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Determine the possible effects on the water level

  1. Jul 2, 2012 #1
    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.
    Read it to me
    The new sphere has density ρ = ρ0 and radius R < R0
    The new sphere has mass M = M0 and density ρ > ρ0
    The new sphere has mass M = M0 and radius R > R0
    The new sphere has radius R > R0 and density ρ < ρ0
    The new sphere has radius R = R0 and density ρ < ρ0
    The new sphere has density ρ = ρ0 and mass M > M0

    My answers:

    F
    F or U
    R or U (this is the one I think may be incorrect...may be just U)
    R or F or U
    F
    R

    I believe the key is whether the mass is increasing (water level will rise) or decreasing (water level will fall). I'm unclear as to which of my answers is wrong.

    Thank you very much for your help.
     
  2. jcsd
  3. Jul 2, 2012 #2

    CWatters

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    I believe the third case is the exact oposite of the second case...

    "The new sphere has mass M = M0 and density ρ > ρ0".... implies R < R0
    "The new sphere has mass M = M0 and radius R > R0.... implies ρ < ρ0
     
  4. Jul 2, 2012 #3
    Yep, that was my line of thinking as well...good to hear. Any idea which is/are incorrect?
     
  5. Jul 2, 2012 #4

    CWatters

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    Perhaps take the examples to extreem and see what happens. For example a very very small dense ball would sink and not raise the water level very much.

    I think the fourth one is easy...

    "The new sphere has radius R > R0 and density ρ < ρ0"

    Sounds like a party balloon. That would have a large radius and low density. what happens to the water level when you float a balloon on it.
     
  6. Jul 2, 2012 #5
    I think I have #3 wrong. I believe it is just U.

    Reasoning: If mass is the same and radius is increasing, density is decreasing. If the density were to decrease only a little tiny bit, the water level would obviously not change. If the density were to decrease as much as possible, it would be a point and the water level would still not change. Is this about right?
     
  7. Jul 2, 2012 #6

    jbriggs444

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    As the sphere gets bigger, the density decreases.
    As the density decreases, the sphere gets bigger.

    It's not going to end up point-like. It's going to end up very big. Like inflating a party
    balloon.

    As long as we neglect atmospheric buoyancy and as long as the enlarged sphere fits in the container, it will still float and will still displace a quantity of water equal to its mass. So yes, the answer I get is "U".
     
  8. Jul 2, 2012 #7

    Chronos

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    As long as the sphere floats, no matter what happen to R and P, the volume of water displaced is always = M. If the sphere sinks, only R matters. A sphere of size R can never displace more than an R volume of water.
     
  9. Jul 3, 2012 #8

    CWatters

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    The mass of water displaced wiil be the same but that's not what the question asks. It asks about the water "level". That's not so easy to visualise. A larger object on the surface means less space around the object. Imagine if the ball had a diameter only slightly less than that of the bowl. Say 1mm less. The water level would have to rise further to displace the same volume of water or would it?

    Perhaps the question is too badly specified to answer? Does it mean the level wrt the object or the bowl? Should we assume the bowl is very large?
     
  10. Jul 3, 2012 #9
    ok. thanks. fogot about volume of water dosent change...
     
    Last edited by a moderator: May 6, 2017
  11. Jul 3, 2012 #10

    sophiecentaur

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    It's like the situation for an enormous ship floating in a 'small' dry dock - almost fitting exactly. If you lifted the ship out with a crane, the water level would drop to nearly zero. Same amount of water displaced as when the ship is on the sea but then the water level wouldn't change detectably
     
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