NEET 2021 Physics Question Paper with Answer and Solution

(D) The formula for escape velocity is $v_{e} = \sqrt{\frac{2GM}{R}}$.
Since mass $M = \text{Volume} \times \text{Density} = \frac{4}{3} \pi R^{3} \rho$,we substitute this into the formula:
$v_{e} = \sqrt{\frac{2G}{R} \times \frac{4}{3} \pi R^{3} \rho} = \sqrt{\frac{8 \pi G \rho}{3} R^{2}} = R \sqrt{\frac{8 \pi G \rho}{3}}$.
This shows that $v_{e} \propto R$ when the density $\rho$ is constant.
Given the radius of the new planet is $R' = 4R$,the new escape velocity $v'$ will be:
$v' = 4 \times v_{e} = 4v$.

(A) When the ball moves with a constant terminal velocity,the net force acting on it is zero.
The forces acting on the ball are:
$1$. Weight of the ball $(W = Mg)$ acting downwards.
$2$. Buoyant force $(F_B)$ acting upwards.
$3$. Viscous force $(F_v)$ acting upwards.
The buoyant force is given by $F_B = V \rho_{liquid} g$,where $V$ is the volume of the ball.
Since the density of the ball is $d$,its volume is $V = \frac{M}{d}$.
Given the density of glycerine is $\frac{d}{2}$,the buoyant force is $F_B = V \left(\frac{d}{2}\right) g = \left(\frac{M}{d}\right) \left(\frac{d}{2}\right) g = \frac{Mg}{2}$.
For constant velocity,the forces must balance:
$F_v + F_B = W$
$F_v + \frac{Mg}{2} = Mg$
$F_v = Mg - \frac{Mg}{2} = \frac{Mg}{2}$.

(C) The displacement equation for a simple harmonic motion $(SHM)$ with frequency $n$ is given by:
$x = A \sin(\omega t) = A \sin(2 \pi n t)$
The potential energy $(U)$ of the body is given by:
$U = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \sin^2(2 \pi n t)$
Using the trigonometric identity $\sin^2 \theta = \frac{1 - \cos(2 \theta)}{2}$,we get:
$U = \frac{1}{2} k A^2 \left[ \frac{1 - \cos(2 \cdot 2 \pi n t)}{2} \right] = \frac{1}{4} k A^2 [1 - \cos(4 \pi n t)]$
The frequency of the potential energy is the coefficient of $2 \pi t$ in the cosine term,which is $2n$.

(B) The input power $P_{\text{input}}$ is given by the rate of change of potential energy: $P_{\text{input}} = \frac{mgh}{t} = \left(\frac{m}{t}\right)gh$.
Given $\frac{m}{t} = 15 \, kg/s$,$g = 10 \, m/s^2$,and $h = 60 \, m$.
$P_{\text{input}} = 15 \times 10 \times 60 = 9000 \, W = 9 \, kW$.
The power loss due to friction is $10 \%$ of the input power.
$\text{Loss} = 10 \% \text{ of } 9 \, kW = 0.1 \times 9 \, kW = 0.9 \, kW$.
The power generated by the turbine is the output power: $P_{\text{output}} = P_{\text{input}} - \text{Loss} = 9 \, kW - 0.9 \, kW = 8.1 \, kW$.

(C) The root mean square speed of gas molecules is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$. Thus, $(A) - (Q)$.
$(B)$ The pressure exerted by an ideal gas is given by $P = \frac{1}{3} n m \bar{v}^{2}$, where $n$ is the number density, $m$ is the mass of a molecule, and $\bar{v}^{2}$ is the mean square speed. Thus, $(B) - (P)$.
$(C)$ The average kinetic energy of a molecule of an ideal gas is given by $E = \frac{3}{2} k_{B} T$. Thus, $(C) - (S)$.
$(D)$ The total internal energy of $1$ mole of a diatomic gas is $U = \frac{f}{2} RT$. For a diatomic gas, the degrees of freedom $f = 5$. Therefore, $U = \frac{5}{2} RT$. Thus, $(D) - (R)$.

(B) The distance travelled by an object in the $n^{th}$ second starting from rest is given by the formula $S_n = u + \frac{a}{2}(2n-1)$.
Since the block starts from rest,the initial velocity $u = 0$.
Therefore,the distance travelled in the $n^{th}$ interval is $S_n = \frac{a}{2}(2n-1)$.
Similarly,the distance travelled in the $(n+1)^{th}$ interval is $S_{n+1} = \frac{a}{2}(2(n+1)-1) = \frac{a}{2}(2n+2-1) = \frac{a}{2}(2n+1)$.
Now,calculating the ratio $\frac{S_n}{S_{n+1}}$:
$\frac{S_n}{S_{n+1}} = \frac{\frac{a}{2}(2n-1)}{\frac{a}{2}(2n+1)} = \frac{2n-1}{2n+1}$.

(D) The Least Count $(L.C.)$ of the screw gauge is calculated as:
$L.C. = \frac{\text{Pitch}}{\text{Total circular scale divisions}} = \frac{1 \, mm}{100} = 0.01 \, mm$.
Since $1 \, mm = 0.1 \, cm$,the $L.C.$ in $cm$ is:
$L.C. = 0.01 \, mm = 0.001 \, cm$.
The diameter $(D)$ is given by the formula:
$D = \text{Main scale reading} + (\text{Circular scale reading} \times L.C.)$
Substituting the given values:
$D = 0 \, mm + (52 \times 0.01 \, mm) = 0.52 \, mm$.
Converting the diameter into $cm$:
$D = \frac{0.52}{10} \, cm = 0.052 \, cm$.

