Three rods of identical cross-sectional area and made from the same metal form the sides of an isosceles triangle $A B C$ , 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

\frac{1}{2 (\sqrt{2} - 1)}

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

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\frac{1}{\sqrt{3} (\sqrt{2} - 1)}

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

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Solution

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., $\frac{Q_{2}}{t} + \frac{Q_{3}}{t} = 0$

$\frac{k A (T_{B} - T_{C})}{l} + \frac{k A (T - T_{C})}{l \sqrt{2}} = 0 (∵ A C = \sqrt{2} B C)$
$\frac{(\sqrt{2} T - T_{C})}{1} + \frac{(T - T_{C})}{\sqrt{2}} = 0$
$2 T - \sqrt{2} T_{C} + T - T_{C} = 0$
which gives $\frac{T_{C}}{T} = \frac{3}{\sqrt{2} + 1}$
, which is choice (b).

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