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(D) The terminal velocity $v_T$ of a spherical drop is given by $v_T = \frac{2}{9} \frac{r^2 g (\rho - \sigma)}{\eta}$,where $r$ is the radius of the

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$27:8$

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(D) The terminal velocity $v_T$ of a spherical drop is given by $v_T = \frac{2}{9} \frac{r^2 g (\rho - \sigma)}{\eta}$,where $r$ is the radius of the drop.Thus,$v_T \propto r^2$.Given the ratio of terminal velocities is $\frac{v_{T_1}}{v_{T_2}} = \frac{9}{4}$.Since $\frac{v_{T_1}}{v_{T_2}} = \frac{r_1^2}{r_2^2}$,we have $\frac{r_1^2}{r_2^2} = \frac{9}{4}$.Taking the square root on both sides,we get $\frac{r_1}{r_2} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.The volume $V$ of a spherical drop is $V = \frac{4}{3} \pi r^3$,so $V \propto r^3$.The ratio of their volumes is $\frac{V_1}{V_2} = \frac{r_1^3}{r_2^3} = \left(\frac{r_1}{r_2}\right)^3 = \left(\frac{3}{2}\right)^3 = \frac{27}{8}$.

Two rain drops reach the earth with different terminal velocities having ratio $9:4$. Then the ratio of their volumes is

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