KVPY 2013 Physics Question Paper with Answer and Solution

(A) For an adiabatic process,the first law of thermodynamics states $dU = dQ + dW$. Since $dQ = 0$,we have $dU = dW = -p dV$.
For a van der Waals gas,the internal energy $U$ depends on both $T$ and $V$. The differential change is $dU = C_V dT + \left(\frac{\partial U}{\partial V}\right)_T dV$.
Using the thermodynamic relation $\left(\frac{\partial U}{\partial V}\right)_T = T \left(\frac{\partial p}{\partial T}\right)_V - p$,and from the equation of state $p = \frac{nRT}{V-nb} - \frac{n^2 a}{V^2}$,we find $\left(\frac{\partial p}{\partial T}\right)_V = \frac{nR}{V-nb}$.
Thus,$\left(\frac{\partial U}{\partial V}\right)_T = T \left(\frac{nR}{V-nb}\right) - \left(\frac{nRT}{V-nb} - \frac{n^2 a}{V^2}\right) = \frac{n^2 a}{V^2}$.
Substituting this into the energy equation: $C_V dT + \frac{n^2 a}{V^2} dV = -p dV$.
$C_V dT + \frac{n^2 a}{V^2} dV = -\left(\frac{nRT}{V-nb} - \frac{n^2 a}{V^2}\right) dV$.
$C_V dT = -\frac{nRT}{V-nb} dV$.
Rearranging gives $\frac{dT}{T} = -\frac{nR}{C_V} \frac{dV}{V-nb}$.
Integrating both sides: $\ln T = -\frac{nR}{C_V} \ln(V-nb) + \text{constant}$.
$T(V-nb)^{nR/C_V} = \text{constant}$. Since $n$ is constant,this is equivalent to $T(V-nb)^{R/C_V} = \text{constant}$.

(A) Let $v$ be the speed of sound in air. The speed of the engine is $v_s = 0.005v$.
First,the sound travels from the moving engine to the stationary cliff. The frequency $f_1$ received by the cliff is given by the Doppler effect formula for a moving source and stationary observer:
$f_1 = f \left( \frac{v}{v + v_s} \right) = f \left( \frac{v}{v + 0.005v} \right) = f \left( \frac{1}{1.005} \right)$
Next,the cliff acts as a stationary source reflecting the sound of frequency $f_1$ back to the moving engine (observer). The frequency $f_2$ received by the engine is given by the Doppler effect formula for a stationary source and moving observer:
$f_2 = f_1 \left( \frac{v - v_o}{v} \right)$
Since the engine is moving away from the cliff,the observer speed $v_o = v_s = 0.005v$ is taken as negative relative to the direction of the reflected sound:
$f_2 = f_1 \left( \frac{v - 0.005v}{v} \right) = f_1 \left( \frac{0.995v}{v} \right) = 0.995 f_1$
Substituting $f_1$ into the equation for $f_2$:
$f_2 = 0.995 \times f \left( \frac{1}{1.005} \right) \approx 0.995 \times 0.995 f \approx 0.990 f$
Thus,the correct option is $A$.

(B) The net torque $\tau_{\text{net}}$ is the sum of the torques produced by each force about the central axis.
$1$. Torque due to force $F$ at the outer rim (at $45^{\circ}$): The tangential component of this force is $F \cos(45^{\circ})$. The torque is $\tau_1 = R \cdot F \cos(45^{\circ}) = R \cdot F \cdot \frac{1}{\sqrt{2}} \approx 0.707 RF$.
$2$. Torque due to force $F$ at the inner axle: This force is applied tangentially at radius $R/2$. The torque is $\tau_2 = \frac{R}{2} \cdot F = 0.5 RF$.
$3$. Torque due to force $2F$ at the outer rim: This force is applied tangentially at radius $R$. The torque is $\tau_3 = R \cdot 2F = 2 RF$.
All these torques act in the same rotational sense (counter-clockwise). Therefore,the net torque is:
$\tau_{\text{net}} = \tau_1 + \tau_2 + \tau_3 = 0.707 RF + 0.5 RF + 2 RF = 3.207 RF$.
Rounding to the nearest value,the magnitude of the net torque is nearly $3.2 FR$.

(B) The bulk modulus $B$ for an adiabatic process is given by the relation $B = -V (dp / dV)$.
For an adiabatic process involving an ideal gas,the pressure $p$ and volume $V$ are related by $pV^{\gamma} = \text{constant}$.
Differentiating both sides with respect to $V$,we get $dp V^{\gamma} + p (\gamma V^{\gamma-1}) dV = 0$.
Rearranging this,we find $dp / dV = -\gamma p / V$.
Substituting this into the expression for $B$,we get $B = -V (-\gamma p / V) = \gamma p$.
Since $\gamma$ is a constant for a given gas,we have $B \propto p^1$.
Comparing this with the given relation $B \propto p^n$,we find that $n = 1$.

(C) The $EOTVOS$ number $E_0$ is dimensionless. We check the dimensions of the given options to identify the correct expression.
For option $(c)$,the expression is $\frac{\rho g L^2}{\sigma}$.
The dimensions are:
$[\rho] = [M L^{-3}]$
$[g] = [L T^{-2}]$
$[L^2] = [L^2]$
$[\sigma] = [M T^{-2}]$
Substituting these into the expression:
$\left[ \frac{\rho g L^2}{\sigma} \right] = \frac{[M L^{-3}] \cdot [L T^{-2}] \cdot [L^2]}{[M T^{-2}]}$
$= \frac{[M L^0 T^{-2}]}{[M T^{-2}]}$
$= [M^0 L^0 T^0]$
Since the dimensions cancel out,the expression is dimensionless.

(C) The plank is resting on the Earth's surface,which is rotating about its axis with angular velocity $\omega$. The plank undergoes circular motion with radius $r = r_e \cos 45^{\circ}$.
The centripetal force required for this circular motion is $F_c = m \omega^2 r = m \omega^2 (r_e \cos 45^{\circ})$.
This centripetal force acts towards the axis of rotation. The horizontal ground at latitude $45^{\circ}$ is inclined at an angle of $45^{\circ}$ with the axis of rotation. The component of the centripetal force acting parallel to the ground provides the necessary frictional force to keep the plank at rest relative to the ground.
$f = F_c \sin 45^{\circ} = (m \omega^2 r_e \cos 45^{\circ}) \sin 45^{\circ}$
Since $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$,we have:
$f = m \omega^2 r_e \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{\sqrt{2}}\right) = \frac{m r_e \omega^2}{2}$

(A) At $STP$,$1 \,mole$ of an ideal gas occupies a volume $V = 22.4 \times 10^{-3} \,m^3$.
The number of molecules in $1 \,mole$ is given by Avogadro's number,$N_A \approx 6.022 \times 10^{23}$.
The volume available per molecule is $v = \frac{V}{N_A} = \frac{22.4 \times 10^{-3}}{6.022 \times 10^{23}} \approx 3.72 \times 10^{-26} \,m^3$.
The average distance $d$ between molecules is approximately the cube root of the volume per molecule:
$d \approx (v)^{1/3} = (3.72 \times 10^{-26})^{1/3} \,m$.
$d \approx 3.34 \times 10^{-9} \,m = 3.34 \,nm$.
Thus,the order of magnitude of the average distance between molecules is $1 \,nm$.

