Physics

Physics

Problem

A rectangular wooden block of height 20 cm20\,\text{cm} and uniform cross-sectional area 50 cm250\,\text{cm}^2 floats upright in fresh water (ρ=1000 kg m−3)(\rho = 1000\,\text{kg\,m}^{-3}). Exactly 15 cm15\,\text{cm} of its height is below the water surface. a. (i) Calculate the density of the wood. b. (ii) The same block is now floated in kerosene of density 800 kg m−3800\,\text{kg\,m}^{-3}. Determine the depth of the block that will be submerged in the kerosene.
Intuitive explanation

The key to floating equilibrium is that the buoyant force (weight of water displaced) must equal the block's weight. Since you know how much of the block is submerged, you can find the block's density by comparing the submerged volume to the total volume—the ratio of submerged height to total height tells you what fraction of the block's density equals water's density. This is a direct application of Archimedes' principle without needing to know the block's mass explicitly.

Intuitive explanation

The key to floating equilibrium is that the buoyant force (weight of water displaced) must equal the block's weight. Since you know how much of the block is submerged, you can find the block's density by comparing the submerged volume to the total volume—the ratio of submerged height to total height tells you what fraction of the block's density equals water's density. This is a direct application of Archimedes' principle without needing to know the block's mass explicitly.

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