Two identical drops of water fall through air with a terminal velocity $2 m s^{- 1} .$ If the drops merge to form a single drop its terminal velocity is

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$2.52 m s^{- 1}$

$3.17 m s^{- 1}$

$1.58 m s^{- 1}$

$4 m s^{- 1}$

answer is A.

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Detailed Solution

When two drops merge let R be the radius, then

$\begin{array}{l} \frac{4}{3} π R^{3} = 2 (\frac{4}{3} π r^{3}) \Rightarrow R = 2^{\frac{1}{3}} r \\ υ_{T} α {(r a d i u s)}^{2} = \frac{υ_{T}}{υ_{T}^{1}} = \frac{r^{2}}{2^{\frac{2}{3}} r^{2}}, \\ υ_{T}^{1} = 4^{\frac{1}{3}} υ_{T} = 4^{\frac{1}{3}} \times 2 = 3.17 m s^{- 1} \end{array}$