TS 6th Class English Guide Unit 1B I Want Peace

Telangana SCERT TS 6th Class English Study Material Pdf Unit 1B I Want Peace Textbook Questions and Answers.

TS 6th Class English Guide Unit 1B I Want Peace

TS 6th Class English Guide Unit 1B I Want Peace 1

I am big and round,
I wonder if there will be peace;
I hear the sound of people who are being killed
I see the people crying for life,
I want to help the poor people.

I am big and round,
I feel the weight of sorrow on me,
I touch the feelings of the people,
I worry about the future of the people on me,
I cry for help from God.
I am big and round.

I understand the problems of the people on me.
I say grace for all people.
I dream about my past,
I try to handle my sorrow,
I hope I won’t be destroyed.
I am big and round.

TS 6th Class English Guide Unit 1B I Want Peace

Now answer the following questions :

Question 1.
What does the earth wish to have and why?
Answer :
The earth wishes to have peace on her to stop the killings and the sorrows of the people who live on her.

Question 2.
“I am big and round.” What do you think is suggested by the repetition of this line?
Answer :
“I am big and round.” is the sentence used by the poet to suggest the condition of the earth. It is repeated in the poem. It is meant that the earth is big and round enough with so many problems and sorrows facing by the people who live on her. ‘Big’ suggests that it is full of the problems.

Question 3.
Was the earth happy or sorrowful ? Give reasons for your opinion.
Answer :
The earth was sorrowful. The earth feels that there is no peace on her and she hears the killings of the innocent people. She sees the crying of the people for saving their lives. She wants to help the poor. All these things tell that the earth is in a sorrowful mood.

TS 6th Class English Guide Unit 1B I Want Peace

Project :

Collect / draw 5 pictures of animals / birds and fill the table given below.

TS 6th Class English Guide Unit 1B I Want Peace 2

Pick your favorite animal/bird from the above list and write about it. Give reasons why you like it present it to the class.
Answer :
TS 6th Class English Guide Unit 1B I Want Peace 3

I like the tiger most as it is a symbol of royalty. It is a carnivorous animal and is the largest living member of the cat family. It lives in Asia, mainly Bhutan, China and Siberia.
There are 49 Tiger forest reservoirs in India.
Tigers are becoming very rare because people hunt them for their skin and destroy the habitats they live in. They are becoming extinct and there is an alarming situation for all of us to wake up and save tigers.

TS 6th Class English Guide Unit 1B I Want Peace

I Want Peace Summary in English

The earth is big and round. The earth feels sorrow that there is no peace and the incidence of killing is very high on her. Here the people are crying to save their lives and the earth wants to help the poor. The earth is anxious about the future of the people and it prays God for his help. The earth understands the problems of the people and it wants the God’s grace for all of them. The earth tries to solve the problem and wishes it won’t be destroyed.

Meanings :

grace (n) : the kindness that God shows towards the human race
dream (v) : a series of events that happen in our mind while we are asleep
handle (v) : to deal with a situation
destroy (v): to damage something

TS 6th Class English Guide Unit 1A Peace and Harmony

Telangana SCERT TS 6th Class English Study Material Pdf Unit 1A Peace and Harmony Textbook Questions and Answers.

TS 6th Class English Guide Unit 1A Peace and Harmony

Look at the picture and discuss the questions that follow:

TS 6th Class English Guide Unit 1A Peace and Harmony 1

Question 1.
What does the picture tell you about the animals ?
Answer :
The animals and birds are discussing seriously something. They might have a problem and to solve the problem, they hold a meeting.

Question 2.
Are they happy or unhappy? How do you know?
Answer :
They are unhappy. I know it from their serious way of holding the meeting.

TS 6th Class English Guide Unit 1A Peace and Harmony

Question 3.
Name the animals you see in the picture.
Answer :
I see the elephant, the giraffe, the tiger, the deer, the rabbit and the monkey in the picture.

Oral Discourse :

Select any one animal from the above picture and describe it.

The elephant

The elephant is the largest animal. It is also strong. Its eyes are very small, but it can see long distances. It can also pick up very small things with its trunk. It has very large ears. Some elephants have tusks. They are called tuskers. Usually, elephants live in forests along with their herds. They are vegetariAnswer : Sometimes they get mad. They destroy everything on their way. They can take revenge also.

Elephants can be tamed. They are used in temple processions and festivals. They are also used to carry heavy logs of wood from the forest. They run very fast. They are trained in circus to do feats.

I. Answer the following questions :

Question 1.
How were the animals before the jackal joined them? Pick out the words which describe their mood.
Answer :
Before the jackal joined them, the animals lived together happily. They roamed about the jungle and played together. They often held a musical evening in a large open field and invited the moon, her friends and the stars to be the guests of honour. The words such as ‘happily’, ‘roamed’, ‘played’, ‘proud’, ‘admired’, ‘enjoyed’, ‘danced’, ‘sang’ describe their mood.

TS 6th Class English Guide Unit 1A Peace and Harmony

Question 2.
What happened to the animals after the jackal came to the forest ?
Answer :
After jackal came to the forest, the animals began suspecting one another. They started to put up boundaries and build fences around their properties. They moved about individually or with their own group.

Question 3.
What was the jackal’s plan? How did it succeed?
Answer :
The jackal’s plan was to bring differences among the creatures. He wanted to disturb the peaceful atmosphere in the forest. It succeeded in his attempts to create the differences among them. Listening to the cunning jackal’s words they began suspecting one another. They put up boundaries and built fences around their properties. Finally there was a chance for the jackal to hunt the animals and the birds and kill them one by one.

Question 4.
What did the animals do to check the jackal’s evil design?
Answer :
The animals held an emergency meeting and decided to approach the Sadhu to kill the cunning jackal. They met the Sadhu and requested him to kill the jackal. He replied that he would not kill the jackal but would help them in a different way.

Question 5.
How was the moon disguised? Why did he choose that form?
Answer :
The moon was disguised as a Sadhu. He wanted to make the creatures in the forest to live in harmony with this deed. He chose that form so as to save all the creatures in the forest from the cunning jackal. He wanted to teach the creatures to live in peace and harmony.

Question 6.
Why did the Sadhu ask the jackal to take him for his meal ? Would you do the same if you were in his place?
Answer :
The jackal came to the house of the hens to take a few of them for his supper. On seeing this, the Sadhu requested the jackal to take himself for his meal and leave the hens. It shows the greatness of sacrificing for others. Yes, I would do the same if I were in his place.

TS 6th Class English Guide Unit 1A Peace and Harmony

Question 7.
What did the Sadhu teach the animals in the forest?
Answer :
The Sadhu taught the animals to live in peace and harmony. He wanted them to break down the fences which they had built.

Question 8.
Do you think the Sadhu sacrificed his life for the sake of animals in the forest? Give reasons for your answer.
Answer :
Yes, I think the Sadhu sacrificed his life for the sake of animals in the forest. He entered the forest for that purpose only. He taught them to live in peace and harmony. When the cunning jackal tried to take the hens as his meal, the Sadhu offered himself as food to the jackal. Though the Sadhu knew that he was going to die, he sacrificed his life for the sake of animals.

II. Read the following paragraph and analyse it in terms of cause and effect/ consequence. One has been done for you.

The animals in the forest were friendly, so the jackal could not eat any animal. The jackal then thought of a plan and implemented it. The animals started suspecting each other as the jackal’s plan worked. The life in the forest became dull because there were no musical evenings. All the animals requested the sadhu for help, so he talked to the animals and the birds and solved their problem. However, the sadhu didn’t want to kill the jackal, but preferred to teach a lesson to it.

TS 6th Class English Guide Unit 1A Peace and Harmony 2
Answer :

CauseConsequence
1. All the animals in the forest were friendly.The jackal couldn’t eat any animal.
2. The jackal’s plan worked.The animals started suspecting each other.
3. There were no musical evenings.The life in the forest became dull.
4. All the animals requested sadhu for help.He talked to the animals and the birds and solved their problem.

TS 6th Class English Guide Unit 1A Peace and Harmony

III. Rearrange the following sentences in the order of their occurrence in the story. Then rewrite them into a meaningful coherent paragraph using appropriate linkers.
1. The animals and the birds requested the moon to help them.
2. The moon came in the form of a Sadhu.
3. The jackal created an unfriendly atmosphere.
4. The peace and harmony in the forest was disturbed.
5. The jackal entered the forest.
6. The peace and harmony in the forest was restored.
7. The jackal did not like this, so it killed the sadhu.
8. The sadhu talked to everyone in the forest and tried to restore peace.
9. They often held musical evenings.
10. The animals and the birds began suspecting each other.
11. The animals and the birds were living happily together.
12. The stomach of the jackal bloated and then burst into pieces.
The correct order of the above sentences : 11,9,5,3,10,4,2,1,8,7,12,6
Answer :
Once there lived the animals and the birds happily together. Every month they had a musical evening. One day the jackal entered the forest. The jackal created an unfriendly atmosphere. So, the animals and the birds began suspecting each

other. Thus the peace and harmony in the forest was disturbed. Seeing all this, the moon came in the form of a Sadhu. As soon as they saw the Sadhu, they requested him to help them. Listening to them, the Sadhu talked to everyone in the forest and tried to restore peace. But the jackal did not like this, and it killed the Sadhu. After taking the Sadhu as his meal, the jackal’s stomach bloated and then burst into pieces. As the Sadhu wished, the peace and harmony in the forest was restored.

Vocabulary :

I. Read the following sentences from the story :

They heard a soft voice.
He was dirty and dangerous.
It was easy for the jackal to hunt the smaller animals and birds.
The jackal told the elephant, “You are the biggest animal in the forest.”
He said to the peacock. “You are the loveliest bird in the forest.”
The underlined words are called Adjectives.
1. They come either before or after a noun (a tall boy)
2. They take intensifiers like very, quite (a very tall boy)
They have degrees of comparison. Positive, Comparative and Superlative.

Here are the three forms of a few adjectives.

TS 6th Class English Guide Unit 1A Peace and Harmony 3

TS 6th Class English Guide Unit 1A Peace and Harmony 12

Interchange of the Degrees of Comparison.

1. Varshini is as tall as Harshini. (Positive)
Harshini is not taller than Varshini. (Comparative)
2. The donkey is bigger than the deer. (Comparative)
The deer is not so big as the donkey. (Positive)
3. Harini is older than Gopi. (Comparative)
Gopi is not as old as Harini. (Positive)
4. Lalitha is not so heavy as Babloo. (Positive)
Babloo is heavier than Lalitha. (Comparative)
5. Tenali is cooler than Jaipur. (Comparative)
Jaipur is not so cool as Tenali. (Positive)

When the comparison is between two, there is no Superlative Degree. When the comparison is among more than two, we can write in all the three degrees of comparison.
The definite article ‘the’ is used before the adjective in the Superlative Degree.
Read the following paragraph carefully, underline all adjectives and write the other degrees of comparison for them.
Answer :
Yesterday we went for a picnic to Nehru Zoological Park which is one of the biggest zoos in the country. The climate was cool and pleasant. We saw many animals, birds, reptiles etc. We also saw a very large elephant. After that, we saw a tall giraffe with a long neck. There were also some small birds which sang sweet songs. There was a beautiful peacock which danced majestically. We went very close to the brown cobra. We felt very happy. We reached home late in the evening.

TS 6th Class English Guide Unit 1A Peace and Harmony 13

II. Read the following words. Each pair has one word the other word is opposite in meaning :

TS 6th Class English Guide Unit 1A Peace and Harmony 4

Fill in the blanks in the sentences given below with the word opposite in meaning to the one in bold letters.

1. The elephant’s eyes are small but its body is ______.
2. The animals were happy in the beginning. After the jackal entered the forest, they became ______.
3. The giraffe’s neck is very long but its tail is ______.
4. Mangoes are sweet but lemons are ______.
5. The coconut is a tall tree but the guava is a ______ tree.
Answer :
1. big
2. unhappy
3. short
4. sour
5. short

TS 6th Class English Guide Unit 1A Peace and Harmony

III. The following pairs of words are similar in meaning. Pick out such pairs from the story you have just read.

TS 6th Class English Guide Unit 1A Peace and Harmony 5
Answer :
1. animals – creatures
2. small – tiny
3. talk – speak

TS 6th Class English Guide Unit 1A Peace and Harmony

Grammar :

I. Read the following sentences.

The elephant is bigger than the donkey.
The donkey is not so big as the elephant.
In the above sentences the elephant and the donkey are compared with respect to their size. The words “big” and “bigger” are adjectives that are used to compare their size.
Here is some information about a group of pupils in 6th class.

TS 6th Class English Guide Unit 1A Peace and Harmony 6

Now make 20 sentences comparing the age, height and weight of the students as shown in the examples given below :

Harini is as tall as Rajesh.
Harini is not so heavy as Raghava.
Harini is older than Gopi.
Siddu is the heaviest boy in the group.

1. Akhila is older than Bhaskar Raju.
2. Gopi is younger than Manoj.
3. Rajesh is not so old as Siddu.
4. Seshagiri is not so old as Vamsi.
5. Raghava is older than Gopi.
6. Bhaskar Raju is younger than Akhila.
7. Siddu is the tallest boy in the group.
8. Vamsi is not so tall as Manoj.
9. Raghava is taller than Bhaskar Raju.
10. Gopi is not so tall as Raghava.
11. Bhaskar Raju is as old as Gopi.
12. Siddu is as old as Vamsi.
13. Manoj is as tall as Seshagiri.
14. Rajesh is as tall as Manoj.
15. Bhaskar Raju is the lightest boy in the class.
16. Gopi is not so heavy as Manoj.
17. Raghava is as heavy as Rajesh.
18. Rajesh is not so heavy as Seshagiri.
19. Seshagiri is heavier than Akhila.
20. Raghava is heavier than Vamsi.

TS 6th Class English Guide Unit 1A Peace and Harmony

II. Look at the following sentences paying special attention to the underlined words.

The tiny ants, which were always busy tickled the dark buffaloes.
They often held musical evenings in a large open field.
The underlined words are adverbs, which tell us how often something the happens.
They are called adverbs of frequency.

Read the following passage and underline the adverbs of frequency.

Sekhar is a busy taxi driver. He never finds time to read books. He often takes food outside. He reaches home early in the evening. But he seldom goes to bed early. So, his children always ask him for a bed time story. Sometimes he takes his children to picnics and buys toys and gifts.
Adverbs of Frequency
An adverb is a word which modifies a verb, an adjective or another adverb. It adds something to the meaning of a verb, adjective or an adverb.
‘Adverbs of frequency’ are the words that tell us how often something happens.
‘Always’, sometimes, never, often, seldom are some of the adverbs of frequency.

Eg:
1) The tiny ants, which are always busy, tickled the dark buffaloes.
2) They often held a musical evening in a large open field.
3) He always drinks milk.
4) I sometimes sip tea.
5) We seldom watch cricket matches.

Look at the conversation between two friends, Murthy and Krishna. Fill in theblanks with always, never, often, seldom, sometimes.

Murthy : Do you like reading books?
Krishna : Yes, very much. I __________ read books, a book a day, How about you?
(Answer : always)
Murthy : I like books too. But I _________ read books, not always.
(Answer : sometimes)
Krishna : How ___________ do you read books?
(Answer : often)
Murthy : Once or twice in a month.
Krishna : Oh, that’s fine. Have you read Asura by Anand Neelakantan?
Murthy : No. I haven’t. I’ll read it next month. How ____________ does your brother read?
(Answer : often)
Krishna : He ___________ reads books. The last he read a book was 10 years ago.
(Answer : seldom)

TS 6th Class English Guide Unit 1A Peace and Harmony

Writing :

I. Here is an invitation card from the animals about a musical programme in the forest. Read it carefully.

TS 6th Class English Guide Unit 1A Peace and Harmony 7

Suppose you are planning to conduct a cultural programme in your school on the occasion of your School Anniversary. Design an invitation card. You may use the model given above.
Answer :
TS 6th Class English Guide Unit 1A Peace and Harmony 14

TS 6th Class English Guide Unit 1A Peace and Harmony

II & III. Imagine that you are the Cultural Secretary of your school. You have been asked to compere the progra mme on the School Anniversary Day. Prepare your script and then give a mock performance before your group.
Answer :
Dear Mr. I.V.Raju, Smt. K. Lakshmi and dear friends,
I welcome you all to our School Anniversary organised at BVS Public School, Vijayawada. As you are aware, we have with us Sri I.V.Raju and Smt. K.Lakshmi as our guests of honour. On behalf of our school and on my behalf, I thank them for sparing some of their valuable time for us. I would request Mr. I.V.Raju, the Joint Commissioner of our city to say a few words on this occasion.
(Mr. I.V.Raju says a few words)
Mr. I.V. Raju Sir,
Thank you sir for saying very kind words for us. Now I would like to request Smt. K.Lakshmi, the ACTO of our city to say a few words on this occasion.
(Mrs. K.Lakshmi says a few words)
Thank you madam for your delivering the speech for us.
Now I will begin our cultural programme with a song called “Punya Bhoomi Na Desam ……”. This will be presented by Vineetha & Suneetha of VI Class B/S.
(A song by Vineetha and Suneetha of VI B / S)
I am sure you have liked the item. The next item in our programme is a dance item called “Pitbull on the floor……” by VI Class C/S Girls.

(Dance item by VI Class C/S Girls)
How was the dance item? Did you like it? Now you will see the play called “Kanyasulkam.” This will be presented by VI Class A/S Girls and Boys.
(A play by VI A / S Girls and Boys)
I am sure you have enjoyed the show. Now you will see a skit called “Laugh if you can……” It will be presented VI B/S Boys.
(Skit by VI B / S Boys)
Now, I invite Master K. Raja Sekhar, CPL of VI B/S to propose a vote of thanks.
(Vote of thanks by K. Raja Sekhar, CPL, VI B/S)
Finally, I invite you all to have a few snacks with a mouthful of juice before you leave.

TS 6th Class English Guide Unit 1A Peace and Harmony

Study Skills :

Read the following Table of Contents of a book :

TS 6th Class English Guide Unit 1A Peace and Harmony 10

Fill in the blanks with the information in the Table of Contents given above.

1. “Fossils” is found on page
Answer :
32

2. If you want to know about crocodiles you must turn to page
Answer :
38

3. “Mosquito” is found in Unit
Answer :
8

TS 6th Class English Guide Unit 1A Peace and Harmony

4. The Unit 7 : “Kites” is found from page
Answer :
47 to 51

5. Unit 4 talks about
Answer :
Vikram Sarabhai Trees (Poem)

Listening and Speaking :

TS 6th Class English Guide Unit 1A Peace and Harmony 11

The Friendly Mongoose

Once, a farmer and his wife lived in a village with their small son. They loved him very much. “We must have a pet,” the farmer said to his wife one day. “When our son grows up, he will need a companion. This pet will be our son’s companion.” His wife liked the idea.

One evening, the farmer brought with him a tiny mongoose. “It’s a baby mongoose,” said his wife, “but will soon be fully grown. He will be a friend to our son.”

Both the baby and the mongoose grew. In five or six months, the mongoose had grown to its full size – a lovely animal with two shining black eyes and a bushy tail. The farmer’s son was still a baby in the cradle, sleeping and crying alternately.

One day, the farmer’s wife wanted to go to the market. She fed the baby and rocked him to sleep in his little cradle. Picking up the basket, she said to her husband, “I’m off to the bazar. The baby is sleeping. Keep an eye on him. Frankly, I don’t like to leave the child alone with the mongoose.”

“You needn’t be afraid,” said the farmer. “The mongoose is a friendly animal. It’s as sweet as our baby and they are the best of friends, you know.”

The wife went away, and the farmer, having nothing to do in the house, decided to go out and take a look at his fields not far away. He ran into some friends on the way back and didn’t return for quite some time.

The farmer’s wife finished her shopping and came back home with a basket full of groceries. She saw the mongoose sitting outside as if waiting for her. On seeing her, he ran to welcome her, as was customary. The farmer’s wife took one look at the mongoose and screamed. “Blood!” she cried. The face and paws of the mongoose were smeared with blood.

“You wicked animal! You have killed my baby,” she screamed hysterically. She was blind with rage and with all her strength brought down the heavy basket full of groceries on the blood-smeared mongoose and ran inside to the child’s cradle.

The baby was fast asleep. But on the floor lay a black snake torn and bleeding. In a flash she realised what had happened. She ran out looking for the mongoose. “Oh! You saved my child! You killed the snake! What have I done?” she cried touching the mongoose, who lay dead and still, unaware of her sobbing. The farmer’s wife, who had acted hastily and rashly, stared long at the dead mongoose. Then she heard the baby crying. Wiping her tears, she went in to feed him.

TS 6th Class English Guide Unit 1A Peace and Harmony

Listen to the story “The Friendly Mongoose” and answer the following questions.

Question 1.
Which character do you like the most in the story and why?
Answer :
I like the character of mongoose the most in the story because it sacrificed its life in saving the child with its faithfulness.

Question 2.
Do you think the mongoose would have bitten the child ?
Answer :
No, the mongoose didn’t bite the child. It bit the snake and saved the child.

Question 3.
Is the woman right in killing the mongoose? What would you have done?
Answer :
No, the woman is not right in killing the mongoose. I woudn’t be hasty in taking the decision. I would have gone into the room where the child slept in the cradle. Then I would have decided what to do.

Question 4.
Why did the husband go to fields leaving the child to the mongoose ?
Answer :
The husband went to fields as he had nothing to do in the house. He ran into some friends on the way. Moreover he had very much confidence in his pet mongoose; So he left the child to the care of mongoose.

Question 5.
Why do you think the mongoose killed the snake?
Answer :
I think the mongoose killed the snake when it tried to bite the little child. Otherwise the child must die with the poison of the snake when it bit him.

TS 6th Class English Guide Unit 1A Peace and Harmony

Question 6.
Suggest one word that describes the character of the mongoose in the story.
Answer :
Loyalty. (Faithfulness)

Question 7.
Is it a good idea to have a mongoose as a pet ? Give reasons.
Answer :
I think it is a good idea to have a mongoose as a pet. We are more pleasant with its presence before us. It defends us from the danger of crawling creatures. It protects us.

Peace and Harmony Summary in English

Long long ago there lived many animals, birds, snakes and insects in a very thick forest. They lived in peace and harmony. They played together happily. They often held a musical evening in a large open field and invited the moon, his friends and the stars to watch their special skills. At that time a cunning jackal entered the forest. He wanted to disturb the peace and kill the animals one by one for his prey. He poisoned their minds and roused enmity among them. They started suspecting one another, put up boundaries and built fences around their properties. They moved about individually.

The cunning jackal, who was waiting for this opportunity, started killing the animals one by one and ate them up. There were no more musical evenings. Then the moon decided to solve the problem and came down to the earth in the form of a Sadhu. When the creatures in the forest saw the Sadhu, they approached him and requested him to solve the problem. The Sadhu met all the creatures personally and advised them to move friendly with other ones. Seeing this the cunning jackal felt very anger with the Sadhu and wanted to kill him.

One day when the jackal tried to attack the hens, the Sadhu offered himself as jackal’s meal and requested the jackal to leave the hens. The jackal, immediately dragged the Sadhu into the forest, tore him into pieces and ate him up. But after sometime, the jackal’s stomach bloated and burst into pieces. At that time, they heard a soft voice wishing that all the creatures should live in peace and harmony. When the jackal’s life was ended in that way, all the animals started again to roam, sing and sleep in peace. The fences were broken down and the moon and the stars were invited for their cultural evenings.

TS 6th Class English Guide Unit 1A Peace and Harmony

Glossary :

1. together (adv) = with each other
Eg: Ramu and Somu go to school together.
2. roam = to travel around an area without any aim
Eg : He roams about the area where different varieties of birds live.
3. admire = to respect somebody for what they have done
Eg : We have to admire the efforts of Vinoba in finding the new ‘Land Gifts Mission’.
4. melodious (adj) = pleasant to listen
Eg: Miss. Chitra sings the songs very melodiously.
5. gymnastics (n) = physical exercises
Eg: We must do gymnastics regularly to make our bodies fit.
6. praise (v) = to express approval for somebody
Eg : The teacher praised Sankar for he had saved a little child’s life.
7. approach (v) = meet
Eg : They approached Gandhiji when he came to their city.
8. whisper (v) = to speak very quietly
Eg : Latha whispered a few words in Vanaja’s ears.
9. suspect = doubt
Eg : I suspected him to be a criminal.
10. fence (n) = A boundary put between two areas of land
Eg : We have a fence around our house.
11. property (n) = objects or things owned by somebody
Eg : We have the most of our properties in our native village.
12. narrate = to tell a story
Eg: Our teacher narrated the story for half an hour.
13. woe (n) = extreme sadness
Eg : Please, listen to my owes.
14. sob (v) = to cry noisily, taking sudden, sharp breaths
Eg : I heard her sobbing loudly.
15. blame = to say that somebody is responsible for something bad.
Eg : You must not blame anybody without knowing the truth.
16. accompany = to go along with somebody
Eg : My father accompanied us on the trip.
17. peace (n) = the situation where there is no violence
Eg : We want peace everywhere.
18. harmony (n) = a state of living together peacefully
Eg : Harmony is the thing we wish for a peaceful life.
19. opportunity (n) = chance
Eg : He is looking for an opportunity to take revenge on his enemy.
20. unanimously (adv) = unitedly
Eg : They selected him as the secretary unanimously.
21. supper (n) = the last meal of the day
Eg : We had a wonderful supper last night.
22. drag (v) = to pull something along with difficulty
Eg : He dragged her from her seat.
23. bloat (v) = become bigger and bigger (swell unpleasantly)

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material Lesson 3(b) Chemical Kinetics Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material Lesson 3(b) Chemical Kinetics

Very Short Answer Questions

Question 1.
Define the speed or rate of a reaction.
Answer:
The change in the concentration of a reactant (or) product in unit time is called the speed or rate of a reaction.
(or)
The decrease in the concentration of a reactant (or) increases in the concentration of product per unit time.

