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A balloon filled with air is weighted, so that it barely floats in water as shown in figure. When it is pushed down so that it gets submerged a short distance in water, then the balloon: -

(A) Will come up again to its former position
(B) Will remain in the position it is left
(C) Will sink to the bottom
(D) Will emerge out of liquid

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Last updated date: 13th Jun 2024
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Answer
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Hint: We know that a body at rest in a fluid is acted upon by a force pushing upward called the buoyant force, which is equal to the weight of the fluid that the body displaces. If the body is completely submerged, the volume of fluid displaced is equal to the volume of the body. If the body is only partially submerged, the volume of the fluid displaced is equal to the volume of the part of the body that is submerged. Based on this concept we have to answer this question.

Complete step by step answer
We know that the buoyancy force is caused by the pressure exerted by the fluid in which an object is immersed. The buoyancy force always points upwards because the pressure of a fluid increases with depth. Buoyancy is the upward force we need from the water to stay afloat, and it's measured by weight. The trapped air weighs much less than the weight of the water it displaces, so the water pushes up harder than the life jacket pushes down, allowing the life jacket to remain buoyant and float.
Assuming that "barely able to float" means that it is still able to float, then its density is still less than water's and it will return to the surface after being pushed below the surface.
However, if by "barely able to float '' the question means that the weighted balloon's density is the same as water, then it will not return to the surface of the water; it will just continue to float at the depth.

Hence Option A is correct.

Note: To answer such a question, it should be known to us that Archimedes' principle is very useful for calculating the volume of an object that does not have a regular shape. The oddly shaped object can be submerged, and the volume of the fluid displaced is equal to the volume of the object. It can also be used in calculating the density or specific gravity of an object.
Let us explain with the help of an example, for an object denser than water, the object can be weighed in air and then weighed when submerged in water. When the object is submerged, it weighs less because of the buoyant force pushing upward. The object's specific gravity is then the object's weight in air divided by how much weight the object loses when placed in water. But most importantly, the principle describes the behaviour of anybody in any fluid, whether it is a ship in water or a balloon in air.