# Does the mass of a beaker filled with water change due to a mass partially submerged in it?

I believe it should be different as there is upthrust acting on the second block and hence by Newton Third's Law it should act back on the liquid adding to more weight. The answer is given isn't so as it highlights that the upthrust is equal to the weight of block which does not seem relevant.

• Hint: What is the mass of the water that the floating block displaces? – PM 2Ring Jun 15 at 7:57
• This should not be a mystery since Archimedes. – Poutnik Jun 15 at 10:39
• By the diagrams you give, water flows out so that the picture on the left has the same water level. That means that by archimedes principle the weight of the water lost is compensated by the amount of weight the block is displacing.en.wikipedia.org/wiki/Archimedes%27_principle – anna v Jun 15 at 14:29

I do not think that Kirtpole is correct.

The total weight of the beaker+water+block in beaker B is equal to the weight of the beaker+water in beaker A. This is because the the weight of the water in beaker B is that of the water in beaker A reduced by the weight of water that was in the space now occuped by the block. But the weight of this displaced water (which overflowed down the sides of thw beaker as the block was introduced) is equal to the weight of the block by Archimedes.

• I agree ... Kirtpole's analysis is not correct. – David White Jun 15 at 18:20

The mass of the beaker and water together should be the same, with or without the floating block. If you measure it on a balance then you will se that the weight increases when you introduce the block but this is because now you’re also measuring the block’s weight. If you subtract the block’s weight from your measurement then you should obtain the same result as without the block.

It is true what you say about the third law bit you must consider that you are also measuring the block’s weight.