To find the average density of a sealed hollow object, divide its total mass by its external volume. To find the density of the shell material, divide the shell’s mass by the volume of the material alone. The two calculations answer different questions.
The 2027 SEC G3 Physics K323 syllabus includes applying density as mass divided by volume. This example uses water displacement for the external volume and a separate cavity measurement for the shell volume.
- Identify whether the question asks about the whole object or the shell material.
- Subtract the initial water reading from the final reading.
- For a sealed submerged object, displacement gives the external volume.
- Subtract cavity volume only when calculating the volume of the shell material.
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Match the mass to the volume
A sealed hollow object keeps water out of its cavity. When it is fully submerged, it displaces water from the space enclosed by its outer surface. That includes the space occupied by the shell and the cavity.
For the shell material alone, use the volume of the solid shell. If the external and cavity volumes are known, subtract the cavity volume from the external volume.
The measurement example assumes a sealed object, full submersion and no attached air bubbles on the outside. It also treats the mass of trapped air as negligible for the shell calculation.
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Worked example | A sealed metal object
A sealed hollow object has a mass of 156 g. When fully submerged, the water reading rises from 90 cm³ to 150 cm³. Its cavity has a volume of 40 cm³.
| Asked for | Volume used | Density |
|---|---|---|
| Average density of sealed object | 60 cm³ | 156 ÷ 60 = 2.60 g/cm³ |
| Density of shell material | 60 − 40 = 20 cm³ | 156 ÷ 20 = 7.80 g/cm³ |
The second result is greater because the same approximate mass is divided by a smaller volume. Write ‘object’ or ‘shell’ beside the number so the reader knows which density you found.
The 150 cm³ reading includes the water already in the container. The rise of 60 cm³ is the displacement due to the object.
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Practise when the cavity fills
Suppose the shell is opened and fully submerged so that water fills the cavity and no air is trapped inside. Water occupies the cavity during the measurement, so the rise now represents the volume of the solid shell.
For the same shell, expect a rise of 20 cm³ if opening it removed a negligible amount of material. If trapped air remains, the rise can be larger than the shell volume.
A useful exam sketch shades the solid and marks the space water cannot enter. Compare that space with the volume needed in the formula.
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Frequently asked questions
Is one millilitre equal to one cubic centimetre?
Yes. A rise of 60 mL is a volume of 60 cm³.
Can I use a floating object’s water rise as its full volume?
The rise gives the submerged volume. A floating object can have part of its volume above the water.
What if the mass of the contents is significant?
Use the total mass for the whole object and the shell mass alone for the density of the shell material.
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