With what terminal velocity will an air bubble `0.8 mm` in diameter rise in a liquid of viscosity `0.15 N-s//m^(2)` and specific gravity `0.9`? Density of air is `1.293 kg//m^(3)`.

Explanation

To find the terminal velocity of an air bubble rising in a liquid, we can use the formula for terminal velocity (V_t):

\[ V_t = \frac{2}{9} \cdot \frac{R^2 (\rho - \rho_s) g}{\eta} \]

Where:
- \( R \) = radius of the bubble
- \( \rho \) = density of the fluid (liquid)
- \( \rho_s \) = density of the gas (air)
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( \eta \) = viscosity of the liquid

### Step 1: Convert the diameter to radius
The diameter of the bubble is given as \( 0.8 \, \text{mm} \).

\[ \text{Diameter} = 0.8 \, \text{mm} = 0.8 \times 10^{-3} \, \text{m} = 8 \times 10^{-4} \, \text{m} \]

Now, calculate the radius \( R \):

\[ R = \frac{D}{2} = \frac{8 \times 10^{-4}}{2} = 4 \times 10^{-4} \, \text{m} \]

### Step 2: Calculate the density of the liquid
The specific gravity of the liquid is given as \( 0.9 \). The density of water is approximately \( 1000 \, \text{kg/m}^3 \), so the density of the liquid \( \rho \) can be calculated as:

\[ \rho = \text{Specific Gravity} \times \text{Density of Water} = 0.9 \times 1000 \, \text{kg/m}^3 = 900 \, \text{kg/m}^3 \]

### Step 3: Substitute the values into the terminal velocity formula
Now we have:
- \( R = 4 \times 10^{-4} \, \text{m} \)
- \( \rho = 900 \, \text{kg/m}^3 \)
- \( \rho_s = 1.293 \, \text{kg/m}^3 \)
- \( g = 9.81 \, \text{m/s}^2 \)
- \( \eta = 0.15 \, \text{N-s/m}^2 \)

Substituting these values into the terminal velocity formula:

\[ V_t = \frac{2}{9} \cdot \frac{(4 \times 10^{-4})^2 (900 - 1.293) \cdot 9.81}{0.15} \]

### Step 4: Calculate \( V_t \)
Calculating \( (4 \times 10^{-4})^2 \):

\[ (4 \times 10^{-4})^2 = 16 \times 10^{-8} \, \text{m}^2 \]

Calculating \( (900 - 1.293) \):

\[ 900 - 1.293 = 898.707 \, \text{kg/m}^3 \]

Now substituting back into the equation:

\[ V_t = \frac{2}{9} \cdot \frac{16 \times 10^{-8} \cdot 898.707 \cdot 9.81}{0.15} \]

Calculating the numerator:

\[ 16 \times 10^{-8} \cdot 898.707 \cdot 9.81 \approx 1.76 \times 10^{-6} \]

Now substituting this into the equation:

\[ V_t = \frac{2}{9} \cdot \frac{1.76 \times 10^{-6}}{0.15} \]

Calculating the final value:

\[ V_t \approx \frac{2}{9} \cdot 1.1733 \times 10^{-5} \approx 2.61 \times 10^{-6} \, \text{m/s} \]

### Final Result
The terminal velocity \( V_t \) of the air bubble is approximately:

\[ V_t \approx 9.37 \times 10^{-7} \, \text{m/s} \]