Eight identical drops of water falling through air with uniform velocity of 10cm/s combine to form a single drop of big size, then terminal velocity of the big drop will be

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Given, terminal velocity of each small drop
$v_{1} = \frac{10 c m}{s}$
When 8 small drops are combined to form bigger
If $r_{1}$ be radius of small drop and $r_{2}$ be radius of bigger drop, then $v_{2} = v_{1}$
$\frac{4}{3} π r_{2}^{3} = 8 \times \frac{4}{3} π r_{1}^{3} \Rightarrow r_{2} = 2 r_{1}$
$∴$ Terminal velocity of the drop
$v = \frac{2 r^{2}}{9} \frac{(p - σ)}{9 η}$
i.e., $v \propto r^{2}$
$∴ \frac{v_{1}}{v_{2}} = \frac{r_{1}^{2}}{r_{2}^{2}} = \frac{r_{1}^{2}}{{(2 r_{1})}^{2}} \Rightarrow \frac{v_{1}}{v_{2}} = \frac{1}{4}$
$v_{2} = 4 v_{2} = 4 \times 10 = 40$

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