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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 liquid  λ*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 

 

 

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