Pressure in fluids and atmospheric pressure (physics only) – GCSE Physics exam questions with mark scheme

1Grade 1–3Triple only4 marks

A concrete paving slab rests flat on a layer of sand.

The weight of the slab is 720 N. The area of the slab in contact with the sand is 0.36 m².

(a) Write down the equation that links area ($A$), force normal to a surface ($F$) and pressure ($p$). [1]

(b) Calculate the pressure exerted on the sand by the slab. Give the unit. [3]

$p = \dfrac{F}{A}$ allow pressure = force normal to a surface ÷ area; allow $F = pA$ or $A = F/p$(1)
$p = \dfrac{720}{0.36}$(1)
$p = 2000$ an answer of 2000 scores the first 2 marks(1)
Pa allow N/m²; do not accept N/m or N m²; unit mark is independent of the numerical answer(1)
Answer: (a) $p = F/A$ (b) 2000 Pa
Examiner tip: The unit mark is easy to lose: write Pa (or N/m²), not N/m.

2Grade 4–5Triple only7 marks

A walker has a weight of 780 N. When she stands in walking boots, the total area of her boots in contact with fresh snow is 0.052 m².

(a) Calculate the pressure on the snow when she stands in her boots. [2]

(b) The fresh snow can support a pressure of 3000 Pa without the walker sinking in.

Calculate the minimum total contact area of snowshoes she needs so that she does not sink into the snow. [3]

(a) $p = \dfrac{780}{0.052}$(1)
(a) $p = 15\,000$ (Pa) allow 1.5 × 10⁴ (Pa); correct answer with no working scores 2 marks(1)
(b) $3000 = \dfrac{780}{A}$(1)
(b) $A = \dfrac{780}{3000}$(1)
(b) $A = 0.26$ (m²) correct answer with no working scores 3 marks(1)
(c) snowshoes have a larger (contact) area and the force (weight) on the snow is the same ignore 'spreads the weight' unless linked to area(1)
(c) so the pressure (on the snow) is smaller allow less pressure; do not accept 'less force'(1)
Answer: (a) 15 000 Pa (b) 0.26 m² (c) same weight over a larger area gives a smaller pressure
Examiner tip: In (c) say the weight stays the same — many answers wrongly claim the snowshoes reduce the force.

3Grade 4–5HigherTriple only7 marks

A tall tank of water has three small holes, A, B and C, in its side. The diagram shows water coming out of the holes. The depth of each hole below the water surface is shown.

(a) Water comes out of hole C with a greater speed than water from hole A.

Explain why. [2]

(b) The water leaves each hole moving horizontally, at right angles to the side of the tank.

Give the reason why. [1]

Density of water = 1000 kg/m³ gravitational field strength = 9.8 N/kg [2]

(d) Calculate the difference in pressure between hole A and hole C. [2]

(a) hole C is deeper (below the surface) / there is a greater height (column) of water above C(1)
(a) so the pressure (at C) is greater allow pressure increases with depth; allow greater weight of water above C for the first mark only(1)
(b) (because) pressure in a fluid acts at right angles (normal) to any surface allow the force from the water pressure acts at 90° to the wall(1)
(c) $p = 0.45 \times 1000 \times 9.8$(1)
(c) $p = 4410$ (Pa) allow 4400 (Pa)(1)
(d) $\Delta p = (0.45 - 0.10) \times 1000 \times 9.8$ or 4410 − 980(1)
(d) $\Delta p = 3430$ (Pa) allow 3400 (Pa); allow ecf from (c)(1)
Answer: (a) C is deeper so pressure there is greater (b) pressure acts at right angles to surfaces (c) 4410 Pa (d) 3430 Pa
Examiner tip: In (d) use the difference in depth (0.45 − 0.10 = 0.35 m), not the depth of hole C alone.

4Grade 4–5Triple only6 marks

This question is about the Earth's atmosphere.

