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Question from Lisa, a student:

Hi! One hundred ball bearings with radius 5 mm are dropped into a cylindrical can, which is half full of oil. The height of the cylinder is 20 cm and the radius is 8 cm. By how much does the level of the oil rise? Many thanks!!!

Hi Lisa.

The question is much harder if the bearings are not fully submerged in the oil, so we have to either assume that or prove it. If the volume of the ball bearings is considerably smaller than the volume of oil, then they'll be submerged.

The volume of oil is the volume of the cylinder up to the top of the liquid. So this is a cylinder that is 4 cm deep. A cylinder's volume is π r2 d, where d is the depth and r is the radius. So the volume of oil is (3.1416)(82)(4) = 804.24 ml.

The volume of a ball bearing is the volume of a sphere. The formula for the volume of a sphere is (4/3) π r2, so 100 ball bearings with radius 0.5 cm (it helps to make sure all measurements are in the same units) is (100)(4/3)(3.1416)(0.52) = 104.72 ml.

So the volume of the ball bearings is much smaller than the volume of oil and they'll be submerged.

So to answer the question, you need to ask yourself whether the ball bearings float. If they float on top of the oil, then the level of oil doesn't rise at all. If they sink, then they will displace as much oil as the space they take up: in other words, the level of the oil has to rise to compensate for the 104.72 ml of ball bearings.

Thus, you need to find the depth d for 104.72 ml of volume V in the cylinder using the formula

V = π r2 d

where you know V is 104.72 ml and r is 8 cm. Solve for d.

Cheers,
Stephen La Rocque.

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