AP EAMCET 2022 Physics Question Paper with Answer and Solution

(A) Let the car cover a distance $s$ from point $A$ to $B$ and then return to point $A$.
The average speed is defined as the total distance covered divided by the total time taken.
Time taken from $A$ to $B$ is $t_1 = \frac{s}{60}$.
Time taken from $B$ to $A$ is $t_2 = \frac{s}{V}$.
The total distance is $s + s = 2s$.
The total time is $t_1 + t_2 = \frac{s}{60} + \frac{s}{V}$.
Given the average speed is $48 \ km \ h^{-1}$,we have:
$48 = \frac{2s}{\frac{s}{60} + \frac{s}{V}}$
Dividing both sides by $s$ (assuming $s \neq 0$):
$48 = \frac{2}{\frac{1}{60} + \frac{1}{V}}$
$\frac{1}{60} + \frac{1}{V} = \frac{2}{48} = \frac{1}{24}$
$\frac{1}{V} = \frac{1}{24} - \frac{1}{60}$
$\frac{1}{V} = \frac{5 - 2}{120} = \frac{3}{120} = \frac{1}{40}$
Therefore,$V = 40 \ km \ h^{-1}$.

(B) The average speed is defined as the total distance divided by the total time taken.
Let the total distance be $L$.
The distance for the first part is $d_1 = \frac{L}{3}$ and the speed is $v_1$.
The time taken for the first part is $t_1 = \frac{d_1}{v_1} = \frac{L/3}{v_1} = \frac{L}{3 v_1}$.
The distance for the second part is $d_2 = \frac{2L}{3}$ and the speed is $v_2$.
The time taken for the second part is $t_2 = \frac{d_2}{v_2} = \frac{2L/3}{v_2} = \frac{2L}{3 v_2}$.
The total time taken is $T = t_1 + t_2 = \frac{L}{3 v_1} + \frac{2L}{3 v_2} = \frac{L}{3} \left( \frac{1}{v_1} + \frac{2}{v_2} \right) = \frac{L}{3} \left( \frac{v_2 + 2 v_1}{v_1 v_2} \right)$.
The average speed $v_{avg} = \frac{\text{Total distance}}{\text{Total time}} = \frac{L}{\frac{L}{3} \left( \frac{2 v_1 + v_2}{v_1 v_2} \right)} = \frac{3 v_1 v_2}{2 v_1 + v_2}$.

(D) The maximum height $(H)$ of a projectile is given by the formula: $H = \frac{u^2 \sin^2 \theta}{2g}$.
Given values are:
Initial speed $u = 50 \ m \ s^{-1}$
Angle of projection $\theta = 60^{\circ}$
Acceleration due to gravity $g = 10 \ m \ s^{-2}$
Substituting these values into the formula:
$H = \frac{(50)^2 \times (\sin 60^{\circ})^2}{2 \times 10}$
$H = \frac{2500 \times (\frac{\sqrt{3}}{2})^2}{20}$
$H = \frac{2500 \times \frac{3}{4}}{20}$
$H = \frac{2500 \times 0.75}{20}$
$H = \frac{1875}{20} = 93.75 \ m$
Therefore,the maximum height reached is $93.75 \ m$.

(C) Velocity is defined as the rate of change of displacement.
$v_y = \frac{dy}{dt}$
$= \frac{d(P t^2 - Q t^3)}{dt}$
$= 2Pt - 3Qt^2$
At maximum height,the vertical velocity of the particle becomes zero.
$2Pt - 3Qt^2 = 0$
$t(2P - 3Qt) = 0$
Since $t=0$ is the start of the motion,the time at maximum height is $t = \frac{2P}{3Q}$.
Now,substituting this time into the displacement equation:
$y_{max} = P(\frac{2P}{3Q})^2 - Q(\frac{2P}{3Q})^3$
$= P(\frac{4P^2}{9Q^2}) - Q(\frac{8P^3}{27Q^3})$
$= \frac{4P^3}{9Q^2} - \frac{8P^3}{27Q^2}$
$= \frac{12P^3 - 8P^3}{27Q^2}$
$= \frac{4P^3}{27Q^2}$

(D) The range of the projectile is given as $R = 90 \,m$.
The formula for the range of a projectile is $R = \frac{u^2 \sin(2\theta)}{g}$.
To hit a target at a fixed distance with the minimum initial velocity $u$, the range must be the maximum possible range for that velocity, which occurs at an angle of projection $\theta = 45^{\circ}$.
Substituting $\theta = 45^{\circ}$ into the range formula: $R = \frac{u^2 \sin(90^{\circ})}{g} = \frac{u^2}{g}$.
Given $R = 90 \,m$ and $g = 10 \,ms^{-2}$, we have $90 = \frac{u^2}{10}$.
Therefore, $u^2 = 900$, which gives $u = 30 \,ms^{-1}$.

(B) For a solid cylinder rolling without slipping,the total kinetic energy $(K.E.)$ is the sum of translational and rotational kinetic energy:
$K.E. = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Since the moment of inertia of a solid cylinder is $I = \frac{1}{2}mR^2$ and the condition for rolling without slipping is $v = R\omega$ (or $\omega = v/R$):
$K.E. = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2$
$K.E. = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$
By the law of conservation of energy,the loss in potential energy $(P.E.)$ equals the gain in kinetic energy:
$mgh = \frac{3}{4}mv^2$
$v^2 = \frac{4gh}{3}$
$v = \sqrt{\frac{4gh}{3}}$

(D) The general equation of the trajectory of a projectile is given by $y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta}$.
Comparing this with the given equation $y = \frac{x}{\sqrt{3}} - \frac{x^2}{60}$:
$1$. Comparing the coefficient of $x$: $\tan \theta = \frac{1}{\sqrt{3}} \Rightarrow \theta = 30^{\circ}$.
$2$. Comparing the coefficient of $x^2$: $\frac{g}{2 u^2 \cos^2 \theta} = \frac{1}{60}$.
Substituting $g = 10 \text{ m s}^{-2}$ and $\theta = 30^{\circ}$:
$\frac{10}{2 u^2 \cos^2 30^{\circ}} = \frac{1}{60} \Rightarrow \frac{5}{u^2 (\sqrt{3}/2)^2} = \frac{1}{60}$.
$\frac{5}{u^2 (3/4)} = \frac{1}{60} \Rightarrow \frac{20}{3 u^2} = \frac{1}{60}$.
$3 u^2 = 1200 \Rightarrow u^2 = 400$.
$u = 20 \text{ m s}^{-1}$.

(C) The horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin(2\theta)}{g} = 90 \,m$.
The maximum height $H$ is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
Dividing $H$ by $R$:
$\frac{H}{R} = \frac{u^2 \sin^2 \theta / 2g}{u^2 (2 \sin \theta \cos \theta) / g} = \frac{\tan \theta}{4}$.
Given $\theta = 45^{\circ}$,$\tan 45^{\circ} = 1$.
Therefore,$H = \frac{R \tan 45^{\circ}}{4} = \frac{90 \times 1}{4} = 22.5 \,m$.

(C) The total range $R$ of the projectile is given by:
$R = \frac{u^2 \sin 2\theta}{g} = \frac{30^2 \sin(2 \times 30^{\circ})}{10} = \frac{900 \times \sin 60^{\circ}}{10} = 90 \times \frac{\sqrt{3}}{2} = 45 \sqrt{3} \,m$.
The time of flight $T$ is given by:
$T = \frac{2u \sin \theta}{g} = \frac{2 \times 30 \times \sin 30^{\circ}}{10} = \frac{60 \times 0.5}{10} = 3 \,s$.
The second player is standing at a distance $d = 21 \sqrt{3} \,m$ from the first player. To catch the ball before it hits the ground, the player must reach the point where the ball lands (or any point along its trajectory). The minimum speed is required to reach the landing point $R$ within the time of flight $T$.
The distance the second player needs to cover is $S = R - d = 45 \sqrt{3} - 21 \sqrt{3} = 24 \sqrt{3} \,m$.
Since the player starts at the same instant as the kick, the time available to cover this distance is the time of flight $T = 3 \,s$.
Therefore, the minimum speed $v$ is:
$v = \frac{S}{T} = \frac{24 \sqrt{3}}{3} = 8 \sqrt{3} \,ms^{-1}$.
Thus, the correct option is $C$.

(C) To calculate the maximum height attained, we consider only the vertical component of the motion.
At the given height $h_1 = 5 \text{ m}$, the vertical component of velocity is $u_y = 5 \text{ m s}^{-1}$.
At the maximum height, the vertical component of velocity becomes $v_y = 0 \text{ m s}^{-1}$.
The acceleration due to gravity is $a_y = -10 \text{ m s}^{-2}$.
Using the kinematic equation $v_y^2 - u_y^2 = 2 a_y s$, where $s$ is the additional height gained:
$0^2 - (5)^2 = 2(-10) s$
$-25 = -20 s$
$s = \frac{25}{20} = 1.25 \text{ m}$
The total maximum height reached by the ball is $H = h_1 + s = 5 \text{ m} + 1.25 \text{ m} = 6.25 \text{ m}$.

(B) We consider the vertical motion of the ball.
Initial vertical velocity: $u_y = u \sin \theta = 20 \times \sin 30^{\circ} = 20 \times \frac{1}{2} = 10 \,m/s$.
Vertical acceleration: $a_y = -g = -10 \,m/s^2$.
Time taken to reach the ground: $t = 3 \,s$.
Using the equation of motion for vertical displacement: $s_y = u_y t + \frac{1}{2} a_y t^2$.
Here, the displacement $s_y = -h$ (where $h$ is the height of the building).
$-h = (10)(3) + \frac{1}{2}(-10)(3)^2$.
$-h = 30 - 5(9) = 30 - 45 = -15 \,m$.
Therefore, $h = 15 \,m$.

(C) At the maximum height,the vertical component of velocity is zero,so the speed of the projectile is $v = u \cos \theta$.
Given that the speed at maximum height is half of the initial speed $u$,we have $u \cos \theta = \frac{1}{2} u$,which implies $\cos \theta = \frac{1}{2}$,so $\theta = 60^{\circ}$.
The range $R$ of a projectile is given by $R = \frac{u^2 \sin 2 \theta}{g}$ and the maximum height $H$ is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
The ratio of range to maximum height is $\frac{R}{H} = \frac{u^2 \sin 2 \theta / g}{u^2 \sin^2 \theta / 2g} = \frac{2 \sin 2 \theta}{\sin^2 \theta}$.
Using the identity $\sin 2 \theta = 2 \sin \theta \cos \theta$,we get $\frac{R}{H} = \frac{2(2 \sin \theta \cos \theta)}{\sin^2 \theta} = 4 \cot \theta$.
Substituting $\theta = 60^{\circ}$,we get $\frac{R}{H} = 4 \cot 60^{\circ} = 4 \times \frac{1}{\sqrt{3}} = \frac{4}{\sqrt{3}}$.

(C) Let $v_{CE}$ be the velocity of the car with respect to the Earth,$v_{RE}$ be the velocity of the rain with respect to the Earth,and $v_{RC}$ be the velocity of the rain with respect to the car.
Given: $v_{CE} = 40 \text{ km h}^{-1}$.
The rain falls vertically,so $v_{RE}$ is along the vertical direction.
The trace of the rain on the side window makes an angle $\theta = 30^{\circ}$ with the vertical.
From the vector triangle of relative velocity,we have $\vec{v}_{RC} = \vec{v}_{RE} - \vec{v}_{CE}$.
In the right-angled triangle formed by these vectors,the side representing the car's velocity $v_{CE}$ is opposite to the angle $\theta = 30^{\circ}$.
Therefore,$\sin \theta = \frac{v_{CE}}{v_{RC}}$.
Rearranging for $v_{RC}$,we get $v_{RC} = \frac{v_{CE}}{\sin 30^{\circ}}$.
Substituting the values,$v_{RC} = \frac{40}{0.5} = 80 \text{ km h}^{-1}$.

(D) When the object is released from rest, its velocity is zero. Therefore, the centripetal acceleration $(a_c = v^2/r)$ is zero.
The forces acting on the object are its weight $(mg)$ acting vertically downwards and the tension $(T)$ in the string acting along the string.
We resolve the weight $(mg)$ into two components: one along the string $(mg \cos 30^{\circ})$ and one perpendicular to the string $(mg \sin 30^{\circ})$.
Since the string remains taut and the object starts from rest, the net force along the string must be zero to satisfy the condition of no radial motion at the instant of release $(T - mg \cos 30^{\circ} = 0)$.
The net force acting on the object is the component perpendicular to the string, which is $F_{net} = mg \sin 30^{\circ}$.
Using Newton's second law, $F_{net} = ma$, we get:
$ma = mg \sin 30^{\circ}$
$a = g \sin 30^{\circ}$
Given $g = 10 \,ms^{-2}$ and $\sin 30^{\circ} = 0.5$:
$a = 10 \times 0.5 = 5.0 \,ms^{-2}$.

