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Tamilnadu State Board New Syllabus Samacheer Kalvi 12th English Guide Pdf Prose Chapter 5 The Chair Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 12th English Solutions Prose Chapter 5 The Chair

12th English Guide The Chair Text Book Back Questions and Answers

Textual Questions:

1. Answer the following questions in one or two sentences each based on your understanding of the story: (Text Book Page No. 148) (Note: IQ – Important Questions)

Question a.
What was put on the family agenda?
Answer:
The idea of buying a chair for the house was put on the family agenda.

Question b.
Who visited the family?
Answer:
Their family friend who was a sub-judge visited the family.

Question c.
Describe the stool that the narrator’s family hpd.
Answer:
The three legged stool measured a mere three – fourth foot. It would topple over if the weight vvps not placed exactly above the legs.

Question d.
What was Pedanna suggested to their father?
Answer:
Pedanna suggested that they could buy a chair from the town.

Question e.
What was offered to Maamanaar by their mother?
Answer:
Their mother offered buttermilk seasoned with asafoetida to Maamanaar.

Question f.
Why were the two chairs compared to Rama-Lakshmana?
Answer:
Rama and Lakshmana were the brothers who were identical in their character. In the same manner the two chairs were identical.

Question g.
When did the children shy away from the chair?
Answer:
The children shied away from the chair as it was used to propup a deadbody in their village.

Question h.
How did Maamanaar handle the chair at home?
Answer:
Maamanaar used to wipe it every morning. While shifting the chair he himself carried it and put it down carefully as if placing down gently a mud pot brimming with waiter.

2. Answer the following questions in three or four sentences each: (Text Book Page No. 149)

Question a.
What happened to the visitor when he sat on the stool?
Answer:
The stool would topple over if one didn’t place one’s weight exactly above the legs. Before the family members tried to caution the sub judge, he sat on it fell down with a thud and rolled over.

Question b.
Why did the family find it difficult to make a chair?
Answer:
The family found it difficult to make a chair because there was no model of a chair in their village. Moreover there was not a single carpenter who knew how to make one.

Question c.
What was grandmother’s suggestion of wood ? Why ?
Answer:
Grandmother suggested to make use of teak wood as it would be of light in weight and could be carried around easily. Moreover it would be strong.

Question d.
How was the chair made and how did the villagers react to it?
Answer:
The chair was made of black wood with a mirror like gleam. It had perfectly shaped front legs and curved back legs. The villagers arrived in groups to see the chair. A few touched it gently. An old man lifted it and fall it to be heavy and strong.

Question e.
When did the children get over the fear of sitting on the chair?
Answer:
One of their neighbour Suganthi came to their home with her baby brother. She put her baby brother on the chair. It was only after that incident the children got over the fear of sitting on the chair.

Question f.
Why did Maamanaar hand over the chair to the villagers to retain it?
Answer:
Maamanaar handed over the chair to the villagers to retain it as he thought that it could be used to prop up dead bodies in the village whenever needed.

3. Answer the following in a paragraph of 100-150 words each: (Text Book Page No. 149)

Question a.
Narrate the humorous incidents that happened in the author’s home before and after the arrival of the chair.
Answer:
There was only a three legged stool in the narrator’s home when the story begins. The woeful thing about the stool was that if one didn’t place one’s weight exactly above the legs, the stool would topple over. Their family friend, a sub-judge paid a visit to their home one day. He was provided with the stool to sit on. Before he was given a caution, he sat on it and fell down with a thud and rolled over.

This incident made the family members giggle for a long time. After the arrival of the chair there arose a different scenario. The chair was asked by the villagers whenever there was a death in the village. It was used to prop up the corpse. Whenever the mourners came for the chair, the family members felt very sad which was misinterpreted by the mourners. The family members had fear to sit on the chair after that incident. They persuaded their visitors to sit on it and made fun that they were rehearsing for their death. When they started to sit on the chair again it would be asked for by the mourners. Thus the witty incidents continued.

Question b.
Write character sketches of Maamanaar and Pedanna.
Answer:
Maamanaar is considered to be a person who is very affectionate and calm. Though his sister’s children tease him to a great extend he remains mum With a smile. He never scolds them for their action. He is the one who is well-mannered and tidy. The way in which he takes care of his betel box and the newly brought chair reveals his perfection in handling things at home.

His cleanliness also comes to lime light when one has a close look at his lime paste dabba. Though he considers and maintains the things in his home as treasures, he is ready to lend his chair to the villagers. He even asks them to retain it for funeral purpose. This act of him highlights his generosity.

Pedanna seems to be a person who disrespects elders. The way in which he ridicules his Maamanaar when the latter pays a visit to his home proves this statement. He is also very good in making tricks. When the mourners come for the chair he diverts them to his Maamanaar’s house, which shows his deceitful act. In other sense, based on the above said qualities we can also consider Pedanna to be a person who likes to lead his life with fun and cheer.

Paragraph:

Introduction:
In the story “The Chair” the narrator, Rajanarayanan amusingly narrates the humorous incidents that took place in his home before and after the arrival of the chair.

Fun with stool:
There was only a three legged stool in the narrator’s home. It would topple over if one didnot place his weight exactly above its legs. Their family friend, a sub judge paid a visit to their home on6 day. He was provided with the. stool to sit on-. Without knowing the wretched thing about the stool, he sat on it and fell down with a thud and rolled over. The family members could not control their giggles.

The arrival of chair:
The narrator’s family decided to make a chair for their home. They had a lengthy discussion about it. Finally they decided to make two chairs. One for them and the other for their maamanaar. The chair arrived. The narrator and his siblings took turns to sit on it and even fought for it often.

Ordeal of the chair:
The village people came in groups to have a look at the chair as it was the first chair in their village. Unfortunately, there was a death of an eminent person in the village and the chair was asked for the propup the corpse of that eminent person. From then the chair was asked for whenever there occurred a death in the village. This act disturbed the sleep of the family members as the chair was asked for mostly during late night hours.

Pedanna’s trick:
The narrator and his siblings were afraid to sit on the chair, once after it was got back after the funeral ceremony. They washed it with many buckets of water. They made each and every visitor who came to their home to sit on it. Only after a long time they started to sit on the chair again. By that time the mourners came for the chair again. In order to overcome it Pedanna diverted the mourners to go to his maamanaar’s house and get the chair.

Maamanaar’s generosity:
Pedanna thought that his Maamanaar would not give the chair to the mourners. Pedanna v and his siblings were of the opinion that their Maamanaar was stingy. But things happened viceversa. Maamanaar was so generous to lend the chair to the villagers and asked them to retain it for funeral purpose.

Conclusion:
Thus the narrator elucidates the incidents that took place because of the chair in a humorous way.

Vocabulary:

a) Find out the synonym of the underlined word in each of the following sentences: (Text Book Page No. 149)

Question 1.
Just as we thought our chuckles had subsided.
a) diminished
b) increased
c) completed
d) submerged.
Answer:
a) diminished

Question 2.
Our father rejected it, saying it wouldn’t be sturdy.
a) weak
b) strong
c) tall
d) good
Answer:
b) strong

Question 3.
Anna would say with an impudent smile.
a) innocent
b) fake
c) disrespectful
d) decent
Answer:
c) disrespectful

Question 4.
A silver tumbler of buttermilk seasoned with asafoetida.
a) mixed
b) garnished
c) filled
d) loosened
Answer:
b) garnished

Question 5.
A few people from the house of bereavement stood outside.
a) rejoice
b) celebration
c) grief
d)war
Answer:
c) grief

b. Find out the antonym of the underlined word in each of the following sentences: (Text Book Page No. 149)

Question 1.
The anticipation of this imagined separation only increases their fondness for the calf.
a) expectancy
b) contemplation
c) outlook
d) ignorance
Answer:
d) ignorance

Question 2.
Fortuitously, a quest visited our house.
a) luckily
b) peacefully
c) unfortunately
d) happily
Answer:
c) unfortunately

Question 3.
There he was sitting in splendour on his chair.
a) magnificence
b) pomp
c) effulgence
d) simplicity
Answer:
d) simplicity

Question 4.
Maamanaar was unmatched.
a) inferior
b) incomparable
c) excellent
d) supreme
Answer:
a) inferior

Question 5.
He greeted me with his usual smile and banter.
a) flattery
b) small talk
c) chitchat
d) repartee.
Answer:
a) flattery

c. Identify the correct combination of the following compound words: (Text Book Page No. 150)

1. Haircut – Noun + Verb
2. waterfall – Noun + Verb
3. drawback – Verb + Adverb (verb+noun)
4. output – Preposition + Verb
5. show case – Verb + Noun
6. headmaster – Noun + Noun
7. swimming pool – Gerund + Verb
8. public speaking – Adjective + Gerund
9. software – Adjective + Noun
10. world famous – Noun + Adjective

d. Blend the following words: (Text Book Page No. 150)

Question 1.
binary + digit
Answer:
bit

Question 2.
electronic + mail
Answer:
email

Question 3.
foreign + exchange
Answer:
forex

Question 4.
motor + pedal
Answer:
moped

Question 5.
parachute + troop
Answer:
paratroop

e. Replace the underlined word/ expressions with possibly polite forms: (Text Book Page No. 150) 

Question a.
The culprit was sent to jail.
Answer:
Jail – Penitentiary

Question b.
Dheeraj is unemployed.
Answer:
unemployed – between jobs

Question c.
I saw a disabled man.
Answer:
disabled – physically challenged

Question d.
Subsidies are given to the poor.
Answer:
poor – needy

Question e.
Elambrathi has a second-hand car.
Answer:
second hand – hand me down

Listening:

After listening to the pledge played on the tape recorder carefully, fill in the following statements with the right options given: (Text Book Page No. 153)

Question a.
The medical graduates take oath to dedicate their _______ to the service of humanity.
i) money
ii) talent
iii) life
iv) nation
Answer:
iii) life

Question b.
The _______ of the patient should be the doctor’s greatest concern.
i) dignity
ii) gratitude
iii) health
iv) honour
Answer:
iii) health

Question c.
The would-be graduates promise to practise their profession with _______ and dignity.
i) conscience
ii) knowledge
iii) understanding
iv) respect
Answer:
i) conscience

Question d.
They should respect the _______ of the patients.
i) age
ii) wealth
iii) background
iv) secrets
Answer:
iv) secrets

Question e.
The pledge is also to treat the patients without any _______.
i) fee
ii) discrimination
iii) interest
iv) hatred
Answer:
ii) discrimination

a. Read the following passage carefully and answer the following questions: (Text Book Page No. 153)
Questions and answer: (for passage please refer Text Book Page No. 152)

Question 1.
What is meant by pollution? Mention the different kinds of pollution.
Answer:
Pollution is the introduction of contaminants into the environment that cause harm to the ecosystem. The different kinds of pollution are air pollution, water pollution, and land pollution.

Question 2.
How does Particulate Matter cause air pollution?
Answer:
Particulate matter (PM), also known as particle pollution, is a complex mixture of extremely small particles and liquid droplets that get into the air. These particles if inhaled can affect health.

Question 3.
Identify three major causes of pollution in air.
Answer:
The release of various gases, finely divided solid particles, or liquid droplets that escape into the atmosphere to disperse and dilute in the environment are the major causes of air pollution.

Question 4.
Name the types of pollution we encounter now.
Answer:
Noise pollution, light pollution, and plastic pollution.

Question 5.
What sort of health issues do people face due to air pollution?
Answer:
Several studies have shown that poor air quality is a cause for many health issues among people with lower respiratory disorders with symptoms like dry cough, breathlessness, wheezing, chest discomfort, serious lung infections, and cardiovascular diseases. Some studies throw light on the fact that about 16 percent of the deaths worldwide in 2015 were due to pollution.

Question 6.
How can we protect ourselves outdoor from air pollution?
Answer:
Air Masks are an option to protect oneself outdoor. Air masks can be used while commuting or while one is exposed to a polluted area.

Question 7.
Suggest a suitable title to the passage.
Answer:
‘Air pollution’ is a suitable title for the passage.

Question 8.
Identify the meaning of the word similar to the one used in the fourth para:
Answer:
a) emerging
b) filtering
c) floating
d) falling.

b) Read the following information given in the table below and answer the questions: (Text Book Page No. 154)

A nice choice from Chennai to the National capital
RAJADHANI EXPRESS TIMETABLE
Shortest Route between Chennai and Hazrat Nizamuddin 8 Halts & 324 intermediate stations in between
Station Name	Departs	Day	Speed
Chennai Central	06.05	1	75
Vijayawada	11.55	1	76
Warangal	14.40	1	77
Balharshah	18.00	1	78
Nagpur	20.45	1	74
Bhopal	02.10	2	89
Jhansi	05.31	2	99
Gwalior	06.32	2	85
Agra Cantt	07.57	2	76
Hazrat Nizamuddin	10.25	2	–

Question a.
The number of stations between Chennai Central and Hazrat Nizamuddin is _______.
i) five
ii) ten
iii) eight
iv) eleven
Answer:
iii) eight

Question b.
The train is expected to reach ______ around 8.45 PM.
i) Warangal
ii) Vijayawada
iii) Bhopal
iv) Nagpur
Answer:
iv) Nagpur

Question c.
Between ____________ the train runs at its maximum speed.
i) Bhopal and Gwalior
ii) Bhopal and Jhansi
iii) Bhopal and Hazarat Nizamudin
iv) Bhopal and Agra
Answer:
ii) Bhopal and Jhansi

Question d.
Almost ____________ the train reaches Vijayawada.
i) the day after
ii) around early morning
iii) late night
iv) around noon
Answer:
iv) around noon

Question e.
People prefer the Rajadhani Express to travel from Chennai to reach the capital because ______.
i) it reaches the destination on the same day
ii) the charge is reasonable
iii) the train halts at ten stations
iv) it is the shortest route from Chennai to New Delhi
Answer:
iv) it is the shortest route from Chennai to New Delhi

Question f.
The destination of Rajadhani express is ______.
i) Hazarat Nizamuddin
ii) New Delhi junction
iii) Old Delhi
iv) Rajkot
Answer:
i) Hazarat Nizamuddin

Grammar:

Non-finite Verbs:

Task 1:
Underline the gerunds in the following sentences: (Text Book Page No. 155)

Question 1.
Boys love playing cricket.
Answer:
playing

Question 2.
I love eating ice creams.
Answer:
eating

Question 3.
Jessie enjoys bothering others.
Answer:
bothering

Question 4.
Painting is an interesting hobby.
Answer:
interesting

Question 5.
Dancing gives me joy.
Answer:
Dancing

Task 2:
Use the gerundial form of the verb in the brackets and fill in the blanks: (Text Book Page No. 156)

Question 1.
______ (exercise) is good for health.
Answer:
Exercising

Question 2.
______ (fly) a kite is fun.
Answer:
Flying

Question 3.
______ (shop) is my favourite hobby.
Answer:
Shopping

Question 4.
My friend waited for the ______ (meet).
Answer:
meeting

Question 5.
Huckleberry Finn was responsible for _______ (signal).
Answer:
signaling

Task 3:
Fill in the blanks with the correct infinitives: (Text Book Page No. 156)

Question 1.
Deva forgot ______ the letter.
Answer:
to post

Question 2.
The doctor advised the patient ______ his medicines without fail.
Answer:
to take

Question 3.
Rajesh went to the airport ______ his friend.
Answer:
to receive

Question 4.
The bear climbed up the tree ______ the honey.
Answer:
to drink

Question 5.
The boys went to the forest ______ birds.
Answer:
to watch

Question 6.
I tried hard ______ both ends meet.
Answer:
to make

Question 7.
The archaeologists are trying ______ the ruins of Keelady.
Answer:
to study

Question 8.
Solar energy is used ______electricity.
Answer:
to generate

Question 9.
______ concession, you have to apply well in advance.
Answer:
To get

Question 10.
We have plans ______ to London during summer vacation.
Answer:
to go

Task 4:
Combine each of the following pairs of sentences using participles. The first one is done for you: (Text Book Page No. 156)

Example: I didn’t know what to do. I phoned the police.
Not knowing what to do, I phoned the police.

Question 1.
The baby cried. She was feeling sleepy.
Answer:
Feeling sleepy, the baby cried.

Question 2.
He lived alone. He had forgotten everybody.
Answer:
Having forgotten everybody, he lived alone.

Question 3.
She walked out. She was smiling.
Answer:
She walked out smiling.

Question 4.
The child says he needs attention. He shouts loudly.
Answer:
Shouting loudly the child says she needs attention.

Question 5.
I threw the pen. It was broken.
Answer:
I threw the broken pen.

Question 6.
His coat is tattered. It needs mending.
Answer:
His tattered coat needs mending.

Question 7.
I heard the noise. I turned around.
Answer:
Hearing the noise I turned around.

Question 8.
He was dissatisfied. He quit his job.
Answer:
Being dissatisfied he quit the job.

Question 9.
The politician entered the campus. He was accompanied by many comrades.
Answer:
Accompanied by many comrades the politician entered the campus.

Question 10.
The girl entered the room. She was singing a song.
Answer:
Singing a song the girl entered the room.

Articles and Determiners:

Articles:

Task 1:
Complete the following exercise using a / an / the / ‘o’ (no article) in the underlined space where appropriate. Change capital letters to lower case letters at the beginning of a sentence if necessary: (Text Book Page No. 156)

According to (1) ______ National Weather Report, cyclones are winds circulating (2) ______ counter clockwise in (3) ______ Northern Hemisphere and clockwise in (4) ______ southern Hemisphere. Cyclones are usually accompanied by (5) ______ stormy weather. Tornadoes and hurricanes are types of cyclones. (6) ______ hurricane is (7) ______ cyclone that forms over (8) ______ tropical oceans and seas. (9) ______ hurricane rotates in (10) ______ shape of (11) ______ oval or a circle. (12) ______ Hurricane Andrew, which hit (13) ______ coasts of Louisiana and Southern Florida in August 1992, caused (14) ______ extreme devastation. It was one of (15) ______ most devasting hurricanes ever to hit (16) ______ U.S.. Fourteen people died of (17) ______ Andrew’s effect.

Answer:

1. the
2. O
3. the
4. the
5. a
6. A
7. a
8. the
9. A
10. the
11. an
12. The
13. the
14. an
15. the
16. the
17. O

Task 2:
Complete the following sentences using appropriate determines: (Text Book Page No. 157)

Question 1.
Only ______ people can afford to buy a flat in Chennai.
Answer:
a few

Question 2.
She earns so ______ that she could not make a decent living.
Answer:
little

Question 3.
______ information that she gave proved false.
Answer:
Every

Question 4.
How ______ sugar do you want?
Answer:
much

Question 5.
I am very tired today, as I had ______ guests today.
Answer:
some

Question 6.
______ of my students have become doctors.
Answer:
Most

Question 7.
______ do I know about his personal life.
Answer:
Little

Question 8.
How ______ pages did you read?
Answer:
many

Question 9.
______ fertilizer used these days spoils the soil.
Answer:
Each

Question 10.
During my student life I used to give ______ trouble to my teachers.
Answer:
much

Degrees of Comparison – Transformation:

Task 1:
Transform each of the following sentences using the comparative degree without changing the meaning: (Text Book Page No. 159)

Question 1.
Very few Indian languages are as ancient as Tamil.
Answer:
Tamil is more ancient than many other Indian languages.

Question 2.
Hurricanes are as dangerous as tornadoes.
Answer:
Tornadoes are not more dangerous than hurricanes.

Question 3.
This is the most challenging task I have ever undertaken.
Answer:
This task is more challenging than any other task, I have ever undertaken.

Question 4.
E-mail is the fastest means of communication.
Answer:
E-mail is faster than any other means of communications.

Question 5.
Compulsive gambling is the worst habit a man can develop.
Answer:
Compulsive gambling is worse than any other gambling a man can develop.

Task 2:
Rewrite each of the following sentences using the superlative degree retaining the meaning: (Text Book Page No. 159)

Question 1.
Shakespeare is greater than many other dramatists of the world.
Answer:
Shakespeare is one of the greatest dramatists in the world.

Question 2.
Some people think that nothing is as important as money in life.
Answer:
Some people think that money is the most important in life.

Question 3.
The peacock is more colourful than any other bird found in India.
Answer:
The peacock is the most colourful bird found in India.

Question 4.
Very few people in this town are as generous as Mr. Mohan.
Answer:
Mr.Mohan is one of the most generous persons in this town.

Question 5.
No other planet in our solar system is as cold as Neptune.
Answer:
Neptune is the coldest planet in the solar system.

Question 6.
I cannot do anything better for you than this.
Answer:
This is the best thing, I can do for you.

Task 3:
Replace the comparative adjectives in the following sentences with their positive forms: (Text Book Page No. 159)

Question 1.
Rural life is certainly more peaceful than urban life.
Answer:
Urban life is certainly not as peaceful as rural life.

Question 2.
The pen is mightier than the sword.
Answer:
The sword is not as mighty as the pen.

Question 3.
Train journey is more comfortable than bus journey.
Answer:
Bus journey is not as comfortable as train journey.

Question 4.
My mother can speak more sweetly than anyone else.
Answer:
No one speaks as sweet as my mother.

Question 5.
Gold is not more useful than Iron.
Answer:
Iron is as useful as Gold.

Writing:

Slogan writing:

Look at the pictures given below, and write slogans to advertise the products. Suggest your own brand name for each of the products. (Text Book Page No. 159)

1. Toothpaste → “Smile with strength”, Everyday protection for sensitive teeth.
2. Water purifier → “Purifying agent”, Perfect quality for perfect living.
3. Camera → “Candid camera”, we capture your memories.
4. Apples → “Orchard apples”, An apple a day keeps the doctor away.

Write slogans to create awareness of the following topics using the tips given above: (Text Book Page No.159)

1. Junk food →“Maintain your weight to just feel great”
2. Labour Day → “There is no success without hard work’’, “No work for labours to labour day”
3. Save Water → “Conserve water, Conserve life”, “Thousand lived without love, but not without water”
4. Yoga → “A boon to people on earth”, “A secret to retaining the beauty of body and mind”
5. Blood Donation → “Bring a life back to power”, “Every blood donor is a lifesaver’.

Write a paragraph of about 150 words, on the following topics: (Text Book Page No. 160)

a) The teacher I like the most:

A teacher is a person who educates a group of students known as his pupils. The teacher instills in his students, the values of discipline, responsibility, honesty, courage, and truthfulness. The teacher guides his students in the right direction and prevents them from falling prey to bad intentions and influences. My favourite teacher is my English teacher. She is an elderly lady with great knowledge. Her class is my favourite of the day.

She is a well-known face in our school. In her class, almost every student pays attention to what she says. She uses the help of charts and new methodology to make us understand even better. She makes sure that her students are comfortable with her and not threatened by her. My English teacher was like a mother to me.

Her affection towards me was like a mother’s towards her child. She is the person who pushed me to try everything in life. Her constant dedication towards my interests and hobbies motivated me to pursue the same with more enthusiasm. It is because of her that I complete my work in time and with sincerity even today.

b) The value of discipline:

Discipline is important in our everyday life. It is also important for the better development of a society. It helps in time management. When one is disciplined, they are better able to manage their time at school. Work and at home. Discipline also helps to build respect. It also helps one to achieve better grades in school.

One is able to concentrate more in their studies and not on other things. Discipline is a virtue that every person should have. There are many benefits associated with discipline. One is able to live a better life when they are disciplined. The absence of discipline brings disorder and chaos. If we do not obey our parents at home, our teachers at school, we can imagine what will happen. So both our parents and the teachers are very anxious to make us see the need for disciplined soldiers.

c) Need for Moral Education in schools:

Education is only complete when it leads to all-round development of the individual, not only mental but also moral development. Moral Education influences the social thinking of the individual and makes him or her distinguish between what is right and what is wrong. Moral Education is important as it teaches diversity, tolerance mutual respect, and moral values.

Moral values are values that express ideas about the good life. Honesty, responsibility, and respect for others is the domain of moral education. The study of morality is vital because we live in a world of rapid change. Education should make children to learn moral values such as truthfulness, honesty, charity, hospitality, tolerance, love, kindness, and simplicity. Moral education should be taught as a separate subject like moral science. Thus moral education is important in schools as it teaches many good things.

d) The importance of Good Health:

Good health is important because a man of health can enjoy great happiness during his lifetime. Without health, we can not do anything in this world. Health is wealth is a common proverb which reveals a very simple meaning by comparing the value of health with wealth. Everyone knows that nothing is important in life than good health. No one can be happy and peaceful without good health.

There is no success in people’s life if they suffer with bad health. In reality, health is more valuable for a person than money because money can not buy good health and happiness intellectually and financially. Maintaining health is not so simple. Good or bad health depends upon several factors including healthy food, environment, lifestyle, sleeping habits, social status, psychological condition, financial condition, and many more things.

e) The importance of Reading:

Reading is one of the most important and priceless activities. If you have ever read a book in life, you will know the pleasure of reading. Reading is the kind of exercise that keeps your mind active and healthy. It is important to develop the habit of reading not only for knowledge but also for personal growth and development. It develops positive thinking and gives you a better perspective of life.

Reading habits develop imagination, knowledge, and vocabulary. The most important reason for reading is that we gain knowledge. Books are a rich source of information and knowledge. It also develops the creative mind of a person. It is the best way to relieve your stress and enhance your mood. It builds your confidence and leads to higher self-esteem. So it is very important to develop a good reading habit.

