Physics - Fluid Question with Solution | TestHub

Let $R$, and $r$ be the radii of the big and small drops respectively. The volume of the big drop $=8\times$ volume of one small drops $\begin{array}{rl}& \therefore \frac{4}{2}\pi R^3=8\times \frac{4}{3}\pi r^3\\ & \therefore R^3=(2r{)}^3\therefore R=2r\end{array}$ The terminal velocity of a drop $v\propto (\text{ radius }{)}^2$ $\begin{array}{rl}& \therefore \frac{V}{v}=\frac{R^2}{r^2}=\frac{(2r{)}^2}{r^3}=4\\ & \therefore V=4v\end{array}$