MHT CET 2023 Physics Question Paper with Answer and Solution

(D) For an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Given $T_1 = T$,$V_1 = V$,$V_2 = 2V$,and $\gamma = 3/2$.
Substituting these values: $T V^{\gamma-1} = T_2 (2V)^{\gamma-1}$.
$T_2 = T \left(\frac{V}{2V}\right)^{\gamma-1} = T \left(\frac{1}{2}\right)^{3/2-1} = T \left(\frac{1}{2}\right)^{1/2} = \frac{T}{\sqrt{2}}$.
The work done in an adiabatic process is $W = \frac{R(T_1 - T_2)}{\gamma - 1}$.
Substituting the values: $W = \frac{R(T - T/\sqrt{2})}{3/2 - 1} = \frac{R T (1 - 1/\sqrt{2})}{1/2}$.
$W = 2 R T \left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) = \sqrt{2} R T (\sqrt{2}-1) = R T (2 - \sqrt{2})$.

(D) Under isochoric (constant volume) conditions,the volume change $dV = 0$.
Since work done $dW = P \cdot dV$,it follows that $dW = 0$.
According to the First Law of Thermodynamics,$dQ = dU + dW$.
Substituting $dW = 0$,we get $dQ = dU$.

(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_C}{T_H}$.
Given initial efficiency $\eta_1 = 50 \% = 0.5$ and source temperature $T_{H1} = 600 \ K$.
$0.5 = 1 - \frac{T_C}{600} \implies \frac{T_C}{600} = 0.5 \implies T_C = 300 \ K$.
Now,the efficiency is increased to $\eta_2 = 70 \% = 0.7$ while keeping the sink temperature $T_C = 300 \ K$ constant.
$0.7 = 1 - \frac{300}{T_{H2}}$.
$\frac{300}{T_{H2}} = 1 - 0.7 = 0.3$.
$T_{H2} = \frac{300}{0.3} = 1000 \ K$.
Thus,the new temperature of the source is $1000 \ K$.

(B) According to Stefan-Boltzmann law,the rate of heat loss $R$ is given by $R = e \sigma A (T^4 - T_0^4)$.
Here,$T$ is the temperature of the sphere and $T_0$ is the temperature of the surroundings.
Given: $T_1 = 627^{\circ} C = 627 + 273 = 900 \ K$,$T_2 = 327^{\circ} C = 327 + 273 = 600 \ K$,and $T_0 = 27^{\circ} C = 27 + 273 = 300 \ K$.
The ratio of the rates of heat loss is $\frac{R_1}{R_2} = \frac{T_1^4 - T_0^4}{T_2^4 - T_0^4}$.
Substituting the values: $\frac{R_1}{R_2} = \frac{900^4 - 300^4}{600^4 - 300^4} = \frac{(300 \times 3)^4 - 300^4}{(300 \times 2)^4 - 300^4}$.
$\frac{R_1}{R_2} = \frac{300^4 (3^4 - 1)}{300^4 (2^4 - 1)} = \frac{81 - 1}{16 - 1} = \frac{80}{15} = \frac{16}{3} \approx 5.33$.
Thus,the ratio is approximately $5.3$.

(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_C}{T_H}$.
For case $(i)$,the sink temperature $T_C = 100 \ K$ and the source temperature $T_H = 600 \ K$.
Therefore,$\eta_1 = 1 - \frac{100}{600} = 1 - \frac{1}{6} = \frac{5}{6}$.
For case $(ii)$,the sink temperature $T_C = T \ K$ and the source temperature $T_H = 960 \ K$.
Therefore,$\eta_2 = 1 - \frac{T}{960}$.
Since the efficiency is the same in both cases,$\eta_1 = \eta_2$.
$\frac{5}{6} = 1 - \frac{T}{960}$.
$\frac{T}{960} = 1 - \frac{5}{6} = \frac{1}{6}$.
$T = \frac{960}{6} = 160 \ K$.

(D) The efficiency of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$
Here,the temperature of the source $T_1 = 227^{\circ}C = 227 + 273 = 500 \ K$.
The temperature of the sink $T_2 = 27^{\circ}C = 27 + 273 = 300 \ K$.
Substituting these values: $\eta = 1 - \frac{300}{500} = 1 - 0.6 = 0.4$ or $\frac{2}{5}$.
We know that efficiency $\eta = \frac{W}{Q_1}$,where $W$ is the work done and $Q_1$ is the heat supplied.
Given $Q_1 = 50 \ kJ$,we have $W = \eta \times Q_1$.
$W = 0.4 \times 50 \ kJ = 20 \ kJ$.

(B) The coefficient of performance of a refrigerator is given by $\beta = \frac{T_2}{T_1 - T_2}$,where $T_1$ is the temperature of the hot reservoir and $T_2$ is the temperature of the cold reservoir.
The efficiency of a heat engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{T_1 - T_2}{T_1}$.
We can rewrite the efficiency as $\eta = \frac{1}{\frac{T_1}{T_1 - T_2}}$.
Since $\frac{T_1}{T_1 - T_2} = \frac{(T_1 - T_2) + T_2}{T_1 - T_2} = 1 + \frac{T_2}{T_1 - T_2} = 1 + \beta$,
Substituting this into the efficiency equation,we get $\eta = \frac{1}{1 + \beta}$.

(B) The coefficient of performance $(COP)$,$\beta$,for an ideal refrigerator is given by the formula: $\beta = \frac{T_2}{T_1 - T_2}$,where $T_2$ is the temperature of the freezer and $T_1$ is the temperature of the surroundings (sink).
Given: $\beta = 5$ and $T_2 = -13^{\circ} C = (-13 + 273) K = 260 K$.
Substituting the values into the formula:
$5 = \frac{260}{T_1 - 260}$
$5(T_1 - 260) = 260$
$5T_1 - 1300 = 260$
$5T_1 = 1560$
$T_1 = 312 K$
Converting back to Celsius: $T_1 = (312 - 273)^{\circ} C = 39^{\circ} C$.

(B) For an adiabatic process,the relationship between temperature and volume is given by $TV^{\gamma-1} = \text{constant}$.
Thus,$T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}$.
Given $V_i = V_0$,$V_f = \frac{V_0}{32}$,and $\gamma = \frac{7}{5}$.
Then $\gamma - 1 = \frac{7}{5} - 1 = \frac{2}{5}$.
Substituting the values: $T_i (V_0)^{2/5} = T_f \left(\frac{V_0}{32}\right)^{2/5}$.
$T_f = T_i \left(\frac{V_0}{V_0/32}\right)^{2/5} = T_i (32)^{2/5}$.
Since $32 = 2^5$,we have $T_f = T_i (2^5)^{2/5} = T_i (2^2) = 4T_i$.
Comparing $T_f = xT_i$ with $T_f = 4T_i$,we get $x = 4$.

(C) Under isothermal conditions,the temperature $T$ remains constant.
Since the process is isothermal,the total number of moles of air inside the bubbles remains constant.
Assuming the pressure inside a soap bubble of radius $r$ is $P = P_0 + \frac{4S}{r}$,where $P_0$ is atmospheric pressure and $S$ is surface tension.
For small bubbles,$P \approx \frac{4S}{r}$.
Using $PV = nRT$,for constant $T$ and $n$,$PV$ is constant.
$P_1 V_1 + P_2 V_2 = P_R V_R$
$\left(\frac{4S}{r_1}\right) \left(\frac{4}{3} \pi r_1^3\right) + \left(\frac{4S}{r_2}\right) \left(\frac{4}{3} \pi r_2^3\right) = \left(\frac{4S}{R}\right) \left(\frac{4}{3} \pi R^3\right)$
$\frac{16}{3} \pi S r_1^2 + \frac{16}{3} \pi S r_2^2 = \frac{16}{3} \pi S R^2$
$r_1^2 + r_2^2 = R^2$
$R = \sqrt{r_1^2 + r_2^2}$

(D) According to the ideal gas equation,$PV = Nk_B T$,where $N$ is the number of molecules and $k_B$ is the Boltzmann constant.
For jar $P$,the equation is: $PV = N_P k_B T$ ... $(i)$
For jar $Q$,the equation is: $(2P) \left( \frac{V}{4} \right) = N_Q k_B (2T)$
Simplifying the left side: $\frac{PV}{2} = N_Q k_B (2T)$
$\Rightarrow PV = 4 N_Q k_B T$ ... $(ii)$
Comparing equations $(i)$ and $(ii)$:
$N_P k_B T = 4 N_Q k_B T$
$N_P = 4 N_Q$
Therefore,the ratio of the number of molecules in jar $P$ to those in jar $Q$ is $\frac{N_P}{N_Q} = 4:1$.

(C) Since the compression is sudden,the process is adiabatic.
For an adiabatic process,the relation between pressure and volume is given by $P_1 V_1^\gamma = P_2 V_2^\gamma$.
Given: $V_2 = \frac{V_1}{4}$,so $\frac{V_1}{V_2} = 4$.
Given: $\gamma = 1.5 = \frac{3}{2}$.
Substituting these values into the equation:
$\frac{P_2}{P_1} = \left( \frac{V_1}{V_2} \right)^\gamma = (4)^{3/2}$.
Calculating the value: $(4)^{3/2} = (2^2)^{3/2} = 2^3 = 8$.
Therefore,the final pressure $P_2 = 8 P_1$,which is $8$ times the initial pressure.

