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. Numerical answer type. Work it out, then open the answer.

Answer: 360. Thermal resistance R = (L)/(Kπ r²), so (RA)/(RB) = (LA)/(LB)·(KB)/(KA)·(rB²)/(rA²) = 1/2· 2·1/4 = 1/4 In series the temperature drop divides in ratio of resistances: drop across A = 200×1/5 = 40 K Interface temperature = 400 - 40 = 360 K

Two cylindrical rods A and B made of different materials, are joined… | JEE Main 2025 (7 Apr, Shift 2)

Q16JEE Main 2025 · 7 APR · SHIFT 2Medium

Heat Conduction

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.

Numerical answer type. Work it out, then open the answer.

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Answer: 360

Thermal resistance $R = \dfrac{L}{K\pi r^2}$ , so $\dfrac{R_A}{R_B} = \dfrac{L_A}{L_B}\cdot\dfrac{K_B}{K_A}\cdot\dfrac{r_B^2}{r_A^2} = \dfrac{1}{2}\cdot 2\cdot\dfrac{1}{4} = \dfrac{1}{4}$
In series the temperature drop divides in ratio of resistances: drop across A $= 200\times\dfrac{1}{5} = 40$ K
Interface temperature $= 400 - 40 = 360$ K

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Heat Conduction question from JEE Main 2025 (7 Apr, Shift 2)

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