HINTS & SOLUTIONS

Answer: (B) $-\frac{2q^{2}}{3\pi\epsilon_{0}}$ Solution: The potential energy of a system of three point charges is given by $U = \frac{1}{4\pi\epsilon_0} \left[ \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_1 q_3}{r_{13}} \right]$. $U = \frac{1}{4\pi\epsilon_0} \left[ \frac{(+q)(-2q)}{|0 - (-1)|} + \frac{(-2q)(+q)}{|2 - 0|} + \frac{(+q)(+q)}{|2 - (-1)|} \right]$ $U = \frac{1}{4\pi\epsilon_0} \left[ \frac{-2q^2}{1} + \frac{-2q^2}{2} + \frac{q^2}{3} \right] = \frac{q^2}{4\pi\epsilon_0} \left[ -2 - 1 + \frac{1}{3} \right] = \frac{q^2}{4\pi\epsilon_0} \left[ -\frac{8}{3} \right] = -\frac{2q^2}{3\pi\epsilon_0}$.

Answer: (C) $V_{1}>V_{2}>V_{3}$ Explanation: Energy of the incident radiation is inversely proportional to its wavelength ($E = hc/\lambda$). Since $\lambda_1 < \lambda_2 < \lambda_3$ (UV has shorter wavelength than visible), the energy order is $E_1 > E_2 > E_3$. Higher energy incident photons result in a higher stopping potential required to halt the emitted photoelectrons. Thus, $V_1 > V_2 > V_3$.

Answer: (A) $\frac{1}{n^{2}}$ Explanation: In the Bohr model of a hydrogen atom, the total energy and potential energy of an electron in the $n^{th}$ orbit are inversely proportional to the square of the principal quantum number. Potential energy $U_n = \frac{-27.2}{n^2}$ eV, so it varies as $1/n^2$.

Answer: (D) The final potential of the system equals $\frac{1}{4\pi\epsilon_{0}}\frac{(q_{1}+q_{2})(r_{1}+r_{2})}{r_{1}r_{2}}$ Explanation: When two conducting spheres are brought into contact, they form a single equipotential surface, and total charge is conserved. Their common potential is $V = \frac{Q_{total}}{C_{total}} = \frac{q_1 + q_2}{4\pi\epsilon_0 (r_1 + r_2)}$. Therefore, statements A, B, and C are correct, making statement D incorrect.

Answer: (C) $\frac{v}{5}$ Solution: The frequency of the emitted photon is proportional to the energy difference: $h\nu = \Delta E = 13.6 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$. For $n=4 \to n=1$: $h\nu = 13.6 \left(\frac{1}{1^2} - \frac{1}{4^2}\right) = 13.6 \left(\frac{15}{16}\right)$. For $n=4 \to n=2$: $h\nu' = 13.6 \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = 13.6 \left(\frac{1}{4} - \frac{1}{16}\right) = 13.6 \left(\frac{3}{16}\right)$. Taking the ratio: $\frac{\nu'}{\nu} = \frac{3/16}{15/16} = \frac{3}{15} = \frac{1}{5} \implies \nu' = \frac{\nu}{5}$.

Answer: (D) Acceleration of the magnet remains uniform as it moves through the loop. Explanation: Because the loop has a small cut, the electrical circuit is incomplete (open). Consequently, no macroscopic induced current can flow around the loop, meaning no opposing magnetic field is generated as per Lenz's Law. The magnet will fall freely with uniform acceleration due to gravity ($a = g$).

Answer: (C) III Solution: The velocity of an electromagnetic wave is given by $v = \frac{\omega}{k}$. In free space, it should equal $c = 3 \times 10^8$ m/s. Let's check the options: I. $v = \frac{1.4 \times 10^{11}}{0.7 \times 10^3} = 2 \times 10^8$ m/s II. $v = \frac{1.5 \times 10^{11}}{0.6 \times 10^3} = 2.5 \times 10^8$ m/s III. $v = \frac{1.5 \times 10^{11}}{0.5 \times 10^3} = 3 \times 10^8$ m/s (Matches $c$) IV. $v = \frac{4.8 \times 10^{11}}{0.2 \times 10^4} = 2.4 \times 10^8$ m/s

Answer: (D) Nickel Explanation: A relative magnetic permeability $\mu_r \gg 1$ implies the substance is ferromagnetic. Among the given options, Aluminium is paramagnetic, Copper and Lead are diamagnetic, and Nickel is ferromagnetic.

