# Depth of lake (Pressure/fluids)

• Elbobo
In summary, a 1.6 m long glass tube, closed at one end, is weighted and lowered to the bottom of a freshwater lake. Upon recovery, an indicator mark shows that water rose to within 0.32 m of the closed end. Using the equation P1*V1 = P2*V2 and the given values of pressure, length, and density, we can determine the depth of the lake to be 0.32 m. The density of the glass and weight of the weight used were not necessary in this calculation.

## 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?

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?

Ah, ok.

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

Thank you!