Question

A solid, incompressible object is lowered into a lake. Why does the buoyant force on the object not increase with depth once the object is totally submerged, even though the surrounding pressure increases?

A. The buoyant force depends on the pressure difference across the object, which stays the same.
B. The object shrinks and displaces less water.
C. The buoyant force increases, but the weight of the object also increases as it is lowered into the lake.
D. The object expands, compensating for the higher pressure.

Answer 1

The buoyant force on a totally submerged object remains constant with depth because it depends on the weight of the fluid displaced, which does not change with depth as long as the object's volume and the fluid's density remain constant.

Step-by-step explanation:

When a solid object is totally submerged in a lake, the buoyant force on the object does not increase with depth even though the surrounding pressure increases. This is because the buoyant force depends on the pressure difference across the object, which remains constant as long as the object is fully submerged. The increase in pressure with depth affects all sides of the submerged object evenly, so the net buoyant force is unchanged. The correct answer to why the buoyant force does not increase with depth once the object is totally submerged is option A: The buoyant force depends on the pressure difference across the object, which stays the same.

According to Archimedes' principle, the buoyant force is equal to the weight of the fluid displaced by the object. Once the object is fully submerged, it displaces a constant volume of fluid, and therefore the weight of the fluid displaced does not change with depth. As a result, the buoyant force remains constant regardless of depth as long as the object does not compress or expand and the density of the fluid remains uniform.