Question

A solid floats with $\left(\begin{array}{c}\frac{1}{4}\end{array}\right)th$ of its volume above the surface of water. Calculate the density of the solid.

Answer 1

Let $V$ be the volume and $d$ be the density of the solid then weight of the body $=Vdg$.
Volume of solid body inside water $=V-\frac{V}{4}=\frac{3V}{4}$
Weight of water displaced by immersed portion of the body
$=\frac{3V}{4}\times {10}^3\times g$
$\therefore$ By the principle of floatation,
$Vdg=\frac{3V}{4}\times {10}^{3}g$
$\Rightarrow d=\frac{3}{4}\times {10}^3=750kg{m}^{-3}$.