Two identical spherical drops of water are falling (vertically downwards) through air with a steady velocity of 5 cm/sec. If both the drops coalsece (combine) to form a new spherical drop, the terminal velocity of the new drop will be (neglect buoyant force on the drops).

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$5 \times \sqrt{2} cm / \sec$

$5 \times {(4)}^{\frac{1}{3}} cm / \sec$

$5 \times 2 cm / \sec$

$\frac{5}{\sqrt{2}} cm / \sec$

answer is C.

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Detailed Solution

When two drops of radius r each combine to form a big drop, the radius of big drop will be given by

$\frac{4}{3} {πR}^{3} = \frac{4 π}{3} r^{2} + \frac{4 π}{3} r^{3}$

or $R^{2} = 2 r^{3} or R = 2^{\frac{1}{3}} r$

Now $\frac{V_{R}}{V_{r}} = {(\frac{R}{r})}^{2} = 2^{\frac{2}{3}} = 4^{\frac{1}{3}}$

$V_{R} = 5 \times 4^{\frac{1}{3}} cm / s$

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