The resistance of the platinum wire of a platinum resistance thermometer at an ice point is \(5 \Omega\) and at steam point is \(5.23 \Omega\). When the thermometer is inserted in a hot bath, the resistance of a platinum wire is \(5.795 \Omega\). Calculate the temperature of the bath.

(C)
Here \(R _0=5 \Omega \quad \theta_0=0^{\circ} C\)
\(\begin{aligned} R _1 =5.23 \Omega \quad \theta_1=100^{\circ} C \\ R _2 =5.795 \Omega \quad \theta_2=? \\ R _\theta = R _0\left[1+\alpha\left(\theta-\theta_0\right)\right]\end{aligned}\)
\(\begin{array}{ll}\therefore R _\theta- R _0= R _0 \alpha\left(\theta-\theta_0\right) \\ \therefore R _\theta- R _0= R _0 \alpha \theta\left(\because \theta_0=0\right)\end{array}\)
\(\therefore \quad R_1-R_0=R_0 \alpha \theta_1 \quad \quad \ldots \ldots(1)\)
\(\therefore \quad R_2-R_\theta=R_0 \alpha \theta_2 \quad \quad \ldots \ldots(2)\)
Taking the ratio of equation (1) and (2),
\(\begin{aligned} \frac{ R _1- R _0}{ R _2- R _0} =\frac{\theta_1}{\theta_2} \\ \therefore \frac{5.23-5}{5.795-5} =\frac{100}{\theta_2} \\ \therefore \frac{0.23}{0.795} =\frac{100}{\theta_2} \\ \therefore \theta_2 =345.65^{\circ} C \end{aligned}\)