A soap bubble of diameter `8 mm` is formed in air. The surface tension of liquid is `3 "dyne"//cm`. Find the excess pressure (`"dyne" //"cm"^(2)`) inside the soap bubble ?

Explanation

To find the excess pressure inside a soap bubble, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Diameter of the soap bubble, \( d = 8 \, \text{mm} \)
- Surface tension of the liquid, \( T = 3 \, \text{dyne/cm} \)

2. **Convert Diameter to Centimeters**:
- Since \( 1 \, \text{mm} = 0.1 \, \text{cm} \), we convert the diameter:
\[ d = 8 \, \text{mm} = 8 \times 0.1 = 0.8 \, \text{cm} \]

3. **Calculate the Radius of the Bubble**:
- The radius \( r \) is half of the diameter:
\[ r = \frac{d}{2} = \frac{0.8 \, \text{cm}}{2} = 0.4 \, \text{cm} \]

4. **Use the Formula for Excess Pressure in a Soap Bubble**:
- The formula for excess pressure \( P \) inside a soap bubble is given by:
\[ P = \frac{4T}{r} \]

5. **Substitute the Values into the Formula**:
- Now substitute \( T = 3 \, \text{dyne/cm} \) and \( r = 0.4 \, \text{cm} \):
\[ P = \frac{4 \times 3 \, \text{dyne/cm}}{0.4 \, \text{cm}} = \frac{12 \, \text{dyne/cm}}{0.4 \, \text{cm}} \]

6. **Calculate the Excess Pressure**:
- Perform the division:
\[ P = 12 \div 0.4 = 30 \, \text{dyne/cm}^2 \]

### Final Answer:
The excess pressure inside the soap bubble is \( 30 \, \text{dyne/cm}^2 \).

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