Heat and Thermal ExpansionHard

Question

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

Options

A.0.25oC
B.0.5oC
C.1oC
D.0.1oC

Solution

Thermoemf of thermocouple = 40μ V/oC
Let θoC be smallest temperature difference that can be measured.
∴ After connecting the thermocouple with the galvanometer, thermoemf,
ε = 10 μ V/oC × θoC = 40θo μV
Smallest potential difference developed across the galvanometer
= IR = 1μA × 10 Ω = 10μV
∴ 40θo μV = 10μV
θ = oC = 0.25o C

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