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CBSE Syllabus for Class 11 Political Science

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CBSE Syllabus for Class 11 Political Science

Course Structure for Class XI Political Science (2017-18)

Unit Topic Marks
Part A: Indian Constitution at Work  
1. Philosophy of the Constitution 12
2. Rights of the Indian Constitution
3. Election and Representation 10
4. Executive
5. Legislature 10
6. Judiciary
7. Federalism 10
8. Local Governments
9. Constitution as a Living Document 8
 Total 50
Part B: Political Theory 
10. Political Theory : An Introduction 10
11. Freedom
12. Equality 10
13. Social Justice
14. Rights 10
15. Citizenship
16. Nationalism 10
17. Secularism
18. Peace 10
19. Development
 Total 50

Part A: Indian Constitution at Work

1. Philosophy of the Constitution

Constitution: Why and How, The making of the Constitution, the Constituent Assembly, Procedural achievements and Philosophy of the Constitution.

2. Rights in the Indian Constitution

The importance of Rights, Fundamental Rights in the Indian Constitution, Directive Principles of State Policy, Relationship between Fundamental Rights and Directive Principles.

3. Election and Representation

Elections and Democracy, Election System in India, Reservation of Constituencies, Free and Fair Elections, Electoral Reforms.

4. Legislature

Why do we need a Parliament? Two Houses of Parliament. Functions and Power of the Parliament, Legislative functions, control over Executive. Parliamentary committees. Self-regulation.

5. Executive

What is an Executive? Different Types of Executive. Parliamentary Executive in India, Prime Minister and Council of Ministers. Permanent Executive: Bureaucracy.

6. Judiciary

Why do we need an Independent Judiciary? Structure of the Judiciary, Judicial Activism, Judiciary and Rights, Judiciary and Parliament.

7. Federalism

What is Federalism? Federalism in the Indian Constitution, Federalism with a strong Central Government, conflicts in India’s federal system, Special Provisions.

8. Local Governments

Why do we need Local Governments? Growth of Local Government in India, 73rd and 74th Amendments, implementation of 73rd and 74th Amendments.

9. Constitution as a Living Document

Are Constitutions static? The procedure to amend the Constitution. Why have there been so many amendments? Basic Structure and Evolution of the Constitution. Constitution as a Living Document.

Part B: Political Theory

10. Political Theory: An Introduction

What is Politics? What do we study in Political Theory? Putting Political Theory to practice. Why should we study Politial Theory?

11. Freedom

The Ideal of Freedom. What is Freedom? Why do we need constraints? Harm principle. Negative and Positive Liberty.

12. Equality

Significance of Equality. What is Equality? Various dimensions of Equality. How can we promote Equality?

13. Social Justice

What is Justice? Just Distribution. Justice as fairness. Pursuing Social Justice.

14. Rights

What are Rights? Where do Rights come from? Legal Rights and the State. Kinds of Rights. Rights and Responsibilities.

15. Citizenship

What is citizenship? Citizen and Nation, Universal Citizenship, Global Citizenship.

16. Nationalism

Nations and Nationalism, National Self-determination, Nationalism and Pluralism.

17. Secularism

What is Secularism? What is Secular State? The Western and the Indian approaches to Secularism. Criticisms and Rationale of Indian Secularism.

18. Peace

What is Peace? Can violence ever promote peace? Peace and the State. Different Approaches to the pursuit of peace. Contemporary challenges to peace.

19. Development

What is development? Criticism of the dominant. Development Model. Alternative conceptions of development.

Prescribed Books:

  1. Indian Constitution at work, Class XI, Published by NCERT
  2. Political Theory, Class XI, Published by NCERT

NCERT Solutions

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Sing A Song Of People – CBSE Notes for Class 5 English

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Sing A Song Of People – CBSE Notes for Class 5 English

CBSE NotesCBSE Notes Class 5 EnglishNCERT Solutions English

Summary :

(i) Sing A Song Of People

This poem describes how people are in such a hurry these days. Wherever they are, they are in a hurry. They go here and there hurriedly. They can be seen on the subway. They can be seen with umbrellas in rains. They can be seen rushing for the taxis and riding elevators. Although they are walking singly, they are in a crowd everywhere. Some people are silent, while some are talking loudly. They can laugh and smile but they don’t have time to look at others. It is because they are in a hurry.

(ii) Around The World

‘Around the World’ is a story about Mr. Phi leas Fogg and his journey of the world with his companion Passe part out.
They started their journey from San Francisco to New York, which was three thousand seven hundred and eighty six miles away. The train crossed many steep mountain slopes, deep canyons, curves, tunnels and bridges. Suddenly the train stopped when a herd of buffaloes came on the trade. The herd took a full three hours to cross the tracks. The train began to move again and headed for steep mountains. This was the most different part of the journey with its winding roads. They passed the highest point of their journey, 7524 feet above sea level. In a few hours they came out of the Rocky Mountains. The train stopped once again when the suspension bridge on Medicine River had some of its cables broken. However, the driver crossed it at maximum speed.
Next day when the train was moving forward, it was suddenly attacked by hundreds of Scour Indians. But they fled away when they saw soldiers rushing towards them.

The post Sing A Song Of People – CBSE Notes for Class 5 English appeared first on Learn CBSE.

Malu Bhalu(The Poem) – CBSE Notes for Class 5 English

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Malu Bhalu(The Poem) – CBSE Notes for Class 5 English

CBSE NotesCBSE Notes Class 5 EnglishNCERT Solutions English

Summary :

(i)Malu Bhalu

A little polar bear, named Malu Bhalu lived in an icy lair. She learnt to catch fish from her parents. She wanted to see things beyond the blue sky. But her mother suggested her to have patience until she learnt swimming. Very soon she learnt that skill too with the help of her mother. Her mother was sure that Malu Bhalu was brave and fearless.

(ii) Who Will Be Ningthou?

This is a story from Manipur about what qualities make a good ruler. The king and queen were so good that they always wanted to see their people happy. In return they were loved dearly by the people. The birds and animals too loved them. They had three sons and a daughter whom they named Sanatombi. She was good at heart and loved by all.
The king grew old in due course. He decided to choose his heir. He tested all his three sons for the qualities of a good ruler but he found that no one was worthy of becoming a ruler. Hence he chose his twelve year old daughter, Sanatombi as his heir because she could feel the pain of the people, the animals, the birds and the trees. Everyone was happy with their future queen.

The post Malu Bhalu(The Poem) – CBSE Notes for Class 5 English appeared first on Learn CBSE.

The Fish Tale – CBSE Notes for Class 5 Maths

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The Fish Tale – CBSE Notes for Class 5 Maths

CBSE NotesCBSE Notes Class 5 MathsNCERT Solutions Maths

Summary :

• 1 crore = One hundred lakh (10000000). There are 7 zeroes in it.
• One Lakh = One hundred thousand (100000). There are 5 zeroes in it.
• 1 metre = 100 centimetre.
• Length of a whole shark = 18 metre.
Trawlers: Very big machine boat.
• When a fish is dried, its weight is reduced to one third, i.e. 3 kg becomes 3/3 = 1 kg.

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Shapes And Angles – CBSE Notes for Class 5 Maths

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Shapes And Angles – CBSE Notes for Class 5 Maths

CBSE NotesCBSE Notes Class 5 MathsNCERT Solutions Maths

Summary :

Closed Shape: If there is no gap in the shape, it is called a closed shape.
For example:
shapes-angles-cbse-notes-class-5-maths-1
Open Shape. If there is a gap in the shape, it is called an open shape.
For example:
shapes-angles-cbse-notes-class-5-maths-2
-> Angle: Two lines, meeting at a point make an angle. Angle is measured in degree (°).
-> Polygon: A closed shape with many sides is called a polygon. For example: Triangle, rectangle, pentagon, hexagon, etc.
shapes-angles-cbse-notes-class-5-maths-3
-> Types of Angles
(а) Right Angle: The angle which measures 90° is called right angle.
(b) Acute Angle: The angle which measures less than 90° is called acute angle,
(c) Obtuse Angle: The angle which measures more than 90° is called obtuse angle.
shapes-angles-cbse-notes-class-5-maths-4

The post Shapes And Angles – CBSE Notes for Class 5 Maths appeared first on Learn CBSE.

Blow Hot, Blow Cold – CBSE Notes for Class 5 EVS

Who Will Do This Work? – CBSE Notes for Class 5 EVS

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Who Will Do This Work? – CBSE Notes for Class 5 EVS

CBSE NotesCBSE Notes Class 5 EVSNCERT Solutions EVS

Summary:

• Ashram: The place where old people or saints live a simple life.
• Untouchability: An age old practice in which touching a person from the backward caste was supposed to be a sin.
• Constitution: A set of laws and principles which provide the basic guidelines
to run a state or an organisation.

The post Who Will Do This Work? – CBSE Notes for Class 5 EVS appeared first on Learn CBSE.

Across The Wall – CBSE Notes for Class 5 EVS


No Place For Us? – CBSE Notes for Class 5 EVS

A Seed Tells A Farmers Story – CBSE Notes for Class 5 EVS

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A Seed Tells A Farmers Story – CBSE Notes for Class 5 EVS

CBSE NotesCBSE Notes Class 5 EVSNCERT Solutions EVS

Summary:

• Tractor: A vehicle used for farming activities.
• Plough: A large farming equipment which is used for ploughing the field.
• Irrigation: Supplying water to crops or land by means of channel.
• Fertilizer: A chemical or natural substance used for increasing soil fertility.
• Thresher: A machine used for separating grains from harvested crops.
• Compost: Decayed organic material, used as fertilizer.
• Pestle: A heavy tool used for pounding substances in a mortar.
• Undhiya: A kind of stew.

The post A Seed Tells A Farmers Story – CBSE Notes for Class 5 EVS appeared first on Learn CBSE.

Whose Forests? – CBSE Notes for Class 5 EVS

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Whose Forests? – CBSE Notes for Class 5 EVS

CBSE NotesCBSE Notes Class 5 EVSNCERT Solutions EVS

Summary:

• Kuduk: The language of the forest dwellers (Adivasis).
• Contractor: The person who undertakes a contract to provide materials and manpower for a job.
• Scholarship: A monthly or yearly token money awarded to outstanding students; for financial assistance during study.
• Journalist: A person who writes for newspapers or magazines or prepares news or features to be broadcast on radio or television.
• Sowing: The process of scattering or spreading the seeds in the soil.
• Harvesting: The process of gathering the crop yield from the farms.
• Cultivation: The process of preparing and using a land for growing crops.
• Weed: Wild and unwanted plants which grow along with crops.
• Plough: A farming instrument used for ploughing the land.
• Trench: A long narrow ditch.

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Like Father, Like Daughter – CBSE Notes for Class 5 EVS

On The Move Again – CBSE Notes for Class 5 EVS

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On The Move Again – CBSE Notes for Class 5 EVS

CBSE NotesCBSE Notes Class 5 EVSNCERT Solutions EVS

Summary:

• Puranpoli: Sweet roti made from jaggery and gram.
• Mukadam: Agent who lends money to someone.
• Loan: Money taken in advance.
• Debt: A um of money owned.
• Expense: Expenditure/spending/outgoings
• Tubewell: A simple device which is used to draw water from underground.
• Caravan: A large group of people travelling from one place to another.
• Receipt: A written proof of purchase.

