Density, pressure and fluid pressure: worked solutions

Answer: V = 33.5 cm³; ρ = 2.61 × 10³ kg/m³. Read the bottom of the water meniscus at eye level. An attached bubble makes the calculated density too low.

1. The displaced volume is V = 71.5 − 38.0 = 33.5 cm³.This equals the stone’s volume because it is fully immersed and does not absorb water.
2. m = 87.5 × 10⁻³ = 0.0875 kg;
   V = 33.5 × 10⁻⁶ = 3.35 × 10⁻⁵ m³.A centimetre is 10⁻² m, so a cubic centimetre is (10⁻²)³ = 10⁻⁶ m³.
3. ρ = m/V = 0.0875/(3.35 × 10⁻⁵) = 2611.94… ≈ 2.61 × 10³ kg m⁻³.
4. Read the bottom of the water meniscus at eye level, with the cylinder upright on a level surface. This avoids viewing the level from above or below.
5. The attached bubble makes the measured displacement larger than the stone’s true volume. The mass is essentially unchanged, so dividing by an overestimated volume gives a density that is too low. Remove bubbles and repeat the measurement.

Check: 87.5/33.5 = 2.61 g/cm³, which is 2610 kg/m³. This is greater than water’s density, so an unsupported stone of this density would sink.

Mark checklist · 5 marks

• 1 mark — Obtains displaced volume 71.5 − 38.0 = 33.5 cm³.
• 1 mark — Converts both mass and volume correctly to kg and m³; accept a correct g/cm³ calculation followed by multiplying density by 1000.
• 1 mark — Uses ρ = m/V to obtain 2.61 × 10³ kg/m³; allow 2.6 × 10³ with correct working. Carry a numerical subtraction error into subsequent correct steps.
• 1 mark — Reads the bottom of the water meniscus at eye level.
• 1 mark — Links the bubble to too large a displaced volume and hence too low a calculated density.

Common mistake: Using 71.5 cm³ includes the original water volume. Subtract the initial reading first. Treating 1 cm³ as 10⁻² m³ also gives a large error: cube the length conversion.

Exam technique: Write each conversion explicitly. The 2022 report’s unit lesson transfers to this displacement task; its original density calculation was for oil in a cylinder.

Related examiner evidence: OCR H156/02, June 2022 — Q1(a)–(b); report p. 5

Answer: d = (12.44 ± 0.04) mm; V = 1.008 × 10⁻⁶ m³; ρ = 7.84 × 10³ kg/m³; percentage uncertainty ≈ 1.22%.

1. d̄ = (12.40 + 12.44 + 12.48 + 12.42 + 12.46 + 12.44)/6 = 12.44 mm.Δd = (12.48 − 12.40)/2 = 0.04 mm.Thus d = (12.44 ± 0.04) mm using the stated estimate.
2. V = π(12.44 × 10⁻³)³/6 = 1.00799… × 10⁻⁶ m³.Alternatively use r = 6.22 mm in 4πr³/3; both give the same volume.
3. ρ = (7.90 × 10⁻³)/(1.00799… × 10⁻⁶) ≈ 7.84 × 10³ kg m⁻³.Retain the unrounded volume for this step.
4. Percentage uncertainty in d = (0.04/12.44) × 100 = 0.3215…%;
   percentage uncertainty in V = 3 × 0.3215… = 0.9646…%.The constant π/6 has no measurement uncertainty.
5. Percentage uncertainty in m = (0.02/7.90) × 100 = 0.2532…%;
   percentage uncertainty in ρ ≈ 0.2532… + 0.9646… = 1.22%.Division still adds percentage uncertainties under this method. They do not subtract.
6. Different directions sample small departures from a perfect sphere and any variation in its diameter. Averaging gives a more representative diameter. Repeating at one orientation alone would miss this variation. Check the micrometer’s zero before use; repeating readings cannot remove a zero error.

Check: The density is 7.84 g/cm³, a plausible value for a dense metal. Using radius also gives Δr = 0.02 mm and Δr/r = Δd/d, so the uncertainty result agrees.