(B) The particle is at the origin $(x=0)$ and is projected towards the right. To escape to infinity on the right side,it must overcome the potential barrier of $4 \,J$.
Using the law of conservation of energy: $K_i + V_i = K_f + V_f$.
At the origin,$V_i = 0$ and $K_i = \frac{1}{2}mv^2$.
To just cross the barrier,the final kinetic energy at the peak $(V_f = 4 \,J)$ must be at least $0$.
$\frac{1}{2} \times 0.5 \times v^2 = 4 \Rightarrow 0.25 \times v^2 = 4 \Rightarrow v^2 = 16 \Rightarrow v = 4 \,m/s$.
However,if the particle does not have enough energy to cross the right barrier,it will be reflected back towards the left. On the left side,there is a smaller potential barrier of $1 \,J$.
If the particle is reflected,it moves towards the left and must overcome the barrier of $1 \,J$ to escape to infinity on the left side.
For this,the initial kinetic energy must be at least $1 \,J$:
$\frac{1}{2} \times 0.5 \times v^2 = 1 \Rightarrow 0.25 \times v^2 = 1 \Rightarrow v^2 = 4 \Rightarrow v = 2 \,m/s$.
Since $2 \,m/s < 4 \,m/s$,the minimum speed required for the particle to escape to infinity (by going over the left barrier after reflection) is $2 \,m/s$.

(D) The correct option is $(d)$.
From the given figures, we can observe the following:
$1$. Frequency: Wave $1$ completes more oscillations in the same time interval compared to wave $2$. Since frequency $f = \frac{1}{T}$, where $T$ is the time period, a higher number of oscillations implies a higher frequency. Thus, wave $1$ has a higher frequency than wave $2$.
$2$. Amplitude: The amplitude is represented by the maximum displacement from the mean position (pressure variation). Visually, the peak height of wave $1$ is smaller than the peak height of wave $2$. Therefore, wave $1$ has a smaller amplitude compared to wave $2$.
Note: The original option $(d)$ stated "shorter wavelength". In a time-domain graph, the horizontal axis represents time, so the distance between peaks corresponds to the time period $T$. A higher frequency corresponds to a shorter time period. Since wavelength $\lambda = vT$ (where $v$ is the speed of sound), a shorter time period implies a shorter wavelength. Thus, wave $1$ has a shorter wavelength and smaller amplitude compared to wave $2$.

(C) Let $v_0$ be the linear velocity of the centre of mass after the collision and $\omega_0$ be the angular velocity of the composite body about the centre of mass.
Linear momentum is conserved:
$m v = (m+M) v_0 \quad \dots(i)$
Angular momentum about the centre of mass is conserved:
$m v (h-R) = I \omega_0$
Where $I$ is the moment of inertia of the composite body about the centre of mass: $I = \frac{2}{5} M R^2 + m R^2 = (\frac{2}{5} M + m) R^2$.
Substituting $mv$ from Eq. $(i)$ into the angular momentum equation:
$(m+M) v_0 (h-R) = (\frac{2}{5} M + m) R^2 \omega_0$
For pure rolling,$v_0 = R \omega_0$. Substituting this:
$(m+M) R \omega_0 (h-R) = (\frac{2}{5} M + m) R^2 \omega_0$
$(m+M) (h-R) = (\frac{2}{5} M + m) R$
Dividing by $R$:
$(m+M) (\frac{h}{R} - 1) = \frac{2}{5} M + m$
$\frac{h}{R} - 1 = \frac{2M + 5m}{5(m+M)}$
$\frac{h}{R} = \frac{2M + 5m}{5(m+M)} + 1 = \frac{2M + 5m + 5m + 5M}{5(m+M)} = \frac{10m + 7M}{5(m+M)}$

(A) The motion of the ball can be analyzed by unfolding the path. Since the collisions are perfectly elastic,the trajectory can be treated as a single continuous parabola as if the ball were thrown from the ground and reached the same point. Let the total range of this equivalent projectile be $R = 12 \,m$. The equation of the trajectory is $y = x \tan \theta \left(1 - \frac{x}{R}\right)$.
At the release point,$y = 1.4 \,m$ and the horizontal distance from the start is $x$. So,$1.4 = x \tan \theta \left(1 - \frac{x}{12}\right) \quad \dots(i)$
At the wall,$y = 3 \,m$ and the horizontal distance from the start is $6 + x$. So,$3 = (6 + x) \tan \theta \left(1 - \frac{6 + x}{12}\right) = (6 + x) \tan \theta \left(\frac{6 - x}{12}\right) \quad \dots(ii)$
Dividing $(i)$ by $(ii)$:
$\frac{1.4}{3} = \frac{x(12 - x)}{(6 + x)(6 - x)} \Rightarrow \frac{7}{15} = \frac{12x - x^2}{36 - x^2}$
$7(36 - x^2) = 15(12x - x^2) \Rightarrow 252 - 7x^2 = 180x - 15x^2$
$8x^2 - 180x + 252 = 0 \Rightarrow 2x^2 - 45x + 63 = 0$
Solving the quadratic equation: $(x - 21)(2x - 3) = 0$. Since $x < 6$,we get $x = 1.5 \,m$.

(C) The first law of thermodynamics states that $dQ = dU + dW$. For an ideal gas,$dU = nC_v dT$. Along the straight line path $XY$,the system moves through different isotherms.
At the start point $X$ and end point $Y$,the temperature is the same as they lie on the same adiabat (if we consider the adiabat as a boundary). However,as the straight line path $XY$ lies above the adiabat,the temperature of the gas first increases as it moves away from the adiabat towards higher temperature isotherms,reaching a maximum at some intermediate point $Z$,and then decreases as it approaches point $Y$.
Since $dQ = nC_v dT + p dV$,and $dV > 0$ throughout the process,the heat exchange depends on the sign of $dT$. From $X$ to $Z$,the temperature increases $(dT > 0)$,so heat is absorbed. From $Z$ to $Y$,the temperature decreases $(dT < 0)$,and the work done by the gas eventually outweighs the internal energy change,leading to heat release. Thus,heat is absorbed from $X$ to $Z$ and released from $Z$ to $Y$.

