Detailed Solutions 📝

Explanation: Superconductors exhibit perfect diamagnetism, a phenomenon known as the Meissner effect. For a superconductor, the relative permeability is zero ($\mu_r=0$), which means its magnetic susceptibility ($\chi = \mu_r - 1$) is exactly $-1$.

Explanation: Heavy doping in a Zener diode makes the depletion region extremely thin (less than a micrometer). This thin region causes the electric field across the junction to be incredibly high ($E=V/d$), which facilitates the internal field emission of electrons, leading to Zener breakdown.

Explanation: Maxwell introduced the concept of displacement current to satisfy the continuity of current in circuits with capacitors. It is given by $I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$, meaning it is produced solely by a time-varying electric flux.

Explanation: A negative total energy implies that the electron-nucleus system is a bound system. If the total energy were zero or positive, the electron would not be bound to the nucleus and would be free.

Explanation: Using the Lens Maker's formula, the focal length depends on the relative refractive index $(\mu_{\text{lens}} / \mu_{\text{medium}}) - 1$. Since $1.5 < 1.6$, the term becomes negative. Thus, the focal length becomes negative, and the converging (convex) lens starts behaving as a diverging (concave) lens.

Explanation: For an ideal gas, internal energy depends only on temperature ($U \propto T$). Since it is an isothermal process, $\Delta T=0$, so $\Delta U=0$. Since the gas expands, the volume increases, meaning the gas does positive work on the surroundings.

Explanation: Initial force $F \propto (Q)(3Q) = 3Q^2$ (attractive). When touched, total charge is $+Q - 3Q = -2Q$. Since they are identical, charge is shared equally: each gets $-Q$. Final force $F' \propto (-Q)(-Q) = Q^2$ (repulsive). The force changed from attractive to repulsive, and the magnitude decreased from $3Q^2$ proportionality to $Q^2$.

Explanation: Resistance of a bulb is given by $R = V^2/P$. Thus, the 60W bulb has a higher resistance than the 100W bulb. In a series circuit, the current $I$ is the same for both. The power consumed (brightness) is $P = I^2R$. Since $R$ is higher for the 60W bulb, it dissipates more power and glows brighter in series.

Explanation: The value of $g$ is affected by both the shape of the earth ($g \propto 1/R^2$, and $R_{\text{equator}} > R_{\text{poles}}$) and the rotation of the earth ($g' = g - R\omega^2\cos^2\lambda$). Both factors contribute to making $g$ minimum at the equator ($\lambda = 0^\circ$) and maximum at the poles ($\lambda = 90^\circ$).

Explanation: At the central position, the path difference is exactly zero for all wavelengths (colors). Since constructive interference condition ($\Delta x = n\lambda$) is satisfied for $n=0$ for all colors simultaneously, they all recombine to form a white central fringe. As you move away, path differences change, and different colors form maxima at different positions.

Explanation: Statement I is incorrect; most particles passed undeflected because most of the atom is empty space, not because charge is uniform (that was Thomson's failed model). Statement II is correct; the impact parameter determines the scattering angle—a small impact parameter leads to a large angle of deviation.

Explanation: As temperature increases, metal ions vibrate with larger amplitudes. This causes electrons to collide more frequently, which decreases the average time between collisions (relaxation time, $\tau$). Since drift velocity $v_d \propto \tau$, the drift velocity actually decreases as temperature increases, making Statement I false.

Explanation: By the Work-Energy theorem (or 3rd equation of motion $v^2 = u^2 + 2as$), the initial kinetic energy is dissipated by the work done by friction: $\frac{1}{2}mv^2 = F \times s$. This means $s \propto v^2$. If speed becomes $3v$, the stopping distance becomes $(3)^2=9$ times the original distance, i.e., $9s$.

Explanation: Angular momentum is conserved: $L_i = L_f \implies I_1 \omega = I_2 \omega'$. Initial moment of inertia $I_1 = \frac{1}{2}MR^2$. Final moment of inertia $I_2 = \frac{1}{2}MR^2 + m(R/2)^2 = \frac{1}{2}MR^2 + \frac{mR^2}{4}$. Equating: $(\frac{1}{2}MR^2)\omega = (\frac{1}{2}MR^2 + \frac{1}{4}mR^2)\omega'$. Cancel $R^2$ and multiply by 4 to clear denominators: $2M\omega = (2M+m)\omega'$. Thus, $\omega' = \frac{2M\omega}{2M + m}$.

Explanation: When capacitors are connected in parallel (positive to positive), the total charge is conserved. $Q_1 = C_1V_1 = 2 \times 10 = 20\,\mu\text{C}$. $Q_2 = C_2V_2 = 4 \times 20 = 80\,\mu\text{C}$. Total charge $Q = 20 + 80 = 100\,\mu\text{C}$. Total capacitance $C = C_1 + C_2 = 2 + 4 = 6\,\mu\text{F}$. Common potential $V = Q/C = 100/6 = 16.67\text{ V}$.

Explanation: From Newton's second law, Force = mass $\times$ acceleration. $F = m \times \frac{v}{T}$. Rearranging for mass: $m = \frac{F \times T}{v} = F v^{-1} T$. So the dimensional formula is $[F v^{-1} T]$.

Explanation: Total initial binding energy = $(20 \times 6) + (30 \times 8) = 120 + 240 = 360\text{ MeV}$. Total mass number of $C = 20+30=50$. Total final binding energy = $50 \times 7.5 = 375\text{ MeV}$. Since final binding energy is greater than initial binding energy, the product nucleus is more tightly bound. The difference $(375 - 360 = 15\text{ MeV})$ is released as Q-value energy.

Explanation: Volume conservation: $\frac{4}{3}\pi R^3 = 8 \times \frac{4}{3}\pi r^3 \implies R = 2r \implies r = R/2$. Terminal velocity $v_t$ is proportional to the square of the radius ($v_t \propto r^2$). New terminal velocity $v' \propto (R/2)^2 \implies v' \propto \frac{R^2}{4}$. Therefore, the new terminal velocity is $v/4$.

Explanation: Time period $T = 2\pi \sqrt{L/g_{\text{eff}}}$. When the lift is in free fall, the downward acceleration $a=g$. The pseudo force acts upwards, perfectly canceling true gravity. Effective gravity $g_{\text{eff}} = g - g = 0$. Substituting $g_{\text{eff}}=0$ into the formula gives $T = \text{Infinity}$. The pendulum simply won't oscillate.

Explanation: Since the block does not slip, it moves with the truck at $2\text{ m/s}^2$. The only force accelerating the block horizontally is the static friction from the truck floor. Force of friction acting on the block $F_f = ma = 10 \times 2 = 20\text{ N}$ (Note: We DO NOT use $\mu mg$ because friction is self-adjusting and limits at 40N, but only 20N is required here). The friction acts in the forward direction (direction of motion). Work done by friction = $F_f \times s = 20 \times 5 = +100\text{ J}$.