Q67 / 120 · NEET 2010 (Mains) · Physics
The thermo emf $E$ in volt of a certain thermo-couple is found to vary with temperature difference $\theta$ in ${ }^{\circ} \mathrm{C}$ between the two junctions according to the relation $E=30 \theta-\frac{\theta^2}{15}$
The neutral temperature for the thermo-couple will be
- A$450^{\circ} \mathrm{C}$
- B$400^{\circ} \mathrm{C}$
- Correct answerC$225^{\circ} \mathrm{C}$
- D$30^{\circ} \mathrm{C}$
Solution
Given, $\mathrm{E}=30 \theta-\frac{\theta^2}{15}$
At neutral temperature
$\begin{aligned}
\frac{\mathrm{dE}}{\mathrm{d} \theta} & =0 \\
\therefore \quad 30-\frac{2 \theta}{15} & =0 \\
\theta & =225^{\circ} \mathrm{C}
\end{aligned}$
Hence, the correct answer is option: 'C'.