Thank you for reporting, we will resolve it shortly
Q.
Neglecting the density of air, the terminal velocity obtained by a raindrop of radius $0.3\, mm$ falling through air of viscosity $1.8 \times 10^{-5}\, N \,s \,m ^{-2}$ will be
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\, mm $
$= 0.3 × 10^{-3} \,m, \rho = 10^{3} \,kg/m^{3}, g = 9.8 \,m/s^{2}$ and
$\eta = 1.8 × 10^{-5} \,N s/m^{2}$, we get
$v_{t} = \frac{2\times\left(0.3\right)^{2}\times10^{-3}\times9.8}{9\times1.8\times10^{-5}}$
$= 10.9 \,m\,s^{-1}$