Question

Three rods made of the same material and having the same cross-section have been joined as shown in the figure. Each rod is of the same length. The left and right ends are kept at $0^{\circ }\text{C}$ and ${90}^{\circ }\text{C}$, respectively. The temperature of junction of the three rods will be

• A. ${45}^{\circ }\text{C}$
• B. ${60}^{\circ }\text{C}$
• C. ${30}^{\circ }\text{C}$
• D. ${20}^{\circ }\text{C}$

Answer 1

The correct option is B ${60}^{\circ }\text{C}$

Let us suppose, $T$ be the temperature of the junction.Thermal resistance $\left(R\right)$ of all the three rods is equal because each rod is identical in every aspect.

Rate of heat flow $\left(Q_3\right)$ through $\text{AB}=$ Rate of heat flow $\left(Q_1\right)$ through $\text{BC}+$ Rate of heat flow $\left(Q_2\right)$ through $\text{BD}$

$\therefore Q_3=Q_1+Q_2$

$\Rightarrow \frac{T-0^{\circ }}{R}=\frac{{90}^{\circ }-T}{R}+\frac{{90}^{\circ }-T}{R}[\because Q=\frac{\Delta T}{R}]$

$\Rightarrow T={60}^{\circ }\text{C}$

Hence, option $\left(b\right)$ is correct.

$\begin{array}{c}\text{Why this question ?}\\ \text{Concept - Rate of heat flow,}Q=\frac{\text{Temperature difference}\left(\Delta T\right)}{\text{Thermal resistance}\left(R\right)}\\ \text{where,}R=\frac{l}{kA},k=\text{Thermal conductivity of the rod.}\\ \text{This is similar to the current flow through a resistance}\left(R\right)\text{where}\\ \text{current (I)= Rate of flow of charge =}\frac{\text{Potential difference}}{\text{Elecrical resistance}}\\ \text{Here,}R=\frac{l}{\sigma A},\sigma =\text{Electrical conductivity}\end{array}$