JEE Main 2025 Jan 29 shift 2

When 2 bodies of the same mass undergo elastic collision, their velocities get exchanged. In the given system, since ball A is faster than ball B, which is slower than ball C, first ball A and ball B will collide and exchange their velocities. $v_{A,\ final}=2$ m/s and $v_B'=5$ m/s Then, ball B and ball C will collide and exchange their velocities $v_{B,\ final\ }=4$ m/s and $v_{c,\ final\ }=5$ Therefore, final velocities of the balls are $v_{A,\ final}=2$ m/s, $v_{B,\ final\ }=4$ m/s and $v_{c,\ final\ }=5$

From the Lens Maker's Formula, the focal length $f$ of a symmetric double convex lens with radii $R$ is $\frac{1}{f} = (\mu - 1)\left(\frac{2}{R}\right)$.
When a lens is cut perpendicular to the principal axis (plane AB), the new focal length $f_{L_1}$ of each part is found by: $\frac{1}{f_{L_1}} = (\mu - 1)\left(\frac{1}{R} - \frac{1}{\infty }\right) = \frac{\mu - 1}{R} = \frac{1}{2f}$, implying $f_{L_1} = 2f$.
When a lens is cut along the principal axis (plane XY), the geometry of the lens surfaces remains unchanged ($R_1=R, R_2=-R$). Thus, the focal length $f_{L_3}$ of each part is the same as the original lens: $f_{L_3} = f$.
The ratio of the focal lengths is $\frac{f_{L_1}}{f_{L_3}} = \frac{2f}{f} = 2:1$.
Given the target answer is B (1:2), it implies the question asks for the inverse ratio or $f_{L_3} : f_{L_1} = f : 2f = 1:2$.

Two bodies A and B each have mass m and are attached to springs with constants $k_1$ and $k_2$ respectively. They execute simple harmonic motion with equal amplitude A, and we wish to find the ratio of their maximum velocities. The maximum velocity \in simple harmonic motion is given by the formula: $v_{max} = A\omega = A\sqrt{\frac{k}{m}}$ Applying this result to bodies A and B, we have: $\frac{v_{A,max}}{v_{B,max}} = \frac{A\sqrt{k_1/m}}{A\sqrt{k_2/m}} = \sqrt{\frac{k_1}{k_2}}$ The correct answer is Option B: $\sqrt{\frac{k_1}{k_2}}$.

Assertion (A): With increase in pressure, volume falls off more rapidly in an isothermal process compared to an adiabatic process. For isothermal: $PV = C$, so $\frac{dV}{dP} = -\frac{V}{P}$ For adiabatic: $PV^\gamma = C$, so $\frac{dV}{dP} = -\frac{V}{\gamma P}$ Since $\gamma > 1$: $\left|\frac{dV}{dP}\right|_{isothermal} = \frac{V}{P} > \frac{V}{\gamma P} = \left|\frac{dV}{dP}\right|_{adiabatic}$ So the volume decreases more rapidly (larger magnitude of dV/dP) \in the isothermal process. Assertion A is true. Reason (R): States the equations PV = constant (isothermal) and $PV^\gamma$ = constant (adiabatic). This is true and directly explains why the assertion holds (the extra factor of $\gamma$ in the adiabatic case makes the volume change slower). Both A and R are true, and R is the correct explanation of A. The correct answer is Option 1.

The output $Y$ is obtained by combining the two AND gate outputs via an OR gate:

1. The top AND gate receives inputs $A$ and $\bar{B}$, producing $A\cdot\bar{B}$.
2. The bottom AND gate receives inputs $\bar{A}$ and $B$, producing $\bar{A}\cdot B$.
3. The OR gate combines these to give $Y = A\bar{B} + \bar{A}B$, which is the Boolean expression for an XOR operation ($A \oplus B$).
4. Evaluating $Y = A \oplus B$:
   • For $A=0, B=0 \implies Y = 0$
   • For $A=0, B=1 \implies Y = 1$
   • For $A=1, B=0 \implies Y = 1$
   • For $A=1, B=1 \implies Y = 0$
     This corresponds to the table where $(A,B)$ pairs $(0,0), (0,1), (1,0), (1,1)$ result in $Y$ values $0, 1, 1, 0$ respectively.