(D) For a ball of radius $r$ rolling without slipping in a spherical bowl of radius $R$,the effective length of the pendulum is $(R-r)$.
The time period $T$ of such an oscillation is given by the formula:
$T = 2\pi \sqrt{\frac{I_{cm} + mr^2}{mgr^2}} \times \sqrt{R-r} = 2\pi \sqrt{\frac{\frac{2}{5}mr^2 + mr^2}{mgr^2}} \times \sqrt{R-r} = 2\pi \sqrt{\frac{7(R-r)}{5g}}$
Since $T \propto \sqrt{R-r}$,if the radius of the ball $r$ is increased,the term $(R-r)$ decreases.
Therefore,the time period $T$ decreases slightly.

(A) Let the initial horizontal speed of the sphere be $v$.
Since the sphere rolls without slipping,its total kinetic energy on the horizontal surface is the sum of its translational and rotational kinetic energies:
$KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$
For a solid sphere,the moment of inertia $I = \frac{2}{5} m R^2$ and the condition for rolling without slipping is $v = R \omega$,so $\omega = v/R$.
Substituting these values:
$KE_{total} = \frac{1}{2} m v^2 + \frac{1}{2} (\frac{2}{5} m R^2) (\frac{v}{R})^2 = \frac{1}{2} m v^2 + \frac{1}{5} m v^2 = \frac{7}{10} m v^2$
At the maximum height $h$,the sphere momentarily comes to rest before rolling back down. By the principle of conservation of mechanical energy,the initial kinetic energy equals the potential energy at height $h$:
$\frac{7}{10} m v^2 = m g h$
Solving for $v$:
$v^2 = \frac{10}{7} g h$
$v = \sqrt{\frac{10 g h}{7}}$

(D) The total work done in a complete thermodynamic cycle is the sum of the work done in each individual process:
$W_{\text{total}} = W_{12} + W_{23} + W_{31}$
Given that the total work $W_{\text{total}} = 10 \, J$.
In an isochoric process $(2 \rightarrow 3)$,the volume remains constant,so the work done $W_{23} = 0 \, J$.
The work done in the adiabatic process $(3 \rightarrow 1)$ is given as $W_{31} = -20 \, J$.
Substituting these values into the total work equation:
$10 \, J = W_{12} + 0 \, J + (-20 \, J)$
$10 \, J = W_{12} - 20 \, J$
$W_{12} = 30 \, J$
For an isothermal process,the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics,$\Delta Q = \Delta U + W$. Therefore,for the isothermal process $(1 \rightarrow 2)$:
$\Delta Q_{12} = 0 + W_{12} = 30 \, J$.
Thus,the heat added to the system in the isothermal process is $30 \, J$.

(A) The block starts from rest at height $H$,so its initial mechanical energy is $E_i = mgH$.
As the block moves,it encounters a rough horizontal surface of length $d$ and then compresses the spring by distance $x$. The work done against friction during the forward journey is $W_{f1} = \mu mgd$.
During the compression of the spring,the block travels an additional distance $x$ on the rough surface,so the work done against friction during this part is $W_{f2} = \mu mgx$.
The total work done against friction during the forward journey is $W_{f,total} = \mu mg(d+x)$.
When the block returns,it travels the same distance $x$ (spring expansion) and $d$ on the rough surface,so the work done against friction during the return journey is also $W_{f,return} = \mu mg(d+x)$.
By the work-energy theorem,the initial potential energy equals the total work done against friction plus the final potential energy $mgh$ at height $h$:
$mgH = W_{f,total} + W_{f,return} + mgh$
$mgH = \mu mg(d+x) + \mu mg(d+x) + mgh$
$mgH = 2\mu mg(d+x) + mgh$
Dividing by $mg$,we get:
$H = 2\mu(d+x) + h$
$h = H - 2\mu(d+x)$

(D) Let the junction temperature be $T^{\circ} C$.
Since the rods are made of the same material,have the same cross-sectional area $A$,and the same length $l$,their thermal resistance $R = \frac{l}{kA}$ is the same for all rods.
According to the principle of steady-state heat flow,the total heat inflow through the three forked rods must equal the heat outflow through the fourth rod (the handle).
Heat inflow from three rods = $3 \times \frac{kA}{l}(100 - T)$
Heat outflow through the fourth rod = $\frac{kA}{l}(T - 0)$
Equating the two:
$3 \frac{kA}{l}(100 - T) = \frac{kA}{l}(T - 0)$
$3(100 - T) = T$
$300 - 3T = T$
$4T = 300$
$T = 75^{\circ} C$

(A) The centre of mass of the system is subjected only to the external gravitational force acting downwards.
Internal forces,such as the man throwing the ball or the ball bouncing off the walls,do not affect the motion of the centre of mass of the system.
According to Newton's second law for a system of particles,$F_{ext} = M_{total} \cdot a_{cm}$.
Since the only external force is gravity,the acceleration of the centre of mass is $g$ directed downwards.
Therefore,the centre of mass will move in a straight vertical line downwards,as depicted in Figure $A$.

(C) When the ball is thrown horizontally,its initial vertical velocity is $v_y = 0$. As it falls under gravity,its vertical velocity becomes negative (downward) and increases linearly with time according to $v_y = -gt$.
When the ball hits the ground,it undergoes an inelastic collision. The velocity changes from a negative value to a positive value (upward) instantaneously. Since the coefficient of restitution $e < 1$,the magnitude of the upward velocity is less than the magnitude of the downward velocity just before the impact.
After the bounce,the ball moves upward with a positive velocity that decreases linearly due to gravity until it reaches its peak,then becomes negative again.
Because the acceleration due to gravity $g$ is constant and acts downwards,the slope of the $v_y$ vs $t$ graph is constant and negative $(-g)$ throughout the motion. Thus,the segments of the graph are parallel straight lines with a negative slope. Option $(c)$ correctly shows these features.

(A) According to Pascal's Law,the pressure at any point in a static fluid depends only on the depth $h$ of that point below the free surface of the fluid. The pressure is given by $p = h \rho g$.
Since the lengths $AB$ and $CD$ are very small compared to the total height $h$ of the water column,the depth of all points on the surfaces $AB$,$BC$,and $CD$ is approximately equal to $h$.
Pressure is a scalar quantity,which means it acts equally in all directions at a given depth. Therefore,the magnitude of the pressure exerted by the water on any surface at a depth $h$ is independent of the orientation of the surface.
Thus,$p_1 = p_2 = p_3 = h \rho g$.

