Physics 9702 · Paper 1

Density

Same size, different mass: a steel cube is heavier than a wooden one. The difference is a property of the material itself.
Density measures how much mass is packed into each unit of volume.

The definition

\rho = \dfrac{m}{V}

Symbols

$\rho$ = density (Greek letter rho) (kg m⁻³)
$m$ = mass (kg)
$V$ = volume (m³)

Density is the mass per unit volume of a substance. That exact phrase is the one-mark definition (9702/21/O/N/24 Q1(a)).
Density belongs to the material, not the object. Half a brick has half the mass and half the volume: the density is unchanged.
Water is about $1000 \text{ kg m}^{-3}$ . The unit conversion to memorise: $1 \text{ g cm}^{-3} = 1000 \text{ kg m}^{-3}$ .

Worked example

Smallest case: a metal cube

A cube has mass 240 g and volume 30 cm³. Find its density.

$\rho = \dfrac{m}{V} = \dfrac{240}{30} = 8.0 \text{ g cm}^{-3} = 8000 \text{ kg m}^{-3}$ .

Answer

\rho = 8000 \text{ kg m}^{-3}

Worked example

Exam version: a glass block

A glass block has mass 0.243 kg and sides 5.41 cm, 11.09 cm and 1.62 cm. Find the density of the glass. (9702/21/O/N/24 Q1)

Change each side to metres before multiplying: $V = 0.0541 \times 0.1109 \times 0.0162 = 9.72 \times 10^{-5} \text{ m}^3$ .
$\rho = \dfrac{0.243}{9.72 \times 10^{-5}} = 2500 \text{ kg m}^{-3}$ .

Answer

\rho = 2500 \text{ kg m}^{-3}

Common mistake

Leaving lengths in centimetres. Convert to metres before you multiply, and remember that cubing a length cubes its conversion:

1 \text{ cm}^3 = 10^{-6} \text{ m}^3

(lesson 1.4).

What the exam pairs it with

Percentage uncertainty (lesson 1.6): for a cube of side $l$ , the volume uses $l$ three times, so $\%\,\rho = \%m + 3 \times \%l$ . The same glass block question continues into an uncertainty of 1.3% (9702/22/F/M/25 Q1(b)).
Estimates (lesson 1.5): a density answer that is far from 1000 kg m⁻³ for a watery object, or far from a few thousand for rock and metal, is a sign of a powers-of-ten slip.

Your turn— tap to reveal the worked answer (9702/12/M/J/24 Q13)

40 cm³ of glycerine (density 1.3 g cm⁻³) is mixed with a volume V of water (1.0 g cm⁻³). The mixture has density 1.1 g cm⁻³. Find V. (9702/12/M/J/24 Q13)

Total mass = total volume × mixture density: $(40)(1.3) + V(1.0) = (40 + V)(1.1)$ .
$52 + V = 44 + 1.1V$ , so $0.1V = 8$ and $V = 80 \text{ cm}^3$ .

Answer: 80 cm³.

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Density

Same size, different mass: a steel cube is heavier than a wooden one. The difference is a property of the material itself.
Density measures how much mass is packed into each unit of volume.

The definition

\rho = \dfrac{m}{V}

Symbols

$\rho$ = density (Greek letter rho) (kg m⁻³)
$m$ = mass (kg)
$V$ = volume (m³)

Density is the mass per unit volume of a substance. That exact phrase is the one-mark definition (9702/21/O/N/24 Q1(a)).
Density belongs to the material, not the object. Half a brick has half the mass and half the volume: the density is unchanged.
Water is about $1000 \text{ kg m}^{-3}$ . The unit conversion to memorise: $1 \text{ g cm}^{-3} = 1000 \text{ kg m}^{-3}$ .

Worked example

Smallest case: a metal cube

A cube has mass 240 g and volume 30 cm³. Find its density.

$\rho = \dfrac{m}{V} = \dfrac{240}{30} = 8.0 \text{ g cm}^{-3} = 8000 \text{ kg m}^{-3}$ .

Answer

\rho = 8000 \text{ kg m}^{-3}

Worked example

Exam version: a glass block

A glass block has mass 0.243 kg and sides 5.41 cm, 11.09 cm and 1.62 cm. Find the density of the glass. (9702/21/O/N/24 Q1)

Change each side to metres before multiplying: $V = 0.0541 \times 0.1109 \times 0.0162 = 9.72 \times 10^{-5} \text{ m}^3$ .
$\rho = \dfrac{0.243}{9.72 \times 10^{-5}} = 2500 \text{ kg m}^{-3}$ .

Answer

\rho = 2500 \text{ kg m}^{-3}

Common mistake

Leaving lengths in centimetres. Convert to metres before you multiply, and remember that cubing a length cubes its conversion:

1 \text{ cm}^3 = 10^{-6} \text{ m}^3

(lesson 1.4).

What the exam pairs it with

Percentage uncertainty (lesson 1.6): for a cube of side $l$ , the volume uses $l$ three times, so $\%\,\rho = \%m + 3 \times \%l$ . The same glass block question continues into an uncertainty of 1.3% (9702/22/F/M/25 Q1(b)).
Estimates (lesson 1.5): a density answer that is far from 1000 kg m⁻³ for a watery object, or far from a few thousand for rock and metal, is a sign of a powers-of-ten slip.

Your turn— tap to reveal the worked answer (9702/12/M/J/24 Q13)

40 cm³ of glycerine (density 1.3 g cm⁻³) is mixed with a volume V of water (1.0 g cm⁻³). The mixture has density 1.1 g cm⁻³. Find V. (9702/12/M/J/24 Q13)

Total mass = total volume × mixture density: $(40)(1.3) + V(1.0) = (40 + V)(1.1)$ .
$52 + V = 44 + 1.1V$ , so $0.1V = 8$ and $V = 80 \text{ cm}^3$ .

Answer: 80 cm³.