Why work done to submerge a ball is $pV$?

Question

In this stack post Mr.mark says the work done to submerge a ball is $pV$. I don't seem to comprehend this .

I believe this would be answered in a comment but and answer could help me show my appreciation as well

Answer 1

Let's assume that the total amount of water is so large that submerging the sphere raises the overall water height by only a negligible amount. Completely submerging a weightless sphere of volume $V$ is then equivalent to elevating a mass of water $m=\rho V$, where $\rho$ is the water density, by a distance $r$, where $r$ is the sphere radius. The reason is that the completely submerged sphere has its center at a depth of $r$; this is also the center of mass of the water that the sphere displaced. That water is now dispersed at the surface, for a net elevation of $r$.

The energy to elevate a mass is $mgh$ (where $h$ is the height), or $\rho Vgr$ in this case. Let $p$ specifically represent the pressure at the depth of the center of the just-submerged sphere. By applying the general formula for hydrostatic pressure, we find that $p=\rho g r$ at that depth. Then, the work done in submerging the sphere/elevating the water is $pV$.