MYP Physics · unit 08
Pressure and fluids
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
A selection from this pathway. Open a topic below for its complete practice and challenge set.
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.
A floating instrument that reaches its loading limit
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. An instrument housing has external volume 5.0×10⁻³ m³ and mass 3.0 kg. It floats upright in fresh water of density 1000 kg m⁻³. Use g = 10 N kg⁻¹. Assume its sealed walls do not deform and any added ballast is inside. The supplied graph gives buoyant force against displaced volume. Supplied data (Displaced volume / litre; Upthrust / N): (0, 0), (1, 10), (3, 30), (5, 50). (a) Find the fraction initially submerged and the maximum ballast mass before complete submersion. [3 marks] (b) It is moved with 1.0 kg ballast into a liquid of density 800 kg m⁻³. Predict its behaviour and explain why density alone does not establish stability. [3 marks]
Fresh-water upthrust
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Upthrust / N
Horizontal axis (x): Displaced volume / litre
Read the exact data as a table
| Displaced volume / litre | Upthrust / N |
|---|---|
| 0 | 0 |
| 1 | 10 |
| 3 | 30 |
| 5 | 50 |
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 fraction initially submerged and the maximum ballast mass before complete submersion.
Reveal the worked solution
- Floating equilibrium requires upthrust equal to the 30 N weight. Archimedes' principle gives Vdisplaced = m/ρ = 0.0030 m³.
- The submerged fraction is 0.0030/0.0050 = 0.60. The maximum upthrust before full submersion is 1000(10)(0.0050) = 50 N.
- The largest supported total mass is 5.0 kg, leaving 2.0 kg ballast. At exactly that limit there is no freeboard, so a safe operating load must be lower.
- The total mass is 3.0+1.0 = 4.0 kg; complete displacement in the new liquid yields 800(0.0050) = 4.0 kg of displaced liquid.
- It therefore has no positive freeboard in the ideal model, and waves or added water would make its operation unreliable.
- Whether a tilted housing returns upright depends on the locations of its centre of mass and the changing centre of buoyancy. Average density predicts floating capacity, not rotational stability.
Answer (a) Initially 60% submerged; limiting ballast mass = 2.0 kg. (b) Its total 4.0 kg mass exactly equals the displaced-liquid mass at full submersion; it is at the neutral-buoyancy limit.
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.