NEET 2019 Physics Question Paper with Answer and Solution

(C) Average speed is defined as the ratio of total distance traveled to the total time taken.
Let the distance between $X$ and $Y$ be $d$.
Time taken to travel from $X$ to $Y$ is $t_1 = \frac{d}{v_1}$.
Time taken to travel from $Y$ to $X$ is $t_2 = \frac{d}{v_2}$.
Total distance $= d + d = 2d$.
Total time $= t_1 + t_2 = \frac{d}{v_1} + \frac{d}{v_2}$.
Average speed $\bar v = \frac{\text{Total distance}}{\text{Total time}} = \frac{2d}{\frac{d}{v_1} + \frac{d}{v_2}}$.
Canceling $d$ from the numerator and denominator,we get $\bar v = \frac{2}{\frac{1}{v_1} + \frac{1}{v_2}}$.
Rearranging this gives $\frac{2}{\bar v} = \frac{1}{v_1} + \frac{1}{v_2}$.

(C) The work done during the phase change at constant pressure is given by $W = P \Delta V$.
Given $P = 1.013 \times 10^5\,Pa$ (atmospheric pressure),$V_{liquid} = 1\,cm^3 = 10^{-6}\,m^3$,and $V_{vapour} = 1671\,cm^3 = 1671 \times 10^{-6}\,m^3$.
$W = 1.013 \times 10^5 \times (1671 - 1) \times 10^{-6} \approx 169\,J$. Using the standard approximation $P = 10^5\,Pa$ as implied by the options: $W = 10^5 \times 1670 \times 10^{-6} = 167\,J$.
The heat supplied is $Q = mL = 1\,g \times 2256\,J/g = 2256\,J$.
From the first law of thermodynamics,$Q = \Delta U + W$,so $\Delta U = Q - W$.
$\Delta U = 2256\,J - 167\,J = 2089\,J$.

(B) In an adiabatic process,the system is thermally insulated from its surroundings.
Therefore,there is no exchange of heat between the system and the surroundings.
This means that the heat absorbed or released by the system is zero.
Mathematically,this is represented as $\Delta Q = 0$.

(B) According to the kinetic theory of gases,the average kinetic energy $(KE)$ of gas molecules is directly proportional to the absolute temperature $(T)$ of the gas.
The relation is given by: $KE = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant.
Therefore,as the temperature of the gas increases,the average kinetic energy of the gas molecules also increases.

(C) For the block to remain stationary,the upward frictional force must balance the downward gravitational force.
$f_s = mg$
The normal force $N$ provided by the wall acts as the centripetal force for the block's circular motion:
$N = mr\omega^2$
The maximum static frictional force is given by:
$f_{L} = \mu N = \mu mr\omega^2$
For the block to stay in equilibrium,the frictional force must be at least equal to the weight:
$f_{L} \geq mg$
$\mu mr\omega^2 \geq mg$
Solving for $\omega$:
$\omega^2 \geq \frac{g}{\mu r}$
$\omega \geq \sqrt{\frac{g}{\mu r}}$
Substituting the given values ($g = 10\; m/s^2$,$\mu = 0.1$,$r = 1\; m$):
$\omega_{\min} = \sqrt{\frac{10}{0.1 \times 1}} = \sqrt{\frac{10}{0.1}} = \sqrt{100} = 10\; rad/s$

(B) For a one-dimensional elastic collision where body $B$ is initially at rest,the final velocity $v_1$ of body $A$ is given by:
$v_1 = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) u$
Substituting $m_1 = 4m$ and $m_2 = 2m$:
$v_1 = \left( \frac{4m - 2m}{4m + 2m} \right) u = \left( \frac{2m}{6m} \right) u = \frac{u}{3}$
Initial kinetic energy of body $A$ is $K_i = \frac{1}{2}(4m)u^2 = 2mu^2$.
Final kinetic energy of body $A$ is $K_f = \frac{1}{2}(4m)v_1^2 = \frac{1}{2}(4m)\left(\frac{u}{3}\right)^2 = \frac{2mu^2}{9}$.
The energy lost by body $A$ is $\Delta K = K_i - K_f = 2mu^2 - \frac{2mu^2}{9} = \frac{16mu^2}{9}$.
The fraction of energy lost is $\frac{\Delta K}{K_i} = \frac{16mu^2 / 9}{2mu^2} = \frac{16}{18} = \frac{8}{9}$.

(A) Let the speed of the swimmer in still water be $v = 20 \; m/s$ and the speed of the river be $u = 10 \; m/s$.
To cross the river along the shortest path (perpendicular to the river flow),the swimmer must swim at an angle $\theta$ with respect to the north (perpendicular to the bank) such that the horizontal component of his velocity cancels the river's velocity.
From the geometry of the velocity vector triangle,we have:
$\sin \theta = \frac{u}{v}$
Substituting the given values:
$\sin \theta = \frac{10}{20} = \frac{1}{2}$
Therefore,$\theta = \arcsin(1/2) = 30^{\circ}$.
Thus,the swimmer should make his strokes at an angle of $30^{\circ}$ west of north.

(C) For a mass $m$ moving in a vertical circle of radius $R$,the tension $T$ in the wire at any angular position $\theta$ from the lowest point is given by the equation:
$T - mg \cos \theta = \frac{mv^2}{R}$
$T = mg \cos \theta + \frac{mv^2}{R}$
At the lowest point,$\theta = 0^{\circ}$,so $\cos 0^{\circ} = 1$. Also,the velocity $v$ is maximum at the lowest point due to the conservation of energy.
Since both $\cos \theta$ and $v$ are at their maximum values at the lowest point,the tension $T$ is maximum at this position.
Therefore,the wire is most likely to break when the mass is at the lowest point.

(B) The given equation is $y = A_{0} + A \sin \omega t + B \cos \omega t$.
To find the amplitude,we express the time-dependent part in the form $R \sin(\omega t + \phi)$.
Let $A = R \cos \phi$ and $B = R \sin \phi$.
Then $A^{2} + B^{2} = R^{2} (\cos^{2} \phi + \sin^{2} \phi) = R^{2}$.
Thus,$R = \sqrt{A^{2} + B^{2}}$.
The equation becomes $y = A_{0} + \sqrt{A^{2} + B^{2}} \sin(\omega t + \phi)$.
Here,$A_{0}$ represents the shift in the mean position,and the coefficient of the trigonometric term,$\sqrt{A^{2} + B^{2}}$,represents the amplitude of the oscillation.

(D) In one complete vibration,the particle starts from a position,moves to the extreme,returns to the starting position,moves to the other extreme,and finally returns to the starting position.
Therefore,the net displacement of the particle in one complete oscillation is $0$.
Since average velocity is defined as the ratio of total displacement to the total time taken,we have:
$\text{Average velocity} = \frac{\text{Total Displacement}}{\text{Total Time}} = \frac{0}{T} = 0$.

(C) The pressure inside a soap bubble is given by $P = P_{0} + \frac{4T}{R}$,where $P_{0}$ is the atmospheric pressure,$T$ is the surface tension,and $R$ is the radius of the bubble.
The pressure at a depth $Z_{0}$ below the free surface of water is given by $P = P_{0} + \rho g Z_{0}$.
Equating the two pressures: $P_{0} + \rho g Z_{0} = P_{0} + \frac{4T}{R}$.
Simplifying,we get $\rho g Z_{0} = \frac{4T}{R}$.
Solving for $Z_{0}$: $Z_{0} = \frac{4T}{\rho g R}$.
Substituting the given values: $T = 2.5 \times 10^{-2}\; N/m$,$R = 1\; mm = 10^{-3}\; m$,$\rho = 10^{3}\; kg/m^{3}$,and $g = 10\; m/s^{2}$.
$Z_{0} = \frac{4 \times 2.5 \times 10^{-2}}{10^{3} \times 10 \times 10^{-3}} = \frac{10^{-1}}{10} = 10^{-2}\; m$.
Converting to centimeters: $Z_{0} = 10^{-2} \times 100\; cm = 1\; cm$.

(D) The problem states that the increase in length for both rods is independent of the increase in temperature,which implies that the change in length $(\Delta \ell)$ for both rods must be equal for any given change in temperature $(\Delta T)$.
Given:
$\ell_{Cu} = 88\; cm$
$\alpha_{Cu} = 1.7 \times 10^{-5}\; K^{-1}$
$\alpha_{Al} = 2.2 \times 10^{-5}\; K^{-1}$
The formula for thermal expansion is $\Delta \ell = \ell \alpha \Delta T$.
Since $(\Delta \ell)_{Cu} = (\Delta \ell)_{Al}$,we have:
$\ell_{Cu} \alpha_{Cu} \Delta T = \ell_{Al} \alpha_{Al} \Delta T$
Canceling $\Delta T$ from both sides:
$\ell_{Cu} \alpha_{Cu} = \ell_{Al} \alpha_{Al}$
Substituting the values:
$88 \times 1.7 \times 10^{-5} = \ell_{Al} \times 2.2 \times 10^{-5}$
Solving for $\ell_{Al}$:
$\ell_{Al} = \frac{88 \times 1.7}{2.2}$
$\ell_{Al} = 40 \times 1.7 = 68\; cm$.

