A $10-cm-diameter$, $20-cm-tall$steel cylinder$\left({\rho }_{\text{steel}}=\right.$$\left.7900\mathrm{kg}/{\mathrm{m}}^3\right)$ floats in mercury. The axis of the cylinder is perpendicular to the surface. What length of steel is above the surface$?$

The cylinder floats in the mercury bath. So, the net force on it is zero. Use the expression for buoyant force and gravitational force and equate them to calculate the length of the steel above the surface.

The expression for buoyant force is,

$F_B={\rho }_f{V}_f\mathrm{g}$

Here, ${\rho }_{\mathrm{f}}$is density of fluid $V_{\mathrm{f}}$is volume of the displaced fluid, and $g$is acceleration due to gravity.

The following figure shows the situation.

Here, $h$is the length of the cylinder above the surface.

The buoyant force on the cylinder submerged in Mercury is,

$F_{\mathrm{B}}={\rho }_{{\mathrm{Hg}}_{}}{V}_{{\mathrm{Hg}}_{}}\mathrm{g}$

From Archimedes principle, the buoyant force acting an the steel cylinder is equal to weight of the cylinder submerged and it is expressed as follows:

$F_{\text{B}}={\rho }_{\text{steel}}{V}_{\text{steel}}g$

Here, $V_{\text{steel}}$is volume of the cylinder.

Combine the above two equations.

${\rho }_{{\mathrm{Hg}}_{}}{V}_{\mathrm{Hg}}g={\rho }_{\text{steel}}{V}_{\text{steel}}g$

${\rho }_{\mathrm{Hg}}{V}_{\mathrm{Hg}}={\rho }_{\text{steel}}{V}_{\text{steel}}$

The volume of the displaced fluid is,

$V_{\mathrm{Hg}}=A(20\mathrm{cm}-h)$.

Replace$V_{\text{Hg}}$by $A(20\mathrm{cm}-h)$,$V_{steel}$by $A(20\mathrm{cm})$in $F_{\mathrm{B}}={\rho }_{\text{steel}}{V}_{\text{steel}}g$and rearrange the expression for $h$.

${\rho }_{\mathrm{Hg}}A(20\mathrm{cm}-h)={\rho }_{\text{steel}}A(20\mathrm{cm})$

$h=20\mathrm{cm}-\frac{{\rho }_{\text{steel}}(20\mathrm{cm})}{{\rho }_{\mathrm{Hg}}}$

Substitute $13600\mathrm{kg}/{\mathrm{m}}^3$for${\rho }_{Hg}$, $7900kg/m^3$for ${\rho }_{steel}$,

$h=(20\mathrm{cm})-\frac{\left(7900\mathrm{kg}/{\mathrm{m}}^3\right)(20\mathrm{cm})}{13600\mathrm{kg}/{\mathrm{m}}^3}$

$h$$=8.38\mathrm{cm}$