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)

$\frac{3}{\sqrt{2}+1}$

Answer 2

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