IIT JEE 1998 Physics Question Paper with Answer and Solution

(C) The work done is given by the line integral $W = \int \overrightarrow F \cdot d\overrightarrow r = \int (-K(y\hat i + x\hat j)) \cdot (dx\hat i + dy\hat j) = -K \int (y\,dx + x\,dy) = -K \int d(xy)$.
Path $1$: From $(0, 0)$ to $(a, 0)$. Here $y = 0$,so $dy = 0$. The work done $W_1 = -K \int_0^a (0) dx = 0$.
Path $2$: From $(a, 0)$ to $(a, a)$. Here $x = a$,so $dx = 0$. The work done $W_2 = -K \int_0^a (a) dy = -K[ay]_0^a = -Ka^2$.
Total work done $W = W_1 + W_2 = 0 + (-Ka^2) = -Ka^2$.

(B) $1$. Dimensional formula of permittivity of free space $(\varepsilon_0)$:
From Coulomb's Law,$F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}$.
Rearranging,$\varepsilon_0 = \frac{q_1 q_2}{4\pi F r^2}$.
Dimensions: $[q] = [IT]$,$[F] = [MLT^{-2}]$,$[r] = [L]$.
Substituting: $[\varepsilon_0] = \frac{[IT][IT]}{[MLT^{-2}][L^2]} = \frac{I^2 T^2}{ML^3 T^{-2}} = M^{-1} L^{-3} T^4 I^2$.
$2$. Dimensional formula of permeability of free space $(\mu_0)$:
From the force between two parallel wires,$F = \frac{\mu_0 I_1 I_2 l}{2\pi r}$.
Rearranging,$\mu_0 = \frac{F \cdot 2\pi r}{I_1 I_2 l}$.
Dimensions: $[F] = [MLT^{-2}]$,$[r] = [L]$,$[I] = [I]$,$[l] = [L]$.
Substituting: $[\mu_0] = \frac{[MLT^{-2}][L]}{[I][I][L]} = MLT^{-2}I^{-2}$.
Comparing with the options,option $B$ is correct.

(D) Let the lowest point be $A$ and the horizontal position be $B$. At $A$, velocity is $\vec{u} = u \hat{i}$.
Using the law of conservation of energy between $A$ and $B$:
$\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgL$
$v^2 = u^2 - 2gL$
$v = \sqrt{u^2 - 2gL}$.
At position $B$, the velocity is directed vertically upward, so $\vec{v} = v \hat{j} = \sqrt{u^2 - 2gL} \hat{j}$.
The change in velocity is $\Delta \vec{v} = \vec{v} - \vec{u} = v \hat{j} - u \hat{i}$.
The magnitude of the change in velocity is $|\Delta \vec{v}| = \sqrt{v^2 + u^2} = \sqrt{(u^2 - 2gL) + u^2} = \sqrt{2u^2 - 2gL} = \sqrt{2(u^2 - gL)}$.

(A) The gravitational force exerted by the Earth on the satellite always acts towards the centre of the Earth. According to Newton's second law,$F = ma$,the acceleration of the satellite is always directed towards the centre of the Earth.
Since the gravitational force is a central force,the torque acting on the satellite about the centre of the Earth is zero. Therefore,the angular momentum $L$ of the satellite remains constant in both magnitude and direction.
According to the law of conservation of energy,in the absence of non-conservative forces,the total mechanical energy of the satellite remains constant throughout its orbit.
Since the distance $r$ of the satellite from the Earth varies in an elliptical orbit,the orbital velocity $v$ must change to conserve angular momentum $(L = mvr \sin \theta)$. Consequently,the linear momentum $p = mv$ is not constant.

(C) As the sphere floats in the liquid,its weight is equal to the total upthrust force acting on it.
Let $V$ be the total volume of the sphere and $\rho$ be its density.
Weight of the sphere = $V \rho g$
Upthrust due to oil = (Volume in oil) $\times$ (Density of oil) $\times g = (V/2) \times 0.8 \times g = 0.4 Vg$
Upthrust due to mercury = (Volume in mercury) $\times$ (Density of mercury) $\times g = (V/2) \times 13.6 \times g = 6.8 Vg$
Total upthrust = $0.4 Vg + 6.8 Vg = 7.2 Vg$
Equating weight and total upthrust:
$V \rho g = 7.2 Vg$
$\rho = 7.2 \; g/cm^3$

