Pressure in a fluid 1 (physics only) - AQA GCSE Physics Exam Questions

Practise 22 exam-style questions on Pressure in a fluid 1 (physics only) for AQA GCSE Physics. Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

22 questions
1–6 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

A rectangular glass tank is filled with water. The water exerts a pressure on the base of the tank and on each of the four vertical side walls.

Describe the direction of the force that the water exerts on one of the vertical side walls.

Model answer

The force acts at right angles (normal) to the wall, so it is horizontal, and it pushes outwards on the wall, away from the water.

Explanation

The pressure in a fluid causes a force normal (at right angles) to any surface it touches, not just downwards. On a vertical wall this means the force is horizontal, pushing the wall outwards. This is why tanks, dams and swimming pool walls must be strong.

Mark scheme

1 mark
at right angles to the wall
allow normal / perpendicular to the wall; allow horizontal
1 mark
outwards (away from the water / pushing the wall away from the inside of the tank)
do not accept downwards for either mark

Question 2

3 marks

A submarine is deep under the sea. The water outside the submarine exerts a pressure of 3.0 × 10⁶ Pa on a hatch. The hatch has an area of 0.25 m².

Calculate the force exerted by the water on the hatch.

Model answer

p = F ÷ A, so F = p × A
F = 3.0 × 10⁶ × 0.25
F = 750 000 N = 7.5 × 10⁵ N

Explanation

Rearranging p = F ÷ A gives F = p × A. Deep water has a very high pressure, so even a small hatch has a force of hundreds of thousands of newtons on it - this is why submarine hulls and hatches must be very strong.

Mark scheme

1 mark
3.0 × 10⁶ = F ÷ 0.25
an answer of 7.5 × 10⁵ (N) scores 3 marks
1 mark
F = 3.0 × 10⁶ × 0.25
rearrangement; allow F = p × A with values
1 mark
7.5 × 10⁵ (N)
allow 750 000 (N)

Question 3

3 marks

A cyclist lets some air out of his bicycle tyres before riding on soft sand. The pressure of the air in the tyres decreases. The weight on each tyre does not change.

Explain why the area of each tyre in contact with the ground increases.

Model answer

The weight on the tyre, and so the force it exerts on the ground, stays the same. Since pressure = force ÷ area, the area = force ÷ pressure. If the air pressure in the tyre is lower, the tyre flattens until a larger area is in contact with the ground, so the same force is still supported.

Explanation

For a fixed force, pressure and area are inversely proportional: halve the pressure and the area must double. A larger contact area also means the tyre presses less hard on each square metre of sand, so it sinks in less.

Mark scheme

1 mark
the force on the ground (the weight) stays the same
1 mark
pressure = force ÷ area / area = force ÷ pressure
allow p = F ÷ A
1 mark
so a lower pressure needs a larger contact area to support the same force
allow area is inversely proportional to pressure for the same force; ignore references to sinking into the sand

Question 4

5 marks

A car has a mass of 1200 kg. Its weight is supported equally by its four tyres. The area of each tyre in contact with the road is 140 cm².

The pressure of the air in a tyre can be taken to be equal to the pressure the tyre exerts on the road.

gravitational field strength = 9.8 N/kg

Estimate the pressure of the air in each tyre.
Give your answer in kilopascals.

Model answer

Weight = m × g = 1200 × 9.8 = 11 760 N
Total contact area = 4 × 140 = 560 cm² = 560 ÷ 10 000 = 0.056 m²
p = F ÷ A = 11 760 ÷ 0.056 = 210 000 Pa
p = 210 kPa

Explanation

The force on the road is the car's weight, found from W = m g, and it is spread over the total contact area of all four tyres. Convert cm² to m² by dividing by 10 000, then divide the pressure in pascals by 1000 to get kilopascals. Real car tyres are inflated to about 200-250 kPa above atmospheric pressure, so this estimate is sensible.

Mark scheme

1 mark
weight = 1200 × 9.8 = 11 760 (N)
an answer of 210 (kPa) scores 5 marks; an answer of 210 000 (Pa) scores 4 marks
1 mark
total area = 4 × 140 = 560 cm² = 0.056 m²
allow force on one tyre = 2940 N with area 0.014 m²
1 mark
p = 11 760 ÷ 0.056
allow p = 2940 ÷ 0.014; allow ecf
1 mark
p = 210 000 (Pa)
allow ecf
1 mark
210 (kPa)
correct conversion of their pressure in Pa to kPa; allow ecf

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