A cube of side 10cm and mass 0.5kg floats in a liquid with only \((1)/(5)\) of its height above the liquid surface. What is the relative density of the liquid?

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$^3$:

$$m_{\text{cube}} = 0.5 \text{ kg} = 500 \text{ g}$$

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:

$$V_{\text{submerged}} = L^2 \times h_{\text{submerged}} = (10 \text{ cm})^2 \times 8 \text{ cm} = 100 \text{ cm}^2 \times 8 \text{ cm} = 800 \text{ cm}^3$$

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:

$$\rho_{\text{liquid}} = \frac{500 \text{ g}}{800 \text{ cm}^3}$$ $$\rho_{\text{liquid}} = \frac{5}{8} \text{ g/cm}^3$$ $$\rho_{\text{liquid}} = 0.625 \text{ g/cm}^3$$

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 $\text{0.625}$.