iit jee | medical |olympiads physics

IIT JEE | Medical |Olympiads

Foundation Curriculum by

IITians

9

PHYSICS

CLASS IX PHYSICS FOUNDATION FOR IIT JEE/ NEET/NTSE/OLYMPIADS, ANY PRESENT

AND FUTURE COMPETITIVE EXAMS

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Foundation Curriculum by IITians for Academic Excellence in Schools

Published by:

Dr. Ravi's Educational Research and Development Center

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[email protected]

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© Dr. Ravi's Educational Research and Development Center

First edition 2012

Version 4

FOR INTERNAL CIRCULATION ONLY

All Rights Reserved

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inadvertently, neither the compiler, publisher nor any of the distributors

take any legal responsibility

In case of any dispute, all matters are subjected to the exclusive

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FOUNDATION FOR SUCCESS BY Dr. RAVI SURA

Dear Students / Parents, Quality education will play a vital role to emerge India as a leader among all developing countries. IITians Propel Academy is designed to enhance the quality education towards excellent training from school level to IIT JEE / NEET. The foundation program is designed exclusively for students who are aiming to attain a top ranks in the competitive examinations. The propel foundation course will focus more on conceptual knowledge. As a result, students are able to devote more time on practicing more complex problems in each subject area to develop a complete command on the key concepts.

The Propel Academy comprising of detailed concept explanations, Introductory Exercises and JEE MAINS / ADVANCED and NEET model questions, which includes previous questions from different competitive exams like International Math’s Olympiad, NTSE, RMO, APAMT, NTSE, IJSO and Ramaiah's SAT examinations. Foundation books are outcome of the in depth research work of IITians and the entire foundation curriculum is well structured by IITians. We strongly believe that nurturing young minds at an early age can lead to great professional careers in future. Contact us for any support or clarification without any hesitation. Wishing you a very bright future!

Dr. Ravi Sura Ph.D, M.S (Swiss & France) M.Tech, M.Sc (IIT Madras) Founder and Executive Director IITians Propel Academy

FOREWORD BY Dr. CHUKKA RAMAIAH

Dear Friends,

While an extra ordinarily gifted student can achieve the goal of attaining a

top rank in competitive examinations with one or two years of focused preparation

during class XI and XII, other students, can achieve the same by starting early and

by following a systematic and focused approach. PROPEL program focuses on

preparing students on each topic to a level of detail that helps them to fully

understand the fundamentals effectively and in a much shorter span of time.

One has to concentrate on, at least three aspects in order to view the best

treatment for teaching and learning. The first one is the ability of a teacher, in using

different methodology to drive home his point. The second one is the teacher’s role

in generating and sustaining the interest of the learners by providing the children

ample hands on experience by motivating him. The third one is the teacher’s skill in

keeping his subject content in the frame work of real life experience. I believe that

Propel Academy can empower our teachers in all these areas. I wish all children

enroll in Propel Academy for a great satisfaction through our informed teaching

methods.

(Dr. Chukka Ramaiah)

Eminent Educationist

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CHAPTER 1

MOTION &

NEWTON’S LAWS

OF MOTION

8 - 31

C 1: Few Basic Concepts S1

Introductory Exercise 1 S2

C 2: Acceleration due to gravity S3


Ex 1: Integer type questions

S5 Ex 2: Matrix type (Only one correct type)

Ex 3: Matrix type (One or more than one possible)

Ex 4: Assertion and Reason

Ex 5: One or More than one correct options

Ex 6: Questions for/from Olympiads and NTSE S6

C 3: Newton’s First Law – Law of Inertia S7

C 4: Newton’s Second Law

C 5: Newton’s Third Law



C 6: Frictional force

S10 C 7: Impulse

Ex 7: Single correct objective type questions

S11 Ex 8: One or More than one correct options

Ex 9: Linked Type Questions


Ex 11: Questions from competitive exams & NTSE S12

CHAPTER 2

WORK, ENERGY

AND POWER

32 - 45

C 1: Work, C 2: Energy S1

C 3: Work - Energy Theorem

C 4: Law Of Conservation Of Energy


C 5: Power


Ex 1: Single correct objective type questions

Ex 2: Assertion and Reason S4

Ex 3: Matrix (One or More than one correct)

Ex 4: One or More than one correct options S5

Ex 5: Questions for/from Olympiads & NTSE S6

CHAPTER 3

REFRACTION

OF LIGHT

46 - 61

C 1: Refraction of light S1

C 2: Refractive Index

C 3: Laws of refraction

C 4: Finding unknown refractive index S2

Introductory Exercise 1

C 5: Bending towards the normal or away? S3

C 6: Total Internal Reflection, C 7: Apparent Shift

Solved Examples S4

Ex 1: Single correct objective type questions S5

Ex 2: Integer Type Questions

S6 Ex 3: Assertion and Reasoning

Ex 4: One or more than one correct options

S7 Ex 5: Matrix type (Only one correct type)

Ex 6: Matrix type (One or more than one possible)

CHAPTER 4

GRAVITATION

62 - 77

C 1: Universal Law of Gravitation

S1 Introductory Exercise 1

C 2: Acceleration due to gravity

S2 C 3: Variation in acceleration due to gravity


C 4: Potential Energy, C 5: Escape velocity S3

C 6: Orbital velocity, C 7: Time period of a satellite

C 8: Total energy of the satellite

S4 C 9: Geostationary Satellite and Polar Satellite


Solved Examples S5


Ex 2: Assertion and Reasoning

S7 Ex 4: Linked Type Questions

Ex 5: Match the following

Ex 6: Questions for/from OLYMPIADS & NTSE S8

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CHAPTER 5

FLOATING

BODIES

78 - 91

C 1: Pressure In A Fluid

S1 C 2: Pressure Exerted by a Liquid Column

C 3: Pascal’s Law, C 4: Equation of Continuity

C 5: Buoyancy And Force Of Buoyancy

S2 C 6: Archimedes Principle, C 7: Law Of Floatation


Solved Examples S3



Ex 3: Matrix type (Only one correct type)

Ex 4: Linked Comprehension Type


S6 Ex 6: Questions for/from OLYMPIADS & NTSE

CHAPTER 6

HEAT

91 - 102

C 1: Heat and Temperature

S1 C 2: Specific Heat


C 3: Latent Heat

S2


C 4: Thermal Equilibrium and Law of Mixtures


C 5: Mechanical Equivalent of heat

S3 Introductory Exercise 4

Ex 1: Single Correct Answer Objective Type

S4 Ex 2: Assertion and Reason


S5

Ex 4: Numerical Type Questions

Ex 5: Linked Comprehension Type

Ex 6: Questions for/from Olympiads & NTSE S6

S1 Means session 1/class 1/period 1 of 50 min

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Propel Academy

CHAPTER 1 Motion and

Laws of motion

Physics

8 © Dr. Ravi’s Educational Research and Development Center

Distance: The total path length covered by body during the motion.

Its magnitude is always positive.

Displacement: The shortest distance between initial and final point.

Its magnitude is may be positive, negative and zero.

Speed v =

=

Velocity v =

=

In motion, time and distance always positive and speed can’t be zero, negative. It is always positive.

In motion, the velocity of a body can be zero, negative or positive.

Speed is a scalar quantity as it has magnitude, but not direction.

Velocity is a vector quantity as it has magnitude and also direction.

S.I. unit of speed is m/s or ms-1

C.G.S. unit of speed is cm/s or cms-1

S.I. unit of velocity is m/s or ms-1

C.G.S. unit of velocity is cm/s or cms-1

Average Speed =

=

Average Velocity=

=

Instantaneous speed is the speed at a particular instant of time.

Speedometer in all vehicles shows instantaneous speed not average speed.

Instantaneous Velocity is the Velocity at a particular instant of time.

Speedometer in all vehicles shows instantaneous Velocity, if it travels in a straight line.

Acceleration =

=

Acceleration is a vector and its SI unit is m/s2. CGS unit is cm/s2. In motion, the acceleration of a body can be zero, negative or positive.

If Acceleration is positive, then final velocity > initial velocity Acceleration is negative then final velocity < initial velocity a = 0, then object is at rest or it is moving with a constant velocity

Whenever acceleration is constant throughout motion, then equations of motion v = u + at s = ut + (1/2) at2 v2 - u2 = 2as (v = final velocity, u = initial velocity, s = displacement, and a = acceleration)

Concept 1 Few Basic concepts

Class IX C1:Kinematics > IITians Propel Academy

© Dr. Ravi’s Educational Research and Development Center 9

JEE MAINS /NEET TYPE Introductory Exercise 1

1) If a train is travelling with a velocity of 25 m/s, how long will it take to travel 700 m?

a. 25 s b. 28 s c. 25 min d. 2.8 s

2) A body moving along a straight line at 20 m/s decelerates at the rate of 4 m/s2. After 2 seconds its speed will be equal to__m/s.

a. 8 b. 12 c. 16 d. - 12

3) If a = 5 + 4t2. Then initial acceleration (in m/s2)of the body is

a. 8 b. 10 c. 5 d. 0

4) A train is travelling at uniform speed of 20 m/s. How long will it take to cover a distance of 40m?

a. 1s b. 2s c. 4s d.6s

5) A bike is moving at 10ms–1. How much distance in kilometers will the bike cover in 25 minutes? (in Km)

a. 30 b. 15 c. 40 d. 50

6) Which of the following cannot be the distance – time graph?

a. b.

c. d.

7) A body moves with a speed of 20kmph in the first 5s and with a speed of 30kmph in the next 5s. Then, the average speed (in kmph) of the body is...

a. 25 b. 30 c. 24 d. 34 8) A bus is beginning to move with an acceleration of 1m/s2. A boy is 12 m behind the bus. He runs with uniform velocity 5 m/s. After what time will the boy be able to catch the bus? a. 2 s b. 4 s c. 6 s d. None 9) The numerical ratio of speed to velocity is

a. < 1 b. > 1 c. = 1 d. 0

10) A submarine uses sonar to measure the depth of water below it. Reflected sound waves are detected 0.40 s after they are transmitted. How deep is the water? (Speed of sound in water = 1500 m/s) a. 400 m b. 50 m c. 20 m d. none

11) A body starts from rest and moves along a straight line path with uniform acceleration. The ratio of velocities at t = 1 sec and t = 2 sec is

a. 2:1 b. 1:2 c. 1:1.4 d. 1.4:1

12) An object starts from rest and travels with uniform acceleration and covers 20 m in 2 s. Acceleration of object is __ m/s2.

a. 45 b. 10 c. 15 d. 20 13) A body starts motion with speed 4 m/s and moves along a straight horizontal path with uniform acceleration 2m/s2, after 5

s, velocity of body will be __ m/s.

a. 10 b. 6 c. 14 d. 4

IITians Propel Academy > Physics Class IX


Objects thrown vertically upwards move up to a certain distance and then fall back to the ground. This is due to Earth’s gravitational force.

Equations of motion of objects under the influence of gravity (neglecting air resistance) can be obtained by replacing 'a' with '-g' and the displacement ‘s’ is substituted by ‘h’.

From the experiments the value of acceleration due to gravity 'g' is found as 9.8 m/s2

and its direction is always towards down.

That’s why we take negative sign all the time in the equations irrespective of either the object moves upward or downward. While solving problems we take g value as -10 m/s2 for easy calculations.

v = u + at becomes v = u - gt or v = u - 10 t

s = ut +

gt2 becomes h = ut -

gt2 or h = ut -

10 t2

v2 - u2 = 2as becomes v2 - u2 = -2gh or v2 - u2 = -2 10 h

Here v is the final velocity (vector), u is initial velocity(vector), g is acceleration due to gravity (vector) and t (Scalar) is time taken. g is chosen to be negative if the body is moving towards the Earth i.e., downward and g is negative if the body is moving in upwards direction, that’s because the Earth’s attraction is always downwards (negative

y axis).

Sign convention Initial velocity (u)

Final velocity (v)

Displacement (h)

Acceleration due to gravity (g)

Object is moving up Positive Positive Positive Negative

Object is moving down Negative Negative Negative Negative

Equations of motion of a body dropped from a certain height: A freely falling body (initial velocity is 0 m/s) moves only under the influence of gravity (i.e. no other force acts on it). Taking u = 0, s=-h and a = -g, we can write the equation of motion for a freely falling body, as

(i) v = u + at or v = - gt Velocity of the object at the time t is – gt

(ii) s = ut + 21at

2 or - h = -

gt2 so h =

gt2

Time taken for the body to reach the ground (time of descent) is 2h

t ,g

(iii) 2 2v u 2as or (-v)2 = 2(-g)(-h) so v2 = 2gh or v =

Velocity of the object after travelling a distance h is

Concept 2 Acceleration due to gravity



Equations of motion for a body projected vertically upwards: Consider a body projected vertically upwards with an initial velocity, u.

As the body moves up its velocity decreases, since the Earth pulls the body downwards with an acceleration of g.

The body moves until its velocity becomes zero (at maximum height). Such bodies are called vertically projected bodies. As the direction of motion is against the direction of acceleration due to gravity (g), the sign of g is taken to be negative.

Taking s = h, a = -g and v at maximum height = 0 we can write the equations of motion for such body as follows.

(i) v = u + at or v = u – gt or 0 = u -gt or u = gt

So if we know time of ascent we can find with what velocity we have thrown up or if we know initial velocity, we can find time of

ascent.

(ii) s = ut –1

2 at2 becomes h = ut 21

gt2

(iii) 2 2v u 2as becomes 2 2v u 2gh

JEE MAINS/ NEET TYPE Introductory Exercise 2

1) A body is projected vertically upwards with a velocity 20m/s. The maximum height reached by the body is _____________ m. a. 10 b. 20 c. 30 d. 40

2) A body dropped from a height reaches the ground in 5s. The velocity with which it reaches the ground is__ m/s.

a. 49 b. 39 c. 89 d. 99

3) A ball dropped freely taken 0.2s to cross the last 6m distance before hitting the ground. Total time of fall in seconds is (g = 10m/s2)

a. 1.1 b. 2.1 c. 3.1 d. 4.1

4) A body is thrown vertically upwards and rises to a maximum height of 10m. The velocity with which the body was thrown upwards is _ m/s.

a. 10 b. 14 c. 20 d. 25

5) A body fall freely from height 19.6m. The body reached on earth surface in time _ sec.

a. 4 b. 3 c. 2 d. 5

6) An objected projected upwards with speed 19.6 m/s. The time after which object reached on the maximum height is _ sec.

a. 4 b. 2 c. 6 d. 3



7) The velocity of a body is given by the equation v = 8 – 0.8t, where t is the time taken. The body is undergoing

a. Uniform retardation b. Uniform acceleration c. Non-uniform acceleration d. Zero-acceleration

8) A freely falling body covers 44.1m in the last second of its journey. The total distance travelled by the body

a. 2.5 b. 22.5 c. 122.5 d. 222.5

9) A body thrown up with some initial velocity reaches a maximum height of 50m. Another body with double the mass thrown up with double the initial velocity will reach a maximum height of __m.

a. 400 b. 150 c. 100 d. 200

10) A body projected up with a velocity reaches a maximum height of 100 m. Another body projected up with double the initial velocity. The maximum height reached will be __m.

a. 400 b. 150 c. 100 d. 200

11) If the distance travelled by a freely falling body in the last second of its journey is equal to the distance travelled in the first 2s, the time of descent of the body is

a. 1.5 s b. 0.5 s c. 2.5 s d. 3.5 s

12) Two balls are dropped to the ground from different heights. One ball is dropped 2s after the other. But both of them strike the ground simultaneously 5s after the first is

dropped. The difference in the heights, when they are dropped is

a. 20 m b. 40 m c. 80 m d. 60 m

13) Two balls are dropped simultaneously from two points separated by a vertical height of 6m. The distance of separation between them after next 2s is

a. 2m b. 1m c. 0m d. 6m

14) A body projected vertically upwards with certain velocity reaches a maximum height at 98m. If it is projected up with the same velocity on the moon, the maximum height reached by it will be.. a. 88 b. 58 c. 588 d. 888

15) A body is thrown vertically up reaches a maximum height of 78.4 m. After what time it will reach the ground from the maximum height....

a. 8s b. 5s c. 4s d. 10s

16) A particle falls from a height then rebounds, which of the following graph is correct

a. b.

c. d.



1) A body starts motion with speed 4m/s and moves along a straight horizontal path with uniform acceleration 2m/s2, after 5 sec the velocity of body will be n m/s. Then n is

2) A body with an initial velocity 3 m/s moves with an acceleration 2m/s2, then the distance traveled in the 4th second is n m. Then n is

3) Water drops fall from a tap of height 1.6m at regular intervals. If the first drop touches the ground, 5th drop detaches from the tap. Then time taken by the first drop to reach the ground is n seconds. Then n is

4) A balloon is rising vertically up with a velocity of 29 ms–1. A stone is dropped from it and it reaches the ground in 10 s. The height of the balloon

when the stone was dropped from it is n/100 m. Then n is

5) An object moving with a velocity 2 m/s and moves with uniform acceleration 2m/s2. After 2sec the velocity of object will be n m/s. Then n is

6) A body is thrown vertically up with a velocity of 20 ms–1. It will fall on the

ground after n seconds. Then n is

JEE MAINS/NEET Exercise 2 Matrix type (Only one correct type)

A) Column I Column II B) Column I Column II

1. Freely falling apple p. Random 1. Translatory motion p. A simple pendulum

2. Motion of Earth q. Non-uniform motion

2. Circular motion q. Firing a bullet from a gun

3. Motion of a bird r. revolution 3. Oscillatory motion r. Rotating fan

JEE ADVANCED Exercise 3 Matrix type (One or more than one possible)

C) Column I Column II D) Column I Column II

1. Acceleration p. m 1. Distance p. Always positive 2. Speed q. scalar 2. Speed q. Can be negative 3. Displacement r. m/s2 3. Acceleration r. Can be zero 4. Velocity s. vector 4. Velocity s. m/s

JEE ADVANCED Exercise 1 Integer type questions



JEE ADVANCED Exercise 4 Assertion and Reason

a. Assertion is True, Reason is True; Reason is a correct explanation for Assertion b. Assertion is True, Reason is True; Reason is not a correct explanation for Assertion

c. Assertion is True, Reason is False d. Assertion is False, Reason is True

1) Assertion: When you drop a mass from a height, is non-uniform motion. Reason: In non-uniform motion, velocity is not constant.

