A silver wire has a resistance of 2.1 `Omega` at `27.5^@C`, and a resistance of 2.7 `Omega` at `100^@C`. Determine the temperature coefficient of resistivity of silver. Also find the resistance of the silver wire at `0^@C`.

Explanation

Given : `R_1 = 2.1 Omega, t_1 = 27.5^@C`
`R_2 = 2.7 Omega, t_2 = 100^@C`
We have, `R_1 = (1 + alpha t_1)` and `R_2 = R_0(1alpha t_2)`
On dividing,
`(R)/(R_2)=(1+ alpha t_1)/(1+alpha t_2)`
or `R_1(1+alpha t_1)=R_2(1+alpha t_2)`
`alpha=(R_2-R_1)/(R_1t_2-R_2t_1)`
`alpha=(2.7-2.1)/((2.1 xx 100)-(2.7xx27.5))`
`alpha=(0.6)/(210-74.25)`
`alpha=(0.6)/(135.75)`
`alpha = 0.0044^@C^(-1)`
Resistance at `0^@C` is
`R_0=(R_1)/(1+alpha t_1)=(2.1)/(1+(0.004xx27.5))`
`R_0=(2.1)/(1.121)`
`R_0=1.87 Omega`