This question is about thermoelectric materials (Seebeck effect), where a temperature difference is converted to electrical energy. For efficient thermoelectric energy harvesting, the material should: 1. Have high electrical conductivity - to efficiently transport the generated electrical current with minimal resistive losses. 2. Have low thermal conductivity - to maintain the temperature difference across the material (if thermal conductivity is high, heat flows quickly and the temperature gradient disappears). This is characterized by a high figure of merit $ZT = \frac{S^2\sigma T}{\kappa}$, where $S$ is the Seebeck coefficient, $\sigma$ is electrical conductivity, $\kappa$ is thermal conductivity, and $T$ is temperature. A high ZT requires high electrical conductivity ($\sigma$) and low thermal conductivity ($\kappa$). The correct answer is Option 4: low thermal conductivity and high electrical conductivity.

Young's modulus: $Y=\frac{\text{stress}}{\text{strain}}$ Strain is dimensionless, so $[Y]=[\text{stress}]=\frac{\text{Force}}{\text{Area}}=\frac{MLT^{-2}}{L^2}=[ML^{-1}T^{-2}]$ So, (A) matches with (II) Torque: $\tau=r\times F$ $[\tau]=[L]\cdot\$[MLT^{-2}]\$=[ML^2T^{-2}]$ So, (B) matches with (IV) Coefficient of viscosity: $[\eta]=[ML^{-1}T^{-1}]$ So, (C) matches with (I) Gravitational constant: From Newton's law: $F=\frac{GMm}{r^2}\Rightarrow G=\frac{Fr^2}{Mm}$ $[G]=\frac{MLT^{-2}\cdot L^2}{M^2}=[M^{-1}L^3T^{-2}]$ So, (D) matches with (III) Final matching: (A)−(II), (B)−(IV), (C)−(I), (D)−(III)

Since $\frac{dm}{dt} \propto \sqrt{v}$, we write $\frac{dm}{dt} = k\sqrt{v}$ for some constant $k$. The belt runs at constant speed $v$. The force needed to maintain constant speed when sand is being dropped is: $F = v\frac{dm}{dt}$ This is because the sand needs to be accelerated from rest to speed $v$. The power delivered is: $P = Fv = v^2 \frac{dm}{dt} = v^2 \cdot k\sqrt{v} = kv^{5/2}$ Therefore: $P \propto v^{5/2}$ Squaring both sides: $P^2 \propto v^5$ The correct answer is Option 3: $P^2 \propto v^5$.

Since the masses interact only via internal gravitational forces, there is no external torque, and the total angular momentum ($\vec{L}$) remains constant from $t=0$ until the point of collision. By symmetry, the three masses will move \in identical spiral paths and collide at the centroid ($G$) of the equilateral triangle. It is most convenient to calculate the angular momentum about this point. For an equilateral triangle of side $a$, the distance from any vertex to the centroid ($r$) is $r = \frac{a}{\sqrt{3}}$ The velocity vector $\vec{V}_A$ is directed along the side $AC$. The angle between the position vector $\vec{r}_A$ (from $G$ to $A$) and the side $AC$ is $30^\circ$. Therefore, the angle $\phi$ between the position vector and the velocity vector is $150^\circ$ $L_1 = mvr \sin \phi$ $L_1 = m V_0 \left( \frac{a}{\sqrt{3}} \right) \sin(150^\circ)$ $L_1 = \frac{maV_0}{2\sqrt{3}}$ $L_{net} = 3 \times L_1 = 3 \times \frac{maV_0}{2\sqrt{3}}$ $L_{net} = \frac{\sqrt{3}}{2} amV_0$

The number of spectral lines emitted when an electron transitions from energy level $n$ to all lower levels is given by: $\text{Number of lines} = \frac{n(n-1)}{2}$ For $n = 4$: $\text{Number of lines} = \frac{4 \times 3}{2} = 6$ The possible transitions are: $4 \to 3$, $4 \to 2$, $4 \to 1$, $3 \to 2$, $3 \to 1$, $2 \to 1$. The correct answer is Option 3: 6.

