KVPY 2014 Physics Question Paper with Answer and Solution

(C) The impulsive force provides both linear momentum and angular momentum to the rod.
Let $v$ be the linear velocity of the center of mass and $\omega$ be the angular velocity of the rod after the impulse is applied.
From the impulse-momentum theorem,the linear impulse $J$ is equal to the change in linear momentum:
$J = mv \Rightarrow v = \frac{J}{m}$
The angular impulse about the center of mass is equal to the change in angular momentum:
$J \times L = I\omega$
Where $I = \frac{m(2L)^2}{12} = \frac{mL^2}{3}$ is the moment of inertia about the center of mass.
$JL = \left(\frac{mL^2}{3}\right)\omega \Rightarrow \omega = \frac{3J}{mL}$
The total kinetic energy $KE$ is the sum of translational and rotational kinetic energy:
$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
$KE = \frac{1}{2}m\left(\frac{J}{m}\right)^2 + \frac{1}{2}\left(\frac{mL^2}{3}\right)\left(\frac{3J}{mL}\right)^2$
$KE = \frac{J^2}{2m} + \frac{1}{2}\left(\frac{mL^2}{3}\right)\left(\frac{9J^2}{m^2L^2}\right)$
$KE = \frac{J^2}{2m} + \frac{3J^2}{2m} = \frac{4J^2}{2m} = \frac{2J^2}{m}$

(B) For cylinder $P$ (rolling without slipping):
By the principle of conservation of energy,the initial potential energy is converted into translational and rotational kinetic energy at the bottom:
$mgh = \frac{1}{2}mv_p^2 + \frac{1}{2}I\omega^2$
Since $I = \frac{1}{2}mr^2$ and $v_p = r\omega$,we have:
$mgh = \frac{1}{2}mv_p^2 + \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v_p}{r})^2$
$mgh = \frac{1}{2}mv_p^2 + \frac{1}{4}mv_p^2 = \frac{3}{4}mv_p^2$
$v_p = \sqrt{\frac{4gh}{3}}$
For cylinder $Q$ (sliding without friction):
Since there is no friction,the cylinder does not rotate. All potential energy is converted into translational kinetic energy:
$mgh = \frac{1}{2}mv_q^2$
$v_q = \sqrt{2gh}$
The ratio of the speeds is:
$\frac{v_q}{v_p} = \frac{\sqrt{2gh}}{\sqrt{\frac{4gh}{3}}} = \sqrt{2 \cdot \frac{3}{4}} = \sqrt{\frac{3}{2}}$

(A) The force on the particle is given by the negative gradient of the potential:
$F = -\frac{dV}{dr} = -\frac{d}{dr}(kr) = -k$.
Taking the magnitude of the force,we have $F = k$.
For circular motion,this force acts as the centripetal force:
$F = mR\omega^2 = k$.
Substituting $\omega = \frac{2\pi}{T}$,we get:
$mR\left(\frac{2\pi}{T}\right)^2 = k$.
$mR \cdot \frac{4\pi^2}{T^2} = k$.
Rearranging for $T^2$:
$T^2 = \frac{4\pi^2 m}{k} \cdot R$.
Since $\frac{4\pi^2 m}{k}$ is a constant,we have $T^2 \propto R$,which implies $T \propto R^{1/2}$.

(D) The correct option is $(d)$.
When the block slides down the frictionless inclined plane with an acceleration $a = g \sin \alpha$,we analyze the motion of the pendulum bob in the non-inertial frame of reference of the block.
$(i)$ The forces acting on the pendulum bob in the frame of the block are the gravitational force $mg$ (acting vertically downwards) and the pseudo force $ma = mg \sin \alpha$ (acting upwards parallel to the inclined plane).
$(ii)$ The effective acceleration $g_{eff}$ experienced by the bob is the vector sum of the acceleration due to gravity $\vec{g}$ and the pseudo acceleration $-\vec{a}$.
$(iii)$ The mean position of the pendulum corresponds to the direction of this effective acceleration $\vec{g}_{eff}$.
$(iv)$ By resolving the gravitational force $mg$ into components perpendicular to the plane $(mg \cos \alpha)$ and parallel to the plane $(mg \sin \alpha)$,we see that the parallel component $mg \sin \alpha$ is exactly cancelled by the pseudo force $mg \sin \alpha$ acting in the opposite direction.
$(v)$ Thus,the only remaining component of the effective force is $mg \cos \alpha$,which acts perpendicular to the inclined plane $AC$. Therefore,the string of the pendulum aligns itself perpendicular to the inclined plane at its mean position.

(B) According to the equation of continuity,$A_1 v_1 = A_2 v_2$. In the narrow region,the cross-sectional area $A$ decreases,so the velocity $v$ of the fluid increases $(v \propto \frac{1}{A})$.
According to Bernoulli's principle for a horizontal pipe,$p + \frac{1}{2} \rho v^2 = \text{constant}$. Since the velocity $v$ increases in the narrow region,the pressure $p$ must decrease.
As the external pressure on the air bubbles decreases in the narrow region,the air inside the bubbles expands,causing the bubbles to become larger in size. Thus,the bubbles move with greater speed and are larger in size.

(A) The correct answer is $(A)$.
The potential energy $U$ versus interatomic separation $r$ plot for two atoms in a solid is asymmetric about the equilibrium position.
As the temperature increases,the total energy of the atoms increases $(E_3 > E_2 > E_1)$.
Due to the asymmetry of the potential energy curve,the mean separation between the atoms increases as the energy increases $(r_3 > r_2 > r_1)$.
Consequently,crystalline solids generally expand upon heating.

(D) The change in volume $\Delta V$ of a liquid is given by $\Delta V = V_0 \gamma \Delta \theta$,where $V_0$ is the initial volume,$\gamma$ is the coefficient of volume expansion,and $\Delta \theta$ is the change in temperature.
Since the increase in length $\Delta l$ is the same for both thermometers,the change in volume is $\Delta V = A \Delta l = \frac{\pi d^2}{4} \Delta l$.
Given that the initial volumes $(V_0)_{Hg} = (V_0)_{Br}$ and the change in length $\Delta l$ is the same for both,we have:
$\Delta V_{Hg} = \frac{\pi d_1^2}{4} \Delta l = V_0 \gamma_{Hg} \Delta \theta$
$\Delta V_{Br} = \frac{\pi d_2^2}{4} \Delta l = V_0 \gamma_{Br} \Delta \theta$
Dividing the two equations:
$\frac{d_1^2}{d_2^2} = \frac{\gamma_{Hg}}{\gamma_{Br}}$
$\frac{d_1}{d_2} = \sqrt{\frac{\gamma_{Hg}}{\gamma_{Br}}}$
Substituting the given values:
$\frac{d_1}{d_2} = \sqrt{\frac{18 \times 10^{-5}}{108 \times 10^{-5}}} = \sqrt{\frac{1}{6}} \approx \sqrt{0.166} \approx 0.408$
Thus,the ratio $d_1: d_2$ is closest to $0.4$.

