Question

The $V-I$ graph for a conductor at temperature $T_1$ and $T_2$ are as shown in the figure. The term $\left(T_2-T_1\right)$ is proportional to

• A. $\cos 2\theta$
• B. $\sin 2\theta$
• C. $\cot 2\theta$
• D. $\tan 2\theta$

Answer 1

Answer: (C)

The slope of $V$ -I curve gives the resistance.
$R_1=\tan \theta =R_0\left(1+\alpha T_1\right)$
and $R_2=\cot \theta =R_0\left(1+\alpha T_2\right)$
$\cot \theta -\tan \theta =R_0\left(1+\alpha T_2\right)-R_0\left(1+\alpha T_1\right)=R_0\alpha \left(T_2-T_1\right)$
or $T_2-T_1=\frac{1}{\alpha R_0}(\cot \theta -\tan \theta )$
$=\frac{1}{\alpha R_0}\left(\frac{\cos \theta }{\sin \theta }-\frac{\sin \theta }{\cos \theta }\right)$
$=\frac{2\cos 2\theta }{\alpha R_0\sin 2\theta }=\frac{2}{\alpha R_0}\cot 2\theta$
or $T_2-T_1\propto \cot 2\theta$