JEE Main 2019 Physics Question Paper with Answer and Solution

(A) The standard equation of the trajectory of a projectile is given by $y = x \tan \theta_0 - \frac{g x^2}{2 v_0^2 \cos^2 \theta_0}$.
Comparing this with the given equation $y = 2x - 9x^2$:
$1$. $\tan \theta_0 = 2$. Since $\tan \theta_0 = \frac{\text{opposite}}{\text{adjacent}} = \frac{2}{1}$,the hypotenuse is $\sqrt{2^2 + 1^2} = \sqrt{5}$. Thus,$\cos \theta_0 = \frac{1}{\sqrt{5}}$ and $\sin \theta_0 = \frac{2}{\sqrt{5}}$.
$2$. $\frac{g}{2 v_0^2 \cos^2 \theta_0} = 9$.
Substituting $g = 10$ and $\cos^2 \theta_0 = \left( \frac{1}{\sqrt{5}} \right)^2 = \frac{1}{5}$:
$\frac{10}{2 v_0^2 (1/5)} = 9 \implies \frac{10}{2 v_0^2 / 5} = 9 \implies \frac{25}{v_0^2} = 9 \implies v_0^2 = \frac{25}{9} \implies v_0 = \frac{5}{3} \, ms^{-1}$.
Therefore,$\theta_0 = \cos^{-1} \left( \frac{1}{\sqrt{5}} \right) = \sin^{-1} \left( \frac{2}{\sqrt{5}} \right)$ and $v_0 = \frac{5}{3} \, ms^{-1}$. Comparing with options,option $A$ is correct.

(D) For a cyclic process,the total change in internal energy is zero: $\Delta U_{ab} + \Delta U_{bc} + \Delta U_{ca} = 0$.
Given $\Delta U_{ca} = -180\, J$,we have $\Delta U_{ab} + \Delta U_{bc} = 180\, J$.
From the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
For path $bc$,the process is isochoric (vertical line in $P-V$ diagram),so $\Delta W_{bc} = 0$. Thus,$\Delta U_{bc} = \Delta Q_{bc} = 60\, J$.
Substituting this into the cyclic equation: $\Delta U_{ab} + 60 = 180 \implies \Delta U_{ab} = 120\, J$.
Now,for path $ab$,$\Delta W_{ab} = \Delta Q_{ab} - \Delta U_{ab} = 250 - 120 = 130\, J$.
The total work done along path $abc$ is $\Delta W_{abc} = \Delta W_{ab} + \Delta W_{bc} = 130 + 0 = 130\, J$.

(A) At the lowest point,the velocity of the person is $V_0 = \omega A = \sqrt{\frac{g}{L}} (\theta_0 L) = \theta_0 \sqrt{gL}$.
When the person stands up,the distance of the center of mass from the pivot changes from $L$ to $L-l$. By the conservation of angular momentum about the pivot point:
$M V_0 L = M V_1 (L-l) \implies V_1 = V_0 \frac{L}{L-l} = V_0 (1 - \frac{l}{L})^{-1} \approx V_0 (1 + \frac{l}{L})$.
The work-energy theorem states: $W_g + W_p = \Delta KE$.
Here,$W_g = -Mgl$ (work done against gravity as the center of mass rises).
$\Delta KE = \frac{1}{2} M (V_1^2 - V_0^2) = \frac{1}{2} M [V_0^2 (1 + \frac{l}{L})^2 - V_0^2] \approx \frac{1}{2} M V_0^2 (1 + \frac{2l}{L} - 1) = M V_0^2 \frac{l}{L}$.
Substituting $V_0^2 = \theta_0^2 gL$:
$\Delta KE = M (\theta_0^2 gL) \frac{l}{L} = Mgl \theta_0^2$.
Thus,$W_p = Mgl + \Delta KE = Mgl + Mgl \theta_0^2 = Mgl(1 + \theta_0^2)$.

(D) The coordinates of the three particles are:
$m_1 = 50\, g$ at $(0, 0)$
$m_2 = 100\, g$ at $(1, 0)$
$m_3 = 150\, g$ at $(0.5, \frac{\sqrt{3}}{2})$
The $x$-coordinate of the center of mass is:
$X_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3} = \frac{50(0) + 100(1) + 150(0.5)}{50 + 100 + 150} = \frac{100 + 75}{300} = \frac{175}{300} = \frac{7}{12}\, m$
The $y$-coordinate of the center of mass is:
$Y_{cm} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{m_1 + m_2 + m_3} = \frac{50(0) + 100(0) + 150(\frac{\sqrt{3}}{2})}{300} = \frac{75\sqrt{3}}{300} = \frac{\sqrt{3}}{4}\, m$
Thus,the coordinates of the center of mass are $\left( \frac{7}{12}\,m, \frac{\sqrt{3}}{4}\,m \right)$.

(D) Since the mass of the object remains constant,the weight of the object is proportional to the acceleration due to gravity $g$.
Given: $\frac{W_{earth}}{W_{planet}} = \frac{9}{4} = \frac{g_{earth}}{g_{planet}}$.
We know that $g = \frac{GM}{R^2}$. Since density $\rho$ is constant,$M = \rho \cdot \frac{4}{3}\pi R^3$,so $g = \frac{G \cdot \rho \cdot \frac{4}{3}\pi R^3}{R^2} = G \cdot \rho \cdot \frac{4}{3}\pi R \propto R$.
However,the problem states the mass of the planet is $\frac{1}{9}$ of the Earth's mass. Let $M_p = \frac{M_e}{9}$.
Using $g = \frac{GM}{R^2}$,we have $\frac{g_e}{g_p} = \frac{M_e}{M_p} \cdot \frac{R_p^2}{R_e^2} = 9 \cdot \frac{R_p^2}{R^2}$.
Given $\frac{g_e}{g_p} = \frac{9}{4}$,so $\frac{9}{4} = 9 \cdot \frac{R_p^2}{R^2}$.
$\frac{1}{4} = \frac{R_p^2}{R^2} \implies \frac{R_p}{R} = \frac{1}{2}$.
Therefore,$R_p = \frac{R}{2}$.

(D) Case $(A)$ (Pushing):
Vertical forces: $N_1 = mg + F \sin 30^o = 5 \times 10 + 20 \times 0.5 = 50 + 10 = 60\, N$.
Horizontal force: $F_x = F \cos 30^o = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}\, N$.
Friction: $f_1 = \mu N_1 = 0.2 \times 60 = 12\, N$.
Acceleration: $a_1 = \frac{F_x - f_1}{m} = \frac{10\sqrt{3} - 12}{5} = 2\sqrt{3} - 2.4 \approx 3.464 - 2.4 = 1.064\, ms^{-2}$.
Case $(B)$ (Pulling):
Vertical forces: $N_2 = mg - F \sin 30^o = 5 \times 10 - 20 \times 0.5 = 50 - 10 = 40\, N$.
Horizontal force: $F_x = F \cos 30^o = 10\sqrt{3}\, N$.
Friction: $f_2 = \mu N_2 = 0.2 \times 40 = 8\, N$.
Acceleration: $a_2 = \frac{F_x - f_2}{m} = \frac{10\sqrt{3} - 8}{5} = 2\sqrt{3} - 1.6 \approx 3.464 - 1.6 = 1.864\, ms^{-2}$.
Difference: $a_2 - a_1 = (2\sqrt{3} - 1.6) - (2\sqrt{3} - 2.4) = 0.8\, ms^{-2}$.

(C) For a diatomic gas with rigid molecules,the molar heat capacity at constant pressure is given by $C_{p} = \frac{7}{2} R$.
The work done in an isobaric process is given by $W = P \Delta V = nR \Delta T = 10 \ J$.
The heat energy absorbed by the gas is given by $\Delta Q = n C_{p} \Delta T$.
Substituting $C_{p} = \frac{7}{2} R$,we get $\Delta Q = n \left( \frac{7}{2} R \right) \Delta T = \frac{7}{2} (nR \Delta T)$.
Since $nR \Delta T = W = 10 \ J$,we have $\Delta Q = \frac{7}{2} \times 10 \ J = 35 \ J$.

(C) The loudness level in decibels $(dB)$ is given by the formula: $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$,where $I$ is the intensity of sound and $I_0 = 10^{-12}\, W/m^2$ is the reference intensity.
Given $L = 120\, dB$,we have: $120 = 10 \log_{10} \left( \frac{I}{10^{-12}} \right)$.
Dividing by $10$,we get $12 = \log_{10} \left( \frac{I}{10^{-12}} \right)$,which implies $\frac{I}{10^{-12}} = 10^{12}$.
Thus,$I = 10^{12} \times 10^{-12} = 1\, W/m^2$.
The intensity $I$ at a distance $r$ from a point source is given by $I = \frac{P}{4 \pi r^2}$,where $P = 2\, W$ is the power.
Substituting the values: $1 = \frac{2}{4 \pi r^2}$.
Solving for $r^2$: $r^2 = \frac{2}{4 \pi} = \frac{1}{2 \pi} \approx \frac{1}{6.28} \approx 0.159\, m^2$.
Taking the square root: $r = \sqrt{0.159} \approx 0.399\, m$.
Converting to centimeters: $r \approx 0.399 \times 100 = 39.9\, cm \approx 40\, cm$.

