Q201 / 225 · JEE Main 2004 · Physics

The thermo emf of a thermocouple varies with the temperature $\theta$ of the hot junction as $\mathrm{E}=\mathrm{a} \theta+\mathrm{b} \theta^2$ in volts where the ratio $\mathrm{a} / \mathrm{b}$ 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
    $350^{\circ} \mathrm{C}$
  • C
    $1400^{\circ} \mathrm{C}$
  • Correct answer
    D
    no neutral temperature is possible for this thermocouple.

Solution

$E=a \theta+b \theta^2$ At neutral temperature $\mathrm{dE} / \mathrm{d} \theta=0$ $\therefore \frac{\mathrm{dE}}{\mathrm{d} \theta}=\mathrm{a}+2 \mathrm{~b} \theta_{\mathrm{n}}=0 ; \theta_{\mathrm{n}}=-\frac{\mathrm{a}}{2 \mathrm{~b}}$ Now $\frac{a}{b}=700^{\circ} \mathrm{C}$ (given) $\theta_n=-700 / 2=-350^{\circ} \mathrm{C}$ Now $\theta_c=0^{\circ} \mathrm{C}$. So, $\theta_n>0^{\circ} \mathrm{C}$ But mathematically $\theta_{\mathrm{n}} < 0^{\circ} \mathrm{C}$.

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

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