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(C) Let the radius of each of the two identical water droplets be $r$. When they coalesce to form a single larger drop of radius $R$,the volume remain

Vedclass · Accepted Answer

$(2)^{2/3} V$

Vedclass · Answer

(C) Let the radius of each of the two identical water droplets be $r$. When they coalesce to form a single larger drop of radius $R$,the volume remains conserved. Thus,$\frac{4}{3} \pi R^3 = 2 	imes \frac{4}{3} \pi r^3$,which gives $R^3 = 2r^3$,or $R = 2^{1/3} r$. The terminal velocity $V$ of a spherical drop falling through a viscous medium is given by $V = \frac{2}{9} \frac{r^2(ho - \sigma)g}{\eta}$. From this expression,we see that $V \propto r^2$. If $V'$ is the terminal velocity of the new drop,then $\frac{V'}{V} = \frac{R^2}{r^2}$. Substituting $R = 2^{1/3} r$,we get $\frac{V'}{V} = \frac{(2^{1/3} r)^2}{r^2} = \frac{2^{2/3} r^2}{r^2} = 2^{2/3}$. Therefore,the new terminal velocity is $V' = 2^{2/3} V$.

Two identical drops of water are falling through air with steady velocity $V$. If the two drops coalesce to form a single drop,what is the new terminal velocity of the single drop?

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