(B) Initial angular speed $\omega_0 = 300 \text{ rpm} = \frac{300 \times 2\pi}{60} \text{ rad/s} = 10\pi \text{ rad/s}$.
Final angular speed $\omega = 0 \text{ rad/s}$.
Time taken $t = 80 \text{ s}$.
Using the equation of motion $\omega = \omega_0 + \alpha t$,we find the angular deceleration $\alpha$:
$0 = 10\pi + \alpha(80) \Rightarrow \alpha = -\frac{10\pi}{80} = -\frac{\pi}{8} \text{ rad/s}^2$.
The total angular displacement $\theta$ is given by $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$:
$\theta = (10\pi)(80) + \frac{1}{2} \left(-\frac{\pi}{8}\right) (80)^2 = 800\pi - \frac{\pi}{16} (6400) = 800\pi - 400\pi = 400\pi \text{ rad}$.
The number of revolutions $n$ is given by $n = \frac{\theta}{2\pi}$:
$n = \frac{400\pi}{2\pi} = 200$.

(A) Given: Radius, $r = 10 \,m$.
Centripetal acceleration, $a_c = 10 \,ms^{-2}$.
We know that the formula for centripetal acceleration is $a_c = \omega^2 r$.
Rearranging the formula to solve for angular speed $\omega$:
$\omega = \sqrt{\frac{a_c}{r}}$
Substituting the given values:
$\omega = \sqrt{\frac{10 \,ms^{-2}}{10 \,m}} = \sqrt{1 \,s^{-2}} = 1 \,rad \,s^{-1}$.
Therefore, the angular speed is $1 \,rad \,s^{-1}$.

(A) Let the velocity at point $A$ be $\vec{v}_1$ and at point $B$ be $\vec{v}_2$. Since the speed is constant,$|\vec{v}_1| = |\vec{v}_2| = v$.
The change in velocity is given by $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1$.
The magnitude of the change in velocity for a particle moving in a circle with constant speed $v$ through an angle $\theta$ is given by the vector subtraction formula:
$|\Delta \vec{v}| = \sqrt{v^2 + v^2 - 2v^2 \cos \theta}$
$|\Delta \vec{v}| = \sqrt{2v^2(1 - \cos \theta)}$
Using the trigonometric identity $1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$,we get:
$|\Delta \vec{v}| = \sqrt{2v^2 \cdot 2 \sin^2 \frac{\theta}{2}}$
$|\Delta \vec{v}| = \sqrt{4v^2 \sin^2 \frac{\theta}{2}}$
$|\Delta \vec{v}| = 2v \sin \frac{\theta}{2}$

(D) The mechanical energy $E$ of a harmonic oscillator is proportional to the square of its amplitude $A$,given by $E = \frac{1}{2} k A^2$.
For a small change in amplitude,the fractional change in energy is given by the differential:
$\frac{dE}{E} = \frac{d(A^2)}{A^2} = \frac{2A dA}{A^2} = 2 \frac{dA}{A}$.
Given that the amplitude decreases by $1.5 \%$,we have $\frac{dA}{A} = 1.5 \%$.
Substituting this value into the expression for fractional change in energy:
$\frac{dE}{E} = 2 \times 1.5 \% = 3 \%$.
Thus,the mechanical energy lost in each cycle is $3 \%$.

(A) The formula for the time period $T$ of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$.
Given: $T = 4 \ s$ and $g = \pi^2 \ ms^{-2}$.
Substituting these values into the formula:
$4 = 2\pi \sqrt{\frac{\ell}{\pi^2}}$
Squaring both sides:
$16 = 4\pi^2 \cdot \frac{\ell}{\pi^2}$
$16 = 4\ell$
$\ell = \frac{16}{4} = 4 \ m$.
Therefore,the length of the pendulum is $4 \ m$.

(C) The frequency of a simple pendulum is given by the formula:
$f = \frac{1}{2 \pi} \sqrt{\frac{g}{\ell}}$
From this formula, we can see that the frequency is inversely proportional to the square root of the length of the pendulum:
$f \propto \frac{1}{\sqrt{\ell}}$
Let the initial length be $\ell_1 = \ell$ and the initial frequency be $f_1 = 8 \,Hz$.
When the string is clamped at its midpoint, the new length becomes $\ell_2 = \frac{\ell}{2}$.
Now, we can find the ratio of the new frequency $f_2$ to the initial frequency $f_1$:
$\frac{f_2}{f_1} = \sqrt{\frac{\ell_1}{\ell_2}} = \sqrt{\frac{\ell}{\ell / 2}} = \sqrt{2}$
Therefore, the new frequency $f_2$ is:
$f_2 = \sqrt{2} \times f_1 = \sqrt{2} \times 8 \,Hz \approx 1.414 \times 8 \,Hz = 11.312 \,Hz$.
Rounding to the nearest provided option, we get $11.28 \,Hz$.

(B) The amplitude of a damped oscillator varies with time as $A(t) = A_0 e^{-\alpha t}$, where $\alpha = \frac{b}{2m}$.
Given that the amplitude drops to half its initial value, $A(t) = \frac{A_0}{2}$.
Substituting this into the equation: $\frac{A_0}{2} = A_0 e^{-\alpha t} \Rightarrow \frac{1}{2} = e^{-\alpha t}$.
Taking the natural logarithm on both sides: $\ln(0.5) = -\alpha t \Rightarrow -\ln 2 = -\alpha t \Rightarrow \ln 2 = \alpha t$.
Given $b = 69.3 \,g \,s^{-1} = 0.0693 \,kg \,s^{-1}$ and $m = 100 \,g = 0.1 \,kg$.
Calculate $\alpha = \frac{b}{2m} = \frac{69.3 \,g \,s^{-1}}{2 \times 100 \,g} = \frac{69.3}{200} \,s^{-1} = 0.3465 \,s^{-1}$.
Using $\ln 2 = 0.693$, we have $0.693 = \alpha t = 0.3465 \times t$.
Solving for $t$: $t = \frac{0.693}{0.3465} = 2 \,s$.

(A) The time period $T$ of a spring-mass system is given by the formula:
$T = 2 \pi \sqrt{\frac{m}{k}}$
Given, mass $m = 2 \,kg$ and spring constant $k = 8 \,N/m$.
Substituting these values into the formula:
$T = 2 \pi \sqrt{\frac{2}{8}} = 2 \pi \sqrt{\frac{1}{4}} = 2 \pi \times \frac{1}{2} = \pi \,s$.
The number of cycles $n$ completed in time $t = 66 \,s$ is given by:
$n = \frac{t}{T} = \frac{66}{\pi}$.
Taking $\pi \approx \frac{22}{7}$, we get:
$n = \frac{66}{22/7} = \frac{66 \times 7}{22} = 3 \times 7 = 21$.
Thus, the object completes $21$ oscillations in the given time.

(C) The potential energy $U$ of a body executing $S.H.M.$ at displacement $x$ is given by $U = \frac{1}{2} k x^2$,where $k$ is the force constant.
Given:
$U_x = \frac{1}{2} k x^2 = 9 \ J$ --- $(1)$
$U_y = \frac{1}{2} k y^2 = 16 \ J$ --- $(2)$
We need to find the potential energy at displacement $(x+y)$,which is $U_{(x+y)} = \frac{1}{2} k (x+y)^2$.
Expanding this,we get:
$U_{(x+y)} = \frac{1}{2} k (x^2 + y^2 + 2xy) = \frac{1}{2} k x^2 + \frac{1}{2} k y^2 + 2 \left( \sqrt{\frac{1}{2} k x^2} \right) \left( \sqrt{\frac{1}{2} k y^2} \right)$.
Substituting the given values:
$U_{(x+y)} = 9 + 16 + 2 \times \sqrt{9} \times \sqrt{16}$
$U_{(x+y)} = 25 + 2 \times 3 \times 4$
$U_{(x+y)} = 25 + 24 = 49 \ J$.

(C) The amplitude of the particle performing $SHM$ is $A = 10 \ cm$.
The instantaneous displacement of the particle is $x = 6 \ cm$.
The kinetic energy $(K)$ of the particle is given by $K = \frac{1}{2} m \omega^2 (A^2 - x^2)$.
The potential energy $(U)$ of the particle is given by $U = \frac{1}{2} m \omega^2 x^2$.
The ratio of kinetic energy to potential energy is $\frac{K}{U} = \frac{\frac{1}{2} m \omega^2 (A^2 - x^2)}{\frac{1}{2} m \omega^2 x^2} = \frac{A^2 - x^2}{x^2}$.
Substituting the values,we get $\frac{K}{U} = \frac{10^2 - 6^2}{6^2} = \frac{100 - 36}{36} = \frac{64}{36}$.
Simplifying the fraction,$\frac{K}{U} = \frac{16}{9}$.
Thus,the ratio of kinetic energy to potential energy is $16: 9$.

(B) In $S.H.M$,the kinetic energy $(K.E)$ and potential energy $(P.E)$ are given by:
$K.E = \frac{1}{2} m \omega^2 (a^2 - y^2)$
$P.E = \frac{1}{2} m \omega^2 y^2$
where $a$ is the amplitude and $y$ is the displacement.
For $K.E = P.E$:
$\frac{1}{2} m \omega^2 (a^2 - y^2) = \frac{1}{2} m \omega^2 y^2$
$a^2 - y^2 = y^2$
$a^2 = 2y^2$
$y = \pm \frac{a}{\sqrt{2}}$
Thus,Assertion $(A)$ is true.
Regarding Reason $(R)$,the potential energy of a particle in $S.H.M$ is indeed periodic and reaches its maximum value at the extreme displacement $(y = \pm a)$. However,this statement does not explain why the kinetic and potential energies are equal at $y = a/\sqrt{2}$. Therefore,both are true,but $(R)$ is not the correct explanation of $(A)$.

(D) The position of the particle is given by $x = 2 \cos (2t)$.
Taking the derivative with respect to time $t$,we get the velocity $v$:
$v = \frac{dx}{dt} = \frac{d}{dt} [2 \cos (2t)] = -2 \sin (2t) \times 2 = -4 \sin (2t) \,m/s$.
The maximum velocity $v_{\max}$ occurs when $|\sin (2t)| = 1$,so $|v_{\max}| = 4 \,m/s$.
The maximum kinetic energy $K_{\max}$ is given by the formula $K_{\max} = \frac{1}{2} m (v_{\max})^2$.
Substituting the given mass $m = 2 \,kg$ and $v_{\max} = 4 \,m/s$:
$K_{\max} = \frac{1}{2} \times 2 \,kg \times (4 \,m/s)^2 = 16 \,J$.

(B) The free body diagrams of the blocks are shown in the figure.
At equilibrium,the net force on a body is zero.
For block $A$ hanging vertically:
$m_A g = k x$
$\Rightarrow k = \frac{m_A g}{x}$
For block $B$ on the inclined plane:
$m_B g \sin 30^{\circ} = k x^{\prime}$
$\Rightarrow x^{\prime} = \frac{m_B g \sin 30^{\circ}}{k} = \frac{m_B g \sin 30^{\circ}}{(m_A g / x)} = \frac{m_B}{m_A} x \sin 30^{\circ}$
Since both blocks are made of the same material,mass is proportional to volume $(m = \rho V)$:
$\Rightarrow x^{\prime} = \frac{V_B}{V_A} x \sin 30^{\circ} = \frac{0.75}{0.25} \times 0.2 \times \sin 30^{\circ}$
$\Rightarrow x^{\prime} = 3 \times 0.2 \times 0.5 = 0.3 \ m$
The total distance of the block from the top is the sum of the unstretched length $l$ and the extension $x^{\prime}$:
$d = l + x^{\prime} = 1.0 \ m + 0.3 \ m = 1.3 \ m$.

(D) The displacement of the particle is given by $x = x_0 \sin \left(\omega t - \frac{\pi}{6}\right)$.
Acceleration $a$ is given by $a = \frac{d^2x}{dt^2} = -\omega^2 x$.
Substituting the expression for $x$:
$a = -\omega^2 x_0 \sin \left(\omega t - \frac{\pi}{6}\right)$.
Using the trigonometric identity $-\sin(\theta) = \sin(\theta + \pi)$:
$a = \omega^2 x_0 \sin \left(\omega t - \frac{\pi}{6} + \pi\right)$.
$a = \omega^2 x_0 \sin \left(\omega t + \frac{5\pi}{6}\right)$.
Comparing this with $a = A \sin(\omega t + \delta)$,we get $A = x_0 \omega^2$ and $\delta = \frac{5\pi}{6}$.

(A) The displacement of a particle in $SHM$ starting from the equilibrium position is given by $x = A \sin(\omega t)$.
Given $A = 10 \,cm$ and $x = 5 \,cm$,we have $5 = 10 \sin(\omega t)$,which implies $\sin(\omega t) = 1/2$.
Thus,$\omega t = \pi/6$. Since $\omega = 2\pi/T$,we get $t = \frac{\pi/6}{2\pi/T} = T/12$.
Given $T = 0.6 \,s$,the time interval is $t = 0.6/12 = 0.05 \,s$.
The mean velocity $V_{\text{mean}}$ is defined as the total displacement divided by the total time interval:
$V_{\text{mean}} = \frac{\Delta x}{\Delta t} = \frac{x_f - x_i}{t_f - t_i}$.
Starting from equilibrium ($x_i = 0$ at $t_i = 0$) to $x_f = 5 \,cm$ at $t_f = 0.05 \,s$:
$V_{\text{mean}} = \frac{5 \,cm - 0}{0.05 \,s} = 100 \,cm/s = 1 \,m/s$.

