- Home
- Standard 11
- Physics
Two thin metallic spherical shells of radii ${r}_{1}$ and ${r}_{2}$ $\left({r}_{1}<{r}_{2}\right)$ are placed with their centres coinciding. A material of thermal conductivity ${K}$ is filled in the space between the shells. The inner shell is maintained at temperature $\theta_{1}$ and the outer shell at temperature $\theta_{2}\left(\theta_{1}<\theta_{2}\right)$. The rate at which heat flows radially through the material is :-
$\frac{4 \pi {Kr}_{1} {r}_{2}\left(\theta_{2}-\theta_{1}\right)}{{r}_{2}-{r}_{1}}$
$\frac{\pi{r}_{1} {r}_{2}\left(\theta_{2}-\theta_{1}\right)}{{r}_{2}-{r}_{1}}$
$\frac{{K}\left(\theta_{2}-\theta_{1}\right)}{{r}_{2}-{r}_{1}}$
$\frac{{K}\left(\theta_{2}-\theta_{1}\right)\left({r}_{2}-{r}_{1}\right)}{4 \pi {r}_{1} {r}_{2}}$
Solution
Thermal resistance of spherical sheet of thicleness $dr$ and radius $r$ is
${d} {R}=\frac{{dr}}{{K}\left(4 \pi {r}^{2}\right)}$
${R}=\int_{{r}_{1}}^{{r}_{2}} \frac{{dr}}{{K}\left(4 \pi {r}^{2}\right)}$
${R}=\frac{1}{4 \pi {K}}\left(\frac{1}{{r}_{1}}-\frac{1}{{r}_{2}}\right)=\frac{1}{4 \pi {K}}\left(\frac{{I}_{2}-{r}_{1}}{{r}_{1} {I}_{2}}\right)$
Thermal current (i) $=\frac{\theta_{2}-\theta_{1}}{R}$
${i} =\frac{4 \pi {Kr}_{1} {r}_{2}\left(\theta_{2}-\theta_{1}\right)}{{r}_{2}-{r}_{1}}$
