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. ` 0.84 m//s `
B. ` 3.5 m//s `
C. `0.48 m//s `
D. `0.35 m//s `

Explanation

To determine the terminal speed of an air bubble rising in water, we can use the formula for terminal velocity derived from Stokes' law. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We are given:
- Radius of the air bubble, \( r = 0.4 \, \text{mm} = 0.4 \times 10^{-3} \, \text{m} \)
- Coefficient of viscosity of water, \( \eta = 1 \times 10^{-3} \, \text{kg/(m \cdot s)} \)
- Density of water, \( \sigma = 1000 \, \text{kg/m}^3 \) (approximately)
- Density of air is negligible.

### Step 2: Write the Formula for Terminal Velocity
The formula for terminal velocity \( V_T \) of a sphere moving through a viscous fluid is given by:
\[ V_T = \frac{2}{9} \cdot \frac{(r^2)(\sigma - \rho)g}{\eta} \]
Where:
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \))
- \( \rho \) is the density of the bubble (which is negligible in this case, so we can consider it as \( 0 \))

### Step 3: Substitute the Values into the Formula
Since the density of air is negligible, we can simplify the equation:
\[ V_T = \frac{2}{9} \cdot \frac{(r^2)(\sigma)g}{\eta} \]
Now substituting the values:
- \( r = 0.4 \times 10^{-3} \, \text{m} \)
- \( \sigma = 1000 \, \text{kg/m}^3 \)
- \( g = 10 \, \text{m/s}^2 \)
- \( \eta = 1 \times 10^{-3} \, \text{kg/(m \cdot s)} \)

Calculating \( r^2 \):
\[ r^2 = (0.4 \times 10^{-3})^2 = 0.16 \times 10^{-6} \, \text{m}^2 = 1.6 \times 10^{-7} \, \text{m}^2 \]

Now substituting these into the terminal velocity formula:
\[ V_T = \frac{2}{9} \cdot \frac{(1.6 \times 10^{-7})(1000)(10)}{1 \times 10^{-3}} \]

### Step 4: Simplify the Expression
Calculating the numerator:
\[ (1.6 \times 10^{-7})(1000)(10) = 1.6 \times 10^{-7} \times 10000 = 1.6 \times 10^{-3} \]

Now substituting back into the formula:
\[ V_T = \frac{2}{9} \cdot \frac{1.6 \times 10^{-3}}{1 \times 10^{-3}} = \frac{2}{9} \cdot 1.6 = \frac{3.2}{9} \]

### Step 5: Calculate the Final Value
Calculating \( \frac{3.2}{9} \):
\[ V_T \approx 0.3556 \, \text{m/s} \]

### Final Answer
The terminal speed of the bubble is approximately:
\[ V_T \approx 0.36 \, \text{m/s} \]