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(B) Let $r$ be the radius of each small drop and $R$ be the radius of the big drop.Since the volume remains constant during coalescence:$n 	imes (\fr

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$5 n^{2/3} \ cm/s$

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(B) Let $r$ be the radius of each small drop and $R$ be the radius of the big drop.Since the volume remains constant during coalescence:$n 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$$R^3 = n r^3 \implies R = n^{1/3} r$The terminal velocity $v_t$ of a drop is given by Stokes' Law: $v_t = \frac{2r^2(ho - \sigma)g}{9\eta}$.Thus,$v_t \propto r^2$.Let $v_1 = 5 \ cm/s$ be the terminal velocity of a small drop and $v_2$ be the terminal velocity of the big drop.$\frac{v_2}{v_1} = \left(\frac{R}{r}ight)^2 = (n^{1/3})^2 = n^{2/3}$.$v_2 = v_1 	imes n^{2/3} = 5 n^{2/3} \ cm/s$.

$n$ small drops of the same size fall through air with a constant terminal velocity of $5 \ cm/s$. If they coalesce to form a single big drop,what is the terminal velocity of the big drop?

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