JEE MAIN - Physics (2003 - No. 9)
The thermo $$e.m.f.$$ of a thermo -couple is $$25$$ $$\mu V/{}^ \circ C$$ at room temperature. A galvanometer of $$40$$ $$ohm$$ resistance, capable of detecting current as low as $${10^{ - 5}}\,A,$$ is connected with the thermo couple. The smallest temperature difference that can be detected by this system is
$${16^0}C$$
$${12^0}C$$
$${8^0}C$$
$${20^0}C$$
Explanation
Let $$\theta $$ be the smallest temperature difference that can be detected by the thermocouple, then
$$I \times R = \left( {25 \times {{10}^{ - 6}}} \right)\theta $$
where $${\rm I}$$ is the smallest current which can be detected by the galvanometer of resistance $$R.$$
$$\therefore$$ $${10^{ - 5}} \times 40 = 25 \times {10^{ - 6}} \times \theta $$
$$\therefore$$ $$\theta = {16^ \circ }C.$$
$$I \times R = \left( {25 \times {{10}^{ - 6}}} \right)\theta $$
where $${\rm I}$$ is the smallest current which can be detected by the galvanometer of resistance $$R.$$
$$\therefore$$ $${10^{ - 5}} \times 40 = 25 \times {10^{ - 6}} \times \theta $$
$$\therefore$$ $$\theta = {16^ \circ }C.$$
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