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

<p>The terminal velocity of a small spherical object moving under the action of gravity through a fluid medium is given by Stokes&#39; Law, which is stated as:</p> <p>$v = \frac{2}{9} \frac{r^2 g (\rho_p - \rho_f)}{\eta}$,</p> <p>where:</p> <ul> <li>$v$ is the velocity of the object (in this case, the air bubble),</li> <li>$r$ is the radius of the object,</li> <li>$g$ is the acceleration due to gravity,</li> <li>$\rho_p$ is the density of the object (negligible in this case, as it&#39;s an air bubble),</li> <li>$\rho_f$ is the density of the fluid (the solution), and</li> <li>$\eta$ is the coefficient of viscosity of the fluid.</li> </ul> <p>Since we are neglecting the density of the air bubble, the formula simplifies to:</p> <p>$v = \frac{2}{9} \frac{r^2 g \rho_f}{\eta}$.</p> <p>Rearranging for $\eta$, we get:</p> <p>$\eta = \frac{2}{9} \frac{r^2 g \rho_f}{v}$.</p> <p>Given that $r = \frac{6 \, \text{mm}}{2} = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m}$, $g = 10 \, \text{ms}^{-2}$, $\rho_f = 1750 \, \text{kg/m}^{3}$, and $v = 0.35 \, \text{cm/s} = 0.35 \times 10^{-2} \, \text{m/s}$, we can substitute these values into the formula to find $\eta$:</p> <p>$\eta = \frac{2}{9} \frac{(3 \times 10^{-3})^2 \times 10 \times 1750}{0.35 \times 10^{-2}} = 10 \, \text{Pas}$.</p> <p>Therefore, the coefficient of viscosity of the solution is $10 \, \text{Pas}$.</p>