(C) The kinetic energy of the gas is converted into internal energy when the container is stopped suddenly.
Let $n$ be the number of moles of the gas. The total mass of the gas is $m_{total} = nM$.
The kinetic energy of the gas is $K = \frac{1}{2} (nM) V^2$.
The change in internal energy of a monoatomic gas is $\Delta U = n C_V \Delta T$.
For a monoatomic gas,the molar heat capacity at constant volume is $C_V = \frac{3}{2} R$.
Equating the kinetic energy to the change in internal energy:
$\frac{1}{2} n M V^2 = n \left( \frac{3}{2} R \right) \Delta T$
Dividing both sides by $n$:
$\frac{1}{2} M V^2 = \frac{3}{2} R \Delta T$
Solving for $\Delta T$:
$\Delta T = \frac{MV^2}{3R}$

(C) An isochoric process is a thermodynamic process in which the volume $(V)$ of the system remains constant.
In a $P-V$ diagram,a constant volume process is represented by a vertical line,because for a single value of $V$,the pressure $(P)$ can vary.
Looking at the given options:
Graph $(A)$ shows $P$ proportional to $V$.
Graph $(B)$ shows constant pressure (isobaric).
Graph $(C)$ shows a vertical line,which indicates that $V$ is constant while $P$ changes.
Graph $(D)$ shows a curve.
Therefore,graph $(C)$ represents an isochoric process.

(A) The root-mean-square velocity of an ideal gas is given by $V_{\text{rms}} = \sqrt{\frac{3RT}{M_0}}$.
Since $V_{\text{rms}} \propto \sqrt{T}$, we have $T \propto V_{\text{rms}}^2$.
If $V_{\text{rms}}$ is reduced by a factor of $3$, then $T_2 = \frac{T_1}{3^2} = \frac{T_1}{9}$, so $\frac{T_1}{T_2} = 9$.
For an adiabatic process, $TV^{\gamma-1} = \text{constant}$, which implies $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Rearranging gives $\frac{V_2}{V_1} = \left(\frac{T_1}{T_2}\right)^{\frac{1}{\gamma-1}}$.
Given $\gamma = 1.5$, then $\gamma - 1 = 0.5 = \frac{1}{2}$.
Substituting the values: $\frac{V_2}{V_1} = (9)^{\frac{1}{1/2}} = (9)^2 = 81$.
Thus, the gas must be expanded $81$ times.

(A) In an adiabatic process,the system is thermally insulated from its surroundings,meaning no heat exchange occurs $(Q = 0)$.
According to the first law of thermodynamics,$\Delta U = Q - W$. Since $Q = 0$,we have $\Delta U = -W$,which means all work done is at the expense of internal energy.
The equation of state for an adiabatic process is $PV^{\gamma} = \text{constant}$,where $\gamma$ is the adiabatic index. The statement $PV = \text{constant}$ describes an isothermal process,not an adiabatic one.
Therefore,the statement '$PV=$ constant' is incorrect.

(B) According to the first law of thermodynamics,the heat supplied to the system $\Delta Q$ is given by $\Delta Q = \Delta U + dW$.
For an adiabatic process,there is no exchange of heat with the surroundings,so $\Delta Q = 0$.
Substituting this into the first law equation: $0 = \Delta U + dW$.
Rearranging the terms,we get $\Delta U = -dW$.
Therefore,the condition $\Delta U = -dW$ corresponds to an adiabatic process.

(B) For an adiabatic process,the relation between pressure $P$ and volume $V$ is $PV^{\gamma} = \text{constant}$.
Since density $\rho = \frac{m}{V}$,we have $V = \frac{m}{\rho}$. Substituting this into the adiabatic equation:
$P \left(\frac{m}{\rho}\right)^{\gamma} = \text{constant}$
$P \rho^{-\gamma} = \text{constant}$
Therefore,$\frac{P^{\prime}}{P} = \left(\frac{\rho^{\prime}}{\rho}\right)^{\gamma}$.
Given $\frac{\rho^{\prime}}{\rho} = 32$ and $\gamma = \frac{7}{5}$,we substitute these values:
$\frac{P^{\prime}}{P} = (32)^{\frac{7}{5}}$
$\frac{P^{\prime}}{P} = (2^5)^{\frac{7}{5}} = 2^7 = 128$.

(D) According to Boyle's Law at constant temperature,$P \propto \frac{1}{V}$,which implies $P_1 V_1 = P_2 V_2$.
Let the initial pressure be $P_1 = P$.
The new pressure is $P_2 = P + 0.05P = 1.05P$.
Substituting these into the equation: $P V_1 = 1.05P V_2$.
Solving for the new volume: $V_2 = \frac{V_1}{1.05} \approx 0.9524 V_1$.
The change in volume is $\Delta V = V_2 - V_1 = 0.9524 V_1 - V_1 = -0.0476 V_1$.
The percentage decrease in volume is $\frac{|\Delta V|}{V_1} \times 100 = 0.0476 \times 100 = 4.76 \%$.

(D) thermodynamic process in which there is no exchange of heat between the system and the surroundings is defined as an adiabatic process.
In this process,the heat exchange $dQ = 0$.
According to the first law of thermodynamics,$dQ = dU + dW$,so for an adiabatic process,$dU = -dW$.

(C) For an isothermal process,the temperature $T$ remains constant.
Since internal energy $U$ of an ideal gas depends only on temperature $(U = f(T))$,the change in internal energy $\Delta U = 0$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$. Since $\Delta U = 0$,$\Delta Q = \Delta W$,meaning all energy exchanged is used to do work.
An isothermal process requires the system to be in perfect thermal equilibrium with the environment to maintain a constant temperature.
The equation of state for an ideal gas is $PV = nRT$. Since $n$,$R$,and $T$ are constant,$PV$ must be constant.
Therefore,the statement '$PV$ is not constant' is incorrect.

(C) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is tension and $\mu$ is mass per unit length.
Since $T$ and $\mu$ are constant for the segments,we have $n \propto \frac{1}{l}$,which implies $nl = k$ (constant).
Given that the total length $l = l_1 + l_2 + l_3$,we can express each segment length as $l_1 = \frac{k}{n_1}$,$l_2 = \frac{k}{n_2}$,and $l_3 = \frac{k}{n_3}$.
The original length is $l = \frac{k}{n}$.
Substituting these into the length equation: $\frac{k}{n} = \frac{k}{n_1} + \frac{k}{n_2} + \frac{k}{n_3}$.
Dividing both sides by $k$,we get $\frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3}$.

(B) The general equation of a simple harmonic progressive wave is $y=A \sin 2 \pi \left( \frac{t}{T} - \frac{x}{\lambda} \right)$.
Given equation: $y=a \sin 2 \pi(b t-c x)$.
Comparing the two,we get amplitude $A=a$,frequency $n = \frac{1}{T} = b$,and wave number $\frac{1}{\lambda} = c$.
The maximum particle velocity is given by $(v_p)_{\max} = A \omega = A(2 \pi n) = a(2 \pi b) = 2 \pi ab$.
The wave velocity is given by $v = n \lambda = \frac{n}{1/\lambda} = \frac{b}{c}$.
According to the problem,$(v_p)_{\max} = \frac{1}{2} v$.
Substituting the values: $2 \pi ab = \frac{1}{2} \left( \frac{b}{c} \right)$.
Canceling $b$ from both sides: $2 \pi a = \frac{1}{2c}$.
Solving for $c$: $c = \frac{1}{4 \pi a}$.

(D) The standard equation of a traveling wave is given by $Y = A \sin(\omega t - kx + \phi)$.
Comparing this with the given equation $Y = 6 \sin(12 \pi t - 0.02 \pi x + \frac{\pi}{3})$:
We have angular frequency $\omega = 12 \pi \ rad/s$ and wave number $k = 0.02 \pi \ rad/m$.
The velocity of the wave $v$ is given by the ratio of angular frequency to wave number:
$v = \frac{\omega}{k}$
$v = \frac{12 \pi}{0.02 \pi}$
$v = \frac{12}{0.02} = \frac{1200}{2} = 600 \ m/s$.

(C) The separation between two consecutive nodes in a standing wave is given by $\frac{\lambda}{2}$.
Given, $\frac{\lambda}{2} = 5 \,cm = 0.05 \,m$.
Therefore, the wavelength $\lambda = 2 \times 0.05 \,m = 0.1 \,m$.
The frequency of vibration $f$ is related to the wave velocity $v$ and wavelength $\lambda$ by the formula $f = \frac{v}{\lambda}$.
Substituting the given values, $f = \frac{2 \,m/s}{0.1 \,m} = 20 \,Hz$.

