Three rods of identical cross-sectional area: Heat Conduction

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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)

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

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).

Three rods of identical cross-sectional area
The rates of heat flow from B to C and from C to A are given by, \begin{align} \frac{\mathrm{d}Q_\text{BC}}{\mathrm{d}t} &=\frac{\kappa A(T_B-T_C)}{x} \\ &=\frac{\kappa A(\sqrt{2}T-T_C)}{x},\\ \end{align} \begin{align} \frac{\mathrm{d}Q_\text{CA}}{\mathrm{d}t}&=\frac{\kappa A(T_C-T_A)}{\sqrt{2}x}\\ &=\frac{\kappa A(T_C-T)}{\sqrt{2}x}. \end{align} In steady state, ${\mathrm{d}Q_\text{BC}}/{\mathrm{d}t}={\mathrm{d}Q_\text{CA}}/{\mathrm{d}t}$. Solve above equations to get, \begin{align} T_C &=\frac{(T_A+\sqrt{2}T_B)}{(1+\sqrt{2})} \\ &=\frac{3T}{(1+\sqrt{2})}.\nonumber \end{align}

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