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

Initial area of the film $A_i=4\times 2=8c{m}^2=8\times {10}^{-4}{m}^2$

Final area of the film $A_f=5\times 4=20c{m}^2=20\times {10}^{-4}{m}^2$

Change in area $∆A=(20-8)\times {10}^{-4}=12\times {10}^{-4}{m}^2$

Work done $W=3\times {10}^{-4} J$

$W=T(2∆A)$

$T=\frac{W}{2∆A}$

=$\frac{3\times {10}^{-4}}{2\times 12\times {10}^{-4}}$

$=\frac{1}{8}$

$=0.125 N{m}^{-1}$