Q. The temperature of hot junction of a thermocouple changes from 80$℃\ $to 100$℃$, the percentage change in thermo electric power is
A) 25%
B) 20%
C) 10%
D) 8%
Explanation
(a) Thermo electric power, $S \propto \theta$
$\therefore\frac{S_{100}}{S_{80}} = \frac{100}{80}\ or\ S_{100} = S_{80} \times \frac{100}{80} = \frac{5}{4}S_{80}$
Therefore % change in thermo electric power
$= \left( \frac{S_{100} - S_{80}}{S_{80}} \right) \times 100$
$= \left( \frac{\frac{5}{4}S_{80} - S_{80}}{S_{80}} \right) \times 100$
= 25%
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.