Density of ice is given as 917kgm' and density of water as 1000kgm³. When a piece of ice is placed on top of pure water, it floats in the water and what fraction of the volume of the piece of ice is above the water?

When a piece of ice floats on water, it does so because its density is less than that of water. The fraction of the ice above the water can be determined using the principle of buoyancy, which states that the weight of the displaced water is equal to the weight of the ice.

Let's denote:
- \(\rho_{\text{ice}}\) as the density of ice, \(917 \, \text{kg/m}^3\).
- \(\rho_{\text{water}}\) as the density of water, \(1000 \, \text{kg/m}^3\).
- \(V_{\text{total}}\) as the total volume of the ice.
- \(V_{\text{submerged}}\) as the volume of ice submerged in water.

### Step-by-Step Solution:

1. **Buoyant Force and Weight Balance:**

   The buoyant force is equal to the weight of the ice. Hence:

   \[
   \rho_{\text{water}} \cdot g \cdot V_{\text{submerged}} = \rho_{\text{ice}} \cdot g \cdot V_{\text{total}}
   \]

   Since \(g\) (acceleration due to gravity) appears on both sides, it cancels out, simplifying to:

   \[
   \rho_{\text{water}} \cdot V_{\text{submerged}} = \rho_{\text{ice}} \cdot V_{\text{total}}
   \]

2. **Finding the Fraction Submerged:**

   The fraction of the ice submerged, \( \frac{V_{\text{submerged}}}{V_{\text{total}}} \), can be found as:

   \[
   \frac{V_{\text{submerged}}}{V_{\text{total}}} = \frac{\rho_{\text{ice}}}{\rho_{\text{water}}}
   \]

   Substituting the values:

   \[
   \frac{V_{\text{submerged}}}{V_{\text{total}}} = \frac{917}{1000} = 0.917
   \]

   This means that 91.7% of the volume of the ice is submerged in the water.

3. **Finding the Fraction Above Water:**

   The fraction of the ice that is above the water is:

   \[
   \text{Fraction above water} = 1 - \frac{V_{\text{submerged}}}{V_{\text{total}}}
   \]

   Substituting the fraction submerged:

   \[
   \text{Fraction above water} = 1 - 0.917 = 0.083
   \]

   Thus, **8.3% of the volume of the ice is above the water**.

### Conclusion:

When a piece of ice is placed on top of pure water, **8.3% of the volume of the ice is above the water**.

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Would you like to see more details or have any questions about this solution?

Here are some related questions for further exploration:

1. What would happen to the fraction of ice above water if salt were added to the water?
2. How does the shape of the ice affect its floating position in the water?
3. How would the fraction above water change if the density of ice changed?
4. What is the buoyant force acting on a 10 kg block of ice floating in water?
5. How would this fraction change in a liquid with a density different from water?

**Tip:** Remember that the lower the density of an object compared to the fluid it is in, the more of the object will remain above the surface.

Density of ice is given as 917 kg/m³ and density of water as 1000 kg/m³. When a piece of ice is placed on top of pure water, it floats in the water. What fraction of the volume of the piece of ice is above the water?

Learn how to calculate the fraction of ice that floats above water, using density values of ice (917 kg/m³) and water (1000 kg/m³). This problem involves key concepts like buoyancy, density, and Archimedes' Principle. Suitable for high school students.

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