Mark checklist · 7 marks

• 1 mark — Mean diameter 12.44 mm.
• 1 mark — Half-range absolute uncertainty 0.04 mm.
• 1 mark — Uses V = πd³/6 correctly with d = 0.01244 m, or the equivalent radius method.
• 1 mark — Density 7.84 × 10³ kg/m³; allow 7.8 × 10³. Carry a prior numerical mean/volume error forward in a correct calculation.
• 1 mark — Triples the diameter percentage uncertainty for volume, obtaining approximately 0.965%.
• 1 mark — Adds mass percentage uncertainty approximately 0.253% to obtain approximately 1.22%; allow 1.2%.
• 1 mark — Explains that different directions account for variation in the sphere’s diameter/shape.

Common mistake: Using πd³ as the sphere volume drops the factor 1/6. Dividing percentage uncertainties because density is m/V is also incorrect: the estimated contributions add, with the diameter contribution multiplied by three.

Exam technique: Identify what each uncertainty describes. Half range is specified for these readings; do not apply it automatically to every instrument measurement. The original 2025 task found mass from known density; here the same volume and uncertainty skills are used to find density.

Related examiner evidence: OCR H156/02, June 2022 — Q1(a)–(b); report p. 5 · OCR H156/02, June 2025 — Q2(a); report pp. 7–8

Answer: Contact pressure = 2.40 × 10⁴ Pa; turning the box increases pressure by a factor of 2.50. Absolute gas pressure = 5.73 × 10⁴ Pa.

1. A = 30.0 × 10⁻⁴ = 3.00 × 10⁻³ m²;
   p = 72.0/(3.00 × 10⁻³) = 2.40 × 10⁴ Pa.This is average normal force per unit contact area.
2. p_new/p_old = A_old/A_new = 30.0/12.0 = 2.50.The new pressure is 6.00 × 10⁴ Pa. The weight and normal force need not increase to make the pressure increase.
3. d = 0.0200 m; r = 0.0100 m;
   A = π(0.0100)² = 3.14159… × 10⁻⁴ m².
4. The piston is stationary. The outward force from the gas balances the inward applied force: p_gasA = 18.0 N;
   p_gas = 18.0/(3.14159… × 10⁻⁴) ≈ 5.73 × 10⁴ Pa.The vacuum pressure is taken as zero, so this is absolute gas pressure.

Check: 1 Pa = 1 N/m². A smaller contact area gives larger pressure. The calculated gas pressure can be below atmospheric pressure because the opposite face is exposed to vacuum.

Mark checklist · 6 marks

• 1 mark — Converts 30.0 cm² to 3.00 × 10⁻³ m².
• 1 mark — Uses p = F/A to obtain 2.40 × 10⁴ Pa; carry an area conversion error into a correct division.
• 1 mark — Pressure increases by factor 30.0/12.0 = 2.50.
• 1 mark — Uses piston radius 0.0100 m or equivalent area πd²/4.
• 1 mark — Obtains piston area approximately 3.14 × 10⁻⁴ m².
• 1 mark — Balances pgasA with 18.0 N to obtain 5.73 × 10⁴ Pa absolute; allow 5.7 × 10⁴ with correct working.

Common mistake: Using the diameter in πr² makes the area four times too large and the pressure four times too small. Square the centimetre-to-metre conversion for an area.

Exam technique: Label the circular face area before substituting. The report supports the p = F/A and diameter-to-area step. Vacuum versus atmosphere is an additional PhysicsUK teaching comparison.

Related examiner evidence: OCR H156/01, June 2023 — Q11; report p. 14

Answer: Gauge pressure = 1.77 × 10⁴ Pa; absolute pressure = 1.19 × 10⁵ Pa. Resultant pressure force = 21.2 N towards the dry compartment.

1. The relevant h is the vertical depth below the free surface: p_gauge = ρgh = 1000 × 9.81 × 1.80 = 17658 Pa ≈ 1.77 × 10⁴ Pa.
2. p_absolute = 101000 + 17658 = 118658 Pa ≈ 1.19 × 10⁵ Pa.Gauge and absolute pressure describe the same location but use different zero references.
3. A = 12.0 × 10⁻⁴ = 1.20 × 10⁻³ m²;
   F_net = (118658 − 101000)(1.20 × 10⁻³) = 21.1896 N ≈ 21.2 N.The water-side pressure is larger, so the force is towards the dry compartment.
4. Multiplying absolute water pressure by area gives about 142 N from the water side alone. The compartment air contributes about 121 N in the opposite direction; their difference is 21.2 N. Retain unrounded values before subtracting.

