If a solid body floating in water has th of the volume above surface, then determine the fraction of its volume that will project upwards if it floats in a liquid of specific gravity 1.2

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Concepts involved : 1) specific gravity = density of liquid / density of water 2) boyancy force = volume immersed *density of liquid (ρ)* g( accn due to gravity) 3) weight of the object = total volume *density (σ)*g Now since 1/6 th of the volume is above liquid surface 5/6 th is below it...
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Concepts involved : 1) specific gravity = density of liquid / density of water 2) boyancy force = volume immersed *density of liquid (ρ)* g( accn due to gravity) 3) weight of the object = total volume *density (σ)*g Now since 1/6 th of the volume is above liquid surface 5/6 th is below it therefore by force balancing ( the upward boyancy force and downward gravity ) in case of water 5/6*V*ρ*g = V*σ*g ----(1) in case of the other liquidλ*V*ρ'*g=V*σ*g ----(2) Therefore from 1 and 2 λ*ρ' =5/6*ρ λ*1.2=5/6*1 λ=25/36 Therefore the fraction of thw volume outside the surface of water = 9/36 read less
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