Question 106 of 126

A large cylindrical rod of length $L$ is made by joining two identical rods of copper and steel of length $\left(\frac{L}{2}\right)$ each. The rods are completely insulated from the surroundings. If the free end of copper rod is maintained at $100^{\circ} \mathrm{C}$ and that of steel at $0^{\circ} \mathrm{C}$ then the temperature of junction is (Thermal conductivity of copper is 9 times that of steel)

A$90^{\circ} \mathrm{C}$B$50^{\circ} \mathrm{C}$C$10^{\circ} \mathrm{C}$D$67^{\circ} \mathrm{C}$

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Answer: (A) $90^{\circ} \mathrm{C}$

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;
$$\begin{aligned} \theta & =\frac{K_{\text {copper }} \theta_{\text {copper }} l_{\text {steel }}+K_{\text {steel }} \theta_{\text {steel }} l_{\text {copper }}}{K_{\text {copper }} l_{\text {steel }}+K_{\text {steel }} l_{\text {copper }}} \\ & =\frac{9 k \times 100 \times \frac{L}{2}+k \times 0 \times \frac{L}{2}}{9 k \times \frac{L}{2}+k \times \frac{L}{2}} \\ & =\frac{\frac{900}{2} k L}{\frac{10 k L}{2}}=90^{\circ} \mathrm{C} \end{aligned}$$