(B) Given equations are:
$y_1 = a_1 \sin \left(\omega t - \frac{2 \pi x}{\lambda}\right)$
$y_2 = a_2 \cos \left(\omega t - \frac{2 \pi x}{\lambda} + \phi\right)$
We know that $\cos \theta = \sin \left(\theta + \frac{\pi}{2}\right)$.
Therefore,$y_2$ can be rewritten as:
$y_2 = a_2 \sin \left(\omega t - \frac{2 \pi x}{\lambda} + \phi + \frac{\pi}{2}\right)$
The phase difference $\Delta \Phi$ between the two waves is the difference in their arguments:
$\Delta \Phi = \left(\omega t - \frac{2 \pi x}{\lambda} + \phi + \frac{\pi}{2}\right) - \left(\omega t - \frac{2 \pi x}{\lambda}\right) = \phi + \frac{\pi}{2}$
The relationship between path difference $\Delta x$ and phase difference $\Delta \Phi$ is given by:
$\Delta x = \frac{\lambda}{2 \pi} \times \Delta \Phi$
Substituting the value of $\Delta \Phi$:
$\Delta x = \frac{\lambda}{2 \pi} \left(\phi + \frac{\pi}{2}\right)$

(D) The given wave equation is $y = A \sin(2\omega t - 2kx)$.
Comparing this with the standard wave equation $y = A \sin(\omega' t - k' x)$,we have angular frequency $\omega' = 2\omega$ and wave number $k' = 2k$.
The velocity of the particle is $v_p = \frac{dy}{dt} = A \cdot (2\omega) \cos(2\omega t - 2kx)$.
The maximum particle velocity is $v_{p,max} = 2A\omega$.
The wave velocity is $v_w = \frac{\omega'}{k'} = \frac{2\omega}{2k} = \frac{\omega}{k}$.
Given that $v_{p,max} = v_w$,we have $2A\omega = \frac{\omega}{k}$.
Thus,$A = \frac{1}{2k}$.
Since the wave number $k = \frac{2\pi}{\lambda}$,we substitute this into the expression for $A$:
$A = \frac{1}{2(2\pi / \lambda)} = \frac{\lambda}{4\pi}$.

(C) The standard equation of a traveling wave is given by $y = A \sin (\omega t - kx + \phi)$ or $y = A \sin (kx - \omega t + \phi)$.
Comparing the given equation $Y = 5 \sin (10 \pi t - 0.02 \pi x + \pi / 3)$ with the standard form,we identify the angular frequency $\omega$ and the wave number $k$:
$\omega = 10 \pi \ rad/s$
$k = 0.02 \pi \ rad/m$
The velocity of the wave $v$ is given by the ratio of angular frequency to wave number:
$v = \frac{\omega}{k}$
Substituting the values:
$v = \frac{10 \pi}{0.02 \pi}$
$v = \frac{10}{0.02} = \frac{1000}{2} = 500 \ m/s$
Therefore,the velocity of the wave is $500 \ m/s$.

(B) The general equation for a progressive wave is $Y = a \sin (\omega t - kx)$.
Comparing this with the given equation $Y = a \sin (2 \pi n t - \frac{2 \pi x}{5})$:
Angular frequency $\omega = 2 \pi n$.
Wave number $k = \frac{2 \pi}{5}$.
Wave velocity $v = \frac{\omega}{k} = \frac{2 \pi n}{2 \pi / 5} = 5n$.
Particle velocity $v_p = \frac{dY}{dt} = a \omega \cos (\omega t - kx)$.
Maximum particle velocity $v_m = a \omega = a (2 \pi n) = 2 \pi n a$.
The ratio of maximum particle velocity to wave velocity is $\frac{v_m}{v} = \frac{2 \pi n a}{5n} = \frac{2 \pi a}{5}$.

(A) When a transverse wave strikes a rigid wall,it undergoes reflection. According to the boundary conditions for a rigid support,the wave experiences a phase reversal of $180^{\circ}$ (or $\pi$ radians). Since the medium remains the same,the speed of the wave does not change,although the direction of propagation is reversed. Therefore,the phase changes by $180^{\circ}$,but the magnitude of the velocity remains the same.

(C) Mass per unit length for wire $PQ$ is $m_{PQ} = \frac{0.06}{4.8} = \frac{1}{80} \ kg/m$.
Mass per unit length for wire $QR$ is $m_{QR} = \frac{0.2}{2.56} = \frac{5}{64} \ kg/m$.
The velocity of the wave in wire $PQ$ is $v_{PQ} = \sqrt{\frac{T}{m_{PQ}}} = \sqrt{\frac{80}{1/80}} = \sqrt{6400} = 80 \ m/s$.
The time taken to travel through wire $PQ$ is $t_1 = \frac{L_{PQ}}{v_{PQ}} = \frac{4.8}{80} = 0.06 \ s$.
The velocity of the wave in wire $QR$ is $v_{QR} = \sqrt{\frac{T}{m_{QR}}} = \sqrt{\frac{80}{5/64}} = \sqrt{\frac{80 \times 64}{5}} = \sqrt{16 \times 64} = 4 \times 8 = 32 \ m/s$.
The time taken to travel through wire $QR$ is $t_2 = \frac{L_{QR}}{v_{QR}} = \frac{2.56}{32} = 0.08 \ s$.
The total time taken is $t = t_1 + t_2 = 0.06 + 0.08 = 0.14 \ s$.

(C) The frequency of a vibrating wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{m}}$.
Since $T$ and $m$ are constant,$f \propto \frac{1}{l}$,which implies $fl = \text{constant}$.
Let $f$ be the frequency of the tuning fork.
Initially,the length $l_1 = 50 \ cm$ and the beat frequency is $3$,so the wire frequency $f_1 = f - 3$ (or $f + 3$).
After shortening the length by $1 \ cm$,$l_2 = 49 \ cm$. The beat frequency remains $3$,so the wire frequency $f_2 = f + 3$ (or $f - 3$).
Since $f_1 l_1 = f_2 l_2$,we have $(f - 3) \times 50 = (f + 3) \times 49$.
$50f - 150 = 49f + 147$.
$50f - 49f = 147 + 150$.
$f = 297 \ Hz$.

(B) For a sonometer wire,the frequency $n$ is inversely proportional to the length $L$ $(n \propto 1/L)$,so $n_1 L_1 = n_2 L_2$.
Given $L_1 = 49 \ cm$ and $L_2 = 48 \ cm$,and the initial frequency $n_1 = n$.
Then $n \times 49 = n_2 \times 48$,which gives $n_2 = \frac{49}{48} n$.
Since the length is decreased,the frequency increases,so $n_2 > n_1$.
The beat frequency is given by $n_2 - n_1 = 6$.
Substituting the value of $n_2$: $\frac{49}{48} n - n = 6$.
$\frac{n}{48} = 6$.
$n = 6 \times 48 = 288 \ Hz$.

(A) According to the Doppler effect,when the listener moves towards the source,the apparent frequency increases,so the numerator is $(V+V_L)$.
When the source moves towards the listener,the apparent wavelength decreases,which increases the frequency,so the denominator is $(V-V_S)$.
Therefore,the observed frequency $n$ is given by the formula:
$n=n_0\left[\frac{V+V_{L}}{V-V_{s}}\right]$

(B) For an observer moving towards a stationary source,the observed frequency is given by $n_1 = n_0 \left[ \frac{v + v_L}{v} \right]$.
For a source moving towards a stationary observer,the observed frequency is given by $n_2 = n_0 \left[ \frac{v}{v - v_s} \right]$.
Substituting the values $v = 330 \,m/s$,$v_L = 50 \,m/s$,and $v_s = 50 \,m/s$:
$n_1 = n_0 \left[ \frac{330 + 50}{330} \right] = n_0 \left[ \frac{380}{330} \right] \approx 1.151 n_0$.
$n_2 = n_0 \left[ \frac{330}{330 - 50} \right] = n_0 \left[ \frac{330}{280} \right] \approx 1.178 n_0$.
Comparing the two results,we find that $n_2 > n_1$.
Therefore,the frequency heard will be more in the second case than in the first case.

(D) The frequency heard by a stationary observer when the source is approaching is given by $n_b = \left( \frac{v}{v - v_s} \right) n$.
When the source is moving away from the stationary observer,the frequency heard is $n_a = \left( \frac{v}{v + v_s} \right) n$.
Taking the ratio of these two frequencies:
$\frac{n_b}{n_a} = \frac{v + v_s}{v - v_s}$.
Given the ratio $\frac{n_b}{n_a} = \frac{11}{9}$,we have:
$\frac{11}{9} = \frac{v + v_s}{v - v_s}$.
Cross-multiplying gives:
$11(v - v_s) = 9(v + v_s)$
$11v - 11v_s = 9v + 9v_s$
$2v = 20v_s$
$v_s = \frac{2v}{20} = \frac{v}{10}$.

