Correct answer is : 40
\(\mathrm{m}=\) mass of small drop
M= mass of bigger drop
\(\mathrm{V}_{\mathrm{t}}=\frac{2}{9} \frac{\mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{\eta}\)
\(8 \propto \mathrm{m}=\mathrm{M}\)
\(8 r^{3}=R^{3} \Rightarrow R=2 R\)
as \(\mathrm{V}_{\mathrm{t}} \times \mathrm{R}^{2} \ \because\) Radius double so \(\mathrm{V}_{\mathrm{t}}\) becomes 4 time
\(\therefore \ 4 \times 10=40 \mathrm{~cm} / \mathrm{s}\)
