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5.5.1.1Pressure in a fluid (p = F/A)

AQA GCSE Physics (8463), Higher tier · Forces › Pressure in fluids

Practise Pressure in a fluid (p = F/A). 12 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.

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Write down the equation that links area (A), force normal to a surface (F) and pressure (p).
\(p = \dfrac{F}{A}\)
A research submarine has a circular hatch of area 0.50 m2. At the depth where the submarine is working, the water pressure on the outside of the hatch is 2.1 MPa. The air pressure inside the submarine is 0.10 MPa. In which direction does this resultant force act?
Into the submarine, at right angles to the hatch.
Write down the equation that links area of a surface (A), force normal to the surface (F) and pressure (p).
\(p = \dfrac{F}{A}\)

Sample questions

Written for this site in the style of AQA exam questions. They are not taken from real past papers.

Question 1Easy5 marks
(a) Which two of these are fluids?
Tick (✓) two boxes.[2]
  • Air
  • Ice
  • Steel
  • Water
(b) The pressure in a fluid causes a force on any surface in contact with the fluid.
In which direction does this force act?
Tick (✓) one box.[1]
  • At right angles (normal) to the surface
  • Parallel to the surface
  • Always vertically downwards
  • Always vertically upwards
(c) A force of 1200 N acts at right angles to a surface of area 0.30 m2.
Calculate the pressure on the surface.
Use the equation:
\(\text{pressure} = \dfrac{\text{force normal to a surface}}{\text{area of that surface}}\)[2]
Show the answer and mark scheme
(a) Answer: Air; Water
(b) Answer: At right angles (normal) to the surface
(c) Answer: 4000 Pa
  • p = 1200 ÷ 0.30
  • 4000 (Pa)
Question 2Medium7 marks
The water in a large aquarium exerts a pressure on a glass viewing window. The area of the window is 0.60 m2.
(a) Write down the equation that links area (A), force normal to a surface (F) and pressure (p).[1]
(b) The average pressure of the water on the window is 18 kPa.
Calculate the force of the water on the window.[3]
(c) In which direction does the force of the water act on the window?[1]
(d) The aquarium is filled to a greater depth.
Suggest what happens to the force of the water on the window. Give a reason for your answer.[2]
Show the answer and mark scheme
(a) Answer: \(p = \dfrac{F}{A}\)
  • p = F ÷ A / pressure = force normal to a surface ÷ area of that surface
(b) Answer: 10 800 N
  • 18 kPa = 18 000 Pa
  • F = 18 000 × 0.60
  • 10 800 (N)
(c) Answer: At right angles to the glass (outwards).
  • at right angles to the window / normal to the glass (outwards)
(d) Answer: It increases, because the pressure of the water on the window increases.
  • the force increases
  • the (average) pressure (on the window) increases (because the water is deeper)
Question 3Hard7 marks
A research submarine has a circular hatch of area 0.50 m2. At the depth where the submarine is working, the water pressure on the outside of the hatch is 2.1 MPa. The air pressure inside the submarine is 0.10 MPa.
(a) Calculate the resultant force on the hatch caused by the water and the air.[3]
(b) In which direction does this resultant force act?[1]
(c) A viewing window can safely withstand a resultant force of 45 kN.
Calculate the maximum area of the viewing window at this depth.[3]
Show the answer and mark scheme
(a) Answer: 1.0 × 106 N
  • pressure difference = 2.1 − 0.10 = 2.0 MPa = 2.0 × 106 Pa
  • F = 2.0 × 106 × 0.50
  • 1.0 × 106 (N)
(b) Answer: Into the submarine, at right angles to the hatch.
  • inwards / into the submarine (at right angles to the hatch)
(c) Answer: 0.0225 m2
  • 45 kN = 45 000 N
  • A = 45 000 ÷ 2.0 × 106
  • 0.0225 (m2)

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