If the resistance of a conductor is `5 Omega` at `50^(@)C` and `7Omega` at `100^(@)C` then the mean temperature coefficient of resistance of the material is

Options

A. `0.008//^(@) C`
B. `0.006//^(@)C`
C. `0.004//^(@)C`
D. `0.01//^(@)C`

Answer

D

Explanation

`R_(1)/R_(2) = (1 + alphaT_(1))/(1 + alphaT_(2)) rArr 5/7 = (1 + alpha50)/(1 + alpha100)`
`rArr 5 + 500alpha = 7+ 350 alpha`
`rArr alpha = 2/150 = 0.01//^(@)C`