A film of water is formed between two straight parallel wires of length 10 cm each separated by `0.5cm` If their separation is increased by `1mm` while still maintaining their parallelism, how much work will have to be done (Surface tension of water `=7.2xx10^-2(N)/(m)`)

Options

A. `7.22xx10^-6Joule`
B. `1.44xx10^-5joule`
C. `2.88xx10^-5joule`
D. `5.76xx10^-5joule`

Explanation

To solve the problem, we need to determine the work done when the separation between two parallel wires is increased while a film of water is maintained between them. The work done is equal to the change in surface energy of the water film.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Length of the wires, \( L = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Initial separation between the wires, \( d_1 = 0.5 \, \text{cm} = 0.005 \, \text{m} \)
- Increase in separation, \( \Delta d = 1 \, \text{mm} = 0.001 \, \text{m} \)
- Final separation, \( d_2 = d_1 + \Delta d = 0.005 \, \text{m} + 0.001 \, \text{m} = 0.006 \, \text{m} \)
- Surface tension of water, \( \gamma = 7.2 \times 10^{-2} \, \text{N/m} \)

2. **Calculate the Change in Area:**
- The area of the film between the wires is given by the product of the length of the wires and the change in separation.
- The change in separation is \( \Delta d = 0.001 \, \text{m} \).
- The change in area \( \Delta A \) can be calculated as:
\[ \Delta A = L \times \Delta d = 0.1 \, \text{m} \times 0.001 \, \text{m} = 0.0001 \, \text{m}^2 \]
- Since there are two surfaces (front and back), the total change in area is:
\[ \Delta A_{\text{total}} = 2 \times \Delta A = 2 \times 0.0001 \, \text{m}^2 = 0.0002 \, \text{m}^2 \]

3. **Calculate the Change in Surface Energy:**
- The change in surface energy \( \Delta U \) is given by:
\[ \Delta U = \gamma \times \Delta A_{\text{total}} \]
- Substituting the values:
\[ \Delta U = 7.2 \times 10^{-2} \, \text{N/m} \times 0.0002 \, \text{m}^2 = 1.44 \times 10^{-5} \, \text{J} \]

4. **Conclusion:**
- The work done \( W \) is equal to the change in surface energy:
\[ W = \Delta U = 1.44 \times 10^{-5} \, \text{J} \]

### Final Answer:
The work done will be \( 1.44 \times 10^{-5} \, \text{J} \).

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