Answer: (C) z-axis Solution: Using the Biot-Savart Law vector form, $d\vec{B} \propto (d\vec{l} \times \vec{r})$. The current element is along the +x axis: $d\vec{l} = dl \hat{i}$. The position vector points to (0, 40cm, 0) on the +y axis: $\vec{r} = y \hat{j}$. The cross product is $\hat{i} \times \hat{j} = \hat{k}$ (positive z-axis).

Answer: (C) 1 mA Solution: Let full-scale deflection current be $I_g$. Galvanometer resistance $G = 27 \text{ }\Omega$. Shunt resistance $S = 3 \text{ }\Omega$. Total current $I = 10 \text{ mA}$. $I_g G = (I - I_g) S \implies I_g(27) = (10 - I_g)(3) \implies 9I_g = 10 - I_g \implies 10I_g = 10 \implies I_g = 1 \text{ mA}$.

Answer: (C) $Wbs^{-2}$, $Wbs^{-1}$ Explanation: Based on the principle of dimensional homogeneity, each term in the equation $\phi = 5At^2 + Bt - 2C$ must have the same unit as magnetic flux (Weber, Wb). For $At^2$: Unit of A $\times$ $s^2$ = Wb $\implies$ Unit of A is Wb s$^{-2}$. For $Bt$: Unit of B $\times$ s = Wb $\implies$ Unit of B is Wb s$^{-1}$.

Answer: Dependent on graphical slope (Explanation provided) Solution: The capacitive reactance is $X_C = \frac{1}{\omega C} = \left(\frac{1}{C}\right) \frac{1}{\omega}$. When plotting $X_C$ versus $\frac{1}{\omega}$, the graph is a straight line passing through the origin with slope $m = \frac{1}{C}$. Therefore, the capacitance is inversely proportional to the slope of the line ($C \propto \frac{1}{m}$). The ratio $C_1/C_2$ equals the inverse ratio of their slopes $m_2/m_1$. (Select the appropriate ratio value from the diagram's angles provided in the actual exam paper, e.g., if angles are 30° and 60°, ratio is $\tan(60^\circ)/\tan(30^\circ) = 3$ or $1/3$ depending on line labels).

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). Explanation: In interference, when a crest of one wave superimposes on the trough of another (destructive interference), the resultant amplitude and intensity drop to zero, producing darkness.

Answer: (C) Assertion (A) is true, but Reason (R) is false. Explanation: The equivalent resistance of heaters in series is $R_s = R_1 + R_2$. The total power consumed is $P_s = \frac{V^2}{R_s}$, which is less than $P_1 = \frac{V^2}{R_1}$. So the assertion is true. However, the reason is false because for devices connected directly across a constant voltage dc source, power $P = V^2/R$, meaning power is inversely (not directly) proportional to resistance.

Answer: (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). Explanation: Increasing intensity means more photons strike the surface per second, emitting more electrons, thus increasing photocurrent (Assertion is true). Stopping potential depends only on the frequency of light (Reason is true). However, the reason does not explain the assertion, which relies on the particle nature/photon flux density.

Answer: (D) Assertion (A) is false and Reason (R) is also false. Explanation: In forward bias, the positive terminal is connected to the p-side and negative to the n-side. The applied external electric field directly opposes the built-in potential barrier electric field, thereby decreasing the height of the potential barrier and allowing current to flow. Both statements are factually incorrect.

