An air bubble of diameter 6 mm rises steadily through a solution of density 1750 kg m$$^{-3}$$ at the rate of 0.35 cm s$$^{-1}$$. The co-efficient of viscosity of the solution (neglect density of air) is ________ Pas (given, $$g = 10$$ m s$$^{-2}$$).

We need to find the coefficient of viscosity using Stokes' law for a rising air bubble.

At terminal velocity, the net force is zero. For a bubble rising in a liquid (neglecting the density of air), the buoyant force equals the viscous drag:

$$6\pi \eta r v = \frac{4}{3}\pi r^3 \rho g$$

Solve for $$\eta$$: $$\eta = \frac{2r^2 \rho g}{9v}$$

Substitute the values: $$r = \frac{6}{2} = 3$$ mm $$= 3 \times 10^{-3}$$ m

$$\rho = 1750$$ kg m$$^{-3}$$

$$g = 10$$ m s$$^{-2}$$

$$v = 0.35$$ cm s$$^{-1}$$ $$= 3.5 \times 10^{-3}$$ m s$$^{-1}$$

$$\eta = \frac{2 \times (3 \times 10^{-3})^2 \times 1750 \times 10}{9 \times 3.5 \times 10^{-3}}$$

$$= \frac{2 \times 9 \times 10^{-6} \times 17500}{9 \times 3.5 \times 10^{-3}}$$

$$= \frac{2 \times 9 \times 17500 \times 10^{-6}}{31.5 \times 10^{-3}}$$

$$= \frac{315000 \times 10^{-6}}{31.5 \times 10^{-3}} = \frac{0.315}{0.0315} = 10 \text{ Pa s}$$

The correct answer is 10 Pa s.