A block of wood of mass 24 kg floats on water. The volume of wood is 0.032 m³.

Find —

(a) the volume of block below the surface of water,

(b) the density of wood.

(Density of water = 1000 kg m^-3)

(a) Given,

Mass = 24 kg

Volume = 0.032 m³

Upthrust = Volume of block below the water x density of liquid x acceleration due to gravity

Hence,

$24 \text {kgf} = v \times 1000 \times g \\[0.5em] v = \dfrac{24}{1000} \\[0.5em] \Rightarrow v = 0.024 \text{ m}^{3} \\[0.5em]$

Hence, volume of block below the surface of water = 0.024 m³

(b) By the principle of floatation,

$\dfrac{\text {Volume of immersed part }}{\text{Total volume}} = \dfrac{\text {Density of wood}}{\text {Density of water}}$

and

Density of water = 1000 kg m^-3

Substituting the values in the formula we get,

$\dfrac{0.024}{0.032} = \dfrac{\text {Density of wood}}{1000} \\[0.5em] \Rightarrow \text{Density of wood} = \dfrac{0.024}{0.032} \times 1000 \\[0.5em] = 7.5 \times 10^{2} \text{ kg m}^{-3} \\[0.5em]$