The excess pressure due to surface tension in a spherical liquid drop of radius r is directly proportional to

Options

A. r
B. `r^(2)`
C. `r^(-1)`
D. `r^(-2)`

Explanation

To solve the question regarding the excess pressure due to surface tension in a spherical liquid drop of radius \( r \), we can follow these steps:

### Step 1: Understand the concept of excess pressure due to surface tension
The excess pressure (\( \Delta P \)) inside a liquid drop due to surface tension is a result of the surface tension acting at the interface between the liquid and the surrounding medium.

### Step 2: Use the formula for excess pressure
The formula for the excess pressure inside a spherical drop is given by:

\[ \Delta P = \frac{4T}{r} \]

where:
- \( \Delta P \) is the excess pressure,
- \( T \) is the surface tension of the liquid,
- \( r \) is the radius of the drop.

### Step 3: Analyze the relationship
From the formula, we can see that the excess pressure (\( \Delta P \)) is directly proportional to the surface tension (\( T \)) and inversely proportional to the radius (\( r \)) of the drop.

### Step 4: Conclusion
Thus, we can conclude that the excess pressure due to surface tension in a spherical liquid drop of radius \( r \) is directly proportional to the surface tension \( T \) and inversely proportional to the radius \( r \).

### Final Answer
The excess pressure due to surface tension in a spherical liquid drop of radius \( r \) is directly proportional to the surface tension \( T \) and inversely proportional to the radius \( r \).

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