A block of wood floats in a liquid with four-fifths of its volume submerged. If the relative density of wood is 0.8, what is the the density of the liquid in units of kg m^-3?

Let the volume of the block be $V{\mathrm{m}}^3$.

Volume of block under liquid $=\frac{4V}{5}{\mathrm{m}}^3$

$\therefore$Volume of liquid displaced $=\frac{4V}{5}{\mathrm{m}}^3$

Now let the density of the liquid be $\rho {\mathrm{kgm}}^{-3}$.

Mass of liquid displaced

= (volume of liquid displaced) x (density of liquid)

                                             $=\frac{4\mathrm{V}}{5}\mathrm{\rho kg}$

Weight of liquid displaced $=\frac{4\mathrm{V}}{5}\times \mathrm{\rho }\times \mathrm{g}\text{ newton }$

Relative density of wood = 0.8

$\therefore$Density of wood $=0.8\times 1000$

                               $=800{\mathrm{kgm}}^{-3}$

$\therefore$Mass of the block $=800\times V\mathrm{kg}$

Weight of the block  $=800\times V\times g\text{ newton }$

From the law of flotation,

Weight of block = weight of liquid displaced

$\begin{array}{c}800\times \mathrm{V}\times \mathrm{g}=\frac{4\mathrm{V}}{5}\times \mathrm{\rho }\times \mathrm{g}\\ \mathrm{\rho }=800\times \frac{5}{4}\\ =1000{\mathrm{kgm}}^{-3}\end{array}$