AP EAMCET 2023 Physics Question Paper with Answer and Solution

(D) Given: Mass of stone, $m = 2 \,kg$.
Length of the string (radius), $r = 2 \,m$.
Maximum tension, $T_{\max} = 64 \,N$.
The centripetal force required for circular motion is provided by the tension in the string: $T_{\max} = \frac{m v_{\max}^2}{r}$.
Substituting the values: $64 = \frac{2 \times v_{\max}^2}{2}$.
Solving for $v_{\max}$: $v_{\max}^2 = 64$, so $v_{\max} = 8 \,m/s$.
We know that $v = r \omega$, where $\omega = 2 \pi f$ and $f$ is the frequency in rotations per second.
So, $v_{\max} = r (2 \pi f) \implies 8 = 2 \times 2 \pi f$.
$4 \pi f = 8 \implies f = \frac{2}{\pi}$ rotations per second.
To find the number of rotations per minute $(N)$, we multiply the frequency by $60$: $N = f \times 60 = \frac{2}{\pi} \times 60 = \frac{120}{\pi}$.
Thus, the permissible maximum number of rotations per minute is $\frac{120}{\pi}$.

(D) Given: Velocity of the car, $v = 30 \,ms^{-1}$. Radius of the circular road, $r = 300 \,m$. Tangential acceleration, $a_t = 4 \,ms^{-2}$.
First, calculate the centripetal acceleration $(a_c)$ using the formula $a_c = \frac{v^2}{r}$.
$a_c = \frac{30^2}{300} = \frac{900}{300} = 3 \,ms^{-2}$.
The net acceleration $(a)$ is the vector sum of the tangential acceleration and the centripetal acceleration, which are perpendicular to each other.
$a = \sqrt{a_t^2 + a_c^2}$.
$a = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \,ms^{-2}$.

(A) For a damped harmonic oscillator,the amplitude at time $t$ is given by $A(t) = A_0 e^{-bt}$,where $A_0$ is the initial amplitude and $b$ is the damping constant.
At $t = 2 \ s$,$A_1 = A_0 e^{-2b} \implies \frac{A_1}{A_0} = e^{-2b}$.
At $t = 4 \ s$,$A_2 = A_0 e^{-4b} \implies \frac{A_2}{A_0} = e^{-4b}$.
We can write $e^{-4b} = (e^{-2b})^2$.
Substituting the expression for $e^{-2b}$,we get $\frac{A_2}{A_0} = \left(\frac{A_1}{A_0}\right)^2$.
$\frac{A_2}{A_0} = \frac{A_1^2}{A_0^2}$.
$A_2 = \frac{A_1^2}{A_0}$.
$A_1^2 = A_0 A_2 \implies A_1 = \sqrt{A_0 A_2}$.

(C) The amplitude of a damped oscillator is given by $A(t) = A_0 e^{-\frac{bt}{2m}}$.
Given that $A(t) = 0.25 A_0 = \frac{A_0}{4}$ at time $t$,we have:
$\frac{A_0}{4} = A_0 e^{-\frac{bt}{2m}} \implies \frac{1}{4} = e^{-\frac{bt}{2m}} \implies 4 = e^{\frac{bt}{2m}}$.
Taking the natural logarithm on both sides: $\ln(4) = \frac{bt}{2m} \implies 2\ln(2) = \frac{bt}{2m} \implies t = \frac{4m \ln(2)}{b} \quad ... (1)$
The mechanical energy of a damped oscillator is given by $E(t) = E_0 e^{-\frac{bt}{m}}$.
Given that $E(t') = 0.5 E_0 = \frac{E_0}{2}$ at time $t'$,we have:
$\frac{E_0}{2} = E_0 e^{-\frac{bt'}{m}} \implies \frac{1}{2} = e^{-\frac{bt'}{m}} \implies 2 = e^{\frac{bt'}{m}}$.
Taking the natural logarithm on both sides: $\ln(2) = \frac{bt'}{m} \implies t' = \frac{m \ln(2)}{b} \quad ... (2)$
Dividing equation $(2)$ by $(1)$:
$\frac{t'}{t} = \frac{m \ln(2) / b}{4m \ln(2) / b} = \frac{1}{4} \implies t' = \frac{t}{4}$.

(B) The mechanical energy of a damped oscillator is given by $E(t) = E_0 e^{-bt/m}$.
At $t = 4 \ s$,$E = E_0/2$.
So,$E_0/2 = E_0 e^{-4b/m} \Rightarrow e^{-4b/m} = 1/2 \Rightarrow e^{4b/m} = 2$.
Taking the natural logarithm on both sides,$4b/m = \ln 2$,so $b/m = (\ln 2)/4$.
At time $T = 4 + t$,the energy is $12.5 \% E_0 = (1/8) E_0$.
So,$E_0/8 = E_0 e^{-b(4+t)/m} \Rightarrow e^{-b(4+t)/m} = 1/8 \Rightarrow e^{b(4+t)/m} = 8 = 2^3$.
Taking the natural logarithm,$b(4+t)/m = 3 \ln 2$.
Substituting $b/m = (\ln 2)/4$,we get $[(\ln 2)/4] \times (4+t) = 3 \ln 2$.
Dividing by $\ln 2$,we get $(4+t)/4 = 3 \Rightarrow 4+t = 12 \Rightarrow t = 8 \ s$.

(C) The damping force $F_d$ is given by the relation $F_d = -bv$,where $b$ is the damping constant and $v$ is the velocity of the oscillator.
Because of the negative sign,the damping force always acts in the direction opposite to the velocity of the body.
Therefore,the statement that the damping force acts in the direction of the velocity is incorrect.

(D) The displacement of a damped harmonic oscillator is given by $x(t) = A(t) \cos(\omega t + \varphi)$,where $A(t) = A_0 e^{-bt/2m}$.
Comparing this with the given equation $x(t) = e^{-0.1 t} \cos(10 \pi t + \varphi)$,the amplitude is $A(t) = A_0 e^{-0.1 t}$,where $A_0 = 1$.
We need to find the time $t$ when the amplitude becomes half of its initial value,i.e.,$A(t) = \frac{A_0}{2}$.
Substituting this into the amplitude equation: $\frac{A_0}{2} = A_0 e^{-0.1 t}$.
Dividing both sides by $A_0$,we get: $\frac{1}{2} = e^{-0.1 t}$.
Taking the natural logarithm on both sides: $\ln(0.5) = -0.1 t$.
Since $\ln(0.5) = -\ln(2) \approx -0.693$,we have: $-0.693 = -0.1 t$.
Solving for $t$: $t = \frac{0.693}{0.1} = 6.93 \ s$.
Rounding to the nearest integer,we get $t \approx 7 \ s$.

(D) The weight of the block is $W = mg = 20 \,N$. Given $g = 10 \,ms^{-2}$, the mass of the block is $m = \frac{20}{10} = 2 \,kg$.
For a block-spring system on a smooth inclined plane, the component of gravity along the plane only shifts the equilibrium position and does not affect the frequency or period of oscillation.
The angular frequency of the oscillation is given by $\omega = \sqrt{\frac{K}{m}}$, where $K = 8 \pi^2 \,Nm^{-1}$ is the spring constant.
$\omega = \sqrt{\frac{8 \pi^2}{2}} = \sqrt{4 \pi^2} = 2 \pi \,rad/s$.
The period of oscillation $T$ is given by $T = \frac{2 \pi}{\omega}$.
Substituting the value of $\omega$, we get $T = \frac{2 \pi}{2 \pi} = 1 \,s$.
Therefore, the period of oscillations is $1 \,s$.

(A) Given mass $m = 8 \,kg$ and extension $x = 20 \,cm = 0.2 \,m$.
Using Hooke's Law, $mg = kx$, we find the spring constant $k = \frac{mg}{x} = \frac{8 \times 10}{0.2} = 400 \,N/m$.
For a mass $M$ oscillating on a spring, the time period $T$ is given by $T = 2\pi \sqrt{\frac{M}{k}}$.
Given $T = \frac{\pi}{5} \,s$, we have $\frac{\pi}{5} = 2\pi \sqrt{\frac{M}{400}}$.
Dividing both sides by $\pi$, we get $\frac{1}{5} = 2 \sqrt{\frac{M}{400}} = 2 \frac{\sqrt{M}}{20} = \frac{\sqrt{M}}{10}$.
Thus, $\sqrt{M} = \frac{10}{5} = 2$.
Squaring both sides, $M = 4 \,kg$.

