The air pressure inside a soap bubbles of radius R exceeds the out side air pressure by 10 Pa. By how much will the pressure inside a bubble of radius 2R exceed the out side air pressure.

Options

A. 20 Pa
B. 40 Pa
C. 2.4 Pa
D. 5 Pa

Explanation

To solve the problem, we need to determine the excess pressure inside a soap bubble of radius \(2R\) given that the excess pressure inside a soap bubble of radius \(R\) is \(10 \, \text{Pa}\).

### Step-by-Step Solution:

1. **Understanding Excess Pressure in a Soap Bubble**:
The excess pressure \(P\) inside a soap bubble is given by the formula:
\[ P = \frac{4S}{R} \]
where \(S\) is the surface tension of the soap solution, and \(R\) is the radius of the bubble.

2. **Given Information**:
For the bubble of radius \(R\), the excess pressure is:
\[ P_1 = \frac{4S}{R} = 10 \, \text{Pa} \]

3. **Finding Surface Tension**:
From the equation above, we can express the surface tension \(S\) in terms of the given excess pressure:
\[ 4S = 10R \implies S = \frac{10R}{4} = \frac{5R}{2} \]

4. **Calculating Excess Pressure for Radius \(2R\)**:
Now, we need to find the excess pressure \(P_2\) for the bubble of radius \(2R\):
\[ P_2 = \frac{4S}{2R} \]

5. **Substituting for Surface Tension**:
Substitute \(S\) into the equation for \(P_2\):
\[ P_2 = \frac{4 \left(\frac{5R}{2}\right)}{2R} \]

6. **Simplifying the Expression**:
Simplifying the equation:
\[ P_2 = \frac{4 \cdot \frac{5R}{2}}{2R} = \frac{20R}{4R} = 5 \, \text{Pa} \]

7. **Conclusion**:
Therefore, the excess pressure inside the bubble of radius \(2R\) is:
\[ P_2 = 5 \, \text{Pa} \]

### Final Answer:
The pressure inside the bubble of radius \(2R\) exceeds the outside air pressure by \(5 \, \text{Pa}\).

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