Q157 / 160 · AP EAMCET 2011 · 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 $a / b$ 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 answerDNone of these
Solution
Given, $E=a \theta+b \theta^2$
$\begin{aligned}
\therefore \quad p & =\frac{d E}{d \theta}=\frac{d}{d \theta}\left(a \theta+b \theta^2\right) \\
& =a+2 b \theta
\end{aligned}$
At $\quad \theta=T_n, p=0$
$\Rightarrow \quad T_n=-\frac{a}{2 b}=-350^{\circ} \mathrm{C}$
No correct option is given.
Hence, the correct answer is option: 'D'.