Consider a rectangular frame (see fig. 11.20), having a sliding wire on one of its arms. Dip the frame in a soap solution and take it out. A soap film is formed on the frame and have two surfaces. Both the surfaces are in contact with the sliding wire. So, surface tension acts on the wire due to both the surface.Let T be the surface tension of the soap solution and L be the length of the wire.The force exerted by each surface on the wire = \\\\(T \ imes L\\\\)So, total force on the wire = \\\\(2TL\\\\) ......(11.23)Let the surfaces contract by \\\\(\\Delta x\\\\).Work done on the film, \\\\(W = F \ imes \\Delta x\\\\)\\\\(= 2T L \\Delta x\\\\) ......(11.24)Here \\\\(2L \ imes \\Delta x \\rightarrow\\\\) total increase in the area of both of the surfaces of film.Let \\\\(2L \ imes \\Delta x = \\Delta A\\\\)\\\\(\ herefore W = T \\Delta A\\\\)or \\\\(T = \\frac{W} {\\Delta A}\\\\) .........(11.25)So, surface tension of a liquid is equal to the work done in increasing the surface area of its free surface by one unit. In other words, surface tension is equal to surface energy per unit area.\\\\(E_s = T \\Delta A\\\\)The unit of T is \\\\(Jm^{-2}\\\\) and its dimensional formula is \\\\(\\left[M^{1} L^{0} T^{-2} \\right]\\\\)