Question 2.
Assuming that the volume of the system is constant, derive the expression for the average rate of the system R → P in terms of R and R [time = ‘t’ sec] [R = reactant, P = product].
Answer:
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 1

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 3.
What are the units of rate of reaction?
Answer:
Units of rate of reaction – moles/Lit × sec – moles. Lit-1 . sec-1

Question 4.
Draw the graphs that relate the concentrations .(C) of the reactants and the reaction times (t) and the concentrations of the products (C) and the reaction times (t) in chemical reactions.
Answer:
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 2

Question 5.
Write the equation for the rate of the reaction.
\(5 \mathrm{Br}_{(\mathrm{aq})}^{-}+\mathrm{BrO}_{3(\mathrm{aq})}^{-}+6 \mathrm{H}_{(\mathrm{aq})}^{+} \longrightarrow 3 \mathrm{Br}_{2(\mathrm{aq})}+3 \mathrm{H}_2 \mathrm{O}_{(\text {) }}\)
Answer:
Given reaction is
\(5 \mathrm{Br}_{(\mathrm{aq})}^{-}+\mathrm{BrO}_{3(\mathrm{aq})}^{-}+6 \mathrm{H}_{(\mathrm{aq})}^{+} \longrightarrow 3 \mathrm{Br}_{2(\mathrm{aq})}+3 \mathrm{H}_2 \mathrm{O}_{(\text {) }}\)
Rate of reaction = \(=-\frac{1}{5} \frac{\Delta\left[\mathrm{Br}^{-}\right]}{\Delta \mathrm{t}}=\frac{-\Delta\left[\mathrm{BrO}_3^{-}\right]}{\Delta \mathrm{t}}=\frac{-1}{6} \frac{\Delta\left[\mathrm{H}^{+}\right]}{\Delta \mathrm{t}}=\frac{-1}{3} \frac{\Delta\left[\mathrm{Br}_2\right]}{\Delta \mathrm{t}}=\frac{1}{3} \frac{\Delta\left[\mathrm{H}_2 \mathrm{O}\right]}{\Delta \mathrm{t}}\)

Question 6.
What is rate law ? Illustrate with an example.
Answer:
The equation that describes mathematically the dependence of the rate of a reaction on the concentration terms of the reactions is known as the rate equation (or) rate law.
Eg : 2A + 3B → 3C
Rate, of the given reaction ∝ [A]2 [B]3

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 7.
Mention a reaction for which the exponents of concentration terms are not the same as their stoichiometric coefficients in the rate equation.
Answer:
The following are the reactions for which the exponents of concentration terms are not the same as their stoichiometric coefficients in the rate equation.
CHCl3 + Cl2 → CCl4 + HCl
rate = k[CHCl3] [Cl2]1/2
CH3COOC2H5 + H2O → CH3COOH + C2H5OH
rate = k[CHCOOC2H5] [H2O]1/2

Question 8.
Define Order of a reaction. Illustrate your answer with an example. [T.S. Mar. 15]
Answer:
Order of a reaction : The sum of the powers of the concentration terms of the reactants present in the rate equation is called order of a reaction.
Order of a reaction can be 0, 1, 2, 3, and even a fraction
Eg. : 1) N2O5 → N2O4 + \(\frac{1}{2}\) O2
rate ∝ [N2O5]
∴ It is a first order reaction.

2) 2N2O → 2N2 + O2
rate ∝ [N2O]2
∴ It is 2nd order reaction.

Question 9.
What are elementary reactions ?
Answer:
The reactions taking place in one step are called elementary reactions.

Question 10.
What are complex reactions ? Name one Complex reaction.
Answer:
A sequence of elementary reactions, reactants give the products, the reactions are called complex reactions.
Eg: Oxidation of Ethane to CO2 and H2O passes through a series of intermediate steps in which alcohol, aldehyde and acid are formed.

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 11.
Give the units of rate constants for Zero, first order and second order reactions.
Answer:
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 3

Question 12.
Define molecularity of a reaction. Illustrate with an example.
Answer:
The number of reacting species (atoms, ions or molecules) taking parts in an elementary reaction, which must colloid simultaneously to bring about a chemical reaction is called molecularity of a reaction.
NH4NO2 → N2 + 2H2O (Unimolecular)
2HI → H2 + I2 (Bimolecular)
2NO + O2 → 2NO2 (Trimolecular)

Question 13.
What is rate determining step in a complex reaction ?
Answer:
The overall rate of a reaction is controlled by the slowest step in a reaction is called the rate determining step in a complex reaction.

Question 14.
Give the mechanism for the decomposition reaction of H2O2 in alkaline medium catalysed by I ions.
Answer:
Chemical equation of decomposition of H2O2 in alkaline medium is
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 4
Mechanism:

  • It is a first order reaction w.r.t. both H2O2, I
  • This reaction takes place is two steps
    1. H2O2 + I → H2O + IO
    2. H2O2 + IO → H2O + I + O2

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 15.
Write the equation relating [R], [R]0 and reaction time ‘t’ for a zero order reaction. [R] = concentration of reactant at time ‘t’ and [R]0 = initial concentration of reactant.
Answer:
Zero order reaction Rate constant k = \(\frac{[R]_0-[R]}{t}\)
Eg : Decomposition of NH3
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 6
Rate = k [NH3]0 = k
∴ \(\frac{\Delta \mathrm{x}}{\Delta \mathrm{t}}\) = k ⇒ ∆x = ∆t × k

Question 16.
Draw the graph that relates the concentration ‘R, of the reactant and ‘t’ the reaction time for a zero Order reaction.
Answer:
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 5

Question 17.
Give two examples for zero Order reactions. [A.P. Mar. 19]
Answer:
Examples for zero order reactions

  1. AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 6
  2. AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 7

Question 18.
Write the Integrated equation for a first order reaction in terms of [R], [R]0 and ‘t’.
Answer:
[Rl = Concentration of reaction after time ‘t
[R]0 = Initial concentrations of reactant
∴ k = \(\frac{2.303}{t} \log \frac{[R]_0}{[R]}\)
This is the integrated Equation for a first order reaction.

Question 19.
Give two examples for gaseous first order reactions. [Mar. 14]
Answer:
The following are the examples for gaseous first order reactions
N2O5(g) → N2O4(g)g + \(\frac{1}{2}\) O2(g)
SO2Cl2(g) → SO2(g) + Cl2(g)

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 20.
For the reaction A(g) → B(g) + C(g), write the integrated rate equation in terms of total pressure ‘P and the partial pressures pApBpC.
Answer:
Given
A(g) → B(g) + C(g)
p = pA + pB + pC
k = (\(\frac{2.303}{t}\)) log (\(\frac{\mathrm{p}_0}{\mathrm{p}_{\mathrm{A}}}\))
k = (\(\frac{2.303}{t}\)) log (\(\frac{\mathrm{p}_0}{2 \mathrm{p}_{\mathrm{i}}-\mathrm{p}_{\mathrm{t}}}\))
p0 = initial pressure
Pi = Total pressure
pA, pB, pC are partial pressures.

Question 21.
What is half-life of a reaction ? Illustrate your answer with an example. .
Answer:
The time required for the initial concentration of the reactants to become half of it’s value during the progress of the reaction is called half life (t1/2) of reaction.
Eg : The radio active of C-14 is exponential with a half life of 5730 years.

Question 22.
Write the equation relating the half-life (t1/2) of a reaction and the rate constant ‘k’ for first order reaction.
Answer:
Half life of first order reaction (t1/2) = \(\frac{0.693}{k}\)
k = rate constant

Question 23.
Write the equation useful to calculate half-life (t1/2) values for zero and first order reactions.
Answer:
Half life (t1/2) of zero order reaction
At half life (t1/2) [R] = \(\frac{[\mathrm{R}]_0}{2}\)
∴ t1/2 = \(\frac{[\mathrm{R}]_0 / 2}{\mathrm{k}}\)
∴ t1/2 = \(\frac{\left[\mathrm{R}_0\right]}{2 \mathrm{k}}\)
Half life of a first order reaction
t1/2 = \(\frac{0.693}{\mathrm{k}}\)
k = rate constant

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 24.
What are pseudo first order reactions ? Give one example.
Answer:
First order reactions whose molecularity is more than one are called pseudo first order reactions.
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 8
Order = 1
molecularity = 2

Question 25.
Write the Arrhenius equation for the rate constant (k) of a reaction.
Answer:
Arrhenius equation is
k = A × e-Ea/RT
k = Rate constant
Ea = activation energy
R = gas constant
T = Temperature (K)

Question 26.
By how many times the rate constant increases for a rise of reaction temperature by 10°C ?
Answer:
For every 10°C rise of temperature rate constant of chemical reactions may be doubled (some times tripled).

Question 27.
Explain the term ‘activation energy’ of a reaction with a suitable diagram.
Answer:
Activation Energy : The energy required to for an intermediate called activated complex (C) during a chemical reaction is called activation energy.
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 9

Question 28.
Write the equation which relates the rate constants k1 and k2 at temperatures T1 and T2 of a reaction.
Answer:
\(\log \left(\frac{\mathrm{k}_2}{\mathrm{k}_1}\right)=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left[\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right]\)
Ea = activation energy; R = Universal gas constant.

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 29.
What is collision frequency (Z) of a reaction ? How is rate related to it for the reaction A + B → Products.
Answer:
The number of collisions per second per unit volume of the reaction mixture is called collision frequency (Z). Fora bimolecular elementary reactions.
A + B → products; Rate = ZAB.e – Ea/RT.

Question 30.
Draw the graphs between potential energy – reaction coordinates for catalysed and uncatalysed reactions.
Answer:
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 10

Question 31.
What is the effect of temperature on the rate constant ?
Answer:
Most of the chemical reactions are accelerated by increase of temperature.

For a chemical reaction with rise of temperature by 10°C the rate constant is nearly doubled k = A.e-Ea/RT

Short Answer Questions

Question 1.
Define average rate of a reaction. How is the rate of reaction expressed in terms of change in the concentration of reactants and products for the following reactions.
1) 2HI(g) → H2(g) + I2(g)
2) Hg(l) + Cl2(g) → HgCl2(g)
3) 5 Br(aq) + \(\mathrm{BrO}_{3(\mathrm{aq})}^{-}\) + \(6 \mathrm{H}_{(\mathrm{aq})}^{+}\) → 3 Br2(aq) + 3 H2O(l)
Answer:
Average rate of a reaction: The change in the concentration of any one of the reactants or products per unit time is called average rate of a reaction.
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 11

Question 2.
What is rate equation ? How is it obtained ? Write the rate equations for
1) 2NO(g) + O2 (g) → 2NO2 (g)
2) CHCl3 + Cl2 → CCl4 + HCl
3) CH3COOC2H5 (l) + H2O(l) → CH3COOH (aq) + C2H5OH(aq)
Answer:
The equation that describes mathematically the dependence of the rate of a reaction on the concentration terms of the reactions is known as the rate equation (or) rate law.
Eg, : 2A + 3B → 3C
Rate of the given reaction ∝ [A]2 [B]3
How to obtain: Each concentration term is raised to some power which may or may not be same as the stoichiometric coefficient of the reacting species
i) 2NO(g) + O2(g) → 2NO2(g)
\(\frac{\Delta[\mathrm{R}]}{\Delta t}\) = k [NO]2 [O2]

ii) CHCl3 + Cl2 → CCl4 + HCl
\(\frac{\Delta[\mathrm{R}]}{\Delta t}\) = k [CHCl3] [Cl2]1/2

iii) CH3COOC2H5(l) + H2O(l) → CH3COOH(aq) + C2H5OH(aq)
\(\frac{\Delta[\mathrm{R}]}{\Delta t}\) = k [CH3COOC2H5].

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 3.
Define and explain the order of a reaction. How is it obtained experimentally?
Answer:
Order of a reaction :. The sum of the powers of the concentration terms of the reactants
present in the rate equation is called order of a reaction.
Order of a reaction can be 0, 1, 2, 3, and even a fraction
Eg. : 1) N2O5 → N2O4 + \(\frac{1}{2}\) O2
rate ∝ [N2O5]
∴ It is a first order reaction.

2) 2N2O → 2N2 + O2
rate ∝ [N2O]2
∴ It is 2nd order reaction.

Order of a reaction can be determined experimentally
Half – Time (t1) method : The time required for the initial concentration (a) of the reactant to become half its value (a/2) during the progress of the reaction is called half-time (t1/2) of the reaction.
A general expression for the half life, (t1/2) is given by
t1/2 ∝ \(\frac{1}{a^{n-1}}\)
Therefore, for a given reaction two half time values (t’1/2and t”1/2) with initial concentrations a’ and a” respectively are determined experimentally and the order is established from the equation.
\(\left(\frac{t_{1 / 2}^{\prime}}{t_{1 / 2}^n}\right)=\left(\frac{a^n}{a^{\prime}}\right)^{n-1}\)
Where ‘n’ is the order of the reaction.

Question 4.
What is “moleculartiy” of a reaction ? How is it different from the ‘order’ of a reaction ? Name one bimolecular and one trimolecular gaseous reactions. [T.S. Mar. 17] [Mar. 14]
Answer:
The number of reacting species (atoms, ions or molecules) taking parts in an elementary reaction, which must colloid simultaneously to bring about a chemical reaction is called molecularity of a reaction.
NH4NO2 → N2 + 2H2O (Unimolecular)
2HI → H2 + I2 (Bimolecular)
2NO + O2 → 2NO2 ( Trimolecular)

  • Molecularity has only integer values (1, 2, 3 )
  • It has nop zero, non fraction values while order has zero, 1, 2, 3 and fractional values.
  • It is determined by reaction mechanism, order is determined experimentally.

Question 5.
Derive the integrated rate equation for a zero order reaction.
Answer:
Zero order reaction is the reaction in which rate of reaction does not depends on the concentration of reactants.
R → P
Rate = \(\frac{-\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}\) = k [R]°
Rate = \(\frac{-\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}\); d[R] = – k. dt
Integrating on both sides
[RJ = -kt + I …………….. (1)
I – Integration constant
At t = 0 → R = [R]0 initial concentration
I = [R]0
Substituting I = [R]0 in the above equation (1)
[R] = – kt + [R]0
k = \(\frac{\left[R_0\right]-[R]}{t}\)
This is the integrated rate equation for a zero order reaction.

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 6.
Derive an integrated rate equation for a first order reaction.
Answer:
In first order reactions rate depends on only one concentration term.
R → P
Rate = k [R]; -M = – k. dt
Integration on both sides
ln [R] = – kt + I
I = Integration constant
At t = 0, [R] = [R]0 ⇒ ln[R]0 = I
Substituting I = ln [R] in the above equation (1)
ln [R] = – kt + ln [R]0
ln \(\frac{[\mathrm{R}]}{\left[\mathrm{R}_0\right]}\) = -kt ………………. (2)
k = \(\frac{1}{\mathrm{t}} \ln \frac{\left[\mathrm{R}_0\right]}{[\mathrm{R}]}\)
Taking antilog on both sides of eq. (2)
R = [R]0. e-kt
This is first order rate equation.

Question 7.
Derive an integrated rate equation in terms of total pressure (P) and the partial pressures pA, pB, pC for the gaseous reaction A(g) → B(g) + C(g).
Answer:
Given first order gas phase equations
A(g) → B(g) + C(g)
Let pi be the initial pressure of A and pt the total pressure at time’t’. Integrated rate equation for such a reaction can be derived as
Total pressure pt = pA + pB + pC (pressure units)
pA, pB and pC are the partial pressures of A, B and C, respectively.
If x atm be the decrease in pressure of A at time t and one mole each of B and C is being formed, the increase in pressure of B and C will also be x atm each.
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 12
Thus the rate expression for 1st order gaseous phase reaction derived.

Question 8.
What is half-life (t1/2) of a reaction ? Derive the equations for the ‘half-life’ value of zero and first order reactions.
Answer:
The time required for the initial concentration of the reactants to become half of it’s value during the progress of the reaction is called half life (t1/2) of reaction.
Eg : The radio active of C-14 is exponential with a half life of 5730 years.
Half life of zero order reaction :
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 13

Question 9.
What is Arrhenius equation ? Derive an equation which describes the effect of rise of temperature (T) on the rate constant (k) of a reaction.
Answer:
The temperature dependence of the rate of reaction can be explained by Arrhenius equation.
k = A.e-Ea/KT
A = Arrhenius factor.
Ea = activation Energy
R = gas constant
T = Temperature. (K)
k = A.e-Ea/KT
ln k = ln A – Ea/RT
2.303 log k = 2.303 log A – Ea/RT
T1, T2 are Temperatures
k1 is rate constant at temperature T1
k2 is rate constant at temperature T2
∴ 2.303 (log k2 – log k1) = \(\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{\mathrm{~T}_2}-\frac{1}{\mathrm{~T}_1}\right)\)
\(\log \frac{\mathrm{k}_2}{\mathrm{k}_1}=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left[\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right]\)
The above equation describes the effect of temperature (T) on rate constant (k)

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 10.
Discuss the effect of catalyst on the kinetics of a chemical reaction with a suitable diagram. [T.S. Mar. 15]
Answer:
The substance which increases the rate of reaction without being consumed in the chemical reaction is called a catalyst.
In presence of catalyst the reaction proceeds in a new path which lowers the activation energy.
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 14

Question 11.
Describe the salient features of the collision theory of reaction rates of bimolecular reactions. [T.S. Mar. 18, 16; A.P. Mar. 17]
Answer:
Collision theory of reaction rate bimolecular reactions salient features.

  • The reaction molecules are assumed to be hard spheres
  • The reaction is postulated to occur when molecules collide with each other.
  • The number of collisions per second per unit volume of the reaction mixture is known as collision frequency (Z).
  • For a bimolecular elementary reaction
    A + B → products
    Rate = ZAB. e-Ea/RT ; ZAB = collision frequency.
  • All collisions do not lead to product formation.
  • The collisions with sufficient kinetic energy (Threshold energy) are responsible for product formation. These are called as effective collisions.
  • To account for effective collisions a factor p called to probability factor or steric factor is introduced.
    Rate = P ZAB. e-Ea/RT

Question 12.
Explain the terms
a) Activation energy (Ea)
b) Collision frequency (Z) :
c) Probability factor (P) with respect to Arrhenius equation.
Answer:
a) Activation Energy: The energy required to for an intermediate called activated complex (C) during a chemical reaction is called activation energy.
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 15
Diagram showing plot of potential energy vs reaction coordinate

b) Collision frequency: The number of collisions per second per unit volume of the reaction mixture is called collision frequency (Z). For a bimolecular elementary reactions
A + B → products
Rate = ZAB. e-Ea/RT

c) Probability factor (P) with respect to Arrhenius equation : To account for effective collisions a factor p called to probability factor or steric factor is introduced.
Rate = P ZAB. e-Ea/RT

Long Answer Questions

Question 1.
Explain the following terms with suitable examples.
a) Average rate of a reaction
b) Slow and fast reactions
c) Order of a reaction
d) Molecularity of a reaction
e) Activation energy of reaction.
Answer:
a) Average rate of a reaction : The change in the concentration of any one of the reactants or products per unit time is called average rate of a reaction.
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 16

b)

  1. Fast Reactions : In case pf ionic compounds reactions takes place fastly i.e., rate is high
    NaCl + AgNO3 → NaNO3 + AgCl
  2. Slow reactions : In case of covalent compounds reactions takes place slowly i.e., rate is low.
    AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 17

c) Order of reaction : The number of reacting species (atoms, ions or molecules) taking parts in an elementary reaction, which must colloid simultaneously to bring about a chemical reaction is called molecularity of a reaction.
NH4NO2 → N2 + 2H2O (Unimolecular)
2HI → H2 + I2 (Bimolecular)
2NO + O2 → 2NO2 (Trimolecular)
Molecularity has only integer values (1, 2, 3 …………..)
It has non zero, non fraction values while order has zero, 1, 2, 3 ……….. and fractional values.
It is determined by reaction mechanism, order is determined experimentally.

d) Molecularity of a reaction: Zero order reaction is the reaction in which rate of reaction does not depends on the concentration of reactants.
R → P
Rate = \(\frac{-\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}\) = k [R]°
Rate = \(\frac{-\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}\); d[R] = – k. dt
Integrating on both sides
[RJ = -kt + I …………….. (1)
I – Integration constant
At t = 0 → R = [R]0 initial concentration
I = [R]0
Substituting I = [R]0 in the above equation (1)
[R] = – kt + [R]0
k = \(\frac{\left[R_0\right]-[R]}{t}\)
This is the integrated rate equation for a zero order reaction.

e) Activation energy of reaction: The energy required to for an intermediate called activated complex (C) during a chemical reaction is called activation energy
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 18
Diagram showing plot of potential energy vs reaction co-ordinate

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 2.
Give two examples for each of zero order and first order reactions. Write the equations for the rate of a reaction in terms of concentration changes of reactants and products for the following reactions.
1) A(g) + B(g) → C(g) + D(g)
2) A(g) → B(g) + C(g) 3) A(g) + B(g) – C(g)
Answer:
Examples of zero order reaction :

  1. AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 6
  2. AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 7

Examples of first order reaction :
N2O5(g) → N2O4(g) + \(\frac{1}{2}\) O2(g)
SO2Cl2(g) → SO2(g) + Cl2(g)
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 19

Question 3.
Discuss the effect of temperature on the rate of a reaction. Derive necessary equations in this context. [T.S. Mar. 15]
Answer:
Most of the chemical reactions are accelerated by increase of temperature.
For a chemical reaction with rise of temperature by 10°C the rate constant is nearly doubled
k = A.e-Ea/RT
The temperature dependence of the rate of reaction can be explained by Arrhenius equation.
k = A.e-Ea/RT
A = Arrhenius factor.
Ea = activation Energy
T = Temperature (°K)
R = gas constant
k = A.e-Ea/RT
In k = ln A – Ea/RT
2.303 log k = 2.303 log A – Ea/RT
T1, T2 are Temperatures
k1 is rate constant at temperature T1
k2 is rate constant at temperature T2
∴ 2.303 (log k2 – log k1) = \(\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{\mathrm{~T}_2}-\frac{1}{\mathrm{~T}_1}\right)\)
\(\log \frac{\mathrm{k}_2}{\mathrm{k}_1}=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left[\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right]\)
The above equation describes the effect of temperature (T) on rate constant (k)

Question 4.
Give a detailed account of the collision theory of reaction rates of bimolecular gaseous reactions. [A.P. Mar. 17, 16]
Answer:
Collision theory of reaction rate bimolecular reactions salient features.

  • The reaction molecules are assumed to be hard spheres
  • The reaction is postulated to occur when molecules collide with each other.
  • The number of collisions per second per unit volume of the reaction mixture is known as collision frequency (Z).
  • For a bimolecular elementary reaction
    A + B → products
  • Rate = ZAB.e-Ea/RT; ZAB = collision frequency.
  • All collisions do not lead to product formation.
  • The collisions with Sufficient kinetic energy (Threshold eriergy) are responsible for product formation. These are called as effective collisions.
  • To account for effective collisions a factor p called to probability factor or steric factor is introduced.
    Rate = PZAB.e-Ea/RT
  • The proper orientation of reactant molecular lead to bond formation where as improper orientation makes them back and no product are formed.
    AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 20

In this theory activation energy and proper orientation of the molecules to gather determine the creteria for an effective collision and hence the rate of a chemical reaction.

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Problems

Numerical Data Based And Concept Oriented Questions

Question 1.
A reaction is 50% completed in 2 hours and 75% completed in 4 hours. What is the order of the reaction ? [T.S. Mar. 16]
Solution:
Given that a reaction is 50% completed in 2 hrs.
75% completed in 4 hrs.
From the data half life is independent of initial concentration so it is a first order reaction.

Question 2.
A reaction has a half – life of 10 minutes. Calculate the rate constant for the first order reaction. [T.S. Mar. 16]
Solution:
In case of fist order reaction t1/2 = \(\frac{0.693}{k}\)
∴ k = \(\frac{0.693}{t_{1 / 2}}=\frac{0.693}{10}\) = 0.0693 min-1

Question 3.
In a first order reaction, the concentration of the reactant is reduced from 0.6 mol/L to 0.2 mol/L in 5 min. Calculate the rate constant (k).
Solution:
a = 0.6mol L-1; a – x = 0.2 mol L-1; t = 5 min.
Since it is a first order reaction.
k = \(\frac{2.303}{t} \log _{10} \frac{a}{(a-x)}\)
k = \(\frac{2.303}{t}\) log \(\frac{0.6}{0.2}\) = 0.2197 min-1.

Question 4.
The rate constant for a zero order reaction in A is 0.0030 mol L-1 s-1. How long it will take for the initial concentration of A to fall from 0.10 M to 0.075 M.
Solution:
In case of zero order reaction
k = \(\frac{1}{t}\) [[A0] – [A]]
[A]0 = Initial concentration
[A] = concentration after time t
k = 0.0030 mol L-1s-1
[A]0 = 0.10 M
[A] = 0.075 M
0.0030 = \(\frac{1}{t}\) [0.10 – 0.075]
t = \(\frac{0.025}{0.0030}\) = 8.33 seconds.