(B) Let the energy $E$ be expressed as $E = k F^{a} A^{b} T^{c}$,where $k$ is a dimensionless constant.
The dimensional formula for energy is $[M^{1} L^{2} T^{-2}]$.
The dimensional formula for force is $[M^{1} L^{1} T^{-2}]$.
The dimensional formula for acceleration is $[L^{1} T^{-2}]$.
The dimensional formula for time is $[T^{1}]$.
Substituting these into the equation:
$[M^{1} L^{2} T^{-2}] = [M^{1} L^{1} T^{-2}]^{a} [L^{1} T^{-2}]^{b} [T^{1}]^{c}$
$[M^{1} L^{2} T^{-2}] = [M^{a} L^{a+b} T^{-2a-2b+c}]$
Comparing the powers of $M, L,$ and $T$ on both sides:
For $M$: $a = 1$
For $L$: $a + b = 2 \Rightarrow 1 + b = 2 \Rightarrow b = 1$
For $T$: $-2a - 2b + c = -2 \Rightarrow -2(1) - 2(1) + c = -2 \Rightarrow -4 + c = -2 \Rightarrow c = 2$
Thus,the dimensions of energy in terms of $F, A,$ and $T$ are $[F^{1} A^{1} T^{2}]$ or $[F][A][T^{2}]$.

(B) The dimensional formula for energy $E$ is $[M L^{2} T^{-2}]$.
The dimensional formula for the gravitational constant $G$ is $[M^{-1} L^{3} T^{-2}]$.
Therefore,the dimensions of $\frac{E}{G}$ are given by:
$\frac{[E]}{[G]} = \frac{[M L^{2} T^{-2}]}{[M^{-1} L^{3} T^{-2}]}$
$= [M^{1 - (-1)}] [L^{2 - 3}] [T^{-2 - (-2)}]$
$= [M^{2}] [L^{-1}] [T^{0}]$

(D) According to Hooke's Law, $F = Kx$, where $F$ is the force, $K$ is the spring constant, and $x$ is the displacement.
Given $F = 10 \,\,N$ and $x = 5 \,\,cm = 0.05 \,\,m$.
$10 = K \times 0.05 \implies K = \frac{10}{0.05} = 200 \,\,N/m$.
The time period $T$ of a spring-mass system is given by $T = 2 \pi \sqrt{\frac{m}{K}}$.
Substituting $m = 2 \,\,kg$ and $K = 200 \,\,N/m$:
$T = 2 \pi \sqrt{\frac{2}{200}} = 2 \pi \sqrt{\frac{1}{100}} = 2 \pi \times \frac{1}{10} = \frac{2 \times 3.14}{10} = 0.628 \,\,s$.

(B) The velocity of the ball just before hitting the ground is given by $v = \sqrt{2gh}$.
Substituting the values: $v = \sqrt{2 \times 10 \times 10} = \sqrt{200} = 10\sqrt{2} \, m/s$.
Since the ball rebounds to the same height,the velocity just after the impact is $v' = -v = -10\sqrt{2} \, m/s$ (taking upward as positive).
The impulse $J$ imparted to the ball is equal to the change in momentum: $J = \Delta p = m(v_{final} - v_{initial})$.
Here,$v_{initial} = -10\sqrt{2} \, m/s$ and $v_{final} = +10\sqrt{2} \, m/s$.
$J = 0.15 \times (10\sqrt{2} - (-10\sqrt{2})) = 0.15 \times 20\sqrt{2} = 3\sqrt{2}$.
Using $\sqrt{2} \approx 1.414$,we get $J = 3 \times 1.414 = 4.242 \, kg \cdot m/s$.
Rounding to the nearest value,the magnitude of the impulse is $4.2 \, kg \cdot m/s$.

(D) At $t=4 \, s$,the velocity of the car (and thus the ball) in the horizontal direction is $v_x = u + at = 0 + 5 \times 4 = 20 \, m/s$.
After the ball is dropped,it is only under the influence of gravity.
At $t=6 \, s$,the time elapsed since the ball was dropped is $\Delta t = 6 - 4 = 2 \, s$.
The horizontal velocity remains constant: $v_x = 20 \, m/s$.
The vertical velocity at $\Delta t = 2 \, s$ is $v_y = u_y + g \Delta t = 0 + 10 \times 2 = 20 \, m/s$.
The resultant velocity is $v = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + 20^2} = 20 \sqrt{2} \, m/s$.
Since the ball is in free fall,its acceleration is equal to the acceleration due to gravity,$a = g = 10 \, m/s^{2}$ downwards.

(D) By the law of conservation of energy, the total energy at the surface equals the total energy at the maximum height $r$ where velocity is zero.
$-\frac{GMm}{R} + \frac{1}{2} m v^2 = -\frac{GMm}{r}$
Given $v = k V_{e}$ and $V_{e} = \sqrt{\frac{2GM}{R}}$, we have $v^2 = k^2 \frac{2GM}{R}$.
Substituting this into the energy equation:
$-\frac{GMm}{R} + \frac{1}{2} m \left( k^2 \frac{2GM}{R} \right) = -\frac{GMm}{r}$
Dividing by $GMm$:
$-\frac{1}{R} + \frac{k^2}{R} = -\frac{1}{r}$
$\frac{1}{r} = \frac{1}{R} - \frac{k^2}{R} = \frac{1-k^2}{R}$
$r = \frac{R}{1-k^2}$
Since $r = R + h$, the maximum height $h$ is:
$h = r - R = \frac{R}{1-k^2} - R = R \left( \frac{1 - (1-k^2)}{1-k^2} \right) = \frac{R k^2}{1-k^2}$.

