MHT CET 2021 Physics Question Paper with Answer and Solution

(C) According to the Stefan-Boltzmann law,the power $P$ radiated by a perfectly black body is proportional to the fourth power of its absolute temperature $T$,given by $P = \sigma A T^4$.
Given the ratio of powers is $\frac{P_2}{P_1} = 16$.
Using the relation $\frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^4$,we substitute the given values:
$16 = \left(\frac{T_2}{T_1}\right)^4$
Taking the fourth root on both sides:
$\left(2^4\right)^{1/4} = \frac{T_2}{T_1}$
$2 = \frac{T_2}{T_1}$
Therefore,$T_2 = 2 \ T_1$.

(A) According to Stefan-Boltzmann Law,the energy emitted per second is given by $E = \sigma A T^4$,where $\sigma$ is the Stefan-Boltzmann constant,$A$ is the area,and $T$ is the absolute temperature in Kelvin.
Initial state: $E = \sigma A T_1^4$,where $T_1 = 27 + 273 = 300 \ K$.
Final state: $E' = \sigma A' T_2^4$,where $T_2 = 327 + 273 = 600 \ K$.
The area $A = \ell \times b$. If length and breadth are reduced to $\frac{1}{3}$ of their initial values,the new area $A' = (\frac{\ell}{3}) \times (\frac{b}{3}) = \frac{A}{9}$.
Taking the ratio: $\frac{E'}{E} = \frac{A'}{A} \times (\frac{T_2}{T_1})^4$.
Substituting the values: $\frac{E'}{E} = \frac{1}{9} \times (\frac{600}{300})^4 = \frac{1}{9} \times (2)^4 = \frac{16}{9}$.
Therefore,$E' = \frac{16 E}{9}$.

(B) According to Wien's displacement law,the product of the wavelength of maximum emission and the absolute temperature is a constant:
$\lambda_m T = \text{constant}$
Therefore,$\lambda_m \propto \frac{1}{T}$.
Given $T_1 = 2000 \ K$ and $T_2 = 3000 \ K$,and the initial wavelength is $\lambda_m$:
$\frac{\lambda_2}{\lambda_m} = \frac{T_1}{T_2}$
$\frac{\lambda_2}{\lambda_m} = \frac{2000}{3000} = \frac{2}{3}$
$\lambda_2 = \frac{2}{3} \lambda_m$
Thus,the corresponding wavelength at $3000 \ K$ is $\frac{2}{3} \lambda_m$.

(D) According to Stefan-Boltzmann Law,the rate of energy emission $R$ from a black body is proportional to the fourth power of its absolute temperature $T$,given by $R \propto T^4$.
Since the distance between the sun and the earth remains constant,the rate of energy received by the earth is directly proportional to the rate of energy emission from the sun.
Let $T_1$ be the initial temperature and $T_2 = 2T_1$ be the final temperature.
The ratio of the rates of energy received is $\frac{R_2}{R_1} = \left(\frac{T_2}{T_1}\right)^4$.
Substituting the values,we get $\frac{R_2}{R_1} = (2)^4 = 16$.
Therefore,the rate of energy received by the earth will increase by a factor of $16$.

(A) The emissive power $E$ is defined as the heat energy $Q$ radiated per unit area $A$ per unit time $t$, given by the formula $E = \frac{Q}{A \cdot t}$.
To find the total heat radiated $Q$, we rearrange the formula: $Q = E \cdot A \cdot t$.
Given values are $E = 0.7 \,kcal \,s^{-1} \,m^{-2}$, $A = 0.04 \,m^2$, and $t = 20 \,s$.
Substituting these values: $Q = 0.7 \times 0.04 \times 20$.
$Q = 0.7 \times 0.8 = 0.56 \,kcal$.
Therefore, the amount of heat radiated is $0.56 \,kcal$.

(C) According to Stefan-Boltzmann law,the rate of radiation $R$ is proportional to the fourth power of the absolute temperature $T$ of the black body: $R \propto T^4$.
Let the initial temperature be $T_1 = T$ and the initial rate of radiation be $R_1$.
The temperature is increased by $50 \%$,so the new temperature $T_2 = T + 0.5T = 1.5T$.
The new rate of radiation $R_2$ is given by:
$\frac{R_2}{R_1} = \left(\frac{T_2}{T_1}\right)^4 = (1.5)^4 = 5.0625 \approx 5$.
Thus,$R_2 \approx 5R_1$.
The percentage increase in the rate of radiation is given by:
$\text{Percentage increase} = \left(\frac{R_2 - R_1}{R_1}\right) \times 100 = \left(\frac{5R_1 - R_1}{R_1}\right) \times 100 = 4 \times 100 = 400 \%$.

(B) According to Newton's law of cooling,the rate of cooling is proportional to the temperature difference between the body and its surroundings: $\frac{dT}{dt} = k(T - T_s)$.
For the first case,$T_1 = 80^{\circ} C$ and $T_s = 30^{\circ} C$. The rate is $1.5^{\circ} C / min$.
$1.5 = k(80 - 30) = k(50) \implies k = \frac{1.5}{50} = 0.03 \ min^{-1}$.
For the second case,$T_2 = 50^{\circ} C$ and $T_s = 30^{\circ} C$. Let the rate be $r$.
$r = k(50 - 30) = k(20)$.
Substituting the value of $k$:
$r = 0.03 \times 20 = 0.6^{\circ} C / min$.

(A) Given: Surface temperature $T = 6000 \,K$,Wien's constant $b = 2.897 \times 10^{-3} \,m-K$.
According to Wien's displacement law,$\lambda_{max} T = b$.
Therefore,$\lambda_{max} = \frac{b}{T} = \frac{2.897 \times 10^{-3}}{6000} \,m$.
$\lambda_{max} = 4.828 \times 10^{-7} \,m$.
Since $1 \,Å = 10^{-10} \,m$,we have $4.828 \times 10^{-7} \,m = 4828 \times 10^{-10} \,m = 4828 \,Å$.

(D) Let the change in temperature be $\Delta T$. The change in length of the first rod is $\Delta L_1 = L_1 \alpha_1 \Delta T$.
The change in length of the second rod is $\Delta L_2 = L_2 \alpha_2 \Delta T$.
Given that the difference $(L_2 - L_1)$ remains constant at all temperatures,the change in length of both rods must be equal.
Therefore,$\Delta L_1 = \Delta L_2$.
Substituting the expressions,we get $L_1 \alpha_1 \Delta T = L_2 \alpha_2 \Delta T$.
Canceling $\Delta T$ from both sides,we obtain the relation $L_1 \alpha_1 = L_2 \alpha_2$.

(C) In the Celsius scale,the freezing point of water is $0^{\circ} C$ and the boiling point is $100^{\circ} C$. The range is $100 - 0 = 100$.
In the given imaginary '$W$' scale,the freezing point is $39^{\circ} W$ and the boiling point is $239^{\circ} W$. The range is $239 - 39 = 200$.
Using the linear conversion formula between two temperature scales:
$\frac{C - C_{freezing}}{C_{boiling} - C_{freezing}} = \frac{W - W_{freezing}}{W_{boiling} - W_{freezing}}$
Substituting the values:
$\frac{C - 0}{100 - 0} = \frac{W - 39}{239 - 39}$
$\frac{C}{100} = \frac{W - 39}{200}$
For $C = 39^{\circ} C$:
$\frac{39}{100} = \frac{W - 39}{200}$
$39 \times 2 = W - 39$
$78 = W - 39$
$W = 78 + 39 = 117^{\circ} W$.

(B) The relation between temperature in Celsius $(C)$ and Kelvin $(K)$ is given by:
$T(K) = T(^{\circ}C) + 273.15$
Given $T(^{\circ}C) = -197^{\circ}C$.
Substituting the value:
$T(K) = -197 + 273 = 76 K$
Therefore,the temperature is $76 K$.

(A) The change in internal energy $\Delta U$ for an ideal gas is given by $\Delta U = n C_v \Delta T$.
Using the relation $C_v = \frac{R}{\gamma-1}$,we get $\Delta U = n \left( \frac{R}{\gamma-1} \right) \Delta T$.
From the ideal gas law $PV = nRT$,at constant pressure $P$,we have $P \Delta V = nR \Delta T$.
Substituting this into the internal energy equation: $\Delta U = \frac{P \Delta V}{\gamma-1}$.
Given that the volume changes from $V$ to $2V$,the change in volume is $\Delta V = 2V - V = V$.
Therefore,$\Delta U = \frac{P(V)}{\gamma-1} = \frac{PV}{\gamma-1}$.

