Two parallel wires each of length 10 cm are 0.5 cm apart. A film of water is formed between them. If surface tension of water is 0.072 N/m, then the work done in increasing the distance between the wires by 1 mm is:

Options

A. ` 1.44 xx 10 ^(-5 ) J`
B. `1.72 xx 10 ^(-5) J`
C. ` 1.44 xx 10 ^(-4) J`
D. ` 1.72 xx 10 ^(-4) J`

Explanation

To solve the problem, we need to calculate the work done in increasing the distance between two parallel wires with a water film between them. The work done is related to the change in surface energy due to the change in the area of the water film.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Length of the wires, \( L = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Initial distance between the wires, \( d = 0.5 \, \text{cm} = 0.005 \, \text{m} \)
- Change in distance, \( \Delta d = 1 \, \text{mm} = 0.001 \, \text{m} \)
- Surface tension of water, \( S = 0.072 \, \text{N/m} \)

2. **Calculate the Change in Area:**
- The change in area \( \Delta A \) due to the increase in distance between the wires can be calculated as:
\[ \Delta A = L \times \Delta d \]
- Substituting the values:
\[ \Delta A = 0.1 \, \text{m} \times 0.001 \, \text{m} = 0.0001 \, \text{m}^2 \]

3. **Calculate the Change in Surface Energy:**
- The change in surface energy \( \Delta U \) is given by:
\[ \Delta U = S \times \Delta A \]
- Substituting the values:
\[ \Delta U = 0.072 \, \text{N/m} \times 0.0001 \, \text{m}^2 = 0.0000072 \, \text{J} \]

4. **Account for Two Surfaces:**
- Since there are two surfaces of the water film (upper and lower), we need to multiply the change in surface energy by 2:
\[ \text{Total Work Done} = 2 \times \Delta U = 2 \times 0.0000072 \, \text{J} = 0.0000144 \, \text{J} \]

5. **Final Result:**
- Therefore, the work done in increasing the distance between the wires by 1 mm is:
\[ \text{Work Done} = 1.44 \times 10^{-5} \, \text{J} \]

### Summary:
The work done in increasing the distance between the wires by 1 mm is \( 1.44 \times 10^{-5} \, \text{J} \).