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(D) soap film has two surfaces, so the total change in area is $2 	imes \Delta A$.Initial area $A_1 = 3 	imes 3 \,cm^2 = 9 	imes 10^{-4} \,m^2$.Fin

Vedclass · Accepted Answer

$80 	imes 10^{-6} \,J$

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(D) soap film has two surfaces, so the total change in area is $2 	imes \Delta A$.Initial area $A_1 = 3 	imes 3 \,cm^2 = 9 	imes 10^{-4} \,m^2$.Final area $A_2 = 5 	imes 5 \,cm^2 = 25 	imes 10^{-4} \,m^2$.Change in area $\Delta A = A_2 - A_1 = (25 - 9) 	imes 10^{-4} \,m^2 = 16 	imes 10^{-4} \,m^2$.Surface tension $T = 2.5 	imes 10^{-2} \,N/m$.Work done $W = T 	imes (2 \Delta A) = 2.5 	imes 10^{-2} 	imes 2 	imes 16 	imes 10^{-4} \,J$.$W = 5 	imes 10^{-2} 	imes 16 	imes 10^{-4} \,J = 80 	imes 10^{-6} \,J$.

Consider a soap film on a rectangular frame of wire of area $3 \times 3 \,cm^2$. If the area of the soap film is increased to $5 \times 5 \,cm^2$, the work done in the process will be (surface tension of soap solution is $2.5 \times 10^{-2} \,N/m$).

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