Height of fluid and Bernoullis equation

[radiator0505]

Homework Statement

An ideal liquid flows horizontally through a pipe of cross sectional area 3cm² with a velocity 1ms^-1. The pipe narrows to a cross sectional area 1cm². 2 vertical pipes are connected to the pipe, one in either region.
Calculate the height difference of the liquid between the 2 vertical pipes.

Homework Equations

p₁ + 0.5ρv₁² + ρgh₁ = p₂ + 0.5ρv₂² + ρgh₂

A₁v₁ = A₂v₂

The Attempt at a Solution

I think I'm a bit confused on what my variables represent in bernoullis equation:
Using the continuity equation gives v₂ = 3ms^-1.
Then I said that for the pipes p₁ = p₂ (because they're both open to the atmosphere), putting everything into bernoullis equation and rearranging gives
h₁ - h₂ = 0.4 m
The book says the answer is 4x10^-3 m.

Am I not allowed to consider the heights of the fluid in the vertical pipe because I calculated the speeds of the fluid in the horizontal pipe? I can't think of any other way to do the problem.

 

[radiator0505]

Oh sorry nevermind, I got it.
Turns out the velocity is initially 0.1 ms^-1, then doing what I said gives the answer.

:)

 

[SteamKing]

IMO, the book's answer is wrong.

I see that the OP had missed a decimal in the initial velocity. Book answer is OK.