Depth of lake (Pressure/fluids)

[Elbobo]

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 = P_A + P_Lake
P_lake = 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 ) = P_A / P_L

(L₂P_A) / L₂ = P_L = 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?

 

[LowlyPion]

Looks like your method is almost OK.

Isn't the key relationship

P₁*V₁ = P₂*V₂

1 atm * 1.6 = P_bottom *.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?

 

[Elbobo]

Ah, ok.

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

Thank you!