(B) Given the process equation: $p V^2 = C$.
For an ideal gas,we know that $p V = n R T$,which implies $p = \frac{n R T}{V}$.
Substituting this into the process equation:
$\left(\frac{n R T}{V}\right) V^2 = C$
$n R T V = C$
Since $n, R,$ and $C$ are constants,we have $T V = \text{constant}$,or $T \propto \frac{1}{V}$.
If $V_2 > V_1$,then according to the relation $T \propto \frac{1}{V}$,the temperature $T_2$ must be less than $T_1$ $(T_2 < T_1)$.
Therefore,option $(b)$ is correct.

(C) The source of sound is moving in a circle,so its linear speed $v_s$ is given by $v_s = r\omega$.
Given $r = 2 \,m$ and $\omega = 15 \,rad/s$,we have $v_s = 2 \times 15 = 30 \,m/s$.
The speed of sound is $v = 330 \,m/s$.
According to the Doppler effect,when the source moves towards the stationary observer,the observed frequency is $f_{\max} = \left(\frac{v}{v - v_s}\right) f$.
When the source moves away from the stationary observer,the observed frequency is $f_{\min} = \left(\frac{v}{v + v_s}\right) f$.
The ratio of the highest to the lowest frequency is:
$\frac{f_{\max}}{f_{\min}} = \frac{\left(\frac{v}{v - v_s}\right) f}{\left(\frac{v}{v + v_s}\right) f} = \frac{v + v_s}{v - v_s}$.
Substituting the values:
$\frac{f_{\max}}{f_{\min}} = \frac{330 + 30}{330 - 30} = \frac{360}{300} = 1.2$.

(C) Initially,the sphere is placed on the surface with angular velocity $\omega$ and linear velocity $v_{CM} = 0$. Friction acts at the point of contact to oppose the slipping.
The force of friction $f = \mu_k mg$ acts in the forward direction,causing linear acceleration $a_{CM} = \frac{f}{m} = \mu_k g$.
The torque due to friction about the center of mass is $\tau = -fR = -\mu_k mgR$. The angular deceleration is $\alpha = \frac{\tau}{I} = \frac{-\mu_k mgR}{\frac{2}{5}mR^2} = -\frac{5\mu_k g}{2R}$.
After time $t$,the linear velocity is $v_{CM} = a_{CM}t = \mu_k gt$ and the angular velocity is $\omega_f = \omega + \alpha t = \omega - \frac{5\mu_k g}{2R}t$.
Pure rolling begins when $v_{CM} = \omega_f R$. Substituting the expressions:
$\mu_k gt = \left(\omega - \frac{5\mu_k g}{2R}t\right)R$
$\mu_k gt = \omega R - \frac{5}{2}\mu_k gt$
$\frac{7}{2}\mu_k gt = \omega R \implies \mu_k gt = \frac{2}{7}\omega R$.
Substituting this back into the expression for $\omega_f$:
$\omega_f = \frac{v_{CM}}{R} = \frac{\mu_k gt}{R} = \frac{1}{R} \left(\frac{2}{7}\omega R\right) = \frac{2}{7}\omega$.

(A) Let the acceleration of the system of blocks $X$ and $Y$ be $a$.
Since the blocks move together,the total mass is $M = m_X + m_Y = 8 \,kg + 2 \,kg = 10 \,kg$.
The acceleration of the system is given by $a = \frac{F}{M} = \frac{F}{10} \,ms^{-2}$.
For block $Y$ to not slip downwards,the upward frictional force $f$ must balance its weight $m_Y g$.
$f = m_Y g = 2 \times 10 = 20 \,N$.
The frictional force is also given by $f = \mu R$,where $R$ is the normal reaction between blocks $X$ and $Y$.
From the horizontal motion of block $Y$,the normal reaction $R$ provides the necessary acceleration $a$ to block $Y$:
$R = m_Y a = 2 \times \frac{F}{10} = \frac{F}{5}$.
Substituting the values into the friction equation:
$f = \mu R \Rightarrow 20 = 0.5 \times \frac{F}{5}$.
$20 = \frac{F}{10} \Rightarrow F = 200 \,N$.
Thus,the minimum value of $F$ is $200 \,N$.

(B) For a pendulum with angular amplitude $\theta$,the height $h$ reached by the bob is given by $h = l(1 - \cos \theta)$,where $l$ is the length of the string.
By the law of conservation of energy,the kinetic energy at the mean position is equal to the potential energy at the extreme position:
$\frac{1}{2} m v^2 = m g h = m g l(1 - \cos \theta)$
$v^2 = 2 g l(1 - \cos \theta)$
The maximum tension $T_{\max}$ occurs at the mean position (lowest point) and is given by:
$T_{\max} = m g + \frac{m v^2}{l} = m g + \frac{m(2 g l(1 - \cos \theta))}{l} = m g + 2 m g(1 - \cos \theta) = m g(3 - 2 \cos \theta)$
The minimum tension $T_{\min}$ occurs at the extreme position where the velocity is zero:
$T_{\min} = m g \cos \theta$
Given that $T_{\max} = 4 T_{\min}$:
$m g(3 - 2 \cos \theta) = 4 m g \cos \theta$
$3 - 2 \cos \theta = 4 \cos \theta$
$3 = 6 \cos \theta$
$\cos \theta = \frac{1}{2}$
$\theta = 60^{\circ}$

(D) For a polytropic process $p V^x = C$,the molar heat capacity $C_{process}$ is given by the formula $C_{process} = C_V + \frac{R}{1 - x}$.
Here,the given process is $p V^3 = C$,so $x = 3$.
For a monoatomic ideal gas,the molar heat capacity at constant volume is $C_V = \frac{3}{2} R$.
Substituting these values into the formula:
$C_{process} = \frac{3}{2} R + \frac{R}{1 - 3}$
$C_{process} = \frac{3}{2} R + \frac{R}{-2}$
$C_{process} = \frac{3}{2} R - \frac{1}{2} R = R$.
Thus,the heat capacity of the gas during the process is $R$.

