KVPY 2011 Physics Question Paper with Answer and Solution

(B) The acceleration due to gravity $g$ decreases as we move away from the earth's surface,given by $g(h) = g_0(1 - 2h/R_e)$.
Let $C$ be the centre of mass of the cabin. The acceleration of any object at a height $h$ is $a = g(h)$.
Since $A$ is above $C$ and $B$ is below $C$,their heights are $h_A > h_C > h_B$.
Consequently,the gravitational accelerations satisfy $a_B > a_C > a_A$.
When viewed from the frame of the cabin (which has acceleration $a_C$),the relative accelerations are:
$a_{A,rel} = a_A - a_C < 0$ (meaning $A$ accelerates upward relative to the cabin).
$a_{B,rel} = a_B - a_C > 0$ (meaning $B$ accelerates downward relative to the cabin).
Thus,$A$ moves slowly upward and $B$ moves slowly downward relative to the cabin. Therefore,option $(b)$ is correct.

(B) For the lower plate of mass $m$ to lift off the ground,the upward spring force must be equal to its weight.
Let $k$ be the spring constant. The condition for the lower plate to lift is $kx = mg$,where $x$ is the extension of the spring from its natural length.
So,$x = \frac{mg}{k}$.
Now,consider the energy conservation between the initial compressed state (position $I$) and the final extended state (position $II$) where the lower plate just lifts off.
Let $h$ be the initial compression of the spring when weight $W$ is placed on the upper plate.
The total energy at position $I$ is the potential energy of the compressed spring: $U_I = \frac{1}{2}kh^2$.
The total energy at position $II$ is the potential energy of the extended spring plus the potential energy of the upper plate of mass $m$: $U_{II} = \frac{1}{2}kx^2 + mgh_{total}$,where $h_{total} = h + x$ is the total height change of the upper plate.
By energy conservation: $\frac{1}{2}kh^2 = mg(h+x) + \frac{1}{2}kx^2$.
Substituting $x = \frac{mg}{k}$:
$\frac{1}{2}kh^2 = mgh + \frac{m^2g^2}{k} + \frac{1}{2}k(\frac{mg}{k})^2 = mgh + \frac{m^2g^2}{k} + \frac{m^2g^2}{2k} = mgh + \frac{3m^2g^2}{2k}$.
Multiplying by $2/k$: $h^2 - \frac{2mgh}{k} - \frac{3m^2g^2}{k^2} = 0$.
Solving for $h$: $h = \frac{\frac{2mg}{k} + \sqrt{(\frac{2mg}{k})^2 + 4(\frac{3m^2g^2}{k^2})}}{2} = \frac{\frac{2mg}{k} + \sqrt{\frac{16m^2g^2}{k^2}}}{2} = \frac{\frac{2mg}{k} + \frac{4mg}{k}}{2} = \frac{3mg}{k}$.
At equilibrium in position $I$,the spring force balances the weight of the upper plate and the weight $W$: $kh = mg + W$.
Substituting $h = \frac{3mg}{k}$: $k(\frac{3mg}{k}) = mg + W \Rightarrow 3mg = mg + W \Rightarrow W = 2mg$.
Thus,the weight $W$ must be just more than $2mg$.

(A) For a simple pendulum undergoing simple harmonic motion,the displacement $x$ is given by $x = A \sin(\omega t)$.
The velocity is $v = \frac{dx}{dt} = A\omega \cos(\omega t)$.
The tangential acceleration is $a = \frac{dv}{dt} = -A\omega^2 \sin(\omega t)$.
From these equations,we have:
$\frac{v}{A\omega} = \cos(\omega t)$ and $\frac{a}{-A\omega^2} = \sin(\omega t)$.
Using the identity $\sin^2(\theta) + \cos^2(\theta) = 1$,we get:
$\left(\frac{v}{A\omega}\right)^2 + \left(\frac{a}{-A\omega^2}\right)^2 = 1$
$\frac{v^2}{A^2\omega^2} + \frac{a^2}{A^2\omega^4} = 1$
This is the equation of an ellipse in the $v-a$ plane,where $v$ is on the horizontal axis and $a$ is on the vertical axis. Thus,the correct graph is $I$.

(A) The expansion occurs in a vacuum,so the work done by the expanding gas is $\Delta W = 0$.
The container is perfectly insulated,so the heat exchanged with the surroundings is $\Delta Q = 0$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Substituting the values,we get $0 = \Delta U + 0$,which implies $\Delta U = 0$.
Therefore,$U_i = U_f$,meaning the initial and final internal energies of the gas are equal.
Since the gas is not ideal,the expansion results in an increase in the intermolecular potential energy. Because the total internal energy remains constant,this increase in potential energy leads to a decrease in the kinetic energy of the molecules,which consequently causes the temperature of the gas to decrease.

(B) Let $n$ be the total number of moles of gas in both bulbs initially. Since the bulbs have identical volumes $V$ and are at the same temperature $T$,the number of moles in each bulb is $n_1 = n_2 = n / 2$.
When bulb $2$ is heated to $2 T$ and bulb $1$ is at $T$,the pressure $p$ in both bulbs must be equal for the system to be in equilibrium.
Using the ideal gas law $p V = n R T$,the number of moles in bulb $1$ $(n_1')$ and bulb $2$ $(n_2')$ are:
$n_1' = \frac{p V}{R T}$
$n_2' = \frac{p V}{R (2 T)} = \frac{p V}{2 R T}$
Since the total number of moles is conserved,$n_1' + n_2' = n = n_1 + n_2 = n / 2 + n / 2 = n$.
$\frac{p V}{R T} + \frac{p V}{2 R T} = n \Rightarrow \frac{3 p V}{2 R T} = n \Rightarrow \frac{p V}{R T} = \frac{2 n}{3}$.
Thus,the final number of moles in bulb $2$ is $n_2' = \frac{1}{2} \times \frac{p V}{R T} = \frac{1}{2} \times \frac{2 n}{3} = \frac{n}{3}$.
The ratio of the final mass (or final moles) in bulb $2$ to the initial mass (or initial moles) in bulb $2$ is:
Ratio $= \frac{n_2'}{n_2} = \frac{n / 3}{n / 2} = \frac{2}{3}$.

(D) Let the junction temperature be $T^{\circ} C$.
In the steady state,the rate of heat flow through the copper rod must be equal to the rate of heat flow through the steel rod,assuming no heat loss from the sides.
The rate of heat flow $H$ is given by $H = \frac{KA(T_1 - T_2)}{l}$.
Equating the heat flow for both rods:
$\frac{K_{\text{copper}} A (100 - T)}{l} = \frac{K_{\text{steel}} A (T - 0)}{l}$
Since the length $l$ and cross-sectional area $A$ are the same for both rods,they cancel out:
$K_{\text{copper}} (100 - T) = K_{\text{steel}} T$
Substituting the given values $K_{\text{copper}} = 385$ and $K_{\text{steel}} = 50$:
$385(100 - T) = 50T$
$38500 - 385T = 50T$
$38500 = 435T$
$T = \frac{38500}{435} \approx 88.5^{\circ} C$.
Rounding to the nearest integer,the junction temperature is $88^{\circ} C$.

