Find depth or density from a liquid-pressure difference
Rearrange the equation after selecting a consistent surface or point reference.
Cross-referenced to the 2028–2029 0653 section using original teaching from the related single-science courses. A 0653-specific objective-by-objective and Core/Supplement review is in progress. Written practice is self-checked and does not demonstrate supervised laboratory competence or award an official Cambridge grade.
Before you begin
- Calculate Δp = ρgΔh.
- Convert kPa to Pa and select a vertical depth reference.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What pressure change separates 118 kPa at a point from 100 kPa at its surface?
What you will learn
- Use Δh = Δp/(ρg) and ρ = Δp/(gΔh).
- Subtract a supplied reference pressure before finding depth.
The numerator must be a pressure change
Rearrange Δp = ρgΔh to Δh = Δp/(ρg). A 2000 Pa increase in water of density 1000 kg/m³ with g = 10 N/kg corresponds to 0.20 m greater depth. If a point's total pressure is 118 kPa while its surface pressure is 100 kPa, use only the 18 kPa liquid increment. In the same water model the depth is 18 000/(1000 × 10) = 1.8 m, not 11.8 m.
When two depths share one unchanged surface reference, subtract their point pressures to obtain the increment between them. Absolute pressure labels need not be guessed or replaced by a familiar value. A reported point pressure below its own surface pressure is inconsistent with a positive below-surface depth in this stationary positive-density model; investigate the data rather than taking the absolute value to hide the inconsistency.
Density needs a non-zero level difference
Density is ρ = Δp/(gΔh). If a stationary liquid's pressure rises by 3600 Pa over a downward 0.40 m interval and g = 10 N/kg, density is 900 kg/m³. The measured pressure and depth differences must refer to the same interval; an unrelated total pressure is not a substitute for the numerator.
For positive g, positive density and a non-zero depth difference, pressure change has the same sign as downward depth change. At equal depth, the zero change carries no information selecting a unique density through 0/0. The liquid equation describes the stated stationary uniform-density model; it does not prove practical measurement quality or a material's complete composition.
Sources for this lesson
Pause and explain
Do equal-level pressure readings determine a unique density through ρ = Δp/(gΔh)?
Worked example
A point has total pressure 118 kPa with surface pressure 100 kPa. Find its water depth for ρ = 1000 kg/m³ and g = 10 N/kg, then find density for an increase of 3600 Pa over downward 0.40 m with g = 10.
Show the worked solution
- The first liquid increment is 18 kPa = 18 000 Pa.
- Depth = 18 000/(1000 × 10) = 1.8 m.
- Second density = 3600/(10 × 0.40) = 900 kg/m³.
Answer 1.8 m; 900 kg/m³.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find water depth for total pressure 118 kPa and surface pressure 100 kPa with ρ = 1000 kg/m³ and g = 10 N/kg.
Give me a hint
Remove the surface contribution before rearranging.
Compare my reasoning
- Δp = 118 − 100 = 18 kPa = 18 000 Pa.
- h = 18 000/(1000 × 10).
- Depth is 1.8 m.
1.8 m.
Look for these in your work
- Use the increment.
- Convert kPa.
Find density for a 3600 Pa increase over downward 0.40 m with g = 10 N/kg; find depth increase for 2000 Pa in density 1000 kg/m³ with the same g.
Give me a hint
Use a consistent positive difference in each rearrangement.
Compare my reasoning
- ρ = 3600/(10 × 0.40) = 900 kg/m³.
- Δh = 2000/(1000 × 10).
- The second increase is 0.20 m.
900 kg/m³; 0.20 m.
Look for these in your work
- Choose the right unknown.
- Use matching differences.
A stated below-surface water point has 102 kPa while the surface is at 104 kPa, with ρ = 1000 kg/m³ and g = 10 N/kg. Check the depth calculation rather than hiding its sign.
Give me a hint
The pressure increment must be positive for a positive downward depth in this model.
Compare my reasoning
- Δp = −2000 Pa.
- The equation gives h = −0.20 m.
- That is inconsistent with the stated below-surface location; check readings, references or model conditions.
Calculated h = −0.20 m; the below-surface claim is inconsistent with the supplied data.
Look for these in your work
- Retain the negative result.
- Identify the inconsistency rather than taking its magnitude.
Common mistakes
- Subtract the supplied reference pressure.
- Use matching pressure and depth differences.
- A zero level difference does not determine density.
What can you explain now?
What depth increase gives 2000 Pa in density 1000 kg/m³ with g = 10 N/kg?
Compare with the explanation
0.20 m.
Divide the pressure increment by ρg.
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Make it stick
Find depth and density from the right pressure difference and detect zero-data or inconsistent-sign cases.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.