What is the weight of an object that is immersed in 100 ft of salt water?

Answer 1

There is really no way to answer the question definitively. It can only be answered in general, theoretical terms. Since you didn't mention the VOLUME of the object, we cannot determine the amount of salt water it displaced. The object's weight would be reduced by an amount equal to the weight of the water displaced. Further, since we don't know the density of the salt water -- salt water is denser than fresh water, and its density varies with salinity -- that further confounds the problem. First, Pascal's principle tells us that fluid pressure on a submerged object is exerted in all directions. But since fluid pressure increases with depth, the pressure on the bottom of the object is greater than on the top, so a net upward force -- buoyancy -- is created. Second, objects of equal volume will have the same buoyancy. (That is somewhat counter-intuitive, since many people assume that a rock will be less buoyant than, say, a similarly sized piece of cork.) The buoyant force opposes the object's weight, so submerged objects have lower effective weight than objects in air. If the buoyant force is greater than an object's real weight, it will rise; if not, it sinks. Third, since buoyancy is proportional to the volume of the submerged object, if the object compresses significantly as it reaches greater depth, its buoyancy will decrease and its effective or apparent weight will increase, which will cause it to sink farther.