Question

Three rods made of 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 $0^{\circ }C$ and ${90}^{\circ }C$, respectively. The temperature of the junction of the three rods will be

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

Answer 1

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

Let ${\theta }^{\circ }C$ be the temperature at $B$. Let $Q$ be the heat flowing per second from $A$ to $B$ on account of temperature difference by conductivity .


$\therefore Q=\frac{kA(90-\theta )}{l}...\left(i\right)$
where $k=$ thermal conductivity of the rod
$A=$ Area of cross section of the rod
$l=$ length of the rod
By symmetry, the same will be the case for heat flow from $C$ to $B$.
$\therefore$ The heat flowing per second from $B$ to $D$ will be
$2Q=\frac{kA(\theta -0)}{l}............\left(ii\right)$
Dividing Eq. $\left(ii\right)$ by Eq. $\left(i\right)$, we get
$2=\frac{\theta }{90-\theta }\Rightarrow \theta ={60}^{\circ }C$