ஆசிரியரைப் பற்றி:

ராஜநாராயணன் என்பவர் எல்லோராலும் தமிழிலில் கி.ரா. என்று அழைக்கப்படுபவர். இவர் ஒரு தமிழ் நாட்டார் வழக்காற்றியலாளர் (Folklorist) மற்றும் ஏராளமாகப் படைத்தளித்த (prolific) ஆசிரியர். “தி சேர்” என்ற சிறுகதை, 1969ல் நாற்காலி என்ற தலைப்பில் எழுதப்பட்டது. நாவல்கள் கோபால கிராமம் மற்றும் கோபாலபுரத்து மக்கள் அனைவராலும் மிகவும் பாராட்டப்பட்டது (acclaimed). இதற்காக 1991ல் அவருக்கு சாகித்ய அகாதமி விருது வழங்கப்பட்டது. நாட்டார் வழக்காற்றியல் துறையில் இருந்ததால், கி.ரா. 10 ஆண்டுகளில் (decade) நாட்டுப்புற கதைகளை கரிசல் காடுகளில் இருந்து சேகரித்து, இதனை மிகப் பிரபலமான இதழ்களில் வெளியிட்டார்.

2007ல் தஞ்சாவூரில் இருக்கும் புத்தக வெளியீட்டு நிறுவனம் (publishing house) “அன்னம்” இந்த நாட்டுப்புறக் கதைகளை, 944 பக்கங்களைக் கொண்ட ஒரு புத்தகமாகத் தொகுத்து “நாட்டுப்புறக் கதைக் களஞ்சியம்” என்ற தலைப்பில் வெளியிடப்பட்டது. 2009ல் கிட்டத்தட்ட அவர் 30 புத்தகங்களை வெளியிட்டார். அதிலிருந்து ஒரு சில புத்தகங்களைத் தேர்ந்தெடுத்து பிரித்தம் K. சக்கரவர்த்தி என்பவர் ஆங்கிலத்தில் மொழிபெயர்த்து, அதனை வேர் ஆர் யு கோயுங், யு மங்கீஸ்? என்ற தலைப்பில் தமிழ்நாட்டு நாட்டுப்புறக் கதைகளாக, புத்தக வடிவில் 2009ல் வெளியிட்டார்.

பாடத்தைப் பற்றி:

கீழே உள்ள கதையில் ஒரு நாற்காலி இல்லாத குடும்பத்தைப் பற்றி பார்ப்போம். (இருப்பினும் அந்த ஒட்டுமொத்த (entire) கிராமமே நாற்காலியைப் பார்த்ததில்லை ) இங்கு கதை சொல்பவர் (narrator) கிராம மக்கள் அந்த திட்டங்களை செயல்படுத்தும் விதம் பற்றியும் மற்றும் நாற்காலி வந்த பிறகு ஏற்படும் பின்விளைவு (aftermath) பற்றியும் ஒரு நகைச்சுவை சம்பவமாக மிகவும் வேடிக்கையாகக் கூறுகிறார். இதைப் பற்றி விரிவாக கீழே காண்போம்.

The Chair Summary in Tamil

தமிழாக்கம் நாற்காலி இல்லாத வீடா?

திடீரென எங்கள் வீட்டில் இருந்த அனைவரும் இவ்வாறு நினைக்கத் தொடங்கினர். அது என்னவெனில் குடும்ப நிகழ்ச்சி நிரலில் போடப்பட்டிருந்தது (agenda) மற்றும் விவாதங்கள் ஆரம்பிக்கப்பட்டன.

ஒரு நாள் முன்பு, ஒரு குடும்ப நண்பர், எங்கள் வீட்டுக்கு விஜயம் செய்தார். அவர் ஒரு துணை நீதிபதி. சட்டை அணிந்து வரவில்லை. மாறாக அவர் கோட்டும், சூட்டும், காலணியும் அணிந்து வந்திருந்தார். எங்கள் வீட்டில் ஒரு முக்காலி மட்டுமே இருந்தது. அது ஒன்றே முக்கால் மீட்டர் அடி அளவு கொண்டதாக இருக்கும். பாட்டி தயிர் சலிக்கும் போது அந்த முக்காலியில் தான் அமர்ந்திருப்பார். பாட்டி கொஞ்சம் எடை கொண்டவர் அதனால், தாத்தா அதனை தச்சரிடம் கொடுத்து கொஞ்சம் பரந்துபட்ட அளவு உடையதாக மாற்றிக் கேட்டார்.

வீட்டுக்கு வந்திருந்த துணை நீதிபதியும் சிறிய உருவம் கொண்டவர் (portly). எங்கள் வீட்டில் எந்த ஒரு இருக்கைகளும் இல்லாத காரணத்தால், அந்த முக்காலியை அவர் உட்கார கொண்டு வந்தோம். அதன் விளிம்பில் ஒருகையைவைத்தவாறு, அவரை அமரச் செய்தோம். அந்த முக்காலியைப் பற்றி ஒரு மோசமான (wretched)

செய்தி: சரியாக அந்த முக்காலியின் கால்கள் மேல் எடையை (உடலை) அமர்த்தவில்லையென்றால் அது கீழே கவிழ்ந்து (topple) விடும். நாங்கள் இது போன்று பலமுறை கவனக்குறைவினால் கீழே விழுந்திருக்கிறோம். அதன் மீது ஏறி, உறியில் இருந்த நெய்யைத் திருடும் போது (rope net). இந்த செய்தியைச் சொல்வதற்கு முன்னால், துணை நீதிபதி கீழே விழுந்தார்.

கீழே “தொப்” (thud) என்ற மெல்லிய சத்தத்தோடு விழுந்த அவர் உருண்டோடினார். நான் எனது தம்பி மற்றும் என் குட்டித் தங்கை, இதனைக் கண்டு இளித்தவர்களாய் (giggles) பின்புறம் உள்ள தோட்டத்துக்கு ஒடினோம். இந்த உள்ளூர சிரிப்பு தணிந்தது என்று தான் நினைத்தோம். என் தங்கை பல குரலில் நன்றாக பேசுவாள். அவள் அந்த துணை நீதிபதி விழுந்தது போன்று கைகளை சாய்த்தும், தரையில் உருண்டும் செய்து காண்பித்தாள். இதைப் பார்த்து எங்களுக்குச் சிரிப்பு நீளத்தான் செய்தது.

இந்த இளிப்புக்கு இன்னொரு காரணமும் உண்டு. இது போன்று மற்ற விருந்தினர்கள் தரையில் உருண்டு விழும்போது, எங்களின் பெற்றோர் சிரிப்பை அடக்கிக் கொண்டு (suppressing) அவர்கள் முன்னே நிற்பது எங்களின் நினைவுக்கு வந்தது.

அதனால், எல்லோருடைய சார்பாகவும் சிரிப்பை முடிவுக்கு கொண்டு வந்து, நாங்கள் மெல்ல வீட்டுக்குள் அடியெடுத்து வைத்தோம் (pussy footed). ஆனால் அந்த துணை நீதிபதி அங்கு இல்லை . அந்த முக்காலியும் அங்கு இல்லை. அவரோடு அதனை எடுத்துச் சென்று விட்டாரா? என்று என் சகோதரி என்னிடம் கேட்டாள்.

இந்த சம்பவத்திற்குப் பிறகு, எங்கள் வீட்டுக்கு ஒரு நாற்காலி செய்ய வேண்டும் என்று முடிவு எடுக்கப்பட்டது. இதில் ஒரு பிரச்சனை என்னவென்றால், அந்த நாற்காலியை செய்வதற்கு, அந்த கிராமத்தில், மாதிரி காண்பிக்க ஒரு நாற்காலி கூட இல்லை. ஒரு தச்சர் கூட அதனை செய்வதற்கு இல்லை.

“பரவாயில்லை. ஒரு நாற்காலியை நகரத்தில் இருந்து வாங்கி விடலாம்” என்று தம்பி பெடன்னா (Pedanna) பரிந்துரைத்தான். ஆனால் என் தந்தை அதனை நிராகரித்தார். காரணம் அது சாத்தியமிக்கது அல்ல (sturdy) என்று கூறினார்.

எங்கள் அத்தை, பக்கத்து கிராமத்தில் ஒரு கை தேர்ந்த (skilled) தச்சர் இருப்பதாகக் கூறினார். ஆனால், அவர் எந்த ஒரு நாற்காலியும் செய்ததில்லை. அந்த ஆளுநர் தாமே அவருக்கு பரிசாகக் கொடுத்திருந்தார்.

அம்மா, அத்தையின் இரண்டாவது வாக்கியத்தைச் கேட்ட போது, தலையைத் திருப்பி, “அவள் எல்லாவற்றையும் பார்த்திருப்பாள்” என்று கூறுவதைப் போல இருந்தது.

அப்பா வேலைக்காரர் ஒருவரை அழைத்து, அந்த தச்சன் இருக்கும் கிராமத்துக்கு அனுப்பி (despatched) அழைத்து வரச் சொன்னார். அவர் அருகில் வந்து அமர்ந்தார். இப்போது, என்ன வகையான மரத்தைப் பயன்படுத்தலாம் என்ற விவாதம் நடந்தது.

“தேக்குதான் சிறந்தது. அது தான் தூக்குவதற்கும், பிற இடங்களுக்கு நகர்த்தி செல்வதற்கும் சாத்தியம் உள்ளதாக இருக்கும்” என்று தன் பக்கவாத கால்களைத் தடவியவாறே கூறினார். பாட்டி (பாட்டி எப்போதும் தன் கால்களை பக்கவாதத்தின் காரணமாக, ஆட்டிக் கொண்டே இருப்பார்)

அந்நேரம், மாமனார் (தாய் வழி மாமன்) உள்ளே நுழைகிறார். பெடன்னா, ஓடிச்சென்று முக்காலியை எடுத்து வருகிறான். சற்று நேரம், ஒட்டுமொத்த குடும்பமே, உணர்ச்சி மிகுதியினால் (spluttered) இளித்தது. எல்லா நிகழ்வுகளும் சீராவதற்கு முன்.

மாமனார், அவரே, தனக்கென ஒரு இருப்பிடத்தை வீட்டுக்கு வரும்போதெல்லாம் தேர்ந்தெடுத்து கொள்வார். தலையை சாய்த்தவாறு (chop off) அந்த இடத்தைத் தவிர வேறு எங்கும் அமர மாட்டார். அந்த தூணில் அவர் சாய்ந்து கொண்டு இருப்பார். அந்த தூண் பொருட்கள் வைக்கும் அறைக்கு தெற்கே உள்ள சுவருக்கும் அடுத்து இருக்கும், முதல் காரியம் என்ன செய்வார் என்றால் உட்கார்ந்த உடனே, குடுமியை அவிழ்த்துவிட்டு, நன்கு குலுக்கி, தலையைத் நன்கு தேய்த்து, பின்பு இறுகலாக கட்டி விடுவார். இந்த சடங்குகளை (ritual) தவறாமல் அவர் செய்வார். பின்பு அந்த தரையினை ஆய்வு செய்வார். “ஏதாவது பணம் தலையில் இருந்து விழுந்ததா என்று பார்த்தீர்களா? அண்ணா ஒரு அடக்கமற்ற சிரிப்போடு சொல்வார்.

அவர் காகித அம்புகள் போல எப்போதும் துளைத்து கொண்டே இருப்பார். எங்களைப் பார்க்கும் போதெல்லாம் ஒரு புன்னகையுடன்கல்லுப்பிள்ளையார்போல இருப்பார். நீங்கள் என் உறவினர்கள். நீங்கள் கிண்டல் செய்யவில்லையென்றால் வேறு யார் செய்வார்கள்? நாங்கள் செய்யும் ஏளனம் எல்லை மீறி போனால், அம்மா எங்களிடம் கோபித்துக் கொண்டு சென்று விடுவார். கடைசியாக வரும் வார்த்தை, “நீங்கள் கழுதைகள்.

மாமனார் உட்கார்ந்த உடன், அம்மா சமையலறைக்குச் சென்றுவிடுவார். ஒரு ஆட்டுக்குட்டியைப் போல அப்பாவும் அவளைப் பின்தொடர்ந்தார். சிறிது நேரத்தில், அம்மா ஒரு வெள்ளி டம்ளரில் மோர் எடுத்து வந்தாள். பெருங்காயமும் கையில் எடுத்து வந்தாள். அப்பா பின்னால் வந்தது அவளுக்கு தெரியாது. நாங்கள் அவளை, “பார், அண்ணா வந்தவுடன், எவ்வாறு அக்கறையோடு கையில் மோர் கொண்டு கொடுக்கிறாள்”. மோரின் வாசனையும், பெருங்காயத்தின் வாசனையும் எங்களையும் குடிக்கத் (asafoetida) தூண்டியது.

நாங்கள் மாமனார் மோர் குடிப்பதற்காகவே வந்திருக்கிறார் என்று நினைத்தோம். எங்கள் பசுவிடம் இருந்து கிடைக்கும் மோர் மிகவும் தெய்வீகம் உடையதாக இருக்கும். இதற்கிடையில் எங்கள் மாமனார், அந்த கிராமத்தில் மிகப்பெரிய கஞ்சன் (stingiest) என்று கேள்விப்பட்டோம். அவரை எதுவும் யாருக்கும் கொடுக்க மாட்டார் என்று நம்பினோம், அவர் தானே கண்ணாவரத்திற்குச் சென்று புகழ்பெற்ற கருப்பு நாக்கு கறவை மாடு, அவரின் தங்கைக்காக வாங்கச் சென்றார். என் தம்பியும் தங்கையும் அந்த கன்றுக்குட்டியுடன் பேரன்பு கொண்டு இருந்தனர்.

மாமனார் இங்கு வரும் போதெல்லாம், அந்த மாட்டின் அருகில் சென்று தட்டிக் கொடுப்பார். (pat) (யாருடைய கண்களும் பட்டுவிடக் கூடாது என நினைப்பார்) சிக்கனமாக நடந்து கொள்வார். இதனால் அந்த குட்டிகள் எப்போதும் பயத்துடனே இருந்தன. மாட்டில் பால் வற்றி விட்டது என்றால் அவர் அந்த மாட்டை, கன்றுகளோடு அவருடைய வீட்டுக்கு அழைத்துச் சென்றுவிடுவார்.

இது நிகழ இருப்பதைக் கண்டு, மாமனாரின் மேல் ஒரு கசப்புத் தன்மையை உருவாக்கியது. அவர் எப்போதெல்லாம் மோரை மிக்க மகிழ்ச்சியோடு (relish) குடிக்கிறாரோ, சிறு பிள்ளைகள் அவரை முறைத்துப் பார்ப்பர்.

மாமனார் அந்த நாற்காலி பற்றி விவாதத்தில் பங்கு கொள்ள ஆர்வம் காட்டினார். தனக்கும் ஒன்று வேண்டும் என்று கூறினார். இதனால் எங்களுக்கும், அவர் உடன் வருவது மகிழ்ச்சி அளித்தது.

மாமனார் வேப்ப மர கட்டையில் நாற்காலி செய்தால் நல்லது என்றார். அது எப்போதும் உடலை குளிர்ச்சியாக வைத்திருக்கும் என்று கூறினார். – அவர் வேப்ப மர கட்டை பற்றி விளக்கும் போது (expounded) அப்பா, வட்டக் கண்களோடு அவரைப் பார்த்து ஆச்சரியப்பட்டார்.

(astonished) ஒரு நாளைக்கு முன்பு தான், அப்பா ஒரு பழமையான வைரம் பொருந்திய வேப்ப மரத்தை வெட்டுவதற்கு ஒரு கூலியாளிடம் (farmhand) பேசிக் கொண்டிருந்தார். பூவரச மரத்தில் நாற்காலி செய்தால் நல்லது. அது ஒரு பளபளப்பான (glossy), நன்றாக தானியங்கள் முடிச்சுகள் இல்லாத கட்டை மற்றும் பலத்திலும் சிறந்தது என்று பெடன்னா கூறினான்.

எங்கள் மூத்த சகோதரி, “இதெல்லாம் மெல்லிய நிற கட்டைகள், பார்ப்பதற்கு அழகாக இருக்காது என்று கூறினாள். சிறிது நேரத்தில் நாம் அவற்றை வெறுத்துவிடுவோம் (detest). பழுத்த கரும்பு அல்லது எள் எண்ணெய் கலரில் இருந்தால் நல்லது அப்புறம் உங்கள் இஷ்டம்” என்று கூறினாள். கருப்பு மரத்தில் கண்ணாடி போன்ற தோற்றத்துடன், நன்றாக வடிவமைக்கப்பட்ட கால்கள், வளைந்த பின் கால்கள், சோர்வு இல்லாத தோற்றம் தான் சாய்வு ஏற்படுத்தக்கூடிய ஒரு நல்ல – நாற்காலி. அந்த ஆடம்பர நாற்காலி தான் எங்கள் கண்களில் முன் நிழலாடியது.

எல்லோரும் அவள் சொல்வது சரி என்று ஆமோதித்தனர். எனவே, இரண்டு நாற்காலிகள் உடனடியாக செய்ய முடிவு. – எடுக்கப்பட்டது. ஒன்று எங்களுக்கு, மற்றொன்று மாமனாருக்கும்.

இரண்டு நாற்காலிகளும் வந்து சேர்ந்த போது, எந்த நாற்காலியை நாங்கள் வைத்துக் கொள்வது என்றும், எதை மாமனாருக்கு அனுப்புவது என்று தெரியவில்லை. ஒன்றைப் பார்த்தால், மற்றொன்றைப் பார்க்க அவசியமில்லை. இரண்டுமே ராமன் லெட்சுமணன் போல உருவத்தில் ஒத்து இருந்தது. நாங்கள் ஒன்றை வைத்துக் கொண்டு மற்றொன்றை. – மாமனாருக்கு அனுப்பிவிட்டோம். ஆனால் ஒரு சந்தேகம். நல்லதை அவருக்கு அனுப்பிவிட்டோமோ என்று?

ஒருவர் பின் ஒருவராக, அந்த புதிய நாற்காலியில் உட்கார நாங்கள் வரிசை அமைத்துக் கொண்டோம். அதிலிருந்து. எழுந்து செல்ல எங்களுக்கு மனமில்லை. ஆனால் மற்றவருக்கும் வாய்ப்புக் கிடைக்க வேண்டும் என்ற நியதிக்காகத் தான் எழ வேண்டியிருந்தது. பெடன்னா உட்கார்ந்து பார்த்துக் கொண்டு, ஆச்சரியத்தோடு ஆஹா ஹா என்று பாராட்டினான்.. இரு கைகளையும் அந்த நாற்காலியின் மேல் தூக்கினான். கால் மேல் கால் போட்டு அமர்ந்தான். “இதற்கு ஒரு உறை தைக்க வேண்டும். இல்லையெனில் அது மண்ணாகி விடும்” என்று அத்தை கூறினார்.

என்னுடைய இளைய சகோதரியும், தம்பியும் அந்நாற்காலிக்காக அடிக்கடி சண்டையிட்டனர். அதனால் அவள் நீ”, ரொம்ப நேரமாக அதில் உட்கார்ந்து கொண்டு இருக்கிறாய். எழுந்திரு. நானும் இப்போது அமர வேண்டும் என்று கத்தினாள்.

‘அய்யோ நான் இப்போது தான் அமர்ந்தேன். அம்மா அவளைப் பாருங்கள்’, என்று கூறுவான். அம்மாவைப் பார்த்து திரைப்புள்ள முகத்தோடு அழுவது போன்று கத்துவான்.

நாற்காலி வீடு வந்து சேர்ந்த விஷயம் காட்டுத்தீ போல் அக்கிராமம் முழுவதும் பரவியது. பெரியவர்கள், குழந்தைகள், முதியவர்கள் என அனைவரும் கூட்டமாக வந்து (hordes) பார்த்துச் சென்றனர். ஒரு சிலர் அதனை பக்கவாட்டில் தட்டிப் பார்த்தனர். ஒரு முதியவர் அதனை தூக்கி, கொஞ்சம் பாரமாகத்தான் இருக்கிறது. நல்ல துணிச்சலோடு செய்து இருக்கிறார்.” என்று அந்த தச்சரைப் பாராட்டினார் (commended).

சில நாட்கள் கழித்தன.

ஓர் இரவு யாரோ ஒருவர் வந்து கதவைத் தட்டினார். உள் திண்ணையில் தூங்கிக் கொண்டிருந்த பெடன்னா, எழுந்து கதவைத் திறந்தான். கிராமத்தில் உள்ள முக்கியமான நபர் ஒருவர் இறந்து விட்டதாகவும், அதற்கு நாற்காலி தேவைப்படுகிறது. என்று வந்தவர் கூறினார்.

இறந்தவர் யார் என்று எங்களுக்கும் தெரியுமாதலால் நாங்களும் அந்த இறுதிச்சடங்கில் கலந்து கொண்டோம். அங்கே எங்கள் நாற்காலியில் தான் இறந்தவரை உட்கார வைத்து முட்டுக்கொடுத்திருந்தனர் (propped).

ஆனால் இதுவரை எங்கள் கிராமத்தில் இறந்தவரை தரையில் தான் படுக்க வைத்திருப்பது வழக்கம். ஒரு பெரிய ஆட்டுக்கல்லை, தரையில் வைத்து, உருண்டு விடாமல் இருக்க முட்டுக் கொடுத்திருப்பர். ஒரு கோணிப்பையை எடுத்துக்கொண்டு அதனுள் தினை வைக்கோலை அடைத்து வைத்திருப்பர். அதன் மீது உயர்த்தி, அந்தப் பிணத்தைச் சாய்த்து வைத்திருப்பர் (bolster).

இதிலிருந்து தான் நகரத்துவாசிகளும் இந்த புதிய முறைக்கு மாறினார் (fad) அவர்களும் பிணத்தை நாற்காலியில் சாய்த்து வைக்கத் தொடங்கினர். எங்களுக்கு எந்த துப்பும் இல்லை. ஆனால் அன்றிலிருந்து தான் எங்கள் நாற்காலிக்கு ஒரு புதிய பிரச்சனை வரத் தொடங்கியது. (தரையில் படுக்க வைத்தவர்கள் இப்போது நாற்காலியில் சாய்த்து வைக்க முன்னேறினர்.

எல்லா நிகழ்வுகளும் அந்த வீட்டில் (துக்க வீட்டில்) முடிந்த பின்னர், அவர்கள் அந்த நாற்காலியினை எங்கள் முன் முற்றத்தில் வைத்துவிட்டனர். வீட்டில் உள்ள குழந்தைகள் யாவரும் அந்த நாற்காலியை பார்க்கப் பயந்தனர். நாங்கள் ஒரு வேலைக்காரனை அமர்த்தி, அதனை கிணற்றுக்குத் தூக்கிக் கொண்டு போய், நன்றாக வைக்கோலினால் தேய்த்து (scrub) அதை மீண்டும் 15 பெரிய வாளிகள் தண்ணீர் கொண்டு தேய்த்து கழுவினோம்.

வெகு நாட்களுக்கு பின்பு கூட, அந்த நாற்காலியில் அமர யாருக்கும் தைரியம் இல்லை (guts) எங்களுக்கு மீண்டும் அந்த நாற்காலியை எப்படி புழக்கத்திற்கு கொண்டு வர வேண்டும் என்று தெரியவில்லை.

ஒரு நாள் அதிர்ஷ்டவசமாக (fortuitously) ஒரு உறவினர் ஒருவர் வீட்டுக்கு வந்தார். அவருக்கு அந்த நாற்காலியை அமருவதற்குக் கொண்டு வந்தோம். இல்லை தம்பி வேண்டாம். நான் இதில் உட்கார்ந்து கொள்கிறேன் என்று அந்த நாற்காலியில் உட்காராமல், அங்கே கிடந்த துணி பாயை (cloth mat) நோக்கி நகர்ந்தார். நாங்கள் அவர் தரையில் உட்கார்ந்து விடுவார் என நினைத்தோம், பயந்தோம். அவரை நாற்காலியில் அமர சம்மதித்தோம்.

அவர் உட்கார்ந்த உடனே சின்னத் தம்பியும், தங்கையும் பின்புறத்தில் உள்ள தோட்டத்திற்கு தப்பி ஒடினர் (fled). சில சமயங்களில் அவர்கள் அங்கிருந்து வந்து நாற்காலியில் உட்கார்ந்து இருப்பவருக்கு எதுவும் நடந்துவிட்டதா என்று உற்றுப் பார்த்தனர்.

மறுநாள், எங்கள் ஊரில் உள்ள ஒரு மூத்தவர் நடந்து செல்லும் போது, தானாகவே, அந்த நாற்காலியில் வந்து அமர நினைத்தார். அது எங்களுக்கு மிகுந்த ஆறுதலைத் தந்தது. (இப்போதே அந்த நாற்காலியில் அமர ஒத்திகை பார்க்கிறார் என்று பெடன்னா என் காதுகளில் ஓதினான்). இப்படித்தான் நாங்கள் அந்த நாற்காலியைப் பயன்படுத்தினோம் (seasoned).

முதலில் அந்த நாற்காலியில் மூத்தவர்கள் அமர்ந்து பார்த்தனர். ஆனால் குழந்தைகள் பயந்து கொண்டு இருந்தனர். என்னுடைய சகோதரி, நீ ஏன் முதலில் போய் அதில் அமரக் கூடாது என்று என் தம்பியை நோக்கி கெஞ்சினாள். அதற்கு அவன், நீ ஏன் முதலில் போய் உட்காரக் கூடாது என எதிருரைத்தான்.

பக்கத்துத் தெருவைச் சேர்ந்தவர் சுகந்தி, அவர் வந்து தனது குழந்தைத் தம்பியை நாற்காலியில் வைத்தார். அன்றிலிருந்து தான், எங்கள் வீட்டில் உள்ள குழந்தைகளும் எவ்வித பயமும் இல்லாமல், அந்த நாற்காலியில் அமரத் தொடங்கினர்.

மீண்டும், ஓர் இரவு, இன்னொருவர் இறந்து போகவே, அந்த நாற்காலியைத் தூக்கிச் சென்றனர். இது அடிக்கடி நிகழ்ந்தது. நாங்கள் ஒவ்வொரு முறையும் சோகத்தில் அனுப்பினோம். துக்க வீட்டினர், எங்கள் சோகத்தை வேறு வழியில்லாமல் புரிந்து கொண்டனர். நாங்கள் இறந்து போனவர்களுக்காக வருந்துறோம் என்று தவறாக நினைத்துக் கொண்டனர்.

