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Three rods of equal length and equal cross-section made of materials with thermal conductivities K1, K2 and K3 are joined end to end into one long rod. One end is held at 100 C and the other at 0 C. In steady state the two junctions are at 70 C and 20 C. Find the correct relation among K1, K2, K3.

1. K1: K3 = 2: 3; K2: K3 = 2: 5
2. K1 < K2 < K3
3. K1: K2 = 5: 2; K1: K3 = 3: 5
4. K1 > K2 > K3

Correct answer: K1: K3 = 2: 3; K2: K3 = 2: 5

Solution

Equal heat current means K1*30 = K2*50 = K3*20, giving K1:K2:K3 = (1/30):(1/50):(1/20) = 10:6:15, i.e. K1:K3 = 2:3 and K2:K3 = 2:5.

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