Check: At zero depth the absolute pressure would equal atmospheric pressure and the net force would be zero. Doubling depth would double the gauge pressure and net force, while the atmospheric contribution stays the same.

Mark checklist · 5 marks

• 1 mark — Uses p = ρgh with h = 1.80 m.
• 1 mark — Gauge pressure approximately 1.77 × 10⁴ Pa.
• 1 mark — Adds atmospheric pressure to obtain absolute pressure approximately 1.19 × 10⁵ Pa.
• 1 mark — Uses the pressure difference and area 1.20 × 10⁻³ m² to obtain 21.2 N; carry an earlier numerical hydrostatic-pressure error forward.
• 1 mark — Force is towards the dry compartment, away from the higher-pressure water side.

Common mistake: Using absolute water pressure alone for the net force ignores the pressure on the reverse side. Also, h is vertical depth; it is not the distance along a sloping wall.

Exam technique: This is original PhysicsUK guidance for the specification’s p = hρg outcome. State whether pressure is gauge, absolute or a pressure difference; no specific examiner-report claim is attached to this calculation.

Answer: Best-fit gradient = 9.80 kPa/m; ρ = 999 kg/m³. The intercept is +0.25 kPa, suggesting a positive zero offset. Density uncertainty ≈ ±41 kg/m³; 1000 kg/m³ is consistent.

1. Use the supplied best-fit points, not individual scattered readings: gradient = (5.15 − 1.23)/(0.500 − 0.100) = 9.80 kPa m⁻¹ = 9800 Pa m⁻¹.
2. ρ = gradient/g = 9800/9.81 = 998.98… ≈ 999 kg m⁻³.The gradient has units Pa/m = N/m³; dividing by g gives kg/m³.
3. intercept = 1.23 − 9.80(0.100) = +0.25 kPa.With a correct depth origin, this suggests a positive additive zero error in a sensor meant to read gauge pressure. Check the reading at the free surface (true gauge pressure zero) and correct the offset; a constant vertical offset does not change the gradient.
4. steepest gradient = (5.25 − 1.17)/(0.500 − 0.100) = 10.2 kPa m⁻¹;
   Δgradient = 10.2 − 9.80 = 0.40 kPa m⁻¹ = 400 Pa m⁻¹.Use the difference from the best fit, not 10.2 itself, as the absolute gradient uncertainty.
5. Δρ = 400/9.81 = 40.77… ≈ 41 kg m⁻³.This gives approximately ρ = (999 ± 41) kg m⁻³, or an interval of about 958–1040 kg m⁻³. Since 1000 lies within it, that density is consistent with the data and stated estimate. This does not prove that the liquid is water.
6. Place a vertical ruler beside the vessel and read both the free-surface level and sensor-opening centre at eye level, then take their vertical difference. Use the opening, rather than an arbitrary point on the casing, as the depth reference.

Check: The liquid gains about 9.8 kPa per metre, close to water’s ρg. The 0.25 kPa offset affects p/h from a single point but cancels when taking Δp/Δh.

Mark checklist · 7 marks

• 1 mark — Best-fit gradient 9.80 kPa/m from well-separated points on the fitted line.
• 1 mark — Converts to 9800 Pa/m and uses gradient = ρg.
• 1 mark — Density approximately 999 kg/m³; accept 1000 with correct working.
• 1 mark — Positive intercept suggests a positive constant zero offset in the gauge sensor if depth origin is correct; accept an equivalent explanation of non-zero reading at zero depth.
• 1 mark — Uses gradient difference 0.40 kPa/m to obtain density uncertainty approximately 41 kg/m³.
• 1 mark — Concludes that 1000 kg/m³ is consistent because it is inside the density interval; accept an equivalent difference-versus-uncertainty comparison.
• 1 mark — Measures the vertical difference from free surface to opening centre using a vertical ruler, viewing both levels at eye level.

Common mistake: Using p/h from one reading folds the zero offset into the density. Another error is dividing 9.80 by 9.81 without converting kPa to Pa, giving a density 1000 times too small.

