The surface tension of a liquid is `5Nm^-1`. If a thin film formed on a loop of area `0.02m^-2` then its surface energy will be

Options

A. `5xx10^-2J`
B. `2.5xx10^-2J`
C. `2xx10^-1J`
D. `3xx10^-1J`

Explanation

To find the surface energy of a thin film formed on a loop, we can use the formula for surface energy, which is given by:

\[ \text{Surface Energy} (E) = \text{Surface Tension} (S) \times \text{Area} (A) \times 2 \]

The factor of 2 is included because a thin film has two surfaces (the top and bottom).

### Step-by-step Solution:

1. **Identify the given values**:
- Surface tension, \( S = 5 \, \text{N/m} \)
- Area of the film, \( A = 0.02 \, \text{m}^2 \)

2. **Calculate the surface energy**:
- Using the formula for surface energy:
\[ E = S \times A \times 2 \]
- Substitute the known values:
\[ E = 5 \, \text{N/m} \times 0.02 \, \text{m}^2 \times 2 \]

3. **Perform the multiplication**:
- First calculate \( 5 \times 0.02 \):
\[ 5 \times 0.02 = 0.1 \, \text{N m} \]
- Now multiply by 2:
\[ 0.1 \times 2 = 0.2 \, \text{J} \]

4. **Final Result**:
- The surface energy of the thin film is:
\[ E = 0.2 \, \text{J} \]

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

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