WBJEE 2011 Physics Question Paper with Answer and Solution

(A) The vehicle moves down a frictionless inclined plane with an acceleration $a = g\sin \alpha$.
In the frame of the vehicle,a pseudo force $F_p = ma = mg\sin \alpha$ acts on the bob of the pendulum in the upward direction along the incline.
The effective acceleration $g_{eff}$ is the vector sum of the acceleration due to gravity $\vec{g}$ and the negative of the vehicle's acceleration $-\vec{a}$.
Resolving the components,the component of $g$ perpendicular to the incline is $g\cos \alpha$ and the component of $g$ parallel to the incline is $g\sin \alpha$. The pseudo force cancels the component $g\sin \alpha$.
Thus,the effective acceleration is $g_{eff} = g\cos \alpha$.
The time period of the simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}} = 2\pi \sqrt{\frac{L}{g\cos \alpha}}$.

(B) The gravitational potential energy of an object of mass $m$ at a distance $r$ from the center of the earth is given by $U = -\frac{GMm}{r}$.
At the surface of the earth,$r = R$,so $U_i = -\frac{GMm}{R}$.
At a height $h = R$ above the surface,the distance from the center is $r = R + h = R + R = 2R$.
Thus,the potential energy at height $h$ is $U_f = -\frac{GMm}{2R}$.
The gain in potential energy is $\Delta U = U_f - U_i = -\frac{GMm}{2R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}$.
Since the acceleration due to gravity on the surface is $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this into the expression for $\Delta U$,we get $\Delta U = \frac{(gR^2)m}{2R} = \frac{mgR}{2}$.

(B) The total pressure inside the container is the sum of the partial pressure of dry air $(p)$ and the saturated vapour pressure of water $(\bar{p})$,so $P_{total} = p + \bar{p}$.
When the volume is compressed to half $(V' = V/2)$ at a constant temperature,the partial pressure of the dry air follows Boyle's Law $(P_1V_1 = P_2V_2)$.
Thus,the new pressure of dry air becomes $p' = p \times (V / (V/2)) = 2p$.
Since the temperature remains constant,the saturated vapour pressure of water $(\bar{p})$ remains unchanged because it depends only on temperature.
Therefore,the new total pressure is $P'_{total} = p' + \bar{p} = 2p + \bar{p}$.

(D) Given: Mass $m = 0.25 \,kg$, initial velocity $u = 10 \,m/s$, final velocity $v = -10 \,m/s$ (since it returns in the opposite direction), and time interval $\Delta t = 0.01 \,s$.
Change in momentum $\Delta P = m(v - u) = 0.25 \times (-10 - 10) = 0.25 \times (-20) = -5 \,kg \cdot m/s$.
The magnitude of change in momentum is $|\Delta P| = 5 \,kg \cdot m/s$.
The force exerted on the ball by the bat is $F = \frac{\Delta P}{\Delta t} = \frac{5}{0.01} = 500 \,N$.
According to Newton's third law, the force exerted on the bat by the ball is equal in magnitude, which is $500 \,N$.

(C) Let $V$ be the volume of the stone,$\sigma$ be the density of the stone,and $\rho$ be the density of water.
The relative density $K$ of the stone is given by $K = \frac{\sigma}{\rho}$,so $\sigma = K\rho$.
The forces acting on the stone as it sinks are:
$1$. Weight of the stone acting downwards: $W = V\sigma g = V(K\rho)g$.
$2$. Buoyant force acting upwards: $F_B = V\rho g$.
According to Newton's second law,the net force $F_{net} = W - F_B = ma$,where $m = V\sigma$ is the mass of the stone.
$V\sigma g - V\rho g = (V\sigma)a$
Dividing by $V\sigma$:
$a = g - \frac{V\rho g}{V\sigma} = g - \frac{\rho}{\sigma}g$
Since $K = \frac{\sigma}{\rho}$,we have $\frac{\rho}{\sigma} = \frac{1}{K}$.
Therefore,the acceleration $a = g - \frac{g}{K} = g\left(1 - \frac{1}{K}\right)$.

