Question
Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity 2 K while that in the middle has thermal conductivity $K$. The left end of the combination is maintained at temperature $3T$ and the right end at $T$. The rods are thermally insulated from outside. In steady state, temperature at the left junction is $T_{1}$ and that at the right junction is $T_{2}$. The ratio $T_{1}/T_{2}$ is
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Solution
In series, $R_{\text{eq.}\text{~}} = R_{1} + R_{2} + R_{3}$
$$\begin{matrix} & \ = \frac{1}{2KA} + \frac{1}{KA} + \frac{1}{2KA} \\ & \ = \frac{4I}{2KA} \\ & R_{\text{eq.}\text{~}} = \frac{2I}{KA} \end{matrix}$$
In series rate of heat flow is same
$$\begin{matrix} & & \ \therefore\ \frac{3T - T_{1}}{R_{1}} = \frac{3T - T}{R_{\text{eq.}\text{~}}} \\ & & \frac{\left( 3T - T_{1} \right)2KA}{I} = \frac{(2T)KA}{2I} \\ & & \ \Rightarrow \ 6T - 2T_{1} = T \\ & & \ \Rightarrow \ 2T_{1} = 5T \\ & \ \Rightarrow \ T_{1} = \frac{5T}{2}\ \ldots(1) & \text{(1)} \end{matrix}$$
Now, equate heat flow rate in $3^{\text{rd}\text{~}}$ section & total section
$$\begin{matrix} & & \frac{T_{2} - T}{R_{3}} = \frac{3T - T}{R_{\text{eq.}\text{~}}} \\ & & \ \Rightarrow \ \frac{\left( T_{2} - T \right)(2KA)}{I} = \frac{2T(KA)}{2I} \\ & & \ \Rightarrow \ 2T_{2} - 2T = T \\ & \ \Rightarrow \ T_{2} = \frac{3T}{2}\ \ldots(2) & \text{(2)} \end{matrix}$$
By equation (1) and equation (2)
$$\frac{T_{1}}{T_{2}} = \frac{5T \times 2}{2 \times 3T} = \frac{5}{3}$$
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