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Question

One end of a thermally insulated rod is kept at a temperature T1 and the other at T2. The rod is composed of two sections of lengths l1 and l2 and thermal conductivities K1 and K2 respectively. The temperature at the interface of two sections is :
26670_a3bc23a909fd4f17a7e5dd4e0cbddee1.png
  1. K1l2T1+K2l1T2K1l2+K2l1
  2. K2l1T1+K1l2T2K2l1+K1l2
  3. K1l1T1+K2l2T2K1l1+K2l2
  4. K2l2T1+K1l1T2K1l1+K2l2

A
K2l1T1+K1l2T2K2l1+K1l2
B
K1l2T1+K2l1T2K1l2+K2l1
C
K1l1T1+K2l2T2K1l1+K2l2
D
K2l2T1+K1l1T2K1l1+K2l2
Solution
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Let T be the temp at the interface.
The rate of flow of heat remains same across the interface.
The equation for heat flow in conduction is:
dQdT=KAdTdx where, A is the cross section area across which heat flows.
K1A(T1T)l1=K2A(TT2)l2
(T1T)l1K1=(TT2)l2K2

T1TTT2=l1K2l2K1

T(K1l2+K2l1)=K1l2T1+K2l1T2

T=K1l2T1+K2l1T2K1l2+K2l1

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