(D) Let $V$ be the volume of the object and $d$ be its density. The apparent weight of the object in a liquid is given by the actual weight minus the buoyant force.
In the first liquid: $m_1 g = V d g - V d_1 g = V g (d - d_1) \implies m_1 = V(d - d_1) \quad (1)$
In the second liquid: $m_2 g = V d g - V d_2 g = V g (d - d_2) \implies m_2 = V(d - d_2) \quad (2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{m_1}{m_2} = \frac{d - d_1}{d - d_2}$
$m_1(d - d_2) = m_2(d - d_1)$
$m_1 d - m_1 d_2 = m_2 d - m_2 d_1$
$d(m_1 - m_2) = m_1 d_2 - m_2 d_1$
$d = \frac{m_1 d_2 - m_2 d_1}{m_1 - m_2}$

(D) Let $V$ be the total volume of the body and $\rho$ be its density.
According to the law of floatation,the weight of the body is equal to the weight of the liquid displaced.
Case $1$: In water (density $\rho_w = 1 \text{ g/cm}^3$):
Volume inside water = $V - 0.4V = 0.6V$.
Weight of body = Weight of water displaced $\implies V \rho g = (0.6V) \rho_w g \implies \rho = 0.6 \rho_w = 0.6 \text{ g/cm}^3$.
Case $2$: In oil (density $\rho_o$):
Volume inside oil = $V - 0.6V = 0.4V$.
Weight of body = Weight of oil displaced $\implies V \rho g = (0.4V) \rho_o g \implies \rho = 0.4 \rho_o$.
Equating the density of the body from both cases: $0.6 \rho_w = 0.4 \rho_o$.
Relative density of oil = $\frac{\rho_o}{\rho_w} = \frac{0.6}{0.4} = 1.5$.

(A) When two soap bubbles coalesce in a vacuum,the total number of moles of air remains constant. Assuming the process is isothermal,we use the ideal gas law $PV = nRT$. Since $n$ is constant and $T$ is constant,$PV$ must be constant.
For a soap bubble,the excess pressure is given by $P_{ex} = \frac{4T}{r}$,where $T$ is the surface tension and $r$ is the radius. In a vacuum,the total pressure inside the bubble is $P = \frac{4T}{r}$.
The volume of a spherical bubble is $V = \frac{4}{3}\pi r^3$.
For the two initial bubbles,$P_1 V_1 = (\frac{4T}{x})(\frac{4}{3}\pi x^3) = \frac{16}{3}\pi T x^2$ and $P_2 V_2 = (\frac{4T}{y})(\frac{4}{3}\pi y^3) = \frac{16}{3}\pi T y^2$.
For the final bubble,$P V = (\frac{4T}{z})(\frac{4}{3}\pi z^3) = \frac{16}{3}\pi T z^2$.
Since the total amount of air is conserved,$P_1 V_1 + P_2 V_2 = PV$.
$\frac{16}{3}\pi T x^2 + \frac{16}{3}\pi T y^2 = \frac{16}{3}\pi T z^2$.
Dividing both sides by $\frac{16}{3}\pi T$,we get $x^2 + y^2 = z^2$,which implies $z = \sqrt{x^2 + y^2}$.

(D) The longitudinal strain is given by $\epsilon_L = 2 \times 10^{-3}$.
Poisson's ratio $\sigma$ is defined as the ratio of lateral strain $\epsilon_d$ to longitudinal strain $\epsilon_L$,i.e.,$\sigma = -\frac{\epsilon_d}{\epsilon_L}$.
Therefore,the lateral strain is $\epsilon_d = -\sigma \epsilon_L = -0.50 \times (2 \times 10^{-3}) = -1 \times 10^{-3}$.
The volumetric strain $\frac{\Delta V}{V}$ is given by the sum of the longitudinal strain and two lateral strains: $\frac{\Delta V}{V} = \epsilon_L + 2\epsilon_d$.
Substituting the values: $\frac{\Delta V}{V} = (2 \times 10^{-3}) + 2(-1 \times 10^{-3}) = 2 \times 10^{-3} - 2 \times 10^{-3} = 0$.
Since the volumetric strain is $0$,the percentage change in volume is $0$.

