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How to Perform in the new G3 SEC Examinations | Learner’s Guide Vol 0108 | Science Physics: Density by Water Displacement — Read the Difference, Not the Final Level

G3 SEC Science Physics | Learner’s Guide Vol 0108. Read the initial and final volume, calculate the displaced difference and explain the density of the same specimen.

A stone sinks into a measuring cylinder, and the water level rises from 25 mL to 34 mL. What is the stone’s volume? It is not 34 mL. That is the final reading for the water and displaced space. The volume attributed to the submerged stone in this simple method is the change: 34 − 25 = 9 mL, equivalent to 9 cm³.

This small subtraction is the foundation of a dependable G3 SEC Physics practical skill. Density is mass divided by volume, but the useful work begins before the formula. You must identify the object whose mass is measured, determine its volume using a suitable method, keep the units compatible and check whether the measurements genuinely belong to the same sample.

Learner’s Guide Vol 0108 shows you how to interpret water-displacement measurements of an irregular solid, read cylinder graduations correctly, convert between common density units, notice why small volume differences can magnify uncertainty and give clear Physics explanations. You will work through original examples and independent questions. The examples describe supervised educational measurements; they are not a set of instructions for unsupervised laboratory work.

Chapter index

1. What density tells you

Density describes mass per unit volume. In a simple school calculation, write density = mass ÷ volume. If an object has mass 72 g and volume 9 cm³, its density is 72/9 = 8 g/cm³. The units matter: grams measure mass, while cubic centimetres measure volume.

A higher density means more mass is contained in a given volume of material. It does not automatically mean an object has more total mass. A small piece of a dense material can have less mass than a very large piece of a less dense material. Size and density answer different questions.

Consider two solid blocks. Block A has mass 40 g and volume 10 cm³, giving density 4 g/cm³. Block B has mass 90 g and volume 30 cm³, giving density 3 g/cm³. Block B is heavier, but Block A is denser. A learner who uses weight or mass alone as a density comparison has missed the denominator.

Density is a ratio, so both parts must refer to the same substance or object under the stated conditions. Mass from one specimen and volume from another cannot produce a meaningful density of either specimen without a valid model connecting them. This is a measurement-ownership issue, not an arithmetic one.

The equation can be rearranged: mass = density × volume, and volume = mass ÷ density. In exam problems you may know two of the three quantities and be asked for the third. Read which is missing before substituting.

A strong first sentence names the property: “The density of the solid is its mass divided by the volume it occupies.” Then calculate using appropriate units. The explanation does not need to be complicated to be precise.

2. Verify the current SEC Physics route

For the 2027 Singapore-Cambridge SEC, the SEAB G3 school-candidate syllabus index lists combined Science (Physics/Chemistry) K326 and combined Science (Physics/Biology) K327. Both contain Physics. Combined Science (Chemistry/Biology) K328 does not include a Physics paper; students taking that combination should follow their own Science content. Pure Physics has its own G3 syllabus K323.

The official K326–K328 combined Science syllabus includes density under Physics Force and Pressure and expects candidates to recall and apply density = mass/volume. Its practical assessment notes explicitly include determining the density of a liquid or of a regularly or irregularly shaped solid that sinks in water. This makes water displacement an appropriately grounded example, not a forecast of exactly what any future practical paper will contain.

For combined Science, Paper 1 lasts 1 hour and carries 40 multiple-choice marks, weighted at 20%. Candidates take the two relevant disciplinary structured-and-free-response papers, each lasting 1 hour 15 minutes and carrying 65 marks, weighted at 32.5% each. Paper 5 is a practical test lasting 1 hour 30 minutes, worth 30 marks and weighted at 15%. Physics students taking K326 or K327 sit the Physics component in Paper 2; K328 students do not.

The official assessment includes observing, measuring, handling data and evaluating methods. Calculating a density is one skill; knowing whether a volume measurement is appropriate and how its uncertainty affects the result is another. Both deserve practice.

Use the Science Hub for the complete subject route and the G3 SEC learner-guide index to follow other chapters. This article owns the narrow concept of identifying an irregular solid’s volume from the change in fluid level.

3. Read initial and final volumes as different measurements

The volume-displacement method relies on the idea that an object submerged in a liquid occupies space that the liquid cannot occupy at the same time. In a suitable, correctly arranged measuring cylinder, the increase in the reading corresponds to the volume of the submerged object, assuming conditions such as full submersion and no water loss are met.

