CBSE Class 9 Syllabus 2027 Science at Advanced Level

FOR CBSE CLASS 9 SYLLABUS EXAM PREPARATION

CBSE Class 9 Syllabus
2027
Question Paper ·
Science
EXAM YEAR TYPE SUBJECT

CBSE Class 9 Syllabus 2027 Question Paper Science
DETAILS

at Advanced Level

Notes · Sample Papers · Previous Year Papers · Mock Tests

.
.co s e m

s em l a
a ag

Science at Advanced Level m
om (Optional) . co
. c e m
em l as
l as ag
ag
Grade 9

m
m .co
s e
l a
ag
2026-27

Academic Unit,
o m
o mCentral Board of Secondary c
Education
.
. c Integrated Office Complex, Sector-23, Phase - 1, Dwarka, e m
e m las
las New Delhi - 110077
ag
ag

m 1 . c
c. o s e m
s em
For more Question Papers, Sample Papers, Notes & Syllabus visit
gl a
Page 1 of 85

Table of Contents

S. No. Chapters Pages

1. Measurement – Foundation of Science 4-8

2. Understanding Motion through Experience 9-14

Newton’s Laws of Motion 15-24
3.

The Geometry of Power – Advanced Simple 25-29
4.
Machines
Work and Energy 30-33
5.

Structure of Atom 34-41
6.

Chemical Bonding 42-48
7.

Mixtures and their Separation 49-53
8.

Microscope and Microscopy 54-69
9.

Engineering Life: Miracles in Biotechnology 69-86
10.

2

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 2 of 85

Content Development Committee

Mr. Aishwary Meet, Physics Expert, Gaur International School, Noida
Ms. Alka Gupta, TGT Science, Mayoor School, Sector-126, Noida
Dr. Amit Sehgal, Professor, Hansraj College, University of Delhi
Dr. Anita Verma, Retired Professor, Kirori Mal College, University of Delhi
Ms. Cenkush Sharma, TGT, Vandana International School, Dwarka, Delhi
Ms. Geetanjali Padhy, Biology Educator, Suncity School, Sector-54, Gurugram
Dr. Girish Choudahry, Retd. Associate Professor, Lady Irwin College,
University of Delhi
Professor K.K Arora, Professor, Zakir Husain College, University of Delhi
Ms. Mridula Arora, Head of School, Navayug School, Sarojini Nagar, New
Delhi
Ms. Pamila Marwaha, PGT Chemistry, Navayug School, Sarojini Nagar, New
Delhi
Dr. Renu Parashar, Professor, Hansraj College, University of Delhi
Dr. Sanjeev, Professor, IGNOU
Ms. Shivani Kaplish, TGT Science, Mayoor School, Sector-126, Noida
Ms. Varsha Krishnan, TGT, Bal Bharati Public School, Rohini, Delhi

3

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 3 of 85

.
.co s e m

s em l a
a ag

Measurement- The Foundation of Science

o m
m
oWhat is Measurement? . c
1.1 Introduction:
Measurement m
. c
is the process of comparing an unknown quantity with a known s
em
s e of the same kind. Physics is based on measurement. Whethergla
la the length of a classroom, the mass of a bag, or the time taken bya a
standard quantity
g
a
we measure
runner, accurate measurement is essential

Examples
● A tailor measures cloth in meters.
● A doctor measures body temperature in degree Celsius.
● A shopkeeper measures rice in kilograms.

m
.co
Without proper units, these measurements would have no meaning.

e m
l a s
Activity 1.1 Measuring the area of the classroom floor

ag :l :l = 1:2:3
Materials: Take three sticks of length l 1 2 3

Procedure:
1. Divide students in 3 groups and hand over one stick to each group.
2. Each group will measure the length and width of the classroom taking stick
as one unit.
Stick 1 Stick 2 Stick 3
Length of wall ……… units* ……… units ……… units
m
m Breadth of wall ……… units ……… units
.co
……… units

m .co Area of floor ……… units2
s em
……… units2 ……… units2

s e g la
*the unit refers to length of the stick used for measurement

g la a
a 3. Compare the length, breadth and area measured by different groups and try
to generate conclusions between unit chosen and numerical values
obtained by all the three groups.

This activity demonstrates that the numerical value of a quantity is inversely
proportional to the size of the unit used. Thus, when a larger unit (longer stick) is
used to measure the classroom floor, the numerical value obtained is smaller, and

om
4
. c
. c e m
m as
se
For more Question Papers, Sample Papers, Notes & Syllabus visit
gl
Page 4 of 85

when a smaller unit is used, the numerical value is larger.

In measurement, the physical quantity remains constant even when the unit changes.
Hence,
Q = n₁u₁ = n₂u₂
Therefore,
n₂ = n₁ (u₁ / u₂)

Activity 1.2: Let’s play an estimation game
Procedure:
1. Estimate the length of the blackboard without measuring.
2. Then measure its length using a meter scale.
3. Compare the estimated and measured values. Now, calculate the
inaccuracy (error) in the measurement.

1.2 Different Systems of Units
In earlier times, different regions/places used their own units of measurement,
which often led to confusion and errors.
(a) CGS System
● Length: centimeter (cm)
● Mass: gram (g)
● Time: second (s)
It is mainly used in laboratory and scientific calculations.
(b) FPS System
● Length: foot (ft)
● Mass: pound (lb)
● Time: second (s)
Commonly used in the United States.
(c) MKS System
● Length: meter (m)
● Mass: kilogram (kg)
● Time: second (s)
This system later developed into the SI (International System of Units or
Système International d'Unités) system.

5

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 5 of 85

Example
● The height of a person is largely measured in centimeters in India, while in
some countries it is measured in feet and inches.

1.3. Need for a Common System of Units
Different systems of units caused difficulties in communication, trade, and scientific
research.
Problems Without a Common System
● Confusion in international trade
● Errors in scientific calculations
● Difficulty in sharing scientific data
Example
If a scientist in India measures length in meters and another in the USA measures
in feet, comparison becomes difficult unless a common unit is used.
Hence, a universal system of units was required.

Activity 1.3: Let us Compare
Materials: Ruler marked in cm and inches
Procedure:
1. Measure the length of a book using both cm and inches.
2. Compare the values and find the relation between them.

1.4. International System of Units (SI)
The International System of Units (SI) is the modern and universally accepted
system of measurement.

SI Base Units

Physical Quantity SI Unit Symbol
Length Meter m
Mass Kilogram kg
Time Second s
Temperature Kelvin K
Electric Current Ampere A

6

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 6 of 85

.
.co s e m

s em l a
a ag
Luminous Intensity Candela Cd
Amount of substance Mole mol
Advantages of SI Units
● Internationally accepted
m
co
● Easy to use and understand
m system .
c. o
● Based on the decimal

s e m
Examples em a
as of vehicles is measured in m/s or km/h g l
l
● Speed
g a
● aMedicines are measured in milligrams (mg)

1.5 Conversion of Units between different Systems
Sometimes, we need to convert a measurement from one unit to another to ensure
comprehension across different systems.
During unit conversion, the numerical value and the unit may change, but the
magnitude of the physical quantity remains the same.
m
Basic Conversions
m .co
● 1 km = 1000 m
s e
l a
ag
● 1 m = 100 cm
● 1 kg = 1000 g
● 1 hour = 3600 s

Examples
Q. Convert 9 km/hr into m/s.
×
Answer: 9 km/hr = = m/s
Q. Convert 1 N into gcm/s2. o m
m . c
. co Answer: 1 kg m/s = 1000g x 100 cm/s = 10 gcm/s = 10 dyne.
2

s
2

e m 5 2 5

s em Quick Check g l a
g la a
a 1. Name any two systems of units.
2. Why is SI system preferred over other systems?
3. Convert 250 N into gcm/s2.
4. Convert 1000 kg/L into kg/m3.

Check Your Understanding

om
7
. c
. c e m
m as
se
For more Question Papers, Sample Papers, Notes & Syllabus visit
gl
Page 7 of 85

1. Which of the following is not an SI unit?
a) Meter
b) Kilogram
c) Second
d) foot
2. The SI unit of mass is:
a) Gram
b) Kilogram
c) Pound
d) tonne

3. Name the system of units used internationally.
4. Why is a common system of units necessary?
5. Why is measurement necessary in physics?
6. Why was there a need for a common system of units?
7. Explain the relation: Magnitude = Numerical value × Unit
8. Why does the same classroom floor give different numerical values when
measured with sticks of different lengths?
Answer questions 9 to 11 that are based on Activity 1.1 (Measuring
Classroom Floor)
Suppose:

Stick Length Length of Wall Breadth of Wall
1 unit 30 units 20 units
2 units 15 units 10 units
3 units 10 units 6.6 units

9. Why are numerical values different?
10. Is the actual size of the classroom different? Why or Why not?
11. What conclusion can you draw about units and measurement from this
activity?
12. Fill in the blanks

a. Measurement is the process of comparing an unknown quantity with a
__________ quantity.
b. The SI unit of mass is __________.

8

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 8 of 85

c. In CGS system, the unit of length is __________.
d. 1 km = __________ m.
e. The modern internationally accepted system of units is called
__________.
13. Match the following:
Column A Column B
CGS Kelvin
FPS Pound
SI International system
MKS Meter-Kilogram-Second

14. What problems might occur if every country used its own system of units for
measurement?
15. A scientist measures length in feet and another in meters. What difficulties
may it lead to?
16. If 1 meter was defined differently in different countries, what would happen
to international trade?
17. A shopkeeper sells rice using kilograms. A foreign customer asks for rice in
pounds.
a. Why is unit conversion necessary here?
b. If 1 kg = 2.2 pounds, how many pounds are there in 5 kg?

************************************************************************************************

9

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 9 of 85

.
.co s e m

s em l a
a ag
Understanding Motion Through Experience

Reflect on the following:
m
co
● Why do we feel pushed backward when a bus suddenly starts moving?