(C) When a stone is thrown vertically upward with initial velocity $u$,the maximum height reached is given by the formula:
$h = \frac{u^2}{2g}$
From this,we can express the square of the initial velocity as:
$u^2 = 2gh$
When the same stone is thrown horizontally (as a projectile) to achieve the maximum horizontal range $R_{\max}$,the angle of projection should be $\theta = 45^{\circ}$. The formula for maximum horizontal range is:
$R_{\max} = \frac{u^2}{g}$
Substituting the value of $u^2$ from the first equation into the second equation:
$R_{\max} = \frac{2gh}{g}$
$R_{\max} = 2h$
Thus,the maximum horizontal distance the person can throw the stone is $2h$.

(C) The given equation is $x = 4(\cos \pi t + \sin \pi t)$.
We can rewrite this in the form $x = A \sin(\omega t + \phi)$ or $x = A \cos(\omega t + \phi)$.
Multiply and divide by $\sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$.
$x = 4\sqrt{2} \left( \frac{4}{4\sqrt{2}} \cos \pi t + \frac{4}{4\sqrt{2}} \sin \pi t \right)$.
$x = 4\sqrt{2} \left( \frac{1}{\sqrt{2}} \cos \pi t + \frac{1}{\sqrt{2}} \sin \pi t \right)$.
Using the trigonometric identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we can write:
$x = 4\sqrt{2} \sin(\pi t + \frac{\pi}{4})$.
Comparing this with the standard equation $x = A \sin(\omega t + \phi)$,the amplitude $A$ is $4\sqrt{2}$.

(B) For configuration $(a)$, the springs are in series. The equivalent spring constant $k_1$ is given by $\frac{1}{k_1} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k}$, so $k_1 = \frac{k}{2}$.
The time period is $T_1 = 2\pi \sqrt{\frac{m}{k_1}} = 2\pi \sqrt{\frac{2m}{k}} = 2 \,s$.
For configuration $(b)$, the springs are in parallel. The equivalent spring constant $k_2$ is $k_2 = k + k = 2k$.
The time period is $T_2 = 2\pi \sqrt{\frac{m}{k_2}} = 2\pi \sqrt{\frac{m}{2k}}$.
Dividing the two time periods: $\frac{T_1}{T_2} = \frac{2\pi \sqrt{2m/k}}{2\pi \sqrt{m/2k}} = \sqrt{\frac{2m}{k} \cdot \frac{2k}{m}} = \sqrt{4} = 2$.
Therefore, $T_2 = \frac{T_1}{2} = \frac{2 \,s}{2} = 1 \,s$.

(B) The potential energy is given by $V(x) = A(1 - \cos px)$.
The force acting on the particle is $F = -\frac{dV}{dx}$.
$F = -\frac{d}{dx} [A(1 - \cos px)] = -A(p \sin px) = -Ap \sin px$.
For small oscillations,$x$ is very small,so $\sin px \approx px$.
Thus,$F \approx -Ap(px) = -Ap^2 x$.
Comparing this with the restoring force equation $F = -kx$,we get the effective spring constant $k = Ap^2$.
The angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{Ap^2}{m}} = p \sqrt{\frac{A}{m}}$.
The time period $T$ is given by $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{Ap^2}}$.

(C) According to the law of conservation of angular momentum,since no external torque acts on the Earth,the angular momentum remains constant: $L_1 = L_2$.
Since $L = I\omega$,we have $I_1 \omega_1 = I_2 \omega_2$.
The moment of inertia of a solid sphere is $I = \frac{2}{5}MR^2$.
Let $R_1 = R$ and $R_2 = \frac{R}{n}$.
Substituting these into the equation: $\frac{2}{5}MR^2 \left(\frac{2\pi}{T_1}\right) = \frac{2}{5}M\left(\frac{R}{n}\right)^2 \left(\frac{2\pi}{T_2}\right)$.
Simplifying the equation: $R^2 \left(\frac{1}{T_1}\right) = \frac{R^2}{n^2} \left(\frac{1}{T_2}\right)$.
Therefore,$T_2 = \frac{T_1}{n^2}$.
Given the initial duration of the day $T_1 = 24 \text{ hr}$,the new duration is $T_2 = \frac{24}{n^2} \text{ hr}$.