(C) For the same range $R$,the angles of projection must be complementary,i.e.,$\theta$ and $(90^\circ - \theta)$.
The range $R$ is given by: $R = \frac{u^2 \sin(2\theta)}{g} = \frac{2u^2 \sin\theta \cos\theta}{g}$.
The maximum heights for the two particles are:
$h_1 = \frac{u^2 \sin^2\theta}{2g}$
$h_2 = \frac{u^2 \sin^2(90^\circ - \theta)}{2g} = \frac{u^2 \cos^2\theta}{2g}$
Multiplying $h_1$ and $h_2$:
$h_1 h_2 = \left( \frac{u^2 \sin^2\theta}{2g} \right) \left( \frac{u^2 \cos^2\theta}{2g} \right) = \frac{u^4 \sin^2\theta \cos^2\theta}{4g^2}$
$h_1 h_2 = \frac{1}{16} \left( \frac{4u^4 \sin^2\theta \cos^2\theta}{g^2} \right) = \frac{1}{16} R^2$
Therefore,$R^2 = 16 h_1 h_2$.

(B) Given the speed $v = b\sqrt{x}$.
We know that $v = \frac{dx}{dt}$,so $\frac{dx}{dt} = b\sqrt{x}$.
Separating the variables,we get $\frac{dx}{\sqrt{x}} = b dt$.
Integrating both sides from $t = 0$ (where $x = 0$) to $t = \tau$ (where $x = x$):
$\int_{0}^{x} x^{-1/2} dx = \int_{0}^{\tau} b dt$
$[2\sqrt{x}]_{0}^{x} = b\tau$
$2\sqrt{x} = b\tau$,which implies $\sqrt{x} = \frac{b\tau}{2}$.
Substituting this into the expression for speed $v = b\sqrt{x}$:
$v = b \left( \frac{b\tau}{2} \right) = \frac{b^2\tau}{2}$.

(C) The total heat $Q$ required to raise the temperature of $1, kg$ of water from $20, ^oC$ to $100, ^oC$ and then to evaporate it is given by:
$Q = mc\Delta T + mL$
Here, $m = 1, kg$, $c = 4200, J/kg, ^oC$, $\Delta T = (100 - 20) = 80, ^oC$, and $L = 2260 \times 10^3, J/kg$.
$Q = (1 \times 4200 \times 80) + (1 \times 2260 \times 10^3) = 336000 + 2260000 = 2596000, J$.
The power $P$ dissipated by the heating element is:
$P = \frac{V_{rms}^2}{R} = \frac{200^2}{20} = \frac{40000}{20} = 2000, W$.
The time $t$ taken is given by $t = \frac{Q}{P}$:
$t = \frac{2596000}{2000} = 1298, s$.
Converting to minutes: $t = \frac{1298}{60} \approx 21.63, min \approx 22, min$.

(C) The length of the cylinder remains unchanged,which means the thermal expansion is exactly compensated by the longitudinal compression.
The thermal expansion strain is given by $\Delta L / L = \alpha T$,where $\alpha$ is the coefficient of linear expansion.
The compressive strain due to force $F$ is given by $\Delta L / L = \text{Stress} / Y = F / (A Y) = F / (\pi r^2 Y)$.
Equating the two strains: $\alpha T = F / (\pi r^2 Y)$.
Therefore,$\alpha = F / (\pi r^2 YT)$.
The coefficient of volume expansion $\gamma$ is related to the coefficient of linear expansion $\alpha$ by the relation $\gamma = 3\alpha$.
Substituting the value of $\alpha$,we get $\gamma = 3F / (\pi r^2 YT)$.

(D) Given: Frequency of sources $f = 660\, Hz$, velocity of sound $v = 330\, m/s$, and beat frequency $f_b = 10\, \text{beats/s}$.
When the listener moves away from $S_1$ with speed $u$, the observed frequency $f_1$ is:
$f_1 = f \left( \frac{v - u}{v} \right)$
When the listener moves towards $S_2$ with speed $u$, the observed frequency $f_2$ is:
$f_2 = f \left( \frac{v + u}{v} \right)$
The beat frequency is the difference between the two observed frequencies:
$f_b = f_2 - f_1 = f \left( \frac{v + u}{v} \right) - f \left( \frac{v - u}{v} \right)$
$f_b = \frac{f}{v} [v + u - (v - u)] = \frac{f}{v} [2u]$
Substituting the given values:
$10 = \frac{660}{330} \times 2u$
$10 = 2 \times 2u$
$10 = 4u$
$u = \frac{10}{4} = 2.5\, m/s$.

(D) Let $m$ be the mass of the bead and $N$ be the normal reaction force exerted by the wire on the bead.
The radius of the circular path of the bead is $R_{path} = r/2$.
The forces acting on the bead are the gravitational force $mg$ (downwards) and the normal reaction $N$ (perpendicular to the wire).
Resolving the forces into horizontal and vertical components:
$N \sin \theta = m R_{path} \omega^2 = m (r/2) \omega^2$ ... $(i)$
$N \cos \theta = mg$ ... (ii)
Dividing $(i)$ by (ii),we get:
$\tan \theta = \frac{(r/2) \omega^2}{g} = \frac{r \omega^2}{2g}$
From the geometry of the circle,the distance from the center $O$ to the vertical axis is $r/2$. The angle $\theta$ that the radius vector $OP$ makes with the vertical is given by $\sin \theta = \frac{r/2}{r} = 1/2$,so $\theta = 30^\circ$.
Thus,$\tan 30^\circ = \frac{1}{\sqrt{3}}$.
Equating the two expressions for $\tan \theta$:
$\frac{1}{\sqrt{3}} = \frac{r \omega^2}{2g}$
$\omega^2 = \frac{2g}{r\sqrt{3}}$

(C) The terminal velocity of a sphere of radius $r$ falling through a viscous fluid is given by Stokes' Law: $v_T = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta}$,where $\rho$ is the density of the sphere and $\sigma$ is the density of the fluid.
From this formula,we see that $v_T \propto r^2$.
Let the radius of the large sphere be $R$ and the radius of each small sphere be $r$. Since the volume remains constant when the sphere is broken into $27$ smaller identical spheres:
$\frac{4}{3} \pi R^3 = 27 \times \frac{4}{3} \pi r^3$
$R^3 = 27 r^3 \Rightarrow R = 3r \Rightarrow r = R/3$.
Now,the ratio of terminal velocities is:
$\frac{\nu_1}{\nu_2} = \frac{R^2}{r^2} = \frac{R^2}{(R/3)^2} = \frac{R^2}{R^2/9} = 9$.
Thus,the ratio $(\nu_1/\nu_2)$ is $9$.

(A) The total number of molecules $N$ is given by the integral of the number density $n(r)$ over the entire volume.
In spherical coordinates,the volume element is $dV = 4\pi r^2 dr$.
Thus,$N = \int_{0}^{\infty} n(r) dV = \int_{0}^{\infty} n_0 e^{-\alpha r^4} (4\pi r^2) dr$.
$N = 4\pi n_0 \int_{0}^{\infty} r^2 e^{-\alpha r^4} dr$.
Let $u = \alpha r^4$,then $r = (u/\alpha)^{1/4}$ and $dr = \frac{1}{4} \alpha^{-1/4} u^{-3/4} du$.
Substituting these into the integral:
$N = 4\pi n_0 \int_{0}^{\infty} (u/\alpha)^{2/4} e^{-u} (\frac{1}{4} \alpha^{-1/4} u^{-3/4}) du$.
$N = \pi n_0 \alpha^{-1/2} \alpha^{-1/4} \int_{0}^{\infty} u^{1/2 - 3/4} e^{-u} du = \pi n_0 \alpha^{-3/4} \int_{0}^{\infty} u^{-1/4} e^{-u} du$.
The integral $\int_{0}^{\infty} u^{-1/4} e^{-u} du$ is a constant (Gamma function $\Gamma(3/4)$).
Therefore,$N \propto n_0 \alpha^{-3/4}$.

(C) In a resonance tube experiment,the speed of sound $\nu$ is related to the frequency $f$ and the difference between two successive resonance lengths $\ell_1$ and $\ell_2$ by the formula: $\nu = 2f(\ell_2 - \ell_1)$.
Given:
Frequency $f = 480\, Hz$
$\ell_1 = 30\, cm = 0.30\, m$
$\ell_2 = 70\, cm = 0.70\, m$
Substituting the values into the formula:
$\nu = 2 \times 480 \times (0.70 - 0.30)$
$\nu = 960 \times 0.40$
$\nu = 384\, m/s$.

