Einstein's Sphere
An Incredible Sphere was found in the laboratory of Einstein. A cylindrical hole, 6 inches long has been drilled through the center of the solid sphere. Remaining volume of the sphere was needed to solve the puzzling equation, written on the black board.
Can you help?
Can you help?
Labels: funphysics, logic, mathemagic
posted by Rajesh Lal at
10:06 PM
8 Comments:
What's the radius of the hole?
1", 2" ??
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I use cm.
I'll convert inches to cm.
1in = 2.54cm
2.54*6 = 15.24
15.24/2 = 7.62
π = pi / 3.141592654
r = radius
h = height
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sphere v = (4/3)*pi*r^3
= 1.33*3.142*(7.62^3)
= 2937.73858
cylinder v = pi*r^2*h
= 3.142*x^2*15.24
x = ?? (radius of cylinder)
To find out the remaining volume of the sphere, first find the radius of the cylinder to figure out the volume of the cylinder.
Deduct the volume of the cylinder from the sphere and there's the answer.
~Cassieadams01
No, there is a hemisphere cap to the cylinder, since it was removed from the sphere. Easiest way would be to drop it in a bucket of water with known volume and markings, then find the difference in the water levels.
you need to know how wide the hole is
i agree with the dropping in the water idea, however, there is caution as well because the sphere must have a higher density than the water.
~Cybersurf~
if volume of a sphere is Vs=(4/3)(pi)(r^3), and the radius is 3, then we know that the volume of the sphere is 36(pi). the volume of the cylinder is Vc=(pi)(r^2), where the height is 6. we get 6(pi)(r^2). we are looking for the volume of the remaining part, which can be laid out in the formula Vr=Vs-Vc. That leaves us with Vr=36(pi) - 6(pi)(r^2). we can manipulate the formula to be Vr= 6(pi)(6-(r^2)). after that, im kinda stumped
First semester calculus solution:
Position the sphere with the Y coordinate axis running vertically through the center. If the sphere has radius R, then the wall of the hole is at Sqrt[R^2-9] and the outer wall is at Sqrt[R^2-y^2] (draw a right triangle from the origin to the wall at y=0 and the wall at y=3 to see this). That means the volume is:
Integral(-3,3)[Pi*(R^2-y^2)-Pi*(R^2-9)]dy. The R^2's cancel out and the result is 36*Pi.
Insight and faith solution:
If you assume the problem has enough data for a solution, then the result must be independent of the radius of the sphere. Then the result is the same, if the hole diameter is zero (i.e. the sphere has R=3), which is 4/3 Pi R^3 or 36 Pi.
If the sphere's density is lighter than the water, simply attach a weight device. Of course, before doing that, measure out the displacment of the weight first before the attaching.
The question begs that the answer is unique. So drill a hole of diameter 0. It's length is 2r = 6".
So, as long as the sphere has diameter greater than 6", the remaining part has the same volume as a 6" diameter sphere.
Otherwise the question is unanswerable.
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