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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