Work of 3.0 times 10^{-4} , J is required to | vedclass

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(B) The initial area of the soap film is $A_1 = 10 \, cm 	imes 6 \, cm = 60 \, cm^2$.The final area of the soap film is $A_2 = 10 \, cm 	imes 11 \,

Vedclass · Accepted Answer

$3 	imes 10^{-2} \, N/m$

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(B) The initial area of the soap film is $A_1 = 10 \, cm 	imes 6 \, cm = 60 \, cm^2$.The final area of the soap film is $A_2 = 10 \, cm 	imes 11 \, cm = 110 \, cm^2$.The increase in area is $\Delta A = A_2 - A_1 = 110 \, cm^2 - 60 \, cm^2 = 50 \, cm^2$.$A$ soap film has two surfaces,so the total increase in surface area is $\Delta A_{total} = 2 	imes 50 \, cm^2 = 100 \, cm^2$.Converting this to $SI$ units: $\Delta A_{total} = 100 	imes 10^{-4} \, m^2 = 10^{-2} \, m^2$.The work done is given by $W = T 	imes \Delta A_{total}$,where $T$ is the surface tension.Therefore,$T = \frac{W}{\Delta A_{total}} = \frac{3.0 	imes 10^{-4} \, J}{10^{-2} \, m^2} = 3.0 	imes 10^{-2} \, N/m$.

Work of $3.0 \times 10^{-4} \, J$ is required to be done in increasing the size of a soap film from $10 \, cm \times 6 \, cm$ to $10 \, cm \times 11 \, cm$. The surface tension of the film is

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