Question

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

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

Answer 1

The correct option is C $\cot 2\theta$

$\text{Step1: Slope of}V-I\text{graph}$

Ohm's Law states that $V=IR$

From the graph

Slope of $V-I$ graph will be resistance

So, $\frac{V_1}{I_1}=\tan \theta =R_1$

$\&$ $\frac{V_2}{I_2}=\tan (90-\theta )=\cot \theta =R_2$

$\text{Step2: Calculation of}$ $T_2-T_1$

Temperature dependence of resistance is given by:

$R_1=R_0(1+\alpha T_1)$

$R_2=R_0(1+\alpha T_2)$

Subtracting above equations, we get

$R_2-R_1=\alpha (T_2-T_1)$

$\therefore T_2-T_1$ is proportional to $R_2-R_1$

$\Rightarrow$ $T_2-T_1\propto (\cot \theta -\tan \theta )=\frac{\cos \theta }{\sin \theta }-\frac{\sin \theta }{\cos \theta }=\frac{{\cos }^2\theta -{\sin }^2\theta }{\sin \theta \cos \theta }$

$T_2-T_1\propto \frac{2\cos 2\theta }{\sin 2\theta }=2\cot 2\theta$

Hence, $T_2-T_1\propto \cot 2\theta$

Therefore, Option C is correct