question on heat conduction

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Three rods of identical cross-sectional area and made from the same metal form the sides of an isosceles triangle ABC, right angeled at B. The points A and B are maintained at temperatures T and (âˆš2)T respectively. In steady state, temperature of point C is T_c. Assuming that only heat conduction takes place, T_c/T is

I am getting 1/(âˆš2 + 1) but ans given is 3/(âˆš2+1)

#Physics #Thermal physics

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4 Answers

262

Aditya Bhutra ·2012-03-29 11:48:44

at equilibrium ,

Q_Bâ†’C = Q_Câ†’A

kA(âˆš2T - T_c)l = kA(T_c-T)âˆš2l

T_c = 3*Tâˆš2+1

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21

mohit_ sharma ·2012-03-29 11:56:16

can we equate Q_Bâ†’A and Q_Aâ†’C to get ans ?

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21

mohit_ sharma ·2012-03-31 01:15:57

please reply to post #3

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262

Aditya Bhutra ·2012-03-31 01:21:40

no.

it is very clear that Ta <Tc<Tb

hence heat will flow from Bâ†’Câ†’A and Bâ†’A

heat will flow from C to A and not the other way.

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