(D) The rate of heat flow is given by the formula: $\frac{dQ}{dt} = -K A \frac{dT}{dx}$.
Here,$\frac{dQ}{dt}$ is the heat flow rate in Watts ($W$ or $J/s$),$A$ is the area in $m^2$,and $\frac{dT}{dx}$ is the temperature gradient in $K/m$.
Rearranging for thermal conductivity $(K)$:
$K = \frac{(dQ/dt)}{A \cdot (dT/dx)}$
Substituting the units:
$K = \frac{W}{m^2 \cdot (K/m)} = \frac{W}{m \cdot K} = W m^{-1} K^{-1}$.
Therefore,the unit of thermal conductivity is $W m^{-1} K^{-1}$.

(C) The elastic potential energy $U$ stored in a stretched wire is given by the formula $U = \frac{1}{2} \times \text{force} \times \text{elongation}$.
Here,the force applied is the weight of the block,which is $F = Mg$.
The elongation produced in the wire is $l$.
Substituting these values into the formula:
$U = \frac{1}{2} (Mg) (l) = \frac{1}{2} Mgl$.
Thus,the elastic potential energy stored in the wire is $\frac{1}{2} Mgl$.

(A) The work required to stop the disc is equal to the change in its total kinetic energy.
Total kinetic energy of a rolling disc is given by $KE = KE_{translational} + KE_{rotational} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
For a disc,the moment of inertia $I = \frac{1}{2}mR^2$ and for pure rolling,$\omega = \frac{v}{R}$.
Substituting these,$KE = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.
Given: $m = 100\; kg$,$v = 20\; cm/s = 0.2\; m/s$.
$KE = \frac{3}{4} \times 100 \times (0.2)^2 = 75 \times 0.04 = 3\; J$.
Since the work done to stop the object is $W = \Delta KE = 0 - KE_{initial} = -3\; J$,the magnitude of work needed is $3\; J$.

(B) Given the expression for $X$ is $X = \frac{A^2 B^{1/2}}{C^{1/3} D^3}$.
The relative error in $X$ is given by the formula: $\frac{\Delta X}{X} = 2 \frac{\Delta A}{A} + \frac{1}{2} \frac{\Delta B}{B} + \frac{1}{3} \frac{\Delta C}{C} + 3 \frac{\Delta D}{D}$.
To find the maximum percentage error,we multiply by $100$:
$\frac{\Delta X}{X} \times 100 = 2 \left( \frac{\Delta A}{A} \times 100 \right) + \frac{1}{2} \left( \frac{\Delta B}{B} \times 100 \right) + \frac{1}{3} \left( \frac{\Delta C}{C} \times 100 \right) + 3 \left( \frac{\Delta D}{D} \times 100 \right)$.
Substituting the given percentage errors $(1\%, 2\%, 3\%, 4\%)$:
$\frac{\Delta X}{X} \times 100 = 2(1\%) + \frac{1}{2}(2\%) + \frac{1}{3}(3\%) + 3(4\%)$.
Calculating the values:
$= 2\% + 1\% + 1\% + 12\% = 16\%$.
Thus,the maximum percentage error in $X$ is $16\%$.

(D) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by the formula: $g' = g(1 - d/R)$,where $g$ is the acceleration due to gravity at the surface and $R$ is the radius of the earth.
Given that the body is halfway down to the center,the depth $d = R/2$.
Substituting this value into the formula:
$g' = g(1 - (R/2)/R) = g(1 - 1/2) = g/2$.
Since weight $W = mg$,the new weight $W'$ at depth $d$ is $W' = mg' = m(g/2) = W/2$.
Given the weight at the surface $W = 200 \; N$,the weight at depth $d = R/2$ is $W' = 200/2 = 100 \; N$.

(A) Given: Mass $m = 2\; kg$,Radius $R = 4\; cm = 0.04\; m$,Initial angular velocity $\omega_0 = 3\; rpm = \frac{3 \times 2\pi}{60} = \frac{\pi}{10}\; rad/s$.
Total angular displacement $\theta = 2\pi \text{ revolutions} = 2\pi \times 2\pi = 4\pi^2\; rad$.
Final angular velocity $\omega = 0$.
Using the kinematic equation $\omega^2 = \omega_0^2 - 2\alpha\theta$:
$0 = (\frac{\pi}{10})^2 - 2\alpha(4\pi^2)$
$8\alpha\pi^2 = \frac{\pi^2}{100} \Rightarrow \alpha = \frac{1}{800}\; rad/s^2$.
The moment of inertia of a solid cylinder is $I = \frac{1}{2}mR^2 = \frac{1}{2} \times 2 \times (0.04)^2 = 0.0016\; kg\cdot m^2$.
The torque required is $\tau = I\alpha = 0.0016 \times \frac{1}{800} = 16 \times 10^{-4} \times \frac{1}{8} \times 10^{-2} = 2 \times 10^{-6}\; Nm$.

(D) From the figure,the radius of the circle is $A = 3 \ m$ and the time period is $T = 4 \ s$.
The angular velocity is given by $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{4} = \frac{\pi}{2} \ rad/s$.
At $t = 0$,the particle $P$ is at the positive $y$-axis,which means its position vector makes an angle $\phi = \frac{\pi}{2}$ with the positive $x$-axis.
The $y$-coordinate of the particle at any time $t$ is given by $y(t) = A \sin(\omega t + \phi)$.
Substituting the values,$y(t) = 3 \sin\left(\frac{\pi t}{2} + \frac{\pi}{2}\right)$.
Using the trigonometric identity $\sin(\theta + \frac{\pi}{2}) = \cos \theta$,we get $y(t) = 3 \cos\left(\frac{\pi t}{2}\right)$.

(C) The work done by a variable force $F$ acting along the $y$-axis is given by the integral: $W = \int_{y_1}^{y_2} F \, dy$.
Given $F = 20 + 10y$,$y_1 = 0 \; m$,and $y_2 = 1 \; m$,we substitute these into the formula:
$W = \int_{0}^{1} (20 + 10y) \, dy$.
Integrating term by term:
$W = [20y + 10 \cdot \frac{y^2}{2}]_{0}^{1}$.
Simplifying the expression:
$W = [20y + 5y^2]_{0}^{1}$.
Applying the limits:
$W = (20(1) + 5(1)^2) - (20(0) + 5(0)^2)$.
$W = 20 + 5 = 25 \; J$.

(D) The angular speed $\omega$ of a particle in uniform circular motion is related to its time period $T$ by the formula $\omega = \frac{2\pi}{T}$.
Given that the time periods of particles $A$ and $B$ are the same,i.e.,$T_A = T_B = T$.
Therefore,the angular speed of particle $A$ is $\omega_A = \frac{2\pi}{T_A} = \frac{2\pi}{T}$.
Similarly,the angular speed of particle $B$ is $\omega_B = \frac{2\pi}{T_B} = \frac{2\pi}{T}$.
Taking the ratio of the angular speeds,we get $\frac{\omega_A}{\omega_B} = \frac{2\pi/T}{2\pi/T} = 1$.
Thus,the ratio of the angular speed of $A$ to that of $B$ is $1: 1$.

(C) Using the third equation of motion,$v^{2} = u^{2} - 2as$. At the highest point,the final velocity $v = 0$.
Therefore,$0 = u^{2} - 2as$,which gives $s = \frac{u^{2}}{2a}$.
For an object on a smooth inclined plane,the acceleration $a = g \sin \theta$.
Thus,the distance traveled is $s = \frac{u^{2}}{2g \sin \theta}$.
Since $u$ and $g$ are constant,$s \propto \frac{1}{\sin \theta}$.
Therefore,$\frac{x_{1}}{x_{2}} = \frac{\sin \theta_{2}}{\sin \theta_{1}} = \frac{\sin 30^{\circ}}{\sin 60^{\circ}}$.
Substituting the values,$\frac{x_{1}}{x_{2}} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}$.
Hence,the ratio $x_{1}: x_{2}$ is $1: \sqrt{3}$.

(A) The velocity of efflux $v$ is given by Torricelli's law: $v = \sqrt{2gh}$.
The volume flow rate $Q$ is given by $Q = A \times v = A \sqrt{2gh}$.
Given: Area $A = 2 \; mm^{2} = 2 \times 10^{-6} \; m^{2}$,height $h = 2 \; m$,and $g = 10 \; m/s^{2}$.
Substituting the values:
$Q = (2 \times 10^{-6}) \sqrt{2 \times 10 \times 2}$
$Q = (2 \times 10^{-6}) \sqrt{40}$
$Q = (2 \times 10^{-6}) \times 2\sqrt{10}$
$Q = 4 \sqrt{10} \times 10^{-6} \; m^{3}/s$.
Since $\sqrt{10} \approx 3.162$,
$Q \approx 4 \times 3.162 \times 10^{-6} \; m^{3}/s = 12.648 \times 10^{-6} \; m^{3}/s$.
Thus,the rate of flow is nearly $12.6 \times 10^{-6} \; m^{3}/s$.

