1. Tardigrade
  2. Question
  3. Physics
  4. 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 √2T , respectively. In steady-state, the temperature of the point C is TC . Assuming that only heat conduction takes place along the lengths of the rods, the value of (TC/T) is

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Q. 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

NTA AbhyasNTA Abhyas 2020Thermal Properties of Matter

Solution:

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

Solution