. c om
● Can an object be at rest for one observer but moving for another?
e m .
em
● How do athletes decide the best angle to throw a ball so that it travels the
l as
l as
maximum distance?
ag
g
● Can we measure motion using simple tools available in the classroom?
a
Discuss your ideas with classmates before beginning the activities.

2.1 What is Motion?
An object is said to be in motion if its position changes with time with respect to a
reference point. Motion can be slow or fast, straight or curved, uniform or
non‑uniform. Understanding motion becomes easier when we observe it directly
and measure it ourselves.
m
Activity 2.1: Let’s observe
m .co
s e
Materials: Notebook, stopwatch (mobile timer), measuring tape
l a
Steps:
ag
1. Mark two points 5 meters apart in the classroom corridor or playground.
2. Ask one student to walk normally from one point to another while another
student measures the time taken using a stopwatch.
3. Repeat the experiment with the student running.
4. Record the distance and time in a table.
Observation: - Compare the time taken for walking and running. - Which motion is
m
mfaster? How can you calculate speed?
c. oCan motion be described using measurable quantities such as distance
m and time? .co
s e
s em 2.2 Frame of Reference g l a
la a
ag A frame of reference that is at rest or moving with constant velocity is called an
inertial frame. Newton’s laws hold without modification in such frames. A frame that
is accelerating is called a non-inertial frame, and we will see in later sections that
special corrections (pseudo forces) become necessary in such frames.
To describe the state of motion of an object, we must specify a reference point.
Without it, we cannot say whether an object is moving or at rest.

om
10
. c
. c e m
m as
se
For more Question Papers, Sample Papers, Notes & Syllabus visit
gl
Page 10 of 85

Activity 2.2: Motion is Relative
Materials: Two students as props
Steps:
1. Let one student stand still while another walks past him.
2. Ask each student to describe the motion of the other student.
3. Now let both students walk in the same direction with the same speed and
describe the motion again.
Discussion: - When both students walk together at the same speed, they appear
at rest relative to each other but moving relative to the classroom.

2.3. Scalars and Vectors
Physical quantities are of two types: -
Scalars: Quantities having magnitude only (distance, time, mass, speed and work)
Vectors: Quantities having both magnitude and direction (displacement, velocity,
force).

Activity 2.3: Direction Matters
Materials: Chalk, measuring tape
Steps:
1. Draw a straight 5‑metre line on the ground and mark the starting point as A
and the end as B.
2. Walk from A to B and note the distance covered.
3. Next walk from A to B and then back to A.
4. Compare the distance travelled and displacement.
Observation: - Distance changes but displacement becomes zero when returning
to the starting point.

2.4 Vector Addition (Graphical Method)
Vectors are physical quantities that have both magnitude and direction, such as
displacement, velocity, and force. When two or more vectors act together, we
combine them to find a single vector called the resultant. This process is known as
vector addition. Vectors can be added graphically using methods like the triangle
method or the parallelogram method.

11

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 11 of 85

Activity 2.4: Graphical Addition of Displacements
Materials: Graph paper, ruler, pencil
Steps:
1. On graph paper, draw a vector representing 4 units towards the east.
2. From the head (end) of this vector, draw another vector representing 3 units
towards the north.
3. Now join the tail (starting point) of the first vector to the head of the second
vector.
Observation:
The line joining the starting point to the final point represents the resultant
displacement.
This is how two vectors can be combined graphically to find a resultant vector, and
how both magnitude and direction are important in describing motion.
Practice Question:

Points A at (1,1), B at (3,1) , C at (3,5) and D at (4,5) (All the values mentioned in
the graph are in km) represent Sita’s House, bus stop, traffic signal and school
respectively. In the morning Sita travels from A to B on foot, then B to D via C in
the school bus. (All the values mentioned in graph are in km) Then calculate:
(a) Distance traveled by Sita on foot,
(b) Distance traveled by Sita by the school bus,
(c) Total displacement of Sita from her house to the school.

12

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 12 of 85

.
.co s e m

s em l a
a ag
2.5. Equations of Motion
When an object moves with constant acceleration, its motion can be described
using equations which are given as:
V = u + at
m
om
S = ut + 2
. co
. c e m
m v² = u² + 2as
as
where, u isseinitial velocity, v is final velocity, a is acceleration, and s is
g l
g l a
displacement. a
a
Activity 2.5: Observing Accelerated Motion Using a Toy Car
Materials: Toy car (or small wheeled object), smooth floor, measuring tape, stopwatch
(mobile timer), chalk/tape
Steps:
1. Mark a straight line on the floor and label the starting point as O.
m
.co
2. Place the toy car at point O and give it a gentle push so that it moves
forward.
e m
l as
3. Use a stopwatch and note the position of the car at equal time intervals

agfloor using chalk or tape.
(every 1 second).
4. Mark these positions on the
5. Measure the distance from the starting point to each marked position and
record it in a table.
Observation: The distance travelled in successive intervals increases, showing
acceleration.
Conclusion:
m
.co
The motion is accelerated motion, as the velocity increases with time.
m
c. o4 equation of motion distance travelled in the
th
second
s e m
e m From the second equation of motion:
la
las 1 ag
ag = +
2
where,
= initial velocity
= acceleration
= time
= displacement in time

om
13
. c
. c e m
m as
se
For more Question Papers, Sample Papers, Notes & Syllabus visit
gl
Page 13 of 85

Distance travelled in seconds
1
= ( )+
2
Distance travelled in (1) second
= ( − 1) + ( − 1)
Distance travelled in the second
1 1
= − = − ( − 1) + ( − 1)
2 2
Solving:
1 1 1
= + − + − ( − 2 + 1) = + [2 − 1]
2 2 2
Final Formula

= + (2 − 1)
2
This is the distance travelled in the second, often called the fourth equation of
motion. Here n must be a positive integer representing the nth second of motion.
The formula gives the displacement specifically during that one-second interval,
not a cumulative displacement.

2.6 Reflect and Discuss
● Why is specifying a reference frame necessary to describe motion?
● How do direction and magnitude together describe displacement?
● Which daily activities around you involve accelerated motion?

2.7 Project-Based Learning
Design a simple experiment using everyday materials to measure the speed of a moving
object (using a bicycle, or a walking student). Present your method, observations,
calculations, and conclusions to the class.
Check Your Understanding
1. Define a frame of reference in your own words.
2. Give two real-life examples where motion depends on the observer.
3. Why does a person sitting in a moving train appear at rest to another
passenger?
4. Classify the following as scalar or vector quantities: speed, velocity,
displacement, distance, acceleration and mass.

14

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 14 of 85

5. Explain the difference between distance and displacement with an activity
diagram.
6. Give two everyday examples of vector quantities.
7. Draw two vectors of 4 units east and 3 units north and find the resultant
using the triangle method.
8. Explain how vector subtraction is performed graphically.
9. Draw two opposite vectors of equal magnitude. Calculate its resultant.
10. A body starts from rest and accelerates at 4 m/s². Find the distance travelled
in the 6th second.
11. A car with initial velocity 8 m/s accelerates at 2 m/s². Find the distance
covered in the 5th second.
************************************************************************************************

15

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 15 of 85

.
.co s e m

s em l a
a ag
Newton’s Laws of Motion

3.1 Limitations of Newton’s Laws in Accelerating Frames
m
Activity 3.1: Let us observe
m . co
c. o situations:
Consider the following
s e m
s em standing in a bus that suddenly
● A passenger
g l a
l a
accelerates
g forward feels pushed backward, a
aeven though no one is actually pushing.
● When a vehicle takes a sharp turn,
passengers feel pushed outward.
Why does this happen? Is there really a force
pushing the passenger backward or outward? Can
these effects be explained only by the usual forces
like gravity or friction?
m
Understanding the Limitations
m .co
e
According to Newton’s First Law of Motion, a body continues to remain at rest or in
s
l a
uniform motion in a straight line unless acted upon by an external force. This law is
ag
strictly valid only in a non-accelerating frame.
However, when the frame itself is accelerating, objects seem to move without any
visible external force acting on them.
To maintain consistency with Newton’s laws, let us examine an additional concept.

Pseudo Force (Fictitious Force)
A force which does not arise due to physical contact or interaction (unlike
m
gravitational, frictional, or tension forces). A pseudo force (also called a fictitious
m .co
.co m
force) is an apparent force that is observed only when motion is described from an
s e
em Now understanding the scenario: What does a passenger fall abackwards when a
accelerating frame of reference.
s g l
g la a
a bus starts suddenly?

Observer 1: Standing on the road (inertial frame)
● Sees the bus accelerate forward
● Sees the passenger trying to remain at rest (inertia)

om
16
. c
. c e m
m as
se
For more Question Papers, Sample Papers, Notes & Syllabus visit
gl
Page 16 of 85

Explanation uses only real physics:
● No backward force exists
● Passenger’s body just resists motion
This follows Newton’s First Law of Motion perfectly.

Observer 2: Inside the accelerating bus (non-inertial frame)
● Sees the passenger “move backward”
● But doesn’t see any real force causing it

To make Newton’s Laws of Motion still work, we introduce the concept of
pseudo force: From the ground (an inertial frame), no force pushes the passenger
backward — the bus simply accelerates away from under them. But if we describe
the situation from inside the accelerating bus (a non-inertial frame), we must add a
pseudo force of magnitude m*a directed backward to make Newton’s First Law
appear valid within that frame. This force has no physical source and no reaction
pair.
When a frame accelerates forward, it exerts an influence on objects inside it. From
within that accelerating frame, we introduce an imaginary force acting in the
opposite direction to explain the observed motion. Thus, in an accelerating frame:
● The frame accelerates in one direction.
● An apparent force (pseudo force) is considered to act on the body in the
opposite direction.
This ensures that Newton’s First Law still appears valid within that frame.

Definition:
A pseudo force is an apparent force observed only in an accelerating frame of
reference.
It is always opposite to the acceleration of the frame and does not arise due to any
physical interaction.
Formula
=−
where:
● = mass of the object
● = acceleration of the frame
● The negative sign indicates that the pseudo force acts opposite to the
acceleration of the frame.

