Animated Solution for Physics - Properties of Solids and Liquids: Temperature difference of $120^{\circ }\text{C}$ is maintained between two ends of a uniform rod $AB$ of length $2L$. Another bent rod $PQ$, of same cross-section as $AB$ and length $\frac{3L}{2}$ is connected across $AB$ (see figure). In steady state, temperature difference between $P$ and $Q$ will be close to

Imagine you are looking at a complex network of water pipes, or better yet, an electrical circuit. Heat transfer through conduction behaves in exactly the same way! Heat flows from a region of higher temperature to a region of lower temperature, driven by a temperature difference ($\Delta T$). This is perfectly analogous to electrical current flowing due to a potential difference (Voltage).

In this problem, the rods act as our thermal resistors. The thermal resistance of any uniform rod is given by the formula:

where $L$ is the length, $K$ is the thermal conductivity, and $A$ is the cross-sectional area. Since both rods are made of the same material and have the same cross-section, their thermal resistance is directly proportional to their length ($R\propto L$).

Let's assign a base resistance to make our calculations elegant. Let the resistance of the middle straight part $PQ$, which has a length of $L$, be $R$.

Now, look at the segments $AP$ and $QB$. Each of these segments has a length of $L/2$. Because resistance is proportional to length, their resistances will be exactly half:

What about the bent rod? It consists of three segments: two vertical segments of length $L/4$ and a horizontal segment of length $L$. Its total length is $L/4+L+L/4=\frac{3L}{2}$. Therefore, its resistance is:

Between points $P$ and $Q$, the heat has two paths to travel: the straight rod and the bent rod. This is a classic parallel combination! To find the equivalent resistance of this section ($R_{P\to Q}$), we use the product-over-sum rule:

Now, let's zoom out and look at the entire rod from $A$ to $B$. The heat must flow through segment $AP$, then through the parallel combination between $P$ and $Q$, and finally through segment $QB$. These three sections are in series. We simply add their resistances to find the total equivalent resistance of the system:

We are given that the total temperature difference across the entire system is $120^{\circ }\text{C}$. Using the thermal equivalent of Ohm's Law ($I=\frac{\Delta T}{R}$), we can find the total thermal current flowing through the main rod:

We are almost there! The question asks for the temperature difference specifically between points $P$ and $Q$.

Just like finding the voltage drop across a specific resistor in a series circuit, we multiply the total thermal current by the equivalent resistance of that specific section:

Notice how beautifully the $R$ cancels out!

And there we have it. By trusting the electrical analogy, a complex thermal geometry collapses into a simple, elegant calculation.