**Problem:**
Three rods of identical cross-sectional area and made from the same metal form the sides of an isosceles triangle ABC, right angled at B. The points A and B are maintained at temperatures $T$ and $\sqrt{2}T$, respectively. In the steady state, the temperature of the point C is $T_c$. Assuming that only heat conduction takes place, $T_c /T$ is, (IIT JEE 1995)

- $\frac{1}{2\left(\sqrt{2}-1\right)}$
- $\frac{3}{\sqrt{2}+1}$
- $\frac{1}{\sqrt{3}\left(\sqrt{2}-1\right)}$
- $\frac{1}{\sqrt{2}+1}$

**Solution:**
Let $T_A=T$, $T_B=\sqrt{2}T$, and the steady state temperature of C is $T_C$. Since $T_B > T_A$, heat flows from B to A. If $T_C < T$ then heat flows to C from A as well as from B leading to a non-steady state. Similarly, if $T_C > \sqrt{2}T$ then heat flows from C to A as well as to B leading to a non-steady state. Thus, in steady state $T < T_C < \sqrt{2} T$ and heat flow from B to C is equal to heat flow from C to A (see figure).

- Heat Conduction
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