Q98 / 160 · JIPMER 2012 · Physics

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

  • A
    $700^{\circ} \mathrm{C}$
  • B
    $1400^{\circ} \mathrm{C}$
  • C
    $390^{\circ} \mathrm{C}$
  • Correct answer
    D
    no neutral temperature is possible for this thermocouple

Solution

$E=a \theta+b \theta^2$ so $\frac{d E}{d \theta}=a+2 b \theta$ For neutral temperature $\theta=\theta_{\mathrm{n}}, \frac{\mathrm{dE}}{\mathrm{d} \theta}=0$ or $a+2 b \theta_n=0$ or $\theta_{\mathrm{n}}=-\mathrm{a} / 2 \mathrm{~b}=-700 / 2=-350^{\circ} \mathrm{C}$ As cold junction is at $0^{\circ} \mathrm{C}$ so negative neutral temperature can not be possible.

Hence, the correct answer is option: 'D'.

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