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 radius of copper wire is r. When it is stretched , then its new radius ,
`r . r - 0.1% ` of r
`rArr" " r = 0.999r `
Now , volume will remain constant.
`rArr " " pi r^(2) l = pi r^(2) l ` [l is its new length]
`rArr " " pir^(2) l = pi (0.99r)^(2) l`
`rArr" " l = (l)/((0.99)^(2))`
If R is new area , then
`(A)/(A) = (pi r^(2))/(pi ( 0.99)^(2)) = (1)/((0.99)^(2))`
`rArr " " A = (0.99)^(2)A`
Percentage increase in resistance
`= (R - R)/(R) xx 100 = ( rho (l)/(A) - (rho l)/(A))/(rho(l)/(A)) xx 100`
` = (1.041 0-1)/(1) xx 100 = 4.1 %`