(A) Given, mass $m = 400 \, g = 0.4 \, kg$.
Force $F = -10 x$.
Comparing with the standard equation of $S.H.M.$, $F = -m \omega^2 x$, we get $m \omega^2 = 10$.
Substituting the value of $m$:
$0.4 \omega^2 = 10$
$\omega^2 = \frac{10}{0.4} = 25$
$\omega = 5 \, rad/s$.
The maximum speed at the centre of oscillation is given by $V_{max} = \omega A$.
Given $V_{max} = 10 \, m/s$, we have:
$10 = 5 \times A$
$A = \frac{10}{5} = 2 \, m$.

(B) In simple harmonic motion $(SHM)$,the maximum velocity of an object is given by:
$v_{\max} = A\omega$
From this,we can express the amplitude $A$ as:
$A = \frac{v_{\max}}{\omega} \quad ...(i)$
where $A$ is the amplitude and $\omega$ is the angular frequency.
The maximum acceleration of an object performing $SHM$ is given by:
$a_{\max} = \omega^2 A$
Substituting the value of $A$ from equation $(i)$ into the expression for $a_{\max}$:
$a_{\max} = \omega^2 \left( \frac{v_{\max}}{\omega} \right)$
$a_{\max} = \omega v_{\max}$

(A) Given,position of particle of mass $m = 3 \ kg$ is $x = 0.3 \cos (\omega t)$.
Velocity of particle,$v = \frac{dx}{dt} = \frac{d}{dt} (0.3 \cos \omega t) = -0.3 \omega \sin (\omega t)$.
Kinetic energy $K(t) = \frac{1}{2} m v^2 = \frac{1}{2} m (-0.3 \omega \sin \omega t)^2 = \frac{1}{2} m (0.09 \omega^2 \sin^2 \omega t)$.
At $t_1 = \frac{\pi}{6 \omega}$,$K\left(\frac{\pi}{6 \omega}\right) = \frac{1}{2} m (0.09 \omega^2 \sin^2 \frac{\pi}{6}) = \frac{1}{2} m (0.09 \omega^2) \left(\frac{1}{2}\right)^2 = \frac{1}{2} m (0.09 \omega^2) \left(\frac{1}{4}\right)$.
At $t_2 = \frac{\pi}{3 \omega}$,$K\left(\frac{\pi}{3 \omega}\right) = \frac{1}{2} m (0.09 \omega^2 \sin^2 \frac{\pi}{3}) = \frac{1}{2} m (0.09 \omega^2) \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{2} m (0.09 \omega^2) \left(\frac{3}{4}\right)$.
Taking the ratio: $\frac{K\left(\frac{\pi}{6 \omega}\right)}{K\left(\frac{\pi}{3 \omega}\right)} = \frac{1/4}{3/4} = \frac{1}{3}$.

(B) Given: $m = 1 \,kg$, $k = 100 \,N \,m^{-1}$, $\theta = 45^{\circ}$, $x = 10 \,cm = 0.1 \,m$, $g = 10 \,m \,s^{-2}$.
According to the work-energy theorem, the work done by gravity minus the work done by friction equals the potential energy stored in the spring when the block comes to rest.
Work done by gravity = $mg \sin \theta \cdot x$
Work done by friction = $f \cdot x = \mu N \cdot x = \mu mg \cos \theta \cdot x$
Potential energy of the spring = $\frac{1}{2} k x^2$
Applying the work-energy principle:
$mg \sin \theta \cdot x - \mu mg \cos \theta \cdot x = \frac{1}{2} k x^2$
$mg \sin \theta - \mu mg \cos \theta = \frac{1}{2} k x$
Substituting the values:
$1 \times 10 \times \sin 45^{\circ} - \mu \times 1 \times 10 \times \cos 45^{\circ} = \frac{1}{2} \times 100 \times 0.1$
$10 \times \frac{1}{\sqrt{2}} - \mu \times 10 \times \frac{1}{\sqrt{2}} = 5$
$\frac{10}{\sqrt{2}} (1 - \mu) = 5$
$1 - \mu = \frac{5 \sqrt{2}}{10} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$
$\mu = 1 - \frac{1}{\sqrt{2}} \approx 1 - 0.707 = 0.293$
Rounding to one decimal place, we get $\mu \approx 0.3$.

(A) Given that the density of the cone $\rho_C$ is half of the density of the water $\rho_W$,so $\rho_C = \frac{1}{2} \rho_W$.
Let $h$ be the height of the cone immersed in water at equilibrium. For floating,the weight of the cone equals the buoyant force:
$\frac{1}{3} \pi R^2 H \rho_C = \frac{1}{3} \pi r^2 h \rho_W$
Since $\frac{r}{h} = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$ and $\frac{R}{H} = \frac{1}{\sqrt{3}}$,we have $r = \frac{h}{\sqrt{3}}$ and $R = \frac{H}{\sqrt{3}}$.
Substituting these into the equilibrium equation:
$\frac{1}{3} \pi (\frac{H}{\sqrt{3}})^2 H \rho_C = \frac{1}{3} \pi (\frac{h}{\sqrt{3}})^2 h \rho_W$
$\frac{H^3}{3} \rho_C = \frac{h^3}{3} \rho_W \Rightarrow \frac{h^3}{H^3} = \frac{\rho_C}{\rho_W} = \frac{1}{2} \Rightarrow h = H (\frac{1}{2})^{\frac{1}{3}}$.
When the cone is displaced by a small distance $\delta$ downward,the additional buoyant force is $F = \pi r^2 \rho_W g \delta$.
The restoring force constant is $k = \pi r^2 \rho_W g$.
The mass of the cone is $M = \frac{1}{3} \pi R^2 H \rho_C = \frac{1}{3} \pi (\frac{H}{\sqrt{3}})^2 H (\frac{1}{2} \rho_W) = \frac{1}{18} \pi H^3 \rho_W$.
The frequency of $SHM$ is $f = \frac{1}{2 \pi} \sqrt{\frac{k}{M}}$.
$k = \pi (\frac{h}{\sqrt{3}})^2 \rho_W g = \frac{\pi h^2 \rho_W g}{3} = \frac{\pi H^2 (1/2)^{2/3} \rho_W g}{3}$.
$f = \frac{1}{2 \pi} \sqrt{\frac{\pi H^2 (1/2)^{2/3} \rho_W g / 3}{\pi H^3 \rho_W / 18}} = \frac{1}{2 \pi} \sqrt{\frac{6 g (1/2)^{2/3}}{H}} = \frac{1}{2 \pi} \sqrt{\frac{6 g}{H} \frac{1}{4^{1/3}}}$.

(B) In simple harmonic motion $(SHM)$,the velocity of a particle is zero at the extreme positions.
Let the displacement be $x = A \sin(\omega t)$. The velocity is $v = \frac{dx}{dt} = A\omega \cos(\omega t)$.
The velocity is zero when $\cos(\omega t) = 0$,which occurs at $t = \frac{T}{4}, \frac{3T}{4}, \dots$
The time interval between two consecutive points of zero velocity (i.e.,between two extreme positions) is the time taken to travel from one extreme to the other,which is half the time period,$\frac{T}{2}$.
Given,$\frac{T}{2} = 0.5 \text{ s}$.
Therefore,$T = 1 \text{ s}$.
The angular frequency $\omega$ is given by $\omega = \frac{2\pi}{T}$.
Substituting $T = 1 \text{ s}$,we get $\omega = \frac{2\pi}{1} = 2\pi \text{ rad s}^{-1}$.

(A) When the hydrometer is floating in equilibrium,its weight is balanced by the buoyant force.
When it is pushed down by a small distance $x$,the additional buoyant force acts as the restoring force.
The additional volume submerged is $V = \pi r^2 x$.
The additional buoyant force is $F = \rho V g = \rho (\pi r^2 x) g$.
Since this force acts in the direction opposite to the displacement,the restoring force is $F_{restoring} = -\rho \pi r^2 g x$.
Using Newton's second law,$m a = -\rho \pi r^2 g x$,which gives $a = -(\frac{\rho \pi r^2 g}{m}) x$.
Comparing this with the standard $SHM$ equation $a = -\omega^2 x$,we get $\omega^2 = \frac{\rho \pi r^2 g}{m}$.
The time period $T$ is given by $T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{m}{\rho \pi r^2 g}}$.

(B) Given,the instantaneous displacement of the particle is $x = A \sin^2(\omega t - \frac{\pi}{4})$.
Using the trigonometric identity $\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$,we can rewrite the expression as:
$x = A \left[ \frac{1 - \cos(2(\omega t - \frac{\pi}{4}))}{2} \right]$
$x = \frac{A}{2} [1 - \cos(2\omega t - \frac{\pi}{2})]$
In simple harmonic motion,the general form is $x = x_0 + A' \cos(\omega' t + \phi)$.
Comparing the expression,the angular frequency of the oscillation is $\omega' = 2\omega$.
The time period $T'$ is given by $T' = \frac{2\pi}{\omega'}$.
Substituting $\omega' = 2\omega$,we get $T' = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$.

(C) The total mechanical energy $E$ of the system is the sum of translational kinetic energy,rotational kinetic energy,and potential energy of the spring:
$E = \frac{1}{2} M V^2 + \frac{1}{2} I \omega^2 + \frac{1}{2} K x^2$
Since the sphere rolls without slipping,$V = R \omega$,so $\omega = V/R$. The moment of inertia of a solid sphere is $I = \frac{2}{5} M R^2$.
Substituting these:
$E = \frac{1}{2} M V^2 + \frac{1}{2} (\frac{2}{5} M R^2) (\frac{V^2}{R^2}) + \frac{1}{2} K x^2 = \frac{1}{2} M V^2 + \frac{1}{5} M V^2 + \frac{1}{2} K x^2 = \frac{7}{10} M V^2 + \frac{1}{2} K x^2$
Since total energy is conserved,$\frac{dE}{dt} = 0$:
$\frac{d}{dt} (\frac{7}{10} M V^2 + \frac{1}{2} K x^2) = 0$
$\frac{7}{10} M (2 V \frac{dV}{dt}) + \frac{1}{2} K (2 x \frac{dx}{dt}) = 0$
Since $V = \frac{dx}{dt}$ and $a = \frac{dV}{dt}$:
$\frac{7}{5} M V a + K V x = 0$
$\frac{7}{5} M a + K x = 0 \implies a = -(\frac{5 K}{7 M}) x$
Comparing with the $SHM$ equation $a = -\omega^2 x$,we get $\omega^2 = \frac{5 K}{7 M}$,so $\omega = \sqrt{\frac{5 K}{7 M}}$.
The time period $T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{7 M}{5 K}}$.

(D) Given: Mass of ice $m_{\text{ice}} = 5 \ g$,initial temperature $T_{\text{ice}} = -30^{\circ} C$. Mass of water $m_w = 20 \ g$,initial temperature $T_w = 35^{\circ} C$.
Specific heat of ice $s_{\text{ice}} = 0.5 \ cal \ g^{-1} {}^{\circ} C^{-1}$,specific heat of water $s_w = 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$,latent heat of fusion $L_f = 80 \ cal \ g^{-1}$.
Heat required to raise ice from $-30^{\circ} C$ to $0^{\circ} C$: $Q_1 = m_{\text{ice}} s_{\text{ice}} \Delta T = 5 \times 0.5 \times 30 = 75 \ cal$.
Heat required to melt ice at $0^{\circ} C$: $Q_2 = m_{\text{ice}} L_f = 5 \times 80 = 400 \ cal$.
Total heat required to convert ice at $-30^{\circ} C$ to water at $0^{\circ} C$ is $Q_{\text{total}} = 75 + 400 = 475 \ cal$.
Heat available from water cooling from $35^{\circ} C$ to $0^{\circ} C$: $Q_{\text{avail}} = m_w s_w \Delta T = 20 \times 1 \times 35 = 700 \ cal$.
Since $Q_{\text{avail}} > Q_{\text{total}}$,the final temperature $T$ will be above $0^{\circ} C$.
Heat lost by water = Heat gained by ice:
$m_w s_w (35 - T) = Q_{\text{total}} + m_{\text{ice}} s_w (T - 0)$
$20 \times 1 \times (35 - T) = 475 + 5 \times 1 \times T$
$700 - 20T = 475 + 5T$
$225 = 25T$
$T = 9^{\circ} C$.

