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When reading about Fluid Dynamics, I saw two questions that presented similar situations but had very different answers, and I'm trying to understand why, since they seem contradictory:

A beach ball filled with air is pushed about 1 m below the surface of a swimming pool and released from rest. Which of the following statements are valid, assuming the size of the ball remains the same? (Choose all correct statements.) (a) As the ball rises in the pool, the buoyant force on it increases. (b) When the ball is released, the buoyant force exceeds the gravitational force, and the ball accelerates upward. (c) The buoyant force on the ball decreases as the ball approaches the surface of the pool. (d) The buoyant force on the ball equals its weight and remains constant as the ball rises. (e) The buoyant force on the ball while it is submerged is approximately equal to the weight of a volume of water that could fill the ball

Here the only correct answer was (b), and it was said that the buoyant force doesn't change because the displaced volume doesn't change. I thought "ok, this makes sense because the ball simply moves to a lower depth, but depth doesn't change the weight of displaced volume, which determines the buoyant force"

Yet there was another task:

A beach ball is made of thin plastic. It has been inflated with air, but the plastic is not stretched. By swimming with fins on, you manage to take the ball from the surface of a pool to the bottom. Once the ball is completely submerged, what happens to the buoyant force exerted on the beach ball as you take it deeper? (a) It increases. (b) It remains constant. (c) It decreases. (d) It is impossible to determine.

I thought the answer would be that it remains constant too, but it actually decreases, and I don't understand why. The textbook answer says that the ball is compressed to a smaller volume, and since the volume is changing the buoyant force changes. But how do we know that? What word in the task indicates that the volume changes? Since they mention "thin plastic", I assume the fact that the plastic is thin has importance, but I don't know what it means. What would change if it were thick plastic? I also don't understand what "the plastic is not stretched" implies.

Why does the buoyant force remain constant in one situation but not in the other?

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The key distinction here is that the plastic is not stretched—this means that the pressure inside of the unstretched ball equilibrates with the external pressure.

Assuming the air behaves like an ideal gas, Boyle’s Law indicates that $p_1 V_1 = p_2 V_2$, so the volume the air occupies decreases as the ball goes down thanks to the pressure increase from the hydrostatic equation.

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I am assuming the ball to be not properly infiltrated.Now because of this when you get the ball to the floor of pool,because of the pressure from the water above the ball the ball gets pressed and assumes the shape of a flat disk.Buoyancy force will affect a body only if the body has some fluid underneath it to push it upward,in all the other cases it will not affect the body.When you do the same experiment with a ball fully infiltrated the amount of liquid underneath the ball is quite high enough to exert the force in the ball upwards.Hope you got this.

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For the first question, (e) is also a correct response. For the second one, you are meant to intuit that the pressure inside the beach ball is no greater than atmospheric pressure—otherwise the “thin” plastic would presumably “stretch”. I’d say that premise is a bit thin, and the reasoning a bit of a stretch, but otherwise the given answer is correct.

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