A drop of water of radius 0.0015 mm is falling in air .If the cofficient of viscosity of air is `2.0 xx 10^(-5) kg m^(-1)s^(-1)` ,the terminal velocity of the drop will be
(The density of water = `10^(3) kg m^(-3)` and g = `10 m s^(-2)` )

Options

A. `1.0 xx 10^(-4) m//s`
B. `2.0 xx 10^(-4) m//s`
C. `2.5 xx 10^(-4) m//s`
D. `5.0 xx 10^(-4) m//s`

Answer

C

Explanation

Here r = 0.0015 mm `= 0.0015 xx 10^(-3)m`
`eta= 2.0 xx 10^(-5) kg//m s, rho = 1.0xx 10^(3) kg//m^(3), g= 10 m//s^(2)`
Neglecting the density of air, the terminal velocity of the water drop is
`v= (2)/(9) (r^(2) rho g)/(eta) = (2 xx (0.0015 xx 10^(-3))^(2) xx 1.0 xx 10^(3) xx 10)/(9 xx 2.0 xx 10^(-5))`
`=2.5 xx 10^(-4) m//s`