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A pot is partially filled with water, in which a plastic cups is floating. Inside the floating cup is a small block of lead. When the lead block is removed from the cup and placed in the water, block sinks to bottom. When this happens, does the water level in the pot: fall, rise, or stay the same?

The answer is apparently fall, but I don't understand why. The solution says "if we take the lead out of the cup and it sinks, the buoyant force acting upon the lead is equal to the weight of the fluid displaced. This means that the water level of the pot falls," but I don't understand how we come to the conclusion that water level falls just by knowing buoyant force is the same as displaced fluid. Any help?

Qmechanic
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Azz Likar
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2 Answers2

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When in the cup, the lead (and cup) displaces a weight of water equal to it's weight. When at the bottom, the lead displaces a volume of water equal to it's volume. (Most of the weight is supported by a normal force from the bottom.) Since the density of lead is several times that of water, it takes more water to support the weight. Since more water is displaced when the lead is in the cup, the height of the surface will be greater in that case.

R.W. Bird
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  • Can i ask why it takes more water to support the weight if the density of lead is greater then water? – Azz Likar Oct 04 '21 at 14:22
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    @AzzLikar let's say you have 10kg of lead which takes up 1 litre of space. To float, it has to displace 10kg of water, which is 10 litres of water. (If it can't do then it sinks). When it sinks it only displaces its actual size which is 1 litre. – user253751 Oct 04 '21 at 15:18
  • (The purpose of the cup is to allow a 1-litre object to displace 10 litres, by keeping water out of the other 9 litres. Otherwise the water would just fill in those 9 litres and you couldn't ever get a 10-litre displacement. This is also how boats work) – user253751 Oct 04 '21 at 15:19
  • That's a good clarification which I failed to make. – R.W. Bird Oct 04 '21 at 19:03
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Let's exaggerate the situation so we can imagine it better. The cup has almost no mass, but the lead is extremely dense and heavy. The pot is only slightly bigger than the cup.

Here are three cases:

Left, just the water.

Middle, we put the small very dense lead mass in and the water rises slightly (displacing only it's own volume of water).

Right, the heavy lead is now in the cup. As you can imagine, the cup has to sink quite far until the weight of the displaced liquid matches the weight of the lead (Archimedes principle). So the liquid rises far relative to the first case.

enter image description here

The question about the lead block being removed from the cup and put in the water is like going from the right hand diagram to the middle one (just imagine the light empty cup floating on top of the water) - so the water level falls.

John Hunter
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