Density of an irregular object

Particle model of matter

GCSE Physics (8463)

ORIGINAL

Volume by displacement, then the unit conversions that decide the mark: g to kg and cm³ to m³.

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Question

A student finds the density of an irregular stone. Its mass is

84\,\text{g}

. When it is lowered into a full displacement can,

30\,\text{cm}^3

of water flows out into a measuring cylinder. Calculate the density of the stone in

\text{kg/m}^3

Every step worked, with the reasoning.

1
Volume of stone $=$ volume of water displaced $= 30\,\text{cm}^3$
The stone takes the place of an equal volume of water, which flows out of the spout.
2
$m = 84\,\text{g} = 0.084\,\text{kg}$
Divide by 1000 to convert grams to kilograms.
3
$V = 30\,\text{cm}^3 = 30 \times 10^{-6}\,\text{m}^3 = 3.0 \times 10^{-5}\,\text{m}^3$
$1\,\text{m} = 100\,\text{cm}$ , so $1\,\text{m}^3 = 100^3 = 1\,000\,000\,\text{cm}^3$ .
4
$\rho = \dfrac{m}{V} = \dfrac{0.084}{3.0 \times 10^{-5}} = 2800\,\text{kg/m}^3$
Check in $\text{g/cm}^3$ : $84 \div 30 = 2.8\,\text{g/cm}^3$ , and $1\,\text{g/cm}^3 = 1000\,\text{kg/m}^3$ .

Answer: $2800\,\text{kg/m}^3$