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Indonesia Spheres In Water – Mind Your Decisions
Thanks to Sunny for this suggestion! This is a tricky question for grade 11 students in Indonesia.
A vertical cylindrical container contains water to a height of 24 cm. Its base has a radius of 10 cm.
Five identical solid spheres, each with a radius of 10 cm, are placed into the cylinder.
By how much does the water level increase? (Assume the cylinder is sufficiently tall to contain the water.)
As usual, watch the video for a solution.
Or keep reading.
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Answer To Indonesia Spheres In Water
(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).
What happens to the water level when 1 sphere is placed into the cylinder and submerged with water? Suppose the water level increases by h. The volume of displaced water, which is a cylinder of height h, must equal the volume of the sphere.
So we can solve for h with the equation:
volume cylinder base r, height h = volume sphere radius r
πr2h = (4/3)πr3
h = (4/3)πr
Substitute r = 10 to get h = 40/3.
Each submerged sphere increases the water level by 40/3, so 5 such spheres would be an increase of 200/3, or about 66.67 cm.
This was the answer every student gave. But unfortunately they all missed the question.
We have made a critical assumption that all spheres would get submerged in water. But we have not checked there is sufficient water for this to happen.
Water to submerge a sphere
Initially the water level is at 24 cm, so the volume of water is equal to:
πr2h
= π(102(24))
= 2400π
If the water in the cylinder just covers the sphere, the volume of water plus the volume of the sphere would equal the volume of a cylinder with a height equal to the diameter of the sphere. So we have:
vol(water) + vol(sphere) = vol(cylinder height 2r)
vol(water) + (4/3)πr3 = πr2(2r)
vol(water) = 2πr3 – (4/3)πr3
Using r = 10 we get:
vol(water) = 2π(1000) – (4/3)π(1000)
vol(water) = 2000π/3
The initial water volume is 2400π. After we submerge 3 spheres, we have used 2000π of water, and only 400π is remaining. This is sufficient to submerge another 0.5 sphere, leaving:
400π – 0.5(2000π)/3
= 200π/3
Since we can submerge 3 entire spheres, we know the water level exceeds that of 3 diameter marks, of 20, 40, 60 cm. Then we know half a sphere can be submerged, so we have another 10 cm to a level of 70.
So all that remains is to calculate the increase h from the last 200π/3 of water.
Spherical segment
This is the hardest part of the question. We need to calculate how much the water level rises in the upper half of the sphere. So we need the volume of water 200π/3 to equal the volume of a cylinder minus that of a spherical segment.
The volume of the cylinder with r = 10 and h height is just πr2h = 100πh. But what’s the volume of this spherical segment?
You could look up the formula and apply it, as follows.
But in grade 11 in Indonesia students could know integral calculus. So the volume can be found by revolving a quarter circle of radius 10 about the y-axis from 0 to h.
In either case, the volume ends up being 100πh – πh3/3. So we can finally calculate:
vol(water) = vol(cylinder) – vol(spherical segment)
200π/3 = 100πh – (100πh – πh3/3)
200π/3 = πh3/3
200 = h3
∛200 = h
The answer
The final water level is 70 + h = 70 + ∛200, and the initial water level was 24 cm.
So the change in water level is:
70 + h – 24
= 46 + h
= 46 + ∛200
≈ 51.848 cm
And that’s the answer!
References
Geogebra interactive
https://www.geogebra.org/m/aRJnx8b5
Spherical segment MathWorld
https://mathworld.wolfram.com/SphericalSegment.html
Spherical segment Wikipedia
https://en.wikipedia.org/wiki/Spherical_segment
Other Indonesia problems
Indonesia logic puzzle
https://www.youtube.com/watch?v=Ka-Ge3PdVOM
Indonesia kissing circles
https://www.youtube.com/watch?v=i0dZukEw1JY
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