Question

A solid cube, with faces either vertical or horizontal, is floating in a liquid of density $6g/cc$. It has two third of its volume submerged. If enough water is added from the top so as to completely cover the cube, what fraction of its volume will remain immersed in the liquid?

Answer 1

Density of liquid $=6g/cc=6000kg/m^3$
Applying the law of floatation, when it is placed only in liquid.
$V{\rho }_{c}g=\left(\frac{2}{3}V\right){\rho }_{l}g..........\left(I\right)$
Let say $x$ fraction of volume is inside the liquid, so $(1-x)$ fraction of volume will be inside water.
Now applying the law of floatation, when it is placed in water and the liquid.
$V{\rho }_{c}g=xV{\rho }_{l}g+(1-x)V{\rho }_{w}g..........\left(II\right)$
Equating $\left(I\right)$ and $\left(II\right)$
$\left(\frac{2}{3}V\right){\rho }_{l}g=xV{\rho }_{l}g+(1-x)V{\rho }_{w}g\Rightarrow \left(\frac{2}{3}\right){\rho }_l=x{\rho }_l+(1-x){\rho }_w\Rightarrow \left(\frac{2}{3}\right)\times 6000=x\times 6000+(1-x)\times 1000\because {\rho }_w=1000kg/m^3\Rightarrow 4000=6000x+1000-1000x\Rightarrow 5000x=3000\Rightarrow x=\frac{3}{5}$