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(C) The terminal velocity $v_T$ of a spherical rain drop of radius $r$ is given by Stokes' Law:$v_T = \frac{2(\sigma - \rho) r^2 g}{9 \eta}$where $\si

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$16: 9$

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(C) The terminal velocity $v_T$ of a spherical rain drop of radius $r$ is given by Stokes' Law:$v_T = \frac{2(\sigma - \rho) r^2 g}{9 \eta}$where $\sigma$ is the density of the drop,$\rho$ is the density of air,$g$ is acceleration due to gravity,and $\eta$ is the coefficient of viscosity.Since $\sigma, \rho, g,$ and $\eta$ are constant for both drops,we have:$v_T \propto r^2$  --- $(i)$The surface area $A$ of a spherical drop is given by:$A = 4 \pi r^2$This implies:$A \propto r^2$  --- $(ii)$Comparing equations $(i)$ and $(ii)$,we see that the surface area is directly proportional to the terminal velocity:$A \propto v_T$Therefore,the ratio of their surface areas is equal to the ratio of their terminal velocities:$\frac{A_1}{A_2} = \frac{v_{T1}}{v_{T2}} = \frac{16}{9}$Thus,the ratio is $16: 9$.

Two spherical rain drops reach the surface of the earth with terminal velocities having ratio $16: 9$. The ratio of their surface area is

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