b
\(\mathrm{m}=\text { mass of small drop }\)

\(\mathrm{M}=\text { mass of bigger drop }\)

\(\mathrm{V}_{\mathrm{t}}=\frac{2}{9} \frac{\mathrm{R}^2(\rho-\sigma) \mathrm{g}}{\eta}\)

\(8 \propto \mathrm{m}=\mathrm{M}\)

\(8 \mathrm{r}^3=\mathrm{R}^3 \Rightarrow \mathrm{R}=2 \mathrm{R}\)

\(\text { as } \mathrm{V}_{\mathrm{t}} \times \mathrm{R}^2 \because \text { Radius double so } \mathrm{V}_{\mathrm{t}} \text { becomes } 4 \text { time }\)

\(\therefore 4 \times 10=40 \mathrm{~cm} / \mathrm{s}\)