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9 February, 23:25

A block of wood floats in fresh water with 0.663 of its volume V submerged and in oil with 0.913 V submerged. Find the density of (a) the wood and (b) the oil.

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  1. 10 February, 03:11
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    (a). 663kg/m³.

    (b). 726.18kg/m³

    Explanation:

    (a). Let V be the volume of the block of wood,

    Then the submerged volume in fresh water is

    V₍s₎=0.663V

    Where V₍s₎ = submerged volume in water.

    Since the block is floating, according to Archimedes principle,

    The weight of the displaced fresh water is equal to the weight of block.

    ∴D₍w₎V₍s₎ = DV ... (1)

    Where D₍w₎ = Density of fresh water, D = Density of Block of wood

    Since V₍s₎=0.663V, and D₍w₎=1000kg/m³.

    We substitute the value of V₍s₎ and D₍w₎ in equation (1) above.

    ⇒ 1000*0.663V = D * V

    ∴ D * V = 1000 * 0.663V

    Dividing both side of the equation by V

    ∴ D = 663kg/m³.

    (b). Let V be the volume of the wood block. and V₍o₎ is the submerged volume in oil.

    V₍o₎ = 0.913V

    Since the wood block is floating, according to Archimedes principle,

    The weight of the displaced oil is equal to the weight of the wood block.

    IF D₍o₎ = Density of oil.

    D₍o₎V₍o₎ = DV ... (2)

    D=663kg/m³, and V₍o₎ = 0.913V,

    Substituting the value of D and V₍o₎ in equation (2)

    D₍o₎ * 0.913V = 663 * V

    Dividing both side of the equation by the coefficient of D₍o₎

    ∴D₍o₎ * 0.913V/0.913V = 663V/0.913V

    ∴ D₍o₎ = 663/0.913

    D₍o₎ = 726.177

    D₍o₎ ≈ 726.18kg/m³
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