Suppose the initial water reading is 25.0 mL. After a solid is appropriately submerged, the reading is 34.0 mL. Volume of the solid = final reading − initial reading = 34.0−25.0=9.0 mL. For volume measurements, 1 mL equals 1 cm³, so the volume is 9.0 cm³.

The subtraction is not a guess about what fraction of the cylinder looks occupied. It is a comparison of two readings from the same measurement setup at different states. One reading represents water alone. The other represents water plus the displacement of the submerged solid. Their difference isolates the change in occupied volume.

If an object is only partly submerged, the change tells you the volume of the submerged portion, not necessarily the whole object. For density of the whole solid, the whole object volume is required. If the method cannot fully submerge the object appropriately, another valid method is needed rather than silently using a partial reading.

If water spills between readings, the final level may not give the intended change. If the object contains open pores that trap air or absorb water, the relation can be more complicated. Practical suitability matters. This is why the syllabus specifies the kind of solid for the simple practical task and why school instructions must be followed.

A correct mathematical subtraction only yields a trustworthy physical volume when the measurement conditions support the inference. Write “volume of solid = rise in water reading” together with the assumptions that make the method valid.

4. Measure the mass of the same object

Imagine a stone whose mass is measured as 76 g on a suitable balance. A supervised cylinder measurement changes from 18 mL to 27.5 mL when that same stone is submerged without loss of liquid. Its measured volume is 27.5−18.0=9.5 cm³. Density = 76/9.5=8 g/cm³.

The phrase “that same stone” matters. If a student records 76 g for one stone but uses the displacement of another stone, the quotient is not automatically the density of either. Two stones may have different sizes or materials. Measurements must be attached to one defined specimen or to a valid shared-material model.

Sometimes a balance displays the total mass of a container and object. If the task asks for the solid’s mass, the container contribution must be removed where appropriate. For example, if the empty dry container has mass 42.0 g and the dry container with the solid is 117.0 g, the solid mass is 117.0−42.0=75.0 g. Dividing 117.0 g by the stone’s volume would wrongly include the container.

You may also be told the specimen is wet after removal from water. Depending on the method, residual water can affect the mass if the balance measurement is made at the wrong stage. A correct practical design identifies when the object’s mass is measured and how unrelated material is excluded.

When reporting density, give the numerical result and unit. “8” alone could be eight grams, eight cubic centimetres or eight grams per cubic centimetre. The unit is part of the scientific answer.

In a calculation with given idealised data, follow the question’s accuracy instructions. If rounding is needed, retain appropriate calculator precision during intermediate work, and avoid implying greater precision than the measurements support.

5. How to read the meniscus carefully

Water in a narrow measuring cylinder commonly forms a curved surface known as a meniscus. For water in ordinary transparent glassware, the correct reading is normally taken at the bottom of that curve with the eye level with the meniscus, according to the method and instrument. Reading from above or below can create a parallax error.

A question may show a picture of the water curve, but this text-only guide describes the decision. First identify the numerical graduations. Then determine the value corresponding to the appropriate level of the curve. Do not assume each printed line represents 1 mL; some cylinders use increments of 2, 5 or other amounts.

Suppose a cylinder has labelled marks at 20 mL and 30 mL with five equal divisions between them. Each division then represents 2 mL. A water level three divisions above 20 mL corresponds to 26 mL. A student who automatically counts each line as one millilitre might write 23 mL and carry a wrong initial value into every later calculation.

If a scale reads downward or contains unusual graduations, read the labels rather than the apparent height alone. A correct interpretation is grounded in how the instrument is marked, not how familiar the drawing looks.

Reading uncertainty matters especially when the measured rise is small. An initial reading of 30 mL and a final reading of 31 mL give a difference of 1 mL, but a coarse scale may make that difference uncertain. A larger appropriate displacement relative to the graduation size may provide a more reliable result, subject to the apparatus and practical constraints.

The goal is not to insist on a fictional number of decimal places. Use the instrument’s displayed resolution, school instructions and the specific question. Precision in a final answer should match the quality of the observations.

6. Why subtracting two readings can magnify uncertainty

Imagine two large readings measured with some uncertainty: initial 40.0 mL and final 42.0 mL. The inferred volume is only 2.0 cm³. If each reading may be uncertain by a small amount, that uncertainty can be a notable fraction of the two-millilitre difference. The relative effect is greater than if the object displaced 20 mL under similar reading conditions.

For a simple illustrative worst-case calculation, suppose each reading can differ from the actual value by up to 0.5 mL. The difference between them could then vary by up to 1.0 mL in the most extreme combined direction. This is not a claim that every result has exactly 1.0 mL uncertainty; it illustrates why the subtraction of two uncertain readings demands care.

