If air bubbles were trapped in the solid under the liquid, would the volume measurement be larger, smaller…?
Assuming that you took a block of solid steel, then hollowed it out so that inside, it was one giant opening (an air bubble), the volume would remain the same, but the density would decrease, because the mass would have decreased, and Density = Mass / Volume. However, if the solid remained heavy enough to continue to rest on the floor of the tank, unchanged, then you would get error because it would still be displacing the same volume and mass of water (or whatever liquid).
