# Two conducting cylinders of equal length: Heat Conduction

Problem: Two conducting cylinders of equal length but different radii are connected in series between two heat baths kept at temperatures T1 = 300 K and T2 = 100 K, as shown in the figure. The radius of the bigger cylinder is twice that of the smaller one and the thermal conductivities of the materials of the smaller and the larger cylinders are K1 and K2 respectively. If the temperature at the junction of the two cylinders in the steady state is 200 K, then K1/K2 = ?  (IIT JEE 2018) Answer: The answer is 4 i.e., ratio of thermal condictivites of $K_1/ K_2=4$.

Solution: The temperature of the source is $T_1=300$ K and that of the sink is $T_{2}=100$ K. The temperature at the junction of two cylinders is $T=200$ K. The radius of the bigger cylinder, $r_2$, is twice the radius of the smaller cylinder, $r_1$ i.e., $r_2=2r_1$.

The rate of heat conduction through a material, with conductivity $K$, cross-section area $A$, length $\Delta x$, and temperature difference between the two ends $\Delta T$, is given by \begin{align} \frac{\Delta Q}{\Delta t}=K A \frac{\Delta T}{\Delta x}. \nonumber \end{align} Thus, the rates of heat conduction through the two cylinders are \begin{align} \frac{\Delta Q_1}{\Delta t}&=K_1 (\pi r_1^2) \frac{T_1-T}{L_1},\nonumber \\ \frac{\Delta Q_2}{\Delta t}&=K_2 (\pi r_2^2) \frac{T-T_2}{L_2}.\nonumber \end{align} There is no heat loss from the cylinders because they are covered with an insulating material. Also, there is no heat accumulation in steady state. Thus, $\Delta Q_1/\Delta t=\Delta Q_2/\Delta t$, which gives \begin{align} \frac{K_1}{K_2}&=\frac{r_2^2 (T-T_2)/L_2}{r_1^2 (T_1-T)/L_1}\\ &=\frac{(2r_1)^2 (200-100)/L}{ r_1^2 (300-200)/L}\\ &=4.\nonumber \end{align}