AP EAMCET 2021 Physics Question Paper with Answer and Solution

(C) From the third equation of kinematics,we have $v^2 - u^2 = 2as$.
Assuming the object starts from rest,the initial velocity $u = 0$.
Substituting this into the equation,we get $v^2 = 2as$,which can be rearranged as $s = \frac{v^2}{2a}$.
Since $a$ is a uniform (constant) acceleration,the relationship between displacement $s$ and velocity $v$ is $s \propto v^2$.
This represents a parabola opening along the $s$-axis.
Graph $C$ shows a parabolic curve where $s$ increases with the square of $v$,starting from the origin $(0, 0)$,which matches the derived relationship $s = \frac{1}{2a} v^2$.

(C) We know that uniform velocity is independent of time. Therefore,in a velocity-time graph,the graph is represented as a straight line parallel to the time axis.
Since the slope of a line is given by $\text{slope} = \tan \theta$,where $\theta$ is the angle the line makes with the horizontal time axis.
From the graph,the line is parallel to the time axis,so $\theta = 0^{\circ}$.
Therefore,$\text{slope} = \tan 0^{\circ} = 0$.

(B) When the object is dropped in a stationary lift from height $h$,the equation of motion is $s = ut + \frac{1}{2}at^2$.
Since $u = 0$ and $a = -g$,we have $-h = -\frac{1}{2}gt_1^2$,which gives $t_1 = \sqrt{\frac{2h}{g}}$.
When the object is dropped from a lift moving upward with constant acceleration $a$,the object experiences a pseudo force in the downward direction relative to the lift frame.
The effective acceleration becomes $g_{eff} = g + a$.
Using the same kinematic equation,$-h = -\frac{1}{2}(g + a)t_2^2$,which gives $t_2 = \sqrt{\frac{2h}{g + a}}$.
Since $(g + a) > g$,it follows that $\frac{2h}{g + a} < \frac{2h}{g}$.
Therefore,$t_2 < t_1$ or $t_1 > t_2$.

(C) The velocity $v$ acquired by a mass $m$ falling freely under gravity from a height $h$ is given by $v = \sqrt{2gh}$.
The linear momentum $p$ is defined as $p = mv$.
For the first body:
$p_1 = m_1 v_1 = m_1 \sqrt{2gh_1}$
For the second body:
$p_2 = m_2 v_2 = m_2 \sqrt{2gh_2}$
The ratio of their linear momenta is:
$\frac{p_1}{p_2} = \frac{m_1 \sqrt{2gh_1}}{m_2 \sqrt{2gh_2}} = \frac{m_1}{m_2} \sqrt{\frac{h_1}{h_2}}$
Given $\frac{m_1}{m_2} = \frac{2}{3}$ and $\frac{h_1}{h_2} = \frac{9}{16}$,we substitute these values:
$\frac{p_1}{p_2} = \frac{2}{3} \times \sqrt{\frac{9}{16}} = \frac{2}{3} \times \frac{3}{4} = \frac{6}{12} = \frac{1}{2}$
Thus,the ratio is $1:2$.

(D) Given, velocity $v = 2t^2 - 8t + 15$.
We know that instantaneous acceleration $a$ is the derivative of velocity with respect to time $t$, i.e., $a = \frac{dv}{dt}$.
Differentiating the given expression for velocity with respect to $t$:
$a = \frac{d}{dt}(2t^2 - 8t + 15) = 4t - 8$.
Now, substitute the value $t = 5 \,s$ into the expression for acceleration:
$a = 4(5) - 8 = 20 - 8 = 12 \,ms^{-2}$.
Therefore, the instantaneous acceleration at $t = 5 \,s$ is $12 \,ms^{-2}$.

(C) Given,the displacement equation is $3 s = 9 t + 5 t^2$.
Dividing the entire equation by $3$,we get $s = 3 t + (5/3) t^2$.
The standard equation of motion for a body with constant acceleration is $s = u t + (1/2) a t^2$,where $u$ is the initial velocity and $a$ is the acceleration.
Comparing the given equation $s = 3 t + (5/3) t^2$ with the standard equation $s = u t + (1/2) a t^2$,we find that $(1/2) a = 5/3$.
Solving for $a$,we get $a = 2 \times (5/3) = 10/3 \ m s^{-2}$.

(C) Given: Length of the train, $l = 100 \,m$.
Time taken to cross the object, $t = 7.2 \,s$.
Velocity of the object moving in the opposite direction, $v_o = 5 \,km/h = 5 \times \frac{5}{18} \,m/s = \frac{25}{18} \,m/s \approx 1.39 \,m/s$.
Let the velocity of the train be $v_t = v$.
Since the objects are moving in opposite directions, the relative velocity is $v_{\text{rel}} = v_t + v_o$.
The distance covered to cross the object is equal to the length of the train, $l = v_{\text{rel}} \times t$.
$100 = (v + \frac{25}{18}) \times 7.2$.
Dividing by $7.2$: $\frac{100}{7.2} = v + \frac{25}{18}$.
$\frac{1000}{72} = v + \frac{25}{18} \Rightarrow \frac{125}{9} = v + \frac{25}{18}$.
$v = \frac{250}{18} - \frac{25}{18} = \frac{225}{18} = 12.5 \,m/s$.
Converting to $km/h$: $v = 12.5 \times \frac{18}{5} = 45 \,km/h$.

(C) Let the speed of the bus be $v_B$ and the speed of the cyclist be $v_C = 20 \,km/h$.
Let $d$ be the distance between two consecutive buses,where $d = v_B \times T$.
Case $1$: Bus moving from $A$ to $B$ (same direction as cyclist).
The relative speed of the bus with respect to the cyclist is $(v_B - v_C)$.
The time interval between buses passing the cyclist is $t_1 = 18 \,min = \frac{18}{60} \,h = 0.3 \,h$.
Thus,$d = (v_B - 20) \times 0.3 \dots (i)$.
Case $2$: Bus moving from $B$ to $A$ (opposite direction to cyclist).
The relative speed of the bus with respect to the cyclist is $(v_B + v_C)$.
The time interval between buses passing the cyclist is $t_2 = 6 \,min = \frac{6}{60} \,h = 0.1 \,h$.
Thus,$d = (v_B + 20) \times 0.1 \dots (ii)$.
Equating $(i)$ and $(ii)$:
$(v_B - 20) \times 0.3 = (v_B + 20) \times 0.1$
$3(v_B - 20) = v_B + 20$
$3v_B - 60 = v_B + 20$
$2v_B = 80 \Rightarrow v_B = 40 \,km/h$.
Now,substitute $v_B$ into $d = v_B \times T$ using equation $(ii)$:
$v_B \times T = (v_B + 20) \times 0.1$
$40 \times T = (40 + 20) \times 0.1$
$40T = 60 \times 0.1 = 6$
$T = \frac{6}{40} \,h = \frac{3}{20} \,h$.

(A) Using the equation of motion $v = u + at$.
Here,$v$ is the final speed,$u$ is the initial speed,and $a$ is the acceleration for time $t$.
Since there is constant retardation,the acceleration is negative,i.e.,$a_{eff} = -a$.
Substituting this into the equation:
$v = u + (-a)t = u - at$.
Since $at > 0$ for $t > 0$,it follows that $v < u$.
Therefore,the speed of the body decreases under constant retardation.

(D) The velocity of the particle is given by the derivative of displacement with respect to time: $v = \frac{ds}{dt}$.
Given $s = 9t^2 - 2t^3$,we differentiate with respect to $t$:
$v = \frac{d}{dt}(9t^2 - 2t^3) = 18t - 6t^2$.
To find the time when the particle attains zero velocity,we set $v = 0$:
$18t - 6t^2 = 0$.
Factoring out $6t$,we get:
$6t(3 - t) = 0$.
This gives two solutions: $t = 0$ and $t = 3$.
Since the particle starts from rest at $t = 0$,the time at which it will attain zero velocity again is $t = 3 \ s$.

(A) The position vector of point $A$ is $\vec{r}_A = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}$.
The position vector of point $B$ is $\vec{r}_B = 6 \hat{i} + 6 \hat{j} + 9 \hat{k}$.
The displacement vector $\vec{d}$ is given by the change in position: $\vec{d} = \vec{r}_B - \vec{r}_A$.
$\vec{d} = (6 - 2) \hat{i} + (6 - 2) \hat{j} + (9 - 3) \hat{k}$.
$\vec{d} = 4 \hat{i} + 4 \hat{j} + 6 \hat{k}$.

(A) Given that, speed of train from city-$A$ to city-$B$ is $v_1 = 18 \,ms^{-1}$.
Speed of train from city-$B$ to city-$A$ is $v_2 = 36 \,ms^{-1}$.
Since the distance travelled from $A$ to $B$ and $B$ to $A$ is the same, we use the formula for average speed over equal distances:
$v_{av} = \frac{2 v_1 v_2}{v_1 + v_2}$
Substituting the values:
$v_{av} = \frac{2 \times 18 \times 36}{18 + 36}$
$v_{av} = \frac{2 \times 18 \times 36}{54}$
$v_{av} = \frac{1296}{54} = 24 \,ms^{-1}$.

(A) Given initial velocity $u = 36 \ km/h = 36 \times \frac{5}{18} \ m/s = 10 \ m/s$.
Since the object comes to rest,the final velocity $v = 0 \ m/s$.
The distance covered $s = 200 \ m$.
Using the third equation of motion,$v^2 - u^2 = 2as$.
Substituting the values: $0^2 - (10)^2 = 2 \times a \times 200$.
$-100 = 400a$.
$a = -\frac{100}{400} = -0.25 \ m/s^2$.
The negative sign indicates retardation.
Therefore,the retardation produced by the brakes is $0.25 \ m/s^2$.

