Question

A film of water is formed between two straight parallel wires each 10cm long & at a separation 0.5 . Calculate the work required to increase the distance between the wires by 1mm . the surface tension of water is 72x10-3N/m

Answer 1

The energy of the surface of a thin film is equal to the surface tension of the film (in this case water) times the area of that film. So we are told that we have a thin film of water formed between two wires 10cm long and 0.5cm wide. Converting these into metres we have a surface area of 0.1x0.005 = 0.0005 m\(^2\). This will have an energy of \[E=0.072\times0.0005=3.6\times10^{-6}\rm{J}\]After stretching, the surface now has an area of 0.1x0.006 =0.0006m\(^2\). This then has an energy of \[E=0.072\times0.0006=4.32\times10^{-6}\rm{J}\]. So the work done is the amount of energy added to the system to do the stretching (i.e. the differences in the two energies) and hence the work done is \[W=4.32\times10^{-6}-3.6\times10^{-6}\rm{J}=72\times10^{-7}\rm{J}\].

BUT, this is just for one side of the film, i.e. one surface. There are two sides to the film, and hence two surfaces, and hence twice the area, so the work done will be twice as much, i.e. \(144\times10^{-7}\) Joules.

The above is the general explanation. The quickest way to calculate it is noting taht the work done is equal to the product of teh surface tension \(T\) and the change in Surface area \(\Delta A\) thus\[W=T\Delta A=72\times10^{-3}\times0.0002=144\times10^{-7}\rm{J}\]