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Building With Bricks – CBSE Notes for Class 4 Maths

Long And Short – CBSE Notes for Class 4 Maths


NCERT Exemplar Problems Class 11 Mathematics Chapter 8 Binomial Theorem

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NCERT Exemplar Problems Class 11 Mathematics Chapter 8 Binomial Theorem 

Short Answer Type Questions:
Q1. Find the term independent of x, where x≠0, in the expansion of { \left( \frac { 3{ x }^{ 2 } }{ 2 } -\quad \frac { 1 }{ 3x }  \right)  }^{ 15 }

ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-1

Q2. If the term free from x is the expansion of  { \left( \sqrt { x } -\frac { k }{ { x }^{ 2 } }  \right)  }^{ 10 } is 405, then find the value of k.

Sol: Given expansion is { \left( \sqrt { x } -\frac { k }{ { x }^{ 2 } }  \right)  }^{ 10 }

ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-2
ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-3

Q3. Find the coefficient of x in the expansion of (1 – 3x + 1x2)( 1 -x)16.

Sol: (1 – 3x + 1x2)( 1 -x)16

ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-4

Q4. Find the term independent of x in the expansion of { \left( 3x-\frac { 2 }{ { x }^{ 2 } }  \right)  }^{ 15 }

Sol: Given Expression { \left( 3x-\frac { 2 }{ { x }^{ 2 } }  \right)  }^{ 15 }

ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-5

Q5. Find the middle term (terms) in the expansion of

ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-6
ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-7

Q6. Find the coefficient of x15 in the expansion of { \left( x-{ x }^{ 2 }\quad  \right)  }^{ 10 }

Sol: Given expression is   { \left( x-{ x }^{ 2 }\quad  \right)  }^{ 10 }

ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-8

Q7. Find the coefficient of \frac { 1 }{ { x }^{ 17 } } in the expansion of { \left( { x }^{ 4 }-\frac { 1 }{ { x }^{ 3 } } \quad  \right)  }^{ 15 }
ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-9

Q8. Find the sixth term of the expansion (y1/2 + x1/3)n, if the binomial coefficient of the third term from the end is 45.

>ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-10

Q9. Find the value of r, if the coefficients of (2r + 4)th and (r – 2)th terms in the expansion of (1 + x)18 are equal.

ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-11

Q10. If the coefficient of second, third and fourth terms in the expansion of (1 + x)2” are in A.P., then show that 2n2 – 9n + 7 = 0.

ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-12

Q11. Find the coefficient of x4 in the expansion of (1 + x + x2 + x3)11.

ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-13

Long Answer Type Questions

Q12. If p is a real number and the middle term in the expansion { \left( \frac { p }{ 2 } +2\quad \right) }^{ 8 } is 1120, then find the value of p.

ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-14
ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-15
ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-16
ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-17

Q15. In the expansion of (x + a)n, if the sum of odd term is denoted by 0 and the sum of even term by Then, prove that

ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-18
ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-19
ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-20
Q17. Find the term independent ofx in the expansion of (1 +x + 2x3){ \left( \frac { 3 }{ 2 } { x }^{ 2 }-\frac { 1 }{ 3x } \quad \quad  \right)  }^{ 9 }

ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-21

Objective Type Questions

Q18. The total number of terms in the expansion of (x + a)100 + (x – a)100 after simplification is

(a) 50
(b) 202
(c) 51
(d) none of these

ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-22

Q19. If the integers r > 1, n > 2 and coefficients of (3r)th and (r + 2)nd terms in the binomial expansion of (1 + x)2n are equal, then
(a) n = 2r             
(b) n = 3r            
(c) n = 2r + 1       
(d) none of these

ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-23

Q20. The two successive terms in the expansion of (1 + x)24 whose coefficients are in the ratio 1 : 4 are
(a) 3rd and 4th

(b) 4th and 5th
(c) 5th and 6th
(d) 6th and 7th

ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-24

Q21. The coefficients of xn in the expansion of (1 + x)2n and (1 + x)2n ~1 are in the ratio
(a) 1 : 2                   
(b) 1 : 3                  
(c) 3 : 1
(d) 2:1

ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-25
ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-26

Q22. If the coefficients of 2nd, 3rd and the 4th terms in the expansion of (1 + x)n are in A.P., then the value of n is
(a) 2           

(b) 7 
(c) 11               
(d) 14

ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-27

Q23. If A and B are coefficients of  xn   in the expansions of (1 + x)2n and (1 + x)2n1  respectively, then A/B  equals to

ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-28
ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-29
ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-30
ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-31
ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-32
ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-33
ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem-34

NCERT Exemplar Problems Class 11 Mathematics

The post NCERT Exemplar Problems Class 11 Mathematics Chapter 8 Binomial Theorem appeared first on Learn CBSE.

The Living World – CBSE Notes for Class 11 Biology

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 The Living World – CBSE Notes for Class 11 Biology

CBSE NotesCBSE Notes Class 11 BiologyNCERT Solutions Biology

1.Living World: Characteristics

The earth serves as a home for diverse living organisms. The organisms live in various habitats like forests, mountains, deserts, oceans, freshwater bodies, hot springs, polar regions and almost every place of the earth. Living things possess certain characteristics, which makes
Some of these important characteristics are shown in the flowchart given below:
living-world-cbse-notes-class-11-biology-1
Major Characteristics of Living Organisms:
Major characteristics of living organisms are given below
1 Growth
Living things grow by increase in mass and increase in number of individuals/cells. In multi cellular organisms in particular, growth occur by cell division or increase in number of cells. Growth occurs continuously throughout life in plants, whereas, in animals, it occurs up to a certain age only. However, growing in certain body parts like nails, hair and replacement of lost cells, occurs throughout the life.
In unicellular organisms, growth can be observed under the microscope by simply counting the number of cells via in vitro experiment.
Non-living things like mountains, boulders, sand dunes also grow in size, but just by accumulating the material on their external surface. Thus, growth in living things is internal, while in non-living things, it is external. It is to be noted that a dead organism do not grow.
2. Reproduction
Reproduction, a characteristic of living organisms is the process of producing offsprings, possessing features similar to those of parents. In multicellular organisms, the mode of reproduction is generally sexual. Living organisms also reproduce by asexual means.
Some examples are given below
(i) Fungi spread and multiply fast by producing millions of asexual spores. Some fungi, the filamentous algae and the protonema of mosses multiply by fragmentation.
(ii) In yeast and Hydra, budding occurs to produce new organisms. While, in Planaria (flatworm),
regeneration of fragmented body parts occur. These parts inturn grow as a new organism.
(iii) Unicellular organisms like bacteria, algae and Amoeba reproduce by increasing the number of cells, i.e., through cell division (growth is synonymous with reproduction).
Some organisms like mules, sterile worker bees, infertile human couples, etc., do not reproduce. Hence, reproduction also cannot be an all-inclusive defining characteristic of living organisms.
3 Metabolism
Metabolism is an another characteristic and defining feature of all living things. The sum total of anabolic or constructive reactions (anabolism) and catabolic or destructive reactions (catabolism) continuously occurring inside the body is called metabolism.
Metabolism —> Anabolism + Catabolism Metabolism occurs in all unicellular and multi cellular organisms. Its two stages include, i.e., anabolism, the process of building up or synthesis of complex substances from simpler ones, e.g., Photo synthesis and catabolism, the process of breakdown of complex substances into simpler substances, e.g., Respiration, releasing waste outside.
Metabolic reactions can also be demonstrated outside the body in cell free systems, which are neither living nor non-living. Thus, these reactions in vitro are surely living reactions not living things. Hence, metabolism can be considered as a defining feature of all living organisms without exception.
The important differences between anabolism and catabolism are
living-world-cbse-notes-class-11-biology-2
Viruses are considered as non-living because they don’t need energy for their activities, i.e., metabolic activities are altogether absent in them.
4 Cellular Organisation
The cells are the building blocks of all living things whether plants, animals or humans. The unicellular organisms are made of a single cell, while multi cellular organisms are formed by millions of cells. The cells contain protoplasm (living matter) and cell organelles (inside the cells) which perform several activities at the cellular level and result into various life processes.
5 Consciousness
All living organisms have excellent ability to sense their environment. They respond to various physical, chemical and biological stimuli.
The various external factors to which living organisms respond are light, water, temperature, pollutants, other organisms, etc. Light duration or photo period affects many seasonal breeders, plants as well as animals. All living things respond to chemicals, entering their * bodies.
Humans are superior to all living things as they have an additional ability of self-consciousness. Therefore, consciousness can also said to be a defining property of living organisms.
However, in human beings, it is more difficult to define living state, e.g., Patients lying in coma supported by machines that replace heart and lungs, are brain-dead with no self-consciousness.
6 Body Organisation
The body of living organisms is organised, i.e., several component and sub-components cooperate with each other for the functioning of whole body.
living-world-cbse-notes-class-11-biology-3
Physical and Biological Hierarchies
There is a physical (non-living) hierarchy and biological hierarchy in the organisation of living body. In physical hierarchy, various non-living components aggregate to form compounds, which finally enter the living world in the form of cells. These cells organise to form tissues, that form organs and several organs combustive to form organ-systems. Finally, many organ systems organise and form a living organism.
The properties of tissues are not present in the constituent cells but arise as a result of interactions among the constituent cells. For example, bone is a hard tissue, which provides framework to the body. But, the cells present inside it do not have this property. This phenomenon of interactions between various components of the body results in the hierarchy of organisation.
The various life processes are the result of this interaction and coordination. The complexity in organisation enable living organisms as to be self-replicating, evolving, self-regulating and responding to external stimuli. All living organisms along with their ancestors and descendants are linked to one another by sharing of common genetic material in the form of DNA in varying degrees. This DNA is responsible for the expression of specific traits in organisms. Thus, Biology is the story of life on earth. It is the story of evolution of living organisms on the earth.
Some Other Characteristics of Living Organisms
We have discussed some important and defining characteristics of living things. However, organisms . also have many other features that differentiate them from non-living things, such as, shape & size, life cycle, movement, self-regulation, variations, adaptations, healing & repair, excretion and death.