ஒவ்வொரு முறையும், எங்கள் தூக்கம் கெடுக்கப்படுவதை நினைத்து வருந்தினோம். அக்காள் ஒரு நாள், எதற்கு இந்த மோசமான (wretched) மனிதர்கள் நிலவுலகில் உரியதாக இயலும் போது, அந்நேரத்தில் ஏன் இறக்க வேண்டும் என்று கடவுளுக்குத் தெரியும் என்று குறிப்பிட்டாள். அண்ணா உற்சாகமாக (exasperatedly) நாம் நல்ல ஒரு நாற்காலியை உருவாக்கினோம். பிணங்களை (corpses) உட்கார வைக்க வேண்டிய நிலை வந்து விட்டதே …. சீ என்றான்.

இதெல்லாம் அந்த நாற்காலியை அமங்கலமான நேரத்தில் செய்ய சொன்னதே தான் காரணம் என்றாள் அத்தை.

பெடன்னா இறுதியாக ஒரு யோசனையை நினைத்தான். ஆனால் நாங்கள் இருவருக்குள்ளும் அதை மூடி மறைத்து வைத்துக் கொண்டோம்.

அம்மா என்னை ஒரு முறை மாமனாரின் வீட்டுக்கு தூது அனுப்பினார் (errand) நான் நுழைந்த வேளை, அங்கே மாமனார், அற்புதமாக நாற்காலியில் அமர்ந்திருந்தார். வெற்றிலையை மடித்து வாயில் போட்டு மென்று கொண்டிருந்தார். இது பார்ப்பதற்கு ஒரு இனிமையான தருணமாக இருந்தது. வெற்றிலை பெட்டியை, அவர் மிகவும் கவனமாக கையாள்வதும், அதற்கு வலிக்காதவாறும் அதனைத் திறந்தார்.

பரந்த அளவில் (span wide) முழங்கை நீளமான, நான்கு விரல்கள் உயரம் கொண்ட அந்த வெற்றிலையை, தினமும் பொன்னைப் போன்று மின்னும் வரை துடைத்தார். அந்த சாதனங்களை பயபக்தியோடு பூஜைப்பெட்டியில் இருந்து எடுப்பதைப் போன்று எடுப்பார்.

அவர் வெற்றிலையைத் தூய்மையாகத் துடைத்தாலும் அதன் தண்டினை (stalk) கிள்ளி எறிய மாட்டார். மிகவும் சிக்கனமானவர். அவர் ஒரு கரடுமுரடான (coarse leaf) இலையைக் கண்டாலும், அதன் நரம்புகளை துண்டு துண்டாக எடுத்துவிடுவார். இது வெற்றிலை பற்றிய ஒரு பழைய புதிர் மந்திரத்தை நினைவூட்டுகிறது.

முத்தப்பனை பிடித்துக் கொண்டு
அவன் முதுகெலும்பை நீக்கு
தூய்மையான வெண்ணெய் கொண்டு பூசு …..

அவர் உடைந்த பாக்கினை நுகர்ந்து பார்ப்பார். இது நுகர்ந்து பார்த்தலின் போது போதை ஏறுவதை விரட்டுகிறது (ward off) பாக்கில் இருக்கும் சின்னச் சின்ன பூச்சிகளை விரட்ட இவ்வாறு செய்யலாம். ஊதுதல் மூலம் ஒளிந்திருக்கும் பூச்சிகளை விரட்டலாம். முதலில் மெதுவாக ஆரம்பித்து பின்பு வேகம் எடுக்கும் பின்பு வாயில் போட்டு மென்று விடுவார்.

ஒருவர் எவ்வாறு சுத்தமாக இருக்கிறார் என்பதை அறிய, அவர் வைத்திருக்கும் சுண்ணாம்பு டப்பாவை பார்த்தாலே போதும். மாமனார் இதற்கு சற்றும் பொருந்தாதவர். விரலிலிருக்கும் மிஞ்சிய சுண்ணாம்பு பசையைக் கூட அடுத்த பொருட்களின் மீது துடைக்க மாட்டார். பயபக்தியோடு அவரின் சுண்ணாம்பு டப்பாவை கண்ணில் ஒற்றி எடுத்துக் கொள்ளலாம். அவரின் எவெரெடி டார்ச் விளக்கு 15 வருடங்களுக்கு முன் வாங்கப்பட்டது.

இன்னும் உபயோகத்தில் உள்ளது. ஆயினும் பளிச் என இருப்பதோடு இப்போது தான் வாங்கியது போன்று இருக்கும். அதோடு எங்கள் குடும்பத்துக்கும் சேர்த்து வாங்கிய டார்ச் லைட், துவாரம் உடையதாகவும், மஞ்சள் நிறத்தில், பரிதாபமிக்க நிலைக்குத் தள்ளப்பட்டு, நாட்பட்ட (chronic) நோயாளி சாகும் தருவாயில் இருப்பதைப் போல் இருக்கிறது.

அவரது வீட்டில் அவரைத் தவிர அந்த நாற்காலியை யாரும் உபயோகப்படுத்த மாட்டார்கள். காலை எழுந்தவுடன் ஒவ்வொரு நாளும் முதல் வேளையாக அதனை நன்கு துடைத்துவிடுவார். ஒரு மண் பானையில் தண்ணீர் உள்ளதைப் போல நினைத்து, அவரே அதனை கவனமாக தூக்குவதும், ஓரிடத்திலிருந்து மற்றொரு இடத்திற்கு மாற்றுவதும் அவரே செய்வார்.

என்னைப் பார்த்து, மாமனார் வாழ்த்து தெரிவித்து வரவேற்றார். மருமகனே, வாருங்கள். வெற்றிலை சாப்பிடுகிறீர்களா? பின்பு அவரே அதற்குப் பதிலளித்தார். பவுளிப்பருவத்து மாணவர்கள் இப்போதே வெற்றிலை. மென்றால், சிறு கோழிக்குஞ்சுகள் கூட பீப்பாய் போல உருமாறிவிடும் (butting).

நான், அம்மா கூறியவற்றை அவரிடம் சொல்லிவிட்டு வீடு திரும்பினேன்.

பின்பு, அன்று ஒரு தேவ பக்தியற்ற இரவில், கதவை யாரோ தட்டினர். எல்லோரும் ஆழ்ந்த தூக்கத்தில் இருந்தனர். நான் பெட்டனாவை எழுப்பினேன். வீட்டுக்கு வெளியே ஒரு சிலர் ஒரு துயரமடைந்த வீட்டில் இருந்து வந்து நின்று கொண்டிருந்தனர். அவர் நாற்காலி வாங்குவதற்காக நின்று கொண்டிருந்தனர். பெடன்னா அவர்களை தெருவுக்குள் அழைத்துச் செல்ல நானும் அவர்களைப் பின்தொடர்ந்தேன். அவர்கள் எதற்காக வந்திருக்கிறார்கள் என்பதை எங்களிடம் சொல்லி முடித்த பின்பு, பொறுமையாக பெடன்னா பதிலளித்தான். நாற்காலி தானே. அது என்னுடைய மாமனார் வீட்டில் உள்ளது. நீங்கள் போய் அவரிடம் கேளுங்கள். தருவார். அவர்களை அனுப்பிய பின் நாங்கள் வீட்டுக்குத் திரும்பினோம். சத்தமில்லாமல் யார் அது? தூக்கத்தில் அப்பா கேட்டார்.

“ஒன்னும் இல்லை ….” சிலர் நமது காளைகளை இதிரடிப்பதற்கு (threshing) கேட்டனர்.

போர்வையை இழுத்துப் போர்த்தியவராய் அப்பா மீண்டும் படுக்கைக்குச் சென்றார். இப்போது மாமனாரின் காட்டில் மழை. சில நாட்கள் கழித்து, மாமனாரை சந்தித்த போது அவர் தரையில் உட்கார்ந்திருந்தார். வெற்றிலை மெல்ல தயார் படுத்திக் கொண்டிருந்தார். அவர் வழக்கமான புன்னகையை உதிர்த்து என்னை வரவேற்றார்.

“இது என்ன? தரையில் அமர்ந்து இருக்கிறீர்கள், நாற்காலி எங்கே?” நான் தேடினேன். சுண்ணாம்பு பசையை வெற்றிலையில் பின்புறத்தில் தடவியவராய் என்னைப் பார்த்து புன்னகைத்தார். பின்பு மெல்லிய குரலில் கூறினார். நான் அவர்களை, அந்த நாற்காலியை அந்த ஒன்றுக்காக மட்டும் பயன்படுத்திக் கொள்ளுங்கள் என்று கூறினேன். இருப்பினும் உனக்கு இன்னொன்று தேவைப்படுகிறது என்றார்.

எனக்கு என்ன சொல்வது என்று தெரியவில்லை வீட்டுக்கு விரைந்தேன். இந்த செய்தியை சொல்வதற்காக பெடன்னாவைத் தேடிச் சென்றேன். ஆனால் என்னுடைய நகர்வுகள் அனைத்தும் சாதாரண நடையாக மாறியது (gait).

Tamilnadu State Board New Syllabus Samacheer Kalvi 9th Science Guide Pdf Chapter 14 Acids, Bases and Salts Text Book Back Questions and Answers, Notes.

Samacheer Kalvi 9th Science Solutions Chapter 14 Acids, Bases and Salts

9th Science Guide Acids, Bases and Salts Text Book Back Questions and Answers

I. Choose the correct answer :

Question 1.
Zn + 2HCl → ZnCl₂ + ….↑ (H₂, O₂, CO₂)
Answer:
H₂

Question 2.
Apple contains malic acid. Orange contains …………………(citric acid, ascorbic acid).
Answer:
ascorbic acid

Question 3.
Acids in plants and animals are organic acids. Whereas Acids in rocks and minerals are …………………. (Inorganic acids, Weak acids).
Answer:
Inorganic acids

Question 4.
Acids turn blue litmus paper to …………….. (green, red, orange).
Answer:
Red

Question 5.
Since metal carbonate and metal bicarbonate are basic, they react with acids to give salt and water with the liberation of ……………….. (NO₂, SO₂, CO₂).
Answer:
CO₂

Question 6.
The hydrated salt of copper sulphate has …………….colour (red, white, blue).
Answer:
Blue

II. Answer in briefly :

Question 1.
Classify the various types of Acids based on their sources.
Answer:
The acids are classified based on their sources and organic and inorganic acids.
Organic acids – acids present in plants and animals.
Inorganic acids – acids prepared from rocks and minerals.

Question 2.
Write any four uses of acids.
Answer:

• Sulphuric acid is called King of Chemicals because it is used in the preparation of many other compounds. It is used in car batteries also.
• Hydrochloric acid is used as a cleansing agent in toilets.
• Citric acid is used in the preparation of effervescent salts and as a food preservative.
• Nitric acid is used in the manufacture of fertilizers, dyes, paints and drugs.
• Oxalic acid is used to clean iron and manganese deposits from quartz crystals. It is also used as bleach for wood and removing black stains.
• Carbonic acid is used in aerated drinks.
• Tartaric acid is a constituent of baking powder.

Question 3.
Give the significance of p^H of .soil in agriculture.
Answer:
In agriculture, the p^H of soil is very important. Citrus fruits require slightly alkaline soil, while rice requires acidic soil and sugarcane requires neutral soil.

Question 4.
What are the various uses of Aquaregia.
Answer:

• It is used chiefly to dissolve metals such as gold and platinum.
• It is used for cleaning and refining gold.

Question 5.
What are the uses of Plaster of Paris?
Answer:

• It is used for plastering bones.
• It is used for making casts for statues.

Question 6.
Two acids ‘A’ and ‘B’ are given. Acid A gives one hydrogen ion per molecule of the acid in solution. Acid B gives two hydrogen ions per molecule of the acid in solution.
(i) Find out acid A and acid B.
(ii) Which acid is called the King of Chemicals?
Answer:
(i) Acid A – HCl – Hydrochloric acid. Acid B – H₂SO₄ – Sulphuric acid.
(ii)Sulphuric acid – H₂SO₄.

Question 7.
Define aquaregia.
Answer:
It is the mixture of hydrochloric acid and nitric acid prepared optimally in a molar ratio of 3 : 1.

Question 8.
Correct the mistakes :
(a) Washing soda is used for making cakes and bread soft, spongy.
(b) Calcium sulphate hemihydrate is used in textile industry.
Answer:
(a) Baking soda (Sodium bicarbonate – NaHCO₃) is used for making cakes and bread soft spongy, (or) Washing soda is used for softening hard water.
(b) Calcium sulphate hemihydrate (CaSO₄, 1/2 H₂O)is used for plastering bones (or) Bleaching powder (Calcium oxy Chloride – CaOC₁₂) is used in textile industry.

Question 9.
What is neutralization reaction? Give an example.
Answer:
Neutralization reaction is a reaction in which an acid reacts with a base to form salt and water and H⁺ ion and OH^– ion combines to generate water. The neutralization of a strong acid and strong base has a p^Hequal 7.

III. Answer in detail :

Question 1.
Differentiate hydrate and anhydrous salts with examples.
Answer:

Hydrated	Anhydrous
1. Hydrons is a term used to explain a substance that contains water as a constituent	Anhydrous is a term used to explain a substance that does not contain water as a constituent.
2. Composed of water molecules.	Not composed of water molecules.
3. These are known as hydrates.	Known as anhydrates
4. Hydroscopic compounds can form hydrous compounds by Absorption of water from the air.	Anhydrous compounds can absorb water from the air.
Ex : CuSO4 5H2O – Blue vitrol.	Ex: NaCl

Question 2.
Give the tests to identify Acids and Bases.
Answer:

• Acids turn blue litmus. Red bases turn red litmus blue.
• In acid phenolphthalein is colourless. In base Phenolphthalein is pink in colour.
• In acid methyl orange is pink. In bases methyl orange is yellow

Question 3.
Write any four uses of bases.
Answer:

• Sodium hydroxide is used in the manufacture of soap.
• Calcium hydroxide is used in the whitewashing of buildings.
• Magnesium hydroxide is used as a medicine for a stomach disorder.
• Ammonium hydroxide is used to remove grease stains from cloth.

Question 4.
Write any five uses of salts.
Answer:
Common Salt (NaCl) :
It is used in our daily food and used as a preservative
Washing Soda (Sodium Carbonate – Na₂CO₃) :

• It is used in softening hard water.
• It is used in glass, soap and paper industries.

Baking Soda (Sodium bicarbonate -NaHCO₃):

• It is used in making of baking powder which is a mixture of baking soda and tartaric acid.
• It is used in soda-acid fire extinguishers.
• Baking powder is used to make cakes and bread, soft and spongy.
• It neutralizes excess acid in the stomach and provides relief.

Bleaching powder (Calcium Oxychloride – CaOCl₂):

• It is used as a disinfectant.
• It is used in textile industry for bleaching cotton and linen.

Plaster of Paris (Calcium Sulphate Hemihydrate – CaSO₄ .1/2 H₂O):

• It is used for plastering bones
• It is used for making casts for statues.

Question 5.
Sulphuric acid is called King of Chemicals. Why is it called so?
Answer:
Sulphuric acid is called King of Chemicals because it is used in the preparation of many other compounds

Intext Activities

ACTIVITY – 2

Take solutions of hydrochloric acid or sulphuric acid. Fix two nails on a cork and place the cork in a 100 ml beaker.
Connect the nails to the two terminals of a 6 V battery through a bulb and a switch as shown in Figure. Now pour some dilute HCl in the beaker and switch on the current. Repeat the activity with dilute sulphuric acid, glucose and alcohol solutions. What do you observe now? Does the bulb glow in all cases?

Answer:

• The bulb glows when sulphuric acid and hydrochloric acid are used.
• The bulb does not glow when the activity is done with alcohol and glucose solution.

ACTIVITY – 3

Collect the following samples from the science laboratory – Hydrochloric acid, Sulphuric acid and Nitric acid, Sodium hydroxide, Potassium hydroxide. Take 2 ml of each solution in a test tube and test with a litmus paper and indicators phenolphthalein and Methyl orange. Tabulate your observations.

Answer:

ACTIVITY-4

Answer:

ACTIVITY – 5
Boil about 100 ml of groundwater in a vessel to dryness. After all the water get evaporated observe the inner wall of the vessel. Can you observe any deposits? This is the deposit of dissolved salts present in water.
Answer:
This is the deposit of dissolved salts present in water.

9th Science Guide Acids, Bases and Salts Additional Important Questions and Answers

I. Choose the correct answer :

Question 1.
Acid secreted in our stomach is ……………….
(a) hydrochloric acid
(b) sulphuric acid
(c) nitric acid
(d) carbonic acid
Answer:
(a) hydrochloric acid

Question 2.
Hydrochloric Acid reacts with metal bicarbonates to give ……………
(a) metal chloride
(b) water
(c) carbon di – oxide
(d) all the above
Answer:
(d) all the above

Question 3.
…………… & ………….. metals do not react with HCl or HNO₃.
(a) Gold & Magnesium
(b) Silver & Magnesium
(c) Gold & Silver
(d) Zinc & Silver
Answer:
(c) Gold & Silver

Question 4.
The molar ratio of hydrochloric acid and nitric acid in aquaregia is ………………..
(a) 1 : 3
(b) 6 : 3
(c) 2 : 3
(d) 3 : 1
Answer:
(d) 3 : 1

Question 5.
Bases ionise in water to form ……………………. ions.
(a) H⁺
(b) H₃O⁺
(c) OH^–
(d) O^2-
Answer:
(c) OH ^–

Question 6.
Which of the following pairs are weak base?
(a) NH₄OH & NaOH ‘
(b) Ca(OH)₂ & KOH
(c) NH₄OH & Ca(OH)₂
(d) NaOH&KOH
Answer:
(c) NH₄OH & Ca(OH)₂

Question 7.
NaOH & KOH are …………….
(a) strong bases
(b) metal Oxides
(c) weak bases
(d) diacidic bases
Answer:
(a) strong bases

Question 8.
Which of the following solution is soapy to touch?
(a) Acidic
(b) Basic
(c) Salt
(a(d) Aquaregia
Answer:
(b) Basic

Question 9.
Which of the following solutions do not conduct electricity?
(a) alcohol
(b) glucose
(c) sulphuric acid
(d) both a & b
Answer:
(d) both a & b

Question 10.
The p^H value of neutral solution is ………………
(a) = 7
(b) <7
(c) none of the above
(d) -7
Answer:
(a) = 7

Question 11.
The p^H of stomach fluid is ……………….
(a) 4
(b) 2
(c) 6
(d) 7
Answer:
(b) 2

Question 12.
A salt which is formed by complete neutralization of an acid and a base is called ………………. salt.
(a) basic
(b) acid
(c) double
(d) normal
Answer:
(d) normal

Question 13.
The number of water molecules present in one molecule of copper sulphate is
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(d) 5

Question 14.
The molecular formula of copper sulphate pentahydrate is ………………..
(a) CuSO₄. H₂O
(b) CuSO₄. 2H₂O
(c) CuSO₄. 5H₂O
(d) CuSO₄. 3H₂O
Answer:
(c) CuSO₄ . 5H₂O

Question 15.
……………is the gas produced when HCl is added with carbonate salt.
(a) H₂
(b) N₂
(C) CO₂
(d) O₂
Answer:
(c)CO₂

Question 16.
The formula of bleaching powder is ……………..
(a) CaCl₂
(b) CaOCl₂
(c) Ca(OH)₂
(d) CaO
Answer:
(b) CaOCl₂

Question 17.
The chemical name of plaster of paris is ……………………..
(a) Calcium sulphate hemihydrate
(b) Calcium sulphate monohydrate
(c) Calcium sulphate dihydrate
(d) Calcium sulphate trihydrate
Answer:
(a) Calcium sulphate hemihydrate

Question 18.
Which of the following metal does not react with sodium hydroxides?
(a) Cu
(b) Ag
(c) Cr
(d) All the above
Answer:
(d) All the above

Question 19.
Curd contains …………….acid.
(a) malic ‘
(b) formic
(c) lactic
(d) ascorbic
Answer:
(c) lactic

Question 20.
Which one of the following acids undergoes complete ionisation?.
(a) HCl
(b) CH₃COOH
(c) H₂SO₄
(d) all the above
Answer:
(a) HCl

II. Fill in the blanks :

1. Acid reacts with base to form a neutral product called ………………
Answer:
Salt

2. The taste of acid is ………………..
Answer:
Sour

3. …………….. contain one or more replaceable hydrogen atoms.
Answer:
Acids

4. ………………..acids have a relatively smaller amount of acids dissolved in a solvent.
Answer:
Dilute

5. Acids react with metallic oxides to produce ……………….
Answer:
salt and water

6. _________ acid is used in aerated drinks.
Answer:
Carbonic acid

7. Chemical formula of aquaregia is ________
Answer:
3HCl + HNO₃

8. Water soluble bases are called ________
Answer:
alkali

9. Non-metallic oxides are ___________ in nature.
Answer:
acid

10. __________ are bitter in taste.
Answer:
Bases

11. __________ alkali has a relatively high percentage of alkali in its aqueous solution.
Answer:
Concentrated

12. Acids turn blue litmus to ………….
Answer:
red

13. Phenolphthalein and methyl Orange are ………………
Answer:
indicators

14. p^H stands for …………….. in a solution.
Answer:
power of hydrogen ion concentration

15. The p^H value of acids are …………….. than 7.
Answer:
lesser

16. White enamel coating of our teeth is…………….
Answer:
Calcium phosphate

17. Salt is ………….. in nature.
Answer:
hygroscopic

18. Salt which is formed by the partial replacement of hydrogen ions of an acid by a metal is called ………………………..
Answer:
acid salt

19 Salts contaíning water of crystallisation are called …………. salts.
Answer:
hydrated

20. Salts that do not contain water of crystallisation is called …………..
Answer:
anhydrous salt

21. p^H value of human blood is ……………..(7.0, 7.4, 7.6).
Answer:
7.4

22. The nature of the toothpaste commonly used is ……………. in nature (acidic, basic, neutral)
Answer:
basic

23. You are given pure water to test the p^H value using p^H paper. It shows colour (White, black, green)
Answer:
green

III. To Match:

Question 1.

Source	Acid present
1) Apple	?
2) ?	Citric acid
3) ?	tartaric acid
4) Tomato	?
5) Vinegar	?

Answer:

Source	Acid present
1) Apple	Malic acid
2) Lemon	Citric acid
3) Grapes	tartaric acid
4) Tomato	Oxalic acid
5) Vinegar	Acetic acid

Question 2.

Answer:

Question 3.

Answer:

Question 4.

Answer:

IV. Complete the following equations.

Question 1.
HCl + H₂O → ? + ?
Answer:
HCl + H₂O → H₃O⁺ + Cl^–

Question 2.
H⁺ + H₂O → ?
Answer:
H⁺ + H₂O → H₃O⁺

Question 3.
Mg+ ? → ? + H₂
Answer:
Mg + H₂SO₄ → MgSO₄ + H₂↑

Question 4.
Na₂CO₃ + 2HCl → ? + ? + CO₂↑
Answer: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂↑

Question 5.
ZnO + 2HCl → ? + ?
Answer:
ZnO + 2HCl → ZnCl₂ + H₂O↑

Question 6.
Zn + ? → Na₂ZnO₂ + H₂ ↑
Answer:
Zn + 2NaOH → Na₂ZnO₂+ H₂↑

Question 7.
CaO + H₂SO₄ → ? + H₂O
Answer:
CaO + H₂SO₄ → CaSO₄ + H₂O

Question 8.
HCl + ? → NaCl + H₂O
Answer:
HCl + NaOH → NaCl + H₂O

Question 9.
Ca(OH)₂ + ? → ? + H₂O
Answer:
Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

Question 10.

Answer:
A = NH₄Cl, B = NH₃↑
C = HCl
D = NaHSO₄,E = H₂O

V. Answer in briefly :

Question 1.
What are organic acids? Given examples.
Answer:
Acids present in plants and animals (living things) are organic acids.
Example: HCOOH, CH₃COOH

Question 2.
How are acids classified based on ionisation? Give examples.
Answer:
Acids get ionise in water (produce H⁺ ions) completely or partially. Based on the extent of ionisation, acids are classified as follows:
Strong Acids:
These are acids that ionise completely in water.
Example: HCl

Weak Acids :
These are acids that ionise partially in the water.
Example: CH₃COOH.

Question 3.
What is Aquaregia? Mention its uses.
Answer:
It is a mixture of hydrochloric acid and nitric acid prepared optimally in a molar ratio of 3:1. It is a yellow-orange fuming liquid. It is a highly corrosive liquid, able to attack gold and other resistant substances.

• Aquaregia is used tp dissolve noble metals such as gold, platinum, and palladium.
• It is used Tor cleaning and refining gold.

Question 4.
What does the acidity of base mean?
Answer:
It means the number of replaceable hydroxyl groups present in one molecule of a base.

Question 5.
What is Potash alum? Write its formula.
Answer:
Potash alum is a mixture of potassium sulphate and aluminium sulphate. KAl(SO₄)₂.12H₂O.

Question 6.
What are double salts? Give an example?
Answer:
Double salts are the salts formed by the combination of the saturated solution of two simple salts in equimolar ratio followed by crystallization. For example : Potash alum.

Question 7.
What are basic salts? Give suitable reaction for this.
Answer:
Basic salts are the product formed by the partial replacement of hydroxide ions of a diacidic or triacidic base with an acid radical.
Pb(OH)₂ + HCl → Pb(OH)Cl + H₂O

Question 8.
Why are tooth pastes basic?
Answer:
Toothpastes which are generally basic and used for cleaning the teeth can neutralise the excess acid and prevent tooth decay.

Question 9.
What is water of crystallisation?
Answer:
Many salts are found as crystals with water molecules they contain. These water molecules are known as water of crystallisation.