(B) The internal energy $U$ of an ideal gas is given by the formula $U = n \frac{f}{2} RT$,where $n$ is the number of moles,$f$ is the degree of freedom,$R$ is the universal gas constant,and $T$ is the absolute temperature.
For $n_1$ moles of hydrogen $(H_2)$ at temperature $T$ with $f_1 = 5$:
$U_1 = n_1 \times \frac{5}{2} \times R \times T$
For $n_2$ moles of helium $(He)$ at temperature $2T$ with $f_2 = 3$:
$U_2 = n_2 \times \frac{3}{2} \times R \times (2T)$
Given that $U_1 = U_2$:
$n_1 \times \frac{5}{2} \times RT = n_2 \times \frac{3}{2} \times R \times 2T$
Canceling $\frac{1}{2}$,$R$,and $T$ from both sides:
$5n_1 = 6n_2$
Therefore,the ratio $\frac{n_1}{n_2} = \frac{6}{5}$.

(C) For an adiabatic process, the relation between pressure and volume is given as $PV^{\gamma} = \text{constant}$. Here, $\gamma = 3/2$.
Using the ideal gas law $PV = nRT$, we can write $P = nRT/V$.
Substituting this into the adiabatic equation: $(nRT/V)V^{\gamma} = \text{constant}$, which simplifies to $TV^{\gamma-1} = \text{constant}$.
Given $\gamma = 3/2$, the relation becomes $TV^{(3/2 - 1)} = TV^{1/2} = \text{constant}$.
Let the initial state be $(T_1, V_1)$ and the final state be $(T_2, V_2)$.
Given $T_1 = T$ and $V_2 = V_1/2$.
Using $T_1 V_1^{1/2} = T_2 V_2^{1/2}$:
$T_2 = T_1 (V_1/V_2)^{1/2} = T (V_1 / (V_1/2))^{1/2} = T (2)^{1/2} = \sqrt{2}T$.

(B) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = \text{constant}$.
Given the condition $T \propto \frac{1}{\sqrt{V}}$,we can write this as $T \propto V^{-1/2}$,which implies $TV^{1/2} = \text{constant}$.
Comparing the two expressions $TV^{\gamma-1} = \text{constant}$ and $TV^{1/2} = \text{constant}$,we get $\gamma - 1 = \frac{1}{2}$.
Solving for $\gamma$,we find $\gamma = 1 + 0.5 = 1.5$.

(D) For a monoatomic gas,the molar heat capacities are $C_p = \frac{5}{2}R$ and $C_v = \frac{3}{2}R$.
Work done at constant pressure is $W = nR \Delta T$.
Heat supplied at constant volume is $Q_v = nC_v \Delta T = n \left( \frac{3}{2}R \right) \Delta T$.
Substituting $nR \Delta T = W$ into the expression for $Q_v$,we get $Q_v = \frac{3}{2} W$.

(D) We know that for an ideal gas,the relationship between specific heats at constant pressure $(C_p)$ and constant volume $(C_v)$ is given by Mayer's relation: $C_p - C_v = R$.
Given that the ratio of specific heats is $\gamma = \frac{C_p}{C_v}$,we can write $C_p = \gamma C_v$.
Substituting this into Mayer's relation: $\gamma C_v - C_v = R$.
Factoring out $C_v$: $C_v(\gamma - 1) = R$.
Therefore,$C_v = \frac{R}{\gamma - 1}$.

(D) An isobaric process is a thermodynamic process in which the pressure of the system remains constant throughout the process.
By definition,if the pressure $P$ is constant,then the change in pressure $\Delta P$ must be equal to zero.
Therefore,the equation that specifies an isobaric process is $\Delta P=0$.

(A) For the isothermal process,the relation is $P_1 V_1 = P_2 V_2$. Given $P_1 = P$,$V_1 = V$,and $V_2 = 2V$.
Substituting these values: $P \times V = P_2 \times 2V$,which gives $P_2 = \frac{P}{2}$.
For the adiabatic process,the relation is $P_2 V_2^\gamma = P_3 V_3^\gamma$. Given $V_2 = 2V$,$V_3 = 16V$,and $\gamma = \frac{5}{3}$.
Substituting these values: $P_3 = P_2 \left(\frac{V_2}{V_3}\right)^\gamma = \frac{P}{2} \left(\frac{2V}{16V}\right)^{5/3}$.
$P_3 = \frac{P}{2} \left(\frac{1}{8}\right)^{5/3} = \frac{P}{2} \times \left(\left(\frac{1}{2}\right)^3\right)^{5/3} = \frac{P}{2} \times \left(\frac{1}{2}\right)^5 = \frac{P}{2} \times \frac{1}{32} = \frac{P}{64}$.

(A) An isochoric process is defined as a process where the volume of the system remains constant $(dV = 0)$.
Since work done $dW = P dV$,if $dV = 0$,then $dW = 0$. Thus,no work is done in the process.
According to the first law of thermodynamics,$dQ = dU + dW$. Since $dW = 0$,the heat exchanged $(dQ)$ is entirely used to change the internal energy $(dU)$ of the system.
Therefore,statement $(A)$ is incorrect because energy exchanged is not used to do work.

(B) According to the first law of thermodynamics,the change in internal energy $\Delta U$ is given by $\Delta Q = \Delta U + dW$,where $\Delta Q$ is the heat supplied to the system and $dW$ is the work done by the system.
In an adiabatic process,there is no exchange of heat with the surroundings,so $\Delta Q = 0$.
Substituting this into the equation,we get $0 = \Delta U + dW$.
Therefore,$\Delta U = -dW$.

(D) For an adiabatic process,the relationship between pressure and volume is given by $P_1 V_1^\gamma = P_2 V_2^\gamma$.
Given that the gas is compressed to $(1/8)^{\text{th}}$ of its initial volume,we have $V_2 = V_1 / 8$,which implies $V_1 / V_2 = 8$.
The adiabatic index is given as $\gamma = 4/3$.
Substituting these values into the adiabatic equation:
$P_2 / P_1 = (V_1 / V_2)^\gamma$
$P_2 / P_0 = (8)^{4/3}$
$P_2 / P_0 = (2^3)^{4/3} = 2^4 = 16$
Therefore,the new pressure is $P_2 = 16 P_0$.

(D) For an isothermal process,the temperature remains constant,so Boyle's Law applies: $P_1 V_1 = P_2 V_2$.
Given that the pressure is decreased by $10 \%$,the new pressure $P_2$ is:
$P_2 = P_1 - 0.10 P_1 = 0.9 P_1 = \frac{9}{10} P_1$.
Substituting this into the equation:
$P_1 V_1 = (\frac{9}{10} P_1) V_2$.
Solving for $V_2$:
$V_2 = \frac{10}{9} V_1 \approx 1.111 V_1$.
The change in volume is $\Delta V = V_2 - V_1 = 1.111 V_1 - V_1 = 0.111 V_1$.
Expressed as a percentage,the volume increases by approximately $11.1 \%$,which is closest to $11 \%$.

(D) For an adiabatic process,the relation between temperature and volume is $TV^{\gamma-1} = \text{constant}$.
Thus,$\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1}$.
Given $T_1 = 27^{\circ} C = 300 \ K$ and $V_2 = \frac{8}{27} V_1$,so $\frac{V_1}{V_2} = \frac{27}{8}$.
Given $\gamma = 5/3$,so $\gamma - 1 = 5/3 - 1 = 2/3$.
Substituting the values: $\frac{T_2}{300} = \left(\frac{27}{8}\right)^{2/3} = \left(\left(\frac{3}{2}\right)^3\right)^{2/3} = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$.
$T_2 = \frac{9}{4} \times 300 = 9 \times 75 = 675 \ K$.
The rise in temperature is $\Delta T = T_2 - T_1 = 675 \ K - 300 \ K = 375 \ K$.

(D) At constant pressure, the work done is given by $W = P \Delta V$.
For an ideal gas, $PV = nRT$, so $P \Delta V = nR \Delta T$.
Thus, $W = nR \Delta T$.
Heat supplied at constant volume is given by $Q = \Delta U = nC_v \Delta T$.
For a monoatomic gas, the molar heat capacity at constant volume is $C_v = \frac{3}{2}R$.
Substituting this into the heat equation, we get $Q = n \left( \frac{3}{2}R \right) \Delta T = \frac{3}{2} (nR \Delta T)$.
Since $W = nR \Delta T$, we substitute $W$ into the equation for $Q$:
$Q = \frac{3}{2} W$.

(B) For isothermal expansion,the process follows Boyle's Law: $P_1 V_1 = P_2 V_2$.
Given $P_1 = P$,$V_1 = V$,and $V_2 = 2V$.
Therefore,$P_i = P \times (V / 2V) = P / 2$.
For adiabatic expansion,the process follows the relation $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$.
Given $P_1 = P$,$V_1 = V$,and $V_2 = 2V$.
Therefore,$P_a = P \times (V / 2V)^{\gamma} = P / 2^{\gamma}$.
Now,calculating the ratio $\frac{P_i}{P_a}$:
$\frac{P_i}{P_a} = \frac{P/2}{P/2^{\gamma}} = \frac{2^{\gamma}}{2} = 2^{\gamma-1}$.

