OCR GCSE Physics A: Liquid pressure, upthrust, floating and sinking (J249) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover statements P1.3h, P1.3i and P1.3j of OCR GCSE (9-1) Physics A (Gateway Science), specification Version 5.0 (July 2026). All three statements are Higher Tier only, so every question is labelled (Higher Tier only). Use g = 10 N/kg and give pressures due to the liquid alone unless told otherwise. Learn the ideas first in the study guide or the revision notes. Course pages: OCR GCSE Physics hub and the printable checklist. For a quick check of your level, try the free 10-minute diagnostics.

Questions

1. (Higher Tier only) State two factors that determine the pressure due to a liquid at a point below its surface. [2]

2. (Higher Tier only) A submarine sensor is 18 m below the surface of fresh water of density 1000 kg/m³. Calculate the pressure due to the water. [3]

3. (Higher Tier only) Two probes hang in seawater of density 1030 kg/m³, at depths of 2.4 m and 9.6 m. Calculate the difference in pressure between the probes. [3]

4. (Higher Tier only) A gauge in a freshwater pond reads a pressure of 45 000 Pa due to the water. The density of the water is 1000 kg/m³. Calculate the depth of the gauge. [3]

5. (Higher Tier only) Explain why the pressure at the bottom of a tank is greater in a deeper tank of the same liquid, and greater in a denser liquid at the same depth. [4]

6. (Higher Tier only) A wooden block floats with its lower part under water. Explain how the pressure of the water produces an upward force on the block. [4]

7. (Higher Tier only) A flat-bottomed wooden block of mass 0.84 kg has a rectangular base measuring 0.25 m by 0.12 m. It floats upright in water of density 1000 kg/m³.

(a) Calculate the weight of the block. [1] (b) Calculate the pressure of the water on the base of the block. [2] (c) Calculate the depth of the base below the water surface. [2]

8. (Higher Tier only) A solid cuboid has a square top face of side 0.30 m and a height of 0.40 m. Its mass is 40 kg. It is held completely under a liquid of density 1200 kg/m³, with the top face 0.50 m below the surface.

(a) Calculate the pressure on the top face and on the bottom face. [2] (b) Calculate the upthrust on the cuboid. [2] (c) The cuboid is released. Calculate the resultant force on it and state its direction. [2]

9. (Higher Tier only) Four objects are placed in water of density 1000 kg/m³.

Object	Mass (kg)	Volume (m³)
A	0.45	0.00050
B	0.90	0.00075
C	1.50	0.0020
D	0.30	0.00030

(a) Calculate the density of each object. [2] (b) State what happens to each object when it is released fully under the water. [2] (c) Calculate the fraction of object A that is under the surface when it floats at rest. [1] (d) State the fraction of object C under the surface when it floats at rest. [1]

10. (Higher Tier only) A flat-bottomed boat has a mass of 840 kg. The hull can hold a maximum of 3.0 m³ of water below the deck line before water spills in. The boat is used in fresh water of density 1000 kg/m³.

(a) Calculate the weight of the empty boat. [1] (b) Calculate the minimum volume of water the boat must displace to float. [2] (c) Calculate the greatest mass of cargo the boat can carry before it sinks to the deck line. [3] (d) Explain why the boat floats even though the steel it is made from is denser than water. [2]

11. (Higher Tier only) A pupil measures the pressure due to a liquid at different depths.

Depth (m)	0.10	0.20	0.30	0.40
Pressure due to the liquid (Pa)	1200	2400	3600	4800

(a) Describe the relationship between pressure and depth shown by the data. [1] (b) Calculate the gradient of a graph of pressure (y-axis) against depth (x-axis). [2] (c) Use the gradient to find the density of the liquid. [2] (d) Predict the pressure at a depth of 0.55 m. [1] (e) A second liquid is less dense. Describe how the graph for it would differ. [2]

Answers

1. The depth below the surface [1]. The density of the liquid [1]. [2] Examiner insight: Naming g or “weight” does not fit “factors of the liquid at a point”; answers gain credit by naming depth and density, and a list of three with one wrong item can lose the mark.

2. p = h x ρ x g [1]. p = 18 x 1000 x 10 [1]. p = 180 000 Pa [1]. [3] Examiner insight: A correct number with no unit, or the wrong unit, is a common way to drop the last mark; write Pa (or 180 kPa) on the answer line.

3. Δh = 9.6 - 2.4 = 7.2 m [1]. Δp = 7.2 x 1030 x 10 [1]. Δp = 74 160 Pa [1]. [3] Examiner insight: Working out both pressures and subtracting gives the same result as the depth difference, so show whichever route you take clearly; do not use only one of the two depths.

4. h = p / (ρ x g) [1]. h = 45 000 / (1000 x 10) [1]. h = 4.5 m [1]. [3] Examiner insight: Rearrange before substituting. Answers that substitute into p = h x ρ x g and then solve for h in one step is riskier; writing each line makes algebra slips easier to spot.

