Calculate the magnitude and direction of the terminal velocity of an air bubble of radius $1\;mm$ passing through an oil viscosity $0.2\;N\,s\, /m^2$ and specific gravity $0.9$ if the density of air is $1.29\;kg/m^3$.

Given,
Radius of the air bubble $(r) = 1\;mm = 1 * 10^{-3}\;m$
Viscosity of the oil $(\eta) = 0.2\; Ns/\;m^2$
Density of the air $(\rho) = 1.92\;kg/m^3$
Density of the oil $(\sigma) = 0.9 * 1000 = 900 \;kg/m^3$
Terminal velocity $(v) = ?$

We have,

$v =$ $\frac{2\;r^2 (\rho \; - \; \sigma)\,g}{9\;\eta} = \frac{2 \; * \; (1 \; * \; 10^{-3})^2 \; * \;(1.26 \; - \; 900 ) \; * \; 10}{9 \; * \; 0.20}$ $= -0.0099 \; m/s$

The $-\;ve$ sign indicates that the ball moves in the upward direction against the force of gravity.

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