A rectangular film of liquid is extended from $$\left( {4\,cm \times 2\,cm} \right)$$   to $$\left( {5\,cm \times 4 \times cm} \right).$$    If the work done is $$3 \times {10^{ - 4}}J,$$   the value of the surface tension of the liquid is

Solution :
$$\eqalign{ & {\text{Increase in surface energy}} = {\text{work done in area}} \times {\text{surface tension}} \cr & \because {\text{Increase in surface area,}} \cr & \Delta A = \left( {5 \times 4 - 4 \times 2} \right) \times 2\,\,\,\left( {\because {\text{film has two surfaces}}} \right) \cr & = \left( {20 - 8} \right) \times 2\,c{m^2} = 24\,c{m^2} = 24 \times {10^{ - 4}}{m^2} \cr} $$
So, work done, $$W = T \cdot \Delta A$$
$$\eqalign{ & 3 \times {10^{ - 4}} = T \times 24 \times {10^{ - 4}} \cr & \therefore T = \frac{1}{8} = 0.125\,N/m \cr} $$