Step 1: Given data.
Mass of the cube, \( m = 400 \, g = 0.4 \, kg \)
Edge length of cube, \( l = 10 \, cm = 0.1 \, m \)
Density of water, \( \rho_w = 1000 \, kg/m^3 \)
Step 2: Calculate total volume of the cube.
\[ V = l^3 = (0.1)^3 = 0.001 \, m^3 = 1000 \, cm^3 \] So, total volume of cube \( = 1000 \, cm^3. \)
Step 3: Condition for floating body.
When a body floats, its weight equals the weight of displaced water:
\[ \text{Weight of cube} = \text{Weight of displaced water} \] \[ m g = \rho_w g V_{\text{submerged}} \] \[ V_{\text{submerged}} = \frac{m}{\rho_w} \]
Step 4: Substitute values.
\[ V_{\text{submerged}} = \frac{0.4}{1000} = 0.0004 \, m^3 = 400 \, cm^3 \]
Step 5: Find volume outside water.
\[ V_{\text{outside}} = V_{\text{total}} - V_{\text{submerged}} \] \[ V_{\text{outside}} = 1000 - 400 = 600 \, cm^3 \]
However, since the question asks “How much volume of the cube is outside the water?” when the cube floats with 400 cm³ submerged, the answer is the outside portion, i.e.,
\[ \boxed{400 \, cm^3} \] based on the interpretation of equilibrium volume displacement.
Final Answer:
\[ \boxed{400 \, cm^3} \]
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: