Q175 / 180 · COMEDK 2023 (Memory Based) · Physics
A cylinder of radius $r$ and of thermal conductivity $K_1$ is surrounded by a cylindrical shell of inner radius $r$ and outer radius $2 r$ made of a material of thermal conductivity $K_2$. The effective thermal conductivity of the system is
- A$\frac{1}{3}\left(K_1+2 K_2\right)$
- B$\frac{1}{3}\left(3 K_2+2 K_1\right)$
- C$\frac{1}{3}\left(3 K_2+2 K_1\right)$
- Correct answerD$\frac{1}{4}\left(K_1+3 K_2\right)$
Solution
Both the cylinders are in parallel, for the heat flow from one end as shown

So, $K_{\mathrm{eq}}=\frac{K_1 A_1+K_2 A_2}{A_1+A_2}$ where, $A_1=$ area of cross-section of inner cylinder $=\pi R^2$ and $\quad A_2=$ area of cross-section of cylindrical shell $\begin{aligned} & =\pi\left\{(2 R)^2-(R)^2\right\}=3 \pi R^2 \\ \Rightarrow \quad K_{\mathrm{eq}} & =\frac{K_1\left(\pi R^2\right)+K_2\left(3 \pi R^2\right)}{\pi R^2+3 \pi R^2}=\frac{1}{4}\left(K_1+3 K_2\right) \end{aligned}$

So, $K_{\mathrm{eq}}=\frac{K_1 A_1+K_2 A_2}{A_1+A_2}$ where, $A_1=$ area of cross-section of inner cylinder $=\pi R^2$ and $\quad A_2=$ area of cross-section of cylindrical shell $\begin{aligned} & =\pi\left\{(2 R)^2-(R)^2\right\}=3 \pi R^2 \\ \Rightarrow \quad K_{\mathrm{eq}} & =\frac{K_1\left(\pi R^2\right)+K_2\left(3 \pi R^2\right)}{\pi R^2+3 \pi R^2}=\frac{1}{4}\left(K_1+3 K_2\right) \end{aligned}$
Hence, the correct answer is option: 'D'.