(A) The process involves three steps:
$1$. Heating ice from $-10^{\circ}C$ to $0^{\circ}C$: $Q_1 = m \cdot S_{ice} \cdot \Delta T = 0.2 \ kg \times 2100 \ J \ kg^{-1} K^{-1} \times 10 \ K = 4200 \ J$
$2$. Melting ice at $0^{\circ}C$ to water at $0^{\circ}C$: $Q_2 = m \cdot L_f = 0.2 \ kg \times 3.35 \times 10^5 \ J \ kg^{-1} = 67000 \ J$
$3$. Heating water from $0^{\circ}C$ to $30^{\circ}C$: $Q_3 = m \cdot S_{water} \cdot \Delta T = 0.2 \ kg \times 4186 \ J \ kg^{-1} K^{-1} \times 30 \ K = 25116 \ J$
Total heat required $Q = Q_1 + Q_2 + Q_3 = 4200 + 67000 + 25116 = 96316 \ J$.

(C) Given: Mass of metal block,$m = 3.3 \ kg = 3300 \ g$.
Temperature change,$\Delta T = 400^{\circ} C$.
Specific heat of metal,$s_m = 0.4 \ J \ g^{-1} \ K^{-1}$.
Latent heat of fusion of ice,$L_f = 330 \ J \ g^{-1}$.
The heat released by the metal block is $Q = m \cdot s_m \cdot \Delta T$.
$Q = 3300 \ g \times 0.4 \ J \ g^{-1} \ K^{-1} \times 400 \ K = 528,000 \ J$.
Let $m'$ be the mass of ice melted. The heat required to melt the ice is $Q = m' \cdot L_f$.
$m' = \frac{Q}{L_f} = \frac{528,000 \ J}{330 \ J \ g^{-1}} = 1600 \ g$.
Therefore,$m' = 1.6 \ kg$.

(D) Given: Mass of water $m = 200 \,g = 0.2 \,kg$, time $t = 200 \,s$, specific heat $s = 4200 \,J \,kg^{-1} \,K^{-1}$, and temperature change $\Delta T = 100^{\circ} C - 40^{\circ} C = 60 \,K$.
The heat energy $Q$ required to raise the temperature of the water is given by $Q = m \cdot s \cdot \Delta T$.
The power $P$ supplied by the heater is $P = \frac{Q}{t} = \frac{m \cdot s \cdot \Delta T}{t}$.
Substituting the values: $P = \frac{0.2 \,kg \times 4200 \,J \,kg^{-1} \,K^{-1} \times 60 \,K}{200 \,s}$.
$P = \frac{50400}{200} \,W = 252 \,W$.

(C) The net rate of energy exchange of the sphere is given by the Stefan-Boltzmann Law: $P = \varepsilon \sigma A (T^4 - T_s^4)$.
Given:
Emissivity $\varepsilon = 0.5$
Stefan-Boltzmann constant $\sigma = 5.67 \times 10^{-8} \ W \ m^{-2} \ K^{-4}$
Surface area $A = 4 \ m^2$
Temperature of the sphere $T = 400 \ K$
Temperature of the environment $T_s = 200 \ K$
Substituting the values into the formula:
$P = 0.5 \times (5.67 \times 10^{-8}) \times 4 \times [400^4 - 200^4]$
$P = 2 \times 5.67 \times 10^{-8} \times [(4 \times 10^2)^4 - (2 \times 10^2)^4]$
$P = 11.34 \times 10^{-8} \times [256 \times 10^8 - 16 \times 10^8]$
$P = 11.34 \times 10^{-8} \times [240 \times 10^8]$
$P = 11.34 \times 240 = 2721.6 \ W$.

(C) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = -k(T - T_s)$,where $T$ is the temperature of the body and $T_s$ is the temperature of the surroundings.
For the first interval,the average temperature is $T_{avg1} = \frac{75+60}{2} = 67.5^{\circ}C$.
For the second interval,the average temperature is $T_{avg2} = \frac{65+55}{2} = 60^{\circ}C$.
Since $T_{avg2} < T_{avg1}$,the temperature difference between the body and the surroundings is smaller in the second case.
Therefore,the rate of cooling $\frac{dT}{dt}$ is lower in the second case.
Since the temperature drop is $10^{\circ}C$ in both cases,a lower rate of cooling implies that it will take more time to cool down in the second case.

(A) The length of the metal tape at temperature $T$ is given by $L = L_0 [1 + \alpha \Delta T]$,where $L_0$ is the length at the calibration temperature $T_0 = 25^{\circ} C$.
The change in length is $\Delta L = L - L_0 = L_0 \alpha (T - T_0)$.
The percentage error in the measurement is given by $|\frac{\Delta L}{L_0} \times 100\%|$.
Substituting the given values: $\alpha = 1 \times 10^{-5} {}^{\circ} C^{-1}$,$T = -15^{\circ} C$,and $T_0 = 25^{\circ} C$.
Percentage error $= |\alpha (T - T_0) \times 100\%|$
$= |1 \times 10^{-5} (-15 - 25) \times 100\%|$
$= |1 \times 10^{-5} (-40) \times 100\%|$
$= |-40 \times 10^{-3}\%| = 0.04\%$.

(D) Given,diameter of sphere $= D$.
Initial surface area,$A = 4 \pi R^2 = 4 \pi (D/2)^2 = \pi D^2$.
Surface area after heating by temperature $\delta T$ is $A' = \pi (D')^2$,where $D'$ is the new diameter.
From the equation of linear expansion,$D' = D(1 + \alpha \delta T)$.
Substituting $D'$ into the expression for $A'$,we get $A' = \pi [D(1 + \alpha \delta T)]^2 = \pi D^2 (1 + 2\alpha \delta T + \alpha^2 \delta T^2)$.
The change in surface area $\Delta A = A' - A = \pi D^2 (1 + 2\alpha \delta T + \alpha^2 \delta T^2) - \pi D^2$.
$\Delta A = \pi D^2 (2\alpha \delta T + \alpha^2 \delta T^2) = \pi D^2 \alpha \delta T (2 + \alpha \delta T)$.

(B) The density $\rho$ is inversely proportional to volume $V$,i.e.,$\rho \propto \frac{1}{V}$.
For a small change in temperature $\Delta \theta$,the volume changes as $V_2 = V_1(1 + \gamma \Delta \theta)$,where $\gamma$ is the coefficient of volume expansion.
Thus,the new density $\rho_2$ is given by $\rho_2 = \frac{m}{V_2} = \frac{m}{V_1(1 + \gamma \Delta \theta)} = \rho_1(1 + \gamma \Delta \theta)^{-1}$.
Using the binomial approximation $(1 + x)^{-1} \approx 1 - x$ for small $x$,we get $\rho_2 \approx \rho_1(1 - \gamma \Delta \theta)$.
The fractional change in density is $\frac{\Delta \rho}{\rho_1} = \frac{\rho_2 - \rho_1}{\rho_1} = -\gamma \Delta \theta$.
Given $\gamma = 5 \times 10^{-4} {^{\circ}C}^{-1}$ and $\Delta \theta = 40^{\circ}C$.
Substituting the values: $\frac{\Delta \rho}{\rho_1} = -(5 \times 10^{-4} {^{\circ}C}^{-1}) \times (40^{\circ}C) = -200 \times 10^{-4} = -0.02$.
The magnitude of the fractional change in density is $0.02$.

(B) Density is given by $\rho = \frac{m}{V}$.
Since mass $m$ remains constant,$\rho \propto \frac{1}{V}$.
Taking the logarithmic derivative,we get $\frac{d\rho}{\rho} = -\frac{dV}{V}$.
For small changes,$\frac{\Delta \rho}{\rho} = -\frac{\Delta V}{V}$.
We know that $\Delta V = V_0 \gamma \Delta T$,where $\gamma$ is the coefficient of volume expansion.
Thus,$\frac{\Delta \rho}{\rho} = -\gamma \Delta T$.
The percentage change in density is $\frac{\Delta \rho}{\rho} \times 100 = -\gamma \Delta T \times 100$.
Given $\gamma = 18.2 \times 10^{-5} \ K^{-1}$ and $\Delta T = 60^{\circ} C - 10^{\circ} C = 50 \ K$.
Percentage change $= 18.2 \times 10^{-5} \times 50 \times 100 = 18.2 \times 10^{-5} \times 5000 = 0.91 \%$.

(A) Given: Work done by the gas, $W = 200 \,J$.
For an ideal monoatomic gas expanding at constant pressure, the heat absorbed is given by $Q = n C_p \Delta T$.
The work done is given by $W = n R \Delta T$.
Taking the ratio of $Q$ to $W$:
$\frac{Q}{W} = \frac{n C_p \Delta T}{n R \Delta T} = \frac{C_p}{R}$.
For a monoatomic gas, the molar heat capacity at constant volume is $C_v = \frac{3}{2} R = 1.5 R$.
Using the relation $C_p = C_v + R$, we get $C_p = 1.5 R + R = 2.5 R$.
Substituting this into the ratio:
$\frac{Q}{W} = \frac{2.5 R}{R} = 2.5$.
Therefore, $Q = 2.5 \times W = 2.5 \times 200 \,J = 500 \,J$.

(B) The internal energy $U$ of $n$ moles of a gas is given by $U = n \frac{f}{2} RT$,where $f$ is the degree of freedom.
For monoatomic gases $(He, Ar)$,$f = 3$.
For diatomic gases $(N_2, H_2)$,$f = 5$ (ignoring vibrational modes).
Total internal energy $U_{total} = U_{He} + U_{Ar} + U_{N_2} + U_{H_2}$.
$U_{total} = (3 \times \frac{3}{2} RT) + (1 \times \frac{3}{2} RT) + (5 \times \frac{5}{2} RT) + (3 \times \frac{5}{2} RT)$.
$U_{total} = (4.5 + 1.5 + 12.5 + 7.5) RT = 26 RT$.

(A) Let the charge on one object be $q$,then the charge on the other object is $(Q - q)$.
According to Coulomb's Law,the electrostatic force $F_e$ between them is given by $F_e = \frac{K q(Q - q)}{r^2}$,where $r$ is the distance between them.
To find the condition for maximum force,we differentiate $F_e$ with respect to $q$ and set it to zero:
$\frac{dF_e}{dq} = \frac{K}{r^2} \frac{d}{dq}(qQ - q^2) = 0$.
This implies $\frac{d}{dq}(qQ - q^2) = 0$.
$Q - 2q = 0$,which gives $q = Q/2$.
The charge on the other object is $(Q - q) = Q - Q/2 = Q/2$.
Therefore,the electrostatic force is maximum when the charges are divided equally as $Q/2$ and $Q/2$.

(A) The four charges are located at $A(1,0,0), B(-1,0,0), C(0,1,0),$ and $D(0,-1,0)$. We need to find the electric field at point $P(0,0,1)$.
The distance $r$ of point $P$ from each charge is $r = \sqrt{1^2 + 1^2} = \sqrt{2} \text{ m}$.
The magnitude of the electric field due to each charge at point $P$ is:
$E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{2}$.
Due to symmetry,the horizontal components of the electric fields from the four charges cancel each other out. Only the vertical components (along the $Z$-axis) contribute to the net electric field.
The angle $\theta$ that the electric field vector from each charge makes with the $Z$-axis is given by $\cos \theta = \frac{1}{r} = \frac{1}{\sqrt{2}}$.
The vertical component of the electric field from each charge is $E_z = E \cos \theta = \left( \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{2} \right) \cdot \frac{1}{\sqrt{2}}$.
Since there are four such charges,the net electric field $E_{\text{net}}$ is:
$E_{\text{net}} = 4 \cdot E_z = 4 \cdot \left( \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{2 \sqrt{2}} \right) = \frac{q}{2 \sqrt{2} \pi \varepsilon_0} \text{ N/C}$.

(A) Let the positions of the charges be $x_1 = 0$ for $+5q$,$x_2 = 2r/3$ for $Q$,and $x_3 = r$ for $-2q$. The distance between $+5q$ and $-2q$ is $r$,and the distance between $Q$ and $-2q$ is $r/3$.
For the net force on charge $-2q$ to be zero,the forces exerted by $+5q$ and $Q$ on $-2q$ must be equal in magnitude and opposite in direction.
Let $F_1$ be the force due to $+5q$ on $-2q$ and $F_2$ be the force due to $Q$ on $-2q$.
$F_1 = \frac{k |5q| |-2q|}{r^2} = \frac{10kq^2}{r^2}$ (Attractive,towards left).
For the net force to be zero,$F_2$ must be repulsive (towards right),so $Q$ must be positive.
$F_2 = \frac{k |Q| |-2q|}{(r/3)^2} = \frac{k |Q| 2q}{r^2/9} = \frac{18k|Q|q}{r^2}$.
Equating the magnitudes: $\frac{10kq^2}{r^2} = \frac{18kQq}{r^2}$.
$10q = 18Q \Rightarrow Q = \frac{10}{18}q = \frac{5}{9}q$.