(D) The speed of the particle in circular motion is $V = \frac{2 \pi R}{T}$.
The maximum height $H$ attained by a projectile is given by $H = \frac{V^2 \sin^2 \theta}{2g}$.
Given $H = 4R$,we substitute $V$ and $H$ into the formula:
$4R = \frac{(\frac{2 \pi R}{T})^2 \sin^2 \theta}{2g}$
$4R = \frac{4 \pi^2 R^2 \sin^2 \theta}{2g T^2}$
$4R = \frac{2 \pi^2 R^2 \sin^2 \theta}{g T^2}$
Solving for $\sin^2 \theta$:
$\sin^2 \theta = \frac{4R \cdot g T^2}{2 \pi^2 R^2} = \frac{2 g T^2}{\pi^2 R}$
Therefore,$\theta = \sin^{-1} \left( \frac{2 g T^2}{\pi^2 R} \right)^{1/2}$.

(A) The moment of inertia of a complete circular ring of mass $M$ and radius $R$ about an axis passing through its centre and perpendicular to its plane is $I = MR^{2}$.
Since the mass is uniformly distributed along the circumference,the mass of the remaining part of the ring after removing a $90^{\circ}$ sector (which is $1/4$ of the total circumference) is $M' = M - \frac{1}{4}M = \frac{3}{4}M$.
The moment of inertia of a point mass $m$ at a distance $R$ from the axis is $I = mR^{2}$. Since every point on the ring is at the same distance $R$ from the central axis,the moment of inertia of any part of the ring with mass $M'$ is simply $I' = M'R^{2}$.
Substituting $M' = \frac{3}{4}M$,we get $I' = \frac{3}{4}MR^{2}$.
Comparing this with $I' = KMR^{2}$,we find $K = \frac{3}{4}$.

(D) The rod is uniform,so its center of mass acts at the midpoint,i.e.,at $100 \, cm$ mark. The mass of the rod is $0.5 \, kg$.
The wedge is at $40 \, cm$. We take the torque about the pivot point (wedge) at $40 \, cm$.
Clockwise torque due to the rod's weight $(0.5 \, kg)$ acting at $100 \, cm$:
$\tau_{cw} = (0.5 \, kg \times g) \times (100 \, cm - 40 \, cm) = 0.5 \times g \times 60 \, cm$.
Clockwise torque due to mass $'m'$ acting at $160 \, cm$:
$\tau_{cw}' = (m \times g) \times (160 \, cm - 40 \, cm) = m \times g \times 120 \, cm$.
Counter-clockwise torque due to $2 \, kg$ mass acting at $20 \, cm$:
$\tau_{ccw} = (2 \, kg \times g) \times (40 \, cm - 20 \, cm) = 2 \times g \times 20 \, cm$.
For equilibrium,$\tau_{ccw} = \tau_{cw} + \tau_{cw}'$:
$2 \times g \times 20 = 0.5 \times g \times 60 + m \times g \times 120$.
Dividing by $g$:
$40 = 30 + 120m$.
$10 = 120m$.
$m = \frac{10}{120} \, kg = \frac{1}{12} \, kg$.

(A) Using the average form of Newton's law of cooling: $\frac{dT}{dt} = k(T_{avg} - T_{room})$.
For the first interval ($90^{\circ} C$ to $80^{\circ} C$):
$\frac{90-80}{t} = k\left(\frac{90+80}{2} - 20\right) \implies \frac{10}{t} = k(85 - 20) = 65k \dots (i)$
For the second interval ($80^{\circ} C$ to $60^{\circ} C$):
$\frac{80-60}{t'} = k\left(\frac{80+60}{2} - 20\right) \implies \frac{20}{t'} = k(70 - 20) = 50k \dots (ii)$
Dividing equation $(i)$ by $(ii)$:
$\frac{10/t}{20/t'} = \frac{65k}{50k}$
$\frac{10}{t} \times \frac{t'}{20} = \frac{65}{50}$
$\frac{t'}{2t} = \frac{13}{10}$
$t' = \frac{13}{10} \times 2t = \frac{13}{5} t$.

(D) Let the height from the surface be $x$. The distance fallen by the particle is $(S - x)$.
Using the equation of motion $v^2 = u^2 + 2as$,where $u = 0$ and $a = g$,we get $v^2 = 2g(S - x)$.
The potential energy at height $x$ is $PE = mgx$.
The kinetic energy at that instant is $KE = \frac{1}{2}mv^2 = \frac{1}{2}m(2g(S - x)) = mg(S - x)$.
According to the problem,$KE = 3 \times PE$.
Substituting the expressions: $mg(S - x) = 3(mgx)$.
Dividing both sides by $mg$: $S - x = 3x$.
$S = 4x \Rightarrow x = \frac{S}{4}$.
Now,substitute $x$ into the velocity equation: $v^2 = 2g(S - \frac{S}{4}) = 2g(\frac{3S}{4}) = \frac{3gS}{2}$.
Therefore,$v = \sqrt{\frac{3gS}{2}}$.
The height is $\frac{S}{4}$ and the speed is $\sqrt{\frac{3gS}{2}}$.