A student may ask why the uncertainty of a difference could be relatively large even though both readings seem close to 40 mL. Because the desired result is not the 40 mL water already present. It is the small change after adding the solid. A small absolute reading uncertainty can be a large percentage of that change.

Suppose another suitable solid displaces 15 mL and is measured with comparable graduation uncertainty. Its relative volume uncertainty may be smaller. That does not automatically mean the specimen is better for every experiment; size, fit, full submersion and other constraints matter. It is a comparison of the measurement signal with instrument limitations.

A practical improvement should address the actual weakness. A measuring cylinder with finer suitable graduations, correct eye position and a repeatable method can support better readings. Saying “be more careful” without specifying what action improves the measurement is less useful.

This continues Volume 0072 on measurement uncertainty and anomalies. That chapter develops uncertainty more generally; here the narrow issue is a small volume inferred from the difference between two larger readings.

7. Density units: grams, centimetres and metres

At school level, density is often expressed in g/cm³ or kg/m³. Both can describe the same physical density. The numbers look different because the units are different sizes. A reliable conversion starts from the definitions rather than an uncertain memory of whether to multiply or divide.

One gram equals 0.001 kg. One cubic centimetre equals 0.000001 cubic metres, because each centimetre is 0.01 m and volume cubes the length conversion. Therefore 1 g/cm³ = 0.001 kg / 0.000001 m³ = 1000 kg/m³.

A density of 2.7 g/cm³ is 2700 kg/m³. A density of 7800 kg/m³ is 7.8 g/cm³. The material has not changed because the numeric value is larger in kg/m³. Only the unit has changed.

Volume is a common source of error. One cubic centimetre equals one millilitre. But 1 cm³ does not equal 1 m³. It takes a million cubic centimetres to fill a cubic metre. Length, area and volume conversion factors grow by different powers of ten.

Try a direct calculation. Mass is 0.054 kg and volume is 0.000006 m³. Density = 0.054/0.000006 = 9000 kg/m³. In grams and cubic centimetres, that is 54 g divided by 6 cm³ = 9 g/cm³. The two answers agree after conversion.

Check the question’s requested unit before presenting the final line. Correct arithmetic in g/cm³ does not answer a task explicitly demanding kg/m³ unless you convert and state it clearly.

8. Regular solids and irregular solids need different volume methods

A rectangular block has a shape that can be described by length, width and height. If those dimensions are given accurately in compatible units, volume = length × width × height. An irregular stone lacks a simple rectangular shape, so a direct formula based on three guessed dimensions may not provide its actual volume.

Suppose a rectangular object measures 2 cm by 3 cm by 4 cm. Its volume is 24 cm³. If its mass is 96 g, density is 96/24=4 g/cm³. There is no need to use water displacement if dimensions are reliable and the intended method is geometrical.

Now suppose an irregular object has mass 96 g and causes a suitable water level to rise from 30 mL to 54 mL. Its volume is also 24 cm³, giving the same density of 4 g/cm³. The numerical answer coincides, but the method for determining volume differs.

A student who tries to approximate an irregular object by measuring its maximum length, width and height is effectively calculating a bounding box. That box may contain empty space not occupied by the object. Density computed from that overestimated volume would be underestimated under the simplified assumptions. A suitable displacement method avoids the need to force the specimen into an invented rectangular shape.

Some objects float, dissolve or react in water, or trap air in ways that invalidate the simple method. A school practical should use an appropriate material and procedure. The official combined Science practical description specifies a solid that sinks in water for the relevant determination. Do not generalise that every unknown object can be measured accurately by dropping it into water.

Choosing a method is part of the physics. The arithmetic comes after the volume measurement is justified.

9. Repeated trials and the same starting conditions

Suppose three suitable displacement trials for the same rigid solid yield volumes of 9.4, 9.6 and 9.5 cm³. The values are close, and their mean is (9.4+9.6+9.5)/3=9.5 cm³. Under appropriate conditions, the close agreement supports the repeatability of the measurement. It does not prove the cylinder calibration is perfect.

A second trial may not be comparable if the water was spilled, the object retained trapped air or a different specimen was used. Repetition only supports confidence when the relevant method and object remain comparable. A tidy set of numbers produced by a biased procedure may be consistently wrong.

If a result is clearly different—say 9.5, 9.4 and 14.0 cm³—do not silently delete the third result. Investigate possible reasons, such as reading a graduation incorrectly, incomplete control of the setup or a recording mistake. Another suitable trial may help, but a decision to exclude a reading needs justification.

