A drop of water of radius 0.0015 mm is falling in air .If the coefficient 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. `1.0xx10^(-4)ms^(-1)`
B. `2.0xx10^(-4)ms^(-1)`
C. `2.5xx10^(-4)ms^(-1)`
D. `5.0xx10^(-4)ms^(-1)`

Answer

C

Explanation

Here
r=0.00115m`=0.0015xx10^(-3)m`
`eta=2.0xx10^(-5)kgm^(-1)s^(-1)`
`rho 1.0xx10^(3)kgm^(-3)`
`g=10ms^(-2)`
Neglecting the density of air, the terminal velocity of the water drop is
`v_(T)=2/9(r^(2)rho g)/(eta)`
`=(2xx(0.0015010^(-3))^(2)xx1.0xx10^(3)xx10)/(9xx2.0xx10^(-5))`
`g=10ms^(-1)`