**Problem:**
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°C and 90°C respectively. The temperature of junction of the three rods will be?
(IIT JEE 2001)

- 45°C
- 60°C
- 30°C
- 20°C

**Answer:** The answer is (B) i.e., temperture of junction of the three rods is 60°C.

**Solution:**
Let T be the temperature of the junction. The rate of heat flow to the junction is
\begin{align}
\frac{\mathrm{d}Q_\text{in}}{\mathrm{d}t}&=\kappa A \frac{(90-T)}{l}+\kappa A \frac{(90-T)}{l}\nonumber\\
&=2\kappa A \frac{(90-T)}{l},
\end{align}
and the rate of heat flow out of the junction is
\begin{align}
\frac{\mathrm{d}Q_\text{out}}{\mathrm{d}t}=\kappa A \frac{(T-0)}{l}=\frac{\kappa AT}{l}.
\end{align}
In steady state, the rate of inflow of heat at the junction is equal to the rate of outflow of heat i.e.,
\begin{align}
\frac{\mathrm{d}Q_\text{in}}{\mathrm{d}t}=\frac{\mathrm{d}Q_\text{out}}{\mathrm{d}t}.
\end{align}
Solve above equations to get temperature of junction of the three rods T = 60°C.

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