Archimedes Principle is a fundamental concept in fluid mechanics, credited to the ancient Greek mathematician and physicist Archimedes. According to Archimedes’ Principle, when an object is immersed in a fluid the object experiences an upward force whose magnitude is equal to the weight of the fluid displaced by the object.

Archimedes’ principle has applications in various fields, from ship design to understanding the behavior of objects in water. Archimedes’ Principle is important for class 9 students.

In this article, we are going to learn about what is Archimedes Principle, its formula, how to derive it, its application, and its experimental verification of Archimedes

What is Archimedes’ Principle?

Archimedes Principle states that when an object is submerged in a fluid, it experiences an upward buoyant force equal to the weight of the fluid it displaces. This principle explains why objects appear to weigh less when placed in a fluid and why some objects float while others sink. The buoyant force is directly proportional to the volume of the fluid displaced by the submerged object, and it counteracts the object’s weight. As a result, if the weight of the fluid displaced is greater than or equal to the weight of the object, it will float, while if the weight of the fluid displaced is less, the object will sink.

It was in the year 246 BC that Archimedes had an idea about the weight displacement by liquids. After a lot of tests about the density and volume of different things, he stated the “Archimedes’ Principle”.

Archimedes’ Principle Definition

“Any object, totally or partially immersed in a fluid or liquid, is buoyed up by a force equal to the weight of the fluid displaced by the object.” – Archimedes

It means that when a body is partially or completely submerged in a liquid, it experiences an apparent loss in because of a force acting in upward direction (Buoyancy), which is equal to the weight of liquid displaced by the submerged part of the body. This force acting in upward direction was later named as Upthrust or force acting vertically upwards.

Archimedes Principle Example

The best example of Archimedes Principle is that when a ship is launched into water it displaces the amount of water equal to its weight.

Experimental Verification of Archimedes’ Principle

We have learnt that as per Archimedes Principle, the upthrust experienced by an object immersed in fluid is equal to the weight of the fluid displaced by the object. Let’s us see how we can verify it experimentally.

Step 1: Take a piece of metal and suspend it from a spring balance using a thread and note its weight.

Step 2: Now take a eureka can fill it with water up to its spout and arrange a measuring cylinder below the spout. (Eureka can is a beaker with a spout near its mouth – also invented by Archimedes)

Step 3: Now gently immerse the solid into the can, such that the the overflown water is collected in the measuring cylinder.

Calculation

In the figure, the solid weighs 300gf in air and 200gf in water, and the water collected in the can is 100ml, i.e. 100cm³

∴ Loss in weight = 300gf – 200gf = 100gf ….. (i)

Volume of water displaced = volume of solid = 100cm³

Since, density of water = 1 g cm^-3

∴ Weight of water displaced = 100gf ….. (ii)

From equations (i) and (ii)

Weight of water displaced = Loss in weight.

Thus, the weight of water displaced by a solid is equal to the apparent loss in weight of the solid. This verifies the Archimedes’ Principle.

Archimedes’ Principle Formula

When the object is immersed in the fluid, the object feels lighter due to the loss of apparent weight which is equal to the weight of the fluid displaced by the liquid.

F_B = ρ × V_L × g

Where,

• F_B is the Upward Thurst or the Buoyant Force
• ρ is the Density of the Fluid
• V_L is the Volume of the Fluid Displaced
• g is the Acceleration Due to Gravity

Archimedes Formula Derivation

Lets take a cylinder ‘PQRS’ with upper surface ‘PQ’ and lower surface ‘RS’, dipped in a liquid of density D_L the height of the cylinder is ‘h’ and it is at a depth ‘d’. Since the depth at PQ is d and at RS is (d+h)

Using the formula for pressure (P = ρ × h × g)

P_PQ = D_L × d × g

P_RS = D_L × (d+h) × g

According to the laws of liquid pressure, pressure at a point inside the liquid increases with the depth from its free surface.

