To find the relative density of the liquid, we will use the principle of flotation, which states that for a floating object, the buoyant force equals the weight of the object. Step 1: Calculate the volume of the cube. The side of the cube is $L = 10 \text{ cm}$. The volume of the cube ($V_{\text{cube}}$) is: $$V_{\text{cube}} = L^3 = (10 \text{ cm})^3 = 1000 \text{ cm}^3$$ Convert the mass of the cube to grams for consistency with cm$^3$: $$m_{\text{cube}} = 0.5 \text{ kg} = 500 \text{ g}$$ Step 2: Determine the volume of the cube submerged in the liquid. The problem states that $\frac{1}{5}$ of the cube's height is above the liquid surface. This means that the fraction of the cube's height submerged in the liquid is $1 - \frac{1}{5} = \frac{4}{5}$. The height submerged ($h_{\text{submerged}}$) is: $$h_{\text{submerged}} = \frac{4}{5} \times L = \frac{4}{5} \times 10 \text{ cm} = 8 \text{ cm}$$ The volume of the cube submerged ($V_{\text{submerged}}$) is the base area multiplied by the submerged height: $$V_{\text{submerged}} = L^2 \times h_{\text{submerged}} = (10 \text{ cm})^2 \times 8 \text{ cm} = 100 \text{ cm}^2 \times 8 \text{ cm} = 800 \text{ cm}^3$$ Step 3: Apply the principle of flotation to find the density of the liquid. For a floating object, the weight of the object is equal to the weight of the displaced liquid. Weight of cube = Weight of displaced liquid $$m_{\text{cube}} \times g = \rho_{\text{liquid}} \times V_{\text{submerged}} \times g$$ We can cancel $g$ from both sides: $$m_{\text{cube}} = \rho_{\text{liquid}} \times V_{\text{submerged}}$$ Now, solve for the density of the liquid ($\rho_{\text{liquid}}$): $$\rho_{\text{liquid}} = \frac{m_{\text{cube}}}{V_{\text{submerged}}}$$ Substitute the values: $$\rho_{\text{liquid}} = \frac{500 \text{ g}}{800 \text{ cm}^3}$$ $$\rho_{\text{liquid}} = \frac{5}{8} \text{ g/cm}^3$$ $$\rho_{\text{liquid}} = 0.625 \text{ g/cm}^3$$ Step 4: Calculate the relative density of the liquid. Relative density (RD) is the ratio of the density of the substance to the density of water. The density of water ($\rho_{\text{water}}$) is $1 \text{ g/cm}^3$. $$\text{RD}_{\text{liquid}} = \frac{\rho_{\text{liquid}}}{\rho_{\text{water}}}$$ $$\text{RD}_{\text{liquid}} = \frac{0.625 \text{ g/cm}^3}{1 \text{ g/cm}^3}$$ $$\text{RD}_{\text{liquid}} = 0.625$$ The final answer is $\boxed{\text{0.625}}$.