(D) For an ideal gas,density $\rho = \frac{m}{V}$. Since the mass $m$ remains constant,$\rho \propto \frac{1}{V}$.
Given that the final density is twice the initial density,$\rho_2 = 2\rho_1$,which implies $V_2 = \frac{V_1}{2}$ or $\frac{V_1}{V_2} = 2$.
For an adiabatic process,the relation between pressure and volume is $P_1 V_1^\gamma = P_2 V_2^\gamma$.
Therefore,the final pressure $P_2$ is given by $P_2 = P_1 \left( \frac{V_1}{V_2} \right)^\gamma$.
Substituting the given values,$P_2 = p \times (2)^{7/5} = p \times (2)^{1.4}$.
Calculating the value,$2^{1.4} \approx 2.639$.
Thus,the final pressure is approximately $2.63p$.

(C) An isobaric process is defined as a thermodynamic process in which the pressure of the system remains constant $(P = \text{constant})$.
According to the ideal gas law, $PV = nRT$. Since $P$ is constant, $V \propto T$.
Therefore, if the volume changes during the process, the temperature of the system must also change.
Statement $(C)$ claims that the temperature remains constant, which is incorrect for an isobaric process; it is a characteristic of an isothermal process.

(C) Given: $\gamma = \frac{C_p}{C_v} = \frac{3}{2}$.
Case $I$: Isothermal process $(T = \text{constant})$
For an isothermal process,$P_1 V_1 = P_2 V_2$.
Given $P_1 = P$,$V_1 = V$,and $V_2 = 9V$.
$P \times V = P_2 \times 9V$
$P_2 = \frac{P}{9}$.
Case $II$: Adiabatic process $(PV^\gamma = \text{constant})$
For an adiabatic process,$P_2 V_2^\gamma = P_3 V_3^\gamma$.
Given $P_2 = \frac{P}{9}$,$V_2 = 9V$,and $V_3 = V$.
$\frac{P}{9} \times (9V)^{3/2} = P_3 \times (V)^{3/2}$
$P_3 = \frac{P}{9} \times \left(\frac{9V}{V}\right)^{3/2}$
$P_3 = \frac{P}{9} \times (9)^{3/2}$
$P_3 = \frac{P}{9} \times (3^2)^{3/2} = \frac{P}{9} \times 3^3 = \frac{P}{9} \times 27 = 3P$.
Therefore,the final pressure is $3P$.

(B) In an isochoric process,the volume of the system remains constant throughout the process.
On a $p-V$ diagram,where pressure $p$ is plotted on the $y$-axis and volume $V$ is plotted on the $x$-axis,a constant volume process is represented by a vertical line.
This vertical line indicates that for any change in pressure,the value of $V$ remains the same.
Looking at the provided diagrams (not fully shown but implied by standard physics problems of this type),the diagram representing a vertical line corresponds to the isochoric process.
Assuming the standard set of diagrams where $II$ represents the vertical line,the correct option is $II$.

(A) Given: Initial volume $V_1$,Final volume $V_2 = V_1/8$,Adiabatic index $\gamma = 5/3$.
For an adiabatic process,the relation between pressure and volume is given by $P_1 V_1^\gamma = P_2 V_2^\gamma$.
Rearranging for the ratio of final pressure to initial pressure: $\frac{P_2}{P_1} = \left(\frac{V_1}{V_2}\right)^\gamma$.
Substituting the values: $\frac{P_2}{P_1} = (8)^{5/3}$.
Since $8 = 2^3$,we have $(2^3)^{5/3} = 2^5 = 32$.
Thus,the ratio of the final pressure to the initial pressure is $32$.

(B) $\text{Density} = \frac{\text{Mass}}{\text{Volume}}$
$\text{The SI unit of mass is kilogram } (kg)$.
$\text{The SI unit of volume is cubic meter } (m^3)$.
$\text{Therefore, the SI unit of density is } \frac{kg}{m^3} \text{ or } kg \cdot m^{-3}$.

(A) Sound waves are mechanical waves that require a medium to travel. In air,particles oscillate parallel to the direction of wave propagation,making them longitudinal waves.
Light waves are electromagnetic waves that do not require a medium. They consist of oscillating electric and magnetic fields perpendicular to the direction of propagation,making them transverse waves.
Therefore,the correct statement is that sound waves in air are longitudinal while light waves in air are transverse.

(D) The given wave equation is $y = 10 \sin \left(\frac{2 \pi t}{30} + \alpha\right)$.
At $t = 0 \ s$,the displacement $y = 5 \ cm$.
Substituting these values into the equation: $5 = 10 \sin \left(\frac{2 \pi (0)}{30} + \alpha\right) \implies 5 = 10 \sin \alpha \implies \sin \alpha = \frac{1}{2}$.
Thus,the initial phase constant $\alpha = \frac{\pi}{6} \ rad$.
The total phase angle $\phi$ at any time $t$ is given by $\phi = \frac{2 \pi t}{30} + \alpha$.
At $t = 7.5 \ s$,the total phase angle is $\phi = \frac{2 \pi (7.5)}{30} + \frac{\pi}{6}$.
$\phi = \frac{15 \pi}{30} + \frac{\pi}{6} = \frac{\pi}{2} + \frac{\pi}{6}$.
$\phi = \frac{3 \pi + \pi}{6} = \frac{4 \pi}{6} = \frac{2 \pi}{3} \ rad$.

(C) The relationship between phase difference $\Delta \phi$ and time interval $\Delta t$ is given by:
$\Delta \phi = \omega \Delta t = (2\pi f) \Delta t$
Given frequency $f = 50 \,Hz$ and time interval $\Delta t = 0.01 \,s$.
Substituting the values:
$\Delta \phi = 2 \times \pi \times 50 \times 0.01$
$\Delta \phi = 100 \pi \times 0.01$
$\Delta \phi = \pi \,rad$.

(B) The given wave equation is $Y = Y_0 \sin(2 \pi n t - \frac{2 \pi x}{\lambda})$.
Comparing this with the standard form $Y = Y_0 \sin(\omega t - kx)$,we get angular frequency $\omega = 2 \pi n$.
The maximum particle velocity $v_{p, \text{max}} = Y_0 \omega = Y_0 (2 \pi n) = 2 \pi n Y_0$.
The wave velocity $v = n \lambda$.
According to the problem,the wave velocity is $(1/8)^{\text{th}}$ of the maximum particle velocity:
$v = \frac{1}{8} v_{p, \text{max}}$
$n \lambda = \frac{1}{8} (2 \pi n Y_0)$
$n \lambda = \frac{\pi n Y_0}{4}$
$\lambda = \frac{\pi Y_0}{4}$.

(D) The standard equation for a wave is $Y = A \sin 2 \pi \left(\frac{t}{T} - \frac{x}{\lambda}\right)$.
For the first wave: $Y_1 = 2 \sin 2 \pi \left(\frac{4t}{0.2} - \frac{4x}{2}\right) = 2 \sin 2 \pi \left(\frac{t}{0.05} - \frac{x}{0.5}\right)$.
Here,$T_1 = 0.05 \text{ s}$ and $\lambda_1 = 0.5 \text{ m}$.
Velocity $v_1 = \frac{\lambda_1}{T_1} = \frac{0.5}{0.05} = 10 \text{ m/s}$.
For the second wave: $Y_2 = 4 \sin 2 \pi \left(\frac{4t}{0.16} - \frac{4x}{1.6}\right) = 4 \sin 2 \pi \left(\frac{t}{0.04} - \frac{x}{0.4}\right)$.
Here,$T_2 = 0.04 \text{ s}$ and $\lambda_2 = 0.4 \text{ m}$.
Velocity $v_2 = \frac{\lambda_2}{T_2} = \frac{0.4}{0.04} = 10 \text{ m/s}$.
Since $v_1 = v_2 = 10 \text{ m/s}$,both waves have the same velocity.

(A) The standard equation of a traveling wave is given by $y=A \sin (kx+\omega t)$.
Comparing this with the given equation $y=4 \sin (3x+60t)$,we get the wave number $k=3 \,m^{-1}$ and the angular frequency $\omega=60 \,rad/s$.
The wave speed $V$ is calculated as $V = \frac{\omega}{k} = \frac{60}{3} = 20 \,m/s$.
The speed of a transverse wave in a stretched string is given by $V = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the mass per unit length.
Squaring both sides,we get $V^2 = \frac{T}{\mu}$,which implies $\mu = \frac{T}{V^2}$.
Substituting the given values $T = 0.4 \,N$ and $V = 20 \,m/s$:
$\mu = \frac{0.4}{(20)^2} = \frac{0.4}{400} = \frac{4 \times 10^{-1}}{4 \times 10^2} = 10^{-3} \,kg \,m^{-1}$.

(A) The relationship between phase difference $\phi$ and path difference $x$ is given by $\phi = \frac{2 \pi}{\lambda} x$.
First,calculate the wavelength $\lambda$ using the formula $\lambda = \frac{v}{f}$,where $v = 330 \,m/s$ and $f = 50 \,Hz$.
$\lambda = \frac{330}{50} = 6.6 \,m$.
Now,rearrange the phase difference formula to solve for $x$: $x = \frac{\phi \lambda}{2 \pi}$.
Substituting the values: $x = \frac{(\pi/3) \times 6.6}{2 \pi} = \frac{6.6}{6} = 1.1 \,m$.
Thus,the distance between the two points is $1.1 \,m$.

