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(B) soap film has two free surfaces,so the total change in area is $2 	imes \Delta A$.The work done $W$ is given by the formula $W = T 	imes 2 \Delt

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$24 	imes 10^{-6} 	ext{ J}$

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(B) soap film has two free surfaces,so the total change in area is $2 	imes \Delta A$.The work done $W$ is given by the formula $W = T 	imes 2 \Delta A$.Initial area $A_1 = 4 	imes 4 = 16 	ext{ cm}^2 = 16 	imes 10^{-4} 	ext{ m}^2$.Final area $A_2 = 4 	imes 5 = 20 	ext{ cm}^2 = 20 	imes 10^{-4} 	ext{ m}^2$.Change in area $\Delta A = A_2 - A_1 = (20 - 16) 	imes 10^{-4} = 4 	imes 10^{-4} 	ext{ m}^2$.Given surface tension $T = 3 	imes 10^{-2} 	ext{ N/m}$.Substituting the values: $W = 3 	imes 10^{-2} 	imes 2 	imes (4 	imes 10^{-4}) = 24 	imes 10^{-6} 	ext{ J}$.

Consider a soap film on a rectangular frame of wire of area $4 \times 4 \text{ cm}^2$. If the area of the soap film is increased to $4 \times 5 \text{ cm}^2$,the work done in the process will be (The surface tension of the soap film is $3 \times 10^{-2} \text{ N/m}$).

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