With what terminal velocity will an air bubble 0.8 mm in a diameter rise in a liquid of viscosity of `0.15 Nsm^(-2)` and specific gravity 0.9? `(rho_("air")=1.2kgm^(-3))`.

Explanation

Given D=0.8mm, `r=0.4 xx 10^(-3) m, eta=0.15Nsm^(-2), rho=0.9 xx 10^(3)kgm^(-3)`
`rho_("air")=1.2kgm^(-3)`
We know that `v=2/9 (r^(2)g)/(eta) (rho-sigma)=2/9 ((4 xx 10^(-4))^(2)xx 9.8)/(0.15) (1.2-900)=-2.1 xx 10^(-3) ms^(-1)`