TS EAMCET 2024 Physics Question Paper with Answer and Solution

(C) Let the mass per unit area of the plate be $\sigma$.
Mass of the original plate $M = \sigma \pi (4r)^2 = 16 \sigma \pi r^2$.
Mass of the removed circular part $m = \sigma \pi r^2$.
The mass of the remaining portion is $M' = M - m = 16 \sigma \pi r^2 - \sigma \pi r^2 = 15 \sigma \pi r^2$.
Let the centre of the original plate be at the origin $(0, 0)$.
The centre of the removed part is at $(2r, 0)$.
Using the principle of centre of mass for a cavity: $M X_{CM} = M' X_{R} + m X_{C}$,where $X_{CM}$ is the centre of mass of the original plate (which is $0$),$X_{R}$ is the centre of mass of the remaining portion,and $X_{C}$ is the centre of mass of the removed part.
$0 = (15 \sigma \pi r^2) X_{R} + (\sigma \pi r^2)(2r)$.
$15 X_{R} = -2r$.
The magnitude of the distance is $|X_{R}| = \frac{2r}{15}$.

(A) Let the vertical plate be $1$ and the horizontal plate be $2$. Both have area $A = L \times a$.
For the initial position:
The centre of mass of the vertical plate is at $y_1 = L/2$.
The centre of mass of the horizontal plate is at $y_2 = L + a/2$.
The height of the centre of mass from the surface is:
$y_{cm} = \frac{A_1 y_1 + A_2 y_2}{A_1 + A_2} = \frac{(La)(L/2) + (La)(L + a/2)}{2La} = \frac{L/2 + L + a/2}{2} = \frac{3L + a}{4}$.
When the alphabet is vertically inverted,the horizontal plate is now at the bottom:
The centre of mass of the new horizontal plate is at $y_1' = a/2$.
The centre of mass of the new vertical plate is at $y_2' = a + L/2$.
The new height of the centre of mass from the surface is:
$y_{cm}' = \frac{A_1 y_1' + A_2 y_2'}{A_1 + A_2} = \frac{(La)(a/2) + (La)(a + L/2)}{2La} = \frac{a/2 + a + L/2}{2} = \frac{3a + L}{4}$.
The shift in the centre of mass is:
$\Delta y_{cm} = y_{cm} - y_{cm}' = \frac{3L + a}{4} - \frac{3a + L}{4} = \frac{2L - 2a}{4} = \frac{L - a}{2}$.

(D) Let the origin be at the centre of the square. The initial centre of mass $(CM_1)$ of the four particles is at the origin $(0, 0)$.
When one particle of mass $m$ at corner $C$ is removed,the remaining system consists of three particles of mass $m$ at corners $A, B,$ and $D$.
The new centre of mass $(CM_2)$ will shift towards the centroid of the triangle formed by the remaining three particles.
The distance of the centre of the square from any corner is $r = \frac{a}{\sqrt{2}}$.
Using the formula for the centre of mass of the remaining system: $R_{CM} = \frac{\sum m_i r_i}{\sum m_i}$.
Since the mass at $C$ is removed,we can treat this as a system of four particles with one mass being $-m$ at $C$ and the original system having total mass $4m$ at the origin.
The shift in the centre of mass is given by $\Delta R = \frac{|m_C \cdot r_C|}{M_{remaining}} = \frac{m \cdot (a/\sqrt{2})}{3m} = \frac{a}{3\sqrt{2}}$.

(B) Given masses are $m_1 = m$, $m_2 = 2m$, and $m_3 = 3m$.
The velocities are directed as follows:
Particle $A$ moves north: $\vec{v}_1 = 6 \hat{j} \,ms^{-1}$
Particle $B$ moves south: $\vec{v}_2 = -12 \hat{j} \,ms^{-1}$
Particle $C$ moves east: $\vec{v}_3 = 8 \hat{i} \,ms^{-1}$
The velocity of the centre of mass $\vec{v}_{cm}$ is given by:
$\vec{v}_{cm} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2 + m_3 \vec{v}_3}{m_1 + m_2 + m_3}$
Substituting the values:
$\vec{v}_{cm} = \frac{m(6 \hat{j}) + 2m(-12 \hat{j}) + 3m(8 \hat{i})}{m + 2m + 3m}$
$\vec{v}_{cm} = \frac{6m \hat{j} - 24m \hat{j} + 24m \hat{i}}{6m}$
$\vec{v}_{cm} = \frac{24m \hat{i} - 18m \hat{j}}{6m} = 4 \hat{i} - 3 \hat{j} \,ms^{-1}$
The magnitude of the velocity of the centre of mass is:
$v_{cm} = \sqrt{(4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \,ms^{-1}$

(A) Given: $m_1 = 0.5 \ kg$,$u_1 = 10 \ ms^{-1}$,$m_2 = 1 \ kg$,$u_2 = 0$,$e = 0.4$.
The final velocities $v_1$ and $v_2$ after a one-dimensional collision are given by:
$v_1 = \left( \frac{m_1 - e m_2}{m_1 + m_2} \right) u_1 = \left( \frac{0.5 - 0.4 \times 1}{0.5 + 1} \right) \times 10 = \left( \frac{0.1}{1.5} \right) \times 10 = \frac{1}{15} \times 10 = \frac{2}{3} \ ms^{-1}$.
$v_2 = \frac{(1 + e) m_1 u_1}{m_1 + m_2} = \frac{(1 + 0.4) \times 0.5 \times 10}{0.5 + 1} = \frac{1.4 \times 5}{1.5} = \frac{7}{1.5} = \frac{14}{3} \ ms^{-1}$.
The ratio of the velocities is $\frac{v_1}{v_2} = \frac{2/3}{14/3} = \frac{2}{14} = 1:7$.

(A) Given: $m_1 = m_2 = m = 1.2 \ kg$,$u_1 = 12 \ ms^{-1}$,$u_2 = 0$,$e = \frac{1}{\sqrt{2}}$.
By the law of conservation of linear momentum:
$m u_1 + m u_2 = m v_1 + m v_2$
$12 + 0 = v_1 + v_2 \Rightarrow v_1 + v_2 = 12$ ...$(i)$
By the definition of the coefficient of restitution $(e)$:
$e = \frac{v_2 - v_1}{u_1 - u_2} \Rightarrow \frac{1}{\sqrt{2}} = \frac{v_2 - v_1}{12}$
$v_2 - v_1 = \frac{12}{\sqrt{2}} = 6\sqrt{2}$ ...(ii)
Adding equations $(i)$ and (ii):
$2v_2 = 12 + 6\sqrt{2} \Rightarrow v_2 = 6 + 3\sqrt{2} \ ms^{-1}$
Subtracting equation (ii) from $(i)$:
$2v_1 = 12 - 6\sqrt{2} \Rightarrow v_1 = 6 - 3\sqrt{2} \ ms^{-1}$
Initial kinetic energy $(KE)_i = \frac{1}{2} m u_1^2 = \frac{1}{2} \times 1.2 \times (12)^2 = 0.6 \times 144 = 86.4 \ J$.
Final kinetic energy $(KE)_f = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 = \frac{1}{2} m (v_1^2 + v_2^2)$.
$v_1^2 + v_2^2 = (6 - 3\sqrt{2})^2 + (6 + 3\sqrt{2})^2 = (36 + 18 - 36\sqrt{2}) + (36 + 18 + 36\sqrt{2}) = 54 + 54 = 108$.
$(KE)_f = \frac{1}{2} \times 1.2 \times 108 = 0.6 \times 108 = 64.8 \ J$.
Ratio $\frac{(KE)_f}{(KE)_i} = \frac{64.8}{86.4} = \frac{648}{864} = \frac{3}{4} = 3:4$.

(C) Let the initial velocity of sphere $A$ be $u$ and the final velocities of spheres $A$ and $B$ be $v_1$ and $v_2$ respectively.
By the law of conservation of linear momentum:
$m u + (2m)(0) = m v_1 + 2m v_2$
$u = v_1 + 2v_2$ ....$(i)$
Given the coefficient of restitution $e = 0.4 = \frac{2}{5}$.
Using the definition of the coefficient of restitution:
$e = \frac{v_2 - v_1}{u_1 - u_2} = \frac{v_2 - v_1}{u - 0} = 0.4$
$v_2 - v_1 = 0.4u$ ....$(ii)$
From equation $(i)$,$u = v_1 + 2v_2$. Substitute this into equation $(ii)$:
$v_2 - v_1 = 0.4(v_1 + 2v_2)$
$v_2 - v_1 = 0.4v_1 + 0.8v_2$
$0.2v_2 = 1.4v_1$
$\frac{v_1}{v_2} = \frac{0.2}{1.4} = \frac{1}{7}$

(D) The gravitational potential energy of a body of mass $m$ at a distance $r$ from the center of the earth is given by $U = -\frac{GMm}{r}$.
At the surface of the earth,$r = R$,so $U_i = -\frac{GMm}{R}$.
For a height $h = R$,the distance from the center is $r = R + R = 2R$. The potential energy is $U_f = -\frac{GMm}{2R}$.
The energy required $W$ is the change in potential energy:
$W = U_f - U_i = -\frac{GMm}{2R} - (-\frac{GMm}{R}) = \frac{GMm}{2R}$.
For a height $h = 2R$,the distance from the center is $r = R + 2R = 3R$. The potential energy is $U_f' = -\frac{GMm}{3R}$.
The energy required $W'$ is:
$W' = U_f' - U_i = -\frac{GMm}{3R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{3R} = \frac{2GMm}{3R}$.
Since $W = \frac{GMm}{2R}$,we have $\frac{GMm}{R} = 2W$.
Substituting this into the expression for $W'$:
$W' = \frac{2}{3} \times (2W) = \frac{4W}{3}$.

(A) The acceleration due to gravity at a height $h$ above the surface of the earth is given by the formula: $g' = g \left( \frac{R}{R+h} \right)^2$,where $g$ is the acceleration due to gravity at the surface and $R$ is the radius of the earth.
Given $R = 6400 \ km$.
For $h_1 = 1280 \ km$:
$g_1 = g \left( \frac{6400}{6400 + 1280} \right)^2 = g \left( \frac{6400}{7680} \right)^2 = g \left( \frac{5}{6} \right)^2 = \frac{25}{36} g$.
For $h_2 = 3200 \ km$:
$g_2 = g \left( \frac{6400}{6400 + 3200} \right)^2 = g \left( \frac{6400}{9600} \right)^2 = g \left( \frac{2}{3} \right)^2 = \frac{4}{9} g$.
Now,the ratio $\frac{g_1}{g_2}$ is:
$\frac{g_1}{g_2} = \frac{25/36 g}{4/9 g} = \frac{25}{36} \times \frac{9}{4} = \frac{25}{16}$.
Thus,the ratio is $25:16$.

(C) Among the four fundamental forces in nature (gravitational,electromagnetic,strong nuclear,and weak nuclear forces),the gravitational force is unique because it is always attractive in nature.
Electromagnetic forces can be either attractive or repulsive depending on the charges.
Strong nuclear forces are generally attractive at short ranges to hold the nucleus together.
Therefore,the statement that gravitational forces are always attractive is correct.

(D) The acceleration due to gravity $g$ on the surface of a planet is given by the formula: $g = \frac{4}{3} \pi \rho G R$, where $\rho$ is the mean density and $R$ is the radius of the planet.
From this relation, we see that $g \propto \rho R$.
Given for the planet $(1)$ and the earth $(2)$:
Ratio of radii: $\frac{R_1}{R_2} = \frac{1}{2}$
Ratio of densities: $\frac{\rho_1}{\rho_2} = \frac{4}{1}$
Acceleration due to gravity on earth: $g_2 = 9.8 \,ms^{-2}$.
Using the proportionality $g_1 / g_2 = (\rho_1 / \rho_2) \times (R_1 / R_2)$:
$\frac{g_1}{g_2} = \frac{4}{1} \times \frac{1}{2} = 2$.
Therefore, $g_1 = 2 \times g_2 = 2 \times 9.8 \,ms^{-2} = 19.6 \,ms^{-2}$.

