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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 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, T_C /T is.

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

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

$\frac{1}{(\sqrt{2} + 1)}$

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

answer is B.

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Detailed Solution

Given that T_B >T_A. So, heat flows from B to A through both the paths. Thus
Heat flowing in BC = Heat flowing in CA
$\frac{KA [T \sqrt{2} - T_{C}]}{l} = \frac{KA [T_{C} - T]}{l \sqrt{2}} ∴ \frac{T_{C}}{T} = \frac{3}{(\sqrt{2} + 1)}$

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