One practical note table might contain initial reading, final reading, calculated difference and an observation about the specimen. It is better than a table containing only three density answers. The fuller record allows you to locate whether disagreement came from mass, volume or calculation.

This connects to Volume 0100 on repeated measurements and drift. That workshop explains why consistency and accuracy differ. In the density case, both instrument and specimen state matter.

A good exam explanation links an improvement to the actual issue: more suitable graduations for a small displacement, consistent reading position, full submersion for an irregular solid or an appropriate repeat to assess variability.

10. A full measurement audit

Consider a hypothetical supervised lab record. A dry solid has mass 61.2 g. The measuring cylinder initially shows 20.0 mL. After full submersion, it shows 27.2 mL. The student calculates density as 61.2/27.2 and writes 2.25 g/cm³.

The calculation is arithmetically plausible but scientifically wrong for the intended object. The 27.2 mL final reading includes the initial water reading. The solid’s inferred volume is 27.2−20.0=7.2 cm³. Density is 61.2/7.2=8.5 g/cm³. The error is not division itself; it is using the wrong denominator.

Now imagine the student realises the mistake but writes 61.2/7.2=8.5 kg/m³. This is a second error. With grams and cubic centimetres, the quotient unit is g/cm³. Converting to kg/m³ gives 8500 kg/m³. A correct number paired with an incorrect unit still misrepresents the quantity.

Suppose another student says the object is “heavier than water” because 8.5 is greater than one. That phrase may be trying to compare densities, not total masses. More precise language is that the object’s density of 8.5 g/cm³ exceeds the approximate density of water under the relevant conditions. A small object can have less mass than a large volume of water while being denser.

Finally, suppose the object includes a sealed hollow cavity. Its overall density based on total external volume may differ from the density of the solid material composing its walls. The question needs to define whether it asks for the average density of the whole object or the density of the material. The measurement boundary affects the meaning of the result.

The full audit therefore checks object identity, mass, initial and final readings, subtraction, units and interpretation. Each stage protects the next.

11. Original practice problems

A. A suitable solid causes the water reading to rise from 18 mL to 26 mL. Find the solid’s volume in cm³.

B. The solid in A has mass 40 g. Calculate its density in g/cm³.

C. A stone has mass 76 g. Its cylinder readings are 18.0 and 27.5 mL. Find its measured volume and density.

D. An empty container has mass 35 g and the same container with a solid has mass 91 g. The solid displaces 8 cm³. Calculate the solid’s density.

E. A solid has mass 0.072 kg and volume 0.000009 m³. Calculate density in kg/m³.

F. Convert 2.5 g/cm³ to kg/m³ and 6400 kg/m³ to g/cm³.

G. A cylinder has marks at 10, 20 and 30 mL with five equal spaces between each consecutive labelled pair. What value does one division represent?

H. A learner reads the final water volume as the object’s volume. Explain the mistake and give the correct mathematical operation.

I. A small object produces a rise from 40.0 to 41.0 mL. Why might graduation uncertainty be particularly important?

J. A specimen is only partly submerged. What physical quantity does the rise represent in the simple method, and why may it be insufficient?

K. A rigid rectangular block measures 2 cm × 3 cm × 4 cm and has mass 48 g. Calculate density.

L. An irregular solid of the same mass displaces 24 mL. Compare its density with the block in K.

M. Repeated volume differences are 9.4, 9.6 and 9.5 cm³. Find their mean and state one limitation of using repeated readings alone as proof of accuracy.

N. Two objects each have mass 50 g, but their volumes are 5 cm³ and 20 cm³. Which is denser, and by what factor?

O. A student calculates 80/10=8 g/cm³. What information is still needed to know whether the result truly describes the density of the intended specimen?

Read the questions without assuming all results are numerical. Some ask you to explain whether a procedure and inference are valid.

12. Worked answers and specific corrections

A. Volume = 26−18=8 mL = 8 cm³.

B. Density=40/8=5 g/cm³. The denominator is the displaced volume, not the final water level.

C. Volume=27.5−18.0=9.5 cm³. Density=76/9.5=8 g/cm³.

D. Solid mass=91−35=56 g. Density=56/8=7 g/cm³. Do not include the container mass.

E. Density=0.072/0.000009=8000 kg/m³. The calculation uses consistent SI units.

F. 2.5 g/cm³=2500 kg/m³; 6400 kg/m³=6.4 g/cm³. One g/cm³ equals 1000 kg/m³.

G. A 10 mL interval divided into five equal spaces means 2 mL per division. Count spaces, not merely labelled numbers.

H. The final reading includes the original liquid level. Subtract initial from final to isolate the volume displaced by the solid.