17

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 17 of 85

Quick Check
1. In which type of reference frame are Newton’s laws valid?
2. Define pseudo force and write its formula.
3. A lift accelerates upward at 4.5 . Calculate the pseudo force
experienced by a 60 kg person inside the lift.
4. Why does pseudo force disappear in an inertial frame?

3.2 Gravitation
Orbital Motion: Why the Earth and Moon Do Not Fall Despite Gravity?

Concept of Centripetal and Centrifugal forces
The Sun and the Earth both have mass, so they attract each other with a
gravitational force. According to Newton’s law of gravitation, the force between
them is equal in magnitude and opposite in direction. However, because the Sun’s
mass is much greater than the Earth’s mass, the Earth experiences a much larger
acceleration as compared to the Sun. That is why the Earth appears to revolve
around the Sun.
The Sun’s gravitational pull on the Earth provides the centripetal force needed to
keep the Earth in its orbit.
Earth is revolving around the Sun and it is moving with very high tangential
velocity. So, due to its inertia, it should tend to continue moving in a straight line.
On the other hand, the gravitational force of the sun is continuously attracting it
towards the centre of the sun. This changes its direction of motion. As a result of
these two aspects, the Earth does not move in a straight line but follows a fixed
curved path called an orbit.
Let us think: Imagine the Earth suddenly slows down. Take a moment to picture
what would happen if its forward (tangential) speed decreases, but the Sun’s
gravitational pull remains just as strong as before. Would the balance still exist?
How would this change affect the Earth’s path? Think about why slowing down
would cause the Earth to drift closer to the Sun instead of continuing smoothly
along its usual orbit.
Try to get the answer with the help of the following activity.

Activity 3.2:
Steps:
1. Tie the ring/bob securely to one end of the thread of length approx. 1 m.
2. Hold the other end of the thread firmly with your finger.
3. Swing the ring/bob in a horizontal circle at a steady speed.
18

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 18 of 85

.
.co s e m

s em l a
a ag
4. Observe how the bob moves in a circular path.
5. Now slowly reduce the speed of rotation.
6. Continue decreasing the speed further and observe what happens to the
circular motion.
Observation:
m
om
At an appropriate speed, the thread provides the centripetal force that pulls the
. co
. c
stone/ bob inward, keeping it in a circular path. At the same time, due to its inertia,
e m
e m
the stone tends to move in a straight line along the tangent. The balance between
l as
as
this outward tendency and the inward centripetal force results in circular motion.
ag
gl speed decreases, the required centripetal force also decreases.
Whenathe
However, the tension in the thread may reduce to the point where it can no longer
keep the stone moving in a circular path. As a result, the string may become slack,
and the motion is no longer circular—the stone begins to move inward or fall.
Conclusion:
Circular motion requires a balance between inward pulling force (centripetal force)
and tangential speed (which provides necessary centrifugal force). If speed
m
.co
decreases too much, the balance is disturbed, and the object can no longer
continue in the same circular path.
e m
s Resistance) on Falling Objects of
Effect of Cross-Sectional Area a(Air
l
Equal Mass ag
Let us imagine two objects with the same mass but different cross-sectional areas These
are dropped from the same height, an important question arises: Will they reach the
ground at the same time?

Case 1: When Air Is Present
Both objects have the same mass, so the gravitational force acting on them is the
same: m
m
c. oF = mg m.co
m s e
s e Since mass (m) is the same, the force of gravity on both objects is equal. This
g la
g la means gravity pulls both objects downward equally.
a
a However, another force also acts on the objects — air resistance (air drag). This
is an upward force that opposes motion.
Air resistance mainly depends on the cross-sectional area of the object (and
shape, speed and density of air). The larger the cross-sectional area, the greater
the air resistance.
Because of this:

om
19
. c
. c e m
m as
se
For more Question Papers, Sample Papers, Notes & Syllabus visit
gl
Page 19 of 85

● The object with a larger cross-sectional area experiences more air
resistance.
● The object with a smaller cross-sectional area faces less opposition.
● Therefore, it has a greater net downward force and falls faster.
Conclusion:
In air, the object with the smaller cross-sectional area reaches the ground first.
Case 2: In a Vacuum (No Air)
Since there is no air in a vacuum, there is no air resistance either.

Conclusion:
In a vacuum, both objects reach the ground at the same time, regardless of their
shape or size.
Variation of acceleration due to Gravity with Altitude and Depth
(without using Binomial Theorem)
● When we throw a ball upward, it comes back down.
● When we jump, we return to the ground.
This happens because the Earth pulls everything toward its centre due to gravity.
But why do astronauts float inside a spacecraft?
Does gravity disappear in space?
Is the value of gravity the same everywhere?
We know that as we go higher above the Earth’s surface, we move farther away
from the centre of the Earth. We know that gravitational force depends on distance.
As distance increases, force decreases.
Astronauts in the International Space Station appear weightless.
So clearly, gravity decreases with height, but it does not become zero.
Now let us derive the expression for the acceleration due to gravity at point A,
which is at a height of h from the surface of the earth.

20

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 20 of 85

Derivation – Acceleration Due to Gravity at Height
Let:
● Mass of object = m
● Radius of Earth = R
● Height above surface = h
● Distance from centre of Earth = (R + h)

Fig: 3.1 Acceleration due to gravity at height h

The acceleration due to gravity at height h is:

=
(ℎ)
Where:
● G = Universal Gravitational Constant
● M = Mass of Earth
On Earth’s surface:

=

Dividing both equations:

=
(ℎ)
This shows clearly that:
<
Example
Calculate acceleration due to gravity at a height of 800 km above Earth.
Given:
R = 6400 km
h = 800 km
g = 9.8 m/s²
21

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 21 of 85

.
.co s e m

s em l a
a ag
6400 64
= = 9.8 = 9.8 = 7.74 /
(ℎ) 7200 72

Acceleration Due to Gravity Below the Surface of Earth
Now, consider a point A located at a depth d inside the Earth. Assuming the Earth has
m
uniform density, let us determine the acceleration due to gravity at that interior point.
Let:
om . co
. c em
e m l as
l as ag
ag

Fig: 3.2 Acceleration due to gravity at depth d
● Radius of Earth = R
m
● Depth below surface = d
m .co
● Distance from centre = (R − d)
s e
● Density of Earth = ρ (uniform)la
g
If density is uniform, then mass ofa the earth can be calculated by
4
ℎ= =
3
At depth d, the object is at a distance (R − d) from the centre, only the mass inside

radius (R − d) contributes to gravity. (This is By Newton’s Shell theorem, the

gravitational effect of a uniform spherical shell on a point inside it is exactly zero.
m
m
Therefore, at depth d, only the sphere of radius (R–d) centred at Earth’s core
c. ocontributes to gravity— the outer shell of thickness d has no net effect)
m .co
s e
s em g l a
la 4
a
ag
= ( − )
3
From Newton’s Law of Gravitation

=
( )
Substitute :

om
22
. c
. c e m
m as
se
For more Question Papers, Sample Papers, Notes & Syllabus visit
gl
Page 22 of 85

4
( − )
= 3
( )
4
= ( − )
3
Compare with gravity at earth’s surface i.e.
At Earth’s surface:
4
=
3
Dividing the two equations:
−
=

Therefore,

=

We conclude from the above derivations that acceleration due to gravity is
maximum at the Earth’s surface and decreases as we go up/down. It will become
zero at the centre of the earth.

Example:
At what depth does g become 1/10th of its surface value?
Given:

=
10
Using the formula:
1 9 9
= 1− = = =
10 10 10 10

Quick Check
1. Where does the acceleration due to gravity reach its maximum value—on
the surface, above, or below the Earth?
2. What happens to g at the centre of the Earth?
3. Calculate g at a height of 400 km if R = 6400 km.
4. At what depth will g become half of its surface value?
5. Why does gravity decrease both above and below the surface of the earth?

23

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 23 of 85

3.3 Turning Forces (Moment of Force/Torque)
Activity 3.3: Let us observe
Look at the picture of a boy trying to enter his classroom.
He pushes the door to open it.
Now, think carefully and answer the following questions:
● Where will the boy apply force to open the door
easily?
(a) Near the handle
(b) Near the hinges
(c) At the centre of the door
● Why are door handles fixed far away from the hinges and not near them?

Now, reflect on the following points:
● The boy applies force in a straight direction, but the door rotates.
Why does this happen?
● Even though the door is heavy, it rotates easily when pushed at the handle.
● How is it possible to rotate such a heavy object by applying force at just one
end?

Fig: 3.3 Some examples of turning effects of forces in our daily life
When we apply a force to an object and it starts to rotate, the force produces a
turning effect.
This turning effect of a force is called the moment of force.
Moment of Force (Torque)τ = F × d × sin θ, where F is the magnitude of the
force, d is the distance from the pivot to the point of application, and θ is the angle
between the force and the line joining the pivot to the point of application. Torque is
maximum when θ = 90° (force perpendicular to the lever arm) and zero when θ =
0° or 180° (force directed toward or away from the pivot).
24

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 24 of 85

.
.co s e m

s em l a
a ag

Since the turning effect of a force depends on both the magnitude of the force and
the distance from the fixed point, its S.I. unit is newton-metre (Nm).
The angle at which force is applied
to a door (and the resulting angle of
m
om
the door itself) is crucial for
. co
. c
controlling the turning, efficiency,
e m
e m
and safety of the opening motion.
l as
as
The fundamental principle is that
l ag
ag
turning is maximized when the force
is applied perpendicular (at a 90-degree angle) to the door surface, making it the
most efficient way to open or close it.

