What will be the surface tension of soap solution if the excess pressure of air in soap bubble of radius 0.4 cm is 10 mm of water column?

Answer

To find the surface tension of the soap solution given the excess pressure in a soap bubble, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Radius of the soap bubble, \( R = 0.4 \, \text{cm} = 0.4 \times 10^{-2} \, \text{m} \)
- Excess pressure, \( \Delta P = 10 \, \text{mm of water column} \)

2. **Convert Excess Pressure to Pascals**:
- We know that \( 1 \, \text{mm of water} \) corresponds to a pressure of \( \frac{1.01 \times 10^5 \, \text{Pa}}{760 \times 13.6} \).
- Therefore, to convert \( 10 \, \text{mm} \) of water column to Pascals:
\[ \Delta P = 10 \, \text{mm} \times \left( \frac{1.01 \times 10^5 \, \text{Pa}}{760 \times 13.6} \right) \]

3. **Calculate the Value of \( \Delta P \)**:
- First, calculate \( 760 \times 13.6 \):
\[ 760 \times 13.6 = 10376 \, \text{Pa} \]
- Now substituting back into the equation for \( \Delta P \):
\[ \Delta P = 10 \times \left( \frac{1.01 \times 10^5}{10376} \right) \approx 97.2 \, \text{Pa} \]

4. **Use the Formula for Excess Pressure in a Soap Bubble**:
- The formula relating excess pressure to surface tension \( T \) is given by:
\[ \Delta P = \frac{4T}{R} \]
- Rearranging this formula to solve for surface tension \( T \):
\[ T = \frac{\Delta P \cdot R}{4} \]

5. **Substitute the Values**:
- Now substitute \( \Delta P \) and \( R \) into the equation:
\[ T = \frac{97.2 \, \text{Pa} \cdot (0.4 \times 10^{-2} \, \text{m})}{4} \]

6. **Calculate Surface Tension**:
- Performing the multiplication:
\[ T = \frac{97.2 \cdot 0.4 \times 10^{-2}}{4} = \frac{38.88 \times 10^{-2}}{4} = 9.72 \times 10^{-2} \, \text{N/m} \]

7. **Final Result**:
- Therefore, the surface tension of the soap solution is approximately:
\[ T \approx 9.72 \times 10^{-2} \, \text{N/m} \]