(D) The bullet exhibits horizontal projectile motion.
Vertical displacement of the bullet,$y = 15 \ cm = 0.15 \ m$.
Horizontal displacement of the bullet,$x = 150 \ m$.
Using the equation of trajectory for horizontal projection,$y = \frac{g x^2}{2 u^2}$.
Substituting the values: $0.15 = \frac{10 \times 150^2}{2 u^2}$.
Rearranging for $u^2$: $u^2 = \frac{10 \times 22500}{2 \times 0.15} = \frac{225000}{0.3} = 750000$.
Taking the square root: $u = \sqrt{750000} = \sqrt{250000 \times 3} = 500 \sqrt{3} \ m \ s^{-1}$.

(D) Given: Speed $v = 30 \,ms^{-1}$, Radius $r = 500 \,m$, Tangential acceleration $a_T = 2 \,ms^{-2}$.
The total acceleration $a$ of a particle in circular motion is the vector sum of tangential acceleration $a_T$ and centripetal (radial) acceleration $a_r$.
First, calculate the centripetal acceleration $a_r = \frac{v^2}{r} = \frac{30^2}{500} = \frac{900}{500} = 1.8 \,ms^{-2}$.
The total acceleration is given by $a = \sqrt{a_T^2 + a_r^2}$.
Substituting the values: $a = \sqrt{2^2 + 1.8^2} = \sqrt{4 + 3.24} = \sqrt{7.24}$.
Calculating the square root: $a \approx 2.69 \,ms^{-2}$, which is approximately $2.7 \,ms^{-2}$.

(A) The time of flight $T$ for a projectile is given by the formula $T = \frac{2u \sin \theta}{g}$.
Given:
Initial velocity $u = 50 \ m s^{-1}$
Angle of projection $\theta = 30^{\circ}$
Acceleration due to gravity $g = 10 \ m s^{-2}$
Substituting the values into the formula:
$T = \frac{2 \times 50 \times \sin 30^{\circ}}{10}$
Since $\sin 30^{\circ} = 0.5$,
$T = \frac{100 \times 0.5}{10} = \frac{50}{10} = 5 \ s$.
Thus,the ball remains in the air for $5 \ s$.

(C) When a ball is projected upwards,the only force acting on it throughout its flight is the gravitational force,which acts vertically downwards. According to Newton's second law of motion,$F = ma$,the acceleration $a$ is in the direction of the net force. At the highest point,the velocity of the ball is momentarily zero,but the gravitational force continues to act on it. Therefore,the acceleration remains $g$ (acceleration due to gravity) and is directed downwards.

(D) The kinetic energy of a projectile is given by $KE = \frac{1}{2}mv^2$,where $v$ is the instantaneous velocity of the particle.
Since the horizontal component of velocity $(v_x = u \cos \theta)$ remains constant throughout the flight,the net velocity $v = \sqrt{v_x^2 + v_y^2}$ depends on the vertical component $v_y$.
The vertical component $v_y = u \sin \theta - gt$ is maximum in magnitude at the point of projection $(t=0)$ and at the point of landing ($t=T$,where $T$ is the total time of flight).
At the point of projection,the horizontal distance covered is $0 \ m$. At the point of landing,the horizontal distance covered is equal to the range $R = 100 \ m$.
Comparing the options,the kinetic energy is maximum at the start and at the end of the trajectory. Since $0 \ m$ is not an option,the correct answer is $100 \ m$.

(A) The horizontal component of velocity remains constant because there is no acceleration in the horizontal direction: $v_x = u_x = 15 \ m \ s^{-1}$.
In the vertical direction,the stone undergoes free fall with initial vertical velocity $u_y = 0$. Using the equation $v_y^2 = u_y^2 + 2gh$:
$v_y^2 = 0 + 2 \times 9.8 \times 490 = 9604$.
$v_y = \sqrt{9604} = 98 \ m \ s^{-1}$.
The final speed $v$ is the magnitude of the resultant velocity vector: $v = \sqrt{v_x^2 + v_y^2}$.
$v = \sqrt{15^2 + 98^2} = \sqrt{225 + 9604} = \sqrt{9829} \approx 99.14 \ m \ s^{-1}$.
Rounding to the nearest integer,the speed is $99 \ m \ s^{-1}$.

(A) Given that, the height of the tower is $h = 19.6 \,m$.
The angle of the line joining the point of projection to the point on the ground with the horizontal is $\theta = 45^{\circ}$.
In the right-angled triangle formed by the height $h$ and horizontal range $R$, we have:
$\tan \theta = \frac{h}{R}$
$\tan 45^{\circ} = \frac{19.6}{R}$
$1 = \frac{19.6}{R} \Rightarrow R = 19.6 \,m$.
The time taken $t$ to reach the ground is given by:
$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 19.6}{9.8}} = \sqrt{4} = 2 \,s$.
The horizontal distance $R$ is given by:
$R = u \times t$
$19.6 = u \times 2$
$u = \frac{19.6}{2} = 9.8 \,m/s$.
Thus, the initial velocity of the ball is $9.8 \,m/s$.

(B) Let $u$ and $\theta$ be the velocity of projection and angle of projection,respectively.
Given that,the horizontal component of velocity is $u_x = u \cos \theta = 3 \ m/s$.
The equation of projectile motion is given by $y = 12x - \frac{3}{4}x^2$ ... $(i)$.
We know the general equation of projectile motion is $y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$ ... (ii).
Comparing equations $(i)$ and (ii),we get $\tan \theta = 12$.
Since $\tan \theta = \frac{\sin \theta}{\cos \theta} = 12$,we have $\sin \theta = 12 \cos \theta$.
Multiplying both sides by $u$,we get $u \sin \theta = 12(u \cos \theta) = 12 \times 3 = 36 \ m/s$.
The range $R$ is given by $R = \frac{2u^2 \sin \theta \cos \theta}{g} = \frac{2(u \sin \theta)(u \cos \theta)}{g}$.
Substituting the values,$R = \frac{2 \times 36 \times 3}{10} = \frac{216}{10} = 21.6 \ m$.

(B) The ball is projected from a height of $20 \,m$ and we need to find the horizontal distance where it returns to the same vertical level ($20 \,m$ from the ground).
This is equivalent to finding the horizontal range of a projectile launched at an angle $\theta$ on a level plane.
The formula for the horizontal range $R$ is given by:
$R = \frac{u^2 \sin(2\theta)}{g}$
Given:
Initial speed $u = 13 \,ms^{-1}$
Angle of projection $\theta = 30^{\circ}$
Acceleration due to gravity $g = 10 \,ms^{-2}$
Substituting the values:
$R = \frac{(13)^2 \times \sin(2 \times 30^{\circ})}{10}$
$R = \frac{169 \times \sin(60^{\circ})}{10}$
$R = \frac{169 \times \frac{\sqrt{3}}{2}}{10}$
$R = \frac{169 \times 1.732}{20}$
$R = \frac{292.708}{20} \approx 14.635 \,m$
Rounding to one decimal place, we get $R = 14.6 \,m$.

(A) Let the velocity of cars $A$ and $B$ be $v_A = v_B = 30 \ km/h$ in the positive direction.
Let the velocity of car $C$ be $v_C$ in the opposite direction,so $v_C = -v$ (where $v$ is the speed of car $C$).
The relative velocity of car $C$ with respect to cars $A$ and $B$ is $v_{rel} = v_A - v_C = 30 - (-v) = 30 + v$.
The distance between car $A$ and $B$ is $d = 10 \ km$.
Car $C$ meets car $B$ and then car $A$ after an interval of $t = 8 \ minutes = \frac{8}{60} \ h = \frac{2}{15} \ h$.
Using the formula $d = v_{rel} \times t$:
$10 = (30 + v) \times \frac{2}{15}$
$10 \times \frac{15}{2} = 30 + v$
$75 = 30 + v$
$v = 75 - 30 = 45 \ km/h$.

(D) Given: Mass $m = 5 \ kg$,Radius $r = 1 \ m$,Angular velocity $\omega = 2 \ rad \ s^{-1}$.
Centripetal force $F_c$ is given by the formula $F_c = m \omega^2 r$.
Substituting the given values:
$F_c = 5 \times (2)^2 \times 1$
$F_c = 5 \times 4 \times 1$
$F_c = 20 \ N$.
Therefore,the centripetal force is $20 \ N$.

(C) The amplitude of a damped oscillator at time $t$ is given by $A(t) = A_0 e^{-\alpha t}$.
Given that at $t_1 = 5 \,s$, $A(5) = 0.9 A_0$.
Substituting this into the equation: $0.9 A_0 = A_0 e^{-5\alpha}$, which implies $e^{-5\alpha} = 0.9$.
We need to find the amplitude after another $20 \,s$, which means at total time $t_2 = 5 \,s + 20 \,s = 25 \,s$.
Let $A'$ be the amplitude at $t_2 = 25 \,s$.
$A' = A_0 e^{-25\alpha} = A_0 (e^{-5\alpha})^5$.
Substituting $e^{-5\alpha} = 0.9$:
$A' = A_0 (0.9)^5 = A_0 \times 0.59049$.
Thus, the amplitude decreases to approximately $0.59$ times its original amplitude.

(C) Using Hooke's law for the spring,$F = kx$.
When a mass of $0.6 \ kg$ is suspended,the spring force balances the weight of the mass,so $F = mg$.
Thus,$k = \frac{mg}{x} = \frac{0.6 \times 10}{0.40} = 15 \ N \ m^{-1}$.
The time period $T$ for a mass $m' = 255 \ g = 0.255 \ kg$ is given by $T = 2\pi \sqrt{\frac{m'}{k}}$.
Substituting the values: $T = 2 \times 3.14 \times \sqrt{\frac{0.255}{15}} = 2 \times 3.14 \times \sqrt{0.017} \approx 6.28 \times 0.13038 \approx 0.8188 \ s$.
Rounding to two decimal places,$T \approx 0.82 \ s$.

