1. Use Fourier’s law for radial heat conduction in a sphere

For steady radial heat flow through a spherical surface of radius $r$,

$\frac{dQ}{dt}=-K A \frac{d\theta}{dr}$

where the area of the spherical surface is

$A=4\pi r^2$

So,

$\frac{dQ}{dt}=-K(4\pi r^2)\frac{d\theta}{dr}$

Let the steady heat current be $H=\dfrac{dQ}{dt}$. Then,

$H=-4\pi K r^2 \frac{d\theta}{dr}$

Since the flow is steady, $H$ is constant.

2. Separate variables and integrate

Rearrange:

$d\theta=-\frac{H}{4\pi K}\frac{dr}{r^2}$

Integrate from inner shell $r=r_1$, temperature $\theta_1$ to outer shell $r=r_2$, temperature $\theta_2$:

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

Thus,

$\theta_2-\theta_1=-\frac{H}{4\pi K}\left[-\frac{1}{r}\right]_{r_1}^{r_2}$

$\theta_2-\theta_1=-\frac{H}{4\pi K}\left(-\frac{1}{r_2}+\frac{1}{r_1}\right)$

$\theta_2-\theta_1=\frac{H}{4\pi K}\left(\frac{1}{r_2}-\frac{1}{r_1}\right)(-1)$

More simply,

$\theta_2-\theta_1=\frac{H}{4\pi K}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$

Now,

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

So,

$\theta_2-\theta_1=\frac{H}{4\pi K}\cdot \frac{r_2-r_1}{r_1 r_2}$

Hence,

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

3. Match with the options

This is exactly:

$\boxed{\frac{4\pi K r_1 r_2(\theta_2-\theta_1)}{r_2-r_1}}$

So the correct option is A.

4. Check options briefly

• A matches the derived expression. ✅
• B misses the factor $4K$. ❌
• C is the plane slab form, not spherical conduction. ❌
• D is the reciprocal form, incorrect. ❌

5. Final Answer

$\boxed{\text{A}}$