The resistance of the wire in the platinum resistance thermometer at ice point is `5Omega` and at steam point is `5.25Omega`. When the thermometer is inserted in an unknown hot bath its resistance is found to be `5.5Omega`. The temperature of the hot bath is

Options

A. `100^(@)C`
B. `200^(@)C`
C. `300^(@)C`
D. `350^(@)C`

Answer

B

Explanation

Here, `R_(0)=5Omega,R_(100)=5.25Omega`
`R_(T)=5.5Omega`
As `R_(T)=R_(0)(1+alphaT) therefore R_(100)=R_(0)(1+alphaxx100)`
`alpha=(R_(100)-R_(0))/(R_(0)xx100)" "...(i)`
Let the temperature of hot bath be `T^(@)C`.
`R_(T)=R_(0)(1+alphaT) therefore alpha=(R_(T)-T_(0))/(R_(0)xxT)" "...(ii)`
Equating equations (i) and (ii), we get
`(R_(100)-R_(0))/(R_(0)xx100)=(R_(T)-R_(0))/(R_(0)xxT)`
`T=(R_(T)-R_(0))/(R_(100)-R_(0))xx100=(5.5-5)/(5.25-5)xx100=200^(@)C`