2) Assertion: When we throw an object vertically up, it’s a = -10 m/s2. Reason: acceleration due to gravity wont depend on the mass of the object what we throw up.

3) Assertion: If you drop 1kg and 2 kg balls from 1 meter, then 2 kg both

will reach the ground at the a same time. Reason: Equations of motion won’t depend on mass.

4) Assertion: Acceleration due to gravity can never be negative. Reason: Acceleration is a vector.

5) Assertion: Time of flight wont depend on mass. Reason: Time of ascend = time of descend.

1) If a body is accelerating, it may

a. Speed up b. Speed down c. Move with same speed d. Move with same velocity

2) A ball is thrown up with a certain velocity. It attains a height of 40m and comes back to thrower. Then

a. Total distance covered is 80 m b. Magnitude of displacement is 80 m c. Displacement is zero d. Average velocity is zero

3) If the velocity of a body is constant

a. | velocity | = speed b. | average velocity | = speed

c. Velocity = average velocity d. Speed = average speed

4) Choose correct statement

a. acc can be nonzero when = 0

b. acc must be zero when = 0

c. acc may be zero when 0 d. The direction of acc must have some

correlation with the direction of .

5) At the maximum height of a body thrown vertically up a. Velocity is not 0 but acceleration is 0 b. Acceleration is not 0 but velocity is 0 c. Both acceleration and velocity are 0 d. Acceleration and velocity are not 0

JEE ADVANCED Exercise 5 One or More than one correct options



JEE ADVANCED Exercise 6 Questions for/from Olympiads and NTSE

1) A racing car has uniform acceleration of 4m/s2. What distance will it cover in 10seconds after the start?

a. 100m b. 300m c. 200m d. 400m

2) A particle moves in a circle of

radius R. In half the period of

revolution its displacement and

distance covered is

a. 2R, R b. R,

c. R, R d. 2R, R

3) The numerical value of the ratio

of average velocity to average

speed is

a. Always less than one

b. Always equal to one

c. Always more than one

d. Equal to or less than one

4) Two stones of different masses are dropped simultaneously from the top of a building. Then which is correct among these?

a. Larger stone hits ground earlier

b. Smaller stone hits ground earlier

c. Both at the same time

d. We can’t say

5) A balloon rises from rest with a constant acceleration g/8. A stone is released from it when it has risen to height h. The time taken by the stone to reach the ground is

a.h

4g

b.h

2g

c.2h

g

d.g

h

6) A stone falls from a balloon that is descending at a uniform rate of 12 m/s. The displacement of the stone from the point of release after 10 sec is _ m.

a. 490 b. 510 c. 610 d. 725

7) Which of the following curve

does not represent motion in one

dimension?

a. b. c. d.

8) Which graph represents uniform

motion?

a. b. c. d.

9) In an imaginary atmosphere, the air exerts a small force F on any particle in the direction of the particle’s motion. A particle of mass ‘m’ projected upward takes a time t1 in reaching the maximum height

and t2 in the return journey to the

original point then a. t1<t2 b. t1>t2 c. t1= t2 d. None

10) A wheel completes two

rotations. Then angle through

which it rotated is

a.2 rad b. 4 rad c. 10 rad d. 20 rad

11) The radius of earth is 6400 km.

The linear velocity of a point at the

equator is ___ km/h.

v

t

v

t

v

t

v

t

x

t

v

t

x

t

v

t



a. 600 b. 800 c. 1000 d. 1600

12) A particle has an initial velocity of 4 m/s and a constant acceleration of 4 m/s2. Its speed after ten seconds will be equal to ___ m/s.

a. 44 b. 10 c.5 2 d. None

13) A stone is released from an elevator going up with acceleration 5 m/s2. The acceleration of the stone, after its release, with respect to the ground, is __ m/s2. a. 5 b. 4.8 upward

c. 4.8 downward d. 9.8 downward

14) A particle is moving eastwards with a velocity of 5 m/s. In 10 s the velocity changes to 5 m/s northwards. Then there is a change in

a. Acceleration b. Velocity

c. Speed d. a and b

15) A person walks along a straight road from his house to a market 2.5kms away with a speed of 5 km/hr and instantly turns back and reaches his house with a speed of 7.5 kms/hr. The average speed of the person during the time interval

0 to 50 minutes is (in m/sec)

a. 3/2 b. 7/3 c. 1/3 d. 5/3

16) A ball is throw vertically upwards from the top of a tower. Velocity at a point ‘h’ m vertically below the point of projection is twice the down ward velocity at a point ‘h’ m vertically above the point of projection. The maximum

height reached by the ball above

the top of the tower is...

a. h/3 b. 2h/3 c. 4h/3 d. 5h/3

17) An athlete completes one round at a circular track of radius R in 40 sec. What will be his displacement at the end of 2 min 20 seconds (2010-E)

a. R b. 4R c. 2R d. 8R

18) A body is projected vertically upwards with a velocity ‘u’. It crosses a point in its journey at a height ‘h’ twice, just after 1 and 7 seconds. The value of u in ms-1 is

a. 10 b. 20 c. 30 d. 40

19) A body is thrown vertically upwards with an initial velocity ‘u’ reaches maximum height in 6s. The ratio of the distance travelled by the body in the first second to the

seventh second is

a. 21:1 b. 1:1 c. 3:1 d. 11:1

20) A ball of mass m1 and an another ball of mass m2 are dropped from equal height. If time taken by the balls are t1 and t2

respectively then

a. t1 = t2 b. t1 = 2t2

c. t1 < t2 d. none

21) A man throws a ball vertically upward and it rises through 20m and returns to his hands. What was initial velocity of the ball and for how much time (T) it remained in

the air (g= 10ms-2)

a. 10, 4 b. 20, 4 c. 3, 4 d. 4, 40

Propel Academy

CHAPTER 4 GRAVITATION

Physics


Kepler do not know what makes planets to revolve as described above. It was Newton who explained it.

Newton's Law of Gravitation states that

F and F

; F

or F G

Where G is proportionality constant called as gravitational constant. This constant is same or will not change in any place of the universe. For this reason, it is also called as universal gravitational constant. Its value was first measured by Cavendish and is now known to be: G = 6.63 × 10-11 N-m2/kg2 From Newton's third law, every action have equal and opposite reaction (F12 = - F21)

The magnitude of the force with which m1 attracts m2 (F12) is G

The magnitude of the force with which m2 attracts m1 (F21) is G

But the direction of force F21 is in the opposite direction of F12.

1) The gravitational force is always an attractive force.

2) The gravitational force between two particles does not depend on the medium.

3) The gravitational force between two particles is along the straight line joining the

particles (called line of centers).

4) The force of attraction between a

solid spherical shell (or spherical sphere)

of uniform density and a point mass

situated outside is just as if the entire

mass of the shell is concentrated at the

centre of the shell.

5) The force of attraction due to a hollow spherical shell of uniform density, on a point mass situated inside it is zero. (Sum of all attractive forces on the point mass from the solid or hollow spherical shell is zero)

Concept 1 Universal Law of Gravitation

Important points

Class IX C4:Gravitation > IITians Propel Academy


1) The force of attractive between two object is 10 N. If we immerse the two objects in to water, then

the force of attraction is

a. >10N b. <10N c. =10N d. none

2) SI unit of gravitational constant is

a. N m2kg2 b. N m2kg-2

c. N m2s-2 d. N mkg-2

3) What is the value of

gravitational constant?

a. 6.63 x 10-11 N mkg2 b. 6.62 x 10-10 N m2kg2 c. 6.17 x 10-11 N m2kg2

d. 6.63 x 10-11 N m2kg2

4) If the distance between two bodies is doubled, the force of attraction F between them will be

a. 0.25 F b. 2 F c. 0.5 F d. F

5) The force of gravitation between two bodies in the universe does not

depend on

a. The distance between them b. The product of their masses c. The sum of their masses

d. The gravitational constant

6) The force of attraction due to a hollow spherical shell of non-uniform density, on a point mass situated inside it is a. Zero b. Non zero c. We cannot tell d. None

7) The fundamental force which holds the planets in their orbits

around the sun is

a. Gravitational force of attraction b. Electrostatic force of attraction c. Nuclear force of attraction

d. Electro static force of repulsion

8) If a pen and pencil separated by 10 cm, then the nature of force between them is

a. Attractive only b. Repulsive only

c. a and b d. Neither a nor b

9) If a proton and electron is separated by a distance of 1 nm, then the nature of force between them is a. Attractive only b. Repulsive only

c. a and b d. Neither a nor b

10) Two uniform spheres r1 and r2 is shown below. What is the distance of separation between

them?

a. r b. r1+r2 c. r1+r+r2 d. r-r1-r2 11) Two uniform spheres r1 and r2 are placed as shown below. What is the distance of separation

between them?

a. 0 b. r1+r2 c. r1+2r2

d. r1-r2

JEE MAINS/NEET TYPE Introductory Exercise 1



If M is the mass of Earth and R is the radius, the Earth attracts a mass m on its surface with a force F given by

F G

But Newton's law F =ma

So G

Then a = G

a = g = Acceleration due to Earth

Acceleration due to gravity is

(i) Directly proportional to the mass of the Earth and

(ii) Inversely proportional to the radius of the Earth.

Example: Calculate the value of acceleration due to gravity (g) on the

Earth and Moon.

(G = 6.673 × 10–11 Nm2kg–2, Mass of Earth = 5.98 × 1024 kg, Radius of Earth = 6.4 ×

106 m. Mass of the moon = 7.4 × 1022 kg, radius of the moon = 1.75 × 106 m)

11 2 2 24

26

6.673 10 Nm kg 5.98 10 kgg

6.4 10 m

= 9.8 Nkg–1 = 9.8 kgms–2kg–1= 9.8 ms–2

In the same way, g on moon=

11 2 2 22

6

6.673 10 Nm kg 7.4 10 kg

1.75 10 m

= 1.6 ms–2

We can see clearly, g on moon =

g on Earth.

Note: g is not depending on the mass of the falling object. It means if we drop two masses m1 and m2 from the same height on Earth surface, both will accelerate with g,

so that both will reach the ground at the same time and with same velocity.

The falling of a body (or object) from a height towards the Earth under the gravitational force of Earth (with no other forces acting on it) is called free fall. A freely falling body has acceleration equal to acceleration due to gravity (g). The product of m and g is called weight of the object. Weight = mg.

Equations of motion for a freely falling object are v = u +

at becomes v = u – 5 t

s = ut + at2 becomes S = ut - 10t2 ; v2 - u2 = 2 as becomes v2 - u2 = -20s

(We took a = g = -10 m/s2, the force of attraction is always towards the Earth)

Concept 2 Acceleration due to gravity (of Earth)



4.1 Above the surface of the Earth: When an object mass m located at a height h above the Earth's surface, then the distance of separation is (R+h).

Then acceleration due to gravity g is given by gh = Acceleration due to gravity at height h above from the surface of the Earth

gh =

We know g = G

Relation between g and gh is gh = g

or gh = g

4.2 Below the surface of the Earth: When an object mass m located at a depth d below the Earth's surface, then the distance from the center is R-d.

Mass of Earth = density of Earth x volume of spherical Earth M = x

R3

When we go down d from Earth surface, then effective mass of the Earth will change as

M = x

d)3. Force between an object mass m and to the Earth is

F G

G

F = G

gd = Acceleration due to gravity at depth d below from the surface of Earth

gd =

G

On the surface g =

=

By dividing gd and g we get

=

=

=

or gd = g

Graphical representation

4.3 Rotation of Earth: Rotation of Earth also effects the value of g. The value of g at a latitude is

= g - R

Concept 3 Variation in acceleration due to gravity



1) The acceleration due to gravity

is zero at the

a. Equator b. Poles c. Center of the Earth d. Sea level

2) If acceleration due to gravity on Earth is 10 m/s2 then, the acceleration due to gravity on

moon is __ m/s2.

a. 1.66 b. 16.6 c. 10 d. 0.166

3) A feather and a coin released simultaneously from the same height do not reach the ground at

the same time because of the

a. Resistance of the air b. Force of gravity c. Force of gravitation d. Difference in mass

4) The weight of an object of mass

10 kg on Earth is

a. 9.8N b. 9.8kg c. 98N d. 98kg

5) The weight of an object of mass

15 kg at the centre of the Earth is

a. 147N b. 147kg c. 0N d. 150N

6) Mass remains ___ throughout

the universe.

a. Varies b. Zero c. Constant d. Negative

7) The mass of an object is __kg,

whose weight on Earth is 196 N.

a. 20 b. 0.2 c. 1960 d. 2

8) As we go from the equator to

the poles, the value of g

a. Remains the same b. Decreases

c. Increases d. None

9) The ratio of SI units to CGS units of g is

a. 102 b. 10 c. 10–1 d. 10–2

10) The value of g on the Earth’s

surface is 980 cm/s2. Its value at a height of 64km from the Earth’s

surface is _ cm/s2

a. 960.4 b. 984.9 c. 982.4 d. 977.5

11) A man weights W on the surface of Earth. What is the

weight at a height equal to R?

a. W b. W/2 c. W/4 d. W/8

12) The value of g is with depth

a. Increases b. Decreases

c. Does not change d. None

13) g __ when we move towards equator to poles.

a. Decreases b. Increases

c. Constant d. None

14) Our weight will increase from

poles to equator.

a. Decreases b. Increases c. Constant d. None

15) Value of acceleration due to moon at center of the moon is _

m/s2

a. 10 b. 8 c. 4 d. 0




If two particles are separated by a distance r, then the

potential energy of the two masses is given as U = - G

.

Here negative symbol represents that, the two masses are bound with that energy. Or

in order to separate two masses, we have to provide G

of energy.

For 3 particles system, as shown below potential energy is

given as U = - (G

+ G

Every object on the surface of Earth will have a potential

energy of

. Here m is mass of the object and M is mass

of Earth and R is radius of Earth.

Escape velocity is the minimum velocity with which a body must be projected from the surface of Earth, so that it will escape the gravitational attraction of the Earth and it will never come back to Earth. In this case Kinetic energy = Potential energy.

or Ve =

=

By substituting the value of g = 9.81 m/s2 and R = 6400 km we get 11.3 km/s. Hence any object thrown with a velocity of 11.3 km/s or more will escape the gravitational field of the Earth and will never come back to the Earth.

Consider a satellite of mass m revolving in a circle at a height h above the Earth's surface. The radius of its orbit is r = R + h, where R is the radius of the Earth. The gravitational force between m and M provides the centripetal force necessary for a

circular motion. The velocity of a satellite in its orbit is called orbital velocity (vo). Then

Force due to gravity = Centripetal force

or Vo =

=

Hence, the orbital velocity of a satellite is decided by the radius of its orbit or its height

above the Earth's surface. For the satellite very close to Earth's surface r = R + h = R.

Vo =

=

=

Relation between orbital velocity and escape velocity:

Ve = and Vo = so Ve = Vo

Concept 4 Potential Energy

Concept 5 Escape velocity

Concept 6 Orbital velocity



The time taken to complete one revolution is called the time period. It is given by

T =

= 2

or T =

or T2 =

or T2

The kinetic energy of the planet is K =

=

The gravitational potential energy of the sun-planet system is U =

The total mechanical energy of the sun - planet system is

Total energy = kinetic energy + potential energy

Total energy = (

The total energy is negative. This is true for any bound system if the potential energy is

taken to be zero at infinite separation.

In another words, we have to provide this much of energy to the satellite so that it will

escape from Earth's orbit.

Geostationary satellites appear to be stationary as the time of rotation of earth = time of rotation of satellite. These are used for telecommunication, weather forecast

and other applications.

According to above equation, T2 =

or

1/32

2

GMTr

4

By substituting the values of G, M (6 × 1024 kg) and T (24 hours), the radius of the

geostationary orbit comes out to be r = 4.2 × 104 km. The height above the surface of

the earth is about 3.6 × 104 km.

Polar satellites will revolve in polar orbits around earth. A polar orbit is that orbit

whose angle of inclination with equatorial plane of earth is 90º.

The altitude of polar satellites is about 500 to 800 km from the surface of the earth. The time period of revaluation of polar satellites is about 100 minutes of few hours. Polar satellites cross any location on earth many times a day.

Concept 7 Time period of a satellite

Concept 8 Total energy of the satellite

Concept 9 Geostationary Satellite and Polar Satellite



1) The ratio of the velocity to the orbital velocity is

a. 2 b. c.

d.