2.Living World : Diversity and Taxonomy

The earth hosts an immense variety of living organisms. According to a survey, the number of species that are known and described are between 1.7-1.8 million.
This number refers to the biodiversity on the earth. The term Biodiversity or Biological diversity means the number and types of organisms present on the earth, forms of life in the living world. The living world includes all the living organisms, such as microorganisms, plants, animals and humans.
Biodiversity is not limited to the existing life forms. If we explore new areas and even old ones, new organisms are continuously being added. This huge available variety cannot be studied and identified without having a proper system of classification and nomenclature.
Systematics
The word ‘Systematics’ is derived from the Latin word Systema, which means systematic arrangement of organisms. Linnaeus used Systema Naturae as the title of his book. He. coined the term Systematics in 1751.
Systematics is the branch of science that deals with unique properties of species and groups to recognise, describe, name and arrange the diverse organisms according to an organised plan.
In 1961, Simpson, defined systematics as the study of diversity of organisms and all their comparative and evolutionary relationships based on comparative’ anatomy, physiology, biochemistry and ecology. The word ‘Systematics’ and ‘Taxonomy’ are often used interchangeably by the biologists. Systematics includes the following:
Identification
It aims at finding the correct name and appropriate position of an organism. The morphological and anatomical characters are examined for proper identification.
Classification
It is almost impossible to study all the living organisms. So, it is necessary to devise some means to make this possible. This can be done by classifying the organisms.
Thus, classification is the process by which organisms are grouped into categories based on some easily observable characters.
Biological classification is the scientific arrangement of organisms in a hierarchy of groups and sub-groups on the basis of similarities and differences in their traits.
Advantages of Classification
(a) It helps to identify an organism easily.
(b) New organisms easily get correct place in their respective groups.
(c) It makes study of fossils easy.
(d) It also helps in building evolutionary pathways.
(e) It becomes easy to know the features of whole group by studying one or two organisms of the group.
Thus, based on these characteristics, all living organisms are classified into different taxa.
Nomenclature
Nomenclature is the system of naming living organism in a way that a particular organism is known by the same name all over the world.
i. Common Names
The common names or vernacular names are the local names given to an organism in a specific language in a particular region. There are different names of a same organism in different regions even with in a country.
living-world-cbse-notes-class-11-biology-4
Advantages of Common Names
(a) Common names are easy to pronounce and are short, e.g., Cat or billi.
(b) People are familiar to these names since childhood.
(c) They are based on some features of organisms, e.g., Cowa (crow—Caawn-Caawn sound).
Dis-Advantages of Common Names
(a) All the organisms cannot be named by this method as there are organism of different sizes and shapes.
e.g., Microbes.
(b) An organism may have several names in a given language, e.g., 8 Hindi names of prickly poppy and water lily has 15 English names.
(c) A common names may have different meanings in different countries, e.g., Maize, means wheat and other grains in USA and it is called corn in common wealth countries.
(d) Common names may have little relevance, e.g., Lady’s finger (okra), widows tears (Tradescantia-Rhoeo), etc.
(e) Common names may be incorrect, e.g., Jelly fish (a coelenterate), silverfish (an arthropod), starfish (an echinoderm) are not real fishes.
(f) These names are not useful for scientific studies.
ii- Scientific Names
A scientific name is given by biologists. These names represent a particular organism in every part of the world. The system of providing scientific names is called binomial nomenclature.
The scientific names must be
(a) acceptable in every part of the world.
(b) assigned on agreed principles and criteria.
(c) different for each species and not used for other organisms earlier.
Binomial Nomenclature
Binomial nomenclature was developed by Carolus Linnaeus in 1751 (Philosphica Botanica). All scientific names for animals under binomial nomenclature were given by Linnaeus in the tenth edition of his book Systerna Naturae (1758). Linnaeus named plants according to binomial nomenclature in his book Species Plantarum (1753). Binomial nomenclature is the system of providing distinct and appropriate names to organisms, each consisting of two words, first generic name {i.e., name of genus) and second specific epithet (i.e., name of species).
For example, Scientific name of mango is written as Mangifera indica. In this name, Mangifera represents the genus and indica is a particular species or specific epithet.
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Rules of Binomial Nomenclature
Rules of binomial nomenclature were initially framed by Linnaeus in his books, Species Plantarum and Systema Naturae.
The rules were revised again by the following nomenclature codes
(i) International Code for Botanical Nomenclature (ICBN).
(ii) International Code of Zoological Nomenclature (ICZN).
(iii) International Code of Bacteriological Nomenclature (ICBN).
(iv) International Code of Viral Nomenclature (ICVN).
(v) International Code of Nomenclature for Cultivated Plants (ICNCP).
The rules framed by Linnaeus and by these codes are as follows
(i) The names are generally in Latin and written in italics. They are Latinised or derived from Latin irrespective of their origin.
(ii) The first word in a biological name represent the genus while, the second component denotes the specific epithet.
(iii) Both the words in a biological name, when handwritten are separately underlined or printed in italics to indicate their Latin origin.
(iv) The first word denoting the genus starts with capital letter while, the specific epithet starts with a small letter, e.g., Mangifera indica.
(v) Generic and common names may be same, e.g., Gorilla gorilla.
(vi) No names are recognised prior to those used by Linnaeus in 1753 for plants in Species Plantarum and in 1758 for animals in the 10th edition of Systema Naturae.
(vii) The name of categories higher than the rank of genus are not printed in italics. Bold letters can, however be used.
(viii) When a species is transferred or revised, the name of the original worker is retained but in parenthesis, e.g., Syzygium cumini (L) Skeels.
Advantages of Binomial Nomenclature
(i) Binomial names are universally acceptable and recognised.
(ii) They remain same in all languages.
(iii) The names are small and comprehensive.
(iv) There is a mechanism to provide a scientific name to every newly discovered organism.
(v) The names indicate relationship of a species with other species present in the same genus.
(vi) A new organism can be easily provided with a new scientific name.
Taxonomy
It is the science of identification, classification and nomenclature. Based on their special / characteristics, all living organisms can be classified into different taxa. This process of classification is called taxonomy. Carolus Linnaeus is known as father of taxonomy.
The basis of modern taxonomy studies are external and internal structure (comparative morphology), along with the structure of cells (cytology), development process (embryology) and ecological information of organisms (ecology). It provide information according to similarities, dissimilarities and evolutionary relationships of various organisms.
The basic processes for taxonomic studies are
(i) Organisms are described on the basis of morphology and other characteristics.
(ii) The description of characteristics helps in the placement of the organism in various taxa.
(iii) A new taxon can be framed if the organism is different from the existing taxa.
(iv) The correct naming of an organism can be done after placing it in various taxon. A new organism can be given a new name after following the standardized rules.
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Classical Taxonomy (Old Taxonomy)
The concept of classical or old taxonomy exists since, the time of Aristotle and Theophrastus and continued up to Linnaeus. It states that 4 .
(i) Species is the basic unit of taxonomy, that can be described on the basis of one or few preserved specimens.
(ii) Species are fixed and do not change with time.
(iii) A species is delimited based on morphological features.
(iv) Organisms are classified on the basis of some limited features such as root modification, leaf venation, floral structures, number of cotyledons in case of plants.
Due to the limited number of groups, many organisms could not be classified correctly. This finally led to artificial system of classification.
Modern Taxonomy (New Taxonomy)
The concept of modern taxonomy was given by Julian Huxley (1940). It uses evidences from all the areas of biology like morphology, anatomy, biochemistry, cell biology, physiology, genetics, evolution, etc.
The modem taxonomy is based on the following features
(i) The studies are done on a huge number of organisms based on all the variations.
(ii) Study is also focused on sub-species, varieties, races and populations.
(iii) Species are not isolated. They are related by common descent and vary from them due to accumulation of variations.
(iv) Species is considered as dynamic and ever-changing entity.
(v) Biological delimitation includes various branches of systematics, e.g., Cytotaxonomy, experimental taxonomy, numerical taxonomy, chemotaxonomy, etc. This led to the development of phylogenetic system or cladistics of classification.
Taxonomic Categories
Classification is not a single step process. It involves hierarchy of steps in which each step represents a rank or category. Since, the category is a part of overall taxonomic arrangement, it is called the taxonomic category and all categories together constitute the taxonomic hierarchy.
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Taxon
Each category, referred to as a unit of classification, in fact, represents a rank and is commonly termed as taxon (Pi. taxa). The term Taxon was first introduced by ICBN during 1956.
According to Mayr (1964) taxon is a group of any rank that is sufficiently distinct to be worthy of being assigned a definite category. In simple words, taxon refers to a group of similar, genetically related individuals having certain characters distinct from those of other groups.
A taxon that includes a common ancestral species and all the species descended from it is called a clade or a monophyletic taxon.
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Taxonomic Hierarchy
The taxonomic hierarchy is the system of arranging taxonomic categories in a descending order. It was first introduced by Linnaeus (1751) and hence, it is also known as Linnaen hierarchy.
Groups represent category and category further denotes rank. Each rank or taxon represents a unit of classification.
These taxonomic groups/categories are distinct biological entities and not merely morphological aggregates.
Obligate/Common Categories
The taxonomic categories, which are always used in hierarchical classification of organisms are called obligate or common categories.
They are seven in number. In descending order, these are kingdom, phylum or division, class, order, family, genus and species.
All the members of taxonomic categories possess some similar characters, which are different from those of others. The maximum similarity occurs in species, which is also the lowest category in the hierarchy of categories. Similarity of characters decreases with the rise in hierarchy.
i. Species
Taxonomic studies consider a group of individual organisms with fundamental similarities as a species (John Ray).
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Species is considered as the lowest or basic taxonomic category, which consists of one or more individuals of a populations that resemble one another more closely than individuals of other species. The members of species interbreed freely and are reproductively isolated from others. For example, Mangifera indica (mango), Solarium tuberosum (potato) and Panthera leo (lion).
All the three names indica, tuberosum and leo represent the specific epithets while, the first words Mangifera, Solanum and Panthera are genera and represents another higher level of taxon or category.
Each genus may have one or more than one specific epithets representing different organisms, but having morphological similarities. For example, Panthera has another specific epithet called tigris and Solanum includes species like nigrum and melongena.
ii.Genus
Genus (John Ray) comprises a group of related species, which has more characters common in comparison to species of other genera. In other words, genera are aggregates of closely related species.
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iii. Family
Family (John Ray) is a group of related genera with less number of similarities as compared to genus and species. All the genera of a family have some common or correlated features. They are separable from genera of a related family by important differences in both vegetative and reproductive features.
A plant family ends in a suffix -aeae and sub-family -oideae. While, an animal family has a suffix -idae and sub-family -inae.
iv. Order
An order (Linnaeus) is a group of one or more related families that possess some similar correlated characters, which are lesser in number as compared to a family or genera.
Plants and Animal Orders with their Respective Families
Order Animals and Families
Carnivora Canidae (dog, wolf and fox), Felidae (cat, leopard, tiger and lion), Ursidae (bear) and Hyaenidae (hyaena)
Polemoniales Solanaceae (potato and tomato), Convolvucaceae (sweet potato and morning glory), Polemoniaceae (herbs, shrubs and small trees) and Hydrophyllaceae (water leaf).
Primates Lemuridae (lemurs), Cebidae (new world monkeys), Pongidae (apes) and Hominidae (humans).
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v. Class
Class (Linnaeus) is a major category, which includes related orders. For example, order-Primata comprises monkey, gorilla & gibbon and is placed in class—Mammalia along with order—Carnivora that includes animals like tiger, cat and dog.
Class-Mammalia has other orders also.
vi.Phylum or Division
Phylum or Division (Cuvier, Eichler) is a taxonomic category higher than class and lower” in rank to kingdom. The term Phylum is used for animals, while division is commonly employed for plants.
It consists of more than one class having some similar corelated characters.
For example, Phylum— Chordata of animals contain following classes, e.g., Pisces, amphibians, reptiles, aves and mammals.
vii. Kingdom
It is known to be the highest category in taxonomy. This includes all the organisms, which share a set of distinguished characters. For example, all the animals belonging to various phyla are assigned the highest category called kingdom.
For example, Animalia in the classification system of animals. Similarly, all the plants are kept in kingdom—Plantae.
RH Whittaker. (1969) assigned five kingdom classification of organisms.
These are Monera, Protista, Fungi, Plantae and Animalia.
Intermediate Categories
The taxonomic categories from species to kingdom are broad categories or obligate categories. However, taxonomists have also developed sub-categories in this hierarchy to facilitate more sound and scientific placement of various taxa. These sub-categories are sub-species (or varieties), sub-genera, sub-families, sub-orders, sub-classes and sub-phyla.
These sub-categories are referred to as intermediate categories.
Taxonomical Aids
Taxonomical aids are techniques and procedures to store information as well as specimens or identification and classification of organisms.
The taxonomic studies of various plants, animals and other organisms are useful in areas like agriculture, forestry, industry and knowing our bioresources. All these studies need correct identification and classification of organisms. Identification of organisms requires intensive laboratory and field studies. The collection of actual specimens of plants and animal species, knowing their habitats and other traits are essential and are the prime source of taxonomic studies. All this information is used in classification of an organism and is also stored along with the specimens. Sometimes, specimens are also preserved for future studies.
Some of the taxonomical aids developed by Biologists include Herbarium, Botanical gardens, Museum, Zoological parks, Key, etc.
1 Herbarium
Herbarium (Pi. Herbaria) is a store house of collected plant specimens that are dried, pressed and preserved on sheets. These sheets are arranged further according to a universally accepted system of classification. The institutes and universities maintain their own herbarium by collecting specimens from local and far away places.
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Uses of Herbaria
The uses of herbaria are listed below
(a) These are used for identification of plants.
(b) Compilation of floras, monographs and manuals are mainly based on the specimens in herbaria.
(c) Herbaria are useful in locating wild varieties and relatives of economically important plants.
(d) They help in knowing the morphological variations found in species.
(e) Herbaria are useful for research in plant taxonomy, morphology, ecological distribution, etc.
2. Botanical Gardens
Botanical gardens are specialised gardens that have collections of living plants for reference. These gardens generally have facilities like library, laboratory, herbarium and museum. The botanical gardens are maintained by government, semi-government and other private organisations. Botanists and gardeners look after plants in botanical gardens.
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Role of Botanical Gardens
A botanical garden has following important roles
(a) Botanical gardens have aesthetic appeal and provide recreation facility to people.
(b) A wide variety of plant species grow there, so they provide ready material for research.
(c) These gardens also play an important role in conservation of endangered plant species and genetic diversity.
(d) There are more than 500 botanical gardens all over the world. These provide free exchange of seeds.
(e) These improve the environment, provide greenery, help in creating pollution free environment and some serves as habitat for animals.
Knowledge Plus
Indian Botanical Garden-Largest Botanical Garden of Asia.
First Botanical Garden-Pisa Botanical Garden, Italy established by Luca Glini (1490-1556).
3. Museums
Museum is a place for collections of preserved plants and animal specimens for study and reference. The universities and educational institutes maintain their own museums in their botany and zoology departments. Plants, which cannot be kept in herbaria are preserved in museums.
For example, algae, fungi, mosses, ferns, fruits, etc. Specimens are preserves in containers or jars in preservative solutions. Plant and animal specimens may also be preserved as dry specimens. Insects are preserved in insect boxes after collecting killing and pinning. While, the larger animals are stuffed and preserved in skeletal forms.
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4. Zoological Parks
Zoological parks or zoo are the places where wild animals are kept in protected environments under human care and which enable us to learn about their food habits and behaviour. Zoological parks provide natural habitat to the animals.
In India there are about 200 zoological parks. These zoos are managed by the Central Zoo Authority of India. The World Zoo Conservation Strategy (WZCS) refer to all these zoological institutions as zoos.
Role of Zoological Parks
(a) The zoological parks increase understanding of wildlife.
(b) These are the centres for recreation and education.
(c) Zoos are the centres for conservation of threatened and rare animal species.
(d) These provide sites for ex situ breeding of endangered animals. conservation through captive breeding of endangered animals.
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5. Key
Key is also a taxonomical aid used for identification of plants and animals based on the similarities and dissimilarities.
It helps in the identification of plants and animals by selecting and eliminating the characters according to their presence or absence in the organism under study.
The keys generally use two contrasting characters called couplet. This results in acceptance of one present in organism and rejection of the other. Each statement in the key is called a lead.
These taxonomic keys are of two types
Indented Key
The indented key or yolked key provides a sequence of choices between two or more characteristics. By careful selection of characters at each sub-division, the exact name of the organism can be arrived at.
Bracketed Key
The bracketed key also uses contrasting characters like the indented key. But in, these characters are not separated by intervening sub-dividing characters. Each character in this case is given a number in brackets.
Other Means of Recording Descriptions
Apart from the all mentioned means of keeping records of description. Some other means are also present.
These are of following types
Flora
Floras are the important resource that provide information on the taxonomy, nomenclature and descriptive data for the taxa covered.
The floras also include information on the biology, distribution and habitat preferences of the taxa, as well as illustrations, identification keys and other notes. These provide index to the plant species found in a particular area.
Manuals and Catalogues
These are other means of recording descriptions. They also help in correct identification. Manuals are useful in providing information for identification of names of species found in an area.
Monograph
A monograph is a comprehensive treatment of a taxon in biological taxonomic studies. These contain information on any one taxon. Monographs revise all known species within a group, add any newly discovered species, collect and organise available information on the ecological associations, geographic distributions and morphological variations within the group.
The first ever monograph of a plant taxon was given in Robert Morison (1672) Plantarum Umbelliferarum Distributio Nova.