(D) For an ideal gas,the speeds are defined as:
${v_{rms}} = \sqrt{\frac{3kT}{m}}$,${v_p} = \sqrt{\frac{2kT}{m}}$,and $\bar{v} = \sqrt{\frac{8kT}{\pi m}}$.
Comparing these values:
${v_p} = \sqrt{2} \sqrt{\frac{kT}{m}} \approx 1.414 \sqrt{\frac{kT}{m}}$
$\bar{v} = \sqrt{\frac{8}{3.14}} \sqrt{\frac{kT}{m}} \approx 1.596 \sqrt{\frac{kT}{m}}$
${v_{rms}} = \sqrt{3} \sqrt{\frac{kT}{m}} \approx 1.732 \sqrt{\frac{kT}{m}}$
Thus,${v_p} < \bar{v} < {v_{rms}}$,which confirms option $(c)$ is correct.
For the average kinetic energy:
${E_{av}} = \frac{1}{2} m v_{rms}^2 = \frac{1}{2} m \left( \frac{3kT}{m} \right) = \frac{3}{2} kT$.
Since ${v_p}^2 = \frac{2kT}{m}$,we have $kT = \frac{1}{2} m v_p^2$.
Substituting this into the energy equation:
${E_{av}} = \frac{3}{2} \left( \frac{1}{2} m v_p^2 \right) = \frac{3}{4} m v_p^2$,which confirms option $(b)$ is correct.
Therefore,both $(b)$ and $(c)$ are correct.

(A) According to the law of equipartition of energy,the average kinetic energy associated with each degree of freedom is $\frac{1}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
Both oxygen $(O_2)$ and nitrogen $(N_2)$ are diatomic molecules.
$A$ diatomic molecule has $2$ degrees of freedom associated with rotational motion.
Therefore,the average rotational kinetic energy for each molecule is $2 \times (\frac{1}{2} k_B T) = k_B T$.
Since both gases are in the same vessel at the same temperature $(T = 300 K)$,the average rotational kinetic energy per molecule for both $O_2$ and $N_2$ is $k_B T$.
Thus,the ratio of the average rotational kinetic energy per $O_2$ molecule to that per $N_2$ molecule is $k_B T : k_B T = 1:1$.

(A) The bulk modulus $B$ is defined as $B = -V \frac{dP}{dV}$.
For an isothermal process,the equation of state for an ideal gas is $PV = \text{constant}$.
Differentiating both sides with respect to $V$,we get $P + V \frac{dP}{dV} = 0$.
This implies $P = -V \frac{dP}{dV}$.
Therefore,the isothermal bulk modulus $B_T = P$.

(D) When ice melts at $273 \, K$ and atmospheric pressure,its volume decreases because the density of water is greater than that of ice at this temperature.
Since work done by the system is given by $W = P \Delta V$,and the change in volume $\Delta V$ is negative,the work done by the ice-water system on the atmosphere is negative.
This implies that positive work is done on the ice-water system by the atmosphere. Thus,option $(B)$ is correct.
According to the first law of thermodynamics,$\Delta U = \Delta Q - W$. During melting,heat is absorbed by the system,so $\Delta Q > 0$. Since $W < 0$,the internal energy change $\Delta U = \Delta Q - W$ becomes $\Delta U = \Delta Q + |W|$,which is positive. Therefore,the internal energy of the system increases. Thus,option $(C)$ is also correct.
Since both $(B)$ and $(C)$ are correct,the final answer is $(D)$.

(C) The process is isothermal,therefore $T = \text{constant}$.
Since $PV = \mu RT$,we have $P = \frac{\mu RT}{V}$.
For chamber $A$,the change in pressure is $\Delta P = P_i - P_f = \frac{\mu_A RT}{V} - \frac{\mu_A RT}{2V} = \frac{\mu_A RT}{2V} \dots (i)$.
For chamber $B$,the change in pressure is $1.5 \Delta P = P_i - P_f = \frac{\mu_B RT}{V} - \frac{\mu_B RT}{2V} = \frac{\mu_B RT}{2V} \dots (ii)$.
Dividing equation $(i)$ by $(ii)$,we get $\frac{\Delta P}{1.5 \Delta P} = \frac{\mu_A}{\mu_B} \implies \frac{1}{1.5} = \frac{\mu_A}{\mu_B} \implies \frac{\mu_A}{\mu_B} = \frac{2}{3}$.
Since $\mu = \frac{m}{M}$,where $M$ is the molar mass,we have $\frac{m_A/M}{m_B/M} = \frac{2}{3} \implies \frac{m_A}{m_B} = \frac{2}{3}$.
Therefore,$3m_A = 2m_B$.

