One column, two liquids, one pressure question
A learner has just become comfortable calculating the pressure contribution of a single stationary liquid. The familiar formula is p = ρgh. Then the next diagram shows two liquids stacked in a vessel: a lighter liquid above a denser liquid, with a sensor below both. What is the density to put into the formula? A rushed answer may choose the lower liquid because the sensor sits there, or average the two densities because there are two of them. Both shortcuts can give the wrong result.
This volume develops a more demanding application: calculate the pressure contributions of separate stationary layers and add them where appropriate. The essential Physics remains familiar. A pressure increase through a liquid column depends on that column’s density, gravitational field strength and vertical thickness. What has changed is that there is no longer one uniform density over the entire depth. We must track each layer as a distinct part of the system.
How we arrived at this problem
Volume 0458 introduced density, depth, liquid pressure and manometers. Volume 0459 practised inverse density and feasible quantities. Volume 0460 turned measurements into careful statements, and Volume 0461 checked two separate uniform liquids against one model. Now we place two different liquid columns inside the same problem.
The distinction matters. In Volume 0461, each vessel contained a single liquid. Here, a lower sensor has a column of one liquid and a column of another above it. This is a new modelling challenge, not a repeat of the previous comparison. Every example below is an invented paper-based exercise. No practical mixing or construction is necessary.
The quick method for a layered column
Draw or describe the upper and lower regions. State the vertical thickness and density of each. Identify the measurement point and the pressure reference at the free surface. Work down from the top, calculating the pressure increase through each layer using density × gravitational field strength × vertical thickness. Add the contributions from the layers above the point. Convert to the required units and check whether the answer is reasonable.
Do not use the full height of the vessel for every layer. Do not assign the same density to the whole column unless that is justified. Do not call the gauge-pressure contribution absolute pressure. If the point is within the upper layer, the lower layer is not part of the fluid column above it. This simple observation is the foundation of the whole guide.
The official 2027 G3 syllabus boundary
The Singapore–Cambridge Secondary Education Certificate begins in 2027. The SEAB G3 school-candidate directory, last updated 26 October 2026, lists standalone Physics K323, replacing the earlier O-Level reference code 6091. Mathematics is K310, previously 4052, and English Language is K300, previously 1184. These are the current identifiers to use when finding the correct official material.
The K323 Physics syllabus specifically requires density = mass/volume and the liquid-column pressure relationship involving density, gravitational field strength and column height. It also covers pressure difference and manometers. Two-layer problems here are a teaching extension that applies those stated principles; “layered-liquid pressure” is not presented as a separately named syllabus topic or a guaranteed examination question. Standalone K323 has Paper 1 multiple choice, one hour, 40 marks, 30%; Paper 2 structured and free response, one hour 45 minutes, 80 marks, 50%; and Paper 3 practical, one hour 50 minutes, 40 marks, 20%. Candidates for Physics-containing Combined Science must consult their own K326/K327/K328 syllabus and actual combination rather than assuming these standalone Physics paper details apply to them.
The fictional layered vessel
Imagine a vertical container open to the atmosphere. It holds an upper 0.30 m layer of Liquid A, with density 800 kg/m³, resting above a lower 0.50 m layer of Liquid B, with density 1000 kg/m³. For these original questions, take gravitational field strength g = 10 N/kg. The liquids are assumed immiscible, at rest, arranged stably with the lower-density liquid above the denser one, and sufficiently uniform for the given density values to apply.
The total liquid depth is 0.80 m. The free surface is our zero-depth position, and pressure due to the liquid column is measured relative to the pressure at that surface. Our statements concern the elementary static model. We have not supplied a vessel wall rating, temperature dependence, real laboratory measurements or a safety certification. Those omissions are not problems to fill with guesses. They define the boundary of the calculation.
