Archimedes’ Principle states that when a body is wholly or partially immersed in a fluid, it experiences an upward buoyant force equal to the weight of the fluid displaced by it. This buoyant force opposes the weight of the object. If the displaced fluid's weight is equal to or greater than the object's, it floats; if not, it sinks.
Example: The best example of Archimedes' principle is that when a ship is launched into water, it displaces an amount of water equal to its weight.
Experimental Verification
We have learned 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.
Step 1: Take a piece of metal and suspend it from a spring balance using a thread. Note its weight in air.
Step 2: Now take an overflow can (a container with a side spout near the top) and fill it with water up to the level of the spout. Place a measuring cylinder below the spout to collect the overflowing water.
Step 3: Gently immerse the solid completely into the can so that the displaced water flows out through the spout and is collected in the measuring cylinder.
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 object.
\boxed{F_B = \rho \times V_L \times g}
Where,
- FB is the Upward Thrust or the buoyant force.
- ρ is the Density of the fluid.
- VL is the Volume of the fluid displaced.
- g is the Acceleration Due to gravity.
Archimedes' Formula Derivation
Let's take a cylinder 'PQRS' with upper surface 'PQ' and lower surface 'RS', dipped in a liquid of density DL. 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)
PPQ = DL × d × g
PRS = DL × (d+h) × g
According to the laws of liquid pressure, pressure at a point inside the liquid increases with the depth below its free surface.
So, PPQ < PRS
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.
FB = P × A, where P is Pressure and A is area.
FB = (PRS - PPQ) × A
FB = ((DL × (d+h) × g) - (DL × d × g)) × A
FB = DL × g × (d + h - d) × A
FB = DL × g × Ah
FB = DL × g × Vo ... Volume object = volume of cylinder = Area × height
FB = DL × VL × g ... 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 FB = Weight of the object, then the object will float in a completely submerged position.
Case 2: If FB > W then the object will float.
Case 3: When FB < 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 an amount of liquid equal to the object's weight. When an object is immersed, it experiences an upward force called the buoyant force. Buoyant force is equal to the weight of the displaced fluid. It equals the weight of the object only when the object is floating in equilibrium.
Applications
- Submarines: They control their depth by adjusting ballast tanks, filling them with water to sink and expelling water to rise.Submarines control their depth by adjusting ballast tanks, changing buoyant force to rise or sink.
- Hot Air Balloons: Heated air inside the balloon becomes less dense than the surrounding air, creating a buoyant force that lifts it upward.
- Life Jackets: They increase the overall volume and buoyant force on a person, helping them float in water.
- Hydrometers: Used to measure the density of liquids based on the level at which they float.
- Oil Separation: Oil floats on water because it is less dense, allowing easy separation in oil extraction processes.
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Solved Problems
Question 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
Question 2: A piece of iron of density 7.8 × 103 kg/m³ and volume 100 cm³ is completely immersed in water (ρ = 1000 kg/m³). Calculate (1) the weight of the iron piece in air, (2) the upthrust, and (3) its apparent weight in water. (Take g = 10 ms⁻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 = VL × ρ × g = 10-4 × 1000 × 10 = 1N
(3) Apparent weight = True weight - Upthrust =7.8N - 1N = 6.8N
Question 3: A metal sphere of radius 7 cm and density 9 g/cm³ is suspended by a thread and is immersed completely in a liquid of density 3 g/cm³. 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
volumesphere = 4/3× π× r3 = 4/3× 22/7× 7× 7× 7 = 4312/3 = 1437.33 cm3
masssphere = V × D = 4312 / 3 × 9 = 12936
(1) weight = 12936g =12936*10 = 129360 dyne
(2) upthrust = weight of liquid displaced = V × D × g = 4312 / 3 × 3 × 10 = 43120 dyne
(3) tension on the thread = total downward force = apparent weight of the sphere = original weight - upthrust = 12936 g - 4312 g = 8624 *10 = 86240 dyne
Unsolved Problems
Q1. A body of volume 100 cm³3 weighs 5 kgf in air. It is completely immersed in a liquid of density 1.8 × 103 kg/m³. Find: (1) upthrust due to the liquid and (2) the weight of the body in the liquid
Q2. A spherical ball of density ρ=7.7 g cm⁻³ has a radius of r=14 cm. 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 13.5 kg, and its volume is 15 × 10⁻³ m. (1) Calculate the upthrust on the block if it is held fully immersed in water. (2) Will the block sink or float on releasing? Give a reason for your answer. (3) What will be the upthrust on the block while floating? take DL = 1000kg m⁻³
Q4. A body of mass 3.5 kg displaces 1 L of water when fully immersed in it. Calculate: (1) the volume of the body, (2) the upthrust on the body, and (3) the apparent weight of the body in water.
Q5. A metal cube with edge 5 cm and density 9.0 g/cm³ is suspended by thread so as to be completely in a liquid of density 1.2 g/cm³. 3. Find the tension in the thread. (take g = 10 m/s⁻¹)