(C) When a block of mass $M$ hangs from a spring in equilibrium,the downward gravitational force is balanced by the upward spring force.
$Mg = kx$,where $k$ is the spring constant and $x$ is the extension in the spring.
Therefore,the extension $x = \frac{Mg}{k}$.
We know that the angular frequency of a spring-mass system is given by $\omega = \sqrt{\frac{k}{M}}$.
Squaring both sides,we get $\omega^2 = \frac{k}{M}$,which implies $k = M\omega^2$.
Substituting the value of $k$ into the expression for $x$:
$x = \frac{Mg}{M\omega^2} = \frac{g}{\omega^2}$.
Thus,when the block is removed,the spring shortens by $\frac{g}{\omega^2}$.

(B) Given: Mass $m = 3 \,kg$, Amplitude $A = \frac{2}{\pi} \,m$, Kinetic energy at mean position $K_{max} = 6 \,J$.
At the mean position, the kinetic energy is equal to the total energy of the simple harmonic oscillator: $K_{max} = \frac{1}{2} m \omega^2 A^2 = 6 \,J$.
Substituting $\omega = \frac{2\pi}{T}$, we get: $\frac{1}{2} m (\frac{2\pi}{T})^2 A^2 = 6$.
$\frac{1}{2} \times 3 \times \frac{4\pi^2}{T^2} \times (\frac{2}{\pi})^2 = 6$.
$\frac{1}{2} \times 3 \times \frac{4\pi^2}{T^2} \times \frac{4}{\pi^2} = 6$.
$6 \times \frac{4}{T^2} = 6$.
$T^2 = 4$, which gives $T = 2 \,s$.

(D) The equation of motion is $x=3 \sin \left(6 t+\frac{\pi}{6}\right)$,where amplitude $A=3 \ m$.
At time $t=0$,the displacement is $x=3 \sin \left(\frac{\pi}{6}\right) = 3 \times \frac{1}{2} = 1.5 \ m$.
The potential energy $V$ is given by $V = \frac{1}{2} k x^2$.
The kinetic energy $K$ is given by $K = \frac{1}{2} k (A^2 - x^2)$.
The ratio of potential energy to kinetic energy is $\frac{V}{K} = \frac{\frac{1}{2} k x^2}{\frac{1}{2} k (A^2 - x^2)} = \frac{x^2}{A^2 - x^2}$.
Substituting the values $x=1.5$ and $A=3$:
$\frac{V}{K} = \frac{(1.5)^2}{3^2 - (1.5)^2} = \frac{2.25}{9 - 2.25} = \frac{2.25}{6.75} = \frac{1}{3}$.
Thus,the ratio is $1:3$.

(D) Given the force equation $F = -75 y$. Comparing this with the standard $SHM$ force equation $F = -ky$, we get the force constant $k = 75 \,N/m$.
Given mass $m = 3 \,kg$, the angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{75}{3}} = \sqrt{25} = 5 \,rad/s$.
The velocity at the mean position is the maximum velocity $V_{\max} = 2.5 \,m/s$.
Since $V_{\max} = A\omega$, the amplitude $A = \frac{V_{\max}}{\omega} = \frac{2.5}{5} = 0.5 \,m$.
The maximum acceleration is given by $a_{\max} = \omega^2 A$.
Substituting the values, $a_{\max} = (5)^2 \times 0.5 = 25 \times 0.5 = 12.5 \,m/s^2$.

(B) When a system is subjected to an external periodic force,it undergoes forced oscillations.
The amplitude of these forced oscillations depends on the driving frequency $\omega_d$ and the natural frequency $\omega$ of the system.
As the driving frequency $\omega_d$ approaches the natural frequency $\omega$,the amplitude of the oscillations increases.
When $\omega_d = \omega$,the system reaches a state known as resonance,where the amplitude of oscillations becomes maximum.
Therefore,the condition for maximum amplitude is $\omega_d = \omega$.

(D) The displacement of a particle executing simple harmonic motion starting from the mean position is given by $y = A \sin(\omega t)$.
Given $\omega = \frac{\pi}{4} \text{ rad s}^{-1}$.
Displacement at $t = 1 \text{ s}$ is $y_1 = A \sin(\frac{\pi}{4} \times 1) = A \sin(\frac{\pi}{4}) = \frac{A}{\sqrt{2}}$.
Displacement at $t = 2 \text{ s}$ is $y_2 = A \sin(\frac{\pi}{4} \times 2) = A \sin(\frac{\pi}{2}) = A$.
Distance travelled in the first second is $d_1 = y_1 = \frac{A}{\sqrt{2}}$.
Distance travelled in the second second is $d_2 = y_2 - y_1 = A - \frac{A}{\sqrt{2}} = A(1 - \frac{1}{\sqrt{2}})$.
The ratio of the distances is $\frac{d_1}{d_2} = \frac{A/\sqrt{2}}{A(1 - 1/\sqrt{2})} = \frac{1/\sqrt{2}}{(\sqrt{2}-1)/\sqrt{2}} = \frac{1}{\sqrt{2}-1}$.
Rationalizing the denominator: $\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} = \frac{\sqrt{2}+1}{2-1} = \sqrt{2}+1$.
Thus,the ratio is $(\sqrt{2}+1): 1$.

(D) Let the displacement of the two particles be $x_1 = A \sin(\omega t + \phi_1)$ and $x_2 = A \sin(\omega t + \phi_2)$.
At the point where they pass each other,$x_1 = x_2 = \frac{A}{\sqrt{2}}$.
For particle $1$,moving away from the mean position,$\sin(\omega t + \phi_1) = \frac{1}{\sqrt{2}}$,which gives the phase $\theta_1 = \frac{\pi}{4}$.
For particle $2$,moving towards the mean position at the same displacement,$\sin(\omega t + \phi_2) = \frac{1}{\sqrt{2}}$. Since it is moving in the opposite direction,its phase must be in the second quadrant,so $\theta_2 = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
The phase difference is $\Delta \phi = |\theta_2 - \theta_1| = |\frac{3\pi}{4} - \frac{\pi}{4}| = \frac{2\pi}{4} = \frac{\pi}{2}$.
Converting to degrees,$\frac{\pi}{2} = 90^{\circ}$.

(D) The given equation of motion is $4 \frac{d^2 y}{dt^2} + \pi^2 y = 0$.
Dividing by $4$,we get $\frac{d^2 y}{dt^2} + \frac{\pi^2}{4} y = 0$.
The general equation for simple harmonic motion is $\frac{d^2 y}{dt^2} + \omega^2 y = 0$.
Comparing the two equations,we find $\omega^2 = \frac{\pi^2}{4}$,which gives $\omega = \frac{\pi}{2} \ rad/s$.
The time period $T$ is given by the formula $T = \frac{2\pi}{\omega}$.
Substituting the value of $\omega$,we get $T = \frac{2\pi}{\pi/2} = 2\pi \times \frac{2}{\pi} = 4 \ s$.

(C) For the particle to move from the extreme position $(x=A)$ to half of the amplitude $(x=A/2)$:
$x = A \cos(\omega t_1) \implies A/2 = A \cos(\omega t_1) \implies \cos(\omega t_1) = 1/2$.
Thus,$\omega t_1 = \pi/3$,which gives $t_1 = \pi / (3\omega)$.
For the particle to move from the mean position $(x=0)$ to half of the amplitude $(x=A/2)$:
$x = A \sin(\omega t_2) \implies A/2 = A \sin(\omega t_2) \implies \sin(\omega t_2) = 1/2$.
Thus,$\omega t_2 = \pi/6$,which gives $t_2 = \pi / (6\omega)$.
Comparing the two times: $t_1 / t_2 = (\pi / 3\omega) / (\pi / 6\omega) = 6/3 = 2$.
Therefore,$t_1 = 2 t_2$.

(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.
At height $h$ above the surface, the acceleration due to gravity is $g' = \frac{g}{(1 + \frac{h}{R})^2}$.
The new time period $T'$ is given by $T' = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{g / (1 + \frac{h}{R})^2}} = T(1 + \frac{h}{R})$.
Given $T' = 2T$, we have $2T = T(1 + \frac{h}{R})$.
$2 = 1 + \frac{h}{R} \Rightarrow \frac{h}{R} = 1$.
Therefore, $h = R = 6400 \text{ km}$.

