Search any question & find its solution
Question:
Answered & Verified by Expert
If the cold junction is held at $0^{\circ} \mathrm{C}$, 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 ${ }^{\circ} \mathrm{C}$. The neutral temperature and the maximum value of thermo emf are respectively :
Options:
Solution:
1795 Upvotes
Verified Answer
The correct answer is:
$200^{\circ} \mathrm{C} ; 1 \mathrm{mV}$
Given, $V=10 \times 10^{-6} t-\frac{1}{40} \times 10^{-6} t^2$
At neutral temperature $\frac{d V}{d t}=0$
$\therefore \quad 10 \times 10^{-6}-\frac{1}{20} \times 10^{-6} t_n=0$
or $\quad t_n=200^{\circ} \mathrm{C}$
Also at neutral temperature, thermo emf is maximum.
Thus,
$V_{\max }=10 \times 10^{-6}(200)-\frac{1}{40} \times 10^{-6}(200)^2$
$=2 \times 10^{-3}-1 \times 10^{-3}$
$=1 \mathrm{mV}$
At neutral temperature $\frac{d V}{d t}=0$
$\therefore \quad 10 \times 10^{-6}-\frac{1}{20} \times 10^{-6} t_n=0$
or $\quad t_n=200^{\circ} \mathrm{C}$
Also at neutral temperature, thermo emf is maximum.
Thus,
$V_{\max }=10 \times 10^{-6}(200)-\frac{1}{40} \times 10^{-6}(200)^2$
$=2 \times 10^{-3}-1 \times 10^{-3}$
$=1 \mathrm{mV}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.