A drop of water of radius `0.0015 mm` is falling in air. If the coefficient of viscosity of air is `1.8 xx 10^(-3)kg//m^(3)`, what will be the terminal velocity of the drop? Density of water `= 1.0 xx 10^(3) kg//m^(3)` and `g = 9.8 N//kg`. Density of air can be neglected.

Options

A. `2.72xx10^-4(m)/(s)`
B. `4.72xx10^-4(m)/(s)`
C. `5.28xx10^-4(m)/(s)`
D. `1.36xx10^-4(m)/(s)`

Answer

A

Explanation

By stokes' law the terminal velocity of a water drop of radius `r` is given by
`v=(2)/(9)(r^2(rho-sigma)g)/(eta)`
Where `rho` is the density of water, `sigma` is the density of air and `eta` the coefficient of viscosity of air. Here `sigma` is negligible and `r=0.0015mm=1.5xx10^-3mm=1.5xx10^-6m`. Substituting values:
`v=(2)/(9)xx((1.5xx10^-6)^2xx(1.0xx10^3)xx9.8)/(1.8xx10^-5)`
`=2.72xx10^-4(m)/(s)`