The thermo e.m.f. of a thermo-couple is 25 μV/^ at room temperature. A galvanometer of 40 Ω resistance, capable of detecting current as low as 10^-5 A - Physics MCQ AIEEE 2003 2026 | ExamTest.live

mediumMCQAIEEE 20032026Physics

1 mark

The thermo e.m.f. of a thermo-couple is 25 μ V/^{\circ} C at room temperature. A galvanometer of 40 Ω resistance, capable of detecting current as low as 1 0^{- 5} A, is connected with the thermocouple. The smallest temperature difference that can be detected by this system is:

Solution & Step-by-step Explanation

Let the smallest temperature difference be Δ θ .The generated thermo-emf is E = 25 \times 1 0^{- 6} \times Δ θ .The minimum detectable voltage V by the galvanometer is: V = I_{min} \times R = 1 0^{- 5} \times 40 = 4 \times 1 0^{- 4} V Equating E and V : 25 \times 1 0^{- 6} \times Δ θ = 4 \times 1 0^{- 4} Δ θ = \frac{4 \times 1 0 ^{- 4}}{25 \times 1 0 ^{- 6}} = \frac{400}{25} = 1 6^{\circ} C

Try it yourself before checking the explanation above.

The thermo e.m.f. of a thermo-couple is 25 μ V/^{\circ} C at room temperature. A galvanometer of 40 Ω resistance, capable of detecting current as low as 1 0^{- 5} A, is connected with the thermocouple. The smallest temperature difference that can be detected by this system is:

1 6^{\circ} C

1 2^{\circ} C

8^{\circ} C

2 0^{\circ} C

Practice with timed mock tests and track your performance across Physics.