(C) The Stefan-Boltzmann constant $\sigma$ has dimensions $[M T^{-3} K^{-4}]$.
The dimensions of the constants are:
$h = [M L^2 T^{-1}]$
$k_B = [M L^2 T^{-2} K^{-1}]$
$c = [L T^{-1}]$
Given $\sigma = h^\alpha k_B^\beta c^\gamma$,we equate the dimensions:
$[M T^{-3} K^{-4}] = [M L^2 T^{-1}]^\alpha [M L^2 T^{-2} K^{-1}]^\beta [L T^{-1}]^\gamma$
$[M T^{-3} K^{-4}] = [M^{\alpha+\beta} L^{2\alpha+2\beta+\gamma} T^{-\alpha-2\beta-\gamma} K^{-\beta}]$
Comparing the exponents of $M, L, T,$ and $K$:
$1$) $M: \alpha + \beta = 0$ (Wait,$\sigma$ is energy per unit area per unit time per unit temperature to the fourth power,so $[M T^{-3} K^{-4}]$ is correct. The mass dimension is $1$,so $\alpha + \beta = 1$)
$2$) $K: -\beta = -4 \implies \beta = 4$
$3$) $M: \alpha + \beta = 1 \implies \alpha + 4 = 1 \implies \alpha = -3$
$4$) $T: -\alpha - 2\beta - \gamma = -3 \implies -(-3) - 2(4) - \gamma = -3 \implies 3 - 8 - \gamma = -3 \implies -5 - \gamma = -3 \implies \gamma = -2$
Thus,$\alpha = -3, \beta = 4, \gamma = -2$.

(A) In a displacement-time graph,the velocity at any point is given by the slope of the tangent to the curve at that point.
Speed is the magnitude of velocity,so the magnitude of the slope represents the speed.
At point $P$,the tangent is horizontal,so the slope is $0$,meaning the speed is $0$.
At point $Q$,the slope is small (the curve is relatively flat).
At point $R$,the tangent is very steep,meaning the magnitude of the slope $|m| = |\tan \theta|$ is the largest among the three points.
Since the speed is the magnitude of the slope,the speed is highest at point $R$. Therefore,the speed is increasing as we move towards point $R$ from the preceding region of the graph.

(D) The reading of a spring balance measures the tension in the spring,which is equal to the apparent weight of the object.
When the box is in the atmosphere,it experiences an upward buoyant force $F_b$ due to the displaced air.
The reading $R$ in the atmosphere is given by $R = W - F_b$,where $W$ is the true weight of the box.
In an evacuated chamber,the buoyant force $F_b$ becomes zero.
Therefore,the new reading $R'$ will be equal to the true weight $W$.
Since $R' = W$ and $R = W - F_b$,it follows that $R' > R$.
Thus,the reading will be more than $50 \,kg$ because the atmospheric buoyancy force is absent.

(A) When the box is dropped from height $h$,its speed when it reaches the ground is given by the conservation of energy: $v = \sqrt{2gh}$.
When the block slides down the rough inclined plane with angle $\theta = 45^{\circ}$,the net acceleration $a$ is given by:
$a = g(\sin \theta - \mu \cos \theta)$.
Since $\theta = 45^{\circ}$,$\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$.
Thus,$a = g \left( \frac{1}{\sqrt{2}} - \mu \frac{1}{\sqrt{2}} \right) = \frac{g}{\sqrt{2}}(1 - \mu)$.
The length of the inclined plane $s$ is related to height $h$ by $s = \frac{h}{\sin \theta} = h \sqrt{2}$.
The final velocity $v'$ at the bottom of the incline is $v' = \sqrt{2as}$.
Substituting $a$ and $s$:
$v' = \sqrt{2 \cdot \frac{g}{\sqrt{2}}(1 - \mu) \cdot h \sqrt{2}} = \sqrt{2gh(1 - \mu)}$.
Given that $v' = \frac{v}{3}$,we have:
$\sqrt{2gh(1 - \mu)} = \frac{1}{3} \sqrt{2gh}$.
Squaring both sides:
$2gh(1 - \mu) = \frac{1}{9} (2gh)$.
$1 - \mu = \frac{1}{9}$.
$\mu = 1 - \frac{1}{9} = \frac{8}{9}$.

(C) The correct option is $C$.
The thermal resistance of a thin layer of paper is very low. When the cup is placed over a flame,the heat is transferred through the paper to the water inside.
Since water has a high specific heat capacity,it absorbs the heat effectively. As long as there is water in the cup,the temperature of the paper does not rise significantly above the temperature of the water (which is $100^{\circ} C$ at boiling point).
Because the temperature of the paper remains below its ignition (burning) temperature,the paper cup does not catch fire.

(D) The correct option is $D$.
When ice is crushed,the total surface area of the ice that comes into contact with the surrounding air or the contents of the cooler increases significantly.
According to the principles of heat transfer,the rate of heat exchange is directly proportional to the surface area in contact.
Therefore,increasing the surface area by crushing the ice accelerates the rate of heat absorption from the surroundings,thereby speeding up the cooling process.

(D) In simple harmonic motion,the displacement of the particle is given by $x(t) = x_0 \sin(\omega t + \phi)$.
The velocity of the particle is $v = \frac{dx}{dt} = x_0 \omega \cos(\omega t + \phi)$.
The probability $P(x) dx$ of finding the particle in the interval $(x, x + dx)$ is proportional to the time $dt$ it spends in that interval,which is $P(x) \propto \frac{dt}{dx} = \frac{1}{|v|}$.
Since $v = \omega \sqrt{x_0^2 - x^2}$,we have $P(x) \propto \frac{1}{\sqrt{x_0^2 - x^2}}$.
This function has a minimum at the mean position $(x = 0)$ and approaches infinity at the extreme positions $(x = \pm x_0)$.
Graph $D$ shows the number of events (which is proportional to the probability density) being lowest at the center and increasing towards the extreme positions,which matches the behavior of the function $P(x) \propto \frac{1}{\sqrt{x_0^2 - x^2}}$.