(C) The torque $\tau$ acting on an electric dipole in an external electric field is given by the formula: $\tau = p E \sin \theta$.
Given:
Dipole moment $p = 5 \times 10^{-7} \text{ C m}$
Electric field $E = 2 \times 10^4 \text{ N C}^{-1}$
Angle $\theta = 60^{\circ}$
Substituting the values:
$\tau = (5 \times 10^{-7}) \times (2 \times 10^4) \times \sin(60^{\circ})$
$\tau = 10 \times 10^{-3} \times \frac{\sqrt{3}}{2}$
$\tau = 10 \times 10^{-3} \times 0.866$
$\tau = 8.66 \times 10^{-3} \text{ N m}$

(C) Let the radius of the semicircle be $R$. The length of the wire is $L = \pi R$,so $R = L / \pi$. The linear charge density is $\lambda = Q / L$.
Consider a small element of the wire of length $dl = R d\theta$ at an angle $\theta$ with the vertical axis. The charge on this element is $dQ = \lambda dl = \lambda R d\theta$.
The electric field due to this element at the centre is $dE = \frac{1}{4 \pi \varepsilon_0} \frac{dQ}{R^2} = \frac{1}{4 \pi \varepsilon_0} \frac{\lambda R d\theta}{R^2} = \frac{1}{4 \pi \varepsilon_0} \frac{\lambda d\theta}{R}$.
By symmetry,the horizontal components of the electric field cancel out. The net electric field is the integral of the vertical components $dE \cos \theta$ from $-\pi / 2$ to $\pi / 2$:
$E = \int_{-\pi / 2}^{\pi / 2} dE \cos \theta = \frac{\lambda}{4 \pi \varepsilon_0 R} \int_{-\pi / 2}^{\pi / 2} \cos \theta d\theta = \frac{\lambda}{4 \pi \varepsilon_0 R} [\sin \theta]_{-\pi / 2}^{\pi / 2} = \frac{\lambda}{4 \pi \varepsilon_0 R} (1 - (-1)) = \frac{2 \lambda}{4 \pi \varepsilon_0 R} = \frac{\lambda}{2 \pi \varepsilon_0 R}$.
Substituting $\lambda = Q / L$ and $R = L / \pi$:
$E = \frac{Q / L}{2 \pi \varepsilon_0 (L / \pi)} = \frac{Q}{2 \pi \varepsilon_0 L} \cdot \frac{\pi}{L} = \frac{Q}{2 \varepsilon_0 L^2} \cdot \pi = \frac{\pi Q}{2 \varepsilon_0 L^2}$.
Wait,checking the options provided,the standard result for a semicircle is $E = \frac{2k\lambda}{R} = \frac{2}{4\pi\varepsilon_0} \cdot \frac{Q/L}{L/\pi} = \frac{2\pi Q}{4\pi\varepsilon_0 L^2} = \frac{Q}{2\varepsilon_0 L^2}$.
Thus,the correct option is $C$.

(C) When the observer is at rest relative to the static charge,they only experience an electric field.
When the observer starts moving away from the charge,there is a relative velocity between the charge and the observer.
According to the principles of electromagnetism,a moving charge creates both an electric field and a magnetic field in the frame of reference of the observer.
Therefore,the observer experiences both an electric field and a magnetic field.

(B) For Assertion $(A)$:
The potential difference is given by $dV = -\vec{E} \cdot d\vec{r}$. If the electric field $\vec{E} = 0$,then $dV = 0$,which implies that the potential $V$ is constant throughout the region.
For Reason $(R)$:
According to Gauss's law,the net electric flux $\phi_E$ through a closed surface is given by $\phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}$.
If $\vec{E} = 0$ everywhere in the region,then the flux $\phi_E = 0$,which implies $q_{enclosed} = 0$. Thus,the reason is correct.
However,the reason explains why the charge must be zero if the field is zero,but it does not explain why the potential is constant in that region (which is derived from the relation $E = -\nabla V$). Therefore,$(R)$ is not the correct explanation of $(A)$.

(A) The charges are placed at $x = 0, \pm 1 \text{ cm}, \pm 2 \text{ cm}, \pm 3 \text{ cm}, \dots$
Since the spherical surface has a radius of $2.5 \text{ cm}$ and is centered at the origin,it encloses the charges located at $x = 0, \pm 1 \text{ cm},$ and $\pm 2 \text{ cm}$.
The total number of charges enclosed is $1$ (at origin) $+ 2$ (at $\pm 1 \text{ cm}$) $+ 2$ (at $\pm 2 \text{ cm}$) $= 5$ charges.
Thus,the total charge enclosed by the sphere is $Q_{\text{enclosed}} = 5q$.
According to Gauss's Law,the total electric flux $\phi$ through the spherical surface is given by $\phi = \frac{Q_{\text{enclosed}}}{\varepsilon_0}$.
Therefore,$\phi = \frac{5q}{\varepsilon_0}$.

(D) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{\text{net}}}{\varepsilon_0}$,where $q_{\text{net}}$ is the net charge enclosed by the surface.
For surface $A$,the charges enclosed are $q$,$3q$,$-2q$,and $-5q$.
Therefore,$(q_{\text{net}})_A = q + 3q - 2q - 5q = -3q$.
Thus,$\phi_A = \frac{-3q}{\varepsilon_0}$.
For surface $B$,the charges enclosed are $3q$,$-q$,and $2q$.
Therefore,$(q_{\text{net}})_B = 3q - q + 2q = 4q$.
Thus,$\phi_B = \frac{4q}{\varepsilon_0}$.
Now,the ratio of the fluxes is:
$\frac{\phi_A}{\phi_B} = \frac{-3q / \varepsilon_0}{4q / \varepsilon_0} = -\frac{3}{4}$.

(A) The electric field $E$ due to a large charged plane with surface charge density $\sigma$ is given by $E = \frac{\sigma}{2 \varepsilon_0}$.
Given $\sigma = 4.9 \times 10^{-6} \text{ C m}^{-2}$.
The electric flux $\phi$ through an area $A$ is given by $\phi = \vec{E} \cdot \vec{A} = EA \cos \theta$,where $\theta$ is the angle between the electric field vector and the area vector (normal to the surface).
Here,the electric field is along the $z$-axis (perpendicular to the $x-y$ plane). The normal to the circular plane makes an angle $\theta = 60^{\circ}$ with the $z$-axis.
Area $A = \pi r^2 = \pi \times (0.01 \text{ m})^2 = \pi \times 10^{-4} \text{ m}^2$.
Using $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$,we have $\frac{1}{2 \varepsilon_0} = 2 \pi \times 9 \times 10^9 = 18 \pi \times 10^9$.
Thus,$\phi = \left( \frac{\sigma}{2 \varepsilon_0} \right) A \cos 60^{\circ} = (\sigma \times 18 \pi \times 10^9) \times (\pi \times 10^{-4}) \times \frac{1}{2}$.
Substituting the values: $\phi = (4.9 \times 10^{-6}) \times (18 \pi^2 \times 10^5) \times 0.5 = 4.9 \times 9 \times \pi^2 \times 10^{-1} = 44.1 \times 9.8696 \times 0.1 \approx 43.56 \text{ N m}^2 \text{ C}^{-1}$.

(A) According to Gauss's Law,the electric flux $\phi$ linked with a closed surface is given by $\phi = \frac{Q_{\text{enclosed}}}{\varepsilon_0}$.
For surface $P_1$,the enclosed charge is $Q_{\text{enclosed}, 1} = \frac{Q}{2}$.
Therefore,the flux linked with $P_1$ is $\phi_1 = \frac{Q/2}{\varepsilon_0} = \frac{Q}{2\varepsilon_0}$.
For surface $P_2$,the enclosed charge is $Q_{\text{enclosed}, 2} = \frac{Q}{2} + 4Q = \frac{9Q}{2}$.
Therefore,the flux linked with $P_2$ is $\phi_2 = \frac{9Q/2}{\varepsilon_0} = \frac{9Q}{2\varepsilon_0}$.
Comparing the two expressions,we get $\phi_2 = 9 \times \left(\frac{Q}{2\varepsilon_0}\right) = 9\phi_1$.

(A) The work done $W$ to move a charge in an electrostatic field is equal to the change in electrostatic potential energy $\Delta U$.
$W = U_f - U_i = k q_1 q_2 (\frac{1}{r_f} - \frac{1}{r_i})$
Given: $q_1 = 10 \times 10^{-6} \ C$,$q_2 = 12 \times 10^{-6} \ C$,$k = 9 \times 10^9 \ N \ m^2/C^2$.
Initial distance $r_i = 10 \ cm = 0.1 \ m$.
Final distance $r_f = 10 \ cm - 6 \ cm = 4 \ cm = 0.04 \ m$.
$W = (9 \times 10^9) \times (10 \times 10^{-6}) \times (12 \times 10^{-6}) \times (\frac{1}{0.04} - \frac{1}{0.1})$
$W = 1.08 \times (25 - 10) = 1.08 \times 15 = 16.2 \ J$.
Note: The calculated result is $16.2 \ J$. None of the provided options match this result. Assuming the question intended to ask for the work done to bring them to a final distance of $6 \ cm$ $(r_f = 0.06 \ m)$:
$W = 1.08 \times (\frac{1}{0.06} - \frac{1}{0.1}) = 1.08 \times (16.67 - 10) = 1.08 \times 6.67 \approx 7.2 \ J$.

(A) Statement $(A)$ is true: Inside a charged hollow metal sphere,the electric field $E$ is $0$ because there is no charge enclosed. However,the electric potential $V$ is constant and equal to the potential at the surface,so $V \neq 0$.
Statement $(B)$ is true: By definition,an equipotential surface is a surface where the potential is the same at all points. Since work done $W = q(V_f - V_i)$ and $V_f = V_i$,the work done is $0$.
Statement $(C)$ is true: When two like charges are brought closer,they repel each other. To move them closer,external work must be done against the electrostatic repulsive force. This work is stored as an increase in the mutual electrostatic potential energy of the system.
Therefore,all three statements are correct.

(C) The electric field is given by $E = 5x \hat{i}$.
The potential difference between two points is given by $V_B - V_A = -\int_{A}^{B} \vec{E} \cdot d\vec{r}$.
Since the electric field is only in the $X$-direction, the $Y$-axis (where $x=0$) is an equipotential line. Thus, the potential at point $A(0, 5)$ is the same as the potential at the origin $C(0, 0)$.
$V_A = V_C$.
Now, we calculate the potential difference between $C(0, 0)$ and $B(2, 0)$:
$V_B - V_C = -\int_{0}^{2} E_x dx = -\int_{0}^{2} 5x dx$.
$V_B - V_C = -5 \left[ \frac{x^2}{2} \right]_{0}^{2} = -5 \left( \frac{4}{2} - 0 \right) = -5(2) = -10 \text{ V}$.
Since $V_A = V_C$, we have $V_B - V_A = -10 \text{ V}$.

(A) The electric field $\vec{E}$ is related to the electric potential $\phi$ by the relation $\vec{E} = -\nabla \phi$.
Since the potential $\phi$ depends only on $x$,the electric field is given by $\vec{E} = -\frac{\partial \phi}{\partial x} \hat{i}$.
Given $\phi(x) = \phi_0 \frac{x_0}{x} = (8 \ V)(5 \ m) \frac{1}{x} = \frac{40}{x} \ V \cdot m$.
Differentiating with respect to $x$: $\frac{\partial \phi}{\partial x} = 40 \frac{d}{dx}(x^{-1}) = 40 (-x^{-2}) = -\frac{40}{x^2}$.
Therefore,$\vec{E} = -(-\frac{40}{x^2}) \hat{i} = \frac{40}{x^2} \hat{i} \ V/m$.
At the point $(10 \ m, 5 \ m, 5 \ m)$,the $x$-coordinate is $10 \ m$.
Substituting $x = 10 \ m$ into the expression for $\vec{E}$: $\vec{E} = \frac{40}{10^2} \hat{i} = \frac{40}{100} \hat{i} = 0.40 \ Vm^{-1} \hat{i}$.

(C) The relation between electric field $\vec{E}$ and potential $V$ is given by $\vec{E} = -\nabla V$,which implies $dV = -\vec{E} \cdot d\vec{r}$.
Given $\vec{E} = A(x \hat{i} + y \hat{j})$,we have $dV = -A(x dx + y dy)$.
Integrating both sides,we get $V = -A \int x dx - A \int y dy = -A \frac{x^2}{2} - A \frac{y^2}{2} + C$,where $C$ is the integration constant.
Substituting $A = 10 \ Vm^{-2}$,we get $V = -5(x^2 + y^2) + C$.
Given that the potential at $(10 \ m, 20 \ m)$ is zero,we have $0 = -5(10^2 + 20^2) + C$.
$0 = -5(100 + 400) + C \Rightarrow 0 = -5(500) + C \Rightarrow C = 2500 \ V$.
Thus,the potential function is $V(x, y) = -5(x^2 + y^2) + 2500$.
At the origin $(0, 0)$,the potential is $V(0, 0) = -5(0^2 + 0^2) + 2500 = 2500 \ V$.

