An air bubble of diameter 2 cm is allowed to rise through a long cylindrical column of viscous liquid and travels at the rate of `0.21 cm s^(-1) `. If the density of the liquid is `1.47 g cm^(-3)`. find the coefficient of viscosity. Density of air is neglected.

Explanation

To find the coefficient of viscosity (η) for the given problem, we will use the formula for terminal velocity of a sphere moving through a viscous fluid. The formula is:

\[ v = \frac{2}{9} \cdot \frac{r^2 \cdot g}{\eta} \cdot (\rho_{body} - \rho_{liquid}) \]

Where:
- \( v \) = terminal velocity (0.21 cm/s)
- \( r \) = radius of the bubble (diameter = 2 cm, so radius = 1 cm)
- \( g \) = acceleration due to gravity (approximately 980 cm/s²)
- \( \eta \) = coefficient of viscosity (what we want to find)
- \( \rho_{body} \) = density of the body (air bubble, which we neglect, so it is 0)
- \( \rho_{liquid} \) = density of the liquid (1.47 g/cm³)

### Step-by-Step Solution:

1. **Convert the diameter to radius**:
\[ \text{Diameter} = 2 \text{ cm} \implies \text{Radius} = \frac{2}{2} = 1 \text{ cm} \]

2. **Substitute known values into the terminal velocity equation**:
\[ 0.21 = \frac{2}{9} \cdot \frac{(1)^2 \cdot 980}{\eta} \cdot (0 - 1.47) \]

3. **Simplify the equation**:
\[ 0.21 = \frac{2}{9} \cdot \frac{980}{\eta} \cdot (-1.47) \]
\[ 0.21 = -\frac{2 \cdot 980 \cdot 1.47}{9 \eta} \]

4. **Rearranging to find η**:
\[ 0.21 \cdot 9 \eta = -2 \cdot 980 \cdot 1.47 \]
\[ 1.89 \eta = -2 \cdot 980 \cdot 1.47 \]
\[ \eta = \frac{-2 \cdot 980 \cdot 1.47}{1.89} \]

5. **Calculate η**:
\[ \eta = \frac{-2892.6}{1.89} \approx 1524.44 \text{ g/cm·s} \]

### Final Answer:
The coefficient of viscosity (η) is approximately **1524.44 g/cm·s**.