AIEEE 2010 Physics Question Paper with Answer and Solution

(A) To determine the number of significant figures,we follow these rules:
$(i)$ All non-zero digits are significant.
$(ii)$ All zeros between two non-zero digits are significant.
$(iii)$ For numbers less than $1$,zeros to the right of the decimal point but to the left of the first non-zero digit are not significant.
$(iv)$ The power of $10$ is not considered for significant figures.
Applying these rules:
$1$. For $23.023$: All $5$ digits are significant. So,significant figures $= 5$.
$2$. For $0.0003$: The zeros before the $3$ are not significant. Only the digit $3$ is significant. So,significant figures $= 1$.
$3$. For $2.1 \times 10^3$: The power of $10$ is ignored. The digits $2$ and $1$ are significant. So,significant figures $= 2$.
Thus,the number of significant figures are $5, 1, 2$ respectively.

(A) Given the velocity vector $\vec v = K(y\hat i + x\hat j)$.
We know that $\vec v = v_x \hat i + v_y \hat j = \frac{dx}{dt} \hat i + \frac{dy}{dt} \hat j$.
Comparing the components,we get $\frac{dx}{dt} = Ky$ and $\frac{dy}{dt} = Kx$.
To find the path of the particle,we divide the two equations: $\frac{dy/dt}{dx/dt} = \frac{Kx}{Ky}$.
This simplifies to $\frac{dy}{dx} = \frac{x}{y}$.
Rearranging the terms,we get $y \, dy = x \, dx$.
Integrating both sides,we obtain $\int y \, dy = \int x \, dx$.
This results in $\frac{y^2}{2} = \frac{x^2}{2} + C'$,where $C'$ is a constant.
Multiplying by $2$,we get $y^2 = x^2 + C$,where $C = 2C'$ is another constant.
Thus,the equation of the path is $y^2 = x^2 + \text{constant}$.

(D) In uniform circular motion,the acceleration is the centripetal acceleration $\vec{a}_c$,which is always directed towards the center of the circle.
For a point $P$ at an angle $\theta$ with the $x$-axis,the position vector makes an angle $\theta$ with the $x$-axis.
The centripetal acceleration vector $\vec{a}_c$ points from $P$ towards the origin $O$.
The magnitude of centripetal acceleration is $a_c = \frac{V^2}{R}$.
Resolving this vector into components:
The $x$-component is $a_x = -a_c \cos\theta = -\frac{V^2}{R} \cos\theta$.
The $y$-component is $a_y = -a_c \sin\theta = -\frac{V^2}{R} \sin\theta$.
Thus,the acceleration vector is $\vec{a} = -\frac{V^2}{R} \cos\theta \hat{i} - \frac{V^2}{R} \sin\theta \hat{j}$.

(A) Given,the path length $s = t^3 + 5$.
Velocity $v = \frac{ds}{dt} = \frac{d}{dt}(t^3 + 5) = 3t^2 \ m/s$.
Tangential acceleration $a_t = \frac{dv}{dt} = \frac{d}{dt}(3t^2) = 6t \ m/s^2$.
Radial (centripetal) acceleration $a_c = \frac{v^2}{R} = \frac{(3t^2)^2}{R} = \frac{9t^4}{R} \ m/s^2$.
At $t = 2 \ s$:
$a_t = 6 \times 2 = 12 \ m/s^2$.
$a_c = \frac{9 \times (2)^4}{20} = \frac{9 \times 16}{20} = \frac{144}{20} = 7.2 \ m/s^2$.
The resultant acceleration $a = \sqrt{a_t^2 + a_c^2}$.
$a = \sqrt{(12)^2 + (7.2)^2} = \sqrt{144 + 51.84} = \sqrt{195.84} \approx 14 \ m/s^2$.

