Question

An ice cube floats with $3/4$ of its volume above the surface of water. What is the density of the solid?

• A. $120kg{m}^{-3}$
• B. $250kg{m}^{-3}$
• C. $200kg{m}^{-3}$
• D. $750kg{m}^{-3}$

Answer 1

Answer: (B)

Let $V$ and $\rho$ is the volume of ice cube and density of cube and $\sigma$ is the density of water weight of ice $=mg=V\rho g$
Volume of ice cube outside the water $=\frac{3}{4}V$
$\therefore$ Volume inside the water $=V-\frac{3}{4}V=\frac{V}{4}$
Weight of water displaced by ice cube $=\frac{V}{4}\times \sigma \times g$
$=\frac{V}{4}\times 10^3\times g$
$\therefore$ Weight of body $=$ weight of water displaced archimedes' principle
$V\rho g=\frac{V}{4}\times 10^3\times g$
$\rho =\frac{1000}{4}$
$=250kg{m}^{-3}$