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Calorimetry

Three rods of...

Question

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 $100^{\circ} C$ and the other at $0^{\circ} C$ (see figure). If the joints of the rod are at $70^{\circ} C$ and $20^{\circ} C$ in steady state and there is no loss of energy from the surface of the rod, the correct relationship between $K_{1}, K_{2} and K_{3}$ is :

A

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

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B

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

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C

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

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D

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

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Solution

The correct option is A $K_{1} : K_{3} = 2 : 3, K_{2} : K_{3} = 2 : 5$
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 $= {(\frac{Δ Q}{Δ t})}_{A B} = {(\frac{Δ Q}{Δ t})}_{B C} = {(\frac{Δ Q}{Δ t})}_{C D}$

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

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

$\Rightarrow K_{1} (30) = K_{2} (50) = K_{3} (20)$

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

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

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

Hence, option $(A)$ is correct.

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