2) The time period of a simple pendulum on a freely moving artificial satellite is

a. Zero b. 3 sec c. 2 sec d. Infinite

3) Orbital velocity of a satellite is_ km/s. a. 5.8 b. 18.4 c. 11.2 d. 8.0

4) The escape velocity of a body from the Earth is u. What is the escape velocity from a planet whose mass and radius are twice those of the Earth?

a. 2u b. u c. 4u d. 16u

5) The atmosphere is held to the Earth by a. Winds b. Clouds c. Gravity d. None

6) Weightlessness of an astronaut moving in a satellite is a situation of a. Zero g b. Zero mass

c. No gravity d. Free fall

7) The escape velocity of a particle o mass m is proportional to

a. m2 b. m c. m0 d. m-1

8) A satellite of mass m revolves round a planet o mass M. Orbital velocity of the satellite will be

a.

b.

c.

d.

9) How much energy will be necessary or making a body of 500 kg escape from the Earth? a. 9.8 × 106J b. 6.4 × 108J

c. 3.1 × 1010J d. 27.4 × 1012J

10) With what velocity, we need to throw a ball so that it won’t come back and fall on the Earth?

a. 1.3 km/s b. 14.3 km/s

c. 10.3 km/s d. 11.3 km/s

11) An artificial satellite moving in a circular orbit around the earth has a total (kinetic + potential)

energy E0. Its potential energy is

a. -E0 b. 1.5 E0 c. 2 E0 d. E0

12) If the distance between the earth and the sun becomes half its present value, the number of days in a year would have been

a. 64.5 b. 129 c. 182.5 d. 730

13) A geostationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the time period of a satellite orbiting a few hundred kilometres above Earth’s

surface will approximately be

a. 0.5 h b. 1 h c. 2h d. 4h

14) Two identical satellites A and B are circulating round the earth at the height of R and 2R respectively, (where R is radius of the earth). The ratio of kinetic energy of A to

that of B is

a. 1/2 b. 2/3 c. 2 d. 3/2

15) The mean radius of the earth's orbit round the sun is 1.5 x 1011m. The mean radius of the orbit of mercury round the sun is 6 x 1010

m. The mercury will rotate around

the sun in

a. 1 yr b. 4 yr c. ¼ yr d. 2.5 yr




1) Calculate the mass from Sun of the following data; distance between the Sun and the Earth = 1.49 x 1011m, G = 6.67 x 10-11 SI

units, one year = 365 days.

Solution: Force of attraction between

the sun and the earth = G. 2SE

ES

d

mm

Considering the orbit of the earth nearly circular the centripetal force acting on

the earth is mE r2

mE r2 = G

2SE

ES

d

mm

mS = 2

2SE

222SE

GT

d4

G

.d

=

211

2112

60x60x24x365x10x67.6

10x49.1x14.3x4

= 19.72 x 1031 kg

2) An artificial satellite of mass 100kg is in circular orbit at 500 km above the earth’s surface. Take the radius of the earth as 6.5 x 106 m. a. Find the acceleration due to gravity at any point along the satellite path. b. What is the centripetal acceleration of the satellite?

Solution: Here, h = 500 km = 0.5 x 106 m R = 6.5 x 106 m r = R+h= 6.5 x 106+0.5x106 =7x 106 m a. Now, g = g

2

6

62

10x0.7

10x5.68.9

hR

R

= 8.45 m/s2

b.Centripetal force on the satellite,

F = r

mv2

Centripetal acceleration,

a =r

rgR

r

v

m

F

22

2

=

2

2

2

2

2

s/m45.8hR

Rg

r

gR

3) Jupiter has mass 318 times that of earth, and its radius is 11.2 times the earth’s radius. Estimate the escape velocity of a body from Jupiter‘s surface given the escape velocity from the earth’s surface as 11.2 km/s.

Solution: Hence, MJ = 318 Me; RJ = 11.2 Re; ve = 11.2 km/s

We know, vJ=J

J

R

GM2

and

ve = e

e

R

GM2

J

e

e

J

e

J

R

Rx

M

M

v

v

vJ = veJ

e

e

J

R

Rx

M

M

vJ =

2

1

e

e

e

e

R2.11

Rx

M

M3182.11

= 11.2 s/km7.592.11

318 2

1

4) A satellite orbits the earth at a height of 3.6 106 m from its surface. Compute its a. kinetic energy b. potential energyc. total energy Mass of satellite = 500 kg, mass of the earth = 6 1024 kg; radius of

Solved Examples



the earth = 6.4 1024 kg; radius of the earth = 6.4 106m; G = 6.67 10–11 Kg–2. Solution: Here, r = R + h = 6.4 106

+ 3.6 106= 107 m Orbital velocity of satellite around the earth is given by

0

GMv

R h

11 24

7

6.67 10 6 10

10

7 16.67 6 6 10 ms

a. K.E. of the satellite

2

0

1mv

2

1500 6.67 6 10

2

= 1010J

b. P. E. of the satellite

GMm

R h

11 24

7

6.67 10 6 10 500

10

= – 2 1010J c. Total energy = KE + PE = 1010 – 2 1010 = – 1010 J. Calculate the escape speed on the surface of a planet of mass 1.7 1025 gram, its radius 1.6 106m. Solution: Here, M = 7.5 1025 g; R = 1.6 108 cm

Escape speed, e

2GMv

R

8 25

8

2 6.67 10 7.5 10

1.6 10

= 2.5006 105 cm/s. 43. An artificial satellite is moving in a circular orbit around the earth with a speed equal to half

the magnitude of escape velocity from the earth. (i) Then the distance of satellite from the surface of earth. (ii) If the satellite is stopped suddenly in its orbit and allowed to fall freely onto the earth, the speed with which it hits the surface of the earth. (iii) The time that it takes to reach the surface of the earth. Solution (i) Let M and R be the mass and radius of the earth respectively. If m be the mass of satellite, then escape velocity

from earth ne =

Velocity of satellite ns = =

Further ns = =

= h = R = 6400 km (ii) Now total energy at height h = total energy at earth's surface (principle of conservation of energy)

0 – G M = mn2 – GM

or mn2 = – (Q h = R)

Solving we get n =

n = = 7.919 km/sec.

)Rg2(

2

e

2/)gR2(

r

GM

hR

gR2

2s hR

gR2

hR

m

R

m

2

1

R

mGM

R2

GMm

)gR(

)8.9x10x6400( 3



1. If R is the radius of earth the height at which the weight of a body becomes one fourths its weight on the surface of earth, is

a. 2R b. R c. 0.5R d. 0.25R

2. When an apple falls from a tree

a. Only earth attracts the apple b. Only apple attracts the earth c. Both the earth and the apple attract each other

d. None attracts each other

3. If the distance between two masses be doubled, then force between them will become _ times.

a.1

4 b. 4 c.

1

2 d. 2

4. The radius vector drawn from the Sun to a planet sweeps out __ Areas in equal time.

a. exponential b. equal c. unequal d. zero

5. The acceleration due to gravity

at the moon’s surface is 1.67 ms2. If the radius of the moon is 1.74 ×

106 m. Then mass of the moon

a. 7.55×1022 kg b. 8.55×1022 kg

c. 7.55×1020 kg d. 7.45×1021 kg

6. In a satellite if the time of revolution is T, then K.E. is

proportional to

a.1

T b. 2

1

T c. 3

1

T d. 2/3T

7. At what height (km) over the Earth's pole, the free fall

acceleration decreases by one

percent?

a. 32 b. 1 c. 80 d. 64

8. When the medium between two bodies changes, force of

gravitation between them

a. Will increase b. Will decrease c. a and b d. Will not change

9. Value of g is

a. Max at poles b. Max at equator c. Same everywhere d. Min at poles

10. At what height above the Earths surface does the acceleration due to gravity fall to 1% of its value at the earths

surface?

a. R b. 5R c. 10R d. 9R

11. The time period of a geostationary satellite is __ hours. a. 24 b. 12 c. 365 d. none 12. The mass of a planet is 6 times that of Earth. The radius of planet is twice that of earth. If the escape velocity from earth is v, then the

escape velocity from the planet is

a. 3 v b. 2 v c. v d. 5 v

13. The radius vector drawn from the Sun to a planet sweeps out

……. Areas in equal time

a. Exponential b. Equal

c. Unequal d. Zero

JEE ADVANCED Exercise 1 Single correct objective type questions



JEE ADVANCED Exercise 2 Assertion and Reasoning

a. Assertion is True, Reason is True; Reason is a correct explanation for Assertion. b. Assertion is True, Reason is True; Reason is not correct explanation for Assertion. c. Assertion is True, Reason is False. d. Assertion is False, Reason is True.

1) Assertion: The moon has no atmosphere. Reason: Escape velocity of moon is smaller than the Earth.

2) Assertion: IF we insert a medium between two bodies, force of attraction will decrease. Reason: Medium has no effect on gravitational attraction.

3) Assertion: When a ball is on the surface of earth then its potential energy is zero. Reason: Distance of separation in the above situation is 0m.

4) Assertion: If we decrease the distance by half then its force increases by four times. Reason: Gravitational force of attraction is inversely proportional to distance of separation.

5) Assertion: Gravitational forces are always attractive.

Reason: The earth attracts the moon with a gravitational force of 1020 N. Then the

moon attracts the earth with a gravitational force of 1010 N. 6) Assertion: The gravitational force between two objects will increases 4 times, if

the mass of one object is doubled.

Reason: The radial acceleration of a planet is zero.

7) Assertion: Gravitational force is a medium dependent force. Reason: Gravitational force is a conservative force.

8) Assertion: g increases with height above the Earth surface. Reason: g affected by the Earth’s rotation.

9) Assertion: g increases with depth from the Earth’s surface.

Reason: The SI unit of g is m/s2.

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The escape velocity is the least velocity required by a body to escape away from the gravitational pull of the earth. In the escaping condition kinetic energy is equal to potential energy, i.e., total energy of the body is zero. The escape velocity is independent of angle of projection. Escape velocity depends on the mass of central body as well as radius of central body. When velocity of orbiting body increases, its kinetic energy increases and hence total energy.

1) A projectile is fired with a velocity less than escape velocity. Then sum of its kinetic and

potential energy is

a. Negative b. Positive c. May be positive or zero d. Zero

2) As the radius of orbiting body decreases. Its time period

a. Increases b. Decreases

c. Remains unchanged d. None

3) For the planets orbiting around the sun, the quantity which

remains constant is

a. Linear speed b. Kinetic energy

c. Angular speed d. Angular momentum

An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of escape velocity from the earth. Total energy is always constant at any point.

1) The height of the satellite above the Earth surface

a. 6400 km b. 3200 km

c. 9600 km d. 4800 km

2) If the satellite is suddenly stopped in its orbit and allowed to fall freely into the earth. Find the

speed with which it hits the surface of the earth:

a. 88 km/h b. 44 km/h

c. 22 km/h d. 66 km/h.

3) Projection kinetic energy will be

a. mgR

2 b.

mgR

4 c. mgR d. 2 mgR.

JEE ADVANCED Exercise 5 Match the following

Column I Column II

1) Kinetic Energy a.

2) Potential Energy b.

3) Total energy c.

Column I Column II

1) g at height a. g 2) g at depth b. g - R

3) g at pole c. g

4) g at equator d. g

JEE ADVANCED Exercise 4 Linked Type Questions



1) At what height from the ground will the value of ‘g’ be the same as that in 10 km deep mine (in km) below the surface of Earth?

a. 20 b. 15 c. 10 d. 5

2) If density of Earth increased 4 times and its radius become half of

what it is, our weight will

a. Becomes 4 times b. Remain same

c. Be doubled d. Be halved

3) At what distance from the centre of the Earth, the value of acceleration due to gravity g will be half that on the surface (R = radius of Earth)

a. 2 R b. 1.4 R c. R d. 0.4 R

4) If the change in the value of ‘g’ at a height h above the surface of the Earth is the same as at a depth x below it, then (both x and h being much smaller than the radius of the

Earth)

a. x = h b. x = h/2

c. x = 2h d. x = h2

5) If radius of Earth is R then the height ‘h’ at which value of ‘g’ becomes one-fourth is

a. R/4 b. R c. 3R/4 d. R/8

6) The weight of a body at the

centre of the Earth is

a. Will not change b. Zero

c. Infinite d. None

7) Force of gravity is least at _

a. The equator b. The poles c. Between equator and any pole d. None

8) A satellite of the Earth is revolving in a circular orbit with a uniform speed v. If the gravitational force suddenly disappears, the satellite will

a. Continue to move with velocity v along the original orbit b. Fall down with increasing velocity c. Move with a velocity v, tangentially to the original orbit d. Ultimately come to rest somewhere on the original orbit

9) If radius of Earth were decreased by 1% its mass remaining the same acceleration due to gravity on the surface of Earth will

a. Increase by 1% b. Decrease by 2%

c. Decrease by 1% d. Increase by 2%

10) Two satellites of masses m1 and m2 are revolving round the Earth in circular orbits of radii r1 and r2 (r1>r2). Then which of following relations is correct regarding their orbital velocity V1 and V2

a. V1 > V2 b. V1 < V2

c. V1 = V2 d. V1 /r1 =V2/r2

11) The Earth revolves round the Sun in one year. If distance between them becomes double

the new period of revolution will be

a. 1/2 b. 2 2 c. 4 d. 8

JEE ADVANCED Exercise 6 Questions for/from OLYMPIADS & NTSE



12) In an artificial satellite a space traveler tries to fill ink in a pen by dipping it ink. Amount of ink filled in the pen as compared to the quantity of ink filled on the Earth’s

surface will be

a. Less b. More c. Same d. Nil 13) The mass o the Earth is 81 times that of the moon and the radius of the Earth is 3.5 times that of the moon. The ratio of the escape velocity on the surface of Earth to that on the surface o moon will be

a. 0.2 b. 2.57 c. 4.81 d. 0.39

14) The period of revolution op a certain planet in an orbit of radius R is T. Its period of revolution in an orbit of radius 4R will be ____

a. 2T b. 2 2 T c. 4T d. 8T

15) If the radius of the Earth were to shrink by 1% its mass remaining the same, the acceleration due to gravity on the Earth’s surface would

a. Decreasing by 1% b. Remain unchanged c. Increase by 1%

d. Increase by 2%

16) The distance of Neptune and Saturn from Sun are respectively 1013 and 1012 meters and their periodic times are respectively TN and TS If their orbits are assumed to be circular the value of TN/TS

a. 100 b. 10 10 c.

1

10 10 d. 10

17) The escape velocity from the Earth’s surface is 11 km/s. If or another planet the radius is twice that of the Earth but the mean density is the same as that of the Earth, the escape velocity from this planet will be

a. 5.5 km/s b. 11 km/s

c. 16.5 km/s d. 22 km/s

18) The mass of planet is 1/9 of the mass of the Earth and its radius is half that of the Earth. If a body weights 9N on the Earth, its weight

on the planet would be,

a. 6 N b. 4.5 N c. 2.25 N d. 1 N

19) Two spheres of mass m and M are situated in air and the gravitational force between them is F. The space around the masses is now filled with a liquid. The gravitational force will now be

a. F b. F/9 c. F/3 d. 3 F

20) Two identical solid copper spheres of radius R placed in contact with each other. The gravitational attraction between

them is proportional to

a.R2 b. R4 c. R-2 d. R-4

21) The diameters of two planets are in the ratio 4:1 and their mean densities in the ratio 1:2. The acceleration due to gravity on the

planets will be in ratio

a. 1:2 b. 2:1 c. 2:3 d. 4:1



IITians

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MATHEMATICS

CLASS IX MATHEMATICS FOUNDATION FOR IIT JEE/ RMO/NTSE/OLYMPIADS, ANY PRESENT



Foundation Curriculum by IITians for

Academic Excellence in Schools

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CHAPTER NO 1: 7-19 NUMBERS AND SURDS C 1: Rational numbers as Decimals C 2: Conversion of Decimals to Rationals Introductory Exercise 1 C 3: Representing Rational, Irrational numbers on the Number Line C 4: Important Identities C5: Rationalization& Rationalizing Factor Introductory Exercise 2 C 6: Surds, C7: Types of Surds C 8: Laws of Radicals C 9: Operations on Surds C 10: Comparison of Surds C 11: Square Roots of a Surd Introductory Exercise 3 Ex 1: Single correct objective questions Ex 2: Match the following Ex 3: One or more than one correct Ex 4: Linked Data Questions Ex 5: Assertion and Reason Ex 6: Questions from NTSE & Olympiads CHAPTER NO 2: 20 - 35 POLYNOMIALS & FACTORIZATION C 1: Degree of the Polynomial C 2: Types of Polynomials Introductory Exercise 1 C 3: Algebraic Identities C 4: Factors of a Polynomial Introductory Exercise 2 C 5: Zeroes or Roots of a Polynomial C 6: Value of a Polynomial C 7: Remainder Theorem C 8: Factor Theorem Introductory Exercise 3 C 9: Factorization Introductory Exercise 4 Ex 1: Single correct objective questions Ex 2: Match the following

Ex 3: One or more than one correct Ex 4: Linked Data Questions Ex 5: Assertion and Reason Ex 6: Questions from NTSE & Olympiads CHAPTER NO 3: 36 - 49 COORDINATE GEOMETRY C 1: Coordinates of a point C 2: Cartesian Plane Introductory Exercise 1 C 3: Distance Between Two Points C 4: Section Formula Introductory Exercise 2 C 5: Area of the Triangle and Centroid Introductory Exercise 3 C 6: Oblique lines and Its Equations Ex 1: Single correct objective questions Ex 2: Match the following Ex 3: Linked Data Questions Ex 4: One or more than one correct Ex 5: Assertion and Reason Ex 6: Questions from NTSE & Olympiads CHAPTER NO 4: 50 - 62 LINEAR EQUATIONS IN 2 VARIABLES C 1: Linear Equation in One Variable C 2: Cross-Multiplication Method Introductory Exercise 1 C 3: Graph of a Linear Equation Introductory Exercise 2 C 4: Linear Equations in Two variables C 5: Applications of Linear Equations C 6: Consistent & Inconsistent Systems Ex 1: Single correct objective questions Ex 2: Linked Data Questions Ex 3: Match the following Ex 4: Assertion and Reason Ex 5: One or more than one correct Ex 6: Questions from NTSE & Olympiads

