A cube of density 250 kg/`m^2` floats in water, then what part of total volume of the cube outside the water?

Options

A. 0.75
B. 0.25
C. 0.333
D. 0.677

Answer

To solve the problem of determining what part of the total volume of a cube floats outside of water, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Density of the cube (\( \rho_{cube} \)) = 250 kg/m³
- Density of water (\( \rho_{water} \)) = 1000 kg/m³

2. **Understand the Concept of Buoyancy**:
- When an object floats, the weight of the object is equal to the weight of the water displaced by the submerged part of the object.
- This can be expressed as:
\[ \text{Weight of the cube} = \text{Weight of the displaced water} \]
- Mathematically, this can be written as:
\[ \rho_{cube} \cdot V_{cube} \cdot g = \rho_{water} \cdot V_{submerged} \cdot g \]
- Here, \( V_{cube} \) is the total volume of the cube, and \( V_{submerged} \) is the volume of the cube that is submerged in water.

3. **Cancel Out Gravity**:
- Since \( g \) (acceleration due to gravity) is present on both sides of the equation, we can cancel it out:
\[ \rho_{cube} \cdot V_{cube} = \rho_{water} \cdot V_{submerged} \]

4. **Express the Volumes**:
- Let’s denote the total volume of the cube as \( V_{cube} \).
- The volume of the cube can be expressed as \( V_{cube} = A^3 \) where \( A \) is the length of a side of the cube.
- For simplicity, we can assume \( A = 1 \) m, thus \( V_{cube} = 1 \, m^3 \).

5. **Substituting the Values**:
- Now substituting the values into the equation:
\[ 250 \cdot 1 = 1000 \cdot V_{submerged} \]
- Rearranging gives:
\[ V_{submerged} = \frac{250}{1000} = 0.25 \, m^3 \]

6. **Calculate the Volume Outside of Water**:
- The volume of the cube that is outside of the water can be calculated as:
\[ V_{outside} = V_{cube} - V_{submerged} = 1 - 0.25 = 0.75 \, m^3 \]

7. **Find the Fraction of Volume Outside**:
- The fraction of the total volume of the cube that is outside of the water is given by:
\[ \text{Fraction outside} = \frac{V_{outside}}{V_{cube}} = \frac{0.75}{1} = 0.75 \]

### Final Answer:
The part of the total volume of the cube that is outside of the water is **0.75**.