Edexcel IGCSE Physics 5.7 Pressure Difference Equation Questions

Apply the pressure-difference equation in manometers, submerged objects and liquid columns, including total pressure above atmosphere.

Exam points

State and rearrange Δp = hρg to relate pressure difference to liquid height, density and gravitational field strength.
Calculate pressure difference or liquid height from liquid data, converting centimetres to metres and using consistent pressure units.
Distinguish pressure difference from total pressure: add atmospheric pressure to find total pressure, or subtract it to find the liquid’s contribution.
For a very large pressure difference in a manometer, explain why a denser liquid gives a smaller, more manageable height difference.

[Maximum number: 3]

The diagram shows a manometer, a device used for measuring differences in pressure.

Figure for Question Edexcel IGCSE Physics 5.7 Pressure Difference Equation Questions question 1 — Edexcel IGCSE Physics

One side of the manometer has some trapped gas. The other side is left open to the atmosphere.

The difference in pressure can be calculated using this formula.
[difference in pressure = height × density × 10 ]
The density of the liquid is $1.3 \times 10^{4} \mathrm{~kg} / \mathrm{m}^{3}$ .
The difference in the levels of the liquid is 3.8 cm.
Calculate the difference in pressure between the atmosphere and the trapped gas.
difference in pressure = Pa

Marking points:
- conversion of cm to m
- substitution into the given formula
- evaluation

Example:
$3.8\ \mathrm{cm}=0.038\ \mathrm{m}$ $\text{pressure difference}=0.038\times1.3\times10^{4}\times10$ $\text{pressure difference}=4900\ \mathrm{Pa}$

Marking guidance:
- power-of-ten error gives a 1 mark penalty
- 494000 gains 2 marks

Accept:
- $4940\ \mathrm{Pa}$
- 4800, 4840, 4846, or $4841\ \mathrm{Pa}$ for use of g=9.8 or 9.81

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Edexcel IGCSE Physics 5.7 Pressure Difference Equation Questions question 1