# Homework Help: Fluid cylindrical bucket Question

1. Sep 25, 2007

### neoking77

A cylindrical bucket, open at the top has height 27.0 cm and diameter 13.0 cm. A circular hole with a cross-sectional area 1.49 cm^2 is cut in the center of the bottom of the bucket. Water flows into the bucket from a tube above it at the rate of 2.28×10−4 m^3/s

my work
let 1 denote the top and 2 denote the bottom

Q = V1A1
V1 = Q/A = 0.01718 m/s

Q = V2A2
V2 = 0.01530 m/s

using bernoulli's
top bottom
P + pgh + 1/2pv^2 = pgh + 1/2pv^2 + P
P's are equal and cancels out
rho's cancel out
there's no pgh on bottom
we are left with
h = [ 1/2(v2)^2-1/2(v1)^2 ] /g

the answer i get is 0.00179m which is 0.1179 cm but that's not the answer...can anyone see where i went wrong? thank you very much.

2. Sep 25, 2007

### olgranpappy

you're probably forgetting to convert something from cm to m or some such thing... I found

h=0.11725 meters

actually, only good to 2 sig fig?

h=0.12 meters

3. Sep 25, 2007

### neoking77

really? ? any idea what i converted wrong?
did you get 0.01718m/s and 0.01530m/s for velocities? ... hm

4. Sep 25, 2007

### olgranpappy

no idea. sorry.

I don't remember off hand, if I can find my scratch-paper I'll let you know.

good luck.

5. Sep 27, 2007

### p0nda

1cm^2 = .0001m^2 so 2.28x10^-4/1.49x10^-4 gives a velocity of 1.53m/s

I just plugged it in to toricelli's equation (v1=sqrt(2g(h2-h1)) and solved for h2 (h1 is 0), which gave the answer .119m = 11.9cm.

6. Sep 27, 2007

### olgranpappy

that neglects the velocity of the water at the top of the bucket... valid if A_top >> A_bottom only.