Solution: Concave mirror focal length $f = -20 \text{ cm}$. Pencil length $L = 5 \text{ cm}$. Nearest end $u_1 = -25 \text{ cm}$. Farthest end $u_2 = -(25+5) = -30 \text{ cm}$. Using mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$: For $u_1$: $\frac{1}{v_1} + \frac{1}{-25} = \frac{1}{-20} \implies \frac{1}{v_1} = \frac{1}{25} - \frac{1}{20} = \frac{4-5}{100} = -\frac{1}{100} \implies v_1 = -100 \text{ cm}$. For $u_2$: $\frac{1}{v_2} + \frac{1}{-30} = \frac{1}{-20} \implies \frac{1}{v_2} = \frac{1}{30} - \frac{1}{20} = \frac{2-3}{60} = -\frac{1}{60} \implies v_2 = -60 \text{ cm}$. Length of image $= |v_1| - |v_2| = |-100| - |-60| = 100 - 60 = 40 \text{ cm}$. Solution (Alternative): Given: $\lambda_1 = 500 \text{ nm}$, $\lambda_2 = 600 \text{ nm}$, $D = 1.8 \text{ m}$, $d = 0.3 \text{ mm} = 0.3 \times 10^{-3} \text{ m}$. Let the $n^{th}$ bright fringe of $\lambda_1$ coincide with the $m^{th}$ bright fringe of $\lambda_2$. Condition for coincidence: $y = n\frac{\lambda_1 D}{d} = m\frac{\lambda_2 D}{d} \implies n\lambda_1 = m\lambda_2$. $\frac{n}{m} = \frac{\lambda_2}{\lambda_1} = \frac{600}{500} = \frac{6}{5}$. For the least distance, choose the minimum integer values: $n = 6, m = 5$. Least distance $y = 6 \times \frac{500 \times 10^{-9} \times 1.8}{0.3 \times 10^{-3}} = 6 \times \left( \frac{900 \times 10^{-9}}{0.3 \times 10^{-3}} \right) = 6 \times 3000 \times 10^{-6} = 18000 \times 10^{-6} \text{ m} = 18 \text{ mm}$ (or 1.8 cm).

Solution: Given: $T_1 = 27^\circ\text{C}$, $\alpha = 2.0 \times 10^{-4} \text{ }^\circ\text{C}^{-1}$. New resistance $R_2$ is 20% more than $R_1$: $R_2 = R_1 + 0.20 R_1 = 1.2 R_1$. Using formula: $R_2 = R_1 [1 + \alpha(T_2 - T_1)]$. $1.2 R_1 = R_1 [1 + 2.0 \times 10^{-4} (T_2 - 27)]$. $1.2 = 1 + 2.0 \times 10^{-4} (T_2 - 27) \implies 0.2 = 2.0 \times 10^{-4} (T_2 - 27)$. $T_2 - 27 = \frac{0.2}{2.0 \times 10^{-4}} = \frac{0.2 \times 10^4}{2} = 1000$. $T_2 = 1000 + 27 = 1027^\circ\text{C}$.

Solution: Given: Angle of prism $A = 60^\circ$, refractive index $\mu = \sqrt{2}$. Ray just suffers TIR at second face, so angle of emergence $e = 90^\circ$ and angle of refraction at second face $r_2 = i_c$ (critical angle). $\sin i_c = \frac{1}{\mu} = \frac{1}{\sqrt{2}} \implies i_c = 45^\circ$. Thus, $r_2 = 45^\circ$. For a prism, $r_1 + r_2 = A \implies r_1 + 45^\circ = 60^\circ \implies r_1 = 15^\circ$. Applying Snell's law at the first face: $\frac{\sin i}{\sin r_1} = \mu \implies \sin i = \sqrt{2} \sin(15^\circ)$. $i = \sin^{-1}(\sqrt{2} \sin 15^\circ)$. (Using $\sin 15^\circ = \frac{\sqrt{3}-1}{2\sqrt{2}}$, $i = \sin^{-1}(\frac{\sqrt{3}-1}{2}) \approx 21.47^\circ$).

Solution: 1. Purity: Intrinsic semiconductors are in their purest form without any doping. Extrinsic semiconductors are impure, formed by adding controlled amounts of trivalent or pentavalent impurities (doping).
2. Charge Carriers & Conductivity: In intrinsic semiconductors, the number density of electrons equals that of holes ($n_e = n_h$), resulting in low electrical conductivity. In extrinsic semiconductors, one type of carrier heavily outnumbers the other ($n_e \neq n_h$), resulting in significantly higher electrical conductivity.

Solution: De Broglie wavelength $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}$. Let mass of proton $= m_p$ and charge $= e$. For alpha particle, mass $m_\alpha \approx 4m_p$ and charge $q_\alpha = 2e$.
(i) Same Kinetic Energy (K): $\frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p}{m_\alpha}} = \sqrt{\frac{m_p}{4m_p}} = \frac{1}{\sqrt{4}} = \frac{1}{2} = 1:2$.
(ii) Same Potential Difference (V): $\frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p q_p}{m_\alpha q_\alpha}} = \sqrt{\frac{m_p \cdot e}{4m_p \cdot 2e}} = \sqrt{\frac{1}{8}} = \frac{1}{2\sqrt{2}} = 1:2\sqrt{2}$.

