A copper wire is stretched to make its radius decreased by 0.1%. Find the percentage increase in resistance.

Explanation

Let l be the length and r be the radius of copper wire. When it is stretched, its new radius becomes
`r.r=-0.1 % of r rarr r. =0.99r`
Now volume will remain constant
`rarr pi r^(2) l =pi(0.99)^(2) l`
`rarr l=(l)/(0.99)^(2)`
if R. is new area then
`(A)/(A.)=(pir^(2))/(pi(0.99r )^(2)=(1)/(0.99)^(2)`
percentage increase in resistance
`=(R.-R)/(R )xx100=(p(l)/(A)-(pl)/(A))/(p(1)/(A))xx100`
`=(1.041-1)/(1)xx100=4.1 %`