IF the length of a copper wire is increased by 0.1%, show that the resistance of the copper wire will increase by 0.2%.

Explanation

IF the initial length be `l_1`, then the find length will be
`l_2timesl_1times(100.1%)=l_1times100.1/100=1.001l_1`
IF the temperature of the wire remains constant,its volume wire remain constant.
Now, volume =length `times` cross sectional area
So if the length increases from `l_1` to 1.001 `l_1` cross sectional are `A_1` decreases to `A_2` where `A_2=A_1/1.001`.
Now, `R_1=rhol_1/A_1`and`R_2=rho.l_2/A_2`
`thereforeR_2/R_1=l_2/l_1timesA_1/A_2=1.001 times1.001=1.002=100.2%`
i.e., increase of resistance =0.2%
Alternative method:
`R=rhol/A`
By logarithmic differentation, `(dR)/R=(dl)/l-(dA)/A` (`rho`= constant)
Now, `V=lA`=constant
By logarithmic differentation ,`(dl)/l=-(dA)/A`
Hence `(dR)/R=(2dl)/l=2times0.1%=0.2%`