A silver wire has a resistance of `2.1Omega` at `27.5^(@)C&2.7Omega` at `100^(@)C` Determine the temperature coefficient of resistivity of silver.

Explanation

If `R_(2)` is the resistance of a material at temperature `t_(2) and R_(1)` is the resistance at temperature `t_(1)`, then the temperature coefficient of resistivity of the material is given by:
`alpha = (R_(2)-R_(1))/(R_(1) (t_(2)-t_(1)))` …(i)
Let `R_(100)` be the resistance of the wire at `100^(@)C and R_(27.5)` be its resistance at `27.5^(@)C`. Substituting `R_(2) = R_(100), R_(1) = R_(27.5), t_(2) = 100 and t_(1) = 27.5` in (i), we get:
`alpha= (R_(100) - R_(27.5))/(R_(27.5) xx (100-27.5))`
Here, `R_(27.5) = 2.1 Omega and R_(100) = 2.7 Omega`
`:. alpha= (2.7 -2.1)/(2.1 xx (100-27.5)) = 0.0039^(@)C^(-1)`