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`

Temperature coefficient of resistance of platinum is `alpha=3.92xx10^(

1.	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`
Answer» `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`

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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`

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