## agdgdgdgwngo 3 years ago An ideal liquid with density ρ is poured into a cylindrical vessel with cross section A1 to a level of height h from the bottom, which has an opening of cross section A2. Find the time it takes for the liquid to flow out

1. agdgdgdgwngo

quick analysis shows there's no need for the density

2. Ishaan94

is it cylindrical or conical?

3. agdgdgdgwngo

says cylindrical right up there :-P

4. Ishaan94

how could a cylinder have different cross sectional area? :-p

5. agdgdgdgwngo

I guess A1 = A2 then :-D

6. Ishaan94

|dw:1324391145896:dw| This is the cylinder right?!

7. agdgdgdgwngo

it's not illustrated inn the problem, but that is a cylinder indeed!

8. agdgdgdgwngo

cmon!! IIT!!

9. Tomas.A

why u posted 2 same questions? and it's physics isn't it?

10. agdgdgdgwngo

it's actually physics.... so yeah :-D

11. agdgdgdgwngo
12. agdgdgdgwngo

im stuck on the very first problem lol

If you could, do try to keep physics questions in the physics group. You can use chat to get people to come to that group if you're not getting traction there :)

14. henkjan

You need to say that you use the 'dimensional analysis' .... It's just about getting the units of time (the unknown) equal to the units on the right side. So all the 'meters' on the right hand side should add up to 0. This is a tool to find a relation that COULD fullfill the requirements. The problem consists of density, h, A1, A2 and gravity g. Now you just write your solution in the form: t = rho^a*h^b*A1^c*A2^d*g^e You see that the only time dependent variable is g. So g should be (m/s^2) to the power of -1/2 to obtain seconds. The other variables like meter work in the same way, try to get rid of all of them by making a couple of easy assumptions

15. agdgdgdgwngo

I don't know how to make valid assumptions though :(