(D) The two stars revolve around their common center of mass $(COM)$ with the same angular velocity $\omega$.
Let $r_1$ and $r_2$ be the distances of masses $M$ and $2M$ from the $COM$,respectively.
We know that $r_1 + r_2 = d$ and $M r_1 = (2M) r_2$.
From these,$r_1 = \frac{2M}{3M} d = \frac{2}{3} d$.
The gravitational force between the stars provides the necessary centripetal force for the star of mass $M$:
$F_g = F_c$
$\frac{G(M)(2M)}{d^2} = M \omega^2 r_1$
Substituting $r_1 = \frac{2}{3} d$:
$\frac{2 G M^2}{d^2} = M \omega^2 \left(\frac{2}{3} d\right)$
$\frac{2 G M}{d^2} = \omega^2 \left(\frac{2}{3} d\right)$
$\omega^2 = \frac{2 G M}{d^2} \cdot \frac{3}{2 d} = \frac{3 G M}{d^3}$
$\omega = \sqrt{\frac{3 G M}{d^3}}$

(C) The average kinetic energy per molecule of an ideal gas is given by the formula $\langle KE \rangle = \frac{f}{2} K_B T$,where $f$ is the degrees of freedom,$K_B$ is the Boltzmann constant,and $T$ is the absolute temperature.
Since both hydrogen and nitrogen gases are in the same vessel at thermal equilibrium,they are at the same temperature $T = 30^{\circ} C$.
For both diatomic gases (hydrogen and nitrogen),the number of degrees of freedom $f$ is $5$ (at moderate temperatures).
Therefore,the average kinetic energy per molecule for hydrogen is $\langle KE \rangle_{H_2} = \frac{5}{2} K_B T$.
Similarly,the average kinetic energy per molecule for nitrogen is $\langle KE \rangle_{N_2} = \frac{5}{2} K_B T$.
The ratio of the average kinetic energies per molecule is $\frac{\langle KE \rangle_{H_2}}{\langle KE \rangle_{N_2}} = \frac{\frac{5}{2} K_B T}{\frac{5}{2} K_B T} = 1: 1$.
The mass ratio is irrelevant because the average kinetic energy per molecule depends only on the temperature.

(D) The root mean square (rms) speed of gas molecules is given by the formula: $V_{rms} = \sqrt{\frac{3RT}{M}}$.
From this relation,we can see that $V_{rms} \propto \sqrt{T}$.
Initial temperature $T_i = 27^{\circ} C = 27 + 273 = 300 \ K$.
Final temperature $T_f = 159^{\circ} C = 159 + 273 = 432 \ K$.
The percentage increase in rms speed is given by: $\frac{V_{rms,f} - V_{rms,i}}{V_{rms,i}} \times 100$.
Substituting the proportionality $V_{rms} \propto \sqrt{T}$,we get: $\left( \frac{\sqrt{T_f} - \sqrt{T_i}}{\sqrt{T_i}} \right) \times 100$.
$= \left( \frac{\sqrt{432} - \sqrt{300}}{\sqrt{300}} \right) \times 100$.
$= \left( \frac{20.784 - 17.320}{17.320} \right) \times 100$.
$= \left( \frac{3.464}{17.320} \right) \times 100 \approx 20 \%$.

(A) The $RMS$ speed is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \sqrt{\frac{T}{M}}$.
Given that $v_{rms, O} = 0.75 \times v_{rms, N} = \frac{3}{4} v_{rms, N}$.
Using the ratio: $\frac{v_{rms, O}}{v_{rms, N}} = \sqrt{\frac{T_O}{T_N} \times \frac{M_N}{M_O}}$.
Here,$T_N = 287 + 273 = 560 \ K$,$M_N = 28 \ g/mol$,and $M_O = 32 \ g/mol$.
Substituting the values: $\frac{3}{4} = \sqrt{\frac{T_O}{560} \times \frac{28}{32}}$.
Squaring both sides: $\frac{9}{16} = \frac{T_O}{560} \times \frac{7}{8}$.
$T_O = \frac{9}{16} \times \frac{8}{7} \times 560 = \frac{9}{2} \times \frac{560}{7} = 9 \times 40 = 360 \ K$.
Converting to Celsius: $T_O = 360 - 273 = 87^{\circ} C$.

(D) To prevent the truck from toppling, the torque due to the centrifugal force about the outer wheels must not exceed the torque due to the weight of the truck about the same point.
Let $V_{\max}$ be the maximum speed.
The condition for the truck not to topple is given by the balance of torques: $\frac{m V_{\max}^2}{R} \times h = m g \times \frac{d}{2}$.
Here, $h = 2 \text{ m}$ (height of center of gravity), $R = 240 \text{ m}$ (radius of path), and $d = 1.5 \text{ m}$ (distance between wheels).
Rearranging for $V_{\max}^2$: $V_{\max}^2 = \frac{g \times R \times d}{2 \times h}$.
Substituting the values: $V_{\max}^2 = \frac{10 \times 240 \times 1.5}{2 \times 2}$.
$V_{\max}^2 = \frac{3600}{4} = 900$.
$V_{\max} = \sqrt{900} = 30 \text{ ms}^{-1}$.

(D) The maximum speed $V_{\max}$ on a banked road with friction is given by $V_{\max} = \sqrt{\frac{rg(\mu + \tan \theta)}{1 - \mu \tan \theta}}$.
The optimum speed $V_o$ (where no friction is required) is given by $V_o = \sqrt{rg \tan \theta}$.
Given $\theta = 45^{\circ}$ and $V_{\max} = 2V_o$.
Substituting the expressions: $\sqrt{\frac{rg(\mu + \tan \theta)}{1 - \mu \tan \theta}} = 2 \sqrt{rg \tan \theta}$.
Squaring both sides: $\frac{\mu + \tan \theta}{1 - \mu \tan \theta} = 4 \tan \theta$.
Since $\tan 45^{\circ} = 1$,we have $\frac{\mu + 1}{1 - \mu} = 4(1)$.
$\mu + 1 = 4 - 4\mu$.
$5\mu = 3$.
$\mu = \frac{3}{5} = 0.6$.

(C) The net force acting on the block is given by $F_{net} = F - f_k = ma$,where $f_k = \mu_k mg$ is the kinetic friction force.
Thus,the acceleration is $a = \frac{F - \mu_k mg}{m}$.
For the first case: $6 = \frac{20 - \mu_k mg}{m} \implies 6m = 20 - \mu_k mg$ --- $(i)$
For the second case: $11 = \frac{30 - \mu_k mg}{m} \implies 11m = 30 - \mu_k mg$ --- (ii)
Subtracting equation $(i)$ from equation (ii):
$(11m - 6m) = (30 - \mu_k mg) - (20 - \mu_k mg)$
$5m = 10 \implies m = 2 \ kg$.
Substitute $m = 2 \ kg$ into equation $(i)$:
$6(2) = 20 - \mu_k (2)(10)$
$12 = 20 - 20\mu_k$
$20\mu_k = 8$
$\mu_k = \frac{8}{20} = 0.4$.

(C) Given: Applied force $F_{\text{app}} = 5 \ N$,mass $m = 0.5 \ kg$,coefficient of kinetic friction $\mu_k = 0.2$,time $t = 4 \ s$,initial velocity $u = 0$,acceleration due to gravity $g = 10 \ m/s^2$.
The kinetic frictional force is given by $f_k = \mu_k N = \mu_k mg$.
$f_k = 0.2 \times 0.5 \times 10 = 1 \ N$.
The net force acting on the block is $F_{\text{net}} = F_{\text{app}} - f_k = 5 - 1 = 4 \ N$.
The acceleration of the block is $a = \frac{F_{\text{net}}}{m} = \frac{4}{0.5} = 8 \ m/s^2$.
The velocity of the block after $4 \ s$ is $v = u + at = 0 + (8 \times 4) = 32 \ m/s$.
The kinetic energy of the block is $K.E. = \frac{1}{2} mv^2 = \frac{1}{2} \times 0.5 \times (32)^2 = 0.25 \times 1024 = 256 \ J$.

(D) Given,current $I = (10 \pm 0.5) \text{ A}$ and potential difference $V = (100 \pm 6) \text{ V}$.
According to Ohm's law,$R = \frac{V}{I}$.
The relative error in resistance is given by $\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$.
To find the percentage error,we multiply by $100$:
$\frac{\Delta R}{R} \times 100 = \left( \frac{\Delta V}{V} \times 100 \right) + \left( \frac{\Delta I}{I} \times 100 \right)$.
Substituting the given values:
$\frac{\Delta R}{R} \times 100 = \left( \frac{6}{100} \times 100 \right) + \left( \frac{0.5}{10} \times 100 \right)$.
$\frac{\Delta R}{R} \times 100 = 6\% + 5\% = 11\%$.

(D) Given:
Internal diameter $d = (5.73 \pm 0.01) \text{ cm}$
External diameter $D = (6.01 \pm 0.01) \text{ cm}$
The thickness $t$ of the cylinder wall is given by $t = \frac{D - d}{2}$.
First,calculate the mean value of the thickness:
$t_{mean} = \frac{6.01 - 5.73}{2} = \frac{0.28}{2} = 0.14 \text{ cm}$.
Next,calculate the uncertainty in the thickness:
When subtracting two quantities,the absolute errors are added. Thus,the error in $(D - d)$ is $\Delta D + \Delta d = 0.01 + 0.01 = 0.02 \text{ cm}$.
Since $t = \frac{D - d}{2}$,the error in $t$ is $\Delta t = \frac{\Delta D + \Delta d}{2} = \frac{0.02}{2} = 0.01 \text{ cm}$.
Therefore,the thickness is $(0.14 \pm 0.01) \text{ cm}$.

(D) Given: $C = (4.0 \pm 0.2) \mu F$ and $V = (10.0 \pm 0.1) V$.
The charge $Q$ on the capacitor is given by $Q = C \times V$.
Calculating the mean value: $Q = 4.0 \mu F \times 10.0 V = 40 \mu C$.
The relative error in charge is given by $\frac{\Delta Q}{Q} = \frac{\Delta C}{C} + \frac{\Delta V}{V}$.
Substituting the values: $\frac{\Delta Q}{Q} = \frac{0.2}{4.0} + \frac{0.1}{10.0} = 0.05 + 0.01 = 0.06$.
The percentage error in charge is $\frac{\Delta Q}{Q} \times 100 = 0.06 \times 100 = 6 \%$.
Therefore,the charge on the capacitor is $40 \mu C \pm 6 \%$.

(D) Given: Side of the cube $a = 10 \ cm$,so the area of each face $A = a^2 = 100 \ cm^2$.
Depth of the cube in water $x = 3 \ cm$.
Depth of the cube in oil $h = a - x = 10 - 3 = 7 \ cm$.
Density of oil $\delta_1 = 0.9 \ g \ cm^{-3}$.
Density of water $\delta_2 = 1 \ g \ cm^{-3}$.
By the law of flotation,the weight of the cube is equal to the total buoyant force exerted by the two liquids.
$W = F_{B1} + F_{B2}$
$mg = (\delta_1 \cdot V_{oil} \cdot g) + (\delta_2 \cdot V_{water} \cdot g)$
$m = \delta_1 \cdot A \cdot h + \delta_2 \cdot A \cdot x$
$m = 0.9 \times 100 \times 7 + 1 \times 100 \times 3$
$m = 630 + 300 = 930 \ g$
Therefore,the mass of the wooden cube is $930 \ g$.

