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4 December, 16:53

A manometer (shown below) that is open to the atmosphere contains mercury (SG = 13.6) with a layer of oil floating on the mercury in the right leg. If the level of the mercury in the left leg is 2 cm above the mercury level in the right leg, and the oil layer on the right is 8 cm above the mercury level in the right leg. What is the density of the oil?

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  1. 4 December, 17:15
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    3400 kg/m³

    Explanation:

    Given:

    Specific gravity of the mercury = 13.6

    therefore, the density of mercury = 13.6 * 1000 = 13600 kg/m³

    Level of mercury in the left limb = 2 cm above the mercury level in right limb

    = 0.02 m

    Level of oil layer in the right limb = 8 cm above the mercury level in right limb

    0.08 m

    Now,

    Pressure in manometer is given as = ρgh

    where,

    ρ is the density

    g is the acceleration due to gravity

    h is the height of the liquid above datum

    also,

    pressure in right limb = pressure in the left limb (for the equilibrium)

    thus,

    ρ₁gh₁ = ρ₂gh₂

    here,

    ρ₁ = density of oil

    ρ₁ * g * 0.08 = 13600 * g * 0.02

    or

    ρ₁ = 3400 kg/m³
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