(B) The compression of air in a compressor is an adiabatic process.
Using the adiabatic relation $P^{1-\gamma} T^{\gamma} = \text{constant}$, we have $P_{\text{in}}^{1-\gamma} T_{\text{in}}^{\gamma} = P_{\text{out}}^{1-\gamma} T_{\text{out}}^{\gamma}$.
Given $P_{\text{in}} = 0.28 \,atm$, $T_{\text{in}} = -40^{\circ} C = 233 \,K$, $P_{\text{out}} = 1 \,atm$, and $\gamma = 1.4 = 7/5$.
Substituting the values: $(0.28)^{1-1.4} (233)^{1.4} = (1)^{1-1.4} (T_{\text{out}})^{1.4}$.
$(0.28)^{-0.4} (233)^{1.4} = (T_{\text{out}})^{1.4}$.
$T_{\text{out}} = 233 \times (0.28)^{-0.4/1.4} = 233 \times (0.28)^{-2/7}$.
Calculating this, $T_{\text{out}} \approx 233 \times 1.48 \approx 345 \,K$.
Since $345 \,K$ is approximately $72^{\circ} C$, which is much higher than the required cabin temperature of $25^{\circ} C$, an air-conditioner is needed to cool the air.

(B) When the observer is stationary and the source is moving towards the observer,the observed frequency is given by the Doppler effect formula:
$f_1 = f_0 \left( \frac{v}{v - u} \right)$
where $v$ is the speed of sound and $u$ is the speed of the source.
When the source is stationary and the observer is moving towards the source,the measured frequency is:
$f_2 = f_0 \left( \frac{v + u}{v} \right)$
To compare $f_1$ and $f_2$,we look at the ratio:
$f_1 = f_0 \left( 1 - \frac{u}{v} \right)^{-1} \approx f_0 \left( 1 + \frac{u}{v} + \frac{u^2}{v^2} + \dots \right)$
$f_2 = f_0 \left( 1 + \frac{u}{v} \right)$
Comparing the two,$f_1 = f_0 \left( \frac{v}{v-u} \right)$ and $f_2 = f_0 \left( \frac{v+u}{v} \right) = f_0 \left( 1 + \frac{u}{v} \right)$.
Since $\frac{v}{v-u} > 1 + \frac{u}{v}$ (because $\frac{v^2}{v(v-u)} > \frac{v^2-u^2}{v(v-u)}$),it follows that $f_1 > f_2$.

(C) For a pipe open at both ends,the fundamental frequency is given by $f = \frac{v}{2L}$,where $v$ is the speed of sound and $L$ is the length of the pipe.
Since the length $L$ of the pipe remains constant,the frequency $f$ is directly proportional to the speed of sound $v$ $(f \propto v)$.
Let $v_1$ and $f_1$ be the speed of sound and frequency at $23^{\circ} C$,and $v_2$ and $f_2$ be the speed of sound and frequency on a hot day.
Given $f_1 = 450 \,Hz$ and $v_2 = v_1 + 0.04 v_1 = 1.04 v_1$.
Using the proportionality $\frac{f_2}{f_1} = \frac{v_2}{v_1}$,we get:
$f_2 = f_1 \times \frac{v_2}{v_1} = 450 \times 1.04 = 468 \,Hz$.

(B) In this collision,the electron transfers a portion of its kinetic energy to the molecule to excite it from the ground state to a metastable excited state.
Since some kinetic energy is converted into the internal potential energy of the molecule,the total kinetic energy of the system decreases. Therefore,the collision is inelastic,and $K^{\prime} < K$.
According to the law of conservation of linear momentum,the total momentum of an isolated system remains constant regardless of whether the collision is elastic or inelastic.
Thus,the total initial momentum $p$ must be equal to the total final momentum $p^{\prime}$,so $p = p^{\prime}$.
Combining these,we get $K^{\prime} < K$ and $p = p^{\prime}$.

(A) The radiative power $P$ of a spherical black body is given by the Stefan-Boltzmann law:
$P = \sigma A T^4$
Since the surface area $A$ of a sphere is $4 \pi R^2$,we have:
$P = \sigma (4 \pi R^2) T^4$
This implies $P \propto R^2 T^4$.
Let the new radius be $R' = 2R$ and the new temperature be $T' = T/2$.
The new radiative power $P'$ is given by:
$P' \propto (R')^2 (T')^4$
$P' \propto (2R)^2 (T/2)^4$
$P' \propto (4 R^2) (T^4 / 16)$
$P' \propto \frac{4}{16} R^2 T^4$
$P' = \frac{1}{4} P$
Therefore,the radiative power becomes $P/4$.

(B) The dimension of resistance $R$ is given by $[R] = [ML^2T^{-3}A^{-2}]$.
Planck's constant $h$ has dimensions $[h] = [ML^2T^{-1}]$.
Electric charge $e$ has dimensions $[e] = [AT]$.
Let the dimension of $R_H$ be proportional to $h^a e^b$.
$[ML^2T^{-3}A^{-2}] = [ML^2T^{-1}]^a [AT]^b = [M^a L^{2a} T^{-a+b} A^b]$.
Comparing the powers of $M, L, T,$ and $A$ on both sides:
For $M$: $a = 1$.
For $A$: $b = -2$.
Substituting these values,we get the dimension of $R_H$ as $[h^1 e^{-2}] = [h/e^2]$.
Thus,the dimension of $R_H$ is the same as $\frac{h}{e^2}$.

(A) Precision refers to the closeness of various measurements for the same quantity. It is determined by the spread or dispersion of the data points.
$A$ smaller spread (narrower distribution curve) indicates higher precision,as the measurements are more consistent with each other.
Looking at the provided graphs,the distribution curve for case $I$ is the narrowest,meaning the measurements are clustered closely together.
Therefore,the measurement with the highest precision is $I$.

