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Q. An air bubble of radius $1\, cm$ rises from the bottom portion through a liquid of density $1.5\, g/cc$ at a constant speed of $0.25\, cm\, s^{-1}$. If the density of air is neglected, the coefficient of viscosity of the liquid is approximately, (In Pa)

AFMCAFMC 2008Mechanical Properties of Fluids

Solution:

$v =\frac{2}{9} \frac{r^{2} \rho g}{\eta}$
$\Rightarrow \eta =\frac{2}{9} \cdot \frac{r^{2} \rho g}{v}$
$=\frac{2}{9} \frac{\left(1 \times 10^{-2}\right)^{2} \times\left(1.5 \times 10^{3}\right) \times 9.8}{0.25 \times 10^{-2}}$
$=130\, Pa - s$