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) $225^{\circ} C$

Show Answer

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 $225^{\circ} C$.