Eight identical small drops of water are falling down vertically through a medium, each with terminal…
Eight identical small drops of water are falling down vertically through a medium, each with terminal velocity $v$. If they combine to form a single drop, then its terminal velocity will be
$3 v$
$6 v$
$5 v$
$4 v$
Solution
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
$\begin{aligned} & \therefore \frac{4}{2} \pi R^3=8 \times \frac{4}{3} \pi r^3 \\ & \therefore R^3=(2 r)^3 \therefore R=2 r\end{aligned}$
The terminal velocity of a drop $v \propto(\text { radius })^2$
$\begin{aligned} & \therefore \frac{V}{v}=\frac{R^2}{r^2}=\frac{(2 r)^2}{r^3}=4 \\ & \therefore V=4 v\end{aligned}$
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