CHAPTER NO 5: 63-79 GEOMETRY (TRIANGLES, QUADRILATERALS AND CIRCLES) C 1: Types of Triangles C 2: Properties of Angles in a Triangle Introductory Exercise 1 C 3: The Comparison Of Two Triangles C 4: Inequalities Of a Triangle Introductory Exercise 2 C 5: Quadrilaterals C 6: Types of Quadrilaterals C 7: Theorems On Quadrilaterals Ex 1: Single correct objective questions Ex 2: Linked Data Questions Ex 3: Match the following Ex 4: One or more than one correct Ex 5: Assertion and Reasoning Ex 6: Questions from NTSE & Olympiads C8: Basics of Circles C9: Theorems on Circles C10: Cyclic Quadrilateral Ex7: Single correct objective questions Ex 8: Match the following Ex 9: Linked Data Questions Ex 10: One or more than one correct Ex 11: Assertion and Reason Ex12:Questions from NTSE & Olympiads CHAPTER NO 6: 80-94 AREAS AND VOLUMES C 1: Area Introductory Exercise 1 C2: Solved examples Ex 1: Single correct objective questions Ex 2: Linked Data Questions Ex 3: Match the following Ex 4: One or more than one correct Ex 5: Assertion and Reason Ex 6: Questions from NTSE & Olympiads CHAPTER NO 7: 95-103 PROBABILITY C 1: Probability

C 2: Probability of rolling a Dice Introductory Exercise 1 C 3: Probability of Tossing C 4: Probability of Drawing a Card Ex 1: Single correct objective questions Ex 2: Linked Data Questions Ex 3: Match the following Ex 4: One or more than one correct Ex 5: Assertion and Reasoning Ex 6: Questions from NTSE & Olympiads CHAPTER NO 8: 104 - 115 LOGARITHMS C 1: Types of Logarithms C 2: Laws of Logarithms Introductory Exercise 1 C 3: Change of Base C 4: Logarithm Equations Introductory Exercise 2 C 5: Characteristic and Mantissa Ex 1: Single correct objective questions Ex 2: Linked Data Questions Ex 3: Match the following Ex 4: One or more than one correct Ex 5: Assertion and Reason Ex 6: Questions from NTSE & Olympiads CHAPTER NO 9: 116 - 224 TRIGONOMETRY C 1: Measurement Of Angles Introductory Exercise 1 C 2: Trigonometric Ratios C 3: Signs of Trigonometric Functions Introductory Exercise 2 C 4: Trigonometric Ratios Introductory Exercise 3 C 5: Trigonometric Identities Ex 1: Single correct objective questions Ex 2: Match the following Ex 3: Linked Data Questions Ex 4: One or more than one correct Ex 5: Assertion and Reasoning Ex 6: Questions from NTSE & Olympiads

Propel Academy

CHAPTER NO 1 NUMBERS AND

SURDS

Mathematics


Numbers that we use in normal day-to-day activities are called Real numbers. The

number of the formp

q, where p and q are integer and q 0 and

p

qis in the lowest

form i.e., p and q have no common factor, is called rational number i.e., all integers

and fraction are rational number. The set of rational numbers is denoted by Q.

Concept 1 Representation of Rational numbers as Decimals

1.1Rational numbers can be represented in the form of decimals. By actual division

We find that1

0.52 ,

20.4

5

30.75

4 ,

111.375

8 ,

10.625

16 .

Let us consider the rational numbers,

1 5 30.3333..... 0.625, 0.428571

3 8 7 .

1.2 Terminating Decimals: When the decimals end with the remainder zero, they

are called as terminating decimals.

Example:5

4 and

1

2

1.3Non – Terminating Decimals: when the decimals end with non zero

remainder, then they are called as non – terminating decimals.

Example:1

3 and

3

7

Concept 2 Conversion of Decimal numbers into Rational Numbers

2.1A rational number can be represented either as either as a termination decimal

or a non-terminating repeating decimal.

Conversion of a Terminating Decimal number of the form p/q.

Step I :Obtain the rational number.

Step II :Determine the number of digits in its decimal part.

Step III :Remove decimal point form the numerator. Write 1 in the

denominator and put as many zeros on the night side of 1 as the

number of digits in the decimal part of the given rational number.

Step IV :Final a common division of the numerator and denominator and

express the rational number and denominator by common divisor.

Introduction

IITians Propel Academy > C1: Number System Class IX


Examples: i. 0.15 = 15

100=

15 5

100 5

=

3

20

ii. 0.675 = 675 675 675 675 25 27

1000 1000 1000 1000 25 40

iii. 15.75 = 1575 1575 25 63

100 100 25 4

iv. –25.6875 = 256875 256875 625 411

10000 1000 625 16

2.2 Irrational Numbers: We have seen that every rational numbers can be but in form of terminating or non-terminating but receiving decimal. On other hand,the

number which cannot be put in form of p

qand neither these are “terminating not

recurring” are known as “Irrational Number”.

Examples: 2, 3,.................etc

IIT JEE MAINS/OLYMPIADS TYPE Introductory Exercise 1

1. If x= 2.16 , then the rational

form of x will be

a. 195

90 b.

198

90 c.

105

95 d.

195

95

2. If x= 0.317 , then the rational

form of x will be

a. 295

900 b.

314

990 c.

314

900 d.

295

990

3. 1.75 in rational form will be

a. 3/4 b. 1/7 c. 7/4 d. 2/5

4. √5 in rational form will be

a. 1/8 b. 2/7 c. 3/5 d. None

5. 3

0in decimal form is

a. 0 b. 3 c. 1 d. Undefined

6. What number should be

subtracted from -1

3 to get

2

6?

a. 2

3 b.

1

3

c.

1

3 d.

2

3

Concept 3 Representing Rational, Irrational numbers on the Number Line

We know that all the number (either rational or irrational) can be represented on

number line.

Example: i. Visualize the position of 5.665 on the number line, through, successive

magnification

Class IX Math’s Foundation > IITians Propel Academy


ii. Represent 3 on the real number line

Draw base OA = 1 unit of length, braw AB OA and cut off AB = 1 unit join OB.

Then 2 2 2 2OB OA AB 1 1 2 . Draw BD OB and cut off BD = 1 unit.

Then OD = 2

2 2OD OB BD 2 1 2 1 3 . Cut off OP = OD = 3

P represents the real number (irrational) 3

Concept 4 Important Identities

Let x and y be positive real numbers, then

2

x y x 2 xy y

2

x y x 2 xy y

x y x y x y

2x y x y x y

x x

y y

xy x. y

x y a b ax ay bx by



Concept 5 Rationalization and Rationalizing Factor

Changing an irrational number into a rational number is called “rationalization” and

the factor by number is called “rationalizing factor”

Rationalizing Factor (R.F) of

a. x is x

b. 2 1

3 3x is x

c. x y is x y

d. x y is x y

e. 1 1

isx y x y

f.

1 1is

a b x a b x

Example: Rationalize 1

4 2 3

1 1 4 2 3

4 2 3 4 2 3 4 2 3

=

4 2 3 4 2 3

16 4 3 4

Note:1.R.F. of a surd is not unique 2. A surd may have any number of R.Fs


1. The Rationalizing factor of the

denominator of 3 3 2

2 5 3 2

is

a. 2 5 3 2 b. 3 3 2

c. 2 5 3 2 d. 3 3 2

2. The Rationalizing factor of the

denominator of 5 3 - 2

5 is

a. 5 3 2 b. 5 3 2

c. 5 d. 5 5 2

3. If If x = 2 + 3 then 1

xx

is

a. 2 b. 4 c. 6 d. 8

4. Rationalizing 1 2

2 3 3 2

gives

a. 3√2-√3 b. 3√2+√3

c. -3√2 –√3 d. -3√2+√3

5. The rationalizing factor of

3 31

2 -2

is

a. 3 23

2

12 1

2 b. 3 3 3

3

12 1

2

c. 3 3 33

12 1

2 d. 3 3 3

3

12 1

2

6. If √x –√12=√4 –√x, then the

value of x will be

a. 4 –√3 b. 4+√3

c. 4+2√3 d. 4 –2√3



Let ‘a’ be a positive rational number and n be a positive integer. If ‘a’ cannot be

expressed as the nth power of some rational number then

is called a Surd (or) Radical.

Example: Some of the surds are3 749, 10, 253..........

Concept 7 Types of Surds

1. Quadratic surd: If the order of a surd is 2, then the surd is called

Quadratic surd.

Example:

2. Cubic Surd: If the order of a surd is 3 then the surd is called cubic surd.

Example:

3. Biquadratic surd: If the order of a surd is 4 then the surd is called

biquadratic surd.

Example:

4. Simple surd: A surd having only one term is called simple surd.

Example:

5. Pure surd (Entire surd): If the rational co – efficient of a surd is 1 then the

surd is called pure surd.(or)If a surd is written entirely inside the radical sign,

then it is called entire surd.

Examples: are pure surds while are not pure

surds as their rational co - efficient are not 1.

6. Mixed Surd: If a is a rational number and is a surd, then and

are called mixed surds.

Concept 6 Surds

n a

2, 3, 8, 20.... are quadratic surds

3 33 32, 3, 4, 9.... are cubic surds

4 4 4 42, 9, 10, 200.... biquadratic are surds

343 5, 6,8 7

54 8, 10, 11 35 3, 2 6, 4 9

n b na bna b



Example:

7. Compound Surd: The sum or difference of two or more dissimilar surds is

called a compound surd.

Example:

8. Binomial Surd: A compound surd consisting of two terms is called a

binomial surd.

Example:

9. Trinomial Surd: A compound surd consisting of three terms is called a

trinomial surd.

Example:3 +

10. Multi nominal surds: A compound surd consisting of more than three terms

is called a multinomial surd.

Example:5 –

11. Like Surds / Similar Surd’s: Two or more surds are said similar if they are

rational multiples of a same surd (or)If the quotient of two surds is a rational

number, then they are said to be like surds.

Example:

12. Unlike surds / dissimilar surds: Surds which are not like are unlike surds.

Example:

Note: i. Every pure surd is a simple surd but the converse need not be true.

Example: are pure surds,

are simple surds

but is a simple surd, but not pure surd.

ii. Every simple surd can be converted into pure surd.

Example:444 42 3 2 3 48x

52 3,3 8,6 2

3 4 46 5,3 8 6, 8 3 5 4 7

35 2, 5 10

345 2, 5 9 8

533 2 5 6,4 3 2 7 8

4 72,2 2,3 2, 2, 2

5 6

5 74 5,3 2, 6, 3

2 5

2, 3

2, 3

3 2



Concept 8 Laws of Radicals 1

.

.

.

.

.

n n

n n n

n

nn

n mnx mx

m nn m mn

i a a

ii ab a b

a aiii

b b

iv a a x R

v a a a

Concept 9 Operations on Surds

9.1 Addition: Two or more surds can be added when they are similar.

Examples: i . The sum of is

ii. The sum of is

9.2 Subtraction: Subtraction among surds is possible when they are similar.

Example: Subtracting gives , = – √3

9.3 Multiplication: Surds of same order can be multiplied.

Example: i .

Since the order is same for both surds, multiplication is possible.

ii.

Since the order is not same, first make the order same by taking their LCM

LCM of 2, 3 = 6

66 3 235 5 and 2 2 6 23 6 65. 2 and 2 125 4 580

9.4 Division: Division is possible between two surds when they are of same order.

Example:i.

Since the order is same, division is possible.

6 3 23

22

ii.

Since the order is not same, first make the order same by taking their LCM

LCM of orders 3, 2, = 6

3 5 7

2 2 2 23 4 3 5 3 10 3

5 3 4 3 2 3 5 3,4 3,2 3

3 4 5 1 4 3 5 4 15

35 2

6 2

3 2 5

6 23

6

6 3

2 2 4

1255 5

3, 5, 7

2 2 23 4 3,5 3



Concept 10 Comparison of Surds and Conjugate Surds

10.1 Comparison of surds is possible only when they are of same order. If so, then

compare the radicands of similar surds.

Example : Arrange in ascending order

Given surds can be written as,

Since 400 < 512 < 729

The ascending order is

10.2 Conjugate Surd: If the sum and product of two surds are rational numbers,

then each is called the conjugate of the other.

Example : Prove that two surds of the form a + , a – are conjugate to each

other?

Sol: Sol: Rational number

Rational number

The sum and product of are rational

Hence are conjugate surds

Note: Every conjugate surd of a surd is the R.F of the surd. But converse need not

be true.

an – bn = (a – b) (an – 1 + an – 2 . b + an – 3 . b2 + ----- + a. bn-2 + bn – 1 )

The R.F of

Concept 11 Square Roots of a Surd

11.1. a b [Positive square root]

Let a b = x y where x > y 2a b x y xy

By comparing corresponding rational and irrational parts on both sides, we get

x + y = a and 2 xy b

------------------ (1)

Now consider (x – y)2 = (x + y)2 – 4xy

= a2 – b2x y a b ------------------ (2)

From 1 & 2, 2x = a + 2a b

2 2

,2 2

a a b a a bx y

643, 8, 20

6 3 212 12 12 12 12 123 , 8 , 20 729, 512, 400

6 420, 8, 3

b b

2a b a b a

,a b a b

,a b a b

2a b a b a b

1 11 2 1 .n n n n nn n nna b is a a b b



11.2. 2 2 2a b c d [Positive square root]

Let 2 2 2a b c d = x y z

2 2 2 2 2 2a b c d x y z xy yz zx

By comparing corresponding rational and irrational parts on both sides we

get

x + y +z = a, xy = b , yz = c, zx = d

Consider x2 = .xy zx bd bd

xyz c c

Consider y2 = . .xy yz b c bc

yxz d d

Consider z2 = . .yz zx c d cd

zxy b b

Note: If [ ]a b x y then a b x y x y

2 2

,2 2

a a b a a bx y

1. Rationalize the numerator of

2 - 3 + x

x -1

a. 2 3

1

x

x

b.

1

2 3 x

c. 1

2 3 x

d.

1

2 3 x

2.If3 + 5

= x + y 53 - 5

, then the

value of x+y is

a. 7/2 b. 3/2 c. 5 d. 2

3. 3 15625 in the simplest form is

a.5 b. 25 c. 125 d. None

4.If a=√5+√6, then the value of

2

2

1a + - 2

a

a. 22 b. 20 c. 24 d. 26

5. If x= 7 –4√3, then 2

x +1

7xis

a. 0 b. 1 c. 2 d. 3

6. If x=√17 – √16, y=√16 –√15

then

a. x>y b. x<y c. x=y d. None




IIT JEE/NTSE Exercise 1 Single correct objective questions

1. The square root of 7 3 2 30 is

a. 4 3 5 2 b. 4 3 5 2

c. 4 3 5 2 d. 4 3 5 2

2. The value of

1 1 1 is

a. 1 5

2

b.

1 5

2

c. 1 5

2

d.

1 5

2

3. If the values of

6 5, 11 10, 22 21x y z ,

then

a. x=y=z b. x>y=z

c. x<y<z d. x<y>z

4. 11 2 7 2 3 84 =

a. 7 3 1 b. 7 3 1

c. 7 3 1 d. 7 3 1

5. If 20 300 a b then the

value of a b

a b

is

a. 2 b. 3 c. 7 d. 11

6.If 7 2 10 a b , then the

value of a + b is

a. 7 b. 5 c. 2 d. 4

7. The value of 4 22 8

5 is

a.24

5 b.

24 5

5 c.

23

5 d.

23 5

5

8. If x = 4 – 3 and y = 4 + 3 ,

then 3(xy)2 – 4xy is

a. 420 b. 435 c. 455 d. 470

9. The positive square root of

10+2√10+2√6+√60 is

a. √2+√3+√5 b. √2 –√3+√5

c.√2 –√3 –√5 d. –√2+√3+√5

10. 0.34 0.34 as a single decimal

will be

a. 0.687 b. 0.6878

c. 0.689 d. 0.6788

11. The value of 5 6

3 2

is

a. 8 2 7 3 b. 7 3 8 2

c.8 2 7 3 d. 8 2 7 3

12. The value of

3 2 7 3 3 7 is

a. 5 5 21 b. 5 21 5

c. 5 5 21 d. 5 5 21



I. Column A Column B II. Column A Column B

1.10 3

5 a.