(a) Describe the Earth's atmosphere. Refer to its thickness and to how the density of the air changes with height. [2]

(b) Explain how air molecules produce atmospheric pressure on a surface. [2]

(c) A climber measures the atmospheric pressure at the bottom and at the top of a mountain. The pressure at the top is lower.

Explain why. [2]

(a) (it is a) thin layer (of air) (compared with the size of the Earth) allow thin layer surrounding the Earth(1)
(a) the density (of the air) decreases as height (altitude) increases allow air gets thinner higher up(1)
(b) (air) molecules collide with the surface allow hit / bump into the surface(1)
(b) each collision exerts a force on the surface (and pressure is the force per unit area) ignore molecules colliding with each other(1)
(c) there is less air above the top of the mountain / the column of air above is shorter allow smaller weight of air above(1)
(c) (air is less dense so) there are fewer molecules (per unit volume) so fewer collisions (with a surface) per second allow fewer collisions with the surface; do not accept 'no air at the top'(1)
Answer: (a) a thin layer whose density decreases with height (b) molecules collide with the surface, exerting a force (c) less air above / fewer molecules so fewer collisions
Examiner tip: Pressure comes from molecules hitting the surface — collisions between molecules do not count.

5Grade 6–7Triple only6 marks

The graph shows how atmospheric pressure changes with altitude (height above sea level).

(a) Use the graph to determine the atmospheric pressure at an altitude of 6.0 km. [1]

(b) Determine the altitude at which atmospheric pressure is half its value at sea level. [2]

One window of the aircraft has an area of 0.060 m².

Calculate the resultant force on the window caused by the difference in pressure between the inside and outside of the aircraft. [3]

(a) 48 kPa allow 46–49 kPa(1)
(b) sea-level pressure read as 101 kPa (allow 100–102) and half of it ≈ 50 kPa identified may be seen as a line drawn on the graph(1)
(b) altitude = 5.5 km allow 5.3–5.8 km(1)
(c) pressure outside at 10 km read as 29 kPa allow 28–30 kPa(1)
(c) pressure difference = 75 − 29 = 46 kPa (= 46 000 Pa) allow ecf from their reading(1)
(c) $F = 46\,000 \times 0.060 = 2760$ N allow 2700–2820 N; allow 2800 N; an answer of 2.76 (kPa not converted to Pa) scores max 2(1)
Answer: (a) ≈ 48 kPa (b) ≈ 5.5 km (c) ≈ 2800 N
Examiner tip: Use the difference in pressure and convert kPa to Pa before multiplying by the area.

6Grade 6–7HigherTriple only8 marks

A solid block of steel sinks when it is placed in water. A ship made of steel floats.

(a) Explain why the solid steel block sinks. [2]

(b) Explain why the steel ship floats. [3]

(c) The ship sails from the sea into a river. The density of sea water is 1030 kg/m³. The density of river water is 1000 kg/m³.

Explain whether the ship floats higher or lower in the river than in the sea. [3]

(a) the weight of the block is greater than the upthrust (on it)(1)
(a) because steel is denser than water / the block cannot displace a weight of water equal to its own weight allow the weight of water displaced is less than the weight of the block(1)
(b) the ship is hollow / contains (a large volume of) air allow the (average) density of the ship is less than that of water(1)
(b) so it displaces a large volume (weight) of water(1)
(b) the ship sinks (into the water) until the upthrust equals its weight / until the weight of water displaced equals the weight of the ship(1)
(c) (floats) lower(1)
(c) the weight of the ship is unchanged so the upthrust needed is the same allow it must displace the same weight of water(1)
(c) river water is less dense so a greater volume of water must be displaced allow pressure at a given depth is less in river water so the ship must sit deeper(1)
Answer: (a) weight > upthrust as steel is denser than water (b) hollow ship displaces enough water for upthrust to equal its weight (c) lower — less dense water, so a larger volume must be displaced
Examiner tip: In (c) do not say the upthrust changes — it still equals the ship's weight; it is the volume displaced that increases.