(D) In the given circuit,let the input terminal be $A$ and the output terminal be $B$.
The two capacitors of capacitance $C$ in the upper and lower branches are connected in series with each other,but they are also connected to a third capacitor $C$ in parallel.
However,looking closely at the circuit,the two capacitors in the top and bottom branches are in series with each other,and this combination is in parallel with the middle capacitor.
Wait,let's re-evaluate: The two capacitors in the top and bottom branches are connected in series,and their combination is in parallel with the middle capacitor.
Actually,the circuit shows two capacitors in series (top and bottom) which are then in parallel with the middle capacitor.
Let the series combination of the two capacitors be $C_s$. Then $1 / C_s = 1 / C + 1 / C = 2 / C$,so $C_s = C / 2$.
Now,this $C_s$ is in parallel with the middle capacitor $C$.
Therefore,the equivalent capacitance $C_{eq} = C_s + C = C / 2 + C = 3 C / 2$.

(D) Polar molecules are those in which the centres of positive and negative charges do not coincide,even in the absence of an external electric field.
Because these centres are separated by a small distance,these molecules possess a permanent electric dipole moment.

(D) The magnetic field $B$ produced by an infinitely long straight conductor at a distance $R$ is given by:
$B = \frac{\mu_{0} I}{2 \pi R}$
Given: $I = 5 \, A$,$R = 20 \, cm = 0.2 \, m$,$v = 10^{5} \, m/s$,$q = e = 1.6 \times 10^{-19} \, C$.
Substituting the values:
$B = \frac{4 \pi \times 10^{-7} \times 5}{2 \pi \times 0.2} = \frac{2 \times 10^{-7} \times 5}{0.2} = 5 \times 10^{-6} \, T$.
The magnetic force $F$ on a moving charge is given by $F = qvB \sin \theta$. Since the electron moves parallel to the conductor,the angle $\theta$ between the velocity vector and the magnetic field (which is perpendicular to the plane of the conductor and the electron's path) is $90^{\circ}$.
$F = qvB \sin 90^{\circ} = qvB$
$F = (1.6 \times 10^{-19} \, C) \times (10^{5} \, m/s) \times (5 \times 10^{-6} \, T)$
$F = 8 \times 10^{-20} \, N$.
Thus,the magnitude of the force is $8 \times 10^{-20} \, N$.

(C) For a thick cylindrical cable of radius $R$ carrying a uniformly distributed current $I$:
$1$. Inside the cable $(r < R)$: Using Ampere's circuital law,the magnetic field is given by $B_{\text{in}} = \frac{\mu_0 I r}{2 \pi R^2}$. This shows that $B \propto r$,which is a linear relationship.
$2$. Outside the cable $(r > R)$: The magnetic field is given by $B_{\text{out}} = \frac{\mu_0 I}{2 \pi r}$. This shows that $B \propto 1/r$,which is a hyperbolic relationship.
$3$. At the surface $(r = R)$: The magnetic field is maximum,$B_0 = \frac{\mu_0 I}{2 \pi R}$.
Therefore,the graph increases linearly from the center to the surface and then decreases hyperbolically as the distance increases beyond the surface.

(A) In a potentiometer, the potential drop per unit length $(\phi)$ is constant.
The balance length $(l)$ is directly proportional to the $EMF$ $(E)$ of the cell, i.e., $E = \phi l$.
Therefore, the ratio is given by: $\frac{E_1}{E_2} = \frac{l_1}{l_2}$.
Given: $E_1 = 1.5\, V$, $l_1 = 36\, cm$, $E_2 = 2.5\, V$.
Substituting the values: $\frac{1.5}{2.5} = \frac{36}{l_2}$.
Simplifying the ratio: $\frac{3}{5} = \frac{36}{l_2}$.
Solving for $l_2$: $l_2 = \frac{36 \times 5}{3} = 12 \times 5 = 60\, cm$.

(B) The direction of propagation of an electromagnetic wave is given by the direction of the Poynting vector,which is $\vec{E} \times \vec{B}$.
Given the wave propagates in the $x$-direction,we must have $\vec{E} \times \vec{B} = \hat{i}$.
Let us check option $B$: $\vec{E} = -\hat{j} + \hat{k}$ and $\vec{B} = -\hat{j} - \hat{k}$.
Calculating the cross product: $(-\hat{j} + \hat{k}) \times (-\hat{j} - \hat{k}) = (-\hat{j} \times -\hat{j}) - (\hat{j} \times -\hat{k}) + (\hat{k} \times -\hat{j}) - (\hat{k} \times -\hat{k})$.
$= 0 + (\hat{j} \times \hat{k}) - (\hat{k} \times \hat{j}) - 0 = \hat{i} - (-\hat{i}) = 2\hat{i}$.
Since the resulting vector is in the $x$-direction,option $B$ is correct.

(D) Given:
Amplitude of current,$I_{0} = 10 \sqrt{2} \, A$.
Potential difference across inductor,$V_{L} = 40 \, V$.
Potential difference across capacitor,$V_{C} = 10 \, V$.
Potential difference across resistor,$V_{R} = 40 \, V$.
Step $1$: Calculate the $RMS$ current $(I_{RMS})$.
$I_{RMS} = \frac{I_{0}}{\sqrt{2}} = \frac{10 \sqrt{2}}{\sqrt{2}} = 10 \, A$.
Step $2$: Calculate the $RMS$ voltage of the source $(V_{RMS})$.
In an $LCR$ series circuit,the total voltage is given by:
$V_{RMS} = \sqrt{V_{R}^{2} + (V_{L} - V_{C})^{2}}$
$V_{RMS} = \sqrt{(40)^{2} + (40 - 10)^{2}}$
$V_{RMS} = \sqrt{40^{2} + 30^{2}} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \, V$.
Step $3$: Calculate the impedance $(Z)$ of the circuit.
$Z = \frac{V_{RMS}}{I_{RMS}} = \frac{50 \, V}{10 \, A} = 5 \, \Omega$.
Therefore,the impedance of the circuit is $5 \, \Omega$.

