Doubling depth doubles the pressure due to the liquid when its density and gravitational field strength stay constant. In water open to the atmosphere, total pressure also includes the pressure at the surface, so doubling depth generally gives less than twice the total pressure.
The 2027 SEC G3 Physics K323 syllabus covers pressure due to a liquid column. This guide uses two points under the same water surface to show when to calculate ρgh, when to add atmospheric pressure and when to subtract two pressures.
- Measure vertical depth from the liquid surface.
- Calculate the liquid contribution with ρgh.
- Add the pressure at the surface when asked for total pressure.
- Use the difference in depths to compare two points in the same liquid.
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Identify the pressure the question asks for
At a depth h below a stationary liquid surface, the pressure increase relative to that surface is ρgh. Here ρ is the liquid density in kg/m³, g is the gravitational field strength in N/kg and h is vertical depth in metres. The result is in pascals.
At the surface of a tank open to the air, the pressure is atmospheric. A point below the surface has this starting pressure plus the increase through the liquid. If a question gives the pressure above a sealed liquid instead, use that stated surface pressure in the same role.
Circle a phrase such as due to the water, total pressure or pressure difference. All three phrases can appear with the same diagram and lead to different numerical answers.
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Compare 2 m and 4 m below the surface
Use water density 1,000 kg/m³, g = 10 N/kg and atmospheric pressure 100,000 Pa, as stated for this practice question. At 2 m the liquid adds 1,000 × 10 × 2 = 20,000 Pa. At 4 m it adds 40,000 Pa.
The total pressures are 100,000 + 20,000 = 120,000 Pa and 100,000 + 40,000 = 140,000 Pa. Their ratio is 140,000 ÷ 120,000 = 7/6. The water contribution doubles while the surface contribution stays at 100,000 Pa.
A quick reasonableness check is to picture both sensors just beneath the surface. At 1 cm and 2 cm, the liquid contributions are in a ratio of 1 to 2, but the total readings are both close to atmospheric pressure.
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CHAPTER 3 OF 4 · Back to contents
Calculate a pressure difference directly
Two sensors in the same body of water are 0.30 m and 0.85 m below the surface. The deeper sensor is 0.55 m below the shallower one, so the pressure difference is 1,000 × 10 × 0.55 = 5,500 Pa. The deeper sensor has the higher reading.
The common surface pressure cancels when you subtract the two total pressures. The shortcut assumes the fluid between the sensors has uniform density. If a diagram shows separate liquid layers, calculate the contribution of each layer the path passes through and add those contributions.
Try this. In the first example, the atmospheric pressure is changed to 98,000 Pa while the water depths and density stay fixed. The new total pressures are 118,000 Pa and 138,000 Pa. Their difference is still 20,000 Pa.
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CHAPTER 4 OF 4 · Back to contents
Frequently asked questions
Which length do I take from a tilted container diagram? Take the vertical difference in height. The sloping distance along the wall has no role in ρgh.
Does the shape of the tank change the pressure at a given depth? The pressure at the same depth is the same for the same stationary liquid under the same surface pressure, regardless of the tank’s shape.
Why is surface pressure present even at h = 0? At h = 0 the liquid contribution is zero. The pressure acting on the surface still applies.
What should I write beside my answer? Label which pressure you found and use the units requested. If you report a difference, state which point has the higher pressure.
Contents · Previous chapter · Read pressure differences in a manometer
Official syllabus and next reading
This independent study guide uses original practice examples. See the official 2027 SEC G3 K323 syllabus and the SEAB syllabus list for G3 school candidates.