(A) The spring constant $k$ of the spring is calculated using Hooke's Law,$F = kx$. Given $F = 15 \ kg \times 10 \ m/s^2 = 150 \ N$ and $x = 0.25 \ m$,we have $k = \frac{150}{0.25} = 600 \ N/m$.
The time period $T$ of a spring-mass system is given by $T = 2 \pi \sqrt{\frac{m}{k}}$.
Substituting the given values,$\frac{2 \pi}{5} = 2 \pi \sqrt{\frac{m}{600}}$.
Dividing both sides by $2 \pi$,we get $\frac{1}{5} = \sqrt{\frac{m}{600}}$.
Squaring both sides,$\frac{1}{25} = \frac{m}{600}$.
Solving for $m$,$m = \frac{600}{25} = 24 \ kg$.

(A) Given: Mass $m = 1 \ kg$,spring constant $k = 100 \ N \ m^{-1}$,amplitude $A = 10 \ cm = 0.1 \ m$,displacement $x = 5 \ cm = 0.05 \ m$.
Angular frequency $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{100}{1}} = 10 \ rad \ s^{-1}$.
Velocity at displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2} = 10 \sqrt{0.1^2 - 0.05^2} = 10 \sqrt{0.01 - 0.0025} = 10 \sqrt{0.0075} = 10 \times 0.0866 = 0.866 \ m \ s^{-1}$.
Kinetic energy $K.E. = \frac{1}{2} m v^2 = \frac{1}{2} \times 1 \times (0.866)^2 = 0.5 \times 0.75 = 0.375 \ J$.
Potential energy $P.E. = \frac{1}{2} k x^2 = \frac{1}{2} \times 100 \times (0.05)^2 = 50 \times 0.0025 = 0.125 \ J$.

(A) simple pendulum consists of a point mass (bob) suspended by a massless,inextensible string.
For a simple pendulum oscillating in a plane,the position of the bob is uniquely determined by a single coordinate,which is the angle $\theta$ that the string makes with the vertical.
Therefore,the number of degrees of freedom for an oscillating simple pendulum is $1$.

(B) Given: Length of pendulum,$l = 0.5 \ m$. Speed at lowest point,$v_1 = 6 \ ms^{-1}$. Angle made by string with vertical,$\theta = 60^{\circ}$.
In $\triangle OBC$,$\cos \theta = \frac{OC}{OB} = \frac{l-h}{l} = \cos 60^{\circ}$.
$\Rightarrow 1 - \frac{h}{l} = 0.5 \Rightarrow \frac{h}{l} = 0.5 \Rightarrow h = 0.5 \times 0.5 = 0.25 \ m$.
Using the law of conservation of mechanical energy:
$K_i + U_i = K_f + U_f$
$\frac{1}{2}mv_1^2 + 0 = \frac{1}{2}mv_2^2 + mgh$
$v_2^2 = v_1^2 - 2gh$
$v_2^2 = (6)^2 - 2 \times 10 \times 0.25 = 36 - 5 = 31$
$v_2 = \sqrt{31} \ ms^{-1}$.
Hence,the final velocity at an angle $60^{\circ}$ is $\sqrt{31} \ ms^{-1}$.

(A) In an evacuated chamber,there is no air resistance or drag force acting on the bob of the simple pendulum.
Since the only force acting on the bob is gravity (which is a conservative force),there is no mechanism to dissipate the mechanical energy of the system.
According to the law of conservation of energy,the total mechanical energy remains constant.
Therefore,the amplitude of the oscillation remains constant over time.

(C) Given that, mass of body $m = 4.9 \, kg$.
Time period of oscillation of the spring $T = 0.5 \, s$.
Acceleration due to gravity $g = 10 \, m/s^2$ and $\pi^2 = 10$.
The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
Squaring both sides, we get $T^2 = 4\pi^2 \frac{m}{k}$.
Rearranging for the ratio $\frac{m}{k}$, we have $\frac{m}{k} = \frac{T^2}{4\pi^2}$.
Substituting the values, $\frac{m}{k} = \frac{(0.5)^2}{4 \times 10} = \frac{0.25}{40} = 0.00625 \, kg/N$.
When the mass is removed, the spring shortens by an amount $x$ equal to the extension produced by the mass in equilibrium.
From Hooke's Law at equilibrium, $kx = mg$, which implies $x = \frac{m}{k} g$.
Substituting the values, $x = 0.00625 \times 10 = 0.0625 \, m$.
Converting to centimeters, $x = 0.0625 \times 100 = 6.25 \, cm$.
Thus, the spring is shortened by $6.25 \, cm$.

(D) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g_{eff}}}$.
When the lift is at rest,the effective acceleration due to gravity is $g_{eff} = g$.
So,$T = 2 \pi \sqrt{\frac{l}{g}}$ ...$(i)$
When the lift moves upwards with an acceleration $a = 3g$,the effective acceleration due to gravity becomes $g_{eff} = g + a = g + 3g = 4g$.
The new time period $T^{\prime}$ is given by $T^{\prime} = 2 \pi \sqrt{\frac{l}{4g}}$.
$T^{\prime} = \frac{1}{2} \times 2 \pi \sqrt{\frac{l}{g}}$.
Substituting the value from Eq. $(i)$,we get $T^{\prime} = \frac{T}{2}$.

(C) The potential energy $U$ of a particle executing simple harmonic motion is given by $U = \frac{1}{2} k x^2$,where $k$ is the force constant and $x$ is the displacement from the mean position.
For the potential energy to be maximum,the displacement $x$ must be at its maximum value.
In simple harmonic motion,the maximum displacement from the mean position is equal to the amplitude $A$.
Therefore,the potential energy is maximum at the extreme positions,which are $x = \pm A$.

(C) Given,amplitude of oscillation,$a = 6 \text{ cm}$.
Let the displacement be $x \text{ cm}$.
When the kinetic energy $(K)$ is equal to the potential energy $(U)$,we have $K = U$.
The kinetic energy of a simple harmonic oscillator is given by $K = \frac{1}{2} m \omega^2 (a^2 - x^2)$.
The potential energy is given by $U = \frac{1}{2} m \omega^2 x^2$.
Equating the two:
$\frac{1}{2} m \omega^2 (a^2 - x^2) = \frac{1}{2} m \omega^2 x^2$
$a^2 - x^2 = x^2$
$2x^2 = a^2$
$x^2 = \frac{a^2}{2}$
$x = \frac{a}{\sqrt{2}}$
Substituting $a = 6 \text{ cm}$:
$x = \frac{6}{\sqrt{2}} = 3 \sqrt{2} \text{ cm}$.

(C) Given that,the force acting on the particle is $F = -kx + F_0$.
We can rewrite this as $F = -k(x - \frac{F_0}{k})$.
Let $y = x - \frac{F_0}{k}$,then the force becomes $F = -ky$.
This is the standard form of the restoring force for Simple Harmonic Motion $(SHM)$,where the mean position is defined by $F = 0$.
Setting $F = 0$,we get $-k(x - \frac{F_0}{k}) = 0$,which implies $x = \frac{F_0}{k}$.
Thus,the particle oscillates about the mean position $x = \frac{F_0}{k}$.
Comparing $F = -ky$ with the standard $SHM$ equation $F = -m\omega^2 y$,we get $m\omega^2 = k$.
Therefore,the angular frequency is $\omega = \sqrt{\frac{k}{m}}$.

(A) For a simple harmonic oscillator $(SHM)$,we know that:
The maximum acceleration is given by $|a_{\max}| = \omega^2 A$.
The maximum velocity is given by $|v_{\max}| = \omega A$.
Here,$\omega$ is the angular frequency and $A$ is the amplitude of oscillation.
Therefore,the ratio of maximum acceleration to maximum velocity is:
$\frac{|a_{\max}|}{|v_{\max}|} = \frac{\omega^2 A}{\omega A} = \omega$.

(B) The equation of a particle executing simple harmonic motion is given by $x = A \cos \left(\omega t + \frac{\pi}{4}\right)$.
The velocity $v$ of the particle is the derivative of displacement with respect to time:
$v = \frac{dx}{dt} = \frac{d}{dt} \left[ A \cos \left(\omega t + \frac{\pi}{4}\right) \right] = -A \omega \sin \left(\omega t + \frac{\pi}{4}\right)$.
The speed $|v|$ is maximum when the magnitude of the sine function is $1$,i.e.,$\sin \left(\omega t + \frac{\pi}{4}\right) = -1$ or $1$.
For the first positive time $t$,we consider $\sin \left(\omega t + \frac{\pi}{4}\right) = -1 = \sin \left(\frac{3\pi}{2}\right)$.
Thus,$\omega t + \frac{\pi}{4} = \frac{3\pi}{2} \Rightarrow \omega t = \frac{3\pi}{2} - \frac{\pi}{4} = \frac{5\pi}{4} \Rightarrow t = \frac{5\pi}{4\omega}$.
However,checking the options provided,if we consider the condition for maximum speed magnitude $|v| = A\omega$,we require $\sin^2(\omega t + \pi/4) = 1$.
If we set $\omega t + \pi/4 = \pi/2$,we get $t = \pi/(4\omega)$,which corresponds to the particle passing through the equilibrium position with maximum speed in the negative direction.

(B) Let the equations of motion for particles $P$ and $Q$ be $x_1 = a \sin(\omega t)$ and $x_2 = a \sin(\omega t + \phi)$,where $\phi$ is the phase difference.
The distance between the particles is given by $d = |x_2 - x_1| = |a \sin(\omega t + \phi) - a \sin(\omega t)|$.
Using the trigonometric identity $\sin A - \sin B = 2 \sin(\frac{A-B}{2}) \cos(\frac{A+B}{2})$,we get:
$d = |2a \sin(\frac{\phi}{2}) \cos(\omega t + \frac{\phi}{2})|$.
The maximum value of this distance occurs when the magnitude of the cosine term is $1$.
Thus,$d_{max} = 2a \sin(\frac{\phi}{2})$.
Given $d_{max} = a \sqrt{2}$,we have $2a \sin(\frac{\phi}{2}) = a \sqrt{2}$.
$\sin(\frac{\phi}{2}) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
$\frac{\phi}{2} = \frac{\pi}{4}$,which implies $\phi = \frac{\pi}{2}$.