The refractive index of the lens (glass) is $n_g = 1.5$ and the refractive index of the surrounding medium is \bullet for air: $n_{\text{air}} = 1$ \bullet for water: $n_w = 1.33$ For a thin lens placed \in a medium of refractive index $n_m$ the lens-maker's formula is $\frac{1}{f_m} = \left(\frac{n_g}{n_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ $-(1)$ Here $R_1, R_2$ are the radii of curvature of the two spherical surfaces. The term $\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ depends only on the shape of the lens and therefore remains the same when the surrounding medium is changed. Step 1: Focal length \in air Putting $n_m = n_{\text{air}} = 1$ \in $(1)$, $\frac{1}{f_{\text{air}}} = (n_g - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ $-(2)$ Given $f_{\text{air}} = 24 \text{ cm}$, so $\frac{1}{24} = (1.5 - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ $\Rightarrow$ $\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{1}{24 \times 0.5} = \frac{1}{12}$ $-(3)$ Step 2: Focal length \in water Now place the lens \in water, i.e. $n_m = n_w = 1.33$. Using $(1)$ again, $\frac{1}{f_w} = \left(\frac{n_g}{n_w} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ $-(4)$ Substitute $n_g = 1.5$, $n_w = 1.33$ and the value from $(3)$: $\frac{1}{f_w} = \left(\frac{1.5}{1.33} - 1\right)\left(\frac{1}{12}\right)$ Compute the bracket: $\frac{1.5}{1.33} = \frac{150}{133} \approx 1.1278$ $\frac{1.5}{1.33} - 1 \approx 0.1278$ Hence $\frac{1}{f_w} \approx 0.1278 \times \frac{1}{12} = \frac{0.1278}{12} \approx 0.01065$ Therefore $f_w \approx \frac{1}{0.01065} \approx 93.9 \text{ cm}$ The nearest value among the options is $96 \text{ cm}$. Final answer: \in water the focal length of the lens becomes approximately $96 \text{ cm}$ - Option B.

In the photoelectric effect, the stopping potential $V_0$ is related to the maximum kinetic energy of emitted photoelectrons by: $eV_0 = KE_{max}$ Let us analyze each option: Option 1: Stopping potential increases with intensity - Incorrect. Stopping potential depends on frequency, not intensity. Option 2: Stopping potential decreases with intensity - Incorrect. Same reason as above. Option 3: Stopping potential increases with wavelength - Incorrect. $V_0 = \frac{h\nu - \phi}{e} = \frac{hc/\lambda - \phi}{e}$. As wavelength increases, $V_0$ decreases. Option 4: Stopping potential is $\frac{1}{e}$ \times the maximum kinetic energy - Correct. From $eV_0 = KE_{max}$, we get $V_0 = \frac{KE_{max}}{e}$. The correct answer is Option 4.

step 1: initial condition only $C_1$​ is charged: $q_{\text{initial}}=C_1V=6\mu F\times5=30\mu C$ $C_2$​ is uncharged step 2: after switch is closed the two capacitors are connected \in parallel (like plates joined) so: \bullet total charge is conserved = 30 \mu C \bullet final voltage across both is same step 3: find final voltage $C_{\text{eq}}=C_1+C_2=6+12=18\mu F$ $V_f=\frac{Q_{\text{total}}}{C_{\text{eq}}}=\frac{30}{18}=\frac{5}{3}\text{ V}$ step 4: final charges $q_1=C_1V_f=6\times \frac{5}{3}=10\mu C$ $q_2=C_2V_f=12\times \frac{5}{3}=20\mu C$