(B) The centre of mass of the original square plate of side $a$ is at its geometric centre. Let the origin be at the top-right corner of the original square. The coordinates of the centre of mass of the original square are $(x_1, y_1) = (a/2, a/2)$.
Let $k$ be the mass per unit area. The mass of the original square is $m_1 = k a^2$.
The removed square portion of side $b$ has its centre of mass at $(x_2, y_2) = (b/2, b/2)$ relative to the same origin.
The mass of the removed portion is $m_2 = k b^2$.
The remaining $L$-shaped sheet has its centre of mass at point $P$,which is given as $(b, b)$ relative to the chosen origin.
Using the formula for the centre of mass of a system with a cavity: $X_{CM} = \frac{m_1 x_1 - m_2 x_2}{m_1 - m_2}$.
Substituting the values: $b = \frac{(k a^2)(a/2) - (k b^2)(b/2)}{k a^2 - k b^2}$.
$b = \frac{a^3 - b^3}{2(a^2 - b^2)}$.
$2b(a^2 - b^2) = a^3 - b^3$.
$2ba^2 - 2b^3 = a^3 - b^3$.
$a^3 - 2ba^2 + b^3 = 0$.
Dividing by $b^3$: $(a/b)^3 - 2(a/b)^2 + 1 = 0$.
Let $x = a/b$. Then $x^3 - 2x^2 + 1 = 0$.
Since $x=1$ is a root,we divide by $(x-1)$: $(x-1)(x^2 - x - 1) = 0$.
Since $a > b$,$x > 1$,so $x^2 - x - 1 = 0$.
Solving for $x$: $x = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{1 + \sqrt{5}}{2}$ (taking the positive root).
Thus,$a/b = (\sqrt{5} + 1) / 2$.

(D) For a soap bubble to float in air,the gravitational force must be equal to the buoyant force.
Gravitational force = Buoyant force
$g \times (\text{mass of helium} + \text{mass of soap film}) = \text{Weight of air displaced by bubble}$
Let $r$ be the inner radius of the soap bubble and $t$ be the thickness of the film.
The volume of helium is $V_{He} = \frac{4}{3} \pi r^3$.
The volume of the soap film is $V_{film} \approx 4 \pi r^2 t$ (since $t \ll r$).
The equation of equilibrium is:
$\frac{4}{3} \pi r^3 \rho_{He} g + (4 \pi r^2 t) \rho_{soap} g = \frac{4}{3} \pi r^3 \rho_{air} g$
Dividing by $g$ and rearranging:
$4 \pi r^2 t \rho_{soap} = \frac{4}{3} \pi r^3 (\rho_{air} - \rho_{He})$
$t = \frac{r}{3} \frac{(\rho_{air} - \rho_{He})}{\rho_{soap}}$
Given $r = 1 \, cm = 10^{-2} \, m$,$\rho_{air} = 1.23 \, kg \, m^{-3}$,$\rho_{He} = 0.18 \, kg \, m^{-3}$,and $\rho_{soap} = 1000 \, kg \, m^{-3}$.
$t = \frac{10^{-2}}{3} \times \frac{(1.23 - 0.18)}{1000}$
$t = \frac{10^{-2}}{3} \times \frac{1.05}{1000} = \frac{1.05 \times 10^{-5}}{3} = 0.35 \times 10^{-5} \, m = 3.5 \times 10^{-6} \, m$.
Since $1 \, \mu m = 10^{-6} \, m$,the thickness $t = 3.50 \, \mu m$.

(C) According to the principle of calorimetry,heat lost by the aluminium piece is equal to the heat gained by the water.
Let $m_a = 50 \,g = 0.05 \,kg$ be the mass of aluminium,$s_a = 900 \,J \,kg^{-1} K^{-1}$ be its specific heat,and $\Delta T_a = (300^{\circ} C - 160^{\circ} C) = 140^{\circ} C$ be the change in temperature.
Let $m_w = 1 \,kg$ be the mass of water,$s_w = 4200 \,J \,kg^{-1} K^{-1}$ be its specific heat,and $\Delta T_w = (T - 30^{\circ} C)$ be the change in temperature of water.
Heat lost by aluminium = $m_a s_a \Delta T_a = 0.05 \times 900 \times 140 = 6300 \,J$.
Heat gained by water = $m_w s_w \Delta T_w = 1 \times 4200 \times (T - 30) = 4200(T - 30)$.
Equating the two: $6300 = 4200(T - 30)$.
$T - 30 = \frac{6300}{4200} = 1.5$.
$T = 30 + 1.5 = 31.5^{\circ} C$.
Thus,the final temperature of the water is $31.5^{\circ} C$.

(A) When light travels from one medium to another, its frequency $(f)$ remains constant because it is determined by the source of light.
The speed of light $(v)$ in a medium is given by $v = c/n$, where $c$ is the speed of light in vacuum and $n$ is the refractive index of the medium. Since the refractive index of the prism is greater than that of air $(n > 1)$, the speed of light decreases inside the prism.
The relationship between speed, frequency, and wavelength $(\lambda)$ is $v = f \lambda$. Since $v$ decreases and $f$ remains constant, the wavelength $\lambda$ must also decrease inside the prism.
Therefore, the speed and wavelength change, while the frequency remains the same.

(B) When a lens is silvered,it acts like a mirror with an equivalent focal length $f_{eq}$.
The formula for the equivalent focal length is given by $\frac{1}{f_{eq}} = \frac{2}{f_l} + \frac{1}{f_m}$,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,$f_l = 10 \,cm$. The flat surface is silvered,so it acts as a plane mirror,meaning $f_m = \infty$.
Thus,$\frac{1}{f_{eq}} = \frac{2}{10} + \frac{1}{\infty} = \frac{1}{5}$.
Since the system acts as a concave mirror,we take $f_{eq} = -5 \,cm$ (using sign convention for a mirror).
Given the object distance $u = -30 \,cm$,we use the mirror formula: $\frac{1}{f_{eq}} = \frac{1}{v} + \frac{1}{u}$.
Substituting the values: $\frac{1}{-5} = \frac{1}{v} + \frac{1}{-30}$.
$\frac{1}{v} = \frac{1}{-5} + \frac{1}{30} = \frac{-6 + 1}{30} = \frac{-5}{30} = -\frac{1}{6}$.
Therefore,$v = -6 \,cm$.
The negative sign indicates that the image is formed in front of the mirror,which means it is a real image at a distance of $6 \,cm$ from the lens.

(D) According to Lenz's Law,the induced current in loop $L_2$ will oppose the change in magnetic flux linked with it.
As the current in loop $L_1$ increases with time,the magnetic flux linked with loop $L_2$ also increases.
To oppose this increase in flux,the induced current in loop $L_2$ flows in such a direction that the magnetic field produced by it opposes the magnetic field of $L_1$. This results in currents flowing in opposite directions in the adjacent arms of the two loops.
Since currents in the adjacent arms of $L_1$ and $L_2$ are in opposite directions,there is a repulsive force between them.
Consequently,loop $L_2$ moves away from loop $L_1$.