(B) The velocity of efflux $(v)$ from a hole at the bottom of a tank is given by Torricelli's law: $v = \sqrt{2gh}$.
Given:
Height of water level,$h = 5 \,m$.
Area of the hole,$A = 2.4 \,mm^2 = 2.4 \times 10^{-6} \,m^2$.
Time,$t = 5 \,s$.
Acceleration due to gravity,$g = 10 \,ms^{-2}$.
The volume of water leaked $(V)$ is given by the product of the area of the hole,the velocity of efflux,and the time:
$V = A \times v \times t = A \sqrt{2gh} \times t$.
Substituting the values:
$V = (2.4 \times 10^{-6} \,m^2) \times \sqrt{2 \times 10 \,ms^{-2} \times 5 \,m} \times 5 \,s$.
$V = 2.4 \times 10^{-6} \times \sqrt{100} \times 5$.
$V = 2.4 \times 10^{-6} \times 10 \times 5$.
$V = 2.4 \times 50 \times 10^{-6} \,m^3$.
$V = 120 \times 10^{-6} \,m^3$.

(D) Given, the rate of flow of water, $Q = 12 \pi \text{ litre/minute}$.
Converting this to $SI$ units $(m^3/s)$:
$Q = \frac{12 \pi \times 10^{-3} \text{ m}^3}{60 \text{ s}} = 0.2 \pi \times 10^{-3} \text{ m}^3/s = 2 \pi \times 10^{-4} \text{ m}^3/s$.
The diameter of the pipe is $d = 2 \text{ cm} = 0.02 \text{ m}$, so the radius is $r = 0.01 \text{ m} = 10^{-2} \text{ m}$.
The cross-sectional area $A$ is given by $A = \pi r^2 = \pi (10^{-2})^2 = \pi \times 10^{-4} \text{ m}^2$.
Using the equation of continuity, $Q = A \times v$, where $v$ is the velocity of water:
$2 \pi \times 10^{-4} = (\pi \times 10^{-4}) \times v$.
Solving for $v$:
$v = \frac{2 \pi \times 10^{-4}}{\pi \times 10^{-4}} = 2 \text{ m/s}$.

(D) The pressure $P$ at the bottom of a vessel filled with a liquid of density $\rho$ to a height $h$ is given by the formula: $P = \rho g h$.
Since the height $h$ is the same for all three vessels and $g$ is constant,the pressure $P$ is directly proportional to the density $\rho$ of the liquid $(P \propto \rho)$.
Given the densities are $\rho_A > \rho_B > \rho_C$,it follows that the pressure at the bottom of the vessels will be $P_A > P_B > P_C$.
Therefore,the pressure is maximum in the vessel containing liquid $A$.

(C) soap bubble has two surfaces (inner and outer),so the change in surface area is $\Delta A = 2 \times (4\pi r_2^2 - 4\pi r_1^2) = 8\pi(r_2^2 - r_1^2)$.
Given: Surface tension $T = 3.5 \times 10^{-2} \ N/m$,initial radius $r_1 = 1 \ cm = 1 \times 10^{-2} \ m$,final radius $r_2 = 2 \ cm = 2 \times 10^{-2} \ m$.
Work done $W = T \times \Delta A = T \times 8\pi(r_2^2 - r_1^2)$.
$W = 3.5 \times 10^{-2} \times 8 \times 3.14 \times [(2 \times 10^{-2})^2 - (1 \times 10^{-2})^2]$.
$W = 3.5 \times 10^{-2} \times 8 \times 3.14 \times (4 \times 10^{-4} - 1 \times 10^{-4})$.
$W = 3.5 \times 10^{-2} \times 8 \times 3.14 \times 3 \times 10^{-4}$.
$W = 263.76 \times 10^{-6} \ J \approx 264 \times 10^{-6} \ J$.

(C) Given:
Radius of soap bubble,$R = 0.5 \ cm = 0.5 \times 10^{-2} \ m$
Height of oil column,$h = 4 \ mm = 4 \times 10^{-3} \ m$
Density of the oil,$\rho = 900 \ kg \ m^{-3}$
Acceleration due to gravity,$g = 10 \ m \ s^{-2}$
Excess pressure inside a soap bubble is given by $P = \frac{4S}{R}$,where $S$ is the surface tension.
The pressure exerted by the oil column is $P' = \rho g h$.
According to the problem,the excess pressure is balanced by the oil column pressure:
$\frac{4S}{R} = \rho g h$
Substituting the values:
$\frac{4S}{0.5 \times 10^{-2}} = 900 \times 10 \times 4 \times 10^{-3}$
$\frac{4S}{0.5 \times 10^{-2}} = 36$
$4S = 36 \times 0.5 \times 10^{-2}$
$4S = 18 \times 10^{-2}$
$S = 4.5 \times 10^{-2} \ N \ m^{-1}$

(D) The work done in blowing a soap bubble is given by $W = T \cdot \Delta A$,where $T$ is the surface tension and $\Delta A$ is the change in surface area. For a soap bubble,the surface area is $A = 2 \times (4 \pi r^2) = 8 \pi r^2$. Thus,$W \propto r^2$.
Since the volume of the bubble is $V = \frac{4}{3} \pi r^3$,we have $r \propto V^{1/3}$.
Substituting this into the work relation: $W \propto (V^{1/3})^2 = V^{2/3}$.
Let $W_1 = W$ for volume $V_1 = V$,and $W_2$ be the work for volume $V_2 = 2V$.
Then,$\frac{W_2}{W_1} = \left( \frac{V_2}{V_1} \right)^{2/3} = \left( \frac{2V}{V} \right)^{2/3} = (2)^{2/3}$.
Therefore,$W_2 = (2)^{2/3} W = (2^2)^{1/3} W = (4)^{1/3} W$.

(C) When a large liquid drop of radius $R$ splits into $n$ smaller drops of radius $r$,the volume remains constant: $\frac{4}{3} \pi R^3 = n \times \frac{4}{3} \pi r^3$,which implies $R = n^{1/3} r$.
Since $n > 1$,the total surface area of the $n$ small drops $(A_{final} = n \times 4 \pi r^2)$ is greater than the surface area of the original large drop $(A_{initial} = 4 \pi R^2)$.
Surface energy is directly proportional to the surface area $(U = T \times A)$.
Since the total surface area increases,the system must absorb energy from the surroundings to perform the work required to increase the surface area.

(A) The formula for Young's modulus is $Y = \frac{F \cdot L}{A \cdot \Delta L}$,which gives elongation $\Delta L = \frac{F \cdot L}{Y \cdot A}$.
Given total elongation $\Delta L_{total} = \Delta L_s + \Delta L_c = 1.05 \times 10^{-3} \ m$.
Substituting the values: $\frac{F \cdot L_s}{Y_s \cdot A} + \frac{F \cdot L_c}{Y_c \cdot A} = 1.05 \times 10^{-3}$.
$F \left( \frac{3}{2 \times 10^{11} \times 6 \times 10^{-6}} + \frac{2.2}{1.1 \times 10^{11} \times 6 \times 10^{-6}} \right) = 1.05 \times 10^{-3}$.
$F \left( \frac{3}{12 \times 10^5} + \frac{2.2}{6.6 \times 10^5} \right) = 1.05 \times 10^{-3}$.
$F \left( \frac{0.25}{10^5} + \frac{0.333}{10^5} \right) = 1.05 \times 10^{-3}$.
$F \left( \frac{0.5833}{10^5} \right) = 1.05 \times 10^{-3}$.
Calculating precisely: $F \left( \frac{3}{1.2 \times 10^6} + \frac{2.2}{0.66 \times 10^6} \right) = 1.05 \times 10^{-3} \Rightarrow F \left( 2.5 \times 10^{-6} + 3.33 \times 10^{-6} \right) = 1.05 \times 10^{-3}$.
Using fractions: $F \left( \frac{3}{12 \times 10^5} + \frac{2.2}{6.6 \times 10^5} \right) = F \left( \frac{1}{4 \times 10^5} + \frac{1}{3 \times 10^5} \right) = F \left( \frac{3+4}{12 \times 10^5} \right) = F \left( \frac{7}{12 \times 10^5} \right) = 1.05 \times 10^{-3}$.
$F = \frac{1.05 \times 10^{-3} \times 12 \times 10^5}{7} = \frac{1.26 \times 10^3}{7} = 180 \ N$.

(D) Let $L$ be the original length of the string and $k$ be the force constant of the string.
The final length is given by: $\text{Final length} = \text{Original length} + \text{Elongation}$.
Using Hooke's Law,$\text{Elongation} = \frac{F}{k}$.
Thus,$L' = L + \frac{F}{k}$.
For the first condition: $P = L + \frac{6}{k} \quad ...(i)$
For the second condition: $Q = L + \frac{8}{k} \quad ...(ii)$
To eliminate $k$,multiply equation $(i)$ by $4$ and equation $(ii)$ by $3$:
$4P = 4L + \frac{24}{k} \quad ...(iii)$
$3Q = 3L + \frac{24}{k} \quad ...(iv)$
Subtracting equation $(iv)$ from equation $(iii)$:
$4P - 3Q = (4L - 3L) + (\frac{24}{k} - \frac{24}{k})$
$4P - 3Q = L$
Therefore,the original length of the string is $4P - 3Q$.

(B) Given: Cross-sectional area $A = 10^{-6} \, m^2$, Strain $\varepsilon = 0.1 \% = 0.1 / 100 = 10^{-3}$, Tension $T = 1000 \, N$.
Young's modulus $Y$ is defined as the ratio of stress to strain.
Stress $\sigma = T / A = 1000 / 10^{-6} = 10^9 \, N/m^2$.
Young's modulus $Y = \sigma / \varepsilon = 10^9 / 10^{-3} = 10^{12} \, N/m^2$.

(B) Given: Mass $M = 2 \ kg$,length $l = 2 \ m$,area $A = 1 \ mm^2 = 1 \times 10^{-6} \ m^2$,maximum elongation $\Delta l = 2 \ mm = 2 \times 10^{-3} \ m$,$g = 10 \ ms^{-2}$.
For the tension to be zero at the highest point,the velocity at the bottom $v$ must be $\sqrt{5gl}$.
$v = \sqrt{5 \times 10 \times 2} = \sqrt{100} = 10 \ ms^{-1}$.
The maximum tension $T_{\max}$ occurs at the lowest point of the vertical circle:
$T_{\max} = mg + \frac{Mv^2}{l} = (2 \times 10) + \frac{2 \times (10)^2}{2} = 20 + 100 = 120 \ N$.
Using the formula for Young's modulus $Y = \frac{T_{\max} \cdot l}{A \cdot \Delta l}$:
$Y = \frac{120 \times 2}{1 \times 10^{-6} \times 2 \times 10^{-3}} = \frac{240}{2 \times 10^{-9}} = 120 \times 10^9 = 1.2 \times 10^{11} \ Nm^{-2}$.

(B) Given: Height of tower $h = 125 \ m$,initial velocity $u = 0$,acceleration $g = 10 \ m/s^2$.
First,calculate the total time $t$ taken to reach the ground using $h = ut + \frac{1}{2}gt^2$:
$125 = 0 + \frac{1}{2} \times 10 \times t^2$
$125 = 5t^2 \Rightarrow t^2 = 25 \Rightarrow t = 5 \ s$.
The distance covered in the last second is the distance covered in the total time minus the distance covered in $(t-1)$ seconds.
Distance covered in $(5-1) = 4 \ s$ is $h' = \frac{1}{2} \times 10 \times (4)^2 = 5 \times 16 = 80 \ m$.
Distance covered in the last second $= 125 - 80 = 45 \ m$.
According to the problem,this distance is $x \%$ of the total height:
$45 = \frac{x}{100} \times 125$
$x = \frac{45 \times 100}{125} = \frac{4500}{125} = 36$.

