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^(-5)kg//ms`, 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.

Answer

` 2.72 `

Explanation

By Stoke’s law, the terminal velocity of a water drop of radius r is given by ` v = (2)/(g) ( r ^2 (rho - sigma ) g ) /(eta )`
Where ` rho ` is the density of water, ` sigma ` is the density of air and ` eta ` the coefficient o viscosity of air. Here ` sigma ` is egligible and ` r = 0.0015 mm = 1.5 xx 10 ^(-3) mm = 1.5 xx 10 ^(-6) m`. Substituting the values:
` v = (2)/(9) xx ((1.5 xx 10 ^(-6)) ^2 xx( 1.0 xx 10 ^3) xx 9.8)/(1.8 xx 10 ^(-5)) = 2.72 xx 10 ^(-4) m//s `