Solution: (i) Mutual Inductance: It is defined as the magnetic flux linked with the secondary coil when a unit current flows through the primary coil ($\Phi_2 = M I_1$). SI Unit is Henry (H).
(ii) Derivation: Let current $I$ flow in the outer solenoid. Magnetic field inside it is uniform: $B = \mu_0 n I$. The magnetic flux linked with the inner circular loop of area $A = \pi r^2$ is $\Phi = B \times A = (\mu_0 n I)(\pi r^2)$. Since $\Phi = MI$, the mutual inductance is $M = \frac{\Phi}{I} = \mu_0 n \pi r^2$. Solution (Alternative): Force between Parallel Conductors: Conductor A carries $I_a$, producing a magnetic field at distance $d$ (where B lies): $B_a = \frac{\mu_0 I_a}{2\pi d}$. Conductor B carries $I_b$ and is placed in $B_a$. By Lorentz force, force on length L of B is $F_b = I_b L B_a \sin(90^\circ) = \frac{\mu_0 I_a I_b L}{2\pi d}$. By Fleming's Left Hand Rule, this force is directed towards A (attractive). Similarly, field due to B at A is $B_b = \frac{\mu_0 I_b}{2\pi d}$. Force on A is $F_a = I_a L B_b = \frac{\mu_0 I_a I_b L}{2\pi d}$, directed towards B. Since $\vec{F}_a = -\vec{F}_b$, it perfectly obeys Newton's third law of motion.

Solution: Ground state energy $E_1 = -13.6 \text{ eV}$. Energy of electron in $n^{th}$ state is $E_n = \frac{-13.6}{n^2} \text{ eV}$. Second excited state implies $n = 3$. Energy $E_3 = \frac{-13.6}{3^2} = -1.51 \text{ eV}$. Kinetic energy $K = -E_n = +1.51 \text{ eV} = 1.51 \times 1.6 \times 10^{-19} \text{ J}$. De Broglie wavelength $\lambda = \frac{h}{\sqrt{2m_e K}}$. $\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9 \times 10^{-31} \times 1.51 \times 1.6 \times 10^{-19}}} = \frac{6.6 \times 10^{-34}}{\sqrt{43.48 \times 10^{-50}}} = \frac{6.6 \times 10^{-34}}{6.59 \times 10^{-25}} \approx 1.0 \times 10^{-9} \text{ m} = 10 \text{ \AA}$.

Solution: (i) Diagnostic tool in medicine: X-Rays. Wavelength range: $10^{-8} \text{ m}$ to $10^{-13} \text{ m}$.
(ii) Remote switches for TV sets: Infrared Waves. Wavelength range: $10^{-3} \text{ m}$ to $7 \times 10^{-7} \text{ m}$.
(iii) Water purifiers: Ultraviolet (UV) Rays. Wavelength range: $4 \times 10^{-7} \text{ m}$ to $10^{-8} \text{ m}$.

Solution: (i) Diagram: For two equal and opposite point charges (an electric dipole), electric field lines originate from the positive charge and terminate on the negative charge in curved paths. The equipotential surfaces are distinct spheres around each charge (bulging away from the center) and a flat vertical plane perfectly bisecting the line joining the two charges.
(ii) Perpendicularity: If the electric field $\vec{E}$ were not perpendicular to the equipotential surface, it would have a non-zero component along the surface. Consequently, work would have to be done to move a test charge along the surface. However, by definition, the potential difference between any two points on an equipotential surface is zero, meaning no work is done ($W = q\Delta V = 0$). Hence, $\vec{E}$ must be strictly perpendicular to the surface.

Solution: (a) Isotopes (same Z, different A): ${}_{6}^{12}\text{C}$ and ${}_{6}^{14}\text{C}$.
Isotones (same Number of Neutrons $N = A - Z$): ${}_{80}^{198}\text{Hg}$ ($198 - 80 = 118$) and ${}_{79}^{197}\text{Au}$ ($197 - 79 = 118$).

(b) Nuclear Density: Experiments show the volume of a nucleus is directly proportional to its mass number $A$. Therefore, radius $R = R_0 A^{1/3}$ (where $R_0 \approx 1.2 \text{ fm}$). Let $m$ be the average mass of a nucleon. Total mass of nucleus $\approx m A$. Volume $V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A$. Density $\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{mA}{\frac{4}{3}\pi R_0^3 A} = \frac{3m}{4\pi R_0^3}$. Since $m$ and $R_0$ are constants, the term $A$ has cancelled out, proving that nuclear density is a universal constant for all nuclei.