(D) For an ideal diatomic gas,the molar heat capacity at constant pressure is $C_p$ and at constant volume is $C_v$. The ratio is given by $\gamma = \frac{C_p}{C_v} = 1.4$.
In cylinder $A$,the piston is free to move,so the process is isobaric (constant pressure). The heat supplied is $\Delta Q = \mu C_p (\Delta T)_A$.
In cylinder $B$,the piston is fixed,so the process is isochoric (constant volume). The heat supplied is $\Delta Q = \mu C_v (\Delta T)_B$.
Since the same amount of heat is given to both,we have $\mu C_p (\Delta T)_A = \mu C_v (\Delta T)_B$.
Rearranging for $(\Delta T)_B$,we get $(\Delta T)_B = \frac{C_p}{C_v} (\Delta T)_A = \gamma (\Delta T)_A$.
Substituting the given values: $(\Delta T)_B = 1.4 \times 30 \ K = 42 \ K$.

(D) According to Wien's displacement law,the wavelength corresponding to the maximum spectral emissive power is given by ${\lambda _m}T = b$.
Substituting the given values,we get ${\lambda _m} = \frac{b}{T} = \frac{2.88 \times {10^6}}{2880} = 1000\;nm$.
This means the peak of the black body radiation curve occurs at $1000\;nm$.
The energy $U$ emitted in a small wavelength interval $\Delta \lambda$ is proportional to the area under the $E_{\lambda} - \lambda$ curve in that interval.
Since the peak of the curve is at $1000\;nm$,the spectral emissive power $E_{\lambda}$ is maximum at $1000\;nm$.
Comparing the intervals,the interval for ${U_2}$ ($999\;nm$ to $1000\;nm$) is closest to the peak wavelength,while the interval for ${U_1}$ ($499\;nm$ to $500\;nm$) is further away.
Therefore,the area under the curve for ${U_2}$ is greater than the area for ${U_1}$,implying ${U_2} > {U_1}$.

(A) The potential energy is given by $U(x) = k|x|^3$.
The force acting on the particle is $F = -\frac{dU}{dx} = -3k|x|^2 \text{sgn}(x)$.
For a particle of mass $m$ oscillating with amplitude $a$,the total energy $E$ is conserved and is equal to the potential energy at the extreme position $(x = a)$:
$E = U(a) = ka^3$.
At any position $x$,the energy is $E = \frac{1}{2}mv^2 + k|x|^3 = ka^3$.
Thus,$v = \frac{dx}{dt} = \sqrt{\frac{2k}{m}(a^3 - |x|^3)}$.
The time period $T$ is given by $T = 4 \int_{0}^{a} \frac{dx}{v} = 4 \int_{0}^{a} \frac{dx}{\sqrt{\frac{2k}{m}(a^3 - x^3)}}$.
Let $x = ay$,then $dx = a dy$. When $x=0, y=0$ and when $x=a, y=1$.
$T = 4 \sqrt{\frac{m}{2k}} \int_{0}^{1} \frac{a dy}{\sqrt{a^3(1 - y^3)}} = 4 \sqrt{\frac{m}{2ka}} \int_{0}^{1} \frac{dy}{\sqrt{1 - y^3}}$.
Since the integral is a constant,$T \propto \frac{1}{\sqrt{a}}$.

(D) The maximum particle velocity of a transverse wave is given by $v_{\max} = a\omega = a(2\pi n).$
Given $v_{\max} = v/10,$ where $v = 10 \ m/s,$ we have $v_{\max} = 10/10 = 1 \ m/s.$
Substituting $a = 10^{-3} \ m,$ we get $10^{-3} \times 2\pi n = 1 \implies n = \frac{10^3}{2\pi} \ Hz.$
Using the wave speed relation $v = n\lambda,$ we find $\lambda = \frac{v}{n} = \frac{10}{10^3 / 2\pi} = 2\pi \times 10^{-2} \ m.$
Thus,both options $(A)$ and $(C)$ are correct.

(D) Since the edges of the square plate are clamped,the displacement $u(x, y)$ must be zero at all boundaries:
$1$. At $y = 0$ (edge $OA$),$u(x, 0) = 0$ for $0 \le x \le L$.
$2$. At $x = L$ (edge $AB$),$u(L, y) = 0$ for $0 \le y \le L$.
$3$. At $y = L$ (edge $BC$),$u(x, L) = 0$ for $0 \le x \le L$.
$4$. At $x = 0$ (edge $OC$),$u(0, y) = 0$ for $0 \le y \le L$.
Evaluating the options:
- For option $(a)$: $u(0, y) = a \cos(0) \cos(\frac{\pi y}{2L}) = a \cos(\frac{\pi y}{2L}) \neq 0$ for all $y$.
- For option $(b)$: $u(x, y) = a \sin(\frac{\pi x}{L}) \sin(\frac{\pi y}{L})$. Here,$u = 0$ at $x=0, x=L, y=0, y=L$. This satisfies the boundary conditions.
- For option $(c)$: $u(x, y) = a \sin(\frac{\pi x}{L}) \sin(\frac{2\pi y}{L})$. Here,$u = 0$ at $x=0, x=L, y=0, y=L/2, y=L$. This also satisfies the boundary conditions.
Thus,both $(b)$ and $(c)$ are valid expressions.

