A film of water is formed between two straight parallel wires each `10 cm` long and at a seperation of `0.5 cm`. Calculate the work required to increase `1mm` distance between the wires. Surface tension of water `=72xx10^(-3)N//m`.

Explanation

To solve the problem, we need to calculate the work required to increase the distance between two parallel wires due to the surface tension of water. Here's a step-by-step solution:

### Step 1: Understand the Given Data
- Length of each wire, \( L = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Initial separation between the wires, \( d = 0.5 \, \text{cm} = 0.005 \, \text{m} \)
- Increase in separation, \( \Delta d = 1 \, \text{mm} = 0.001 \, \text{m} \)
- Surface tension of water, \( \gamma = 72 \times 10^{-3} \, \text{N/m} \)

### Step 2: Calculate the Change in Area
When the distance between the wires increases by \( \Delta d \), the change in area \( \Delta A \) of the water film can be calculated as:
\[ \Delta A = L \times \Delta d \]
Substituting the values:
\[ \Delta A = 0.1 \, \text{m} \times 0.001 \, \text{m} = 0.0001 \, \text{m}^2 \]

### Step 3: Calculate the Work Done
The work done \( W \) to increase the distance between the wires is given by the formula:
\[ W = \gamma \times \Delta A \]
Since there are two surfaces (one on each wire), we need to multiply the change in area by 2:
\[ W = 2 \times \gamma \times \Delta A \]
Substituting the values:
\[ W = 2 \times (72 \times 10^{-3} \, \text{N/m}) \times (0.0001 \, \text{m}^2) \]
Calculating this:
\[ W = 2 \times 72 \times 10^{-3} \times 0.0001 = 1.44 \times 10^{-5} \, \text{J} \]

### Final Answer
The work required to increase the distance between the wires by \( 1 \, \text{mm} \) is:
\[ W = 1.44 \times 10^{-5} \, \text{J} \]

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