A wooden block of relative density 0.4 floats in a liquid of density 1600 kg m\(^{-3}\). What fraction of its volume is immersed?

Answer Details

A floating body sinks until the upthrust equals its weight. By Archimedes' principle the upthrust equals the weight of liquid displaced, so for a block of volume \(V\) with a fraction \(f\) of that volume submerged in a liquid of density \(\rho_L\):

\[\rho_b V g = \rho_L (fV) g \quad\Rightarrow\quad f = \frac{\rho_b}{\rho_L}.\]

The fraction immersed is simply the ratio of the density of the body to the density of the liquid.

Relative density is a density compared with that of water, so a relative density of \(0.4\) means

\[\rho_b = 0.4\times 1000 = 400\ \text{kg m}^{-3}.\]

Therefore

\[f = \frac{400}{1600} = 0.25.\]

A quarter of the block's volume is below the liquid surface, and the other three quarters stay above it.

The mistake this question is designed to catch is using the relative density \(0.4\) directly as though it were the density in \(\text{kg m}^{-3}\), or dividing \(0.4\) by \(1600\), both of which give far too small a fraction. Relative density has no unit, so it must be multiplied by \(1000\ \text{kg m}^{-3}\) before it is compared with a liquid density given in \(\text{kg m}^{-3}\). Also remember the sanity check: since the block floats, the fraction immersed must lie between \(0\) and \(1\), and a denser liquid means less of the block is submerged.