Question

Two equal drops of water are falling through air with. steady velocity v. If the drops coalesced, what will be the new velocity ?

Answer 1

When the two equal drops coalesces, volume of water remains same and a bigger drop is formed. Lets say $r$ is radius of drop before these combine and $R$ is radius after these combine.

We have

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

and we know that terminal velocity $V_T\propto$ radius${}^2$, so we have

$\frac{V_T^{\prime }}{V_T}=\frac{R^2}{r^2}$

$\Rightarrow V_T^{\prime }={\left(\frac{R}{r}\right)}^2{V}_T$

$=(2^{1/3}{)}^{2}v$

$=2^{2/3}v\text{m/s}$