
Total restoring force $F_{net} = F_1 + 2 F_2 = kx + 2(\frac{kx}{2}) = kx + kx = 2kx$.
Thus,the equivalent spring constant is $K_{eq} = 2k$.
The time period of oscillation is $T = 2\pi \sqrt{\frac{m}{K_{eq}}} = 2\pi \sqrt{\frac{m}{2k}}$.

(A) The process of heating ice from $-20^{\circ} C$ to steam at $100^{\circ} C$ involves several stages:
$1$. Heating ice from $-20^{\circ} C$ to $0^{\circ} C$: The temperature increases linearly with heat supplied.
$2$. Melting ice at $0^{\circ} C$ to water at $0^{\circ} C$: The temperature remains constant (phase change).
$3$. Heating water from $0^{\circ} C$ to $100^{\circ} C$: The temperature increases linearly with heat supplied.
$4$. Boiling water at $100^{\circ} C$ to steam at $100^{\circ} C$: The temperature remains constant (phase change).
Therefore,the temperature-heat graph should show two linear increases and two horizontal constant-temperature segments. Option $A$ correctly represents this behavior.

(B) According to Wien's displacement law, the product of the wavelength corresponding to maximum intensity $(\lambda_m)$ and the absolute temperature $(T)$ of the black body is a constant $(b).$
Formula: $\lambda_m T = b$
Given values:
$\lambda_m = 480 \, nm = 480 \times 10^{-9} \, m$
$b = 2.88 \times 10^{-3} \, mK$
Calculation:
$T = \frac{b}{\lambda_m}$
$T = \frac{2.88 \times 10^{-3}}{480 \times 10^{-9}}$
$T = \frac{2.88}{480} \times 10^6$
$T = 0.006 \times 10^6 = 6000 \, K$
Therefore, the surface temperature of the sun is $6000 \, K$.

(D) According to Stefan-Boltzmann Law,the rate of loss of heat $E$ from a black body at temperature $T$ in an environment at temperature $T_0$ is given by $E \propto (T^4 - T_0^4)$.
Given temperatures are $T_1 = 327^{\circ} C = 600 \ K$,$T_2 = 427^{\circ} C = 700 \ K$,and $T_0 = 27^{\circ} C = 300 \ K$.
The ratio of the rates of loss of heat is $\frac{E_1}{E_2} = \frac{T_1^4 - T_0^4}{T_2^4 - T_0^4}$.
Substituting the values: $\frac{E_1}{E_2} = \frac{(600)^4 - (300)^4}{(700)^4 - (300)^4}$.
Factoring out $(100)^4$: $\frac{E_1}{E_2} = \frac{6^4 - 3^4}{7^4 - 3^4} = \frac{1296 - 81}{2401 - 81}$.
$\frac{E_1}{E_2} = \frac{1215}{2320}$.
Dividing both numerator and denominator by $5$,we get $\frac{243}{464}$.

(A) The standard equation of a progressive wave is given by $y = A \sin(\omega t - kx + \phi)$.
Comparing this with the given equation $y = 4 \sin(4 \pi t - 0.04 x + \pi / 3)$,we identify the angular frequency $\omega = 4 \pi \ rad/s$ and the wave number $k = 0.04 \ rad/m$.
The velocity of the wave $(v)$ is given by the ratio of angular frequency to the wave number: $v = \frac{\omega}{k}$.
Substituting the values: $v = \frac{4 \pi}{0.04} = \frac{400 \pi}{4} = 100 \pi \ m/s$.

(C) The given wave equation is $x = x_0 \sin 2 \pi (nt - x/\lambda)$.
Comparing this with the standard form $x = A \sin (\omega t - kx)$,we have amplitude $A = x_0$,angular frequency $\omega = 2 \pi n$,and wave number $k = 2 \pi / \lambda$.
The maximum particle velocity $V_p$ is given by $V_p = A \omega = x_0 (2 \pi n) = 2 \pi n x_0$.
The wave velocity $V_w$ is given by $V_w = \frac{\omega}{k} = \frac{2 \pi n}{2 \pi / \lambda} = n \lambda$.
According to the problem,$V_p = 4 V_w$.
Substituting the expressions,we get $2 \pi n x_0 = 4 (n \lambda)$.
Dividing both sides by $n$,we get $2 \pi x_0 = 4 \lambda$.
Therefore,$\lambda = \frac{2 \pi x_0}{4} = \frac{\pi x_0}{2}$.

