The V – I graph for a conductor at temperature ${\mathrm{T}}_1 \mathrm{and} {\mathrm{T}}_2$ are as shown in figure. Then (${\mathrm{T}}_2– {\mathrm{T}}_1$) is equal to [$\alpha$ = temperature coefficient of resistivity]

 

According to Ohm’s Law, V = R.I. It means that ‘R’ is the slope of V – I graph.

$$\begin{gathered} \therefore {\mathrm{R}}_1=\mathrm{tan\theta } \mathrm{and} \therefore {\mathrm{R}}_2={\mathrm{R}}_1\left[1+\mathrm{\alpha }\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)\right] \\ \therefore \mathrm{cot\theta }=\mathrm{tan\theta }\left[1+\mathrm{\alpha }\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)\right] \\ \Rightarrow \mathrm{\alpha }\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)=\left(\frac{\mathrm{cot\theta }}{\mathrm{tan\theta }}-1\right)\Rightarrow \left({\mathrm{T}}_2-{\mathrm{T}}_1\right)=\frac{1}{\mathrm{\alpha }}.\frac{\cos 2\mathrm{\theta }}{{\sin }^2\mathrm{\theta }}=\frac{1}{\mathrm{\alpha }}\frac{\cos {60}^0}{{\sin }^2{30}^0}=\frac{2}{\mathrm{\alpha }} \end{gathered}$$