How deep must a diver go in fresh water before the pressure due to the water's weight alone equals 1.00 atm, that is 1.01 x 10^5 Pa? Take the density of water as 1.00 x 10^3 kg/m^3 and g as 9.80 m/s^2.
A1.03 m
B10.3 m
C103 m
D20.6 m
Answer & Solution
Correct answer: B. 10.3 m
1. Pressure due to the weight of a fluid is pressure = h rho g, so the depth is h = pressure / (rho g).
2. The target pressure is 1.00 atm, which is 1.01 x 10^5 N/m^2.
3. Denominator = (1.00 x 10^3 kg/m^3)(9.80 m/s^2) = 9.80 x 10^3 kg/(m^2 s^2).
4. h = (1.01 x 10^5 N/m^2) / (9.80 x 10^3 kg/(m^2 s^2)) = 10.3 m.
5. So just over ten metres of water weighs as much per unit area as 120 km of air.
6. The values 1.03 m and 103 m are the same digits with the power of ten slipped, which is what happens if the density is entered as 10^4 or 10^2 kg/m^3.
7. The value 20.6 m doubles the answer, the trap for anyone who adds the atmosphere's own pressure before dividing.
_Source: OpenStax College Physics (CC BY 4.0), Ch 11 "Fluid Statics", section 11.4 Variation of Pressure with Depth in a Fluid_
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