Submerged Cone Pressure Ratio

We know that the buoyant force is $F=\rho gV$ , where $\rho$ is the density of water. At the same time the buoyant force is the net upward force acting on the body, and in our case it can be expressed as $F=F_B-F_S$ , since $F_B$ and $F_S$ are both vertical forces. Therefore we have

$\frac{F_B}{F_S}=\frac{5}{4}$ and

$F_B-F_S= \rho gV$ .

We solve this for $F_B$

$F_B=5\rho gV$

On the other hand the pressure at the bottom of the cone is $p=\rho g d$ , where $d$ is the distance between the water surface and the base plane. We can calculate $d$ by using $F_B=p r^2\pi=r^2\pi \rho gd$ and combining it with our result for $F_B$

$5\rho g r^2 \pi \frac{h}{3} = r^2\pi \rho gd$

where we expressed the volume with the height $h$ and base radius $r$ . We get

$d=\frac{5}{3} h$

and therefore the top is at the depth of

$d-h=\frac{2}{3} h= 2.0m$

I don't know if it's easy to make sense out of this but here is how I solved it -

Something interesting to note is that the density of liquid has no effect on the depth of the pyramid.