Explanation

(d) Let a bubble of radius $r$ and density$\rho$ is rising up in a liquid whose density is$\sigma$ and coefficient of viscosity $ɳ$.
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Then effective force acting downward
$= V(\rho - \sigma)g = \frac{4}{3}{\pi r}^{3}(\rho - \sigma)g$
Viscous force acting upward$= 6\pi ɳ\ rv.$
Since bubble is moving up with constant velocity $v$, there is no acceleration in it, the net force acting on it must be zero.
$\therefore\ 6\pi ɳ\ rv = \frac{4}{3}\ {\pi r}^{3}(\rho - \sigma)g$
$\Longrightarrow \ ɳ = \frac{2}{9}\frac{r^{2}(\rho - \sigma)}{v}g$
$Given,\ v = - 2 \times 10^{- 3}\ {ms}^{- 1},\ r = 10^{- 2}m$
$\ \ \ \rho = 0,\sigma = 1.5 \times 10^{3}\ {kgm}^{- 3},\ g = 10^{- 2}$
$\therefore\ ɳ = \frac{2 \times (10^{- 2})^{2} \times ( - 1.5 \times 10^{3}) \times 10}{9 \times \left( - 2 \times 10^{- 3} \right)}$
$\ \ \ \ = \frac{3}{18 \times 10^{- 3}} = \frac{1}{6} \times 10^{3}$
$\ \ \ \ \ = 0.167 \times 10^{3} = 167\ units.$