Thermal Properties of Matter

JEE Main level, single correct, medium

111 of 126

Question 111 of 126

The thermo emf of a thermocouple varies with the temperature θ\theta of the hot junction as E=aθ+bθ2\mathrm{E}=\mathrm{a} \theta+\mathrm{b} \theta^2 in volts where the ratio a/b\mathrm{a} / \mathrm{b} is 700∘C700^{\circ} \mathrm{C}. If the cold junction is kept at 0∘C0^{\circ} \mathrm{C}, then the neutral temperature is

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Answer: (D) no neutral temperature is possible for this thermocouple.

E=aθ+bθ2E=a \theta+b \theta^2
At neutral temperature dE/dθ=0\mathrm{dE} / \mathrm{d} \theta=0
∴dEdθ=a+2 bθn=0;θn=−a2 b\therefore \frac{\mathrm{dE}}{\mathrm{d} \theta}=\mathrm{a}+2 \mathrm{~b} \theta_{\mathrm{n}}=0 ; \theta_{\mathrm{n}}=-\frac{\mathrm{a}}{2 \mathrm{~b}}
Now ab=700∘C\frac{a}{b}=700^{\circ} \mathrm{C} (given)
θn=−700/2=−350∘C\theta_n=-700 / 2=-350^{\circ} \mathrm{C}
Now θc=0∘C\theta_c=0^{\circ} \mathrm{C}.
So, θn>0∘C\theta_n\gt 0^{\circ} \mathrm{C}
But mathematically θn<0∘C\theta_{\mathrm{n}} \lt 0^{\circ} \mathrm{C}.

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