(B) The linear mass density of the string is $\mu = \frac{M}{L} = \frac{10^{-2}}{0.4} = 2.5 \times 10^{-2}\, kg/m$.
The velocity of the wave pulse in the string is $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{1.6}{2.5 \times 10^{-2}}} = \sqrt{\frac{160}{2.5}} = \sqrt{64} = 8\, m/s$.
For constructive interference,the pulse must return to the starting point in the same phase. When a pulse reflects from a fixed end,it undergoes a phase change of $\pi$. After two reflections (traveling a total distance of $2L$),the pulse undergoes a total phase change of $2\pi$,returning to its original phase.
Therefore,the time taken for the pulse to travel $2L$ is $\Delta t_{\min} = \frac{2L}{v} = \frac{2 \times 0.4}{8} = \frac{0.8}{8} = 0.1\, s$.

(D) Since there are no external torques acting on the system,the angular momentum of the system remains constant.
Initially,when the beads are at the center of the rod,the moment of inertia of the system is $I_1 = \frac{ML^2}{12}$.
The initial angular momentum is $L_1 = I_1{\omega _0} = \left( \frac{ML^2}{12} \right){\omega _0}$.
When the beads reach the ends of the rod,the moment of inertia of the system becomes $I_2 = \frac{ML^2}{12} + m\left( \frac{L}{2} \right)^2 + m\left( \frac{L}{2} \right)^2 = \frac{ML^2}{12} + \frac{mL^2}{4} + \frac{mL^2}{4} = \frac{ML^2}{12} + \frac{mL^2}{2} = \frac{ML^2 + 6mL^2}{12} = \frac{L^2(M + 6m)}{12}$.
The final angular momentum is $L_2 = I_2\omega ' = \left( \frac{L^2(M + 6m)}{12} \right)\omega '$.
By the principle of conservation of angular momentum,$L_1 = L_2$:
$\left( \frac{ML^2}{12} \right){\omega _0} = \left( \frac{L^2(M + 6m)}{12} \right)\omega '$
$\omega ' = \frac{M{\omega _0}}{M + 6m}$.

(A) For a uniform square plate,the moment of inertia about any axis passing through the centre and lying in the plane of the plate is the same,provided the axis is symmetric with respect to the square's geometry.
Let $I_Z$ be the moment of inertia about an axis passing through the centre and perpendicular to the plane of the square.
According to the perpendicular axis theorem,$I_Z = I_{AB} + I_{A'B'}$,where $AB$ and $A'B'$ are two mutually perpendicular axes in the plane of the plate.
Since the square is symmetric,$I_{AB} = I_{A'B'} = I$. Thus,$I_Z = 2I$.
Similarly,for any two mutually perpendicular axes $CD$ and $C'D'$ in the plane of the plate passing through the centre,$I_Z = I_{CD} + I_{C'D'}$.
Due to the rotational symmetry of the square,the moment of inertia about any axis in the plane passing through the centre is constant and equal to $I$. Therefore,$I_{CD} = I$.

