Q. If the cold junction is held at $0℃,$ the same thermo-emf $V$ of a thermocouple varies as $V = 10 \times 10^{- 6}\ t - \frac{1}{40} \times 10^{- 6}\ t^{2},\ $ where $t$ is the temperature of the hot junction in $℃.$ The neutral temperature and the maximum value of thermo-emf are respectively
A) $200℃;2\ mV$
B) $400℃;2\ mV$
C) $100℃;1\ mV$
D) $200℃;1\ mV$
Explanation
(d) Given, $V = 10 \times 10^{- 6}t - \frac{1}{40} \times 10^{- 6}t^{2}$
At neutral temperature
$\frac{dV}{dt} = 0$
$\therefore\ \ 10 \times 10^{- 6} - \frac{1}{20} \times 10^{- 6}\ t_{n} = 0$
or $t_{n} = 200℃$
Also at neutral temperature, thermo-emf is maximum.
Thus,
$V_{\max} = 10 \times 10^{- 6}(200) - \frac{1}{40} \times 10^{- 6}(200)^{2}$
$\ \ \ \ \ \ \ \ \ \ = 2 \times 10^{- 3} - 1 \times 10^{- 3} = 1\ mv$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.