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A thermocouple of negligible resistance produces an e.m.f. of 40 μV/°C in the linear range of temperature. A galvanometer of resistance 10 ohm whose sensitivity is 1 μA/div, is employed with the thermocouple. The smallest value of temperature difference that can be detected by the system will be

1. 0.5°C
2. 1°C
3. 0.1°C
4. 0.25°C

Correct answer: 0.5°C

Solution

The smallest current that the galvanometer can detect is 1 μA (1 μA/div sensitivity). Using Ohm's law, the smallest voltage detectable is V = IR = (1 × 10⁻⁶ A)(10 Ω) = 10 μV. Since the thermocouple produces 40 μV/°C, the smallest temperature difference is ΔT = V / (40 μV/°C) = 10 μV / 40 μV/°C = 0.25°C.

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