(D) Given the torque $\overrightarrow{\tau} = \overrightarrow{A} \times \overrightarrow{L}$. Since $\overrightarrow{\tau} = \frac{d\overrightarrow{L}}{dt}$,we have $\frac{d\overrightarrow{L}}{dt} = \overrightarrow{A} \times \overrightarrow{L}$.
$1$. By the definition of the cross product,$\frac{d\overrightarrow{L}}{dt}$ is perpendicular to both $\overrightarrow{A}$ and $\overrightarrow{L}$. Thus,$\frac{d\overrightarrow{L}}{dt} \perp \overrightarrow{L}$,making option $(a)$ correct.
$2$. To check the magnitude of $\overrightarrow{L}$,consider $L^2 = \overrightarrow{L} \cdot \overrightarrow{L}$. Differentiating with respect to time $t$: $\frac{d}{dt}(L^2) = 2\overrightarrow{L} \cdot \frac{d\overrightarrow{L}}{dt}$. Since $\frac{d\overrightarrow{L}}{dt} \perp \overrightarrow{L}$,the dot product $\overrightarrow{L} \cdot \frac{d\overrightarrow{L}}{dt} = 0$. Therefore,$\frac{d}{dt}(L^2) = 0$,which means the magnitude $L$ is constant. Thus,option $(c)$ is correct.
$3$. Since $\frac{d\overrightarrow{L}}{dt} = \overrightarrow{A} \times \overrightarrow{L}$,the rate of change of $\overrightarrow{L}$ is always perpendicular to $\overrightarrow{A}$. The component of $\overrightarrow{L}$ along $\overrightarrow{A}$ is given by $L_A = \overrightarrow{L} \cdot \hat{A}$. Differentiating with respect to time: $\frac{dL_A}{dt} = \frac{d\overrightarrow{L}}{dt} \cdot \hat{A} = (\overrightarrow{A} \times \overrightarrow{L}) \cdot \hat{A}$. Since the cross product $(\overrightarrow{A} \times \overrightarrow{L})$ is perpendicular to $\overrightarrow{A}$,the dot product with $\hat{A}$ is zero. Thus,$\frac{dL_A}{dt} = 0$,meaning the component of $\overrightarrow{L}$ along $\overrightarrow{A}$ is constant. Thus,option $(b)$ is correct.
Since all statements are correct,the answer is $(d)$.

(D) The electric potential $V$ at the origin due to a system of point charges is given by the sum $V = \sum \frac{1}{4\pi\varepsilon_0} \frac{q_i}{r_i}$.
Substituting the given positions and charges:
$V = \frac{1}{4\pi\varepsilon_0} \left[ \left( \frac{q}{x_0} + \frac{q}{3x_0} + \frac{q}{5x_0} + \dots \right) - \left( \frac{q}{2x_0} + \frac{q}{4x_0} + \frac{q}{6x_0} + \dots \right) \right]$
$V = \frac{q}{4\pi\varepsilon_0 x_0} \left[ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \dots \right]$
Using the Taylor series expansion for $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$, for $x=1$, we get $\ln(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$
Therefore, $V = \frac{q \ln 2}{4\pi\varepsilon_0 x_0}$.

(D) The electric field on the axis of a charged ring at a distance $z_0$ from the center is given by $E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Q z_0}{(R^2 + z_0^2)^{3/2}}$.
Since the particle $P$ has a negative charge $-q$,the force acting on it is $F = -qE = -\frac{1}{4\pi \varepsilon_0} \cdot \frac{Q q z_0}{(R^2 + z_0^2)^{3/2}}$.
The negative sign indicates that the force is always directed towards the origin $O$. As the particle crosses the origin,the force reverses direction,always pulling it back towards the center. Thus,the motion is periodic for all $z_0 > 0$.
If $z_0 \ll R$,we can approximate $(R^2 + z_0^2)^{3/2} \approx (R^2)^{3/2} = R^3$.
Then the force becomes $F \approx -\left( \frac{Q q}{4\pi \varepsilon_0 R^3} \right) z_0$.
Since $F \propto -z_0$,the motion is approximately simple harmonic for small $z_0$.

(D) For a non-conducting solid sphere,the electric field $E$ is given by:
Inside the sphere $(r < R)$: $E = \frac{kQr}{R^3}$,which means $E \propto r$. Thus,as $r$ increases,$E$ increases.
Outside the sphere $(r > R)$: $E = \frac{kQ}{r^2}$,which means $E \propto \frac{1}{r^2}$. Thus,as $r$ increases,$E$ decreases.
Therefore,both statements $(a)$ and $(c)$ are correct.

(D) The negative plate is at $x = 0$ and the positive plate is at $x = 3d$. The electric field lines are always directed from the positive plate to the negative plate. Therefore,the direction of the electric field is from $x = 3d$ towards $x = 0$ (i.e.,in the negative $x$-direction) throughout the region. Thus,the direction remains the same.
The magnitude of the electric field in the air gap is $E_{air} = \frac{\sigma}{\varepsilon_0}$ and in the dielectric slab is $E_{dielectric} = \frac{\sigma}{K\varepsilon_0}$. Since $K > 1$,the magnitudes are different.
Electric potential $V$ is related to the electric field $E$ by $E = -\frac{dV}{dx}$. Since the electric field is directed towards the negative $x$-axis,$E_x < 0$. Thus,$-\frac{dV}{dx} < 0$,which implies $\frac{dV}{dx} > 0$. This means the electric potential $V$ increases continuously as we move from $x = 0$ (negative plate) to $x = 3d$ (positive plate).
Therefore,both statements $(b)$ and $(c)$ are correct.