Question 10.
Why do blue colour copper sulphate becomes white on heating?
Answer:
On heating, blue colour copper sulphate loses its water molecules and becomes white.

Question 11.
Acidic or basic solutions are good conductors of electricity. Justify your answer.
Answer:
Acidic and basic solutions in water conduct electricity because they produce hydrogen and hydroxide ions respectively.

Question 12.
What are hygroscopic substances?
Answer:
Substance which absorbs water from the surroundings are called hygroscopic substances.

Question 13.
Define indicator. Give examples
Answer:
Chemical substances used to find out whether the given solution is acid or base are called indicators. Eg: Phenolphthalein, methyl orange.

Question 14.
Define ionisation.
Answer:
Ionisation is the condition of being dissociated into ions by heat or radiation or chemical reactions or electrical discharge.

Question 15.
How is normal salt obtained? Give a suitable reaction.
Answer:
A normal salt is obtained by complete neutralization of an acid by a base.
NaOH + HCl → NaCl + H₂O

Question 16.
How are bases classified based on ionisation?
Answer:
Based on Ionisation
(a) Strong Bases :
These are bases which ionise completely in aqueous solution.
Example: NaOH, KOH

(b) Weak Bases:
These are bases that ionise partially in aqueous solution.
Example: NH₄OH, Ca(OH)₂

VI. To Interpret:

Question 1.
CH₄ and NH₃, are not acids.
Answer:
CH₄ and NH₃ do not produce hydrogen ion (H⁺) in its aqueous solution.

Question 2.
Acetic acid (CH₃COOH) is a mono basic acid.
Answer:
Though acetic acid contains four hydrogen atoms only one hydrogen can be replaced in its aqueous solution. So CH₃COOH is a mono basic acid.
CH₃COOH → H⁺+ + CH₃COO^–

Question 3.
Al(OH)₃ & Zn(OH)₂ are not alkalis.
Answer:
Al(OH)₃ & Zn(OH)₂ are water insoluble bases. So they are bases not alkalies.

Question 4.
NaOH & KOH are strong bases.
Answer:
These are bases which ionise completely in aqueous solution.

Question 5.
NaHSO₄ is an acid salt.
Answer:
It is formed by the partial replacement of hydrogen ion of sulphuric acid (H₂ SO₄ ) by a metal present in sodium hydroxide (NaOH).
NaOH + H₂ SO₄ → NaHSO₄ + H₂O

Question 6.
Non-metallic oxides are acidic is nature.
Answer:
When non – metallic oxides react with bases, they can form a salt and water similar to the reaction of base with acids. So non-metallic oxides are acidic in nature.

VII. Complete the following table:

Question 1.

Answer:

Question 2.

Answer:

Question 3.

1) Hydrochloric acid	a) Fertilizer
2) Citric acid	b) Baking powder
3) Nitric acid	c) Car batteries
4) Oxalic acid	d) Cleansing agent in toilets
5) Sulphuric acid	e) Removing black stains
	f) Effervescent salts

Answer:
1. – d, 2. – f, 3. – a, 4. – e, 5. – c

Question 4.

1) Hardest substance in our body	a) Aluminium phosphate
2) Citrus fruits	b) Potash alum
3) Rice plant	c) Insoluble in water
4) Double salts	d) Acidic soil
5) Silver chloride	e) Calcium phophate
	f) Alkaline soil

Answer:
l.-e,2.-f,3.-d,4-b,5.-c

VIII. Find odd one out & give reason :

Question 1.
HCl, HNO₃, HCOOH, H₃PO₄
Answer:
H₃PO₄,
It is a tribasic acid whereas other three are monobasic acids.

Question 2.
Acetic acid, formic acid, tartaric acid, sulphuric acid.
Answer:
Sulphuric acid.
It is a mineral acid whereas other three are organic acids.

Question 3.
CaO, Na₂O, ZnO, NaOH
Answer:
NaOH.
It is a base containing replaceable hydroxyl ion. whereas other three bases contain replaceable oxide ions.

Question 4.
Bitter taste, soupy to touch, turns red litmus to blue, produce pink colour with methyl orange.
Answer:
Produce pink colour with methyl orange.
It is the property of acids whereas other three are properties of bases.

Question 5.
Litmus paper, phenolphthalein, methyl orange, Aquaregia.
Answer:
Aquaregia.
It is a mixture of two acids namely, hydrochloric acid and nitric acid, used to dissolve metals such as gold and platinum whereas other three are indicators used to identify the nature of the solution.

IX. Spot the error / Correct the wrong statement given below :

Question 1.
An acid is the compound which are capable of forming hydroxyl ions (OH^– ) in aqueous solution. .
Answer:
An acid is the compound which are capable of forming hydrogen ions (H+) in aqueous , solution, (or) A base is the compound which are capable of forming hydroxyl ion (OH-) in aqueous solution.

Question 2.
Nitric Acid is a constituent of baking powder.
Answer:
Tartaric acid is a constituent of baking powder.

Question 3.
The p^H value of the base in lesser than 7.
Answer:
The p^H value of the base is greater than 7. (or) p^H value of an acid is lesser than 7.

Question 4.
Ca(OH)₂ is a triacidic base.
Answer:
Ca(OH)₂ is a diacidic base (or) Al(OH)₃ is a triacidic base.

Question 5.
Magnesium hydroxide is used in whitewashing of buildings.
Answer:
Magnesium hydroxide is used as an antacid. (or) Calcium hydroxide is used in whitewashing of buildings.

X. Answer in detail :

Question 1.
Explain the classification of acids based on their basicity.
Answer:
Monobasic Acid :
Acid that contain only one replaceable hydrogen atom per molecule is called monobasic acid. It gives one hydrogen ion per molecule of the acid in solutions.
Example: HCl, HNO₃
.
Dibasic Acid :
An acid which gives two hydrogen ions per molecule of the acid in solution.
Example : H₂SO₄, H₂CO₃
‘
Tribasic Acid :
An acid which gives three hydrogen ions per molecule of the acid in solution.
Example: H₃PO₄.

Question 2.
Write notes on the properties of acids.
Answer:
(a) They have sour taste.
(b) Their aqueous solutions conduct electricity since they contain ions
(c) Acids turns blue litmus red.
(d) Acids react with active metals to give hydrogen gas.
Mg + H₂SO₄ → MgSO₄ + H ₂↑
Zn + 2HCl → ZnCl₂ + H₂↑

(e) Acids react with metal carbonate and metal hydrogen carbonate to give carbon dioxide.
Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂↑
NaHCO₃ + HCl → NaCl + H₂O + CO₂↑

(f) Acids react with metallic oxides to give salt and water.
CaO + H₂SO₄ → CaSO₄+ H₂O

(g) Acids react with bases to give salt and water.
HCl + NaOH → NaCl + H₂O.

Question 3.
Write notes on the properties of bases. .
Answer:
(a) They have bitter taste.
(b) Their aqueous solutions have soapy touch.
(c) They turn red litmus blue.
(d)Their aqueous solutions conduct electricity.
(e) Bases react with metals to form salt with the liberation of hydrogen gas.
Zn + 2 NaOH → Na₂ZnO₂ + H₂↑

(f) Bases react with non-metallic oxides to produce salt and water. Since this is similar to the reaction between a base and an acid, we can conclude that non-metallic oxides are acidic in nature,
Ca(OH)₂ + CO₂ → CaCO₃+ H₂O

(g) Bases react with acids to form salt and water.
KOH + HCl → KCl + H₂O
The above reaction between a base and an acid is known as Neutralisation reaction.

(h) On heating with ammonium salts, bases give ammonia gas.
NaOH + NH₄Cl → NaCl + H₂O+ NH₃↑

Question 4.
Describe the classification of bases based on their acidity.
Answer:
(a) Monoacidic Base :
It is a base that ionises in water to give one hydroxide ion per molecule.
Example: NaOH, KOH

(b) Diacidic Base:
It is a base that ionises in water to give two hydroxide ions per molecule.
Example: Ca(OH)₂, Mg(OH)₂

(c) Triacidic Base :
It is a base that ionises in water to give three hydroxide ions per molecule.
Example: Al(OH)₃, Fe(OH)₃

Question 5.
Write notes on importance of pH in everyday life.
Answer:
p^H in our digestive system :
Hydrochloric acid produced in our stomach helps in the digestion of food without harming the stomach. During indigestion, the stomach produces too much acid and this causes pain and irritation. pH of stomach fluid is approximately 2.0

p^H changes is the cause of tooth decay :
White enamel coating of our teeth is calcium phosphate, the hardest substance in our body. Toothpaste which are generally basic and used for cleaning the teeth can neutralise the excess acid and prevent tooth decay.

p^H of soil:
In agriculture, the pH of soil is very important. Citrus fruits require slightly alkaline soil, while rice requires acidic soil and sugarcane requires neutral soil.

p^H of rain water:
The pH of rainwater is approximately 7 which means that it is neutral and also represents its high purity. If the atmospheric air is polluted with oxides of sulphur and nitrogen, they get dissolved in rainwater and make its pH less than 7. Thus, if the pH of rainwater is less than 7, then it is called acid rain. When acid rain flows into the rivers it lowers the pH of the river water. The survival of aquatic life in such rivers becomes difficult.

Question 6.
List the properties of salts.
Answer:

•  Salts are mostly solids which melt as well as boil at high temperature.
• Most of the salts are soluble in water.
  For example, chloride salts of potassium and sodium are soluble in water. But silver chloride is insoluble in water.
• They are odourless, mostly white, cubic crystals or crystalline powder with a salty taste.
• Salt is hygroscopic in nature.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5

Choose the correct or the most suitable answer:

Question 1.
The value of 2 + 4 + 6 + …………… + 2n is
(1) \(\frac{n(n-1)}{2}\)
(2) \(\frac{n(n+1)}{2}\)
(3) \(\frac{2 n(2 n+1)}{2}\)
(4) n(n + 1)
Answer:
(4) n(n + 1)

Explaination:
2 + 4 + 6 + ……… + 2n = 2(1 + 2 + 3 + ………….. + n)
= 2 × \(\frac{n(n+1)}{2}\)
= n(n + 1)

Question 2.
The coefficient of x⁶ in (2 + 2x)¹⁰ is
(1) 10C₆
(2) 2₆
(3) 10C₆2⁶
(4) 10C₆2¹⁰
Answer:
(4) 10C₆2¹⁰

Explaination:

Question 3.
The coefficient of x⁸ y¹² in the expansion of (2x + 3y)²⁰ is
(1) 0
(2) 2⁸ 3¹²
(3) 2⁸ 3¹² + 2¹² 3⁸
(4) 20C₈ 2⁸ 3¹²
Answer:
(4) 20C₈ 2⁸ 3¹²

Explaination:

Question 4.
If nC₁₀ > nC_r for all possible r then a value of n is
(1) 10
(2) 21
(3) 19
(4) 20
Answer:
(4) 20

Explaination:
Out of ¹⁰C₁₀, ²¹C₁₀, ¹⁹C₁₀ and ²⁰C₁₀, ²⁰C₁₀ is larger.

Question 5.
If a is the Arithmetic mean and g is the Geometric mean of two numbers then
(1) a ≤ g
(2) a ≥ g
(3) a = g
(4) a > g
Answer:
(2) a ≥ g

Explaination:
Given Arithmetic mean = a,
Geometric mean = g
We have A. M ≥ G. M
∴ a ≥ g

Question 6.
If (1 + x²)² (1 + x)ⁿ = a₀ + a₁x + a₂x² + ………… + x^{n + 4} and if a₀, a₁, a₂ are in A. P then n is
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(3) 3

Explaination:

n² – 5n + 6 = 0
(n – 2) (n – 3) = 0
n = 2 or n = 3

Question 7.
If a, 8, b are in A .P , a, 4 , b are in G. P and if a, x ,b are in H . P then x is
(1) 2
(2) 1
(3) 4
(4) 16
Answer:
(1) 2

Explaination:
Given a, 8, b are in A. P ∴ 2 × 8 = a + b ⇒ a + b = 16 ——— (1)
Also a, 4, b are in G.P ∴ 4² = a . b ⇒ ab = 16 ——- (2)
Also a, x, b are in H.P. ∴ \(\frac{1}{a}, \frac{1}{x}, \frac{1}{b}\) are in A.P

Question 8.

(1) A. P
(2) G.P
(3) H.P
(4) AGP
Answer:
(3) H.P

Explaination:

Question 9.
The H.M of two positive numbers whose A.M and G.M are 16,8 respectively is
(1) 10
(2) 6
(3) 5
(4) 4
Answer:
(4) 4

Explaination:
Let a, b be the two numbers. Given A. M = \(\frac{a+b}{2}\) = 16
G.M = \(\sqrt{a b}\) = 8
a+b = 16 × 2 = 32
ab = 8² = 64
H.M = \(\frac{2 a b}{a+b}\)
= \(\frac{2 \times 64}{32}\) = 2 × 2 = 4

Question 10.
If S denote the sum of n terms of an A. P whose common difference is d, the value of S_n – 2S_{n- 1} + S_{n – 2} is
(1) d
(2) 2d
(3) 4d
(4) d²
Answer:
(1) d

Explaination:

= a + (n – 1) d – (a + (n – 2)d)
= a + (n – 1) d – a – (n – 2)d
= a + nd – d – a – nd + 2d = d

Question 11.
The remainder when 38¹⁵ is divided by 13 is
(1) 12
(2) 1
(3) 11
(4) 5
Answer:
(1) 12

Explaination:
38¹⁵ = (39 – 1)¹⁵ = 39¹⁵ – 15C₁ 39¹⁴(1) + 15C₂ (39)¹³(1)² – 15C₃ (39)¹²(1)³ ….. + 15C₁₄ (39)¹(1) – 15C₁₅(1)
Except -1 all other terms are divisible by 13.
∴ When 1 is added to it the number is divisible by 13. So the remainder is 13 – 1 = 12.

Question 12.
The nth term of the sequence 1, 2, 4, 7, 11, ………….. is
(1) n² + 3n² + 2n
(2) n³ – 3n² + 3n
(3) \(\frac{n(n+1)(n+2)}{3}\)
(4) \(\frac{n^{2}-n+2}{2}\)
Answer:
(4) \(\frac{n^{2}-n+2}{2}\)

Explaination:

Question 13.
The sum up to n terms of the series

(1) \(\sqrt{2 n+1}\)
(2) \(\frac{\sqrt{2 n+1}}{2}\)
(3) \(\sqrt{2 n+1}-1\)
(4) \(\frac{\sqrt{2 n+1}-1}{2}\)
Answer:
(4) \(\frac{\sqrt{2 n+1}-1}{2}\)

Explaination:

Question 14.
The n^th term of the sequence

(1) 2ⁿ – n – 1
(2) 1 – 2^-n
(3) 2^-n + n – 1
(4) 2^n-1
Answer:
(2) 1 – 2^-n

Explaination:

Question 15.
The sum up to n terms of the series

(1) \(\frac{\mathbf{n}(\mathbf{n}+1)}{2}\)
(2) 2n (n + 1)
(3) \(\frac{\mathbf{n}(\mathbf{n}+1)}{\sqrt{2}}\)
(4) 1
Answer:
(3) \(\frac{\mathbf{n}(\mathbf{n}+1)}{\sqrt{2}}\)

Explaination:

Question 16.
The value of the series

(1) 14
(2) 7
(3) 4
(4) 6
Answer:
(1) 14

Explaination:

Question 17.
The sum of an infinite G.P is 18. If the first term is 6 the common ratio is
(1) \(\frac{1}{3}\)
(2) \(\frac{2}{3}\)
(3) \(\frac{1}{6}\)
(4) \(\frac{3}{4}\)
Answer:
(2) \(\frac{2}{3}\)

Explaination:
Let the geometric series be a, ar, ar², …………… ar^n-1
Given a = 6, S_∞ = 18

Question 18.
The coefficient of x⁵ in the series e^-2x is
(1) \(\frac{2}{3}\)
(2) \(\frac{3}{2}\)
(3) \(-\frac{4}{15}\)
(4) \(\frac{4}{15}\)
Answer:
(3) \(-\frac{4}{15}\)

Explaination:

Question 19.
The value of

(1) \(\frac{e^{2}+1}{2 e}\)
(2) \(\frac{(e+1)^{2}}{2 e}\)
(3) \(\frac{(e-1)^{2}}{2 e}\)
(4) \(\frac{e^{2}+1}{2 e}\)
Answer:
(3) \(\frac{(e-1)^{2}}{2 e}\)

Explaination:

Question 20.
The value of

(1) log \(\left(\frac{5}{3}\right)\)
(2) \(\frac{3}{2}\) log \(\left(\frac{5}{3}\right)\)
(3) \(\frac{5}{3}\) log \(\left(\frac{5}{3}\right)\)
(4) \(\frac{2}{3}\) log \(\left(\frac{2}{3}\right)\)
Answer:
(2) \(\frac{3}{2}\) log \(\left(\frac{5}{3}\right)\)

Explaination:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Question 1.
Expand the following in ascending powers of x and find the condition on x for which the binomial expansion is valid.
(i) \(\frac{1}{5+x}\)
Answer:

(ii) \(\frac{2}{(3+4 x)^{2}}\)
Answer:

(iii) (5 + x²)^2/3
Answer:

(iv) \((x+2)^{-\frac{2}{3}}\)
Answer:

Question 2.
Find \(\sqrt[3]{1001}\) approximately. (two decimal places)
Answer:

Question 3.
Prove that \(\sqrt[3]{x^{3}+6}-\sqrt[3]{x^{3}+3}\) is approximately equal to \(\frac{1}{x^{2}}\) when x is sufficiently large.
Answer:

Question 4.
Prove that \(\sqrt{\frac{1-x}{1+x}}\) is approximately equal to 1 – x + \(\frac{x^{2}}{2}\) when x is very small.
Answer:

Question 5.
Write the first 6 terms of the exponential series
(i) e^5x
Answer:

(ii) e^-2x
Answer:

(iii) e^x/2
Answer:

Question 6.
Write the first 4 terms of the logarithmic series.
(i) log (1 + 4x)
(ii) log (1 – 2x)
(iii) log \(\left(\frac{1+3 x}{1-3 x}\right)\)
(iv) log \(\left(\frac{1-2 x}{1+2 x}\right)\)
Find the intervals on which the expansions are valid.
Answer:
(i) log ( 1 + 4x )

(ii) log (1 – 2x)

(iii) log \(\left(\frac{1+3 x}{1-3 x}\right)\)

(iv) log \(\left(\frac{1-2 x}{1+2 x}\right)\)

Question 7.

Answer:

Question 8.

Answer:

Question 9.
Find the coefficient of x⁴ in the expansion of \(\frac{3-4 x+x^{2}}{e^{2 x}}\)
Answer:

Question 10.
Find the value of
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3

Question 1.
Find the sum of the first 20 terms of the arithmetic progression having the sum of first 10 terms as 52 and the sum of the first 15 terms as 77 .
Answer:
Sum of the first n terms of an Arithmetic progression is S_n = \(\) [2a + (n – 1)d]
Given S₁₀ = 52
52 = \(\frac{10}{2}\) [ 2a + (10 – 1) d ]
52 = 5 [2a + 9d]
52 = 10a + 45d ……….. (1)
Also given S₁₅ = 77
77 = \(\frac{15}{2}\) [2a + (15 – 1)d]
77 = \(\frac{15}{2}\) [2a + 14d]
77 = 15 [a + 7d]
77 = 15a + 105d ………….. (2)

Substituting the value of d in (1)

Question 2.
Find the sum up to the 17^th term of the series
Answer:

Question 3.
Compute the sum of first n terms of the following series.
(i) 8 + 88 + 888 + 8888 + . . . . . . .
Answer:

(ii) 6 + 66 + 666 + 6666 + . . . . . . .
Answer:

Question 4.
Compute the sum of first n terms of 1 + (1 + 4) + (1 + 4 + 4²) + (1 + 4 + 4² + 4³) + …………..
Answer:
The given series is 1 + (1 + 4) + (1 + 4 + 4²) + (1 + 4 + 4² + 4³) + …………..
n^th term of the series is T_n = 1 + 4 + 4² + 4³ + ………………. + 4^n-1

Sum to n terms of the series
S_n = T₁ + T₂ + ………….. + T_n

Question 5.
Find the general terms and sum to n terms of the sequence 1, \(\frac{4}{3}\), \(\frac{7}{9}\), \(\frac{10}{27}\), ………….
Answer:
The given sequence as 1, \(\frac{4}{3}\), \(\frac{7}{9}\), \(\frac{10}{27}\), ………….
Consider the terms in the numerator 1, 4, 7, 10, ……………….. which is an Arithmetic progression
with first term a = 1, common difference d = 4 – 1 = 3
a_n = a + (n – 1)d = 1 + (n – 1)3 = 1 + 3n – 3 = 3n – 2
The given sequence can be written as 1, (1 + 3)\(\left(\frac{1}{3}\right)\), (1 + 2 × 3)\(\left(\frac{1}{3}\right)^{2}\) (1 + 3 × 3) \(\left(\frac{1}{3}\right)^{3}\), …………….
where \(1, \frac{1}{3}, \frac{1}{3^{2}}, \frac{1}{3^{3}}, \ldots \ldots\) is a G. P with first term a = 1 , common ratio r = \(\frac{1}{3}\)
∴ The given sequence is an Arithmetico – Geometric sequence.


which is an arithmetic – geometric sequence.
∴ The sum of first n terms of the arithmetico – geometric sequence is given by

Question 6.
Find the value of n if the sum to n terms of the series \(\sqrt{3}+\sqrt{75}+\sqrt{243}+\ldots \ldots . . . \text { is } 435 \sqrt{3}\)
Answer:

Question 7.
Show that the sum of ( m + n)^th and ( m – n)^th term of an A.P is equal to twice the m^th term.
Answer:
Let the A.P. be a, a + d, a + 2d, ……..
t_{m + n} = a + (m + n – 1)d
t_{m – n} = a + (m – n – 1)d
t_m = a + (m – 1)d
2t_m = 2[a + (m – 1)d]
To prove t_{m + n} + t_{m – n} = 2t_m
LHS t_{m + n} + t_{m – n} = a + (m + n – 1)d + a + (m – n – 1)d

= 2a + d [2m – 2]
= 2[a + (m – 1)d) = 2 t_m = RHS

Question 8.
A man repays an amount of Rs.3250 by paying Rs.20 in the first month and then increases the payment by Rs. 15 per month. How long will it take him to clear the amount?
Answer:
Amount of loan = 3250
Let n be the number of months taken to clear the loan
Amount paid in the first month a = 20
Increased payment in every month d = 15
∴ Amount paid in the second month = 20 + 15 = 35
Amount paid in the third month = 35 + 15 = 50
∴ The sequence of amount paid in every month is 20, 35, 50, …………. which is an A.P with first term a = 20 and common difference = 15
Given S_n = \(\frac{n}{2}\) [2a + (n – 1) d]
S_n = 3250
3250 = \(\frac{n}{2}\) [2 × 20 + (n – 1) 15]
6500 = n[40 + 15n – 15]
6500 = n[25 + 15n]
6500 = 25n + 15n²
1300 = 5n + 3n²
3n² + 5n – 1300 = 0
3n² + 65n – 60n – 1300 = 0
n(3n + 65) – 20 (3n + 65) = 0
(n – 20) (3n + 65) = 0
n = 20 = 0 or 3n + 65 = 0
n = 20 or n = \(-\frac{65}{3}\)
n = \(-\frac{65}{3}\) is not possible ∴ n = 20
Thus, in 20 months the loan is cleared.

Question 9.
In a race, 20 balls are placed in a line at intervals of 4 meters with the first ball, 24 meters away from the starting point. A contestant is required to bring the balls back to the starting place one at a time. How far would the contestant run to bring back all balls?
Answer:
Number of balls placed in a line = 20
Let A be the starting point.
The distance of the first ball from A = 24 m
The distance of the second ball from the first ball = 4 m

A contestant starts from point A, travels a distance of 24 m and picks the first ball and brings it back to starting point A. This is continued for each ball.
Distance travelled by the contestant to bring the first ball = 24 + 24 = 2 × 24 = 48 m
Distance travelled to bring the second ball = 2 ( 24 + 4 ) = 2 × 28 = 56 m
Distance travelled to bring the third ball = 2 × (24 + 4 + 4) = 64 m

∴ The sequence of distances travelled are 48, 56, 64 ……………
This is an Arithmetic progression with first term a = 48 ,
common difference d = 56 – 48 = 8 ,
number of terms n = 20.
The sum of 20 terms of this A.P gives the total distance travelled by the contestant in bringing all balls to the starting place
S_n = \(\frac{\mathrm{n}}{2}\) [2a + (n – 1)d]
S₂₀ = \(\frac{\mathrm{20}}{2}\) [2 × 48 + (20 – 1)8] = 10 [96 + 19 × 8]
= 10 [ 96 + 152] = 10 × 248 = 2480
Total distance traveled = 2480 m

Question 10.
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally how many bacteria will be present at the end of the 2nd hour, 4th hour, and nth hour?
Answer:
Number of bacteria at the beginning = 30
Number of bacteria after 1 hour = 30 × 2 = 60
Number of bacteria after 2 hours = 30 × 2² = 120
Number of bacteria after 4 hours = 30 × 24 = 30 × 16 = 480
∴ Number of bacteria after n^th hour = 30 × 2ⁿ

Question 11.
What will Rs. 500 amounts to in 10 years after its deposit in a bank which pays an annual interest rate of 10 % compounded annually?
Answer:

Question 12.
In a certain town, a viral disease caused severe health hazards, upon its people disturbing their normal life. It was found that on each day, the virus which caused the disease spread in Geometric Progression. The amount of infected virus particle gets doubled each day, being 5 particles on the first day. Find the day when the infections virus particles just grow over 1,50,000 units?
Answer:
The number of viruses present at the beginning = 5, Given virus, gets doubled each day
∴ The sequence of a number of viruses in each day is 5, 10, 20, 40, 80, ………………. which is a G. P with First-term a = 5, Common ratio r = \(\frac{10}{5}\) = 2
n^th term t_n = ar^n-1

∴ On the 15^th day, the infectious virus grows over 1,50,000 units.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 5 Binomial Theorem, Sequences and Series Ex 5.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.2

Question 1.
Write the first 6 terms of the sequence whose nth terms are given below and classify them as Arithmetic progression Geometric progression, Arithmetic – geometric progression, Harmonic progression and none of them.
(i) \(\frac{1}{2^{n+1}}\)
Answer:

(ii)
Answer:

(iii) \(4\left(\frac{1}{2}\right)^{n}\)
Answer:

(iv) \(\frac{(-1)^{n}}{n}\)
Answer:

(v) \(\frac{2 n+3}{3 n+4}\)
Answer:

(vi) 2018
Answer:
The n^th term a_n = 2018
a₁ = 2018,
a₂ = 2018,
a₃ = 2018,
a₄ = 2018,
a₅ = 2018,
a₆ = 2018,
∴ The given sequence is 2018, 2018, 2018, 2018, 2018, 2018, ………….
This is a œnstant sequence which has same common ratio and common difference.
Hence this is an A. P, G . P and AGP.

