Q.
Three rods made of same material and having same cross-section have been joined as shown in figure. Each rod is of same length. The left and right ends are kept at 0$℃$ and 90$℃$ respectively. The temperature of the junction of the three rods will be
A) 45$℃$
B) 60$℃$
C) 30$℃$
D) 20$℃$
Explanation
(b) Let the temperature of junction be$Q$. In equilibrium, rate of flow of heat through rod 1= sum of rate of flow of heat through rods 2 and 3.
$\left( \frac{dQ}{dt} \right)_{1} = \left( \frac{dQ}{dt} \right)_{2} + \left( \frac{dQ}{dt} \right)_{3}$
$KA\frac{(\theta - 0)}{l} = \ \frac{KA(90{^\circ} - \theta)}{l} + \frac{KA(90{^\circ} - \theta)}{l}$
$\theta\ \ = 2(90{^\circ} - \theta)$
3$\theta = 180{^\circ}\ ,\ \theta = \frac{180{^\circ}}{3} = 60{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.