Sixty four rain drops of radius 1 mm each falling down with a terminal velocity of $10 \mathrm{~cm}

Question

Sixty four rain drops of radius 1 mm each falling down with a terminal velocity of $10 \mathrm{~cm} / \mathrm{s}$ coalesce to form a bigger drop. The terminal velocity of bigger drop is $\_\_\_\_$ $\mathrm{cm} / \mathrm{s}$.

Professor PC Thomas Classes · Answer

Let $r$ be the radius of a small drop and $R$ be the radius of the coalesced bigger drop. By conservation of volume, $64 	imes \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \implies R^3 = 64r^3 \implies R = (64)^{1/3} r = 4r$. According to Stokes' law, terminal velocity is proportional to the square of the radius, $v_t \propto r^2$. Therefore, the terminal velocity of the bigger drop is $V = v_t \left(\frac{R}{r}ight)^2 = v_t (4)^2 = 16 v_t$. Given the initial terminal velocity $v_t = 10 	ext{ cm/s}$, we get $V = 16 	imes 10 = 160 	ext{ cm/s}$.