Let's do some cool math:

A tank in the form of a right-circular cylinder standing on end is leaking water through a circular hole in its bottom. Suppose the tank is $10$ feet in height and has radius $2$ feet and the circular hole has radius $1/2$ inch. If the tank is initially full, how long will it take to empty?

Whoever gets this right first gets a cookie.

Question

Let's do some cool math:

A tank in the form of a right-circular cylinder standing on end is leaking water through a circular hole in its bottom. Suppose the tank is $10$ feet in height and has radius $2$ feet and the circular hole has radius $1/2$ inch. If the tank is initially full, how long will it take to empty?

Whoever gets this right first gets a cookie.

across · Answer

Hint 1:

Torricelli's law states that$$v=\sqrt{2gh},$$where $v$ is the speed of the water leaving at the bottom of a tank of height $h$. $g$ stands for the acceleration due to gravity, as usual.

anonymous · Answer

I really want to work this out... but, unless I'm missing something, using Bernoulli's, it'd take too long, and I don't think I have the attention span...

anonymous · Answer

~30 min

across · Answer

@Algebraic!, how did you get that? ;P

anonymous · Answer

math&stuff

across · Answer

That's not a valid explanation! ;O

anonymous · Answer

where's my cookie?

anonymous · Answer

Roughly 30.275 mins.

anonymous · Answer

idk differential equations, but i can try setting up one :p

anonymous · Answer

Yeah, I forgot about Torricelli's law... I had started trying to derive it from Bernoulli's and was like... nope.

anonymous · Answer

flowrate outside = $\pi (1/2)^2\sqrt{2gh} $  cubic inch per sec

hartnn · Answer

using this differential equation : 
dh/dt = -Ah/Aw sqrt (2gh)
take g= 32 ft/s^2
Ah=Area of hole , Aw is area of water ,so Aw/Ah = 576
so (h^(-1/2))dh = -(8/576)dt
so 2 (sqrt h) = -(8/576) t
put h= 10 to get t = 30.36 minutes.

across · Answer

You guys are right. The answer is roughly $30$ minutes. I guess I'll have to start posting Millennium Prize problems. That way, you guys will take at least five minutes longer to answer correctly. ;P

anonymous · Answer

I mean, it's not hard to compute, just a bunch of numbers and converting feet to metres. So, converting everything, we get the very 'elegant' expressions of:
$$
f_\text{rate}=(0.0127)^2\pi\sqrt{2\cdot3.048g}\approx0.003918\text{ m}^3/\text{s}\
vf_\text{rate}\approx30.275\text{ mins.}
$$

anonymous · Answer

I meant:
$$
\frac{v}{f_\text{rate}}
$$... oops.

anonymous · Answer

still waiting...