Physics - Fluid Question with Solution | TestHub

Let $V$ be the volume of the cube and ${\rho }_c$ be its density.

When the cube floats in water, the volume submerged is $V-\frac{1}{3}V=\frac{2}{3}V$.

By the principle of flotation, the weight of the cube equals the buoyant force:

$V{\rho }_{c}g=\left(\frac{2}{3}V\right){\rho }_{w}g$

${\rho }_c=\frac{2}{3}{\rho }_w$

When the cube floats in the liquid, the volume submerged is $V-\frac{3}{4}V=\frac{1}{4}V$.

Applying the principle of flotation again:

$V{\rho }_{c}g=\left(\frac{1}{4}V\right)\rho g$

$\rho =4{\rho }_c$

Substituting the value of ${\rho }_c$:

$\rho =4\left(\frac{2}{3}{\rho }_w\right)=\frac{8}{3}{\rho }_w$

Given ${\rho }_w=1$ g/cc, we get:

$\rho =\frac{8}{3}$ g/cc

Answer: $\frac{8}{3}$