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A lead brick with volume of 7*10^-4m^3 is floating in bath of liquid mercury.

a) What fraction of the lead brick’s volume is above the surface of mercury?
A student uses a stick to push the lead brick below the mercury surface so that it is completely submerged. What force is required to hold the lead brick below the mercury surface?

  • physics -

    You need to know the specific gravity (or density) of both lead and mercury to answer this question.

    Lead has a specific gravity of 11.35, so it is 11.35 times as dense as water. It's density in SI units is
    rho1 = 11.35*10^3 kg/m^3

    Mercury has a specific gravity of 13.56 and density
    rho2 = 13.56*10^3 kg/m^3

    The buoyancy force is (rho2)*g*V'
    where V' is the displaced volume of mercury. The brick's weight is W = (rho1*g*V)
    where V is the volume of the brick.

    Setting them equal,
    V' = (rho1/rho2)*V = 0.837

    The fraction above the surface must be 0.163 or 16.3%

    To hold it all beneath the surface, the force required is (rho2 - rho1)*g*V =
    2.21*10^3*9.8*7*10^-4 = 15.2 N

  • physics -

    Thanks a lot

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