Physics - Fluid Question with Solution | TestHub

The terminal velocity of the spherical raindrop of radius $r$ is given by $v_t=\frac{2r^2\rho g}{9\eta }$ where $\rho$ is the density of water and $\eta$ the viscosity of air. Substituting $r=0.3\mathrm{mm}$ $=0.3\times 10^{-3}\mathrm{m},\rho =10^3\mathrm{kg}/{\mathrm{m}}^3,g=9.8\mathrm{m}/{\mathrm{s}}^2$ and $\eta =1.8\times 10^{-5}\mathrm{N}\mathrm{s}/{\mathrm{m}}^2$, we get $\begin{array}{rl}& v_t=\frac{2\times (0.3{)}^2\times 10^{-3}\times 9.8}{9\times 1.8\times 10^{-5}}\\ & =10.9\mathrm{m}{\mathrm{s}}^{-1}\end{array}$