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Question

(A) The work done $W$ in increasing the area of a soap film is given by $W = T 	imes \Delta A 	imes 2$,where $T$ is the surface tension and the fact

Vedclass · Accepted Answer

$2 	imes 10^{-2}\;N/m$

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(A) The work done $W$ in increasing the area of a soap film is given by $W = T 	imes \Delta A 	imes 2$,where $T$ is the surface tension and the factor $2$ accounts for the two surfaces of the soap film.Initial area $A_1 = 10\;cm 	imes 6\;cm = 60\;cm^2 = 60 	imes 10^{-4}\;m^2$.Final area $A_2 = 10\;cm 	imes 11\;cm = 110\;cm^2 = 110 	imes 10^{-4}\;m^2$.Change in area $\Delta A = A_2 - A_1 = (110 - 60) 	imes 10^{-4}\;m^2 = 50 	imes 10^{-4}\;m^2$.Given $W = 2 	imes 10^{-4}\;J$.Using the formula $W = 2T \Delta A$,we get $T = \frac{W}{2 \Delta A}$.$T = \frac{2 	imes 10^{-4}}{2 	imes (50 	imes 10^{-4})} = \frac{1}{50} = 0.02\;N/m = 2 	imes 10^{-2}\;N/m$.

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

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