(D) To find the equivalent resistance,we simplify the circuit from right to left.
First,the last branch has a $2\,\Omega$ resistor in series with a $4\,\Omega$ resistor,which is in parallel with an $8\,\Omega$ resistor. The resistance of this part is $R_1 = \frac{(2+4) \times 8}{(2+4) + 8} = \frac{6 \times 8}{14} = \frac{48}{14} = \frac{24}{7}\,\Omega$.
Adding the $2\,\Omega$ resistor in series with this combination gives $R_2 = 2 + \frac{24}{7} = \frac{38}{7}\,\Omega$.
This is in parallel with the next $8\,\Omega$ resistor: $R_3 = \frac{(\frac{38}{7}) \times 8}{(\frac{38}{7}) + 8} = \frac{304}{38 + 56} = \frac{304}{94} = \frac{152}{47}\,\Omega$.
Adding the remaining series resistors ($3\,\Omega$ and $2\,\Omega$),the total equivalent resistance is $R_{eq} = 3 + 2 + \frac{152}{47} = 5 + 3.23 = 8.23\,\Omega$ (approx).
However,using the provided solution image logic: The main current is $I = 1\,A$.
At the first junction,the current splits. Following the circuit reduction,the current through the $4\,\Omega$ resistor is $0.25\,A$.

(A) The current $i$ produced by the rotating charge $q$ is given by $i = \frac{q}{T} = \frac{q}{2\pi / \omega} = \frac{q\omega}{2\pi}$.
The magnetic moment $M$ of the system is $M = iA$,where $A = \pi R^2$ is the area of the circle traced by the particles.
$M = \left( \frac{q\omega}{2\pi} \right) (\pi R^2) = \frac{q\omega R^2}{2}$.
The angular momentum $L$ of the system about the centre is $L = I\omega$,where $I$ is the moment of inertia.
For two particles of mass $m$ at distance $R$ from the axis,$I = mR^2 + mR^2 = 2mR^2$.
Thus,$L = (2mR^2)\omega = 2mR^2\omega$.
The ratio of the magnitudes is $\frac{M}{L} = \frac{q\omega R^2 / 2}{2mR^2\omega} = \frac{q}{4m}$.
Wait,re-evaluating the current: Since there are two charges,the total current $i = \frac{2q}{T} = \frac{2q\omega}{2\pi} = \frac{q\omega}{\pi}$.
Then $M = iA = (\frac{q\omega}{\pi})(\pi R^2) = q\omega R^2$.
Then $\frac{M}{L} = \frac{q\omega R^2}{2mR^2\omega} = \frac{q}{2m}$.

(D) Let the two wires be placed in the $xy$-plane,parallel to the $y$-axis,at $x = -d/2$ and $x = d/2$. The currents are in opposite directions (e.g.,$+I$ in the $y$-direction and $-I$ in the $y$-direction).
At the midpoint (origin),the magnetic field $B_1$ due to the first wire is directed into the plane ($-z$ direction),and the magnetic field $B_2$ due to the second wire is also directed into the plane ($-z$ direction).
The resultant magnetic field $B = B_1 + B_2$ is perpendicular to the plane of the wires (along the $-z$ direction).
The velocity $v$ of the charge $q$ is given as perpendicular to the plane of the wires (along the $z$ direction).
The magnetic force on a moving charge is given by $F = q(v \times B)$.
Since the velocity vector $v$ and the magnetic field vector $B$ are collinear (both along the $z$-axis),the cross product $v \times B = 0$.
Therefore,the magnitude of the magnetic force acting on the charge is $0$.

(B) When a metal rod of length $l$ moves with a velocity $v$ in a uniform magnetic field $B$ perpendicular to both the length of the rod and the velocity,a motional electromotive force (emf) is induced across the ends of the rod.
According to the Lorentz force law,the free electrons in the rod experience a magnetic force $F_m = q(v \times B)$. Due to this force,electrons accumulate at one end of the rod,creating a potential difference between the ends.
This separation of charge creates an internal electric field $E$ within the rod,which exerts an electric force $F_e = qE$ on the charges,opposing the magnetic force.
At equilibrium,the magnetic force is balanced by the electric force,i.e.,$qE = qvB$,which implies $E = vB$.
Since an electric field $E$ exists within the rod,the potential varies along the length of the rod. Therefore,statement $(b)$ is correct.