(D) The electric field $E$ produced by an infinite non-conducting sheet is given by $E = \frac{\sigma}{2 \varepsilon_0}$.
Given $\sigma = 7 \times 10^{-7} \text{ C m}^{-2}$ and $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ SI units}$.
We know that $\varepsilon_0 = \frac{1}{4 \pi \times 9 \times 10^9} \approx 8.85 \times 10^{-12} \text{ F m}^{-1}$.
The relation between electric field and potential difference $\Delta V$ for a distance $\Delta r$ is $|E| = \frac{\Delta V}{\Delta r}$.
Substituting the values:
$\frac{\sigma}{2 \varepsilon_0} = \frac{\Delta V}{\Delta r}$
$\Delta r = \frac{\Delta V \times 2 \varepsilon_0}{\sigma} = \frac{19.8 \times 2 \times 8.85 \times 10^{-12}}{7 \times 10^{-7}}$
$\Delta r = \frac{19.8 \times 17.7 \times 10^{-12}}{7 \times 10^{-7}} \approx 5 \times 10^{-4} \text{ m} = 0.5 \text{ mm}$.

(A) The electric potential $V$ at the center of a square of side $a$ with charges $q$ at each of the $4$ corners is given by $V = 4 \times \frac{Kq}{r}$,where $r$ is the distance from the center to any corner.
For a square of side $a$,the diagonal is $a\sqrt{2}$,so the distance $r$ from the center to a corner is $r = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.
For the first square,$a_1 = 1 \ m$,so $r_1 = \frac{1}{\sqrt{2}} \ m$. The potential is $V_1 = 4 \times \frac{K \times 2}{1/\sqrt{2}} = 8\sqrt{2}K$.
For the second square,$a_2 = 2 \ m$,so $r_2 = \frac{2}{\sqrt{2}} = \sqrt{2} \ m$. The potential is $V_2 = 4 \times \frac{K \times 2}{\sqrt{2}} = \frac{8K}{\sqrt{2}} = 4\sqrt{2}K$.
Therefore,the ratio is $\frac{V_2}{V_1} = \frac{4\sqrt{2}K}{8\sqrt{2}K} = \frac{1}{2}$.

(C) The electrostatic potential energy $U$ of a system of point charges is given by the sum of the potential energies of all pairs of charges: $U = \sum \frac{k q_i q_j}{r_{ij}}$.
For the given system with charges $-q$,$+q$,and $Q$ at the vertices of a right-angled triangle with sides $x, x$,and hypotenuse $\sqrt{x^2 + x^2} = \sqrt{2}x$,the total potential energy is:
$U = k \left[ \frac{(-q)(q)}{x} + \frac{(q)(Q)}{x} + \frac{(-q)(Q)}{\sqrt{2}x} \right]$
For the system to have zero potential energy,we set $U = 0$:
$0 = k \left[ -\frac{q^2}{x} + \frac{qQ}{x} - \frac{qQ}{\sqrt{2}x} \right]$
Dividing by $\frac{kq}{x}$ (assuming $q, x \neq 0$):
$0 = -q + Q - \frac{Q}{\sqrt{2}}$
$q = Q \left( 1 - \frac{1}{\sqrt{2}} \right)$
$q = Q \left( \frac{\sqrt{2} - 1}{\sqrt{2}} \right)$
$Q = \frac{\sqrt{2} q}{\sqrt{2} - 1}$
To match the options,multiply the numerator and denominator by $(\sqrt{2} + 1)$ or simplify differently:
$Q = \frac{\sqrt{2} q (\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{(2 + \sqrt{2})q}{2 - 1} = (2 + \sqrt{2})q$
Alternatively,looking at option $C$: $\frac{2q}{2-\sqrt{2}} = \frac{2q(2+\sqrt{2})}{4-2} = \frac{2q(2+\sqrt{2})}{2} = (2+\sqrt{2})q$. Thus,option $C$ is correct.

(D) There are four fundamental forces in nature: $1$. Gravitational force,$2$. Electromagnetic force,$3$. Weak nuclear force,and $4$. Strong nuclear force.
Normal force,frictional force,and spring force are derived forces that arise from electromagnetic interactions between atoms and molecules. Therefore,the strong nuclear force is the only fundamental force among the given options.

(C) The magnetic force on a current-carrying conductor of length $\vec{L}$ in a uniform magnetic field $\vec{B}$ is given by $\vec{F} = I(\vec{L} \times \vec{B})$.
For any closed current-carrying loop placed in a uniform magnetic field,the net magnetic force is always zero.
Mathematically,$\vec{F}_{net} = I \oint (d\vec{l} \times \vec{B}) = I (\oint d\vec{l}) \times \vec{B}$.
Since the loop is closed,the vector sum of all length elements $\oint d\vec{l} = 0$.
Therefore,$\vec{F}_{net} = 0$.

(A) The torque $\vec{\tau}$ on a magnetic dipole $\vec{M}$ in a magnetic field $\vec{B}$ is given by $\vec{\tau} = \vec{M} \times \vec{B}$.
Let the magnitude of the magnetic moment be $M$. The unit vector in the direction of the needle is $\hat{u} = \frac{\sqrt{3}\hat{i} + \hat{j}}{2}$. Thus,$\vec{M} = \frac{M}{2}(\sqrt{3}\hat{i} + \hat{j})$.
Given $\vec{B}_1 = B\hat{i}$,the torque is $\vec{\tau}_1 = \frac{M}{2}(\sqrt{3}\hat{i} + \hat{j}) \times B\hat{i} = \frac{MB}{2}(\sqrt{3}(\hat{i} \times \hat{i}) + (\hat{j} \times \hat{i})) = \frac{MB}{2}(0 - \hat{k}) = -\frac{MB}{2}\hat{k}$.
Given $|\vec{\tau}_1| = 0.06 \ N-m$,we have $\frac{MB}{2} = 0.06$,so $MB = 0.12 \ N-m$.
In the second case,the needle is oriented in the direction $(\hat{i} + \sqrt{3}\hat{j})$,so $\vec{M}_2 = \frac{M}{2}(\hat{i} + \sqrt{3}\hat{j})$. The magnetic field is $\vec{B}_2 = 2B\hat{j}$.
The torque is $\vec{\tau}_2 = \vec{M}_2 \times \vec{B}_2 = \frac{M}{2}(\hat{i} + \sqrt{3}\hat{j}) \times 2B\hat{j} = MB(\hat{i} \times \hat{j} + \sqrt{3}(\hat{j} \times \hat{j})) = MB(\hat{k} + 0) = MB\hat{k}$.
Thus,$|\vec{\tau}_2| = MB = 0.12 \ N-m$.

(A) When two short magnets of equal dipole moments $M$ are fastened perpendicularly at their centres,their resultant dipole moment $M^{\prime}$ is given by the vector sum:
$M^{\prime} = \sqrt{M^2 + M^2 + 2 M M \cos 90^{\circ}} = M \sqrt{2}$
This resultant dipole moment $M^{\prime}$ acts along the bisector of the right angle.
The point at distance $d$ on the bisector lies on the axial line of this resultant dipole $M^{\prime}$.
The magnetic field $B$ on the axial position of a short magnetic dipole is given by:
$B = \frac{\mu_0}{4 \pi} \frac{2 M^{\prime}}{d^3}$
Substituting $M^{\prime} = M \sqrt{2}$ into the formula:
$B = \frac{\mu_0}{4 \pi} \frac{2 (M \sqrt{2})}{d^3} = \frac{\mu_0}{4 \pi} \frac{2 \sqrt{2} M}{d^3}$

(D) The mean radius $r$ of the toroid is given by the average of the inner and outer radii:
$r = \frac{r_1 + r_2}{2} = \frac{24 \ cm + 25 \ cm}{2} = 24.5 \ cm = 24.5 \times 10^{-2} \ m$
Given number of turns $N = 4900$ and current $I = 12 \ A$.
The magnetic field $B$ inside the core of a toroid is given by the formula:
$B = \frac{\mu_0 N I}{2 \pi r}$
Substituting the values:
$B = \frac{(4 \pi \times 10^{-7} \ T \cdot m/A) \times 4900 \times 12}{2 \pi \times 24.5 \times 10^{-2} \ m}$
$B = \frac{2 \times 10^{-7} \times 4900 \times 12}{24.5 \times 10^{-2}}$
$B = \frac{2 \times 4900 \times 12}{24.5} \times 10^{-5}$
$B = \frac{117600}{24.5} \times 10^{-5} = 4800 \times 10^{-5} \ T = 48 \times 10^{-3} \ T = 48 \ mT$.

(A) The magnetic field $B$ produced by a long straight current-carrying wire at a distance $r$ is given by the formula:
$B = \frac{\mu_0 i}{2 \pi r}$
From this expression,we can see that $B \propto \frac{1}{r}$.
Given that at distance $r$,$B = 1 \ T$.
$(a)$ At distance $\frac{r}{2}$:
$B_{\frac{r}{2}} = B \times \frac{r}{\frac{r}{2}} = 1 \times 2 = 2 \ T$.
$(b)$ At distance $2r$:
$B_{2r} = B \times \frac{r}{2r} = 1 \times \frac{1}{2} = \frac{1}{2} \ T$.
$(c)$ At distance $3r$:
$B_{3r} = B \times \frac{r}{3r} = 1 \times \frac{1}{3} = \frac{1}{3} \ T$.
Thus,the values are $2 \ T, \frac{1}{2} \ T, \text{ and } \frac{1}{3} \ T$.

(D) Let the point $P$ be at a distance $r$ from wire $A$ where the resultant magnetic field is zero.
Since the currents are in the same direction,the magnetic fields produced by the two wires at a point between them will be in opposite directions.
For the resultant magnetic field to be zero at point $P$,the magnitudes of the magnetic fields produced by wires $A$ and $B$ must be equal.
$B_A = B_B$
$\frac{\mu_0 i_1}{2 \pi r} = \frac{\mu_0 i_2}{2 \pi (d - r)}$
Where $i_1 = 12 \,A$,$i_2 = 36 \,A$,and $d = 50 \,cm = 0.5 \,m$.
$\frac{12}{r} = \frac{36}{0.5 - r}$
$\frac{1}{r} = \frac{3}{0.5 - r}$
$0.5 - r = 3r$
$4r = 0.5$
$r = \frac{0.5}{4} = 0.125 \,m = 12.5 \,cm$
Thus,the point is at a distance of $12.5 \,cm$ from wire $A$.

(A) The wires are located at $A(1, 1)$ and $B(1, -1)$. The distance of each wire from the origin $O(0, 0)$ is $r = \sqrt{1^2 + 1^2} = \sqrt{2} \text{ cm}$.
Since both wires carry current in the same direction (perpendicular to the $xy$-plane),the magnetic field vectors at the origin are perpendicular to the position vectors $OA$ and $OB$.
The angle of $OA$ with the $x$-axis is $45^\circ$ and $OB$ is $-45^\circ$. The magnetic field $B_A$ due to wire $A$ is perpendicular to $OA$ at $135^\circ$ to the $x$-axis,and $B_B$ due to wire $B$ is perpendicular to $OB$ at $45^\circ$ to the $x$-axis.
The angle between $B_A$ and $B_B$ is $90^\circ$.
The magnitude of the field due to one wire is $B_0 = \frac{\mu_0 I}{2 \pi r}$.
The net magnetic field magnitude is $|B| = \sqrt{B_0^2 + B_0^2 + 2 B_0 B_0 \cos(90^\circ)} = \sqrt{2 B_0^2} = \sqrt{2} B_0$.
Therefore,$\frac{|B|}{B_0} = \sqrt{2}$.

(A) The distance of wire $I$ from the origin is $d = \sqrt{1^2 + 1^2} = \sqrt{2} \ cm = \sqrt{2} \times 10^{-2} \ m$.
The magnetic field due to wire $I$ at the origin is $B_1 = \frac{\mu_0 I}{2 \pi d}$. Let $B_0 = \frac{\mu_0 I}{2 \pi d}$.
Using the right-hand rule,the direction of $B_1$ is perpendicular to the position vector $(1, 1)$ and the current direction $(+z)$. The unit vector for the position is $\frac{\hat{i} + \hat{j}}{\sqrt{2}}$. The field direction is $\frac{\hat{i} + \hat{j}}{\sqrt{2}} \times \hat{k} = \frac{\hat{j} - \hat{i}}{\sqrt{2}}$. Thus,$\vec{B}_1 = B_0 \left( \frac{\hat{j} - \hat{i}}{\sqrt{2}} \right)$.
For wire $II$ at $y = 1 \ cm$ carrying current in $+x$ direction,the distance is $d_2 = 1 \ cm = 10^{-2} \ m$. The field magnitude is $B_2 = \frac{\mu_0 I}{2 \pi d_2} = \sqrt{2} B_0$.
The direction of $B_2$ using the right-hand rule for current in $+x$ at $y=1$ is $-\hat{k}$. Thus,$\vec{B}_2 = -\sqrt{2} B_0 \hat{k}$.
The total field is $\vec{B} = \vec{B}_1 + \vec{B}_2 = B_0 \left( -\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} - \sqrt{2} \hat{k} \right)$.
Therefore,$\frac{\vec{B}}{B_0} = \left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, -\sqrt{2} \right)$. Note: The provided options suggest a sign convention difference; based on standard vector analysis,the ratio is $\left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, -\sqrt{2} \right)$,which matches option $A$ in magnitude and components.

