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(C) The length of the wire is $l = 10 	ext{ cm} = 0.1 	ext{ m}$.The initial distance between the wires is $d_1 = 0.5 	ext{ cm} = 0.005 	ext{ m}$.T

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$1.44 	imes 10^{-5} 	ext{ J}$

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(C) The length of the wire is $l = 10 	ext{ cm} = 0.1 	ext{ m}$.The initial distance between the wires is $d_1 = 0.5 	ext{ cm} = 0.005 	ext{ m}$.The increase in distance is $\Delta d = 1 	ext{ mm} = 0.001 	ext{ m}$.$A$ water film has two surfaces,so the change in area is $\Delta A = 2 	imes (l 	imes \Delta d)$.$\Delta A = 2 	imes (0.1 	ext{ m} 	imes 0.001 	ext{ m}) = 2 	imes 10^{-4} 	ext{ m}^2$.The surface tension of water is $T = 72 	ext{ mN/m} = 72 	imes 10^{-3} 	ext{ N/m}$.The work done is $W = T 	imes \Delta A$.$W = (72 	imes 10^{-3} 	ext{ N/m}) 	imes (2 	imes 10^{-4} 	ext{ m}^2) = 144 	imes 10^{-7} 	ext{ J} = 1.44 	imes 10^{-5} 	ext{ J}$.

$A$ water film is formed between two parallel wires of $10 \text{ cm}$ length. The distance of $0.5 \text{ cm}$ between the wires is increased by $1 \text{ mm}$. The work done in the process is (surface tension of water $= 72 \text{ mN/m}$).

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