So, P_PQ < P_RS

According to the laws of liquid pressure, liquid pressure is the same in all directions about a point in a liquid. So, there will be an upward force acting on the bottom surface RS, which is the Upthrust or Buoyant Force

F_B = P × A where P is Pressure and A is Area

F_B = (P_RS – P_PQ) × A

F_B = ((D_L × (d+h) × g) – (D_L × d × g)) × A

F_B = D_L × g × ( d + h – d) × A

F_B = D_L × g × Ah

F_B = D_L × g × V_o …………….. Volume_object = volume of cylinder = Area × height

F_B = D_L × V_L × g …………….. Volume_object = Volume _{liquid displaced}

Archimedes Principle Cases

There are three possible cases as per Archimedes Principle, these cases are mentioned below:

Case 1: If the F_B = Weight_object then the object will float in completely submerged position;

Case 2: If F_B > W then the object will float;

Case 3: When F_B < W then the object will sink.

Archimedes Principle and Buoyancy

Archimedes Principle in simple words states that when an object is immersed in water it displaces the amount of liquid weight of the object. When object is immersed it experiences an upward force called buoyant force. The magnitude of this buoyant force is equal to the weight of the object immersed.

Applications of Archimedes’ Principle

Archimedes’ principle is crucial in designing ships, submarines, and other watercraft. By understanding how the buoyant force works, engineers can design vessels that can float and maintain stability. This principle is also used to calculate the maximum cargo capacity of ships and to ensure they don’t exceed their weight limits. Let’s see some of the applications of Archimedes Principle

• Submarines : Submarines use Archimedes’ principle to control their depth. By adjusting their ballast tanks to change the amount of water they displace, submarines can either rise to the surface or dive to greater depths.
• Hot Air Balloons : The principle is applied in hot air balloons, where the heated air inside the balloon is less dense than the surrounding air, creating a buoyant force that allows the balloon to rise.
• Life Jackets and Buoyancy Aids : Life jackets and buoyancy aids are designed to help people float in water. They work by increasing the buoyant force acting on the person, thus preventing them from sinking.
• Density Measurement : Archimedes’ principle is used to determine the density of irregularly shaped objects. By measuring the weight of an object in air and then in a fluid (usually water), the volume of the object can be calculated, which, in turn, allows for the determination of its density.
• Hydrometers : A hydrometer is an instrument that measures the density of a liquid, typically used in applications like determining the sugar content in a liquid (e.g., in winemaking or brewing).
• Oil Extraction : In the oil industry, Archimedes’ principle is employed to separate oil from water by using large tanks. The less dense oil rises to the surface, making it easier to collect.

Also, Check

• Difference Between Buoyant Force and Buoyancy
• Equations of Motion
• Laws of Motion
• Real Life Applications of Archimedes’ Principle

Solved Examples of Archimedes’ Principle

Example 1: A body weighs 400 gf in air and 280 gf when completely immersed in water. Calculate: (1)the loss in weight of the body. (2) Calculate the volume of water displaced. (3) The upthrust on the body.

Solution:

Given: Weight of the object in air = 400 gf

Weight of the body in water = 280 gf

(1)Loss in weight in water = 400 gf – 280 gf = 120 gf

(2)Weight of water displaced = weight of object reduced = 120 gf

(3)Upthrust = loss of weight in water = 120 gf

Example 2: A piece of iron of density 7.8× 10³ kg m^-3 and volume 100cm³ is completely immersed in water(ρ = 1000kgm^-3). Calculate : (1) the weight of iron piece in air, (2) the upthrust and, (3) its apparent weight in water.(Take g = 10ms^-2)

Solution:

Given: Volume of iron piece = 100cm³= 10^-4m³

(1)Weight of iron piece in air = Volume × Density × g = 10^-4 × 7.8 × 10³ × 10 = 7.8N

(2) Upthrust = V_L × ρ × g = 10^-4 × 1000 × 10 = 1N

(3) Apparent weight = True weight – Upthrust =7.8N – 1N = 6.8N

Example 3: A metal sphere of radius 7cm³ and density 9gcm^-3 is suspended by a thread and is immersed completely in a liquid of density 3gcm^-3. Find: (1) the weight of the sphere, (2)the upthrust on the sphere and, (3) the tension on the thread.