(B) The magnetic field due to a current-carrying coil of radius $R$ on its axis at a distance $x$ from its centre is given by:
$B_{\text{axis}} = \frac{\mu_0 N I R^2}{2(x^2 + R^2)^{3/2}}$
Given: $x = 8 \text{ cm} = 8 \times 10^{-2} \text{ m}$,$R = 6 \text{ cm} = 6 \times 10^{-2} \text{ m}$,and $B_{\text{axis}} = 27 \mu \text{T} = 27 \times 10^{-6} \text{ T}$.
Substituting these values into the formula:
$27 \times 10^{-6} = \frac{\mu_0 N I (6 \times 10^{-2})^2}{2((8 \times 10^{-2})^2 + (6 \times 10^{-2})^2)^{3/2}}$
$27 \times 10^{-6} = \frac{\mu_0 N I (36 \times 10^{-4})}{2(64 \times 10^{-4} + 36 \times 10^{-4})^{3/2}}$
$27 \times 10^{-6} = \frac{\mu_0 N I (36 \times 10^{-4})}{2(100 \times 10^{-4})^{3/2}} = \frac{\mu_0 N I (36 \times 10^{-4})}{2(10^{-2})^{3/2}}$
$27 \times 10^{-6} = \frac{\mu_0 N I (36 \times 10^{-4})}{2 \times 10^{-3}}$
$\mu_0 N I = \frac{27 \times 10^{-6} \times 2 \times 10^{-3}}{36 \times 10^{-4}} = \frac{54 \times 10^{-9}}{36 \times 10^{-4}} = 1.5 \times 10^{-5} \text{ T} \cdot \text{m}$.
Now,the magnetic field at the centre is $B_{\text{centre}} = \frac{\mu_0 N I}{2R}$.
$B_{\text{centre}} = \frac{1.5 \times 10^{-5}}{2 \times 6 \times 10^{-2}} = \frac{1.5 \times 10^{-5}}{12 \times 10^{-2}} = 0.125 \times 10^{-3} \text{ T} = 125 \mu \text{T}$.

(D) The magnetic field due to a long straight wire is given by $B = \frac{\mu_0 I}{2\pi r}$.
For the wire along the $x$-axis carrying current $I_1 = 8 \,A$, the point $P(2, 4)$ is at a perpendicular distance $r_1 = 4$ units. Using the right-hand rule, the magnetic field $B_1$ at $P$ is directed into the plane $(\otimes)$.
$B_1 = \frac{\mu_0 I_1}{2\pi r_1} = \frac{(4\pi \times 10^{-7}) \times 8}{2\pi \times 4} = 4 \times 10^{-7} \,T$ (into the plane).
For the wire along the $y$-axis carrying current $I_2 = 6 \,A$, the point $P(2, 4)$ is at a perpendicular distance $r_2 = 2$ units. Using the right-hand rule, the magnetic field $B_2$ at $P$ is directed out of the plane $(\odot)$.
$B_2 = \frac{\mu_0 I_2}{2\pi r_2} = \frac{(4\pi \times 10^{-7}) \times 6}{2\pi \times 2} = 6 \times 10^{-7} \,T$ (out of the plane).
The net magnetic field $B_{net} = |B_2 - B_1| = |6 \times 10^{-7} - 4 \times 10^{-7}| = 2 \times 10^{-7} \,T$.

(A) The magnetic field $B$ inside a long solenoid is given by the formula $B = \mu_0 n I$, where $n$ is the number of turns per unit length.
Given:
Length $L = 1 \,m$
Number of layers $= 5$
Turns per layer $= 500$
Total number of turns $N = 5 \times 500 = 2500$
Number of turns per unit length $n = N / L = 2500 / 1 = 2500 \,m^{-1}$
Magnetic field $B = 4.4 \,mT = 4.4 \times 10^{-3} \,T$
Permeability of free space $\mu_0 = 4\pi \times 10^{-7} \,T \cdot m/A$
Substituting the values into the formula:
$4.4 \times 10^{-3} = (4\pi \times 10^{-7}) \times 2500 \times I$
$4.4 \times 10^{-3} = (4 \times 3.14159 \times 10^{-7}) \times 2500 \times I$
$4.4 \times 10^{-3} = 3.14159 \times 10^{-3} \times I$
$I = 4.4 / 3.14159 \approx 1.4 \,A$
Therefore, the current carried is $1.4 \,A$.

(B) The magnetic field due to a long straight wire at a distance $d$ is given by $B = \frac{\mu_0 I}{2 \pi d}$.
At point $P(0, 2, 0) \, m$:
$1$. Magnetic field due to wire $A$ (along $X$-axis, $I_1 = 40 \, A$): The distance from the wire to point $P$ is $d_1 = 2 \, m$. The direction of the field is given by the right-hand rule, which is in the $+z$-direction $(\hat{k})$.
$B_1 = \frac{\mu_0 (40)}{2 \pi (2)} = \frac{20 \mu_0}{\pi} \, T$.
$2$. Magnetic field due to wire $B$ (passing through $(0, 3, 0) \, m$, parallel to $Z$-axis, current $I_2$): The distance from the wire to point $P$ is $d_2 = |3 - 2| = 1 \, m$. The direction of the field is in the $+x$-direction $(\hat{i})$.
$B_2 = \frac{\mu_0 I_2}{2 \pi (1)} = \frac{\mu_0 I_2}{2 \pi} \, T$.
Since $B_1$ and $B_2$ are perpendicular, the resultant magnetic field $R$ is $R = \sqrt{B_1^2 + B_2^2}$.
Given $R = 5 \times 10^{-6} \, T$ and $\frac{\mu_0}{2 \pi} = 2 \times 10^{-7} \, T \cdot m/A$:
$R = \frac{\mu_0}{2 \pi} \sqrt{20^2 + I_2^2} = 2 \times 10^{-7} \sqrt{400 + I_2^2} = 5 \times 10^{-6}$.
$\sqrt{400 + I_2^2} = \frac{5 \times 10^{-6}}{2 \times 10^{-7}} = 25$.
$400 + I_2^2 = 625$.
$I_2^2 = 225 \Rightarrow I_2 = 15 \, A$.

(D) The straight wire segments $ab$ and $cd$ do not produce any magnetic field at point $O$ because the point $O$ lies on the axis of these wires.
For the arc $bc$ (radius $R_1 = 2 \, cm = 2 \times 10^{-2} \, m$), the magnetic field $B_1$ at $O$ is directed inward $(\otimes)$ and is given by:
$B_1 = \frac{\mu_0 I}{4 \pi R_1} \theta = \frac{10^{-7} \times I \times \theta}{R_1}$
$B_1 = \frac{10^{-7} \times (1.2 / \pi) \times (30^\circ \times \pi / 180^\circ)}{2 \times 10^{-2}} = \frac{10^{-7} \times 1.2 \times (1/6)}{2 \times 10^{-2}} = 10^{-6} \, T = 1000 \, nT$ (inward).
For the arc $ad$ (radius $R_2 = 1 \, cm = 1 \times 10^{-2} \, m$), the magnetic field $B_2$ at $O$ is directed outward $(\odot)$ and is given by:
$B_2 = \frac{\mu_0 I}{4 \pi R_2} \theta = \frac{10^{-7} \times (1.2 / \pi) \times (30^\circ \times \pi / 180^\circ)}{1 \times 10^{-2}} = \frac{10^{-7} \times 1.2 \times (1/6)}{1 \times 10^{-2}} = 2 \times 10^{-6} \, T = 2000 \, nT$ (outward).
The net magnetic field at $O$ is $B_{net} = B_2 - B_1 = 2000 \, nT - 1000 \, nT = 1000 \, nT = 1 \, \mu T$.

(C) The magnetic moment of a straight wire of length $l$ and pole strength $m$ is given by $M = m \times l$.
When the wire is bent into a semicircular arc of radius $R$,the length of the wire corresponds to the circumference of the semicircle: $l = \pi R$.
Therefore,the radius of the arc is $R = \frac{l}{\pi}$.
The new magnetic moment $M'$ is the product of the pole strength $m$ and the straight-line distance between the two ends (the diameter of the semicircle).
The diameter is $d = 2R = \frac{2l}{\pi}$.
Thus,$M' = m \times d = m \times \frac{2l}{\pi}$.
Since $M = ml$,we substitute $ml$ with $M$ to get $M' = \frac{2}{\pi} M$.

(A) When a charged particle moves perpendicular to a magnetic field,it follows a circular path.
The magnetic Lorentz force provides the necessary centripetal force: $q v B = \frac{m v^2}{r}$.
This simplifies to $\frac{v}{r} = \frac{q B}{m}$.
Since the angular frequency $\omega = \frac{v}{r}$,we have $\omega = \frac{q B}{m}$.
The frequency of rotation $f$ is given by $f = \frac{\omega}{2 \pi} = \frac{q B}{2 \pi m}$.
As the frequency $f$ depends only on the charge $q$,magnetic field $B$,and mass $m$,it is independent of the velocity or kinetic energy of the particle.
Therefore,for both electrons $e_1$ and $e_2$,the frequencies are equal,i.e.,$f_1 = f_2$.

(A) The magnetic field at the origin due to the wire at $x=1 \ cm$ is $B_0 = \frac{\mu_0 i}{2 \pi (0.01)}$ in the $+z$ direction (using the right-hand rule,$\hat{k}$).
The magnetic field at the origin due to the wire at $x=-2 \ cm$ is $B' = \frac{\mu_0 i}{2 \pi (0.02)}$ in the $-z$ direction $(-\hat{k})$.
The net magnetic field at the origin is $\vec{B}_{net} = B_0 \hat{k} - \frac{B_0}{2} \hat{k} = \frac{B_0}{2} \hat{k}$.
The velocity of the electron is $\vec{v} = U \cos(45^{\circ}) \hat{i} + U \sin(45^{\circ}) \hat{j} = \frac{U}{\sqrt{2}} \hat{i} + \frac{U}{\sqrt{2}} \hat{j}$.
The force on the electron is $\vec{F} = q(\vec{v} \times \vec{B}) = (-e) \left( \frac{U}{\sqrt{2}} \hat{i} + \frac{U}{\sqrt{2}} \hat{j} \right) \times \left( \frac{B_0}{2} \hat{k} \right)$.
Calculating the cross product: $\hat{i} \times \hat{k} = -\hat{j}$ and $\hat{j} \times \hat{k} = \hat{i}$.
$\vec{F} = -e \left( \frac{U B_0}{2 \sqrt{2}} (-\hat{j}) + \frac{U B_0}{2 \sqrt{2}} (\hat{i}) \right) = \frac{-e U B_0}{2 \sqrt{2}} (\hat{i} - \hat{j})$.

(A) The kinetic energy of the electron is $K = 100 eV = 100 \times 1.6 \times 10^{-19} J = 1.6 \times 10^{-17} J$.
The radius of the circular path is $r = 10 cm = 0.1 m$.
The mass of the electron is $m = 0.5 MeV/c^2 = \frac{0.5 \times 10^6 \times 1.6 \times 10^{-19} J}{(3 \times 10^8 m/s)^2} \approx 8.89 \times 10^{-31} kg$.
The radius of the circular path of a charged particle in a magnetic field is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$.
Rearranging for $B$,we get $B = \frac{\sqrt{2mK}}{rq}$.
Substituting the values: $B = \frac{\sqrt{2 \times 8.89 \times 10^{-31} \times 1.6 \times 10^{-17}}}{0.1 \times 1.6 \times 10^{-19}}$.
$B = \frac{\sqrt{28.448 \times 10^{-48}}}{1.6 \times 10^{-20}} = \frac{5.33 \times 10^{-24}}{1.6 \times 10^{-20}} \approx 3.33 \times 10^{-4} T$.

(C) The radius of curvature $r$ of a charged particle moving in a uniform magnetic field $B$ is given by the formula:
$r = \frac{mv}{qB} = \frac{\sqrt{2m(K.E.)}}{qB}$
Since $m$,$q$,and $B$ are constant,we have $r \propto \sqrt{K.E.}$.
Let the initial kinetic energy be $K_1$ and the final kinetic energy be $K_2 = \frac{K_1}{2}$.
The ratio of the radii is given by:
$\frac{r_2}{r_1} = \sqrt{\frac{K_2}{K_1}} = \sqrt{\frac{K_1/2}{K_1}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Therefore,the radius of curvature is reduced to $\frac{1}{\sqrt{2}}$ times its initial value.