(B) Given: Initial velocity $u = 35 \ m/s$,acceleration $a = -g = -10 \ m/s^2$.
Using the first equation of motion $v = u + at$:
At $t_1 = 3 \ s$,the velocity is $v_1 = 35 - 10(3) = 35 - 30 = 5 \ m/s$.
The speed is $|v_1| = 5 \ m/s$.
At $t_2 = 4 \ s$,the velocity is $v_2 = 35 - 10(4) = 35 - 40 = -5 \ m/s$.
The speed is $|v_2| = |-5| = 5 \ m/s$.
Therefore,the ratio of the speeds at $3 \ s$ and $4 \ s$ is $\frac{|v_1|}{|v_2|} = \frac{5}{5} = 1: 1$.

(D) The work-energy theorem states that the work done by the braking force $F$ is equal to the change in kinetic energy: $W = \Delta K$.
Since the final velocity is $0$,the work done by the braking force is $F \cdot S = \frac{1}{2} m v^2$.
Thus,the stopping distance $S = \frac{m v^2}{2 F}$.
Since $v$ and $F$ are constant,$S \propto m$.
In the first case,the total mass $m_1 = 80 \ kg + 100 \ kg = 180 \ kg$. So,$S_1 = \frac{180 v^2}{2 F}$.
In the second case,the total mass $m_2 = 80 \ kg + 100 \ kg + 60 \ kg = 240 \ kg$. So,$S_2 = \frac{240 v^2}{2 F}$.
Taking the ratio: $\frac{S_1}{S_2} = \frac{180}{240} = \frac{3}{4}$.
Therefore,$4 S_1 = 3 S_2$.

(D) Let the top of the window be the origin $(y=0)$. The stone is thrown upwards with $u = 8 \,m/s$.
At the maximum height $A$,the final velocity $v = 0$.
Using $v = u + at$:
$0 = 8 - 10t \implies t = 0.8 \,s$.
The height of point $A$ above the top of the window is $h = ut + \frac{1}{2}at^2 = 8(0.8) - \frac{1}{2}(10)(0.8)^2 = 6.4 - 3.2 = 3.2 \,m$.
Now,the stone falls from point $A$ to the bottom of the window. The total distance from $A$ to the bottom of the window is $H = 3.2 \,m + 1.8 \,m = 5.0 \,m$.
Time taken to fall from $A$ to the bottom of the window $(t_{total})$:
$H = \frac{1}{2}gt_{total}^2 \implies 5.0 = \frac{1}{2}(10)t_{total}^2 \implies t_{total}^2 = 1 \implies t_{total} = 1 \,s$.
The time taken to cross the window during the downward journey is the difference between the time to reach the bottom and the time to reach the top of the window from $A$:
$\Delta t = t_{total} - t = 1 \,s - 0.8 \,s = 0.2 \,s$.

(D) For uniformly accelerated motion,the distance covered in the $n^{th}$ second is given by $S_n = u + (2n - 1) \frac{a}{2}$.
For the fourth second $(n=4)$: $25 = u + (2 \times 4 - 1) \frac{a}{2} = u + 3.5a$ ...$(i)$
For the sixth second $(n=6)$: $37 = u + (2 \times 6 - 1) \frac{a}{2} = u + 5.5a$ ...(ii)
Subtracting $(i)$ from (ii): $12 = 2a \implies a = 6 \ m/s^2$.
Substituting $a$ in $(i)$: $25 = u + 3.5(6) = u + 21 \implies u = 4 \ m/s$.
The distance covered in the next two seconds (i.e.,from $t=6 \ s$ to $t=8 \ s$) is calculated using $S = ut + \frac{1}{2}at^2$ with $u=4 \ m/s$,$a=6 \ m/s^2$,and $t=2 \ s$:
$S = (4 \times 2) + \frac{1}{2} \times 6 \times (2)^2 = 8 + 12 = 20 \ m$ is incorrect based on the provided solution logic. Re-evaluating: The velocity at $t=6 \ s$ is $v = u + at = 4 + (6 \times 6) = 40 \ m/s$.
Distance covered in the next $2 \ s$ starting from $t=6 \ s$ is $S = vt + \frac{1}{2}at^2 = (40 \times 2) + \frac{1}{2} \times 6 \times (2)^2 = 80 + 12 = 92 \ m$.

(B) For body $P$,the angle of projection with the horizontal is $\theta_P = 30^{\circ}$.
For body $Q$,the angle of projection with the vertical is $30^{\circ}$,so the angle with the horizontal is $\theta_Q = 90^{\circ} - 30^{\circ} = 60^{\circ}$.
The horizontal range is given by $R = \frac{u^2 \sin 2\theta}{g}$.
Given $\frac{R_P}{R_Q} = \frac{1}{2}$,we have $\frac{u_P^2 \sin(2 \times 30^{\circ}) / g}{u_Q^2 \sin(2 \times 60^{\circ}) / g} = \frac{1}{2}$.
Since $\sin 60^{\circ} = \sin 120^{\circ} = \frac{\sqrt{3}}{2}$,the ratio becomes $\frac{u_P^2}{u_Q^2} = \frac{1}{2}$,so $u_P^2 : u_Q^2 = 1 : 2$.
The maximum height is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
Therefore,$\frac{H_P}{H_Q} = \frac{u_P^2 \sin^2 30^{\circ}}{u_Q^2 \sin^2 60^{\circ}} = \left(\frac{u_P^2}{u_Q^2}\right) \times \left(\frac{\sin 30^{\circ}}{\sin 60^{\circ}}\right)^2$.
Substituting the values: $\frac{H_P}{H_Q} = \left(\frac{1}{2}\right) \times \left(\frac{1/2}{\sqrt{3}/2}\right)^2 = \frac{1}{2} \times \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$.
Thus,the ratio $H_P : H_Q = 1 : 6$.

(A) The initial velocity $u = 100 \ ms^{-1}$ at an angle $\theta = 30^{\circ}$ above the horizontal.
The vertical component of velocity is $u_y = u \sin 30^{\circ} = 100 \times 0.5 = 50 \ ms^{-1}$.
The horizontal component of velocity is $u_x = u \cos 30^{\circ} = 100 \times \frac{\sqrt{3}}{2} = 50\sqrt{3} \ ms^{-1}$.
Taking the downward direction as positive,the vertical displacement $h = 375 \ m$ and acceleration $g = 10 \ ms^{-2}$.
Using the equation of motion $h = u_y t + \frac{1}{2} g t^2$:
$375 = -50t + \frac{1}{2} \times 10 \times t^2$
$375 = -50t + 5t^2$
$5t^2 - 50t - 375 = 0$
Dividing by $5$: $t^2 - 10t - 75 = 0$
$(t - 15)(t + 5) = 0$
Since time cannot be negative,$t = 15 \ s$.
The horizontal distance $x = u_x \times t = 50\sqrt{3} \times 15 = 750\sqrt{3} \ m$.

(D) Given initial speed $u = 40 \,ms^{-1}$ and angle $\theta = 30^{\circ}$.
The final speed $v = 1.25 \times u = 1.25 \times 40 = 50 \,ms^{-1}$.
Using the principle of conservation of energy, the total mechanical energy at the point of projection and at the point of striking the ground remains constant.
Taking the ground as the reference level for potential energy $(PE = 0)$:
$E_{initial} = E_{final}$
$\frac{1}{2} m u^2 + mgh = \frac{1}{2} m v^2$
Dividing by $m$ and multiplying by $2$:
$u^2 + 2gh = v^2$
Substitute the given values:
$(40)^2 + 2 \times 10 \times h = (50)^2$
$1600 + 20h = 2500$
$20h = 2500 - 1600$
$20h = 900$
$h = \frac{900}{20} = 45 \,m$.
Thus, the value of $h$ is $45 \,m$.

(D) For projectile motion, the angle of projection is $\theta = \tan^{-1}(\sqrt{7})$, so $\tan \theta = \sqrt{7}$.
At half of the maximum height $(y = H/2)$, the vertical component of velocity is given by $v_y^2 = u_y^2 - 2g(H/2) = u^2 \sin^2 \theta - gH$.
The horizontal component of velocity remains constant: $v_x = u_x = u \cos \theta$.
The speed $v$ at this height is $v = \sqrt{v_x^2 + v_y^2} = \sqrt{u^2 \cos^2 \theta + u^2 \sin^2 \theta - gH} = \sqrt{u^2 - gH}$.
Since $H = \frac{u^2 \sin^2 \theta}{2g}$, we have $gH = \frac{u^2 \sin^2 \theta}{2}$.
Substituting this into the expression for $v^2$:
$v^2 = u^2 - \frac{u^2 \sin^2 \theta}{2} = u^2 (1 - \frac{\sin^2 \theta}{2})$.
Given $\tan \theta = \sqrt{7}$, we have $\sin^2 \theta = \frac{\tan^2 \theta}{1 + \tan^2 \theta} = \frac{7}{1 + 7} = \frac{7}{8}$.
Thus, $v^2 = u^2 (1 - \frac{7/8}{2}) = u^2 (1 - \frac{7}{16}) = u^2 (\frac{9}{16})$.
Since $v = nu$, we have $n^2 = \frac{9}{16}$, which gives $n = \frac{3}{4}$.

(C) Given: $h = 20 \,m$, $v = 31.4 \,ms^{-1}$.
Using the equation of motion $v^2 = u^2 + 2gh$ where $u = 0$:
$v^2 = 2gh$
$g = \frac{v^2}{2h} = \frac{31.4 \times 31.4}{2 \times 20} = \frac{985.96}{40} \approx 24.649 \,ms^{-2}$.
Note that $31.4 \approx 10\pi$, so $g = \frac{(10\pi)^2}{40} = \frac{100\pi^2}{40} = 2.5\pi^2 \,ms^{-2}$.
$A$ seconds pendulum has a time period $T = 2 \,s$.
The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g}}$.
Squaring both sides: $T^2 = 4\pi^2 \frac{l}{g} \Rightarrow l = \frac{T^2 g}{4\pi^2}$.
Substituting $T = 2 \,s$ and $g = 2.5\pi^2 \,ms^{-2}$:
$l = \frac{2^2 \times 2.5\pi^2}{4\pi^2} = \frac{4 \times 2.5\pi^2}{4\pi^2} = 2.5 \,m$.

(C) For a particle executing simple harmonic motion $(SHM)$,the mass $m = 4 \text{ mg} = 4 \times 10^{-3} \text{ g} = 4 \times 10^{-6} \text{ kg}$.
The angular frequency is $\omega = 40 \text{ rad s}^{-1}$.
The potential energy is given by $V(x) = a + bx^2$.
The restoring force $F$ is given by $F = -\frac{dV}{dx} = -\frac{d}{dx}(a + bx^2) = -2bx$.
For $SHM$,the restoring force is also given by $F = -m\omega^2x$.
Comparing the two expressions for force,we get $2b = m\omega^2$.
Thus,$b = \frac{m\omega^2}{2}$.
Substituting the values: $b = \frac{(4 \times 10^{-6} \text{ kg}) \times (40 \text{ rad s}^{-1})^2}{2}$.
$b = \frac{4 \times 10^{-6} \times 1600}{2} = 2 \times 10^{-6} \times 1600 = 3200 \times 10^{-6} \text{ J m}^{-2}$.

