Q. The emf of a thermocouple, cold junction of which is kept at $- 300℃$ is given by $E = 40t + \frac{1}{10}t^{2}.$ The temperature of inversion of thermocouple will be
A) $200℃$
B) $400℃$
C) $- 200℃$
D) $- 100℃$
Explanation
(c) $E = 40t + \frac{1}{10}t^{2}\ at$
Inversion, $E$ will be minimum
$\frac{dE}{dt} = 0$
$\frac{d}{dt}\left( 40t + \frac{1}{10}\ t^{2} \right) = 0$
$40 + \frac{1}{5}\ t = 0$
$t = - 40 \times 5$
$t = - 200℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.