(B) Given: Mass $m = 2 \,kg$, inclination $\theta = \sin^{-1}(3/5)$, so $\sin \theta = 3/5$ and $\cos \theta = 4/5$. Distance $s = 80 \,cm = 0.8 \,m$. Specific heat $c = 420 \,J kg^{-1} K^{-1}$. Acceleration due to gravity $g = 10 \,ms^{-2}$.
Since the block slides at a constant speed, the work done by friction equals the loss in potential energy. The work done by friction $W_f = f_k \cdot s$, where $f_k = \mu_k N = \mu_k mg \cos \theta$. Since the speed is constant, $mg \sin \theta = f_k$.
Thus, the heat generated $Q = W_f = (mg \sin \theta) \cdot s$.
Equating heat generated to heat absorbed: $Q = mc \Delta T$.
$mc \Delta T = (mg \sin \theta) s$.
$\Delta T = \frac{g \sin \theta \cdot s}{c} = \frac{10 \times (3/5) \times 0.8}{420}$.
$\Delta T = \frac{10 \times 0.6 \times 0.8}{420} = \frac{4.8}{420} \approx 0.0114^{\circ} C$.

(C) Specific heat capacity is an intrinsic property of a material,which means it depends only on the nature of the substance and not on the amount or mass of the substance.
Since the material remains copper,its specific heat capacity remains constant regardless of the change in mass.
Therefore,if the mass is doubled,the specific heat capacity will remain $s$.

(B) In a standard phase diagram (Pressure vs Temperature):
$1$. The curve separating the Solid and Liquid phases is the Fusion (or Melting) curve. In the figure,$AO$ separates the Solid and Liquid regions,so $AO$ is the Fusion curve.
$2$. The curve separating the Solid and Vapour phases is the Sublimation curve. In the figure,$BO$ separates the Solid and Vapour regions,so $BO$ is the Sublimation curve.
$3$. The curve separating the Liquid and Vapour phases is the Vaporization (or Boiling) curve. In the figure,$CO$ separates the Liquid and Vapour regions,so $CO$ is the Vaporization curve.
Therefore,$AO =$ Fusion curve,$BO =$ Sublimation curve,and $CO =$ Vaporization curve.

(C) The energy lost by the ball during the bounce is equal to the difference in potential energy at the two heights.
Energy lost $\Delta E = mg(h_1 - h_2)$.
Given $m = 200 \,g = 0.2 \,kg$, $h_1 = 20 \,m$, $h_2 = 10.8 \,m$, $g = 10 \,ms^{-2}$, and $c = 460 \,Jkg^{-1} K^{-1}$.
Assuming the lost energy is absorbed by the ball, $\Delta E = mc \Delta T$.
Equating the two: $mg(h_1 - h_2) = mc \Delta T$.
Canceling $m$ from both sides: $g(h_1 - h_2) = c \Delta T$.
Substituting the values: $10 \times (20 - 10.8) = 460 \times \Delta T$.
$10 \times 9.2 = 460 \times \Delta T$.
$92 = 460 \times \Delta T$.
$\Delta T = \frac{92}{460} = 0.2^{\circ} C$.

(D) Given: Mass of water,$m = 2 \,kg$
Change in temperature,$\Delta T = 10^{\circ} C$
Specific heat capacity of water,$c = 4200 \,J \,kg^{-1} \,K^{-1}$
The heat energy $Q$ released is calculated using the formula:
$Q = m c \Delta T$
Substituting the values:
$Q = 2 \,kg \times 4200 \,J \,kg^{-1} \,K^{-1} \times 10 \,K$
$Q = 84000 \,J$

(B) Step $1$: Calculate heat required to bring ice from $-10^{\circ} C$ to $0^{\circ} C$ (ice): $Q_1 = m_i c_i \Delta T = 50 \times 0.5 \times 10 = 250 \,cal$.
Step $2$: Calculate heat required to melt $50 \,g$ of ice at $0^{\circ} C$: $Q_2 = m_i L_f = 50 \times 80 = 4000 \,cal$.
Total heat required by ice to become water at $0^{\circ} C$ is $Q_{total} = 250 + 4000 = 4250 \,cal$.
Step $3$: Calculate heat released by $200 \,g$ of water cooling from $30^{\circ} C$ to $0^{\circ} C$: $Q_{release} = m_w c_w \Delta T = 200 \times 1 \times 30 = 6000 \,cal$.
Since $Q_{release} > Q_{total}$,the ice melts completely and the final temperature $T_f$ will be above $0^{\circ} C$.
Step $4$: Apply the principle of calorimetry: Heat lost by water = Heat gained by ice.
$200 \times 1 \times (30 - T_f) = 4250 + 50 \times 1 \times (T_f - 0)$.
$6000 - 200 T_f = 4250 + 50 T_f$.
$1750 = 250 T_f$.
$T_f = 7^{\circ} C$.

(C) According to Stefan-Boltzmann law, the power radiated per unit area $E$ is proportional to the fourth power of the absolute temperature $T$ of the black body:
$E = \sigma T^4$
Given that the final emissivity $E_2$ is $16$ times the initial emissivity $E_1$:
$E_2 = 16 E_1$
Substituting the Stefan-Boltzmann law into the ratio:
$\frac{E_2}{E_1} = \frac{\sigma T_2^4}{\sigma T_1^4} = \left( \frac{T_2}{T_1} \right)^4$
$16 = \left( \frac{T_2}{T} \right)^4$
Taking the fourth root on both sides:
$\sqrt[4]{16} = \frac{T_2}{T}$
$2 = \frac{T_2}{T}$
$T_2 = 2 \,T$

(D) According to Wien's displacement law, the wavelength corresponding to the maximum intensity of emission $(\lambda_m)$ is inversely proportional to the absolute temperature $(T)$ of the body, given by $\lambda_m T = b$, where $b$ is Wien's constant.
As the temperature of the iron rod increases, the wavelength of the emitted radiation decreases.
Initially, at lower temperatures, the emitted radiation is in the red part of the visible spectrum.
As the temperature increases further, the peak wavelength shifts towards shorter wavelengths (yellow, then blue, and finally white), which is a combination of all visible wavelengths.
Thus, the change in colour is explained by Wien's displacement law.

(D) Let $L_0$ be the length of the rod at $0^{\circ} C$ and $\alpha$ be the coefficient of linear expansion.
The length at temperature $T$ is given by $L_T = L_0(1 + \alpha T)$.
Length at $10^{\circ} C$: $L_{10} = L_0(1 + 10\alpha)$.
Length at $40^{\circ} C$: $L_{40} = L_0(1 + 40\alpha)$.
Given that $L_{40} - L_{10} = 0.05 \ cm$.
Substituting the expressions: $L_0(1 + 40\alpha) - L_0(1 + 10\alpha) = 0.05$.
$L_0(40\alpha - 10\alpha) = 0.05$.
$30 L_0 \alpha = 0.05$.
$L_0 = \frac{0.05}{30 \alpha}$.
Substituting $\alpha = 1.5 \times 10^{-5} \ {}^{\circ} C^{-1}$:
$L_0 = \frac{0.05}{30 \times 1.5 \times 10^{-5}} = \frac{0.05}{45 \times 10^{-5}} = \frac{0.05 \times 10^5}{45} = \frac{5000}{45} \approx 111.1 \ cm$.

(A) Let $L_0 = 300 \,cm$ be the length of the steel tape at $27^{\circ} C$.
At $50^{\circ} C$, the length of the tape becomes $L_T = L_0(1 + \alpha \Delta T)$.
Here, $\Delta T = 50^{\circ} C - 27^{\circ} C = 23^{\circ} C$.
$L_T = 300(1 + 1.2 \times 10^{-5} \times 23) = 300(1 + 0.000276) = 300.0828 \,cm$.
The measured length of the rod is $L_m = 110 \,cm$.
The actual length $L_a$ is given by $L_a = L_m \times \frac{L_T}{L_0}$.
$L_a = 110 \times \frac{300.0828}{300} = 110 \times (1 + 0.000276) = 110 + 0.03036 = 110.03036 \,cm$.
Rounding to two decimal places, the actual length is $110.03 \,cm$.

(A) Given: Heat supplied at constant pressure, $Q_P = 210 \,J$.
For a diatomic gas, the molar heat capacity at constant volume is $C_V = \frac{5}{2}R$.
The molar heat capacity at constant pressure is $C_P = C_V + R = \frac{5}{2}R + R = \frac{7}{2}R$.
We know that $Q_P = n C_P \Delta T = 210 \,J$.
Substituting $C_P$, we get $n (\frac{7}{2}R) \Delta T = 210$.
Thus, $n R \Delta T = 210 \times \frac{2}{7} = 60 \,J$.
The work done by the gas at constant pressure is given by $W = P \Delta V = n R \Delta T$.
Therefore, $W = 60 \,J$.

