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(B) The work done in increasing the area of a film is given by $W = T 	imes \Delta A$,where $T$ is the surface tension and $\Delta A$ is the change i

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$22.4$

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(B) The work done in increasing the area of a film is given by $W = T 	imes \Delta A$,where $T$ is the surface tension and $\Delta A$ is the change in surface area.Since a thin film has two surfaces,the total change in area is $\Delta A = 2 	imes l 	imes (d_2 - d_1)$.Given: $l = 8 \ cm = 0.08 \ m$,$d_1 = 0.6 \ cm = 0.006 \ m$,$d_2 = 0.8 \ cm = 0.008 \ m$,and $T = 0.07 \ N/m$.Change in distance $\Delta d = d_2 - d_1 = 0.8 \ cm - 0.6 \ cm = 0.2 \ cm = 0.002 \ m$.$\Delta A = 2 	imes 0.08 \ m 	imes 0.002 \ m = 0.00032 \ m^2$.Work done $W = 0.07 \ N/m 	imes 0.00032 \ m^2 = 0.0000224 \ J$.$W = 22.4 	imes 10^{-6} \ J = 22.4 \ \mu J$.

$A$ thin film of water is formed between two straight parallel wires each of length $8 \ cm$ separated by a distance of $0.6 \ cm$. The work done to increase the distance between the wires to $0.8 \ cm$ is (Surface tension of water $= 0.07 \ N/m$) (in $\mu J$)

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