(C) The power $P$ emitted by a source is given by the product of the number of photons per second $n$ and the energy of a single photon $E = \frac{hc}{\lambda}$.
$P = n \frac{hc}{\lambda} \Rightarrow n = \frac{P \lambda}{hc}$
Given: $P = 3.3 \times 10^{-3} \, W$,$\lambda = 600 \times 10^{-9} \, m$,$h = 6.6 \times 10^{-34} \, Js$,and $c = 3 \times 10^8 \, m/s$.
Substituting the values:
$n = \frac{3.3 \times 10^{-3} \times 600 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8}$
$n = \frac{3.3 \times 600 \times 10^{-12}}{19.8 \times 10^{-26}}$
$n = \frac{1980 \times 10^{-12}}{19.8 \times 10^{-26}} = 100 \times 10^{14} = 10^{16}$
Thus,the number of photons emitted per second is $10^{16}$.

(C) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = K_{\max} + \phi$.
Given that the work function $\phi$ is negligible,we have $\frac{hc}{\lambda} = K_{\max}$.
The de-Broglie wavelength of the emitted photoelectron is given by $\lambda_{d} = \frac{h}{\sqrt{2mK_{\max}}}$.
Squaring both sides,we get $\lambda_{d}^{2} = \frac{h^{2}}{2mK_{\max}}$,which implies $K_{\max} = \frac{h^{2}}{2m\lambda_{d}^{2}}$.
Substituting the expression for $K_{\max}$ into the photoelectric equation: $\frac{hc}{\lambda} = \frac{h^{2}}{2m\lambda_{d}^{2}}$.
Solving for $\lambda$,we get $\lambda = \frac{hc \cdot 2m\lambda_{d}^{2}}{h^{2}} = \left(\frac{2mc}{h}\right) \lambda_{d}^{2}$.

(A) The drift velocity is given by $v_{d} = \frac{e E}{m} \tau$. Thus, $(A) \rightarrow (R)$.
The electrical resistivity $\rho$ is defined as the ratio of electric field to current density, $\rho = \frac{E}{J}$. Thus, $(B) \rightarrow (S)$.
The relaxation period $\tau$ is related to resistivity by $\rho = \frac{m}{n e^{2} \tau}$, which implies $\tau = \frac{m}{n e^{2} \rho}$. Thus, $(C) \rightarrow (P)$.
The current density $J$ is given by $J = n e v_{d}$. Thus, $(D) \rightarrow (Q)$.
Therefore, the correct matching is $(A)-(R), (B)-(S), (C)-(P), (D)-(Q)$.

(B) The electric field lines are closer together near the $+q$ charge,which means the magnitude of the electric field is stronger at the position of the $+q$ charge than at the position of the $-q$ charge. Let $E_1$ be the field at $+q$ and $E_2$ be the field at $-q$. Thus,$|E_1| > |E_2|$.
The force on the positive charge is $F_+ = qE_1$ (towards the right) and the force on the negative charge is $F_- = qE_2$ (towards the left).
Since $|E_1| > |E_2|$,the net force $F_{net} = q(E_1 - E_2)$ acts towards the right.
$A$ physical system naturally moves in a direction that decreases its potential energy. Therefore,the dipole will move towards the right as its potential energy decreases.

(A) The charge on the capacitor is given by $q = CV$.
Substituting the given voltage $V = V_{0} \sin \omega t$,we get $q = C V_{0} \sin \omega t$.
The displacement current $I_{d}$ is equal to the rate of change of charge on the capacitor plates,which is $I_{d} = \frac{dq}{dt}$.
$I_{d} = \frac{d}{dt} (C V_{0} \sin \omega t)$.
Since $C$ and $V_{0}$ are constants,$I_{d} = C V_{0} \frac{d}{dt} (\sin \omega t)$.
Using the derivative of $\sin \omega t$,which is $\omega \cos \omega t$,we get $I_{d} = C V_{0} \omega \cos \omega t$.
Therefore,$I_{d} = V_{0} \omega C \cos \omega t$.

(D) Let the resistance of each wire be $R$. Since all four wires have equal length,equal area of cross-section,and are made of the same material,their resistances are identical.
When $n$ identical resistors are connected in parallel,the equivalent resistance $R_p$ is given by $R_p = \frac{R}{n}$.
Given $n = 4$ and $R_p = 0.25\, \Omega$,we have $0.25 = \frac{R}{4}$,which implies $R = 0.25 \times 4 = 1\, \Omega$.
When these four resistors are connected in series,the equivalent resistance $R_s$ is given by $R_s = n \times R$.
Therefore,$R_s = 4 \times 1 = 4\, \Omega$.

