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

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Aicharya · Accepted Answer

$v=\frac{2}{9} \frac{r^2 ho g}{\eta}$ $\Rightarrow \eta=\frac{2}{9} \cdot \frac{r^2 ho g}{v}$ $=\frac{2}{9} \frac{\left(1 	imes 10^{-2}ight)^2 	imes\left(1.5 	imes 10^3ight) 	imes 9.8}{0.25 	imes 10^{-2}}$ $=130$ pa-s.

An air bubble of radius $1 \mathrm{~cm}$ rises from the bottom portion through a liquid of density $1.5…

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