(B) The relationship between velocity $(v)$,frequency $(n)$,and wavelength $(\lambda)$ is given by $v = n \lambda$.
Since both sound sources are at the same temperature,the velocity of sound $(v)$ is the same for both waves.
Let the frequencies be $n_1$ and $n_2$ corresponding to wavelengths $\lambda_1 = 50 \ cm = 0.5 \ m$ and $\lambda_2 = 50.5 \ cm = 0.505 \ m$.
We have $v = n_1 \lambda_1 = n_2 \lambda_2$.
Thus,$n_1 (0.5) = n_2 (0.505)$,which implies $\frac{n_1}{n_2} = \frac{0.505}{0.5} = 1.01$.
Given $|n_1 - n_2| = 6 \ Hz$,we can write $n_1 = n_2 + 6$.
Substituting this into the ratio: $\frac{n_2 + 6}{n_2} = 1.01$.
$1 + \frac{6}{n_2} = 1.01 \implies \frac{6}{n_2} = 0.01$.
$n_2 = \frac{6}{0.01} = 600 \ Hz$.
Now,calculate the velocity: $v = n_2 \lambda_2 = 600 \times 0.505 \ m = 303 \ m/s$.

(A) The Doppler effect depends on the relative motion between the source of sound and the observer.
In this case, the passenger (observer) and the engine (source) are both in the same train.
Since they are moving together at the same velocity, there is no relative motion between the source and the listener.
Therefore, the apparent frequency $f$ heard by the passenger is equal to the actual frequency $n$ of the whistle.
Thus, $f = n$.

(A) The apparent frequency $n$ heard by a stationary observer when the source moves towards them with speed $v_S$ is given by the Doppler effect formula: $n = n_0 \left( \frac{v}{v - v_S} \right)$,where $n_0$ is the real frequency and $v$ is the speed of sound.
Given that $v_S = \frac{v}{10}$,we substitute this into the formula:
$\frac{n}{n_0} = \frac{v}{v - \frac{v}{10}}$
$\frac{n}{n_0} = \frac{v}{\frac{9v}{10}}$
$\frac{n}{n_0} = \frac{10}{9}$
Thus,the ratio of apparent to real frequency is $10:9$.

(B) For a closed organ pipe,the frequency of the first overtone is given by $f_1 = \frac{3v_1}{4L_1}$.
For an open organ pipe,the frequency of the first overtone is given by $f_2 = \frac{2v_2}{2L_2} = \frac{v_2}{L_2}$.
Given that the frequencies are equal,we have $\frac{3v_1}{4L_1} = \frac{v_2}{L_2}$.
The speed of sound in a gas is $v = \sqrt{\frac{1}{\rho K}}$,where $K$ is the compressibility. Since $K$ is the same for both gases,$v \propto \frac{1}{\sqrt{\rho}}$.
Thus,$\frac{v_1}{v_2} = \sqrt{\frac{\rho_2}{\rho_1}}$.
Substituting this into the frequency equation: $\frac{3}{4L_1} \sqrt{\frac{\rho_2}{\rho_1}} = \frac{1}{L_2}$.
Solving for $L_2$,we get $L_2 = \frac{4L_1}{3} \sqrt{\frac{\rho_1}{\rho_2}}$.

(B) For an open pipe, the end correction $e$ is given by the formula $e = 0.6 \times d$, where $d$ is the inner diameter of the pipe.
Given that the end correction $e = 0.8 \,cm$.
Substituting the value in the formula: $0.8 = 0.6 \times d$.
Therefore, $d = \frac{0.8}{0.6} = \frac{4}{3} \,cm$.
Since the inner radius $r$ is half of the diameter $d$, we have $r = \frac{d}{2}$.
$r = \frac{4/3}{2} = \frac{4}{6} = \frac{2}{3} \,cm$.
Thus, the inner radius of the pipe is $\frac{2}{3} \,cm$.

(B) For an open organ pipe of length $l$, the fundamental frequency is given by $f_0 = \frac{v}{2l}$.
Given $v = 333 \,m/s$ and $l = 33.3 \,cm = 0.333 \,m$.
$f_0 = \frac{333}{2 \times 0.333} = \frac{333}{0.666} = 500 \,Hz$.
In an open organ pipe, all harmonics are present, so the $n^{\text{th}}$ overtone is the $(n+1)^{\text{th}}$ harmonic.
The $5^{\text{th}}$ overtone corresponds to the $6^{\text{th}}$ harmonic.
Frequency of $5^{\text{th}}$ overtone $f_5 = 6 \times f_0 = 6 \times 500 \,Hz = 3000 \,Hz$.

(C) The height of a liquid column in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.
Given for liquid '$A$': $h_1 = 5 \,cm$,surface tension = $T_1$,density = $\rho_1$.
Given for liquid '$B$': surface tension $T_2 = 2T_1$,density $\rho_2 = 2\rho_1$,and the radius $r$ and angle of contact $\theta$ remain the same.
Substituting these into the formula for liquid '$B$':
$h_2 = \frac{2T_2 \cos \theta}{r \rho_2 g} = \frac{2(2T_1) \cos \theta}{r(2\rho_1) g} = \frac{4T_1 \cos \theta}{2r \rho_1 g} = \frac{2T_1 \cos \theta}{r \rho_1 g}$.
Since $h_1 = \frac{2T_1 \cos \theta}{r \rho_1 g} = 5 \,cm$,it follows that $h_2 = h_1 = 5 \,cm$.
Converting to meters: $5 \,cm = 0.05 \,m$.

(D) For a stretched string,the fundamental frequency (first harmonic) is $n = \frac{v}{2l} = \frac{1}{2l} \sqrt{\frac{T}{m}}$.
The frequency of the second harmonic is $2n = \frac{1}{l} \sqrt{\frac{T}{m}}$.
The first overtone is the next possible frequency above the fundamental,which is $n_1 = \frac{2v}{2l} = \frac{v}{l} = \frac{1}{l} \sqrt{\frac{T}{m}}$.
Comparing these,the frequency of the first overtone is equal to the frequency of the second harmonic.
Therefore,option $D$ is correct.

(B) The fundamental frequency of an open organ pipe of length $L$ is given by $n_{\text{open}} = \frac{v}{2L} = n$ ... $(i)$
Since the pipe is in unison with the string,the frequency of the string is $n_s = n = \frac{v}{2L}$.
When the tube is dipped in water such that $75 \%$ of its length is inside,the remaining length of the air column is $l_1 = 25 \% \times L = \frac{L}{4}$.
The pipe now acts as a closed organ pipe (closed at one end).
The fundamental frequency of this closed pipe is $n_{\text{closed}} = \frac{v}{4l_1} = \frac{v}{4(L/4)} = \frac{v}{L}$.
We need the ratio of the fundamental frequency of the air column of the dipped tube $(n_{\text{closed}})$ to that of the string $(n_s)$:
Ratio $= \frac{n_{\text{closed}}}{n_s} = \frac{v/L}{v/2L} = \frac{2}{1}$.

(A) The fundamental frequency of a closed pipe $A$ of length $L_1$ is given by $n_1 = \frac{v}{4L_1}$.
For an open pipe $B$ of length $L_2$,the frequencies are $n = \frac{mv}{2L_2}$ where $m = 1, 2, 3, \dots$. The first overtone is $m=2$ and the second overtone is $m=3$.
Thus,the frequency of the second overtone of pipe $B$ is $n_2 = \frac{3v}{2L_2}$.
Given that the frequencies are in unison,$n_1 = n_2$.
Therefore,$\frac{v}{4L_1} = \frac{3v}{2L_2}$.
Rearranging the terms to find the ratio $\frac{L_1}{L_2}$,we get $\frac{L_1}{L_2} = \frac{2}{4 \times 3} = \frac{2}{12} = \frac{1}{6}$.
So,the ratio is $1:6$.

(C) Let $l$ be the length of the open pipe and $L$ be the length of the closed pipe. Let $v$ be the speed of sound in air.
For an open pipe,the frequency of the $p$-th overtone is given by $f_p = (p+1) \frac{v}{2l}$. The second overtone corresponds to $p=2$,so $f_{o} = 3 \frac{v}{2l}$.
For a closed pipe,the frequency of the $p$-th overtone is given by $f_p = (2p+1) \frac{v}{4L}$. The first overtone corresponds to $p=1$,so $f_{c} = 3 \frac{v}{4L}$.
Given that the frequencies are equal: $f_{o} = f_{c}$.
$\frac{3v}{2l} = \frac{3v}{4L}$.
Canceling $3v$ from both sides,we get $\frac{1}{2l} = \frac{1}{4L}$.
Therefore,$2l = 4L$,which simplifies to $l = 2L$.

(A) The velocity of the wave is given by $v = f \lambda$.
Given $v = 50 \,m/s$ and $f = 200 \,Hz$.
Calculating the wavelength $\lambda$:
$\lambda = \frac{v}{f} = \frac{50}{200} = 0.25 \,m$.
In a stationary wave, the distance between two consecutive antinodes is equal to half of the wavelength $(\frac{\lambda}{2})$.
Distance $= \frac{0.25 \,m}{2} = 0.125 \,m$.