6 15 9

17

1. 3√7 a. Pure surd

2. 20 5 20 5 b.7

6

2. 5 15 b. Simple surd

3. 174 27

4 2

c. 15 3. 43 8 c. Cubic surd

4. 3 3

2 5 3

d. 2 15 4. 3 11 d. Compound surd

IIT JEE ADVANCED TYPE Ex 3 One or more than one correct option

1. If 2 28 8

7 4 3 7 4 3 14a a

then the value of a is a. ±3 b. ±7 c. ±5 d. ±9

2. The root of 91+52 3 is

a. (2+√3)√13 b. (2 –√3)√13 c. (4 –√3)√13 d. (4+√3)√13

3. If 1 2a then the value of

a + 2 a -1 - a - 2 a -1 is

a. 11 + 11 b. 2 c. 21+21 d. 4 4. Factors of x in the expression

2013= x x x is

a. 2012 b. 2013 c. 2014 d. 2015 5. Which of the following is not the value of the expression

1 3 4

11 2 30 7 2 10 8 4 3

is

a. –1 b. 0 c. 1 d. 2

6. If 4 3 3

7 4 3x y

then the

product of x and y will be a. 12 b. –12 c. 1 x 12 d. –1 x 12

7. If x = 2+ 3 , then 1

xx

is

a. 2 b. 4 c. 6 d. 8

8. π is not a

a. Rational number b. Irrational number c. Transcendental number d. Integer number 9. The square root of

21 4 5 8 3 4 15 is

a. –√5+√4+√12 b. √5 –√4 –√12 c. √5+√4+√12 d. –√5 –√4 –√12

10. If a = 1- 2 then the value of

31

a -a

will be

a. 2x12 –4o +3o b. 8

c. 2 d. 1o+24-31x31

IIT JEE MAINS/OLYMPIADS Exercise 2 Match the following



I. If the value of x=1 2 3

1 2 3

, II. If x= 23+a 10 = 18 + 5

1. The product of x and inverse of 1. The value of a is

x will be

a. 7

21 b.

21

7 c. 1 d. 2 a. 6 b.5 c. 1 d. 2

2. The value of

2

2

2

1 1x+ - x +

x xis 2. The value of

2a -1

ais

a.228

121 b. 1 c.

301

434 d. 2 a.

30

36 b.

35

36 c.

30

6 d.

35

6

a. Assertion is True; Reason is True; Reason is a correct explanation for Assertion.

b. Assertion is True; Reason is True; Reason is not a correct explanation for Assertion.

c. Assertion is True, Reason is False.

d. Assertion is False, Reason is True.

1. Assertion: If 1 1x x x , then x= 16

25

Reason: x value can be found by squaring twice the given equation.

2. Assertion: If x= 1/3 2/33 3 3 , then x3 –9x2 +18x –12 = 0

Reason: A surd of order 2 is called a quadratic surd.

3. Assertion: The fourth root of 128 – 32√15 is √5+√3

Reason: The square root of 8+2√15 is √3+√5

4. Assertion: The value 1 1 1

.....................3 3 3

of is 1

3

Reason: The value of 1 1 1

.....................3 3 3 is

1

6

5. Assertion: 5√3 and 7√3 are dissimilar surds

Reason: The Rationalizing factor of 1 1

3 37 7

is2/3 2/37 1 7

IIT JEE ADVANCED TYPE Exercise 4 Linked Data Questions

IIT JEE ADVANCED TYPE Exercise 5 Assertion and Reason



Exercise 6 Questions from NTSE, Olympiads and Competitive Exams

1. If2

x =3 + 7

, then the value of

(x –3)2 is

a. 6 b. 7 c. 8 d. 9

2. If2 + 3

x =2 - 3

, then the value of

x2(x –4)2 is

a. 0 b. –1 c. 1 d. 2

3. If the values of

x = 7 - 5,y = 5 - 3,z = 3 - 7 ,

then the value of x3 + y3 +z3 –

2xyz is

a. 0 b. xyz c. 1 d. –xyz

4. The value of 12- 68+ 48 2 is

a. –2 –√2 b. 2+√2

c. 2√2 d. 2 –√2

5. The value of

2 + 5 - 6 - 3 5 + 14 - 6 5 is

a. 1 b. 2 c. –1 d. –2

6. If

1 1x = 7 +

2 7then

2

2

x -1

x - x -1

will be

a. 2 b. 3 c. 4 d. 5

7. The value of the expression

36 15- 2 56 7 + 2 2 is

a. 0 b. 1 c. 2 d. 3

7. If x = 3 + 2 2 then the value of

1x -

x is

a.3 3 b.3 3 c.1 d.2

8. The value of x in 3 4x - 5 - 5 = 0

is

a. 31.3 b. 32.5

c.33.7 d. 34.9

9. If a + a - 1 - a = 1 then value

of a will be

a. 15

16 b.

16

23 c.

15

23 d.

16

25

10. If

1 2

3 3k = 3 + 3 + 3 then k3 -9k2

+18k will be

a. 11 b. -11 c. -12 d. 12

11. If 1 1

3 3m = 4 - 15 + 4 + 15

then m3-3m-8 will be

a. 0 b. -1 c. 2 d. 3

Propel Academy

CHAPTER NO 2 POLYNOMIALS AND

FACTORIZATION Mathematics


An algebraic expression f(x) of the form f(x) = a0 + a1x + a2x2 + ........+ anxn,

where a0,a1,a2 .......,an are real numbers and all the index of ‘x’ are non-negative

integers is called a polynomials in x.

Concept 1 Degree of the Polynomial

Highest Index of x in a term of a polynomial is called the degree of the polynomial,

here a0, a1x, a2x2..... anxn are called the terms of the polynomial and a0, a1,

a2......, an are called coefficients of the polynomial f(x).

A polynomial in x is said to be in standard form when the terms are written either in increasing order or decreasing order of the indices of x in various terms.

Concept 2 Types of Polynomials

Generally, we divide the polynomials based on degree and number of terms as

shown below.

2.1 Based on degree: There are four types of polynomials based on degree.

i. Linear Polynomials: A polynomials of degree one is called a linear

polynomial. The general form of a linear polynomial is ax + b, where

a and b are any real numbers and a 0.

ii. Quadratic Polynomials: A polynomial of degree two is called a quadratic

polynomial. The general form of a quadratic polynomial is ax2 + b + c,

where a 0.

iii. Cubic Polynomials: A polynomial of degree three is called a cubic

polynomial. The general form of a cubic polynomial is ax3 + bx2 + cx + d,

where a 0.

iv. Bi-quadratic Polynomials: A polynomial of degree four is called a biquadratic

(quadratic) polynomial. The general form of a biquadratic polynomial is

ax4 + bx3 + cx2 + dx + e , where a 0.

Introduction

Class IX Maths Foundation > IITians Propel Academy


2.2 Based on number of terms: There are three types of polynomials based on

number of terms. They are

i. Monomial: A polynomial is said to be monomial if it has only one term.

Example: x, 9x2, 5x3 all are monomials.

ii. Binomial: A polynomial is said to be binomial if it contains only two terms

Example: 2x2 + 3x, 3 x + 5x3, -8x3 + 3, all are binomials.

iii. Trinomial: A polynomial is said to be a trinomial it if contains exactly three

terms.

Example: 3x3 - 8 + ,2

5107x 8x4 - 3x2, 5 - 7x + 8x9, are all trinomials.

Note:

i.A polynomial having four or more than four terms does not have particular Name.

These are simply called as polynomials.

ii. A polynomial of degree five or more than five does not have any particular name.

Such a polynomial usually called a polynomial of degree five or six or ....etc.

1. The given expression x2 –5x

+3x4+9 is a

a. Monomial b. Binomial

c. Trinomial d. Polynomial

2. The degree of the polynomial

4x – x2 + 6x3 –x5 +3 is

a. 5 b. –5 c. 1 d. 4

3. The degree of the polynomial

4x2 is

a. 1 b. 2 c. 3 d. 4

4. We divide the polynomials

based on

a. Number of terms b. Degree

c. Neither a nor b d. Both a and b

5. The general form of a cubic

polynomial is px3 + qx2 + rx + s

when

a. p 0 b. p = 0 c. q 0 d. p =

0

6. The general form of linear

polynomial is

a. ax+b b. ax2+bx+c

c. ax2+bx d. All of these


IITians Propel Academy > C2: Polynomials Class IX


Concept 3 Algebraic Identities

An identity is an equality which is true for all values of the variables with given

specifications

Some important identities are

(i) (a + b)2 = a2 + 2ab + b2

(ii) (a - b)2 = a2 - 2ab + b2

(iii) a2 - b2 = (a + b) (a - b)

(iv) a3 + b3 = (a + b) (a2 - ab + b2)

(v) a3 - b3 = (a - b) (a2 + ab + b2)

(vi) (a + b)3 = a3 + b3 + 3ab (a + b)

(vii) (a - b)3 = a3 - b3 – 3ab (a - b)

(viii) a4 + a2b2 + b4 = (a2 + ab + b2) (a2 – ab + b2)

(ix) a3 + b3 + c3 –3abc = (a + b + c) (a2 + b2 + c2 –ab – bc –ac)

Special case: If a + b + c = 0 then a3 + b3 + c3 = 3abc.

(a) Value Form:

(i) a2 + b2 = (a + b)2 - 2ab, if a + b and ab are given

(ii) a2 + b2 = (a - b)2 + 2ab if a - b and ab are given

(iii) a + b = ab4ba 2 if a - b and ab are given

(iv) a -b = ab4ba 2 if a + b and ab are given

(v) a2 + 2a

1a

a

12

2

if a +

a

1 is given

(vi) a2 + 2a

1a

a

12

2

if a -

a

1 is given

(vii) a3 + b3 = (a + b)3 - 3ab(a + b) if (a + b) and ab are given

(viii) a3 - b3 = (a - b)3 + 3ab(a - b) if (a - b) and ab are given

(ix)

a

1a3

a

1a

a

1x

3

33

ifa

1a is given

(x)

3

3

3

1 1 13 ,a a a

a a a

if

a

1a is given

(xi) a4+b4 = (a2+b2)2 -2a2b2 = [(a + b)2 - 2ab]2 - 2a2b2, (a + b),ab are given

(xii) a4 - b4 = (a2 + b2) (a2 - b2) = [(a + b)2 -2ab](a + b) (a - b)

(xiii) a5 + b5 = (a3 + b3) (a2 + b2) - a2b2 (a + b)



Concept 4 Factors of a Polynomial

If a polynomial f(x) can be written as a product of two or more polynomials f1(x),

f2(x), f3(x),..... then each of the polynomials f1(x), f2(x),..... is called a factor of

polynomial f(x). The method of finding the factors of a polynomial is called

factorization.

4.1 Factorization by Making a Perfect Square:

Example 1: 81a2b2c2 + 64a6b2 - 144a4b2c

Sol. 81a2b2c2 + 64a6b2 –144a4b2c= [9abc]2 –2 [9abc][8a3b] + [8a3b]2

= [9abc – 8a3b]2

=a2b2[9c – 8a2]2

Example 2:

3

b

1a3a2

b

1c9

b

1a3

b

1a3

2

Sol.

3

b

1a3a2

b

1c9

b

1a36

b

1a3

2

3

b

1a3a2

b

1c)3(

b

1a33.2

b

1a3 2

2

3

b

1a3a2

b

1c3

b

1a3

2

a2

b

13

b

1a33

b

1a3 ]3ca[3

b

1a3

4.2 Factorization by Using the Formula for the Difference of Two Squares:

a2 - b2 = (a + b) (a - b):

Example : Factorize 4(2a + 3b - 4c)2 - (a - 4b + 5c).2

Sol. = 4(2a + 3b - 4c)2 - (a - 4b + 5c)2

= [2(2a + 3b - 4c)]2 - (a - 4b + 5c)2

= [4a + 6b - 8c + a - 4b + 5c] [4a + 6b - 8c - a + 4b - 5c]

= [5a + 2b - 3c] [3a + 10b - 13c]

4.3 Factorization by Using Formula of a3 + b3 and a3 - b3:

Example 1: Factorize 64a13b + 343ab13.

Sol. 64a13b + 343ab13 = ab[64a12 + 343b12] = ab[(4a4)3 + (7b4)3]



= ab[4a4 + 7b4] [(4a4)2 - (4a4) (7b4) + (7b4)2]

= ab[4a4 + 7b4][16a8 - 28a4b4 + 49b8]

Example 2: Factorize p3q2x4 + 3p2qx3 + 3px2 + q

x - q2r3x

Sol. p3q2x4 + 3p2qx3 + 3px2 + q

x - q2r3x

= q

x[p3q3x3 + 3p2q2x2 + 3pqx + 1 - q3r3]

= q

x[(pqx)3 + 3(pqx)2 .1 + 3pqx . (1)2 + (1)3 - q3r3]

Let pqx = A & 1 = B

= q

x [A3 + 3A2B + 3AB2 + B3 - q3r3]

= q

x[(pqx + 1)3 - (qr)3] =

q

x[pqx + 1 - qr][(pqx + 1)2 + (pqx + 1) qr + (qr)2]

= q

x[pqx + 1 - qr][p2q2x2 + 1 + 2pqx + pq2xr + qr + q2r2]

1. The factors of the polynomial

2 2

2

14 9 2

4x y

x is

a. 1 1

2 3 2 32 2

x y x yx x

b.1 1

2 3 2 32 2

x y x yx x

c. 1 1

2 3 2 32 2

x y x yx x

d. 1 1

2 3 2 32 2

x y x yx x

2. On Factorization of x3 - 6x2 +

32, the factors are

a.(x + 2) (x - 4)2

b. (x - 2) (x - 4)2

c. (x + 2) (x + 4)2

d. (2-x) (x - 4)2

3. The factors of the expression

4

4

13a

a is

a. 2 2

2 2

1 11 1a a

a a

b. 2 2

2 2

1 11 1a a

a a

c. 2 2

2 2

1 11 1a a

a a

d. 2 2

2 2

1 11 1a a

a a

4. (x2+y2+xy) (x2+y2- xy) gives

a. x4 – x2y2 – y4

b. x4 + x2y2 – y4

c. x4 + x2y2 + y4

d. x4 – x2y2 + y4




Concept 5 Zeroes or Roots of a Polynomial

A real number is a root or zero of polynomial

f(x) = ,axa......xaxaxa 012n

2n1n

1nn

n

where a0,a1,----an are

real and an is not zero if f )( = 0. i.e

.0aa......aaa 012n

2n1nn

n

Example: x=3 is root of the polynomial f(x) = x3 - 6x2 + 11x - 6, because

Sol.f(3) = (3)3 - 6(3)2 + 11(3) - = 27 - 54 + 33 - 6 = 0.

but x = -2 is not a root of the above polynomial,

f(-2) = (-2).3 - 6(-2)2 + 11(-2) - 6

f(-2) = -8 -24 -22 – 6 = –60 0

Concept 6 Value of a Polynomial

The value of a polynomial f(x) at x= is obtained by substituting x = in the

given polynomial and is denoted by f )( .

Example 1:Show that x = 2 is a root of 2x3 + x2 - 7x - 6.

Sol. p(x) = 2x3 + x2 - 7x - 6 then,

p(2) = 2(2)3 + (2)2 7(2) - 6 = 16 + 4 - 14 - 6 = 0

Hence x = 2 is a root of p(x).

Example 2: If f(x) = 2x3 - 13x2 + 17x + 12 then its value at x = 1 is.

Sol. f(1) = 2(1)3 - 13(1)2 + 17(1) + 12

= 2 - 13 + 17 + 12 = 18.

Concept 7 Remainder Theorem

Let ‘p(x)’ be any polynomial of degree greater than or equal to one and a be any

real number and If p(x) is divided by (x - a), then the remainder is equal to p(a).

Let q(x) be the quotient and r(x) be the remainder when p(x) is divided by (x - a)

then

Dividend = Divisor × Quotient + Remainder

p(x) = (x - a) × q(x) + r(x) , where r(x) = 0 or degree of r(x) < degree of (x - 1 ).

But (x - a) is a polynomial of degree 1 and a polynomial of degree less than 1 is a

constant. Therefore, either r(x) = 0 or r(x) = Constant.



Let r(x) = r, then p(x) = (x - a)q(x) + r,putting x = a in above equation

p(a) = (a - a)q(a) + r = 0. q(a) + r

p(a) = 0 + r = r

This shows that the remainder is p(a) when p(x) is divided by (x - a).

REMARK: If a polynomial p(x) is divided by (x + a), (ax - b), (x + b), (b - ax) then

the remainder in the value of p(x) at x = a

b,a

b,a

b,a i.e. p(-a),

a

bp,

a

bp,

a

bp

respectively.

Example 1: Find the remainder when f(x) = x3 - 6x2 + 2x - 4 is divided by

g(x) = 1- 2x.

Sol. 1 - 2x = 0 2x = 1 x = 2

1

42

12

2

16

2

1

2

1f

23

412

3

8

1

8

35

8

328121

Example 2: The polynomials ax3 + 3x2 - 13 and 2x3 - 5x + a are divided by x + 2

if the remainder in each case is the same, find the value of a.

Sol. p(x) = ax3 + 3x2 - 13 and q(x) = 2x3 - 5x + a

when p(x) & q(x) are divided by x + 2 = 0 then x = -2

p (-2) = q(-2)

a(-2)3 + 3(-2)2 - 13 = 2 (-2)3 - 5(-2) + a

-8a + 12 - 13 = - 16 + 10 + a

-9a = - 59

5a

Concept 8 Factor Theorem

Let p(x) be a polynomial of degree greater than or equal to 1 and ‘a’ be a real

number such that p(a) = 0, than (x - a) is a factor of p(x). Conversely, if (x - a) is a

factor of p(x), then p(a) = 0.

Example 1: Show that x + 1 an d 2x - 3 are factors of 2x3 - 9x2 + x + 12.

Sol. To prove that (x + 1) and (2x - 3) are factors of 2x3 - 9x2 + x + 12 ,it is

sufficient to show that p(-1) and

2

3p both are equal to zero.



p(-1) = 2(-1)3 - 9(-1)2 + (-1) + 12 = - 2 - 9 - 1 + 12 = - 12 + 12 = 0

And, 122

3

2

39

2

32

2

3p

3

04

8181

4

486812712

2

3

4

81

4

27

Hence, (x + 1) and (2x – 3) are the factors 2x3 - 9x2 + x + 12.

