IGCSE · 1 · extended response
Two manometers, one gas pressure
Motion, forces and energy · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
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.
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
Ready to self-mark?Reveal the detailed answer guide
(a)
Find gauge pressure, absolute gas pressure, and the expected level difference in the oil manometer.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
The lower liquid surface on the gas side shows that gas pressure exceeds atmospheric pressure.
- 2
Gauge pressure is ρgh = 1000(10)(0.24) = 2400 Pa. Absolute pressure is 100000+2400 = 102400 Pa.
- 3
For the same pressure difference, hoil = 2400/(800×10) = 0.30 m. The less dense liquid requires a taller column.
(b)
If the tube is tilted, explain which height is required and propose a measurement that reduces uncertainty.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Hydrostatic pressure depends on vertical depth because the gravitational-energy change depends on vertical height.
- 2
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.
- 3
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.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- 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.
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