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 length L₁ and L₂ and thermal conductivities K₁ and K₂ respectively. The temperature at the interface of the two section is (a) ((K₁L₁T₁+K₂L₂T₂))/((K₁L₁+K₂L₂)) (b) ((K₂L₂T₁+K₁L₁T₂))/((K₁L₁+K₂L₂)) (c) ((K₂L₁T₁+K₁L₂T₂))/((K₂L₁+K₁L₂)) (d) ((K₁L₂T₁+K₂L₁T₂))/((K₁L₂+K₂L₁))

Answer: (d). Same heat current: (K₁A(T₁ - T))/(L₁) = (K₂A(T - T₂))/(L₂) K₁L₂(T₁ - T) = K₂L₁(T - T₂) T = (K₁L₂T₁ + K₂L₁T₂)/(K₁L₂ + K₂L₁), so the answer is (d).

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Heat Conduction

One end of a thermally insulated rod is kept at a temperature $T_1$ and the other at $T_2$ . The rod is composed of two sections of length $L_1$ and $L_2$ and thermal conductivities $K_1$ and $K_2$ respectively. The temperature at the interface of the two section is

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Answer: (d)

Same heat current: $\frac{K_1A(T_1 - T)}{L_1} = \frac{K_2A(T - T_2)}{L_2}$
$K_1L_2(T_1 - T) = K_2L_1(T - T_2)$
$T = \frac{K_1L_2T_1 + K_2L_1T_2}{K_1L_2 + K_2L_1}$ , so the answer is (d).

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Heat Conduction question from JEE Main 2007 (AIEEE)

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