A large cylindrical rod of length $L$ is made by joining two identical rods of copper and steel of length…

1. $90^{\circ} \mathrm{C}$
2. $50^{\circ} \mathrm{C}$
3. $10^{\circ} \mathrm{C}$
4. $67^{\circ} \mathrm{C}$

Solution

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} $

Asked in: JEE Main 2012 (19 May Online)

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