A solid, homogeneous sphere with a mass of m0, a radius of r0 and a density of 0

Question

A solid, homogeneous sphere with a mass of m0, a radius of r0 and a density of 0 is placed in a container of water. Initially the sphere floats and the water level is marked on the side of the container. What happens to the water level, when the original sphere is replaced with a new sphere which has different physical parameters? Notation: r means the water level rises in the container, f means falls, s means stays the same. Combination answers like 'r or f or s' are possible answers in some of the cases.

1.The new sphere has a radius of r > r0 and a mass of m < m0.

2.The new sphere has a radius of r > r0 and a density of < 0.

3.The new sphere has a mass of m < m0 and a density of > 0.

Select r / f / s / r or s / f or s / r or f or s for the three questions.

Explanation / Answer

Here, first of all you should keep in mind is that the amount of water displaced equals the weight of the sphere (as long as it floats).

Now come on case - (1):

Same mass, so same weight, same amount of water displaced, therefore notation: s

For case - (2):

Less mass, so less weight, less water displaced, therefore notation: f

And for case (3):
Clearly, the volume is the same (same r0) but the mass increases because of the increased density. Clearly then, more water needs to be displaced to keep it afloat. If it doesn't float but sinks, also more water will be displaced because the radius (volume) of the sphere is the same, so going under displaces more water than staying afloat.

therefore notation: r.

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