Example 2: Find , if x + 1 and x + 2 are factors of p(x) =x3 + 3x2 - x2

Sol. When we put x + 1 =0 or x =-1 and x+2 =0 or x=-2 in p(x)

Then, p(-1) = 0 & p(-2) = 0

Therefore, p(-1) = (-1)3 + 3(-1)2 - 2 (-1) + = 0

- 1 + 3 + 2 + = 0 = - 2 - 2 ....(i)

And, p(-2) = (-2)3 + 3(-2)2 - 2 (-2) + = 0

- 8 + 12 + 4 + = 0 = - 4 - 4 .....(ii)

From equation (i) and (ii)

- 2 - 2 = - 4 - 4

2 = - 2 = - 1 Put = -1 in equation (i) = -2(-1) - 2 = 2 - 2 = 0.

Hence, = - 1 = 0.

Example 3: What must be added to 3x3 + x2 - 22x + 9 so that the result is exactly

divisible by 3x2 + 7x - 6.

Sol. Let p(x) = 3x3 + x2 - 22x + 9 and q(x) = 3x2 + 7x - 6.

We know if p(x) is divided by q(x) which is quadratic polynomial therefore if

p(x) is not exactly divisible by q(x) then the remainder be r(x) and degree of

r(x) is less than q(x) (or Divisor)

By long division method

2 3 2

3 2

2

2

2

3 7 6 3 22 9

3 7 6

( ) ( ) ( )

_______________

6 16 9

6 14 12

( ) ( ) ( )

________________

2 3

x

x x x x x

x x x

x x

x x

x

Hence, -2x -3 must be added to the given polynomial so that the result is exactly

divisible by 3x2 + 7x – 6



1. If x = 4/3 is a root of the

polynomial f(x) = 6x3-11x2 +kx -

20 then the value of k is

a. 19 b. –19 c. 20 d. –20

2. If x =2, x =0 are two roots of

the polynomial f(x) = 2x3 - 5x2 +

ax +b. then the values of a and b

are

a. 0, 2 b. 2, 0 c. 1, 1 d. 3, –1

3. The factors of the polynomial

9x2 –3x –2 are

a.(3x+1)(3x –2) b.(2x+3)(2x-3)

c. (x+1)(x-2) d. (3x–1)(3x+2)

4. If x –3 is a factor of x2 –7x+12,

then the remainder is

a. 0 b. 1 c. 2 d. 3

5. If x2 –x –72 is exactly divided

by x –9, then the exact factor will

be

a. x+8 b. x+6

c. x –9 d. x+9

6. The remainder obtained when

x3 +1 is divided by x+1 is

a. x2 +x +1 b. x2 –x –1

c. 0 d. 1

7. If (2x+k)3 + (x+1)7 is exactly

divisible by x+2, then the value of

k is

a. 4 b. 5 c. 6 d. 2

8. If x –2 is a factor of x2 –6x +8

then the other factor is

a. 4 –x b. x+4

c. x –4 d. –x –4

Concept 9 Factorization of a Quadratic polynomial

For factorization of a quadratic expression ax2 + bx + a where a 0

9.1 By Method of Completion of Square: In the form ax2 + bx + c where a

0, firstly we take ‘a’ common in the whole expression then factorize by converting

the expression

a

cx

a

bxa 2

as the difference of two squares.

Example 1: Factorize x2 - 31x + 220.

Sol. x2 - 31a + 220

= 2202

31

2

31x.

2

31.2x

222




= 4

81

2

31x220

4

961

2

31x

22

2

9

2

31x

2

9

2

31x

2

9

2

31x

2

= (x - 11)(x - 20)

Example 2: Factorize - 10x2 + 31x - 24

Sol. -10x2 + 31x - 24

= -[10x2 - 31x + 24] = - 10

10

24

10

31x2

10

24

20

31

20

31x.

20

31.2x10

222

400

1

20

31x10

10

24

400

961

20

31x10

22

20

1

20

31x

20

1

20

31x10

20

1

20

31x10

22

)8x5)(x23()8x5)(3x2(5

8x5

2

3x210

9.2 By Splitting the Middle Term: In the quadratic expression ax2 + bx + c,

where a is the coefficient of x2, b is the coefficient of x and c is the constant term.

In the quadratic expression of the form x2 + bx + c, a = 1 is the multiple of x2 and

another terms are the same as above.

Example 1: Factorize 2x2 + 12 x2 + 35.

Sol. 2x2 + 12 x2 + 35

Product ac = 70 & b = 12 2

Split the middle term as 25&27

2x2 + 12 x2 + 35 = 2x2 + 7 x25x2 + 35

= 7x257x2x2

= 7x25x2

Example 2: Factorize x2 - 14x + 24.

Sol. Product ac = 24 & b = -14

Split the middle term as - 12 & - 2

x2 - 14x + 24 = x2 - 12 - 2x + 24

x(x - 12) - 2 (x - 12) = (x - 12)(x - 2)



Example 3: Factorize .12

1x

24

13x2

Sol. ]2x13x24[24

1

12

1x

24

13x 22

Product ac = - 48 & b = - 13 We split the middle term as - 16x + 3x.

=24

1[24x2 - 16x + 3x - 2]

=24

1[8x(3x - 2) + 1(3x - 2)]

=24

1(3x - 2)(8x + 1)

Example 4: Factorize 2

35x8x

2

3 2

Sol. )35x5x21x3(2

1)35x16x3(

2

1

2

35x8x

2

3 222

)5x3)(7x(2

1)]7x(x5)7x(x3[

2

1

9.3 Integral Root Theorem: If f(x) is a polynomial with integral coefficient and

the leading coefficient is 1, then any integer root of f(x) is a factor of the constant

term. Thus if f(x) = x3 - 6x2 + 11x - 6 has an Integral root, then it is one of the

factors of 6 which are .6,3,2,1 .

In fact, f(1) = (1)3 - 6(1)2 + 11(1) - 6 = 1 - 6 + 11 - 6 =0

f(2) = (2)3 - 6(2)2 + 11(2) - 6 = 8 - 24 + 22 - 6 = 0

f(3) = (3)3 - 6(3)2 + 11(3) - 6 = 27 - 54 + 33 - 6 = 0

Therefore Integral roots of f(X) are 1, 2, 3.

9.4Rational Root Theorem: Let c

b be a rational fraction in lowest terms. If

c

b is a

root of the polynomial f(x) = 0a,axa....xaxa n011n

1nn

n with integral

coefficients. Then b is a factor of constant term a0 and c is a factor of the leading

coefficient an.

Example: If c

b is a rational root of the polynomial f(x) = 6x3 + 5x2 - 3x - 2, then

the values of b are limited to the factors of -2 which are 2,1 and the value of c

are limited to the factors of 6, which are 6,3,2,1

Hence, the possible rational roots of f(x) are .3

2,6

1,3

1,2

1,2,1 .6,3,2,1

Infact -1 is an Integral root,2

1,3

2 are the rational roots of f(x) = 6x3 + 5x2 - 3x - 2.



NOTE:

(i) An nth degree polynomial can have at most n real roots.

(ii) Finding a zero or root of polynomial f(x) means solving the polynomial equation

f(x) = 0. It follows from the above discussion that if f(x) = ax + b, 0a is a linear

polynomial, then it has only one root given by f(x) = 0 i.e. f(x) = ax + b = 0

ax = - b x = a

b

Thusa

ba is the only root of f(x) = ax + b.

1. If 2x3 + 4x2 + 4x + 2 is divided

by 2x + 2, then the quotient is

a.x2 – x + 1 b. x2 + x + 1

c. x2 + x – 1 d. –x2 – x – 1

2. On factorization x3 +y3+x+y

gives

a. (x –y)(x2 –xy +y2 +1)

b. (x+y)(x2 + xy +y2 +1)

c. (x – y)(x2 –xy –y2 +1)

d. (x+y)(x2 – xy +y2 +1)

3.(1+a-b)p = 1-(a-b)2, then p is

a. 1 a b b. 1 a b

c. 1 a b d. 1 a b

4. If x2 – 4 is a factor of ax4 +

2x3 – 3x2 + bx – 4, then a and b

are

a. 1, –8 b. 3, 5 c. 6, 7 d. –1, 5

5. If ax3 + bx2 + cx + d is exactly

divisible by (x+3), (x+4) and

(x+5), then the polynomial is

a.x3 + 11x2+ 47x + 60

b. x3 + 12x2 + 47x + 60

c. x3 + 12x2 + 48x + 60

d. x3 +12x2 + 47x + 62

6. The remainder when x3 –

7x2a+8xa2+15a3 is divided by

(x+2a) is

a. -35a3b. -37 a3c. 35a3d. 37a3

7. If f(x)=x3– 6x2+ 5x + 8, then

f(x+2) is

a. x3 + 7x + 2 b. x3 + 7x –2

c. x3 – 7x + 2 d. x3 – 7x – 2

8. The polynomial 2

11a -12 2a + 2on factorization

gives

a. 2 11 2a a

b. a+ 2 11a+ 2

c. 2 11 2a a

d. 2 11 2a a




JEE MAINS/Olympiads Exercise 1 Single correct objective questions

1. The sum of the zeroes of the polynomial 4a2 – 2007 is

a. 0 b. –1 c. 2007

4

d.

2007

4

2. If x – 1 is a factor of x3 + 10x2 + px – q, then the condition is a. 11 – p – q = 0 b. 11 + p + q = 0 c. 11 + p – q = 0 d.–11 – p – q = 0

3. If ax4 + bx3 + cx2 + dx + e is a polynomial where a + b + c + d + e = 0, then one of its factor is ] a. x – 1 b. x + 1 c. x – 2 d. x + 2

4. The factors of 2x 6 3x 48 are

a. x 8 x 2 3 b. x 8 3 x 2 3

c. x 8 3 x 2 3 d. x 8 3 x 2

5. The expression whose roots are –1, 0, 1 is a. x3+x2+x b. x3 –x c. x3 +x d. x3 –x2 +x 6. The coefficient of mn in

2 249m ?mn 100n , a perfect

square trinomial will be

a. 70 b. 140 c. 490 d.- 70 7. If 100a2 +20a +25 is divided by 5a –1, then the remainder is a. 31 b. 34 c. 33 d. 30 8. If a, b are zeroes of the

expansion 2 3 2x x then the

values of a and b are a. 1,2 b. 2,-1 c. 1, –2 d. –2, 1 9. If (x+a) be a common zero of x2 +px +q and x2+lx+m, then the value of a will be

a.

m q

l p b.

m q

l p c.

m q

l p d.

m q

l p 10. The degree of the polynomial f(x) is 6, the degree of the polynomial g(x) is 8, then degree of the polynomial f(x) + g(x) is a. 6 b. 8 c. 14 d.48

IIT JEE MAINS TYPE Exercise 2 Match the following

I. Column A Column B II. Column A Column B

1. 2x 5x 6 a. x 10 (x 3) 1. 3 2x 6x 11x 6 a. x 1 x 1 3x 1

2. 2x 7x 30 b. x 18 2x 1 2. 3 2x 2x x 2 b. x 1 x 2 x 5

3. 22x 35 18 c. x 2 x 18 3. 3 2x 6x 3x 10 c. x 1 x 1 x 2

4. 2x 16x 36 d. x 2 x 3 4. 3 23x x 3x 1 d. x 1 x 2 x 3



IIT JEE ADVANCED TYPE Ex 3 One or more than one correct option

1. If ax3 + bx2 + cx + d is divided by x–2, the coefficients of quotient are 2, –1, 3 and the remainder is 4, then the polynomial is

a. 2x3 + 5x2 + 5x – 2

b. 2x3 – 5x2 + 5x – 2

c. 2x3 – 5x2 + 5x + 2

d. 2x3 + 5x2 + 5x + 2 2. The degree of the polynomial

(2x5+3x4 +2x2 –10x + 1)3 is a. 3x5 b. 3+5 c. 8 d. 15 3. Sum of the values of the

polynomial p(x) = x4 – 3x2 + 2x – 1 at x = 1 and x = 2 is [ ] a. 9 b. 11 c.15 d. 6 4. The factors of the polynomial

15a2 –26a +8

a.(3a – 4)(5a –2) b.(3a+4)(5a+2)

c. (–3a+4)(–5a+2) d. (–3a–4)(–5a–2) 5. If ‘–2’ is a zero of the

polynomial ax3 + bx2 + x – b and the value of the polynomial is 4 at x = 2, then the values of a and b respectively are

a. 0 b.2

3 c.

2

3

d.

1

3

6. Which of the following number is a multiple of the zero of the

polynomial x3 – x2 – x – 2 ?

a. 16 b. 35 c. 36 d. 72

7. The polynomials kx3 + 3x2 – 3

and 2x3 – 5x + k where divided by (x – 4) leave the same remainder in each case, then which of the following is not the value of k?

a.1 b. 2 c. 3 d. 4

8. For what value of a is 2x3 +

ax2 +11x+a+3 is exactly divisible by (2x–1) a. –5 b. –19 c. –7 d. –16

9. What must be added to 3x3 +

x2 – 22x + 9 so that the result is

exactly divisibleby 3x2 + 7x – 6 is a.–2x–3 b. 2x–1 c.2x+3 d. 2x+1

10. Sum of the zeroes of the polynomial(x-1)(x-2)(x-3)…….. (x-10) is

a. 45 b. 55 c.10x11

2 d.

10x9

2

IIT JEE ADVANCED TYPE Exercise 4 Linked Data Questions

I. If ax3 + bx2 + cx + d is divided by x + 2 the coefficients of quotient are 4, 5, 6 and the remainder is zero, then 1. The polynomial is

a. 4x3 + 13x2 + 16x + 12

b. 4x3 + 13x2 + 17x + 12

c. 4x3 + 13x2 + 14x + 12

d. 4x3 + 13x2 + 18x + 12 2. The values of a, b, c and d are a. 4, 13, 16, 12 b. 4, 13, 17, 12 c. 4, 13, 14, 12 d. 4, 13, 18, 12



II. If ax2 +bx+c is a quadratic expression and p, q are the roots of expression, then

1. The sum of the roots is a. b/a b. –b/a c. c/a d. –c/a 2. The product of the roots is a. b/a b. –b/a c. c/a d. –c/a

IIT JEE ADVANCED TYPE Exercise 5 Assertion and Reason

a. Assertion is True, Reason is True; Reason is a correct explanation for Assertion b. Assertion is True, Reason is True; Reason is not a correct explanation for Assertion

c. Assertion is True, Reason is False d. Assertion is False, Reason is True

1. Assertion: The square root of 24 ( ) ( )abc a b ab ac bc is ab bc ca

Reason: (a – b)2 +4ab = (a+b)2

2. Assertion: If a+b+c =0, then

2 2 2

3a b c

bc ac ab

Reason: On factorization 3 3 3

2 2 2 2 2 2a b b c c a gives

2 2 2 2 2 23 a b b c c a

3. Assertion: If (2a+b)=12 and ab=15, then the value of 8a3+b3 = 650

Reason: If a+b+c =11 andab+bc+ca =25 then a2+b2+c2=71

4. Assertion: The polynomial x3 + ax2 + bx + 6 has x – 2 as a factor and leaves a

remainder 3, when divided by x – 3. Then the values of a and b

respectively are -3 and -1.

Reason: Let R1 and R2 are the remainders when the polynomials

x3 + 2x2 – 5ax – 7and x3 + ax2 – 12x + 6 are divided by 2x + 1 and

2x – 3 respectively. If 2R1 + R2 = 2, then the value of a is 191

158

5. Assertion: .If a linear polynomial is multiplied with cubic polynomial, then the

obtained polynomial have degree 4

Reason: The remainder when x4 – 4x3 + 4x2 – 2 is divided by (x – 3) is 7



JEE MAINS/NTSE Ex 6 Questions from NTSE & Olympiads

1. If px2 + qx + r is exactly

divisible by (4x–3) and (3x+2),

then the values of p, q and r is

a. 12, 1, 6 b. 12, –1, 6

c. –12, –1, –6 d. 12, –1, –6

2. The product of zeroes of the

polynomial

2 2 22 2( 1)( )a c ax bx c x ac b

a.

2 2

2 2

2

2

c b ac

a b ac

b.

2 2

2 2

2

2

c b ac

a b ac

c.

2 2

2 2

2

2

c b ac

a b ac

d. Can’t say

3. If the ratios of the zeroes of the

expression x2+px+q is equal to

the ratio of zeroes of x2 +lx +m

then

a. p2m=lq2 b. p2m=l2q

c. pm2=lq2 d. p2l=q2m

4.Without actual substitution, the

remainder when x9 – 3x8 + 5x7 –

15x6 + 13x5 – 39x4 + 7x3 – 21x2

+ 17x – 15 divided by x – 3

a. 46 b. 0 c. 36 d. 56

5. If ax3 + bx2 + x – b has x + 2

as a factor and leaves a remainder

4 when divided by x – 2, then the

values of a and b respectively are

a. 0 and 2

3

b. 0 and

1

3

c. 0 and 2

3 d. 0 and

1

3

6. If p, q are the zeroes of the

expression ax2 +bx+c, then the

value of 55 8 8

p q +p q is

a. 5 3

8

3c abc b

a

b.

5 3

8

3c abc b

a

c. 5 3

8

3c abc b

a

d.

