If a copper wire is stretched to make it `0.1%` longer wha is the percentage change in its resistance?

Explanation

`(R_f - R_i)/(R_i) = (rho(I_f)/(A_f) - rho(I_i)/(A_f))/(rho(I_i)/(A_i)) = ((I_f)/(A_f) - (I_i)/(A_i)/(I_i)/(A_i) )` (1)
Let the initial of the wire be 100 cm, then the new length is
`I_f = 100 +(0.1)/(100)xx100 = 100.1cm (ii)`
Let `A_i and A-f` be the initial and final ares of cross section. Then
`100xxA_f = 100.1A_f`
`or A_f = (100)/(100.1)A_i (iii)`
From Eqs. (i), (ii), and (iii) we get
`(R_f- R_i)/(R_i) xx100 =((100.1)^2/(100A_i) -(100)/(A_i))/((100)/(A_i)) xx100`
`=((100.1)^2 - (100)^2)/((100)^2) xx100 `
`=(200.1xx0.1)/(100xx100)xx100 = 0.2%`