Final answer:
In a vacuum, two objects that have the same apparent weight when submerged in water would weigh the same, regardless of their volumes. This is because the buoyant forces that affect apparent weight in water are not present in a vacuum.
Step-by-step explanation:
If two objects have different volumes but the same apparent weight when submerged in water, this suggests that their densities are different because they displace the same weight of water, as per Archimedes' principle. However, in a vacuum, there are no buoyant forces acting on the objects, as there is no fluid to displace. Therefore, their weights would be solely determined by their masses and the acceleration due to gravity. The apparent weight loss they experience in water is irrelevant in a vacuum. For objects with the same apparent weight in water, if their volumes are different, it typically implies that their densities are different, with the less dense object having a larger volume. The differences in density and volume do not affect their actual weight in a vacuum. So, the answer to the question is (a) they weigh the same in a vacuum, as the apparent weights in water were the same, indicating that their actual weights are also the same.