1) The document discusses Archimedes' principle and buoyancy. It provides examples measuring the weight and buoyant force of objects in air and water. 2) Archimedes' principle states that the buoyant force on an object immersed in a fluid is equal to the weight of the fluid the object displaces. 3) The document uses examples to demonstrate how to calculate buoyant force and determine if an object will float or sink based on its density compared to the fluid.

1. Buoyancy and Archimedes’
principle
For class -IX

2. Introduction
Measurement in air by
spring scale
Measurement in water by
spring scale
Mass of gold brick 1.5 kg 1.5 kg
Weight of gold brick 14.7 N 13.9 N
Buoyant force 9.1x10-4 N 0.8 N
The difference in weight in air and water = 0.8 N
It is weight of the fluid displaced by the body, and it is called force of buoyancy
Archimedes’ principle talks of this buoyancy force
Buoyancy force depends upon the density of fluid
Air has very less density, so buoyancy force is negligible in air
Some Data:
Density of gold brick= 19.3 x 103 kg/m3
Density of Water= 1000 kg/m3
Density of air = 1.2 kg/m3

3. Archimedes’ challenge
• Whether the gold crown of the king was pure or impure?
• Step 1: Measurement of spilled water from the water filled vessel when
pure gold (of similar mass as of gold crown) was immersed
• Step 2: Measurement of spilled water from the water filled vessel when
gold crown was immersed
• Step 3: Measurement of difference between quantity of spilled water
• Test: No difference in quantity of spilled water: gold crown pure
• Actual finding: gold crown displaced more water than pure gold.
Hence, gold crown impure

4. Learning objective
• What is Buoyancy and Buoyant force?
• When a body will sink or float in fluid?
• How to calculate Force of Buoyancy?

5. Archimedes’ Principle
• When a body is completely or partially
immersed in a fluid, the fluid exerts an upward
force on the body equal to the weight of the
fluid displaced by the body

6. Archimedes’ Principle-part I
• A body immersed in a fluid experiences a
vertical buoyant force equal to the weight of
the fluid it displaces
• Buoyant Force, FB =weight of fluid equivalent
to body volume

7. Archimedes’ Principle -part II
• A floating body displaces its own weight in the
fluid in which it floats
• Floating bodies are special case
• Only a portion of the body is submerged
• Buoyant Force for floating body, FB =weight
of fluid equivalent to displaced volume

8. Examples
• In what direction does the buoyant force on an
object immersed in a liquid act?
• Why does a block of plastic released under
water come up to the surface of water?

9. Example
• The volume of 50 g of substance is 20 cm3 . If the
density of water is 1 g cm-3 , will the substance float or
sink?
Ans) Body Volume = 20 cm3
Mass of body = 50 g
Density of body = Mass of body/Body Volume
= 50/20 g/cm3 = 2.5 g/cm3
2.5g/cm3 > 1 g/cm3
or Density of body > Density of water
Hence, body will sink

10. Example
• The volume of a 500 g sealed packet is 350 cm3 . Will the packet float or sink
in water if the density of water is 1 g cm-3 ? What will be the mass of the water
displaced by this packet? (given g = 1000 cm s-2 )
Ans) Volume of body = 350cm3
FB = weight of liquid equivalent to body volume
FB = liquid x Vbody x g
FB = (1 g/cm3 )(350 cm3)(1000 cm s-2)
FB = 350000 Dynes
Wbody = mg = 500 x1000 = 500000 Dynes
Since Wbody > FB , object will sink
Amount of water displaced = mass of liquid= density of liquid x body volume
= 1 g cm-3 x 350 cm3
= 350 g
FB
mg

11. Example
• A ball weighing 4 kg of density 4000 kg m-3 is
completely immersed in water of density 103 kg m-3 .
Find the force of buoyancy on it. (given g= 10 ms-2 )
FB = weight of fluid equivalent to body volume
=mass of fluid x g
=density of fluid x body volume x g
= 103 kg m-3 x (4/4000 m3 ) x 10 ms-2
= 10 kg ms-2 = 10 N
mg
FB