(D) Given: Mass of the particle $m = 0.4 \text{ kg}$,Amplitude $A = 0.4 \text{ m}$,Initial phase $\phi = \pi / 4$.
The kinetic energy at the mean position is equal to the total energy of the oscillator,given by $KE = \frac{1}{2} m \omega^2 A^2$.
Substituting the given values: $256 \times 10^{-3} = \frac{1}{2} \times 0.4 \times \omega^2 \times (0.4)^2$.
$0.256 = 0.2 \times \omega^2 \times 0.16$.
$0.256 = 0.032 \times \omega^2$.
$\omega^2 = \frac{0.256}{0.032} = 8$.
$\omega = \sqrt{8} = 2 \sqrt{2} \text{ rad/s}$.
The general equation for simple harmonic motion is $x = A \sin(\omega t + \phi)$.
Substituting the values,we get $x = 0.4 \sin \left((2 \sqrt{2}) t + \frac{\pi}{4}\right)$.

(B) Resonance is a phenomenon where the amplitude of oscillations increases significantly when the frequency of an externally applied periodic force matches the natural frequency of the system.
In a forced oscillation system,the sharpness of the resonance curve is determined by the damping present in the system.
The sharpness of the resonance is inversely proportional to the damping coefficient.
Therefore,when the dampening force is small,the energy loss per cycle is minimal,leading to a very sharp and high-amplitude resonance peak.
Thus,the correct option is $B$.

(C) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$,where $l$ is the effective length of the pendulum measured from the point of suspension to the center of mass of the bob.
Initially,the center of mass of the water-filled sphere is at its geometric center.
As water drains out from the hole at the bottom,the center of mass of the system shifts downwards. This increases the effective length $l$,causing the time period $T$ to increase.
As the water level continues to drop,the center of mass reaches its lowest point and then begins to rise back towards the geometric center of the sphere as the bob becomes empty.
Once the water is completely drained,the center of mass returns to the geometric center of the hollow sphere,restoring the original effective length $l$ and thus the original time period $T$.

(A) Given,maximum speed of $SHM$ is $v_{\max} = A\omega = 40 \,ms^{-1}$ ... $(i)$
Maximum acceleration of $SHM$ is $a_{\max} = A\omega^2 = 60 \,ms^{-2}$ ... (ii)
Dividing Eq. (ii) by Eq. $(i)$,we get:
$\frac{A\omega^2}{A\omega} = \frac{60}{40}$
$\omega = 1.5 \,rad/s = \frac{3}{2} \,rad/s$
We know that the time period $T$ is related to angular frequency $\omega$ by the formula:
$T = \frac{2\pi}{\omega}$
Substituting the value of $\omega$:
$T = \frac{2\pi}{3/2} = \frac{4\pi}{3} \,s$
Hence,the time period of the oscillation is $\frac{4\pi}{3} \,s$.

(C) Specific heat $s$ of a substance is defined as the amount of heat $Q$ required to raise the temperature of a unit mass $m$ of the substance by a unit degree $\Delta T$.
The formula is given by $s = \frac{Q}{m \Delta T}$.
During the process of boiling water changing into steam,the substance undergoes a phase change.
In a phase change process,the temperature of the substance remains constant,meaning $\Delta T = 0$.
Substituting this into the formula,we get $s = \frac{Q}{m \times 0} = \infty$.
Therefore,the specific heat of boiling water during this phase change is infinite.

(A) The molar specific heat capacity $C$ is defined as the amount of heat required to raise the temperature of $1 \text{ mole}$ of a gas by $1 \text{ K}$.
Mathematically,$C = \frac{\Delta Q}{n \Delta T}$.
In an adiabatic process,the system is thermally insulated from its surroundings,meaning no heat is exchanged with the environment.
Therefore,the heat change $\Delta Q = 0$.
Substituting this into the formula,we get $C = \frac{0}{n \Delta T} = 0$.
Thus,the specific heat of a gas undergoing an adiabatic change is $0$.

(A) Given: Mass of steam,$m_s = 5 \text{ g}$,Temperature of steam,$T_s = 100^{\circ} \text{C}$.
Mass of ice,$m_i = 5 \text{ g}$,Temperature of ice,$T_i = 0^{\circ} \text{C}$.
Latent heat of vaporization,$L_v = 540 \text{ cal/g}$.
Latent heat of fusion,$L_f = 80 \text{ cal/g}$.
Specific heat of water,$s = 1 \text{ cal/g}^{\circ} \text{C}$.
Step $1$: Calculate heat released by steam to condense into water at $100^{\circ} \text{C}$:
$Q_1 = m_s \times L_v = 5 \times 540 = 2700 \text{ cal}$.
Step $2$: Calculate heat required by ice to melt and reach $100^{\circ} \text{C}$:
Heat to melt ice: $Q_{melt} = m_i \times L_f = 5 \times 80 = 400 \text{ cal}$.
Heat to raise water temperature from $0^{\circ} \text{C}$ to $100^{\circ} \text{C}$: $Q_{rise} = m_i \times s \times \Delta T = 5 \times 1 \times 100 = 500 \text{ cal}$.
Total heat required by ice: $Q_2 = 400 + 500 = 900 \text{ cal}$.
Step $3$: Compare $Q_1$ and $Q_2$:
Since $Q_1 > Q_2$,the steam has more than enough energy to melt the ice and heat the resulting water to $100^{\circ} \text{C}$.
Therefore,the final mixture will be at $100^{\circ} \text{C}$ with some steam remaining.

(A) The correct matches are as follows:
$(A)$ The process of converting a liquid into a solid is called freezing or fusion (in some contexts,fusion refers to melting,but here it corresponds to the phase change process).
$(B)$ The process of converting a liquid into vapour is known as vaporisation.
$(C)$ The process of converting a solid directly into vapour without passing through the liquid state is known as sublimation.
$(D)$ The phenomenon of melting of ice due to the application of pressure is known as regelation.
Therefore,the correct matching is $A-3, B-4, C-2, D-1$.

(D) The density of water is maximum at $4^{\circ} C$.
As the temperature increases above $4^{\circ} C$,the density of water decreases,causing the volume to increase.
Similarly,as the temperature decreases below $4^{\circ} C$,the density of water also decreases due to anomalous expansion,which again causes the volume to increase.
Since the beaker is already full,any increase in volume leads to overflow.
Therefore,water overflows when heated above $4^{\circ} C$ or cooled below $4^{\circ} C$.
Statement $D$ is incorrect because water does overflow when cooled below $4^{\circ} C$.

(B) reversible process is one that can be reversed without leaving any trace on the surroundings.
Melting of ice,isothermal expansion,and adiabatic expansion (if performed quasi-statically) can be considered reversible processes.
Conduction of heat is an irreversible process because it occurs due to a finite temperature difference between two bodies.
Heat always flows spontaneously from a higher temperature to a lower temperature,and this process cannot be reversed without external work,thus increasing the entropy of the universe.

(C) Given that two plates have the same surface area $A$.
Let $K_1, K_2$ be their thermal conductivities and $t_1, t_2$ be their thicknesses respectively.
According to the problem,the ratios are:
$\frac{K_1}{K_2} = \frac{2}{3}$ and $\frac{t_1}{t_2} = \frac{2}{3}$.
Since the plates are in series,the rate of heat flow through both plates must be the same in steady state:
$\frac{dQ}{dt} = \frac{K_1 A (T_1 - T)}{t_1} = \frac{K_2 A (T - T_2)}{t_2}$
Where $T$ is the temperature of the common surface.
Substituting the given values $T_1 = 10^{\circ} C$ and $T_2 = 0^{\circ} C$:
$\frac{K_1}{t_1} (10 - T) = \frac{K_2}{t_2} (T - 0)$
$\frac{K_1}{K_2} \cdot \frac{t_2}{t_1} (10 - T) = T$
Substituting the ratios $\frac{K_1}{K_2} = \frac{2}{3}$ and $\frac{t_2}{t_1} = \frac{3}{2}$:
$\left( \frac{2}{3} \right) \cdot \left( \frac{3}{2} \right) (10 - T) = T$
$1 \cdot (10 - T) = T$
$10 - T = T$
$2T = 10$
$T = 5^{\circ} C$

(B) Given that,two charges $q_1 = 20 \mu C$ and $q_2 = 10 \mu C$ are separated by a distance $d = 50 \ m$.
Work done $W$ in bringing a unit positive charge $q = 1 \ C$ from infinity to the midpoint $M$ is given by $W = qV$,where $V$ is the electric potential at point $M$.
The potential at $M$ due to charges $q_1$ and $q_2$ is $V = \frac{K q_1}{(d/2)} + \frac{K q_2}{(d/2)} = \frac{2K}{d} (q_1 + q_2)$.
Substituting the values:
$W = 1 \times \frac{2 \times (9 \times 10^9)}{50} \times (20 + 10) \times 10^{-6} \ J$
$W = \frac{18 \times 10^9}{50} \times 30 \times 10^{-6} \ J$
$W = \frac{540 \times 10^3}{50} \ J = 10.8 \times 10^3 \ J$.

(B) By definition,an equipotential surface is a surface where the electric potential is constant at all points.
Since the work done in moving a charge $q$ along the surface is $W = 0$,and $W = \int \vec{F} \cdot d\vec{l} = q \int \vec{E} \cdot d\vec{l} = 0$,it implies that the electric field $\vec{E}$ must be perpendicular to the displacement vector $d\vec{l}$ at every point on the surface.
Therefore,the electric lines of force are always normal (perpendicular) to the equipotential surface.

