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(B) When two drops of radius $r$ coalesce to form a single drop of radius $R$,the volume remains conserved.$2 	imes (\frac{4}{3} \pi r^3) = \frac{4}{

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$(4)^{1/3}v$

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(B) When two drops of radius $r$ coalesce to form a single drop of radius $R$,the volume remains conserved.$2 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$$R^3 = 2r^3 \implies R = 2^{1/3}r$The terminal velocity $v_t$ of a spherical drop is given by Stokes' Law: $v_t = \frac{2r^2g(ho - \sigma)}{9\eta}$,which implies $v_t \propto r^2$.Let $v$ be the terminal velocity of the individual drop and $v'$ be the terminal velocity of the coalesced drop.$\frac{v'}{v} = \frac{R^2}{r^2} = \frac{(2^{1/3}r)^2}{r^2} = (2^{1/3})^2 = 2^{2/3} = (2^2)^{1/3} = 4^{1/3}$.Therefore,the new terminal velocity is $v' = 4^{1/3}v$.

Two drops of same radius are falling through air with steady velocity of $v \ cm/s$. If the two drops coalesce,what would be the terminal velocity?

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