(D) The given equation of the transverse wave is $y=2 \sin (0.01 x+30 t)$.
Comparing this with the standard wave equation $y=A \sin (kx+\omega t)$,we identify the wave number $k=0.01 \ rad/cm$ and the angular frequency $\omega=30 \ rad/s$.
The wave speed $v$ is calculated as $v = \frac{\omega}{k} = \frac{30}{0.01} = 3000 \ cm/s$.
Converting the speed to $SI$ units,$v = 30 \ m/s$.
The time taken to travel the length of the string is $t = 0.5 \ s$.
Therefore,the length of the string $L = v \times t = 30 \ m/s \times 0.5 \ s = 15 \ m$.

(A) Given: Frequency $f = 160 \,Hz$, Velocity $v = 320 \,m/s$.
First, calculate the wavelength $\lambda$ using the formula $\lambda = v/f$:
$\lambda = 320 / 160 = 2 \,m = 200 \,cm$.
The relationship between phase difference $\phi$ and path difference $x$ is given by $\phi = \frac{2\pi x}{\lambda}$.
Given phase difference $\phi = 90^{\circ} = \pi/2$ radians.
Substituting the values: $\frac{\pi}{2} = \frac{2\pi x}{200}$.
Solving for $x$: $x = \frac{\pi}{2} \cdot \frac{200}{2\pi} = \frac{200}{4} = 50 \,cm$.

(C) The wavelength $\lambda$ of a sound wave is given by the formula $\lambda = \frac{V}{f}$,where $V$ is the velocity of sound and $f$ is the frequency.
For the first tuning fork,$f_1 = 320 \,Hz$,so $\lambda_1 = \frac{320 \,ms^{-1}}{320 \,Hz} = 1 \,m$.
For the second tuning fork,$f_2 = 480 \,Hz$,so $\lambda_2 = \frac{320 \,ms^{-1}}{480 \,Hz} = \frac{2}{3} \,m \approx 0.67 \,m$.
The difference between the wavelengths is $\Delta\lambda = \lambda_1 - \lambda_2 = 1 \,m - 0.67 \,m = 0.33 \,m$.
Converting to centimeters,$0.33 \,m = 33 \,cm$.

(A) Given: Frequency $f = 220 \,Hz$, Velocity $v = 330 \,m/s$.
First, calculate the wavelength $\lambda$ using the formula $v = f \lambda$:
$\lambda = \frac{v}{f} = \frac{330}{220} = 1.5 \,m$.
In one vibration, the sound wave travels a distance equal to one wavelength $\lambda$.
Therefore, for $80$ vibrations, the total distance $d$ is:
$d = 80 \times \lambda = 80 \times 1.5 = 120 \,m$.

(A) The speed of a transverse wave in a string is given by $v = f \lambda = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Since the frequency $f$ of the pulse remains constant as it travels,we have $\lambda \propto \sqrt{T}$.
Therefore,the ratio of wavelengths is $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_2}{T_1}}$.
At the lower end (bottom),the tension $T_1$ is due to the block of mass $2 \ kg$: $T_1 = 2g$.
At the top of the rope,the tension $T_2$ supports both the block $(2 \ kg)$ and the rope $(6 \ kg)$: $T_2 = (2 + 6)g = 8g$.
Given $\lambda_1 = 0.06 \ m$,we calculate $\lambda_2$ as:
$\lambda_2 = \lambda_1 \sqrt{\frac{T_2}{T_1}} = 0.06 \times \sqrt{\frac{8g}{2g}} = 0.06 \times \sqrt{4} = 0.06 \times 2 = 0.12 \ m$.

(A) The general equation for a wave is given by $y = a \sin(2 \pi n t)$,where $n$ is the frequency.
For the first wave,$2 \pi n_1 = 2000 \pi$,which gives $n_1 = 1000 \text{ Hz}$.
For the second wave,$2 \pi n_2 = 2008 \pi$,which gives $n_2 = 1004 \text{ Hz}$.
The beat frequency is the difference between the two frequencies: $f_{\text{beat}} = |n_2 - n_1| = |1004 - 1000| = 4 \text{ Hz}$.
Therefore,the number of beats heard per second is $4$.

(C) Given the frequencies of the tuning forks are $n_A, n_B, n_C$ such that $n_A > n_B > n_C$.
When forks $A$ and $B$ are sounded together,the beat frequency is $n_1 = n_A - n_B$ (Equation $i$).
When forks $A$ and $C$ are sounded together,the beat frequency is $n_2 = n_A - n_C$ (Equation $ii$).
We want to find the beat frequency when $B$ and $C$ are sounded together,which is $n_B - n_C$.
Subtracting Equation $i$ from Equation $ii$:
$(n_A - n_C) - (n_A - n_B) = n_2 - n_1$
$n_A - n_C - n_A + n_B = n_2 - n_1$
$n_B - n_C = n_2 - n_1$.
Thus,the number of beats produced per second is $n_2 - n_1$.

(A) Given: Wavelengths $\lambda_1 = 5.0 \ m$ and $\lambda_2 = 5.5 \ m$. Velocity of sound $v = 300 \ m/s$.
The frequency $n$ of a wave is given by $n = \frac{v}{\lambda}$.
Frequency of the first wave: $n_1 = \frac{300}{5.0} = 60 \ Hz$.
Frequency of the second wave: $n_2 = \frac{300}{5.5} = \frac{3000}{55} \approx 54.55 \ Hz$.
The number of beats produced per second is the difference in frequencies: $n_{beats} = |n_1 - n_2| = |60 - 54.55| = 5.45 \ Hz$.
Rounding to the nearest integer,the number of beats is approximately $5 \ Hz$ or $6 \ Hz$ depending on the precision. Given the options,$6 \ Hz$ is the closest match.

(A) The frequency of the tuning fork is $n = 264 \,Hz$ and the speed of sound is $V = 330 \,m/s$.
For a pipe closed at one end,the resonance occurs at lengths given by $\ell = (2k-1) \frac{\lambda}{4}$,where $k = 1, 2, 3, \dots$.
The wavelength is $\lambda = \frac{V}{n} = \frac{330}{264} = 1.25 \,m = 125 \,cm$.
Thus,the possible lengths are $\ell = (2k-1) \frac{125}{4} = (2k-1) \times 31.25 \,cm$.
For $k=1$,$\ell = 31.25 \,cm$.
For $k=2$,$\ell = 3 \times 31.25 = 93.75 \,cm$.
For $k=3$,$\ell = 5 \times 31.25 = 156.25 \,cm$.
Comparing these with the given options,$62.50 \,cm$ is not an odd multiple of $31.25 \,cm$,hence it is not possible.

(D) In a pipe closed at one end, the allowed frequencies are odd multiples of the fundamental frequency $(f_0)$: $f_n = (2n-1)f_0$, where $n = 1, 2, 3, ...$.
These frequencies are $f_1 = f_0$, $f_2 = 3f_0$, $f_3 = 5f_0$, $f_4 = 7f_0$, and so on.
The difference between two consecutive harmonics in a closed pipe is $2f_0$.
Given the consecutive frequencies are $150 \,Hz$ and $250 \,Hz$, their difference is $250 \,Hz - 150 \,Hz = 100 \,Hz$.
Therefore, $2f_0 = 100 \,Hz$, which gives $f_0 = 50 \,Hz$.

(C) The time period of one vibration is $T = \frac{1}{n} \ s$.
For $x$ vibrations,the total time taken is $t = x \times T = \frac{x}{n} \ s$.
The distance traveled by the wave in time $t$ is given by $d = V \times t$.
Substituting the value of $t$,we get $d = V \times \frac{x}{n} = \frac{xV}{n} \ m$.

(C) The potential energy $U$ of a system of point charges is given by the sum of the potential energies of all pairs of charges: $U = \sum \frac{k q_i q_j}{r_{ij}}$.
Given the charges $-q, Q, -q$ are placed at distances $x$ apart,the pairs are: $(-q, Q)$ at distance $x$,$(Q, -q)$ at distance $x$,and $(-q, -q)$ at distance $2x$.
The total potential energy is:
$U = \frac{1}{4 \pi \varepsilon_0} \left( \frac{(-q)(Q)}{x} + \frac{(Q)(-q)}{x} + \frac{(-q)(-q)}{2x} \right) = 0$
Dividing by $\frac{1}{4 \pi \varepsilon_0 x}$ (assuming $x \neq 0$):
$-qQ - qQ + \frac{q^2}{2} = 0$
$-2qQ + \frac{q^2}{2} = 0$
$2qQ = \frac{q^2}{2}$
Dividing both sides by $2q$ (assuming $q \neq 0$):
$Q = \frac{q}{4}$
Therefore,the ratio $\frac{Q}{q} = \frac{1}{4}$.

