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 Vm3.Volume of block under liquid =4V5m3∴Volume of liquid displaced =4V5m3Now let the density of the liquid be ρkgm−3.Mass of liquid displaced= (volume of liquid displaced) x (density of liquid)                                             =4V5ρkgWeight of liquid displaced =4V5×ρ×g newton Relative density of wood = 0.8∴Density of wood =0.8×1000                               =800kgm−3∴Mass of the block =800×VkgWeight of the block  =800×V×g newton From the law of flotation,Weight of block = weight of liquid displaced800×V×g=4V5×ρ×gρ=800×54=1000kgm−3