The language of the diagram
Even without a drawing, locate the important heights. The top is at depth 0.00 m. The boundary between liquids is at depth 0.30 m. The bottom is at depth 0.80 m. When the sensor is at depth 0.60 m, it has 0.30 m of Liquid A above the interface and 0.30 m of Liquid B between the interface and the sensor. When the sensor is at depth 0.20 m, it lies inside Liquid A, so the column above it is only 0.20 m of Liquid A.
Keep “total depth” separate from “thickness of each layer above the point”. The total depth is useful for locating the measurement, but a pressure calculation must account for the material actually above that measurement point. This prevents the classic mistake of putting 0.60 m into both terms and counting the same vertical distance twice.
Start at the free surface
At the top of the open container, the liquid-column gauge-pressure contribution is zero by our chosen reference. That does not mean there is no absolute pressure. The air around the vessel exerts atmospheric pressure, but its numerical value is not given. The calculation begins with the difference attributable to the liquid column, so zero at the reference surface is appropriate.
This reference matters throughout the article. A gauge pressure of 2400 Pa at the interface means the liquid column above contributes 2400 Pa relative to the pressure at the free surface. It does not mean that a physically exposed surface has an absolute pressure of only 2400 Pa. If a question explicitly supplies atmospheric pressure and asks for absolute pressure, the surface reference must be included.
Find the pressure at the interface
The boundary is 0.30 m below the surface, entirely beneath Liquid A. The model gives p_interface = ρ_A g h_A = 800 × 10 × 0.30 = 2400 Pa. In kilopascals that is 2.40 kPa. The lower liquid has not been crossed to reach this point, so its own lower column contributes nothing to the pressure increase from the surface to the interface.
That result is a checkpoint. Any full-depth calculation should include the 2400 Pa contributed by the upper layer. If a learner’s final answer is smaller than 2400 Pa at the bottom despite adding a positive depth of lower liquid, something is wrong with the method or arithmetic. Physical monotonicity, within the stated static model, is a useful check before looking at the mark scheme.
Move 0.30 m into the lower liquid
A sensor at depth 0.60 m lies 0.30 m below the interface. The upper 0.30 m layer contributes 2400 Pa. The lower 0.30 m of Liquid B contributes 1000 × 10 × 0.30 = 3000 Pa. Add them: 2400 + 3000 = 5400 Pa, or 5.40 kPa. This calculation shows why one density cannot be used blindly over the whole 0.60 m.
If the learner mistakenly calculates 800 × 10 × 0.60, the answer is 4800 Pa and neglects the denser lower region. If the learner calculates 1000 × 10 × 0.60, the answer is 6000 Pa and treats the upper layer as if it were the denser liquid. Both expressions use a familiar formula. Their failure lies in the mapping between the physical regions and the values assigned to them.
Calculate the bottom pressure contribution
At the bottom, the sensor lies beneath all 0.30 m of Liquid A and all 0.50 m of Liquid B. The pressure contribution from A is 800 × 10 × 0.30 = 2400 Pa. The contribution from B is 1000 × 10 × 0.50 = 5000 Pa. Their sum is 7400 Pa, or 7.40 kPa, relative to the free-surface pressure. That is our central worked result.
Label each term before adding. The statement “p = 2400 + 5000” is easier to check when the preceding line identifies the two columns. A final “7.4” without unit or reference is incomplete. The reader should know that the calculation concerns gauge-pressure contribution at the bottom under the stated model, not the vessel’s total weight, the total downward force on its base or the atmospheric pressure.
Check with a different route
Suppose we express the entire 0.80 m depth as an effective uniform column only for the purpose of checking the total pressure. An appropriate thickness-weighted average density is (800 × 0.30 + 1000 × 0.50)/0.80. The numerator is 240 + 500 = 740, and dividing by 0.80 gives 925 kg/m³. Then 925 × 10 × 0.80 = 7400 Pa, agreeing with the direct sum of contributions.