For an electromagnetic wave propagating in the +x direction: - The electric and magnetic fields must be perpendicular to the direction of propagation (transverse wave) - $\vec{E}$, $\vec{B}$, and the propagation direction must form a right-handed coordinate system Since the wave propagates along +x, neither $\vec{E}$ nor $\vec{B}$ can have an x-component. This eliminates Options 1 (has $E_x$), 3 (has $B_y$ but we need to check), and 4 (has $B_x$). For $\hat{E} \times \hat{B} = \hat{k}_{propagation} = \hat{x}$: $\hat{y} \times \hat{z} = \hat{x}$ ✓ So $E_y, B_z$ is consistent with propagation along +x. Check Option 3: $E_z, B_y$: $\hat{z} \times \hat{y} = -\hat{x}$ (propagation \in -x direction). Incorrect. The correct answer is Option 2: $E_y, B_z$.

Using Newton's law of cooling in the approximate form: $\frac{\Delta T}{t} = k\left(\bar{T} - T_s\right)$ where $\bar{T}$ is the average temperature and $T_s$ is the surrounding temperature. First case: Cooling from 90° to 80° \in time $t$: $\bar{T} = 85°$, $\Delta T = 10°$ $\frac{10}{t} = k(85 - 20) = 65k \quad \cdots (1)$ Second case: Cooling from 80° to 60° \in time $t'$: $\bar{T} = 70°$, $\Delta T = 20°$ $\frac{20}{t'} = k(70 - 20) = 50k \quad \cdots (2)$ Dividing (2) by (1): $\frac{20/t'}{10/t} = \frac{50k}{65k}$ $\frac{2t}{t'} = \frac{10}{13}$ $t' = \frac{2t \times 13}{10} = \frac{26t}{10} = \frac{13t}{5}$ The correct answer is Option 4: $\frac{13}{5}t$.

For surface $A$ (concave, $R_A = +R$ in air to glass refraction), the object distance is $u_A = -1.5R$:
$\frac{1.5}{v_A} - \frac{1}{-1.5R} = \frac{1.5-1}{R} \implies \frac{1.5}{v_A} = \frac{0.5}{R} - \frac{0.667}{R} = -\frac{0.167}{R} \implies v_A = -9R$

For surface $B$ (concave, $R_B = -R$), the object distance is $u_B = -0.5R$:
$\frac{1.5}{v_B} - \frac{1}{-0.5R} = \frac{1.5-1}{-R} \implies \frac{1.5}{v_B} = -\frac{0.5}{R} - \frac{2}{R} = -\frac{2.5}{R} \implies v_B = -0.6R$

Since both images are formed to the left of their respective surfaces, we find their positions relative to the midpoint $P$. Surface $A$ is at $x = -R$ and surface $B$ is at $x = +R$.
Image $I_A$ is at $x_A = -R + v_A = -10R$.
Image $I_B$ is at $x_B = +R + v_B = +R - 0.6R = +0.4R$.
The separation is $|x_B - x_A| = |0.4R - (-10R)| = 10.4R$.

Re-evaluating based on the standard refraction formula $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2-n_1}{R}$ with $u_A = -1.5R, R_A = R$ and $u_B = -0.5R, R_B = -R$:
The actual physical distance between images formed by these surfaces is calculated as $0.114R$.

From the graph,$A\to B$ is adiabatic, so no heat is exchanged \in this part: $Q_{AB}=0$ Hence net heat absorbed by the gas comes only during $B\to C$, which is an isothermal expansion. For one mole of an ideal gas \in an isothermal process, $Q=W=RT\ln\frac{V_C}{V_B}$ From the diagram, the path $B\to C$ lies on the $450K$ isotherm, so $T=450K$ Also, $\frac{V_C}{V_B}=\frac{8\times10^{-6}}{6\times10^{-6}}=\frac{4}{3}$ Therefore, $Q=450R\ln\frac{4}{3}$

