The surface tension the surface tension of a liquid is 10 N/m. If a film is held on a ring of area 0.01 m square, its surface energy is about Newton

Options

A. `2 * 10^-3` J
B. `2*10^-1` J
C. `2*10^-2` J
D. `2*10^-4 `J

Answer

To solve the problem, we need to calculate the surface energy of a liquid film held on a ring. The surface energy (U) can be calculated using the formula:

\[ U = T \times A \]

where:
- \( U \) is the surface energy,
- \( T \) is the surface tension of the liquid,
- \( A \) is the area of the film.

### Step-by-step Solution:

1. **Identify the given values**:
- Surface tension \( T = 10 \, \text{N/m} \)
- Area of the film \( A = 0.01 \, \text{m}^2 \)

2. **Consider the film on both sides of the ring**:
- Since the film is held on both sides of the ring, the effective area \( A' \) that contributes to the surface energy is:
\[ A' = 2 \times A = 2 \times 0.01 \, \text{m}^2 = 0.02 \, \text{m}^2 \]

3. **Calculate the surface energy**:
- Now, substitute the values into the surface energy formula:
\[ U = T \times A' = 10 \, \text{N/m} \times 0.02 \, \text{m}^2 \]
- Performing the multiplication:
\[ U = 0.2 \, \text{N} \cdot \text{m} = 0.2 \, \text{J} \]

4. **Final answer**:
- The surface energy is approximately \( 0.2 \, \text{J} \) or \( 2 \times 10^{-1} \, \text{J} \).

### Conclusion:
The surface energy of the liquid film held on the ring is \( 0.2 \, \text{J} \).

---