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Question

(B) 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 = 1

Vedclass · Accepted Answer

$3.0 	imes 10^{-2} \, N/m$

Vedclass · Answer

(B) 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 $\Delta A = A_2 - A_1 = (110 - 60) 	imes 10^{-4} \, m^2 = 50 	imes 10^{-4} \, m^2$. $A$ soap film has two surfaces,so the work done $W$ is given by $W = T 	imes 2 	imes \Delta A$. Given $W = 3 	imes 10^{-4} \, J$. Substituting the values: $3 	imes 10^{-4} = T 	imes 2 	imes (50 	imes 10^{-4})$. $3 	imes 10^{-4} = T 	imes 100 	imes 10^{-4}$. $T = \frac{3 	imes 10^{-4}}{100 	imes 10^{-4}} = 0.03 \, N/m = 3.0 	imes 10^{-2} \, N/m$.

The work done in increasing the size of a rectangular soap film from $10 \, cm \times 6 \, cm$ to $10 \, cm \times 11 \, cm$ is $3 \times 10^{-4} \, J$. The surface tension of the film is:

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