How Does Replacing a Sphere Affect Water Level in a Bowl?

AI Thread Summary
Replacing a solid sphere in a bowl of water affects the water level based on the sphere's density and mass. If the new sphere has the same mass but a larger radius, the water level rises due to increased displacement. Conversely, if the new sphere has a smaller radius and the same density, the water level falls as it results in less displacement. For cases where the density increases while mass remains constant, the outcome depends on whether the sphere sinks or floats, leading to varying effects on the water level. Overall, the relationship between mass, volume, and density is crucial in determining the water level changes.
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A small solid sphere of mass M0, of radius R0, and of uniform density ρ0 is placed in a large bowl containing water. It floats and the level of the water in the dish is L. Given the information below, determine the possible effects on the water level L, (R-Rises, F-Falls, U-Unchanged), when that sphere is replaced by a new solid sphere of uniform density.
https://s3.lite.msu.edu/enc/74/b3c49e2ca8cb7c50641a94dd509544cd403157aec9507f20a52bdc5947f0cb19184f2a219a924c3481bb57610950b65b7dd0f547f2f8b7a8b89c27e8be908981d875d7cfcda8738d36912b4c413a1f8c3b2a53a7dd9ca143fd48be3e50ded0c9.mp3
The new sphere has radius R > R0 and mass M = M0
The new sphere has radius R < R0 and density ρ = ρ0
The new sphere has mass M = M0 and density ρ > ρ0
The new sphere has mass M > M0 and density ρ = ρ0
The new sphere has radius R = R0 and mass M > M0
The new sphere has radius R < R0 and density ρ > ρ0

Homework Equations


B=mass/volume

The Attempt at a Solution


1. U
If it is not submerged then the displacement is equal to mass which doesn't change
2. F
This means less mass therefore less displacement
3. U
If it is not submerged then the displacement is equal to mass which doesn't change
4. R
More displacement
5. R
More displacement
6. R or F or U
It depends how it effects the mass[/B]
 
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ScrubTier said:
A small solid sphere of mass M0, of radius R0, and of uniform density ρ0 is placed in a large bowl containing water. It floats and the level of the water in the dish is L. Given the information below, determine the possible effects on the water level L, (R-Rises, F-Falls, U-Unchanged), when that sphere is replaced by a new solid sphere of uniform density.
https://s3.lite.msu.edu/enc/74/b3c49e2ca8cb7c50641a94dd509544cd403157aec9507f20a52bdc5947f0cb19184f2a219a924c3481bb57610950b65b7dd0f547f2f8b7a8b89c27e8be908981d875d7cfcda8738d36912b4c413a1f8c3b2a53a7dd9ca143fd48be3e50ded0c9.mp3
The new sphere has radius R > R0 and mass M = M0
The new sphere has radius R < R0 and density ρ = ρ0
The new sphere has mass M = M0 and density ρ > ρ0
The new sphere has mass M > M0 and density ρ = ρ0
The new sphere has radius R = R0 and mass M > M0
The new sphere has radius R < R0 and density ρ > ρ0

Homework Equations


B=mass/volume

The Attempt at a Solution


1. U
If it is not submerged then the displacement is equal to mass which doesn't change
2. F
This means less mass therefore less displacement
3. U
If it is not submerged then the displacement is equal to mass which doesn't change
4. R
More displacement
5. R
More displacement
6. R or F or U
It depends how it effects the mass[/B]
Not bad, but in a couple of places you wrote "if it is not submerged". You need to decide whether it can be submerged and if so how that affects the answer.
 
1. R or U
Floating: U mass doesn't change
Submerged: Radius increase means more displacement
2. F
To keep p constant both must fall
3. F or U
Floating: U mass doesn't change
Submerged: To get p to be larger with m constant then volume must decrease
4. R
If Mass increases but p stays constant then volume must also increase. Displacement increases in both situations
5. R or U
Floating: R Mass increases therefore more displacement
Submerged: U volume remains constant no change
6. R or F or U
Floating: Mass may or may not increase could be Rise or U
Submerged: Volume decreases therefore F

This is still wrong. I really can't tell why.
 
ScrubTier said:
1. R or U
Floating: U mass doesn't change
Submerged: Radius increase means more displacement
2. F
To keep p constant both must fall
3. F or U
Floating: U mass doesn't change
Submerged: To get p to be larger with m constant then volume must decrease
4. R
If Mass increases but p stays constant then volume must also increase. Displacement increases in both situations
5. R or U
Floating: R Mass increases therefore more displacement
Submerged: U volume remains constant no change
6. R or F or U
Floating: Mass may or may not increase could be Rise or U
Submerged: Volume decreases therefore F

This is still wrong. I really can't tell why.
You are told it is floating initially. What you have to consider is which of the proposed changes might result in the sphere sinking, and how that might affect the change in water level.
 
In 1, 2,4 It would not submerge
so 1. Would be no change because not submerging and mass no change
2. Fall because p being the same, R getting smaller, than M must also get smaller
4. Rise because p is the same, M gets larger
?
 
ScrubTier said:
In 1, 2,4 It would not submerge
so 1. Would be no change because not submerging and mass no change
2. Fall because p being the same, R getting smaller, than M must also get smaller
4. Rise because p is the same, M gets larger
?
Right.
What about 3, 5 and 6?
 
Well doesn't it depend on if it actual sinks? Like how much the density changes which I don't know?
 
ScrubTier said:
Well doesn't it depend on if it actual sinks? Like how much the density changes which I don't know?
That's right. For the cases where the density increases, consider both possibilities. If this leads to different answers for R, F, U, list all those that may occur.
(For at least one of 3, 5, 6, not all of R, F, U are possible.)
 
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