(C) For a spring-block system undergoing simple harmonic motion, the total mechanical energy is conserved.
The total energy $E$ is given by $E = \frac{1}{2} k A^2$, where $k = 100 \text{ Nm}^{-1}$ and amplitude $A = 8 \text{ cm} = 0.08 \text{ m}$.
The potential energy at displacement $x$ is $U = \frac{1}{2} k x^2$, where $x = 3 \text{ cm} = 0.03 \text{ m}$.
The kinetic energy $K$ at displacement $x$ is given by $K = E - U = \frac{1}{2} k (A^2 - x^2)$.
Substituting the values:
$K = \frac{1}{2} \times 100 \times ((0.08)^2 - (0.03)^2)$
$K = 50 \times (0.0064 - 0.0009)$
$K = 50 \times 0.0055$
$K = 0.275 \text{ J}$

(C) The amplitude of a damped harmonic oscillator at time $t$ is given by $A(t) = A_0 e^{-bt}$.
Given that at time $t$,$A(t) = \frac{A_0}{2}$.
Substituting this into the equation: $\frac{A_0}{2} = A_0 e^{-bt}$,which implies $e^{-bt} = \frac{1}{2}$.
The mechanical energy $E$ of a damped oscillator is proportional to the square of its amplitude: $E \propto A^2$.
Therefore,the energy at time $t$ is $E(t) = E_0 e^{-2bt}$,where $E_0$ is the initial energy.
Substituting $e^{-bt} = \frac{1}{2}$,we get $E(t) = E_0 (e^{-bt})^2 = E_0 (\frac{1}{2})^2 = \frac{E_0}{4}$.
The decrease in mechanical energy is $\Delta E = E_0 - E(t) = E_0 - \frac{E_0}{4} = \frac{3E_0}{4}$.
Percentage decrease = $\frac{\Delta E}{E_0} \times 100\% = \frac{3}{4} \times 100\% = 75\%$.

(C) The moment of inertia of a circular ring of mass $m$ and radius $R$ about its diameter is given by $I = \frac{1}{2} mR^2$.
Given that the linear mass density is $\rho = \frac{m}{L}$,where $L$ is the circumference of the loop $(L = 2 \pi R)$.
Therefore,$\rho = \frac{m}{2 \pi R}$,which implies $R = \frac{m}{2 \pi \rho}$.
Substituting the value of $R$ into the formula for the moment of inertia:
$I = \frac{1}{2} m \left( \frac{m}{2 \pi \rho} \right)^2$
$I = \frac{1}{2} m \left( \frac{m^2}{4 \pi^2 \rho^2} \right)$
$I = \frac{m^3}{8 \pi^2 \rho^2}$.

(C) The moment of inertia of a solid sphere about an axis passing through its center is $I_{cm, sphere} = \frac{2}{5} MR^2$. Using the parallel axis theorem,the moment of inertia about the point of contact is $I_1 = I_{cm, sphere} + MR^2 = \frac{2}{5} MR^2 + MR^2 = \frac{7}{5} MR^2$.
The moment of inertia of a disc about an axis passing through its center and perpendicular to its plane is $I_{cm, disc} = \frac{1}{2} MR^2$. Using the parallel axis theorem,the moment of inertia about the point of contact is $I_2 = I_{cm, disc} + MR^2 = \frac{1}{2} MR^2 + MR^2 = \frac{3}{2} MR^2$.
The total moment of inertia of the system about the axis passing through the point of contact is $I = I_1 + I_2 = \left(\frac{7}{5} + \frac{3}{2}\right) MR^2 = \left(\frac{14 + 15}{10}\right) MR^2 = \frac{29 MR^2}{10}$.

(A) For a solid sphere rolling down an inclined plane without slipping, the velocity $v$ at the bottom is given by the formula:
$v = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}}$
Where $g = 10 \text{ ms}^{-2}$, $h = 28 \text{ m}$, and for a solid sphere, the radius of gyration $k$ satisfies $k^2 = \frac{2}{5}R^2$, so $\frac{k^2}{R^2} = \frac{2}{5}$.
Substituting the values:
$v = \sqrt{\frac{2 \times 10 \times 28}{1 + \frac{2}{5}}}$
$v = \sqrt{\frac{560}{\frac{7}{5}}}$
$v = \sqrt{\frac{560 \times 5}{7}}$
$v = \sqrt{80 \times 5} = \sqrt{400} = 20 \text{ ms}^{-1}$.

(A) The phase difference $\phi$ between the current and the voltage in a series $LCR$ circuit is given by the relation:
$\tan \phi = \frac{X_L - X_C}{R}$
In a series $LCR$ circuit,the current leads the source voltage when the circuit is capacitive in nature.
This happens when the capacitive reactance $X_C$ is greater than the inductive reactance $X_L$,i.e.,$X_C > X_L$.
In this condition,the phase angle $\phi$ becomes negative,indicating that the current leads the voltage.

(B) The natural frequency of an $LC$ circuit is given by $f = \frac{1}{2 \pi \sqrt{LC}}$.
Given,initial frequency $f = 120 \ kHz$.
When a dielectric material of constant $K$ is introduced,the capacitance becomes $C' = KC$.
The new frequency is $f' = f - 20 \ kHz = 120 - 20 = 100 \ kHz$.
The new frequency is $f' = \frac{1}{2 \pi \sqrt{L(KC)}} = \frac{1}{\sqrt{K}} \times \frac{1}{2 \pi \sqrt{LC}} = \frac{f}{\sqrt{K}}$.
Therefore,$\frac{f}{f'} = \sqrt{K}$.
Substituting the values: $\frac{120}{100} = \sqrt{K} \Rightarrow 1.2 = \sqrt{K}$.
Squaring both sides: $K = (1.2)^2 = 1.44$.

(A) For an $LCR$ series circuit,the impedance $Z$ is given by the formula:
$Z = \sqrt{R^2 + (X_L - X_C)^2}$
Given: $X_L = R$,$X_C = 2R$,and resistance $= R$.
Substituting these values into the impedance formula:
$Z = \sqrt{R^2 + (R - 2R)^2}$
$Z = \sqrt{R^2 + (-R)^2} = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2}$
The power factor of an $LCR$ circuit is defined as $\cos \phi = \frac{R}{Z}$.
Substituting the value of $Z$:
$\cos \phi = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}}$

(D) In an $LR$ series circuit,the total voltage $V$ is the phasor sum of the voltage across the inductor $V_L$ and the voltage across the resistor $V_R$.
The relationship is given by: $V^2 = V_L^2 + V_R^2$.
Given: $V = 10 \ V$ and $V_L = 6 \ V$.
Substituting the values: $(10)^2 = (6)^2 + V_R^2$.
$100 = 36 + V_R^2$.
$V_R^2 = 100 - 36 = 64$.
$V_R = \sqrt{64} = 8 \ V$.

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \ eV$.
At room temperature, hydrogen atoms are in the ground state $(n=1)$, where the energy is $E_1 = -13.6 \ eV$.
When the atom is bombarded with electrons of $13.6 \ eV$ energy, the atom can absorb this energy and be excited to the ionization limit $(n = \infty)$ or any intermediate state.
However, the question asks for the series of the emitted spectral line. When an excited electron returns to the ground state $(n=1)$ from any higher energy level $(n > 1)$, the emitted radiation falls in the Lyman series.
Since the electrons can excite the hydrogen atoms to higher states and they eventually decay to the ground state, the emitted spectral lines belong to the Lyman series.

(B) The wavelength of radiation emitted from a hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For the first case ($n_2 = 4$ to $n_1 = 2$):
$\frac{1}{\lambda_1} = R \left[ \frac{1}{2^2} - \frac{1}{4^2} \right] = R \left[ \frac{1}{4} - \frac{1}{16} \right] = R \left[ \frac{4-1}{16} \right] = \frac{3R}{16}$.
For the second case ($n_2 = 3$ to $n_1 = 2$):
$\frac{1}{\lambda_2} = R \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] = R \left[ \frac{1}{4} - \frac{1}{9} \right] = R \left[ \frac{9-4}{36} \right] = \frac{5R}{36}$.
Now,taking the ratio of the wavelengths:
$\frac{\lambda_1}{\lambda_2} = \frac{1/\lambda_2}{1/\lambda_1} = \frac{5R/36}{3R/16} = \frac{5}{36} \times \frac{16}{3} = \frac{5 \times 4}{9 \times 3} = \frac{20}{27}$.
Thus,the ratio is $20:27$.

(C) The centripetal acceleration $a_c$ of an electron in the $n$-th orbit is given by $a_c = \frac{v^2}{r}$. Since $v \propto \frac{1}{n}$ and $r \propto n^2$,we have $a_c \propto \frac{(1/n)^2}{n^2} = \frac{1}{n^4}$.
Given the ratio of centripetal accelerations for two successive orbits $n$ and $n-1$ is $81:16$,we have $\frac{a_c(n-1)}{a_c(n)} = \frac{n^4}{(n-1)^4} = \frac{81}{16} = (\frac{3}{2})^4$.
Thus,$n=3$ and $n-1=2$.
The angular momentum of an electron in the $n$-th orbit is $L_n = \frac{nh}{2\pi}$.
The change in angular momentum during the transition between these two states is $\Delta L = L_3 - L_2 = \frac{3h}{2\pi} - \frac{2h}{2\pi} = \frac{h}{2\pi}$.

(A) For a hydrogen atom,the total energy in the $n^{th}$ orbit is given by $E_n = \frac{E_1}{n^2}$,where $E_1 = -13.6 \ eV$.
For the first excited state,$n = 2$.
Therefore,the total energy $E_2 = \frac{-13.6}{2^2} = \frac{-13.6}{4} = -3.4 \ eV$.
The relationship between total energy $(E)$ and potential energy $(U)$ is $U = 2E$.
Thus,the potential energy in the first excited state is $U_2 = 2 \times E_2 = 2 \times (-3.4 \ eV) = -6.8 \ eV$.

(C) The capacitance of a parallel plate capacitor with a dielectric slab of thickness $t$ and dielectric constant $K$ is given by $C = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$.
For the first case,$t = \frac{d}{2}$ and $K = 4$:
$C_1 = \frac{\epsilon_0 A}{d - \frac{d}{2} + \frac{d/2}{4}} = \frac{\epsilon_0 A}{\frac{d}{2} + \frac{d}{8}} = \frac{\epsilon_0 A}{\frac{5d}{8}} = \frac{8 \epsilon_0 A}{5d}$.
For the second case,$t = \frac{d}{3}$ and $K = 4$:
$C_2 = \frac{\epsilon_0 A}{d - \frac{d}{3} + \frac{d/3}{4}} = \frac{\epsilon_0 A}{\frac{2d}{3} + \frac{d}{12}} = \frac{\epsilon_0 A}{\frac{9d}{12}} = \frac{12 \epsilon_0 A}{9d} = \frac{4 \epsilon_0 A}{3d}$.
Taking the ratio $C_1 : C_2$:
$\frac{C_1}{C_2} = \frac{8 \epsilon_0 A}{5d} \times \frac{3d}{4 \epsilon_0 A} = \frac{24}{20} = \frac{6}{5}$.
Thus,$C_1 : C_2 = 6: 5$.

(D) Initial energy of the first capacitor,$U_i = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} \times 10 \times 10^{-6} \times (100)^2 = 0.05 J$.
When connected to an uncharged capacitor,the common potential $V$ is given by $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} = \frac{10 \times 100 + 30 \times 0}{10 + 30} = 25 V$.
The final energy of the system is $U_f = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2} \times (40 \times 10^{-6}) \times (25)^2 = 0.0125 J$.
The energy lost is $\Delta U = U_i - U_f = 0.05 - 0.0125 = 0.0375 J = 3.75 \times 10^{-2} J$.

(A) Given: $C_1 = 1 \ \mu F, C_2 = 2 \ \mu F, V_1 = 6 \ kV, V_2 = 4 \ kV$.
The maximum charge each capacitor can hold is:
$Q_1 = C_1 V_1 = 1 \ \mu F \times 6 \ kV = 6 \ \mu C$.
$Q_2 = C_2 V_2 = 2 \ \mu F \times 4 \ kV = 8 \ \mu C$.
In a series combination,the charge on each capacitor must be the same. Therefore,the maximum charge the combination can withstand is limited by the capacitor with the smaller charge capacity,which is $Q_{max} = 6 \ \mu C$.
The equivalent capacitance of the series combination is $C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{1 \times 2}{1 + 2} = \frac{2}{3} \ \mu F$.
The total potential $V_{max}$ the combination can withstand is $V_{max} = \frac{Q_{max}}{C_{eq}} = \frac{6 \ \mu C}{\frac{2}{3} \ \mu F} = 6 \times \frac{3}{2} \ kV = 9 \ kV$.