(A) The force on a current-carrying conductor due to a long straight wire is given by $F = \frac{\mu_0 I_1 I_2 L}{2 \pi r}$.
For the rectangular coil $ABCD$,the forces on the horizontal segments $BC$ and $AD$ are equal and opposite,so they cancel each other out.
The force on segment $AB$ (at distance $r_1 = 2 \text{ cm} = 0.02 \text{ m}$) is attractive (towards the wire) because the currents are parallel:
$F_{AB} = \frac{\mu_0 (2 \text{ A})(1 \text{ A})(0.15 \text{ m})}{2 \pi (0.02 \text{ m})} = \frac{2 \times 10^{-7} \times 2 \times 1 \times 0.15}{0.02} = 30 \times 10^{-7} \text{ N}$.
The force on segment $CD$ (at distance $r_2 = 2 \text{ cm} + 10 \text{ cm} = 12 \text{ cm} = 0.12 \text{ m}$) is repulsive (away from the wire) because the currents are anti-parallel:
$F_{CD} = \frac{\mu_0 (2 \text{ A})(1 \text{ A})(0.15 \text{ m})}{2 \pi (0.12 \text{ m})} = \frac{2 \times 10^{-7} \times 2 \times 1 \times 0.15}{0.12} = 5 \times 10^{-7} \text{ N}$.
The net force is $F_{net} = F_{AB} - F_{CD} = 30 \times 10^{-7} - 5 \times 10^{-7} = 25 \times 10^{-7} \text{ N}$.
Since $F_{AB} > F_{CD}$,the net force is directed towards the wire.

(A) The magnetic dipole moment of the coil is given by $M = N A I$. Substituting the given values: $M = 10 \times (2 \times 10^{-4} \ m^2) \times 0.5 \ A = 10^{-3} \ A \ m^2$.
The magnetic field inside a long solenoid is given by $B = \mu_0 n I_s$. Substituting the values: $B = (4 \pi \times 10^{-7} \ T \ m/A) \times (10^3 \ m^{-1}) \times (3 \ A) = 12 \pi \times 10^{-4} \ T$.
The torque on a magnetic dipole in a magnetic field is $\tau = M B \sin(\theta)$. Since the axis of the coil is perpendicular to the axis of the solenoid,the angle between the magnetic moment vector and the magnetic field is $\theta = 90^\circ$,so $\sin(90^\circ) = 1$.
Therefore,$\tau = (10^{-3} \ A \ m^2) \times (12 \pi \times 10^{-4} \ T) \times 1 = 12 \pi \times 10^{-7} \ N \ m$.

(B) Given that, for a moving coil galvanometer:
Effective area, $A = 4 \times 10^{-2} \, m^2$
Magnetic field, $B = 5 \times 10^{-2} \, Wb \, m^{-2}$
Angle of deflection, $\phi = 0.2 \, rad$
Electric current, $I = 5 \, mA = 5 \times 10^{-3} \, A$
Assuming the number of turns $N = 1$ (as not specified, we calculate the product $NAB$):
The torque balance equation is $NAB I = k \phi$, where $k$ is the torsional constant.
$k = \frac{NAB I}{\phi} = \frac{(1) \times (4 \times 10^{-2}) \times (5 \times 10^{-2}) \times (5 \times 10^{-3})}{0.2}$
$k = \frac{100 \times 10^{-7}}{0.2} = 500 \times 10^{-7} = 5 \times 10^{-5} \, N \, m \, rad^{-1}$.
Current sensitivity $S_I = \frac{\phi}{I} = \frac{0.2}{5 \times 10^{-3}} = 40 \, rad \, A^{-1}$.
Thus, the current sensitivity is $40 \, rad \, A^{-1}$.

(B) The torque $\tau$ acting on a magnetic needle of magnetic moment $M$ in a uniform magnetic field $B$ is given by the formula: $\tau = M B \sin \theta$.
Here,$\theta$ is the angle between the magnetic moment and the magnetic field.
It is given that the needle remains perpendicular to the magnetic field,so $\theta = 90^{\circ}$.
Since $\sin 90^{\circ} = 1$,the torque becomes $\tau = M B$.
For the two different places with magnetic fields $B_1$ and $B_2$,the torques are $\tau_1 = M B_1$ and $\tau_2 = M B_2$.
Taking the ratio of the two torques: $\frac{\tau_1}{\tau_2} = \frac{M B_1}{M B_2} = \frac{B_1}{B_2}$.
Therefore,the ratio $B_1 : B_2$ is equal to $\tau_1 : \tau_2$.

(A) According to Ampere's circuital law,$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$.
For a point inside the hollow cylinder $(r < R)$,the enclosed current $I_{\text{enclosed}} = 0$. Therefore,$B(2\pi r) = 0$,which implies $B = 0$ for $r < R$.
For a point outside the hollow cylinder $(r \geq R)$,the enclosed current $I_{\text{enclosed}} = i$. Therefore,$B(2\pi r) = \mu_0 i$,which implies $B = \frac{\mu_0 i}{2\pi r}$.
This shows that the magnetic field is zero inside the cylinder and decreases as $1/r$ outside the cylinder. Graph $(i)$ correctly represents this behavior.

(C) The magnetic field at the center of a circular coil with $n$ turns, radius $a$, and current $i$ is given by $B = \frac{\mu_0 n i}{2a}$.
Since the total length of the wire $L$ is constant, $L = n(2\pi a)$.
For a single turn $(n=1)$, $L = 2\pi a$, so $a = \frac{L}{2\pi}$.
For two turns $(n'=2)$, $L = 2(2\pi a')$, so $a' = \frac{L}{4\pi} = \frac{a}{2}$.
Since $B \propto \frac{n}{a}$, we have $\frac{B'}{B} = \frac{n'}{n} \times \frac{a}{a'}$.
Substituting the values, $\frac{B'}{B} = \frac{2}{1} \times \frac{a}{a/2} = 2 \times 2 = 4$.
Therefore, $B' = 4B$.

(A) The magnetic field at the center of a circular coil of radius $r$ carrying current $i$ is given by $B = \frac{\mu_0 i}{2r}$.
Given $r = 4 \pi \text{ cm} = 4 \pi \times 10^{-2} \text{ m}$.
For the first coil with current $i_1 = 10 \text{ A}$:
$B_1 = \frac{\mu_0 \times 10}{2 \times 4 \pi \times 10^{-2}} = \frac{4 \pi \times 10^{-7} \times 10}{8 \pi \times 10^{-2}} = 0.5 \times 10^{-4} \text{ T} = 5 \times 10^{-5} \text{ T}$.
For the second coil with current $i_2 = 24 \text{ A}$:
$B_2 = \frac{\mu_0 \times 24}{2 \times 4 \pi \times 10^{-2}} = \frac{4 \pi \times 10^{-7} \times 24}{8 \pi \times 10^{-2}} = 1.2 \times 10^{-4} \text{ T} = 12 \times 10^{-5} \text{ T}$.
Since the coils are at right angles,the resultant magnetic field $B$ is given by $B = \sqrt{B_1^2 + B_2^2}$.
$B = \sqrt{(5 \times 10^{-5})^2 + (12 \times 10^{-5})^2} = \sqrt{25 + 144} \times 10^{-5} \text{ T} = \sqrt{169} \times 10^{-5} \text{ T} = 13 \times 10^{-5} \text{ Wb m}^{-2}$.

(B) Given: $R = 5 \ cm = 0.05 \ m$,$r = 12 \ cm = 0.12 \ m$,$B_{axis} = 250 \ \mu T$.
The magnetic field due to a current-carrying circular loop at a point on its axis is given by:
$B_{axis} = \frac{\mu_0 I R^2}{2(R^2 + r^2)^{3/2}}$
The magnetic field at the centre of the loop is:
$B_{centre} = \frac{\mu_0 I}{2R}$
Dividing $B_{centre}$ by $B_{axis}$:
$\frac{B_{centre}}{B_{axis}} = \frac{\mu_0 I}{2R} \times \frac{2(R^2 + r^2)^{3/2}}{\mu_0 I R^2} = \frac{(R^2 + r^2)^{3/2}}{R^3}$
Substituting the values:
$B_{centre} = B_{axis} \times \frac{(R^2 + r^2)^{3/2}}{R^3}$
$B_{centre} = 250 \ \mu T \times \frac{(0.05^2 + 0.12^2)^{3/2}}{0.05^3}$
$B_{centre} = 250 \ \mu T \times \frac{(0.0025 + 0.0144)^{3/2}}{0.000125}$
$B_{centre} = 250 \ \mu T \times \frac{(0.0169)^{3/2}}{0.000125}$
$B_{centre} = 250 \ \mu T \times \frac{(0.13)^3}{0.000125}$
$B_{centre} = 250 \ \mu T \times \frac{0.002197}{0.000125} = 250 \ \mu T \times 17.576 = 4394 \ \mu T$.

(B) For the square frame of side $a$,the magnetic field at the centre $O$ is the sum of the fields produced by its four sides.
Each side acts as a finite wire of length $a$ at a perpendicular distance $r = a/2$ from the centre.
The magnetic field due to one side is $B_1 = \frac{\mu_0 I}{4 \pi r} (\sin 45^{\circ} + \sin 45^{\circ}) = \frac{\mu_0 I}{4 \pi (a/2)} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 I}{2 \pi a} (\frac{2}{\sqrt{2}}) = \frac{\sqrt{2} \mu_0 I}{\pi a}$.
Since there are $4$ sides,the total magnetic field is $B = 4 \times B_1 = \frac{4 \sqrt{2} \mu_0 I}{\pi a}$.
For the circular coil,the perimeter is equal to the perimeter of the square,so $2 \pi R = 4 a$,which gives $R = \frac{2 a}{\pi}$.
The magnetic field at the centre of the circular coil is $B^{\prime} = \frac{\mu_0 I}{2 R} = \frac{\mu_0 I}{2 (2 a / \pi)} = \frac{\mu_0 \pi I}{4 a}$.
Taking the ratio,$\frac{B}{B^{\prime}} = \frac{4 \sqrt{2} \mu_0 I / \pi a}{\mu_0 \pi I / 4 a} = \frac{16 \sqrt{2}}{\pi^2}$.

