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Current Electricity question

2003 · Shift 0 · Q127
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Current Electricity question

2003 · Shift 0 · Q127

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
The thermo e.m.f.e.m.f.e.m.f. of a thermo -couple is 25μV/∘C25\mu V/{}^ \circ C25μV/∘C at room temperature. A galvanometer of 40ohm40ohm40ohm resistance, capable of detecting current as low as 10−5 A,{10^{ - 5}}\,A,10−5A, is connected with the thermo couple. The smallest temperature difference that can be detected by this system is
  1. A
    160C{16^0}C160C
  2. B
    120C{12^0}C120C
  3. C
    80C{8^0}C80C
  4. D
    200C{20^0}C200C
View written solutionFree

Correct answer: A

  1. Given data
  • Thermo e.m.f. sensitivity of thermocouple: 25 μV/∘C=25×10−6 V/∘C25\,\mu V/^\circ C = 25 \times 10^{-6}\,V/^\circ C25μV/∘C=25×10−6V/∘C
  • Galvanometer resistance: R=40 ΩR = 40\,\OmegaR=40Ω
  • Minimum detectable current: Imin⁡=10−5 AI_{\min} = 10^{-5}\,AImin​=10−5A
  1. Minimum e.m.f. needed to produce detectable current

Using Ohm’s law, Vmin⁡=Imin⁡RV_{\min} = I_{\min}RVmin​=Imin​R Vmin⁡=10−5×40=4×10−4 VV_{\min} = 10^{-5} \times 40 = 4 \times 10^{-4}\,VVmin​=10−5×40=4×10−4V

  1. Relate thermo e.m.f. to temperature difference

If the thermocouple gives 25×10−6 V25 \times 10^{-6}\,V25×10−6V per degree Celsius, then for temperature difference ΔT\Delta TΔT, E=25×10−6ΔTE = 25 \times 10^{-6} \Delta TE=25×10−6ΔT

For the smallest detectable temperature difference, 25×10−6ΔT=4×10−425 \times 10^{-6} \Delta T = 4 \times 10^{-4}25×10−6ΔT=4×10−4

So, ΔT=4×10−425×10−6\Delta T = \frac{4 \times 10^{-4}}{25 \times 10^{-6}}ΔT=25×10−64×10−4​ ΔT=425×102=16\Delta T = \frac{4}{25} \times 10^2 = 16ΔT=254​×102=16

Thus, ΔT=16∘C\Delta T = 16^\circ CΔT=16∘C

  1. Check options
  • A: 16∘C16^\circ C16∘C ✅
  • B: 12∘C12^\circ C12∘C ❌
  • C: 8∘C8^\circ C8∘C ❌
  • D: 20∘C20^\circ C20∘C ❌

Therefore, the correct option is A.

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