Question

A solid of density 3000 kg/m^3 weighs 0.3kgf in air. 1/4th of this solid is immersed in a liquid of density 8000 kg/m^3. Calculate the apparent wegiht of the solid.

Answer 1

Density of the solids is, ρ = 3000 kg/m³
Its weight in air is, W = 0.3 kgf = 0.3g N
If V be its volume then,
Vρg = 0.3g
=> V = 0.3/3000
=> V = 0.0001 m³
¼ th of V is immersed in a liquid of density 8000 kg/m³.
The buoyant force is, B = ¼ V(8000)g = ¼ (0.0001)(8000)g = 0.2g N
Thus, the apparent weight of the solid is = W – B = 0.3g – 0.2g = 0.1g N = 0.1 kgf