(B) The rate of heat conduction is given by the formula: $\frac{dQ}{dt} = \frac{KA \Delta T}{x}$
Given:
Heat $Q = 1.56 \times 10^5 \ J$
Area $A = 2 \ m^2$
Thickness $x = 12 \ cm = 0.12 \ m$
Time $t = 1 \ hour = 3600 \ s$
Temperature difference $\Delta T = 20^{\circ} C$
Substituting the values:
$\frac{1.56 \times 10^5}{3600} = \frac{K \times 2 \times 20}{0.12}$
$K = \frac{1.56 \times 10^5 \times 0.12}{3600 \times 40}$
$K = \frac{1.56 \times 10^5 \times 12 \times 10^{-2}}{3600 \times 40}$
$K = \frac{1.56 \times 10^3 \times 12}{3600 \times 40} = \frac{1.56 \times 12000}{144000} = \frac{1.56}{12} = 0.13 \ W \ m^{-1} \ K^{-1}$

(C) Given the relationship between the two temperature scales is $\frac{A-42}{110} = \frac{B-72}{220}$.
To find the temperature where both scales have the same reading,we set $A = B = T$.
Substituting $T$ into the equation: $\frac{T-42}{110} = \frac{T-72}{220}$.
Simplifying the equation by dividing the denominators by $110$: $\frac{T-42}{1} = \frac{T-72}{2}$.
Cross-multiplying gives: $2(T-42) = T-72$.
Expanding the terms: $2T - 84 = T - 72$.
Rearranging to solve for $T$: $2T - T = 84 - 72$.
Thus,$T = 12$.
Therefore,the two scales have the same reading at $12^{\circ}$.

(A) Let the initial state of the gas be $(P_i, V, T)$.
Step $1$: Isothermal compression.
The gas is compressed until the pressure doubles. Since the process is isothermal,$P_i V = P_f V_f$. Given $P_f = 2P_i$,we have $P_i V = (2P_i) V_f$,which gives $V_f = V/2$.
The state after isothermal compression is $(2P_i, V/2, T)$.
Step $2$: Adiabatic expansion.
The gas expands adiabatically to regain its original volume $V$. Let the final pressure be $P_f'$.
Using the adiabatic process equation $P_1 V_1^\gamma = P_2 V_2^\gamma$:
$(2P_i) (V/2)^\gamma = P_f' V^\gamma$
$P_f' = (2P_i) \left(\frac{V/2}{V}\right)^\gamma = (2P_i) \left(\frac{1}{2}\right)^\gamma = 2P_i \times 2^{-\gamma}$
$P_f' = P_i \times 2^{1-\gamma}$
Given $\gamma = 1.4$,we have $P_f' = P_i \times 2^{1-1.4} = P_i \times 2^{-0.4}$.
Wait,the problem states $2^{-1.4} = 0.38$. Let's re-evaluate: $P_f' = 2P_i \times (1/2)^{1.4} = 2P_i \times 2^{-1.4}$.
Substituting the given value: $P_f' = 2P_i \times 0.38 = 0.76 P_i$.
Therefore,the ratio of the final to initial pressure is $P_f'/P_i = 0.76: 1$.

(A) The standard equation of a plane progressive wave is given by $y = A \cos(2\pi(\nu t - \frac{x}{\lambda}))$,where $\nu$ is the frequency.
Given equation: $y = 2 \cos 6.284(330t - x)$.
Since $6.284 \approx 2\pi$,we can rewrite the equation as $y = 2 \cos 2\pi(330t - x)$.
Comparing this with the standard form,the frequency $\nu = 330 \text{ Hz}$.
The time period $T$ is the reciprocal of the frequency,$T = \frac{1}{\nu}$.
Therefore,$T = \frac{1}{330} \text{ s}$.