Solution: Consider a circular coil of radius $r$ carrying current $I$. We need the field at point P on the axis at distance $x$. Consider a current element $Id\vec{l}$ at the top of the loop. Distance to P is $s = \sqrt{r^2 + x^2}$. By Biot-Savart law, $d\vec{B} = \frac{\mu_0}{4\pi} \frac{Id\vec{l} \sin 90^\circ}{s^2}$. Resolve $d\vec{B}$ into two components: $dB \cos\phi$ perpendicular to the axis, and $dB \sin\phi$ along the axis. Due to diametrically opposite current elements, all perpendicular components cancel out. Only axial components add up. Net field $B = \int dB \sin\phi$. From geometry, $\sin\phi = \frac{r}{s} = \frac{r}{\sqrt{r^2 + x^2}}$. $B = \int \frac{\mu_0 I dl}{4\pi (r^2 + x^2)} \frac{r}{\sqrt{r^2 + x^2}} = \frac{\mu_0 I r}{4\pi (r^2 + x^2)^{3/2}} \int dl$. For a coil of $N$ turns, total length $\int dl = N(2\pi r)$. $B = \frac{\mu_0 I r}{4\pi (r^2 + x^2)^{3/2}} \times N(2\pi r) = \frac{\mu_0 N I r^2}{2(r^2 + x^2)^{3/2}}$.

Solution: (a) The direction of an arrow over a wire merely indicates the geometric path the flow of charge is taking in the macroscopic circuit. Current does not obey the mathematical laws of vector addition (like the triangle or parallelogram law). For instance, when two currents meet at a junction, they add algebraically ($I_1 + I_2$), regardless of the physical angle between the wires. Thus, it is a scalar quantity.
(b) Circuit Calculation (Methodology): Apply Kirchhoff's Junction Rule (KCL) at the nodes to define loop currents $I_1$ and $I_2$. Apply Kirchhoff's Voltage Rule (KVL) to two independent closed loops, setting the algebraic sum of potential drops across resistors and batteries to zero. Solve the two simultaneous linear equations to find the specific current passing through the 3 $\Omega$ resistor branch.

Solution: (i) (A) $\sqrt{D(D-4f)}$ Reasoning: Let $u$ be object distance and $v$ be image distance. $u + v = D$ and $v - u = d$. Solving gives $u = (D-d)/2$ and $v = (D+d)/2$. Using lens formula $f = \frac{uv}{u+v} = \frac{D^2 - d^2}{4D}$. Rearranging for $d$: $d^2 = D^2 - 4Df \implies d = \sqrt{D(D-4f)}$.

(ii) (C) enlarged, reduced (Assuming standard order of $u$ increasing) OR (D) enlarged, reduced Reasoning: In the first position (closer to object), $u$ is small and $v$ is large. Magnification $m = v/u > 1$ (enlarged). In the second position (closer to screen), $u$ is large and $v$ is small. $m = v/u < 1$ (reduced).

(iii) (B) 18.75 cm Reasoning: Using the formula $f = \frac{D^2 - d^2}{4D}$. Here $D = 80$ cm, $d = 20$ cm. $f = \frac{80^2 - 20^2}{4 \times 80} = \frac{6400 - 400}{320} = \frac{6000}{320} = 18.75 \text{ cm}$.

(iv) (D) 65 cm Reasoning: For two sharp image positions to exist, the discriminant must be real and positive: $D^2 - 4Df > 0 \implies D > 4f$. Here $f = 15$ cm, so $4f = 60$ cm. The only option where $D > 60$ cm is 65 cm. (iv) (A) -15 cm Reasoning: Equivalent power $P = P_1 + P_2$. $f_1 = 10 \text{ cm} = 0.1 \text{ m} \implies P_1 = \frac{1}{0.1} = +10 \text{ D}$. Given $P = 10/3 \text{ D}$. $P_2 = P - P_1 = \frac{10}{3} - 10 = -\frac{20}{3} \text{ D}$. $f = \frac{1}{P_2} = -\frac{3}{20} \text{ m} = -15 \text{ cm}$.

