This document provides information about a physics class covering physical quantity and measurement, density, and floatation. It defines density as the mass per unit volume and provides conversions between kg/m3 and g/cm3. It describes methods to measure the density of irregular solids and liquids using a measuring cylinder or Eureka can. The principles of floatation and applications such as ships, swimming, and submarines are also summarized. Numerical problems are included for practice calculations.

1. WELCOME TO MY CLASS
Mr. Souvik Chatterjee
M.Sc (Physics ) B.Ed

2. PHYSICAL QUANTITY &
MEASUREMENT
Measurement of density of irregular solids
using : Eureka can, Measuring cylinder
Measurement of density of fluids: Basic
Concept, Concept of floatation and sinking of
substance (relate to density)
Comparision of densities in the three states of
matter

3. DENSITY
Density of a substance is defined as amount of
mass per unit of volume of that substance
 S.I unit of density kilogram/metre3 (kg/ m3 )
 C.G.S unit of density gram/centimetre3(g/cm3)
1 g/cm3=1000 kg/m3
 1 kg/m3=1/1000 g/cm3

4. 1 kg/m3 = 1kg/ 1m3 =1000g/(100)3 cm3 =
1g/1000 cm3 = 1/1000 g/cm3
1 g/cm3 = 1000 kg/m3
 1 g/litre= 1g/1 litre= 1g/1000 ml = 1/1000
g/ml= 1/1000 g/cm3 = (1/1000 X
1000) kg/ 1m3 = 1kg/ 1m3
 Therefore, 1 g/litre = 1/1000 g/cm3 &
1g/litre = 1kg/1m3

5. NUMERICALS ON CONCISE
PHYSICS
1.Density of air is 1.28 g/litre. Express it in: (a)
g/cm3 (b) kg/m3 .
(a)1.28 g/litre= 1.28/1000=0.00128 g/cm3
(b)1.28 g/litre= 1.28 kg/m3
 2.The dimension of the hall are 10m X 7m X 5m.
If the density of air is 1.11 kg/m3, Find the mass
of air in the hall.
 Volume=(10 X 7 X 5)=350 m3
 Mass= 350 X 1.11 = 388.5 kg

6. 3. Density of aluminium is 2.7 g/cm3 . Express
it in kg/m3
 2.7 g/cm3 = (2.7 X 1000) kg/m3 = 2700
kg/m3
4. Density of alcohol is 600 kg/m3. Express it
in g/cm3 .
600 kg/m3 = 600/1000= 0.6 g/cm3
5. A piece of zinc of mass 438.6 g has a
volume of 86 cm3.Calculate the density of
zinc.
Density= 438.6/86= 5.1 g/cm3

7. 6. A piece of wood of mass 150 g has a volume of
200 cm3. Find density of wood in (a) C.G.S Unit
(b) S.I unit
(a) Density in C.G.S unit = 150/200 = 0.75 g/cm3
(b) Density in S.I unit = (0.75 X 1000) =
750 kg/m3
7. Calculate the volume of wood of mass 6000 kg
if the density is 0.8 g/cm3.
0.8 g/cm3 = (0.8 X 1000) kg/m3= 800 kg/m3
 Volume = 6000/800 = 7.5 m3

8. KEY POINTS
 The density of a substance does not change with any
change in its shape or size.
 Almost all substances expand on heating and contract
on cooling, but their mass does not change. So the
density of a substance decreases with the increase in
temperature and increases with the decrease in
temperature.
 Exception in water which contracts on heating above 4
degree C. So density of water increases from 0 degree
C to 4 degree C and then decreases above 4 degree C.
 The density of water is maximum at 4 degree C which
is equal to 1000 kg/ m3 or 1 g/ cm3

9. VESSELS FOR MEASURING
VOLUME
Measuring Cylinder

10. MEASURING CYLINDER
It is made up of glass or plastic and is graduated
in mililitre (ml) with zero mark at the bottom.
The graduations are increasing upwards.
We have measuring cylinders available of
different capacities, such as 50 ml, 100 ml, 200
ml, 500 ml etc.
The capacity of measuring cylinder is marked on
it.

11. Measuring Beaker

12. MEASURING BEAKER
A measuring beaker is made up of glass,
plastic or metal like aluminium.
 It is used to take out a fixed volume of liquids
(say milk, oil etc) such as 50 ml, 100 ml, 200
ml, 500 ml, 1 litre from a large container.
 The capacity of measuring braker is marked
on it.

13. Eureka Can

14. EUREKA CAN
A Eureka can is a glass (or polythene) beaker
with a side opening near its mouth which is
known as spout.
Thus, the beaker can contain a volume of
liquid up to the spout.
 Any excess of liquid overflows through the
spout.