(B) To pump the water out of the vessel, we need to lift the center of mass of the water to the top level of the vessel.
Let the side length of the cube be $L = 1 \,m$.
The volume of the water is $V = L^3 = 1^3 = 1 \,m^3$.
The density of water is $\rho = 1000 \,kg/m^3$.
The mass of the water is $m = \rho V = 1000 \times 1 = 1000 \,kg$.
The center of mass of the water in a full cubical vessel is at a height of $h_{cm} = L/2 = 0.5 \,m$ from the bottom.
To pump the water out, the center of mass must be raised to the top of the vessel, which is at a height $h = L/2 = 0.5 \,m$ from the initial center of mass position.
The work done is $W = mgh = 1000 \times 10 \times 0.5 = 5000 \,J$.

(C) Given that power $P$ is constant.
We know that $P = F \cdot v = m \cdot a \cdot v = m \cdot \frac{dv}{dt} \cdot v$.
Rearranging the terms,we get $v \cdot dv = \frac{P}{m} \cdot dt$.
Integrating both sides,$\int v \cdot dv = \int \frac{P}{m} \cdot dt$,which gives $\frac{v^2}{2} = \frac{P}{m} \cdot t$.
Thus,$v = \sqrt{\frac{2P}{m}} \cdot t^{\frac{1}{2}}$.
Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \sqrt{\frac{2P}{m}} \cdot t^{\frac{1}{2}}$.
Integrating with respect to time $t$,$x = \int \sqrt{\frac{2P}{m}} \cdot t^{\frac{1}{2}} \cdot dt = \sqrt{\frac{2P}{m}} \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}}$.
Therefore,$x = \frac{2}{3} \sqrt{\frac{2P}{m}} \cdot t^{\frac{3}{2}}$.
Since $P$ and $m$ are constants,$x \propto t^{\frac{3}{2}}$.

(D) Given: Mass $m = 6 \,kg$,displacement $x = \frac{t^2}{4} \,m$.
Velocity $v = \frac{dx}{dt} = \frac{d}{dt}(\frac{t^2}{4}) = \frac{2t}{4} = \frac{t}{2} \,m/s$.
At $t = 0 \,s$,velocity $v_i = \frac{0}{2} = 0 \,m/s$.
At $t = 2 \,s$,velocity $v_f = \frac{2}{2} = 1 \,m/s$.
According to the Work-Energy Theorem,the work done $W$ is equal to the change in kinetic energy $\Delta K$.
$W = K_f - K_i = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$.
$W = \frac{1}{2} \times 6 \times (1)^2 - \frac{1}{2} \times 6 \times (0)^2$.
$W = 3 - 0 = 3 \,J$.

(B) The path difference $\Delta x$ at a point $P$ on the screen at a distance $y$ from the center is given by $\Delta x = d \sin \theta \approx d \tan \theta = d \left( \frac{y}{D} \right)$.
For a dark fringe,the path difference must be an odd multiple of half the wavelength: $\Delta x = (2n - 1) \frac{\lambda}{2}$,where $n = 1, 2, 3, \dots$.
The first dark fringe corresponds to $n = 1$,so $\Delta x = \frac{\lambda}{2}$.
The problem states that the first dark fringe is observed directly opposite to one of the slits. The distance of the slits from the center is $d/2$. Thus,$y = d/2$.
Substituting these values into the path difference formula:
$\frac{d}{2} \cdot \frac{d}{D} = \frac{\lambda}{2}$
$\frac{d^2}{D} = \lambda$
Therefore,the wavelength of the light is $\lambda = \frac{d^2}{D}$.

(B) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6 \ eV}{n^2}$.
For the ground state $(n=1)$,$E_1 = -13.6 \ eV$.
For the $M$-shell $(n=3)$,the energy is $E_3 = -\frac{13.6 \ eV}{3^2} = -\frac{13.6}{9} \ eV \approx -1.51 \ eV$.
To ionize the hydrogen atom from the $M$-state,the atom must be provided with enough energy to reach the ionization limit $(E_{\infty} = 0 \ eV)$.
The energy required is $\Delta E = E_{\infty} - E_3 = 0 - (-1.51 \ eV) = +1.51 \ eV$.
Therefore,the second electron must transfer at least $+1.51 \ eV$ of energy to the atom.

