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A system consist of a cylinder surrounded by a cylindrical shell. A cylinder is a radius $R$ and is made of material of thermal conductivity $K$, whereas a cylindrical shell has inner radius $R$ and outer radius $2 R$ and is made of material of thermal conductivity twice as that of cylinder. Assuming the system in steady state and negligible heat loss across the cylindrical surface, find the effective thermal conductivity of the system, if the two ends of the combined system are maintained at two different temperatures.
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The correct answer is:
$\frac{7 K}{4}$
$\therefore \quad R_1=\frac{L}{K_1 \cdot \pi R^2}$
and $\begin{aligned} R_2 & =\frac{L}{K_2\left(4 \pi R^2-\pi R^2\right)}=\frac{L}{3 K_2 \pi R^2} \\ R_{\text {eq }} & =\frac{L}{K_{\text {eq }} \cdot 4 \pi R^2}\end{aligned}$
$R_1$ and $R_2$ are in parallel.
$\therefore \quad \frac{1}{R_{\text {eq }}}=\frac{1}{R_1}+\frac{1}{R_2}$
$\begin{aligned} \Rightarrow & & \frac{4 \pi R^2 K_{\text {eq }}}{L} & =\frac{\pi R^2 K_1}{L}+\frac{3 \pi R^2 K_2}{L} \\ & \Rightarrow & 4 K_{\text {eq }} & =K_1+3 K_2\end{aligned}$
On putting values of $K_1$ and $K_2$, we get
$\begin{aligned} 4 K_{\text {eq }} & =K+3(2 K) \quad\left[\because K_1=K, K_2=2 K\right] \\ K_{\text {eq }} & =\frac{K+6 K}{4}=\frac{7 K}{4}\end{aligned}$
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