Solution:

Given: radius of the sphere = 7cm³

volume_sphere = 4/3× π× r³ = 4/3× 22/7× 7× 7× 7 = 4312/3 = 1437.33 cm³

mass_sphere = V × D = 4312 / 3 × 9 = 12936

(1) weight = 12936 gf

(2) upthrust = weight of liquid displaced = V × D × g = 4312 / 3 × 3 × g = 4312 gf

(3) tension on the thread = total downward force = apparent weight of the sphere = original weight – upthrust = 12936 gf – 4312 gf = 8624 gf

Practice Problems on Archimedes’ Principle

Q1. A body of volume 100cm³ weighs 5kgf in air. It is completely immersed in a liquid of density 1.8 × 10³ kg m^-3. Find : (1) upthrust due to the liquid, (2) the weight of the body in the liquid

Q2. A spherical ball of density ρ=7.7g cm^-3 has a radius of r=14cm. If the ball is placed on the surface of water and released, how much of the ball becomes submerged in the water? (g=10ms^-2)

Q3. The mass of a block made of a certain material is 13.5kg and its volume is 15 × 10^-3m³.(1) Calculate the upthrust on the block if it is held full immersed in water, (2)Will the block sink or float on releasing? Give reason for your answer, (3)What will be the upthrust on the block while floating? take D_L = 1000kg m^-3

Q4. A body of mass 3.5kg displaces 1L of water when fully immersed in it. Calculate: (1)The volume of the body, (2)the upthrust on the body, (3) the apparent weight of the body in water.

Q5. A metal cube with edge 5cm and density 9.0 gcm^-3 is suspended by thread so as to be completely in a liquid of density 1.2 g cm^-3. Find the the tension in the thread. (take g = 10 m s^-2)

Q6. A block of wood is floating with tis dimension 75 cm × 75 cm × 75cm inside water. Calculate the buoyant force acting on the the block. Take g = 9.8 N kg^-1.

Q7. You are provided with a hollow iron ball A of volume 15 cm^-3 and mass 12 g and a sold iron ball B of mass 12g. Both are placed on the surface of water contained in a large tub. (1) FInd the upthrust on each ball, (2) Which ball will sink? Give reason for your answer. (Density_iron = 8.0 g cm^-1)

Q8. Two spheres A and B, each of volume 100cm³ are placed on water (density = 1.0 g cm^-3). The sphere A is made of wood of density = 0.3 g cm^-3 and the sphere B is made of iron of density 8.9 g cm^-3. Find: (1) the weight of each sphere, (2) the upthrust on each sphere. Which sphere will float? Give reason.

Q9. State Archimedes Principle.

FAQs on Archimedes Principle

1. What is Archimedes Principle Class 9? Who stated it?

Archimedes’ principle is a property of fluids that was discovered by the ancient Greek mathematician and physicist Archimedes. It states that when an object is partially or completely submerged in a fluid i.e. liquid or gas, it experiences an upward buoyant force that is equal to the weight of the fluid that the object displaces.

2. What is Buoyancy?

Buoyancy is the phenomena due to which an object when immersed in water experiences an upward thrust. The upward force experienced by the object is called Buoyant Force.

3. How can we Experimentally Verify Archimedes’ Principle.

We can experimentally verify Archimedes Principle by immerisng an object in a liquid and observe change in the volume of the water. The detail has been discussed in the article above

4. State 2 factors on which Upthrust depends.

The two factors on which upthrust depends are: (1) The volume of the object submerged, (2)The density of the liquid in which the object is submerged.

5. When does the Object Float?

According to the statement of Archimedes’ principle, the object will float when the weight of the water is displaced equal to the weight of the object.

6. What is an Example of Archimedes’ Principle in our daily life?

One common example of Archimedes’ principle in our daily life is the experience of floating in a swimming pool or bathtub. When you get into the water, your body displaces an amount of water equal to your own volume.

8. What is Archimedes’ Principle of Density?

The Archimedes principle can also be related with density of an object to derive a ratio between the volume of the of liquid and the total volume of the solid along with the densities of the liquid and solid. When an object is placed in water and is floating: V_L ×D_L× g = V_O ×D_O ×g [W = F_B] ⇒ V_L×D_L = V_O×D_O ⇒ D_L/D_O = V_O/V_L. Thus we can conclude the ratio D_L : D_O :: V_O : V_L, for floating bodies. For sinking bodies W > F_B and D_O > D_L.