(D) Let the square be $ABCD$ with side length $2L$. Charges $+q$ are at $A$ and $B$,and charges $-q$ are at $D$ and $C$. Point $P$ is the midpoint of $AB$.
Distance $AP = PB = L$.
The distance from $P$ to $D$ and $C$ can be found using the Pythagorean theorem in $\triangle ADP$ and $\triangle BCP$:
$PD = PC = \sqrt{(2L)^2 + L^2} = \sqrt{4L^2 + L^2} = \sqrt{5}L$.
The electric potential $V$ at point $P$ is the sum of potentials due to all four charges:
$V = \frac{1}{4 \pi \epsilon_0} \left( \frac{q}{AP} + \frac{q}{BP} + \frac{-q}{PD} + \frac{-q}{PC} \right)$
$V = \frac{1}{4 \pi \epsilon_0} \left( \frac{q}{L} + \frac{q}{L} - \frac{q}{\sqrt{5}L} - \frac{q}{\sqrt{5}L} \right)$
$V = \frac{1}{4 \pi \epsilon_0} \left( \frac{2q}{L} - \frac{2q}{\sqrt{5}L} \right)$
$V = \frac{1}{4 \pi \epsilon_0} \frac{2q}{L} \left( 1 - \frac{1}{\sqrt{5}} \right)$

(C) According to Ampere's circuital law,the line integral of the magnetic field $\overrightarrow{B}$ around any closed loop is equal to $\mu_0$ times the net current $I_{\text{enclosed}}$ passing through the loop.
$\oint \overrightarrow{B} \cdot d\overrightarrow{l} = \mu_0 I_{\text{enclosed}}$
In the given figure,two wires carrying currents of $5 \ A$ and $2 \ A$ are inside the loop. Since these currents are in opposite directions,the net current enclosed by the loop is $I_{\text{enclosed}} = 5 \ A - 2 \ A = 3 \ A$.
The wire carrying $3 \ A$ current is situated outside the loop,so it does not contribute to the net enclosed current.
Therefore,the value of the line integral is $\oint \overrightarrow{B} \cdot d\overrightarrow{l} = \mu_0 \times 3 \ A = 3 \mu_0$.

(C) The force per unit length between two parallel long conductors is given by the formula: $F = \frac{\mu_0 I_1 I_2}{2 \pi d}$.
Initially,the force is $F = \frac{\mu_0 I_1 I_2}{2 \pi d}$.
When the current in one conductor is doubled $(I_1' = 2I_1)$ and its direction is reversed,the new current becomes $-2I_1$. The distance is increased to $d' = 3d$.
The new force $F'$ is given by: $F' = \frac{\mu_0 (-2I_1) I_2}{2 \pi (3d)}$.
Simplifying this,we get: $F' = -\frac{2}{3} \left( \frac{\mu_0 I_1 I_2}{2 \pi d} \right)$.
Substituting the initial force $F$,we get: $F' = -\frac{2F}{3}$.

(D) The force per unit length between two parallel conductors carrying currents $I_1$ and $I_2$ separated by distance $r$ is given by $f = \frac{\mu_0 I_1 I_2}{2 \pi r}$.
$1$. Force $F_1$ exerted by $B$ on $A$:
Since currents in $A$ and $B$ are in the same direction,the force is attractive (towards $B$).
$F_1 = \frac{\mu_0 I \cdot I}{2 \pi x} \cdot L = \frac{\mu_0 I^2 L}{2 \pi x}$.
$2$. Force $F_2$ exerted by $C$ on $A$:
Since currents in $A$ and $C$ are in opposite directions,the force is repulsive (away from $C$).
The distance between $A$ and $C$ is $2x$.
$F_2 = \frac{\mu_0 I \cdot 2I}{2 \pi (2x)} \cdot L = \frac{\mu_0 I^2 L}{2 \pi x}$.
Comparing the magnitudes,we see that $F_1 = F_2$. However,since the forces are in opposite directions (one is attractive towards $B$ and the other is repulsive away from $C$),we have $F_1 = -F_2$ in vector notation.

(C) The torque $\tau$ acting on a current-carrying loop in a uniform magnetic field is given by the formula: $\tau = N i A B \sin \theta$.
Here,$N$ is the number of turns,$i$ is the current,$A$ is the area of the loop,$B$ is the magnetic field strength,and $\theta$ is the angle between the normal to the loop and the magnetic field.
From the formula,it is evident that the torque depends on $N, i, A, B,$ and $\theta$.
It does not depend on the shape of the loop,as long as the area $A$ remains constant.

(A) The potential energy $U$ of a magnetic dipole in an external magnetic field is given by the formula $U = -\vec{M} \cdot \vec{B} = -MB \cos \theta$,where $\theta$ is the angle between the magnetic moment $\vec{M}$ and the magnetic field $\vec{B}$.
For the minimum potential energy,$\cos \theta$ must be maximum,which occurs at $\theta = 0^{\circ}$. Thus,$U_{\text{min}} = -MB \cos 0^{\circ} = -MB(1) = -MB$.
For the maximum potential energy,$\cos \theta$ must be minimum,which occurs at $\theta = 180^{\circ}$. Thus,$U_{\text{max}} = -MB \cos 180^{\circ} = -MB(-1) = +MB$.
Therefore,the minimum and maximum values are $-MB$ and $+MB$ respectively.

(C) For a long solenoid,the magnetic field $B$ at the centre is given by $B = \mu_0 n I$,where $\mu_0$ is the permeability of free space,$n$ is the number of turns per unit length,and $I$ is the current.
The magnetic field intensity $H$ is defined as $H = B / \mu_0$.
Substituting the value of $B$,we get $H = (\mu_0 n I) / \mu_0 = n I$.
Therefore,the magnetic field intensity at the centre of a long solenoid is $n I$.

(A) The magnetic field $B$ at the center of a circular current loop is given by the formula: $B = \frac{\mu_0 I}{2r}$.
Here,the current $I$ is defined as the charge per unit time,$I = \frac{e}{T}$.
The time period $T$ for one revolution is $T = \frac{2 \pi r}{V}$.
Substituting $T$ into the current equation,we get $I = \frac{e}{(2 \pi r / V)} = \frac{eV}{2 \pi r}$.
Now,substitute the value of $I$ into the magnetic field formula:
$B = \frac{\mu_0}{2r} \left( \frac{eV}{2 \pi r} \right) = \frac{\mu_0 eV}{4 \pi r^2}$.

(D) The magnetic field $B$ inside a long solenoid is given by the formula:
$B = \mu_0 n I$
where $n$ is the number of turns per unit length and $I$ is the current flowing through the solenoid.
Given that the initial magnetic field is $B = \mu_0 n I$.
According to the problem,the new number of turns per unit length $n' = 2n$ and the new current $I' = \frac{1}{3}I$.
The new magnetic field $B'$ is given by:
$B' = \mu_0 n' I'$
$B' = \mu_0 (2n) \left(\frac{1}{3}I\right)$
$B' = \frac{2}{3} (\mu_0 n I)$
$B' = \frac{2}{3} B$
Therefore,the new value of the magnetic field is $\frac{2B}{3}$.

(A) The magnetic field $B$ at the center of a circular current loop is given by $B = \frac{\mu_0 I}{2r}$.
Since the electron revolves with period $T$, the equivalent current is $I = \frac{e}{T}$.
Given $T = \frac{2\pi r}{v}$, we have $I = \frac{ev}{2\pi r}$.
Substituting this into the magnetic field formula: $B = \frac{\mu_0 (ev/2\pi r)}{2r} = \frac{\mu_0 ev}{4\pi r^2}$.
For a hydrogen atom in the ground state $(n=1)$, the velocity $v$ and radius $r$ are given by Bohr's theory:
$v = \frac{e^2}{2\epsilon_0 h}$ and $r = \frac{\epsilon_0 h^2}{\pi m e^2}$.
Substituting these into the expression for $B$:
$B = \frac{\mu_0 e}{4\pi} \cdot \left( \frac{e^2}{2\epsilon_0 h} \right) \cdot \left( \frac{\pi m e^2}{\epsilon_0 h^2} \right)^2$
$B = \frac{\mu_0 e^3}{8\pi \epsilon_0 h} \cdot \frac{\pi^2 m^2 e^4}{\epsilon_0^2 h^4} = \frac{\mu_0 e^7 \pi m^2}{8 \epsilon_0^3 h^5}$.

