Calculating Submergence of a Floating Cube

This question asks us to determine how much of a cubical block is submerged when it floats in water. This involves understanding the principle of buoyancy, which is a fundamental concept in fluid mechanics.

Understanding Buoyancy and Flotation

When an object is placed in a fluid, it experiences an upward force called the buoyant force. Archimedes' principle states that this buoyant force is equal to the weight of the fluid displaced by the object. For an object floating on the surface of a fluid, a special condition applies: the buoyant force is equal to the weight of the object itself. This balance is what allows the object to float.

In this specific problem:

• The object is a cubical block.
• The fluid is water.
• The block is floating in water.

Therefore, the weight of the cubical block must be equal to the weight of the water displaced by the submerged part of the block.

Step-by-Step Calculation

Let's break down the calculation:

1. Calculate the Weight of the Block:
   • Mass of the block, \( m = 600 \, \text{g} = 0.6 \, \text{kg} \).
   • Weight of the block, \( W = mg \). We can use \( g = 9.8 \, \text{m/s}^2 \) or just work with mass and density ratios. Let's use mass and density, as it's often simpler for floating problems. The weight is proportional to mass.
2. Calculate the Volume of the Block:
   • Side of the cube, \( s = 10 \, \text{cm} = 0.1 \, \text{m} \).
   • Total volume of the block, \( V_{\text{total}} = s^3 = (0.1 \, \text{m})^3 = 0.001 \, \text{m}^3 \).
   • In cm³, \( V_{\text{total}} = (10 \, \text{cm})^3 = 1000 \, \text{cm}^3 \).
3. Relate Weight to Buoyant Force:
   • For flotation, Weight of block = Buoyant force.
   • Buoyant force = Weight of displaced water.
   • Weight of displaced water = Volume of displaced water \(\times\) density of water \(\times\) \( g \).
   • The volume of displaced water is equal to the submerged volume of the block, let's call it \( V_{\text{submerged}} \).
   • Density of water, \( \rho_{\text{water}} \approx 1000 \, \text{kg/m}^3 \) or \( 1 \, \text{g/cm}^3 \).
4. Equate and Solve for Submerged Volume:
   • Weight of block = \( m \times g \).
   • Buoyant force = \( V_{\text{submerged}} \times \rho_{\text{water}} \times g \).
   • So, \( m \times g = V_{\text{submerged}} \times \rho_{\text{water}} \times g \).
   • We can cancel \( g \) from both sides: \( m = V_{\text{submerged}} \times \rho_{\text{water}} \).
   • Using mass in grams and density in g/cm³: \( 600 \, \text{g} = V_{\text{submerged}} \times 1 \, \text{g/cm}^3 \).
   • Therefore, \( V_{\text{submerged}} = 600 \, \text{cm}^3 \).
5. Calculate the Percentage Submerged:
   • Percentage submerged \( = \left( \frac{V_{\text{submerged}}}{V_{\text{total}}} \right) \times 100\% \).
   • Percentage submerged \( = \left( \frac{600 \, \text{cm}^3}{1000 \, \text{cm}^3} \right) \times 100\% \).
   • Percentage submerged \( = 0.6 \times 100\% = 60\% \).

So, 60% of the cubical block is submerged in water.

Alternatively, we can consider the densities. Let \( \rho_{\text{block}} \) be the density of the block.

• Density of the block, \( \rho_{\text{block}} = \frac{\text{Mass}}{\text{Volume}} = \frac{600 \, \text{g}}{1000 \, \text{cm}^3} = 0.6 \, \text{g/cm}^3 \).
• Density of water, \( \rho_{\text{water}} = 1 \, \text{g/cm}^3 \).

For a floating object, the ratio of the submerged volume to the total volume is equal to the ratio of the object's density to the fluid's density:

\( \frac{V_{\text{submerged}}}{V_{\text{total}}} = \frac{\rho_{\text{block}}}{\rho_{\text{water}}} \)

\( \frac{V_{\text{submerged}}}{V_{\text{total}}} = \frac{0.6 \, \text{g/cm}^3}{1 \, \text{g/cm}^3} = 0.6 \)

Percentage submerged \( = 0.6 \times 100\% = 60\% \).

Both methods yield the same result, confirming that 60% of the cube is submerged.

Revision Table: Floating Objects

Additional Information: Factors Affecting Submergence

The percentage of an object submerged when floating depends primarily on the relative densities of the object and the fluid it is floating in. Specifically:

• A denser object (compared to the fluid) will sink further, meaning a larger percentage of its volume will be submerged.
• A less dense object will float higher, with a smaller percentage submerged.
• If the object's density is equal to the fluid's density, it will be fully submerged but will not sink (neutral buoyancy).
• If the object's density is greater than the fluid's density, it will sink completely.

In this problem, the density of the block (0.6 g/cm³) is 60% of the density of water (1 g/cm³), which directly corresponds to 60% of the volume being submerged.