(D) As the ball falls through a vertical height $h = l \sin \theta$ and the string turns by an angle $\theta$ from the horizontal,the velocity $v$ of the ball is obtained by the conservation of mechanical energy:
$\frac{1}{2} m v^2 = m g h = m g l \sin \theta$
$\Rightarrow v^2 = 2 g l \sin \theta$
The radial acceleration $a_r$ is given by:
$a_r = \frac{v^2}{l} = \frac{2 g l \sin \theta}{l} = 2 g \sin \theta$
The tangential acceleration $a_t$ is produced by the component of the gravitational force acting along the tangent to the circular path. At an angle $\theta$ from the horizontal,the angle with the vertical is $(90^\circ - \theta)$. Thus,the component of weight along the tangent is $m g \cos \theta$:
$a_t = \frac{F_t}{m} = g \cos \theta$
The magnitude of the total acceleration $a$ is:
$a = \sqrt{a_r^2 + a_t^2}$
$a = \sqrt{(2 g \sin \theta)^2 + (g \cos \theta)^2}$
$a = \sqrt{4 g^2 \sin^2 \theta + g^2 \cos^2 \theta}$
$a = g \sqrt{4 \sin^2 \theta + (1 - \sin^2 \theta)}$
$a = g \sqrt{3 \sin^2 \theta + 1}$

(D) Given,internal energy $U = a V^3$.
For an ideal gas,$U = \frac{f}{2} n R T$. Since $n = 1$ and assuming a monatomic gas $(f = 3)$,we have $U = \frac{3}{2} R T$.
Equating the two expressions: $\frac{3}{2} R T = a V^3$.
Using the ideal gas law $P V = R T$,we substitute $R T = P V$:
$\frac{3}{2} P V = a V^3 \Rightarrow P = \frac{2 a}{3} V^2$.
The work done $W$ is given by $\int_{V_i}^{V_f} P dV$.
Here,$V_i = V$ and $V_f = 2V$.
$W = \int_{V}^{2V} \frac{2 a}{3} V^2 dV = \frac{2 a}{3} \left[ \frac{V^3}{3} \right]_{V}^{2V} = \frac{2 a}{9} (8V^3 - V^3) = \frac{2 a}{9} (7V^3) = \frac{14 a V^3}{9}$.
Since $a V^3 = \frac{3}{2} R T$ at the initial state,we substitute $a V^3 = \frac{3}{2} R T$:
$W = \frac{14}{9} \times \left( \frac{3}{2} R T \right) = \frac{7}{3} R T$.

(C) The work done in a cyclic process is equal to the area enclosed by the cycle on the $p-V$ diagram. The cycle is $ABCA$.
$1$. Work done in process $AB$ (isobaric expansion):
$W_{AB} = p_A(V_B - V_A) = 200 \times (V_B - 2)$.
Since $BC$ is isothermal,$p_B V_B = p_C V_C$.
$200 \times V_B = 500 \times 2 \Rightarrow V_B = 5 \, m^3$.
$W_{AB} = 200 \times (5 - 2) = 600 \, kJ$.
$2$. Work done in process $BC$ (isothermal compression):
$W_{BC} = \int_{V_B}^{V_C} p \, dV = \int_{5}^{2} \frac{p_B V_B}{V} \, dV = 1000 \ln(2/5) = 1000 \times (-0.916) \approx -916 \, kJ$.
$3$. Work done in process $CA$ (isochoric compression):
$W_{CA} = 0$ (since volume is constant).
Total work done $W = W_{AB} + W_{BC} + W_{CA} = 600 - 916 + 0 = -316 \, kJ$.
Rounding to the nearest option,the work done is approximately $-300 \, kJ$.

(A) The path of the particle is given by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Differentiating with respect to time $t$,we get $\frac{2x}{a^2} \frac{dx}{dt} + \frac{2y}{b^2} \frac{dy}{dt} = 0$,which simplifies to $\frac{x}{a^2} v_x + \frac{y}{b^2} v_y = 0 \dots (i)$.
Differentiating again with respect to time $t$,we get $\frac{1}{a^2} (v_x^2 + x a_x) + \frac{1}{b^2} (v_y^2 + y a_y) = 0 \dots (ii)$.
At point $(0, b)$,we have $x = 0$ and $y = b$. Given $v_x = u$ at this point.
Substituting $x = 0$ into equation $(i)$,we get $\frac{0}{a^2} u + \frac{b}{b^2} v_y = 0$,which implies $v_y = 0$.
Now,substituting $x = 0, y = b, v_x = u, v_y = 0$ into equation $(ii)$,we get $\frac{1}{a^2} (u^2 + 0 \cdot a_x) + \frac{1}{b^2} (0^2 + b \cdot a_y) = 0$.
This simplifies to $\frac{u^2}{a^2} + \frac{b a_y}{b^2} = 0$,or $\frac{u^2}{a^2} + \frac{a_y}{b} = 0$.
Therefore,$a_y = -\frac{b u^2}{a^2}$.

(A) The correct option is $A$.
In a simple pendulum,the total mechanical energy is conserved in the absence of friction. The total energy of the pendulum at its extreme positions is purely potential energy,given by $U = mgh$,where $h$ is the vertical height of the bob from the lowest point of the oscillation.
Since the pendulum starts from point $A$ with zero initial velocity,its total energy is $E = mgh_A$. At the other extreme point $D$,the velocity is again zero,so its total energy is $E = mgh_D$.
By the law of conservation of energy,$mgh_A = mgh_D$,which implies $h_A = h_D$. Since $A$ and $B$ lie on the same horizontal line,$h_A = h_B$. Therefore,$h_D = h_B$,which means the extreme point $D$ must lie on the same horizontal line $A B$.

(A) Since the toy is in equilibrium (it does not move),the net force acting on it must be zero.
The forces acting on the toy are:
$1$. The pushing force $F$ applied by the child.
$2$. The weight of the toy $(W = mg)$ acting downwards.
$3$. The normal force $(N)$ exerted by the ground acting upwards.
$4$. The frictional force $(f)$ exerted by the ground acting horizontally to oppose the motion.
According to Newton's first law,for an object in equilibrium,the vector sum of all external forces acting on it must be zero.
Therefore,$\vec{F} + \vec{W} + \vec{N} + \vec{f} = 0$.
This means the resultant of these four forces is zero.
Option $(a)$ is correct.

(C) Let the first ball reach a maximum height $H = \frac{u^2}{2g}$.
At the instant the first ball reaches height $H$,it starts falling. Let the balls collide at a height $h$ from the ground after time $t$ from the moment the second ball is thrown.
The first ball falls a distance $(H - h)$ in time $t$. Using the equation of motion for the first ball:
$H - h = \frac{1}{2}gt^2 \quad \dots(i)$
The second ball rises to height $h$ in time $t$. Using the equation of motion for the second ball:
$h = ut - \frac{1}{2}gt^2 \quad \dots(ii)$
Adding equations $(i)$ and $(ii)$:
$(H - h) + h = \frac{1}{2}gt^2 + ut - \frac{1}{2}gt^2$
$H = ut \implies t = \frac{H}{u}$
Since $H = \frac{u^2}{2g}$,we have $u = \sqrt{2gH}$. Thus,$t = \frac{H}{\sqrt{2gH}} = \sqrt{\frac{H}{2g}}$.
Substituting $t$ into equation $(ii)$:
$h = u\left(\frac{H}{u}\right) - \frac{1}{2}g\left(\frac{H}{u}\right)^2 = H - \frac{1}{2}g\left(\frac{H^2}{u^2}\right)$
Since $u^2 = 2gH$,we get:
$h = H - \frac{1}{2}g\left(\frac{H^2}{2gH}\right) = H - \frac{H}{4} = \frac{3H}{4}$.