Exam technique: Use fitted-line points and state gradient units. This pressure experiment is an original application of OCR’s graph and uncertainty outcomes; it is not a reconstruction of the density experiments discussed in the reports.

Answer: Top gauge pressure = 3.53 kPa; bottom gauge pressure = 4.47 kPa. Upthrust = 2.35 N upwards; U = ρgAH = ρgVdisplaced.

1. The bottom depth is h_bottom = 0.300 + 0.0800 = 0.3800 m.p_top,gauge = 1200 × 9.81 × 0.300 = 3531.6 Pa ≈ 3.53 kPa;
   p_bottom,gauge = 1200 × 9.81 × 0.3800 = 4473.36 Pa ≈ 4.47 kPa.
2. A = 25.0 × 10⁻⁴ = 2.50 × 10⁻³ m².The gauge-pressure contribution at the top is 3531.6A = 8.829 N downwards; at the bottom it is 4473.36A = 11.1834 N upwards.
3. U = (p_bottom − p_top)A = (4473.36 − 3531.6)(2.50 × 10⁻³) = 2.3544 N ≈ 2.35 N upwards.Side-pressure forces are horizontal and cancel in pairs in this uniform, static liquid.
4. For a top face at depth h and bottom face at h + H, the absolute pressures are p₀ + ρgh and p₀ + ρg(h + H). Hence U = [(p₀ + ρg(h + H)) − (p₀ + ρgh)]A = ρgHA.The common p₀ cancels.
5. The block displaces volume V = AH, so U = ρgV = (mass of displaced liquid) × g.This is Archimedes’ principle: upthrust equals the weight of fluid displaced. The block’s density is not needed to find this upthrust.

Check: The displaced volume is AH = 2.00 × 10⁻⁴ m³. Direct use of 1200 × 9.81 × 2.00 × 10⁻⁴ gives the same 2.35 N.

Mark checklist · 6 marks

• 1 mark — Top gauge pressure approximately 3.53 kPa.
• 1 mark — Bottom depth 0.3800 m and gauge pressure approximately 4.47 kPa.
• 1 mark — Uses the difference in pressure forces with area 2.50 × 10⁻³ m².
• 1 mark — Obtains 2.35 N upwards; carry earlier numerical pressure errors into a correct pressure difference.
• 1 mark — Derives U = ρgHA = ρgV and identifies ρVg as displaced-liquid weight.
• 1 mark — Shows or explains cancellation of equal atmospheric-pressure contributions on top and bottom.

Common mistake: Adding the top and bottom force magnitudes ignores their opposite directions. Using the block’s density in ρVg would calculate its weight, rather than the weight of liquid displaced.

Exam technique: Name the displaced fluid and show the intermediate volume or pressure-force calculation. The 2022 report’s sphere calculation illustrates the same Archimedes principle; this block derivation is a different teaching example.

Related examiner evidence: OCR H156/02, June 2022 — Q1(c); report p. 6

Answer: Weight is downwards; tension and upthrust are upwards. U = 1.47 N; V = 1.50 × 10⁻⁴ m³; ρobject = 4.01 × 10³ kg/m³. The deeper reading remains 4.42 N.

1. The three forces are weight W downwards, tension T upwards and upthrust U upwards. The object is at rest, so the vertical resultant is zero: T + U = W; U = 5.89 − 4.42 = 1.47 N.
2. V = U/(ρ_waterg) = 1.47/(1000 × 9.81) = 1.49847… × 10⁻⁴ m³ ≈ 1.50 × 10⁻⁴ m³.Because the object is fully immersed, the displaced-water volume equals its full volume.
3. m = W/g = 5.89/9.81 = 0.600408… kg;
   ρ_object = m/V = 4006.80… ≈ 4.01 × 10³ kg m⁻³.Combining the equations also gives ρ_object = ρ_waterW/(W − T).
4. Moving deeper does not change the displaced volume or water density in the stated model. Therefore upthrust stays at 1.47 N and T = W − U = 4.42 N.The object can remain at rest even though U < W, because the string supplies the remaining upward force.

Check: The inferred object density is greater than water’s. Without the string the object would not float; its weight is greater than the available fully immersed upthrust.