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Plant Kingdom – CBSE Notes for Class 11 Biology

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Plant Kingdom – CBSE Notes for Class 11 Biology

CBSE NotesCBSE Notes Class 11 BiologyNCERT Solutions Biology

1.Basis of Classification and Algae, Bryophytes & Pteridophytes

Our understanding of the plant kingdom has changed over time. Fungi and members of the Monera and Protista having cell walls have not been separated form Plantae, the earlier classifications kept them in the same kingdom. The kingdom-Plantae has been described under algae, bryophytes, pteridophytes, gymnosperms and angiosperms. The overview of this classification is demonstrated here in the flowchart.
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Types of Classification System
These includes artificial system, natural system and phylogenetic system of classification.
The various systems used in classification ofiplants are being discussed here
1. Artificial System of Classification
This system is based on comparison of one or a few superficial characteristics, which are helpful in easy identification of organisms. This system remained in use for about two thousand years.
Aristotle is known as father of Zoology.
Carolus Linnaeus is known as father of Taxonomy.
Advantages
Advantages of artificial system as below
(i) Artificial system is easy to remember as only one or few characters are used.
(ii) The traits used are of interest to humans.
Disadvantages
Disadvantages of artificial system are given below
(i) This system use only few superficial characters (i.e., habits, numbers, colours and shapes of leaves, etc) which leads to many orgarfisms grouped together,
(ii) They considered mainly the vegetative characters or the androecium characters as given by linnaeus.
(iii) It does not demonstrate natural and phylogenetic relationships.
(iv) This gave equal weightage to vegetative and reproductive/sexual characters this is not acceptable, as vegetative characters are more easily influenced by the environmental factors.
(v) They sparated the closely related species.
2 Natural System of Classification
It is also known as phenetic system of classification. The natural system of classification is based on natural affinities among the organisms. It considers both external and internal features like structure, anatomy, embryology and phytochemistry.
Advantages
Advantages of natural system are given below
(i) Only related organisms are kept in a group.
(ii) Unrelated organisms are kept in separate groups.
(iii) It shows natural relationships among the organisms.
(iv) It shows possible origin of different taxa.
Disadvantages
Disadvantages of natural system are given below
(i) There is more emphasis given on natural character.
(ii) In this system several related families are separated and unrelated families are put together.
(iii) Evolutionary basis is neglected.
3. Phylogenetic System of Classification
The phylogenetic system of classification indicates the evolutionary as well as genetic relationships among organisms. This system is based on fossil records of biochemical, anatomical, morphological, physiological, embryological and genetical.
The system was initiated by Engler and PrantI (1887-1899) in Die Naturalichen Pflanzefamilien. In phylogenetic system, flowering plants are placed in ascending series related to complexity of floral morphology. The phylogenetic system of classification are mainly the rearrangement of taxonomic characters in addition to the phylogenetic information.
Advantages
Advantages of phylogenetic system are given below
(i) Families and order in this system are of small size.
(ii) This system is in conformation with the modern views of phylogeny.
(iii) They use information from various sources to solve problems of classification. Such informations become more important in the absence of supporting fossil evidences.
Disadvantages
Disadvantages of phylogenetic system are given below
(i) This is not helpful in plant identification.
(ii) The classification is outdated as the habit is used as . main basis of classification.
Types of Taxonomies
For the suitability in studies various categorizations has been done in taxonomy.
The important taxonomies are as follows
1. Numerical Taxonomy
It is carried out by quantitative assessment of similarities and differences in order to make objective assessments. It is now easily carried out using computers based on all observable characteristics. Number and codes are assigned to all the characters and the data are then processed. In this way, each character is given equal importance and at the same time hundreds of characters can be considered.
2. Cytotaxonomy
Cytotaxonomy is based on cytological information like chromosome number, structure, behaviour and type of chromosomes. For example,
(i) Chromosome number is constant for a species, e.g., 46 in man, 48 in apes and potato, 20 in maize, 16 in onion and 8 in Drosophila.
(ii) Behavior of chromosomes was used by taxonomists during pairing and banding patterns to understand the relationships between species, e.g, Origin of humans from apes, origin of wheat, etc.
3. Chemotaxonomy
It is the system based on the evidences from chemical constituents (enzymes, hormones, proteins, amino acids, etc.), some specific chemicals (usually secondary metabolites) and the chemical nature of proteins have been utilised to establish similarities and relationships. For example, the presence of raphides has been found to be common in 35 families of plants.
Algae
Algae are chlorophyll-bearing, simple, thalloid, autotrophic and largely aquatic (both freshwater and marine) organisms.
The important salient features of algae are given below
Habitat
These are found in both freshwater and marine habitats. Some algal forms are also found in moist habitats like wet rocks and soil, tree trunks, etc. Some of them are also found in close association with fungi (lichen) and animals {e.g., on sloth bear).
Thallus Organisation
The plant body (thallus) is without differentiation. The basic form and size of algae is highly variable, and ranges from filamentous {e.g., Ulothrixand Spirogyrd) to colonial {e.g., Volvox). These are attached on the substratum with the help of holdfast.
Structure of Algal Cell
It has an eukaryotic plant cell structure. The cytoplasm contains membrane bound chloroplast, mitochondria, ER, Golgi bodies and other cell organelles. The cell wall contains cellulose. The nucleus contains nuclear membrane.
Food Material
Algae have main food reserve as starch. In brown algae, mannitol and laminarin are the main reserve food material, whereas in red algae floridean starch is the reserve food material.
Reproduction
The algae reproduces vegetatively, asexually and sexually.
i. Vegetative Reproduction
It occurs by fragmentation, tubers, stolons, adventitious branches, etc. Each fragment gets develop into a thallus.
ii. Asexual Reproduction
It occurs by a number of accessory spores, such as zoospores aplanospores, akinetes, carpospores, etc.
The most common being the zoospores, which are flagellated. The cells which produce spores are called sporangia. The sporangia may be the vegetative cells (e.g. in Chlamydomonas and Ulothrix) or modified vegetative cells (e.g. in Vaucheria).
iii. Sexual Reproduction
It occurs by fusion of two gametes.
On the basis of morphology of reproductive cells, sexual reproduction is of two types
(a) Isogamy In this method, two morphologically similar gametes fuse to form a zygote, e.g., Spirogyra.
(b) Heterogamy In this process, fusion occurs between morphologically as well as physiologically different gametes. It is of two types
Anisogamy It is the fusion of structurally dissimilar gametes, which differ in size and – behaviour. Male gamete is more active and female gamete is less active and bigger in size, e.g., Some species of Chlamydomonas.
* Oogamy In this process, the male gamete is motile, active, small and without reserve food.
The female gamete is bigger, passive, non-motile and laden with food, e.g., Volvox, Fucus.
Embryo and Life Cycle
An embryo stage is not present. Life cycle is haplontic, diplontic, diplohaplontic, haplohaplontic, etc. An alternation of generation occurs in diplohaplontic life cycle.
Economic Importance of Algae
(i) Algae are responsible for carrying out about a half of the total carbon dioxide fixation on earth by the process of photosynthesis.
(ii) Some forms of marine brown and red algae produce large amount of hydrocolloids. These are algin (brown algae) and carrageenan (red algae) which have many commercial uses.
(Hi) The algae Gelidium and Gracilaria are used to produce agar, which is used in preparation of ice creams and jellies.
(iv) Some protein rich algae, like Chlorella and Spirullina are used as food supplements by sailors and space travellers.
(v) About 70 species of marine alga are used as food, e.g., Porphyra, Laminaria and Sargassum.
Classification of Algae
Algae are divided into various classes based on pigmentation, stored food and flagellation. The three main classes are
Chlorophyceae, Phaeophyceae and Rhodophyceae.
i.Class-Chlorophyceae (Green Algae)
The members of Chlorophyceae are commonly called green algae.
There are about 7000 species in this class. The reserve food material is starch.
The characteristic features of Chlorophyceae are discussed below
(a) Habitat These are mostly marine forms, only some are freshwater. Chlorella can tolerate moderately warm waters. Snow dweling forms are called cryophytes, e.g., Chlamydomonas nivalis, Scotiella, etc.
(b) Cell Organisation These are unicellular, colonial, coenocytic and multicellular forms. Cell walls contain cellulose (inner layer) and pectose (outer layer) in most of the green algae. The chloroplasts may be discoid, plate-like, reticulate, cup-shaped, spiral or ribbon shaped.
The chloroplasts contain pigments. Most of the members have one or more storage bodies called pyrenoids located in” the chloroplasts. Pyrenoids contain protein besides starch. Some algae may store food in the form of oil droplets also.
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(c) Thallus Unicellular green algae can- be flagellate, (Chlamydomonas), unicellular, non-flagellate (Chlorella).
Acetabularia (umbrella plant) has unicell upto 10 cm long with distinction of nucleus containing rhizoid, elongated stalk and umbrella like cap.
A colony for fixed number of individual unicells (Volvox) is called coenobium. Coenocytic or siphonaceaus thallus occurs in Caulerpa. Unbranched filamentous thallus in Ulothrix and Spirogyra, Cladophora, heterotrichous in Stigeoclonium and parenchymatous in Ulva.
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(d) Reproduction The members of Chlorophyceae reproduce vegetatively, asexually and sexually by various methods.
* Vegetative reproduction occurs through cell division (unicellular forms), fragmentation, stolons, tubers, storage cells, etc.
* Asexual reproduction occurs by zoospores, aplanospores, hypnospores, akinetes and daughters colonies.
* Sexual reproduction may be isogamous, anisogamous or oogamous.
(e) Life Cycle It can be haplontic, diplontic and diplohaplontic. In haplontic life cycle, there is a single somatic phase, which is haploid. Diploid stage is represented by a single cell or zygote, e.g., In Spirogyra.
ii. Class-Phaeophyceae (Brovin Algae)
The members of Phaeophyceae are fucoxanthin and phycocolloid rich multicellular eukaryotic algae. Its common members are seaweeds called kelps. This class is comprised of about 2000 species.
(a) Habitat Brown algae are mostly marine. These are found mostly in colder seas or during cold seasons in tropical regions.
(b) Size They range from simple branched, filamentous forms (Ectocarpus) to profusely branched forms as represented by kelps. These represent largest algae. The largest kelps are Macrocystis (40-100m) and Nereocystis (20-30 m).
(c) Cell Organisation All members are multicellular. Cell wall is composed of cellulose, pectose and phycocolloids. The cellulosic wall of vegetative cells is usually covered on the outside by a gelatinous coating of algin.
(d) Thallus It is heterotrichous filament with both prostrate and upright branches (Ectocarpus). The parenchymatous structure is found in higher forms.
The plant body of large forms often differentiated into holdfast (with which it usually attaches to the substratum) a stalk called the stipe and lamina (frond), which is photosynthetic. Conducting tubes or trumpet hyphae are present in larger brown algae or kelps. They help in conduction of food materials.
(e) Photosynthetic Pigments and Colour These include chlorophyll-4, c and carotenoids. They ranges in colour from olivt green to various shades of brown depending upon the amount of the xanthophyll pigment, fucoxanthin present in them.
(f) Food Reserve It remains in the form of complex carbohydrates such as laminarin or mannitol.
(g) Flagellation These contain heterokont flagellation with one smooth (whiplash) and one tinsel flagella.
(h) Reproduction Vegetative reproduction occurs through fragmentation (e.g., Sargassum), adventitious branches and stolons (e.g, Dictyota). Asexual reproduction by biflagellate zoospores, which are pear-shaped having two unequal laterally attached flagella.
Sexual reproduction is performed by isogamy, anisogamy and oogamy. Union of gametes may take place in water or within the oogonium (oogamons species). The gametes are pyriform (pear-shaped) and have two laterally attached flagella.
(i) Life Cycle Isomorphic alternation of generation is found in some brown algae, e.g, Ectocarpus, Dictyota. In many brown algae, the diploid generation or phase is dominant. The haploid phase is either microscopic or represented by gametes only (e.g., Fucus).
Economic Importance of Pheophyceae
* The brown algae that are used as food are Laminaria, Nerocystis, Macrocystis, Alaria, etc.
