The Physics Question Bank

Course: Separate Physics only  |  Tier: Higher only  |  Equation sheet: given in the exam

The equation

pressure due to a column of liquid = height of column × density of liquid × gravitational field strength    p = h ρ g

Quantity	Symbol	Unit
pressure	p	pascals (Pa)
height (depth)	h	metres (m)
density	ρ	kilograms per cubic metre (kg/m³)
gravitational field strength	g	newtons per kilogram (N/kg)

Rearranged: h = p ÷ (ρ g)  |  ρ = p ÷ (h g)  |  Fresh water ρ = 1000 kg/m³, sea water ρ = 1030 kg/m³, g = 9.8 N/kg

Use FIFA for every answer

• F – Formula: write the equation as it appears on the sheet.
• I – Insert: put in the numbers (convert to standard units first).
• F – Fix: rearrange to make the unknown the subject.
• A – Answer: calculate and always give the unit.

Higher tier only: work through all 10. Most questions need rearranging.

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Questions

Q1 A swimming pool is 2.0 m deep. Calculate the pressure due to the water at the bottom.

Show solution

F: p = h ρ g
I: p = 2.0 × 1000 × 9.8
F: p is already the subject
A: 19 600 Pa

Q2 The water pressure at the bottom of a garden water butt is 29 400 Pa. How deep is the water?

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F: p = h ρ g
I: 29 400 = h × 1000 × 9.8
F: h = 29 400 ÷ 9800
A: 3.0 m

Q3 A 0.50 m deep tank of cooking oil exerts a pressure of 4508 Pa at the bottom. Calculate the density of the oil.

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F: p = h ρ g
I: 4508 = 0.50 × ρ × 9.8
F: ρ = 4508 ÷ 4.9
A: 920 kg/m³

Q4 A diver in a freshwater lake feels a water pressure of 98 kPa. How deep is she?

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Convert first: 98 kPa = 98 000 Pa
F: p = h ρ g
I: 98 000 = h × 1000 × 9.8
F: h = 98 000 ÷ 9800
A: 10 m

Q5 The water pressure at the bottom of a fish tank is 3.92 kPa. How deep is the water in cm?

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Convert first: 3.92 kPa = 3920 Pa
F: p = h ρ g
I: 3920 = h × 1000 × 9.8
F: h = 3920 ÷ 9800 = 0.4 m
A: 40 cm

Q6 The water pressure at the base of a dam is 441 kPa. How deep is the reservoir?

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Convert first: 441 kPa = 441 000 Pa
F: p = h ρ g
I: 441 000 = h × 1000 × 9.8
F: h = 441 000 ÷ 9800
A: 45 m

Q7 In a barometer, atmospheric pressure of 1.0 × 10⁵ Pa holds up a column of mercury (ρ = 13 600 kg/m³). Calculate the height of the column in mm, to 2 significant figures.

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F: p = h ρ g
I: 1.0 × 10⁵ = h × 13 600 × 9.8
F: h = 100 000 ÷ 133 280 = 0.750… m
A: 750 mm (2 s.f.)

Q8 A deep-sea submarine feels a seawater pressure of 5.0 × 10⁷ Pa. Calculate its depth, in standard form to 2 significant figures.

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F: p = h ρ g
I: 5.0 × 10⁷ = h × 1030 × 9.8
F: h = 5.0 × 10⁷ ÷ 10 094 = 4953… m
A: 5.0 × 10³ m

Q9 A scuba diver is 12 m deep in the sea. Her mask has an area of 0.020 m². Calculate the force of the water on the mask, to 2 significant figures.

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Step 1 – pressure
F: p = h ρ g
I: p = 12 × 1030 × 9.8 = 121 128 Pa
Step 2 – force
F: p = F / A
I: 121 128 = F ÷ 0.020
F: F = 121 128 × 0.020 = 2422… N
A: 2400 N (2 s.f.)

Q10 Atmospheric pressure is 1.0 × 10⁵ Pa. At what depth in a freshwater lake is the total pressure double atmospheric pressure? Give your answer to 3 significant figures.

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Think: the water must add another 1.0 × 10⁵ Pa on top of the atmosphere
F: p = h ρ g
I: 100 000 = h × 1000 × 9.8
F: h = 100 000 ÷ 9800 = 10.20… m
A: 10.2 m (3 s.f.)