Final answer:
When a steel ball and a hollow plastic ball of the same volume are submerged in water on a double-pan mechanical balance, the balance will not move because both displace equal volumes of water and the fastening of the hollow ball transmits any net force to the beaker.
Step-by-step explanation:
The question relates to the concept of buoyant force and Archimedes' principle in physics. The scenario involves comparing the effect of submerging objects of different densities, but identical volumes, in water using a double-pan mechanical balance. When a steel ball, which is denser than water, is submerged in one container, it displaces a volume of water equal to its own volume, exerting a force due to the weight of the displaced water. However, since the steel ball is denser than water, it has a greater weight than the weight of the displaced water, resulting in no net change in the balance. On the other hand, submerging a hollow plastic ball, which has the same volume but is less dense than the steel ball, will displace the same volume of water, but its weight is less than the weight of the displaced water. In this case, the force on the bottom of the beaker due to the displaced water would not be counteracted by the weight of the hollow plastic ball, potentially causing an imbalance. However, because the plastic ball is fastened to the bottom, any net upward buoyant force is transmitted to the beaker, maintaining the balance. Hence, the balance will not move in either direction if both objects have the same volume, and the situation is set up as described.