(D) The force constant $k$ of a spring is inversely proportional to its unstretched length,i.e.,$k \propto \frac{1}{\ell}$,which implies $k\ell = C$ (a constant).
For the two pieces,we have $k_1 \ell_1 = C$ and $k_2 \ell_2 = C$.
Therefore,$k_1 \ell_1 = k_2 \ell_2$.
Rearranging for the ratio,we get $\frac{k_1}{k_2} = \frac{\ell_2}{\ell_1}$.
Given that $\ell_1 = n\ell_2$,we substitute this into the ratio:
$\frac{k_1}{k_2} = \frac{\ell_2}{n\ell_2} = \frac{1}{n}$.

(A) Let time $T \propto c^{x} G^{y} h^{z}$.
$\Rightarrow T = k c^{x} G^{y} h^{z}$.
Taking dimensions on both sides: $[M^{0} L^{0} T^{1}] = [L T^{-1}]^{x} [M^{-1} L^{3} T^{-2}]^{y} [M L^{2} T^{-1}]^{z}$.
$[M^{0} L^{0} T^{1}] = [M^{-y+z} L^{x+3y+2z} T^{-x-2y-z}]$.
Equating powers of $M, L, T$ on both sides:
$-y + z = 0 \implies z = y \quad \dots(1)$
$x + 3y + 2z = 0 \quad \dots(2)$
$-x - 2y - z = 1 \quad \dots(3)$
Adding $(2)$ and $(3)$: $(x + 3y + 2z) + (-x - 2y - z) = 0 + 1 \implies y + z = 1$.
Since $z = y$,we have $2y = 1 \implies y = 1/2$.
Thus,$z = 1/2$.
Substituting into $(2)$: $x + 3(1/2) + 2(1/2) = 0 \implies x + 3/2 + 1 = 0 \implies x = -5/2$.
Therefore,$[T] = [G^{1/2} h^{1/2} c^{-5/2}]$.

(A) From the relation $q = CV$, the slope of the $q-V$ graph represents the capacitance $C = q/V$.
For line $A$, the capacitance is $C_A = 500\,\mu C / 10\,V = 50\,\mu F$.
For line $B$, the capacitance is $C_B = 80\,\mu C / 10\,V = 8\,\mu F$.
Since the parallel combination has a higher equivalent capacitance than the series combination, we have $C_{parallel} = 50\,\mu F$ and $C_{series} = 8\,\mu F$.
Let the two capacitors be $C_1$ and $C_2$. Then $C_1 + C_2 = 50$ and $(C_1 C_2) / (C_1 + C_2) = 8$.
Substituting $C_1 + C_2 = 50$ into the series formula: $(C_1 C_2) / 50 = 8$, so $C_1 C_2 = 400$.
Solving the quadratic equation $x^2 - 50x + 400 = 0$, we get $(x - 40)(x - 10) = 0$.
Thus, the capacitances are $40\,\mu F$ and $10\,\mu F$.

(B) The force per unit length on a wire carrying current $I$ due to another parallel wire carrying current $I'$ at a distance $r$ is given by $F = \frac{\mu_0 I I'}{2 \pi r}$.
For the net force on wire $C$ to be zero,the forces exerted by wires $A$ and $B$ must be equal in magnitude and opposite in direction.
Let wire $C$ be at a distance $x$ from $A$ and $(d-x)$ from $B$. The force per unit length due to $A$ is $F_A = \frac{\mu_0 I_1 I}{2 \pi x}$ (repulsive if currents are in the same direction,attractive if opposite).
The force per unit length due to $B$ is $F_B = \frac{\mu_0 I_2 I}{2 \pi (d-x)}$.
Since the currents $I_1$ and $I_2$ are in opposite directions (as shown in the figure),the forces will be in the same direction if $C$ is between $A$ and $B$. Thus,the equilibrium point must lie outside the region between $A$ and $B$.
Equating the magnitudes of the magnetic fields produced by $A$ and $B$ at the position of $C$:
$\frac{\mu_0 I_1}{2 \pi x} = \frac{\mu_0 I_2}{2 \pi |x - d|}$
$\frac{I_1}{x} = \frac{I_2}{|x - d|}$
Case $1$: $x > d$,then $x - d = x - d$,so $I_1(x - d) = I_2 x \Rightarrow x(I_1 - I_2) = I_1 d \Rightarrow x = \frac{I_1 d}{I_1 - I_2}$.
Case $2$: $x < 0$,then $|x - d| = d - x$,so $I_1(d - x) = I_2 x \Rightarrow I_1 d = x(I_1 + I_2) \Rightarrow x = \frac{I_1 d}{I_1 + I_2}$ (This is not possible as $x$ is distance from $A$).
Re-evaluating the geometry for equilibrium: The point where the net magnetic field is zero is $x = \frac{I_1 d}{I_1 - I_2}$.

(D) The radius of a charged particle moving in a magnetic field is given by $r = \frac{mv}{qB}$.
Since kinetic energy $K = \frac{1}{2}mv^2$,we have $mv = \sqrt{2mK}$.
Substituting this into the radius formula,we get $r = \frac{\sqrt{2mK}}{qB}$.
Given that $K$ and $B$ are the same for all particles,$r \propto \frac{\sqrt{m}}{q}$.
For a proton $(p)$: $m_p = m, q_p = e$,so $r_p \propto \frac{\sqrt{m}}{e}$.
For an electron $(e)$: $m_e \approx \frac{m}{1836}, q_e = e$,so $r_e \propto \frac{\sqrt{m/1836}}{e} = \frac{r_p}{\sqrt{1836}}$. Thus,$r_e < r_p$.
For a Helium nucleus $(He^{2+})$: $m_{He} \approx 4m, q_{He} = 2e$,so $r_{He} \propto \frac{\sqrt{4m}}{2e} = \frac{2\sqrt{m}}{2e} = \frac{\sqrt{m}}{e} = r_p$.
Therefore,$r_e < r_p = r_{He}$.

(B) The circuit consists of two cells of $1.5\,V$ each in series,so the total $EMF$ is $E_{eq} = 1.5 + 1.5 = 3\,V$. The total internal resistance is $r_{eq} = r + r = 2r$.
The external circuit has a $15\,\Omega$ and $10\,\Omega$ resistor in parallel,which are in series with a $2\,\Omega$ resistor.
The equivalent resistance of the parallel part is $R_p = \frac{15 \times 10}{15 + 10} = \frac{150}{25} = 6\,\Omega$.
The total external resistance is $R_{ext} = 6 + 2 = 8\,\Omega$.
The total resistance of the circuit is $R_{total} = 8 + 2r$.
The total current in the circuit is $i = \frac{E_{eq}}{R_{total}} = \frac{3}{8 + 2r}$.
The voltmeter is connected across the $10\,\Omega$ resistor,which is part of the parallel combination. The voltage across the parallel combination is $V_p = i \times R_p = i \times 6$.
Given $V_p = 2\,V$,we have $2 = \frac{3}{8 + 2r} \times 6$.
$2 = \frac{18}{8 + 2r} \Rightarrow 16 + 4r = 18 \Rightarrow 4r = 2 \Rightarrow r = 0.5\,\Omega$.

(A) The electric field is $\vec E = E_0 \hat i \cos(kz) \cos(\omega t)$.
Since the wave propagates in the $+z$ direction and $\vec E$ is along the $x$-axis,the magnetic field $\vec B$ must be along the $y$-axis.
Using Faraday's Law in differential form: $\nabla \times \vec E = -\frac{\partial \vec B}{\partial t}$.
For a wave propagating in the $z$-direction,$\frac{\partial E_x}{\partial z} = -\frac{\partial B_y}{\partial t}$.
Calculating the derivative of $E_x$ with respect to $z$: $\frac{\partial}{\partial z} [E_0 \cos(kz) \cos(\omega t)] = -k E_0 \sin(kz) \cos(\omega t)$.
Thus,$\frac{\partial B_y}{\partial t} = k E_0 \sin(kz) \cos(\omega t)$.
Integrating with respect to $t$: $B_y = \int k E_0 \sin(kz) \cos(\omega t) dt = \frac{k E_0}{\omega} \sin(kz) \sin(\omega t)$.
Since $c = \frac{\omega}{k}$,we have $\frac{k}{\omega} = \frac{1}{c}$.
Therefore,$\vec B = \frac{E_0}{c} \hat j \sin(kz) \sin(\omega t)$.

(B) The power gain in $dB$ is given by $G_p = 10 \log_{10} \left( \frac{P_{out}}{P_{in}} \right) = 60\, dB$.
Therefore,$\frac{P_{out}}{P_{in}} = 10^{(60/10)} = 10^6$.
Power gain is also expressed as $A_p = \beta^2 \times \frac{R_{out}}{R_{in}}$,where $\beta$ is the current gain,$R_{out} = 10\, k\Omega = 10^4\,\Omega$,and $R_{in} = 100\,\Omega$.
Substituting the values: $10^6 = \beta^2 \times \frac{10^4}{100}$.
$10^6 = \beta^2 \times 10^2$.
$\beta^2 = \frac{10^6}{10^2} = 10^4$.
$\beta = \sqrt{10^4} = 100$.

