A large cylindrical rod of length L is made by joining two...

Let conductivity of steel $K_{\text{steel }}=k$ then from question Conductivity of copper $K_{\text{copper }}=9 k$$\begin{aligned}& \theta_{\text{copper }}=100^{{}^{\circ}} \mathrm{C}\\ & \theta_{\text{steel }}=0^{{}^{\circ}} \mathrm{C}\\ & l_{\text{steel }}=l_{\text{copper }}=\frac{L}{2} \end{aligned}$$$From formula temperature of junction;$$\begin{aligned}\theta& =\frac{K_{\text{copper }} \theta_{\text{copper }} l_{\text{steel }}+K_{\text{steel }} \theta_{\text{steel }} l_{\text{copper }}}{K_{\text{copper }} l_{\text{steel }}+K_{\text{steel }} l_{\text{copper }}}\\ & =\frac{9 k \times 100 \times \frac{L}{2}+k \times 0 \times \frac{L}{2}}{9 k \times \frac{L}{2}+k \times \frac{L}{2}}\\ & =\frac{\frac{900}{2} k L}{\frac{10 k L}{2}}=90^{{}^{\circ}} \mathrm{C} \end{aligned}$$