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 = \(1000 { kg/m}^3\))
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The floating condition follows Archimedes’ principle: the buoyant force equals the weight of the displaced liquid.
The total volume of the cube is:
\[V_{{total}} = (10 { cm})^3 = 1000 { cm}^3\]
The mass of the cube is:
\[m = 400 { g} = 0.4 { kg}\]
The density of the cube is:
\[\rho_{{cube}} = \frac{m}{V_{{total}}} = \frac{0.4}{1000 \times 10^{-6}} = 400 { kg/m}^3\]
As the cube floats, the submerged volume is calculated as:
\[V_{{submerged}} = V_{{total}} \times \frac{\rho_{{cube}}}{\rho_{{water}}}\]
\[V_{{submerged}} = 1000 \times \frac{400}{1000} = 600 { cm}^3\]
Consequently, the volume of the cube that is not submerged in water is:
\[V_{{outside}} = V_{{total}} - V_{{submerged}}\]
\[V_{{outside}} = 1000 - 600 = 400 { cm}^3\]
Therefore, the correct answer is (4) 400 cm³.