(B) The relationship between kinetic energy $K$ and momentum $P$ is given by $K = \frac{P^2}{2m}$.
Let the initial momentum be $P$ and the initial kinetic energy be $K = \frac{P^2}{2m}$.
If the momentum is increased by $20 \%$,the new momentum $P'$ becomes $P' = P + 0.20P = 1.2P$.
The new kinetic energy $K'$ is given by $K' = \frac{(P')^2}{2m} = \frac{(1.2P)^2}{2m} = \frac{1.44P^2}{2m}$.
Substituting $K = \frac{P^2}{2m}$,we get $K' = 1.44K$.
The percentage increase in kinetic energy is given by $\frac{K' - K}{K} \times 100 \% = \frac{1.44K - K}{K} \times 100 \% = 0.44 \times 100 \% = 44 \%$.

(D) Let the capacitance of each capacitor be $C = 3 \mu F$.
Looking at the circuit,we can identify the arrangement of capacitors.
There are two capacitors in parallel in the middle branch. Their equivalent capacitance is $C_p = C + C = 2C = 2 \times 3 = 6 \mu F$.
This combination is in series with another capacitor $C$ in that same branch. The equivalent capacitance of this branch is $C_s = \frac{C \times C_p}{C + C_p} = \frac{C \times 2C}{C + 2C} = \frac{2C^2}{3C} = \frac{2}{3}C = \frac{2}{3} \times 3 = 2 \mu F$.
Finally,this branch is in parallel with the top capacitor $C$. The total effective capacitance between $A$ and $B$ is $C_{eq} = C + C_s = 3 + 2 = 5 \mu F$.

(A) Let $r$ be the radius of each small droplet and $R$ be the radius of the large drop formed by coalescing $n$ droplets.
Since the volume remains conserved:
$n \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$
$R^3 = n r^3 \implies R = r n^{1/3}$
Each small droplet has a charge $q = C_0 V = (4 \pi \varepsilon_0 r) V$.
The total charge on the large drop is $Q = n q = n (4 \pi \varepsilon_0 r) V$.
The potential of the large drop is $V' = \frac{Q}{4 \pi \varepsilon_0 R}$.
Substituting the values of $Q$ and $R$:
$V' = \frac{n (4 \pi \varepsilon_0 r) V}{4 \pi \varepsilon_0 (r n^{1/3})}$
$V' = n \times \frac{1}{n^{1/3}} V = n^{2/3} V$.

(C) From the circuit diagram, the resistors $2 \Omega$, $3 \Omega$, and $6 \Omega$ are connected in parallel.
Let their equivalent resistance be $R_p$. Then, $\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = 1 \Omega^{-1}$.
Thus, $R_p = 1 \Omega$.
This parallel combination is in series with the $4 \Omega$ resistor and the $2 \text{ V}$ battery.
Therefore, the total equivalent resistance of the circuit is $R_{\text{eq}} = R_p + 4 \Omega = 1 \Omega + 4 \Omega = 5 \Omega$.
The current $I$ measured by the ammeter is given by Ohm's law: $I = \frac{V}{R_{\text{eq}}} = \frac{2 \text{ V}}{5 \Omega} = 0.4 \text{ A}$.

(B) The resistance of a wire is given by $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity,$L$ is the length,and $A$ is the cross-sectional area.
When the wire is elongated $n$-fold,the new length $L' = nL$.
Since the volume of the wire remains constant,$V = A \times L = A' \times L'$.
Therefore,$A' = \frac{A \times L}{L'} = \frac{A \times L}{nL} = \frac{A}{n}$.
The new resistance $R'$ is given by $R' = \rho \frac{L'}{A'} = \rho \frac{nL}{A/n} = n^2 \left( \rho \frac{L}{A} \right) = n^2 R$.

(B) The induced $emf$ $(e)$ in a coil is given by the formula $e = -L \frac{dI}{dt}$.
For the $emf$ to be zero,the rate of change of current $\frac{dI}{dt}$ must be zero.
Given $I = t^2 e^{-t}$.
Using the product rule for differentiation: $\frac{dI}{dt} = \frac{d}{dt}(t^2) \cdot e^{-t} + t^2 \cdot \frac{d}{dt}(e^{-t})$.
$\frac{dI}{dt} = 2t e^{-t} + t^2 (-e^{-t}) = e^{-t} (2t - t^2) = e^{-t} t(2 - t)$.
Setting $\frac{dI}{dt} = 0$,we get $e^{-t} t(2 - t) = 0$.
Since $e^{-t} \neq 0$ for finite $t$,the solutions are $t = 0$ or $t = 2 \ s$.
At $t = 0$,the current is zero,but the $emf$ becomes zero at $t = 2 \ s$ as the current reaches its maximum value.

