JEE Main 2025 Jan 24 Shift 1, Physics Q7: Properties of Fluids

Work done against surface tension equals the increase in surface area times the surface tension $S$. Splitting a drop of radius $R$ into $n$ equal droplets conserves volume:

$\frac{4}{3}\pi R^3=n\cdot \frac{4}{3}\pi r^3\Rightarrow r=R{n}^{-1/3}$

Change in surface area:

$\Delta A=n\cdot 4\pi r^2-4\pi R^2=4\pi R^2\left(n^{1/3}-1\right)$

So the work scales as $\left(n^{1/3}-1\right)$. Taking the ratio of the two cases:

$\frac{W_{64}}{W_{27}}=\frac{64^{1/3}-1}{27^{1/3}-1}=\frac{4-1}{3-1}=\frac{3}{2}$

$W_{64}=\frac{3}{2}\times 10=15$ J.

$\therefore$ Option (1).