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Q.
An ice cube floats with $3 / 4$ of its volume above the surface of water. What is the density of the solid?
Mechanical Properties of Fluids
Solution:
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
$V \rho g =\frac{ V }{4} \times 10^{3} \times g$
$\rho=\frac{1000}{4}$
$=250 kgm ^{-3}$