Calculate Viscosity Coefficient from Air Bubble Motion

To calculate the coefficient of viscosity ($\eta$) of the liquid, we use Stokes' Law for the terminal velocity of a spherical body (or air bubble) moving through a viscous fluid.

The magnitude of the upward viscous force at terminal velocity equals the net upward buoyant force acting on the air bubble: $6 \pi \eta r v = \frac{4}{3} \pi r^3 (\rho_l - \rho_a) g$

Since the density of air ($\rho_a$) is negligible compared to the density of the liquid ($\rho_l = 2000 \text{ kg/m}^3$), we can approximate $\rho_l - \rho_a \approx \rho_l$.

Rearranging the expression for the coefficient of viscosity ($\eta$): $\eta = \frac{2}{9} \frac{r^2 \rho_l g}{v}$

Given values in SI units:

• Radius of the bubble, $r = \frac{\text{diameter}}{2} = \frac{2 \text{ mm}}{2} = 1 \text{ mm} = 10^{-3} \text{ m}$
• Density of the liquid, $\rho_l = 2000 \text{ kg/m}^3$
• Acceleration due to gravity, $g = 10 \text{ m/s}^2$
• Terminal speed, $v = 0.5 \text{ cm/s} = 0.5 \times 10^{-2} \text{ m/s} = 5 \times 10^{-3} \text{ m/s}$

Substitute these values into the equation for $\eta$: $\eta = \frac{2}{9} \times \frac{(10^{-3})^2 \times 2000 \times 10}{5 \times 10^{-3}}$

$\eta = \frac{2}{9} \times \frac{10^{-6} \times 20000}{5 \times 10^{-3}}$

$\eta = \frac{2}{9} \times \frac{2 \times 10^{-2}}{5 \times 10^{-3}} = \frac{2}{9} \times 4 = \frac{8}{9} \text{ Pa}\cdot\text{s}$

To convert the viscosity from SI units ($\text{Pa}\cdot\text{s}$) to CGS units ($\text{Poise}$), we use the conversion factor $1 \text{ Pa}\cdot\text{s} = 10 \text{ Poise}$:

$\eta = \frac{8}{9} \times 10 \text{ Poise} = \frac{80}{9} \text{ Poise} \approx 8.8 \text{ Poise}$

Hence, the correct option is B.