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(B) The work done $(W)$ in increasing the surface area of a soap film is given by $W = T 	imes \Delta A_{total}$.Since a soap film has two surfaces,t

Vedclass · Accepted Answer

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

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(B) The work done $(W)$ in increasing the surface area of a soap film is given by $W = T 	imes \Delta A_{total}$.Since a soap film has two surfaces,the total change in area is $\Delta A_{total} = 2 	imes (A_{final} - A_{initial})$.Initial area $A_i = 10 \, cm 	imes 6 \, cm = 60 \, cm^2 = 60 	imes 10^{-4} \, m^2$.Final area $A_f = 10 \, cm 	imes 11 \, cm = 110 \, cm^2 = 110 	imes 10^{-4} \, m^2$.Change in area $\Delta A = A_f - A_i = (110 - 60) 	imes 10^{-4} \, m^2 = 50 	imes 10^{-4} \, m^2$.Total change in area $\Delta A_{total} = 2 	imes 50 	imes 10^{-4} \, m^2 = 100 	imes 10^{-4} \, m^2 = 10^{-2} \, m^2$.Given $W = 3 	imes 10^{-4} \, J$.Using $W = T 	imes \Delta A_{total}$,we get $T = \frac{W}{\Delta A_{total}} = \frac{3 	imes 10^{-4}}{10^{-2}} = 3 	imes 10^{-2} \, N/m$.

The work done in increasing the size of a 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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