Molecular forces exist between the molecules of a liquid in a container. The molecules on the surface have unequal force leading to a tension on the surface. If this is not compensated by a force, the equilibrium of the liquid will be a difficult task. This leads to an excess pressure on the surface. The nature of the meniscus can inform us of the direction of the excess pressure. The angle of contact of the liquid decided by the forces between the molecules, air and container can make the angle of contact.
In a meniscus of radius `r`, with excess pressure p in atmospheric pressure `p_(0)`, the force experienced is

Options

A. `(p-p_(0))pr^(2)`
B. `(p-p_(0))2pir`
C. `ppir^(2)`
D. `p_(0)2pir`

Explanation

To solve the problem regarding the force experienced by a liquid in a container with a meniscus, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Pressure**:
- We know that pressure (P) is defined as force (F) per unit area (A). Mathematically, this is expressed as:
\[ P = \frac{F}{A} \]
- Rearranging this gives us:
\[ F = P \times A \]

2. **Identify the Excess Pressure**:
- The problem states that there is an excess pressure \( P \) acting on the surface of the liquid, in addition to the atmospheric pressure \( P_0 \).
- Therefore, the total pressure acting on the liquid surface can be expressed as:
\[ P_{\text{total}} = P_0 + P \]

3. **Calculate the Change in Pressure**:
- The change in pressure (\( \Delta P \)) can be calculated as the difference between the total pressure and the atmospheric pressure:
\[ \Delta P = (P_0 + P) - P_0 = P \]

4. **Determine the Area**:
- The area \( A \) of the meniscus can be considered as the cross-sectional area of the liquid surface. For a circular meniscus with radius \( r \), the area is given by:
\[ A = \pi r^2 \]

5. **Calculate the Force**:
- Now substituting the change in pressure and the area into the force equation:
\[ F = \Delta P \times A = P \times (\pi r^2) \]
- Thus, the force experienced by the liquid due to the excess pressure is:
\[ F = P \cdot \pi r^2 \]

6. **Conclusion**:
- Therefore, the final expression for the force experienced by the liquid in the meniscus of radius \( r \) with excess pressure \( P \) is:
\[ F = P \cdot \pi r^2 \]