(C) According to the triangle law of vector addition,if three forces are represented by the sides of a triangle taken in the same order,their resultant is zero.
In the given triangle $PQR$,the forces are $\overrightarrow{PQ}$,$\overrightarrow{QR}$,and $\overrightarrow{RP}$.
Since they are in the same order,the net force $\overrightarrow{F}_{net} = \overrightarrow{PQ} + \overrightarrow{QR} + \overrightarrow{RP} = \overrightarrow{0}$.
According to Newton's second law,$\overrightarrow{F}_{net} = m\overrightarrow{a}$.
Since $\overrightarrow{F}_{net} = \overrightarrow{0}$,the acceleration $\overrightarrow{a} = \overrightarrow{0}$.
Therefore,the velocity $\overrightarrow{V}$ of the particle remains constant.

(C) The gravitational potential energy at the surface of the earth is $U_i = -\frac{GMm}{R}$.
The gravitational potential energy at height $h$ is $U_f = -\frac{GMm}{R+h}$.
The work done $W$ is the change in potential energy: $W = U_f - U_i = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right)$.
Given $h = R$,we substitute this into the equation:
$W = GMm \left( \frac{1}{R} - \frac{1}{R+R} \right) = GMm \left( \frac{1}{R} - \frac{1}{2R} \right) = GMm \left( \frac{1}{2R} \right) = \frac{GMm}{2R}$.
Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting $GM = gR^2$ into the expression for $W$:
$W = \frac{(gR^2)m}{2R} = \frac{1}{2}mgR$.

(D) In $SHM$,a particle oscillates between the extreme positions $+A$ and $-A$.
Starting from the mean position $(x=0)$,the particle moves to $+A$ (distance $= A$).
Then it moves from $+A$ to $-A$ (distance $= 2A$).
Finally,it moves from $-A$ back to the mean position $(x=0)$ (distance $= A$).
Total distance covered in one complete oscillation (one time period) $= A + 2A + A = 4A$.

(D) The time of fall $t$ for an object falling from height $h$ is given by $t = \sqrt{\frac{2h}{g}}$. Thus,$t \propto \frac{1}{\sqrt{g}}$.
The time period $T$ of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$. Thus,$T \propto \frac{1}{\sqrt{g}}$.
Taking the ratio,we get $\frac{t}{T} = \frac{k_1 / \sqrt{g}}{k_2 / \sqrt{g}} = \text{constant}$.
Since the ratio $\frac{t}{T}$ is independent of the acceleration due to gravity $g$,the relationship remains the same on any planet.
Given that on Earth $t = 2T$,it follows that on the other planet $t' = 2T'$.

(C) For a resonance column tube closed at one end,the resonance lengths $l_1, l_2, l_3, \dots$ are given by $l_n = (2n-1) \frac{\lambda}{4}$.
The difference between two successive resonance lengths is $\Delta l = l_{n+1} - l_n = \frac{\lambda}{2}$.
Given $l_1 = 9.75 \; cm$ and $l_2 = 31.25 \; cm$,the difference is $\Delta l = 31.25 \; cm - 9.75 \; cm = 21.50 \; cm = 0.215 \; m$.
Thus,$\frac{\lambda}{2} = 0.215 \; m$,which implies $\lambda = 0.43 \; m$.
The speed of sound $v$ is given by $v = f \lambda$.
Substituting the values $f = 800 \; Hz$ and $\lambda = 0.43 \; m$,we get $v = 800 \times 0.43 = 344 \; m/s$.

(A) According to the law of conservation of linear momentum,the total momentum before the break is equal to the total momentum after the break.
Let the total mass be $M = m_1 + m_2$. Given the ratio $m_1 : m_2 = 1 : 5$,we have $m_1 = \frac{M}{6}$ and $m_2 = \frac{5M}{6}$.
Initial momentum: $P_i = M \vec{v}_0 = M(20 \hat{i} + 25 \hat{j} - 12 \hat{k})$.
Final momentum: $P_f = m_1 \vec{v}_1 + m_2 \vec{v}_2 = \frac{M}{6}(100 \hat{i} + 35 \hat{j} + 8 \hat{k}) + \frac{5M}{6} \vec{v}_2$.
Equating $P_i = P_f$:
$M(20 \hat{i} + 25 \hat{j} - 12 \hat{k}) = \frac{M}{6}(100 \hat{i} + 35 \hat{j} + 8 \hat{k}) + \frac{5M}{6} \vec{v}_2$
Dividing by $M$ and multiplying by $6$:
$6(20 \hat{i} + 25 \hat{j} - 12 \hat{k}) = (100 \hat{i} + 35 \hat{j} + 8 \hat{k}) + 5 \vec{v}_2$
$120 \hat{i} + 150 \hat{j} - 72 \hat{k} = 100 \hat{i} + 35 \hat{j} + 8 \hat{k} + 5 \vec{v}_2$
$5 \vec{v}_2 = (120 - 100) \hat{i} + (150 - 35) \hat{j} + (-72 - 8) \hat{k}$
$5 \vec{v}_2 = 20 \hat{i} + 115 \hat{j} - 80 \hat{k}$
$\vec{v}_2 = 4 \hat{i} + 23 \hat{j} - 16 \hat{k}$.

(D) According to Newton's Law of Cooling: $\frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_s \right)$
For the first interval ($80^{\circ}C$ to $70^{\circ}C$):
$\frac{80 - 70}{12} = K \left( \frac{80 + 70}{2} - 25 \right)$
$\frac{10}{12} = K (75 - 25) = 50K \implies K = \frac{10}{12 \times 50} = \frac{1}{60} \ldots(1)$
For the second interval ($70^{\circ}C$ to $60^{\circ}C$):
$\frac{70 - 60}{t} = K \left( \frac{70 + 60}{2} - 25 \right)$
$\frac{10}{t} = K (65 - 25) = 40K \ldots(2)$
Substituting $K = \frac{1}{60}$ into equation $(2)$:
$\frac{10}{t} = 40 \times \frac{1}{60} = \frac{2}{3}$
$t = \frac{10 \times 3}{2} = 15 \text{ minutes}$.

(D) The terminal velocity $v_{T}$ of a spherical ball of radius $r$ and density $\sigma$ falling in a medium of density $\rho$ and viscosity $\eta$ is given by $v_{T} = \frac{2 r^{2}(\sigma - \rho) g}{9 \eta}$.
Given that the masses of the two balls are equal,$m_{1} = m_{2}$.
Since $m = \text{volume} \times \text{density} = \frac{4}{3} \pi r^{3} \sigma$,we have $\frac{4}{3} \pi r_{1}^{3} \rho_{1} = \frac{4}{3} \pi r_{2}^{3} \rho_{2}$.
Given $r_{1} = 1 \; mm$,$r_{2} = 2 \; mm$,and $\rho_{1} = 8 \rho_{2}$,we check the mass condition: $1^{3} \times 8 \rho_{2} = 8 \rho_{2}$ and $2^{3} \times \rho_{2} = 8 \rho_{2}$. The condition holds.
The ratio of terminal velocities is $\frac{v_{1}}{v_{2}} = \left(\frac{r_{1}}{r_{2}}\right)^{2} \frac{(\rho_{1} - \rho_{medium})}{(\rho_{2} - \rho_{medium})}$.
Here,$\rho_{medium} = 0.1 \rho_{2}$.
Substituting the values: $\frac{v_{1}}{v_{2}} = \left(\frac{1}{2}\right)^{2} \frac{(8 \rho_{2} - 0.1 \rho_{2})}{(\rho_{2} - 0.1 \rho_{2})} = \frac{1}{4} \times \frac{7.9 \rho_{2}}{0.9 \rho_{2}} = \frac{1}{4} \times \frac{79}{9} = \frac{79}{36}$.

(C) The particle starts from rest,so initial angular velocity $\omega_{0} = 0$.
After $n$ rounds,the total angular displacement is $\theta = 2 \pi n$.
The final linear velocity is $V_{0}$,so the final angular velocity is $\omega = \frac{V_{0}}{r}$.
Using the kinematic equation $\omega^{2} = \omega_{0}^{2} + 2 \alpha \theta$,we get:
$\alpha = \frac{\omega^{2} - \omega_{0}^{2}}{2 \theta} = \frac{(V_{0}/r)^{2} - 0}{2(2 \pi n)} = \frac{V_{0}^{2}/r^{2}}{4 \pi n} = \frac{V_{0}^{2}}{4 \pi n r^{2}} \; rad/s^{2}$.

