Diameter

[Guest]

A lead ball is formed into a circular sheet. Determine diameter of sheet.

Diameter of ball - 2 in.

Thickness of sheet - 0.01 in.

I have the answer. How do I solve ? Thanks.

[Guest]

Please reply. Thanks again.

[Guest]

s ( vol of sphere=4/3 X pi X r sq ) 4/3 X 3.14 X 1 = 4.186
c ( vol of cylinder=pi X r sq X h ) 3.14 X 0.01 X r sq = 4.186
Ds = 2 : rs = 1 4.186 / ( 3.14 X 0.01 ) = r sq
vol s = vol c 133.31 = r sq
rs = 1 r = 11.54
h = .01 _____________________________
solve for rc

do you think that's it ?
KC

[Guest]

s ( vol of sphere=4/3 X pi X r sq )
c ( vol of cylinder=pi X r sq X h )
Ds = 2 : rs = 1
vol s = vol c
rs = 1
h = .01
__________________________

4/3 X 3.14 X 1 = 4.186
3.14 X 0.01 X r sq = 4.186
4.186 / ( 3.14 X 0.01 ) = r sq
133.31 = r sq
r = 11.54
do you think that's it ?
KC

[Guest]

"r = 11.54"

diameter = r^2 = 11.54 * 11.54 = 133.71 = 134

That is too high.

[Guest]

4/3 x Pi x 1^3 = Pi x R^2 x 0.01
Pi on each side cancels
1.3333 / 0.01 = R^2
133.3333 = R^2
11.547 = R
23.094 = Diameter

[Guest]

Correct - answer given approx. 23 in. diameter.