Before you start
You need these ideas and skills for this practical. If any of them are new, read this section first.
Key words
- Mass (m): the amount of matter in an object, measured in grams (g) with a balance.
- Volume (V): the space an object takes up, measured in cm³. 1 cm³ = 1 ml.
- Density (ρ): the mass of each cubic centimetre of a substance: ρ = m ÷ V.
- Irregular solid: an object with no simple shape (a stone, a key, a bolt), so you cannot find its volume with a ruler and a formula.
- Displacement: when an object goes under water it pushes aside (displaces) its own volume of water, so the water level rises.
- Sinker: a heavy object tied to something that floats, to hold it completely under water.
The big idea
An object under water takes up space that the water used to fill. The water has nowhere to go but up, so the level rises by exactly the volume of the object.
V₁ is the reading before the object goes in, and V₂ is the reading with the object completely under water.
Reading the measuring cylinder
Put your eye level with the water surface and read the bottom of the meniscus. Eye above: the reading looks too high. Eye below: too low (parallax error).
Between two marks? Record the nearest mark, or estimate to half a division (e.g. 57.5 cm³). Do not always take the lower mark: that is a systematic error.
Because V is the difference of two readings, each with an uncertainty of about ±0.5 cm³, the volume can be out by up to ±1 cm³. That is a big fraction of a small object's volume.
Skills you need
- Zeroing and reading a top-pan balance to 0.1 g.
- Reading a measuring cylinder to the nearest 0.5 cm³.
- Recording results in a table with units in the headings, e.g. V₁ / cm³.
- Calculating density with ρ = m ÷ V, and converting 1 g/cm³ = 1000 kg/m³.
Aim and apparatus
Aim
To find the volume of irregularly shaped solids by the displacement of water, and so find their densities. To use a sinker to find the volume and density of objects that float.
Syllabus link: Cambridge IGCSE Physics 0625 / 0972, topic 1.4 Density: describe how to determine the density of an irregularly shaped solid by the method of displacement. Supplement: predict whether an object will float, using density data.
Apparatus
- 100 cm³ measuring cylinder, with 1 cm³ divisions
- Top-pan balance, reading to 0.1 g
- Beaker of water and a dropper (pipette)
- Thread, to lower each object gently
- Objects that fit inside the cylinder: stone, glass marble, aluminium block, brass weight, steel bolt, rubber stopper, a mystery metal
- Objects that float: cork, a wooden block, candle wax, and a steel sinker
- Paper towels, to dry each object
Too big for the cylinder? Use a displacement (Eureka) can: fill it to the spout, lower the object in, and collect the water that overflows in a measuring cylinder. Its volume is the object's volume.
Theory
Volume by displacement
This works for any shape, as long as the object is completely under water, does not dissolve, and does not soak up water.
Density
With m in g and V in cm³, ρ is in g/cm³. Multiply by 1000 for kg/m³.
Density is a property of the material, not the object: a small and a large piece of brass have the same density. On a graph of mass against volume they lie on the same straight line through the origin, and the gradient is the density.
Objects that float: the sinker method
An object less dense than water (ρ < 1.00 g/cm³) floats with only part of it under water. A floating object pushes aside its own mass of water, not its own volume, so V₂ − V₁ would be too small.
Tie a heavy sinker to it:
- V₁ = water + sinker (sinker under water)
- V₂ = water + sinker + object (both under water)
The sinker is under water in both readings, so its volume cancels out.
Densities of some solids
| Material | ρ / g/cm³ | ρ / kg/m³ |
|---|---|---|
| Cork | 0.24 | 240 |
| Wood (pine) | 0.5 – 0.7 | 500 – 700 |
| Candle wax | 0.90 | 900 |
| Water (for comparison) | 1.00 | 1000 |
| Rubber | 1.1 – 1.5 | 1100 – 1500 |
| Glass | 2.5 | 2500 |
| Stone (granite) | 2.65 | 2650 |
| Aluminium | 2.70 | 2700 |
| Titanium | 4.51 | 4510 |
| Steel | 7.85 | 7850 |
| Brass | 8.5 | 8500 |
| Copper | 8.96 | 8960 |
| Lead | 11.3 | 11 300 |
Use this table to identify the Mystery object in the lab. Anything denser than water sinks; anything less dense floats.
