Physics - Fluid Question with Solution | TestHub

Let the total volume of the sphere be $V$ and its density be $\rho$.

According to the principle of flotation, the weight of the sphere is balanced by the total buoyant force exerted by the two liquids.

Weight of the sphere, $W=V\rho g$

The sphere has half of its volume in mercury and the other half in oil. The total buoyant force $F_B$ is the sum of the buoyant forces from both liquids:
$F_B=F_{B,\text{mercury}}+F_{B,\text{oil}}$
$F_B=\left(\frac{V}{2}\right){\rho }_{\text{Hg}}g+\left(\frac{V}{2}\right){\rho }_{\text{oil}}g$

Equating the weight of the sphere to the total buoyant force:
$V\rho g=\frac{V}{2}{\rho }_{\text{Hg}}g+\frac{V}{2}{\rho }_{\text{oil}}g$

Dividing both sides by $Vg$, we get:
$\rho =\frac{{\rho }_{\text{Hg}}+{\rho }_{\text{oil}}}{2}$

Given ${\rho }_{\text{Hg}}=13.6{\text{ gm/cm}}^3$ and ${\rho }_{\text{oil}}=0.8{\text{ gm/cm}}^3$, substituting these values:
$\rho =\frac{13.6+0.8}{2}$
$\rho =\frac{14.4}{2}=7.2{\text{ gm/cm}}^3$

The density of the material of the sphere is $7.2{\text{ gm/cm}}^3$.

Answer: $7.2$