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 5.
A first order decomposition reaction takes 40 min. for 30% decomposition. Calculate it’s t1/2 value.
Solution:
Given t = 40 min, a = 100
a – x = 100 – 30 = 70
In case of first order reaction
k = \(\frac{2.303}{t} \log \frac{a}{(a-x)}\)
k = \(\frac{2.303}{40} \log \frac{100}{70}\) = 0.0576 [2.0000 – 1.8451]
= 0.0576 (0.1549) = 8.922 × 10-3
\(\mathrm{t}_{\frac{1}{2}}=\frac{0.693}{\mathrm{k}}=\frac{0.693}{8.922 \times 10^{-3}}\) = 77.673 min.

Question 6.
Calculate the half-life of first order réaction whose rate constant is 200 s-1.
Solution:
Half – life period for first order reaction
\(\mathrm{t}_{\frac{1}{2}}=\frac{0.693}{\mathrm{k}}\)
\(\mathrm{t}_{\frac{1}{2}}=\frac{0.693}{200 \mathrm{~s}^{-1}}\) = 0.346 × 10-2s = 3.46 × 10-3s

Question 7.
The thermal decomposition of HCOOH is a first order reaction. The rate constant is 2.4 × 10-3 s-1 at a certain temperature. Calculate how long will it take for 3/4 of initial quantity of HCOOH to decompose. ‘
Solution:
Given reaction is a first order reaction.
\(\mathrm{t}_{\frac{1}{2}}=\frac{0.693}{\mathrm{k}}=\frac{0.693}{2.4 \times 10^{-3}}\)
= 288.75 sec.
To convert into \(\frac{3}{4}\) of the original, two half lifes are required.
Time to decompose = 2 × 288.75
= 577.5 sec.
= 5.775 × 102 sec.

Question 8.
The decomposition of a compound is found to follow first order rate law. If it takes 15 minutes for 20% of original material to react, calculate the rate constant.
Solution:
Given t = 15 min., a = 100
a – x = 100 – 20 = 80
k = \(\frac{2.303}{t} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\)
k = \(\frac{2.303}{15} \log \frac{100}{80}\) = 0.1535 [2.0000 – 1.9031] = 0.1535 × 0.0969
k = 0.0148 min-1.

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 9.
In a pseudo first order hydrolysis of ester in water, the following results are obtained
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 21
Calculate the average rate of reaction between the time interval 30 to 60 s.
Solution:
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 22

Question 10.
The half-life for a first order reaction is 5 × 10-6s. What percentage of the initial reactant will react in 2 hours ?
Solution:
Given t1/2 = 5 × 10-6 sec.
k = \(\frac{0.693}{1_{1 / 2}}=\frac{0.693}{5 \times 10^{-6}}\) = 0.1386 × 106 sec-1
k = 0.1386 × 106 sec-1 = 1.386 × 105 sec-1
Here a = 100, t = 2 hrs. = 2 × 60 × 60 sec, (a – x) = ?
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 23

Question 11.
H2O2(aq) decomposes to H2O(l) and O2(g) in a first order reaction w.r.t. H2O2. The rate constant is k = 1.06 × 10-3 min-1. How long it will take 15% of the sample to decompose?
Solution:
k = 1.06 × 10-3 min-1
a = 100 .
a – x = 100 – 15 = 85
k = \(\frac{2.303}{t} \log \frac{a}{a-x}\)
1.06 x 10 = \(\frac{2.303}{t} \log \frac{100}{85}\)
t = \(\frac{2.303}{1.06 \times 10^{-3}} \log \frac{100}{85}\) = 153.4 min

Question 12.
Show that in the case of first order reaction, the time required for 99.9% completion of the reaction is 10 times that required for 50% completion. (log 2 = 0.30 10)
Solution:
t1/2 = \(\frac{0.693}{k}\)
For 99.9% → a = 100
a – x = 100 – 99.9 = 0.1
k = \(\frac{2.303}{\mathrm{k}} \log \frac{100}{0.1}\)
But k = \(\frac{0.693}{\mathrm{t}_{1 / 2}}=\frac{2.303}{0.693}\) × t1/2 log 1000 = 3.33 × t1/2 × 3 = 9.99 × t1/2
t99.9% is 10 times t1/2

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 13.
The rate constant of a reaction is doubled when the temperature is raised from 298 K to 308 K. Calculate the activation energy.
Solution:
Ea = ? k2 = 2k1; T1 = 298 K; T2 = 308 K
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 24

Question 14.
The first order rate constant ‘k’ for the reaction C2H5I(g) → C2H4 (g) + HI(g) at 600 K is 1.60 × 10-5 s-1. The energy of activation is 209 kJ/mol. Calculate ‘k’ at 700 K.
Solution:
k1 = 1.60 × 10-5 s-1
k2 = ?
T1 = 600 K
T2 = 700 K
Ea = 209 kJ mol-1 = 209000 J mol-1
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 25

Question 15.
The activation energy for the reaction 2HI(g) → H2(g) + I2(g) at 581 K is 209.5 kJ/mol. Calculate the fraction of molecules having energy equal to or greater than activation energy. [R = 8.31 JK-1 mol-1] .
Solution:
Fraction of molecules [x] having energy equal to or more than activation energy may be calculated as fallows:
x – n/N = e-Ea/RT
In x = \(\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}\) or log x = \(\frac{-\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}\)
or log x = –\(\frac{209.5 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}}{2.303 \times\left[8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right] \times 581 \mathrm{~K}}\) = -18.8323
x = Antilog [-18.8324] = Antilog \(\overline{19} .1677\) = 1.471 × 10-19
Fraction of molecules = 1.471 × 10-19

Question 16.
For the reaction R → P, the concentration of a reactant changes from 0.03M to 0.02M in 25 minutes. Calculate the average rate of reaction using units seconds.
Solution:
R→ P
0.03M to 0.02 M in 25 minutes.
0.03M – 0.02M in 25 × 60 sec.
Average rate \(\frac{0.03-0.02}{25 \times 60}=\frac{0.01}{25 \times 60}\)
= \(\frac{0.01}{1500}\) = 6.66 × 10-6 ms-1

Question 17.
In a reaction 2A → Products, the concentration of A decreases from 0.5 mol L-1 to 0.4 mol L-1 in 10 minutes. Calculate the rate during this interval.
Solution:
2A → products
0.5 – 0.4 moI lit rate ∝ [A]2
rate = \(\frac{-1}{2} \frac{\mathrm{d}}{\mathrm{dt}}\) [A] = \(\frac{1}{2}\left[\frac{(0.5-0.4)}{10}\right]\) = 0.5 × 10-2
rate = 5 × 10-3 mol Lit-1 min-1

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 18.
For a reaction, A + B → Product : the rate law is given by r = k [A]1/2 [B]2 What is the order of the reaction?
Solution:
A + B → product
r = k [A]1/2 [B]2
Rate of the reaction r = \(\frac{1}{2}\) + 2 = 2.5

Question 19.
The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased by three times, how will it affect the rate of formation of Y.
Solution:
x → y It is a second order reaction
r ∝ [x]2
If x = 1 r = 1
If x = 3 r = 32 = 9
The rate of formation of y increases by 9 times.

Question 20.
A first order reaction has a rate constant 1.15 × 10-3 s-1. How long will 5g of this reactant take to reduce to 3 g?
Solution:
Rate constant [k] = 1.15 × 10-3 s-1
Initial amount [R]0 = 5g; Final amount [R] = 3g
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 26

Question 21.
Time required to decompose SO2Cl2 to half of its initial amount Is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.
Solution:
For the first order reaction
Rate constant (k) = \(\frac{0.693}{60 \mathrm{~min}}=\frac{0.693}{(60 \times 60) \mathrm{s}}\) = 1.925 × 10-4 s-1
k = 1.925 × 10-4 s-1

Question 22.
From the rate expression for the following reactions, determine their order of reaction . and the dimensions of the rate constants.
i) 3NO(g) → N2O(g) Rate = k[NO]2
ii) H2O2 (aq) + 3I (aq) + 2H+ → 2H2O(l) + I3 Rate = k[H2O2] [I]
iii) CH3CHO(g) → CH4 (g) + CO(g) Rate = k[CH3CHO]3/2
iv) C2H5Cl(g) → C2H2 .(g) + HCl(g) Rate = k[C2H5Cl]
Solution:
i) Rate = k [NO]2
order of reaction = 2 .
units (dimensions) of rate constant
[k] = \(\frac{\text { Rate }}{[\mathrm{NO}]^2}=\frac{\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}}{\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)^2}\) = mol-1 L s-1

ii) Rate = k[H2O2] [I]
order of reaction = 1 + 1 = 2
Dimensions of k = \(\) = mol-1 L s-1

iii) Rate = k[CH3CHO]3/2
order of reaction = \(\frac{3}{2}\)
Dimensions of k = \(\frac{\text { Rate }}{\left[\mathrm{CH}_3 \mathrm{CHO}\right]^{3 / 2}}=\frac{\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}}{\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)^{3 / 2}}\) = mol-1/2 L1/2 s-1

iv) Rate = k[C2H5Cl]
order of reaction = 1
Dimensions of k = \(\frac{\text { Rate }}{\mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}}=\frac{\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}}{\mathrm{~mol} \mathrm{~L}^{-1}}\) = s-1

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 23.
For the reaction 2A + B → A, B, the rate = K[A] [B]2 with k = 2.0 × 10-6 v mol-2 L2 s-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L-1, [B] = 0.2 mol L-1, Calculate the rate of reaction after [A] is reduced to 0.06 mol L-2.
Solution:
i) Case I:
Rate = k [A] [B]2
= [2.0 × 10-6 mol2 L-2 s-1] × (0.1 mol L-1) × (0.2 mol L-1)2
= 8.0 × mol L-1 s-1

ii) Case II:
Concentration of A at a particular time = 0.06 mol L-1
Amount of A reacted = (0.1 – 0.06) = 0.04 mol L-1
Amount of B reacted = \(\frac{1}{2}\) × 0.04 mol L-1 = 0.02 mol L-1
Concentration of B at a particular time = [0.2 – 0.02] mol L-1 = 0.18 mol L-1.
Rate = k [A] [B]2
= [2.0 × 10-6 mol-2 L2 s-1] × [0.06 mol L-1] × (0.18 mol L-1)2
= 3.89 × 10-9 mol L-1 s-1.

Question 24.
The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10-4 mol-1 Ls-1.
Solution:
For a zero order reaction
rate = \(\frac{\mathrm{dx}}{\mathrm{dt}}\) = k.
dx = change in the concentration
dt = difference in time
k = rate constant
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 27
Rate of production of N2 = k = 2.5 × 10-14 mol L-1 s-1
Rate of production of H2 = 3 × 2.5 × 10-14 = 7.5 × 10-14 mol L-1 s-1

Question 25.
The rate expression for the decomposition of dimethyl ether in terms of partial pressures is given as Rate = k (pCH3 O CH3)3/2. If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constant ?
Solution:
Unit of rate = bar min-1
unit of k = \(\frac{\text { Rate }}{\left[\left(\mathrm{PCH}_3 \mathrm{OCH}_3\right]^{3 / 2}\right.}\)
= \(\frac{\text { bar } \min ^{-1}}{[\mathrm{bar}]^{3 / 2}}\) = bar-1/2 min-1

Question 26.
A reaction is second order with respect to a reactant. How is the rate of reaction is affected if the concentration of the reactant is
i) doubled
ii) reduced to half
Solution:
For a reaction A → products
Rate = k[A]2 = ka2
i) When concentration of A is doubled
i.e., [A] = 2a
Rate = k[2a]2 = 4ka2
Rate of reaction becomes 4 times

ii) When concentration of A is reduced to \(\frac{1}{2}\) i.e., [A] = \(\frac{1}{2}\) a
Rate = k [latex]\frac{a}{2}[/latex]2 = \(\frac{1}{4}\) ka2
Rate of reaction becomes \(\frac{1}{2}\) times i.e reduced to one fourth.

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 27.
A reaction is first order in A and second order in B.

  1. Write the differential rate equation
  2. How is the rate affected on increasing the concentrations of B there times?
  3.  How is the rate affected when the concentrations of both A and B are doubled ?

Solution:

  1. Rate k [A] [B]2
  2. Rate = k [A] [3B]2 = 9k [A] [B]2 Rate of reaction becomes 9 times
  3. Rate = k [2A] [2B]2 = 8k [A] [B]2 Rate of reaction becomes 8 times

Question 28.
In a reaction between A and B, the initial rate of reaction (ro) was measured for different initial concentrations of A and B as given below:
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 28
What is the order of the reaction with respect to A and B ?
Solution:
Rate law states that
Rate = k [A]x [B]y
(Rate)1 = k[0.20]x [0.30]y = 5.07 × 10-5 ………………. (i)
(Rate)2 = k[0.20]x [0.10]y = 5.07 × 10-5 …………….. (ii)
(Rate)3 = k[0.40]x [0.05]y = 1.43 × 10-4 …………… (iii)
Dividing equation (i) by equation (ii)
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 29
x log 2 = log 2.82
x = 1.4957 = 1.5
∴ order with respect to A = 1.5
order with respect to B = 0.

Question 29.
The following results have been obtained during the kinetic studies of the reaction: 2A + B → C + D
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 30
Determine the rate law and rate constant for the reaction.
Solution:
Rate law may be expressed. as
Rate = k [A]x [B]y
[Rate]1 = 6.0 × 10-3 = k (0.1)x (0.1)y …………….. (i)
[Rate]2 = 7.2 × 10-2 = k (0.3)x (0.2)x …………….. (ii)
[Rate]3 = 2.88 × 10-1 = k (0.3)x (0.4)x ……………. (iii)
[Rate]4 = 2.40 × 10-2 = k (0.4)x (0.l)x …………….. (iv)
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 31
∴ Rate law expression is given by Rate = k [A] [B]2
Rate constant k, can be determined by placing the values of A,B and rate of formation of D.
By taking the values from experimant II.
Rate k[A] [B]2
k = \(\frac{\text { Rate }}{[\mathrm{A}][\mathrm{B}]^2}=\frac{7.2 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}}{\left(0.3 \mathrm{molL}^{-1}\right)\left(0.2 \mathrm{molL}^{-1}\right)^2}\) = 6.0 mol-2 L2 min-1
∴ k = 6.0 mol-2 L2 min-1

Question 30.
The rate constant for a first order reaction is 60 s-1. How much time will it take to reduce. the initial concentration of the reactant to its 1/ 16th value?
Solution:
For 1st order reaction
t = \(\frac{2.303}{k} \log \frac{a}{(a-x)}\) …………….. (i)
Given (a – x) \(\frac{a}{16}\), k = 60 s-1
Placing the values in eq. (i)
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 32

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 31.
For a first order reaction, show that the time required for 99% completion is twice the time required for completion of 90% of reaction.
Solution:
Case I:
If a = 100; (a – x) = (100 – 99) = 1
For 99% completion of the reaction
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 33
It means that time required for 99% completion of the reaction is twice the time required to complete 90% of the reaction.

Question 32.
Here the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 34
Calculate the rate constant.
Solution:
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 35
Total pressure after time t,
i e., [pt] = [pi – p] + p + p = pi + p
Or p = pt – pi
a = pi [a – x] = pi – p on substituting the values of pi
[a – x] = pi – [pt – pi]
i. e., [a – x] = 2pi – pi
The decomposition reaction is of gaseous nature and the rate constant k can be calculate as :
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 36
Average rate constant k = \(\frac{(2.17+2.24) \times 10^{-3} \mathrm{~s}^{-1}}{2}\)
k = 2.21 × 10-3 s-1

Question 33.
The following data were obtained during the first order thermal decomposition of SO2C2 at a constant volume.
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 37
Calculate the rate of reaction when total pressure is 0.65 atm.
Solution:
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 38
Total presstce after time t
i.e., pt = pi – p + p + p = pi + p
So, a = pi
a – x = pi – (pt – pi) pi – pt + pi = 2pi – pt
Substitutions of the value of a and (a – x) gives
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 39
b. Calculation of reaction rate when total pressure is 0.65 atm.
PSO2Cl2 = 0.5 – (0.65 – 0.50) = (1 – 0.65) = 0.35atm
k = 2.23 × 10-3 s-1
Rate = k × PSO2Cl2 = (2.23 × 10-3 s-1) × (0.35 atm)
Rate = 7.3 × 10-4 atm s-1

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 34.
The rate constant for the decomposition of hydrocarbons is 2.418 × 10-5 s-1 at 546 K. If the energy of activation is 179.9 kJ/mol. What will be the value of pre-exponential factor ?
Solution:
According to Arrhenius equation,
log k = log A – \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}\)
k = 2.418 × 10-5 s-1
Ea = 179.9 KJ mol-1 or 179900 J mol-1
R = 8.314 JK-1 mol-1
T = 546K
log A = log k + \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}\)
= log (2.418 × 10-5 s-1) + \(\frac{179900 \mathrm{Jmol}^{-1}}{2.303 \times\left(8.314 \mathrm{Jk}^{-1} \mathrm{~mol}^{-1}\right) \times 546 \mathrm{k}}\)
= -4.6184 + 17.21 = 12.5916
A = Antilog 12.5916 = 3.9 × 1012s-1
A = 3.9 × 1012 s-1

Question 35.
Consider a certain reaction A → Products with k = 2.0 × 10-2s-1. Calculate the concentration of A remaining after 100 s If the initial concentration of A is 1.0 mol L-1.
Solution:
For the first order reaction
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 40

Question 36.
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t\(\frac{1}{2}\) = 3.00 hours. What fraction of sample of sucrose remains 8 after 8 hours?
Solution:
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 41

Question 37.
The decomposition of hydrocarbon follows the equation K = (4.5 × 1011 s-1) e-28000K/T. Calculate Ea.
Solution:
According to Arrhenius equation
k = Ae-Ea/RT ………………. (i)
According to the available data
k = (4.5 × 10 s-1)e-28000k/T ………………….. (ii)
on comparing both equations
\(-\frac{E_{\mathrm{a}}}{\mathrm{RT}}=\frac{-28000 \mathrm{k}}{\mathrm{T}}\)
Ea = (28000 k) × R
= (28000 k) × (8.314 K-1 J mol-1)
= 232792 Jk mol-1
Ea = 232.792 kJ-1 mol-1

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 38.
The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34 – 1.25 × 104 K/T. Calculate Ea for this reaction and at what temperature will its half-life period be 256 minutes?
Solution:
a) Calculation of activation energy Ea
According to Arrhenius equation; k = Ae-Ea/RT
log k = log A – \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}\) …………………. (i)
log K = 14.34 – \(\frac{1.25 \times 10^4 \mathrm{~K}}{\mathrm{~T}}\) …………………. (ii)
on comparing both equations.
\(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}=\frac{1.25 \times 10^4 \mathrm{~K}}{\mathrm{~T}}\)
Ea = 1.25 × 104 K × 2.303 × 8.314 (JK-1 mol-1)
= 23.93 × 104 J mol-1 = 239.3 kJ mol-1

b) Calculation of required temperature
If t1/2 = 256 min. for 1st order reaction;
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 42

Question 39.
The decomposition of A into product has value of k as 4.5 × 103 s-1 at 10°C and energy of activation 60 kJ mol-1. At what temperature would k be 1.5 × 104 s-1?
Solution:
According to Arrhenius equation
\(\log \frac{\mathrm{k}_2}{\mathrm{k}_1}=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}} \times \frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{~T}_1 \mathrm{~T}_2}\)
k1 = 4.5 × 103 s-1
k2 = 1.5 × 104 s-1
T1 = 10°C = 283 K
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 43
T2 = 297.19 K = (297.19 – 273.0) = 24.19°C
Temperature = 24.19°C

Question 40.
The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 × 1010 s-1. calculate k at 318K andE .
Solution:
Calculation of activation energy (Ea)
For 1st order reaction :
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 44
According to Arrhenius theory
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 45

b) Calculation of rate constant (k)
According to Arrhenius equation
log k = log A \(\frac{-\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}\)
log k = log (4 × 1010) – \(\frac{76640 \mathrm{~J} \mathrm{~mol}^{-1}}{2.303 \times\left(8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right) \times(318 \mathrm{~K})}\)
log k = 10.6021 – 12.5870 = – 1.9849
k = Antilog (-1.9849) = Antilog \((\overline{2} .0151)\) = 1.035 × 10-2 s-1
Ea = 76.640 kJ mol-1
k = 1.035 × 10-2 s-1

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 41.
The rate of a reaction quadruples when the temperature changes from 293 K to 313K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Solution:
According to Arrhenius equation
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 46

Textual Examples

Question 1.
From the concentrations of C4H9Cl (butyl chloride) at different times given below, calculate the average rate of the reaction :
C4H9Cl + H2O → C4H9OH + HCl
during different intervals of time.
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 47
Solution:
We can determine the difference in concentration over different intervals of time and thus determine the average rate by dividing ∆ [R] by ∆t
Average rates of hydrolysis of butyl chloride
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 48

Question 2.
The decomposition of N2O5 in CCl4 at 318K has been studied by monitoring the concentration of N2O5 in the solution. Initially the concentration of N2O5 is 2.33 mol L-1 and after 184 minutes, it is reduced to 2.08 mol L-1. The reaction takes place according to the equation
2 N2O5 (g) → 4 NO2 (g) + O2 (g)
Calculate the average rate of this reaction in terms of hours, minutres and seconds. What is the rate of production of NO2 during this period?
Solution:
Average Rate = \(\frac{1}{2}\left\{-\frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta \mathrm{t}}\right\}=-\frac{1}{2}\left[\frac{(2.08-2.33) \mathrm{mol} \mathrm{L}^{-1}}{184 \mathrm{~min}}\right]\)
= 6.79 × 10-4 mol L-1/min
= (6.79 × 10-4 mol L-1 min-1) × (60 min / 1h)
=4.07 × 10-2 mol L-1/h .
= 6.79 × 10-4 mol L-1 × 1mm / 60s = 1.13 × 10-5 mol L-1s-1
It may be remembered that
Rate = \(\frac{1}{4}\left\{\frac{\Delta\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}}\right\}\)
\(\frac{\Delta\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}}\) = 6.79 × 10-4 × 4 mol L-1 min-1
= 2.72 × 10-3 mol L-1 min-1

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 3.
Calculate the overall order of a reaction which has the rate expression
a) Rate = k [A]1/2 [B]3/2
b) Rate = k [A]3/2 [B]-1
Solution:
a) Rate = k [A]x [B]y
order = x + y
So order = 1/2 + 3/2 = 2, i.e., second order

b) order = 3/2 + (- 1) = 1/2, i.e., half order.
A balanced chemical equation never gives us a true picture of how a reaction takes place since rarely a reaction gets completed in one step. The reactions taking place in one step are called elementary reactions. If a sequence of elementary reactions (called mechanism), reactants give in the products, the reactions are called complex reactions. These may be consecutive reactions (e.g., oxidation of ethane to CO2 and H2O passes through a series of intermediate steps in which alcohol, aldehyde and acid are formed), reversible reactions and parallel reactions (e.g., nitration of phenol yields o-nitrophenol and p-nitrophenol).

Question 4.
Identify the reaction order from each of the following rate constants.

  1. k = 2.3 × 10-5 L mol-1 s-1
  2. k = 3 × 10-4 s-1

Solution:

  1. The unit of second order rate constant is L mol-1 s-1, therefore k = 2.3 × 10-5 L mol-1 s-1represents a second order reaction.
  2. The unit of a first order rate constant is s-1 therefore k = 3 × 10-4 s-1 represents a first order reaction.

Question 5.
The initial concentration of N2O5 in the following first order reaction
N2O5(g) → 2 NO2(g) + \(\frac{1}{2 \mathrm{O}_2}\) (g) was 124 × 10-2 mol L-1 at 318 K. The concentration of N2O5 after 60 minutes was 0.20 × 10-2 mol L-1 . Calculate the rate constant of the reaction at 318 K.
Solution:
For a first order reaction
\(\log \frac{[\mathrm{R}]_1}{[\mathrm{R}]_2}=\frac{\mathrm{k}\left(\mathrm{t}_2-\mathrm{t}_1\right)}{2.303}\)
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 49
k = 0.0304 min-1.

