**Problem:**
A cylinder of radius R made of a material of thermal conductivity K_{1} is surrounded by a cylindrical shell of inner radius R and outer radius 2R made of material of thermal conductivity K_{2}. The two ends of the combined system are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system is?
(IIT JEE 1988)

- $K_1+K_2$
- ${K_1 K_2}/{(K_1 +K_2)}$
- ${(K_1 +3 K_2)}/{4}$
- ${(3K_1 + K_2)}/{4}$

**Answer:** The answer is (C) i.e., effective thermal conductivity of the system is ${(K_1 +3 K_2)}/{4}$.

**Solution:**
Let $\Delta T=(T_h-T_l)$ be the temperature difference between the two ends and $x$ be the length of the cylinders.

The rate of heat flow through the cylinder of radius $R$ and thermal conductivity $K_1$ is \begin{align} \frac{\mathrm{d}Q_1}{\mathrm{d}t}&=K_1A_1 \frac{\Delta T}{x}\\ &=K_1(\pi R^2)\Delta T/x.\nonumber \end{align} The rate of heat flow through the cylindrical shell of inner radius $R$, outer radius $2R$, and thermal conductivity $K_2$ is \begin{align} \frac{\mathrm{d}Q_2}{\mathrm{d}t}&=K_2A_2 \frac{\Delta T}{x}\\ &=K_2(4\pi R^2- \pi R^2)\Delta T/x \nonumber\\ &=3K_2\pi R^2\Delta T/x.\nonumber \end{align} The rate of heat flow through the system from one end to the other end is \begin{align} \label{moa:eqn:1} \frac{\mathrm{d}Q}{\mathrm{d}t}&=\frac{\mathrm{d}Q_1}{\mathrm{d}t}+\frac{\mathrm{d}Q_2}{\mathrm{d}t}\nonumber \nonumber\\ & = (K_1+3K_2)(\pi R^2) \Delta T/x\nonumber\\ & =\left(\tfrac{K_1+3K_2}{4}\right)(4\pi R^2)\Delta T/x \nonumber\\ &=K_\text{eff}A_\text{eff}\Delta T/x. \end{align} From above equation, the effective thermal conductivity is $K_\text{eff}=(K_1+3K_2)/4$.

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