The figure shows a system of two concentric spheres of radii $$r_1$$ and $$r_2$$ and kept at temperatures $$T_1$$ and $$T_2$$ respectively. The radial rate of flow of heat in a substance between the two concentric sphere is proportional to

For steady-state radial heat conduction through a material of thermal conductivity $$k$$ located between two concentric spheres, the heat current $$\dot Q$$ is given by Fourier’s law.

At a radius $$r$$ the area of the spherical surface is $$A = 4\pi r^{2}$$.
Fourier’s law (in one dimension) is $$\dot Q = -kA \frac{dT}{dr}$$, so

$$\dot Q = -k \,4\pi r^{2}\,\frac{dT}{dr} \qquad -(1)$$

Because the system is in steady state, $$\dot Q$$ is the same through every spherical surface between $$r_1$$ and $$r_2$$. Rearrange $$-(1)$$ to separate variables:

$$\frac{dT}{dr} = -\frac{\dot Q}{4\pi k}\,\frac{1}{r^{2}}$$

Integrate from the inner sphere (radius $$r_1$$, temperature $$T_1$$) to the outer sphere (radius $$r_2$$, temperature $$T_2$$):

$$\int_{T_1}^{T_2} dT = -\frac{\dot Q}{4\pi k}\int_{r_1}^{r_2}\frac{dr}{r^{2}}$$

$$T_2 - T_1 = -\frac{\dot Q}{4\pi k}\left[-\frac{1}{r}\right]_{r_1}^{r_2} = -\frac{\dot Q}{4\pi k}\left(-\frac{1}{r_2}+\frac{1}{r_1}\right)$$

Simplify the bracket:

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

Solving for the heat current,

$$\dot Q = 4\pi k\,(T_1 - T_2)\,\frac{r_1 r_2}{r_2 - r_1}$$

Thus the radial rate of heat flow is directly proportional to $$\dfrac{r_1 r_2}{r_2 - r_1}$$.

Option C which is: $$\dfrac{r_1 r_2}{r_2 - r_1}$$