Two cylindrical rods A and B made of different materials, are joined in a straight line. The ratios of lengths, radii and thermal conductivites of these rods are: frac L A L B =1/2, frac r A r B =2 and frac K A K B =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.

Physics · Heat Transfer

JEE Main 2025 — 7 April, Evening Shift — Question 67

Two cylindrical rods $A$ and $B$ made of different materials, are joined in a straight line. The ratios of lengths, radii and thermal conductivites 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 $400 \mathrm{~K}, 200 \mathrm{~K}$ , respectively. The temperature of rods interface is $\qquad$ K, when equilibrium is established.

Answer: 360

Numerical answer — enter this value.

Step-by-step solution

$\frac{K_{A} \pi r_{A}^{2}(400-T)}{L_{A}}=\frac{K_{B} \pi r_{B}^{2}(T-200)}{L_{B}}$

$\frac{K_{A}}{K_{B}}\left(\frac{r_{A}}{r_{B}}\right)^{2}(400-T)=\frac{L_{A}}{L_{B}}(T-200)$

$T=360 K$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 7 April, Evening Shift
Subject: Physics
Chapter: Heat Transfer
Topic: Convection and Radiation