Lesson 6, Topic 4
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# Density

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Density of a substance is the ratio of mass to a volume of a substance.

OR

Density is mass per unit volume of a substance.

Density = $$\frac {mass}{volume}$$

= $$\frac {kilogramme}{m^3}$$ = Kgm-3

Its unit is kgm-3.

Different materials have different densities as shown;

### Determination of Density of a Solid:

(i) Regular Shaped Object: The mass of a regular object is measured with a chemical balance or beam balance and its volume measured by metre rule, Vernier Calliper or micrometre screw gauge, depending on the shape of the body. Applying the formula,

Density = $$\frac {mass}{volume}$$

$$\scriptsize \rho = \normalsize \frac {m}{v}$$

(ii) Irregular Shaped Object: The volume of the object is determined by displacement method using Eureka can.

The volume of water displaced by suspending the object in the water is equal to the volume of the object.

The Formula for density $$\frac {mass}{volume}$$ is used to determine the density of the body.

### Relative Density:

Relative density of a substance is the ratio of mass of a given volume of a substance to the mass of an equal volume of water.

Relative density = $$\frac {mass\: of \: substance}{mass \: of \:equal \:volume}$$

Relative density = $$\frac {weight\: of \: substance}{weight \: of \:equal \:volume \: of \: water}$$

(Because mass is directly proportional to the weight of a body)

Relative density = $$\frac {density\: of \: a\:substance}{density \: of \:equal \:volume \: of \: water}$$

Relative density has no unit.

### Example 1:

A solid weighs 1.80N in air and 0.90N in a liquid of density 900kgm-3.

Calculate;
(i) the upthrust of liquid in solid
(ii) the volume of the liquid.

Solution:

(i) Upthrust = Weight in air – weight in liquid

= 1.80 – 0.90

= 0.90N

(ii) mass = 0.09kg (1kg = 10N) = $$\frac {W}{g} = \frac {0.9}{10}$$

= 0.09kg

= 90g

Volume of liquid = $$\frac {mass}{density} \\ = \frac {m}{\rho} \\ = \frac{0.09}{900}$$

= 0.0001m3 = 10-4m3

= 10cm3

### Example 2:

A solid material of volume 2 x 10-5m3 and density 2.5 x 103kgm3 is suspended from a spring balance with half of the volume of the solid immersed in water. What is the reading on the spring balance? (g = 10ms-2).

Solution:

Relative density = $$\frac {weight\: of \: substance}{weight \: of \:equal \:volume \: of \: water}$$

Weight = Vρg

= 2.5 x 103 x 2.0 x 10-5 x 10

= 0.5N

Upthrust in water = weight of water displaced

= Vρg = $$\frac {V}{2} \scriptsize \rho g$$

density f water = 1 x 103 kgm-3

= $$\normalsize \frac {2 \:\times \: 10^{-2}}{2} \scriptsize \: \times \: 1.0 \times 10^3 \: \times \: 10$$

= 0.1N

Spring balance reading = Weight in air – Upthrust

= 0.5 – 0.1

= 0.4N

Evaluation Questions:

1. A solid weighs 0.04N in air and 0.024 in a liquid of density 800kgm-3. Find the volume of the solid. (g = 10ms-3).
2. State Archimedes principle
3. Describe an experiment to verify Archimedes principle
4. State the principle of floatation
5. Briefly explain why a solid metal sinks in water while a ship made of the same material does not sink.

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