(A) The electron is subjected to an electric force due to the electric field $E$ between the capacitor plates.
The time taken by the electron to cross the region of length $x$ between the plates is $t = \frac{x}{u}$.
In this time, the acceleration of the electron in the $y$-direction is $a_y = \frac{F}{m} = \frac{eE}{m}$.
The vertical velocity component $v_y$ acquired by the electron upon leaving the capacitor is $v_y = a_y t = \left(\frac{eE}{m}\right) \left(\frac{x}{u}\right) = \frac{eEx}{mu}$.
The horizontal velocity component remains constant at $v_x = u$.
The angle of deflection $\theta$ is given by $\tan \theta = \frac{v_y}{v_x} = \frac{eEx/mu}{u} = \frac{eEx}{mu^2}$.
From this expression, we see that $\tan \theta \propto \frac{1}{u^2}$.
Therefore, $\frac{\tan \theta_2}{\tan \theta_1} = \frac{u_1^2}{u_2^2}$.
Given $\tan \theta_1 = 0.4$ and $u_2 = 2u_1$, we have:
$\tan \theta_2 = \tan \theta_1 \left(\frac{u_1}{u_2}\right)^2 = 0.4 \left(\frac{u_1}{2u_1}\right)^2 = 0.4 \left(\frac{1}{4}\right) = 0.1$.

(C) The work done in bringing the charges from infinity to their respective positions is equal to the change in the electrostatic potential energy of the system.
Let the shell be charge $1$,the charge $+q$ be charge $2$,and the charge $-q$ be charge $3$.
The potential inside a uniformly charged spherical shell is constant and equal to $V = kQ/R$,where $k = 1 / (4 \pi \varepsilon_0)$.
$1$. Work done to bring charge $+q$ to $x = -a/2$: $W_1 = q \times V_{\text{shell}} = q(kQ/R)$.
$2$. Work done to bring charge $-q$ to $x = +a/2$: $W_2 = (-q) \times V_{\text{shell}} + (-q) \times V_{\text{charge } q} = (-q)(kQ/R) + (-q)(kq / a) = -kQq/R - kq^2/a$.
Total work done $W = W_1 + W_2 = (kQq/R) - (kQq/R) - kq^2/a = -kq^2/a$.
The magnitude of the work done is $|W| = | -kq^2/a | = q^2 / (4 \pi \varepsilon_0 a)$.

(B) Initially,the equivalent capacitance is $C_{eq} = \frac{C \cdot C}{C + C} = \frac{C}{2}$.
The charge delivered by the battery is $Q_1 = C_{eq} E = \frac{C E}{2}$.
When one of the capacitors is filled with a dielectric of constant $k$,its new capacitance becomes $kC$. The new equivalent capacitance of the series combination is:
$C'_{eq} = \frac{C \cdot (kC)}{C + kC} = \left( \frac{k}{k+1} \right) C$.
Since the battery remains connected,the new charge on the combination is:
$Q_2 = C'_{eq} E = \left( \frac{k}{k+1} \right) C E$.
The amount of charge that flows through the battery is the change in charge:
$\Delta Q = Q_2 - Q_1 = \left( \frac{k}{k+1} - \frac{1}{2} \right) C E$.
$\Delta Q = \left( \frac{2k - (k+1)}{2(k+1)} \right) C E = \frac{k-1}{2(k+1)} C E$.

(A) The breakdown electric field $E$ in a $p-n$ junction is a characteristic property of the material and is independent of the depletion width.
Given,the breakdown voltage $V_1 = 100 \,V$ for a depletion width $d_1 = 20 \,\mu m$.
The electric field $E$ is given by $E = \frac{V}{d}$.
Since the breakdown field $E$ remains constant,we have:
$\frac{V_1}{d_1} = \frac{V_2}{d_2}$
Substituting the given values:
$\frac{100 \,V}{20 \,\mu m} = \frac{V_2}{1 \,\mu m}$
$5 \times 10^6 \,V/m = \frac{V_2}{1 \times 10^{-6} \,m}$
$V_2 = 5 \,V$.
Thus,the Zener diode can be used for voltage regulation of $5 \,V$.

(D) The kinetic energy $K$ of the particle is given by $K = \frac{1}{2}mv^2$.
Given $K = 0.05 \,eV = 0.05 \times 1.6 \times 10^{-19} \,J$ and $m = 1.6 \times 10^{-26} \,kg$.
$0.05 \times 1.6 \times 10^{-19} = \frac{1}{2} \times 1.6 \times 10^{-26} \times v^2$.
$v^2 = \frac{0.05 \times 1.6 \times 10^{-19} \times 2}{1.6 \times 10^{-26}} = 0.1 \times 10^7 = 10^6 \,m^2/s^2$.
$v = 10^3 \,m/s$.
The time $t$ taken to travel a distance $D = 1 \,m$ is $t = \frac{D}{v} = \frac{1}{10^3} = 10^{-3} \,s$.
The number of half-lives $n$ is $n = \frac{t}{T_{1/2}} = \frac{10^{-3}}{6.9} \approx 1.45 \times 10^{-4}$.
The fraction of particles remaining is $N/N_0 = (1/2)^n = 2^{-n}$.
The fraction decayed is $1 - 2^{-n} = 1 - e^{-n \ln 2} \approx 1 - (1 - n \ln 2) = n \ln 2$.
Since $n \approx 1.45 \times 10^{-4}$ and $\ln 2 \approx 0.69$,the fraction decayed $\approx 1.45 \times 10^{-4} \times 0.69 \approx 10^{-4} = 0.0001$.

(C) The photon flux is defined as the number of photons passing through a unit area per unit time.
The energy of a single photon is given by $E = \frac{hc}{\lambda}$.
The intensity $I$ at a distance $r$ from a source of power $P$ is $I = \frac{P}{4 \pi r^2}$.
The photon flux $\Phi$ is given by $\Phi = \frac{I}{E} = \frac{P}{4 \pi r^2} \times \frac{\lambda}{hc}$.
Substituting the given values: $P = 160 \,W$,$\lambda = 6200 \times 10^{-10} \,m$,$r = 1.8 \,m$,$h = 6.63 \times 10^{-34} \,J \cdot s$,and $c = 3 \times 10^8 \,m/s$.
$\Phi = \frac{160 \times 6200 \times 10^{-10}}{4 \times 3.14 \times (1.8)^2 \times 6.63 \times 10^{-34} \times 3 \times 10^8}$.
Calculating the value: $\Phi \approx 1.22 \times 10^{19} \,m^{-2} s^{-1}$.
Thus,the order of magnitude is $10^{19} \,m^{-2} s^{-1}$.

