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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 $E = 30\theta - \frac{\theta^2}{15}$. The neutral temperature for the thermocouple will be
A
$450^\circ C$
B
$400^\circ C$
C
$225^\circ C$
D
$30^\circ C$
Detailed Solution
The neutral temperature is the temperature at which the thermo emf is maximum, i.e., $\frac{dE}{d\theta} = 0$
$E = 30\theta - \frac{\theta^2}{15}$
$\frac{dE}{d\theta} = 30 - \frac{2\theta}{15}$
Setting it to zero: $30 - \frac{2\theta}{15} = 0$
$\theta = \frac{30\times15}{2} = 225^\circ C$
$E = 30\theta - \frac{\theta^2}{15}$
$\frac{dE}{d\theta} = 30 - \frac{2\theta}{15}$
Setting it to zero: $30 - \frac{2\theta}{15} = 0$
$\theta = \frac{30\times15}{2} = 225^\circ C$