Understanding Buoyancy and Floating Objects

When an object like a rectangular block floats in water, it does so because of the principle of buoyancy. This principle states that an object immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced by the object.

For a floating object, a specific condition holds true: the buoyant force acting upwards is exactly equal to the weight of the object acting downwards. This balance of forces is why the object remains suspended at a certain depth without sinking or rising.

Therefore, to find the weight of the rectangular block, we need to calculate the buoyant force acting on it, which is equal to the weight of the water it displaces.

Calculating the Weight of the Displaced Water

The volume of water displaced by the floating block is the volume of the part of the block that is submerged below the water surface. We are given the dimensions of the block and its depth of immersion.

• Length of the block (L) = 2 m
• Width of the block (W) = 1 m
• Depth of the block (H) = 1 m
• Depth of immersion (h) = 0.5 m
• Weight density of water ($\rho_w g$) = 10 kN/m³

The submerged part of the block forms a rectangular volume with the given length, width, and the depth of immersion.

Volume of displaced water (V_displaced) = Length × Width × Depth of immersion

Let's perform the calculation:

V_displaced = L \times W \times h

V_displaced = 2\ \text{m} \times 1\ \text{m} \times 0.5\ \text{m}

V_displaced = 1\ \text{m}^3

Now that we have the volume of the displaced water, we can calculate its weight using the given weight density of water.

Weight of displaced water = Volume of displaced water × Weight density of water

\text{Weight of displaced water} = V_{displaced} \times (\rho_w g)

\text{Weight of displaced water} = 1\ \text{m}^3 \times 10\ \text{kN/m}^3

\text{Weight of displaced water} = 10\ \text{kN}

Determining the Weight of the Rectangular Block

According to the principle of flotation, the weight of the floating object (the rectangular block) is equal to the weight of the fluid it displaces (the water).

Weight of block = Weight of displaced water

\text{Weight of block} = 10\ \text{kN}

Therefore, the weight of the rectangular block is 10 kN.

Summary of Calculation Steps

1. Identify the given dimensions of the rectangular block and the depth of immersion.
2. Recall the principle of buoyancy and flotation: Weight of block = Buoyant force = Weight of displaced fluid.
3. Calculate the volume of the displaced fluid (water), which is the volume of the submerged part of the block.
4. Calculate the weight of the displaced fluid using its volume and the given weight density.
5. Equate the weight of the block to the weight of the displaced fluid.

Final Answer Determination

The calculated weight of the rectangular block is 10 kN. We check this against the provided options.

• Option 1: 5 kN
• Option 2: 20 kN
• Option 3: 15 kN
• Option 4: 10 kN

Our calculated value matches Option 4.

Revision Table: Buoyancy and Floating

Additional Information on Buoyancy and Density

The depth of immersion for a floating object depends on its density relative to the fluid density. If the object's density ($\rho_{object}$) is less than the fluid's density ($\rho_{fluid}$), it floats. The fraction of the object submerged is equal to the ratio of the object's density to the fluid's density.

\text{Fraction submerged} = \frac{V_{submerged}}{V_{total}} = \frac{\rho_{object}}{\rho_{fluid}}

In this problem, the volume of the block is $2\ \text{m} \times 1\ \text{m} \times 1\ \text{m} = 2\ \text{m}^3$. The submerged volume is $1\ \text{m}^3$. The fraction submerged is $1\ \text{m}^3 / 2\ \text{m}^3 = 0.5$.

Using the formula, the ratio of densities should be 0.5. Since the weight density of water is 10 kN/m³, its mass density is $\rho_w = (\rho_w g) / g$. Assuming $g \approx 9.8\ \text{m/s}^2$, $\rho_w \approx 10000\ \text{N/m}^3 / 9.8\ \text{m/s}^2 \approx 1020\ \text{kg/m}^3$. Using the given weight density directly, we can find the weight density of the block.

Weight of block = Volume of block × Weight density of block

10\ \text{kN} = 2\ \text{m}^3 \times (\rho_{block} g)

(\rho_{block} g) = \frac{10\ \text{kN}}{2\ \text{m}^3} = 5\ \text{kN/m}^3

The weight density of the block is 5 kN/m³, which is half the weight density of water (10 kN/m³). This confirms why exactly half the block (0.5 m out of 1 m depth) is submerged when floating in water.