The surface tension of a liquid is `5` Newton per metre. If a film is held on a ring of area `0.02 metres^(2)`, its surface energy is about :

Explanation

To find the surface energy of a liquid film held on a ring, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Surface tension (T) = 5 Newton/meter
- Area (A) = 0.02 square meters

2. **Understand the Concept of Surface Energy**:
- Surface energy (E) is calculated using the formula:
\[ E = T \times 2A \]
- This is because the film has two surfaces (front and back) in contact with the air.

3. **Calculate the Effective Area**:
- Since the film is held on both sides of the ring, the effective area becomes:
\[ 2A = 2 \times 0.02 \, \text{m}^2 = 0.04 \, \text{m}^2 \]

4. **Substitute the Values into the Surface Energy Formula**:
- Now substitute the values of T and 2A into the surface energy formula:
\[ E = T \times 2A = 5 \, \text{N/m} \times 0.04 \, \text{m}^2 \]

5. **Perform the Calculation**:
- Calculate the surface energy:
\[ E = 5 \times 0.04 = 0.2 \, \text{Joules} \]

6. **Final Answer**:
- Therefore, the surface energy of the liquid film is:
\[ E = 0.2 \, \text{Joules} \]

### Summary:
The surface energy of the liquid film held on the ring is **0.2 Joules**.