(D) The induced electromotive force $e$ is given by $e = \frac{d\phi}{dt} = L \frac{dI}{dt}$.
From this,the unit of inductance $L$ is $[L] = \frac{[\text{Weber}]}{[\text{Ampere}]}$.
Since $e = \frac{d\phi}{dt}$,we have $[\text{Volt}] = \frac{[\text{Weber}]}{[\text{Second}]}$,which implies $[\text{Weber}] = [\text{Volt} \cdot \text{Second}]$.
Substituting this,we get $[L] = \frac{[\text{Volt} \cdot \text{Second}]}{[\text{Ampere}]}$.
Also,the energy stored in an inductor is $U = \frac{1}{2} L I^2$. Thus,$[\text{Joule}] = [L] \cdot [\text{Ampere}]^2$,which implies $[L] = \frac{[\text{Joule}]}{[\text{Ampere}]^2}$.
Therefore,all the given expressions are equivalent to the henry.

(B) The magnetic field $B$ at the centre of a large square loop of side $L$ carrying current $i$ is given by:
$B = \frac{\mu_0}{4\pi} \frac{8\sqrt{2}i}{L}$
Since the smaller loop of side $l$ is placed at the centre and $L \gg l$, we can assume the magnetic field is approximately uniform over the area of the smaller loop.
The magnetic flux $\phi$ linked with the smaller loop is:
$\phi = B \times (\text{Area of smaller loop}) = B \times l^2$
$\phi = \left( \frac{\mu_0}{4\pi} \frac{8\sqrt{2}i}{L} \right) l^2$
By definition, $\phi = Mi$, where $M$ is the mutual inductance.
$Mi = \left( \frac{\mu_0}{4\pi} \frac{8\sqrt{2}l^2}{L} \right) i$
$M = \frac{\mu_0}{4\pi} \frac{8\sqrt{2}l^2}{L}$
Therefore, $M \propto \frac{l^2}{L}$.

(C) The work function $W_0$ is given as $4.0 \,eV$.
The threshold wavelength $\lambda_0$ is the longest wavelength capable of causing photoelectric emission.
It is calculated using the formula: $\lambda_0 = \frac{hc}{W_0}$.
Using the approximation $hc \approx 12400 \,eV \cdot \mathring{A}$:
$\lambda_0 = \frac{12400 \,eV \cdot \mathring{A}}{4.0 \,eV} = 3100 \,\mathring{A}$.
Since $1 \,nm = 10 \,\mathring{A}$,we have $\lambda_0 = 310 \,nm$.

(B) The continuous $X$-ray spectrum is produced due to the deceleration of electrons hitting the target. The maximum energy of a photon emitted is equal to the kinetic energy of the incident electron,given by $E_{\max} = eV = \frac{hc}{\lambda_{\min}}$.
Thus,the minimum wavelength is $\lambda_{\min} = \frac{hc}{eV}$.
Since electrons can lose any amount of energy less than their total kinetic energy,photons of all wavelengths greater than $\lambda_{\min}$ are emitted.
Therefore,the range of wavelengths for the continuous $X$-ray spectrum is from $\lambda_{\min}$ to $\infty$.

(D) According to the Bohr model,the time period $T$ of an electron in an orbit with principal quantum number $n$ is given by $T = \frac{2\pi r}{v}$.
Since $r \propto n^2$ and $v \propto \frac{1}{n}$,we have $T \propto n^3$.
Given that the time period in the initial state is eight times that in the final state: $T_{n_1} = 8 T_{n_2}$.
Substituting the proportionality: $n_1^3 = 8 n_2^3$.
Taking the cube root on both sides: $n_1 = 2 n_2$.
Checking the options:
For option $(a)$: $n_1 = 4, n_2 = 2$. Here $4 = 2(2)$,which is correct.
For option $(b)$: $n_1 = 6, n_2 = 3$. Here $6 = 2(3)$,which is correct.
Thus,both $(a)$ and $(b)$ are valid possibilities.

(D) Radioactive decay is a stochastic (probabilistic) process.
The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$,where $\lambda$ is the decay constant.
This equation implies that the probability of a nucleus decaying is distributed over time from $t = 0$ to $t = \infty$.
Therefore,it is impossible to predict exactly when a specific nucleus will decay. $A$ given nucleus has a non-zero probability of decaying at any instant $t > 0$.
Thus,option $(D)$ is correct.

(D) The mass of a nucleus is always less than the sum of the masses of its constituent nucleons due to the mass defect,which accounts for the binding energy of the nucleus.
For the $_{10}^{20}Ne$ nucleus,which consists of $10$ protons and $10$ neutrons,the mass ${M_1}$ is given by ${M_1} < 10({m_p} + {m_n})$. This confirms option $(a)$.
For the $_{20}^{40}Ca$ nucleus,which consists of $20$ protons and $20$ neutrons,the binding energy per nucleon is higher than that of $_{10}^{20}Ne$. Since the binding energy per nucleon increases with the mass number for lighter nuclei,the mass defect per nucleon is greater for $_{20}^{40}Ca$ than for $_{10}^{20}Ne$.
Therefore,the mass of the $_{20}^{40}Ca$ nucleus ${M_2}$ is less than twice the mass of the $_{10}^{20}Ne$ nucleus ${M_1}$,i.e.,${M_2} < 2{M_1}$. This confirms option $(c)$.
Thus,both $(a)$ and $(c)$ are correct.