15. DETERMINATION OF DENSITY OF AN
IRREGULAR SOLID
Using Mesuring Cylinder

16.  Measure the mass of the given solid using a common
balance/ spring balance.
 Take a cylinder. Fill it partly with water. Note the level
of water, let it be V1 ml.
 Now tie the solid with a thread and gently lower the
solid in water contained in the measuring cylinder. Take
care that no water splashes out. Note the level of water
again. Let it be V2 ml.
 Find the difference, V2−V1. It gives the volume V of
the solid. i.e
 V = V2−V1 cm3
 Then calculate the density of the solid by
using the following formula
Density=Mass/Volume= M/V

17. PROBLEM FROM CONCISE
PHYSICS
8.Calculate the density of solid from the
following data:
(a) Mass of solid = 72 g
(b) Initial volume of water in measuring cylinder
= 24 ml
(c) Final volume of water when solid is
completely immersed in water = 42 ml
Volume of solid = (42 − 24) = 18 ml = 18
cm3
 Density = 72/18 = 4 g/cm3

18. Using Eureka Pan

19.  Pour water into the can until it starts overflowing through the
spout. When the water has stopped dripping, remove the
measuring cylinder. Empty it, dry it and again place it under
the spout.
 Now tie the irregular solid by a thread. Immerse the solid
gently into the water contained in the Eureka can. The solid
displaces water equal to its own volume which overflows
through the spout and gets collected in measuring cylinder.
When water stops dripping out through the spout, note the
volume of water collected in the measuring cylinder [100
cm3]
 Dry the solid. Measure the mass M of the given solid with a
bam balance/ spring balance.
 Let the mass be 300 g.
 Volume of the solid (V) = Volume of water collected in the
measuring cylinder= 100 cm3
 Density = M/V = 300/100 = 3 g/cm3

20. DETERMINATION OF DENSITY OF A
LIQUID
To determine the density of a liquid (milk or oil
etc.), its mass M is measured by a common
beam balance and its volume V is measured by
a measuring cylinder. Then density is
calculated by using the relation
D=M/V

21. DENSITY BOTTLE

22. DENSITY BOTTLE
 Density bottle is specially designed bottle which is used to
determine the density of a liquid.
 It is a small glass bottle having glass stopper at its neck. The
bottle can store a fixed volume of a liquid.
 Generally the volume of bottle is 25 ml or 50 ml.
The stopper has a narrow hole through it.
 When the bottle is filled with the liquid and the stopper is
inserted, the excess liquid rises through the hole and drains
out. Thus, the bottle contains the same volume of liquid
each time when it is filled.
 Since the density of water is 1 g/cm3, so the mass(in g) of
water needed to fill the bottle completely, will give the
volume (in ml) of bottle.
 Now a days it is named as specific gravity bottle.

23. RELATIVE DENSITY
The relative density of a substance is defined
as the ratio of the density of the substance to
the density of water at 4 degree C

24. Or,
Or,
Density of water at 4 degree C =1 g/cm3

25. Density of water at 4 degree C S.I UNIT= 1000
kg/m3

26. But density of a substance is the mass of 1 cm3 of that
substance, therefore we can also express the relative
density of a substance as follows:
Thus, relative density of a substance can also be defined
as the ratio of the mass of any volume of the substance to
the mass of an equal volume of water

27. Determination of the relative density
using a density bottle
• R.D of liquid

28. NUMERICALS FROM CONCISE
PHYSICS
 9. The mass of an empty density bottle is 21.8 g, when filled
completely with water it is 41.8 g and when filled completely
with liquid it is 40.6 g. Find:
(a) the volume of density bottle
(b) the relative density of liquid
Ans.(a) Mass of water having volume same as density bottle =
(M2 − M1) = (41.8 − 21.8) = 20 g
Volume of water = Volume of density bottle = (Mass of
water)/(Density of water in C.G.S unit) = 20 g/ (1 g/cm3) = 20
cm3 = 20 ml [ As, 1 cm3 = 1 ml]
(b) Mass of liquid having same volume as density bottle = (M3 −
M1) = (40.6 − 21.8) = 18.8 g
Relative Density = (M3 − M1)/ (M2 − M1) = 18.8/20.0 = 0.94

29. 10.From the following observations, calculate
the density and relative density of a brine
solution.
Mass of empty bottle = 22 g
Mass of bottle + water = 50 g
Mass of bottle + brine solution = 54 g
Ans. Density in C.G.S unit = Relative density =
(M3 − M1)/ (M2 − M1) = (54 − 22)/ (50 − 22) =
32/28 = 1.14
Density = 1.14 g/cm3

30. 11.The mass of an empty density bottle is 30 g, it is 75 g when
filled completely with water and 65 g when filled completely
with liquid. Find :
(a)Volume of density bottle
(b)Density of liquid, &
(c)Relative density
Ans. Mass of water having same volume as density bottle = (75
− 30) = 45 g
Volume of density bottle = Volume of water = Mass/ Density in
C.G.S unit = 45 g/1g/cm3 = 45 cm3 = 45 ml [As,1cm3=1 ml]
Mass of liquid having same volume as density bottle = (65 −
30) = 35 g
Density of liquid = Mass/ Volume of liquid =
Mass/ Volume of density bottle = 35/45 = 0.77 g/cm3
R.D = Density in C.G.S unit = 0.77

31. FLOATING AND SINKING
A body floats on a liquid if its density is less
than the density of liquid.
A body sinks in a liquid if its density is more
than the density of liquid.
Example:- A piece of cork(density = 0.25
g/cm3 ) floats and an iron nail(density = 7.86
g/cm3) sinks in water.