Planning the investigation
Variables
| Type | Variable | How |
|---|---|---|
| Measured | Mass m | Top-pan balance, zeroed first |
| Measured | Volumes V₁ and V₂ | Measuring cylinder, eye level, bottom of the meniscus |
| Calculated | V = V₂ − V₁, ρ = m ÷ V | For each object |
| Control | The water | Enough to cover the object, not so much it goes above 100 cm³ |
| Control | The object | Dry, clean, and with no air bubbles stuck to it |
How much water?
Start with about 50 cm³. It must cover the object completely once it is in, but leave room for the level to rise without going past the top of the scale.
Making it more accurate
Use the largest object that fits, so that the ±1 cm³ uncertainty in V₂ − V₁ is a smaller fraction of V. A narrower measuring cylinder with smaller divisions also helps. Repeat with the same object and take a mean.
Method in a real lab
- Find the mass. Zero the balance with nothing on the pan, then place the dry object on it and record m.
- Half-fill the measuring cylinder with water (about 50 cm³). Stand it on a flat, level bench.
- Read V₁ with your eye level with the water surface, at the bottom of the meniscus.
- Tie a thread to the object and lower it gently into the water until it is completely covered. Do not drop it: water could splash out, and the glass could crack.
- Check for air bubbles on the object; tap the cylinder gently to free them. Then read V₂.
- Calculate V = V₂ − V₁ and ρ = m ÷ V. Repeat for each object.
- For an object that floats: lower a sinker into the water and read V₁. Tie the object to the sinker, lower both so they are completely under water, and read V₂. The object's volume is V₂ − V₁.
How to use this virtual lab
- Choose an object at the top of the panel, and Small or Large.
- Step 1, Weigh: press Zero the balance, then Put object on balance. The mass is saved for you.
- Step 2, Fill: press and hold “Hold to pour” and let go at about 50 cm³. Use + a few drops to fine-tune. Press Next.
- Step 3, V₁: the magnifier shows the scale. Keep your eye Level, type the volume you read and press Save V₁. Stuck? Help me read it draws a red line at the bottom of the meniscus.
- Step 4, V₂: press Lower it gently on the thread. When the object is under water, read V₂ in the magnifier and press Record result. The lab works out V₂ − V₁ and the density.
- Repeat with other objects and sizes. Open Table & graph to compare densities and plot mass against volume.
- Try the mistakes: What if I drop it in? splashes water out; choosing Cork shows what happens when an object floats; Above/Below shows parallax error. Results with a mistake are saved with ⚠.
- More to explore: switch on the sinker method to measure cork, wood and wax.
Results
Results table
Sample results with the large objects:
| Object | m / g | V₁ / cm³ | V₂ / cm³ | V / cm³ | ρ / g/cm³ |
|---|---|---|---|---|---|
| Stone | 19.8 | 50.0 | 57.5 | 7.5 | 2.64 |
| Glass marble | 13.7 | 50.0 | 55.5 | 5.5 | 2.49 |
| Aluminium block | 26.2 | 50.0 | 59.5 | 9.5 | 2.76 |
| Steel bolt | 46.3 | 50.0 | 56.0 | 6.0 | 7.72 |
| Brass weight | 88.7 | 50.0 | 60.5 | 10.5 | 8.45 |
| Rubber stopper | 12.9 | 50.0 | 60.0 | 10.0 | 1.29 |
| Wooden block (with sinker) | 6.3 | 53.0 | 62.5 | 9.5 | 0.66 |
Worked example
Brass weight: m = 88.7 g, V₁ = 50.0 cm³, V₂ = 60.5 cm³.
This is close to the book value for brass, 8.5 g/cm³.
Analysis and conclusion
Comparing materials
Compare each density with the table in the Theory section. The metals (steel, brass) are about three times denser than stone, glass and aluminium. The rubber stopper is only a little denser than water, so it sinks slowly; the wood is less dense than water, so it floats.
Mass against volume
Plot mass (y-axis) against volume (x-axis) for the small and large piece of each material. Each material gives a straight line through the origin, and its gradient is its density. The two sizes of one material lie on the same line: density does not depend on how big the object is.
