The weight of a piece of iron in water is 1.67N, find its volume.

The weight of a piece of iron in water is equal to the difference between two directly opposite forces: the force of gravity and the force of Archimedes.
Р = Fт – Fa = m * g – ρк * g * V, where m is the mass of the piece (m = ρl * V), g is the acceleration of gravity (g = 9.8 m / s ^ 2), ρ is the density water (ρ = 1000 kg / m ^ 3), V is the volume of the piece (V = m / ρst., where ρst. is the density of iron (ρl. = 7800 kg / m ^ 3)).
P = m * g – ρw * g * V = ρl * V * g – ρw * g * V = V * (ρl * g – ρw * g);
V = P / (ρl * g – ρw * g) = 1.67 / (7800 * 9.8 – 1000 * 9.8) = 2.5 * 10 ^ -5 m ^ 3 = 25 cm ^ 3.
Answer: The volume of a piece of iron is 25 cm ^ 3.