The effective-density route is a check, not an excuse to erase the separate layers. It works for the total pressure contribution here because the same g applies and the pressure contributions add linearly. At an intermediate depth, the relevant effective density depends on which portions of the layers lie above that position. The separate-column method remains clearer and safer for solving the actual layered problem.
Why the simple average density is wrong
One student writes (800 + 1000)/2 = 900 kg/m³ and multiplies by 10 × 0.80, obtaining 7200 Pa. The arithmetic is correct for the expression chosen, but the expression is unjustified: the two layers do not have equal thickness. A simple arithmetic mean gives them equal influence regardless of how much of the depth each occupies.
If the upper and lower layers were equal in thickness, then the simple mean would match the thickness-weighted mean in this special linear calculation. They are not equal here. The broader lesson is to ask which quantity is being averaged and which weights are appropriate. Mathematics can produce a plausible-looking number from an inappropriate model. Physics tells us which model represents the actual arrangement.
Understand pressure continuity at a boundary
In the ideal static model, the pressure at the shared horizontal interface is the same when approached immediately from above or below, neglecting special effects such as curved interfaces or surface tension. The gradient of pressure with depth changes as the fluid density changes; pressure itself does not jump merely because the material changes. At depth 0.30 m, the contribution is 2400 Pa on both sides of the idealised interface.
Imagine a graph that leapt suddenly from 2400 Pa to 3000 Pa at the boundary with no additional depth. That would not represent our chosen simple equilibrium model. The transition is better understood as a continuous line whose slope becomes steeper in the denser lower layer. This is a valuable conceptual check when working with unfamiliar graphs.
Read the pressure-depth relationship in two parts
Let h denote vertical depth below the top surface in metres. For 0 ≤ h ≤ 0.30, the sensor is in Liquid A, so p = 800 × 10 × h = 8000h Pa. For 0.30 < h ≤ 0.80, the upper layer contributes 2400 Pa and the lower portion has thickness h − 0.30. Therefore p = 2400 + 10000(h − 0.30) Pa. Simplifying gives p = 10000h − 600 Pa in that lower region.
These formulas agree at h = 0.30: 8000 × 0.30 = 2400, and 10000 × 0.30 − 600 = 2400. They give 7400 at h = 0.80. The graph is piecewise linear, with a change in gradient at the interface. This mathematical representation is not a new physical law. It is two applications of the same familiar pressure-column relationship.
A graph tells you more than a single answer
The upper segment starts at (0, 0) and reaches (0.30, 2400) when depth is measured in metres and pressure in pascals. Its gradient is 8000 Pa/m. The lower segment begins at that same point and reaches (0.80, 7400), with gradient (7400 − 2400)/(0.80 − 0.30) = 10000 Pa/m. The lower segment is steeper because the lower liquid has the greater density under the same g.
Do not confuse the steeper gradient with a sudden increase at the interface. The graph’s value is continuous. The gradient is what changes. A sentence such as “pressure increased sharply at the boundary” can suggest an instantaneous jump; prefer “the pressure increased more rapidly per metre of additional depth below the boundary”. The wording preserves the actual mathematical relationship.
Test a position inside the upper layer
At depth 0.20 m, the sensor lies above the interface. Only Liquid A contributes above that point. The model gives 800 × 10 × 0.20 = 1600 Pa, or 1.60 kPa. The denser lower liquid is present in the container but does not add a column above the sensor at 0.20 m. A learner who includes the entire bottom layer has reversed the geometry.
This is the quickest test of whether the student understands the arrangement. Ask them to name which parts of the column physically lie above the sensor before they calculate. If they can answer that in words, the correct equation is usually close. If they cannot, one more memorised formula will not repair the problem.
Test a position just below the interface
At depth 0.32 m, the upper liquid contributes 2400 Pa, and the lower liquid contributes only 0.02 m × 1000 kg/m³ × 10 N/kg = 200 Pa. The total is 2600 Pa. Compare it with the 2400 Pa at 0.30 m. Two centimetres of additional lower liquid give a 200 Pa increase in this model. That small step is consistent with a lower gradient of 10000 Pa per metre.
