ICSE Class 8 Physics Chapter 08 Archimedes Principle Including Density and Relative Density

Archimedes' Principle

Including Density and Relative Density

Density

Density of a substance is defined as the mass per unit volume of that substance.

\[\text{Density} = \frac{\text{Mass}}{\text{Volume}}\]

Units

In M.K.S. (i.e. S.I.) system, the unit of mass is kg, the unit of volume is m³ and the unit of density is kg m⁻³.

\[\text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{1 \text{ kg}}{1 \text{ m}^3} = 1 \text{ kg m}^{-3}\]

In C.G.S. system, the unit of mass is g (gram), the unit of volume is cm³ and the unit of density is g cm⁻³.

\[\text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{1 \text{ g}}{1 \text{ cm}^3} = 1 \text{ g cm}^{-3}\]

The density of water at 4°C is 1 g cm⁻³. This means, at 4°C, the mass of 1 cm³ of water is 1 g.

The density of mercury is 13.6 x 10³ kg m⁻³. This means, the mass of 1 m³ of mercury is 13.6 x 10³ kg.

Also, note that:

\[x \text{ g cm}^{-3} = x \times 1000 \text{ kg m}^{-3}\]

(i) 1 g cm⁻³ = 1000 kg m⁻³

(ii) 7.8 g cm⁻³ = 7.8 x 10³ kg m⁻³

Conversely,

\[x \text{ kg m}^{-3} = \frac{x}{1000} \times \text{g cm}^{-3}\]

(i) 13.6 x 10³ kg m⁻³ = \(\frac{13.6 \times 10^3}{1000}\) g cm⁻³ = 13.6 g cm⁻³

(ii) 800 kg m⁻³ = \(\frac{800}{1000}\) g cm⁻³ = 0.8 g cm⁻³

Teacher's Note

Density helps us understand why some objects sink while others float - the same material can be rearranged to change how much space it takes up, which is why a solid iron nail sinks but an iron ship floats.

Relative Density

Relative density (R.D.) of a substance is defined as the ratio of density of the substance to the density of water at 4°C. Relative density is also known as specific gravity of the substance.

\[\text{R.D.} = \frac{\text{Density of substance}}{\text{Density of water at 4°C}}\]

\[= \frac{\text{Mass of unit volume of substance}}{\text{Mass of unit volume of water at 4°C}} \times \frac{v}{v}\]

\[= \frac{\text{Mass of certain volume of substance}}{\text{Mass of same volume of water at 4°C}}\]

R.D. can also be defined as the ratio of mass of a substance to the mass of same volume of water at 4°C.

Since R.D. is the ratio of two densities, it has no unit.

Relation Between Density and Relative Density

We know density of water at 4°C is 1 g cm⁻³.

Therefore, R.D. of a substance

\[= \frac{\text{Density of the substance}}{\text{Density of water at 4°C}} = \frac{\text{Density of the substance in g cm}^{-3}}{1 \text{ g cm}^{-3}}\]

Thus, if the density of a substance is 3.6 g cm⁻³, its R.D. is 3.6. Conversely, if R.D. of a substance is 1.4, its density is 1.4 g cm⁻³.

If relative density of a substance is less than one it will float in water. For example, a piece of ice, which has a relative density of about 0.9, will float on water.

A substance with a relative density greater than 1 will sink in water.

Uses

(1) Relative density is often used by geologists and mineralogists to determine the mineral content of a rock or other sample.

(2) Gemologists use it as an aid for identification of gemstones.

Note: The relative density of a liquid can be measured using a hydrometer.

Teacher's Note

Relative density explains why salt water feels easier to float in than fresh water - the higher density of salt water provides more buoyant force on your body.

Example 1

A piece of iron of volume 30 cm³ has a mass of 234 g. Find:

(i) the density of iron

(ii) relative density (R.D.) of iron

Solution

Since, mass of iron = 234 g and its volume = 30 cm³.

(i) \[\text{Density of iron} = \frac{\text{Mass}}{\text{Volume}} = \frac{234 \text{ g}}{30 \text{ cm}^3} = 7.8 \text{ g cm}^{-3}\]

(ii) Since, density of iron = 7.8 g cm⁻³

Its R.D. = 7.8

Example 2

10 cm³ of silver weighs 103 g. Find its:

(i) density in C.G.S.

