Calculating Wood Density Using Buoyancy

This problem requires us to determine the density of a block of wood based on how it floats in water. The key physical principle involved here is buoyancy, specifically Archimedes' principle and the principle of flotation.

Understanding Buoyancy and Flotation

When an object floats in a fluid, the upward buoyant force exerted by the fluid on the object is equal to the weight of the object. Archimedes' principle states that the buoyant force is equal to the weight of the fluid displaced by the submerged part of the object.

• Weight of the object: This depends on the object's volume and density.
• Buoyant force: This depends on the volume of the submerged part of the object and the density of the fluid.

Applying the Principle of Flotation

Let \(V\) be the total volume of the block of wood.

We are given that 30% of the volume is above water. This means the volume of the wood submerged in water is:

Volume submerged (\(V_{submerged}\)) = Total volume - Volume above water

\(V_{submerged} = V - 0.30V = 0.70V\)

The density of water (\(\rho_{water}\)) is given as \(10^3 \, \text{kg/m}^3\), which is \(1000 \, \text{kg/m}^3\). Let the density of the wood be \(\rho_{wood}\).

According to the principle of flotation:

Weight of the wood block = Buoyant force

The weight of the wood block is given by:

Weight = mass \(\times\) acceleration due to gravity (\(g\))

mass = density \(\times\) volume = \(\rho_{wood} \times V\)

Weight of wood = \((\rho_{wood} \times V) \times g\)

The buoyant force is equal to the weight of the water displaced, which is the weight of the water occupying the submerged volume:

Weight of displaced water = mass of displaced water \(\times g\)

mass of displaced water = density of water \(\times\) volume of displaced water

mass of displaced water = \(\rho_{water} \times V_{submerged}\)

Buoyant force = \((\rho_{water} \times V_{submerged}) \times g\)

Equating the weight and the buoyant force for flotation:

\((\rho_{wood} \times V) \times g = (\rho_{water} \times V_{submerged}) \times g\)

We can cancel \(g\) from both sides (assuming \(g > 0\)):

\(\rho_{wood} \times V = \rho_{water} \times V_{submerged}\)

Substitute the value of \(V_{submerged}\) in terms of \(V\):

\(\rho_{wood} \times V = \rho_{water} \times (0.70V)\)

Assuming \(V > 0\), we can cancel \(V\) from both sides:

\(\rho_{wood} = \rho_{water} \times 0.70\)

Now, substitute the given density of water:

\(\rho_{wood} = (10^3 \, \text{kg/m}^3) \times 0.70\)

\(\rho_{wood} = 1000 \, \text{kg/m}^3 \times 0.70\)

\(\rho_{wood} = 700 \, \text{kg/m}^3\)

The density of the wood is \(700 \, \text{kg/m}^3\). This can also be written in scientific notation as \(7 \times 10^2 \, \text{kg/m}^3\).

Verifying the Options

Let's compare our calculated density with the given options:

• Option 1: \(6 \times 10^3 \, \text{kg/m}^3 = 6000 \, \text{kg/m}^3\)
• Option 2: \(5 \times 10^2 \, \text{kg/m}^3 = 500 \, \text{kg/m}^3\)
• Option 3: \(7 \times 10^2 \, \text{kg/m}^3 = 700 \, \text{kg/m}^3\)
• Option 4: \(3 \times 10^2 \, \text{kg/m}^3 = 300 \, \text{kg/m}^3\)

Our calculated density, \(700 \, \text{kg/m}^3\), matches Option 3.

Density Calculation Summary

Revision Table: Buoyancy and Density Concepts

Additional Information: Why Objects Float or Sink

An object's ability to float or sink in a fluid depends on the relationship between its density and the density of the fluid. This is directly explained by the principle of flotation we used to find the wood density.

• If \(\rho_{object} < \rho_{fluid}\): The object will float. It will sink until the weight of the displaced fluid (buoyant force) equals its own weight. The fraction of the object's volume submerged is equal to the ratio of the object's density to the fluid's density (\(V_{submerged}/V_{total} = \rho_{object}/\rho_{fluid}\)). In our case, \(\rho_{wood}/\rho_{water} = 700/1000 = 0.7\), meaning 70% is submerged, leaving 30% above water, which matches the problem statement.
• If \(\rho_{object} > \rho_{fluid}\): The object will sink. The buoyant force when fully submerged is less than the object's weight.
• If \(\rho_{object} = \rho_{fluid}\): The object will be in equilibrium anywhere within the fluid; it neither sinks nor floats to the surface (neutrally buoyant).

Understanding density and buoyancy is fundamental in fluid mechanics and has applications in shipbuilding, submarine design, and many other areas.