The terminal velocity of a rain drop is $30 \mathrm{cm}/\mathrm{s}$. If the viscosity of air is $1.8\times {10}^{-5}{\mathrm{Nsm}}^{-2}$. The radius of rain drop is

${\mathrm{v}}_{\mathrm{t}}=\frac{2}{9}\frac{{\mathrm{r}}^2(\mathrm{\rho }-\mathrm{\sigma })\mathrm{g}}{\mathrm{\eta }}$

Ignoring the density of air $\mathrm{\sigma }$ we have,

$$\begin{gathered} {\mathrm{y}}_{\mathrm{T}}=\frac{2}{9}\frac{{\mathrm{r}}^2\mathrm{\rho g}}{\mathrm{\eta }} \\ \therefore \mathrm{r}=\sqrt{\frac{9{\mathrm{\eta y}}_{\mathrm{T}}}{2\mathrm{\rho g}}} \\ =\sqrt{\frac{9\times 1.8\times {10}^{-5}\times 0.3}{2\times {10}^3\times 9.8}} \\ =0.49\times {10}^{-4} \mathrm{m} \\ =0.05 \mathrm{mm} \end{gathered}$$