Mark checklist · 6 marks

• 1 mark — Names weight downwards, tension upwards and upthrust upwards; accept an accurately labelled free-body diagram.
• 1 mark — Uses equilibrium to obtain upthrust 5.89 − 4.42 = 1.47 N.
• 1 mark — Uses U = ρwaterVg to obtain volume approximately 1.50 × 10⁻⁴ m³.
• 1 mark — Uses W = mg to obtain mass approximately 0.600 kg.
• 1 mark — Uses m/V, or the combined force-reading ratio, to obtain density approximately 4.01 × 10³ kg/m³; carry an earlier numerical volume error forward.
• 1 mark — Reading remains 4.42 N because upthrust is unchanged at the same fully immersed volume and fluid density.

Common mistake: Calling 4.42 N the object’s new weight confuses tension with gravitational force. Its mass and W = mg have not changed. Upthrust need not equal weight when an additional support force acts.

Exam technique: Draw the forces before using Archimedes’ principle. The report lesson identifies displaced-fluid weight; the force-meter context and equilibrium reasoning here are original PhysicsUK guidance.

Related examiner evidence: OCR H156/02, June 2022 — Q1(c); report p. 6

Answer: ρblock = 800 kg/m³; fraction submerged = 0.800. Maximum extra mass = 0.600 kg. With a 0.300 kg load, x = 1.50 cm in fresh water; x increases in sea water.

1. ρ_block = 2.40/(3.00 × 10⁻³) = 800 kg m⁻³.For the unloaded floating block, ρ_waterV_subg = ρ_blockV_totalg, so V_sub/V_total = 800/1000 = 0.800.Thus 80.0% of its volume is submerged, not 80.0% of its weight balanced.
2. At the top-face loading limit the full block volume is submerged: m_total,max = ρ_waterV_total = 1000 × 3.00 × 10⁻³ = 3.00 kg;
   m_load,max = 3.00 − 2.40 = 0.600 kg.Upthrust then is 29.43 N. Extra mass beyond this cannot be supported by the block’s buoyancy in the stated model.
3. With the 0.300 kg load, use total mass m_total = 2.40 + 0.300 = 2.70 kg;
   V_sub = 2.70/1000 = 2.70 × 10⁻³ m³;
   h = V_sub/A = (2.70 × 10⁻³)/(2.00 × 10⁻²) = 0.135 m.
4. x = H − h = 0.150 − 0.135 = 0.0150 m = 1.50 cm.This is the height of the top face above the surface, sometimes called freeboard.
5. In sea water the weight is unchanged. The same upthrust therefore requires a smaller displaced volume because ρ is larger. Less of the block is submerged, so x increases. An optional numerical check is x = 0.150 − 2.70/(1025 × 0.0200) = 0.0182927… m ≈ 1.83 cm.The upthrust still equals 2.70 × 9.81 = 26.487 N; it does not increase for the new floating equilibrium.

Check: The loaded displaced volume is 90.0% of the total, below the 100% loading limit. The total supported weight is 26.5 N in either liquid; the equilibrium submerged volume adjusts.

Mark checklist · 7 marks

• 1 mark — Block density 800 kg/m³.
• 1 mark — Uses floating equilibrium to find submerged fraction 0.800 or 80.0%.
• 1 mark — At full immersion the maximum total supported mass is 3.00 kg.
• 1 mark — Subtracts block mass to obtain maximum additional mass 0.600 kg.
• 1 mark — Includes the load and finds submerged depth 0.135 m; accept equivalent displaced-volume working.
• 1 mark — Subtracts from total height to obtain x = 0.0150 m or 1.50 cm; carry a prior numerical displaced-volume error forward if the method remains physically applicable.
• 1 mark — States x increases and links unchanged weight/upthrust with a smaller displaced volume in denser sea water.

Common mistake: Using only the wooden block’s mass omits the load. Substituting total volume for submerged volume gives maximum upthrust, not the actual upthrust of a partly immersed floating block. Saying “sea water gives more upthrust” misses the new equilibrium: it floats higher until upthrust again equals the same weight.

Exam technique: Include every supported mass, distinguish submerged depth from height above the surface, and connect a density comparison to displaced volume. These are the skills emphasised in the 2025 loaded-cylinder report; the block, loading-limit calculation and data here are original.

Related examiner evidence: OCR H156/02, June 2025 — Q2(b); report pp. 10–11