A copper wire is stretched to make it 0.1% longer .What is the percentage increase In resistance?

Explanation

Data supplied,
Original length=`l_(1)=l`
Origoinal area of cross section`A_(1)=A`
New length =`l_(2)=l+(0.0l)/(100)= 1.001 l `
New area of cross section`=A_(2)`
As `A_(1) l_(1)=A_(2) l_(2)" " A_(2)=(A_(1) l_(1))/(l_(2))=(A)/(1.001)`
Let `R_(2)=R_(1)` be the original and new resistance.Then resistivity `p=(R_(1)l_(1))/(l_(1))=(R_(2)l_(2))/(l_(2))`
`R_(2)=((A_(1))/(A_(2))).((l_(1))/(l_(2))). R_(1)=(A)/(("A/1.00."))xx((1.001l)/(l))xxR_(1)`
`=(1.001)^(2)R_(1) =(1+.001)^(2)R_(1)`
`=[1+2xx.001+"higher terma of"0.001]R_(1)=(1+0.002)R_(1)`
`R_(2)=R_(1)+0.002R_(1)`(Explained by Bonomial Theorem and neglected hihger power beiong very small)
`therefore R_(2)-R_(1)=0.002R_(1)`
Hence percentage increase in resistence `=(R_(2)-R_(1))/(R_(1))xx100=(0.002R_(1))/(R_(1))xx100=0.2%`