The gas bubble is rising through the liquid of densily $1.75 \mathrm{g} / \mathrm{cm}^{3}$ with constant speed. The acting forces on bubble are


Buoyancy force of liquid $=F_{0}$ (in upward direction) Viscous force due to liquid = $F_{V}$, (in downward direction)
$$
\therefore \quad F_{V}=F_{b}
$$
For free falling in a viscous medium
$$
F_{V}=6 \pi \eta {V_{T}}
$$
$\eta$ = viscosity coefficient of the liquid
$r=$ radius of bubble
$$
\begin{aligned}
F_{D} &=m g=v \times \rho g \\
&=\frac{4}{3} \pi r^{3} \rho_{e} \cdot g
\end{aligned}
$$
$\begin{aligned} 6 \pi \eta r V_{t} &=\frac{4}{3} \pi r^{3} \rho_{e} \cdot g \\ \eta &=\frac{24 \pi \cdot r^{3} \rho_{e} \cdot g}{3 \times 6 \pi rV_{T}}=\frac{2}{9} \cdot \frac{r^{2} \rho_{e} \cdot g}{v_{T}} \end{aligned}$
$\begin{aligned} D &=2 \mathrm{cm}, r=1 \mathrm{cm}=1 \times 10^{-2} \mathrm{m} \\ \rho_{e} &=1.75 \mathrm{g} / \mathrm{cm}^{3}=1.75 \times 10^{* 3} \mathrm{kg} / \mathrm{m}^{3} \\ g &=10 \mathrm{m} / \mathrm{s}^{2} \\ \mathrm{v}_{T} &=\text { terminal velocity } \\ &=0.35 \times 10^{-2} \mathrm{m} / \mathrm{s} \\ \eta &=\frac{2}{9} \times \frac{\left(10^{-2}\right)^{2} \times 1.75 \times 10^{+3} \times 10}{0.35 \times 10^{-2}} \\ &=\frac{2}{9} \times \frac{10^{-4} \times 1.75 \times 10^{3} \times 10}{0.35 \times 10^{-2}} \\ &=\frac{2 \times 175 \times 10 \times 10}{9 \times 35} \times 10 \text { poise } \\ &=1089 \text { poise } \end{aligned}$