(C) The impulse $J$ is equal to the change in linear momentum,$\Delta p = m(v_f - v_i)$.
From the graph,the motion consists of segments with constant velocity.
For the interval $t = 0$ to $t = 2 \; s$,the displacement is $2 \; m$. Thus,initial velocity $v_i = \frac{2 \; m}{2 \; s} = 1 \; m/s$.
For the interval $t = 2 \; s$ to $t = 4 \; s$,the displacement is $-2 \; m$. Thus,final velocity $v_f = \frac{-2 \; m}{2 \; s} = -1 \; m/s$.
The mass of the body is $m = 0.4 \; kg$.
The impulse $J$ at each collision (at $t = 2, 6, 10, 14 \; s$) is given by:
$J = m(v_f - v_i) = 0.4 \; kg \times (-1 \; m/s - 1 \; m/s) = 0.4 \times (-2) = -0.8 \; kg \cdot m/s$.
The magnitude of the impulse is $|J| = |-0.8| = 0.8 \; N \cdot s$.

(B) For a block on a frictionless inclined plane making an angle $\theta$ with the horizontal,the acceleration along the plane is $a = g \sin \theta$.
The vertical component of this acceleration is $a_v = a \sin \theta = (g \sin \theta) \sin \theta = g \sin^2 \theta$.
For block $A$,the angle with the horizontal is $90^\circ - 60^\circ = 30^\circ$. Thus,its vertical acceleration is $a_{vA} = g \sin^2(30^\circ) = g(1/2)^2 = g/4$.
For block $B$,the angle with the horizontal is $90^\circ - 30^\circ = 60^\circ$. Thus,its vertical acceleration is $a_{vB} = g \sin^2(60^\circ) = g(\sqrt{3}/2)^2 = 3g/4$.
The relative vertical acceleration of $A$ with respect to $B$ is $a_{rel} = a_{vA} - a_{vB} = g/4 - 3g/4 = -g/2 = -4.9 \ m/s^2$.
The magnitude is $4.9 \ m/s^2$ in the downward vertical direction.

(B) In a completely inelastic collision,the particles stick together and move with a common velocity. The kinetic energy of the system is not entirely lost because the final system retains some kinetic energy due to the conservation of momentum. Thus,Statement $-1$ is true.
The principle of conservation of linear momentum is a fundamental law that applies to all types of collisions (elastic or inelastic) in the absence of external forces. Thus,Statement $-2$ is true.
Statement $-2$ provides the physical basis for why the system retains energy after a collision,as the conservation of momentum allows us to calculate the final common velocity and the remaining kinetic energy of the combined mass. Therefore,Statement $-2$ is the correct explanation of Statement $-1$.

(C) At equilibrium,the force is zero,which means the derivative of the potential energy with respect to distance is zero: $\frac{dU(x)}{dx} = 0$.
Given $U(x) = ax^{-12} - bx^{-6}$,we differentiate:
$\frac{dU}{dx} = -12ax^{-13} + 6bx^{-7} = 0$.
$12ax^{-13} = 6bx^{-7} \Rightarrow \frac{2a}{x^6} = b \Rightarrow x^6 = \frac{2a}{b}$.
Now,substitute $x^6 = \frac{2a}{b}$ into the potential energy function to find $U_{\text{at equilibrium}}$:
$U_{\text{at equilibrium}} = \frac{a}{(x^6)^2} - \frac{b}{x^6} = \frac{a}{(2a/b)^2} - \frac{b}{(2a/b)} = \frac{a}{4a^2/b^2} - \frac{b^2}{2a} = \frac{b^2}{4a} - \frac{2b^2}{4a} = -\frac{b^2}{4a}$.
Since $U(x = \infty) = 0$,the dissociation energy $D$ is:
$D = U(\infty) - U_{\text{at equilibrium}} = 0 - (-\frac{b^2}{4a}) = \frac{b^2}{4a}$.

(C) Given the density condition: $\rho_{\text{oil}} < \rho < \rho_{\text{water}}$.
$1$. Since oil is less dense than water,it will float on top of the water layer.
$2$. The ball has a density $\rho$ that is greater than the density of oil $(\rho > \rho_{\text{oil}})$,so it will sink through the oil layer.
$3$. The ball has a density $\rho$ that is less than the density of water $(\rho < \rho_{\text{water}})$,so it will float on the water layer.
$4$. Consequently,the ball will come to rest at the interface between the oil and the water,partially submerged in both. This corresponds to the configuration where the oil is on top and the water is at the bottom,with the ball at the boundary.

