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(C) The terminal velocity $v_t$ of a spherical drop falling through a viscous medium is given by Stokes' Law as $v_t = \frac{2r^2(\rho - \sigma)g}{9\e

Vedclass · Accepted Answer

$64$

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(C) The terminal velocity $v_t$ of a spherical drop falling through a viscous medium is given by Stokes' Law as $v_t = \frac{2r^2(ho - \sigma)g}{9\eta}$.From this,we see that $v_t \propto r^2$.Let $r$ be the radius of each small drop and $R$ be the radius of the big drop.Since the volume remains constant when $8$ small drops combine to form one big drop:$8 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$$R^3 = 8r^3 \Rightarrow R = 2r$.Now,using the proportionality $v_t \propto r^2$:$\frac{v_{big}}{v_{small}} = \frac{R^2}{r^2} = \frac{(2r)^2}{r^2} = 4$.Given $v_{small} = 16 \ cm/s$,the terminal velocity of the big drop is:$v_{big} = 4 	imes 16 \ cm/s = 64 \ cm/s$.

Eight equal drops of water,each of radius $r = 2 \ mm$,are falling through air with a terminal velocity of $16 \ cm/s$. The eight drops combine to form a single big drop. The terminal velocity of the bigger drop will be $.... \ cm/s$.

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