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## What are the Applications of Archimedes’ Principle?

A body appears lighter when It is immersed totally or partly in a liquid or a gas. For example, a pitcher full of water appears lighter when immersed in water. On drawing a bucket full of water from a well, it appears lighter as long as

it is inside water. But once it rises above water, it appears heavier. In no case does the actual weight of a body decrease, but the body experiences an apparent loss of weight.

Definition: The ability of a liquid or a gas at rest to exert an upward force on a body immersed in that fluid is called buoyancy.

When a body is immersed in a liquid or a gas, the liquid or the gas exerts pressure that is different on different parts of the body. The pressure at any point on the body depends on the depth of that point from the free surface or the upper surface of that liquid or gas—as explained in Section 2.5.3—characteristics (iii). Since the immersed body experiences different forces at different points, we can say that the body experiences a resultant force or thrust which acts in the upward direction. This resultant upward thrust balances a part of the downward force due to the weight of the body and hence the body suffers an apparent loss of weight.

Definition: The upward thrust exerted by a liquid or a gas at rest on a body partly or totally immersed in it is called buoyant force.

The terms buoyancy and buoyant force usually have the same meaning.

Magnitude of buoyoncy on a body immersed in a liquid: Let a cube ABCDEFGH be immersed in a liquid of density ρ [Fig.]. The depths of the surfaces ABCD and EFGH of the cube from the upper surface of the liquid are h_{1} and h_{2} respectively. Each side of the cube is of length l.

The surfaces ABCD and EFGII of the cube are horizontal but the other surfaces are vertical. The liquid exerts a normal thrust on each surface of the cube. The lateral thrusts acting horizontally on the mutually opposite vertical faces ABHE and CDFG, being equal and opposite, balance each other.

Similarly, the lateral thrusts acting on the side faces AEFD and BCGH balance each other. So, no lateral resultant thrust acts on the cube in the horizontal direction. But, due to the difference in the depths of the surfaces ABCD and EFGH from the upper surface of the Liquid, unequal thrusts act on these two surfaces.

The downward pressure of the liquid at any point on the surface ABCD = h1pg

∴ The downward thrust on the surface ABCD = pressure × area = h_{1}ρg ᐧ l^{2}

The upward pressure of the liquid at any point on the surface EFGH = h_{2}ρg

∴ The upward thrust on the surface EFGH = h_{2}ρg ᐧ l^{2}

Since h_{2} > h_{1}, the upward thrust acting on the cube becomes greater than the downward thrust.

∴ The net upward thrust on the cube

= h_{2}ρgl^{2} – h_{1}ρgl^{2} = l^{2}ρg(h_{2} – h_{1}) = l^{3}ρg

[∵ h_{2} – h_{1} = l].

But, l^{3} = V = volume of the cube

∴ l^{3}ρg = Vρg = weight of the liquid of a volume equal to that of the cube.

So, the buoyancy on the cube due to the liquid = weight of the liquid displaced by the cube.

This inference is applicable to any object of any shape whatsoever.

Hence, we can say that when a body is immersed partly or totally in a liquid, then the body feels an upward thrust and this upthrust or buoyant force is equal to the weight of the liquid displaced by the body.

From the presence of the factor g in the expression of the buoyant force, it is evident that buoyancy is a direct consequence of the effect of gravity on fluid pressure.

### Characteristics of buoyant force:

i) The buoyant force depends on

- the volume of the immersed portion of the body
- the density of the displaced liquid and
- the acceleration due to gravity

ii) The buoyant force acts in the direction opposite to the weight of the body.

iii) For a totally immersed body, the buoyant force does not depend on the depth up to which the body is immersed

inside the liquid.

iv) If the volume of a body is kept unaltered, then

- the buoyant force on a totally immersed body does not depend on the size, physical state (i.e., solid, liquid or gas) and mass of the body; but
- for a partly immersed body, the buoyant force on it does not depend upon its size or physical state, but depends on its mass.

It is clear that there will be no buoyant force acting on a body if both the body and the liquid are in a weightless condition. This situation arises in an artificial satellite or for a freely falling system.

### Units and dimension of buoyant force:

The dimension of the buoyant force is MLT^{-2}.

Centre of buoyancy: The point where the centre of gravity of the liquid or gas is located before it is displaced by the immersed body is the centre of buoyancy or the centre of floatation of the immersed body.

The buoyant force on an immersed body acts through the centre of buoyancy. The centre of gravity and the centre of buoyancy coincide for a totally immersed body in a fluid. But for a partly immersed body, these two points are different.

Reaction of buoyant force: When a body is immersed in a liquid, the displaced liquid exerts an upthrust or buoyant force on the body, and according to Newton’s third law of motion, the body also exerts an equal but opposite reaction on the liquid. The existence of this reaction can be realised from the following experiment.

