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 ne 0`

Answer

To solve the problem, we need to determine the excess pressure inside an air bubble just below the surface of water and compare it with the excess pressure inside a drop of the same radius just outside the surface of water.

### Step-by-Step Solution:

1. **Understanding Excess Pressure in a Bubble**:
- The formula for excess pressure inside a bubble is given by:
\[ P = \frac{4T}{r} \]
- This is because a bubble has two surfaces (the inner and outer surfaces).

2. **Understanding Excess Pressure in a Drop**:
- The formula for excess pressure inside a drop is given by:
\[ P = \frac{2T}{r} \]
- A drop has only one surface (the outer surface).

3. **Identifying the Conditions**:
- In the case of the air bubble just below the surface of water, we consider the pressure exerted by the water on the bubble. The excess pressure \( P_1 \) inside the bubble is:
\[ P_1 = \frac{4T}{r} \]
- In the case of the drop just outside the surface of water, the excess pressure \( P_2 \) inside the drop is:
\[ P_2 = \frac{2T}{r} \]

4. **Comparing the Two Pressures**:
- Since the question states that both the bubble and the drop have the same radius \( r \), we can analyze the two expressions for excess pressure:
- For the bubble:
\[ P_1 = \frac{4T}{r} \]
- For the drop:
\[ P_2 = \frac{2T}{r} \]

5. **Conclusion**:
- The excess pressure \( P_1 \) inside the bubble is greater than the excess pressure \( P_2 \) inside the drop. Therefore, we can conclude that:
\[ P_1 \neq P_2 \]
- However, the problem states that we need to find a relationship between \( P_1 \) and \( P_2 \). Since both depend on the same surface tension \( T \) and radius \( r \), we can express the relationship as:
\[ P_1 = 2P_2 \]

### Final Answer:
The relationship between the excess pressures is:
\[ P_1 = 2P_2 \]