(B) For an open pipe of length '$l$' with end correction '$e$' at both ends,the effective length is $L_{eff} = l + 2e$.
For an open pipe,the fundamental frequency is $n_1 = \frac{v}{2L_{eff}}$.
The harmonics are given by $n_p = p \cdot n_1$,where $p = 1, 2, 3, \dots$.
The first overtone is $p=2$,and the second overtone is $p=3$.
Thus,the frequency of the second overtone is $n_3 = \frac{3v}{2(l+2e)}$.
Using the relation $v = n_3 \lambda_3$,we get $\lambda_3 = \frac{v}{n_3}$.
Substituting the value of $n_3$,we have $\lambda_3 = \frac{v}{\frac{3v}{2(l+2e)}} = \frac{2(l+2e)}{3}$.

(A) Let $l_1$ and $l_2$ be the lengths for the first and second resonance,and $x$ be the end correction.
For the first resonance: $l_1 + x = \frac{\lambda}{4}$
For the second resonance: $l_2 + x = \frac{3\lambda}{4}$
Dividing the two equations: $\frac{l_2 + x}{l_1 + x} = 3 \implies l_2 + x = 3l_1 + 3x \implies 2x = l_2 - 3l_1$
$2x = 70.2 - 3(22.7) = 70.2 - 68.1 = 2.1 \,cm \implies x = 1.05 \,cm$
For the third resonance: $l_3 + x = \frac{5\lambda}{4} = 5(l_1 + x)$
$l_3 = 5l_1 + 4x = 5(22.7) + 4(1.05) = 113.5 + 4.2 = 117.7 \,cm$

(D) The magnetic moment of the straight rod is $M = m \times L$,where $m$ is the pole strength.
When the rod is bent into a semicircular arc,the length $L$ becomes the arc length of the semicircle.
Thus,$L = \pi r$,which gives the radius $r = \frac{L}{\pi}$.
The new magnetic moment $M_{\text{new}}$ is the product of the pole strength $m$ and the effective length (the distance between the two ends,which is the diameter $2r$).
$M_{\text{new}} = m \times (2r) = m \times \left( \frac{2L}{\pi} \right)$.
Since $m = \frac{M}{L}$,we substitute this into the equation:
$M_{\text{new}} = \left( \frac{M}{L} \right) \times \left( \frac{2L}{\pi} \right) = \frac{2M}{\pi}$.

(D) The magnetization $M$ is defined as the magnetic dipole moment per unit volume,$M = \frac{m_{net}}{V}$.
Therefore,the net magnetic dipole moment is $m_{net} = M \times V$.
The volume $V$ of a cylindrical rod is given by $V = \pi r^2 l = \pi \left(\frac{d}{2}\right)^2 l$.
Given: $M = 5.3 \times 10^3 \ A/m$,$l = 5 \ cm = 0.05 \ m$,$d = 1 \ cm = 0.01 \ m$.
Radius $r = \frac{d}{2} = 0.005 \ m$.
$V = 3.142 \times (0.005)^2 \times 0.05 = 3.142 \times 25 \times 10^{-6} \times 0.05 = 3.9275 \times 10^{-6} \ m^3$.
$m_{net} = (5.3 \times 10^3) \times (3.9275 \times 10^{-6}) \approx 2.08 \times 10^{-2} \ J/T$.

(C) The orbital magnetic moment $M_0$ of an electron orbiting in a circular path is given by the relation:
$M_0 = \frac{-e}{2m_e} L$
where '$e$' is the charge of the electron,'$m_e$' is the mass of the electron,and '$L$' is the angular momentum.
From this expression,it is clear that the orbital magnetic moment $M_0$ is directly proportional to the angular momentum $L$ of the electron.

(D) The magnetic moment $(m)$ of a current-carrying coil is given by the formula:
$m = nIA$
where $n$ is the number of turns,$I$ is the current,and $A$ is the area of the coil.
For a circular coil of radius $r$,the area $A = \pi r^2$.
Substituting this into the formula,we get:
$m = nI(\pi r^2)$
Thus,the magnetic moment depends on the number of turns $(n)$,the current $(I)$,and the radius $(r)$ of the coil.

(A) Let the point where the electric field is zero be at a distance $x$ from the centre of the dipole with moment $P$. The distance from the dipole with moment $27P$ is $(24 - x)$.
For an axial point of a dipole, the electric field magnitude is given by $E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{2p}{r^3}$.
At the point where the net electric field is zero, the magnitudes of the electric fields due to both dipoles must be equal:
$\frac{1}{4 \pi \varepsilon_0} \cdot \frac{2P}{x^3} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{2(27P)}{(24 - x)^3}$
$\frac{1}{x^3} = \frac{27}{(24 - x)^3}$
Taking the cube root on both sides:
$\frac{1}{x} = \frac{3}{24 - x}$
$24 - x = 3x$
$4x = 24$
$x = 6 \,cm$.

(B) The length of the wire remains constant. Let $N_1 = n$ and $N_2$ be the number of turns in the new coil.
Since the total length of the wire $L = N_1 (2 \pi R) = N_2 (2 \pi R_2)$,where $R_2 = \frac{R}{3}$:
$N_1 (2 \pi R) = N_2 (2 \pi \frac{R}{3})$
$N_1 = \frac{N_2}{3} \implies N_2 = 3 N_1$.
The magnetic moment $\mu$ of a coil is given by $\mu = N I A$,where $A = \pi R^2$.
For the original coil: $\mu_1 = N_1 I (\pi R^2)$.
For the new coil: $\mu_2 = N_2 I (\pi R_2^2) = (3 N_1) I \pi (\frac{R}{3})^2 = 3 N_1 I \pi \frac{R^2}{9} = \frac{1}{3} N_1 I \pi R^2$.
Therefore,the ratio $\frac{\mu_2}{\mu_1} = \frac{\frac{1}{3} N_1 I \pi R^2}{N_1 I \pi R^2} = \frac{1}{3}$.

(D) The magnetic field at the center of a circular loop is $B = \frac{\mu_0 i}{2r}$.
The magnetic moment of the loop is $M = i \pi r^2$.
The ratio is $x = \frac{B}{M} = \frac{\mu_0 i / 2r}{i \pi r^2} = \frac{\mu_0}{2 \pi r^3}$.
When the current $i$ becomes $2i$ and the radius $r$ becomes $2r$,the new magnetic field $B'$ is $B' = \frac{\mu_0 (2i)}{2(2r)} = \frac{\mu_0 i}{2r} = B$.
The new magnetic moment $M'$ is $M' = (2i) \pi (2r)^2 = (2i) \pi (4r^2) = 8(i \pi r^2) = 8M$.
The new ratio $x'$ is $x' = \frac{B'}{M'} = \frac{B}{8M} = \frac{1}{8} \left( \frac{B}{M} \right) = \frac{x}{8}$.

(C) The magnetic moment $M$ of a circular loop is given by the formula $M = I A$,where $I$ is the current and $A$ is the area of the loop.
For a circular loop of radius $R$,the area is $A = \pi R^2$.
Thus,$M = I (\pi R^2)$.
The length of the wire $L$ forms the circumference of the loop,so $L = 2 \pi R$,which implies $R = \frac{L}{2 \pi}$.
Substituting the value of $R$ into the magnetic moment equation:
$M = I \pi \left( \frac{L}{2 \pi} \right)^2$
$M = I \pi \left( \frac{L^2}{4 \pi^2} \right)$
$M = \frac{I L^2}{4 \pi}$
Rearranging to solve for $L$:
$L^2 = \frac{4 M \pi}{I}$
$L = \sqrt{\frac{4 M \pi}{I}}$ or $\left[ \frac{4 M \pi}{I} \right]^{\frac{1}{2}}$.

(B) For a charged particle moving in a uniform magnetic field,the magnetic Lorentz force provides the necessary centripetal force for circular motion.
$qvB = \frac{mv^2}{R}$
$\therefore R = \frac{mv}{qB}$ ... $(i)$
Given the new speed $v' = \frac{v}{2}$ and the new magnetic field $B' = 2B$.
The new radius $R'$ is given by:
$R' = \frac{mv'}{qB'} = \frac{m(v/2)}{q(2B)}$
$R' = \frac{mv}{4qB}$
Substituting equation $(i)$ into the expression for $R'$:
$R' = \frac{R}{4}$

(C) The magnetic force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $h$ is given by $F/l = \frac{\mu_0 I_1 I_2}{2 \pi h}$.
Since the currents are in opposite directions,the force is repulsive.
At equilibrium,the repulsive magnetic force balances the gravitational force acting on wire $PQ$:
$F = mg$
$\frac{\mu_0 I^2 l}{2 \pi h} = mg$
Given: $I = 25 \,A$,$l = 1 \,m$,$m = 2.5 \,g = 2.5 \times 10^{-3} \,kg$,$g = 9.8 \,m/s^2$,$\mu_0 = 4 \pi \times 10^{-7} \,T \cdot m/A$.
Substituting the values:
$h = \frac{\mu_0 I^2 l}{2 \pi m g} = \frac{(4 \pi \times 10^{-7}) \times (25)^2 \times 1}{2 \pi \times (2.5 \times 10^{-3}) \times 9.8}$
$h = \frac{2 \times 10^{-7} \times 625}{2.5 \times 10^{-3} \times 9.8} = \frac{1250 \times 10^{-7}}{24.5 \times 10^{-3}} = \frac{1250}{24.5} \times 10^{-4} \,m \approx 51 \times 10^{-4} \,m \approx 5.1 \,mm$.
Rounding to the nearest given option,the height is $5 \,mm$.

