Number of 1.5" balls that can be put into a glass?

Question

How many 1.5" diameter balls can we put in a glass that has the following dimensions.
Height: 72"
Diameter of base: 18"
Diameter of rim (top) 30"

Kindly show the working. Thanks!!!

Answer 1

This glass geometrically is the frustum of an inverted cone, the height from the bottom of this imaginary inverted cone to top of the glass rim is 180".

Thus the volume of the glass = π*(15)^2*180/3 - π*(9)^2*108/3
= π*10584in^3

The volume of each ball = 4/3*π*(.75)^3 = π*.5625in^3

So volume wise π*10584/π*.5625 = 18,816 balls could fit in the glass. This is a good theoretical upper limit.

Because the conical nature of the glass, and the property of similiar triangles. One can equate the radius to the height h above the bottom of the glass as r = 9+(h/12).

Suppose these balls were cubes instead, each side 1.5" long. Each side would have a surface area of 1.5^2= 2.25in^2.

Suppose we considered the cross sections of the glass every 1.5" up (48 sections) and the number of squares that would make a the largest embedded square (sqrt 2*r x sqrt 2*r) at the top of each section. This gives a theoretical lower limit.

= Σ(n=1 to 48) 2*(9+1.5*n/12)^2/(1.5^2)
= 2/2.25 * Σ(n=1 to 48) (9+n/8)^2
~ 6336

The answer is between 6336 and 18,816 balls.
If I had to give an estimate I'd take the average and say
~12,500 balls

Answer 2

7

Answer 3

This is a good question. The answer lies in the volume of the interstices. I'd consult Isaac Newton on this one...

Answer 4

I'd suggest doing your own homework in the first place. You're going to have to anyway...because I certainly wouldn't trust most of the idiots that answer questions here.