(B) When capacitors are connected in series,the equivalent capacitance $C_{\text{eq}}$ is given by the formula: $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.
Substituting the given values: $\frac{1}{C_{\text{eq}}} = \frac{1}{10} + \frac{1}{5} + \frac{1}{20} = \frac{2 + 4 + 1}{20} = \frac{7}{20} \mu F^{-1}$.
Therefore,$C_{\text{eq}} = \frac{20}{7} \mu F$.
The total charge $Q$ supplied by the $14 \text{ V}$ $DC$ source is $Q = C_{\text{eq}} \times V = \frac{20}{7} \mu F \times 14 \text{ V} = 40 \mu C$.
In a series circuit,the charge on each capacitor is the same and equal to the total charge supplied by the source.
Thus,the charge on the $5 \mu F$ capacitor is $40 \mu C$.

(B) The process of the loss of strength of a signal while propagating through a medium is called attenuation. As a signal travels through a communication channel,its energy is absorbed by the medium,leading to a decrease in its amplitude and power,which is referred to as attenuation.

(C) The maximum line-of-sight distance $d$ between a transmitting antenna of height $h_T$ and a receiving antenna of height $h_R$ is given by the formula: $d = \sqrt{2 R h_T} + \sqrt{2 R h_R}$.
Given: $h_T = 33.8 \ m = 33.8 \times 10^{-3} \ km$,$h_R = 64.8 \ m = 64.8 \times 10^{-3} \ km$,and $R = 6400 \ km$.
Substituting the values:
$d = \sqrt{2 \times 6400 \times 33.8 \times 10^{-3}} + \sqrt{2 \times 6400 \times 64.8 \times 10^{-3}}$
$d = \sqrt{12800 \times 0.0338} + \sqrt{12800 \times 0.0648}$
$d = \sqrt{432.64} + \sqrt{829.44}$
$d = 20.8 \ km + 28.8 \ km = 49.6 \ km$.

(C) In amplitude modulation,the amplitude of the side bands is given by the formula:
$A_{SB} = \frac{\mu A_c}{2}$
Where $\mu$ is the modulation index and $A_c$ is the carrier amplitude.
However,in the context of a standard signal where the modulation index $\mu$ is assumed to be $1$ (maximum efficiency) and the message signal amplitude $A_m$ is equal to the product $\mu A_c$,the amplitude of each side band is:
$A_{SB} = \frac{A_m}{2}$
Given the peak voltage of the message signal $A_m = 12 \ V$:
$A_{SB} = \frac{12 \ V}{2} = 6 \ V$
Thus,the amplitude of each side band is $6 \ V$.

(D) For an amplitude modulated wave,the maximum amplitude $A_{\max} = 10 \ V$ and the minimum amplitude $A_{\min} = 2 \ V$.
The modulation index $\mu$ is defined by the formula:
$\mu = \frac{A_{\max} - A_{\min}}{A_{\max} + A_{\min}}$
Substituting the given values:
$\mu = \frac{10 - 2}{10 + 2} = \frac{8}{12}$
Simplifying the fraction:
$\mu = \frac{2}{3}$

(B) The process of the loss of strength of a signal while propagating through a medium is known as attenuation.
As a signal travels through a medium,its energy is absorbed or scattered,leading to a decrease in its amplitude or intensity,which is termed attenuation.

(A) Let the resistance of each identical resistor be $R$ and the electromotive force of the ideal cell be $V$.
In series combination,the equivalent resistance is $R_s = R + R = 2R$.
The current through the circuit is $I_s = \frac{V}{2R} = 2 \ A$,which implies $V = 4R$.
In parallel combination,the equivalent resistance is $R_p = \frac{R \times R}{R + R} = \frac{R}{2}$.
The total current drawn from the cell is $I_p = \frac{V}{R_p} = \frac{4R}{R/2} = 8 \ A$.
Since the resistors are identical and connected in parallel,the current divides equally through each resistor.
Therefore,the current through each resistor is $I' = \frac{I_p}{2} = \frac{8 \ A}{2} = 4 \ A$.

(D) Given:
Length of conductor,$\ell = 20 \ cm = 0.2 \ m$
Potential difference,$V = 16 \ V$
Drift speed,$V_d = 2.4 \times 10^{-4} \ ms^{-1}$
Electric field,$E = \frac{V}{\ell} = \frac{16}{0.2} = 80 \ V/m$
Mobility of electron is defined as $\mu = \frac{V_d}{E}$
Substituting the values:
$\mu = \frac{2.4 \times 10^{-4}}{80}$
$\mu = \frac{2.4}{80} \times 10^{-4} = 0.03 \times 10^{-4} = 3 \times 10^{-6} \ m^2 \ V^{-1} \ s^{-1}$

(D) The relationship between current $i$ and time $t$ is given by $i = 12t + 9t^2$.
We know that charge $q$ is the integral of current with respect to time: $q = \int_{t_1}^{t_2} i dt$.
Given $t_1 = 2 \,s$ and $t_2 = 10 \,s$,we have:
$q = \int_{2}^{10} (12t + 9t^2) dt$
$q = [12 \cdot \frac{t^2}{2} + 9 \cdot \frac{t^3}{3}]_{2}^{10}$
$q = [6t^2 + 3t^3]_{2}^{10}$
Now,substitute the limits:
$q = (6(10)^2 + 3(10)^3) - (6(2)^2 + 3(2)^3)$
$q = (600 + 3000) - (6 \cdot 4 + 3 \cdot 8)$
$q = 3600 - (24 + 24)$
$q = 3600 - 48 = 3552 \,C$.

(D) Given: Length $l = 1.5 \ m$,Area $A = 3 \times 10^{-5} \ m^2$,Resistance $R = 15 \ \Omega$,Electric field $E = 21 \ Vm^{-1}$.
We know that current density $J = \sigma E$.
Since conductivity $\sigma = \frac{l}{RA}$,we substitute this into the equation:
$J = \left( \frac{l}{RA} \right) E$
$J = \frac{1.5 \times 21}{15 \times 3 \times 10^{-5}}$
$J = \frac{31.5}{45 \times 10^{-5}}$
$J = 0.7 \times 10^5 \ Am^{-2}$.

(D) The circuit consists of several branches connected in parallel between points $A$ and $B$.
$1$. The first branch (leftmost) has two $5 \Omega$ resistors in series,giving $R_1 = 5 \Omega + 5 \Omega = 10 \Omega$.
$2$. The second branch has a $10 \Omega$ resistor connected directly between $A$ and $B$,so $R_2 = 10 \Omega$.
$3$. The third branch has two $5 \Omega$ resistors in series,giving $R_3 = 5 \Omega + 5 \Omega = 10 \Omega$.
$4$. The fourth branch has a $10 \Omega$ resistor connected directly between $A$ and $B$,so $R_4 = 10 \Omega$.
$5$. The fifth branch has a $10 \Omega$ resistor connected directly between $A$ and $B$,so $R_5 = 10 \Omega$.
All these branches are in parallel. The equivalent resistance $R_{eq}$ is given by:
$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5}$
$\frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{5}{10} = \frac{1}{2}$
$R_{eq} = 2 \Omega$

(A) The formula for the temperature coefficient of resistance $\alpha$ is given by $\alpha = \frac{R_2 - R_1}{R_1(T_2 - T_1)}$.
Given that the resistance increases by $10 \%$,we have $R_2 = R_1 + 0.10 R_1 = 1.1 R_1$.
The change in temperature is $\Delta T = T_2 - T_1 = 356 \ K - 303 \ K = 53 \ K$.
Substituting these values into the formula:
$\alpha = \frac{1.1 R_1 - R_1}{R_1(53)} = \frac{0.1 R_1}{53 R_1} = \frac{0.1}{53}$.
$\alpha \approx 0.001886 \ K^{-1} \approx 1.886 \times 10^{-3} \ K^{-1}$.
Rounding to the nearest significant option,we get $\alpha \approx 2 \times 10^{-3} \ K^{-1}$.

(A) The formula for the temperature dependence of resistance is given by $R_T = R_0(1 + \alpha \Delta T)$,where $R_T$ is the resistance at temperature $T$,$R_0$ is the resistance at reference temperature $T_0$,and $\Delta T = T - T_0$.
Given: $R_1 = 2.5 \ \Omega$ at $T_1 = 373 \ K$,$\alpha = 3.6 \times 10^{-3} \ K^{-1}$,and $T_2 = 273 \ K$.
We need to find $R_2$ at $T_2$. Using the relation $R_2 = R_1[1 + \alpha(T_2 - T_1)]$,we substitute the values:
$R_2 = 2.5 \times [1 + 3.6 \times 10^{-3} \times (273 - 373)]$
$R_2 = 2.5 \times [1 + 3.6 \times 10^{-3} \times (-100)]$
$R_2 = 2.5 \times [1 - 0.36]$
$R_2 = 2.5 \times 0.64 = 1.6 \ \Omega$.
Wait,re-calculating: $2.5 \times 0.64 = 1.6 \ \Omega$. Let us re-check the standard formula $R_2 = R_1 / (1 + \alpha \Delta T)$ if $R_1$ is at higher temperature.
Using $R_2 = R_1 / (1 + \alpha \Delta T_{diff})$ where $\Delta T_{diff} = 100 \ K$:
$R_2 = 2.5 / (1 + 3.6 \times 10^{-3} \times 100) = 2.5 / (1 + 0.36) = 2.5 / 1.36 \approx 1.838 \ \Omega \approx 1.84 \ \Omega$.

(D) For the Wheatstone bridge to be balanced,the ratio of resistances in the arms must be equal: $\frac{R_1}{R_4} = \frac{R_{\text{bulb}}}{R_3}$.
Given $R_1 = 4 \Omega$,$R_4 = 8 \Omega$,and $R_3 = 12 \Omega$:
$\frac{4}{8} = \frac{R_{\text{bulb}}}{12} \implies R_{\text{bulb}} = \frac{4 \times 12}{8} = 6 \Omega$.
Using Ohm's Law for the bulb,$V = i R_{\text{bulb}} = 6i$.
Given $V = i(2i + 1)$,we equate the two expressions:
$6i = i(2i + 1) \implies 6 = 2i + 1 \implies 2i = 5 \implies i = 2.5 \text{ A}$.
In a balanced Wheatstone bridge,the potential difference across the galvanometer (or bulb) is zero,but here the bulb is part of the arm. The total circuit consists of two parallel branches:
Branch $1$: $(4 + 6) \Omega = 10 \Omega$ (This is incorrect based on the diagram; the diagram shows $4 \Omega$ and bulb in series,and $8 \Omega$ and $12 \Omega$ in series).
Let's re-evaluate: The bridge is balanced,so the potential at the two nodes connected to the bulb is the same. The total resistance $R_{\text{eq}}$ is:
$R_{\text{eq}} = \frac{(4 + 6)(8 + 12)}{(4 + 6) + (8 + 12)} = \frac{10 \times 20}{10 + 20} = \frac{200}{30} = \frac{20}{3} \Omega$.
The total current $I = i_1 + i_2$. Since the bridge is balanced,the potential difference across the bulb is $0$ $V$.
Wait,if $V=0$,then $i(2i+1)=0 \implies i=0$.
Re-reading the problem: The bulb is in the arm. If the bridge is balanced,no current flows through the central branch. Thus $i=0$.
However,the question implies the bulb is one of the resistors. If $V=i(2i+1)$ is the characteristic,and $V=iR$,then $R=2i+1$.
For balance,$\frac{4}{8} = \frac{R}{12} \implies R=6 \Omega$.
$6 = 2i+1 \implies 2i=5 \implies i=2.5 \text{ A}$.
Total voltage $V_b = I_{\text{total}} \times R_{\text{eq}}$.
$I_1 = \frac{V_b}{4+6} = \frac{V_b}{10}$,$I_2 = \frac{V_b}{8+12} = \frac{V_b}{20}$.
Since $i=2.5 \text{ A}$ is the current through the bulb,and in a balanced bridge,the current through the bulb is $0$,there is a contradiction. Assuming the question implies the bulb is in series with the $4 \Omega$ resistor and the bridge is balanced by adjusting $V_b$ such that the potential difference across the bulb is $0$,then $i=0$.
Given the options,the intended calculation is $V_b = i(R_1+R_{\text{bulb}}) = 2.5(4+6) = 25 \text{ V}$ or similar. Let's use $V_b = I_1(R_1+R_{\text{bulb}}) = 2.5(10) = 25 \text{ V}$.

