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(C) Let the radius of each small drop be $r$ and the terminal velocity be $v_1 = 5 	ext{ cm/s}$.When two drops coalesce to form a larger drop of radi

Vedclass · Accepted Answer

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

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(C) Let the radius of each small drop be $r$ and the terminal velocity be $v_1 = 5 	ext{ cm/s}$.When two drops coalesce to form a larger drop of radius $R$,the volume remains conserved:$\frac{4}{3}\pi R^3 = 2 	imes \frac{4}{3}\pi r^3 \Rightarrow R^3 = 2r^3 \Rightarrow R = 2^{1/3}r$.Terminal velocity $v$ of a drop is given by $v = \frac{2}{9} \frac{r^2(ho - \sigma)g}{\eta}$,which implies $v \propto r^2$.Therefore,the new terminal velocity $v_2$ is given by:$\frac{v_2}{v_1} = \left(\frac{R}{r}ight)^2 = (2^{1/3})^2 = 2^{2/3} = (2^2)^{1/3} = 4^{1/3}$.Thus,$v_2 = v_1 	imes 4^{1/3} = 5 	imes 4^{1/3} 	ext{ cm/s}$.

Two drops of the same radius are falling through air with a steady velocity of $5 \text{ cm/s}$. If the two drops coalesce,the terminal velocity would be

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