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JEE MAIN - Physics (2025 - 7th April Evening Shift - No. 21)

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: $\frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}}=\frac{1}{2}, \frac{\mathrm{r}_{\mathrm{A}}}{\mathrm{r}_{\mathrm{B}}}=2$ and $\frac{\mathrm{K}_{\mathrm{A}}}{\mathrm{K}_{\mathrm{B}}}=\frac{1}{2}$. 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.
Answer
360

Explanation

JEE Main 2025 (Online) 7th April Evening Shift Physics - Heat and Thermodynamics Question 11 English Explanation

$$\begin{aligned} & \mathrm{R}_1=\frac{\ell_1}{\mathrm{~K}_1 \mathrm{~A}_1}, \mathrm{R}_2=\frac{\ell_2}{\mathrm{~K}_2 \mathrm{~A}_2} \\ & \frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\Delta \mathrm{T}}{\mathrm{R}} \\ & \left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_1=\left(\frac{\mathrm{dQ}}{\mu}\right)_2 \\ & \frac{400-\mathrm{T}}{\mathrm{R}_1}=\frac{\mathrm{T}-200}{\mathrm{R}_2} \\ & \frac{400-\mathrm{T}}{\mathrm{~T}-200}=\frac{\mathrm{R}_1}{\mathrm{R}_2}=\left(\frac{\ell_1}{\ell_2}\right)\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right)^2 \times \frac{\mathrm{K}_2}{\mathrm{~K}_1} \\ & =\frac{1}{2} \times\left(\frac{1}{2}\right)^2 \times 2 \\ & =\left(\frac{1}{4}\right) \end{aligned}$$

$$\begin{aligned} & \frac{400-\mathrm{T}}{\mathrm{~T}-200}=\frac{1}{4} \\ & 1600-4 \mathrm{~T}=\mathrm{T}-200 \\ & 5 \mathrm{~T}=1800 \\ & \mathrm{~T}=360 \mathrm{~K} \end{aligned}$$

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