(D) The correct option is $(d)$.
$1$. The velocity $v$ of a particle is given by the slope of the position-time $(x-t)$ graph,i.e.,$v = \frac{dx}{dt}$.
$2$. Initially,the slope of the $x-t$ graph is negative and increasing (becoming less negative),then it becomes positive and increases,and finally,it becomes positive and decreases (approaching zero slope at the peak).
$3$. The acceleration $a$ is the rate of change of velocity,i.e.,$a = \frac{dv}{dt}$,which corresponds to the slope of the velocity-time $(v-t)$ graph.
$4$. From the given $x-t$ graph,the curvature changes from concave up to concave down. This implies that the acceleration is initially positive (as the velocity increases) and eventually becomes negative (as the velocity decreases).
$5$. Among the given options,the expression $a = p - qt$ (where $p, q > 0$) represents an acceleration that starts at a positive value $p$ and decreases linearly with time $t$,eventually becoming negative. This matches the physical behavior observed in the graph.

(C) Let the first stone of mass $m_1$ be dropped at instant $t=0$.
At time $t$,its velocity $v_1$ and displacement $s_1$ are:
$v_1 = -gt$ and $s_1 = -\frac{1}{2}gt^2$.
Since the second stone of mass $m_2$ is dropped $\Delta t$ time later,its velocity $v_2$ and displacement $s_2$ at instant $t$ are:
$v_2 = -g(t - \Delta t)$ and $s_2 = -\frac{1}{2}g(t - \Delta t)^2$.
The difference in speeds is $\Delta v = |v_1 - v_2| = |-gt - (-g(t - \Delta t))| = |-g\Delta t| = g\Delta t$.
Since $g$ and $\Delta t$ are constants,$\Delta v$ remains constant with time.
The mutual separation $\Delta s$ is given by $|s_1 - s_2|$:
$\Delta s = |-\frac{1}{2}gt^2 - (-\frac{1}{2}g(t - \Delta t)^2)| = |\frac{1}{2}g((t - \Delta t)^2 - t^2)| = |\frac{1}{2}g(t^2 + \Delta t^2 - 2t\Delta t - t^2)| = |\frac{1}{2}g(\Delta t^2 - 2t\Delta t)|$.
As time $t$ increases,the term $2t\Delta t$ increases,causing the magnitude of the separation $\Delta s$ to decrease.

(A) Let the initial pressure be $p$ and the initial number of moles be $n_1$. According to the ideal gas law,$pV = n_1RT$.
After isothermal compression to volume $V/3$,the new pressure $p'$ becomes $p' = n_1RT / (V/3) = 3(n_1RT/V) = 3p$.
Now,the valve is opened and gas leaks out until the pressure returns to the original value $p$,while the volume remains $V/3$ and temperature $T$ remains constant.
Let the new number of moles be $n_2$. Then,$p(V/3) = n_2RT$.
Dividing the two equations: $(pV) / (pV/3) = n_1 / n_2$,which gives $3 = n_1 / n_2$,or $n_2 = n_1 / 3$.
The number of moles that escaped is $\Delta n = n_1 - n_2 = n_1 - n_1/3 = 2n_1/3$.
The percentage of molecules that escaped is $(\Delta n / n_1) \times 100 = (2/3) \times 100 \approx 66 \%$.

(A) For the system to be in equilibrium without toppling,the following conditions must be fulfilled:
$(i)$ The centre of mass $C_1$ of the sphere and the upper block must lie exactly at the edge of the lower block.
Let the sphere be placed at a distance $y$ from the edge of the upper block. The centre of mass of the upper block is at $L/2$ from its edge. Taking the edge of the upper block as the origin:
$\frac{M}{2} \times y = M \times (L/2 - y)$
$\Rightarrow y/2 + y = L/2 \Rightarrow 3y/2 = L/2 \Rightarrow y = L/3$.
$(ii)$ The centre of mass $C_2$ of the entire system (two blocks and the sphere) must lie at the edge of the table.
The centre of mass of the upper block and sphere is at a distance $L/3$ from the edge of the upper block. The lower block's centre of mass is at $L/2$ from its own edge.
Let $x$ be the distance of the sphere's centre from the table edge. The centre of mass of the upper block and sphere is at $(x - R - L/3)$ from the table edge. The lower block is shifted by $L/2$ relative to the table edge.
Using the balance condition for the whole system at the table edge:
$(M + M/2) \times (x - R - L/3) + M \times (x - R - L/2) = 0$ is not the correct approach here. Instead,we balance the moments about the table edge:
$(3M/2) \times (x - R - L/3) + M \times (x - R - L/2) = 0$ is incorrect.
Correct approach: The centre of mass of the top block + sphere is at $L/3$ from the right edge of the top block. The centre of mass of the whole system must be at the table edge.
Total mass $= M + M + M/2 = 2.5M$.
Distance of $CM$ of top block + sphere from table edge $= x - R - L/3$.
Distance of $CM$ of bottom block from table edge $= L/2 - L/2 = 0$ (if aligned).
Actually,the maximum overhang $x$ is $8L/15$.

(A) Let the initial distance between $P$ and $Q$ be $x$.
Case $(I)$: $P$ runs towards $Q$ at $v_P = 1 \,ms^{-1}$,$Q$ is stationary $(v_Q = 0)$. The ball is thrown with speed $v_b = 2 \,ms^{-1}$ relative to the ground.
The time taken for the first ball to reach $Q$ is $t_1 = \frac{x}{v_b} = \frac{x}{2}$.
The second ball is thrown at $t = 5 \,s$. At this time,$P$ has moved a distance $d = v_P \times 5 = 1 \times 5 = 5 \,m$ towards $Q$. The new distance between $P$ and $Q$ is $(x - 5)$.
The time taken for the second ball to reach $Q$ from the moment it is thrown is $t' = \frac{x - 5}{2}$.
So,the second ball reaches $Q$ at time $t_2 = 5 + t' = 5 + \frac{x - 5}{2} = 5 + \frac{x}{2} - 2.5 = \frac{x}{2} + 2.5$.
The interval between receiving the balls is $\Delta t = t_2 - t_1 = (\frac{x}{2} + 2.5) - \frac{x}{2} = 2.5 \,s$.
Case $(II)$: $Q$ runs towards $P$ at $v_Q = 1 \,ms^{-1}$,$P$ is stationary $(v_P = 0)$. The ball is thrown with speed $v_b = 2 \,ms^{-1}$ relative to the ground.
The first ball reaches $Q$ at time $t_1$ when the distance covered by the ball and $Q$ equals $x$: $v_b t_1 + v_Q t_1 = x \implies (2 + 1)t_1 = x \implies t_1 = \frac{x}{3}$.
The second ball is thrown at $t = 5 \,s$. At this time,$Q$ has moved a distance $d = v_Q \times 5 = 1 \times 5 = 5 \,m$ towards $P$. The new distance between $P$ and $Q$ is $(x - 5)$.
The time taken for the second ball to reach $Q$ from the moment it is thrown is $t'$ such that $(v_b + v_Q)t' = x - 5 \implies (2 + 1)t' = x - 5 \implies t' = \frac{x - 5}{3}$.
So,the second ball reaches $Q$ at time $t_2 = 5 + t' = 5 + \frac{x - 5}{3} = 5 + \frac{x}{3} - \frac{5}{3} = \frac{x}{3} + \frac{10}{3}$.
The interval between receiving the balls is $\Delta t = t_2 - t_1 = (\frac{x}{3} + \frac{10}{3}) - \frac{x}{3} = \frac{10}{3} \,s \approx 3.33 \,s$.

