Three rods of identical cross-section and lengths are made of three different materials of thermal conductivity $K_{1}, K_{2}$ and $K_{3}$ respectively. They are joined together at their ends to make a long rod (see figure). One end of the long rod is maintained at 100°C and the other at 0°C (see figure). If the joints of the rod are at 70°C and 20°C in steady state and there is no loss of energy from the surface of the rod, the correct relationship between $K_{1}, K_{2}$ and $K_{3}$

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$K_{1} > K_{2} > K_{3}$

$_{K_{1} : K_{3} = 3 : 5}^{K_{1} : K_{2} = 5 : 2,}$

$_{K_{2} : K_{3} = 2 : 5}^{K_{1} : K_{3} = 2 : 3,}$

$K_{1} < K_{2} < K_{3}$

answer is C.

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Detailed Solution

Let cross-section area and length of each rod is A and l respectively. Then current (thermal) through each of them are equal.

Using $\dot{Q} = \frac{△ T A K}{L}$

$\begin{array}{l} S o \frac{(100 - 70) A K_{1}}{l} = \frac{(70 - 20) A K_{2}}{l} = \frac{(20 - 0) A K_{3}}{l} \\ \Rightarrow 30 K_{1} = 50 K_{2} = 20 K_{3} \\ \Rightarrow \frac{K_{1}}{K_{3}} = \frac{2}{3} a n d \frac{K_{2}}{K_{3}} = \frac{2}{5} \end{array}$