(B) The magnetic field at the centre of a circular coil is given by $B = \frac{\mu_0 I}{2r}$.
Given that the magnetic field at the centre is the same for both coils,we have:
$\frac{\mu_0 I_1}{2(2r)} = \frac{\mu_0 I_2}{2(r)} \Rightarrow \frac{I_1}{I_2} = 2 \quad ...(i)$
Since the coils are made from the same wire,their resistance $R$ is proportional to their length $l$ $(R = \rho \frac{l}{A})$.
The lengths are $l_1 = 2\pi(2r) = 4\pi r$ and $l_2 = 2\pi(r) = 2\pi r$.
Thus,the ratio of resistances is $\frac{R_1}{R_2} = \frac{l_1}{l_2} = \frac{4\pi r}{2\pi r} = 2 \quad ...(ii)$
Using Ohm's law,$V = IR$,we have $I = V/R$.
Substituting this into equation $(i)$:
$\frac{V_1/R_1}{V_2/R_2} = 2 \Rightarrow \frac{V_1}{V_2} = 2 \times \frac{R_1}{R_2}$
Substituting the ratio from equation $(ii)$:
$\frac{V_1}{V_2} = 2 \times 2 = 4$.

(A) Given: Inner radius $r_1 = 20 \,cm = 0.2 \,m$, Outer radius $r_2 = 25 \,cm = 0.25 \,m$, Number of turns $N = 800$, Current $I = 12 \,A$.
The magnetic field $B$ inside a toroid at a distance $r$ from the center is given by $B = \frac{\mu_0 N I}{2 \pi r}$.
Since $B \propto \frac{1}{r}$, the magnetic field is maximum at the minimum radius $(r_1)$ and minimum at the maximum radius $(r_2)$.
Maximum magnetic field: $B_{\max} = \frac{4 \pi \times 10^{-7} \times 800 \times 12}{2 \pi \times 0.2} = \frac{2 \times 10^{-7} \times 9600}{0.2} = 9.6 \times 10^{-3} \,T = 9.6 \,mT$.
Minimum magnetic field: $B_{\min} = \frac{4 \pi \times 10^{-7} \times 800 \times 12}{2 \pi \times 0.25} = \frac{2 \times 10^{-7} \times 9600}{0.25} = 7.68 \times 10^{-3} \,T = 7.68 \,mT$.

(B) The magnetic moment $M$ of a current loop is given by the product of the current $I$ and the area $A$ of the loop,$M = I A$.
When a wire of length $L$ is bent into a circle,its circumference is $2 \pi r = L$,which implies the radius $r = \frac{L}{2 \pi}$.
The area of this circular loop is $A = \pi r^2 = \pi \left( \frac{L}{2 \pi} \right)^2 = \frac{L^2}{4 \pi}$.
Substituting the area into the magnetic moment formula,we get $M = I \times \frac{L^2}{4 \pi} = \frac{L^2 I}{4 \pi}$.

(B) Given that, magnetic moment, $M = 2 \,J \,T^{-1}$.
Magnetic field, $B = 0.1 \,T$.
Since the magnetic moment is aligned in the direction of the magnetic field, the initial angle is $\theta_1 = 0^{\circ}$.
When the magnet is normal to the field, the final angle is $\theta_2 = 90^{\circ}$.
The work done $W$ in rotating a magnet in a magnetic field is given by the formula:
$W = MB(\cos \theta_1 - \cos \theta_2)$
Substituting the values:
$W = 2 \times 0.1 \times (\cos 0^{\circ} - \cos 90^{\circ})$
$W = 0.2 \times (1 - 0)$
$W = 0.2 \,J$.
Hence, the net work done is $0.2 \,J$.

(C) toroid is essentially a solenoid bent into a circular shape. The magnetic field lines produced by a toroid are confined entirely within its core,forming concentric circles.
Because the magnetic field is confined within the toroid,the net magnetic field outside the toroid is zero.
If the magnetic moment $m$ were non-zero,the toroid would act like a magnetic dipole at large distances,producing a magnetic field that falls off as $\frac{1}{r^3}$.
Since the magnetic field outside an ideal toroid is zero,its magnetic moment $m$ must be zero.

(C) The magnetic force (Lorentz force) on a moving charge is given by the expression $\vec{F} = q(\vec{v} \times \vec{B})$.
From this formula,it is evident that the force depends on the velocity $\vec{v}$ of the charge.
If the charge is at rest,its velocity $\vec{v} = 0$.
Substituting this into the equation,we get $\vec{F} = q(0 \times \vec{B}) = 0$.
Therefore,a magnetic field exerts no force on a charge that is at rest.

(C) Let $\phi$ be the true dip angle and $\theta$ be the magnetic declination. When the dip circle is in the geographic meridian,the angle between the plane and the magnetic meridian is $\theta$. The apparent dip $\delta_1$ is given by $\tan \delta_1 = \frac{\tan \phi}{\cos \theta}$.
When the plane is perpendicular to the geographic meridian,the angle between the plane and the magnetic meridian is $(90^\circ - \theta)$. The apparent dip $\delta_2$ is given by $\tan \delta_2 = \frac{\tan \phi}{\cos(90^\circ - \theta)} = \frac{\tan \phi}{\sin \theta}$.
From these two equations,we have $\tan \phi = \tan \delta_1 \cos \theta$ and $\tan \phi = \tan \delta_2 \sin \theta$.
Equating the two expressions for $\tan \phi$: $\tan \delta_1 \cos \theta = \tan \delta_2 \sin \theta$.
Rearranging gives $\frac{\sin \theta}{\cos \theta} = \frac{\tan \delta_1}{\tan \delta_2}$,which implies $\tan \theta = \frac{\tan \delta_1}{\tan \delta_2}$.
Therefore,$\theta = \tan ^{-1}\left(\frac{\tan \delta_1}{\tan \delta_2}\right)$.

(B) According to Curie's law, the magnetization $M$ of a paramagnetic material is given by $M = C \frac{B}{T}$, where $B$ is the external magnetic field, $T$ is the absolute temperature, and $C$ is Curie's constant.
Given $M_1 = 0.8 \text{ A m}^{-1}$, $B_1 = 0.8 \text{ T}$, and $T_1 = 5 \text{ K}$.
Substituting these values into the formula: $0.8 = C \frac{0.8}{5} \Rightarrow C = 5 \text{ K}$.
Now, for the new temperature $T_2 = 20 \text{ K}$ with the same magnetic field $B_2 = 0.8 \text{ T}$, the new magnetization $M_2$ is:
$M_2 = C \frac{B_2}{T_2} = 5 \times \frac{0.8}{20} = \frac{4}{20} = 0.2 \text{ A m}^{-1}$.

(D) Let the magnetic flux density outside the material be $B_0$.
Given that the magnetic flux density inside the material is $B = 4 B_0$.
The relationship between the magnetic flux density inside a material and the external magnetic field is given by $B = \mu_r B_0$,where $\mu_r$ is the relative magnetic permeability of the material.
Substituting the given values,we get $4 B_0 = \mu_r B_0$.
Therefore,$\mu_r = 4$.

(A) The mass of a stable nucleus is always less than the sum of the masses of its constituent protons and neutrons. This difference in mass is known as the mass defect $(\Delta m)$.
According to Einstein's mass-energy equivalence principle $(E = \Delta m c^2)$, this mass defect is converted into binding energy, which holds the nucleus together.
Since energy is released when a nucleus is formed, the system reaches a lower energy state, meaning the mass of the bound nucleus must be less than the sum of the masses of its individual nucleons.
Therefore, option $A$ is the correct statement.

(C) nuclear reaction is possible if it satisfies the laws of conservation of charge (atomic number) and mass number.
For option $C$: ${ }_{93} Np^{239} \longrightarrow{ }_{94} Pu^{239}+\beta^{-}+\bar{\nu}$.
Charge conservation: $93 = 94 + (-1) + 0 = 93$. This is satisfied.
Mass number conservation: $239 = 239 + 0 + 0 = 239$. This is satisfied.
This reaction represents the $\beta^{-}$-decay of Neptunium-$239$ into Plutonium-$239$,which is a physically possible process.

(B) Nuclear fusion is a process in which two light nuclei combine to form a heavier nucleus. This process requires overcoming the strong electrostatic repulsion between the positively charged nuclei. To achieve this,the nuclei must have extremely high kinetic energy,which is provided by maintaining a very high temperature (on the order of $10^7$ to $10^8 \ K$). Therefore,fusion reactions are initiated with the help of high temperature.

(D) Nuclear density is defined as the ratio of the mass of the nucleus to its volume.
The mass of a nucleus with mass number $A$ is approximately $M = A \cdot m$,where $m$ is the average mass of a nucleon $(m \approx 1.67 \times 10^{-27} \text{ kg})$.
The radius of a nucleus is given by $R = R_0 A^{1/3}$,where $R_0 \approx 1.2 \times 10^{-15} \text{ m}$.
The volume of the nucleus is $V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A$.
Nuclear density $\rho = \frac{M}{V} = \frac{A \cdot m}{\frac{4}{3} \pi R_0^3 A} = \frac{3m}{4 \pi R_0^3}$.
Since $\rho$ is independent of the mass number $A$,it is constant for all nuclei.
Substituting the values,$\rho \approx 2.3 \times 10^{17} \text{ kg m}^{-3}$ to $3.0 \times 10^{17} \text{ kg m}^{-3}$.
Thus,the correct order of magnitude is $10^{17} \text{ kg m}^{-3}$.

