Page 86 - Physics

Physics for Secondary Schools

Example 4.1

What would be the density of a glass block calculated based on its dimensions of
0.12 m × 0.04 m × 0.10 m and a mass of 1.2 kg?

Solution
mass
Density = . Whereby;
volume

Volume of the block = 0.12 m × 0.04 m × 0.10 m = 0.00048 m 3
Volume of the block = 0.12 m × 0.04 m × 0.10 m = 0.00048 m 3

Mass of the block = 1.2 kg
m
Density of block, ρ =
V
1.2 kg
∴ Densityof block = = 2 500 kg/m 3
0.00048 m 3

Therefore, the density of the block is 2 500 kg/m .
3

Example 4.2 where m is the mass of glass, ρ is
g
g
Using Table 4.1, calculate the mass of the density of the glass, and V is the
g
the glass that has the same volume as volume of the glass.
5.4 g of aluminium. Then,
m =2.5 g/cm × 2 cmm =5 g
3
3
Solution g g
Density of aluminium = 2 700 kg/m 3 Therefore, the mass of glass is 5 g.
= 2.7 g/cm 3
Mass of aluminium = 5.4 g Determinati on of density of a
Density of glass = 2.5 g/cm 3 substance
Required: Mass of the glass A technique used in the determination
of density of a substance depends on the
m
From the formula of density, V = properties of the substance. For a given
ρ substance, the appropriate method for
measuring its mass and volume must be
The volume of aluminium will be: used. Once the measured values of mass
5.4 g
V = 2.7 g/cm 3 = 2 cm 3 and volume are known, the density of a

substance can be calculated.
This volume is the same as that of glass, It is important to note that different

Thus, m = ρ × V g techniques are used to determine the
g
g
density of solids, liquids and gases.
80
Student’s Book Form One

Physics Form 1 Final.indd 80 16/10/2024 20:56