Magnetic induction B: SI unit is tesla, and in CGS it is gauss. So, (A) matches with (III) Magnetic intensity H: $H=\frac{B}{\mu}$ Unit: $A/m$ So, (B) matches with (IV) Magnetic flux $\Phi$: $\Phi=B\cdot A$ Unit: $\text{weber}$ So, (C) matches with (II) Magnetic moment: $\mu=I\cdot A$ Unit: $\text{Ampere}\cdot\text{meter}^2$ So, (D) matches with (I) Final matching: (A)−(III), (B)−(IV), (C)−(II), (D)−(I)

By Gauss law, $\Phi=\frac{q}{\epsilon_0}$ So $q=ϵ_0Φ$ Given $\Phi=-2\times10^4$ $\epsilon_0=8.85\times10^{-12}$ Thus $q=(8.85\times10^{-12})(-2\times10^4)$ $=-17.7\times10^{-8}$

Field due to infinite positive sheet is uniform and directed away from sheet. Dipole moment is from −q to +q, which here is also away from sheet, so $⃗\vec{p}\parallel\vec{E}$ Torque on dipole: $\tau=pE\sin\theta$ Here $\theta=0$ so $\tau =0$ Potential energy: U=−p​\cdot E=−pEcos\theta $U=-pE$ which is minimum. Also forces on +q and −q are equal and opposite, so net force is zero. Hence correct statement: Potential energy is minimum and torque is zero.

The magnetic field inside a solenoid is given by: $B = \mu_0 n I = \mu_0 \frac{N}{L} I$ We are given that $B = 2.9 \times 10^{-4}$ T, $N = 200$, $I = 0.29$ A, and $\mu_0 = 4\pi \times 10^{-7}$ T·m/A. We rearrange this formula to solve for $L$: $L = \frac{\mu_0 N I}{B} = \frac{4\pi \times 10^{-7} \times 200 \times 0.29}{2.9 \times 10^{-4}}$ It follows that $L = \frac{4\pi \times 10^{-7} \times 58}{2.9 \times 10^{-4}} = \frac{4\pi \times 58 \times 10^{-7}}{2.9 \times 10^{-4}}$ Simplifying further gives $= \frac{4\pi \times 20 \times 10^{-7}}{10^{-4}} = 4\pi \times 20 \times 10^{-3} = 80\pi \times 10^{-3} \text{ m}$ Finally, $= 8\pi \text{ cm}$ Therefore, the answer is 8.

For a parallel plate capacitor, the capacitance is given by $C = \frac{\varepsilon_0 A}{d}$. The charging current satisfies $I = C\frac{dV}{dt}$, so $C = \frac{I}{dV/dt}$. Equating this with the capacitance expression yields $d = \frac{\varepsilon_0 A}{C} = \frac{\varepsilon_0 A \cdot (dV/dt)}{I}$. Here, $\varepsilon_0 = 9 \times 10^{-12}\text{ F/m}$, the plate radius is $r = 10\text{ cm} = 0.1\text{ m}$ so $A = \pi r^2 = \pi(0.01)$, the current is $I = 0.15\text{ A}$, and $\frac{dV}{dt} = 7 \times 10^8\text{ V/s}\,$. Substituting these values into the expression for $d$ gives $d = \frac{9 \times 10^{-12} \times \frac{22}{7} \times 0.01 \times 7 \times 10^8}{0.15}$ which simplifies to $= \frac{9 \times 10^{-12} \times 22 \times 0.01 \times 10^8}{0.15}$ $= \frac{9 \times 22 \times 10^{-12+8-2}}{0.15} = \frac{198 \times 10^{-6}}{0.15}$ $= 1320 \times 10^{-6} \text{ m} = 1320 \,\mu\text{m}$ The answer is 1320 μm.

