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 Time to run out water (Posted on 2006-07-13)
Consider a solid sphere (capable of withstanding full vacuum) of 3m diameter filled completely with water resting at sea level. It has a 10cm hole at the bottom with a cork on it. If you open the cork, what is the time taken for water to completely drain out.

What happens for higher diameter spheres?

 No Solution Yet Submitted by Salil Rating: 1.2500 (4 votes)

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 solution (maybe) | Comment 3 of 11 |
let a be the area of the drainage hole and v be the velocity of water out of that hole, then av is the rate at which the water is leaving the sphere and V be the volume of the sphere then

(1) DV/Dt=-av

now when the sphere is filled to hight h the volume is equal to

V=(Pi*(2.25-h)*h^3)/3

(2) DV/dh=Pi*(3-h)*h

now using (1), (2), and the chain rule

DV/dt=(DV/dh)(Dh/Dt)=Pi*(3-h)*h*(Dh/Dt)=-av
Dh/Dt=-av/(Pi*(3-h)*h)
a=Pi*(.1)^2=.01*Pi
Dh/Dt=-.01*v/(h*(3-h))

v(h)=Sqrt(2*g*h)  where g is acceleration due to gravity on earth (9.8m/s^2)

Dh/Dt=-.01*Sqrt(2*g*h)/(h*(3-h))
100*h*(3-h)/Sqrt[2*g*h] dh = -Dt
(3) 20*(5-h)*sqrt(2*g*h)/g=-t+c
when t=0 h=3
thus
c=20*2*Sqrt(2*g*3)/g=40*sqrt(6*g)/g

now the sphere is empty when h=0 thus we can find how long it takes to drain by setting h=0 in (3) and solving for t thus
t=40*sqrt(6*g)/g

now that seems a little quick for that much water to be able to drain so I think I may have made a mistake somewhere, if someone could please point it out I would be greatly appreciate it :-D

 Posted by Daniel on 2006-07-14 01:41:54

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