32. PRINCIPLE OF FLOATATION
When the body is immersed in a liquid, the following
two forces act on it:
(1)The weight of the body W acting vertically
downwards. This force has a tendency to sink the body.
(2) The buoyant force of the liquid FB acting vertically
upwards. The buoyant force is equal to the weight of
the liquid displaced by the immersed part of the body.
This force has a tendency to move the body up. This is
why buoyant force is also called upthrust.

35. • (1)In this case, the resultant force on the
body is (W−FB) which acts downwards.
The body will sink in the liquid to the
bottom under the influence of the
resultant force (W−FB). This is shown in
the central diagram. This happens when
the density of solid is greater than the
density of liquid.

36. • (2)The weight of the body W is equal to
the buoyant force FB. In the case, the
resultant force on the body is zero i.e the
apparent weight of the body is zero. The
body will float just inside the surface of
liquid. This is shown the right most
diagram. This happens when the density
of solid is equal to the density of liquid.

37. • (3)The weight of the body W is less than the
buoyant force, if the body is completely immersed
in liquid. In this case, the resultant force acts on
the body upwards. The body will float partially
above the surface of liquid. Only that much
portion of the body will immerse inside the liquid
by which the weight of the of liquid displaced FB
balances the total weight of the body. This is
shown in left most diagram. This happens when
density of solid is less than density of liquid. So,
now while floating, FB=W, so apparent weight is
zero.

38. LAW OF FLOATATION
When a body floats in a liquid, the weight of the liquid
displaced by its immersed part is equal to the total
weight of the body. This is the law of floatation, i.e
while floating
Weight of the floating body=Weight of the liquid
displaced by its immersed part(i.e, buoyant force).
In other words, according to the law of floatation, the
apparent weight of a floating body is zero. A body
floats with its more portion outside the surface of liquid
of more density than that of low density.

39. SOME APPLICATION OF
FLOATATION
• (1)Floatation of an iron ship: A nail made of iron
sinks in water, but a ship made of iron does not.
The reason that a nail is solid and the density of
iron is greater than that of water. The weight of
the nail is more than the buoyant force of water
on it. So the nail sinks in water. On the other
hand, the ship is hollow and the empty space
contains air. This makes the average density of
ship less than that of water. Therefore, a ship
floats on water.

41. (2)Floatation of man: It is easier for a person
to swim in sea water than in river water. The
reason is that sea water contains salt and so its
density is more than the density of river water.
The weight of a man gets balanced by the less
immersed part of his body in sea water as
compared to that in river water. Thus, it is
easier to swim in sea water than in river water.

43. • (3)Floatation of ice in water: A piece of ice
floats on water with its 9/10 part of inside the
water and only 1/10 part of it outside the water.
The reason is that the density of ice is 0.9
g/cm3 while the density of water is 1 g/cm3.
Hence, the weight of water displaced by 9/10
th part of ice immersed inside water becomes
equal to the total weight of the ice piece.

45. (4) Submarine: A submarine can be made to dive or to
rise to the surface of water as and when desired. The
reason is that a submarine is a water-tight boat which
can travel under water like a ship. A submarine is
provided with water tanks. To make the submarine
dive, the tanks are filled with water so that the average
density of the submarine becomes greater than the
density of sea water and it sinks. To make the
submarine rise to the surface of water , these tanks are
emptied. This makes the average density of the
submarine less than the density of sea water, so the
submarine rises up to the surface of water.

47. • (5)Iceburgs: Very huge and large pieces of ice
floating on sea water, are called iceburgs. They
are dangerous for ship. The reason is that the
density of ice is less than the density of sea water.
The density of ice is 0.9 g/cm3 and the density of
sea water 1.02 g/cm3. Hence, an iceburg floats in
sea water with its large portion submerged inside
the water and only a little portion of it is above
the surface of water. Thus, a ship can collide with
the invisible part of iceburg under the surface of
water, hence it is dangerous for ships

49. • (6)Whales: The whales can sink or rise at their
will. Whales are sea animals, They have special
organ in their body which is called swim bladder.
In order to come to the surface of water, they fill
the bladder with air. This decreases the average
density of the whale and so rises to the surface. To
dive into the sea, they empty the bladder. This
increases the average density of the whale and so
it sinks.

51. • (7)Balloons: A hydrogen or helium filled
balloon rises in air. The reason is that the
density of these gases is less than the density
of air. Therefore, the buoyant force
experienced by the balloon due to air, becomes
greater than the weight the weight of the
balloon. Hence, the balloon rises up under the
influence of the net upward force.