(vii) \(\frac{3 n-2}{3^{n-1}}\)
Answer:

Question 2.
Write the first 6 terms of the sequences whose n^th term a_n is given below
(i)
n = 1, a_n = n + 1, a₁ = 1 + 1 = 2
n = 2, a_n = n, a₂ = 2
n = 3, a_n = n + 1, a₃ = 3 + 1 = 4
n = 4, a_n = n, a₄ = 4
n = 5, a_n = n + 1, a₅ = 5 + 1 = 6
n = 6, a_n = n, a₆ = 6
∴ The first six terms are 2, 2, 4, 4, 6, 6

(ii)
n = 1, a₁ = 1, n = 2, a₂ = 2
n = 3, a_n = a_n+1 + a_n-2, a₃ = a_{3 – 1} + a_{3 – 2} = a₂ + a₁ = 2 + 1 = 3

n = 4, a_n = a_n+1 + a_n-2, a₄ = a_{4 – 1} + a_{4 – 2} = a₃ + a₂ = 3 + 2 = 5

n = 5, a_n = a_n+1 + a_n-2, a₅ = a_{5 – 1} + a_{5 – 2} = a₄ + a₃ = 5 + 3 = 8

n = 6, a_n = a_n+1 + a_n-2, a₆ = a_{6 – 1} + a_{6 – 2} = a₅ + a₄ = 8 + 5 = 13
∴ The first six terms are 1, 2, 3, 5, 8, 13

(iii)
Answer:
n = 1, a_n = n, a₁ = 1
n = 2, a_n = n, a₂ = 1
n = 3, a_n = n, a₃ = 1

n = 4, a_n = a_n-1 + a_n-2 + a_n-3
a₄ = a_4-1 + a_4-2 + a_4-3
a₄ = a₃ + a₂ + a₁
a₄ = 3 + 2 + 1 = 6

n = 5, a_n = a_n-1 + a_n-2 + a_n-3
a₅ = a_5-1 + a_5-2 + a_5-3
a₅ = a₄ + a₃ + a₂
a₅ = 6 + 3 + 2 = 11

n = 6, a_n = a_n-1 + a_n-2 + a_n-3
a₆ = a_6-1 + a_6-2 + a_6-3
a₆ = a₅ + a₄ + a₃
a₆ = 11 + 6 + 3 = 20
∴ The first six terms are 1, 2, 3, 6, 11, 20

Question 3.
Write the n^th term of the following sequences.
(i) 2, 2, 4, 4, 6, 6, ……………..
Answer:
The odd terms are a₁ = 2, a₃ = 4, a₅ = 6
The even terms are a₂ = 2, a₄ = 4, a₆ = 6

(ii)
Answer:

The terms in the numerator are 1 , 2 , 3, 4
a = 1 , d = 2 – 1 = 1
a_n = a + (n – 1) d
a_n = 1 + (n – 1)(1) = 1 + n – 1 = n
a_n = n
The terms in the denominator are 2 , 3 , 4 , 5 , 6 .
a = 2, d = 3 – 2 = 1
a_n = a + (n – 1) d
a_n = 2 + (n – 1) (1) = 2 + n – 1 = n + 1
a_n = n + 1
∴ The n^th term of the given sequence is a_n = \(\frac{n}{n+1}\) for all n ∈ N

(iii)
Answer:

The terms in the numerator are 1, 3, 5, 7, 9, ………….
a = 1 , d = 3 – 1 = 2
a_n = a + (n – 1) d
a_n = 1 + (n – 1)2
a_n = 1 + 2n – 2 = 2n – 1
The terms in the denominator are 2, 4, 6, 8, 10, …………..
a = 2, d = 4 – 2 = 2
a_n = a + (n – 1) d
a_n = 2 + (n – 1)(2)
a_n = 2 + 2n – 2 = 2n
∴ The n^th term of the given sequence is a_n = \(\frac{2 n-1}{2 n}\) for all n ∈ N

(iv) 6, 10, 4, 12, 2, 14, 0, 16, – 2 …………………
Answer:
The given sequence is 6, 10, 4, 12, 2, 14, 0, 16, – 2 ……………..
The odd terms are a₁ = 6, a₃ = 4 , a₅ = 2 , a₇ = 0, a₉ = – 2
∴ a_n = n – 7, n is odd
The even terms are a₂ = 10, a₄ = 12 , a₆ = 14 , a₈ = 16
∴ a_n = 8 + n, n is even.

Question 4.
The product of three increasing numbers in a G.P is 5832 . If we add 6 to the second number and 9 to the third number, then resulting numbers form an A.P. Find the numbers in G.P.
Answer:
Let the increasing numbers in G.P be \(\), a, ar.
Given \(\frac{a}{r}\) × a × ar = 5832 ⇒ a³ = 5832 = 18³ ⇒ a = 18
Also given \(\frac{a}{r}\), a + 6, ar + 9 form an A.P.
∴ 2(a + b) = \(\frac{a}{r}\) + (ar + 9)
⇒ (a + 6) + (a + 6) = \(\frac{a}{r}\) + (ar + 9)
⇒ (a + 6) – \(\frac{a}{r}\) = (ar + 9) – (a + 6)
⇒ a + 6 – \(\frac{a}{r}\) = ar + 9 – a – 6
⇒ a + 6 – \(\frac{a}{r}\) = ar – a + 3
Substituting the value of a = 18, we get


39r = 18r² + 18
18r² – 39r + 18 = 0
(2r – 3)(3r -2) = 0
2r – 3 = 0 or 3r – 2 = 0
r = \(\frac{3}{2}\) or r = \(\frac{2}{3}\)

Case (i) When a = 18, r = \(\frac{3}{2}\) the numbers in G.P are

Case (ii) When a = 18, r = \(\frac{2}{3}\)

Question 5.
Write the n^th term of the sequence \(\frac{3}{1^{2} \cdot 2^{2}}, \frac{5}{2^{2} \cdot 3^{2}}, \frac{7}{3^{2} \cdot 4^{2}}\), …………….. as a difference of two terms.
Answer:

The terms in the numerator are 3,5, 7
which forms an A. P with first term a = 3 and common difference d = 5 – 3 = 2
n^th term t_n = a + (n – 1) d
= 3 + (n – 1)(2)
= 3 + 2n – 2 = 2n + 1
t_n = 2n + 1
The terms in the denominator are 1² . 2², 2² . 3², 3² . 4² ……………….
n^th term t_n = n² . (n + 1)²

Question 6.
If t_k is the k^th term of a G.P then show that t_{n – k}, t_k, t_{n + k} also form a G.F for any positive integer k.
Answer:
Given t^k is the k^th term of a G.P. We have n^th term of a G.P is t_n = ar^n-1

Question 7.
If a, b, c are in geometric progression and if a^1/x = b^1/y = c^1/z are in Arithmetic progression.
Answer:

Question 8.
The A.M of two numbers exceeds their G.M by 10 and H.M by 16. Find the numbers.
Answer:
Let the numbers be a and b

Question 9.
If the roots of the equation (q – r)x² + (r – p)x + (p – q) = 0 are equal then show that p , q and r are in A. P.
Answer:
The roots are equal ⇒ ∆ = 0
(i.e.) b² – 4ac = 0
Hence, a = q – r ; b = r – p ; c = p – q
b² – 4ac = 0
⇒ (r – p)² – 4(q – r)(p – q) = 0
r² + p² – 2pr – 4[qr – q² – pr + pq] = 0
r² + p² – 2pr – 4qr + 4q² + 4pr – 4pq = 0
(i.e.) p² + 4q² + r² – 4pq – 4qr + 2pr = 0
(i.e.) (p – 2q + r)² = 0
⇒ p – 2q + r = 0
⇒ p + r = 2q
⇒ p, q, r are in A.P.

Question 10.
If a , b , c are respectively the p^th, q^th and r^th terms of a G . P show that (q – r) log a + (r – p) log b + (p – q) log c = 0
Answer:
Let A be first term and R be the jmnon ratio of the G.P.
Given a = p^th term of the G.P
General term of a G. P with first term A and common ratio R is t_n = AR^{n – 1}
∴ a = t_p = AR^{P – 1}
log a = log AR^p-1 = log A + log R^p-1 = log A + (p – 1) log R

b = q^th term of the G.P
b = t_q = AR^q-1
log b = log AR^q-1 = log A + log R^q-1 = log A + (q – r)log R

c = r^th term of the G.P
c = t_r = AR^r-1
log c = log AR^r-1 = log A + log R^r-1 = log A + (r – 1) log R

(q – r) log a + (r – p) log b + (p – q) log c
= (q – r) [log A + (p – 1) log R] + (r – p) [ log A + (q – 1) log R ] + (P – q) [ log A + (r – 1) log R]
= (q – r) log A + (q – r) (p – 1 ) log R + (r – p) log A + (r – p) (q – 1) log R + (P – q) log A + (p – q) (r – 1 ) log R
= [ q – r + r – p + p – q ] log A + [ (q – r) (p – 1) + (r – p) (q – 1) + (p – q)(r – 1)] log R
= 0 × log A + [pq – q – rp + r + rq – r – pq + p + pr – p – rq + q] log R
= 0 × log R = 0
∴ (q – r) log a + (r – p) log b + (p – q) log c = 0

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th English Guide Pdf Prose Chapter 4 The Summit Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 12th English Solutions Prose Chapter 4 The Summit

12th English Guide The Summit Text Book Back Questions and Answers

Textual Questions:

1. Based on your reading of the text, answer the following questions in one or two sentences each. (Text Book Page No. 116) (Note: IQ → Important Questions)

Question a.
What did Hillary do with his wet boots? (IQ)
Answer:
Hillary cooked his wet boots over the fierce flame of the Primus and managed to soften them.

Question b.
Name equipment and a tool carried by the climbers during their expedition. (IQ)
Answer:
They carried oxygen bottles, ice-ax, and crampons during their expedition.

Question c.
Why did Hillary become clumsy-fingered and slow-moving?
Answer:
After reaching the peak, Hillary ran out of oxygen, he was becoming clumsy-fingered and slow-moving. So, he quickly replaced his oxygen set.

Question d.
What did Hillary find in a tiny hollow? (IQ)
Answer:
Hillary found two oxygen bottles left by Evan and Bourdillon, who had their earlier attempt to reach Mt. Everest.

Question e.
When did Hillary feel a sense of freedom and well being?
Answer:
Their first partly-full bottle of oxygen got exhausted. They had only one oxygen bottle to cany. With reduced load of 20 litre bottle, Hillary cut steps down off the South Summit. So, he felt a sense of freedom and well-being.

Question f.
What did Hillary mean by saying “We had had enough to do the job, but by no means too much”?
Answer:

1. Hillary meant that they carried an exact number of oxygen bottles needed for their expedition.
2. They didn’t even carry a single bottle more.

2. Answer the following questions in two or three sentences each:(Text Book Page No. 116)

Question a.
How did the mountaineers belay?
Answer:
Belaying refers to a variety of techniques used by mountaineers to exert tension on a climbing rope so that a climber does not fall very far. A climbing partner typically applies tension at the other end of the rope whenever the climber is not climbing and removes the tension from the rope when the climber needs a rope to continue to climb.

Question b.
Why was the original zest fading away?
Answer:
Their original zest was fading away as they had to struggle a lot to find the top of Mt. Everest.

Question c.
What did Edmund Hillary do to escape the ’ large overhanging ice cornices?
Answer:
In a number of places, the overhanging ice cornices were very large. In order to escape them, Hillary cut a line of steps down to where the snow met the rocks on the west.

Question d.
What did Tenzing and Edmond Hillary gift to the God of Softy Summit? How did they do it? (IQ)
Answer:

1. Tenzing gifted a bar of chocolate, a packet of biscuits and a handful of lollies whereas Hillary placed a small crucifix given by Colonel Hunt, besides Tenzing’s gifts.
2. They did so by making a little hole in the snow.

Question e.
What did the photograph portray?
Answer:
The photograph portrayed North Col and the old route which had been made famous by the struggles of those great climbers in 1920’s and 1930’s.

Question f.
The soft snow was difficult and dangerous. Why?
Answer:
The soft snow was difficult and dangerous as it sometimes held Hillary’s weight but often fell down suddenly.

Question g.
How did the firm snow at the higher regions fill them with hope?
Answer:
They were a little perturbed by slippery soft snow. But as they reached firmer snow higher up, they felt better. As one bottle of oxygen got exhausted, their load was now less. As Hillary’s ax bit into the first steep slope of the ridge, his high hopes were realized. The snow was crystalline and firm. They were able to make comfortable belays to haul themselves up slowly.

3. Based on the text, answer the following questions in a paragraph of about 100 – 150 words each: (Text Book Page No. 116)

Question a.
How did Hillary and Tenzing prepare themselves before they set off to the summit?
Answer:
They started up their cookers and drank lots of lemon juice and sugar. Then they took sardines and biscuits. Hillary cleaned the ice off the oxygen sets. He rechecked and tested them. He had removed his boots which had become wet the day before. They were now frozen solid. It would be very challenging to start climbing the ice-cold Himalayas with such wet and chilling boots. So, he cooked them over the fierce flame of Primus and managed to soften them up. They were also conscious of the probability of braving snowstorms during the ascent. They fortified their clothing with windproof and also pulled three pairs of gloves silk, woolen, and windproof on to their hands. At 6.30 am they crawled out of their tent into the snow. They hoisted their 30 lb. of oxygen gear on their backs. Connecting their oxygen masks they turned on the valves to bring life-giving oxygen into their lungs. Taking a few deep breaths, They got ready to go.

Question b.
Give an account of the journey to the South Coi from 28,000 feet.
Answer:
They reached a crest of a ridge at about 28000 feet from where the ridge narrowed to a knife-edge. Hillary took over the lead. The route on top of the ridge was filled with soft snow which was both difficult and dangerous. After several hundred feet they reached a tiny hollow where they found two oxygen bottles left by the earlier expeditors. They took it for their use. Hillary took the lead for nearly 400 feet. They made frequent changes of the lead. When Hillary was stamping a trail in deep soft snow, a portion of it gave way and he slipped back. After this, they had a dilemma to proceed further. Still, without losing hope they proceeded further and climbed on to the south peak.

Question c.
Describe the feelings of Edmund Hillary and Tenzing as they reached the top of the Summit.
Answer:
Hillary’s first feelings were of relief on reaching Everest. There were no more ridges to traverse and no more humps to tease them off with the hope of success. He looked at Tenzing. In spite of the balaclava helmet, goggles, and oxygen mask, all encrusted with long icicles, that concealed his face, his delight was visible. He looked around with a grin of delight. They shook hands. Tenzing hugged Hillary and thumped each other till they were breathless. It was 1 .30 am. The ridge had taken them two and a half hours but it seemed like a lifetime.

Question d.
The ridge had taken us two and a half hours, but it seemed like a lifetime. Why?
Answer:
Tenzing and Hillary struggled hard to reach the top of the ridge. They helped each other belaying. After crossing the crack with great struggle, Hillary signaled to Tenzing to come on up who wriggled his way up the crack and collapsed at the top like a giant fish hailing from the sea. Again the ridge curved away to the right and they were unable to find its top. They felt that the ridge seemed never-ending. At this juncture, they lost their original enthusiasm but didn’t give up their try.

Hillary felt at one spot that the ridge seemed to be dropped sharply away instead of rising. Later he found a narrow snow ridge running up to a snowy summit. With the help of the ice-axe they paved the way in the firm snow and reached the top. Though it took two and a half hours for them to reach the top of the ridge, they felt that it took their whole lifetime just because of the struggle they had to reach it.

Question e.
Describe the view from the top. What was the most important photograph?
Answer:
On reaching the peak, Tenzing and Hillary felt a great relief. To the east was their giant neighbour Makalu, unexplored and unclimbed. Far away across the clouds, the great bulk of Kanchenjunga loomed on the horizon. To the west, they could see the unexplored ranges of Nepal stretching off into the distance. The most important photograph was a shot down at the north ridge showed the North Col and the old route. It had been made famous by the famous climbers of 1920’s and 1930’s. It was the breath-taking view of the snow’ clad peak all round.

Question f.
‘There is no height, no depth that the spirit of man, guided by higher Spirit cannot attain’. Discuss the above statement in the context of the achievement of Edmund Hillary and Tenzing.
Answer:
Tenzing and Hillary started with their ascent from 28,000 feet on 29th May. They had undergone many hardships before they reach the summit. The ridge was narrow from 28,000 feet and they had to make their way through soft snow which was both difficult and dangerous. The soft snow gave way suddenly when Hillary had his trail in the deep snow. In spite of his slipping back, he did not give up his effort to move forward. They also found it very difficult to cross the ridge when the snow was firm.

Crossing the narrow crack between the rock and the cornice was really thrilling. It was done by them with fierce determination. Though their energy level went down when they found the ridge to be never-ending, they did not lose hope and proceeded further. It was this strong determination and undying spirit that made them reach the summit successfully. Thus the statement, ‘There is no height, no depth that the spirit of man, guided by the higher spirit cannot attain gets proved.

Paragraph:

Introduction:
Edmund Hillary in his “The Summit” tells how the summit of Everest was reached.

Preparation:
Hillary and his friend Tenzing got ready for the climb from camp-8 on May 29. They drank large quantities of lemon juice and had some biscuits. He softened his boots which became frozen solid. They also wore three pairs of gloves.

Journey:
They reached a crest of a ridge at about 28000 feet which narrowed to a knife-edge. The route on top of the ridge was both difficult and dangerous as it was filled with soft snow without losing hope they proceeded further and crampons on to the south peak.

The struggle:
The snow was crystalline and firm. With the help of his ice-axe he made away and they proceeded further belaying each other. The overhanging ice cornices threatened them. They struggled hard to cross a great cornice which had a forty feet narrow crack between the cornice and the rock. Hillary found a narrow snow ridge running up to a snowy summit. Thus they reached the top. It took nearly two and a half hours for them to/each the top of the ridge.

Sense of fulfillment:
They thumped each other on their back till they became breathless. As they had undergone a huge risk they felt that they had risked their whole lifetime to reach the top. The most important photograph was that of the North ridge. They made a little hole in the snow and placed some offerings to the Gods. From there they safely started descending to the South Col.

Conclusion:
Thus their ascent of Everest depicts the value of teamwork which helps one to overcome any difficulties. The adventure of Hillary and Tenzing enriches the fact that there is no height, no depth, that the spirit of man, guided by a higher spirit cannot attain.

Vocabulary:

Idioms:

I. Given below are some Idiomatic expressions with their meanings. Understand the meaning: (Text Book Page No.117)

1. wait for the dust to settle – to wait for a situation to become clear or certain.
2. get/have all your ducks in a row – to have made all the preparations needed to do something / to be well-organized
3. fetch and carry (for somebody) – to do a lot of little jobs for somebody as if you were their servant
4. do the math – to think carefully about something before doing it, so that you know all the relevant facts or figures
5. round the corner – very near

II. Fill in the blanks with the right idioms. Choose from the above-given idioms: (Text Book Page No.117)

Question a.
The Sherpas are cheerful, gallant men, who________________ tents, oxygen, food etc., for climbers during their ascent of the summit.
Answer:
fetch and carry

Question b.
The team _________________ carefully so as to reach the summit successfully.
Answer:
does the math

Question c.
When they had to climb through deep new snow the party sometimes had to _________________.
Answer:
wait for the dust to settle

Question d.
Each member of the team had all their _________________.
Answer:
get/have all your ducks in a row

Question e.
We could not believe that with a few more whacks of the ice axe in the firm snow we were _______________ to the top.
Answer:
round the corner

III. Understand the meaning of the given idiomatic expression and choose the right one to complete the sentence:
(Text Book Page No.117)

• the icing on the cake — something extra and not essential, but is added to make it even better
• break the ice — to make people more relaxed, especially at the beginning of the meeting

Question a.
The conference room was silent though packed. The chairman introduced an interactive session to _________.
Answer:
break the ice

Question b.
Our headmistress not only promised us to take us for an excursion but also announced that on return we would get a holiday. It was like _________.
Answer:
the icing on the cake

Phrasal Verbs: (Text Book Page No.118)

i. Given below are the phrasal verbs with their meanings. Use the given phrasal verbs in sentences of your own (Text Book Page No.118)

1. I want to turn on the television.
2. Mr. Prem took over charge as manager.
3. The children set off for school.
4.  The meeting was put off to next week.

ii. Given below are some Phrasal Verbs which are frequently used in connection with travelling. Guess the meaning and match: (Text Book Page No.118)

Question a.

Answer:

see off	to go to the station or airport to say goodbye to someone
stopover	to stay at a place for a short period of time when travelling to another place
set off	start off / to begin a journey

Question b.

Answer:

get in	arrive inside the train, bus, etc.
get off	leave a bus or train etc.
get on	enter a bus, train place
getaway	to go away from home for a vacation.

Question c.

Answer:

check-in	arrive and register at the airport or hotel
check out	pay the bill when leaving a hotel

Compound words: (Text Book Page No. 118)

i. Here are some compound words chosen from the text:

I. Match the following with their right field, choosing appropriately from the box given: (Text Book Page No. 119)

Question 1.
snow-board
Answer:
Sports

Question 2.
snow-mobile
Answer:
transportation

Question 3.
snow-chains
Answer:
machinery

Question 4.
snow-storm
Answer:
weather

Question 5.
snow-bird
Answer:
travel

Question 6.
snow-belt
Answer:
geography

Listening:

First, read the following statements. Then, listen to the passage read aloud by your teacher or played on the recorder and complete the statements. You may listen to it again if required. Complete the following: (Text Book Page No. 119)

Question a.
List any three aspects which contributed to the success of the ascent of the summit.

1. ___________
2. __________
3. __________.

Answer:

1. Knowledge gained from other climbers
2. Careful planning
3. Excellence of equipment

Question b.
Without the help of ____________ nothing would have been possible.
Answer:
Teamwork / Sherpas

Question c.
The main idea of the passage is _________
Answer:
reasons for the success of the summit

Question d.
The biggest thing of all is _________.
Answer:
togetherness

Question e.
_________ were cheerful and gallant men.
Answer:
Sherpas

Speaking:

a) Group Activity: (Text Book Page No. 119)

Question i.
Have you ever been on an adventurous trip? If so, share your success story with your friends.
Answer:
I with my brother went on a trip from Tirunelveli to Chennai on a bike. When we were just nearing Chennai, the tire of the bike burst, we did not know what to do. Luckily we managed not to hurt ourselves. We were waiting for someone to help us. Finally, a car came that way. We pleaded with them to help us. They promised to send a mechanic to that place. We waited for nearly two hours. Then the mechanic came and repaired our bike. Later we reached our uncle’s home in Chennai safely.

Question ii.
How will you organize or plan for a trip or an event? Do you have the habit of preparing a check-list? Discuss.
Answer:
I organize a trip with my family to Mysore for four days to visit all the important places. I planned to go by train. So I will book the to and fro tickets to Mysore. I will book a room to stay there in advance, I will prepare a checklist a day before my travel one toothbrush, towel, soap, money, my Id card, medicines, first aid kit. A tourist guide book of Mysore. I will plan everything properly to make my trip safe and pleasant.

b) Individual Activity: (Text Book Page No. 119)

Given below are a few proverbs. Prepare a short speech of two minutes on one of the proverbs:

a) Nothing is impossible:
The word ‘impossible’ means it is possible. You can do anything. “The word impossible is found only in the dictionary of fools,” says Napoleon Bonaparte. In this word, everything is possible on the basis of will power, determination, and sacrifice. To accomplish the most difficult tasks, you need to put in a lot of hard work, extra perseverance, and concentration on a simple objective. You should have determination, dedication, and devotion to attain success. Never bother about the results keep on going even if the pace is slow. Just ensure it remains steady.

b) Where there is a will there’s away.
“Where there is a will there’s a way” is a message of hope and encouragement. Perseverance leads to success. Most of us are endowed with a fair share of intelligence, and all that is needed for success is diligence and perseverance. Some may not be blessed with good health. Some may not be blessed with a sharp intelligence or a retentive memory. Some may be dull. Yet, it is in the will power of all to attain to a certain standard of success. Thus, we may conclude with another saying “Strong Determination is the key to success”.

c) Together we can achieve more.
This means Unity is strength. If people work together, the work is easier and is completed more quickly. This proverb means that you should take advantage of a favourable situation before it changes. It is best to always be honest and tell the truth. By doing so, you will win the trust and respect of others. A country is strong whose citizens are united, a family is strong whose members stay together. Thus unity is very important in each and every sphere of our life.