The pressure change is not 10000 Pa simply because the lower layer’s gradient is 10000 Pa/m. A gradient is a change per unit depth. Multiply it by the actual depth change, 0.02 m. This distinction helps prevent a common graph-reading mistake: treating the numerical gradient as though it were the vertical-axis value at each point.
Compare a uniform-water vessel
Now imagine a separate vessel with the same total depth of 0.80 m, filled entirely with water of density 1000 kg/m³. Under the same g = 10 N/kg, its bottom liquid-column contribution is 1000 × 10 × 0.80 = 8000 Pa. Our layered vessel gave 7400 Pa. The difference is 600 Pa. That result follows because 0.30 m of the denser liquid has been replaced by an equal thickness of the lighter liquid.
The comparison is not a statement about the overall safety, buoyancy or stability of real containers. It is a comparison of pressure contributions at equal total depth under explicit model conditions. A learner who reaches for “more mass” should return to the column calculation and check whether the mass in question has actually been defined.
Change the upper-layer thickness while keeping total depth fixed
Let the total depth remain 0.80 m, but call the lighter liquid’s thickness x metres. The lower liquid then has thickness 0.80 − x. The bottom pressure contribution is 800 × 10 × x + 1000 × 10 × (0.80 − x). Simplify: p = 8000x + 8000 − 10000x = 8000 − 2000x Pa. The physical domain is 0 ≤ x ≤ 0.80 for this idealised arrangement.
At x = 0, all 0.80 m is denser liquid, so p = 8000 Pa. At x = 0.80, all 0.80 m is lighter liquid, so p = 6400 Pa. At x = 0.30, p = 8000 − 600 = 7400 Pa, our original result. The formula is a clear model of the stated replacement, not a universal expression for any layered-liquid system.
A reverse problem with a target pressure
Suppose a question asks how thick the upper lighter layer must be so that the bottom gauge-pressure contribution is at most 7500 Pa, keeping the total depth and densities fixed. Use 8000 − 2000x ≤ 7500. Subtract 8000: −2000x ≤ −500. Divide by the negative number and reverse the inequality, giving x ≥ 0.25 m. The interpretation is that at least 0.25 m of the column must be the lighter liquid within the stated model and domain.
If a student forgets to reverse the inequality, they would conclude x ≤ 0.25, exactly the wrong direction. Physics offers a check: increasing the lighter layer’s thickness replaces denser liquid and reduces the model’s bottom pressure contribution. The algebraic answer must agree with that physical direction.
The importance of the domain
The expression p = 8000 − 2000x is meaningful only under the stated arrangement with fixed total depth and densities. If someone substitutes x = 1.2 m, the lower thickness becomes negative, which is physically impossible for a vessel of total liquid depth 0.80 m. A calculator will still produce a number. The model will no longer describe the problem.
Before solving, write the range of feasible values. In this case 0 ≤ x ≤ 0.80. If a computed threshold fell outside the range, the appropriate conclusion might be that no permitted arrangement meets the condition, or that every permitted arrangement does. The equation is not an invitation to invent an impossible layer.
Why buoyancy is not the immediate answer
A student recognises different densities and begins discussing floating objects and upthrust. Those ideas can be interesting, but the question asks for pressure at a depth inside stationary liquid layers. The necessary model is the pressure contribution of the columns above that point. Introducing a second physical system before solving the first can distract the learner and create unsupported assumptions.
The official K323 pressure outcomes name density, pressure, hydraulics, liquid columns, atmospheric-pressure measurement and manometers. Our layered-column exercise deliberately builds on those specified outcomes. Treat additional topics as enrichment only when they are useful and correctly introduced; do not claim that every extra concept is a separately assessed requirement.