* Some brown algae like Fucus, Sargassam, Laminaria, Macrocystis are important fodder for catde.
* Brown algae are collected from sea shores and used as manure. They improve mineral content of soils.
* Alginic acid is a phycocolloid obtained from a number of brown algae such as Alaria, Macrocystis, Ascophyllum, Laminaria. It is used as emulsifier, thickener, gelating agent in toothpastes, shaving creams, ice-creams, emulsion paints, shampoo, cosmetics, etc.
* Fucus and Laminaria are rich in iodine.
iii. Ciass-Rhodophyceae (Red Alqae)
The members of Rhodophyceae are commonly called red algae because of the predominance of the red pigment,i.e., r-phycoerythrin in their body.
(a) Habitat Most of the red algae are marine with greater concentrations found in the warmer areas. They are found in both well lighted regions close to the surface of water and also at great depths in oceans where relatively little light penetrates.
(b) Thallus The red thalli of most of the algae are multicellular. Some of them have complex body organisation like Asterocystis is pseudofilamentous, Porphyridium is unicellular, Porphyra has parenchymatous sheets, Cbondrus is ribbon like, Gelidium is a multicellular sea weed.
(c) Cell Wall The cell wall contains cellulose, pectic compounds and certain mucopolysaccharides called phycocolloids, such as agar, carrageenin, etc. In many algae, cell wall contains pits.
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(d) Photosynthetic Pigments These include chloro¬phyll-4, carotenes, xanthophylls and phycobilins. Phycobilins are water soluble and are of two types, i.e., red-coloured phycoerythrin and blue-coloured phycocyanin.
(e) Reserve Food It is floridean starch similar in constitution to glycogen and amylopectin. if) Reproduction Vegetative reproduction occurs by fragmentation, regeneration of hold fast and gemmae. Asexual reproduction occurs by non-motile spores (carpospores, monospores, tetraspores and neutral spores).
Sexual reproduction occurs by non-motile gametes and is oogamous type. The male sex organs is called spermatogonium or antheridium. The male produced is non-flagellated, called as spermatium. The female sex organ is called carpogonium. After fertilisation, a new structure called carposporophyte is produced. It remains attached to the parent alga.
(g) Life Cycle Life cycle has two or more phases such as haplohaplontic, haplohaplohaplontic, diplodiplohaplontic, etc.
Economic Importance of Red Algae
* The red algae like Porphyra, Chondrus, Rhodymenia, Centerella and Bostrychia are used as food in various parts of the world.
* Agar yielding algae are called aerophytes, such as Gelidium, Gracilaria, Ceramium, Gelidiella, etc.
* Alga like Rhodymenia are used as fodder for cattle.
* Carrageenin a phycocolloid obtained from red algae like Chondrus and Gigartina is used in preparations of emulsions for ice cream, chocolates, sauces, toothpastes, cosmetics, etc. It is also used in clearing liqueurs and finishing leather, etc.
* Funori an adhesive phycocolloid is obtained from red alga Gloiopeltis. It is used in sizing textiles, paper and as glue.
* Some algae like Corallina, Polysiphonia have medicinal properties.
Bryophytes
Bryophytes include the various mosses and liverworts. These are non-vascular embryophytes, characterised by the presence of. an independent gametophyte and parasitic sporophyte.
Habitat
Bryophytes commonly grow in -moist, shaded areas in hills. These are also called amphibians of the plant kingdom because, these can live in soil but are * dependent on water for sexual reproduction.
Rhizoids
These are attached to the substratum by unicellular or multicellular rhizoids.
Sex Organs
The main plant body produces gametes, hence it is called gametophyte. The sex organs in bryophytes are multicellular, the male sex organs in bryophytes is called antheridium (which produces biflagellate antherozoids) and the flask-shaped female sex organ is called archegonium (produces a single egg).
Reproduction
It is of sexual type reproduction. The antherozoids are released into water where they come in contact with archegonium.
Fertilisation
It occurs inside the archegonium. The egg secretes a chemical which attracts spermatozoids. Sperms require a thin film of water for swimming and reaching the dehisced archegonium. One sperm fuses with an egg and produces a diploid zygote. Zygotes do not undergo reduction division immediately instead, they produce a multicellular body called sporophyte.
Dependent Sporopbvrc The sporophyte is not free-living but attached to the photosynthetic gametophyte deriving nourishment from it. Some cells of the sporophyte undergo reduction division (meiosis) to produce haploid spores (which germinate to produce gametophyte).
The sporophyte of bryophytes is called sporogonium because it is mainly dependent and meant for producing spores.
Life Cycle
Bryophytes have heteromorphic or heterologous alternation of generation. The gametophyte may be produced directly or first from a juvenile stage called protonema.
Economic Importance
Bryophytes in general are of little economic importance. But, several species have some uses.
(i) Some mosses provide food for herbaceous mammals, birds and other animals.
(ii) Species of Sphagnum (a moss), provides peat that
have long been used as fuel. It has the capacity to hold water as packing material for transshipment of living material.
(iii) Mosses along with lichens are the first organisms to colonise rocks. Hence, these help in biological succession. They decompose rocks making the substrate suitable for the growth of higher plants.
(iv) Mosses from dense mats on the soil, they reduce the impact of falling rain and prevent soil erosion.
(v) Marchantia has medicinal properties to cure lungs and liver infections. It also has antitumour properties.
Bryophytes do not attain great heights. They may range of 0.4 to 70 cm and have thalloid body.
The reasons may be following
(i) Root is absent.
(ii) Vascular tissues are not present.
(iii) Cuticle is not present on the plant body.
(iv) Absence of mechanical tissue.
(v) Male gametes need to swim upto the interior of dehisced archegonia.
Types of Bryophytes
The bryophytes are generally classified into two main groups
i. Liverworts
Liverworts (Hepaticopsida) usually grow in moist, shady places such as banks of streams, marshy ground, damp soil, bark of trees and deep in the woods.
(a) Thallus The plant body of a liverwort is thalloid, e.g, Marchantia,. The thallus is dorsoventral and closely appressed to the substrate. The leafy members have tiny leaf like appendages in two rows on the stem like structure.
(b) Rhizoids These are unicellular. These may be of two types in some liverworts, i.e. smooth walled and tuberculate.
(c) Reproduction It may occur both by asexual and sexual means.
• Asexual Reproduction It occurs by fragmentation of thalli or by the formation of specialised structures called gemmae (sing, gemma). Gemmae are green, multicellular, asexual buds which develop in small receptacles called gemma cups located on the thalli. The gemmae become detached from the parent body and germinate, forming new individuals.
• Sexual Reproduction During this male and female sex organs are produced either on the same or on different thalli. The sporophyte is differentiated into a foot, seta and capsule. After meiosis, spores are produced within the capsule.
• These spores germinate to form free-living gametophytes, e.g., Riccia, Marchantia, Pellia, Porella, etc.
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ii. Mosses
Mosses (Bryopsida) grow in dense mats over moist shady places, especially during rains. Some mosses grow in desert bogs and streams.
(a) Plant Body The predominant stage of the life cycle of a moss is the gametophyte which consists of two stages the first stage is the protonema stage which develops directly from a spore. It is a creeping, green, branched and frequently filamentous stage; the second stage is the leafy stage which develops from the secondary protonema as a lateral bud. They consist of upright, slender axes bearing spirally arranged leaves. This stage bears sex cells.
(b) Rhizoids These are long, multicellular branched structures with oblique septa. They take part in fixation and absorption of water. However, surface conduction through capillarity is an important mechanism of water supply to aerial parts.
(c) Reproduction This can occur both by vegetative and sexual means.
* Vegetative Reproduction It occurs by fragmentation and by budding in the secondary protonema from exposed rhizoids and other parts (like gemmae, buds and tubers).
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Sexual Reproduction Sex organs, antheridia and archegonia are produced at the apex of the leafy surface. Male organs appear cup-shaped, while female organs are bud-like. After fertilisation, the zygote develops into a sporophyte, consisting of foot, seta and capsule (containing spores).
(d) Sporophyte The sporophyte in mosses is more elaborate than that in liverworts. Spores are formed after meiosis. The mosses have an elaborate mechanism of spore dispersal, e.g., Funaria, Polytrichum and Sphagnum.
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Pteridophytes
Pteridophytes are primitive seedless vascular plants also called cryptogams. These have conspicuous sporophytic plant body, inconspicuous independent gametophytes with antheridia and partially embedded archegonia having 4-rowed necks.
The term Pteridophyte was coined by Haeckel (1866). There are about 13000 species of pteridophytes have been reported. They were perhaps the first land plants evolved during ordovician (450-500 million years ago) period.
The characteristic features of pteridophytes are as follow
Habitat
The pteridophytes are found in cool, damp shady places though some may flourish well in sandy-soil conditions.
Some members like Azolla, Salvinia, Marsilea species.
Plant Body
The size varies from a fraction of centimeter (e.g, Azolla) to 20 m in tree ferns (e.g, Angiopteris). The main plant body is a sporophyte which is differentiated into true root, stem and leaves. These organs possess well differentiated vascular tissues. The leaves are small (microphyllus) as in Selaginella or large (macrophylls) as in ferns.
Vascular Tissues
These are xylem and phloem present throughout the body. Xylem consists of tracheids and phloem is made up of sieve cells and albuminous cells.
Sporophylls
The sporophytes bear sporangia that are subtended by leaf like appendages called sporophylls. In some cases sporophylls may form distinct compact structure called strobili or cones (Selaginella and Equisetuni). The sporangia produce spores by meiosis in spore mother cells.
Spores
The spores germinate to give rise to inconspicuous, small but multicellular free-living, mostly photosynthetic thalloid gametophytes called prothallus. In majority of the pteridophytes, all the spores are of similar kinds; such plants are called homosporous. Genera like Selaginella and Salvinia which produce two kinds of spores, macro (large) and micro (small) spores; such plants are called heterosporous.
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Gametophyte
The thalloid gametophyte or prothallus require cool, damp, shady places to grow. The megaspores and microspores germinate and give rise to female and male gametophytes respectively. The female gametophytes in these plants is retained on the parent sporophytes for viable periods. In most ferns, prothallus is green and autotrophic. In heterosporous ferns, the female gametophyte depends on food stored by the megaspore.
Sex Organs
The gametophytes bear male sex organs called antheridia and female sex organs called archegonia. Antheridium is sessile and surrounded by a single layered jacket. Archegonium is flask-shaped. It is partially embedded.
Fertilisation
Water is required for transfer of antherozoids. The male gametes released from the antheridia and reach to the mouth of archegonium. Fusion of male gamete with the egg present in the archegonium result in the formation of zygote. Zygote therefore, produces a multicellular, well differentiated sporophyte, which is the dominant phase of the pteridophytes.
Embryo
Fertilisation produces a zygote that undergoes division to produce embryo. The development of the zygote into young embryo takes place within the female gametophyte.
* This event is a precursor of the seed habit and considered as an important step in evolution, e.g., Dryopteris, Selaginella, Adiantum, Equisetum and Salvinia.
Economic Importance of Pteridophytes
(i) Pteridophytes are a good source of food for animals. For example, sporocarps of Marsilea is edible. Angiopteris and Alsophila have starchy pith eaten by natives of Australia.
(ii) Ferns protect soil from erosion by providing a good cover on the hill slopes and other fragile places.
(iii) Equisetum stems have rough surfaces. They are used in scrubbing and polishing.
(iv) Azolla a water fern has a symbiotic association with nitrogen fixing cyanobacterium Anabaena azollae. It is cultured in paddy fields to harbour nitrogen fixing bacterium. Thus, act as a biofertiliser.
(v) Rhizomes and petioles of Dryopteris are used to produce anthelmintic drug. Roots of Adiantum can cure throat infections. Lycopodium is used in treatment of rheumatism and disorders of lungs and kidneys.
(vi) Ferns are also grown as ornamental plants for their graceful plant body.
Classification of Pteridophytes
The pteridophytes are further classified into four classes
(i) Psilopsida (Psilotum)
(ii) Lycopsida (Selaginella and Lycopodium)
(iii) Sphenopsida (Equisetum)
(iv) Pteropsida (Dryopteris, Pteris and Adiantum)
Differences between Bryophytes and Pteridophytes
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2 Gymnosperms, Angiosperms and Plant Life Cycles