(A) For a monatomic gas,the degrees of freedom $f = 3$.
$C_p = \frac{5}{2}R$ and $C_v = \frac{3}{2}R$.
Total heat supplied at constant pressure is $Q = n C_p \Delta T = \frac{5}{2} nR \Delta T$.
Heat used to increase internal energy is $\Delta U = n C_v \Delta T = \frac{3}{2} nR \Delta T$.
Percentage of heat used to increase internal energy $= \frac{\Delta U}{Q} \times 100 = \frac{\frac{3}{2} nR \Delta T}{\frac{5}{2} nR \Delta T} \times 100 = 60\%$.
Heat used to do external work is $W = P \Delta V = nR \Delta T$.
Percentage of heat used to do external work $= \frac{W}{Q} \times 100 = \frac{nR \Delta T}{\frac{5}{2} nR \Delta T} \times 100 = 40\%$.
Thus,the percentages are $40\%$ and $60\%$ respectively.

(D) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$,where $\Delta U$ is the change in internal energy and $\Delta W$ is the work done by the gas.
For an isothermal process,the temperature remains constant,so the change in internal energy $\Delta U = 0$.
For an adiabatic process,there is no heat exchange,so $\Delta Q = 0$,which implies $\Delta U = -\Delta W$.
Let the total change in internal energy be $\Delta U_{total} = \Delta U_a + \Delta U_b + \Delta U_c = -20 \,J$.
For the isothermal steps $(a)$ and $(c)$,$\Delta U_a = 0$ and $\Delta U_c = 0$.
Therefore,the total change in internal energy is due to the adiabatic step $(b)$: $\Delta U_b = -20 \,J$.
Since $\Delta U_b = -W$ for the adiabatic process,we have $-W = -20 \,J$.
Thus,$W = 20 \,J$.

(B) Number of moles,$n = 3$.
Initial volume,$V_i = V$.
Final volume,$V_f = 3V$.
Ratio of specific heats,$\gamma = \frac{C_p}{C_v}$.
Change in internal energy is given by $\Delta U = n C_v \Delta T$.
We know that $C_v = \frac{R}{\gamma - 1}$.
At constant pressure,according to Charles's Law,$\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Substituting the values,$\frac{V}{T} = \frac{3V}{T_2}$,which gives $T_2 = 3T$.
Change in temperature,$\Delta T = T_2 - T_1 = 3T - T = 2T$.
Now,substituting these into the internal energy formula:
$\Delta U = n \left( \frac{R}{\gamma - 1} \right) \Delta T = 3 \left( \frac{R}{\gamma - 1} \right) (2T) = \frac{6RT}{\gamma - 1}$.

(B) Initial internal energy,$U_i = 80 cal = 80 \times 4.2 J = 336 J$.
Work done on the gas,$W = -18 cal = -18 \times 4.2 J = -75.6 J$ (Work done on the system is negative in the convention $\Delta Q = \Delta U + \Delta W$).
Heat released by the gas,$Q = -42 J$ (Heat released is negative).
According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
$-42 = (U_f - 336) + (-75.6)$.
$-42 = U_f - 336 - 75.6$.
$-42 = U_f - 411.6$.
$U_f = 411.6 - 42 = 369.6 J$.

(C) Temperature of the sink, $T_2 = 300 \,K$. Efficiency of the engine, $\eta_1 = 0.25$.
The formula for efficiency is $\eta = 1 - \frac{T_2}{T_1}$.
Substituting the values: $0.25 = 1 - \frac{300}{T_1} \Rightarrow \frac{300}{T_1} = 0.75 \Rightarrow T_1 = \frac{300}{0.75} = 400 \,K$.
When the source temperature is increased by $100 \,K$, the new source temperature $T_1' = 400 + 100 = 500 \,K$.
The new efficiency $\eta_2 = 1 - \frac{T_2}{T_1'} = 1 - \frac{300}{500} = 1 - 0.6 = 0.40$.
The increase in efficiency is $\Delta\eta = \eta_2 - \eta_1 = 0.40 - 0.25 = 0.15$.

(C) In a Carnot engine,the sink is defined as a thermal energy reservoir. $A$ thermal reservoir is a system with an infinite heat capacity,meaning that it can absorb or reject any amount of heat without undergoing a change in its temperature. Therefore,as the gas gives heat energy to the sink,the temperature of the sink remains constant.

(C) Carnot engine operates on an idealized reversible cycle. The source is defined as a thermal reservoir with an infinite heat capacity. Therefore,when the working substance (gas) absorbs heat energy from the source during the isothermal expansion process,the temperature of the source remains constant.

(A) The initial efficiency of a Carnot engine is given by $\eta_1 = 1 - \frac{T_2}{T_1} = \frac{T_1 - T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature $(T_1 > T_2)$.
When both temperatures are decreased by $100 \ K$,the new temperatures are $T_1' = T_1 - 100$ and $T_2' = T_2 - 100$.
The new efficiency is $\eta_2 = 1 - \frac{T_2 - 100}{T_1 - 100} = \frac{(T_1 - 100) - (T_2 - 100)}{T_1 - 100} = \frac{T_1 - T_2}{T_1 - 100}$.
Since $T_1 - 100 < T_1$,the denominator of the new efficiency is smaller than the original,while the numerator remains the same.
Therefore,$\eta_2 > \eta_1$,which means the efficiency of the engine increases.

(D) The heat energy required at constant volume is given by the formula $Q = n C_v \Delta T$.
For a monatomic gas,the molar heat capacity at constant volume is $C_v = \frac{3}{2} R$.
Given $n_1 = 2 \text{ moles}$,$\Delta T_1 = 40^{\circ} C - 30^{\circ} C = 10 \text{ K}$.
So,$Q = 2 \times \frac{3}{2} R \times 10 = 30 R$.
For a diatomic gas,the molar heat capacity at constant volume is $C_v = \frac{5}{2} R$.
Given $n_2 = 4 \text{ moles}$,$\Delta T_2 = 33^{\circ} C - 28^{\circ} C = 5 \text{ K}$.
Let the required heat be $Q'$.
$Q' = n_2 C_v \Delta T_2 = 4 \times \frac{5}{2} R \times 5 = 50 R$.
Now,taking the ratio: $\frac{Q'}{Q} = \frac{50 R}{30 R} = \frac{5}{3}$.
Therefore,$Q' = \frac{5}{3} Q$.

(C) Initial state: $P_1 = P$,$V_1 = V$.
Step $1$: Isothermal expansion to $V_2 = 27 V$.
For an isothermal process,$P_1 V_1 = P_2 V_2$.
$P_2 = \frac{P_1 V_1}{V_2} = \frac{P \times V}{27 V} = \frac{P}{27}$.
Step $2$: Adiabatic compression to $V_3 = V$.
For a monatomic gas,the adiabatic index $\gamma = \frac{5}{3}$.
For an adiabatic process,$P_2 V_2^\gamma = P_3 V_3^\gamma$.
$P_3 = P_2 \left( \frac{V_2}{V_3} \right)^\gamma = \frac{P}{27} \left( \frac{27 V}{V} \right)^{5/3}$.
$P_3 = \frac{P}{27} \times (27)^{5/3} = \frac{P}{27} \times (3^3)^{5/3} = \frac{P}{27} \times 3^5$.
$P_3 = \frac{P}{27} \times 243 = 9 P$.
The final pressure of the gas is $9 P$.

(D) For a monatomic gas, the degrees of freedom $f=3$.
The adiabatic index $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3}$.
For an adiabatic process, the relationship between temperature $T$ and volume $V$ is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Given $T_1 = 630 \,K$, $V_1 = V$, and $V_2 = 27V$.
Substituting these values:
$630 \times V^{\frac{5}{3}-1} = T_2 \times (27V)^{\frac{5}{3}-1}$
$630 = T_2 \times (27)^{\frac{2}{3}}$
$630 = T_2 \times (3^3)^{\frac{2}{3}}$
$630 = T_2 \times 3^2$
$630 = T_2 \times 9$
$T_2 = \frac{630}{9} = 70 \,K$.

(C) The work done by a gas during expansion is given by the area under the $P-V$ curve,$W = \int_{V_1}^{V_2} P \, dV$.
For a given change in volume from $V_1$ to $V_2$,the pressure $P$ must be as high as possible throughout the process to maximize the integral.
In an isobaric process,the pressure remains constant at its initial maximum value.
In other processes like isothermal or adiabatic,the pressure decreases as the volume increases.
Therefore,the area under the $P-V$ curve is largest for an isobaric expansion,making the work done maximum.

(D) From the given $P-V$ diagram,the cyclic process forms an ellipse.
The length of the axis along the $V$-axis is $2b = V_2 - V_1$,so $b = \frac{V_2 - V_1}{2}$.
The length of the axis along the $P$-axis is $2a = P_2 - P_1$,so $a = \frac{P_2 - P_1}{2}$.
The work done in a cyclic process is equal to the area enclosed by the loop in the $P-V$ diagram.
Area of an ellipse = $\pi ab$.
Therefore,Work done = $\pi \times \left(\frac{P_2 - P_1}{2}\right) \times \left(\frac{V_2 - V_1}{2}\right) = \frac{\pi}{4}(P_2 - P_1)(V_2 - V_1)$.

