Question

An iron piece of density 7.8*103 kg/m3 and the volume100 cm3 is totally immersed in water. Find:-
i. weight of the iron piece in air.
ii. upthrust
iii.apparent weight of object in water.

Answer 1

density of iron (Dr) = 7.8×10³ kg/m³
volume of iron piece(V) = 100cm³ = 10^(-4) m³
density of water (Dw) = 10³ kg/m³

(a) weight in air = Dr×V×g = 7.8×10³ × 10^(-4) ×10 = 7.8 N

(b) upthrust is the buoyant force exerted by water
    upthrust = Dw×V×g = 10³ × 10^(-4) ×10 = 1N

(c) apparent weight in water = weight in air - upthrust
                                         = 7.8N - 1N = 6.8N