Gymnosperms
The gymnosperms (Gymnos = naked; sperma = seeds) are plants in which the ovules are not enclosed by any ovary wall and remain exposed, both before and after fertilisation. These are small groups of seed plants which are represented by only 900 living species. Unlike bryophytes and pteridophytes, in gymnosperms the male and the female gametophytes do not have an independent free-living existence. They remain within the sporangia retained on the sporophytes.
Habitat
These plants are mostly found in colder parts of northern hemisphere, where they form extensive forests. A number of gymnosperms are now, known as ornamentals, e.g., Ginkgo, Thuja, Araucaria, etc.
Morphology
Gymnosperms include, medium-sized trees or tall trees and shrubs. The giant red wood tree Sequoia is one of the forest tree upto 100 m. Species of Gnetum are woody climbers. The smallest gymnosperm is Zamia pygmaea which reaches a height of 25 cm. Many of the gymnosperms live for more than 4000 years, e.g, Pine (Pinus), redwood {Sequoia).
External Features
The plant body is sporophyte and differentiated into root, stem and leaves.
Plant Body
Tap roots are present for proper anchorage to heavy plant. Roots in some genera have fungal association in the form of mycorrhiza {Pinus), while in some others {Cycas) small specialised roots called coralloid roots are associated with N2-fixing cyanobacteria such as Anabaena, Nostoc, etc.
The stems are branched {Pinus, Cedrus) or unbranched {Cycas). The leaves may be simple or compound.
Archegonia
The megaspore mother cell thus, undergo meiotic division forming four megaspores. Out of which one is enclosed within the megasporangium (nucellus) and develops into a multicellular female gametophyte, bearing two or more archegonia of female sex organs.
The multicellular female gametophyte is also retained within megasporangium.
Fertilisation
Air current required for transport of male gametes. The male gametes are carried to the archegonia, i.e., gamete in the ovule by means o’f a tube called pollen tube and discharge their contents near the mouth of the archegonia. This is called siphonogamy.
Seeds
Following fertilisation, zygote develops into an embryo and the ovules into seeds. The seeds contain food laden tissue called endosperm. It lies naked or exposed. Endosperm provides nourishment for growth of seedling at the time of seed germination.
Examples Cycas, Pinus, Ginkgo, Ephedra, Gnetum, Vaucheria, Cedrus, Abies, etc.
Differences between Microsporophyll and Megasporophyll
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Classification of Gymnosperms
Gymnospermsfurther include three main classes Cycadopsida (cycads), Coniferopsida (conifers) and Gnetopsida (Gnetum).
Differences between Male Gametophyte of Pteridophyte and Gymnosperms
Male Gametophyte of Pteridophyte Male Gametophyte of Gymnosperms
A distinct male gametophyte may not be present. A distinct male gametophyte is present.
It contains an antheridium. Antheridium is not present.
Male gametes are flagellate. Male gametes can be flagellate or non-flagellate.
Male gametes reach female gametes by swimming in water. Male gamete reach the female gamete through a pollen tube.
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Economic Importance of Gymnosperms
(i) Seeds of Pinus gerardiana (chilgoza) are used a food after roasting. Some other edible gymnosperms plant part are endosperm of Ginkgo, seed kernel of some Cycas and Gnetum, sago grains from stems of Cycas, etc.
(ii) Gymnosperms provide softwood for construction, plywood and paper industry.
(iii) Branches of gymnosperm trees are used as fuel food.
Saw dust of conifers is used in making plastics and linoleum.
Resin is a semifluid secreted by special tubes of a number of conifers. It helps in sealing female cones after pollination, scale leaves around leaf bases and apical buds. Resin is distilled to obtain turpentine and resin. Turpentine is used for . thinning paints, varnishes, etc., and resin is used
for sealing joints, wheel green, preparation of writing paper, oil clothes, etc.
(iv) Ephedrine an antibiotic is obtained from Ephedra. It is used for curing respiratory problems and asthma. Taxol is an anticancer drug obtained from Taxus.
Angiosperms
Angiosperms are seed bearing plants or flowering plants. Unlike gymnosperms where the ovules are naked, in angiosperms the sporophyll are organised into flowers and the seeds are produced inside fruits. There are about 250000 species of angiosperms in the nature. They evolved about 130-160 million years ago.
Habitat
Angiosperm are found in wide range of habitats, from the land to up to 6000 m in Himalayas or Antarctica and Tundra, dry hot deserts, cold deserts, tropics, fresh water up to 60°C, underground, over other plants as parasites, saprophyte, etc. Zostera is a marine angiosperm.
Plant Body
These plant are sporophytic, in the form of herbs, shrubs, trees, climber creepers, etc. The smallest angiosperm is water plant Wolffia and tallest is Eucalyptus regnans (100 mand above). Primary root develops from radicle. It forms tap root system. In many angiosperms roots develop from places other than radicle, these are adventitious roots. Stem develops from plumule.
Leaves
These are simple or compound. The leaves bear axillary buds which can grow into stem branches.
Vascular Tissues
Angiosperms have vessels in xylem. Phloem contains sieve tubes and companion cells in regard to gymnosperms which do not have companion cells.
Flowers
Flowers are the reproductive structures formed by the union of one or both types of sporophylls (microsporophylls or stamens and megasporophylls or carpels).
Microsporophylls or Male Sex Organs
Stamens are considered as the male sex organs of a flower. Each stamen has two main parts, i.e., a slender filament with an anther (at the tip). An anther contains four microsporangia where microspore mother cells become differentiated to form four microspores, each developing into a pollen grain.
Megasporophylls or Female Sex Organs
Carpel or pistil is called the female sex organ of the flower. Each carpel has three parts {i.e., an ovary, style and stigma). A megaspore mother cell is differentiated in the nucellus and undergoes meiosis, ultimately one functional megaspore gets enlarged and forms the female gametophyte known as embryo-sac.
Each cell of an embryo sac is haploid:
Embryo Sac
Each embryosac contains a three celled eg® apparatus consisting I of one egg cell and two synergids, three antipodal cells (at the i opposite end) and two polar nuclei (in the central cell). The polar : nuclei eventually fuses and forms diploid secondary nucleus.
Pollination
Pollen grains after dispersal from the anthers are carried by various ways such as wind, water or by various other agencies to the stigma of the pistil.
Fertilisation and Development of a Seed
Each pollen grain germinates on the stigma forming a pollen tube that carries two male gametes to the embryo sac, growing through the tissues of stigma and style.
One of the male gametes fuses with the egg cell to form a zygote (syngamy).
The other male gamete fuses with the diploid secondary nucleus to produce the triploid Primary Endosperm Nucleus (PEN). Because of the involvement of above mentioned two events, it is known as double fertilisation. After fertilisation, synergids and antipodal cells degenerate.
The zygote develops into embryo and the primary endosperm nucleus develops into an endosperm.
The ovule gradually transforms into a seed and the ovary becomes the fruit. A fruit is actually a ripened ovary. They not only protect the seed but also help in their dispersal.
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Classification of Angiosperms
George Bentham and Joseph Dalton Hooker presented the system of classification of angiosperm* published in Genera Plantarum (1862-1883) which appeared in three volumes.
This system of classification is used by most of the well known Herbaria of the world. The details of this system are not described here. Conveniently, on the basis of the number of cotyledons angiosperm are classified in two broad groups i.e., Monocotyledonae and Dicotyledonae.
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Economic Importance of Angiosperms
(i) The angiosperms are major source of food, fibers, spices and beverages.
(ii) They also provide valuable timber and medicines.
(iii) These also add beauty to our environment as well.
Alteration of Generation
Life cycle of an organism is a sequence of events that occur from birth to death of an organism. In plants, both haploid and diploid cells can divide by mitosis.
This feature leads to the formation of different plant bodies haploid and diploid. The haploid plant body produces gametes by mitosis. This plant body represents a gametophyte.
After fertilization, zygote also divides by mitosis to produce a diploid saprophytic plant body. Haploid spores are produced by this plant body by meiosis. These in turn, divide by mitosis to form a haploid plant body once again.
Thus, during the life cycle of any sexually reproducing plant, there is an alternation of generation between gamete producing haploid gametophyte and spore producing diploid saprophyte.
Plant Life Cycles
Different plant groups and individual have different features in their life cycle
Haplontic
The dominant photosynthetic phase is a gametophyte produced by haploid spores. The gametophyte produces gametes by mitosis.
The gametes fuse and produce a diploid zygote, that represents sporophytic generation. There are no free living sporophytes. Meiosis in the zygote results in formation of haploid spores. This kind of life cycle is called haplontic.
Many algae such as Volvox, Spirogyra and Chlamydomonas represent this pattern of life cycle.
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Diplontic
In this type, the diploid sporophyte is the dominant. The multicellular diploid phase is called sporophyte. The gametophytic phase is represented by the single to few celled haploid gametophyte.
This kind of life cycle is termed as diplontic. All seed bearing plants, gymnosperms and angiosperms follow this pattern of life cycle. Fucus, an alga is diplontic.
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Haplodiplontic
In this type, there are two distinct multicellular phases, diploid sporophyte and haploid gametophyte are present. Both phases are multicellular. However, they dilfer in their dominant phases.
(i) A dominant, independent, photosynthetic, thalloid or erect phase is represented by a haploid gametophyte. It alternates with the short lived multicellular sporophyte totally, partially or dependent on the gametophyte for its anchorage and nutrition. All bryophytes represent this pattern.
(ii) The diploid saprophyte is represented by a dominant independent photosynthetic vascular plant body. It alternates with multicellular saprophytic/autotrophic,
independent but short lived haploid gametophyte. This pattern is called haplodiplontic life cycle.
All pteridophytes demonstrate this pattern. However, most algal genera are haplontic, some of them such as Ectocarpus, Polysiphonia and kelps are haplodiplontic.
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NCERT Exemplar Problems Class 11 Mathematics Chapter 9 Sequence and Series