(D) The time taken for an object to fall a distance $h$ under constant acceleration $a$ is given by $t = \sqrt{\frac{2h}{a}}$.
When the elevator is at rest,its acceleration is $0$,so the effective acceleration of the coin relative to the floor is $g$.
When the elevator is moving uniformly (constant velocity),its acceleration is also $0$. Therefore,the effective acceleration of the coin relative to the floor remains $g$.
Since the effective acceleration is the same in both cases $(a_{real} = g)$,the time taken for the coin to reach the floor is the same.
Thus,$t_{1} = t_{2}$.

(B) When the truck accelerates to the right with acceleration $a$,a pseudo force $F_{\text{pseudo}} = ma$ acts on the bob in the opposite direction (i.e.,to the left) in the frame of the truck.
Let $\theta$ be the angle the string makes with the vertical.
In the equilibrium position of the bob in the truck's frame,the forces acting on the bob are the tension $T$,the gravitational force $mg$ acting downwards,and the pseudo force $ma$ acting to the left.
Resolving the forces,we get:
$T \sin \theta = ma$
$T \cos \theta = mg$
Dividing the two equations,we get:
$\tan \theta = \frac{ma}{mg} = \frac{a}{g}$
Therefore,$\theta = \tan^{-1}\left(\frac{a}{g}\right)$.
Since the pseudo force acts to the left,the pendulum tilts to the left.

(B) At the same horizontal level in a continuous static fluid,the pressure is the same.
Let the pressure at the bottom of the $U$-tube be $P$.
The pressure exerted by the water column on the left side is $P_{\text{left}} = P_0 + \rho_{\text{water}} g h_{\text{water}}$,where $P_0$ is atmospheric pressure.
The pressure exerted by the oil column on the right side is $P_{\text{right}} = P_0 + \rho_{\text{oil}} g h_{\text{oil}}$.
Since the pressures at the bottom are equal,we have:
$\rho_{\text{water}} g h_{\text{water}} = \rho_{\text{oil}} g h_{\text{oil}}$
$\rho_{\text{oil}} = \frac{\rho_{\text{water}} h_{\text{water}}}{h_{\text{oil}}}$
Given $\rho_{\text{water}} = 1000 \; kg/m^3$,$h_{\text{water}} = 15 \; cm$,and $h_{\text{oil}} = 20 \; cm$:
$\rho_{\text{oil}} = \frac{1000 \times 15}{20} = 750 \; kg/m^3$.

(A) The heat conducted through the ice layer of thickness $x$ in time $dt$ is given by $dQ = \frac{KA(T_2 - T_1)}{x} dt$.
Here,$T_2 = 0^{\circ}C$ (temperature at the water-ice interface) and $T_1 = -26^{\circ}C$ (outside air temperature).
So,$dQ = \frac{KA(0 - (-26))}{x} dt = \frac{26KA}{x} dt$.
This heat causes an additional thickness $dx$ of water to freeze. The mass of this new ice layer is $dm = A \cdot dx \cdot \rho$.
The heat released during this phase change is $dQ = dm \cdot L = A \cdot dx \cdot \rho \cdot L$.
Equating the two expressions for $dQ$:
$\frac{26KA}{x} dt = A \cdot dx \cdot \rho \cdot L$.
Rearranging to find the rate of increase of thickness $\frac{dx}{dt}$:
$\frac{dx}{dt} = \frac{26K}{\rho x L}$.

(A) The adiabatic index is given by $\gamma = 1 + \frac{2}{f}$,where $f$ is the number of degrees of freedom.
$1$. Hydrogen $(H_2)$ is a diatomic gas at room temperature. Its degrees of freedom $f = 5$ ($3$ translational + $2$ rotational). Thus,$\gamma = 1 + \frac{2}{5} = \frac{7}{5}$.
$2$. Helium $(He)$ is a monoatomic gas. Its degrees of freedom $f = 3$ ($3$ translational). Thus,$\gamma = 1 + \frac{2}{3} = \frac{5}{3}$.
$3$. Gas $X$ is a diatomic gas with an additional vibrational mode. $A$ vibrational mode contributes $2$ degrees of freedom ($1$ kinetic + $1$ potential). So,$f = 5 + 2 = 7$. Thus,$\gamma = 1 + \frac{2}{7} = \frac{9}{7}$.
Therefore,the values are $\frac{7}{5}, \frac{5}{3}, \text{and } \frac{9}{7}$.

(A) The horizontal distance between the two buildings is $d = 100 \; m$. The bullets are fired towards each other with a horizontal velocity $v = 25 \; m/s$ each.
The relative horizontal velocity is $v_{rel} = v_1 + v_2 = 25 + 25 = 50 \; m/s$.
The time taken for the bullets to collide is $t = \frac{d}{v_{rel}} = \frac{100}{50} = 2 \; s$.
In this time,both bullets fall vertically due to gravity. The vertical displacement $s_y$ is given by $s_y = -\frac{1}{2} gt^2$.
Substituting the values: $s_y = -\frac{1}{2} \times 10 \times (2)^2 = -5 \times 4 = -20 \; m$.
The height of the collision from the ground is $H_{collision} = H_{initial} + s_y = 200 - 20 = 180 \; m$.
Thus,the bullets collide after $2 \; s$ at a height of $180 \; m$.

(A) In a stress-strain curve,the ultimate strength point represents the maximum stress a material can withstand. The fracture point is the point where the material breaks.
For a brittle material,the material breaks soon after the elastic limit or ultimate strength point,meaning the fracture point is very close to the ultimate strength point.
For a ductile material,the material undergoes significant plastic deformation after the ultimate strength point before it finally fractures,meaning the fracture point is far from the ultimate strength point.
Since material $X$ has these points close together,it is brittle.
Since material $Y$ has these points far apart,it is ductile.
Therefore,$X$ and $Y$ are brittle and ductile,respectively.

(C) The normal reaction $N$ acting on the body is $N = mg$.
The frictional force $f$ acting on the body is a static frictional force,which balances the applied horizontal force. Since the body does not move,$f \leq \mu N = \mu mg$.
The resultant force $F$ of the normal reaction $N$ and the frictional force $f$ is given by the vector sum $\overrightarrow{F} = \overrightarrow{N} + \overrightarrow{f}$.
Since $N$ and $f$ are perpendicular to each other,the magnitude of the resultant is $|\overrightarrow{F}| = \sqrt{N^{2} + f^{2}}$.
Substituting $N = mg$ and $f \leq \mu mg$,we get $|\overrightarrow{F}| = \sqrt{(mg)^{2} + f^{2}}$.
Since $f^{2} \leq (\mu mg)^{2}$,it follows that $|\overrightarrow{F}| \leq \sqrt{(mg)^{2} + (\mu mg)^{2}} = mg \sqrt{1 + \mu^{2}}$.

(D) Let the velocity of the third fragment of mass $3m$ be $\vec{v}'$. According to the law of conservation of linear momentum,the initial momentum is zero,so the final momentum must also be zero.
$3m \vec{v}' + m v \hat{i} + m v \hat{j} = 0$
$3m \vec{v}' = -mv \hat{i} - mv \hat{j}$
$\vec{v}' = -\frac{v}{3} \hat{i} - \frac{v}{3} \hat{j}$
The magnitude of the velocity of the third fragment is $|\vec{v}'| = \sqrt{(-\frac{v}{3})^2 + (-\frac{v}{3})^2} = \sqrt{\frac{2v^2}{9}} = \frac{\sqrt{2}}{3} v$.
The energy released is equal to the total kinetic energy of the fragments:
$K = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 + \frac{1}{2} (3m) |\vec{v}'|^2$
$K = m v^2 + \frac{3}{2} m (\frac{2v^2}{9}) = m v^2 + \frac{1}{3} m v^2 = \frac{4}{3} m v^2$.

(C) The work done by the force is equal to the change in kinetic energy: $W = \Delta KE = \frac{1}{2} m v^2$. The mass $m = 500 \; g = 0.5 \; kg$.
$1$. Work done up to $X = 8 \; m$:
$W_8 = (20 \; N \times 5 \; m) + (10 \; N \times 3 \; m) = 100 + 30 = 130 \; J$.
Using $W = \frac{1}{2} m v^2$: $130 = \frac{1}{2} (0.5) v_8^2 \Rightarrow v_8^2 = 520 \Rightarrow v_8 = \sqrt{520} \approx 22.8 \; m/s \approx 23 \; m/s$.
$2$. Work done up to $X = 12 \; m$:
$W_{12} = W_8 + (\text{Area from } 8 \; m \text{ to } 10 \; m) + (\text{Area from } 10 \; m \text{ to } 12 \; m)$.
Area from $8 \; m$ to $10 \; m = (2 \; m) \times (-25 \; N) = -50 \; J$.
Area from $10 \; m$ to $12 \; m = (2 \; m) \times (10 \; N) = 20 \; J$.
$W_{12} = 130 - 50 + 20 = 100 \; J$.
Using $W = \frac{1}{2} m v^2$: $100 = \frac{1}{2} (0.5) v_{12}^2 \Rightarrow v_{12}^2 = 400 \Rightarrow v_{12} = 20 \; m/s \approx 20.6 \; m/s$ (closest option).
Thus, the velocities are approximately $23 \; m/s$ and $20.6 \; m/s$.