Check Your Understanding
1. Why is it easier to open a door when you push at the handle rather than
near the hinges?

m
2. A force is applied to a wrench at different angles. At which angle will the

.co
rotating force be maximum? What happens to the turning effect when the
m
e
force is applied parallel to the wrench?
s force to open a gate. One pushes
3. Two students apply the same
l a
perpendicular to the gategat 20cm from the hinge. The other pushes
a
perpendicular to the gate at 80 cm. Who produces greater torque? Justify
4. Is it possible for a force to act on a body and still produce zero turning about
a given fixed point? Give a real-life example.
5. Two forces act on a rod pivoted at its centre:
I. 10 N downward at 0.5 m on the left
II. 10 N downward at 0.5 m on the right
m
.co
Will the rod rotate? Explain your reasoning
m How can a mechanic loosen a tight bolt
c. o 6. using s e m
e m a long spanner instead of applying a
la
las very large force? Explain using the torque
ag
ag formula.
7. A force of 20 N is applied to a door at 0.8 m from the hinge. Calculate the
torque when the force is applied at (a) 90°, (b) 60° (c) 30° to the door
surface.

****************************************************************************************

om
25
. c
. c e m
m as
se
For more Question Papers, Sample Papers, Notes & Syllabus visit
gl
Page 25 of 85

The Geometry of Power- Advanced Simple
Machines

4.1 Introduction
Welcome to the study of Mechanical Advantage. While simple machines such as
levers and inclined planes form the foundational concepts of physics, the machines
that shape our modern world—like cranes, trucks, and bicycles—apply these same
principles in more advanced and integrated ways through systems of wheels,
axles, and pulleys.
In this chapter, we will examine how the principles of geometry and force
distribution allow a relatively small input force to be transformed into a much larger
output force.
Activity 4.1:
● A truck driver turning a massive vehicle using only two hands.
● A crane lifting heavy concrete beams smoothly.
● A cyclist moving very fast by pedaling lightly.
Now think carefully:
● Is the driver extremely strong?
● Does the crane create extra force?
● Does the cyclist get “free” speed?
In all these cases, machines are helping us multiply force or increase speed.
This multiplication is called Mechanical Advantage (MA).

4.2 Wheel and Axle – The Steering Mastery
Activity 4.2: Think and Answer
“Think about a steering wheel and axle (steering column) and their respective
radius.”
● The steering wheel is large. The steering column connected to it is small.
Why is this so?
● Why not make both of equal size?

A wheel and axle consist of:
● a large wheel
● a smaller axle fixed at the center

26

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 26 of 85

Both rotate together. When effort is applied on the wheel, torque increases at the
axle.
The Mechanical Advantage is calculated using the formula:
M.A =

Example
A driver needs to maneuver the truck on muddy ground, requiring a resistance
force of 1,200 N to turn the steering axle. The steering wheel has a radius of 30cm,
and the steering axle has a radius of 3cm.
a) Calculate the Mechanical Advantage (MA) of the steering system.
b) How much effort (E) must the driver apply to the rim of the steering wheel to
turn the truck?
Solution:
a) M.A =

= = 10

b) Effort (E) =
.

= =120N

The driver needs to apply an effort of 120N to the rim of the steering wheel to turn
the truck.
Note: In practice, some input work is lost to friction within the machine. The
efficiency of a machine is defined as η = (useful output work / total input
work) × 100%. A real machine always has η < 100%. Mechanical Advantage
as calculated here assumes an ideal (frictionless) machine.

Quick Check
In a mechanical watch, a single power source (a spring or motor) must move three
different hands at three different speeds. This is achieved through a Gear Train,
where the "output" of one gear becomes the "input" for the next. The seconds-to-
minutes gear ratio is 60:1 and the minutes-to-hours ratio is 60:1;

1. If the seconds gear is 2 mm, how large would the hour gear be in meters?
2. Which of the three hands gear should be directly connected to the motor?
Why?

27

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 27 of 85

.
.co s e m

s em l a
a ag
4.3 Tension
Activity 4.3:
Hang a thread from an iron stand as shown in figure.
Observe its natural length.
m
Now attach a small bob to the lower end of the
om . co
c
thread. Notice how the thread stretches slightly.
. e m
e
Replace the smallm bob with a heavier bob. Does the l as
l as
stretch increase or decrease?
ag
agobserve that the thread stretches more when
You will
a heavier bob is attached. This shows that a greater
pulling force is acting on the thread.
Further,
Pass the thread over a pulley. Attach a weight to one side
and observe.
Now attach equal weights (equal bobs) on both sides of the m
pulley. Does the rope move, or does it only stretch?
m .co
s e
Replace one of the equal bobs with a heavier bob on the
l a
ag
left side. Observe carefully the direction in which the
system moves.
When both sides have equal weights, the system remains
at rest because the forces are balanced. When one side is
heavier, the system moves toward the heavier side.
Tension
Tension is the pulling force/ stretch force that travels
m
.co
through a stretched string, thread, rope, or cable. When you
m
.co
hang an object using a thread, the object pulls the thread
e m
e m
downward because of its weight. In response, the thread pulls the object upward.
las
las This pulling force inside the thread is called tension. If the weight attached to the
string increases, the tension in the string also increases. ag
ag It always acts along the length of the string and pulls away from the object to which
it is attached.
S.I unit of tension is Newton.
When the forces acting on an object are balanced (for example, the upward
tension is equal to the downward weight), the object remains at rest or moves with
constant speed. This state is called equilibrium. In a pulley system, if equal
m 28
. c
c. o s e m
s em
For more Question Papers, Sample Papers, Notes & Syllabus visit
gl a
Page 28 of 85

weights are placed on both sides, the tensions balance and the system does not
move. But if one side is heavier, the forces become unbalanced, and the system
moves toward the heavier side. This is by Newton’s First Law of Motion.
Let us Calculate: Tension and acceleration are produced when two unequal
masses are connected over a pulley.
1. Do the setup of weights, string and simple pulley as shown.

2. Since 0.55 kg > 0.5 kg, the 0.55 kg mass will move downward. The 0.5 kg
mass will move upward. Both masses will move with the same acceleration
because they are connected by the same string.
3. For 0.55 kg mass (moving downward):
a. Downward force = Weight = ___________
b. Upward force = Tension (T)
4. Net force: 0.55 − = 0.55
5. For 0.5 kg mass (moving upward):
a. Downward force = Weight = ___________
b. Upward force = Tension (T)
6. Net force: − 0.5 = 0.5
7. Add both equations: ________________________________
8. Acceleration of the system: ________________
9. Find Tension: __________________

Examples:
1. A 5kg object is suspended stationary from a rope. Calculate the tension.
Ans: The weight of the object is: = × = 5 × 9.8 = 49
2. A 4 kg mass is lifted upward with an acceleration of 2 m/s².
Calculate the tension.
Ans: Using Newton’s Second Law
− = = ( + ) = 4(9.8 + 2) = 47.2

29

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 29 of 85

Check Your Understanding :

1. Show the direction of weight and tension for both objects m 1 and m2.
2. An 8 kg mass hangs freely from a single fixed pulley. The system is at rest.
Find the tension in the rope.
3. Observe the given diagram. Find out in which direction the rope will move?
What will be the net downward force?

4. A 6 kg mass hangs freely from a single fixed pulley. The system is at rest.
Find the tension in the rope.
5. Two objects having masses 2 kg and 6 kg are connected over a frictionless
pulley with the help of rope. Find acceleration and tension in the rope.

*************************************************************************************************

30

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 30 of 85

.
.co s e m

s em l a
a ag
Work and Energy

5.1 CONSERVATIVE AND NON-CONSERVATIVE FORCES
Recollect these common occurrences: m
● A ball throwno m comes back down to your hand. . co
c upward
. rubber band returns to its original shape. e m
e
● A stretchedm l as
as
l book on a table finally stops.
● A sliding ag
g
a carefully:
Now think
● Why does the ball come back?
● Why does the rubber band regain its shape?
● Why does the book stop moving?
In all these cases, forces are acting. But are all these forces the same?
Conservative Forces
o m
. c
m
● If the work done by the force doesn’t depend on the path.
e path is always zero.
a s
● Work done by the force on the closed
● For a conservative force, a glwork done by the force equals the decrease in
the
potential energy: W = –ΔU = –(U_final – U_initial) = U_initial – U_final.
Equivalently, ΔU = U_final – U_initial = –W.
Examples: Gravitational force (Earth pulling objects downward) and Spring force
(stretched or compressed spring)
Non-Conservative Forces: If the work done by the force depends on the path
taken.
m
mExamples: Friction (solid and drag)
c. oWhen you slide a book across a table, it eventually stops becausesefriction
m converts .co
s em its kinetic energy into heat. This lost energy cannot be fully recovered.
g l a
la a
ag That is why friction is a non-conservative force.
Reflect on the following:
● If there were no friction, would a moving object ever stop?
● Why do pendulums slowly stop after some time?
● Why do machines require lubrication?

om
31
. c
. c e m
m as
se
For more Question Papers, Sample Papers, Notes & Syllabus visit
gl
Page 31 of 85

Quick Check
1. Define a conservative force with one example.
2. Why is gravitational force called a conservative force?
3. Why is friction called a non-conservative force?
4. What happens to energy when a non-conservative force acts on an object?
5. If there were no friction on Earth, how would motion be different? Explain.

