Internal pressure inside a liquid drop of radius r and surface tension T is

Options

A. `(2T)/r-P_(0)`
B. `(4T)/r + P_(0)`
C. `P_(0)+(2T)/r`
D. `T/(4r)-P_(0)`

Answer

To find the internal pressure inside a liquid drop of radius \( r \) and surface tension \( T \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Pressure Difference**:
The internal pressure \( P \) inside the liquid drop and the external pressure \( P_0 \) outside the drop create a pressure difference. This difference is referred to as the excess pressure.

2. **Formula for Excess Pressure**:
For a liquid drop, the excess pressure \( \Delta P \) is given by the formula:
\[ \Delta P = P - P_0 = \frac{2T}{r} \]
where:
- \( T \) is the surface tension of the liquid,
- \( r \) is the radius of the liquid drop.

3. **Relate Internal and External Pressure**:
From the formula for excess pressure, we can express the internal pressure \( P \) in terms of the external pressure \( P_0 \):
\[ P = P_0 + \Delta P \]
Substituting the expression for excess pressure:
\[ P = P_0 + \frac{2T}{r} \]

4. **Final Expression for Internal Pressure**:
Therefore, the internal pressure inside the liquid drop can be expressed as:
\[ P = P_0 + \frac{2T}{r} \]

### Conclusion:
The internal pressure inside a liquid drop of radius \( r \) and surface tension \( T \) is given by:
\[ P = P_0 + \frac{2T}{r} \]