Question

A block of wood floats in brine solution of density 1.2 g/cm3, , such that 3/8th of its volume is above brine. Calculate the density of wood?

Answer 1

The density of the brine solution,
$$\begin{gathered} {\rho }_b=1.2g/c{m}^3=\frac{1.2\times {10}^{-3}}{({10}^{-2}{)}^3}=1.2\times {10}^3{\text{kg/m}}^3 \\ \text{Let the volume of the wood be}V. \\ \text{The volume of the wood inside the brime solution}, \\ V\text{'}=\left(1-\frac{3}{8}\right)V=\frac{5}{8}V \\ \text{The weight of the wood is balanced by the weight of the displaced volume brime solution.} \\ \text{Therefore,} \\ \text{weight of wood}=\text{weight of displaced volume of brime solution} \\ \Rightarrow {\rho }_{w}Vg={\rho }_{b}V\text{'}g \\ \Rightarrow {\rho }_{w}Vg={\rho }_b\times \frac{5}{8}Vg \\ \Rightarrow {\rho }_w=1.2\times {10}^3\times \frac{5}{8}=0.75\times {10}^3\text{kg}/{\text{m}}^3 \end{gathered}$$