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.
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?