 Pressure in a fluid-mechanical properties of fluids
Question

# 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?

Moderate
Solution

## Let the volume of the block be $V{\mathrm{m}}^{3}$.Volume of block under liquid $=\frac{4V}{5}{\mathrm{m}}^{3}$$\therefore$Volume of liquid displaced $=\frac{4V}{5}{\mathrm{m}}^{3}$Now let the density of the liquid be $\rho {\mathrm{kgm}}^{-3}$.Mass of liquid displaced= (volume of liquid displaced) x (density of liquid)                                             $=\frac{4\mathrm{V}}{5}\mathrm{\rho kg}$Weight of liquid displaced Relative density of wood = 0.8$\therefore$Density of wood $=0.8×1000$                               $=800{\mathrm{kgm}}^{-3}$$\therefore$Mass of the block $=800×V\mathrm{kg}$Weight of the block  From the law of flotation,Weight of block = weight of liquid displaced$\begin{array}{c}800×\mathrm{V}×\mathrm{g}=\frac{4\mathrm{V}}{5}×\mathrm{\rho }×\mathrm{g}\\ \mathrm{\rho }=800×\frac{5}{4}\\ =1000{\mathrm{kgm}}^{-3}\end{array}$

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