Density of ice is `900 kg//m^(3)`. A piece of ice is floating in water of density `1000kg//m^(3)`. Find the fraction of volume of the picec of ice outside the water.

Explanation

To solve the problem of finding the fraction of the volume of a piece of ice that is outside of water when it is floating, we can follow these steps:

### Step 1: Understand the problem
We have a piece of ice with a density of \( \rho_{\text{ice}} = 900 \, \text{kg/m}^3 \) floating in water with a density of \( \rho_{\text{water}} = 1000 \, \text{kg/m}^3 \). We need to find the fraction of the volume of the ice that is above the water.

### Step 2: Set up the equations
When the ice is floating, it is in equilibrium. The weight of the ice is equal to the buoyant force acting on it.

- Let \( V \) be the total volume of the ice.
- Let \( V_i \) be the volume of the ice that is immersed in water.

The weight of the ice can be expressed as:
\[ \text{Weight of ice} = m_{\text{ice}} \cdot g = \rho_{\text{ice}} \cdot V \cdot g \]

The buoyant force (upthrust) can be expressed as:
\[ \text{Buoyant force} = \text{Weight of displaced water} = \rho_{\text{water}} \cdot V_i \cdot g \]

### Step 3: Set the forces equal
At equilibrium, the weight of the ice is equal to the buoyant force:
\[ \rho_{\text{ice}} \cdot V \cdot g = \rho_{\text{water}} \cdot V_i \cdot g \]

We can cancel \( g \) from both sides:
\[ \rho_{\text{ice}} \cdot V = \rho_{\text{water}} \cdot V_i \]

### Step 4: Relate the immersed volume to the total volume
From the equation above, we can express \( V_i \) in terms of \( V \):
\[ V_i = \frac{\rho_{\text{ice}}}{\rho_{\text{water}}} \cdot V \]

### Step 5: Substitute the densities
Substituting the given densities:
\[ V_i = \frac{900}{1000} \cdot V = 0.9 V \]

### Step 6: Find the volume outside the water
The volume of the ice that is outside the water \( V_o \) is given by:
\[ V_o = V - V_i \]
Substituting for \( V_i \):
\[ V_o = V - 0.9 V = 0.1 V \]

### Step 7: Find the fraction of volume outside the water
The fraction of the volume of the ice that is outside the water is:
\[ \text{Fraction outside} = \frac{V_o}{V} = \frac{0.1 V}{V} = 0.1 \]

### Conclusion
Thus, the fraction of the volume of the piece of ice that is outside the water is \( 0.1 \) or \( 10\% \).

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