Giving instructions: (Text Book Page No. 119)
Here are a few instructions are given by a Health Inspector to a group of students, in order to prevent malaria and dengue. Complete the series by adding some more important instructions. (Text Book Page No. 120)

1. Do not allow water to stagnate in and around your house.
2. Keep your surroundings clean.
3. Wear long-sleeved shirts/blouses and long pants/skirts that cover your arms and legs.
4. Do not litter the place with wastes
5. Divide the waste into perishable and non¬perishable and put them in the proper place

Now, write a set of 8 to 10 instructions for the following situations:

1. A doctor instructing a patient regarding a healthy diet and proper care after surgery.

• Eat food at regular intervals.
• Do not take fatty food
• Eat more green vegetables
• Go for a walk daily
• Do not take more sugar and salt

2. A traffic police personnel to the public, as to how to move around in safety, in crowded public places during festival seasons.

• Do not wear lots of jewels
• Take care of personal belongings.
• Do not allow children to stray away.
• Always follow the queue

3. A mother to her children, on safety measures to be taken before leaving home on vacation.

• Do not go alone
• Carry your belongings safely
• Do not eat from roadside shops
• Do not waste time unnecessarily

Reading:

On the basis of your understanding of the given passage, make notes in any appropriate format:

The Sherpas were nomadic people who first migrated from Tibet approximately 600 years ago, through the Nangpa La pass and settled in the Solukhumbu District, Nepal. These nomadic people then gradually moved westward along salt trade routes. During the 14th century, Sherpa ancestors migrated from Kham. The group of people from the Kham region, east of Tibet, was called “ShyarKhamba”.

The inhabitants of ShyarKhumbu were called Sherpa. Sherpa migrants travelled through O and Tsang, before crossing the Himalayas. According to Sherpa’s oral history, four groups migrated out of Solukhumbu at different times, giving rise to the four fundamental Sherpa clAnswer: Minyagpa, Thimmi, Sertawa, and Chawa. These four groups have since split into the more than 20 different clans that exist today.

Sherpas had little contact with the world beyond the mountains and they spoke their own language. Ang Dawa, a 76-year-old former mountaineer recalled “My first expedition was to Makalu [the world’s fifth highest mountain] with Sir Edmund Hillary’ We were not allowed to got really heavy when wet, and we only got a little salary, but we danced the Sherpa dance, and we were able to buy firewood and make campfires, and we spent a lot of the time dancing and singing and drinking. Today Sherpas get good pay and good equipment, but they don’t have good entertainment. My one regret is that I never got to the top of Everest. I got to the South Summit, but I never got a chance to go for the top.

The transformation began when the Sherpa Tenzing Norgay and the New in 1953. Edmund Hillary took efforts to
build schools and health clinics to raise the living standards of the Sherpas. Thus life in Khumbu improved due to the efforts taken by Edmund Hillary and hence he was known as ‘Sherpa King’.

Sherpas working on the Everest generally tend to perish one by one, casualties of crevasse falls, avalanches, and altitude sickness. Some have simply disappeared on the mountain, never to be seen again. Apart from the bad seasons in 1922, 1970, and 2014 they do not die en masse. Sherpas carry the heaviest loads and pay the highest prices on the world’s tallest mountain. In some ways, Sherpas have benefited from the commercialization of Everest more than any group, earning income from thousands of climbers and trekkers drawn to the mountain. While interest in climbing Everest grew gradually over the decades after the first ascent, it wasn’t until the 1990s that the economic motives of commercial guiding on Everest began.

This leads to an eclipse of the amateur impetus of traditional mountaineering. Climbers looked after each other for the love of adventure and “the brotherhood of the rope” now are tending to mountain businesses. Sherpas have taken up jobs as guides to look after clients for a salary. Commercial guiding agencies promised any reasonably fit person a shot at Everest.

Note Making:

Notes:
Life of Sherpas

I. Nomadic Sherpas migrated from Tibet to Nepal
a) 600 years ago
b) Shyarkhamba
c) 4 Sherpa clans – Minyaagpa, Thimmi, Sertawa, Chawa.

II. Little contact beyond mountains
a) have their own language
b) have no opportunity to reach the top
c) carry things for others
d) Edmund Hillary the ‘Sherpa King’

III. Sherpas die in mountain casualties:
a) carry heaviest loads
b) earn money from climbers
c) 1990s commercial guiding on Everest

Grammar:

Kinds of sentences — Simple, Complex, and Compound:

a) Simple sentence: (Text Book Page No.121)

Task 1:
Pick out the finite verbs in the following sentences: (Text Book Page No. 121)

Question a.
You can solve this problem in different ways.
Answer:
can solve

Question b.
The professor has been working on the last chapter of the book since March.
Answer:
as been working

Question c.
Despite being a celebrity, Ravi mingles easily with everyone.
Answer:
mingles

Question d.
You must speak clearly to make yourself understood.
Answer:
must speak

Question e.
The chairman being away, the clerk is unable to approve the proposal.
Answer:
is

Question f.
Getting down from the car, the Chief Guest walked towards the dais amidst applause.
Answer:
walked

Question g.
The old man struggled to walk without support.
Answer:
struggled

Question h.
In case of emergency, please contact this number.
Answer:
contact

Question i.
The sun having set, the temperature fell rapidly.
Answer:
fell

Question j.
But for your help, I could not have completed the assignment.
Answer:
completed

Task 2:

II. Read the following passage and identify the simple sentences: (Text Book Page No. 121)

Question  1.
Sunflowers turn according to the position of the sun. In other words, they ‘chase the light’. Have you ever wondered what happens on cloudy, rainy days when the sun is completely covered by clouds? If you think the sunflower withers or turns its head towards the ground, you are completely mistaken. Do you know what happens? Sunflowers turn to each other to share their energy. Learning from Nature, we too should support and empower each other.
Answer:
Sunflowers turn according to the position of the sun. In other words, they ‘chase the light’. Have you ever wondered what happens on cloudy, rainy days when the sun is completely covered by clouds? If you think the sunflower withers or turns its head towards the ground, you are completely mistaken. Do you know what happens? Sunflowers turn to each other to share their energy. Learning from Nature, we too should support and empower each other.

b) Complex sentence: (Text Book Page No.122)

Task 1:

Look at the following complex sentences. Circle the Main clauses and underline the Subordinate clauses: (Text Book Page No. I23) 

Question a.
Nobody knows when the power supply will resume.
Answer:
Nobody knows when the power supply will resume.
M.C                                              S.C

Question b.
Please tell me what the time is.
Answer:
Please tell me             what the time is.
M.C                                   S.C

Question c.
The man who directed the film was my schoolmate.
Answer:
The man             who directed the film          was my schoolmate.
M.C                             S.C                                          M.C

Question d.
I believe that all men are basically good.
Answer:
I believe                  that all men are basically good.
M.C                                           S.C

Question e.
No one knows when he will return.
Answer:
No one knows               when he will return.
M.C                                           S.C

Task 2:

Pick out the complex sentences in the following passage: (Text Book Page No. 123)

Question  1.
A man saw a lion in the bush, as he was walking through the forest. He did not know what to do. He was helpless. He was too scared to turn around and run. He just knelt down as if he were getting ready to pray. He closed his eyes, thinking that the lion would pounce on him anytime. Out of the corner of his eye, he saw the lion on its knees too. Shocked, he asked the lion what it was doing. The lion replied that he was praying before he started his meal.
Answer:
A man saw a lion in the bush, as he was walking through the forest. He did not know what to do. He was helpless. He was too scared to turn around and run. He just knelt down as if he were getting ready to pray. He closed his eyes, thinking that the lion would pounce on him anytime. Out of the corner of his eye, he saw the lion on its knees too. Shocked, he asked the lion what it was doing. The lion replied that he was praying before he started his meal.

c) Compound sentence: (Text Book Page No. 123)
Identify the two Main clauses and conjunction in each of the following sentences:

Question a.
It started raining suddenly and people ran for shelter.
Answer:
It started raining suddenly            and                              people ran for shelter.
M.C                                            conjunction                                    M.C

Question b.
Understand the concept well, otherwise, you cannot solve the problem.
Answer:
Understand the concept well,              otherwise,                          you cannot solve the problem.
M.C                                                      conjunction                                            M.C

Question c.
Fifty candidates appeared for the interview, but only five were selected.
Answer:
Fifty cT’rlidates appeared for the interview,                 but                   only five were selected.
M.C                                                                         conjunction                          M.C

Question d.
Ramesh did not know Spanish, so he wanted a translator.
Answer:
Ramesh did not know Spanish,                                so                                        he wanted a translator.
M.C                                                                     conjunction                                           M.C

Question e.
He is a good actor, still, he is not popular.
Answer:
He is a good actor,                   Still,                         he is not popular.
M.C                                       conjunction                          M.C

Task 2:

Pick out the compound sentences in the following passage: (Text Book Page No. 123) 

Question  1.
The food we eat has to be digested and then thrown out of the body. The air we breathe in, has to be thrown out, to help us survive. But we hold negative emotions like insecurity, anger, and jealousy within ourselves for years. If these negative emotions are not eliminated, the mind grows corrupt and diseased. Let us do away with hatred and lead a healthy life filled with peace and joy.
Answer:
The food we eat has to be digested and then thrown out of the body. The air we breathe in, has to be thrown out, to help us survive. But we hold negative emotions like insecurity, anger, and jealousy within ourselves for years. If these negative emotions are not eliminated, the mind grows corrupt and diseased. Let us do away with hatred and lead a healthy life filled with peace and joy.

Task 3:

Complete the sentences choosing the right endings: (Text Book Page No. 124)


Answer:

1. We were thoroughly disappointed	a. since our team did not get a prize
2. Hardly had he stepped out	b. when it began to rain
3. They wanted	c. to find out his address
4. Since we had run out of petrol	d. we could not go further
5. The cause of his injury	e. was his reckless driving

Conditional clause:

Task 1:

Read the following sentences and fill in the blanks: (Text Book Page No. 124)

Question a.
If I _______ (be) a spider, I would ______ (weave) webs.
Answer:
were, weave

Question b.
If Raj ______ (be) a sculptor, he would ______ (make) beautiful idols.
Answer:
were, make

Question c.
If Mary had an umbrella, she would ______ (lend) it to me.
Answer:
lend

Question d.
Rex would have played with me if he ______ (has) time.
Answer:
had

Question e.
If I were you, I would ______ (accept) this offer.
Answer:
accept

Question f.
We will ______ (select) storybooks for kids if we allow time for storytelling.
Answer:
select

Question g.
The Education Minister will ______ (visit) our school tomorrow if he goes by this way.
Answer:
visit

Question h.
You will be rewarded by the wise if you ______ (stand) for truth.
Answer:
stand

Question i.
If my mother ______ (know) of my poor performance in the exam, she will not allow me to watch a movie.
Answer:
knows

Question j.
If I had won the lottery, I would have ______ (donate) relief materials for the flood victims.
Answer:
donated

Task 2:

Rewrite the following sentences using ‘If’ without changing the meaning: (Text Book Page No. 125)

e.g. Unless you go for a walk regularly, you cannot reduce your weight. (Use ‘If’)
If you do not go for a walk regularly, you cannot reduce your weight.

Question a.
Sindhu would not have won the world championship unless she had single-minded devotion.
Answer:
If Sindhu had not had single-minded devotion, she would not have won the world championship.

Question b.
You will not reach your goal unless you chase your dream.
Answer:
If you do not chase your dream, you will not reach your goal.

Question c.
Unless we plant more trees, we cannot save our planet.
Answer:
If we plant more trees, we can save our planet.

Question d.
The rescue team would not have saved the victims unless they had received the call in time.
Answer:
If the rescue team had received the call in time, they would have saved the victims.

Question e.
The palace cannot be kept clean unless we appoint more people.
Answer:
If we do not appoint more people, the palace cannot be kept clean.

Question f.
The portraits would not have been so natural unless the artist had given his best.
Answer:
If the artist had not given his best, the portraits would not have been so natural.

Question g.
The manager would not have selected Nithiksha unless she exhibited good accounting skills.
Answer:
If she had not exhibited good accounting skills, the manager would not have selected Nithiksha.

Question h.
The policeman would not have arrested the man unless he had violated the rules.
Answer:
If he had not violated the rules, the policeman would not have arrested the man.

Question i.
Mr. Kunaal would not sponsor my higher education unless I studied well.
Answer:
If I studied well, Mr. Kunaal would sponsor my higher education.

Question j.
Kavin will not stop flying kites unless he understands the risk involved in it.
Answer:
If kavin understands the risk involved in it, he will stop flying kites.

Question k.
Tanya would not know the answer unless she referred to the answer key.
Answer:
If she referred to the answer key, Tanya would know the answer.

Question l.
My village cannot achieve a 100 % literacy rate unless the elders of the village cooperate with the education department.
Answer:
If the elders of the village do not cooperate with the education department, my village cannot achieve 100% literacy.

Writing:

Summarizing is to briefly sum up the various points from the notes made from the original passage.

Refer to the reading passage (Text Book Page No. 120). You must have completed reading. Now go through the passage once again and refer to the notes made and do the summarizing.

Rough Copy:
The Sherpas were nomadic people who migrated from Tibet about 600 years ago. They moved to the west. The ancestors from the Kham region were called Shayar khamba. The inhabitants were called Sherpas. There were fundamental clans, sherpas lived beyond the mountains. An old Sherpa Ang Dawa accompanied Edwin Hillary – He was called the Sherpa king, because he took efforts to build schools and health clinics to raise their living standards, Sherpas working on the Everest got perished due to Crevasse falls, Avalanches, and altitude sickness. They can carry heavy loads. They have benefited from the commercialization. Sherpas have taken up to look after clients for a salary. Sherpas’ life is entangled with the mountains.

Fair Copy:

The Sherpas

The Sherpas were nomadic people who migrated from Tibet about 600 years ago. They moved to the west. The ancestors from the Kham region were called Shayar Khamba. The inhabitants were called Sherpas. There were fundamental clans. Sherpas lived beyond the mountains. An old Sherpa Ang Dawa accompanied Edwin Hillary – He was called the Sherpa king because he took efforts to build schools and health clinics to raise their living standards. Sherpas working on the Everest got perished due to Crevasse falls, Avalanches, and altitude sickness. They can cany heavy loads. They have benefited by the commercialization. Sherpas have taken up jobs as guides to look after clients for a salary. Sherpas’ life is entangled with the mountains.

Interpreting non-verbal presentation:

Reading a map:

Let us together scale the summit. Here is a drawing of Everest showing the way to the summit, and the position of the camps with their heights. Trace the trekking trail to reach the summit with the given detail and write an interesting paragraph in about 100 words. (Text Book Page No. 126)

Paragraph about trekking trail to reach the summit:

We started our journey at the base camp. We have sufficient oxygen with us. At the 3rd camp, we suffered a lot due to ice fail. We didn’t come out till the weather was clear. Only three of our team could reach the 8th camp. From here the ridge narrowed to a knife-edged and it was a steep slope. So we had to cut steps and then moved. Since there was a lot of firm snow, we could do it. With great efforts. We reached the 9th camp. One of my friends became sick, I started my challenging journey with my other friend. Having fierce determination, nothing could stop us from reaching the top. When we realized that the ridge ahead of us dropped sharply away, we stood on the summit. It was a splendid joy we shared with each other and shouted, at the height of 28,700 feet.

ஆசிரியரைப் பற்றி:

சர் எட்மெண்ட் பெர்சிவல் ஹிலாரி (Sir Edmund Percival Hillary) (20 ஜீலை 1919 – 11 ஜனவரி 2008) நியூசிலாந்து நாட்டைச் சேர்ந்த மலை ஏறுபவர் (mountaineer), கண்டுபிடிப்பாளர் (explorer) மற்றும் தாராளமான நன்கொடையாளர் (philanthropist). இவர் இரண்டாம் உலகப்போரின் போது “நியூசிலாந்து ராயல் விமானப் படையில்” மாலுமியாக (navigator) பணியாற்றினார்.

எவரெஸ்ட் மலையில் ஏறியதற்கு பின்பு, அவர் தான் உருவாக்கிய “ஹிமாலயன் டிரஸ்ட்” மூலம் நேபாள நாட்டில் உள்ள “ஷெர்பா” (Sherpa) மக்களுக்கு சேவை செய்வதில் தன்னை ஈடுபடுத்திக் கொண்டார். “High Adventure, No Latitude for Error, Nothing Venture, Nothing Win. View from the Summit, The Remarkable Memoir by the First Person to Conquer Everest” ஆகியவை அவருடைய பிரபலமான (famous) புத்தகங்களாகும்.

பாடத்தைப் பற்றி:

இப்பாடத்தில் எவரெஸ்ட் சிகரத்தை தொட்ட உலக புகழ் பெற்ற மலைஏறும் வீரர் எட்மண்ட் ஹிலாரி தன் அனுபவத்தை கொடுத்திருக்கிறார். குறிப்பாக மே மாதம் 29ம் தேதி மலையேறும் இறுதி நாளில் நடந்த மெய் சிலிர்க்கும் நிகழ்வை பற்றியும் காலை 4.30 மணி முதல் பிற்பகல் 2 மணி வரை டென்சிங்குடன் சேர்ந்து அவர் எதிர்கொண்ட அபாயங்கள், புத்திசாலித்தனங்கள் பற்றியும் விரிவாக சொல்லப்பட்டு இருக்கிறது. அவர் புரிந்த இந்த அபாயகரமான நிகழ்வை, பாடத்தைப் படித்து விரிவாகக் காண்போம். சாகசங்கள், பயணங்கள் மற்றும் கண்டுபிடிப்புகள் எப்போதும் உற்சாகம் தரக்கூடியவை.

குறிப்பாக அவை எப்பொழுதும் மெய்யானதாகவும் மற்றும் முதன்முதலில் நடப்பதாகவும் இருந்தால் மெய்சிலிப்பேற்றக்கூடியவை. ஒருவர் மனதில் எழும் ஒரே கேள்வி என்னவென்றால், இவ்வாறு அதிகப்படியான ஆபத்துக்கள் அடங்கிய காரியங்களை ஒருவரை செய்யத் தூண்டுவது எது என்பதே ஆகும். இது வலிமைமிக்க துணிகரச் செயலை புரியக்கூடிய மனநிலை மற்றும் தனிச்சிறப்புமிக்க பண்புகளே அவர்களை அத்தகைய உயரங்களை தொடச் செய்கிறது.

The Summit Summary in Tamil

“எவரெஸ்ட்டிற்கு ஏறும் வழி” (The Ascent of Everest) என்னும் ஜான் ஹன்ட் (John Hunt) எழுதிய நூலை சற்று தழுவி இப்பாடப்பகுதி எடுக்கப்பட்டது. எவரெஸ்ட் சிகரம் எவ்வாறு அடையப்பட்டது என்பது சர் எட்மெண்ட் ஹிலாரியின் (Sir Edmund Hillary) சொந்த வார்த்தைகளால் சொல்லப்படுகிறது.

மே 28ல் தெற்கு கணவாயில் போடப்பட்ட முகாமில் 8 பேர் இருந்தனர், முறையே எட்மெண்ட் ஹிலாரி (Edmund Hillary), டென்சிங் (Tenzing), ஜார்ஜ் லோ (George Lowe), அல்பிரட் கிரிகோரி (Albert Gregory) மற்றும் இரண்டு செர்பாக்கள் (Sherpas), பெம்பா (Pemba), அங்க் நியிமா (Ang Nyima). ஆனால் பெம்பா ஏற முடியாத அளவிற்கு உடல்நிலை சரியில்லாமல் இருந்தார். மற்றவர்கள் அதிக சுமையுடன் அன்று 27900 அடி ஏறினர். அங்கே ஹிலாரியும், டென்சிங்கும் சிறிய கூடாரத்தில் இருந்து அவர்களுடன் வந்த மூன்று நபர்கள் முகடுகளில் சறுக்கி தெற்கு கணவாய்க்கு திரும்புவதை கண்டார்கள்.

சூரியன் மறையும் போது, ஹிலாரி மற்றும் டென்சிங் தங்கள் கூடாரத்திற்குள் சென்று, வெப்பம் தரும் (warm clothing) ஆடைகளை அணிந்து, தூங்கும் பைகளுக்குள் நெளிந்து சென்றார்கள். மறுநாள், மே 29ல் காலை 4 மணிக்கு ஏறுவதற்கு அவர்கள் தயாரானார்கள் (ready).

நாங்கள் எங்களது சூட்டடுப்பை (cooker) பற்ற வைத்து, அதிக அளவு எலுமிச்சை சாறையும் (lemon juice), சர்க்கரையும் (sugar) கலந்து குடித்தோம், அதன்பின் எங்களது கடைசி தகரக்குவளையில் உள்ள சார்டின் (sardines) எனப்படும் கடல் உணவையும், பிஸ்கெட்டையும் எடுத்துக் கொண்டோம். நான் எனது ஆக்சிஜன் தொகுப்பு பொருட்களை கூடாரத்தின் உள்ளே இழுத்து, அதன் மேலிருந்த பனிக்கட்டிகளை சுத்தம் செய்து அதனை பரிசோதனை செய்து சோதித்து பார்த்தேன்.

முந்தைய நாள் ஈரமாயிருந்து. இன்று உறைந்து போன (frozen) எனது காலணிகளை கழற்றினேன். எனவே அவைகளை நான் அடுப்பில் இருந்து வரும் புகையினால் சூடேற்றி மென்மையாக்கினேன். எங்களது கீழ் ஆடைக்கு மேலே நாங்கள் காற்று சீராக்கும் ஆடையை அணிந்தும், எங்கள் கைகளில் பட்டு, கம்பளி மற்றும் காற்று சீராக்கும், கையுறைகளையும் (gloves) அணிந்துக் கொண்டோம்.

காலை 6.30 மணியளவில் நாங்கள் எங்கள் கூடராத்தை விட்டு பனிக்குள் (snow) நகர்ந்து வந்து, 30 அளவுடைய ஆக்சிஜன் (oxygen) உருளையை (gear) முதுகில் சுமந்து, முகமூடியை அணிந்து, ஆக்சிஜன் உருளையில் குழாயைத் திறந்து எங்களது நுரையீரலுக்குள் ஆக்சிஜன் வாழ்வை அனுப்பினோம். ஒரு சில சிறந்த ஆழமான சுவாசத்திற்குப் பிறகு நாங்கள் புறப்பட தயாரானோம். எனது உறைந்து போன பாதங்களைப் பற்றி நான் சிறிது கவலைப்பட்டாலும், நான் டென்சிங்கை நகரச் சொன்னேன்.

டென்சிங் நீண்ட மலைமுகட்டை நோக்கி (traverse) செல்லக் கூடிய குறுக்கு பாதையில் நடந்தார். நாங்கள் 28000 அடி பனிக்குவியலை உண்டாக்கிய மலைமுகட்டின் உச்சியை அடைந்தோம். அதன் பிறகு அந்த முகடானது கத்தி போல் கூர்மையான பாதையாக காணப்பட்டது. என் காலகள் இப்போது வெப்பமானதால், நான் டென்சிங்கை முந்தினேன்.

கடினமானதாகவும், ஆபத்தாகவும் இருக்கக்கூடிய மலை முகட்டிற்கு செல்லக்கூடிய பாதையானது மென்பனியால் உருவாகயிருந்தது. அது சில நேரங்களில் என் நிலையிலேயே இருக்க செய்தாலும், அடிக்கடி மேலே செல்ல வழி விட்டது. பல நூறு அடிகளுக்கு அப்புறம் நாங்கள் சிறிய காலியான இடத்திற்கு வந்தோம். அங்கே எவான் மற்றும் போர்டில்லான் ஆகியோரின் முந்தைய முயற்சியில் விட்டுச் சென்ற இரண்டு ஆக்சிஜன் குடுவைகளை கண்டோம்.

மதிப்பு அளவீட்டிலிருந்த பனியை நான் விலக்கி, நாங்கள் அதனை சிக்கனமாக உபயோகித்தால் திரும்ப தெற்கு கணவாய் வரை செல்லத் தேவையான பல நூறு லிட்டர் ஆக்சிஜனை அவை இன்னும் கொண்டுள்ளன என்பதை கண்டறிந்து நிம்மதியானேன்.

நான் தெற்கு உச்சியின் கடைசி 400 அடிக்கு முன்னால் உள்ள மலைமுகட்டிற்கு (ridge) செல்லக்கூடிய பாதையில் தொடர்ந்து பயணமானோம்.

முன்னேறி செல்வதில் நாங்கள் அடிக்கடி (frequent) மாற்றங்களை கொண்டு வந்தோம். நான் எனது அடியை ஆழமான பாதையின் மீதிருந்த ஆழமான பனியில் வைத்த போது, ஒரு பகுதி விலகி சென்றதனால், நான் மூன்று நான்கு அடிகள் பின்னோக்கி வழுக்கினேன். நாங்கள் முன்னேறி செல்வது குறித்து டென்சிங்கிடம் நான் அறிவுரைகளை (advisability) கேட்ட போது, அவர் பனியைக் குறித்து மகிழ்ச்சியற்றுக் காணப்பட்டாலும் அவருடைய வழக்கமான கூற்றான “நீங்கள் விரும்பிய படியே” என்று உரையை முடித்தார்.

நான் முன்னேறிச் செல்வதற்கு முடிவெடுத்தேன் (decided). கடைசியாக நாங்கள் மேலே இருந்த உறுதியான பனியை அடைந்த போது நாங்கள் கடைசியா இருந்த செங்குத்தான சரிவினில் பனியை உடைக்கும் கம்பி கொண்டு பனியை வெட்டி படிகளை உருவாக்கினோம். அப்பொழுது நேரம் காலை 9 மணி.

தெற்கு சிகரத்திற்கு (south summit) கீழே நாங்கள் அமர்வதற்கு ஒரு இருக்கையை வெட்டினோம், பின்பு எங்களது பாதி நிரப்பப்பட்ட முதல் ஆக்சிஜன் குடுவை தீர்ந்திருந்த போது எங்களிடம் ஒரே ஒரு ஆக்சிஜன் நிரம்பிய குடுவை மட்டுமே மீதம் இருந்தது. இப்போது எங்களது உபகரணங்கள் சுமப்பதற்கு இலகுவாகவும், சுமார் 20 க்கு சற்று அதிகமாகவும் தான் எடை கொண்டிருந்தன.

