Two cylindrical rods A and B made of different materials, are joined in a straight line. The ratio of lengths, radii and thermal conductivities 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.

For steady one-dimensional conduction through a rod, the heat current is
$$Q=\frac{K\,A}{L}\,\left(T_{\text{hot}}-T_{\text{cold}}\right)$$

When two rods are joined in series, the same heat current $$Q$$ flows through each of them in the steady state. Let $$T$$ be the interface temperature (at the junction of rods A and B).

Given ratios
$$\frac{L_A}{L_B}=\frac12 \;\Rightarrow\; L_A=\frac{L_B}{2}$$
$$\frac{r_A}{r_B}=2 \;\Rightarrow\; r_A=2r_B$$
$$\frac{K_A}{K_B}=\frac12 \;\Rightarrow\; K_A=\frac{K_B}{2}$$

Cross-sectional areas: $$A=\pi r^{2}$$, so
$$A_A=\pi r_A^{2}=\pi\,(2r_B)^{2}=4\pi r_B^{2}=4A_B$$
thus $$\frac{A_A}{A_B}=4$$.

Heat current through rod A:
$$Q=\frac{K_AA_A}{L_A}\,(400-T)$$

Heat current through rod B:
$$Q=\frac{K_BA_B}{L_B}\,(T-200)$$

Equate the two heat currents:

$$\frac{K_AA_A}{L_A}\,(400-T)=\frac{K_BA_B}{L_B}\,(T-200)$$

Substitute the ratio values:
$$K_A=\frac{K_B}{2},\;A_A=4A_B,\;L_A=\frac{L_B}{2}$$

$$\frac{\left(\tfrac{K_B}{2}\right)\,(4A_B)}{\tfrac{L_B}{2}}\,(400-T)=\frac{K_BA_B}{L_B}\,(T-200)$$

Simplify the fraction on the left:
$$\frac{(\tfrac{K_B}{2})\,(4A_B)}{\tfrac{L_B}{2}}=\frac{2K_BA_B}{\tfrac{L_B}{2}}=\frac{2K_BA_B\cdot2}{L_B}=\frac{4K_BA_B}{L_B}$$

Hence
$$\frac{4K_BA_B}{L_B}\,(400-T)=\frac{K_BA_B}{L_B}\,(T-200)$$

Cancel the common factor $$\frac{K_BA_B}{L_B}$$ from both sides:

$$4(400-T)=T-200$$

Solve for $$T$$:
$$1600-4T=T-200$$
$$1600+200=5T$$
$$1800=5T$$
$$T=360\,\text{K}$$

The temperature at the interface of the two rods is $$\mathbf{360\;K}$$.