The temperature of inversion of a thermocouple is $620^{\circ} \mathrm{C}$ and the neutral temperature is…
- $320^{\circ}$
- $20^{\circ} \mathrm{C}$
- $-20^{\circ} \mathrm{C}$
- $40^{\circ} \mathrm{C}$
Solution
$\begin{array}{l}
T_n=\frac{T_1+T_C}{2} \\
\text {or } 600 =620+T_C \\
\Rightarrow T_C =-20^{\circ} \mathrm{C}
\end{array}$ ^
Asked in: NEET 2005
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