(D) Since there is no external horizontal force acting on the system,the position of the centre of mass of the system remains unchanged.
Let the initial position of the girl be at the origin $(x = 0)$ and the box be at $x = 20 \,m$.
The mass of the girl is $m_1 = 36 \,kg$ and the mass of the box is $m_2 = 9 \,kg$.
The initial position of the centre of mass $(X_{CM})$ is:
$X_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \frac{36 \times 0 + 9 \times 20}{36 + 9} = \frac{180}{45} = 4 \,m$.
Let the girl move a distance $x$ towards the box and the box move a distance $(20 - x)$ towards the girl. When they meet,they will both be at the position of the centre of mass,$X_{CM} = 4 \,m$.
Therefore,the girl has travelled $4 \,m$ from her initial position to reach the centre of mass,and the box has travelled $20 - 4 = 16 \,m$ to reach the same point.
Thus,the girl has travelled $4 \,m$.

(A) When objects are left in a closed room for a long duration (overnight),they exchange heat with the surrounding air until they reach thermal equilibrium.
According to the Zeroth Law of Thermodynamics,if two systems are in thermal equilibrium with a third system,they are in thermal equilibrium with each other.
Since all three objects are in the same room,they will eventually reach the same temperature as the room's ambient temperature.
Therefore,$T_1 = T_2 = T_3$.

(B) The correct answer is $(B)$.
Gas molecules in a room are in constant,random thermal motion. Due to the high speed of these molecules and the relatively small height of a room,the effect of gravity on the density distribution of the air inside the room is negligible.
According to the kinetic theory of gases,pressure is exerted by the collisions of gas molecules with the walls of the container. Since the distribution of gas molecules is uniform throughout the room,the frequency and force of these collisions are essentially the same on the floor,the walls,and the ceiling.
Therefore,the air pressure is nearly the same on the floor,the walls,and the ceiling.

(B) Given:
Input power $(P_{\text{in}})$ = $750 \,W$
Volume of water $(V)$ = $300 \,L$,so mass $(m)$ = $300 \,kg$ (since density of water is $1 \,kg/L$)
Height $(h)$ = $6 \,m$
Time $(t)$ = $1 \,minute = 60 \,s$
Acceleration due to gravity $(g)$ = $10 \,m/s^2$
The useful power output $(P_{\text{out}})$ is the rate of doing work against gravity:
$P_{\text{out}} = \frac{mgh}{t} = \frac{300 \times 10 \times 6}{60} = \frac{18000}{60} = 300 \,W$
The efficiency $(\eta)$ is defined as the ratio of useful power output to input power:
$\eta = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{300}{750} = 0.4$
Converting to percentage:
$\eta = 0.4 \times 100 = 40 \%$

(C) The volume of the room is $V = 5 \,m \times 5 \,m \times 4 \,m = 100 \,m^3$.
At standard temperature and pressure $(STP)$,the pressure $p = 1.013 \times 10^5 \,Pa \approx 10^5 \,Pa$ and temperature $T = 273 \,K$.
The Boltzmann constant $k_B = 1.38 \times 10^{-23} \,J/K$.
Using the ideal gas equation $pV = Nk_BT$,the number of molecules $N$ is given by $N = \frac{pV}{k_BT}$.
Substituting the values: $N = \frac{10^5 \times 100}{1.38 \times 10^{-23} \times 273} \approx \frac{10^7}{3.7674 \times 10^{-21}} \approx 2.65 \times 10^{27}$.
Rounding to the nearest order of magnitude,$N \approx 3 \times 10^{27}$ molecules.

(B) For the equilibrium of the block hung from the spring,the sum of the spring force and the buoyant force must equal the weight of the block.
Let $V$ be the volume of the object and $k$ be the spring constant.
The weight of the object is $W = \rho V g$.
The buoyant force in a liquid of density $\rho_f$ is $F_B = \rho_f V g$.
The spring force is $F_s = kx$.
For equilibrium: $kx + \rho_f V g = \rho V g$.
In the first liquid (density $\rho_1$): $k x_1 + \rho_1 V g = \rho V g \quad \dots(i)$
In the second liquid (density $\rho_2$): $k x_2 + \rho_2 V g = \rho V g \quad \dots(ii)$
From $(i)$,$k = \frac{(\rho - \rho_1) V g}{x_1}$.
From $(ii)$,$k = \frac{(\rho - \rho_2) V g}{x_2}$.
Equating the two expressions for $k$:
$\frac{(\rho - \rho_1) V g}{x_1} = \frac{(\rho - \rho_2) V g}{x_2}$
$(\rho - \rho_1) x_2 = (\rho - \rho_2) x_1$
$\rho x_2 - \rho_1 x_2 = \rho x_1 - \rho_2 x_1$
$\rho (x_2 - x_1) = \rho_1 x_2 - \rho_2 x_1$
$\rho = \frac{\rho_1 x_2 - \rho_2 x_1}{x_2 - x_1}$

(D) According to the work-kinetic energy theorem,the work done by the force is equal to the change in kinetic energy of the body.
$W = \Delta K = K_f - K_i = \frac{1}{2} m (v_f^2 - v_i^2)$
The work done is equal to the area under the force-displacement graph from $x=4 \,m$ to $x=8 \,m$. From the graph,at $x=4 \,m$,$F=1.5 \,N$ and at $x=8 \,m$,$F=3 \,N$.
The area is a trapezoid (or a rectangle plus a triangle) with height $h = (8-4) = 4 \,m$ and parallel sides $F_1 = 1.5 \,N$ and $F_2 = 3 \,N$.
$W = \text{Area} = \frac{1}{2} \times (F_1 + F_2) \times \Delta x = \frac{1}{2} \times (1.5 + 3) \times 4 = 2 \times 4.5 = 9 \,J$.
Given $m = 0.5 \,kg$ and $v_i = 3.16 \,ms^{-1}$.
$9 = \frac{1}{2} \times 0.5 \times (v_f^2 - (3.16)^2)$
$9 = 0.25 \times (v_f^2 - 9.9856)$
$36 = v_f^2 - 9.9856$
$v_f^2 = 45.9856$
$v_f \approx 6.78 \,ms^{-1} \approx 6.8 \,ms^{-1}$.