(D) The magnetic field produced by a circular current-carrying loop at any point on its axis is directed along the axis of the loop.
Since the electron is projected along the same axis,its velocity vector $\vec{v}$ is parallel or anti-parallel to the magnetic field vector $\vec{B}$.
Therefore,the angle $\theta$ between the velocity and the magnetic field is either $0^{\circ}$ or $180^{\circ}$.
The magnetic force on a moving charge is given by the Lorentz force formula: $F = qvB \sin \theta$.
Substituting $\theta = 0^{\circ}$ or $180^{\circ}$,we get $\sin(0^{\circ}) = 0$ or $\sin(180^{\circ}) = 0$.
Thus,$F = 0$. The electron experiences no force.

(B) The motional e.m.f. induced in a conductor moving through a magnetic field is given by the formula: $\varepsilon = B \cdot l \cdot v$.
Given values:
Magnetic field $B = 5 \times 10^{-5} \ T$,
Wing span (length) $l = 40 \ m$,
Speed $v = 500 \ m/s$.
Substituting these values into the formula:
$\varepsilon = (5 \times 10^{-5} \ T) \times (40 \ m) \times (500 \ m/s)$
$\varepsilon = 5 \times 10^{-5} \times 20000$
$\varepsilon = 5 \times 10^{-5} \times 2 \times 10^4$
$\varepsilon = 10 \times 10^{-1} \ V$
$\varepsilon = 1 \ V$.
Therefore,the e.m.f. generated is $1 \ V$.

(B) According to Curie's law,the magnetization $M$ of a paramagnetic material is directly proportional to the applied magnetic field $B$ and inversely proportional to the absolute temperature $T$.
Mathematically,this is expressed as $M \propto \frac{B}{T}$.
Introducing Curie's constant $C$,we get the relation $M = C \frac{B}{T}$,which can be rewritten as $M = \frac{CB}{T}$.
Therefore,the correct option is $B$.

(B) Retentivity or remanence is the ability of a magnetic substance to retain magnetism even in the absence of an external magnetizing field.
When the magnetizing force $(H)$ is reduced to zero,the residual magnetic induction $(B)$ present in the material is known as retentivity.
Therefore,retentivity corresponds to the value of $B$ when $H = 0$.

(C) Given: $N = 100$, $A = 2 \, m^2$, $\omega = 30 \, rad/s$, $B = 2 \times 10^{-2} \, T$, $R = 600 \, \Omega$.
The peak electromotive force $(E_0)$ induced in the coil is given by:
$E_0 = N A B \omega$
$E_0 = 100 \times 2 \times (2 \times 10^{-2}) \times 30$
$E_0 = 100 \times 2 \times 0.02 \times 30 = 120 \, V$.
The maximum power dissipated in an $A.C.$ circuit is given by:
$P_{\max} = \frac{E_0^2}{2R}$
Substituting the values:
$P_{\max} = \frac{120 \times 120}{2 \times 600}$
$P_{\max} = \frac{14400}{1200} = 12 \, W$.

(D) When a charge '$q$' moves with velocity '$\overrightarrow{v}$' in a region where both electric field '$\overrightarrow{E}$' and magnetic field '$\overrightarrow{B}$' are present,it experiences two forces:
$1$. Electric force: $\overrightarrow{F}_e = q\overrightarrow{E}$
$2$. Magnetic force (Lorentz force): $\overrightarrow{F}_m = q(\overrightarrow{v} \times \overrightarrow{B})$
The total force acting on the charge is the vector sum of these two forces,known as the Lorentz force:
$\overrightarrow{F} = \overrightarrow{F}_e + \overrightarrow{F}_m = q\overrightarrow{E} + q(\overrightarrow{v} \times \overrightarrow{B})$
Therefore,the correct option is $D$.

(B) As the bar magnet falls through the hollow metal pipe,the magnetic flux linked with the pipe changes continuously.
According to Faraday's law of electromagnetic induction,this change in magnetic flux induces an electromotive force $(emf)$ in the pipe.
Since the pipe is a conductor,this induced $emf$ causes eddy currents to flow within the body of the pipe.
According to Lenz's law,these eddy currents create a magnetic field that opposes the motion of the falling magnet.
This opposing force acts upwards,reducing the net downward force on the magnet.
Consequently,the net acceleration of the magnet is less than the acceleration due to gravity $(g)$.

(A) The magnetic susceptibility $(\chi)$ of diamagnetic substances is small and negative. It arises due to the orbital motion of electrons and is essentially independent of temperature. In contrast, the magnetic susceptibility of paramagnetic and ferromagnetic substances depends on temperature according to Curie's Law and the Curie-Weiss Law, respectively.

(A) Magnetic shielding is achieved by enclosing the sensitive instrument within a shell made of a material with high magnetic permeability,such as soft iron.
When an external magnetic field is applied,the magnetic field lines prefer to pass through the high-permeability material rather than the air or the interior space.
This effectively diverts the magnetic flux around the protected region,thereby shielding the instrument from the external magnetic field.

(D) When the wire attains terminal velocity,the net force acting on it becomes zero.
Therefore,the magnetic force equals the gravitational force.
$iBl = mg$
Since the induced current $i = \frac{e}{R}$ and the induced electromotive force $e = Bvl$,we substitute these into the equation:
$\frac{Bvl}{R} \cdot Bl = mg$
$\frac{B^2 l^2 v}{R} = mg$
Solving for terminal velocity $v$:
$v = \frac{mgR}{B^2 l^2}$

(B) According to the radioactive disintegration law,the number of atoms $N$ at time $t$ is given by $N = N_0 e^{-\lambda t}$.
Here,$N_0$ is the initial number of atoms.
We need to find the ratio $\frac{N_0}{N}$ at time $t = \frac{1}{2 \lambda}$.
Substituting the value of $t$ in the equation:
$\frac{N}{N_0} = e^{-\lambda \times \frac{1}{2 \lambda}}$
$\frac{N}{N_0} = e^{-\frac{1}{2}}$
Taking the reciprocal to find $\frac{N_0}{N}$:
$\frac{N_0}{N} = e^{\frac{1}{2}}$
$\frac{N_0}{N} = \sqrt{e}$

(C) The activity $A$ of a radioactive sample is given by the formula $A = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of undecayed atoms.
The decay constant $\lambda$ is related to the half-life $T$ by the relation $\lambda = \frac{\ln 2}{T}$.
For the two radioactive elements,the activities are:
$A_1 = \lambda_1 N_1 = \frac{\ln 2}{T_1} N_1$
$A_2 = \lambda_2 N_2 = \frac{\ln 2}{T_2} N_2$
The ratio of their activities is:
$\frac{A_1}{A_2} = \frac{(\frac{\ln 2}{T_1}) N_1}{(\frac{\ln 2}{T_2}) N_2}$
Simplifying the expression:
$\frac{A_1}{A_2} = \frac{N_1}{T_1} \times \frac{T_2}{N_2} = \frac{N_1 T_2}{N_2 T_1}$

(B) Let the original nucleus be ${ }_{Z} X^{A}$.
When one $\alpha$ particle $({ }_{2} He^{4})$ is emitted,the nucleus becomes ${ }_{Z-2} Y^{A-4}$.
When one $\beta^{-}$ particle $({ }_{-1} e^{0})$ is emitted,the atomic number increases by $1$ $(Z-2+1 = Z-1)$.
When two $\beta^{-}$ particles are emitted,the atomic number increases by $2$ $(Z-2+2 = Z)$.
Thus,the final nucleus is ${ }_{Z} X^{A-4}$,which is an isotope of the original nucleus ${ }_{Z} X^{A}$ because they have the same atomic number $Z$ but different mass numbers.

(C) Given:
Total time,$T = 10$ years
Half-life,$T_{1/2} = 5$ years
Number of half-lives,$n = \frac{T}{T_{1/2}} = \frac{10}{5} = 2$
Using the radioactive decay law,the fraction remaining is given by $\frac{N}{N_0} = \left(\frac{1}{2}\right)^n$
Substituting the value of $n$,we get $\frac{N}{N_0} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$
This means $\frac{1}{4}$ of the original substance remains after $10$ years.
The fraction decayed is $1 - \frac{N}{N_0} = 1 - \frac{1}{4} = \frac{3}{4}$
To express this as a percentage,we multiply by $100$: $\frac{3}{4} \times 100 = 75 \%$
Therefore,$75 \%$ of the sample will decay in $10$ years.