Separate the layer arrangement from chemistry
Our two liquids are imagined as uniform, immiscible and at rest. That is a modelling assumption. A real pair of liquids may mix, react, form an emulsion, or change density with temperature or composition. Those possibilities do not belong in the numerical answer unless the question specifies them. If the liquid densities are not constant or a stable interface is absent, the simple two-layer representation may not apply.
This is why questions often include words such as “uniform”, “stationary” and “immiscible”. They establish the conditions under which the calculation is intended. Underline them. A reader who treats these words as background decoration risks applying a formula to a different physical situation than the one described.
A height is not always a depth
Suppose a diagram labels the lower layer as 0.50 m tall and marks the sensor 0.20 m above the bottom. The amount of lower liquid above the sensor is 0.50 − 0.20 = 0.30 m, not 0.20 m. The upper layer remains 0.30 m. Thus the pressure contribution is 2400 + 3000 = 5400 Pa. The same position is at total depth 0.60 m below the free surface.
The distinction between height from the base and depth from the surface is a frequent source of wrong substitution. Before calculating, convert every position into one shared coordinate convention. Use either depth downward from the surface or height upward from the base, but do not mix the two halfway through your working.
Force on a surface is a separate question
Pressure describes force per unit area at a point, or locally over a small area. To find a force on a horizontal base at uniform bottom pressure, one may use force = pressure × area, with the appropriate pressure difference and conditions. But a vertical wall spans different depths, so the pressure varies along it; a single bottom-pressure value does not describe the pressure at every height on that wall.
Do not turn a bottom reading of 7400 Pa into a total wall force without being given the relevant geometry and a suitable method. The presence of a wide container does not alter our point-pressure calculation at fixed depth and fluid arrangement, but area matters when changing the question to force. Distinguishing pressure and force is central to avoiding plausible yet invalid generalisations.
A manometer is not the same as this layered vessel
A U-tube manometer uses height differences in an appropriate fluid to compare pressures. Our layered vertical vessel involves pressure contributions from two different fluids arranged in one vertical column. Both use hydrostatic reasoning, but the diagrams and reference levels may differ. If a manometer has fluids of different densities in its arms, the reasoning requires careful identification of connected levels and each relevant segment; a memorised one-liquid shortcut can fail.
For a G3 learner, the safest approach is to identify known pressures at suitable reference points and account for pressure changes when moving vertically through each fluid. Do not infer a manometer result from the horizontal distance between the arms. The difference in levels and the fluid properties are the essential quantities in the simple cases described by the syllabus.
Units that protect the calculation
Density in kg/m³, g in N/kg and vertical thickness in m produce pressure in N/m², or Pa. To express 7400 Pa in kPa, divide by 1000: 7.40 kPa. The upper-layer density 800 kg/m³ equals 0.800 g/cm³. A thickness of 30 cm equals 0.30 m. If a task gives 500 mm for the lower layer, convert it to 0.50 m before substitution.
A quick unit check can expose an error earlier than a full recalculation. If a student uses 30 rather than 0.30, the upper-layer contribution becomes 240000 Pa, one hundred times the intended value. If the final result is unreasonably large for the given short column, return to the units before questioning the physical law.
Decide how to round
In our fictional calculation, the values are chosen to produce exact results with the given round numbers. A real question may specify significant figures or a different value of g. Follow the paper’s instructions. For K310 Mathematics, SEAB notes that non-exact numerical answers should generally be given to three significant figures, and angles in degrees to one decimal place, unless a different accuracy is specified. Physics questions may state their own conventions and input precisions. Do not impose an unrelated Mathematics rule on every scientific answer.
Keep enough digits in intermediate calculations to avoid rounding error. Write units beside the final quantity. Explain any chosen approximation when it matters to a conclusion. If a problem explicitly gives g = 9.8 N/kg, do not silently replace it with 10 merely because an earlier example did so. The gravitational field strength belongs to the question, not to your preference.
