Question

a solid body floats in water 1/6th of its volume above the surface . what fraction of its volume will project out if it floats in liquid of density 1020 kq/m/m/m ?

Answer 1

Using law of flotation  

Vwater  displaced = V-V/6 = 5V/6 

Vsolid xPSolidxg = v water displaced x p water x g 

V x P solid x g = (5V/6) x p water x g  

Psolid = (5000/6) kg/m3 

When liquid of 1020 kg/m3 is used 

V solid x P solid x g = V liquid displaced x p liquid x g 

V (5000/6) = V liquid displaced x 1020 

V liquid displaced = .81V 

So .81V is inside liquid  

V- 0.81V = 0.19V will project upward 

Or (19/100) V 

Answer 2

Explanation:

Pls refer to the attached file