Question 6.
The following data were obtained during the first order thermal decomposition of N2O5 (g) at constant volume:
2N2O5 (g) → 22 O4 (g) + O2 (g)
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 50
Calculate the rate constant.
Solution:
Let the pressure of N2O5 (g) decrease by 2x atm. As two moles of N2O5 decompose to give two moles of N2O4 (g) and one mole of O2 (g), the pressure of N2O4 (g) increases by 2x atm and that of O2 (g) increases by x atm.
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 51

Question 7.
A first order reaction is found to have a rate constant, k = 5.5 × 10-14 s-1. Find the half¬life of the reaction.
Solution:
Half-life for a first order reaction is
t1/2 = \(\frac{0.693}{k}\)
t1/2 = \(\frac{0.693}{5.5 \times 10^{-14} \mathrm{~S}^{-1}}\) = 1.26 × 1013 s

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 8.
Show that in a first order reaction, time required for completion of 99.9% is 10 times of half-life (t1/2) of the reaction.
Solution:
When reaction is completed 99.9%, [R]n = [R]0 – 0.999 [R]0
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 52

Question 9.
Hydrolysis of methyl acetate in aqueous solution has been studied by titrating the liberated acetic acid against sodium hydroxide. The concentration of the ester at different times is given below.
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 53
Show that it follows a pseudo first order reaction, as the concentration of water remains nearly constant (55 mol L-1), during the course of the reaction. What is the value of k’ in this equation ?
Rate = k’ [CH3 COOCH3] [H2O]
Solution:
For pseudo first order reaction, the reaction should be first order with respect to ester when [H2O] is constant. The rate constant k for pseudo first order reaction is
k = \(\frac{2.303}{t} \log \frac{C_0}{C}\) where k = k’ [H2O]
From the above data we note
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 54
It can be seen that k’ [H2O] is constant and equal to 2.004 × 10-3 min-1 and hence, it is pseudo first order reaction. We can now determine k from
k’ [H2O] = 2.004 × 10-3 min-1
k'[55 mol L-1] = 2.004 × 10-3 min-1
k’ = 3.64 × 10-5 mol-1 L min-1

Question 10.
The rate constants of a reaction at 500 K and 700 K are 0.02s-1 and 0.07s-1 respectively. Calculate the values of Ea and A.
Solution:
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 55

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 11.
The first order rate constant for the decomposition of ethyl iodide by the reaction.
C2H5I(g) → C2H4 (g) + HI(g) at 600K is 1.60 × 10-5 s-1. It is energy of activation is 209 kj/mol. Calculate the rate constant of the reaction at 700 K.
Solution:
We know that
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 56

Intext Questions

Question 1.
For the reaction, R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 min. Calculate the average rate of reaction using units of time both in minutes and seconds.
Solution:
For the reaction R → P
AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics 57

Question 2.
In a reaction, 2A → products, concentration of A decreases from 0.5 mol L-1 to 0.4 to mol L-1 in 10 min. Calculate the rate during this interval.
Solution:
Rate of reaction = rate of disappearance of A
= \(-\frac{1}{2} \frac{\Delta[\mathrm{A}]}{\Delta \mathrm{t}}\)
= \(-\frac{1}{2} \frac{(0.4-0.5) \mathrm{mol} \mathrm{L}^{-1}}{10 \mathrm{~min}}\)
= 0.005 mol L-1 m-1

Question 3.
For a reaction, A + B → product; the rate law is given by, r = k[A]1/2 [B]2. What is the order of the reaction ?
Solution:
Order of reaction = \(\frac{1}{2}\) + 2 = \(\frac{1}{2}\)
Order = \(\frac{5}{2}\)

AP Inter 2nd Year Chemistry Study Material Chapter 3(b) Chemical Kinetics

Question 4.
The conversion of molecules X to Y follows second order kinetics. If concentration of x is increased to three times how will it affect the rate of formation of y ?
Solution:
For the reaction x → y
Reaction rate (r) = k[x]2 …………….. (i)
If the concentration of x is increased three times, then
Reaction rate (r’) = k[3x]2 = kx [9x]2 ……….. (ii)
Dividing Eq. (ii) by Eq. (i) .
\(\frac{\mathrm{r}^{\prime}}{\mathrm{r}}=\frac{\mathrm{k} \times[9 \mathrm{x}]^2}{\mathrm{k} \times[\mathrm{x}]^2}=9\)
It means that the rate of formation of y will increase by nine times.

Question 5.
What will be the effect of temperature on rate constant ?
Solution:
The rate constant for a reaction is nearly doubled with about 10° rise in temperature. The reason is that the number of effective collision becomes almost double. The exact dependence of the rate of reaction can be given by Arrhenius equation; k = Ae-Ea/RT. Where, A is called the frequency factor and Ea is the activation energy of the reaction.

Inter 2nd Year Maths 2A Quadratic Expressions Solutions Ex 3(b)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Quadratic Expressions Solutions Exercise 3(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Quadratic Expressions Solutions Exercise 3(b)

I.

Question 1.
If the quadratic equations ax2 + 2bx + c = 0 and ax2 + 2cx + b = 0, (b ≠ c) have a common root, then show that a + 4b + 4c = 0
Solution:
Let α be the common roots of the equations ax2 + 2bx + c = 0 and ax2 + 2cx + b = 0
2 + 2bα + c = 0
2 + 2cα + b = 0
on Subtracting,
2α(b – c) + c – b = 0
2α(b – c) = b – c
2α = 1 (b ≠ c)
α = \(\frac{1}{2}\)
Substitute α = \(\frac{1}{2}\) in ax2 + 2bx + c = 0 is
\(a\left(\frac{1}{4}\right)+2 b \frac{1}{2}+c=0\)
⇒ a + 4b + 4c = 0
∴ a + 4b + 4c = 0

Inter 2nd Year Maths 2A Quadratic Expressions Solutions Ex 3(b)

Question 2.
If x2 – 6x + 5 = 0 and x2 – 12x + p = 0 have a common root, then find p.
Solution:
Given x2 – 6x + 5 = 0, x2 – 12x + p = 0 have a common root.
If α is the common root then
α2 – 6α + 5 = 0, α2 – 12α + p = 0
α2 – 6α + 5 = 0
⇒ (α – 1) (α – 5) = 0
⇒ α = 1 or 5
If α = 1 then α2 – 12α + p = 0
⇒ 1 – 12 + p = 0
⇒ p = 11
If α = 5 then α2 – 12α + p = 0
⇒ 25 – 60 + p = 0
⇒ p = 35
∴ p = 11 or 35

Question 3.
If x2 – 6x + 5 = 0 and x2 – 3ax + 35 = 0 have a common root, then find a.
Solution:
The roots of the equation x2 – 6x + 5 = 0 are
(x – 1) (x – 5) = 0
⇒ x = 1, x = 5
Case (i): x = 1 is a common root then it is also root for the equation x2 – 3ax + 35 = 0
⇒ 1 – 3a(1) + 35 = 0
⇒ a = 12
Case (ii): x = 5 is a common root then
(5)2 – 3a(5) + 35 = 0
⇒ 60 – 15a = 0
⇒ a = 4
∴ a = 12 (or) a = 4

Question 4.
If the equation x2 + ax + b = 0 and x2 + cx + d = 0 have a common root and the first equation has equal roots, then prove that 2(b + d) = ac.
Solution:
Let α be the common root.
∴ x2 + ax + b = 0 has equal roots.
Its roots are α, α
α + α = -a
⇒ α = \(-\frac{a}{2}\)
α . α = b
⇒ α2 = b
∴ α is a root of x2 + cx + d = 0
⇒ α2 + cα + d = 0
⇒ b + c(\(-\frac{a}{2}\)) + d = 0
⇒ 2(b + d) = ac

Inter 2nd Year Maths 2A Quadratic Expressions Solutions Ex 3(b)

Question 5.
Discuss the signs of the following quadratic expressions when x is real.
(i) x2 – 5x + 4
Solution:
x2 – 5x + 4 = (x – 1) (x – 4)
a = 1 > 0
The expression x2 – 5x + 2 is positive if x < 1 or x > 4 and is negative if 1 < x < 4

(ii) x2 – x + 3
Solution:
∆ = b2 – 4ac
= (-1)2 – 4 (1) (3)
= 1 – 12
= -11 < 0
a = 1 > 0, ∆ < 0
⇒ The given expression is positive for all real x.

Question 6.
For what values of x, the following expressions are positive?
(i) x2 – 5x + 6
Solution:
x2 – 5x + 6 = (x – 2) (x – 3)
Roots of x2 – 5x + 6 = 0 are 2, 3 which are real.
The expression x2 – 5x + 6 is positive if x < 2 or x > 3
∴ a = 1 > 0

(ii) 3x2 + 4x + 4
Solution:
Here a = 3, b = 4, c = 4
∆ = b2 – 4ac
= 16 – 48
= -32 < 0
∴ 3x2 + 4x + 4 is positive ∀ x ∈ R
∴ a = 3 > 0 and ∆ < 0
ax2 + bx + c and ‘a’ have same sign ∀ x ∈ R, if ∆ < 0

(iii) 4x – 5x2 + 2
Solution:
Roots of 4x – 5x2 + 2 = 0 are \(\frac{-4 \pm \sqrt{16+40}}{-10}\)
i.e., \(\frac{2 \pm \sqrt{14}}{5}\) which is real
∴ The expression 4x – 5x2 + 2 is positive when
\(\frac{2-\sqrt{14}}{5}<x<\frac{2+\sqrt{14}}{5}\) [∵ a = -5 < 0]

(iv) x2 – 5x + 14
Solution:
Here a = 1, b = -5, c = 14
∆ = b2 – 4ac
= 25 – 56
= -31 < 0
∴ ∆ < 0 ∵ a = 1 > 0 and ∆ < 0
⇒ x2 – 5x + 14 is positive ∀ x ∈ R.

Question 7.
For what values of x, the following expressions are negative?
(i) x2 – 7x + 10
Solution:
x2 – 7x + 10 = (x – 2)(x – 5)
Roots of x2 – 7x + 10 = 0 are 2, 5 which are real.
∴ The expression x2 – 7x + 10 is negative if 2 < x < 5, ∵ a = 1 > 0

(ii) 15 + 4x – 3x2
Solution:
The roots of 15 + 4x – 3x2 = 0 are \(\frac{-4 \pm \sqrt{16+180}}{-6}\)
i.e., \(\frac{-5}{3}\), 3
∴ The expression 15 + 4x – 3x2 is negative if
-5x < \(\frac{-5}{3}\) or x > 3, ∵ a = -3 < 0

(iii) 2x2 + 5x – 3
Solution:
The roots of 2x2 + 5x – 3 = 0 are \(\frac{-5 \pm \sqrt{25+24}}{4}\)
i.e., -3, \(\frac{1}{2}\)
∴ The expression 2x2 + 5x – 3 is negative if -3 < x < \(\frac{1}{2}\), ∵ a = 2 > 0

(iv) x2 – 5x – 6
Solution:
x2 – 5x – 6 = (x – 6) (x + 1)
Roots of x2 – 5x – 6 = 0 are -1, 6 which are real.
∴ The expression x2 – 5x – 6 is negative if -1 < x < 6, ∵ a = 1 > 0

Inter 2nd Year Maths 2A Quadratic Expressions Solutions Ex 3(b)

Question 8.
Find the changes in the sign of the following expressions and find their extreme values.
Hint: Let α, β are the roots of ax2 + bx + c = 0 and α < β
(1) If x < α or x > β, ax2 + bx + c and ‘a’ have same sign.
(2) If α < x < β, ax2 + bx + c and ‘a’ have opposite sign.

(i) x2 – 5x + 6
Solution:
x2 – 5x + 6 = (x – 2) (x – 3)
(1) If 2 < x < 3, the sign of x2 – 5x + 6 is negative, ∵ a = 1 > 0
(2) If x < 2 or x > 3, the sign of x2 – 5x + 6 is positive, ∵ a = 1 > 0
Since a > 0, the minimum value of x2 – 5x + 6 is \(\frac{4 a c-b^{2}}{4 a}\)
= \(\frac{4(1)(6)-(-5)^{2}}{4(1)}\)
= \(\frac{24-25}{4}\)
= \(-\frac{1}{4}\)
Hence the extreme value of the expression x2 – 5x + 6 is \(-\frac{1}{4}\)

(ii) 15 + 4x – 3x2
Solution:
15 + 4x – 3x2 = 15 + 9x – 5x – 3x2
= 3(5 + 3x) – x(5 + 3x)
= (3 – x) (5 + 3x)
(1) If \(-\frac{5}{3}\) < x < 3 the sign of 15 + 4x – 3x2 is positive, ∵ a = -3 < 0
(2) If x < \(-\frac{5}{3}\) or x > 3, the sign of 15 + 4x – 3x2 is negative, ∵ a = -3 < 0
Since a < 0, the maximum value of 15 + 4x – 3x2 is \(\frac{4 a c-b^{2}}{4 a}\)
= \(\frac{4(-3)(15)-16}{4(-3)}\)
= \(\frac{49}{3}\)
Hence the extreme value of the expression 15 + 4x – 3x2 is \(\frac{49}{3}\)

Question 9.
Find the maximum or minimum of the following expressions as x varies over R.
(i) x2 – x + 7
Solution:
a = 1 > 0,
minimum value = \(\frac{4 a c-b^{2}}{4 a}\)
= \(\frac{28-1}{4}\)
= \(\frac{27}{4}\)

(ii) 12x – x2 – 32
Solution:
a = -1 < 0,
maximum value = \(\frac{4 a c-b^{2}}{4 a}\)
= \(\frac{128-144}{-4}\)
= 4

(iii) 2x + 5 – 3x2
Solution:
a = -3 < 0,
maximum value = \(\frac{4 a c-b^{2}}{4 a}\)
= \(\frac{(4)(-3)(5)-(2)^{2}}{4 \times-3}\)
= \(\frac{16}{3}\)

(iv) ax2 + bx + a
Solution:
If a < 0, then maximum value = \(\frac{4 a \cdot a-b^{2}}{4 a}\) = \(\frac{4 a^{2}-b^{2}}{4 a}\)
If a > 0, then minimum value = \(\frac{4 a \cdot a-b^{2}}{4 a}\) = \(\frac{4 a^{2}-b^{2}}{4 a}\)

II.

Question 1.
Determine the range of the following expressions.
(i) \(\frac{x^{2}+x+1}{x^{2}-x+1}\)
Solution:
Let y = \(\frac{x^{2}+x+1}{x^{2}-x+1}\)
⇒ x2y – xy + y = x2 + x + 1
⇒ x2y – xy + y – x2 – x – 1 = 0
⇒ x2 (y – 1) – x(y + 1) + (y – 1) = 0
∴ x is real ⇒ b2 – 4ac ≥ 0
⇒ (y + 1 )2 – 4(y – 1 )2 ≥ 0
⇒ (y + 1)2 – (2y – 2)2 ≥ 0
⇒ (y + 1 + 2y – 2) (y + 1 – 2y + 2) ≥ 0
⇒ (3y – 1) (-y + 3) ≥ 0
⇒ -(3y – 1) (y – 3) ≥ 0
a = Coeff of y2 = -3 < 0
But The expression ≥ 0
⇒ y lies between \(\frac{1}{3}\) and 3
∴ The range of \(\frac{x^{2}+x+1}{x^{2}-x+1}\) is [\(\frac{1}{3}\), 0]

(ii) \(\frac{x+2}{2 x^{2}+3 x+6}\)
Solution:
Let y = \(\frac{x+2}{2 x^{2}+3 x+6}\)
Then 2yx2 + 3yx + 6y = x + 2
⇒ 2yx2 + (3y – 1)x + (6y – 2) = 0
∴ x is real ⇒ discriminant ≥ 0
⇒ (3y – 1)2 – 4(2y)(6y – 2) ≥ 0
⇒ 9y2 + 1 – 6y – 48y2 + 16y ≥ 0
⇒ -39y2 + 10y + 1 ≥ 0
⇒ 39y2 – 10y – 1 < 0
⇒ 39y2 – 13y + 3y – 1 < 0
⇒ 13y(3y – 1) + 1(3y – 1) ≤ 0
⇒ (3y – 1) (13y + 1) ≤ 0
∴ a = Coeff of y2 = 39 > 0 and the exp ≤ 0
⇒ y lies between \(\frac{-1}{13}\) and \(\frac{1}{3}\)
∴ Range of \(\frac{x+2}{2 x^{2}+3 x+6}\) is \(\left[-\frac{1}{13}, \frac{1}{3}\right]\)

Inter 2nd Year Maths 2A Quadratic Expressions Solutions Ex 3(b)

(iii) \(\frac{(x-1)(x+2)}{x+3}\)
Solution:
Let y = \(\frac{(x-1)(x+2)}{x+3}\)
⇒ yx + 3y = x2 + x – 2
⇒ x2 + (1 – y)x – 3y – 2 = 0
x ∈ R ⇒ (1 – y2) – 4(-3y – 2) ≥ 0
⇒ 1 + y2 – 2y + 12y + 8 ≥ 0
⇒ y2 + 10y + 9 ≥ 0
y2 + 10y + 9 = 0
⇒ (y + 1) (y + 9) = 0
⇒ y = -1, -9
y2 + 10y + 9 ≥ 0
∴ a = Coeff of y2 = 1 > 0 and exp ≥ 0
⇒ y ≤ -9 or y ≥ -1
∴ Range = (-∞, -9] ∪ [-1, ∞)

(iv) \(\frac{2 x^{2}-6 x+5}{x^{2}-3 x+2}\)
Solution:
Let y = \(\frac{2 x^{2}-6 x+5}{x^{2}-3 x+2}\)
⇒ yx2 – 3yx + 2y = 2x2 – 6x + 5
⇒ (y – 2)x2 + (6 – 3y)x + (2y – 5) = 0
x ∈ R ⇒ (6-3y)2 – 4(y – 2) (2y – 5) ≥ 0
⇒ 36 + 9y2 – 36y – 4(2y2 – 9y + 10) ≥ 0
⇒ 36 + 9y2 – 36y – 8y2 + 36y – 40 ≥ 0
⇒ y2 – 4 ≥ 0
y2 – 4 = 0
⇒ y2 = 4
⇒ y = ±2
y2 – 4 ≥ 0
⇒ y ≤ -2 or y ≥ 2
⇒ y does not lie between -2, 2,
∵ y2 Coeff is > 0 and exp is also ≥ 0
∴ Range of \(\frac{2 x^{2}-6 x+5}{x^{2}-3 x+2}\) is (-∞, -2] ∪ [2, ∞)

Question 2.
Prove that \(\frac{1}{3 x+1}+\frac{1}{x+1}-\frac{1}{(3 x+1)(x+1)}\) does not lie between 1 and 4, if x is real.
Solution:
Let y = \(\frac{1}{3 x+1}+\frac{1}{x+1}-\frac{1}{(3 x+1)(x+1)}\)
⇒ y = \(\frac{x+1+3 x+1-1}{(3 x+1)(x+1)}\)
⇒ y = \(\frac{4 x+1}{3 x^{2}+4 x+1}\)
⇒ 3yx2 + 4yx + y = 4x + 1
⇒ 3yx2 + (4y – 4)x + (y – 1) = 0
x ∈ R ⇒ (4y – 4)2 – 4(3y)(y – 1) ≥ 0
⇒ 16y2 + 16 – 32y – 12y2 + 12y ≥ 0
⇒ 4y2 – 20y + 16 ≥ 0
4y2 – 20y + 16 = 0
⇒ y2 – 5y + 4 = 0
⇒ (y – 1)(y – 4) = 0
⇒ y = 1, 4
4y2 – 20y + 16 ≥ 0
⇒ y ≤ 1 or y ≥ 4
⇒ y does not lie between 1 and 4
Since y2 Coeff is the and exp ≥ 0.

Inter 2nd Year Maths 2A Quadratic Expressions Solutions Ex 3(b)

Question 3.
If x is real, prove that \(\frac{x}{x^{2}-5 x+9}\) lies between 1 and \(\frac{-1}{11}\).
Solution:
Let y = \(\frac{x}{x^{2}-5 x+9}\)
⇒ yx2 + (-5y – 1)x + 9y = 0
x ∈ R ⇒ (5y – 1)2 – 4y(9y) ≥ 0
⇒ 25y2 + 1 + 10y – 36y2 ≥ 0
⇒ -11y2 + 10y + 1 ≥ 0 ……….(1)
-11y2 + 10y + 1 = 0
⇒ -11y2 + 11y – y + 1 = 0
⇒ 11y(-y + 1) + 1(-y + 1) = 0
⇒ (-y + 1) (11y + 1) = 0
⇒ y = 1, \(\frac{-1}{11}\)
-11y2 + 10y + 1 ≥ 0
∴ y2 Coeff is -ve, but the exp is ≥ 0 from (1)
⇒ \(\frac{-1}{11}\) ≤ y ≤ 1
⇒ y lies between 1 and \(\frac{-1}{11}\)

Question 4.
If the expression \(\frac{x-p}{x^{2}-3 x+2}\) takes all real value for x ∈ R, then find the bounts for p.
Solution:
Let y = \(\frac{x-p}{x^{2}-3 x+2}\), given y is real
Then yx2 – 3yx + 2y = x – p
⇒ yx2 + (-3y – 1)x + (2y + p) = 0
∵ x is real ⇒ (-3y – 1)2 – 4y(2y + p) ≥ 0
⇒ 9y2 + 6y + 1 – 8y2 – 4py ≥ 0
⇒ y2 + (6 – 4p)y + 1 ≥ 0
∵ y is real ⇒ y2 + (6 – 4p)y + 1 ≥ 0
⇒ The roots are imaginary or real and equal
⇒ ∆ ≤ 0
⇒ (6 – 4p)2 – 4 ≤ 0
⇒ 4(3 – 2p)2 – 4 ≤ 0
⇒ (3 – 2p)2 – 1 ≤ 0
⇒ 4p2 – 12p + 8 ≤ 0
⇒ p2 – 3p + 2 ≤ 0
⇒ (p – 1)(p – 2) ≤ 0
If p = 1 or p = 2 then \(\frac{x-p}{x^{2}-3 x+2}\) is not defined.
∴ 1 < p < 2

Inter 2nd Year Maths 2A Quadratic Expressions Solutions Ex 3(b)

Question 5.
If c2 ≠ ab and the roots of (c2 – ab)x2 – 2(a2 – bc)x + (b2 – ac) = 0 are equal, then show that a3 + b3 + c3 = 3abc or a = 0.
Solution:
Given equation is (c2 – ab)x2 – 2(a2 – bc)x + (b2 – ac) = 0
Discriminant = 4(a2 – bc)2 – 4(c2 – ab) (b2 – ac)
= 4[(a2 – bc)2 – (c2 – ab) (b2 – ac)]
= 4(a4 + b2c2 – 2a2bc – b2c2 + ac3 + ab3 – a2bc)
= 4(a4 + ab3 + ac3 – 3a2bc)
= 4a(a3 + b3 + c3 – 3abc)
The roots are equal ⇒ discriminant = 0
4a(a3 + b3 + c3 – 3abc) = 0
a = 0 or a3 + b3 + c3 – 3abc = 0
i.e., a = 0 or a3 + b3 + c3 = 3abc

AP Inter 1st Year Economics Study Material Chapter 5 Theory of Value

Andhra Pradesh BIEAP AP Inter 1st Year Economics Study Material 5th Lesson Theory of Value Textbook Questions and Answers.

AP Inter 1st Year Economics Study Material 5th Lesson Theory of Value

Essay Questions

Question 1.
Explain the classification of Markets. [March 18, 17]
Answer:
Edwards defiried “Market , as a mechanism by which buyers and sellers are brought together”. Hence market means where selling and buying transactions take place. The classification of markets is’ based on three factors.

  1. On the basis of area
  2. On the basis of time
  3. On the basis of competition.

I. On the basis of area : According to the area, markets can be of three types.

  1. Local market : When a commodity is sold at particular locality. It is called a local market. Ex : Vegetables, flowers, fruits etc.
  2. National market : When a commodity is demanded and supplied throughout the country is called national market. Ex : Wheat, rice etc.
  3. International market: When a commodity is demanded and supplied all over the world is called international market. Ex : Gold, silver etc.-

II. On the basis of time : It can be further classified into three types.

  1. Market period or very short period : In this period where producer cannot make any changes in supply of a commodity. Here supply remains constant. Ex : Perishable goods. .
  2. Short period : In this period supply can be change to some extent by changing the variable factors of production.
  3. Long period : In this period-supply can be adjusted in according change in demand. In long run all factors will become variable.

III. On the basis of competition : This can be classified into two types.

  1. Perfect market: A perfect market is one in which the number of buyers and sellers is very large, all engaged in buying and selling a homogeneous products without any restrictions.
  2. Imperfect market: In this market, competition is imperfect among the buyers and sellers. These markets are divided into
    1. Monopoly
    2. Duopoly
    3. Oligopoly
    4. Monopolistic competition.

AP Inter 1st Year Economics Study Material Chapter 5 Theory of Value

Question 2.
Elucidate the features of perfect competition.
Answer:
Perfect competitive market is one in which the number of buyers and sellers is very large, all engaged in buying and selling a homogeneous products without any restrictions.
The following are the features of perfect competition :
1) Large number of buyers and sellers : Under perfect competition the number of buyers and sellers are large. The share of each seller and buyer in total supply or total demand is small. So no buyer and seller cannot influence the price. The price is determine only demand and supply. Thus the firm is price taker.

2) Homogeneous product: The commodities produced by all the firm of an industry are homogeneous or identical. The cross elasticity of products of sellers is infinite. As a result, single price will rule in the industry.

3) Free entry and exit: In this competition there is a freedom of free entry and exit. If existing firms are getting profits. New firms enter into the market. But when a firm getting losses, it would leave to the market.

4) Perfect mobility of factors of production : Under perfect competition the factors of production are freely mobile between the firms. This is useful for free entry and exits of firms.

5) Absence of transport cost: There are no transport cost. Due to this, price of the commodity will be the same throughout the market.

6) Perfect knowledge of the economy : All the buyers and sellers have full information regarding the prevailing and future prices and availability of the commodity. Information regarding market conditions is availability of commodity.

