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(B) For a small drop falling through a viscous medium at terminal velocity,the terminal velocity $v$ is given by $v = \frac{2}{9} \frac{r^2 (\rho - \s

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) For a small drop falling through a viscous medium at terminal velocity,the terminal velocity $v$ is given by $v = \frac{2}{9} \frac{r^2 (ho - \sigma) g}{\eta}$,which implies $v \propto r^2$.Let $r$ be the radius of each small drop and $R$ be the radius of the large drop formed by coalescing $8$ small drops.Since the volume is conserved,$8 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$,which gives $R^3 = 8r^3$,so $R = 2r$.Using the proportionality $v \propto r^2$,we have $\frac{v_1}{v_2} = (\frac{r}{R})^2$.Substituting the values,$\frac{10}{v_2} = (\frac{r}{2r})^2 = (\frac{1}{2})^2 = \frac{1}{4}$.Therefore,$v_2 = 10 	imes 4 = 40\,cm/s$.

Eight equal drops of water are falling through air with a steady speed of $10\,cm/s$. If the drops coalesce,the new velocity is $.........\,cm/s$.

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