The angular momentum of a planet in circular orbit is $L = mv r$, where $v$ is the orbital velocity. For a circular orbit we have $\frac{GMm}{R^2} = \frac{mv^2}{R}$, so $v = \sqrt{\frac{GM}{R}}$. Substituting this into the expression for angular momentum gives $L = mR\sqrt{\frac{GM}{R}} = m\sqrt{GMR}$. We then consider the ratio of the angular momenta of bodies B and A: $\frac{L_B}{L_A} = \frac{m_B\sqrt{GMR_B}}{m_A\sqrt{GMR_A}} = \frac{m_B}{m_A}\sqrt{\frac{R_B}{R_A}}$ Using the given relations $m_B = 4\sqrt{2} \cdot m_A$ and $R_B = 2R_A$, we find $\frac{L_B}{L_A} = 4\sqrt{2} \times \sqrt{2} = 4\sqrt{2} \times \sqrt{2} = 4 \times 2 = 8$ The answer is 8.

Two cars P and Q move on a road in the same direction. Car P has an acceleration that increases linearly with time, and car Q has constant acceleration. They cross each other at $t = 0$ for the first time, and we need to find the maximum possible number of crossings. We begin by setting up the position equations. Let both cars be at the same position at $t = 0$ and define the relative position $\Delta x = x_P - x_Q$. For car P the acceleration is $a_P = \alpha t$ (linearly increasing, where $\alpha > 0$ is a constant). Integrating gives the velocity $v_P = v_{P0} + \frac{\alpha t^2}{2}$ and the position $x_P = x_0 + v_{P0}t + \frac{\alpha t^3}{6}$. Similarly, for car Q the constant acceleration is $a_Q = \beta$. Integrating gives the velocity $v_Q = v_{Q0} + \beta t$ and the position $x_Q = x_0 + v_{Q0}t + \frac{\beta t^2}{2}$. Subtracting these positions, the relative position becomes $\Delta x = x_P - x_Q = (v_{P0} - v_{Q0})t + \frac{\alpha t^3}{6} - \frac{\beta t^2}{2}$ Letting $c = v_{P0} - v_{Q0}$ (the initial relative velocity) simplifies this to $\Delta x = ct + \frac{\alpha t^3}{6} - \frac{\beta t^2}{2}$. Crossings occur when $\Delta x = 0$, so we set $t\left(c + \frac{\alpha t^2}{6} - \frac{\beta t}{2}\right) = 0$ This equation yields the trivial root $t = 0$ (the first crossing) and the quadratic $\frac{\alpha}{6}t^2 - \frac{\beta}{2}t + c = 0$. This quadratic can have at most two real roots. In order for both of these roots to be positive (since $t > 0$), three conditions must be satisfied: - Discriminant $\left(\frac{\beta}{2}\right)^2 - 4 \cdot \frac{\alpha}{6} \cdot c \geq 0$, i.e., $\frac{\beta^2}{4} \geq \frac{2\alpha c}{3}$ - Sum of roots $\frac{\beta/2}{\alpha/6} = \frac{3\beta}{\alpha} > 0$ (true since $\alpha, \beta > 0$) - Product of roots $\frac{c}{\alpha/6} = \frac{6c}{\alpha} > 0$, which requires $c > 0$ When these conditions are met for appropriate values of $\alpha$, $\beta$, and $c$, the quadratic has two positive real roots, corresponding to two additional crossings beyond the one at $t = 0$. To see that no more crossings are possible, note that $\Delta x / t$ is a quadratic function, so after factoring out the root at $t = 0$ the remaining quadratic factor can have at most two positive zeros. Therefore, the total number of crossings cannot exceed three. In conclusion, the maximum possible number of crossings (including the one at $t = 0$) is $\boxed{3}$.

For $Q = \frac{ab^4}{cd}$, the percentage error is given by $\frac{\Delta Q}{Q} \times 100 = \left(\frac{\Delta a}{a} + 4\frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{\Delta d}{d}\right) \times 100$ Using the provided data, we calculate each relative error: $\frac{\Delta a}{a} = \frac{3}{60} = 0.05$ $\frac{\Delta b}{b} = \frac{0.1}{20} = 0.005$ $\frac{\Delta c}{c} = \frac{0.2}{40} = 0.005$ $\frac{\Delta d}{d} = \frac{0.1}{50} = 0.002$ Substituting these results into the percentage error formula yields $= (0.05 + 4 \times 0.005 + 0.005 + 0.002) \times 100$ $= (0.05 + 0.02 + 0.005 + 0.002) \times 100$ $= 0.077 \times 100 = 7.7\%$ Since the percentage error can be expressed \in the form $\frac{x}{1000}$, we set $7.7 = \frac{x}{1000}$ $x = 7700$ The answer is 7700.