Question 3.
Describe Price determination in the imperfect competition.
Answer:
Monopoly is one of the market in the imperfect competition. The word Mono’ means single and Poly means seller. Thus monopoly means single seller market.
In the words of Bilas “Monopoly is represented by a market situation in which there is a single seller of a product for which there are no close substitutes, this single seller is unaffected by and does not affect, the prices and outputs of other products sold in the economy”. Monopoly exists under the following conditions.

  1. There is a single seller of product.
  2. There are no close substitutes.
  3. Strong barriers to entry into the industry exist.

Features of monopoly :

  1. There is no single seller in the market.
  2. No close substitutes.
  3. There is no difference between firm and industry.
  4. The monopolist either fix the price or output.

AP Inter 1st Year Economics Study Material Chapter 5 Theory of Value

Price determination : Under monopoly the monopolist has complete control over the supply of the product. He is price maker who can set the price to attain maximum profit. But he cannot do both things simultaneously. Either he can fix the price and leave the output to be determined by consumer demand at a particular price. Or he can fix the output to be produced and leave the price to be determined by the consumer demand for his product. This can be shown in the diagram.
AP Inter 1st Year Economics Study Material Chapter 5 Theory of Value 1
In the above diagram on ‘OX’ axis measures output and OY axis measures cost. AR is Average Revenue curve, AC is Average Cost curve. In the above diagram at E point where MC = MR at that point the monopolist determine the output. Price is determine where this output line touches the AR line. In the above diagram for producing OQ quantity cost of production is OCBQ and revenue is OPAQ.
Profit = Revenue – Cost
= PACB shaded area is profit under monopoly.

Short Answer Questions

Question 1.
What are the main features of perfect competition ?
Answer:
Perfect competitive market is one in which the number of buyers and sellers is very large, all engaged in buying and selling a homogeneous products without any restrictions.
The following are the features of perfect competition :
1) Large number of buyers and sellers : Under perfect competition the number of buyers and sellers are large. The share of each seller and buyer in total supply or total demand is small. So no buyer and seller cannot influence the price. The price is determine only demand and supply. Thus the firm is price taker.

2) Homogeneous product: The commodities produced by all the firm of an industry are homogeneous or identical. The cross elasticity of products of sellers is infinite. As a result, single price will rule in the industry.

3) Free entry and exit: In this competition there is a freedom of free entry and exit. If existing firms are getting profits. New firms enter into the market. But when a firm getting losses, it would leave to the market.

4) Perfect mobility of factors of production : Under perfect competition the factors of production are freely mobile between the firms. This is useful for free entry and exits of firms.

5) Absence of transport cost: There are no transport cost. Due to this, price of the commodity will be the same throughout the market.

6) Perfect knowledge of the economy : All the buyers and sellers have full information regarding the prevailing and future prices and availability of the commodity. Information regarding market conditions is availability of commodity.

Question 2.
What is meant by price discrimination ? Explain various methods of price discrimination.
Answer:
Price discrimination refers to the practice of a monopolist charging different prices for different customers of. the same product.
In the words of Joan Robinson “The act of selling the same article, produced under single control at different prices to different buyers is known as price discrimination”. Price discrimination is of three types. 1. Personal 2. Local 3. Use or trade discrimination.

  1. Personal discrimination : When a seller charges different prices from different persons.
    Ex : A book is sold ₹ 15/- to one person and other person at discount rate of ?
  2. Local discrimination : When a seller charges different prices from people of different localities or places.
    Ex : Dumping.
  3. Use discrimination : When different prices of commodity are charged according to the uses to which the commodity is put is known discrimination is according to use.
    Ex : Electricity is sold at a cheaper rate for uses of domestic purposes than for industrial purposes.

AP Inter 1st Year Economics Study Material Chapter 5 Theory of Value

Question 3.
Define Oligopoly.
Answer:
The term ‘Oligopoly’ is derived from two Greek word “Oligoi” meaning a few and “Pollein” means to sell. Oligopoly refers to a market situation in which the number of sellers dealing in a homogeneous or differentiated product is small. It is called competition among the few. The main features of oligopoly are the following.

  1. Few sellers of the product.
  2. There is interdependence in the determination of price.
  3. Presence of monopoly power.
  4. There is existence of price rigidity.
  5. There is excessive selling cost or advertisement cost.

Question 4.
Compare Perfect competition and Monopoly.
Answer:
Perfect competition

  1. There is large number of sellers.
  2. All products are homogeneous.
  3. There is freedom of free entry and exists.
  4. There is difference between industry and firm.
  5. Industry determines the price and firm receives the price.
  6. There is universal price.
  7. The AR, MR curves are parallel to ‘X, axis.

Monopoly

  1. There is only one seller.
  2. No close substitutes.
  3. There is no freedom of free entry and exists.
  4. Industry and firm both are same.
  5. Firm alone determine the price.
  6. Price discrimination is possible.
  7. The AR, MR curves are different and slopes downs from left to right.

Additional Questions

Question 5.
Define Monopolistic competition. Explain the important characteristics of Monopolistic competition.
Answer:
It is a market with many sellers for a product but the products are different in certain respects. It is mid way of monopoly and perfect competition. Prof. E.H. Chamberlin and Mrs. Joan Robinson pioneered this market analysis.
Characteristics of Monopolistic competition :

  1. Relatively small number of firms : The number of firms in this market are less than that of perfect competition. No one should not control the output in the market as a result of high competition.
  2. Product differentiation : One of the features of monopolistic competition is product
    differentiation. It take the form of brand names, trade marks etc. It cross elasticity of demand is very high.
  3. Entry and exit: Entry into the industry is unrestricted. New firms are able to commence production of very close substitutes for the existing brands of the product.
  4. Selling cost: Advertisement or sales promotion technique is the important feature of Monopolitic competition. Such costs are called selling costs.
  5. More Elastic Demand : Under this competition the demand curve slopes downwards from left to the right. It is highly elastic.

AP Inter 1st Year Economics Study Material Chapter 5 Theory of Value

Question 6.
Explain the price determination under perfect competition ?
Answer:
Perfect competition is a competition in which the number of buyers and sellers is very large. All enged in buying and selling a homogeneous without any restrictions.

Under this competition there are large no. of by buyers and sellers no buyer is.a seller can’t influence market price all products are homogeneous there is a freedom of free entry and exit. There is a perfect mobility of factors are production. There is no transport cost these are the main features are perfect competition.

Price determination:
Under perfect competition sellers and buyers can’t decide the price industry decides the price of the good in the supply and demand determined the price these can shown in the following table.
AP Inter 1st Year Economics Study Material Chapter 5 Theory of Value 2
In this competition where demand supply both are equal at that point price and output determine the table changes in price always lead to a change in supply and demand as price increases there is a fall in the quantity demanded. The relation between price and demand is negative. The relation between price and supply is positive. It can be observe the table price 1 ₹ market demand 60 and supply is 20. When price increases 5 ₹ supply increases’ 60 and demand decreases 20. When the price is 3 ₹ the demand supply are equal that is 40 these price called equilibrium price. This process is explain with help of the diagram.
AP Inter 1st Year Economics Study Material Chapter 5 Theory of Value 3
In the diagram ‘OX’ axis shown demand and supply OY’ axis represented price. DD demand curve, ‘SS’ is supply curve. In the diagram the demand curve and supply curve intersect at point E. Where the price is ‘OP’ and output is ‘OQ’.

Very Short Answer Questions

Question 1.
Market
Answer:
Market is place where commodities are brought and sold and where buyers and sellers meet. Communication facilities help us today to purchase and sell without sell without going to the market. All the activities take place is now called as market.

AP Inter 1st Year Economics Study Material Chapter 5 Theory of Value

Question 2.
Local Market
Answer:
A product is said to have local market, when buyers and sellers of the product carry on the business in a particular locality. These goods are highly perishable cannot to take to distant places. They cannot even stored for a longer time.
Ex : Vegetables, milk, fruits etc.

Question 3.
National Market
Answer:
A National Market is said to exist when a commodity is demanded and supplies all over our country.
Ex : Rice, wheat, sugar etc. .

Question 4.
Monopoly
Answer:
Mono means single, Poly means seller. In this market single seller and there is no close substitutes. The monopolist is a price maker.

Question 5.
Monopolistic Competitions
Answer:
It is a market where several firms produce same commodity with small differences is called monopolistic competition. In this market producers to produce close substitute goods.
Ex : Soaps, cosmetics etc.

Question 6.
Oligopoly
Answer:
A market with a small number of producer is called oligopoly. The product may be homogeneous or may be differences. This market exists in automobiles, electricals etc.

AP Inter 1st Year Economics Study Material Chapter 5 Theory of Value

Question 7.
Duopoly [march 16]
Answer:
When there are only two sellers of a product, there exist -duopoly. Each seller under duopoly must consider the other firms reactions to any changes that he make in price or output. They make decisions either independently or together.

Question 8.
Equilibrium Price
Answer:
Equilibrium price is that price where demand and supply are equal in the market.

Question 9.
Price discrimination [March 18, 17, 16]
Answer:
Monopolist will charge different prices for the same commodity or service in the market. This is known as discriminating monopoly or price discrimination.

Question 10.
Selling Costs [March 18, 17]
Answer:
An important feature of monopolistic market is every firm makes expenditure to sell more output. Advertisements through newspapers, journals, electronic media etc., these methods are used to attract more consumers by each firm.

Additional Question

Question 11.
Perfect competition
Answer:
In this market large number of buyers and sellers who promote competition. In this market goods are homogeneous. There is no transport fares and publicity costs. So price is uniform of any market.

AP Board 7th Class Maths Solutions Chapter 7 Ratio and Proportion Ex 7.4

SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion Ex 7.4 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion Ex 7.4

Question 1.
If 5 men take 8 days to type 10 books, then how many days does it take for 8 men to type 2 books?
Answer:

Number of MenNumber of books typedNumber of Days
5108
82x

Here the 3 quantities are men, books typed and number of days.
Let the number days taken to type 2 books by 8 men be x.

From the above,

  • Number of days and men are in inverse proportion.
  • Number of days and typed books are in direct proportion.

Here number of days depends on both men and number of books typed.

So, we have to take compound ratio of 8 : 5 and 10:2.
8 : x = 8 × 10 : 5 × 2
Since the ratios are equal,
The product of means = The product of extremes
⇒ 8 : x = 80 : 10
⇒ x × 80 = 8 × 10
⇒ \(\frac{80 x}{80}=\frac{80}{80}\)
⇒ x= 1 day
Or
x = \(\frac{5}{8} \times \frac{2}{10}\) × 8 = 1

∴ Number of days taken to type 2 books by 8 men is 1.

AP Board 7th Class Maths Solutions Chapter 7 Ratio and Proportion Ex 7.4

Question 2.
If 5 men take 9 days to plough 18 acres of land, then find how long it takes 25 men for to plough 30 acres?
AP Board 7th Class Maths Solutions Chapter 7 Ratio and Proportion Ex 7.4 1
Answer:
Here we have three quantities number of men, jj number of days and acres of land ploughed.

Number of MenNumber of LandNumber of Days
5189
2530X

Let the number of days taken to plough the 30 acres of land by 25 men is x.

From the above,

  • Number of days and number of men are in increase proportion.
  • Number of days and acres of land ploughed in direct proportion.

Here number of days depends on both men and number of acres of land ploughed.

So, we have to take compound ratio of 25 : 5 and 18 : 30 is
25 × 18 : 5 × 30
∴ 9 : x = 25 × 18 : 5 × 30
Since the ratios are equal,
The product of means = The product of extremes
⇒ x × 25 × 18 = 9 × 5 × 30
AP Board 7th Class Maths Solutions Chapter 7 Ratio and Proportion Ex 7.4 2
Or
x = \(\frac{5}{25} \times \frac{30}{18}\) × 9 = 3 days
∴ Number of days taken to plough 30 acres of land by 25 men is 3 days.

Question 3.
Rice cost ₹ 480 is needed for 8 members for 20 days. What is the cost of rice required for 12 members for 15 days?
Answer:
Here we have three quantities cost of rice, needed members and number of days.

Number of MembersNumber of DaysCost of Rice
820₹ 480
1215X

Let the Cost of rice needed for 12 members for 15 days is x.

From the above,
• Number of members and number of days are in inverse proportion.
• Number of days and cost of rice are in direct proportion.

Here the cost of rice depends on both number of member needed and number of days.
So, we have to take compound ratio of 8 : 12 = 20 : 15 is
8 × 20: 12 × 15
480 : x = 8 × 20 : 12 × 15 Since the ratios are equal,
The product of means = The product of extremes
⇒ x × 8 × 20 = 480 × 12 × 15
AP Board 7th Class Maths Solutions Chapter 7 Ratio and Proportion Ex 7.4 3
∴ The cost of rice needed for 12 members for 15 days is ₹ 540.

AP Board 7th Class Maths Solutions Chapter 7 Ratio and Proportion Ex 7.4

Question 4.
24 men working at 8 hours per day can dp a piece of work in 15 days. In how many days can 20 men working at 9 hours per day do the same work?
Answer:
Here we have three quantities number of men, working hours per day and number of days.

Number of MenWorking hoursNumber of Days
24815
209x

Let the number of days needed for 20 men working at 9 hours is x.
From the above,
• Number of men and members are in inversely proportional:
• Number of men and their working hours per day are in direct proportion.
Here the number of days depends on both number of men and working hours per day.
So, we have to take compound ratio of 24 : 20 = 8 : 9 is
24 × 8 : 20 × 9
∴ x : 15 = 24 × 8 : 20 × 9

Since the ratios are equal,
The product of means = The product of extremes
⇒ x × 20 × 9 = 15 × 24 × 8
AP Board 7th Class Maths Solutions Chapter 7 Ratio and Proportion Ex 7.4 4
∴ The number of days needed fur 20 men working at 9 hours is 16 days.

Question 5.
12 painters can paint a wall of 180 meters long in 3 days. How many painters are required to paint 200 meters long wall in 5 days?
Answer:
Here we have three quantities number of painters, length of wall and number of days required.

Length of wallNumber of working daysNumber of Painters
180312
2005X

Let the number of painters needed to paint 200 m long wall in 5 days is x.

From the above,

  • Number of working days and number of painters are in inversely proportional.
  • Number of working days and length of wall are in direct proportion.

Here the number of painters depends on both number of working days and length of wall.
So, we have to take compound ratio of 180 :,200 = 5 : 3 is 180 × 5 : 200 × 3
∴ 12 : x = 180 × 5 : 200 × 3
Since the ratios are equal,
The product of means = The product of extremes
⇒ x × 180 × 5 = 12 × 200 × 3
AP Board 7th Class Maths Solutions Chapter 7 Ratio and Proportion Ex 7.4 5
⇒ x = 8
The number of painters needed to paint 200 m long wall in 5 days is 8.

TS 6th Class Social 21th Lesson Questions and Answers Telangana – Greenery in Telangana

Telangana SCERT 6th Class Social Study Material Telangana 21th Lesson – Greenery in Telangana Textbook Questions and Answers.

Greenery in Telangana – TS 6th Class Social 21th Lesson Questions and Answers Telangana

Question 1.
Why is it necessary to increase the greenery in Telangana ?
Answer:

  1. According to National Forest Policy of India to maintain ecological balance 33% of tree cover is necessary.
  2. But in Telangana the present tree cover area is only 24%.
  3. Due to this Telangana state receives very less rainfall and ground water level also reduces.
  4. Total wildlife is in endangered position due to less greenery.
  5. So it is necessary to increase the greenery.

Question 2.
What are the uses of forests ?
Answer:

  1. If we grow more plants and trees, it will be useful in getting rainfall, improving the surface and ground water level.
  2. Tanks and wells will be with full of water and so better irrigation’will be provided for agriculture.
  3. Animals such as sheep, goats, cows will have abundant fodder.
  4. Grasslands, green meadows will be available to these animals for grazing.
  5. The pollution caused due to emissions from industries, vehicles and fossil fuel burning will be reduced. •
  6. Forest produce like timber, bamboo, leaves for medicines and beedi leaves, fire wood, honey, fruits, nuts, etc. will be available from these forests.
  7. Birds and wild animals will have a proper shelter and will not poach on to the habitations.

TS 6th Class Social 21th Lesson Questions and Answers Telangana - Greenery in Telangana

Question 3.
Suggest a few measures to protect the plants.
Answer:
Measures to protect the plants :

  1. Use natural pesticides
  2. Try Botanical insecticides
  3. Provide shelter and warmth
  4. Add some mulch
  5. Water then daily
  6. Plant more trees
  7. Prevent excess deforestation

Question 4.
How are the trees on both sides of the roads helpful to us ?
Answer:

  1. The trees on both sides of the roads in the city make greenery.
  2. They provide us shelter.
  3. They make temperature cool.
  4. They prevent air pollution.
  5. Abundant oxygen is supplied by taking carbondioxide.

Question 5.
Collect newspaper clippings, showing the news related to planting saplings and make an album.
Answer:
Plant saplings:
TS 6th Class Social 21th Lesson Questions and Answers Telangana - Greenery in Telangana 1

Note : Students can collect some more images.

Question 6.
How is Vanamahotsavam conducted recently in your locality ?
Answer:

  1. Vanamahotsavam is being held in our country since 1950 during the first week of July.
  2. We are residing in Bhupalpally. We have celebrated Vanamahotsavam 1st July.
  3. Students, teachers, SMC members, president and some famous persons in our area, participated in this programme.
  4. We selected non-greenery areas and plant trees like neem. kanuga, peepal. banyan, etc. on either side of roads.
  5. After that we conducted a cultural programme on greenery.

TS 6th Class Social 21th Lesson Questions and Answers Telangana - Greenery in Telangana

Question 7.
Create a few slogans on afforestation.
Answer:

  1. Save trees, Save the world, Save yourself.
  2. Live green, Love green, n think green.
  3. You want to smile, so let the Earth smile.
  4. T.R.E.E.S : Treat Our Existing Ecosystem Sacredly.

I. Conceptual Understandings

Question 1.
How was Vanamahotsav programme held in your school ?
Answer:
It is our primary duty to take care of these plants. In my school Vanamahotsav programme was held on August 4th. For this programme my headmaster invite the local leader to participate in it. At first our H.M. Chief guest and other teachers gave their valuable speeches about importance of greenery.

After that they distributed each student one plant and make them to plant surroundings of my school. Remaining students went into my village and plant the saplings in open places. According to this programme we known very much about the importance of greenery.

Question 2.
Make a list of different saplings planted in your school.
Answer:
We have sappled the following plants in our school zone.

  1. Bougenvellia
  2. Jasmine
  3. Marigold
  4. Rose
  5. Hybiscus
  6. Peepal
  7. Banyan
  8. Neredu
  9. Neem
  10. Mango
  11. Indian goosberry
  12. Guava
  13. Caster plants
  14. Coconut
  15. Ashoka
  16. Eucalyptus
  17. Survey plants
  18. Subabul etc.

Question 3.
How can you say that increase in forest cover will help in high rainfall ?
Answer:

  1. Increase forest may leads high rainfall to transportation by the plants / trees.
  2. High transpiration into the atmosphere allows for the formation of clouds in that region and may lead to abundant rainfall and increasing oxygen levels.
  3. The carbondioxide levels also will decrease in that area and reduce global warming affect.

Analysation:

Question 1.
Imagine the situations in your village / town or your region, if it is fully covered with trees and greenery.
Answer:
Due to the continuous efforts made our local leaders, now my village turned into self-sufficient in increasing plants and trees. The Vanamahotsav programme has been continuing for the last four years, we made our village full greenery and the village also involving in the waste lands and converted into parks. This greenery in our village gives us recreation and also habitat for various new animals surviving, it controls endangering of animals in our regions. It leads to increasing oxygen levels, agriculture works and heavy rainfall. Now our village has become a role model to all other surrounding villages and cities to bring awareness about the greenery importance in the people.

Discussion:

Question 1.
Discuss in the classroom, how the plantation has taken place in your locality.
Answer:
The ecofriendly technology method is insisted by the environmental scientist, nation leaders, state leaders as well as village leaders, we started the plantation programme in our city areAnswer:

Ours is a small school in our locality, my panchayat leaders, local MLA’s have conducted a weekend programme of increasing greenery in our village. Before that we all save trees save country campaign programme. Later our local leaders and our school principal have donated a plant a each student and asked them to plant in your house or near by areas and to take care of them, hence they have become trees. As we do this programme every year, the oxygen levels in the atmosphere will increase and carbondioxide levels will decrease. This helps in sustaining of life.

TS 6th Class Social 21th Lesson Questions and Answers Telangana - Greenery in Telangana

Question 2.
Discuss about the trees, gardens, parks available in your village / town or region,
Answer:
Ours is a small town and we have so many trees along the road side, and our leaders also have taken care of growing plants in our area and a few parts are created. Because of this our small town is cool in hot seasons when compared will surrounding towns and village. If we continue maintaining. This type of situations in every village. Global temperature will definitely come down and we can lead the happiest.

II. Learning and Thinking Questions for Examination Purpose

Question 1.
Where do Telangana lies ?
Answer:
Telangana lies between 15°46′ and 19°47! North latitude and 77°16’ and 81°43′ Eastern longitude. It is situated in the Deccan Plateau.

Question 2.
Write about NFPI (1988) ?
Answer:
According to National Forest Policy of India (1988) the total geographical area under tree cover should be a minimum of 33% to maintain environmental stability and ecological balance. It is very essential for the sustenance of all life forms, human beings and animals.

Question 3.
What are the two folds of Telangana massive plantation programme ?
Answer:

  1. Initiatives in notified areas.
  2. Initiatives in areas outside the notified.

Question 4.
Read the following para and answer the given questions.

Massive planting activities are taken up in areas such as road-side avenues, river and canal bank, barren hills, tank bunds and foreshore areas, schools and colleges, religious places, housing colonies, community lands, municipalities, industrial parks and farm lands, etc.

Question 1.
What activities are taken up ?
Answer:
Massive planting activities are taken up.

Question 2.
Name the areas that are related to students.
Answer:
Schools and colleges.

TS 6th Class Social 21th Lesson Questions and Answers Telangana - Greenery in Telangana

Question 5.
Read the following para and answer the given questions.

Medicinal plants like neem, tulsi, etc. Fruit and nut bearing like almonds, mango, j sapota, guava, custard apple, etc. Flowering plants like jasmine, marigold, rose.! gulmohar, etc and several other plants like silver oak, raavi. marri, neredu. etc. are! being planted in the above mentioned areas.

1. Name any medicinal plant.
Answer:
Neem/Tulasi.

2. Name the nut bearing plants.
Answer:
Almonds, mango, sapota, guava, custard apple etc.

Question 6.
Read the given para and comment on it.

Planting of saplings is a passion to many of us, but the problem is with regard to safeguarding and protecting of the plants. Normally, more saplings are planted, but not many of them are grown due to our own negligence.
Answer:
Now-a-days, on many oecassions many of us plantives, saplings and posting the photos in social mediAnswer: This become a great activity to many of us. And we feel it as our pride. But nobody takes care of them. Only a few saplings will grow to plants and then to trees. As the plants are main source of our livelyhood. we should not ignore their growth.

Question 7.
In Telangana the present tree cover area is only about 24%. Discuss the sensitivity in this issue ? *
Answer:

  1. According to National Forest Policy of India (1988). the total geographical area under tree cover should be a minimum of 33% to maintain environmental stability and ecological balance.
  2. But it is only 24% in Telangana.
  3. This leads to many problems to the life in Telangana and India.
  4. So steps to be taken to fill the gap between two percentages.

Question 8.
What are the uses of growing more plants and trees ? Write an essay.
Answer:
If we grow more plants and trees, it will be useful in getting rainfall, improving the surface and ground water level. Tanks arid wells will be with full of water and so better irrigation will be provided for agriculture. Animals such as sheep, goats, cows will have abundant fodder. Grasslands, green meadows will be available to these animals for grazing.

The pollution caused due to emissions from industries, vehicles and fossil fuel burning will be reduced. Forest produce like timber, bamboo, leaves for medicines and beedi leaves, fire wood, honey, fruits, nuts, etc. will be available from these forests. Birds and wild animals will have a proper shelter and will not poach on to the habitations.

TS 6th Class Social 21th Lesson Questions and Answers Telangana - Greenery in Telangana

Question 9.
Appreciate the Forest Man of India.
Answer:
The Forest Man of India :
Jadav Molai Payeng was a forest worker in Jorhat district of Assom. At the age of 16, he participated in the social forestry program in 1979 as a labourer in the project which continued for 5 years. After completion of the project, all the workers left but he continued to plant more trees on his own.

He planted and cared number of trees in an area of about 1360 acres over a period of 20 years on a sand bar of the river Brahmaputra. With his work, the whole place turned into a forest, called the Molai forest, after him. He is called as “the Forest Man of India” and awarded Padmashree, the fourth highest civilian honour in the country.

Question 10.
Name the programme of greenery in some other schools.
Answer:
Each one, Plant one.

II. Reading the Text (given), Understanding and Interpretation

Question 1.
Can you think of the reasons for the condition in the first picture ?
Answer:

  1. Due IO lack of rains the area has turned into desertification.
  2. It also includes acid rains and usage of nuclear weapons
  3. Excessive use of underground water created this situation.
  4. Excessive use of bio-chemicals, fertilizers and pesticides, the land turned into barren.
  5. Global warming is also main reason for this condition.

IV. Reflection on Contemporary Issues and Questioning
TS 6th Class Social 20th Lesson Questions and Answers Telangana - Sculptures and Buildings 4

Question 1.
With regard to your region compare the situations of the above pictures.
Answer:

  1. I am living in Telangana area, naturally our area is upland area.
  2. There are scarcity of rivers and rocky soil.
  3. Cultivation is not possible due to lack of water.
  4. So the first picture situation is also available in same parts of my area.