(B) The magnetic field at the center of a circular loop of radius $r$ carrying current $i$ is given by $B = \frac{\mu_0 i}{2r}$.
For the two loops,the magnetic fields are $B_1 = \frac{\mu_0 i_1}{2r_1}$ and $B_2 = \frac{\mu_0 i_2}{2r_2}$.
Since the currents are in opposite directions,the resultant magnetic field at the center is $B = B_1 - B_2$ (assuming $B_1 > B_2$).
Given that $B = \frac{B_1}{2}$,we have:
$\frac{B_1}{2} = B_1 - B_2$
$B_2 = \frac{B_1}{2}$
Substituting the expressions for $B_1$ and $B_2$:
$\frac{\mu_0 i_2}{2r_2} = \frac{1}{2} \left( \frac{\mu_0 i_1}{2r_1} \right)$
$\frac{i_2}{r_2} = \frac{i_1}{2r_1}$
Given $r_2 = 2r_1$,we substitute this into the equation:
$\frac{i_2}{2r_1} = \frac{i_1}{2r_1}$
$\frac{i_2}{i_1} = 1$.

(D) When a battery is connected across the diagonal of a square frame,the current splits into two equal paths.
Each path consists of two sides of the square.
Due to the symmetry of the circuit,the current flowing through each side creates a magnetic field at the center of the square.
For any side of the square,the magnetic field produced at the center is equal in magnitude but opposite in direction to the field produced by the diametrically opposite side.
Since the currents in these opposite segments are equal and flow in such a way that their magnetic field contributions at the center cancel each other out,the net magnetic field at the center is zero.

(A) The magnetic field $B$ at the center of a circular loop carrying current $I$ is given by $B = \frac{\mu_0 I}{2r}$.
Here,the charge $q = 100e$ is revolving with a frequency $f = 1 \ r.p.s$.
The equivalent current $I$ is given by $I = qf = 100e \times 1 = 100e \ A$.
Given radius $r = 0.8 \ m$ and $e = 1.6 \times 10^{-19} \ C$.
Substituting these values into the formula:
$B = \frac{\mu_0 \times 100e}{2 \times 0.8} = \frac{\mu_0 \times 100 \times 1.6 \times 10^{-19}}{1.6}$.
$B = \frac{\mu_0 \times 100 \times 1.6 \times 10^{-19}}{1.6} = 100 \times 10^{-19} \mu_0 = 10^{-17} \mu_0 \ T$.

(B) The magnetic field $B$ inside a toroidal solenoid is given by the formula: $B = \mu_0 n I$,where $\mu_0$ is the permeability of free space,$n$ is the number of turns per unit length,and $I$ is the current flowing through the solenoid.
Since $B \propto I$,if the current is tripled $(I_2 = 3I_1)$,the magnetic field will also become three times the original value.
Given $B_1 = 0.2 \ mT$ and $I_2 = 3I_1$,we have:
$B_2 = 3 \times B_1 = 3 \times 0.2 \ mT = 0.6 \ mT$.

(C) The magnetic field $B$ at the center of a circular loop carrying current $I$ is given by the formula: $B = \frac{\mu_0 I}{2 R}$.
Since the ring has a charge $q$ rotating with frequency $f$ (revolutions per second), the equivalent current $I$ is defined as the charge passing a point per unit time.
Thus, $I = q \times f$.
Substituting this value of $I$ into the magnetic field formula, we get:
$B = \frac{\mu_0 (qf)}{2 R}$.
Therefore, the correct option is $C$.

(A) The magnetic field inside a long solenoid is given by $B = \mu_0 n I$,where $n$ is the number of turns per unit length and $I$ is the current.
Initially,$B_1 = \mu_0 n_1 I_1$.
Given that the current is reduced to $20 \%$,the new current $I_2 = 0.2 I_1$.
The number of turns per $cm$ is increased five times,so the new turn density $n_2 = 5 n_1$.
The new magnetic field is $B_2 = \mu_0 n_2 I_2$.
Substituting the values: $B_2 = \mu_0 (5 n_1) (0.2 I_1) = \mu_0 n_1 I_1 (5 \times 0.2) = \mu_0 n_1 I_1 (1) = B_1$.
Therefore,the new magnetic field $B_2$ is equal to $B_1$.

(B) According to Ampere's Circuital Law, the magnetic field $B$ at a distance $R$ from the axis of a long straight cylindrical wire carrying current $I$ is given by $B = \frac{\mu_0 I}{2 \pi R}$ for points outside the wire $(R \ge \text{radius of wire})$.
Since the current $I$ is the same in both cases and the distance $R$ is the same, the magnetic field depends only on the current and the distance from the wire.
Therefore, the diameter of the wire does not affect the magnetic field at a distance $R$ outside the wire.
Thus, $B_1 = B_2$.

(C) The magnetic field $B$ at the centre of a long solenoid is given by the formula $B = \mu_0 n I$.
Here,$n$ is the number of turns per unit length.
Total number of turns $N = 4 \times 1000 = 4000$.
Length of the solenoid $L = 2 \,m$.
Therefore,$n = \frac{N}{L} = \frac{4000}{2} = 2000 \text{ turns/m}$.
The current $I = 5 \,A$.
Substituting the values into the formula:
$B = (4 \pi \times 10^{-7} \,Wb/Am) \times (2000 \text{ turns/m}) \times (5 \,A)$.
$B = 4 \pi \times 10^{-7} \times 10000$.
$B = 4 \pi \times 10^{-3} \,T$.

(A) The area of the circular coil is given by $A = \pi R^2$,where $R$ is the radius of the coil.
From this,the radius is $R = \sqrt{\frac{A}{\pi}}$.
The magnetic field at the centre of a circular coil carrying current $I$ is $B = \frac{\mu_0 I}{2 R}$.
Rearranging for current $I$,we get $I = \frac{2 B R}{\mu_0}$.
The magnetic moment $M$ of the coil is defined as $M = I A$.
Substituting the expression for $I$ and $R$ into the formula for $M$:
$M = \left( \frac{2 B R}{\mu_0} \right) A = \frac{2 B A}{\mu_0} \sqrt{\frac{A}{\pi}}$.
Simplifying this,we get $M = \frac{2 B A^{3/2}}{\mu_0 \sqrt{\pi}}$.

(B) According to Bohr's theory,the angular momentum '$L$' of an electron in the '$n^{th}$' orbit is given by $L = \frac{nh}{2\pi}$.
The magnetic moment '$M$' associated with a revolving electron is given by $M = \frac{e}{2m_e} L$.
Substituting the value of '$L$' into the equation for '$M$':
$M = \frac{e}{2m_e} \left( \frac{nh}{2\pi} \right) = \frac{enh}{4\pi m_e}$.
Since '$e$','$h$','$m_e$',and '$\pi$' are constants,we have $M \propto n$.

(D) The gyromagnetic ratio is defined as the ratio of the magnetic moment $(\mu_L)$ to the angular momentum $(L)$ of an orbital electron.
$\text{Gyromagnetic ratio} = \frac{\mu_L}{L} = \frac{e}{2m}$.
Given that the charge to mass ratio of an electron is $\frac{e}{m} = A \ C/kg$.
Substituting this into the expression,we get:
$\text{Gyromagnetic ratio} = \frac{1}{2} \times \left(\frac{e}{m}\right) = \frac{A}{2} \ C/kg$.

(C) Consider an electron of charge $e$ and mass $m$ revolving in a circular orbit of radius $r$ with a speed $v$ and time period $T$.
The current $I$ associated with this revolving electron is $I = \frac{e}{T} = \frac{e}{2\pi r / v} = \frac{ev}{2\pi r}$.
The magnetic moment $M$ is given by $M = I \times A = I \times (\pi r^2)$.
Substituting the value of $I$,we get $M = \left(\frac{ev}{2\pi r}\right) \times (\pi r^2) = \frac{evr}{2}$.
The angular momentum $L$ of the electron is $L = mvr$.
Thus,$vr = \frac{L}{m}$.
Substituting this into the expression for $M$,we get $M = \frac{e}{2} \times \left(\frac{L}{m}\right) = \frac{eL}{2m}$.

(D) The length of the conductor $L$ forms the circumference of the circular loop.
$L = 2 \pi r$,where $r$ is the radius of the loop.
Therefore,$r = \frac{L}{2 \pi}$.
The area $A$ of the circular loop is given by $A = \pi r^2$.
Substituting the value of $r$,we get $A = \pi \left( \frac{L}{2 \pi} \right)^2 = \pi \left( \frac{L^2}{4 \pi^2} \right) = \frac{L^2}{4 \pi}$.
The magnetic moment $M$ of a current-carrying loop is given by $M = I \times A$.
Substituting the value of $A$,we get $M = I \left( \frac{L^2}{4 \pi} \right) = \frac{IL^2}{4 \pi}$.

(D) The current $i$ produced by an electron moving in a circular orbit is given by $i = \frac{e}{T}$,where $T$ is the time period of revolution. Since $T = \frac{2 \pi r}{v}$,we have $i = \frac{ev}{2 \pi r}$.
The magnetic moment $M$ of a current loop is $M = iA$,where $A = \pi r^2$ is the area of the orbit.
Substituting the values,$M = \left(\frac{ev}{2 \pi r}\right) (\pi r^2) = \frac{evr}{2}$.
Multiplying and dividing by the mass of the electron $m_{e}$,we get $M = \frac{e}{2 m_{e}} (m_{e}vr)$.
According to Bohr's quantization condition,the angular momentum $L = m_{e}vr = \frac{nh}{2 \pi}$.
Substituting this into the expression for $M$,we get $M = \frac{e}{2 m_{e}} \left(\frac{nh}{2 \pi}\right)$.