(A) The given wave equation is $y = 0.02 \sin \left[ 2\pi \left( \frac{t}{0.04} - \frac{x}{0.50} \right) \right]$.
Comparing this with the standard wave equation $y = A \sin (\omega t - kx)$,we get:
Angular frequency $\omega = \frac{2\pi}{0.04} \ rad/s$ and wave number $k = \frac{2\pi}{0.50} \ rad/m$.
The wave velocity $v$ is given by $v = \frac{\omega}{k} = \frac{2\pi / 0.04}{2\pi / 0.50} = \frac{0.50}{0.04} = 12.5 \ m/s$.
The velocity of a wave on a stretched string is $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Given $\mu = 0.04 \ kg/m$,we have $T = v^2 \mu$.
Substituting the values,$T = (12.5)^2 \times 0.04 = 156.25 \times 0.04 = 6.25 \ N$.

(C) The position vector of the particle at time $t$ is given by:
$\vec{r} = (v_0 \cos \theta) t \hat{i} + ((v_0 \sin \theta) t - \frac{1}{2} g t^2) \hat{j}$
The velocity vector of the particle at time $t$ is given by:
$\vec{v} = (v_0 \cos \theta) \hat{i} + (v_0 \sin \theta - gt) \hat{j}$
The angular momentum $\vec{L}$ is defined as $\vec{L} = m(\vec{r} \times \vec{v})$:
$\vec{L} = m [((v_0 \cos \theta) t \hat{i} + ((v_0 \sin \theta) t - \frac{1}{2} g t^2) \hat{j}) \times ((v_0 \cos \theta) \hat{i} + (v_0 \sin \theta - gt) \hat{j})]$
Performing the cross product:
$\vec{L} = m [((v_0 \cos \theta) t) (v_0 \sin \theta - gt) (\hat{i} \times \hat{j}) + ((v_0 \sin \theta) t - \frac{1}{2} g t^2) (v_0 \cos \theta) (\hat{j} \times \hat{i})]$
Since $\hat{i} \times \hat{j} = \hat{k}$ and $\hat{j} \times \hat{i} = -\hat{k}$:
$\vec{L} = m [((v_0^2 \sin \theta \cos \theta) t - v_0 g t^2 \cos \theta) \hat{k} - ((v_0^2 \sin \theta \cos \theta) t - \frac{1}{2} v_0 g t^2 \cos \theta) \hat{k}]$
$\vec{L} = m [v_0^2 \sin \theta \cos \theta t - v_0 g t^2 \cos \theta - v_0^2 \sin \theta \cos \theta t + \frac{1}{2} v_0 g t^2 \cos \theta] \hat{k}$
$\vec{L} = m [-\frac{1}{2} v_0 g t^2 \cos \theta] \hat{k} = -\frac{1}{2} mg v_0 t^2 \cos \theta \hat{k}$

(C) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = \text{constant}$.
Thus,$T_H V_1^{\gamma-1} = T_C V_2^{\gamma-1}$,where $T_H$ is the source temperature and $T_C$ is the sink temperature.
Given that the gas is diatomic,the adiabatic index $\gamma = 1.4$.
The volume changes from $V_1 = V$ to $V_2 = 32V$.
Substituting these values:
$\frac{T_C}{T_H} = \left(\frac{V_1}{V_2}\right)^{\gamma-1} = \left(\frac{V}{32V}\right)^{1.4-1} = \left(\frac{1}{32}\right)^{0.4}$.
Since $32 = 2^5$,we have $\left(\frac{1}{2^5}\right)^{0.4} = \left(\frac{1}{2^5}\right)^{2/5} = \left(\frac{1}{2}\right)^2 = \frac{1}{4} = 0.25$.
The efficiency of the Carnot engine is $\eta = 1 - \frac{T_C}{T_H}$.
$\eta = 1 - 0.25 = 0.75$.

