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 $T$ 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, $\frac{T_C}{T}$ is

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

Answer 1

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


Let $a$ be the length of AB & BC.
Since temperature at point B is higher than temperature at point A, heat flows from B to A. Let us assume the heat then flows from A to C and C to B.
For AC, $\frac{\Delta Q}{\Delta t}=KA\left(\frac{T-T_C}{\sqrt{2}a}\right)$
For CB, $\frac{\Delta Q}{\Delta t}=KA\left(\frac{T_C-\sqrt{2}T}{a}\right)$
$\left[\begin{array}{c}K=\text{Thermal conductivity of rod}\\ A=\text{cross -sectional area of rod}\end{array}\right]$

Equating the two equations for steady state,
$KA\left(\frac{T-T_C}{\sqrt{2}a}\right)=KA\left(\frac{T_C-\sqrt{2}T}{a}\right)\Rightarrow \frac{T_C}{T}=\frac{3}{(\sqrt{2}+1)}$
i.e $T_C=\frac{3}{\sqrt{2}+1}T$
or $\frac{T_C}{T}=\frac{3}{\sqrt{2}+1}$