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What's the weight of the part?


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What's the "work instruction?" Is this homework?

 

It depends on how you're measuring how long the hole is - how surface to surface on the original sphere, or surface to surface on the remainder. Since you only have enough information to answer if it's the latter, then go with that.

 

And the hint is: think about what the remainder looks like as the original sphere gets smaller and the hole gets narrower.

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No prob. It is simple math. Here isone way and here is another.

 

But there is a short solution for that. Try to find out.

 

Here's a more accurate description of the problem (yours didn't specify whether the hole was through the center nor what was 6 inches):

 

A sphere has a cylindrical hole drilled right through its center. The hole is six inches long -- that is, from one point on one edge of the hole to the closest point on the other edge is a distance of six inches.

 

What is the volume of the remaining part of the sphere outside the hole?

 

It may sound like you don't have enough information to solve this puzzle, but you do.

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Here's a more accurate description of the problem (yours didn't specify whether the hole was through the center nor what was 6 inches):

 

A sphere has a cylindrical hole drilled right through its center. The hole is six inches long -- that is, from one point on one edge of the hole to the closest point on the other edge is a distance of six inches.

 

What is the volume of the remaining part of the sphere outside the hole?

 

It may sound like you don't have enough information to solve this puzzle, but you do.

Or the thread title/question should be changed to "What's the weight of a solid steel sphere with a 6" long cylindrical hole through?".

Thank you, Mr Skeptic.

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