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:
\[ \boxed{400 \, cm^3} \]
For a given reaction \( R \rightarrow P \), \( t_{1/2} \) is related to \([A_0]\) as given in the table. Given: \( \log 2 = 0.30 \). Which of the following is true?
| \([A]\) (mol/L) | \(t_{1/2}\) (min) |
|---|---|
| 0.100 | 200 |
| 0.025 | 100 |
A. The order of the reaction is \( \frac{1}{2} \).
B. If \( [A_0] \) is 1 M, then \( t_{1/2} \) is \( 200/\sqrt{10} \) min.
C. The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 M to 0.500 M.
D. \( t_{1/2} \) is 800 min for \( [A_0] = 1.6 \) M.