Explanation

(a) If $r$ is the radius of small droplet and $R$ is the radius of big drop, then according to question,
$\frac{4}{3}\pi R^{3} = 10^{6} \times \frac{4}{3}\pi r^{3}\ $ or $r = \frac{R}{100} = 0.01\ R$
$= 0.01 \times 10^{- 2}m = 10^{- 4}m$
Work done = surface tension $\times$ increase in area
$= 35 \times 10^{- 2}\left\lbrack 10^{6} \times 4\pi \times \left( 10^{- 4} \right)^{2} - 4\pi \times \left( 10^{- 3} \right)^{2} \right\rbrack$
$= 4.35 \times 10^{- 2}$ J