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(D) The work done in stretching a liquid film is given by $W = T 	imes \Delta A_{total}$.Since a liquid film has two surfaces,the change in area is $

Vedclass · Accepted Answer

$0.125$

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(D) The work done in stretching a liquid film is given by $W = T 	imes \Delta A_{total}$.Since a liquid film has two surfaces,the change in area is $\Delta A_{total} = 2 	imes (A_2 - A_1)$.Initial area $A_1 = 4 \, cm 	imes 2 \, cm = 8 \, cm^2 = 8 	imes 10^{-4} \, m^2$.Final area $A_2 = 5 \, cm 	imes 4 \, cm = 20 \, cm^2 = 20 	imes 10^{-4} \, m^2$.Change in area $\Delta A = A_2 - A_1 = 12 \, cm^2 = 12 	imes 10^{-4} \, m^2$.Total change in area $\Delta A_{total} = 2 	imes 12 	imes 10^{-4} \, m^2 = 24 	imes 10^{-4} \, m^2$.Given work done $W = 3 	imes 10^{-4} \, J$.Using $W = T 	imes \Delta A_{total}$,we get $T = \frac{W}{\Delta A_{total}} = \frac{3 	imes 10^{-4}}{24 	imes 10^{-4}} = \frac{1}{8} = 0.125 \, N \, m^{-1}$.

$A$ rectangular film of liquid is extended from $(4 \, cm \times 2 \, cm)$ to $(5 \, cm \times 4 \, cm).$ If the work done is $3 \times 10^{-4} \, J,$ the value of the surface tension of the liquid is ............ $N \, m^{-1}$

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