(A) The law of radioactive decay states that the fraction of the sample remaining undecayed is given by the formula: $\frac{N}{N_0} = \left(\frac{1}{2}\right)^{\frac{t}{T}}$
Here,the total time elapsed is $t = 6400$ years and the half-life is $T = 1600$ years.
Substituting these values into the formula:
$\frac{N}{N_0} = \left(\frac{1}{2}\right)^{\frac{6400}{1600}}$
$\frac{N}{N_0} = \left(\frac{1}{2}\right)^4$
$\frac{N}{N_0} = \frac{1}{16}$
Thus,the fraction of the sample remaining undecayed after $6400$ years is $\frac{1}{16}$.

(D) When light passes through a lens,the refractive index of the lens material depends on the wavelength of the light. Since different colours have different wavelengths,they are refracted by different amounts. Consequently,rays of different colours do not focus at the same point after passing through a lens. This phenomenon is known as chromatic aberration.

(D) For a thin prism,the angle of deviation is given by $\delta = (\mu - 1)A$,where $\mu$ is the relative refractive index of the prism material with respect to the surrounding medium.
$1$. When the prism is in air:
$\mu_1 = \frac{\mu_{\text{glass}}}{\mu_{\text{air}}} = \frac{3/2}{1} = \frac{3}{2}$
$\delta_1 = (\frac{3}{2} - 1)A = \frac{1}{2}A$
$2$. When the prism is in water:
$\mu_2 = \frac{\mu_{\text{glass}}}{\mu_{\text{water}}} = \frac{3/2}{4/3} = \frac{9}{8}$
$\delta_2 = (\frac{9}{8} - 1)A = \frac{1}{8}A$
$3$. Calculating the ratio:
$\frac{\delta_1}{\delta_2} = \frac{\frac{1}{2}A}{\frac{1}{8}A} = \frac{1}{2} \times 8 = 4$
Therefore,the ratio $\delta_1 : \delta_2$ is $4 : 1$.

(C) The power of a lens is given by $P = \frac{1}{f} = (n_l - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a convex lens,$R_1 = +40 \ cm$ and $R_2 = -40 \ cm$.
In air $(n_a = 1)$: $P_{air} = (1.5 - 1) \left( \frac{1}{40} - \frac{1}{-40} \right) = 0.5 \times \frac{2}{40} = \frac{0.5}{20} = \frac{1}{40} \ cm^{-1}$.
In liquid $(n_l = 1.25)$: The effective refractive index is $n_{rel} = \frac{n_g}{n_l} = \frac{1.5}{1.25} = 1.2$.
$P_{liquid} = (1.2 - 1) \left( \frac{1}{40} - \frac{1}{-40} \right) = 0.2 \times \frac{2}{40} = \frac{0.4}{40} = \frac{1}{100} \ cm^{-1}$.
The ratio of power in liquid to air is $\frac{P_{liquid}}{P_{air}} = \frac{1/100}{1/40} = \frac{40}{100} = \frac{2}{5}$.

(B) The power of a lens is given by $P = \frac{1}{f}$, where $f$ is the focal length in meters.
For a combination of thin lenses in contact, the equivalent power $P$ is the algebraic sum of the individual powers:
$P = P_1 + P_2$
Given:
$P_1 = -15 D$
$P_2 = +5 D$
Substituting the values:
$P = -15 D + 5 D = -10 D$
Now, calculate the focal length $f$:
$f = \frac{1}{P} = \frac{1}{-10} \,m$
$f = -0.1 \,m$
Converting to centimeters $(1 \,m = 100 \,cm)$:
$f = -0.1 \times 100 \,cm = -10 \,cm$
Thus, the focal length of the combination is $-10 \,cm$.

(C) For a real image formed by a convex lens,the magnification $m$ is given by $m = \frac{v}{u}$,where $v$ is the image distance and $u$ is the object distance. Since the image is real,$m$ is negative. Let the magnitude of magnification be $M = |m|$. Then $v = M u$.
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{F}$.
Substituting $u = \frac{v}{M}$ (taking magnitudes): $\frac{1}{v} - \frac{M}{v} = \frac{1}{F}$.
$\frac{1-M}{v} = \frac{1}{F} \implies v = F(1-M)$.
However,in standard convention where $m$ is defined as the ratio of image size to object size,for a real image,$v = F(1+m)$ is the standard result derived from $\frac{1}{v} - \frac{1}{u} = \frac{1}{F}$ with $m = \frac{v}{u}$ (where $u$ is negative).
Thus,$v = F(1+m)$.

(A) For a double convex lens, the lens maker's formula is given by $\frac{1}{F} = (\mu - 1) \left( \frac{1}{R_1} + \frac{1}{R_2} \right)$.
Assuming $R_1 = R_2 = R$, we have $\frac{1}{F} = (\mu - 1) \left( \frac{2}{R} \right)$.
When the lens is cut into two equal parts along the vertical axis, each part becomes a plano-convex lens.
For a plano-convex lens, one radius of curvature is $R$ and the other is $\infty$.
Applying the lens maker's formula for the new focal length $F'$:
$\frac{1}{F'} = (\mu - 1) \left( \frac{1}{R} + \frac{1}{\infty} \right) = \frac{\mu - 1}{R}$.
Comparing the two equations:
$\frac{1/F}{1/F'} = \frac{(\mu - 1) \times (2/R)}{(\mu - 1) / R} = 2$.
Therefore, $\frac{F'}{F} = 2$, which gives $F' = 2 F$.

(A) The ability of a lens or mirror to converge or diverge light rays is defined as its power.
For a lens,the power $P$ is given by the reciprocal of the focal length $f$ in meters,i.e.,$P = 1/f$.
Therefore,the correct term for this ability is focal power (or simply power).

(D) From the lens maker's formula,$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a convex lens with equal radii of curvature $R$,we have $R_1 = R$ and $R_2 = -R$. Thus,$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (\mu - 1) \left( \frac{2}{R} \right)$.
So,$\frac{1}{f} = \frac{2(\mu - 1)}{R}$.
When one surface is made plane by grinding,the new radii are $R_1 = R$ and $R_2 = \infty$.
The new focal length $f'$ is given by $\frac{1}{f'} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right) = \frac{\mu - 1}{R}$.
Comparing the two equations,we see that $\frac{1}{f'} = \frac{1}{2} \left( \frac{1}{f} \right)$,which implies $f' = 2f$.
Since focal power $P = \frac{1}{f}$,the new power $P' = \frac{1}{f'} = \frac{1}{2f} = \frac{P}{2}$.
Therefore,the new focal length is $2f$ and the new power is $\frac{P}{2}$.

(D) Let the powers of the two lenses be $P_1$ and $P_2$. When the lenses are in contact,the equivalent power is given by $P = P_1 + P_2 = 10 D$.
When the lenses are separated by a distance $d = 0.25 m$,the equivalent power is given by $P = P_1 + P_2 - d P_1 P_2 = 6 D$.
Substituting $P_1 + P_2 = 10$ into the second equation: $10 - 0.25 P_1 P_2 = 6$.
This simplifies to $0.25 P_1 P_2 = 4$,so $P_1 P_2 = 16$.
We know that $(P_1 - P_2)^2 = (P_1 + P_2)^2 - 4 P_1 P_2$.
$(P_1 - P_2)^2 = (10)^2 - 4(16) = 100 - 64 = 36$.
Thus,$P_1 - P_2 = 6 D$.
Solving the system $P_1 + P_2 = 10$ and $P_1 - P_2 = 6$,we get $2 P_1 = 16 \Rightarrow P_1 = 8 D$ and $P_2 = 2 D$.

(B) The resolving power $(P)$ of a compound microscope is given by the formula: $P = \frac{2 \mu \sin \theta}{1.22 \lambda}$.
Here,$\mu$ is the refractive index of the medium between the object and the objective lens,$\theta$ is the half-angle of the cone of light from the object,and $\lambda$ is the wavelength of light used.
To improve the resolving power,one must increase the numerator or decrease the denominator.
Therefore,increasing the refractive index $(\mu)$ of the medium between the object and the objective lens increases the resolving power.

(C) The formation of a secondary rainbow involves two refractions,two total internal reflections,and angular dispersion of light within the water droplets.
Interference is a wave phenomenon that is not a primary mechanism in the geometric optics description of rainbow formation.
Therefore,interference is $NOT$ involved in the formation of a secondary rainbow.

(D) For the light ray to retrace its path after reflection from the silvered surface,it must strike the silvered surface normally.
In the prism,the angle of refraction at the first surface is '$r_1$'. From the geometry of the prism,the angle of incidence at the second surface is '$r_2$'. Since the ray strikes normally,'$r_2 = 0$'.
Using the relation '$A = r_1 + r_2$',we get '$A = r_1 + 0$',so '$r_1 = A$'.
Applying Snell's law at the first surface: '$\mu = \frac{\sin i}{\sin r_1}$'.
Given the angle of incidence '$i = 2A$',we have '$\mu = \frac{\sin 2A}{\sin A}$'.
Using the trigonometric identity '$\sin 2A = 2 \sin A \cos A$',we get '$\mu = \frac{2 \sin A \cos A}{\sin A} = 2 \cos A$'.
Thus,the refractive index of the material of the prism is '$2 \cos A$'.

