With what terminal velocity will an air bubble `0.8 mm` in diameter rise in a liquid of viscosity `0.15 N-s//m^(2)` and specific gravity `0.9`? Density of air is `1.293 kg//m^(3)`.

Options

A. `-0.21 cms^(-1)`
B. `+0.21cms^(-1)`
C. `-0.41 cms^(-1)`
D. `+0.41cms^(-1)`

Answer

A

Explanation

The terminal velocity of the bubble is given by ,
`v_T=2/9(r^2(rho-sigma)g)/(eta)`
Here, `r=0.4xx10^(-3) m,`
`sigma = 0.9xx10^3 kgm^(-3) rho= 1.293kgm^(-3)`
`eta=0.15 Nsm^(-2) " and " g=9.8 ms^(-2)`
Substituting the values , we have
`v_T=2/9xx((0.4xx10^(-3))^2(1.293-0.9xx10^3)xx9.8)/(0.15)`
`=-0.0021 ms^(-1)`
or `v_T=-0.21 cms^(-1)`