If an air bubble of diameter 2 mm rises steadily through a liquid of density 2000 kg/m$$^3$$ at a rate of 0.5 cm/s, then the coefficient of viscosity of liquid is _________ Poise. (Take $$g = 10$$ m/s$$^2$$)

At steady (terminal) speed the upward buoyant force on the bubble is exactly balanced by the downward viscous drag exerted by the liquid.

Buoyant force (because the weight of air in the bubble is negligible) is equal to the weight of the displaced liquid:
$$F_{\text{buoy}} = \rho V g = \rho\left(\frac{4}{3}\pi r^{3}\right)g$$

For creeping (Stokes-flow) conditions, the viscous drag on a sphere of radius $$r$$ moving with speed $$v$$ in a liquid of viscosity $$\eta$$ is
$$F_{\text{drag}} = 6\pi \eta r v$$

At terminal velocity, $$F_{\text{drag}} = F_{\text{buoy}}$$, hence
$$6\pi \eta r v = \rho\left(\frac{4}{3}\pi r^{3}\right)g$$

Simplifying (cancel $$\pi$$ and one factor of $$r$$):
$$6 \eta v = \frac{4}{3}\rho r^{2} g$$
$$\eta = \frac{2}{9}\,\frac{\rho g r^{2}}{v}$$ $$-(1)$$

Insert the data:
Radius     $$r = \frac{2\text{ mm}}{2} = 1\text{ mm} = 1\times10^{-3}\text{ m}$$
Speed     $$v = 0.5\text{ cm s}^{-1} = 5\times10^{-3}\text{ m s}^{-1}$$
Density  $$\rho = 2000\text{ kg m}^{-3}$$
Acceleration $$g = 10\text{ m s}^{-2}$$

Compute the numerator of $$(1)$$:
$$\rho g r^{2} = 2000 \times 10 \times (1\times10^{-3})^{2} = 20000 \times 10^{-6} = 2\times10^{-2}$$

Divide by $$v$$:
$$\frac{\rho g r^{2}}{v} = \frac{2\times10^{-2}}{5\times10^{-3}} = 4$$

Finally, from $$(1)$$:
$$\eta = \frac{2}{9} \times 4 = \frac{8}{9}\text{ Pa·s} \approx 0.888\text{ Pa·s}$$

Conversion to Poise (1 Pa·s = 10 Poise):
$$\eta \approx 0.888 \times 10 = 8.88\text{ Poise}$$

Rounded to two significant figures, $$\eta \approx 8.8\text{ Poise}$$.

Option B which is: $$8.8$$