5.2 Potential Energy of a Spring
Activity 5.2:
Collect the following items: A spring, a stand, a
weight hanger, slotted weights, a ruler.
1. Suspend a spring vertically from a rigid
support.
2. Attach a weight hanger to the free end of
the spring and note the initial length of the
spring.
3. Add a known weight to the hanger and
measure the extension produced in the
spring.
4. Increase the weight gradually and note the corresponding extension each
time.
5. Repeat the experiment using springs made of different materials or
thickness.
Observation:
● As more weight is added, the extension of the spring increases.
Different springs show different extensions for the same applied weight.
Conclusion:
The extension of a spring is directly proportional to the applied force (weight),
provided the elastic limit is not exceeded. This relationship can be expressed as:
=
where
= applied force,
= extension produced,
= spring constant, which depends on the nature of the spring.
32

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 32 of 85

This law is called Hooke’s law and is mathematically stated as F=−kx. The
negative sign indicates the force is a restoring force acting against the direction of
displacement (elongation or compression), aiming to return the spring to its original
length.
Its unit is N m-1. The spring is said to be stiff if k is large and soft if k is small.
Derivation
● The spring obeys Hooke’s Law
∝
=
Prepare a graph between Force and extension in the spring with the help of data
observed in the activity by taking
● X-axis → Extension (m or cm)

● Y-axis → Force (N)
Observation Table
These values give a straight-line graph passing
through origin.
Sample Values for k= 100 N/m
Force (F) in N Extension (x) Extension (x)
(in cm) (in m)
0 0 cm 0m
20 20 cm 0.2 m
40 40 cm 0.40 m
60 60 cm 0.60 m
80 80 cm 0.80 m
100 100 cm 1m

Calculation of Average Force:
For a spring stretched from 0 to maximum force:
+
=
2
When the spring is stretched gradually from zero extension to a maximum
extension , the force acting on it does not remain constant.
● At the beginning, force = 0
● At extension , force =
So, the average force (spring force changes linearly from 0 to maximum as
extension increases.) acting on the spring is given by:

33

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 33 of 85

.
.co s e m

s em l a
a ag
0+
= =
2 2
Work done in stretching the spring = Average force × Extension
1
= × =
2 2
o m
Conclusion
m . c
. co
The work done in stretching the spring is stored in it as elastic potential energy.
s e m
s em =
1
g l a
g laA spring obeys Hooke’s law with a spring constant of 30 N m⁻¹. If a force
2
a
a
Example:
of 100 N is applied to the spring, calculate the extension produced in the spring.
The Force-extension graph of a spring of spring constant 100 N m⁻¹ is given in
figure:
(a) Using the graph, determine the work done in stretching the spring from 2 cm
to 6 cm.
(b) If the spring is released from the stretched position of 6 cm, calculate the
m
maximum speed of a body of mass 0.5 kg attached to the spring, assuming
no loss of energy.
m .co
s e
11
l a
ag
10
9
8
7
Force (N)

6
5
4
3
2

m
.co
1

m 0

.co m
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

m Extension (m)
s e
s e g la
g la Solution: a
a (a) Work done = area under force–extension graph

om
34
. c
. c e m
m as
se
For more Question Papers, Sample Papers, Notes & Syllabus visit
gl
Page 34 of 85

(b) k = 100 N/m, x = 0.06 m. Elastic PE = ½ × 100 × 0.06² = 0.18 J. At maximum

speed, all PE converts to KE: 0.18 = ½ × 0.5 × v², so v² = 0.72, v ≈ 0.85 m/s.
Check Your Understanding
1) Explain the conversion of potential energy to kinetic energy when a ball is
thrown upward.
2) Why is gravitational potential energy considered a conservative force?
3) Calculate the potential energy of a 5 kg object kept on the top of a 30m high
building. (Considering potential energy to be zero at the base of the
building.)
4) What is the increment in its potential energy?
5) A 10 kg weight is hung from a 5 m wire, causing it to stretch by 1 mm.
Calculate the energy stored.
6) Calculate the work done by an external force to lift a 2 m long rod from a
horizontal to a vertical position.

***********************************************************************************************

35

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 35 of 85

Structure Of Atom

6.1 Discovery of Subatomic Particles
You have learnt about the development of models of the structure of the atom. You
would recall that in 1803, Dalton proposed that atoms are the smallest indivisible
particles of matter. However, this idea could not explain the results of several
experiments. For example, it was observed that substances like glass or ebonite,
when rubbed with silk or fur, acquire electric charge. This and many other
experiments on electrical discharge through gases showed that atoms are not
indivisible. They are made up of smaller particles called subatomic particles.
J. J. Thomson, in 1897, discovered the electron as a constituent of the atom and
confirmed that the atom is not the smallest particle of matter. He proposed the so-
called plum-pudding model of the atom, in which electrons are embedded in a
sphere of positive charge. This model was later shown to be incorrect by
Rutherford, as it could not explain the results of the gold foil experiment. Rutherford
then proposed a model in which electrons revolve around a small, positively
charged nucleus. However, this model could not explain the stability of the atom.
Thereafter, another model was proposed by Niels Bohr.
Here, you will learn about the discovery of the subatomic particles—electron,
proton, and neutron—which contribute to our understanding of atomic structure.
Before discussing these exploration we must recall a basic principle: like charges
repel each other, while unlike charges attract each other.
6.1.1 Discovery of Electron
In the late nineteenth century, many scientists, including Michael Faraday, William
Crookes, and others, studied electrical discharge in partially evacuated tubes
known as cathode ray discharge tubes. J.J Thomson carried out experiments by
taking gases at low pressure in discharge tube which is a long glass tube in which
two metal plates connected to oppositely charged poles of battery (Fig 1)

Fig. 6.1: A schematic representation of cathode ray tube showing the cathode rays going from the cathode to
anode in a straight line
36

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 36 of 85

.
.co s e m

s em l a
a ag
When a sufficiently high voltage is applied across the electrodes, rays are
observed to travel from the negatively charged electrode (cathode) towards the
positively charged electrode (anode). (Fig 2)
These are called cathode rays. The presence of these rays can be detected by
allowing them to pass through a hole in the anode and strike at screen coated with
m
a special material placed behind it. A bright spot is observed on the screen,
om . co
c
indicating that the rays travel in straight lines.
. em
e m l as
l as ag
ag

Fig.6.2: A schematic representation of the deflection of cathode rays to positive plate of the applied electric
field.

m
Further to determine the nature of these rays, Thomson carried out experiments

.co
by applying electric and magnetic fields in the path of the rays. He observed that
m
e
the rays were deflected towards the positively charged plate.
s
l
This showed that the particles in athe rays carry negative charge on further
ag that cathode rays consist of tiny negatively
experimentation. Thomson concluded
charged particles, later called electrons.
When these experiments were repeated using different gases (such as hydrogen,
nitrogen, neon, etc.) and different electrode materials, it was found that the
properties of cathode rays remained unchanged. This showed that electrons are
present in all atoms.
The main characteristics of cathode rays are as follows:
m
m  They originate from the cathode and move towards the anode.
.co
m .co 
s e
They are not visible themselves but produce bright spot when they strike m
s e certain materials.
g la
g la a fields.
 They travel in straight lines in the absence of external
a  They are deflected by electric and magnetic fields in such a way that
indicates them to be negatively charged.
 Their properties do not depend on the nature of the gas or the electrode
material.
Thus, electrons are a fundamental constituent of all atoms.

om
37
. c
. c e m
m as
se
For more Question Papers, Sample Papers, Notes & Syllabus visit
gl
Page 37 of 85

6.1.2 Discovery of Protons
After the discovery of the electron, it was realised that since electrons are
negatively charged, atoms must also contain positive charge to maintain electrical
neutrality.
Eugen Goldstein, in 1886, performed experiments using a discharge tube similar to
that used for cathode rays, but with a cathode having holes in it (perforated
cathode). When high voltage was applied, a faint glow was observed behind the
cathode. The rays responsible for this glow passed through the holes (or canals) in
the cathode and were therefore called canal rays.
Further studies showed that these rays were deflected towards the negatively
charged plate in electric and magnetic fields, indicating that they consist of
positively charged particles.
However, it is important to note that canal rays are not made up of a single type of
particle. They consist of positively charged ions of the gas present in the tube.
Therefore, their properties depend on the nature of the gas used.
When hydrogen gas was used in the discharge tube, the positively charged
particles obtained were the lightest known and were identified as hydrogen ions
(H⁺). These particles were later recognised as protons. The proton was finally
established as a fundamental particle by Rutherford in 1919.
The main characteristics of canal rays are:
 They are positively charged.
 Their behaviour in electric and magnetic fields is opposite to that of electrons.
 Their properties depend on the nature of the gas present.
 The lightest positive particle was obtained from hydrogen and is called the
proton.
6.1.3 Discovery of Neutrons
Once electrons and protons were known, it appeared that the structure of the atom
was understood. However, another problem arose when atomic masses were
measured. The mass of atoms was found to be greater than the sum of the
masses of their protons and electrons. For example, helium contains two protons,
yet its mass is about four times that of hydrogen. This indicated the possibility of
the presence of another particle contributing to the mass of the atom.
It was proposed that there must be a neutral particle present in the atom. This
particle was discovered by James Chadwick in 1932. He bombarded a thin sheet
of beryllium with alpha particles and observed the emission of powerful neutral

38

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 38 of 85

radiation. This radiation consisted of particles having no charge and a mass nearly
equal to that of the proton. These particles were called neutrons.
Neutrons are present in the nuclei of almost all atoms. The most common isotope
of hydrogen that does not contain a neutron is protium (1H1) but its heavier
isotopes Deuterium (1H2) and Tritium (1H3) contains neutrons.
Thus, the presence of neutrons explains the mass of atoms. For example, helium
contains two protons and two neutrons, which accounts for its mass being
approximately four times that of hydrogen.
Chadwick was awarded the Nobel Prize in Physics in 1935 for the discovery of the
Subatomic particle neutron.
From these discoveries, it became clear that atoms are composed of three
subatomic particles:
 Electrons (negative charge)
 Protons (positive charge)
 Neutrons (no charge)
These particles together determine the structure and properties of atoms.
Quick Check:
1. Why do cathode rays bend towards the positive plate?
2. What conclusion did Thomson draw from using different gases in discharge
tubes?
3. Why are canal rays different from cathode rays in nature?
4. Why was the discovery of neutron necessary?
5. In a cathode ray experiment, it was observed that the rays bend towards a
positively charged plate. What can we conclude about the nature of these
rays?
6. In a discharge tube experiment, the gas is changed from hydrogen to neon,
but the behaviour of cathode rays remains unchanged. What does this
observation tell us about electrons?
7. If cathode rays were neutral instead of being negatively charged, how would
their behaviour differ in an electric field?
8. In an experiment with canal rays, different gases are used and different
masses of particles are observed. What conclusion can be drawn about the
nature of canal rays?

