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

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

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

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

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

answer is A.

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Detailed Solution

As the rods are identical, so they have same length (l) and area of cross-section (A). They are connected in series. So, heat current will be same for all rods.

Heat current $= {(\frac{Δ Q}{Δ t})}_{A B} = {(\frac{Δ Q}{Δ t})}_{B C} = {(\frac{Δ Q}{Δ t})}_{C D}$

$\Rightarrow \frac{(100 - 70) K_{1} A}{l} = \frac{(70 - 20) K_{2} A}{l} = \frac{(20 - 0) K_{3} A}{l}$

$\Rightarrow K_{1} (100 - 70) = K_{2} (70 - 20) = K_{3} (20 - 0) \Rightarrow K_{1} (30) = K_{2} (50) = K_{3} (20)$

$\Rightarrow \frac{K_{1}}{10} = \frac{K_{2}}{6} = \frac{K_{3}}{15} \Rightarrow K_{1} : K_{2} : K_{3} = 10 : 6 : 15$

$\Rightarrow K_{1} : K_{3} = 2 : 3.$

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