A piece of metal of relative density 5.0 weighs 60N in air. Calculate its weight when fully immersed in water

The weight of an object is equal to the force of gravity acting on it. When an object is submerged in a fluid, it experiences a buoyant force, which is equal to the weight of the fluid displaced by the object. To calculate the weight of the metal when fully immersed in water, we need to first determine the volume of water displaced by the metal. We can do this by using the principle of Archimedes, which states that the buoyant force on an object is equal to the weight of the fluid displaced by the object. The weight of the metal in air is 60N, which is equal to its mass times the acceleration due to gravity. We can calculate the volume of the metal using its density and mass. Density is defined as mass per unit volume. Since the relative density of the metal is 5.0, its density is 5 times that of water, which has a density of 1000 kg/m^3. Density of metal = 5 x Density of water = 5 x 1000 kg/m^3 = 5000 kg/m^3 Mass of metal = Weight of metal / Acceleration due to gravity = 60N / 9.81 m/s^2 = 6.11 kg Volume of metal = Mass of metal / Density of metal = 6.11 kg / 5000 kg/m^3 = 0.001222 m^3 Since the metal is fully submerged in water, the volume of water displaced is equal to the volume of the metal. Weight of water displaced = Density of water x Volume of water displaced x Acceleration due to gravity = 1000 kg/m^3 x 0.001222 m^3 x 9.81 m/s^2 = 12N Therefore, the weight of the metal when fully immersed in water is 60N - 12N = 48N. Answer: 48N