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(C) Let the radius of the small drop be $r$ and the radius of the big drop be $R$.Since the volume is conserved, the volume of the big drop is equal t

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$24 \,cm \,s^{-1}$

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(C) Let the radius of the small drop be $r$ and the radius of the big drop be $R$.Since the volume is conserved, the volume of the big drop is equal to the sum of the volumes of the eight small drops:$\frac{4}{3} \pi R^3 = 8 	imes \frac{4}{3} \pi r^3$$R^3 = 8r^3 \implies R = 2r$The terminal velocity $v_t$ of a spherical drop is given by Stokes' Law:$v_t = \frac{2}{9} \frac{r^2}{\eta} (ho - \sigma) g$Since the buoyancy of air is neglected, $\sigma \approx 0$, so $v_t \propto r^2$.Let $v_1 = 6 \,cm \,s^{-1}$ be the terminal speed of the small drop and $v_2$ be the terminal speed of the big drop.$\frac{v_2}{v_1} = \frac{R^2}{r^2} = \frac{(2r)^2}{r^2} = 4$$v_2 = 4 	imes v_1 = 4 	imes 6 \,cm \,s^{-1} = 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)

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