Two thin metallic spherical shells of radii $r_1$ and $r_2$ $(r_1 < r_2)$ 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 $\theta_2$ $(\theta_1 < \theta_2)$. The rate at which heat flows radially through the material is:
Answer: (D) $\dfrac{4\pi Kr_1r_2(\theta_2 - \theta_1)}{r_2 - r_1}$
Through a thin shell of radius $r$ and thickness $dr$: $H = -K(4\pi r^2)\dfrac{d\theta}{dr}$ (same $H$ for every shell in steady state).
$$H\int_{r_1}^{r_2}\frac{dr}{r^2} = 4\pi K(\theta_2 - \theta_1) \Rightarrow H\left(\frac{1}{r_1} - \frac{1}{r_2}\right) = 4\pi K(\theta_2 - \theta_1)$$
$$H = \frac{4\pi Kr_1r_2(\theta_2 - \theta_1)}{r_2 - r_1}$$
Solution by Sreeraj P, M.Sc Physics