Explanation

(c) If two drops of same radius $r$ coalesce then radius of new drop is given by $R$
$\frac{4}{3}\pi R^{3} = \frac{4}{3}\pi r^{3} + \frac{4}{3}\pi r^{3} \Rightarrow R^{3} = 2r^{3} \Rightarrow R = 2^{1/3}r$
If drop of radius $r$ is falling in viscous medium then it acquire a critical velocity $v$ and $v \propto r^{2}$
$\frac{v_{2}}{v_{1}} = \left( \frac{R}{r} \right)^{2} = \left( \frac{2^{1/3}r}{r} \right)^{2}$
$\Rightarrow v_{2} = 2^{2/3} \times v_{1} = 2^{2/3} \times (5) = 5 \times (4)^{1/3}m/s$