(C) In the steady state,the capacitor acts as an open circuit,meaning no current flows through the branch containing the capacitor.
Therefore,the current $I$ flows only through the branch containing the resistor $R_2$ and the internal resistance $r$ of the battery.
Using Ohm's law,the current in the circuit is $I = \frac{E}{R_2 + r}$.
The potential difference $V$ across the capacitor is equal to the potential difference across the resistor $R_2$,because they are connected in parallel.
Thus,$V = I R_2 = \frac{E R_2}{R_2 + r}$.
The charge $Q$ on the capacitor is given by $Q = CV$.
Substituting the value of $V$,we get $Q = C \left( \frac{E R_2}{R_2 + r} \right) = \frac{C E R_2}{R_2 + r}$.

(C) The total e.m.f. of the two cells connected in series is $E + E = 2E$.
The total resistance of the circuit is $R + r_1 + r_2$.
The current $I$ flowing through the circuit is given by $I = \frac{2E}{R + r_1 + r_2}$.
The potential difference $V$ across the first cell (with internal resistance $r_1$) is given by $V = E - Ir_1$.
We are given that the potential difference across the first cell is zero,so $E - Ir_1 = 0$,which implies $E = Ir_1$.
Substituting the value of $I$ into this equation:
$E = \left( \frac{2E}{R + r_1 + r_2} \right) r_1$
$1 = \frac{2r_1}{R + r_1 + r_2}$
$R + r_1 + r_2 = 2r_1$
$R = 2r_1 - r_1 - r_2$
$R = r_1 - r_2$

(C) Let the nodes be labeled. The first node is $A$. The wire connects the start of the first resistor to the end of the second resistor. The second wire connects the start of the second resistor to the end of the third resistor (point $B$).
By analyzing the circuit,we see that all three resistors of resistance $R$ are connected in parallel between points $A$ and $B$.
For resistors connected in parallel,the equivalent resistance $R_{eq}$ is given by:
$\frac{1}{R_{eq}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R}$
Therefore,$R_{eq} = \frac{R}{3}$.

(C) From the circuit diagram,we can see that the resistors are arranged in a bridge-like structure.
By symmetry,the potential at point $F$ and point $E$ can be analyzed.
When a voltage $V$ is applied between $A$ and $B$,the circuit simplifies such that the equivalent resistance between $A$ and $B$ is $R_{eq} = R$.
Thus,the total current drawn from the battery is $I = \frac{V}{R}$.
Due to the symmetric nature of the circuit,this current splits equally into the two parallel branches connected to $F$.
Therefore,the current flowing through the resistor $FC$ is $I_{FC} = \frac{I}{2} = \frac{V}{2R}$.

(B) Let the side of the square be $a$. The distance of the center from each corner is $r = \frac{a}{\sqrt{2}}$.
For the system to be in equilibrium,the net force on any charge at the corner must be zero.
Consider a charge $-Q$ at one corner. The forces acting on it are:
$1$. Repulsive forces from the other three charges $-Q$ at the corners.
Let $F = \frac{1}{4 \pi \epsilon_0} \frac{Q^2}{a^2}$.
The resultant of the two forces along the sides is $F_{res} = \sqrt{F^2 + F^2} = \sqrt{2} F$.
The force from the diagonal charge is $F_{diag} = \frac{1}{4 \pi \epsilon_0} \frac{Q^2}{(\sqrt{2}a)^2} = \frac{F}{2}$.
Total repulsive force $F_{total} = \sqrt{2} F + \frac{F}{2} = F(\sqrt{2} + \frac{1}{2})$.
$2$. Attractive force from the central charge $q$: $F_q = \frac{1}{4 \pi \epsilon_0} \frac{q Q}{r^2} = \frac{1}{4 \pi \epsilon_0} \frac{q Q}{(a/\sqrt{2})^2} = \frac{2 q Q}{4 \pi \epsilon_0 a^2}$.
For equilibrium,$F_q = F_{total} \implies \frac{2 q Q}{4 \pi \epsilon_0 a^2} = \frac{Q^2}{4 \pi \epsilon_0 a^2} (\sqrt{2} + \frac{1}{2})$.
$2q = Q(\sqrt{2} + 0.5) = Q(\frac{2\sqrt{2}+1}{2})$.
$q = \frac{Q}{4}(1 + 2\sqrt{2})$.
Since the force must be attractive,$q$ must have a sign opposite to $-Q$,so $q$ is positive.