(B) Work done on the system,$W = 166.28 \ J$.
Increase in temperature,$\Delta T = 8^{\circ} C = 8 \ K$.
For an adiabatic process,the work done on the gas is given by $W = \frac{nR\Delta T}{\gamma - 1}$.
Given $n = 1 \ mol$ and $R = 8.314 \ J \ mol^{-1} \ K^{-1}$.
Substituting the values: $166.28 = \frac{1 \times 8.314 \times 8}{\gamma - 1}$.
$\gamma - 1 = \frac{66.512}{166.28} = 0.4$.
$\gamma = 1.4$.
Since the adiabatic index $\gamma = 1.4$ corresponds to a diatomic gas,the gas is diatomic.

(B) For an adiabatic process,the relation between pressure $(P)$ and volume $(V)$ is given by $PV^{\gamma} = \text{constant}$.
From the ideal gas equation,we know that $PV = nRT$,which implies $V = \frac{nRT}{P}$.
Substituting the expression for $V$ into the adiabatic equation:
$P \left( \frac{nRT}{P} \right)^{\gamma} = \text{constant}$.
Since $n$ and $R$ are constants,we can write:
$P \cdot \frac{T^{\gamma}}{P^{\gamma}} = \text{constant}$.
$P^{1-\gamma} T^{\gamma} = \text{constant}$.

(D) According to the first law of thermodynamics,the heat supplied to a system $(\Delta Q)$ is equal to the sum of the change in internal energy $(\Delta U)$ and the work done by the system $(\Delta W)$:
$\Delta Q = \Delta U + \Delta W$
In an adiabatic process,there is no exchange of heat between the system and the surroundings,so $\Delta Q = 0$.
Substituting this into the first law equation:
$0 = \Delta U + \Delta W$
Therefore,$\Delta U = -\Delta W$.

(D) An adiabatic process is a thermodynamic process in which there is no exchange of heat between the system and its surroundings $(dQ = 0)$.
For an ideal gas undergoing a reversible adiabatic process,the relationship between pressure $(P)$ and volume $(V)$ is given by $PV^{\gamma} = \text{constant}$,where $\gamma$ is the adiabatic index (ratio of specific heats $C_p/C_v$).
Using the ideal gas law $PV = nRT$,we can express pressure as $P = \frac{nRT}{V}$.
Substituting this into the adiabatic equation: $(\frac{nRT}{V})V^{\gamma} = \text{constant}$.
Since $nR$ is a constant,we get $T V^{\gamma-1} = \text{constant}$.

(D) For an ideal gas,the equation of state is $PV = nRT$.
Rearranging this,we get $P = (nR/V)T$.
Comparing this with the equation of a straight line $y = mx$,where $y = P$ and $x = T$,the slope $m$ of the graph is given by $m = nR/V$.
Since $n$ and $R$ are constants,the slope is inversely proportional to the volume,i.e.,$m \propto 1/V$.
From the given graph,the slope of the line for $V_1$ is the largest,and the slope of the line for $V_4$ is the smallest.
Therefore,$m_1 > m_2 > m_3 > m_4$.
Since $V \propto 1/m$,it follows that $V_1 < V_2 < V_3 < V_4$.

(D) Mass of gas,$m = 20 \,g$.
Change in temperature,$\Delta T = 35^{\circ} C - 25^{\circ} C = 10^{\circ} C$.
Specific heat capacity of the gas at constant volume,$C_{v} = 0.2 \,cal \,g^{-1} {}^{\circ} C^{-1}$.
Since the process occurs at constant volume,the work done $W = 0$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + W$.
Since $W = 0$,the change in internal energy is $\Delta U = \Delta Q = m C_{v} \Delta T$.
Using the conversion factor $1 \,cal = 4.2 \,J$:
$\Delta U = 20 \,g \times 0.2 \,cal \,g^{-1} {}^{\circ} C^{-1} \times 10^{\circ} C \times 4.2 \,J/cal$.
$\Delta U = 40 \times 4.2 \,J = 168 \,J$.

(D) The dimensional formula for speed of light $c$ is $[L T^{-1}]$.
Given $c = 3 \times 10^8 \,m/s$ ... $(1)$
The dimensional formula for acceleration due to gravity $g$ is $[L T^{-2}]$.
Given $g = 10 \,m/s^2$ ... $(2)$
To find the unit of time $[T]$, we divide the dimensions of speed by the dimensions of acceleration:
$\frac{[L T^{-1}]}{[L T^{-2}]} = [T]$
Substituting the given values:
$[T] = \frac{3 \times 10^8}{10} = 3 \times 10^7 \,s$
Therefore, the unit of time in this system is $3 \times 10^7 \,s$.

(B) For a charged hollow sphere,the electric field $(E)$ inside the sphere is zero because there is no enclosed charge $(q_{enclosed} = 0)$.
According to the relation between electric field and potential,$E = -\frac{dV}{dr}$.
Since $E = 0$,it implies that $\frac{dV}{dr} = 0$,which means the potential $(V)$ must be constant throughout the interior of the sphere.
The value of this constant potential is equal to the potential at the surface of the sphere,given by $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$.

(C) The electric field $\vec{E}$ is given by the negative gradient of the potential $V$: $\vec{E} = -\vec{\nabla} V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} \right)$.
Given $V = \frac{1}{2} y^2 - 2x$.
Calculating partial derivatives:
$\frac{\partial V}{\partial x} = -2$
$\frac{\partial V}{\partial y} = y$
Substituting these into the electric field formula:
$\vec{E} = -(-2 \hat{i} + y \hat{j}) = 2 \hat{i} - y \hat{j}$.
At the point $(x = 1 \text{ m}, y = 1 \text{ m})$:
$\vec{E} = 2 \hat{i} - (1) \hat{j} = 2 \hat{i} - \hat{j} \text{ Vm}^{-1}$.

(C) The potential energy $U$ of a system of point charges is given by $U = k \sum \frac{q_i q_j}{r_{ij}}$.
Initial potential energy $U_i$ with side $r_1 = 1 \text{ m}$:
$U_i = k \left[ \frac{1 \times 2}{1} + \frac{2 \times 3}{1} + \frac{3 \times 1}{1} \right] = k [2 + 6 + 3] = 11k$.
Final potential energy $U_f$ with side $r_2 = 0.5 \text{ m}$:
$U_f = k \left[ \frac{1 \times 2}{0.5} + \frac{2 \times 3}{0.5} + \frac{3 \times 1}{0.5} \right] = k [4 + 12 + 6] = 22k$.
The work done $W$ is the change in potential energy:
$W = U_f - U_i = 22k - 11k = 11k$.
Given $k = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$,
$W = 11 \times 9 \times 10^9 = 99 \times 10^9 \text{ J}$.

(B) The potential at point $A$ due to $+q$ is $V_{A(+q)} = \frac{1}{4\pi\epsilon_0} \frac{q}{x}$ and due to $-q$ is $V_{A(-q)} = \frac{1}{4\pi\epsilon_0} \frac{-q}{x+y}$.
Thus,$V_A = \frac{q}{4\pi\epsilon_0} \left( \frac{1}{x} - \frac{1}{x+y} \right) = \frac{q}{4\pi\epsilon_0} \left( \frac{y}{x(x+y)} \right)$.
Similarly,the potential at point $B$ is $V_B = \frac{1}{4\pi\epsilon_0} \frac{q}{x+y} + \frac{1}{4\pi\epsilon_0} \frac{-q}{x} = \frac{q}{4\pi\epsilon_0} \left( \frac{1}{x+y} - \frac{1}{x} \right) = -V_A$.
Therefore,$V_A - V_B = V_A - (-V_A) = 2V_A = 2 \times \frac{q}{4\pi\epsilon_0} \left( \frac{y}{x(x+y)} \right)$.
Given $q = 10^{-6} \text{ C}$,$x = 0.02 \text{ m}$,$y = 0.03 \text{ m}$,and $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ Nm}^2/\text{C}^2$.
$V_A - V_B = 2 \times 9 \times 10^9 \times 10^{-6} \times \frac{0.03}{0.02(0.02+0.03)} = 18 \times 10^3 \times \frac{0.03}{0.02 \times 0.05} = 18 \times 10^3 \times \frac{0.03}{0.001} = 18 \times 10^3 \times 30 = 5.4 \times 10^5 \text{ V}$.

