## across Group Title 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. one year ago one year ago

1. across Group Title

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.

2. LolWolf Group Title

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...

3. Algebraic! Group Title

~30 min

4. across Group Title

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

5. Algebraic! Group Title

math&stuff

6. across Group Title

That's not a valid explanation! ;O

7. Algebraic! Group Title

8. LolWolf Group Title

Roughly 30.275 mins.

9. sara12345 Group Title

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

10. LolWolf Group Title

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

11. sara12345 Group Title

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

12. hartnn Group Title

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.

13. across Group Title

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

14. LolWolf Group Title

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: $f_\text{rate}=(0.0127)^2\pi\sqrt{2\cdot3.048g}\approx0.003918\text{ m}^3/\text{s}\\ vf_\text{rate}\approx30.275\text{ mins.}$

15. LolWolf Group Title

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

16. Algebraic! Group Title

still waiting...