(B) The magnetic field at point $P$ is produced by the segments $BC$ and $B^{\prime} C^{\prime}$.
Segments $AB$ and $A^{\prime} B^{\prime}$ are directed towards and away from the axis containing $P$,so the magnetic field due to these segments at point $P$ is zero because $P$ lies on the line of these current elements.
For the semi-infinite wire $BC$,the magnetic field at distance $r$ from the end $B$ is given by $B_1 = \frac{\mu_0 I}{4 \pi r}$.
Similarly,for the semi-infinite wire $B^{\prime} C^{\prime}$,the magnetic field at distance $r$ from the end $B^{\prime}$ is $B_2 = \frac{\mu_0 I}{4 \pi r}$.
Using the right-hand rule,both fields at $P$ point in the same direction (into the page).
Therefore,the total magnetic field is $B = B_1 + B_2 = \frac{\mu_0 I}{4 \pi r} + \frac{\mu_0 I}{4 \pi r} = \frac{\mu_0}{4 \pi} \left( \frac{2 I}{r} \right)$.

(D) The magnetic field due to a straight wire of finite length at a perpendicular distance $r$ is given by $B = \frac{\mu_0 I}{4 \pi r} (\sin \theta_1 + \sin \theta_2)$.
For a square loop of side $L$,the perpendicular distance from the center to any side is $r = L/2$.
The angles subtended by the ends of each side at the center are $\theta_1 = 45^{\circ}$ and $\theta_2 = 45^{\circ}$.
Since there are $4$ identical sides,the total magnetic field $B_{total}$ is $4$ times the field due to one side:
$B_{total} = 4 \times \left[ \frac{\mu_0 I}{4 \pi (L/2)} (\sin 45^{\circ} + \sin 45^{\circ}) \right]$
$B_{total} = 4 \times \left[ \frac{\mu_0 I}{2 \pi L} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) \right]$
$B_{total} = 4 \times \left[ \frac{\mu_0 I}{2 \pi L} (\frac{2}{\sqrt{2}}) \right]$
$B_{total} = 4 \times \left[ \frac{\mu_0 I}{\pi L} \times \frac{1}{\sqrt{2}} \times 2 \right] = \frac{8 \mu_0 I}{2 \sqrt{2} \pi L} = \frac{4 \mu_0 I}{\sqrt{2} \pi L} = \frac{2 \sqrt{2} \mu_0 I}{\pi L}$.

(D) In a nuclear reaction,both the total atomic number $(Z)$ and the total mass number $(A)$ must be conserved on both sides of the equation.
Let $X$ be represented as ${ }_{Z'}^{A'} X$.
The reaction is: ${ }_{7}^{14} N + { }_{Z'}^{A'} X \rightarrow { }_{6}^{14} C + { }_{1}^{1} H$.
Conservation of mass number $(A)$: $14 + A' = 14 + 1 \Rightarrow A' = 1$.
Conservation of atomic number $(Z)$: $7 + Z' = 6 + 1 \Rightarrow 7 + Z' = 7 \Rightarrow Z' = 0$.
$A$ particle with mass number $1$ and atomic number $0$ is a neutron,denoted as ${ }_{0}^{1} n$.

(C) According to the law of conservation of linear momentum,the initial momentum of the system is zero because the nucleus is initially at rest.
Let $M$ be the mass of the parent nucleus,which is proportional to the mass number $A$. Thus,$M = kA$.
The mass of the $\alpha$-particle is $4$ (in atomic mass units),and the mass of the daughter nucleus is $(A-4)$.
Let $V_r$ be the recoil speed of the daughter nucleus.
By conservation of momentum: $P_{\text{initial}} = P_{\text{final}}$
$0 = m_{\alpha} v + m_{\text{daughter}} V_r$
$0 = 4v + (A-4) V_r$
$V_r = -\frac{4v}{A-4}$
The magnitude of the recoil speed is $\frac{4v}{A-4}$.