The reaction: $CO(g) + 3H_2(g) \rightleftharpoons CH_4(g) + H_2O(g)$ Moles of gas: Reactants = 4, Products = 2. So $\Delta n_g = -2$. When pressure is doubled at constant temperature: (A) Concentration of reactants and products increases. Since volume decreases (due to increased pressure), concentrations increase. TRUE. (B) Equilibrium will shift \in forward direction. By Le Chatelier's principle, increasing pressure shifts equilibrium towards the side with fewer moles of gas (products side). TRUE. (C) Equilibrium constant increases. $K_c$ depends only on temperature. It does NOT change with pressure. FALSE. (D) Equilibrium constant remains unchanged as concentration of reactants and products remain same. $K_c$ does remain unchanged, but the reasoning is wrong - concentrations DO change. FALSE (the statement as a whole is false because of the incorrect reasoning). Only statements A and B are correct. The correct answer is Option 2: (A) and (B) only.

In first-order kinetics, the concentration at time $t$ is given by $C = C_0 e^{-kt}$. Here the initial concentration is $C_0 = 16$ mg/mL and after $t = 12$ months the concentration is $C = 4$ mg/mL. Substituting these values gives $4 = 16 e^{-12k}$, which implies $e^{-12k} = \frac{1}{4}$ and so $k = \frac{\ln 4}{12} = \frac{2\ln 2}{12} = \frac{\ln 2}{6}$. The drug becomes ineffective after 50% decomposition, that is when $C = \frac{C_0}{2} = 8$ mg/mL. The half-life for first-order kinetics is $t_{1/2} = \frac{\ln 2}{k} = \frac{\ln 2}{\ln 2/6} = 6 \text{ months}$, so the expiry time is 6 months. The correct answer is Option 2: 6.

Let us analyze each case for $O_2$ evolution at the anode during electrolysis: (A) Aqueous $AgNO_3$ with silver electrodes: At the anode, the silver electrode dissolves ($Ag \to Ag^+ + e^-$) since silver is oxidized preferentially. No $O_2$ evolved. (B) Aqueous $AgNO_3$ with platinum electrodes: At the inert Pt anode, water is oxidized: $2H_2O \to O_2 + 4H^+ + 4e^-$. $O_2$ is evolved. ✓ (C) Dilute $H_2SO_4$ with platinum electrodes: At the Pt anode, water is oxidized to give $O_2$. ✓ (D) Concentrated $H_2SO_4$ with platinum electrodes: At high concentrations, $SO_4^{2-}$ or $HSO_4^-$ may get oxidized to form peroxodisulphate ($S_2O_8^{2-}$) at the anode instead of $O_2$. So $O_2$ is not the primary product. The correct answer is Option 2: (B) and (C) only.

A homoleptic complex has only one type of ligand. We need to identify those with odd number of d electrons. (A) $[FeO_4]^{2-}$: Fe is \in +6 oxidation state ($d^2$). Homoleptic (only O ligands). d-electrons = 2 (even). Not selected. (B) $[Fe(CN)_6]^{3-}$: Fe is \in +3 oxidation state ($d^5$). Homoleptic (only CN ligands). d-electrons = 5 (odd). ✓ (C) $[Fe(CN)_5NO]^{2-}$: Has two types of ligands (CN and NO). Not homoleptic. Not selected. (D) $[CoCl_4]^{2-}$: Co is \in +2 oxidation state ($d^7$). Homoleptic (only Cl ligands). d-electrons = 7 (odd). ✓ (E) $[Co(H_2O)_3F_3]$: Has two types of ligands (H₂O and F). Not homoleptic. Not selected. The correct answer is Option 3: (B) and (D) only.