Question

A piece of wood is floating in water kept in a bottle. The bottle is connected to an air pump. Neglect the compressibility of water. When more air is pushed into the bottle from the pump, the piece of wood will float with
(a) larger part in the water
(b) lesser part in the water
(c) same part in the water
(d) it will sink.

Answer 1

Answer ⇒ Option (c). Same Part in water.

Explanation ⇒ When the piece of wood is floating in water, then,

Upthrust = Weight of fluid displaced.

Now, whether we increase the pressure, or decrease the pressure by pushing the air, it will not affect it.  

Since, Pushing air inside the bottle will increase the air pressure but when the pressure of air inside the bottle is increased it does not increase the weight of the wood nor the density of the water.

So there will be no change in the submerged part.  

Hence, the submerged part will remains the same.

Hence, Option (c). is correct.

Hope it helps.  

Answer 2

A particular piece of wood floats in water that is held in a bottle. The bottle is plugged into an air compressor. Neglect the water is compressible. If more air is pumped out of the pump into the tank, the single piece of wood would float with similar part in the water

Explanation:

Option (c) is correct.

• A particular piece of wood floats in water that is held in a bottle. The bottle is plugged into an air compressor. Neglect the water is compressible. If more air is pumped out of the pump into the tank, the single piece of wood would float with similar part in the water
• If more air from the pump is pumped into the container, the pressure of air rises with the same volume on the wood as on the surface of the water. So, the water and wood levels don't change. Thus, the single piece of natural wood floats in water with the same part. Therefore, option (c) is correct.

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