Q. If $E = at + bt^{2},$ what is the neutral temperature
A) $- \frac{a}{2b}$
B) $+ \frac{a}{2b}$
C) $- \frac{a}{b}$
D) $+ \frac{a}{b}$
Explanation
(a) $E = \alpha t + bt^{2}.$ At temperature of inversion $E$ is minimum
$i.e.,\ E = 0$
$\therefore\alpha t_{i} + bt_{i}^{2} = 0,\ i.e.,\ t_{1} = - \frac{a}{b}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.