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 step by step, we will calculate the work done in increasing the distance between the two parallel wires by 1 mm, given the surface tension of water.

### Step 1: Understand the given data
- Length of each wire (L) = 10 cm = 0.1 m
- Initial distance between the wires (d) = 0.5 cm = 0.005 m
- Increase in distance (Δd) = 1 mm = 0.001 m
- Surface tension of water (T) = 0.072 N/m

### Step 2: Calculate the change in area (ΔA)
When the distance between the wires is increased, the area of the water film between them changes. The change in area can be calculated using the formula:
\[ \Delta A = L \times \Delta d \]
Since there are two surfaces (one on each wire), the total change in area will be:
\[ \Delta A = 2 \times (L \times \Delta d) \]

Substituting the values:
\[ \Delta A = 2 \times (0.1 \, \text{m} \times 0.001 \, \text{m}) = 2 \times 0.0001 \, \text{m}^2 = 0.0002 \, \text{m}^2 \]

### Step 3: Calculate the work done (W)
The work done in increasing the distance between the wires is given by the formula:
\[ W = T \times \Delta A \]
Substituting the values:
\[ W = 0.072 \, \text{N/m} \times 0.0002 \, \text{m}^2 \]
Calculating this gives:
\[ W = 0.072 \times 0.0002 = 0.0000144 \, \text{J} \]

### Step 4: Convert the work done to standard form
To express the work done in standard form:
\[ W = 1.44 \times 10^{-5} \, \text{J} \]

### Final Answer
The work done in increasing the distance between the wires by 1 mm is:
\[ \boxed{1.44 \times 10^{-5} \, \text{J}} \]