Two equal drops are falling through air with | vedclass

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Question

(A) Let the radius of each small drop be $r$ and the terminal velocity be $v_t = 5\, cm/s$.When two drops coalesce to form a single large drop of radi

Vedclass · Accepted Answer

$5 	imes (4)^{1/3}\, cm/s$

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(A) Let the radius of each small drop be $r$ and the terminal velocity be $v_t = 5\, cm/s$.When two drops coalesce to form a single large 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$.Terminal velocity $v_t$ is given by Stokes' Law: $v_t = \frac{2r^2(ho - \sigma)g}{9\eta}$,which implies $v_t \propto r^2$.Let $V_T$ be the terminal velocity of the large drop:$\frac{V_T}{v_t} = \frac{R^2}{r^2} = \frac{(2^{1/3}r)^2}{r^2} = 2^{2/3} = (2^2)^{1/3} = 4^{1/3}$.Therefore,$V_T = 5 	imes 4^{1/3}\, cm/s$.

Two equal drops are falling through air with a steady velocity of $5\, cm/s$. If two drops coalesce to form one drop,then the new terminal velocity will be:

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