Derive and use hydrostatic pressure
Build the pressure-depth relationship from a fluid column's weight and calculate pressure differences in a uniform fluid.
AS Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 4. These original explanations and numerical practice support theory study; they do not demonstrate supervised practical competence. Full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use SI units, resolve perpendicular components, and distinguish force from pressure.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which distance is Δh in the hydrostatic equation?
What you will learn
- Derive Δp = ρgΔh from pressure and density definitions
- Use the hydrostatic pressure equation
- Interpret the depth interval and state assumptions
A fluid column links density to pressure change
Consider a vertical fluid column with cross-sectional area A and height Δh. Its volume is AΔh and its mass is ρAΔh, so its weight is ρAΔh g.
The pressure difference across the column supports this weight: Δp = weight/A. Substitution gives Δp = ρgΔh. The area cancels, so the pressure difference depends on fluid density, gravitational field strength and vertical depth change.
Apply the equation to a depth difference
The relation gives a pressure difference between two levels in a fluid of uniform density. A greater depth below the surface corresponds to a greater hydrostatic pressure. It does not require the container to have a particular shape.
Use the vertical depth difference, not a sloping path length. The equation models a fluid at rest in a uniform gravitational field and uses a consistent density and g value. Absolute pressure may also include the pressure at the surface.
Pause and explain
In the same resting fluid and field, what happens to hydrostatic pressure difference if depth doubles?
Worked example
Find the pressure increase 2.0 m below the surface of water with density 1000 kg m⁻³. Take g = 10 N kg⁻¹.
Show the worked solution
- Use Δp = ρgΔh.
- Δp = 1000 × 10 × 2.0.
- The result is a pressure difference in pascals.
Answer 2.0 × 10⁴ Pa above the surface pressure.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
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.
Calculate the pressure increase 0.80 m below a liquid surface. Use ρ = 800 kg m⁻³ and g = 10 N kg⁻¹.
Give me a hint
Multiply density, field strength and vertical depth difference.
Compare my reasoning
- Δp = ρgΔh.
- Δp = 800 × 10 × 0.80.
- Give the pressure difference in Pa.
6400 Pa.
Look for these in your work
- Equation used
- Depth is vertical
- Pressure unit correct
A diver moves from 1.0 m to 4.0 m below the surface in seawater of density 1030 kg m⁻³. Take g = 9.8 N kg⁻¹. Find the pressure increase.
Give me a hint
Use the change in depth, not the final depth alone.
Compare my reasoning
- Δh = 4.0 − 1.0 = 3.0 m.
- Δp = 1030 × 9.8 × 3.0.
- Calculate the difference in pressure.
Δp ≈ 3.03 × 10⁴ Pa.
Look for these in your work
- Depth difference found
- Equation substituted consistently
- Pressure increase reported
A pressure gauge shows an increase of 24 kPa between two points 3.0 m apart vertically in a resting fluid. With g = 10 N kg⁻¹, find the fluid density.
Give me a hint
Rearrange Δp = ρgΔh and convert kilopascals to pascals.
Compare my reasoning
- Δp = 24 000 Pa and Δh = 3.0 m.
- ρ = Δp/(gΔh) = 24 000/(10 × 3.0).
- Give density in kg m⁻³.
ρ = 800 kg m⁻³.
Look for these in your work
- Pressure unit converted
- Equation rearranged
- Density unit stated
Common mistakes
- Using the length along a sloping container wall instead of vertical depth
- Forgetting density or g in the product
- Calling the pressure difference the absolute pressure without considering surface pressure
What can you explain now?
Which distance is Δh in the hydrostatic equation?
Compare with the explanation
Vertical depth difference
Pressure changes with vertical height in the gravitational field.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Why does container shape not appear in the hydrostatic pressure-difference equation?
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.