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\(^2\))

The terminal velocity of an air bubble rising in a liquid is given by Stokes' law:

\(v = \dfrac{2}{9} \dfrac{r^2 \rho g}{\eta}\)

Where:
\(r = \dfrac{D}{2} = 1 \text{ mm} = 10^{-3} \text{ m}\)
\(\rho = 2000 \text{ kg/m}^3\)
\(g = 10 \text{ m/s}^2\)
\(v = 0.5 \text{ cm/s} = 5 \times 10^{-3} \text{ m/s}\)

Rearranging the formula for the coefficient of viscosity \(\eta\):

\(\eta = \dfrac{2}{9} \dfrac{r^2 \rho g}{v}\)

Substituting the given values:

\(\eta = \dfrac{2}{9} \times \dfrac{(10^{-3})^2 \times 2000 \times 10}{5 \times 10^{-3}}\)

\(\eta = \dfrac{2}{9} \times \dfrac{10^{-6} \times 20000}{5 \times 10^{-3}}\)

\(\eta = \dfrac{2}{9} \times \dfrac{2 \times 10^{-2}}{5 \times 10^{-3}}\)

\(\eta = \dfrac{2}{9} \times 4 = \dfrac{8}{9} \text{ Pa}\cdot\text{s}\)

To convert the viscosity from SI units (Pa\(\cdot\)s) to CGS units (Poise), we multiply by \(10\):

\(\eta = \dfrac{8}{9} \times 10 = \dfrac{80}{9} \text{ Poise}\)

\(\eta \approx 8.88 \text{ Poise}\)

The closest given option is \(8.8\).

Answer: \(8.8\)