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:

We begin with the well-known Fourier law of heat conduction in its differential form. For any medium, the instantaneous heat current $$dQ/dt$$ (which we shall denote simply by $$Q$$ for convenience) is related to the temperature gradient by

$$Q = -K A\,\frac{d\theta}{dr},$$

where $$K$$ is the thermal conductivity of the material, $$A$$ is the area normal to the direction of flow and $$\dfrac{d\theta}{dr}$$ is the spatial rate of fall of temperature. Because we are dealing with a spherical geometry, every point at a distance $$r$$ from the common centre lies on a sphere of area

$$A = 4\pi r^{2}.$$

So, for purely radial (i.e. spherically symmetric) flow we can write

$$Q = -K(4\pi r^{2})\,\frac{d\theta}{dr} \;=\; -\,4\pi K r^{2}\,\frac{d\theta}{dr}.$$

Under steady-state conditions the same quantity of heat per unit time must cross every spherical surface, hence $$Q$$ is independent of $$r$$. We now isolate the derivative:

$$\frac{d\theta}{dr} \;=\; -\,\frac{Q}{4\pi K}\,\frac{1}{r^{2}}.$$

Next, we integrate this relation from the inner radius $$r_1$$ (at temperature $$\theta_1$$) to the outer radius $$r_2$$ (at temperature $$\theta_2$$). Writing the integral explicitly, we have

$$\int_{\theta_1}^{\theta_2} d\theta \;=\; -\,\frac{Q}{4\pi K}\,\int_{r_1}^{r_2} \frac{dr}{r^{2}}.$$

The left-hand side integrates to

$$\theta_2 - \theta_1.$$

For the right-hand side, we use the elementary integral $$\displaystyle\int r^{-2}dr = -\,\dfrac{1}{r}$$, giving

$$-\,\frac{Q}{4\pi K}\;\Bigl[-\frac{1}{r}\Bigr]_{r_1}^{r_2} \;=\; -\,\frac{Q}{4\pi K}\;\Bigl(-\frac{1}{r_2}+\frac{1}{r_1}\Bigr) \;=\; -\,\frac{Q}{4\pi K}\;\Bigl(\frac{1}{r_1}-\frac{1}{r_2}\Bigr).$$

Equating both results, we obtain

$$\theta_2 - \theta_1 \;=\; -\,\frac{Q}{4\pi K}\;\Bigl(\frac{1}{r_1}-\frac{1}{r_2}\Bigr).$$

Since $$r_1 < r_2$$, the bracketed term $$\Bigl(\dfrac{1}{r_1}-\dfrac{1}{r_2}\Bigr)$$ is positive. For clarity we take the magnitude of the heat current (the direction being implicitly from the hotter to the colder surface) and remove the minus sign:

$$Q \;=\; 4\pi K\,(\theta_2 - \theta_1)\,\frac{1}{\dfrac{1}{r_1}-\dfrac{1}{r_2}}.$$

Combining the fractions in the denominator,

$$\frac{1}{r_1}-\frac{1}{r_2} = \frac{r_2 - r_1}{r_1 r_2}.$$

Substituting this back, we obtain

$$Q = 4\pi K\,(\theta_2 - \theta_1)\, \frac{r_1 r_2}{\,r_2 - r_1\,}.$$

Thus the rate at which heat flows radially through the material separating the two concentric spherical shells is

$$Q = \frac{4\pi K\,r_1 r_2\,(\theta_2 - \theta_1)}{\,r_2 - r_1\,}.$$

This expression exactly matches Option D.

Hence, the correct answer is Option D.