The resistances corresponding to the lower and the uppper fixed points, in a platinum resistance thermometer are `3.50Omegaand3.65Omega`. What would be the resistance at a temperature equal to the freezing point of mercury `(-37^(@)C)`?

Explanation

Given : `R_(0) = 3.50 Omega, R_(100) = 3.65 Omega` and t = `-37^(@)C, R_(t) ` = ?
Using t = `[ (R_(t) - R_(0))/(R_(100) - R_(0)) ] 100^(@)C `
We have , - `37^(@)C = [ (R_(t) - 3.50)/(3.65 - 3.50) ] 100^(@)` C
`rArr R_(t) - 3.50 = - 0.056 " " R_(t) = 3.55 Omega` .
`therefore ` The required resistance is `3.55 Omega`