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Two equal drops are falling through air with a steady velocity of 5 cm/s. If the drops coalesce, then the new terminal velocity will be.

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a

$5 \times {(4)}^{1 / 3} c m / s$

b

$5 / \sqrt{2} c m / s$

c

$5 \times 2 c m / s$

d

$5 \times \sqrt{2} c m / s$

answer is C.

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Detailed Solution

$\begin{array}{l} R = n^{\frac{1}{3}} r \\ R = (2)^{1 / 3} r \\ V \propto R^{2} \\ \frac{V_{1}}{V_{2}} = {(\frac{R_{1}}{R_{2}})}^{2} = {(\frac{1}{2^{\frac{1}{3}}})}^{2} \\ \frac{5}{V_{2}} = \frac{1}{2^{2 / 3}} \\ V_{2} = 5 \times (2)^{2 / 3} = 5 \times (4)^{1 / 3} cm / s \end{array}$

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