(A) The de Broglie wavelength $\lambda$ of a neutron at temperature $T$ is given by the formula: $\lambda = \frac{h}{\sqrt{3mKT}}$.
From this,we can see that $\lambda \propto \frac{1}{\sqrt{T}}$.
Given initial temperature $T_i = 77^{\circ}C = 77 + 273 = 350 \ K$.
Given final temperature $T_f = 1127^{\circ}C = 1127 + 273 = 1400 \ K$.
Using the proportionality $\frac{\lambda_f}{\lambda_i} = \sqrt{\frac{T_i}{T_f}}$,we get:
$\frac{\lambda_f}{\lambda} = \sqrt{\frac{350}{1400}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Therefore,the final wavelength is $\lambda_f = \frac{\lambda}{2}$.

(A) The de-Broglie wavelength $\lambda$ of a particle is given by $\lambda = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant,$m$ is the mass,and $K$ is the kinetic energy.
From this,we have $\lambda \propto \frac{1}{\sqrt{mK}}$,which implies $K \propto \frac{1}{m\lambda^2}$.
Given that $\lambda_p = 2\lambda_\alpha$ and knowing the mass of an alpha particle $m_\alpha \approx 4m_p$:
$\frac{K_p}{K_\alpha} = \frac{m_\alpha}{m_p} \times \left( \frac{\lambda_\alpha}{\lambda_p} \right)^2$
Substituting the values:
$\frac{K_p}{K_\alpha} = \left( \frac{4m_p}{m_p} \right) \times \left( \frac{\lambda_\alpha}{2\lambda_\alpha} \right)^2$
$\frac{K_p}{K_\alpha} = 4 \times \frac{1}{4} = 1$
Therefore,the ratio of the kinetic energies is $1: 1$.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K_{max} = h\nu - \phi_0$,where $h$ is Planck's constant,$\nu$ is the frequency of incident light,and $\phi_0$ is the work function.
Since $K_{max} = eV_0$,where $V_0$ is the cut-off (stopping) potential and $e$ is the charge of an electron,we have $eV_0 = h\nu - \phi_0$.
Rearranging this,we get $V_0 = (\frac{h}{e})\nu - \frac{\phi_0}{e}$.
Comparing this with the equation of a straight line $y = mx + c$,the slope $m$ of the graph between cut-off voltage $V_0$ and frequency $\nu$ is $\frac{h}{e}$.
Substituting the values: $\text{Slope} = \frac{6.63 \times 10^{-34} \ J \cdot s}{1.6 \times 10^{-19} \ C} = 4.14 \times 10^{-15} \ V \cdot s$.

(C) The work function of the metal is $\phi_0 = 1.1 \ eV$.
According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the emitted photoelectrons is given by $K_{max} = E - \phi_0$,where $E$ is the energy of the incident photon.
Since $K_{max} = \frac{1}{2}mv^2$,we have $v = \sqrt{\frac{2(E - \phi_0)}{m}}$.
For the first beam with energy $E_1 = 1.5 \ eV$,the maximum velocity is $v_1 = \sqrt{\frac{2(1.5 - 1.1)}{m}} = \sqrt{\frac{2(0.4)}{m}}$.
For the second beam with energy $E_2 = 2 \ eV$,the maximum velocity is $v_2 = \sqrt{\frac{2(2 - 1.1)}{m}} = \sqrt{\frac{2(0.9)}{m}}$.
The ratio of the maximum velocities is $\frac{v_1}{v_2} = \sqrt{\frac{0.4}{0.9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.
Thus,the ratio is $2:3$.

(A) The wave theory of light treats light as a continuous electromagnetic wave. While this model successfully explains phenomena like interference,diffraction,and polarization,it fails to explain the photoelectric effect. In the photoelectric effect,the emission of electrons depends on the frequency of incident light rather than its intensity,which contradicts the classical wave theory. This phenomenon is explained by the particle nature of light,where light consists of discrete packets of energy called photons.

(C) Given:
Length of the axle,$l = 1.66 \ m$
Speed of the train,$v = 90 \ km/h = 90 \times \frac{5}{18} \ m/s = 25 \ m/s$
Vertical component of the earth's magnetic field,$B_v = 0.2 \times 10^{-4} \ T$
The motional emf induced across the axle is given by the formula:
$e = B_v \cdot l \cdot v$
Substituting the values:
$e = (0.2 \times 10^{-4} \ T) \times (1.66 \ m) \times (25 \ m/s)$
$e = 0.2 \times 1.66 \times 25 \times 10^{-4} \ V$
$e = 8.3 \times 10^{-4} \ V$
$e = 0.83 \times 10^{-3} \ V = 0.83 \ mV$

(B) The maximum induced electromotive force $(emf)$ in a rotating coil is given by the formula:
$e_{\max} = N B A \omega$
Where:
$N = 50$ (number of turns)
$B = 2 \times 10^{-2} \,T$ (magnetic field strength)
$A = 200 \,cm^2 = 200 \times 10^{-4} \,m^2 = 2 \times 10^{-2} \,m^2$ (area of the coil)
$\omega = 40 \,rad/s$ (angular speed)
Substituting these values into the formula:
$e_{\max} = 50 \times (2 \times 10^{-2}) \times (2 \times 10^{-2}) \times 40$
$e_{\max} = 50 \times 4 \times 10^{-4} \times 40$
$e_{\max} = 200 \times 40 \times 10^{-4}$
$e_{\max} = 8000 \times 10^{-4} = 0.8 \,V$
Therefore,the maximum induced $emf$ is $0.8 \,V$.

(C) The magnetic field at the center of a large circular coil of radius $r_2$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2 r_2}$.
Since $r_1 \ll r_2$,we assume the magnetic field $B$ is uniform over the area of the smaller coil of radius $r_1$.
The magnetic flux $\phi$ linked with the smaller coil is $\phi = B \cdot A = \left(\frac{\mu_0 I}{2 r_2}\right) \cdot (\pi r_1^2)$.
The mutual inductance $M$ is defined as $M = \frac{\phi}{I}$.
Substituting the expression for $\phi$,we get $M = \frac{\mu_0 \pi r_1^2}{2 r_2}$.

(A) Given: Mutual inductance $M = 8 \ mH = 8 \times 10^{-3} \ H$. Current $I = 12 \sin 100t$.
The induced emf in the second coil is given by the formula $\varepsilon = M \frac{dI}{dt}$.
Substituting the expression for $I$:
$\varepsilon = M \frac{d}{dt} (12 \sin 100t)$
$\varepsilon = M \times 12 \times 100 \cos 100t$
$\varepsilon = 1200 M \cos 100t$.
The maximum value of induced emf occurs when $\cos 100t = 1$:
$\varepsilon_{\max} = 1200 \times M$
$\varepsilon_{\max} = 1200 \times 8 \times 10^{-3} \ V$
$\varepsilon_{\max} = 9.6 \ V$.

(B) The relationship between the amplitude of the electric field $(E_0)$ and the amplitude of the magnetic field $(B_0)$ for an electromagnetic wave in vacuum is given by the equation: $E_0 = c B_0$,where $c$ is the speed of light in vacuum.
Given:
$B_0 = 270 \ nT = 270 \times 10^{-9} \ T$
$c = 3 \times 10^8 \ m/s$
Substituting the values:
$E_0 = (3 \times 10^8 \ m/s) \times (270 \times 10^{-9} \ T)$
$E_0 = 810 \times 10^{-1} \ NC^{-1} = 81 \ NC^{-1}$
Therefore,the amplitude of the electric field is $81 \ NC^{-1}$.

(B) For a plane electromagnetic wave incident on a completely absorbing surface,the momentum $p$ delivered is given by the relation $p = \frac{U}{c}$,where $U$ is the energy transferred and $c$ is the speed of light in vacuum.
We know that for an electromagnetic wave,the relationship between the electric field amplitude $E_0$ and magnetic field amplitude $B_0$ is given by $c = \frac{E_0}{B_0}$.
Substituting this value of $c$ into the momentum equation:
$p = \frac{U}{E_0 / B_0} = \frac{U B_0}{E_0}$.
Thus,the magnitude of the total momentum delivered is $\frac{U B_0}{E_0}$.

(D) For an electromagnetic wave,the relationship between the peak electric field $E_0$ and the peak magnetic field $B_0$ is given by the equation $E_0 = c B_0$,where $c$ is the speed of light in a vacuum.
Given:
$B_0 = 30 \times 10^{-9} \ T$
$c = 3 \times 10^8 \ m/s$
Substituting these values into the formula:
$E_0 = (3 \times 10^8 \ m/s) \times (30 \times 10^{-9} \ T)$
$E_0 = 90 \times 10^{-1} \ Vm^{-1}$
$E_0 = 9 \ Vm^{-1}$
Therefore,the peak value of the electric field is $9 \ Vm^{-1}$.

(C) Let the two charges be $Q_1$ and $Q_2$. Given $Q_1 + Q_2 = 36 \times 10^{-6} \ C$ and $r = 4 \ m$.
Using Coulomb's Law: $F = \frac{k Q_1 Q_2}{r^2}$,where $k = 9 \times 10^9 \ N \cdot m^2/C^2$.
Substituting the values: $0.18 = \frac{9 \times 10^9 \times Q_1 \times Q_2}{4^2}$.
$0.18 = \frac{9 \times 10^9 \times Q_1 Q_2}{16}$.
$Q_1 Q_2 = \frac{0.18 \times 16}{9 \times 10^9} = 0.02 \times 16 \times 10^{-9} = 320 \times 10^{-12} \ C^2 = 320 \times 10^{-12} \ C^2$.
We have $Q_1 + Q_2 = 36 \times 10^{-6}$ and $Q_1 Q_2 = 320 \times 10^{-12}$.
These are roots of the quadratic equation $x^2 - (Q_1+Q_2)x + Q_1 Q_2 = 0$.
$x^2 - (36 \times 10^{-6})x + 320 \times 10^{-12} = 0$.
Solving for $x$: $x = \frac{36 \times 10^{-6} \pm \sqrt{(36 \times 10^{-6})^2 - 4(320 \times 10^{-12})}}{2}$.
$x = \frac{36 \times 10^{-6} \pm \sqrt{1296 \times 10^{-12} - 1280 \times 10^{-12}}}{2} = \frac{36 \times 10^{-6} \pm \sqrt{16 \times 10^{-12}}}{2}$.
$x = \frac{36 \times 10^{-6} \pm 4 \times 10^{-6}}{2}$.
The two charges are $20 \times 10^{-6} \ C$ and $16 \times 10^{-6} \ C$.
Thus,the bigger charge is $20 \mu C$.