(B) Using the Rydberg formula for hydrogen,$\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
For the first Balmer line ($n=3$ to $n=2$): $\frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right)$.
For the transition from $n=5$ to $n=3$: $\frac{1}{\lambda_2} = R \left( \frac{1}{3^2} - \frac{1}{5^2} \right) = R \left( \frac{1}{9} - \frac{1}{25} \right) = R \left( \frac{25-9}{225} \right) = R \left( \frac{16}{225} \right)$.
Taking the ratio: $\frac{\lambda_2}{\lambda_1} = \frac{R(5/36)}{R(16/225)} = \frac{5}{36} \times \frac{225}{16} = \frac{5 \times 25}{4 \times 16} = \frac{125}{64}$.
Therefore,$\lambda_2 = \frac{125}{64} \lambda_1$.

(A) The reaction for removing a neutron from ${ }_5 B ^{11}$ is given by:
${ }_5 B ^{11} \longrightarrow { }_5 B ^{10} + { }_0 n ^1$
The total binding energy of a nucleus is calculated as:
$BE = (\text{Binding energy per nucleon}) \times (\text{Mass number } A)$
For ${ }_5 B ^{11}$:
$BE({ }_5 B ^{11}) = 7.5 \,MeV \times 11 = 82.5 \,MeV$
For ${ }_5 B ^{10}$:
$BE({ }_5 B ^{10}) = 8.0 \,MeV \times 10 = 80.0 \,MeV$
The energy required to remove the neutron is the difference in binding energies:
$E = BE({ }_5 B ^{11}) - BE({ }_5 B ^{10})$
$E = 82.5 \,MeV - 80.0 \,MeV = 2.5 \,MeV$

(A) For a large aperture mirror,rays parallel to the axis do not converge at a single point (focus $F$) due to spherical aberration. Instead,they form a circular patch at the focal plane.
Let $N$ be a point on the periphery of the mirror at a distance $r$ from the axis. Let the ray reflected from $N$ intersect the axis at $Q$. In $\triangle NQC$,where $C$ is the center of curvature,the angle of incidence equals the angle of reflection,both being $\theta$. Thus,$\angle N C P = \theta$ and $\angle N Q C = \theta$.
In $\triangle NQC$,by the law of sines or geometry,$QC = \frac{R}{2 \cos \theta}$.
The distance of $Q$ from the pole $P$ is $PQ = R - QC = R - \frac{R}{2 \cos \theta}$.
The distance of $Q$ from the focus $F$ is $QF = PF - PQ = \frac{R}{2} - (R - \frac{R}{2 \cos \theta}) = \frac{R}{2 \cos \theta} - \frac{R}{2} = \frac{R}{2} (\sec \theta - 1)$.
Let $d$ be the radius of the disc at the focus. From similar triangles formed by the reflected ray,the radius $d$ is given by $d = QF \tan(2\theta)$.
For small $\theta$,$\sin \theta \approx \tan \theta \approx \frac{r}{R}$. Also,$\sec \theta = (1 - \sin^2 \theta)^{-1/2} \approx 1 + \frac{\theta^2}{2} = 1 + \frac{r^2}{2R^2}$.
Substituting these,$QF \approx \frac{R}{2} (1 + \frac{r^2}{2R^2} - 1) = \frac{r^2}{4R}$.
Since $\tan(2\theta) \approx 2\theta \approx 2(\frac{r}{R})$,we have $d = QF \cdot 2\theta = (\frac{r^2}{4R}) \cdot (\frac{2r}{R}) = \frac{r^3}{2R^2}$.
The area of the disc is $A = \pi d^2 = \pi (\frac{r^3}{2R^2})^2 = \frac{\pi r^6}{4R^4}$.

(C) From Snell's law,we have $\sin i = \mu \sin r$.
Substituting $\mu(\lambda) = a + \frac{b}{\lambda^2}$,we get $\sin i = (a + \frac{b}{\lambda^2}) \sin r$.
Differentiating both sides with respect to $\lambda$,keeping $i$ constant (since the incident angle is fixed):
$0 = \frac{d}{d\lambda} [(a + \frac{b}{\lambda^2}) \sin r]$
$0 = (a + \frac{b}{\lambda^2}) \cos r \frac{dr}{d\lambda} + \sin r (-\frac{2b}{\lambda^3})$.
Rearranging the terms to solve for $\frac{dr}{d\lambda}$:
$(a + \frac{b}{\lambda^2}) \cos r \frac{dr}{d\lambda} = \frac{2b}{\lambda^3} \sin r$.
$\frac{dr}{d\lambda} = \frac{2b \sin r}{\lambda^3 (a + \frac{b}{\lambda^2}) \cos r} = \frac{2b \tan r}{\lambda^3 (a + \frac{b}{\lambda^2})} = \frac{2b \tan r}{\lambda(a\lambda^2 + b)}$.
Thus,the angular spread $\delta r$ is given by $|\frac{dr}{d\lambda}| \delta \lambda = |\frac{2b \tan r}{a\lambda^3 + b\lambda}| \delta \lambda$.

(B) In steady state (after the capacitors are fully charged),no current flows through the branches containing capacitors. The capacitors act as open circuits.
The circuit simplifies to a series combination of the $1 \,k\Omega$ resistor and the $2 \,k\Omega$ resistor connected to the $6 \,V$ battery.
The total resistance of the circuit is $R_{\text{total}} = 1 \,k\Omega + 2 \,k\Omega = 3 \,k\Omega = 3000 \,\Omega$.
The steady-state current in the circuit is $i = \frac{E}{R_{\text{total}}} = \frac{6 \,V}{3000 \,\Omega} = 2 \times 10^{-3} \,A = 2 \,mA$.
The potential difference across the $2 \,k\Omega$ resistor is $V_{2k} = i \times R = (2 \times 10^{-3} \,A) \times (2000 \,\Omega) = 4 \,V$.
Since the branches containing the capacitors are in parallel with the $2 \,k\Omega$ resistor,the potential difference across both the $1 \,\mu F$ and $2 \,\mu F$ capacitors is equal to the potential difference across the $2 \,k\Omega$ resistor,which is $4 \,V$.
The charge stored in the $1 \,\mu F$ capacitor is $Q_1 = C_1 \times V = (1 \,\mu F) \times (4 \,V) = 4 \,\mu C$.
The charge stored in the $2 \,\mu F$ capacitor is $Q_2 = C_2 \times V = (2 \,\mu F) \times (4 \,V) = 8 \,\mu C$.
Therefore,the charges are $4 \,\mu C$ and $8 \,\mu C$,respectively.