Conclusion
“The volume of each irregular object was found from the rise in the water level, V₂ − V₁. The densities were close to the book values: for example, brass 8.45 g/cm³ (book value 8.5 g/cm³). The wooden block had to be held under water with a sinker, because its density (0.66 g/cm³) is less than that of water.”
Evaluation
| Source of error | Effect | How to reduce it |
|---|---|---|
| Parallax: eye not level with the meniscus | V₁ or V₂ too high (eye above) or too low (eye below) | Eye level with the water surface; read the bottom of the meniscus. |
| Small object: V₂ − V₁ is only a few cm³ | The ±0.5 cm³ uncertainty in each reading is a large percentage of V | Use a larger object, or a narrower cylinder with finer divisions; repeat and average. |
| Dropping the object in | Water splashes out, so V₂ is too small and the density too large; the glass may crack | Lower it gently on a thread, or tilt the cylinder and let the object slide down. |
| Air bubbles stuck to the object | V₂ too large, so the density is too small | Tap the cylinder or shake the object gently on its thread. |
| Object not completely covered | V₂ − V₁ is too small | Use more water to start with. |
| Object floats | Only part of it is under water; V₂ − V₁ is too small | Use a sinker to hold it completely under water. |
| Object absorbs water (wood, cork, chalk) | The water level falls slowly, and the mass increases | Read V₂ quickly, or seal the object with a thin layer of varnish or wax. |
| Wet object weighed | Mass too large | Always weigh the object before putting it in the water, or dry it well. |
| The thread | Its volume adds a tiny amount to V₂ | Use thin thread, and lower only a short length into the water. |
Safety
- Lower objects gently: a heavy object dropped into a glass measuring cylinder can crack it.
- Stand the cylinder on a flat bench away from the edge, or use a plastic cylinder.
- Wipe up spilled water straight away, and keep water away from the balance's power socket.
- Wash your hands after handling lead or other metal samples.
Exam practice
Practical-style questions like those in Paper 5 (Practical Test) and Paper 6 (Alternative to Practical). Try each one before opening the answer.
1. A stone has a mass of 52.0 g. The water level rises from 40.0 cm³ to 60.0 cm³ when the stone is lowered in. Calculate the density of the stone in g/cm³ and kg/m³.
V = 60.0 − 40.0 = 20.0 cm³. ρ = 52.0 ÷ 20.0 = 2.6 g/cm³ = 2600 kg/m³.
2. Describe how to find the volume of an irregularly shaped stone using a measuring cylinder.
Partly fill the measuring cylinder with water and read the volume V₁ at eye level, at the bottom of the meniscus. Tie the stone to a thread and lower it gently into the water until it is completely covered. Read the new volume V₂. The volume of the stone is V₂ − V₁.
3. Why should the object be lowered on a thread instead of being dropped in?
Dropping it could splash water out of the cylinder, so V₂ would be too small (and the density too large). A heavy object could also crack the glass.
4. A student finds the volume of a cork by putting it in water. Explain why her value is too small, and how she could improve the method.
The cork floats, so only part of it is under water and it displaces less water than its own volume. She should tie a sinker to the cork: read V₁ with only the sinker under water, then V₂ with the sinker and cork both under water. The cork's volume is V₂ − V₁.
5. Explain why a student should use a large object rather than a small one in this experiment.
Each volume reading has an uncertainty of about ±0.5 cm³. For a large object, V₂ − V₁ is larger, so this uncertainty is a smaller percentage of the volume, and the density is more accurate.
6. An object of mass 45.0 g has a volume of 5.0 cm³. Use the table of densities to suggest what it is made of. Will it float in water?
ρ = 45.0 ÷ 5.0 = 9.0 g/cm³, close to copper (8.96 g/cm³). It is much denser than water (1.00 g/cm³), so it sinks.
7. Some air bubbles stick to an object under water. What effect does this have on the density the student calculates?
The bubbles push aside water too, so V₂ (and V₂ − V₁) is too large. The calculated density m ÷ V is therefore too small.
8. A large irregular object does not fit into any measuring cylinder. Describe how its volume could be found.
Use a displacement (Eureka) can. Fill it with water until water just runs out of the spout, and wait for it to stop. Place a measuring cylinder under the spout. Lower the object gently into the can on a thread. The volume of water collected in the measuring cylinder equals the volume of the object.