(C) When a $PN-$ junction is formed,electrons diffuse from the $N-$ region to the $P-$ region,and holes diffuse from the $P-$ region to the $N-$ region.
This diffusion leaves behind ionized donors in the $N-$ region (positive charge) and ionized acceptors in the $P-$ region (negative charge) near the junction,creating a depletion layer.
Due to this charge distribution,a potential barrier is established such that the $N-$ side is at a higher potential than the $P-$ side.
Since the electric field $E$ is directed from higher potential to lower potential,the electric field at the junction is directed from the $N-$ type side to the $P-$ type side.

(D) In a common emitter $(CE)$ amplifier configuration,the transistor must be operated in the active region.
For the active region,the base-emitter junction must be forward-biased,and the collector-base junction must be reverse-biased.
Therefore,statement $(a)$ is correct.
Furthermore,in a $CE$ amplifier,the input signal is applied to the base-emitter circuit,which is in series with the $DC$ bias voltage applied to the base-emitter junction.
Therefore,statement $(c)$ is also correct.
Thus,both $(a)$ and $(c)$ are correct.

(D) For total internal reflection $(TIR)$ to occur,the angle of incidence $(i)$ must be greater than the critical angle $(C)$.
$i > C$
Taking the sine of both sides,we get $\sin i > \sin C$.
Given $i = 45^\circ$,we have $\sin 45^\circ > \frac{1}{n}$.
Since $\sin 45^\circ = \frac{1}{\sqrt{2}}$,the inequality becomes $\frac{1}{\sqrt{2}} > \frac{1}{n}$.
This implies $n > \sqrt{2}$.
Since $\sqrt{2} \approx 1.414$,we require $n > 1.414$.
Comparing this with the given options:
Option $(a)$ $1.3 < 1.414$ (Incorrect).
Option $(b)$ $1.6 > 1.414$ (Correct).
Option $(c)$ $1.5 > 1.414$ (Correct).
Therefore,both $(b)$ and $(c)$ are possible values for $n$.

(A) When a light ray travels from $\mu_{1}$ to $\mu_{2}$ after refraction at a single spherical surface,the formula is given by:
$\frac{\mu_{2}}{v} - \frac{\mu_{1}}{u} = \frac{\mu_{2} - \mu_{1}}{R}$
Given that the object is in air $(\mu_{1} = 1.0)$ and the image is in glass $(\mu_{2} = 1.5)$.
According to the sign convention,the object distance $u = -PO = -x$ and the image distance $v = +OQ = +x$ (since $PO = OQ = x$). The radius of curvature $R$ is positive as the center of curvature lies in the glass.
Substituting these values into the formula:
$\frac{1.5}{x} - \frac{1}{-x} = \frac{1.5 - 1.0}{R}$
$\frac{1.5}{x} + \frac{1}{x} = \frac{0.5}{R}$
$\frac{2.5}{x} = \frac{0.5}{R}$
$x = \frac{2.5}{0.5} R = 5 R$
Therefore,the distance $PO$ is $5 R$.

(D) Initially,the light rays from the object at $C$ strike the mirror normally and retrace their path,forming an image at $C$. When the mirror is filled with water,the light rays from the object at $C$ travel through air and then enter the water. Due to refraction at the water surface,the rays bend towards the normal. These rays then strike the concave mirror and reflect. After reflection,they pass through the water again and refract away from the normal at the water surface. As a result,the rays converge at a point $I$ located between $C$ and $O$. Thus,a real image is formed between $C$ and $O$.

(D) The path difference between the rays coming from the edges of a slit of width $d$ at an angle $\theta$ is given by $\Delta x = d \sin \theta$.
For a single slit diffraction pattern,the condition for the $n^{th}$ secondary maximum is given by $d \sin \theta = (n + \frac{1}{2}) \lambda$,where $n = 1, 2, 3, \dots$.
For the first secondary maximum,$n = 1$,so $d \sin \theta = \frac{3\lambda}{2}$.
The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula $\phi = \frac{2\pi}{\lambda} \Delta x$.
Substituting the value of path difference for the first secondary maximum: $\phi = \frac{2\pi}{\lambda} \times \frac{3\lambda}{2} = 3\pi$.
Therefore,the phase difference is $3\pi$.