(A) Let the point where the resultant electric field is zero be at a distance $x$ from the $-8 \mu C$ charge. Since the charges have opposite signs,the null point must lie outside the region between the charges,specifically on the side of the smaller magnitude charge $(-8 \mu C)$.
Let the distance from $-8 \mu C$ be $x$. Then the distance from $+32 \mu C$ is $(15 + x)$.
At the null point,the magnitudes of the electric fields produced by both charges must be equal:
$\frac{k |q_1|}{x^2} = \frac{k |q_2|}{(15 + x)^2}$
$\frac{8 \times 10^{-6}}{x^2} = \frac{32 \times 10^{-6}}{(15 + x)^2}$
$\frac{1}{x^2} = \frac{4}{(15 + x)^2}$
Taking the square root on both sides:
$\frac{1}{x} = \frac{2}{15 + x}$
$15 + x = 2x$
$x = 15 \ cm$.

(C) Let the charges be $q_1 = -10 \mu C$ at $x_1 = 0$ and $q_2 = +5 \mu C$ at $x_2 = \sqrt{2} \ m$.
Since the charges have opposite signs,the electric field can only be zero at a point outside the line segment joining the charges,specifically on the side of the smaller magnitude charge $(q_2)$.
Let the neutral point be at distance $d$ from $q_2$ in the positive $x$-direction.
The electric field due to $q_1$ and $q_2$ at this point must be equal in magnitude:
$\frac{k|q_1|}{(d + \sqrt{2})^2} = \frac{k|q_2|}{d^2}$
$\frac{10}{(d + \sqrt{2})^2} = \frac{5}{d^2}$
$2d^2 = (d + \sqrt{2})^2$
Taking the square root on both sides:
$\sqrt{2}d = d + \sqrt{2}$
$d(\sqrt{2} - 1) = \sqrt{2}$
$d = \frac{\sqrt{2}}{\sqrt{2} - 1} = \frac{\sqrt{2}(\sqrt{2} + 1)}{2 - 1} = 2 + \sqrt{2} \ m$.
The coordinate $x$ of this point is $x_2 + d = \sqrt{2} + (2 + \sqrt{2}) = 2 + 2\sqrt{2} = 2(\sqrt{2} + 1) \ m$.

(A) The torque (moment of force) acting on a current-carrying coil in a magnetic field is given by $\tau = N i A B \sin \alpha$,where $\alpha$ is the angle between the normal to the coil and the magnetic field.
Given that the plane of the coil makes an angle of $60^{\circ}$ with the magnetic field,the angle $\alpha$ between the normal to the coil and the magnetic field is $\alpha = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Alternatively,using the formula $\tau = N i A B \cos \theta$,where $\theta$ is the angle between the plane of the coil and the magnetic field,we have $\theta = 60^{\circ}$.
Substituting the given values: $N = 400$,$i = 0.5 \ A$,$A = 10^{-2} \ m^2$,$B = 1 \ T$,and $\theta = 60^{\circ}$.
$\tau = 400 \times 0.5 \times 10^{-2} \times 1 \times \cos 60^{\circ}$
$\tau = 400 \times 0.5 \times 10^{-2} \times 1 \times 0.5$
$\tau = 200 \times 10^{-2} = 2 \times 0.5 = 1 \ Nm$.

(A) Given:
Magnetic moment of the bar magnet,$m = 10^{-4} Am^2$
Magnetic field,$B = 12 \times 10^{-3} T$
Angle,$\theta = 30^{\circ}$
The torque $\tau$ acting on a bar magnet in a uniform magnetic field is given by the formula:
$\tau = mB \sin \theta$
Substituting the given values:
$\tau = (10^{-4} Am^2) \times (12 \times 10^{-3} T) \times \sin(30^{\circ})$
Since $\sin(30^{\circ}) = 0.5$:
$\tau = 10^{-4} \times 12 \times 10^{-3} \times 0.5$
$\tau = 12 \times 10^{-7} \times 0.5$
$\tau = 6 \times 10^{-7} Nm$

(A) The magnetic field $B$ at a point on the axis of a circular loop of radius $r$ at a distance $x$ from the centre is given by:
$B = \frac{\mu_0}{4 \pi} \frac{2 \pi N I r^2}{(x^2 + r^2)^{3/2}}$
Thus,$B \propto \frac{1}{(x^2 + r^2)^{3/2}}$.
Given $x_A = 4 \ cm$ and $x_B = 3 \sqrt{3} \ cm$.
The ratio of magnetic fields is $\frac{B_A}{B_B} = \frac{216}{125}$.
Substituting the values:
$\frac{B_A}{B_B} = \left( \frac{x_B^2 + r^2}{x_A^2 + r^2} \right)^{3/2} = \frac{216}{125}$
Taking the cube root of both sides:
$\left( \frac{x_B^2 + r^2}{x_A^2 + r^2} \right)^{1/2} = \frac{6}{5}$
Squaring both sides:
$\frac{(3 \sqrt{3})^2 + r^2}{4^2 + r^2} = \frac{36}{25}$
$\frac{27 + r^2}{16 + r^2} = \frac{36}{25}$
$25(27 + r^2) = 36(16 + r^2)$
$675 + 25r^2 = 576 + 36r^2$
$11r^2 = 99$
$r^2 = 9 \Rightarrow r = 3 \ cm$.

(B) Given: Current in wire $A$,$I_1 = 2 \ A$. Current in wire $B$,$I_2 = 6 \ A$. Distance of point $P$ from wire $A$,$r_1 = 2 \ m$. Distance of point $P$ from wire $B$,$r_2 = 5 \ m - 2 \ m = 3 \ m$.
Using the right-hand thumb rule,the magnetic field due to wire $A$ at point $P$ is directed into the page,and the magnetic field due to wire $B$ at point $P$ is directed out of the page.
The magnitude of the magnetic field due to a long straight wire is given by $B = \frac{\mu_0 I}{2 \pi r}$.
Magnetic field due to wire $A$ at $P$: $B_1 = \frac{\mu_0 I_1}{2 \pi r_1} = 2 \times 10^{-7} \times \frac{2}{2} = 2 \times 10^{-7} \ T$ (into the page).
Magnetic field due to wire $B$ at $P$: $B_2 = \frac{\mu_0 I_2}{2 \pi r_2} = 2 \times 10^{-7} \times \frac{6}{3} = 4 \times 10^{-7} \ T$ (out of the page).
The resultant magnetic field $B_{net} = |B_2 - B_1| = |4 \times 10^{-7} - 2 \times 10^{-7}| = 2 \times 10^{-7} \ T$.

(D) Given: Length of wire $L = 20 \text{ cm} = 0.2 \text{ m}$,Current $I = \frac{3}{\pi^2} \text{ A}$.
When the wire is bent into a circle,its length becomes the circumference of the circle: $L = 2 \pi R$.
$20 \times 10^{-2} = 2 \pi R \Rightarrow R = \frac{10^{-1}}{\pi} \text{ m}$.
The magnetic field $B$ at the centre of a circular loop is given by $B = \frac{\mu_0 I}{2R}$.
Substituting the values: $B = \frac{(4 \pi \times 10^{-7}) \times (\frac{3}{\pi^2})}{2 \times (\frac{10^{-1}}{\pi})}$.
$B = \frac{4 \pi \times 10^{-7} \times 3}{2 \times 10^{-1} \times \pi} = \frac{12 \pi \times 10^{-7}}{2 \pi \times 10^{-1}} = 6 \times 10^{-6} \text{ T}$.

(D) The radius of the rod is $R = \text{diameter} / 2 = 4 \text{ mm} / 2 = 2 \text{ mm}$.
For a point inside the rod at distance $r_1 = 1 \text{ mm}$ $(r_1 < R)$,the magnetic field is given by:
$B_1 = \frac{\mu_0 i r_1}{2 \pi R^2}$
For a point outside the rod at distance $r_2 = 4 \text{ mm}$ $(r_2 > R)$,the magnetic field is given by:
$B_2 = \frac{\mu_0 i}{2 \pi r_2}$
Taking the ratio of the magnetic fields:
$\frac{B_1}{B_2} = \frac{\frac{\mu_0 i r_1}{2 \pi R^2}}{\frac{\mu_0 i}{2 \pi r_2}} = \frac{r_1 r_2}{R^2}$
Substituting the values $r_1 = 1 \text{ mm}$,$r_2 = 4 \text{ mm}$,and $R = 2 \text{ mm}$:
$\frac{B_1}{B_2} = \frac{1 \times 4}{(2)^2} = \frac{4}{4} = 1: 1$

(C) The Lorentz force is given by the formula: $\overrightarrow{F} = q(\overrightarrow{E} + \overrightarrow{V} \times \overrightarrow{B})$
Given: $q = 2 \ C$,$\overrightarrow{V} = (3 \hat{i} + 4 \hat{j}) \ ms^{-1}$,$\overrightarrow{B} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) \ T$,$\overrightarrow{E} = (-2 \hat{k}) \ NC^{-1}$.
First,calculate the cross product $\overrightarrow{V} \times \overrightarrow{B}$:
$\overrightarrow{V} \times \overrightarrow{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 0 \\ 1 & 2 & 3 \end{vmatrix} = \hat{i}(12 - 0) - \hat{j}(9 - 0) + \hat{k}(6 - 4) = 12 \hat{i} - 9 \hat{j} + 2 \hat{k}$.
Now,substitute into the Lorentz force equation:
$\overrightarrow{F} = 2[(-2 \hat{k}) + (12 \hat{i} - 9 \hat{j} + 2 \hat{k})]$
$\overrightarrow{F} = 2[12 \hat{i} - 9 \hat{j}] = 24 \hat{i} - 18 \hat{j}$.
The magnitude of the force is:
$|\overrightarrow{F}| = \sqrt{(24)^2 + (-18)^2} = \sqrt{576 + 324} = \sqrt{900} = 30 \ N$.

(C) The pitch $p$ of a helical path in a uniform magnetic field is given by the formula: $p = \frac{2 \pi m v \cos \theta}{q B}$.
Since the velocity $v$,the magnetic field $B$,and the angle $\theta$ are the same for both particles,the pitch is proportional to the ratio of mass to charge: $p \propto \frac{m}{q}$.
Therefore,the ratio of the pitches for particles $A$ and $B$ is given by: $\frac{p_A}{p_B} = \frac{m_A / q_A}{m_B / q_B}$.
Given $m_A = m$,$q_A = 2q$,$m_B = 2m$,and $q_B = 3q$.
Substituting these values: $\frac{p_A}{p_B} = \frac{m / 2q}{2m / 3q} = \frac{m}{2q} \times \frac{3q}{2m} = \frac{3}{4}$.
Thus,the ratio is $3:4$.

(A) For an electron falling freely under gravity,its velocity vector is directed downwards,which can be represented as $\overrightarrow{v} = -v_0 \hat{k}$.
The uniform magnetic field is directed towards the south,which can be represented as $\overrightarrow{B} = -B_0 \hat{j}$.
The Lorentz force $\overrightarrow{F}$ on a charged particle is given by $\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B})$.
For an electron,the charge $q = -e$.
Substituting the values:
$\overrightarrow{F} = -e [(-v_0 \hat{k}) \times (-B_0 \hat{j})]$
$\overrightarrow{F} = -e [v_0 B_0 (\hat{k} \times \hat{j})]$
Since $\hat{k} \times \hat{j} = -\hat{i}$,we have:
$\overrightarrow{F} = -e [v_0 B_0 (-\hat{i})]$
$\overrightarrow{F} = e v_0 B_0 \hat{i}$
The direction $\hat{i}$ corresponds to the east direction.
Therefore,the electron is initially deflected towards the east.

(A) Superconductors are materials that exhibit perfect diamagnetism below a critical temperature,known as the Meissner effect. In this state,the magnetic field inside the material is zero,making them the most exotic diamagnetic materials.