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

Answer

B

Explanation

`(R_f-R_i)/(R_i) xx 100 = (((rho(l_(f)/A_(f))- rho (l_i)/(A_i))/(rho (l_i/A_i))) xx 100 `
`(((l_f/A_f)-(l_i/A_i))/(l_i/A_i)) xx 100 ` ………….(i)
Let the initial length of the wire be 100 cm, then the new
length is `100 + 0.1/100 xx 100 `
`1 = 100.1 cm` ......(ii)
Let `A_(i) and A_(f)` be the initial and final area of cross-section.
Then
`100 xx A_i = 100.1 A_(f)`
`rArr A_(f) = 100/100.1 A_i` ......(iii)
From (i), (ii) and (iii)
`(R_(f) -R_(i))/(R_i) xx 100 = ((100.1)^2/(100A_i) - (100/A_i))/(100/A_i) xx 100`
`=((100.1)^2 - (100)^2)/(100)^2 xx 100 = (200.1 xx 0.1)/(100 xx 100) xx 100 `
`= 0.2%`
Thus the resistance increases by 0.2%.
Alternatively for small chage `(DeltaR)/R = (DeltaA)/A + (Deltal)/l` .