7Grade 6–7HigherTriple only7 marks

A student lowered a pressure sensor into a tall cylinder of salt solution. She recorded the pressure due to the liquid at different depths.

Depth in m	Pressure due to the liquid in Pa
0.10	1080
0.20	2160
0.30	3550
0.40	4310
0.50	5390

(a) One of the results is anomalous. Identify the anomalous result. [1]

(b) Describe the relationship between the pressure due to the liquid and the depth. Use data from the table in your answer. [2]

gravitational field strength = 9.8 N/kg [3]

(d) Predict the pressure that the student should have recorded at a depth of 0.30 m. [1]

(a) 3550 Pa (at 0.30 m)(1)
(b) pressure increases as depth increases(1)
(b) (directly) proportional, supported by data, e.g. doubling the depth from 0.20 m to 0.40 m doubles the pressure from 2160 Pa to (about) 4310 Pa allow pressure increases by about 1080 Pa for every 0.10 m(1)
(c) $5390 = 0.50 \times \rho \times 9.8$ allow any non-anomalous pair of values(1)
(c) $\rho = \dfrac{5390}{0.50 \times 9.8}$(1)
(c) $\rho = 1100$ (kg/m³) allow 1090–1110; use of 3550 Pa and 0.30 m (gives 1210) scores max 2(1)
(d) 3230 Pa allow 3200–3250 Pa(1)
Answer: (a) 3550 Pa (b) directly proportional (c) 1100 kg/m³ (d) ≈ 3230 Pa
Examiner tip: Never use the anomalous result in a calculation — pick a pair that fits the pattern.

8Grade 8–9HigherTriple only6 marks

A diver descends from the surface of the sea to a depth of 40 m. A climber climbs from sea level to an altitude of 4000 m.

Compare how the pressure on the diver changes with depth with how the atmospheric pressure on the climber changes with altitude.

Your answer should explain both changes in terms of the water and the air. [6]

pressure in the sea increases with depth because there is a greater height (weight) of water above the diver(1)
water is (almost) incompressible so its density is (about) the same at all depths(1)
so pressure due to the water increases at a constant rate / is directly proportional to depth ($p = h\rho g$)(1)
e.g. each 10 m of sea water adds about $10 \times 1030 \times 9.8 \approx 1 \times 10^{5}$ Pa (about one atmosphere)(1)
total pressure on the diver = atmospheric pressure + pressure due to the water(1)
atmospheric pressure is caused by air molecules colliding with surfaces(1)
atmospheric pressure decreases with altitude because there is less air (a smaller weight of air) above the climber(1)
the atmosphere is a thin layer whose density decreases with altitude / air is compressed most near the ground(1)
so there are fewer molecules per unit volume and fewer collisions per second with a surface higher up(1)
so atmospheric pressure does not decrease at a constant rate — it falls by less for each extra metre at higher altitude(1)
water is far denser than air, so pressure changes much more over a given distance in the sea: about 4 × 10⁵ Pa increase over 40 m, compared with a decrease of about 4 × 10⁴ Pa over 4000 m of air(1)

Level	Marks	What it looks like
Level 3	5–6	A detailed, coherent comparison of both situations. Explains the change in each fluid using the weight of fluid above and/or particle collisions, and links the different patterns to water having (nearly) constant density while air density decreases with altitude. May include supporting values.
Level 2	3–4	Describes the change in pressure in both situations with some explanation for at least one of them. The comparison may be incomplete or the link to density may be missing or unclear.
Level 1	1–2	Simple, mainly descriptive statements, e.g. pressure increases as the diver goes deeper and/or decreases as the climber goes higher. Little or no explanation.
No relevant content	0	Nothing worthy of credit.