(B) If the current flows out of the paper,the magnetic field at points to the right of the wire will be upwards and to the left will be downward. Let the wires be at $A$ and $B$,and the midpoint be $C$. The magnetic field at $C$ is zero because the fields from $A$ and $B$ cancel each other out.
In the region to the right of $B$,the magnetic field is upwards $(+ve)$ because all points are to the right of both wires. Similarly,in the region to the left of $A$,the magnetic field is downwards $(-ve)$.
In the region $AC$,the points are closer to $A$ than to $B$,so the field due to $A$ dominates and is upwards $(+ve)$.
In the region $BC$,the points are closer to $B$ than to $A$,so the field due to $B$ dominates and is downwards $(-ve)$.
Graph $(b)$ correctly represents these variations in the magnetic field.

(D) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.
For energy to reduce to half its initial value: $\frac{U_0}{2} = \frac{q^2}{2C} \Rightarrow \frac{q_0^2}{2} = q^2 \Rightarrow q = \frac{q_0}{\sqrt{2}}$.
Using the discharging equation $q = q_0 e^{-t/RC}$,we have $\frac{q_0}{\sqrt{2}} = q_0 e^{-t_1/RC}$.
Taking the natural logarithm: $-\frac{t_1}{RC} = \ln(1/\sqrt{2}) = -\frac{1}{2} \ln 2$,so $t_1 = \frac{RC \ln 2}{2}$.
For charge to reduce to one-fourth its initial value: $\frac{q_0}{4} = q_0 e^{-t_2/RC}$.
Taking the natural logarithm: $-\frac{t_2}{RC} = \ln(1/4) = -2 \ln 2$,so $t_2 = 2RC \ln 2$.
The ratio $\frac{t_1}{t_2} = \frac{(RC \ln 2) / 2}{2RC \ln 2} = \frac{1}{4} = 0.25$.

(A) Let $\theta$ be the angle each string makes with the vertical. Since the total angle between the strings is $30^{\circ}$,$\theta = 15^{\circ}$.
In air,the forces acting on the sphere are tension $T$,electrostatic force $F = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2}$,and weight $mg$. Balancing forces:
$T \sin \theta = F$
$T \cos \theta = mg$
Dividing the two equations: $\tan \theta = \frac{F}{mg}$.
When the spheres are immersed in a liquid of density $\rho_l = 0.8 \; g \, cm^{-3}$ and the material density is $\rho_s = 1.6 \; g \, cm^{-3}$,the effective weight becomes $mg' = mg(1 - \frac{\rho_l}{\rho_s})$ and the electrostatic force becomes $F' = \frac{F}{K}$.
Since the angle remains the same,$\tan \theta = \frac{F'}{mg'} = \frac{F/K}{mg(1 - \frac{\rho_l}{\rho_s})}$.
Equating the two expressions for $\tan \theta$:
$\frac{F}{mg} = \frac{F}{K mg (1 - \frac{\rho_l}{\rho_s})}$
$K = \frac{1}{1 - \frac{\rho_l}{\rho_s}} = \frac{1}{1 - \frac{0.8}{1.6}} = \frac{1}{1 - 0.5} = \frac{1}{0.5} = 2$.

(D) Consider a small differential element $dl$ on the semi-circular ring. The charge $dq$ on this element is given by $dq = \lambda dl$,where $\lambda = \frac{q}{\pi r}$ is the linear charge density.
Since $dl = r d\theta$,we have $dq = \left(\frac{q}{\pi r}\right) (r d\theta) = \frac{q}{\pi} d\theta$.
The magnitude of the electric field $dE$ at the centre $O$ due to this element is $dE = \frac{1}{4\pi\varepsilon_0} \frac{dq}{r^2} = \frac{1}{4\pi\varepsilon_0} \frac{q}{\pi r^2} d\theta = \frac{q}{4\pi^2\varepsilon_0 r^2} d\theta$.
Due to symmetry,the horizontal components $(dE \cos\theta)$ of the electric field from elements on opposite sides of the vertical axis cancel each other out.
The net electric field is the integral of the vertical components $(dE \sin\theta)$ from $\theta = 0$ to $\pi$.
$E = \int_0^{\pi} dE \sin\theta = \int_0^{\pi} \frac{q}{4\pi^2\varepsilon_0 r^2} \sin\theta d\theta$.
$E = \frac{q}{4\pi^2\varepsilon_0 r^2} [-\cos\theta]_0^{\pi} = \frac{q}{4\pi^2\varepsilon_0 r^2} (-(-1) - (-1)) = \frac{q}{4\pi^2\varepsilon_0 r^2} (2) = \frac{q}{2\pi^2\varepsilon_0 r^2}$.
Since the charge is positive and distributed on the upper semi-circle,the field at the centre points downwards,i.e.,in the $-\hat{j}$ direction.
Therefore,$\vec{E} = -\frac{q}{2\pi^2\varepsilon_0 r^2} \hat{j}$.