(D) The total kinetic energy of a rolling body is $K = \frac{1}{2}mv^2(1 + \frac{K^2}{R^2})$.
For a solid cylinder,the radius of gyration $K$ satisfies $K^2 = \frac{1}{2}R^2$,so $\frac{K^2}{R^2} = \frac{1}{2}$.
By the law of conservation of energy,the initial kinetic energy equals the final potential energy at the maximum height $h$:
$\frac{1}{2}mv^2(1 + \frac{1}{2}) = mgh$
$\frac{1}{2}v^2(\frac{3}{2}) = gh$
Given $v = 4 \; m/s$ and taking $g = 10 \; m/s^2$:
$\frac{1}{2} \times 16 \times \frac{3}{2} = 10h$
$12 = 10h \Rightarrow h = 1.2 \; m$.
The distance travelled along the incline $\ell$ is related to height $h$ by $h = \ell \sin 30^{\circ}$.
$\ell = \frac{h}{\sin 30^{\circ}} = \frac{1.2}{0.5} = 2.4 \; m$.

(A) According to Kepler's Third Law of planetary motion,the square of the time period $(T)$ is proportional to the cube of the orbital radius $(r)$: $T^2 \propto r^3$ or $T \propto r^{3/2}$.
The orbital radius $r$ is the distance from the center of the Earth,given by $r = R_{E} + h$,where $h$ is the height above the surface.
For the first satellite: $r_1 = R_{E} + 6 R_{E} = 7 R_{E}$ and $T_1 = 24 \; h$.
For the second satellite: $r_2 = R_{E} + 2.5 R_{E} = 3.5 R_{E}$.
Using the ratio: $\frac{T_2}{T_1} = \left( \frac{r_2}{r_1} \right)^{3/2} = \left( \frac{3.5 R_{E}}{7 R_{E}} \right)^{3/2} = \left( \frac{1}{2} \right)^{3/2} = \frac{1}{2 \sqrt{2}}$.
Therefore,$T_2 = T_1 \times \frac{1}{2 \sqrt{2}} = \frac{24}{2 \sqrt{2}} = \frac{12}{\sqrt{2}} = 6 \sqrt{2} \; h$.

(C) Given that the main scale has $n$ divisions per $cm$,the value of $1 \text{ MSD}$ (Main Scale Division) is $\frac{1}{n} \text{ cm}$.
According to the problem,$n \text{ VSD}$ (Vernier Scale Divisions) coincide with $(n-1) \text{ MSD}$.
Therefore,$1 \text{ VSD} = \frac{n-1}{n} \text{ MSD}$.
The least count of a vernier calliper is defined as $1 \text{ MSD} - 1 \text{ VSD}$.
Least Count $= 1 \text{ MSD} - \left( \frac{n-1}{n} \right) \text{ MSD} = \left( 1 - \frac{n-1}{n} \right) \text{ MSD} = \frac{1}{n} \text{ MSD}$.
Substituting the value of $1 \text{ MSD} = \frac{1}{n} \text{ cm}$,we get:
Least Count $= \frac{1}{n} \times \frac{1}{n} \text{ cm} = \frac{1}{n^2} \text{ cm}$.

(B) The gravitational potential energy $U$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.
Initial distance from the center is $r_i = R$.
Final distance from the center is $r_f = R + h$.
The change in potential energy is $\Delta U = U_f - U_i$.
$\Delta U = \left( -\frac{GMm}{R+h} \right) - \left( -\frac{GMm}{R} \right)$.
$\Delta U = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right)$.
$\Delta U = GMm \left( \frac{R+h-R}{R(R+h)} \right)$.
$\Delta U = \frac{GMmh}{R(R+h)}$.

(C) The work function $W_0$ is given as $4.0 \,eV$.
The threshold wavelength $\lambda_0$ is the longest wavelength capable of causing photoelectric emission.
It is calculated using the formula: $\lambda_0 = \frac{hc}{W_0}$.
Using the approximation $hc \approx 12400 \,eV \cdot \mathring{A}$:
$\lambda_0 = \frac{12400 \,eV \cdot \mathring{A}}{4.0 \,eV} = 3100 \,\mathring{A}$.
Since $1 \,nm = 10 \,\mathring{A}$,we have $\lambda_0 = 310 \,nm$.

(B) The relationship between half-life $T_{1/2}$ and decay constant $\lambda$ is given by $T_{1/2} = \frac{\ln 2}{\lambda} \approx \frac{0.693}{\lambda}$.
Given $T_{1/2} = 2.2 \times 10^9 \; s$,we find $\lambda = \frac{0.693}{2.2 \times 10^9} \approx 3.15 \times 10^{-10} \; s^{-1}$.
The rate of disintegration $R$ is related to the number of radioactive atoms $N$ by the formula $R = \lambda N$.
Given $R = 10^{10} \; s^{-1}$,we calculate $N = \frac{R}{\lambda} = \frac{10^{10}}{3.15 \times 10^{-10}} \approx 3.17 \times 10^{19}$ atoms.

(C) The total energy $(TE)$ of an electron in an orbit is given as $-3.4 \; eV$.
For an electron in a hydrogen-like atom,the relationship between total energy $(TE)$,kinetic energy $(KE)$,and potential energy $(PE)$ is:
$KE = -TE$
$PE = 2 \times TE$
Substituting the given value of $TE = -3.4 \; eV$:
$KE = -(-3.4 \; eV) = +3.4 \; eV$
$PE = 2 \times (-3.4 \; eV) = -6.8 \; eV$
Therefore,the kinetic energy is $3.4 \; eV$ and the potential energy is $-6.8 \; eV$.

(C) In the given circuit,the switches $A$ and $B$ are connected in parallel with the ground. When a switch is at position $0$,it is open,and when it is at position $1$,it is closed (connected to ground).
$1$. If both $A=0$ and $B=0$,both switches are open. The current flows through the $LED$ $(Y)$,so $Y=1$.
$2$. If $A=0$ and $B=1$,switch $B$ is closed (grounded). The current flows through the $LED$ $(Y)$,so $Y=1$.
$3$. If $A=1$ and $B=0$,switch $A$ is closed (grounded). The current flows through the $LED$ $(Y)$,so $Y=1$.
$4$. If $A=1$ and $B=1$,both switches are closed (grounded). The current bypasses the $LED$ through the switches to the ground,so $Y=0$.
The truth table is:

$A$	$B$	$Y$
$0$	$0$	$1$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$0$

This truth table corresponds to the $NAND$ gate operation.

(D) Given: Number of turns $N = 800$,Area $A = 0.05\; m^{2}$,Magnetic field $B = 5 \times 10^{-5}\; T$,Time interval $\Delta t = 0.1\; s$.
Initial magnetic flux $\phi_{1} = N B A \cos(0^{\circ}) = N B A$.
Final magnetic flux $\phi_{2} = N B A \cos(90^{\circ}) = 0$.
Induced $emf$ $e = -\frac{\Delta \phi}{\Delta t} = -\frac{\phi_{2} - \phi_{1}}{\Delta t} = \frac{N B A}{\Delta t}$.
Substituting the values: $e = \frac{800 \times 5 \times 10^{-5} \times 0.05}{0.1}$.
$e = \frac{800 \times 5 \times 10^{-5} \times 5 \times 10^{-2}}{10^{-1}} = 800 \times 5 \times 5 \times 10^{-7} \times 10^{1} = 20000 \times 10^{-6} = 0.02\; V$.

(A) The visible spectrum consists of colours ranging from violet to red.
According to the electromagnetic spectrum, the wavelength $(\lambda)$ increases as we move from violet to red.
Therefore, red light has the longest wavelength, approximately $700 \ nm$, while violet light has the shortest wavelength, approximately $400 \ nm$.
Thus, the correct option is $A$.

(B) For a hollow metal sphere,the charge resides only on the outer surface.
According to Gauss's Law,for any point inside the sphere $(r < R)$,the enclosed charge is $q_{enc} = 0$,therefore the electric field $E = 0$.
For any point outside the sphere $(r > R)$,the sphere acts as a point charge located at the centre. The electric field is given by $E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}$,which means $E \propto \frac{1}{r^2}$.
As $r$ increases for $r > R$,the value of $E$ decreases.
Thus,the electric field is zero for $r < R$ and decreases as $r$ increases for $r > R$.

