1. Heat conduction through a spherical shell

For steady radial heat flow through a spherical surface of radius $r$, the area is $A = 4\pi r^2.$

Using Fourier’s law, $\frac{dQ}{dt} = -kA\frac{dT}{dr} = -k(4\pi r^2)\frac{dT}{dr}.$

Since the heat current is constant in steady state, let $\frac{dQ}{dt} = H.$

Then, $H = -4\pi k r^2 \frac{dT}{dr}.$

2. Rearrange and integrate

$\frac{dT}{dr} = -\frac{H}{4\pi k r^2}.$

Integrate from $r_1$ to $r_2$ and corresponding temperatures from $T_1$ to $T_2$: $\int_{T_1}^{T_2} dT = -\frac{H}{4\pi k} \int_{r_1}^{r_2} \frac{dr}{r^2}.$

Now, $T_2 - T_1 = -\frac{H}{4\pi k}\left[-\frac{1}{r}\right]_{r_1}^{r_2}.$

$T_2 - T_1 = -\frac{H}{4\pi k}\left(-\frac{1}{r_2} + \frac{1}{r_1}\right).$

So, $T_1 - T_2 = \frac{H}{4\pi k}\left(\frac{1}{r_1} - \frac{1}{r_2}\right).$

3. Solve for heat current

$H = 4\pi k \frac{T_1-T_2}{\left(\frac{1}{r_1}-\frac{1}{r_2}\right)}.$

Simplify: $\frac{1}{r_1}-\frac{1}{r_2} = \frac{r_2-r_1}{r_1r_2}.$

Hence, $H = 4\pi k (T_1-T_2)\frac{r_1r_2}{r_2-r_1}.$

Therefore, the radial rate of flow of heat is proportional to $\frac{r_1r_2}{r_2-r_1}.$

4. Match with options

• A: $\ln\left(\frac{r_2}{r_1}\right)$ — incorrect
• B: $\frac{r_2-r_1}{r_1r_2}$ — inverse of required expression
• C: $(r_2-r_1)$ — incorrect
• D: $\frac{r_1r_2}{r_2-r_1}$ — correct

So the correct option is D.