Two cylindrical rods A and B made of different materials, are joined in a straight line. The ratios of lengths, radii and thermal conductivities of these rods are: LA/LB = 1/2, rA/rB = 2 and KA/KB = 1/2. The free ends of rods A and B are maintained at 400 K, 200 K, respectively. The temperature of rods interface is _______ K, when equilibrium is established.

The correct answer is Option 360. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q47 – JEE-MAIN 2025

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Home Physics Thermal PropertiesQuestion #495

Two cylindrical rods A and B made of different materials, are joined in a straight line. The ratios of lengths, radii and thermal conductivities of these rods are:

\frac{L_{A}}{L_{B}} = \frac{1}{2}

\frac{r_{A}}{r_{B}} = 2

and

\frac{K_{A}}{K_{B}} = \frac{1}{2}

. The free ends of rods A and B are maintained at

400 K

200 K

, respectively. The temperature of rods interface is _______ K, when equilibrium is established.

Enter Calculated Numerical Value:

Clue 1 of 3: Steady State Heat Flow

In a series combination of rods, at thermal equilibrium, the rate of heat flow through each rod is the same.

🧠 Full Solution Path

Step 1: Identify the principle of heat transfer✦ Active

When two rods are joined in series and thermal equilibrium is established, the rate of heat flow through both rods must be equal.

Q_{A} = Q_{B}

Step 2: Apply the formula for heat conduction○ Expand

The rate of heat flow $Q$ through a rod is given by $Q = \frac{K A Δ T}{L}$ . For cylindrical rods, the cross-sectional area is $A = π r^{2}$ . Thus, for rods A and B, we have:

\frac{K_{A} π r_{A}^{2} (T_{A} - T_{i})}{L_{A}} = \frac{K_{B} π r_{B}^{2} (T_{i} - T_{B})}{L_{B}}

Step 3: Substitute given ratios and solve for interface temperature○ Expand

Given $\frac{L_{A}}{L_{B}} = \frac{1}{2}$ , $\frac{r_{A}}{r_{B}} = 2$ , and $\frac{K_{A}}{K_{B}} = \frac{1}{2}$ . Also, $T_{A} = 400 K$ and $T_{B} = 200 K$ . Substituting these into the equation from Step 2 and simplifying:

\frac{K_{A} (2 r_{B})^{2} (400 - T_{i})}{L_{A}} = \frac{(2 K_{A}) r_{B}^{2} (T_{i} - 200)}{2 L_{A}}

\frac{K_{A} \cdot 4 r_{B}^{2} (400 - T_{i})}{L_{A}} = \frac{K_{A} r_{B}^{2} (T_{i} - 200)}{L_{A}}

4 (400 - T_{i}) = (T_{i} - 200)

1600 - 4 T_{i} = T_{i} - 200

1800 = 5 T_{i}

T_{i} = 360 K

💡 Teacher's Secret Hint

Ensure careful substitution of ratios and algebraic simplification to avoid errors.

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Physics Question 47 – JEE-MAIN 2025

Clue 1 of 3: Steady State Heat Flow

🧠 Full Solution Path