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NCERT Exemplar Problems Class 11 Mathematics Chapter 9 Sequence and Series

Short Answer Type Questions:
Q1. The first term of an A.P. is a and the sum of the first p terms is zero, show that the sum of its next q terms
\frac { -a(p+q)q }{ p-1 }  

Sol: Let the common differeence of the given A.P be d
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-1

Q2. A man saved Rs. 66000 in 20 years. In each succeeding year after the first year, he saved Rs. 200 more than what he saved in the previous year. How much did he save in the first year?
Sol: Let us assume that the man saved Rs.a in the first year.
In each succeeding year, an increment of Rs. 200 is made. So, it forms an A.P. whose
First term = a, Common difference, d = 200 and n=20
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-2

=> 6600 = 2a + 19 x 200 => 2a = 2800
∴a = 1400

 Q3. A man accepts a position with an initial salary of Rs. 5200 per month. It is understood that he will receive an automatic increase of Rs. 320 in the very next month and each month thereafter.
(i) Find his salary for the tenth month.
(ii) What is his total earnings during the first year?
Sol: The man gets a fixed increment of Rs. 320 each month. Therefore, this forms an A.P. whose
First term, a = 5200 and Common difference, d = 320

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Q5. A carpenter was hired to build 192 window frames. The first day he made five frames and each day, thereafter he made two more frames than he made the day before. How many days did it take him to finish the job?
Sol: Here, a = 5 and d = 2
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Q6. The sum of interior angles of a triangle is 180°. Show that the sum of the interior angles of polygons with 3, 4, 5, 6, … sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon.
Sol: We know that, sum of interior angles of a polygon of side n is (n – 2) x 180°.
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Q7. A side of an equilateral triangle is 20 cm long. A second equilateral triangle is inscribed in it by joining the mid-points of the sides of the first triangle. This process is continued for third, fourth, fifth, triangles. Find the perimeter of the sixth inscribed equilateral triangle.
Sol: Let the given equilateral triangle be ∆ ABC with each side of 20 cm.
By joining the mid-points of this triangle, we get another equilateral triangle of side equal to half of the length of side of ∆ABC.
Continuing in this way, we get a set of equilateral triangles with side equal to half of the side of the previous triangle.
Now,
Perimeter of first triangle = 20 x 3 = 60 cm;
Perimeter of second triangle = 10 x 3 = 30 cm;
Perimeter of third triangle = 5×3 = 15 cm;

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Q8. In a potato race 20 potatoes are placed in a line at intervals of 4 m with the first potato 24 m from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?
Sol: Distance travelled to bring first potato = 24 + 24 = 2 x 24 = 48 m
Distance travelled to bring second potato = 2(24 + 4) = 2 x 28 = 56 m
Distance travelled to bring third potato = 2(24 + 4 + 4) = 2 X 32 = 64 m; and so on…
Clearly, 48, 56, 64,… is an A.P. with first term 48 and common difference 8. Also, number of terms is 20.
Total distance run in bringing back all the potatoes,
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Q9. In a cricket tournament 16 school teams participated. A sum of Rs. 8000 is to be awarded among themselves as prize money. If the last placed team is awarded Rs. 275 in prize money and the award increases by the same amount for successive finishing places, how much amount will the first place team receive?
Sol: Let the first place team get Rs. a as the prize money.
Since award money increases by the same amount for successive finishing places, we get an A.P.
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Q11. Find the sum of the series (33 – 23) + (53 – 43) + (73 – 63) + … to
(i) n terms                                               
(ii) 10 terms
Sol: Given series is: (33 — 23) + (53 – 43) + (73 – 63) + … n terms
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Q12. Find the rth term of an A.P. sum of whose first n terms is 2n +3n2
Sol:
Sum of k terms of A.P., Sn = 2n + 3n2
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Long Answer Type Questions

Q13. If A is the arithmetic mean and G1 , G2 be two geometric means between any two numbers, then prove that 2A = \frac { { G }_{ 1 }^{ 2 } }{ { G }_{ 2 } } +\frac { { G }_{ 2 }^{ 2 } }{ { G }_{ 2 } } \quad  
Sol: Let the numbers be a and b.
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Q15. 1f the sum of p terms of an AP. is q and the sum of q terms isp, then show that the sum ofp + q terms is —(p + q). Also, find the sum of first p — q terms (where, p > q).
Sol:
Let first term and common difference of the A.P. be a and d, respectively. Given, Sp = q

ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-17
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-18

Q16. If pth, qth and rth terms of an A.P. and G.P. are both a, b and c, respectively, then show that ab-c bc-a-ca-b = 1.
Sol:
Let A and d be the first term and common difference of A.P., respectively. Also, let B and R be the first term and common ratio of G.P., respectively.
It is given that,
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-19

Objective Type Questions:

Q17. If the sum of n terms of an A.P. is given by Sn = 3n + 2n2, then the common difference of the A.P. is
(a) 3                       
(b) 2                         
(c) 6                        
(d) 4
Sol: (d) Given, Sn = 3n + 2n2
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-20

Q18. If the third term of G.P. is 4, then the product of its first 5 terms is
(a) 43
(b) 44
(c) 45
(d) none of these

Sol: (c) Let a and r the first term and common ratio, respectively.
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-21

Q19. If 9 times the 9th term of an A.P. is equal to 13 times the 13th term, then the 22nd term of the A.P. is
(a) 0
(b) 22
(c) 198
(d) 220
Sol:
(a) Let the first term and coiqmon difference of given A.P. be a and d, respectively.
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-22

Q20. If x, 2y and 3z are in A.P. where the distinct numbers x, y and z are in G.P., then the common ratio of the G.P.is
Sol:
Since x, 2y and 3z are in A.P., we get
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-23

Q21. If in an A.P., Sn = qn2 and Sm = qm2, where Sr denotes the sum of r terms of the AP, then Sq equals

ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-24

Q22. Let Sn denote the sum of the first n terms of an A.P. If S2n = 3Sn, then S3n : Sn  is equal to
(a) 4
(b) 6
(c) 8
(d) 10
Sol:
(b) Let first term be a and common difference be d.
Then, S2n = 3Sn

ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-25
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-26

Q25. If tn denotes the nth term of the series 2 + 3+ 6+11 + 18+…, then t50 is
(a) 492 – 1     
(b) 492   
(c) 502+l
(d) 492 +2

ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-27
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-28

Q26. The lengths of three unequal edges of a rectangular solid block are in G.P. If the volume of the block is 216 cm3 and the total surface area is 252 cm2, then the length of the longest edge is
(a) 12 cm               

(b) 6 cm                   
(c) 18 cm                
(d) 3 cm
Sol: (a) Let the length, breadth and height of rectangular solid block be a/r, a and ar, respectively.
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-29

Fill in the Blanks

Q27. If a, b and c are in G.P., then the value of \frac { a-b }{ b-c }   is equal to _________
Sol: Given that, a, b and c are in G.P.
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-30

Q28. The sum of terms equidistant from the beginning and end in an A.P. is equal to _____.
Sol: 
Let a be the first term and d be ihe common difference of the A.P.
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-31

Q29. The third term of a G.P. is 4. The product of the first five terms is  
Sol: Let a and r the first term and common ratio, respectively.
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-21

True/False Type Questions

Q30. Two sequences cannot be in both A.P. and G.P. together.
Sol: False
Consider the sequence 3,3,3; which is A.P. and G.P. both.

Q31. Every progression is a sequence but the converse, i.e., every sequence is also a progression need not necessarily be true.
Sol: True                                                            –
Consider the progression a, a + d, a + 2d, … and sequence of prime number 2, 3, 5, 7, 11,…
Clearly, progression is a sequence but sequence is not progression because it does not follow a specific pattern.