(ii) density in kg m⁻³

(iii) relative density

Solution

(i) Since, mass = 103 g and volume = 10 cm³

\[\text{Density in C.G.S.} = \frac{103 \text{ g}}{10 \text{ cm}^3} = 10.3 \text{ g cm}^{-3}\]

(ii) Density in kg m⁻³ = Density in C.G.S. x 1000 = 10.3 x 1000 kg m⁻³ = 10300 kg m⁻³

(iii) Density in C.G.S. = 10.3 g cm⁻³

R.D. = 10.3

Example 3

The relative density of a substance is 2.7 and its volume is 100 cm³. Find its mass.

Solution

Since, R.D. = 2.7, therefore density = 2.7 g cm⁻³

Now, \[\text{Density} = \frac{\text{Mass}}{\text{Volume}}\]

\[\Rightarrow \text{Mass} = \text{Volume} \times \text{Density}\]

\[\Rightarrow \text{Mass} = 100 \text{ cm}^3 \times 2.7 \text{ g cm}^{-3} = 270 \text{ g}\]

Example 4

A stone of mass 6000 kg has a volume of 3 m³. Find its:

(i) density in kg m⁻³

(ii) density in g cm⁻³

(iii) relative density

Solution

(i) \[\text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{6000 \text{kg}}{3 \text{ m}^3} = 2000 \text{ kg m}^{-3}\]

(ii) \[\text{Density in g cm}^{-3} = \frac{\text{Density in kg m}^{-3}}{1000} = \frac{2000}{1000} \text{ g cm}^{-3} = 2 \text{ g cm}^{-3}\]

(iii) Since, density in C.G.S. = 2 g cm⁻³

Relative density = 2

Teacher's Note

Understanding density conversions is practical in engineering and construction, where materials must be specified in different measurement systems depending on international or local standards.

Buoyant Force/Upthrust

Whenever a body is immersed partially or completely inside a fluid, it experiences an upward force. This upward force is called the buoyant force or upthrust, whereas the property of the fluids to apply upward force on the body immersed in it is called buoyancy. It is due to the buoyant force (upthrust) that a body submerged partially or wholly in a fluid appears to lose its weight i.e., appears to be lighter.

When a body is immersed in a liquid, two forces act on it.

1. W₁ = Weight of the body which acts vertically downward = True weight of the body

2. W₂ = Upthrust = Buoyant force = Upward force applied by the liquid on the body

Due to upthrust, the body appears to be lighter and its weight appears to be its true weight minus the buoyant force.

Apparent weight of the body = Its true weight - buoyant force = W₁ - W₂

We must note that a body experiences an upthrust in gas too. For example, a balloon filled with hydrogen gas moves in upward direction due to the upthrust of air present underneath it.

It is due to the upward force (buoyant force or upthrust) exerted by water a ship floats in water, although it is made of iron.

Factors On Which The Buoyant Force Depends

The buoyant force primarily depends on two factors:

(a) The volume of the body immersed, i.e., volume of the fluid displaced.

(b) The density of the fluid.

To understand these two factors, let us do the following simple experiments.

(1) Take a hollow metal cube and attach it to the hook of a spring balance.

Note the weight (W₁) shown on the scale of the spring balance.

Immerse the cube, hanging on the spring balance, into water in a beaker and note the reading of the weight (W₂) shown on the scale of the spring balance.

Now take the cube out and hammer it several times, to make it a solid piece. You will notice that it has become smaller by volume after you have hammered it. Now again, put it back into water and note the weight (W₃) in the spring balance. With smaller volume, you will notice that the weight shown is more than the second case when the volume was bigger, i.e., W₃ is more than W₂.

Therefore we conclude, the more is the volume of the body immersed, more is the upthrust because more volume of the body immersed means more amount (volume) of liquid is displaced. This is the reason why a solid iron nail submerges in water but the same nail when beaten into a thin sheet and given some shape, floats in water.

(2) Take a metal cube and suspend it to the hook of the spring balance with the help of a thread. Record the weight (W₁) of the metal cube as shown by the scale of the balance.

Immerse the cube into water in a beaker and record the new weight (W₂) of the cube.

Add salt to the water. Adding salt increases the density of water. Now immerse the object again and note the weight (W₃) shown by the spring balance.

You will notice that weight (W₂) recorded in the second case will be more than in the third case (W₃). It means that the buoyant force in the third case is more than the second case. Hence, the denser is the liquid, the more is the buoyant force.

Note:

W₁ - W₂ = Buoyant force when cube is immersed in water and

W₁ - W₃ = Buoyant force when cube is immersed in salt solution.

Teacher's Note

The buoyant force principle explains everyday experiences like why objects feel lighter in swimming pools and why it's harder to push a beach ball underwater - the water pushes back with greater force the more you try to submerge it.

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