A flim of water if found between two straight parallel wires of length 0.10m each separated by `5 xx 10^(-3)`m. If their separation is increased by `1 xx 10^(-3)m`, while maintaining their parallelism, how much work will have to be done? (S.T. of water is `7.2 xx 10^(-2) Nm^(-2))`.

Explanation

Given `d=5 xx 10^(-3) m, l=0.10m`
`d=6 xx 10^(-3)m, l=0.10m`
`A=ld =0.10 xx 5 xx 10^(-3) =5 xx 10^(-4)m^(2)`
`A.=ld. =0.1 xx 6 xx 10^(-3) =6 xx 10^(-4) m^(2)`
`DeltaA=1 xx 10^(-4) m^(2)`.
Net area `=2DeltaA=2 xx 10^(-4)m^(2)`
Work done `=(T)(2DeltaA)=0.072 xx 2 xx 1 xx 10^(-4)=0.144 xx 10^(-4) =1.44 xx 10^(-5)J`