(D) Eddy currents are loops of electrical current induced within conductors by a changing magnetic field. They are utilized in induction furnaces for heating,magnetic braking systems in trains,and speedometers. An electric heater,however,operates on the principle of the Joule heating effect $(H = I^2Rt)$,where heat is produced by the resistance of a conductor to the flow of electric current,not by eddy currents.

(B) Let the resistance of each bulb be $R$.
Case $(i)$: All six bulbs are glowing. Section $A$ has $3$ bulbs in parallel,and section $B$ has $3$ bulbs in parallel. These two sections are in series. The equivalent resistance is $R_{eq_1} = (R/3) + (R/3) = 2R/3$. The power consumed is $P_1 = E^2 / R_{eq_1} = 3E^2 / 2R$.
Case $(ii)$: Two bulbs from section $A$ are in parallel (resistance $R/2$) and one bulb from section $B$ is in series with this combination. The equivalent resistance is $R_{eq_2} = R/2 + R = 3R/2$. The power consumed is $P_2 = E^2 / R_{eq_2} = 2E^2 / 3R$.
The ratio of power consumption is $P_1 : P_2 = (3E^2 / 2R) : (2E^2 / 3R) = 9 : 4$.

(C) The angle of dip (or magnetic inclination) is defined as the angle that the total magnetic field of the earth makes with the surface of the earth.
By convention, the angle of dip is taken as positive $(+ve)$ in the northern hemisphere, where the north pole of a magnetic needle points downwards.
Conversely, the angle of dip is taken as negative $(-ve)$ in the southern hemisphere, where the north pole of a magnetic needle points upwards.
Since point $A$ has a positive dip $(\delta = +25^{\circ})$, it is located in the northern hemisphere.
Since point $B$ has a negative dip $(\delta = -25^{\circ})$, it is located in the southern hemisphere.

(C) The correct condition to observe a rainbow is that the observer must have their back towards the sun. Therefore,the statement that an observer can see a rainbow when their front is towards the sun is incorrect. Thus,option $C$ is the wrong answer.

(C) According to Ampere's circuital law,the magnetic field $B$ at a distance $d$ from the axis of a long cylindrical conductor of radius $R$ carrying a current $I$ is given by:
For $d \leq R$ (inside the conductor),$B = \frac{\mu_{0}Id}{2 \pi R^{2}}$,which shows that $B \propto d$ (a linear relationship).
For $d > R$ (outside the conductor),$B = \frac{\mu_{0} I}{2 \pi d}$,which shows that $B \propto \frac{1}{d}$ (a hyperbolic relationship).
At the surface $(d = R)$,the magnetic field is maximum,$B_{max} = \frac{\mu_{0} I}{2 \pi R}$.
Comparing this with the given options,the graph that shows a linear increase up to $d = R$ and a hyperbolic decrease for $d > R$ is represented by Figure $C$.

(B) For two thin lenses in contact,the equivalent focal length $F_{1}$ is given by $\frac{1}{F_{1}} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}$,which implies $F_{1} = \frac{f}{2}$.
When the space between the lenses is filled with a liquid of the same refractive index as the glass,the combination acts as a single block of glass. Since the outer surfaces are convex and the inner space is filled,the system effectively becomes a single convex lens with the same radii of curvature as the original lenses. However,the refractive index of the liquid is $\mu = 1.5$,which is equal to the glass. Thus,the entire system behaves as a single lens with focal length $F_{2} = f$.
Therefore,the ratio $\frac{F_{1}}{F_{2}} = \frac{f/2}{f} = \frac{1}{2}$.

(D) The critical angle $(i_c)$ is defined as the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is $90^o$.
When the light ray travels from a denser medium to a rarer medium at an angle of incidence equal to the critical angle,the refracted ray grazes the interface between the two media.
Therefore,the angle of refraction is $90^o$.

(C) The electric field due to an infinite line charge at a distance $r$ is given by $E = \frac{\lambda}{2\pi\epsilon_0 r}$.
Let the two line charges be $1$ and $2$,separated by a distance $2R$. The mid-point $P$ is at a distance $R$ from both lines.
The electric field $\overrightarrow{E}_1$ due to the positive line charge at $P$ points away from it (towards the negative line charge),so $E_1 = \frac{\lambda}{2\pi\epsilon_0 R}$.
The electric field $\overrightarrow{E}_2$ due to the negative line charge at $P$ points towards it (in the same direction as $\overrightarrow{E}_1$),so $E_2 = \frac{|-\lambda|}{2\pi\epsilon_0 R} = \frac{\lambda}{2\pi\epsilon_0 R}$.
Since both fields are in the same direction,the net electric field is $E = E_1 + E_2 = \frac{\lambda}{2\pi\epsilon_0 R} + \frac{\lambda}{2\pi\epsilon_0 R} = \frac{2\lambda}{2\pi\epsilon_0 R} = \frac{\lambda}{\pi\epsilon_0 R}\; N/C$.

(B) In a $p-$type semiconductor,the intrinsic semiconductor (like $Si$ or $Ge$) is doped with trivalent impurity atoms (such as $Boron$,$Aluminum$,or $Indium$).
These trivalent atoms create vacancies in the crystal lattice,which are known as holes.
Since these holes are present in a much higher concentration than the thermally generated electrons,holes act as the majority charge carriers in $p-$type semiconductors.

(D) fuse is a safety device used in electrical circuits to protect against overcurrent or short-circuit conditions. When the current exceeds a safe limit,the fuse wire melts due to the heating effect of the electric current,thereby breaking the circuit and preventing damage to appliances.

(B) The charge $Q$ on a capacitor is given by $Q = CV$, where $C$ is the capacitance and $V$ is the potential difference.
The conduction current $i_c$ in the connecting wires is the rate of change of charge: $i_c = \frac{dQ}{dt} = C \frac{dV}{dt}$.
Given $C = 20 \; \mu F$ and $\frac{dV}{dt} = 3 \; V/s$, we have $i_c = 20 \; \mu F \times 3 \; V/s = 60 \; \mu A$.
According to Maxwell's modification of Ampere's law, the displacement current $i_d$ between the plates of a capacitor is equal to the conduction current in the connecting wires.
Therefore, $i_d = i_c = 60 \; \mu A$.
Thus, the conduction current is $60 \; \mu A$ and the displacement current is $60 \; \mu A$.

(C) In Circuit $1$, the voltmeter $V_{1}$ is connected in parallel with the $10 \; \Omega$ resistor. Since an ideal voltmeter has infinite resistance, it draws no current. Thus, the voltage across the $10 \; \Omega$ resistor is $10 \; V$, so $V_{1} = 10 \; V$. The current $I_{1}$ is given by $I_{1} = \frac{10 \; V}{10 \; \Omega} = 1 \; A$.
In Circuit $2$, the voltmeter $V_{2}$ is connected in series with a $10 \; \Omega$ resistor, and this combination is in parallel with the main $10 \; \Omega$ resistor. An ideal voltmeter has infinite resistance, so the branch containing the voltmeter and the $10 \; \Omega$ resistor draws no current. The voltage across this branch is still $10 \; V$. Since the voltmeter is ideal, the entire potential drop of $10 \; V$ appears across the voltmeter itself, so $V_{2} = 10 \; V$. The current $I_{2}$ flows only through the main $10 \; \Omega$ resistor, so $I_{2} = \frac{10 \; V}{10 \; \Omega} = 1 \; A$.
Therefore, $V_{1} = V_{2} = 10 \; V$ and $I_{1} = I_{2} = 1 \; A$.

(A) An $\alpha$-particle is a helium nucleus,denoted as $\alpha = {}_2^4 \text{He}^{2+}$.
It consists of $2$ protons and $2$ neutrons,having a total mass number of $4$ and an atomic number of $2$.

(B) The de Broglie wavelength $\lambda$ for an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{12.27}{\sqrt{V}} \; \mathring{A}$.
Given $V = 10,000 \; V = 10^4 \; V$.
Substituting the value of $V$ in the formula:
$\lambda = \frac{12.27}{\sqrt{10^4}} \; \mathring{A} = \frac{12.27}{100} \; \mathring{A} = 0.1227 \; \mathring{A}$.
Since $1 \; \mathring{A} = 10^{-10} \; m$,we have $\lambda = 0.1227 \times 10^{-10} \; m = 12.27 \times 10^{-12} \; m$.
Rounding to the nearest option,the value is approximately $12.2 \times 10^{-12} \; m$.

