In This post you will get answer of Q.
Thermo e.m.f. $E$ volt of certain thermo-couple varies with temperature $theta$ of a hot junction as
$E=40,theta-frac{theta^2}{20}$
The neutral temperature of the thermo-couple is
Solution:
When $theta = theta_n , frac{dE}{dtheta} $ = 0
$therefore$ $40 – frac{1}{10} theta_n $ = 0 or $frac{1}{10}theta_n$ = 40
or $theta_n$ = 400$^circ$C
Solution:
When $theta = theta_n , frac{dE}{dtheta} $ = 0
$therefore$ $40 – frac{1}{10} theta_n $ = 0 or $frac{1}{10}theta_n$ = 40
or $theta_n$ = 400$^circ$C