The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:
`R = R_(0) [1 + alpha (T - T_(0))]`
The resistance is `101.6 Omega` at the triple-point of water `273.16 K , " and " 165.5 Omega` at the normal melting point of lead `(600.5 K)`. What is the temperature when the resistance is `123.4 Omega`?

Explanation

Here, `R_(0)=101.6Omega,T_(0)=273.16K`
Case (i) `R_(1)=165.5Omega,T_(1)=600.5K`
Case (ii) `R_(2)=123.4Omega,T_(1)=?`
Using the relation `R=R_(0)[1+alpha(T-T_(0))]`
Case (i) `165.5=101.6[1+alpha(600.6-273.16)]`
`alpha=(165.5-101.6)/(101.6xx(600.5-273.16))=63.9/(101.6xx327.34)`
Case (ii) `123.4=101.6[1+alpha(T_(2)-273.16)]`
or `123.4=101.6[1+63.9/(101.6xx327.34)(T_(2)-273.16)]`
= `101.6+63.9/327.34(T_(2)-273.16)`
or `T_(2)=((123.4-101.6)xx327.34)/63.9+273.16`
= `111.67+273.16=384.83K`