(C) The refractive index $\mu$ of a diamond is very high (approximately $2.42$).
Since the critical angle $\theta_c$ is given by the relation $\sin \theta_c = \frac{1}{\mu}$,a high refractive index results in a very small critical angle.
When light enters a well-cut diamond,it strikes the internal surfaces at angles greater than the critical angle.
This leads to multiple total internal reflections within the diamond.
Consequently,the light is trapped and reflected multiple times,making the diamond appear bright.

(A) The refractive index of glass with respect to water is given by the ratio of their absolute refractive indices (or refractive indices with respect to air).
Given: Refractive index of water,$\mu_w = \frac{4}{3}$.
Refractive index of glass,$\mu_g = \frac{5}{3}$.
The relative refractive index of glass with respect to water is $\mu = \frac{\mu_g}{\mu_w} = \frac{5/3}{4/3} = \frac{5}{4}$.
The critical angle $C$ is defined by the relation $\sin C = \frac{1}{\mu}$.
Substituting the value of $\mu$,we get $\sin C = \frac{1}{5/4} = \frac{4}{5}$.
Therefore,the critical angle $C = \sin ^{-1}\left(\frac{4}{5}\right)$.

(D) The focal length $f$ of a lens is given by the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
According to Cauchy's equation,the refractive index $\mu$ of a material depends on the wavelength $\lambda$ of light,where $\mu$ increases as $\lambda$ decreases.
Since the wavelength of blue light is shorter than that of red light,the refractive index for blue light is greater than that for red light $(\mu_{blue} > \mu_{red})$.
Consequently,the focal length $f$ decreases when red light is replaced by blue light.
Therefore,the assertion $(A)$ is false and the reason $(R)$ is also false.

(C) Given,refractive index of the lens,$\mu = 1.5$.
Radius of the curved face,$R_1 = 12 \,cm$.
Radius of the plane face,$R_2 = \infty$.
Using the lens maker's formula:
$\frac{1}{f} = (\mu - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$
Substituting the values:
$\frac{1}{f} = (1.5 - 1) \left[ \frac{1}{12} - \frac{1}{\infty} \right]$
Since $\frac{1}{\infty} = 0$,we have:
$\frac{1}{f} = 0.5 \times \frac{1}{12}$
$\frac{1}{f} = \frac{0.5}{12} = \frac{1}{24}$
Therefore,$f = 24 \,cm$.
Thus,the focal length of the given plano-convex lens is $24 \,cm$.

(C) The focal length of the convex lens is given as $f$.
The object is placed at a distance $x$ from the first focal point.
Since the first focal point is at a distance $f$ from the optical center,the total object distance $u$ from the optical center is $u = -(f + x)$.
Using the lens formula,$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Substituting $u = -(f + x)$,we get $\frac{1}{v} - \frac{1}{-(f + x)} = \frac{1}{f}$.
$\frac{1}{v} = \frac{1}{f} - \frac{1}{f + x} = \frac{f + x - f}{f(f + x)} = \frac{x}{f(f + x)}$.
Thus,$v = \frac{f(f + x)}{x}$.
The magnification $m$ is given by $m = \frac{v}{u}$.
$m = \frac{\frac{f(f + x)}{x}}{-(f + x)} = -\frac{f}{x}$.
The ratio of the size of the real image to the object is the magnitude of magnification,which is $|m| = \frac{f}{x}$.

(A) Given: Refractive index of the lens,$\mu_g = 1.5$.
Refractive index of the medium,$\mu_m = 1.6$.
Power of the lens in air,$P = -5 \ D$.
Refractive index of air,$\mu_a = 1$.
Using the Lens Maker's Formula:
$P = \frac{1}{f} = (\mu_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$ (in air).
For the medium,the power $P'$ is given by:
$P' = \frac{1}{f'} = \left( \frac{\mu_g}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Dividing $P'$ by $P$:
$\frac{P'}{P} = \frac{(\frac{\mu_g}{\mu_m} - 1)}{(\mu_g - 1)} = \frac{(\frac{1.5}{1.6} - 1)}{(1.5 - 1)} = \frac{(\frac{1.5 - 1.6}{1.6})}{0.5} = \frac{-0.1}{1.6 \times 0.5} = \frac{-0.1}{0.8} = -\frac{1}{8}$.
Therefore,$P' = P \times (-\frac{1}{8}) = -5 \times (-0.125) = 0.625 \ D$.

(A) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2meV}}$.
Alternatively, using the simplified relation: $\lambda \approx \frac{12.27}{\sqrt{V}} \text{ Å}$.
Given $\lambda = 0.50 \text{ Å}$, we substitute this into the equation:
$0.50 = \frac{12.27}{\sqrt{V}}$
$\sqrt{V} = \frac{12.27}{0.50} = 24.54$
$V = (24.54)^2 \approx 602.2 \text{ V}$.
Thus, the required voltage is approximately $602 \text{ V}$.

(A) The speed of light in medium $A$ is given by $v_A = f \lambda_A = (5 \times 10^{14} \ Hz) \times (300 \times 10^{-9} \ m) = 1.5 \times 10^8 \ m/s$.
The absolute refractive index of medium $A$ is $\mu_A = \frac{c}{v_A} = \frac{3 \times 10^8 \ m/s}{1.5 \times 10^8 \ m/s} = 2$.
We are given the ratio of speeds $\frac{v_A}{v_B} = \frac{4}{5}$.
Since the refractive index $\mu$ is inversely proportional to the speed of light $v$ $(\mu = \frac{c}{v})$,we have $\frac{\mu_B}{\mu_A} = \frac{v_A}{v_B}$.
Substituting the values: $\frac{\mu_B}{2} = \frac{4}{5}$.
Therefore,$\mu_B = 2 \times \frac{4}{5} = \frac{8}{5} = 1.6$.

(A) In a $p-n$ junction,the depletion region is formed due to the diffusion of charge carriers.
On the $p$-side of the junction,there are negatively charged acceptor ions,which create a negative charge density.
On the $n$-side of the junction,there are positively charged donor ions,which create a positive charge density.
The charge density $\rho$ is negative on the $p$-side and positive on the $n$-side.
At the junction interface $(r = 0)$,the charge density transitions from negative to positive.
Therefore,the graph that shows a negative charge density on the $p$-side and a positive charge density on the $n$-side,passing through the origin,is the correct representation.
This corresponds to the graph where the curve is below the $r$-axis for the $p$-region and above the $r$-axis for the $n$-region.

(B) The input voltage $V_i$ is an alternating voltage with a peak value of $10 \, V$.
For the positive half-cycle of $V_i$, the top terminal is at a higher potential than the bottom terminal. In this case, diode $D_2$ is forward-biased and conducts, while diode $D_1$ is reverse-biased and acts as an open circuit.
The circuit simplifies to a voltage divider consisting of the $2 \, k\Omega$ resistor (where $V_0$ is measured) in series with another $2 \, k\Omega$ resistor connected to the negative terminal.
The total resistance in the path is $2 \, k\Omega + 2 \, k\Omega = 4 \, k\Omega$.
Using the voltage divider rule, the output voltage $V_0$ across the $2 \, k\Omega$ resistor is:
$V_0 = V_i \times \frac{2 \, k\Omega}{2 \, k\Omega + 2 \, k\Omega} = V_i \times \frac{2}{4} = \frac{V_i}{2}$.
Since the maximum input voltage is $10 \, V$, the maximum output voltage is:
$V_{0, \text{max}} = \frac{10 \, V}{2} = 5 \, V$.

(B) Given that,$R_1 = 1 \text{ k}\Omega = 1000 \text{ } \Omega$,and Zener breakdown voltage $V_z = 6 \text{ V}$.
Let $I_1$ be the current through $R_1$ and $I_z$ be the current through the Zener diode.
Since $R_1$ is in parallel with the Zener diode,the voltage across $R_1$ is $V_z = 6 \text{ V}$.
Therefore,$I_1 = \frac{V_z}{R_1} = \frac{6 \text{ V}}{1000 \text{ } \Omega} = 6 \times 10^{-3} \text{ A}$.
According to the given condition,the Zener current is $I_z = 5 I_1 = 5 \times (6 \times 10^{-3} \text{ A}) = 30 \times 10^{-3} \text{ A}$.
The total current $I$ drawn from the source $V_s = 30 \text{ V}$ is $I = I_1 + I_z = 6 \times 10^{-3} + 30 \times 10^{-3} = 36 \times 10^{-3} \text{ A}$.
Also,the voltage drop across $R$ is $V_s - V_z = 30 \text{ V} - 6 \text{ V} = 24 \text{ V}$.
Using Ohm's law,$R = \frac{V_s - V_z}{I} = \frac{24}{36 \times 10^{-3}} = \frac{24000}{36} \text{ } \Omega = \frac{2000}{3} \text{ } \Omega$.

(A) Given,collector resistance $R_C = 800 \Omega$ and voltage drop across it $V_R = 0.5 V$.
Collector current $I_C = \frac{V_R}{R_C} = \frac{0.5}{800} = 6.25 \times 10^{-4} A = 0.625 \times 10^{-3} A$.
We know that the current gain $\alpha = \frac{I_C}{I_E}$,so emitter current $I_E = \frac{I_C}{\alpha} = \frac{0.625 \times 10^{-3}}{0.96} \approx 6.51 \times 10^{-4} A$.
The base current $I_B = I_E - I_C = I_C \left( \frac{1}{\alpha} - 1 \right) = I_C \left( \frac{1 - \alpha}{\alpha} \right)$.
Substituting the values: $I_B = 0.625 \times 10^{-3} \times \left( \frac{1 - 0.96}{0.96} \right) = 0.625 \times 10^{-3} \times \left( \frac{0.04}{0.96} \right) = 0.625 \times 10^{-3} \times \frac{1}{24} \approx 0.02604 \times 10^{-3} A = 2.6 \times 10^{-5} A$.