(D) The magnetic moment $M$ of a current loop is given by $M = I \times A$.
Here, the current $I$ due to the revolving electron is $I = qf = ef$, where $f$ is the frequency of revolution.
The area $A$ of the circular orbit is $A = \pi r^2$.
Substituting these into the formula, we get $M = (ef)(\pi r^2)$.
Given: $r = 0.05 \,nm = 0.05 \times 10^{-9} \,m = 5 \times 10^{-11} \,m$, $f = 10^{16} \,Hz$, and $e = 1.6 \times 10^{-19} \,C$.
$M = (1.6 \times 10^{-19} \,C) \times (10^{16} \,s^{-1}) \times (3.14) \times (5 \times 10^{-11} \,m)^2$.
$M = 1.6 \times 10^{-3} \times 3.14 \times 25 \times 10^{-22}$.
$M = 1.6 \times 3.14 \times 25 \times 10^{-25}$.
$M = 125.6 \times 10^{-25} = 1.256 \times 10^{-23} \,A-m^2 \approx 1.26 \times 10^{-23} \,A-m^2$.

(B) The force $F$ acting on a charge $q$ in an electric field $E$ is given by $F = qE$.
Given charge $q = 2e$ and electric field $E$,the force is $F = 2eE$.
The acceleration $a$ is given by Newton's second law,$a = \frac{F}{m_{total}}$.
Given mass $m_{total} = 4m$,the acceleration is $a = \frac{2eE}{4m}$.
Simplifying the expression,we get $a = \frac{eE}{2m}$.

(C) The force acting on a moving charge in the presence of both electric and magnetic fields is known as the Lorentz force.
The force due to the electric field is given by $\overrightarrow{F}_{e} = q\overrightarrow{E}$.
The force due to the magnetic field is given by $\overrightarrow{F}_{m} = q(\overrightarrow{V} \times \overrightarrow{B})$.
Therefore,the total force $\overrightarrow{F}$ acting on the charge is the vector sum of these two forces:
$\overrightarrow{F} = \overrightarrow{F}_{e} + \overrightarrow{F}_{m} = q\overrightarrow{E} + q(\overrightarrow{V} \times \overrightarrow{B})$.

(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.

(C) The frequency of oscillation for a bar magnet in a uniform magnetic field is given by $f = \frac{1}{2 \pi} \sqrt{\frac{MB}{I}}$.
For magnet $P$: $f_{P} = \frac{1}{2 \pi} \sqrt{\frac{M_{P} B}{I_{P}}}$.
For magnet $Q$: $f_{Q} = \frac{1}{2 \pi} \sqrt{\frac{M_{Q} B}{I_{Q}}}$.
Given $f_{P} = 2 f_{Q}$ and $I_{P} = 2 I_{Q}$.
Substituting these into the ratio: $\frac{f_{P}}{f_{Q}} = \sqrt{\frac{M_{P} I_{Q}}{M_{Q} I_{P}}} = 2$.
Squaring both sides: $\frac{M_{P} I_{Q}}{M_{Q} I_{P}} = 4$.
Since $I_{P} = 2 I_{Q}$,we have $\frac{I_{Q}}{I_{P}} = \frac{1}{2}$.
Therefore,$\frac{M_{P}}{M_{Q}} \cdot \frac{1}{2} = 4$,which implies $\frac{M_{P}}{M_{Q}} = 8$,or $M_{P} = 8 M_{Q}$.

(C) The relationship between relative permeability $\mu_{r}$ and magnetic susceptibility $\chi$ is given by the formula: $\mu_{r} = 1 + \chi$.
Given that the relative permeability $\mu_{r} = 0.85$.
Substituting the value into the formula: $0.85 = 1 + \chi$.
Therefore,$\chi = 0.85 - 1 = -0.15$.
Thus,the magnetic susceptibility of the medium is $-0.15$.

(A) Given,permeability of the metal $\mu = 0.1256 \ TmA^{-1}$.
We know that the permeability of free space $\mu_0 = 4 \pi \times 10^{-7} \ TmA^{-1}$.
Substituting $\pi = 3.14$,we get $\mu_0 = 4 \times 3.14 \times 10^{-7} = 12.56 \times 10^{-7} \ TmA^{-1}$.
The relative permeability $\mu_r$ is defined as the ratio of the permeability of the medium to the permeability of free space:
$\mu_r = \frac{\mu}{\mu_0} = \frac{0.1256}{12.56 \times 10^{-7}}$.
$\mu_r = \frac{12.56 \times 10^{-2}}{12.56 \times 10^{-7}} = 10^{-2} \times 10^7 = 10^5$.
Thus,the relative permeability is $10^5$.

(C) The absolute permeability $\mu$ is given by the formula $\mu = \mu_r \mu_0$.
Given that the relative permeability $\mu_r = 2000$ and the permeability of free space $\mu_0 = 4 \pi \times 10^{-7} \text{ T m/A}$.
Substituting these values into the formula:
$\mu = 2000 \times (4 \pi \times 10^{-7}) \text{ T m/A}$.
$\mu = 8000 \pi \times 10^{-7} \text{ T m/A}$.
$\mu = 8 \pi \times 10^{-4} \text{ T m/A}$.

(D) Given: Length $L = 3 \,cm = 3 \times 10^{-2} \,m$, Cross-sectional area $A = 2 \,cm^2 = 2 \times 10^{-4} \,m^2$, Magnetic moment $M = 3 \,Am^2$.
The intensity of magnetization $I$ (or $M_z$) is defined as the magnetic moment per unit volume.
Volume $V = L \times A = (3 \times 10^{-2} \,m) \times (2 \times 10^{-4} \,m^2) = 6 \times 10^{-6} \,m^3$.
Intensity of magnetization $I = \frac{M}{V} = \frac{3 \,Am^2}{6 \times 10^{-6} \,m^3}$.
$I = 0.5 \times 10^6 \,A/m = 5 \times 10^5 \,A/m$.

(C) By the law of conservation of momentum,the two parts will have equal and opposite momentum,assuming the nucleus was initially at rest.
$m_1 v_1 = m_2 v_2$
$\frac{m_1}{m_2} = \frac{v_2}{v_1} = \frac{1}{2}$
Since the nuclear mass $m$ is proportional to the volume,which is proportional to the cube of the radius $r^3$ (i.e.,$m \propto r^3$):
$\frac{m_1}{m_2} = \left(\frac{r_1}{r_2}\right)^3$
Substituting the mass ratio:
$\left(\frac{r_1}{r_2}\right)^3 = \frac{1}{2}$
$\frac{r_1}{r_2} = \left(\frac{1}{2}\right)^{1/3} = \frac{1}{2^{1/3}}$

(A) Given: Initial number of nuclei $N_0 = 8 \times 10^{16}$,half-life $T = 15 \text{ days}$,and total time $t = 60 \text{ days}$.
First,calculate the number of half-lives $n$:
$n = \frac{t}{T} = \frac{60}{15} = 4$.
The number of nuclei remaining $N$ after $n$ half-lives is given by:
$N = N_0 \left(\frac{1}{2}\right)^n = 8 \times 10^{16} \times \left(\frac{1}{2}\right)^4 = 8 \times 10^{16} \times \frac{1}{16} = 0.5 \times 10^{16}$.
The number of nuclei decayed is the difference between the initial and remaining nuclei:
$\text{Decayed nuclei} = N_0 - N = 8 \times 10^{16} - 0.5 \times 10^{16} = 7.5 \times 10^{16}$.

(D) Let $N_0$ be the initial number of nuclei for both materials.
After time $t$,the number of nuclei remaining for $X_1$ is $N_1 = N_0 e^{-5 \lambda t}$.
After time $t$,the number of nuclei remaining for $X_2$ is $N_2 = N_0 e^{-\lambda t}$.
The ratio of the number of nuclei is given by $\frac{N_1}{N_2} = \frac{N_0 e^{-5 \lambda t}}{N_0 e^{-\lambda t}} = e^{-5 \lambda t + \lambda t} = e^{-4 \lambda t}$.
Given that $\frac{N_1}{N_2} = \frac{1}{e} = e^{-1}$.
Equating the exponents: $-4 \lambda t = -1$.
Therefore,$t = \frac{1}{4 \lambda}$.

(C) Let the initial amount of the radioactive substance be $N_i = 100 \%$.
At $40 \%$ decay, the remaining amount is $N_1 = 100 \% - 40 \% = 60 \%$.
At $85 \%$ decay, the remaining amount is $N_2 = 100 \% - 85 \% = 15 \%$.
We know that the amount remaining after time $t$ is given by $N(t) = N_i \left( \frac{1}{2} \right)^{t/T_{1/2}}$, where $T_{1/2} = 30 \text{ minutes}$.
For the interval between $N_1$ and $N_2$, the ratio of remaining nuclei is $\frac{N_2}{N_1} = \frac{15 \%}{60 \%} = \frac{1}{4}$.
Since $\frac{1}{4} = \left( \frac{1}{2} \right)^2$, the time taken corresponds to two half-lives.
Therefore, the time taken is $t = 2 \times T_{1/2} = 2 \times 30 \text{ minutes} = 60 \text{ minutes}$.