VI. Appreciation and Sensitivity

Question 1.
What will we do to make our regions like the second picture ?
Answer:

  1. We encourage afforestation and planting the saplings in the open places and even in dry waste lands.
  2. We discourage afforestation and encourage tribal people to use alternate source for their fire wood.
  3. In every function, in our region a new habit has inculcated among the people to give each and every couple a plant to develop greenery.

TS 6th Class Social 21th Lesson Notes – Greenery in Telangana

  • Ecological balance : The ecological balance is the equilibrium between harmonious co-existence of organisms and their environment.
  • Pollution : The presence in or introduction into the environment of a substance which has harmful or poisonous effects.
  • Ground water level : It is a term that is used in a relatively loose way, normally referring to the level, either below ground or above ordnance datum at which soil or rock is saturated, (water table)
  • Greenery : Green foliage, growing plants or vegetation.
  • Predominently : Mainly, mostly, for the most part.
  • Scanty : Small or insufficient in quantity or amount.
  • Erratic : Not even or regular in pattern or movement, unpredicable,
  • Scrub lands : it is a plant community characterised by vegetation dominated by vegetation dominated by shrubs.
  • Degradation : The process by which something changes to a worse condition.
  • Erosion : The process by which rock or soil is gradually destroyed by wind, rain, or the seAnswer:
  • Massive : Large and heavy or solid.
  • Prominent : Important, famous.
  • Meadows : A piece of grassland, especially one used for hay.
  • Emmissions : The production and discharge of something, especially gas or radiation.
  • Saplings : A young tree, especially one with a slender trunk.
  • Intention : A thing intended, an aim or plan.
  • Transformation : A marked change in form, nature or appearance.
  • Desperate : Feeling or showing a hopeless sense that a situation is so bad as to be impossible to deal with.

Mind Mapping:

TS 6th Class Social 21th Lesson Questions and Answers Telangana - Greenery in Telangana 2

TS 6th Class Social 21th Lesson Questions and Answers Telangana - Greenery in Telangana 3

 

TS 6th Class Social 20th Lesson Questions and Answers Telangana – Sculptures and Buildings

Telangana SCERT 6th Class Social Study Material Telangana 20th Lesson –Sculptures and Buildings Textbook Questions and Answers.

Sculptures and Buildings – TS 6th Class Social 20th Lesson Questions and Answers Telangana

Question 1.
Why do you think Ashoka chose lions and bulls or his pillars rather than cows or parrots ? What animal would you have chosen if you were in his place ? Give reasons for your choice.
Answer:
The animals chosen by Ashoka are believed to symbolize different’ steps of Lord Buddha’s life.

  • Elephant: The elephant represents the Buddha’s idea in reference to the dream of Queen Maya of a white elephant entering her womb.
  • Bull : The bull represents desire during the life of the Buddha as a prince.
  • Horse : The horse represents Buddha’s departure from palatial life.
  • Lion : The lion represents the accomplishment of BuddhAnswer: If I were in his place 1 would have chosen Cows/Bulls, Dogs and Horses.

Reasons for my choice :

  1. Cows are sacred.
  2. Bulls do hardwork for us.
  3. Dogs are our best friends.
  4. Horses represent our strength.

TS 6th Class Social 20th Lesson Questions and Answers Telangana - Sculptures and Buildings

Question 2.
Many stupas of Andhra Pradesh are on the banks of rivers (like Salihundam, Nagaijunakonda, Amaravati, etc). Why do you think the monks selected these places to build stupas?
Answer:
The civilizations are flourished on the banks of rivers. Many people settled there. The monks received donations from the people to construct stupas. The stupas are also should be visited by people to offer prayers. Any religion can come into wide practice if it receives the affection of people and support of the kings. So these are plenty near the banks of the rivers. So they constructed stupas near banks.

Question 3.
Look at the sculpture showing the calming of the elephant. Who are the people watching the event from a height ?
Answer:
These people were Queen and her servants.

Question 4.
Why do you think only the rock cut viharas and chaityas have survived till today ?
Answer:
a) Viharas:

  1. Viharas were the monasteries or living places of Buddhist monks where religious education was imparted.
  2. A vihara usually had small rooms around an open courtyard and had a small shrine containing a stupa or an image of the Buddha at one end.
  3. This shrine was called a chaitya which was also a prayer hall for monks. There are cave viharas excavated on hill sides like in Nashik and Karle. The other viharas were built with brick or stone blocks like in Takshasila, Nagarjunakonda and Nalanda which became great place of learnings. Viharas were made of rock using chisel and hammer, so they were survived till today.

b) Chaityas:

  1. Traders, artisans and kings of the time of the Satavahanas donated generally to build a unique kind of chaityas and viharas in several places in Maharashtra like Karie, Bhaja, Kamheri and Nashik.
  2. These were built like caves on hill sides. Stone workers usipg chisel and hammer carved living rocks and reproduced structures which carpenters and wood workers had made.
  3. They made large prayer halls, stupas and small rooms for monks – all in the rock. So they were also survived and had attracted the pilgrims from other countries.

TS 6th Class Social 20th Lesson Questions and Answers Telangana - Sculptures and Buildings

Question 5.
Compare the features of a Chaitya and a temple. Do you think the mode of worship in a Chaitya and a temple would be different ?
Answer:

  1. The temple was the greatest socio-religious institution created in ancient IndiAnswer: The whole South India is dotted with temples. Tamil Nadu alone has more than 30,000 temples. Chaitya was a prayer hall for the monks.
  2. The temple was not merely a peace of worship, it was also the hub of the social, religious, cultural, economic and political life of the community. Chaityas were the great places of learning.
  3. The temple was conceived as a place not only for worship, but as a centre of learning also.
    Apart from imparting education in chaityas and viharas monks were als.o treating the physical ailments of people.
  4. The temple was a great seat of intercourse. The life of man from the cradle to the grave was wrapped in rituals and ceremonies to be performed in temple. Buddhist monks taught Buddhist scriptures to disciples.
  5. Temples were served as Ghatikas and enjoyed patronage of wealthy merchants
    and ruling prince. Ordinary women-men, farmers, traders, soldiers and artisans made liberal donations for chaityas. .
  6. The Gopuram tower acted as a beacon or pointer for the location of village, acting like a light house during the night. They also maintain hospitals and made provisions for medicines.

In chaityas we can see simple structure – a small room, Mandapa, Shikara, etc..

Question 6.
Why are the Buddhist Stupas and Chaityas are sacred ?
Answer:

  1. Buddhist Stupas and Chaityas are considered to be sacred.
  2. In the centre of the stupa, relics of Buddha or other important monks were buried.
  3. That is why it is considered to be sacred.
  4. Pilgrims worship the stupa by offering flowers, etc. and going around it in pradakshina and meditating in front of it.
  5. Chaitya is a small shrine (in viharas) containing a stupa or an image of Buddha and is used as prayer hall for monks.
  6. That is why it is also considered to be sacred.

Question 7.
How did the Buddhist monks probably use the Viharas and Chaityas ?
Answer:

  1. Viharas were the monasteries or living places of Buddhist monks where religious education was imparted.
  2. A vihara usually had small rooms around an open courtyard and had a small shrine containing a stupa or an image of the Buddha at one end.
  3. In viharas, monks taught Buddhist scriptures to disciples.
  4. Apart from imparting education in viharas, monks were also treating the physical ailments of people.
  5. ‘Ordinary men and women, farmers, soldiers and artisans made liberal donations for building and expenses of viharas.
  6. Traders, artisans and kings of the time donated generally to build a unique kind of chaityas and viharas.

TS 6th Class Social 20th Lesson Questions and Answers Telangana - Sculptures and Buildings

Question 8.
Locate Buddhist and Jain sites on the outline map of TelanganAnswer:
Answer:
Nelakondapalli – Khammam District
TS 6th Class Social 20th Lesson Questions and Answers Telangana - Sculptures and Buildings 1

Question 9.
Read the first para of Text Page : 170 (178) (Vihara’s were the for the monks) and comment on it.
Answer:
Viharas were the monasteries or living places of Buddhist monks where religious education vyas imparted..A vihara usually had small rooms around an open courtyard, and had a small shrine containing a stupa or an image of the Buddha at one end. This shrine was called a chaitya, which was also a prayer hall for the monks.
Answer:
Viharas were the Buddhist monasteries. Buddhist religious teaching were taught there. These viharas had small rooms and a ‘Chait.ya’ (stupa) with a open court yard. These were used for both education and prayers. Mostly the wandering monks used to stay here for several months for education. These play a key role in spreading the Buddhism.

I. Conceptual Understanding

Question 1.
Do you feel that the lions in Sarnath Lion Capital look natural or do they look artificially posing for the sculpture ?
Answer:
I feel that the lions look natural. These four majestic lions facing the four directions hold on their head the wheel of DharmAnswer: These pillars and the lion capital represent the power and majesty of the Mauryan Emperor.

Question 2.
Mention the structural details of Nagaijunakonda Stupa.
Answer:

  1. The Nagarjunakonda Stupa is one of the famous Buddhist stupas.
  2. It was the capital city of Ikshvaku kings.
  3. It was on the bank of the Krishna river.
  4. Vijayapuri had a large number of Stupas, Viharas and palaces.
  5. It also had a beautifully designed ampitheatre and ghats leading to the river.
  6. Unfortunately today the entire city lies under water – submerged in the Nagarjuna- sagar dam.
  7. A large number of sculptures and other remains have been removed and kept in a museum nearby.
  8. These sculpture panels represent the earliest sculptural activity in South India.

TS 6th Class Social 20th Lesson Questions and Answers Telangana - Sculptures and Buildings

Question 3.
Explain earliest surviving stone temples at Sanchi in Madhya Pradesh.
Answer:
This is a very simple structure, with just a small room (called garbhagriha) in which the idol of the Buddha was kept, and a small open porch or mandapa with no walls but only pillars. The door and the pillars are beautifully carved. However, the temple has a flat roof without any tower or shikhara on it. This was built around 1600 years ago.

Question 4.
What did Hu-en-Tsang and others describe about “Nalanda”?
Answer:
Hu-En-Tsang and others Chinese pilgrims spent time studying in Nalanda (Bihar) the most famous Buddhist monastery of the period. This is how he describes it:

“The teachers are men of the highest ability and talent. They follow the teachings of the Buddha in all sincerity. The rulers of the monastery are strict, and everyone has to follow them. Discussions are held throughout the day, and the old and the young , mutually help one another. Learned men from different cities come here to clear their doubts. The gate keeper asks new entrants difficult questions. They are allowed to enter only after they have to answered them. Seven or eight out Of every ten are not able to answer.”

Question 5.
Look ait the sculpture panel from Kanaganahalli and identify the features that distinguish the king from other people.
Answer:
The king has a turban like crown on his head. He has some ear rings and some metal ornaments around his wrist and hands. He has some mark of ‘Vibhudhi’ or ‘Tilak’ on his forehead. There is a ‘belt’ like thing around his neck and chest. No other person in the picture has such things.

TS 6th Class Social 20th Lesson Questions and Answers Telangana - Sculptures and Buildings

Question 6.
What are the salient features of ‘Harappan culture’ ?
Answer:

  1. ‘Harappan culture or ‘Indus Valley civilisation progressed in the plains watered by river Indus and its tributaries.
  2. It was predominantly city cultured i.e. people do not depend on agriculture or hunting gathering for livelihoods.
  3. Buildings here were of baked bricks of standard size and shape.’
  4. They used seals and weights and a pictographic language.
  5. They developed an organised drainage system.
  6. They used objects made of different kinds of metals like copper, silver, tin etc. and bronze.
  7. They prayed Mother Goddess, Pashupati, etc.
  8. They organised rituals society as a whole.

Question 7.
Do you think Telangana was a part of the Harappan culture ? Would the people living in Telangana have known about the Harappan people ? How ?
Answer:

  1. I don’t think Telangana was a part of the Harappan culture.
  2. It was not drained by either Indus or its tributaries.
  3. The people living in Telangana might have known about Harappan culture.
  4. There were trade relations between Northern and Southern parts-of India from ancient times.
  5. This might have made them to know about the Harappan culture.

II. Reading the Text (given), Understanding and Interpretation

Can you compare a stupa with a temple and tell what the similarities and the differences are between them and how people worship in them?
Temples are the greatest socio-religious institutions created in ancient indiAnswer: The whole of South India is dotted with temples. The following are the differences between stupa and temple.

TempleStupa
1. The temple was not merely a place of worship, it was also the hub of the social, religious cultural, economic and political life of the community.1. It is supposed to represent the BuddhAnswer: At the same time it is believed that the dome of stupa represents the universe with the ‘ central pillar being the axis around which everything revolves and which connects the world below the earth and the heavens.
2. The temple was conceived as a place not only for worship, but as a centre of learning also.2. Pilgrims worship the stupa by offering flowers etc., and going round it in Pradakshina and meditating in front of it.
3. The Sabha Mandapa or the Vyakhyana Mandapa was for the exposition of ethical texts.3. A stupa is built on a platform. In the centre of stupa were buried relics of the Buddha or important monks.
4. The daily recitation of the sacred texts imparted ethical education to the congregation which gathered of learning various branches were taught.4. Apart from imparting education in viharas, monks were also treating the physical ailments of people.
5. It was the great seat of social intercourse.5. Discussions are held throughout the day, and the old and the young mutually help one another.
6. The temples also played a major role in economic life of the community.6. The stupas also played a major role at the time of Ashoka.

Similarities:

  1. Both of them represent the universe and worship the god.
  2. Both of them play a major role in the economic life of the community.

Question 2.
Why did the Stupa of Amaravathi in Guntur district become ‘famous among other Stupas?
Answer:

  1. Amaravati Stupa in Guntur district is the most famous one in Andhra Pradesh.
  2. This was built during the rule of Satavahanas about 1900 years ago.
  3. The flying figures on the top of the stupa are supposed to be gods from heavens who came to worship the BuddhAnswer:
  4. At the bottom, you can see some men and women also bowing down and worshipping.
  5. We can also see the four lions on the gateway.
  6. Today, the Amaravathi stupa is just a mound of rubble.
  7. All that we have to imagine is how this great stupa was with the help of the panels which show us how it must have looked 1900 years ago.
  8. As you can see from the picture the stupa was covered with panels of sculptures depicting the Buddha and his teachings.
  9. These panels were found by the British and were taken away by them to London.
  10. Some of the panels that could not’be transported werekept in Madras museum.

III. Information Skills

1. Read the following passage.
Traders, artisans and kings of the time of the Satavahanas donated generously to build unique kinds of chaityas and viharas in several places in Maharashtra like Karle, Bhaja, Kanheri and Nasik. These were built like caves on hill sides. Stone workers, carved rocks and reproduced the structures made by the carpenters with wood. They made large prayer halls, stupas and small rooms for monks – all made in rock.They also carved beautiful images of people and animals and some times of the Buddha The Karle Chaitya was built about 2100 years ago.

Answer the following questions:
1) Who donated generously to build a unique kind of chaityas and viharas ?
Answer:
Traders, artisans and kings of the time Satavahanas donated generously to build a unique kind of chaityas and viharas.

2) Name the chaityas and viharas in MaharashtrAnswer:
Answer:
Karle, Bhaja, Kanheri and Nasik.

TS 6th Class Social 20th Lesson Questions and Answers Telangana - Sculptures and Buildings

3) What did the Stone workers use to carve living rocks ?
Answer:
Chisel and hammer.

4) What did they make with the rock?
Answer:
They made large prayer halls, stupas and small rooms for monks.

5) When was Karle Chaitya built ?
Answer:
It was built about 2100 years ago.

IV. Reflection on Contemporary Issues and Questioning

Question 1.
Who were the Chinese pilgrims attracted by the monasteries ?
Answer:
Chinese pilgrims Fa-hi-an, Itsing and Hu-En-Tsang came to visit places associated with the life of Buddha as well as famous monasteries.
Pilgrims were the people who took up journeys to holy places to offer worship. Hu-En-Tsang and others spent time studying in Nalanda (Bihar) the most famous • monastery of the period.

Question 2.
Explain the sculpture panels of Nagaijunakonda and Amaravathi.
Answer:
Introduction : As time went on sculptors started making large and complete figures instead of just the reliefs. Large statues of the Buddha tried to give people an idea of the peaceful, calm and serene personality of the BuddhAnswer: The most famous sculptures of the Buddha come from Gandhara in the Northwest, Mathura and Saranath in Uttar Pradesh. We also can see some of these sclptures in NagarjunakondAnswer: These sculpture panels represent the earliest sculptural acitivity in South India You can get a glimpse of people of those times from them.

  1. Nagarjuna Konda Stupa was the part of Vijayapuri, the capital city of Ikshavaku kings. It was on the banks of the Krishna river. Vijayapuri, had a large number of stupas and viharas and palaces.
  2. Amaravati Stupa in Guntur district is the most famous one in Andhra Pradesh. This was built during the rule of Satavahanas about 1900 years ago. The flying figures of Stupa are supported to be gods from heavens who came to worship the Buddha

Question 3.
Where did the archaeologists find remains of stupas ?
Answer:
Archaeologists have found remains of stupas from very early times in several places of Andhra Pradesh like Amaravati, Bhattiprolu, Ramatirtham, Salihundam etc. At Bhattiprolu stupa was found a crystal casket containing of the relics of the BuddhAnswer: Amaravati Stupa in Guntur district, Nagarjunakonda Stupa at Vijayapuri are the most famous stupas in Andhra Pradesh.

TS 6th Class Social 20th Lesson Questions and Answers Telangana - Sculptures and Buildings

Question 4.
Observe the dresses of men and women in the sculptures and draw them in a notebook. How are they different from the dresses we wear today ?
Answer:
They wore tight, short clothes to cover their lower body. But today we wear clothes to cover our full body.

VI. Appreciation and Sensitivitve

Question 1.
Explain the meanings of stupa.
Answer:
The stupa has many meanings. It is supposed to represent the BuddhAnswer: At the same time it is believed that the dome of the stupa represents the universe with the central pillar being the axis around which everything revolves and which connects the world below, the earth and the heavens. Pilgrims worship the stupa by offering flowers etc., and going round it in pradakshina and meditating in front of it.

Project:

Question 1.
Visit a place of worship in your village or town and draw a sketch of the structure. – Try to get the name of each part, its meaning and use. Prepare a report about this place of worship and conduct an exhibition in your class.
Answer:
Student’s Self Exercise.

TS 6th Class Social 20th Lesson Notes – Sculptures and Buildings

  • Chaitya : A Buddhist hall containing a stupa as the object of veneration. It is the prayer hall for the monks.
  • Vihara : It was the residence of Buddhist / Jain monks. These are the monasteries or living places of Buddhist monks where religious education was imparted. Relics : Relics are remains of body like teeth, bone, hair etc.
  • Monks : Members of a religious group of men who often live apart from other
    people in a monastery and who don’t marry or have personal possessions.
  • Stupa : A Stupa is built on a platform. In the centre of a stupa were buried reljcs of the Buddha or important monks. It is usally surrounded by a stone fence with decorated gateways.
  • Monastery: It is a building in which Buddhist monks live together.
  • Sculpture : A work of art that is solid figure or object made by carving or shaping wood, stone, clay, metal etc.
  • Brahmi : It is the script used by AshokAnswer: Most of the scripts we use in India are in Brahmi script.
  • Script : A written text of a play, film/movie, broadcast, talk etc.
  • Pilgrims : Pilgrims are the people who take up journeys to holy places to offer worship.

Mind Mapping:

TS 6th Class Social 20th Lesson Questions and Answers Telangana - Sculptures and Buildings 3

TS 6th Class Social 19th Lesson Questions and Answers Telangana – Language, Writing and Great Books

Telangana SCERT 6th Class Social Study Material Telangana 19th Lesson – Language, Writing and Great Books Textbook Questions and Answers.

Language, Writing and Great Books – TS 6th Class Social 18th Lesson Questions and Answers Telangana

Question 1.
What is the importance of languages ?
Answer:
Language has become so much a part of our lives that we seldom stop to think about it. We think and understand with the help of language.
The importance of language :

  1. First of all we use language to help us to organise and plan our own work.
  2. Secondly we use language to attract the attention of others and also ask them to do
    something on the other. .
  3. Thirdly, we use language to create new things and just to have fun.
  4. Fourthly, we try to understand the world around us with the help of language.
  5. Fifthly, we use language to tell others about what we feel and experience.
  6. Language has helped people to pass to their children information about what they saw, what they learnt and what they felt.
  7. Language brings together people who are far away from each other.

TS 6th Class Social 19th Lesson Questions and Answers Telangana - Language, Writing and Great Books

Question 2.
How can you say that Aryabhatta was the father of Astronomy ?
Answer:
Many people felt that the stars, planets sun and moon contained secrets about the entire universe and we should understand them better by observing them more carefully calculating their movements. One such person was Aryabhatta who wrote his book called Aryabhattiyam in which he said that the earth rotates causing day and night and that actually the sun did not go around the earth people did not agree with him in those days.

Question 3.
Differentiate between Charaka Samhita and Sushruta Samhita.
Answer:

  1. In earlier times there were many people who studied why people fell ill, and how they can be cursed how people hurt in wars could be healed.
  2. They collected herbs and other things and prepared medicines and 4Tso wrote about these medicines and how to lead healthy lives.
Charaka SamhitaSushruta Samhita
1. It denotes suitable medicine prepared by herbs for illness.1. It focuses on Surgery,
2. It denotes why people fell ill, and how they can be cured.2. It reveals the illness through surgery,

Question 4.
Mention a few inventions in Mathematics.
Answer:

  1. Our ancestors were fascinated by the heavenly bodies like the sun and moon, the stars and the planets.
  2. They observed them day after day and noted the things in the sky kept changing over time. *
  3. They noted all these carefully and found out that the skies had two or three distinct kind of bodies.
  4. Aryabhatta wrote his book called Aryabhattiyam in which he said that the earth rotates causing day and night and that actually the sun didn’t go around the earth.
  5. Aryabhatta and another mathematicians of ancient India also adopted the decimal system of numbers and place value system writing numbers the way we do today.

Question 5.
Look at a currency note and write down the different scripts on them. Identify the language. Is the same script used for different languages ? Which are they ?
Answer:
Different languages written or printed on,the currency note are 15.

  1. Assamese,
  2. Bengali,
  3. Gujarati,
  4. Kannada,
  5. Kashmiri,
  6. Konkani,
  7. Malayalam,
  8. Marathi,
  9. Maithili,
  10. Oriya,
  11. Punjabi,
  12. Sanskrit,
  13. Tamil,
  14. Telugu
  15. Urdu.

Apart from these Hindi is also printed. Different languages using same script are
a) Assamese, Bengali,
b) Hindi, Sanskrit, Konkani, Marathi, Maithili
c) Kashmiri, Urdu.

Question 6.
Refer to any general knowledge book, list out and tabulate five great books in Telugu language and other languages.
Answer:

TeluguSanskritTamil
1) Mahabharatam1) Mahabharata1) Tolkappiyam
2) Ramayanam2) Ramayana2) Tirukkural
3) Bhagavatam3) Asthadhyayi3) Silappadikaram
4) Shringara Naishadam4) Amarakosha4) Manimekalai
5) Dashakumaracharitam5) Buddhacharita5) Ramayan

TS 6th Class Social 19th Lesson Questions and Answers Telangana - Language, Writing and Great Books

Question 7.
How did the work done in various fields by the ancient Indian scientists inspire the present time scientists ?

  1. The work done in various fields by the ancient Indian scientists inspired the present time scientists in many ways.
  2. The books of Charaka Samhita and Sushruta Sainhita laid foundations for. Ayurveda and many scientists are working in the field.
  3. Books on astronomy and mathematics by Aryabhatta, Varahamihira, etc. are of great help to present day scientists for their inventions.
  4. Apart from these there are many inventions in botany, physical science, organic and inorganic chemistry at present, vow their roots to ancient Indian scientists works.

Question 8.
Locate the following in the map India.
a) Bihar
b) Tamil Nadu
c) Uttar Pradesh
d) Maharashtra
e) West Bengal
Answer:
TS 6th Class Social 19th Lesson Questions and Answers Telangana - Language, Writing and Great Books 2

Question 9.
Read the para under the title The great Epics’ of page 146 (165) and comment on it.
Epics are deeds of heroic and legendary figures. They describe conflicts between good and evil, dharma ’ and ‘adharmaEvery country has its own epics and in India, we have two important epics – the Rarnayana and the Mahabharata. The Ramayana is the story of Rama and Sita and the fight between Rama and Ravana. The Ramayana tries to portray an ideal human being – an ideal son, an ideal wife, an ideal brother, an ideal ruler etc. This epic, composed by Valmiki, is considered to be the first Kavya in Sanskrit literature. Initially, it was recited orally by wandering preachers and later on edited and written down.
Answer:
The Ramayana and the Mahabharata are the two great epics of our country. The Ramayana is the very soul of IndiAnswer:It is a complete guide to God realization, the path of nighteousness. The ideas of man are beautifully portrayed in it. Sage Valmiki wrote Ramayana

The Mahabharata was written by famous sage Vedavyas. This epic talks about the importance of following one’s assigned duties in life. It is a brilliant portrayal of ancient politics. It also tells us about the Lord Krishna.

