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(C) The terminal velocity $v$ of a spherical drop is given by $v = \frac{2}{9} \frac{r^2(\rho-\sigma) g}{\eta}$.Since the buoyant force is ignored,$\s

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$\sqrt[3]{4} v$

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(C) The terminal velocity $v$ of a spherical drop is given by $v = \frac{2}{9} \frac{r^2(ho-\sigma) g}{\eta}$.Since the buoyant force is ignored,$\sigma \approx 0$,so $v \propto r^2$.When two drops of radius $r$ combine to form a bigger drop of radius $R$,the volume is conserved:$\frac{4}{3} \pi R^3 = 2 	imes \frac{4}{3} \pi r^3$$R^3 = 2r^3 \Rightarrow R = 2^{1/3} r$.Let the terminal velocity of the bigger drop be $v'$.$\frac{v'}{v} = \frac{R^2}{r^2} = \frac{(2^{1/3} r)^2}{r^2} = 2^{2/3} = \sqrt[3]{4}$.Therefore,$v' = \sqrt[3]{4} v$.

The terminal velocity of a liquid drop of radius $r$ falling through air is $v$. If two such drops are combined to form a bigger drop,the terminal velocity with which the bigger drop falls through air is (ignore any buoyant force due to air).

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