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## 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 * r

^{2}* L

_{1}(L

_{1}= 1.6 m)

## The Attempt at a Solution

So I tried setting up a proportion:

L

_{1}/ L

_{2}(L

_{2}= 0.32 m ) = P

_{A}/ P

_{L}

(L

_{2}P

_{A}) / L

_{2}= 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?