(B) Initially,the charges are $q_A = Q$ and $q_B = -Q$. The force between them at a distance $r$ is given by Coulomb's Law:
$F = \frac{k Q (-Q)}{r^2} = -\frac{k Q^2}{r^2}$
When $25 \%$ of the charge from $A$ is transferred to $B$,the amount transferred is $\Delta q = 0.25 Q = \frac{Q}{4}$.
The new charges are:
$q_A' = Q - \frac{Q}{4} = \frac{3Q}{4}$
$q_B' = -Q + \frac{Q}{4} = -\frac{3Q}{4}$
The new force $F'$ between the charges is:
$F' = \frac{k q_A' q_B'}{r^2} = \frac{k (\frac{3Q}{4})(-\frac{3Q}{4})}{r^2}$
$F' = -\frac{9}{16} \frac{k Q^2}{r^2}$
Since $F = -\frac{k Q^2}{r^2}$,we substitute this into the expression for $F'$:
$F' = \frac{9}{16} F$

(A) The radius of a charged particle moving in a magnetic field is given by $r = \frac{mv}{qB} = \frac{p}{qB}$,where $p$ is the momentum,$q$ is the charge,and $B$ is the magnetic field.
Given that the momenta $p$ and the magnetic field $B$ are the same for both particles,the radius is inversely proportional to the charge: $r \propto \frac{1}{q}$.
For an ionized hydrogen atom (proton),the charge is $q_{H} = +e$.
For an $\alpha$-particle (helium nucleus),the charge is $q_{\alpha} = +2e$.
Therefore,the ratio of the radii is $\frac{r_{H}}{r_{\alpha}} = \frac{q_{\alpha}}{q_{H}} = \frac{2e}{e} = \frac{2}{1}$.
Thus,the ratio $r_{H}: r_{\alpha}$ is $2:1$.

(B) The angular width of the fringes in a Young's double slit experiment is given by $\theta = \frac{\lambda}{d}$.
When the apparatus is immersed in a medium of refractive index $\mu$,the wavelength of light changes to $\lambda^{\prime} = \frac{\lambda}{\mu}$.
Consequently,the new angular width $\theta^{\prime}$ becomes $\theta^{\prime} = \frac{\lambda^{\prime}}{d} = \frac{\lambda}{\mu d} = \frac{\theta}{\mu}$.
Given $\theta = 0.2^{\circ}$ and $\mu = 4/3$,we have $\theta^{\prime} = \frac{0.2^{\circ}}{4/3} = 0.2^{\circ} \times \frac{3}{4} = 0.15^{\circ}$.

(D) Initial charges on the spheres are $Q_1 = \sigma (4 \pi R^2)$ and $Q_2 = \sigma (4 \pi (2R)^2) = 16 \pi R^2 \sigma$.
Total charge $Q_{total} = Q_1 + Q_2 = 4 \pi R^2 \sigma + 16 \pi R^2 \sigma = 20 \pi R^2 \sigma$.
When the spheres are brought into contact,the total charge is redistributed such that they reach the same potential $V = \frac{k Q_1'}{R} = \frac{k Q_2'}{2R}$.
This implies $Q_2' = 2 Q_1'$.
Since $Q_1' + Q_2' = 20 \pi R^2 \sigma$,we have $Q_1' + 2 Q_1' = 20 \pi R^2 \sigma$,which gives $3 Q_1' = 20 \pi R^2 \sigma$,so $Q_1' = \frac{20}{3} \pi R^2 \sigma$.
Then $Q_2' = 2 \times \frac{20}{3} \pi R^2 \sigma = \frac{40}{3} \pi R^2 \sigma$.
The new surface charge densities are $\sigma_1' = \frac{Q_1'}{4 \pi R^2} = \frac{20/3 \pi R^2 \sigma}{4 \pi R^2} = \frac{5}{3} \sigma$.
And $\sigma_2' = \frac{Q_2'}{4 \pi (2R)^2} = \frac{40/3 \pi R^2 \sigma}{16 \pi R^2} = \frac{40}{48} \sigma = \frac{5}{6} \sigma$.

(B) The energy of the emitted photon is given by $E = \frac{hc}{\lambda}$.
Given the energy gap $E_g = 1.9\; eV$.
The wavelength $\lambda$ is calculated using the formula $\lambda = \frac{hc}{E_g}$.
Using the approximation $hc \approx 1240\; eV \cdot nm$,we get:
$\lambda = \frac{1240\; eV \cdot nm}{1.9\; eV} \approx 652.6\; nm$.
Rounding to the nearest provided option,we get $\lambda \approx 654\; nm$.

(A) In the given circuit,switches $A$ and $B$ are connected in parallel. The output $Y$ is the state of the $LED$.
When both switches $A$ and $B$ are at position $0$ (connected to ground),the $LED$ is connected across the $+6V$ supply through resistor $R$,so the $LED$ glows $(Y=1)$.
If either switch $A$ or $B$ is moved to position $1$,the circuit is shorted to ground,and the potential difference across the $LED$ becomes zero,so the $LED$ does not glow $(Y=0)$.
The truth table is:

$A$	$B$	$Y$
$0$	$0$	$1$
$0$	$1$	$0$
$1$	$0$	$0$
$1$	$1$	$0$

This truth table corresponds to the $NOR$ gate.

(D) The power of a lens is given by the lens maker's formula: $P = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
When a lens is cut by a plane containing the principal axis,the radii of curvature $R_1$ and $R_2$ of the lens surfaces remain unchanged.
Since the refractive index $\mu$ and the radii of curvature $R_1$ and $R_2$ remain the same for the cut part,the focal length $f$ remains unchanged.
As power $P = \frac{1}{f}$,the power of each part remains $P$.

(C) The condition for destructive interference (minima) in a Young's double slit experiment is given by the path difference $\Delta x = (2n - 1) \frac{\lambda}{2}$,where $n = 1, 2, 3, \dots$ represents the order of the minimum.
For the fifth minimum,we substitute $n = 5$ into the formula:
$\Delta x = (2(5) - 1) \frac{\lambda}{2}$
$\Delta x = (10 - 1) \frac{\lambda}{2}$
$\Delta x = 9 \frac{\lambda}{2}$.

(C) Using the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Given: $f = 25\; cm$,$\mu = 1.5$. For a double convex lens,$R_1 = R$ and $R_2 = -2R$.
Substituting the values: $\frac{1}{25} = (1.5 - 1) \left( \frac{1}{R} - \frac{1}{-2R} \right)$.
$\frac{1}{25} = 0.5 \left( \frac{1}{R} + \frac{1}{2R} \right) = 0.5 \left( \frac{3}{2R} \right) = \frac{1.5}{2R} = \frac{3}{4R}$.
$4R = 25 \times 3 = 75$.
$R = \frac{75}{4} = 18.75\; cm$.
Therefore,the radii are $R_1 = 18.75\; cm$ and $R_2 = 2R = 37.5\; cm$.

(A) The magnetic field $B$ inside a toroid is given by the formula $B = \frac{\mu_0 N i}{2 \pi r}$,where $N$ is the total number of turns,$i$ is the current,and $r$ is the average radius.
Given for toroid $1$: $N_1 = 200$,$r_1 = 40 \; cm$.
Given for toroid $2$: $N_2 = 100$,$r_2 = 20 \; cm$.
Since the current $i$ is the same for both,the ratio of the magnetic fields is:
$\frac{B_1}{B_2} = \frac{\frac{\mu_0 N_1 i}{2 \pi r_1}}{\frac{\mu_0 N_2 i}{2 \pi r_2}} = \frac{N_1}{N_2} \times \frac{r_2}{r_1}$
Substituting the values:
$\frac{B_1}{B_2} = \left( \frac{200}{100} \right) \times \left( \frac{20}{40} \right) = 2 \times \frac{1}{2} = 1$
Therefore,the ratio is $1:1$.

(A) The current $i$ splits into $i_1$ and $i_2$ at the junction. Since the two arcs are in parallel,the potential difference across them is the same. Thus,$i_1 R_1 = i_2 R_2$,where $R_1$ and $R_2$ are the resistances of the arcs. Since $R \propto \text{length} \propto \theta$,we have $i_1 \theta_1 = i_2 \theta_2$. Given $\theta_1 = 90^\circ = \pi/2$ and $\theta_2 = 270^\circ = 3\pi/2$,we get $i_1(\pi/2) = i_2(3\pi/2)$,which implies $i_2 = 3i_1$. Since $i_1 + i_2 = i$,we have $i_1 + 3i_1 = i$,so $i_1 = i/4$ and $i_2 = 3i/4$.
The magnetic field at the center due to an arc of angle $\theta$ is $B = \frac{\mu_0 i \theta}{4 \pi R}$.
For the upper arc (angle $270^\circ = 3\pi/2$): $B_1 = \frac{\mu_0 i_1 (3\pi/2)}{4 \pi R} = \frac{3 \mu_0 i_1}{8 R} = \frac{3 \mu_0 (i/4)}{8 R} = \frac{3 \mu_0 i}{32 R}$ (inward).
For the lower arc (angle $90^\circ = \pi/2$): $B_2 = \frac{\mu_0 i_2 (\pi/2)}{4 \pi R} = \frac{\mu_0 i_2}{8 R} = \frac{\mu_0 (3i/4)}{8 R} = \frac{3 \mu_0 i}{32 R}$ (outward).
Since $B_1$ and $B_2$ are equal in magnitude and opposite in direction,the net magnetic field at the center $P$ is $B_{net} = B_1 - B_2 = 0$.