(A) In a meter bridge, the unknown resistance $X$ is given by the formula $X = R \frac{l_2}{l_1} = R \frac{(100 - l)}{l}$, where $l$ is the balancing length from the left end.
For reading $1$: $X = 1000 \times \frac{(100 - 60)}{60} = 1000 \times \frac{40}{60} \approx 666.67 \, \Omega$.
For reading $2$: $X = 100 \times \frac{(100 - 13)}{13} = 100 \times \frac{87}{13} \approx 669.23 \, \Omega$.
For reading $3$: $X = 10 \times \frac{(100 - 1.5)}{1.5} = 10 \times \frac{98.5}{1.5} \approx 656.67 \, \Omega$.
For reading $4$: $X = 1 \times \frac{(100 - 1)}{1} = 1 \times 99 = 99 \, \Omega$.
Comparing the calculated values of $X$, the values for readings $1, 2,$ and $3$ are consistent (around $660 \, \Omega$), whereas the value for reading $4$ is significantly different. Thus, reading $4$ is inconsistent.

(A) Given: Number of turns in primary $N_{p} = 300$,Number of turns in secondary $N_{s} = 150$,Output power $P_{s} = 2.2\, kW = 2200\, W$,Secondary current $I_{s} = 10\, A$.
First,calculate the secondary voltage $V_{s}$ using $P_{s} = V_{s} I_{s}$:
$2200 = V_{s} \times 10 \Rightarrow V_{s} = 220\, V$.
Using the transformer ratio $\frac{V_{p}}{V_{s}} = \frac{N_{p}}{N_{s}}$:
$\frac{V_{p}}{220} = \frac{300}{150} = 2 \Rightarrow V_{p} = 440\, V$.
Assuming an ideal transformer,input power $P_{p} = P_{s} = 2200\, W$:
$P_{p} = V_{p} I_{p} \Rightarrow 2200 = 440 \times I_{p} \Rightarrow I_{p} = \frac{2200}{440} = 5\, A$.
Thus,the input voltage is $440\, V$ and the input current is $5\, A$.

(D) The potential at a distance $z$ on the axis of a ring of radius $R$ and charge $q$ is $V = \frac{1}{4\pi \epsilon_0} \frac{q}{\sqrt{R^2 + z^2}}$.
Here,$R = 3a$. At $z = 4a$,the potential energy is $U_i = qV = \frac{q^2}{4\pi \epsilon_0 \sqrt{(3a)^2 + (4a)^2}} = \frac{q^2}{4\pi \epsilon_0 (5a)}$.
At the origin $(z = 0)$,the potential energy is $U_f = qV = \frac{q^2}{4\pi \epsilon_0 (3a)}$.
By the law of conservation of energy,$K_i + U_i = K_f + U_f$. For the charge to just cross the origin,$K_f = 0$.
$\frac{1}{2}mv^2 + \frac{q^2}{4\pi \epsilon_0 (5a)} = 0 + \frac{q^2}{4\pi \epsilon_0 (3a)}$.
$\frac{1}{2}mv^2 = \frac{q^2}{4\pi \epsilon_0 a} (\frac{1}{3} - \frac{1}{5}) = \frac{q^2}{4\pi \epsilon_0 a} (\frac{2}{15})$.
$v^2 = \frac{2}{m} \cdot \frac{2}{15} \cdot \frac{q^2}{4\pi \epsilon_0 a}$.
$v = \sqrt{\frac{2}{m}} \left( \frac{2}{15} \frac{q^2}{4\pi \epsilon_0 a} \right)^{1/2}$.

(B) The mobility $\mu$ is defined as the ratio of drift velocity $V_d$ to the electric field $E$: $\mu = \frac{V_d}{E}$.
From Ohm's law in vector form,$E = \rho J$,where $\rho$ is resistivity and $J$ is current density.
Current density $J = \frac{I}{A} = \frac{I}{\pi r^2}$.
Given: $I = 5\, A$,$\rho = 1.7 \times 10^{-8}\, \Omega \, m$,$r = 5\, mm = 5 \times 10^{-3}\, m$,and $V_d = 1.1 \times 10^{-3}\, m/s$.
First,calculate the electric field $E$:
$E = \rho \times \frac{I}{\pi r^2} = 1.7 \times 10^{-8} \times \frac{5}{\pi \times (5 \times 10^{-3})^2} = 1.7 \times 10^{-8} \times \frac{5}{\pi \times 25 \times 10^{-6}} = \frac{1.7 \times 10^{-8} \times 5}{78.54 \times 10^{-6}} \approx 1.08 \times 10^{-3}\, V/m$.
Now,calculate mobility $\mu$:
$\mu = \frac{1.1 \times 10^{-3}}{1.08 \times 10^{-3}} \approx 1.01\, m^2/Vs$.
Rounding to the nearest given option,the correct value is $1.0\, m^2/Vs$.

(A) Given:
Message signal frequency $f_m = 100\, MHz = 10^8\, Hz$.
Peak voltage of message signal $V_m = 100\, V$.
Carrier wave frequency $f_c = 300\, GHz$.
Peak voltage of carrier wave $V_c = 400\, V$.
The modulation index $\mu$ is given by the formula $\mu = \frac{V_m}{V_c} = \frac{100}{400} = 0.25$.
The two sideband frequencies are $(f_c + f_m)$ and $(f_c - f_m)$.
The difference between the two sideband frequencies is $(f_c + f_m) - (f_c - f_m) = 2f_m$.
Substituting the value of $f_m$,we get $2 \times 10^8\, Hz$.

(D) According to Snell's law,$n_1 \sin i = n_2 \sin r$.
Here,$n_1 = 1$ (vacuum),$i = 60^o$,$n_2 = \mu$ (glass),and $r = 30^o$.
$1 \cdot \sin 60^o = \mu \cdot \sin 30^o$
$\frac{\sqrt{3}}{2} = \mu \cdot \frac{1}{2} \Rightarrow \mu = \sqrt{3}$.
The optical path length is defined as the sum of the products of refractive index and geometric path length for each medium.
Optical path length $= n_{vacuum} \cdot AO + n_{glass} \cdot OB$.
From the geometry of the figure:
$AO = \frac{a}{\cos 60^o} = \frac{a}{1/2} = 2a$.
$OB = \frac{b}{\cos 30^o} = \frac{b}{\sqrt{3}/2} = \frac{2b}{\sqrt{3}}$.
Optical path length $= 1 \cdot (2a) + \sqrt{3} \cdot \left( \frac{2b}{\sqrt{3}} \right) = 2a + 2b$.

(C) Let $N_0$ be the initial number of nuclei for both materials.
For material $A$,the number of nuclei at time $t$ is $N_A = N_0 e^{-10\lambda t}$.
For material $B$,the number of nuclei at time $t$ is $N_B = N_0 e^{-\lambda t}$.
We are given that the ratio $\frac{N_A}{N_B} = \frac{1}{e}$.
Substituting the expressions,we get $\frac{N_0 e^{-10\lambda t}}{N_0 e^{-\lambda t}} = e^{-1}$.
This simplifies to $e^{-10\lambda t + \lambda t} = e^{-1}$,which means $e^{-9\lambda t} = e^{-1}$.
Equating the exponents,we have $-9\lambda t = -1$.
Therefore,$t = \frac{1}{9\lambda}$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by:
$K_{\max} = E - \phi$
where $E$ is the energy of the incident photon and $\phi$ is the work function.
Given $E = \frac{1237}{\lambda}$ and $\phi = \frac{1237}{\lambda_0}$,where $\lambda_0 = 380 \, nm$ and $\lambda = 260 \, nm$.
$K_{\max} = \frac{1237}{260} - \frac{1237}{380}$
$K_{\max} = 1237 \times \left( \frac{380 - 260}{380 \times 260} \right)$
$K_{\max} = 1237 \times \left( \frac{120}{98800} \right)$
$K_{\max} \approx 1.5 \, eV$.

(C) For the $1^{\text{st}}$ lens (plano-convex),using the lens maker's formula: $\frac{1}{f_1} = (\mu_1 - 1) \left( \frac{1}{\infty} - \frac{1}{-R} \right) = \frac{\mu_1 - 1}{R}$.
For the $2^{\text{nd}}$ lens (plano-concave),using the lens maker's formula: $\frac{1}{f_2} = (\mu_2 - 1) \left( \frac{1}{-R} - \frac{1}{\infty} \right) = -\frac{\mu_2 - 1}{R}$.
The equivalent focal length $f_{eq}$ of the combination is given by $\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2}$.
Substituting the values: $\frac{1}{f_{eq}} = \frac{\mu_1 - 1}{R} - \frac{\mu_2 - 1}{R} = \frac{\mu_1 - 1 - \mu_2 + 1}{R} = \frac{\mu_1 - \mu_2}{R}$.
Therefore,$f_{eq} = \frac{R}{\mu_1 - \mu_2}$.

