Question

The electrical resistance of pure platinum increases linearly with increasing temperature over a small range of temperature. This property is used in a Platinum resistance thermometer. The relation between $R_{\theta }$ (Resistance at $\theta K$) and $R_0$ (Resistance at ${\theta }_{0}K)$ is given by $R_{\theta }=R_0[1+\alpha (\theta -{\theta }_0\left)\right]$, where $\alpha =$ temperature coefficient of resistance. Now, if a Platinum resistance thermometer reads $0^{\circ }C$ when its resistance is $80\Omega$ and ${100}^{\circ }C$ when its resistance is $90\Omega$, find the temperature at which its resistance is $86\Omega$.

Answer 1

Using the given relationship
we have $(90-80)\Omega =\alpha \left(80\Omega \right)\left(100\right).....\left(i\right)$
$(86-80)\Omega =\alpha \left(80\Omega \right)\left(\theta \right).....\left(ii\right)$
Where $\theta$ is the desired temperature.
Taking the ratio of (i) and (ii)
$\frac{10}{6}=\frac{100}{\theta }\Rightarrow =60K$.