A cube of ice of side length 10 cm is floating in water of density 1000 `kg//m^(3)`. Then pick up the correct statement (density of ice = 900 `kg//m^(3)`)

Options

A. a. 1 cm of the cube will be out of water
B. b'. 9 cm of the cube will be out of water
C. c. 9 cm of the cube will be in water
D. d. 1 cm of the cube will be in water

Answer

To solve the problem of a cube of ice floating in water, we can follow these steps:

### Step 1: Understand the Given Data
- Side length of the ice cube, \( a = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Density of water, \( \rho_{\text{water}} = 1000 \, \text{kg/m}^3 \)
- Density of ice, \( \rho_{\text{ice}} = 900 \, \text{kg/m}^3 \)

### Step 2: Calculate the Volume of the Ice Cube
The volume \( V \) of the cube can be calculated using the formula:
\[ V = a^3 \]
Substituting the value of \( a \):
\[ V = (0.1 \, \text{m})^3 = 0.001 \, \text{m}^3 \]

### Step 3: Calculate the Mass of the Ice Cube
The mass \( m \) of the ice cube can be calculated using the formula:
\[ m = V \cdot \rho_{\text{ice}} \]
Substituting the values:
\[ m = 0.001 \, \text{m}^3 \cdot 900 \, \text{kg/m}^3 = 0.9 \, \text{kg} \]

### Step 4: Apply the Principle of Buoyancy
According to Archimedes' principle, the buoyant force \( F_b \) acting on the ice cube is equal to the weight of the water displaced by the submerged part of the cube:
\[ F_b = \rho_{\text{water}} \cdot V_{\text{submerged}} \cdot g \]
Where \( V_{\text{submerged}} \) is the volume of the ice cube that is submerged in water.

### Step 5: Set Up the Equation
Since the ice cube is floating, the weight of the ice cube is equal to the buoyant force:
\[ m \cdot g = \rho_{\text{water}} \cdot V_{\text{submerged}} \cdot g \]
Cancelling \( g \) from both sides:
\[ m = \rho_{\text{water}} \cdot V_{\text{submerged}} \]

### Step 6: Substitute the Mass of the Ice Cube
Substituting the mass we calculated:
\[ 0.9 \, \text{kg} = 1000 \, \text{kg/m}^3 \cdot V_{\text{submerged}} \]
Solving for \( V_{\text{submerged}} \):
\[ V_{\text{submerged}} = \frac{0.9 \, \text{kg}}{1000 \, \text{kg/m}^3} = 0.0009 \, \text{m}^3 \]

### Step 7: Calculate the Height of the Submerged Part
The volume submerged can also be expressed in terms of height \( h \) of the submerged part:
\[ V_{\text{submerged}} = a^2 \cdot h \]
Substituting \( a = 0.1 \, \text{m} \):
\[ 0.0009 \, \text{m}^3 = (0.1 \, \text{m})^2 \cdot h \]
\[ 0.0009 \, \text{m}^3 = 0.01 \, \text{m}^2 \cdot h \]
Solving for \( h \):
\[ h = \frac{0.0009 \, \text{m}^3}{0.01 \, \text{m}^2} = 0.09 \, \text{m} = 9 \, \text{cm} \]

### Step 8: Determine the Height Above Water
The height of the cube that is above water is:
\[ \text{Height above water} = a - h = 10 \, \text{cm} - 9 \, \text{cm} = 1 \, \text{cm} \]

### Conclusion
Thus, the correct statement is that **1 cm of the cube will be out of the water**.

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