Two drops of same radius are falling through air with steady speed $\mathrm{V}$ . If the two drops coalesce, what would be the terminal speed?

${\mathrm{v}}_{\mathrm{t}}=\frac{2{\mathrm{r}}^2(\rho -\sigma )\mathrm{g}}{\vartheta \eta }\text{ or }{\mathrm{v}}_{\mathrm{t}}\propto {\mathrm{r}}^2$

Let R be the radius of bigger drop. Equating the volumes, we have

$\frac{4}{3}\pi {\mathrm{R}}^3=2\left(\frac{4}{3}\pi {\mathrm{r}}^3\right)\therefore \mathrm{R}=(2{)}^{1/3}\mathrm{r}$

Hence, the terminal velocity of big drop ${\mathrm{v}}^1=(2{)}^{2/3}\mathrm{v}$