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 following formula:

\[ v_t = \frac{2}{9} \cdot r^2 \cdot (ρ - σ) \cdot g \div η \]

Where:
- \( v_t \) = terminal velocity
- \( r \) = radius of the bubble
- \( ρ \) = density of the liquid
- \( σ \) = density of the air bubble
- \( g \) = acceleration due to gravity
- \( η \) = viscosity of the liquid

### Step 1: Convert the diameter to radius
The diameter of the bubble is given as \( 0.8 \, \text{mm} \). To find the radius, we divide the diameter by 2.

\[ r = \frac{0.8 \, \text{mm}}{2} = 0.4 \, \text{mm} = 0.4 \times 10^{-3} \, \text{m} \]

### Step 2: Convert specific gravity to density
The specific gravity of the liquid is given as \( 0.9 \). To convert this to density (in kg/m³), we multiply by the density of water (\( 1000 \, \text{kg/m}^3 \)).

\[ σ = 0.9 \times 1000 \, \text{kg/m}^3 = 900 \, \text{kg/m}^3 \]

### Step 3: Write down the known values
- Density of air (\( ρ \)) = \( 1.293 \, \text{kg/m}^3 \)
- Viscosity of the liquid (\( η \)) = \( 0.15 \, \text{N-s/m}^2 \)
- Acceleration due to gravity (\( g \)) = \( 9.8 \, \text{m/s}^2 \)

### Step 4: Substitute the values into the formula
Now we can substitute all the values into the terminal velocity formula.

\[ v_t = \frac{2}{9} \cdot (0.4 \times 10^{-3})^2 \cdot (1.293 - 900) \cdot 9.8 \div 0.15 \]

### Step 5: Calculate the terminal velocity
Calculating each part step-by-step:

1. Calculate \( r^2 \):
\[ (0.4 \times 10^{-3})^2 = 0.16 \times 10^{-6} \, \text{m}^2 \]

2. Calculate \( ρ - σ \):
\[ 1.293 - 900 = -898.707 \, \text{kg/m}^3 \]

3. Now substitute into the formula:
\[ v_t = \frac{2}{9} \cdot 0.16 \times 10^{-6} \cdot (-898.707) \cdot 9.8 \div 0.15 \]

4. Calculate the numerator:
\[ 2 \cdot 0.16 \times 10^{-6} \cdot (-898.707) \cdot 9.8 = -2.944 \times 10^{-3} \]

5. Divide by \( 0.15 \):
\[ v_t = \frac{-2.944 \times 10^{-3}}{0.15} = -0.0196 \, \text{m/s} \]

### Step 6: Convert to cm/s
To convert from m/s to cm/s, multiply by 100:
\[ v_t = -0.0196 \times 100 = -1.96 \, \text{cm/s} \]

### Final Result
The terminal velocity of the air bubble is approximately \( -1.96 \, \text{cm/s} \). The negative sign indicates the direction of motion (upwards).

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