An air bubble of radius $$1\,cm$$  rises with terminal velocity $$0.21\,cm/s$$   in liquid column. If the density of liquid is $$1.47 \times {10^3}\,kg/{m^3}.$$    Then the value of coefficient of viscosity of liquid ignoring the density of air, will be

Solution :
Using the formula of the terminal velocity of a body falling through a viscous medium,
$$V = \frac{{2{r^2}\left( {\rho - \sigma } \right)}}{{9\eta }} \Rightarrow \eta = \frac{{2{r^2}\left( {\rho - \sigma } \right)g}}{{9v}}$$
Where $$\rho $$ is the density of material of body and $$\sigma $$ is the density of medium.
In case of the air bubble $$\rho = 1$$  and $$\sigma = 1.47 \times {10^3}\,kg/ms$$     and the air bubble rises up.
$$\eqalign{ & \eta = \frac{{2{r^2}\sigma g}}{{9V}} = \frac{{2 \times {{\left( {{{10}^{ - 2}}} \right)}^2} \times 1.47 \times {{10}^3} \times 9.8}}{{9 \times 0.21 \times {{10}^{ - 2}}}} \cr & = 1.52 \times {10^3}\,{\text{decapoise}} \cr & = 1.52 \times {10^4}\,{\text{poise}} \cr} $$