(D) Let $G$ be the resistance of the galvanometer and $I_g = 10^{-4} \ A$ be the full-scale deflection current.
For a voltmeter of range $V = 5 \ V$ with series resistance $R_s = 2 \ M\Omega = 2 \times 10^6 \ \Omega$:
$V = I_g(R_s + G)$
$5 = 10^{-4}(2 \times 10^6 + G)$
$5 \times 10^4 = 2 \times 10^6 + G$
$G = 50000 - 2000000 = -1950000 \ \Omega$.
Since the resistance $G$ cannot be negative,the problem statement contains inconsistent values. However,assuming the intended series resistance was $R_s = 48 \ k\Omega$ (which would yield $G = 2000 \ \Omega$):
For an ammeter of range $I = 10 \ mA = 10^{-2} \ A$ with shunt resistance $S$:
$S = \frac{I_g G}{I - I_g} = \frac{10^{-4} \times 2000}{10^{-2} - 10^{-4}} = \frac{0.2}{0.0099} \approx 20.2 \ \Omega$.
Given the provided parameters lead to a physical impossibility,the correct choice is $D$.

(D) The magnetic field $B$ due to a straight wire of length $L$ at a perpendicular distance $r$ is given by $B = \frac{\mu_0 i}{4\pi r} (\sin \theta_1 + \sin \theta_2)$.
For an equilateral triangle of side $a = 1\,m$,the distance $r$ from the center to the side is $r = \frac{a}{2\tan(60^\circ)} = \frac{a}{2\sqrt{3}}$.
For each side,$\theta_1 = \theta_2 = 60^\circ$,so $\sin \theta_1 + \sin \theta_2 = 2 \sin(60^\circ) = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$.
The magnetic field due to one side is $B_1 = \frac{\mu_0 i}{4\pi (a/2\sqrt{3})} \times \sqrt{3} = \frac{\mu_0 i \sqrt{3}}{2\pi a} \times \sqrt{3} = \frac{3\mu_0 i}{2\pi a}$.
Since there are $3$ sides,the total magnetic field at the center is $B = 3 \times B_1 = 3 \times \frac{3\mu_0 i}{2\pi a} = \frac{9\mu_0 i}{2\pi a}$.
Substituting the values: $B = \frac{9 \times (4\pi \times 10^{-7}) \times 10}{2\pi \times 1} = 18 \times 10^{-6}\,T = 18\,\mu T$.

(D) Let $N_0$ be the initial number of nuclei for both substances $A$ and $B$.
The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.
For substance $A$: $N_A(t) = N_0 e^{-5\lambda t}$.
For substance $B$: $N_B(t) = N_0 e^{-\lambda t}$.
The ratio of the number of nuclei is $\frac{N_A(t)}{N_B(t)} = \frac{N_0 e^{-5\lambda t}}{N_0 e^{-\lambda t}} = e^{-4\lambda t}$.
We are given that this ratio is $(1/e)^2 = e^{-2}$.
So,$e^{-4\lambda t} = e^{-2}$.
Equating the exponents: $-4\lambda t = -2$.
Solving for $t$: $t = \frac{-2}{-4\lambda} = \frac{1}{2\lambda}$.

(A) The number of spectral lines emitted when an electron transitions from an excited state $n$ to the ground state $(n=1)$ is given by the formula: $N = \frac{n(n-1)}{2}$.
Given $N = 6$,we have $\frac{n(n-1)}{2} = 6$,which implies $n^2 - n - 12 = 0$. Solving this quadratic equation,we get $(n-4)(n+3) = 0$. Since $n > 0$,the electron is excited to the $n = 4$ level.
The energy of the photon absorbed to excite the electron from $n=1$ to $n=4$ in a hydrogen-like ion with atomic number $Z$ is given by: $\Delta E = 13.6 \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) eV$.
For $Li^{++}$,$Z = 3$. Thus,$\Delta E = 13.6 \times 3^2 \left( \frac{1}{1^2} - \frac{1}{4^2} \right) = 13.6 \times 9 \times \left( 1 - \frac{1}{16} \right) = 13.6 \times 9 \times \frac{15}{16} \approx 114.75 \, eV$.
The wavelength $\lambda$ is given by $\lambda = \frac{hc}{\Delta E}$. Using $hc \approx 1240 \, eV \cdot nm$,we get $\lambda = \frac{1240}{114.75} \approx 10.8 \, nm$.

(D) For a thin lens,the magnification $m$ is given by $m = \frac{v}{f} - 1$,where $v$ is the image distance and $f$ is the focal length.
This is a linear equation of the form $y = mx + c$,where the slope is $\frac{1}{f}$.
From the graph,the slope is the change in magnification divided by the change in image distance:
Slope $= \frac{\Delta m}{\Delta v} = \frac{c}{b}$.
Since the slope is also equal to $\frac{1}{f}$,we have $\frac{1}{f} = \frac{c}{b}$.
Therefore,the focal length is $f = \frac{b}{c}$.

(C) The energy flux $I$ is given as $25\,W/cm^2.$ The total power $P$ incident on the surface is $P = I \times A = 25\,W/cm^2 \times 25\,cm^2 = 625\,W.$
For a completely absorbing surface,the radiation pressure exerts a force $F = \frac{P}{c},$ where $c$ is the speed of light $(3 \times 10^8\,m/s)$.
$F = \frac{625}{3 \times 10^8}\,N.$
The momentum $p$ transferred to the surface in time $t$ is given by $p = F \times t.$
Given time $t = 40\,min = 40 \times 60\,s = 2400\,s.$
$p = \left( \frac{625}{3 \times 10^8} \right) \times 2400 = \frac{625 \times 2400}{3 \times 10^8} = \frac{1500000}{3 \times 10^8} = 500000 \times 10^{-8} = 5.0 \times 10^{-3}\,Ns.$

(D) Consider a spherical shell of radius $x$ and thickness $dx$ between the two spheres.
The area of this shell is $A = 4\pi x^2$.
The resistance $dR$ of this thin shell is given by $dR = \rho \frac{dx}{A} = \rho \frac{dx}{4\pi x^2}$.
To find the total resistance $R$ between the spheres,we integrate from $r = a$ to $r = b$:
$R = \int_a^b \frac{\rho}{4\pi x^2} dx$.
$R = \frac{\rho}{4\pi} \int_a^b x^{-2} dx$.
$R = \frac{\rho}{4\pi} \left[ -\frac{1}{x} \right]_a^b$.
$R = \frac{\rho}{4\pi} \left( -\frac{1}{b} - (-\frac{1}{a}) \right) = \frac{\rho}{4\pi} \left( \frac{1}{a} - \frac{1}{b} \right)$.

(D) Given: Breakdown voltage of Zener diode $V_Z = 6\,V,$ load resistance $R_L = 4\,k\Omega,$ series resistance $R_i = 1\,k\Omega.$
The current through the load resistance is constant because the Zener diode maintains a constant voltage $V_Z$ across it:
$I_L = \frac{V_Z}{R_L} = \frac{6\,V}{4\,k\Omega} = 1.5\,mA.$
The total current $I_i$ supplied by the battery through the series resistor $R_i$ is given by $I_i = \frac{V_B - V_Z}{R_i}.$
For minimum battery voltage $V_B = 8\,V$:
$I_{i,min} = \frac{8\,V - 6\,V}{1\,k\Omega} = \frac{2\,V}{1\,k\Omega} = 2\,mA.$
The Zener current $I_Z = I_i - I_L = 2\,mA - 1.5\,mA = 0.5\,mA.$
For maximum battery voltage $V_B = 16\,V$:
$I_{i,max} = \frac{16\,V - 6\,V}{1\,k\Omega} = \frac{10\,V}{1\,k\Omega} = 10\,mA.$
The Zener current $I_Z = I_i - I_L = 10\,mA - 1.5\,mA = 8.5\,mA.$
Thus,the minimum and maximum currents through the Zener diode are $0.5\,mA$ and $8.5\,mA$ respectively.

