A soap bubble A of radius 0.03 m and another bubble B of radius 0.04 m are brought together, so that the combined bubble has a common interface of radius r, then the value of r is

Options

A. `0.24 m`
B. `0.48 m`
C. `0.12 m`
D. None of these

Explanation

To solve the problem of finding the radius \( r \) of the common interface of two soap bubbles with given radii, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Radius of bubble A, \( r_1 = 0.03 \, \text{m} \)
- Radius of bubble B, \( r_2 = 0.04 \, \text{m} \)

2. **Understand the Concept of Excess Pressure**:
- The excess pressure inside a soap bubble is given by the formula:
\[ \Delta P = \frac{4S}{r} \]
where \( S \) is the surface tension and \( r \) is the radius of the bubble.

3. **Write the Excess Pressure for Both Bubbles**:
- For bubble A:
\[ \Delta P_1 = \frac{4S}{r_1} = \frac{4S}{0.03} \]
- For bubble B:
\[ \Delta P_2 = \frac{4S}{r_2} = \frac{4S}{0.04} \]

4. **Set Up the Pressure Difference Equation**:
- When the two bubbles combine, the pressure difference between them can be expressed as:
\[ \Delta P' = \Delta P_1 - \Delta P_2 = \frac{4S}{r} \]
- Therefore, we have:
\[ \Delta P_1 - \Delta P_2 = \frac{4S}{r} \]

5. **Substituting the Excess Pressure Values**:
- Substitute the expressions for \( \Delta P_1 \) and \( \Delta P_2 \):
\[ \frac{4S}{0.03} - \frac{4S}{0.04} = \frac{4S}{r} \]

6. **Simplifying the Equation**:
- Factor out \( 4S \):
\[ 4S \left(\frac{1}{0.03} - \frac{1}{0.04}\right) = \frac{4S}{r} \]
- Cancel \( 4S \) from both sides (assuming \( S \neq 0 \)):
\[ \frac{1}{0.03} - \frac{1}{0.04} = \frac{1}{r} \]

7. **Finding a Common Denominator**:
- The common denominator for \( 0.03 \) and \( 0.04 \) is \( 0.12 \):
\[ \frac{4}{0.12} - \frac{3}{0.12} = \frac{1}{r} \]
- This simplifies to:
\[ \frac{1}{0.12} = \frac{1}{r} \]

8. **Solving for \( r \)**:
- Therefore, we find:
\[ r = 0.12 \, \text{m} \]

### Final Answer:
The radius \( r \) of the common interface is \( 0.12 \, \text{m} \).

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