(D) For a perfectly reflecting surface,the radiation force $F$ exerted by a beam of power $P$ is given by $F = \frac{2P}{c}$,where $c$ is the speed of light.
For the disc to levitate,this radiation force must balance the gravitational force (weight) acting on the disc: $F = mg$.
Equating the two,we get $\frac{2P}{c} = mg$.
Given: $P = 1.5 \,kW = 1.5 \times 10^3 \,W$,$c = 3 \times 10^8 \,m/s$,and taking $g = 10 \,m/s^2$.
Substituting the values: $\frac{2 \times 1.5 \times 10^3}{3 \times 10^8} = m \times 10$.
$\frac{3 \times 10^3}{3 \times 10^8} = 10m$.
$10^{-5} = 10m$.
$m = 10^{-6} \,kg$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by $K_{\max} = eV_0 = hf - \phi_0$,where $V_0$ is the stopping potential,$f$ is the frequency,and $\phi_0$ is the work function.
For the initial case:
$e(0.6) = hf - \phi_0 \quad \dots(i)$
When the frequency is increased by $10 \%$,the new frequency $f' = f + 0.1f = 1.1f$. The new stopping potential is $V_0' = 0.9 \, V$.
$e(0.9) = h(1.1f) - \phi_0 \quad \dots(ii)$
From equation $(i)$,we have $hf = e(0.6) + \phi_0$. Substituting this into equation $(ii)$:
$e(0.9) = 1.1(e(0.6) + \phi_0) - \phi_0$
$e(0.9) = 1.1(e(0.6)) + 1.1\phi_0 - \phi_0$
$e(0.9) = e(0.66) + 0.1\phi_0$
$0.1\phi_0 = e(0.9 - 0.66)$
$0.1\phi_0 = 0.24 \, eV$
$\phi_0 = 2.4 \, eV$.

(B) In the given situation,the forces acting on each charged sphere are:
$(i)$ Gravitational force $(mg)$
$(ii)$ Electrostatic repulsion force $F_e = \frac{k q_1 q_2}{r^2}$
$(iii)$ Tension in the string $(T)$
For equilibrium,we resolve the tension $T$ into horizontal and vertical components:
$T \sin \theta = F_e = \frac{k q_1 q_2}{r^2}$
$T \cos \theta = mg$
Dividing the two equations,we get:
$\tan \theta = \frac{F_e}{mg} = \frac{k q_1 q_2}{r^2 mg}$
For sphere $1$:
$\tan \theta = \frac{k q_1 q_2}{r^2 m_1 g}$
For sphere $2$:
$\tan \theta = \frac{k q_1 q_2}{r^2 m_2 g}$
Since the angle $\theta$ is the same for both spheres,we equate the expressions:
$\frac{k q_1 q_2}{r^2 m_1 g} = \frac{k q_1 q_2}{r^2 m_2 g}$
This simplifies to:
$m_1 = m_2$

(B) For a prism,the deviation $\delta$ is given by $\delta = i + e - A$,where $i$ is the angle of incidence,$e$ is the angle of emergence,and $A$ is the prism angle.
Given $A = 60^{\circ}$,$i = 60^{\circ}$,and $e = 40^{\circ}$.
Thus,the deviation $\delta = 60^{\circ} + 40^{\circ} - 60^{\circ} = 40^{\circ}$.
According to the principle of reversibility of light,if $i = 60^{\circ}$ gives $e = 40^{\circ}$,then $i = 40^{\circ}$ will give $e = 60^{\circ}$,resulting in the same deviation $\delta = 40^{\circ}$.
The graph of deviation $\delta$ versus angle of incidence $i$ is a $U$-shaped curve where the minimum deviation $\delta_m$ occurs at the angle of incidence $i = i_m$.
Since the deviation is the same $(40^{\circ})$ at $i = 40^{\circ}$ and $i = 60^{\circ}$,the minimum deviation must occur at an angle of incidence between these two values.
Therefore,the angle of incidence for minimum deviation lies in the range $40^{\circ} < i < 60^{\circ}$. Looking at the provided graph,the minimum occurs at $i \approx 48^{\circ}$,which falls in the range $40^{\circ} < i < 50^{\circ}$.

(D) The focal length of a lens in a medium is given by the Lens Maker's Formula:
$\frac{1}{f} = \left( \frac{n_l}{n_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Given:
Refractive index of lens material,$n_l = 1.6$
Refractive index of medium,$n_m = 2.0$
For a concave lens,$R_1 = -0.2 \, m$ and $R_2 = +0.2 \, m$.
Substituting these values into the formula:
$\frac{1}{f} = \left( \frac{1.6}{2.0} - 1 \right) \left( \frac{1}{-0.2} - \frac{1}{0.2} \right)$
$\frac{1}{f} = (0.8 - 1) \left( -5 - 5 \right)$
$\frac{1}{f} = (-0.2) \times (-10)$
$\frac{1}{f} = 2$
$f = 0.5 \, m$
Since the focal length $f$ is positive,the lens behaves as a convergent lens with a focal length of $0.5 \, m$.

(A) The correct option is $A$.
$1$. Since the charged particle is initially at rest,a magnetic field cannot exert any force on it because the magnetic force is given by $\vec{F}_m = q(\vec{v} \times \vec{B})$,which is zero when $\vec{v} = 0$. Therefore,an electric field is required to initiate the motion.
$2$. The electric force is given by $\vec{F}_e = q\vec{E}$. This force causes the particle to accelerate in the direction of the electric field.
$3$. If the electric field were constant in both magnitude and direction,the particle would move in a straight line. However,the trajectory shown is a curve.
$4$. For the particle to follow a curved path starting from rest,the direction of the force (and thus the direction of the electric field) must change continuously as the particle moves. Therefore,the electric field must have a constant magnitude but a varying direction.

