Q. The expression for thermo $e$.m.f. in a thermocouple is given by the relation $E = 40\theta - \frac{\theta^{2}}{20},$ where $\theta$ is the temperature difference of two junctions. For this, the neutral temperature will be
A) $100℃$
B) $200℃$
C) $300℃$
D) $400℃$
Explanation
(d) Comparing the given equation with standard equation
$E = \alpha t + \frac{1}{2}\beta t^{2}$
$\alpha = 40\ and\frac{1}{2}\beta = - \frac{1}{20} \Rightarrow \beta = - \frac{1}{10}$
Hence neutral temperature $t_{n} = - \frac{\alpha}{\beta} = \frac{- 40}{- 1/10}$
$\Rightarrow t_{n} = 400℃$
Alternate solution : We know at neutral temperature
$\frac{dE}{d\theta} = 0$
$E = 40\theta - \frac{\theta^{2}}{20} \Rightarrow 40 - \frac{2\theta}{20} = 0 \Rightarrow \theta = 400℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.