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Depth of lake (Pressure/fluids)

  • Thread starter Elbobo
  • Start date
  • #1
145
0

Homework Statement


A 1.6 m long glass tube closed at one end
is weighted and lowered to the bottom of a
freshwater lake. When the tube is recovered,
an indicator mark shows that water rose to
within 0.32 m of the closed end.
Assume the temperature is constant, the
acceleration of gravity is 9.8 m/s2, the density
of the glass is 3400 kg/m3 , the density of the
lake water is 1000 kg/m3, and air pressure is
1.013 × 105 Pa.

Determine the depth of the lake. Answer in
units of m.


Homework Equations


P = PA + PLake
Plake = rho * g * h (h = depth of lake)
rho = m / V
V = pi * r2 * L1 (L1 = 1.6 m)

The Attempt at a Solution


So I tried setting up a proportion:

L1 / L2 (L2 = 0.32 m ) = PA / PL

(L2PA) / L2 = PL = rho * g * h
From there, h seems to be easy to figure out, but I don't think my process is right because I ignored the density of the glass given.

Can someone help me?
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
4
Looks like your method is almost OK.

Isn't the key relationship

P1*V1 = P2*V2

1 atm * 1.6 = Pbottom *.32

ρ *g * h = 1.6 / .32 atm

They gave you the density of glass but not the weight of the weight used to submerge it? To what purpose? Maybe not all givens are always necessary?
 
  • #3
145
0
Ah, ok.

I forgot P(bottom) was 1 atm + pgh.

Thank you!
 

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