Q. The $V - i$ graph for a conductor at temperatures $T_{1}$ and $T_{2}$ are as shown in the figure. $(T_{2} - T_{1})$ is proportional to

A) $\cos{2\theta}$
B) $\sin\theta$
C) $\cot{2\theta}$
D) $\tan\theta$
Explanation

(c) As we know, for conductors, resistance $\propto$ temperature.
From figure $R_{1} \propto T_{1} \Rightarrow \tan\theta \propto T_{1} \Rightarrow \tan\theta = kT_{1}$ …(i)
and $R_{2} \propto T_{2} \Rightarrow \tan{(90{^\circ} - \theta) \propto T_{2} \Rightarrow \cot{\theta = kT_{2}}}$ …(ii)
From equation (i) and (ii), $k\left( T_{2} - T_{1} \right) = (\cot{\theta - \tan\theta})$
$\left( T_{2} - T_{1} \right) = \left( \frac{\cos\theta}{\sin\theta} - \frac{\sin\theta}{\cos\theta} \right) = \frac{(\cos^{2}{\theta - \sin^{2}\theta})}{\sin{\theta\cos\theta}} = 2\cot{2\theta}$
$\Rightarrow (T_{2} - T_{1}) \propto \cot{2\theta}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.