Question

A block of wood floats in a liquid of density $0.8\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$c{m}^3$}\right.$ with one-fourth of its volume submerged. In oil, the block floats with $60\%$of its volume submerged. Find the density of the oil.

Answer 1

Given quantities:

$$\begin{gathered} {\rho }_l=0.8\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$c{m}^3$}\right. \\ Volumesubmerged=\frac{3V}{5} \end{gathered}$$

Formula Used:

$\rho Vg={\rho }_1{V}_{1}g$

The calculation for the density of wood:

$$\begin{gathered} \rho ={\rho }_{oil}\frac{\frac{3V}{5}}{V} \\ \rho =\frac{3{\rho }_{oil}}{5} \\ {\rho }_{oil}=\frac{0.2\times 5}{3} \\ {\rho }_{oil}=0.33\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$c{m}^3$}\right. \end{gathered}$$

Hence, the density of the wood is$0.33\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$c{m}^3$}\right.$.