Consider a spherical air bubble of diameter 1 mm raised in a liquid of viscosity `0.15 N//m_2`. If the specific gravity of air is 0.001 and the specific gravity of the liquid is 0.901, Then find the terminal velocity in cm /s:

Answer

`00.33`

Explanation

Terminal velocity of air bubble,
`v=((2r^2(rho-sigma)g)/(9 eta)`
Given: specific gravity of liquid
= 0.901
`:. "Density of liquid" = 901 kg /m^3` and specific gravity of air = 0.001
`:. "Density of air" = 1 kg/m^3`
`:. upsilon=(2 xx (5 xx 10^-4)^2 xx (1-901) xx 10)/(9 xx 0.15)`
`=(2 xx 25 xx 10^-8 xx -900 xx 10)/(9 xx 0.15)`
`-(10 xx 10^-5)/0.03 = (-1)/3 xx 10^-2`
`=-0.33 xx 10^-2 m//s`
`=-0.33 xx 10^-2 m//s`
`:. v=-0.33 cm//s`