I\'ve already posted this question, but didn\'t get a correct answer (and accide

Question

I've already posted this question, but didn't get a correct answer (and accidentally hit the wrong rating button so now I can't go back and reopen it for answers) Any help is appreciated.

A container of liquid of density has a surface area A exposed to normal atmospheric pressure. A distance h below the surface of the liquid is a tiny hole of area A0. The depth of the liquid in the container is H.

http://imageshack.us/f/192/phy21.png/

If A0 A, which expression correctly gives the velocity v0 (in terms of only g, h, H, A0 and A) at which the uid escapes through the hole of area A0?

1/((1-x)n 1+nx if x <<1

1) (gh)
2) (A-A0)gh
3)not enough info
5) (2gh)(1-(1/2)(A0/A)2)
6) (2g(H+h)
7) (2gh)(1+2(A0/A)2)
9) (2gh)(1+(1/2)A02)
10) (2gh)(1+(1/2)(A0/A)2)

Options 4 and 8 are not listed because they are not correct.

Explanation / Answer

Using Bernoulli’s equation
P1 + 1/2v1^2 = P2 + 1/2v2^2+pgh.

We will assume that the pressure at the top of the tank is atmospheric, so that P1 = P2. Using the
equation of continuity,
A1v1 = A2v2 so v2 = (A1/A2)v1.
Substituting this expression into the first,
1/2(v1^2) = 1/2 (A1/A2)(v1^2) + gh.

Solving for v1, we conclude that
v1 = v2gh(1 - A0^2/A^2)

Since 1/(1-x)^n ~ 1+ nx,

We know choice 10) v(2gh)(1+(1/2)(A0/A)2) is correct.

I took PHY303K before and got an A :)

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