(A) In nature,there are four fundamental forces: gravitational force,electromagnetic force,weak nuclear force,and strong nuclear force. Among these,the strong nuclear force is the most powerful force,acting between nucleons (protons and neutrons) to hold the nucleus together. Therefore,the strongest force in nature is the nuclear force.

(D) According to the Right-Hand Thumb Rule for a circular current-carrying loop,if the fingers of the right hand are curled in the direction of the current,the thumb points in the direction of the magnetic field lines inside the loop.
If the current flows in an anticlockwise direction when viewed from the top,the magnetic field lines point upwards (towards the observer).
Since magnetic field lines emerge from the North Pole,the face of the loop where the current appears anticlockwise acts as a North Pole.
Conversely,if the current flows in a clockwise direction,the face acts as a South Pole.
Therefore,the upper side of the loop acts as a North Pole if the current is anticlockwise.

(A) When two parallel conductors carry steady currents, each conductor produces a magnetic field in the space around it, which is described by the Biot-Savart law.
These currents then experience a magnetic force due to the magnetic field produced by the other, which is described by the Lorentz force law $(F = I \vec{L} \times \vec{B})$.
According to Newton's third law, the force exerted by the first conductor on the second is equal in magnitude and opposite in direction to the force exerted by the second conductor on the first.
Thus, the combination of the Biot-Savart law and the Lorentz force law explains the interaction between parallel conductors in accordance with Newton's third law.

(B) Ampere's circuital law states that the line integral of the magnetic field $B$ around any closed loop is equal to $\mu_0$ times the total current $I$ passing through the surface enclosed by the loop.
Mathematically,this is expressed as $\oint B \cdot dl = \mu_0 I$.

(C) The magnetic force on a moving charge is given by the Lorentz force formula: $F = q(v \times B)$.
By the definition of the cross product of two vectors,the resulting vector $F$ is always perpendicular to the plane containing the two vectors $v$ and $B$.
Therefore,the magnetic force $F$ is perpendicular to both the velocity vector $v$ and the magnetic field vector $B$.

(C) The force per unit length on a current-carrying conductor in a magnetic field is given by the formula: $\frac{F}{l} = iB \sin \theta$.
Here,the current $i = 1 \,A$ flows from south to north.
The magnetic field $B = 3 \times 10^{-5} \,T$ also points from south to north.
Since the current and the magnetic field are in the same direction,the angle $\theta$ between them is $0^\circ$.
Therefore,the force per unit length is: $\frac{F}{l} = 1 \times (3 \times 10^{-5}) \times \sin(0^\circ) = 1 \times (3 \times 10^{-5}) \times 0 = 0 \,N/m$.

(B) Current sensitivity is defined as the deflection produced per unit current flowing through the galvanometer.
It is given by the formula: $I_{S} = \frac{\theta}{I}$
Where $\theta$ is the deflection in divisions and $I$ is the current in amperes.
Given: $\theta = 25 \ div$ and $I = 0.1 \ A$.
Therefore,$I_{S} = \frac{25}{0.1} = 250 \ div/A$.

(D) The torque $\tau$ on a current-carrying coil in a magnetic field is given by $\tau = N i A B \sin \theta$, where $\theta$ is the angle between the normal to the coil and the magnetic field.
Given:
$N = 30$
$r = 8 \,cm = 0.08 \,m$
$i = 6 \,A$
$B = 1.0 \,T$
$\theta = 20^{\circ}$
Area $A = \pi r^2 = 3.14 \times (0.08)^2 = 3.14 \times 0.0064 = 0.020096 \,m^2$
Calculating the torque:
$\tau = 30 \times 6 \times 0.020096 \times 1.0 \times \sin(20^{\circ})$
$\tau = 180 \times 0.020096 \times 0.342$
$\tau \approx 1.236 \,Nm$
Note: Based on the provided options and standard textbook problem variations where $\theta$ is often $30^{\circ}$, if $\theta = 30^{\circ}$, then $\tau = 30 \times 6 \times 3.14 \times (0.08)^2 \times 1.0 \times 0.5 = 1.808 \,Nm$. Given the options, the intended angle was likely $30^{\circ}$.

(B) When a flexible wire loop carrying current is placed in an external magnetic field,it experiences magnetic forces that tend to expand the loop.
To maximize the magnetic flux linked with the loop,the loop tends to enclose the maximum possible area for a given perimeter.
According to geometric principles,for a fixed perimeter,a circle encloses the maximum area.
Therefore,the loop changes its shape to a circular shape with its plane normal to the magnetic field.

(C) Given: Magnetic moment,$m = 0.4 \,Am^2$
Distance,$r = 50 \,cm = 0.5 \,m$
The formula for the magnitude of the axial magnetic field of a short bar magnet is:
$B_{\text{axial}} = \frac{\mu_0}{4\pi} \left( \frac{2m}{r^3} \right)$
Substituting the values:
$B_{\text{axial}} = (10^{-7}) \times \frac{2 \times 0.4}{(0.5)^3}$
$B_{\text{axial}} = 10^{-7} \times \frac{0.8}{0.125}$
$B_{\text{axial}} = 10^{-7} \times 6.4 = 6.4 \times 10^{-7} \,T$

(A) For a short bar magnet of magnetic moment $m$ at a distance $r$ from its center:
$1$. The magnetic field on the axial line is given by: $B_{A} = \frac{\mu_{0}}{4 \pi} \frac{2m}{r^{3}}$
$2$. The magnetic field on the equatorial line is given by: $B_{E} = \frac{\mu_{0}}{4 \pi} \frac{m}{r^{3}}$
$3$. Comparing the two expressions,we can see that $B_{A} = 2 \times \left( \frac{\mu_{0}}{4 \pi} \frac{m}{r^{3}} \right) = 2 B_{E}$.
Therefore,the correct relation is $B_{A} = 2 B_{E}$.

(D) The magnetizing field (or magnetic intensity) $H$ of a solenoid is given by the formula $H = nI$,where $n$ is the number of turns per unit length and $I$ is the current flowing through the windings.
Given:
$n = 500 \,m^{-1}$
$I = 4 \,A$
Substituting these values into the formula:
$H = 500 \times 4 = 2000 \,Am^{-1}$
$H = 2 \times 10^3 \,Am^{-1}$
Comparing this result with the given options,none of the options match $2000 \,Am^{-1}$. Therefore,the correct option is $D$.

(D) The magnetic field at the centre of a circular coil is given by $B = \frac{\mu_0 n I}{2r}$,where $n$ is the number of turns,$I$ is the current,and $r$ is the radius of the coil.
Let $L$ be the total length of the wire. For a coil with $n$ turns and radius $r$,$L = n(2\pi r)$,which implies $r = \frac{L}{2\pi n}$.
Substituting $r$ into the magnetic field formula: $B = \frac{\mu_0 n I}{2(L / 2\pi n)} = \frac{\mu_0 n^2 I \pi}{L}$.
Since the length of the wire $L$ and the current $I$ are constant,$B \propto n^2$.
Therefore,the ratio of the magnetic fields is $\frac{B_1}{B_2} = \left( \frac{n_1}{n_2} \right)^2$.
Given $n_1 = 5$ and $n_2 = 10$,the ratio is $\frac{B_1}{B_2} = \left( \frac{5}{10} \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.

(D) The formula for the magnetic field intensity $B$ at the centre of a circular loop is given by $B = \frac{\mu_0 i}{2r}$.
Given values are: current $i = 0.2 \,A$ and radius $r = 0.1 \,m$.
The permeability of free space is $\mu_0 = 4 \pi \times 10^{-7} \,T \cdot m/A$.
Substituting these values into the formula:
$B = \frac{(4 \pi \times 10^{-7} \,T \cdot m/A) \times (0.2 \,A)}{2 \times (0.1 \,m)}$
$B = \frac{4 \pi \times 10^{-7} \times 0.2}{0.2}$
$B = 4 \pi \times 10^{-7} \,T$.

(C) Given:
Number of turns,$n = 100$
Radius,$r = 10 \,cm = 0.1 \,m$
Current,$I = 2 \,A$
Permeability of free space,$\mu_0 = 4\pi \times 10^{-7} \,T \cdot m/A$
The magnetic field $B$ at the centre of a circular coil is given by the formula:
$B = \frac{\mu_0 n I}{2r}$
Substituting the values:
$B = \frac{(4\pi \times 10^{-7}) \times 100 \times 2}{2 \times 0.1}$
$B = \frac{4\pi \times 10^{-5} \times 2}{0.2}$
$B = \frac{8\pi \times 10^{-5}}{0.2} = 40\pi \times 10^{-5} = 4\pi \times 10^{-4} \,T$
Using $\pi \approx 3.14$:
$B = 4 \times 3.14 \times 10^{-4} \,T = 12.56 \times 10^{-4} \,T$

(D) According to the Biot-Savart Law, the magnetic field $d\vec{B}$ due to a current element $i d\vec{l}$ at a point with position vector $\vec{r}$ is given by $d\vec{B} = \frac{\mu_0}{4\pi} \frac{i d\vec{l} \times \vec{r}}{r^3}$.
For the straight segments of the wire, the current element $d\vec{l}$ and the position vector $\vec{r}$ (pointing from the element to the center $O$) are collinear (either parallel or anti-parallel).
Therefore, the cross product $d\vec{l} \times \vec{r} = 0$.
Consequently, the magnetic field at the center $O$ due to the straight segments is zero.

