A block of wood floats in a liquid with four-fifths of its volume submerged. If 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 m³
.
      Volume of block under liquid $=\frac{4V}{5}{m}^3$
          Volume of liquid displaced $=\frac{4V}{5}{m}^3$
Now let the density of the liquid be $\rho kg m^{-3}$ Mass of liquid displaced
= (volume of liquid displaced) x ( density of liquid)

                                                    $=\frac{4V}{5}\rho kg$
   Weight of liquid displaced$=\frac{4V}{5}\times \rho \times g$  newton
      Relative density of wood $=0.8$

 $$\begin{gathered} Density of wood=0.8\times 1000 \\ =800 kg m^{-3} \end{gathered}$$

$$\begin{gathered} \therefore Mass of the block=800\times V kg \\ Weight of the block=800\times V\times g newton \end{gathered}$$

From the law of flotation,

     Weight of.block = weight of liquid displaced

$$\begin{gathered} or 800\times V\times g=\frac{4V}{5}\times \rho \times g \\ or \rho =800\times \frac{5}{4} \\ =1000 kg m^{-3} \end{gathered}$$