(C) $1$. In the first step,${}_{Z}^{A}X \rightarrow {}_{Z-1}B$,the atomic number decreases by $1$. This corresponds to $\beta^{+}$ decay (positron emission).
$2$. In the second step,${}_{Z-1}B \rightarrow {}_{Z-3}C$,the atomic number decreases by $2$. This corresponds to $\alpha$ decay.
$3$. In the third step,${}_{Z-3}C \rightarrow {}_{Z-2}D$,the atomic number increases by $1$. This corresponds to $\beta^{-}$ decay (electron emission).
$4$. Therefore,the sequence of decay particles is $\beta^{+}, \alpha, \beta^{-}$.

(C) The energy density $u$ (energy per unit volume) in a parallel plate capacitor is given by the formula: $u = \frac{1}{2} \varepsilon_{0} E^{2}$.
The volume $V$ of the space between the plates is the product of the area $A$ and the distance $d$,so $V = Ad$.
The total energy $U$ stored in the capacitor is the product of the energy density and the volume:
$U = u \times V$
$U = (\frac{1}{2} \varepsilon_{0} E^{2}) \times (Ad)$
$U = \frac{1}{2} \varepsilon_{0} E^{2} Ad$.

(C) The current $I$ in a semiconductor is given by the formula $I = n e A v_d$,where $v_d = \mu E$.
Thus,$I = n e A \mu E$.
In an $n$-type semiconductor,the majority charge carriers are electrons,and in a $p$-type semiconductor,the majority charge carriers are holes.
Given that the carrier concentrations are the same $(n_e = n_h)$,the current depends on the mobility $(\mu)$ of the charge carriers.
Since the mobility of electrons $(\mu_e)$ is greater than the mobility of holes $(\mu_h)$ in a semiconductor,it follows that $I_n > I_p$.
Therefore,the current in the $n$-type semiconductor is greater than the current in the $p$-type semiconductor.

(B) For a parallel beam of light to remain parallel after passing through a combination of two lenses,the second focal point of the first lens must coincide with the first focal point of the second lens.
Let $f_1 = 20 \, cm$ (convex lens) and $f_2 = -5 \, cm$ (concave lens).
The distance $d$ between the lenses is given by the formula $d = f_1 + f_2$.
Substituting the values,we get $d = 20 \, cm + (-5 \, cm) = 15 \, cm$.
Thus,the distance $d$ is $15 \, cm$.

(A) From the geometry of the prism,the light ray enters the prism normally to the hypotenuse face. Thus,the angle of incidence at the first surface is $i_1 = 0^{\circ}$,which implies the angle of refraction $r_1 = 0^{\circ}$.
Given the prism angle $A = 30^{\circ}$,we use the relation $r_1 + r_2 = A$. Substituting $r_1 = 0^{\circ}$,we get $r_2 = 30^{\circ}$.
Applying Snell's law at the second surface (the vertical face): $\mu \sin r_2 = 1 \times \sin e$,where $\mu = \sqrt{3}$ and $e$ is the angle of emergence.
$\sqrt{3} \sin 30^{\circ} = \sin e$
$\sqrt{3} \times \frac{1}{2} = \sin e$
$\sin e = \frac{\sqrt{3}}{2}$
Therefore,$e = 60^{\circ}$.

(C) Statement $(A)$ is correct: $A$ Zener diode is specifically designed to operate in the reverse breakdown region,making it an effective voltage regulator.
Statement $(B)$ is incorrect: The potential barrier for a Germanium $(Ge)$ $p-n$ junction is approximately $0.3 \, V$,while for a Silicon $(Si)$ $p-n$ junction,it is approximately $0.7 \, V$. Therefore,the range $0.1 \, V$ to $0.3 \, V$ is not representative of all common $p-n$ junctions,as it excludes the widely used Silicon diodes.

(B) When two conductors are connected by a wire,charge flows until their potentials become equal.
Let $V_{1}$ and $V_{2}$ be the potentials of the spheres. Since they are connected,$V_{1} = V_{2}$.
The potential of a charged spherical conductor is given by $V = \frac{kQ}{R}$.
Thus,$\frac{kQ_{1}}{R_{1}} = \frac{kQ_{2}}{R_{2}}$,which implies $\frac{Q_{1}}{Q_{2}} = \frac{R_{1}}{R_{2}}$.
The surface charge density $\sigma$ is defined as $\sigma = \frac{Q}{A} = \frac{Q}{4\pi R^{2}}$.
Therefore,the ratio of surface charge densities is:
$\frac{\sigma_{1}}{\sigma_{2}} = \frac{Q_{1} / (4\pi R_{1}^{2})}{Q_{2} / (4\pi R_{2}^{2})} = \frac{Q_{1}}{Q_{2}} \times \frac{R_{2}^{2}}{R_{1}^{2}}$.
Substituting $\frac{Q_{1}}{Q_{2}} = \frac{R_{1}}{R_{2}}$,we get:
$\frac{\sigma_{1}}{\sigma_{2}} = \frac{R_{1}}{R_{2}} \times \frac{R_{2}^{2}}{R_{1}^{2}} = \frac{R_{2}}{R_{1}}$.

(D) The magnifying power of an astronomical telescope is given by $MP = \frac{f_o}{f_e}$, where $f_o$ is the focal length of the objective and $f_e$ is the focal length of the eyepiece.
The resolving power $(R.P.)$ of a telescope is given by $R.P. = \frac{a}{1.22 \lambda}$, where $a$ is the aperture of the objective lens and $\lambda$ is the wavelength of light.
$1$. $A$ large aperture $(a)$ increases the resolving power, allowing the telescope to distinguish between two closely spaced objects.
$2$. $A$ large aperture increases the light-gathering power of the objective, which is proportional to the area of the lens $(\pi r^2)$. This ensures that faint objects are visible and the image quality is improved.
Since all these factors contribute to the performance of an astronomical telescope, the correct answer is $D$.