(B) An alternating current $(AC)$ is defined as a current that periodically reverses its direction and changes its magnitude with time.
In the given graphs,the $EMF$ (and consequently the current) in all four cases $(a), (b), (c),$ and $(d)$ crosses the time axis,meaning the $EMF$ changes its polarity (sign) periodically.
Since the polarity of the $EMF$ changes in all four graphs,all of them represent alternating $EMF$ or $AC$ waveforms.
Therefore,the correct option is $(b)$.

(C) The Bohr radius is given by $r = \frac{n^2 h^2 \epsilon_0}{\pi m e^2}$, which implies $r \propto \frac{1}{m}$.
Given $m_{\mu} = 207 m_e$, the new radius $r_{\mu} = \frac{r_e}{207} = \frac{0.53 \times 10^{-10} \; m}{207} \approx 2.56 \times 10^{-13} \; m$.
The ground state energy is given by $E = -\frac{m e^4}{8 \epsilon_0^2 n^2 h^2}$, which implies $E \propto m$.
Given $m_{\mu} = 207 m_e$, the new energy $E_{\mu} = E_e \times 207 = -13.6 \; eV \times 207 = -2815.2 \; eV \approx -2.8 \; keV$.

(D) The circuit consists of two parallel branches connected to a $2 \text{ V}$ battery.
In the upper branch,the resistors $20 \ \Omega$ and $30 \ \Omega$ are in series. The potential at the junction $V_1$ is given by the voltage divider rule: $V_1 = 2 \text{ V} \times \frac{30 \ \Omega}{20 \ \Omega + 30 \ \Omega} = 2 \times \frac{30}{50} = 1.2 \text{ V}$.
In the lower branch,the resistors $30 \ \Omega$ and $20 \ \Omega$ are in series. The potential at the junction $V_2$ is given by: $V_2 = 2 \text{ V} \times \frac{20 \ \Omega}{30 \ \Omega + 20 \ \Omega} = 2 \times \frac{20}{50} = 0.8 \text{ V}$.
The reading of the ideal voltmeter is the potential difference between the two junctions: $V = |V_1 - V_2| = |1.2 \text{ V} - 0.8 \text{ V}| = 0.4 \text{ V}$.

(D) According to the principle of the Wheatstone bridge,the condition for balance is independent of the positions of the cell and the galvanometer. This is known as the reciprocity theorem in electrical networks. If the galvanometer and the cell are interchanged,the bridge will still work,and the balance condition will remain the same,i.e.,$\frac{P}{Q} = \frac{l_{1}}{l_{2}}$.

(B) The Earth's total magnetic field $B_{E}$ can be resolved into two rectangular components:
$1$. The horizontal component $H$,which acts along the horizontal direction,is given by $H = B_{E} \cos \delta$.
$2$. The vertical component $V$,which acts along the vertical direction,is given by $V = B_{E} \sin \delta$.
Here,$\delta$ is the angle of dip (or magnetic inclination).

(A) For the $A.C.$ source,the impedance $Z = \frac{V}{I_{AC}} = \frac{12 \; V}{0.2 \; A} = 60 \; \Omega$.
For the $D.C.$ source,the resistance $R = \frac{V}{I_{DC}} = \frac{12 \; V}{0.4 \; A} = 30 \; \Omega$.
Since $Z > R$,the circuit must contain an inductor $(L)$ in addition to the resistor $(R)$.
In a $D.C.$ circuit,an ideal capacitor acts as an open circuit (infinite resistance),but since current flows in the $D.C.$ case,the circuit cannot contain a capacitor in series. Therefore,the circuit is a series $LR$ circuit.

(B) The motional $EMF$ induced in a rotating rod (or spoke of a wheel) of length $R$ in a uniform magnetic field $B$ rotating with angular velocity $\omega$ is given by the formula:
$E = \frac{1}{2} B \omega R^2$
Given values:
$B = 0.1 \; T$
$\omega = 10 \; rad/s$
$R = 0.5 \; m$
Substituting these values into the formula:
$E = \frac{1}{2} \times 0.1 \times 10 \times (0.5)^2$
$E = 0.5 \times 0.25$
$E = 0.125 \; V$
Therefore,the $EMF$ generated between the centre and the rim is $0.125 \; V$.

(A) The velocity of light in a medium is given by the formula $v = \frac{1}{\sqrt{\mu \epsilon}}$.
Substituting $\mu = \mu_{r} \mu_{0}$ and $\epsilon = \epsilon_{r} \epsilon_{0}$,we get $v = \frac{1}{\sqrt{\mu_{r} \epsilon_{r} \mu_{0} \epsilon_{0}}}$.
Since the speed of light in vacuum is $c = \frac{1}{\sqrt{\mu_{0} \epsilon_{0}}} = 3 \times 10^{8} \;m/s$,the formula becomes $v = \frac{c}{\sqrt{\mu_{r} \epsilon_{r}}}$.
Given $\mu_{r} = 1.0$ and $\epsilon_{r} = 1.44$,we have $v = \frac{3 \times 10^{8}}{\sqrt{1.0 \times 1.44}}$.
$v = \frac{3 \times 10^{8}}{\sqrt{1.44}} = \frac{3 \times 10^{8}}{1.2}$.
$v = 2.5 \times 10^{8} \;m/s$.

(B) According to Gauss's Law, the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{\text{enclosed}}}{\epsilon_0}$.
An electric dipole consists of two equal and opposite charges, $+q$ and $-q$.
Here, the charges are $+3 \times 10^{-6} \; C$ and $-3 \times 10^{-6} \; C$.
The total charge enclosed by the sphere is $q_{\text{enclosed}} = (+3 \times 10^{-6}) + (-3 \times 10^{-6}) = 0 \; C$.
Therefore, the total electric flux $\phi = \frac{0}{\epsilon_0} = 0 \; \text{N m}^2 / \text{C}$.

(C) Initially,capacitor $C_{1}$ is charged to a potential $V$ by the battery. The initial energy stored in $C_{1}$ is $U_{i} = \frac{1}{2} C V^{2}$.
When the battery is disconnected and $C_{1}$ is connected in parallel with an uncharged capacitor $C_{2}$ (where $C_{1} = C_{2} = C$),charge is redistributed.
The common potential $V'$ after connection is given by $V' = \frac{C_{1}V + C_{2}(0)}{C_{1} + C_{2}} = \frac{CV}{2C} = \frac{V}{2}$.
The final energy stored in the system is $U_{f} = \frac{1}{2} (C_{1} + C_{2}) (V')^{2} = \frac{1}{2} (2C) (\frac{V}{2})^{2} = C \cdot \frac{V^{2}}{4} = \frac{1}{4} C V^{2}$.
The loss in energy is $\Delta U = U_{i} - U_{f} = \frac{1}{2} C V^{2} - \frac{1}{4} C V^{2} = \frac{1}{4} C V^{2}$.
The percentage loss of energy is $\frac{\Delta U}{U_{i}} \times 100 = \frac{\frac{1}{4} C V^{2}}{\frac{1}{2} C V^{2}} \times 100 = 50 \%$.

(B) The angular width of the central maxima in Fraunhofer diffraction is given by $\theta = \frac{2\lambda}{d}$,where $\lambda$ is the wavelength and $d$ is the slit width.
Since $\theta \propto \lambda$,we have the ratio $\frac{\theta_2}{\theta_1} = \frac{\lambda_2}{\lambda_1}$.
Given that the angular width decreases by $30 \%$,the new angular width $\theta_2 = \theta_0 - 0.30\theta_0 = 0.70\theta_0$.
Substituting the values: $\frac{0.70\theta_0}{\theta_0} = \frac{\lambda_2}{6000 \; \mathring{A}}$.
$0.70 = \frac{\lambda_2}{6000 \; \mathring{A}}$.
$\lambda_2 = 0.70 \times 6000 \; \mathring{A} = 4200 \; \mathring{A}$.

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mE_k}}$,where $h$ is Planck's constant,$m$ is the mass of the particle,and $E_k$ is the kinetic energy.
Since both particles have the same kinetic energy $E_k$,we have $\lambda \propto \frac{1}{\sqrt{m}}$.
Therefore,$\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha}}{m_p}}$.
We know that the mass of an $\alpha$-particle is approximately $4$ times the mass of a proton,i.e.,$m_{\alpha} = 4m_p$.
Substituting this value,we get $\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{4m_p}{m_p}} = \sqrt{4} = 2$.
Thus,the ratio $\lambda_p : \lambda_{\alpha}$ is $2:1$.