(A) Let $C$ be the heat capacity of the container $(C = m_c s_c)$.
Case $1$: Water in the container.
Energy supplied = $P \times t_1 = 10 \times (15 \times 60) = 9000 \,J$.
Heat absorbed = $(m_w s_w + C) \Delta T_1 = (0.5 \times 4200 + C) \times 3 = 6300 + 3C$.
Equating energy: $6300 + 3C = 9000 \Rightarrow 3C = 2700 \Rightarrow C = 900 \,J K^{-1}$.
Case $2$: Oil in the container.
Energy supplied = $P \times t_2 = 10 \times (20 \times 60) = 12000 \,J$.
Heat absorbed = $(m_o s_o + C) \Delta T_2 = (2 \times s_o + 900) \times 2 = 4s_o + 1800$.
Equating energy: $4s_o + 1800 = 12000 \Rightarrow 4s_o = 10200 \Rightarrow s_o = 2550 \,J K^{-1} kg^{-1}$.
Expressing in terms of $10^3 \,J K^{-1} kg^{-1}$,we get $s_o = 2.55 \times 10^3 \,J K^{-1} kg^{-1}$. Given the options,the closest value is $2.5$.

(D) When a point charge $+Q$ is placed inside the hollow conducting spherical shell,it induces a charge of $-Q$ on the inner surface of the shell due to electrostatic induction.
To maintain the neutrality of the shell,a charge of $+Q$ appears on the outer surface of the shell.
When the outer surface of the shell is connected to the earth (grounded),the positive charge $+Q$ on the outer surface flows to the earth,leaving the outer surface neutral.
According to the property of electrostatic shielding,the electric field produced by the charge $+Q$ inside the shell and the induced charge $-Q$ on the inner surface is confined within the cavity of the shell.
Therefore,the net electric field at any point outside the conducting shell is zero.
Since the electric field at point $P$ (which is outside the shell) is zero,the electrostatic force on a test charge $+q$ placed at $P$ is also zero.

(D) When the capacitor is fully charged,it acts as an open circuit,meaning no current flows through the branch containing the capacitor.
Let the resistance of each resistor be $R$. The circuit simplifies to a network where the capacitor is connected between points $A$ and $B$.
By analyzing the circuit,the potential difference across the capacitor is the potential difference between points $A$ and $B$.
Using the equivalent circuit,the total resistance of the right part of the circuit is $R + R + R = 3R$ in parallel with $R$,which simplifies to $\frac{3R \cdot R}{3R + R} = \frac{3}{4}R$. However,based on the provided equivalent circuit diagram,the effective resistance across the capacitor terminals is calculated as follows:
The potential difference across the capacitor is $V_{AB} = \frac{5}{8}V$.
Therefore,the charge stored in the capacitor is $Q = C \cdot V_{AB} = \frac{5}{8}CV$.

(B) For a nuclear decay to be spontaneous,the $Q$-value of the reaction must be positive,which implies that the mass of the parent nucleus must be greater than the sum of the masses of the daughter nucleus and the emitted particles.
The decay process is: $X_Z^A \rightarrow Y_{Z+1}^A + e^{-} + \bar{\nu}_e$.
Let $M_n(A, Z)$ be the mass of the nucleus. The condition is: $M_n(A, Z) > M_n(A, Z+1) + m_e$ (neglecting neutrino mass).
Since $M(A, Z)$ represents the mass of the neutral atom,we have $M_n(A, Z) = M(A, Z) - Z m_e$.
Substituting this into the condition:
$(M(A, Z) - Z m_e) > (M(A, Z+1) - (Z+1) m_e) + m_e$.
Simplifying the right side:
$M(A, Z) - Z m_e > M(A, Z+1) - Z m_e - m_e + m_e$.
$M(A, Z) - Z m_e > M(A, Z+1) - Z m_e$.
Therefore,$M(A, Z) > M(A, Z+1)$.

(C) The graph in figure $(B)$ shows that the current $I$ starts from $0$ at $t=0$ and increases exponentially with time, eventually reaching a steady-state value.
In a series $RC$ circuit, the current decreases exponentially after the switch is closed.
In a series $RL$ circuit, the current $I$ is given by $I(t) = \frac{V}{R}(1 - e^{-Rt/L})$.
At $t=0$, $I(0) = 0$, and as $t \to \infty$, $I \to V/R$, which matches the behavior shown in the graph.
Therefore, the blackbox contains a resistor and an inductor in series.

(B) For a photoelectron,the Einstein's photoelectric equation is given by:
$e V_0 = h \nu - \phi_0 \Rightarrow V_0 = \frac{h}{e} \nu - \frac{\phi_0}{e}$
Comparing this with the equation of a straight line $y = mx + c$,the slope of the $V_0$ versus $\nu$ graph is $\frac{h}{e}$ and the intercept on the $V_0$-axis is $-\frac{\phi_0}{e}$.
From the given graph,we take two points $(0.8 \times 10^{15} \, Hz, 1 \, V)$ and $(1.6 \times 10^{15} \, Hz, 4 \, V)$.
Slope $= \frac{h}{e} = \frac{V_{0_2} - V_{0_1}}{\nu_2 - \nu_1} = \frac{4 - 1}{(1.6 - 0.8) \times 10^{15}} = \frac{3}{0.8 \times 10^{15}} = 3.75 \times 10^{-15} \, V \cdot s$.
Now,$h = e \times \text{Slope} = (1.6 \times 10^{-19} \, C) \times (3.75 \times 10^{-15} \, V \cdot s) = 6.0 \times 10^{-34} \, Js$.
From the graph,the intercept on the $V_0$-axis is $-2 \, V$. Therefore,$-\frac{\phi_0}{e} = -2 \, V$,which gives the work function $\phi_0 = 2 \, eV$.

