Two drops of the same radius are falling through air with a steady velocity of 5 cm per sec. If the two drops coalesce, the terminal velocity would be

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

$\frac{4}{3}{\mathrm{\pi R}}^3=\frac{4}{3}{\mathrm{\pi r}}^3+\frac{4}{3}{\mathrm{\pi r}}^3\Rightarrow {\mathrm{R}}^3=2{\mathrm{r}}^3\Rightarrow \mathrm{R}=2^{1/3}\mathrm{r}$

If drop of radius r is falling in viscous medium then it acquire a critical velocity v and $\mathrm{v}\propto {\mathrm{r}}^2$

$\frac{{\mathrm{v}}_2}{{\mathrm{v}}_1}={\left(\frac{\mathrm{R}}{\mathrm{r}}\right)}^2={\left(\frac{2^{1/3}\mathrm{r}}{\mathrm{r}}\right)}^2\Rightarrow {\mathrm{v}}_2=2^{2/3}{\mathrm{xv}}_1=2^{2/3}\mathrm{x}\left(5\right) = 5 \mathrm{x} {\left(4\right)}^{1/3} \mathrm{m}/\mathrm{s}$