(B) Given: $\Delta I_b = 20 \ \mu A = 20 \times 10^{-6} \ A$,$\Delta I_c = 2 \ mA = 2 \times 10^{-3} \ A$,and $\Delta V_{BE} = 0.04 \ V$.
The input resistance $R_{\text{input}}$ is defined as the ratio of the change in base-emitter voltage to the change in base current:
$R_{\text{input}} = \frac{\Delta V_{BE}}{\Delta I_b} = \frac{0.04}{20 \times 10^{-6}} = \frac{0.04 \times 10^6}{20} = 2 \times 10^3 \ \Omega = 2 \ k\Omega$.
The $AC$ current gain $\beta$ is defined as the ratio of the change in collector current to the change in base current:
$\beta = \frac{\Delta I_c}{\Delta I_b} = \frac{2 \times 10^{-3}}{20 \times 10^{-6}} = \frac{2000}{20} = 100$.
Thus,the input resistance is $2 \ k\Omega$ and the $AC$ current gain is $100$.

(B) Given that, $V_{CE}=7 \, V$, $\beta=100$, $R_C=2 \, k\Omega = 2000 \, \Omega$, and $V_{CC}=15 \, V$.
We know that the current gain in a common emitter circuit is given by:
$\beta = \frac{I_C}{I_B} \dots (i)$
Applying Kirchhoff's Voltage Law $(KVL)$ in the output loop (collector-emitter loop):
$V_{CC} - I_C R_C - V_{CE} = 0$
Substituting the given values:
$15 - I_C(2000) - 7 = 0$
$8 = I_C(2000)$
$I_C = \frac{8}{2000} \, A = 4 \times 10^{-3} \, A = 4 \, mA \dots (ii)$
Now, substituting the value of $I_C$ from equation $(ii)$ into equation $(i)$:
$100 = \frac{4 \, mA}{I_B}$
$I_B = \frac{4}{100} \, mA = 0.04 \, mA$.

(A) Given, current gain, $\alpha = 0.96$.
Emitter current, $I_E = 7.2 \,mA = 7.2 \times 10^{-3} \,A$.
We know that, $\alpha = I_C / I_E$.
Therefore, $I_C = \alpha \times I_E = 0.96 \times 7.2 \,mA = 6.912 \,mA$.
Also, the relation between emitter, collector, and base current is $I_E = I_C + I_B$.
Thus, $I_B = I_E - I_C = 7.2 \,mA - 6.912 \,mA = 0.288 \,mA \approx 0.29 \,mA$.

(B) The truth table shows that the output $X$ is $1$ if either input $A$ or input $B$ (or both) is $1$.
This behavior is represented by the Boolean expression $X = A + B$.
The logic gate that performs this operation is the $OR$ gate.
Therefore,the correct option is $B$.

(D) The truth table for the waveform is given by:

Time Interval	$A$	$B$	$Y$
$t_1-t_2$	$0$	$0$	$1$
$t_2-t_3$	$0$	$1$	$1$
$t_3-t_4$	$1$	$0$	$1$
$t_4-t_5$	$1$	$1$	$0$

From the truth table,we observe that the output $Y$ is $0$ only when both inputs $A$ and $B$ are $1$. In all other cases,the output is $1$.
This behavior corresponds to the Boolean expression $Y = \overline{A \cdot B}$.
This is the characteristic truth table of a $NAND$ gate.

(C) From the given figure,it is evident that the rectifier is a full-wave rectifier.
For a full-wave rectifier,the output current is given by $I = I_0 \sin(\omega t)$ for the first half-cycle and $I = I_0 \sin(\omega t)$ for the second half-cycle (due to rectification).
The average value of the current over one complete cycle $T$ is given by:
$I_{\text{avg}} = \frac{1}{T} \int_{0}^{T} I(t) dt$
For a full-wave rectifier,the period is $T/2$. The average value is calculated as:
$I_{\text{avg}} = \frac{1}{T/2} \int_{0}^{T/2} I_0 \sin(\omega t) dt$
$I_{\text{avg}} = \frac{2}{T} \cdot I_0 \left[ -\frac{\cos(\omega t)}{\omega} \right]_{0}^{T/2}$
Since $\omega = \frac{2\pi}{T}$,we have:
$I_{\text{avg}} = \frac{2 I_0}{T} \cdot \frac{T}{2\pi} [-\cos(\pi) + \cos(0)]$
$I_{\text{avg}} = \frac{I_0}{\pi} [1 + 1] = \frac{2 I_0}{\pi}$

(B) The conductivity of an intrinsic semiconductor is given by $\sigma = n_i e (\mu_e + \mu_h)$.
Given $n_i = 2.5 \times 10^{19} \ m^{-3}$,$e = 1.6 \times 10^{-19} \ C$,$\mu_e = 0.39 \ m^2/V\cdot s$,and $\mu_h = 0.19 \ m^2/V\cdot s$.
$\sigma = (2.5 \times 10^{19}) \times (1.6 \times 10^{-19}) \times (0.39 + 0.19) = 4 \times 0.58 = 2.32 \ \Omega^{-1}m^{-1}$.
The resistivity $\rho = \frac{1}{\sigma} = \frac{1}{2.32} \ \Omega\cdot m$.
The resistance $R = \rho \frac{L}{A}$.
Given $L = 0.925 \ cm = 9.25 \times 10^{-3} \ m$ and $A = 1 \ mm^2 = 10^{-6} \ m^2$.
$R = \frac{1}{2.32} \times \frac{9.25 \times 10^{-3}}{10^{-6}} = \frac{9250}{2.32} \approx 3987 \ \Omega \approx 4.0 \ k\Omega$.

(C) The electromagnetic force and the weak nuclear force are two of the four fundamental forces in nature.
It was proposed that these two forces can be unified into a single interaction known as the electroweak interaction.
This theoretical unification was successfully achieved by three physicists: Sheldon Glashow,Steven Weinberg,and Abdus Salam.
Therefore,among the given options,Salam is the correct answer.

(C) Sir $C$.$V$. Raman was awarded the Nobel Prize in Physics in $1930$ for his pioneering work on the scattering of light,a phenomenon now known as the Raman Effect.
He discovered that when a beam of light traverses a transparent medium,a small fraction of the light is scattered in directions other than the incident direction,and this scattered light undergoes a change in wavelength and frequency.

(C) Given that atomic mass $= 1 \text{ amu} = 1.66 \times 10^{-27} \text{ kg}$.
By using Einstein's mass-energy equivalence relation,$E = mc^2$,where $c = 3 \times 10^8 \text{ m/s}$ is the speed of light.
$E = (1.66 \times 10^{-27} \text{ kg}) \times (3 \times 10^8 \text{ m/s})^2 = 1.494 \times 10^{-11} \text{ J}$.
We know that $1 \text{ MeV} = 1.602 \times 10^{-13} \text{ J}$.
Therefore,$E = \frac{1.494 \times 10^{-11} \text{ J}}{1.602 \times 10^{-13} \text{ J/MeV}} \approx 931 \text{ MeV}$.
Thus,$1 \text{ amu}$ is equivalent to $931 \text{ MeV}$ of energy.

(C) The generation,propagation,and detection of electromagnetic waves are the fundamental principles behind wireless communication systems. Radio and television broadcasting rely entirely on the transmission of electromagnetic waves through space,which are then detected by receivers to convert them back into audio and visual signals. Therefore,radio and television are based on these principles.

(B) The book “The Evolution of Physics” (often referred to in the context of physics literature) was co-authored by Albert Einstein and Leopold Infeld. Given the options provided,Albert Einstein is the correct author.

(B) For a single slit of width $a$,the condition for the $n^{th}$ minimum is given by $a \sin \theta = n \lambda$.
For the first minimum $(n=1)$,the angle is $\theta = 30^{\circ}$.
Thus,$a \sin 30^{\circ} = 1 \cdot \lambda \Rightarrow a(0.5) = \lambda \Rightarrow a = 2 \lambda$.
The condition for the $n^{th}$ secondary maximum is given by $a \sin \theta' = (n + \frac{1}{2}) \lambda$.
For the first secondary maximum,we take $n=1$.
Substituting $a = 2 \lambda$ and $n=1$ into the equation:
$(2 \lambda) \sin \theta' = (1 + \frac{1}{2}) \lambda$
$2 \sin \theta' = \frac{3}{2}$
$\sin \theta' = \frac{3}{4}$
$\theta' = \sin^{-1} \left( \frac{3}{4} \right)$.

(C) For a plane transmission grating,the condition for the $n$th order diffraction maxima is given by $d \sin \theta = n \lambda$,where $d$ is the grating element,$\theta$ is the angle of diffraction,and $\lambda$ is the wavelength of light.
From the relation $\sin \theta = \frac{n \lambda}{d}$,it is clear that for a fixed order $n$ and grating constant $d$,the angle of diffraction $\theta$ is directly proportional to the wavelength $\lambda$ (i.e.,$\theta \propto \lambda$ for small angles).
When the source is replaced by another source of shorter wavelength,the value of $\lambda$ decreases. Consequently,$\sin \theta$ decreases,which means the angle of diffraction $\theta$ for all orders $(n=1, 2, 3)$ decreases.
The direct image at $O$ (the $0$th order) remains stationary because $\sin \theta = 0$ for $n=0$,regardless of the wavelength. However,the diffracted images $A, B$,and $C$ will move closer to the central maximum $O$ as their respective diffraction angles decrease.
Therefore,the images $C, B$,and $A$ will shift towards $O$.