Submerge The Ball
Suppose you have a very large steel box that is open topped.It measures 50 feet x 50 feet x 50 feet.Inside this box is a steel ball-bearing that has a diameter of 42 feet.If you started filling this box with mercury,how many cubic feet would you have to pour in, in order to completely submerge the ball?
posted by Knightmare at
4:15 PM
19 Comments:
8622.8417404197 cu feet
(4/3)pi(21)^3=116377.1582595803
50^3=125000
125000-116377.1582595803=8622.8417404197
Anonymous...that isn't the answer i have
My bad ( iPod calculator is annoying @ times)
actuall instead of 116377...., it's 38792.38609(according to a graphing calculator)
so it's 125000-38792.38609=86207.61931
That was my dad sry
This post has been removed by the author.
The ball bearing would float (unless you fixed it down), so it can't be done.
66,207.53 cu ft.
doint htis while on the phone ... probably wrong
yea ...didn't consider it floating
Chris...you got it(but too fast-i wanted to mess with Anonymous some more) :{
Yeah that's a bit too advanced for me
if it did not float ...
Anonymous used the entire 50ft height of the box and to cover the non-floating ball ...would only need 42ft. in height.
Sorry Knightmare.
If it didn't float, then it would be 66207.6139 approx.
I reckon that's the best part $10,000,000 worth of mercury.
i get the stuff at discount
According to me the answer comes out to be 92652 cubic feet.
Required volume =
= 50X50X42 - 4/3X21X21X21
= 42 (2500 - 294)
= 92652.
sorry, missed the PI in the post..
Required volume =
= 50X50X42 - (4/3)X(22/7)X21X21X21
= 42 (2500 - 924)
= 66192
Thats the answer!
Hi RiDiM. If you'd used π instead of 22/7, you'd have got 66207.6139...
This is a classic trick question, Chris got it right, stell floats in mercury so it's not possible.
oops meant STEEL floats in mercury :)
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