39

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 39 of 85

.
.co s e m

s em l a
a ag
9. Why did scientists feel the need to propose the existence of neutral particles
even after discovering electrons and protons? Explain using the example of
helium.
10. In Chadwick’s experiment, the emitted particles were not deflected by
electric or magnetic fields. What does this observation indicate about the
o m
nature of these particles?
m . c
6.2. Spectrum .co e m
em passes through a glass prism, what do we observe? We see ga la
When whiteslight
s
a consists of a continuous spread of many colours from violet to red
rainbow, lthat
g a
a Such a spread is called a continuous spectrum in which the colours
(VIBGYOR).
are present without any gap.

m
.co
Fig 6.3 Continuous Spectrum

e m
Now, suppose we take a sodium vapour lamp, which you would have seen at
as
street lights or in parks, and pass the ‘yellow light’ given out by it, through a prism.
l
ag
We find that we do not get all the colours but only a few bright coloured lines. Two
of these are intense yellow lines. We may do a similar experiment with mercury
vapour lamp and observe another set of distinct lines. Such a spectrum that
contains distinct lines is called a line spectrum.

m
m .co
m .co s e m
s e g la
g la a
a Fig 6.4 (a) Sodium vapour lamp Fig 6.4 (b) Lines Spectrum of Sodium

Line spectrum means only specific radiation are emitted, not all. It is important to
know that the line spectrum is characteristic of the element causing it. This fact is
used to identify elements present in stars by studying their spectra. Even without
going to the star, we can know what it is made of.

om
40
. c
. c e m
m as
se
For more Question Papers, Sample Papers, Notes & Syllabus visit
gl
Page 40 of 85

6.3 Line Spectrum of Hydrogen
Now, when the radiation from a discharge tube containing hydrogen gas in it is
passed through a prism, it also gives line spectrum called hydrogen atom
spectrum.
The Spectral lines for atomic hydrogen are:
Series ni nf
Lyman 1 2,3_ _ _ _ _
Balmer 2 3,4_ _ _ _ _
Paschen 3 4,5_ _ _ _ _
Brackett 4 5,6_ _ _ _ _
Pfund 5 6,7_ _ _ _ _

The energies of the distinct spectral lines observed in hydrogen atom spectrum
could be expressed empirically in terms of mathematical expressions involving two
sets of integers. This equation is known as Rydberg Equation.

RE =109,677 cm-1
Where, RE is the Rydberg constant expressed in terms of energy and Z is the
atomic number. These empirical formulas worked very well but could not be
explained until Bohr’s model came.

6.4 Limitation of Rutheford Model of Atom
 Rutherford could not explain stability as the electron continuously loses
energy when it moves around the nucleus.
 As the electron in the atom is allowed to have continuous energies, therefore
the emitted radiation is expected to give a continuous set of radiation.
However, we observe a line spectrum. Therefore, we say that Rutherford’s
model fails to explain the existence of line spectrum of hydrogen.

6.5 Bohr’s model
Neils Bohr, a student of Rutherford, in 1913 proposed his model for an atom. He
combined Rutherford’s nuclear model with the new quantum idea introduced by
Max Planck. He made two revolutionary assumptions which are as below:

41

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 41 of 85

 Electrons can move only in certain allowed circular orbits without radiating
energy. That is, they have a fixed energy as long as they are in a given orbit.
 The radiation is emitted or absorbed only when an electron jumps from one
allowed orbit to another,

Fig 6.5:Energy change is electron jump

6.5.1 Achivement of Bohr Model
“When an electron jumps from an orbit of higher energy to that of a lower energy it
releases energy in the form of radiation. The amount of energy released depends
on the difference in the energies of the two levels. Since the orbits of only certain
energies exist, only of fixed quantity energy differences are possible. Therefore, we
get a line spectrum. Bohr’s model could explain the observed line spectrum of
hydrogen fairly well.
6.5.2 Limitations of Bohr’s model
Bohr model was unable to explain:
 Finer details (that is closely spaced lines) of hydrogen atom spectrum
observed by sophisticated spectroscopic techniques.
 The spectrum of atom other than hydrogen.
 The splitting of spectral lines in presence of magnetic field (Zeeman effect)
or an electric field (Stark’s effect).

6.6 Check Your Understanding
1. The hydrogen spectrum consists of only a few sharp spectral lines instead
of a continuous spectrum. What information does it provide about the
energy of electrons in an atom.
2. Explain why would Rutherford’s model predict a continuous spectrum rather
than a line spectrum.
3. A discharge tube filled with an unknown gas produces a line spectrum
identical to hydrogen. What can you conclude about the gas? Give reason.
4. If electrons in an atom were allowed to have a continuous set of energy
values, what kind of spectrum would you expect? Why is this not observed?
5. “Bohr’s model solved all problems of atomic structure.” Comment.
6. How does the concept of fixed energy levels explain the stability of atoms?
42

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 42 of 85

.
.co s e m

s em l a
a ag
7. Why do different elements produce different line spectra? Give a conceptual
explanation.
8. Explain why Bohr’s model works well for hydrogen but not for multi-electron
atoms.
9. State two limitations of Rutherford’s model.
m
co
10. Rutherford’s model explained the structure of the atom but failed to explain

. c om
atomic stability and spectra. Discuss.
e m .
11. What was the main drawback of Rutherford’s model regarding electron
em l as
l as
motion? What assumption was made by Bohr to overcome this problem.
12. How does Bohr’s model explain line spectrum of hydrogen? ag
ag
13. Outline the limitations of Bohr’s model.
14. Define line spectrum and continuous spectrum with one example each.
15. Write two main postulates of Bohr’s model.
16. What is meant by fine structure in hydrogen spectrum?
17. What is the significance of Rydberg equation?

m
m .co
s e
l a
ag

m
m .co
m .co s e m
s e g la
g la a
a

om
43
. c
. c e m
m as
se
For more Question Papers, Sample Papers, Notes & Syllabus visit
gl
Page 43 of 85

Chemical Bonding

7.1 Octet Rule
You have learnt that the atoms having eight electrons in its valence shell are
stable. The atoms other than hydrogen tends to form bonds until it is surrounded
by eight valence electrons. They do so by gaining, losing or sharing electrons. It is
called Octet rule and is quite useful in describing the formation of simple
molecules. It is important to note that the octet rule is just a guiding principle and
not a law. In case of hydrogen, the valence shell attains the electron configuration
of helium, i.e., a total of two electrons.

7.1.1 Lewis Approach
Lewis Symbol :- The Valance electron of an atom in terms of dot is written around
the atom.
For Example:
Fluorine (F=9) Electronic configuration=2,7
Valence electron=7
Hence is represented as:

Bonding in molecule:
Duing bonding the valance electrons are written around the atoms and then
electrons are shared in such a way so as to complete the octet of each atom.
For Example the Lewis formula for Hydrogen Fluoride is

or

Here the pair of dots (representing electrons) placed between the symbols of the
combining atoms represent the bonding electrons. The remaining dots represent
the non-bonding electrons. As the name suggests, the non-bonding electrons do
not contribute to the bonding. You may note that the hydrogen atom in this
molecule has only two electrons (a duplet) around it. The line here indicates the
bond between hydrogen and fluorine atom.

44

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 44 of 85

7.1.2 Exceptions of Octet Rule
Many stable molecules do not follow the octet rule. These are called exceptions to
the octet rule. Let us discuss about these exceptions.
Molecules with incomplete octets
In case of some elements the valence shell has less than four valence electrons. In
these cases, their atoms cannot form four bonds to complete the octet. Also, these
do not have sufficient lone pairs that can complete the octet. As a result, the octet
remains incomplete in such cases. For example, in case of lithium, beryllium and
boron there are only 1, 2 and 3 valence electrons respectively. Therefore, these
can form 1, 2 and 3 bonds respectively. In such cases the central atom would have
2, 4 and 6 electrons respectively on forming the molecule. These represent
molecules which do not complete the octet and yet are stable. One common
example is that of boron trifluoride. In this molecule, one boron atom makes bonds
with three fluorine atoms and is represented as

The lines here indicate the bond between boron and fluorine atom.
Molecules with expanded octets
Another exception to the octet rule is observed in the formation of molecules
having more than eight valence electrons around central atom. Such molecules are
formed by the atoms of the elements having more than four electrons in their
valence shell. For example, in case of sulfur hexafluoride one atom of sulphur
combines with six atoms of fluorine. The Central sulphur atom has 12 electrons in
its valence shell representing an expanded octet. We can represent its structure as

You will learn about the formation of such compounds in higher classes.
Molecules with odd number of electrons
Certain molecules have an odd number of electrons. For example, an atom of
nitrogen (having 5 valence electrons) makes two bonds with an atom of oxygen
(having six valence electrons) to form a molecule of NO. It has a total of 11 valence
electrons, five from N and six from O atom. The Lewis structure for this molecule
can be represented as

45

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 45 of 85

.
.co s e m

s em l a
a ag
The two lines here indicate two bonds. Whenever there are odd number of
electrons in a molecule then at least one atom would have an incomplete octet.
Secondly in such a molecule there would always be an unpaired electron.
Quick Check
1. What is meant by the octet rule? m
om
2. Why does hydrogen not follow the octet rule? . co
. c
3. Give one example each of the molecule with e m
em
a) incomplete octet
l as
l as
b) expanded octet ag
ag
c) an odd electron
4. Why can boron form compounds with only six electrons around it?
5. What is meant by a duplet configuration?
6. Why is NO considered an exception to the octet rule?
7. Draw the Lewis dot structure of BF₃ and explain why boron does not
complete its octet.
8. Assertion: SF₆ violates the octet rule.
Reason: Sulphur can accommodate more than eight electrons.
m
explanation of the assertion. m .co
A. Assertion and reason, both are correct and reason is the correct

s e
a
B. Assertion and reason, both are correct but reason is not the correct
l
explanation of the assertion.
ag
C. Assertion is correct but reason is a wrong statement.
D. Assertion is wrong but the reason is a correct statement .