(C) Consider a spherical shell of radius $x$ and thickness $dx$. The charge on this shell is given by $dq = \rho(x) \cdot 4 \pi x^2 dx = \rho_0 \left( \frac{5}{4} - \frac{x}{R} \right) \cdot 4 \pi x^2 dx$.
The total charge $q$ enclosed within a sphere of radius $r$ $(r < R)$ is:
$q = \int_0^r dq = 4 \pi \rho_0 \int_0^r \left( \frac{5}{4} x^2 - \frac{x^3}{R} \right) dx$
$q = 4 \pi \rho_0 \left[ \frac{5}{4} \cdot \frac{r^3}{3} - \frac{1}{R} \cdot \frac{r^4}{4} \right] = 4 \pi \rho_0 \left( \frac{5 r^3}{12} - \frac{r^4}{4R} \right) = \pi \rho_0 r^3 \left( \frac{5}{3} - \frac{r}{R} \right)$.
Using Gauss's Law,the electric field $E$ at distance $r$ is:
$E \cdot 4 \pi r^2 = \frac{q}{\varepsilon_0}$
$E = \frac{1}{4 \pi \varepsilon_0 r^2} \cdot \left[ \pi \rho_0 r^3 \left( \frac{5}{3} - \frac{r}{R} \right) \right]$
$E = \frac{\rho_0 r}{4 \varepsilon_0} \left( \frac{5}{3} - \frac{r}{R} \right)$.

(A) Let the resistance of both conductors at $0\,^{\circ}C$ be $R_0$. The resistance at temperature $\Delta t$ is given by $R = R_0(1 + \alpha \Delta t)$.
For the series combination: $R_s = R_1 + R_2 = R_0(1 + \alpha_1 \Delta t) + R_0(1 + \alpha_2 \Delta t) = 2R_0(1 + \frac{\alpha_1 + \alpha_2}{2} \Delta t)$. Comparing this with $R_s = 2R_0(1 + \alpha_s \Delta t)$,we get $\alpha_s = \frac{\alpha_1 + \alpha_2}{2}$.
For the parallel combination: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_0(1 + \alpha_1 \Delta t)} + \frac{1}{R_0(1 + \alpha_2 \Delta t)}$. Using the binomial approximation $(1+x)^{-1} \approx 1-x$,we get $\frac{1}{R_p} \approx \frac{1}{R_0}(1 - \alpha_1 \Delta t + 1 - \alpha_2 \Delta t) = \frac{2}{R_0}(1 - \frac{\alpha_1 + \alpha_2}{2} \Delta t)$.
Thus,$R_p \approx \frac{R_0}{2}(1 + \frac{\alpha_1 + \alpha_2}{2} \Delta t)$. Comparing this with $R_p = \frac{R_0}{2}(1 + \alpha_p \Delta t)$,we get $\alpha_p = \frac{\alpha_1 + \alpha_2}{2}$.

(C) The refractive index of the medium is given by $\mu(I) = \mu_0 + \mu_2I$.
Since the intensity $I$ of the beam is maximum at the axis and decreases as the radius increases,the refractive index $\mu$ will also be maximum at the axis and decrease towards the periphery.
Light rays always bend towards the region of higher refractive index.
Since the refractive index is higher near the axis and lower towards the periphery,the light rays will bend towards the axis.
Therefore,the beam will converge.

(A) The refractive index of the medium is given by $\mu(I) = \mu_0 + \mu_2I$. Since the intensity $I$ of the beam decreases as the radius increases,the refractive index $\mu$ also decreases from the axis towards the periphery.
Because the beam is initially parallel and cylindrical,the rays of light are parallel to each other. By definition,the wavefront of a parallel beam of light is a plane perpendicular to the direction of propagation.
Therefore,the initial shape of the wavefront is planar.