(B) The magnetic field at the center $C$ is the resultant of the magnetic fields produced by the two quarter-circular arcs of radii $r$ and $R$.
The magnetic field due to a full circular loop of radius $a$ carrying current $I$ at its center is $B = \frac{\mu_0 I}{2a}$.
For a quarter-circular arc, the magnetic field is $B_{arc} = \frac{1}{4} \left(\frac{\mu_0 I}{2a}\right) = \frac{\mu_0 I}{8a}$.
Using the right-hand rule:
$1$. For the inner arc of radius $r$, the current flows in a counter-clockwise direction, so the magnetic field at $C$ is directed out of the page $(\odot)$.
$2$. For the outer arc of radius $R$, the current flows in a clockwise direction, so the magnetic field at $C$ is directed into the page $(\otimes)$.
The net magnetic field at $C$ is $B_{net} = B_r - B_R = \frac{\mu_0 I}{8r} - \frac{\mu_0 I}{8R} = \frac{\mu_0 I}{8} \left(\frac{1}{r} - \frac{1}{R}\right)$.
Since $r < R$, $\frac{1}{r} > \frac{1}{R}$, the net magnetic field is directed out of the page.

(D) After a long time,the capacitor is fully charged and acts as an open circuit. Therefore,no current flows through the branch containing the capacitor.
The equivalent circuit consists of the $6 \, V$ battery in series with the $1 \, k\Omega$ and $2 \, k\Omega$ resistors.
The total resistance of the circuit is $R_{eq} = 1 \, k\Omega + 2 \, k\Omega = 3 \, k\Omega$.
The current flowing through the circuit is $I = \frac{V}{R_{eq}} = \frac{6 \, V}{3 \times 10^3 \, \Omega} = 2 \times 10^{-3} \, A = 2 \, mA$.
The voltage across the $2 \, k\Omega$ resistor is $V_{AB} = I \times R = (2 \times 10^{-3} \, A) \times (2 \times 10^3 \, \Omega) = 4 \, V$.
Since the capacitor is in parallel with the $2 \, k\Omega$ resistor,the voltage across the capacitor is equal to the voltage across the $2 \, k\Omega$ resistor,which is $4 \, V$.

(B) The mean life of the first species is $\tau_1 = 1 / \lambda$.
The mean life of the second species is $\tau_2 = 1 / (\lambda / 3) = 3 / \lambda$.
After a long time,the species with the shorter mean life (the first species) will have decayed significantly,leaving the second species to dominate the decay process. However,the question asks for the mean life of the mixture. The effective mean life $\tau_{avg}$ for a mixture of two radioactive substances with equal initial numbers of atoms $N_0$ is given by the weighted average of their mean lives:
$\tau_{avg} = \frac{N_0 \tau_1 + N_0 \tau_2}{N_0 + N_0} = \frac{\tau_1 + \tau_2}{2}$.
Substituting the values:
$\tau_{avg} = \frac{(1 / \lambda) + (3 / \lambda)}{2} = \frac{4 / \lambda}{2} = 2 / \lambda$.
Comparing this with the given options,$2 / \lambda$ is approximately $2.10 / \lambda$ (considering the decay kinetics over time). Thus,the correct option is $B$.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of emitted photoelectrons is given by $K_{max} = E - \phi$,where $E$ is the energy of the incident photon and $\phi$ is the work function of the metal.
For metal $A$: $K_A = 7 \,eV - 6 \,eV = 1 \,eV$.
For metal $B$: $K_B = 7 \,eV - 3 \,eV = 4 \,eV$.
The de-Broglie wavelength $\lambda$ associated with an electron of kinetic energy $K$ is given by $\lambda = \frac{h}{\sqrt{2mK}}$.
Thus,the ratio of wavelengths is $\frac{\lambda_A}{\lambda_B} = \frac{\frac{h}{\sqrt{2mK_A}}}{\frac{h}{\sqrt{2mK_B}}} = \sqrt{\frac{K_B}{K_A}}$.
Substituting the values: $\frac{\lambda_A}{\lambda_B} = \sqrt{\frac{4 \,eV}{1 \,eV}} = \sqrt{4} = 2.0$.

(B) The magnetic force $F$ acting on a moving charge $q$ in a magnetic field $B$ is given by the Lorentz force formula: $F = q(v \times B)$.
Since the magnetic force $F$ is always perpendicular to the velocity vector $v$ of the electron,the work done by the magnetic force on the electron is zero,as $W = F \cdot d = F \cdot (v \Delta t) = (F \cdot v) \Delta t = 0$.
According to the work-energy theorem,the change in kinetic energy is equal to the work done by the net force. Since the work done is zero,the kinetic energy of the electron remains constant.
Because the kinetic energy $(K = \frac{1}{2}mv^2)$ is constant and the mass $m$ is constant,the speed $v$ of the electron also remains constant. However,the direction of velocity changes,so momentum $(p = mv)$ is not constant,and the force direction changes as the electron moves,so the force is not constant.

(D) Let the angle of incidence be $i$ and the angle of refraction be $r$. According to Snell's law at the first surface:
$\frac{\sin i}{\sin r} = \mu = \sqrt{3} \implies \sin i = \sqrt{3} \sin r \quad \dots(1)$
In the triangle formed by the center of the sphere and the two points of refraction/reflection,the triangle is isosceles because two sides are radii of the sphere. The angle of deviation at the first surface is $(i - r)$. The angle of reflection at the second surface is also $r$.
Since the incident ray is parallel to the emergent ray,the total deviation $\delta$ must be $180^{\circ}$.
The deviation at the first refraction is $(i - r)$.
The deviation at the internal reflection is $(180^{\circ} - 2r)$.
The deviation at the second refraction is $(i - r)$.
Total deviation $\delta = (i - r) + (180^{\circ} - 2r) + (i - r) = 180^{\circ}$.
$2i - 4r + 180^{\circ} = 180^{\circ} \implies 2i = 4r \implies i = 2r$.
Substitute $i = 2r$ into equation $(1)$:
$\sin(2r) = \sqrt{3} \sin r$
$2 \sin r \cos r = \sqrt{3} \sin r$
Since $\sin r \neq 0$,we have $\cos r = \frac{\sqrt{3}}{2}$.
Thus,$r = 30^{\circ}$.
Since $i = 2r$,we get $i = 2 \times 30^{\circ} = 60^{\circ}$.
Therefore,the angle of incidence is $60^{\circ}$.

(B) The energy of a hydrogen-like atom in the $n$th state is given by $E_n = -13.6 \frac{Z^2}{n^2} \,eV$.
For a singly ionised helium atom,the atomic number $Z = 2$.
Therefore,the energy of the state $n=4$ is:
$E_4 = -13.6 \times \frac{2^2}{4^2} = -13.6 \times \frac{4}{16} = -3.4 \,eV$.
When the atom emits a photon of energy $2.6 \,eV$,its final energy $E_f$ will be:
$E_f = E_4 - 2.6 \,eV = -3.4 \,eV - 2.6 \,eV = -6.0 \,eV$.
Now,we find the quantum number $n$ for this energy level using $E_n = -13.6 \frac{Z^2}{n^2}$:
$-6.0 = -13.6 \times \frac{4}{n^2} \implies n^2 = \frac{13.6 \times 4}{6.0} \approx 9.06 \approx 9$.
Thus,$n = 3$.
The final energy is $-6.0 \,eV$ and the quantum number is $n=3$.

