A vessel contains oil (density $= 0.8\,\text{gm/cm}^{3}$) over mercury (density $= 13.6\,\text{gm cm}^{3}$). A homogeneous sphere floats with half its volume immersed in mercury and the other half in oil. The density of the material of the sphere in $\text{gm/cm}^{3}$ is

Step 1: Identify the given parameters.

Let $V$ be the total volume of the homogeneous sphere, and $\rho$ be its density.

$$\rho_{\text{oil}} = 0.8\,\text{g/cm}^{3}$$ $$\rho_{\text{Hg}} = 13.6\,\text{g/cm}^{3}$$

The sphere floats with half of its volume in mercury and half in oil.

$$V_{\text{Hg}} = \frac{V}{2}$$ $$V_{\text{oil}} = \frac{V}{2}$$

Step 2: Set up the equilibrium condition.

For the sphere to float in equilibrium, its total weight must be balanced by the sum of the buoyant forces exerted by the mercury and the oil.

$$W = F_{\text{Hg}} + F_{\text{oil}}$$

Step 3: Express the forces in terms of density and volume.

The weight of the sphere and the buoyant forces are defined by the following relations.

$$W = \rho V g$$ $$F_{\text{Hg}} = \rho_{\text{Hg}} \left(\frac{V}{2}\right) g$$ $$F_{\text{oil}} = \rho_{\text{oil}} \left(\frac{V}{2}\right) g$$

Step 4: Solve for the density of the sphere.

Substitute the force expressions into the equilibrium equation.

$$\rho V g = \rho_{\text{Hg}} \left(\frac{V}{2}\right) g + \rho_{\text{oil}} \left(\frac{V}{2}\right) g$$

Divide both sides by $Vg$.

$$\rho = \frac{\rho_{\text{Hg}} + \rho_{\text{oil}}}{2}$$

Substitute the given densities into the equation.

$$\rho = \frac{13.6 + 0.8}{2}$$ $$\rho = 7.2\,\text{g/cm}^{3}$$

Answer: The density of the material of the sphere is $7.2\,\text{g/cm}^{3}$.