Question

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 point $C$ is $T_C$.
Assuming that only heat conduction takes place, the ratio $T_C/T$ is

• A. $\frac{1}{2(\sqrt{2}-1)}$
• B. $\frac{3}{\sqrt{2}+1}$
• C. $\frac{1}{\sqrt{3}(\sqrt{2}-1)}$
• D. $\frac{1}{\sqrt{2}+1}$

Answer 1

The correct option is B $\frac{3}{\sqrt{2}+1}$

Refer to Fig. Since $T_B>T_A$, heat flows from $B$ to $A$.
In the steady state, rate of flow of heat from $B$ to $C+$ rate of flow of heat from $A$ to $C=0$, i.e., $\frac{Q_2}{t}+\frac{Q_3}{t}=0$

$\frac{kA(T_B-T_C)}{l}+\frac{kA(T-T_C)}{l\sqrt{2}}=0(\because AC=\sqrt{2}BC)$
$\frac{(\sqrt{2}T-T_C)}{1}+\frac{(T-T_C)}{\sqrt{2}}=0$
$2T-\sqrt{2}{T}_C+T-T_C=0$
which gives $\frac{T_C}{T}=\frac{3}{\sqrt{2}+1}$
, which is choice (b).