Neglecting the density of air, the terminal velocity obtained by a raindrop of radius `0.3 mm` falling through the air of viscosity `1.8 xx10^(-5) N//m^(2)` will be

Options

A. `10.9m//s`
B. `8.3m//s`
C. `9.2m//s`
D. `7.6m//s`

Answer

A

Explanation

The terminal velocity of the spherical raindrop of radius `r` is given by
`v_(t)=(2r^(2)rhog)/(9eta)`
where `rho` is the density of water and `eta` the viscosity of air. Substituting `r=0.3mm=0.3x10^(-3)m, rho=10^(3)kg//m^(3)`
`g=9.8ms^(-12)` and `eta=1.8xx10^(-5)Ns//m^(2)`
We get `v_(t)=(2xx(0.3)^(2)xx10^(-3)xx9.8)/(9xx1.8xx10^(5))=10.9m//s`