Solution: (i) (C) Boron Reasoning: To obtain a p-type semiconductor, intrinsic silicon must be doped with a trivalent impurity (Group 13), creating a hole. Boron is a trivalent impurity.

(ii) (D) $4.5\times10^{9}m^{-3}$ Reasoning: By the mass-action law for semiconductors in thermal equilibrium, $n_e n_h = n_i^2$. $n_h = \frac{n_i^2}{n_e} = \frac{(1.5 \times 10^{16})^2}{5 \times 10^{22}} = \frac{2.25 \times 10^{32}}{5 \times 10^{22}} = 0.45 \times 10^{10} = 4.5 \times 10^9 \text{ m}^{-3}$.

(iii) Current is mainly due to diffusion of majority carriers. (Note: Select corresponding option from original paper if misprinted in text snippet) Reasoning: Forward biasing lowers the built-in potential barrier, making it easier for majority carriers to overcome the junction. This massively increases the diffusion current across the junction, which dominates the overall forward current.

(iv) (C) 0.7 V Reasoning: The typical knee voltage (or threshold/cut-in voltage) required to overcome the barrier potential in a Silicon p-n junction diode is approximately 0.7 V. (iv) (A) 0.01 eV Reasoning: The fifth valence electron of a pentavalent donor atom in Germanium is very weakly bound to its parent nucleus. The thermal ionization energy required to free it into the conduction band is approximately $\approx 0.01 \text{ eV}$ for Ge (and $\approx 0.05 \text{ eV}$ for Si).

(a) Solution: LCR Circuit Impedance: Let $I = I_0 \sin(\omega t)$. Voltage across Resistor $V_R = I R$ (in phase with $I$). Voltage across Inductor $V_L = I X_L$ (leads $I$ by $90^\circ$). Voltage across Capacitor $V_C = I X_C$ (lags $I$ by $90^\circ$). In the phasor diagram, $V_L$ and $V_C$ are anti-parallel. Assuming $V_L > V_C$, the net reactive voltage is $(V_L - V_C)$, leading $V_R$ by $90^\circ$. Applying Pythagoras theorem: $V^2 = V_R^2 + (V_L - V_C)^2$. $V = \sqrt{(IR)^2 + (I X_L - I X_C)^2} = I \sqrt{R^2 + (X_L - X_C)^2}$. Impedance $Z = \frac{V}{I} = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$. Phase angle $\phi$: $\tan\phi = \frac{V_L - V_C}{V_R} = \frac{X_L - X_C}{R}$.

(b) Solution: (i) Impedance is minimum: At resonance, when $X_L = X_C \implies \omega L = \frac{1}{\omega C}$. Here $Z = R$. (ii) Wattless current flows: When average power consumed is zero ($P = VI\cos\phi = 0$). This happens when the phase difference $\phi = 90^\circ$, which occurs in purely inductive or purely capacitive circuits ($R = 0$). (Alternative) Solution: (i) A.C. Generator: Principle: Electromagnetic Induction. When a coil is continuously rotated in a uniform magnetic field, the magnetic flux linked with it changes, inducing an alternating emf. Construction & Working: Contains an armature coil rotating between strong magnet poles. The ends connect to slip rings and carbon brushes. As the coil rotates, the angle $\theta$ between the area vector and B-field changes continuously, inducing a sinusoidal current in the external circuit via the brushes.
(ii) EMF Expression: Flux $\Phi = N A B \cos(\omega t)$. By Faraday's law, $e = -\frac{d\Phi}{dt} = -\frac{d}{dt} [NAB \cos(\omega t)] = NAB\omega \sin(\omega t)$.
(iii) Maximum EMF: $e$ is maximum when $\sin(\omega t) = \pm 1$. This occurs when $\omega t = \pi/2, 3\pi/2$. Since $\omega = 2\pi/T$, substituting yields: $(\frac{2\pi}{T})t = \frac{\pi}{2} \implies t = T/4$. And similarly $t = 3T/4$.

(a) Solution: Coherent Sources: Two sources of light are coherent if they emit light waves of the same frequency and maintain a constant phase difference over time. They are strictly necessary for stable interference because if phase difference fluctuates rapidly, the maxima and minima will rapidly shift, resulting in average uniform illumination and no observable pattern.
Graph: A plot of Intensity $I$ vs path difference $\Delta x$ (or angle). It shows equal-height, equally-spaced peaks at $0, \pm\lambda, \pm2\lambda$, dipping to exactly $0$ intensity halfway between peaks.

