(i)A block of mass 7 kg and volume 0.07 m^3 floats in a liquid of density 140 kg m^-3 Calculate volume of block above surface of liquid
(ii)A solid of density 5000 kg m^-3 weighs 0.5 kg in air It is completely immersed in liquid of density 8000 kg m^-3.Calculate apparent weight of solid in liquid
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Weight of block = buoyancy force = weight of liquid displaced

Volume of liquid displaced = V m³

7 * 10 Newtons = V * 140 * 10 Newtons

V = 0.05 m³

Volume above the surface of liquid = 0.07 - 0.05 = 0.02 m³

mass of solid = 0.5 kg       density = 5000 kg /m³

mass of liquid displaced by solid = volume of solid * density of liquid
              = (0.5/5, 000) * 8, 000 = 0.8 kg
Buoyancy force on solid = 0.8 kg * 10 m/s² = 8 Newtons

When the solid is completely immersed in the liquid, the buoyancy force is more than its own weight.  Hence, it will float immediately.  It looks as if its apparent weight is negative and is -3 Newtons or -0.3 kg.  ( 0.5g - 0.8g)

When it reaches the surface of the liquid, it will displace less volume of liquid such that buoyancy force = weight of solid. Then it floats.   Hence apparent weight = 0.

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