Question

Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of $6cm{s}^{-1}$. If they coalesce to form one big drop, what will be its terminal speed? Neglect the buoyancy due to air:

• A. $1.5cm{s}^{-1}$
• B. $6cm{s}^{-1}$
• C. $24cm{s}^{-1}$
• D. $32cm{s}^{-1}$

Answer 1

Answer: (C)

$24cm{s}^{-1}$

When eight 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
$8\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3\Rightarrow R=2r$
and we know that terminal velocity $V_T\propto$ radius${}^2$, so we have
$\frac{V_T^{\prime }}{V_T}=\frac{R^2}{r^2}\Rightarrow V_T^{\prime }={\left(\frac{R}{r}\right)}^2{V}_T={\left(2\right)}^2\times 6cm{s}^{-1}=24cm{s}^{-1}$