Since the battery is disconnected, charge $Q$ is constant:

$Q = C_0 V_0 = 12 \times 10^{-12} \times 10 = 120\ \text{pC}$

Initial energy: $U_i = \frac{Q^2}{2C_0} = \frac{(120\times10^{-12})^2}{2\times12\times10^{-12}} = 600\ \text{pJ}$

New capacitance: $C = KC_0 = 6.5 \times 12 = 78\ \text{pF}$

Final energy: $U_f = \frac{Q^2}{2C} = \frac{(120)^2 \times 10^{-24}}{2\times78\times10^{-12}} \approx 92.3\ \text{pJ}$

Work done by capacitor on slab $= U_i - U_f = 600 - 92.3 \approx 508\ \text{pJ}$

Let total distance $= 2d$.

Total time $= \dfrac{d}{v_1} + \dfrac{d}{v_2} = \dfrac{d(v_1+v_2)}{v_1 v_2}$

Average speed $= \dfrac{2d}{\dfrac{d(v_1+v_2)}{v_1 v_2}} = \mathbf{\dfrac{2v_1 v_2}{v_1+v_2}}$

(This is the harmonic mean of $v_1$ and $v_2$.)

Strong electrolytes dissociate completely. KI is an ionic salt → strong electrolyte. Acetic acid, NH₄OH, H₂CO₃ are weak electrolytes.

Using the future value formula: \(FV = PV \times (1+r)^n\). So \(1000 = 250\times(1+r)^{16}\). \((1+r)^{16} = 4\). \(1+r = 4^{1/16} = 4^{0.0625}\). \(\ln(1+r) = \frac{\ln 4}{16} = \frac{1.386}{16} = 0.0866\). \(1+r = e^{0.0866} \approx 1.0905\). So \(r \approx 9\%\). Wait: \(4^{1/16}\): \(\log(4)/16 = 0.602/16=0.0376\), antilog \(=1.089\), so \(r\approx 9\%\) = option (B). Re-examining: \(4^{0.0625}=1.0905\Rightarrow r\approx9\%\). Answer: option (B) = 9 percent.