(D) The initial nucleus has a mass number $A = 240$ and binding energy per nucleon $BE_{initial} = 7.6 \, MeV$.
The total binding energy of the initial nucleus is $240 \times 7.6 = 1824 \, MeV$.
After the process,the nucleus breaks into two fragments,each with mass number $A = 120$ and binding energy per nucleon $BE_{final} = 8.5 \, MeV$.
The total binding energy of the two fragments is $2 \times (120 \times 8.5) = 240 \times 8.5 = 2040 \, MeV$.
The gain in binding energy is the difference between the total binding energy of the products and the reactants:
$Q = BE_{products} - BE_{reactants} = 2040 \, MeV - 1824 \, MeV = 216 \, MeV$.

(B) The activity $A$ of a radioactive sample at time $t$ is given by the formula $A = A_{0} \left(\frac{1}{2}\right)^{t/T_{H}}$,where $A_{0}$ is the initial activity and $T_{H}$ is the half-life.
Given $T_{H} = 100 \, hours$ and $t = 150 \, hours$.
The fraction of activity remaining is $\frac{A}{A_{0}} = \left(\frac{1}{2}\right)^{150/100}$.
This simplifies to $\left(\frac{1}{2}\right)^{1.5} = \left(\frac{1}{2}\right)^{3/2} = \frac{1}{2^{3/2}} = \frac{1}{2 \sqrt{2}}$.

(C) Given: $L = 5.0 \, H$,$C = 80 \, \mu F = 80 \times 10^{-6} \, F$,$R = 40 \, \Omega$.
First,calculate the resonant angular frequency $\omega_0$:
$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{5.0 \times 80 \times 10^{-6}}} = \frac{1}{\sqrt{400 \times 10^{-6}}} = \frac{1}{0.02} = 50 \, rad/s$.
Next,calculate the bandwidth $\Delta \omega$:
$\Delta \omega = \frac{R}{L} = \frac{40}{5.0} = 8 \, rad/s$.
The frequencies at which the power is half the maximum power (half-power frequencies) are given by $\omega = \omega_0 \pm \frac{\Delta \omega}{2}$.
$\omega_1 = \omega_0 - \frac{\Delta \omega}{2} = 50 - \frac{8}{2} = 50 - 4 = 46 \, rad/s$.
$\omega_2 = \omega_0 + \frac{\Delta \omega}{2} = 50 + \frac{8}{2} = 50 + 4 = 54 \, rad/s$.
Thus,the angular frequencies are $46 \, rad/s$ and $54 \, rad/s$.

(D) The magnetic field $B_{1}$ at the center of a circular loop of radius $R_{1}$ carrying current $I_{1}$ is given by $B_{1} = \frac{\mu_{0} I_{1}}{2 R_{1}}$.
The magnetic flux $\phi_{2}$ linked with the smaller loop of radius $R_{2}$ is $\phi_{2} = B_{1} A_{2}$,where $A_{2} = \pi R_{2}^{2}$ is the area of the smaller loop.
Since $R_{1} >> R_{2}$,we can assume the magnetic field $B_{1}$ is uniform over the area of the smaller loop.
The mutual inductance $M$ is defined as $M = \frac{\phi_{2}}{I_{1}}$.
Substituting the expressions,we get $M = \frac{B_{1} A_{2}}{I_{1}} = \frac{(\frac{\mu_{0} I_{1}}{2 R_{1}}) (\pi R_{2}^{2})}{I_{1}} = \frac{\mu_{0} \pi R_{2}^{2}}{2 R_{1}}$.
Thus,$M \propto \frac{R_{2}^{2}}{R_{1}}$.

(D) $1$. First,consider the image formation by the lens:
Given $u = -60\, cm$ and $f = +30\, cm$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{30} - \frac{1}{60} = \frac{2-1}{60} = \frac{1}{60}$
So,$v = +60\, cm$ from the lens.
$2$. This real image acts as an object for the plane mirror. The mirror is at $40\, cm$ from the lens. Since the image is at $60\, cm$ from the lens,it is $60 - 40 = 20\, cm$ behind the mirror.
$3$. The plane mirror forms a virtual image at the same distance behind it as the object is in front of it. Here,the object is $20\, cm$ behind the mirror,so the mirror forms a real image $20\, cm$ in front of it.
$4$. This image acts as a virtual object for the second refraction through the lens. The distance of this object from the lens is $40 - 20 = 20\, cm$. Thus,$u' = -20\, cm$ (as it is on the same side as the light source).
$5$. Using the lens formula again:
$\frac{1}{v'} = \frac{1}{f} + \frac{1}{u'} = \frac{1}{30} - \frac{1}{20} = \frac{2-3}{60} = -\frac{1}{60}$
So,$v' = -60\, cm$.
The negative sign indicates that the final image is virtual and formed $60\, cm$ in front of the lens,which is $20\, cm$ behind the plane mirror.

