Buoyancy of a floating wooden block A wooden block (density = 0.75 relative to water) is 5 m long, 2 m wide, and 3 m high. When it floats in fresh water, what volume of water is displaced?

Introduction / Context:
Archimedes’ principle governs floating bodies: the weight of the displaced fluid equals the weight of the body. For a wooden block with given relative density, the submerged volume is a fixed fraction of its total volume.

Given Data / Assumptions:

• Block dimensions: 5 m × 2 m × 3 m.
• Relative density of wood = 0.75 (i.e., rho_wood = 0.75 * rho_water).
• Floating at equilibrium in fresh water; no additional loads.

Concept / Approach:

For floating equilibrium: Weight of block = Weight of displaced water. Since weights are proportional to densities times volumes, the submerged volume fraction equals the relative density. Thus, V_displaced = RD * V_block.

Step-by-Step Solution:

Verification / Alternative check:

Because RD < 1, the block must float with part above water. The displaced volume (22.5 m^3) is less than the block volume (30 m^3), which is physically consistent.

Why Other Options Are Wrong:

17.5, 20.0, and 25.0 m^3 do not equal RD * V_block; 30.0 m^3 implies full submergence (RD = 1), which is not the case.

Common Pitfalls:

Mixing density ratio with mass fraction; forgetting that volume fraction submerged equals relative density for floating bodies.

Final Answer:

22.5 m^3