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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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The correct answer is:
130
$v=\frac{2}{9} \frac{r^2 \rho g}{\eta}$
$\Rightarrow \eta=\frac{2}{9} \cdot \frac{r^2 \rho g}{v}$
$=\frac{2}{9} \frac{\left(1 \times 10^{-2}\right)^2 \times\left(1.5 \times 10^3\right) \times 9.8}{0.25 \times 10^{-2}}$
$=130$ pa-s.
$\Rightarrow \eta=\frac{2}{9} \cdot \frac{r^2 \rho g}{v}$
$=\frac{2}{9} \frac{\left(1 \times 10^{-2}\right)^2 \times\left(1.5 \times 10^3\right) \times 9.8}{0.25 \times 10^{-2}}$
$=130$ pa-s.
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