Exam-style task one: interface pressure
Question: A top layer has density 700 kg/m³ and thickness 0.25 m. Take g = 10 N/kg. What is its contribution to pressure at the interface? Answer: 700 × 10 × 0.25 = 1750 Pa. A learner who uses the density of a separate lower liquid has included irrelevant information. The point is reached before the lower layer is crossed.
Next question: a denser lower layer of density 1100 kg/m³ extends another 0.40 m downward. What is the contribution at the bottom? Calculate 1750 + 1100 × 10 × 0.40 = 1750 + 4400 = 6150 Pa. State the reference and conditions. The two linked questions test whether the learner can build the full answer from a known intermediate position.
Exam-style task two: inverse depth
Our original layered vessel has a gauge-pressure contribution of 6400 Pa at some point. Is that point within the upper or lower layer? The interface pressure is only 2400 Pa, so 6400 Pa must occur in the lower region. Solve 2400 + 10000(h − 0.30) = 6400. This gives 10000(h − 0.30) = 4000, hence h − 0.30 = 0.40 and h = 0.70 m.
Check the depth is between 0.30 and 0.80 m. Substitute into the piecewise formula: 10000 × 0.70 − 600 = 6400 Pa. This double check combines a physical-region decision with algebra. Solving a correct equation on the wrong branch would be a model error even if the arithmetic were flawless.
Exam-style task three: a mistaken graph
A graph is described as a straight line from (0, 0) to (0.80, 7400), with no change in gradient at 0.30 m. Does it represent the pressure at every depth in the layered column? It passes through the correct endpoints but does not reproduce the interface value of 2400 Pa: the straight-line estimate at 0.30 m would be 2775 Pa. Therefore it is not the correct pointwise model for the distinct layers, even though it matches the bottom pressure.
This is an important warning. Matching one final answer does not establish that the intermediate reasoning is valid. A model should reproduce the relevant conditions and known intermediate quantities, not merely the final endpoint. It is possible to fit a single number and still misunderstand the system.
Exam-style task four: reverse the layer order
Suppose someone draws the denser liquid above the lighter liquid while still assuming the system remains permanently at rest with a perfectly stable horizontal interface. In ordinary gravity, that arrangement may not be stable to disturbances if the liquids are freely able to move. A thoughtful reader should notice the physical assumption rather than blindly copying the earlier equilibrium solution. A question may deliberately impose an idealised arrangement, but the learner should state the assumption under which the model is used.
For our main calculations, we deliberately chose the lower-density liquid above the higher-density liquid and assumed a stable static configuration. The reason to mention the physical setting is not to turn the guide into an advanced fluid-instability chapter. It is to remind the student that equations must describe a plausible or explicitly idealised situation.
Exam-style task five: compare models rather than slogans
A student says, “The pressure at the bottom is the density of the lower liquid multiplied by the total depth and g.” In our model, that gives 1000 × 0.80 × 10 = 8000 Pa. The correct value is 7400 Pa. The shortcut overestimates because it assigns the denser liquid’s density to the entire upper layer. A clear explanation names the substituted quantity that was wrong, not merely the numerical difference.
Another says, “Use the upper density over the whole depth.” That gives 6400 Pa and underestimates because the lower layer is denser. These two wrong answers form a useful bracket around 7400 Pa. If the densities lie between 800 and 1000 and each layer has positive thickness, the effective density should lie between the two. A bottom contribution outside 6400–8000 Pa is a reason to check the working.
A text-only worked graph description
At a depth of 0.00 m, liquid-column gauge pressure is 0 Pa. At 0.10 m it is 800 Pa. At 0.20 m it is 1600 Pa. At the interface, 0.30 m, it is 2400 Pa. At 0.40 m it is 3400 Pa. At 0.60 m it is 5400 Pa. At the bottom, 0.80 m, it is 7400 Pa. Notice that a further 0.10 m in the upper liquid adds 800 Pa, while a further 0.10 m in the lower liquid adds 1000 Pa.
