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

To find the work done in increasing the separation between the two wires, we can use the concept of surface energy and surface tension of a liquid film.

1. Identify the given parameters:
Length of each parallel wire, ^l = 10 cm = 0.1 m
Initial separation, ^d₁ = 0.5 cm = 0.005 m
Increase in separation, $\mathrm{\Delta d}=1\text{mm}={10}^{-3}\text{m}$
Surface tension of water, $T=7.2\times {10}^{-2}\text{N/m}$

2. Calculate the change in surface area:
A film of water formed between two parallel wires has two free surfaces (an upper surface and a lower surface) in contact with the air. Therefore, any change in the physical area of the film results in twice the change in the surface area of the water-air interface.
The physical change in area of the film is given by:
$\text{Physical Area Change}=l\times \mathrm{\Delta d}$
Since there are two free surfaces, the total increase in surface area ($\mathrm{\Delta A}$) is:
$\mathrm{\Delta A}=2\times l\times \mathrm{\Delta d}$

Substituting the given values:
$\mathrm{\Delta A}=2\times 0.1\text{m}\times {10}^{-3}\text{m}$
$\mathrm{\Delta A}=2\times {10}^{-4}{\text{m}}^2$

3. Calculate the work done:
The work done (W) in increasing the surface area is equal to the product of the surface tension (T) and the total increase in surface area ($\mathrm{\Delta A}$):
$W=T\times \mathrm{\Delta A}$

Substituting the values of T and $\mathrm{\Delta A}$:
$W=(7.2\times {10}^{-2}\text{N/m})\times (2\times {10}^{-4}{\text{m}}^2)$
$W=14.4\times {10}^{-6}\text{J}$
$W=1.44\times {10}^{-5}\text{J}$

Thus, the work that will have to be done is 1.44 x 10⁻⁵ J, which corresponds to the second option.