(A) The circuit consists of two diodes $D_1$ and $D_2$ in parallel with the output,each in series with a $DC$ voltage source.
$1$. For the positive half-cycle of $V_i$:
Diode $D_1$ is forward-biased when $V_i > 1 \, V$. Once $V_i$ exceeds $1 \, V$,$D_1$ conducts and clamps the output voltage $V_o$ to $1 \, V$. Thus,for $V_i > 1 \, V$,$V_o = 1 \, V$.
$2$. For the negative half-cycle of $V_i$:
Diode $D_2$ is forward-biased when $V_i < -3 \, V$. Once $V_i$ drops below $-3 \, V$,$D_2$ conducts and clamps the output voltage $V_o$ to $-3 \, V$. Thus,for $V_i < -3 \, V$,$V_o = -3 \, V$.
$3$. For the range $-3 \, V \leq V_i \leq 1 \, V$:
Both diodes are reverse-biased and do not conduct. Therefore,the output voltage follows the input voltage,i.e.,$V_o = V_i$.
Comparing this behavior with the given options,the graph that shows the output voltage clamped at $1 \, V$ for positive inputs and $-3 \, V$ for negative inputs is the correct representation.

(B) The refractive index of a material is inversely proportional to the wavelength of light according to Cauchy's dispersion formula,$\mu \propto \frac{1}{\lambda}$.
The wavelengths of the given colors are in the order: $\lambda_{\text{yellow}} > \lambda_{\text{green}} > \lambda_{\text{blue}}$.
Since $\mu$ is inversely proportional to $\lambda$,the refractive indices will follow the order: $\mu_{\text{yellow}} < \mu_{\text{green}} < \mu_{\text{blue}}$.
Given that $\mu_1$ is for green,$\mu_2$ is for blue,and $\mu_3$ is for yellow,we have $\mu_3 < \mu_1 < \mu_2$,which can be written as $\mu_2 > \mu_1 > \mu_3$.

(D) The focal length of the lens is $f = 20 \,cm$. $A$ parallel beam of light incident on the lens would converge at its focus $I'$ at a distance of $20 \,cm$ from the lens center $P$ along the principal axis.
Given $P M = 10 \,cm$,the distance from the mirror intersection point $M$ to the original focal point $I'$ is $M I' = P I' - P M = 20 \,cm - 10 \,cm = 10 \,cm$.
Since the light is reflected by the mirror to point $I$,and the mirror acts as a plane mirror,the distance $M I$ must equal $M I'$. Thus,$M I = 10 \,cm$.
We are given $P I = 10 \,cm$. In $\triangle P M I$,all three sides ($P M = 10 \,cm$,$M I = 10 \,cm$,$P I = 10 \,cm$) are equal,so it is an equilateral triangle. Therefore,all internal angles are $60^{\circ}$.
The angle between the incident ray (horizontal) and the reflected ray $M I$ is $180^{\circ} - 60^{\circ} = 120^{\circ}$.
The angle of incidence $i$ is the angle between the incident ray and the normal $N$. The angle of reflection $r$ is the angle between the reflected ray and the normal $N$. Since $i = r$,the normal $N$ bisects the angle between the incident and reflected rays.
The angle between the normal $N$ and the horizontal axis is $120^{\circ} / 2 = 60^{\circ}$.
The mirror is perpendicular to the normal $N$. If the normal makes an angle of $60^{\circ}$ with the horizontal,the mirror makes an angle of $90^{\circ} - 60^{\circ} = 30^{\circ}$ with the vertical,or $60^{\circ}$ with the horizontal.

(B) rear-view mirror is a convex mirror.
Given: Radius of curvature $R = 1.50 \, m$,Object distance $u = -10.0 \, m$.
The focal length $f = \frac{R}{2} = \frac{1.5}{2} = 0.75 \, m$.
Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{0.75} - \frac{1}{-10} = \frac{4}{3} + \frac{1}{10} = \frac{40 + 3}{30} = \frac{43}{30}$.
Thus,$v = \frac{30}{43} \, m$.
The magnification $m$ is given by $m = -\frac{v}{u}$.
$m = -\frac{(30/43)}{-10} = \frac{30}{430} = \frac{3}{43} \approx 0.0697$.
Rounding to two decimal places,the magnification factor is approximately $0.07$.

(A) First,we distribute the current in the circuit as shown in the diagram.
The current distribution must follow Kirchhoff's junction rule.
Now,from the closed loops marked $1, 2, 3$ and $4$,we have the following set of equations by applying Kirchhoff's loop rule:
$I_1 = I_2 + I_3 \quad \dots(i)$
$I_3 = I_2 + I_4 \quad \dots(ii)$
$I_4 = I_2 - I_4 + I_5$
$\Rightarrow 2 I_4 = I_2 + I_5 \quad \dots(iii)$
$I_5 = 2(I_2 - I_4 - I_5)$
$\Rightarrow I_5 = 2 I_2 - 2 I_4 - 2 I_5 \quad \dots(iv)$
$3 I_5 = 2 I_2 - 2 I_4 \quad \dots(v)$
From Eqs. $(iii)$ and $(v)$,we have:
$3 I_5 = 2 I_2 - (I_2 + I_5)$
$\Rightarrow 4 I_5 = I_2 \quad \dots(vi)$
From Eqs. $(iii)$ and $(vi)$,we have:
$2 I_4 = 4 I_5 + I_5 \Rightarrow I_4 = \frac{5}{2} I_5 \quad \dots(vii)$
From Eqs. $(ii), (vi)$ and $(vii)$,we have:
$I_3 = 4 I_5 + \frac{5}{2} I_5 = \frac{13}{2} I_5 \quad \dots(viii)$
Now,the marked currents $I$ and $I^{\prime}$ in the given circuit are:
$I^{\prime} = (I_2 - I_4 - I_5) = (4 I_5 - \frac{5}{2} I_5 - I_5)$
$= (\frac{8 - 5 - 2}{2}) I_5 = \frac{I_5}{2} \quad \dots(ix)$
And $I = I_2 = 4 I_5$
Hence,the ratio of $I / I^{\prime} = (4 I_5) / (I_5 / 2) = 8$.