என்னுடைய பனிக் கோடாரி (ice-axe) அந்த முகட்டின் முதல் செங்குத்தான சாய்வினை வெட்டிய போது, என்னுடைய உயர்ந்த நம்பிக்கைகளை (high hopes) நான் உணர்ந்தேன். பனியாகவும் படிகமாகவும் (crystalline), உறுதியானதாகவும் இருந்தது. பனிக்கோடாரியின் இரண்டு அல்லது மூன்று அடிகள் எங்களது பெரிய படிகட்டுகளை (large steps) உண்டு பண்ணியது. பனிக்கோடாரியின் ஒரு உறுதியான அடியானது கைப்பிடியை பாதி வரை மூழ்கச் செய்து, உறுதியான வசதியான மலை ஏற்றத்துக்கு வழி வகுக்கும்.

நாங்கள் ஒரு நேரத்தில் ஒருவர் மட்டுமே நகர்ந்தோம். என்னால் 40 படிகட்டுகளை வெட்ட முடியும், நான் அவ்வாறு வெட்டும் போது என் கயிரை டென்சிங் பிடித்துக் கொண்டு என் எடையைத் தாங்குவார் (belaying). அதன்பின் நான் என் கைப்பிடியை மூழ்கச் செய்து, அதனைச் சுற்றி கயிரால் கண்ணிகள் (loops) போட்டேன். டென்சிங்கோ , உடையும் படிக்கட்டுகளுக்கு எதிராக என்னை நோக்கி நகர்ந்து வந்தார். பின்னர் மறுபடியும் அவர் என் கயிரை பிடித்துக் கொண்டு என் எடையத்ை தாங்கும் போது, நான் வெட்டினேன்.

அதிக இடங்களில் தொங்குகின்ற பனிப்பாறைகள் மிகப் பெரியதாக இருந்ததால், அவற்றிலிருந்து தப்பிக்க நான் பனியின் மேற்பரப்பில் இருந்து பாறை வரையிலும் படிக்கட்டுகளை வெட்டினேன். இந்தப் பெரிய பாறையிலிருந்து நேராக 8000 அடிக்கு கீழாக மேற்கு பள்ளத்தாக்கில் உள்ள முகாம் 4ன் சிறிய கூடாரங்களைப் (tiny tents) பார்த்து சிலிர்ப்பேற்றியது. பாறைகளில் மேலே தவழ்ந்ததாலும் (crawl) பனியில் கைப்பிடிகளை வெட்டியதாலும் எங்களால் இந்த கடினமான பகுதிகளை வேகமாக கடந்து செல்ல முடிந்தது.

அதன் கிழக்குப் பகுதியில் (east side) இன்னொரு பெரிய பனிப்பாதை இருந்தது. அது 40-அடி உயர படியில் உருண்டோடி வந்தது, அங்கே பனிப்பாறைக்கும் பாறைக்கும் இடையே ஒரு விரிசல் (crack) இருந்தது. டென்சிங்கை எவ்வளவு சிறப்பாக என் எடையைத் தாங்க முடியுமோ, அவ்வாறு செய்து விட்டு, நான் அந்த விரிசலுக்குள் ஏறினேன். பின்னர், பின்னோக்கி உதைத்தவாறு என் மலையேறும் கருவியில் இருந்த ஆணிகளை உறைந்த பனியில் முழுவதுமாக மூழ்கச் செய்து நான் தரையில் இருந்து நெம்பியவாறு மேலே எழுந்தேன்.

அங்கிருந்து அனைத்து சிறிய பாறைப்பிடிகளின் துணையோடு, என் மூட்டு, தோள்பட்டை, கைகளால் முடிந்த அளவு பலத்தை திரட்டிக்கொண்டு, பனிப்பாறையுடன் ஒட்டிக்கொள்ள வேண்டும் என்று ஜெபித்துக் (praying) கொண்டே, நான் விரிசலில் மலையேறும் கருவியின் மூலம் பின்புறமாக நகர்ந்தேன்.

எனது முன்னேற்றம் (progress) மெதுவாக இருந்தாலும், நிலையானதாக இருந்தது. டென்சிங் கயிரை விட்டுவிட்டதால், நான் மேல் நோக்கி மெதுமெதுவாக பாறையின் உச்சியைத் தொடும் வரை நகர்ந்து, விரிசலிலிருந்து வெளிவந்து நான் தெற்கு சிகரத்திற்கு படிகளை வெட்டிக் கொண்டிருந்த போது, நான் சுதந்திரமாகவும், நன்றாக இருப்பது போன்றும் உணர்ந்தேன்.

ஒரு சில நிமிடங்களுக்கு நான் மூச்சுவிடுவதற்காக படுத்திருந்தேன். முதன் முறையாக நாங்கள் மலைச் சிகரத்தை அடைவதை எதுவும் தடுக்க முடியாது என்று மிக உறுதியாக உணர்ந்தோம், நான் அந்த தொங்கும் பாறையில் நிலையாக நின்று கொண்டு டென்சிங்கை நோக்கி மேலே வருமாறு கையசைத்தேன் (signed).

நான் அந்த கயிறில் நிம்மதி பெருமூச்சு விட்டபோது, டென்சிங்கோ அந்த விரிசலை நோக்கி மேலே நெளிந்து (wriggled) வந்து கடைசியாக உச்சத்தை அடைந்த போது, மிகப்பெரிய போராட்டத்திற்கு பின், கடலில் இருந்து வெளியே இழுத்து வரப்பட்ட பெரிய இராட்சச மீன் போல் நிலைகுலைந்தார்.

மலை முகடுகள் முன்பு போலவே தொடர்ந்தன, பெரிய பனிப்பாறையின் (giant cornices) வலது புறமும், கூர்மையான மலைச் சரிவுகள் இடது புறமும் இருந்தன. அந்த மலைமுகடானது வலது புறமாக வளைந்து (curved) சென்றதால் எங்களுக்கு மலை உச்சி எங்கு இருக்கிறது என்று தெரியாமல் போனது. நான் (கூனல்) திமிலின் பின்புறத்தை வெட்டும் போது அடுத்த திமில் கண்ணில் தென்படும். நேரம் செலவழிந்து கொண்டே இருந்தது, மலைமுகடுகளும் முடிவது போலத் தோன்றவில்லை.

எங்களது உண்மையான பேரார்வம் சிறிது விலகியது. இது ஒரு கடுமையான போராட்டமாகவே மாறியது. எங்களுக்கு முன்னே இருந்த பனிப்பாறை (snowy summit) எழுவதற்குப் பதிலாக, கிழே விழ ஆரம்பித்ததை நான் அதன்பின் உணர்ந்தேன். ஒரு குறுகலான பனிமுகடு (ice-axe) ஒரு மலை உச்சியை நோக்கி ஓடுவதைப் பார்க்க நான் மேல் நோக்கி பார்த்தேன்.

இதிலிருந்து விடுபட வேண்டும் (relief) என்பதே என் முதல் உணர்வாக இருந்தது. மேலும் இனிமேல் வெட்டுவதற்கு படிகட்டுகளோ, கடப்பதற்கு மலை முகடுகளோ, வென்றுவிடலாம் என்ற நம்பிக்கைக்கு ஏமாற்றம் அளிக்கும் மலை அங்கே காணப்படவில்லை. நான் டென்சிங்கை நோக்கினேன்.

மறைக்கக்கூடிய தலைகவசத்தையும் (balaclavahelmet), பனிக் கண்ணாடியையும், ஆக்சிஜன் வாயு முகமூடியையும் அணிந்திருந்தால் அனைத்தும் ஈட்டி போன்ற கூர்மையான வடிவுடைய பனிக்கட்டியும் சேர்த்து அவருடைய முகத்தை மறைத்தாலும், அவரைச் சுற்றிலும் அவர் பார்த்து விட்டு சந்தோஷத்தினால் சிரித்ததை எதையும் மறைக்கவில்லை.

நாங்கள் இருவரும் கை குலுக்கிக் கொண்டோம். டென்சிங் அவரது கையை என் தோள்மேல் போட்டுக் கொண்டார், கிட்டத்தட்ட மூச்சு முட்டும் வரை நாங்கள் இருவரும் முதுகில் தட்டிக் கொண்டோம். அப்போது நேரம் காலை 11,30 மணியாய் இருந்தது. இந்த மலை முகடு 2 1/2 மணி நேரத்தை எடுத்துக் கொண்டது, அது எங்களுக்கு ஒரு ஆயுட்காலமாகவே தோன்றியது.

எங்களுக்கு கிழக்கில் இதுவரை கண்டறியப்படாத ஏற்படாத பெரிய இராட்சச (giant) அண்டையரான மக்காலு (makalu) இருந்தது. மேகங்களுக்கிடையில் மிகத் தூரத்தில் கஞ்சன்சங்காவின் (Kanchenjunga) மிகப்பெரும் பகுதி அடிவானத்தில் நிழல் போல அச்சுறுத்தும் விதமாக காணப்பட்டது. மேற்கில் இதுவரை கண்டறியப்படாத நேபாளத்தின் மலைப்பகுதிகள் தூரத்திற்கு நீண்டு கிடந்ததை எங்களால் பார்க்க முடிந்தது.

வடக்கு மலைமுகட்டுக்கு கிழே உள்ள வடக்கு கணவாயையும் (North Col), 1920 மற்றும் 1930 களில் மலை ஏறுபவர்களால் போராடி பிரபலமாக்கப்பட்ட பழைய பாதையையும் (old route) காட்டும் மிக முக்கியமான புகைப்படமாக (photograph) நான் கருதும் புகைப்படம் அப்போது எடுக்கப்பட்டது. பத்து மணித்துளிகளுக்கும் பின், நான் மெதுவாக நகர்பவனாகவும் என் விரல்கள் வளைவுத் தன்மையற்றும் மாறின அதனால் நான் உடனே என் ஆக்சிஜன் உபகரணத்தை மாற்றினேன்.

இந்த வேளையில் (meanwhile), டென்சிங் பனியின் நடுவே ஒரு சிறிய துளையிட்டு (hole), அதனுள் – பலதரப்பட்ட சிறிய உணவுப் பொருட்களான (articles), சாக்லேட், பிஸ்கட், குச்சிமிட்டாய் (lollies) போன்றவற்றை வைத்தார். உண்மையில் இவை சிறிய படைப்புகள் தான் ஆனாலும் இந்த உயரமான மலைச்சிகரத்தை இருப்பிடமாக கொண்டுள்ளதாக நம்பும் அனைத்து பக்திமயமான புத்தமதத்தவர்களின் கடவுள்களுக்கு (devout Buddhists) ஒரு சிறு அடையாளப் பரிசாக இருக்கும்.

2 நாட்கள் முன்னதாக நாங்கள் தெற்கு கணவாயில் ஒன்றாக இருந்த போது, கர்னல் ஹன்ட் (Colonel Hunt) என்னிடம் ஒரு சிறிய சிலுவையை (crucifix) கொடுத்து, அதனை மலை உச்சிக்கு எடுத்துச் செல்லுமாறு கேட்டுக் கொண்டார். நானும் பனியில் சிறிய துளையிட்டு, டென்சிங்கின் பரிசுப்பொருட்களுக்கு அருகில் அந்த சிலுவையை வைத்தேன்.

15 நிமிடங்களுக்கு பிறகு, நான் உச்சியிலிருந்து எங்களது படிக்கட்டுகளை நோக்கி கீழே இறங்கினேன். ஆக்ஸிஜன் ‘ குறைந்து கொண்டிருப்பதன் அவசரத்தால் தூண்டப்பட்ட நாங்கள் நேரத்தை வீணாக்காமல் எங்கள் பாதையில் முன்னேறி சென்றோம் (cramponed). நாங்கள் மிக கவனமாக பாறையின் குறுக்கே தவழ்ந்து கடந்து, வழுக்கக்கூடிய பனிப்பகுதியில் ஒருவர் பின் ஒருவராக நகர்ந்து, கடைசியாக எங்களின் படிக்கட்டுகளில் மலையேறினோம் மேலும் திரும்பவும் தெற்கு வந்தோம்.

அப்போது நாங்கள் மிகவும் களைப்பாக இருந்தோம், ஆனாலும், நாங்கள் முகட்டில் சேமித்து வைத்திருந்த உருளைகளை (cylinders) நோக்கி கீழே நகர்ந்தோம். நாங்கள் எங்கள் முகாமை விட்டு மிக குறுகிய தூரத்தில் இருந்தாலும், எங்களது சொந்த குடுவைகளில் மிகச் சிறிய அளவே ஆக்சிஜன் மீதமிருந்ததாலும், நாங்கள் கூடுதல் உருளைகளை சுமந்து எதிர்பார்க்க முடியாத தரைத்தளத்தை (crazy platform) உடைய எங்களது கூடாரத்தை மதியம் 2 மணி அளவில் அடைந்தோம்.

எங்களுக்கு இதுவரை எல்லாவித தேவைகளுக்கும் நன்றாக உதவிய முகாமை கடைசியாக ஒருமுறை பார்த்த படி, நாங்கள் கீழ்நோக்கி திரும்பி, எங்கள் கால்களை இழுத்த படியே, மலை முகட்டிலிருந்து தெற்கு கணவாய்க்கு பாதுகாப்பாக கீழறங்கும் வேலையில் எங்களை ஈடுபடுத்திக் கொண்டோம்.

கனவைப் (dream) போலவே நேரம் கடந்தது முகாமிற்கு 200 அடிக்கு மேலே எங்களை நோக்கி வந்த இருவர் எங்களை சந்தித்தன. அவர்கள் சூடான சூப்பையும், நெருக்கடி கால ஆக்சிஜனையும் சுமந்து கொண்டிருந்த ஜரர்ஜ் லோ மற்றும் வில்பிரட் நாய்ஸ் ஆவார்கள். உயரத்திற்கு மிக அருகில் வந்த போது எனது ஆக்ஸிஜன் தீர்ந்தது. எங்களது வேலையை முடிப்பதற்கு போதுமான அளவு ஆக்ஸிஜனை நாங்கள் வைத்திருந்தோம், ஆனால் கொஞ்சம் கூட தேவைக்கு அதிகமாக இல்லை.

நாங்கள் எங்கள் கூடாரத்திற்குள் தவழ்ந்து வந்து, மகிழ்ச்சிப் பெருமூச்சு விட்டு (delight), எங்கள் தூங்கும் பைகளின் (sleeping bags) மேல் நிலை குலைந்து விழுந்த போது எங்களின் கூடாரங்கள் முடிவில்லாத தெற்கு கணவாயில் காற்றினால் சிறகடித்து (flapped) விழுந்து எங்களை பொட்டலம் போல் பொதிந்தது.

எங்களது பிரயாணத்தை வழிநடத்திய ஜான் ஹண்ட் என்பார் இவ்வாறு கூறுகிறார், “இது ஒரு மறக்க முடியாத நாளாக இருந்தது”. அவர்கள் உச்சியை ஏறி அடைந்தார்கள். அங்கே சந்தோஷ கூச்சலும், கைகலக்கல்களும், அரவணைப்புகள் இந்த இரண்டு கதாநாயகர்களுக்காக காணப்பட்டன. அவர்களது சந்தோஷமும் பெருமிதமும் டென்சிங் மற்றும் ஹிலாரியால் மிகச் கச்சிதமாக முடிக்கப்பட்ட இந்த சாலையை அவர்கள் எவ்வாறு பகிர்ந்து கொண்டார்கள், என்பதை காண்பித்தது. சாகசமானது (adventure) நிறைவுக்கு (concluded) வந்தது.

ஆபத்து மற்றும் கடினமான சூழ்நிலைகளில் உருவாக்கப்பட்ட குழுப்பணியும் (teamwork), தோழமை உணர்வும், ஒன்றையொன்று சந்தித்து, ஒன்றாக வென்றது தான் எவரெஸ்ட் சிகரத்தை ஏறிய கதையாகும், அவர்கள் இந்த பெரிய மலையின் உச்சியில் ஏறிய பிறகும், மற்றவர்களுக்கு அவர்களுடைய சொந்த “எவரெஸ்ட்களை” கண்டறியும் வண்ணம்.

சாகசத்திற்கான நிறைய சந்தர்ப்பங்கள் இன்னனமும் உள்ளன. அவற்றுள் சில, கைக்கு மிக அருகாமையிலும், மற்றவை வேறு நிலங்களில் வெகு தூரத்திலும் உள்ளன. அனைத்து சாகசங்களும் உற்சாகமளிப்பதில்லை. அதுபோல மலையில், மட்டுமே சாகசங்கள் புரியப்படுவதில்லை. நம்முடைய அன்றாட வாழ்க்கையிலும், கடவுளின் உன்னத சக்தியால் மேலே ஏறுவதற்கு எவெரெஸ்ட்கள் உள்ளன.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1

Question 1.
Expand
(i) \(\left(2 x^{2}-\frac{3}{x}\right)^{3}\)
(ii) \(\left(2 x^{2}-3 \sqrt{1-x^{2}}\right)^{4}+\left(2 x^{2}+3 \sqrt{1-x^{2}}\right)^{4}\)
Answer:
(i) \(\left(2 x^{2}-\frac{3}{x}\right)^{3}\)

(ii) \(\left(2 x^{2}-3 \sqrt{1-x^{2}}\right)^{4}+\left(2 x^{2}+3 \sqrt{1-x^{2}}\right)^{4}\)

Question 2.
Compute
(i) 102⁴
(ii) 99⁴
(iii) 9⁷
Answer:
(i) 102⁴
102⁴ = (100 + 2)⁴

= 100000000 + 8 × 1000000 + 24 × 10000 + 3200 + 16
= 100000000 + 8000000 + 240000 + 3216
= 108243216

(ii) 99⁴
99⁴ = (100 – 1)⁴

= 100000000 – 4 × 1000000 + 6 × 10000 – 400 + 1
= 100000000 – 4000000 + 60000 -400+1
= 96059601

(iii) 9⁷
9⁷ = (10 – 1)⁷


= 10³ (10⁴ – 7 × 10³ + 21 × 10² – 35 × 10 + 35) – 21 × 100 + 70 – 1
= 10³ (10000 – 7000 + 2100 – 350 + 35 ) – 2100 + 70 – 1
= 10³ (12135 – 7350) – 2031
= 10³ × 4785 – 2031
= 4785000 – 2031
= 4782969

Question 3.
Using binomial theorem, indicate which of the following two number is larger. (l.Ol)1000000, 10000 Answer:

= 1000000C₀ + 1000000C₁( . 01) + …………………….
= 1 + 1000000 × 0 . 01 + ……………………..
= 1 + 10000 + other positive terms
= 10001 + other positive terms
(1.01)^1000000 – 10000 = 10001 + other positive terms – 10000 > 0
∴ (1.01)^1000000 > 10000 ⇒ (1.01)^1000000 is larger.

Question 4.
Find the coefficient of x¹⁵ in \(\left(x^{2}+\frac{1}{x^{3}}\right)^{10}\)
Answer:
General term T_r+1 = nC_r x^n-r . a^r
∴ In the expansion \(\left(x^{2}+\frac{1}{x^{3}}\right)^{10}\)

To find the coefficient of x¹⁵, Put 20 – 5r = 15
20 – 15 = 5r ⇒ 5r = 5 ⇒ r = 1
∴ T_{1 + 1} = 10C₁x^{20 – 5} ⇒ T₂ = 10 . x¹⁵
∴ The coefficient of x¹⁵ is 10

Question 5.
Find the coefficient of x6 and the coefficient of x² in \(\left(x^{2}-\frac{1}{x^{3}}\right)^{6}\)
Answer:
The general terms is T_r+1 = nC_ra^n-r . a^r
∴ The general term in the expansion of \(\left(x^{2}-\frac{1}{x^{3}}\right)^{6}\) is

To find the coefficient of x⁶ , Put 12 – 5r = 6
12 – 6 = 5r ⇒ 5r = 6 ⇒ r = \(\frac{6}{5}\) which is impossible.
∴ There is no x⁶ term in the expansion.
To find the coefficient of x²,
Put 12 – 5r = 2
⇒ 12 – 2 = 5r
⇒ 10 = 5r
⇒ r = \(\frac{10}{5}\) = 5
Substituting in (1) we have

T₃ = 15
∴ The coefficient of x² is 15

Question 6.
Find the coefficient of x⁴ in the expansion of (1 + x³)⁵⁰ \(\left(x^{2}+\frac{1}{x}\right)^{5}\)
Answer:


= 10x⁴ × 1 + 5 × 1225 x⁴ + 19600 x⁴ + 500 x⁴
= x⁴ (10 + 6125 + 19600 + 500) = 26235 . x⁴
∴ The coefficient of x⁴ is 26235.

Question 7.
Find the constant term of \(\left(x^{3}-\frac{1}{3 x^{2}}\right)^{5}\)
Answer:


To get the constant term,
Put 15 – 5r = 0
⇒ 5r = 15
⇒ r = 3

Question 8.
Find the last two digits of the number 3⁶⁰⁰.
Answer:
Consider 3⁶⁰⁰
3⁶⁰⁰ = (3²)³⁰⁰ = 9³⁰⁰ = (1o – 1)³⁰⁰

= 10³⁰⁰ – 300 (10)²⁹⁹ + ………………. + 300 C₁ × 10 × – 1 + 1 × 1 × 1
= 10³⁰⁰ – 300 (1o)²⁹⁹ + …………….. – 300 × 10 + 1
= 10³⁰⁰ – 300 × 10²⁹⁹ + ……………… – 3000 + 1
All the terms except the last are multiples of 100 and hence divisible by 100.
∴ The last two digits will be 01.

Question 9.
If n is a positive integer show that 9ⁿ⁺¹ – 8n – 9 is always divisible by 64.
Answer:

which is divisible by 64 for all positive integer n.
∴ 9ⁿ – 8n – 1 is divisible by 64 for all positive integer n.
Put n = n + 1 we get
9^{n + 1} – 8 (n + 1) – 1 is divisible by 64 for all possible integer n
(9^{n + 1} – 8n – 8 – 1) is divisible by 64
∴ 9^{n + 1} – 8n – 9 is always divisible by 64

Question 10.
If n is an odd positive integer, prove that the coefficients of the middle terms in the expansion of (x + y)ⁿ are equal.
Answer:
Given (x + y)ⁿ
If n is odd the middle term in the expansion of (x + y)ⁿ are \(\frac{T_{n-1}}{2}+1\) and \(\mathrm{T}_{\frac{\mathrm{n}+1}{2}+1}\)

Question 11.
If n is a positive integer and r is a non-negative integer, prove that the coefficients of x^r and x^n-r in the expansion of (1 + x)ⁿ are equal.
Answer:
Given (1 + x)ⁿ.
General term T_r+1 = nC_r x^{n – r} . a^r
∴ The general term in the expansion of (1 + x)ⁿ is T_r+1 = nC_r . (1)^{n – r} . x^r
T_r+1 = nC_r . x^r ……….. (1)
∴ Coefficient of x^r is nC_r,
Put r = n – r in (1)
T_{n – r + 1} = nC_{n – r} . x^n-r
∴ The coefficient of x^n-r is nC_n-r …………. (2)
we know nC_r = nC^n-r
∴ The coefficient of x^r and coefficient of x^n-r are equal, (by (1) & (2))

Question 12.
If a and b are distinct integers, prove that a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer.
Answer:
Let a = a + b – b
= b + (a – b)
aⁿ = [b + (a – b)]ⁿ
Using binomial expansion

Question 13.
In the binomial expansion of (a + b )n , the coefficients of the 4th and 13th terms are equal to each other , find n.
Answer:
In (a + b)ⁿ general term is t_{r + 1} = ⁿC_r a^{n – r} b^r
So, t₄ = t_{3 + 1} = ⁿC₃ = ⁿC₁₂
⇒ n = 12 + 3 = 15
We are given that their coefficients are equal ⇒ ⁿC₃ = ⁿC₁₂ ⇒ n = 12 + 3 = 15
[ⁿC_x = ⁿC_y ⇒ x = y (or) x + y = n]

Question 14.
If the binomial co – efficients of three consecutive terms in the expansion of (a + x)ⁿ are in the ratio 1 : 7 : 42, then find n.
Answer:
The general term in the expansion of (a + b)ⁿ is T_r+1 = nC_r . a^n-r . b^r
∴ The general term in the expansion of (a + x)ⁿ is T_r+1 = nC_r . a^n-r . x^r
Let the three consecutive terms be r^th term, (r + 1 )^th term, (r + 2 )^th term.

