The resistance of a platinum wire is `15 Omega` at `20^(@)C`. This wire is put in hot furnace and lthe resistance of the wire is found to be `40 Omega`. Find the temperature of the hot furnace if temperature coefficient of resistance of platinum is `3.6 xx 10^(-3)^(0)C^(-1)`

Explanation

`R_(1) = R_(0)(l + alpha t) rArr (R_(2))/(R_(1)) = ((l + alphat))/((l + alphat_(1))`
`(40)/(15)`
`= ((l + alphat_(2))/(l + alphat_(1))) rArr 40 - 15 = alpha(15t_(2) - 40alpha_(1))`
`15t_(2) = (25)/(3.6 xx 10^(-3)) + 40 xx 20 approx`
`Rightarrowt_(2)=516^(@)C`