Floating Wood Density Calculation

To find the density of a floating object, we apply the principle of buoyancy. An object floats when its weight is balanced by the upward buoyant force. The buoyant force equals the weight of the fluid displaced by the object's submerged volume.

Applying the Buoyancy Principle

The condition for flotation is:

\( \text{Weight of object} = \text{Weight of displaced fluid} \)

Expressed using density (\( \rho \)) and volume (\( V \)):

\( \rho_{object} \times V_{object} \times g = \rho_{fluid} \times V_{submerged} \times g \)

Where:

• \( \rho_{object} \) is the density of the wood.
• \( V_{object} \) is the total volume of the wood block.
• \( \rho_{fluid} \) is the density of water.
• \( V_{submerged} \) is the volume of the wood submerged in water.
• \( g \) is the acceleration due to gravity.

Since \( g \) is constant, it can be cancelled out:

\( \rho_{object} \times V_{object} = \rho_{fluid} \times V_{submerged} \)

Calculating the Object's Density

The problem states that 65% of the wood's volume is submerged:

\( V_{submerged} = 0.65 \times V_{object} \)

Substitute this into the simplified buoyancy equation:

\( \rho_{object} \times V_{object} = \rho_{fluid} \times (0.65 \times V_{object}) \)

We can cancel \( V_{object} \) from both sides:

\( \rho_{object} = \rho_{fluid} \times 0.65 \)

The density of water (\( \rho_{fluid} \)) is approximately \( 1000 \text{ kg/m}^3 \). Plugging this value in:

\( \rho_{object} = 1000 \text{ kg/m}^3 \times 0.65 \)

\( \rho_{object} = 650 \text{ kg/m}^3 \)

Matching the Result with Options

We need to express the calculated density (\( 650 \text{ kg/m}^3 \)) in the scientific notation used in the options:

\( 650 \text{ kg/m}^3 = 0.65 \times 1000 \text{ kg/m}^3 = 0.65 \times 10^3 \text{ kg/m}^3 \)

This value matches Option 2.