The leaking container

If $h$ is the height of the water level above the bottom of the sphere, and $V$ the volume of water, then we have $V \; = \; \tfrac13\pi h^2(3R-h)$ where $R$ is the radius of the container, and Bernoulli's Principle tells us that $\dot{V} \; = \; -\pi r^2\sqrt{2gh}$ where $r$ is the radius of the outflow. Thus $\begin{aligned} \pi h(2R-h)\dot{h} & = \; \dot{V} \; = \; -\pi r^2\sqrt{2gh} \\ \sqrt{h}(2R-h) \dot{h} & = \; -r^2\sqrt{2g} \end{aligned}$ and so, if $T$ is the time for the water level to drop from $\tfrac12R$ above the centre of the sphere to $\tfrac12R$ below the centre, then $T \; = \; \int_{\frac12R}^{\frac32R} \frac{\sqrt{h}(2R-h)}{r^2\sqrt{2g}}\,dh \; = \; \frac{R^{\frac52}}{r^2\sqrt{2g}}\int_{\frac12}^{\frac32} \sqrt{k}(2-k)\,dk \; = \; \frac{33\sqrt{3}-17}{60}\left(\frac{R}{r}\right)^2 \sqrt{\frac{R}{g}}$ With $R = 0.3$ , $r = 0.003$ , $g = 9.81$ we deduce that $T = 1170.424628$ seconds, giving the answer $\boxed{11704}$ .

Torricelli's Law is a consequence of Bernoulli's principle, and it is useful here (second line of math below). In addition to the parameters already defined in the problem, let $a$ be the cross-sectional area of the hole, let $h$ be the height of the water above the hole, let $u$ be the flow speed at the hole, let $y$ be the height of the water above the sphere center, and let $V$ be the volume of water in the sphere. Conceptualize the infinitesimal change in volume $dV$ as a cylinder of water at the surface, with height $dy$ and radius $x = \sqrt{R^2 - y^2}$ .

$h = y - (-R) \\ u = \sqrt{2 g h} \\ \frac{dV}{dt} = - a u \\ dV = \pi \, x^2 \, dy = \pi \, (R^2 - y^2) \, dy \\ \frac{\pi \, (R^2 - y^2) \, dy}{dt} = - a u \\ dy = \frac{- a u \, dt }{\pi \, (R^2 - y^2)}$

I numerically integrated $dy$ to arrive at a final time of $\approx 1170.4$

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35

import math

g = 9.81

D = 0.6
R = D/2.0

d = 0.006
r = d/2.0

a = math.pi*(r**2.0)

####################################

dt = 10.0**(-4.0)

t = 0.0
y = 0.15

while y >= -0.15:

    h = y - (-R)
    u = math.sqrt(2.0*g*h)

    xsq = R**2.0 - y**2.0

    dy = -a*u*dt/(math.pi*(xsq))

    y = y + dy
    t = t + dt

####################################

print dt
print t

Let the water surface be point A, the hole be point B, the datum is at the sphere's center. See the picture below:

Surface is a circle with area: ${ A }_{ A }=\pi ({ 0.3 }^{ 2 }-{ h }_{ A }^{ 2 })$

Hole area: ${ A }_{ B }=\pi { 0.003 }^{ 2 }$

Volume flow rate is constant, so: ${ A }_{ A }{ v }_{ A }={ A }_{ B }{ v }_{ B }$

$\equiv \frac { { v }_{ B } }{ { v }_{ A } } =\frac { { 0.3 }^{ 2 }-{ h }_{ A }^{ 2 } }{ { 0.003 }^{ 2 } }$

Substituting this into Bernoulli's equation gives: $\frac { { v }_{ A }^{ 2 } }{ 2 } +g{ h }_{ A }=-0.3g+\frac { { ({ 0.3 }^{ 2 }-{ h }_{ A }^{ 2 }) }^{ 2 } }{ { 0.003 }^{ 4 } } { v }_{ A }^{ 2 }$ (Gauge pressure at both points is 0, then you can cancel density)

A bit of algebraic manipulations give ${ v }_{ A }=-\sqrt { 2g\frac { { h }_{ A }+0.3 }{ \frac { { ({ 0.3 }^{ 2 }-{ h }_{ A }^{ 2 }) }^{ 2 } }{ { 0.003 }^{ 4 } } -1 } }$ (water level is going down so it's negative)

The rest is calculus:

${ v }_{ A }=\frac { d{ h }_{ A } }{ dt }$

$\equiv dt=\frac { d{ h }_{ A } }{ { v }_{ A } }$

$\equiv t=\int _{ h=0.15 }^{ h=-0.15 }{ \frac { dh }{ v } }$

$\equiv t=-\int _{ h=-0.15 }^{ h=0.15 }{ \frac { dh }{ v } }$

Now you basically solve this integral and get answer:

(I changed the variables so it's easier to see)

Now if you can solve this by hand, please tell me. I had to use Simpson's rule and programming.

Put these codes into python:

The program gives 1170.42... so the answer is 11704.