(D) The bob of the pendulum is subjected to two perpendicular forces: the gravitational force $mg$ acting downwards and the electric force $qE$ acting horizontally.
The effective acceleration $g_{eff}$ experienced by the bob is the vector sum of the gravitational acceleration $g$ and the electric acceleration $a_e = \frac{qE}{m}$.
Since these two accelerations are perpendicular to each other,the magnitude of the effective acceleration is given by:
$g_{eff} = \sqrt{g^2 + a_e^2} = \sqrt{g^2 + \left(\frac{qE}{m}\right)^2}$
The time period $T$ of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$.
Substituting the value of $g_{eff}$,we get:
$T = 2\pi \sqrt{\frac{L}{\sqrt{g^2 + \left(\frac{qE}{m}\right)^2}}}$

(D) Given: $q_A = 1\,\mu C = 10^{-6}\,C$,$q_B = 1\,\mu C = 10^{-6}\,C$,$m_B = 4\,\mu g = 4 \times 10^{-9}\,kg$,$r_1 = 1\,mm = 10^{-3}\,m$,$r_2 = 9\,mm = 9 \times 10^{-3}\,m$.
By the law of conservation of energy,the loss in electrostatic potential energy equals the gain in kinetic energy:
$\Delta U = \Delta K$
$k q_A q_B \left( \frac{1}{r_1} - \frac{1}{r_2} \right) = \frac{1}{2} m_B v^2$
Substituting the values:
$9 \times 10^9 \times (10^{-6}) \times (10^{-6}) \left( \frac{1}{10^{-3}} - \frac{1}{9 \times 10^{-3}} \right) = \frac{1}{2} \times 4 \times 10^{-9} \times v^2$
$9 \times 10^{-3} \left( 1000 - \frac{1000}{9} \right) = 2 \times 10^{-9} \times v^2$
$9 \times 10^{-3} \times 1000 \left( 1 - \frac{1}{9} \right) = 2 \times 10^{-9} \times v^2$
$9 \times \frac{8}{9} = 2 \times 10^{-9} \times v^2$
$8 = 2 \times 10^{-9} \times v^2$
$v^2 = 4 \times 10^9$
$v = \sqrt{40 \times 10^8} = 2 \times 10^4 \times \sqrt{10} \approx 6.32 \times 10^4\,m/s$.
Since this value is not among the options,the correct choice is $D$.

(D) The power $P$ of the laser is given by $P = n \cdot E_{photon},$ where $n$ is the number of photons emitted per second and $E_{photon} = \frac{hc}{\lambda}.$
Given $P = 2\,mW = 2 \times 10^{-3}\,W,$ $\lambda = 500\,nm = 500 \times 10^{-9}\,m,$ $h = 6.6 \times 10^{-34}\,Js,$ and $c = 3.0 \times 10^8\,m/s.$
Rearranging for $n$: $n = \frac{P \cdot \lambda}{h \cdot c}$
Substituting the values:
$n = \frac{2 \times 10^{-3} \times 500 \times 10^{-9}}{6.6 \times 10^{-34} \times 3.0 \times 10^8}$
$n = \frac{1000 \times 10^{-12}}{19.8 \times 10^{-26}}$
$n = \frac{10^{-9}}{19.8 \times 10^{-26}} \approx 0.0505 \times 10^{17} = 5.05 \times 10^{15} \approx 5 \times 10^{15}.$
Thus,the number of photons emitted per second is $5 \times 10^{15}.$

(A) Given: Self-inductance $L = 10\,mH = 10 \times 10^{-3}\,H$,coil resistance $r_1 = 0.1\,\Omega$,internal resistance $r_2 = 0.9\,\Omega$.
The total resistance of the circuit is $R = r_1 + r_2 = 0.1 + 0.9 = 1.0\,\Omega$.
The time constant of the $LR$ circuit is $\tau = \frac{L}{R} = \frac{10 \times 10^{-3}}{1.0} = 10^{-2}\,s$.
The current at any time $t$ is given by $i = i_0(1 - e^{-t/\tau})$,where $i_0$ is the saturation current.
We want to find $t$ when $i = 0.8 i_0$.
$0.8 i_0 = i_0(1 - e^{-t/\tau})$
$0.8 = 1 - e^{-t/\tau}$
$e^{-t/\tau} = 0.2 = \frac{1}{5}$
$e^{t/\tau} = 5$
Taking natural logarithm on both sides: $\frac{t}{\tau} = \ln 5$.
Given $\ln 5 = 1.6$,so $t = 1.6 \times \tau = 1.6 \times 10^{-2}\,s = 0.016\,s$.

(A) The intensity of light $I$ is directly proportional to the slit width $w$,so $I_1/I_2 = w_1/w_2 = 4/1$.
Let $I_1 = 4I_0$ and $I_2 = I_0$.
The amplitudes are $a_1 = \sqrt{I_1} = 2\sqrt{I_0}$ and $a_2 = \sqrt{I_2} = \sqrt{I_0}$.
The ratio of maximum to minimum intensity is given by $\frac{I_{\max}}{I_{\min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2$.
Substituting the values: $\frac{I_{\max}}{I_{\min}} = \left( \frac{2\sqrt{I_0} + \sqrt{I_0}}{2\sqrt{I_0} - \sqrt{I_0}} \right)^2 = \left( \frac{3\sqrt{I_0}}{\sqrt{I_0}} \right)^2 = (3)^2 = 9/1$.

(C) For a square loop of side $l$,the perimeter is $P = 4l$ and the area is $A_s = l^2$. The magnetic dipole moment is $m = I A_s = I l^2$.
When reshaped into a circular loop of radius $r$,the perimeter remains the same: $2\pi r = 4l$,which gives $r = \frac{2l}{\pi}$.
The area of the circular loop is $A_c = \pi r^2 = \pi \left(\frac{2l}{\pi}\right)^2 = \frac{4l^2}{\pi}$.
The new magnetic dipole moment $m'$ is $m' = I A_c = I \left(\frac{4l^2}{\pi}\right)$.
Since $m = I l^2$,we substitute this into the expression for $m'$:
$m' = \frac{4}{\pi} (I l^2) = \frac{4m}{\pi}$.

(C) The resolving power of a microscope is defined by the minimum separation $d$ between two points that can be resolved.
The formula for the minimum separation $d$ is given by:
$d = \frac{0.61 \lambda}{\text{NA}}$
Given:
$\lambda = 5000\,\mathring{A} = 5000 \times 10^{-10}\,\text{m} = 5 \times 10^{-7}\,\text{m}$
$\text{NA} = 1.25$
Substituting the values:
$d = \frac{0.61 \times 5 \times 10^{-7}}{1.25}$
$d = \frac{3.05 \times 10^{-7}}{1.25}$
$d = 2.44 \times 10^{-7}\,\text{m}$
Converting to micrometers $(\mu m)$:
$d = 2.44 \times 10^{-7} \times 10^6\,\mu m = 0.244\,\mu m \approx 0.24\,\mu m$
Thus, the correct option is $C$.

(A) First,we calculate the equivalent resistance of the network.
$1$. The first parallel combination of two $4 \, R$ resistors is $R_{p1} = \frac{4R \times 4R}{4R + 4R} = 2R$.
$2$. The second parallel combination of $6 \, R$ and $12 \, R$ resistors is $R_{p2} = \frac{6R \times 12R}{6R + 12R} = \frac{72R^2}{18R} = 4R$.
$3$. The total equivalent resistance $R_{eq}$ is the sum of these in series with the other two $R$ resistors: $R_{eq} = 2R + R + 4R + R = 8R$.
$4$. The power consumed is given by $P = \frac{V^2}{R_{eq}}$.
$5$. Substituting the given values: $4 = \frac{16^2}{8R} \implies 4 = \frac{256}{8R} \implies 4 = \frac{32}{R}$.
$6$. Therefore,$R = \frac{32}{4} = 8 \, \Omega$.

(D) $1$. Analyze the resistance network: The network is a Wheatstone bridge with arms $1\, \Omega, 2\, \Omega, 1\, \Omega, 2\, \Omega$ and a central $3\, \Omega$ resistor. Since $\frac{1}{1} = \frac{2}{2}$,it is a balanced Wheatstone bridge. The central $3\, \Omega$ resistor carries no current. The equivalent resistance of the bridge is $R_{eq} = \frac{(1+2)(1+2)}{(1+2)+(1+2)} = \frac{3 \times 3}{3+3} = 1.5\, \Omega$.
$2$. Calculate the induced electromotive force $(EMF)$: The motional $EMF$ is given by $\varepsilon = B \ell v$. Here,$B = 1\, T$,$\ell = 5\, cm = 0.05\, m$,and $v = 1\, cm/s = 0.01\, m/s$. Thus,$\varepsilon = 1 \times 0.05 \times 0.01 = 5 \times 10^{-4}\, V$.
$3$. Calculate the total resistance: The total resistance $R_{total} = R_{loop} + R_{eq} = 1.7\, \Omega + 1.5\, \Omega = 3.2\, \Omega$.
$4$. Calculate the current: The current $i = \frac{\varepsilon}{R_{total}} = \frac{5 \times 10^{-4}}{3.2} \approx 1.5625 \times 10^{-4}\, A = 156.25\, \mu A$. The closest option is $150\, \mu A$ or $170\, \mu A$. Re-evaluating the bridge: $R_{eq} = \frac{3 \times 3}{3+3} = 1.5\, \Omega$. Total $R = 1.7 + 1.5 = 3.2\, \Omega$. $i = 156.25\, \mu A$. Given the options,$150\, \mu A$ is the closest value.

