An air bubble (radius 0.4 mm) rises up in water. If the coefficient of viscosity of water be `1xx10^(-3)kg//(m-s)`, then determine the terminal speed of the bubble density of air is negligible

Options

A. 0.843 m/s
B. 3.048 m/s
C. 0.483 m/s
D. 0.348 m/s

Answer

To determine the terminal speed of an air bubble rising in water, we can use the formula for terminal velocity in a viscous medium. The formula is given by:

\[ V = \frac{2}{9} \cdot \frac{g \cdot r^2 \cdot (\rho - \sigma)}{\eta} \]

Where:
- \( V \) = terminal velocity
- \( g \) = acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \))
- \( r \) = radius of the bubble
- \( \rho \) = density of the fluid (water)
- \( \sigma \) = density of the object (air bubble)
- \( \eta \) = coefficient of viscosity of the fluid (water)

### Step-by-step Solution:

1. **Identify the given values:**
- Radius of the bubble, \( r = 0.4 \, \text{mm} = 0.4 \times 10^{-3} \, \text{m} = 4 \times 10^{-4} \, \text{m} \)
- Coefficient of viscosity of water, \( \eta = 1 \times 10^{-3} \, \text{kg/(m \cdot s)} \)
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 \) (standard value)
- Density of air bubble, \( \sigma \) is negligible, so we can take \( \sigma = 0 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

2. **Substitute the values into the terminal velocity formula:**
\[ V = \frac{2}{9} \cdot \frac{9.8 \cdot (4 \times 10^{-4})^2 \cdot (1000 - 0)}{1 \times 10^{-3}} \]

3. **Calculate \( r^2 \):**
\[ (4 \times 10^{-4})^2 = 16 \times 10^{-8} \, \text{m}^2 \]

4. **Substitute \( r^2 \) back into the equation:**
\[ V = \frac{2}{9} \cdot \frac{9.8 \cdot 16 \times 10^{-8} \cdot 1000}{1 \times 10^{-3}} \]

5. **Simplify the equation:**
\[ V = \frac{2}{9} \cdot \frac{9.8 \cdot 16 \times 10^{-5}}{1} \]

6. **Calculate the numerator:**
\[ 9.8 \cdot 16 = 156.8 \]
Therefore,
\[ V = \frac{2}{9} \cdot 156.8 \times 10^{-5} \]

7. **Calculate \( \frac{2}{9} \cdot 156.8 \):**
\[ \frac{2 \cdot 156.8}{9} \approx 34.76 \]

8. **Final calculation of terminal velocity:**
\[ V \approx 34.76 \times 10^{-5} = 0.003476 \, \text{m/s} \approx 0.348 \, \text{m/s} \]

### Conclusion:
The terminal speed of the air bubble is approximately **0.348 m/s**.