(B) For a thermally isolated system,the total internal energy remains constant because $\Delta Q = 0$ and $\Delta W = 0$.
Since the initial temperatures of both boxes are the same $(T)$,the final equilibrium temperature $T_f$ will also be $T$.
Using the ideal gas law $PV = nRT$,we can find the number of moles in each box:
$n_1 = \frac{P_1 V_1}{RT} = \frac{1 \times V}{RT} = \frac{V}{RT}$
$n_2 = \frac{P_2 V_2}{RT} = \frac{0.5 \times 4V}{RT} = \frac{2V}{RT}$
When the valve is opened,the total number of moles $n_{total} = n_1 + n_2$ occupies the total volume $V_{total} = V_1 + V_2 = V + 4V = 5V$ at temperature $T$.
$n_{total} = \frac{V}{RT} + \frac{2V}{RT} = \frac{3V}{RT}$
Using the ideal gas law for the final state:
$P_f V_{total} = n_{total} R T$
$P_f (5V) = \left(\frac{3V}{RT}\right) RT$
$P_f (5V) = 3V$
$P_f = \frac{3}{5} \, atm = 0.6 \, atm$.

(C) Let the intensity of the incident light be $I_0$. According to Malus' law,the intensity of light after passing through a polariser is $I = I_{in} \cos^2 \theta$,where $\theta$ is the angle between the incident light's polarisation plane and the polariser's axis.
Since the first polariser's axis is parallel to the plane of polarisation of the incident light,the angle $\theta_1 = 0^{\circ}$. Thus,the intensity after the first polariser is $I_1 = I_0 \cos^2(0^{\circ}) = I_0$.
For the subsequent polarisers,each is rotated by $30^{\circ}$ relative to the previous one. Therefore,the angle between the light's polarisation plane and the next polariser is $30^{\circ}$.
Intensity after the second polariser: $I_2 = I_1 \cos^2(30^{\circ}) = I_0 (\frac{\sqrt{3}}{2})^2 = \frac{3}{4} I_0$.
Intensity after the third polariser: $I_3 = I_2 \cos^2(30^{\circ}) = (\frac{3}{4} I_0) \times \frac{3}{4} = (\frac{3}{4})^2 I_0$.
Intensity after the fourth polariser: $I_4 = I_3 \cos^2(30^{\circ}) = (\frac{3}{4})^2 I_0 \times \frac{3}{4} = (\frac{3}{4})^3 I_0$.
Intensity after the fifth polariser: $I_5 = I_4 \cos^2(30^{\circ}) = (\frac{3}{4})^3 I_0 \times \frac{3}{4} = (\frac{3}{4})^4 I_0$.
Calculating the value: $I_5 = (0.75)^4 I_0 = 0.3164 I_0 \approx 31.6 \% I_0$.
Comparing with the given options,the intensity is closest to $30 \%$ of the incident light.

(B) In Young's double slit experiment,the fringe width (fringe separation) $\beta$ is given by the formula:
$\beta = \frac{\lambda D}{d}$
Since the wavelength $\lambda$ is related to the frequency $f$ by $\lambda = \frac{c}{f}$,where $c$ is the speed of light,we can substitute this into the formula:
$\beta = \frac{cD}{fd}$
Assuming the distance between the slits $d$ and the distance to the screen $D$ remain constant,the fringe width is inversely proportional to the frequency:
$\beta \propto \frac{1}{f}$
Given that the initial fringe separation $\beta_1 = 1.0 \,mm$ at frequency $f_1 = f$,when the frequency is doubled $(f_2 = 2f)$,the new fringe separation $\beta_2$ will be:
$\beta_2 = \beta_1 \times \frac{f_1}{f_2} = 1.0 \,mm \times \frac{f}{2f} = 0.5 \,mm$
Therefore,the new separation of the bright fringes is $0.5 \,mm$.

(A) For a domestic $AC$ supply,the given voltage $220 \,V$ represents the root-mean-square $(V_{\text{rms}})$ value.
The peak voltage $(V_{\text{max}})$ is calculated as $V_{\text{max}} = V_{\text{rms}} \times \sqrt{2} = 220 \sqrt{2} \,V$.
The frequency $f$ is $50 \,cps$ (or $50 \,Hz$). The angular frequency $\omega$ is given by $\omega = 2 \pi f = 2 \pi \times 50 = 100 \pi \,rad/s$.
The instantaneous potential difference $V(t)$ is expressed as $V(t) = V_{\text{max}} \cos(\omega t)$.
Substituting the values,we get $V(t) = 220 \sqrt{2} \cos(100 \pi t) \,V$.

(A) First,simplify the circuit. The resistors $R_3 = 60 \, \Omega$ and $R_4 = 30 \, \Omega$ are in parallel. Their equivalent resistance $R_p$ is given by:
$R_p = \frac{60 \times 30}{60 + 30} = \frac{1800}{90} = 20 \, \Omega$.
This combination is in series with $R_5 = 20 \, \Omega$,so the equivalent resistance of this branch is $R_{branch} = 20 + 20 = 40 \, \Omega$.
Now,this branch is in parallel with $R_2 = 40 \, \Omega$. The equivalent resistance of this parallel part is $R_{eq'} = \frac{40 \times 40}{40 + 40} = 20 \, \Omega$.
Finally,this is in series with $R_1 = 40 \, \Omega$. The total resistance of the circuit is $R_{total} = 40 + 20 = 60 \, \Omega$.
The total current from the battery is $I = \frac{V}{R_{total}} = \frac{3.0}{60} = 0.05 \, A$.
Power dissipated in $R_1$ is $P_1 = I^2 R_1 = (0.05)^2 \times 40 = 0.0025 \times 40 = 0.1 \, W$.
The current $I$ splits equally into the two parallel branches ($R_2$ and the branch containing $R_3, R_4, R_5$) because both have a resistance of $40 \, \Omega$. Thus,$I_2 = I_{branch} = 0.025 \, A$.
Power in $R_2$ is $P_2 = (0.025)^2 \times 40 = 0.000625 \times 40 = 0.025 \, W$.
In the branch with $R_3, R_4, R_5$,the current is $0.025 \, A$. Power in $R_5$ is $P_5 = (0.025)^2 \times 20 = 0.0125 \, W$.
The voltage across the parallel combination of $R_3$ and $R_4$ is $V_p = I_{branch} \times R_p = 0.025 \times 20 = 0.5 \, V$.
Power in $R_3$ is $P_3 = \frac{V_p^2}{R_3} = \frac{0.5^2}{60} = \frac{0.25}{60} \approx 0.0042 \, W$.
Power in $R_4$ is $P_4 = \frac{V_p^2}{R_4} = \frac{0.5^2}{30} = \frac{0.25}{30} \approx 0.0083 \, W$.
Comparing all powers,$P_1 = 0.1 \, W$ is the maximum. Therefore,$R_1$ dissipates the most power.

(D) The correct answer is $IV$.
At $t=0$,when the switch is closed,the current in the circuit is zero because the inductor opposes any instantaneous change in current. Therefore,the potential drop across the resistor $(V_R = iR)$ is zero.
By Kirchhoff's voltage law,the entire source voltage of $10 \, V$ appears across the inductor at $t=0$. Thus,the initial voltage across the inductor is $10 \, V$.
As time passes,the current $i$ in the circuit grows exponentially according to the equation $i(t) = \frac{V}{R} (1 - e^{-Rt/L})$.
The voltage across the inductor is given by $V_L = L \frac{di}{dt} = V e^{-Rt/L}$.
As $t$ increases,the current increases,and the potential drop across the resistor increases,while the potential drop across the inductor decreases exponentially from $10 \, V$ to $0 \, V$.
When the current reaches its maximum steady value,the rate of change of current becomes zero,and the potential drop across the inductor tends to zero. This exponential decay is best represented by graph $IV$.

