What will happen to the water level of the swimming pool?

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When a pebble is thrown from a boat into a swimming pool, the water level remains unchanged. While the pebble is in the boat, it displaces water equal to its weight, which is greater than its volume due to its higher density. Once the pebble is thrown into the pool, it displaces only its own volume. The boat rises slightly as it loses the weight of the pebble, but this does not affect the overall water level since the displaced volume remains constant. Thus, the conclusion is that the water level stays the same regardless of whether the pebble is in the boat or at the bottom of the pool.
  • #31
Try some example numbers, like the mass weighs 1,000 kg, and has a volume of .05 m^3 (1/20th cubic meter). When the mass is in the boat, how much water is displaced? When the mass is resting at the bottom of the pool, how much water is displaced?
 
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  • #32
hsdrop said:
...but you're saying that if the mass was submerged the water level would stay the same no matter what the volume of the mass is(provided it would sink and not float on it own)??

Check again. Did he really say that?
 
  • #33
ok now I'm starting to see what the water level would do. it just the way the question was given had me a little mixed up
also i am a veary slow at read (i have a savear reading disablety with decoding) and i just now got to the boddom of the https://en.wikipedia.org/wiki/Archimedes'_principle page that gives a detaled explanchen on how and why the principle works witch help emancely
 
  • #34
CWatters said:
Check again. Did he really say that?
and you're right i looked and no he did not say that

thank you everyone for being so patient with me
 
  • #35
Interesting problem. I don't know if the OP's question ever got answered, but when the rock is in the boat it will cause an extra displacement of water equal to its weight. If the density of the rock is greater than the density of water, the amount of water displaced by the rock when it is in the boat will be greater than the volume of the rock. When the rock is thrown in the water, it sinks to the bottom and the rock only displaces its volume. Thereby the water level of the pool holding the boat drops a small amount. Alternatively, if the rock is less dense than water, it will float (it won't be entirely submerged) when tossed out of the boat and the amount of water displaced by the rock will be the same whether it was in the boat or out of the boat.
 
  • #36
Now what I understood reading the complete thread. Boat is assumed weightless and stone is of finite density & some finite volume. Now only weight of stone will contribute in displacing certain volume of water. The proportion of density between water and pebble will decide how much volume (multiple of stone's volume) water will be displaced. As the weight of stone will be equal to the weight of displaced water I.e. Archimedes principle. Let's say the density of stone is 1000 times that of water. So when pebble is in the boat it will displace 1000 times more volume compared to the pebble sinking to the bottom of the pool. Because when pebble is sinking it is only displacing its own volume. So level of swimming pool will fall after throwing the pebble.
 
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