I. Conceptual Understanding

Question 1.
What is the Indo Aryan family ?
Answer:
There is another language family in India called the Indo – Aryan family. The famous languages of this family are Sanskrit, Hindi, Bengali, Marathi etc. There are some other language families in IndiAnswer:As people travelled and mixed with each other their languages too mixed with each other and people adopted several words from each other’s languages. Thus in Teiugu today uses many words taken from Sanskrit, Marathi, Arabic, Persian and English.

Question 2.
Write about Sangam literature in Tamil.
Answer:

  1. Sangam refers to the assembly of scholars and poets in Madurai, the capital of the Pandyas.
  2. Tamil was the language that was spoken in a large part of South India in the early days.
  3. Sangam literature is a collection of poems composed by a large number of poets and poetesses.
  4. They write about heroes who bravely fought wars in defence of their tribes, about the love of brave heroes and beautiful girls, about the beauty of the landscapes and seasons and their effect on human beings.
  5. This Sangam literature refers to the history and culture of ancient kingdoms of Cheras, Cholas and Pandyas.

Question 3.
Narrate the two great Epics – Ramayana and Mahabharata,
Answer:
Epics are deeds of great heroes and heroines. They describe conflicts between heroes and evil forces. Every country lias its epic and in India we have two important epics – Ramayana and MahabharatAnswer:Ramayana is the story of Rama and Sita and the fight between Rama and Ravan who had carried off Sita to LankAnswer:This epic is considered the first kavya in Sanskrit literature and composed by Vaimiki. Initially it was related orally by wandering preachers and later on edited and written down.

Mahabharata is the story of the conflict between two sets of cousins (Kauravas and Pandavas) over who should rule the kingdom, ft finally ended in a great war which was won by Pandavas with the help of Krishna. It is one of the largest poems in the world and contains hundreds of small stories. This was composed by Vyasa and again recited orally by wandering story tellers and eventually written down.

TS 6th Class Social 19th Lesson Questions and Answers Telangana - Language, Writing and Great Books

Question 4.
What did the people write it on ?
Answer:

  1. In the beginning people probably wrote on cloth, leaves, barks, etc. which decayed and got destroyed with time.
  2. Some people even wrote on pots by scratching on them.
  3. Some of these pots have survived.
  4. However, the earliest waiting of a long text that Has survived to our own time in Ashoka’s inscriptions which were engraved on rocks or stone pillars.
  5. In many parts of South India people wrote on palm leaves which were cut to a definite size.
  6. They used a pin like pen to scratch on the dried leaf and coloured it with black ink.
  7. In North India they used the barks of Bhoja trees which grew in the Himalayas.
  8. These barks peeled off just like paper and one could easily write on them using ink. In those days paper was not in use.

Question 5.
What are the different types of scripts mentioned in this lesson ?
Answer:

  1. We write and read languages with the help of scripts.
  2. We use many kinds of scripts. We use a) Roman script b) Devanagari script c) Arabic script d) Telugu script and e) Tamil script.
  3. Actually it took human beings thousands of years to develop scripts like the ones we use today.
  4. In the beginning they just drew pictures instead of writing. To say that goat is running they would make a picture of a goat and another picture of two running legs.
  5. Slowly the scripts as we know them developed.
  6. Four thousand years ago the people of Indus Valley used scripts, but we don’t know how to read them.

II. Reading the Text (given), Understanding and Interpretation

Question 1.
Why are there so many languages ?
Answer:

  1. Language is developed by a group of people living together. They decide to call the drink that quenches their thirst as water.
  2. Another group of people who may be living far away from the first group may decide to call the same drink as NEELLU. Another group may call it as THANNI.
  3. They mean the same but they use different sounds or symbols for it.
  4. That is why people developed different languages such as English, Sanskrit, Persian, Chinese, Swahili, Hindi, Telugu etc.
  5. Often it also happened that one group of people who spoke a language got separated into different sub groups and each group in turn developed different languages based on the old language.
  6. We say that all these languages belong to one family.
  7. Thus, originally a group of people spoke a language which today we call early Dravidian.
  8. When the people who spoke this language began to live in distant places and mixed with other people, they developed new languages.
  9. Among them are Tamil, Telugu, Kannada, Malayalam, gondi etc. These are today mainly spoken in South India.

Question 2.
Explain about oral literature.
Answer:

  1. Many of stories were not written down but told by one generation to another generation to another just like our parents told us. These are called oral literature.
  2. That is literature which has been passed from mouth to mouth.
  3. This was the earliest form of literature.
  4. The earliest people composed poems, songs and stories and told them to their children who in turn told their children.
  5. They also passed on their understanding of the world through sayings and proverbs which too were similarly taught.

Question 3.
Why were Jataka tales delightful stories which we may like to read ?
Answer:
Gautama Buddha was the founder of Buddhism. His followers wrote many stories about these births of Buddha and compiled them in the form of Jataka stories. These are delightful stories which we may like to read. These too were collected and written down around 1600-1800 years ago.

III. Information Skills

1. Read the following passage and answer the following questions.

Historians believe that even though these epics were composed much earlier they were given their final form about 1600 years ago.
You had read about Gautama Buddha in an earlier lesson. His followers believed that he had been born several times earlier to preach dharma to fellow human beings and animals.They wrote many stories about Buddha’s previous births and compiled them in the form of Jataka stories. These are delightful stories, which you might enjoy reading. These too were collected and written down around 1600-1800 years ago.

Anwer the following questions :

1) When were the epics given final form ?
Answer:
They were given their final form about 1600 years ago.

2) Who was the founder of Buddhism ?
Answer:
Gautama Buddha was the founder of Buddhism.

3) What was the belief of Buddha’s followers ?
Answer:
His followers, believed that he had taken several births earlier to preach dharma to fellow human beings and animals. .

4) When were Jataka tales written down ?
Answer:
Jataka stories were collected and written down around 1600-1800 years ago.

IV. Reflection on Contemporary Issues and Questioning

Question 1.
What was the Buddha trying to teach sorrowful mother ? (By the story of KisagotamJ)
Answer:

  1. Once there was a woman named Kisagotami, whose son had died.
  2. She was so sad that she roamed through the streets of the city carrying the child with her asking for help to bring him back to life.
  3. A kind man took her to BuddhAnswer:Then Buddha asked her to bring him a handful of mustard seeds, and he will bring her child back to life.
    4) He also said the seeds must come from the, house of a family where nobody has died. ‘
    5) She couldn’t bring seeds any where.
    6) She returned with sad. Then Buddha taught her death is common thing as birth. No one can escape from dfeath. Man is mortal.

TS 6th Class Social 19th Lesson Questions and Answers Telangana - Language, Writing and Great Books

Question 2.
Why is Ramayana considered the first Kavya ?
Answer:

  1. Every country has its epic and in India we have two important epics Ramayana and Mahabharata.
  2. Ramayana is the story of Rama and Sita and the fight between Rama and Ravan who had carried off Sita to Lanka.
  3. This epic is considered the first Kavya in Sanskrit literature.
  4. And it was composed by Valmiki.
  5. Initially it was related orally by wandering preachers and later on edited and written down.

VI. Appreciation and Sensitivity

Question 1.
Explain, how human beings were developed scripts like the ones we use today?
Answer:
In the beginning they just drew pictures instead of writing. To say that the goat is ‘ running, they would make a picture of a goat and another picture of two running legs! Slowly the scripts as we know them developed. Four thousand years ago the people of Indus Valley used scripts, but we don’t’know how to read them.
मेरा नाम रघु – Hindi
నా పేరు రఘు – Telugu

Project:

Question 1.
Listen to different languages on TV or radio and try to think of the similarities and differences.
Answer:
Student’s Self Exercise.

TS 6th Class Social 19th Lesson Notes – Language, Writing and Great Books

  • Epic – Epics are deeds of great heroes and heroines. They describe conflicts between heroes and evil forces. Every country has its epic and in India we have two important epics – Ramayana and Mahabharata.
  • inscription – a thing inscribed, as on a monumental or in a book.
  • Ayurveda – Ayurveda is a type of traditional Hindu medicine that treats illness using a combination of foods, herbs and breathing exercises.
  • Sangam Literature – Sangam literature is a collection of poems composed by a large number of poets and poetesses. They write about heroes who bravely fought was indefence of their tribes.
  • Decimal system – It is based on or counted in tens or tenths etc.
  • Language – The system of communication in speech and writing that is used by people of particular country or area.

Mind Mapping:

TS 6th Class Social 19th Lesson Questions and Answers Telangana - Language, Writing and Great Books 1

 

TS 6th Class Social 18th Lesson Questions and Answers Telangana – Devotion and Love towards God

Telangana SCERT 6th Class Social Study Material Telangana 18th Lesson – Devotion and Love towards God Textbook Questions and Answers.

Devotion and Love towards God – TS 6th Class Social 18th Lesson Questions and Answers Telangana

Question 1.
Imagine a discussion between a Buddhist monk and a hunter – gatherer and how they learnt from each other. Write a short dialogue between them.
Answer:
Buddhist Monk : Buddham Saranam Gachchami
Dharmam Saranami
Saranam Gachchami

Hunter-gatherer : Sami, move aside, I have to hit that bird.

B.M : No man, it is sin. You should not do such‘himsa’.

H.G : Then how can we get our daily food ? Some birds and animals are born for our food. And it is my

B.M : The God gave us many plants, fruits, vegetables and grains for our food. So we should not kill them. Even though, they cannot express their feelings, they curse us in their thoughts.

H.G : Let them curse. Nothing will happen to us. Leave me Sami, let me do my work. B.M : Let me complete my son. We believe in one universal rule – “what we sow, so we reap.” If we do harm to others, in return we have to receive it in double. So don’t kill them. Let them be happy in the nature. Then you will be happy in the world.

H.G : Is it true Sami ? Oh! I am doing mistakes for all these years. How can I come out from this ?

B.M : Your repentance is enough. Say the three jewels
Buddham Saranam Gachchami
Dharmam Saranam Gachchami
Sangham Saranam Gachchami
Every sin will be vanished: But remember one thing, repeating the mistakes is not good for you.

H.G : No Sami, I will never do it again. I will work in some other field and feed my family. Namaste Sami Namaste.
Buddham Saranam Gachchami
Dharmam Saranam Gachchami
Sangham Saranam Gachchami
In this conversation the Buddhist monk learnt that ‘need’ leads the human behaviour. Hunter-gatherer learnt that ‘one should bear one’s sins.’

TS 6th Class Social 18th Lesson Questions and Answers Telangana - Devotion and Love towards God

Question 2.
Can you see any similarities and differences between the early and the Vedic reli¬gion ?
Answer:
Similarities:

  1. Both religions believed in rebirth.
  2. Both worshipped dead ancestors.
  3. Both believed in one supreme God.
  4. Both worshipped animals and plants.
  5. Both performed idol worship.
Early religionVedic religion
1. They worshipped mother goddess, . snakes, animals, trees and dead ancestors.1. They performed yagnas and yagas with prayers and pithru karmas.
2. They do not know any indestructible being like AtmAnswer:2. They perform Tapasya to know ‘atma’ or brahman.
3. The religious beliefs were carried to next generations by practicing.3. They had many discussions regarding several questions and improved their religion.

Question 3.
In what ways were the early Bhagavata and Shaiva thinking different from that of the Buddhists and Jains ?
Answer:

  1. Buddhists or Jains or Upanishadic thinkers did not emphasise worship of any god as the road to salvation or mokshAnswer:
  2. They advocated meditation, abandoning desires or purifying oneself through penance etc.
  3. But the Bhagavatas believed that intense devotion and worship of Vishnu alone could relieve us from all problems and would lead to salvation.
  4. They thought that yagnas or giving alms to Brahmins etc., or even reciting the Vedas was not necessary. They built temples and worshipped images of KrishnAnswer:
  5. Shaivas too built small temples with images of Shiva or a lingam. Some people also worshipped goddess Durga in the same manner.
  6. Even as Vishnu or Shiva or Durga was being worshipped as Supreme God, the Buddhists and Jainas also began worshipping the Buddha and the Tirthankaras.
  7. They made beautiful images of the Buddha and the Tirthankaras and set them up in stupas or temples called Chaityas. Eg : Amaravati, Nagarjuna Konda,Phanigiri and Nelakondapalli.
  8. Even though Buddha or Mahavira did not claim to be gods, they began to be treated as supreme god’s representatives by now.

TS 6th Class Social 18th Lesson Questions and Answers Telangana - Devotion and Love towards God

Question 4.
Discuss how the Puranas helped in bringing together different religious practices in the country.
Answer:

  1. The stories about Vishnu or Shiva were written down as Puranas.
  2. These Puranas, besides combining various religious traditions also extolled one or the other god as the Supreme God who was all powerful, all knowing etc.
  3. They advocated worshipping of the God as the easy way to solve one’s problems and attain salvation.

Question 5.
Can you point out the most important difference between the early beliefs and the beliefs of the Bhaktas of Tamilnadu ?
Answer:

  1. About 550 CE in Tamilnadu there arose a new movement of devotion to God.
  2. They were the followers of Vishnu and ShivAnswer:
  3. They did not believe in performance of any elaborate ritual like yagnas or animal sacrifices.
  4. They did not pray for solving problems or for riches or power, but for union with God.
  5. These above two are important differences between the early beliefs and the beliefs of the Bhaktas of Tamilnadu

Question 6.
In what way did the prophet Mohammad explain the equality of all human beings?
Answer:

  1. The founder of Islam was Prophet Mohammad.
  2. He was born in Arabia at Mecca in 570 CE.
  3. He opined that the way all the children are equal before their parents, all human beings are equal before god.
  4. God or Allah has no shape and therefore it is wrong to worship idols.
  5. He emphasised the importance of love for whole of humanity.

Question 7.
Read the para under the title ‘The belief in a Supreme God’ of page 136 (160) and write your comment.

The Belief in a Supreme God

Can you see some common ideas emerging among the Hindu Bhaktas, Christian and Muslim religions ? They all believed in the existence of one Supreme God and the need to worship Him. They also believed in equality, love and respect for all human beings irrespective of their wealth, education or social status. These ideas were increasingly adopted by a very large number of people. Kings and rulers also adopted these ideas and started encouraging their subjects to practice these religions. They built vast temples, churches or mosques in which people could worship.
Answer:
The belief in a ‘Supreme God’. This can also be called as Monotheism. This belief was developed in all the religions. Many rulers and people adopted this belief. This means that there is only one supreme reality. To practice this many temples, churches and mosques were built.

TS 6th Class Social 18th Lesson Questions and Answers Telangana - Devotion and Love towards God

Question 8.
Write your view on the religious programme held recently in your areAnswer:
Answer:
In our village ‘Sripalle’ a religious programme was held on Karthika pournami. My father Sir., Yellomraju Madhava Rao and my mother Smt. Rajyalakshmi celebrated this with our relatives and villagers. We performed ‘Abishekams’ and ‘Lakshapathri Pooja’ to Lord SivAnswer: On the next day we performed ‘Chandi Homam’ in the temple. All the villagers gathered there and performed Gowri Pooja also. It was really a wonderful occasion. The temple was decorated with leaves, flowers and lights. Vedic Chantings were continued for two days. To say… 1 feel the presence of the Goddess in the ‘Homam’. At the end ‘Santarpana’ was organized with the help of the committee members and villagers.

Question 9.
Point out the following in the world map.
a) Jerusalem
b) Mecca
c) Kerala State
d) Chennai
e) Amaravati
TS 6th Class Social 18th Lesson Questions and Answers Telangana - Devotion and Love towards God 2

I. Conceptual Understanding

Question 1.
Who are Alvars and Navanars ?
Answer:
a) Alvars : Twelve main devotees of Vishnu were called the Alvars – the most important among them being Periyalvar and Nammalvar. Andal the daughter of Periyalvar is the only woman among them. Most of them travelled to different temples of Vishnu and composed and sang songs called Pasurams. Nammalvar considers Vishnu to be the creator of the universe,

b) Nayanars : The Navanars were 63 in number and were devotees of ShivAnswer: They came from different castes, including Kannappa, a hunter and Nandanar, a Pulaiya, or an outcaste. Some of them like Appar, Sambandar, Sundarar and Manikkavasagar went from one temple to another and composed songs in praise of ShivAnswer: A couple of them like Karaikkal Arraiyar were women.

TS 6th Class Social 18th Lesson Questions and Answers Telangana - Devotion and Love towards God

Question 2.
Why does Nammalvar feel that God is playing hide and seek ?
Answer:
In his song “You dwell in heaven ………… will you always play hide and seek?”
Nammalvar is pointing out that the Lord is present everywhere but he is not appearing before him. He is desperately seeking a vision of God, but the God is not showing himself to Nammalvar. He is unable to have a glimpse of God.

Question 3.
Write a short notes on ‘Christianity’.
Answer:

  1. Just as people in the Indian subcontinent started worshipping a Supreme God and believed that love and devotion to God is the path for salvation, people in other parts of the world too developed similar ideas.
  2. In the Middle East in the region now called Israel and Palestine developed a new religion called Christianity started for Jesus Christ.
  3. He was born at Bethlehem near Jerusalem about 2000 years ago.
  4. Jesus Christ preached that all people are equal. He taught us to love one another. He taught the importance of peace, love and compassion.
  5. Saint Thomas a follower of Jesus Christ came to India with Roman traders and brought with him the teachings of Christ. St. Thomas propagated Christianity in South India

Question 4.
Give a brief account of Islam.
Answer:
The Islam religion was founded by Prophet Mohammad. Prophet Mohammad was born in Arabia at Mecca in 570 CE. Prophet taught that there is only one God and all human beings are his creation. The way all the children are equal before their parents all human beings are equal before god. God or Allah has no shape and therefore it is wrong to worship idols. He taught that all men are brothers. He emphasised the importance of love for the whole of humanity. Mohammad is considered as a Prophet or messenger of Allah. The teachings of Allah are written in a book called Quran. It is the holy book of Islam.

Question 5.
What are the common ideas emerging among the Hindus, Christians and Muslims?
Answer:

  1. They all believed in the existence of one Supreme God and the need to worship him.
  2. They also believed in equality, love and respect for all human beings irrespective of their wealth, education or social status.
  3. These ideas were increasingly adopted by a very large number of people.
  4. Kings and rulers also adopted these ideas and started encouraging their subjects to practise these religions.
  5. They built vast temples, churches or mosques in which people could worship.
  6. Kings and emperors also encouraged such ideas and often tried to claim to be representatives of those Supreme Gods themselves.

II. Reading the Text (given), Understanding and Interpretation

Question 1.
Can you compare the religious beliefs of the Vedic period and the Bhakti movement ?
Answer:
The religious beliefs of the Vedic period :

  1. The Aryans believed in the existence of one supreme power as the creator of the entire universe. The gods were represented as great and powerful.
  2. The earliest religious ideas of the Aryans were based on nature worship. Indra, Agni. Surya, Varuna etc., are their gods.
  3. The Gods were worshipped by means of sacrifice and prayers. So yagnas occupied a very important place in the religion of early Vedic period.
  4. In later Vedic period Indra and Varuna lost their pre-eminence. Praja Pathi Brahma came to occupy a supreme position as the creator of the universe. Vishnu came to be regarded as the protector of life. Rama and Krishna were believed to be the incarnations of Vishnu.
    5) Under the influence of the brahman as the cult of sacrifices (yagnas) became the corner stone of Aryan culture. Simple ceremonial worship gave place to elaborate rituals and sacrifices.

The religious beliefs of the Bhakti movement:

  1. The Bhaktas were the followers of Vishnu or Shiva.
  2. They believed in one God “Vishnu”.
  3. Bhaktas did not believe in performance of any elaborate ritual like yagnas or sacrifices.
  4. They believed that intense love for Shiva or Vishnu and a desire for union with him alone was the path for salvation.
  5. All Bhaktas together sang songs of love for the God and danced in jey.

TS 6th Class Social 18th Lesson Questions and Answers Telangana - Devotion and Love towards God

Question 2.
In what way was Nammalvar expressing his intense love for Vishnu ?
Answer:
Nammalvar considers Vishnu to be the creator of the universe, but at the same time talks of him as being his own father and lord. Thus all powerful God is as close as father to his son. In his verse in that original moment (Vishnu), our lord and fath …….. And the gods of rain.” He expressed his intense love for Vishnu.

III. Information Skills

1. Read the following passage and answer the following questions.

About the same time, some other people began to worship Shiva in a similar manner. They too built small temples with images of Shiva or a Lingam. Some people also worshipped goddess Durga in the same manner.

As Vishnu, Shiva and Durga were being worshipped as Supreme Gods, the Buddhists and Jainas also began worshipping the Buddha and the Tirthankaras (Earliest Jaina gurus).They made beautiful images of the Buddha and the Tirthankaras and worshipped , them.These are found in many places like Phanigiri and Nelakondapalli in Telangana and at Amaravati, Nagarjunakonda, Jaggayyapeta, Bhattiprolu in Andhra Pradesh.

1) Where does the images of Buddha found ?
Answer:
They were found in Phanigiri and Nelakondapalli in Telangana and at Amaravathi, Nagarjunakonda, etc. in Andhra Pradesh

2) Who was being worshipped as Supreme God ?
Answer:
Vishnu, Shiva and Durga vas being worshipped as Supreme God.

IV. Reflection on Contemporary Issues and Questioning

Question 1.
What were the important teachings of Jesus Christ ?
Answer:

  1. The Lord Jesus Christ founded the religion “Christianity”.
  2. This religion preached love, kindness and mercy.
  3. It preached repentence, non-violence and truth.
  4. Christ preached that love and faith in God is an early way to the kingdom of God.
  5. God is all powerful in the universe.
  6. Love thy neighbour as thyself.

V. Mapping skills

A) Map Pointing:
Locate the following on the world map :
1) Europe
2) Asia
3) Africa
4) India
5) Mecca
TS 6th Class Social 18th Lesson Questions and Answers Telangana - Devotion and Love towards God 3

B) Map Reading:
Observe the following world map and answer the following questions. African, Asian and European Territories Conquered by Arabs in 7th Century
Answer:
TS 6th Class Social 18th Lesson Questions and Answers Telangana - Devotion and Love towards God 4

1) Where was Prophet Mohammad born ?
Answer:
In Arabia at Mecca
2) Where was Madina ?
Answer:
To the north of Mecca

3) In which continents was the Islam spread ?
Answer: Asia Africa and Europe

4) Where was Spain ?
Answer:
In Europe

5) Where is Sindh now ?
Answer:
In Pakistan

VI. Appreciation and Sensitivity

1. Can you see the similarities between the Bhagavatas, the Shaivas and the Buddhists ?

BhagavatasShaivasBuddhists
They worshipped images of Krishna1. They worshipped with the images of Shiva or a lingam.1. They worshipped Buddha and the Tirthankaras.
They believed that intense devotion of Vishnu alone could relieve us from all problems. .2. They believe that Shiva could relieve us from all problems.2. Buddhists believe that the worship of God as the road to salvation.
Bhagavatas built temples some 2000 years ago in Vidisha in Madhya Pradesh.3. The earliest lingam images of Shiva probably carved more than 2000 years ago.3. Many monasteries were organised in £ dramatic way. Buddhism was essentially a missionary religion.
They built temples in Vidisha in Madhya Pradesh.4. Gudimallam in Chittoor district.4. eg: Amaravati, Nagarjuna Konda, Bhattiprolu, Phanigiri and Nelakondapalli in Telangana.

TS 6th Class Social 18th Lesson Questions and Answers Telangana - Devotion and Love towards God

Question 2.
Can you compare the religious belief of hunter-gatherers and the Bhakti movement and point out the similarities and differences?
Answer:
a) Similarities:

  1. Painting and dance were important parts of the religious lives of the hunter- gatherers. They probably painted hunting scenes or danced to imitate a hunting scene in the belief that this will give them a good hunt. Bhaktas also believed that anyone irrespective of their caste or community could love god. All devotees together sang songs of love for the God and danced in joy.
  2. Hunter-gatherers believed that life continues even after death and hence made there arrangements for the after life of the dead persons.
  3. Instead Bhaktas believed that intense love for Shiva or Vishnu and a desire for union with him alone was the path for salvation.

b) Differences:

Hunter gatherersBhakti Movement
1. They wandered from place to place in search of food.1. They stayed out one place.
2. They believed that the forest and wild animals were sacred and had to be worshipped.2. They believed devotion to a particular deity.
3. They offer diety meat, honey, fruits, flowers to the gods and goddesses of the forest.3. They did not believe in performance of any elaborate ritual like yagnas or animal sacrifices.

3. Can you list the similarities and differences between the religious feelings ?
Answer:
Similarities:

  1. They all believed in the existence of one Supreme God and the need to worship him.
  2. They also believed in equality, love and respect for all human beings irrespective of their wealth education social status.
  3. They built vast temples, churches or mosques in which people could worship.

Differences:

  1. Many people were very unhappy about this growing inequality and the suppression of the poor and the powerless.
  2. Some historians feel that Buddhism and Jainism which emphasised meditation and control over one’s own desires lost their popularity.
  3. Some in fact did not believe in the existence of an all powerful God at all.