(C) When a diver is inside the water,they see the outside world through a cone of light due to the phenomenon of total internal reflection.
The semi-vertical angle of this cone is equal to the critical angle $c$ for the water-air interface.
The formula for the critical angle is given by $\sin c = \frac{1}{\mu}$.
Given the refractive index of water $\mu = \frac{4}{3}$,we have:
$\sin c = \frac{1}{4/3} = \frac{3}{4}$.
Therefore,the semi-vertical angle is $c = \sin ^{-1}\left(\frac{3}{4}\right)$.

(A) The power $P$ of a lens is given by $P = \frac{100}{f}$ where $f$ is the focal length in $cm$.
For the first lens,$P_1 = \frac{100}{20} = 5 \ D$.
For the second lens,$P_2 = \frac{100}{25} = 4 \ D$.
When two thin lenses are placed in contact,the effective power $P_{eff}$ of the combination is the sum of their individual powers: $P_{eff} = P_1 + P_2$.
Therefore,$P_{eff} = 5 \ D + 4 \ D = 9 \ D$.

(A) Given: Focal length $f = +30 \,cm$ (for a convex lens).
Magnification $m = -5$ (since the image is real).
We know that magnification $m = \frac{v}{u}$, so $v = mu = -5u$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{30} = \frac{1}{-5u} - \frac{1}{u}$.
$\frac{1}{30} = \frac{-1 - 5}{5u} = \frac{-6}{5u}$.
$5u = -180$.
$u = -36 \,cm$.
The magnitude of the object distance is $36 \,cm$.

(B) The power $P$ of a lens is given by the lens maker's formula: $P = \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a plano-concave lens, the first surface is plane $(R_1 = \infty)$ and the second surface is concave $(R_2 = -100 \,cm = -1 \,m)$.
Given refractive index $\mu = 1.5$.
Substituting the values: $P = (1.5 - 1) \left( \frac{1}{\infty} - \frac{1}{-1} \right)$.
$P = (0.5) (0 + 1) = 0.5 \,D$.
Wait, for a concave lens, the power must be negative. Let's re-evaluate the sign convention: For a plano-concave lens, light enters the plane surface first $(R_1 = \infty)$ and exits the concave surface $(R_2 = -R = -1 \,m)$.
$P = (1.5 - 1) \left( \frac{1}{\infty} - \frac{1}{-1} \right) = 0.5 \times 1 = +0.5 \,D$ (This is for a plano-convex).
For a plano-concave lens, the curved surface is the first surface $(R_1 = -1 \,m)$ and the second is plane $(R_2 = \infty)$.
$P = (1.5 - 1) \left( \frac{1}{-1} - \frac{1}{\infty} \right) = 0.5 \times (-1) = -0.5 \,D$.

(C) The magnifying power $(m)$ of an astronomical telescope in normal adjustment is given by the formula: $m = -\frac{f_o}{f_e}$,where $f_o$ is the focal length of the objective lens and $f_e$ is the focal length of the eyepiece.
If the focal length of the eyepiece is doubled,the new focal length becomes $f_e' = 2f_e$.
The new magnifying power $(m')$ will be: $m' = -\frac{f_o}{f_e'} = -\frac{f_o}{2f_e} = \frac{1}{2} \left( -\frac{f_o}{f_e} \right) = \frac{m}{2}$.
Therefore,the magnifying power becomes half of its original value.

(D) The truth table provided is as follows:
| Input $A$ | Input $B$ | Output $Q$ |
| :--- | :--- | :--- |
| $0$ | $0$ | $1$ |
| $0$ | $1$ | $1$ |
| $1$ | $0$ | $1$ |
| $1$ | $1$ | $0$ |
Analyzing the truth table:
$1$. When both inputs $A$ and $B$ are $1$, the output $Q$ is $0$.
$2$. For all other input combinations ($0,0$; $0,1$; $1,0$), the output $Q$ is $1$.
This behavior corresponds to the $NAND$ gate, which is equivalent to an $AND$ gate followed by a $NOT$ gate. The Boolean expression for this gate is $Q = \overline{A \cdot B}$.