7.2 Metallic Bonding
You all are familiar with metals like iron, copper, aluminium, and so on. They are
hard, they can be beaten into sheets, drawn into wires, and they conduct
electricity. You may be wondering how can we explain these properties of metals.
m
.co
We can explain these in terms of a simple model known as the electron sea model
m
.co m
for metals. You have learnt about bonding in case of ionic and covalent
m compounds. In these, the atoms bind by transfer or sharing of electrons between s e
s e specific atoms. The electron sea model, in fact, is a simple model for bonding in
g la
g la metals. It involves bonding between a very large number of atoms of the metal. Leta
a us understand this model and learn how we can explain the properties of metals by
using this model.
7.2.1 Electron Sea Model
A metal atom has a few electrons in its outermost shell. These outer electrons are
not held very tightly by the nucleus. Because of this, when many metal atoms
come together to form a solid, these outer electrons do not remain attached to any
m 46
. c
c. o s e m
s em
For more Question Papers, Sample Papers, Notes & Syllabus visit
gl a
Page 46 of 85

one atom. Instead, they become delocalised and are free to move throughout the
entire piece of metal.
You know that when an atom loses an electron it becomes a cation. In metals, the
atoms can be thought of as forming positive metal ions arranged in a regular
pattern. These ions form a kind of fixed structure. The free electrons move
continuously and randomly in all directions around and between these ions. This
collection of freely moving electrons is called a “sea of electrons”.
Thus, according to the electron sea model, a metal can be seen as a structure in
which positive metal ions are fixed in place, and a “sea” of mobile electrons moves
around them. This is why the model is called the electron sea model as shown in
Fig.7.1.

Fig. 7.1: Schematic representation of Electron sea model

These ions and electrons together form a stable structure. The attraction between
the positive ions and the sea of electrons is what holds the metal together. This
attraction is called metallic bonding. So, we can say that metallic bonding is the
force of attraction between positive metal ions and the sea of electrons. It is
important to note that unlike covalent bonds, metallic bonds are not localised. The
electrons are shared by all the atoms collectively, forming a non-directional bond
that can adjust to shifting positions of the metal ions.

7.2.1: Electron sea model and properties of metal
Now let us see how this simple electron sea model helps us to understand the
properties of metals.

Electrical conductivity
Since electrons are free to move, when we apply an electric field across a metal,
these electrons start moving in a particular direction. This movement of electrons is
what we call electric current. That is why metals are good conductors of electricity.

47

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 47 of 85

Fig.7.2: Schematic representation of electrical conduction by metals in terms of electron sea model

Thermal conductivity
When one part of a metal is heated, the electrons in that region gain energy and
start moving faster. As they move, they transfer this energy to other parts of the
metal. At the same time, the metal ions also vibrate more and help in passing the
heat along. In this way, heat spreads quickly. This is why metals are good
conductors of heat.

Malleability and Ductility
You know that malleability refers to the ability of metals to be beaten into thin
sheets. In the electron sea model, the positive metal ions can slide over one
another without breaking the structure. This is possible because the electrons are
not fixed; they continue to move and hold the ions together. So, even when layers
of the metal ions shift, the metal does not break.
Similarly, you know that ductility refers to the ability of metals to be drawn into
wires. When we stretch a metal, these metal ions slide past each other without
breaking the non-directional metallic bonds, allowing the metal to stretch into wires.
The free electrons help maintain the attraction between ions even when the shape
changes.
You must remember that the electron sea model gives a simple picture and
explains many basic properties of metals. The electrons are not completely lost as
in ionic bonding; rather, they are shared collectively by all atoms in the metal. You
will learn more detailed models of metallic bonding in your higher classes.

Check Your Understanding
1. What is meant by the term “electron sea” in metals?
2. What type of particles are in a fixed position in a metal according to the
Electron sea model?
3. Define metallic bonding.
4. Why are metallic bonds called non-directional?
48

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 48 of 85

.
.co s e m

s em l a
a ag
5. Name two properties of metals explained by the electron sea model.
6. Explain how the electron sea model accounts for electrical conductivity in
metals.
7. How does the electron sea model explain thermal conductivity in metals?
8. Why can metals be beaten into thin sheets? Explain using the Electron sea m
model.
om . co
. c e m
9. What is meant by ductility? How is it explained by the electron sea model?
me l as
as
10. How is metallic bonding different from covalent bonding?
l the structure of a metal according to the electron sea model. ag
a g
11. Explain
12. If electrons in a metal were not free to move, which property would be most
affected? Explain.
13. Explain why metals do not break when hammered but instead change
shape.
14. Copper is used for electrical wiring, while rubber is not. Explain using the
electron sea model.
m
.co
15. Why are metals generally good conductors of heat as compared to non-
metals?
e m
l as
19. Assertion (A): Metals are good conductors of electricity.

ag
Reason (R): Metals contain free electrons that can move under an electric
field.
A. Assertion and reason, both are correct and reason is the correct
explanation of the assertion.
B. Assertion and reason, both are correct but reason is not the correct
explanation of the assertion.
C. Assertion is correct but reason is a wrong statement.
D. Assertion is wrong but the reason is a correct statement.
m
.co
20. Assertion (A): Metallic bonds are non-directional.
m
.co m
Reason (R): Electrons in metals are localised between two atoms.

m s e
la
A. Assertion and reason, both are correct and reason is the correct
s e g
la
explanation of the assertion.
g a
B. Assertion and reason, both are correct but reason is not the correct
a explanation of the assertion.
C. Assertion is correct but reason is a wrong statement.
D. Assertion is wrong but the reason is a correct statement.
21. Assertion (A): Metals are malleable.
Reason (R): Layers of metal ions can slide while electrons continue to hold
them together.

om
49
. c
. c e m
m as
se
For more Question Papers, Sample Papers, Notes & Syllabus visit
gl
Page 49 of 85

A. Assertion and reason, both are correct and reason is the correct
explanation of the assertion.
B. Assertion and reason, both are correct but reason is not the correct
explanation of the assertion.
C. Assertion is correct but reason is a wrong statement.
D. Assertion is wrong but the reason is a correct statement.

************************************************************************************************

50

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 50 of 85

Mixtures And Separation Of Mixtures

8.1 Chromatography
Chromatography was first developed by the Russian botanist Mikhail Tswett in 1906
while studying plant pigments. He used this technique to separate the different coloured
constituents of chlorophyll. This method was named chromatography components of
chlorophyll. The name chromatography comes from the Greek words chroma (colour) and
graphein (to write). The technique is used to separate the mixtures into components,
purification of compounds and also to test the purity of compounds.
Principle: The technique is based on the difference in the rates at which the components
move through a stationary medium under the influence of moving phase..
Application: Today, chromatography is widely used in chemistry, biology, and medicine to
identify and separate different substances in a mixture.
Column Chromatography
Modern method for the separation of mixtures into its components. The selective removal
of the components may be due to adsorption or partition process.
When a mobile phase is allowed to move over a stationary phase, the components of the
mixture move by varying distances over the stationary phase because of different
adsorption tendencies. In this case the stationary phase can be held on a cylindrical
column of solid . hence it is called column chromatography.
Principle: It is based on the fact that different compounds are adsorbed on an adsorbernt
to different degrees
Procedure: In this technique, a long glass tube having a stop cock near the bottom, called
a column is used. First a plug of cotton or glass wool is placed at the bottom of the column.
Then it is filled with a solid material such as silica gel or alumina, which acts as the
stationary phase (fixed in a place). The mixture to be separated is placed at the top of
silica gel in the column, and then a little amount of glass wool is placed above the mixture.
After this a suitable liquid solvent is poured from above and allowed to flow through the
column under the influence of gravity. This is called an eluent or the mobile phase (which
moves). The solvent coming out from the column is collected in different fractions. In this
way, the components of the mixture get separated and are collected separately. Column
chromatography is widely used in chemistry laboratories to purify compounds as well as to
separate them from the mixtures.

51

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 51 of 85

.
.co s e m

s em l a
a ag

m
om . co
. c e m
e m l as
l a s ag
ag Fig 8.1: Column chromatography: Different stages of separation of components of a mixture.

Application: The method has been used:
 To separate blue and red dyes
 To separate and purify plant pigments

8.2.1 Fractional Distillation
This method is used for the purification of liquids which boil without decomposition
and contain non volatile impurities.
o m
Principle: Fractional Distillation is a technique c
. to separate a mixture of two miscible
m
liquids whose boiling points differ by lessethan 25 C shown in Fig. 8.2 (a).
s
0

g
In this process, the mixture of liquids l ais heated in a distillation flask which is fitted
a
with a fractionating column before the condenser as shown in figure 8.2(b).
Fractionating Column: The fractionating column is a long tube provided with
obstructions to the passage of vapours moving upwards and liquid moving
downwards. It increases the cooling surface area.

m
m .co
m .co s e m
s e g la
g la a
a

(a) (b)
Fig 8.2: (a) Fractional distillation apparatus and (b) A sample fractional distillation column

om
52
. c
. c e m
m as
se
For more Question Papers, Sample Papers, Notes & Syllabus visit
gl
Page 52 of 85

Procedure: When the mixture is added to distillation flask and the flask is heated
the vapours of more volatile liquid having low boiling point rises up in the
fractionating column.Due to the obstruction in the fractionating column,some of the
vapours condense and fall back in the column.Some of the condensing liquid in the
fractionating column gets heat from the ascending vapours and revaporizes. As a
result the vapours become richer in low boiling component. These rise up in the
fractionating column and condense while passing through condenser and collected
in the receiver. The same process will occur again and again. This repeated
condensation and vaporization helps in better separation of the liquids.By carefully
controlling the temperature, different liquids in the mixture can be separated one
after another according to their increasing boiling points.
8.2.2 How is it different from simple distillation?
 In fractional distillation, a fractionating column is placed between the
distillation flask and the condenser. The column provides many surfaces
where repeated condensation and vaporization occur. This allows better
separation of liquids whose boiling points are close to each other.
 Simple Distillation is used to separate miscible liquid which differ in boiling
point by at least 25oC but in fractional distillation the boiling point differ by
less than 25oC
8.2.3 Application
Crude oil is a complex mixture of many hydrocarbons with different boiling
points. In a refinery, the crude oil is heated, and the vapours enter a tall
fractionating column. As the vapours rise in the column, they cool and
condense at different levels according to their boiling points. Different fractions
are collected at different heights of the column as shown in the figure below
along with the temperature range. The various important fractions used in our
daily life or industry are:
 Petroleum gas (LPG, contains butane and propane)
 Petrol (gasoline)
 Kerosene
 Diesel
 Fuel oils
 Lubricating oil and heavy oils