TS 6th Class Social 18th Lesson Notes – Devotion and Love towards God

  • Sculptures : Work of art that is solid figure or object made by carving or shaping wood, stone, clay, metal, etc.
  • Incarnations : A period of life in a particular form. This is called avatarAnswer: Vishnu took birth on the Earth in various forms of incarnations to save the world from evil.
  • Salvation : The state of being saved from the power or evils.
  • Jataka Stories : The Buddhist stories were written down as the Jataka stories.
  • Sramanas : The Sramanas of Buddhist and Jaina established Viharas and Stupas and preached the teachings of Buddha or Jaina.
  • Bhakti Movement : About 550 Common Era, in Tamil Nadu there arose a new movement of devotion to God. They were the followers of Vishnu or ShivAnswer: They didn’t believe in performance of any elaborate ritual like yagnas or animal sacrifices. Ramanuja, Ramananda, Kabir, Gurunanak, etc., were prominent saints.
  • Bhagavatas : A new kind of religious belief called BhagavatAnswer: The Bhagavatas worshipped Vishnu or NarayanAnswer: They believed that Vishnu was the ultimate God and creator of universe
  • Tirthankaras : The earliest Jaina gurus are called Tirthankaras. They are spiritual leaders. Vardhamana Mahavira was the 24th Tirthankara
  • Bhaktas : Bhaktas were the followers Vishnu or ShivAnswer: They didn’t believe in performance of any elaborate ritual like yagnas or animal sacrifices.
  • Puranas : The stories about.Vishnu or Shiva were written down as Puranas.
  • Alvars : Tweleve main devotees of Vishnu were called the Alvars – the most important among them being Periyalvar and Nammalvar.
  • Nayanars : The Nayanars were 63 in number and were devotees of ShivAnswer: They came from different castes including Kannappa, a hunter and Nandanar, a Pulaiya or an outcaste
  • Holy books : These are connected with God ora particular religion. The Bible is the holy book of the Christians. Quran is the holy book of Muslims.

Mind Mapping:

TS 6th Class Social 18th Lesson Questions and Answers Telangana - Devotion and Love towards God 1

Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(a)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(a)

I.

Question 1.
If z1 = (2, -1), z2 = (6, 3), find z1 – z2.
Solution:
z1 = (2, -1), z2 = (6, 3)
∴ z1 – z2 = (2 – 6, -1 – 3) = (-4, -4)

Question 2.
If z1 = (3, 5) and z2 = (2, 6), find z1 . z2
Solution:
Given z1 = (3, 5) = 3 + 5i
and z2 = (2, 6) = 2 + 6i
z1 . z2 = (3 + 5i) . (2 + 6i)
= 6 + 10i + 18i + 30i2
= 6 + 28i + 30(-1) [since i2 = -1]
= -24 + 28i
= (-24, 28)

Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(a)

Question 3.
Write the additive inverse of the following complex numbers.
(i) (√3, 5)
(ii) (-6, 5) + (10, -4)
(iii) (2, 1) (-4, 6)
Solution:
The additive inverse of (a, b) is (-a, -b)
(i) The additive inverse of (√3, 5) is (-√3, -5)
(ii) (-6, 5) + (10, -4)
= (-6 + 10, 5 + (-4))
= (4, 1)
∴ The additive inverse of (4, 1) is (-4, -1)
(iii) (2, 1) . (-4, 6)
= ((2 × -4 – 1 × 6), (1 × -4 + 2 × 6))
= (-8 – 6, -4 + 12)
= (-14, 8)
∴ The additive inverse of (-14, 8) is (14, -8)

II.

Question 1.
If z1 = (6, 3); z2 = (2, -1), find z1/z2.
Solution:
Given z1 = (6, 3) = 6 + 3i
and z2 = (2, -1) = 2 – i

Question 2.
If z = (cos θ, sin θ), find (z – \(\frac{1}{z}\))
Solution:
Given z = (cos θ, sin θ) = cos θ + i sin θ
⇒ \(\frac{1}{z}\) = cos θ – i sin θ
∴ z – \(\frac{1}{z}\) = (cos θ + i sin θ) – (cos θ – i sin θ)
= 2 i sin θ
= 0 + i (2 sin θ)
= (0, 2 sin θ)

Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(a)

Question 3.
Write the multiplicative inverse of the following complex numbers.
(i) (3, 4)
(ii) (sin θ, cos θ)
(iii) (7, 24)
(iv) (-2, 1)
Solution:
The multiplicative inverse of the complex number (a, b) is \(\left(\frac{a}{a^{2}+b^{2}}, \frac{-b}{a^{2}+b^{2}}\right)\)
(i) Multiplicative inverse of (3, 4) = \(\left(\frac{3}{3^{2}+4^{2}} \cdot \frac{-4}{3^{2}+4^{2}}\right)\) = \(\left(\frac{3}{25}, \frac{-4}{25}\right)\)
(ii) Multiplicative inverse of (sin θ, cos θ) = \(\left(\frac{\sin \theta}{\sin ^{2} \theta+\cos ^{2} \theta}, \frac{-\cos \theta}{\sin ^{2} \theta+\cos ^{2} \theta}\right)\) = (sin θ, -cos θ)
(iii) Multiplicative inverse of (7, 24) = \(\left(\frac{7}{7^{2}+24^{2}}, \frac{-24}{7^{2}+24^{2}}\right)\) = \(\left(\frac{7}{625}, \frac{-24}{625}\right)\)
(iv) Multiplicative inverse of (-2, 1) = \(\left(\frac{-2}{(-2)^{2}+(1)^{2}}, \frac{-1}{(-2)^{2}+(1)^{2}}\right)\) = \(\left(-\frac{2}{5},-\frac{1}{5}\right)\)

Inter 2nd Year Maths 2B System of Circles Solutions Ex 2(a)

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B System of Circles Solutions Exercise 2(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B System of Circles Solutions Exercise 2(a)

I.

Question 1.
Find k if the following pairs of circles are orthogonal.
i) x² + y² + 2by – k = 0, x² + y² + 2ax + 8=0.
Solution:
g1 = 0; f1 = b; c1 = -k
g2 = a; f2 = 0; c2 = 8
Two circles are said to be orthogonal
2g1g2 + 2f1f2 = c1 + c2
2(0)(a) +2(b)(0)= -k + 8
0 = – k + 8
k = 8

ii) x² + y² – 6x – 8y + 12 = 0;
x² + y² – 4x + 6y + k = 0
Solution:
g1 = -3; f1 = -4; c1 = 12
g2 = -2; f2 = 3; c2 = k
Two circles are said to be orthogonal
2g1g2 + 2f1f2 = c1 + c2
2(-3)(-2) + 2(3)(-4) = 12 + k
+ 12 – 24 = 12+ k ⇒ k = -24

iii) x² + y² – 5x – 14y – 34 = 0,
x² + y² + 2x + 4y + k = 0
Solution:
g1 = \(\frac{-5}{2}\) ; f1 = -7 ; c1 = -34
g2 = 1; f2 = 2; c2 = k
Two circles are said to be orthogonal
2g1g2 + 2f1f2 = c1 + c2
2(\(\frac{-5}{2}\)(1) + 2(-7)(2) = -34 + k
-5 – 28 = -34 + k
– 33 = – 34 + k
k = 34 – 33
⇒ k = 1

iv) x² + y² + 4x + 8 = 0, x² + y² – 16y +k = 0
Solution:
g1 =2 ; f1 = 0 ; c1 = 8
g2 = 0 ; f2 = – 4; c2 = k
Two circles are said to be orthogonal
2g1g2 + 2f1f2 = c1 + c2
2(2)(0) + 2(0)(-8) = 8 + k
0 + 0 = 8 + k
⇒ k = -8

Inter 2nd Year Maths 2B System of Circles Solutions Ex 2(a)

Question 2.
Find the angle between the circles given by the equations.
i) x² + y² – 12x – 6y + 41 = 0,
x² + y² + 4x + 6y – 59= 0.
Solution:
Inter 2nd Year Maths 2B System of Circles Solutions Ex 2(a) 1

ii) x² + y²+ 6x – 10y – 135 = 0, x² + y² – 4x + 14y – 116 = 0.
Solution:
Inter 2nd Year Maths 2B System of Circles Solutions Ex 2(a) 2
Inter 2nd Year Maths 2B System of Circles Solutions Ex 2(a) 3

Question 3.
Show that the angle between the circles x + y = a , x + y = ax + ay is \(\frac{3 \pi}{4}\).
Solution:
Equations of the circles are
S ≡ x² + y² – a² = 0
S’ ≡ x² + y² – ax – ay = 0
Inter 2nd Year Maths 2B System of Circles Solutions Ex 2(a) 4

Question 4.
Show that the circles given by the following equations intersect each other orthogonally.
i) x² + y² – 2x – 2y – 7 = 0,
3x² + 3y² – 8x + 29y = 0.
Solution:
C1 = (1, 1)
g = -1, f = -1, c =-7
g’ = \(\frac{-4}{3}\), f’ = \(\frac{29}{6}\) ; c’ = o
Condition that two circles are orthogonal is
Inter 2nd Year Maths 2B System of Circles Solutions Ex 2(a) 5
∴ -7 = -7
Hence both circles cut orthogonally.

ii) x² + y² + 4x – 2y – 11 = 0,
x² + y² – 4x – 8y + 11 =0.
Solution:
g1 = 2, f1 = -1, c1 = -11
g2 = -2, f2 = -4, c2 = 11
Two circles are said to be orthogonal
2g1g2 + 2f1f2 = c1 + c2
2(2)(-2) + 2(-1)(-4) = -11 + 11
-8 + 8 = 0
∴ Two circles are orthogonal.

iii) x² + y² – 2x + 4y + 4 = 0, .
x² + y² + 3x + 4y + 1 = 0.
Solution:
g = -1, f = 2, c = 4
g’ = \(\frac{3}{2}\), f’= 2, c’= 1
2gg’ + 2ff’ = c + c’
2(-1). \(\frac{3}{2}\) + 2 × 2 × 2 = 4 + 1
-3 + 8 = 5
5 = 5
Hence circles cut orthogonally.

iv) x² + y² – 2lx + g = 0, x² + y² + 2my – g = 0.
Solution:
g1 = -l ; f1 = 0, c1 = g, g2 = 0, f2 = m, c2= -g
2g1g2 + 2f1f2 = c1 + c2 is condition for two circles be orthogonal
2(-1)(0) + 2(0)(m) = g – g
0 = 0
∴ Two circles are orthogonal.

II.

Question 1.
Find the equation of the circle which passes through the origin and intersects the circles below, orthogonally.
i) x² + y² – 4x + 6y + 10 = 0, x² + y² + 12y + 6 = 0.
Solution:
Let equation of circle be
x² + y² + 2gx + 2fy + c = 0 ……… (i)
Above circle passes through origin
∴ c = 0
Circle (i) is orthogonal to
x² + y² – 4x + 6y + 10 = 0 then
2g(-2) + 2f(3) = 0 + 10
-4g + 6f = 10 ………… (ii)
Circle (i) is orthogonal to
x² + y² + 12y + 6 = 0
2g(0) + 2f(6) = 6 + 0
12f = 6
f = \(\frac{1}{2}\) …………… (iii)
Solving (ii) and (iii) we get
– 4g + 6 × \(\frac{1}{2}\) = 10
-4g = 10 – 3
g = –\(\frac{7}{4}\)
∴ Equation of circle be
x² + y² – \(\frac{7}{2}\)x + y = 0
2x² + 2y² – 7x + 2y = 0.

ii) x² + y² – 4x – 6y – 3 = 0, x² + y² – 8y + 12 = 0.
Solution:
Let the equation of the circle be
x² + y² + 2gx + 2fy + c = 0
It cuts
x² + y² – 4x – 6y – 3 = 0; x² + y² – 8y + 12 = 0
g1 = -2, f1 = -3, c1 = -3
g2 = 0 ; f2 = -4, c2 = 12
Let g,f, c be constants of required circle.
Required circle passes through origin
∴ c = 0
Requited circle is orthogonal to both circles.
∴ 2g(-2) +2f(-3) = -3 + 0 …………… (i)
2g(0) + 2f(-4) = 12 + 0 …………….. (ii)
Solving (i) and (ii) we get
2g1g2 + 2f1f2 = c1 + c2
Condition of orthogonality
f = –\(\frac{3}{2}\), g = \(\frac{6}{2}\)
Required circle be x² + y² + 6x – 3y = 0

Inter 2nd Year Maths 2B System of Circles Solutions Ex 2(a)

Question 2.
Find the equation of the circle which passes through the point (0, -3) and intersects the circles given by the equations x² + y² – 6x + 3y + 5 = 0 and x² + y² – x – 7y = 0 orthogonally.
Solution:
Let circle be x² + y² + 2gx + 2fy + c = 0 ……… (i)
(i) is orthogonal with x² + y² – 6x + 3y + 5 = 0
then 2g(-3) + 2f\(\frac{+3}{2}\) = c + 5
-6g + 3f = c + 5 ……… (ii)
(i) is orthogonal with x² + y² – x – 7y = 0
-g – 7f = c ……… (iii)
Circle passes through (0, -3)
0 + 9 – 6f + c = 0 ……… (iv)
(iii) – (ii)
5g – 10f = -5
g – 2f = -1
(iii) + (iv)
Inter 2nd Year Maths 2B System of Circles Solutions Ex 2(a) 6
x² + y² + \(\frac{4}{3}\)y + \(\frac{2}{3}\)x – 5 = 0
3x² + 3y² + 4y + 2x – 15 = 0
(or) 3x² + 3y² + 2x + 4y – 15 = 0

Question 3.
Find the equation of the circle passing through the origin, having its centre on the line x + y = 4 and intersecting the circle x² + y² – 4x + 2y + 4 = 0 orthogonally.
Solution:
Let circle be x² + y²+ 2gx + 2fy + c = 0
Equation is passing through (0, 0)
0 + 0 + 2g.0 + 2f.0 + c = 0 ⇒ c = 0
x² + y² + 2gx + 2fy = 0
Centre passes through x + y = 4
∴ -g – f = 4 ………. (i)
Circle is orthogonal to
x² + y² – 4x + 2y + 4 = 0
-4g + 2f = 4 + 0
f – 2g = 2 ……… (ii)
Solving (i) and (ii) we get
– 3g = 6
9 = -2
f = -2
Equation of circle be x² + y² – 4x – 4y = 0

Question 4.
Find the equation of the circle which passes through the points (2, 0), (0, 2) and orthogonal to the circle 2x² + 2y² + 5x – 6y + 4 = 0.
Solution:
Let equation of circle be
x² + y² + 2gx + 2fy + c = 0
Passes through (2, 0), (0, 2) then
Inter 2nd Year Maths 2B System of Circles Solutions Ex 2(a) 7
Orthogonal to x + y + \(\frac{5}{2}\)x – \(\frac{6}{2}\)y + 2 = 0
2g\(\frac{5}{4}\) + 2f(-\(\frac{2}{2}\)) = 2 + c
\(\frac{5}{2}\)g – 3f = 2 + c
But g = f
\(\frac{5}{2}\)g – 3g = 2 + c
⇒ -g = 4 + 2C
Putting value of g in equation (i)
-16 – 8c + c = -4
c = –\(\frac{12}{7}\)
– g = 4 – \(\frac{24}{7}\) = +\(\frac{4}{7}\)
g = \(\frac{-4}{7}\) = f
Equation of the circle is
x + y – \(\frac{8x}{7}\) – \(\frac{8y}{7}\) – \(\frac{12}{7}\) = 0
⇒ 7(x² + y²) – 8x – 8y – 12 = 0

Question 5.
Find the equation of the circle which cuts orthogonally the circle x² + y² – 4x + 2y- 7 = 0 and having a centre at (2, 3).
Solution:
The given circle is
x² + y² – 4x + 2y – 7 = 0 ……. (1)
Let the required circle which, cuts orthogonally the circle (1) is
x² + y² + 2gx + 2fy + c = 0 ………… (2)
Its centre is (-g, -f) = (2, 3) given
g = -2, f = -3
The two circles (1) and (2) are cutting each other orthogonally.
So 2g1g2 + 2f1f2 = c1 + c2
2(-2)(-2) + 2(-3)(1) = – 7 + c
8 – 6 = -7 + c
+ 2 = -7 + c
c = 7 + 2 = 9 ⇒ c = 9
Hence the required circle is,
x² + y² – 4x – 6y + 9 = 0

III.

Question 1.
Find the equation of the circle which intersects the circle
x² + y² – 6x + 4y – 3 = 0 orthogonally and passes through the point (3, 0) and touches Y-axis.
Solution:
Let circle be (x – h)² + (y – k)² = r²
If circle touches Y-axis then co-ordinate of centre (h, k); radius = |h|
(x – h)² + (y – k)² = h²
x² – 2hx + h² + y²
-2ky + k² = h²
Orthogonal to
x² + y² – 6x + 4y – 3 = 0
2(-h) (-3) + 2(-k) (2) = -3 + k²
6h – 4k = -3 + k²
x² – 2hx + y²
-2ky + k² = h
Passes through (3, 0) 6h – 4k + 3 – k² = 0
9 – 6h + k² = 0 ……… (ii)
Adding .(i) and (ii) we get
c = 9
12 – 4k = 0 or k = 3, h = 3,
Equation of circle be y² + x² – 6x – 6y + 9 =0.

Inter 2nd Year Maths 2B System of Circles Solutions Ex 2(a)

Question 2.
Find the equation of the circle which cuts the circles x² + y² – 4x – 6y + 11 =0 and x² + y² – 10x – 4y + 2t = 0 orthogonally and has the diameter along the straight line 2x + 3y = 7.
Solution:
Let circle be x² + y² + 2gx + 2fy + c = 0 ………. (i) Orthogonal to circle
2g(-2) + 2f(-3) = 11 + c ……… (ii)
2g (-5) + 2f(-2) = 21 + c ……….. (iii)
Subtracting it we get
-6g + 2f = 10 ………. (iv)
Circles centre is on 2x + 3y = 7
∴ -2g – 3f = 7 …….. (v)
Solving (iv) and (v)
f = -1, g = -2, c = 3
Equation of circle be x² + y² – 4x – 2y + 3 = 0

Question 3.
If P, Q are conjugate points with respect to a circle S ≡ x² + y² + 2gx + 2fy + c = 0 then prove that the circle PQ as diameter cuts the circles S = 0 orthogonally.
Solution:
Let P = (x1, y1) and Q(x2, y2) be the conjugate points w.r.t. the circle
S ≡ x² + y² – a² = 0 …….. (i)
Polar of P w.r.t.(1) is xx1 + yy1 – a² = 0 ………… (ii)
Given P and Q are conjugate points ⇒ Q lies on (ii)
x1x2 + y1y2 – a² = 0 …….. (iii)
Equation of the circle on PQ as diameter is
(x – x1) (x – x2) + (y – y1) (y – y2) = 0
⇒ x² + y² – (x1 + x2) x – (y1 + y2)y + (x1x2 + y1y2) = 0 …………… (iv)
(i) and (iv) are orthogonal.
2g1g2 + 2f1f2 = 2[0(\(\frac{-x_{1}+x_{2}}{2}\)) + 0(\(\frac{-y_{1}+y_{2}}{2}\))]
c1 + c2 = -a² + a²
⇒ 2g1g2 + 2f1f2 = c1 + c2
∴ The circle on PQ as diameter cuts S orthogonally.

Inter 2nd Year Maths 2B System of Circles Solutions Ex 2(a)

Question 4.
If the equations of two circles whose radii are a, a’ are S = 0 and S’ = 0, then show that the circles \(\frac{S}{a}+\frac{S’}{a’}\) = 0 and \(\frac{S}{a}-\frac{S’}{a’}\) = 0 intersect orthogonally.
Solution:
Let 2d be the distance between the centres of the circles S = 0 and S’ =. 0. Take the line joining the centres as X-axis and the point midway between the centres as origin.
Then the equations of the circles are
S ≡ (x – d)² + y² – a² = 0
S’ ≡ (x + d)² + y² – a² = 0
∴ \(\frac{S}{a}+\frac{S’}{a’}\) =0 becomes Sa’ + S’a = 0
(or) [(x – d)² + y² -a²]a’ + [(x + d)² + y² – a’²]a = 0
(or) x² + y² + 2{\(\frac{(a-a’)d}{(a+a’)}\)}x + (d² – aa’) = 0 ………… (i)
Putting -a’ for a’ in we get \(\frac{S}{a}-\frac{S’}{a’}\) = 0
reduces to x² + y² + 2{\(\frac{(a+a’)d}{(a-a’)}\)}x + (d² + aa’) = 0 ………… (ii)
If (i) and (ii) cut orthogonally then
L.H.S. = 2gg’ + 2ff = c + c’
= 2{\(\frac{(a+a’)d}{(a-a’)}\)}{\(\frac{(a+a’)d}{(a-a’)}\)} + 2(0). (0) = 2d²
= (d² – aa’) + (d² + aa’) = 2d²
Which being true the circles (i) and (ii) the circles \(\frac{S}{a}±\frac{S’}{a’}\) = 0 cut each other at right angles.

Question 5.
Find the equation of the circle which intersects each of the following circles orthognally
i) x² + y² + 2x + 4y + 1 = 0, x² + y² – 2x + 6y – 3 = 0, 2(x² + y²) +6x + 8y – 3 = 0.
Solution:
Let equation of circle be
x² + y² + 2gx + 2fy + c = 0
Given circle is orthogonal to all 3 circles then
2g(1) + 2f(2) = c + 1 ……….. (i)
2g(\(\frac{3}{2}\)) = 2f(2) = c – \(\frac{3}{2}\) ………… (ii)
2g (-1) +2f(3) = c – 3 ………… (iii)
(iii) – (ii)
-5g + 2f = \(\frac{-3}{2}\) (or) – 10g + 4f = -3 ……….. (iv)
(iii) – (i)
-4g + 2f = – 4
f – 2g = -2 ……….. (v)
Solving (iv) and (v) we get
f = -7, g = -5/2, c = -34
∴ Equation of circle be
x² + y² – 5x – 14y – 34 = 0

ii) x² + y² + 4x + 2y + 1 = 0, 2(x² + y²) + 8x + 6y – 3 = 0, x² + y² + 6x – 2y – 3 = 0.
Solution:
Let required circle equation be
x² + y² + 2gx + 2fy + c = 0 this circle is orthogonal to above three circles.
∴ 2g(2) + 2f(1) = c + 1 ……….. (i)
2g(2) + 2f(\(\frac{3}{2}\)) = c – \(\frac{3}{2}\) ……….. (ii)
2g(3) + 2f(-1) = c – 3 ……….. (iii)
(i) – (ii) we get (ii) – (iii) we get
– f = \(\frac{5}{2}\) then – 2g + 5f = \(\frac{3}{2}\)
We get
g = -7 and f = \(\frac{-5}{2}\)
Substituting ‘g’ and ‘f’ in (i) we get
4(-7) + 2(\(\frac{-5}{2}\)) = c + 1
c = -34
Required equation of circle be
x² +y² – 5y – 14x – 34 = 0

Inter 2nd Year Maths 2B System of Circles Solutions Ex 2(a)

Question 6.
If the straight line 2x + 3y = 1 intersects the circle x² + y² = 4 at the points A and B, then find the equation of the circle having AB as diameter.
Solution:
Equation of circle passing through x² + y² – 4 and 2x + 3y – 1 =0 can be written as
(x² + y² – 4) + λ(2x + 3y – 1) = 0
x² + y² + 2λx + 3λy – 4 – λ = 0
Centre : (-λ, \(\frac{-3 \lambda}{2}\))
Centre lies on 2x + 3y – 1 =0
∴ 2(-λ) + 3(\(\frac{-3 \lambda}{2}\)) – 1 = 0
λ = \(\frac{-2}{13}\)
∴ Equation of circle be
13 (x² + y²) – 4 x 13 – 2(2x + 3y- 1) = 0
13 (x² + y²) – 4x – 6y – 50 = 0.

Question 7.
If x + y = 3 is the equation of the chord AB of the circle x² + y² – 2x + 4y – 8 = 0, find the equation of the circle having \(\overline{\mathrm{AB}}\) as diameter.
Solution:
Required equation of circle passing through intersection S = 0 and L = 0 is S + λL = 0
(x² + y² – 2x + 4y – 8) + λ(x + y – 3) = 0
(x² + y² + x(-2 + λ) + y (4 + λ) – 8 – 3λ = 0 ………… (i)
x² + y² + 2gx + 2fy + c = 0 ………… (ii)
Comparing (i) and (ii) we get
Inter 2nd Year Maths 2B System of Circles Solutions Ex 2(a) 8
2 – λ – 4 – λ = 6
-2λ = 8 ⇒ λ = -4
Required equation of circle be
(x² + y² – 2x + 4y – 8) – 4(x + y – 3) = 0
x² + y² – 6x + 4 = 0

Inter 2nd Year Maths 2B System of Circles Solutions Ex 2(a)

Question 8.
Find the equation of the circle passing through the intersection of the circles x² + y² = 2ax and x² + y² = 2by and having its centre on the line \(\frac{x}{a}-\frac{y}{b}\) = 2.
Solution:
Equation of circle passes through
x² + y² – 2ax = 0 and x² + y² – 2by = 0 can be written as
x² + y² – 2ax + X (x² + y² – 2by) = 0
x²(1 + λ) + y²(1 + λ) + x(-2a) – (2bλ)y = 0
Inter 2nd Year Maths 2B System of Circles Solutions Ex 2(a) 9
1- λ = (1 + λ)
λ = -1/3
Equation of circle be
3x² + 3y² – 6ax – x² – y² + 2by = 0
⇒ 2x² + 2y² – 6ax + 2by = 0
⇒ x² + y² – 3ax + by = 0