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Eight drops of water, each of radius 2 mm are falling through air at a terminal velocity of 8 cm $s^{- 1}$ . If they coalesce to form a single drop, then the terminal velocity of combined drop will be

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30 cm $s^{- 1}$

24 cm $s^{- 1}$

32 $cm s^{- 1}$

28 cm $s^{- 1}$

answer is A.

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Detailed Solution

Let the radius of bigger drop is R is smaller drop is r then

$\frac{4}{3} {πR}^{3} = 8 \times \frac{4}{3} \times {πR}^{3}$

or R = 2r ------(i)

Terminal velocity, v $\infty$ $r^{2}$

$\frac{v^{'}}{v} = \frac{R^{2}}{r^{2}} = {(\frac{2 r}{r})}^{2} = 4$

or $v^{'} = 4 v = 4 \times 8 = 32 cm s^{- 1}$

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