(C) The refractive index of glass is $\mu_g = 1.5 = \frac{3}{2}$.
The refractive index of water is $\mu_w = 1.33 = \frac{4}{3}$.
The relative refractive index of glass with respect to water is given by ${}_w\mu_g = \frac{\mu_g}{\mu_w} = \frac{3/2}{4/3} = \frac{3}{2} \times \frac{3}{4} = \frac{9}{8}$.
The formula for the critical angle $C$ is $\sin C = \frac{1}{{}_w\mu_g}$.
Substituting the value,$\sin C = \frac{1}{9/8} = \frac{8}{9}$.
Therefore,the critical angle is $C = \sin^{-1}\left(\frac{8}{9}\right)$.

(C) In the first case,$\theta$ is the critical angle for the glass-air interface.
According to the definition of the critical angle,$\sin \theta = \frac{1}{\mu}$.
In the second case,light travels from air to glass. Using Snell's law,$\frac{\sin i}{\sin r} = \mu$,where $i = \theta$.
Therefore,$\sin r = \frac{\sin \theta}{\mu}$.
Substituting $\sin \theta = \frac{1}{\mu}$ into the equation,we get $\sin r = \frac{1/\mu}{\mu} = \frac{1}{\mu^2}$.
Thus,the angle of refraction is $r = \sin^{-1}\left(\frac{1}{\mu^2}\right)$.

(D) The 'Circle of least confusion' is a term used in optics to describe the smallest cross-sectional area of a light beam that has passed through a lens or mirror system.
It is specifically associated with 'Spherical aberration',where rays of light passing through the edges of a lens focus at a different point than rays passing through the center.
Because the rays do not converge at a single point,the image appears blurred,and the region of minimum blur is known as the 'Circle of least confusion'.

(D) The focal length of a lens in air is given by the Lens Maker's Formula:
$\frac{1}{f} = (n_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \quad \dots(1)$
where $n_g = \frac{3}{2}$ and $f = 15 \,cm$.
The focal length of the same lens in a liquid of refractive index $n_l$ is given by:
$\frac{1}{f'} = \left( \frac{n_g}{n_l} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \quad \dots(2)$
Dividing equation $(1)$ by equation $(2)$, we get:
$\frac{f'}{f} = \frac{(n_g - 1)}{\left( \frac{n_g}{n_l} - 1 \right)}$
Substituting the given values $n_g = \frac{3}{2}$ and $n_l = \frac{9}{5}$:
$\frac{f'}{15} = \frac{(\frac{3}{2} - 1)}{(\frac{3/2}{9/5} - 1)} = \frac{1/2}{(\frac{3}{2} \times \frac{5}{9} - 1)} = \frac{1/2}{(\frac{5}{6} - 1)} = \frac{1/2}{-1/6} = -3$
Therefore, $f' = -3 \times 15 = -45 \,cm$.

(B) Magnification,$m = -\frac{1}{4}$ (Since the image is real and inverted,magnification is negative).
Using the magnification formula,$m = \frac{v}{u} = -\frac{1}{4}$,which gives $v = -\frac{u}{4}$.
Applying the lens formula,$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Substituting the value of $v$: $\frac{1}{-u/4} - \frac{1}{u} = \frac{1}{f}$.
$-\frac{4}{u} - \frac{1}{u} = \frac{1}{f}$.
$-\frac{5}{u} = \frac{1}{f}$.
Therefore,$u = -5f$.
The magnitude of the object distance is $5f$.

(A) The image is real and hence inverted.
Therefore,the magnification $m = \frac{v}{u} = -n$,which implies $u = -\frac{v}{n}$.
Using the lens formula,$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Substituting the value of $u$,we get $\frac{1}{v} - (-\frac{n}{v}) = \frac{1}{f}$.
This simplifies to $\frac{1+n}{v} = \frac{1}{f}$.
Therefore,the image distance $v = f(n+1)$.

(C) Given: Focal length $f = 8 \,cm$. The mean position of the particle is $u_0 = -14 \,cm$. The amplitude of the particle is $A_p = 1 \,cm$.
First, find the position of the image $v_0$ when the particle is at the mean position:
$\frac{1}{v_0} = \frac{1}{f} + \frac{1}{u_0} = \frac{1}{8} - \frac{1}{14} = \frac{7-4}{56} = \frac{3}{56}$
$v_0 = \frac{56}{3} \approx 18.67 \,cm$.
Next, find the position of the image $v_1$ when the particle is at the extreme position $u_1 = -14 - 1 = -15 \,cm$:
$\frac{1}{v_1} = \frac{1}{f} + \frac{1}{u_1} = \frac{1}{8} - \frac{1}{15} = \frac{15-8}{120} = \frac{7}{120}$
$v_1 = \frac{120}{7} \approx 17.14 \,cm$.
The amplitude of the oscillating image is the difference between the image positions:
$A_i = |v_0 - v_1| = |18.67 - 17.14| = 1.53 \,cm$.
Rounding to the nearest integer, the amplitude is approximately $2 \,cm$.

(C) The focal length of a lens is given by the Lens Maker's Formula: $\frac{1}{f} = (\mu_{rel} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$,where $\mu_{rel} = \frac{\mu_{lens}}{\mu_{liquid}}$.
Given that the refractive index of the liquid is equal to the refractive index of the lens material,we have $\mu_{lens} = \mu_{liquid}$,which implies $\mu_{rel} = 1$.
Substituting this into the formula: $\frac{1}{f} = (1 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = 0$.
Therefore,$\frac{1}{f} = 0$,which means $f = \infty$ (the focal length becomes infinite).

(C) For a biconvex lens,the lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Given $R_1 = 30 \ cm$ and $R_2 = -30 \ cm$ (by sign convention for biconvex lens),and $\mu = 1.6$.
Substituting the values: $\frac{1}{f} = (1.6 - 1) \left( \frac{1}{30} - \frac{1}{-30} \right) = 0.6 \times \left( \frac{2}{30} \right) = \frac{1.2}{30} = \frac{1}{25}$.
Thus,the focal length of the lens is $f = 25 \ cm$.
Since the focal length of the concave mirror is equal to the focal length of the lens,$f_{mirror} = 25 \ cm$.
The radius of curvature of a concave mirror is $R = 2f = 2 \times 25 \ cm = 50 \ cm$.

(D) For a convex lens,to obtain an image of equal size to the object,the magnification $m$ must be $-1$.
This occurs when the object is placed at $2f$ from the lens,and the image is formed at $2f$ on the other side.
Given that the image is formed on a wall at a distance $d$ from the lens,we have the image distance $v = d$.
Since the image size equals the object size,the object distance $u$ must also be $d$.
Thus,the total distance between the two walls is $u + v = d + d = 2d$.
For a real image of equal size,the distance between the object and the screen must be at least $4f$.
Therefore,$4f = 2d$,which gives $f = \frac{d}{2}$.

(B) The lens maker's formula is given by $P = 1/f = (\mu - 1)(1/R_1 - 1/R_2)$.
For a convex lens with equal radii of curvature,$R_1 = R$ and $R_2 = -R$.
Thus,$P = (\mu - 1)(1/R + 1/R) = (\mu - 1)(2/R)$.
When one surface is made plane,the new radii are $R_1 = R$ and $R_2 = \infty$.
The new power $P' = (\mu - 1)(1/R - 1/\infty) = (\mu - 1)/R$.
Comparing $P'$ with $P$,we get $P' = P/2$.
Since $f' = 1/P'$,we have $f' = 1/(P/2) = 2/P = 2f$.
Therefore,the new focal length is $2f$ and the new power is $P/2$.

(A) For the plano-convex lens,the focal length $f_1$ is given by the lens maker's formula: $\frac{1}{f_1} = (\mu_1 - 1)(\frac{1}{R} - \frac{1}{\infty}) = \frac{\mu_1 - 1}{R}$.
For the plano-concave lens,the focal length $f_2$ is given by: $\frac{1}{f_2} = (\mu_2 - 1)(\frac{1}{-\infty} - \frac{1}{-R}) = \frac{\mu_2 - 1}{-R} = -\frac{\mu_2 - 1}{R}$.
The focal length $f$ of the combination is given by $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$.
Substituting the values: $\frac{1}{f} = \frac{\mu_1 - 1}{R} - \frac{\mu_2 - 1}{R} = \frac{\mu_1 - 1 - \mu_2 + 1}{R} = \frac{\mu_1 - \mu_2}{R}$.
Therefore,$f = \frac{R}{\mu_1 - \mu_2}$.