(A) The dispersive power $\omega$ of a prism is given by the formula $\omega = \frac{\delta_v - \delta_r}{\delta_y}$,where $\delta_v$ is the deviation for violet light,$\delta_r$ is the deviation for red light,and $\delta_y$ is the mean deviation,calculated as $\delta_y = \frac{\delta_v + \delta_r}{2}$.
For the first prism:
$\delta_v = 11^{\circ}$,$\delta_r = 9^{\circ}$.
$\delta_{y1} = \frac{11 + 9}{2} = 10^{\circ}$.
$\omega_1 = \frac{11 - 9}{10} = \frac{2}{10} = \frac{1}{5}$.
For the second prism:
$\delta_v = 13^{\circ}$,$\delta_r = 11^{\circ}$.
$\delta_{y2} = \frac{13 + 11}{2} = 12^{\circ}$.
$\omega_2 = \frac{13 - 11}{12} = \frac{2}{12} = \frac{1}{6}$.
The ratio of the dispersive power of the second prism to the first prism is $\frac{\omega_2}{\omega_1} = \frac{1/6}{1/5} = \frac{5}{6}$.
Thus,the ratio is $5: 6$.

(C) Given: The ray emerges normally from the second surface,so the angle of emergence $e = 0$.
Since $e = 0$,the angle of refraction at the second surface $r_2 = 0$.
For a prism,the angle of the prism $A = r_1 + r_2$. Substituting $r_2 = 0$,we get $A = r_1$.
According to Snell's law at the first surface,$\mu = \frac{\sin i}{\sin r_1}$.
Since the prism angle $A$ is small,$i$ and $r_1$ are also small,so $\sin i \approx i$ and $\sin r_1 \approx r_1$.
Thus,$\mu = \frac{i}{r_1}$.
Substituting $r_1 = A$,we get $\mu = \frac{i}{A}$,which implies $i = \mu A$.

(B) For an equilateral prism,the angle of the prism is $A = 60^{\circ}$.
Given that the angle of incidence $(i)$ is equal to the angle of emergence $(e)$,and $e = \frac{3}{4} A$.
Substituting the value of $A$:
$e = \frac{3}{4} \times 60^{\circ} = 45^{\circ}$.
Since $i = e$,we have $i = 45^{\circ}$.
The formula for the angle of deviation $(\delta)$ is $\delta = i + e - A$.
Substituting the values:
$\delta = 45^{\circ} + 45^{\circ} - 60^{\circ} = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Therefore,the angle of deviation is $30^{\circ}$.

(C) According to the law of reflection,the angle of incidence $i$ is equal to the angle of reflection $r$,so $i = r$.
According to Snell's Law,for light traveling from a denser medium to a rarer medium with refractive index $n$ (where $n$ is the refractive index of the denser medium relative to the rarer medium,$n > 1$):
$\frac{\sin i}{\sin r^{\prime}} = \frac{1}{n}$
Since $i = r$,we have $\frac{\sin r}{\sin r^{\prime}} = \frac{1}{n}$.
Given that the reflected and refracted rays are at right angles to each other,the sum of the angles on the straight line is $r + 90^{\circ} + r^{\prime} = 180^{\circ}$.
Therefore,$r^{\prime} = 90^{\circ} - r$.
Substituting this into Snell's Law:
$\frac{\sin r}{\sin(90^{\circ} - r)} = \frac{1}{n}$
$\frac{\sin r}{\cos r} = \frac{1}{n}$
$\tan r = \frac{1}{n}$
The critical angle $i_c$ is defined by $\sin i_c = \frac{1}{n}$.
Comparing the two equations,we get $\sin i_c = \tan r$.
Therefore,$i_c = \sin^{-1}(\tan r)$.

(D) The number of images $n$ formed by two plane mirrors inclined at an angle $\theta$ is given by the formula $n = \left( \frac{360^{\circ}}{\theta} - 1 \right)$ if $\frac{360^{\circ}}{\theta}$ is an even integer.
Given that the number of images $n = 3$, we substitute this into the formula:
$3 = \frac{360^{\circ}}{\theta} - 1$
$4 = \frac{360^{\circ}}{\theta}$
$\theta = \frac{360^{\circ}}{4} = 90^{\circ}$
Thus, the angle between the two plane mirrors should be $90^{\circ}$.

(D) Total internal reflection $(TIR)$ occurs when light travels from a denser medium to a rarer medium.
For $TIR$ to occur,two conditions must be met:
$1$. The light must travel from an optically denser medium to an optically rarer medium.
$2$. The angle of incidence $(i)$ must be greater than the critical angle $(i_{c})$ for the given pair of media.
Therefore,the correct condition is $i > i_{c}$.

(B) The normal shift $x$ produced by a glass slab of thickness $t$ and refractive index $\mu$ is given by the formula:
$x = t \left(1 - \frac{1}{\mu}\right)$
Rearranging the formula to solve for $t$:
$x = t \left(\frac{\mu - 1}{\mu}\right)$
$t = \frac{x \cdot \mu}{\mu - 1}$
Therefore,the thickness of the glass slab is $t = \frac{\mu x}{\mu - 1}$.

(A) Let the distance of the air bubble from the first surface be $l_1$ and from the second surface be $l_2$. The total length of the cube is $L = 24 \,cm$. Thus, $l_1 + l_2 = 24 \,cm$, which implies $l_2 = 24 - l_1$.
The formula for apparent depth is given by $\mu = \frac{\text{Real depth}}{\text{Apparent depth}}$.
For the first surface:
$\mu = \frac{l_1}{10} \quad \dots(i)$
For the second surface:
$\mu = \frac{l_2}{6} = \frac{24 - l_1}{6} \quad \dots(ii)$
Equating $(i)$ and $(ii)$:
$\frac{l_1}{10} = \frac{24 - l_1}{6}$
$6l_1 = 10(24 - l_1)$
$6l_1 = 240 - 10l_1$
$16l_1 = 240$
$l_1 = \frac{240}{16} = 15 \,cm$
Therefore, the distance of the air bubble from the first surface is $15 \,cm$.

(B) The refractive index of water is $\mu_w = 1.33$ and the refractive index of air is $\mu_a = 1.0$.
According to the Lens Maker's Formula,the focal length $f$ is given by $\frac{1}{f} = (\frac{\mu_l}{\mu_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$,where $\mu_l$ is the refractive index of the lens material and $\mu_m$ is the refractive index of the surrounding medium.
For a double convex air bubble,$R_1 > 0$ and $R_2 < 0$,so $(\frac{1}{R_1} - \frac{1}{R_2}) > 0$.
Here,$\mu_l = \mu_a = 1.0$ and $\mu_m = \mu_w = 1.33$.
Since $\frac{\mu_a}{\mu_w} = \frac{1.0}{1.33} < 1$,the term $(\frac{\mu_a}{\mu_w} - 1)$ is negative.
Therefore,the focal length $f$ is negative,which indicates that the air bubble behaves as a divergent lens.

(D) Given: $u = -x$ (object is in air),$v = +x$ (image is in glass),$n_1 = 1$ (refractive index of air),$n_2 = 1.5$ (refractive index of glass),and $R$ is the radius of curvature.
Using the refraction formula for a spherical surface:
$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$
Substituting the given values:
$\frac{1.5}{x} - \frac{1}{-x} = \frac{1.5 - 1}{R}$
$\frac{1.5}{x} + \frac{1}{x} = \frac{0.5}{R}$
$\frac{2.5}{x} = \frac{0.5}{R}$
$x = \frac{2.5}{0.5} R = 5 R$
Therefore,the distance $x$ is equal to $5 R$.

(C) For a prism,the angle of deviation is given by $\delta = i + e - A$. When the refracted ray inside the prism is parallel to the base,the prism is in the position of minimum deviation. In the position of minimum deviation,the angle of incidence is equal to the angle of emergence,i.e.,$i = e$. Therefore,the correct option is $C$.

(B) The zener diode operates at a constant voltage $V_{Z} = 5.0 \, V$.
Given the input voltage $V_{S} = 12 \, V$ and the maximum power rating $P_{Z} = 2.0 \, W$.
The maximum current $I_{Z_{\max}}$ that the zener diode can handle is calculated as:
$I_{Z_{\max}} = \frac{P_{Z}}{V_{Z}} = \frac{2.0 \, W}{5.0 \, V} = 0.4 \, A = 400 \, mA$.
The series resistance $R_{S}$ is used to drop the excess voltage $(V_{S} - V_{Z})$ and limit the current to $I_{Z_{\max}}$.
$R_{S} = \frac{V_{S} - V_{Z}}{I_{Z_{\max}}} = \frac{12 \, V - 5.0 \, V}{0.4 \, A} = \frac{7}{0.4} \, \Omega = 17.5 \, \Omega$.

(B) When a $p-n$ junction diode is reverse biased,the positive terminal of the battery is connected to the $n$-region and the negative terminal to the $p$-region.
This configuration pulls the majority charge carriers (holes in $p$-region and electrons in $n$-region) away from the junction.
As a result,the depletion layer is depleted of charge carriers even further,causing its width to increase.
Therefore,the correct option is $B$.