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    A student devised a simple experiment to demonstrate the change in position of a ship when in water of different salt concentrations. The student placed a straw, sealed and weighted at the bottom, into a test tube containing distilled water. The length of straw below the surface of the water was then measured. 13.3 cm The straw settled in a position so that the length below the surface of the water was 13.3 cm. The experiment was then repeated using water containing 20% salt. Calculate the length of straw now below the surface of the water. density of distilled water = 998 kg m−3 density of distilled water containing 20% salt = 1150 kg m−3 .
    Can someone solve it plzz
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    (Original post by mgy2000)
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    Use Archimedes' principle and the fact that the buoyancy force must be the same for the two cases.

    You should not neglect the inclusion of the added weight (although your calculations will show that it is irrelevant).
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    (Original post by pleasedtobeatyou)
    Use Archimedes' principle and the fact that the buoyancy force must be the same for the two cases.

    You should not neglect the inclusion of the added weight (although your calculations will show that it is irrelevant).
    Can u plzz show me ur working ?
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    (Original post by mgy2000)
    Can u plzz show me ur working ?
    The question should be well within your ability if you are doing AS Physics.

    Here's a hint for Archimedes' principle with the buoyancy force B:

    B = \rho_{fluid} g V_{fluid \ displaced}
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    (Original post by pleasedtobeatyou)
    Use Archimedes' principle
    This isn't nessesary in it's full form, or possible as dimensions are not given.
    Simply take a force balance of the 2 situations. What do you notice about them? Is it possible to eliminate an unknown you don't know?
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    (Original post by EwanWest)
    This isn't nessesary in it's full form, or possible as dimensions are not given.
    Simply take a force balance of the 2 situations. What do you notice about them? Is it possible to eliminate an unknown you don't know?
    It is possible if you equate the buoyancy force for the two scenarios since:

    V \propto L

    because the cross-sectional area of the straw remains constant and will cancel.
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    We're essentially saying the same thing, Achermedes principle is merely the generalised 3D approach, where as this example only needs a 1D analysis
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    Also, you've made the assumption that the cross-sectional area remains constant, which you're not explicitly told
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    (Original post by EwanWest)
    We're essentially saying the same thing, Achermedes principle is merely the generalised 3D approach, where as this example only needs a 1D analysis
    Yes but OP is simply trying to learn the fundamentals of Archimedes' principle, rather than apply it in optimum fashion. It would be worth their time starting with Archimedes' principle in its most-general form.


    (Original post by EwanWest)
    Also, you've made the assumption that the cross-sectional area remains constant, which you're not explicitly told
    I think we can assume that the straw being used has a constant cross-section

    If you simplify the problem to 1D, you are assuming a constant cross-section (!)
 
 
 
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