(B) Given: Magnetic flux $\phi = 5t^2 - 4t + 1 \text{ Wb}$ and resistance $R = 10 \Omega$.
According to Faraday's law of induction, the induced electromotive force $(EMF)$ is $e = -\frac{d\phi}{dt}$.
Differentiating $\phi$ with respect to $t$: $\frac{d\phi}{dt} = \frac{d}{dt}(5t^2 - 4t + 1) = 10t - 4$.
The induced current $I$ is given by $I = \frac{|e|}{R} = \frac{|-d\phi/dt|}{R} = \frac{|-(10t - 4)|}{10} = \frac{|4 - 10t|}{10}$.
At $t = 0.2 \text{ s}$, the current is $I = \frac{|4 - 10(0.2)|}{10} = \frac{|4 - 2|}{10} = \frac{2}{10} = 0.2 \text{ A}$.

(A) According to Gauss's Law,the total electric flux $\phi$ through any closed surface is equal to the net charge enclosed by the surface divided by the permittivity of free space $\varepsilon_0$.
Mathematically,$\phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}$.
In this problem,the point charge $+q$ is placed at the centre of the cube,which is a closed surface.
Therefore,the total electric flux emerging from the cube is $\phi = \frac{q}{\varepsilon_0}$.

(A) The magnetic field $B$ at the centre of a circular loop of radius $a$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2a}$.
The magnetic moment $M$ of the loop is given by $M = I A = I(\pi a^2)$.
The ratio $x$ is given by:
$x = \frac{B}{M} = \frac{\mu_0 I}{2a} \times \frac{1}{I \pi a^2} = \frac{\mu_0}{2 \pi a^3}$.
When the current $I$ is doubled $(I' = 2I)$ and the radius $a$ is doubled $(a' = 2a)$,the new ratio $x'$ is:
$x' = \frac{\mu_0}{2 \pi (a')^3} = \frac{\mu_0}{2 \pi (2a)^3} = \frac{\mu_0}{2 \pi (8a^3)} = \frac{1}{8} \left( \frac{\mu_0}{2 \pi a^3} \right) = \frac{x}{8}$.

(D) The energy released per fission is $E_{fission} = 200 MeV$.
Converting this energy into Joules: $E_{fission} = 200 \times 10^6 \times 1.6 \times 10^{-19} J = 3.2 \times 10^{-11} J$.
The power required is $P = 3.2 W$,which means $3.2 J$ of energy is required per second.
The number of fissions per second $(n)$ is given by the ratio of total power to energy per fission:
$n = \frac{P}{E_{fission}} = \frac{3.2 J/s}{3.2 \times 10^{-11} J} = 10^{11} \text{ fissions/second}$.

(C) The radioactive decay law is given by $N(t) = N_0 e^{-\lambda t}$, where $N(t)$ is the amount remaining at time $t$.
For $20 \%$ decay, the amount remaining is $N_1 = 80 \% \text{ of } N_0 = 0.8 N_0$. Thus, $0.8 N_0 = N_0 e^{-\lambda t_1} \Rightarrow e^{-\lambda t_1} = 0.8$.
For $80 \%$ decay, the amount remaining is $N_2 = 20 \% \text{ of } N_0 = 0.2 N_0$. Thus, $0.2 N_0 = N_0 e^{-\lambda t_2} \Rightarrow e^{-\lambda t_2} = 0.2$.
Dividing the two equations: $\frac{e^{-\lambda t_1}}{e^{-\lambda t_2}} = \frac{0.8}{0.2} = 4$.
This simplifies to $e^{\lambda(t_2 - t_1)} = 4$.
Taking the natural logarithm on both sides: $\lambda(t_2 - t_1) = \ln(4) = 2 \ln(2)$.
Since $\lambda = \frac{\ln 2}{T_{1/2}}$, we have $\frac{\ln 2}{T_{1/2}}(t_2 - t_1) = 2 \ln 2$.
Therefore, $t_2 - t_1 = 2 \times T_{1/2} = 2 \times 20 \text{ min} = 40 \text{ min}$.