A learner can reconstruct the entire piecewise graph from these verbal coordinate pairs. There is no need for an illustration to understand the relationship. The pressure rises continuously and changes slope at the interface. If the problem asks you to describe the graph, use that sentence before launching into arithmetic.
A careful experimental limitation
The case assumes two perfectly uniform stationary layers. We did not measure their actual densities, the temperature distribution, atmospheric pressure or sensor calibration. A real investigation would need appropriate apparatus and school supervision. The example should therefore be described as an idealised model rather than a verified reading from a laboratory or a claim about actual industrial vessels.
This is not an excuse to avoid the calculation. The question supplies an idealised system so the learner can apply a specific principle. Solve that system accurately. Then recognise the difference between the model and a real-world engineering decision. Strong Physics combines confidence inside the stated assumptions with caution outside them.
Why this matters to English explanations
A scientifically correct answer can become misleading in ordinary language. “Pressure jumps at the interface” suggests a discontinuity that our model does not contain. Better: “Pressure rises more rapidly with depth in the denser lower layer, while the value remains continuous at the boundary.” That sentence preserves both the graph shape and the physical interpretation. It uses ordinary prose to carry a mathematical distinction.
Likewise, “the bottom is under 7400 Pa of pressure” can be misread as total absolute pressure. Name the quantity: “the liquid-column contribution to gauge pressure at the bottom is 7400 Pa relative to the free-surface pressure.” The extra words are worth using when the reference matters. For general audiences, explain the reference in simpler terms without dropping it.
A fifteen-minute practice sequence
Spend three minutes drawing the layers with depths and densities in words. Spend four minutes finding the interface and bottom contributions. Spend three minutes constructing the two pressure-depth expressions. Spend three minutes solving an inverse-depth question and checking the branch. Use the final two minutes to explain the pattern without claiming the interface causes a sudden pressure jump. These are suggested practice stages, not mandated national-examination timing.
If the learner makes an error, do not simply repeat the calculation with the same numbers. Find the first wrong decision: incorrect depth reference, wrong density, inappropriate averaging, missing unit conversion or equation used outside its domain. Then change the layer thicknesses and test again. The goal is a method that survives a new diagram.
Build a transfer problem on your own
Choose two positive densities and two positive thicknesses. Place the lower-density liquid on top under an assumed stable static arrangement. Give a value of g and calculate the interface and bottom contributions. Create one intermediate sensor position within each layer. Now calculate the corresponding pressures and describe the graph. Finally, explain what would change if the lower layer became thicker while total depth remained fixed.
Check your constructed problem using three rules: pressure contribution should be zero at the top reference; it should increase with depth for positive density and gravity; and it should remain continuous at the interface in this simple model. These checks make the practice independent of a remembered final numerical answer.
Subject owners and the primary-to-secondary bridge
For the wider Physics concepts, use the eduKate Sengkang Physics hub and Science Hub. For equations, graphs, ratios and the next inverse problem, use the Secondary Mathematics capability map. For explaining an intermediate result accurately, use the Secondary English Learning Hub.
A student still struggling to identify quantities can return to the PSLE Learning Guide or the PSLE Science Learning Guide. Those guides reinforce describing a situation before choosing a method. Upper-secondary work adds more variables, units and abstraction, but the old foundation remains useful. The subject of this volume is the layered hydrostatic model; the other hubs remain owners of the broader English and Mathematics teaching.
What to carry into the next volume
Keep three results in mind. The upper layer contributes 2400 Pa at the interface. The complete column contributes 7400 Pa at the bottom. The pressure-depth graph changes gradient at the interface but remains continuous in the stated model. Those facts are different views of the same system. If a later calculation contradicts them, review the geometry before seeking a more complicated formula.
Use the 2027 SEAB G3 syllabus directory and K323 Physics syllabus for the actual examination standard and scope. The next Mathematics volume will turn this piecewise Physics model into an inverse-graph and constraint problem. Our immediate learning target remains simple: know which liquid lies above the point, add the proper column contributions, and let the units and boundary values check the result.