(A) For an electron and nucleus pair,the potential energy is given by $V(r) = \frac{K(-e)(+Ze)}{r} = -\frac{KZe^2}{r}$.
Since the potential energy is negative and approaches zero as $r$ increases,graph $(3)$ represents the electron.
For a neutron,there is no electrostatic force outside the nucleus $(r > r_0)$. Thus,the potential energy is zero for $r > r_0$. Graph $(1)$ represents the neutron.
For a proton,the electrostatic force is repulsive for $r > r_0$ because both the nucleus and the proton are positively charged. Thus,the potential energy is positive and decreases as $r$ increases. Graph $(2)$ represents the proton.
Therefore,the correct pairing is $(1, n), (2, p^{+}), (3, e^{-})$,which corresponds to option $A$.

(D) The energy of a photon emitted during a transition is given by $E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the wavelength.
For the first three energy levels $(n=1, 2, 3)$,the possible transitions are:
$1$. $n=3 \rightarrow n=1$ (Energy $E_1$,wavelength $\lambda_1$)
$2$. $n=2 \rightarrow n=1$ (Energy $E_2$,wavelength $\lambda_2$)
$3$. $n=3 \rightarrow n=2$ (Energy $E_3$,wavelength $\lambda_3$)
Since energy is inversely proportional to wavelength $(E \propto \frac{1}{\lambda})$,the transition with the largest energy change corresponds to the shortest wavelength. Thus,$\lambda_1$ is the shortest wavelength $(n=3 \rightarrow n=1)$,and $\lambda_3$ is the longest wavelength $(n=3 \rightarrow n=2)$.
From the conservation of energy,the energy of the transition from $n=3$ to $n=1$ is equal to the sum of the energies of the transitions from $n=3$ to $n=2$ and $n=2$ to $n=1$:
$E_{3 \rightarrow 1} = E_{3 \rightarrow 2} + E_{2 \rightarrow 1}$
Substituting $E = \frac{hc}{\lambda}$:
$\frac{hc}{\lambda_1} = \frac{hc}{\lambda_3} + \frac{hc}{\lambda_2}$
Dividing by $hc$ on both sides,we get:
$\frac{1}{\lambda_1} = \frac{1}{\lambda_2} + \frac{1}{\lambda_3}$

(B) For a solid sphere of radius $R$ with a uniform volume charge density $\rho$,the electric field $E$ at a distance $r$ from the centre is given by:
$1$. Inside the sphere $(r < R)$: Using Gauss's Law,$E \cdot 4\pi r^2 = \frac{q_{enclosed}}{\epsilon_0} = \frac{\rho \cdot \frac{4}{3}\pi r^3}{\epsilon_0}$. This simplifies to $E = \frac{\rho r}{3\epsilon_0}$,which means $E \propto r$. Thus,the graph is a straight line passing through the origin.
$2$. Outside the sphere $(r \geq R)$: The sphere behaves like a point charge at its centre. The electric field is $E = \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{r^2}$,which means $E \propto \frac{1}{r^2}$. Thus,the graph is a hyperbola.
At the surface $(r = R)$,the electric field is maximum,with a value of $E_{max} = \frac{kq}{R^2}$.
Comparing these characteristics with the given options,curve $II$ correctly represents this variation.

(C) For an uncharged particle in projectile motion,the range is $L = \frac{u^2 \sin 2\theta}{g} \quad \dots(i)$
For a charged particle,the effective acceleration becomes $g' = g + \frac{qE}{m}$.
Given that the range for a particle with charge $q$ is $L/2$,we have:
$\frac{L}{2} = \frac{u^2 \sin 2\theta}{g + \frac{qE}{m}}$
Substituting $u^2 \sin 2\theta = Lg$ from equation $(i)$:
$\frac{L}{2} = \frac{Lg}{g + \frac{qE}{m}}$
$\Rightarrow g + \frac{qE}{m} = 2g \Rightarrow \frac{qE}{m} = g \quad \dots(ii)$
Now,for a particle of mass $m$ and charge $2q$,the effective acceleration is $g'' = g + \frac{2qE}{m}$.
Using equation $(ii)$,$g'' = g + 2g = 3g$.
The new range $R$ is:
$R = \frac{u^2 \sin 2\theta}{3g} = \frac{1}{3} \left( \frac{u^2 \sin 2\theta}{g} \right) = \frac{L}{3}$.

(B) For the light beam to pass through the material without undergoing total internal reflection at the interface,the condition for the critical angle must be satisfied at the boundary between the material and the glass.
From Snell's Law at the interface,$n_1 \sin i = n_2 \sin r$. For the limiting case of total internal reflection,the angle of refraction $r = 90^{\circ}$.
Here,the refractive index of glass is $n_1 = 1.5$ and the minimum refractive index of the material is $n_2 = 1.2$.
Thus,$\sin i_{\max} = \frac{n_2}{n_1} = \frac{1.2}{1.5} = 0.8$.
Therefore,$i_{\max} = \sin^{-1}(0.8) = 53.1^{\circ}$.

(A) Let the linear charge density of the wire be $\lambda$. The potential difference between two points at distances $r_1$ and $r_2$ from a long charged wire is given by $V_1 - V_2 = \frac{\lambda}{2\pi\varepsilon_0} \ln\left(\frac{r_2}{r_1}\right)$.
Applying the principle of conservation of energy between point $A$ (at distance $l$) and point $B$ (at distance $2l$):
$qV_A + \frac{1}{2}mu^2 = qV_B + \frac{1}{2}m(\sqrt{2}u)^2$
$q(V_A - V_B) = \frac{1}{2}m(2u^2 - u^2) = \frac{1}{2}mu^2$
$q \frac{\lambda}{2\pi\varepsilon_0} \ln\left(\frac{2l}{l}\right) = \frac{1}{2}mu^2$
$q \frac{\lambda}{2\pi\varepsilon_0} \ln(2) = \frac{1}{2}mu^2 \quad \dots(i)$
Now, applying energy conservation between point $A$ (at distance $l$) and point $C$ (at distance $4l$):
$qV_A + \frac{1}{2}mu^2 = qV_C + \frac{1}{2}mv^2$
$\frac{1}{2}mv^2 = q(V_A - V_C) + \frac{1}{2}mu^2$
$\frac{1}{2}mv^2 = q \frac{\lambda}{2\pi\varepsilon_0} \ln\left(\frac{4l}{l}\right) + \frac{1}{2}mu^2$
$\frac{1}{2}mv^2 = q \frac{\lambda}{2\pi\varepsilon_0} \ln(4) + \frac{1}{2}mu^2$
Since $\ln(4) = 2\ln(2)$, we have:
$\frac{1}{2}mv^2 = 2 \left(q \frac{\lambda}{2\pi\varepsilon_0} \ln(2)\right) + \frac{1}{2}mu^2$
Substituting from equation $(i)$:
$\frac{1}{2}mv^2 = 2 \left(\frac{1}{2}mu^2\right) + \frac{1}{2}mu^2 = \frac{3}{2}mu^2$
$v^2 = 3u^2 \Rightarrow v = \sqrt{3}u$.