5 3

8

3c abc b

a

7. If 3+√2 is a zero of x3 –11x2

+37x –35, then the rational zero

is

a. 5 b. –5 c. 11 d. –11

8. The quadratic expression

whose zeroes are a

a ± a - b

will be

a. 2

2 2a a ax x

b b

b. 2

2 2a a ax x

b b

c. 2

2 2a a ax x

b b

d. 2

2 2a a ax x

b b

9. If the expression (a2 –4a+3)x2

+ (a–1)x +a2 has infinite zeroes,

then the value of a will be

a. 2 b. 7 c. 1 d. 5

10. If ax3 + bx2y + cxy2 + dy3 is a homogeneous expression of third degree in x andy, where a = p + 1, b = q + 2, c = r + 3, d = s + 4 and p = 2, q = 3, r = 4, s = 5, then the expression is

a. 2x3 + 3x2y + 4xy2 + 5y3

b. 5x3 + 4x2y + 3x2y + 2y3

c. 3x3 + 5x2y + 7xy2 + 9y3

d. 9x3 + 5x2y + 7xy2 + 3y3



IITians

9

CHEMISTRY

CLASS IX CHEMISTRY FOUNDATION FOR IIT JEE/ NEET/NTSE/OLYMPIADS, ANY PRESENT



Foundation Curriculum by IITians for Academic Excellence in Schools


Dear Students / Parents, Quality education will play a vital role to emerge India as a leader among all developing countries. IITians Propel Academy is designed to enhance the quality education towards excellent training from school level to IIT JEE / NEET. The foundation program is designed exclusively for students who are aiming to attain a top ranks in the competitive examinations. The Propel Academy course will focus more on conceptual knowledge. As a result, students are able to devote more time on practicing more complex problems in each subject area to develop a complete command on the key concepts. The Propel Academy comprising of detailed concept explanations, Introductory Exercises and JEE MAINS / ADVANCED and NEET model questions, which includes previous questions from different competitive exams like International Math’s Olympiad, NTSE, RMO, APAMT, NTSE, IJSO and Ramaiah's SAT examinations.

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methods.




CHAPTER 1

7 - 27

WHAT IS INSIDE

THE ATOM

C1: Ideas about atom S1

C2: Fundamental particles of an atom

C3: Dalton’s atomic theory S2

C4: Important models of atom

Introductory Exercise 1 C5: Atomic structure S3


C6: Electronic configuration of an Atom S4

C7: Important terms in the atomic structure S5


Ex 1: Single correct objective questions S6

Ex 2: Match the following


Ex 4: One or more than one correct options S7

Ex 5: Linked Data Questions

Ex 6: Questions from/for Olympiads, NTSE S8

CHAPTER 2

28 - 47

PERIODIC TABLE

C1: Mendeleev’s Periodic Law S1

C2: Modern & Present form of Periodic table S2


C3: Periodic Properties & their Trends S4

C4: Ionization Energy S5


C5: Electron Affinity S7

C6: Electronegativity S8 Introductory Exercise 3



Ex 3: Match the Following

Ex 4: Linked comprehension type



C1: Theory of Chemical Bonding S1

C2: Compounds which Don’t Obey Octect Rule

C3: Types of Bonds & their Properties S2 Introductory Exercise 1

CHAPTER 3

48 - 63

CHEMICAL BONDING

C4: Conditions favorable for the Formation of Ionic Bonds S3 C5: Covalent bond-Orbital Concept


C 6: Types of Covalent Bonds S4 C 7: Polar and Non-polar Covalent Bonds



Ex 2: Match the following S6

Ex 3: Assertion and reason



Ex 6: Questions from/for Olympiads& NTSE S8

CHAPTER 4

64 - 79

ACIDS, BASES

AND SALTS

C1: Strong and Weak Acids, C2: Bases S1


C3: Basicity and Acidity, C4: pH scale and Numericals S2


C5: pH in Everyday Life, C6: Salt S3

C7: Classification of Salt S4


C8: Some Common Salts, C9: Gypsum & Plaster of Paris S5



Ex 2: Match the following S7



Ex 5: Linked Comprehension type


CHAPTER 5

80 - 101 CARBON AND

ITS COMPOUNDS

C1: Allotrope of Carbon, C2: Versatile Nature of Carbon S1


C3: Satured & Unsatured Carbon Compounds S3

C4: Homologous Series S4


C5: Isomerism, C6: Soaps and Detergents S6



Ex 2: Assertion and reason S9 Ex 3: Match the following



Ex 6: Questions from/for Olympiads & NTSE S11

S1 Means Session 1 / Class one of 45 min to 1 hr session

Propel Academy

CHAPTER 1 WHAT IS INSIDE

THE ATOM

Chemistry


We know that an element is a pure substance which cannot be subdivided into two or

more new substance by any means. Chemists have found 118 elements which exist in

nature. The smallest unit of any element is called an atom. The atoms of different

elements combine with one another to form new substance called

compounds.

Concept 1 Ideas about atom

Ancient time onwards people started searching about atom. (i) In sixth century BC, the Indian philosopher Maharshi Kanad came forward with paramanu. Matter is not continuous, and made up of tiny particles. Kanad further said that two or more paramanus combine to form bigger particles. (ii)In the fifth century BC, the Greek philosophers Democritus and Leucippus came up with a similar idea. They thought that on dividing a piece of a substance, one would ultimately get a particle that could not be divided further. They gave the name atomos (in Greek, atomos means indivisible) to these ultimate particles.

(iii) John Dalton, in 1808 published theory of atom assuming that atoms are the

ultimate indivisible particles of matter.

(iv) Later the works of William Crookes (1878), J.J. Thomson (1897) and Goldstein proved that atom of any element contains smaller particles which are either positively charged or negatively charged.

(v) Work of Rutherford and Neils Bohr confirmed that an atom consists of three

subatomic particles, which are electrons, protons and neutrons.

(vi) It has been established that the central core of an atom consists of protons and

neutrons and is commonly called nucleus. The electrons revolve around the nucleus.

(vii) The atom as a whole is electrically neutral as the number of protons in it, is equal to the number of electrons.

The smallest indivisible particle or unit of an element is called an atom, which can take

part in a chemical reaction and may or may not exist independently.

An element is a pure substance which cannot be subdivided into two or more new

substances by any means.

Introduction

Class IX Chemistry > IITians Propel Academy


(a) Electron: Electron has a negative charge on it; its mass is 1/1837

times the mass of one atom of hydrogen. It is denoted by the symbol,

1

0

, where 0 denotes it’s mass and -1 denotes its charge. Electrons in the outer most shell

are called valence electrons.

(b) Proton: Proton has a unit positive charge, it is denoted by the symbol ,p

1

1

where 1

denotes it atomic mass and +1 denotes its charge.

(c) Neutron: Neutron has no electric charge on it. Its mass is almost equal to the

mass of one atom of hydrogen. it is denoted by the symbol,n

0

1

, where 1 denotes its

atomic mass and 0 denotes its charge.

In the neutral atom the total number of protons in the nucleus is equal to the number of electrons revolving round the nucleus.

Property Electron Proton Neutron

1. Discovery J.J. Thomson E. Goldstein James Chadwick

2. Symbol E p n

3. Nature Negatively charged Positively charged Neutral

4. Relative charge -1 +1 0

5. Absolute charge 1.602 × 10-19 C 1.602 × 10-19 C 0

6. Relative mass 1837

1

1 1

7. Absolute mass 9.109 × 10-28 g 1.6725 × 10-24 g 1.6748 × 10-24 g

Concept 3 Dalton’s atomic theory

In 1803, an English chemist, John Dalton, put forward his atomic theory. The main

points in that theory are

a. Composition of matter: Elements are made up of very small particles of matter,

called atoms (derived from the Greek word atomos).

b. Indivisibility of atoms: Atoms are indivisible. They cannot be further broken

down. Atoms can be neither be created nor be destroyed in a chemical reaction, cannot

be converted to that of another element.

c. Atoms of similar elements: The atoms of an element are alike in all the respects.

Concept 2 Fundamental particles of an atom

IITians Propel Academy > C1: Atomic Structure Class IX


The atoms of different elements are different in all the respects.

d. Combination of atoms: Atoms combine in small whole numbers to

form compound (molecules).

e.Role of atom in a chemical reaction: An atom is defined as the smallest part of

an element that takes part in a chemical reaction.

Comparison of DALTON’S atomic theory by modern atomic theory:

Though the Dalton’s theory is one of successful attempts in understanding about nature

of atom, it is contradicted by the modern research.

Similarity:Similarity between these two theories is: Atom is the smallest unit of matter

which takes part in a chemical reaction.

Concept 4 Important models of atom

(i) Thomson’s Model of an Atom: After the discovery of electrons and protons J.J. Thomson (1898) tried to explain the arrangement of electrons and protons within the atom. He proposed that an atom consists of a sphere of positive electricity in which electrons are embedded like plum in pudding or seeds evenly distributed in red spongy

mass in watermelon.

The radius of the sphere is that the other 10-8 cm which is equal to the size of the atom. Although, Thmoson’s model could explain the electrical neutrality of an atom but this model could not satisfy experimental facts proposed by Rutherford and hence was

discarded.

S.No. Dalton's atomic theory Modern atomic theory

1 Matter consists of small indivisible particles called atoms

Atom is no longer indivisible, but consists of neutrons, protons and electrons

2 Atoms of same element are alike in all respects

All atoms have isotopes. It means some of the atoms of an element have different atomic weights

3 Atoms of different elements are different in all respects.

Atoms of different elements are sometimes similar in some respects. For example, atoms of argon and calcium have same atomic weight.

4 Atoms combine in small whole numbers to form compound atoms (molecules)

Atoms in organic compounds do not combine in small whole number ratio



“Proton is a sub - atomic particle having a unit positive charge (+1.602 × 10-19 kg) & mass (1.6725 × 10-27 kg) which is about 1837 times greater than the mass of an electron.”

(ii) Rutherford’s Model of an Atom

(a) Rutherford’s Alpha Scattering Experiment: Ernest Rutherford and his coworkers performed numerous experiments in which particles emitted from a radioactive element such as polonium were allowed to strike the thin sheets of metals

such as gold and platinum.

(i) A beam of -particles (He2+) we obtained by placing polonium in a lead box and letting the alpha particles come out of a pinhole in the lead box. This beam of -rays directed against a thin gold foil (0.00004 cm). A circular screen coated with zinc

sulphide was placed on the other side of the foil.

(ii) About 99% of the particles passed undeflected through the gold foil and caused

illumination of zinc sulphide screen.

(iii) Very few -particles underwent small and large deflections after passing through

the gold foil.

(iv) A very few (about 1 in 20,000) were deflected backward on their path at an angle of 1800.



Rutherford was able to explain these observations as follows :

(i) Since a large number of -particles pass through the atom undeflected, hence, there

must be large empty space within the atom.

(ii) As some of the - particles got deflected, therefore, there must be something massive and positively charged present in the atom.

(iii) The number of -particles which got deflected is very small, therefore, the whole positive charge in the atom is concentrated in a very small space.

(iv) Some of the -particles retracted their path i.e. came almost straight back towards

the sources as a result of their direct collisions with the heavy mass.

- Particles are made up of two protons and two neutrons and are He nuclei.

The positively charged heavy mass which occupies only a small space as compared to the total space occupied by the atom is called nucleus.

(b) Rutherford’s Nuclear Model of Atom: Rutherford proposed a new picture of the

structure of the atom. Main feature of this model are as follows -

(i) The atom of an element consists of a small positively charged “Nucleus” which is situated at the centre of the atom and which carries almost the entire mass of the

atom.

(ii) The electrons are distributed in the empty space of the atom around the nucleus in different concentric circular paths (orbits).

(iii) The number of electrons in the orbits is equal to the total number of protons in the

nucleus.

(iv) Volume of nucleus is very small as compared to the volume of atom.

(v) Most of the space in the atom is empty.

The stability of such a system in which negatively charged electrons surround a positively charged nucleus was explained by proposing that the electrons revolve around the nucleus at very high speed in circular orbits. This arrangement is just like



our solar system. The high speed of the moving electrons given them centrifugal force acting away from the nucleus. The centrifugal force balance the electrostatic force of attraction acting between the nucleus and the electrons.

(c) Defects in Rutherford’s Model: The main defects in the Rutherford’s atomic

model are briefly explained below:

(i) Rutherford did not specify the number of electrons in each orbit.

(ii) According to electromagnetic theory, if a charged particle (electron) is accelerated around another charged particle (nucleus) then there would be continuous radiation of energy. This loss of energy would slow down the speed of electron and eventually the electron would fall into the nucleus. But such a collapse does not occur. Rutherford’s

model could not explain this theory.

1) Where are protons located in an

atom?

a. Around the nucleus b. Inside the

nucleus c. Both (a) & (b) d. None

2) Which of the following statements is true? A proton is

a. 1837 times heavier than an electron b. 1/1837 times heavier than an electron c. 1/1837 times lighter than an electron d. having the same mass as an electron

3) When alpha particles are sent through a thin metal foil, most of them go straight through the foil because–

a. Alpha particles are much heavier than electrons

b. Alpha particles are positively charged

c. Most part of the atom is empty d. Alpha particles move with high

velocity

4) Rutherford’s scattering experiment is related to the size of

a. Nucleus b. Atom

c. Electron d. Neutron

5) The mass of a proton is -

a. 1.00728 amub. 1.673 × 10-24 gmc.

1.673 × 10-27 kg d. All

6) Rutherford performed his alpha

scattering experiment using -

a. Silver b. Gold c. Mercury d. Diamond 7) A proton is usually represented as -

a. 1

1p b.

1

1H

c. 4

2He d. a and b

JEE MAINS / NEET MODEL Introductory Exercise 1



(iii) Bohr’s Model of an Atom: Rutherford’s model of the atom was unable to explain certain observations with regard to the atom i.e. stability of the atom and the occurrance of the atomic spectra. Neils Bohr accepted Rutherford’s idea that the positive charge and most of the mass of the atom is concentrated in its nucleus with

the electrons present at some distance away.

According to Bohr’s theory -

(i) Electrons revolve around the nucleus in well defined orbits or shells, each shell having a definite amount of energy associated with the electrons in it. Therefore, these shells are also called energy levels.

(ii) The energy associated with the electrons in an orbit increases as the radius of the orbit increases. These shells are known as K, L, M, N etc. starting from the one closest to the nucleus.

(iii) An electron in a shell can move to a higher or lower energy shell by absorbing or

releasing a fixed amount of energy.

(iv) The amount of energy absorbed or emitted is given by the difference of energies

associated with the two energy levels.

Energy absorbed, ΔE = E2 - E1 = h

Energy emitted, ΔE = E2 - E1 = h

Where h is Plank’s constant (h = 6.62 × 10-34Js) and is the frequency of the radiation.

(v) By applying the concept of quantization of energy, Bohr calculated the radius and

energy of the nth orbit of hydrogen atom

2 2 2 4

n 2 2 2 2

2r = ;

4n

n h meE

me n h

Where m is mass of electron, n is orbit number, h is planck’s constant, e is the energy



Concept 5 Atomic structure

An atom consists of two parts: (a) Nucleus: Nucleus is situated in the centre of an atom. All the protons & neutrons are situated in the nucleus; therefore, the entire mass of an atom is almost concentrated in the nucleus. The overall charge of nucleus is positive due to the presence of positively charged protons (neutrons present have no charge). The protons & neutrons are collectively called nucleons. The radius of the nucleus of an atom is of the order of 10-15 m.

(b) Extra nuclear region: In extra nuclear part or in the region outside the nucleus, electrons are present which revolve around the nucleus in orbits of fixed energies. These orbits are called energy levels. These energy levels are designated as K, L, M, N and so on.

The maximum number of electrons that can be accommodated in a shell is given by the

formula 2n2 (where n = number of the shell i.e. 1, 2, 3 ….)

Shell n 2n2 Max. no. of electrons

K 1 2(1)2 2 L 2 2(2)2 8 M 3 2(3)2 18 N 4 2(4)2 32

Each energy level is further divided into subshells designated as s, p, d, f.

1st shell (K) contains 1 subshell (s) 2nd shell (L) contains 2 subshells (s,p)

3rd shell ( M) contains 3 subshells (s,p,d) 4 th shell (N) contains 4 subshells (s,p,d,f).

Orbitals: Like shells are divided into subshells, subshells further contain orbitals. An orbital may be defined as a Region in the three - dimensional space around the nucleus where the probability of finding an electron is maximum. The maximum capacity of

each orbital is of two electrons.

The total number of nucleons is equal to the mass number (a) of the atom.

Subshell Orbital (s) Max. no. of electrons

s 1 2 p 3 6 d 5 10 f 7 14



Filling of electron in sub-shell

Electrons are filled in the order

1s, 2s, 2p, 3s, 3p, 4s, 3d..............


1) The formula that gives the maximum number of electrons in a particular shell is

a. n2 b. 2n2 c. 2n d. n2/2

2) The radius of an atomic nucleus

is of the order of

a. 10-10 cm b. 10-13 cm

c. 10-15 cm d. 10-8 cm

3) A p-orbital can accommodate

upto

a. 4 electrons b. 2 electrons

c. 6 electrons d. 3 electrons

4) Energy levels are designated as

a. K, L, M, N and so on b. k, l, m, n and so on c. I, II, III, IV

and so on d. All

of these

5) A neutron is represented as

a.0

0n b.

1

1n c.

1

0n d.

