Cambridge IGCSE Physics · topic 1.8
Pressure
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
A hydraulic lift below the reservoir surface
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. An ideal hydraulic system contains oil of density 800 kg m⁻³. Its small piston has area 2.0 cm² and its large piston has area 40 cm². The large piston is 0.50 m below the small piston. A 60 N force is applied downward to the small piston; atmospheric pressure acts on both external faces. Use g = 10 N kg⁻¹ and neglect piston weights and friction. Supplied data (Depth below small piston / m; Additional pressure / Pa): (0, 0), (0.25, 2000), (0.5, 4000). (a) Find the gauge pressure immediately below each piston and the greatest downward load the large piston can balance. [3 marks] (b) For a short motion, the small piston moves down 0.10 m. Find the large-piston rise and explain why force amplification does not create energy. [3 marks]
Hydrostatic gauge-pressure contribution below the small piston
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Additional pressure / Pa
Horizontal axis (x): Depth below small piston / m
Read the exact data as a table
| Depth below small piston / m | Additional pressure / Pa |
|---|---|
| 0 | 0 |
| 0.25 | 2000 |
| 0.5 | 4000 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Find the gauge pressure immediately below each piston and the greatest downward load the large piston can balance.
Reveal the worked solution
- Convert 2.0 cm² to 2.0×10⁻⁴ m². Applied gauge pressure is F/A = 60/(2.0×10⁻⁴) = 300000 Pa.
- The lower piston has an additional ρgh = 800(10)(0.50) = 4000 Pa, giving 304000 Pa gauge.
- Its area is 40 cm² = 0.0040 m², so the balanced load is pA = 304000(0.0040) = 1216 N. Atmospheric contributions cancel across each piston.
- Incompressibility gives equal displaced volumes: As xs = AL xL, so xL = (2/40)(0.10) = 0.0050 m.
- The large force acts through a smaller displacement. For equal piston heights the ideal input and output work are equal.
- Here the initial pressure difference also involves gravitational potential energy of the oil. For a finite stroke the height difference changes; a full energy audit must include the oil and the varying pressure, rather than multiplying the initial force by any arbitrary stroke.
Answer (a) Small-piston gauge pressure = 300 kPa; large = 304 kPa; load = 1216 N. (b) Large-piston rise = 0.0050 m; input work and gravitational-energy changes must both be included.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Two manometers, one gas pressure
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. An open U-tube connects a gas vessel to the atmosphere. Water of density 1000 kg m⁻³ is 0.24 m lower on the gas side than on the atmospheric side. Atmospheric pressure is 100 kPa; use g = 10 N kg⁻¹. A second manometer uses oil of density 800 kg m⁻³. The graph gives the water calibration. Supplied data (Vertical level difference / m; Pressure difference / Pa): (0, 0), (0.1, 1000), (0.2, 2000), (0.3, 3000). (a) Find gauge pressure, absolute gas pressure, and the expected level difference in the oil manometer. [3 marks] (b) If the tube is tilted, explain which height is required and propose a measurement that reduces uncertainty. [3 marks]
Water manometer calibration
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Pressure difference / Pa
Horizontal axis (x): Vertical level difference / m
Read the exact data as a table
| Vertical level difference / m | Pressure difference / Pa |
|---|---|
| 0 | 0 |
| 0.1 | 1000 |
| 0.2 | 2000 |
| 0.3 | 3000 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Find gauge pressure, absolute gas pressure, and the expected level difference in the oil manometer.
Reveal the worked solution
- The lower liquid surface on the gas side shows that gas pressure exceeds atmospheric pressure.
- Gauge pressure is ρgh = 1000(10)(0.24) = 2400 Pa. Absolute pressure is 100000+2400 = 102400 Pa.
- For the same pressure difference, hoil = 2400/(800×10) = 0.30 m. The less dense liquid requires a taller column.
- Hydrostatic pressure depends on vertical depth because the gravitational-energy change depends on vertical height.
- Read both surfaces from the same horizontal reference with a vertical scale, avoiding parallax. Add the two surface-reading uncertainties when estimating a conservative uncertainty in their difference.
- An inclined tube can spread a small vertical difference over a longer scale, improving readability only when its angle is measured and the vertical component is used.
Answer (a) Gauge = +2.4 kPa; absolute = 102.4 kPa; oil difference = 0.30 m. (b) Use vertical separation of free surfaces, not liquid length along the tilted tube.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.