JEE Advanced 2018
LEVELJEE Main
Visualized Solution
The Sigma Insight: Heat Transfer
Imagine you are an engineer tasked with designing a thermal bridge between two extreme environments. On one side, a scorching heat bath at . On the other, a freezing bath at .
Between these two extremes, you place two conducting cylinders in series.
This isn't just a textbook problem; it's a classic scenario in thermodynamics that tests your understanding of steady-state heat flow.
I know these types of problems can sometimes look intimidating with all the variables, but let's take a breath and break it down logically.
Analyzing the Setup
Let's look closely at what we have.
We have a smaller cylinder with thermal conductivity and a larger cylinder with thermal conductivity .
They are connected end-to-end, meaning they are in series.
The problem tells us that the system has reached a steady state. This is a crucial piece of information!
In a steady state, the temperature profile doesn't change with time. More importantly, the rate of heat flow—the heat current —must be exactly the same through every cross-section of our series combination.
Think of it like water flowing through a pipe; if there are no leaks, the amount of water entering one end must equal the amount exiting the other.
The Master Equation
To solve this, we need our trusty tool: Fourier's Law of Heat Conduction.
The law states that the heat current is given by:
Here, is the thermal conductivity, is the cross-sectional area, is the temperature difference across the conductor, and is its length.
This equation is the bridge that connects the physical dimensions of our cylinders to the thermal properties we want to find.
Equating the Heat Currents
Since the cylinders are in series and in a steady state, the heat current through the first cylinder () must equal the heat current through the second cylinder ().
Let's set up our master equation:
Substituting Fourier's Law for both sides, we get:
This might look like a lot of variables, but watch what happens next.
The Elegance of Cancellation
Let's calculate the temperature differences for each cylinder.
For the first cylinder, the temperature drops from the hot bath at to the junction at .
For the second cylinder, the temperature drops from the junction at to the cold bath at .
Notice something beautiful? The temperature differences are exactly the same!
Furthermore, the problem explicitly states that the cylinders are of equal length, so .
When we plug these identical values back into our equation, they perfectly cancel out from both sides.
This leaves us with a wonderfully simple relationship:
The Final Calculation
We are looking for the ratio of their thermal conductivities, .
From our simplified equation, we can rearrange the terms:
Now, we just need the ratio of their areas. The problem gives us a vital clue: the radius of the bigger cylinder is twice that of the smaller one ().
Since the cross-sectional area of a cylinder is proportional to the square of its radius (), we can write:
Substituting this back into our conductivity ratio, we arrive at our final destination:
And there we have it! The thermal conductivity of the smaller cylinder must be four times that of the larger one to maintain this specific steady state.
Food for Thought
Before we wrap up, let's do a quick thought experiment.
What if these two cylinders were connected in parallel instead of series, between the same two heat baths?
In a parallel setup, the temperature difference across both cylinders would be the same, but the total heat current would be the sum of the individual heat currents.
How would you calculate the equivalent thermal conductivity then?
Exploring these "what if" scenarios is what truly builds your intuition for JEE Physics. Keep questioning, keep exploring, and the physics will always reveal its secrets to you!
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