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JIPMER 2015
Exam
JIPMER
Year
2015
Subject
Physics
Chapter
Thermal Properties of Matter

JIPMER 2015 · Physics · Thermal Properties of Matter

The temperature of the cold junction of thermocouple is 0^ C and...

The temperature of the cold junction of thermocouple is 0∘C0^{\circ} \mathrm{C} and the temperature of hot junction is T∘CT^{\circ} \mathrm{C}. The emf is E=16T−0.04T2μVE=16 T-0.04 T^2 \quad \mu \mathrm{V}. The inversion temperature TiT_i is

Options

  1. A300∘C300^{\circ} \mathrm{C}
  2. B200∘C200^{\circ} \mathrm{C}
  3. C500∘C500^{\circ} \mathrm{C}
  4. D400∘C400^{\circ} \mathrm{C}

Correct answer

D. 400∘C400^{\circ} \mathrm{C}

Step-by-step solution

As, E=16T−0.04T2E=16 T-0.04 T^2 At inversion temperature, E=0E=0⇒16T−0.04T2=0⇒Ti=160.04=400∘C\begin{aligned} \Rightarrow \quad 16 T-0.04 T^2 & =0 \\ \Rightarrow \quad T_i & =\frac{16}{0.04}=400^{\circ} \mathrm{C}\end{aligned}$$
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