LEVELJEE Main
Visualized Solution
The Sigma Insight: Heat Transfer
Welcome to a classic problem in thermal physics! This question tests your fundamental understanding of steady-state heat conduction and your ability to manipulate algebraic expressions cleanly.
Imagine you are standing in front of a composite rod—two different materials welded together end-to-end. One side is exposed to a hot reservoir, and the other to a cold one. Our mission is to find the exact temperature at the junction where these two materials meet.
Analyzing the Setup
Let's break down the physical reality of the situation. We have a thermally insulated rod, which means heat cannot escape from the sides. It can only flow straight through from one end to the other, much like water flowing through a pipe.
The left section has a length and a thermal conductivity . Its outer end is maintained at a temperature .
The right section has a length and a thermal conductivity . Its outer end is maintained at a temperature .
We are looking for the temperature at the interface, which we will call .
The most crucial physical insight here is the concept of a "steady state." In a steady state, the temperature profile of the rod is no longer changing with time. Because no heat is being stored or lost, the rate at which heat flows through the first section must be exactly equal to the rate at which it flows through the second section.
The Master Equation
To translate this physical insight into mathematics, we use Fourier's Law of Heat Conduction.
Fourier's Law states that the heat current (the rate of heat flow ) is proportional to the cross-sectional area and the temperature difference , and inversely proportional to the length .
Mathematically, it is written as:
Since the heat current is the same in both sections, we can set up our master equation by equating and :
Algebraic Manipulation
Now, we enter the execution phase. Our goal is to isolate .
First, notice that the cross-sectional area is the same for both sections. We can immediately cancel it out from both sides, simplifying our equation:
Next, to eliminate the fractions, we cross-multiply the lengths and :
Let's expand the brackets carefully to separate the terms containing our unknown :
Final Calculation
We are almost there! We need to group all the terms containing on one side of the equation. Let's move the negative term from the left to the right, and the negative term from the right to the left:
Now, factor out on the right side:
Finally, isolate by dividing both sides by the bracketed term:
This perfectly matches option (c).
The Electrical Analogy
If you are familiar with current electricity, this result should look incredibly familiar!
Heat conduction is mathematically identical to electrical conduction. Temperature difference is analogous to voltage difference, and heat current is analogous to electrical current.
The thermal resistance of a rod is given by .
If you have two electrical resistors in series, the voltage at the node between them is given by the exact same weighted average formula:
By substituting the thermal resistances into this electrical formula, you can arrive at the exact same answer in seconds! This is a powerful mental model that can save you precious time in competitive exams like JEE.
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