Question

The mass of a tile is 500 g. If the density of the tile is 2.5 g/cm³ , what will be the weight of the tile when it is completely immersed in water? (g = 9.8 m/s² , ρ(water) = 1000 kg/m³ )

Answer 1

Data: m = 500, g = 0.5 kg, ρ 

(tile) = 2.5 g/cm³ = 2500 kg/m³ , g = 9.8 m/s² , ρ (water) = 1000 kg/m³ , weight of the tile when completely immersed in water (also called the apparent weight)  = ?

= \(\rho\) = \(\cfrac{m}{V}\)

∴ Volume of water displaced by the tile

= 2 × 10^-4 m³

∴ Mass of water displaced by the tile

m’ = ρ (water) V = 1000 kg/m³ × 2 × 10^-4 m³

= 0.2 kg

∴ Magnitude of the weight of this water 

= mg = 0.2 kg × 9.8 m/s² = 1.96 N

∴ Buoyant force exerted on the tile = 1.96 N

Magnitude of the weight of the tile = mg = 0.5 kg × 9.8 m/s² = 4.9 N

∴ Weight of the tile when completely immersed in water (apparent weight) = weight of the tile in airbuoyant force on the tile 

= 4.9 N – 1.96 N = 2.94 N(downward)