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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

Qmechanic
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Glowingbluejuicebox
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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 $pV$.

Chemomechanics
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    Nice answer. Just to emphasize, the work is only pV if the pressure p is taken at the centroid of the submerged volume. – Ben51 Feb 26 '21 at 14:00
  • Thank you sir ,but I did not understand the first paragraph. Doesn't the elevation of the water depend on the area of the container ? You say the water level raises by r . – Glowingbluejuicebox Feb 26 '21 at 18:57
  • I added an initial sentence for clarification. The overall water height is assumed not to change, which is valid when the total water volume is far greater than the sphere volume. – Chemomechanics Feb 26 '21 at 19:32
  • "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." Sir this part of the answer only confuses me I've understood the rest.How can the center of mass of the displaced water be at r if it's negligible and I'm also confused by "net elevation of r" – Glowingbluejuicebox Feb 27 '21 at 03:54
  • Where the sphere is submerged, we originally had a spherical volume of water at depth $r$ (on average). That water is now displaced by the sphere and is floating on the surrounding water at zero depth. Therefore, the water was raised by a height $r$. Hope this clarifies. If not, please try drawing a diagram of the problem before and after the sphere is submerged, including where the water is originally and where it ends up. – Chemomechanics Feb 27 '21 at 05:40