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(C) Let the radius of each small drop be $r$ and the terminal velocity be $v_1 = 5 \, cm/s$.When two drops coalesce,the volume is conserved: $\frac{4}

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the radius of each small drop be $r$ and the terminal velocity be $v_1 = 5 \, cm/s$.When two drops coalesce,the volume is conserved: $\frac{4}{3}\pi R^3 = 2 	imes \frac{4}{3}\pi r^3$,where $R$ is the radius of the new drop.Thus,$R^3 = 2r^3$,which gives $R = 2^{1/3}r$.The terminal velocity $v$ of a drop is given by Stokes' Law: $v = \frac{2}{9} \frac{r^2(ho - \sigma)g}{\eta}$,which implies $v \propto r^2$.Therefore,the ratio of the terminal velocities is $\frac{v_2}{v_1} = \left(\frac{R}{r}ight)^2$.Substituting $R = 2^{1/3}r$,we get $\frac{v_2}{v_1} = (2^{1/3})^2 = 2^{2/3} = (4)^{1/3}$.So,$v_2 = v_1 	imes (4)^{1/3} = 5 	imes (4)^{1/3} \, cm/s$.

Two drops of equal radius are falling with a terminal velocity of $5 \, cm/s$. If they coalesce to form a single larger drop,what will be the terminal velocity of the new drop?

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