Temperature coefficient of resistance of platinum is `alpha=3.92xx10^(-3)K^(-1)` at `0^(@)C`. Find the temperature at which the increase in the resistance of platinum wire is `10%` of its value at `0^(@)C`

Explanation

`R_(2)=(110R_(1))/(100)=1.1R_(1),alpha=3.92xx10^(-3)K^(-1)`
`Deltat=(R_(2)-R_(1))/(R_(1)alpha)implies(1.1R_(1)-R_(1))/(R_(1)alpha)`
`=(R_(1)(1.1-1))/(R_(1)alpha)=(0.1R_(1))/(R_(1)alpha)=(0.1)/(3.92xx10^(-3))`
`Deltat=25.51^(@)C,t_(2)=25.51+20=45.51^(@)C`