Archimedes' Principle And Floatation

Question

A wooden block of density $600\,	ext{kg/m}^3$ floats freely in water of density $1000\,	ext{kg/m}^3$. What fraction of the block's total volume remains submerged beneath the water surface in equilibrium?

RankGuru · Accepted Answer

Floatation is governed by Archimedes' principle, which states that the upward buoyant force on a body equals the weight of the fluid it displaces. For a freely floating block in equilibrium the buoyant force exactly balances the weight, so $\rho_{block}V_{total}g = \rho_{water}V_{sub}g$. Cancelling $g$ and rearranging gives the submerged fraction $\frac{V_{sub}}{V_{total}} = \frac{\rho_{block}}{\rho_{water}}$. Substituting the densities, $\frac{V_{sub}}{V_{total}} = \frac{600}{1000} = 0.6$, meaning sixty percent of the block lies below the surface. The value 0.4 mistakenly gives the fraction above the water rather than below. The value 0.5 would require the block density to be half that of water. The value 0.8 corresponds to a denser block. This is the classic NCERT result that the submerged fraction equals the ratio of densities for a floating body, and it explains why icebergs, only slightly less dense than seawater, hide most of their bulk below the surface. The result is independent of the block's shape or total volume because both the weight and the buoyant force scale together with that volume. As a plausibility check, since the wood is less dense than water it must float with part of it above the surface, and a submerged fraction of 0.6 leaving 0.4 exposed is physically consistent.

Archimedes' Principle And Floatation

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