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A piece of gold (\(\rho = 19.3 \, \text{g/cm}^3\)) has a cavity in it. It weighs 38.2 g in air and 36.2 g in water.

The volume of the cavity in the gold is:

a) 1.0 cm³
b) 2.0 cm³
c) 3.0 cm³
d) 4.0 cm³

Answer :

Final answer:

The volume of the cavity in the gold piece is 2.0 cm³, as this is the volume of water that the gold displaces due to the difference in weight when submerged in water compared to in air. b) 2.0 cm³

Explanation:

To calculate the volume of the cavity inside a piece of gold, we will use the principle of buoyancy that is provided by Archimedes' principle. This principle states that the buoyant force on a submerged object is equal to the weight of the fluid that the object displaces. The weight loss of the gold piece in water (38.2 g in air - 36.2 g in water) equals the weight of the water displaced by the submerged piece of gold. Since the density of water is 1 g/cm³, the weight loss of 2.0 g in water corresponds to a volume displacement of 2.0 cm³.

Therefore, the volume of the cavity in the gold piece is equal to the volume of water displaced. Since the weight loss in water is 2.0 g and this is the weight of the water that has been displaced, the volume of the cavity is 2.0 cm³.

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Rewritten by : Jeany

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