(b) Solution: Resultant intensity $I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi = 2I_0 + 2I_0\cos\phi = 2I_0(1+\cos\phi) = 4I_0\cos^2(\phi/2)$. Relation between path diff $\Delta x$ and phase diff $\phi = \frac{2\pi}{\lambda}\Delta x$. (i) $\Delta x = \frac{\lambda}{4} \implies \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$. $I = 2I_0 (1 + \cos 90^\circ) = 2I_0 (1 + 0) = 2I_0$. (ii) $\Delta x = \frac{\lambda}{3} \implies \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3} (120^\circ)$. $I = 2I_0 (1 + \cos 120^\circ) = 2I_0 (1 - 0.5) = I_0$. (Alternative) Solution: (a) Refracting Telescope: Ray Diagram: Parallel rays from a distant object enter the objective lens (large $f_o$) and form a real, inverted image at the focal plane $f_o$. The eyepiece (small $f_e$) is adjusted so this image falls exactly on its focal plane, sending parallel rays to the eye (final image at infinity). Derivation: Magnifying power $M = \beta/\alpha$. Since angles are small, $\alpha \approx \frac{h'}{f_o}$ and $\beta \approx \frac{h'}{-f_e}$ (where $h'$ is image height at focal plane). $M = \frac{h' / -f_e}{h' / f_o} = -\frac{f_o}{f_e}$.
(b) (i) Large Aperture: A large objective gathers more light to make faint distant objects visible, and crucially increases the resolving power of the telescope.
(ii) Advantages of Reflecting Telescope: (1) No chromatic aberration because mirrors reflect all colors equally. (2) Easier mechanical support (large heavy mirrors can be supported across their entire back, unlike lenses which must be supported only at thin edges).

(a) Solution: Capacitance of Parallel Plate Capacitor: Consider two parallel plates of area $A$ separated by distance $d$. Charge densities are $+\sigma$ and $-\sigma$, where $\sigma = Q/A$. The electric field between the plates is uniform and given by $E = \frac{\sigma}{\epsilon_0} = \frac{Q}{A\epsilon_0}$. The potential difference $V$ across the plates is work done in moving a unit charge: $V = E \times d$. Substitute $E$: $V = \frac{Qd}{A\epsilon_0}$. By definition, capacitance $C = \frac{Q}{V} = \frac{Q}{\frac{Qd}{A\epsilon_0}} = \frac{\epsilon_0 A}{d}$.

(b) Solution: Because the DC battery remains connected, the potential difference ($V$) across both capacitors remains constant. Initial capacitance $C$; New capacitance $C' = KC$. (i) Charge: $Q' = C'V = (KC)V = K(CV) = K Q$. The charge on each capacitor increases by a factor of K. (ii) Energy Stored: $U' = \frac{1}{2} C' V^2 = \frac{1}{2} (KC) V^2 = K (\frac{1}{2} C V^2) = K U$. The energy stored increases by a factor of K. (Alternative) Solution: (a) Drift Velocity ($v_d$): When E-field is established, free electrons experience force $\vec{F} = -e\vec{E}$, resulting in acceleration $\vec{a} = -e\vec{E}/m$. Because electrons frequently collide with fixed lattice ions, they lose this accelerated velocity and re-accelerate. The average steady velocity acquired between collisions is the drift velocity. $v_d = a \tau = \frac{eE\tau}{m}$ (independent of time due to continuous resetting by collisions). Relation to current: In time $dt$, electrons move distance $v_d dt$. Volume swept is $A v_d dt$. Total electrons in this volume is $N = n A v_d dt$. Charge crossing area $dQ = e N = n e A v_d dt$. Current $I = \frac{dQ}{dt} = n e A v_d$.

(b) (i) Instant Current: While physical electrons drift very slowly, the electric field (the signal) propagates through the entire wire at near the speed of light ($3\times10^8$ m/s). This field instantly sets all free electrons everywhere in the circuit into motion simultaneously, establishing current instantly.
(ii) Ratio of Drift Velocities: In series, the current $I$ is identical in both wires. $I = n e A v_d \implies v_d = \frac{I}{neA} \implies v_d \propto \frac{1}{A} \propto \frac{1}{r^2}$. Therefore, $\frac{v_{d1}}{v_{d2}} = \left(\frac{r_2}{r_1}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$.