(B) Given: mass $m = 1 \times 10^{-27} \ kg$,charge $q = 1 \times 10^{-16} \ C$,speed $v = 1000 \ m/s$,angle $\theta = 60^{\circ}$,number of turns $N = 5000$,current $I = 5 \ A$.
The velocity components are:
$v_{\parallel} = v \cos 60^{\circ} = 1000 \times 0.5 = 500 \ m/s$ (along the axis)
$v_{\perp} = v \sin 60^{\circ} = 1000 \times \frac{\sqrt{3}}{2} = 500\sqrt{3} \ m/s$ (perpendicular to the axis)
The magnetic field inside the solenoid is $B = \mu_0 n I = \mu_0 (N/L) I$.
The time taken to travel the length $L$ of the solenoid is $t = \frac{L}{v_{\parallel}} = \frac{L}{500}$.
The time period of one revolution in the magnetic field is $T = \frac{2\pi m}{qB}$.
The number of revolutions $n'$ is given by $n' = \frac{t}{T} = \frac{L/v_{\parallel}}{2\pi m / qB} = \frac{L \cdot q \cdot B}{v_{\parallel} \cdot 2\pi m}$.
Substituting $B = \frac{\mu_0 N I}{L}$:
$n' = \frac{L \cdot q \cdot (\mu_0 N I / L)}{v_{\parallel} \cdot 2\pi m} = \frac{q \mu_0 N I}{v_{\parallel} \cdot 2\pi m}$.
Plugging in the values:
$n' = \frac{10^{-16} \times (4\pi \times 10^{-7}) \times 5000 \times 5}{500 \times 2\pi \times 10^{-27}}$
$n' = \frac{10^{-16} \times 4\pi \times 10^{-7} \times 25000}{1000\pi \times 10^{-27}}$
$n' = \frac{10^{-16} \times 10^{-7} \times 10^5}{10^{-27} \times 10^2} = \frac{10^{-18}}{10^{-25}} = 10^7 / 10 = 10^6$.
Thus,the number of revolutions is $1 \times 10^6$.

(D) The magnetic force on a charged particle moving in a magnetic field is given by the Lorentz force formula: $F = q(v \times B)$.
Given:
$q = 1.0 \times 10^{-16} \ C$
$v = (2 \hat{i} + 4 \hat{j}) \ ms^{-1}$
$B = B_0(\hat{i} + 4 \hat{j}) \ T$
$F = 3 \times 10^{-16} \hat{k} \ N$
Substituting these values into the formula:
$3 \times 10^{-16} \hat{k} = 1.0 \times 10^{-16} \times [(2 \hat{i} + 4 \hat{j}) \times B_0(\hat{i} + 4 \hat{j})]$
$3 \hat{k} = B_0 \times [2 \hat{i} \times \hat{i} + 8 \hat{i} \times \hat{j} + 4 \hat{j} \times \hat{i} + 16 \hat{j} \times \hat{j}]$
Using cross product rules $(\hat{i} \times \hat{i} = 0, \hat{j} \times \hat{j} = 0, \hat{i} \times \hat{j} = \hat{k}, \hat{j} \times \hat{i} = -\hat{k})$:
$3 \hat{k} = B_0 \times [0 + 8 \hat{k} - 4 \hat{k} + 0]$
$3 \hat{k} = B_0 \times (4 \hat{k})$
$4 B_0 = 3$
$B_0 = \frac{3}{4} = 0.75 \ T$.

(D) The frequency of oscillation of a magnetic needle in the Earth's magnetic field is given by $f = \frac{1}{2 \pi} \sqrt{\frac{\mu B_H}{I}}$,where $B_H = B \cos \theta$ is the horizontal component of the Earth's magnetic field,$\mu$ is the magnetic moment,and $I$ is the moment of inertia.
Since the same needle is used,$\mu$ and $I$ are constant. Thus,$f \propto \sqrt{B \cos \theta}$.
Given $f_1 = 20$ oscillations/min at $\theta_1 = 45^{\circ}$ and $f_2 = 30$ oscillations/min at $\theta_2 = 30^{\circ}$.
Taking the ratio: $\frac{f_1}{f_2} = \sqrt{\frac{B_1 \cos 45^{\circ}}{B_2 \cos 30^{\circ}}} \Rightarrow \frac{20}{30} = \sqrt{\frac{B_1 (1/\sqrt{2})}{B_2 (\sqrt{3}/2)}}$.
Squaring both sides: $\frac{4}{9} = \frac{B_1}{B_2} \times \frac{1}{\sqrt{2}} \times \frac{2}{\sqrt{3}} = \frac{B_1}{B_2} \times \frac{\sqrt{2}}{\sqrt{3}}$.
Therefore,$\frac{B_1}{B_2} = \frac{4}{9} \times \frac{\sqrt{3}}{\sqrt{2}} = \frac{2 \times 2 \times \sqrt{3}}{3 \times 3 \times \sqrt{2}} = \frac{2 \sqrt{2} \times \sqrt{2} \times \sqrt{3}}{3 \sqrt{3} \times \sqrt{3} \times \sqrt{2}} = \frac{2 \sqrt{2}}{3 \sqrt{3}}$.
Thus,$B_1: B_2 = 2 \sqrt{2}: 3 \sqrt{3}$.

(A) The angle of dip $\delta$ is the angle that the total magnetic field of the earth makes with the horizontal direction. Here,$\delta = 30^{\circ}$.
Given,the horizontal component of the earth's magnetic field is $B_H = 0.3 \ G$.
The relationship between the total magnetic field $B$,the horizontal component $B_H$,and the angle of dip $\delta$ is given by $B_H = B \cos \delta$.
Substituting the given values: $0.3 = B \cos 30^{\circ}$.
Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,we have $0.3 = B \times \frac{\sqrt{3}}{2}$.
Solving for $B$: $B = \frac{0.3 \times 2}{\sqrt{3}} = \frac{0.6}{\sqrt{3}} = \frac{0.6 \times \sqrt{3}}{3} = 0.2 \sqrt{3} \ G$.
Alternatively,$B = \frac{0.6}{\sqrt{3}} = \frac{6}{10 \sqrt{3}} = \frac{3}{5 \sqrt{3}} = \frac{\sqrt{3}}{5} \ G$.

(A) The time period of a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MB_{H}}}$, where $I$ is the moment of inertia, $M$ is the magnetic moment, and $B_{H}$ is the horizontal component of the earth's magnetic field.
From the formula, we see that $T \propto \frac{1}{\sqrt{B_{H}}}$.
Given that $T_{A} = 2T_{B}$, we can write the ratio as $\frac{T_{A}}{T_{B}} = 2$.
Using the proportionality, $\frac{T_{A}}{T_{B}} = \sqrt{\frac{B_{HB}}{B_{HA}}}$.
Substituting the values, $2 = \sqrt{\frac{32 \times 10^{-6}}{B_{HA}}}$.
Squaring both sides, $4 = \frac{32 \times 10^{-6}}{B_{HA}}$.
Therefore, $B_{HA} = \frac{32 \times 10^{-6}}{4} = 8 \times 10^{-6} \,T$.

(C) Each magnet has a magnetic moment vector $\vec{M_1}$ and $\vec{M_2}$ with magnitude $M = ml$.
Since the magnets are placed at right angles to each other,the angle between their magnetic moment vectors is $90^{\circ}$.
The resultant magnetic moment of the system is given by the vector sum $\vec{M_{net}} = \vec{M_1} + \vec{M_2}$.
The magnitude of the resultant magnetic moment is $M_{net} = \sqrt{M_1^2 + M_2^2 + 2M_1M_2 \cos(90^{\circ})}$.
Since $\cos(90^{\circ}) = 0$,we have $M_{net} = \sqrt{M^2 + M^2} = \sqrt{2M^2} = \sqrt{2}M$.
Substituting $M = ml$,we get $M_{net} = \sqrt{2} ml$.

(A) The torque $\tau$ experienced by a magnetic dipole in a uniform magnetic field $B$ is given by the formula: $\tau = MB \sin \theta$.
Given:
Torque $\tau = 3.6 \times 10^{-5} \,J$
Magnetic field $B = 28.3 \times 10^{-3} \,T$
Angle $\theta = 45^{\circ}$
Rearranging the formula to solve for the magnetic moment $M$:
$M = \frac{\tau}{B \sin \theta}$
Substituting the values:
$M = \frac{3.6 \times 10^{-5}}{28.3 \times 10^{-3} \times \sin 45^{\circ}}$
Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}} \approx 0.707$:
$M = \frac{3.6 \times 10^{-5}}{28.3 \times 10^{-3} \times 0.707} \approx \frac{3.6 \times 10^{-5}}{20.008 \times 10^{-3}} \approx 0.1799 \times 10^{-2} \approx 1.8 \times 10^{-3} \,J \,T^{-1}$.

(B) At the null point on the axis,the magnetic field of the bar magnet $(B_{axis})$ is equal and opposite to the Earth's horizontal magnetic field $(B)$.
$B_{axis} = \frac{\mu_0}{4 \pi} \frac{2M}{d^3} = B$
On the equatorial line (normal bisector) at the same distance $d$,the magnetic field of the bar magnet $(B_{eq})$ is given by:
$B_{eq} = \frac{\mu_0}{4 \pi} \frac{M}{d^3} = \frac{B_{axis}}{2} = \frac{B}{2}$
Since the angle of dip is $0^{\circ}$,the Earth's magnetic field is entirely horizontal. On the equatorial line,the magnetic field of the magnet is in the same direction as the Earth's magnetic field.
Therefore,the total magnetic field $(B_{total})$ at this point is:
$B_{total} = B_{eq} + B = \frac{B}{2} + B = \frac{3B}{2}$
Given that $B_{total} = 0.6 \ G$,we have:
$0.6 = \frac{3B}{2}$
$B = \frac{0.6 \times 2}{3} = 0.4 \ G$

(C) The magnetic field produced by a short bar magnet on its axis $(B_{\text{axis}})$ and on its equatorial line $(B_{\text{equator}})$ at a distance $r$ from the centre is given by:
$B_{\text{axis}} = \frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3}$
$B_{\text{equator}} = \frac{\mu_0}{4\pi} \cdot \frac{M}{r^3}$
Thus,$B_{\text{axis}} = 2 \times B_{\text{equator}}$ for the same distance $r$.
Given:
$B_{\text{equator}} = 6.4 \times 10^{-5} \,T$ at $r_2 = 20 \,cm = 0.2 \,m$
We need to find $B_{\text{axis}}$ at $r_1 = 40 \,cm = 0.4 \,m$.
Using the general formula $B \propto \frac{1}{r^3}$:
$\frac{B_{\text{axis}}(r_1)}{B_{\text{equator}}(r_2)} = \frac{\frac{\mu_0}{4\pi} \cdot \frac{2M}{r_1^3}}{\frac{\mu_0}{4\pi} \cdot \frac{M}{r_2^3}} = 2 \times \left(\frac{r_2}{r_1}\right)^3$
$B_{\text{axis}} = 2 \times B_{\text{equator}} \times \left(\frac{20}{40}\right)^3$
$B_{\text{axis}} = 2 \times (6.4 \times 10^{-5}) \times \left(\frac{1}{2}\right)^3$
$B_{\text{axis}} = 2 \times (6.4 \times 10^{-5}) \times \frac{1}{8}$
$B_{\text{axis}} = \frac{6.4 \times 10^{-5}}{4} = 1.6 \times 10^{-5} \,T$

(B) When a bar magnet is cut perpendicular to its length,each piece becomes a new,smaller bar magnet.
Specifically,the cut creates a new South pole $(S)$ on the left piece and a new North pole $(N)$ on the right piece at the point of the cut.
As shown in the figure,the left piece will have its original North pole $(N)$ on the left and a new South pole $(S)$ on the right.
The right piece will have a new North pole $(N)$ on the left and its original South pole $(S)$ on the right.
When these two pieces are held close to each other,the new South pole $(S)$ of the left piece faces the new North pole $(N)$ of the right piece.
Since opposite magnetic poles attract each other,the two pieces will attract each other.

(A) The relationship between the radius $R$ of a nucleus and its mass number $A$ is given by the formula:
$R = R_0 A^{1/3} \dots(1)$
Taking the natural logarithm on both sides:
$\ln(R) = \ln(R_0 A^{1/3}) = \ln(R_0) + \frac{1}{3} \ln(A)$
Rearranging the terms to isolate the ratio:
$\ln(R) - \ln(R_0) = \frac{1}{3} \ln(A)$
$\ln \left(\frac{R}{R_0}\right) = \frac{1}{3} \ln(A)$
This equation is of the form $y = mx$,where $y = \ln \left(\frac{R}{R_0}\right)$,$x = \ln(A)$,and the slope $m = \frac{1}{3}$.
Since this represents a linear equation passing through the origin,the graph is a straight line.

(D) The radius $R$ of an atomic nucleus is related to its mass number $A$ by the formula $R = R_0 A^{1/3}$,where $R_0$ is a constant.
This implies that $R \propto A^{1/3}$ or $A \propto R^3$.
Given for the first nucleus: $A_1 = 64$ and $R_1 = 4.8 \text{ fermi}$.
For the second nucleus: $R_2 = 6 \text{ fermi}$ and we need to find $A_2$.
Using the ratio: $\frac{A_2}{A_1} = \left(\frac{R_2}{R_1}\right)^3$.
Substituting the values: $\frac{A_2}{64} = \left(\frac{6}{4.8}\right)^3$.
Simplifying the fraction: $\frac{6}{4.8} = \frac{60}{48} = \frac{5}{4} = 1.25$.
Therefore,$A_2 = 64 \times (1.25)^3 = 64 \times \frac{125}{64} = 125$.