Two equal drops are falling through air with a steady velocity 5 cm/s. If the drops coalesce, the new terminal velocity will be

When two equal droplets combine, the volume of water stays the same but a larger drop is created. Let's suppose that R is the radius after these combine and that r is the drop's radius before these mix. There are

$$\begin{gathered} 2\times \frac{4}{3}{\mathrm{\pi r}}^3=\frac{4}{3}{\mathrm{\pi R}}^3 \\ \mathrm{R}=2^{\frac{1}{3}}\mathrm{r} \end{gathered}$$

Also we know that terminal velocity ${\mathrm{V}}_{\mathrm{T}}\propto {\mathrm{R}}^2,$

$$\begin{gathered} \frac{\mathrm{V}{\text{'}}_{\mathrm{T}}}{{\mathrm{V}}_{\mathrm{T}}}=\frac{{\mathrm{R}}^2}{{\mathrm{r}}^2} \\ \mathrm{V}{\text{'}}_{\mathrm{T}}={\left(\frac{\mathrm{R}}{\mathrm{r}}\right)}^2{\mathrm{V}}_{\mathrm{T}}={\left(2^{\frac{1}{3}}\right)}^2\times 5=5\times 4^{\frac{1}{3}}\mathrm{cm}/\mathrm{s} \end{gathered}$$

Hence the correct answer is $5\times 4^{\frac{1}{3}}\mathrm{cm}/\mathrm{s}.$