A rowboat holding a large rock is floating at equilibrium in a swimming pool. If the rock is tossed out of the rowboat and sinks to the bottom of the pool will the water level of the pool rise, fall or remain the same?
2007-06-20 12:44:24 · 9 answers · asked by MT 1 in Science & Mathematics ➔ Physics
When the rock is in the boat, the water level is higher by the amount of water it takes to weigh the same as the rock (which is a greatr volume of water than the volume of the rock since the rock is denser than the water). So after the rock is in the pool, two things have changed: The boat is not as heavy, so the water level goes down, and the rock is now in the water, so it goes up. But the first effect, as noted above, is greater, since the amount of water that equals the rocks weight is more than the amount of water that equals the rocks volume. So overall the water level falls.
2007-06-20 12:55:33 · answer #1 · answered by Anonymous · 3⤊ 0⤋
Steve and PJ are both correct. The volume of water dispaced will decrease by the volume of water equal to the mass of the rock, less the volume of the rock. If the rock is small enough to lift and be tossed overboard and the pool is big enough to float a rowboat, the drop in the water level will probably go unnoticed. This drop in level could be computed if the mass and volume of the rock, the surface area of the pool, and the area of the rowboat at the waterline are known.
2007-06-20 20:25:29 · answer #2 · answered by devilsadvocate1728 6 · 1⤊ 0⤋
I'll offer my easy way to analyze this. Exchange the rock with an equal volume of water while holding the boat fixed, which will leave the pool water level unaffected. Then since the boat with the blob of water will be lighter or more buoyant, the boat will rise when released, and the pool water level will fall.
2007-06-20 20:46:30 · answer #3 · answered by Scythian1950 7 · 1⤊ 0⤋
Actually it will fall.
Initially the rock is effectively floating on the water, therefore the boat is displacing a volume of water equivalent to the weight of the rock (and the boat but this is irrelevant). That volume of water is greater than the volume of the rock (because generally rocks sink). When you toss the rock in the volume of water displaced is only equal to the volume of the rock (which is less that the volume of the water displaced to keep the rock floating). Hence the level falls (but only very slightly obviously).
2007-06-20 19:57:59 · answer #4 · answered by PJ 3 · 1⤊ 0⤋
It will stay the same -- The rock in the boat displaces the volume of the water equal to the mass of the rock so when the rock is in the water out of the baot the same amount of the volume of water equal to the same mass of the rock. This was discovered by Archimedes in a bath tub one night.
2007-06-20 20:55:46 · answer #5 · answered by 4xmom 2 · 0⤊ 1⤋
The water level will fall. I got this as a test question during JC.
2007-06-21 21:32:05 · answer #6 · answered by cipher_01 1 · 0⤊ 0⤋
stay the same. It will fall when you pick up the rock and rise when you drop it back in the water.
2007-06-20 20:38:27 · answer #7 · answered by Idaly M 2 · 0⤊ 1⤋
Rise
2007-06-20 19:49:07 · answer #8 · answered by eric l 6 · 0⤊ 3⤋
uhm... can u repeat the question?
2007-06-20 19:52:03 · answer #9 · answered by Bree 2 · 0⤊ 2⤋