What happens to the water level in the container?

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SUMMARY

The discussion centers on the behavior of a Styrofoam piece with a lead weight when inverted in water. When the Styrofoam is turned upside down, it continues to float with its top surface flush with the water line. The water level in the container drops because the buoyant force required is less when the lead weight is suspended below the Styrofoam, resulting in reduced water displacement. This conclusion is based on the principles of buoyancy and the relationship between weight and displaced volume.

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Homework Statement


Answer both of the questions below using the following information: A lead weight is fastened on top of a large piece of Styrofoam that floats in a container of water. Because of the weight of the lead, the water line is flush with the top surface of the Styrofoam. If the piece of Styrofoam is turned upside down so that the lead weight is now suspended underneath it, . . .
(a) what happens to the arrangement? Does it sink, float with its top surface flush with the water line, or float with its top surface above the water line?
(b) what happens to the water level in the container? Does it rise, drop, or stay the same?

Homework Equations





The Attempt at a Solution


for (a), I said that it would float with the top surface flush to the water line; (b) drop because there is less buoyant force needed, so there is less water displaced

I don't know if I'm thinking about this the right way, so any help would be greatly appreciated
 
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Remember that the buoyant force is provided by the weight of the volume of water displaced. So if originally the styrofoams volume was enough to support the weight of the styrofoam and the lead, then what would happen if you turned it upside down and the lead now contributed to the volume of water displaced?

For b) why do you think there is less buoyant force needed? If you turn the system upside down it still has the same weight and that means it still needs the same support.
 

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