A 400 g solid cube having an edge of length 10 cm floats in water. How much volume of the cube is outside the water ? (Given : density of water $$= 1000 kg m^{-3}$$)

Mass of the cube is given as $$m = 400\text{ g} = 0.4\text{ kg}$$ and edge length is $$a = 10\text{ cm} = 0.1\text{ m}$$.

Total volume of the cube is$$V_{\text{total}} = a^3 = (0.1)^3 = 1.0\times 10^{-3}\text{ m}^3\;. $$

For floating equilibrium, weight of cube = buoyant force. We state the formula:

Buoyant force = $$\rho_{\text{water}}\;V_{\text{submerged}}\;g$$
Weight = $$m\,g$$
Equating, we get$$\rho_{\text{water}}\;V_{\text{submerged}}\;g = m\,g\quad-(1)$$

Cancel $$g$$ on both sides of $$(1)$$ to obtain the submerged volume:

$$V_{\text{submerged}} = \frac{m}{\rho_{\text{water}}} \quad-(2)$$

Substitute $$m = 0.4\text{ kg}$$ and $$\rho_{\text{water}} = 1000\text{ kg/m}^3$$ into $$(2)$$:

$$V_{\text{submerged}} = \frac{0.4}{1000} = 4.0\times 10^{-4}\text{ m}^3\;. $$

Convert $$V_{\text{submerged}}$$ to cubic centimetres using $$1\text{ m}^3 = 10^6\text{ cm}^3$$:

$$V_{\text{submerged}} = 4.0\times 10^{-4}\times 10^6 = 400\text{ cm}^3\;. $$

Therefore, the volume of the cube outside the water is

$$V_{\text{outside}} = V_{\text{total}} - V_{\text{submerged}} = 1000\text{ cm}^3 - 400\text{ cm}^3 = 600\text{ cm}^3\;. $$

Final Answer: Option B, $$600\text{ cm}^3\;.$$