(B) The magnetic moment $\mu$ of an electron revolving in an orbit is given by $\mu = I A$,where $I$ is the current and $A$ is the area of the orbit.
For an electron revolving in an orbit of radius $r$ with velocity $v$,the current $I = \frac{e}{T} = \frac{ev}{2 \pi r}$.
The area $A = \pi r^2$.
Thus,$\mu = \left( \frac{ev}{2 \pi r} \right) (\pi r^2) = \frac{evr}{2}$.
According to Bohr's quantization condition,the angular momentum $L = mvr = \frac{nh}{2 \pi}$.
Therefore,$vr = \frac{nh}{2 \pi m}$.
Substituting this into the expression for $\mu$:
$\mu = \frac{e}{2} \left( \frac{nh}{2 \pi m} \right) = \frac{neh}{4 \pi m}$.
Since $e$,$h$,and $m$ are constants,we have $\mu \propto n$.

(C) The angular velocity is given by $\omega = 4 \pi \times 10^6 \text{ rad s}^{-1}$.
The velocity component parallel to the magnetic field is $v_{\parallel} = 3 \times 10^5 \text{ m s}^{-1}$.
The time period of one complete revolution is $T = \frac{2 \pi}{\omega}$.
The pitch of the helix is the distance traveled along the magnetic field direction in one time period: $\text{Pitch} = v_{\parallel} \times T$.
Substituting the values: $\text{Pitch} = (3 \times 10^5) \times \left( \frac{2 \pi}{4 \pi \times 10^6} \right)$.
$\text{Pitch} = 3 \times 10^5 \times \frac{1}{2 \times 10^6} = \frac{3}{20} \text{ m} = 0.15 \text{ m}$.
Converting to centimeters: $0.15 \text{ m} = 15 \text{ cm}$.

(D) In a cyclotron,the radius $R$ of the path of an ion is given by the relation $R = \frac{mv}{qB}$,where $v$ is the velocity of the ion.
Rearranging this for velocity,we get $v = \frac{qBR}{m}$.
The kinetic energy $(K.E.)$ of the ion is given by the formula $K.E. = \frac{1}{2}mv^2$.
Substituting the expression for $v$ into the kinetic energy formula:
$K.E. = \frac{1}{2}m \left( \frac{qBR}{m} \right)^2$
$K.E. = \frac{1}{2}m \left( \frac{q^2B^2R^2}{m^2} \right)$
$K.E. = \frac{q^2B^2R^2}{2m}$.

(A) When a charged particle moves in a magnetic field with a velocity component $v_{\perp}$ perpendicular to the field and $v_{\parallel}$ parallel to the field,it follows a helical path.
The time period $T$ for one complete rotation is determined by the perpendicular component of velocity:
$T = \frac{2 \pi m}{q B}$
The distance moved along the magnetic field in one rotation is called the pitch $(p)$.
$p = v_{\parallel} \times T$
Substituting the value of $T$:
$p = v_{\parallel} \times \frac{2 \pi m}{q B}$
If the total velocity $v$ is considered as the parallel component,the distance is $\frac{2 \pi m v}{q B}$.

(D) The cyclotron frequency is given by the formula $f = \frac{qB}{2\pi m}$.
Rearranging the formula to solve for the magnetic field $B$,we get $B = \frac{2\pi mf}{q}$.
Given values are:
Frequency $f = 20 MHz = 20 \times 10^6 Hz$
Charge of proton $q = 1.6 \times 10^{-19} C$
Mass of proton $m = 1.67 \times 10^{-27} kg$
Substituting these values into the formula:
$B = \frac{2 \times 3.14 \times 1.67 \times 10^{-27} \times 20 \times 10^6}{1.6 \times 10^{-19}}$
$B = \frac{209.536 \times 10^{-21}}{1.6 \times 10^{-19}}$
$B \approx 1.31 T$.

(C) The radius $R$ of the circular path of a charged particle moving perpendicular to a magnetic field is given by the formula: $R = \frac{mv}{qB}$.
Given values are:
Mass of electron,$m = 9 \times 10^{-31} \ kg$
Velocity of electron,$v = 3.2 \times 10^7 \ m/s$
Charge of electron,$q = 1.6 \times 10^{-19} \ C$
Magnetic field,$B = 6 \times 10^{-4} \ T$
Substituting these values into the formula:
$R = \frac{(9 \times 10^{-31}) \times (3.2 \times 10^7)}{(1.6 \times 10^{-19}) \times (6 \times 10^{-4})}$
$R = \frac{28.8 \times 10^{-24}}{9.6 \times 10^{-23}}$
$R = \frac{28.8}{9.6} \times 10^{-1} \ m$
$R = 3 \times 10^{-1} \ m = 0.3 \ m = 30 \ cm$.

(C) Diamagnetic substances are weakly repelled by magnetic fields.
When placed in a non-uniform magnetic field,they experience a force that pushes them from the stronger region of the field toward the weaker region.
This behavior is a characteristic property of diamagnetic materials.

(A) Ferromagnetic materials are classified based on their retentivity and coercivity.
Soft ferromagnetic materials have low retentivity and low coercivity,meaning they are easily magnetized and demagnetized.
Therefore,when the external magnetic field is removed,their magnetization disappears.
In contrast,hard ferromagnetic materials retain their magnetization even after the external field is removed.

(C) The Curie temperature $T_{C}$ is the critical temperature above which a ferromagnetic material loses its spontaneous magnetization and behaves as a paramagnetic material.
When a ferromagnetic substance is heated above its Curie temperature,the thermal agitation of the atoms overcomes the alignment of magnetic dipoles,resulting in a transition to a paramagnetic state.
Therefore,the Curie temperature represents the transition from ferromagnetic to paramagnetic.

(C) In nature,electric charges can exist as isolated monopoles (positive or negative). However,magnetic fields are produced by current loops or intrinsic spin,which always form dipoles. Magnetic monopoles have never been observed experimentally,and Gauss's Law for magnetism states that the net magnetic flux through any closed surface is zero,implying that magnetic monopoles do not exist.

(D) Magnetic field lines are continuous closed loops. Outside the magnet,they emerge from the north pole and enter the south pole,while inside the magnet,they travel from the south pole to the north pole to complete the loop.

(B) The binding energy per nucleon is a measure of the stability of a nucleus.
Experimental observations show that the binding energy per nucleon is maximum for nuclei with mass numbers in the range of $30 < A < 170$.
Among the given options,${ }_{26}^{56} Fe$ (Iron-$56$) has a mass number of $56$,which falls within this range.
It is well-established that ${ }_{26}^{56} Fe$ possesses the highest binding energy per nucleon,approximately $8.8 \text{ MeV/nucleon}$,making it one of the most stable nuclei.

(D) Nuclear fuels are materials that can be consumed by nuclear fission or fusion to produce energy. Uranium,thorium,and plutonium are well-known radioactive elements used in nuclear reactors. Titanium is a transition metal known for its high strength-to-weight ratio and corrosion resistance,but it is not a radioactive material capable of sustaining a nuclear chain reaction. Therefore,it is not used as a nuclear fuel.

(B) Deep inside the core of stars,protons collide with each other at extremely high speeds due to high temperature and pressure. These collisions result in the formation of a helium nucleus,releasing a tremendous amount of energy in the process. This specific nuclear reaction is known as nuclear fusion.

(A) The half-life $T_{1/2}$ of a radioactive substance is given by the formula $T_{1/2} = \frac{\ln 2}{\lambda}$,where $\lambda$ is the decay constant.
The mean life time $\tau$ is defined as the reciprocal of the decay constant,i.e.,$\tau = \frac{1}{\lambda}$.
Substituting $\lambda = \frac{1}{\tau}$ into the half-life formula,we get:
$T_{1/2} = \tau \ln 2$.
Therefore,the correct relation is $T_{1/2} = \tau \ln 2$.

(B) positron,also known as an antielectron,is the antiparticle of the electron. It has the same mass as an electron but carries an opposite electric charge ($+e$ instead of $-e$).

