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(C) The length of the wires is $l = 10 	ext{ cm} = 0.1 	ext{ m}$.The increase in separation is $\Delta x = 1 	ext{ mm} = 10^{-3} 	ext{ m}$.$A$ wat

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The length of the wires is $l = 10 	ext{ cm} = 0.1 	ext{ m}$.The increase in separation is $\Delta x = 1 	ext{ mm} = 10^{-3} 	ext{ m}$.$A$ water film has two surfaces,so the change in area $\Delta A$ is given by $2 	imes (l 	imes \Delta x)$.$\Delta A = 2 	imes (0.1 	ext{ m} 	imes 10^{-3} 	ext{ m}) = 2 	imes 10^{-4} 	ext{ m}^2$.The work done $W$ is given by $W = T 	imes \Delta A$,where $T$ is the surface tension.$W = (7.2 	imes 10^{-2} 	ext{ N/m}) 	imes (2 	imes 10^{-4} 	ext{ m}^2) = 14.4 	imes 10^{-6} 	ext{ J} = 1.44 	imes 10^{-5} 	ext{ J}$.

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