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(D) soap film has two surfaces,so the work done $W$ is given by $W = 2 	imes T 	imes \Delta A$,where $T$ is the surface tension and $\Delta A$ is th

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None of these

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(D) soap film has two surfaces,so the work done $W$ is given by $W = 2 	imes T 	imes \Delta A$,where $T$ is the surface tension and $\Delta A$ is the change in area.Initial area $A_1 = 10\, cm 	imes 6\, cm = 60\, cm^2 = 60 	imes 10^{-4}\, m^2$.Final area $A_2 = 60\, cm 	imes 11\, cm = 660\, cm^2 = 660 	imes 10^{-4}\, m^2$.Change in area $\Delta A = A_2 - A_1 = (660 - 60) 	imes 10^{-4}\, m^2 = 600 	imes 10^{-4}\, m^2 = 6 	imes 10^{-2}\, m^2$.Given $W = 2 	imes 10^{-4}\, J$.Substituting the values: $2 	imes 10^{-4} = 2 	imes T 	imes (6 	imes 10^{-2})$.$T = \frac{2 	imes 10^{-4}}{2 	imes 6 	imes 10^{-2}} = \frac{1}{6} 	imes 10^{-2} = 0.166 	imes 10^{-2} = 1.66 	imes 10^{-3}\, N/m$.Since this value is not among the options,the correct answer is $D$.

If the work done in increasing the size of a soap film from $10\, cm \times 6\, cm$ to $60\, cm \times 11\, cm$ is $2 \times 10^{-4}\, J$,what is the surface tension?

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