(B) When a fish is at a depth $h$ below the surface of water,it sees the outside world through a circular window due to total internal reflection.
The radius $r$ of this circle is given by the formula $r = \frac{h}{\sqrt{\mu^2 - 1}}$.
Given: depth $h = 12 \ m$ and refractive index $\mu = 4/3$.
Substituting the values into the formula:
$r = \frac{12}{\sqrt{(4/3)^2 - 1}}$
$r = \frac{12}{\sqrt{16/9 - 1}}$
$r = \frac{12}{\sqrt{7/9}}$
$r = \frac{12 \times 3}{\sqrt{7}}$
$r = \frac{36}{\sqrt{7}} \ m$.

(D) When a lens is silvered, it acts as a mirror. The equivalent power $P_{eq}$ of the system is given by $P_{eq} = 2P_L + P_M$, where $P_L$ is the power of the lens and $P_M$ is the power of the mirror.
Given, focal length of the lens $f = 20 \,cm$. Power of the lens $P_L = \frac{1}{f} = \frac{1}{20} \,cm^{-1}$.
The plane surface is silvered, so it acts as a plane mirror. The power of a plane mirror $P_M = 0$.
Thus, $P_{eq} = 2 \times P_L + 0 = 2 \times \frac{1}{20} = \frac{1}{10} \,cm^{-1}$.
The equivalent focal length $F$ of the system is given by $P_{eq} = -\frac{1}{F}$ (negative sign because it acts as a concave mirror).
So, $-\frac{1}{F} = \frac{1}{10}$, which gives $F = -10 \,cm$.
The magnitude of the focal length is $10 \,cm$.

(C) The number of images $n$ formed by two plane mirrors inclined at an angle $\theta$ is given by $n = \frac{360^{\circ}}{\theta} - 1$ if $\frac{360^{\circ}}{\theta}$ is an even integer,or if it is an odd integer and the object is placed symmetrically.
Given $n = 5$,we have $5 = \frac{360^{\circ}}{\theta} - 1$,which implies $\frac{360^{\circ}}{\theta} = 6$.
Thus,$\theta = \frac{360^{\circ}}{6} = 60^{\circ}$.
When the angle is decreased by $30^{\circ}$,the new angle $\theta' = 60^{\circ} - 30^{\circ} = 30^{\circ}$.
The new number of images $n'$ is given by $n' = \frac{360^{\circ}}{30^{\circ}} - 1 = 12 - 1 = 11$.

(D) Given:
$\beta = 50$
Input resistance $R_i = 1 \text{ k}\Omega = 10^3 \Omega$
Peak input voltage $V_i = 0.01 \text{ V}$
Step $1$: Calculate the peak value of base current $(\Delta I_B)$:
$\Delta I_B = \frac{V_i}{R_i} = \frac{0.01 \text{ V}}{10^3 \Omega} = 10^{-5} \text{ A}$
Step $2$: Calculate the peak value of collector current $(\Delta I_C)$:
Using the relation $\beta = \frac{\Delta I_C}{\Delta I_B}$,we get:
$\Delta I_C = \beta \times \Delta I_B$
$\Delta I_C = 50 \times 10^{-5} \text{ A}$
$\Delta I_C = 5 \times 10^{-4} \text{ A} = 500 \times 10^{-6} \text{ A} = 500 \mu\text{A}$

(D) The fringe width $\beta$ in Young's double slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength of light, $D$ is the distance between the slits and the screen, and $d$ is the distance between the two slits.
When the entire arrangement is placed in a liquid of refractive index $n$, the wavelength of light changes to $\lambda' = \frac{\lambda}{n}$.
Since $D$ and $d$ remain unchanged, the new fringe width $\beta'$ becomes $\beta' = \frac{\lambda' D}{d} = \frac{(\lambda / n) D}{d} = \frac{1}{n} \left( \frac{\lambda D}{d} \right) = \frac{\beta}{n}$.
Therefore, the new fringe width is $\frac{\beta}{n}$.

(B) Given the ratio of intensities $I_1 / I_2 = 9 / 1$.
Since intensity $I \propto A^2$,the ratio of amplitudes is $A_1 / A_2 = \sqrt{I_1 / I_2} = \sqrt{9 / 1} = 3 / 1$.
Let $A_1 = 3k$ and $A_2 = k$.
The intensity of maxima is given by $I_{\max} = (A_1 + A_2)^2 = (3k + k)^2 = (4k)^2 = 16k^2$.
The intensity of minima is given by $I_{\min} = (A_1 - A_2)^2 = (3k - k)^2 = (2k)^2 = 4k^2$.
Therefore,the ratio of intensities of maxima and minima is $I_{\max} / I_{\min} = 16k^2 / 4k^2 = 4 / 1$.