Current Electricity - Result Question 47
50. The thermo e.m.f $E$ in volts of a certain thermocouple is found to vary with temperature difference $\theta$ in ${ }^{\circ} C$ between the two junctions according to the relation
[2010]
$E=30 \theta-\frac{\theta^{2}}{15}$
The neutral temperature for the thermocouple will be
(a) $30^{\circ} C$
(b) $450^{\circ} C$
(c) $400^{\circ} C$
(d) $225^{\circ} C$
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Answer:
Correct Answer: 50. (d)
Solution:
(d) $E=30 \theta-\frac{\theta^{2}}{15}$
For neutral temperature, $\frac{d E}{d \theta}=0$
$0=30-\frac{2}{15} \theta$
$\therefore \theta=15 \times 15$
$=225^{\circ} C$
Hence, neutral temperature is $225^{\circ} C$.
