# Solving a depth pressure problem?

Working on an assignment, I get 30m but that gets marked as wrong. Suggestions?

Water has a density of 1000 kg/m3. The pressure of the atmosphere is 1x105 Pa. Approximately what distance below the surface of a lake is the total pressure three times the atmospheric pressure at the surface?

### 3 Answers

- Anonymous4 weeks ago
It is 20m. You have forgotten to allow for the pressure from the atmosphere at the surface.

At depth h, the absolute (total) pressure is three times the atmospheric pressure. This is made up of the pressure due to the water (ρgh) *plus* atmospheric pressure acting at the surface.

Total pressure = ρgh + atmospheric pressure at surface

That means ρgh = is twice atmospheric pressure (because 3 = 1+2).

ρgh = 2x1x10^5

h = 2x10^5 /(ρh)

Taking g = 10m/s^2

h = 2x10^5 /(1000x10) = 20m

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- Anonymous4 weeks ago
A bit higher than 20 meters (20.6 to be precise, since 10.3 meters of fresh water column are worth 1.00 atm) l

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- Anonymous4 weeks ago
You want the depth where the pressure is 3x atmospheric or where the water column is adding 2 atm pressure to the atmosphere pressure due to air. You are calculating the depth where the water pressure is an additional 3 atm, not 2 atm.

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