Question

$V-I$ graph of a conductor of temperature $T_1$ and $T_2$ are shown in the figure. Then, $(T_2-T_1)$ is proportional to

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

Answer 1

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


Slope of line gives ,

$\therefore R_1=\tan \theta =R_0[1+\alpha T_1]------\left(1\right)$

$R_2=\tan (90-\theta )=R_0[1+\alpha T_2]$

$\cot \theta =R_0[1+\alpha T_2]------\left(2\right)$

$\left(2\right)-\left(1\right)$ gives us,

$\cot \theta -\tan \theta =R_0\alpha [T_2-T_1]$

$\frac{\cos \theta }{\sin \theta }-\frac{\sin \theta }{\cos \theta }=R_0\alpha (T_2-T_1)$

$R_0\alpha (T_2-T_1)=\frac{{\cos }^2\theta -{\sin }^2\theta }{\sin \theta \cos \theta }=\frac{\cos 2\theta }{\left(\frac{\sin 2\theta }{2}\right)}=2\cot 2\theta$

$\therefore (T_2-T_1)\propto \cot 2\theta$