NEET 2025

Question

Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity $2K$ while that in the middle has thermal conductivity $K$ . The left end is at $3T$ and right end at $T$ . In steady state, the temperatures at the junctions are $T_1$ (left) and $T_2$ (right). The ratio $T_1/T_2$ is:

Three rods in series: 2K | K | 2K, left end at 3T, right end at T, junctions T₁ and T₂

![Three rods in series: 2K | K | 2K, left end at 3T, right end at T, junctions T₁ and T₂](https://www.edvaya.com/neet/physics/q116-three-rods-series-conductivity-2025.png) A). $\dfrac{5}{4}$ B). $\dfrac{3}{2}$ C). $\dfrac{4}{3}$ D). $\dfrac{5}{3}$

Edvaya · Accepted Answer

The correct answer is $\dfrac{5}{3}$. In steady state, the same heat flux passes through all three rods.
For identical cross-sections and lengths:
$2K(3T - T_1) = K(T_1 - T_2) = 2K(T_2 - T)$
From the third equality:
$K(T_1-T_2) = 2K(T_2-T) \Rightarrow T_1-T_2 = 2T_2-2T \Rightarrow T_1 = 3T_2-2T$ ...(i)
From the first equality:
$2(3T-T_1) = T_1-T_2 \Rightarrow 6T-2T_1 = T_1-T_2 \Rightarrow 6T = 3T_1-T_2$ ...(ii)
Substituting (i) into (ii):
$6T = 3(3T_2-2T)-T_2 = 9T_2-6T-T_2 = 8T_2-6T$ 
$12T = 8T_2 \Rightarrow T_2 = \dfrac{3T}{2}$ 
$T_1 = 3 	imes \dfrac{3T}{2} - 2T = \dfrac{9T}{2} - 2T = \dfrac{5T}{2}$ 
$\dfrac{T_1}{T_2} = \dfrac{5T/2}{3T/2} = \dfrac{5}{3}$

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