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)`.

Explanation

Here, `r= 0.8//2 = 0.4 mm`
`= 4 xx 10^(-4)m`,
For liquid, `eta = 0.15 Nm^(-2)s` ,
sp. Gravity = 0.9
Therefore, density of liquid , `sigma = 0.9 g//c c`
`=0.9 xx 10^(3) kg m^(-3)`
Specific gravity of air = 0.001293
Density of air, `rho = 0.001293 g//c c`
`= 0.001293 xx 10^(3) kg//m^(3) = 1.293 kg//m^(3)`,
Terminal velocity of air bubble in liquid is
`upsilon = (2 r^(2) (rho-sigma)g)/(9 eta)`
`=(2 xx (4 xx 10^(-4))^(2)(1.293 - 0.9 xx10^(3))xx9.8)/(9xx0.15)`
` = -2.09 xx 10^(-3)ms^(-1)`
The negative value of `upsilon` shows that the bubble will be moving upward and not downward.
For water, `density` `sigma = 1000 kg//m^(3)`,
`eta ' = 10^(-3) Nm^(-2)s, rho = 1.293 kg//m^(3)`
Terminal velocity of air bubble in water is
`upsilon = (2 r^(2) (rho-sigma')g)/(9 eta')`
`=(2 xx (4 xx 10^(-4))^(2)(1.293 - 1000)xx9.8)/(9xx10^(-3))`
`-0.348 ms^(-1)`.