53

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 53 of 85

Fig 3: (a) Schematic representation of separation of different components of crude oil by fractional distillation

Questions
1. What is chromatography? Mention its two main phases.
2. Who discovered chromatography and in which year?
3. What is meant by stationary phase and mobile phase?
4. Name two common adsorbents used in column chromatography.
5. What is an eluent in column chromatography?
6. Why do different substances move at different speeds in column
chromatography?
7. What is fractional distillation?
8. When is fractional distillation preferred over simple distillation?
9. What is the role of the fractionating column?
10. In column chromatography, a mixture of two compounds A and B is
separated. A comes out first. What can you say about its interaction with the
stationary phase?
11. A mixture of ethanol (b.p. 78°C) and water (b.p. 100°C) is to be separated.
Which method will you use and why?
12. Explain why repeated condensation and vaporization improve separation in
fractional distillation.
13. In a fractional distillation column, why does temperature decrease from
bottom to top?
14. Why is simple distillation not suitable for separating liquids with close boiling
points?
15. Assertion: In chromatography, separation occurs due to difference in boiling
points.
Reason: Components move at different speeds in the column.
A. Assertion and reason, both are correct and reason is the correct
explanation of the assertion.

54

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 54 of 85

.
.co s e m

s em l a
a ag
B. Assertion and reason, both are correct but reason is not the correct
explanation of the assertion.
C. Assertion is correct but reason is a wrong statement.
D. Assertion is wrong but the reason is a correct statement.
16. Assertion: Fractional distillation gives better separation than simple
m
co
distillation.

. c om
Reason: It involves repeated condensation and vaporization.
e m .
A. Assertion and reason, both are correct and reason is the correct
em l as
l as
explanation of the assertion.
B. Assertion and reason, both are correct, but reason is not the correct ag
ag explanation of the assertion.
C. Assertion is correct, but reason is a wrong statement.
D. Assertion is wrong, but the reason is a correct statement.
17. Difference in which property forms the basis for separating components in
fractional distillation?
A. Solubility
B. Boiling points
C. Particle size
m
D. Chemical reactivity
m .co
e
18. What is the main purpose of the "fractionating column" in fractional
s
distillation?
l a
A. To heat the mixture faster.
ag
B. To cool the vapours at fast rate.
C. To provide more surface area for vapours.
D. To let the vapours of two liquids mix properly
19. In column chromatography, the solid substance that is filled in the column is
called the:
A. Mobile phase
B. Solvent
m
.co
C. Stationary phase
m
.co m
D. Mixture

m 20. That component of a mixture moves down the column at a faster rate which s e
s e is
g la
g la A. most attracted to the stationary phase. a
a B. having the highest boiling point.
C. The one most soluble in the mobile phase (solvent)
D. The one with the largest particle
********************************************************************************

om
55
. c
. c e m
m as
se
For more Question Papers, Sample Papers, Notes & Syllabus visit
gl
Page 55 of 85

Microscope and Microscopy

9.1. What is a Microscope?
You have read in Grade 8 that a special instrument called the microscope (micro
– small; + skopion - "means of viewing") is required to observe tiny living
organisms or their parts which cannot be seen through naked eyes by magnifying
them. With a microscope, you can see small specimens such as onion cells, cheek
cells, bacteria and even dust particles etc. It helps doctors to see germs and study
cells in living organisms.
What do we call the ability of a human eye to see two very close objects as
separate and distinct? Imagine two tiny dots drawn on a piece of paper. As the
dots are moved closer, there comes a point at which they can no longer appear as
separate. When viewed from about 25 cm (the near point of the eye), two points
separated by about 0.1 mm (100 µm) can be observed as distinct, otherwise, they
appear as a single point. This defines the limit of resolution of the human eye.
A cell is generally too tiny to be observed by an unaided eye. This raises an
important question - how do cell biologists study the structure and functioning of
cells, which are much smaller than the limit of resolution of the human eye?
When Robert Hooke observed ‘cork’ under the microscope developed by him in
1665. He examined thin slices of bark of an oak tree and observed tiny hexagonal
box-like spaces just like the patterns of honeycomb and called them cells. Around
the same time, Antony van Leeuwenhoek made tiny, powerful lenses and saw
“animalcules” – what we now know as bacteria and protozoa. Those simple lenses
opened the door to a completely new world.

Figure 9.1: A. Drawings of Cork cells as published in the ‘Micrographia’; B. Microscope developed by Robert
Hooke,

56

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 56 of 85

Activity 9.1: Let us think and write:
If you could shrink yourself and travel inside a leaf, what would you see? Write 3 –
4 lines imagining that journey.
9.2. A Quick historical Journey of Microscopes
Let us walk through time and see how microscopes evolved:
 13th–15th century: Simple magnifying glasses used by spectacle makers.

 1590 - Hans and Zacharias Janssen: A Dutch father-and-son duo of
spectacle makers developed an early compound microscope by combining
two lenses within a single tube.

 1665 – Robert Hooke: Coined the term ‘cell’ for empty, hexagonal, box-like
structures by examining the cork of an oak tree under the microscope
developed by him. He published his findings in a book called “Micrographia”.

 1670s – Antony van Leeuwenhoek: He worked with a simple, single-lens
microscope capable of magnifying up to about 300 times, which allowed him
to observe tiny living organisms he called “animalcules,” including bacteria
and protozoa. He was the first to study living microorganisms and is widely
regarded as the Father of Microscopy.

 1878 Ernst Abbe: Postulated a mathematical theory linking resolution to the
wavelength.

 19th–20th century: Better lenses and illumination improved the compound
light microscope.

 1930s onwards: Electron microscopes (TEM and SEM) were invented
where viruses, cell organelles and cell surfaces could be observed.

 1938 Ernst Ruska: developed the first electron microscope, which operated
on the principle using electrons as the illumination source (instead of light)
that provides shorter wavelengths and thereby significantly enhancing the
resolving power.

 1953: Frits Zernike received the Nobel Prize in Physics for inventing and
demonstrating the phase-contrast microscope.

Activity 9.2: A Timeline Strip
Draw a horizontal line. Mark at least 5 important dates in microscopy and add a
tiny sketch or symbol for each (e.g., cork cells, bacteria, electron beam etc.).

57

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 57 of 85

.
.co s e m

s em l a
a ag
9.3. How Does a Microscope Work?
An important parameter in microscopy is the resolution, contrast and
magnification of the object that is viewed under the lens, which makes it
appear several times larger to the human eye. The operating principle varies
with the type of microscope, which can be broadly classified by whether they
m
use multiple lenses or electron beams. In each case, a system of lenses or
om . co
c
electromagnetic fields is used to produce an enlarged, detailed image of a
. e m
em
specimen that cannot be clearly seen with the naked eye.
l as
as
9.3.1 Types of Light microscope
l ag
ag
Light (Optical) microscopes rely on visible light and glass lenses to enlarge and
view specimens.
Basic Classification -
 A simple microscope utilizes a single lens to magnify an object, similar to
how a magnifying glass works. For example, dissecting microscope is used
for 3D viewing of small objects.
Compound Microscope: Most commonly used laboratory microscope
m


.co
utilizes at least two sets of lenses - the objective lens (near the specimen)
m
and the eyepiece (ocular lens) - to achieve high magnification.
e

l as
Advanced optical microscopes
Beyond the standard compound g
a microscope, a diverse family of advanced light
microscopes exists—such as Phase-Contrast and Fluorescence, each using
unique optical technique to reveal hidden cellular secrets that would otherwise
remain invisible to the naked eyes. Fluorescence microscopy uses high-intensity
light to excite specialized dyes in a specimen, causing specific cellular structures to
glow brilliantly against a dark background like stars in the night sky. Phase-
Contrast microscopy is used for viewing living cells in their natural state because
it enhances contrast without the need for chemical stains that would otherwise kill
m
.co
the specimen.
m
m .co s e m
s e g la
g la a
a

Figure 9.2: A. Fluorescence microscope; B. Phase contrast microscope

om
58
. c
. c e m
m as
se
For more Question Papers, Sample Papers, Notes & Syllabus visit
gl
Page 58 of 85

Figure 9.3: Cells imaged with – A. Traditional optical microscope (Mag. 40X) B. Phase-contrast
microscope (Mag. 40X) and C. Fluorescence microscope (Mag. 20X)

9.3.2 Parts of a Compound microscope
Core Components and their Roles
1. Light Source; Provides illumination (LED or halogen lamp).
2. Condenser Lens: Focuses light onto the specimen to optimize numerical
aperture and contrast.
3. Specimen Stage: Holds the slide containing the specimen.
4. Objective Lens: The primary magnifying lens (e.g., 4X, 10X, 40X, 100X).
It forms an enlarged, inverted image of the specimen that is real in nature.
5. Eyepiece (Ocular Lens): It enhances the magnification of the image
already produced by the objective lens.

Activity 9.3: Let us examine a compound microscope
Key parts include eyepiece, objectives, nosepiece, stage with clips, coarse/ fine
focus, condenser, iris diaphragm, illuminator, arm, and base

Figure 9.4: Light (compound) microscope

 Objective lenses: Main magnifying lenses (with different magnification
power (X) – 4, 10, 15, 20, 40 etc.) close to the slide. These form a real,
magnified image of the object.

59

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 59 of 85