I. The desired change is only 1.0 mL, so a small absolute reading uncertainty may be large relative to the inferred volume. Finer suitable graduations or a more appropriate method could help.

J. The reading rise corresponds to displaced volume of the submerged portion. The whole object’s density needs the object’s entire volume, not only part of it.

K. Volume=2×3×4=24 cm³. Density=48/24=2 g/cm³.

L. The irregular solid has volume 24 cm³ by displacement. Its density is also 48/24=2 g/cm³, assuming the measurements are valid and refer to the same object.

M. Mean=(9.4+9.6+9.5)/3=9.5 cm³. Repeatability does not by itself establish correct calibration or freedom from systematic bias.

N. First density=50/5=10 g/cm³; second density=50/20=2.5 g/cm³. The first is 10/2.5=4 times as dense.

O. We need to know that 80 g is the mass of the intended specimen and 10 cm³ is its correctly measured whole volume under appropriate conditions. Arithmetic alone cannot validate measurement ownership.

For errors, write a precise label: failed subtraction, included container mass, mixed units, partial submersion, uncertain cylinder reading or unsupported interpretation.

13. How to write a practical explanation

A short but complete method explanation can name the measured mass, measured initial water level, measured final level, inferred displaced volume and calculated density. In a supervised practical question, also follow the specific apparatus and safety instructions supplied. The essential reasoning is that the difference between water readings gives volume under the appropriate conditions.

If asked for an improvement, identify a specific weakness. A coarse scale may be addressed by a suitable cylinder with clearer graduations. A parallax problem requires viewing the liquid level appropriately at eye level. A risk of incomplete submersion requires an arrangement suitable for fully submerging the specified solid without invalidating the measurement.

If asked whether a result is reliable, repetition can help assess consistency. But do not claim the value is accurate solely because three readings agree. Reference and calibration may still matter. Likewise, an unexplained anomalous trial should be investigated, not discarded automatically.

For a conclusion, state what the result refers to. “The specimen’s measured density is approximately 8 g/cm³” is precise. “This object must be made of a particular metal” may be unjustified if the task supplies no material identification evidence and different materials or composites could have overlapping density values.

Avoid vague phrases such as “the water goes up because it is heavier”. The water-level rise is about displacement of volume, not a direct comparison of total masses. The object must also satisfy the physical requirements of the method.

A good response allows another learner to trace exactly where the denominator came from. If that explanation is clear, the density equation becomes more than a memorised formula.

14. Seven days from measurements to independent reasoning

Day 1: distinguish properties. Practise mass, volume and density definitions and their units. Compare a heavy large object with a small dense one to avoid using mass and density interchangeably.

Day 2: read graduated scales. Work with text or teacher-provided measuring-cylinder diagrams. Identify division values and meniscus readings before performing subtraction.

Day 3: volume difference. Solve five initial/final water-reading pairs. Explain why the final reading alone cannot be used as solid volume.

Day 4: mass and containers. Practise subtracting container mass from assembly mass, then combining the result with the correct displaced volume.

Day 5: unit conversion. Convert between g/cm³ and kg/m³, and between mL and cm³. Check every conversion by a second route.

Day 6: measurement quality. Compare small and large displacements under similar graduation uncertainty. Identify possible errors from parallax, bubbles, water loss or incomplete submersion as appropriate to the given method.

Day 7: independent clinic. Solve Questions A–O, then mark the first step where your reasoning went wrong. Revisit that exact issue with new numerical values a few days later.

Parents can ask, “What changed in the cylinder?” followed by “What exactly is being divided by that change?” These questions test the model beneath the calculation, not merely recall of density = mass/volume.

15. The correct difference makes the formula meaningful

A good density answer is a chain of matched evidence. The balance supplies the mass of the intended object. The difference between two suitable cylinder readings supplies its volume. The quotient supplies density in an appropriate unit. Each link must belong to the same specimen and the same defined conditions.

This connects to Volume 0104 on system boundaries, because defining what is inside the measurement matters. It also connects to Volume 0100 on measurement drift. For full curriculum continuity, return to the Science Hub and the G3 SEC learner-guide route.

Official content and assessment details are grounded in the SEAB 2027 G3 school-candidate syllabus index and combined Science K326–K328 syllabus. All numerical tasks are original practice examples, not official SEC examination questions.