(A) Total length of wire $L = 12 a$.
Case $(i)$: Equilateral triangle of side $a$.
Perimeter of one turn $= 3 a$.
Number of turns $N_{1} = \frac{12 a}{3 a} = 4$.
Area of one triangle $A_{1} = \frac{\sqrt{3}}{4} a^{2}$.
Magnetic dipole moment $\mu_{1} = N_{1} I A_{1} = 4 \times I \times \frac{\sqrt{3}}{4} a^{2} = \sqrt{3} I a^{2}$.
Case $(ii)$: Square of side $a$.
Perimeter of one turn $= 4 a$.
Number of turns $N_{2} = \frac{12 a}{4 a} = 3$.
Area of one square $A_{2} = a^{2}$.
Magnetic dipole moment $\mu_{2} = N_{2} I A_{2} = 3 \times I \times a^{2} = 3 I a^{2}$.
Thus,the magnetic dipole moments are $\sqrt{3} I a^{2}$ and $3 I a^{2}$.

(B) In the given circuit,the current $i_{1}$ splits into two parallel branches containing resistors $r_{2}$ and $r_{3}$.
According to the current divider rule,the current $i_{3}$ flowing through resistor $r_{3}$ is given by:
$i_{3} = i_{1} \left( \frac{r_{2}}{r_{2} + r_{3}} \right)$
Dividing both sides by $i_{1}$,we get the ratio:
$\frac{i_{3}}{i_{1}} = \frac{r_{2}}{r_{2} + r_{3}}$

(A) Given that the transformer is ideal (ignoring power losses),the input power is equal to the output power.
$P_{\text{in}} = P_{\text{out}}$
Since $P_{\text{out}} = 44 \, W$ and $V_p = 220 \, V$,we use the relation $P_{\text{in}} = V_p \times I_p$.
$220 \times I_p = 44$
$I_p = \frac{44}{220} \, A$
$I_p = \frac{1}{5} \, A = 0.2 \, A$
Therefore,the current in the primary circuit is $0.2 \, A$.

(A) Let $r$ be the radius of each small drop and $q$ be the charge on each small drop.
The potential of each small drop is $V_S = \frac{kq}{r} = 220 \, V$.
When $N = 27$ drops combine to form a bigger drop of radius $R$ and charge $Q$,the volume remains constant.
$\frac{4}{3} \pi R^3 = N \times \frac{4}{3} \pi r^3 \implies R = N^{1/3} r = (27)^{1/3} r = 3r$.
The total charge on the bigger drop is $Q = Nq = 27q$.
The potential of the bigger drop is $V_B = \frac{kQ}{R} = \frac{k(Nq)}{N^{1/3}r} = N^{2/3} \frac{kq}{r} = N^{2/3} V_S$.
Substituting the values: $V_B = (27)^{2/3} \times 220 = (3^3)^{2/3} \times 220 = 3^2 \times 220 = 9 \times 220 = 1980 \, V$.

(D) The circuit consists of an $AND$ gate and a $NAND$ gate whose outputs are fed into an $OR$ gate. Let the inputs be $A, B, C$. The output of the $AND$ gate is $A \cdot B$. The output of the $NAND$ gate is $\overline{B \cdot C}$. The final output $Y$ of the $OR$ gate is $Y = (A \cdot B) + \overline{(B \cdot C)}$.
Analyzing the truth table for the intervals:
- Interval $0-t_1$: $A=0, B=0, C=1$. $Y = (0 \cdot 0) + \overline{(0 \cdot 1)} = 0 + 1 = 1$.
- Interval $t_1-t_2$: $A=1, B=0, C=1$. $Y = (1 \cdot 0) + \overline{(0 \cdot 1)} = 0 + 1 = 1$.
- Interval $t_2-t_3$: $A=0, B=1, C=0$. $Y = (0 \cdot 1) + \overline{(1 \cdot 0)} = 0 + 1 = 1$.
- Interval $t_3-t_4$: $A=1, B=1, C=0$. $Y = (1 \cdot 1) + \overline{(1 \cdot 0)} = 1 + 1 = 1$.
- Interval $t_4-t_5$: $A=0, B=0, C=1$. $Y = (0 \cdot 0) + \overline{(0 \cdot 1)} = 0 + 1 = 1$.
- Interval $t_5-t_6$: $A=1, B=0, C=1$. $Y = (1 \cdot 0) + \overline{(0 \cdot 1)} = 0 + 1 = 1$.
Since the output is $1$ $(5 \text{ V})$ for all intervals,the correct representation is a constant $5 \text{ V}$ signal.

(B) Given the Lorentz force equation: $\overrightarrow{F} = q(\vec{v} \times \overrightarrow{B})$.
Since $q = 1$,we have $\overrightarrow{F} = \vec{v} \times \overrightarrow{B}$.
Let $\overrightarrow{B} = B \hat{i} + B \hat{j} + B_0 \hat{k}$.
The cross product is given by the determinant:
$\vec{v} \times \overrightarrow{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 6 \\ B & B & B_0 \end{vmatrix} = \hat{i}(4B_0 - 6B) - \hat{j}(2B_0 - 6B) + \hat{k}(2B - 4B) = 4 \hat{i} - 20 \hat{j} + 12 \hat{k}$.
Comparing the components:
$1$) $4B_0 - 6B = 4$
$2$) $-(2B_0 - 6B) = -20 \implies 2B_0 - 6B = 20$
$3$) $2B - 4B = 12 \implies -2B = 12 \implies B = -6$.
Substitute $B = -6$ into equation $(2)$:
$2B_0 - 6(-6) = 20 \implies 2B_0 + 36 = 20 \implies 2B_0 = -16 \implies B_0 = -8$.
Thus,$\overrightarrow{B} = -6 \hat{i} - 6 \hat{j} - 8 \hat{k}$.