Question

Three rods made from the same material and having the same cross-sectional area, form the sides of an isosceles triangle $ABC$ , right-angled at $B$ . The points $A$ and $B$ are maintained at temperature $T$ and $\sqrt{2}T$ , respectively. In steady-state, the temperature of the point $C$ is $T_C$ . Assuming that only heat conduction takes place along the lengths of the rods, the value of $\frac{T_C}{T}$ is

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

Answer 1

Answer: (B)

${\left(\frac{\Delta \text{Q}}{\Delta \text{t}}\right)}_{\text{BC}}={\left(\frac{\Delta \text{Q}}{\Delta \text{t}}\right)}_{\text{CA}}$
$\Rightarrow \frac{\mathrm{kA}\left(\sqrt{2}\mathrm{T}-{\mathrm{T}}_{\mathrm{C}}\right)}{\mathrm{a}}=\frac{\mathrm{kA}\left({\mathrm{T}}_{\mathrm{C}}-\mathrm{T}\right)}{\sqrt{2}\mathrm{a}}$
Solve to get
$\frac{{\text{T}}_{\text{C}}}{\text{T}}=\frac{3}{\sqrt{2}+1}$