Answer: Pressure in the sea increases linearly with depth (water's density is constant); atmospheric pressure decreases with altitude but not linearly (air density falls with height); water's much greater density gives much larger pressure changes over a short distance.
Examiner tip: Top-band answers explain why the two patterns differ: constant density of water versus density of air decreasing with height.

9Grade 6–7HigherTriple only7 marks

A heating-oil tank has a pressure sensor fixed to its flat base. The sensor shows that the pressure due to the oil at the base is 6.3 kPa.

Density of the oil = 860 kg/m³ gravitational field strength = 9.8 N/kg

(a) Calculate the depth of oil in the tank. [3]

(b) The base of the tank has an area of 1.6 m².

Calculate the force exerted on the base of the tank by the oil. [2]

Calculate the decrease in pressure at the base of the tank. [2]

(a) $6300 = h \times 860 \times 9.8$ allow $6.3 = h \times 860 \times 9.8$ for this mark(1)
(a) $h = \dfrac{6300}{860 \times 9.8}$(1)
(a) $h = 0.75$ (m) allow 0.7475 (m); this mark needs 6.3 kPa converted to 6300 Pa; an answer of 0.00075 (m) scores 2 marks(1)
(b) $F = 6300 \times 1.6$(1)
(b) $F = 10\,080$ (N) allow 10 000 (N); allow 1.0 × 10⁴ (N)(1)
(c) pressure at 0.40 m: $p = 0.40 \times 860 \times 9.8 = 3371$ (Pa) allow alternative: $\Delta p = (0.75 - 0.40) \times 860 \times 9.8$(1)
(c) decrease = 6300 − 3371 = 2929 (Pa) allow 2900–2950 (Pa); allow ecf from (a)(1)
Answer: (a) 0.75 m (b) 10 080 N (≈ 10 000 N) (c) ≈ 2900 Pa
Examiner tip: Convert 6.3 kPa to 6300 Pa before rearranging $p = h\rho g$ — otherwise the depth comes out 1000 times too small.

10Grade 8–9HigherTriple only8 marks

A cube is held completely under water by a string, as shown in the diagram. The top face of the cube is horizontal and is 0.30 m below the water surface. Each side of the cube is 0.20 m long.

Density of water = 1000 kg/m³ gravitational field strength = 9.8 N/kg

(a) Explain why the water exerts a resultant upward force (upthrust) on the cube. [2]

(b) Calculate the upthrust on the cube. [4]

State what happens to the upthrust on the cube. Give a reason for your answer. [2]

(a) the bottom face is deeper than the top face so the pressure on the bottom face is greater allow pressure increases with depth(1)
(a) so the upward force on the bottom face is greater than the downward force on the top face ignore references to forces on the sides; allow 'forces on the sides cancel'(1)
(b) pressure on top face $= 0.30 \times 1000 \times 9.8 = 2940$ Pa or pressure difference $= 0.20 \times 1000 \times 9.8 = 1960$ Pa scores first 2 marks(1)
(b) pressure on bottom face $= 0.50 \times 1000 \times 9.8 = 4900$ Pa(1)
(b) area of face $= 0.040$ m² and a force calculated using $F = pA$, e.g. $2940 \times 0.040 = 117.6$ N or $4900 \times 0.040 = 196$ N(1)
(b) upthrust $= 196 - 117.6 = 78.4$ N allow 78 N; correct answer with no working scores 4 marks; allow 78.4 N from weight of water displaced: $0.008 \times 1000 \times 9.8$(1)
(c) (upthrust) stays the same no mark for the reason if this is wrong(1)
(c) the difference in depth (and so pressure difference) between the top and bottom faces is still the same / still 0.20 m allow the volume of water displaced is unchanged(1)
Answer: (a) greater pressure on the deeper bottom face gives a bigger upward force (b) 78.4 N (c) unchanged — the pressure difference across the cube is the same
Examiner tip: Upthrust depends on the difference in pressure between the top and bottom faces, not on how deep the cube is.

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