A solid of density 5000 kg/m^3 weighs 0.5 kgf in air.It is completely immersed in water of density 1000 kg/m^3. Calculate the apparent weight of the solid in water.

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  1. volume=.5kg/5000kg/m3= 100ml
    mwass water displaced=100g

    apparent mass=origianal-displaced=400g

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  2. We have,

    Density of solid, d = 5000 kg/m3

    Weight in air, W = 0.5 kgf = 0.5g N

    Density of water, q = 1000 kg/m3

    Let, V be the volume of the solid.

    So, W = Vdg

    => 0.5g = V × 5000 × g

    => V = 10-4 m-3

    So, Buoyant force is, B = Vqg = 10-4 × 1000 × g = 0.1 g N = 0.1 kgf

    Thus, apparent weight = W – B = 0.5 – 0.1 = 0.4 kgf

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  3. 45kg

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  4. A solid of density of 5000kg m^3 weighs 0.5 kg f in air is completely immersed in the liquid of density 8000 kg m^3 calculate the apparent weight of solid in liquid

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  5. 0.5g

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