Two identical solid pieces, one of gold and other of silver, when immersed completely in water exhibit equal weights. When weighed in air (given that density of gold is greater than that of silver)

According to Archimedes' Principle, an object immersed in a liquid experiences an upward buoyant force equal to the weight of the liquid displaced [1]. The question states the gold and silver pieces are 'identical,' implying they have the same volume. Since they have the same volume, they displace the same amount of water and thus experience the same buoyant force [1]. The apparent weight in water is calculated as: Apparent Weight = Real Weight (in air) - Buoyant Force. If their apparent weights in water are equal and the buoyant force is identical for both, their real weights in air must also be equal. However, the problem specifies that the density of gold is greater than silver. For two pieces of the same volume to have equal weight in air, they must have the same mass, which contradicts the density premise unless 'identical' refers to their external dimensions. If they have the same volume, the silver piece, being less dense, would actually require more volume to match gold's weight, or if volumes are equal, gold is heavier. Given the standard physics interpretation of this specific problem, the silver piece must weigh more in air to compensate for its larger volume (and thus larger buoyancy) if they were to result in equal submerged weights; however, since the pieces are 'identical' in volume, the silver piece weighs more in air to result in equal submerged weight.

Sources

1. [1] Science ,Class VIII . NCERT(Revised ed 2025) > Chapter 5: Exploring Forces > A step further > p. 76