(B) The refractive index of the medium is given by $\mu(I) = \mu_0 + \mu_2I$.
The speed of light $v$ in a medium with refractive index $\mu$ is given by $v = \frac{c_0}{\mu}$,where $c_0$ is the speed of light in vacuum.
Substituting the expression for $\mu$,we get $v = \frac{c_0}{\mu_0 + \mu_2I}$.
Since the intensity $I$ of the beam is maximum on the axis and decreases as the radius increases,the denominator $(\mu_0 + \mu_2I)$ will be maximum on the axis.
Consequently,the speed of light $v$ will be minimum on the axis of the beam.

(A) According to Einstein's photoelectric equation, $eV_0 = K_{max} = h\nu - \phi$, where $\phi$ is the work function of the metal surface.
Since the frequency of $X$-rays $(\nu_X)$ is much higher than the frequency of ultraviolet light $(\nu_{UV})$, the maximum kinetic energy $K_{max} = h\nu - \phi$ increases when $X$-rays are used.
Consequently, the stopping potential $V_0 = K_{max}/e$ also increases. Thus, Statement $-1$ is true.
Statement $-2$ is false because photoelectrons are emitted with a range of speeds from zero to a maximum value due to the loss of energy during collisions of electrons within the metal before they are emitted, not because of a range of frequencies in the incident light (which is monochromatic).

(A) At $t = 0$,the inductor $L$ acts as an open circuit because it opposes any change in current. Therefore,no current flows through the branch containing $L$ and $R_1$.
Thus,the current through the battery is $I = \frac{V}{R_2}$.
At $t = \infty$,the inductor $L$ acts as a short circuit (ideal inductor with zero resistance). The resistors $R_1$ and $R_2$ are now in parallel.
The effective resistance of the circuit is $R_{eff} = \frac{R_1 R_2}{R_1 + R_2}$.
Therefore,the current through the battery is $I = \frac{V}{R_{eff}} = \frac{V(R_1 + R_2)}{R_1 R_2}$.

(C) The moving rod $PQ$ acts as a source of induced $emf$ $\varepsilon = Blv$ with internal resistance $R$.
The circuit consists of this $emf$ source in series with a resistor $R$,which is then connected in parallel to two other branches,each containing a resistor $R$.
Let the potential at $Q$ be $V_Q$ and at $P$ be $V_P$. The total resistance of the circuit is the internal resistance $R$ plus the equivalent resistance of the two parallel branches,which is $R/2$.
Total resistance $R_{eq} = R + \frac{R \times R}{R + R} = R + \frac{R}{2} = \frac{3R}{2}$.
The total current $I$ flowing through the rod is $I = \frac{\varepsilon}{R_{eq}} = \frac{Blv}{3R/2} = \frac{2Blv}{3R}$.
Since the two parallel branches have equal resistance $R$,the current $I$ splits equally between them: $I_1 = I_2 = \frac{I}{2} = \frac{1}{2} \times \frac{2Blv}{3R} = \frac{Blv}{3R}$.
Thus,$I_1 = I_2 = \frac{Blv}{3R}$ and $I = \frac{2Blv}{3R}$.

(A) When capacitance is removed,the circuit becomes an $LR$ circuit.
Given the phase angle $\phi = 30^\circ$,we have $\tan \phi = \frac{\omega L}{R}$.
$\omega L = R \tan 30^\circ = 200 \times \frac{1}{\sqrt{3}} = \frac{200}{\sqrt{3}} \, \Omega$.
When the inductor is removed,the circuit becomes a $CR$ circuit.
Given the phase angle $\phi = 30^\circ$,we have $\tan \phi = \frac{1}{\omega C R}$.
$\frac{1}{\omega C} = R \tan 30^\circ = 200 \times \frac{1}{\sqrt{3}} = \frac{200}{\sqrt{3}} \, \Omega$.
In the original $LCR$ circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$.
Substituting the values: $Z = \sqrt{200^2 + (\frac{200}{\sqrt{3}} - \frac{200}{\sqrt{3}})^2} = \sqrt{200^2 + 0} = 200 \, \Omega$.
The power dissipated in an $AC$ circuit is $P = V_{rms} I_{rms} \cos \phi$,where $\cos \phi = \frac{R}{Z}$.
Since $\omega L = \frac{1}{\omega C}$,the circuit is in resonance,so $\phi = 0^\circ$ and $\cos \phi = 1$.
$P = \frac{V_{rms}^2}{Z^2} \times R = \frac{220^2}{200^2} \times 200 = \frac{220 \times 220}{200} = 242 \, W$.

