A cylinder of radius R is surrounded by a cylindrical shell of inner radius R and outer radius 2R. The thermal conductivity of the material of the inner cylinder is $$K_1$$ and that of the outer cylinder is $$K_2$$. Assuming no loss of heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is:

We have a solid cylinder of radius $$R$$ made of a material of thermal conductivity $$K_1$$. All around it, in perfect contact, there is a coaxial hollow cylindrical shell whose inner radius is also $$R$$ and whose outer radius is $$2R$$. The shell is made of another material whose thermal conductivity is $$K_2$$. Heat is allowed to flow only along the axis (length) of the composite cylinder, and we assume that there is no loss of heat sideways. Our aim is to find one single number $$K_{\text{eff}}$$ that can replace the two conductors so that the heat flow calculated with $$K_{\text{eff}}$$ is exactly the same as the actual heat flow through the two materials taken together.

First, recall Fourier’s law of steady-state heat conduction along a length:

$$Q = \dfrac{K\,A\,\Delta T}{L},$$

where

$$Q$$ = heat current (rate of heat flow),
$$K$$ = thermal conductivity of the material,
$$A$$ = cross-sectional area perpendicular to the heat flow,
$$\Delta T$$ = temperature difference applied between the two ends, and
$$L$$ = length of the conductor in the direction of heat flow.

For axial (longitudinal) flow, each cylindrical region has its flat circular cross-section perpendicular to the flow. Because the two regions extend from the same hot face to the same cold face, the temperature difference $$\Delta T$$ and the length $$L$$ are identical for both materials. Therefore both regions behave as parallel thermal conductors. In a parallel arrangement, individual heat currents simply add:

$$Q_{\text{total}} = Q_1 + Q_2.$$

Using Fourier’s law for each part we write

$$Q_1 = \dfrac{K_1\,A_1\,\Delta T}{L}, \qquad Q_2 = \dfrac{K_2\,A_2\,\Delta T}{L}.$$

Substituting these into the addition rule gives

$$Q_{\text{total}} = \dfrac{K_1\,A_1\,\Delta T}{L} + \dfrac{K_2\,A_2\,\Delta T}{L}.$$

Now we define the effective conductivity $$K_{\text{eff}}$$ through

$$Q_{\text{total}} = \dfrac{K_{\text{eff}}\,A_{\text{total}}\,\Delta T}{L},$$

where $$A_{\text{total}}$$ is the total cross-sectional area of the whole composite cylinder.

Equating the two expressions for $$Q_{\text{total}}$$ and cancelling the common factors $$\Delta T/L$$ on both sides, we obtain the relation

$$K_{\text{eff}}\,A_{\text{total}} = K_1\,A_1 + K_2\,A_2.$$

All that remains is to insert the actual areas. Because the cross-sections are circular, each area equals $$\pi \times (\text{radius})^2$$.

Area of the inner solid cylinder:

$$A_1 = \pi R^2.$$

Area of the outer hollow shell (annulus):

We subtract the inner area from the outer area: $$A_2 = \pi (2R)^2 \;-\; \pi R^2 = \pi (4R^2 - R^2) = 3\pi R^2.$$

Total cross-sectional area of the composite cylinder:

$$A_{\text{total}} = \pi (2R)^2 = 4\pi R^2.$$

Substituting $$A_1, A_2$$ and $$A_{\text{total}}$$ into the earlier relation gives

$$K_{\text{eff}}\,(4\pi R^2) = K_1\,(\pi R^2) + K_2\,(3\pi R^2).$$

We now cancel the common factor $$\pi R^2$$ from every term:

$$4\,K_{\text{eff}} = K_1 + 3K_2.$$

Solving for $$K_{\text{eff}}$$ we divide both sides by $$4$$:

$$K_{\text{eff}} = \dfrac{K_1 + 3K_2}{4}.$$

Therefore the effective thermal conductivity of the entire assembly is

$$K_{\text{eff}} = \frac{K_1 + 3K_2}{4}.$$

Hence, the correct answer is Option D.