Excess pressure in a soap bubble of radius r is proportional to:

Options

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

Explanation

To determine the relationship between excess pressure in a soap bubble and its radius, we can use the formula for excess pressure in a soap bubble. The excess pressure \( P \) inside a soap bubble is given by the equation:

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

where:
- \( P \) is the excess pressure,
- \( T \) is the surface tension of the soap solution,
- \( r \) is the radius of the bubble.

### Step-by-Step Solution:

1. **Understanding the Formula**: The formula \( P = \frac{4T}{r} \) indicates that the excess pressure \( P \) is inversely proportional to the radius \( r \) of the soap bubble.

2. **Identifying Proportionality**: From the equation, we can see that as the radius \( r \) increases, the excess pressure \( P \) decreases. This means that:
- If \( r \) increases, \( P \) decreases.
- If \( r \) decreases, \( P \) increases.

3. **Expressing Proportionality**: We can express the proportionality as:
\[ P \propto \frac{1}{r} \]
This means that the excess pressure is proportional to the reciprocal of the radius.

4. **Evaluating the Options**: Given the options in the question:
- \( \frac{1}{r} \) (this is correct),
- \( \frac{1}{r^2} \) (this is incorrect),
- \( r \) (this is incorrect),
- \( r^2 \) (this is incorrect).

5. **Conclusion**: The correct answer is that the excess pressure in a soap bubble of radius \( r \) is proportional to \( \frac{1}{r} \).

### Final Answer:
The excess pressure in a soap bubble of radius \( r \) is proportional to \( \frac{1}{r} \).