Pressure in fluids Cambridge International AS & A Level Physics revision
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In plain words
Pressure is force spread over an area. In a liquid it comes from the weight of everything above you, so it grows steadily with depth. That difference between the bottom and the top of a submerged object is exactly what pushes it upwards.
Three things to know
- Pressure p = F ÷ A, in pascals (1 Pa = 1 N m⁻²).
- The difference in pressure between two depths in a fluid is Δp = ρgΔh.
- Upthrust is caused by that difference: the fluid pushes up on the bottom of an object harder than it pushes down on the top.
Worked example
Find the pressure due to 10 m of water of density 1000 kg/m³.
- Δp = ρgΔh.
- = 1000 × 9.81 × 10.
- = 9.8 × 10⁴ Pa, about one atmosphere.
Tips and tricks
- ρgh gives the pressure due to the liquid alone. Add atmospheric pressure for the total.
- Pressure in a liquid depends on depth and density only, not on the shape or width of the container.
It lands in your notebook with its questions as flashcards.
Pressure in fluids: 5 questions and answers
These are the quiz’s questions. Do the quiz first, then come back here for the ones that got you.
What is the pressure due to a column of liquid given by?
Density × g × depth.
What are the base units of pressure?
N m⁻², with the newton as kg m s⁻².
Two containers of different shapes hold water to the same depth. How do the pressures at the bottom compare?
Only the depth and the density matter.
A force of 200 N acts on an area of 0.050 m². What is the pressure?
200 ÷ 0.050.
Why is there an upthrust on a submerged object?
Pressure increases with depth.
Quiz
5 questions
Tap an answer and you’ll see straight away whether it’s right, and why.
Worksheet
4 questions, 10 marks. Write your answers on paper, then check them.
Pressure in fluids
Cambridge International AS & A Level Physics 9702 · 10 marks · papermunch.org
Name ______________________________ Date ______________
A force of 600 N acts on an area of 0.030 m². Find the pressure.[2]
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2.0 × 10⁴ Pa.
Atmospheric pressure is 1.01 × 10⁵ Pa. Find the total pressure 25 m below the surface of sea water of density 1030 kg/m³.[3]
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3.5 × 10⁵ Pa. 1030 × 9.81 × 25 = 2.53 × 10⁵ Pa, plus 1.01 × 10⁵ Pa.
Show that the pressure due to a column of liquid of height h and density ρ is ρgh.[3]
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Take a column of cross-sectional area A. Its volume is Ah, so its mass is ρAh and its weight is ρAhg. Pressure = weight ÷ area = ρAhg ÷ A = ρgh.
Explain, in terms of pressure, why a submerged object experiences an upthrust.[2]
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The pressure in the fluid increases with depth, so the upward force on the bottom of the object is greater than the downward force on the top.
Answers: Pressure in fluids
- 1. 2.0 × 10⁴ Pa.
- 2. 3.5 × 10⁵ Pa. 1030 × 9.81 × 25 = 2.53 × 10⁵ Pa, plus 1.01 × 10⁵ Pa.
- 3. Take a column of cross-sectional area A. Its volume is Ah, so its mass is ρAh and its weight is ρAhg. Pressure = weight ÷ area = ρAhg ÷ A = ρgh.
- 4. The pressure in the fluid increases with depth, so the upward force on the bottom of the object is greater than the downward force on the top.