1

1n

6) The different subshells in an

atom are represented as

a.s,p,d, f b. S,P,D, F

c. 1,2,3, 4 d. All of these

7) The maximum number of

electrons in N shell is

a. 2 b. 8 c. 18 d. 32

8) The maximum number of electrons in f - subshell is

a. 5 b. 6 c. 14 d. 10

Class IX Chemistry >IITians Propel Academy


Concept 6 Electronic configuration of an Atom

(i) Each of the orbits can accommodate a fixed number of electrons. Maximum number

of electrons in an orbit is equal to 2n2, where ‘n’ is the number of the orbit.

If n = 1 then 2n2 = 2 , n = 2 then 2n2 = 8

n = 3 then 2n2 = 18, n = 4 then 2n2 = 32

(ii) In the outermost shell of any atom, the maximum possible number of electrons is 8,

except in the first shell which can have at the most 2 electrons.

(iii) The arrangement of the electrons is different shells is known as the electronic configuration of the element. The pictorial representations of Bohr’s model of hydrogen, helium, carbon, sodium and calcium atoms having 1, 2, 6, 11 and 20 electrons respectively are shown in the figure where the centre of the circle represents the

nucleus.

If the outermost shell has 8 electrons it is said to be an octet. If the first shell has its

full quota of 2 electrons, it is said to be duplet.

Electronic Configuration of Elements Upto Atomic Number 30

Atomic number

Symbols of the element

Name of the element Electronic configuration

1 H Hydrogen 1 2 He Helium 2 3 Lo Lithium 2,1

4 Be Beryllium 2,2 5 B Boron 2,3 6 C Carbon 2,4

7 N Nitrogen 2,5 8 O Oxygen 2,6

9 F Fluorine 2,7 10 Ne Neon 2,8 11 Na Sodium 2,8,1 12 Mg Magnesium 2,8,2 13 Al Aluminium 2, 8,3



Concept 7 Important terms in the atomic structure

(i) Valency: Valency of an element is the combining capacity of the atoms of the element with atoms of the same or different elements. The combining capacity of the atoms of other elements was explained in terms of their tendency to attain a fully - filled outermost shell (stable octet or duplet). The number of electrons gained, lost or contributed for sharing by an atom of the

element gives us directly the combining capacity or valency of the element.

Valency of an element is determined by the number of valence electrons in an atom of

the element.

The valency of an element = number of valence electrons

(Whennumber of valence electrons is from 1 to 4)

The valency of an element = 8 - number of valence electrons.

(Whennumbers of valence electrons are more than 4)

eg. Na has 1 valence electron, thus, its valency is 1.

Cl has 7 valence electrons, thus, its valency is 8 - 7 = 1.

(ii) Atomic Number (Z): The number of protons is the nucleus of an atom of a given

element is called the atomic number of that element. Or

14 Si Silicon 2 ,8, 4

15 P Phosphorus 2, 8,5 16 S Sulphur 2,8,6 17 Cl Chlorine 2,8,7 18 Ar Argon 2,8,8 19 K Potassium 2,8,8,1 20 Ca Calcium 2,8,8,2 21 Sc Scandium 2,8,9,2 22 Ti Titanium 2,8,10,2 23 V Vanadium 2, 8,11,2 24 Cr Chromium 2,8,12,2 25 Mn Manganese 2,8,13,2 26 Fe Iron 2,8,14,2 27 Co Cobalt 2,8,15,2

28 Ni Nickel 2,8,16,2 29 Cu Copper 2,8,17,2

30 Zn Zinc 2,8,18,2



Atomic number is the number of protons present in the atom of an element. It is

denoted by “Z”

Atomic number = Number of protons = Number of electrons

(In a neutral atom)

Atomic number = Number of protons

(In an ion)

Example: How many number of electrons are revolving round the nucleus of sodium?

Atomic number of sodium is 11; nucleus of sodium has 11 protons. Nucleus of sodium has 11 units of positive charge. There are 11 electrons, revolving round the nucleus of sodium.

The atomic number is represented on the LHS (Left Hand Side) to the symbol of the

element as subscript.

(iii) Mass Number (A):Mass number is the number of protons and neutrons present in the atom of an element. It is denoted by “A”. Each element has a unique atomic number which is its identity. The mass number is represented either on the left hand side (LHS) or on the right hand side (RHS) of the symbol of the element as superscript.

Mass number = Number of protons + Number of neutrons.

e.g.27

13Al

Mass number of Aluminum is 27.

The total number of protons and neutrons is the nucleus of aluminum is 27.

Number of protons is 13, Number of neutrons is = 27 - 13 = 14.

(a) Relation between Z, A and N is A = Z + N

Z = Number of Protons, N = Number of neutrons

A = Mass number N = A - Z

(iv) Isotopes: Atoms of same element having the same chemical properties, but differing in mass are known as isotopes. All the isotopes of an element have identical

chemical properties.

The isotopes of an element have the same atomic number but different atomic masses. Isotopes have the same electrical charges means the same number of protons. The difference in their masses is due to the presence of different number of neutrons.



(a) Isotopes of hydrogen

Hydrogen Characteristics isotopes

Protium

H1

1

Deuterium

H1

2

Tritium

H1

3

1. Atomic number 1 1 1 2. No. of protons 1 1 1 3. No. of electrons 1 1 1 4. No. of neutrons 0 1 2

5. Mass number 1 2 3

(b) Isotopes of oxygen

Oxygen Characteristics isotopes

O8

16

O8

17

O8

18

1. Atomic number 8 8 8 2. No. of protons 8 8 8 3. No. of electrons 8 8 8 4. No. of neutrons 8 9 10

5. Mass number 16 17 18

Characteristics of Isotopes:

(i) The physical properties of the isotopes of an element are different because the number of neutrons in their nuclei differs. Hence mass, density and other physical

properties of the isotopes of an element are different.

(ii) All the isotopes of an element contain the same number of electrons. So, they have the same electronic configuration with the same number of valence electrons.

Since the chemical properties of an element are determined by the number of valence

electrons in its atom, all the isotopes of an element have identical chemical properties.

(d) Reason for the fractional atomic masses of elements:

The atomic masses of many elements are in fraction and not whole number. The fractional atomic masses of elements are due to the existence of their isotopes having different masses.e.g. : The atomic mass of chlorine is 35.5 u. Chlorine has two isotopes



35

17Cl and

37

17Cl with abundance of 75% and 25% respectively. Thus the average mass

of a chlorine atom will be 75% of Cl - 35 and 25% of Cl-37, which is 35.5 u.

i.e., average atomic mass of chlorine = 75 25

35 37100 100

= 2625 925

100 100 = 26.25 + 9.25 = 35.5 u.

Thus, the average atomic mass of chlorine is 35.5 u.

Similarly, average atomic mass of copper is 63.5 u.

Applications of Radioactive isotopes:

(i) In agriculture: Certain elements such as Boron, Cobalt, Copper, Manganese, Zinc

and Molybdenum are necessary is very minute quantities for plant nutrition.

By radioactive isotopes we can identify the presence and requirements of these element in the nutrition of plants.

(ii) In industry: Coating on the arm of the clock to see in the dark and to identify the

cracks in the metal casting industry.

(iii) In medicine: Thyroid, bone diseases, brain tumors and cancer and diagnosed,

controlled or destroyed with the help of radioactive isotopes like 60

27Co ,

24

21Na , Iodine,

phosphorus etc.

(iv) Determination of the mechanism of chemical reaction: by replacing an

atom or molecule by its isotope.

(v) In carbon dating: Will and Libby (1960) developed the technique of radiocarbon

dating to determine the age of plants, fossils and archeological samples.

Isotopes (Like Uranium - 238) are used in nuclear reactor to produce energy and

power.

(v) Isobars:The atoms of different elements with different atomic numbers, but same mass number are called isobars.

E.g.14

6C and

14

7N are isobars

40

20Ca and

40

18Ar are isobars



(vi) Isotones: The isotones may be defined as the atoms of different elements

containing same number of neutrons.

e.g. 13

6C and

14

7N

Number of neutrons (N) = A – Z For 13

6C N = 13 - 6 For

14

7N N = 14 - 7

Other example 30 31

,14 15

Si P and 32

16S

(vii) Isoelectric: Ion or atom or molecule which has the same number of electrons is

called as isoelectronic species.

e.g. Cl- Ar K+ Ca+2

No. of electrons 18 18 18 18

Isobars contain different number of electrons, protons and neutrons.


1) Which of the following has the same number of protons, electrons & neutrons?

a. 54

27X b.

155

27X

c.54

26X d.

55

28X

2) In an atom there are four orbits, the maximum number of electrons in this atom will be -

a. 30 b. 36 c. 32 d. 62

3) Isobars contain different

number of

a. electrons b. protons c. neutrons d. all

4) Isoelectric will have same number of

a. electrons b. protons c. neutrons d. all

Characteristics Isobars

Ar18

40

Ca

20

40

1. Atomic number 18 20

2. Mass number 40 40

3. No. of electrons 18 20

4. No. of protons 18 20

5. No. of neutrons 22 20

6. Electronic configuration

2.8.8 2,8,8,2



1) The number of valence electrons

in Na is

a. 1 b. 2 c. 3 d. 4

2) The valency of Ne is

a. 10 b. 8 c. 2 d. 0

3) An isotone of 76

32Ge is -

a. 77

32Ge b.

77

33As

c.77

34Se d.

79

34Se

4) Two atoms of the same element

are found to have different number

of neutrons in their nuclei. These

two atoms are

a. Isomers b. Isotopes

c. Isobars d. Allotropes

5) The electronic configuration of

Mn+2 is

a. 2, 8, 13 b. 2, 8, 11, 2

c. 2, 8, 13, 2 d. None of these

6) The number of valance electron

in Na is

a. 1 b. 2 c. 3 d. 4

7) Neutron is a fundamental

particle which has?

a. + 1 unit charge and mass

b. No charge and 1 unit mass

c. Have no charge and mass

d. Have – 1 unit charge and 1 unit mass

8) Number of unpaired electron in

1s2, 2s2,2p4

a. 4 b. 2 c. 0 d. 1

9) The atomic size is nearly –

a. 10-10 cm b. 10-6 m

c. 10-7 m d. 10-10 m

10) An atom which has a mass

number of 14 or 8 neutron is

a. Isotope of oxygen

b. Isotope of oxygen

c. Isotope of carbon

d. Isobar of carbon

11) The maximum number of

electrons that can be accommodate

in the nth level is

a. n2 b. 2n2 c. 4n2 d. n

12) Which isotope is generally used

as a nuclear fuel?

a. U – 235 b. Co – 60

c. P-32 d.Iodine

JEE MAINS/NEET Exercise 1 Single correct objective questions



Column A Column B

1) Atomic theory a. Maharshi Kannada

2) Electron b. Goldstein

3) Proton c. John Dalton

4)Paramanu d. J.J. Thomson

Element Configuration

1) Hydrogen a. 2, 4

2) Boron b. 2

3) Helium c. 1

4) Carbon d. 2, 3

a. Assertion is true; Reason is true; Reason is the correct explanation for Assertion. b. Assertion is true; Reason is true; Reason is not the correct explanation for Assertion. c. Assertion is true; Reason is false. d. Assertion is false; Reason is true. 1) Assertion: Electron has unit negative charge.

Reason: Electron mass is higher than a proton.

2) Assertion: In a neutral atom the number of protons and electrons are same.

Reason: Neutron has no electric charge.

3) Assertion: Atom is mostly empty.

Reason: Most of - particles went straight through the foil.

4) Assertion: The atomic number of an element is 17.

Reason: Number of neutrons are 16.

5) Assertion: In Na+ number of e is more than the number of protons

Reason: Isotopes of iodine are used for making tincture of iodine n medicine.

6) Assertion: In Bohr’s structure of atom, 5th shell has more energy than the 4th shell

Reason: with the increase in the number of the shell, the energy of the shell also

increases.

JEE ADVANCED TYPE Exercise 2 Match the following

JEE ADVANCED TYPE Exercise 3 Assertion and Reason



1) Pick out the isoelectronic structures

from the following

a. CH3+ b. H3O+ c. NH3 d. CH3-

2) Members of which of the following

have similar chemical properties?

a. Isotopes b. Isobars

c. Allotropes d. None

3) The number of electrons in the L

shell of phosphorus are equal to that

in the

a. L - shell of neon

b. M - shell of potassium

c. M - shell of chromium

d. M - shell of argon

4) Nucleus consists of

a. Proton b. Electron

c. Neutrons d. Ions

5) 35Cl17 and 35Cl 17 have the same

number of

a. Electron b. Proton

c. Neutron d. Atoms

6) 40 Ar18 and 40 Ca20 have the same

number of

a. Electron b. Proton

c. Nucleons d. Atoms

7) Difference between Cl and Cl- ion is

of

a. Proton b. Electron

c. Neutron d. None

8) 35

17 Cl Indicates, nuclide contains

a. 17 protons b. 17 electrons

c. 18 neutrons d. 16 protons

9) The orbits of atom can be

represented by which shell?

a. K b. L c. M d. N (d) N shell

10) Which one of the following is a

correct electronic configuration of

chromium?

a. 2, 8, 8, 6 b. 2, 8, 8, 2, 4

c. 2, 8, 8, 1, 5 d. 2, 8, 8, 8,2,5

11) The protons and neutrons are

collectively called -

a. Deuterons b. Positrons

c. Mesons d. Nucleons

12) 23

11Na contains?

a. 22 protons b. 22 neutrons c. 12 neutrons d. None

13) The credit of discovering neutron

goes to -

a. Rutherford b. Thomson

c. Goldstein d. Chadwick

14) The maximum number of electrons that can be accommodated

in the valence shell of an atom is

a. 5 b. 6 c. 7 d. 8

JEE ADVANCED TYPE Exercise 4 One or more than one correct options



I. The equation given by Bohr to calculate the radius of nth orbit of

hydrogen atom is 2 2

n 2 2

n hr =

4π me

The equation given by Bohr to calculate the energy of nth orbit of

hydrogen atom is2 4

n 2 2

-2π meE =

n h

From the above given data, answer

the following

1) The ratio of radii of 1st and 2nd orbits of Hydrogen atom is

a. 1:2 b. 1:4 c. 2:1 d. 4:1

2) The ratio of energies of 1st and 2nd

orbits of Hydrogen atom is

a. 1:2 b. 1:4 c. 2:1 d. 4:1

II. Read the following passage

The electrons are filled from the lowest energy level to the highest energy level of an atom. The number of electrons occupying an orbit can be calculated by the formula 2n2 where n is the number of an orbit. Based on the given information, answer the

following

1) The number of electrons present in the valence shell of an atom with atomic number 38 is

a. 1 b. 2 c. 8 d. 10

2) The n value of the last shell of the atom with atomic number 38 is

a. 2 b. 3 c. 4 d. 5

1) Number of electrons in CONH2 is

a. 24 b. 28 c. 22 d. 20

2) The size of nucleus is of the

order of

a. 10-15 m b. 10-12 m c. 10-8

m d. 10-10 m

3) If the atomic weight of an element is 23 times that of the lightest element and it has 11

protons, then it contains

a. 11 protons, 11 neutrons, 23 electrons b. 11 protons, 23 neutrons, 11 electrons c. 11 protons, 12 neutrons, 11 electrons

d. 11 protons, 11 neutrons, 11 electrons

4) The number of unpaired

electrons in the Fe+ ion is

a. 4 b. 3 c. 0 d. 6

5) The nucleus of the element having atomic number 25 and atomic weight 55 will contain

a. 25 neutrons and 30 protons b. 55 neutrons c. 25 protons and

30 neutrons d. 55 protons

6) The number of electrons in one

molecule of Oxygen is

a. 44 b. 88 c. 2 d.none

7) 35

17 Cl and

37

17 Cl are________

JEE ADVANCED TYPE Exercise 5 Linked Data Questions

JEE MAINS Exercise 6 Questions from/for Olympiads, NTSE




c.Isotones d. Can’t say

8) Rutherford’s model of an atom

could not explain its stability. This

was overcome by another atomic

model. The postulates of the new

model of the atom are

(i) An atom consists of a positively

charged sphere and the electrons are

embedded in it.

(ii) Only discrete orbits of electrons are

allowed inside the atom

(iii) While revolving in orbits, electrons do not radiate.

a. (i) and (ii) b. (ii) and (iii) b. (ii) and (iii)

c. (i) and (iii) d. (i),(ii) & (iii) d. (i), (ii) and (iii)

9) “All matter is made up of very small particles which cannot be further broke down. These particles are called atoms”. This statements is one of the assumptions of _____

a. Rutherford’s nuclear theory b. Bohr’s theory c. Dalton’s atomic theory d. Kinetic theory of gases

10) In Rutherford gold foil

experiment most of the -particles

pass through the gold foil without any deviation from their paths. This

indicates that

a. The atom is spherical b. There is a positively charged nucleus at the center of the atom c. The entire mass of the atom is concentrated at the nucleus of the atom

d. Most portion of the atom is empty space 11) 18Ar40,19K40,20Ca40 are


c. A & B both d. None

12) A monovalent anions has 10

electron and 10 neutrons. The

atomic number and mass number

of the element of respectively

a. 10, 20 b. 9, 18

c. 10, 20 d. 9, 19

13)A trivalent cation of an element

contains 10 electrons. The atomic

number of the element is

a. 10 b. 7 c. 13 d. Can’t say

14) In which of the following pairs

of shells, energy difference

between two adjacent orbits is

minimum?

a. K,L b. L,M c. M, N d. N, O

15) The electron revolves only in

the orbits in which

a. mvr > nh/2 b. mvr < nh/2

c. mvr = nh/2 d. mvr ≤ nh/2

16) The mass number of an atom

whose uni positive ion has 10

electrons and 12 neutrons is

a. 20 b. 21 c. 22 d. 23

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