(C) Isotopes are atoms of the same element that have the same atomic number $(Z)$ but different mass numbers $(A)$.
Let the original nucleus be ${ }_{Z}^{A} X$.
After the emission of one $\alpha$ particle $({ }_{2}^{4} He)$,the atomic number decreases by $2$ and the mass number decreases by $4$:
${ }_{Z}^{A} X \rightarrow { }_{Z-2}^{A-4} Y + { }_{2}^{4} He$.
To return to the original atomic number $Z$,we need to increase the atomic number by $2$. This is achieved by the emission of two $\beta^-$ particles $({ }_{-1}^{0} e)$:
${ }_{Z-2}^{A-4} Y + 2({ }_{-1}^{0} e) \rightarrow { }_{Z}^{A-4} Y$.
Thus,the emission of one $\alpha$ particle and two $\beta$ particles results in an isotope of the original nucleus with a mass number reduced by $4$.

(C) Given: Half-life period $T_{1/2} = 2$ years,Initial mass $N_0 = 1 \,g$,Total time $t = 4$ years.
Using the formula for radioactive decay: $N = N_0 \left( \frac{1}{2} \right)^n$,where $n$ is the number of half-lives.
The number of half-lives $n = \frac{t}{T_{1/2}} = \frac{4}{2} = 2$.
Substituting the values: $N = 1 \times \left( \frac{1}{2} \right)^2 = 1 \times \frac{1}{4} = 0.25 \,g$.
Therefore,the amount of remaining radioactive material is $0.25 \,g$.

(D) The critical angle $i_c$ is given by the formula $\sin(i_c) = \frac{1}{n}$,where $n$ is the refractive index of the material with respect to air.
For diamond,the refractive index $n \approx 2.42$.
Therefore,$\sin(i_c) = \frac{1}{2.42} \approx 0.413$.
Taking the inverse sine,$i_c = \arcsin(0.413) \approx 24.4^{\circ}$.

(C) The critical angle $C$ is given by the formula $\sin C = \frac{1}{n}$,where $n$ is the refractive index of the medium.
Since the refractive index $n$ is inversely proportional to the wavelength $\lambda$ (Cauchy's relation: $n \approx A + \frac{B}{\lambda^2}$),as the wavelength decreases,the refractive index increases.
For red light,the wavelength is $\lambda_1$. For yellow light,the wavelength is $\lambda_2$. Since $\lambda_1 > \lambda_2$,the refractive index for red light $(n_r)$ is less than the refractive index for yellow light $(n_y)$.
Therefore,$n_r < n_y$.
Since $\sin C = \frac{1}{n}$,a higher refractive index results in a smaller critical angle.
Thus,the critical angle for yellow light will be less than the critical angle for red light $(\theta)$.

(B) For a convex mirror, the focal length $f = +30 \,cm$.
Magnification $m = \frac{h_i}{h_o} = +\frac{1}{4}$ (since the image formed by a convex mirror is always virtual and erect).
Using the magnification formula $m = -\frac{v}{u}$, we get $\frac{1}{4} = -\frac{v}{u}$, which implies $v = -\frac{u}{4}$.
Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$:
$\frac{1}{30} = \frac{1}{-u/4} + \frac{1}{u}$
$\frac{1}{30} = -\frac{4}{u} + \frac{1}{u}$
$\frac{1}{30} = -\frac{3}{u}$
$u = -30 \times 3 = -90 \,cm$.
The distance of the object from the mirror is $90 \,cm$.

(A) For a convex lens,the radius of curvature of the first surface $R_1$ is positive $(+40 \,cm)$ and the radius of curvature of the second surface $R_2$ is negative $(-40 \,cm)$.
Using the Lens Maker's formula:
$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Given $\mu = 1.5$,$R_1 = 40 \,cm$,and $R_2 = -40 \,cm$.
Substituting the values:
$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{40} - \frac{1}{-40} \right)$
$\frac{1}{f} = (0.5) \left( \frac{1}{40} + \frac{1}{40} \right)$
$\frac{1}{f} = (0.5) \left( \frac{2}{40} \right) = 0.5 \times 0.05 = 0.025$
$\frac{1}{f} = \frac{1}{40}$
Therefore,the focal length $f = 40 \,cm$.

(D) The blue colour of the sky is due to the scattering of light.
According to Rayleigh's law of scattering,the intensity of scattered light is inversely proportional to the fourth power of its wavelength,i.e.,$I \propto \frac{1}{\lambda^4}$.
Among the visible colours,blue light has a shorter wavelength compared to red light.
Because the wavelength of blue light is smaller,it is scattered much more strongly by the atmospheric particles (molecules of air) than other colours.
Therefore,the sky appears blue to our eyes.

(D) In a prism,let the angle of the prism be $A$.
When a light ray enters the first face,it refracts at an angle $r_1$.
When it reaches the second face,it strikes at an angle $r_2$ with respect to the normal.
From the geometry of the quadrilateral formed by the two normals and the two refracting faces of the prism,the sum of the angle of the prism $A$ and the angle between the two normals is $180^{\circ}$.
Also,in the triangle formed by the refracted ray inside the prism and the two refracting faces,the sum of the angles is $180^{\circ}$,which gives $A + r_1 + r_2 = 180^{\circ}$ (considering the interior angles of the triangle).
Thus,the relationship between the angle of the prism and the angles of refraction is $A = r_1 + r_2$.

(C) The angle of incidence $i$ is the angle between the incident ray and the normal. The problem states the light enters at an angle of $45^{\circ}$ with the interface. Therefore,the angle of incidence is $i = 90^{\circ} - 45^{\circ} = 45^{\circ}$.
The angle of deviation $D$ is given as $15^{\circ}$. The relationship between the angle of incidence $i$,the angle of refraction $r$,and the angle of deviation $D$ is $D = i - r$.
Substituting the given values: $15^{\circ} = 45^{\circ} - r$,which gives $r = 45^{\circ} - 15^{\circ} = 30^{\circ}$.
Using Snell's law,$n_1 \sin i = n_2 \sin r$,where $n_1 = 1$ (for air) and $n_2 = \mu$ (refractive index of the medium):
$1 \times \sin 45^{\circ} = \mu \times \sin 30^{\circ}$
$\frac{1}{\sqrt{2}} = \mu \times \frac{1}{2}$
$\mu = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.414$.
Thus,the refractive index of the medium is $1.414$.

(C) The reverse saturation current $I$ in a $p-n$ junction diode under reverse bias is given by Ohm's law:
$I = \frac{V}{R}$
Given:
Voltage $V = 8 \, V$
Resistance $R = 4 \times 10^7 \, \Omega$
Substituting the values:
$I = \frac{8}{4 \times 10^7} \, A$
$I = 2 \times 10^{-7} \, A$
To convert this into microamperes $(\mu A)$, we multiply by $10^6$:
$I = 2 \times 10^{-7} \times 10^6 \, \mu A$
$I = 2 \times 10^{-1} \, \mu A = 0.2 \, \mu A$
Therefore, the correct option is $C$.

(C) When a Zener diode is reverse biased and operated in the breakdown region,the voltage across it remains constant even if the current through it changes significantly.
Due to this property,it is widely used as a voltage regulator to maintain a steady output voltage across a load.

(A) In a $p-n-p$ transistor,the emitter is heavily doped to provide a large number of charge carriers for current flow. The base is very thin and lightly doped to allow most carriers to pass through to the collector. The collector is moderately doped and has a larger physical area compared to the emitter to dissipate heat generated during operation.

(D) The relationship between emitter current $(I_E)$,collector current $(I_C)$,and base current $(I_B)$ in a transistor is given by the equation:
$I_E = I_B + I_C$
Given:
$I_E = 9.85 \,mA$
$I_C = 9.5 \,mA$
To find the base current $(I_B)$,we rearrange the formula:
$I_B = I_E - I_C$
Substituting the given values:
$I_B = 9.85 \,mA - 9.5 \,mA = 0.35 \,mA$
Therefore,the base current is $0.35 \,mA$.

(A) Given: Emitter current, $I_{E} = 6.6 \, mA$.
Current amplification factor, $\beta = 59$.
We know that the relationship between emitter current $(I_{E})$, collector current $(I_{C})$, and base current $(I_{B})$ is $I_{E} = I_{C} + I_{B}$.
Since $I_{C} = \beta I_{B}$, we can write $I_{E} = \beta I_{B} + I_{B} = I_{B}(\beta + 1)$.
Therefore, the base current is $I_{B} = \frac{I_{E}}{\beta + 1}$.
Substituting the values: $I_{B} = \frac{6.6 \, mA}{59 + 1} = \frac{6.6 \, mA}{60} = 0.11 \, mA$.