Experiment: A beaker containing water is placed on one pan of a common balance and by placing the required counterpoising weights on the other pan, the balance beam is made horizontal. Now, a piece of iron is tied by a thread and is immersed in water by holding the free end of the thread so that the piece of iron does not touch the wall of the beaker. It is seen that the pan with the beaker immediately goes down. As the liquid exerts an upthrust on the piece of iron, the iron piece at the same time applies a downward reaction on the liquid. This reaction ultimately acts on the pan and hence it goes down.

Apparent weight of a body immersed in a liquid: Let the real weight of a body be W_{1}; this weight acts vertically downwards through the centre of gravity of the body When the body is immersed in a liquid, then the buoyancy W_{2} acts in the upward direction through the centre of buoyancy.

So, apparent weight of the body immersed in a liquid = real weight of the body – buoyant force acting on the body (i.e., weight of liquid displaced by the body)

= W_{1} – W_{2}

Hence, the apparent weight of the body < its real weight. So, a body suffers an apparent loss of weight due to the upthrust of the liquid when h is immersed in that liquid.

### Archimedes’ Principle

‘When a body is totally or partly immersed in a liquid or ; gas at rest, it appears to lose a part of its weight. This apparent loss of weight is equal to the weight of the liquid or gas displaced by the body. This is Archimedes’ principle.

It is to be noted, Archimedes’ principle is related to the weight of a body. So, this principle is not applicable for a body in a weightless condition. The weight of the body in an artificial satellite or in a freely falling situation is zero. So in these cases, Archimedes’ principle is not applicable.

### Application of Archimedes’ Principle

We can determine the following with the help of Archimedes principle:

- The volume of a solid of any shape
- The density of a substance
- The amount of the constituent elements in a piece of alloy made of two elements

For the determination of the above things, the choice of the liquid used should be such that

- the body should not be soluble in that liquid and
- the body under experiment should not react chemically with the liquid.

1. Determination of the volume of a solid of any shape: The method described below is the simplest one to find the volume of a body of any shape.

The body is heavier than the liquid: Let the weight of the body in air be W_{1} and its weight when totally immersed in the liquid be W_{2}.

According to Archimedes’ principle,

W_{1} – W_{2} = apparent loss in the weight of the body

= weight of the liquid displaced by the body

= weight of the liquid equal to the volume of the body

If the density of the liquid is p, then volume of the displaced liquid = \(\frac{W_1-W_2}{\rho g}\)

= volume of the body ………. (1)

The body is lighter than the liquid: In this case, a sinker is used to immerse the lighter body under experiment completely in the liquid.

Let the weight body in air be W_{1},

the weight of the sinker inside the liquid be W_{2}

and the weight of the body with the sinker inside the liquid = W_{3}

According to Archimedes’ principle, W_{1} – ( W_{3} – W_{2})

= apparent weight loss of the body

= weight of the liquid displaced by the body

= weight of equal volume of the liquid

If the density of the liquid is ρ, then the volume of the dis-placed liquid = \(\frac{W_1-\left(W_3-W_2\right)}{\rho g}\)

= volume of the body ………. (2)

2. Determination of density of a substance: Let the weight of a body in air be W_{1}; when it is immersed in a

liquid of density ρ, it weighs W_{2}.

According to Archimedes’ principle, from equation (1) we get the volume of the body, V = \(\frac{W_1-W_2}{\rho g}\)

∴ Density of the material of the body,

D \(=\frac{\text { mass of the body }}{\text { volume of the body }}\) = \(\frac{W_1 / g}{V}\) = \(\frac{W_1 \rho}{W_1-W_2}\) …….. (3)

From the measured value of density, the purity of a metal can be known. If the measured value of the density is equal to the actual density of that metal, then we can say that the metal is pure; otherwise it is impure.

3. Determination of the amounts of constituent elements in a piece of alloy made of two elements: Let us assume that an alloy is made of two metals A and B.

Let the mass of the alloy in air be W_{1} and its weight when immersed completely in water be W_{2}.

According to Archimedes’ principle, volume of the alloy V = \(\frac{W_1-W_2}{\rho g}\)

Let us assume that the amount of masses and the densities of metals A and B in the alloy are m_{a}, m_{b} and ρ_{a}, ρ_{b}

∴ Volume of metal A = \(\frac{m_a}{\rho_a}\)

and volume of metal B = \(\frac{m_b}{\rho_b}\) = \(\frac{\left(W_1 / g\right)-m_a}{\rho_b}\)

So, if the values of W_{1}, W_{2}, ρ, ρ_{a} and ρ_{b} are known, then the amount of metal A can be determined from equation (4) and from this, the amount of metal B can be determined.