
Which is true about an inflatable beach ball as it is pushed under water?

A) When the ball is under water, the pressure on the ball is the same at all places on the surface of the ball.
B) The buoyant force on the ball increases the farther below the surface of the water as you push the ball.
C) The buoyant force on the ball increases until the entire ball is underwater.
D) The ball experiences pressure from the water only in the vertical direction.
Answer
123.6k+ views
Hint: Using the Archimedes Principle, this states that when an object is introduced in a liquid half or full, there is loss in the weight which is also known as apparent weight which is equal to the weight of the liquid displaced by the object introduced. The formula for Buoyant Force is:
\[F=\rho gh\]
where \[\rho \] is the density, \[d\] is the gravity and \[h\] is the height of the liquid.
Complete step by step solution:
Now as said before, the Archimedes principle is the buoyant force which is equal to the liquid weight displaced by the object when dipped in the liquid completely or half. Now in the diagram, the user dropped the balloon in the liquid partially and the user will experience a buoyant force which is equal to the weight of the water displaced meaning the force is not static but volatile and depends on the depth the object is introduced in the water and is entirely dependent upon the properties of the object or in this case the balloon.
If we check the options both (A) and (D) are wrong as pressure is not same and is applied on all direction and not vertically only and for option (B) and (C), the option (B) is correct but as it doesn't specify the application of buoyancy force till the ball is under which is mentioned in the option (C).
Therefore, the correct option is (C).
Note: The buoyancy force of the object (solid not hollow and density greater than water) is dynamic in nature until the ball is not dipped completely it will show resistance in form of force of buoyancy but the moment it is completely dipped in the water all the pressure from the object is gone and it sunk to the bottom.
\[F=\rho gh\]
where \[\rho \] is the density, \[d\] is the gravity and \[h\] is the height of the liquid.
Complete step by step solution:
Now as said before, the Archimedes principle is the buoyant force which is equal to the liquid weight displaced by the object when dipped in the liquid completely or half. Now in the diagram, the user dropped the balloon in the liquid partially and the user will experience a buoyant force which is equal to the weight of the water displaced meaning the force is not static but volatile and depends on the depth the object is introduced in the water and is entirely dependent upon the properties of the object or in this case the balloon.
If we check the options both (A) and (D) are wrong as pressure is not same and is applied on all direction and not vertically only and for option (B) and (C), the option (B) is correct but as it doesn't specify the application of buoyancy force till the ball is under which is mentioned in the option (C).
Therefore, the correct option is (C).
Note: The buoyancy force of the object (solid not hollow and density greater than water) is dynamic in nature until the ball is not dipped completely it will show resistance in form of force of buoyancy but the moment it is completely dipped in the water all the pressure from the object is gone and it sunk to the bottom.
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