JEE Main 2025 — 24 January, Morning Shift — Question 53

$\quad \mathrm{W}=\Delta \mathrm{U}=\mathrm{S} \Delta \mathrm{A}$ One drop to n drop $\frac{4}{3} \lambda R^{3}=n \frac{4}{3} \lambda r^{3}$ $r=\frac{R}{n^{\frac{1}{3}}}$ So

$W=S\left(n 4 \pi r^{2}-4 \pi R^{2}\right)$ $=S 4 \pi R^{2}\left(n^{\frac{1}{3}}-1\right)$ For on drop to 27 drops $\mathrm{W}=\mathrm{S} 4 \pi \mathrm{R}^{2}\left(27^{\frac{1}{3}}-1\right)=10$ For one drop to 64 drops $\mathrm{W}^{\prime}=S 4 \pi \mathrm{R}^{2}\left(64^{\frac{1}{3}}-1\right) \ldots$ (ii) (ii)/(i) $\frac{\mathrm{W}^{\prime}}{\mathrm{W}}=\frac{4-1}{3-1}=\frac{3}{2}$ $\mathrm{W}^{\prime}=\frac{3}{2} \mathrm{~W}=155$