5. A column of liquid lies above the point and has weight [1]. At greater depth there is a taller column, so more weight presses on each unit of area [1]. A denser liquid has more mass in each cubic metre [1], so the same height of column has more weight and gives more pressure [1]. [4] Examiner insight: Four marks need four separate statements. Writing “more water so more pressure” without mentioning weight or area is a typical partial answer.

6. The bottom of the block is deeper than the top [1]. The pressure at the bottom is greater than the pressure at the top [1]. This gives a larger upward force on the bottom than the downward force on the top [1]. The difference is a resultant upward force, the upthrust [1]. [4] Examiner insight: For a floating block, remember that the top is in air, not water. The key phrase is a pressure difference between lower and upper surfaces, not “water pushes up”.

7. (a) W = m x g = 0.84 x 10 = 8.4 N [1] (b) Base area = 0.25 x 0.12 = 0.030 m² [1]. The block floats, so the upward force on the base equals the weight: p = 8.4 / 0.030 = 280 Pa [1] (c) h = p / (ρ x g) = 280 / (1000 x 10) [1] = 0.028 m [1] [5] Examiner insight: Link the weight and the base force in words (“upthrust equals weight because it floats”). Part (c) builds on the pressure from part (b), so write that value clearly and reuse it.

8. (a) Top: p = 0.50 x 1200 x 10 = 6 000 Pa [1]. Bottom depth = 0.50 + 0.40 = 0.90 m, so p = 0.90 x 1200 x 10 = 10 800 Pa [1] (b) Pressure difference = 10 800 - 6 000 = 4 800 Pa [1]. Upthrust = 4 800 x (0.30 x 0.30) = 4 800 x 0.090 = 432 N [1] (c) Weight = 40 x 10 = 400 N [1]. Resultant = 432 - 400 = 32 N upwards [1] [6] Examiner insight: The bottom face depth is top depth plus the height of the block, a step often missed. State the direction as well as the size for the final mark.

9. (a) ρ = m / V. A: 0.45 / 0.00050 = 900 kg/m³; B: 0.90 / 0.00075 = 1200 kg/m³ [1]. C: 1.50 / 0.0020 = 750 kg/m³; D: 0.30 / 0.00030 = 1000 kg/m³ [1] (b) A floats, because 900 is less than 1000 [1]. C floats, because 750 is less than 1000. B sinks, because 1200 is greater than 1000. D has the same density as water, so it hangs at any depth [1] (c) Fraction = 900 / 1000 = 0.90 [1] (d) 750 / 1000 = 0.75 [1] [6] Examiner insight: A verdict needs a comparison of densities; “A is light” does not earn the mark. Object D is the case where the densities are equal, so “floats” or “sinks” would be wrong.

10. (a) W = 840 x 10 = 8 400 N [1] (b) Upthrust must equal weight, so V = 8 400 / (1000 x 10) [1] = 0.84 m³ [1] (c) Greatest upthrust = 3.0 x 1000 x 10 = 30 000 N [1]. Greatest total mass = 30 000 / 10 = 3000 kg [1]. Cargo = 3000 - 840 = 2160 kg [1] (d) The hull is hollow and wide, so the boat displaces a large volume of water [1]. Its average density (mass of boat divided by the volume it occupies) is less than that of water, and the upthrust can equal its weight [1] [8] Examiner insight: Longer questions reward a clear chain: weight, then required upthrust, then volume. In part (d) the mark is for average density or displaced volume, not for “steel is light”.

11. (a) Pressure is directly proportional to depth [1] (b) Gradient = (4800 - 1200) / (0.40 - 0.10) [1] = 12 000 Pa/m [1] (c) Gradient = ρ x g, so ρ = 12 000 / 10 [1] = 1200 kg/m³ [1] (d) p = 0.55 x 1200 x 10 = 6 600 Pa [1] (e) The graph would still be a straight line through the origin [1] but with a smaller gradient [1] [8] Examiner insight: Take the gradient from two widely spaced points on the line, with units of Pa/m. In (a), “increases” alone is weaker than a statement that the increase is in proportion.

Where marks are usually lost

• Leaving the unit off a pressure answer, or using N or N/m instead of Pa.
• Using density in g/cm³ in the equation without converting to kg/m³.
• Using the total height of the container instead of the depth of the point.
• Forgetting to add the height of an object to find the depth of its bottom face.
• Explaining floating by “light things float” instead of comparing densities or forces.
• Describing upthrust as water “pushing up” with no pressure difference between top and bottom.
• Saying “same density” objects float or sink, rather than hang at any depth.
• Stopping after the upthrust and not comparing it with the weight.

Next steps

• Revise the key points in the revision notes.
• Re-read the worked examples in the study guide.
• Course hub: OCR GCSE Physics. Printable checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Source: OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026), first assessment 2018, published by OCR. Statements P1.3h to P1.3j.