(B) The magnetic field $B$ produced by the long wire carrying current $I$ is directed into the page (inward) in the region of the loop. As the loop is pulled to the right,it moves into a region where the magnetic field strength decreases,thus the magnetic flux through the loop decreases.
According to Lenz's law,the induced current in the loop will flow in a direction to oppose this decrease in flux,which means it will create its own magnetic field directed into the page. By the right-hand rule,this corresponds to a clockwise induced current.
For the left side of the loop (closest to the wire),the current flows upward,which is in the same direction as the current $I$ in the long wire. Parallel currents attract,so the force on the left side is directed towards the long wire (to the left).
For the right side of the loop,the current flows downward,which is opposite to the current $I$ in the long wire. Anti-parallel currents repel,so the force on the right side is directed away from the long wire (to the right).

(D) Using nodal analysis at the junction point above resistor $R$,let the potential be $V_x$. The current $I$ flowing through $R$ is $I = V_x / R$.
Applying Kirchhoff's Current Law $(KCL)$ at the junction:
$(V_x - V_1) / R_1 + (V_x - V_2) / R_2 + V_x / R = 0$
$V_x (1/R_1 + 1/R_2 + 1/R) = V_1/R_1 + V_2/R_2$
$V_x = \frac{V_1/R_1 + V_2/R_2}{1/R_1 + 1/R_2 + 1/R} = \frac{V_1 R_2 + V_2 R_1}{R_1 R_2 / R_1 + R_1 R_2 / R_2 + R_1 R_2 / R} = \frac{V_1 R_2 + V_2 R_1}{R_2 + R_1 + R_1 R_2 / R}$
Since $I = V_x / R$,we have $I = \frac{V_1 R_2 + V_2 R_1}{R(R_1 + R_2) + R_1 R_2}$.
Case $1$: $V_1 = 2 \,V, V_2 = 0 \,V, I = 3 \,mA \Rightarrow 3 = \frac{2 R_2}{R(R_1 + R_2) + R_1 R_2} \quad \dots(1)$
Case $2$: $V_1 = 0 \,V, V_2 = 4 \,V, I = 4 \,mA \Rightarrow 4 = \frac{4 R_1}{R(R_1 + R_2) + R_1 R_2} \quad \dots(2)$
Dividing $(1)$ by $(2)$: $3/4 = (2 R_2) / (4 R_1) \Rightarrow 3/4 = R_2 / (2 R_1) \Rightarrow R_2 / R_1 = 3/2 \Rightarrow R_2 = 1.5 R_1$.
Substitute $R_2 = 1.5 R_1$ into $(1)$: $3 = \frac{2(1.5 R_1)}{R(R_1 + 1.5 R_1) + R_1(1.5 R_1)} = \frac{3 R_1}{2.5 R R_1 + 1.5 R_1^2} = \frac{3}{2.5 R + 1.5 R_1}$.
So,$2.5 R + 1.5 R_1 = 1$.
Case $3$: $V_1 = 10 \,V, V_2 = 10 \,V$. Then $I = \frac{10 R_2 + 10 R_1}{R(R_1 + R_2) + R_1 R_2} = \frac{10(R_1 + R_2)}{R(R_1 + R_2) + R_1 R_2}$.
Using $R_2 = 1.5 R_1$,$I = \frac{10(2.5 R_1)}{R(2.5 R_1) + 1.5 R_1^2} = \frac{25 R_1}{2.5 R R_1 + 1.5 R_1^2} = \frac{25}{2.5 R + 1.5 R_1}$.
Since $2.5 R + 1.5 R_1 = 1$,$I = 25 / 1 = 25 \,mA$.

(C) To reach point $Q$ in the shortest time,the girl must follow a path that minimizes the total travel time. This is analogous to Fermat's principle in optics,which states that light follows the path that takes the least time.
Since the girl's speed on the beach $(v_1 = 6 \, kmh^{-1})$ is greater than her speed in the sea $(v_2 = 4 \, kmh^{-1})$,she should travel a longer distance on the beach and a shorter distance in the sea to minimize the total time. According to Snell's law,which is derived from Fermat's principle,the path should bend towards the normal when entering a medium with a lower speed.
Comparing the given paths,the path $P C Q$ represents the optimal refraction-like trajectory that minimizes the total time taken to reach $Q$.

(C) For total internal reflection $(TIR)$ to occur at the face $BC$,the angle of incidence $i$ must be greater than or equal to the critical angle $C$.
From the geometry of the isosceles right-angled prism,the light ray enters face $AB$ normally and strikes face $BC$ at an angle of incidence $i = 45^{\circ}$.
For $TIR$ to occur,the condition is $i \geq C$,where $C$ is the critical angle.
Therefore,$45^{\circ} \geq C$.
Taking the sine on both sides,$\sin 45^{\circ} \geq \sin C$.
We know that $\sin C = \frac{1}{\mu}$,where $\mu$ is the refractive index of the prism.
So,$\sin 45^{\circ} \geq \frac{1}{\mu}$.
$\frac{1}{\sqrt{2}} \geq \frac{1}{\mu}$.
$\mu \geq \sqrt{2}$.
Since $\sqrt{2} \approx 1.414$,the minimum value of the refractive index $\mu$ is approximately $1.42$.

(D) When a lens is used to form an image,every point on the lens contributes to the formation of the entire image.
If the upper half of the lens is blackened,the light rays passing through that portion are blocked.
However,the remaining lower half of the lens still allows light rays from all parts of the object to pass through and converge at the same focal point.
Consequently,the complete image is still formed at the same position on the screen.
Since the amount of light reaching the screen is reduced (as half the aperture is blocked),the intensity or brightness of the image decreases.