Question 15.
In the binomial coefficients of (1 + x)ⁿ, the coefficients of the 5^th, 6^th, and 7^th terms are in A.P. Find all values of n.
Answer:
Given (1 + x)ⁿ

12(n – 4) = 30 + n² – 5n – 4n + 20 ⇒ 12n – 48 = 30 + n² – 9n + 20
⇒ n² – 9n – 12n + 50 + 48 = 0 ⇒ n² – 21n + 98 = 0
⇒ n² – 14n – 7n + 98 = 0 ⇒ n(n – 14) – 7(n – 14) = 0
⇒ (n – 7) (n – 14) = 0 ⇒ n = 7 or n = 14

Question 16.
prove that

Answer:

This relation is true for all values of n. Equating coefficient of xⁿ on both sides, we have
The general term in the expansion of (1 + x)²ⁿ is T_{r + 1} = 2nC_r (1)^{2n – r} . x^r
Put r = n we get, T_n+1 = 2nC_n . xⁿ
∴ The coefficient of xⁿ in the expansion of (1 + x)²ⁿ is 2nC_n

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 4 Combinatorics and Mathematical Induction Ex 4.5 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.5

Choose the correct or the most suitable answer:

Question 1.
The sum if the digits at the 10^th place of all numbers formed with the help of 2, 4, 5, 7 taken all at a time is
(1) 432
(2) 108
(3) 36
(4) 18
Answer:
(2) 108

Explaination:

Number of digits given = 2, 4, 5, 7
Number of 4 digit numbers formed = 4! =24
So each digit occur \(\frac{24}{4}\) = 6 times
Sum of the digits = 2 + 4 + 5 + 7 = 18
So sum of the digits in each place = 18 × 6 = 108

Question 2.
In an examination, there are three multiple-choice questions and each question has 5 choices. The number of ways in which a student can fail to get all answer correct is
(1) 125
(2) 124
(3) 64
(4) 63
Answer:
(2) 124

Explanation:
Number of multiple-choice questions = 3
Number of choices in each question = 5
Number of ways of answering = 5 × 5 × 5 = 5³ = 125
Number of ways of getting all answer correct = 1
∴ Number of ways of getting incorrect answer = 125 – 1 = 124

Question 3.
The number of ways in which the following prize be given to a class of 30 boys first and second in mathematics, first and second in Physics, first in Chemistry and first in English is
(1) 30⁴ × 29²
(2) 30² × 29³
(3) 30² × 29⁴
(4) 30 × 29⁵
Answer:
(1) 30⁴ × 29²

Explanation:
I and II in maths can be given in 30 × 29 ways.
I and II in physics can be given in 30 × 29 ways.
I and chemistry can be given in 30 ways.
I in English can be given in 30 ways.
So total number of ways = 30 × 29 × 30 × 29 × 30 × 30 = 30⁴ × 29²

Question 4.
The number of 5 digit numbers all digits of which are odd is
(1) 25
(2) 5⁵
(3) 5⁶
(4) 625
Answer:
(2) 5⁵

Explanation:
The odd digit numbers = 1, 3, 5, 7, 9
Since repetition is allowed, each digit can be filled by 1, 3, 5, 7, 9
Unit’s place can be filled in 5 ways
Ten’s place can be filled in 5 ways etc.
∴ The number of 5 digit numbers with an odd digit in each place = 5 × 5 × 5 × 5 × 5 = 5⁵ ways

Question 5.
In 3 fingers, the number of ways four rings can be worn is ————- ways.
(1) 4³ – 1
(2) 3⁴
(3) 68
(4) 64
Answer:
(4) 64

Explanation:
Each letter can be ported in 3 ways
∴ 4 letter is 3⁴ ways

Question 6.

then the value of n are
(1) 7 and 11
(2) 6 and 7
(3) 2 and 11
(4) 2 and 6
Answer:
(2) 6 and 7

Explaination:

( n + 5 ) ( n + 4) = 2 × 11 × (n – 1)
n² + 5n + 4n + 20 = 22n – 22
n² + 9n + 20 – 22n + 22 = 0
n² – 13n + 42 = 0
n (n – 6) – 7 (n – 6) = 0
(n – 7) (n – 6) = 0
n – 7 = 0 or n – 6 = 0
n = 7 and n = 6

Question 7.
The product of r consecutive positive integers is divisible by
(1) r !
(2) (r – 1) !
(3) ( r + 1 ) !
(4) r^r
Answer:
(1) r !

Explanation:
1(2) (3) ….. (r) = r! which is ÷ by r!

Question 8.
The number of five-digit telephone numbers having at least one of their digits repeated is
(1) 90000
(2) 10000
(3) 30240
(4) 69760
Answer:
(4) 69760

Explaination:
Case (i) When zero is allowed in the first place.
The number of five-digit telephone numbers which can
be formed using the digits 0, 1, 2, …………,9 is 10⁵.
The number of five-digit telephone numbers which have none of their digits repeated is 10P⁵
= 10 × 9 × 8 × 7 × 6 = 30240
∴ The required number of telephone number
= 10⁵ – 30240
= 100000 – 30240
= 69,760

Case (ii) When zero is not allowed in the first place,
(a) Repetition allowed:

The number of ways of filling the first box, second box, …………….. , fifth box
= 9 × 10 × 10 × 10 × 10
= 90000

(b) Repetition riot allowed:
Number of ways of filling 1 box = 9
Number of ways of filling 2^nd box = 9
Number of ways of filling 3^rd box = 8
Number of ways of filling 4^th box = 7
Number of ways of filling 5^th box = 6
∴ Number of numbers with no digit is repeated = 9 × 9 × 8 × 7 × 6 = 37216
∴ Number of numbers having atleast one of the digits repeated = 90000 – 37216 = 52,784

Question 9.
If (a² – a ) C₂ = ( a² – a) C₄ then the value of a is
(1) 2
(2) 3
(3) 4
(4) 5
Answer:
(2) 3

Explaination:
(a² – a)C₂ = (a² – a) C₄
(a² – a)C₂ = (a² – a) C_{(a² – a) – 4} [nC_r = nC_n-r]
(a² – a)C₂ = (a² – a) C_{a² – a – 4}
∴ 2 = a² – a – 4
a² – a – 4 – 2 = 0 ⇒ a² – a – 6 = 0
a² – 3a + 2a – 6 = 0
a (a – 3) + 2 (a – 3) = 0
⇒ (a + 2)(a – 3) = 0 ⇒ a = – 2 or 3

Question 10.
There are 10 points in a plane and 4 of them are collinear. The number of straight lines joining any two points is
(1) 45
(2) 40
(3) 39
(4) 38
Answer:
(2) 40

Explaination:
Total number of points = 10
The number of straight lines joining any two points is same as the number ways of selecting 2 points from 10 points.
∴ The number of straight lines = 10C₂
Given 4 points are collinear.
∴ The number of straight lines that these 4 points contribute = 4C₂
Since 4 points are collinear they lie on a line.
∴ Required number of lines = 10C₂ – 4C₂ + 1

= 5 × 9 – 2 × 3 + 1
= 45 – 6 + 1 = 40

Question 11.
The number of ways in which a host lady invite for a party of 8 out of 12 people of whom two do not want to attend the party together is
(1) 2 × 11C₇ + 10C₈
(2) 11C₇ + 10C₈
(3) 12C₈ – 10C₆
(4) 10C₆ + 2!
Answer:
(3) 12C₈ – 10C₆

Explaination:
Number of people = 12
Number of people invited for the party = 8
Number of people not willing to attend the party = 2
Number of ways of selecting 8 people from 12 people is 12C₈
Two people do not attend party out of 10, remaining people 8 can attend in 10C₆ ways.
∴ Number of ways in which two of them do not attend together = 12C₈ – 10C₆

Question 12.
The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines.
(1) 6
(2) 9
(3) 12
(4) 18
Answer:
(4) 18

Explaination:
Number of first set of parallel lines = 4
Number of second set of parallel lines = 3
Number of parallelograms = 4C₂ × 3C₂
= \(\frac{4 \times 3}{1 \times 2} \times \frac{3 \times 2}{1 \times 2}\)
= 6 × 3
= 18

Question 13.
Everybody in a room shakes hands with everybody else. The total number of shake hands is 66. The number of persons in the room is
(1) 11
(2) 12
(3) 10
(4) 6
Answer:
(2) 12

Explaination:
Number of shake hands = 66
Let the number of persons = n
First person shakes hands with the remaining n – 1 persons.
Second person shakes hands with the remaining n – 2 persons.

∴ Total number of shake hands
= (n – 1) + (n – 2) + …………. + 2 + 1
= \(\frac{(\mathrm{n}-1)(\mathrm{n}-1+1)}{2}\)
Given 66 = \(\frac{(\mathrm{n}-1) \mathrm{n}}{2}\)
n² – n = 132
n² – n – 132 = 0
(n – 12) (n + 11) = 0
n = 12 or n = – 11
n = – 11 is not possible.
∴ n = 12

Question 14.
Number of sides of a polygon having 44 diagonals is
(1) 4
(2) 4!
(3) 11
(4) 22
Answer:
(3) 11

Explaination:
Number of diagonals of a polygon with n sides
= nC₂ – n
= \(\frac{n(n-1)}{2}\)
Given \(\frac{n(n-1)}{2}\) – n = 44
n² – n + 2n = 88
n² – 3n – 88 = 0
(n – 11 ) (n + 8) = 0
n = 11 and n = – 8
n = – 8 is not possible.
∴ n = 11

Question 15.
If 10 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, then the total number of point of intersection are
(1) 45
(2) 40
(3) 10!
(4) 210
Answer:
(1) 45

Explaination:
Number of non-parallel lines = 10
Two lines will interest.
∴ Number of points of intersection
= 10C₂ = \(\frac{10 \times 9}{1 \times 2}\) = 45

Question 16.
In a plane there are 10 points are there out of which 4 points are collinear, then the number of triangles formed is
(1) 110
(2) 10C₃
(3) 120
(4) 116
Answer:
(4) 116

Explaination:
Number of points = 10
Number of triangles formed by using 10 points is same as number of ways of choosing 3 points from 10 points = 10C₃
Also given 4 points are collinear.
∴ These 4 points do not contribute triangles.
Thus, total number of triangles = 10C₃ – 4C₃
= \(\frac{10 \times 9 \times 8}{1 \times 2 \times 3}\) – 4
= 5 × 3 × 8 – 4
= 120 – 4
= 116

Question 17.
In 2nC₃ : nC₃ = 11 : 1 then
(1) 5
(2) 6
(3) 11
(4) 7
Answer:
(2) 6

Explaination:
Given 2nC₃ : nC₃ = 11 :

4(2n – 1) = 11 (n – 2)
8n – 4 = 11 n – 22
-4 + 22 = 11 n – 8n
18 = 3n
⇒ n = 6

Question 18.
(n – 1) C_r + ( n – 1 ) C_{r – 1} is
(1) (n + 1) C_r
(2) (n – 1) C_r
(3) nC_r
(4) nC_{r – 1}
Answer:
(3) nC_r

Explaination:
[nC_r + nC_{r – 1} = (n + 1 ) C_r
(n – 1)C_r + (n – 1)C_{r – 1} = (n – 1 + 1) C_r
= nC_r

Question 19.
The number of ways of choosing 5 cards out of a deck of 52 cards which include atleast one king is
(1) 52C₅
(2) 48 C₅
(3) 52C₅ + 48C₅
(4) 52C₅ – 48C₅
Answer:
(4) 52C₅ – 48C₅

Explaination:
Number of cards = 52,
Number of kings = 4
Number of ways of choosing 5 cards out of 52 cards = 52 C₅
Number of cards without kings = 48
Number of ways of choosing 5 cards with no kings from 48 cards = 48 C₅
∴ Number of ways of choosing 5 cards out of 52 cards
= 52C₅ – 48C₅

Question 20.
The number of rectangles that a chessboard has
(1) 81
(2) 9⁹
(3) 1296
(4) 6561
Answer:
(3) 1296

Explaination:
Number of rectangles in a chessboard is

= 36 × 36
= 1296

Question 21.
The number of 10 digit number that can be written by using digits 2 and 3 is
(1) 10C₂ + 9C₂
(2) 2¹⁰
(3) 2¹⁰ – 2
(4) 10!
Answer:
(2) 2¹⁰

Explaination:

Each box can be filled in two ways using digits 2 and 3.
∴ A number of 10 digit number that can be written by using the digits 2 and 3 is 2 × 2 × 2 × …………. 10 times. = 210

Question 22.
If Pr stands for rPr, then the sum of the series 1 + P₁ + 2P₃ + 3P₃ + ……… + nP_n is
(1) P_{n + 1}
(2) P_{n + 1} – 1
(3) P_{n + 1} + 1
(4) (n + 1)P_{n – 1}
Answer:
(1) P_{n + 1}

Explaination:
Given P, = rP_r = r!
1 + 1 × P₁! + 2 × P₂ + 3 × P₃ + …………… + n × P_n!
= 1 + (1 × 1! + 2 × 2! + 3 × 3! + …………. + n × n!)
= (n + 1 )!
= P_{n + 1}
[(1 × 1 !) + (2 × 2!) + (3× 3 !) + ….. + (n × n !) = (n + 1)! – 1]

Question 23.
The product of first n odd natural numbers equals
(1) 2nC_n × nP_n
(2) \(\left(\frac{1}{2}\right)^{n}\) 2nC_n × nP_n
(3) \(\left(\frac{1}{4}\right)^{n}\) × 2nC_n × 2nP_n
(4) nC_n × nP_n
Answer:
(2) \(\left(\frac{1}{2}\right)^{n}\) 2nC_n × nP_n

Explaination:
1 . 3 . 5 …………. (2n – 1)

Question 24.
If nC₄, nC₅, nC₆ are in A. P. the value of n can be
(1) 14
(2) 11
(3) 9
(4) 5
Answer:
(1) 14

Explaination:
Given nC_n, nC₅, nC₆ are in A. P.
∴ 2 × nC₅ = nC₄ + nC₆

12(n – 4) = 30 + (n – 4) (n – 5)
12n – 48 = 30 + n² – 5n – 4n + 20
12n – 48 = 50 + n² – 9n
n² – 9n – 50 – 12n + 48 = 0
n² – 21n + 98 = 0
n² – 14n – 7n + 98 = 0
n(n – 14) – 7(n – 14) = 0
(n – 7) (n – 14) = 0
n – 7 = 0 or n – 14 = 0
n = 7 or n = 14

Question 25.
1 + 3 + 5 + 7 + …………. + 17 is equal to
(1) 101
(2) 81
(3) 71
(4) 61
Answer:
(2) 81

Explaination:
Given 1 + 3 + 5 + 7 + ………… + 17
This is an A.P. with first term a = 1
Common difference d = 2
Last term t_n = 17
a + (n – 1)d = 17
1 + (n – 1)2 = 17
(n – 1)2 = 17 – 1
(n – 1)2 = 16
n – 1 = \(\frac{16}{2}\) = 8
n = 9

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.4

Question 1.
By the principle of mathematical induction, prove that, for n ≥ 1
1² + 2² + 3² + ……….. + n³ = \(\left(\frac{\mathbf{n}(\mathbf{n}+\mathbf{1})}{2}\right)^{2}\)
Answer:
Let P(n) = 1² + 2² + 3² + ……….. + n³
p(n) = \(\left(\frac{\mathbf{n}(\mathbf{n}+\mathbf{1})}{2}\right)^{2}\)

Step 1:
Let us verify the statement for n = 1
p(1) = 1³ = \(\left(\frac{1(1+1)}{2}\right)^{2}\)

P(1) = 1
∴ The given statement is true for n = 1

Step 2 :
Let us assume that the given statement is true for n = k
P(k) = 1³ + 2³ + 3³ + …………… + k³

Step 3 :
Let us prove the statement is true for n = k + 1
P(k + 1 ) = 1³ + 2³ + 3³ + …………… + k³ + (k + 1)³

This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
1² + 2² + 3² + ……….. + n³ = \(\left(\frac{\mathbf{n}(\mathbf{n}+\mathbf{1})}{2}\right)^{2}\)
is true for all natural numbers n.

Question 2.
By the principle of mathematical induction, prove that, for n ≥ 1

Answer:

Step 1 :
Let us verify the statement for n = 1

∴ The given statement is true for n = 1

Step 2 :
Let us assume that the given statement is true for n = k

Step 3:
Let us prove the statement is true for n = k + 1


This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.

is true for all natural numbers n.

Question 3.
Prove that the sum of the first ’n’ non-zero even numbers is n² + n.
Answer:
Let P(n) : 2 + 4 + 6 +…+2n = n² + n, ∀ n ∈ N

Step 1:
P( 1) : 2 = 1² + 1 = 2 which is true for P( 1)

Step 2:
P(k): 2 + 4 + 6+ …+ 2k = k² + k. Let it be true.

Step 3:
P(k + 1) : 2 + 4 + 6 + … + 2k + (2k + 2)
= k²+ k + (2k + 2) = k² + 3k + 2
= k² + 2k + k + 1 + 1
= (k+ 1)²+ (k + 1)
Which is true for P(k + 1)
So, P(k + 1) is true whenever P(k) is true.

Question 4.
By the principle of mathematical induction, prove that, for n ≥ 1, 1 . 2 + 2 . 3 + 3 . 4 + …………. + n . (n + 1) = \(\frac{\mathbf{n}(\mathbf{n}+1)(\mathbf{n}+2)}{3}\)
Answer:
Let P(n) = 1 . 2 + 2 . 3 + 3 . 4 + ………… + n . (n + 1) = \(\frac{\mathbf{n}(\mathbf{n}+1)(\mathbf{n}+2)}{3}\)

Step 1:
Let us verify the statement for n = 1
P(1) = 1 . 2 = \(\frac{1(1+1)(1+2)}{3}\)
= 1 . 2 = \(\frac{1 \cdot 2 \cdot 3}{3}\)
∴ The result is true for n = 1

Step 2:
Let us assume that the result is true for n = k
P (k) = 1-2 + 2-3 + 3- 4 + + k (k + 1 ) =

Step 3:
Let us prove the result for n = k + 1
P(k + 1) = 1 . 2 + 2 . 3 + 3 . 4 + …………… + k (k + 1) (k + 1) (k + 1 + 1)
P (k + 1) = 1 . 2 + 2 . 3 + 3 . 4 + ………….. + k (k + 1) + (k + 1 ) (k + 2)
P (k + 1 ) = P(k) + (k + 1) (k + 2)

This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1, Hence by the principle of mathematical induction, the result is true for all natural numbers n.
1 . 2 + 2 . 3 + 3 . 4 + ………. + n (n + 1) = \(\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}{3}\)
is true for all natural numbers n ≥ 1

Question 5.
Using the Mathematical induction, show that for any natural number n ≥ 2,

Answer:
Let

The first stage is n = 2

Step 1:
Let us verify the result for n = 2

∴ The given result is true for n = 2.

Step 2:
Let us assume that the result is true for n = k.

Step 3:
Let us prove the result for n = k + 1


∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n ≥ 2.

is true for all natural numbers n ≥ 2.

Question 6.
Using the Mathematical induction, show that for any natural number n ≥ 2,

Answer:

Step 1:
Since n ≥ 2 the first stage is n = 2
Let us verify the result for n = 2

∴ P (2 ) is true.
The result is true for n = 2.

Step 2:
Let us assume that the result is true for n = k.

Step 3:
Let us prove the result for n = k + 1


This implies p(k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n ≥ 2.

is true for all natural numbers n ≥ 2

Question 7.
Using the mathematical induction, show that for any natural number n,

Answer:

Step 1:
First, let us verify the result for n = 1

∴ The given result is true for n = 1

Step 2:
Let us assume the result for n = k

Step 3:
Let us prove the result for n = k + 1

Factorizing k³ + 6k² + 9k + 4
f (k) = k³ + 6k² + 9k + 4
f(- 1) = (- 1)³ + 6(- 1)² + 9(- 1) + 4
f(- 1) = – 1 + 6 – 9 + 4 = 0
∴ (k + 1) is a factor of f(k)

k³ + 6k² + 9k + 4 = (k + 1) (k² + 5k + 4).
= (k + 1) (k² + 4k + k + 4)
= (k + 1) [k(k + 4) + 1(k + 4)]
= (k + 1) (k + 4) (k + 1)

This implies P(k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.

is true for all natural numbers n

Question 8.
Using the Mathematical induction, show that for any natural number n,

Answer:

Step 1:
First, let us verify the result for n = 1


∴ The result is true for n = 1

Step 2:
Let us assume that the result is true for n = k

Step 3:
Let us prove the result for n = k + 1


This implies P(k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.

Question 9.
Prove by mathematical induction that
1! + (2 × 2 !) + (3 × 3!) +…. .+ (n × n!) = (n + 1)! – 1
Answer:
P(n) is the statement
1! + (2 × 2!) + (3 × 3!) + ….. + (n × n!) = (n + 1)! – 1
To prove for n = 1
LHS = 1! = 1
RHS = (1 + 1)! – 1 = 2! – 1 = 2 – 1 = 1
LHS = RHS ⇒ P(1) is true
Assume that the given statement is true for n = k
(i.e.) 1! + (2 × 2!) + (3 × 3!) + … + (k × k!) = (k + 1)! – 1 is true
To prove P(k + 1) is true
p(k + 1) = p(k) + t_{(k + 1)}
P(k + 1) = (k + 1)! – 1 + (k + 1) × (k + 1)!
= (k + 1)! + (k + 1) (k + 1)! – 1
= (k + 1)! [1 + k + 1] – 1
= (k + 1)! (k + 2) – 1
= (k + 2)! – 1
= (k + 1 + 1)! – 1
∴ P(k + 1) is true
⇒ P(k) is true, So by the principle of mathematical induction P(n) is true.

Question 10.
Using the Mathematical induction, show that for any natural number n, x²ⁿ – y²ⁿ is divisible by x + y.
Answer:
Let P(n) = x²ⁿ – y²ⁿ is divisible by (x + y)
For n = 1
P(1) = x^{2 × 1} – y^{2 × 1} is divisible by (x + y)
⇒ (x + y) (x – y) is divisible by (x + y)
∴ P(1) is true
Let P(n) be true for n = k
∴ P(k) = x^2k – y^2k is divisible by (x + y)
⇒ x^2k – y^2k = λ(x + y) …… (i)
For n = k + 1
⇒ P(k + 1) = x^{2(k + 1)} – y^{2(k + 1)} is divisible by (x + y)
Now x^{2(k + 2)} – y^{2(k + 2)}
= x^{2k + 2} – x^2ky² + x^2ky² – y^{2k + 2}
= x^2k.x² – x^2ky² + x^2ky² – y^2ky²
= x^2k (x² – y²) + y²λ (x + y) [Using (i)]
⇒ x^{2k + 2} – y^{2k + 2} is divisible by (x + y)
∴ P(k + 1) is true.
Thus P(k) is true ⇒ P(k + 1) is true.
Hence by the principle of mathematical induction, P(n) is true for all n ∈ N

Question 11.
By the principle of mathematical induction, prove that, for n ≥ 1, 1² + 2² + 3² + ………….. + n² > \(\frac{n^{3}}{3}\)
Answer:
Let P (n) = 1² + 2² + 3² + ………….. + n² > \(\frac{n^{3}}{3}\)

Step 1:
Let us first verify the result for n = 1

∴ The result is true for n = 1

Step 2:
Let us assume that the result is true for n = k.
P(k) = 1² + 2² + 3² + …………… + k² > \(\frac{\mathrm{k}^{3}}{3}\)

Step 3:
Let us prove the result for n = k + 1
P(k + 1) = 1² + 2² + 3² + …………… + k² + (k + 1)²
P(k + 1) = P(k) + ( k + 1)²

This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
1² + 2² + 3² + ………….. + n² > \(\frac{n^{3}}{3}\)

Question 12.
Use induction to prove that n³ – 7n + 3, is divisible by 3, for all natural numbers n.
Answer:
Let P(n) : n³ – 7n + 3
Step 1:
P(1) = (1)³ – 7(1) + 3
= 1 – 7 + 3 = -3 which is divisible by 3
So, it is true for P(1).

Step 2:
P(k) : k³ – 7k + 3 = 3λ. Let it be true
⇒ k³ = 3λ + 7k – 3

Step 3:
P(k + 1) = (k + 1)³ – 7(k + 1) + 3
= k³ + 1 + 3k² + 3k – 7k – 7 + 3
= k³ + 3k² – 4k – 3
= (3λ + 7k – 3) + 3k² – 4k – 3 (from Step 2)
= 3k² + 3k + 3λ – 6
= 3(k² + k + λ – 2) which is divisible by 3.
So it is true for P(k + 1).
Hence, P(k + 1) is true whenever it is true for P(k).

Question 13.
Use induction to prove that 5^{n + 1} + 4 × 6ⁿ when divided by 20 leaves a remainder 9 , for all natural numbers n.
Answer:
To prove 5^{n + 1} + 4 × 6ⁿ when divided by 20 leaves a remainder 9
That is to prove 5^{n + 1} + 4 × 6ⁿ – 9 is divisible by 20
Let P (n) be the statement 5^{n + 1} + 4 × 6ⁿ
which is divisible by 20
P (n) = 5^{n + 1} + 4 × 6ⁿ – 9 = 20λ

Step 1 :
Let us verify the statement for n = 1
P (1) = 5^{i + 1} + 4 × 6¹ – 9
= 5² + 4 × 6 – 9
= 25 + 24 – 9
= 49 – 9 = 40
which is divisible by 20
∴ The statement is true for n = 1

Step 2:
Let us assume that the statement is true for n = k
P(k) = 5^{k + 1} + 4 × 6^k – 9 is divisible by 20
P(k) = 5^{k + 1} + 4 × 6^k – 9 = 20λ …………… (1)

Step 3:
Let us prove the result for n = k + 1

Hence, P(k + 1) is divisible by 20.
This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
Thus, 5^{n + 1} + 4 × 6ⁿ when divided by 20 leaves a remainder 9, for all natural numbers n.

Question 14.
Use induction to prove that 10ⁿ + 3 × 4^{n + 2} + 5 is divisible by 9, for all natural numbers n.
Answer:
Let P( n) = 10ⁿ + 3 × 4^{n + 2} + 5 is divisible by 9

Step 1:
Let us verify the result for n = 1
P ( 1 ) = 10¹ + 3 × 4^{1 + 2} + 5
= 15 + 3 × 4³
= 15 + 3 × 64
= 15 + 192 = 207
which is divisible by 9
Thus we have verified the result for n = 1

Step 2:
Let us assume the result is true for n = k
P(k) = 10^k + 3 × 4^{k + 2} + 5 is divisible by 9
∴ 10^k + 3 × 4^{k + 2} + 5 = 9m for some m
10^k = 9m – 5 – 3 × 4^{k + 2} for some m

Step 3:
Let us prove the result for n = k + 1

which is divisible by 9
This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
10ⁿ + 3 × 4^{n + 2} + 5 is divisible by 9,
for all natural numbers n.

Question 15.
Prove that using mathematical induction

Answer:

This implies P (k + 1) is true. Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.