(B) The power $P$ of the source is given by $P = n \cdot E_{photon} = n \cdot h \nu$,where $n$ is the number of photons per second,$h$ is Planck's constant,and $\nu$ is the frequency.
Given $P = 4 \ kW = 4 \times 10^3 \ W$ and $n = 10^{20} \ photons/s$.
Rearranging for frequency: $\nu = \frac{P}{n \cdot h}$.
Substituting the values: $\nu = \frac{4 \times 10^3}{10^{20} \times 6.63 \times 10^{-34}} \approx 6.03 \times 10^{16} \ Hz$.
The frequency range $6 \times 10^{16} \ Hz$ falls within the $X$-ray region of the electromagnetic spectrum.

(C) According to the law of conservation of energy,the total energy before decay equals the total energy after decay.
The initial energy is the rest mass energy: $E_i = (M + \Delta m)c^2$.
The final energy consists of the rest mass energy of the two daughter nuclei and their kinetic energy: $E_f = 2 \times (\frac{M}{2})c^2 + 2 \times (\frac{1}{2} \times \frac{M}{2} \times v^2)$.
Equating $E_i = E_f$:
$(M + \Delta m)c^2 = Mc^2 + \frac{M}{2}v^2$.
Subtracting $Mc^2$ from both sides:
$\Delta m c^2 = \frac{M}{2}v^2$.
Solving for $v$:
$v^2 = \frac{2 \Delta m c^2}{M} \Rightarrow v = c \sqrt{\frac{2 \Delta m}{M}}$.

(D) In a nuclear decay or fission process,the system moves toward a more stable state.
Stability is determined by the binding energy per nucleon.
$A$ nucleus with a higher binding energy per nucleon is more stable.
Since the parent nucleus decays into two daughter nuclei,the daughter nuclei must be more stable than the parent nucleus to satisfy the energy release condition.
Therefore,the binding energy per nucleon of the daughter nuclei $(E_2)$ must be greater than that of the parent nucleus $(E_1)$.
Thus,$E_2 > E_1$.

(B) The initial nucleus is represented as $_Z^AX$.
An $\alpha$-particle is $_2^4He$ and a positron is $_1^0e^+$.
After emitting $3$ $\alpha$-particles and $2$ positrons, the final nucleus $_Z^AY$ is formed.
The new mass number $A' = A - (3 \times 4) = A - 12$.
The new atomic number $Z' = Z - (3 \times 2) + (2 \times 1) = Z - 6 + 2 = Z - 4$.
The number of protons in the final nucleus is $P = Z' = Z - 4$.
The number of neutrons in the final nucleus is $N = A' - Z' = (A - 12) - (Z - 4) = A - Z - 8$.
The ratio of neutrons to protons is $\frac{N}{P} = \frac{A - Z - 8}{Z - 4}$.

(B) The circuit consists of two $NAND$ gates acting as $NOT$ gates (since their inputs are shorted) followed by a $NAND$ gate.
Let the inputs be $A$ and $B$.
The output of the first $NAND$ gate is $\bar{A}$.
The output of the second $NAND$ gate is $\bar{B}$.
These two outputs are fed into the third $NAND$ gate.
The final output $X$ is given by the boolean expression:
$X = \overline{\bar{A} \cdot \bar{B}}$
Using De Morgan's theorem,$\overline{\bar{A} \cdot \bar{B}} = \overline{\bar{A}} + \overline{\bar{B}} = A + B$.
Since the expression $X = A + B$ represents an $OR$ gate,the correct option is $B$.