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 $\dfrac{L_A}{L_B} = \dfrac12$, $\dfrac{r_A}{r_B} = 2$ and $\dfrac{K_A}{K_B} = \dfrac12$. The free ends of rods A and B are maintained at $400\ \text{K}$ and $200\ \text{K}$ respectively. The temperature of the rods' interface is ______ K, when equilibrium is established.
Numerical value type. Enter your answer.
Answer: 360
Thermal resistance $R = \dfrac{L}{KA}$ with $A \propto r^2$:
$$\frac{R_A}{R_B} = \frac{L_A}{L_B}\cdot\frac{K_B}{K_A}\cdot\left(\frac{r_B}{r_A}\right)^2 = \frac12\times2\times\frac14 = \frac14$$
The same heat current flows through both rods:
$$\frac{400 - T}{R_A} = \frac{T - 200}{R_B} \Rightarrow 400 - T = \frac{T - 200}{4} \Rightarrow 5T = 1800 \Rightarrow T = 360\ \text{K}$$
Solution by Sreeraj P, M.Sc Physics