Question

A block of wood of mass $24kg$ floats in water. The volume of the wood is $0.032m^3$. Taking the density of water as $1000kg/m^3$, calculate the;

1. Volume of the block below the surface of the water, and
2. The density of the wood.

Answer 1

Step 1: Given data
Mass of the wooden block = $24kg$
The volume of the wooden block = $0.032m^3$
The density of water = $1000kg/m^3$

Step 2: Finding the density of the wood
$$\begin{gathered} \mathrm{Density}\mathrm{of}\mathrm{the}\mathrm{wooden}\mathrm{block}=\frac{\mathrm{mass}}{\mathrm{volume}} \\ =\frac{24\mathrm{kg}}{0.032{\mathrm{m}}^3}=750\mathrm{kg}/{\mathrm{m}}^3 \end{gathered}$$
Hence, the density of the wooden block is $\mathbf{750}\mathbf{}\mathit{k}\mathit{g}\mathbf{/}{\mathit{m}}^{\mathbf{3}}$.

Step 3: Finding the volume of the block below the surface of water
$$\begin{gathered} \mathrm{Relative}\mathrm{density}\mathrm{of}\mathrm{the}\mathrm{wooden}\mathrm{block}=\frac{\mathrm{density}\mathrm{of}\mathrm{the}\mathrm{wooden}\mathrm{block}}{\mathrm{density}\mathrm{of}\mathrm{water}} \\ =\frac{750\mathrm{kg}/{\mathrm{m}}^3}{1000\mathrm{kg}/{\mathrm{m}}^3}=0.75 \\ \mathrm{Also},\mathrm{relative}\mathrm{density}\mathrm{of}\mathrm{the}\mathrm{wooden}\mathrm{block}=\frac{\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{block}\mathrm{immersed}\mathrm{in}\mathrm{water}}{\mathrm{total}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{block}} \\ \mathrm{So},\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{block}\mathrm{immersed}\mathrm{in}\mathrm{water}=\mathrm{relative}\mathrm{density}\mathrm{of}\mathrm{the}\mathrm{wooden}\mathrm{block}\times \mathrm{total}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{block} \\ =0.75\times 0.032{\mathrm{m}}^3=0.024{\mathrm{m}}^3 \end{gathered}$$

Hence, the volume of the wooden block below the surface of water is $\mathbf{0}\mathbf{.}\mathbf{024}\mathbf{}{\mathit{m}}^{\mathbf{3}}$.