Question

1. if a ball that has a density of 1.73 g/cm3 is placed in the beaker, where will the ball come to rest? explain! 8. if the ball was taken out of the beaker and then cut in half and both pieces were then placed back into the beaker, where will both balls come to rest and what will the density of both balls be?

Explanation:

Step1: Recall density - buoyancy principle

If an object's density ($
ho_{object}$) is greater than the density of the liquid ($
ho_{liquid}$) it is placed in, it sinks; if $
ho_{object}<
ho_{liquid}$, it floats; if $
ho_{object}=
ho_{liquid}$, it is in neutral - buoyancy. But we are not given the density of the liquid in the beaker. Assuming the beaker contains water (density $
ho_{water}=1\ g/cm^{3}$). Since the density of the ball $
ho_{ball}=1.73\ g/cm^{3}>1\ g/cm^{3}$, the ball will sink to the bottom of the beaker.

Step2: Understand density property

Density is an intensive property, which means it does not depend on the amount of the substance. When the ball is cut in half, the density of each half - piece remains the same as the original ball, i.e., $1.73\ g/cm^{3}$. And since $1.73\ g/cm^{3}>1\ g/cm^{3}$ (assuming water in the beaker), both half - balls will sink to the bottom of the beaker.

Answer:

1. The ball will sink to the bottom of the beaker because its density ($1.73\ g/cm^{3}$) is greater than the density of water ($1\ g/cm^{3}$).
2. Both balls will sink to the bottom of the beaker and the density of both balls will be $1.73\ g/cm^{3}$ because density is an intensive property and does not change with the amount of the substance.