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(C) Let the radius of each small drop be $r$ and the radius of the big drop be $R$. Since the volume is conserved,the volume of $8$ small drops equals

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$24$

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(C) Let the radius of each small drop be $r$ and the radius of the big drop be $R$. Since the volume is conserved,the volume of $8$ small drops equals the volume of the big drop: $8 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$.This simplifies to $R^3 = 8r^3$,which gives $R = 2r$.The terminal velocity $V_T$ of a spherical drop is given by Stokes' Law as $V_T = \frac{2}{9} \frac{r^2 g (ho - \sigma)}{\eta}$.Since the density $ho$ and viscosity $\eta$ are constant,$V_T \propto r^2$.Therefore,$\frac{V_{T_{big}}}{V_{T_{small}}} = (\frac{R}{r})^2 = (\frac{2r}{r})^2 = 4$.Given $V_{T_{small}} = 6 \ cm \ s^{-1}$,the terminal velocity of the big drop is $V_{T_{big}} = 4 	imes 6 = 24 \ cm \ s^{-1}$.

Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of $6 \ cm \ s^{-1}$. If they coalesce to form one big drop,what will be the terminal speed of the bigger drop? (Neglect the buoyancy of the air) ....... $cm \ s^{-1}$

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