(B) For capacitor $(I)$,the dielectrics are in series. The capacitance of each part is $C_i = \frac{3 \varepsilon_0 A K_i}{d}$.
Since they are in series,$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{d}{3 \varepsilon_0 A} (\frac{1}{K_1} + \frac{1}{K_2} + \frac{1}{K_3}) = \frac{d(K_2 K_3 + K_3 K_1 + K_1 K_2)}{3 \varepsilon_0 A K_1 K_2 K_3}$.
Thus,$C_{eq} = \frac{3 \varepsilon_0 A K_1 K_2 K_3}{d(K_1 K_2 + K_2 K_3 + K_3 K_1)}$.
For capacitor $(II)$,the dielectrics are in parallel. The capacitance of each part is $C_i' = \frac{\varepsilon_0 (A/3) K_i}{d} = \frac{\varepsilon_0 A K_i}{3d}$.
Since they are in parallel,$C_{eq}' = C_1' + C_2' + C_3' = \frac{\varepsilon_0 A}{3d} (K_1 + K_2 + K_3)$.
The energy stored is $E = \frac{1}{2} C V^2$. Therefore,$\frac{E_1}{E_2} = \frac{C_{eq}}{C_{eq}'} = \frac{3 \varepsilon_0 A K_1 K_2 K_3}{d(K_1 K_2 + K_2 K_3 + K_3 K_1)} \cdot \frac{3d}{\varepsilon_0 A (K_1 + K_2 + K_3)} = \frac{9 K_1 K_2 K_3}{(K_1 + K_2 + K_3)(K_1 K_2 + K_2 K_3 + K_3 K_1)}$.

(B) According to Einstein's photoelectric equation:
$h\nu = \phi + eV_0$
Rearranging for the stopping potential $V_0$:
$V_0 = \frac{h\nu}{e} - \frac{\phi}{e}$
From the given graph,the threshold frequency $(\nu_0)$ is the frequency at which the stopping potential $V_0$ becomes zero. Looking at the graph,$V_0 = 0$ when $\nu = 4 \times 10^{14} \, Hz$.
At this point:
$0 = \frac{h\nu_0}{e} - \frac{\phi}{e}$
$\Rightarrow \phi = h\nu_0$
Substituting the values:
$\phi = (6.63 \times 10^{-34} \, Js) \times (4 \times 10^{14} \, Hz)$
$\phi = 26.52 \times 10^{-20} \, J$
To convert the work function into electron-volts $(eV)$:
$\phi (eV) = \frac{26.52 \times 10^{-20} \, J}{1.6 \times 10^{-19} \, C}$
$\phi = 1.6575 \, eV \approx 1.66 \, eV$

(D) The frequency of oscillation of a magnetic needle in a magnetic field $B$ is given by $f = \frac{1}{2\pi} \sqrt{\frac{\mu B_H}{I}}$,where $B_H = B \cos \delta$ is the horizontal component of the earth's magnetic field and $\delta$ is the angle of dip.
Given $f_1 = 30 \text{ oscillations/min}$ at $\delta_1 = 45^{\circ}$ and $f_2 = 40 \text{ oscillations/min}$ at $\delta_2 = 30^{\circ}$.
Thus,$f_1^2 \propto B_1 \cos 45^{\circ}$ and $f_2^2 \propto B_2 \cos 30^{\circ}$.
Taking the ratio: $\frac{f_1^2}{f_2^2} = \frac{B_1 \cos 45^{\circ}}{B_2 \cos 30^{\circ}}$.
Substituting the values: $(\frac{30}{40})^2 = \frac{B_1 (1/\sqrt{2})}{B_2 (\sqrt{3}/2)}$.
$\frac{9}{16} = \frac{B_1}{B_2} \times \frac{1}{\sqrt{2}} \times \frac{2}{\sqrt{3}} = \frac{B_1}{B_2} \times \sqrt{\frac{2}{3}}$.
$\frac{B_1}{B_2} = \frac{9}{16} \times \sqrt{\frac{3}{2}} = 0.5625 \times 1.2247 \approx 0.689 \approx 0.7$.

(C) The energy of a photon is given by $E = \frac{1240}{\lambda} \, eV$.
For the first transition, $\lambda_1 = 108.5 \, nm$, so $E_1 = \frac{1240}{108.5} \approx 11.43 \, eV$.
For the second transition, $\lambda_2 = 30.4 \, nm$, so $E_2 = \frac{1240}{30.4} \approx 40.79 \, eV$.
The total energy released is $E_{total} = E_1 + E_2 = 11.43 + 40.79 = 52.22 \, eV$.
The energy of an electron in the $n$-th state of a hydrogen-like ion is $E_n = -13.6 \, Z^2 \frac{1}{n^2}$.
For $He^+$, $Z = 2$, so $E_n = -13.6 \times 4 \times \frac{1}{n^2} = -54.4 \frac{1}{n^2} \, eV$.
The transition is from state $n$ to ground state $(n=1)$, so $E_{total} = E_n - E_1 = -54.4 \left(\frac{1}{n^2} - 1\right) = 54.4 \left(1 - \frac{1}{n^2}\right)$.
Setting $54.4 \left(1 - \frac{1}{n^2}\right) = 52.22$, we get $1 - \frac{1}{n^2} = \frac{52.22}{54.4} \approx 0.96$.
$\frac{1}{n^2} = 1 - 0.96 = 0.04$.
$n^2 = \frac{1}{0.04} = 25$, so $n = 5$.

(A) Given: Input resistance $R_{\text{in}} = 100\,\Omega$,Output resistance $R_{\text{out}} = 100\,k\Omega = 10^5\,\Omega$.
From the graph,the current gain $\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{(20 - 5) \times 10^{-3} \text{ A}}{(400 - 100) \times 10^{-6} \text{ A}} = \frac{15 \times 10^{-3}}{300 \times 10^{-6}} = \frac{15000}{300} = 50$.
Voltage gain $A_V = \beta \times \frac{R_{\text{out}}}{R_{\text{in}}} = 50 \times \frac{10^5}{100} = 50 \times 10^3 = 5 \times 10^4$.
Power gain $A_P = \beta \times A_V = 50 \times (5 \times 10^4) = 250 \times 10^4 = 2.5 \times 10^6$.

(B) The total charge of the dipole is $0$,so the net charge induced on the inner surface is $0$. Since the shell is a conductor,the total charge $Q$ resides on the outer surface.
Because the dipole is at the centre,it creates a non-uniform electric field inside the cavity. This induces a non-uniform charge distribution on the inner surface of the shell.
However,for any point outside the shell $(r > b)$,the electric field produced by the induced charges on the inner surface and the dipole itself cancels out due to the principle of superposition and the properties of a conductor in electrostatic equilibrium.
Thus,the electric field outside the shell is solely due to the total charge $Q$ distributed uniformly on the outer surface,which is equivalent to a point charge $Q$ at the centre: $E = \frac{kQ}{r^2}$.

(D) The dimensional formula for permittivity of vacuum is $[\varepsilon_0] = M^{-1}L^{-3}T^4A^2$.
The dimensional formula for permeability of vacuum is $[\mu_0] = MLT^{-2}A^{-2}$.
The dimensional formula for electrical resistance is $[R] = ML^2T^{-3}A^{-2}$.
Now,consider the expression $\sqrt{\frac{\mu_0}{\varepsilon_0}}$:
$\sqrt{\frac{\mu_0}{\varepsilon_0}} = \sqrt{\frac{MLT^{-2}A^{-2}}{M^{-1}L^{-3}T^4A^2}} = \sqrt{M^2L^4T^{-6}A^{-4}} = ML^2T^{-3}A^{-2}$.
This matches the dimension of electrical resistance $[R]$.
Therefore,the correct option is $D$.

(D) The full-scale deflection current of the galvanometer is $I_g = 50 \times 20 \times 10^{-6} \, A = 10^{-3} \, A = 1 \, mA$.
The resistance of the galvanometer is $G = 100 \, \Omega$.
For a voltmeter of range $V$,the total resistance required is $R_{total} = V / I_g$.
The series resistance to be added is $R = R_{total} - G$.
For $0-2 \, V$ range: $R_{total} = 2 / 10^{-3} = 2000 \, \Omega$. So,$R_1 = 2000 - 100 = 1900 \, \Omega$.
For $0-10 \, V$ range: $R_{total} = 10 / 10^{-3} = 10000 \, \Omega$. The resistance added in series with $R_1$ is $R_2 = 10000 - 2000 = 8000 \, \Omega$.
For $0-20 \, V$ range: $R_{total} = 20 / 10^{-3} = 20000 \, \Omega$. The resistance added in series with $R_1 + R_2$ is $R_3 = 20000 - 10000 = 10000 \, \Omega$.
Thus,the circuit must have $R_1 = 1900 \, \Omega$,$R_2 = 8000 \, \Omega$,and $R_3 = 10000 \, \Omega$ in series.

