A copper wire iis stretched to make it `0.1%` longer. What is the percentage change in its resistance?

Explanation

`R= rho (l)/(A)`
`V=Al`
By differentiation, `0=ldA +Adl" "…(1)`
By differentiation, `dR=(rho(Adl-ldA))/(A^(2))" "…(2)`
By putting value of `-ldA` in equation (2), we get
`dR=(rho^(2)Adl)/(A^(2))`
`dR=(2rho dl)/(A) or (dR)/(R )=(2dl)/(l)" " ( :. rho=(RA)/(l))`
So, `(dR)/(R )%= 2xx (dl)/(l)%`
`(dR)/(R )%=2xx0.1%`
Hence, `(dR)/(R )% =0.2%`