A small solid sphere of mass M0, of radius R0, and of uniform density 0 is place

Question

A small solid sphere of mass M0, of radius R0, and of uniform density 0 is placed in a large bowl containing water. It floats and the level of the water in the dish is L. Given the information below, determine the possible effects on the water level L, (R-Rises, F-Falls, U-Unchanged), when that sphere is replaced by a new solid sphere of uniform density.

1.) The new sphere has mass M=M0 and density p<p0

2.) The new sphere has radius R=R0 and density p>p0

3.) The new sphere has density p=p0 and mass M<M0

4.) The new sphere has mass M=M0 and radius R<R0

5.) The new sphere has a radius R<R0 and density p>p0

6.) The new sphere has a density p=p0 and mass M>M0

Explanation / Answer

We know: Vsub = [ / w] * V = M / w

where Vsub = submerged volume of sphere, V = total volume of sphere, = density of sphere, w = density of water, M = mass of sphere

1) Since mass of sphere does not change, submerged volume will remain same. Hence, water level will remain unchanged..

2) Since volume of sphere is same and its density more, mass of sphere will be greater. So, submerged volume will be more. Hence, water level will rise.

3) Since mass of sphere is less than before, submerged volume will be less. Hence, water level will fall.

4) Since mass of sphere is same, submerged volume of sphere will be same. Hence, water level will remain unchanged.

5) Sphere volume is less and its density is more. So, we cannot comment on its mass. Thus, we cannot comment on the submerged volume and the water level.

6) Since mass of sphere is greater, submerged volume will increase. Hence, water level will rise.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.