(C) The material of the rod remains the same,so the resistivity $\rho$ is constant for both rods.
Since the volume of the material remains constant during melting and reshaping,we have:
$V_1 = V_2$
$A_1 L_1 = A_2 L_2$
Given $L_1 = L$ and $L_2 = 2L$,we get:
$A_1 L = A_2 (2L) \Rightarrow A_2 = \frac{A_1}{2}$
Now,using the resistance formula $R = \rho \frac{l}{A}$,the new resistance $R_2$ is:
$R_2 = \rho \frac{L_2}{A_2} = \rho \frac{2L}{A_1 / 2} = 4 \left( \frac{\rho L}{A_1} \right)$
Since $R = \rho \frac{L}{A_1}$,we find:
$R_2 = 4R$

(B) Let the distance between charges $A$ and $B$ be $d$. Let the point where the electric field is zero be at a distance $x$ from charge $A$ on the line joining them.
For the electric field to be zero,the magnitudes of the electric fields due to both charges must be equal,and their directions must be opposite.
$1$. Between $A$ and $B$: The fields due to $+Q$ and $-2 Q$ are in the same direction,so the net field cannot be zero.
$2$. Right of $B$: The field due to $-2 Q$ is always stronger than the field due to $+Q$ because $-2 Q$ is larger in magnitude and closer to any point in this region. Thus,the net field cannot be zero.
$3$. Left of $A$: Let the point be at a distance $x$ from $A$. The electric field due to $+Q$ is $E_A = \frac{kQ}{x^2}$ (directed to the left) and the field due to $-2 Q$ is $E_B = \frac{k(2Q)}{(d+x)^2}$ (directed to the right).
Setting $E_A = E_B$,we get $\frac{kQ}{x^2} = \frac{2kQ}{(d+x)^2}$,which simplifies to $(d+x)^2 = 2x^2$,or $d+x = \sqrt{2}x$. This gives $x = \frac{d}{\sqrt{2}-1}$,which is a finite distance.
Therefore,the electric field is zero at a point to the left of $A$.

(D) To find the current in the portion $PQ$,we apply Kirchhoff's current law $(KCL)$ at the junctions.
$1$. At the junction where the $2 \, A$ and $8 \, A$ currents meet,the total incoming current is $2 \, A + 8 \, A = 10 \, A$. This $10 \, A$ current flows towards the next junction.
$2$. At the junction below $Q$,the total outgoing current is $4 \, A + 2 \, A = 6 \, A$. Since $10 \, A$ enters this junction and $6 \, A$ leaves downwards,the remaining $10 \, A - 6 \, A = 4 \, A$ must flow upwards towards $Q$.
$3$. At junction $Q$,the incoming currents are $3 \, A$ (from the top branch) and $4 \, A$ (from the lower branch). The outgoing current is $1 \, A$ (to the right). Let the current in $PQ$ be $I_{PQ}$ flowing from $Q$ to $P$.
$4$. Applying $KCL$ at junction $Q$: $I_{incoming} = I_{outgoing}$.
$3 \, A + 4 \, A = 1 \, A + I_{PQ}$
$7 \, A = 1 \, A + I_{PQ}$
$I_{PQ} = 6 \, A$.
Since the result is positive,the current $6 \, A$ flows in the assumed direction,i.e.,from $Q$ to $P$.

(A) The decay process is as follows:
$1$. Initial nucleus: $Pb_{82}^{214}$.
$2$. Emission of two electrons ($\beta^-$-decay): Each $\beta^-$-decay increases the atomic number by $1$ and keeps the mass number constant.
$Pb_{82}^{214} \rightarrow A_{84}^{214} + 2_{-1}^{0}e$.
$3$. Emission of an $\alpha$-particle: An $\alpha$-particle $(He_{2}^{4})$ emission decreases the atomic number by $2$ and the mass number by $4$.
$A_{84}^{214} \rightarrow X_{82}^{210} + He_{2}^{4}$.
$4$. The resulting nucleus is $X_{82}^{210}$.
$5$. Number of protons $(Z)$ = $82$.
$6$. Number of neutrons $(N)$ = $A - Z = 210 - 82 = 128$.
Therefore,the resulting nucleus has $82$ protons and $128$ neutrons.

(C) The ray from the bottom centre $C$ travels to the top edge and then refracts into the air towards the observer.
Using Snell's law at the water-air interface: $n_1 \cdot \sin i = n_2 \cdot \sin r$.
Here,the angle of incidence $i$ in water is the angle the ray makes with the normal. Since the ray makes an angle $\theta$ with the horizontal,the angle with the vertical normal is $i = 90^{\circ} - \theta$.
The angle of refraction $r$ is the angle the ray makes with the normal in air. From the geometry,$\tan r = \frac{x/2}{h} = \frac{x}{2h}$.
Given $\frac{h}{x} = \frac{7}{4}$,we have $\tan r = \frac{1}{2(h/x)} = \frac{1}{2(7/4)} = \frac{1}{3.5} = \frac{2}{7}$.
Thus,$\sin r = \frac{2}{\sqrt{2^2 + 7^2}} = \frac{2}{\sqrt{4 + 49}} = \frac{2}{\sqrt{53}}$.
Applying Snell's law: $n_{water} \cdot \sin i = n_{air} \cdot \sin r$.
$\frac{4}{3} \cdot \sin(90^{\circ} - \theta) = 1 \cdot \sin r$.
$\frac{4}{3} \cdot \cos \theta = \frac{2}{\sqrt{53}}$.
$\cos \theta = \frac{2}{\sqrt{53}} \cdot \frac{3}{4} = \frac{6}{4\sqrt{53}} = \frac{3}{2\sqrt{53}}$.
Wait,re-evaluating the geometry: The ray from $C$ to the edge makes an angle $r$ with the vertical. $\tan r = \frac{x/2}{h} = \frac{x}{2h} = \frac{1}{2(7/4)} = \frac{2}{7}$. Correct.
Snell's law: $1 \cdot \sin(90^{\circ}-\theta) = \frac{4}{3} \sin r \Rightarrow \cos \theta = \frac{4}{3} \cdot \frac{2}{\sqrt{53}} = \frac{8}{3\sqrt{53}}$.

(A) In the given circuit,the $3 \,\Omega$ and $1 \,\Omega$ resistors are in series with a $10 \,V$ battery.
The equivalent resistance of the circuit is $R_{eq} = 3 \,\Omega + 1 \,\Omega = 4 \,\Omega$.
The current flowing through the circuit is $i = \frac{V}{R_{eq}} = \frac{10 \,V}{4 \,\Omega} = 2.5 \,A$.
The power dissipated in the $1 \,\Omega$ resistor is $P = i^2 R = (2.5)^2 \times 1 = 6.25 \,W$.
When the $1 \,\Omega$ resistor is replaced by a $9 \,\Omega$ resistor,the new equivalent resistance becomes $R'_{eq} = 3 \,\Omega + 9 \,\Omega = 12 \,\Omega$.
The new current in the circuit is $i' = \frac{10 \,V}{12 \,\Omega} = \frac{5}{6} \,A$.
The power dissipated in the $9 \,\Omega$ resistor is $P' = (i')^2 \times 9 = \left(\frac{5}{6}\right)^2 \times 9 = \frac{25}{36} \times 9 = \frac{25}{4} \,W = 6.25 \,W$.
Since $P = 6.25 \,W$ and $P' = 6.25 \,W$,we have $P' = P$.