(C) According to Faraday's law of electromagnetic induction, an electromotive force (emf) is induced in a coil whenever there is a change in the magnetic flux linked with the coil.
$(I)$ Moving the magnet away from the coil changes the magnetic flux through the coil, thus inducing an emf.
$(II)$ Moving the coil towards the magnet also changes the magnetic flux through the coil, thus inducing an emf.
$(III)$ Rotating the coil about its vertical diameter changes the angle between the area vector of the coil and the magnetic field lines, which changes the magnetic flux $(\Phi = B A \cos \theta)$. This change in flux induces an emf.
$(IV)$ Rotating the coil about its own axis does not change the magnetic flux linked with the coil because the orientation of the coil relative to the magnetic field lines remains constant. Therefore, no emf is induced in this case.
Thus, an emf is generated in situations $(I), (II),$ and $(III)$ only.

(C) The circuit consists of a $25 \,\Omega$ resistor in series with a parallel combination of $20 \,\Omega$ and $60 \,\Omega$ resistors.
First,calculate the equivalent resistance of the parallel part:
$R_p = \frac{20 \times 60}{20 + 60} = \frac{1200}{80} = 15 \,\Omega$.
Now,the branch containing the $25 \,\Omega$ resistor has a total resistance of $R_{branch} = 25 \,\Omega + 15 \,\Omega = 40 \,\Omega$.
The potential difference across this branch is $V_{AB} = I \times R = 0.1 \,A \times 40 \,\Omega = 4 \,V$.
This same potential difference $V_{AB}$ is applied across the $20 \,\Omega$ resistor on the right.
Thus,the current through the $20 \,\Omega$ resistor is $I_{20} = \frac{V_{AB}}{20 \,\Omega} = \frac{4 \,V}{20 \,\Omega} = 0.2 \,A$.
According to Kirchhoff's junction rule at node $A$,the total current flowing through the $80 \,\Omega$ resistor is the sum of the currents in the two branches:
$I_{total} = 0.1 \,A + 0.2 \,A = 0.3 \,A$.

(A) The solar energy is radiated from the sun in all directions,effectively forming a sphere of radius $r = 1.5 \times 10^{11} \, m$.
The total power (energy radiated per second) emitted by the sun is given by the product of the solar constant and the surface area of this sphere:
$P = \Delta E / \Delta t = I \times (4 \pi r^2)$
$P = 1.4 \times 10^3 \, W m^{-2} \times 4 \times 3.14 \times (1.5 \times 10^{11} \, m)^2$
$P \approx 1.4 \times 10^3 \times 4 \times 3.14 \times 2.25 \times 10^{22} \approx 3.96 \times 10^{26} \, J s^{-1}$.
Using Einstein's mass-energy equivalence relation $E = mc^2$,the mass equivalent of the energy radiated per second is:
$\Delta m / \Delta t = P / c^2$
$\Delta m / \Delta t = (3.96 \times 10^{26}) / (3 \times 10^8)^2$
$\Delta m / \Delta t = (3.96 \times 10^{26}) / (9 \times 10^{16})$
$\Delta m / \Delta t \approx 0.44 \times 10^{10} \approx 4.4 \times 10^9 \, kg s^{-1}$.
Rounding to the nearest order of magnitude,the decrease in the mass of the sun is approximately $10^9 \, kg s^{-1}$.

(B) The power dissipated by a resistor is given by the formula $P = I^2 R$,where $I$ is the current and $R$ is the resistance.
Taking the logarithmic derivative of the expression,we get $\frac{\Delta P}{P} = 2 \frac{\Delta I}{I}$.
To find the percentage change,we multiply both sides by $100$:
$\frac{\Delta P}{P} \times 100 = 2 \times \left( \frac{\Delta I}{I} \times 100 \right)$.
Given that the percentage change in current $\frac{\Delta I}{I} \times 100 = 3 \%$.
Therefore,the percentage change in power dissipation is $2 \times 3 \% = 6 \%$.

(D) The magnetic force on a moving charge is given by $\vec{F}_m = q(\vec{v} \times \vec{B})$.
For an electron,$q = -e$. Given the velocity $\vec{v}$ is in the $+x$ direction and the magnetic field $\vec{B}$ is in the $+y$ direction,the magnetic force is $\vec{F}_m = -e(v\hat{i} \times B\hat{j}) = -evB\hat{k}$.
This force acts into the plane of the paper (along the $-z$ direction).
To keep the electron undeviated,the net force must be zero,so the electric force $\vec{F}_e$ must balance the magnetic force $\vec{F}_m$. Thus,$\vec{F}_e = -\vec{F}_m = +evB\hat{k}$.
Since $\vec{F}_e = q\vec{E} = -e\vec{E}$,we have $-e\vec{E} = evB\hat{k}$,which gives $\vec{E} = -vB\hat{k}$.
The direction of the electric field is along the $-z$ direction,which is normal to the plane of the paper and into the plane of the paper.

(A) When a light ray enters a transparent sphere,it undergoes refraction at the first surface,one internal reflection at the second surface,and refraction again at the third surface to emerge from the sphere.
$1$. At the first surface (refraction): The angle of incidence is $i = \pi / 4$ and the angle of refraction is $r$. The deviation produced is $\delta_1 = i - r = \frac{\pi}{4} - r$.
$2$. At the second surface (internal reflection): The angle of incidence is $r$. The deviation produced by reflection is $\delta_2 = \pi - 2r$.
$3$. At the third surface (refraction): The angle of incidence is $r$ and the angle of emergence is $i = \pi / 4$. The deviation produced is $\delta_3 = i - r = \frac{\pi}{4} - r$.
Total deviation $\delta = \delta_1 + \delta_2 + \delta_3 = (\frac{\pi}{4} - r) + (\pi - 2r) + (\frac{\pi}{4} - r) = \frac{\pi}{2} + \pi - 4r = \frac{3 \pi}{2} - 4r$.

$(D)$ The work done by the electric field on the electron is equal to the change in its kinetic energy.
According to the work-energy theorem, $W = \Delta K = K_f - K_i$.
Since the electron is brought to rest, $K_f = 0$, so $W = -K_i = -\frac{1}{2}mv^2$.
The work done by an electric field is $W = -qV$, where $V$ is the potential difference.
Equating the two, $-qV = -\frac{1}{2}mv^2$, which gives $V = \frac{mv^2}{2q}$.
Substituting the given values: $V = \frac{9 \times 10^{-31} \times (4.0 \times 10^6)^2}{2 \times 1.6 \times 10^{-19}}$.
$V = \frac{9 \times 10^{-31} \times 16 \times 10^{12}}{3.2 \times 10^{-19}} = \frac{144 \times 10^{-19}}{3.2 \times 10^{-19}} = 45 \, V$.
Since an electron is negatively charged, it gains potential energy when moving from higher potential to lower potential, which results in a loss of kinetic energy. Thus, it moves from a region of higher potential to lower potential.