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(B) The length of the wires is $l = 10\,cm = 0.1\,m$.Initially,the separation is $d_1 = 0.5\,cm = 0.005\,m$.Finally,the separation is $d_2 = 0.5\,cm +

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$1.44 	imes 10^{-5}\,J$

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(B) The length of the wires is $l = 10\,cm = 0.1\,m$.Initially,the separation is $d_1 = 0.5\,cm = 0.005\,m$.Finally,the separation is $d_2 = 0.5\,cm + 0.1\,cm = 0.6\,cm = 0.006\,m$.$A$ water film has two surfaces,so the change in area $\Delta A$ is given by $2 	imes l 	imes (d_2 - d_1)$.$\Delta A = 2 	imes 0.1\,m 	imes (0.006\,m - 0.005\,m) = 0.2 	imes 0.001 = 2 	imes 10^{-4}\,m^2$.The work done $W$ is equal to the change in surface energy: $W = T 	imes \Delta A$.Given $T = 7.2 	imes 10^{-2}\,N/m$.$W = 7.2 	imes 10^{-2} 	imes 2 	imes 10^{-4} = 14.4 	imes 10^{-6} = 1.44 	imes 10^{-5}\,J$.

$A$ film of water is formed between two straight parallel wires of length $10\,cm$ each,separated by $0.5\,cm$. If their separation is increased by $1\,mm$ while still maintaining their parallelism,how much work will have to be done? (Surface tension of water = $7.2 \times 10^{-2}\,N/m$)

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