The excess pressure inside an air bubble of radius `r` just below the surface of water is `P_(1)`. The excess pressure inside a drop of the same radius just outside the surface is `P_(2)`. If `T` is surface tension then

Options

A. `P_(1)=2P_(2)`
B. `P_(1)=P_(2)`
C. `P_(2)=2P_(1)`
D. `P_(2)=0,P_(1)ne0`

Explanation

To solve the problem, we need to determine the relationship between the excess pressure inside an air bubble (P1) and the excess pressure inside a drop of the same radius (P2) using the concept of surface tension.

### Step-by-Step Solution:

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

2. **Calculate Excess Pressure Inside the Air Bubble (P1)**:
- For an air bubble just below the surface of water, the excess pressure \( P_1 \) is:
\[ P_1 = \frac{4T}{r} \]

3. **Calculate Excess Pressure Inside the Drop (P2)**:
- For a liquid drop just outside the surface, the excess pressure \( P_2 \) is given by the same formula:
\[ P_2 = \frac{2T}{r} \]
(Note: The formula for a drop is \( \frac{2T}{r} \) because the pressure difference across the surface of a drop is half that of a bubble due to the curvature being different.)

4. **Relate P1 and P2**:
- Now we can compare \( P_1 \) and \( P_2 \):
\[ P_1 = \frac{4T}{r} \]
\[ P_2 = \frac{2T}{r} \]
- From this, we can see that:
\[ P_1 = 2P_2 \]

5. **Conclusion**:
- Therefore, the relationship between the excess pressures is:
\[ P_1 = 2P_2 \]
- This means that the correct option is that the excess pressure inside the air bubble (P1) is twice that of the excess pressure inside the drop (P2).

### Final Answer:
The correct relationship is:
\[ P_1 = 2P_2 \]