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Q.

Three rods of same dimensions have thermal conductivity 3K,2K and K. They are arranged as shown in figure. Then the temperature of the junction in steady state is

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a

20030C

b

10030C

c

75°C

d

5030C

answer is A.

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Detailed Solution

Qt3K=Qt2K+QtK3KA(100−θ)l=2KA(θ−50)l+KA(θ−0)lθ is temperature of the junction.
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