JEE Mains · Physics · STD 11 - 10.2 transmission of heat
Two cylindrical rods A and B made of different materials, are joined in a straight line. The ratio of lengths, radii and thermal conductivities 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 \(400 \mathrm{~K}, 200 \mathrm{~K}\), respectively. The temperature of rods interface is ________ K, when equilibrium is established.
- A 340
- B 360
- C 380
- D 400
Answer & Solution
Correct Answer
(B) 360
Step-by-step Solution
Detailed 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}} \\ &…
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