If two drops of same radius r coalesce then radius of new drop is given by R

$\frac{4}{3} \pi R^{3} = \frac{4}{3} \pi r^{3} + \frac{4}{3} \pi r^{3} \Rightarrow R^{3} = 2 r^{3} \Rightarrow R = 2^{1/3} r \text{If drop of radius } r \text{ is falling in viscous medium then it acquire a critical velocity } v \text{ and } \mathbf{v} \propto r^{2} \frac{v_{2}}{v_{1}} = \left(\frac{R}{r}\right)^{2} = \left(\frac{2^{1/3} r}{r}\right)^{2}$

⇒ ⇒ $\mathbf{v}_2 = 2^{2/3} \times \mathbf{v}_1 = 2^{2/3} \times (5) = 5 \times (4)^{1/3} \mathrm{m/s}$