- Home
- Standard 11
- Physics
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
$K_1+K_2$
$\frac {K_1K_2}{K_1+K_2}$
$\frac {K_1+3K_2}{4}$
$\frac {3K_1+K_2}{4}$
Solution
Both the cylinders are in parallel, for the heat flow from one end as shown.
$\text { Hence } \mathrm{K}_{\mathrm{eq}}=\frac{\mathrm{K}_{1} \mathrm{A}_{1}+\mathrm{K}_{2} \mathrm{A}_{2}}{\mathrm{A}_{1}+\mathrm{A}_{2}} ; \text { where } \mathrm{A}_{1}=\text { Area of }$
cross section of inner cylinder $=\pi \mathrm{R}^{2}$ and $A_{2}=$ Area of cross section of cylindrical shell
$=\pi\left\{(2 R)^{2}-(R)^{2}\right\}=3 \pi R^{2}$
$\Rightarrow \mathrm{K}_{\mathrm{eq}}=\frac{\mathrm{K}_{1}\left(\pi \mathrm{R}^{2}\right)+\mathrm{K}_{2}\left(3 \pi \mathrm{R}^{2}\right)}{\pi \mathrm{R}^{2}+3 \pi \mathrm{R}^{2}}=\frac{\mathrm{K}_{1}+3 \mathrm{K}_{2}}{4}$