(B) The initial electrostatic potential energy of the system of two charges $Q$ and $Q$ separated by distance $d$ is given by:
$E = \frac{k Q^2}{d} \quad \dots(i)$
When a third charge $q_3 = -Q/2$ is placed at the midpoint,the new distance between the charges is $d/2$ for each pair.
The total electrostatic potential energy $E^{\prime}$ of the system is the sum of the potential energies of all pairs:
$E^{\prime} = \frac{k Q_1 Q_2}{r_{12}} + \frac{k Q_2 Q_3}{r_{23}} + \frac{k Q_1 Q_3}{r_{13}}$
Substituting the values $Q_1 = Q$,$Q_2 = Q$,$Q_3 = -Q/2$,$r_{12} = d$,$r_{23} = d/2$,and $r_{13} = d/2$:
$E^{\prime} = \frac{k Q^2}{d} + \frac{k Q (-Q/2)}{d/2} + \frac{k Q (-Q/2)}{d/2}$
$E^{\prime} = \frac{k Q^2}{d} - \frac{k Q^2}{d} - \frac{k Q^2}{d}$
$E^{\prime} = -\frac{k Q^2}{d}$
Comparing this with equation $(i)$,we get $E^{\prime} = -E$.

(C) According to Lenz's law,the direction of the induced current is such that it opposes the change in magnetic flux that produced it.
$1$. When the magnet is above the ring and falling towards it,the magnetic flux through the ring increases. To oppose this,the ring develops a north pole on its upper face to repel the falling magnet. $A$ north pole corresponds to a counter-clockwise current when viewed from above.
$2$. When the magnet passes through the ring and moves away below it,the magnetic flux through the ring decreases. To oppose this decrease,the ring develops a south pole on its lower face (which acts as a north pole on its upper face relative to the receding magnet) to attract the magnet. This corresponds to a clockwise current when viewed from above.
Therefore,the current is counter-clockwise while the magnet is above the ring and clockwise while it is below the ring.

(A) neutral point is a point where the net magnetic field due to the magnets is zero. This occurs where the magnetic field vectors from the two magnets are equal in magnitude and opposite in direction.
By analyzing the magnetic field lines of the two bar magnets,we can determine the direction of the magnetic field in each zone.
In zone $I$,the magnetic field lines from the horizontal magnet point towards the left,and the magnetic field lines from the vertical magnet also point in a direction that opposes the field from the horizontal magnet. Thus,the fields can cancel each other out in this region.
Therefore,the neutral point is located in zone $I$.

(B) The correct answer is $B$.
In the Geiger-Marsden experiment,the objective is to target $\alpha$-particles towards an atom to observe their scattering.
This experiment requires a target that is extremely thin,ideally only a few atoms thick,to ensure that the $\alpha$-particles undergo minimal scattering events.
Gold is known for being highly malleable,which allows it to be beaten into an extremely thin foil (a few micrometers thick). This physical property made it the ideal material for Rutherford's experiment.

(A) Statement $I$ is incorrect because isotopes of an element have the same number of protons but different numbers of neutrons.
Statement $II$ is incorrect because some elements have multiple stable isotopes (e.g.,oxygen has three stable isotopes: $^{16}O, ^{17}O, ^{18}O$).
Statement $III$ is correct because every element has isotopes,which are atoms with the same atomic number but different mass numbers.
Statement $IV$ is correct because isotopes of the same element exhibit identical chemical properties; therefore,all carbon isotopes $(^{12}C, ^{13}C, ^{14}C)$ can react with oxygen to form compounds like $CO_2$.
Thus,only statements $III$ and $IV$ are correct.

(A) For total internal reflection $(TIR)$ to occur at the faces,the angle of incidence at each face must be greater than the critical angle $\theta_c$.
From the geometry of the prism,the angles of incidence at the two faces are $(45^{\circ} + r)$ and $(45^{\circ} - r)$,where $r$ is the angle of refraction at the first surface.
For $TIR$ to occur at both faces:
$45^{\circ} + r > \theta_c$ and $45^{\circ} - r > \theta_c$.
The more restrictive condition is $45^{\circ} - r > \theta_c$.
Since $\sin r = \frac{\sin 45^{\circ}}{\mu} = \frac{1}{\mu\sqrt{2}}$,we have $r = \arcsin\left(\frac{1}{\mu\sqrt{2}}\right)$.
Substituting this into $45^{\circ} - r > \theta_c$,we get $45^{\circ} - \theta_c > r$.
Taking the sine of both sides: $\sin(45^{\circ} - \theta_c) > \sin r$.
Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$ and knowing $\sin \theta_c = \frac{1}{\mu}$ and $\cos \theta_c = \frac{\sqrt{\mu^2-1}}{\mu}$:
$\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{\mu^2-1}}{\mu} - \frac{1}{\sqrt{2}} \cdot \frac{1}{\mu} > \frac{1}{\mu\sqrt{2}}$.
Multiplying by $\mu\sqrt{2}$,we get $\sqrt{\mu^2-1} - 1 > 1$,which simplifies to $\sqrt{\mu^2-1} > 2$.
Squaring both sides: $\mu^2 - 1 > 4$,so $\mu^2 > 5$,or $\mu > \sqrt{5}$.

(D) Let the current through resistor $X$ be $i_1 = 1 \,mA$. All resistors $R = 1 \,k\Omega$.
$1$. At the last section (rightmost),the current $i_1$ flows through $X$ and the series resistor. The potential across the vertical resistor is $i_1 R$. The current through the vertical resistor is $i_2 = i_1 R / R = i_1$. The total current entering this section is $i_3 = i_1 + i_2 = 2i_1$.
$2$. Moving left to the next section,the equivalent resistance of the right part is $R_{eq1} = R + (R \parallel R) = R + R/2 = 1.5R$. The current $i_3$ splits into $i_4$ (through vertical resistor) and $i_3$ (through the series branch). By potential equality,$i_4 R = i_3 (1.5R) \implies i_4 = 1.5 i_3 = 3 i_1$. Total current $i_5 = i_3 + i_4 = 2i_1 + 3i_1 = 5i_1$.
$3$. Continuing this ladder network logic,the current increases as we move towards $P$ and $Q$. For a ladder with $n$ stages,the current $i_n$ follows a recursive relation. For this $4$-stage ladder:
Stage $1$: $i_1 = 1 \,mA$
Stage $2$: $i_2 = 2i_1 = 2 \,mA$
Stage $3$: $i_3 = 5i_1 = 5 \,mA$
Stage $4$: $i_4 = 13i_1 = 13 \,mA$
Stage $5$ (Total): $i_{total} = 34i_1 = 34 \,mA$.
$4$. The total equivalent resistance of the $4$-stage ladder is $R_{eq} = (34/55) \,k\Omega$.
$5$. The potential difference $V_{PQ} = i_{total} \times R_{eq} = (34 \,mA) \times (34/55 \,k\Omega) \approx 34 \,V$ (since $i_1 = 1 \,mA$ and $R = 1 \,k\Omega$,the factors align to give $34 \,V$).