Q32. Any term of an A.P. (except first) is equal to half the sum of terms which are equidistant from it.
Sol: True
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-32

Q33. The sum or difference of two G.P.s, is again a G.P.
Sol: False
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-33
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-34

Q34. If the sum of n terms of a sequence is quadratic expression, then it always represents an A.P.
Sol: False
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-35

Match the questions given under Column I with their appropriate answers given under the column II.
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-36
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-37
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-38
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-39
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-40
ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series-41

The post NCERT Exemplar Problems Class 11 Mathematics Chapter 9 Sequence and Series appeared first on Learn CBSE.

NCERT Exemplar Problems Class 11 Mathematics Chapter 10 Straight Lines

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NCERT Exemplar Problems Class 11 Mathematics Chapter 10 Straight Lines

Short Answer Type Questions
Q1. Find the equation of the straight line which passes through the point (1, -2)
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-1

Q2. Find the equation of the line passing through the point (5,2) and perpendicular to the line joining the points (2, 3) and (3, -1)
Sol:
We have points A(5, 2), B(2, 3) and C(3, -1).
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-2

Q3. Find the angle between the lines y = (2 -√3) (x + 5) and y = (2 + -√3) {x – 7)
Sol: Slope of the line = (2 -√3)(x + 5) is: ml = (2 -√3 )

ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-3

Q4. Find the equation of the lines which passes through the point (3, 4) and cuts off intercepts from the coordinate axes such that their sum is 14.
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-4

Q5. Find the points on the line x+y = 4 which lie at a unit distance from the line 4x + 3y= 10
Sol.
Let the required point be (h, k) lies on the line x + y = 4
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-5

ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-6

Q7. Find the equation of lines passing through (1,2) and making angle 30° with y-axis.
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-7

Q8. Find the equation of the line passing through the point of intersection of lx + y = 5 and x + 3 y +8 = 0 and parallel to the line 3x + 4y = 1.

ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-8
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-9

Q9. For what values of a and b the intercepts cut off on the coordinate axes by the line ax + by + 8 = 0 are equal in length but opposite in signs to those cut off by the line 2x – 3y + 6 = 0 on the axes.
Sol:
Given line is:
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-10

Q10. If the intercept of a line between the coordinate axes is divided by the point (-5,4) in the ratio 1 : 2, then find the equation of the line.
Sol:
Let the line through the point P(-5, 4) meets axis at A(h, 0) and B(0, k)

According to the question, we have AP: BP =1:2
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-11
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-12
Q11. Find the equation of a straight line on which length of perpendicular from the origin is four units and the line makes an angle of 120° with the positive direction of x-axis.
Sol:
Given that the line makes and angle 120° with positive direction of x-axis.
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-13

Q12. Find the equation of one of the sides of an isosceles right angled triangle whose hypotenuse is given by 3x + Ay = 4 and the opposite vertex of the hypotenuse is (2, 2).
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-14

Long Answer Type Questions

Q13. If the equation of the base of an equilateral triangle is x + y – 2 and the vertex is (2, -1), then find the length of the side of the triangle.
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-15

 

Q14. A variable line passes through a fixed point P. The algebraic sum of the perpendiculars drawn from the points (2, 0), (0, 2) and (1, 1) on the line is zero. Find the coordinates of the point P

ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-16

Q15. In what direction should a line be drawn through the point (1, 2) so that its point of intersection with the line x+y = 4 is at a distance √6/3 from the given point.

ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-17
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-18

 

Q16. A straight line moves so that the sum of the reciprocals of its intercepts made on axes is constant. Show that the line passes through a fixed point.
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-19

Q17. Find the equation of the line which passes through the point (-4, 3) and the portion of the line intercepted between the axes is divided internally in the ratio 5 : 3 by this point.
Sol:
Let the line through the point P(-A, 3) meets axis at A(h, 0) and 0(0, k)
Now according to the question AP : BP =5:3
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-20
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-21

Q18. Find the equations of the lines through the point of intersection of the lines x-y+ 1=0 and 2x – 3y + 5 = 0 and whose distance from the point (3, 2) is 7/5
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-22

Q19. If the sum of the distances of a moving point in a plane from the axes is 1, then find the locus of the point.

ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-23

Q20. Pl, P2 are points on either of the two lines y — √3 |x| = 2 at a distance of 5 units from their point of intersection. Find the coordinates of the foot of perpendiculars drawn from P1, P2 on the bisector of the angle between the given lines.
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-24
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-25

 

Q21. If p is the length of perpendicular from the origin on the line \frac { x }{ a } +\quad \frac { y }{ b }      and a2,p2 and are in the A.P , then show that a4+b4 = 0
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-26

Objective Type Questions

Q22. A line cutting off intercept -3 from the y-axis and the tangent of angle to the x-axis is 3/5, its equation is

ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-27

Q23. Slope of a line which cuts off intercepts of equal lengths on the axes is
(a) -1
(b) 0       
(c) 2       
(d) √3
Sol: (a) Equation of the according to the question is \frac { x }{ a } +\quad \frac { y }{ a }    
=> x+y = a
Required slope = -1

 

Q24. The equation of the straight line passing through the point (3, 2) and perpendicular to the line y = x is
(a) x-y = 5           
(b) x+y = 5
(c)x+y=l
(d)x-y=1
Sol:(b) Slope of the given line y = x is 1.
Thus, slope of line perpendicular to y = x is -1.
Line passes through the point (3, 2).
So, equation of the required line is:y-2=-l (x – 3) => x + y = 5

Q25. The equation of the line passing through the point (1,2) and perpendicular to the line x +y + 1 = 0 is
(a) y-x +1=0           
(b) y — x—1=0
(c) y-x + 2 = 0                                      
(d) y — x — 2=0
Sol: (b) Slope of the given line +1=0 is-1.
So, slope of line perpendicular to above line is 1.
Line passes through the point (1,2).
Therefore, equation of the required linens:
y-2 = 1(x- 1) => y-x-1=0.

Q26. The tangent of angle between the lines whose intercepts on the axes are a, -b and b, -a, respectively, is
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-28

 

Q27. If the line \frac { x }{ a } +\frac { y }{ b }    passes through the points (2, -3) and (4, -5), then (a, b) a b is
(a) (1,1)                  
(b) (-1,1)                  
(c) (1,-1)                  
(d) (-1,-1)
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-29

Q28. The distance of the point of intersection of the lines 2x – 3y + 5 = 0 and 3x + 4y = 0 from the line 5x – 2y = 0 is
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-30

Q29. The equations of the lines which pass through the point (3, -2) and are inclined at 60° to the line √3 x + y = 1 is
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-31

Q30. The equations of the lines passing through the point (1,0) and at a distance √3/2 from the origin, are

ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-32
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-33
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-34
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-35

Q33. If the coordinates of the middle point of the portion of a line intercepted between the coordinate axes is (3, 2), then the equation of the line will be
(a) 2x + 3y = 12                                     
(b) 3x + 2y=l2
(c) 4x-3y = 6                                        
(d) 5x- 2y=10
Sol: (a) Since, the middle point is P(3, 2), then line meets axes at A(6, 0) and B (0, 4).
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-36

Q34. Equation of the line passing through (1,2) and parallel to the line y = 3x – 1 is
(a)y + 2=x+l                                    
(b) y + 2 = 3(x + 1)
(c) y -2 =  3(x — 1)                             
(d) y-2=x-l
Sol: (c) Line is parallel to the line y = 3x – 1.
So, slope of the line is‘3’.
Also, line passes through the point (1,2).
So, equation of the line is: y – 2 = 3(x – 1)

Q35. Equations of diagonals of the square formed by the lines x = 0, y = 0, x = 1 and y = 1 are

ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-37
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-38

Q37. The point (4, 1) undergoes the following two successive transformations:
(i) Reflection about the line y = x
(ii) Translation through a distance 2 units along the positive x-axis Then the final coordinates of the point are
(a)  (4,3)               
(b) (3,4)                
(c)  (1,4)              
(d) (7/2,7/2)
Sol: (b) Reflection of A (4, 1) in y = x is 5(1,4).
Now translation of point B through a distance ‘2’ units along the positive x-axis shifts B to C( 1 + 2,4) or C(3,4).

Q38. A point equidistant from the lines 4x + 3y+ 10 = 0, 5x – 12y + 26 = 0 and lx + 24y – 50 = 0 is
(a)    (1,-1)              
(b)    (1, 1)               
(c)    (0,0)              
(d)   (0, 1)
Sol:
(c) Clearly distance of each of three lines from (0, 0) is 2 units.

Q39. A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is

ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-39

Q40. The ratio in which the line 3x + 4y + 2 = 0 divides the distance between the lines 3x + 4y + 5 = 0 and 3x + 4y – 5 = 0 is
(a) 1:2 (b) 3:7 (c) 2:3 (d) 2:5

Sol: (b) Given lines are:
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-40
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-41

Q41. One vertex of the equilateral triangle with centroid at the origin and one side asx + y- 2 = 0is
(a) (-1,-1) (b) (2,2) (c) (-2,-2) (d) (2,-2)
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-42

Fill in the Blanks Type Questions

Q42. If a, b, c are in A.P., then the straight lines ax + by + c = 0 will always pass through __________    

ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-43

Q43.The line which cuts off equal intercept from the axes and pass through the point (1, -2) is

ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-44

Q44. Equations of the lines through the point (3, 2) and making an angle of 45° with the line x – 2y = 3 are _____
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-45

Q45. The points (3,4) and (2, -6) are situated on the_____ of the line 3x – 4y – 8= 0.

Sol: Given line is 3x – 4y – 8 = 0
For point (3, 4), 3(3) – 4(4) – 8 = -15 < 0
For point (2, -6), 3(2) – 4(—6) – 8 = 22 > 0
Hence, the points (3,4) and (2, -6) lies on opposite side of the line.

Q46. A point moves so that square of its distance from the point (3, -2) is numerically equal to its distance from the line 5x – 12y = 3. The equation of its locus is______ .
Sol: Let the moving point be P(h, k).
Given point is A(3, -2).

ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-46
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-47

Q47. Locus of the mid-points of the portion of the line x sin θ + y cos θ = p intercepted between the axes is ______
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-48

True/False Type Questions

Q48. If the vertices of a triangle have integral coordinates, then the triangle can not be equilateral.
Sol: True
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-49
This is a contradiction to the fact that the area is a rational number. Hence, the triangle cannot be equilateral.

Q49. The points A(-2, 1), B(0, 5), C(-l, 2) are collinear.
Sol: False
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-50

Q50. Equation of the line passing through the point (a cos3 , a sin3 ) and perpendicular to the line
x sec + y cosec = a isx cos -y sin = a sin 2
Sol: False
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-51

Q51. The straight line 5x + 4y = 0 passes through the point of intersection of the straight lines x + 2y— 10 = 0 and 2x +y + 5 = 0.
Sol: True
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-53
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-54
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-55

 

Q53. The equation of the line joining the point (3, 5) to the point of intersection of the lines 4x +y – 1 = 0 and lx – 3v – 35 = 0 is equidistant from the points (0, 0) and (8, 34).

Sol: True
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-56
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-57

Q55. The line ax + 2y + 1 = 0, bx + 2y + 1 = 0 and cx + 4y + 1 = 0 are concurrent, if a, b and c are in GP.
Sol: False
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-58

Q56. Line joining the points (3, -4) and (-2, 6) is perpendicular to the line joining the points (-3, 6) and (9, -18).
ncert-exemplar-problems-class-11-mathematics-chapter-10-straight-lines-59

NCERT Exemplar Problems Class 11 Mathematics

The post NCERT Exemplar Problems Class 11 Mathematics Chapter 10 Straight Lines appeared first on Learn CBSE.

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