Q. A thermocouple of resistance $1.6\Omega$ is connected in series with a galvanometer of $8\Omega$ resistance. The thermocouple develops an $e.$m.f. of $10\mu V$ per degree temperature difference between two junctions. When one junction is kept at $0℃$ and the other in a molten metal, the galvanometer reads $8\ millivolt.$ The temperature of molten metal, when $e$.m.f. varies linearly with temperature difference, will be
A) $960℃$
B) $1050℃$
C) $1275℃$
D) $1545℃$
Explanation
(a) Let the temperature of molten metal is $t℃$
The thermo-$e$mf $e = 10 \times 10^{- 6}t\ volt$
Current in the circuit $\frac{e}{R + R_{G}} = \frac{10^{- 5}t}{8 + 1.6} = \frac{10^{- 5}t}{9.6}amp$
But$\ i = \frac{V}{R_{G}} = \frac{8 \times 10^{- 3}}{8}$
$\therefore\frac{10^{- 5}t}{9.6} = \frac{8 \times 10^{- 3}}{8}\ or\ t = \frac{9.6 \times 10^{- 3}}{10^{- 5}} = 960℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.