(B) When a thin film of thickness $t$ and refractive index $\mu$ is introduced in front of one of the slits in a Young's double slit experiment,the path difference introduced is $\Delta x = (\mu - 1)t$.
The shift in the fringe pattern is given by $\Delta y = \frac{D}{d} \Delta x = \frac{D}{d} (\mu - 1)t$.
The fringe width is $\beta = \frac{\lambda D}{d}$.
Given that the shift is equal to one fringe width,we have $\Delta y = \beta$.
Therefore,$\frac{D}{d} (\mu - 1)t = \frac{\lambda D}{d}$.
Simplifying this,we get $(\mu - 1)t = \lambda$.
Thus,$t = \frac{\lambda}{\mu - 1}$.

(D) The potential $V$ at any point on the equatorial plane (the $y$-axis in this case) of a dipole is zero because the potential due to the positive and negative charges of the dipole cancels out at every point on this plane.
$V = 0$
The electric field $\vec{E}$ at a point on the equatorial plane at a distance $d$ from the dipole is given by the formula:
$\vec{E} = -\frac{1}{4\pi\varepsilon_0} \frac{\vec{p}}{d^3}$
Substituting the given dipole moment $\vec{p} = -p_0\hat{x}$,the electric field is directed along the positive $x$-axis,which is consistent with the direction of $-\vec{p}$.
Thus,the potential is $0$ and the electric field is $-\frac{\vec{p}}{4\pi\varepsilon_0 d^3}$.

(A) The general form of an electromagnetic wave is $\vec{E} = E_0 \hat{n} \sin(\omega t + \vec{k} \cdot \vec{r})$.
Comparing this with the given equation $\vec{E} = E_0 \hat{n} \sin[\omega t + (6y - 8z)]$,we have the wave vector $\vec{k} = 6\hat{j} - 8\hat{k}$.
The direction of propagation $\hat{s}$ is given by the unit vector in the direction of $-\vec{k}$.
Thus,$\hat{s} = -\frac{\vec{k}}{|\vec{k}|} = -\frac{6\hat{j} - 8\hat{k}}{\sqrt{6^2 + (-8)^2}} = -\frac{6\hat{j} - 8\hat{k}}{\sqrt{36 + 64}} = -\frac{6\hat{j} - 8\hat{k}}{10} = \frac{-6\hat{j} + 8\hat{k}}{10} = \frac{-3\hat{j} + 4\hat{k}}{5}$.

(D) The magnetic field $B$ at the centre of a rotating charged ring is given by $B = \frac{\mu_0 i}{2R}$.
Here,the equivalent current $i$ is $i = \frac{q}{T} = \frac{q \omega}{2 \pi}$.
Substituting $i$ into the magnetic field formula: $B = \frac{\mu_0 q \omega}{2R(2 \pi)} = \frac{\mu_0 q \omega}{4 \pi R}$.
Given: $R = 0.1 \, m$,$\omega = 40 \pi \, rad/s$,$B = 3.8 \times 10^{-9} \, T$,and $\mu_0 = 4 \pi \times 10^{-7} \, N/A^2$.
Rearranging for $q$: $q = \frac{B \cdot 4 \pi R}{\mu_0 \omega}$.
$q = \frac{3.8 \times 10^{-9} \times 4 \pi \times 0.1}{4 \pi \times 10^{-7} \times 40 \pi}$.
$q = \frac{3.8 \times 10^{-10}}{40 \pi \times 10^{-7}} = \frac{3.8 \times 10^{-3}}{40 \pi} \approx \frac{3.8 \times 10^{-3}}{125.66} \approx 3.02 \times 10^{-5} \, C$.
Thus,the charge is approximately $3 \times 10^{-5} \, C$.

(D) The given circuit consists of an $OR$ gate and a $NAND$ gate. Let the output of the $OR$ gate be $C$. Then $C = A + B$.
The inputs to the $NAND$ gate are $A$ and $C$. The output $Y$ of the $NAND$ gate is given by $Y = \overline{A \cdot C}$.
Substituting $C = A + B$ into the expression for $Y$, we get $Y = \overline{A \cdot (A + B)} = \overline{A \cdot A + A \cdot B} = \overline{A + A \cdot B} = \overline{A(1 + B)} = \overline{A}$.
Let us verify this with a truth table:

$A$	$B$	$C = A + B$	$A \cdot C$	$Y = \overline{A \cdot C}$
$0$	$0$	$0$	$0$	$1$
$0$	$1$	$1$	$0$	$1$
$1$	$0$	$1$	$1$	$0$
$1$	$1$	$1$	$1$	$0$

Comparing this with the given options, the truth table matches option $D$.

(B) Light incident from particle $P$ will be reflected at the mirror.
For the mirror,the object distance $u = -5\, cm$ and the focal length $f = -R/2 = -20\, cm$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} + \frac{1}{-5} = \frac{1}{-20}$
$\frac{1}{v} = \frac{1}{5} - \frac{1}{20} = \frac{4-1}{20} = \frac{3}{20}$
$v = +\frac{20}{3}\, cm$.
This image acts as a virtual object for the light getting refracted at the water surface.
The distance of this virtual object from the water surface is $d_{obj} = 5 + \frac{20}{3} = \frac{35}{3}\, cm$.
Since the light travels from water $(\mu_1 = 4/3)$ to air $(\mu_2 = 1)$,the apparent depth $d'$ is given by $d' = d_{obj} \times (\mu_2 / \mu_1)$.
$d' = \left( \frac{35}{3} \right) \times \left( \frac{1}{4/3} \right) = \frac{35}{3} \times \frac{3}{4} = \frac{35}{4} = 8.75\, cm$.
Thus,$d \approx 8.8\, cm$.

(C) The terminal potential difference $V$ across a battery is given by $V = E - Ir$,where $E$ is the $emf$ and $r$ is the internal resistance.
From the graph,when the current $I = 0$,the voltage $V = E = 1.5 \, V$.
When the current $I = 1000 \, mA = 1 \, A$,the voltage $V = V_0 \approx 0 \, V$.
Substituting these values into the equation $V = E - Ir$:
$0 = 1.5 - (1 \, A) \times r$
$r = 1.5 \, \Omega$.
Thus,the $emf$ of the battery is $1.5 \, V$ and its internal resistance is $1.5 \, \Omega$.

(B) The Zener diode acts as a voltage regulator, maintaining a constant voltage of $V_Z = 6\,V$ across the load resistance $R_L = 4\,k\Omega$ when it is in the breakdown region.
The load current $I_L$ is constant and is given by:
$I_L = \frac{V_Z}{R_L} = \frac{6\,V}{4\,k\Omega} = 1.5\,mA$
The total source current $I_S$ is given by:
$I_S = \frac{V_{in} - V_Z}{R_S}$
The Zener current $I_Z$ is given by:
$I_Z = I_S - I_L = \frac{V_{in} - 6}{2\,k\Omega} - 1.5\,mA$
To maximize $I_Z$, we must use the maximum input voltage $V_{in} = 16\,V$:
$I_{S,max} = \frac{16\,V - 6\,V}{2\,k\Omega} = \frac{10\,V}{2\,k\Omega} = 5\,mA$
Therefore, the maximum Zener current is:
$I_{Z,max} = I_{S,max} - I_L = 5\,mA - 1.5\,mA = 3.5\,mA$

(B) First,simplify the circuit. The $1\, \mu F$ and $5\, \mu F$ capacitors are in parallel,so their equivalent capacitance is $C_p = 1\, \mu F + 5\, \mu F = 6\, \mu F$.
This $6\, \mu F$ capacitor is in series with the $4\, \mu F$ capacitor. Their equivalent capacitance is $C_s = \frac{4 \times 6}{4 + 6} = 2.4\, \mu F$.
This $2.4\, \mu F$ branch is in parallel with the $3\, \mu F$ capacitor across the $10\, V$ battery.
The charge on the $2.4\, \mu F$ branch is $q = C_s \times V = 2.4\, \mu F \times 10\, V = 24\, \mu C$.
Since the $4\, \mu F$ and $6\, \mu F$ (equivalent of $1\, \mu F$ and $5\, \mu F$) capacitors are in series,the charge on each of them is the same as the total charge on the branch,which is $24\, \mu C$.

(B) For an electron in a hydrogen atom,the condition for a stable orbit according to Bohr's postulate is given by the relation $2 \pi r_n = n \lambda_n$,where $n$ is the principal quantum number,$r_n$ is the radius of the $n^{th}$ orbit,and $\lambda_n$ is the de-Broglie wavelength.
The ground state corresponds to $n=1$,the first excited state to $n=2$,and the second excited state to $n=3$.
Given that the electron is in the second excited state,we have $n=3$ and the radius $r_3 = 4.65 \, \mathring{A}$.
Substituting these values into the formula:
$3 \lambda_3 = 2 \pi r_3$
$\lambda_3 = \frac{2 \pi (4.65 \, \mathring{A})}{3}$
$\lambda_3 = 2 \times 3.14159 \times 1.55 \, \mathring{A}$
$\lambda_3 \approx 9.738 \, \mathring{A}$.
Rounding to one decimal place,we get $\lambda_3 = 9.7 \, \mathring{A}$.