The excess pressure inside a liquid drop of water will be maximum for the one having radius

Options

A. R
B. 2 R
C. `(R)/(2)`
D. `(R)/(4)`

Answer

To solve the problem regarding the excess pressure inside a liquid drop of water and its dependence on the radius, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Excess Pressure**: The excess pressure (\( \Delta P \)) inside a liquid drop is given by the formula:
\[ \Delta P = \frac{2s}{r} \]
where \( s \) is the surface tension of the liquid and \( r \) is the radius of the drop.

2. **Identify the Relationship**: From the formula, we can see that the excess pressure is inversely proportional to the radius (\( r \)). This means that as the radius decreases, the excess pressure increases.

3. **Determine the Condition for Maximum Pressure**: To maximize the excess pressure, we need to minimize the radius \( r \). Therefore, the smaller the radius, the greater the excess pressure.

4. **Evaluate the Given Options**: The problem states that we need to find the radius for which the excess pressure is maximum. The options provided include different values of \( r \).

5. **Find the Minimum Radius**: Among the options, the minimum radius will yield the maximum excess pressure. If one of the options is \( \frac{R}{4} \) (where \( R \) is a reference radius), this would be the smallest value compared to other options.

6. **Conclusion**: Therefore, the excess pressure inside a liquid drop of water will be maximum for the drop having a radius of \( \frac{R}{4} \).

### Final Answer:
The excess pressure inside a liquid drop of water will be maximum for the one having a radius of \( \frac{R}{4} \).